that, with $\ell_{\max}=\lceil\frac{1}{2}\log_{2}N\rceil$,

$\displaystyle\mathsf{Drop}^{N}_{j}(a,\delta,\eta)\subseteq\bigcup_{\ell=0}^{\ell_{\max}}(\mathsf{E}_{\ell})^{c}.$
(10.19)

Indeed, it holds that $(x_{i+1,\ell_{\max}}-x_{i,\ell_{\max}})N^{2/3}=\delta
N^{2/3-1/2}=\delta N^{1/6}$; since $L^{N}_{j}$ is a Bernoulli path, this
implies that, on the complement of the righthand side of the previous display,
i.e., on $\cap_{\ell=0}^{\ell_{\max}}\mathsf{E}_{\ell}$, and for all large
enough $N$ (depending on $\eta$, $\delta$),

$\displaystyle\inf_{xN^{-2/3}\in[0,\delta]}\overline{L}^{N}_{j}(aN^{2/3}+x)$
$\displaystyle\geq\inf_{\begin{subarray}{c}i=1,\ldots,2^{\ell}\\\
\ell=0,\ldots,\ell_{\max}\end{subarray}}\overline{L}^{N}_{j}(x_{i,\ell}N^{2/3})-\delta
N^{1/6}$ $\displaystyle>\inf_{\begin{subarray}{c}i=1,\ldots,2^{\ell}\\\
\ell=0,\ldots,\ell_{\max}\end{subarray}}\overline{L}^{N}_{j}(x_{i,\ell}N^{2/3})-\tfrac{1}{2}\eta
N^{1/3}.$

For any $i$ and $\ell$, setting $i_{m+1}:=\lfloor i_{m}/2\rfloor$ for $m\geq
1$ and $i_{0}=i$, again on $\cap_{\ell=0}^{\ell_{\max}}\mathsf{E}_{\ell}$,

$\displaystyle\overline{L}^{N}_{j}(x_{i,\ell}N^{2/3})$
$\displaystyle=\sum_{m=0}^{\ell}\overline{L}^{N}_{j}(x_{i_{m},\ell-m}N^{2/3})-\overline{L}^{N}_{j}(x_{i_{m+1},\ell-m-1}N^{2/3})+\overline{L}^{N}_{j}(aN^{2/3})$
$\displaystyle\geq\overline{L}^{N}_{j}(aN^{2/3})-\sum_{m=0}^{\ell}\alpha_{\ell-m}\eta
N^{1/3}\geq\overline{L}^{N}_{j}(aN^{2/3})-\tfrac{1}{4}\eta N^{1/3}$

by the definition of $\alpha_{\ell}$. So, on
$\cap_{\ell=0}^{\ell_{\max}}E_{\ell}$, it indeed holds that
$\inf_{xN^{-2/3}\in[0,\delta]}\smash{\overline{L}^{N}_{j}(aN^{2/3}+x)}>\smash{\overline{L}^{N}_{j}(aN^{2/3})}-\eta
N^{1/3}$, i.e., (10.19) holds.

Let
$\mathsf{SingSep}^{N}_{j-1}(2\eta_{0})=\mathsf{SingSep}^{N}_{j-1}(2\eta_{0},[-2T,2T],L^{N}_{j})$
be as defined in (10.8). Let $\eta_{0}>0$ be such that
$\mathbb{P}(\mathsf{SingSep}^{N}_{j-1}(2\eta_{0}))$, as well as the
probability of the second event in (10.17), is at least
$\smash{1-\frac{1}{8}\varepsilon}$, which is possible by the assumptions in
Lemma 10.5. By (10.18), let $\delta_{0}>0$ be such that

$\mathbb{P}\left(\mathsf{Drop}^{N}_{j-1}(a,\delta_{0},\tfrac{1}{2}\eta_{0})\cup\mathsf{Drop}^{N}_{j-1}(2T,\delta_{0},\tfrac{1}{2}\eta_{0})\right)\leq\tfrac{1}{8}\varepsilon.$

Let $\mathsf{Fav}(M,\eta_{0},\delta_{0})$ be as in Lemma 10.6, i.e., (10.17).
Now, by the assumptions in Lemma 10.5 and these choices, performing a union
bound shows that we may choose $M$ such that
$\mathbb{P}(\mathsf{Fav}(M,\eta_{0},\delta_{0})^{c}\cup\mathsf{SingSep}^{N}_{j-1}(2\eta_{0})^{c})\leq\frac{1}{2}\varepsilon$.
Reduce $\eta$ if necessary so that it is less than $\frac{1}{2}\eta_{0}$.
Then, by (10.19),

$\displaystyle\mathbb{P}\left(\mathsf{Drop}^{N}_{j}(a,\delta,\eta)\right)$
$\displaystyle\leq\mathbb{P}\left(\mathsf{Drop}^{N}_{j}(a,\delta,\eta),\mathsf{Fav}(M,\eta_{0},\delta_{0}),\mathsf{SingSep}^{N}_{j-1}(2\eta_{0})\right)+\tfrac{1}{2}\varepsilon$
$\displaystyle\leq\sum_{\ell=0}^{\ell_{\max}}\mathbb{P}\left(\mathsf{E}_{\ell}^{c},\mathsf{Fav}(M,\eta_{0},\delta_{0}),\mathsf{SingSep}^{N}_{j-1}(2\eta_{0})\right)+\tfrac{1}{2}\varepsilon.$

By Lemma 10.6 and a union bound over the $2^{\ell}$ events in
$\mathsf{E}_{\ell}^{c}$, we know that the sum in the previous display is upper
bounded, for $\delta>0$, by

$\displaystyle\sum_{\ell=0}^{\infty}2^{\ell}C\exp\left(-c\alpha_{\ell}^{2}\eta^{2}/(2^{-\ell}\delta)\right)=\sum_{\ell=0}^{\infty}2^{\ell}C\exp\left(-c\delta^{-1}\rho^{2}2^{-\ell/2+\ell}\eta^{2}\right).$

This is upper bounded by $C\exp(-c\delta^{-1}\eta^{2})$. Setting $\delta>0$
small enough such that this is upper bounded by $\frac{1}{2}\varepsilon$
completes the proof. ∎

Next we turn to establishing the one-point statement of there being no quick
drops (Lemma 10.6).

###### Proof of Lemma 10.6.

Note that $a$ is measurable with respect to
$(\smash{L^{N}_{k},L^{N}_{k+1}},\ldots)$ by hypothesis, so
$[aN^{2/3},2TN^{2/3}]$ forms a stopping domain (Definition 6.1) with respect
to $(\smash{L^{N}_{1},\ldots,L^{N}_{k-1}})$. Let
$\mathcal{F}=\mathcal{F}_{\mathrm{ext}}(\\{k-1\\},\llbracket
aN^{2/3},2TN^{2/3}\rrbracket,\bm{L}^{N})$ be as in that definition, i.e., it
is the $\sigma$-algebra generated by
$\\{L^{N}_{i}(x):(i,x)\not\in\\{k-1\\}\times\llbracket
aN^{2/3}+1,TN^{2/3}-1\rrbracket\\}$. Note also that $\mathsf{Fav}$ (10.17) is
$\mathcal{F}$-measurable, and recall $\mathsf{SingSep}^{N}_{k-2}$ from (10.8).
Let $B$ be a Bernoulli random walk bridge from
$(aN^{2/3},L^{N}_{k-1}(aN^{2/3}))$ to $(2TN^{2/3},L^{N}_{k-1}(2TN^{2/3}))$ and
$\overline{B}(x)=B(x)-px$. By the strong Hall-Littlewood Gibbs property
(Proposition 6.2), the probability in the statement of the lemma equals

$\displaystyle\mathbb{E}\left[\frac{\mathbb
E_{\mathcal{F}}\Bigl{[}\mathbbm{1}_{\overline{B}((a+\delta)N^{2/3})<\overline{L}^{N}_{k-1}(aN^{2/3})-\eta
N^{1/3},\mathsf{SingSep}^{N}_{k-2}(2\eta_{0},B)}W(B,L^{N}_{k-2},L^{N}_{k})\Bigr{]}}{\mathbb
E_{\mathcal{F}}[W(B,L^{N}_{k-2},L^{N}_{k})]}\mathbbm{1}_{\mathsf{Fav}}\right]$
$\displaystyle\leq\mathbb{E}\left[\frac{\mathbb
E_{\mathcal{F}}[\mathbbm{1}_{\overline{B}((a+\delta)N^{2/3})<\overline{L}^{N}_{k-1}(aN^{2/3})-\eta
N^{1/3},}W(B,L^{N}_{k-2},L^{N}_{k})\mid\mathsf{SingSep}^{N}_{k-2}(2\eta_{0},B)]}{\mathbb
E_{\mathcal{F}}[W(B,L^{N}_{k-2},L^{N}_{k})\mid\mathsf{SingSep}^{N}_{k-2}(2\eta_{0},B)]}\mathbbm{1}_{\mathsf{Fav}}\right]$
$\displaystyle=\mathbb{E}\biggl{[}\frac{\mathbb
E_{\mathcal{F}}[W(B,L^{N}_{k-2},L^{N}_{k})\mid\mathsf{SingSep}^{N}_{k-2}(2\eta_{0},B),\overline{B}((a+\delta)N^{2/3})<\overline{L}^{N}_{k-1}(aN^{2/3})-\eta
N^{1/3}]}{\mathbb
E_{\mathcal{F}}[W(B,L^{N}_{k-2},L^{N}_{k})\mid\mathsf{SingSep}^{N}_{k-2}(2\eta_{0},B)]}$
$\displaystyle\qquad\qquad\times\mathbb
P_{\mathcal{F}}\left(\overline{B}((a+\delta)N^{2/3})<\overline{L}^{N}_{k-1}(aN^{2/3})-\eta
N^{1/3}\mid\mathsf{SingSep}^{N}_{k-2}(2\eta_{0},B)\right)\mathbbm{1}_{\mathsf{Fav}}\biggr{]}.$

Now, on $\mathsf{SingSep}^{N}_{k-2}(2\eta_{0},B)$, it holds by Lemma 9.1 that
$W_{\mathrm{up}}(B,L^{N}_{k-2})\geq\frac{1}{2}$ (recall from Definition 3.15)
for all large enough $N$, and $W_{\mathrm{up}}(B,L^{N}_{k-2})\leq 1$
trivially. Thus the previous display is upper bounded by

$\displaystyle 2\cdot\mathbb{E}\biggl{[}\frac{\mathbb
E_{\mathcal{F}}[W(B,L^{N}_{k})\mid\mathsf{SingSep}^{N}_{k-2}(2\eta_{0},B),\overline{B}((a+\delta)N^{2/3})<\overline{L}^{N}_{k-1}(aN^{2/3})-\eta
N^{1/3}]}{\mathbb
E_{\mathcal{F}}[W(B,L^{N}_{k})\mid\mathsf{SingSep}^{N}_{k-2}(2\eta_{0},B)]}$
$\displaystyle\times\mathbb
P_{\mathcal{F}}\left(\overline{B}((a+\delta)N^{2/3})<\overline{L}^{N}_{k-1}(aN^{2/3})-\eta
N^{1/3}\mid\mathsf{SingSep}^{N}_{k-2}(2\eta_{0},B)\right)\mathbbm{1}_{\mathsf{Fav}}\biggr{]}.$

By Corollary 8.5, the first factor in the previous line is upper bounded by
$c(q)^{-1}$, yielding that the previous display is upper bounded by

$\displaystyle 2c(q)^{-1}\mathbb{E}\biggl{[}\mathbb
P_{\mathcal{F}}\Bigl{(}\overline{B}\bigl{(}(a+\delta)N^{2/3}\bigr{)}<\overline{L}^{N}_{k-1}(aN^{2/3})-\eta
N^{1/3}\ \Big{|}\
\mathsf{SingSep}^{N}_{k-2}(2\eta_{0},B)\Bigr{)}\mathbbm{1}_{\mathsf{Fav}}\biggr{]}.$

We wish to remove the conditioning event, for which we first lower bound its
probability, on $\mathsf{Fav}$. Now, on $\mathsf{Fav}$, the separation between
$\smash{\overline{B}}$ and $\smash{\overline{L}^{N}_{k-2}}$ at $aN^{2/3}$ and
$2TN^{2/3}$ is at least $4\eta_{0}N^{1/3}$. Further,
$\smash{\overline{L}^{N}_{k-2}}$ drops by at most
$\smash{\frac{1}{2}\eta_{0}N^{1/3}}$ in $[aN^{2/3},(a+\delta_{0})N^{2/3}]$ and
$[(2T-\delta_{0})N^{2/3},2TN^{2/3}]$ and remains above $-MN^{1/3}$ on all of
$[-2TN^{2/3},2TN^{2/3}]$. Thus for $\overline{B}$ to maintain a separation of
$2\eta_{0}N^{1/3}$ with $\smash{\overline{L}^{N}_{k-2}}$ on
$[-2TN^{2/3},2TN^{2/3}]$, it is sufficient if it does the following (similar
to Figure 21 in the proof of Lemma 10.3): drops to or below
$-(M+4\eta_{0})N^{1/3}$ in $[aN^{2/3},(a+\delta_{0})N^{2/3}]$ and never goes
above $\smash{\overline{L}^{N}_{k-1}}(aN^{2/3})+\frac{1}{2}\eta_{0}N^{1/3}$ in
the same interval, stays below $-(M+3\eta_{0})N^{1/3}$ on
$[(a+\delta_{0})N^{2/3},(2T-\delta_{0})N^{2/3}]$, and jumps up to
$\smash{\overline{L}^{N}_{k-1}}(2TN^{2/3})$ at $2TN^{2/3}$ while staying below
$\overline{L}^{N}_{k-1}(2TN^{2/3})+\frac{1}{2}\eta_{0}N^{1/3}$ on
$[(2T-\delta_{0})N^{2/3},2TN^{2/3}]$. The probability that $\overline{B}$
satisfies these conditions is lower bounded by some deterministic
$\alpha=\alpha(T,\delta_{0},M,\eta_{0})>0$ (since the same is true for the
Brownian bridge weak limit of the rescaled $\overline{B}$), using also that
$\smash{\overline{L}^{N}_{k-1}}(aN^{2/3}),\smash{\overline{L}^{N}_{k-1}}(2TN^{2/3})\in[-MN^{1/3},MN^{1/3}]$
on $\mathsf{Fav}$. This yields that the previous display is upper bounded by

$\displaystyle 2\alpha^{-1}c(q)^{-1}\mathbb{E}\left[\mathbb
P_{\mathcal{F}}\Bigl{(}\overline{B}((a+\delta)N^{2/3})<\overline{L}^{N}_{k-1}(aN^{2/3})-\eta
N^{1/3}\Bigr{)}\mathbbm{1}_{\mathsf{Fav}}\right].$

Since $\overline{B}(2TN^{2/3}),\overline{B}(aN^{2/3})\in[-MN^{1/3},MN^{1/3}]$
(so that the endpoints of $B$ define a line of slope $p+O(1)MN^{-1/3}$), it
follows by one-point bounds on Bernoulli random walk bridges (Lemma 6.13)
that, for all $\delta>0$, the previous display is upper bounded by

$\displaystyle C\alpha^{-1}c(q)^{-1}\exp\left(-c\delta^{-1}\eta^{2}\right).$

This completes the proof. ∎

##  11\. Partition function and non-intersection probability estimates

In this section we prove Proposition 6.12. The argument is fairly standard
given what we have already done; it first appeared in [CH14], and has been
used in a number of subsequent works [CH16, CD18, DFF+21, DFF+21, BCD23]. It
relies on sampling from a measure on paths on a larger interval where the
interaction between paths is turned off in the interval of interest and
considering the Radon-Nikodym derivative of this measure with the original
one; if one has a probability lower bound on the partition function of the new
measure, it can be made into a high probability lower bound on the partition
function of the original measure, by an argument that can be understood as a
form of “size-biased” sampling.

However, our arguments are slightly more complicated than in earlier works as
the larger interval we work on is $[U_{1},U_{2}]$ (defined as in Lemma 10.3
such that there is control on the growth of the lower curve near $U_{i}$) and
so is random. This is because we need some such control on the lower curve at
the boundaries of the interval to obtain a lower bound on the partition
function for the new measure mentioned above. In the earlier works, this lower
bound was achieved on a deterministic interval, using the stochastic
monotonicity and other features of the Gibbs property that were available in
those models.

As in the last two sections, we assume here that $L^{N}_{k}$ satisfies
Assumptions 1 and 2 without explicitly mentioning it in the statements.

###  11.1. Lower bounding the partition function

Fix $T\geq 2$. Let $U_{1}\in[-2T,-\frac{3}{2}T]$ be the largest point and
$U_{2}\in[\frac{3}{2}T,2T]$ be the smallest point such that

$\begin{split}\overline{L}^{N}_{k+1}(x)&\leq\overline{L}^{N}_{k+1}(U_{1}N^{2/3})+(x-U_{1}N^{2/3})PN^{-1/3}\quad\text{for}\quad
xN^{-2/3}\in[U_{1},U_{1}+1]\\\
\overline{L}^{N}_{k+1}(x)&\leq\overline{L}^{N}_{k+1}(U_{2}N^{2/3})+(U_{2}N^{2/3}-x)PN^{-1/3}\quad\text{for}\quad
xN^{-2/3}\in[U_{2}-1,U_{2}],\end{split}$ (11.1)

for a $P$ to be specified later. In the event that such a $U_{1}$ does not
exist, we set it to $-2T$, and in the event that such a $U_{2}$ does not
exist, we set it to $2T$.

Fix an interval $[-\frac{3}{2}T,\frac{3}{2}T]\subseteq[a,b]\subseteq[-2T,2T]$
and $\bm{x},\bm{y}\in\\{\bm{z}\in\mathbb{R}^{k}:z_{1}>\ldots>z_{k}\\}$. We
define a discrete line ensemble (Definition 3.6)
$\smash{\tilde{\bm{L}}^{N}=(\tilde{L}^{N}_{1},\ldots,\tilde{L}^{N}_{k})}$,
where each $\tilde{L}^{N}_{i}:\llbracket
aN^{2/3},bN^{2/3}\rrbracket\to\mathbb{R}$ with
$\tilde{L}^{N}_{i}(aN^{2/3})=x_{i}$ and $\tilde{L}^{N}_{i}(bN^{2/3})=y_{i}$,
as follows. The law $\tilde{\mathbb}{P}^{[a,b],\bm{x},\bm{y}}_{N}$ of
$\tilde{\bm{L}}^{N}$ is specified via its Radon-Nikodym derivative with
respect to the law $\mathbb{P}^{k,[a,b],\bm{x},\bm{y}}_{\mathrm{free},N}$ of
$k$ independent Bernoulli random walk bridges on $\llbracket
aN^{2/3},bN^{2/3}\rrbracket$, the $j$th from $(aN^{2/3},x_{j})$ to
$(bN^{2/3},y_{j})$ for each $j$: for $\bm{\gamma}:\llbracket
1,k\rrbracket\times\llbracket aN^{2/3},bN^{2/3}\rrbracket\to\mathbb{Z}$ a
collection of $k$ Bernoulli paths with endpoints $(aN^{2/3},x_{j})$ and
$(bN^{2/3},y_{j})$ for $j=1,\ldots,k$,

$\displaystyle\frac{\mathrm{d}\tilde{\mathbb}{P}^{[a,b],\bm{x},\bm{y}}_{N}}{\mathrm{d}\mathbb{P}^{k,[a,b],\bm{x},\bm{y}}_{\mathrm{free},N}}(\bm{\gamma})\propto\prod_{i=1}^{k}\prod_{x\in
I_{N,T}}\left(1-q^{\Delta_{i}(x)}\mathbbm{1}_{\Delta_{i}(x)=\Delta_{i}(x-1)-1}\right)$
(11.2)

where $I_{N,T}=\llbracket
U_{1}N^{2/3},U_{2}N^{2/3}\rrbracket\setminus\llbracket-
TN^{2/3},TN^{2/3}\rrbracket$ and $\Delta_{i}(x)=\gamma_{i}(x)-\gamma_{i+1}(x)$
with $\gamma_{k+1}=L^{N}_{k+1}$ by convention. In words,
$\tilde{\mathbb}{P}^{[a,b],\bm{x},\bm{y}}_{N}$ is the usual Hall-Littlewood
Gibbs measure but where the interaction is absent on $\llbracket-
TN^{2/3},TN^{2/3}\rrbracket$.

Writing in the just introduced notation, the definition (6.5) reads

$\displaystyle
Z^{k,[-T,T],\bm{x},\bm{y},g}_{N}=\mathbb{E}^{k,[-T,T],\bm{x},\bm{y}}_{\mathrm{free},N}[W(\bm{B},g)]$
(11.3)

(recall $W$ from (3.9)), where, under
$\mathbb{E}^{k,[-T,T],\bm{x},\bm{y}}_{\mathrm{free}}$,
$\bm{B}=(B_{1},\ldots,B_{k})$ is a collection of $k$ independent Bernoulli
random walk bridges, with $B_{i}$ from $(-TN^{2/3},x_{i})$ to
$(TN^{2/3},y_{i})$.

######  Lemma 11.1.

Suppose $g:[aN^{2/3},bN^{2/3}]\to\mathbb{R}$ satisfies (with
$\overline{g}(x):=g(x)-px$)

$\begin{split}\overline{g}(x)&\leq\overline{g}(aN^{2/3})+(x-aN^{2/3})PN^{-1/3}\quad\text{for}\quad
xN^{-2/3}\in[a,a+1]\\\
\overline{g}(x)&\leq\overline{g}(bN^{2/3})+(bN^{2/3}-x)PN^{-1/3}\quad\text{for}\quad
xN^{-2/3}\in[b-1,b],\end{split}$ (11.4)

Suppose also that $M$ is such that $\smash{\sup_{xN^{-2/3}\in[a,b]}|g(x)|\leq
MN^{1/3}}$, and that there exists $\eta>0$ such that
$x_{i}>x_{i+1}+\eta\smash{N^{1/3}}$ and $y_{i}>y_{i+1}+\eta\smash{N^{1/3}}$
for each $i=1,\ldots,k$, with $x_{k+1}=g(a\smash{N^{2/3}})$ and
$y_{k+1}=g(bN^{2/3})$. Then there exist $\delta>0$, $\varepsilon>0$, and
$N_{0}$, all depending on $k,M,T,\eta$, such that, for all $N\geq N_{0}$,

$\displaystyle\tilde{\mathbb}{P}^{[a,b],\bm{x},\bm{y}}_{N}\left(Z^{k,[-T,T],\tilde{\bm{L}}^{N}(-TN^{2/3}),\tilde{\bm{L}}^{N}(TN^{2/3}),g}_{N}>\delta\right)>\varepsilon,$

where $\tilde{\bm{L}}^{N}(TN^{2/3})$ and $\tilde{\bm{L}}^{N}(-TN^{2/3})$ in
the superscript respectively indicate
$(\tilde{L}^{N}_{i}(TN^{2/3}))_{i=1}^{k}$ and
$(\tilde{L}^{N}_{i}(-TN^{2/3}))_{i=1}^{k}$.

###### Proof.

We adopt the shorthand
$Z^{k,T,g}_{N}=Z^{k,[-T,T],\tilde{\bm{L}}^{N}(-TN^{2/3}),\tilde{\bm{L}}^{N}(TN^{2/3}),g}_{N}$.
With some positive probability $2\varepsilon>0$ (independent of $N$) it holds
that $k$ independent Bernoulli random walk bridges
$\bm{B}=(B_{1},\ldots,B_{k})$ starting from $(aN^{2/3},\bm{x})$ and going to
$(bN^{2/3},\bm{y})$ reach points at $(a+1)N^{2/3}$ and $(b-1)N^{2/3}$ which
are all above $2MN^{1/3}$ and are separated by at least $N^{1/3}$; reach
points at $\pm TN^{2/3}$ which have the same properties; and do so while
staying separated (from each other as well as from $g$) by at least
$\frac{1}{2}\eta N^{1/3}$ on $[aN^{2/3},-TN^{2/3}]$ and $[TN^{2/3},bN^{2/3}]$;
this uses the condition (11.4) controlling $g$ and the initial separation of
$\eta N^{1/3}$ between the endpoints of $\bm{B}$. Call this event
$\mathsf{Corr}$ (for corridors).

On $\mathsf{Corr}$, it holds that the righthand side of (11.5) is lower
bounded by $\frac{1}{2}$ for all large enough $N$ by lower bounding it as in
the proof of Lemma 9.1; and the implicit normalization factor in (11.5) is
trivially upper bounded by $1$. Thus with
$\smash{\tilde{\mathbb}{P}^{[a,b],\bm{x},\bm{y}}_{N}}$-probability at least
$\varepsilon$, we have that the entries of
$\smash{\tilde{\bm{L}}^{N}(-TN^{2/3})}$,
$\smash{\tilde{\bm{L}}^{N}(TN^{2/3})}$ are separated by at least $N^{1/3}$ and
are all above $2MN^{1/3}$.

We claim that there exists $\delta>0$ such that, on the event that the entries
of $\tilde{\bm{L}}^{N}(\pm TN^{2/3})$ have the just mentioned properties,
$\smash{Z^{k,T,g}_{N}}>\delta$. Indeed, with some positive probability
$2\delta$, $k$ independent Bernoulli random walk bridges from
$(-TN^{2/3},\tilde{\bm{L}}^{N}(-TN^{2/3}))$ to
$(TN^{2/3},\tilde{\bm{L}}^{N}(TN^{2/3}))$ will maintain separation at least
$\frac{1}{2}N^{1/3}$ throughout $[-TN^{2/3},TN^{2/3}]$ while staying above
$MN^{1/3}$. On this event, the weight function $W$ is again lower bounded by
$\smash{\frac{1}{2}}$ for all large enough $N$ by Lemma 9.1. This completes
the proof. ∎

###### Proof of Proposition 6.12.

We adopt the shorthand
$Z^{k,T}_{N}=Z^{k,[-T,T],\bm{L}^{N}(-TN^{2/3}),\bm{L}^{N}(TN^{2/3}),L^{N}_{k+1}}_{N}$.
By Theorem 6.10 (lower tail control) and Proposition 10.2 (upper tail
control), there exists $M=M(\varepsilon,k,T)$ such that
$\mathbb{P}(\mathsf{RangeCtrl})\geq 1-\frac{1}{4}\varepsilon$, where

$\mathsf{RangeCtrl}=\left\\{\sup_{xN^{-2/3}\in[-2T,2T]}|L^{N}_{k+1}(x)|\leq
MN^{1/3}\right\\}.$

Let $U_{1}$ and $U_{2}$ be as in (11.1). On $\mathsf{RangeCtrl}$, by Lemma
10.1, there exists $P=P(\varepsilon,k,T,M)=P(\varepsilon,k,T)$ such that
$U_{1}\neq-2T$ and $U_{2}\neq 2T$, i.e., (11.1) holds. Next, by Theorem 9.2,
let $\eta=\eta(\varepsilon,k,T)>0$ be such that
$\mathbb{P}(\mathsf{Sep}_{k}^{N}(\eta,\\{U_{1},U_{2}\\},\bm{L}^{N}))>1-\frac{1}{4}\varepsilon$.
Let
$\mathsf{Fav}=\mathsf{RangeCtrl}\cap\mathsf{Sep}_{k}^{N}(\eta,\\{U_{1},U_{2}\\},\bm{L}^{N})$,
so that $\mathbb{P}(\mathsf{Fav}^{c})\leq\frac{1}{2}\varepsilon$.

Let $\mathcal{F}=\mathcal{F}_{\mathrm{ext}}(k,\llbracket
U_{1}N^{2/3},U_{2}N^{2/3}\rrbracket,\bm{L}^{N})$, and note that $\llbracket
U_{1}N^{2/3},U_{2}N^{2/3}\rrbracket$ forms a stopping domain with respect to
$(L^{N}_{1},\ldots,L^{N}_{k})$ (Definition 6.1), since $U_{1},U_{2}$ are
measurable with respect to $L^{N}_{k+1}$. Conditionally on $\mathcal{F}$, let
$\mathbb{P}_{N}$ be the law of $\smash{\bm{L}^{N}|_{\llbracket
1,k\rrbracket\times\llbracket U_{1}N^{2/3},U_{2}N^{2/3}\rrbracket}}$ and adopt
the shorthand $\smash{\tilde{\mathbb}{P}_{N}}$ for the law defined by (11.2).
Let $\smash{\tilde{\bm{L}}^{N}\sim\tilde{\mathbb}{P}_{N}}$. Again
conditionally on $\mathcal{F}$, define $\tilde{\bm{L}}^{N,\mathrm{res}}$ and
$\bm{L}^{N,\mathrm{res}}$ (“res” short for restricted) by

$\tilde{L}^{N,\mathrm{res}}_{i}:=\tilde{L}^{N}_{i}|_{\llbracket
U_{1}N^{2/3},U_{2}N^{2/3}\rrbracket\setminus\llbracket-
TN^{2/3},TN^{2/3}\rrbracket}$

and $L^{N,\mathrm{res}}_{i}:=L^{N}_{i}|_{\llbracket
U_{1}N^{2/3},U_{2}N^{2/3}\rrbracket\setminus\llbracket-
TN^{2/3},TN^{2/3}\rrbracket}$ for $i\in\llbracket 1,k\rrbracket$, and let
$\tilde{\mathbb}{P}^{\mathrm{res}}_{N}$ and $\mathbb{P}^{\mathrm{res}}_{N}$,
respectively, be their laws. Note that, for any collection
$\bm{\gamma}:\llbracket 1,k\rrbracket\times\llbracket
U_{1}N^{2/3},U_{2}N^{2/3}\rrbracket\setminus\llbracket-
TN^{2/3},TN^{2/3}\rrbracket\to\mathbb{Z}$ of Bernoulli paths,

$\displaystyle\frac{\mathrm{d}\mathbb{P}^{\mathrm{res}}_{N}}{\mathrm{d}\tilde{\mathbb}{P}^{\mathrm{res}}_{N}}(\bm{\gamma})=(\tilde{Z}^{k,T}_{N})^{-1}Z^{k,T}_{N}(\bm{\gamma}),$
(11.5)

where
$Z^{k,T}_{N}(\bm{\gamma}):=Z^{k,[-T,T],\bm{\gamma}(-TN^{2/3}),\bm{\gamma}(TN^{2/3}),L^{N}_{k+1}}_{N}$
as in (11.3) and
$\tilde{Z}^{k,T}_{N}:=\tilde{\mathbb}{E}^{\mathrm{res}}_{N}[Z^{k,T}_{N}(\bm{\gamma})]$.
Since $Z^{k,T}_{N}(\bm{\gamma})$ is a function of $\bm{\gamma}(\pm TN^{2/3})$
($L^{N}_{k+1}$ is conditioned on), which have the same joint distribution when
$\bm{\gamma}$ is distributed as
$\smash{\tilde{\mathbb}{P}^{\mathrm{res}}_{N}}$ or
$\smash{\tilde{\mathbb}{P}_{N}}$, it holds that
$\tilde{Z}^{k,T}_{N}=\tilde{\mathbb}{E}_{N}[Z^{k,T}_{N}(\bm{\gamma})]$.
Similarly, $\smash{Z^{k,T}_{N}=Z^{k,T}_{N}(\bm{L}^{N})}$ has the same
distribution under $\mathbb{P}_{N}$ and $\mathbb{P}^{\mathrm{res}}_{N}$. Thus,
since $\mathsf{Fav}$ is $\mathcal{F}$-measurable,

$\displaystyle\mathbb{P}\left(Z^{k,T}_{N}<\delta\right)\leq\mathbb{E}\left[\mathbb
P_{\mathcal{F}}\left(Z^{k,T}_{N}<\delta\right)\mathbbm{1}_{\mathsf{Fav}}\right]+\mathbb{P}\left(\mathsf{Fav}^{c}\right)$
$\displaystyle=\mathbb{E}\left[\mathbb{P}^{\mathrm{res}}_{N}\left(Z^{k,T}_{N}<\delta\right)\mathbbm{1}_{\mathsf{Fav}}\right]+\tfrac{1}{2}\varepsilon.$

Now by (11.5), the previous display equals

$\displaystyle\mathbb{E}\left[(\tilde{Z}^{k,T}_{N})^{-1}\tilde{\mathbb}{E}^{\mathrm{res}}_{N}\left[\mathbbm{1}_{Z^{k,T}_{N}<\delta}Z^{k,T}_{N}\right]\mathbbm{1}_{\mathsf{Fav}}\right]+\tfrac{1}{2}\varepsilon.$
(11.6)

On $\mathsf{Fav}$, $(L^{N}_{i}(\pm TN^{2/3}))_{i=1}^{k}$ and $L^{N}_{k+1}$
satisfy the conditions in Lemma 11.1. So we have that, for some
$\delta^{\prime},\varepsilon^{\prime}>0$ (the $\delta$, $\varepsilon$ provided
by Lemma 11.1),

$\displaystyle\tilde{Z}^{k,T}_{N}=\tilde{\mathbb}{E}[Z^{k,T}_{N}]\geq\rho:=\delta^{\prime}\varepsilon^{\prime},$

This yields that (11.6) is upper bounded by

$\displaystyle\rho^{-1}\cdot\mathbb{E}\left[\tilde{\mathbb}{E}^{\mathrm{res}}_{N}\left[\mathbbm{1}_{Z^{k,T}_{N}<\delta}Z^{k,T}_{N}\right]\right]+\tfrac{1}{2}\varepsilon\leq\delta\rho^{-1}+\tfrac{1}{2}\varepsilon.$

Picking $\delta=\frac{1}{2}\varepsilon\rho$ completes the proof. ∎

##  12\. Brownian Gibbs property in the limit

In this section we prove that all weak limit points of $\bm{\mathfrak{L}}^{N}$
as defined in (3.13) possess the Brownian Gibbs property, thus establishing
Proposition 6.5. The first step is to compare the partition function
associated to the Hall-Littlewood Gibbs property with the probability of
separation (its analog for the Brownian Gibbs property); we do this in the
next section. Then we will give the proof of Proposition 6.5 in Section 12.2.
As in the last few sections, we assume, without explicitly stating it again,
that $\bm{L}^{N}$ satisfies Assumptions 1 and 2.

###  12.1. Comparing the partition function and the separation probability

To emphasize the separation required with the lower boundary condition, in
this section we extend the notation (9.1) as follows: for a set
$A\subseteq\mathbb{R}$, an integer $k\in\mathbb{N}$,
$\Lambda\subseteq\mathbb{N}$ a (possibly infinite) interval with $\llbracket
1,k\rrbracket\subseteq\Lambda$, a process
$\bm{X}:\Lambda\times\mathbb{R}\to\mathbb{R}$, a process
$Y:\mathbb{R}\to\mathbb{R}$, and $\delta>0$, we define the event

$\begin{split}\mathsf{Sep}^{N}_{k}(\delta,A,\bm{X},Y)&=\left\\{\min_{i=1,\ldots,k-1}\inf_{xN^{-2/3}\in
A}\bigl{(}X_{i}(x)-X_{i+1}(x)\bigr{)}\geq\delta N^{1/3}\right\\}\\\
&\qquad\cap\left\\{\inf_{xN^{-2/3}\in
A}\bigl{(}X_{k}(x)-Y(x)\bigr{)}\geq\delta N^{1/3}\right\\},\end{split}$ (12.1)

i.e., in the notation (9.1), if we define $\tilde{\bm{X}}:\llbracket
1,k+1\rrbracket\times\mathbb{R}\to\mathbb{R}$ by
$\tilde{X}_{i}(\bm{\cdot})=X_{i}(\bm{\cdot})$ for $i=1,\ldots,k$ and
$\tilde{X}_{k+1}(\bm{\cdot})=Y(\bm{\cdot})$, then
$\mathsf{Sep}^{N}_{k}(\delta,A,\bm{X},Y)=\mathsf{Sep}^{N}_{k}(\delta,A,\tilde{\bm{X}})$.
At times we will drop the set $A$ from the notation, but this will be
explicitly stated.

For the next statement, recall $W$ from Definition 3.15.

######  Lemma 12.1.

Fix $a>b$ and $k\in\mathbb{N}$. Let
$\mathcal{F}=\mathcal{F}_{\mathrm{ext}}(k,\llbracket
aN^{2/3},bN^{2/3}\rrbracket,\bm{L}^{N})$ and $\bm{B}=(B_{1},\ldots,B_{k})$ be
a collection of $k$ independent Bernoulli random walk bridges, with $B_{i}$
from $(aN^{2/3},L^{N}_{i}(aN^{2/3}))$ to $(bN^{2/3},L^{N}_{i}(bN^{2/3}))$. For
any $\varepsilon>0$, there exist $N_{0}=N_{0}(\varepsilon,k,b-a)$ and
$\delta=\delta(\varepsilon,k,b-a)>0$ such that, for all $N\geq N_{0}$,

$\displaystyle\mathbb{P}\left(\frac{\mathbb
E_{\mathcal{F}}[W(\bm{B},L^{N}_{k+1},\llbracket
aN^{2/3},bN^{2/3}\rrbracket)]}{\mathbb
P_{\mathcal{F}}(\mathsf{Sep}^{N}_{k}(\delta,[a,b],\bm{B},L^{N}_{k+1}))}\in[1-\varepsilon,1+\varepsilon]\right)\geq
1-\varepsilon.$

###### Proof.

Adopt the shorthand
$\mathsf{Sep}^{N}_{k}(\delta,\bm{B},L^{N}_{k+1})=\mathsf{Sep}^{N}_{k}(\delta,[a,b],\bm{B},L^{N}_{k+1})$
and drop the interval $\llbracket aN^{2/3},bN^{2/3}\rrbracket$ from the $W$
notation. We first pick $\delta>0$ such that, by Theorem 9.2,

$\displaystyle\mathbb{P}(\mathsf{Sep}^{N}_{k}(\delta,\bm{L}^{N},L^{N}_{k+1}))\geq
1-\varepsilon^{2}.$

Since the lefthand side of this inequality equals $\mathbb{E}[\mathbb
P_{\mathcal{F}}(\mathsf{Sep}^{N}_{k}(\delta,\bm{L}^{N},L^{N}_{k+1}))]$, and
the conditional probability is a random variable taking values in $[0,1]$, it
follows that, with probability at least $1-\varepsilon$,

$\displaystyle\mathbb
P_{\mathcal{F}}(\mathsf{Sep}^{N}_{k}(\delta,\bm{L}^{N},L^{N}_{k+1}))\geq
1-\varepsilon.$

We work on the event that the previous inequality holds. We have the following
string of inequalities, using $W(\bm{B},L^{N}_{k+1})\leq 1$ in the first
inequality, the Hall-Littlewood Gibbs property to go from the second to the
third line, and the previous display for the inequality in the last line:

$\displaystyle\mathbb
P_{\mathcal{F}}(\mathsf{Sep}^{N}_{k}(\delta,\bm{B},L^{N}_{k+1}))$
$\displaystyle\geq\mathbb
E_{\mathcal{F}}[\mathbbm{1}_{\mathsf{Sep}^{N}_{k}(\delta,\bm{B},L^{N}_{k+1})}W(\bm{B},L^{N}_{k+1})]$
$\displaystyle=\frac{\mathbb
E_{\mathcal{F}}[\mathbbm{1}_{\mathsf{Sep}^{N}_{k}(\delta,\bm{B},L^{N}_{k+1})}W(\bm{B},L^{N}_{k+1})]}{\mathbb
E_{\mathcal{F}}[W(\bm{B},L^{N}_{k+1})]}\cdot\mathbb
E_{\mathcal{F}}[W(\bm{B},L^{N}_{k+1})]$ $\displaystyle=\mathbb
P_{\mathcal{F}}(\mathsf{Sep}^{N}_{k}(\delta,\bm{L}^{N},L^{N}_{k+1}))\cdot\mathbb
E_{\mathcal{F}}[W(\bm{B},L^{N}_{k+1})]\geq(1-\varepsilon)\mathbb
E_{\mathcal{F}}[W(\bm{B},L^{N}_{k+1})].$

It also trivially holds that $\mathbb
E_{\mathcal{F}}[W(\bm{B},L^{N}_{k+1})]\geq\mathbb
E_{\mathcal{F}}[W(\bm{B},\smash{L^{N}_{k+1}})\mathbbm{1}_{\mathsf{Sep}^{N}_{k}(\delta,\bm{B},\smash{L^{N}_{k+1}})}]$
and that, on the event
$\smash{\mathsf{Sep}^{N}_{k}}(\delta,\bm{B},\smash{L^{N}_{k+1}})$ and by Lemma
9.1, $W(\bm{B},\smash{L^{N}_{k+1}})\geq 1-\varepsilon$ for all large enough
$N$. This yields that, for such $N$,

$\displaystyle\mathbb
E_{\mathcal{F}}[W(\bm{B},L^{N}_{k+1})]\geq(1-\varepsilon)\mathbb
P_{\mathcal{F}}(\mathsf{Sep}^{N}_{k}(\delta,\bm{B},L^{N}_{k+1})).$

This completes the proof after relabeling $\varepsilon$. ∎

######  Corollary 12.2.

Fix $\varepsilon\in(0,1)$, $k\in\mathbb{N}$, $M>0$, and $a<b$. Let
$\mathcal{F}=\mathcal{F}_{\mathrm{ext}}(k,[aN^{2/3},bN^{2/3}]$, $\bm{L}^{N})$
and $\bm{B}=(B_{1},\ldots,B_{k})$ be a collection of $k$ independent Bernoulli
random walk bridges with $B_{i}$ from $(aN^{2/3},L^{N}_{i}(aN^{2/3}))$ to
$(bN^{2/3},L^{N}_{i}(bN^{2/3}))$. There exist
$N_{0}=N_{0}(\varepsilon,k,b-a,M)$ and $\delta=\delta(\varepsilon,k,b-a,M)>0$
such that, for any $N\geq N_{0}$, the following holds with probability at
least $1-\varepsilon$. For any bounded measurable function
$F:\mathcal{C}(\llbracket
1,k\rrbracket\times[aN^{2/3},bN^{2/3}],\mathbb{R})\to\mathbb{R}$ such that
$\sup|F|\leq M$,

$\displaystyle\left|\frac{\mathbb
E_{\mathcal{F}}\left[F(\bm{B})W(\bm{B},L^{N}_{k+1},\llbracket
aN^{2/3},bN^{2/3}\rrbracket)\right]}{\mathbb
E_{\mathcal{F}}[W(\bm{B},L^{N}_{k+1},\llbracket
aN^{2/3},bN^{2/3}\rrbracket)]}-\mathbb
E_{\mathcal{F}}\left[F(\bm{B})\mid\mathsf{Sep}^{N}_{k}(\delta,[a,b],\bm{B},L^{N}_{k+1})\right]\right|\leq\varepsilon.$

###### Proof.

We again adopt the shorthand
$\mathsf{Sep}^{N}_{k}(\delta,\bm{B},L^{N}_{k+1})=\mathsf{Sep}^{N}_{k}(\delta,[a,b],\bm{B},L^{N}_{k+1})$
and drop the interval $\llbracket aN^{2/3},bN^{2/3}\rrbracket$ from the $W$
notation. We may assume that $F\geq 0$ as we have assumed $F$ is bounded and
adding a constant to $F$ does not change the lefthand side of the inequality
we are proving. Let $\varepsilon^{\prime}=\varepsilon M^{-1}$. Pick $\delta>0$
from Lemma 12.1 with $\varepsilon$ there equal to $\varepsilon^{\prime}$, and
let $N_{0}=N_{0}(\varepsilon,k,b-a,M)$ be such that $W(\bm{B},L^{N}_{k+1})\geq
1-\varepsilon^{\prime}$ on $\mathsf{Sep}^{N}_{k}(\delta,\bm{B},L^{N}_{k+1})$
for $N>N_{0}$ (by Lemma 9.1). Then

$\displaystyle\frac{\mathbb
E_{\mathcal{F}}\left[F(\bm{B})W(\bm{B},L^{N}_{k+1})\right]}{\mathbb
E_{\mathcal{F}}[W(\bm{B},L^{N}_{k+1})]}$ $\displaystyle\geq\frac{\mathbb
E_{\mathcal{F}}\left[F(\bm{B})\mathbbm{1}_{\mathsf{Sep}^{N}_{k}(\delta,\bm{B},L^{N}_{k+1})}W(\bm{B},L^{N}_{k+1})\right]}{\mathbb
E_{\mathcal{F}}[W(\bm{B},L^{N}_{k+1})]}$
$\displaystyle\geq(1-\varepsilon^{\prime})\frac{\mathbb
E_{\mathcal{F}}\left[F(\bm{B})\mathbbm{1}_{\mathsf{Sep}^{N}_{k}(\delta,\bm{B},L^{N}_{k+1})}\right]}{\mathbb
E_{\mathcal{F}}[W(\bm{B},L^{N}_{k+1})]}.$

Now invoking Lemma 12.1 and our choice of $\delta$, with probability at least
$1-\varepsilon$, the previous display is lower bounded by

$\displaystyle\frac{1-\varepsilon^{\prime}}{1+\varepsilon^{\prime}}\cdot\frac{\mathbb
E_{\mathcal{F}}\left[F(\bm{B})\mathbbm{1}_{\mathsf{Sep}^{N}_{k}(\delta,\bm{B},L^{N}_{k+1})}\right]}{\mathbb
P_{\mathcal{F}}(\mathsf{Sep}^{N}_{k}(\delta,\bm{B},L^{N}_{k+1}))}=\frac{1-\varepsilon^{\prime}}{1+\varepsilon^{\prime}}\cdot\mathbb
E_{\mathcal{F}}[F(\bm{B})\mid\mathsf{Sep}^{N}_{k}(\delta,\bm{B},L^{N}_{k+1})].$

Rearranging this gives one side of the claimed inequality after relabeling
$\varepsilon$, since $F$ is bounded by $M$, and the other is obtained by
replacing $F$ with $M-F$, and rearranging appropriately again. ∎

###  12.2. Brownian Gibbs property in the limit

Throughout this section, we recall the definition of the Brownian Gibbs
property from Definition 6.4.

######  Lemma 12.3.

For each $N\in\mathbb{N}$, let $B^{N}$ be a Bernoulli random walk bridge from
$(0,0)$ to $(N,z_{N})$, and suppose $z_{N}/N\to p\in(-1,0)$ as $N\to\infty$.
Then

$\displaystyle N^{-1/2}(B^{N}(Nt)-z_{N}t)\stackrel{{\scriptstyle
d}}{{\to}}B^{\mathrm{Br}}(t)$

as $N\to\infty$ as processes on $\mathcal{C}([0,1],\mathbb{R})$ with the
topology of uniform convergence, where $B^{\mathrm{Br}}$ is a Brownian bridge
on $[0,1]$ with zero endpoints and variance $p(1-p)$.

###### Proof.

This follows from the KMT coupling Lemma 6.14 (with the $p$ there equal to
$z_{N}/N$ here). ∎

###### Proof of Proposition 6.5.

It is sufficient to prove the resampling property when resampling the top $k$
curves for any $k\in\mathbb{N}$, i.e, prove Definition 6.4 holds with $j=1$,
as the conditional distribution in the case of $j>1$ is obtained from the
$j=1$ case along with the fact that non-intersecting Brownian bridges have the
Brownian Gibbs property themselves [DM21, Lemma 2.13] (this is stated in the
case of no lower boundary condition, but the proof also applies if there is
one). More precisely, given an interval $[a,b]$ and knowledge that the
Brownian Gibbs property holds for a line ensemble $\bm{\mathfrak{L}}$ when
$j=1$ (i.e., when conditioned on
$\mathcal{F}_{\mathrm{ext}}(k,[a,b],\bm{\mathfrak{L}})$), to obtain the same
for $j>1$, one further conditions on $\mathcal{F}_{\mathrm{ext}}(\llbracket
j,k\rrbracket,[a,b],\bm{\mathfrak{L}})$ and makes use of [DM21, Lemma 2.13]
along with the fact that iterative conditioning is commutative [Kal21, Theorem
8.15].

Now we turn to prove the $j=1$ case. For $k\in\mathbb{N}$ and $a<b$, let
$\bm{\mathfrak{B}}:\llbracket 1,k\rrbracket\times[a,b]\to\mathbb{R}$ with
$\bm{\mathfrak{B}}=(\mathfrak{B}_{1},\ldots,\mathfrak{B}_{k})$ be a collection
of $k$ rate two independent Brownian bridges, with
$\mathfrak{B}_{i}(x)=\mathfrak{L}_{i}(x)$ for $i\in\llbracket 1,k\rrbracket$
and $x\in\\{a,b\\}$. Similar to (12.1), we extend the definition (9.17) to
emphasize the lower boundary condition: for an interval $I$, define

$\displaystyle\mathsf{NonInt}_{k}(I,\bm{\mathfrak{B}},\mathfrak{L}_{k+1}):=\Bigl{\\{}\mathfrak{B}_{1}(x)>\mathfrak{B}_{2}(x)>\ldots>\mathfrak{B}_{k}(x)>\mathfrak{L}_{k+1}(x)\text{
for all }x\in I\Bigr{\\}}.$

In the rest of the proof we take $I=[a,b]$ and omit it from the notation. Let
$\mathcal{F}=\mathcal{F}_{\mathrm{ext}}(k,[a,b],\bm{\mathfrak{L}})$. For any
$k\in\mathbb{N}$, $a<b$, and $F:\mathcal{C}(\llbracket
1,k\rrbracket\times[a,b],\mathbb{R})\to\mathbb{R}$ bounded and continuous, we
have to show that

$\displaystyle\mathbb{E}\left[F(\bm{\mathfrak{L}})\mid\mathcal{F}\right]=\mathbb
E_{\mathcal{F}}[F(\bm{\mathfrak{B}})\mid\mathsf{NonInt}(\bm{\mathfrak{B}},\mathfrak{L}_{k+1})],$

where $F(\bm{X})$ is shorthand for $F(\bm{X}|_{\llbracket
1,k\rrbracket\times[a,b]})$.

By the definition of conditional expectations, and a monotone class argument,
the previous display will be proven if we show that, for any $j\in\mathbb{N}$,
$c<d$, and $G:\mathcal{C}(\llbracket
1,j\rrbracket\times[c,d]\setminus(\llbracket
1,k\rrbracket\times[a,b]),\mathbb{R})\to\mathbb{R}$ bounded and continuous
(with $G(\bm{X})$ shorthand for $G(\bm{X}|_{\llbracket
1,j\rrbracket\times[c,d]\setminus(\llbracket 1,k\rrbracket\times[a,b])})$),

$\displaystyle\mathbb{E}\left[F(\bm{\mathfrak{L}})\cdot
G(\bm{\mathfrak{L}})\right]=\mathbb{E}\bigl{[}\mathbb
E_{\mathcal{F}}[F(\bm{\mathfrak{B}})\mid\mathsf{NonInt}(\bm{\mathfrak{B}},\mathfrak{L}_{k+1})]\cdot
G(\bm{\mathfrak{L}})\bigr{]}.$ (12.2)

By Skorokhod’s representation theorem and by the weak convergence of
$\bm{\mathfrak{L}}^{N}$ to $\bm{\mathfrak{L}}$, we may assume that we work on
a probability space such that $\bm{\mathfrak{L}}^{N}\to\bm{\mathfrak{L}}$
uniformly on compact sets almost surely. Then we obtain, with
$\mathcal{F}^{N}:=\mathcal{F}_{\mathrm{ext}}^{N}(k,\llbracket\beta^{-1}aN^{2/3},\beta^{-1}bN^{2/3}\rrbracket,\bm{L}^{N})$,
by the dominated convergence theorem and the Hall-Littlewood Gibbs property,
that

$\displaystyle\mathbb{E}\left[F(\bm{\mathfrak{L}})\cdot
G(\bm{\mathfrak{L}})\right]$
$\displaystyle=\lim_{N\to\infty}\mathbb{E}\left[F(\bm{\mathfrak{L}}^{N})\cdot
G(\bm{\mathfrak{L}}^{N})\right]$
$\displaystyle=\lim_{N\to\infty}\mathbb{E}\left[\mathbb{E}[F(\bm{\mathfrak{L}}^{N})\mid\mathcal{F}^{N}]\cdot
G(\bm{\mathfrak{L}}^{N})\right]$
$\displaystyle=\lim_{N\to\infty}\mathbb{E}\left[\frac{\mathbb{E}_{\mathcal{F}^{N}}[F(\bm{\mathfrak{B}}^{N})W(\bm{B}^{N},L^{N}_{k+1})]}{\mathbb{E}_{\mathcal{F}^{N}}[W(\bm{B}^{N},L^{N}_{k+1})]}\cdot
G(\bm{\mathfrak{L}}^{N})\right],$ (12.3)

the last line using the Hall-Littlewood Gibbs property and with
$\bm{\mathfrak{B}}^{N}=(B^{N}_{1},\ldots,B^{N}_{k}):\llbracket
1,k\rrbracket\times[a,b]\to\mathbb{R}$ given by (recall $p$ from Assumption 2,
$\beta$ and $\sigma$ from (3.12), and the definition of
$\bm{\mathfrak{L}}^{N}$ in terms of $\bm{L}^{N}$ from (3.13))
$\bm{\mathfrak{B}}^{N}=T_{p,\lambda,N}(\bm{B}^{N})$, i.e.,

$\mathfrak{B}^{N}_{i}(x)=\sigma^{-1}N^{-1/3}\left(B^{N}_{i}(\beta
xN^{2/3})-p\beta xN^{2/3}\right)$

with ${\bm{B}}^{N}=(B^{N}_{1},\ldots,B^{N}_{k})$ a family of $k$ independent
Bernoulli random walk bridges, with $B^{N}_{i}$ from
$(\beta^{-1}aN^{2/3},L^{N}_{i}(\beta^{-1}aN^{2/3}))$ to
$(\beta^{-1}bN^{2/3},L^{N}_{i}(\beta^{-1}bN^{2/3}))$. Now, by Lemma 12.3, it
holds that $\bm{\mathfrak{B}}^{N}$ converges weakly as $N\to\infty$ to
$\bm{\mathfrak{B}}=(\mathfrak{B}_{1},\ldots,\mathfrak{B}_{k})$, as above, a
collection of $k$ independent rate two Brownian bridges with
$\mathfrak{B}_{i}$ going from $(a,\mathfrak{L}_{i}(a))$ to
$(b,\mathfrak{L}_{i}(b))$; the fact that the rate of the Brownian bridges is
$2$ is a consequence of the relation (3.12) between $\beta$ and $\sigma$ in
the definitions of $\bm{\mathfrak{L}}^{N}$ and thus $\bm{\mathfrak{B}}^{N}$.

Recall the extended definition of $\mathsf{Sep}$ from (12.1). Because it
requires separation of order $N^{1/3}$, which should not be present for the
limiting objects, we define an event $\mathsf{ScSep}$ (scaled separation) for
an interval $[a,b]\subset\mathbb{R}$, an integer $k\in\mathbb{N}$, a (possibly
infinite) interval $\Lambda\subseteq\mathbb{N}$ such that $\llbracket
1,k\rrbracket\subseteq\Lambda$, and processes
$\bm{X}:\Lambda\times\mathbb{R}\to\mathbb{R}$ and $Y:\mathbb{R}\to\mathbb{R}$,
by

$\displaystyle\mathsf{ScSep}_{k}(\delta,[a,b],\bm{X},Y)$
$\displaystyle:=\left\\{\min_{i=1,\ldots,k-1}\inf_{x\in[a,b]}\left(X_{i}(x)-X_{i+1}(x)\right)\geq\delta\right\\}$
$\displaystyle\qquad\cap\left\\{\inf_{x\in[a,b]}\left(X_{k+1}(x)-Y(x)\right)\geq\delta\right\\}.$

Then observe that, by the definitions of $\bm{\mathfrak{B}}^{N}$ and
$\bm{\mathfrak{L}}^{N}$, it holds that

$\displaystyle\mathsf{Sep}^{N}_{k}(\delta\sigma,[\beta^{-1}a,\beta^{-1}b],\bm{B}^{N},L^{N}_{k+1})=\mathsf{ScSep}_{k}(\delta,[a,b],\bm{\mathfrak{B}}^{N},\mathfrak{L}^{N}_{k+1}).$

By Corollary 12.2 and the previous display, we obtain that there is a sequence
$\varepsilon_{N,\delta}\downarrow\varepsilon_{\delta}$ as $N\to\infty$, where
$\varepsilon_{\delta}$ is a sequence which goes to $0$ as $\delta\downarrow
0$, such that (12.3) equals

$\displaystyle\lim_{N\to\infty}\mathbb{E}\left[\left(\mathbb{E}_{\mathcal{F}^{N}}[F(\bm{\mathfrak{B}}^{N})\mid\mathsf{ScSep}_{k}(\delta,[a,b],\bm{\mathfrak{B}}^{N},\mathfrak{L}^{N}_{k+1})]+\varepsilon_{N,\delta}\right)\cdot
G(\bm{\mathfrak{L}}^{N})\right]$
$\displaystyle=\mathbb{E}\left[\lim_{N\to\infty}\left(\mathbb{E}_{\mathcal{F}^{N}}[F(\bm{B}^{N})\mid\mathsf{ScSep}_{k}(\delta,[a,b],\bm{\mathfrak{B}}^{N},\mathfrak{L}^{N}_{k+1})]+\varepsilon_{N,\delta}\right)\cdot\lim_{N\to\infty}G(\bm{\mathfrak{L}}^{N})\right]$
$\displaystyle=\mathbb{E}\left[(\mathbb{E}_{\mathcal{F}}[F(\bm{\mathfrak{B}})\mid\mathsf{ScSep}_{k}(\delta,[a,b],\bm{\mathfrak{B}},\mathfrak{L}_{k+1})]+\varepsilon_{\delta})\cdot
G(\bm{\mathfrak{L}})\right];$

the penultimate equality used the dominated convergence theorem since $F$ and
$G$ are bounded and continuous, and the final equality used the Portmanteau
theorem along with the fact that
$\mathbb{P}(\mathsf{ScSep}_{k}(\delta,[a,b],\bm{\mathfrak{B}},\mathfrak{L}_{k+1}))$
$>0$ (since $\mathfrak{L}_{k+1}$ is a continuous function and with probability
$1$ there is strictly positive separation between the endpoints of
$\mathfrak{B}_{i}$ and $\mathfrak{B}_{i+1}$ as well as between those of
$\mathfrak{B}_{k}$ and $\mathfrak{B}_{k+1}$, as follows from Theorem 9.2) to
also take the limit of the conditioning.

Finally taking $\delta\to 0$ and using that
$\mathbb{E}_{\mathcal{F}}[F(\bm{\mathfrak{B}})\mid\mathsf{ScSep}_{k}(\delta,[a,b],\bm{\mathfrak{B}},\mathfrak{L}_{k+1})]\to\mathbb{E}_{\mathcal{F}}[F(\bm{\mathfrak{B}})\mid\mathsf{NonInt}(\bm{\mathfrak{B}},\mathfrak{L}_{k+1})]$
completes the proof. ∎

##  Appendix A The Yang-Baxter equation, S6V, and the $q$-Boson model

In this section we give a self-contained proof of the partition function
expression in (3.1) and of Proposition 3.12 on the distributional relation of
the top line of the colored Hall-Littlewood line ensemble and the colored
height function of the colored S6V model. First, recall the description of the
colored $q$-Boson model from Section 3.1 and the colored S6V model from
Section 2.3, with the domain restriction performed as in Section 3.2.1. In
particular, in both models we have a spectral parameter $z\in(0,1)$ fixed and
a boundary condition $\sigma:\llbracket 1,N\rrbracket\to\llbracket
1,N\rrbracket$ where the arrow horizontally entering $(1,k)$ (for colored S6V)
and the arrow horizontally entering $(-K,k)$ for all large enough $K$ (for
colored $q$-Boson) are both of color $\sigma(k)$, for each $k\in\llbracket
1,N\rrbracket$. To explain the relation between the colored $q$-Boson model
and the colored S6V model, we need the Yang-Baxter relation enjoyed by these
models, which we introduce next. Then in Section A.2 we establish the
expression for the partition function of the colored $q$-Boson model and in
Section A.3 we give the proof of Proposition 3.12.

###  A.1. Yang-Baxter equation

The Yang-Baxter equation is an equality of partition functions of
configurations consisting of three vertices. In our case, the first vertex has
vertex weight given by the weights of the colored S6V model (Figure 8) and the
other two vertices have weights given by the
$\scalebox{1.05}{\mbox{$\mathtt{L}$}}_{u}$ weights (Figure 10). Let us write
the weights from Figure 8 with spectral parameter $u\in(0,1)$ as
$\scalebox{1.05}{\mbox{$\mathtt{R}$}}_{u}(a,i;b,j)$ with $a,i,b,j\in\llbracket
0,N\rrbracket$, where recall $a,i,b,j$ respectively represent the colors of
the arrows vertically entering, horizontally entering, vertically exiting, and
horizontally exiting from the vertex. Unlike the convention in Section 2.3,
here we will take $0$ to represent the absence of an arrow, so as to match the
setup of the colored $q$-Boson model. In this notation, the Yang-Baxter
equation can be written as follows; it was originally due to [KRS81, Baz85,
Jim86], but the notation we have adopted is from [BW22] (it can also be
verified directly from the formulas for the weights).

######  Lemma A.1 (Yang-Baxter equation, [BW22, Proposition 2.3.1]).

Fix $a_{1},i_{1},b_{2},j_{2}\in\llbracket 0,N\rrbracket$ and
$\bm{I},\bm{J}\in\mathbb{Z}_{\geq 0}^{N}$. Let $x,y\in(0,1)$ and let $z=y/x$.
Then it holds that

$\displaystyle\sum_{\begin{subarray}{c}b_{1},j_{1}\in\llbracket
0,N\rrbracket,\\\ \bm{K}\in\mathbb{Z}_{\geq
0}^{N}\end{subarray}}\scalebox{1.05}{\mbox{$\mathtt{R}$}}_{y/x}(a_{1},i_{1};b_{1},j_{1})\,\scalebox{1.05}{\mbox{$\mathtt{L}$}}_{x}(\bm{I},j_{1};\bm{K},j_{2})\,\scalebox{1.05}{\mbox{$\mathtt{L}$}}_{y}(\bm{K},b_{1};\bm{J},b_{2})$
$\displaystyle=\sum_{\begin{subarray}{c}b_{1},j_{1}\in\llbracket
0,N\rrbracket,\\\ \bm{K}\in\mathbb{Z}_{\geq
0}^{N}\end{subarray}}\,\scalebox{1.05}{\mbox{$\mathtt{L}$}}_{y}(\bm{I},a_{1};\bm{K},b_{1})\,\scalebox{1.05}{\mbox{$\mathtt{L}$}}_{x}(\bm{K},i_{1};\bm{J},j_{1})\,\scalebox{1.05}{\mbox{$\mathtt{R}$}}_{y/x}(b_{1},j_{1};b_{2},j_{2});$

here $\scalebox{1.05}{\mbox{$\mathtt{L}$}}_{u}$ and
$\scalebox{1.05}{\mbox{$\mathtt{R}$}}_{u}$ are zero if their arguments do not
satisfy colored arrow conservation.

Diagrammatically, the Yang-Baxter equation can be written as follows (we adopt
a style of depiction similar to [BW22]).

$\displaystyle\sum_{\begin{subarray}{c}b_{1},j_{1}\in\llbracket
0,N\rrbracket,\\\ \bm{K}\in\mathbb{Z}_{\geq 0}^{N}\end{subarray}}\vbox{\hbox{
\leavevmode\hbox to100.28pt{\vbox
to80.05pt{\pgfpicture\makeatletter\hbox{\hskip 70.10686pt\lower-30.0663pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }{{}}{
{}{{}}{}{{}{}{}{{}}{{{}{}{}{}}}}{} {} {} {}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setdash{0.4pt,1.0pt}{0.0pt}\pgfsys@invoke{
}{}\pgfsys@moveto{-24.89604pt}{4.9792pt}\pgfsys@curveto{-22.14621pt}{7.72903pt}{-22.14621pt}{12.18756pt}{-24.89604pt}{14.9374pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}{{}}{{}} {}{}
{}{}{{}}{{}} {}{}{{}}{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{1.0pt}\pgfsys@invoke{
}{}{}{}{}{{}}\pgfsys@moveto{-49.7921pt}{19.91684pt}\pgfsys@lineto{-39.83368pt}{19.91684pt}\pgfsys@lineto{-19.91684pt}{0.0pt}\pgfsys@lineto{17.01683pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
}{{}{{}}{}{}{{}}{{{}}{{{}}{\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{17.01683pt}{0.0pt}\pgfsys@invoke{
}\pgfsys@invoke{ \lxSVG@closescope }\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope}}{{}}}}\hbox{\hbox{\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{
}{{}{}{{ {}{}}}{ {}{}} {{}{{}}}{{}{}}{}{{}{}}{}{}{}{}{} {
}{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{0.7}{0.0}{0.0}{0.7}{-51.97787pt}{23.86304pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{$i_{1}$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope}}}\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{ {}{}}}{
{}{}} {{}{{}}}{{}{}}{}{{}{}}{}{}{}{}{} {
}{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{0.7}{0.0}{0.0}{0.7}{-22.53848pt}{-7.29974pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{$j_{1}$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope}}}\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{ {}{}}}{
{}{}} {{}{{}}}{{}{}}{}{{}{}}{}{}{}{}{} {
}{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{0.7}{0.0}{0.0}{0.7}{22.59993pt}{-1.62778pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{$j_{2}$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}{{}}{{}} {}{}
{}{}{{}}{{}} {}{}{{}}{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{1.0pt}\pgfsys@invoke{
}{}{}{}{}{{}}\pgfsys@moveto{-49.7921pt}{0.0pt}\pgfsys@lineto{-39.83368pt}{0.0pt}\pgfsys@lineto{-19.91684pt}{19.91684pt}\pgfsys@lineto{17.01683pt}{19.91684pt}\pgfsys@stroke\pgfsys@invoke{
}{{}{{}}{}{}{{}}{{{}}{{{}}{\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{17.01683pt}{19.91684pt}\pgfsys@invoke{
}\pgfsys@invoke{ \lxSVG@closescope }\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope}}{{}}}}\hbox{\hbox{\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{
}{{}{}{{ {}{}}}{ {}{}} {{}{{}}}{{}{}}{}{{}{}}{}{}{}{}{} {
}{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{0.7}{0.0}{0.0}{0.7}{-52.62213pt}{3.9462pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{$a_{1}$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope}}}\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{ {}{}}}{
{}{}} {{}{{}}}{{}{}}{}{{}{}}{}{}{}{}{} {
}{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{0.7}{0.0}{0.0}{0.7}{-22.3989pt}{23.86304pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{$b_{1}$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope}}}\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{ {}{}}}{
{}{}} {{}{{}}}{{}{}}{}{{}{}}{}{}{}{}{} {
}{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{0.7}{0.0}{0.0}{0.7}{22.59993pt}{18.11786pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{$b_{2}$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}{{}}{{}}
{}{}{{}}{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{2.0pt}\pgfsys@invoke{ }{}{{ {\pgfsys@beginscope{} {} {}
{}
\pgfsys@moveto{4.40002pt}{0.0pt}\pgfsys@lineto{-2.64001pt}{3.52002pt}\pgfsys@lineto{0.0pt}{0.0pt}\pgfsys@lineto{-2.64001pt}{-3.52002pt}\pgfsys@fill\pgfsys@endscope}}
}{}{}{{}}\pgfsys@moveto{0.0pt}{-19.91684pt}\pgfsys@lineto{0.0pt}{35.43365pt}\pgfsys@stroke\pgfsys@invoke{
}{{}{{}}{}{}{{}}{{{}}{{{}}{\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{0.0}{1.0}{-1.0}{0.0}{0.0pt}{35.43365pt}\pgfsys@invoke{
}\pgfsys@invoke{ \lxSVG@closescope }\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope}}{{}}}}\hbox{\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{
}{{}{}{{ {}{}}}{ {}{}} {{}{{}}}{{}{}}{}{{}{}}{}{}{}{}{} {
}{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{0.7}{0.0}{0.0}{0.7}{-1.81317pt}{-27.73322pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{$\bm{I}$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope}}}\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{ {}{}}}{
{}{}} {{}{{}}}{{}{}}{}{{}{}}{}{}{}{}{} {
}{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{0.7}{0.0}{0.0}{0.7}{-2.2774pt}{42.86676pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{$\bm{J}$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{2.0pt}\pgfsys@invoke{
}{}{}{}{}{{}}{{}{{}}{}{}{{}}{{{}}{{{}}{\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{0.0}{1.0}{-1.0}{0.0}{0.0pt}{9.5417pt}\pgfsys@invoke{
}\pgfsys@invoke{ \lxSVG@closescope }\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope}}{{}}}} \pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{{}}{{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{
}{{}{}{{ {}{}}}{ {}{}} {{}{{}}}{{}{}}{}{{}{}}{}{}{}{}{} {
}{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{0.7}{0.0}{0.0}{0.7}{6.73553pt}{7.56677pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{$\bm{K}$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
{{}}{{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{{}{}}}{{}{}}
{{}{{}}}{{}{}}{}{{}{}}{}{}{}{}{} { }{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{0.9}{0.0}{0.0}{0.9}{-67.10718pt}{16.70187pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{$x\rightarrow$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
{{}}{{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{{}{}}}{{}{}}
{{}{{}}}{{}{}}{}{{}{}}{}{}{}{}{} { }{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{0.9}{0.0}{0.0}{0.9}{-66.69888pt}{-2.33998pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{$y\rightarrow$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} }
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}} }
}=\sum_{\begin{subarray}{c}b_{1},j_{1}\in\llbracket 0,N\rrbracket,\\\
\bm{K}\in\mathbb{Z}_{\geq 0}^{N}\end{subarray}}\vbox{\hbox{ \leavevmode\hbox
to100.28pt{\vbox to80.05pt{\pgfpicture\makeatletter\hbox{\hskip
70.10686pt\lower-30.0663pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }{{}}{
{}{{}}{}{{}{}{}{{}}{{{}{}{}{}}}}{} {} {} {}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setdash{0.4pt,1.0pt}{0.0pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{4.9792pt}\pgfsys@curveto{2.74983pt}{7.72903pt}{2.74983pt}{12.18756pt}{0.0pt}{14.9374pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}{{}}{{}}
{}{}{{}}{{}} {}{} {}{}{{}}{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{1.0pt}\pgfsys@invoke{
}{}{}{}{}{{}}\pgfsys@moveto{-49.7921pt}{19.91684pt}\pgfsys@lineto{-14.93762pt}{19.91684pt}\pgfsys@lineto{4.9792pt}{0.0pt}\pgfsys@lineto{17.01683pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
}{{}{{}}{}{}{{}}{{{}}{{{}}{\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{17.01683pt}{0.0pt}\pgfsys@invoke{
}\pgfsys@invoke{ \lxSVG@closescope }\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope}}{{}}}}\hbox{\hbox{\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{
}{{}{}{{ {}{}}}{ {}{}} {{}{{}}}{{}{}}{}{{}{}}{}{}{}{}{} {
}{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{0.7}{0.0}{0.0}{0.7}{-51.97787pt}{23.86304pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{$i_{1}$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope}}}\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{ {}{}}}{
{}{}} {{}{{}}}{{}{}}{}{{}{}}{}{}{}{}{} {
}{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{0.7}{0.0}{0.0}{0.7}{-17.55927pt}{23.96101pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{$j_{1}$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope}}}\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{ {}{}}}{
{}{}} {{}{{}}}{{}{}}{}{{}{}}{}{}{}{}{} {
}{{{{}}\pgfsys@beginscope\pgfsys@invoke{
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}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{$j_{2}$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}{{}}{{}}
{}{}{{}}{{}} {}{} {}{}{{}}{{}}\pgfsys@beginscope\pgfsys@invoke{
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}{}{}{}{}{{}}\pgfsys@moveto{-49.7921pt}{0.0pt}\pgfsys@lineto{-14.93762pt}{0.0pt}\pgfsys@lineto{4.9792pt}{19.91684pt}\pgfsys@lineto{17.01683pt}{19.91684pt}\pgfsys@stroke\pgfsys@invoke{
}{{}{{}}{}{}{{}}{{{}}{{{}}{\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{17.01683pt}{19.91684pt}\pgfsys@invoke{
}\pgfsys@invoke{ \lxSVG@closescope }\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope}}{{}}}}\hbox{\hbox{\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{
}{{}{}{{ {}{}}}{ {}{}} {{}{{}}}{{}{}}{}{{}{}}{}{}{}{}{} {
}{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{0.7}{0.0}{0.0}{0.7}{-52.62213pt}{3.9462pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{$a_{1}$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope}}}\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{ {}{}}}{
{}{}} {{}{{}}}{{}{}}{}{{}{}}{}{}{}{}{} {
}{{{{}}\pgfsys@beginscope\pgfsys@invoke{
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}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{$b_{1}$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope}}}\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{ {}{}}}{
{}{}} {{}{{}}}{{}{}}{}{{}{}}{}{}{}{}{} {
}{{{{}}\pgfsys@beginscope\pgfsys@invoke{
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}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}{{}}{{}}
{}{}{{}}{{}}\pgfsys@beginscope\pgfsys@invoke{
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}{}{}{}{}{{}}\pgfsys@moveto{-34.85448pt}{-19.91684pt}\pgfsys@lineto{-34.85448pt}{35.43365pt}\pgfsys@stroke\pgfsys@invoke{
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}\pgfsys@invoke{ \lxSVG@closescope }\pgfsys@invoke{\lxSVG@closescope
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}{{}{}{{ {}{}}}{ {}{}} {{}{{}}}{{}{}}{}{{}{}}{}{}{}{}{} {
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}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope
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{}{}} {{}{{}}}{{}{}}{}{{}{}}{}{}{}{}{} {
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\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{{}}{{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{
}{{}{}{{ {}{}}}{ {}{}} {{}{{}}}{{}{}}{}{{}{}}{}{}{}{}{} {
}{{{{}}\pgfsys@beginscope\pgfsys@invoke{
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}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{$\bm{K}$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} {}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{2.0pt}\pgfsys@invoke{
}{}{}{}{}{{}}{{}{{}}{}{}{{}}{{{}}{{{}}{\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{0.0}{1.0}{-1.0}{0.0}{-34.85448pt}{9.5417pt}\pgfsys@invoke{
}\pgfsys@invoke{ \lxSVG@closescope }\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope}}{{}}}} \pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{{}}{{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{
}{{}{}{{{}{}}}{{}{}} {{}{{}}}{{}{}}{}{{}{}}{}{}{}{}{} {
}{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{0.9}{0.0}{0.0}{0.9}{-67.10718pt}{16.70187pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{$x\rightarrow$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
{{}}{{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{{}{}}}{{}{}}
{{}{{}}}{{}{}}{}{{}{}}{}{}{}{}{} { }{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{0.9}{0.0}{0.0}{0.9}{-66.69888pt}{-2.33998pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{$y\rightarrow$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} }
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}} } }$ (A.1)

Here, the left and righthand sides mean the sum over the weights of the
depicted configurations of the three vertices, and the arc between two edges
indicates which of those two edges corresponds to exiting vertically (upper
end of arc) and horizontally (lower end of arc). The weights of the two
vertices at the intersection of the two horizontal lines and the thick
vertical line are the $\scalebox{1.05}{\mbox{$\mathtt{L}$}}_{u}$ weights (so
they can hold an arbitrary number of arrows) with spectral parameter $u$ as
indicated by $x$ or $y$ at the beginning of the line (which should be followed
straight through the cross), while the weight of the vertex at the center of
the cross is the $\scalebox{1.05}{\mbox{$\mathtt{R}$}}_{z}$ weights with
spectral parameter $z=y/x$.

$\cdots$$1$$1$$1$$z$$z$$z$$z$$\sigma(1)$$\sigma(2)$$\sigma(3)$$M$$N$$\mathcal{R}_{1}$$\mathcal{R}_{2}$
Figure 22. The model with the $M\times N$ rectangle of
$\scalebox{1.05}{\mbox{$\mathtt{R}$}}_{z}$-weight vertices attached to the
left. If any arrow tries to deviate from the configuration depicted except for
finitely many columns at the right end (i.e., the portion to the right of the
$\bm{\cdots}$), some arrow must travel infinitely far horizontally in the top
$M$ rows, where the weight of a single arrow moving horizontally (with no
other arrows incident) across a vertex is $z\in(0,1)$; thus any such
configuration has weight zero and the depicted configuration must occur in all
but finitely many columns (that is, to the left of the $\bm{\cdots}$). The
arrows paths have not been shown in the region to the right of the
$\bm{\cdots}$, where their trajectories are random.

###  A.2. Partition function of the colored $q$-Boson model

With the Yang-Baxter equation in hand, we may proceed to derive the expression
for the partition function of the colored $q$-Boson model given in (3.1). To
do this we consider an augmented vertex model, as in Figure 22, by the
following informal description (see the next paragraph for a more formal one).
We attach a copy of an $M\times N$ rectangular grid (which we denote
$\mathcal{R}_{1}$), each of whose vertices has vertex weights given by the
$\scalebox{1.05}{\mbox{$\mathtt{R}$}}_{z}$-weights (i.e., with spectral
parameter $z$ as that of the colored $q$-Boson model), to the left of the
original domain $\mathbb{Z}_{\leq 0}\times\llbracket 1,N+M\rrbracket$ (which
we denote $\mathcal{R}_{2}$). The augmented model has boundary conditions
given by an arrow of color $\sigma(i)$ horizontally entering at the $i$th row
of the just added rectangle, no arrows vertically entering, no arrows
horizontally exiting, and all arrows vertically exiting at $(0,N+M)$ in the
original domain.

We are being somewhat informal in saying to attach the $M\times N$ rectangle
$\mathcal{R}_{1}$ to the “left” of the semi-infinite domain $\mathcal{R}_{2}$;
more formally, a configuration of the augmented model (with boundary condition
$\sigma$) consists of an assignment of arrow configurations to the vertices of
$\mathcal{R}_{1}$ as well as $\mathcal{R}_{2}$ such that (i) consistency and
colored arrow conservation are satisfied within each, (ii) the entering
boundary condition for $\mathcal{R}_{1}$ is given by $\sigma$, (iii) the
exiting boundary condition for $\mathcal{R}_{2}$ is that
$(\\#\sigma^{-1}(\\{j\\}))_{j=1}^{N}$ vertically exits from $(0,N+M)$, and
(iv) the exiting arrow configurations of $\mathcal{R}_{1}$ equals the boundary
condition of $\mathcal{R}_{2}$ as shown in Figure 22, i.e., for all large
enough $K$ and all $y\in\llbracket 1,N+M\rrbracket$, the arrow configuration
at $(-K,y)\in\mathcal{R}_{2}$ equals $(0^{N},\tau_{y};0^{n},\tau_{y})$, where
$\tau_{y}\in\llbracket 0,N\rrbracket$ is the color of the arrow exiting
$\mathcal{R}_{1}$ from the $y$th vertex on its right and top boundaries
(counted counterclockwise).

We first observe that the horizontally entering arrows to $\mathcal{R}_{1}$
are forced to go horizontally straight through: if any arrow does not pass
straight through, at least one arrow will enter $\mathcal{R}_{2}$ through one
of the top $M$ lines and then travel an infinite horizontal distance; any such
configuration of arrows has weight zero since, under the
$\scalebox{1.05}{\mbox{$\mathtt{L}$}}_{z}$ weights, the individual vertex
weight of passing horizontally through a given vertex in the top $M$ rows is
$z<1$. Thus the partition function $\mathcal{Z}^{\mathrm{aug}}_{\sigma}$ of
the augmented model is the partition function
$\mathcal{Z}^{\mathrm{cB}}_{\sigma}$ of the original colored $q$-Boson model
(with boundary condition $\sigma$) multiplied by the weight of the
configuration of arrows passing horizontally straight through
$\mathcal{R}_{1}$; by Figure 8, the latter is
$\left((1-z)/(1-qz)\right)^{NM}$, so that

$\displaystyle\mathcal{Z}^{\mathrm{aug}}_{\sigma}=\left(\frac{1-z}{1-qz}\right)^{NM}\cdot\mathcal{Z}^{\mathrm{cB}}_{\sigma}.$
(A.2)

Next, observe that there are $MN$ vertices present in the rectangle
$\mathcal{R}_{1}$ in the augmented model. We move each of these vertices to
the right, one at a time (starting at the rightmost corner vertex of the
rectangle and ending at the leftmost one), all the way across
$\mathcal{R}_{2}$. By the Yang-Baxter equation Lemma A.1 (see also (A.1)),
this does not change the partition function
$\mathcal{Z}_{\sigma}^{\mathrm{aug}}$. Upon doing so, we obtain the situation
depicted in Figure 23. Observe that again the only configuration of arrows
with non-zero weight is the one depicted, in which the arrows pass
horizontally straight through (this uses that an arrow is entering at every
one of the top $N$ rows). By Figure 10, the configuration in $\mathcal{R}_{2}$
has weight $1$, and from Figure 8, the empty configuration in
$\mathcal{R}_{1}$ also has weight $1$. Thus, for any $\sigma$,

$\displaystyle\mathcal{Z}^{\mathrm{aug}}=1.$ (A.3)

Note that the partition function was not changed between Figures 22 and 23. In
other words, we can conclude from (A.2) and (A.3) that
$\mathcal{Z}^{\mathrm{cB}}_{\sigma}=\left((1-qz)/(1-z)\right)^{NM}$ for any
$\sigma$, completing the proof of the claim in (3.1).

$\cdots$$1$$1$$1$$z$$z$$z$$z$$\sigma(1)$$\sigma(2)$$\sigma(3)$$N$$M$$\mathcal{R}_{2}$$\mathcal{R}_{1}$
Figure 23. The configuration after all the crosses of $\mathcal{R}_{1}$ have
been pulled all the way through $\mathcal{R}_{2}$ to the right by applying
(A.1) repeatedly. We note that now, due to the location where all the arrows
must exit and by the condition that at most one arrow can be present on each
horizontal edge, the arrows are forced into the single configuration of moving
horizontally straight through till the very last column, where they move
vertically.

###  A.3. Proof of Proposition 3.12

Fix $M,N\in\mathbb{N}$ and a boundary condition $\sigma:\llbracket
1,N\rrbracket\to\llbracket 1,N\rrbracket$, and recall $h^{\mathrm{S6V}}$ from
(2.17). We define a variant height function $\tilde{h}^{\mathrm{S6V}}$ as
follows. Let
$(j^{\mathrm{S6V}}_{v},b^{\mathrm{S6V}}_{v})_{v\in\mathcal{R}_{1}}$ be sampled
from the colored S6V measure on $\mathcal{R}_{1}$ with boundary condition
$\sigma$, where (as in Section 2.3.1) $j^{\mathrm{S6V}}_{v}$ and
$b^{\mathrm{S6V}}_{v}$ are the colors of the arrows exiting horizontally and
vertically, respectively, from the vertex $v$. For $k\in\llbracket
1,N\rrbracket$ and $y\in\llbracket 0,N-1\rrbracket$, define

$\tilde{h}^{\mathrm{S6V}}(k;y):=\\#\left\\{i>y:j^{\mathrm{S6V}}_{(M,i)}\geq
k\right\\}=h^{\mathrm{S6V}}(k,0;y,M).$ (A.4)

For $k\in\llbracket 1,N\rrbracket$ and $y\in\llbracket N,N+M\rrbracket$,
define

$\displaystyle\tilde{h}^{\mathrm{S6V}}(k;y):=\\#\left\\{i<N+M-y:b^{\mathrm{S6V}}_{(i,N)}\geq
k\right\\}.$

In words, $\tilde{h}^{\mathrm{S6V}}(k;y)$ counts the number of arrows of color
at least $k$ which exit from the right and top boundaries of $\llbracket
1,M\rrbracket\times\llbracket 1,N\rrbracket$ from a site which is $(y+1)$st or
later when counted counterclockwise from the bottom right corner.

Consider a sequence of Bernoulli paths $L^{(k)}:\llbracket
0,N+M\rrbracket\to\mathbb{Z}$ such that, (i)
$L^{(k)}(0)=\\#\sigma^{-1}(\llbracket k,N\rrbracket)$ and $L^{(k)}(N+M)=0$,
and (ii) $L^{(k)}(\bm{\cdot})-L^{(k+1)}(\bm{\cdot})$ is also a Bernoulli path.
We will show that

$\begin{split}&\mathbb{P}\left((\tilde{h}^{\mathrm{S6V}}(k;y))_{k\in\llbracket
1,N\rrbracket,y\in\llbracket 0,N+M\rrbracket}=(L^{(k)}(y))_{k\in\llbracket
1,N\rrbracket,y\in\llbracket 0,N+M\rrbracket}\right)\\\
&=\mathbb{P}\left((L^{\mathrm{cHL},(k)}_{1}(y))_{k\in\llbracket
1,N\rrbracket,y\in\llbracket 0,N+M\rrbracket}=(L^{(k)}(y))_{k\in\llbracket
1,N\rrbracket,y\in\llbracket 0,N+M\rrbracket}\right);\end{split}$ (A.5)

this implies Proposition 3.12 by a projection of $y$ onto $\llbracket
1,N-1\rrbracket$ and (A.4).

To establish (A.5), we first define an exiting boundary condition
$(j^{\prime}_{(-1,y)})_{y\in\llbracket 1,N+M\rrbracket}$ in terms of
$\smash{(L^{(k)}(y))_{k\in\llbracket 1,N\rrbracket,y\in\llbracket
0,N+M\rrbracket}}$ as follows: letting $\Delta(L,y)=L(y-1)-L(y)\in\\{0,1\\}$
for a Bernoulli path $L:\llbracket 0,N+M\rrbracket\to\mathbb{Z}$, for
$y\in\llbracket 1,N+M\rrbracket$,

$\displaystyle j^{\prime}_{(-1,y)}:=\min\left\\{k\in\llbracket
1,N\rrbracket:\Delta(L^{(k+1)},y)=\Delta(L^{(k)},y)-1\right\\},$

where by convention $\Delta(L^{(N+1)},y)=0$ for all $y$ and $\min\emptyset=0$.

Let $(j^{\mathrm{cB}}_{v})_{v\in\mathcal{R}_{2}}$ be sampled according to the
colored $q$-Boson measure (as in Section 3.1.1 and Definition 3.1) with
boundary condition $\sigma$. It follows from Definition 3.7 of
$\bm{L}^{\mathrm{cHL},(k)}$ that the righthand side of (A.5) equals

$\displaystyle\mathbb{P}\left(j^{\mathrm{cB}}_{(-1,y)}=j^{\prime}_{(-1,y)}\text{
for all }y\in\llbracket 1,N+M\rrbracket\right).$

This in turn equals the ratio of two partition functions,
$\mathcal{Z}^{\mathrm{cB}}_{\sigma}(\bm{j}^{\prime})/\mathcal{Z}^{\mathrm{cB}}_{\sigma}$
(see the left panel of Figure 24): the numerator is the partition function
$\smash{\mathcal{Z}^{\mathrm{cB}}_{\sigma}(\bm{j}^{\prime})}$ of the colored
$q$-Boson measure with (incoming) boundary condition $\sigma$ as well as the
(outgoing) boundary condition $(j^{\prime}_{(-1,y)}:y\in\llbracket
1,N+M\rrbracket)$ at $\\{-1\\}\times\llbracket 1,N+M\rrbracket$, and the
denominator is the partition function
$\smash{\mathcal{Z}^{\mathrm{cB}}_{\sigma}=((1-qz)/(1-z))^{NM}}$ of the whole
measure, as derived in Section A.2.

$\mathcal{Z}^{\mathrm{cB}}(\bm{j}^{\prime})=$$\displaystyle\sum$$\cdots$$z$$z$$z$$z$$1$$1$$1$$\sigma(1)$$\sigma(2)$$\sigma(3)$$M$$N$$\widetilde{\mathcal{R}}_{2}$$j^{\prime}_{1}$$j^{\prime}_{2}$$j^{\prime}_{N+M-1}$$j^{\prime}_{N+M}$...$\mathcal{Z}^{\mathrm{S6V}}(\tilde{\bm{j}},\tilde{\bm{b}})=$$\displaystyle\sum$$\tilde{b}_{1}$$\tilde{b}_{N}$$\tilde{j}_{1}$$\tilde{j}_{M}$......$1$$1$$1$$z$$z$$z$$z$$\sigma(1)$$\sigma(2)$$\sigma(3)$$M$$N$
Figure 24. The definitions of the partition functions
$\mathcal{Z}^{\mathrm{cB}}_{\sigma}(\bm{j}^{\prime})$ (left) and
$\mathcal{Z}^{\mathrm{S6V}}_{\sigma}(\tilde{\bm{b}},\tilde{\bm{j}})$ (right);
here the summation means to add the weights of all configurations which
satisfy the depicted boundary conditions defined by $\sigma$ and
$\bm{j}^{\prime}$ in the left panel and $\sigma$, $\tilde{\bm{b}}$, and
$\tilde{\bm{j}}$ in the right panel.

Recall that $(b^{\mathrm{S6V}}_{(x,N)},j^{\mathrm{S6V}}_{(M,y)}:x\in\llbracket
1,M\rrbracket,y\in\llbracket 1,N\rrbracket)$ are sampled according to the
colored S6V model with boundary condition $\sigma$. Let
$\smash{\tilde{b}_{x}:=j^{\prime}_{(-1,N+M+1-x)}}$ for $x\in\llbracket
1,M\rrbracket$ and $\smash{\tilde{j}_{y}:=j^{\prime}_{(-1,y)}}$ for
$y\in\llbracket 1,N\rrbracket$. Then the lefthand side of (A.5) equals

$\displaystyle\mathbb{P}\left(b^{\mathrm{S6V}}_{(x,N)}=\tilde{b}_{x},j^{\mathrm{S6V}}_{(M,y)}=\tilde{j}_{y}\text{
for all }x\in\llbracket 1,M\rrbracket,y\in\llbracket 1,N\rrbracket\right),$

which in turn equals the partition function
$\mathcal{Z}^{\mathrm{S6V}}_{\sigma}(\tilde{\bm{b}},\tilde{\bm{j}})$ (see the
right panel of Figure 24) for colored S6V with (incoming) boundary condition
$\sigma$ and the (outgoing) boundary condition
$((\tilde{b}_{(x,N)},\tilde{j}_{(M,y)}):x\in\llbracket
1,M\rrbracket,y\in\llbracket 1,N\rrbracket)$ (we do not divide by the
normalization constant of the model as a whole since it is $1$).

Thus, to establish (A.5), we must show that

$\frac{\mathcal{Z}^{\mathrm{cB}}_{\sigma}(\bm{j}^{\prime})}{\mathcal{Z}^{\mathrm{cB}}_{\sigma}}=\mathcal{Z}^{\mathrm{S6V}}_{\sigma}(\tilde{\bm{b}},\tilde{\bm{j}}).$
(A.6)

We establish this by using the Yang-Baxter equation (A.1) via a similar
argument as in the last section. We start from an augmented model with
$\mathcal{R}_{1}$ to the left of
$\smash{\widetilde{\mathcal{R}}_{2}:=\mathbb{Z}_{\leq-1}\times\llbracket
1,N+M\rrbracket}$, with (incoming) boundary conditions $\sigma$ and (outgoing)
boundary conditions $(j^{\prime}_{(-1,y)})_{y\in\llbracket 1,N+M\rrbracket}$;
see the left side of Figure 25. By reasoning as in Section A.2, the partition
function of this model is
$\smash{((1-z)/(1-qz))^{NM}\cdot\mathcal{Z}^{\mathrm{cB}}_{\sigma}(\bm{j}^{\prime})=\mathcal{Z}^{\mathrm{cB}}_{\sigma}(\bm{j}^{\prime})/\mathcal{Z}^{\mathrm{cB}}_{\sigma}}$.

$\cdots$$1$$1$$1$$z$$z$$z$$z$$\sigma(1)$$\sigma(2)$$\sigma(3)$$M$$N$$\mathcal{R}_{1}$$\widetilde{\mathcal{R}}_{2}$$\stackrel{{\scriptstyle\eqref{e.yang-
baxter}}}{{\Longrightarrow}}$$\cdots$$1$$1$$1$$z$$z$$z$$z$$\sigma(1)$$\sigma(2)$$\sigma(3)$$N$$M$$\mathcal{R}_{1}$$\widetilde{\mathcal{R}}_{2}$
Figure 25. Left: The augmented model with outgoing boundary conditions. Right:
the same when $\mathcal{R}_{1}$ is moved all the way through
$\widetilde{\mathcal{R}}_{2}$ using repeated applications of the Yang-Baxter
equation (A.1); again, only a single configuration of arrows till
$\mathcal{R}_{1}$ has non-zero weight.

Now we move the crosses in $\mathcal{R}_{1}$ to the right of
$\widetilde{\mathcal{R}}_{2}$ using (A.1) again and obtain the situation in
the right side of Figure 25. Now, as depicted, there is a single configuration
of non-zero weight for the arrows in $\widetilde{\mathcal{R}}_{2}$: they go
horizontally straight through and are incident on $\mathcal{R}_{1}$ in the
same way as the latter having incoming boundary condition $\sigma$, and
outgoing boundary condition given by $(\tilde{\bm{b}},\tilde{\bm{j}})$. Thus
the partition function of the right side of Figure 25 is simply
$\mathcal{Z}^{\mathrm{S6V}}_{\sigma}(\tilde{b},\tilde{j})$, since the weight
of the fixed configuration in the $\smash{\widetilde{\mathcal{R}}_{2}}$ region
is $1$. Since the partition function is not modified by applying (A.1), we
obtain (A.6) and the proof is complete.

##  Appendix B Proofs of miscellaneous lemmas

In this section we prove various miscellaneous statements in the main text
whose proofs were deferred. Lemmas 5.3 and 5.4 are proved in Section B.1;
Lemma 6.13 in Section B.2; Lemma 8.6 in Section B.4; Lemma 9.7 in Section B.5;
and Lemma 10.4 in Section B.6. An alternative proof of Theorem 2.8 when
$\alpha=0$ (as mentioned in Remark 6.7) is given in Section B.3.

###  B.1. Finite speed of propagation bounds for colored ASEP and colored S6V

###### Proof of Lemma 5.3.

The lemma follows from the claim that, with probability at least
$1-\exp(-ct)$, no particle in colored ASEP (with packed initial condition) of
color greater than $-\lfloor 2t\rfloor$ is at any location to the right of
$\lfloor 4t\rfloor$ by time $t$. To see this, we merge all colors greater than
or equal to $-\lfloor 2t\rfloor$ to be particles and all particles of color
less than $-\lfloor 2t\rfloor$ to be holes, resulting in uncolored ASEP with
step initial condition from location $\lfloor 2t\rfloor$, in which setting we
must show the probability that the rightmost particle at time $t$ is at
location greater than $\lfloor 4t\rfloor$ is at most $\exp(-ct)$. By ignoring
left jump attempts, and letting $\smash{(X_{i})_{i=1}^{\lfloor 2t\rfloor-1}}$
be a collection of independent and identically distributed exponential rate 1
random variables, this probability is upper bounded by
$\smash{\mathbb{P}(\sum_{i=1}^{\lfloor 2t\rfloor-1}X_{i}<t)}$, which is indeed
upper bounded by $\exp(-ct)$ by standard concentration results (e.g., [Jan18,
Theorem 5.1]). ∎

###### Proof of Lemma 5.4.

By monotonicity of $h^{\mathrm{S6V}}(x,0;\bm{\cdot},t)$ and since
$h^{\mathrm{S6V}}(x,0;\bm{\cdot},t)\leq\lfloor\delta^{-1}t\rfloor-x+1$ for all
$x$, it is sufficient to prove the case $z=\lfloor 4t\rfloor$. The event in
the lemma with $z=\lfloor 4t\rfloor$ contains the event that no arrow of color
at least $-\lfloor 2t\rfloor$ exits horizontally from a vertex in
$I_{t}:=\\{\lfloor\delta^{-1}t\rfloor\\}\times\llbracket-\lfloor
2t\rfloor,\lfloor\delta^{-1}t\rfloor-\lfloor 4t\rfloor\rrbracket$. To lower
bound the probability of this event, we first merge all arrows of color at
least $-\lfloor 2t\rfloor$ to color 1, and all other arrows to color $0$,
obtaining an uncolored S6V model with step initial condition at $-\lfloor
2t\rfloor$. Then, equivalently, we need to upper bound the probability that
the lowest arrow, i.e., the one starting at $(1,-\lfloor 2t\rfloor)$, exits
horizontally from $I_{t}$; we call this event $\mathsf{E}_{t}$. It can be
described as the bottom-most arrow traveling at least
$\lfloor\delta^{-1}t\rfloor$ horizontally before traveling
$\lfloor\delta^{-1}t\rfloor-\lfloor 2t\rfloor$ vertically. The trajectory of
the bottom-most arrow can be broken up into alternating intervals of
horizontal travel and intervals of vertical travel. Note that the $i$th
interval of horizontal travel begins with a vertex configuration as in the
fourth panel of Figure 8 and is followed by a random number $X_{i}$ of
configurations as in the third panel of Figure 8, where $X_{i}$ is a geometric
random variable, with distribution given by
$\mathbb{P}(X_{i}=k)=\delta^{k}(1-\delta)$ for $k\in\mathbb{Z}_{\geq 0}$; thus
the $i$th interval of horizontal travel has length $1+X_{i}$. The $X_{i}$ are
independent for different $i$, and there are at most
$\lfloor\delta^{-1}t\rfloor-\lfloor 2t\rfloor$ many intervals of horizontal
travel, so
$\mathbb{P}(\mathsf{E}_{t})\leq\mathbb{P}(\sum_{i=1}^{\smash{\lfloor\delta^{-1}t\rfloor-\lfloor
2t\rfloor}}(1+X_{i})>\lfloor\delta^{-1}t\rfloor)=\mathbb{P}(\sum_{i=1}^{\smash{\lfloor\delta^{-1}t\rfloor-\lfloor
2t\rfloor}}X_{i}>\lfloor 2t\rfloor)$. Using that
$\smash{\mathbb{E}[\sum_{i=1}^{\lfloor\delta^{-1}t\rfloor-\lfloor
2t\rfloor}X_{i}]=t(1-2\delta)/(1-\delta)<2t}$, this is upper bounded by
$\exp(-ct)$ for a $c>0$ independent of $\delta$, again by standard
concentration inequalities (e.g., [Jan18, Theorem 2.1]). ∎

###  B.2. Two-point fluctuation bound for Bernoulli random walk bridge

###### Proof of Lemma 6.13.

It is enough to prove the two-point bound without the supremum, i.e., to prove
that there exist $C$, $c>0$ such that for all $M>0$,

$\displaystyle\mathbb{P}\left(|B(x)-B(y)-p(x-y)|>M|x-y|^{1/2}\right)\leq
C\exp(-cM^{2}).$

From here the claim is proved by a standard chaining argument, i.e., doing a
union bound over dyadically decreasing scales and using the above two point
bound at each scale; e.g., we may apply [Dau23, Lemma 2.3] or imitate the
proofs of Proposition 9.3 or Lemma 10.5.

Assume without loss of generality that $x>y$. Observe that
$\\{B(w)-B(w-1)\\}_{w=1}^{n}$ is a collection of $pn$ many $-1$s and $(1-p)n$
many 0s in uniformly random order. Clearly, $B(x)-B(y)$ is the number of $-1$s
lying in positions $y+1,\ldots,x$. Observe that this has the same distribution
as the number of $-1$s in the first $k:=x-y$ positions, which is the number
$X$ of $-1$s drawn without replacement in the first $k$ draws from a
collection of $n$ numbers of which $K:=pn$ are $-1$s. In other words, $X$ is a
hypergeometric distribution with parameters $n$, $K$, and $k$. By a theorem of
Hoeffding [Hoe63, Theorems 1 and 4], it follows that

$\displaystyle\mathbb{P}\left(|X-pk|\geq Mk^{1/2}\right)\leq 2\exp(-2M^{2}),$

which completes the proof. ∎

###  B.3. Alternative proof of ASEP sheet to Airy sheet convergence at
$\bm{\alpha=0}$

In this subsection we provide the alternate proof of Theorem 2.8 in the case
of $\alpha=0$, that was mentioned in Remark 6.7.

###### Alternative proof of Theorem 2.8 for $\alpha=0$.

As in the original proof in Section 7.1, we again take
$\mathcal{S}^{N}(x;y)=\mathcal{S}^{\mathrm{ASEP},\theta,\varepsilon}(-x;-y)$
as processes in $(x,y)$ with $N=\varepsilon^{-1}$, and
$\bm{L}^{(j),N}=\bm{L}^{\mathrm{ASEP},(j),\theta,\varepsilon}$. As in the
proof of Theorem 2.8 in Section 7.1, (3.14), (7.3), and (7.4) hold and
$\bm{L}^{(1),N}$ satisfies Assumptions 1 and 2. So by Theorem 3.18 and
Proposition 6.5, $\bm{\mathfrak{L}}^{(1),N}$ is tight and all subsequential
limits satisfy the Brownian Gibbs property; call such a subsequential limit
$\bm{\mathfrak{L}}^{(1)}$. Combining Lemma 5.2 with [QS22, Theorem 2.2 (2)]
yields that
$\mathfrak{L}^{\smash{(1)}}_{1}\stackrel{{\scriptstyle\smash{d}}}{{=}}\mathcal{P}_{1}$
(this uses $\alpha=0$), and, since $\bm{\mathfrak{L}}^{\smash{(1)}}$ has the
Brownian Gibbs property, [DM21, Theorem 1.1] implies that
$\bm{\mathfrak{L}}^{(1)}\stackrel{{\scriptstyle\smash{d}}}{{=}}\bm{\mathcal{P}}$.
Since every subsequential limit is the same, we obtain that
$\bm{\mathfrak{L}}^{(1),N}\xrightarrow{\smash{d}}\bm{\mathcal{P}}$. Invoking
Theorem 7.1 and the fact that
$\mathcal{S}(x;y)\stackrel{{\scriptstyle\smash{d}}}{{=}}\mathcal{S}(-x;-y)$,
as processes on $\mathbb{R}^{2}$, from [DV21a, Proposition 1.23] completes the
proof. ∎

###  B.4. Quantitative monotonicity for non-intersecting Bernoulli random walk
bridge

###### Proof of Lemma 8.6.

We construct a Markov chain whose value at any time $t$ is a pair of Bernoulli
paths $(B^{\shortuparrow}_{t},B^{\downarrow}_{t})$ such that

$\displaystyle B^{\downarrow}_{t}(x)\leq B^{\shortuparrow}_{t}(x)\leq
B^{\downarrow}_{t}(x)+M$ (B.1)

holds almost surely and the large $t$ marginal distributions of the chain are
respectively the marginal distributions of $B^{\shortuparrow}$ and
$B^{\downarrow}$. Any subsequential limiting coupling so produced will
establish the lemma. We may assume $M\geq 1$ as otherwise there is nothing to
prove.

For $*\in\\{\shortuparrow,\downarrow\\}$, define $B^{*}_{0}$ to be the minimal
Bernoulli path from $(a,x^{*})$ to $(b,y^{*})$ which satisfies $g(x)\leq
B^{*}_{0}(x)\leq f(x)$ for all $x\in\llbracket a,b\rrbracket$ (the assumptions
guarantee at least one such path exists). We claim that (B.1) holds at $t=0$.
That the first inequality holds is immediate by minimality of the paths.

For the second inequality, suppose to the contrary that there exists
$x_{0}\in\llbracket a,b\rrbracket$ such that
$\smash{B^{\shortuparrow}_{0}(x_{0})=B^{\downarrow}_{0}(x_{0})+M+1}$, and we
take $x_{0}$ to be the smallest such point. Let $y_{0}>x_{0}$ be the first
such that $\smash{B^{\shortuparrow}_{0}(y_{0})=B^{\downarrow}_{0}(y_{0})+M}$;
it exists since the final separation of the paths at $b$ is at most $M$ and
the separation changes by an element of $\\{-1,0,1\\}$ at each step. Now, by
minimality of $x_{0}$ and $y_{0}$, $B^{\shortuparrow}$ has a flat-step
starting at $x_{0}$ and a down-step starting at $y_{0}-1$. Let
$\smash{\tilde{B}^{\shortuparrow}_{0}}$ be the path obtained from $B_{0}$ by
flipping the flat-step at $x_{0}$ to a down-step and flipping the down-step at
$y_{0}-1$ to a flat-step; then $\smash{\tilde{B}^{\shortuparrow}_{0}}$ is
lower than $B^{\shortuparrow}_{0}$ and does not intersect $g$ (as
$\smash{\tilde{B}^{\shortuparrow}_{0}(x)\geq B^{\downarrow}_{0}(x)\geq g(x)}$
for all $x\in\llbracket x_{0},y_{0}\rrbracket$), contradicting the minimality
of $\smash{B^{\downarrow}_{0}}$. Thus our claim is established, i.e., (B.1)
holds at $t=0$.

Now we specify the Markov dynamics. At each time $t\geq 1$, an $X\in\llbracket
a,b\rrbracket$ and $\sigma\in\\{0,-1\\}$ are chosen independently and
uniformly at random. For each $*\in\\{\shortuparrow,\downarrow\\}$, if
$B^{*}_{t-1}(X+1)=B^{*}_{t-1}(X-1)-1$ (i.e., there is a down-step followed by
a flat-step or vice versa from $X-1$ to $X+1$ in $B^{*}_{t-1}$), we set
$B^{*}_{t}(X)=B^{*}_{t-1}(X-1)+\sigma$ and $B^{*}_{t}(x)=B^{*}_{t-1}(x)$ for
all $x\neq X$, as long as the resulting $B^{*}_{t}$ satisfies $g\leq
B^{*}_{t}\leq f$; if not, no change is made.

It is easy to check that (B.1) is satisfied under these dynamics, and that the
dynamics are aperiodic and irreducible. It is also straightforward that the
marginal distributions of $B^{\shortuparrow}$ and $B^{\downarrow}$ are
stationary for the marginal Markov chains; thus the limiting stationary
distribution has the correct marginals. This completes the proof. ∎

###  B.5. Gibbs property on a certain random interval

###### Proof of Lemma 9.7.

For a stochastic process $X:\llbracket a,b\rrbracket\to\mathbb{Z}$ taking
values in $\Omega_{\llbracket a,b\rrbracket}$ (the space of Bernoulli paths on
$\llbracket a,b\rrbracket$, as defined before Lemma 9.7), $t\in\llbracket
a,b\rrbracket$, and $p\in(-1,0)$, let $\overline{X}(x)=X(x)-px$,
$a^{\prime}(X):=\max\\{x\in\llbracket a,t\rrbracket:\overline{X}(x)\geq m\\}$,
and $b^{\prime}(X):=\min\\{x\in\llbracket t,b\rrbracket:\overline{X}(x)\geq
m\\}$. With the notational convention for event names from Remark 6.16 in
mind, for such a process $X$, define
$\mathsf{E}(x,y,z_{1},z_{2},X):=\mathsf{BdyVal}(x,y,z_{1},z_{2},X)\cap\mathsf{IntVal}(x,y,X)$,
where

$\displaystyle\mathsf{BdyVal}(x,y,z_{1},z_{2},X)$
$\displaystyle:=\\{X(x)=z_{1},X(y)=z_{2}\\},\quad\text{and}$
$\displaystyle\mathsf{IntVal}(x,y,X)$
$\displaystyle:=\\{a^{\prime}(X)=x,b^{\prime}(X)=y\\}.$

Let $\mathcal{F}_{x,y}:=\mathcal{F}_{\mathrm{ext}}(\\{k\\},\llbracket
x,y\rrbracket,\bm{L})$, $a^{\prime}:=a^{\prime}(L_{k})$,
$b^{\prime}:=b^{\prime}(L_{k})$, and
$\mathsf{E}(x,y,z_{1},z_{2}):=\mathsf{E}(x,y,z_{1},z_{2},L_{k})$ for short. In
the case that $a^{\prime}=b^{\prime}=t$, there is nothing to prove, so we my
assume that this does not hold. By the definition of $\mathcal{F}$ from Lemma
9.7, we must show that, for any $x\in\llbracket a,t\rrbracket$,
$y\in\llbracket t,b\rrbracket$, $z_{1},z_{2}\in\mathbb{Z}$, and
$\mathsf{A}\in\mathcal{F}_{x,y}$ (recalling the notation
$\mathbb{E}^{z_{1},z_{2},\llbracket x,y\rrbracket}$ introduced just before
Lemma 9.7),

$\begin{split}&\mathbb{E}\left[F(L_{k}|_{\llbracket
a^{\prime},b^{\prime}\rrbracket},a^{\prime},b^{\prime})\mathbbm{1}_{\mathsf{A}\cap\mathsf{E}(x,y,z_{1},z_{2})}\right]\\\
&=\mathbb{E}\left[\frac{\mathbb{E}^{z_{1},z_{2},\llbracket
x,y\rrbracket}\left[F(B,x,y)W(B,L_{k-1},L_{k+1})\mid\overline{B}(s)<m\ \forall
s\in\llbracket x+1,y-1\rrbracket\right]}{\mathbb{E}^{z_{1},z_{2},\llbracket
x+1,y-1\rrbracket}\left[W(B,L_{k-1},L_{k+1})\mid\overline{B}(s)<m\ \forall
s\in\llbracket
x+1,y-1\rrbracket\right]}\mathbbm{1}_{\mathsf{A}\cap\mathsf{E}(x,y,z_{1},z_{2})}\right],\end{split}$
(B.2)

where the $W$ factors are evaluated on the interval $\llbracket
x,y\rrbracket$. First, if $z_{1}-px<m$ or $z_{2}-py<m$,
$\mathsf{E}(x,y,z_{1},z_{2})=\emptyset$ by definition of $a^{\prime}$ and
$b^{\prime}$, in which case the equality trivially holds. So we assume
$z_{1}-px,z_{2}-py\geq m$. By conditioning on $\mathcal{F}_{x,y}$, we see that
the lefthand side of the previous display equals

$\displaystyle\mathbb{E}\Bigl{[}\mathbb{E}_{\mathcal{F}_{x,y}}\bigl{[}F(L_{k}|_{\llbracket
x,y\rrbracket},x,y)\mathbbm{1}_{\mathsf{E}(x,y,z_{1},z_{2})}\bigr{]}\mathbbm{1}_{\mathsf{A}}\Bigr{]}.$
(B.3)

Adopt the shorthand
$\mathsf{BdyVal}(x,y,z_{1},z_{2}):=\mathsf{BdyVal}(x,y,z_{1},z_{2},L_{k})$ and
recall that
$\mathsf{E}(x,y,z_{1},z_{2})=\mathsf{BdyVal}(x,y,z_{1},z_{2})\cap\mathsf{IntVal}(x,y,z_{1},z_{2},L_{k})$.
Applying the Hall-Littlewood Gibbs property shows that the (inner) conditional
expectation in (B.3) equals

$\displaystyle\frac{\mathbb{E}^{z_{1},z_{2},\llbracket
x,y\rrbracket}\left[F(B,x,y)\mathbbm{1}_{\mathsf{IntVal}(x,y,B)}W(B,L_{k-1},L_{k+1})\right]}{\mathbb{E}^{z_{1},z_{2},\llbracket
x,y\rrbracket}\left[W(B,L_{k-1},L_{k+1})\right]}\mathbbm{1}_{\mathsf{BdyVal}(x,y,z_{1},z_{2})}.$

Note that $\mathsf{IntVal}(x,y,B)=\\{\overline{B}(s)<m\ \forall s\in\llbracket
x+1,y-1\rrbracket\\}$ when $\mathsf{BdyVal}(x,y,z_{1},z_{2})$ holds, since
$z_{1}-px,z_{2}-py\geq m$. Thus we can rewrite the previous display as (with
the shorthand $\overline{B}<m$ for $\overline{B}(s)<m\ \forall s\in\llbracket
x+1,y-1\rrbracket$)

$\displaystyle\frac{\mathbb{E}^{z_{1},z_{2},\llbracket
x,y\rrbracket}\left[F(B,x,y)W(B,L_{k-1},L_{k+1})\mid\overline{B}<m\right]}{\mathbb{E}^{z_{1},z_{2},\llbracket
x,y\rrbracket}\left[W(B,L_{k-1},L_{k+1})\right]}\cdot\mathbb
P_{\mathcal{F}}\left(\overline{B}<m\right)\mathbbm{1}_{\mathsf{BdyVal}(x,y,z_{1},z_{2})}$
$\displaystyle=\frac{\mathbb{E}^{z_{1},z_{2},\llbracket
x,y\rrbracket}\left[F(B,x,y)W(B,L_{k-1},L_{k+1})\mid\overline{B}<m\right]}{\mathbb{E}^{z_{1},z_{2},\llbracket
x,y\rrbracket}\left[W(B,L_{k-1},L_{k+1})\mid\overline{B}<m\right]}$
$\displaystyle\qquad\qquad\times\frac{\mathbb{E}^{z_{1},z_{2},\llbracket
x,y\rrbracket}\left[W(B,L_{k-1},L_{k+1})\mathbbm{1}_{\overline{B}<m}\right]}{\mathbb{E}^{z_{1},z_{2},\llbracket
x,y\rrbracket}\left[W(B,L_{k-1},L_{k+1})\right]}\mathbbm{1}_{\mathsf{BdyVal}(x,y,z_{1},z_{2})}$
$\displaystyle=\frac{\mathbb{E}^{z_{1},z_{2},\llbracket
x,y\rrbracket}\left[F(B,x,y)W(B,L_{k-1},L_{k+1})\mid\overline{B}<m\right]}{\mathbb{E}^{z_{1},z_{2},\llbracket
x,y\rrbracket}\left[W(B,L_{k-1},L_{k+1})\mid\overline{B}<m\right]}$
$\displaystyle\qquad\qquad\times\mathbb{P}_{\mathcal{F}_{x,y}}\left(\overline{L}_{k}(s)<m\
\forall s\in\llbracket
x,y\rrbracket\right)\mathbbm{1}_{\mathsf{BdyVal}(x,y,z_{1},z_{2})}$
$\displaystyle=\frac{\mathbb{E}^{z_{1},z_{2},\llbracket
x,y\rrbracket}\left[F(B,x,y)W(B,L_{k-1},L_{k+1})\mid\overline{B}<m\right]}{\mathbb{E}^{z_{1},z_{2},\llbracket
x,y\rrbracket}\left[W(B,L_{k-1},L_{k+1})\mid\overline{B}<m\right]}\cdot\mathbb{P}_{\mathcal{F}_{x,y}}\left(\mathsf{IntVal}(x,y,L_{k})\right)\mathbbm{1}_{\mathsf{BdyVal}(x,y,z_{1},z_{2})}.$

Since all the factors outside the $\mathbb P_{\mathcal{F}}$ factor are
$\mathcal{F}_{x,y}$-measurable, and since by definition
$\mathsf{E}(x,y,z_{1},z_{2})=\mathsf{IntVal}(x,y,L_{k})\cap\mathsf{BdyVal}(x,y,z_{1},z_{2})$,
the tower property of conditional expectation yields that the previous display
equals

$\displaystyle\mathbb{E}_{\mathcal{F}_{x,y}}\left[\frac{\mathbb{E}^{z_{1},z_{2},\llbracket
x,y\rrbracket}\left[F(B,x,y)W(B,L_{k-1},L_{k+1})\mid\overline{B}<m\right]}{\mathbb{E}^{z_{1},z_{2},\llbracket
x,y\rrbracket}\left[W(B,L_{k-1},L_{k+1})\mid\overline{B}<m\right]}\mathbbm{1}_{\mathsf{E}(x,y,z_{1},z_{2})}\right].$

Substituting this in (B.3), using the tower property again, and that
$\mathsf{A}$ is $\mathcal{F}_{x,y}$ measurable, yields (B.2). This completes
the proof. ∎

###  B.6. Uniform separation for Brownian and random walk bridges

In this section we establish Lemma 10.4 which, recall, states that conditional
on a Bernoulli random walk bridge staying above a lower boundary curve $f$, it
will maintain some on-scale uniformly positive separation from $f$ with high
probability. The argument goes by proving a similar statement for Brownian
bridges and then using the KMT coupling to obtain it for the Bernoulli case.

The Brownian bridge case is the following.

######  Lemma B.1 (Brownian bridge uniform separation).

Fix $M>0$, $\sigma>0$, $p>0$ and $T\geq 2$. Let $U_{1}\in[-2T,-\frac{3}{2}T]$
and $U_{2}\in[\frac{3}{2}T,2T]$. Let $B^{\mathrm{Br}}$ be a Brownian bridge
from $(U_{1}N^{2/3},z_{1})$ to $(U_{2}N^{2/3},z_{2})$ of variance
$\sigma^{2}>0$, for some $z_{i}$ such that
$z_{i}-pU_{i}N^{2/3}\in[-MN^{1/3},MN^{1/3}]$ for each $i\in\\{1,2\\}$. Also
let $f:[U_{1}N^{2/3},U_{2}N^{2/3}]\to\mathbb{R}$ be a function with
$f(U_{i}N^{2/3})\leq z_{i}$. Let $\\{B^{\mathrm{Br}}\geq f\\}$ denote
$\\{B^{\mathrm{Br}}(x)\geq f(x)\ \forall x\in[U_{1},U_{2}]\\}$ and suppose,
for some $\rho>0$, that

$\displaystyle\mathbb{P}\left(B^{\mathrm{Br}}\geq f\right)\geq\rho.$

Then for any $\varepsilon>0$ there exists
$\delta=\delta(\varepsilon,\rho,M,\sigma,T)$ such that

$\displaystyle\mathbb{P}\left(\inf_{x\in[-TN^{2/3},TN^{2/3}]}\left(B^{\mathrm{Br}}(x)-f(x)\right)\leq\delta
N^{1/3}\ \Big{|}\ B^{\mathrm{Br}}\geq f\right)\leq\varepsilon.$

Using the previous lemma and Lemma 6.14, we give the straightforward proof of
Lemma 10.4.

###### Proof of Lemma 10.4.

We will eventually set $\delta<\omega$ and will use this fact in the coming
arguments. By Lemma 6.14, there exists $N_{0}=N_{0}(\varepsilon,\delta,T,M)$
such that there is a coupling between $B^{\mathrm{RW}}$ and $B^{\mathrm{Br}}$
(of variance lower bounded by a positive constant depending on only $p$) such
that, for $N\geq N_{0}$, with probability at least $1-\varepsilon\rho$,
$\sup_{[U_{1}N^{2/3},U_{2}N^{2/3}]}|B_{k}-B^{\mathrm{Br}}|\leq\delta N^{1/3}$.
Thus

$\displaystyle\mathbb{P}\left(\inf_{x\in[-TN^{2/3},TN^{2/3}]}\left(B^{\mathrm{RW}}(x)-f(x)\right)\leq\delta
N^{1/3}\ \Big{|}\ B^{\mathrm{RW}}\geq f\right)$
$\displaystyle=\frac{\mathbb{P}\left(\inf_{x\in[-TN^{2/3},TN^{2/3}]}\left(B^{\mathrm{RW}}(x)-f(x)\right)\leq\delta
N^{1/3},B^{\mathrm{RW}}\geq f\right)}{\mathbb{P}\left(B^{\mathrm{RW}}\geq
f\right)}$
$\displaystyle\leq\frac{\mathbb{P}\left(\inf_{x\in[-TN^{2/3},TN^{2/3}]}\left(B^{\mathrm{Br}}(x)-f(x)\right)\leq
2\delta N^{1/3},B^{\mathrm{Br}}\geq f-\delta
N^{1/3}\right)+\varepsilon\rho}{\mathbb{P}\left(B^{\mathrm{Br}}\geq f+\delta
N^{1/3}\right)-\varepsilon\rho}.$

Since $\delta<\omega$, $\mathbb{P}\left(B^{\mathrm{Br}}\geq f+\delta
N^{1/3}\right)\geq\rho$ by hypothesis. So (assuming $\varepsilon<\frac{1}{2}$
without loss of generality) the previous display is upper bounded by

$\displaystyle
2\rho^{-1}\cdot\mathbb{P}\left(\inf_{x\in[-TN^{2/3},TN^{2/3}]}\left(B^{\mathrm{Br}}(x)-f(x)\right)\leq
2\delta N^{1/3},B^{\mathrm{Br}}\geq f-\delta N^{1/3}\right)+2\varepsilon.$

By Lemma B.1 we may pick $\delta<\omega$ with
$\delta=\delta(\varepsilon,\rho,M,p)$ such that the probability is less than
$\rho\varepsilon$, which completes the proof after relabeling $\varepsilon$. ∎

Next we prove Lemma B.1 on the uniform on-scale separation of a Brownian
bridge from its lower boundary condition. The idea is to condition on the
entire bridge except for its value at zero; more precisely, one conditions on
the shape of the curve so that, given the value of $B^{\mathrm{Br}}(0)$, the
whole process is determined by an affine shift. The conditional distribution
of $B^{\mathrm{Br}}(0)$ is explicit and the proof comes down to a calculation
about the probability of this random variable being close to the lower edge of
its support.

Similar ideas were introduced in [Ham22] and have been used in a number of
subsequent works, including [CHH23, CHHM23, GHZ23].

###### Proof of Lemma B.1.

We assume that $\sigma^{2}=1$ and $p=0$ without loss of generality. Recall
that $B^{\mathrm{Br}}$ is a Brownian bridge defined on
$[U_{1}N^{2/3},U_{2}N^{2/3}]$. For an interval
$[a,b]\subseteq[U_{1}N^{2/3},U_{2}N^{2/3}]$, we adopt the notation
$B^{\mathrm{Br},[a,b]};[a,b]\to\mathbb{R}$ to be $B^{\mathrm{Br}}$ affinely
shifted to equal zero at $a$ and $b$, i.e., for $x\in[a,b]$,

$\displaystyle
B^{\mathrm{Br},[a,b]}(x):=B^{\mathrm{Br}}(x)-\frac{x-a}{b-a}B^{\mathrm{Br}}(b)-\frac{b-x}{b-a}B^{\mathrm{Br}}(a).$
(B.4)

We let $\tilde{\mathbb}{P}$ be the probability measure
$\mathbb{P}(\bm{\cdot}\mid B^{\mathrm{Br}}\geq f)$, so that under
$\tilde{\mathbb}{P}$, $B^{\mathrm{Br}}$ is distributed as a Brownian bridge
conditioned to stay above $f$. We consider the $\sigma$-algebra $\mathcal{F}$
generated by $B^{\mathrm{Br},[U_{1}N^{2/3},0]}(\bm{\cdot})$ and
$B^{\mathrm{Br},[0,U_{2}N^{2/3}]}(\bm{\cdot})$. Then, conditionally on
$\tilde{\mathbb}{P}_{\mathcal{F}}$, the only remaining randomness is in the
value $B^{\mathrm{Br}}(0)$. The distribution of this random variable is that
of a normal random variable of mean $\mu=\frac{1}{2}(z_{1}+z_{2})$ and
variance $\widetilde{\sigma}^{2}=|U_{1}|\cdot U_{2}N^{2/3}/(U_{2}+|U_{1}|)$
conditioned to stay above an $\mathcal{F}$-measurable random variable
$\mathrm{Cor}$ (short for corner, a name given to a similar quantity in
[Ham22]) defined by

$\displaystyle\mathrm{Cor}:=\inf\left\\{w:B^{\mathrm{Br},w}(x)\geq f(x)\text{
for all }x\in[U_{1}N^{2/3},U_{2}N^{2/3}]\right\\},$

where $B^{\mathrm{Br},w}$ is the reconstruction of $B^{\mathrm{Br}}$ from
$B^{\mathrm{Br},[U_{1}N^{2/3},0]}$ and $B^{\mathrm{Br},[0,U_{2}N^{2/3}]}$ when
$B(0)=w$, i.e.,

$\displaystyle
B^{\mathrm{Br},w}(x):=\begin{cases}\frac{U_{1}-xN^{-2/3}}{U_{1}}w+\frac{xN^{-2/3}}{U_{1}}z_{1}+B^{\mathrm{Br},[U_{1}N^{2/3},0]}(x)&x\in[U_{1}N^{2/3},0]\\\
\frac{U_{2}-xN^{-2/3}}{U_{2}}w+\frac{xN^{-2/3}}{U_{2}}z_{2}+B^{\mathrm{Br},[0,U_{2}N^{2/3}]}(x)&x\in[0,U_{2}N^{2/3}].\end{cases}$

Notice that for any $R$,

$\displaystyle\tilde{\mathbb}{P}\left(\mathrm{Cor}>RN^{1/3}\right)\leq\tilde{\mathbb}{P}\left(B^{\mathrm{Br}}(0)>RN^{1/3}\right)$
$\displaystyle=\mathbb{P}\left(B^{\mathrm{Br}}(0)>RN^{1/3}\ \Big{|}\
B^{\mathrm{Br}}\geq f\right)$
$\displaystyle\leq\rho^{-1}\mathbb{P}\left(B^{\mathrm{Br}}(0)>RN^{1/3}\right)$
$\displaystyle=\rho^{-1}\mathbb{P}\left(\mathcal{N}(\mu,\widetilde{\sigma}^{2})>RN^{1/3}\right).$

Since $\mu\geq-MN^{1/3}$ and $\widetilde{\sigma}^{2}\leq TN^{2/3}$, there is
$R=R(\varepsilon,\rho,M,T)$ (and since $T\geq 1$) such that the previous
display is at most $\varepsilon$. We fix such a value of $R$ for the rest of
the proof.

Now observe that, if $w\geq\mathrm{Cor}+3\delta N^{1/3}$, then

$\displaystyle\inf_{x\in[-TN^{2/3},TN^{2/3}]}\left(B^{\mathrm{Br},w}(x)-f(x)\right)\geq\delta
N^{1/3}.$

This is because the lefthand side is at least $0$ at $w=\mathrm{Cor}$ by
definition of $\mathrm{Cor}$, $B^{\mathrm{Br},w}(x)$ increases with $w$, and
$\frac{U_{1}-xN^{-2/3}}{U_{1}}\cdot 3\delta N^{1/3}\geq\delta N^{1/3}$ when
$x\in[-TN^{2/3},TN^{2/3}]$ since $U_{1}\in[-2T,-\frac{3}{2}T]$, and similarly
for $U_{2}$. Thus it follows that (recall $\mathcal{F}$ from right after
(B.4))

$\displaystyle\tilde{\mathbb}{P}\left(\inf_{x\in[-TN^{2/3},TN^{2/3}]}\left(B^{\mathrm{Br}}(x)-f(x)\right)\leq\delta
N^{1/3}\right)$
$\displaystyle=\tilde{\mathbb}{E}\left[\tilde{\mathbb}{P}_{\mathcal{F}}\left(\inf_{x\in[-TN^{2/3},TN^{2/3}]}\left(B^{\mathrm{Br}}(x)-f(x)\right)\leq\delta
N^{1/3}\right)\right]$
$\displaystyle\leq\tilde{\mathbb}{E}\left[\tilde{\mathbb}{P}_{\mathcal{F}}\left(B^{\mathrm{Br}}(0)\leq\mathrm{Cor}+3\delta
N^{1/3}\right)\right].$

Letting $X\sim\mathcal{N}(0,1)$, the previous display equals

$\displaystyle\tilde{\mathbb}{E}\left[\tilde{\mathbb}{P}_{\mathcal{F}}\left(\widetilde{\sigma}X+\mu\leq\mathrm{Cor}+3\delta
N^{1/3}\mid\widetilde{\sigma}X+\mu\geq\mathrm{Cor}\right)\right]$
$\displaystyle\leq\tilde{\mathbb}{E}\left[\tilde{\mathbb}{P}_{\mathcal{F}}\left(X\leq\widetilde{\sigma}^{-1}(\mathrm{Cor}-\mu)+3\widetilde{\sigma}^{-1}\delta
N^{1/3}\mid
X\geq\widetilde{\sigma}^{-1}(\mathrm{Cor}-\mu)\right)\mathbbm{1}_{\mathrm{Cor}\leq
RN^{1/3}}\right]+\varepsilon.$

Now by Lemma 8.1, the conditional probability in the previous display is upper
bounded by the same with $\widetilde{\sigma}^{-1}(\mathrm{Cor}-\mu)$ replaced
by an upper bound on it. Since $\widetilde{\sigma}^{2}\geq cTN^{2/3}$ and
$\mu\geq-MN^{1/3}$, on the event $\mathrm{Cor}\leq RN^{1/3}$, it holds that
$\widetilde{\sigma}^{-1}(\mathrm{Cor}-\mu)\leq c^{-1/2}T^{-1/2}(R+M)$. By the
upper bound on $\widetilde{\sigma}^{-1}$, we also see that
$3\widetilde{\sigma}^{-1}\delta N^{1/3}\leq 3c^{-1/2}T^{-1/2}\delta$. Thus,
the previous display is upper bounded (using the bound on
$3\widetilde{\sigma}^{-1}\delta N^{1/3}$ for the first line and Lemma 8.1 for
the second) by

$\displaystyle\tilde{\mathbb}{E}\left[\tilde{\mathbb}{P}_{\mathcal{F}}\left(X\leq\widetilde{\sigma}^{-1}(\mathrm{Cor}-\mu)+3c^{-1/2}T^{-1/2}\delta\
\Big{|}\
X\geq\widetilde{\sigma}^{-1}(\mathrm{Cor}-\mu)\right)\mathbbm{1}_{\mathrm{Cor}\leq
RN^{1/3}}\right]+\varepsilon$
$\displaystyle\leq\tilde{\mathbb}{E}\left[\tilde{\mathbb}{P}_{\mathcal{F}}\left(X\leq
c^{-1/2}T^{-1/2}(R+M+3\delta)\ \Big{|}\ X\geq
c^{-1/2}T^{-1/2}(R+M)\right)\right]+\varepsilon.$

It is now clear using standard estimates on normal probabilities that, for
$\delta$ small enough depending on $R$, $M$, and $T$ (therefore on
$\varepsilon$, $\rho$, $M$, and $T$), the previous display is upper bounded by
$2\varepsilon$. This completes the proof. ∎

##  Appendix C LPP maximizer location asymptotics

In this section, we give the proof of Proposition 7.2. Before doing so, we set
up some preliminary lemmas that will be needed. Recall the definition of
$Z^{N}_{k}$ as the maximizer of the maximum over $z$ of (7.7).

###  C.1. Monotonicity properties

We need a monotonicity property of $Z^{N}_{k}$ and of LPP values. We start
with the first; its proof is the same as that of [DOV22, Lemma 3.6] and is
omitted.

######  Lemma C.1.

For fixed $N$, $x$, and $y$, it almost surely holds that $Z^{N}_{k}(x,y)\geq
Z^{N}_{k+1}(x,y)$ for all $k\in\llbracket 1,N^{1/6}\rrbracket$.

######  Lemma C.2 ([DOV22, Lemma 3.3]).

Let $f=(f_{1},\ldots,f_{n})$ be a sequence of continuous functions with
$f_{i}:\mathbb{R}\to\mathbb{R}$. Then for any $x\in\mathbb{R}$,

$\displaystyle z\mapsto f[(z,k)\to(x,1)]+f_{k}(z)$

is non-increasing.

###  C.2. Prelimiting quantities

Define $F^{N}_{k}$, $G^{N}_{k}$ and $H^{N}$ by

$\displaystyle F^{N}_{k}(x,z)$ $\displaystyle:=\sup_{w\leq
z}\left(\mathfrak{L}^{N}_{N^{1/6}}(x;w)+\bm{\mathfrak{L}}^{(1),N}[(w,N^{1/6}-1)\to(z,k)]\right)-\mathfrak{L}^{(1),N}_{k}(z)$
$\displaystyle G^{N}_{k}(z,y)$
$\displaystyle:=\bm{\mathfrak{L}}^{(1),N}[(z,k)\to(y,1)]+\mathfrak{L}^{(1),N}_{k}(z)$
$\displaystyle H^{N}(x,y)$
$\displaystyle:=\mathcal{S}^{N}(x;y)=\mathfrak{L}^{N}_{1}(x;y);$

note that the definition of $G^{N}_{k}(z,y)$ requires $z\leq y$. Also define

$\begin{split}G_{k}(z,y)&:=\bm{\mathcal{P}}[(z,k)\to(y,1)]+\mathcal{P}_{k}(z)\\\
H(x,y)&:=\mathcal{S}(x;y).\end{split}$ (C.1)

Observe that $G^{N}_{k}\to G^{k}$ (by Theorem 3.19). The quantities in (C.1)
are defined in this way to obtain the monotonicity stated in Lemma C.2, i.e.,
$G_{k}(z,y)$ is non-increasing in $z$ by Lemma C.2, a property not possessed
by $\bm{\mathcal{P}}[(z,k)\to(y,1)]$.

The proof of Proposition 7.2 will also need some asymptotic information about
$G_{k}$, which we record in the following lemma. We defer its proof to Section
C.3 as it requires the introduction of the prelimiting model of Brownian LPP.
Recall the meaning of the notation $\mathfrak{o}$ from (7.5).

######  Lemma C.3.

Fix $z<y$. Then it holds that

$\displaystyle
G_{k}\left(z\sqrt{k},y\sqrt{k}\right)=-y^{2}k-\frac{\sqrt{k}}{y-z}+\mathfrak{o}(\sqrt{k}).$

Now we give the proof of Proposition 7.2, which is an adaptation of the
argument from [DOV22, Lemma 7.1].

###### Proof of Proposition 7.2.

Note that $H^{N}(x,y)\stackrel{{\scriptstyle
d}}{{=}}\mathfrak{L}^{(1),N}_{1}(y-x)$ for fixed $x,y$ by (7.3) and
$\mathfrak{L}^{(1),N}_{1}\stackrel{{\scriptstyle d}}{{\to}}\mathcal{P}_{1}$ as
$N\to\infty$ by Theorem 3.19. We added $\mathfrak{L}^{(1),N}_{k}(z)$ in the
definition of $G^{N}_{k}$ because we will need $G^{N}_{k}$ to converge to
$G^{k}$. We subtracted the same quantity from $F^{n}_{k}$ so that the
following holds, which follows from (7.7):

$\displaystyle H^{N}(x,y)$ $\displaystyle\geq
F^{N}_{k}(x,z)+G^{N}_{k}(z,y)+\mathfrak{o}(N^{-1/11}).$ (C.2)

Now the left hand side is tight in $N$ and independent of $k$ by the tightness
of $\mathfrak{L}^{(1)}_{N}$ (Theorem 3.19). For any $\varepsilon>0$, we claim
that if $|z+\sqrt{k/2x}|>\varepsilon\sqrt{k}$, then the right hand side is at
most $-\varepsilon^{2}\sqrt{kx}/2$. By (C.2) and the tightness of
$H^{N}(x,y)$, thus would imply that $Z^{N}_{k}$, the maximizer over $z$, must
lie within $\varepsilon\sqrt{k}$ of $-\sqrt{k/2x}$, establishing the
proposition.

For $z$ and $y$ in a fixed compact subset independent of $N$ (but which may
depend on $k$), we know from Theorem 3.19 that
$\smash{G^{N}_{k}(z,y)\stackrel{{\scriptstyle d}}{{\to}}G_{k}(z,y)}$ as
$N\to\infty$. Further, we know from Lemma C.3 and by the convergence of
$G^{N}_{k}$ to $G_{k}$ that for each fixed $z$ and $y$

$\displaystyle
G^{N}_{k}\left(z\sqrt{k},y\sqrt{k}\right)=-y^{2}k-\frac{\sqrt{k}}{y-z}+\mathfrak{o}(\sqrt{k}).$
(C.3)

We can use this information to obtain an upper bound on
$F^{N}_{k}(x,z\sqrt{k})$. First, we know from (7.7) and the definitions of
$F^{N}_{k}$, $G^{N}_{k}$, and $H^{N}$ that, for all $x,y,z$,

$\displaystyle F^{N}_{k}(x,z\sqrt{k})\leq
H^{N}(x,y\sqrt{k})-G^{N}_{k}(z\sqrt{k},y\sqrt{k})+\mathfrak{o}(N^{-1/11}).$

Define

$\tilde{F}^{N}_{k}(x,z\sqrt{k}):=\inf_{y}\left(H^{N}(x,y\sqrt{k})-G^{N}_{k}(z\sqrt{k},y\sqrt{k})\right),$
(C.4)

so that $z\mapsto\tilde{F}(x,z\sqrt{k})$ is non-decreasing, since, by Lemma
C.2, $G^{N}_{k}(z\sqrt{k},y\sqrt{k})$ is monotonically non-increasing in $z$.
Note also that

$F^{N}_{k}(x,z\sqrt{k})\leq\tilde{F}^{N}_{k}(x,z\sqrt{k})+\mathfrak{o}(N^{-1/11})$
(C.5)

for all $x$ and $z$. Our next step is to upper bound
$\tilde{F}^{N}_{k}(x,z\sqrt{k})$.

Since for fixed $x,y$ we know $H^{N}(x,y)\stackrel{{\scriptstyle
d}}{{\to}}\mathcal{P}_{1}(y-x)$ as $N\to\infty$,
$x\mapsto\mathcal{P}_{1}(x)+x^{2}$ is stationary, and exponentially decaying
tail bounds for $\mathcal{P}_{1}(0)$ (i.e., the GUE Tracy-Widom distribution)
are well-known (e.g., [AGZ10, Exercise 3.8.3]), it follows that
$H^{N}(x,y\sqrt{k})=-(x-y\sqrt{k})^{2}+\mathfrak{o}(\sqrt{k})$. Combining this
with the estimate from (C.3) of $G^{N}_{k}(z\sqrt{k},y\sqrt{k})$ yields

$\displaystyle\tilde{F}^{N}_{k}(x,z\sqrt{k})$
$\displaystyle=\inf_{y}\left(-(x-y\sqrt{k})^{2}-\left(-y^{2}k-\frac{\sqrt{k}}{y-z}\right)\right)+\mathfrak{o}(\sqrt{k})$
$\displaystyle=\inf_{y}\left(2xy\sqrt{k}+\frac{\sqrt{k}}{y-z}\right)+\mathfrak{o}(\sqrt{k}),$

where in the last equality we used that $x$ is uniformly bounded,
independently of $k$. The optimizer of the infimum is
$y=z+\frac{1}{\sqrt{2x}}$. Substituting this into the previous display yields
that, for each fixed $x$ and $z$,

$\displaystyle\tilde{F}^{N}_{k}(x,z\sqrt{k})\leq
2\sqrt{2kx}+2xz\sqrt{k}+\mathfrak{o}(\sqrt{k}).$ (C.6)

With this information, we can show that the maximizer in (7.7) equals
$-\sqrt{k/2x}+\mathfrak{o}(\sqrt{k})$. Indeed, (7.7) can be written as

$\displaystyle H^{N}(x,y)=\sup_{z\leq
y}\left(F^{N}_{k}(x,z)+G^{N}_{k}(z,y)\right)+\mathfrak{o}(N^{-1/11}).$

Now, since $x$ and $y$ are bounded independently of $k$, the lefthand side of
the previous display is $\mathfrak{o}(\sqrt{k})$ (by Theorem 3.19). So at
$z=Z^{N}_{k}$ the maximizer, we must have

$\displaystyle
F^{N}_{k}(x,Z^{N}_{k})+G^{N}_{k}(Z^{N}_{k},y)=\mathfrak{o}(\sqrt{k}).$

We will show that if $|z+\sqrt{k/2x}|>\varepsilon\sqrt{k}$, then
$F^{N}_{k}(x,z)+G^{N}_{k}(z,y)$ is smaller then $-c\varepsilon^{2}\sqrt{k}$.
More precisely, we will show that for some absolute constant $c>0$,

$\displaystyle\sup_{\begin{subarray}{c}z:z\leq y\\\
|z+\sqrt{k/2x}|>\varepsilon\sqrt{k}\end{subarray}}\tilde{F}^{N}_{k}(x,z)+G^{N}_{k}(z,y)\leq-c\varepsilon^{2}\sqrt{k}+\mathfrak{o}(\sqrt{k});$

this suffices to establish our claim since, by (C.5),
$F^{N}_{k}(x,z)\leq\tilde{F}^{N}_{k}(x,z)+\mathfrak{o}(N^{-1/11})$. Recall
from just after (C.4) that $\tilde{F}^{N}_{k}(x,z\sqrt{k})$ is monotonically
non-decreasing in $z$. We first reduce the previous display to a supremum over
a finite set using the monotonicity in $z$ of $\tilde{F}^{N}_{k}$ and
$G^{N}_{k}$. Letting

$\displaystyle\lfloor x\rfloor_{N,k}$
$\displaystyle:=\max\left\\{w<x:w\in\tfrac{1}{4}\sqrt{\frac{k}{x}}\varepsilon^{2}\mathbb{Z}\right\\}$
$\displaystyle\lceil x\rceil_{N,k}$
$\displaystyle:=\min\left\\{w>x:w\in\tfrac{1}{4}\sqrt{\frac{k}{x}}\varepsilon^{2}\mathbb{Z}\right\\},$

the monotonicity in $z$ of $\tilde{F}^{N}_{k}$ and $G^{N}_{k}$ implies that it
is sufficient to show that

$\displaystyle\sup_{\begin{subarray}{c}z:z\leq y\\\
|z+\sqrt{k/2x}|>\varepsilon\sqrt{k}\end{subarray}}\tilde{F}^{N}_{k}(x,\lceil
z\rceil_{N,k})+G^{N}_{k}(\lfloor
z\rfloor_{N,k},y)\leq-c\varepsilon^{2}\sqrt{k}.$ (C.7)

Combining the bounds on $G^{N}_{k}$ and $\tilde{F}^{N}_{k}$ from (C.3) and
(C.6) respectively (taking $y$ in the $G^{N}_{k}$ bound to be $yk^{-1/2}$
here, and using the fact that $y$ is bounded independently of $k$), we obtain,
for each fixed $z$,

$\displaystyle\tilde{F}^{N}_{k}(x,\lceil z\rceil_{N,k})+G^{N}_{k}(\lfloor
z\rfloor_{N,k},y)\leq
2\sqrt{2kx}+2x(z+\tfrac{1}{4}\varepsilon^{2}x^{-1/2}\sqrt{k})+\frac{k}{z-\tfrac{1}{4}\varepsilon^{2}\sqrt{k/x}}+\mathfrak{o}(\sqrt{k}).$
(C.8)

If one substitutes $\varepsilon=0$, the righthand side is maximized at
$z=-\sqrt{k/2x}$ (under the condition $z\leq y$), at which value it equals
$0$. Further, by the concavity of the righthand side as a function in $z$ (and
the negativity of the third derivative) and using Taylor’s theorem, if
$|z+\sqrt{k/2x}|>\varepsilon\sqrt{k}$, the righthand side is at most
$-c\varepsilon^{2}\sqrt{k}$. Since these bounds also apply simultaneously for
all the finite number of $z$’s in (C.7), we obtain the asymptotic for
$Z^{N}_{k}$.

The tightness of $\\{Z^{N}_{k}(x,y)\\}_{N=1}^{\infty}$ for each $k$ follows
from the asymptotic and the monotonicity property $y\geq Z^{N}_{1}(x,y)\geq
Z^{N}_{2}(x,y)\geq\ldots\geq Z^{N}_{N^{1/6}}$ from Lemma C.1. ∎

######  Remark C.4.

Let remark briefly on the difference of the above argument compared to
[DOV22]. There, an analog of the $F^{N}_{k}$, $G^{N}_{k}$ and $H^{N}_{k}$
quantities were defined in a model known as Brownian last passage percolation
(introduced here in Section C.3 ahead). In their context, asymptotics for
$F^{N}_{k}$ were available (via estimates for a limiting analog $F_{k}$), and
these were used to derive an upper bound on $G^{N}_{k}$ to then upper bound
the analog of (C.8). Here, we do not _a priori_ have such estimates for
$F^{N}_{k}$ as we do not know it weakly converges to $F_{k}$; instead, we know
the convergence of $G^{N}_{k}$ to its analog in the parabolic Airy line
ensemble, where the previously mentioned estimates for $F_{k}$ apply. These
will lead to the estimates on $G_{k}$ recorded in Lemma C.3, yielding the
estimates for $G^{N}_{k}$ that were used. We essentially reversed the argument
in [DOV22] to obtain estimates for $F^{N}_{k}$ that allowed us to then bound
(C.8).

######  Remark C.5.

Recall from (7.2) and (7.3) that $L_{1}^{N,(k)}$ is equal in law to a
translation of $L_{1}^{N,(1)}$, which was important for our arguments in
Section 7. This statement is false for the lower-indexed curves, namely,
$L_{j}^{N,(k)}$ does not have the same law as that of the corresponding
translation of $\smash{L_{j}^{N,(1)}}$ for $j>1$. This is closely related to
the fact that, as noted in Remark 3.3, we cannot merge colors 1 and 0 in the
colored $q$-Boson model. It can also be seen from the fact that, if such an
equality in law were true, then it would imply precise asymptotics for
$F_{k}^{N}$ (since the asymptotics of $\smash{L_{j}^{(N)}}$, converging to the
$j$th curve of the Airy line ensemble, are well understood, e.g., [DV21,
Corollary 5.3]), which are actually in disagreement with those shown in (C.5)
and (C.6).

###  C.3. $\bm{G_{k}}$ asymptotics

Here we give the proof of Lemma C.3. In order to access a certain variational
formula, we prove its analog in the model of Brownian LPP, and use the
convergence of the latter to the Airy sheet to obtain the result for $G_{k}$
itself. We briefly recall the model.

An up-right path from $(x,j)$ to $(y,k)$ for $x<y$ and $j>k$ is a function
$\gamma:[x,y]\to\llbracket k,j\rrbracket$ which is non-increasing and càdlàg
(right continuous with left-limits) with $\gamma(x)=j$ and $\gamma(y)=k$. The
jump times $(t_{1},\ldots,t_{N-1})$ of a path from line $N$ to line $1$
defined by $t_{i}$ being the time $\gamma$ jumps from line $i+1$ to line $i$,
i.e., the time $t$ such that $\gamma(t^{-})=i+1$ and $\gamma(t)=i$. A pair of
up-right paths $\pi$ and $\pi^{\prime}$ from $(x,j)$ to $(y,k)$ and
$(x^{\prime},j^{\prime})$ to $(y^{\prime},k^{\prime})$, respectively, are said
to be disjoint if either $\pi(t)>\pi^{\prime}(t)$ for all
$t\in(x,y)\cap(x^{\prime},y^{\prime})$ (i.e., excluding endpoints), or
$\pi(t)<\pi^{\prime}(t)$ for all $t\in(x,y)\cap(x^{\prime},y^{\prime})$.

Let $\bm{B}=(B_{1},\ldots,B_{N})$ be a collection of $N$ independent and
identically distributed Brownian motions on $[0,\infty)$. The weight of a path
$\gamma$ from $(x,N)$ to $(y,1)$ with jump times $(t_{1},\ldots,t_{N-1})$ is
given by (with the convention $t_{0}=y$ and $t_{N}=x$)

$\displaystyle\bm{B}[\gamma]=\sum_{i=1}^{N}\bigl{(}B_{i}(t_{i-1})-B_{i}(t_{i})\bigr{)}.$

Let $\bm{W}\\!\bm{B}=(W\\!B_{1},\ldots,W\\!B_{N})$ be the curves of the
_watermelon_ , defined by

$\displaystyle\sum_{i=1}^{k}(W\\!B)_{i}(t)=\bm{B}[(0,N)^{k}\to(t,1)^{k}]$

for each $k=1,\ldots,N$, where $\bm{B}[(0,N)^{k}\to(t,1)^{k}]$ is the maximum
weight over all collections of $k$ disjoint paths
$(\gamma_{1},\ldots,\gamma_{k})$, with $\gamma_{i}$ from $(0,N-i+1)$ to
$(t,1)$, with the weight of a collection of paths being the sum of the
weights.

Recall the LPP notation from (2.2) (which agrees with the above convention
with $k=1$). Let $x_{N}=2xN^{2/3}$, and similarly for $y_{N}$ and $z_{N}$.
Define

$\displaystyle F^{N,\mathrm{Br}}_{k}(x,z)$
$\displaystyle=N^{-1/3}\left(\bm{W}\\!\bm{B}[(x_{N},N)\to(N+z_{N},k)]-(W\\!B)_{k}(N+z_{N})+2xN^{2/3}\right)$
$\displaystyle G^{N,\mathrm{Br}}_{k}(z,y)$
$\displaystyle=N^{-1/3}\left(\bm{W}\\!\bm{B}[(N+z_{N},k)\to(N+y_{N},1)]+(W\\!B)_{k}(N+z_{N})-2N-2yN^{2/3}\right)$
$\displaystyle H^{N,\mathrm{Br}}(x,y)$
$\displaystyle=N^{-1/3}\left(\bm{W}\\!\bm{B}[(x_{N},N)\to(N+y_{N},1)]-2N-2(y-x)N^{2/3}\right).$

It follows from [DOV22, Proposition 2.2] or [CH14, Proposition 4.1] that
$G^{N,\mathrm{Br}}_{k}(z,y)\stackrel{{\scriptstyle d}}{{\to}}G_{k}(z,y)$ for
any fixed $z$ and $y$ (which may depend on $k$).

We will also need a certain stationarity property of $G_{k}$:

######  Lemma C.6.

Fix $\varepsilon\in\mathbb{R}$ and $z<y$. Then

$\displaystyle G_{k}(z\sqrt{k},y\sqrt{k})\stackrel{{\scriptstyle
d}}{{=}}G_{k}((z+\varepsilon)\sqrt{k},(y+\varepsilon)\sqrt{k}))+(\varepsilon^{2}+2\varepsilon
y)k.$

###### Proof.

Recall that $G_{k}(z,y)=\bm{\mathcal{P}}[(z,k)\to(y,1)]+\mathcal{P}_{k}(z)$
and that $\mathcal{P}_{i}(x)=\mathcal{A}_{i}(x)-x^{2}$ for all
$i\in\mathbb{N}$ and $x\in\mathbb{R}$, where $\bm{\mathcal{A}}$ is the Airy
line ensemble and is stationary. Also note that for any up-right path $\gamma$
from $(z,k)$ to $(y,1)$,
$\bm{\mathcal{P}}[\gamma]=\bm{\mathcal{A}}[\gamma]-y^{2}+z^{2}$. Thus it
follows that

$\displaystyle G_{k}(z,y)$
$\displaystyle=\bm{\mathcal{A}}[(z,k)\to(y,1)]-y^{2}+z^{2}+\mathcal{A}_{k}(z)-z^{2}$
$\displaystyle\stackrel{{\scriptstyle
d}}{{=}}\bm{\mathcal{A}}[(z+\varepsilon,k)\to(y+\varepsilon,1)]-y^{2}+\mathcal{A}_{k}(z+\varepsilon)$
$\displaystyle=\bm{\mathcal{P}}[(z+\varepsilon,k)\to(y+\varepsilon,1)]-y^{2}+(y+\varepsilon)^{2}+\mathcal{P}_{k}(z+\varepsilon).$

This equals $G(z+\varepsilon,y+\varepsilon)+\varepsilon^{2}+2\varepsilon y$.
Replacing $z$ by $z\sqrt{k}$, $y$ by $y\sqrt{k}$, and $\varepsilon$ by
$\varepsilon\sqrt{k}$ completes the proof. ∎

###### Proof of Lemma C.3.

In the proof we will prove the statement with $z_{0}$ in place of $z$ as we
will need to use $z$ as a generic variable. As mentioned, we will prove that

$G^{N,\mathrm{Br}}_{k}(z_{0}\sqrt{k},y\sqrt{k})=-y^{2}k-\sqrt{k}/(y-z_{0})+\mathfrak{o}(\sqrt{k});$

then since
$G^{N,\mathrm{Br}}_{k}(z_{0}\sqrt{k},y\sqrt{k})\stackrel{{\scriptstyle
d}}{{\to}}G_{k}(z_{0}\sqrt{k},y\sqrt{k})$ as $N\to\infty$, we will be done.

We know from [DOV22, Proposition 6.1] that
$F^{N,\mathrm{Br}}_{k}(x,z\sqrt{k})=(2\sqrt{2x}+2xz)\sqrt{k}+\mathfrak{o}(k^{1/2})$
for any $0<x<z$, and from [DOV22, Theorem 1.4, Proposition 4.1] that
$\smash{H^{N,\mathrm{Br}}(x,y)\stackrel{{\scriptstyle
d}}{{\to}}\mathcal{S}(x;y)}=-(x-y)^{2}+\mathfrak{o}(\sqrt{k})$. By the
variational definition of the LPP values, we also know that

$\displaystyle
H^{N,\mathrm{Br}}(x,y)=\sup_{z\in\mathbb{R}}F^{N,\mathrm{Br}}_{k}(x,z)+G^{N,\mathrm{Br}}_{k}(z,y).$
(C.9)

Let $x_{0}=1/(2(y-z_{0})^{2})$. The previous display implies that there is a
random $z^{*,N}_{k}$ such that

$F^{N,\mathrm{Br}}_{k}(x_{0},z^{*,N}_{k}\sqrt{k})+G^{N,\mathrm{Br}}_{k}(z^{*,N}_{k}\sqrt{k},y\sqrt{k})=H^{N,\mathrm{Br}}(x_{0},y\sqrt{k})=-(x_{0}-y\sqrt{k})^{2}+\mathfrak{o}(\sqrt{k}).$
(C.10)

We claim that

$z^{*,N}_{k}=-\sqrt{\frac{1}{2x_{0}}}+y+\mathfrak{o}(1)=z_{0}+\mathfrak{o}(1)$
(C.11)

and that $\\{z^{*,N}_{k}\\}_{N=1}^{\infty}$ is tight. By (C.9), we know

$G^{N,\mathrm{Br}}_{k}(z\sqrt{k},y\sqrt{k})\leq
H^{N,\mathrm{Br}}(x,y\sqrt{k})-F^{N,\mathrm{Br}}_{k}(x,z\sqrt{k})$

for all $x,y,z$. Taking $x=1/(2(y-z)^{2})$ and the asymptotics
$H^{N,\mathrm{Br}}(x,y\sqrt{k})=-(x-y\sqrt{k})^{2}+\mathfrak{o}(\sqrt{k})$,
$F^{N,\mathrm{Br}}_{k}(x,z\sqrt{k})=2\sqrt{2kx}+2\sqrt{k}xz+\mathfrak{o}(\sqrt{k})$
yields that

$G^{N,\mathrm{Br}}_{k}(z\sqrt{k},y\sqrt{k})\leq-y^{2}k-\frac{\sqrt{k}}{y-z}+\mathfrak{o}(\sqrt{k});$
(C.12)

by taking $N\to\infty$ and using
$G^{N,\mathrm{Br}}_{k}(z\sqrt{k},y\sqrt{k})\to G_{k}(z\sqrt{k},y\sqrt{k})$ in
distribution we also obtain the upper bound half of the overall equality we
are aiming to prove. Using this upper bound and the same asymptotics on
$H^{N,\mathrm{Br}}$ and $F^{N,\mathrm{Br}}_{k}$ in (C.10) yields that the
latter cannot hold if the claimed (C.11) does not hold. As with the tightness
of $\smash{Z^{N}_{k}}$ in Proposition 7.2, the tightness of
$\smash{z^{*,N}_{k}}$ follows from the just established asymptotic and the
monotonicity of the sequence in $k$ from Lemma C.1.

It remains to prove a lower bound on $G_{k}(z\sqrt{k},y\sqrt{k})$ matching the
upper bound in (C.12). With (C.11) we see that

$\displaystyle G^{N,\mathrm{Br}}_{k}(z^{*,N}_{k}\sqrt{k},y\sqrt{k})$
$\displaystyle=H^{N,\mathrm{Br}}(x_{0},y\sqrt{k})-F^{N,\mathrm{Br}}_{k}(x_{0},z^{*,N}_{k}\sqrt{k})$
$\displaystyle=-(x_{0}-y\sqrt{k})^{2}-(2\sqrt{2x_{0}}+2x_{0}z^{*,N}_{k})\sqrt{k}+\mathfrak{o}(\sqrt{k})$
$\displaystyle=-y^{2}k+2x_{0}y\sqrt{k}-(2\sqrt{2x_{0}}+2x_{0}(-(2x_{0})^{-1/2}+y))\sqrt{k}+\mathfrak{o}(\sqrt{k}).$

By simplifying, the previous line equals

$\displaystyle
G^{N,\mathrm{Br}}_{k}(z^{*,N}\sqrt{k},y\sqrt{k})=-y^{2}k-\sqrt{2x_{0}k}+\mathfrak{o}(\sqrt{k})$
$\displaystyle=-y^{2}k-\frac{\sqrt{k}}{y-z_{0}}+\mathfrak{o}(\sqrt{k}).$
(C.13)

By using the tightness of $z^{*,N}_{k}$ and the weak convergence of
$G^{N,\mathrm{Br}}_{k}$ to $G_{k}$, we obtain the existence of a $z^{*}_{k}$
satisfying (C.11) and the previous display with $G_{k}$ and $z^{*}_{k}$ in
place of their prelimiting analogs.

To go from $G_{k}(z^{*}_{k}\sqrt{k},y\sqrt{k})$ to
$G_{k}(z_{0}\sqrt{k},y\sqrt{k})$, we use that $G_{k}(\bm{\cdot},y)$ is
monotonically decreasing by Lemma C.2, as well as the stationarity property
from Lemma C.6. In the case that $z^{*}_{k}\geq z_{0}$, the monotonicity of
$G_{k}(\bm{\cdot},y\sqrt{k})$ along with (C.13) immediately yields that

$\displaystyle G_{k}(z_{0}\sqrt{k},y\sqrt{k})\geq
G_{k}(z^{*}_{k}\sqrt{k},y\sqrt{k})\geq-y^{2}k-\frac{\sqrt{k}}{y-z_{0}}+\mathfrak{o}(\sqrt{k}).$
(C.14)

In the case that $z^{*}_{k}<z_{0}$, we know from above that
$z^{*}_{k}>z_{0}-\mathfrak{o}(1)$, i.e, for any $\varepsilon>0$ it holds that
$\sum_{k}\mathbb{P}\left(z^{*}_{k}<z_{0}-\varepsilon\right)<\infty$. We fix
$\varepsilon>0$, and apply Lemma C.6 along with monotonicity of
$G_{k}(\bm{\cdot},y\sqrt{k})$ again to obtain that, for any deterministic
sequence $M_{k}$,

$\displaystyle\mathbb{P}\left(G_{k}(z_{0}\sqrt{k},y\sqrt{k})<M_{k}-\varepsilon\sqrt{k}\right)$
$\displaystyle=\mathbb{P}\left(G_{k}((z_{0}-\varepsilon)\sqrt{k},(y-\varepsilon)\sqrt{k})<M_{k}-(\varepsilon^{2}-2\varepsilon
y)k-\varepsilon\sqrt{k}\right)$
$\displaystyle\leq\mathbb{P}\left(G_{k}(z^{*}\sqrt{k},(y-\varepsilon)\sqrt{k})<M_{k}-(\varepsilon^{2}-2\varepsilon
y)k-\varepsilon\sqrt{k}\right)$
$\displaystyle\qquad+\mathbb{P}\left(z^{*}_{k}<z_{0}-\varepsilon\right).$

Now by (C.13), when $M_{k}=-(y-\varepsilon)^{2}k-\sqrt{k}/(y-\varepsilon-
z_{0})+(\varepsilon^{2}-2\varepsilon y)k$, the first term in the previous
display is summable, and we already noted that the second is also summable. We
can write

$\displaystyle
M_{k}=-y^{2}k-\frac{\sqrt{k}}{y-z_{0}}-\varepsilon\sqrt{k}\cdot\left(\frac{1}{(y-z_{0})(y-z_{0}-\varepsilon)}\right),$

and the parenthetical term is positive and $O(1)$. Thus we have established
that
$\mathbb{P}(G_{k}(z_{0}\sqrt{k},y\sqrt{k})<-y^{2}k-\sqrt{k}/(y-z_{0})-\varepsilon\sqrt{k})$
is summable for every $\varepsilon>0$, i.e.,
$G_{k}(z_{0}\sqrt{k},y\sqrt{k})\geq-y^{2}k-\sqrt{k}/(y-z_{0})+\mathfrak{o}(\sqrt{k})$.
This, along with (C.14) and (C.12), completes the proof. ∎

##  Appendix D Directed landscape and KPZ fixed point convergence

In this section we prove Corollaries 2.12 (ASEP) and 2.22 (S6V), using an idea
communicated to us by Shalin Parekh; it was also used by him in the work
[Par23] establishing joint convergence of ASEP in the weakly asymmetric regime
under multiple initial conditions with the basic coupling to the KPZ equation
with multiple initial conditions driven by the same noise. We prove Corollary
2.12 in Section D.1 and Corollary 2.22 in Section D.2. The latter relies on an
approximate monotonicity result (Proposition D.3) which is proved in Section
D.3, and a finite speed of discrepancy bound (Lemma D.4) proved in Section
D.4.

###  D.1. ASEP to directed landscape

We start by giving a lemma on height monotonicity for ASEP under the basic
coupling. Recall the definition from below (2.10) of the ASEP height function
$h^{\mathrm{ASEP}}(h_{0},0;\bm{\cdot},t)$ when started from an arbitrary
height function $h_{0}$.

######  Lemma D.1 (Height monotonicity of ASEP).

Let $H\in\mathbb{Z}$ be an integer and
$h_{0},h_{0}^{\prime}:\mathbb{Z}\to\mathbb{Z}$ be Bernoulli paths such that
$h_{0}(x)+H\geq h_{0}^{\prime}(x)$ for all $x\in\mathbb{Z}$. Then, under the
basic coupling, almost surely, $h^{\mathrm{ASEP}}(h_{0},0;x,t)+H\geq
h^{\mathrm{ASEP}}(h_{0}^{\prime},0;x,t)$ for all $x\in\mathbb{Z}$ and $t>0$.

###### Proof.

This follows immediately from the definitions of the basic coupling in Section
2.2.6 and $h^{\mathrm{ASEP}}(h_{0},0;\bm{\cdot},t)$ from below (2.10). ∎

We will also need to use the following basic observation, whose proof is
trivial and omitted.

######  Lemma D.2.

Let $X,Y$ be random variables such that
$\mathbb{E}[|X|],\mathbb{E}[|Y|]<\infty$. Suppose $X\geq Y$ almost surely and
$\mathbb{E}[X]=\mathbb{E}[Y]$. Then $X=Y$ almost surely.

###### Proof of Corollary 2.12 (1) & (2).

Corollary 2.12 (2) follows immediately from (1), so we only need to establish
the latter. The convergence of
$(\mathcal{L}^{\mathrm{ASEP},\varepsilon}(\bm{\cdot},s;\bm{\cdot},t):(s,t)\in\mathcal{T}^{2}_{<})$
to $(\mathcal{L}(\bm{\cdot},s;\bm{\cdot},t):(s,t)\in\mathcal{T}^{2}_{<})$ in
the space
$\smash{\mathcal{C}(\mathbb{R}^{2},\mathbb{R})^{\mathcal{T}^{2}_{<}}}$ is
equivalent to the convergence

$\displaystyle\left(\mathcal{L}^{\mathrm{ASEP},\varepsilon}(\bm{\cdot},s;\bm{\cdot},t):(s,t)\in\mathcal{U}^{2}_{<}\right)\xrightarrow{\smash{d}}\left(\mathcal{L}(\bm{\cdot},s;\bm{\cdot},t):(s,t)\in\mathcal{U}^{2}_{<}\right)$
(D.1)

for every finite subset $\mathcal{U}\subseteq\mathcal{T}$. Fix such a
$\mathcal{U}$ and write $\mathcal{U}=\\{s_{i}:i\in\llbracket 1,n\rrbracket\\}$
with $0\leq s_{1}<\ldots<s_{n}$ for some $n\in\mathbb{N}$.

We first observe that
$(\mathcal{L}^{\mathrm{ASEP},\varepsilon}(\bm{\cdot},s;\bm{\cdot},t):(s,t)\in\mathcal{U}^{2}_{<})$
is a tight sequence in $\varepsilon$. This follows from combining the fact
that, by Definitions 2.7 and 2.10,

$\mathcal{L}^{\mathrm{ASEP},\varepsilon}(x,s;y,t)\stackrel{{\scriptstyle\smash{d}}}{{=}}(t-s)^{1/3}\mathcal{S}^{\mathrm{ASEP},(t-s)^{-1}\varepsilon}((t-s)^{-2/3}x;(t-s)^{-2/3}y)$
(D.2)

as processes in $(x,y)$, with Theorem 2.8 and the relation of $\mathcal{L}$
with $\mathcal{S}$ from Definition 2.4 (i), namely, for any $t>0$, and as
processes in $x$ and $y$,

$\displaystyle\mathcal{L}(x,s;y,t)\stackrel{{\scriptstyle
d}}{{=}}(t-s)^{1/3}\mathcal{S}((t-s)^{-2/3}x,(t-s)^{-2/3}y).$ (D.3)

Let
$(\tilde{\mathcal{L}}(\bm{\cdot},s;\bm{\cdot},t):(s,t)\in\mathcal{U}^{2}_{<})$
be any subsequential limit point of the lefthand side of (D.1). Then we
further see from Theorem 2.8, (D.2), and (D.3) that, for each
$(s,t)\in\mathcal{U}^{2}_{<}$,
$\tilde{\mathcal{L}}(\bm{\cdot},s;\bm{\cdot},t)\stackrel{{\scriptstyle\smash{d}}}{{=}}\mathcal{L}(\bm{\cdot},s;\bm{\cdot},t)$.
We next observe that
$(\tilde{\mathcal{L}}(\bm{\cdot},s_{i};\bm{\cdot},s_{i+1}):i\in\llbracket
1,n-1\rrbracket)\stackrel{{\scriptstyle\smash{d}}}{{=}}(\mathcal{L}(\bm{\cdot},s_{i};\bm{\cdot},s_{i+1}):i\in\llbracket
1,n-1\rrbracket)$, since each of the components of the lefthand vector are
independent, and the same for the righthand vector.

With these facts in hand, (D.1) is reduced to showing that, for any
$r,s,t\in\mathcal{U}$ with $r<s<t$ fixed, almost surely, for all
$x,y\in\mathbb{R}^{2}$,

$\displaystyle\tilde{\mathcal{L}}(x,r;y,t)=\sup_{z\in\mathbb{R}}\left(\tilde{\mathcal{L}}(x,r;z,s)+\tilde{\mathcal{L}}(z,s;y,t)\right);$
(D.4)

this suffices since the righthand side is a deterministic function of the
independent quantities $\tilde{\mathcal{L}}(\bm{\cdot},r;\bm{\cdot},s)$ and
$\tilde{\mathcal{L}}(\bm{\cdot},s;\bm{\cdot},t)$ and the same relation holds
with $\mathcal{L}$ in place of $\tilde{\mathcal{L}}$. We show (D.4) by
returning to the prelimit. Fix $r,s,t$ as above. We first claim that, almost
surely, for all $x,y,z\in\mathbb{Z}$ (recall $\gamma=1-q$),

$\begin{split}&h^{\mathrm{ASEP}}(x,2\gamma^{-1}r\varepsilon^{-1};z,2\gamma^{-1}s\varepsilon^{-1})+h^{\mathrm{ASEP}}(z,2\gamma^{-1}s\varepsilon^{-1};y,2\gamma^{-1}s\varepsilon^{-1})\\\
&\geq
h^{\mathrm{ASEP}}(x,2\gamma^{-1}r\varepsilon^{-1};y,2\gamma^{-1}s\varepsilon^{-1}).\end{split}$
(D.5)

This holds for $z\geq y$ as both sides are equal at $z=y$, the first term on
the lefthand side is non-increasing and decreases by at most 1 in going from
$z$ to $z+1$, and the second term on the lefthand side is

$h^{\mathrm{ASEP}}(z,2\gamma^{-1}s\varepsilon^{-1};y,2\gamma^{-1}s\varepsilon^{-1})=(z-y)\mathbbm{1}_{y\leq
z}$

and so increases by $1$ on going from $z$ to $z+1$. Similarly, for $z\leq y$,
we have equality in (D.5) at $z=y$, the first term on the lefthand side weakly
increases on going from $z$ to $z-1$, and the second term on the lefthand side
is $0$ for all $z\leq y$.

We now view (D.5) as an ordering of initial conditions: for each $x,z$ fixed,
we regard the first term
$h^{\mathrm{ASEP}}(x,2\gamma^{-1}r\varepsilon^{-1};z,2\gamma^{-1}s\varepsilon^{-1})$
as a (random) constant, and
$h^{\mathrm{ASEP}}(z,2\gamma^{-1}s\varepsilon^{-1};\bm{\cdot},2\gamma^{-1}s\varepsilon^{-1})$
and
$h^{\mathrm{ASEP}}(x,2\gamma^{-1}r\varepsilon^{-1};\bm{\cdot},2\gamma^{-1}s\varepsilon^{-1})$
both as initial conditions. Evolving both for time
$2\gamma^{-1}(t-s)\varepsilon^{-1}$ under the basic coupling, Lemma D.1
guarantees that, almost surely, for all $x,y,z\in\mathbb{Z}$,

$\displaystyle
h^{\mathrm{ASEP}}(x,2\gamma^{-1}r\varepsilon^{-1};z,2\gamma^{-1}s\varepsilon^{-1})+h^{\mathrm{ASEP}}(z,2\gamma^{-1}s\varepsilon^{-1};y,2\gamma^{-1}t\varepsilon^{-1})$
$\displaystyle\geq
h^{\mathrm{ASEP}}(x,2\gamma^{-1}r\varepsilon^{-1};y,2\gamma^{-1}t\varepsilon^{-1}).$

Centering and rescaling according to Definition 2.10, and taking
$\varepsilon\to 0$, the previous display yields that, almost surely, for all
$x,y\in\mathbb{R}$,

$\displaystyle\tilde{\mathcal{L}}(x,r;y,t)\geq\sup_{z\in\mathbb{R}}\left(\tilde{\mathcal{L}}(x,r;z,s)+\tilde{\mathcal{L}}(z,s;y,t)\right).$
(D.6)

However, we know that both sides of this inequality are marginally distributed
as the corresponding expression with $\mathcal{L}$ in place of
$\tilde{\mathcal{L}}$ and, further, we know that, almost surely, for all
$x,y\in\mathbb{R}$,