1 Introduction
The interpolation problem has occupied a central position in mathematics for several millennia. Roughly speaking, it asks:
Question. When can a curve of a given type be drawn through a given collection of points?
The first results on the interpolation problem date to classical antiquity. Two such results appear in Euclid’s Elements (circa 300 B.C.): A line can be drawn through any two distinct points in the plane (the first postulate), and a circle can be drawn through any three noncollinear points in the plane (Proposition 5 of Book IV).

The study of the interpolation problem in antiquity culminated in the work of Pappus (circa 340 A.D.), who showed in his Collection that a conic section can be drawn through any five points in the plane (Proposition 14 of Book VIII).
 The introduction of algebraic techniques to geometry enabled a second wave of results in the 18th century, and cast the interpolation problem firmly in the then-emerging field of algebraic geometry. For example, Cramer generalized Pappus’s result to plane curves of arbitrary degree n, which he showed can pass through 
 $n(n + 3)/2$
 general points in 1750 [Reference Cramer11]. Then Waring solved the interpolation problem for graphs of polynomial functions in 1779 [Reference Waring32]. (Lagrange independently rediscovered this result in 1795 [Reference Lagrange21] and thus it is often known as the ‘Lagrange interpolation formula’.) Cauchy [Reference Cauchy6], Hermite and Borchardt [Reference Hermite19] and Birkhoff [Reference Birkhoff5], all subsequently generalized Waring’s result in several different directions. These results are of interest far outside algebraic geometry and even outside of mathematics. For example, they play essential roles in the Newton–Cotes method for numerical integration, in Shamir’s cryptographic secret sharing protocol [Reference Shamir28], and in Reed–Solomon error-correcting codes [Reference Reed and Solomon26] (which currently power most digital storage media).
$n(n + 3)/2$
 general points in 1750 [Reference Cramer11]. Then Waring solved the interpolation problem for graphs of polynomial functions in 1779 [Reference Waring32]. (Lagrange independently rediscovered this result in 1795 [Reference Lagrange21] and thus it is often known as the ‘Lagrange interpolation formula’.) Cauchy [Reference Cauchy6], Hermite and Borchardt [Reference Hermite19] and Birkhoff [Reference Birkhoff5], all subsequently generalized Waring’s result in several different directions. These results are of interest far outside algebraic geometry and even outside of mathematics. For example, they play essential roles in the Newton–Cotes method for numerical integration, in Shamir’s cryptographic secret sharing protocol [Reference Shamir28], and in Reed–Solomon error-correcting codes [Reference Reed and Solomon26] (which currently power most digital storage media).
The key prerequisite to the modern study of the interpolation problem was the development of Brill–Noether theory in the 20th century, which studies maps from general curves to projective space, and thus identifies the most natural class of curves for which to study the interpolation problem. Namely, let C be a general curve of genus g. From our perspective here, the two key facts are:
- 
1. There exists a nondegenerate map  $C \to \mathbb {P}^r$
 of degree d if and only if the quantity satisfies $C \to \mathbb {P}^r$
 of degree d if and only if the quantity satisfies  $\rho \geq 0$
. [Proven in 1980 by Griffiths and Harris [Reference Griffiths and Harris17].] $\rho \geq 0$
. [Proven in 1980 by Griffiths and Harris [Reference Griffiths and Harris17].]
- 
2. In this case, the universal space of such maps has a unique component dominating  $\overline {M}_g$
. [Proven in the 1980s by Fulton and Lazarsfeld [Reference Fulton and Lazarsfeld15], Gieseker [Reference Gieseker16] and Eisenbud and Harris [Reference Eisenbud and Harris12].] $\overline {M}_g$
. [Proven in the 1980s by Fulton and Lazarsfeld [Reference Fulton and Lazarsfeld15], Gieseker [Reference Gieseker16] and Eisenbud and Harris [Reference Eisenbud and Harris12].]
We call stable maps 
 $f \colon C \to \mathbb {P}^r$
 corresponding to points in this unique component Brill–Noether curves (BN-curves). (The general such curve is an embedding of a smooth curve for
$f \colon C \to \mathbb {P}^r$
 corresponding to points in this unique component Brill–Noether curves (BN-curves). (The general such curve is an embedding of a smooth curve for 
 $r \geq 3$
.) This language then gives us a precise and natural formulation of the interpolation problem:
$r \geq 3$
.) This language then gives us a precise and natural formulation of the interpolation problem:
Question. Let 
 $d, g, r, n$
 be nonnegative integers with
$d, g, r, n$
 be nonnegative integers with 
 $\rho (d, g, r) \geq 0$
. When can we pass a BN-curve of degree d and genus g through n general points in
$\rho (d, g, r) \geq 0$
. When can we pass a BN-curve of degree d and genus g through n general points in 
 $\mathbb {P}^r$
?
$\mathbb {P}^r$
?
 Equivalently, writing 
 $M^\circ _{g, n}(\mathbb {P}^r, d)$
 for the component corresponding to BN-curves, this question is asking when the evaluation map
$M^\circ _{g, n}(\mathbb {P}^r, d)$
 for the component corresponding to BN-curves, this question is asking when the evaluation map 
 $M^\circ _{g, n}(\mathbb {P}^r, d) \to (\mathbb {P}^r)^n$
 is dominant. It is evidently necessary for:
$M^\circ _{g, n}(\mathbb {P}^r, d) \to (\mathbb {P}^r)^n$
 is dominant. It is evidently necessary for: 
 $$\begin{align*}rn = \dim (\mathbb{P}^r)^n \leq \dim M^\circ_{g, n}(\mathbb{P}^r, d) = (r + 1)d - (r - 3)(g - 1) + n,\end{align*}$$
$$\begin{align*}rn = \dim (\mathbb{P}^r)^n \leq \dim M^\circ_{g, n}(\mathbb{P}^r, d) = (r + 1)d - (r - 3)(g - 1) + n,\end{align*}$$
or upon rearrangement,
 $$\begin{align*}(r - 1)n \leq (r + 1)d - (r - 3)(g - 1).\end{align*}$$
$$\begin{align*}(r - 1)n \leq (r + 1)d - (r - 3)(g - 1).\end{align*}$$
Despite cases where this is not sufficient, it is a folklore conjecture that it usually suffices:
Conjecture 1.1. Let 
 $d, g, r, n$
 be nonnegative integers with
$d, g, r, n$
 be nonnegative integers with 
 $\rho (d, g, r) \geq 0$
. Then there is a BN-curve of degree d and genus g through n general points in
$\rho (d, g, r) \geq 0$
. Then there is a BN-curve of degree d and genus g through n general points in 
 $\mathbb {P}^r$
 if and only if
$\mathbb {P}^r$
 if and only if 
 $$\begin{align*}(r - 1)n \leq (r + 1)d - (r - 3)(g - 1),\end{align*}$$
$$\begin{align*}(r - 1)n \leq (r + 1)d - (r - 3)(g - 1),\end{align*}$$
apart from finitely many exceptions.
 This conjecture has been studied intensely in recent years. As in previous eras, this attention has been motivated by both intrinsic interest, as well as by striking applications to a wide range of other interesting geometric problems. Recent examples of such problems include smoothing curve singularities [Reference Stevens30], constructing moving curves in 
 $\overline {M}_g$
 [Reference Atanasov3, Reference Chen, Farkas and Morrison7], the first author’s resolution of Severi’s 1915 maximal rank conjecture [Reference Larson22], as well as various generalizations thereof [Reference Ballico4].
$\overline {M}_g$
 [Reference Atanasov3, Reference Chen, Farkas and Morrison7], the first author’s resolution of Severi’s 1915 maximal rank conjecture [Reference Larson22], as well as various generalizations thereof [Reference Ballico4].
 The easiest cases of this conjecture are when d is large relative to g and r, and such cases have therefore been the focus of significant work. For example, Sacchiero proved Conjecture 1.1 for rational curves in 1980 [Reference Sacchiero27]; Ran later gave an independent proof in this case in 2007 [Reference Ran25]. Subsequently, the first author, in joint work with Atanasov and Yang, proved Conjecture 1.1 when 
 $d \geq g + r$
 in characteristic zero [Reference Atanasov, Larson and Yang2]. Another case of interest is the proof of Conjecture 1.1 for canonical curves in characteristic zero in a pair of papers by Stevens from 1989 and 1996 [Reference Stevens29, Reference Stevens30]. Many authors have also considered this conjecture in low dimensions. For example, Ellingsrud and Hirschowitz in 1984 [Reference Ellingsrud and Hirschowitz13], Perrin in 1987 [Reference Perrin24] and later Atanasov in 2014 [Reference Atanasov3], all made significant progress on Conjecture 1.1 for space curves, but their analysis left infinitely many cases unsolved. This effort culminated in the proof of Conjecture 1.1 for space curves in characteristic zero by the second author in 2018 [Reference Vogt31], and for curves in
$d \geq g + r$
 in characteristic zero [Reference Atanasov, Larson and Yang2]. Another case of interest is the proof of Conjecture 1.1 for canonical curves in characteristic zero in a pair of papers by Stevens from 1989 and 1996 [Reference Stevens29, Reference Stevens30]. Many authors have also considered this conjecture in low dimensions. For example, Ellingsrud and Hirschowitz in 1984 [Reference Ellingsrud and Hirschowitz13], Perrin in 1987 [Reference Perrin24] and later Atanasov in 2014 [Reference Atanasov3], all made significant progress on Conjecture 1.1 for space curves, but their analysis left infinitely many cases unsolved. This effort culminated in the proof of Conjecture 1.1 for space curves in characteristic zero by the second author in 2018 [Reference Vogt31], and for curves in 
 $\mathbb {P}^4$
 in characteristic zero by both authors in 2021 [Reference Larson and Vogt23].
$\mathbb {P}^4$
 in characteristic zero by both authors in 2021 [Reference Larson and Vogt23].
Nevertheless, despite this significant interest, fundamental limitations of previous techniques have prevented the resolution of Conjecture 1.1 in general and limited even partial results largely to characteristic zero. Our main result gives the first comprehensive answer to the interpolation problem.
Theorem 1.2. Conjecture 1.1 holds in full generality and in any characteristic. More precisely: Let 
 $d, g, r, n$
 be nonnegative integers with
$d, g, r, n$
 be nonnegative integers with 
 $\rho (d, g, r) \geq 0$
. There is a BN-curve of degree d and genus g through n general points in
$\rho (d, g, r) \geq 0$
. There is a BN-curve of degree d and genus g through n general points in 
 $\mathbb {P}^r$
 if and only if
$\mathbb {P}^r$
 if and only if 
 $$ \begin{align} (r - 1)n \leq (r + 1)d - (r - 3)(g - 1), \end{align} $$
$$ \begin{align} (r - 1)n \leq (r + 1)d - (r - 3)(g - 1), \end{align} $$
except in the following four exceptional cases:
 $$\begin{align*}(d, g, r) \in \{(5, 2, 3), (6, 4, 3), (7, 2, 5), (10, 6, 5)\}.\end{align*}$$
$$\begin{align*}(d, g, r) \in \{(5, 2, 3), (6, 4, 3), (7, 2, 5), (10, 6, 5)\}.\end{align*}$$
 Since the normal bundle 
 $N_C$
 controls the deformation theory of C, Conjecture 1.1 is closely related to a certain property, also known as interpolation, for
$N_C$
 controls the deformation theory of C, Conjecture 1.1 is closely related to a certain property, also known as interpolation, for 
 $N_C$
.
$N_C$
.
Definition 1.3. A vector bundle E on a curve C satisfies interpolation if 
 $H^1(E) = 0$
, and for every
$H^1(E) = 0$
, and for every 
 $n> 0$
, there exists an effective divisor D of degree n such that
$n> 0$
, there exists an effective divisor D of degree n such that 
 $$ \begin{align} H^0(E(-D)) = 0 \quad \text{or} \quad H^1(E(-D)) = 0. \end{align} $$
$$ \begin{align} H^0(E(-D)) = 0 \quad \text{or} \quad H^1(E(-D)) = 0. \end{align} $$
For C irreducible, 
 $\operatorname {Sym}^n C$
 is also irreducible, so interpolation is equivalent to (1.2) for D general.
$\operatorname {Sym}^n C$
 is also irreducible, so interpolation is equivalent to (1.2) for D general.
 Given 
 $p_1, \dots , p_n \in C\subset \mathbb {P}^r$
, a standard argument in deformation theory (see [Reference Perrin24, Theorem 1.5]) implies that the evaluation map
$p_1, \dots , p_n \in C\subset \mathbb {P}^r$
, a standard argument in deformation theory (see [Reference Perrin24, Theorem 1.5]) implies that the evaluation map 
 $M^\circ _{g, n}(\mathbb {P}^r, d) \to (\mathbb {P}^r)^n$
 is smooth at the point
$M^\circ _{g, n}(\mathbb {P}^r, d) \to (\mathbb {P}^r)^n$
 is smooth at the point 
 $(C, p_1, \dots , p_n)$
, and hence dominant, if
$(C, p_1, \dots , p_n)$
, and hence dominant, if 
 $H^1(N_C(-p_1 - \cdots -p_n))=0$
. Since
$H^1(N_C(-p_1 - \cdots -p_n))=0$
. Since 
 $\chi (N_C(-p_1-\cdots -p_n)) = (r+1)d - (r-3)(g-1) - (r-1)n$
, we have
$\chi (N_C(-p_1-\cdots -p_n)) = (r+1)d - (r-3)(g-1) - (r-1)n$
, we have 
 $\chi (N_C(-p_1-\cdots -p_n)) \geq 0$
 precisely when equation (1.1) is satisfied. Therefore, interpolation for
$\chi (N_C(-p_1-\cdots -p_n)) \geq 0$
 precisely when equation (1.1) is satisfied. Therefore, interpolation for 
 $N_C$
 implies that
$N_C$
 implies that 
 $H^1(N_C(-p_1 - \cdots -p_n))=0$
 when equation (1.1) is satisfied, which implies Conjecture 1.1. In fact, interpolation for
$H^1(N_C(-p_1 - \cdots -p_n))=0$
 when equation (1.1) is satisfied, which implies Conjecture 1.1. In fact, interpolation for 
 $N_C$
 is a slightly stronger condition, the essential differences being:
$N_C$
 is a slightly stronger condition, the essential differences being: 
- 
1. It implies an analog of Conjecture 1.1 where the general points are replaced by general linear spaces: There is a BN-curve of degree d and genus g incident to general linear spaces  $\Lambda _i$
 of dimension $\Lambda _i$
 of dimension $\lambda _i$
 if and only if (1.3)(This implication can be deduced from [Reference Atanasov3, Theorem 8.1], cf. the introduction to loc. cit.) $\lambda _i$
 if and only if (1.3)(This implication can be deduced from [Reference Atanasov3, Theorem 8.1], cf. the introduction to loc. cit.) $$ \begin{align} \sum (r - 1 - \lambda_i) \leq (r + 1)d - (r - 3)(g - 1). \end{align} $$ $$ \begin{align} \sum (r - 1 - \lambda_i) \leq (r + 1)d - (r - 3)(g - 1). \end{align} $$
- 
2. It implies that  $M^\circ _{g, n}(\mathbb {P}^r, d) \to (\mathbb {P}^r)^n$
 is generically smooth, rather than merely dominant. (This is of course equivalent in characteristic zero but is a stronger statement in positive characteristic.) $M^\circ _{g, n}(\mathbb {P}^r, d) \to (\mathbb {P}^r)^n$
 is generically smooth, rather than merely dominant. (This is of course equivalent in characteristic zero but is a stronger statement in positive characteristic.)
Theorem 1.2 is a consequence of our main theorem, which asserts:
Theorem 1.4. Let 
 $d, g, r$
 be nonnegative integers with
$d, g, r$
 be nonnegative integers with 
 $\rho (d, g, r) \geq 0$
, and
$\rho (d, g, r) \geq 0$
, and 
 $C \subset \mathbb {P}^r$
 be a general BN-curve of degree d and genus g. Then
$C \subset \mathbb {P}^r$
 be a general BN-curve of degree d and genus g. Then 
 $N_C$
 satisfies interpolation if and only if neither of the following hold:
$N_C$
 satisfies interpolation if and only if neither of the following hold: 
- 
1. The tuple  $(d, g, r)$
 is one of the following five exceptions: (1.4) $(d, g, r)$
 is one of the following five exceptions: (1.4) $$ \begin{align} (d, g, r) \in \{(5,2,3), (6,4,3), (6,2,4), (7,2,5),(10,6,5)\}. \end{align} $$ $$ \begin{align} (d, g, r) \in \{(5,2,3), (6,4,3), (6,2,4), (7,2,5),(10,6,5)\}. \end{align} $$
- 
2. The characteristic is  $2$
, and $2$
, and $g=0$
, and $g=0$
, and $d \not \equiv 1$
 mod $d \not \equiv 1$
 mod $r - 1$
. $r - 1$
.
There are several exceptions in Theorem 1.4 that are not exceptions for Theorem 1.2:
- 
1. The case  $(d, g, r) = (6, 2, 4)$
: Such curves have the expected behavior for passing through points, but not for incidence to linear spaces of arbitrary dimension. More precisely, a naive dimension count suggests that they can pass through nine general points and meet a general line, but this is not true. $(d, g, r) = (6, 2, 4)$
: Such curves have the expected behavior for passing through points, but not for incidence to linear spaces of arbitrary dimension. More precisely, a naive dimension count suggests that they can pass through nine general points and meet a general line, but this is not true.
- 
2. The cases in characteristic  $2$
: In these cases, the evaluation map $2$
: In these cases, the evaluation map $M^\circ _{g, n}(\mathbb {P}^r, d) \to (\mathbb {P}^r)^n$
 is dominant but not generically smooth. $M^\circ _{g, n}(\mathbb {P}^r, d) \to (\mathbb {P}^r)^n$
 is dominant but not generically smooth.
We discuss these two cases in more depth in Sections 2 and 9.
 Our approach to Theorem 1.4 will be via degeneration to reducible curves 
 $X \cup Y$
. In general, although the restrictions
$X \cup Y$
. In general, although the restrictions 
 $N_{X \cup Y}|_X$
 and
$N_{X \cup Y}|_X$
 and 
 $N_{X \cup Y}|_Y$
 admit nice descriptions, fitting these together to describe
$N_{X \cup Y}|_Y$
 admit nice descriptions, fitting these together to describe 
 $N_{X \cup Y}$
 is extremely challenging outside a handful of special cases. This fundamental obstacle has limited previous attempts to study the interpolation problem. For example, the key innovation of [Reference Atanasov, Larson and Yang2] was an essentially complete description of
$N_{X \cup Y}$
 is extremely challenging outside a handful of special cases. This fundamental obstacle has limited previous attempts to study the interpolation problem. For example, the key innovation of [Reference Atanasov, Larson and Yang2] was an essentially complete description of 
 $N_{X \cup Y}$
 in the special case that Y was a one- or two-secant line. Considering only such degenerations leads to two severe limitations:
$N_{X \cup Y}$
 in the special case that Y was a one- or two-secant line. Considering only such degenerations leads to two severe limitations: 
- 
1. Only nonspecial curves can be obtained by successively adding one- and two-secant lines. 
- 
2. Since the set of degenerations used is so limited, only a few types of elementary modifications to the normal bundle appear. Because there are only a few types of modifications, it is difficult to produce desired modifications by combining them, in a way reminiscent of the Frobenius coin problem. Circumventing this difficulty requires additional tools that work only in characteristic zero. 
Previous attempts to overcome these difficulties were limited to ad-hoc constructions in low-dimensional projective spaces. The present paper introduces two key innovations:
- 
1. We consider a third degeneration, where Y is an  $(r + 1)$
-secant rational curve of degree $(r + 1)$
-secant rational curve of degree $r - 1$
 contained in a hyperplane H, which allows us to obtain any BN-curve. Describing how $r - 1$
 contained in a hyperplane H, which allows us to obtain any BN-curve. Describing how $N_{X \cup Y}|_X$
 and $N_{X \cup Y}|_X$
 and $N_{X \cup Y}|_Y$
 fit together to give $N_{X \cup Y}|_Y$
 fit together to give $N_{X \cup Y}$
 is intractable even in a degeneration of this complexity. Nevertheless, thanks to our detailed study of this setup in Sections 5.3 and 13.1, we are able to reduce interpolation for $N_{X \cup Y}$
 is intractable even in a degeneration of this complexity. Nevertheless, thanks to our detailed study of this setup in Sections 5.3 and 13.1, we are able to reduce interpolation for $N_{X \cup Y}$
 to interpolation for a certain modification of $N_{X \cup Y}$
 to interpolation for a certain modification of $N_X$
. $N_X$
.
- 
2. Although Y does not have many interesting degenerations in H, we show in Section 7 that as H becomes tangent to C, a plethora of such degenerations appear. As in the Frobenius coin problem, this plethora of additional degenerations makes it possible to produce the desired modifications by combining them. This eliminates the restriction to characteristic zero that plagued previous work. 
The techniques we develop here hold promise of application to other problems about the geometry of normal bundles. Indeed, in [Reference Coskun, Larson and Vogt10], they have already been applied to prove semistability of normal bundles of canonical curves, thereby establishing a (slightly weaker version of a) conjecture of Aprodu, Farkas and Orgeta [Reference Aprodu, Farkas and Ortega1]. (A vector bundle of integral slope which satisfies interpolation is necessarily semistable, c.f. [Reference Vogt31, Remark 1.6], but in these cases the slope is not integral. Nevertheless, our techniques may be applied.)
Structure of the paper
We begin, in Section 2, by discussing the various exceptional cases appearing in Theorem 1.4. Then in Section 3, we introduce the notation we shall use for the remainder of the paper and discuss a few other preliminary points.
In Section 4, we explain the basic strategy of proof for the hard direction of Theorem 1.4, that is, that there are no other exceptional cases besides those mentioned in the statement of Theorem 1.4 above. After explaining the basic strategy, we give a roadmap to the proof, which occupies the remainder of the paper.
2 Counterexamples
2.1 Counterexamples in all characteristics
We start with the five counterexamples to Theorem 1.4 that occur in all characteristics:
 $$\begin{align*}(d, g, r) \in \{(5,2,3), (6,4,3), (6,2,4), (7,2,5),(10,6,5)\}.\end{align*}$$
$$\begin{align*}(d, g, r) \in \{(5,2,3), (6,4,3), (6,2,4), (7,2,5),(10,6,5)\}.\end{align*}$$
In each of these cases, we will construct a certain surface S containing C, and see that S prevents Theorem 1.2 (or the generalization (1.3) thereof) from holding. Indeed, if S cannot be made to pass through the requisite number of points (or be made incident to the requisite linear spaces), then C cannot either. Since Theorem 1.4 implies Theorem 1.2 (and the generalization (1.3)), this implies that these five cases must also be counterexamples to Theorem 1.4.
Remark 2.1. An alternative approach, the details of which we leave to the interested reader, would be to see directly that the geometry of S obstructs Theorem 1.4. The basic idea is that, for any effective divisor D on C, we have 
 $h^0(N_C(-D)) \geq h^0(N_{C/S}(-D))$
; in the five exceptional cases, this inequality will prevent
$h^0(N_C(-D)) \geq h^0(N_{C/S}(-D))$
; in the five exceptional cases, this inequality will prevent 
 $N_C$
 from satisfying interpolation.
$N_C$
 from satisfying interpolation.
2.1.1 The family 
 $(d,g,r)=(r+2, 2, r)$
$(d,g,r)=(r+2, 2, r)$
 Let C be a curve of genus 
 $2$
 and L be a line bundle of degree
$2$
 and L be a line bundle of degree 
 $r+2$
 on C. Write
$r+2$
 on C. Write 
 $f \colon C \to \mathbb {P}^1$
 for the hyperelliptic map. Then
$f \colon C \to \mathbb {P}^1$
 for the hyperelliptic map. Then 
 is a vector bundle of rank
 is a vector bundle of rank 
 $2$
 on
$2$
 on 
 $\mathbb {P}^1$
 with
$\mathbb {P}^1$
 with 
 $$\begin{align*}\chi(E) = \chi(L) = r+1.\end{align*}$$
$$\begin{align*}\chi(E) = \chi(L) = r+1.\end{align*}$$
By Riemann–Roch, 
 $c_1(E) = r-1$
. The inclusion
$c_1(E) = r-1$
. The inclusion 
 $L \to f^*f_* L$
 embeds C in the projective bundle
$L \to f^*f_* L$
 embeds C in the projective bundle 
 $\mathbb {P} E^\vee $
 so that
$\mathbb {P} E^\vee $
 so that 
 $O_{\mathbb {P} E^\vee }(1)|_C \simeq L$
. Therefore, the image of C in
$O_{\mathbb {P} E^\vee }(1)|_C \simeq L$
. Therefore, the image of C in 
 $\mathbb {P}^r$
 under the complete linear series for L lies on the image S of
$\mathbb {P}^r$
 under the complete linear series for L lies on the image S of 
 $\mathbb {P} E^\vee $
 under the complete linear series for
$\mathbb {P} E^\vee $
 under the complete linear series for 
 $\mathcal {O}_{\mathbb {P} E^\vee }(1)$
. The surface S is a scroll of degree equal to
$\mathcal {O}_{\mathbb {P} E^\vee }(1)$
. The surface S is a scroll of degree equal to 
 $$\begin{align*}[\mathcal{O}_{\mathbb{P} E^\vee}(1)]^2 = - c_1(E^\vee) \cdot \mathcal{O}_{\mathbb{P} E^\vee}(1) = r-1.\end{align*}$$
$$\begin{align*}[\mathcal{O}_{\mathbb{P} E^\vee}(1)]^2 = - c_1(E^\vee) \cdot \mathcal{O}_{\mathbb{P} E^\vee}(1) = r-1.\end{align*}$$
By [Reference Coskun8, Lemma 2.6], the dimension of the space of such scrolls is 
 $r^2 + 2r - 6$
.
$r^2 + 2r - 6$
.
- 
If  $\boldsymbol{(d, g, r) = (5, 2, 3)}$
 Then $\boldsymbol{(d, g, r) = (5, 2, 3)}$
 Then $r^2 + 2r - 6 = 9$
. Since it is $r^2 + 2r - 6 = 9$
. Since it is $1$
 condition for a surface in $1$
 condition for a surface in $\mathbb {P}^3$
 to pass through a point, S cannot pass through more than nine general points. This contradicts (1.1), which predicts that C should be able to pass through $\mathbb {P}^3$
 to pass through a point, S cannot pass through more than nine general points. This contradicts (1.1), which predicts that C should be able to pass through $10$
 general points. $10$
 general points.
- 
If  $ \boldsymbol{(d, g, r) = (6, 2, 4)}$
 Then $ \boldsymbol{(d, g, r) = (6, 2, 4)}$
 Then $r^2 + 2r - 6 = 18$
. Since it is two conditions for a surface in $r^2 + 2r - 6 = 18$
. Since it is two conditions for a surface in $\mathbb {P}^4$
 to pass through a point and one condition to meet a line, S cannot pass through nine general points while meeting a general line. This contradicts equation (1.3), which predicts that C should be able to pass through nine general points points while meeting a general line. $\mathbb {P}^4$
 to pass through a point and one condition to meet a line, S cannot pass through nine general points while meeting a general line. This contradicts equation (1.3), which predicts that C should be able to pass through nine general points points while meeting a general line.
- 
If  $ \boldsymbol{(d, g, r) = (7, 2, 5)} $
 Then $ \boldsymbol{(d, g, r) = (7, 2, 5)} $
 Then $r^2 + 2r - 6 = 29$
. Since it is three conditions for a surface in $r^2 + 2r - 6 = 29$
. Since it is three conditions for a surface in $\mathbb {P}^5$
 to pass through a point, S cannot pass through more than nine general points. This contradicts equation (1.1), which predicts that C should be able to pass through $\mathbb {P}^5$
 to pass through a point, S cannot pass through more than nine general points. This contradicts equation (1.1), which predicts that C should be able to pass through $10$
 general points. $10$
 general points.
2.1.2 The case 
 $(d,g,r)=(6, 4, 3)$
$(d,g,r)=(6, 4, 3)$
 A general canonical curve of genus 
 $4$
 is a cubic section of a quadric surface S. There is a nine-dimensional family of quadric surfaces, and it is one condition for a surface in
$4$
 is a cubic section of a quadric surface S. There is a nine-dimensional family of quadric surfaces, and it is one condition for a surface in 
 $\mathbb {P}^3$
 to pass through a point, so S cannot pass through more than two points. This contradicts equation (1.1), which predicts that C should be able to pass through
$\mathbb {P}^3$
 to pass through a point, so S cannot pass through more than two points. This contradicts equation (1.1), which predicts that C should be able to pass through 
 $12$
 general points.
$12$
 general points.
2.1.3 The case 
 $(d,g,r)=(10, 6, 5)$
$(d,g,r)=(10, 6, 5)$
 A general canonical curve of genus 
 $6$
 is a quadric section of a quintic del Pezzo surface S. There is a
$6$
 is a quadric section of a quintic del Pezzo surface S. There is a 
 $35$
-dimensional family of quintic del Pezzo surfaces, and it is three conditions for a surface in
$35$
-dimensional family of quintic del Pezzo surfaces, and it is three conditions for a surface in 
 $\mathbb {P}^5$
 to pass through a point, so S cannot pass through more than
$\mathbb {P}^5$
 to pass through a point, so S cannot pass through more than 
 $11$
 points. This contradicts (1.1), which predicts that C should be able to pass through
$11$
 points. This contradicts (1.1), which predicts that C should be able to pass through 
 $12$
 general points.
$12$
 general points.
2.2 Rational curves in characteristic 
 $2$
$2$
 In this section, we explain the final infinite family of counterexamples to Theorem 1.4 that occurs only in characteristic 
 $2$
. This phenomenon was already observed for space curves in [Reference Coskun, Larson and Vogt9] in relation to semistability. We begin by describing which vector bundles on a rational curve satisfy interpolation in terms of the Birkhoff–Grothendieck classification.
$2$
. This phenomenon was already observed for space curves in [Reference Coskun, Larson and Vogt9] in relation to semistability. We begin by describing which vector bundles on a rational curve satisfy interpolation in terms of the Birkhoff–Grothendieck classification.
Lemma 2.2. The bundle 
 $E = \bigoplus _i \mathcal {O}_{\mathbb {P}^1}(e_i)$
 satisfies interpolation if and only if for all
$E = \bigoplus _i \mathcal {O}_{\mathbb {P}^1}(e_i)$
 satisfies interpolation if and only if for all 
 $i, j$
,
$i, j$
, 
 $$\begin{align*}|e_i - e_j| \leq 1\quad \text{and} \quad e_i \geq -1.\end{align*}$$
$$\begin{align*}|e_i - e_j| \leq 1\quad \text{and} \quad e_i \geq -1.\end{align*}$$
Proof. For any 
 $n \geq 0$
,
$n \geq 0$
, 
 $$ \begin{align*} h^0(E(-n)) = 0 \quad &\Leftrightarrow\quad n \geq 1 + \max(e_i) \\ h^1(E(-n)) = 0 \quad &\Leftrightarrow \quad n \leq 1 + \min(e_i). \end{align*} $$
$$ \begin{align*} h^0(E(-n)) = 0 \quad &\Leftrightarrow\quad n \geq 1 + \max(e_i) \\ h^1(E(-n)) = 0 \quad &\Leftrightarrow \quad n \leq 1 + \min(e_i). \end{align*} $$
One of these two conditions holds for all 
 $n \geq 0$
 if and only if
$n \geq 0$
 if and only if 
 $|e_i - e_j| \leq 1$
 for all i and j, and the second of these holds for
$|e_i - e_j| \leq 1$
 for all i and j, and the second of these holds for 
 $n=0$
 if and only if
$n=0$
 if and only if 
 $e_i \geq -1$
 for all i.
$e_i \geq -1$
 for all i.
As a consequence of the Euler sequence, the conormal bundle of C sits in the exact sequence
 $$\begin{align*}0 \to N_C^\vee(1) \to \mathcal{O}_C^{\oplus r+1} \to \mathscr{P}^1(\mathcal{O}_C(1)) \to 0, \end{align*}$$
$$\begin{align*}0 \to N_C^\vee(1) \to \mathcal{O}_C^{\oplus r+1} \to \mathscr{P}^1(\mathcal{O}_C(1)) \to 0, \end{align*}$$
where 
 $\mathscr {P}^1(\mathcal {O}_C(1))$
 is the bundle of first principle parts of
$\mathscr {P}^1(\mathcal {O}_C(1))$
 is the bundle of first principle parts of 
 $\mathcal {O}_C(1)$
. If the characteristic is
$\mathcal {O}_C(1)$
. If the characteristic is 
 $2$
, and we write
$2$
, and we write 
 $\pi \colon C \to C^{(2)}$
 for the (relative) Frobenius morphism, then we have
$\pi \colon C \to C^{(2)}$
 for the (relative) Frobenius morphism, then we have 
 $$\begin{align*}\mathscr{P}^1(\mathcal{O}_C(1)) \simeq \pi^*\pi_*\mathcal{O}_C(1).\end{align*}$$
$$\begin{align*}\mathscr{P}^1(\mathcal{O}_C(1)) \simeq \pi^*\pi_*\mathcal{O}_C(1).\end{align*}$$
Therefore,
 $ N_C^\vee (1)$
 is isomorphic to the pullback of a bundle under the Frobenius morphism, so every entry of its splitting type is even.
$ N_C^\vee (1)$
 is isomorphic to the pullback of a bundle under the Frobenius morphism, so every entry of its splitting type is even.
Lemma 2.3. Assume that the characteristic of the ground field is 
 $2$
. Let
$2$
. Let 
 $C \subset \mathbb {P}^r$
 be a rational curve of degree d. Then
$C \subset \mathbb {P}^r$
 be a rational curve of degree d. Then 
 $N_C$
 satisfies interpolation only if
$N_C$
 satisfies interpolation only if 
 $$\begin{align*}d \equiv 1 \pmod{r-1}.\end{align*}$$
$$\begin{align*}d \equiv 1 \pmod{r-1}.\end{align*}$$
Proof. Suppose that 
 $N_C \simeq \bigoplus _{i = 1}^{r - 1} \mathcal {O}_{\mathbb {P}^1}(e_i)$
. Since
$N_C \simeq \bigoplus _{i = 1}^{r - 1} \mathcal {O}_{\mathbb {P}^1}(e_i)$
. Since 
 $N_C^\vee (1) \simeq \bigoplus _{i = 1}^{r - 1} \mathcal {O}_{\mathbb {P}^1}(d - e_i)$
, and every entry of its splitting type is even, every
$N_C^\vee (1) \simeq \bigoplus _{i = 1}^{r - 1} \mathcal {O}_{\mathbb {P}^1}(d - e_i)$
, and every entry of its splitting type is even, every 
 $e_i$
 satisfies
$e_i$
 satisfies 
 $e_i \equiv d$
 mod
$e_i \equiv d$
 mod 
 $2$
. Applying Lemma 2.2, we conclude that
$2$
. Applying Lemma 2.2, we conclude that 
 $N_C$
 can only satisfy interpolation if all
$N_C$
 can only satisfy interpolation if all 
 $e_i$
 are equal. This implies
$e_i$
 are equal. This implies 
 $$\begin{align*}(r + 1)d - 2 = c_1(N_C) \equiv (r - 1)d \pmod{2(r - 1)},\end{align*}$$
$$\begin{align*}(r + 1)d - 2 = c_1(N_C) \equiv (r - 1)d \pmod{2(r - 1)},\end{align*}$$
and therefore 
 $d \equiv 1$
 mod
$d \equiv 1$
 mod 
 $r - 1$
 as desired.
$r - 1$
 as desired.
3 Preliminaries
3.1 Elementary modifications of vector bundles
In this section, we give a brief overview of the key properties of elementary modifications of vector bundles. Our presentation will roughly follow the more detailed exposition given in Sections 2–4 of [Reference Atanasov, Larson and Yang2].
Definition 3.1. Let E be a vector bundle on a scheme X, and 
 $D \subset X$
 be a Cartier divisor and
$D \subset X$
 be a Cartier divisor and 
 $F \subset E|_D$
 be a subbundle of the restriction of E to D. We define the negative elementary modification of E along D towards F
 by
$F \subset E|_D$
 be a subbundle of the restriction of E to D. We define the negative elementary modification of E along D towards F
 by 

We then define the (positive) elementary modification of E along D towards F as

Remark 3.2. This notation differs slightly from [Reference Atanasov, Larson and Yang2], in which negative modifications were denoted by 
 $E[D \to F]$
 (and no separate notation was given for positive modifications).
$E[D \to F]$
 (and no separate notation was given for positive modifications).
 By construction, a modification of E along D is naturally isomorphic to E when restricted to the complement of D. If 
 $D_1$
 and
$D_1$
 and 
 $D_2$
 are disjoint, then we may easily make sense of multiple modifications such as
$D_2$
 are disjoint, then we may easily make sense of multiple modifications such as 
 by working locally. However, if
 by working locally. However, if 
 $D_1$
 and
$D_1$
 and 
 $D_2$
 meet, then we do not have enough data to even define multiple modifications: For example, if
$D_2$
 meet, then we do not have enough data to even define multiple modifications: For example, if 
 $D_1 = D_2 = D$
 and
$D_1 = D_2 = D$
 and 
 $F_1 = F_2 = F$
, then we should have
$F_1 = F_2 = F$
, then we should have 
 , so we must know how F extends over
, so we must know how F extends over 
 $2D$
. To sidestep these issues, we suppose when defining multiple modifications – at least along divisors that meet – that we are given not just a subbundle
$2D$
. To sidestep these issues, we suppose when defining multiple modifications – at least along divisors that meet – that we are given not just a subbundle 
 $F_i$
 of
$F_i$
 of 
 $E|_{D_i}$
, but a subbundle
$E|_{D_i}$
, but a subbundle 
 $F_i$
 of
$F_i$
 of 
 $E|_{U_i}$
 where
$E|_{U_i}$
 where 
 $U_i \subset X$
 is an open neighborhood of
$U_i \subset X$
 is an open neighborhood of 
 $D_i$
.
$D_i$
.
 We first construct the modification 
 , which is naturally isomorphic to E on
, which is naturally isomorphic to E on 
 $X {\smallsetminus} D_1$
, and so in particular on
$X {\smallsetminus} D_1$
, and so in particular on 
 $U_2 {\smallsetminus} D_1$
. If the data of a subbundle
$U_2 {\smallsetminus} D_1$
. If the data of a subbundle 
 $F_2 \subset E|_{U_2 {\smallsetminus} D_1}$
 extend to
$F_2 \subset E|_{U_2 {\smallsetminus} D_1}$
 extend to 
 $U_2$
, then it does so uniquely, and we may modify along
$U_2$
, then it does so uniquely, and we may modify along 
 $D_2$
 towards this extension. However, this subbundle may not extend to
$D_2$
 towards this extension. However, this subbundle may not extend to 
 $U_2$
. The following situation where it does will include all situations we shall need in this paper.
$U_2$
. The following situation where it does will include all situations we shall need in this paper.
Definition 3.3. Let 
 $M = \{(D_i, U_i, F_i)\}_{i \in I}$
 be a collection of modification data. For each point
$M = \{(D_i, U_i, F_i)\}_{i \in I}$
 be a collection of modification data. For each point 
 $x \in X$
, define
$x \in X$
, define 
 $I_x \subseteq I$
 to be the set of indices for which
$I_x \subseteq I$
 to be the set of indices for which 
 $x \in D_i$
. We say that M is tree-like if for all
$x \in D_i$
. We say that M is tree-like if for all 
 $x \in X$
, and all subsets
$x \in X$
, and all subsets 
 $I' \subset I_x$
, the following condition holds: Whenever the fibers
$I' \subset I_x$
, the following condition holds: Whenever the fibers 
 $\{F_i|_x\}_{i \in I'}$
 are dependent, there exist indices
$\{F_i|_x\}_{i \in I'}$
 are dependent, there exist indices 
 $i,j \in I'$
 and an open
$i,j \in I'$
 and an open 
 $U \subseteq U_i \cap U_j$
 containing x such that
$U \subseteq U_i \cap U_j$
 containing x such that 
 $F_i|_U \subseteq F_j|_U$
.
$F_i|_U \subseteq F_j|_U$
.
 By [Reference Atanasov, Larson and Yang2, Proposition 2.17], we can transfer modification data as above when it is treelike. That is, given modification data M for E such that 
 $\{(D, U, F)\} \cup M$
 is treelike, we obtain modification data
$\{(D, U, F)\} \cup M$
 is treelike, we obtain modification data 
 $M'$
 for
$M'$
 for 
 . In this way, we inductively define the multiple modification
. In this way, we inductively define the multiple modification 
 $E[M]$
 for treelike modification data M. This is independent of the order in which the modifications from M are applied [Reference Atanasov, Larson and Yang2, Proposition 2.20].
$E[M]$
 for treelike modification data M. This is independent of the order in which the modifications from M are applied [Reference Atanasov, Larson and Yang2, Proposition 2.20].
Example 3.4. A simplifying special case is when F is a direct summand of E. Writing 
 $E \simeq F \oplus E'$
,
$E \simeq F \oplus E'$
, 

In nice cases, a short exact sequence of vector bundles induces a short exact sequence of modifications. To make this more precise, consider a short exact sequence
 $$ \begin{align} 0 \to S \to E \to Q \to 0. \end{align} $$
$$ \begin{align} 0 \to S \to E \to Q \to 0. \end{align} $$
For example, first suppose that 
 $F \cap S$
 is flat over the base X. Then equation (3.1) induces the short exact sequence
$F \cap S$
 is flat over the base X. Then equation (3.1) induces the short exact sequence 

 A more interesting example is when the base is a curve 
 $X = C$
, and
$X = C$
, and 
 $F \subset E$
 is a line subbundle, and
$F \subset E$
 is a line subbundle, and 
 $D = np$
, where
$D = np$
, where 
 $p \in C$
 is a smooth point. Then we obtain an induced sequence for the modification
$p \in C$
 is a smooth point. Then we obtain an induced sequence for the modification 
 , as follows. Define
, as follows. Define 
 $k'$
 to be the order to which F is contained in S in a neighborhood of p. In other words, if F is not contained in S, this is the length of the subscheme
$k'$
 to be the order to which F is contained in S in a neighborhood of p. In other words, if F is not contained in S, this is the length of the subscheme 
 $\mathbb {P} S \cap \mathbb {P} F$
 in
$\mathbb {P} S \cap \mathbb {P} F$
 in 
 $\mathbb {P} E$
; if F is contained in S, this is
$\mathbb {P} E$
; if F is contained in S, this is 
 $\infty $
. Let
$\infty $
. Let 
 $k = \max (k', n)$
. Then equation (3.1) induces the short exact sequence
$k = \max (k', n)$
. Then equation (3.1) induces the short exact sequence 

where 
 $\overline {F}$
 is the saturation of the image of F in Q. When
$\overline {F}$
 is the saturation of the image of F in Q. When 
 $k' = \infty $
 or
$k' = \infty $
 or 
 $k' = 0$
, this agrees with equation (3.2).
$k' = 0$
, this agrees with equation (3.2).
3.2 Pointing bundles
 Given an unramified map 
 $f \colon C \to \mathbb {P}^r$
, the sheaf
$f \colon C \to \mathbb {P}^r$
, the sheaf 
 $N_f = \ker (f^*\Omega _{\mathbb {P}^r} \to \Omega _X)^\vee $
 is a vector bundle, which we refer to as the normal bundle of the map f
. In almost all cases that we shall consider, f will be an embedding, in which case
$N_f = \ker (f^*\Omega _{\mathbb {P}^r} \to \Omega _X)^\vee $
 is a vector bundle, which we refer to as the normal bundle of the map f
. In almost all cases that we shall consider, f will be an embedding, in which case 
 $N_f = N_C$
 coincides with the normal bundle of the image.
$N_f = N_C$
 coincides with the normal bundle of the image.
 We will primarily deal with modifications of 
 $N_f$
 towards pointing subbundles
$N_f$
 towards pointing subbundles 
 $N_{f \to \Lambda }$
, whose definition we now recall. Let
$N_{f \to \Lambda }$
, whose definition we now recall. Let 
 $\Lambda \subset \mathbb {P}^r$
 be a linear space of dimension
$\Lambda \subset \mathbb {P}^r$
 be a linear space of dimension 
 $\lambda $
. Let
$\lambda $
. Let 
 $\pi _{\Lambda } \circ f$
 denote the composition of f with the projection map
$\pi _{\Lambda } \circ f$
 denote the composition of f with the projection map 
 $$\begin{align*}\pi_{\Lambda} \colon \mathbb{P}^r \dashrightarrow \mathbb{P}^{r-\lambda -1}. \end{align*}$$
$$\begin{align*}\pi_{\Lambda} \colon \mathbb{P}^r \dashrightarrow \mathbb{P}^{r-\lambda -1}. \end{align*}$$
Let 
 $U =U_{\Lambda }$
 denote the open locus of
$U =U_{\Lambda }$
 denote the open locus of 
 $C {\smallsetminus} (\Lambda \cap C)$
 where
$C {\smallsetminus} (\Lambda \cap C)$
 where 
 $\pi _{\Lambda } \circ f$
 is unramified; explicitly this is the locus of points of C whose tangent space does not meet
$\pi _{\Lambda } \circ f$
 is unramified; explicitly this is the locus of points of C whose tangent space does not meet 
 $\Lambda $
. Assuming that U is dense and contains the singular locus of C, we may define
$\Lambda $
. Assuming that U is dense and contains the singular locus of C, we may define 
 $N_{f \to \Lambda }$
 as the unique subbundle of
$N_{f \to \Lambda }$
 as the unique subbundle of 
 $N_f$
 whose restriction to U is
$N_f$
 whose restriction to U is 
 $$\begin{align*}\ker(N_f|_U \to N_{\pi_{\Lambda} \circ f}|_U).\end{align*}$$
$$\begin{align*}\ker(N_f|_U \to N_{\pi_{\Lambda} \circ f}|_U).\end{align*}$$
The notation 
 $N_{f \to \Lambda }$
 is evocative of the geometry of sections of
$N_{f \to \Lambda }$
 is evocative of the geometry of sections of 
 $N_{f \to \Lambda }$
: Informally speaking, they ‘point towards’ the subspace
$N_{f \to \Lambda }$
: Informally speaking, they ‘point towards’ the subspace 
 $\Lambda \subset \mathbb {P}^r$
. When f is an embedding, we write
$\Lambda \subset \mathbb {P}^r$
. When f is an embedding, we write 
 $N_{C \to \Lambda } = N_{f \to \Lambda }$
. If the projection
$N_{C \to \Lambda } = N_{f \to \Lambda }$
. If the projection 
 $(\pi _\Lambda \circ f) \colon C \to \mathbb {P}^{r-\lambda -1}$
 is unramified, then
$(\pi _\Lambda \circ f) \colon C \to \mathbb {P}^{r-\lambda -1}$
 is unramified, then 
 $N_{f \to \Lambda }$
 sits in the pointing bundle exact sequence
$N_{f \to \Lambda }$
 sits in the pointing bundle exact sequence 
 $$ \begin{align} 0 \to N_{f \to \Lambda} \to N_f \to N_{\pi_\Lambda \circ f} (\Lambda \cap C) \to 0. \end{align} $$
$$ \begin{align} 0 \to N_{f \to \Lambda} \to N_f \to N_{\pi_\Lambda \circ f} (\Lambda \cap C) \to 0. \end{align} $$
The same definitions work for families of curves in a projective bundle. For a treatment in this more general setting, see [Reference Atanasov, Larson and Yang2, Section 5].
 The simplest case, and our primary interest, is when 
 $\Lambda = p$
 is a point in
$\Lambda = p$
 is a point in 
 $\mathbb {P}^r$
. In this case, by [Reference Atanasov, Larson and Yang2, Propositions 6.2 and 6.3], we have the following explicit descriptions:
$\mathbb {P}^r$
. In this case, by [Reference Atanasov, Larson and Yang2, Propositions 6.2 and 6.3], we have the following explicit descriptions: 
- 
• If  $p \in \mathbb {P}^r$
 is a general point (in which case $p \in \mathbb {P}^r$
 is a general point (in which case $U_p = C$
), then $U_p = C$
), then $N_{f \to p} \simeq f^*\mathcal {O}_{\mathbb {P}^r}(1)$
. $N_{f \to p} \simeq f^*\mathcal {O}_{\mathbb {P}^r}(1)$
.
- 
• If  $p \in C$
 is a general point (in which case $p \in C$
 is a general point (in which case $U_p = C {\smallsetminus} p$
), then $U_p = C {\smallsetminus} p$
), then $N_{f \to p} \simeq f^*\mathcal {O}_{\mathbb {P}^r}(1)(2p)$
. $N_{f \to p} \simeq f^*\mathcal {O}_{\mathbb {P}^r}(1)(2p)$
.
When modifying towards a pointing bundle, we use the simpler notation

For two points 
 $p, q \in C$
, we also define the more compact notation
$p, q \in C$
, we also define the more compact notation 

 We now restate a foundational result of Hartshorne–Hirschowitz, which describes the normal bundle of a nodal curve in projective space, in this language of pointing bundles. Let 
 $X \cup _{\Gamma } Y$
 be a reducible nodal curve. For each point
$X \cup _{\Gamma } Y$
 be a reducible nodal curve. For each point 
 $p_i \in \Gamma $
, let
$p_i \in \Gamma $
, let 
 $q_i$
 denote any point on
$q_i$
 denote any point on 
 $T_{p_i}Y {\smallsetminus} p_i$
. For simplicity, we introduce the following notation. For any subset
$T_{p_i}Y {\smallsetminus} p_i$
. For simplicity, we introduce the following notation. For any subset 
 $\Gamma ' = \{p_1, \dots p_n\} \subseteq \Gamma $
, we write
$\Gamma ' = \{p_1, \dots p_n\} \subseteq \Gamma $
, we write 

When 
 $\Gamma ' = \Gamma $
 is the full set of points where
$\Gamma ' = \Gamma $
 is the full set of points where 
 $X $
 and Y meet, we simplify further and write
$X $
 and Y meet, we simplify further and write 

(We analogously define
 
and
 
.)
Proposition 3.5 [Reference Hartshorne and Hirschowitz18, Corollary 3.2]
 As above, let 
 $X \cup Y \subseteq \mathbb {P}^r$
 be a reducible nodal curve. Then
$X \cup Y \subseteq \mathbb {P}^r$
 be a reducible nodal curve. Then 

3.3 Interpolation for vector bundles
3.3.1 Interpolation for bundles on nodal curves
 Let C be a nodal curve, and let E be a nonspecial vector bundle on C. Twisting down by a point can only decrease 
 $h^0$
 and can only increase
$h^0$
 and can only increase 
 $h^1$
. Therefore, E satisfies interpolation provided that there exist two divisors
$h^1$
. Therefore, E satisfies interpolation provided that there exist two divisors 
 $D_+$
 and
$D_+$
 and 
 $D_-$
 for which
$D_-$
 for which 
 $$ \begin{align} h^0(E(-D_+)) = 0, \quad h^1(E( - D_-)) = 0, \quad \text{and} \quad \deg D_+ - \deg D_- \leq 1. \end{align} $$
$$ \begin{align} h^0(E(-D_+)) = 0, \quad h^1(E( - D_-)) = 0, \quad \text{and} \quad \deg D_+ - \deg D_- \leq 1. \end{align} $$
 Alternatively, if E has rank r, then twisting down by a point either decreases 
 $h^0$
 by r or increases
$h^0$
 by r or increases 
 $h^1$
. Thus, E satisfies interpolation if and only if, for every
$h^1$
. Thus, E satisfies interpolation if and only if, for every 
 $n> 0$
, some collection of n points impose the expected number of conditions:
$n> 0$
, some collection of n points impose the expected number of conditions: 
 $$\begin{align*}h^0(E(-p_1 - \cdots - p_n)) = \max(0, h^0(E) - rn) \qquad \text{for some}\ p_1, \dots, p_n.\end{align*}$$
$$\begin{align*}h^0(E(-p_1 - \cdots - p_n)) = \max(0, h^0(E) - rn) \qquad \text{for some}\ p_1, \dots, p_n.\end{align*}$$
More generally, we can use this idea to define interpolation for a space of sections of a vector bundle. Given 
 $V \subseteq H^0(E)$
, write
$V \subseteq H^0(E)$
, write 

We say that 
 $V \subseteq H^0(E)$
 satisfies interpolation if
$V \subseteq H^0(E)$
 satisfies interpolation if 
 $H^1(E) = 0$
 and, for every
$H^1(E) = 0$
 and, for every 
 $n> 0$
, there are n points
$n> 0$
, there are n points 
 $p_1, \dots , p_n$
 such that
$p_1, \dots , p_n$
 such that 
 $$\begin{align*}\dim V(-p_1 - \cdots - p_n) = \max(0,\dim V - rn).\end{align*}$$
$$\begin{align*}\dim V(-p_1 - \cdots - p_n) = \max(0,\dim V - rn).\end{align*}$$
 The following basic result allows us to reduce interpolation for a vector bundle E on a reducible curve 
 $X \cup _{\Gamma } Y$
 to interpolation for a space of sections on one component.
$X \cup _{\Gamma } Y$
 to interpolation for a space of sections on one component.
Lemma 3.6 [Reference Larson and Vogt23, Lemma 2.10]
 Let E be a vector bundle on 
 $X \cup _{\Gamma } Y$
. If the restriction map on Y
$X \cup _{\Gamma } Y$
. If the restriction map on Y 
 $$\begin{align*}\operatorname{res}_{Y,\Gamma} \colon H^0(Y, E|_Y) \to E|_\Gamma\end{align*}$$
$$\begin{align*}\operatorname{res}_{Y,\Gamma} \colon H^0(Y, E|_Y) \to E|_\Gamma\end{align*}$$
is injective and the space of sections 
 $\{\sigma \in H^0(X, E|_X) : \sigma |_\Gamma \in \operatorname {Im}(\operatorname {res}_{Y, \Gamma })\}$
 has dimension
$\{\sigma \in H^0(X, E|_X) : \sigma |_\Gamma \in \operatorname {Im}(\operatorname {res}_{Y, \Gamma })\}$
 has dimension 
 $\chi (E)$
 and satisfies interpolation, then E satisfies interpolation.
$\chi (E)$
 and satisfies interpolation, then E satisfies interpolation.
 The main case of interest in this paper is when 
 $Y = R$
 is a rational curve and
$Y = R$
 is a rational curve and 
 $E|_R$
 is perfectly balanced, that is,
$E|_R$
 is perfectly balanced, that is, 
 $E|_R \simeq \mathcal {O}_{\mathbb {P}^1}(a)^{\oplus r}$
 for some
$E|_R \simeq \mathcal {O}_{\mathbb {P}^1}(a)^{\oplus r}$
 for some 
 $a \in \mathbb {Z}$
.
$a \in \mathbb {Z}$
.
Lemma 3.7. Let 
 $C \cup _{\Gamma } R$
 be a nodal curve with R rational, and let E be a vector bundle on
$C \cup _{\Gamma } R$
 be a nodal curve with R rational, and let E be a vector bundle on 
 $C \cup R$
 with
$C \cup R$
 with 
 $E|_R$
 perfectly balanced of slope at least
$E|_R$
 perfectly balanced of slope at least 
 $\#\Gamma - 1$
. If
$\#\Gamma - 1$
. If 
 $E|_C$
 satisfies interpolation, then E satisfies interpolation.
$E|_C$
 satisfies interpolation, then E satisfies interpolation.
Proof. Let D be an effective divisor of degree 
 $\mu (E|_R) - \#\Gamma + 1$
 supported on
$\mu (E|_R) - \#\Gamma + 1$
 supported on 
 $R {\smallsetminus} \Gamma $
. It suffices to prove that
$R {\smallsetminus} \Gamma $
. It suffices to prove that 
 $E(-D)$
 satisfies interpolation, as we now explain. Let
$E(-D)$
 satisfies interpolation, as we now explain. Let 
 $D_+$
 and
$D_+$
 and 
 $D_-$
 be the two divisors satisfying equation (3.5) for the vector bundle
$D_-$
 be the two divisors satisfying equation (3.5) for the vector bundle 
 $E(-D)$
. Then the divisors
$E(-D)$
. Then the divisors 
 $D_+ + D$
 and
$D_+ + D$
 and 
 $D_-+D$
 satisfy equation (3.5) for the vector bundle E. The bundle
$D_-+D$
 satisfy equation (3.5) for the vector bundle E. The bundle 
 $E(-D)|_R$
 is perfectly balanced of slope
$E(-D)|_R$
 is perfectly balanced of slope 
 $\#\Gamma - 1$
, and so
$\#\Gamma - 1$
, and so 
 $\operatorname {res}_{R, \Gamma }$
 is an isomorphism. The result now follows from Lemma 3.6.
$\operatorname {res}_{R, \Gamma }$
 is an isomorphism. The result now follows from Lemma 3.6.
3.3.2 Interpolation and twists
 If E satisfies interpolation, then as in the proof of Lemma 3.7, the twist 
 $E(D)$
 by any effective divisor D also satisfies interpolation. Conversely, we have the following.
$E(D)$
 by any effective divisor D also satisfies interpolation. Conversely, we have the following.
Lemma 3.8 [Reference Atanasov, Larson and Yang2, Proposition 4.12]
Suppose that E is a vector bundle on a genus g curve such that
 $$\begin{align*}\chi(E) \geq \operatorname{rk}(E) \cdot g.\end{align*}$$
$$\begin{align*}\chi(E) \geq \operatorname{rk}(E) \cdot g.\end{align*}$$
If there exists an effective divisor D for which 
 $E(D)$
 satisfies interpolation, then E also satisfies interpolation.
$E(D)$
 satisfies interpolation, then E also satisfies interpolation.
3.3.3 Interpolation and modifications
 Consider a vector bundle E and its modification 
 . Given sufficient generality of either p or F, or if the slope
. Given sufficient generality of either p or F, or if the slope 
 $\mu (E) \in \mathbb {Z}$
, it is sometimes possible to deduce that
$\mu (E) \in \mathbb {Z}$
, it is sometimes possible to deduce that 
 satisfies interpolation from the assumption that E satisfies interpolation.
 satisfies interpolation from the assumption that E satisfies interpolation.
Lemma 3.9. Let E be a vector bundle on C, let 
 $p \in C$
 be a smooth point and let
$p \in C$
 be a smooth point and let 
 $F \subseteq E|_p$
. If E satisfies interpolation and
$F \subseteq E|_p$
. If E satisfies interpolation and 
 $\mu (E) \in \mathbb {Z}$
, then
$\mu (E) \in \mathbb {Z}$
, then 
 satisfies interpolation.
 satisfies interpolation.
Proof. Since 
 $\mu (E) \in \mathbb {Z}$
, there is an effective divisor D with
$\mu (E) \in \mathbb {Z}$
, there is an effective divisor D with 
 $$\begin{align*}H^0(E(-D)) = 0 \quad \text{and} \quad H^1(E(-D)) = 0.\end{align*}$$
$$\begin{align*}H^0(E(-D)) = 0 \quad \text{and} \quad H^1(E(-D)) = 0.\end{align*}$$
Then

Definition 3.10. We say that a collection of subspaces 
 $\{W_b\}_{b \in B}$
 of a vector space V are linearly general if, for any subspace
$\{W_b\}_{b \in B}$
 of a vector space V are linearly general if, for any subspace 
 $U \subset V$
, there is some
$U \subset V$
, there is some 
 $b \in B$
 so that
$b \in B$
 so that 
 $W_b$
 is transverse to U.
$W_b$
 is transverse to U.
Lemma 3.11 [Reference Atanasov, Larson and Yang2, Proposition 4.10]
 Let E be a vector bundle on C, and let 
 $p \in C$
 be a smooth point. Let
$p \in C$
 be a smooth point. Let 
 $\{F_b\}_{b \in B}$
 be a collection of subspaces of
$\{F_b\}_{b \in B}$
 be a collection of subspaces of 
 $E|_p$
 that all contain a fixed subspace
$E|_p$
 that all contain a fixed subspace 
 $F_0$
. Suppose that both E and
$F_0$
. Suppose that both E and 
 $E[p \to F_0]$
 satisfy interpolation. If the collection
$E[p \to F_0]$
 satisfy interpolation. If the collection 
 $\{F_b/F_0\}_{b \in B}$
 is linearly general in
$\{F_b/F_0\}_{b \in B}$
 is linearly general in 
 $E|_p/F_0$
, then for some
$E|_p/F_0$
, then for some 
 $b\in B$
, the positive modification
$b\in B$
, the positive modification 
 satisfies interpolation.
 satisfies interpolation.
Lemma 3.12 [Reference Atanasov, Larson and Yang2, Proposition 4.21 for 
 $n=1$
]
$n=1$
]
 Suppose E satisfies interpolation, 
 $L \subset E$
 is a nonspecial line subbundle, and the quotient
$L \subset E$
 is a nonspecial line subbundle, and the quotient 
 $Q = E/L$
 also satisfies interpolation. If
$Q = E/L$
 also satisfies interpolation. If 
 $\mu (L) \leq \mu (H)$
, then
$\mu (L) \leq \mu (H)$
, then 
 satisfies interpolation.
 satisfies interpolation.
Remark 3.13. While [Reference Atanasov, Larson and Yang2] assumes characteristic zero, none of the specific results we quote from [Reference Atanasov, Larson and Yang2] use this assumption – except for Proposition 4.21. This proposition states that if 
 $\mu (L) \leq \mu (H) + n - 1$
, then
$\mu (L) \leq \mu (H) + n - 1$
, then 
 satisfies interpolation. The proof uses that vanishing at
 satisfies interpolation. The proof uses that vanishing at 
 $np$
 imposes n conditions on sections of any linear series. This is true in any characteristic when
$np$
 imposes n conditions on sections of any linear series. This is true in any characteristic when 
 $n=1$
, but fails in positive characteristic for
$n=1$
, but fails in positive characteristic for 
 $n> 1$
. Since we will use Proposition 4.21 of [Reference Atanasov, Larson and Yang2] only when
$n> 1$
. Since we will use Proposition 4.21 of [Reference Atanasov, Larson and Yang2] only when 
 $n = 1$
, we do not need a restriction on the characteristic.
$n = 1$
, we do not need a restriction on the characteristic.
Lemma 3.14. Let E be a vector bundle on an irreducible curve C. Let 
 $p_1, \ldots , p_n \in C$
 be points, and let
$p_1, \ldots , p_n \in C$
 be points, and let 
 $L_i \subseteq E|_{p_i}$
 be one-dimensional subspaces. Suppose that both E and
$L_i \subseteq E|_{p_i}$
 be one-dimensional subspaces. Suppose that both E and 
 
 satisfy interpolation. Then for any 
 $0 <m <n$
, there is a collection of distinct indices
$0 <m <n$
, there is a collection of distinct indices 
 $i_1, \dots , i_m$
 such that
$i_1, \dots , i_m$
 such that 

satisfies interpolation.
Proof. By induction on n we reduce to the case 
 $m=n-1$
. Write
$m=n-1$
. Write 
 and
 and 
 $N = \lceil \chi (E') / \operatorname {rk} E'\rceil $
. Let
$N = \lceil \chi (E') / \operatorname {rk} E'\rceil $
. Let 
 $D_N$
 and
$D_N$
 and 
 $D_{N-1}$
 be general divisors of degrees N and
$D_{N-1}$
 be general divisors of degrees N and 
 $N - 1$
, respectively. Since E and
$N - 1$
, respectively. Since E and 
 $E'$
 both satisfy interpolation and
$E'$
 both satisfy interpolation and 
 $\chi (E) < \chi (E')$
, we have
$\chi (E) < \chi (E')$
, we have 
 $$\begin{align*}h^0(E'(-D_N)) = 0, \quad h^0(E'(-D_{N-1})) \neq 0, \quad \text{and} \quad h^0(E(-D_{N-1})) < h^0(E'(-D_{N-1})).\end{align*}$$
$$\begin{align*}h^0(E'(-D_N)) = 0, \quad h^0(E'(-D_{N-1})) \neq 0, \quad \text{and} \quad h^0(E(-D_{N-1})) < h^0(E'(-D_{N-1})).\end{align*}$$
Let 
 . Since
. Since 
 $\chi (E_i) = \chi (E') - 1$
, it suffices to show
$\chi (E_i) = \chi (E') - 1$
, it suffices to show 
 $h^0(E_i(-D_{N-1})) < h^0(E'(-D_{N-1}))$
 for some i. This follows from the fact that
$h^0(E_i(-D_{N-1})) < h^0(E'(-D_{N-1}))$
 for some i. This follows from the fact that 
 $$\begin{align*}\bigcap_i H^0(E_i(-D_{N-1})) = H^0(E(-D_{N-1})) \subsetneq H^0(E'(-D_{N-1})).\\[-49pt] \end{align*}$$
$$\begin{align*}\bigcap_i H^0(E_i(-D_{N-1})) = H^0(E(-D_{N-1})) \subsetneq H^0(E'(-D_{N-1})).\\[-49pt] \end{align*}$$
3.3.4 Interpolation and short exact sequences
Lemma 3.15. Consider an exact sequence
 $$\begin{align*}0 \to S \to E \to Q \to 0\end{align*}$$
$$\begin{align*}0 \to S \to E \to Q \to 0\end{align*}$$
of vector bundles on an irreducible curve C. Suppose that S and Q satisfy interpolation and
 $$ \begin{align}\mu(S) \leq \lfloor \mu(Q) \rfloor + 1 \quad \text{and} \quad \mu(Q) \leq \lfloor \mu(S) \rfloor +1.\end{align} $$
$$ \begin{align}\mu(S) \leq \lfloor \mu(Q) \rfloor + 1 \quad \text{and} \quad \mu(Q) \leq \lfloor \mu(S) \rfloor +1.\end{align} $$
Then E also satisfies interpolation.
Proof. Since S and Q are nonspecial, E is nonspecial. By (3.6), there exists an integer 
 $n \in \mathbb {Z}$
 such that
$n \in \mathbb {Z}$
 such that 
 $\mu (S)$
 and
$\mu (S)$
 and 
 $\mu (Q)$
 are contained in the closed interval
$\mu (Q)$
 are contained in the closed interval 
 $[n, n+1]$
. Since (3.5) is satisfied for
$[n, n+1]$
. Since (3.5) is satisfied for 
 $D_+$
 a general divisor of degree
$D_+$
 a general divisor of degree 
 $n + 1$
, and
$n + 1$
, and 
 $D_-$
 a general divisor of degree n, we conclude that E satisfies interpolation as desired.
$D_-$
 a general divisor of degree n, we conclude that E satisfies interpolation as desired.
We will most often use this result in the special case in which S is a line subbundle of E.
Corollary 3.16. Suppose that 
 $S \subset E$
 is a nonspecial line subbundle and
$S \subset E$
 is a nonspecial line subbundle and 
 $|\mu (S) - \mu (E)| < 1$
. If the quotient
$|\mu (S) - \mu (E)| < 1$
. If the quotient 
 $Q = E/S$
 satisfies interpolation, then E does as well.
$Q = E/S$
 satisfies interpolation, then E does as well.
Proof. By assumption 
 $-1 < \frac {\deg (Q) - \operatorname {rk}(Q)\deg (S)}{\operatorname {rk}(Q)+1} < 1.$
 Hence, we have strict inequalities
$-1 < \frac {\deg (Q) - \operatorname {rk}(Q)\deg (S)}{\operatorname {rk}(Q)+1} < 1.$
 Hence, we have strict inequalities 
 $$\begin{align*}\deg(S) < \mu(Q) + \frac{\operatorname{rk}(Q) + 1}{\operatorname{rk}(Q)} \quad \text{and} \quad \mu(Q) < \deg(S) + \frac{\operatorname{rk}(Q) + 1}{\operatorname{rk}(Q)},\end{align*}$$
$$\begin{align*}\deg(S) < \mu(Q) + \frac{\operatorname{rk}(Q) + 1}{\operatorname{rk}(Q)} \quad \text{and} \quad \mu(Q) < \deg(S) + \frac{\operatorname{rk}(Q) + 1}{\operatorname{rk}(Q)},\end{align*}$$
which imply the required inequalities in Lemma 3.15, since 
 $\mu (Q)$
 is a
$\mu (Q)$
 is a 
 $[1/\operatorname {rk}(Q)]$
-integer.
$[1/\operatorname {rk}(Q)]$
-integer.
4 Overview
4.1 Base cases
 We can reduce the number of base cases by extending Theorem 1.4 to 
 $r=1$
 and
$r=1$
 and 
 $r=2$
. For
$r=2$
. For 
 $r=2$
, we replace
$r=2$
, we replace 
 $N_C$
 with the normal sheaf
$N_C$
 with the normal sheaf 
 $N_f$
, where
$N_f$
, where 
 $f \colon C \to \mathbb {P}^2$
 is a general BN-curve. In this case, adjunction implies that
$f \colon C \to \mathbb {P}^2$
 is a general BN-curve. In this case, adjunction implies that 
 $N_f = K_C \otimes f^*\mathcal {O}_{\mathbb {P}^2}(3)$
 is a nonspecial line bundle, and therefore satisfies interpolation. For
$N_f = K_C \otimes f^*\mathcal {O}_{\mathbb {P}^2}(3)$
 is a nonspecial line bundle, and therefore satisfies interpolation. For 
 $r=1$
, we only consider the case where
$r=1$
, we only consider the case where 
 $f \colon C \to \mathbb {P}^1$
 is an isomorphism, so
$f \colon C \to \mathbb {P}^1$
 is an isomorphism, so 
 $N_f = 0 $
 satisfies interpolation.
$N_f = 0 $
 satisfies interpolation.
4.2 first strategy: degeneration of C
 The first inductive strategy we will use is degeneration of C to reducible curves 
 $X \cup Y$
. In Section 5 we will study certain such degenerations, for which Y has a prescribed form, and we can thus relate interpolation for
$X \cup Y$
. In Section 5 we will study certain such degenerations, for which Y has a prescribed form, and we can thus relate interpolation for 
 $N_C$
 to interpolation for certain modifications of
$N_C$
 to interpolation for certain modifications of 
 $N_X$
.
$N_X$
.
 Since the sectional monodromy group of a general BN-curve always contains the alternating group [Reference Kadets20], and in particular is 
 $(r+1)$
-transitive, it makes sense to talk of a general
$(r+1)$
-transitive, it makes sense to talk of a general 
 $(r+1)$
-secant rational curve of degree
$(r+1)$
-secant rational curve of degree 
 $r-1$
 in a hyperplane. While the following hypothesis does not encompass all modifications that might appear using this method, it includes those modifications that will play the most central role in our inductive argument:
$r-1$
 in a hyperplane. While the following hypothesis does not encompass all modifications that might appear using this method, it includes those modifications that will play the most central role in our inductive argument:
Hypothesis 4.1 (
 $I(d, g, r, \ell , m)$
)
$I(d, g, r, \ell , m)$
)
 Let 
 $C \subset \mathbb {P}^r$
 be a general BN-curve of degree d and genus g. Let
$C \subset \mathbb {P}^r$
 be a general BN-curve of degree d and genus g. Let 
 $u_1, v_1, \dots , u_\ell , v_\ell $
 be
$u_1, v_1, \dots , u_\ell , v_\ell $
 be 
 $\ell $
 pairs of general points on C. Let
$\ell $
 pairs of general points on C. Let 
 $R_1, \dots , R_m$
 be m general
$R_1, \dots , R_m$
 be m general 
 $(r+1)$
-secant rational curves of degree
$(r+1)$
-secant rational curves of degree 
 $r - 1$
 (contained in hyperplanes transverse to C). Then the modification
$r - 1$
 (contained in hyperplanes transverse to C). Then the modification 

of the normal bundle of C satisfies interpolation.
 A central complicating factor is that the inductive hypothesis 
 $I(d, g, r, \ell , m)$
 is not always true. The following definition describes a set of tuples
$I(d, g, r, \ell , m)$
 is not always true. The following definition describes a set of tuples 
 $(d, g, r, \ell , m)$
 for which we will prove that it holds.
$(d, g, r, \ell , m)$
 for which we will prove that it holds.
Definition 4.2. A tuple 
 $(d, g, r, \ell , m)$
 is called good if it satisfies all the following conditions:
$(d, g, r, \ell , m)$
 is called good if it satisfies all the following conditions: 
- 
• The following inequalities hold:  $$\begin{align*}d \geq g + r, \quad 0 \leq \ell \leq \frac{r}{2}, \quad \text{and} \quad 0\leq m \leq \rho(d,g,r).\end{align*}$$ $$\begin{align*}d \geq g + r, \quad 0 \leq \ell \leq \frac{r}{2}, \quad \text{and} \quad 0\leq m \leq \rho(d,g,r).\end{align*}$$
- 
• If  $g=m=0$
, then where for integers a and b we write $g=m=0$
, then where for integers a and b we write $$\begin{align*}2 \ell \geq (1-d) \% (r-1),\end{align*}$$ $$\begin{align*}2 \ell \geq (1-d) \% (r-1),\end{align*}$$ $a \% b$
 for the reduced residue of a modulo b, and $a \% b$
 for the reduced residue of a modulo b, and
- 
• It is not the following set: (XEx) $$ \begin{align} \left\{ \begin{array}{cccccc} (5, 2, 3, 0, 0), &(4, 1, 3, 1, 0), & (4, 1, 3, 0, 1), & (4, 1, 3, 1, 1), \\ (6, 2, 4, 0, 0), & (5, 1, 4, 1, 0),&& (5, 1, 4, 1, 1), & (5, 1, 4, 2, 1), & (6, 2, 4, 1, 1), \\ (7, 2, 5, 0, 0), &&(6, 1, 5, 0, 1), &(6, 1, 5, 1, 1). \end{array} \right\} \end{align} $$ $$ \begin{align} \left\{ \begin{array}{cccccc} (5, 2, 3, 0, 0), &(4, 1, 3, 1, 0), & (4, 1, 3, 0, 1), & (4, 1, 3, 1, 1), \\ (6, 2, 4, 0, 0), & (5, 1, 4, 1, 0),&& (5, 1, 4, 1, 1), & (5, 1, 4, 2, 1), & (6, 2, 4, 1, 1), \\ (7, 2, 5, 0, 0), &&(6, 1, 5, 0, 1), &(6, 1, 5, 1, 1). \end{array} \right\} \end{align} $$
Remark 4.3. By Clifford’s theorem, the first inequality 
 $d \geq g + r$
 follows from
$d \geq g + r$
 follows from 
 $g \leq r$
.
$g \leq r$
.
 We conclude Section 5 by using this first strategy to show that if 
 $I(d, g, r, 0, m)$
 holds for every good tuple
$I(d, g, r, 0, m)$
 holds for every good tuple 
 $(d, g, r, 0, m)$
, then Theorem 1.4 holds except possibly for rational curves or canonical curves of even genus.
$(d, g, r, 0, m)$
, then Theorem 1.4 holds except possibly for rational curves or canonical curves of even genus.
4.3 Second Strategy: Limits of Modifications and Projection
The basic issue with the first strategy described above is that every time we apply it, we get more modifications. In order to make an inductive argument work, we need a second inductive strategy that decreases the number of modifications.
 Hypothesis 
 $I(d,g,r,\ell , m)$
 asserts that
$I(d,g,r,\ell , m)$
 asserts that 
 
satisfies interpolation. Let p be a general point on C. The pointing bundle exact sequence induces the exact sequence

where 
 $\overline {C}$
 and
$\overline {C}$
 and 
 $\overline {R}_j$
 denote the images of C and
$\overline {R}_j$
 denote the images of C and 
 $R_j$
, respectively under projection from p. In order to apply Corollary 3.16 to relate interpolation for the original bundle
$R_j$
, respectively under projection from p. In order to apply Corollary 3.16 to relate interpolation for the original bundle 
 $N_C'$
 to interpolation for the quotient bundle, the sequence must be close to balanced. The failure of the sequence to be balanced is related to the quantity
$N_C'$
 to interpolation for the quotient bundle, the sequence must be close to balanced. The failure of the sequence to be balanced is related to the quantity 

We first apply these ideas in Section 6 to treat the family of good tuples 
 $(d, g,r, 0, 0)$
 with
$(d, g,r, 0, 0)$
 with 
 ${\delta (d, g, r, 0,0)=1}$
, which are difficult from the perspective of our more uniform inductive arguments.
${\delta (d, g, r, 0,0)=1}$
, which are difficult from the perspective of our more uniform inductive arguments.
 More generally, in order to make this sequence sufficiently close to balanced, we will appropriately specialize the points on C at which the modifications occur. To illustrate this idea in the simplest possible case, assume here that 
 $\ell \geq \lfloor \delta \rfloor $
. Since the points
$\ell \geq \lfloor \delta \rfloor $
. Since the points 
 $v_1, \dots , v_{\lfloor \delta \rfloor }$
 are general on C, we may specialize them all to the point p. This induces the specialization of
$v_1, \dots , v_{\lfloor \delta \rfloor }$
 are general on C, we may specialize them all to the point p. This induces the specialization of 
 $N_C'$
 to
$N_C'$
 to 

Using equation (3.1), the pointing bundle exact sequence becomes

By our auspicious choice to specialize exactly 
 ${\lfloor \delta \rfloor }$
 points to p, this sequence is now close enough to balanced to reduce to proving interpolation for the quotient bundle. Furthermore, since
${\lfloor \delta \rfloor }$
 points to p, this sequence is now close enough to balanced to reduce to proving interpolation for the quotient bundle. Furthermore, since 
 $u_1, \dots , u_{\lfloor \delta \rfloor }$
 are general, the modification at p in the quotient is linearly general and we can erase it by Lemma 3.12. It therefore suffices to prove interpolation for
$u_1, \dots , u_{\lfloor \delta \rfloor }$
 are general, the modification at p in the quotient is linearly general and we can erase it by Lemma 3.12. It therefore suffices to prove interpolation for 

which evidently has fewer modifications. However, there are two basic issues with this argument:
- 
1. In general, we might not have  $\ell \geq \lfloor \delta \rfloor $
. $\ell \geq \lfloor \delta \rfloor $
.
- 
2. Since  $\overline {R}_i$
 is still an $\overline {R}_i$
 is still an $(r+1)$
-secant curve of degree $(r+1)$
-secant curve of degree $r-1$
, the argument does not reduce to another case of our inductive hypothesis. $r-1$
, the argument does not reduce to another case of our inductive hypothesis.
To surmount both of these two difficulties, we will need to specialize the 
 $R_i$
 as well.
$R_i$
 as well.
 This second strategy will be fleshed out in Sections 7 and 8: In Section 7, we will study how to specialize the 
 $R_i$
 so that they can also contribute modifications to
$R_i$
 so that they can also contribute modifications to 
 $N_{C \to p}$
. Then in Section 8, we will refine the basic argument outlined above to use these degenerations of the
$N_{C \to p}$
. Then in Section 8, we will refine the basic argument outlined above to use these degenerations of the 
 $R_i$
 as well, and also explain further degenerations that will be necessary to reduce to another case of our inductive hypothesis.
$R_i$
 as well, and also explain further degenerations that will be necessary to reduce to another case of our inductive hypothesis.
4.4 Outline of the remainder of paper
 Section 9 is a brief interlude in which we use the inductive arguments of Section 8 to treat the case of rational curves not implied by 
 $I(d,g,r,\ell ,m)$
 for good tuples. We also explain the counterexamples to Theorem 1.4 that are not counterexamples to Theorem 1.2. At this point, we will have reduced both Theorem 1.4 and Theorem 1.2 to
$I(d,g,r,\ell ,m)$
 for good tuples. We also explain the counterexamples to Theorem 1.4 that are not counterexamples to Theorem 1.2. At this point, we will have reduced both Theorem 1.4 and Theorem 1.2 to 
 $I(d,g,r,\ell , m)$
 for good tuples, as well as Theorem 1.4 for canonical curves of even genus, which we treat at the end of the paper in Section 13. The intervening sections 10–12 inductively prove
$I(d,g,r,\ell , m)$
 for good tuples, as well as Theorem 1.4 for canonical curves of even genus, which we treat at the end of the paper in Section 13. The intervening sections 10–12 inductively prove 
 $I(d,g,r,\ell ,m)$
 for good tuples.
$I(d,g,r,\ell ,m)$
 for good tuples.
 In Section 10, we complete a purely combinatorial analysis, in which we show that the inductive arguments of Section 8 can be applied to reduce 
 $I(d, g, r, \ell , m)$
 for all good tuples to
$I(d, g, r, \ell , m)$
 for all good tuples to 
- 
•  $I(d, g, r, \ell , m)$
 for a certain large but finite list of sporadic cases $I(d, g, r, \ell , m)$
 for a certain large but finite list of sporadic cases $(d, g, r, \ell , m)$
 with $(d, g, r, \ell , m)$
 with $r \leq 13$
. $r \leq 13$
.
- 
• The infinite family of tuples  $(d, g, r, 0, 0)$
 with $(d, g, r, 0, 0)$
 with $\delta = 1$
, which was already treated in Section 6. $\delta = 1$
, which was already treated in Section 6.
In Section 11, we give a more complicated, yet more flexible, inductive argument in the style of those in Section 8 and verify by exhaustive computer search that it reduces the finitely many sporadic cases identified above to a managable list of 
 $30$
 base cases. These base cases are treated by ad-hoc techniques in Section 12.
$30$
 base cases. These base cases are treated by ad-hoc techniques in Section 12.
5 Basic degenerations
 In this section, we discuss the three basic degenerations of BN-curves, to reducible curves 
 $C \cup D$
, that we will use in the proof of Theorem 1.4. In each subsection, we will first show that these degenerations lie in the Brill–Noether component. We will then relate interpolation for
$C \cup D$
, that we will use in the proof of Theorem 1.4. In each subsection, we will first show that these degenerations lie in the Brill–Noether component. We will then relate interpolation for 
 $N_{C \cup D}$
, or a modification thereof, to interpolation for a particular modification of
$N_{C \cup D}$
, or a modification thereof, to interpolation for a particular modification of 
 $N_C$
.
$N_C$
.
 In what follows, write 
 $N_{C \cup D}'$
 for a modification of
$N_{C \cup D}'$
 for a modification of 
 $N_{C \cup D}$
 away from D. In other words,
$N_{C \cup D}$
 away from D. In other words, 
 $N_{C \cup D}'$
 is a vector bundle on
$N_{C \cup D}'$
 is a vector bundle on 
 $C \cup D$
, equipped with an isomorphism to
$C \cup D$
, equipped with an isomorphism to 
 $N_{C \cup D}$
 over a dense open subset of
$N_{C \cup D}$
 over a dense open subset of 
 $C \cup D$
 containing the entire curve D, and in particular containing a neighborhood U of
$C \cup D$
 containing the entire curve D, and in particular containing a neighborhood U of 
 $C \cap D$
 in C. Write
$C \cap D$
 in C. Write 
 $N_C'$
 for the bundle obtained by making the same modifications to
$N_C'$
 for the bundle obtained by making the same modifications to 
 $N_C$
. In other words,
$N_C$
. In other words, 
 $N_C'$
 is obtained from
$N_C'$
 is obtained from 
 $N_{C \cup D}'|_{C {\smallsetminus} (C \cap D)}$
 by gluing along
$N_{C \cup D}'|_{C {\smallsetminus} (C \cap D)}$
 by gluing along 
 $U {\smallsetminus} (C \cap D)$
 via our given isomorphism to
$U {\smallsetminus} (C \cap D)$
 via our given isomorphism to 
 $N_C|_U$
.
$N_C|_U$
.

5.1 Peeling off one-secant lines
 Our most basic degeneration will be when 
 $D = L$
 is a quasitransverse one-secant line. (Recall that two subschemes X and Y of a scheme Z are transverse (respectively, quasitransverse) at a point
$D = L$
 is a quasitransverse one-secant line. (Recall that two subschemes X and Y of a scheme Z are transverse (respectively, quasitransverse) at a point 
 $p \in X \cap Y$
 if the natural map of tangent spaces
$p \in X \cap Y$
 if the natural map of tangent spaces 
 $T_pX \oplus T_pY \to T_pZ$
 is surjective (respectively, either injective or surjective).) If C has degree d and genus g, then
$T_pX \oplus T_pY \to T_pZ$
 is surjective (respectively, either injective or surjective).) If C has degree d and genus g, then 
 $C \cup L$
 has degree
$C \cup L$
 has degree 
 $d+1$
 and genus g. Write
$d+1$
 and genus g. Write 
 $v \in L {\smallsetminus} \{u\}$
 for any other point on the line L.
$v \in L {\smallsetminus} \{u\}$
 for any other point on the line L.
Lemma 5.1. If C is a BN-curve, then 
 $C \cup L$
 is also a BN-curve.
$C \cup L$
 is also a BN-curve.
Proof. Generalizing C, we may suppose C is a general BN-curve. We will show 
 $H^1(T_{\mathbb {P}^r}|_{C \cup L}) =0$
, which implies that the map
$H^1(T_{\mathbb {P}^r}|_{C \cup L}) =0$
, which implies that the map 
 $C \cup L \to \mathbb {P}^r$
 may be lifted as
$C \cup L \to \mathbb {P}^r$
 may be lifted as 
 $C \cup L$
 is deformed to a general curve.
$C \cup L$
 is deformed to a general curve.
 Since C is a general BN-curve, 
 $H^1(T_{\mathbb {P}^r}|_C)=0$
 by the Gieseker–Petri theorem. Furthermore, we have
$H^1(T_{\mathbb {P}^r}|_C)=0$
 by the Gieseker–Petri theorem. Furthermore, we have 
 $H^1(T_{\mathbb {P}^r}|_L(-u)) =0$
 because
$H^1(T_{\mathbb {P}^r}|_L(-u)) =0$
 because 
 $T_{\mathbb {P}^r}|_L \simeq \mathcal {O}_{\mathbb {P}^1}(2) \oplus \mathcal {O}_{\mathbb {P}^1}(1)^{\oplus (r-1)}$
. This implies
$T_{\mathbb {P}^r}|_L \simeq \mathcal {O}_{\mathbb {P}^1}(2) \oplus \mathcal {O}_{\mathbb {P}^1}(1)^{\oplus (r-1)}$
. This implies 
 $H^1(T_{\mathbb {P}^r}|_{C \cup L}) = 0$
 as desired, using
$H^1(T_{\mathbb {P}^r}|_{C \cup L}) = 0$
 as desired, using 
 $$\begin{align*}0 \to T_{\mathbb{P}^r}|_L(-u) \to T_{\mathbb{P}^r}|_{C \cup L} \to T_{\mathbb{P}^r}|_C \to 0.\\[-38pt] \end{align*}$$
$$\begin{align*}0 \to T_{\mathbb{P}^r}|_L(-u) \to T_{\mathbb{P}^r}|_{C \cup L} \to T_{\mathbb{P}^r}|_C \to 0.\\[-38pt] \end{align*}$$
Lemma 5.2 (Lemma 8.5 of [Reference Atanasov, Larson and Yang2])
 If 
 satisfies interpolation, then so does
 satisfies interpolation, then so does 
 $N_{C \cup L}'$
.
$N_{C \cup L}'$
.
 When C is nonspecial with genus small relative to r, we can combine Lemma 5.2 with Lemma 3.8 to reduce to a positive modification of 
 $N_C'$
.
$N_C'$
.
Corollary 5.3. Suppose that 
 $N_C'$
 is a positive modification of
$N_C'$
 is a positive modification of 
 $N_C$
. If
$N_C$
. If 
 $d \geq g + r$
 and
$d \geq g + r$
 and 
 $g \leq r + 6$
 and
$g \leq r + 6$
 and 
 satisfies interpolation, then
 satisfies interpolation, then 
 $N_{C \cup L}'$
 satisfies interpolation.
$N_{C \cup L}'$
 satisfies interpolation.
5.2 Peeling off one-secant lines
 Our next basic degeneration will be to the union of a curve C and a quasitransverse one-secant line L, meeting C at points u and v. If C has degree d and genus g, then 
 $C \cup L$
 has degree
$C \cup L$
 has degree 
 $d+1$
 and genus
$d+1$
 and genus 
 $g+1$
.
$g+1$
.
Lemma 5.4. If C is a BN-curve, then 
 $C \cup L$
 is also a BN-curve.
$C \cup L$
 is also a BN-curve.
Proof. As in the proof of Lemma 5.1, we have that 
 $H^1(T_{\mathbb {P}^r}|_{C \cup L})=0$
 by combining
$H^1(T_{\mathbb {P}^r}|_{C \cup L})=0$
 by combining 
 $H^1(T_{\mathbb {P}^r}|_C)=0$
 (from the Gieseker–Petri theorem) and
$H^1(T_{\mathbb {P}^r}|_C)=0$
 (from the Gieseker–Petri theorem) and 
 $H^1(T_{\mathbb {P}^r}|_L(-u-v)) = 0$
 (from
$H^1(T_{\mathbb {P}^r}|_L(-u-v)) = 0$
 (from 
 $T_{\mathbb {P}^r}|_L \simeq \mathcal {O}_{\mathbb {P}^1}(2) \oplus \mathcal {O}_{\mathbb {P}^1}(1)^{\oplus r-1}$
).
$T_{\mathbb {P}^r}|_L \simeq \mathcal {O}_{\mathbb {P}^1}(2) \oplus \mathcal {O}_{\mathbb {P}^1}(1)^{\oplus r-1}$
).
 We generalize Lemma 5.2 to one-secant lines. In slightly greater generality, let 
 $p \in L {\smallsetminus} \{u, v\}$
 be a point, and
$p \in L {\smallsetminus} \{u, v\}$
 be a point, and 
 $\Lambda $
 be a linear space disjoint from the span of the tangent lines to C at u and v.
$\Lambda $
 be a linear space disjoint from the span of the tangent lines to C at u and v.
Lemma 5.5 (Slight generalization of Lemma 8.8 of [Reference Atanasov, Larson and Yang2])
 If 
 satisfies interpolation, then so does
 satisfies interpolation, then so does 
 .
.
Proof. Imitate the proof Lemma 8.8 of [Reference Atanasov, Larson and Yang2], mutatis mutandis.
 (In the notation of [Reference Atanasov, Larson and Yang2]: Take T instead to have dimension 
 $r - 5 - \dim \Lambda $
, where by convention
$r - 5 - \dim \Lambda $
, where by convention 
 $\dim \emptyset = -1$
, and use instead the decomposition
$\dim \emptyset = -1$
, and use instead the decomposition 
 $N_L \simeq N_{L \to x} \oplus N_{L \to y} \oplus N_{L \to \Lambda } \oplus N_{L \to T}$
.)
$N_L \simeq N_{L \to x} \oplus N_{L \to y} \oplus N_{L \to \Lambda } \oplus N_{L \to T}$
.)
 As in Corollary 5.3, we can reduce to a positive modification of 
 $N_C'$
.
$N_C'$
.
Corollary 5.6. Suppose that 
 $N_C'$
 is a positive modification of
$N_C'$
 is a positive modification of 
 $N_C$
. If
$N_C$
. If 
 $d \geq g + r$
 and
$d \geq g + r$
 and 
 $g \leq r + 6$
 and
$g \leq r + 6$
 and 
 satisfies interpolation, then
 satisfies interpolation, then 
 $N_{C \cup L}'$
 satisfies interpolation.
$N_{C \cup L}'$
 satisfies interpolation.
5.3 Peeling off rational normal curves in hyperplanes
 The final of our basic degenerations is to the union of a BN-curve C of degree at least 
 $r+1$
 in
$r+1$
 in 
 $\mathbb {P}^r$
, and an
$\mathbb {P}^r$
, and an 
 $(r+1)$
-secant rational curve R of degree
$(r+1)$
-secant rational curve R of degree 
 $r-1$
 contained in a hyperplane H. If C has degree d and genus g, then
$r-1$
 contained in a hyperplane H. If C has degree d and genus g, then 
 $C \cup R$
 has degree
$C \cup R$
 has degree 
 $d + r-1$
 and genus
$d + r-1$
 and genus 
 $g + r$
. Observe that
$g + r$
. Observe that 
 $$\begin{align*}\rho(d,g,r) = \rho(d + r -1, g+r, r) + 1.\end{align*}$$
$$\begin{align*}\rho(d,g,r) = \rho(d + r -1, g+r, r) + 1.\end{align*}$$
Lemma 5.7. If 
 $\rho (d, g, r) \geq 1$
, and C is a BN-curve, then
$\rho (d, g, r) \geq 1$
, and C is a BN-curve, then 
 $C \cup R$
 is also a BN-curve.
$C \cup R$
 is also a BN-curve.
Proof. Generalizing C, we may suppose C is a general BN-curve.
 If 
 $g> 0$
, then we can specialize C to the union
$g> 0$
, then we can specialize C to the union 
 $C' \cup L$
, where
$C' \cup L$
, where 
 $C'$
 is a general BN-curve of degree
$C'$
 is a general BN-curve of degree 
 $d-1$
 and genus
$d-1$
 and genus 
 $g-1$
, and L is a general one-secant line. Otherwise, if
$g-1$
, and L is a general one-secant line. Otherwise, if 
 $g = 0$
, then we can specialize C to the union
$g = 0$
, then we can specialize C to the union 
 $C' \cup L$
, where
$C' \cup L$
, where 
 $C'$
 is a general BN-curve of degree
$C'$
 is a general BN-curve of degree 
 $d-1$
 and genus g, and L is a general one-secant line. Either way, we can arrange for one of the points p of
$d-1$
 and genus g, and L is a general one-secant line. Either way, we can arrange for one of the points p of 
 $C \cap R$
 to specialize onto L, and the rest to specialize onto
$C \cap R$
 to specialize onto L, and the rest to specialize onto 
 $C'$
.
$C'$
.

 Write 
 . Note that this is a set of
. Note that this is a set of 
 $r + 1$
 or
$r + 1$
 or 
 $r+2$
 points on
$r+2$
 points on 
 $C'$
. We will show the following:
$C'$
. We will show the following: 
- 
(a) The curve  $C' \cup L \cup R$
 is a smooth point of the Hilbert scheme. $C' \cup L \cup R$
 is a smooth point of the Hilbert scheme.
- 
(b) The curve  $L \cup R$
 can be smoothed to a rational normal curve M while preserving the points of incidence with $L \cup R$
 can be smoothed to a rational normal curve M while preserving the points of incidence with $C'$
. $C'$
.
- 
(c) The curve  $C' \cup M$
 is in the Brill–Noether component. $C' \cup M$
 is in the Brill–Noether component.
By part (a), the curves 
 $C \cup R$
 and
$C \cup R$
 and 
 $C' \cup M$
 are both generalizations of a smooth point of the Hilbert scheme. Hence, they are in the same component, which must be the Brill–Noether component by part (c).
$C' \cup M$
 are both generalizations of a smooth point of the Hilbert scheme. Hence, they are in the same component, which must be the Brill–Noether component by part (c).
 Beginning with part (b), write N for the subsheaf of 
 $N_{L \cup R}$
 whose sections fail to smooth the node p. It suffices to show that
$N_{L \cup R}$
 whose sections fail to smooth the node p. It suffices to show that 
 $H^1(N(-\Gamma ))=0$
. This follows from the exact sequence
$H^1(N(-\Gamma ))=0$
. This follows from the exact sequence 
 $$\begin{align*}0 \to N_{R \cup L}|_R(-p - \Gamma) \to N(-\Gamma) \to N|_L(-\Gamma) \to 0,\end{align*}$$
$$\begin{align*}0 \to N_{R \cup L}|_R(-p - \Gamma) \to N(-\Gamma) \to N|_L(-\Gamma) \to 0,\end{align*}$$
together with the isomorphisms
 $$ \begin{align*} N_{R \cup L}|_R &\simeq N_{R/H} \oplus \mathcal{O}_R(1)(p) \simeq \mathcal{O}_{\mathbb{P}^1}(r+2)^{\oplus r-2} \oplus \mathcal{O}_{\mathbb{P}^1}(r), \\ N|_L &\simeq N_L \simeq \mathcal{O}_{\mathbb{P}^1}(1)^{\oplus r-1}. \end{align*} $$
$$ \begin{align*} N_{R \cup L}|_R &\simeq N_{R/H} \oplus \mathcal{O}_R(1)(p) \simeq \mathcal{O}_{\mathbb{P}^1}(r+2)^{\oplus r-2} \oplus \mathcal{O}_{\mathbb{P}^1}(r), \\ N|_L &\simeq N_L \simeq \mathcal{O}_{\mathbb{P}^1}(1)^{\oplus r-1}. \end{align*} $$
 Since 
 $N_{L \cup R}(-\Gamma )$
 is a positive modification of
$N_{L \cup R}(-\Gamma )$
 is a positive modification of 
 $N(-\Gamma )$
, we also have
$N(-\Gamma )$
, we also have 
 $H^1(N_{L \cup R}(-\Gamma )) = 0$
. Similarly,
$H^1(N_{L \cup R}(-\Gamma )) = 0$
. Similarly, 
 $N_{C' \cup L \cup R}|_{C'}$
 is a positive modification of
$N_{C' \cup L \cup R}|_{C'}$
 is a positive modification of 
 $N_{C'}$
. As
$N_{C'}$
. As 
 $C'$
 is a general BN-curve, the Gieseker–Petri theorem implies that
$C'$
 is a general BN-curve, the Gieseker–Petri theorem implies that 
 $H^1(N_{C' \cup L \cup R}|_{C'})=0$
. Hence, using the exact sequence
$H^1(N_{C' \cup L \cup R}|_{C'})=0$
. Hence, using the exact sequence 
 $$\begin{align*}0 \to N_{L \cup R}(-\Gamma) \to N_{C' \cup L \cup R} \to N_{C' \cup L \cup R}|_{C'} \to 0, \end{align*}$$
$$\begin{align*}0 \to N_{L \cup R}(-\Gamma) \to N_{C' \cup L \cup R} \to N_{C' \cup L \cup R}|_{C'} \to 0, \end{align*}$$
we see that 
 $H^1(N_{C' \cup L \cup R}) = 0$
, and part (a) follows.
$H^1(N_{C' \cup L \cup R}) = 0$
, and part (a) follows.
 Finally, for part (c), we will show that 
 $H^1(T_{\mathbb {P}^r}|_{C' \cup M})= 0$
, and hence the map from
$H^1(T_{\mathbb {P}^r}|_{C' \cup M})= 0$
, and hence the map from 
 $C' \cup M$
 to
$C' \cup M$
 to 
 $\mathbb {P}^r$
 can be lifted as
$\mathbb {P}^r$
 can be lifted as 
 $C' \cup M$
 is smoothed to a general curve. This vanishing follows from the exact sequence
$C' \cup M$
 is smoothed to a general curve. This vanishing follows from the exact sequence 
 $$\begin{align*}0 \to T_{\mathbb{P}^r}|_M(-\Gamma) \to T_{\mathbb{P}^r}|_{C' \cup M} \to T_{\mathbb{P}^r}|_{C'} \to 0, \end{align*}$$
$$\begin{align*}0 \to T_{\mathbb{P}^r}|_M(-\Gamma) \to T_{\mathbb{P}^r}|_{C' \cup M} \to T_{\mathbb{P}^r}|_{C'} \to 0, \end{align*}$$
the isomorphism 
 $T_{\mathbb {P}^r}|_M \simeq \mathcal {O}_{\mathbb {P}^r}(r+1)^{\oplus r} $
, and the Gieseker–Petri theorem (
$T_{\mathbb {P}^r}|_M \simeq \mathcal {O}_{\mathbb {P}^r}(r+1)^{\oplus r} $
, and the Gieseker–Petri theorem (
 $H^1(T_{\mathbb {P}^r}|_{C'}) = 0$
).
$H^1(T_{\mathbb {P}^r}|_{C'}) = 0$
).
 Our next goal is to study the restricted normal bundle 
 , which is of slope
, which is of slope 
 $r + 2$
. In most cases, this bundle is perfectly balanced (equivalently semistable):
$r + 2$
. In most cases, this bundle is perfectly balanced (equivalently semistable):
Lemma 5.8. Unless r is odd and C is an elliptic normal curve, 
 is perfectly balanced, that is
 is perfectly balanced, that is 

If r is odd and C is an elliptic normal curve, then 
 is ‘almost balanced’, that is, is isomorphic to one of the two bundles:
 is ‘almost balanced’, that is, is isomorphic to one of the two bundles: 
 $$\begin{align*}\mathcal{O}_{\mathbb{P}^1}(r + 2)^{\oplus (r - 1)} \quad \text{or} \quad \mathcal{O}_{\mathbb{P}^1}(r + 3) \oplus \mathcal{O}_{\mathbb{P}^1}(r + 2)^{\oplus (r - 3)} \oplus \mathcal{O}_{\mathbb{P}^1}(r + 1).\end{align*}$$
$$\begin{align*}\mathcal{O}_{\mathbb{P}^1}(r + 2)^{\oplus (r - 1)} \quad \text{or} \quad \mathcal{O}_{\mathbb{P}^1}(r + 3) \oplus \mathcal{O}_{\mathbb{P}^1}(r + 2)^{\oplus (r - 3)} \oplus \mathcal{O}_{\mathbb{P}^1}(r + 1).\end{align*}$$
Proof. Write d and g for the degree and genus of C. First, we reduce to the cases where C is nonspecial, that is, where 
 $d \geq g + r$
. To do this, when
$d \geq g + r$
. To do this, when 
 $d < g + r$
, we inductively specialize C to a union
$d < g + r$
, we inductively specialize C to a union 
 $C' \cup D$
, where
$C' \cup D$
, where 
 $C'$
 is a general BN-curve of degree
$C'$
 is a general BN-curve of degree 
 $d' \geq r + 2$
. Since in particular
$d' \geq r + 2$
. Since in particular 
 $d' \geq r + 1$
, we may specialize the points where R meets C onto
$d' \geq r + 1$
, we may specialize the points where R meets C onto 
 $C'$
, thereby replacing C by
$C'$
, thereby replacing C by 
 $C'$
, which is not an elliptic normal curve since
$C'$
, which is not an elliptic normal curve since 
 $d' \geq r + 2$
. To find such a specialization, we break into cases as follows:
$d' \geq r + 2$
. To find such a specialization, we break into cases as follows: 
- 
1. If  $d < g + r$
 and $d < g + r$
 and $\rho (d, g, r)> 0$
 (which forces $\rho (d, g, r)> 0$
 (which forces $d \geq 2r + 1 \geq r + 3$
), we apply Lemma 5.4 to degenerate C to the union of a general BN-curve $d \geq 2r + 1 \geq r + 3$
), we apply Lemma 5.4 to degenerate C to the union of a general BN-curve $C'$
 of degree $C'$
 of degree $d - 1$
 and genus $d - 1$
 and genus $g - 1$
, with a one-secant line D. $g - 1$
, with a one-secant line D.
- 
2. If  $d < g + r$
 and $d < g + r$
 and $\rho (d, g, r) = 0$
, but C is not a canonical curve (which forces $\rho (d, g, r) = 0$
, but C is not a canonical curve (which forces $d \geq 3r \geq 2r + 2$
), we claim that we may specialize C to the union of a general BN-curve $d \geq 3r \geq 2r + 2$
), we claim that we may specialize C to the union of a general BN-curve $C'$
 of degree $C'$
 of degree $d - r$
 and genus $d - r$
 and genus $g - r - 1$
, with an $g - r - 1$
, with an $(r + 2)$
-secant rational normal curve D. Indeed, as in the proof of Lemma 5.1, we have that $(r + 2)$
-secant rational normal curve D. Indeed, as in the proof of Lemma 5.1, we have that $H^1(T_{\mathbb {P}^r}|_{C' \cup D})=0$
 by combining $H^1(T_{\mathbb {P}^r}|_{C' \cup D})=0$
 by combining $H^1(T_{\mathbb {P}^r}|_{C'})=0$
 (from the Gieseker–Petri theorem) and $H^1(T_{\mathbb {P}^r}|_{C'})=0$
 (from the Gieseker–Petri theorem) and $H^1(T_{\mathbb {P}^r}|_D(-D \cap C')) = 0$
 (from $H^1(T_{\mathbb {P}^r}|_D(-D \cap C')) = 0$
 (from $T_{\mathbb {P}^r}|_D \simeq \mathcal {O}_{\mathbb {P}^1}(r + 1)^{\oplus r}$
). $T_{\mathbb {P}^r}|_D \simeq \mathcal {O}_{\mathbb {P}^1}(r + 1)^{\oplus r}$
).
- 
3. If C is a canonical curve, we claim that we may specialize C to the union of a general BN-curve  $C'$
 of degree $C'$
 of degree $r + 2$
 and genus $r + 2$
 and genus $2$
, with a r-secant rational curve of degree $2$
, with a r-secant rational curve of degree $r - 2$
. Indeed, we glue an abstract curve of genus $r - 2$
. Indeed, we glue an abstract curve of genus $2$
 to $2$
 to $\mathbb {P}^1$
 at r general points, and map it to projective space via the complete linear series for the dualizing sheaf. $\mathbb {P}^1$
 at r general points, and map it to projective space via the complete linear series for the dualizing sheaf.
 For C nonspecial, we will prove the lemma by induction on r. Let 
 $\Gamma $
 be a collection of
$\Gamma $
 be a collection of 
 $r + 1$
 points where C meets R; for
$r + 1$
 points where C meets R; for 
 $r \geq 3$
, this is exactly the intersection
$r \geq 3$
, this is exactly the intersection 
 $C \cap R$
, and so
$C \cap R$
, and so 
 . However, we can extend the lemma to cover the case
. However, we can extend the lemma to cover the case 
 $r = 2$
 as well, by replacing equation (5.3) with the assertion that
$r = 2$
 as well, by replacing equation (5.3) with the assertion that 
 . With this formulation, the base case of
. With this formulation, the base case of 
 $r=2$
 is clear since
$r=2$
 is clear since 
 is a line bundle of degree
 is a line bundle of degree 
 $4$
. The base case
$4$
. The base case 
 $r=3$
 is [Reference Coskun, Larson and Vogt9, Lemma 4.2].
$r=3$
 is [Reference Coskun, Larson and Vogt9, Lemma 4.2].
 For the inductive step, we suppose 
 $r \geq 4$
. Let
$r \geq 4$
. Let 
 $H'$
 be a general hyperplane transverse to H. We will degenerate C to a union
$H'$
 be a general hyperplane transverse to H. We will degenerate C to a union 
 $C' \cup L$
, where
$C' \cup L$
, where 
 $C' \subset H'$
 is a general BN-curve of degree
$C' \subset H'$
 is a general BN-curve of degree 
 $d-1$
 and genus g. By hypothesis,
$d-1$
 and genus g. By hypothesis, 
 $d\geq r+1$
, and so
$d\geq r+1$
, and so 
 $C'$
 meets H in at least r points, which are in linear general position in
$C'$
 meets H in at least r points, which are in linear general position in 
 $H \cap H' \simeq \mathbb {P}^{r - 2}$
 since the sectional monodromy group of a general curve always contains the alternating group [Reference Kadets20]. Since
$H \cap H' \simeq \mathbb {P}^{r - 2}$
 since the sectional monodromy group of a general curve always contains the alternating group [Reference Kadets20]. Since 
 $\operatorname {Aut} \mathbb {P}^{r-2}$
 acts transitively on r-tuples of points in linear general position, we can apply an automorphism so that
$\operatorname {Aut} \mathbb {P}^{r-2}$
 acts transitively on r-tuples of points in linear general position, we can apply an automorphism so that 
 $r-1$
 of these points are on R and the final point p lies on a one-secant line L to R.
$r-1$
 of these points are on R and the final point p lies on a one-secant line L to R.

 We claim that 
 $C' \cup L$
 is a BN-curve of degree d and genus g, as
$C' \cup L$
 is a BN-curve of degree d and genus g, as 
 $H^1(T_{\mathbb {P}^r}|_{C' \cup L})=0$
. Indeed,
$H^1(T_{\mathbb {P}^r}|_{C' \cup L})=0$
. Indeed, 
 $H^1(T_{\mathbb {P}^r}|_{C'}) = 0$
 because
$H^1(T_{\mathbb {P}^r}|_{C'}) = 0$
 because 
 $C'$
 is nonspecial (and
$C'$
 is nonspecial (and 
 $T_{\mathbb {P}^r}$
 is a quotient of
$T_{\mathbb {P}^r}$
 is a quotient of 
 $\mathcal {O}_{\mathbb {P}^r}(1)^{\oplus r+1}$
). Furthermore, since
$\mathcal {O}_{\mathbb {P}^r}(1)^{\oplus r+1}$
). Furthermore, since 
 $T_{\mathbb {P}^r}|_L \simeq \mathcal {O}_{\mathbb {P}^1}(2) \oplus \mathcal {O}_{\mathbb {P}^r}(1)^{\oplus r}$
, we have
$T_{\mathbb {P}^r}|_L \simeq \mathcal {O}_{\mathbb {P}^1}(2) \oplus \mathcal {O}_{\mathbb {P}^r}(1)^{\oplus r}$
, we have 
 $H^1(T_{\mathbb {P}^r}|_L(-p)) = 0$
. The result now follows by considering
$H^1(T_{\mathbb {P}^r}|_L(-p)) = 0$
. The result now follows by considering 
 $$\begin{align*}0 \to T_{\mathbb{P}^r}|_L(-p) \to T_{\mathbb{P}^r}|_{C' \cup L} \to T_{\mathbb{P}^r}|_{C'} \to 0.\end{align*}$$
$$\begin{align*}0 \to T_{\mathbb{P}^r}|_L(-p) \to T_{\mathbb{P}^r}|_{C' \cup L} \to T_{\mathbb{P}^r}|_{C'} \to 0.\end{align*}$$
Call u and v the points on R where L is one-secant. Projecting from L induces an exact sequence

The curve 
 $\overline {R}$
 is again a rational curve of degree
$\overline {R}$
 is again a rational curve of degree 
 $r-3$
 in a hyperplane that is incident to
$r-3$
 in a hyperplane that is incident to 
 $\overline {C'}$
 at
$\overline {C'}$
 at 
 $r-1$
 points. Furthermore, if C is not an elliptic normal curve, then neither is
$r-1$
 points. Furthermore, if C is not an elliptic normal curve, then neither is 
 $\overline {C'}$
. Applying our inductive hypothesis (for
$\overline {C'}$
. Applying our inductive hypothesis (for 
 $\mathbb {P}^{r - 2}$
), in combination with the above exact sequence, completes the proof.
$\mathbb {P}^{r - 2}$
), in combination with the above exact sequence, completes the proof.
 It is natural to ask what happens when C is an elliptic normal curve and r is odd. This case is only necessary to treat the special family of canonical curves of even genus. When we treat that case in Section 13, we will show that 
 is not perfectly balanced in this case. Moreover, we will give a geometric construction of its Harder–Narasimhan filtration.
 is not perfectly balanced in this case. Moreover, we will give a geometric construction of its Harder–Narasimhan filtration.
5.4 Reduction to good tuples
 In this section, we show that, apart from rational curves and canonical curves of even genus, all other cases of Theorem 1.4 follow from 
 $I(d,g,r, 0, m)$
 for good tuples with
$I(d,g,r, 0, m)$
 for good tuples with 
 $\ell = 0$
.
$\ell = 0$
.
Lemma 5.9. Suppose that 
 $\rho (d,g,r) \geq 0$
 and that
$\rho (d,g,r) \geq 0$
 and that 
 $(d, g,r) \neq (2r, r+1, r)$
 if r is odd. If
$(d, g,r) \neq (2r, r+1, r)$
 if r is odd. If 
 $g \geq r$
 and
$g \geq r$
 and 
 $I(d-r+1, g-r, r,\ell ,m+1)$
 holds, then
$I(d-r+1, g-r, r,\ell ,m+1)$
 holds, then 
 $I(d,g,r,\ell ,m)$
 holds.
$I(d,g,r,\ell ,m)$
 holds.
Proof. Let C be a general BN-curve of degree d and genus g in 
 $\mathbb {P}^r$
 with
$\mathbb {P}^r$
 with 
 $g \geq r$
. Let
$g \geq r$
. Let 
 $u_1, v_1, \ldots , u_\ell , v_\ell $
 be general points on C. Let
$u_1, v_1, \ldots , u_\ell , v_\ell $
 be general points on C. Let 
 $R_1, \dots , R_m$
 be general
$R_1, \dots , R_m$
 be general 
 $(r+1)$
-secant rational curves of degree
$(r+1)$
-secant rational curves of degree 
 $r-1$
. The statement
$r-1$
. The statement 
 $I(d, g, r, \ell , m)$
 asserts that
$I(d, g, r, \ell , m)$
 asserts that 

satisfies interpolation. Combining the assumptions that 
 $\rho (d,g,r)\geq 0$
 and
$\rho (d,g,r)\geq 0$
 and 
 $g \geq r$
, we see that
$g \geq r$
, we see that 
 $$\begin{align*}d \geq r + \frac{rg}{r+1} \geq r + \frac{r^2}{r + 1}> 2r - 1.\end{align*}$$
$$\begin{align*}d \geq r + \frac{rg}{r+1} \geq r + \frac{r^2}{r + 1}> 2r - 1.\end{align*}$$
We may therefore prove 
 $I(d, g, r, \ell , m)$
 by peeling off an additional
$I(d, g, r, \ell , m)$
 by peeling off an additional 
 $(r+1)$
-secant rational curve
$(r+1)$
-secant rational curve 
 $R_{m+1}$
 of degree
$R_{m+1}$
 of degree 
 $r-1$
. That is, we specialize the curve C as in Lemma 5.7 to the union of a general BN-curve
$r-1$
. That is, we specialize the curve C as in Lemma 5.7 to the union of a general BN-curve 
 $C'$
 of degree
$C'$
 of degree 
 $d-r+1$
 and genus
$d-r+1$
 and genus 
 $g-r$
 and a rational curve
$g-r$
 and a rational curve 
 $R_{m+1}$
, in such a way that the points of
$R_{m+1}$
, in such a way that the points of 
 $C \cap (R_1 \cup \cdots \cup R_m)$
 and the
$C \cap (R_1 \cup \cdots \cup R_m)$
 and the 
 $u_i$
 and
$u_i$
 and 
 $v_i$
 specialize onto
$v_i$
 specialize onto 
 $C'$
. By virtue of specializing the auxilliary points onto
$C'$
. By virtue of specializing the auxilliary points onto 
 $C'$
, we have
$C'$
, we have 

Since we assume that 
 $(d,g,r)\neq (2r, r+1, r)$
 if r is odd, Lemma 5.8 implies that
$(d,g,r)\neq (2r, r+1, r)$
 if r is odd, Lemma 5.8 implies that 
 $N_{C' \cup R_{m+1}}|_{R_{m+1}}$
 is perfectly balanced. By Lemma 3.7, it suffices to prove that
$N_{C' \cup R_{m+1}}|_{R_{m+1}}$
 is perfectly balanced. By Lemma 3.7, it suffices to prove that 
 $N_{C' \cup R_{m+1}}|_{C'}$
 satisfies interpolation. This restriction is
$N_{C' \cup R_{m+1}}|_{C'}$
 satisfies interpolation. This restriction is 

which satisfies interpolation by our assumption that 
 $I(d-r+1, g-r, r,\ell ,m+1)$
 holds.
$I(d-r+1, g-r, r,\ell ,m+1)$
 holds.
Proposition 5.10. Suppose that 
 $I(d,g,r, 0, m)$
 holds for all good
$I(d,g,r, 0, m)$
 holds for all good 
 $(d,g,r,0,m)$
. Then
$(d,g,r,0,m)$
. Then 
 $I(d, g, r, 0, 0)$
 holds whenever
$I(d, g, r, 0, 0)$
 holds whenever 
 $\rho (d, g, r) \geq 0$
, except if
$\rho (d, g, r) \geq 0$
, except if 
- 
•  $(d,g,r)$
 is in the list (1.4), or $(d,g,r)$
 is in the list (1.4), or
- 
•  $(d,g,r)= (2r, r+1, r)$
 and r is odd, or $(d,g,r)= (2r, r+1, r)$
 and r is odd, or
- 
•  $g=0$
 and $g=0$
 and $d\not \equiv 1 \pmod {r-1}$
. $d\not \equiv 1 \pmod {r-1}$
.
Proof. When 
 $g = 0$
, the tuple
$g = 0$
, the tuple 
 $(d, g, r, 0, 0)$
 is good when
$(d, g, r, 0, 0)$
 is good when 
 $d \equiv 1$
 mod
$d \equiv 1$
 mod 
 $r - 1$
, so the result is a tautology. We therefore suppose
$r - 1$
, so the result is a tautology. We therefore suppose 
 $g \geq 1$
.
$g \geq 1$
.
 We will prove by induction on g that 
 $I(d,g,r,0,m)$
 holds for
$I(d,g,r,0,m)$
 holds for 
 $g \geq 1$
 subject to the conditions that
$g \geq 1$
 subject to the conditions that 
 $m \leq \rho (d, g, r)$
 and
$m \leq \rho (d, g, r)$
 and 
 $(d, g, r, 0, m)$
 is not in the list (XEx). (If
$(d, g, r, 0, m)$
 is not in the list (XEx). (If 
 $(d, g, r, 0, 0)$
 is in the list (XEx), then
$(d, g, r, 0, 0)$
 is in the list (XEx), then 
 $(d, g, r)$
 is in the list (1.4), so this is sufficient.) Our base cases will be
$(d, g, r)$
 is in the list (1.4), so this is sufficient.) Our base cases will be 
 $g \leq r$
; in these cases,
$g \leq r$
; in these cases, 
 $d \geq g + r$
 by Remark 4.3, and so
$d \geq g + r$
 by Remark 4.3, and so 
 $(d, g, r, 0, m)$
 is good. For the inductive step, we apply Lemma 5.9 to reduce from
$(d, g, r, 0, m)$
 is good. For the inductive step, we apply Lemma 5.9 to reduce from 
 $I(d,g,r,0,m)$
 to
$I(d,g,r,0,m)$
 to 
 $I(d-r+1, g-r, r, 0, m + 1)$
.
$I(d-r+1, g-r, r, 0, m + 1)$
.
6 
The family with 
 $\boldsymbol {\delta } = \mathbf {1}$
 and
$\boldsymbol {\delta } = \mathbf {1}$
 and 
 $\boldsymbol {\ell } = \mathbf {m} = \mathbf {0}$
$\boldsymbol {\ell } = \mathbf {m} = \mathbf {0}$
 In this section, we establish 
 $I(d, g, r, 0, 0)$
 for good tuples with
$I(d, g, r, 0, 0)$
 for good tuples with 
 $\delta = 1$
. When
$\delta = 1$
. When 
 $\ell = m = 0$
, the condition
$\ell = m = 0$
, the condition 
 $\delta = 1$
 is equivalent to
$\delta = 1$
 is equivalent to 
 $$ \begin{align} 2d + 2g = 3r - 1, \end{align} $$
$$ \begin{align} 2d + 2g = 3r - 1, \end{align} $$
and 
 $I(d, g, r, 0, 0)$
 asserts interpolation for
$I(d, g, r, 0, 0)$
 asserts interpolation for 
 $N_C$
 (with no modifications). Our argument will be by induction on
$N_C$
 (with no modifications). Our argument will be by induction on 
 $d - g - r$
.
$d - g - r$
.
 When 
 $d - g - r = 0$
, then by equation (6.1), we have
$d - g - r = 0$
, then by equation (6.1), we have 
 $(d, g, r) = (5g + 1, g, 4g + 1)$
. We may therefore conclude interpolation by [Reference Atanasov, Larson and Yang2, Lemma 11.3].
$(d, g, r) = (5g + 1, g, 4g + 1)$
. We may therefore conclude interpolation by [Reference Atanasov, Larson and Yang2, Lemma 11.3].
 To complete the inductive step, observe that if 
 $g = 0$
, then
$g = 0$
, then 
 $d = (3r - 1)/2$
 by equation (6.1), and so such tuples
$d = (3r - 1)/2$
 by equation (6.1), and so such tuples 
 $(d, 0, r, 0, 0)$
 are never good. It thus suffices to prove the following proposition in the case
$(d, 0, r, 0, 0)$
 are never good. It thus suffices to prove the following proposition in the case 
 $g> 0$
. (The
$g> 0$
. (The 
 $g = 0$
 case is included here too since it will be useful later, when establishing Theorem 1.4 for rational curves, and it will follow via the same argument.)
$g = 0$
 case is included here too since it will be useful later, when establishing Theorem 1.4 for rational curves, and it will follow via the same argument.)
Proposition 6.1. Suppose that 
 $\ell = m = 0 $
, and
$\ell = m = 0 $
, and 
 $d> g + r$
, and equation (6.1) holds. If
$d> g + r$
, and equation (6.1) holds. If 
 $I(d-3, g, r-2, 0, 0)$
 holds, and
$I(d-3, g, r-2, 0, 0)$
 holds, and 
 $g>0$
 or the characteristic is not
$g>0$
 or the characteristic is not 
 $2$
, then
$2$
, then 
 $I(d,g,r,0,0)$
 also holds.
$I(d,g,r,0,0)$
 also holds.
In order to prove this proposition, we first establish the following lemmas.
Lemma 6.2. Let L be a line meeting C quasitransversely at a smooth point x. For any points 
 $y, y' \in L {\smallsetminus} x$
, the sections of
$y, y' \in L {\smallsetminus} x$
, the sections of 
 $\mathbb {P} N_C$
 corresponding to
$\mathbb {P} N_C$
 corresponding to 
 $N_{C \to y}$
 and
$N_{C \to y}$
 and 
 $N_{C \to y'}$
 are tangent over x.
$N_{C \to y'}$
 are tangent over x.
Proof. We prove this by a calculation in local coordinates. We may choose an affine neighborhood of x in 
 $\mathbb {P}^r$
, and a local coordinate t on C so that
$\mathbb {P}^r$
, and a local coordinate t on C so that 
 $x = C(0) = 0$
, and C is given parametrically by the power series
$x = C(0) = 0$
, and C is given parametrically by the power series 
 $C(t) = tC_1 + t^2C_2 + O(t^3)$
. By assumption,
$C(t) = tC_1 + t^2C_2 + O(t^3)$
. By assumption, 
 $y' = a \cdot y$
 for some invertible scalar a. It suffices to show that the three vectors
$y' = a \cdot y$
 for some invertible scalar a. It suffices to show that the three vectors 
 $C(t)-y$
, and
$C(t)-y$
, and 
 $C(t) - y'$
, and
$C(t) - y'$
, and 
 $C'(t)$
, are dependent mod
$C'(t)$
, are dependent mod 
 $t^2$
. The explicit dependence is
$t^2$
. The explicit dependence is 
 $$\begin{align*}-a(C(t) - y) + (C(t) - y') + t(a-1)C'(t) \equiv -a (tC_1 - y) + (tC_1 - y') + t(a-1)C_1 = 0 \pmod{t^2}. \end{align*}$$
$$\begin{align*}-a(C(t) - y) + (C(t) - y') + t(a-1)C'(t) \equiv -a (tC_1 - y) + (tC_1 - y') + t(a-1)C_1 = 0 \pmod{t^2}. \end{align*}$$
Lemma 6.3. Let u, v, and x be general points on a BN-curve C of degree d and genus g with
 $$ \begin{align} 2(d+1) + 2g = 3r - 1 \quad \text{and} \quad d \geq g+r. \end{align} $$
$$ \begin{align} 2(d+1) + 2g = 3r - 1 \quad \text{and} \quad d \geq g+r. \end{align} $$
Let y be a general point on the one-secant line 
 $\overline {uv}$
. Then the bundle
$\overline {uv}$
. Then the bundle 

satisfies interpolation if and only if 
 $g> 0$
 or the characteristic is not
$g> 0$
 or the characteristic is not 
 $2$
.
$2$
.
Proof. Subtracting the first equation in equation (6.2) from 
 $6$
 times the second inequality implies
$6$
 times the second inequality implies 
 $2d + 4 - 2g \geq 2g$
. By Lemma 3.8, interpolation for
$2d + 4 - 2g \geq 2g$
. By Lemma 3.8, interpolation for 
 
is thus equivalent to interpolation for

 By Lemma 6.2, the subbundles 
 $N_{C \to y}$
 and
$N_{C \to y}$
 and 
 $N_{C \to v}$
 agree to second order at u. Therefore, the composition
$N_{C \to v}$
 agree to second order at u. Therefore, the composition 
 $N_{C \to y} \to N_{C \to u} \oplus N_{C \to v} \to N_{C \to u}$
 vanishes to order
$N_{C \to y} \to N_{C \to u} \oplus N_{C \to v} \to N_{C \to u}$
 vanishes to order 
 $2$
 at u. Similarly,
$2$
 at u. Similarly, 
 $N_{C \to y} \to N_{C \to v}$
 vanishes to order
$N_{C \to y} \to N_{C \to v}$
 vanishes to order 
 $2$
 at v. Therefore, under the isomorphisms
$2$
 at v. Therefore, under the isomorphisms 
 $$ \begin{align} N_{C\to y} \simeq \mathcal{O}_C(1), \quad N_{C \to u}\simeq \mathcal{O}_C(1)(2u), \quad \text{and} \quad N_{C \to v}\simeq \mathcal{O}_C(1)(2v), \end{align} $$
$$ \begin{align} N_{C\to y} \simeq \mathcal{O}_C(1), \quad N_{C \to u}\simeq \mathcal{O}_C(1)(2u), \quad \text{and} \quad N_{C \to v}\simeq \mathcal{O}_C(1)(2v), \end{align} $$
the map 
 $N_{C \to y} \to N_{C \to u} \oplus N_{C \to v}$
 is a diagonal inclusion
$N_{C \to y} \to N_{C \to u} \oplus N_{C \to v}$
 is a diagonal inclusion 
 $$\begin{align*}\mathcal{O}_C(1) \to \mathcal{O}_C(1)(2u) \oplus \mathcal{O}_C(1)(2v)\end{align*}$$
$$\begin{align*}\mathcal{O}_C(1) \to \mathcal{O}_C(1)(2u) \oplus \mathcal{O}_C(1)(2v)\end{align*}$$
given by constant sections of 
 $\mathcal {O}_C(2u)$
 and
$\mathcal {O}_C(2u)$
 and 
 $\mathcal {O}_C(2v)$
 that vanish at
$\mathcal {O}_C(2v)$
 that vanish at 
 $2u$
 and
$2u$
 and 
 $2v$
, respectively. Such a section is indexed by two nonzero constants c and d. In other words,
$2v$
, respectively. Such a section is indexed by two nonzero constants c and d. In other words, 

where 
 $O_C(1)$
 is the diagonal subbundle identified above.
$O_C(1)$
 is the diagonal subbundle identified above.
 Since 
 $\mu (N)=d + 1$
, it satisfies interpolation if and only if it has no cohomology when twisted down by a general line bundle of degree
$\mu (N)=d + 1$
, it satisfies interpolation if and only if it has no cohomology when twisted down by a general line bundle of degree 
 $d + 2 - g$
. Write such a line bundle as
$d + 2 - g$
. Write such a line bundle as 
 $L^\vee (1)(2u+2v)$
, where L is a general line bundle of degree
$L^\vee (1)(2u+2v)$
, where L is a general line bundle of degree 
 $g+2$
. We therefore want
$g+2$
. We therefore want 

to have no global sections, where the diagonal subbundle 
 $L(-2u-2v)$
 is indexed by
$L(-2u-2v)$
 is indexed by 
 $[c:d] \in \mathbb {G}_m$
 as above.
$[c:d] \in \mathbb {G}_m$
 as above.
 As L is a general line bundle of degree 
 $g+2$
 and u and v are general points,
$g+2$
 and u and v are general points, 
 $h^0(L(-2v)) = h^0(L(-2u)) = 1$
; write
$h^0(L(-2v)) = h^0(L(-2u)) = 1$
; write 
 $\sigma $
 and
$\sigma $
 and 
 $\tau $
 for the unique (up to scaling) sections of L vanishing to order
$\tau $
 for the unique (up to scaling) sections of L vanishing to order 
 $2$
 at u and v, respectively. Every section of
$2$
 at u and v, respectively. Every section of 
 $L(-2u) \oplus L(-2v)$
 is a linear combination
$L(-2u) \oplus L(-2v)$
 is a linear combination 
 $a \sigma \oplus b \tau $
, viewed as a section of
$a \sigma \oplus b \tau $
, viewed as a section of 
 $L \oplus L$
. Such a global section comes from the subsheaf
$L \oplus L$
. Such a global section comes from the subsheaf 
 when it is dependent with the constant diagonal section
 when it is dependent with the constant diagonal section 
 $c \oplus d$
 at
$c \oplus d$
 at 
 $2x$
, that is, when the section
$2x$
, that is, when the section 
 $ad \sigma - bc \tau $
 of L vanishes at
$ad \sigma - bc \tau $
 of L vanishes at 
 $2x$
. Hence, equation (6.4) has no cohomology if x is not a ramification point of the map
$2x$
. Hence, equation (6.4) has no cohomology if x is not a ramification point of the map 
 $\varphi \colon C \to \mathbb {P}^1$
 determined by
$\varphi \colon C \to \mathbb {P}^1$
 determined by 
 $\langle \sigma , \tau \rangle \subseteq H^0(L)$
.
$\langle \sigma , \tau \rangle \subseteq H^0(L)$
.
 As x was a general point, this holds if and only if 
 $\varphi $
 is separable. If
$\varphi $
 is separable. If 
 $\varphi $
 is not separable, then the characteristic p of the ground field is positive and
$\varphi $
 is not separable, then the characteristic p of the ground field is positive and 
 $\varphi $
 factors through the Frobenius morphism F:
$\varphi $
 factors through the Frobenius morphism F: 
 $$\begin{align*}C \xrightarrow{F} C' \to \mathbb{P}^1.\end{align*}$$
$$\begin{align*}C \xrightarrow{F} C' \to \mathbb{P}^1.\end{align*}$$
In this case, L and the linear system 
 $\langle \sigma , \tau \rangle $
 are necessarily pulled back under F. In other words,
$\langle \sigma , \tau \rangle $
 are necessarily pulled back under F. In other words, 
 $L \simeq F^*M$
 for a line bundle M (necessarily general because L is general), of degree
$L \simeq F^*M$
 for a line bundle M (necessarily general because L is general), of degree 
 $(g+2)/p$
 with
$(g+2)/p$
 with 
 $h^0(M) \geq 2$
. Therefore,
$h^0(M) \geq 2$
. Therefore, 
 $$\begin{align*}\frac{g + 2}{p} + 1 - g = h^0(M) \geq 2,\end{align*}$$
$$\begin{align*}\frac{g + 2}{p} + 1 - g = h^0(M) \geq 2,\end{align*}$$
or upon rearrangement,
 $$\begin{align*}p \leq \frac{g+2}{g+1}.\end{align*}$$
$$\begin{align*}p \leq \frac{g+2}{g+1}.\end{align*}$$
Thus 
 $g=0$
 and
$g=0$
 and 
 $p=2$
.
$p=2$
.
 Conversely, when 
 $g = 0$
, there is a choice of coordinates
$g = 0$
, there is a choice of coordinates 
 $[t : s]$
 on C so that
$[t : s]$
 on C so that 
 $\langle \sigma , \tau \rangle = \langle t^2, s^2 \rangle $
. If in addition
$\langle \sigma , \tau \rangle = \langle t^2, s^2 \rangle $
. If in addition 
 $p = 2$
, then the map
$p = 2$
, then the map 
 $\varphi $
 is inseparable.
$\varphi $
 is inseparable.
Proof of Proposition 6.1
 Specialize C to the union 
 $C' \cup L$
, where
$C' \cup L$
, where 
 $C'$
 is a general BN-curve of degree
$C'$
 is a general BN-curve of degree 
 $d - 1$
 and genus g in
$d - 1$
 and genus g in 
 $\mathbb {P}^r$
, and L is a one-secant line
$\mathbb {P}^r$
, and L is a one-secant line 
 $\overline {xy}$
 meeting
$\overline {xy}$
 meeting 
 $C'$
 at x. It suffices to show interpolation for
$C'$
 at x. It suffices to show interpolation for 

Let 
 $u,v \in C'$
 be general points, and specialize y to a general point on the line
$u,v \in C'$
 be general points, and specialize y to a general point on the line 
 $\overline {uv}$
. Projection from
$\overline {uv}$
. Projection from 
 $\overline {uv}$
 induces a pointing bundle exact sequence
$\overline {uv}$
 induces a pointing bundle exact sequence 

By Lemma 6.3, the subbundle
 
 satisfies interpolation. By hypothesis, 
 $N_{\overline {C'}}$
, and hence the quotient
$N_{\overline {C'}}$
, and hence the quotient 
 $N_{\overline {C'}}(u+v)$
, also satisfies interpolation. Finally, by equation (6.1),
$N_{\overline {C'}}(u+v)$
, also satisfies interpolation. Finally, by equation (6.1), 

Thus,
 
satisfies interpolation by Lemma 3.15.
7 
Specializations of the 
 $\mathbf {R}_{\mathbf {i}}$
$\mathbf {R}_{\mathbf {i}}$
 In this section, for some integers 
 $n=n_i$
, we construct a specialization of one of the rational curves
$n=n_i$
, we construct a specialization of one of the rational curves 
 $R = R_i$
 so that exactly n modifications point towards a point p on C. We then show that this specialization plays well with projection from p.
$R = R_i$
 so that exactly n modifications point towards a point p on C. We then show that this specialization plays well with projection from p.
7.1 Setup
 Let n be an integer satisfying 
 $0 \leq n \leq r - 1$
 and
$0 \leq n \leq r - 1$
 and 
 $n \equiv r - 1$
 mod
$n \equiv r - 1$
 mod 
 $2$
. Let
$2$
. Let 
 $p, q_1, q_2, \ldots , q_{r - 1} \in C$
 be points such that
$p, q_1, q_2, \ldots , q_{r - 1} \in C$
 be points such that 
 $2p + q_1 + \cdots + q_{r - 1}$
 lies in a hyperplane H. Assume that
$2p + q_1 + \cdots + q_{r - 1}$
 lies in a hyperplane H. Assume that 
 $2p + q_1 + \cdots + q_{r - 1}$
 is otherwise in linear general position, that is,
$2p + q_1 + \cdots + q_{r - 1}$
 is otherwise in linear general position, that is, 
 $p + q_1 + \cdots + q_{r - 1}$
 and each
$p + q_1 + \cdots + q_{r - 1}$
 and each 
 $2p + q_1 + \cdots + q_{i - 1} + q_{i + 1} + \cdots + q_{r - 1}$
 spans H. When C is a general nonspecial BN-curve, we claim this assumption is satisfied if p is general and H is a general hyperplane containing
$2p + q_1 + \cdots + q_{i - 1} + q_{i + 1} + \cdots + q_{r - 1}$
 spans H. When C is a general nonspecial BN-curve, we claim this assumption is satisfied if p is general and H is a general hyperplane containing 
 $2p$
. Indeed, in this case, the projection
$2p$
. Indeed, in this case, the projection 
 $\overline {C}$
 from p is a general BN-curve, and p remains a general point on
$\overline {C}$
 from p is a general BN-curve, and p remains a general point on 
 $\overline {C}$
. Since the sectional monodromy group of a general curve always contains the alternating group [Reference Kadets20], the corresponding points of
$\overline {C}$
. Since the sectional monodromy group of a general curve always contains the alternating group [Reference Kadets20], the corresponding points of 
 $\overline {C}$
 are in linear general position in the projection of H.
$\overline {C}$
 are in linear general position in the projection of H.
 For i between 
 $1$
 and n, write
$1$
 and n, write 
 $L_i$
 for the line joining p and
$L_i$
 for the line joining p and 
 $q_i$
. For j from
$q_i$
. For j from 
 $1$
e to
$1$
e to 
 , let
, let 
 $Q_j$
 be a plane conic passing through p,
$Q_j$
 be a plane conic passing through p, 
 $q_{n + 2j - 1}$
 and
$q_{n + 2j - 1}$
 and 
 $q_{n + 2j}$
. The following diagram illustrates the
$q_{n + 2j}$
. The following diagram illustrates the 
 $L_i$
 and
$L_i$
 and 
 $Q_j$
:
$Q_j$
:

Define

 We will study when 
 $R^\circ $
 is a limit of
$R^\circ $
 is a limit of 
 $(r + 1)$
-secant rational normal curves
$(r + 1)$
-secant rational normal curves 
 $R^t$
 in hyperplanes that are
$R^t$
 in hyperplanes that are 
 $(r + 1)$
-secant to C, in such a way that exactly two points of secancy limit together to p while the remaining points of secancy limit to
$(r + 1)$
-secant to C, in such a way that exactly two points of secancy limit together to p while the remaining points of secancy limit to 
 $q_1, q_2, \ldots , q_{r - 1}$
. For this, it is evidently necessary to have a containment of Zariski tangent spaces
$q_1, q_2, \ldots , q_{r - 1}$
. For this, it is evidently necessary to have a containment of Zariski tangent spaces 
 $T_p C \subset T_p R^\circ $
. In what follows, we will show that this condition is sufficient.
$T_p C \subset T_p R^\circ $
. In what follows, we will show that this condition is sufficient.
 Suppose that 
 $T_p C \subset T_p R^\circ $
 and
$T_p C \subset T_p R^\circ $
 and 
 
. Then the tangent line to 
 $L_i$
 (respectively to
$L_i$
 (respectively to 
 $Q_j$
) at p gives a distinguished point
$Q_j$
) at p gives a distinguished point 
 $a_i$
 (respectively
$a_i$
 (respectively 
 $b_j$
) in
$b_j$
) in 

Write 
 $\Gamma = \{a_1, \ldots , a_n, b_1, \ldots , b_{n'}\}$
, which is a collection of m points in
$\Gamma = \{a_1, \ldots , a_n, b_1, \ldots , b_{n'}\}$
, which is a collection of m points in 
 $\Lambda $
. Our linear generality assumption on
$\Lambda $
. Our linear generality assumption on 
 $2p + q_1 + \cdots + q_{r - 1}$
 implies that
$2p + q_1 + \cdots + q_{r - 1}$
 implies that 
 $\Gamma $
 is linearly general in
$\Gamma $
 is linearly general in 
 $\Lambda $
. Indeed, a linear dependence between
$\Lambda $
. Indeed, a linear dependence between 
 $a_{i_1}, \ldots , a_{i_\alpha }$
,
$a_{i_1}, \ldots , a_{i_\alpha }$
, 
 $b_{j_1}, \ldots , b_{j_\beta }$
 in
$b_{j_1}, \ldots , b_{j_\beta }$
 in 
 $\Lambda $
 implies a linear dependence between
$\Lambda $
 implies a linear dependence between 
 $2p, q_{i_1}, \ldots , q_{i_\alpha }, q_{n + 2j_1 - 1}, q_{n + 2j_1}, \ldots , q_{n + 2j_\beta - 1}, q_{n + 2j_\beta }$
 in H. Let
$2p, q_{i_1}, \ldots , q_{i_\alpha }, q_{n + 2j_1 - 1}, q_{n + 2j_1}, \ldots , q_{n + 2j_\beta - 1}, q_{n + 2j_\beta }$
 in H. Let 
 $T \subset \Lambda $
 be a general rational normal curve in
$T \subset \Lambda $
 be a general rational normal curve in 
 $\Lambda $
 passing through
$\Lambda $
 passing through 
 $\Gamma $
, and let M be a general one-secant line to T. Our argument will furthermore show that we can choose the family
$\Gamma $
, and let M be a general one-secant line to T. Our argument will furthermore show that we can choose the family 
 $R^t$
 so that the modifications along
$R^t$
 so that the modifications along 
 $R^t$
 at the points approaching p limit to M, that is, so that
$R^t$
 at the points approaching p limit to M, that is, so that 
 
fits into a flat family whose central fiber is
 
.
7.2 The construction
 To construct the desired family 
 $R^t$
, let B be the spectrum of a DVR, with special point
$R^t$
, let B be the spectrum of a DVR, with special point 
 $t = 0$
. Consider the blowup of
$t = 0$
. Consider the blowup of 
 $\mathbb {P}^r \times B$
 along
$\mathbb {P}^r \times B$
 along 
 $C \times 0$
. The special fiber X over
$C \times 0$
. The special fiber X over 
 $0$
 has two components: The first is isomorphic to the blowup
$0$
 has two components: The first is isomorphic to the blowup 
 $\operatorname {Bl}_C \mathbb {P}^r$
, and contains the proper transform
$\operatorname {Bl}_C \mathbb {P}^r$
, and contains the proper transform 
 $\hat {R}^\circ = \hat {L}_1 \cup \cdots \cup \hat {L}_n \cup \hat {Q}_1 \cup \cdots \cup \hat {Q}_{n'}$
 of
$\hat {R}^\circ = \hat {L}_1 \cup \cdots \cup \hat {L}_n \cup \hat {Q}_1 \cup \cdots \cup \hat {Q}_{n'}$
 of 
 $R^\circ $
. The second is isomorphic to the normal cone
$R^\circ $
. The second is isomorphic to the normal cone 
 $\mathbb {P}(N_C \oplus \mathcal {O}_C)$
, and contains the special fiber of the proper transform of
$\mathbb {P}(N_C \oplus \mathcal {O}_C)$
, and contains the special fiber of the proper transform of 
 $C \times B$
, which coincides with
$C \times B$
, which coincides with 
 $\mathbb {P} \mathcal {O}_C \subset \mathbb {P}(N_C \oplus \mathcal {O}_C)$
 and is isomorphic to C. The two components meet along
$\mathbb {P} \mathcal {O}_C \subset \mathbb {P}(N_C \oplus \mathcal {O}_C)$
 and is isomorphic to C. The two components meet along 
 $\mathbb {P} N_C$
, and the intersection
$\mathbb {P} N_C$
, and the intersection 
 $\hat {R}^\circ \cap \mathbb {P} N_C$
 is the finite set of points
$\hat {R}^\circ \cap \mathbb {P} N_C$
 is the finite set of points 
 $\Gamma \cup \Gamma '$
, where
$\Gamma \cup \Gamma '$
, where 
 $\Gamma = \{a_1, \ldots , a_n, b_1, \ldots , b_{n'}\}$
 lies in the fiber over p, and
$\Gamma = \{a_1, \ldots , a_n, b_1, \ldots , b_{n'}\}$
 lies in the fiber over p, and 
 $\Gamma ' = \{q_1', q_2', \ldots , q_{r - 1}'\}$
 contains one point
$\Gamma ' = \{q_1', q_2', \ldots , q_{r - 1}'\}$
 contains one point 
 $q_i'$
 in the fiber over each
$q_i'$
 in the fiber over each 
 $q_i$
. The following diagram illustrates the central fiber X:
$q_i$
. The following diagram illustrates the central fiber X:

 Let 
 $\ell _k \subset \mathbb {P}(N_C \oplus \mathcal {O}_C)|_{q_k}$
 denote the line joining
$\ell _k \subset \mathbb {P}(N_C \oplus \mathcal {O}_C)|_{q_k}$
 denote the line joining 
 $q_k$
 to
$q_k$
 to 
 $q_k'$
. These lines are pictured as dotted vertical lines in the above diagram.
$q_k'$
. These lines are pictured as dotted vertical lines in the above diagram.
 Write 
 $\Delta = M \cap T$
. The linear series
$\Delta = M \cap T$
. The linear series 
 defines a map
 defines a map 
 $\mathbb {P}^1 \simeq T \to \mathbb {P} V \simeq \mathbb {P}^{m-1}$
 of degree m, which identifies the two points of
$\mathbb {P}^1 \simeq T \to \mathbb {P} V \simeq \mathbb {P}^{m-1}$
 of degree m, which identifies the two points of 
 $\Delta $
 to a common point. Fix an embedding
$\Delta $
 to a common point. Fix an embedding 
 $\mathbb {P} V \hookrightarrow \mathbb {P}(N_C \oplus \mathcal {O}_C)$
, which agrees with the identification
$\mathbb {P} V \hookrightarrow \mathbb {P}(N_C \oplus \mathcal {O}_C)$
, which agrees with the identification 
 $\mathbb {P} H^0(\mathcal {O}_T(1)) \simeq \Lambda $
, and sends this common point to
$\mathbb {P} H^0(\mathcal {O}_T(1)) \simeq \Lambda $
, and sends this common point to 
 . Composing these maps, we obtain a map
. Composing these maps, we obtain a map 
 $f \colon \mathbb {P}^1 \simeq T \to \mathbb {P}(N_C \oplus \mathcal {O}_C)$
, which is pictured as the dotted curve in the above diagram. By construction, f passes through
$f \colon \mathbb {P}^1 \simeq T \to \mathbb {P}(N_C \oplus \mathcal {O}_C)$
, which is pictured as the dotted curve in the above diagram. By construction, f passes through 
 $\Gamma $
 and is nodal at p. Moreover, composing projection from p with f is the identity on T, and projection from p sends the Zariski tangent space of the image of f at p to M.
$\Gamma $
 and is nodal at p. Moreover, composing projection from p with f is the identity on T, and projection from p sends the Zariski tangent space of the image of f at p to M.
 We now glue 
 $\hat {R}^\circ $
 to f and the
$\hat {R}^\circ $
 to f and the 
 $\ell _i$
, that is, we consider the map
$\ell _i$
, that is, we consider the map 
 $F \colon \hat {R}^\circ \cup T \cup \ell _1 \cup \cdots \cup \ell _{r - 1} \to X$
 defined by the natural inclusions on
$F \colon \hat {R}^\circ \cup T \cup \ell _1 \cup \cdots \cup \ell _{r - 1} \to X$
 defined by the natural inclusions on 
 $\hat {R}^\circ $
 and the
$\hat {R}^\circ $
 and the 
 $\ell _i$
, and by f on T. To complete the argument, it suffices to deform F to the general fiber in a way that preserves its incidence to C. To check that this is possible, we just need to check that the corresponding obstruction space vanishes, that is, that
$\ell _i$
, and by f on T. To complete the argument, it suffices to deform F to the general fiber in a way that preserves its incidence to C. To check that this is possible, we just need to check that the corresponding obstruction space vanishes, that is, that 
 .
.
7.3 the normal space 
 $\mathbb {P} N_C|_p$
$\mathbb {P} N_C|_p$
 One tool that we will use – both to show in the next section that 
 , and in the following section to analyze the transformation
, and in the following section to analyze the transformation 
 – is the natural identification of
 – is the natural identification of 
 $\mathbb {P} N_C|_p$
 with the projection of
$\mathbb {P} N_C|_p$
 with the projection of 
 $\mathbb {P}^r$
 from
$\mathbb {P}^r$
 from 
 $T_p C$
. Under this projection,
$T_p C$
. Under this projection, 
 $\overline {q}_1, \overline {q}_2, \ldots , \overline {q}_{r - 1}$
 are a collection of
$\overline {q}_1, \overline {q}_2, \ldots , \overline {q}_{r - 1}$
 are a collection of 
 $r - 1$
 points, which are general subject to the condition of being contained in a hyperplane H. The conics
$r - 1$
 points, which are general subject to the condition of being contained in a hyperplane H. The conics 
 $Q_j$
 project to the lines through
$Q_j$
 project to the lines through 
 $\overline {q}_{n + 2j - 1}$
 and
$\overline {q}_{n + 2j - 1}$
 and 
 $\overline {q}_{n + 2j}$
. The image
$\overline {q}_{n + 2j}$
. The image 
 $\overline {p}$
 of p is identified with the osculating
$\overline {p}$
 of p is identified with the osculating 
 $2$
-plane, which coincides with
$2$
-plane, which coincides with 
 $\mathbb {P} N_{C\to p}|_p \in \mathbb {P} N_C|_p$
. Under this identification, the points
$\mathbb {P} N_{C\to p}|_p \in \mathbb {P} N_C|_p$
. Under this identification, the points 
 $a_i$
 are identified with
$a_i$
 are identified with 
 $\overline {q}_i$
, and the points
$\overline {q}_i$
, and the points 
 $b_j$
 lie on
$b_j$
 lie on 
 $\overline {Q}_j$
. The following diagram illustrates this setup:
$\overline {Q}_j$
. The following diagram illustrates this setup:

7.4 Vanishing of 
 
 Given a vector bundle E on a reducible curve 
 $X \cup _{\Gamma } Y$
, recall that the Mayer–Vietoris sequence is
$X \cup _{\Gamma } Y$
, recall that the Mayer–Vietoris sequence is 
 $$\begin{align*}0 \to E \to E|_X \oplus E|_Y \to E|_{\Gamma} \to 0.\end{align*}$$
$$\begin{align*}0 \to E \to E|_X \oplus E|_Y \to E|_{\Gamma} \to 0.\end{align*}$$
Applying this to 
 , we obtain
, we obtain 

For each of the direct summands in the middle term, we both show that 
 $H^1$
 vanishes, and extract information about the image of its global sections in
$H^1$
 vanishes, and extract information about the image of its global sections in 
 $T_{\mathbb {P} N_C}|_{\Gamma \cup \Gamma '}$
. We then combine this information to show that the rightmost map is surjective on global sections.
$T_{\mathbb {P} N_C}|_{\Gamma \cup \Gamma '}$
. We then combine this information to show that the rightmost map is surjective on global sections.
Lemma 7.1. We have 
 , and
, and 
 surjects onto
 surjects onto 
 $T_{\mathbb {P} N_C}|_{q_k'}$
.
$T_{\mathbb {P} N_C}|_{q_k'}$
.
Proof. Both statements follow from
 
, which can in turn be deduced from

using the isomorphisms
 $$\begin{align*}N_{\ell_k / \mathbb{P}(N_C \oplus \mathcal{O}_C)|_{q_k}} \simeq \mathcal{O}_{\ell_k}(1)^{r-1} \quad \text{and} \quad N_{\mathbb{P}(N_C \oplus \mathcal{O}_C)|_{q_k} / \mathbb{P}(N_C \oplus \mathcal{O}_C)} \simeq \mathcal{O}_{\mathbb{P}(N_C \oplus \mathcal{O}_C)|_{q_k}}.\\[-41pt] \end{align*}$$
$$\begin{align*}N_{\ell_k / \mathbb{P}(N_C \oplus \mathcal{O}_C)|_{q_k}} \simeq \mathcal{O}_{\ell_k}(1)^{r-1} \quad \text{and} \quad N_{\mathbb{P}(N_C \oplus \mathcal{O}_C)|_{q_k} / \mathbb{P}(N_C \oplus \mathcal{O}_C)} \simeq \mathcal{O}_{\mathbb{P}(N_C \oplus \mathcal{O}_C)|_{q_k}}.\\[-41pt] \end{align*}$$
Lemma 7.2. We have 
 . Moreover, the image
. Moreover, the image 
 consists of those deformations of
 consists of those deformations of 
 $\Gamma $
 that can be lifted to deformations of
$\Gamma $
 that can be lifted to deformations of 
 $\Lambda $
, that is, this image coincides with the full preimage in
$\Lambda $
, that is, this image coincides with the full preimage in 
 $T_{\mathbb {P} N_C}|_\Gamma $
 of the image of
$T_{\mathbb {P} N_C}|_\Gamma $
 of the image of 
 $H^0(N_{\Lambda /\mathbb {P} N_C}) \to N_{\Lambda /\mathbb {P} N_C}|_\Gamma $
.
$H^0(N_{\Lambda /\mathbb {P} N_C}) \to N_{\Lambda /\mathbb {P} N_C}|_\Gamma $
.
Proof. The map 
 $\pi _p \circ f \colon T \to T$
 is the identity map, so the pointing bundle exact sequence yields a surjection
$\pi _p \circ f \colon T \to T$
 is the identity map, so the pointing bundle exact sequence yields a surjection 

which we may further compose with the surjection 
 $N_T \to N_{\Lambda / \mathbb {P} N_C}|_T$
 coming from the inclusion
$N_T \to N_{\Lambda / \mathbb {P} N_C}|_T$
 coming from the inclusion 
 $T \subset \Lambda $
. Define K via the exact sequence
$T \subset \Lambda $
. Define K via the exact sequence 

By considering the diagram

 and noting that 
 $H^0(N_\Lambda ) \simeq H^0(N_\Lambda |_T)$
, it suffices to show that
$H^0(N_\Lambda ) \simeq H^0(N_\Lambda |_T)$
, it suffices to show that 
 $$\begin{align*}H^1(K(-\Gamma)) = H^1(N_{\Lambda / \mathbb{P} N_C}|_T) = 0.\end{align*}$$
$$\begin{align*}H^1(K(-\Gamma)) = H^1(N_{\Lambda / \mathbb{P} N_C}|_T) = 0.\end{align*}$$
The first vanishing statement follows from the pointing bundle sequence
 $$\begin{align*}0 \to N_{f \to C}(-\Delta) \simeq \mathcal{O}_T(\Delta + \Gamma) \to K \to N_{T/\Lambda} \simeq \mathcal{O}_T(\Gamma)^{\oplus (m - 3)} \to 0.\end{align*}$$
$$\begin{align*}0 \to N_{f \to C}(-\Delta) \simeq \mathcal{O}_T(\Delta + \Gamma) \to K \to N_{T/\Lambda} \simeq \mathcal{O}_T(\Gamma)^{\oplus (m - 3)} \to 0.\end{align*}$$
The second vanishing statement follows from the sequence:
 $$\begin{align*}0 \to N_{\Lambda / \mathbb{P} N_C|_p}|_T \simeq \mathcal{O}_T(1)^{r - m} \to N_{\Lambda / \mathbb{P} N_C}|_T \to N_{\mathbb{P} N_C|_p / \mathbb{P} N_C}|_T \simeq \mathcal{O}_T \to 0.\\[-42pt] \end{align*}$$
$$\begin{align*}0 \to N_{\Lambda / \mathbb{P} N_C|_p}|_T \simeq \mathcal{O}_T(1)^{r - m} \to N_{\Lambda / \mathbb{P} N_C}|_T \to N_{\mathbb{P} N_C|_p / \mathbb{P} N_C}|_T \simeq \mathcal{O}_T \to 0.\\[-42pt] \end{align*}$$
We finally consider the bundle
 $$\begin{align*}N_{\hat{R}^\circ / \operatorname{Bl}_C \mathbb{P}^r} \simeq \bigoplus_{i = 1}^n N_{\hat{L}_i / \operatorname{Bl}_C \mathbb{P}^r} \oplus \bigoplus_{j = 1}^{n'} N_{\hat{Q}_j / \operatorname{Bl}_C \mathbb{P}^r}.\end{align*}$$
$$\begin{align*}N_{\hat{R}^\circ / \operatorname{Bl}_C \mathbb{P}^r} \simeq \bigoplus_{i = 1}^n N_{\hat{L}_i / \operatorname{Bl}_C \mathbb{P}^r} \oplus \bigoplus_{j = 1}^{n'} N_{\hat{Q}_j / \operatorname{Bl}_C \mathbb{P}^r}.\end{align*}$$
To describe the images of
 $$\begin{align*}H^0(N_{\hat{L}_i / \operatorname{Bl}_C \mathbb{P}^r}) \to T_{\mathbb{P} N_C}|_{a_i} \quad \text{and} \quad H^0(N_{\hat{Q}_j / \operatorname{Bl}_C \mathbb{P}^r}) \to T_{\mathbb{P} N_C}|_{b_j},\end{align*}$$
$$\begin{align*}H^0(N_{\hat{L}_i / \operatorname{Bl}_C \mathbb{P}^r}) \to T_{\mathbb{P} N_C}|_{a_i} \quad \text{and} \quad H^0(N_{\hat{Q}_j / \operatorname{Bl}_C \mathbb{P}^r}) \to T_{\mathbb{P} N_C}|_{b_j},\end{align*}$$
we define
 $$\begin{align*}A_i = T_{q_i} C \quad \text{and} \quad B_j = \langle T_{q_{n+2j - 1}} C, T_{q_{n+2j}} C\rangle,\end{align*}$$
$$\begin{align*}A_i = T_{q_i} C \quad \text{and} \quad B_j = \langle T_{q_{n+2j - 1}} C, T_{q_{n+2j}} C\rangle,\end{align*}$$
for the tangent line or span of tangent lines, and 
 $\overline {A}_i$
 and
$\overline {A}_i$
 and 
 $\overline {B}_j$
 for their projections from
$\overline {B}_j$
 for their projections from 
 $T_p C$
.
$T_p C$
.
Lemma 7.3. We have 
 $H^1(N_{\hat {L}_i / \operatorname {Bl}_C \mathbb {P}^r}) = 0$
. Moreover, the image of
$H^1(N_{\hat {L}_i / \operatorname {Bl}_C \mathbb {P}^r}) = 0$
. Moreover, the image of 
 $H^0(N_{\hat {L}_i / \operatorname {Bl}_C \mathbb {P}^r}) \to T_{\mathbb {P} N_C}|_{a_i}$
 has the following two properties:
$H^0(N_{\hat {L}_i / \operatorname {Bl}_C \mathbb {P}^r}) \to T_{\mathbb {P} N_C}|_{a_i}$
 has the following two properties: 
- 
1. It surjects onto  $T_C|_{p}$
. $T_C|_{p}$
.
- 
2. The kernel of the map from the image to  $T_C|_{p}$
 is precisely $T_C|_{p}$
 is precisely $T_{a_i} \overline {A}_i$
. $T_{a_i} \overline {A}_i$
.
Proof. We have an exact sequence

Using this, the kernel in equation (2) is the image of
 
 in 
 $T_{\mathbb {P} N_C}|_{a_i}$
. Moreover, the normal bundle exact sequence for
$T_{\mathbb {P} N_C}|_{a_i}$
. Moreover, the normal bundle exact sequence for 
 $L_i$
 in the span
$L_i$
 in the span 
 $\overline {L_i A_i}$
 gives
$\overline {L_i A_i}$
 gives 

and 
 $N_{L_i / \overline {L_i A_i}}(-p)|_p$
 is identified with
$N_{L_i / \overline {L_i A_i}}(-p)|_p$
 is identified with 
 $T_{a_i}\overline {A}_i$
 under projection from
$T_{a_i}\overline {A}_i$
 under projection from 
 $T_pC$
.
$T_pC$
.

The same diagram chase as in Lemma 7.2 implies that it suffices to show:
 $$\begin{align*}H^1(N_{L_i / \overline{L_i A_i}}(-2p)) = H^0(N_{\overline{L_i A_i} / \mathbb{P}^r}|_{L_i} (- q_i-p)) = H^1(N_{\overline{L_i A_i} / \mathbb{P}^r}|_{L_i} (- q_i-p)) = 0.\end{align*}$$
$$\begin{align*}H^1(N_{L_i / \overline{L_i A_i}}(-2p)) = H^0(N_{\overline{L_i A_i} / \mathbb{P}^r}|_{L_i} (- q_i-p)) = H^1(N_{\overline{L_i A_i} / \mathbb{P}^r}|_{L_i} (- q_i-p)) = 0.\end{align*}$$
These statements follow from the isomorphisms
 $$\begin{align*}N_{L_i / \overline{L_i A_i}}(-2p) \simeq \mathcal{O}_{L_i}(-1) \quad \text{and} \quad N_{\overline{L_i A_i} / \mathbb{P}^r}|_{L_i} (-p - q_i) \simeq \mathcal{O}_{L_i}(-1)^{r - 2}. \end{align*}$$
$$\begin{align*}N_{L_i / \overline{L_i A_i}}(-2p) \simeq \mathcal{O}_{L_i}(-1) \quad \text{and} \quad N_{\overline{L_i A_i} / \mathbb{P}^r}|_{L_i} (-p - q_i) \simeq \mathcal{O}_{L_i}(-1)^{r - 2}. \end{align*}$$
Lemma 7.4. We have 
 $H^1(N_{\hat {Q}_j / \operatorname {Bl}_C \mathbb {P}^r}) = 0$
. Moreover, the image of
$H^1(N_{\hat {Q}_j / \operatorname {Bl}_C \mathbb {P}^r}) = 0$
. Moreover, the image of 
 $H^0(N_{\hat {Q}_j / \operatorname {Bl}_C \mathbb {P}^r}) \to T_{\mathbb {P} N_C}|_{b_j}$
 has the following two properties:
$H^0(N_{\hat {Q}_j / \operatorname {Bl}_C \mathbb {P}^r}) \to T_{\mathbb {P} N_C}|_{b_j}$
 has the following two properties: 
- 
1. It surjects onto  $T_C|_{p}$
. $T_C|_{p}$
.
- 
2. The kernel of the map from the image to  $T_C|_{p}$
 is precisely $T_C|_{p}$
 is precisely $T_{b_j} \overline {B}_j$
. $T_{b_j} \overline {B}_j$
.
Proof. We will imitate the proof of Lemma 7.3. We have an exact sequence

Moreover, the normal bundle exact sequence for 
 $Q_j$
 in the span
$Q_j$
 in the span 
 $\overline {Q_j B_j}$
 gives
$\overline {Q_j B_j}$
 gives 

and
 
 is identified with 
 $T_{b_j}\overline {B}_j$
 under projection from
$T_{b_j}\overline {B}_j$
 under projection from 
 $T_pC$
. Our goal is therefore to show both
$T_pC$
. Our goal is therefore to show both 

and
 $$\begin{align*}H^0(N_{\overline{Q_j B_j} / \mathbb{P}^r}|_{Q_j} (- q_{n+2j-1} - q_{n+2j}-p)) = H^1(N_{\overline{Q_j B_j} / \mathbb{P}^r}|_{Q_j} (- q_{n+2j-1} - q_{n+2j}-p)) = 0.\end{align*}$$
$$\begin{align*}H^0(N_{\overline{Q_j B_j} / \mathbb{P}^r}|_{Q_j} (- q_{n+2j-1} - q_{n+2j}-p)) = H^1(N_{\overline{Q_j B_j} / \mathbb{P}^r}|_{Q_j} (- q_{n+2j-1} - q_{n+2j}-p)) = 0.\end{align*}$$
The first vanishing statement follows from the exact sequence:

The second vanishing statement follows from the isomorphism
 $$\begin{align*}N_{\overline{Q_j B_j} / \mathbb{P}^r}|_{Q_j} (- q_{n+2j-1} - q_{n+2j}-p) \simeq O_{\mathbb{P}^1}(-1)^{\oplus (r - 4)}. \end{align*}$$
$$\begin{align*}N_{\overline{Q_j B_j} / \mathbb{P}^r}|_{Q_j} (- q_{n+2j-1} - q_{n+2j}-p) \simeq O_{\mathbb{P}^1}(-1)^{\oplus (r - 4)}. \end{align*}$$
 Combining these lemmas, we immediately see that 
 $H^1$
 of the middle terms in the sequence (7.1) vanish. We now see that the rightmost map of equation (7.1) is surjective on global sections, as follows. First, we apply Lemma 7.1 to handle the points of
$H^1$
 of the middle terms in the sequence (7.1) vanish. We now see that the rightmost map of equation (7.1) is surjective on global sections, as follows. First, we apply Lemma 7.1 to handle the points of 
 $\Gamma '$
; this reduces our problem to showing the surjectivity of
$\Gamma '$
; this reduces our problem to showing the surjectivity of 

Applying Lemmas 7.3 and 7.4, we see that the composition to 
 $(T_C|_p)^m$
 is surjective. It thus suffices to show that the image contains the kernel of
$(T_C|_p)^m$
 is surjective. It thus suffices to show that the image contains the kernel of 
 $T_{\mathbb {P} N_C}|_\Gamma \to (T_C|_p)^m$
, that is,
$T_{\mathbb {P} N_C}|_\Gamma \to (T_C|_p)^m$
, that is, 
 $T_{\mathbb {P} N_C|_p}|_\Gamma $
.
$T_{\mathbb {P} N_C|_p}|_\Gamma $
.
 Since removing any point from 
 $\Gamma $
 yields a linearly independent collection of points, any deformation in
$\Gamma $
 yields a linearly independent collection of points, any deformation in 
 $\mathbb {P} N_C|_p$
 of all but one point of
$\mathbb {P} N_C|_p$
 of all but one point of 
 $\Gamma $
 lifts to a deformation of
$\Gamma $
 lifts to a deformation of 
 $\Lambda $
. Combining Lemma 7.2 with Lemma 7.3, we therefore conclude that the image contains each
$\Lambda $
. Combining Lemma 7.2 with Lemma 7.3, we therefore conclude that the image contains each 
 $T_{\langle \Lambda , \overline {A}_i\rangle }|_\Gamma $
. Similarly, combining Lemma 7.2 with Lemma 7.4, we conclude that the image contains each
$T_{\langle \Lambda , \overline {A}_i\rangle }|_\Gamma $
. Similarly, combining Lemma 7.2 with Lemma 7.4, we conclude that the image contains each 
 $T_{\langle \Lambda , \overline {B}_j\rangle }|_\Gamma $
. Since the
$T_{\langle \Lambda , \overline {B}_j\rangle }|_\Gamma $
. Since the 
 $T_{\langle \Lambda , \overline {A}_i\rangle }|_\Gamma $
 and
$T_{\langle \Lambda , \overline {A}_i\rangle }|_\Gamma $
 and 
 $T_{\langle \Lambda , \overline {B}_j\rangle }|_\Gamma $
 span
$T_{\langle \Lambda , \overline {B}_j\rangle }|_\Gamma $
 span 
 $T_{\mathbb {P} N_C|_p}|_\Gamma $
, the desired result follows.
$T_{\mathbb {P} N_C|_p}|_\Gamma $
, the desired result follows.
7.5 The transformation 
 
 We next show that 
 $ M$
 is ‘suitably generic’ in
$ M$
 is ‘suitably generic’ in 
 $\mathbb {P} N_C|_p$
.
$\mathbb {P} N_C|_p$
.
Lemma 7.5. Fix a general BN-curve C and a general point 
 $p \in C$
.
$p \in C$
. 
- 
1. If  $n \geq 2$
: As $n \geq 2$
: As $q_1, q_2, \ldots , q_{r-1}$
 vary, M is linearly general in $q_1, q_2, \ldots , q_{r-1}$
 vary, M is linearly general in $\mathbb {P} N_C|_p$
. $\mathbb {P} N_C|_p$
.
- 
2. If  $n \geq 3$
: This remains true if we fix $n \geq 3$
: This remains true if we fix $q_{n + 1}, q_{n + 2}, \ldots , q_{r - 1} \in C$
 to be general. In other words, as just the remaining points $q_{n + 1}, q_{n + 2}, \ldots , q_{r - 1} \in C$
 to be general. In other words, as just the remaining points $q_1, q_2, \ldots , q_n$
 vary, M is still linearly general in $q_1, q_2, \ldots , q_n$
 vary, M is still linearly general in $\mathbb {P} N_C|_p$
. $\mathbb {P} N_C|_p$
.
Proof. Fix 
 $\Lambda \subset \mathbb {P} N_C|_p$
 of codimension
$\Lambda \subset \mathbb {P} N_C|_p$
 of codimension 
 $2$
; we want to show that M can be disjoint from
$2$
; we want to show that M can be disjoint from 
 $\Lambda $
. In the first case,
$\Lambda $
. In the first case, 
 $\overline {q}_1$
 and
$\overline {q}_1$
 and 
 $\overline {q}_2$
 are general points on
$\overline {q}_2$
 are general points on 
 $\overline {C}$
. In the second case, since any
$\overline {C}$
. In the second case, since any 
 $(r - 1 - n) + 2 = r + 1 - n\leq r-2$
 points in
$(r - 1 - n) + 2 = r + 1 - n\leq r-2$
 points in 
 $\mathbb {P}^{r-2}$
 lie in a hyperplane, the points
$\mathbb {P}^{r-2}$
 lie in a hyperplane, the points 
 $\overline {q}_1$
 and
$\overline {q}_1$
 and 
 $\overline {q}_2$
 remain general even as
$\overline {q}_2$
 remain general even as 
 $q_{n + 1}, q_{n + 2}, \ldots , q_{r - 1} \in C$
 are fixed. In either case, the line between
$q_{n + 1}, q_{n + 2}, \ldots , q_{r - 1} \in C$
 are fixed. In either case, the line between 
 $\overline {q}_1$
 and
$\overline {q}_1$
 and 
 $\overline {q}_2$
 is therefore disjoint from
$\overline {q}_2$
 is therefore disjoint from 
 $\Lambda $
. This completes the proof because M can be specialized to the line between
$\Lambda $
. This completes the proof because M can be specialized to the line between 
 $\overline {q}_1$
 and
$\overline {q}_1$
 and 
 $\overline {q}_2$
.
$\overline {q}_2$
.
 Lemma 7.5(2) is sharp, in the sense that the conclusion is always false if 
 $n = 2$
 (the subspace M is never transverse to
$n = 2$
 (the subspace M is never transverse to 
 $\Lambda = \overline {q_3 q_4 \cdots q_{r - 1}}$
). Nevertheless, there is a variant that does hold for
$\Lambda = \overline {q_3 q_4 \cdots q_{r - 1}}$
). Nevertheless, there is a variant that does hold for 
 $n = 2$
. By part (1), the general such M is disjoint from
$n = 2$
. By part (1), the general such M is disjoint from 
 $\overline {p} = \mathbb {P} N_{C \to p}|_p \in \mathbb {P} N_C|_p$
. We may therefore ask for the weaker conclusion that the image of M is linearly general in the quotient
$\overline {p} = \mathbb {P} N_{C \to p}|_p \in \mathbb {P} N_C|_p$
. We may therefore ask for the weaker conclusion that the image of M is linearly general in the quotient 
 $\mathbb {P} (N_C / N_{C \to p})|_p$
, that is, that M is transverse to any
$\mathbb {P} (N_C / N_{C \to p})|_p$
, that is, that M is transverse to any 
 $\Lambda $
 containing
$\Lambda $
 containing 
 $\overline {p}$
. In this case, the analog of Lemma 7.5(2) holds apart from a single counterexample.
$\overline {p}$
. In this case, the analog of Lemma 7.5(2) holds apart from a single counterexample.
Lemma 7.6. Suppose 
 $n = 2$
, and fix a general BN-curve C and general points
$n = 2$
, and fix a general BN-curve C and general points 
 $p, q_3, q_4, \ldots , q_{r - 1} \in C$
. If C is not an elliptic normal curve, then as
$p, q_3, q_4, \ldots , q_{r - 1} \in C$
. If C is not an elliptic normal curve, then as 
 $q_1, q_2$
 vary, M is linearly general in
$q_1, q_2$
 vary, M is linearly general in 
 $\mathbb {P} (N_C / N_{C \to p})|_p$
.
$\mathbb {P} (N_C / N_{C \to p})|_p$
.
Proof. By assumption, 
 $r-1\equiv n=2$
 mod
$r-1\equiv n=2$
 mod 
 $2$
; hence, r is odd. If C is not an elliptic curve, then since
$2$
; hence, r is odd. If C is not an elliptic curve, then since 
 $d \geq r + 1$
, either
$d \geq r + 1$
, either 
 $d \geq r + 2$
 or
$d \geq r + 2$
 or 
 $(d, g) = (r + 1, 0)$
. We consider these two cases separately.
$(d, g) = (r + 1, 0)$
. We consider these two cases separately.
 
Case 1: 
 ${\boldsymbol{d}} \geq {\boldsymbol{r}} + \mathbf {2}$
. Let
${\boldsymbol{d}} \geq {\boldsymbol{r}} + \mathbf {2}$
. Let 
 $\Lambda \subset \mathbb {P} N_C|_p$
 be any codimension
$\Lambda \subset \mathbb {P} N_C|_p$
 be any codimension 
 $2$
 plane containing
$2$
 plane containing 
 $\overline {p}$
. We will show that M can be chosen disjoint from
$\overline {p}$
. We will show that M can be chosen disjoint from 
 $\Lambda $
. Since any
$\Lambda $
. Since any 
 $(r - 1 - n) + 1 = r - n=r-2$
 points lie in a hyperplane,
$(r - 1 - n) + 1 = r - n=r-2$
 points lie in a hyperplane, 
 $\overline {q}_1$
 is a general point on
$\overline {q}_1$
 is a general point on 
 $\overline {C}$
 and is therefore not contained in
$\overline {C}$
 and is therefore not contained in 
 $\Lambda $
. Let
$\Lambda $
. Let 
 $H \simeq \mathbb {P}^{r-3}$
 be a general hyperplane containing
$H \simeq \mathbb {P}^{r-3}$
 be a general hyperplane containing 
 $\overline {q}_3, \overline {q}_4, \ldots , \overline {q}_{r - 1}$
. Since
$\overline {q}_3, \overline {q}_4, \ldots , \overline {q}_{r - 1}$
. Since 
 $\overline {p} \notin H$
 and
$\overline {p} \notin H$
 and 
 $\overline {p} \in \Lambda $
, it follows that
$\overline {p} \in \Lambda $
, it follows that 
 $\Lambda $
 is transverse to H. As
$\Lambda $
 is transverse to H. As 
 $d \geq r + 2$
, the hyperplane section
$d \geq r + 2$
, the hyperplane section 
 $H \cap \overline {C}$
 contains two points
$H \cap \overline {C}$
 contains two points 
 $\{x, y\}$
 distinct from
$\{x, y\}$
 distinct from 
 $\overline {q}_1, \overline {q}_3, \overline {q}_4, \ldots , \overline {q}_{r - 1}$
. Since the sectional monodromy group of a general curve always contains the alternating group [Reference Kadets20], the points
$\overline {q}_1, \overline {q}_3, \overline {q}_4, \ldots , \overline {q}_{r - 1}$
. Since the sectional monodromy group of a general curve always contains the alternating group [Reference Kadets20], the points 
 $\{x, y, \overline {q}_1, \overline {q}_3, \overline {q}_4, \ldots , \overline {q}_{r - 1}\}$
 are in linear general position. For any
$\{x, y, \overline {q}_1, \overline {q}_3, \overline {q}_4, \ldots , \overline {q}_{r - 1}\}$
 are in linear general position. For any 
 $k \geq 0$
, there is a unique k-plane in
$k \geq 0$
, there is a unique k-plane in 
 $\mathbb {P}^{2k+2}$
 meeting each of
$\mathbb {P}^{2k+2}$
 meeting each of 
 $k+2$
 lines in linear general position. Applying this with
$k+2$
 lines in linear general position. Applying this with 
 $k = (r-7)/2$
, we see that there is a unique
$k = (r-7)/2$
, we see that there is a unique 
 $[(r - 3)/2]$
-plane
$[(r - 3)/2]$
-plane 
 $\Lambda _x\subset H$
 containing
$\Lambda _x\subset H$
 containing 
 $\overline {q}_1$
 and x, and meeting each of the lines
$\overline {q}_1$
 and x, and meeting each of the lines 
 $\overline {Q}_j$
. If
$\overline {Q}_j$
. If 
 $\overline {q}_2 = x$
, then by this uniqueness,
$\overline {q}_2 = x$
, then by this uniqueness, 
 $\Lambda _x$
 coincides with the projection of
$\Lambda _x$
 coincides with the projection of 
 $T_p R^\circ $
. Similarly define
$T_p R^\circ $
. Similarly define 
 $\Lambda _y$
. Because M can be linearly general in either
$\Lambda _y$
. Because M can be linearly general in either 
 $\Lambda _x$
 or
$\Lambda _x$
 or 
 $\Lambda _y$
, it suffices to show that one of
$\Lambda _y$
, it suffices to show that one of 
 $\Lambda _x$
 or
$\Lambda _x$
 or 
 $\Lambda _y$
 contains a line disjoint from
$\Lambda _y$
 contains a line disjoint from 
 $\Lambda $
.
$\Lambda $
.
 Note that 
 $\Lambda _x \cap \overline {Q}_1$
 is the projection of x from
$\Lambda _x \cap \overline {Q}_1$
 is the projection of x from 
 $\langle \overline {q}_1, \overline {Q}_2, \ldots , \overline {Q}_{n'}\rangle $
 onto
$\langle \overline {q}_1, \overline {Q}_2, \ldots , \overline {Q}_{n'}\rangle $
 onto 
 $\overline {Q}_1$
, and similarly for
$\overline {Q}_1$
, and similarly for 
 $\Lambda _y \cap \overline {Q}_1$
. It follows that
$\Lambda _y \cap \overline {Q}_1$
. It follows that 
 $\Lambda _x \cap \overline {Q}_1 \neq \Lambda _y \cap \overline {Q}_1$
, and thus that
$\Lambda _x \cap \overline {Q}_1 \neq \Lambda _y \cap \overline {Q}_1$
, and thus that 
 $\langle \Lambda _x, \Lambda _y \rangle $
 contains
$\langle \Lambda _x, \Lambda _y \rangle $
 contains 
 $\overline {Q}_1$
. Similarly,
$\overline {Q}_1$
. Similarly, 
 $\langle \Lambda _x, \Lambda _y \rangle $
 contains all other
$\langle \Lambda _x, \Lambda _y \rangle $
 contains all other 
 $\overline {Q}_i$
. By inspection,
$\overline {Q}_i$
. By inspection, 
 $\langle \Lambda _x, \Lambda _y \rangle $
 contains
$\langle \Lambda _x, \Lambda _y \rangle $
 contains 
 $\overline {q}_1$
, x, and y. Therefore,
$\overline {q}_1$
, x, and y. Therefore,
 $\langle \Lambda _x, \Lambda _y \rangle = H$
. In particular,
$\langle \Lambda _x, \Lambda _y \rangle = H$
. In particular, 
 $\langle \Lambda _x, \Lambda _y \rangle $
 is a distinct hyperplane from
$\langle \Lambda _x, \Lambda _y \rangle $
 is a distinct hyperplane from 
 $\langle \Lambda , \overline {q}_1\rangle $
. Without loss of generality,
$\langle \Lambda , \overline {q}_1\rangle $
. Without loss of generality, 
 $\Lambda _x$
 contains a point
$\Lambda _x$
 contains a point 
 $z \notin \langle \Lambda , \overline {q}_1\rangle $
. Then
$z \notin \langle \Lambda , \overline {q}_1\rangle $
. Then 
 $\langle z, \overline {q}_1 \rangle $
 gives the desired line contained in
$\langle z, \overline {q}_1 \rangle $
 gives the desired line contained in 
 $\Lambda _x$
 and disjoint from
$\Lambda _x$
 and disjoint from 
 $\Lambda $
.
$\Lambda $
.
 
Case 2: 
 $({\boldsymbol{d}}, {\boldsymbol{g}}) = ({\boldsymbol{r}} + \mathbf {1}, \mathbf {0})$
. Since
$({\boldsymbol{d}}, {\boldsymbol{g}}) = ({\boldsymbol{r}} + \mathbf {1}, \mathbf {0})$
. Since 
 $\overline {f} \colon \mathbb {P}^1 \simeq C \to \mathbb {P}^{r - 2}$
 is a general rational curve of degree
$\overline {f} \colon \mathbb {P}^1 \simeq C \to \mathbb {P}^{r - 2}$
 is a general rational curve of degree 
 $r - 1$
, it suffices to verify that M is linearly general for a particular choice of
$r - 1$
, it suffices to verify that M is linearly general for a particular choice of 
 $\overline {f}$
. We may therefore take
$\overline {f}$
. We may therefore take 
 $$\begin{align*}\overline{f}(t) = \left[t^2 + 1 : t : \frac{t - p_3}{t - q_3} : \frac{t - p_4}{t - q_4} : \cdots : \frac{t - p_{r - 1}}{t - q_{r - 1}}\right],\end{align*}$$
$$\begin{align*}\overline{f}(t) = \left[t^2 + 1 : t : \frac{t - p_3}{t - q_3} : \frac{t - p_4}{t - q_4} : \cdots : \frac{t - p_{r - 1}}{t - q_{r - 1}}\right],\end{align*}$$
where 
 $p_i \in \mathbb {P}^1$
 are general. For
$p_i \in \mathbb {P}^1$
 are general. For 
 $3 \leq i \leq r-1$
, we have
$3 \leq i \leq r-1$
, we have 
 $\overline {f}(q_i) = [0: \cdots : 0 : 1 : 0 : \cdots : 0]$
, where the
$\overline {f}(q_i) = [0: \cdots : 0 : 1 : 0 : \cdots : 0]$
, where the 
 $1$
 occurs in the ith position. (The interested reader may verify that this is not actually a specialization, that is, the general rational curve of degree
$1$
 occurs in the ith position. (The interested reader may verify that this is not actually a specialization, that is, the general rational curve of degree 
 $r - 1$
 in
$r - 1$
 in 
 $\mathbb {P}^{r - 2}$
 is of this form after applying automorphisms of the source and target.)
$\mathbb {P}^{r - 2}$
 is of this form after applying automorphisms of the source and target.)
 Let 
 $H = H_s$
 be a generic hyperplane passing through
$H = H_s$
 be a generic hyperplane passing through 
 $\overline {q}_3, \overline {q}_4, \ldots , \overline {q}_{r - 1}$
, defined by the ratio of the first two coordinates being equal to s. Note that H meets
$\overline {q}_3, \overline {q}_4, \ldots , \overline {q}_{r - 1}$
, defined by the ratio of the first two coordinates being equal to s. Note that H meets 
 $\overline {f}(\mathbb {P}^1)$
 at two other points
$\overline {f}(\mathbb {P}^1)$
 at two other points 
 $\overline {q}_1 = \overline {f}(q_1)$
 and
$\overline {q}_1 = \overline {f}(q_1)$
 and 
 $\overline {q}_2 = \overline {f}(q_2)$
. The parameters
$\overline {q}_2 = \overline {f}(q_2)$
. The parameters 
 $q_1$
 and
$q_1$
 and 
 $q_2$
 are the solutions of the equation
$q_2$
 are the solutions of the equation 
 $t + t^{-1} = (t^2 + 1)/t = s$
. The projection
$t + t^{-1} = (t^2 + 1)/t = s$
. The projection 
 $\Lambda _s$
 of
$\Lambda _s$
 of 
 $T_p R^\circ $
 is the unique
$T_p R^\circ $
 is the unique 
 $[(r - 3)/2]$
-plane
$[(r - 3)/2]$
-plane 
 $\Lambda _s$
 containing
$\Lambda _s$
 containing 
 $\overline {q}_1$
 and
$\overline {q}_1$
 and 
 $\overline {q}_2$
 and meeting each of the lines
$\overline {q}_2$
 and meeting each of the lines 
 $\overline {Q}_i$
. We will show that, for
$\overline {Q}_i$
. We will show that, for 
 $s \in \mathbb {P}^1$
 generic,
$s \in \mathbb {P}^1$
 generic, 
 $\Lambda _s$
 is transverse to any fixed subspace
$\Lambda _s$
 is transverse to any fixed subspace 
 $\Lambda $
 of codimension
$\Lambda $
 of codimension 
 $2$
 containing
$2$
 containing 
 $\overline {p}$
. Hence, a general line
$\overline {p}$
. Hence, a general line 
 $M \subseteq \Lambda _s$
 is disjoint from
$M \subseteq \Lambda _s$
 is disjoint from 
 $\Lambda $
.
$\Lambda $
.
 To show this, we calculate 
 $\Lambda _s$
 explicitly. Since
$\Lambda _s$
 explicitly. Since 
 $\Lambda _s$
 is unique, it suffices to exhibit a particular
$\Lambda _s$
 is unique, it suffices to exhibit a particular 
 $[(r - 3)/2]$
-plane containing
$[(r - 3)/2]$
-plane containing 
 $\overline {q}_1$
 and
$\overline {q}_1$
 and 
 $\overline {q}_2$
 and meeting each of the lines
$\overline {q}_2$
 and meeting each of the lines 
 $\overline {Q}_i$
. We claim that we may take:
$\overline {Q}_i$
. We claim that we may take: 
 $$\begin{align*}\Lambda_s = \langle \alpha(s), \beta_1(s), \beta_2(s), \ldots, \beta_{(r - 3)/2}(s)\rangle,\end{align*}$$
$$\begin{align*}\Lambda_s = \langle \alpha(s), \beta_1(s), \beta_2(s), \ldots, \beta_{(r - 3)/2}(s)\rangle,\end{align*}$$
where
 $$ \begin{align*} \alpha(s) &= \bigg[s : 1 : \frac{p_3 q_4 s - p_3 - q_4}{q_3 q_4 - 1} : \frac{p_4 q_3 s - p_4 - q_3}{q_3 q_4 - 1} : \\ &\qquad\qquad\qquad\qquad \cdots : \frac{p_{r-2} q_{r-1} s - p_{r-2} - q_{r-1}}{q_{r-2} q_{r-1} - 1} : \frac{p_{r-1} q_{r-2} s - p_{r-1} - q_{r-2}}{q_{r-2} q_{r-1} - 1} \bigg] \\ \beta_i(s) &= \left[0 : 0 : \cdots : 0 : 0 : \frac{p_{2i+1}}{q_{2i + 1}} \cdot \frac{s - q_{2i+1} - p_{2i + 1}^{-1}}{s - q_{2i+1} - q_{2i+1}^{-1}} : \frac{p_{2i+2}}{q_{2i + 2}} \cdot \frac{s - q_{2i+2} - p_{2i + 2}^{-1}}{s - q_{2i+2} - q_{2i+2}^{-1}} : 0 : 0 : \cdots : 0 : 0\right]. \end{align*} $$
$$ \begin{align*} \alpha(s) &= \bigg[s : 1 : \frac{p_3 q_4 s - p_3 - q_4}{q_3 q_4 - 1} : \frac{p_4 q_3 s - p_4 - q_3}{q_3 q_4 - 1} : \\ &\qquad\qquad\qquad\qquad \cdots : \frac{p_{r-2} q_{r-1} s - p_{r-2} - q_{r-1}}{q_{r-2} q_{r-1} - 1} : \frac{p_{r-1} q_{r-2} s - p_{r-1} - q_{r-2}}{q_{r-2} q_{r-1} - 1} \bigg] \\ \beta_i(s) &= \left[0 : 0 : \cdots : 0 : 0 : \frac{p_{2i+1}}{q_{2i + 1}} \cdot \frac{s - q_{2i+1} - p_{2i + 1}^{-1}}{s - q_{2i+1} - q_{2i+1}^{-1}} : \frac{p_{2i+2}}{q_{2i + 2}} \cdot \frac{s - q_{2i+2} - p_{2i + 2}^{-1}}{s - q_{2i+2} - q_{2i+2}^{-1}} : 0 : 0 : \cdots : 0 : 0\right]. \end{align*} $$
Here, the nonzero entries of 
 $\beta _i(s)$
 occur in the
$\beta _i(s)$
 occur in the 
 $(2i + 1)$
st and
$(2i + 1)$
st and 
 $(2i + 2)$
nd coordinates. Indeed,
$(2i + 2)$
nd coordinates. Indeed, 
 $\Lambda _s$
 meets
$\Lambda _s$
 meets 
 $\overline {Q}_i$
 at
$\overline {Q}_i$
 at 
 $\beta _i(s)$
, so it suffices to check that
$\beta _i(s)$
, so it suffices to check that 
 $\Lambda _s$
 contains
$\Lambda _s$
 contains 
 $\overline {f}(t)$
 when
$\overline {f}(t)$
 when 
 $s = t + t^{-1}$
. This follows from the following identity, which may be verified by separately considering the first coordinate, the second coordinate, the
$s = t + t^{-1}$
. This follows from the following identity, which may be verified by separately considering the first coordinate, the second coordinate, the 
 $(2i + 1)$
st coordinate, and the
$(2i + 1)$
st coordinate, and the 
 $(2i + 2)$
nd coordinate:
$(2i + 2)$
nd coordinate: 
 $$\begin{align*}\overline{f}(t) = t \cdot \alpha(t + t^{-1}) - \sum_i \frac{(q_{2i + 1} t - 1)(q_{2i + 2} t - 1)}{q_{2i + 1} q_{2i + 2} - 1} \cdot \beta_i(t + t^{-1}).\end{align*}$$
$$\begin{align*}\overline{f}(t) = t \cdot \alpha(t + t^{-1}) - \sum_i \frac{(q_{2i + 1} t - 1)(q_{2i + 2} t - 1)}{q_{2i + 1} q_{2i + 2} - 1} \cdot \beta_i(t + t^{-1}).\end{align*}$$
This establishes that 
 $\Lambda _s$
 is given by the above explicit formula, as claimed.
$\Lambda _s$
 is given by the above explicit formula, as claimed.
 From the above explicit formulas for 
 $\alpha $
 and the
$\alpha $
 and the 
 $\beta _i$
, it is evident that
$\beta _i$
, it is evident that 
 $\alpha $
 is an isomorphism from
$\alpha $
 is an isomorphism from 
 $\mathbb {P}^1$
 onto a line L, and the
$\mathbb {P}^1$
 onto a line L, and the 
 $\beta _i$
 are quadratic maps from
$\beta _i$
 are quadratic maps from 
 $\mathbb {P}^1$
 onto lines
$\mathbb {P}^1$
 onto lines 
 $M_i$
 such that
$M_i$
 such that 
 $L, M_1, M_2, \ldots , M_{(r - 3)/2}$
 are linearly independent and span
$L, M_1, M_2, \ldots , M_{(r - 3)/2}$
 are linearly independent and span 
 $\mathbb {P}^{r - 2}$
. In fact, the above formulas for the
$\mathbb {P}^{r - 2}$
. In fact, the above formulas for the 
 $\beta _i$
 imply that, up to changing coordinates on the
$\beta _i$
 imply that, up to changing coordinates on the 
 $M_i$
, the
$M_i$
, the 
 $\beta _i$
 are independently general quadratic maps – so in particular distinct (from themselves and from
$\beta _i$
 are independently general quadratic maps – so in particular distinct (from themselves and from 
 $\alpha $
). Since the image of
$\alpha $
). Since the image of 
 $\overline {f}$
 does not lie in any union of proper linear subspaces, and
$\overline {f}$
 does not lie in any union of proper linear subspaces, and 
 $\Lambda $
 must meet
$\Lambda $
 must meet 
 $\overline {p}$
 (which is a general point on the image of
$\overline {p}$
 (which is a general point on the image of 
 $\overline {f}$
), all that remains is to prove Lemma 7.7 below.
$\overline {f}$
), all that remains is to prove Lemma 7.7 below.
Lemma 7.7. Let 
 $L_1, L_2, \ldots , L_k \subset \mathbb {P}^{2k - 1}$
 be linearly independent lines, and
$L_1, L_2, \ldots , L_k \subset \mathbb {P}^{2k - 1}$
 be linearly independent lines, and 
 $\beta _i \colon \mathbb {P}^1 \to L_i$
 be maps which are pairwise distinct (under every possible identification of
$\beta _i \colon \mathbb {P}^1 \to L_i$
 be maps which are pairwise distinct (under every possible identification of 
 $L_i$
 with
$L_i$
 with 
 $L_j$
).
$L_j$
).
 If 
 $\Lambda \subset \mathbb {P}^{2k + 1}$
 is a fixed codimension
$\Lambda \subset \mathbb {P}^{2k + 1}$
 is a fixed codimension 
 $2$
 subspace that is not transverse to
$2$
 subspace that is not transverse to 
 $\langle \beta _1(s), \beta _2(s), \ldots , \beta _k(s) \rangle $
 for
$\langle \beta _1(s), \beta _2(s), \ldots , \beta _k(s) \rangle $
 for 
 $s \in \mathbb {P}^1$
 general, then
$s \in \mathbb {P}^1$
 general, then 
 $\Lambda $
 is the span of
$\Lambda $
 is the span of 
 $k - 1$
 of the k given lines
$k - 1$
 of the k given lines 
 $L_1, L_2, \ldots , L_k$
.
$L_1, L_2, \ldots , L_k$
.
Proof. We argue by induction on k. For the base case, we take 
 $k = 1$
, which is vacuous.
$k = 1$
, which is vacuous.
 For the inductive step, we suppose 
 $k \geq 2$
, and we divide into cases based on how
$k \geq 2$
, and we divide into cases based on how 
 $\Lambda $
 meets
$\Lambda $
 meets 
 $\langle L_1, L_2, \ldots , L_{k - 1}\rangle $
. If
$\langle L_1, L_2, \ldots , L_{k - 1}\rangle $
. If 
 $\Lambda = \langle L_1, L_2, \ldots , L_{k - 1}\rangle $
, then the desired conclusion evidently holds.
$\Lambda = \langle L_1, L_2, \ldots , L_{k - 1}\rangle $
, then the desired conclusion evidently holds.
 Next, consider the case when 
 $\Lambda $
 meets
$\Lambda $
 meets 
 $\langle L_1, L_2, \ldots , L_{k - 1}\rangle $
 in codimension
$\langle L_1, L_2, \ldots , L_{k - 1}\rangle $
 in codimension 
 $1$
. Fix
$1$
. Fix 
 $s \in \mathbb {P}^1$
 general. Then the intersection
$s \in \mathbb {P}^1$
 general. Then the intersection 
 $\Lambda \cap \langle L_1, L_2, \ldots , L_{k - 1}\rangle $
 does not contain, and hence is transverse to,
$\Lambda \cap \langle L_1, L_2, \ldots , L_{k - 1}\rangle $
 does not contain, and hence is transverse to, 
 $\langle \beta _1(s), \beta _2(s), \ldots , \beta _{k-1}(s) \rangle $
 inside of
$\langle \beta _1(s), \beta _2(s), \ldots , \beta _{k-1}(s) \rangle $
 inside of 
 $\langle L_1, L_2, \ldots , L_{k - 1}\rangle \simeq \mathbb {P}^{2k - 3}$
. Also, we have
$\langle L_1, L_2, \ldots , L_{k - 1}\rangle \simeq \mathbb {P}^{2k - 3}$
. Also, we have 
 $\beta _k(s) \notin \Lambda + \langle L_1, L_2, \ldots , L_{k - 1}\rangle $
, since
$\beta _k(s) \notin \Lambda + \langle L_1, L_2, \ldots , L_{k - 1}\rangle $
, since 
 $\langle L_1, L_2, \ldots , L_k\rangle = \mathbb {P}^{2k-1}$
. Combining these,
$\langle L_1, L_2, \ldots , L_k\rangle = \mathbb {P}^{2k-1}$
. Combining these, 
 $\Lambda $
 is transverse to
$\Lambda $
 is transverse to 
 $\langle \beta _1(s), \beta _2(s), \ldots , \beta _k(s) \rangle $
 in violation of our assumption.
$\langle \beta _1(s), \beta _2(s), \ldots , \beta _k(s) \rangle $
 in violation of our assumption.
 Finally, consider the case when 
 $\Lambda $
 is transverse to
$\Lambda $
 is transverse to 
 $\langle L_1, L_2, \ldots , L_{k - 1}\rangle $
. Applying our inductive hypothesis,
$\langle L_1, L_2, \ldots , L_{k - 1}\rangle $
. Applying our inductive hypothesis, 
 $\Lambda \cap \langle L_1, L_2, \ldots , L_{k - 1}\rangle $
 is the span of
$\Lambda \cap \langle L_1, L_2, \ldots , L_{k - 1}\rangle $
 is the span of 
 $k - 2$
 of the
$k - 2$
 of the 
 $k - 1$
 given lines
$k - 1$
 given lines 
 $L_1, L_2, \ldots , L_{k-1}$
. If
$L_1, L_2, \ldots , L_{k-1}$
. If 
 $k \geq 3$
, then
$k \geq 3$
, then 
 $\Lambda $
 contains some
$\Lambda $
 contains some 
 $L_i$
, and projecting from this
$L_i$
, and projecting from this 
 $L_i$
 and applying our inductive hypothesis completes the proof.
$L_i$
 and applying our inductive hypothesis completes the proof.
 It thus remains only to rule out the case when 
 $k = 2$
 and
$k = 2$
 and 
 $\Lambda $
 is transverse to
$\Lambda $
 is transverse to 
 $L_1$
; exchanging the roles of
$L_1$
; exchanging the roles of 
 $L_1$
 and
$L_1$
 and 
 $L_2$
, we may also suppose
$L_2$
, we may also suppose 
 $\Lambda $
 is transverse to
$\Lambda $
 is transverse to 
 $L_2$
. Projection from
$L_2$
. Projection from 
 $\Lambda $
 then defines an isomorphism
$\Lambda $
 then defines an isomorphism 
 $L_1 \simeq L_2$
. By assumption,
$L_1 \simeq L_2$
. By assumption, 
 $\beta _1 \neq \beta _2$
 with respect to this identification of
$\beta _1 \neq \beta _2$
 with respect to this identification of 
 $L_1$
 with
$L_1$
 with 
 $L_2$
, that is,
$L_2$
, that is, 
 $\Lambda $
 is disjoint from
$\Lambda $
 is disjoint from 
 $\langle \beta _1(s), \beta _2(s)\rangle $
 for
$\langle \beta _1(s), \beta _2(s)\rangle $
 for 
 $s \in \mathbb {P}^1$
 generic, in violation of our assumption.
$s \in \mathbb {P}^1$
 generic, in violation of our assumption.
8 Inductive arguments
 In this section, we suppose that 
 $(d,g,r,\ell ,m)$
 is good and give several inductive arguments that reduce
$(d,g,r,\ell ,m)$
 is good and give several inductive arguments that reduce 
 $I(d, g, r, \ell , m)$
 to cases where d is smaller or where d is the same and m is smaller. In the next section, we will show that these arguments reduce all allowed instances
$I(d, g, r, \ell , m)$
 to cases where d is smaller or where d is the same and m is smaller. In the next section, we will show that these arguments reduce all allowed instances 
 $I(d, g, r, \ell , m)$
 to the already considered infinite family of cases with
$I(d, g, r, \ell , m)$
 to the already considered infinite family of cases with 
 $(\delta , \ell , m) = (1, 0, 0)$
, plus finitely many sporadic base cases in small projective spaces.
$(\delta , \ell , m) = (1, 0, 0)$
, plus finitely many sporadic base cases in small projective spaces.
8.1 Outline of inductive arguments
 In order to indicate the specializations and projections of the original BN-curve C, we introduce the following notation. Write 
 $C(0,0;0) = C$
 for our original general BN-curve of degree d and genus g in
$C(0,0;0) = C$
 for our original general BN-curve of degree d and genus g in 
 $\mathbb {P}^r$
. More generally, the notation
$\mathbb {P}^r$
. More generally, the notation 
 $C(a, b; c)$
 will denote a curve obtained from
$C(a, b; c)$
 will denote a curve obtained from 
 $C(0,0;0)$
 by peeling off a one-secant lines (as described in equation (1) below), peeling off b two-secant lines (as described in (2) below), and projecting from c general points on the curve (as described in equation (4) below). In particular,
$C(0,0;0)$
 by peeling off a one-secant lines (as described in equation (1) below), peeling off b two-secant lines (as described in (2) below), and projecting from c general points on the curve (as described in equation (4) below). In particular, 
 $C(a,b;c)$
 is a BN-curve of degree
$C(a,b;c)$
 is a BN-curve of degree 
 $d-a-b-c$
 and genus
$d-a-b-c$
 and genus 
 $g-b$
 in
$g-b$
 in 
 $\mathbb {P}^{r-c}$
. The inductive arguments we will give will make use of the following six key ingredients:
$\mathbb {P}^{r-c}$
. The inductive arguments we will give will make use of the following six key ingredients: 
- 
1. (cf. Section 5.1) We peel off a one-secant line, that is, we degenerate  $C(a, b; c)$
 to $C(a, b; c)$
 to $C(a + 1, b; c) \cup L$
, where L is a one-secant line to $C(a + 1, b; c) \cup L$
, where L is a one-secant line to $C(a + 1, b; c)$
, meeting $C(a + 1, b; c)$
, meeting $C(a + 1, b; c)$
 at a point we will call x. In this case, we write y for some point in $C(a + 1, b; c)$
 at a point we will call x. In this case, we write y for some point in $L {\smallsetminus} \{x\}$
. We always do this specialization so that all marked points determining the modification data specialize onto $L {\smallsetminus} \{x\}$
. We always do this specialization so that all marked points determining the modification data specialize onto $C(a + 1, b; c) {\smallsetminus} \{x\}$
. $C(a + 1, b; c) {\smallsetminus} \{x\}$
.
- 
2. (cf. Section 5.2) We peel off a one-secant line, that is, we degenerate  $C(a, b; c)$
 to $C(a, b; c)$
 to $C(a, b + 1; c) \cup L$
, where L is one-secant to $C(a, b + 1; c) \cup L$
, where L is one-secant to $C(a, b + 1; c)$
, meeting $C(a, b + 1; c)$
, meeting $C(a, b + 1; c)$
 at points we will denote $C(a, b + 1; c)$
 at points we will denote $\{z, w\}$
. We always do this specialization so that all marked points determining the modification data specialize onto $\{z, w\}$
. We always do this specialization so that all marked points determining the modification data specialize onto $C(a, b + 1; c) {\smallsetminus} \{z, w\}$
. $C(a, b + 1; c) {\smallsetminus} \{z, w\}$
.
- 
3. We specialize the modification data. For the modifications  , we use the technology developed in Section 7. For the remaining modifications, we specialize the marked points determining the modification data (which start out general). , we use the technology developed in Section 7. For the remaining modifications, we specialize the marked points determining the modification data (which start out general).
- 
4. We project from a point  $p \in C(a, b; c)$
. Namely, if we write $p \in C(a, b; c)$
. Namely, if we write $C(a, b; c + 1)$
 for the projection of $C(a, b; c + 1)$
 for the projection of $C(a, b; c)$
 from p, then the pointing bundle exact sequence induces (cf. equation (3.3)) an exact sequence If the number n of modifications towards p satisfies $C(a, b; c)$
 from p, then the pointing bundle exact sequence induces (cf. equation (3.3)) an exact sequence If the number n of modifications towards p satisfies $$\begin{align*}\qquad 0 \to N_{C(a, b; c) \to p}(\text{mods to}\ p) \to N_{C(a, b; c)}[\text{mods}] \to N_{C(a, b; c + 1)}[\text{residual mods}](p) \to 0.\end{align*}$$ $$\begin{align*}\qquad 0 \to N_{C(a, b; c) \to p}(\text{mods to}\ p) \to N_{C(a, b; c)}[\text{mods}] \to N_{C(a, b; c + 1)}[\text{residual mods}](p) \to 0.\end{align*}$$ $|n - \delta | < 1$
, then by Corollary 3.16 interpolation for $|n - \delta | < 1$
, then by Corollary 3.16 interpolation for $N_{C(a, b; c)}[\text {mods}]$
 follows from interpolation for $N_{C(a, b; c)}[\text {mods}]$
 follows from interpolation for $N_{C(a, b; c + 1)}[\text {residual mods}]$
. More generally, if n satisfies $N_{C(a, b; c + 1)}[\text {residual mods}]$
. More generally, if n satisfies $|n - \delta | \leq 1 - \frac {\epsilon }{r - 1}$
, then we may iterate this construction (i.e., first specialize as desired and then project) a total of $|n - \delta | \leq 1 - \frac {\epsilon }{r - 1}$
, then we may iterate this construction (i.e., first specialize as desired and then project) a total of $\epsilon $
 times. $\epsilon $
 times.
- 
5. We erase modifications that are linearly general. Namely, suppose that one of our modifications  is linearly general. Then interpolation for is linearly general. Then interpolation for follows from interpolation for N by Lemma 3.11. More generally, if M is not linearly general, but contains some subspace follows from interpolation for N by Lemma 3.11. More generally, if M is not linearly general, but contains some subspace $M_0$
 and is linearly general in the quotient $M_0$
 and is linearly general in the quotient $N|_p / M_0$
, then interpolation for $N|_p / M_0$
, then interpolation for follows from interpolation for N and follows from interpolation for N and by Lemma 3.11. by Lemma 3.11.
- 
6. We specialize any remaining  $R_i$
 to pass through the center of projection. In more detail, suppose that we projected from a point p, and that prior to this step, $R_i$
 to pass through the center of projection. In more detail, suppose that we projected from a point p, and that prior to this step, $R_i$
 remains general; write $R_i$
 remains general; write $\overline {R}_i$
 for the projection of $\overline {R}_i$
 for the projection of $R_i$
 from p. Specializing $R_i$
 from p. Specializing $R_i$
 to pass through p then induces the specialization of $R_i$
 to pass through p then induces the specialization of $\overline {R}_i$
 to a union $\overline {R}_i$
 to a union $R^{\prime }_i \cup L$
, where $R^{\prime }_i \cup L$
, where $R^{\prime }_i$
 is an r-secant rational normal curve in a hyperplane (the projection from p of the hyperplane containing $R^{\prime }_i$
 is an r-secant rational normal curve in a hyperplane (the projection from p of the hyperplane containing $R_i$
), and L is a line passing through p and a point $R_i$
), and L is a line passing through p and a point $t \in R^{\prime }_i$
. This has the effect of replacing the modification $t \in R^{\prime }_i$
. This has the effect of replacing the modification with the modifications with the modifications . By Lemma 8.1 below, . By Lemma 8.1 below, $R^{\prime }_i$
 is a general r-secant rational normal curve in a hyperplane, and $R^{\prime }_i$
 is a general r-secant rational normal curve in a hyperplane, and $t \in R_i'$
 is a general point. The modification $t \in R_i'$
 is a general point. The modification is therefore in a linearly general direction, and can be erased as above. In other words, at least when no other modifications are made at p, the combined effect of these steps is to replace is therefore in a linearly general direction, and can be erased as above. In other words, at least when no other modifications are made at p, the combined effect of these steps is to replace with with (which fits well with our inductive hypothesis). (which fits well with our inductive hypothesis).
Lemma 8.1. Let 
 $p, q_1, \ldots , q_{r} \in \mathbb {P}^{r-1}$
 be a general set of points, and write
$p, q_1, \ldots , q_{r} \in \mathbb {P}^{r-1}$
 be a general set of points, and write 
 $\overline {q}_i \in \mathbb {P}^{r - 2}$
 for the projection of
$\overline {q}_i \in \mathbb {P}^{r - 2}$
 for the projection of 
 $q_i$
 from p. Let
$q_i$
 from p. Let 
 $\overline {R} \subset \mathbb {P}^{r - 2}$
 be a general rational normal curve passing through
$\overline {R} \subset \mathbb {P}^{r - 2}$
 be a general rational normal curve passing through 
 $\overline {q}_1, \overline {q}_2, \ldots , \overline {q}_{r}$
, and
$\overline {q}_1, \overline {q}_2, \ldots , \overline {q}_{r}$
, and 
 $x \in \overline {R}$
 be a general point. Then there exists a rational normal curve R through
$x \in \overline {R}$
 be a general point. Then there exists a rational normal curve R through 
 $p, q_1, \ldots , q_{r}$
 whose tangent direction at p corresponds to x, and whose projection from p is
$p, q_1, \ldots , q_{r}$
 whose tangent direction at p corresponds to x, and whose projection from p is 
 $\overline {R}$
.
$\overline {R}$
.
Proof. Such a rational curve, if it exists, is unique. We can therefore simply compare the dimension of the space of rational curves through 
 $p, q_1, \ldots , q_{r}$
, to the dimension of the space of rational curves through
$p, q_1, \ldots , q_{r}$
, to the dimension of the space of rational curves through 
 $\overline {q}_1, \overline {q}_2, \ldots , \overline {q}_{r}$
 together with a choice of point on that rational curve. Visibly both are equal to
$\overline {q}_1, \overline {q}_2, \ldots , \overline {q}_{r}$
 together with a choice of point on that rational curve. Visibly both are equal to 
 $r - 2$
.
$r - 2$
.
8.2 Main inductive arguments
 We begin with the following proposition, which applies this method without utilizing specialization (2) (peeling off a one-secant line) and which specializes the 
 $R_i$
 as in Section 7. Since this is the first application of the method described above, we include some additional explanations which serve to clarify this method, and will be omitted in subsequent applications.
$R_i$
 as in Section 7. Since this is the first application of the method described above, we include some additional explanations which serve to clarify this method, and will be omitted in subsequent applications.
Proposition 8.2. Let 
 $\ell '$
 and
$\ell '$
 and 
 $m'$
 be integers satisfying
$m'$
 be integers satisfying 
 $0 \leq \ell ' \leq \ell $
 and
$0 \leq \ell ' \leq \ell $
 and 
 $0 \leq m' \leq m$
, with
$0 \leq m' \leq m$
, with 
 $m' = 0$
 if
$m' = 0$
 if 
 $r = 3$
. Let
$r = 3$
. Let 
 $d'$
 be an integer satisfying
$d'$
 be an integer satisfying 
 $g + r \leq d' \leq d$
, with
$g + r \leq d' \leq d$
, with 
 $d'> g + r$
 if both
$d'> g + r$
 if both 
 $g = 0$
 and
$g = 0$
 and 
 $m \neq 0$
. For
$m \neq 0$
. For 
 $1 \leq i \leq m'$
, let
$1 \leq i \leq m'$
, let 
 $n_i$
 be an integer satisfying
$n_i$
 be an integer satisfying 
 $n_i \equiv r - 1$
 mod
$n_i \equiv r - 1$
 mod 
 $2$
 and
$2$
 and 
 $2 \leq n_i \leq r - 1$
, with
$2 \leq n_i \leq r - 1$
, with 
 $n_i \neq 2$
 if
$n_i \neq 2$
 if 
 $(d', g) = (r + 1, 1)$
. Define
$(d', g) = (r + 1, 1)$
. Define 
 $$\begin{align*}\overline{\ell} = \ell - \ell' + \frac{(r - 1)m' - \sum n_i}{2} \quad \text{and} \quad \overline{m} = m - m'.\end{align*}$$
$$\begin{align*}\overline{\ell} = \ell - \ell' + \frac{(r - 1)m' - \sum n_i}{2} \quad \text{and} \quad \overline{m} = m - m'.\end{align*}$$
If
 $$\begin{align*}2m' + \ell' \leq r - 2 \quad \text{and} \quad \left|\delta - \left[\ell' + 2(d - d') + \sum n_i \right] \right| \leq 1 - \frac{1}{r - 1},\end{align*}$$
$$\begin{align*}2m' + \ell' \leq r - 2 \quad \text{and} \quad \left|\delta - \left[\ell' + 2(d - d') + \sum n_i \right] \right| \leq 1 - \frac{1}{r - 1},\end{align*}$$
and 
 $I(d' - 1, g, r - 1, \overline {\ell }, \overline {m})$
 holds, then so does
$I(d' - 1, g, r - 1, \overline {\ell }, \overline {m})$
 holds, then so does 
 $I(d, g, r, \ell , m)$
.
$I(d, g, r, \ell , m)$
.
Proof. Our goal is to establish 
 $I(d, g, r, \ell , m)$
, which asserts interpolation for
$I(d, g, r, \ell , m)$
, which asserts interpolation for 

Our assumption that 
 $g + r \leq d' \leq d$
, with
$g + r \leq d' \leq d$
, with 
 $d'> g + r$
 if both
$d'> g + r$
 if both 
 $g = 0$
 and
$g = 0$
 and 
 $m \neq 0$
, implies that we may peel off
$m \neq 0$
, implies that we may peel off 
 $d - d'$
 one-secant lines. (Recall from the discussion at the beginning of the section that this means we specialize
$d - d'$
 one-secant lines. (Recall from the discussion at the beginning of the section that this means we specialize 
 $C(0, 0; 0)$
 to the union of a BN-curve
$C(0, 0; 0)$
 to the union of a BN-curve 
 $C(d - d', 0; 0) \subset \mathbb {P}^r$
 of degree
$C(d - d', 0; 0) \subset \mathbb {P}^r$
 of degree 
 $d'$
 and genus g, with
$d'$
 and genus g, with 
 $d - d'$
 one-secant lines, in such a way that all
$d - d'$
 one-secant lines, in such a way that all 
 $u_i$
 and
$u_i$
 and 
 $v_i$
, and all points of intersection with the
$v_i$
, and all points of intersection with the 
 $R_i$
, specialize onto
$R_i$
, specialize onto 
 $C(d - d', 0; 0)$
.) This reduces our problem to showing interpolation for
$C(d - d', 0; 0)$
.) This reduces our problem to showing interpolation for 

 For 
 $1 \leq i \leq m'$
, write
$1 \leq i \leq m'$
, write 
 $n_i' = (r-1-n_i)/2$
, and degenerate
$n_i' = (r-1-n_i)/2$
, and degenerate 
 $R_i$
 as in Section 7 to the union
$R_i$
 as in Section 7 to the union 
 $R_i^\circ $
, of
$R_i^\circ $
, of 
 $n_i$
 lines
$n_i$
 lines 
 $L_{i,j}$
 meeting C at
$L_{i,j}$
 meeting C at 
 $p_i$
 and
$p_i$
 and 
 $q_{i,j}$
, and
$q_{i,j}$
, and 
 $n_i'$
 conics
$n_i'$
 conics 
 $Q_{i,j}$
 meeting C at
$Q_{i,j}$
 meeting C at 
 $p_i$
 and
$p_i$
 and 
 $q_{i,n_i + 2j-1}$
 and
$q_{i,n_i + 2j-1}$
 and 
 $q_{i,n_i + 2j}$
. This induces a specialization of N to
$q_{i,n_i + 2j}$
. This induces a specialization of N to 

Now, fix a general point 
 $p \in C$
, and specialize
$p \in C$
, and specialize 
 $p_1, p_2, \ldots , p_{m'}, v_1, v_2, \ldots , v_{\ell '}, y_1, y_2, \ldots , y_{d - d'}$
 all to p. Because
$p_1, p_2, \ldots , p_{m'}, v_1, v_2, \ldots , v_{\ell '}, y_1, y_2, \ldots , y_{d - d'}$
 all to p. Because 
 $2m' + \ell ' \leq r - 2 \leq r-1$
 by assumption, the limiting directions
$2m' + \ell ' \leq r - 2 \leq r-1$
 by assumption, the limiting directions 
 $M_1, \dots , M_{m'}, u_1, \dots , u_{\ell '}$
 are linearly independent in
$M_1, \dots , M_{m'}, u_1, \dots , u_{\ell '}$
 are linearly independent in 
 $ \mathbb {P} N_C|_p$
, and the limit is therefore treelike (cf. Defintion 3.3). Hence, this induces a further specialization of N to
$ \mathbb {P} N_C|_p$
, and the limit is therefore treelike (cf. Defintion 3.3). Hence, this induces a further specialization of N to 

where 
 $M = \operatorname {Span}(M_1, \dots , M_{m'}, u_1, \dots , u_{\ell '}) \subset \mathbb {P} N_C|_p$
. Furthermore, M is disjoint from
$M = \operatorname {Span}(M_1, \dots , M_{m'}, u_1, \dots , u_{\ell '}) \subset \mathbb {P} N_C|_p$
. Furthermore, M is disjoint from 
 $\mathbb {P} N_{C \to p}|_p$
 by combining the assumption
$\mathbb {P} N_{C \to p}|_p$
 by combining the assumption 
 $2m' + \ell ' \leq r - 2 $
 with Lemma 7.5(1). Finally, M is linearly general in
$2m' + \ell ' \leq r - 2 $
 with Lemma 7.5(1). Finally, M is linearly general in 
 $\mathbb {P} (N_C / N_{C \to p})|_p$
 by Lemmas 7.5(2) and 7.6.
$\mathbb {P} (N_C / N_{C \to p})|_p$
 by Lemmas 7.5(2) and 7.6.
 It remains to see that 
 $N^\circ $
 satisfies interpolation. For this, we project from p. In other words, as described at the beginning of the section, we use the following pointing bundle exact sequence:
$N^\circ $
 satisfies interpolation. For this, we project from p. In other words, as described at the beginning of the section, we use the following pointing bundle exact sequence: 
 $$ \begin{align} &0 \to N_{C(d - d', 0; 0) \to p}(u_1 + \cdots + u_{\ell'} + 2x_1 + \cdots + 2x_{d-d'} + (q_{1,1} + \cdots + q_{1, n_1})\nonumber\\ &\qquad + \cdots + (q_{m',1} + \cdots + q_{m', n_{m'}})) \to N^\circ \to Q(p) \to 0, \end{align} $$
$$ \begin{align} &0 \to N_{C(d - d', 0; 0) \to p}(u_1 + \cdots + u_{\ell'} + 2x_1 + \cdots + 2x_{d-d'} + (q_{1,1} + \cdots + q_{1, n_1})\nonumber\\ &\qquad + \cdots + (q_{m',1} + \cdots + q_{m', n_{m'}})) \to N^\circ \to Q(p) \to 0, \end{align} $$
where

The number of transformations towards p is 
 $\ell ' + 2(d - d') + \sum n_i$
. (These transformations occur at
$\ell ' + 2(d - d') + \sum n_i$
. (These transformations occur at 
 $u_1, \ldots , u_{\ell '}, x_1, \ldots , x_{d - d'}, q_{1,1}, \ldots , q_{1, n_1}, \ldots , q_{m',1}, \ldots , q_{m', n_{m'}}$
, c.f. equation (8.1). In particular, this specialization does not produce a positive transformation at p in the direction of p, because M is disjoint from
$u_1, \ldots , u_{\ell '}, x_1, \ldots , x_{d - d'}, q_{1,1}, \ldots , q_{1, n_1}, \ldots , q_{m',1}, \ldots , q_{m', n_{m'}}$
, c.f. equation (8.1). In particular, this specialization does not produce a positive transformation at p in the direction of p, because M is disjoint from 
 $\mathbb {P} N_{C \to p}|_p$
 as explained above.) Our assumption that
$\mathbb {P} N_{C \to p}|_p$
 as explained above.) Our assumption that 
 $|\delta - [\ell ' + 2(d - d') + \sum n_i]| \leq 1 - \frac {1}{r - 1}$
 therefore implies that interpolation for
$|\delta - [\ell ' + 2(d - d') + \sum n_i]| \leq 1 - \frac {1}{r - 1}$
 therefore implies that interpolation for 
 $N^\circ $
 follows from interpolation for Q by Corollary 3.16.
$N^\circ $
 follows from interpolation for Q by Corollary 3.16.
 We next erase the transformation at p. In other words, the only way that Q depends on the points 
 $u_1, \dots , u_{\ell '}, q_{1,1}, \dots , q_{1, n_1}, \dots , q_{m', 1}, \dots , q_{m', n_{m'}}$
 is via the dependence of
$u_1, \dots , u_{\ell '}, q_{1,1}, \dots , q_{1, n_1}, \dots , q_{m', 1}, \dots , q_{m', n_{m'}}$
 is via the dependence of 
 $\overline {M}$
 on these points. As only these points vary,
$\overline {M}$
 on these points. As only these points vary, 
 $\overline {M}$
 is linearly general. Thus interpolation for Q follows, by Lemma 3.11, from interpolation for
$\overline {M}$
 is linearly general. Thus interpolation for Q follows, by Lemma 3.11, from interpolation for 

 Finally, we specialize the remaining 
 $R_i$
, for
$R_i$
, for 
 $m' + 1 \leq i \leq m$
, to pass through p. Namely, we first specialize
$m' + 1 \leq i \leq m$
, to pass through p. Namely, we first specialize 
 $R_{m' + 1}$
 to pass through p, which induces the specialization of
$R_{m' + 1}$
 to pass through p, which induces the specialization of 
 $\overline {R}_{m' + 1}$
 to a union
$\overline {R}_{m' + 1}$
 to a union 
 $R^{\prime }_{m' + 1} \cup L$
 as described in Subsection 8.1(6). The effect of this specialization on the above bundle is to replace the modification
$R^{\prime }_{m' + 1} \cup L$
 as described in Subsection 8.1(6). The effect of this specialization on the above bundle is to replace the modification 
 
with the modifications
 
, where t is a general point on 
 $R^{\prime }_{m'+1}$
. In other words, the above bundle specializes to
$R^{\prime }_{m'+1}$
. In other words, the above bundle specializes to 

The modification
 
is in a linearly general direction and may therefore be erased by Lemma 3.11. In other words, interpolation for this bundle follows from interpolation for

Similarly specializing 
 $R_{m' + 2}$
, then
$R_{m' + 2}$
, then 
 $R_{m' + 3}$
, and so on until
$R_{m' + 3}$
, and so on until 
 $R_m$
, we reduce to interpolation for the bundle
$R_m$
, we reduce to interpolation for the bundle 

which is just the assertion 
 $I(d' - 1, g, r - 1, \overline {\ell }, \overline {m})$
.
$I(d' - 1, g, r - 1, \overline {\ell }, \overline {m})$
.
 The 
 $n_i$
 appearing in Proposition 8.2 are constrained mod
$n_i$
 appearing in Proposition 8.2 are constrained mod 
 $2$
. It is thus often difficult to apply Proposition 8.2 in situations where
$2$
. It is thus often difficult to apply Proposition 8.2 in situations where 
 $\delta $
 is an integer with the ‘wrong’ parity. We introduce the following variant, which has the advantage that its difficult parity is the opposite of the difficult parity for Proposition 8.2.
$\delta $
 is an integer with the ‘wrong’ parity. We introduce the following variant, which has the advantage that its difficult parity is the opposite of the difficult parity for Proposition 8.2.
Proposition 8.3. Let 
 $\ell '$
,
$\ell '$
, 
 $m'$
,
$m'$
, 
 $d'$
, the
$d'$
, the 
 $n_i$
,
$n_i$
, 
 $\overline {\ell }$
 and
$\overline {\ell }$
 and 
 $\overline {m}$
 be as in Proposition 8.2. If
$\overline {m}$
 be as in Proposition 8.2. If 
 $$\begin{align*}m' < m, \quad 2m' + \ell' < r - 2, \quad \text{and} \quad \left|\delta - \left[1 + \ell' + 2(d - d') + \sum n_i\right]\right| \leq 1 - \frac{1}{r - 1},\end{align*}$$
$$\begin{align*}m' < m, \quad 2m' + \ell' < r - 2, \quad \text{and} \quad \left|\delta - \left[1 + \ell' + 2(d - d') + \sum n_i\right]\right| \leq 1 - \frac{1}{r - 1},\end{align*}$$
and
 $$\begin{align*}I(d' - 1, g, r - 1, \overline{\ell}, \overline{m}), \quad I(d' - 1, g, r - 1, \overline{\ell}, \overline{m} - 1), \quad \text{and} \quad I(d' - 2, g, r - 2, \overline{\ell}, \overline{m})\end{align*}$$
$$\begin{align*}I(d' - 1, g, r - 1, \overline{\ell}, \overline{m}), \quad I(d' - 1, g, r - 1, \overline{\ell}, \overline{m} - 1), \quad \text{and} \quad I(d' - 2, g, r - 2, \overline{\ell}, \overline{m})\end{align*}$$
all hold, then so does 
 $I(d, g, r, \ell , m)$
.
$I(d, g, r, \ell , m)$
.
Proof. As in the proof of Proposition 8.2, it suffices to show that 
 $N^\circ $
 satisfies interpolation, where
$N^\circ $
 satisfies interpolation, where 

Write 
 $R_{m' + 1} \cap C = \{s_0, s_1, s_2, \ldots , s_{r - 1}, s_r\}$
. We first specialize
$R_{m' + 1} \cap C = \{s_0, s_1, s_2, \ldots , s_{r - 1}, s_r\}$
. We first specialize 
 $R_{m' + 1}$
 to a union
$R_{m' + 1}$
 to a union 
 $R \cup L$
, where L is the line through
$R \cup L$
, where L is the line through 
 $s_0$
 and
$s_0$
 and 
 $s_r$
, and R is a rational curve of degree
$s_r$
, and R is a rational curve of degree 
 $r - 2$
 passing through
$r - 2$
 passing through 
 $s_1, s_2, \ldots , s_{r - 1}$
 and meeting L at a single point. We then specialize
$s_1, s_2, \ldots , s_{r - 1}$
 and meeting L at a single point. We then specialize 
 $s_r$
 to p. These specializations induce a specialization of
$s_r$
 to p. These specializations induce a specialization of 
 $N^\circ $
 to
$N^\circ $
 to 

where 
 $M' = \langle M, \mathbb {P} N_{C \to s_0}|_p\rangle $
. We then project from p, thereby reducing to interpolation for
$M' = \langle M, \mathbb {P} N_{C \to s_0}|_p\rangle $
. We then project from p, thereby reducing to interpolation for 

Here, we have written out the modification
 
 because 
 $\overline {R}$
 also meets
$\overline {R}$
 also meets 
 $C(d-d',0;1)$
 at
$C(d-d',0;1)$
 at 
 $s_0$
. Note that
$s_0$
. Note that 
 $\overline {M}'$
 is not linearly general since it contains the fixed direction
$\overline {M}'$
 is not linearly general since it contains the fixed direction 
 $\mathbb {P} N_{C(d-d', 0;1) \to s_0}|_p$
; however, it is linearly general in the quotient by
$\mathbb {P} N_{C(d-d', 0;1) \to s_0}|_p$
; however, it is linearly general in the quotient by 
 $\mathbb {P} N_{C(d-d', 0;1) \to s_0}|_p$
. Using Lemma 3.11, we reduce to interpolation for the pair of bundles
$\mathbb {P} N_{C(d-d', 0;1) \to s_0}|_p$
. Using Lemma 3.11, we reduce to interpolation for the pair of bundles 

Specializing 
 $R_{m' + 2}, R_{m' + 3}, \ldots , R_m$
 to pass through p, interpolation for these two bundles follows from interpolation for the two bundles
$R_{m' + 2}, R_{m' + 3}, \ldots , R_m$
 to pass through p, interpolation for these two bundles follows from interpolation for the two bundles 

 By Lemma 3.14, interpolation for 
 $Q^-$
 follows from interpolation for the two closely related vector bundles where all (respectively none) of the transformations along
$Q^-$
 follows from interpolation for the two closely related vector bundles where all (respectively none) of the transformations along 
 $\overline {R}$
 are performed:
$\overline {R}$
 are performed: 


But these are the assertions 
 $I(d' - 1, g, r - 1, \overline {\ell }, \overline {m})$
 and
$I(d' - 1, g, r - 1, \overline {\ell }, \overline {m})$
 and 
 $I(d' - 1, g, r - 1, \overline {\ell }, \overline {m} - 1)$
, respectively, which hold by assumption.
$I(d' - 1, g, r - 1, \overline {\ell }, \overline {m} - 1)$
, respectively, which hold by assumption.
 It remains to see that 
 $Q^+$
 satisfies interpolation. Applying Lemma 3.12 and noting that we have already established interpolation for
$Q^+$
 satisfies interpolation. Applying Lemma 3.12 and noting that we have already established interpolation for 
 $Q^-$
 above, it suffices to check interpolation for
$Q^-$
 above, it suffices to check interpolation for 
 $Q^- / N_{C(d - d', 0; 1) \to s_0}$
, which after twisting down by
$Q^- / N_{C(d - d', 0; 1) \to s_0}$
, which after twisting down by 
 $s_0$
 is isomorphic to
$s_0$
 is isomorphic to 

Specializing 
 $R^{\prime }_{m' + 2}, R^{\prime }_{m' + 3}, \ldots , R^{\prime }_m$
 to pass through
$R^{\prime }_{m' + 2}, R^{\prime }_{m' + 3}, \ldots , R^{\prime }_m$
 to pass through 
 $s_0$
, we reduce to interpolation for
$s_0$
, we reduce to interpolation for 

But this is just the assertion 
 $I(d' - 2, g, r - 2, \overline {\ell }, \overline {m})$
, which holds by assumption.
$I(d' - 2, g, r - 2, \overline {\ell }, \overline {m})$
, which holds by assumption.
8.3 Large parameters
 Both of the main inductive arguments above impose upper bounds on 
 $2m' + \ell '$
 (depending on r). It is thus difficult to apply them when any of the remaining parameters, that is, d, g, or m, is large. (Note that
$2m' + \ell '$
 (depending on r). It is thus difficult to apply them when any of the remaining parameters, that is, d, g, or m, is large. (Note that 
 $\ell $
 is already bounded in terms of r by construction.) We therefore next give three inductive arguments that apply for large values of d, g and m, respectively.
$\ell $
 is already bounded in terms of r by construction.) We therefore next give three inductive arguments that apply for large values of d, g and m, respectively.
Proposition 8.4. Suppose that 
 $d \geq g + 2r - 1$
. If
$d \geq g + 2r - 1$
. If 
 $I(d - (r - 1), g, r, \ell , m)$
 holds, then so does
$I(d - (r - 1), g, r, \ell , m)$
 holds, then so does 
 $I(d, g, r, \ell , m)$
.
$I(d, g, r, \ell , m)$
.
Proof. We want to show interpolation for

Peeling off 
 $r - 1$
 one-secant lines, it suffices to show interpolation for
$r - 1$
 one-secant lines, it suffices to show interpolation for 

Specializing 
 $x_1, x_2, \ldots , x_{r - 1}$
 to a common point
$x_1, x_2, \ldots , x_{r - 1}$
 to a common point 
 $x \in C$
 (while leaving
$x \in C$
 (while leaving 
 $y_1, y_2, \ldots , y_{r - 1}$
 general) reduces to interpolation for
$y_1, y_2, \ldots , y_{r - 1}$
 general) reduces to interpolation for 

Removing the twist, this bundle satisfies interpolation provided that

satisfies interpolation, which is the assertion 
 $I(d - (r - 1), g, r, \ell , m)$
 that holds by assumption.
$I(d - (r - 1), g, r, \ell , m)$
 that holds by assumption.
Proposition 8.5. Suppose that 
 $g \geq r$
. If
$g \geq r$
. If 
 $I(d - (r - 1), g - r, r, \ell , m + 1)$
 holds, then so does
$I(d - (r - 1), g - r, r, \ell , m + 1)$
 holds, then so does 
 $I(d, g, r, \ell , m)$
.
$I(d, g, r, \ell , m)$
.
Proof. Since 
 $(d,g,r,\ell , m)$
 is good,
$(d,g,r,\ell , m)$
 is good, 
 $(d,g,r) \neq (2r, r+1, r)$
, and so this is a special case of Lemma 5.9.
$(d,g,r) \neq (2r, r+1, r)$
, and so this is a special case of Lemma 5.9.
Proposition 8.6. Suppose that 
 $m \geq r - 1$
. If
$m \geq r - 1$
. If 
 $I(d, g, r, \ell , m - (r - 1))$
 holds, then so does
$I(d, g, r, \ell , m - (r - 1))$
 holds, then so does 
 $I(d, g, r, \ell , m)$
.
$I(d, g, r, \ell , m)$
.
Proof. We want to show interpolation for

Fix points 
 $q_1, q_2, \ldots , q_{r + 1}$
 lying in a general hyperplane section of C. For
$q_1, q_2, \ldots , q_{r + 1}$
 lying in a general hyperplane section of C. For 
 $m - (r - 2) \leq i \leq m$
, specialize
$m - (r - 2) \leq i \leq m$
, specialize 
 $R_i$
 to a general rational curve of degree
$R_i$
 to a general rational curve of degree 
 $r - 1$
 meeting C at
$r - 1$
 meeting C at 
 $q_1, q_2, \ldots , q_{r + 1}$
. This induces a specialization of the above bundle to
$q_1, q_2, \ldots , q_{r + 1}$
. This induces a specialization of the above bundle to 

Removing the twist, this bundle satisfies interpolation provided that

satisfies interpolation, which is the assertion 
 $I(d, g, r, \ell , m - (r - 1))$
 that holds by assumption.
$I(d, g, r, \ell , m - (r - 1))$
 that holds by assumption.
8.4 Small parameters
 The remaining cases where our main inductive arguments do not apply are when various parameters are small (which deprives us of flexibility in choosing 
 $\ell '$
 and the
$\ell '$
 and the 
 $n_i$
). Some of the arguments we give here readily generalize to larger values of various parameters, but since we will not need them in that regime, we opt to simplify the exposition as far as possible. We first consider two cases where
$n_i$
). Some of the arguments we give here readily generalize to larger values of various parameters, but since we will not need them in that regime, we opt to simplify the exposition as far as possible. We first consider two cases where 
 $\ell = 0$
 and
$\ell = 0$
 and 
 $m = 1$
.
$m = 1$
.
Proposition 8.7. Suppose that 
 $\ell = 0$
 and
$\ell = 0$
 and 
 $m = 1$
. Let
$m = 1$
. Let 
 $\epsilon $
 be an integer satisfying
$\epsilon $
 be an integer satisfying 
 $0 \leq \epsilon \leq (d - g - r) / 2$
, with
$0 \leq \epsilon \leq (d - g - r) / 2$
, with 
 $\epsilon < (d - g - r)/2$
 if
$\epsilon < (d - g - r)/2$
 if 
 $g = 0$
. If
$g = 0$
. If 
 $$\begin{align*}\left| \delta - (2\epsilon + 1)\right| \leq 1 - \frac{2}{r - 1},\end{align*}$$
$$\begin{align*}\left| \delta - (2\epsilon + 1)\right| \leq 1 - \frac{2}{r - 1},\end{align*}$$
and 
 $I(d - 2\epsilon - 2, g, r - 2, 0, 1)$
 holds, then so does
$I(d - 2\epsilon - 2, g, r - 2, 0, 1)$
 holds, then so does 
 $I(d, g, r, 0, 1)$
.
$I(d, g, r, 0, 1)$
.
Proof. We want to show that
 
 satisfies interpolation. Peeling off 
 $2\epsilon $
 one-secant lines, we reduce to interpolation for
$2\epsilon $
 one-secant lines, we reduce to interpolation for 

Write 
 $R_1 \cap C = \{s_0, s_1, s_2, \ldots , s_{r - 1}, s_r\}$
. As in the proof of Proposition 8.3, specialize
$R_1 \cap C = \{s_0, s_1, s_2, \ldots , s_{r - 1}, s_r\}$
. As in the proof of Proposition 8.3, specialize 
 $R_1$
 to a union
$R_1$
 to a union 
 $R \cup L$
, where L is the line through
$R \cup L$
, where L is the line through 
 $s_0$
 and
$s_0$
 and 
 $s_r$
, and R is a rational curve of degree
$s_r$
, and R is a rational curve of degree 
 $r - 2$
 passing through
$r - 2$
 passing through 
 $s_1, s_2, \ldots , s_{r - 1}$
 and meeting L at a single point. Then specialize
$s_1, s_2, \ldots , s_{r - 1}$
 and meeting L at a single point. Then specialize 
 $y_1, y_2, \ldots , y_\epsilon $
 to
$y_1, y_2, \ldots , y_\epsilon $
 to 
 $s_0$
, and
$s_0$
, and 
 $y_{\epsilon + 1}, y_{\epsilon + 2}, \ldots , y_{2\epsilon }$
 to
$y_{\epsilon + 1}, y_{\epsilon + 2}, \ldots , y_{2\epsilon }$
 to 
 $s_r$
. This reduces our problem to interpolation for
$s_r$
. This reduces our problem to interpolation for 

Projecting from 
 $s_0$
 and then from
$s_0$
 and then from 
 $s_r$
, we reduce to interpolation for
$s_r$
, we reduce to interpolation for 
 
, which is the assertion 
 $I(d - 2\epsilon - 2, g, r - 2, 0, 1)$
 that holds by assumption.
$I(d - 2\epsilon - 2, g, r - 2, 0, 1)$
 that holds by assumption.
Proposition 8.8. If 
 $I(4k - 3, 2k - 2, 2k - 1, k - 3, 0)$
 holds, then so does
$I(4k - 3, 2k - 2, 2k - 1, k - 3, 0)$
 holds, then so does 
 $I(4k + 1, 2k - 1, 2k + 1, 0, 1)$
, provided that
$I(4k + 1, 2k - 1, 2k + 1, 0, 1)$
, provided that 
 $k \geq 3$
.
$k \geq 3$
.
Proof. Note that 
 $\delta (4k + 1, 2k - 1, 2k + 1, 0, 1) = 5$
. Our goal is to show interpolation for
$\delta (4k + 1, 2k - 1, 2k + 1, 0, 1) = 5$
. Our goal is to show interpolation for 
 
. Peeling off a one-secant line and a one-secant line, we reduce to interpolation for

Degenerate 
 $R_1$
 as in Section 7 to the union
$R_1$
 as in Section 7 to the union 
 $R_1^\circ $
, of four lines
$R_1^\circ $
, of four lines 
 $L_j$
 meeting C at p and
$L_j$
 meeting C at p and 
 $q_j$
, and
$q_j$
, and 
 $k - 2$
 conics
$k - 2$
 conics 
 $Q_j$
 meeting C at p and
$Q_j$
 meeting C at p and 
 $q_{2j + 3}$
 and
$q_{2j + 3}$
 and 
 $q_{2j + 4}$
. This induces a specialization of the above bundle to
$q_{2j + 4}$
. This induces a specialization of the above bundle to 

where M is linearly general as 
 $q_1, q_2, q_3, q_4$
 vary. Specializing z to p, we reduce to interpolation for
$q_1, q_2, q_3, q_4$
 vary. Specializing z to p, we reduce to interpolation for 

Projecting from p, we reduce to interpolation for

Specializing y and 
 $q_5$
 to w, we reduce to interpolation for
$q_5$
 to w, we reduce to interpolation for 

Projecting from w, we reduce to interpolation for

Erasing the transformation
 
, and then
 
, we reduce to interpolation for

which is the assertion 
 $I(4k - 3, 2k - 2, 2k - 1, k - 3, 0)$
 that holds by assumption.
$I(4k - 3, 2k - 2, 2k - 1, k - 3, 0)$
 that holds by assumption.
 We finally consider several arguments that are adapted to the case 
 $m = 0$
.
$m = 0$
.
Proposition 8.9. Suppose 
 $m = 0$
, and
$m = 0$
, and 
 $g \geq 3$
, and
$g \geq 3$
, and 
 $r \geq 6$
. Let
$r \geq 6$
. Let 
 $\epsilon $
 be an integer with
$\epsilon $
 be an integer with 
 $0 \leq \epsilon \leq (d - g - r) / 3$
. If
$0 \leq \epsilon \leq (d - g - r) / 3$
. If 
 $$\begin{align*}\left|\delta - (2\epsilon + 3)\right| \leq 1 - \frac{3}{r - 1},\end{align*}$$
$$\begin{align*}\left|\delta - (2\epsilon + 3)\right| \leq 1 - \frac{3}{r - 1},\end{align*}$$
and 
 $I(d - 3\epsilon - 6, g - 3, r - 3, \ell + 1, 0)$
 and
$I(d - 3\epsilon - 6, g - 3, r - 3, \ell + 1, 0)$
 and 
 $I(d - 3\epsilon - 6, g - 3, r - 3, \ell , 0)$
 hold, then so does
$I(d - 3\epsilon - 6, g - 3, r - 3, \ell , 0)$
 hold, then so does 
 $I(d, g, r, \ell , 0)$
.
$I(d, g, r, \ell , 0)$
.
Proof. Our goal is to show interpolation for

Peeling off three two-secant lines and 
 $3\epsilon $
 one-secant lines, we reduce to interpolation for
$3\epsilon $
 one-secant lines, we reduce to interpolation for 

Specializing 
 $y_1, y_2, \ldots , y_\epsilon , w_1$
 to
$y_1, y_2, \ldots , y_\epsilon , w_1$
 to 
 $z_2$
, and
$z_2$
, and 
 $y_{\epsilon + 1}, y_{\epsilon + 2}, \ldots , y_{2\epsilon }, w_2$
 to
$y_{\epsilon + 1}, y_{\epsilon + 2}, \ldots , y_{2\epsilon }, w_2$
 to 
 $z_3$
, we reduce to interpolation for
$z_3$
, we reduce to interpolation for 

Projecting from 
 $z_2$
, and then
$z_2$
, and then 
 $z_3$
, we reduce to interpolation for
$z_3$
, we reduce to interpolation for 

Specializing 
 $y_{2\epsilon + 1}, y_{2\epsilon + 2}, \ldots , y_{3\epsilon }, w_3$
 to
$y_{2\epsilon + 1}, y_{2\epsilon + 2}, \ldots , y_{3\epsilon }, w_3$
 to 
 $z_2$
, we reduce to interpolation for
$z_2$
, we reduce to interpolation for 

Projecting from 
 $z_2$
 (again), we reduce to interpolation for
$z_2$
 (again), we reduce to interpolation for 

Erasing the transformation
 
, we reduce to interpolation for the pair of bundles

The first is our assumption 
 $I(d - 3\epsilon - 6, g - 3, r - 3, \ell + 1, 0)$
. For the second, we erase the transformation
$I(d - 3\epsilon - 6, g - 3, r - 3, \ell + 1, 0)$
. For the second, we erase the transformation 
 
to reduce to interpolation for

which is our assumption 
 $I(d - 3\epsilon - 6, g - 3, r - 3, \ell , 0)$
.
$I(d - 3\epsilon - 6, g - 3, r - 3, \ell , 0)$
.
Proposition 8.10. Suppose 
 $m = 0$
 and
$m = 0$
 and 
 $g \geq 1$
 and that
$g \geq 1$
 and that 
 $$\begin{align*}|\delta - 2| \leq 1 - \frac{1}{r - 1}.\end{align*}$$
$$\begin{align*}|\delta - 2| \leq 1 - \frac{1}{r - 1}.\end{align*}$$
If 
 $I(d - 2, g - 1, r - 1, \ell + 1, 0)$
 holds, then so does
$I(d - 2, g - 1, r - 1, \ell + 1, 0)$
 holds, then so does 
 $I(d, g, r, \ell , 0)$
.
$I(d, g, r, \ell , 0)$
.
Proof. Our goal is to show interpolation for
 
. Peeling off a one-secant line, we reduce to interpolation for

Projecting from w, we reduce to interpolation for

which is our assumption 
 $I(d - 2, g - 1, r - 1, \ell + 1, 0)$
.
$I(d - 2, g - 1, r - 1, \ell + 1, 0)$
.
Proposition 8.11. Suppose 
 $m = 0$
 and
$m = 0$
 and 
 $g \geq 3$
 and
$g \geq 3$
 and 
 $r \geq 6$
 and that
$r \geq 6$
 and that 
 $$\begin{align*}|\delta - 4| \leq 1 - \frac{2}{r - 1}.\end{align*}$$
$$\begin{align*}|\delta - 4| \leq 1 - \frac{2}{r - 1}.\end{align*}$$
If 
 $I(d - 5, g - 3, r - 2, \ell + 1, 0)$
 and
$I(d - 5, g - 3, r - 2, \ell + 1, 0)$
 and 
 $I(d - 5, g - 3, r - 2, \ell , 0)$
 hold, then so does
$I(d - 5, g - 3, r - 2, \ell , 0)$
 hold, then so does 
 $I(d, g, r, \ell , 0)$
.
$I(d, g, r, \ell , 0)$
.
Proof. Our goal is to show interpolation for
 
. Peeling off three two-secant lines, we reduce to interpolation for

Specializing 
 $w_2$
 to
$w_2$
 to 
 $w_1$
, we reduce to interpolation for
$w_1$
, we reduce to interpolation for 

Projecting from 
 $w_1$
, we reduce to interpolation for
$w_1$
, we reduce to interpolation for 

Specializing 
 $w_3$
 to
$w_3$
 to 
 $w_1$
, we reduce to interpolation for
$w_1$
, we reduce to interpolation for 

Projecting from 
 $w_1$
 (again), we reduce to interpolation for
$w_1$
 (again), we reduce to interpolation for 

Erasing the transformation
 
, we reduce to interpolation for the pair of bundles

The first is our assumption 
 $I(d - 5, g - 3, r - 2, \ell + 1, 0)$
. For the second, we erase the transformation
$I(d - 5, g - 3, r - 2, \ell + 1, 0)$
. For the second, we erase the transformation 
 
to reduce to interpolation for

which is our assumption 
 $I(d - 5, g - 3, r - 2, \ell , 0)$
.
$I(d - 5, g - 3, r - 2, \ell , 0)$
.
9 
Interlude: some cases not implied by 
 $\mathbf {I}(\mathbf {d, g, r,} \ell , \mathbf {m})$
$\mathbf {I}(\mathbf {d, g, r,} \ell , \mathbf {m})$
 As explained in Section 4, our main inductive argument will establish 
 $I(d, g, r, \ell , m)$
 for all good tuples. We have already seen that:
$I(d, g, r, \ell , m)$
 for all good tuples. We have already seen that: 
- 
•  $I(d, g, r, \ell , m)$
 for all good tuples implies Theorem 1.4 except in a couple of cases. $I(d, g, r, \ell , m)$
 for all good tuples implies Theorem 1.4 except in a couple of cases.
- 
• Theorem 1.4 implies Theorem 1.2 except in a couple of cases. 
Of course, we must also check Theorem 1.4 and Theorem 1.2, respectively, in these couple of cases. The most difficult of these is Theorem 1.4 for canonical curves of even genus 
 $g \geq 8$
, which we defer to Section 13. Here, we quickly take care of all the others.
$g \geq 8$
, which we defer to Section 13. Here, we quickly take care of all the others.
9.1 Theorem 1.4 for rational curves
 Consulting Proposition 5.10, we may assume 
 $d \not \equiv 1$
 mod
$d \not \equiv 1$
 mod 
 $r - 1$
. By assumption, this implies the characteristic is distinct from
$r - 1$
. By assumption, this implies the characteristic is distinct from 
 $2$
. It thus suffices to argue that Theorem 1.4 holds for rational curves in characteristic distinct from
$2$
. It thus suffices to argue that Theorem 1.4 holds for rational curves in characteristic distinct from 
 $2$
, which we do by induction on d as follows:
$2$
, which we do by induction on d as follows:
- 
If  $\boldsymbol{\delta < 1} $
: We apply Proposition 8.2 with $\boldsymbol{\delta < 1} $
: We apply Proposition 8.2 with $d' = d$
. $d' = d$
.
- 
If  $\boldsymbol{\delta = 1}$
: We apply Proposition 6.1 (the characteristic assumption enters here). $\boldsymbol{\delta = 1}$
: We apply Proposition 6.1 (the characteristic assumption enters here).
- 
If  $\boldsymbol{1 < \delta < 2}$
: We apply Proposition 8.2 with $\boldsymbol{1 < \delta < 2}$
: We apply Proposition 8.2 with $d' = d - 1$
. $d' = d - 1$
.
- 
If  $\boldsymbol{2 \leq \delta} $
: Upon rearrangement this implies $\boldsymbol{2 \leq \delta} $
: Upon rearrangement this implies $d \geq 2r - 1$
. We may thus apply Proposition 8.4. $d \geq 2r - 1$
. We may thus apply Proposition 8.4.
9.2 Theorem 1.2 for rational curves
 Using Theorem 1.4, we deduce Theorem 1.2 for rational curves when the characteristic is distinct from 
 $2$
. Here, we show that Theorem 1.2 also holds for rational curves in characteristic
$2$
. Here, we show that Theorem 1.2 also holds for rational curves in characteristic 
 $2$
.
$2$
.
Lemma 9.1. Suppose the evaluation map 
 $\overline {M}_{g, n}(\mathbb {P}^r, d) \to (\mathbb {P}^r)^n$
 is dominant in characteristic
$\overline {M}_{g, n}(\mathbb {P}^r, d) \to (\mathbb {P}^r)^n$
 is dominant in characteristic 
 $0$
. Then it is dominant in all characteristics.
$0$
. Then it is dominant in all characteristics.
Proof. Because 
 $\overline {M}_{g, n}(\mathbb {P}^r, d)$
 is proper over
$\overline {M}_{g, n}(\mathbb {P}^r, d)$
 is proper over 
 $\operatorname {Spec} \mathbb {Z}$
, and the evaluation map is dominant in characteristic
$\operatorname {Spec} \mathbb {Z}$
, and the evaluation map is dominant in characteristic 
 $0$
, the evaluation map is therefore surjective over
$0$
, the evaluation map is therefore surjective over 
 $\operatorname {Spec} \mathbb {Z}$
.
$\operatorname {Spec} \mathbb {Z}$
.
 If 
 $g = 0$
, then
$g = 0$
, then 
 $\overline {M}_{0, n}(\mathbb {P}^r, d)$
 is irreducible in any characteristic, and so we conclude the truth of Theorem 1.2 in characteristic
$\overline {M}_{0, n}(\mathbb {P}^r, d)$
 is irreducible in any characteristic, and so we conclude the truth of Theorem 1.2 in characteristic 
 $2$
 from the truth of Theorem 1.2 in characteristic
$2$
 from the truth of Theorem 1.2 in characteristic 
 $0$
.
$0$
.
Remark 9.2. The reader might hope to apply Lemma 9.1 to higher genus curves. Unfortunately, all we learn is that some component of 
 $\overline {M}_{g, n}(\mathbb {P}^r, d)$
 dominates
$\overline {M}_{g, n}(\mathbb {P}^r, d)$
 dominates 
 $(\mathbb {P}^r)^n$
 in positive characteristic. This is a fatal flaw when the genus is positive, because there are other components, not corresponding to BN-curves, which would tell us nothing about the interpolation problem for positive-genus curves. For example, consider the component containing those stable maps which contract a smooth curve of genus g to a point and map a rational tail to
$(\mathbb {P}^r)^n$
 in positive characteristic. This is a fatal flaw when the genus is positive, because there are other components, not corresponding to BN-curves, which would tell us nothing about the interpolation problem for positive-genus curves. For example, consider the component containing those stable maps which contract a smooth curve of genus g to a point and map a rational tail to 
 $\mathbb {P}^r$
 with degree d.
$\mathbb {P}^r$
 with degree d.
9.3 Theorem 1.2 for 
 $(d, g, r) = (6, 2, 4)$
$(d, g, r) = (6, 2, 4)$
 We want to show such a BN-curve can pass through nine general points. It suffices to show 
 $H^1(N_C(-D)) = 0$
 when D is a general divisor of degree
$H^1(N_C(-D)) = 0$
 when D is a general divisor of degree 
 $9$
 on C. Peeling off a one-secant line and specializing one of the points of D onto the one-secant line, this reduces to
$9$
 on C. Peeling off a one-secant line and specializing one of the points of D onto the one-secant line, this reduces to 

where 
 $D'$
 is now a general divisor of degree
$D'$
 is now a general divisor of degree 
 $8$
 on
$8$
 on 
 $C(0, 1; 0)$
. This follows in turn from
$C(0, 1; 0)$
. This follows in turn from 

because this is a subsheaf with punctual quotient. Since 
 $v + D'$
 is a general divisor of degree
$v + D'$
 is a general divisor of degree 
 $9$
 on
$9$
 on 
 $C(0, 1; 0)$
, this follows from interpolation for
$C(0, 1; 0)$
, this follows from interpolation for 

Projecting from u, we reduce to interpolation for 
 $N_{C(0, 1; 1)}$
, which is Theorem 1.4 for
$N_{C(0, 1; 1)}$
, which is Theorem 1.4 for 
 $(d, g, r) = (4, 1, 3)$
.
$(d, g, r) = (4, 1, 3)$
.
10 Combinatorics
 In this section, we show, by a purely combinatorial argument, that the inductive arguments in Section 8 apply to all good tuples 
 $(d, g,r, \ell , m)$
 except for:
$(d, g,r, \ell , m)$
 except for: 
- 
• The infinite family  $(d, g,r, 0,0)$
 with $(d, g,r, 0,0)$
 with $\delta = 1$
 already treated in Section 6; $\delta = 1$
 already treated in Section 6;
- 
• A finite number of other cases. 
We begin by showing that these inductive arguments apply to all but finitely many tuples for each projective space, that is, for each value of r. To reduce casework, define:
 $$\begin{align*}\epsilon_0 = \epsilon_0(g) = \begin{cases} 1 & \text{if}\ g = 0; \\ 0 & \text{if}\ g \neq 0. \end{cases} \quad \text{and} \quad \epsilon_1 = \epsilon_1(d, g) = \begin{cases} 1 & \text{if}\ d> g + r; \\ 0 & \text{if}\ d = g + r. \end{cases} \end{align*}$$
$$\begin{align*}\epsilon_0 = \epsilon_0(g) = \begin{cases} 1 & \text{if}\ g = 0; \\ 0 & \text{if}\ g \neq 0. \end{cases} \quad \text{and} \quad \epsilon_1 = \epsilon_1(d, g) = \begin{cases} 1 & \text{if}\ d> g + r; \\ 0 & \text{if}\ d = g + r. \end{cases} \end{align*}$$
Proposition 10.1. Let 
 $(d, g, r, \ell , m)$
 be a good tuple. Then one of the arguments of Section 8.3 may be applied unless
$(d, g, r, \ell , m)$
 be a good tuple. Then one of the arguments of Section 8.3 may be applied unless 
 $$ \begin{align} d \leq g + 2r - 1, \quad g \leq r - 1, \quad \text{and} \quad m \leq r - 2 + \epsilon_0, \end{align} $$
$$ \begin{align} d \leq g + 2r - 1, \quad g \leq r - 1, \quad \text{and} \quad m \leq r - 2 + \epsilon_0, \end{align} $$
or unless
 $$ \begin{align} (d, g, r, \ell, m - (r - 1)) \ \text{lies in equation (XEx)}. \end{align} $$
$$ \begin{align} (d, g, r, \ell, m - (r - 1)) \ \text{lies in equation (XEx)}. \end{align} $$
Proof. If 
 $g \geq r$
, then we may apply Lemma 8.5. If
$g \geq r$
, then we may apply Lemma 8.5. If 
 $m \geq r - 1 + \epsilon _0$
, then we may apply Lemma 8.6, unless
$m \geq r - 1 + \epsilon _0$
, then we may apply Lemma 8.6, unless 
 $(d, g, r, \ell , m - (r - 1))$
 lies in equation (XEx).
$(d, g, r, \ell , m - (r - 1))$
 lies in equation (XEx).
 We may thus assume 
 $m \leq r - 2 + \epsilon _0 \leq r - 1$
. For any
$m \leq r - 2 + \epsilon _0 \leq r - 1$
. For any 
 $d'> g + r$
, this implies
$d'> g + r$
, this implies 
 $\rho (d', g, r) \geq r + 1 \geq m$
. Therefore, if
$\rho (d', g, r) \geq r + 1 \geq m$
. Therefore, if 
 $d \geq g + 2r$
, we may apply Lemma 8.4.
$d \geq g + 2r$
, we may apply Lemma 8.4.
 For any fixed r, conditions (10.1) and (10.2) describe a finite set of tuples 
 $(d, g, r, \ell , m)$
 as promised. It therefore suffices to prove:
$(d, g, r, \ell , m)$
 as promised. It therefore suffices to prove:
Theorem 10.2. If 
 $r \geq 14$
, one of the arguments in Section 8 may be applied, unless
$r \geq 14$
, one of the arguments in Section 8 may be applied, unless 
 $\ell = m = 0$
 and
$\ell = m = 0$
 and 
 $\delta = 1$
.
$\delta = 1$
.
 The remainder of this section is devoted to a proof of Theorem 10.2, which is a purely combinatorial exercise. Since all tuples in equation (XEx) have 
 $r \leq 5$
, by Proposition 10.1, we may suppose equation (10.1) is satisfied.
$r \leq 5$
, by Proposition 10.1, we may suppose equation (10.1) is satisfied.
10.1 The cases with 
 $m \neq 0$
$m \neq 0$
Our first step will be to show that Proposition 8.2 by itself handles the majority of these cases. This consists of showing that we may assign integer values to the various parameters appearing in Proposition 8.2 that satisfy the desired inequalities. We shall accomplish this using the following lemma, which gives a sufficient criterion for a system of inequalities to have an integer solution.
Lemma 10.3. Let 
 $a_i/b_i$
 and
$a_i/b_i$
 and 
 $c_j/d_j$
 be rational numbers. There is an integer n satisfying
$c_j/d_j$
 be rational numbers. There is an integer n satisfying 
 $$\begin{align*}n \geq \frac{a_i}{b_i} \ \text{for all}\ i \quad \text{and} \quad n \leq \frac{c_j}{d_j} \ \text{for all}\ j,\end{align*}$$
$$\begin{align*}n \geq \frac{a_i}{b_i} \ \text{for all}\ i \quad \text{and} \quad n \leq \frac{c_j}{d_j} \ \text{for all}\ j,\end{align*}$$
provided that, for all i and j, we have
 $$\begin{align*}\frac{a_i}{b_i} \leq \frac{c_j}{d_j} - \frac{(b_i - 1)(d_j - 1)}{b_i d_j}.\end{align*}$$
$$\begin{align*}\frac{a_i}{b_i} \leq \frac{c_j}{d_j} - \frac{(b_i - 1)(d_j - 1)}{b_i d_j}.\end{align*}$$
Proof. The collection of intervals 
 $[a_i/b_i, c_j/d_j]$
 is closed under intersection, so it suffices to check that there is an integer n satisfying
$[a_i/b_i, c_j/d_j]$
 is closed under intersection, so it suffices to check that there is an integer n satisfying 
 $$ \begin{align} \frac{a}{b} \leq n \leq \frac{c}{d} \end{align} $$
$$ \begin{align} \frac{a}{b} \leq n \leq \frac{c}{d} \end{align} $$
provided that
 $$ \begin{align} \frac{a}{b} \leq \frac{c}{d} - \frac{(b - 1)(d - 1)}{bd}. \end{align} $$
$$ \begin{align} \frac{a}{b} \leq \frac{c}{d} - \frac{(b - 1)(d - 1)}{bd}. \end{align} $$
For this, we note that equation (10.3) is equivalent to
 $$\begin{align*}\frac{a - 1}{b} < n < \frac{c + 1}{d}.\end{align*}$$
$$\begin{align*}\frac{a - 1}{b} < n < \frac{c + 1}{d}.\end{align*}$$
Since any interval of length greater than 
 $1$
 contains an integer, it suffices to have
$1$
 contains an integer, it suffices to have 
 $$\begin{align*}\frac{c + 1}{d} - \frac{a - 1}{b}> 1,\end{align*}$$
$$\begin{align*}\frac{c + 1}{d} - \frac{a - 1}{b}> 1,\end{align*}$$
or equivalently,
 $$\begin{align*}\frac{c + 1}{d} - \frac{a - 1}{b} \geq 1 + \frac{1}{bd}.\end{align*}$$
$$\begin{align*}\frac{c + 1}{d} - \frac{a - 1}{b} \geq 1 + \frac{1}{bd}.\end{align*}$$
Upon rearrangement this yields equation (10.4) as desired.
The following simple observations will be used repeatedly in what follows.
Lemma 10.4. If r is even and 
 $\delta $
 is an integer, then
$\delta $
 is an integer, then 
 $\delta \equiv m$
 mod
$\delta \equiv m$
 mod 
 $2$
.
$2$
.
Proof. This follows directly from examining the formula
 $$\begin{align*}\delta = \frac{2d + 2g - 2r + 2\ell + (r + 1)m}{r - 1}. \end{align*}$$
$$\begin{align*}\delta = \frac{2d + 2g - 2r + 2\ell + (r + 1)m}{r - 1}. \end{align*}$$
Lemma 10.5. In Propositions 8.2 and 8.3, suppose that 
 $d' \neq g + r$
 if
$d' \neq g + r$
 if 
 $d \neq g + r$
. Then
$d \neq g + r$
. Then 
 $$\begin{align*}\overline{m} \leq \rho(d' - 1, g, r - 1) \quad \text{and} \quad \overline{m} \leq \rho(d' - 2, g, r - 2).\end{align*}$$
$$\begin{align*}\overline{m} \leq \rho(d' - 1, g, r - 1) \quad \text{and} \quad \overline{m} \leq \rho(d' - 2, g, r - 2).\end{align*}$$
Proof. We divide into cases as follows.
 
Case 1: 
 ${\boldsymbol{d}} = {\boldsymbol{g}} + {\boldsymbol{r}}$
. This implies
${\boldsymbol{d}} = {\boldsymbol{g}} + {\boldsymbol{r}}$
. This implies 
 $d' = d = g + r$
; thus,
$d' = d = g + r$
; thus, 
 $g = \rho (d, g, r) = \rho (d' - 1, g, r - 1) = \rho (d' - 2, g, r - 2)$
. On the other hand, because
$g = \rho (d, g, r) = \rho (d' - 1, g, r - 1) = \rho (d' - 2, g, r - 2)$
. On the other hand, because 
 $m \leq \rho (d, g, r)$
, we have
$m \leq \rho (d, g, r)$
, we have 
 $\overline {m} = m - m' \leq m \leq \rho (d, g, r)$
.
$\overline {m} = m - m' \leq m \leq \rho (d, g, r)$
.
 
Case 2: 
 ${\boldsymbol{d}}> {\boldsymbol{g}} + {\boldsymbol{r}}$
. This implies
${\boldsymbol{d}}> {\boldsymbol{g}} + {\boldsymbol{r}}$
. This implies 
 $d' \geq g + r + 1$
; thus,
$d' \geq g + r + 1$
; thus, 
 $\rho (d' - 1, g, r - 1) \geq g + r$
 and
$\rho (d' - 1, g, r - 1) \geq g + r$
 and 
 $\rho (d' - 2, g, r - 2) \geq g + r - 1$
. On the other hand, because
$\rho (d' - 2, g, r - 2) \geq g + r - 1$
. On the other hand, because 
 $m \leq r - 2 + \epsilon _0$
, we have
$m \leq r - 2 + \epsilon _0$
, we have 
 $\overline {m} = m - m' \leq m \leq r - 1$
.
$\overline {m} = m - m' \leq m \leq r - 1$
.
The first main step of our combinatorial analysis is the following.
Proposition 10.6. Let 
 $(d, g, r, \ell , m)$
 be a good tuple satisfying (10.1) with
$(d, g, r, \ell , m)$
 be a good tuple satisfying (10.1) with 
 $m \neq 0$
 and
$m \neq 0$
 and 
 $r \geq 14$
. Then the conditions of Proposition 8.2 can be satisfied unless one of the following holds:
$r \geq 14$
. Then the conditions of Proposition 8.2 can be satisfied unless one of the following holds: 
- 
1.  $\ell = 0$
, and $\ell = 0$
, and $\delta $
 is an integer with the same parity as r, and $\delta $
 is an integer with the same parity as r, and $\delta < r$
 if r is even. $\delta < r$
 if r is even.
- 
2.  $\ell < \delta < \ell + 2$
 and $\ell < \delta < \ell + 2$
 and $g> 0$
. $g> 0$
.
- 
3.  $(d, g, r, \ell , m) = (3k + 1, k, 2k, 0, 2k - 3)$
 for some k. $(d, g, r, \ell , m) = (3k + 1, k, 2k, 0, 2k - 3)$
 for some k.
- 
4.  $(d, g, r, \ell , m) = (k + 1, 0, k, 0, 1)$
 for some k. $(d, g, r, \ell , m) = (k + 1, 0, k, 0, 1)$
 for some k.
Proof. We will show a slightly stronger statement: The conditions of Proposition 8.2 can be satisfied, together with the additional conditions that
 $$\begin{align*}m' \neq m \ \text{if} \ g = 0, \quad \text{and} \quad d' \neq g + r \ \text{if} \ d \neq g + r,\end{align*}$$
$$\begin{align*}m' \neq m \ \text{if} \ g = 0, \quad \text{and} \quad d' \neq g + r \ \text{if} \ d \neq g + r,\end{align*}$$
unless either one of the above-mentioned conditions holds or
 $$\begin{align*}(d, g, r, \ell, m) = (4k - 2, 0, 2k, 0, 1) \ \text{or} \ (4k + 1, 2k - 1, 2k, 0, 2k - 3) \quad \text{for some}\ k.\end{align*}$$
$$\begin{align*}(d, g, r, \ell, m) = (4k - 2, 0, 2k, 0, 1) \ \text{or} \ (4k + 1, 2k - 1, 2k, 0, 2k - 3) \quad \text{for some}\ k.\end{align*}$$
 This is indeed a stronger statement because if 
 $(d, g, r, \ell , m) = (4k - 2, 0, 2k, 0, 1)$
, then the conditions of Proposition 8.2 can be satisfied by taking:
$(d, g, r, \ell , m) = (4k - 2, 0, 2k, 0, 1)$
, then the conditions of Proposition 8.2 can be satisfied by taking: 
 $$\begin{align*}\ell' = 0, \quad m' = m = 1, \quad d' = d = 4k - 2, \quad n_1 = 3,\end{align*}$$
$$\begin{align*}\ell' = 0, \quad m' = m = 1, \quad d' = d = 4k - 2, \quad n_1 = 3,\end{align*}$$
and if 
 $(d, g, r, \ell , m) = (4k + 1, 2k - 1, 2k, 0, 2k - 3)$
, then the conditions of Proposition 8.2 can be satisfied by taking:
$(d, g, r, \ell , m) = (4k + 1, 2k - 1, 2k, 0, 2k - 3)$
, then the conditions of Proposition 8.2 can be satisfied by taking: 
 $$\begin{align*}\ell' = 0, \quad m' = 1, \quad d' = 4k - 1, \quad n_1 = 2k - 1.\end{align*}$$
$$\begin{align*}\ell' = 0, \quad m' = 1, \quad d' = 4k - 1, \quad n_1 = 2k - 1.\end{align*}$$
 The advantage of this first additional condition is that 
 $m' \neq m$
 implies
$m' \neq m$
 implies 
 $\overline {m} \neq 0$
. In combination with Lemma 10.5 (which applies because of the second additional condition), these conditions therefore imply that
$\overline {m} \neq 0$
. In combination with Lemma 10.5 (which applies because of the second additional condition), these conditions therefore imply that 
 $(d' - 1, g, r - 1, \overline {\ell }, \overline {m})$
 is good provided only that
$(d' - 1, g, r - 1, \overline {\ell }, \overline {m})$
 is good provided only that 
 $$\begin{align*}\overline{\ell} \leq \frac{r - 1}{2}.\end{align*}$$
$$\begin{align*}\overline{\ell} \leq \frac{r - 1}{2}.\end{align*}$$
 A further advantage of this second additional condition is that 
 $\sum n_i$
 can be any integer of the form
$\sum n_i$
 can be any integer of the form 
 $(r - 1)m' - 2n$
 where
$(r - 1)m' - 2n$
 where 

 We next write down a system of inequalities such that an integer solution (for 
 $\ell '$
,
$\ell '$
, 
 $m'$
,
$m'$
, 
 $d'$
, and n) to this system guarantees that the conditions of Proposition 8.2 can be satisfied:
$d'$
, and n) to this system guarantees that the conditions of Proposition 8.2 can be satisfied: 
 $$ \begin{align} 0 \leq m' \leq m - \epsilon_0\end{align} $$
$$ \begin{align} 0 \leq m' \leq m - \epsilon_0\end{align} $$
 $$ \begin{align} 2m' + \ell' \leq r - 2 \end{align} $$
$$ \begin{align} 2m' + \ell' \leq r - 2 \end{align} $$
 $$ \begin{align} g + r + \epsilon_1 \leq d' \leq d \end{align} $$
$$ \begin{align} g + r + \epsilon_1 \leq d' \leq d \end{align} $$
 $$ \begin{align} 0 \leq n \leq \kappa m' \end{align} $$
$$ \begin{align} 0 \leq n \leq \kappa m' \end{align} $$
 $$ \begin{align} 0 \leq \ell' \leq \ell \end{align} $$
$$ \begin{align} 0 \leq \ell' \leq \ell \end{align} $$
 $$ \begin{align} \left|\delta - \left[\ell' + 2(d - d') + (r - 1)m' - 2n \right] \right| \leq 1 - \frac{1}{r - 1} \end{align} $$
$$ \begin{align} \left|\delta - \left[\ell' + 2(d - d') + (r - 1)m' - 2n \right] \right| \leq 1 - \frac{1}{r - 1} \end{align} $$
 $$ \begin{align} \ell - \ell' + n \leq \frac{r - 1}{2}. \end{align} $$
$$ \begin{align} \ell - \ell' + n \leq \frac{r - 1}{2}. \end{align} $$
Using equation (10.9), the inequality (10.6) follows from 
 $2m' + \ell \leq r - 2$
. We introduce a new variable
$2m' + \ell \leq r - 2$
. We introduce a new variable 
 $s = d' + n$
. Replacing equation (10.6) with
$s = d' + n$
. Replacing equation (10.6) with 
 $2m' + \ell \leq r - 2$
 and rewriting the resulting system in terms of s and n, we obtain:
$2m' + \ell \leq r - 2$
 and rewriting the resulting system in terms of s and n, we obtain: 
 $$ \begin{align} 0 \leq m' \leq m - \epsilon_0 \end{align} $$
$$ \begin{align} 0 \leq m' \leq m - \epsilon_0 \end{align} $$
 $$ \begin{align} 2m' + \ell \leq r - 2 \end{align} $$
$$ \begin{align} 2m' + \ell \leq r - 2 \end{align} $$
 $$ \begin{align} s - d \leq n \leq s - g - r - \epsilon_1 \end{align} $$
$$ \begin{align} s - d \leq n \leq s - g - r - \epsilon_1 \end{align} $$
 $$ \begin{align} 0 \leq n \leq \kappa m' \end{align} $$
$$ \begin{align} 0 \leq n \leq \kappa m' \end{align} $$
 $$ \begin{align} 0 \leq \ell' \leq \ell \end{align} $$
$$ \begin{align} 0 \leq \ell' \leq \ell \end{align} $$
 $$ \begin{align} \delta - \left[2d - 2s + (r - 1)m' \right] - \frac{r - 2}{r - 1} \leq \ell' \leq \delta - \left[2d - 2s + (r - 1)m'\right] + \frac{r - 2}{r - 1} \end{align} $$
$$ \begin{align} \delta - \left[2d - 2s + (r - 1)m' \right] - \frac{r - 2}{r - 1} \leq \ell' \leq \delta - \left[2d - 2s + (r - 1)m'\right] + \frac{r - 2}{r - 1} \end{align} $$
 $$ \begin{align} n \leq \frac{r - 1}{2} + \ell' - \ell. \end{align} $$
$$ \begin{align} n \leq \frac{r - 1}{2} + \ell' - \ell. \end{align} $$
We use Lemma 10.3 to eliminate the variable n. In other words, equations (10.14), (10.15) and (10.18) involve n. Applying Lemma 10.3, there is such an integer n provided that:
 $$ \begin{align} s - d &\leq s - g - r - \epsilon_1\end{align} $$
$$ \begin{align} s - d &\leq s - g - r - \epsilon_1\end{align} $$
 $$ \begin{align} s - d &\leq \kappa m' \end{align} $$
$$ \begin{align} s - d &\leq \kappa m' \end{align} $$
 $$ \begin{align} s - d &\leq \frac{r - 1}{2} + \ell' - \ell \end{align} $$
$$ \begin{align} s - d &\leq \frac{r - 1}{2} + \ell' - \ell \end{align} $$
 $$ \begin{align} 0 &\leq s - g - r - \epsilon_1 \end{align} $$
$$ \begin{align} 0 &\leq s - g - r - \epsilon_1 \end{align} $$
 $$ \begin{align} 0 &\leq \kappa m' \end{align} $$
$$ \begin{align} 0 &\leq \kappa m' \end{align} $$
 $$ \begin{align} 0 &\leq \frac{r - 1}{2} + \ell' - \ell. \end{align} $$
$$ \begin{align} 0 &\leq \frac{r - 1}{2} + \ell' - \ell. \end{align} $$
Inequalities (10.19) and (10.23) are immediate (they follow from 
 $d \geq g + r + \epsilon _1$
 and
$d \geq g + r + \epsilon _1$
 and 
 $m' \geq 0$
 respectively). Rearranging the remaining inequalities, and including the inequalities (10.12), (10.13), (10.16) and (10.17) that do not involve n, it therefore suffices to show that there is an integer solution to the following system:
$m' \geq 0$
 respectively). Rearranging the remaining inequalities, and including the inequalities (10.12), (10.13), (10.16) and (10.17) that do not involve n, it therefore suffices to show that there is an integer solution to the following system: 
 $$ \begin{align*}\begin{array}{rl} \displaystyle 0 \leq m' &\leq m - \epsilon_0 \\[4pt] \displaystyle 2m' + \ell &\leq r - 2 \\[4pt] \displaystyle 0 \leq \ell' &\leq \ell \\[4pt] \displaystyle\delta - \left[2d - 2s + (r - 1)m' \right] - \frac{r - 2}{r - 1} &\leq \ell' \leq \delta - \left[2d - 2s + (r - 1)m'\right] + \frac{r - 2}{r - 1} \\[4pt] \displaystyle s &\leq d + \kappa m' \\[4pt] \displaystyle s - d + \ell - \frac{r - 1}{2} &\leq \ell' \\[4pt] \displaystyle g + r + \epsilon_1 &\leq s \\[4pt] \displaystyle\ell - \frac{r - 1}{2} &\leq \ell'. \end{array}\end{align*} $$
$$ \begin{align*}\begin{array}{rl} \displaystyle 0 \leq m' &\leq m - \epsilon_0 \\[4pt] \displaystyle 2m' + \ell &\leq r - 2 \\[4pt] \displaystyle 0 \leq \ell' &\leq \ell \\[4pt] \displaystyle\delta - \left[2d - 2s + (r - 1)m' \right] - \frac{r - 2}{r - 1} &\leq \ell' \leq \delta - \left[2d - 2s + (r - 1)m'\right] + \frac{r - 2}{r - 1} \\[4pt] \displaystyle s &\leq d + \kappa m' \\[4pt] \displaystyle s - d + \ell - \frac{r - 1}{2} &\leq \ell' \\[4pt] \displaystyle g + r + \epsilon_1 &\leq s \\[4pt] \displaystyle\ell - \frac{r - 1}{2} &\leq \ell'. \end{array}\end{align*} $$
Using Lemma 10.3 to eliminate the variable 
 $\ell '$
 replaces the inequalities involving
$\ell '$
 replaces the inequalities involving 
 $\ell '$
 with:
$\ell '$
 with: 
 $$ \begin{align} 0 &\leq \ell,\end{align} $$
$$ \begin{align} 0 &\leq \ell,\end{align} $$
 $$ \begin{align} 0 &\leq \delta - \left[2d - 2s + (r - 1)m'\right] + \frac{r - 2}{r - 1} \end{align} $$
$$ \begin{align} 0 &\leq \delta - \left[2d - 2s + (r - 1)m'\right] + \frac{r - 2}{r - 1} \end{align} $$
 $$ \begin{align} \delta - \left[2d - 2s + (r - 1)m' \right] - \frac{r - 2}{r - 1} &\leq \ell \end{align} $$
$$ \begin{align} \delta - \left[2d - 2s + (r - 1)m' \right] - \frac{r - 2}{r - 1} &\leq \ell \end{align} $$
 $$ \begin{align} \delta - \left[2d - 2s + (r - 1)m' \right] - \frac{r - 2}{r - 1} &\leq \delta - \left[2d - 2s + (r - 1)m'\right] + \frac{r - 2}{r - 1} - \frac{(r - 2)^2}{(r - 1)^2} \end{align} $$
$$ \begin{align} \delta - \left[2d - 2s + (r - 1)m' \right] - \frac{r - 2}{r - 1} &\leq \delta - \left[2d - 2s + (r - 1)m'\right] + \frac{r - 2}{r - 1} - \frac{(r - 2)^2}{(r - 1)^2} \end{align} $$
 $$ \begin{align} s - d + \ell - \frac{r - 1}{2} &\leq \ell \end{align} $$
$$ \begin{align} s - d + \ell - \frac{r - 1}{2} &\leq \ell \end{align} $$
 $$ \begin{align} s - d + \ell - \frac{r - 1}{2} &\leq \delta - \left[2d - 2s + (r - 1)m'\right] + \frac{r - 2}{r - 1} - \frac{r - 2}{2r - 2} \end{align} $$
$$ \begin{align} s - d + \ell - \frac{r - 1}{2} &\leq \delta - \left[2d - 2s + (r - 1)m'\right] + \frac{r - 2}{r - 1} - \frac{r - 2}{2r - 2} \end{align} $$
 $$ \begin{align} \ell - \frac{r - 1}{2} &\leq \ell \end{align} $$
$$ \begin{align} \ell - \frac{r - 1}{2} &\leq \ell \end{align} $$
 $$ \begin{align} \ell - \frac{r - 1}{2} &\leq \delta - \left[2d - 2s + (r - 1)m'\right] + \frac{r - 2}{r - 1} - \frac{r - 2}{2r - 2}. \end{align} $$
$$ \begin{align} \ell - \frac{r - 1}{2} &\leq \delta - \left[2d - 2s + (r - 1)m'\right] + \frac{r - 2}{r - 1} - \frac{r - 2}{2r - 2}. \end{align} $$
Inequalities (10.25), (10.28) and (10.31) are immediate. Simplifying the remaining ones and including the inequalities that do not involve 
 $\ell '$
, we obtain:
$\ell '$
, we obtain: 
 $$ \begin{align} s &\leq d + \frac{(r - 1)m' - \delta}{2} + \frac{\ell}{2} + \frac{r - 2}{2r - 2}\end{align} $$
$$ \begin{align} s &\leq d + \frac{(r - 1)m' - \delta}{2} + \frac{\ell}{2} + \frac{r - 2}{2r - 2}\end{align} $$
 $$ \begin{align} s &\leq d + \frac{r - 1}{2} v \end{align} $$
$$ \begin{align} s &\leq d + \frac{r - 1}{2} v \end{align} $$
 $$ \begin{align} s &\leq d + \kappa m' \end{align} $$
$$ \begin{align} s &\leq d + \kappa m' \end{align} $$
 $$ \begin{align} s &\geq d + \frac{(r - 1)m' - \delta}{2} - \frac{r - 2}{2r - 2} \end{align} $$
$$ \begin{align} s &\geq d + \frac{(r - 1)m' - \delta}{2} - \frac{r - 2}{2r - 2} \end{align} $$
 $$ \begin{align} s &\geq d + (r - 1)m' - \delta + \ell - \frac{r^2 - r - 1}{2r - 2} \end{align} $$
$$ \begin{align} s &\geq d + (r - 1)m' - \delta + \ell - \frac{r^2 - r - 1}{2r - 2} \end{align} $$
 $$ \begin{align} s &\geq d + \frac{(r - 1)m' - \delta}{2} + \frac{\ell}{2} - \frac{r^2 - r - 1}{4r - 4} \end{align} $$
$$ \begin{align} s &\geq d + \frac{(r - 1)m' - \delta}{2} + \frac{\ell}{2} - \frac{r^2 - r - 1}{4r - 4} \end{align} $$
 $$ \begin{align} s &\geq g + r + \epsilon_1 \end{align} $$
$$ \begin{align} s &\geq g + r + \epsilon_1 \end{align} $$
 $$ \begin{align} 0 \leq m' &\leq m - \epsilon_0 \end{align} $$
$$ \begin{align} 0 \leq m' &\leq m - \epsilon_0 \end{align} $$
 $$ \begin{align} 2m' + \ell &\leq r - 2. \end{align} $$
$$ \begin{align} 2m' + \ell &\leq r - 2. \end{align} $$
We now eliminate the variable s. Mostly, we will accomplish this by using Lemma 10.3, except we will compare equations (10.33) and (10.36) by ad-hoc methods. Namely, for equations (10.33) and (10.36), we want there to be an integer between
 $$\begin{align*}d + \frac{(r - 1)m' - \delta}{2} - \frac{r - 2}{2r - 2} \quad \text{and} \quad d + \frac{(r - 1)m' - \delta}{2} + \frac{\ell}{2} + \frac{r - 2}{2r - 2}.\end{align*}$$
$$\begin{align*}d + \frac{(r - 1)m' - \delta}{2} - \frac{r - 2}{2r - 2} \quad \text{and} \quad d + \frac{(r - 1)m' - \delta}{2} + \frac{\ell}{2} + \frac{r - 2}{2r - 2}.\end{align*}$$
By direct inspection, such an integer exists if and only if
 $$ \begin{align} \ell \neq 0 \quad \text{or} \quad (r - 1)m' - \delta \ \text{is not an odd integer}. \end{align} $$
$$ \begin{align} \ell \neq 0 \quad \text{or} \quad (r - 1)m' - \delta \ \text{is not an odd integer}. \end{align} $$
Eliminating s, we therefore have condition (10.42) plus the following system of inequalities:
 $$ \begin{align} d + \frac{(r - 1)m' - \delta}{2} - \frac{r - 2}{2r - 2} &\leq d + \frac{r - 1}{2} - \frac{2r - 3}{4r - 4} \end{align} $$
$$ \begin{align} d + \frac{(r - 1)m' - \delta}{2} - \frac{r - 2}{2r - 2} &\leq d + \frac{r - 1}{2} - \frac{2r - 3}{4r - 4} \end{align} $$
 $$ \begin{align} d + \frac{(r - 1)m' - \delta}{2} - \frac{r - 2}{2r - 2} &\leq d + \kappa m' \end{align} $$
$$ \begin{align} d + \frac{(r - 1)m' - \delta}{2} - \frac{r - 2}{2r - 2} &\leq d + \kappa m' \end{align} $$
 $$ \begin{align} d + (r - 1)m' - \delta + \ell - \frac{r^2 - r - 1}{2r - 2} &\leq d + \frac{(r - 1)m' - \delta}{2} + \frac{\ell}{2} + \frac{r - 2}{2r - 2} - \frac{(2r - 3)^2}{(2r - 2)^2} \end{align} $$
$$ \begin{align} d + (r - 1)m' - \delta + \ell - \frac{r^2 - r - 1}{2r - 2} &\leq d + \frac{(r - 1)m' - \delta}{2} + \frac{\ell}{2} + \frac{r - 2}{2r - 2} - \frac{(2r - 3)^2}{(2r - 2)^2} \end{align} $$
 $$ \begin{align} d + (r - 1)m' - \delta + \ell - \frac{r^2 - r - 1}{2r - 2} &\leq d + \frac{r - 1}{2} - \frac{2r - 3}{4r - 4} \end{align} $$
$$ \begin{align} d + (r - 1)m' - \delta + \ell - \frac{r^2 - r - 1}{2r - 2} &\leq d + \frac{r - 1}{2} - \frac{2r - 3}{4r - 4} \end{align} $$
 $$ \begin{align} d + (r - 1)m' - \delta + \ell - \frac{r^2 - r - 1}{2r - 2} &\leq d + \kappa m' \end{align} $$
$$ \begin{align} d + (r - 1)m' - \delta + \ell - \frac{r^2 - r - 1}{2r - 2} &\leq d + \kappa m' \end{align} $$
 $$ \begin{align} d + \frac{(r - 1)m' - \delta}{2} + \frac{\ell}{2} - \frac{r^2 - r - 1}{4r - 4} &\leq d + \frac{(r - 1)m' - \delta}{2} + \frac{\ell}{2} + \frac{r - 2}{2r - 2} - \frac{(4r - 5)(2r - 3)}{(4r - 4)(2r - 2)} \end{align} $$
$$ \begin{align} d + \frac{(r - 1)m' - \delta}{2} + \frac{\ell}{2} - \frac{r^2 - r - 1}{4r - 4} &\leq d + \frac{(r - 1)m' - \delta}{2} + \frac{\ell}{2} + \frac{r - 2}{2r - 2} - \frac{(4r - 5)(2r - 3)}{(4r - 4)(2r - 2)} \end{align} $$
 $$ \begin{align} d + \frac{(r - 1)m' - \delta}{2} + \frac{\ell}{2} - \frac{r^2 - r - 1}{4r - 4} &\leq d + \frac{r - 1}{2} - \frac{4r - 5}{8r - 8} \end{align} $$
$$ \begin{align} d + \frac{(r - 1)m' - \delta}{2} + \frac{\ell}{2} - \frac{r^2 - r - 1}{4r - 4} &\leq d + \frac{r - 1}{2} - \frac{4r - 5}{8r - 8} \end{align} $$
 $$ \begin{align} d + \frac{(r - 1)m' - \delta}{2} + \frac{\ell}{2} - \frac{r^2 - r - 1}{4r - 4} &\leq d + \kappa m' \end{align} $$
$$ \begin{align} d + \frac{(r - 1)m' - \delta}{2} + \frac{\ell}{2} - \frac{r^2 - r - 1}{4r - 4} &\leq d + \kappa m' \end{align} $$
 $$ \begin{align} g + r + \epsilon_1 &\leq d + \frac{(r - 1)m' - \delta}{2} + \frac{\ell}{2} + \frac{r - 2}{2r - 2} \end{align} $$
$$ \begin{align} g + r + \epsilon_1 &\leq d + \frac{(r - 1)m' - \delta}{2} + \frac{\ell}{2} + \frac{r - 2}{2r - 2} \end{align} $$
 $$ \begin{align} g + r + \epsilon_1 &\leq d + \frac{r - 1}{2} \end{align} $$
$$ \begin{align} g + r + \epsilon_1 &\leq d + \frac{r - 1}{2} \end{align} $$
 $$ \begin{align} g + r + \epsilon_1 &\leq d + \kappa m' \end{align} $$
$$ \begin{align} g + r + \epsilon_1 &\leq d + \kappa m' \end{align} $$
 $$ \begin{align} 0 \leq m' &\leq m - \epsilon_0 \end{align} $$
$$ \begin{align} 0 \leq m' &\leq m - \epsilon_0 \end{align} $$
 $$ \begin{align} 2m' + \ell &\leq r - 2. \end{align} $$
$$ \begin{align} 2m' + \ell &\leq r - 2. \end{align} $$
Inequalities (10.48), (10.52) and (10.53) are immediate. Moreover, equations (10.43), (10.45), (10.46) and (10.49) all follow from
 $$\begin{align*}(r - 1)m' - \delta + \ell \leq \frac{r^2 - 2r}{r - 1},\end{align*}$$
$$\begin{align*}(r - 1)m' - \delta + \ell \leq \frac{r^2 - 2r}{r - 1},\end{align*}$$
and equations (10.47) and (10.50) follow from
 $$\begin{align*}(r - 1)m' - \delta + \ell \leq \kappa m' + \frac{r^2 - r - 2}{2r - 2}.\end{align*}$$
$$\begin{align*}(r - 1)m' - \delta + \ell \leq \kappa m' + \frac{r^2 - r - 2}{2r - 2}.\end{align*}$$
The above system of inequalities therefore follows from the following system:
 $$ \begin{align} (r - 1)m' - \delta &\leq 2\kappa m' + \frac{r - 2}{r - 1} \end{align} $$
$$ \begin{align} (r - 1)m' - \delta &\leq 2\kappa m' + \frac{r - 2}{r - 1} \end{align} $$
 $$ \begin{align} (r - 1)m' - \delta + \ell &\leq \frac{r^2 - 2r}{r - 1} \end{align} $$
$$ \begin{align} (r - 1)m' - \delta + \ell &\leq \frac{r^2 - 2r}{r - 1} \end{align} $$
 $$ \begin{align} (r - 1)m' - \delta + \ell &\leq \kappa m' + \frac{r^2 - r - 2}{2r - 2} \end{align} $$
$$ \begin{align} (r - 1)m' - \delta + \ell &\leq \kappa m' + \frac{r^2 - r - 2}{2r - 2} \end{align} $$
 $$ \begin{align} g + r + \epsilon_1 &\leq d + \frac{(r - 1)m' - \delta}{2} + \frac{\ell}{2} + \frac{r - 2}{2r - 2} \end{align} $$
$$ \begin{align} g + r + \epsilon_1 &\leq d + \frac{(r - 1)m' - \delta}{2} + \frac{\ell}{2} + \frac{r - 2}{2r - 2} \end{align} $$
 $$ \begin{align} 0 &\leq m' \leq m - \epsilon_0 \end{align} $$
$$ \begin{align} 0 &\leq m' \leq m - \epsilon_0 \end{align} $$
 $$ \begin{align} 2m' + \ell &\leq r - 2. \end{align} $$
$$ \begin{align} 2m' + \ell &\leq r - 2. \end{align} $$
All that remains is therefore to show that there is an integer 
 $m'$
 satisfying equations10.56–10.61 plus condition (10.42). For this, we divide into three cases as follows.
$m'$
 satisfying equations10.56–10.61 plus condition (10.42). For this, we divide into three cases as follows.
 
Case 1: 
 $\boldsymbol {\ell } = \mathbf {0}$
 and
$\boldsymbol {\ell } = \mathbf {0}$
 and 
 ${\boldsymbol{r}}$
 is even and
${\boldsymbol{r}}$
 is even and 
 $\boldsymbol {\delta }$
 is an even integer. In this case, we will take
$\boldsymbol {\delta }$
 is an even integer. In this case, we will take 
 $m' = 2$
, which evidently satisfies equation (10.42). Substituting
$m' = 2$
, which evidently satisfies equation (10.42). Substituting 
 $\ell = 0$
 and
$\ell = 0$
 and 
 $m' = 2$
 into equations 10.56–10.61, it remains only to verify:
$m' = 2$
 into equations 10.56–10.61, it remains only to verify: 
 $$ \begin{align} 2(r - 1) - \delta &\leq \frac{2r^2 - 9r + 6}{r - 1} \end{align} $$
$$ \begin{align} 2(r - 1) - \delta &\leq \frac{2r^2 - 9r + 6}{r - 1} \end{align} $$
 $$ \begin{align} 2(r - 1) - \delta &\leq \frac{r^2 - 2r}{r - 1} \end{align} $$
$$ \begin{align} 2(r - 1) - \delta &\leq \frac{r^2 - 2r}{r - 1} \end{align} $$
 $$ \begin{align} 2(r - 1) - \delta &\leq \frac{3r^2 - 11r + 6}{2r - 2} \end{align} $$
$$ \begin{align} 2(r - 1) - \delta &\leq \frac{3r^2 - 11r + 6}{2r - 2} \end{align} $$
 $$ \begin{align} g + r + \epsilon_1 &\leq d + \frac{2(r - 1) - \delta}{2} + \frac{r - 2}{2r - 2} \end{align} $$
$$ \begin{align} g + r + \epsilon_1 &\leq d + \frac{2(r - 1) - \delta}{2} + \frac{r - 2}{2r - 2} \end{align} $$
 $$ \begin{align} 0 &\leq 2 \leq m - \epsilon_0 \end{align} $$
$$ \begin{align} 0 &\leq 2 \leq m - \epsilon_0 \end{align} $$
 $$ \begin{align} 4 &\leq r - 2. \end{align} $$
$$ \begin{align} 4 &\leq r - 2. \end{align} $$
Note that 
 $\delta \geq r$
 by our exclusion of the cases
$\delta \geq r$
 by our exclusion of the cases 
 $\delta < r$
 in Proposition 10.6(1). This implies equations (10.62), (10.63) and (10.64). Since
$\delta < r$
 in Proposition 10.6(1). This implies equations (10.62), (10.63) and (10.64). Since 
 $d \geq g + r + \epsilon _1$
, inequality (10.65) follows from
$d \geq g + r + \epsilon _1$
, inequality (10.65) follows from 
 $\delta \leq 2r - 2$
. Inequality (10.66) follows from
$\delta \leq 2r - 2$
. Inequality (10.66) follows from 
 $m \geq 2 + \epsilon _0$
, and equation (10.67) is immediate. All that remains is therefore to check the following pair of inequalities:
$m \geq 2 + \epsilon _0$
, and equation (10.67) is immediate. All that remains is therefore to check the following pair of inequalities: 
 $$ \begin{align} \delta &\leq 2r - 2 \end{align} $$
$$ \begin{align} \delta &\leq 2r - 2 \end{align} $$
 $$ \begin{align} m &\geq 2 + \epsilon_0. \end{align} $$
$$ \begin{align} m &\geq 2 + \epsilon_0. \end{align} $$
For equation (10.68), we note that
 $$\begin{align*}\delta = \frac{2(d - g - 2r + 1) + 4g + (r + 1)m + 2r - 2}{r - 1} \leq \frac{4(r - 1) + (r + 1)(r - 1) + 2r - 2}{r - 1} \leq 2r - 2.\end{align*}$$
$$\begin{align*}\delta = \frac{2(d - g - 2r + 1) + 4g + (r + 1)m + 2r - 2}{r - 1} \leq \frac{4(r - 1) + (r + 1)(r - 1) + 2r - 2}{r - 1} \leq 2r - 2.\end{align*}$$
For equation (10.69), we note that m is even by Lemma 10.4; in particular, 
 $m \geq 2$
. Inequality (10.69) thus holds unless
$m \geq 2$
. Inequality (10.69) thus holds unless 
 $g = 0$
 and
$g = 0$
 and 
 $m = 2$
. But in this case,
$m = 2$
. But in this case, 
 $$\begin{align*}\delta = \frac{2d + 2}{r - 1} \leq \frac{2(2r - 1) + 2}{r - 1} < r,\end{align*}$$
$$\begin{align*}\delta = \frac{2d + 2}{r - 1} \leq \frac{2(2r - 1) + 2}{r - 1} < r,\end{align*}$$
contradicting our assumption that 
 $\delta \geq r$
.
$\delta \geq r$
.
 
Case 2: 
 $\boldsymbol {\ell } = \mathbf {0}$
 and
$\boldsymbol {\ell } = \mathbf {0}$
 and 
 ${\boldsymbol{r}}$
 is even and
${\boldsymbol{r}}$
 is even and 
 $\boldsymbol {\delta }$
 is an odd integer. In this case, we will take
$\boldsymbol {\delta }$
 is an odd integer. In this case, we will take 
 $m' = 1$
, which again evidently satisfies equation (10.42). Substituting
$m' = 1$
, which again evidently satisfies equation (10.42). Substituting 
 $\ell = 0$
 and
$\ell = 0$
 and 
 $m' = 1$
 into equations 10.56–10.61, it remains only to verify:
$m' = 1$
 into equations 10.56–10.61, it remains only to verify: 
 $$ \begin{align} (r - 1) - \delta &\leq \frac{r^2 - 4r + 2}{r - 1}\end{align} $$
$$ \begin{align} (r - 1) - \delta &\leq \frac{r^2 - 4r + 2}{r - 1}\end{align} $$
 $$ \begin{align} (r - 1) - \delta &\leq \frac{r^2 - 2r}{r - 1} \end{align} $$
$$ \begin{align} (r - 1) - \delta &\leq \frac{r^2 - 2r}{r - 1} \end{align} $$
 $$ \begin{align} (r - 1) - \delta &\leq \frac{r^2 - 3r + 1}{r - 1} \end{align} $$
$$ \begin{align} (r - 1) - \delta &\leq \frac{r^2 - 3r + 1}{r - 1} \end{align} $$
 $$ \begin{align} g + r + \epsilon_1 &\leq d + \frac{(r - 1) - \delta}{2} + \frac{r - 2}{2r - 2} \end{align} $$
$$ \begin{align} g + r + \epsilon_1 &\leq d + \frac{(r - 1) - \delta}{2} + \frac{r - 2}{2r - 2} \end{align} $$
 $$ \begin{align} 0 &\leq 1 \leq m - \epsilon_0 \end{align} $$
$$ \begin{align} 0 &\leq 1 \leq m - \epsilon_0 \end{align} $$
 $$ \begin{align} 2 &\leq r - 2. \end{align} $$
$$ \begin{align} 2 &\leq r - 2. \end{align} $$
Inequalities (10.70), (10.71) and (10.72) follow from 
 $\delta \geq 3$
, inequality (10.73) from the inequality
$\delta \geq 3$
, inequality (10.73) from the inequality 
 $\delta \leq 2(d - g - r - \epsilon _1) + (r - 1)$
, inequality (10.74) from
$\delta \leq 2(d - g - r - \epsilon _1) + (r - 1)$
, inequality (10.74) from 
 $m \geq 1 + \epsilon _0$
, and equation (10.75) is immediate. All that remains is therefore to check the following system of inequalities:
$m \geq 1 + \epsilon _0$
, and equation (10.75) is immediate. All that remains is therefore to check the following system of inequalities: 
 $$ \begin{align} \delta &\geq 3\end{align} $$
$$ \begin{align} \delta &\geq 3\end{align} $$
 $$ \begin{align} \delta &\leq 2(d - g - r - \epsilon_1) + (r - 1) \end{align} $$
$$ \begin{align} \delta &\leq 2(d - g - r - \epsilon_1) + (r - 1) \end{align} $$
 $$ \begin{align} m &\geq 1 + \epsilon_0. \end{align} $$
$$ \begin{align} m &\geq 1 + \epsilon_0. \end{align} $$
 For equation (10.76), since 
 $m \geq 1$
, we have
$m \geq 1$
, we have 
 $\delta> 1$
. Since
$\delta> 1$
. Since 
 $\delta $
 is an odd integer,
$\delta $
 is an odd integer, 
 $\delta \geq 3$
 as desired. For equation (10.78), since
$\delta \geq 3$
 as desired. For equation (10.78), since 
 $m \geq 1$
, the inequality holds unless
$m \geq 1$
, the inequality holds unless 
 $g = 0$
 and
$g = 0$
 and 
 $m = 1$
. But in this case,
$m = 1$
. But in this case, 
 $$\begin{align*}\delta = \frac{2d}{r - 1} - 1 \leq \frac{2(2r - 1)}{r - 1} - 1 < 5,\end{align*}$$
$$\begin{align*}\delta = \frac{2d}{r - 1} - 1 \leq \frac{2(2r - 1)}{r - 1} - 1 < 5,\end{align*}$$
and so 
 $\delta = 3$
, and so
$\delta = 3$
, and so 
 $d = 2r - 2$
. In other words, writing
$d = 2r - 2$
. In other words, writing 
 $r = 2k$
, we have
$r = 2k$
, we have 
 $$\begin{align*}(d, g, r, \ell, m) = (4k - 2, 0, 2k, 0, 1),\end{align*}$$
$$\begin{align*}(d, g, r, \ell, m) = (4k - 2, 0, 2k, 0, 1),\end{align*}$$
which is one of the cases excluded by assumption.
 All that remains is to verify equation (10.77). Note that m is odd by Lemma 10.4; in particular, since 
 $m \leq r-2+\epsilon _0$
 by equation (10.1), we have one of:
$m \leq r-2+\epsilon _0$
 by equation (10.1), we have one of: 
 $$\begin{align*}m \leq r - 5, \quad m = r - 3, \quad \text{or} \quad m = r - 1,\end{align*}$$
$$\begin{align*}m \leq r - 5, \quad m = r - 3, \quad \text{or} \quad m = r - 1,\end{align*}$$
where the final case can only occur if 
 $g = 0$
. Our argument will be via casework as follows.
$g = 0$
. Our argument will be via casework as follows.
 
Subcase 2.1: 
 $d \geq g + r + 2$
. By separately considering the cases
$d \geq g + r + 2$
. By separately considering the cases 
 $g = 0$
 (in which case
$g = 0$
 (in which case 
 $m \leq r - 1$
) and
$m \leq r - 1$
) and 
 $g> 0$
 (in which case
$g> 0$
 (in which case 
 $m \leq r - 3$
), we have
$m \leq r - 3$
), we have 
 $4g + (r + 1)m \leq r^2 + 2r - 7$
, with equality only if
$4g + (r + 1)m \leq r^2 + 2r - 7$
, with equality only if 
 $g = r - 1$
 and
$g = r - 1$
 and 
 $m = r - 3$
. Therefore,
$m = r - 3$
. Therefore, 
 $$ \begin{align*} \delta &= 2(d - g - r - 1) + (r + 1) - \frac{(r^2 + 2r - 7) - 4g - (r + 1)m + (2r - 4)(d - g - r - 2)}{r - 1} \\ &\leq 2(d - g - r - \epsilon_1) + (r + 1), \end{align*} $$
$$ \begin{align*} \delta &= 2(d - g - r - 1) + (r + 1) - \frac{(r^2 + 2r - 7) - 4g - (r + 1)m + (2r - 4)(d - g - r - 2)}{r - 1} \\ &\leq 2(d - g - r - \epsilon_1) + (r + 1), \end{align*} $$
with equality only if 
 $g = r - 1$
 and
$g = r - 1$
 and 
 $m = r - 3$
 and
$m = r - 3$
 and 
 $d = g + r + 2 = 2r + 1$
. Since
$d = g + r + 2 = 2r + 1$
. Since 
 $\delta $
 is an odd integer, we therefore have
$\delta $
 is an odd integer, we therefore have 
 $\delta \leq 2(d - g - r - \epsilon _1) + (r - 1)$
 unless, writing
$\delta \leq 2(d - g - r - \epsilon _1) + (r - 1)$
 unless, writing 
 $r = 2k$
, we have
$r = 2k$
, we have 
 $$\begin{align*}(d, g, r, \ell, m) = (2r + 1, r - 1, r, 0, r - 3) = (4k + 1, 2k - 1, 2k, 0, 2k - 3),\end{align*}$$
$$\begin{align*}(d, g, r, \ell, m) = (2r + 1, r - 1, r, 0, r - 3) = (4k + 1, 2k - 1, 2k, 0, 2k - 3),\end{align*}$$
which is again one of the cases excluded by assumption.
 
Subcase 2.2: 
 $d \leq g + r + 1$
 and
$d \leq g + r + 1$
 and 
 $m \leq r - 5$
. We have
$m \leq r - 5$
. We have 
 $$\begin{align*}\delta = \frac{2(d - g - r - 1) + 4g + 2 + (r + 1)m}{r - 1} \leq \frac{4(r - 1) + 2 + (r + 1)(r - 5)}{r - 1} = r + 1 - \frac{6}{r - 1} < r + 1.\end{align*}$$
$$\begin{align*}\delta = \frac{2(d - g - r - 1) + 4g + 2 + (r + 1)m}{r - 1} \leq \frac{4(r - 1) + 2 + (r + 1)(r - 5)}{r - 1} = r + 1 - \frac{6}{r - 1} < r + 1.\end{align*}$$
Since 
 $\delta $
 is an odd integer, this implies
$\delta $
 is an odd integer, this implies 
 $\delta \leq r - 1$
 as desired.
$\delta \leq r - 1$
 as desired.
 
Subcase 2.3: 
 $d = g + r$
 and
$d = g + r$
 and 
 $m = r - 3$
. We have
$m = r - 3$
. We have 
 $$\begin{align*}\delta = r - 1 + \frac{4(g - 1)}{r - 1}.\end{align*}$$
$$\begin{align*}\delta = r - 1 + \frac{4(g - 1)}{r - 1}.\end{align*}$$
Since 
 $\delta $
 is an integer, and
$\delta $
 is an integer, and 
 $r - 1$
 is odd, this implies
$r - 1$
 is odd, this implies 
 $g \equiv 1$
 mod
$g \equiv 1$
 mod 
 $r - 1$
, which since
$r - 1$
, which since 
 $0 \leq g \leq r - 1$
 in turn implies
$0 \leq g \leq r - 1$
 in turn implies 
 $g = 1$
, and so
$g = 1$
, and so 
 $\delta = r - 1$
.
$\delta = r - 1$
.
 
Subcase 2.4: 
 $d = g + r + 1$
 and
$d = g + r + 1$
 and 
 $m = r - 3$
. We have
$m = r - 3$
. We have 
 $$\begin{align*}\delta = r - 1 + \frac{2(2g - 1)}{r - 1}.\end{align*}$$
$$\begin{align*}\delta = r - 1 + \frac{2(2g - 1)}{r - 1}.\end{align*}$$
Since 
 $\delta $
 is an integer, and
$\delta $
 is an integer, and 
 $r - 1$
 is odd, this implies
$r - 1$
 is odd, this implies 
 $2g \equiv 1 \equiv r$
 mod
$2g \equiv 1 \equiv r$
 mod 
 $r - 1$
, which since
$r - 1$
, which since 
 $0 \leq g \leq r - 1$
 in turn implies
$0 \leq g \leq r - 1$
 in turn implies 
 $g = r/2$
. Writing
$g = r/2$
. Writing 
 $r = 2k$
, we therefore have
$r = 2k$
, we therefore have 
 $g = k$
 and
$g = k$
 and 
 $d = g + r + 1 = 3k + 1$
, that is, we have
$d = g + r + 1 = 3k + 1$
, that is, we have 
 $$\begin{align*}(d, g, r, \ell, m) = (3k + 1, k, 2k, 0, 2k - 3),\end{align*}$$
$$\begin{align*}(d, g, r, \ell, m) = (3k + 1, k, 2k, 0, 2k - 3),\end{align*}$$
which is again one of the cases excluded by assumption.
 
Subcase 2.5: 
 $d \leq g + r + 1$
 and
$d \leq g + r + 1$
 and 
 $m = r - 1$
. Since
$m = r - 1$
. Since 
 $m \leq r - 2 + \epsilon _0$
, we would have
$m \leq r - 2 + \epsilon _0$
, we would have 
 $g = 0$
, and thus
$g = 0$
, and thus 
 $d = r + 1$
. But this would imply
$d = r + 1$
. But this would imply 
 $\delta = (r^2 + 1)/(r - 1) \notin \mathbb {Z}$
, in contradiction to our assumption that
$\delta = (r^2 + 1)/(r - 1) \notin \mathbb {Z}$
, in contradiction to our assumption that 
 $\delta $
 is an odd integer.
$\delta $
 is an odd integer.
 
Case 3: 
 $\boldsymbol {\ell \neq } \mathbf {0}$
 or
$\boldsymbol {\ell \neq } \mathbf {0}$
 or 
 ${\boldsymbol{r}}$
 is odd or
${\boldsymbol{r}}$
 is odd or 
 $\boldsymbol {\delta }$
 is not an integer. If
$\boldsymbol {\delta }$
 is not an integer. If 
 $\ell \neq 0$
 or
$\ell \neq 0$
 or 
 $\delta $
 is not an integer, then equation (10.42) holds. Otherwise, the current assumption implies r is odd, so
$\delta $
 is not an integer, then equation (10.42) holds. Otherwise, the current assumption implies r is odd, so 
 $\delta $
 is even (the cases where
$\delta $
 is even (the cases where 
 $\delta $
 is also odd are excluded), and so equation (10.42) again holds. We conclude that equation (10.42) is automatic.
$\delta $
 is also odd are excluded), and so equation (10.42) again holds. We conclude that equation (10.42) is automatic.
 All that remains is therefore to check that there exists an integer 
 $m'$
 satisfying equations 10.56–10.61. Rearranging to make the bounds on
$m'$
 satisfying equations 10.56–10.61. Rearranging to make the bounds on 
 $m'$
 explicit, this is the system:
$m'$
 explicit, this is the system: 
 $$ \begin{align} m' &\leq \frac{1}{r - 1 - 2\kappa} \cdot \left(\delta + \frac{r - 2}{r - 1}\right) \end{align} $$
$$ \begin{align} m' &\leq \frac{1}{r - 1 - 2\kappa} \cdot \left(\delta + \frac{r - 2}{r - 1}\right) \end{align} $$
 $$ \begin{align} m' &\leq \frac{1}{r - 1} \cdot \left(\delta - \ell + \frac{r^2 - 2r}{r - 1}\right) \end{align} $$
$$ \begin{align} m' &\leq \frac{1}{r - 1} \cdot \left(\delta - \ell + \frac{r^2 - 2r}{r - 1}\right) \end{align} $$
 $$ \begin{align} m' &\leq \frac{1}{r - 1 - \kappa} \cdot \left(\delta - \ell + \frac{r^2 - r - 2}{2r - 2}\right) \end{align} $$
$$ \begin{align} m' &\leq \frac{1}{r - 1 - \kappa} \cdot \left(\delta - \ell + \frac{r^2 - r - 2}{2r - 2}\right) \end{align} $$
 $$ \begin{align} m' &\leq m - \epsilon_0 \end{align} $$
$$ \begin{align} m' &\leq m - \epsilon_0 \end{align} $$
 $$ \begin{align} m' &\leq \frac{r - 2 - \ell}{2} \end{align} $$
$$ \begin{align} m' &\leq \frac{r - 2 - \ell}{2} \end{align} $$
 $$ \begin{align} m' &\geq \frac{1}{r - 1} \cdot \left(\delta - 2(d - g - r - \epsilon_1) - \ell - \frac{r - 2}{r - 1}\right) \end{align} $$
$$ \begin{align} m' &\geq \frac{1}{r - 1} \cdot \left(\delta - 2(d - g - r - \epsilon_1) - \ell - \frac{r - 2}{r - 1}\right) \end{align} $$
 $$ \begin{align} m' &\geq 0. \end{align} $$
$$ \begin{align} m' &\geq 0. \end{align} $$
We will compare the inequalities (10.79) and (10.81) to equation (10.84) using ad-hoc methods. But first we handle all of the other comparisons using Lemma 10.3 by verifying the following system of inequalities:
 $$ \begin{align*} 0 &\leq \frac{1}{r - 1 - 2\kappa} \cdot \left(\delta + \frac{r - 2}{r - 1}\right) \\ 0 &\leq \frac{1}{r - 1} \cdot \left(\delta - \ell + \frac{r^2 - 2r}{r - 1}\right) \\ 0 &\leq \frac{1}{r - 1 - \kappa} \cdot \left(\delta - \ell + \frac{r^2 - r - 2}{2r - 2}\right) \\ 0 &\leq m - \epsilon_0 \\ 0 &\leq \frac{r - 2 - \ell}{2} \\ \frac{1}{r - 1} \cdot \left(\delta - 2(d - g - r - \epsilon_1) - \ell - \frac{r - 2}{r - 1}\right) &\leq \frac{1}{r - 1} \cdot \left(\delta - \ell + \frac{r^2 - 2r}{r - 1}\right) - \frac{(r - 2)^4}{(r - 1)^4} \\ \frac{1}{r - 1} \cdot \left(\delta - 2(d - g - r - \epsilon_1) - \ell - \frac{r - 2}{r - 1}\right) &\leq m - \epsilon_0 \\ \frac{1}{r - 1} \cdot \left(\delta - 2(d - g - r - \epsilon_1) - \ell - \frac{r - 2}{r - 1}\right) &\leq \frac{r - 2 - \ell}{2} - \frac{(r - 1)^2 - 1}{2(r - 1)^2}. \end{align*} $$
$$ \begin{align*} 0 &\leq \frac{1}{r - 1 - 2\kappa} \cdot \left(\delta + \frac{r - 2}{r - 1}\right) \\ 0 &\leq \frac{1}{r - 1} \cdot \left(\delta - \ell + \frac{r^2 - 2r}{r - 1}\right) \\ 0 &\leq \frac{1}{r - 1 - \kappa} \cdot \left(\delta - \ell + \frac{r^2 - r - 2}{2r - 2}\right) \\ 0 &\leq m - \epsilon_0 \\ 0 &\leq \frac{r - 2 - \ell}{2} \\ \frac{1}{r - 1} \cdot \left(\delta - 2(d - g - r - \epsilon_1) - \ell - \frac{r - 2}{r - 1}\right) &\leq \frac{1}{r - 1} \cdot \left(\delta - \ell + \frac{r^2 - 2r}{r - 1}\right) - \frac{(r - 2)^4}{(r - 1)^4} \\ \frac{1}{r - 1} \cdot \left(\delta - 2(d - g - r - \epsilon_1) - \ell - \frac{r - 2}{r - 1}\right) &\leq m - \epsilon_0 \\ \frac{1}{r - 1} \cdot \left(\delta - 2(d - g - r - \epsilon_1) - \ell - \frac{r - 2}{r - 1}\right) &\leq \frac{r - 2 - \ell}{2} - \frac{(r - 1)^2 - 1}{2(r - 1)^2}. \end{align*} $$
Substituting in the definition of 
 $\delta $
 and rearranging, these inequalities are equivalent to:
$\delta $
 and rearranging, these inequalities are equivalent to: 
 $$ \begin{align} 2(d - g - r) + 4g + 2\ell + (r + 1)m &\geq -r + 2 \end{align} $$
$$ \begin{align} 2(d - g - r) + 4g + 2\ell + (r + 1)m &\geq -r + 2 \end{align} $$
 $$ \begin{align} 4(d - g - r) + 8g + (r - 3)(r-2\ell) + 2(r + 1)m &\geq -r^2 + r \end{align} $$
$$ \begin{align} 4(d - g - r) + 8g + (r - 3)(r-2\ell) + 2(r + 1)m &\geq -r^2 + r \end{align} $$
 $$ \begin{align} 4(d - g - r) + 8g + (r - 3)(r-2\ell) + 2(r + 1)m &\geq -2r + 2 \end{align} $$
$$ \begin{align} 4(d - g - r) + 8g + (r - 3)(r-2\ell) + 2(r + 1)m &\geq -2r + 2 \end{align} $$
 $$ \begin{align} m - 1 &\geq \epsilon_0 - 1 \end{align} $$
$$ \begin{align} m - 1 &\geq \epsilon_0 - 1 \end{align} $$
 $$ \begin{align} r-2\ell &\geq -r + 4 \end{align} $$
$$ \begin{align} r-2\ell &\geq -r + 4 \end{align} $$
 $$ \begin{align} 2(r - 1)^3 (d - g - r - \epsilon_1) &\geq -5r^3 + 23r^2 - 35r + 18 \end{align} $$
$$ \begin{align} 2(r - 1)^3 (d - g - r - \epsilon_1) &\geq -5r^3 + 23r^2 - 35r + 18 \end{align} $$
 $$ \begin{align} (2r - 4)(d - g - r - \epsilon_1) + 4(r - 1 - g) + (r - 3)\ell \qquad & \end{align} $$
$$ \begin{align} (2r - 4)(d - g - r - \epsilon_1) + 4(r - 1 - g) + (r - 3)\ell \qquad & \end{align} $$
 $$ \begin{align} \phantom{.} + (r^2 - 3r)(m - 1) + 2(1 - \epsilon_1) &\geq (r - 1)^2 \epsilon_0 - r^2 + 6r\notag\\ (8r - 16)(d - g - r) + 16(r - 1 - g) + (r^2 - 4r + 7)(r - 2\ell) \qquad & \end{align} $$
$$ \begin{align} \phantom{.} + (r^2 - 3r)(m - 1) + 2(1 - \epsilon_1) &\geq (r - 1)^2 \epsilon_0 - r^2 + 6r\notag\\ (8r - 16)(d - g - r) + 16(r - 1 - g) + (r^2 - 4r + 7)(r - 2\ell) \qquad & \end{align} $$
 $$\begin{align*} \phantom{.} + (4r + 4)(r - 1 - m) + 8(1 - \epsilon_1) &\geq -r^3 + 10r^2 + 5r. \end{align*}$$
$$\begin{align*} \phantom{.} + (4r + 4)(r - 1 - m) + 8(1 - \epsilon_1) &\geq -r^3 + 10r^2 + 5r. \end{align*}$$
From these expressions, we see that all but equation (10.92) is immediate and that equation (10.92) holds when 
 $\epsilon _0 = 0$
. But when
$\epsilon _0 = 0$
. But when 
 $\epsilon _0 = 1$
, then
$\epsilon _0 = 1$
, then 
 $g = 0$
 and
$g = 0$
 and 
 $\epsilon _1 = 1$
, and so equation (10.92) becomes
$\epsilon _1 = 1$
, and so equation (10.92) becomes 
 $$\begin{align*}(2r - 4)(d - r - 1) + (r^2 - 3r)(m - 1) + (r - 3)\ell \geq 5,\end{align*}$$
$$\begin{align*}(2r - 4)(d - r - 1) + (r^2 - 3r)(m - 1) + (r - 3)\ell \geq 5,\end{align*}$$
which holds unless 
 $d = r + 1$
 and
$d = r + 1$
 and 
 $m = 1$
 and
$m = 1$
 and 
 $\ell = 0$
, or equivalently unless
$\ell = 0$
, or equivalently unless 
 $$\begin{align*}(d, g, r, \ell, m) = (k + 1, 0, k, 0, 1),\end{align*}$$
$$\begin{align*}(d, g, r, \ell, m) = (k + 1, 0, k, 0, 1),\end{align*}$$
which is again one of the cases excluded by assumption.
All that remains is our promised ad-hoc comparison of equations (10.79) and (10.81) to equation (10.84). That is, we want to show that there are integers between:
 $$ \begin{align*} \frac{1}{r - 1} \cdot \left(\delta - 2(d - g - r - \epsilon_1) - \ell - \frac{r - 2}{r - 1}\right) \quad &\text{and} \quad \frac{1}{r - 1 - 2\kappa} \cdot \left(\delta + \frac{r - 2}{r - 1}\right) \\ \frac{1}{r - 1} \cdot \left(\delta - 2(d - g - r - \epsilon_1) - \ell - \frac{r - 2}{r - 1}\right) \quad &\text{and} \quad \frac{1}{r - 1 - \kappa} \cdot \left(\delta - \ell + \frac{r^2 - r - 2}{2r - 2}\right). \end{align*} $$
$$ \begin{align*} \frac{1}{r - 1} \cdot \left(\delta - 2(d - g - r - \epsilon_1) - \ell - \frac{r - 2}{r - 1}\right) \quad &\text{and} \quad \frac{1}{r - 1 - 2\kappa} \cdot \left(\delta + \frac{r - 2}{r - 1}\right) \\ \frac{1}{r - 1} \cdot \left(\delta - 2(d - g - r - \epsilon_1) - \ell - \frac{r - 2}{r - 1}\right) \quad &\text{and} \quad \frac{1}{r - 1 - \kappa} \cdot \left(\delta - \ell + \frac{r^2 - r - 2}{2r - 2}\right). \end{align*} $$
 
Remark 
 $\ast $
: If
$\ast $
: If 
 $(d, g) = (r + 1, 1)$
 then
$(d, g) = (r + 1, 1)$
 then 
 $\delta < \ell + 2$
. Indeed, if
$\delta < \ell + 2$
. Indeed, if 
 $(d, g) = (r + 1, 1)$
, then
$(d, g) = (r + 1, 1)$
, then 
 $m \leq \rho (d, g, r) = 1$
, and so
$m \leq \rho (d, g, r) = 1$
, and so 
 $$\begin{align*}\delta = \frac{4 + 2\ell + (r + 1)m}{r - 1} \leq \frac{r + 5 + 2\ell}{r - 1} < \ell + 2.\end{align*}$$
$$\begin{align*}\delta = \frac{4 + 2\ell + (r + 1)m}{r - 1} \leq \frac{r + 5 + 2\ell}{r - 1} < \ell + 2.\end{align*}$$
 
Subcase 3.1: 
 $\delta < \ell + 1 + 2(d - g - r - \epsilon _1)$
. In this case, the lower bound is nonpositive. We have already shown that both upper bounds are nonnegative above, so there is nothing more to check.
$\delta < \ell + 1 + 2(d - g - r - \epsilon _1)$
. In this case, the lower bound is nonpositive. We have already shown that both upper bounds are nonnegative above, so there is nothing more to check.
 
Subcase 3.2: 
 $\delta \geq \ell + 3$
. By Remark
$\delta \geq \ell + 3$
. By Remark 
 $\ast $
 above,
$\ast $
 above, 
 $\kappa \geq \frac {r - 4}{2}$
. It suffices to show that there are integers between
$\kappa \geq \frac {r - 4}{2}$
. It suffices to show that there are integers between 
 $$ \begin{align*} \frac{1}{r - 1} \cdot \left(\delta - \ell - \frac{r - 2}{r - 1}\right) \quad &\text{and} \quad \frac{1}{r - 1 - 2 \cdot \frac{r - 4}{2}} \cdot \left(\delta - \ell + \frac{r - 2}{r - 1}\right) = \frac{1}{3} \cdot \left(\delta - \ell + \frac{r - 2}{r - 1}\right)\\ \frac{1}{r - 1} \cdot \left(\delta - \ell - \frac{r - 2}{r - 1}\right) \quad &\text{and} \quad \frac{1}{r - 1 - \frac{r - 4}{2}} \cdot \left(\delta - \ell + \frac{r^2 - r - 2}{2r - 2}\right) = \frac{1}{r + 2} \cdot \left(2(\delta - \ell) + \frac{r^2 - r - 2}{r - 1}\right). \end{align*} $$
$$ \begin{align*} \frac{1}{r - 1} \cdot \left(\delta - \ell - \frac{r - 2}{r - 1}\right) \quad &\text{and} \quad \frac{1}{r - 1 - 2 \cdot \frac{r - 4}{2}} \cdot \left(\delta - \ell + \frac{r - 2}{r - 1}\right) = \frac{1}{3} \cdot \left(\delta - \ell + \frac{r - 2}{r - 1}\right)\\ \frac{1}{r - 1} \cdot \left(\delta - \ell - \frac{r - 2}{r - 1}\right) \quad &\text{and} \quad \frac{1}{r - 1 - \frac{r - 4}{2}} \cdot \left(\delta - \ell + \frac{r^2 - r - 2}{2r - 2}\right) = \frac{1}{r + 2} \cdot \left(2(\delta - \ell) + \frac{r^2 - r - 2}{r - 1}\right). \end{align*} $$
By Lemma 10.3, this follows from the following inequalities:
 $$ \begin{align*} \frac{1}{r - 1} \cdot \left(\delta - \ell - \frac{r - 2}{r - 1}\right) &\leq \frac{1}{3} \cdot \left(\delta - \ell + \frac{r - 2}{r - 1}\right) - \frac{(3(r - 1) - 1) ((r - 1)^2 - 1)}{3(r - 1)^3} \\ \frac{1}{r - 1} \cdot \left(\delta - \ell - \frac{r - 2}{r - 1}\right) &\leq \frac{1}{r + 2} \cdot \left(2(\delta - \ell) + \frac{r^2 - r - 2}{r - 1}\right) - \frac{((r + 2)(r - 1) - 1)((r - 1)^2 - 1)}{(r + 2)(r - 1)^3} \end{align*} $$
$$ \begin{align*} \frac{1}{r - 1} \cdot \left(\delta - \ell - \frac{r - 2}{r - 1}\right) &\leq \frac{1}{3} \cdot \left(\delta - \ell + \frac{r - 2}{r - 1}\right) - \frac{(3(r - 1) - 1) ((r - 1)^2 - 1)}{3(r - 1)^3} \\ \frac{1}{r - 1} \cdot \left(\delta - \ell - \frac{r - 2}{r - 1}\right) &\leq \frac{1}{r + 2} \cdot \left(2(\delta - \ell) + \frac{r^2 - r - 2}{r - 1}\right) - \frac{((r + 2)(r - 1) - 1)((r - 1)^2 - 1)}{(r + 2)(r - 1)^3} \end{align*} $$
But these are immediate for 
 $\delta - \ell \geq 3$
, using the assumption
$\delta - \ell \geq 3$
, using the assumption 
 $r \geq 14$
.
$r \geq 14$
.
 
Subcase 3.3: 
 $\ell + 1 + 2(d - g - r - \epsilon _1) \leq \delta < \ell + 3$
. These inequalities force
$\ell + 1 + 2(d - g - r - \epsilon _1) \leq \delta < \ell + 3$
. These inequalities force 
 $d = g + r + \epsilon _1$
 or equivalently
$d = g + r + \epsilon _1$
 or equivalently 
 $$ \begin{align} d = g + r \quad \text{or} \quad d = g + r + 1. \end{align} $$
$$ \begin{align} d = g + r \quad \text{or} \quad d = g + r + 1. \end{align} $$
The inequality 
 $\delta < \ell + 3$
 also implies
$\delta < \ell + 3$
 also implies 
 $$\begin{align*}\frac{1}{r - 1} \cdot \left(\delta - \ell - \frac{r - 2}{r - 1}\right) \leq 1.\end{align*}$$
$$\begin{align*}\frac{1}{r - 1} \cdot \left(\delta - \ell - \frac{r - 2}{r - 1}\right) \leq 1.\end{align*}$$
It therefore suffices to show
 $$\begin{align*}\frac{1}{r - 1 - 2\kappa} \cdot \left(\delta + \frac{r - 2}{r - 1}\right) \geq 1 \quad \text{and} \quad \frac{1}{r - 1 - \kappa} \cdot \left(\delta - \ell + \frac{r^2 - r - 2}{2r - 2} \right) \geq 1,\end{align*}$$
$$\begin{align*}\frac{1}{r - 1 - 2\kappa} \cdot \left(\delta + \frac{r - 2}{r - 1}\right) \geq 1 \quad \text{and} \quad \frac{1}{r - 1 - \kappa} \cdot \left(\delta - \ell + \frac{r^2 - r - 2}{2r - 2} \right) \geq 1,\end{align*}$$
or upon rearrangement
 $$\begin{align*}\delta \geq r - 2 - 2\kappa + \frac{1}{r - 1} \quad \text{and} \quad \delta \geq \ell + \frac{r}{2} - 1 - \kappa + \frac{1}{r - 1}.\end{align*}$$
$$\begin{align*}\delta \geq r - 2 - 2\kappa + \frac{1}{r - 1} \quad \text{and} \quad \delta \geq \ell + \frac{r}{2} - 1 - \kappa + \frac{1}{r - 1}.\end{align*}$$
 Subsubcase 3.3.1: 
 $g = 0$
. By equation (10.94), we have
$g = 0$
. By equation (10.94), we have 
 $d = r + 1$
. Since
$d = r + 1$
. Since 
 $g = 0$
, we have
$g = 0$
, we have 
 $(d, g) \neq (r + 1, 1)$
. Our goal is thus to show
$(d, g) \neq (r + 1, 1)$
. Our goal is thus to show 
 $$\begin{align*}\delta \geq \begin{cases} 1 + \frac{1}{r - 1} & \text{if}\ r\text{ is odd,} \\ 2 + \frac{1}{r - 1} & \text{if}\ r\text{ is even;} \end{cases} \quad \text{and} \quad \delta \geq \ell + \begin{cases} \frac{1}{2} + \frac{1}{r - 1} & \text{if}\ r\text{ is odd,} \\ 1 + \frac{1}{r - 1} & \text{if}\ r\text{ is even.} \end{cases}\end{align*}$$
$$\begin{align*}\delta \geq \begin{cases} 1 + \frac{1}{r - 1} & \text{if}\ r\text{ is odd,} \\ 2 + \frac{1}{r - 1} & \text{if}\ r\text{ is even;} \end{cases} \quad \text{and} \quad \delta \geq \ell + \begin{cases} \frac{1}{2} + \frac{1}{r - 1} & \text{if}\ r\text{ is odd,} \\ 1 + \frac{1}{r - 1} & \text{if}\ r\text{ is even.} \end{cases}\end{align*}$$
When 
 $\ell = 0$
, the first inequality implies the second. In this case, recall that
$\ell = 0$
, the first inequality implies the second. In this case, recall that 
 $(d, g, r, \ell , m) = (k + 1, 0, k, 0, 1)$
 is excluded by assumption. Therefore,
$(d, g, r, \ell , m) = (k + 1, 0, k, 0, 1)$
 is excluded by assumption. Therefore,
 $m \geq 2$
, which implies the first inequality because
$m \geq 2$
, which implies the first inequality because 
 $$\begin{align*}\delta = \frac{2 + (r + 1)m}{r - 1} \geq \frac{2 + 2(r + 1)}{r - 1} \geq 2 + \frac{1}{r - 1}.\end{align*}$$
$$\begin{align*}\delta = \frac{2 + (r + 1)m}{r - 1} \geq \frac{2 + 2(r + 1)}{r - 1} \geq 2 + \frac{1}{r - 1}.\end{align*}$$
 Now, suppose 
 $\ell \geq 1$
. Note that
$\ell \geq 1$
. Note that 
 $\delta \geq \ell + 1 \geq 2$
, which implies the first inequality unless r is even and
$\delta \geq \ell + 1 \geq 2$
, which implies the first inequality unless r is even and 
 $\delta = \ell + 1$
. Similarly,
$\delta = \ell + 1$
. Similarly, 
 $\delta \geq \ell + 1$
 implies the second inequality unless r is even and
$\delta \geq \ell + 1$
 implies the second inequality unless r is even and 
 $\delta = \ell + 1$
. It thus remains only to show that it is impossible to have
$\delta = \ell + 1$
. It thus remains only to show that it is impossible to have 
 $\delta = \ell + 1$
 when r is even. To see this, observe that
$\delta = \ell + 1$
 when r is even. To see this, observe that 
 $$\begin{align*}\ell + 1 = \delta = \frac{2 + 2\ell + (r + 1)m}{r - 1}\end{align*}$$
$$\begin{align*}\ell + 1 = \delta = \frac{2 + 2\ell + (r + 1)m}{r - 1}\end{align*}$$
implies
 $$\begin{align*}(r - 3)(\ell + 1) = (r + 1)m.\end{align*}$$
$$\begin{align*}(r - 3)(\ell + 1) = (r + 1)m.\end{align*}$$
But if r were even, then this would imply 
 $(r + 1) \mid (\ell + 1)$
, which forces
$(r + 1) \mid (\ell + 1)$
, which forces 
 $\ell + 1 \geq r + 1$
, contradicting our assumption that
$\ell + 1 \geq r + 1$
, contradicting our assumption that 
 $\ell \leq r/2$
.
$\ell \leq r/2$
.
 Subsubcase 3.3.2: 
 $g> 0$
. We excluded the cases
$g> 0$
. We excluded the cases 
 $\ell < \delta < \ell + 2$
 in Proposition 10.6(2). Since we have
$\ell < \delta < \ell + 2$
 in Proposition 10.6(2). Since we have 
 $\ell + 1 \leq \delta < \ell + 3$
, we therefore have
$\ell + 1 \leq \delta < \ell + 3$
, we therefore have 
 $\ell + 2 \leq \delta < \ell + 3$
. Moreover, by Remark
$\ell + 2 \leq \delta < \ell + 3$
. Moreover, by Remark 
 $\ast $
, we have
$\ast $
, we have 
 $(d, g) \neq (r + 1, 1)$
. As in the previous subsubcase, our goal thus is to show
$(d, g) \neq (r + 1, 1)$
. As in the previous subsubcase, our goal thus is to show 
 $$\begin{align*}\delta \geq \begin{cases} 1 + \frac{1}{r - 1} & \text{if}\ r\text{ is odd,} \\ 2 + \frac{1}{r - 1} & \text{if}\ r\text{ is even;} \end{cases} \quad \text{and} \quad \delta \geq \ell + \begin{cases} \frac{1}{2} + \frac{1}{r - 1} & \text{if}\ r\text{ is odd,} \\ 1 + \frac{1}{r - 1} & \text{if}\ r\text{ is even.} \end{cases}\end{align*}$$
$$\begin{align*}\delta \geq \begin{cases} 1 + \frac{1}{r - 1} & \text{if}\ r\text{ is odd,} \\ 2 + \frac{1}{r - 1} & \text{if}\ r\text{ is even;} \end{cases} \quad \text{and} \quad \delta \geq \ell + \begin{cases} \frac{1}{2} + \frac{1}{r - 1} & \text{if}\ r\text{ is odd,} \\ 1 + \frac{1}{r - 1} & \text{if}\ r\text{ is even.} \end{cases}\end{align*}$$
 Since 
 $\delta \geq \ell + 2$
, the second inequality is immediate in all cases. Also, the first inequality is immediate if r is odd. To see the first inequality when r is even, note that
$\delta \geq \ell + 2$
, the second inequality is immediate in all cases. Also, the first inequality is immediate if r is odd. To see the first inequality when r is even, note that 
 $\delta \geq \ell + 2 \geq 2$
, so the first inequality holds unless we have equality everywhere, that is, unless
$\delta \geq \ell + 2 \geq 2$
, so the first inequality holds unless we have equality everywhere, that is, unless 
 $\ell = 0$
 and
$\ell = 0$
 and 
 $\delta = 2$
. But this possibility is excluded by assumption (recall that in Case 3 we have
$\delta = 2$
. But this possibility is excluded by assumption (recall that in Case 3 we have 
 $\ell \neq 0$
 or r odd or
$\ell \neq 0$
 or r odd or 
 $\delta $
 is not an integer).
$\delta $
 is not an integer).
The majority of the remaining cases are handled by Proposition 8.3. More precisely:
Proposition 10.7. Let 
 $(d, g, r, \ell , m)$
 be a good tuple satisfying (10.1) with
$(d, g, r, \ell , m)$
 be a good tuple satisfying (10.1) with 
 $m \neq 0$
 and
$m \neq 0$
 and 
 $r \geq 14$
. Suppose in addition that either condition (1) or (2) of Lemma 10.6 is satisfied. Then the conditions of Proposition 8.3 can be satisfied unless one of the following holds:
$r \geq 14$
. Suppose in addition that either condition (1) or (2) of Lemma 10.6 is satisfied. Then the conditions of Proposition 8.3 can be satisfied unless one of the following holds: 
- 
1.  $(d, g, r, \ell , m) = (4k, 0, 2k + 1, 0, 1)$
 for some k. $(d, g, r, \ell , m) = (4k, 0, 2k + 1, 0, 1)$
 for some k.
- 
2.  $(d, g, r, \ell , m) = (4k + 1, 2k - 1, 2k + 1, 0, 1)$
 for some k. $(d, g, r, \ell , m) = (4k + 1, 2k - 1, 2k + 1, 0, 1)$
 for some k.
Proof. We separately consider the following three cases.
 
Case 1: 
 $\boldsymbol {\ell < \delta < \ell } + \mathbf {2}$
 and
$\boldsymbol {\ell < \delta < \ell } + \mathbf {2}$
 and 
 ${\boldsymbol{g}}> \mathbf {0}$
. We take
${\boldsymbol{g}}> \mathbf {0}$
. We take 
 $m' = 0$
,
$m' = 0$
, 
 $\ell ' = \ell $
, and
$\ell ' = \ell $
, and 
 $d' = d$
, which satisfy the conditions of Proposition 8.3 by Lemma 10.5.
$d' = d$
, which satisfy the conditions of Proposition 8.3 by Lemma 10.5.
 
Case 2: 
 $\boldsymbol {\ell } = \mathbf {0}$
, and
$\boldsymbol {\ell } = \mathbf {0}$
, and 
 $\boldsymbol {\delta }$
 and
$\boldsymbol {\delta }$
 and 
 ${\boldsymbol{r}}$
 are even integers with
${\boldsymbol{r}}$
 are even integers with 
 $\boldsymbol {\delta } < {\boldsymbol{r}}$
. We take
$\boldsymbol {\delta } < {\boldsymbol{r}}$
. We take 
 $$\begin{align*}m' = 1, \quad \ell' = 0, \quad d' = d, \quad \text{and} \quad n_1 = \delta - 1.\end{align*}$$
$$\begin{align*}m' = 1, \quad \ell' = 0, \quad d' = d, \quad \text{and} \quad n_1 = \delta - 1.\end{align*}$$
Applying Lemma 10.5, the conditions of Proposition 8.3 are satisfied provided that
 $$\begin{align*}m> 1 \quad \text{and}\end{align*}$$
$$\begin{align*}m> 1 \quad \text{and}\end{align*}$$
 $$\begin{align*}2 \leq n_1 = \delta - 1 \leq r - 1 \quad \text{with}\ n_1 = \delta - 1 \neq 2\ \text{if}\ (d', g) = (r + 1, 1).\end{align*}$$
$$\begin{align*}2 \leq n_1 = \delta - 1 \leq r - 1 \quad \text{with}\ n_1 = \delta - 1 \neq 2\ \text{if}\ (d', g) = (r + 1, 1).\end{align*}$$
Since 
 $\delta $
 is an even integer,
$\delta $
 is an even integer, 
 $\delta - 1 \neq 2$
; since
$\delta - 1 \neq 2$
; since 
 $\delta < r$
, we have
$\delta < r$
, we have 
 $\delta - 1 \leq r - 1$
. All that remains to check is therefore that
$\delta - 1 \leq r - 1$
. All that remains to check is therefore that 
 $m \geq 2$
 and that
$m \geq 2$
 and that 
 $2 \leq \delta - 1$
, which since
$2 \leq \delta - 1$
, which since 
 $\delta $
 is an even integer is equivalent to
$\delta $
 is an even integer is equivalent to 
 $\delta> 2$
.
$\delta> 2$
.
 To see this, we first apply Lemma 10.4 to conclude that m is even. Since 
 $m \neq 0$
, this implies
$m \neq 0$
, this implies 
 $m \geq 2$
 as desired. This in turn implies
$m \geq 2$
 as desired. This in turn implies 
 $\delta> 2$
 because
$\delta> 2$
 because 
 $$\begin{align*}\delta = \frac{2d + 2g - 2r + 2\ell + (r + 1)m}{r - 1} \geq \frac{2(r + 1)}{r - 1}> 2.\end{align*}$$
$$\begin{align*}\delta = \frac{2d + 2g - 2r + 2\ell + (r + 1)m}{r - 1} \geq \frac{2(r + 1)}{r - 1}> 2.\end{align*}$$
 
Case 3: 
 $\boldsymbol {\ell } = \mathbf {0}$
, and
$\boldsymbol {\ell } = \mathbf {0}$
, and 
 $\boldsymbol {\delta }$
 and
$\boldsymbol {\delta }$
 and 
 ${\boldsymbol{r}}$
 are odd integers. As in the proof of Proposition 10.6, we show a slightly stronger statement: The conditions of Proposition 8.3 can be satisfied, together with the additional conditions that
${\boldsymbol{r}}$
 are odd integers. As in the proof of Proposition 10.6, we show a slightly stronger statement: The conditions of Proposition 8.3 can be satisfied, together with the additional conditions that 
 $$\begin{align*}m' \neq m - 1 \ \text{if} \ g = 0, \quad \text{and} \quad d' \neq g + r \ \text{if} \ d \neq g + r,\end{align*}$$
$$\begin{align*}m' \neq m - 1 \ \text{if} \ g = 0, \quad \text{and} \quad d' \neq g + r \ \text{if} \ d \neq g + r,\end{align*}$$
unless either one of the above-mentioned conditions holds or
 $$\begin{align*}(d, g, r, \ell, m) = (3k + 1, k - 1, 2k + 1, 0, 1) \quad \text{for some}\ k.\end{align*}$$
$$\begin{align*}(d, g, r, \ell, m) = (3k + 1, k - 1, 2k + 1, 0, 1) \quad \text{for some}\ k.\end{align*}$$
This is indeed a stronger statement because if 
 $(d, g, r, \ell , m) = (3k + 1, k - 1, 2k + 1, 0, 1)$
, then the conditions of Proposition 8.3 can be satisfied by taking:
$(d, g, r, \ell , m) = (3k + 1, k - 1, 2k + 1, 0, 1)$
, then the conditions of Proposition 8.3 can be satisfied by taking: 
 $$\begin{align*}\ell' = 0, \quad m' = 0, \quad \text{and} \quad d' = d - 1 = 3k.\end{align*}$$
$$\begin{align*}\ell' = 0, \quad m' = 0, \quad \text{and} \quad d' = d - 1 = 3k.\end{align*}$$
 Again as in the proof of Proposition 10.6, Lemma 10.5 guarantees that the tuples 
 $(d' - 1, g, r - 1, \overline {\ell }, \overline {m})$
,
$(d' - 1, g, r - 1, \overline {\ell }, \overline {m})$
, 
 $(d' - 1, g, r - 1, \overline {\ell }, \overline {m} - 1)$
, and
$(d' - 1, g, r - 1, \overline {\ell }, \overline {m} - 1)$
, and 
 $(d' - 2, g, r - 2, \overline {\ell }, \overline {m})$
, are all good provided only that
$(d' - 2, g, r - 2, \overline {\ell }, \overline {m})$
, are all good provided only that 
 $$\begin{align*}\overline{\ell} \leq \frac{r - 3}{2}.\end{align*}$$
$$\begin{align*}\overline{\ell} \leq \frac{r - 3}{2}.\end{align*}$$
 With 
 $\kappa $
 as in the proof of Proposition 10.6, our task is thus to show that the following system of inequalities can be satisfied for integers
$\kappa $
 as in the proof of Proposition 10.6, our task is thus to show that the following system of inequalities can be satisfied for integers 
 $\ell '$
,
$\ell '$
, 
 $m'$
,
$m'$
, 
 $d'$
, and n:
$d'$
, and n: 
 $$ \begin{align} 0 \leq m' \leq m - 1 - \epsilon_0 \end{align} $$
$$ \begin{align} 0 \leq m' \leq m - 1 - \epsilon_0 \end{align} $$
 $$ \begin{align} 2m' + \ell' \leq r - 3 \end{align} $$
$$ \begin{align} 2m' + \ell' \leq r - 3 \end{align} $$
 $$ \begin{align} g + r + \epsilon_1 \leq d' \leq d \end{align} $$
$$ \begin{align} g + r + \epsilon_1 \leq d' \leq d \end{align} $$
 $$ \begin{align} 0 \leq n \leq \kappa m' \end{align} $$
$$ \begin{align} 0 \leq n \leq \kappa m' \end{align} $$
 $$ \begin{align} 0 \leq \ell' \leq \ell = 0 \end{align} $$
$$ \begin{align} 0 \leq \ell' \leq \ell = 0 \end{align} $$
 $$ \begin{align} \left|\delta - \left[1 + \ell' + 2(d - d') + (r - 1)m' - 2n \right] \right| \leq 1 - \frac{1}{r - 1} \end{align} $$
$$ \begin{align} \left|\delta - \left[1 + \ell' + 2(d - d') + (r - 1)m' - 2n \right] \right| \leq 1 - \frac{1}{r - 1} \end{align} $$
 $$ \begin{align} \ell - \ell' + n \leq \frac{r - 3}{2}. \end{align} $$
$$ \begin{align} \ell - \ell' + n \leq \frac{r - 3}{2}. \end{align} $$
Inequalities (10.99) and (10.100) are satisfied by taking
 $$\begin{align*}\ell' = 0 \quad \text{and} \quad n = d - d' + \frac{r - 1}{2} m' - \frac{\delta - 1}{2}.\end{align*}$$
$$\begin{align*}\ell' = 0 \quad \text{and} \quad n = d - d' + \frac{r - 1}{2} m' - \frac{\delta - 1}{2}.\end{align*}$$
Substituting these into the remaining inequalities and rearranging, we reduce to the system of inequalities:
 $$ \begin{align} 0 \leq m' \leq m - 1 - \epsilon_0 \end{align} $$
$$ \begin{align} 0 \leq m' \leq m - 1 - \epsilon_0 \end{align} $$
 $$ \begin{align} m' \leq \frac{r - 3}{2} \end{align} $$
$$ \begin{align} m' \leq \frac{r - 3}{2} \end{align} $$
 $$ \begin{align} g + r + \epsilon_1 \leq d' \leq d \end{align} $$
$$ \begin{align} g + r + \epsilon_1 \leq d' \leq d \end{align} $$
 $$ \begin{align} d + \frac{r - 1}{2} m' - \frac{\delta - 1}{2} - \kappa m' \leq d' \leq d + \frac{r - 1}{2} m' - \frac{\delta - 1}{2} \end{align} $$
$$ \begin{align} d + \frac{r - 1}{2} m' - \frac{\delta - 1}{2} - \kappa m' \leq d' \leq d + \frac{r - 1}{2} m' - \frac{\delta - 1}{2} \end{align} $$
 $$ \begin{align} d + \frac{r - 1}{2} m' - \frac{\delta - 1}{2} - \frac{r - 3}{2} \leq d'. \end{align} $$
$$ \begin{align} d + \frac{r - 1}{2} m' - \frac{\delta - 1}{2} - \frac{r - 3}{2} \leq d'. \end{align} $$
All bounds on 
 $d'$
 are integers because r and
$d'$
 are integers because r and 
 $\delta $
 are both odd integers. Using Lemma 10.3 to eliminate
$\delta $
 are both odd integers. Using Lemma 10.3 to eliminate 
 $d'$
 replaces equations 10.104–10.106 with
$d'$
 replaces equations 10.104–10.106 with 
 $$ \begin{align} g + r + \epsilon_1 &\leq d\end{align} $$
$$ \begin{align} g + r + \epsilon_1 &\leq d\end{align} $$
 $$ \begin{align} g + r + \epsilon_1 &\leq d + \frac{r - 1}{2} m' - \frac{\delta - 1}{2} \end{align} $$
$$ \begin{align} g + r + \epsilon_1 &\leq d + \frac{r - 1}{2} m' - \frac{\delta - 1}{2} \end{align} $$
 $$ \begin{align} d + \frac{r - 1}{2} m' - \frac{\delta - 1}{2} - \kappa m' &\leq d \end{align} $$
$$ \begin{align} d + \frac{r - 1}{2} m' - \frac{\delta - 1}{2} - \kappa m' &\leq d \end{align} $$
 $$ \begin{align} d + \frac{r - 1}{2} m' - \frac{\delta - 1}{2} - \kappa m' &\leq d + \frac{r - 1}{2} m' - \frac{\delta - 1}{2} \end{align} $$
$$ \begin{align} d + \frac{r - 1}{2} m' - \frac{\delta - 1}{2} - \kappa m' &\leq d + \frac{r - 1}{2} m' - \frac{\delta - 1}{2} \end{align} $$
 $$ \begin{align} d + \frac{r - 1}{2} m' - \frac{\delta - 1}{2} - \frac{r - 3}{2} &\leq d \end{align} $$
$$ \begin{align} d + \frac{r - 1}{2} m' - \frac{\delta - 1}{2} - \frac{r - 3}{2} &\leq d \end{align} $$
 $$ \begin{align} d + \frac{r - 1}{2} m' - \frac{\delta - 1}{2} - \frac{r - 3}{2} &\leq d + \frac{r - 1}{2} m' - \frac{\delta - 1}{2}. \end{align} $$
$$ \begin{align} d + \frac{r - 1}{2} m' - \frac{\delta - 1}{2} - \frac{r - 3}{2} &\leq d + \frac{r - 1}{2} m' - \frac{\delta - 1}{2}. \end{align} $$
Inequalities (10.107) and (10.112) always hold, while equation (10.110) is implied by 
 $m' \geq 0$
. Rearranging the others, and including equations (10.102) and (10.103), we arrive at the system
$m' \geq 0$
. Rearranging the others, and including equations (10.102) and (10.103), we arrive at the system 
 $$ \begin{align*} m' &\geq 0 \\ m' &\geq \frac{\delta - 1 - 2(d - g - r - \epsilon_1)}{r - 1} \\ m' &\leq \frac{\delta - 1}{r - 1 - 2\kappa} \\ m' &\leq \frac{\delta + r - 4}{r - 1} \\ m' &\leq m - 1 - \epsilon_0 \\ m' &\leq \frac{r - 3}{2}. \end{align*} $$
$$ \begin{align*} m' &\geq 0 \\ m' &\geq \frac{\delta - 1 - 2(d - g - r - \epsilon_1)}{r - 1} \\ m' &\leq \frac{\delta - 1}{r - 1 - 2\kappa} \\ m' &\leq \frac{\delta + r - 4}{r - 1} \\ m' &\leq m - 1 - \epsilon_0 \\ m' &\leq \frac{r - 3}{2}. \end{align*} $$
Applying Lemma 10.3 to eliminate 
 $m'$
, we reduce to the system
$m'$
, we reduce to the system 
 $$ \begin{align} 0 &\leq \frac{\delta - 1}{r - 1 - 2\kappa} \end{align} $$
$$ \begin{align} 0 &\leq \frac{\delta - 1}{r - 1 - 2\kappa} \end{align} $$
 $$ \begin{align} 0 &\leq \frac{\delta + r - 4}{r - 1} \end{align} $$
$$ \begin{align} 0 &\leq \frac{\delta + r - 4}{r - 1} \end{align} $$
 $$ \begin{align} 0 &\leq m - 1 - \epsilon_0 \end{align} $$
$$ \begin{align} 0 &\leq m - 1 - \epsilon_0 \end{align} $$
 $$ \begin{align} 0 &\leq \frac{r - 3}{2} \end{align} $$
$$ \begin{align} 0 &\leq \frac{r - 3}{2} \end{align} $$
 $$ \begin{align} \frac{\delta - 1 - 2(d - g - r - \epsilon_1)}{r - 1} &\leq \frac{\delta - 1}{r - 1 - 2\kappa} - \frac{\left(\frac{r - 1}{2} - 1\right)\left(\frac{r - 1 - 2\kappa}{2} - 1\right)}{\left(\frac{r - 1}{2}\right)\left(\frac{r - 1 - 2\kappa}{2}\right)} \end{align} $$
$$ \begin{align} \frac{\delta - 1 - 2(d - g - r - \epsilon_1)}{r - 1} &\leq \frac{\delta - 1}{r - 1 - 2\kappa} - \frac{\left(\frac{r - 1}{2} - 1\right)\left(\frac{r - 1 - 2\kappa}{2} - 1\right)}{\left(\frac{r - 1}{2}\right)\left(\frac{r - 1 - 2\kappa}{2}\right)} \end{align} $$
 $$ \begin{align} \frac{\delta - 1 - 2(d - g - r - \epsilon_1)}{r - 1} &\leq \frac{\delta + r - 4}{r - 1} - \frac{\left(\frac{r - 1}{2} - 1\right)^2}{\left(\frac{r - 1}{2}\right)^2} \end{align} $$
$$ \begin{align} \frac{\delta - 1 - 2(d - g - r - \epsilon_1)}{r - 1} &\leq \frac{\delta + r - 4}{r - 1} - \frac{\left(\frac{r - 1}{2} - 1\right)^2}{\left(\frac{r - 1}{2}\right)^2} \end{align} $$
 $$ \begin{align} \frac{\delta - 1 - 2(d - g - r - \epsilon_1)}{r - 1} &\leq m - 1 - \epsilon_0 \end{align} $$
$$ \begin{align} \frac{\delta - 1 - 2(d - g - r - \epsilon_1)}{r - 1} &\leq m - 1 - \epsilon_0 \end{align} $$
 $$ \begin{align} \frac{\delta - 1 - 2(d - g - r - \epsilon_1)}{r - 1} &\leq \frac{r - 3}{2}. \end{align} $$
$$ \begin{align} \frac{\delta - 1 - 2(d - g - r - \epsilon_1)}{r - 1} &\leq \frac{r - 3}{2}. \end{align} $$
Since 
 $\delta $
 is an odd integer, we have
$\delta $
 is an odd integer, we have 
 $\delta \geq 1$
, which implies equations (10.113) and (10.114). The inequality (10.116) is immediate. Inequality (10.118) follows from
$\delta \geq 1$
, which implies equations (10.113) and (10.114). The inequality (10.116) is immediate. Inequality (10.118) follows from 
 $d \geq g + r + \epsilon _1$
, which holds by construction. For the remaining inequalities (10.115), (10.117), (10.119) and (10.120), we use the inequality
$d \geq g + r + \epsilon _1$
, which holds by construction. For the remaining inequalities (10.115), (10.117), (10.119) and (10.120), we use the inequality 
 $\epsilon _1 \leq 1$
 to reduce to the system
$\epsilon _1 \leq 1$
 to reduce to the system 
 $$ \begin{align} m &\geq 1 + \epsilon_0 \end{align} $$
$$ \begin{align} m &\geq 1 + \epsilon_0 \end{align} $$
 $$ \begin{align} \frac{\delta - 1 - 2(d - g - r - 1)}{r - 1} &\leq \frac{\delta - 1}{r - 1 - 2\kappa} - \frac{(r - 3)(r - 3 - 2\kappa)}{(r - 1)(r - 1 - 2\kappa)} \end{align} $$
$$ \begin{align} \frac{\delta - 1 - 2(d - g - r - 1)}{r - 1} &\leq \frac{\delta - 1}{r - 1 - 2\kappa} - \frac{(r - 3)(r - 3 - 2\kappa)}{(r - 1)(r - 1 - 2\kappa)} \end{align} $$
 $$ \begin{align} \frac{\delta - 1 - 2(d - g - r - 1)}{r - 1} &\leq m - 1 - \epsilon_0 \end{align} $$
$$ \begin{align} \frac{\delta - 1 - 2(d - g - r - 1)}{r - 1} &\leq m - 1 - \epsilon_0 \end{align} $$
 $$ \begin{align} \frac{\delta - 1 - 2(d - g - r - 1)}{r - 1} &\leq \frac{r - 3}{2}. \end{align} $$
$$ \begin{align} \frac{\delta - 1 - 2(d - g - r - 1)}{r - 1} &\leq \frac{r - 3}{2}. \end{align} $$
We divide our analysis as follows.
 
Inequality (10.121): This inequality asserts that we do not simultaneously have 
 $g = 0$
 (hence,
$g = 0$
 (hence, 
 $\epsilon _0 = 1$
) and
$\epsilon _0 = 1$
) and 
 $m = 1$
. So assume
$m = 1$
. So assume 
 $g = 0$
 and
$g = 0$
 and 
 $m = 1$
. Then
$m = 1$
. Then 
 $$\begin{align*}\delta = \frac{2d + 2g - 2r + 2\ell + (r + 1)m}{r - 1} = \frac{2d - r + 1}{r - 1}.\end{align*}$$
$$\begin{align*}\delta = \frac{2d + 2g - 2r + 2\ell + (r + 1)m}{r - 1} = \frac{2d - r + 1}{r - 1}.\end{align*}$$
Since 
 $r = g + r \leq d \leq g + 2r - 1 = 2r - 1$
, we would have
$r = g + r \leq d \leq g + 2r - 1 = 2r - 1$
, we would have 
 $$\begin{align*}1 < \frac{2r - r + 1}{r - 1} \leq \delta \leq \frac{2(2r - 1) - r + 1}{r - 1} < 5.\end{align*}$$
$$\begin{align*}1 < \frac{2r - r + 1}{r - 1} \leq \delta \leq \frac{2(2r - 1) - r + 1}{r - 1} < 5.\end{align*}$$
Since 
 $\delta $
 is an odd integer,
$\delta $
 is an odd integer, 
 $\delta = 3$
, and so
$\delta = 3$
, and so 
 $d = 2r - 2$
. In other words, writing
$d = 2r - 2$
. In other words, writing 
 $r = 2k + 1$
, we would have
$r = 2k + 1$
, we would have 
 $(d, g, r, \ell , m) = (4k, 0, 2k + 1, 0, 1)$
. But this case is excluded by assumption.
$(d, g, r, \ell , m) = (4k, 0, 2k + 1, 0, 1)$
. But this case is excluded by assumption.
 
Inequality (10.122) when 
 $(d, g) = (r + 1, 1)$
: In this case,
$(d, g) = (r + 1, 1)$
: In this case, 
 $\kappa = (r - 5)/2$
 so upon rearrangement, equation (10.122) becomes
$\kappa = (r - 5)/2$
 so upon rearrangement, equation (10.122) becomes 
 $$\begin{align*}\delta \geq \frac{3r - 3}{r - 5}.\end{align*}$$
$$\begin{align*}\delta \geq \frac{3r - 3}{r - 5}.\end{align*}$$
However,
 $$\begin{align*}\delta = \frac{2d + 2g - 2r + 2\ell + (r + 1)m}{r - 1} = \frac{(r + 1)m + 4}{r - 1};\end{align*}$$
$$\begin{align*}\delta = \frac{2d + 2g - 2r + 2\ell + (r + 1)m}{r - 1} = \frac{(r + 1)m + 4}{r - 1};\end{align*}$$
since 
 $\delta $
 is an integer, this implies
$\delta $
 is an integer, this implies 
 $2m + 4 \equiv (r + 1)m + 4 \equiv 0$
 mod
$2m + 4 \equiv (r + 1)m + 4 \equiv 0$
 mod 
 $r - 1$
, and so
$r - 1$
, and so 
 $m \equiv -2$
 mod
$m \equiv -2$
 mod 
 $(r - 1)/2$
, which implies
$(r - 1)/2$
, which implies 
 $m \geq (r - 1)/2 - 2 = (r - 5)/2$
. Therefore,
$m \geq (r - 1)/2 - 2 = (r - 5)/2$
. Therefore, 
 $$\begin{align*}\delta = \frac{(r + 1)m + 4}{r - 1} \geq \frac{(r + 1) \cdot (r - 5)/2 + 4}{r - 1} = \frac{r - 3}{2}.\end{align*}$$
$$\begin{align*}\delta = \frac{(r + 1)m + 4}{r - 1} \geq \frac{(r + 1) \cdot (r - 5)/2 + 4}{r - 1} = \frac{r - 3}{2}.\end{align*}$$
As 
 $r \geq 14$
, this implies
$r \geq 14$
, this implies 
 $\delta \geq (3r - 3)/(r - 5)$
 as desired.
$\delta \geq (3r - 3)/(r - 5)$
 as desired.
 
Inequality (10.123) when 
 $m \leq 1 + \epsilon _0$
: As we have already established
$m \leq 1 + \epsilon _0$
: As we have already established 
 $m \geq 1 + \epsilon _0$
, this implies
$m \geq 1 + \epsilon _0$
, this implies 
 $m = 1 + \epsilon _0$
. In this case, equation (10.123) asserts
$m = 1 + \epsilon _0$
. In this case, equation (10.123) asserts 
 $$\begin{align*}\delta \leq 2d - 2g - 2r - 1.\end{align*}$$
$$\begin{align*}\delta \leq 2d - 2g - 2r - 1.\end{align*}$$
By definition,
 $$ \begin{align*} \delta &= \frac{2d + 2g - 2r + (r + 1)(1 + \epsilon_0)}{r - 1} \\ &= 2d - 2g - 2r + 5 - \frac{(2r - 4)(d - g - r) + (4r - 6) - (4g + (r + 1)\epsilon_0)}{r - 1}. \end{align*} $$
$$ \begin{align*} \delta &= \frac{2d + 2g - 2r + (r + 1)(1 + \epsilon_0)}{r - 1} \\ &= 2d - 2g - 2r + 5 - \frac{(2r - 4)(d - g - r) + (4r - 6) - (4g + (r + 1)\epsilon_0)}{r - 1}. \end{align*} $$
 Since 
 $d \geq g + r$
, we have
$d \geq g + r$
, we have 
 $(2r - 4)(d - g - r) \geq 0$
, with
$(2r - 4)(d - g - r) \geq 0$
, with 
 $(2r - 4)(d - g - r) \geq 2r - 4 \geq 4$
 unless equality holds. Similarly, since
$(2r - 4)(d - g - r) \geq 2r - 4 \geq 4$
 unless equality holds. Similarly, since 
 $g \leq r - 1$
, we have
$g \leq r - 1$
, we have 
 $4g + (r + 1)\epsilon _0 \leq 4r - 4$
, with
$4g + (r + 1)\epsilon _0 \leq 4r - 4$
, with 
 $4g + (r + 1)\epsilon _0 \leq 4r - 8$
 unless equality holds. Putting this together, we have
$4g + (r + 1)\epsilon _0 \leq 4r - 8$
 unless equality holds. Putting this together, we have 
 $\delta \leq 2d - 2g - 2r + 5 + 2/(r - 1)$
, with
$\delta \leq 2d - 2g - 2r + 5 + 2/(r - 1)$
, with 
 $\delta \leq 2d - 2g - 2r + 5 - 2/(r - 1)$
 unless equality holds. As
$\delta \leq 2d - 2g - 2r + 5 - 2/(r - 1)$
 unless equality holds. As 
 $\delta $
 is an odd integer, this implies
$\delta $
 is an odd integer, this implies 
 $\delta \leq 2d - 2g - 2r + 3$
.
$\delta \leq 2d - 2g - 2r + 3$
.
 If 
 $g = 0$
, then
$g = 0$
, then 
 $4g + (r + 1)\epsilon _0 = r + 1$
. Therefore
$4g + (r + 1)\epsilon _0 = r + 1$
. Therefore 
 $\delta \leq 2d - 2g - 2r + 5 - (3r - 7)/(r - 1)$
, with
$\delta \leq 2d - 2g - 2r + 5 - (3r - 7)/(r - 1)$
, with 
 $\delta \leq 2d - 2g - 2r + 5 - (5r - 11)/(r - 1) < 2d - 2g - 2r + 1$
 unless equality holds. As
$\delta \leq 2d - 2g - 2r + 5 - (5r - 11)/(r - 1) < 2d - 2g - 2r + 1$
 unless equality holds. As 
 $\delta $
 is an odd integer, this implies
$\delta $
 is an odd integer, this implies 
 $\delta \leq 2d - 2g - 2r - 1$
 as desired.
$\delta \leq 2d - 2g - 2r - 1$
 as desired.
 It thus remains only to rule out the cases where 
 $g> 0$
 and
$g> 0$
 and 
 $\delta = 2d - 2g - 2r + 3$
 or
$\delta = 2d - 2g - 2r + 3$
 or 
 $\delta = 2d - 2g - 2r + 1$
. Upon rearrangement, this is equivalent to
$\delta = 2d - 2g - 2r + 1$
. Upon rearrangement, this is equivalent to 
 $$\begin{align*}(r - 2)(d - g - r) - 2g = -(r - 2) \ \text{or} \ 1.\end{align*}$$
$$\begin{align*}(r - 2)(d - g - r) - 2g = -(r - 2) \ \text{or} \ 1.\end{align*}$$
Since r is odd, considering the above equation mod 
 $2$
 implies that
$2$
 implies that 
 $d - g - r$
 must also be odd. If
$d - g - r$
 must also be odd. If 
 $d - g - r \geq 3$
, then since
$d - g - r \geq 3$
, then since 
 $g \leq r - 1$
, the left-hand side is at least
$g \leq r - 1$
, the left-hand side is at least 
 $3(r - 2) - 2(r - 1) = r - 4$
, which is impossible. Therefore,in this case we must have
$3(r - 2) - 2(r - 1) = r - 4$
, which is impossible. Therefore,in this case we must have 
 $d - g - r = 1$
. Solving for g we obtain
$d - g - r = 1$
. Solving for g we obtain 
 $g = r - 2$
 or
$g = r - 2$
 or 
 $g = (r - 3)/2$
. In other words, writing
$g = (r - 3)/2$
. In other words, writing 
 $r = 2k + 1$
, we would have
$r = 2k + 1$
, we would have 
 $(d, g, r, \ell , m) = (4k + 1, 2k - 1, 2k + 1, 0, 1)$
 or
$(d, g, r, \ell , m) = (4k + 1, 2k - 1, 2k + 1, 0, 1)$
 or 
 $(d, g, r, \ell , m) = (3k + 1, k - 1, 2k + 1, 0, 1)$
. But these cases are excluded by assumption.
$(d, g, r, \ell , m) = (3k + 1, k - 1, 2k + 1, 0, 1)$
. But these cases are excluded by assumption.
 
Inequalities (10.122) when 
 $(d, g) \neq (r + 1, 1)$
, and (10.123) when
$(d, g) \neq (r + 1, 1)$
, and (10.123) when 
 $m \geq 2 + \epsilon _0$
, and (10.124) (in all cases): Since
$m \geq 2 + \epsilon _0$
, and (10.124) (in all cases): Since 
 $(d, g) \neq (r + 1, 1)$
 for equation (10.122), we may substitute
$(d, g) \neq (r + 1, 1)$
 for equation (10.122), we may substitute 
 $\kappa = (r - 3)/2$
. Substituting in the definition of
$\kappa = (r - 3)/2$
. Substituting in the definition of 
 $\delta $
 and rearranging equations (10.122), (10.123) and (10.124), we obtain
$\delta $
 and rearranging equations (10.122), (10.123) and (10.124), we obtain 
 $$ \begin{align*} (6r - 10)(d - g - r) + (r^2 - 2r - 3)(m - 1 - \epsilon_0) + (4r - 12)(g + \epsilon_0 - 1) \qquad & \\ \qquad \qquad \phantom{.} + (r^2 - 6r + 9)\epsilon_0 + 2r - 14 &\geq 0\\ (2r - 4)(d - g - r) + (r^2 - 3r)(m - 2 - \epsilon_0) + 4(r - 1 - g) + (r + 1)(1 - \epsilon_0) + r^2 - 10r + 3 &\geq 0 \\ (4r - 8)(d - g - r) + (2r + 2)(r - 1 - m) - 8(r - 1 - g) + r^3 - 7r^2 - 3r + 9 &\geq 0. \end{align*} $$
$$ \begin{align*} (6r - 10)(d - g - r) + (r^2 - 2r - 3)(m - 1 - \epsilon_0) + (4r - 12)(g + \epsilon_0 - 1) \qquad & \\ \qquad \qquad \phantom{.} + (r^2 - 6r + 9)\epsilon_0 + 2r - 14 &\geq 0\\ (2r - 4)(d - g - r) + (r^2 - 3r)(m - 2 - \epsilon_0) + 4(r - 1 - g) + (r + 1)(1 - \epsilon_0) + r^2 - 10r + 3 &\geq 0 \\ (4r - 8)(d - g - r) + (2r + 2)(r - 1 - m) - 8(r - 1 - g) + r^3 - 7r^2 - 3r + 9 &\geq 0. \end{align*} $$
This establishes the desired inequalities (using that 
 $m \geq 2 + \epsilon _0$
 for equation (10.123)).
$m \geq 2 + \epsilon _0$
 for equation (10.123)).
 Finally, we complete our analysis of the case 
 $m \neq 0$
 by verifying the desired result in the four remaining one-parameter infinite families of cases:
$m \neq 0$
 by verifying the desired result in the four remaining one-parameter infinite families of cases: 
- 
Case (3) of Lemma 10.6: This follow from Proposition 8.3 with the following parameters:  $$\begin{align*}\ell' = 0, \quad m' = 2, \quad d' = d = 3k + 1, \quad \text{and} \quad (n_1, n_2) = (3, 2k - 3).\end{align*}$$ $$\begin{align*}\ell' = 0, \quad m' = 2, \quad d' = d = 3k + 1, \quad \text{and} \quad (n_1, n_2) = (3, 2k - 3).\end{align*}$$
- 
Case (4) of Lemma 10.6: This follows from Proposition 8.7 with  $\epsilon = 0$
. $\epsilon = 0$
.
- 
Case (1) of Lemma 10.7: This follows from Proposition 8.7 with  $\epsilon = 1$
. $\epsilon = 1$
.
10.2 The cases with 
 $m = 0$
 and
$m = 0$
 and 
 $g \neq 0$
$g \neq 0$
 As in the case 
 $m \neq 0$
, we will begin by showing that Proposition 8.2 handles ‘most’ of the cases by itself.
$m \neq 0$
, we will begin by showing that Proposition 8.2 handles ‘most’ of the cases by itself.
Proposition 10.8. Let 
 $(d, g, r, \ell , 0)$
 be a good tuple (with
$(d, g, r, \ell , 0)$
 be a good tuple (with 
 $m = 0$
) satisfying equation (10.1), such that
$m = 0$
) satisfying equation (10.1), such that 
 $g \neq 0$
 and
$g \neq 0$
 and 
 $r \geq 14$
. Then the conditions of Proposition 8.2 can be satisfied unless one of the following holds:
$r \geq 14$
. Then the conditions of Proposition 8.2 can be satisfied unless one of the following holds: 
- 
1.  $\delta \geq \ell + 1 + 2(d - g - r)$
. $\delta \geq \ell + 1 + 2(d - g - r)$
.
- 
2.  $\ell = 0$
 and $\ell = 0$
 and $\delta $
 is an odd integer. $\delta $
 is an odd integer.
Proof. Our goal is to show the existence of integers 
 $d'$
 and
$d'$
 and 
 $\ell '$
, such that
$\ell '$
, such that 
 $(d' - 1, g, r - 1, \ell - \ell ', 0)$
 is good (which is equivalent to
$(d' - 1, g, r - 1, \ell - \ell ', 0)$
 is good (which is equivalent to 
 $\ell - \ell ' = \overline {\ell } \leq (r - 1)/2$
), and the inequalities of Proposition 8.2 are satisfied:
$\ell - \ell ' = \overline {\ell } \leq (r - 1)/2$
), and the inequalities of Proposition 8.2 are satisfied: 
 $$ \begin{align} \ell - \ell' \leq \frac{r - 1}{2} \end{align} $$
$$ \begin{align} \ell - \ell' \leq \frac{r - 1}{2} \end{align} $$
 $$ \begin{align} 0 \leq \ell' \leq \ell \end{align} $$
$$ \begin{align} 0 \leq \ell' \leq \ell \end{align} $$
 $$ \begin{align} g + r \leq d' \leq d \end{align} $$
$$ \begin{align} g + r \leq d' \leq d \end{align} $$
 $$ \begin{align} |\delta - [\ell' + 2(d - d')]| \leq 1 - \frac{1}{r - 1} \end{align} $$
$$ \begin{align} |\delta - [\ell' + 2(d - d')]| \leq 1 - \frac{1}{r - 1} \end{align} $$
 $$ \begin{align} \ell' \leq r - 2. \end{align} $$
$$ \begin{align} \ell' \leq r - 2. \end{align} $$
Inequality (10.129) follows from equation (10.126) and the hypothesis 
 $\ell \leq r/2$
. Rewriting equations 10.125–10.128, we obtain the system
$\ell \leq r/2$
. Rewriting equations 10.125–10.128, we obtain the system 
 $$ \begin{align*} \ell - \frac{r - 1}{2} \leq \ell' \\ 0 \leq \ell' \leq \ell \\ \delta - 2(d - d') - \frac{r - 2}{r - 1} \leq \ell' \leq \delta - 2(d - d') + \frac{r - 2}{r - 1}\\ g + r \leq d' \leq d. \end{align*} $$
$$ \begin{align*} \ell - \frac{r - 1}{2} \leq \ell' \\ 0 \leq \ell' \leq \ell \\ \delta - 2(d - d') - \frac{r - 2}{r - 1} \leq \ell' \leq \delta - 2(d - d') + \frac{r - 2}{r - 1}\\ g + r \leq d' \leq d. \end{align*} $$
Applying Lemma 10.3 to eliminate 
 $\ell '$
, it suffices to show there is an integer solution
$\ell '$
, it suffices to show there is an integer solution 
 $d'$
 to the system:
$d'$
 to the system: 
 $$ \begin{align} \ell - \frac{r - 1}{2} &\leq \ell \end{align} $$
$$ \begin{align} \ell - \frac{r - 1}{2} &\leq \ell \end{align} $$
 $$ \begin{align} \ell - \frac{r - 1}{2} &\leq \delta - 2(d - d') + \frac{r - 2}{r - 1} - \frac{r - 2}{2r - 2} \end{align} $$
$$ \begin{align} \ell - \frac{r - 1}{2} &\leq \delta - 2(d - d') + \frac{r - 2}{r - 1} - \frac{r - 2}{2r - 2} \end{align} $$
 $$ \begin{align} 0 &\leq \ell \end{align} $$
$$ \begin{align} 0 &\leq \ell \end{align} $$
 $$ \begin{align} 0 &\leq \delta - 2(d - d') + \frac{r - 2}{r - 1} \end{align} $$
$$ \begin{align} 0 &\leq \delta - 2(d - d') + \frac{r - 2}{r - 1} \end{align} $$
 $$ \begin{align} \delta - 2(d - d') - \frac{r - 2}{r - 1} &\leq \ell \end{align} $$
$$ \begin{align} \delta - 2(d - d') - \frac{r - 2}{r - 1} &\leq \ell \end{align} $$
 $$ \begin{align} \delta - 2(d - d') - \frac{r - 2}{r - 1} &\leq \delta - 2(d - d') + \frac{r - 2}{r - 1} - \frac{(r - 2)^2}{(r - 1)^2} \end{align} $$
$$ \begin{align} \delta - 2(d - d') - \frac{r - 2}{r - 1} &\leq \delta - 2(d - d') + \frac{r - 2}{r - 1} - \frac{(r - 2)^2}{(r - 1)^2} \end{align} $$
 $$ \begin{align} g + r &\leq d' \leq d. \end{align} $$
$$ \begin{align} g + r &\leq d' \leq d. \end{align} $$
Inequalities (10.130), (10.132) and (10.135) are immediate. Rearranging the remaining inequalities, we obtain
 $$ \begin{align} d' &\geq d - \frac{\delta}{2} + \frac{\ell}{2} - \frac{r^2 - r - 1}{4r - 4}\end{align} $$
$$ \begin{align} d' &\geq d - \frac{\delta}{2} + \frac{\ell}{2} - \frac{r^2 - r - 1}{4r - 4}\end{align} $$
 $$ \begin{align} d' &\geq d - \frac{\delta}{2} - \frac{r - 2}{2r - 2} \end{align} $$
$$ \begin{align} d' &\geq d - \frac{\delta}{2} - \frac{r - 2}{2r - 2} \end{align} $$
 $$ \begin{align} d' &\geq g + r \end{align} $$
$$ \begin{align} d' &\geq g + r \end{align} $$
 $$ \begin{align} d' &\leq d - \frac{\delta}{2} + \frac{\ell}{2} + \frac{r - 2}{2r - 2} \end{align} $$
$$ \begin{align} d' &\leq d - \frac{\delta}{2} + \frac{\ell}{2} + \frac{r - 2}{2r - 2} \end{align} $$
 $$ \begin{align} d' &\leq d. \end{align} $$
$$ \begin{align} d' &\leq d. \end{align} $$
We next eliminate 
 $d'$
. Comparing equation (10.138) to equation (10.140), we want there to be an integer between
$d'$
. Comparing equation (10.138) to equation (10.140), we want there to be an integer between 
 $$\begin{align*}d - \frac{\delta}{2} - \frac{r - 2}{2r - 2} \quad \text{and} \quad d - \frac{\delta}{2} + \frac{\ell}{2} + \frac{r - 2}{2r - 2}.\end{align*}$$
$$\begin{align*}d - \frac{\delta}{2} - \frac{r - 2}{2r - 2} \quad \text{and} \quad d - \frac{\delta}{2} + \frac{\ell}{2} + \frac{r - 2}{2r - 2}.\end{align*}$$
By inspection, such an integer exists unless 
 $\delta $
 is an odd integer and
$\delta $
 is an odd integer and 
 $\ell = 0$
, which is excluded by assumption Proposition 10.8(2). Applying Lemma 10.3 for the remaining pairs of inequalities, we reduce to verifying
$\ell = 0$
, which is excluded by assumption Proposition 10.8(2). Applying Lemma 10.3 for the remaining pairs of inequalities, we reduce to verifying 
 $$ \begin{align} d - \frac{\delta}{2} + \frac{\ell}{2} - \frac{r^2 - r - 1}{4r - 4} &\leq d - \frac{\delta}{2} + \frac{\ell}{2} + \frac{r - 2}{2r - 2} - \frac{(2r - 3)(4r - 5)}{(2r - 2)(4r - 4)} \end{align} $$
$$ \begin{align} d - \frac{\delta}{2} + \frac{\ell}{2} - \frac{r^2 - r - 1}{4r - 4} &\leq d - \frac{\delta}{2} + \frac{\ell}{2} + \frac{r - 2}{2r - 2} - \frac{(2r - 3)(4r - 5)}{(2r - 2)(4r - 4)} \end{align} $$
 $$ \begin{align} d - \frac{\delta}{2} + \frac{\ell}{2} - \frac{r^2 - r - 1}{4r - 4} &\leq d \end{align} $$
$$ \begin{align} d - \frac{\delta}{2} + \frac{\ell}{2} - \frac{r^2 - r - 1}{4r - 4} &\leq d \end{align} $$
 $$ \begin{align} d - \frac{\delta}{2} - \frac{r - 2}{2r - 2} &\leq d \end{align} $$
$$ \begin{align} d - \frac{\delta}{2} - \frac{r - 2}{2r - 2} &\leq d \end{align} $$
 $$ \begin{align} g + r &\leq d - \frac{\delta}{2} + \frac{\ell}{2} + \frac{r - 2}{2r - 2} \end{align} $$
$$ \begin{align} g + r &\leq d - \frac{\delta}{2} + \frac{\ell}{2} + \frac{r - 2}{2r - 2} \end{align} $$
 $$ \begin{align} g + r &\leq d. \end{align} $$
$$ \begin{align} g + r &\leq d. \end{align} $$
Upon rearrangement, equation (10.145) is equivalent to
 $$\begin{align*}\delta \leq \ell + 1 + 2(d - g - r) - \frac{1}{r - 1},\end{align*}$$
$$\begin{align*}\delta \leq \ell + 1 + 2(d - g - r) - \frac{1}{r - 1},\end{align*}$$
which holds by our assumption Proposition 10.8(1). The remaining inequalities rearrange to
 $$ \begin{align*} 2r^3 - 8r^2 + 10r - 5 &\geq 0\\[5pt](r - 3)(r - 2\ell) + 4(d - g - r) + 8g + 2r - 1 &\geq 0\\[5pt]2\ell + 2(d - g - r) + 4g + r - 2 &\geq 0\\[5pt] d - g - r &\geq 0, \end{align*} $$
$$ \begin{align*} 2r^3 - 8r^2 + 10r - 5 &\geq 0\\[5pt](r - 3)(r - 2\ell) + 4(d - g - r) + 8g + 2r - 1 &\geq 0\\[5pt]2\ell + 2(d - g - r) + 4g + r - 2 &\geq 0\\[5pt] d - g - r &\geq 0, \end{align*} $$
which hold because 
 $(d, g, r, \ell , 0)$
 is good and
$(d, g, r, \ell , 0)$
 is good and 
 $r \geq 14$
.
$r \geq 14$
.
Lemma 10.9. Suppose that condition (1) or (2) of Proposition 10.8 is satisfied, but 
 $(\ell , \delta ) \neq (0, 1)$
 and
$(\ell , \delta ) \neq (0, 1)$
 and 
 $r \geq 14$
. Then one of the following two conditions holds:
$r \geq 14$
. Then one of the following two conditions holds: 
 $$ \begin{align} \ell \leq 3, \quad g \geq 4, \quad \text{and} \quad 1 + \frac{1}{r - 1} \leq \delta \leq 5 - \frac{2}{r - 1} \end{align} $$
$$ \begin{align} \ell \leq 3, \quad g \geq 4, \quad \text{and} \quad 1 + \frac{1}{r - 1} \leq \delta \leq 5 - \frac{2}{r - 1} \end{align} $$
or
 $$ \begin{align} \ell = 0, \quad g \geq 4, \quad d \geq g + r + 3, \quad \text{and} \quad \delta = 5. \end{align} $$
$$ \begin{align} \ell = 0, \quad g \geq 4, \quad d \geq g + r + 3, \quad \text{and} \quad \delta = 5. \end{align} $$
Proof. We divide into cases according to whether equation (1) or equation (2) is satisfied.
 
Case 1: equation (1) holds. In this case, we establish equation (10.147). Because 
 $\ell $
 and g are integers, the first two inequalities follow from
$\ell $
 and g are integers, the first two inequalities follow from 
 $\ell < 4$
 and
$\ell < 4$
 and 
 $g> 3$
 respectively. Upon rearrangement, these inequalities become
$g> 3$
 respectively. Upon rearrangement, these inequalities become 
 $$ \begin{align*} (r - 1)(\delta - (\ell + 1 + 2(d - g - r))) + (2r - 4)(d - g - r) + 4(r - 1 - g) + (r - 9) &> 0 \\ (r - 1)(\delta - (\ell + 1 + 2(d - g - r))) + (2r - 4)(d - g - r) + (r - 3)\ell + (r - 13) &> 0, \end{align*} $$
$$ \begin{align*} (r - 1)(\delta - (\ell + 1 + 2(d - g - r))) + (2r - 4)(d - g - r) + 4(r - 1 - g) + (r - 9) &> 0 \\ (r - 1)(\delta - (\ell + 1 + 2(d - g - r))) + (2r - 4)(d - g - r) + (r - 3)\ell + (r - 13) &> 0, \end{align*} $$
and therefore hold for 
 $r \geq 14$
 as desired. Since
$r \geq 14$
 as desired. Since 
 $\ell \geq 0$
 and
$\ell \geq 0$
 and 
 $d \geq g + r$
, we have
$d \geq g + r$
, we have 
 $$\begin{align*}\delta \geq \ell + 1 + 2(d - g - r) \geq 1,\end{align*}$$
$$\begin{align*}\delta \geq \ell + 1 + 2(d - g - r) \geq 1,\end{align*}$$
with equality only if 
 $\ell = 0$
. But equality is excluded by assumption (as
$\ell = 0$
. But equality is excluded by assumption (as 
 $(\ell , \delta ) \neq (0, 1)$
). Finally, the inequality
$(\ell , \delta ) \neq (0, 1)$
). Finally, the inequality 
 $\delta \leq 5 - 2/(r - 1)$
 becomes upon rearrangement
$\delta \leq 5 - 2/(r - 1)$
 becomes upon rearrangement 
 $$\begin{align*}(2r - 2)(\delta - (\ell + 1 + 2(d - g - r))) + (2r - 2)(d - g - r) + (4r - 4)(r - 1 - g) + (r^2 - 12r + 15) \geq 0.\end{align*}$$
$$\begin{align*}(2r - 2)(\delta - (\ell + 1 + 2(d - g - r))) + (2r - 2)(d - g - r) + (4r - 4)(r - 1 - g) + (r^2 - 12r + 15) \geq 0.\end{align*}$$
Case 2: equation (2) holds. In this case,
 $$\begin{align*}(r - 1)(7 - \delta) = 2(g + 2r - 1 - d) + 4(r - 1 - g) + (r - 1)> 0.\end{align*}$$
$$\begin{align*}(r - 1)(7 - \delta) = 2(g + 2r - 1 - d) + 4(r - 1 - g) + (r - 1)> 0.\end{align*}$$
Since 
 $\delta $
 is an odd integer, but
$\delta $
 is an odd integer, but 
 $\delta \neq 1$
 (because
$\delta \neq 1$
 (because 
 $\ell = 0$
 so
$\ell = 0$
 so 
 $\delta = 1$
 is excluded by assumption), we therefore have
$\delta = 1$
 is excluded by assumption), we therefore have 
 $\delta = 3$
 or
$\delta = 3$
 or 
 $\delta = 5$
. In particular,
$\delta = 5$
. In particular, 
 $$\begin{align*}4(g - 3) = (r - 1)(\delta - 3) + 2(g + 2r - 1 - d) + (r - 13)> 0;\end{align*}$$
$$\begin{align*}4(g - 3) = (r - 1)(\delta - 3) + 2(g + 2r - 1 - d) + (r - 13)> 0;\end{align*}$$
since g is an integer, this implies 
 $g \geq 4$
.
$g \geq 4$
.
 
Subcase 2.1: 
 $\delta = 3$
. Then equation (10.147) is satisfied.
$\delta = 3$
. Then equation (10.147) is satisfied.
 
Subcase 2.2: 
 $\delta = 5$
. In this case,
$\delta = 5$
. In this case, 
 $2(d - g - r - 3) = (r - 1)(\delta - 5) + 4(r - 1 - g) + (r - 7) \geq 0$
, so equation (10.148) is satisfied.
$2(d - g - r - 3) = (r - 1)(\delta - 5) + 4(r - 1 - g) + (r - 7) \geq 0$
, so equation (10.148) is satisfied.
 Recall that the case 
 $\delta = 1$
 and
$\delta = 1$
 and 
 $\ell = m = 0$
 is excluded in Theorem 10.2. Therefore, to complete our analysis of the case
$\ell = m = 0$
 is excluded in Theorem 10.2. Therefore, to complete our analysis of the case 
 $m = 0$
 and
$m = 0$
 and 
 $g \neq 0$
, we just have to handle the following two cases:
$g \neq 0$
, we just have to handle the following two cases: 
- 
If equation (10.147) holds Then we apply one of the following propositions according to the value of  $\delta $
: $\delta $
:- 
• If  $1 + \frac {1}{r - 1} \leq \delta \leq 3 - \frac {1}{r - 1}$
: Proposition 8.10. $1 + \frac {1}{r - 1} \leq \delta \leq 3 - \frac {1}{r - 1}$
: Proposition 8.10.
- 
• If  $2 + \frac {3}{r - 1} \leq \delta \leq 4 - \frac {3}{r - 1}$
: Proposition 8.9 with $2 + \frac {3}{r - 1} \leq \delta \leq 4 - \frac {3}{r - 1}$
: Proposition 8.9 with $\epsilon = 0$
. $\epsilon = 0$
.
- 
• If  $3 + \frac {2}{r - 1} \leq \delta \leq 5 - \frac {2}{r - 1}$
: Proposition 8.11. $3 + \frac {2}{r - 1} \leq \delta \leq 5 - \frac {2}{r - 1}$
: Proposition 8.11.
 Note that the union of these intervals covers the entire interval for  $\delta $
 given by the final inequality of equation (10.147). Moreover, the conditions $\delta $
 given by the final inequality of equation (10.147). Moreover, the conditions $\ell \leq 3$
 and $\ell \leq 3$
 and $g \geq 4$
 imply that all tuples appearing in these lemmas are good (they have positive genus and at most $g \geq 4$
 imply that all tuples appearing in these lemmas are good (they have positive genus and at most $\ell + 1 \leq 5$
 lines in a projective space of dimension at least $\ell + 1 \leq 5$
 lines in a projective space of dimension at least $r - 3 \geq 11$
). $r - 3 \geq 11$
).
- 
- 
If equation (10.148) holds Then we apply Proposition 8.9 with  $\epsilon = 1$
. $\epsilon = 1$
.
10.3 The cases with 
 $m = g = 0$
$m = g = 0$
 Since 
 $m = 0$
, Lemma 8.4 can be applied unless
$m = 0$
, Lemma 8.4 can be applied unless 
 $d \leq 2r - 2$
. In the remaining cases, we will show that Proposition 8.2 always applies.
$d \leq 2r - 2$
. In the remaining cases, we will show that Proposition 8.2 always applies.
Proposition 10.10. Let 
 $(d, 0, r, \ell , 0)$
 be a good tuple (with
$(d, 0, r, \ell , 0)$
 be a good tuple (with 
 $m = g = 0$
) satisfying
$m = g = 0$
) satisfying 
 $d \leq 2r - 2$
 and
$d \leq 2r - 2$
 and 
 $r \geq 14$
. Then the conditions of Proposition 8.2 can be satisfied.
$r \geq 14$
. Then the conditions of Proposition 8.2 can be satisfied.
Proof. As in the proof of Lemma 10.8, our goal is to show the existence of certain integers 
 $d'$
 and
$d'$
 and 
 $\ell '$
 which in particular must satisfy:
$\ell '$
 which in particular must satisfy: 
 $$ \begin{align} \ell - \frac{r - 1}{2} \leq \ell' \end{align} $$
$$ \begin{align} \ell - \frac{r - 1}{2} \leq \ell' \end{align} $$
 $$ \begin{align} 0 \leq \ell' \leq \ell \end{align} $$
$$ \begin{align} 0 \leq \ell' \leq \ell \end{align} $$
 $$ \begin{align}\delta - 2(d - d') - \frac{r - 2}{r - 1} \leq \ell' \leq \delta - 2(d - d') + \frac{r - 2}{r - 1}\end{align} $$
$$ \begin{align}\delta - 2(d - d') - \frac{r - 2}{r - 1} \leq \ell' \leq \delta - 2(d - d') + \frac{r - 2}{r - 1}\end{align} $$
 $$ \begin{align} r \leq d' \leq d, \end{align} $$
$$ \begin{align} r \leq d' \leq d, \end{align} $$
plus possibly some additional conditions to guarantee that 
 $(d' - 1, 0, r - 1, \ell - \ell ', 0)$
 is good. For this, we divide into cases as follows:
$(d' - 1, 0, r - 1, \ell - \ell ', 0)$
 is good. For this, we divide into cases as follows:
 
Case 1: 
 ${\boldsymbol{d}} = {\boldsymbol{r}}$
. In this case, we take
${\boldsymbol{d}} = {\boldsymbol{r}}$
. In this case, we take 
 $d' = r$
. With this choice
$d' = r$
. With this choice 
 $(1 - (d' - 1)) \% ((r - 1) - 1) = 0$
, and so equations (10.149)–(10.152) are sufficient for
$(1 - (d' - 1)) \% ((r - 1) - 1) = 0$
, and so equations (10.149)–(10.152) are sufficient for 
 $(d' - 1, g, r - 1, \ell - \ell ', 0)$
 to be good. Substituting
$(d' - 1, g, r - 1, \ell - \ell ', 0)$
 to be good. Substituting 
 $d = d' = r$
 and
$d = d' = r$
 and 
 $\delta = 2\ell / (r - 1)$
, our goal is thus to show that there is an integer
$\delta = 2\ell / (r - 1)$
, our goal is thus to show that there is an integer 
 $\ell '$
 satisfying
$\ell '$
 satisfying 
 $$ \begin{align*} \ell - \frac{r - 1}{2} \leq \ell' \\ 0 \leq \ell' \leq \ell \\ \frac{2\ell - r + 2}{r - 1} \leq \ell' \leq \frac{2\ell + r - 2}{r - 1}. \end{align*} $$
$$ \begin{align*} \ell - \frac{r - 1}{2} \leq \ell' \\ 0 \leq \ell' \leq \ell \\ \frac{2\ell - r + 2}{r - 1} \leq \ell' \leq \frac{2\ell + r - 2}{r - 1}. \end{align*} $$
Applying Lemma 10.3, it suffices to verify
 $$ \begin{align} 0 &\leq \ell \end{align} $$
$$ \begin{align} 0 &\leq \ell \end{align} $$
 $$ \begin{align} 0 &\leq \frac{2\ell + r - 2}{r - 1} \end{align} $$
$$ \begin{align} 0 &\leq \frac{2\ell + r - 2}{r - 1} \end{align} $$
 $$ \begin{align} \frac{2\ell - r + 2}{r - 1} &\leq \ell \end{align} $$
$$ \begin{align} \frac{2\ell - r + 2}{r - 1} &\leq \ell \end{align} $$
 $$ \begin{align} \frac{2\ell - r + 2}{r - 1} &\leq \frac{2\ell + r - 2}{r - 1} - \frac{(r - 2)^2}{(r - 1)^2} \end{align} $$
$$ \begin{align} \frac{2\ell - r + 2}{r - 1} &\leq \frac{2\ell + r - 2}{r - 1} - \frac{(r - 2)^2}{(r - 1)^2} \end{align} $$
 $$ \begin{align} \ell - \frac{r - 1}{2} &\leq \ell \end{align} $$
$$ \begin{align} \ell - \frac{r - 1}{2} &\leq \ell \end{align} $$
 $$ \begin{align} \ell - \frac{r - 1}{2} &\leq \frac{2\ell + r - 2}{r - 1} - \frac{r - 2}{2r - 2}. \end{align} $$
$$ \begin{align} \ell - \frac{r - 1}{2} &\leq \frac{2\ell + r - 2}{r - 1} - \frac{r - 2}{2r - 2}. \end{align} $$
Inequalities (10.153)–(10.155) follow from 
 $\ell \geq 0$
, and equations (10.156) and (10.157) are automatic, and equation (10.158) follows from
$\ell \geq 0$
, and equations (10.156) and (10.157) are automatic, and equation (10.158) follows from 
 $\ell \leq r/2$
.
$\ell \leq r/2$
.
 
Case 2: 
 ${\boldsymbol{d}} \geq {\boldsymbol{r}} + \mathbf {1}$
. Since
${\boldsymbol{d}} \geq {\boldsymbol{r}} + \mathbf {1}$
. Since 
 $r + 1 \leq d \leq 2r - 2$
, we have
$r + 1 \leq d \leq 2r - 2$
, we have 
 $(1 - d) \% (r - 1) = (1 - d) + 2(r - 1) = 2r - 1 - d$
. Therefore, as
$(1 - d) \% (r - 1) = (1 - d) + 2(r - 1) = 2r - 1 - d$
. Therefore, as 
 $(d, g, r, \ell , 0)$
 is good by assumption,
$(d, g, r, \ell , 0)$
 is good by assumption, 
 $\ell $
 satisfies
$\ell $
 satisfies 
 $$ \begin{align} \frac{2r - 1 - d}{2} \leq \ell \leq \frac{r}{2}. \end{align} $$
$$ \begin{align} \frac{2r - 1 - d}{2} \leq \ell \leq \frac{r}{2}. \end{align} $$
Similarly, because 
 $d' - 1 \leq 2r - 3$
, we have
$d' - 1 \leq 2r - 3$
, we have 
 $(1 - (d' - 1)) \% ((r - 1) - 1) \leq (2 - d') + 2(r - 2) = 2r - 2 - d'$
. Therefore,
$(1 - (d' - 1)) \% ((r - 1) - 1) \leq (2 - d') + 2(r - 2) = 2r - 2 - d'$
. Therefore,
 $(d' - 1, g, r - 1, \ell - \ell ', 0)$
 is good provided equations (10.149)–(10.152) are satisfied, and also
$(d' - 1, g, r - 1, \ell - \ell ', 0)$
 is good provided equations (10.149)–(10.152) are satisfied, and also 
 $2(\ell - \ell ') \geq 2r - 2 - d'$
. In other words, we want to show that there are integers
$2(\ell - \ell ') \geq 2r - 2 - d'$
. In other words, we want to show that there are integers 
 $\ell '$
 and
$\ell '$
 and 
 $d'$
 satisfying the following system (here we have substituted in
$d'$
 satisfying the following system (here we have substituted in 
 $\delta = (2d - 2r + 2\ell )/(r - 1)$
):
$\delta = (2d - 2r + 2\ell )/(r - 1)$
): 
 $$ \begin{align} \ell - \frac{r - 1}{2} \leq \ell' \end{align} $$
$$ \begin{align} \ell - \frac{r - 1}{2} \leq \ell' \end{align} $$
 $$ \begin{align} 0 \leq \ell' \leq \ell \end{align} $$
$$ \begin{align} 0 \leq \ell' \leq \ell \end{align} $$
 $$ \begin{align} \frac{2d - 2r + 2\ell}{r - 1} - 2(d - d') - \frac{r - 2}{r - 1} \leq \ell' \leq \frac{2d - 2r + 2\ell}{r - 1} - 2(d - d') + \frac{r - 2}{r - 1} \end{align} $$
$$ \begin{align} \frac{2d - 2r + 2\ell}{r - 1} - 2(d - d') - \frac{r - 2}{r - 1} \leq \ell' \leq \frac{2d - 2r + 2\ell}{r - 1} - 2(d - d') + \frac{r - 2}{r - 1} \end{align} $$
 $$ \begin{align} r \leq d' \leq d \end{align} $$
$$ \begin{align} r \leq d' \leq d \end{align} $$
 $$ \begin{align} \ell' \leq \ell - \frac{2r - 2 - d'}{2}. \end{align} $$
$$ \begin{align} \ell' \leq \ell - \frac{2r - 2 - d'}{2}. \end{align} $$
 
Subcase 2.1: 
 $d = r + 1$
. In this case, equation (10.159) becomes
$d = r + 1$
. In this case, equation (10.159) becomes 
 $r/2 - 1 \leq \ell \leq r/2$
. In other words,
$r/2 - 1 \leq \ell \leq r/2$
. In other words, 
 $(d, r, \ell )$
 is of one of the following forms:
$(d, r, \ell )$
 is of one of the following forms: 
 $$\begin{align*}(d, r, \ell) = (2k + 1, 2k, k - 1), \quad (d, r, \ell) = (2k + 1, 2k, k), \quad \text{or} \quad (d, r, \ell) = (2k + 2, 2k + 1, k).\end{align*}$$
$$\begin{align*}(d, r, \ell) = (2k + 1, 2k, k - 1), \quad (d, r, \ell) = (2k + 1, 2k, k), \quad \text{or} \quad (d, r, \ell) = (2k + 2, 2k + 1, k).\end{align*}$$
These cases may satisfy equations 10.160–10.164 by taking
 $$\begin{align*}(d', \ell') = (2k, 0), \quad (d', \ell') = (2k + 1, 1), \quad \text{respectively} \quad (d', \ell') = (2k + 1, 0).\end{align*}$$
$$\begin{align*}(d', \ell') = (2k, 0), \quad (d', \ell') = (2k + 1, 1), \quad \text{respectively} \quad (d', \ell') = (2k + 1, 0).\end{align*}$$
 
Subcase 2.2: 
 $d \geq r + 2$
. In this case, we take
$d \geq r + 2$
. In this case, we take 
 $\ell ' = 2 - 2(d - d')$
, in which case equations 10.160–10.164 become
$\ell ' = 2 - 2(d - d')$
, in which case equations 10.160–10.164 become 
 $$ \begin{align*} d' \geq \frac{4d + 2\ell - r - 3}{4} \\ d - 1 \leq d' \leq d - 1 + \frac{\ell}{2} \\ 3r \leq 2d + 2\ell \leq 5r - 4 \\ r \leq d' \leq d \\ d' \leq \frac{4d + 2\ell - 2r - 2}{3}. \end{align*} $$
$$ \begin{align*} d' \geq \frac{4d + 2\ell - r - 3}{4} \\ d - 1 \leq d' \leq d - 1 + \frac{\ell}{2} \\ 3r \leq 2d + 2\ell \leq 5r - 4 \\ r \leq d' \leq d \\ d' \leq \frac{4d + 2\ell - 2r - 2}{3}. \end{align*} $$
The inequality 
 $d' \geq r$
 follows from
$d' \geq r$
 follows from 
 $d' \geq d - 1$
 since
$d' \geq d - 1$
 since 
 $d = r + 2$
. Deleting the inequality
$d = r + 2$
. Deleting the inequality 
 $d' \geq r$
 and eliminating
$d' \geq r$
 and eliminating 
 $d'$
 via Lemma 10.3, we reduce to the system
$d'$
 via Lemma 10.3, we reduce to the system 
 $$ \begin{align} \frac{4d + 2\ell - r - 3}{4} &\leq d - 1 + \frac{\ell}{2} - \frac{3}{8}\end{align} $$
$$ \begin{align} \frac{4d + 2\ell - r - 3}{4} &\leq d - 1 + \frac{\ell}{2} - \frac{3}{8}\end{align} $$
 $$ \begin{align} \frac{4d + 2\ell - r - 3}{4} &\leq d \end{align} $$
$$ \begin{align} \frac{4d + 2\ell - r - 3}{4} &\leq d \end{align} $$
 $$ \begin{align} \frac{4d + 2\ell - r - 3}{4} &\leq \frac{4d + 2\ell - 2r - 2}{3} - \frac{1}{2} \end{align} $$
$$ \begin{align} \frac{4d + 2\ell - r - 3}{4} &\leq \frac{4d + 2\ell - 2r - 2}{3} - \frac{1}{2} \end{align} $$
 $$ \begin{align} d - 1 &\leq d - 1 + \frac{\ell}{2} \end{align} $$
$$ \begin{align} d - 1 &\leq d - 1 + \frac{\ell}{2} \end{align} $$
 $$ \begin{align} d - 1 &\leq d \end{align} $$
$$ \begin{align} d - 1 &\leq d \end{align} $$
 $$ \begin{align} d - 1 &\leq \frac{4d + 2\ell - 2r - 2}{3} \end{align} $$
$$ \begin{align} d - 1 &\leq \frac{4d + 2\ell - 2r - 2}{3} \end{align} $$
 $$ \begin{align} 3r &\leq 2d + 2\ell \end{align} $$
$$ \begin{align} 3r &\leq 2d + 2\ell \end{align} $$
 $$ \begin{align} 2d + 2\ell &\leq 5r - 4. \end{align} $$
$$ \begin{align} 2d + 2\ell &\leq 5r - 4. \end{align} $$
Inequalities (10.165), (10.168) and (10.169) are immediate. Rearranging the others, we obtain
 $$ \begin{align*} (r - 2\ell) + 3 &\geq 0 \\ (2\ell - 2r + 1 + d) + 3(d - r - 2) &\geq 0 \\ 2\ell - 2r + 1 + d &\geq 0 \\ (2\ell - 2r + 1 + d) + (d - r - 2) + 1 &\geq 0 \\ (r - 2\ell) + 2(2r - 2 - d) + 8 &\geq 0, \end{align*} $$
$$ \begin{align*} (r - 2\ell) + 3 &\geq 0 \\ (2\ell - 2r + 1 + d) + 3(d - r - 2) &\geq 0 \\ 2\ell - 2r + 1 + d &\geq 0 \\ (2\ell - 2r + 1 + d) + (d - r - 2) + 1 &\geq 0 \\ (r - 2\ell) + 2(2r - 2 - d) + 8 &\geq 0, \end{align*} $$
which all follow from equation (10.159), our assumption 
 $r + 2 \leq d \leq 2r - 2$
, and the hypotheses in (10.1).
$r + 2 \leq d \leq 2r - 2$
, and the hypotheses in (10.1).
11 Most of the sporadic cases
The finite set of sporadic cases identified in the previous section is unfortunately rather large. Our next task is to introduce an additional argument that, in combination with the arguments of Section 8, applies to handle most of the sporadic cases, that is, all but a list that is short enough to write down explicitly.
 This argument will, essentially, be a variant on Proposition 8.2, but where we allow transformations to come together at p in a less restricted way. In particular, we will weaken the hypothesis ‘
 $2m' + \ell \leq r-2$
’ in the statement of Proposition 8.2 by allowing more modifications to limit to the point p than the rank of the normal bundle. In this regime, the limiting bundle can depend on how the points are specialized into p. We will be able to give a description of some possible limits by limiting the marked points into p one at a time inductively. At each step, we will be able to identify what the limiting modifications are at p. Suppose that, after limiting some collection of marked points into p, we have a transformation at p of the form
$2m' + \ell \leq r-2$
’ in the statement of Proposition 8.2 by allowing more modifications to limit to the point p than the rank of the normal bundle. In this regime, the limiting bundle can depend on how the points are specialized into p. We will be able to give a description of some possible limits by limiting the marked points into p one at a time inductively. At each step, we will be able to identify what the limiting modifications are at p. Suppose that, after limiting some collection of marked points into p, we have a transformation at p of the form 

where 
 $\Lambda _1$
 and
$\Lambda _1$
 and 
 $\Lambda _2$
 are linear spaces in
$\Lambda _2$
 are linear spaces in 
 $\mathbb {P} N_C|_p$
. Let S and W be sets of parameters varying in irreducible bases with
$\mathbb {P} N_C|_p$
. Let S and W be sets of parameters varying in irreducible bases with 
 $S \subseteq W$
. (For us, W will be the collection of all corresponding marked points, and S will be those marked points at which the projected normal bundle is not modified.) Assume that
$S \subseteq W$
. (For us, W will be the collection of all corresponding marked points, and S will be those marked points at which the projected normal bundle is not modified.) Assume that 
 $\Lambda _2$
 is linearly general as the parameters S vary, and assume that
$\Lambda _2$
 is linearly general as the parameters S vary, and assume that 
 $\Lambda _1$
 is either:
$\Lambda _1$
 is either:
- 
• Linearly general as the parameters W vary (‘weakly general’); 
- 
• Weakly general and its image in  $\mathbb {P} (N_C / N_{C \to p})|_p$
 is linearly general as only the parameters S vary (‘strongly general’). $\mathbb {P} (N_C / N_{C \to p})|_p$
 is linearly general as only the parameters S vary (‘strongly general’).
 We will summarize this situation by three pieces of data: the linear dimensions 
 $t_1 = \operatorname {rk} \Lambda _1 = \dim \Lambda _1 + 1$
 and
$t_1 = \operatorname {rk} \Lambda _1 = \dim \Lambda _1 + 1$
 and 
 $t_2 = \operatorname {rk} \Lambda _2 = \dim \Lambda _2 + 1$
, and whether we are in the weak or strong case. Note that we always have
$t_2 = \operatorname {rk} \Lambda _2 = \dim \Lambda _2 + 1$
, and whether we are in the weak or strong case. Note that we always have 
 $t_2 \leq t_1 \leq r - 2$
. (We do not need to keep track of the integer n since this is can be deduced from the Euler characteristic of the limit bundle, which is the same as the original bundle.)
$t_2 \leq t_1 \leq r - 2$
. (We do not need to keep track of the integer n since this is can be deduced from the Euler characteristic of the limit bundle, which is the same as the original bundle.)
Modifications of type (†) occur naturally when considering the degenerations that make up the key inductive argument outlined in Section 8.1. We review them here (and add one additional argument that we will use in this section) in order to motivate the shape of equation (†).
- 
1. If we peel off a one-secant line  $\overline {xy}$
 and limit the point x to p, we obtain $\overline {xy}$
 and limit the point x to p, we obtain at p. This is of the form (†) with at p. This is of the form (†) with $\Lambda _1 = \Lambda _2 = y$
; hence, we have $\Lambda _1 = \Lambda _2 = y$
; hence, we have $t_1 = t_2 = 1$
. Since y is a general point, and no modifications occur at y in the quotient by projection from p, both $t_1 = t_2 = 1$
. Since y is a general point, and no modifications occur at y in the quotient by projection from p, both $\Lambda _1$
 and $\Lambda _1$
 and $\Lambda _2$
 are strongly linearly general. $\Lambda _2$
 are strongly linearly general.
- 
2. If we peel off a one-secant line  $\overline {zw}$
 and limit the point z to p, we obtain $\overline {zw}$
 and limit the point z to p, we obtain at p. This is of the form (†) with at p. This is of the form (†) with $\Lambda _1 = T_wC$
 and $\Lambda _1 = T_wC$
 and $\Lambda _2 = w$
; hence, we have $\Lambda _2 = w$
; hence, we have $(t_1, t_2) = (2, 1)$
. Since w is a general point on the curve, at which no modifications occur in the quotient by projection from p, both $(t_1, t_2) = (2, 1)$
. Since w is a general point on the curve, at which no modifications occur in the quotient by projection from p, both $\Lambda _1$
 and $\Lambda _1$
 and $\Lambda _2$
 are strongly linearly general. $\Lambda _2$
 are strongly linearly general.
- 
3. If we specialize R as in Section 7 to contain n lines through p, then at p we obtain the modification  , where , where $\operatorname {rk} \Lambda = 2$
. This is of the form (†) with $\operatorname {rk} \Lambda = 2$
. This is of the form (†) with $(t_1, t_2) = (2, 0)$
. By Lemmas 7.5 and 7.6, the subspace $(t_1, t_2) = (2, 0)$
. By Lemmas 7.5 and 7.6, the subspace $\Lambda $
 is strongly linearly general if $\Lambda $
 is strongly linearly general if $$\begin{align*}n \geq \begin{cases} 3 & \text{if}\ C\text{ is an elliptic normal curve} \\ 2 & \text{otherwise.} \end{cases}\end{align*}$$ $$\begin{align*}n \geq \begin{cases} 3 & \text{if}\ C\text{ is an elliptic normal curve} \\ 2 & \text{otherwise.} \end{cases}\end{align*}$$
- 
(3′) If we specialize one of the  $v_i$
 to p, we obtain modification $v_i$
 to p, we obtain modification at p. This is of the form (†) with at p. This is of the form (†) with $(t_1, t_2) = (1, 0)$
 and is strongly linearly general since $(t_1, t_2) = (1, 0)$
 and is strongly linearly general since $u_i$
 is a general point, at which no modifications occur in the quotient by projection from p. $u_i$
 is a general point, at which no modifications occur in the quotient by projection from p.
- 
7. We allow ourselves one new degeneration in our more general inductive step, which is similar to equation (6) from Section 8.1 but crucially different in that we specialize R to pass through p before we project. We first specialize R to the union of a line L through  $2$
 points $2$
 points $s_0, s_r$
 on C and a rational curve $s_0, s_r$
 on C and a rational curve $R'$
 of degree $R'$
 of degree $r-2$
 through $r-2$
 through $r-1$
 points on C and meeting L at one point. Then we specialize $r-1$
 points on C and meeting L at one point. Then we specialize $s_0$
 to p. This results in the modification $s_0$
 to p. This results in the modification at p, which is of type at p, which is of type $(t_1, t_2) = (1,0)$
. This modification is linearly general as all the points of contact between C and $(t_1, t_2) = (1,0)$
. This modification is linearly general as all the points of contact between C and $R'$
 move. However, $R'$
 move. However, $s_r$
 is constrained to be one of the points at which the r-secant rational curve $s_r$
 is constrained to be one of the points at which the r-secant rational curve $\overline {R'}$
 meets $\overline {R'}$
 meets $\overline {C}$
, and modifications occur at the remainder of these points, so it is only weakly linearly general. $\overline {C}$
, and modifications occur at the remainder of these points, so it is only weakly linearly general.
 Our first goal is to understand what happens when we limit into p another point 
 $p'$
, at which we have another transformation
$p'$
, at which we have another transformation 
 of the same form (†) (depending on sets of parameters
 of the same form (†) (depending on sets of parameters 
 $S' \subseteq W'$
 disjoint from W). In the following five cases, which we consider separately, we will see that in the limit we obtain another transformation of the form (†) (depending on parameters
$S' \subseteq W'$
 disjoint from W). In the following five cases, which we consider separately, we will see that in the limit we obtain another transformation of the form (†) (depending on parameters 
 $S \cup S' \subseteq W \cup W'$
).
$S \cup S' \subseteq W \cup W'$
).
 Most of the subspaces whose generality we must assess are of the form 
 $\Lambda + \Lambda '$
 (the span of
$\Lambda + \Lambda '$
 (the span of 
 $\Lambda $
 and
$\Lambda $
 and 
 $\Lambda '$
). If
$\Lambda '$
). If 
 $\Lambda $
 and
$\Lambda $
 and 
 $\Lambda '$
 are both linearly general as independent parameters X and
$\Lambda '$
 are both linearly general as independent parameters X and 
 $X'$
 vary, then their span is linearly general, as we now show. Let M be a fixed subspace; there is a choice of the parameters X for which the corresponding subspace
$X'$
 vary, then their span is linearly general, as we now show. Let M be a fixed subspace; there is a choice of the parameters X for which the corresponding subspace 
 $\Lambda $
 meets M transversely. Then there is a choice of the parameters
$\Lambda $
 meets M transversely. Then there is a choice of the parameters 
 $X'$
 for which the corresponding subspace
$X'$
 for which the corresponding subspace 
 $\Lambda '$
 meets
$\Lambda '$
 meets 
 $M + \Lambda $
 transversely. For this choice of
$M + \Lambda $
 transversely. For this choice of 
 $X \cup X'$
, the subspace M meets
$X \cup X'$
, the subspace M meets 
 $\Lambda + \Lambda '$
 transversely. The only case where the resulting modification is not of this form is (c) below.
$\Lambda + \Lambda '$
 transversely. The only case where the resulting modification is not of this form is (c) below.
- 
(a) If  $t_1 + t_1' < r - 1$
: In this case, the limiting transformation is This transformation is of the desired form. The subspace $t_1 + t_1' < r - 1$
: In this case, the limiting transformation is This transformation is of the desired form. The subspace  $\Lambda _1 + \Lambda _1'$
 is strongly general if $\Lambda _1 + \Lambda _1'$
 is strongly general if $\Lambda _1$
 and $\Lambda _1$
 and $\Lambda _1'$
 are both strongly general and weakly general otherwise. $\Lambda _1'$
 are both strongly general and weakly general otherwise.
- 
(b) If  $t_2 + t_2' < t_1 + t_1' = r - 1$
: In this case, the limiting transformation is This transformation is of the desired form, and the subspace $t_2 + t_2' < t_1 + t_1' = r - 1$
: In this case, the limiting transformation is This transformation is of the desired form, and the subspace  $\Lambda _2 + \Lambda _2'$
 is always strongly general. $\Lambda _2 + \Lambda _2'$
 is always strongly general.
- 
(c) If  $t_2' = 0$
 and $t_2' = 0$
 and $t_1' + t_2 \leq r - 1 \leq t_1' + t_1$
: In this case, the limiting transformation is We now show that $t_1' + t_2 \leq r - 1 \leq t_1' + t_1$
: In this case, the limiting transformation is We now show that  $\Lambda _2 + (\Lambda _1 \cap \Lambda _1')$
 is strongly linearly general if both $\Lambda _2 + (\Lambda _1 \cap \Lambda _1')$
 is strongly linearly general if both $\Lambda _1$
 and $\Lambda _1$
 and $\Lambda _1'$
 are strongly general and weakly linearly general otherwise. Indeed, let M be any fixed subspace; we want to show that M is transverse to $\Lambda _1'$
 are strongly general and weakly linearly general otherwise. Indeed, let M be any fixed subspace; we want to show that M is transverse to $\Lambda _2 + (\Lambda _1 \cap \Lambda _1')$
. Both $\Lambda _2 + (\Lambda _1 \cap \Lambda _1')$
. Both $\Lambda _1$
 and $\Lambda _1$
 and $\Lambda _2$
 are linearly general; since the parameters W vary in an irreducible base, there is a single choice of the parameters W for which M is simultaneously transverse to both $\Lambda _2$
 are linearly general; since the parameters W vary in an irreducible base, there is a single choice of the parameters W for which M is simultaneously transverse to both $\Lambda _1$
 and $\Lambda _1$
 and $\Lambda _2$
. Since $\Lambda _2$
. Since $\Lambda _1$
 is transverse to M for our choice of parameters W, and $\Lambda _1$
 is transverse to M for our choice of parameters W, and $\Lambda _2 \subset \Lambda _1$
, we can restrict to $\Lambda _2 \subset \Lambda _1$
, we can restrict to $\Lambda _1$
 and consider transversality as subspaces of $\Lambda _1$
 and consider transversality as subspaces of $\Lambda _1$
. The subspace $\Lambda _1$
. The subspace $M \cap \Lambda _1$
 is transverse to $M \cap \Lambda _1$
 is transverse to $\Lambda _2$
 for our choice of parameters W. The subspace $\Lambda _2$
 for our choice of parameters W. The subspace $\Lambda _1' \cap \Lambda _1$
 is transverse to $\Lambda _1' \cap \Lambda _1$
 is transverse to $\Lambda _2$
 since $\Lambda _2$
 since $\Lambda _1'$
 is linearly general, varying with independent parameters. Therefore,the transversality of $\Lambda _1'$
 is linearly general, varying with independent parameters. Therefore,the transversality of $M \cap \Lambda _1$
 and $M \cap \Lambda _1$
 and $\Lambda _2 + (\Lambda _1' \cap \Lambda _1)$
 is equivalent to the transversality of $\Lambda _2 + (\Lambda _1' \cap \Lambda _1)$
 is equivalent to the transversality of $\Lambda _1 \cap \Lambda _1'$
 and $\Lambda _1 \cap \Lambda _1'$
 and $\Lambda _2 + (M \cap \Lambda _1)$
, which again follows by the linear generality of $\Lambda _2 + (M \cap \Lambda _1)$
, which again follows by the linear generality of $\Lambda _1'$
. $\Lambda _1'$
.
- 
(d) If  $t_1' + t_2 < t_1 + t_2' = r - 1$
: In this case, the limiting transformation is This transformation is of the desired form, with $t_1' + t_2 < t_1 + t_2' = r - 1$
: In this case, the limiting transformation is This transformation is of the desired form, with  $\Lambda _2 + \Lambda _1'$
 strongly general if $\Lambda _2 + \Lambda _1'$
 strongly general if $\Lambda _1'$
 is strongly general, and weakly general otherwise. $\Lambda _1'$
 is strongly general, and weakly general otherwise.
- 
(e) If  $t_1 + t_2' = t_1' + t_2 = r - 1$
: In this case, the limiting transformation is This transformation is of the desired form, with $t_1 + t_2' = t_1' + t_2 = r - 1$
: In this case, the limiting transformation is This transformation is of the desired form, with  $\emptyset $
 always strongly general. $\emptyset $
 always strongly general.
Definition 11.1. For integers 
 $0 \leq i,j < r-1$
, let
$0 \leq i,j < r-1$
, let 
 $\{s_{ij}\}$
 and
$\{s_{ij}\}$
 and 
 $\{w_{ij}\}$
 be collections of nonnegative integers, and consider
$\{w_{ij}\}$
 be collections of nonnegative integers, and consider 
 $s_{ij}$
 (respectively
$s_{ij}$
 (respectively 
 $w_{ij}$
) marked points decorated with modifications of type (†) with
$w_{ij}$
) marked points decorated with modifications of type (†) with 
 $(t_1, t_2) = (i, j)$
 and
$(t_1, t_2) = (i, j)$
 and 
 $\Lambda _1$
 strongly (respectively weakly) general. Consider all ways of limiting these marked points into p, one at a time in some order, such that at every step of the process, we are in one of the five cases discussed above. If there is such an order for which the final resulting transformation at p satisfies
$\Lambda _1$
 strongly (respectively weakly) general. Consider all ways of limiting these marked points into p, one at a time in some order, such that at every step of the process, we are in one of the five cases discussed above. If there is such an order for which the final resulting transformation at p satisfies 
 $t_2 = 0$
 and
$t_2 = 0$
 and 
 $\Lambda _1$
 is strongly general, then we say that
$\Lambda _1$
 is strongly general, then we say that 
 $\{s_{ij}\}$
 and
$\{s_{ij}\}$
 and 
 $\{w_{ij}\}$
 is erasable.
$\{w_{ij}\}$
 is erasable.
We are now ready to state our more flexible variant on Proposition 8.2. The high-level overview is that, in some order, we do the following specializations:
- 
• Peel off  $g - g'$
 two-secant lines. Specialize all of them into p as in equation (1). $g - g'$
 two-secant lines. Specialize all of them into p as in equation (1).
- 
• Peel off  $\epsilon _{\text {in}} + \epsilon _{\text {out}} = d - g - d' + g'$
 one-secant lines. Specialize $\epsilon _{\text {in}} + \epsilon _{\text {out}} = d - g - d' + g'$
 one-secant lines. Specialize $\epsilon _{\text {in}}$
 of these into p as in the proof of Proposition 8.2 and the remaining $\epsilon _{\text {in}}$
 of these into p as in the proof of Proposition 8.2 and the remaining $\epsilon _{\text {out}}$
 of them into p as in equation (2). $\epsilon _{\text {out}}$
 of them into p as in equation (2).
- 
• Specialize  $m'$
 of the rational curves $m'$
 of the rational curves $R_i$
 as in Section 7 to lines and conics through $R_i$
 as in Section 7 to lines and conics through $p_i$
. Specialize all of the $p_i$
. Specialize all of the $p_i$
 to p as in equation (3). $p_i$
 to p as in equation (3).
- 
• Specialize  $\ell '$
 of the points $\ell '$
 of the points $v_i$
 to p as in (3′). $v_i$
 to p as in (3′).
- 
• Specialize  $m"$
 of the rational curves $m"$
 of the rational curves $R_i$
 to the union $R_i$
 to the union $L_i \cup R_i'$
 as in equation (7). Specialize one of the points where $L_i \cup R_i'$
 as in equation (7). Specialize one of the points where $L_i$
 meets C into p. $L_i$
 meets C into p.
- 
• Specialize the remaining  $m - m' - m"$
 rational curves $m - m' - m"$
 rational curves $R_i$
 to pass through p as in Section 8.1(6). $R_i$
 to pass through p as in Section 8.1(6).
After projecting from p, we will reduce to a case of our inductive hypothesis plus a single linearly general modification at p precisely when the modifications at p above are erasable.
Proposition 11.2. Let 
 $\ell '$
,
$\ell '$
, 
 $m'$
, and
$m'$
, and 
 $m"$
 be nonnegative integers satisfying
$m"$
 be nonnegative integers satisfying 
 $\ell ' \leq \ell $
 and
$\ell ' \leq \ell $
 and 
 $m' + m" \leq m$
, with
$m' + m" \leq m$
, with 
 $m' = 0$
 if
$m' = 0$
 if 
 $r = 3$
. Let
$r = 3$
. Let 
 $d'$
 and
$d'$
 and 
 $g'$
 be integers satisfying
$g'$
 be integers satisfying 
 $0 \leq g' \leq g$
 and
$0 \leq g' \leq g$
 and 
 $g' + r \leq d' \leq d - g + g'$
, with
$g' + r \leq d' \leq d - g + g'$
, with 
 $d'> g' + r$
 if both
$d'> g' + r$
 if both 
 $g' = 0$
 and
$g' = 0$
 and 
 $m \neq 0$
. Let
$m \neq 0$
. Let 
 $\epsilon _{\text {in}}$
 and
$\epsilon _{\text {in}}$
 and 
 $\epsilon _{\text {out}}$
 be nonnegative integers with
$\epsilon _{\text {out}}$
 be nonnegative integers with 
 $\epsilon _{\text {in}} + \epsilon _{\text {out}} = d - g - d' + g'$
. For
$\epsilon _{\text {in}} + \epsilon _{\text {out}} = d - g - d' + g'$
. For 
 $1 \leq i \leq m'$
, let
$1 \leq i \leq m'$
, let 
 $n_i$
 be an integer satisfying
$n_i$
 be an integer satisfying 
 $n_i \equiv r - 1$
 mod
$n_i \equiv r - 1$
 mod 
 $2$
 and
$2$
 and 
 $2 \leq n_i \leq r - 1$
, with
$2 \leq n_i \leq r - 1$
, with 
 $n_i \neq 2$
 if
$n_i \neq 2$
 if 
 $(d', g') = (r + 1, 1)$
. Define
$(d', g') = (r + 1, 1)$
. Define 
 $$\begin{align*}\overline{\ell} = \ell - \ell' + \frac{(r - 1)m' - \sum n_i}{2} \quad \text{and} \quad \overline{m}_{\text{max}} = m - m' \quad \text{and} \quad \overline{m}_{\text{min}} = m - m' - m".\end{align*}$$
$$\begin{align*}\overline{\ell} = \ell - \ell' + \frac{(r - 1)m' - \sum n_i}{2} \quad \text{and} \quad \overline{m}_{\text{max}} = m - m' \quad \text{and} \quad \overline{m}_{\text{min}} = m - m' - m".\end{align*}$$
Suppose that the following collection is erasable:
 $$ \begin{align} s_{10} &= \ell' + m - m' - m" \notag \\ s_{11} &= \epsilon_{\text{out}} \notag \\ s_{20} &= m' \\ s_{21} &= g - g' \notag \\ w_{10} &= m". \notag \end{align} $$
$$ \begin{align} s_{10} &= \ell' + m - m' - m" \notag \\ s_{11} &= \epsilon_{\text{out}} \notag \\ s_{20} &= m' \\ s_{21} &= g - g' \notag \\ w_{10} &= m". \notag \end{align} $$
If
 $$\begin{align*}\left|\delta - \left[2\epsilon_{\text{in}} + g - g' + m" + \ell' + \left \lfloor\frac{2 \epsilon_{\text{out}} + 3 (g - g') + m + m' + \ell'}{r - 1} \right \rfloor + \sum n_i \right] \right| \leq 1 - \frac{1}{r - 1},\end{align*}$$
$$\begin{align*}\left|\delta - \left[2\epsilon_{\text{in}} + g - g' + m" + \ell' + \left \lfloor\frac{2 \epsilon_{\text{out}} + 3 (g - g') + m + m' + \ell'}{r - 1} \right \rfloor + \sum n_i \right] \right| \leq 1 - \frac{1}{r - 1},\end{align*}$$
and 
 $I(d' - 1, g', r - 1, \overline {\ell }, \overline {m})$
 holds for all
$I(d' - 1, g', r - 1, \overline {\ell }, \overline {m})$
 holds for all 
 $\overline {m}$
 with
$\overline {m}$
 with 
 $\overline {m}_{\text {min}} \leq \overline {m} \leq \overline {m}_{\text {max}}$
, then so does
$\overline {m}_{\text {min}} \leq \overline {m} \leq \overline {m}_{\text {max}}$
, then so does 
 $I(d, g, r, \ell , m)$
.
$I(d, g, r, \ell , m)$
.
Proof. Our goal is to show interpolation for

Peeling off 
 $\epsilon _{\text {in}} + \epsilon _{\text {out}} = d - g - d' + g'$
 one-secant lines and
$\epsilon _{\text {in}} + \epsilon _{\text {out}} = d - g - d' + g'$
 one-secant lines and 
 $g - g'$
 two-secant lines reduces to interpolation for
$g - g'$
 two-secant lines reduces to interpolation for 

 For 
 $1 \leq i \leq m'$
, write
$1 \leq i \leq m'$
, write 
 $n_i' = (r-1-n_i)/2$
, and degenerate
$n_i' = (r-1-n_i)/2$
, and degenerate 
 $R_i$
 as in Section 7 to the union
$R_i$
 as in Section 7 to the union 
 $R_i^\circ $
, of
$R_i^\circ $
, of 
 $n_i$
 lines
$n_i$
 lines 
 $L_{i,j}$
 meeting C at
$L_{i,j}$
 meeting C at 
 $p_i$
 and
$p_i$
 and 
 $q_{i,j}$
, and
$q_{i,j}$
, and 
 $n_i'$
 conics
$n_i'$
 conics 
 $Q_{i,j}$
 meeting C at
$Q_{i,j}$
 meeting C at 
 $p_i$
 and
$p_i$
 and 
 $q_{i,n_i + 2j-1}$
 and
$q_{i,n_i + 2j-1}$
 and 
 $q_{i,n_i + 2j}$
. For
$q_{i,n_i + 2j}$
. For 
 $m' + 1 \leq i \leq m$
, write
$m' + 1 \leq i \leq m$
, write 
 $R_i \cap C = \{s^i_0, s^i_1, s^i_2, \ldots , s^i_{r - 1}, s^i_r\}$
. For
$R_i \cap C = \{s^i_0, s^i_1, s^i_2, \ldots , s^i_{r - 1}, s^i_r\}$
. For 
 $m' + 1 \leq i \leq m' + m"$
, specialize
$m' + 1 \leq i \leq m' + m"$
, specialize 
 $R_i$
 to a union
$R_i$
 to a union 
 $R_i^- \cup L_i$
, where
$R_i^- \cup L_i$
, where 
 $L_i$
 is the line through
$L_i$
 is the line through 
 $s^i_0$
 and
$s^i_0$
 and 
 $s^i_r$
, and
$s^i_r$
, and 
 $R_i^-$
 is a rational curve of degree
$R_i^-$
 is a rational curve of degree 
 $r - 2$
 passing through
$r - 2$
 passing through 
 $s^i_1, s^i_2, \ldots , s^i_{r - 1}$
 and meeting
$s^i_1, s^i_2, \ldots , s^i_{r - 1}$
 and meeting 
 $L_i$
 at a single point. This induces a specialization of the above bundle to
$L_i$
 at a single point. This induces a specialization of the above bundle to 

Fix a general point 
 $p \in C$
, and specialize
$p \in C$
, and specialize 
 $p_1, p_2, \ldots , p_{m'}, v_1, v_2, \ldots , v_{\ell '}, y_1, y_2, \ldots , y_{\epsilon _{\text {in}}}, x_{\epsilon _{\text {in}} + 1}, x_{\epsilon _{\text {in}} + 2}, \ldots , x_{\epsilon _{\text {in}} + \epsilon _{\text {out}}}, z_1, z_2, \ldots , z_{g - g'}, s^{m' + 1}_0, s^{m' + 2}_0, \ldots , s^m_0$
 all to p in some order. Our assumption that equation (11.1) is erasable implies that we may choose the order so that the limiting bundle is
$p_1, p_2, \ldots , p_{m'}, v_1, v_2, \ldots , v_{\ell '}, y_1, y_2, \ldots , y_{\epsilon _{\text {in}}}, x_{\epsilon _{\text {in}} + 1}, x_{\epsilon _{\text {in}} + 2}, \ldots , x_{\epsilon _{\text {in}} + \epsilon _{\text {out}}}, z_1, z_2, \ldots , z_{g - g'}, s^{m' + 1}_0, s^{m' + 2}_0, \ldots , s^m_0$
 all to p in some order. Our assumption that equation (11.1) is erasable implies that we may choose the order so that the limiting bundle is 

for some integer n and subspace 
 $\Lambda \subset \mathbb {P} N_{C(d - g - d' + g', g - g';0)}|_p$
, disjoint from
$\Lambda \subset \mathbb {P} N_{C(d - g - d' + g', g - g';0)}|_p$
, disjoint from 
 $\mathbb {P} N_{C(d - g - d' + g', g - g';0) \to p}|_p$
 and whose image
$\mathbb {P} N_{C(d - g - d' + g', g - g';0) \to p}|_p$
 and whose image 
 $\overline {\Lambda }$
 in
$\overline {\Lambda }$
 in 
 $\mathbb {P} (N_{C(d - g - d' + g', g - g';0)}/N_{C(d - g - d' + g', g - g';0) \to p})|_p$
 is linearly general. Computing the Euler characteristic, we obtain
$\mathbb {P} (N_{C(d - g - d' + g', g - g';0)}/N_{C(d - g - d' + g', g - g';0) \to p})|_p$
 is linearly general. Computing the Euler characteristic, we obtain 
 $$\begin{align*}(r - 1) n + \operatorname{rk} \Lambda = 2 \epsilon_{\text{out}} + 3 (g - g') + m + m' + \ell',\end{align*}$$
$$\begin{align*}(r - 1) n + \operatorname{rk} \Lambda = 2 \epsilon_{\text{out}} + 3 (g - g') + m + m' + \ell',\end{align*}$$
and so
 $$\begin{align*}n = \left\lfloor \frac{2 \epsilon_{\text{out}} + 3 (g - g') + m + m' + \ell'}{r - 1}\right\rfloor.\end{align*}$$
$$\begin{align*}n = \left\lfloor \frac{2 \epsilon_{\text{out}} + 3 (g - g') + m + m' + \ell'}{r - 1}\right\rfloor.\end{align*}$$
Projecting from p, we reduce to interpolation for

Erasing the transformation at p, we reduce to interpolation for

By Lemma 3.14, this follows in turn from interpolation for the bundles

with 
 $m' + 1 \leq i_1 < i_2 < \cdots < i_j \leq m' + m"$
. Finally, specializing
$m' + 1 \leq i_1 < i_2 < \cdots < i_j \leq m' + m"$
. Finally, specializing 
 $R_{m' + m" + 1}, R_{m' + m" + 2}, \ldots , R_m$
 to pass through p (as in the proof of Proposition 8.2), we reduce to interpolation for
$R_{m' + m" + 1}, R_{m' + m" + 2}, \ldots , R_m$
 to pass through p (as in the proof of Proposition 8.2), we reduce to interpolation for 

But these are precisely our assumptions 
 $I(d' - 1, g', r - 1, \overline {\ell }, \overline {m})$
 for
$I(d' - 1, g', r - 1, \overline {\ell }, \overline {m})$
 for 
 $\overline {m}_{\text {min}} \leq \overline {m} \leq \overline {m}_{\text {max}}$
.
$\overline {m}_{\text {min}} \leq \overline {m} \leq \overline {m}_{\text {max}}$
.
 We then write a computer program in python [14] (see Appendix 1) which iterates over all of the finitely many sporadic cases identified in the previous section, that is, those tuples 
 $(d, g, r, \ell , m)$
 satisfying
$(d, g, r, \ell , m)$
 satisfying 
 $r \leq 13$
 and equation (10.1) or (10.2), but excluding those tuples with
$r \leq 13$
 and equation (10.1) or (10.2), but excluding those tuples with 
 $(\delta , \ell , m) = (1, 0, 0)$
. In each case, all possible parameters for every inductive argument in Section 8, as well as all possible parameters for Proposition 11.2, are tried. In all but the following
$(\delta , \ell , m) = (1, 0, 0)$
. In each case, all possible parameters for every inductive argument in Section 8, as well as all possible parameters for Proposition 11.2, are tried. In all but the following 
 $30$
cases, one of these arguments applies:
$30$
cases, one of these arguments applies:

 Our remaining task is therefore to verify 
 $I(d, g, r, \ell , m)$
 in these
$I(d, g, r, \ell , m)$
 in these 
 $30$
 base cases (as well as to prove Theorem 1.4 for canonical curves of even genus).
$30$
 base cases (as well as to prove Theorem 1.4 for canonical curves of even genus).
12 The remaining sporadic cases
12.1 The cases 
 $(d, g, r, \ell , m) = (7, 1, 6, 2, 1), (7, 1, 6, 3, 1)$
, and
$(d, g, r, \ell , m) = (7, 1, 6, 2, 1), (7, 1, 6, 3, 1)$
, and 
 $(8, 2, 6, 2, 0)$
$(8, 2, 6, 2, 0)$
 In these three cases, our previous arguments apply provided that 
 $I(6, 1, 5, 3, 0)$
 holds. (Note that
$I(6, 1, 5, 3, 0)$
 holds. (Note that 
 $(6, 1, 5, 3, 0)$
 is, however, not good, which is why we were not able to deal with these cases in the previous section and need to separately consider them here.) Indeed:
$(6, 1, 5, 3, 0)$
 is, however, not good, which is why we were not able to deal with these cases in the previous section and need to separately consider them here.) Indeed:
- 
If  $ \boldsymbol{(d, g, r, \ell , m) = (7, 1, 6, 2, 1)} $
 We apply Proposition 8.2 with parameters $ \boldsymbol{(d, g, r, \ell , m) = (7, 1, 6, 2, 1)} $
 We apply Proposition 8.2 with parameters $$\begin{align*}\ell' = 0, \quad m' = 1, \quad d' = 7, \quad \text{and} \quad n_1 = 3.\end{align*}$$ $$\begin{align*}\ell' = 0, \quad m' = 1, \quad d' = 7, \quad \text{and} \quad n_1 = 3.\end{align*}$$
- 
If  $ \boldsymbol{(d, g, r, \ell , m) = (7, 1, 6, 3, 1)} $
 We apply Proposition 8.2 with parameters $ \boldsymbol{(d, g, r, \ell , m) = (7, 1, 6, 3, 1)} $
 We apply Proposition 8.2 with parameters $$\begin{align*}\ell' = 1, \quad m' = 1, \quad d' = 7, \quad \text{and} \quad n_1 = 3.\end{align*}$$ $$\begin{align*}\ell' = 1, \quad m' = 1, \quad d' = 7, \quad \text{and} \quad n_1 = 3.\end{align*}$$
- 
If  $ \boldsymbol{(d, g, r, \ell , m) = (8, 2, 6, 2, 0)} $
 We apply Proposition 8.10. $ \boldsymbol{(d, g, r, \ell , m) = (8, 2, 6, 2, 0)} $
 We apply Proposition 8.10.
 It thus remains to check 
 $I(6, 1, 5, 3, 0)$
. For this, we simply apply Proposition 8.2 with parameters
$I(6, 1, 5, 3, 0)$
. For this, we simply apply Proposition 8.2 with parameters 
 $$\begin{align*}\ell' = 3, \quad m' = 0, \quad \text{and} \quad d' = 6,\end{align*}$$
$$\begin{align*}\ell' = 3, \quad m' = 0, \quad \text{and} \quad d' = 6,\end{align*}$$
thereby reducing 
 $I(6, 1, 5, 3, 0)$
 to
$I(6, 1, 5, 3, 0)$
 to 
 $I(5, 1, 4, 0, 0)$
, which suffices because
$I(5, 1, 4, 0, 0)$
, which suffices because 
 $(5, 1, 4, 0, 0)$
 is good.
$(5, 1, 4, 0, 0)$
 is good.
12.2 The cases 
 $ \boldsymbol{(d, g, r, \ell , m) = (4, 0, 3, 1, 1), (5, 1, 3, 1, 1)} $
, and
$ \boldsymbol{(d, g, r, \ell , m) = (4, 0, 3, 1, 1), (5, 1, 3, 1, 1)} $
, and 
 $(5, 2, 3, 1, 1)$
$(5, 2, 3, 1, 1)$
In each of these cases, we want to show interpolation for

Write 
 $R \cap C = \{q_1, q_2, q_3, q_4\}$
. Specializing u to
$R \cap C = \{q_1, q_2, q_3, q_4\}$
. Specializing u to 
 $q_1$
 and v to
$q_1$
 and v to 
 $q_2$
 induces a specialization of this bundle to
$q_2$
 induces a specialization of this bundle to 

Removing the twists at 
 $q_1$
 and
$q_1$
 and 
 $q_2$
, we reduce to interpolation for
$q_2$
, we reduce to interpolation for 

Specializing R to the union of the lines 
 $\overline {q_1 q_2} \cup \overline {q_3 q_4}$
 induces a specialization of this bundle to
$\overline {q_1 q_2} \cup \overline {q_3 q_4}$
 induces a specialization of this bundle to 

Interpolation for this bundle is the assertion 
 $I(d, g, 3, 1, 0)$
, and
$I(d, g, 3, 1, 0)$
, and 
 $(d, g, 3, 1, 0)$
 is good in each of these cases.
$(d, g, 3, 1, 0)$
 is good in each of these cases.
12.3 The cases 
 $(d, g, r, \ell , m) = (5, 1, 3, 0, 1)$
 and
$(d, g, r, \ell , m) = (5, 1, 3, 0, 1)$
 and 
 $(6, 2, 3, 0, 1)$
$(6, 2, 3, 0, 1)$
In both of these cases, we we want to show interpolation for
 
. Peeling off a one-secant line, we reduce to interpolation for

Removing the twist at z, we reduce to interpolation for

Interpolation for this bundle is the assertion 
 $I(d - 1, g - 1, 3, 1, 1)$
, and
$I(d - 1, g - 1, 3, 1, 1)$
, and 
 $(d - 1, g - 1, 3, 1, 1)$
 is good in both of these cases.
$(d - 1, g - 1, 3, 1, 1)$
 is good in both of these cases.
12.4 The cases 
 $(d, g, r, \ell , m) = (4, 0, 3, 0, 2)$
 and
$(d, g, r, \ell , m) = (4, 0, 3, 0, 2)$
 and 
 $(5, 2, 3, 0, 2)$
$(5, 2, 3, 0, 2)$
 Let 
 $x_1, y_1, x_2, y_2 \in C$
 be four general points. Projection from
$x_1, y_1, x_2, y_2 \in C$
 be four general points. Projection from 
 $\overline {x_i y_i}$
 defines a general map
$\overline {x_i y_i}$
 defines a general map 
 $\pi _i \colon C \to \mathbb {P}^1$
 of degree
$\pi _i \colon C \to \mathbb {P}^1$
 of degree 
 $d - 2$
, which is in particular separable. Since
$d - 2$
, which is in particular separable. Since 
 $x_2$
 and
$x_2$
 and 
 $y_2$
 are general,
$y_2$
 are general, 
 $\overline {x_2 y_2}$
 does not meet the tangent line to C at either
$\overline {x_2 y_2}$
 does not meet the tangent line to C at either 
 $x_1$
,
$x_1$
, 
 $y_1$
, or any of the ramification points of
$y_1$
, or any of the ramification points of 
 $\pi _1$
. Thus,
$\pi _1$
. Thus, 
 $(\pi _1, \pi _2) \colon C \to \mathbb {P}^1 \times \mathbb {P}^1$
 is birational onto its image, an isomorphism near
$(\pi _1, \pi _2) \colon C \to \mathbb {P}^1 \times \mathbb {P}^1$
 is birational onto its image, an isomorphism near 
 $x_1$
 and
$x_1$
 and 
 $y_1$
 (and by symmetry near
$y_1$
 (and by symmetry near 
 $x_2$
 and
$x_2$
 and 
 $y_2$
), and its image is nodal.
$y_2$
), and its image is nodal.
 The number of nodes is the difference between the arithmetic and geometric genus, which is 
 $(d - 3)^2 - g \neq 0$
. Therefore, there is a pair of points
$(d - 3)^2 - g \neq 0$
. Therefore, there is a pair of points 
 $z, w \in C$
, distinct from each other and
$z, w \in C$
, distinct from each other and 
 $x_1, x_2, y_1, y_2$
, with
$x_1, x_2, y_1, y_2$
, with 
 $\pi _i(z) = \pi _i(w)$
 for both i. Geometrically,
$\pi _i(z) = \pi _i(w)$
 for both i. Geometrically, 
 $x_i, y_i, z, w$
 are four distinct coplanar points. Since
$x_i, y_i, z, w$
 are four distinct coplanar points. Since 
 $x_i$
 and
$x_i$
 and 
 $y_i$
 are general,
$y_i$
 are general, 
 $\overline {x_i y_i}$
 is not a trisecant to C, so in particular,
$\overline {x_i y_i}$
 is not a trisecant to C, so in particular, 
 $(x_i, y_i, z)$
 and
$(x_i, y_i, z)$
 and 
 $(x_i, y_i, w)$
 are not collinear. Because
$(x_i, y_i, w)$
 are not collinear. Because 
 $x_i$
 and
$x_i$
 and 
 $y_i$
 can be exchanged via monodromy, this implies no three of
$y_i$
 can be exchanged via monodromy, this implies no three of 
 $x_i, y_i, z, w$
 are collinear.
$x_i, y_i, z, w$
 are collinear.
Our goal is to show interpolation for
 
. Specializing 
 $R_i$
 to meet C at
$R_i$
 to meet C at 
 $x_i, y_i, z, w$
, this bundle specializes to
$x_i, y_i, z, w$
, this bundle specializes to 

Removing the twists at z and w, we reduce to interpolation for

Specializing 
 $R_i$
 to the union of lines
$R_i$
 to the union of lines 
 $\overline {x_i y_i} \cup \overline {zw}$
, this bundle specializes to
$\overline {x_i y_i} \cup \overline {zw}$
, this bundle specializes to 

Interpolation for this bundle is the assertion 
 $I(d, g, 3, 2, 0)$
. Although the
$I(d, g, 3, 2, 0)$
. Although the 
 $(d, g, 3, 2, 0)$
 are not good, our previous arguments still apply in these cases:
$(d, g, 3, 2, 0)$
 are not good, our previous arguments still apply in these cases:
- 
If  $ \boldsymbol{(d, g) = (4, 0)} $
 We apply Proposition 8.2 with parameters: $ \boldsymbol{(d, g) = (4, 0)} $
 We apply Proposition 8.2 with parameters: $$\begin{align*}\ell' = 1, \quad m' = 0, \quad \text{and} \quad d' = 3.\end{align*}$$ $$\begin{align*}\ell' = 1, \quad m' = 0, \quad \text{and} \quad d' = 3.\end{align*}$$
- 
If  $ \boldsymbol{(d, g) = (5, 2)} $
 We apply Proposition 11.2 with parameters: (The required erasability of $ \boldsymbol{(d, g) = (5, 2)} $
 We apply Proposition 11.2 with parameters: (The required erasability of $$\begin{align*}\ell' = 1, \quad m' = m" = \epsilon_{\text{in}} = \epsilon_{\text{out}} = 0, \quad d' = 3, \quad \text{and} \quad g' = 0.\end{align*}$$ $$\begin{align*}\ell' = 1, \quad m' = m" = \epsilon_{\text{in}} = \epsilon_{\text{out}} = 0, \quad d' = 3, \quad \text{and} \quad g' = 0.\end{align*}$$ $(s_{10}, s_{11}, s_{20}, s_{21}, w_{10}) = (1, 0, 0, 2, 0)$
 can be checked by specializing the points in any order.) $(s_{10}, s_{11}, s_{20}, s_{21}, w_{10}) = (1, 0, 0, 2, 0)$
 can be checked by specializing the points in any order.)
12.5 The cases 
 $(d, g, r, \ell , m) = (4, 0, 3, 0, 1)$
 and
$(d, g, r, \ell , m) = (4, 0, 3, 0, 1)$
 and 
 $(5, 0, 3, 0, 1)$
$(5, 0, 3, 0, 1)$
In both of these cases, we want to show interpolation for
 
. Write 
 $C \cap R = \{q_1, q_2, q_3, q_4\}$
. Peel off a one-secant line, that is, degenerate C to
$C \cap R = \{q_1, q_2, q_3, q_4\}$
. Peel off a one-secant line, that is, degenerate C to 
 $C(1, 0; 0) \cup L$
 – but in such a way that
$C(1, 0; 0) \cup L$
 – but in such a way that 
 $q_4$
 specializes onto L, while
$q_4$
 specializes onto L, while 
 $q_1$
,
$q_1$
, 
 $q_2$
, and
$q_2$
, and 
 $q_3$
 specialize onto
$q_3$
 specialize onto 
 $C(1, 0; 0)$
. The restriction of the modified normal bundle to L is perfectly balanced of slope
$C(1, 0; 0)$
. The restriction of the modified normal bundle to L is perfectly balanced of slope 
 $2$
, so by Lemma 3.7, this reduces interpolation for
$2$
, so by Lemma 3.7, this reduces interpolation for 
 
to interpolation for

Erasing the transformation
 
, we reduce to interpolation for

Specializing R to the union of lines 
 $\overline {q_1 q_2} \cup \overline {q_3 q_4}$
 induces a specialization of this bundle to
$\overline {q_1 q_2} \cup \overline {q_3 q_4}$
 induces a specialization of this bundle to 

Erasing the transformation at 
 $q_3$
, we reduce to interpolation for
$q_3$
, we reduce to interpolation for 

which is the assertion 
 $I(d - 1, 0, 3, 1, 0)$
. Both
$I(d - 1, 0, 3, 1, 0)$
. Both 
 $(3, 0, 3, 1, 0)$
 and
$(3, 0, 3, 1, 0)$
 and 
 $(4, 0, 3, 1, 0)$
 are good.
$(4, 0, 3, 1, 0)$
 are good.
12.6 The case 
 $(d, g, r, \ell , m) = (5, 2, 3, 0, 1)$
$(d, g, r, \ell , m) = (5, 2, 3, 0, 1)$
In this case, we want to show interpolation for
 
. Write 
 $C \cap R = \{q_1, q_2, q_3, q_4\}$
. Peel off a one-secant line, that is, degenerate C to
$C \cap R = \{q_1, q_2, q_3, q_4\}$
. Peel off a one-secant line, that is, degenerate C to 
 $C(0, 1; 0) \cup L$
 – but in such a way that
$C(0, 1; 0) \cup L$
 – but in such a way that 
 $q_3$
 and
$q_3$
 and 
 $q_4$
 specialize onto L, while
$q_4$
 specialize onto L, while 
 $q_1$
 and
$q_1$
 and 
 $q_2$
 specialize onto
$q_2$
 specialize onto 
 $C(0, 1; 0)$
. The restriction of the modified normal bundle to L is perfectly balanced of slope
$C(0, 1; 0)$
. The restriction of the modified normal bundle to L is perfectly balanced of slope 
 $3$
, so by Lemma 3.7, this reduces interpolation for
$3$
, so by Lemma 3.7, this reduces interpolation for 
 
to interpolation for

Let Q be the unique quadric containing 
 $C(0, 1; 0)$
 and the line
$C(0, 1; 0)$
 and the line 
 $\overline {zw}$
. Then interpolation for this bundle follows from the balanced exact sequence
$\overline {zw}$
. Then interpolation for this bundle follows from the balanced exact sequence 

12.7 The case 
 $(d, g, r, \ell , m) = (5, 0, 4, 2, 0)$
$(d, g, r, \ell , m) = (5, 0, 4, 2, 0)$
In this case, we want to show interpolation for

Peel off a one-secant line, that is, degenerate C to 
 $C(1, 0; 0) \cup L$
 – but in such a way that
$C(1, 0; 0) \cup L$
 – but in such a way that 
 $v_1$
 and
$v_1$
 and 
 $v_2$
 specialize onto L, while
$v_2$
 specialize onto L, while 
 $u_1$
 and
$u_1$
 and 
 $u_2$
 specialize onto
$u_2$
 specialize onto 
 $C(1, 0; 0)$
. By Lemma 3.7, this reduces to interpolation for
$C(1, 0; 0)$
. By Lemma 3.7, this reduces to interpolation for 

Specializing 
 $v_1$
 to z, we reduce to interpolation for
$v_1$
 to z, we reduce to interpolation for 

Projecting from z, we reduce to interpolation for

Specializing 
 $v_2$
 onto the line
$v_2$
 onto the line 
 $\overline {zu_2}$
, we reduce to interpolation for
$\overline {zu_2}$
, we reduce to interpolation for 

This is the assertion 
 $I(3, 0, 3, 1, 0)$
, and
$I(3, 0, 3, 1, 0)$
, and 
 $(3, 0, 3, 1, 0)$
 is good.
$(3, 0, 3, 1, 0)$
 is good.
12.8 The cases 
 $(d, g, r, \ell , m) = (6, 2, 4, 0, 2)$
 and
$(d, g, r, \ell , m) = (6, 2, 4, 0, 2)$
 and 
 $(7, 3, 4, 0, 1)$
$(7, 3, 4, 0, 1)$
In these cases, we want to show interpolation for
 
. Write 
 $R_i \cap C = \{q_{i1}, q_{i2}, q_{i3}, q_{i4}, q_{i5}\}$
. Note that
$R_i \cap C = \{q_{i1}, q_{i2}, q_{i3}, q_{i4}, q_{i5}\}$
. Note that 
 $m \leq g$
 in both cases, so we may peel off m two-secant lines, that is, degenerate C to
$m \leq g$
 in both cases, so we may peel off m two-secant lines, that is, degenerate C to 
 $C(0, m; 0) \cup L_1 \cup \cdots \cup L_m$
 – but in such a way that
$C(0, m; 0) \cup L_1 \cup \cdots \cup L_m$
 – but in such a way that 
 $q_{i5}$
 specializes onto
$q_{i5}$
 specializes onto 
 $L_i$
, while the remaining
$L_i$
, while the remaining 
 $q_{ij}$
 specialize onto
$q_{ij}$
 specialize onto 
 $C(0, m; 0)$
. By Lemma 3.7, this reduces interpolation for
$C(0, m; 0)$
. By Lemma 3.7, this reduces interpolation for 
 
to interpolation for

Specialize 
 $R_i$
 to the union of lines
$R_i$
 to the union of lines 
 $\overline {q_{i1} q_{i2}}$
,
$\overline {q_{i1} q_{i2}}$
, 
 $\overline {q_{i3}q_{i4}}$
 and the unique line through
$\overline {q_{i3}q_{i4}}$
 and the unique line through 
 $q_{i5}$
 meeting both of these two lines. This induces a specialization of the above bundle to
$q_{i5}$
 meeting both of these two lines. This induces a specialization of the above bundle to 

In other words, all that remains is to check the assertion 
 $I(d - m, g - m, 4, 3m, 0)$
.
$I(d - m, g - m, 4, 3m, 0)$
.
- 
If  $\boldsymbol{(d, g, m) = (6, 2, 2)}$
 In this case, writing C for a curve of degree $\boldsymbol{(d, g, m) = (6, 2, 2)}$
 In this case, writing C for a curve of degree $d - m = 4$
 and genus $d - m = 4$
 and genus $g - m = 0$
, we want to establish interpolation for Specializing ‘to a tetrahedron’, that is, specializing $g - m = 0$
, we want to establish interpolation for Specializing ‘to a tetrahedron’, that is, specializing  $u_1, u_2$
 to $u_1, u_2$
 to $u_3$
, and $u_3$
, and $u_4, v_6$
 to $u_4, v_6$
 to $v_1$
, and $v_1$
, and $v_4, u_5$
 to $v_4, u_5$
 to $v_2$
, and $v_2$
, and $v_5, u_6$
 to $v_5, u_6$
 to $v_3$
, this bundle specializes to Removing the twists at $v_3$
, this bundle specializes to Removing the twists at $$\begin{align*}N_C(u_3 + v_1 + v_2 + v_3).\end{align*}$$ $$\begin{align*}N_C(u_3 + v_1 + v_2 + v_3).\end{align*}$$ $u_3, v_1, v_2$
, and $u_3, v_1, v_2$
, and $v_3$
, we reduce to interpolation for $v_3$
, we reduce to interpolation for $N_C$
, which is the assertion $N_C$
, which is the assertion $I(4, 0, 4, 0, 0)$
. Note that $I(4, 0, 4, 0, 0)$
. Note that $(4, 0, 4, 0, 0)$
 is good. $(4, 0, 4, 0, 0)$
 is good.
- 
If  $\boldsymbol{(d, g, m) = (7, 3, 1)}$
 In this case, writing C for a curve of degree $\boldsymbol{(d, g, m) = (7, 3, 1)}$
 In this case, writing C for a curve of degree $d - m = 6$
 and genus $d - m = 6$
 and genus $g - m = 2$
, we want to establish interpolation for Note that $g - m = 2$
, we want to establish interpolation for Note that  $\delta = 4\frac {2}{3}$
. Peeling off two one-secant lines, we reduce to interpolation for Limiting $\delta = 4\frac {2}{3}$
. Peeling off two one-secant lines, we reduce to interpolation for Limiting  $w_1$
 to $w_1$
 to $w_2$
, this bundle specializes to Projecting from $w_2$
, this bundle specializes to Projecting from  $w_2$
, we reduce to interpolation for Limiting $w_2$
, we reduce to interpolation for Limiting  $v_3$
 to $v_3$
 to $w_2$
, and $w_2$
, and $v_2$
 to $v_2$
 to $v_1$
, we reduce to interpolation for Projecting from $v_1$
, we reduce to interpolation for Projecting from  $w_2$
 again, we reduce to interpolation for which is a nonspecial line bundle and therefore satisfies interpolation. $w_2$
 again, we reduce to interpolation for which is a nonspecial line bundle and therefore satisfies interpolation. $$\begin{align*}N_{C(0, 2; 2)}(u_1 + u_2 + w_2),\end{align*}$$ $$\begin{align*}N_{C(0, 2; 2)}(u_1 + u_2 + w_2),\end{align*}$$
12.9 The cases 
 $(d, g, r, \ell , m) = (5, 0, 4, 0, 1)$
, and
$(d, g, r, \ell , m) = (5, 0, 4, 0, 1)$
, and 
 $(7, 3, 4, 1, 1)$
$(7, 3, 4, 1, 1)$
 We want to show interpolation for a vector bundle of rank 
 $3$
, and degree
$3$
, and degree 
 $28$
 and
$28$
 and 
 $46$
, respectively. By Lemma 3.9, it suffices to check interpolation for corresponding vector bundles of rank
$46$
, respectively. By Lemma 3.9, it suffices to check interpolation for corresponding vector bundles of rank 
 $27$
 and
$27$
 and 
 $45$
 where one positive transformation is omitted.
$45$
 where one positive transformation is omitted.
 For 
 $(d, g, r, \ell , m) = (5, 0, 4, 0, 1)$
, the bundle of degree
$(d, g, r, \ell , m) = (5, 0, 4, 0, 1)$
, the bundle of degree 
 $28$
 for which we want to show interpolation is
$28$
 for which we want to show interpolation is 
 . Write
. Write 
 $C \cap R = \{q_1, q_2, q_3, q_4, q_5\}$
. It suffices to establish interpolation for the degree
$C \cap R = \{q_1, q_2, q_3, q_4, q_5\}$
. It suffices to establish interpolation for the degree 
 $27$
 vector bundle
$27$
 vector bundle 
 . Specialize R to the union of the lines
. Specialize R to the union of the lines 
 $\overline {q_1q_2}$
,
$\overline {q_1q_2}$
, 
 $\overline {q_3q_4}$
 and the unique line through
$\overline {q_3q_4}$
 and the unique line through 
 $q_5$
 meeting both of these two lines. This induces a specialization of this bundle to
$q_5$
 meeting both of these two lines. This induces a specialization of this bundle to 
 , which is the assertion
, which is the assertion 
 $I(5, 0, 4, 2, 0)$
. Observe that
$I(5, 0, 4, 2, 0)$
. Observe that 
 $(5, 0, 4, 2, 0)$
 was already considered above in Section 12.7.
$(5, 0, 4, 2, 0)$
 was already considered above in Section 12.7.
 For 
 $(d, g, r, \ell , m) = (7, 3, 4, 1, 1)$
, our bundle of degree
$(d, g, r, \ell , m) = (7, 3, 4, 1, 1)$
, our bundle of degree 
 $46$
 is
$46$
 is 
 , and we can reduce to interpolation for the degree
, and we can reduce to interpolation for the degree 
 $45$
 vector bundle
$45$
 vector bundle 
 . Erasing the transformation at u, this reduces to interpolation for
. Erasing the transformation at u, this reduces to interpolation for 
 , which is the assertion
, which is the assertion 
 $I(7, 3, 4, 0, 1)$
. Observe that
$I(7, 3, 4, 0, 1)$
. Observe that 
 $(7, 3, 4, 0, 1)$
 was already considered above in Section 12.8.
$(7, 3, 4, 0, 1)$
 was already considered above in Section 12.8.
12.10 The case 
 $(d, g, r, \ell , m) = (7, 1, 5, 0, 1)$
$(d, g, r, \ell , m) = (7, 1, 5, 0, 1)$
In this case we want to show interpolation for
 
. Write 
 $R\cap C = \{q_1, q_2, q_3, q_4, q_5, q_6\}$
. Peel off a one-secant line, that is, degenerate C to
$R\cap C = \{q_1, q_2, q_3, q_4, q_5, q_6\}$
. Peel off a one-secant line, that is, degenerate C to 
 $C(0, 1; 0) \cup L$
 — but in such a way that
$C(0, 1; 0) \cup L$
 — but in such a way that 
 $q_2$
 and
$q_2$
 and 
 $q_4$
 specialize onto L and the remaining points specialize onto
$q_4$
 specialize onto L and the remaining points specialize onto 
 $C(0,1;0)$
. By Lemma 3.7, this reduces interpolation for
$C(0,1;0)$
. By Lemma 3.7, this reduces interpolation for 
 
to interpolation for

Specializing R to the union of the three lines 
 $\overline {q_1 q_2}$
,
$\overline {q_1 q_2}$
, 
 $\overline {q_3 q_4}$
,
$\overline {q_3 q_4}$
, 
 $\overline {q_5 q_6}$
, and the unique fourth line in
$\overline {q_5 q_6}$
, and the unique fourth line in 
 $\mathbb {P}^5$
 meeting these three lines, we reduce to interpolation for
$\mathbb {P}^5$
 meeting these three lines, we reduce to interpolation for 

Limiting 
 $q_2$
 to u and
$q_2$
 to u and 
 $q_4$
 to v, we reduce to proving interpolation for
$q_4$
 to v, we reduce to proving interpolation for 

Projecting from u and then v, we reduce to interpolation for

This is the assertion 
 $I(4, 0, 3, 1, 0)$
, and
$I(4, 0, 3, 1, 0)$
, and 
 $(4, 0, 3, 1, 0)$
 is good.
$(4, 0, 3, 1, 0)$
 is good.
12.11 The case 
 $(d, g, r, \ell , m) = (7, 2, 5, 0, 1)$
$(d, g, r, \ell , m) = (7, 2, 5, 0, 1)$
We want to show that
 
satisfies interpolation. Peeling off a one-secant line, we reduce to interpolation for

We now specialize R as in Section 7 to the union of two lines 
 $\overline {q_1p}$
 and
$\overline {q_1p}$
 and 
 $\overline {q_2 p}$
, and a three-secant conic through
$\overline {q_2 p}$
, and a three-secant conic through 
 $q_3, q_4, p$
. Then limit w to p. This induces a specialization of our bundle to
$q_3, q_4, p$
. Then limit w to p. This induces a specialization of our bundle to 

Projecting from p we reduce to interpolation for

Limiting 
 $q_4$
 to p, we reduce to interpolation for
$q_4$
 to p, we reduce to interpolation for 

Projecting from p, we reduce to interpolation for
 
. Erasing the transformation
 
, this reduces to 
 $I(4, 1, 3, 0, 0)$
, and
$I(4, 1, 3, 0, 0)$
, and 
 $(4, 1, 3, 0, 0)$
 is good.
$(4, 1, 3, 0, 0)$
 is good.
12.12 The case 
 $(d, g, r, \ell , m) = (7, 2, 5, 2, 2)$
$(d, g, r, \ell , m) = (7, 2, 5, 2, 2)$
This case asserts interpolation for

We first specialize each 
 $R_i$
 as in Section 7 to the union of two lines
$R_i$
 as in Section 7 to the union of two lines 
 $\overline {q_{i1}p_i}, \overline {q_{i2}p_i}$
 and a three-secant conic through
$\overline {q_{i1}p_i}, \overline {q_{i2}p_i}$
 and a three-secant conic through 
 $\{p_i, q_{i3}, q_{i4}\}$
. We then specialize
$\{p_i, q_{i3}, q_{i4}\}$
. We then specialize 
 $p_1$
 and
$p_1$
 and 
 $p_2$
 together to a common point p. This induces a specialization of our bundle to
$p_2$
 together to a common point p. This induces a specialization of our bundle to 

Limiting 
 $u_1$
 to p and removing the overall twist at p reduces us to interpolation for
$u_1$
 to p and removing the overall twist at p reduces us to interpolation for 

Projecting from p, we reduce to interpolation for

Erasing the transformation
 
and peeling off two one-secant lines, we reduce to

Limiting 
 $w_1$
 and
$w_1$
 and 
 $w_2$
 to p, we reduce to interpolation for
$w_2$
 to p, we reduce to interpolation for 

Projecting from p, we reduce to interpolation for

Limiting 
 $q_{24}$
 to
$q_{24}$
 to 
 $q_{14}$
, and
$q_{14}$
, and 
 $v_2$
 to p, and removing the resulting twist at
$v_2$
 to p, and removing the resulting twist at 
 $q_{14}$
, we reduce to
$q_{14}$
, we reduce to 

Projecting from p, we reduce to interpolation for 
 $N_{C(0,2;3)}$
, which is a nonspecial line bundle.
$N_{C(0,2;3)}$
, which is a nonspecial line bundle.
12.13 The case 
 $(d, g, r, \ell , m) = (9, 2, 5, 0, 0)$
$(d, g, r, \ell , m) = (9, 2, 5, 0, 0)$
Peeling off two one-secant lines reduces to interpolation for

Limit the points 
 $z_1$
 and
$z_1$
 and 
 $w_2$
 to a common point p. This induces the specialization of our bundle to
$w_2$
 to a common point p. This induces the specialization of our bundle to 

Projection from p reduces to interpolation for

Erasing the transformation
 
 and then projecting from p, we reduce to interpolation for 
 $N_{C(0,2;2)}$
. This is
$N_{C(0,2;2)}$
. This is 
 $I(5, 0, 3, 0, 0)$
, and
$I(5, 0, 3, 0, 0)$
, and 
 $(5, 0, 3, 0, 0)$
 is good.
$(5, 0, 3, 0, 0)$
 is good.
12.14 The cases 
 $(d, g, r, \ell , m) = (9, 4, 5, 0, 0)$
 and
$(d, g, r, \ell , m) = (9, 4, 5, 0, 0)$
 and 
 $(9, 4, 5, 1, 0)$
$(9, 4, 5, 1, 0)$
 We want that both 
 $N_C$
 and
$N_C$
 and 
 
satisfy interpolation. Peeling off four one-secant lines, we reduce to interpolation for

and

Specializing 
 $w_2$
 to
$w_2$
 to 
 $w_1$
, and
$w_1$
, and 
 $w_4$
 to
$w_4$
 to 
 $w_3$
, we reduce to interpolation for
$w_3$
, we reduce to interpolation for 

Projecting from 
 $w_1$
 and then
$w_1$
 and then 
 $w_3$
, we reduce to interpolation for
$w_3$
, we reduce to interpolation for 

Specializing v to 
 $w_1$
, we reduce to interpolation for
$w_1$
, we reduce to interpolation for 

Projecting from 
 $w_1$
, we reduce to interpolation for
$w_1$
, we reduce to interpolation for 
 $N_{C(0,4;3)}$
, which is a nonspecial line bundle.
$N_{C(0,4;3)}$
, which is a nonspecial line bundle.
12.15 The cases 
 $(d, g, r, \ell , m) = (8, 3, 5, 2, 0)$
 and
$(d, g, r, \ell , m) = (8, 3, 5, 2, 0)$
 and 
 $(11, 5, 6, 0, 0)$
$(11, 5, 6, 0, 0)$
We first reduce interpolation for both of these bundles to the same statement.
- 
For  $\boldsymbol{(d, g, r, \ell , m) = (11, 5, 6, 0, 0)}$
 Note that $\boldsymbol{(d, g, r, \ell , m) = (11, 5, 6, 0, 0)}$
 Note that $\delta = 4$
. Our goal is to establish interpolation for $\delta = 4$
. Our goal is to establish interpolation for $N_C$
. We first peel off one two-secant lines, which reduces our problem to interpolation for Limiting $N_C$
. We first peel off one two-secant lines, which reduces our problem to interpolation for Limiting  $w_2$
 to $w_2$
 to $w_1$
 induces a specialization of this bundle to Projecting from $w_1$
 induces a specialization of this bundle to Projecting from  $w_1$
, we reduce to interpolation for $w_1$
, we reduce to interpolation for 
- 
For  $\boldsymbol{(d, g, r, \ell , m) = (8, 3, 5, 2, 0)}$
, our goal is to establish interpolation for Limiting $\boldsymbol{(d, g, r, \ell , m) = (8, 3, 5, 2, 0)}$
, our goal is to establish interpolation for Limiting  $v_2$
 to $v_2$
 to $v_1$
, we reduce to interpolation for $v_1$
, we reduce to interpolation for 
 To finish the argument, let C be a general BN-curve of degree 
 $8$
 and genus
$8$
 and genus 
 $3$
 in
$3$
 in 
 $\mathbb {P}^5$
, and
$\mathbb {P}^5$
, and 
 $p, q_1, q_2 \in C$
 be general points. Above, we have shown that both of the desired assertions reduce to interpolation for the modified normal bundle
$p, q_1, q_2 \in C$
 be general points. Above, we have shown that both of the desired assertions reduce to interpolation for the modified normal bundle 

We next peel off two one-secant lines, that is, degenerate C to 
 $C \cup L_1 \cup L_2$
, where
$C \cup L_1 \cup L_2$
, where 
 $L_1$
 and
$L_1$
 and 
 $L_2$
 are one-secant lines to C – but in such a way that
$L_2$
 are one-secant lines to C – but in such a way that 
 $q_i$
 limits onto
$q_i$
 limits onto 
 $L_i$
, and p limits onto C. Applying Lemma 5.5, we reduce to interpolation for
$L_i$
, and p limits onto C. Applying Lemma 5.5, we reduce to interpolation for 

Over the function field of the moduli space of unordered pairs of triples 
 $\{(z_1, w_1, q_1), (z_2, w_2, q_2)\}$
, the transformation
$\{(z_1, w_1, q_1), (z_2, w_2, q_2)\}$
, the transformation 
 
 is linearly general as just 
 $q_1$
 and
$q_1$
 and 
 $q_2$
 vary. Indeed, geometrically, it is transverse to any subspace of the normal space at p except for the two subspaces
$q_2$
 vary. Indeed, geometrically, it is transverse to any subspace of the normal space at p except for the two subspaces 
 $N_{C \to L_1}|_p$
 and
$N_{C \to L_1}|_p$
 and 
 $N_{C \to L_2}|_p$
 – but neither of these subspaces is rational over this function field. Therefore, we may erase the transformation at p, thereby reducing to interpolation for
$N_{C \to L_2}|_p$
 – but neither of these subspaces is rational over this function field. Therefore, we may erase the transformation at p, thereby reducing to interpolation for 

Note that 
 $\delta = 3\frac {1}{2}$
 for this bundle. Peeling off a one-secant line, we reduce to interpolation for
$\delta = 3\frac {1}{2}$
 for this bundle. Peeling off a one-secant line, we reduce to interpolation for 

Specializing 
 $w_3$
 to p, and
$w_3$
 to p, and 
 $w_2$
 to
$w_2$
 to 
 $w_1$
, we reduce to interpolation for
$w_1$
, we reduce to interpolation for 

Projecting from 
 $w_1$
 and then p, we reduce to interpolation for
$w_1$
 and then p, we reduce to interpolation for 

Finally, projecting from 
 $w_1$
 again, we reduce to interpolation for
$w_1$
 again, we reduce to interpolation for 
 $N_{C(0, 3; 3)}$
, which is a nonspecial line bundle.
$N_{C(0, 3; 3)}$
, which is a nonspecial line bundle.
12.16 The case 
 $(d, g, r, \ell , m) = (7, 0, 6, 0, 1)$
$(d, g, r, \ell , m) = (7, 0, 6, 0, 1)$
 Arguing as in the proof of Proposition 8.3, it suffices to show that 
 $Q^-$
 and
$Q^-$
 and 
 $Q^+$
 satisfy interpolation, where
$Q^+$
 satisfy interpolation, where 

As in the proof of Proposition 8.3, interpolation for 
 $Q^+$
 follows from interpolation for
$Q^+$
 follows from interpolation for 
 $Q^-$
 given the assertion
$Q^-$
 given the assertion 
 $I(5, 0, 4, 0, 1)$
. Since
$I(5, 0, 4, 0, 1)$
. Since 
 $(5, 0, 4, 0, 1)$
 is good, it suffices to prove interpolation for
$(5, 0, 4, 0, 1)$
 is good, it suffices to prove interpolation for 
 $Q^-$
. By Lemma 3.8, this follows in turn from interpolation for
$Q^-$
. By Lemma 3.8, this follows in turn from interpolation for 

where 
 $\Lambda \subset Q^-|_{s_0}$
 is codimension
$\Lambda \subset Q^-|_{s_0}$
 is codimension 
 $1$
. By Lemma 3.9, since
$1$
. By Lemma 3.9, since 
 
, interpolation for 
 $Q^-(s_0)$
 follows from interpolation for
$Q^-(s_0)$
 follows from interpolation for 

This is the assertion 
 $I(6, 0, 5, 0,1)$
, and
$I(6, 0, 5, 0,1)$
, and 
 $(6,0,5,0,1)$
 is good.
$(6,0,5,0,1)$
 is good.
12.17 The cases 
 $(d, g, r, \ell , m) = (8, 1, 7, 0, 1)$
 and
$(d, g, r, \ell , m) = (8, 1, 7, 0, 1)$
 and 
 $(8, 1, 7, 1, 1)$
$(8, 1, 7, 1, 1)$
In both of these cases, our goal is to show interpolation for

We specialize 
 $R_1$
 to the union of the lines
$R_1$
 to the union of the lines 
 $q_1 q_2$
,
$q_1 q_2$
, 
 $q_3 q_4$
,
$q_3 q_4$
, 
 $q_5 q_6$
,
$q_5 q_6$
, 
 $q_7 q_8$
, together with a plane conic meeting each of these four lines. This induces a specialization of this bundle to
$q_7 q_8$
, together with a plane conic meeting each of these four lines. This induces a specialization of this bundle to 

Note that the points 
 $q_1, q_2, q_3, q_4, q_5, q_6, q_7, q_8$
 are not general, as they are constrained to lie in a hyperplane.
$q_1, q_2, q_3, q_4, q_5, q_6, q_7, q_8$
 are not general, as they are constrained to lie in a hyperplane.
 Let 
 $p_1, p_2, p_3, p_4 \in C$
 be points with
$p_1, p_2, p_3, p_4 \in C$
 be points with 
 $\mathcal {O}_C(1) = 2p_1 + 2p_2 + 2p_3 + 2p_4$
 (such points exist by Riemann–Roch because C is an elliptic curve). By construction,
$\mathcal {O}_C(1) = 2p_1 + 2p_2 + 2p_3 + 2p_4$
 (such points exist by Riemann–Roch because C is an elliptic curve). By construction, 
 $H^0(\mathcal {O}_C(1)(-2p_1 - 2p_2 - 2p_3 - 2p_4)) = 1$
; as C is embedded by a complete linear series, we conclude that the tangent lines to C at
$H^0(\mathcal {O}_C(1)(-2p_1 - 2p_2 - 2p_3 - 2p_4)) = 1$
; as C is embedded by a complete linear series, we conclude that the tangent lines to C at 
 $p_1, p_2, p_3, p_4$
 span a hyperplane H. Specializing the hyperplane containing
$p_1, p_2, p_3, p_4$
 span a hyperplane H. Specializing the hyperplane containing 
 $q_1, q_2, q_3, q_4, q_5, q_6, q_7, q_8$
 to H, in such a way that
$q_1, q_2, q_3, q_4, q_5, q_6, q_7, q_8$
 to H, in such a way that 
 $q_1$
 and
$q_1$
 and 
 $q_8$
 specialize to
$q_8$
 specialize to 
 $p_1$
, and
$p_1$
, and 
 $q_2$
 and
$q_2$
 and 
 $q_3$
 specialize to
$q_3$
 specialize to 
 $p_2$
, and
$p_2$
, and 
 $q_4$
 and
$q_4$
 and 
 $q_5$
 specialize to
$q_5$
 specialize to 
 $p_3$
, and
$p_3$
, and 
 $q_6$
 and
$q_6$
 and 
 $q_7$
 specialize to
$q_7$
 specialize to 
 $p_4$
, we obtain a further specialization of the above bundle to
$p_4$
, we obtain a further specialization of the above bundle to 

Note that, since 
 $H^0(\mathcal {O}_C(1)(-p_1 - p_2 - p_3 - p_4)) = 4$
, the points
$H^0(\mathcal {O}_C(1)(-p_1 - p_2 - p_3 - p_4)) = 4$
, the points 
 $p_1, p_2, p_3, p_4$
 are linearly independent. We claim interpolation for (12.1) reduces to interpolation for
$p_1, p_2, p_3, p_4$
 are linearly independent. We claim interpolation for (12.1) reduces to interpolation for 
 $N_{C(0, 0; 4)}$
. To see this, we divide into cases as follows.
$N_{C(0, 0; 4)}$
. To see this, we divide into cases as follows.
- 
If  $\boldsymbol{\ell = 0}$
 Note that $\boldsymbol{\ell = 0}$
 Note that $\delta = 2$
. Our goal is to show interpolation for Projecting from each of $\delta = 2$
. Our goal is to show interpolation for Projecting from each of  $p_1, p_2, p_3, p_4$
 in turn, we reduce to interpolation for $p_1, p_2, p_3, p_4$
 in turn, we reduce to interpolation for $N_{C(0, 0; 4)}$
. $N_{C(0, 0; 4)}$
.
- 
If  $\boldsymbol{\ell = 1}$
 Note that $\boldsymbol{\ell = 1}$
 Note that $\delta = 2\frac {1}{3}$
. Our goal is to show interpolation for Specializing $\delta = 2\frac {1}{3}$
. Our goal is to show interpolation for Specializing  $u_1$
 to $u_1$
 to $p_1$
 and $p_1$
 and $v_1$
 to $v_1$
 to $p_3$
, we reduce to interpolation for Projecting from $p_3$
, we reduce to interpolation for Projecting from  $p_1$
, $p_1$
, $p_2$
, $p_2$
, $p_3$
, and $p_3$
, and $p_4$
, we reduce to interpolation for $p_4$
, we reduce to interpolation for $N_{C(0, 0; 4)}$
. $N_{C(0, 0; 4)}$
.
 It remains to check interpolation for 
 $N_{C(0, 0; 4)}$
, which is the assertion
$N_{C(0, 0; 4)}$
, which is the assertion 
 $I(4, 1, 3, 0, 0)$
, and
$I(4, 1, 3, 0, 0)$
, and 
 $(4, 1, 3, 0, 0)$
 is good.
$(4, 1, 3, 0, 0)$
 is good.
12.18 The case 
 $(d, g, r, \ell , m) = (11, 4, 7, 1, 0)$
$(d, g, r, \ell , m) = (11, 4, 7, 1, 0)$
Our goal is to show interpolation for
 
. Note that 
 $\delta = 3$
. Peeling off a one-secant line, we reduce to interpolation for
$\delta = 3$
. Peeling off a one-secant line, we reduce to interpolation for 

Specializing v to w, we reduce to interpolation for

Projecting from w, we reduce to interpolation for

Erasing the transformation
 
, we reduce to interpolation for the two bundles

The first of these statements is the assertion 
 $I(9, 3, 6, 1, 0)$
. For the second, erasing the transformation at z reduces it to interpolation for
$I(9, 3, 6, 1, 0)$
. For the second, erasing the transformation at z reduces it to interpolation for 
 $N_{C(0, 1; 1)}$
, which is the assertion
$N_{C(0, 1; 1)}$
, which is the assertion 
 $I(9, 3, 6, 0, 0)$
. Note that both
$I(9, 3, 6, 0, 0)$
. Note that both 
 $(9, 3, 6, 0, 0)$
 and
$(9, 3, 6, 0, 0)$
 and 
 $(9, 3, 6, 1, 0)$
 are good.
$(9, 3, 6, 1, 0)$
 are good.
13 Canonical curves of even genus
 In this section, we prove interpolation for the normal bundle of a general canonical curve of even genus 
 $g \geq 8$
, which is the last remaining case (cf. Proposition 5.10 and Section 9). These cases are difficult, in part, because interpolation does not hold for canonical curves of genus
$g \geq 8$
, which is the last remaining case (cf. Proposition 5.10 and Section 9). These cases are difficult, in part, because interpolation does not hold for canonical curves of genus 
 $4$
 and
$4$
 and 
 $6$
, that is, when
$6$
, that is, when 
 $r=3$
 or
$r=3$
 or 
 $r=5$
.
$r=5$
.
 We will do this via degeneration to 
 $E \cup R$
, as in Section 5.3. That is, E is an elliptic normal curve in
$E \cup R$
, as in Section 5.3. That is, E is an elliptic normal curve in 
 $\mathbb {P}^r$
, and R is a general
$\mathbb {P}^r$
, and R is a general 
 $(r+1)$
-secant rational curve of degree
$(r+1)$
-secant rational curve of degree 
 $r-1$
, where
$r-1$
, where 
 $r = g - 1$
 is odd.
$r = g - 1$
 is odd.
13.1 Reduction to a bundle on E
 Recall that, due to the exceptional case of elliptic normal curves in odd-dimensional projective spaces in Lemma 5.8, we cannot reduce interpolation for 
 $N_{E \cup R}$
 to interpolation for
$N_{E \cup R}$
 to interpolation for 
 $N_{E \cup R}|_E$
. Instead, we will reduce interpolation for
$N_{E \cup R}|_E$
. Instead, we will reduce interpolation for 
 $N_{E \cup R}$
 to interpolation for a certain modification of
$N_{E \cup R}$
 to interpolation for a certain modification of 
 $N_{E \cup R}|_E$
. Our first step will be to show that
$N_{E \cup R}|_E$
. Our first step will be to show that 
 $N_{E \cup R}|_R$
 is not perfectly balanced, and give a geometric description of its Harder–Narasimhan (HN) filtration.
$N_{E \cup R}|_R$
 is not perfectly balanced, and give a geometric description of its Harder–Narasimhan (HN) filtration.
Lemma 13.1. Let 
 $q_1, \dots , q_{2n+2}$
 be a general collection of points on
$q_1, \dots , q_{2n+2}$
 be a general collection of points on 
 $\mathbb {P}^1$
. Let
$\mathbb {P}^1$
. Let 
 $p_1, \dots , p_{2n+2}$
 be a general collection of points on a general elliptic curve E. Then there exist exactly two maps of degree
$p_1, \dots , p_{2n+2}$
 be a general collection of points on a general elliptic curve E. Then there exist exactly two maps of degree 
 $n+1$
 from E to
$n+1$
 from E to 
 $\mathbb {P}^1$
 that send
$\mathbb {P}^1$
 that send 
 $p_i$
 to
$p_i$
 to 
 $q_i$
.
$q_i$
.
Proof. If 
 $f \colon E \to \mathbb {P}^1$
 is a general map of degree
$f \colon E \to \mathbb {P}^1$
 is a general map of degree 
 $n+1$
, then
$n+1$
, then 
 $f^*T_{\mathbb {P}^1}(-p_1 - \cdots -p_{2n+2})$
 has vanishing cohomology. Therefore, deformations of f are in bijection with deformations of the
$f^*T_{\mathbb {P}^1}(-p_1 - \cdots -p_{2n+2})$
 has vanishing cohomology. Therefore, deformations of f are in bijection with deformations of the 
 $f(p_i)$
. The number of maps of degree
$f(p_i)$
. The number of maps of degree 
 $n+1$
 from E to
$n+1$
 from E to 
 $\mathbb {P}^1$
 that send
$\mathbb {P}^1$
 that send 
 $p_i$
 to
$p_i$
 to 
 $q_i$
 is therefore finite and nonzero.
$q_i$
 is therefore finite and nonzero.
 To calculate this number, we degenerate the target 
 $\mathbb {P}^1$
 to a binary curve, with
$\mathbb {P}^1$
 to a binary curve, with 
 $q_1, q_2, \ldots , q_{n+1}$
 on the left
$q_1, q_2, \ldots , q_{n+1}$
 on the left 
 $\mathbb {P}^1$
, and
$\mathbb {P}^1$
, and 
 $q_{n + 2}, q_{n + 3}, \ldots , q_{2n+2}$
 on the right
$q_{n + 2}, q_{n + 3}, \ldots , q_{2n+2}$
 on the right 
 $\mathbb {P}^1$
. This degeneration is illustrated in the following diagram:
$\mathbb {P}^1$
. This degeneration is illustrated in the following diagram:

 Such a map 
 $E \to \mathbb {P}^1$
 then degenerates to an admissible cover from a marked curve whose stable model is
$E \to \mathbb {P}^1$
 then degenerates to an admissible cover from a marked curve whose stable model is 
 $(E, p_1, p_2, \ldots , p_{2n+2})$
. One can construct two such admissible covers (both with no infinitesimal deformations sending
$(E, p_1, p_2, \ldots , p_{2n+2})$
. One can construct two such admissible covers (both with no infinitesimal deformations sending 
 $p_i$
 to
$p_i$
 to 
 $q_i$
): In one such cover, E maps to the left component, with
$q_i$
): In one such cover, E maps to the left component, with 
 $p_1, p_2, \ldots , p_{n+1}$
 mapping to
$p_1, p_2, \ldots , p_{n+1}$
 mapping to 
 $q_1, q_2, \ldots , q_{n+1}$
, and
$q_1, q_2, \ldots , q_{n+1}$
, and 
 $p_{n + 2}, p_{n + 3}, \ldots , p_{2n+2}$
 mapping to the node; each of the points
$p_{n + 2}, p_{n + 3}, \ldots , p_{2n+2}$
 mapping to the node; each of the points 
 $p_{n + 2}, p_{n + 3}, \ldots , p_{2n+2}$
 is then attached to a rational tail mapping isomorphically onto the right component. Similarly, in the other such cover, E maps to the right component, with
$p_{n + 2}, p_{n + 3}, \ldots , p_{2n+2}$
 is then attached to a rational tail mapping isomorphically onto the right component. Similarly, in the other such cover, E maps to the right component, with 
 $p_{n + 2}, p_{n + 3}, \ldots , p_{2n+2}$
 mapping to
$p_{n + 2}, p_{n + 3}, \ldots , p_{2n+2}$
 mapping to 
 $q_{n + 2}, q_{n + 3}, \ldots , q_{2n+2}$
, and
$q_{n + 2}, q_{n + 3}, \ldots , q_{2n+2}$
, and 
 $p_1, p_2, \ldots , p_{n+1}$
 mapping to the node and attached to a rational tail mapping isomorphically onto the left component. These covers are pictured in the following diagrams:
$p_1, p_2, \ldots , p_{n+1}$
 mapping to the node and attached to a rational tail mapping isomorphically onto the left component. These covers are pictured in the following diagrams:

 In fact, these are the only two such admissible covers. Indeed, the curve E must map to one of the two components of the above degeneration of the target 
 $\mathbb {P}^1$
, say without loss of generality to the left component. Then
$\mathbb {P}^1$
, say without loss of generality to the left component. Then 
 $p_{n+2}, \dots , p_{2n+2}$
 must map to the node, which we normalize to
$p_{n+2}, \dots , p_{2n+2}$
 must map to the node, which we normalize to 
 $[1:0]$
. Hence, the map
$[1:0]$
. Hence, the map 
 $E \to \mathbb {P}^1$
 is given by
$E \to \mathbb {P}^1$
 is given by 
 $[s:1]$
 for a section
$[s:1]$
 for a section 
 $s \in H^0(\mathcal {O}(p_{n+2} + \cdots + p_{2n+2}))$
. Since
$s \in H^0(\mathcal {O}(p_{n+2} + \cdots + p_{2n+2}))$
. Since 
 $p_1, \ldots , p_{n+1}$
 are general, the evaluation map
$p_1, \ldots , p_{n+1}$
 are general, the evaluation map 
 $H^0(\mathcal {O}(p_{n+2} + \cdots + p_{2n+2})) \to \bigoplus _{i = 1}^{n+1} \mathcal {O}(p_{n+2} + \cdots + p_{2n+2})|_{p_i}$
 is an isomorphism. Hence, s is uniquely determined. We conclude that, when
$H^0(\mathcal {O}(p_{n+2} + \cdots + p_{2n+2})) \to \bigoplus _{i = 1}^{n+1} \mathcal {O}(p_{n+2} + \cdots + p_{2n+2})|_{p_i}$
 is an isomorphism. Hence, s is uniquely determined. We conclude that, when 
 $q_1, q_2, \ldots , q_{2n + 2} \in \mathbb {P}^1$
 are general, there are exactly two such maps.
$q_1, q_2, \ldots , q_{2n + 2} \in \mathbb {P}^1$
 are general, there are exactly two such maps.
 Write 
 $f_i \colon E \to \mathbb {P}^1$
 (for
$f_i \colon E \to \mathbb {P}^1$
 (for 
 $i \in \{1, 2\}$
) for these two maps, and
$i \in \{1, 2\}$
) for these two maps, and 
 $\overline {f} = (f_1, f_2) \colon E \to \mathbb {P}^1 \times \mathbb {P}^1$
 for the resulting map.
$\overline {f} = (f_1, f_2) \colon E \to \mathbb {P}^1 \times \mathbb {P}^1$
 for the resulting map.
Lemma 13.2. In the setup of Lemma 13.1, the map 
 $\overline {f}$
 is a general map from E to
$\overline {f}$
 is a general map from E to 
 $\mathbb {P}^1\times \mathbb {P}^1$
 of bidegree
$\mathbb {P}^1\times \mathbb {P}^1$
 of bidegree 
 $(n+1, n+1)$
. In particular:
$(n+1, n+1)$
. In particular:
- 
•  $\overline {f}$
 is birational onto its image, and its image is nodal. $\overline {f}$
 is birational onto its image, and its image is nodal.
- 
•  $\overline {f}^* \mathcal {O}_{\mathbb {P}^1\times \mathbb {P}^1}(1, -1) \in \operatorname {Pic}^0(E)$
 is general (and thus nontrivial). $\overline {f}^* \mathcal {O}_{\mathbb {P}^1\times \mathbb {P}^1}(1, -1) \in \operatorname {Pic}^0(E)$
 is general (and thus nontrivial).
Proof. Fix a general elliptic curve E. Let 
 $f \colon E \to \mathbb {P}^1 \times \mathbb {P}^1$
 be a general map. Write
$f \colon E \to \mathbb {P}^1 \times \mathbb {P}^1$
 be a general map. Write 
 $\Delta \subset \mathbb {P}^1 \times \mathbb {P}^1$
 for the diagonal. Then the
$\Delta \subset \mathbb {P}^1 \times \mathbb {P}^1$
 for the diagonal. Then the 
 $\{p_1, p_2, \ldots , p_{2n+2}\} = f^{-1}(\Delta ) \subset E$
, and their images
$\{p_1, p_2, \ldots , p_{2n+2}\} = f^{-1}(\Delta ) \subset E$
, and their images 
 $\{q_1, q_2, \ldots , q_{2n+2}\} \subset \mathbb {P}^1$
 under the composition of either projection with f, satisfy all the hypotheses of Lemma 13.1 except for possibly genericity.
$\{q_1, q_2, \ldots , q_{2n+2}\} \subset \mathbb {P}^1$
 under the composition of either projection with f, satisfy all the hypotheses of Lemma 13.1 except for possibly genericity.
 To complete the proof, it remains to check that there is no obstruction to deforming f so that these points become general. In other words, we must check 
 $H^1(f^* T_{\mathbb {P}^1 \times \mathbb {P}^1}(-\Delta )) = 0$
. But this is true because f is general and
$H^1(f^* T_{\mathbb {P}^1 \times \mathbb {P}^1}(-\Delta )) = 0$
. But this is true because f is general and 
 $f^* T_{\mathbb {P}^1 \times \mathbb {P}^1}(-\Delta ) \simeq f^*\mathcal {O}_{\mathbb {P}^1 \times \mathbb {P}^1}(1, -1) \oplus f^*\mathcal {O}_{\mathbb {P}^1 \times \mathbb {P}^1}(-1, 1)$
.
$f^* T_{\mathbb {P}^1 \times \mathbb {P}^1}(-\Delta ) \simeq f^*\mathcal {O}_{\mathbb {P}^1 \times \mathbb {P}^1}(1, -1) \oplus f^*\mathcal {O}_{\mathbb {P}^1 \times \mathbb {P}^1}(-1, 1)$
.
Lemma 13.3. In the setup of Lemma 13.1, we have
 $$\begin{align*}\overline{f}^* \mathcal{O}_{\mathbb{P}^1 \times \mathbb{P}^1}(1, 1) \simeq \mathcal{O}_E(p_1 + p_2 + \cdots + p_{2n+2}).\end{align*}$$
$$\begin{align*}\overline{f}^* \mathcal{O}_{\mathbb{P}^1 \times \mathbb{P}^1}(1, 1) \simeq \mathcal{O}_E(p_1 + p_2 + \cdots + p_{2n+2}).\end{align*}$$
Proof. The isomorphism class of the line bundle 
 $\overline {f}^* \mathcal {O}_{\mathbb {P}^1 \times \mathbb {P}^1}(1, 1)$
 is independent of the moduli of the points
$\overline {f}^* \mathcal {O}_{\mathbb {P}^1 \times \mathbb {P}^1}(1, 1)$
 is independent of the moduli of the points 
 $q_1, \dots , q_{2n+2}$
 because they vary in a rational base. Hence, we may calculate it in the degeneration of Lemma 13.1, where the result clearly holds.
$q_1, \dots , q_{2n+2}$
 because they vary in a rational base. Hence, we may calculate it in the degeneration of Lemma 13.1, where the result clearly holds.
Lemma 13.4. In the setup of Lemma 13.1, the pushforward of 
 $\mathcal {O}_E(p_1 + p_2 + \cdots + p_{2n+2})$
 along either map is perfectly balanced, that is,
$\mathcal {O}_E(p_1 + p_2 + \cdots + p_{2n+2})$
 along either map is perfectly balanced, that is, 
 $$\begin{align*}(f_i)_* \mathcal{O}_E(p_1 + p_2 + \cdots + p_{2n+2}) \simeq \mathcal{O}_{\mathbb{P}^1}(1)^{\oplus(n+1)}.\end{align*}$$
$$\begin{align*}(f_i)_* \mathcal{O}_E(p_1 + p_2 + \cdots + p_{2n+2}) \simeq \mathcal{O}_{\mathbb{P}^1}(1)^{\oplus(n+1)}.\end{align*}$$
Proof. Since 
 $f_i$
 is of degree
$f_i$
 is of degree 
 $n+1$
, the pushforward
$n+1$
, the pushforward 
 $(f_i)_* \mathcal {O}_E(p_1 + p_2 + \cdots + p_{2n+2})$
 is a rank
$(f_i)_* \mathcal {O}_E(p_1 + p_2 + \cdots + p_{2n+2})$
 is a rank 
 $n+1$
 vector bundle on
$n+1$
 vector bundle on 
 $\mathbb {P}^1$
, that is, we can write
$\mathbb {P}^1$
, that is, we can write 
 $$\begin{align*}(f_i)_* \mathcal{O}_E(p_1 + p_2 + \cdots + p_{2n+2}) \simeq \bigoplus_{j = 1}^{n+1} \mathcal{O}_{\mathbb{P}^1}(a_j).\end{align*}$$
$$\begin{align*}(f_i)_* \mathcal{O}_E(p_1 + p_2 + \cdots + p_{2n+2}) \simeq \bigoplus_{j = 1}^{n+1} \mathcal{O}_{\mathbb{P}^1}(a_j).\end{align*}$$
The integers 
 $a_j$
 satisfy
$a_j$
 satisfy 
 $$\begin{align*}\sum a_j = \chi\left(\bigoplus_{j = 1}^{n+1} \mathcal{O}_{\mathbb{P}^1}(a_j)\right) - (n+1) &= \chi(\mathcal{O}_E(p_1 + p_2 + \cdots + p_{2n+2})) - (n+1)\\ &= 2n+2 - (n+1) = n+1,\end{align*}$$
$$\begin{align*}\sum a_j = \chi\left(\bigoplus_{j = 1}^{n+1} \mathcal{O}_{\mathbb{P}^1}(a_j)\right) - (n+1) &= \chi(\mathcal{O}_E(p_1 + p_2 + \cdots + p_{2n+2})) - (n+1)\\ &= 2n+2 - (n+1) = n+1,\end{align*}$$
so to see that 
 $a_j = 1$
 for all j, it suffices to see that
$a_j = 1$
 for all j, it suffices to see that 
 $a_j < 2$
 for all j, that is, that
$a_j < 2$
 for all j, that is, that 
 $$\begin{align*}0 = H^0\left(\bigoplus_{j = 1}^{n+1} \mathcal{O}_{\mathbb{P}^1}(a_j - 2)\right) = H^0(\mathcal{O}_E(p_1 + p_2 + \cdots + p_{2n+2}) \otimes f_i^* \mathcal{O}_{\mathbb{P}^1}(-2)),\end{align*}$$
$$\begin{align*}0 = H^0\left(\bigoplus_{j = 1}^{n+1} \mathcal{O}_{\mathbb{P}^1}(a_j - 2)\right) = H^0(\mathcal{O}_E(p_1 + p_2 + \cdots + p_{2n+2}) \otimes f_i^* \mathcal{O}_{\mathbb{P}^1}(-2)),\end{align*}$$
or equivalently that
 $$\begin{align*}\mathcal{O}_E(p_1 + p_2 + \cdots + p_{2n+2}) \not\simeq \overline{f}^* \mathcal{O}_{\mathbb{P}^1 \times \mathbb{P}^1}(2, 0) = \overline{f}^* \mathcal{O}_{\mathbb{P}^1 \times \mathbb{P}^1}(1, 1) \otimes \overline{f}^* \mathcal{O}_{\mathbb{P}^1 \times \mathbb{P}^1}(1, -1),\end{align*}$$
$$\begin{align*}\mathcal{O}_E(p_1 + p_2 + \cdots + p_{2n+2}) \not\simeq \overline{f}^* \mathcal{O}_{\mathbb{P}^1 \times \mathbb{P}^1}(2, 0) = \overline{f}^* \mathcal{O}_{\mathbb{P}^1 \times \mathbb{P}^1}(1, 1) \otimes \overline{f}^* \mathcal{O}_{\mathbb{P}^1 \times \mathbb{P}^1}(1, -1),\end{align*}$$
 Write 
 $r = 2n + 1$
. Applying Lemma 13.1, there are exactly two maps
$r = 2n + 1$
. Applying Lemma 13.1, there are exactly two maps 
 $f_i \colon E \to \mathbb {P}^1 \simeq R$
, of degree
$f_i \colon E \to \mathbb {P}^1 \simeq R$
, of degree 
 $n + 1$
, sending
$n + 1$
, sending 
 $\Gamma |_E$
 to
$\Gamma |_E$
 to 
 $\Gamma |_R$
, where
$\Gamma |_R$
, where 
 $\Gamma = E \cap R$
. Write
$\Gamma = E \cap R$
. Write 
 $\overline {f} = (f_1, f_2) \colon E \to \mathbb {P}^1 \times \mathbb {P}^1$
. Let S denote the blowup of
$\overline {f} = (f_1, f_2) \colon E \to \mathbb {P}^1 \times \mathbb {P}^1$
. Let S denote the blowup of 
 $\mathbb {P}^1 \times \mathbb {P}^1$
 at the nodes of the image of E under
$\mathbb {P}^1 \times \mathbb {P}^1$
 at the nodes of the image of E under 
 $\overline {f}$
, so that the
$\overline {f}$
, so that the 
 $f_i$
 give rise to an embedding
$f_i$
 give rise to an embedding 
 $f \colon E \hookrightarrow S$
. Writing
$f \colon E \hookrightarrow S$
. Writing 
 $F_1, \dots , F_{n^2-1}$
 for the exceptional divisors, define
$F_1, \dots , F_{n^2-1}$
 for the exceptional divisors, define 

 By adjunction on S and Lemma 13.3, we have 
 $L|_E \simeq \mathcal {O}_S(1, 1)|_E \simeq \mathcal {O}_E(1)$
. Let
$L|_E \simeq \mathcal {O}_S(1, 1)|_E \simeq \mathcal {O}_E(1)$
. Let 
 $\pi _1\colon S \to \mathbb {P}^1$
 and
$\pi _1\colon S \to \mathbb {P}^1$
 and 
 $\pi _2\colon S \to \mathbb {P}^1$
 denote the two projections. By intersection theory, for
$\pi _2\colon S \to \mathbb {P}^1$
 denote the two projections. By intersection theory, for 
 $x \in \mathbb {P}^1$
, the restriction of the line bundle
$x \in \mathbb {P}^1$
, the restriction of the line bundle 
 $L(-E)$
 to the corresponding fiber
$L(-E)$
 to the corresponding fiber 
 $\pi _i^{-1}(x)$
 is the (unique) line bundle of total degree
$\pi _i^{-1}(x)$
 is the (unique) line bundle of total degree 
 $-1$
 that is isomorphic to
$-1$
 that is isomorphic to 
 $\mathcal {O}_{\mathbb {P}^1}(-1)$
 on any exceptional divisors lying over x. In particular, it has vanishing cohomology, so by the theorem on cohomology and base-change,
$\mathcal {O}_{\mathbb {P}^1}(-1)$
 on any exceptional divisors lying over x. In particular, it has vanishing cohomology, so by the theorem on cohomology and base-change, 
 $(\pi _i)_* L(-E) = R^1 (\pi _i)_* L(-E) = 0$
. Combining this with Lemma 13.4, we therefore have a natural identification
$(\pi _i)_* L(-E) = R^1 (\pi _i)_* L(-E) = 0$
. Combining this with Lemma 13.4, we therefore have a natural identification 
 $$ \begin{align}(\pi_i)_* L \simeq (f_i)_* \mathcal{O}_E(1) \simeq \mathcal{O}_{\mathbb{P}^1}(1)^{n + 1}.\end{align} $$
$$ \begin{align}(\pi_i)_* L \simeq (f_i)_* \mathcal{O}_E(1) \simeq \mathcal{O}_{\mathbb{P}^1}(1)^{n + 1}.\end{align} $$
 The map 
 $S \to \mathbb {P}^{2n + 1}$
 via
$S \to \mathbb {P}^{2n + 1}$
 via 
 $|L|$
 thus factors through a (uniquely defined) embedding of
$|L|$
 thus factors through a (uniquely defined) embedding of 
 $\mathbb {P} [(\pi _i)_* L] \simeq \mathbb {P}^1 \times \mathbb {P}^n$
, via the complete linear system of the relative
$\mathbb {P} [(\pi _i)_* L] \simeq \mathbb {P}^1 \times \mathbb {P}^n$
, via the complete linear system of the relative 
 $\mathcal {O}(1)$
 on
$\mathcal {O}(1)$
 on 
 $\mathbb {P} [(\pi _i)_* L]$
, which corresponds to
$\mathbb {P} [(\pi _i)_* L]$
, which corresponds to 
 $|\mathcal {O}_{\mathbb {P}^1 \times \mathbb {P}^n}(1, 1)|$
 on
$|\mathcal {O}_{\mathbb {P}^1 \times \mathbb {P}^n}(1, 1)|$
 on 
 $\mathbb {P}^1 \times \mathbb {P}^n$
. In particular, since
$\mathbb {P}^1 \times \mathbb {P}^n$
. In particular, since 
 $S \to \mathbb {P} [(\pi _i)_* L]$
 is an embedding, so is
$S \to \mathbb {P} [(\pi _i)_* L]$
 is an embedding, so is 
 $S \to \mathbb {P}^{2n + 1}$
. Write
$S \to \mathbb {P}^{2n + 1}$
. Write 
 $\Sigma _i \subset \mathbb {P}^{2n + 1}$
 for the scroll obtained as the image of the map
$\Sigma _i \subset \mathbb {P}^{2n + 1}$
 for the scroll obtained as the image of the map 
 $\mathbb {P}^1 \times \mathbb {P}^n \to \mathbb {P}^{2n+1}$
.
$\mathbb {P}^1 \times \mathbb {P}^n \to \mathbb {P}^{2n+1}$
.
Putting all of this together, we can summarize this situation with the following diagram of inclusions:

Lemma 13.5. The intersection 
 $\Sigma _1 \cap \Sigma _2$
 coincides with S.
$\Sigma _1 \cap \Sigma _2$
 coincides with S.
Proof. Let 
 $x_1, x_2 \in \mathbb {P}^1$
 be any two points, and write
$x_1, x_2 \in \mathbb {P}^1$
 be any two points, and write 
 $\Lambda _i$
 for the fiber of
$\Lambda _i$
 for the fiber of 
 $\Sigma _i$
 over
$\Sigma _i$
 over 
 $x_i$
. Note that
$x_i$
. Note that 
 $\Lambda _i$
 is the span of the divisor
$\Lambda _i$
 is the span of the divisor 
 $f_i^{-1}(x_i)$
.
$f_i^{-1}(x_i)$
.
 First, suppose that 
 $(x_1, x_2)$
 is not a node of
$(x_1, x_2)$
 is not a node of 
 $\overline {f}(E)$
. If
$\overline {f}(E)$
. If 
 $(x_1, x_2)$
 does not lie on
$(x_1, x_2)$
 does not lie on 
 $\overline {f}(E)$
, then the span of
$\overline {f}(E)$
, then the span of 
 $\Lambda _1$
 and
$\Lambda _1$
 and 
 $\Lambda _2$
 is the span of the divisor
$\Lambda _2$
 is the span of the divisor 
 $f_1^{-1}(x_1) + f_2^{-1}(x_2)$
. Since this divisor is linearly equivalent to
$f_1^{-1}(x_1) + f_2^{-1}(x_2)$
. Since this divisor is linearly equivalent to 
 $O_E(1)$
, the span is a hyperplane. Otherwise, if
$O_E(1)$
, the span is a hyperplane. Otherwise, if 
 $(x_1, x_2) = \overline {f}(y)$
 lies on
$(x_1, x_2) = \overline {f}(y)$
 lies on 
 $\overline {f}(E)$
, then the span of
$\overline {f}(E)$
, then the span of 
 $\Lambda _1$
 and
$\Lambda _1$
 and 
 $\Lambda _2$
 is the span of the divisor
$\Lambda _2$
 is the span of the divisor 
 $f_1^{-1}(x_1) + f_2^{-1}(x_2) - y$
. Since any
$f_1^{-1}(x_1) + f_2^{-1}(x_2) - y$
. Since any 
 $2n + 1$
 points on E are linearly general, this span is again a hyperplane. Either way, since
$2n + 1$
 points on E are linearly general, this span is again a hyperplane. Either way, since 
 $\Lambda _1$
 and
$\Lambda _1$
 and 
 $\Lambda _2$
 span a hyperplane, they must meet at the single point that is the image of
$\Lambda _2$
 span a hyperplane, they must meet at the single point that is the image of 
 $(x_1, x_2)$
 on S.
$(x_1, x_2)$
 on S.
 Next, suppose that 
 $(x_1, x_2)$
 is a node of
$(x_1, x_2)$
 is a node of 
 $\overline {f}(E)$
, say
$\overline {f}(E)$
, say 
 $(x_1, x_2) = \overline {f}(y_1) = \overline {f}(y_2)$
. Then the span of
$(x_1, x_2) = \overline {f}(y_1) = \overline {f}(y_2)$
. Then the span of 
 $\Lambda _1$
 and
$\Lambda _1$
 and 
 $\Lambda _2$
 is the span of the divisor
$\Lambda _2$
 is the span of the divisor 
 $f_1^{-1}(x_1) + f_2^{-1}(x_2) - y_1 - y_2$
. Since any
$f_1^{-1}(x_1) + f_2^{-1}(x_2) - y_1 - y_2$
. Since any 
 $2n$
 points on E are linearly general, this span is codimension
$2n$
 points on E are linearly general, this span is codimension 
 $2$
, and so
$2$
, and so 
 $\Lambda _1$
 and
$\Lambda _1$
 and 
 $\Lambda _2$
 meet along the line that is the image of the exceptional divisor over
$\Lambda _2$
 meet along the line that is the image of the exceptional divisor over 
 $(x_1, x_2)$
.
$(x_1, x_2)$
.
 Combining these two cases, we see that 
 $\Sigma _1 \cap \Sigma _2$
 coincides with S set-theoretically. To upgrade this to a scheme-theoretic equality, we must show that
$\Sigma _1 \cap \Sigma _2$
 coincides with S set-theoretically. To upgrade this to a scheme-theoretic equality, we must show that 
 $\Sigma _1$
 and
$\Sigma _1$
 and 
 $\Sigma _2$
 are quasi-transverse along S, meaning that the tangent spaces to
$\Sigma _2$
 are quasi-transverse along S, meaning that the tangent spaces to 
 $\Sigma _1$
 and
$\Sigma _1$
 and 
 $\Sigma _2$
 at points of S span a hyperplane. (They cannot span all of
$\Sigma _2$
 at points of S span a hyperplane. (They cannot span all of 
 $\mathbb {P}^r$
, because
$\mathbb {P}^r$
, because 
 $\Sigma _1 \cap \Sigma _2$
 is pure of dimension
$\Sigma _1 \cap \Sigma _2$
 is pure of dimension 
 $2$
.) Away from the exceptional divisors, this is straightforward: The tangent space to
$2$
.) Away from the exceptional divisors, this is straightforward: The tangent space to 
 $\Sigma _i$
 contains
$\Sigma _i$
 contains 
 $\Lambda _i$
, so it suffices to note that
$\Lambda _i$
, so it suffices to note that 
 $\Lambda _1$
 and
$\Lambda _1$
 and 
 $\Lambda _2$
 span a hyperplane.
$\Lambda _2$
 span a hyperplane.
 It remains to consider an exceptional divisor 
 $M = \Lambda _1 \cap \Lambda _2$
. Write
$M = \Lambda _1 \cap \Lambda _2$
. Write 
 $\Lambda $
 for the span of
$\Lambda $
 for the span of 
 $\Lambda _1$
 and
$\Lambda _1$
 and 
 $\Lambda _2$
. As in the previous case, the span of the tangent spaces to the
$\Lambda _2$
. As in the previous case, the span of the tangent spaces to the 
 $\Sigma _i$
 contains
$\Sigma _i$
 contains 
 $\Lambda $
; however, in this case
$\Lambda $
; however, in this case 
 $\Lambda $
 is codimension
$\Lambda $
 is codimension 
 $2$
. It thus remains to show that the natural map
$2$
. It thus remains to show that the natural map 
 $N_{M / \Sigma _1} \oplus N_{M / \Sigma _2} \to N_{\Lambda /\mathbb {P}^r}|_M$
 is everywhere nonzero along M. Recall that we write
$N_{M / \Sigma _1} \oplus N_{M / \Sigma _2} \to N_{\Lambda /\mathbb {P}^r}|_M$
 is everywhere nonzero along M. Recall that we write 
 $y_i$
 for the points on E lying over the node, that is, satisfying
$y_i$
 for the points on E lying over the node, that is, satisfying 
 $\overline {f}(y_1) = \overline {f}(y_2) = (x_1, x_2)$
 so that M is the line spanned by
$\overline {f}(y_1) = \overline {f}(y_2) = (x_1, x_2)$
 so that M is the line spanned by 
 $y_1$
 and
$y_1$
 and 
 $y_2$
. Let
$y_2$
. Let 
 $y_i^j$
 be nontrivial first-order deformations of the
$y_i^j$
 be nontrivial first-order deformations of the 
 $y_i$
 satisfying
$y_i$
 satisfying 
 $f_j(y_1^j) = f_j(y_2^j)$
. Such deformations are pictured in the following diagram:
$f_j(y_1^j) = f_j(y_2^j)$
. Such deformations are pictured in the following diagram:

 The lines joining 
 $y_1^j$
 and
$y_1^j$
 and 
 $y_2^j$
 give first-order deformations of M in
$y_2^j$
 give first-order deformations of M in 
 $\Sigma _j$
, that is, sections
$\Sigma _j$
, that is, sections 
 $\sigma _j$
 of
$\sigma _j$
 of 
 $N_{M / \Sigma _j}$
. It suffices to see that the images of the
$N_{M / \Sigma _j}$
. It suffices to see that the images of the 
 $\sigma _j$
 in
$\sigma _j$
 in 
 $N_{\Lambda / \mathbb {P}^r}|_M$
 do not simultaneously vanish anywhere along M.
$N_{\Lambda / \mathbb {P}^r}|_M$
 do not simultaneously vanish anywhere along M.
 The span of 
 $\Lambda $
 and the tangent line to E at
$\Lambda $
 and the tangent line to E at 
 $y_1$
 is the span of the divisor
$y_1$
 is the span of the divisor 
 $f_1^{-1}(x_1) + f_2^{-1}(x_2) - y_2$
. Since any
$f_1^{-1}(x_1) + f_2^{-1}(x_2) - y_2$
. Since any 
 $2n + 1$
 points on E are linearly general, this span is a hyperplane. Similarly, the span of
$2n + 1$
 points on E are linearly general, this span is a hyperplane. Similarly, the span of 
 $\Lambda $
 and the tangent line to E at
$\Lambda $
 and the tangent line to E at 
 $y_2$
 is a hyperplane. Moreover, the span of
$y_2$
 is a hyperplane. Moreover, the span of 
 $\Lambda $
 and the tangent lines to E at both
$\Lambda $
 and the tangent lines to E at both 
 $y_1$
 and
$y_1$
 and 
 $y_2$
 is the span of the divisor
$y_2$
 is the span of the divisor 
 $f_1^{-1}(x_1) + f_2^{-1}(x_2)$
, which is linearly equivalent to
$f_1^{-1}(x_1) + f_2^{-1}(x_2)$
, which is linearly equivalent to 
 $\mathcal {O}_E(1)$
, and therefore again spans a hyperplane. Since this hyperplane contains the first two of these hyperplanes, all three of these hyperplanes must be equal. Write
$\mathcal {O}_E(1)$
, and therefore again spans a hyperplane. Since this hyperplane contains the first two of these hyperplanes, all three of these hyperplanes must be equal. Write 
 $\Lambda '$
 for this hyperplane.
$\Lambda '$
 for this hyperplane.
 By construction, the images of both 
 $\sigma _j$
 are nonzero sections in the subspace
$\sigma _j$
 are nonzero sections in the subspace 
 $N_{\Lambda / \Lambda '}|_M \simeq \mathcal {O}_M(1)$
. Because
$N_{\Lambda / \Lambda '}|_M \simeq \mathcal {O}_M(1)$
. Because 
 $\overline {f}(E)$
 is nodal, the deformations
$\overline {f}(E)$
 is nodal, the deformations 
 $(y_1^1, y_2^1), (y_1^2, y_2^2)$
 form a basis of
$(y_1^1, y_2^1), (y_1^2, y_2^2)$
 form a basis of 
 $T_{y_1} E \oplus T_{y_2} E$
. The images of these two sections thus form a basis of
$T_{y_1} E \oplus T_{y_2} E$
. The images of these two sections thus form a basis of 
 $H^0(N_{\Lambda / \Lambda '}|_M) = H^0(\mathcal {O}_M(1))$
 and so do not simultaneously vanish anywhere along M as desired.
$H^0(N_{\Lambda / \Lambda '}|_M) = H^0(\mathcal {O}_M(1))$
 and so do not simultaneously vanish anywhere along M as desired.
 The upshot of this is that we have a natural filtration of 
 $N_{E \cup R}$
, whose successive quotients are vector bundles of ranks
$N_{E \cup R}$
, whose successive quotients are vector bundles of ranks 
 $1$
,
$1$
, 
 $2n - 2$
 and
$2n - 2$
 and 
 $1$
, respectively:
$1$
, respectively: 
 $$ \begin{align} 0 \subset N_{E \cup R / S} \subset N_{E \cup R / \Sigma_1} + N_{E \cup R / \Sigma_2} \subset N_{E \cup R}. \end{align} $$
$$ \begin{align} 0 \subset N_{E \cup R / S} \subset N_{E \cup R / \Sigma_1} + N_{E \cup R / \Sigma_2} \subset N_{E \cup R}. \end{align} $$
Lemma 13.6. We have 
 $c_1(N_{E \cup R / \Sigma _j}|_R) = n(2n + 3) + 1$
.
$c_1(N_{E \cup R / \Sigma _j}|_R) = n(2n + 3) + 1$
.
Proof. The Picard group of 
 $\Sigma _j$
 is spanned by the class
$\Sigma _j$
 is spanned by the class 
 $\gamma $
 of one n-plane and the restriction of the hyperplane class h from
$\gamma $
 of one n-plane and the restriction of the hyperplane class h from 
 $\mathbb {P}^{2n+1}$
. One computes that
$\mathbb {P}^{2n+1}$
. One computes that 
 $K_{\Sigma _j} = (n-1)\gamma - (n+1) h$
 for such a scroll. By adjunction we have
$K_{\Sigma _j} = (n-1)\gamma - (n+1) h$
 for such a scroll. By adjunction we have 
 $$ \begin{align*} c_1(N_{R/\Sigma_j}) &= c_1(K_R) - c_1(K_S|_R) \\ &= -2 - ((n-1) \gamma - (n+1) h) \cdot R\\ &= -2 - (n-1) (\gamma \cdot R) + (n + 1) (h \cdot R) \\ &= -2 - (n-1) \cdot 1 + (n + 1) \cdot 2n \\ &= 2n^2 + n - 1. \end{align*} $$
$$ \begin{align*} c_1(N_{R/\Sigma_j}) &= c_1(K_R) - c_1(K_S|_R) \\ &= -2 - ((n-1) \gamma - (n+1) h) \cdot R\\ &= -2 - (n-1) (\gamma \cdot R) + (n + 1) (h \cdot R) \\ &= -2 - (n-1) \cdot 1 + (n + 1) \cdot 2n \\ &= 2n^2 + n - 1. \end{align*} $$
Therefore,
 $$\begin{align*}c_1(N_{E \cup R / \Sigma_j}|_R) = c_1(N_{R/S}) + \# \Gamma = (2n^2 + n - 1) + (2n + 2) = n(2n + 3) + 1. \end{align*}$$
$$\begin{align*}c_1(N_{E \cup R / \Sigma_j}|_R) = c_1(N_{R/S}) + \# \Gamma = (2n^2 + n - 1) + (2n + 2) = n(2n + 3) + 1. \end{align*}$$
Proposition 13.7. The vector bundle 
 is isomorphic to
 is isomorphic to 
 $$\begin{align*}\mathcal{O}_{\mathbb{P}^1}(2n + 4) \oplus \mathcal{O}_{\mathbb{P}^1}(2n + 3)^{\oplus (2n - 2)} \oplus \mathcal{O}_{\mathbb{P}^1}(2n + 2).\end{align*}$$
$$\begin{align*}\mathcal{O}_{\mathbb{P}^1}(2n + 4) \oplus \mathcal{O}_{\mathbb{P}^1}(2n + 3)^{\oplus (2n - 2)} \oplus \mathcal{O}_{\mathbb{P}^1}(2n + 2).\end{align*}$$
Moreover, its HN-filtration is precisely the restriction of the filtration (13.2) to R.
Proof. By Lemma 5.8, we have either 
 or
 or 
 . Lemma 13.6 rules out the second case since
. Lemma 13.6 rules out the second case since 
 $\mathcal {O}_{\mathbb {P}^1}(2n + 3)^{\oplus 2n}$
 admits no subbundle of rank n and first Chern class
$\mathcal {O}_{\mathbb {P}^1}(2n + 3)^{\oplus 2n}$
 admits no subbundle of rank n and first Chern class 
 $n(2n + 3) + 1$
.
$n(2n + 3) + 1$
.
 Moreover, any subbundle of 
 $\mathcal {O}_{\mathbb {P}^1}(2n + 4) \oplus \mathcal {O}_{\mathbb {P}^1}(2n + 3)^{\oplus (2n - 2)} \oplus \mathcal {O}_{\mathbb {P}^1}(2n + 2)$
 of rank n and first Chern class
$\mathcal {O}_{\mathbb {P}^1}(2n + 4) \oplus \mathcal {O}_{\mathbb {P}^1}(2n + 3)^{\oplus (2n - 2)} \oplus \mathcal {O}_{\mathbb {P}^1}(2n + 2)$
 of rank n and first Chern class 
 $n(2n + 3) + 1$
 contains
$n(2n + 3) + 1$
 contains 
 $O_{\mathbb {P}^1}(2n + 4)$
 and is contained in
$O_{\mathbb {P}^1}(2n + 4)$
 and is contained in 
 $O_{\mathbb {P}^1}(2n + 4) \oplus \mathcal {O}_{\mathbb {P}^1}(2n + 3)^{\oplus (2n - 2)}$
. Since the graded pieces of equation (13.2) are the intersection and span of the
$O_{\mathbb {P}^1}(2n + 4) \oplus \mathcal {O}_{\mathbb {P}^1}(2n + 3)^{\oplus (2n - 2)}$
. Since the graded pieces of equation (13.2) are the intersection and span of the 
 $N_{E \cup R / \Sigma _j}$
, they must therefore coincide with the HN-filtration.
$N_{E \cup R / \Sigma _j}$
, they must therefore coincide with the HN-filtration.
 This provides the promised determination of 
 , and the promised geometric construction of its HN-filtration. This geometric description of the HN-filtration of
, and the promised geometric construction of its HN-filtration. This geometric description of the HN-filtration of 
 $N_{E \cup R}|_R$
 allows us to reduce interpolation for
$N_{E \cup R}|_R$
 allows us to reduce interpolation for 
 $N_{E \cup R}$
 to interpolation for a modification of
$N_{E \cup R}$
 to interpolation for a modification of 
 as follows.
 as follows.
Lemma 13.8. Let p and q be two distinct points of 
 $E \cap R$
. Then
$E \cap R$
. Then 
 $N_{E \cup R}$
 satisfies interpolation provided that
$N_{E \cup R}$
 satisfies interpolation provided that 

satisfies interpolation.
Proof. We imitate the basic idea of the proof of [Reference Atanasov, Larson and Yang2, Lemma 8.8]. Write
 
, which has size 
 $r+1$
. Write
$r+1$
. Write 
 $x, y, z$
 for three general points on R. Twisting down, we have
$x, y, z$
 for three general points on R. Twisting down, we have 
 $$\begin{align*}N_{E \cup R}(-x-y-z)|_R \simeq \mathcal{O}_{\mathbb{P}^1}(r-2) \oplus \mathcal{O}_{\mathbb{P}^1}(r-1)^{\oplus r-3} \oplus \mathcal{O}_{\mathbb{P}^1}(r).\end{align*}$$
$$\begin{align*}N_{E \cup R}(-x-y-z)|_R \simeq \mathcal{O}_{\mathbb{P}^1}(r-2) \oplus \mathcal{O}_{\mathbb{P}^1}(r-1)^{\oplus r-3} \oplus \mathcal{O}_{\mathbb{P}^1}(r).\end{align*}$$
Therefore,the evaluation map
 $$\begin{align*}\operatorname{ev}_{R,\Gamma} \colon H^0(N_{E \cup R}|_R)\to N_{E \cup R}|_{\Gamma} \end{align*}$$
$$\begin{align*}\operatorname{ev}_{R,\Gamma} \colon H^0(N_{E \cup R}|_R)\to N_{E \cup R}|_{\Gamma} \end{align*}$$
is injective when restricted to the subspace 
 $H^0(N_{E \cup R}|_R(-x-y-z))$
. Our aim is to suitably specialize the points
$H^0(N_{E \cup R}|_R(-x-y-z))$
. Our aim is to suitably specialize the points 
 $x,y,z$
 so as to be able to identify the subspace of sections of
$x,y,z$
 so as to be able to identify the subspace of sections of 
 $N_{E \cup R}|_E$
$N_{E \cup R}|_E$
 

that glues to the image of 
 $H^0(N_{E \cup R}|_R(-x-y-z))$
 under
$H^0(N_{E \cup R}|_R(-x-y-z))$
 under 
 $\operatorname {ev}_{R,\Gamma }$
. By Lemma 3.6, it suffices to show that this subspace of sections has the correct dimension and satisfies interpolation to conclude that
$\operatorname {ev}_{R,\Gamma }$
. By Lemma 3.6, it suffices to show that this subspace of sections has the correct dimension and satisfies interpolation to conclude that 
 $N_{E \cup R}(-x-y-z)$
 satisfies interpolation, which implies that
$N_{E \cup R}(-x-y-z)$
 satisfies interpolation, which implies that 
 $N_{E \cup R}$
 satisfies interpolation.
$N_{E \cup R}$
 satisfies interpolation.
 Since 
 $\# \Gamma = r + 1$
, the evaluation map
$\# \Gamma = r + 1$
, the evaluation map 
 $\operatorname {ev}_{R, \Gamma }$
 restricted to the sections of the largest factor
$\operatorname {ev}_{R, \Gamma }$
 restricted to the sections of the largest factor 
 $\mathcal {O}_{\mathbb {P}^1}(r)$
 is already an isomorphism. The evaluation on the other factors is not an isomorphism: On the
$\mathcal {O}_{\mathbb {P}^1}(r)$
 is already an isomorphism. The evaluation on the other factors is not an isomorphism: On the 
 $\mathcal {O}_{\mathbb {P}^1}(r-1)$
 factors, the image is a codimension
$\mathcal {O}_{\mathbb {P}^1}(r-1)$
 factors, the image is a codimension 
 $1$
 subspace of
$1$
 subspace of 
 $\mathcal {O}_{\mathbb {P}^1}(r-1)|_\Gamma $
, and on the
$\mathcal {O}_{\mathbb {P}^1}(r-1)|_\Gamma $
, and on the 
 $\mathcal {O}_{\mathbb {P}^1}(r-2)$
 factor, the image is a codimension
$\mathcal {O}_{\mathbb {P}^1}(r-2)$
 factor, the image is a codimension 
 $2$
 subspace of
$2$
 subspace of 
 $\mathcal {O}_{\mathbb {P}^1}(r-2)|_\Gamma $
. We will appropriately specialize so as to force these subspaces to be ‘coordinate’ planes.
$\mathcal {O}_{\mathbb {P}^1}(r-2)|_\Gamma $
. We will appropriately specialize so as to force these subspaces to be ‘coordinate’ planes.
 First, limit x to p. The gluing data across the nodes (in particular at p) is fixed, and therefore the limiting codimension 
 $1$
 subspace of
$1$
 subspace of 
 $\mathcal {O}_{\mathbb {P}^1}(r-1)|_\Gamma $
 contains the subspace
$\mathcal {O}_{\mathbb {P}^1}(r-1)|_\Gamma $
 contains the subspace 
 $\mathcal {O}_{\mathbb {P}^1}(r-1)|_{\Gamma {\smallsetminus} p}\oplus 0|_p$
 of sections vanishing at p. Since this subspace has the correct dimension, it must be the flat limit. Then limit y to q. In the limit, the codimension
$\mathcal {O}_{\mathbb {P}^1}(r-1)|_{\Gamma {\smallsetminus} p}\oplus 0|_p$
 of sections vanishing at p. Since this subspace has the correct dimension, it must be the flat limit. Then limit y to q. In the limit, the codimension 
 $2$
 subspace of
$2$
 subspace of 
 $\mathcal {O}_{\mathbb {P}^1}(r-2)|_{\Gamma }$
 must contain the subspace
$\mathcal {O}_{\mathbb {P}^1}(r-2)|_{\Gamma }$
 must contain the subspace 
 $\mathcal {O}_{\mathbb {P}^1}(r-2)|_{\Gamma {\smallsetminus} \{p, q\}}\oplus 0|_p\oplus 0|_q$
 of sections vanishing at p and q. Since this has the correct dimension, it must be the flat limit. Since the HN-filtration on
$\mathcal {O}_{\mathbb {P}^1}(r-2)|_{\Gamma {\smallsetminus} \{p, q\}}\oplus 0|_p\oplus 0|_q$
 of sections vanishing at p and q. Since this has the correct dimension, it must be the flat limit. Since the HN-filtration on 
 $N_{E \cup R}|_R$
 is the restriction of equation (13.2) to R, this flat limit is
$N_{E \cup R}|_R$
 is the restriction of equation (13.2) to R, this flat limit is 

Since this subspace is the space of sections of a vector bundle, it suffices to show that this bundle satisfies interpolation. To complete the proof, we note that
 
, and so it suffices by Lemma 3.8 to prove interpolation after twisting up by p.
13.2 The case 
 $r \geq 9$
$r \geq 9$
By Lemma 13.8, it suffices to show equation (13.3) satisfies interpolation. On an elliptic curve, we can characterize which bundles satisfy interpolation in terms of the Atiyah classification.
Lemma 13.9. Let 
 $\mathscr {E}$
 be a vector bundle on an elliptic curve E. Then
$\mathscr {E}$
 be a vector bundle on an elliptic curve E. Then 
 $\mathscr {E}$
 satisfies interpolation if and only if there is a nonnegative integer a for which every Jordan–Holder (JH) factor
$\mathscr {E}$
 satisfies interpolation if and only if there is a nonnegative integer a for which every Jordan–Holder (JH) factor 
 $\mathscr {F}$
 of
$\mathscr {F}$
 of 
 $\mathscr {E}$
 satisfies
$\mathscr {E}$
 satisfies 
 $$ \begin{align} a \leq \mu(\mathscr{F}) \leq a+1 \quad \text{and} \quad \mathscr{F} \not\simeq \mathcal{O}_E.\end{align} $$
$$ \begin{align} a \leq \mu(\mathscr{F}) \leq a+1 \quad \text{and} \quad \mathscr{F} \not\simeq \mathcal{O}_E.\end{align} $$
Proof. By the Atiyah classification, every JH-factor of 
 $\mathscr {E}$
 is both a subbundle and quotient of
$\mathscr {E}$
 is both a subbundle and quotient of 
 $\mathscr {E}$
.
$\mathscr {E}$
.
 First, suppose 
 $\mathscr {E}$
 satisfies interpolation. Then
$\mathscr {E}$
 satisfies interpolation. Then 
 $\mathscr {E}$
 is nonspecial, so every JH-factor
$\mathscr {E}$
 is nonspecial, so every JH-factor 
 $\mathscr {F}$
 is nonspecial or equivalently satisfies
$\mathscr {F}$
 is nonspecial or equivalently satisfies 
 $\mu (\mathscr {F}) \geq 0$
 and
$\mu (\mathscr {F}) \geq 0$
 and 
 $\mathscr {F} \not \simeq \mathcal {O}_E$
. If no such nonnegative integer a exists, then there would be a positive integer b and JH-factors
$\mathscr {F} \not \simeq \mathcal {O}_E$
. If no such nonnegative integer a exists, then there would be a positive integer b and JH-factors 
 $\mathscr {F}_1$
 and
$\mathscr {F}_1$
 and 
 $\mathscr {F}_2$
 with
$\mathscr {F}_2$
 with 
 $\mu (\mathscr {F}_1) < b < \mu (\mathscr {F}_2)$
. This is a contradiction, since for general points
$\mu (\mathscr {F}_1) < b < \mu (\mathscr {F}_2)$
. This is a contradiction, since for general points 
 $p_1, p_2, \ldots , p_b \in E$
, we would have
$p_1, p_2, \ldots , p_b \in E$
, we would have 
 $$\begin{align*}H^0(\mathscr{E}(-p_1 - \cdots - p_b)) \neq 0 \quad \text{and} \quad H^1(\mathscr{E}(-p_1 - \cdots - p_b)) \neq 0.\end{align*}$$
$$\begin{align*}H^0(\mathscr{E}(-p_1 - \cdots - p_b)) \neq 0 \quad \text{and} \quad H^1(\mathscr{E}(-p_1 - \cdots - p_b)) \neq 0.\end{align*}$$
 In the other direction, suppose there is a nonnegative integer a for which every JH-factor 
 $\mathscr {F}$
 satisfies equation (13.4). Then for general points
$\mathscr {F}$
 satisfies equation (13.4). Then for general points 
 $p_1, p_2, \ldots , p_{a + 1} \in E$
,
$p_1, p_2, \ldots , p_{a + 1} \in E$
, 
 $$\begin{align*}H^0(\mathscr{E}(-p_1 - \cdots - p_{a + 1})) = 0 \quad \text{and} \quad H^1(\mathscr{E}(-p_1 - \cdots - p_a)) = 0.\end{align*}$$
$$\begin{align*}H^0(\mathscr{E}(-p_1 - \cdots - p_{a + 1})) = 0 \quad \text{and} \quad H^1(\mathscr{E}(-p_1 - \cdots - p_a)) = 0.\end{align*}$$
Therefore,
 $\mathscr {E}$
 satisfies interpolation.
$\mathscr {E}$
 satisfies interpolation.
Lemma 13.10. Let 
 $\mathscr {E}$
 be a vector bundle on an elliptic curve E, and let a and b be integers. For two points
$\mathscr {E}$
 be a vector bundle on an elliptic curve E, and let a and b be integers. For two points 
 $p, q \in E$
, consider subspaces
$p, q \in E$
, consider subspaces 
 $\Delta \subseteq \mathscr {E}|_p$
 of rank
$\Delta \subseteq \mathscr {E}|_p$
 of rank 
 $1$
 and
$1$
 and 
 $\Lambda \subseteq \mathscr {E}|_q$
 of corank
$\Lambda \subseteq \mathscr {E}|_q$
 of corank 
 $1$
. If every JH-factor
$1$
. If every JH-factor 
 $\mathscr {F}$
 of
$\mathscr {F}$
 of 
 $\mathscr {E}$
 satisfies
$\mathscr {E}$
 satisfies 
 $a < \mu (\mathscr {F}) < b$
, then every JH-factor
$a < \mu (\mathscr {F}) < b$
, then every JH-factor 
 $\mathscr {F}'$
 of
$\mathscr {F}'$
 of 
 satisfies
 satisfies 
 $a \leq \mu (\mathscr {F}') \leq b$
.
$a \leq \mu (\mathscr {F}') \leq b$
.
Proof. Up to replacing 
 $\mathscr {E}'$
 with its dual, it suffices to show that every JH-factor
$\mathscr {E}'$
 with its dual, it suffices to show that every JH-factor 
 $\mathscr {F}'$
 of
$\mathscr {F}'$
 of 
 $\mathscr {E}'$
 satisfies
$\mathscr {E}'$
 satisfies 
 $\mu (\mathscr {F}') \leq b$
. Since every JH-factor is a subbundle, and
$\mu (\mathscr {F}') \leq b$
. Since every JH-factor is a subbundle, and 
 $\mathscr {E}'$
 is a subsheaf of
$\mathscr {E}'$
 is a subsheaf of 
 , it suffices to show that every subsheaf
, it suffices to show that every subsheaf 
 $\mathscr {F}'$
 of
$\mathscr {F}'$
 of 
 satisfies
 satisfies 
 $\mu (\mathscr {F}') \leq b$
.
$\mu (\mathscr {F}') \leq b$
.
 If 
 $\mathscr {F}'$
 is a subsheaf of
$\mathscr {F}'$
 is a subsheaf of 
 $\mathscr {E}$
, we are done by assumption. Otherwise, write x for the degree of
$\mathscr {E}$
, we are done by assumption. Otherwise, write x for the degree of 
 $\mathscr {F}'$
 and y for the rank of
$\mathscr {F}'$
 and y for the rank of 
 $\mathscr {F}'$
 so that
$\mathscr {F}'$
 so that 
 $\mu (\mathscr {F}') = x/y$
. Write
$\mu (\mathscr {F}') = x/y$
. Write 
 $\mathscr {F} = \mathscr {F}' \cap \mathscr {E}$
 for the corresponding subsheaf of
$\mathscr {F} = \mathscr {F}' \cap \mathscr {E}$
 for the corresponding subsheaf of 
 $\mathscr {E}$
. Then
$\mathscr {E}$
. Then 
 $(x - 1)/y = \mu (\mathscr {F}) < b$
. Since b is an integer,
$(x - 1)/y = \mu (\mathscr {F}) < b$
. Since b is an integer, 
 $\mu (\mathscr {F}') = x/y \leq b$
 as desired.
$\mu (\mathscr {F}') = x/y \leq b$
 as desired.
Combining Lemmas 13.9 and 13.10, it suffices to prove:
Proposition 13.11. If 
 $r \geq 9$
 is odd, every JH-factor
$r \geq 9$
 is odd, every JH-factor 
 $\mathscr {F}$
 of
$\mathscr {F}$
 of 
 satisfies
 satisfies 
 $r+4 < \mu (\mathscr {F}) < r+5$
.
$r+4 < \mu (\mathscr {F}) < r+5$
.
Our proof of Proposition 13.11 will be by induction on r.
 When 
 $r = 7$
, we have
$r = 7$
, we have 
 . Since
. Since 
 $I(8, 1, 7, 0, 1)$
 holds, we deduce from Lemma 13.9 that every JH-factor
$I(8, 1, 7, 0, 1)$
 holds, we deduce from Lemma 13.9 that every JH-factor 
 $\mathscr {F}$
 of
$\mathscr {F}$
 of 
 has slope exactly
 has slope exactly 
 $r + 5$
 in this case. Although Proposition 13.11 does not hold in this case, it is close enough that we will be able to leverage it to establish the case
$r + 5$
 in this case. Although Proposition 13.11 does not hold in this case, it is close enough that we will be able to leverage it to establish the case 
 $r = 9$
.
$r = 9$
.
In general, our strategy will be to use our inductive hypothesis to show that Proposition 13.11 is close enough to holding that naturality of the HN-filtration forces it to hold exactly.
Definition 13.12. Let E be a genus 
 $1$
 curve. We say that a map
$1$
 curve. We say that a map 
 $$\begin{align*}\operatorname{Pic}^aE \xrightarrow{f} \operatorname{Pic}^bE\end{align*}$$
$$\begin{align*}\operatorname{Pic}^aE \xrightarrow{f} \operatorname{Pic}^bE\end{align*}$$
is natural if for any automorphism 
 $\theta \colon E \to E$
, the following diagram commutes:
$\theta \colon E \to E$
, the following diagram commutes:

Proposition 13.13. If 
 $\operatorname {Pic}^aE \to \operatorname {Pic}^bE$
 is natural, then
$\operatorname {Pic}^aE \to \operatorname {Pic}^bE$
 is natural, then 
 $a $
 divides b.
$a $
 divides b.
Proof. Translation by an a-torsion point acts as the identity on 
 $\operatorname {Pic}^aE$
, and so it must also act as the identity on
$\operatorname {Pic}^aE$
, and so it must also act as the identity on 
 $\operatorname {Pic}^bE$
.
$\operatorname {Pic}^bE$
.
Proof of Proposition 13.11
 It suffices to prove that 
 has no subbundles of slope
 has no subbundles of slope 
 $r+5$
 or more and no quotient bundles of slope
$r+5$
 or more and no quotient bundles of slope 
 $r+4$
 or less. We will prove this by induction on r, using the case
$r+4$
 or less. We will prove this by induction on r, using the case 
 $r = 7$
 discussed above as our base case. Our argument will consist of two steps:
$r = 7$
 discussed above as our base case. Our argument will consist of two steps:
- 
1. We specialize so that the statement of the proposition becomes false but still close enough to true that we can gather information about the possible limits of subbundles of large slope (respectively quotient bundles of small slope). 
- 
2. Leveraging this information, we apply Proposition 13.13 to the general fiber. 
 Our specialization will be of R to the union 
 $R^\circ = \overline {pq} \cup R^-$
, of a one-secant line
$R^\circ = \overline {pq} \cup R^-$
, of a one-secant line 
 $\overline {pq}$
 and an
$\overline {pq}$
 and an 
 $(r-1)$
-secant rational curve
$(r-1)$
-secant rational curve 
 $R^-$
 of degree
$R^-$
 of degree 
 $r-2$
 meeting
$r-2$
 meeting 
 $\overline {p q}$
 at a single point. Projection from
$\overline {p q}$
 at a single point. Projection from 
 $\overline {pq}$
 induces an exact sequence
$\overline {pq}$
 induces an exact sequence 

The bundle S is perfectly balanced of slope 
 $r+4$
. The bundle Q is a twist of another instance of our problem in
$r+4$
. The bundle Q is a twist of another instance of our problem in 
 $\mathbb {P}^{r-2}$
. If
$\mathbb {P}^{r-2}$
. If 
 $r \geq 11$
, then by induction, every JH-factor of Q has slope strictly between
$r \geq 11$
, then by induction, every JH-factor of Q has slope strictly between 
 $r+4$
 and
$r+4$
 and 
 $r+5$
; if
$r+5$
; if 
 $r=9$
, then every JH-factor of Q has slope exactly
$r=9$
, then every JH-factor of Q has slope exactly 
 $r+5$
.
$r+5$
.
 We begin by showing 
 has no quotient bundles of slope
 has no quotient bundles of slope 
 $r+4$
 or less. Since Q has no quotient bundles of slope
$r+4$
 or less. Since Q has no quotient bundles of slope 
 $r + 4$
 or less, any such quotient must specialize to a quotient of S, and therefore must have slope exactly
$r + 4$
 or less, any such quotient must specialize to a quotient of S, and therefore must have slope exactly 
 $r + 4$
 and rank at most
$r + 4$
 and rank at most 
 $2$
. Let G be the maximal such quotient of
$2$
. Let G be the maximal such quotient of 
 (i.e., on the general fiber). Our above specialization of R shows that
 (i.e., on the general fiber). Our above specialization of R shows that 
 $\mu (G) = r + 4$
 and
$\mu (G) = r + 4$
 and 
 . The determinant
. The determinant 
 $\det G$
 depends on the following data:
$\det G$
 depends on the following data:
- 
• A line bundle  $\mathcal {O}_E(1)$
. $\mathcal {O}_E(1)$
.
- 
• A basis for  $H^0(\mathcal {O}_E(1))$
. $H^0(\mathcal {O}_E(1))$
.
- 
• A hyperplane  $H \subset \mathbb {P} H^0(\mathcal {O}_E(1))^\vee $
. $H \subset \mathbb {P} H^0(\mathcal {O}_E(1))^\vee $
.
- 
• A rational curve  $R \subset H$
 of degree $R \subset H$
 of degree $r - 1$
 passing through the $r - 1$
 passing through the $r + 1$
 points of $r + 1$
 points of $E \cap H$
. $E \cap H$
.
 Except for the choice of line bundle 
 $\mathcal {O}_E(1)$
, all of these data vary in a rational family. Since any map from a rational variety to an abelian variety is constant,
$\mathcal {O}_E(1)$
, all of these data vary in a rational family. Since any map from a rational variety to an abelian variety is constant, 
 $\det G$
 depends only on the choice of line bundle
$\det G$
 depends only on the choice of line bundle 
 $\mathcal {O}_E(1)$
. Extracting the determinant of G therefore gives a natural map
$\mathcal {O}_E(1)$
. Extracting the determinant of G therefore gives a natural map 
 $$\begin{align*}\operatorname{Pic}^{r+1} E \to \operatorname{Pic}^{n(r+4)}E.\end{align*}$$
$$\begin{align*}\operatorname{Pic}^{r+1} E \to \operatorname{Pic}^{n(r+4)}E.\end{align*}$$
Hence, by Proposition 13.13, we have 
 $(r+1) \mid n(r+4)$
, and therefore
$(r+1) \mid n(r+4)$
, and therefore 
 $(r + 1) \mid 3n$
. Since we have
$(r + 1) \mid 3n$
. Since we have 
 $3n \leq 6 < 10 \leq r + 1$
, we must have
$3n \leq 6 < 10 \leq r + 1$
, we must have 
 $3n = 0$
, that is,
$3n = 0$
, that is, 
 $n = 0$
 as desired.
$n = 0$
 as desired.
 We next show that 
 has no subbundles of slope
 has no subbundles of slope 
 $r+5$
 or more. If
$r+5$
 or more. If 
 $r> 9$
, then the specialization
$r> 9$
, then the specialization 
 has no such subbundle because S and Q do not. It therefore remains only to consider the case
 has no such subbundle because S and Q do not. It therefore remains only to consider the case 
 $r=9$
, in which every JH-factor of S has slope
$r=9$
, in which every JH-factor of S has slope 
 $13$
 and every JH-factor of Q has slope
$13$
 and every JH-factor of Q has slope 
 $14$
. Let G be the maximal such subbundle of
$14$
. Let G be the maximal such subbundle of 
 (i.e., on the general fiber). Our above specialization of R shows that
 (i.e., on the general fiber). Our above specialization of R shows that 
 $\mu (G) = 14$
 and
$\mu (G) = 14$
 and 
 . Extracting the determinant of G therefore gives a natural map
. Extracting the determinant of G therefore gives a natural map 
 $$\begin{align*}\operatorname{Pic}^{10} E \to \operatorname{Pic}^{14n} E.\end{align*}$$
$$\begin{align*}\operatorname{Pic}^{10} E \to \operatorname{Pic}^{14n} E.\end{align*}$$
Hence, by Proposition 13.13, we have that 
 $10 \mid 14n$
, so
$10 \mid 14n$
, so 
 $5 \mid n$
. If
$5 \mid n$
. If 
 $n = 0$
, we are done, so suppose that
$n = 0$
, we are done, so suppose that 
 $n = 5$
.
$n = 5$
.
 To obtain a contradiction, we analyze what happens in our specialization, in which G specializes to a subbundle 
 $G^\circ $
 of Q with slope
$G^\circ $
 of Q with slope 
 $14$
 and rank
$14$
 and rank 
 $5$
. We consider the determinant
$5$
. We consider the determinant 
 $\det [G^\circ (-p-q)]$
. A priori this depends only on
$\det [G^\circ (-p-q)]$
. A priori this depends only on 
 $\mathcal {O}_E(1)$
, p, and q (the remaining data vary in a rational family). In fact, we claim it depends only on
$\mathcal {O}_E(1)$
, p, and q (the remaining data vary in a rational family). In fact, we claim it depends only on 
 $$\begin{align*}\mathcal{O}_{E(0,0;2)}(1) = \mathcal{O}_E(1) - p - q.\end{align*}$$
$$\begin{align*}\mathcal{O}_{E(0,0;2)}(1) = \mathcal{O}_E(1) - p - q.\end{align*}$$
Indeed, 
 $\det [G^\circ (-p-q)]$
 is a product of JH-factors of
$\det [G^\circ (-p-q)]$
 is a product of JH-factors of 
 , which is a discrete set of possibilities once we fix
, which is a discrete set of possibilities once we fix 
 $\mathcal {O}_{E(0,0;2)}(1)$
 and some additional data varying in a rational family. As we fix
$\mathcal {O}_{E(0,0;2)}(1)$
 and some additional data varying in a rational family. As we fix 
 $\mathcal {O}_{E(0,0;2)}(1)$
 and these additional data, we may allow
$\mathcal {O}_{E(0,0;2)}(1)$
 and these additional data, we may allow 
 $\{p, q\}$
 to vary arbitrarily in
$\{p, q\}$
 to vary arbitrarily in 
 $E \times E$
 by Lemma 8.1. In this way, we obtain a map from
$E \times E$
 by Lemma 8.1. In this way, we obtain a map from 
 $E \times E$
 to this discrete set, which must therefore be constant because
$E \times E$
 to this discrete set, which must therefore be constant because 
 $E \times E$
 is connected. Therefore,
$E \times E$
 is connected. Therefore,
 $\det [G^\circ (-p-q)]$
 depends only on
$\det [G^\circ (-p-q)]$
 depends only on 
 $\mathcal {O}_{E(0,0;2)}(1)$
 plus these additional data varying in a rational family and thus only on
$\mathcal {O}_{E(0,0;2)}(1)$
 plus these additional data varying in a rational family and thus only on 
 $\mathcal {O}_{E(0,0;2)}(1)$
. The determinant of
$\mathcal {O}_{E(0,0;2)}(1)$
. The determinant of 
 $G^\circ (-p-q)$
 therefore gives a natural map
$G^\circ (-p-q)$
 therefore gives a natural map 
 $$\begin{align*}\operatorname{Pic}^8 E \to \operatorname{Pic}^{60} E.\end{align*}$$
$$\begin{align*}\operatorname{Pic}^8 E \to \operatorname{Pic}^{60} E.\end{align*}$$
This is a contradiction by Proposition 13.13 since 
 $8\nmid 60$
.
$8\nmid 60$
.
13.3 The case 
 $r = 7$
$r = 7$
 To handle this case, we will first have to study the restriction of equation (13.2) to E, which we do for arbitrary odd r. This is a filtration of 
 whose successive quotients are:
 whose successive quotients are: 
 $$\begin{align*}N_{E/S}(1), \quad N_{S/\Sigma_1}|_E \oplus N_{S/\Sigma_2}|_E, \quad \text{and} \quad \frac{N_E}{N_{E / \Sigma_1} + N_{E / \Sigma_2}}.\end{align*}$$
$$\begin{align*}N_{E/S}(1), \quad N_{S/\Sigma_1}|_E \oplus N_{S/\Sigma_2}|_E, \quad \text{and} \quad \frac{N_E}{N_{E / \Sigma_1} + N_{E / \Sigma_2}}.\end{align*}$$
We write 
 for the corresponding hyperplane class.
 for the corresponding hyperplane class.
Proposition 13.14. We have 
 $N_{E/S} \simeq \mathcal {O}_E(3 - n)$
 (and so
$N_{E/S} \simeq \mathcal {O}_E(3 - n)$
 (and so 
 $N_{E/S}(1) \simeq \mathcal {O}_E(4 - n)$
).
$N_{E/S}(1) \simeq \mathcal {O}_E(4 - n)$
).
Proof. We consider the sequence (not exact) of maps
 $$\begin{align*}[N_{E/S} \simeq N_f] \to N_{\overline{f}} \to \overline{f}^* N_{\overline{f}(E) / \mathbb{P}^1 \times \mathbb{P}^1}.\end{align*}$$
$$\begin{align*}[N_{E/S} \simeq N_f] \to N_{\overline{f}} \to \overline{f}^* N_{\overline{f}(E) / \mathbb{P}^1 \times \mathbb{P}^1}.\end{align*}$$
By inspection, both of these maps drop rank exactly at the points of E lying over the nodes of 
 $\overline {f}$
. Therefore,their Chern classes lie in a linear progression, that is,
$\overline {f}$
. Therefore,their Chern classes lie in a linear progression, that is, 
 $$ \begin{align*} c_1(N_{E/S}) &= 2 \cdot c_1(N_{\overline{f}}) - c_1(\overline{f}^* N_{\overline{f}(E) / \mathbb{P}^1 \times \mathbb{P}^1}) \\ &= -2 c_1(\overline{f}^* K_{\mathbb{P}^1 \times \mathbb{P}^1}) - c_1(\overline{f}^* \mathcal{O}_{\mathbb{P}^1 \times \mathbb{P}^1}(n + 1, n + 1)) \\ &= -2 (-2 H_1 - 2H_2) - (n + 1)(H_1 + H_2) \\ &= (3 - n)(H_1 + H_2). \end{align*} $$
$$ \begin{align*} c_1(N_{E/S}) &= 2 \cdot c_1(N_{\overline{f}}) - c_1(\overline{f}^* N_{\overline{f}(E) / \mathbb{P}^1 \times \mathbb{P}^1}) \\ &= -2 c_1(\overline{f}^* K_{\mathbb{P}^1 \times \mathbb{P}^1}) - c_1(\overline{f}^* \mathcal{O}_{\mathbb{P}^1 \times \mathbb{P}^1}(n + 1, n + 1)) \\ &= -2 (-2 H_1 - 2H_2) - (n + 1)(H_1 + H_2) \\ &= (3 - n)(H_1 + H_2). \end{align*} $$
Proposition 13.15. We have 
 $c_1(N_{S/\Sigma _1}|_E \oplus N_{S/\Sigma _2}|_E) = \mathcal {O}_E(3n - 3)$
.
$c_1(N_{S/\Sigma _1}|_E \oplus N_{S/\Sigma _2}|_E) = \mathcal {O}_E(3n - 3)$
.
Proof. As in the proof of Proposition 13.7, define the classes 
 $\gamma _j$
 and h in
$\gamma _j$
 and h in 
 $\operatorname {Pic} \Sigma _j$
 to be the class of one n-plane, and the restriction of the hyperplane class from
$\operatorname {Pic} \Sigma _j$
 to be the class of one n-plane, and the restriction of the hyperplane class from 
 $\mathbb {P}^{2n+1}$
, respectively. By adjunction,
$\mathbb {P}^{2n+1}$
, respectively. By adjunction, 
 $$\begin{align*}c_1(N_{E/\Sigma_j}) = -c_1(K_{\Sigma_j}|_E) = - \left((n - 1)\gamma_j - (n + 1) h\right) \cdot E.\end{align*}$$
$$\begin{align*}c_1(N_{E/\Sigma_j}) = -c_1(K_{\Sigma_j}|_E) = - \left((n - 1)\gamma_j - (n + 1) h\right) \cdot E.\end{align*}$$
Therefore,
 $$ \begin{align*} c_1(N_{S/\Sigma_1}|_E \oplus N_{S/\Sigma_2}|_E) &= c_1(N_{E/\Sigma_1}) + c_1(N_{E/\Sigma_2}) - 2c_1(N_{E/S}) \\ &= -(n - 1)(\gamma_1 + \gamma_2) \cdot E + 2(n + 1) (h \cdot E) - 2(3 - n)(H_1 + H_2) \\ &= -(n - 1)(H_1 + H_2) + 2(n + 1) (H_1 + H_2) - 2(3 - n)(H_1 + H_2) \\ &= (3n - 3)(H_1 + H_2). \end{align*} $$
$$ \begin{align*} c_1(N_{S/\Sigma_1}|_E \oplus N_{S/\Sigma_2}|_E) &= c_1(N_{E/\Sigma_1}) + c_1(N_{E/\Sigma_2}) - 2c_1(N_{E/S}) \\ &= -(n - 1)(\gamma_1 + \gamma_2) \cdot E + 2(n + 1) (h \cdot E) - 2(3 - n)(H_1 + H_2) \\ &= -(n - 1)(H_1 + H_2) + 2(n + 1) (H_1 + H_2) - 2(3 - n)(H_1 + H_2) \\ &= (3n - 3)(H_1 + H_2). \end{align*} $$
Proposition 13.16. We have 
 $\frac {N_E}{N_{E / \Sigma _1} + N_{E / \Sigma _2}} \simeq \mathcal {O}_E(2)$
.
$\frac {N_E}{N_{E / \Sigma _1} + N_{E / \Sigma _2}} \simeq \mathcal {O}_E(2)$
.
Proof. Since 
 $K_{\mathbb {P}^{2n + 1}} = \mathcal {O}_{\mathbb {P}^{2n + 1}}(-(2n + 2))$
, we have
$K_{\mathbb {P}^{2n + 1}} = \mathcal {O}_{\mathbb {P}^{2n + 1}}(-(2n + 2))$
, we have 
 $c_1(N_E) = \mathcal {O}_E(2n + 2)$
. Combined with the previous two propositions, this implies the statement of the proposition (
$c_1(N_E) = \mathcal {O}_E(2n + 2)$
. Combined with the previous two propositions, this implies the statement of the proposition (
 $(2n + 2) - (3 - n) - (3n - 3) = 2$
).
$(2n + 2) - (3 - n) - (3n - 3) = 2$
).
 We now take 
 $r = 7$
 (equivalently
$r = 7$
 (equivalently 
 $n = 3$
) and let p and q be points of
$n = 3$
) and let p and q be points of 
 $E \cap R$
. By Lemma 13.8, it suffices to show interpolation for
$E \cap R$
. By Lemma 13.8, it suffices to show interpolation for 

This bundle has slope 
 $12$
, so it suffices to show that for a general effective divisor D of degree
$12$
, so it suffices to show that for a general effective divisor D of degree 
 $12$
,
$12$
, 

Furthermore, 
 $I(8, 1, 7, 0, 1)$
 holds, so
$I(8, 1, 7, 0, 1)$
 holds, so 
 
, which implies that
 
. Call the unique section 
 $\sigma $
. If there is any point
$\sigma $
. If there is any point 
 $q \in E \cap R {\smallsetminus} \{p\}$
 for which
$q \in E \cap R {\smallsetminus} \{p\}$
 for which 
 $\sigma |_q \notin (N_{E /\Sigma _1} + N_{E/\Sigma _2})|_q$
, then we have proved the desired vanishing (13.6). We may therefore assume that at all points of
$\sigma |_q \notin (N_{E /\Sigma _1} + N_{E/\Sigma _2})|_q$
, then we have proved the desired vanishing (13.6). We may therefore assume that at all points of 
 $E \cap R {\smallsetminus} \{p\}$
, the value of
$E \cap R {\smallsetminus} \{p\}$
, the value of 
 $\sigma $
 lies in the subbundle
$\sigma $
 lies in the subbundle 
 $N_{E /\Sigma _1} + N_{E/\Sigma _2}$
. By Proposition 13.16, we have the exact sequence
$N_{E /\Sigma _1} + N_{E/\Sigma _2}$
. By Proposition 13.16, we have the exact sequence 

Since 
 $\deg \left (\mathcal {O}_E(2)(-D)(-E\cap R +p)\right ) = -3$
, we must have that
$\deg \left (\mathcal {O}_E(2)(-D)(-E\cap R +p)\right ) = -3$
, we must have that 
 $\sigma $
 comes from a section of
$\sigma $
 comes from a section of 

It therefore suffices to show that for some 
 $p \in E \cap R$
, this bundle has no global sections (or, equivalently since the degree is
$p \in E \cap R$
, this bundle has no global sections (or, equivalently since the degree is 
 $-3$
 such that
$-3$
 such that 
 $h^1 = 3$
). Using equation (13.7), we have
$h^1 = 3$
). Using equation (13.7), we have 
 
, so such a point 
 $p \in E \cap R$
 exists unless making positive modifications towards all points of
$p \in E \cap R$
 exists unless making positive modifications towards all points of 
 $ E \cap R$
 does not decrease the
$ E \cap R$
 does not decrease the 
 $h^1$
:
$h^1$
: 

Equivalently, we are done unless

Taking the sum of the two normal bundle exact sequences for 
 $S \hookrightarrow \Sigma _1$
 and
$S \hookrightarrow \Sigma _1$
 and 
 $S\hookrightarrow \Sigma _2$
 along E yields
$S\hookrightarrow \Sigma _2$
 along E yields 

Twisting down by D, the line subbundle 
 $N_{E/S}(2(E \cap R))(-D)$
 has degree
$N_{E/S}(2(E \cap R))(-D)$
 has degree 
 $4$
, and hence four global sections and vanishing
$4$
, and hence four global sections and vanishing 
 $H^1$
. The quotient twisted down by D (which has degree
$H^1$
. The quotient twisted down by D (which has degree 
 $0$
) must therefore also have four global section in order for equation (13.8) to hold. Furthermore, the two scrolls
$0$
) must therefore also have four global section in order for equation (13.8) to hold. Furthermore, the two scrolls 
 $\Sigma _1$
 and
$\Sigma _1$
 and 
 $\Sigma _2$
 are exchanged by monodromy because the two maps
$\Sigma _2$
 are exchanged by monodromy because the two maps 
 $f_i \colon E \to \mathbb {P}^1$
 in Lemma 13.1 have degree
$f_i \colon E \to \mathbb {P}^1$
 in Lemma 13.1 have degree 
 $n+1$
, which is not a multiple of
$n+1$
, which is not a multiple of 
 $2n + 2$
, and hence by Proposition 13.13 they cannot be individually naturally defined. Thus, the two rank
$2n + 2$
, and hence by Proposition 13.13 they cannot be individually naturally defined. Thus, the two rank 
 $2$
 bundles
$2$
 bundles 
 $N_{S/\Sigma _i}|_E(-D)$
 necessarily both have
$N_{S/\Sigma _i}|_E(-D)$
 necessarily both have 
 $2$
 sections. If
$2$
 sections. If 
 $N_{S/\Sigma _i}|_E$
 were indecomposable, then by the Atiyah classification, it would necessarily be an extension of a degree
$N_{S/\Sigma _i}|_E$
 were indecomposable, then by the Atiyah classification, it would necessarily be an extension of a degree 
 $12$
 line bundle M by itself. As long as
$12$
 line bundle M by itself. As long as 
 $\mathcal {O}_E(D) \not \simeq M$
, we would have
$\mathcal {O}_E(D) \not \simeq M$
, we would have 
 $h^0(N_{S/\Sigma _i}|_E(-D)) = 0$
. Therefore,
$h^0(N_{S/\Sigma _i}|_E(-D)) = 0$
. Therefore,
 $N_{S/\Sigma _i}|_E$
 is a direct sum of line bundles. Since
$N_{S/\Sigma _i}|_E$
 is a direct sum of line bundles. Since 
 $h^0(N_{S/\Sigma _i}|_E(-D)) = 2$
:
$h^0(N_{S/\Sigma _i}|_E(-D)) = 2$
: 
 $$\begin{align*}N_{S/\Sigma_i}|_E \simeq L_{i1} \oplus L_{i2}\qquad \text{where} \qquad \deg(L_{i1}) = 14 \ \text{and} \ \deg(L_{i2}) = 10.\end{align*}$$
$$\begin{align*}N_{S/\Sigma_i}|_E \simeq L_{i1} \oplus L_{i2}\qquad \text{where} \qquad \deg(L_{i1}) = 14 \ \text{and} \ \deg(L_{i2}) = 10.\end{align*}$$
Then 
 $\det [L_{11} \oplus L_{21}]$
 gives a natural map
$\det [L_{11} \oplus L_{21}]$
 gives a natural map 
 $$\begin{align*}\operatorname{Pic}^8 E \to \operatorname{Pic}^{28} E,\end{align*}$$
$$\begin{align*}\operatorname{Pic}^8 E \to \operatorname{Pic}^{28} E,\end{align*}$$
which is a contradiction by Proposition 13.13, since 
 $8 \nmid 28$
.
$8 \nmid 28$
.
A Code for Section 11

Acknowledgements
We thank Dan Abramovich, Atanas Atanasov, Izzet Coskun, Joe Harris, Gavril Farkas, Dave Jensen, Ravi Vakil, David Yang and the anonymous referees for many stimulating discussions about normal bundles of curves and/or comments on an earlier version of this manuscript. We particularly thank Izzet Coskun for many helpful discussions about the normal bundle of a canonical curve and, in particular, the Harder–Narasimhan filtration in the degeneration used in Section 13.
Competing interest
The authors have no competing interest to declare.
Financial support
During the preparation of this article, E.L. was supported by NSF grants DMS-1802908 and DMS-2200641, and I.V. was supported by NSF grants DMS-1902743 and DMS-2200655.
 
 



 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 


 
 

 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

 
 
 
 

 
 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

 
 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

 

 
 

 

 
 
 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

 
 

 

 

 
 

 
 
 
 
 

 
 
 
 

 
 
 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
