Proof of Lemma 3.6.

We may assume throughout that $|t - t_0|$ and $jn^{-1}$ are sufficiently small, for otherwise $|t - t_0|^3 + jn^{-1}$ is of order $1$ (and thus of $\operatorname {\mathrm {diam}} (\mathfrak {P}) \geq \gamma _{K - j} (t)$ ). Then, observe that $\big ( \gamma _K (s), s \big ) \in \mathfrak {A} (\mathfrak {P})$ for each s in a neighborhood of $t_0$ . Indeed, since $\nabla H^* (w, s) = (0, 0)$ for all $(w, s) \in \mathfrak {P} \setminus \mathfrak {L} (\mathfrak {P})$ sufficiently close to $(x_0, t_0)$ , we have $n H^* (w, s) = n H (x_0, t_0) = K$ for each $(w, s) \in \mathfrak {A} (\mathfrak {P})$ in a neighborhood of $(x_0, t_0)$ , implying $\big ( \gamma _K (s), s \big ) = (w, s) \in \mathfrak {A} (\mathfrak {P})$ .

Throughout this proof, set $\widetilde {x} = \gamma _{K - j} (t)$ . We first consider the case $t = t_0$ . Fix some $x \in [\widetilde {x}, x_0]$ , and abbreviate $f_0 = f_{t_0} (x_0)$ and $f = f_{t_0} (x)$ . We will approximately express f in terms of $f_0$ , and then we will use this with (3.1) to compare the classical locations $\gamma _{K - j} (t_0)$ and $\gamma _K (t_0) = x_0$ . To that end, the second part of Proposition 3.3 implies

$$ \begin{align*} Q_0 (f_0) = x_0 (f_0 + 1) - t_0 f_0; \qquad Q_0 (f) = x (f + 1) - t_0 f. \end{align*} $$

Subtracting these and applying a Taylor expansion yields

$$ \begin{align*} (f - f_0) Q_0' (f_0) + \displaystyle\frac{(f - f_0)^2}{2} Q_0" (f_0) + \mathcal{O} \big( |f - f_0|^3 \big) & = Q_0 (f) - Q_0 (f_0) \\ & = (f + 1) (x - x_0) + (x_0 - t_0) (f - f_0), \end{align*} $$

where the error depends on the first three derivatives of $Q_0$ at f, which is uniformly bounded if $(x_0, t_0)$ is bounded away from a singularity or tangency location of $\mathfrak {A} (\mathfrak {P})$ . Since the third part of Proposition 3.3 gives $Q_0' (f_0) = x_0 - t_0$ , we find that

$$ \begin{align*} (f - f_0)^2 = \displaystyle\frac{2 (f_0 + 1)}{Q_0" (f_0)} (x - x_0) + \mathcal{O} \big( |f - f_0|^3 \big). \end{align*} $$

In particular, $|f_0 - f| = \mathcal {O} \big ( |x - x_0|^{1/2} \big )$ and, more specifically,

(3.6) $$ \begin{align} f - f_0 = \bigg( \displaystyle\frac{2 (f_0 + 1)}{Q_0" (f_0)} \bigg)^{1/2} (x - x_0)^{1/2} + \mathcal{O} \big( |x - x_0|^{3 / 2} \big). \end{align} $$

Since $\gamma _{K - j} (t_0) = \widetilde {x} \leq x \leq x_0 \leq \gamma _K (t_0)$ , we have $(x - x_0)^{1/2} \in \mathrm {i} \mathbb {R}$ . Since moreover $f_0 \in \mathbb {R}$ , which implies $Q_0" (f_0) \in \mathbb {R}$ , we deduce

$$ \begin{align*} \arg^* f & = f_0^{-1} \operatorname{\mathrm{Im}} (f - f_0) + \mathcal{O} \big( |f - f_0|^3 + |x - x_0|^{3/2} \big) \\ & = \bigg( \displaystyle\frac{2 |f_0 + 1|}{f_0^2 \big| Q_0" (f_0) \big|} \bigg)^{1/2} (x_0 - x)^{1/2} + \mathcal{O} \big( |x_0 - x|^{3 / 2} \big). \end{align*} $$

In particular, since $n H^* (x_0, t_0) = K$ and $n H^* (\widetilde {x}, t_0) = K - j$ , this implies by Equation (3.1) that

$$ \begin{align*} \displaystyle\frac{j}{n} = H^* (x_0, t_0) - H^* (\widetilde{x}, t_0) & = \displaystyle\int_{\widetilde{x}}^{x_0} \partial_x H^* (w, t_0) \mathrm{d}w \\ & = \displaystyle\frac{1}{\pi} \displaystyle\int_{\widetilde{x}}^{x_0} \arg^* f_{t_0} (w) \mathrm{d} w \\ & = \bigg( \displaystyle\frac{8 |f_0 + 1|}{9 \pi^2 f_0^2 \big| Q_0" (f_0) \big|} \bigg)^{1 / 2} (x_0 - \widetilde{x})^{3/2} + \mathcal{O} \big( |x_0 - \widetilde{x}|^{5 / 2} \big). \end{align*} $$

Hence,

(3.7) $$ \begin{align} \gamma_K (t_0) - \gamma_{K - j} (t_0) = x_0 - \widetilde{x} = \bigg( \displaystyle\frac{f_0^2 \big| Q_0" (f_0) \big|}{2 |f_0 + 1|} \bigg)^{1/3} \bigg( \displaystyle\frac{3 \pi j}{2 n} \bigg)^{2/3} + \mathcal{O} (j n^{-1}), \end{align} $$

which by Lemma 3.7 and the definition of $\mathfrak {s}$ from (2.9) implies the lemma when $t = t_0$ .

If $t \ne t_0$ , then set $\widehat {x}_0 = \gamma _K (t) \in \mathfrak {A} (\mathfrak {P})$ and $\widehat {f}_0 = f_t (\widehat {x}_0)$ . Then, the same reasoning as used to deduce Equation (3.7) implies

$$ \begin{align*} \gamma_K (t) - \gamma_{K - j} (t) & = \Bigg( \displaystyle\frac{\widehat{f}_0^2 \big| Q_0" (\widehat{f}_0)\big|}{2 |\widehat{f}_0 + 1|} \Bigg)^{1/3} \bigg( \displaystyle\frac{3 \pi j}{2 n} \bigg)^{2/3} + \mathcal{O} (jn^{-1}) \\ & = \bigg( \displaystyle\frac{f_0^2 \big| Q_0" (f_0) \big|}{2 |f_0 + 1|} \bigg)^{1/3} \bigg( \displaystyle\frac{3 \pi j}{2 n} \bigg)^{2/3} + \mathcal{O} \big( |t - t_0| j^{2/3} n^{-2/3} + jn^{-1} \big), \end{align*} $$

where in the last equality we used the fact that $f_t$ is uniformly smooth in t around $t_0$ along $\mathfrak {A} (\mathfrak {P})$ . By Equation (2.7), it follows that

$$ \begin{align*} \gamma_{K - j} (t) & = \gamma_K (t) - \bigg( \displaystyle\frac{f_0^2 \big| Q_0" (f_0) \big|}{2 |f_0 + 1|} \bigg)^{1/3} \bigg( \displaystyle\frac{3 \pi j}{2 n} \bigg)^{2/3} + \mathcal{O} \big( |t - t_0| j^{2/3} n^{-2/3} + jn^{-1} \big) \\ & = \gamma_K (t_0) + \mathfrak{l} (t - t_0) + \mathfrak{q} (t - t_0)^2 - \bigg( \displaystyle\frac{f_0^2 \big| Q_0" (f_0) \big|}{2 |f_0 + 1|} \bigg)^{1/3} \bigg( \displaystyle\frac{3 \pi j}{2 n} \bigg)^{2/3} + \mathcal{O} \big( jn^{-1} + |t - t_0|^3 \big), \end{align*} $$

which implies the lemma due to Lemma 3.7 and Equation (2.9) again.

Proof. We first show that Theorem 3.10 implies, for any t in a sufficiently small (independent of n) neighborhood of $t_0$ , that with overwhelming probability we have

(3.8) $$ \begin{align} \gamma_{K - j - n^{\delta} + 1} (t) - n^{-1} \leq n^{-1} \mathsf{x}_{K - j + 1} (tn) \leq \min \big\{ \gamma_{K - j + n^{\delta} + 1} (t), \gamma_K (t) + n^{\delta / 2 - 2/3} \}, \end{align} $$

where we recall the classical locations $\gamma _i(t)$ from Definition 3.5 (and we assume that $tn \in \mathbb {Z}$ for notational convenience). Let us only show the second bound in Equation (3.8), for the proof of the first is entirely analogous. Then, from the bijection between tilings and nonintersecting Bernoulli walk ensembles described in Section 2.2, we have $n^{-1} (\mathsf {x}_{K - j + 1} (tn)+1) \leq x$ if and only if $\mathsf {H} (xn, tn) \geq K - j + 1$ .

So, setting $\gamma = \gamma _{K - j + n^{\delta } + 1} (t)$ , the first part of Theorem 3.10 implies with overwhelming probability that $\mathsf {H} (\gamma n, tn) \geq n H^* (\gamma n, tn) - n^{\delta } = K - j + 1$ . Hence, $\mathsf {x}_{K - j + 1} (tn) \leq \gamma _{K - j + n^{\delta } + 1} (t)$ holds with overwhelming probability. Moreover, denoting $x^{\prime } = \gamma _K (t) + N^{\delta / 2 - 2/3}$ , we have by the second part of Theorem 3.10 that $\mathsf {H} (x^{\prime } n, tn) = n H^* (x^{\prime }, t) = n H^* \big ( \gamma _K (t), t \big ) = K$ with overwhelming probability, where in the second equality we used the fact that $\nabla H^* (x, t) = (0, 0)$ for $(x, t)$ in a neighborhood of $(x_0, t_0)$ to the right of $\mathfrak {A}$ . Hence, $n^{-1} \mathsf {x}_{K - j + 1} (t) \leq n^{-1} \mathsf {x}_K (t) \leq x^{\prime } = \gamma _K (t) + N^{\delta / 2 - 2/3}$ with overwhelming probability. This confirms Equation(3.8).

Now, Equation (3.8) and Lemma 3.6 together imply that

(3.9) $$ \begin{align} n^{-1} \mathsf{x}_{K - j + 1} (tn) & = x_0 + \mathfrak{l} (t - t_0) + \mathfrak{q} (t - t_0)^2 - \mathfrak{s}^{3 / 2} \Big( \displaystyle\frac{3 \pi j}{2n} \Big)^{2/3} \nonumber\\ & \quad + \mathcal{O} \big( j n^{-1} + |t - t_0|^3 + j^{-1/3} n^{\delta - 2/3} \big) \end{align} $$

holds for each $j \in \mathbb {Z}$ and $t \in \mathbb {R}$ , with overwhelming probability. Here, we have also used the fact that Lemma 3.6 implies the classical locations $\gamma _j (t)$ (from Definition 3.5) with respect to $\mathfrak {P}$ satisfy $\gamma _j (t) - \gamma _{j - n^{\delta }} (t) = \mathcal {O} (j^{-1/3} n^{- 2/3})$ . Since $\delta \in \big ( 0, 1/100 \big )$ and $j \in [1, 2 n^{10 \delta }]$ , we have $jn^{-1} + |t - t_0|^3 + j^{-1/3} n^{\delta - 2/3} \leq 3 j^{-1/3} n^{\delta - 2/3}$ for $j \in [1, 2n^{10 \delta }]$ and $|t - t_0| \leq n^{20 \delta - 2/3}$ . Setting $s = (t - t_0) n$ in (3.9) then yields the corollary.

Proof. Throughout, we assume $\mathfrak {q}> 0$ , as the proof when $\mathfrak {q} < 0$ is entirely analogous. This proposition will follow from a comparison between the random Bernoulli walk ensemble $\mathsf {X}$ and the one associated with a random tiling of a suitably chosen hexagon. So, we begin by identifying real numbers $a, b, c> 0$ and a point $(x_0, t_0) \in \mathfrak {A} (a, b, c) = \partial \mathfrak {L}_{a, b, c}$ on the ellipse inscribed in the hexagon $\mathfrak {E}_{a, b, c}$ (recall Definition 3.12) whose curvature parameters are given by $(\mathfrak {l}, \mathfrak {q})$ . To that end, first observe that there are two points on $\mathfrak {A} (1, 1, 1)$ at which a line with inverse slope $\mathfrak {l}$ is tangent. Let $(\widetilde {x}, \widetilde {t}) \in \mathfrak {A} (1, 1, 1)$ be the one whose curvature parameters $(\widetilde {\mathfrak {l}}, \widetilde {\mathfrak {q}}) = (\mathfrak {l}, \widetilde {\mathfrak {q}})$ are such that $\operatorname {\mathrm {sgn}} \widetilde {\mathfrak {q}} = \operatorname {\mathrm {sgn}} \mathfrak {q} = 1$ . Since $\widetilde {\mathfrak {l}} = \mathfrak {l}$ is bounded away from $0$ and $\infty $ , as is $\widetilde {\mathfrak {q}}$ . Then setting $r = \widetilde {\mathfrak {q}} \mathfrak {q}^{-1}$ , $(a, b, c) = (r, r, r)$ and $(x_0, t_0) = (r \widetilde {x}, r \widetilde {t}) \in \mathfrak {A} (a, b, c)$ , the curvature parameters at $(x_0, t_0)$ are $(\mathfrak {l}, \mathfrak {q})$ .

We will now perturb the quadratic curvature parameter of $(x_0, t_0)$ with respect to ${\mathfrak {A}} (\mathfrak {E}_{a, b, c})$ through scaling by $\kappa $ and $\nu $ , where

$$ \begin{align*} \kappa = 1 + n^{-20 \delta}; \qquad \nu = 1 - n^{-20 \delta}. \end{align*} $$

Set $(a^{\prime }, b^{\prime }, c^{\prime }) = (\kappa ^{-1} a, \kappa ^{-1} b, \kappa ^{-1} c)$ and $(a^{\prime \prime }, b^{\prime \prime }, c^{\prime \prime }) = (\nu ^{-1} a, \nu ^{-1} b, \nu ^{-1} c)$ ; further, let $(x_0^{\prime }, t_0^{\prime }) = (\kappa ^{-1} x_0, \kappa ^{-1} t_0)$ and $(x_0^{\prime \prime }, t_0^{\prime \prime }) = (\nu ^{-1} x_0, \nu ^{-1} t_0)$ . Observe that the curvature parameters at $(x_0^{\prime }, t_0^{\prime })$ with respect to $\mathfrak {A} (a^{\prime }, b^{\prime }, c^{\prime })$ and at $(x_0^{\prime \prime }, t_0^{\prime \prime })$ with respect to $\mathfrak {A} (a^{\prime \prime }, b^{\prime \prime }, c^{\prime \prime })$ are given by $(\mathfrak {l}^{\prime }, \mathfrak {q}^{\prime }) = (\mathfrak {l}, \kappa \mathfrak {q})$ and $(\mathfrak {l}^{\prime \prime }, \mathfrak {q}^{\prime \prime }) = (\mathfrak {l}, \nu \mathfrak {q})$ , respectively.

Next, define $(\mathsf {a}^{\prime }, \mathsf {b}^{\prime }, \mathsf {c}^{\prime }) = (na^{\prime }, nb^{\prime }, nc^{\prime })$ and $(\mathsf {a}^{\prime \prime }, \mathsf {b}^{\prime \prime }, \mathsf {c}^{\prime \prime }) = (na^{\prime \prime }, nb^{\prime \prime }, nc^{\prime \prime })$ , which we assume for notational simplicity are integers. Denote the hexagons $\mathsf {P}^{\prime } = \mathfrak {E}_{\mathsf {a}^{\prime }, \mathsf {b}^{\prime }, \mathsf {c}^{\prime }}$ and $\mathsf {P}^{\prime \prime } = \mathfrak {E}_{\mathsf {a}^{\prime \prime }, \mathsf {b}^{\prime \prime }, \mathsf {c}^{\prime \prime }}$ ; let $\mathscr {M}^{\prime }$ and $\mathscr {M}^{\prime \prime }$ denote uniformly random tilings of $\mathsf {P}^{\prime }$ and $\mathsf {P}^{\prime \prime }$ , which are associated with nonintersecting Bernoulli walk ensembles $\mathsf {Y}^{\prime } = (\mathsf {y}_1^{\prime }, \mathsf {y}_2^{\prime }, \ldots , \mathsf {y}_{\mathsf {a}^{\prime }}^{\prime })$ and $\mathsf {Y}^{\prime \prime } = (\mathsf {y}_1^{\prime \prime }, \mathsf {y}_1^{\prime \prime }, \ldots , \mathsf {y}_{\mathsf {a}^{\prime \prime }}^{\prime \prime })$ , respectively. Define the Bernoulli walk ensembles $\mathsf {X}^{\prime } = (\mathsf {x}_1^{\prime }, \mathsf {x}_2^{\prime }, \ldots , \mathsf {x}_m^{\prime })$ and $\mathsf {X}^{\prime \prime } = (\mathsf {x}_1^{\prime \prime }, \mathsf {x}_2^{\prime \prime }, \ldots , \mathsf {x}_m^{\prime \prime })$ through a spacial and index shift of $\mathsf {Y}^{\prime }$ and $\mathsf {Y}^{\prime \prime }$ , respectively; specifically, for each j and t, set

$$ \begin{align*} \mathsf{x}_j^{\prime} (t) = \mathsf{y}_{\mathsf{a}^{\prime} + j - m }^{\prime} (t + t_0^{\prime} n) - x_0^{\prime} n + 3 n^{1/3 - \delta}; \qquad \mathsf{x}_j^{\prime\prime} (t) = \mathsf{y}_{\mathsf{a}^{\prime\prime} + j - m}^{\prime\prime} (t + t_0^{\prime\prime} n) - x_0^{\prime\prime} n - 3 n^{1/3 - \delta}. \end{align*} $$

Given this notation, we will first use Lemma 3.15 to bound the ensemble $\mathsf {X}$ between $\mathsf {X}'$ and $\mathsf {X}^{\prime \prime }$ ; see Figure 8. Then, we will apply Proposition 3.13 to show that $(\mathsf {X}', \mathsf {X}^{\prime \prime })$ converges to the same Airy line ensemble under the normalization (3.13). To implement the former, define the sequences $\mathsf {d}', \mathsf {e}', \mathsf {d}^{\prime \prime }, \mathsf {e}^{\prime \prime } \in \mathbb {Z}^m$ by

$$ \begin{align*} & \mathsf{d}' = \big( \mathsf{x}_1' (-\mathsf{T}), \mathsf{x}_2' (-\mathsf{T}), \ldots , \mathsf{x}_m' (-\mathsf{T}) \big); \quad \mathsf{e}' = \big( \mathsf{x}_1' (\mathsf{T}), \mathsf{x}_2' (\mathsf{T}), \ldots , \mathsf{x}_m' (\mathsf{T}) \big); \\ & \mathsf{d}^{\prime\prime} = \big( \mathsf{x}_1^{\prime\prime} (-\mathsf{T}), \mathsf{x}_2^{\prime\prime} (-\mathsf{T}), \ldots , \mathsf{x}_m^{\prime\prime} (-\mathsf{T}) \big); \quad \mathsf{e}^{\prime\prime} = \big( \mathsf{x}_1^{\prime\prime} (\mathsf{T}), \mathsf{x}_2^{\prime\prime} (\mathsf{T}), \ldots , \mathsf{x}_m^{\prime\prime} (\mathsf{T}) \big), \end{align*} $$

Figure 8 Shown above trajectories for the paths $\mathsf {x}_j' \leq \mathsf {x}_j \leq \mathsf {x}_j"$ in the proof of Proposition 3.18; they approximately coincide in the shaded region.

and the functions $\mathsf {f}', \mathsf {f}^{\prime \prime } : [-\mathsf {T}, \mathsf {T}] \rightarrow \mathbb {Z}$ by

$$ \begin{align*} \mathsf{f}' (t) = \mathsf{x}_0' (t); \qquad \mathsf{f}^{\prime\prime} (t) = \mathsf{x}_0^{\prime\prime} (t). \end{align*} $$

Then, by the Gibbs property described in Remark 3.17, $\mathsf {X}'$ is a uniformly random nonintersecting Bernoulli walk ensemble with entrance data $\mathsf {d}'$ , exit data $\mathsf {e}'$ and boundary conditions $(\mathsf {x}_0'; \infty )$ ; a similar statement holds for $\mathsf {X}^{\prime \prime }$ . Let us show with overwhelming probability that

(3.14) $$ \begin{align} \mathsf{d}^{\prime\prime} \leq \mathsf{d} \leq \mathsf{d}'; \qquad \mathsf{e}^{\prime\prime} \leq \mathsf{e} \leq \mathsf{e}'; \qquad \mathsf{f}^{\prime\prime} \leq \mathsf{f} \leq \mathsf{f}'. \end{align} $$

To verify the first statement of Equation (3.14), observe for any $j \in [1, m]$ that Equations (3.11) and (3.12) imply

(3.15) $$ \begin{align} \mathsf{d}_j \leq \mathfrak{q} n^{-1} \mathsf{T}^2 - \mathfrak{l} \mathsf{T} + 5 (\mathfrak{s}^2 + 1) n^{1/3 + 10 \delta} \leq \mathfrak{q}' n^{-1} \mathsf{T}^2 - \mathfrak{l} \mathsf{T} - 5 (\mathfrak{s}^2 + 1) n^{1/3 + 10 \delta}. \end{align} $$

To deduce the first bound, we used the facts that $n^{1/3 - \delta } \leq \mathfrak {s}^{-3/2} K_0 \leq 3 m^{2/3} n^{1/3} \leq n^{1/3 + 10 \delta }$ since $m = \lfloor n^{10 \delta } \rfloor $ ; to deduce the second, we used the facts that $\mathfrak {q}' = \kappa \mathfrak {q} = (1 + n^{-20 \delta }) \mathfrak {q}$ and $\mathsf {T} = \lfloor n^{2/3 + 20 \delta } \rfloor $ .

Moreover, Corollary 3.11 (applied with the $\mathsf {P}$ there equal to $\mathsf {P}'$ here) implies with overwhelming probability that

(3.16) $$ \begin{align} \mathsf{x}_{m - j}^{\prime} (-\mathsf{T}) = \mathsf{y}_{\mathsf{a}^{\prime} - j}^{\prime} (t_0 n - \mathsf{T}) - x_0^{\prime} n + 3 n^{1/3 - \delta} \geq \mathfrak{q}^{\prime} n^{-1} \mathsf{T}^2 - \mathfrak{l} \mathsf{T} - 5 (\mathfrak{s}^2 + 1) n^{1/3 + 10 \delta}, \end{align} $$

where we have again used the facts that $n^{1/3 - \delta } \leq \mathfrak {s}^{-3/2} \leq K_0 \leq 3 m^{2/3} n^{1/3} \leq n^{1/3 + 10 \delta }$ . Combining Equations (3.15) and (3.16), it follows that $\mathsf {d} \leq \mathsf {d}^{\prime }$ . The proof that $\mathsf {d}^{\prime \prime } \leq \mathsf {d}$ is entirely analogous, thereby establishing the first statement of Equation (3.14); the second is shown similarly.

The third statement of Equation (3.14) follows from the fact that for any $t \in [-\mathsf {T}, \mathsf {T}]$ we have

(3.17) $$ \begin{align} \mathsf{f}^{\prime} (t) = \mathsf{x}_0^{\prime} (t) & = \mathsf{y}_{\mathsf{a}^{\prime}- m}^{\prime} (t + t_0^{\prime} n) - x_0^{\prime} n + 3 n^{1/3 - \delta} \nonumber\\ & \geq \mathfrak{l} t + \mathfrak{q}^{\prime} n^{-1} t^2 - K_0^{\prime} - m^{-1/3} n^{1/3 + \delta} + 3 n^{1/3 - \delta} \\ & \nonumber\geq \mathsf{f} (t) + K_0 - K_0^{\prime} + (\mathfrak{q}^{\prime} - \mathfrak{q}) n^{-1} t^2 + n^{1/3 - \delta} \geq \mathsf{f} (t), \end{align} $$

where we have set

$$ \begin{align*} K_0^{\prime} = \mathfrak{s}^{\prime3/2} n^{1/3} \Big( \displaystyle\frac{3 \pi m}{2} \Big)^{2/3} = \kappa^{-1/2} K_0, \end{align*} $$

and we have denoted

(3.18) $$ \begin{align} \mathfrak{s}^{\prime} = \Big| \displaystyle\frac{\mathfrak{l}^{2/3} (1 - \mathfrak{l})^{2/3}}{4^{1/3} \mathfrak{q}^{\prime1/3}} \Big|; \quad \mathfrak{r}^{\prime} = \Big| \displaystyle\frac{\mathfrak{l}^{1/3} (1 - \mathfrak{l})^{1/3}}{2^{1/3} \mathfrak{q}^{\prime1/3}} \Big|. \end{align} $$

The first and second statements of Equation (3.17) follow from the definitions of $\mathsf {f}^{\prime }$ and $\mathsf {x}^{\prime }$ , the third from Corollary 3.11 (applied with j there equal to m here), the fourth from Equation (3.11) and the definition $m = \lfloor n^{10 \delta } \rfloor $ and the fifth from the facts that $\mathfrak {q}^{\prime } = \kappa \mathfrak {q} \geq \mathfrak {q}$ and that

$$ \begin{align*} |K_0 - K_0^{\prime}| \leq |1 - \kappa^{-1/2}| K_0 \leq n^{-10 \delta} K_0 \leq 5 \mathfrak{s}^{3/2} n^{1/3 -10 \delta} m^{2/3} = \mathrm{o}(n^{1/3 - \delta}), \end{align*} $$

as $m \leq n^{10 \delta }$ . Hence, $\mathsf {f} \leq \mathsf {f}^{\prime }$ ; similarly, $\mathsf {f}^{\prime \prime } \leq \mathsf {f}$ . This verifies that Equation (3.14) holds with overwhelming probability. So, it follows from Lemma 3.15 that there exists a coupling between $(\mathsf {X}, \mathsf {X}^{\prime }, \mathsf {X}^{\prime \prime })$ such that $\mathsf {x}_j^{\prime \prime } \leq \mathsf {x}_j \leq \mathsf {x}_j^{\prime }$ holds for each $j \in [1, m]$ , with overwhelming probability.

Define normalizations of these Bernoulli walk ensembles, denoted $\mathcal {X}_n^{\prime } = (\mathsf {X}_1^{\prime }, \mathsf {X}_2^{\prime }, \ldots , \mathsf {X}_m^{\prime })$ , $\mathcal {X}_n^{\prime \prime } = (\mathsf {X}_1^{\prime \prime }, \mathsf {X}_2^{\prime \prime }, \ldots , \mathsf {X}_n^{\prime \prime })$ , $\mathcal {W}_n' = (\mathsf {W}_1', \mathsf {W}_2', \ldots , \mathsf {W}_m')$ and $\mathcal {W}_n^{\prime \prime } = (\mathsf {W}_1^{\prime \prime }, \mathsf {W}_2^{\prime \prime }, \ldots , \mathsf {W}_m^{\prime \prime })$ by (recall the notation from Equation (3.18)) setting

$$ \begin{align*} \mathsf{X}_i^{\prime} (t) &= \mathfrak{s}^{-1} n^{-1/3} \Big( \mathsf{x}_{m - i + 1}' (\mathfrak{r} n^{2/3} t) - \mathfrak{l} n^{2/3} t\Big); \qquad \mathsf{X}_i^{\prime\prime} (t) = \mathfrak{s}^{-1} n^{-1/3} \Big( \mathsf{x}_{m -i + 1} (\mathfrak{r} n^{2/3} t) - \mathfrak{l} n^{2/3} t \Big); \\ \mathsf{W}_i^{\prime} (t) &= \mathfrak{s}^{\prime-1} {n^{-1/3}} \Big( \mathsf{x}_{m - i + 1}' (\mathfrak{r}^{\prime} {n^{2/3}} t) - {\mathfrak{l}} {n^{2/3}} t\Big); \qquad \mathsf{W}_i^{\prime\prime} (t) = \mathfrak{s}^{\prime\prime-1} {n^{-1/3}} \Big( \mathsf{x}_{m - i + 1} (\mathfrak{r}^{\prime\prime} {n^{2/3}} t) - \mathfrak{l} {n^{2/3}} t \Big). \end{align*} $$

Then, Proposition 3.13 implies that $\mathcal {W}_n^{\prime }$ and $\mathcal {W}_n^{\prime \prime }$ converge to $\mathcal {R}$ , uniformly on compact subsets of $\mathbb {Z}_{> 0} \times \mathbb {R}$ , as n tends to $\infty $ . Since Equation (3.18) and the facts that $(\mathfrak {q}, \mathfrak {q}') = (\kappa \mathfrak {q}, \nu \mathfrak {q})$ and $\kappa - 1 = 1 - \nu = n^{-20 \delta } = \mathrm {o}(1)$ imply $\mathfrak {s}' = \mathfrak {s} \big ( 1 + \mathrm {o}(1) \big )$ , $\mathfrak {r}' = \mathfrak {r} \big ( 1 + \mathrm {o}(1) \big )$ , $\mathfrak {s}^{\prime \prime } = \mathfrak {s} \big ( 1 + \mathrm {o}(1) \big )$ and $\mathfrak {r}^{\prime \prime } = \mathfrak {r} \big ( 1 + \mathrm {o}(1) \big )$ , we deduce that $\big | \mathsf {W}_j^{\prime } (t) - \mathsf {X}_j' (t) \big |$ and $\big | \mathsf {W}_j^{\prime \prime } (t) - \mathsf {X}_j^{\prime \prime } (t) \big | = \mathrm {o}(1)$ for each $j \in [1, m]$ , uniformly on compact subsets of $\mathbb {Z}_{> 0} \times \mathbb {R}$ . Hence, $\mathcal {X}_n'$ and $\mathcal {X}_n^{\prime \prime }$ both converge to $\mathcal {R}$ . Since $\mathsf {x}_j^{\prime \prime } \leq \mathsf {x}_j \leq \mathsf {x}_j^{\prime \prime }$ , it follows that $\mathsf {X}_j^{\prime \prime } \leq \mathsf {X}_j \leq \mathsf {X}_j'$ , and thus the same convergence holds for $\mathcal {X}_n$ .

Proof of Theorem 2.10.

This will follow from Proposition 3.18. Fix a real number $\delta \in \big ( 0, 1/100 \big )$ , and define $m = n^{10 \delta }$ and $\mathsf {T} = n^{2/3 + 20 \delta }$ (as in Proposition 3.18), which we assume for notational convenience are integers. Define the nonintersecting Bernoulli walk ensemble $\mathsf {Y} = (\mathsf {y}_1, \mathsf {y}_2, \ldots , \mathsf {y}_m)$ by

(3.19) $$ \begin{align} \mathsf{y}_i (t) = \mathsf{x}_{K + i - m} (t_0 n + t) - x_0 n. \end{align} $$

Denoting the sequences $\mathsf {d}, \mathsf {e} \in \mathbb {Z}^m$ and function $\mathsf {f} : [-\mathsf {T}, \mathsf {T}] \rightarrow \mathbb {Z}$ by

$$ \begin{align*} \mathsf{d} = \big( \mathsf{y}_1 (- \mathsf{T}), \mathsf{y}_2 (-\mathsf{T}), & \ldots , \mathsf{y}_m (-\mathsf{T}) \big); \qquad \mathsf{e} = \big( \mathsf{y}_1 (\mathsf{T}), \mathsf{y}_2 (\mathsf{T}), \ldots , \mathsf{y}_m (\mathsf{T}) \big); \\ & \mathsf{f} (t) = \mathsf{x}_{K - m} (t_0 n + t) - x_0 n, \end{align*} $$

it follows from the Gibbs property described in Remark 3.17 that $\mathsf {Y}$ is a uniformly random nonintersecting Bernoulli walk ensemble with entrance and exit data $(\mathsf {d}; \mathsf {e})$ and boundary conditions $(\mathsf {f}; \infty )$ . Let us verify that $(\mathsf {d}, \mathsf {e}, \mathsf {f})$ satisfy Equations (3.12) and (3.11).

To that end, since $m = n^{10 \delta }$ , the $j = m$ case of Corollary 3.11 implies with overwhelming probability that

(3.20) $$ \begin{align} \displaystyle\max_{s \in [-\mathsf{T}, \mathsf{T}]} \bigg| \mathsf{f} (s) - \mathfrak{l} s - \mathfrak{q} n^{-1} s^2 + \mathfrak{s}^{3/2} \Big( \displaystyle\frac{3 \pi m}{2} \Big)^{3/2} n^{1/3} \bigg| < m^{-1/3} n^{1/3 + \delta} < n^{1/3 - \delta}, \end{align} $$

and so $\mathsf {f}$ satisfies Equation (3.11). Applying Corollary 3.11 with $j \in [0, m - 1]$ and also using Equation (3.20) yields

$$ \begin{align*} \big| \mathsf{y}_1 (-\mathsf{T}) - f(-\mathsf{T}) \big| \leq 2 \mathfrak{s}^{3/2} \Big( \displaystyle\frac{3 \pi m}{2} \Big)^{2/3} n^{1/3} + 2n^{1/3} \leq n^{1/3 + 10 \delta}, \end{align*} $$

where to deduce the last inequality we used the fact that $m = n^{10 \delta }$ . This verifies that $\mathsf {d}$ satisfies (3.12) with overwhelming probability; the proof that $\mathsf {e}$ does as well is very similar and thus omitted.

Hence, Proposition 3.18 applies and gives that $\mathcal {Y}_n = (\mathsf {Y}_1, \mathsf {Y}_2, \ldots , \mathsf {Y}_m)$ converges to $\mathcal {R}$ , uniformly on compact subsets of $\mathbb {Z}_{> 0} \times \mathbb {R}$ , as n tends to $\infty $ , where

$$ \begin{align*} \mathsf{Y}_i (t) = \mathfrak{s}^{-1} n^{-1/3} \Big( \mathsf{y}_{m - i + 1} (\mathfrak{r} n^{2/3} t) - \mathfrak{l} n^{2/3} t \Big). \end{align*} $$

By Equation (3.19), $\mathsf {Y}_i (t) = \mathsf {X}_i (t)$ , meaning that $\mathcal {Y}_n = \mathcal {X}_n$ , so the same convergence holds for $\mathcal {X}_n$ .

Assumption 4.1. Assume the following constraints hold.

1. (1) The boundary height function h is constant along both $\partial _{\operatorname {\mathrm {ea}}} (\mathfrak {D})$ and $\partial _{\operatorname {\mathrm {we}}} (\mathfrak {D})$ .

2. (2) Either $\partial _{\operatorname {\mathrm {no}}} (\mathfrak {D})$ is packed with respect to h or there exists $\mathfrak {t}'> \mathfrak {t}_2$ such that $H^*$ admits an extension to time $\mathfrak {t}'$ .

3. (3) There exists $\widetilde {\mathfrak {t}} \in [\mathfrak {t}_1, \mathfrak {t}_2)$ such that the following holds. For $t \in [\mathfrak {t}_1, \widetilde {\mathfrak {t}})$ , the set $I_t$ consists of two nonempty disjoint intervals, and for $t \in [\widetilde {\mathfrak {t}}, \mathfrak {t}_2]$ , the set $I_t$ consists of one nonempty interval.Footnote ⁸

4. (4) Any tangency location along $\partial \overline {\mathfrak {L}}$ is of the form $\min I_t$ or $\max I_t$ , for some $t \in (\mathfrak {t}_1, \mathfrak {t}_2)$ . At most one tangency location is of the form $\min I_t$ , and at most one is of the form $\max I_t$ .

5. (5) There exists an algebraic curve Q such that, for any $(x, t) \in \overline {\mathfrak {L}}$ , we have

   $$ \begin{align*} Q \bigg( f_t (x), x - \displaystyle\frac{t f_t (x)}{f_t (x) + 1} \bigg) = 0. \end{align*} $$

   Furthermore, the curve Q ‘approximately comes from a polygonal domain’ in the following sense. There exists a polygonal domain $\mathfrak {P}$ satisfying Assumption 2.8 with liquid region $\mathfrak {L} (\mathfrak {P})$ and a real number $\alpha = \alpha _n \in \mathbb {R}$ with $|\alpha - 1| < n^{-\delta }$ such that, if $Q_{\mathfrak {L} (\mathfrak {P})}$ is the algebraic curve associated with $\mathfrak {L} (\mathfrak {P})$ from Proposition 3.3, then

   $$ \begin{align*} Q (u, v) = \displaystyle\frac{u + 1}{\alpha^{-1} u + 1} Q_{\mathfrak{L} (\mathfrak{P})} (\alpha^{-1} u, v). \end{align*} $$

Assumption 4.2. Adopt Assumption 4.1, and assume the following on how $\mathsf {h}$ converges to h.

1. (1) For each $v \in \partial \mathfrak {D}$ , we have $\big | \mathsf {h} (nv) - n h(v) \big | < n^{\delta / 2}$ .

2. (2) For each $v \in \partial _{\operatorname {\mathrm {ea}}} (\mathfrak {D}) \cup \partial _{\operatorname {\mathrm {we}}} (\mathfrak {D})$ , and each $v \in \partial _{\operatorname {\mathrm {no}}} \mathfrak {D}\cup \partial _{\operatorname {\mathrm {so}}} \mathfrak {D}$ such that $v \notin \mathfrak {L}_+^{\delta / 2} (\mathfrak {D})$ , we have $\mathsf {h} (nv) = n h(v)$ .

Proof of Proposition 4.6.

This follows from Lemma 4.11, Lemma 4.10 and Proposition 4.9 (the last applied with the $\varepsilon $ there equal to $\frac {e^{-2A}}{32A^4}$ here).

Proof. Throughout this proof, we recall the interpretation of $M_{\operatorname {\mathrm {cwf}}}$ as a lazy version of $M_{\operatorname {\mathrm {fl}}}$ from Remark 4.13; we also recall the notation of that remark and set $T_0 = 4A^3 (\log \varepsilon ^{-1})^2 \cdot \big ( t_{\operatorname {\mathrm {mix}}} (\varepsilon; M_{\operatorname {\mathrm {fl}}}) \big ) + 2 \le 4 A^3 (\log \varepsilon ^{-1})^2 \cdot \big ( t_{\operatorname {\mathrm {mix}}} (\varepsilon; M_{\operatorname {\mathrm {fl}}}) + 1 \big )$ . By Chernoff’s inequality, with probability at least $1 - \varepsilon $ , the number of active steps in this walk after some time $T \ge T_0 \ge 4A^3 (\log \varepsilon ^{-1})^2$ is at least

$$ \begin{align*} \displaystyle\frac{|\mathsf{R} \setminus \partial \mathsf{R}|}{\mathfrak{m} \cdot \mathfrak{M}} \cdot T - T^{1/2} \log \varepsilon^{-1} \ge \displaystyle\frac{T}{\mathfrak{M}} - T^{1/2} \log \varepsilon \ge A^{-3/2} T - (\log \varepsilon^{-1}) T^{1/2} \ge \displaystyle\frac{T_0}{2A^{3/2}}, \end{align*} $$

where we used the fact that $\mathfrak {m} \le |\mathsf {R} \setminus \partial \mathsf {R}|$ (as $\mathfrak {m}(\mathsf {v}) \ge 1$ for each $\mathsf {v}\in \mathsf {R} \setminus \partial \mathsf {R}$ ) in the first bound, the fact that $\mathfrak {M} \le k \cdot |\mathsf {R} \setminus \mathsf {R}| \le (\operatorname {\mathrm {diam}} \mathsf {R})^3 = A^{3/2}$ (as $\mathfrak {m} (\mathsf {v}) \le k \le \operatorname {\mathrm {diam}} \mathsf {R}$ for each $\mathsf {V} \in \mathsf {R} \setminus \partial \mathsf {R}$ ) in the second and the fact that $T \ge T_0 \ge 4 A^3 (\log \varepsilon )^2$ in the third. It follows for $T \ge T_0$ that the number of active steps in these dynamics is at least $t_{\operatorname {\mathrm {mix}}} (\varepsilon; M_{\operatorname {\mathrm {fl}}})$ , with probability at least $1 - \varepsilon $ . Conditioning on this event, we find for any $\mathsf {H}_0 \in \mathscr {G} (\mathsf {h})$ that we may couple $M_{\operatorname {\mathrm {cwf}}}^{\lceil T_0 \rceil } \mathsf {H}_0$ to coincide with a uniformly random element $\mathsf {F} \in \mathscr {G} (\mathsf {h})$ with probability $1 - \varepsilon $ . Hence, by a union bound, we may couple $M_{\operatorname {\mathrm {cwf}}}^{\lceil T_0 \rceil } \mathsf {H}_0$ to coincide with $\mathsf {F}$ with probability $1 - 2 \varepsilon $ , from which the lemma follows, as $T_0 \le 4A^3 (\log \varepsilon ^{-1})^2 \cdot \big ( t_{\operatorname {\mathrm {mix}}} (\varepsilon; M_{\operatorname {\mathrm {fl}}}) + 1 \big )$ .

Proof. We only establish the second statement of Equation (4.6), as the proof of the first is entirely analogous. First, observe that there exists a grand coupling the region-flip dynamics $M_{\operatorname {\mathrm {rf}}}$ over all choices of initial data in $\mathscr {G} (\mathsf {h})$ by running them under the same choices of sequences $\mathcal {X} = (X_1, X_2, \ldots )$ and vertices $\mathsf {v}$ (at which each flip is made) from Definition 4.15. It is quickly verified (see [Reference GorinGor21, Proposition 25.7], for example) that this coupling is monotone, meaning that if for some $\mathsf {H}_1, \mathsf {H}_2 \in \mathscr {G} (\mathsf {h})$ we have $\mathsf {H}_1 (\mathsf {v}) \le \mathsf {H}_2 (\mathsf {v})$ for each $\mathsf {v} \in \mathsf {R}$ , then it holds that $M_{\operatorname {\mathrm {rf}}}^t \mathsf {H}_1 (\mathsf {v}) \le M_{\operatorname {\mathrm {rf}}}^t \mathsf {H}_2 (\mathsf {v})$ for each $t \ge 0$ and $v \in \mathsf {R}$ .

Observe that it suffices to show under these coupled dynamics that, with probability at least $1 - \varepsilon $ , the models started at $\mathsf {H}^{\operatorname {\mathrm {top}}}$ and at $\mathsf {H}^{\operatorname {\mathrm {btm}}}$ coincide after time $R \big ( \frac {\varepsilon }{4A^2}, M_{\operatorname {\mathrm {rf}}} \big ) + S \big ( \frac {\varepsilon }{4A^2}, M_{\operatorname {\mathrm {rf}}} \big )$ , that is,

(4.7) $$ \begin{align} \mathbb{P} [ M_{\operatorname{\mathrm{rf}}}^T \mathsf{H}^{\operatorname{\mathrm{top}}} = M_{\operatorname{\mathrm{rf}}}^T \mathsf{H}^{\operatorname{\mathrm{btm}}}] \ge 1 - \varepsilon, \qquad \text{if}\; T \ge R \Big( \displaystyle\frac{\varepsilon}{4A^2}; M_{\operatorname{\mathrm{rf}}} \Big) + S \Big( \displaystyle\frac{\varepsilon}{4A^2}; M_{\operatorname{\mathrm{rf}}} \Big). \end{align} $$

Indeed, given Equation (4.7), it follows since $\mathsf {H}^{\operatorname {\mathrm {btm}}} \le \mathsf {F} \le \mathsf {H}^{\operatorname {\mathrm {top}}}$ for each $\mathsf {F} \in \mathscr {G}(\mathsf {h})$ that with probability at least $1 - \varepsilon $ the $M_{\operatorname {\mathrm {rf}}}^T \mathsf {F}$ all, over every $\mathsf {F} \in \mathscr {G} (\mathsf {h})$ , coincide for $T \ge R \big ( \frac {\varepsilon }{4A^2}, M_{\operatorname {\mathrm {rf}}} \big ) + S \big ( \frac {\varepsilon }{4A^2}, M_{\operatorname {\mathrm {rf}}} \big )$ . In particular, sampling $\mathsf {F}$ under the stationary measure $\rho $ for $M_{\operatorname {\mathrm {rf}}}$ , we deduce for any $\mathsf {H} \in \mathscr {G}(\mathsf {h})$ that one can couple $M_{\operatorname {\mathrm {rf}}}^T \mathsf {H}$ to coincide with a height function sampled under $\rho $ , with probability $1 - \varepsilon $ ; hence, $t_{\operatorname {\mathrm {mix}}} (\varepsilon; M_{\operatorname {\mathrm {rf}}}) \le R \big ( \frac {\varepsilon }{4A^2}, M_{\operatorname {\mathrm {rf}}} \big ) + S \big ( \frac {\varepsilon }{4A^2}, M_{\operatorname {\mathrm {rf}}} \big )$ , confirming the lemma.

It remains to verify Equation (4.7). Since $T \ge R \big ( \frac {\varepsilon }{4A^2}; M_{\operatorname {\mathrm {rf}}} \big )$ , we have by Equation (4.4) that it is possible to couple $M_{\operatorname {\mathrm {rf}}}^T \mathsf {H}^{\operatorname {\mathrm {top}}}$ with a height function $\mathsf {F}$ sampled under the stationary measure $\rho $ of $M_{\operatorname {\mathrm {rf}}}$ such that $M_{\operatorname {\mathrm {rf}}}^T \mathsf {H}^{\operatorname {\mathrm {top}}} = \mathsf {F}$ with probability at least $1 - \frac {\varepsilon }{4A^2}$ . Moreover, since $M_{\operatorname {\mathrm {rf}}}^T \mathsf {H}^{\operatorname {\mathrm {top}}} (\mathsf {u}) = \mathsf {h} (\mathsf {u}) = \mathsf {F}(\mathsf {u})$ for each $\mathsf {u} \in \partial \mathsf {R}$ and since $\operatorname {\mathrm {diam}} \mathsf {R} \le A$ , it follows (as $\mathsf {H}$ is $1$ -Lipschitz) that $\big | M_{\operatorname {\mathrm {rf}}}^T \mathsf {H}^{\operatorname {\mathrm {top}}} (\mathsf {v}) - \mathsf {F} (\mathsf {v}) \big | \le 2A$ . Combining these two statements, we deduce that $\mathbb {E} \big [ M_{\operatorname {\mathrm {rf}}}^T \mathsf {H}^{\operatorname {\mathrm {top}}} (\mathsf {v}) \big ] \le \mathbb {E} \big [ \mathsf {F} (\mathsf {v}) \big ] + \frac {\varepsilon }{2A}$ , for each $\mathsf {v} \in \mathsf {R}$ . Similarly, we have $\mathbb {E} \big [ M_{\operatorname {\mathrm {rf}}}^T \mathsf {H}^{\operatorname {\mathrm {btm}}} (\mathsf {v}) \big ] \le \mathbb {E} \big [ \mathsf {F} (\mathsf {v}) \big ] - \frac {\varepsilon }{2A}$ .

Therefore, $\mathbb {E} \big [ M_{\operatorname {\mathrm {rf}}}^T \mathsf {H}^{\operatorname {\mathrm {top}}} (\mathsf {v}) \big ] \le \mathbb {E} \big [ M_{\operatorname {\mathrm {rf}}}^T \mathsf {H}^{\operatorname {\mathrm {btm}}} (\mathsf {v})| \big ] + \frac {\varepsilon }{A}$ for any $\mathsf {v} \in \mathsf {R}$ . Together with the above grand coupling satisfying $M_{\operatorname {\mathrm {rf}}}^T \mathsf {H}^{\operatorname {\mathrm {btm}}} \le M_{\operatorname {\mathrm {rf}}}^T \mathsf {H}^{\operatorname {\mathrm {top}}}$ , the fact that any height function is integer-valued and a Markov inequality, we deduce that $\mathbb {P} \big [ M_{\operatorname {\mathrm {rf}}}^T \mathsf {H}^{\operatorname {\mathrm {top}}} (\mathsf {v}) \ne M_{\operatorname {\mathrm {rf}}}^T \mathsf {H}^{\operatorname {\mathrm {btm}}} (\mathsf {v}) \big ] \le \frac {\varepsilon }{A}$ , under this grand coupling. A union bound over all $|\mathsf {R}| \le A$ vertices $\mathsf {v} \in \mathsf {R}$ then yields Equation (4.7) and thus the lemma.

Proof. We first bound the mixing time of $M_{\operatorname {\mathrm {crf}}}$ in terms of that of $M_{\operatorname {\mathrm {wf}}}$ . To that end, recall that the state $\mathsf {H}_t$ at time $t \geq 1$ under $M_{\operatorname {\mathrm {wf}}}$ is defined from $\mathsf {H}_{t - 1}$ by performing a random flip at a vertex $\mathsf {v} \in \mathsf {R} \setminus \partial \mathsf {R}$ chosen with probability $\mathfrak {m} (\mathsf {v}) \cdot \mathfrak {M}^{-1}$ . Observe that we equivalently sample $\mathsf {v}$ by first selecting an index $i \in [1, k]$ with probability $p_i$ and then selecting $\mathsf {v} \in \mathsf {R}_i \setminus \partial \mathsf {R}_i$ uniformly at random. Recalling the random sequence $\mathcal {X} = (X_1, X_2, \ldots )$ from Definition 4.15, this is in turn equivalent to sampling $\mathsf {v} \in \mathsf {R}_{X_t} \setminus \partial \mathsf {R}_{X_t}$ uniformly at random, where we have denoted $\mathsf {R}_j = \mathsf {R}_i$ for $i \in [1, k]$ the integer such that k divides $j - i$ .

It follows that $M_{\operatorname {\mathrm {fl}}}$ and $M_{\operatorname {\mathrm {crf}}}$ can be coupled so that the former at time t coincides with the latter at time $(X_t - 1) A^{5} \le k t A^{5}$ , where in the last equality we used the fact that $X_t \leq kt$ (as $X_t - X_{t - 1} \leq k$ ). Hence,

(4.8) $$ \begin{align} t_{\operatorname{\mathrm{mix}}} (8A^2 \varepsilon; M_{\operatorname{\mathrm{crf}}}) \leq k A^{5} t_{\operatorname{\mathrm{mix}}} (8A^2 \varepsilon; M_{\operatorname{\mathrm{wf}}}) \le A^6 t_{\operatorname{\mathrm{mix}}} (8 A^2 \varepsilon; M_{\operatorname{\mathrm{wf}}}). \end{align} $$

Moreover, we have

$$ \begin{align*} t_{\operatorname{\mathrm{mix}}} (8 A^2 \varepsilon; M_{\operatorname{\mathrm{wf}}}) \le R (2 \varepsilon; M_{\operatorname{\mathrm{wf}}}) + S(2 \varepsilon; M_{\operatorname{\mathrm{wf}}}) & \le 2 t_{\operatorname{\mathrm{mix}}} (2\varepsilon; M_{\operatorname{\mathrm{cwf}}}) \\ & \le 8A^3 (\log \varepsilon^{-1})^2 \cdot \big( t_{\operatorname{\mathrm{mix}}} (\varepsilon; M_{\operatorname{\mathrm{fl}}}) + 1 \big), \end{align*} $$

where in the first inequality we applied Lemma 4.17, in the second we applied Proposition 4.16 and in the third we applied Lemma 4.14. Combining this with Equation (4.8) yields the lemma.

Proof of Lemma 4.10.

By Lemma 4.17, Proposition 4.16, and Lemma 4.18, we have for sufficiently large A that

$$ \begin{align*} t_{\operatorname{\mathrm{mix}}} (e^{-2A}; M_{\operatorname{\mathrm{rf}}}) &\leq R \Big(\frac{e^{-2A}}{4A^2}, M_{\operatorname{\mathrm{rf}}} \Big) + S \Big(\frac{e^{-2A}}{4A^2}, M_{\operatorname{\mathrm{rf}}} \Big)\\ &\leq R \Big(\frac{e^{-2A}}{4A^2}, M_{\operatorname{\mathrm{crf}}} \Big) + S \Big(\frac{e^{-2A}}{4A^2}, M_{\operatorname{\mathrm{crf}}} \Big)\\ &\leq 2t_{\operatorname{\mathrm{mix}}} \Big(\frac{e^{-2A}}{4A^2}, M_{\operatorname{\mathrm{crf}}} \Big)\leq 16A^9 (3A)^2 \cdot \bigg( t_{\operatorname{\mathrm{mix}}} \Big(\frac{e^{-2A}}{32A^4}; M_{\operatorname{\mathrm{fl}}} \Big) + 1 \bigg), \end{align*} $$

which yields the lemma.

Proof of Lemma 4.11.

First, observe that for sufficiently large A we have $t_{\operatorname {\mathrm {mix}}} \big (\frac {e^{-2A}}{4A^2}; M_{\operatorname {\mathrm {fl}}} \big ) \leq \frac {A^5}{200}$ , by Proposition 4.9; thus, Lemma 4.10 implies that $t_{\operatorname {\mathrm {mix}}} (e^{-2A}; M_{\operatorname {\mathrm {rf}}}) \leq A^{16}$ . So, to establish the lemma it suffices to couple the dynamics $M_{\operatorname {\mathrm {alt}}}$ at time t to coincide with $M_{\operatorname {\mathrm {rf}}}$ at time $A^5 t$ for each $t \in [0, A^{11}]$ , away from an event of probability at most $A^{11} e^{-2A} \leq e^{-A} - e^{-2A}$ .

To that end, let $\mathsf {H}_t$ denote the state after $t \geq 0$ steps of the dynamics $M_{\operatorname {\mathrm {rf}}}$ . Furthermore, for any integer $s \geq 0$ and $i \in [1, k]$ such that k divides $s - i + 1$ , set $\mathsf {H}_s' = \mathsf {H}_{s A^5}$ and $\mathsf {h}_{s+1}' = \mathsf {H}_{s A^5} |_{\partial \mathsf {R}_i}$ . Then, Definition 4.8 implies $\mathsf {H}_{s+1}' \in \mathscr {G} (\mathsf {h}_{s+1}')$ is obtained from $\mathsf {H}_s' \in \mathscr {G} (\mathsf {h}_s')$ from applying the flip dynamics $M_{\operatorname {\mathrm {fl}}}$ on $\mathsf {R}_i$ for time $A^5$ . Hence, since $t_{\operatorname {\mathrm {mix}}} (e^{-2A}; M_{\operatorname {\mathrm {fl}}})\leq t_{\operatorname {\mathrm {mix}}} \big ( \frac {e^{-2A}}{4A^2}; M_{\operatorname {\mathrm {fl}}} \big ) \leq A^5$ , we may couple $\mathsf {H}_s' |_{\mathsf {R}_i}$ with a uniformly random element of $\mathscr {G} (\mathsf {h}_s')$ , away from an event of probability at most $e^{-2A}$ .

It follows that the sequence $\{ \mathsf {H}_0', \mathsf {H}_1', \ldots , \mathsf {H}_s' \}$ can be coupled with s steps of the alternating dynamics $M_{\operatorname {\mathrm {alt}}}$ with initial data $\mathsf {H}_0$ , away from an event of probability at most $s e^{-2A}$ . Taking $s = t_{\operatorname {\mathrm {mix}}} (e^{-2A}; M_{\operatorname {\mathrm {fl}}}) \leq A^{11}$ and recalling that $\mathsf {H}_s' = \mathsf {H}_{s A^5}$ , we deduce the lemma.

Assumption 5.2. Adopt the notation of Theorem 2.10, and suppose $\mathfrak {D} \subseteq \mathfrak {P}$ , with $\mathsf {D} = n \mathfrak {D} \subseteq \mathbb {T}$ . Assume that the second, third and fourth constraints listed in Assumption 4.1 hold for $\mathfrak {D}$ (with respect to $H^*$ ). Further, suppose that either $\partial _{\operatorname {\mathrm {ea}}} (\mathfrak {D})$ is disjoint with $\overline {\mathfrak {L}(\mathfrak {P})}$ or that $\partial _{\operatorname {\mathrm {ea}}} (\mathfrak {D}) \subset \partial \mathfrak {P}$ ; similarly, suppose that either $\partial _{\operatorname {\mathrm {we}}} (\mathfrak {D})$ is disjoint with $\overline {\mathfrak {L}(\mathfrak {P})}$ or that $\partial _{\operatorname {\mathrm {we}}} (\mathfrak {D}) \subset \partial \mathfrak {P}$ . We denote the liquid region inside $\mathfrak D$ as $\mathfrak {L}=\mathfrak {L}(\mathfrak P)\cap \mathfrak D$ . Additionally, fix a real number $\mathfrak {s} \in [\mathfrak {t}_1, \mathfrak {t}_2]$ , and assume that no cusp or tangency location of $\mathfrak {A}(\mathfrak {P})$ in $\overline {\mathfrak {D}}$ is of the form $(x, y)$ , with $x \in \mathbb {R}$ and $y \in \{ \mathfrak {t}_1, \mathfrak {s}, \mathfrak {t}_2 \}$ .