Limiting Current Distribution for a Two Species Asymmetric Exclusion Process

Abstract

We study current fluctuations of a two-species totally asymmetric exclusion process, known as the Arndt–Heinzel–Rittenberg model. For a step-Bernoulli initial condition with finite number of particles, we provide an explicit multiple integral expression for a certain joint current probability distribution. By performing an asymptotic analysis we prove that the joint current distribution is given by a product of a Gaussian and a GUE Tracy–Widom distribution in the long time limit, as predicted by non-linear fluctuating hydrodynamics.

Appendices

A. Bethe Wave Function

We consider the AHR model with n plus and m minus particles. Define a set of coordinates by ${\mathbb {W}}^k\,{:=}\, \{\mathbf {x}=(x_1,\dots ,x_k)\in {\mathbb {Z}}^k: x_1<x_2<\dots <x_k\}$, and let $\mathbf {x}=(x_1,\dots ,x_n)\in {\mathbb {W}}^n$ and $\mathbf {y}=(y_1,\dots ,y_m)\in {\mathbb {W}}^m$ be the positions of plus and minus particles, respectively. To each set of coordinates we associate a unit basis vector $|\mathbf {x};\mathbf {y}\rangle$, and construct the state space as $S=\mathrm{Span}\{|\mathbf {x};\mathbf {y}\rangle \}$. Let

$$\begin{aligned} |P(t)\rangle \,{:=}\, \sum _{\mathbf {x},\mathbf {y}} P(\mathbf {x},\mathbf {y};t) |\mathbf {x};\mathbf {y}\rangle \end{aligned}$$

be the vector of probabilities to observe a configuration $|\mathbf {x},\mathbf {y}\rangle$ of the AHR model at time t with plus particles at positions $\mathbf {x}$ and minus particles at positions $\mathbf {y}$. The vector $|P(t)\rangle$ satisfies the master equation

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} |P(t)\rangle = M |P(t)\rangle , \end{aligned}$$

(A.1)

where M is the transition matrix encoding the jumping rates of the AHR model. The $\big ((\mathbf {x};\mathbf {y}),(\mathbf {x}';\mathbf {y}')\big )$ entry of M is the transition rate from state $(\mathbf {x}';\mathbf {y}')$ to $(\mathbf {x};\mathbf {y})$, given by

$$\begin{aligned} M\big ((\mathbf {x};\mathbf {y}),(\mathbf {x}';\mathbf {y}')\big ) = \left\{ \begin{array}{cl} \beta , &{} \text { if }(\mathbf {x}';\mathbf {y}') = (\mathbf {x}_i^-;\mathbf {y}), \forall i \in [1,n] \\ \alpha , &{} \text { if }(\mathbf {x}';\mathbf {y}') = (\mathbf {x};\mathbf {y}_j^+), \forall j \in [1,m] \\ -\beta n - \alpha m &{} \text { if }(\mathbf {x}';\mathbf {y}') = (\mathbf {x};\mathbf {y}) \\ 0, &{} \text { otherwise}, \end{array} \right. \end{aligned}$$

(A.2)

where $\mathbf {x}_i^{\pm }\,{:=}\,(x_1,\dots ,x_i {\pm } 1,\dots ,x_n)$. (A.1) is equivalent to (2.2).

To derive an expression for $P(\mathbf {x},\mathbf {y};t)$, or more precisely the transition probability (2.7) considered in Sect. 2, we first look at the eigenvalue problem of the master equation with eigenvector

$$\begin{aligned} |\psi \rangle = \sum _{\mathbf {x},\mathbf {y}} \psi (\mathbf {x};\mathbf {y}) |\mathbf {x};\mathbf {y}\rangle ,\qquad M|\psi \rangle = \varLambda _{n,m} |\psi \rangle , \end{aligned}$$

where $\varLambda _{n,m}$ is the eigenvalue.

We will follow Bethe’s method in [10] to obtain the Bethe wave function $\psi (\mathbf {x};\mathbf {y})$ for the AHR model. First we consider the case of single species particles only, before considering the two species case. Then for each case we start with two particles and then generalise to the many particles case.

Such method of Bethe ansatz is first introduced in [10], and applied to many particles systems in [65, 99, 100]. The Bethe ansatz was especially used to solve the master equation of ASEP and TASEP in [43, 87]. A more complicated formula would follow from the standard nested Bethe ansatz [25].

1.1 A.1. Single species

First we consider the case without minus particles and two plus particles, i.e., $n=2,m=0$. The corresponding eigenvalue equation for $x_1+1<x_2$ is

$$\begin{aligned} \varLambda _{2,0} \psi (x_1,x_2) = \beta \psi (x_1-1,x_2) + \beta \psi (x_1,x_2-1) - 2\beta \psi (x_1,x_2), \end{aligned}$$

(A.3)

while for $x_1=x=x_2-1$ it is

$$\begin{aligned} \varLambda _{2,0} \psi (x,x+1) = \beta \psi (x-1,x+1)-\beta \psi (x,x+1). \end{aligned}$$

(A.4)

These two equations can be solved by the trial wave function

$$\begin{aligned} \psi (x_1,x_2)=A_{12}z_1^{x_1}z_2^{x_2}+A_{21}z_1^{x_2}z_2^{x_1}, \end{aligned}$$

for which (A.3) results in the eigenvalue expression $\varLambda _{2,0}=\beta (z_1^{-1}+z_2^{-1}-2),$ while (A.4) gives the scattering relation $A_{12}/A_{21}=-(1-z_1)/(1-z_2)$.

The number of equations increases as the number n of plus particles increases. To solve the resulting system of equations it is convenient to replace (A.4) by

$$\begin{aligned} \psi (x,x)=\psi (x,x+1), \end{aligned}$$

(A.5)

because by imposing (A.5) on (A.3) for $x_1=x=x_2-1$, equation (A.4) is automatically satisfied. In other words, (A.3) along with the boundary condition (A.5) gives the same solution as that of the eigenvalue problem. This boundary condition (A.5) is the same as in [87], since the AHR model reduced to TASEP when there are only one species particles.

It can thus be seen that the eigenvalue problem corresponding to $m=0$ and general n is equivalent to

$$\begin{aligned} \varLambda _{n,0}\psi (\mathbf {x})&= \beta \sum _{i=1}^n\psi (\mathbf {x}_i^-) - n \beta \psi (\mathbf {x}),\\ \psi (x_1,\dots ,x_i,x_{i+1}=x_{i},\dots ,x_n)&= \psi (x_1,\dots ,x_i,x_{i+1}=x_{i}+1,\dots ,x_n),\!\! \quad \!\! i\in [1,n-1]. \end{aligned}$$

These equations are solved by the wave function

$$\begin{aligned} \psi (\mathbf {x})= \sum _{\pi \in S_n}A_{\pi }\prod _{i=1}^nz_{\pi _i}^{x_i}, \end{aligned}$$

with eigenvalue $\varLambda _{n,0}=\beta \sum _{i=1}^{n}(z_i^{-1}-1)$, and with scattering relation

$$\begin{aligned} \frac{A_{\pi }}{A_{s_i \pi }} = -\frac{1-z_{\pi _i}}{1-z_{\pi _{i+1}}}, \end{aligned}$$

where $s_i\; (i=1,\ldots , n-1)$ are simple transposition, i.e., the generators of the symmetric group $S_n$. This is the same ratio of amplitudes obtained in [87]. This ratio is satisfied if $A_{\pi }$ is of the form

$$\begin{aligned} A_{\pi } = {{\,\mathrm{sign}\,}}(\pi ) \prod _{i=1}^n \left( \frac{1}{1-z_{\pi _i}} \right) ^i. \end{aligned}$$

The derivation of $\psi (\mathbf {y})$, when $n=0$ and m general, is very similar to the case when $m=0$ and n general. The corresponding eigenvalue problem is

$$\begin{aligned} \varLambda _{0,m}\psi (\mathbf {y}) =&\alpha \sum _{j=1}^m\psi (\mathbf {y}_j^+) - m \alpha \psi (\mathbf {y}),\\ \psi (y_1,\dots ,y_{i-1}=y_{i},y_i,\dots ,y_m) =&\psi (y_1,\dots ,y_{i-1}=y_{i}-1,y_i,\dots ,y_m), \quad i\in [2,m], \end{aligned}$$

and we will not repeat analogous details of derivation here. Naively this observation would suggest that a wave function for two species particles is of the form

$$\begin{aligned} \sum _{\pi \in S_n}{{\,\mathrm{sign}\,}}(\pi ) \prod _{i=1}^{n} \left( \frac{1}{1-z_{\pi _i}}\right) ^i z_{\pi _i}^{x_i} \sum _{\sigma \in S_m}{{\,\mathrm{sign}\,}}(\sigma ) \prod _{j=1}^{m} \left( \frac{1}{1-w_{\sigma _j}}\right) ^{-j} w_{\sigma _j}^{-y_j}, \end{aligned}$$

(A.6)

with eigenvalue $\varLambda _{n,m} = \beta \sum _{j=1}^n (z_j^{-1}-1) + \alpha \sum _{k=1}^m (w_k^{-1}-1).$ Such a trial wave function would only work when there is no interaction between the plus and the minus particles. In the following, we shall consider a modification of this wave function so that the interaction between different types of particles can be taken into account.

As we have seen, the Bethe wave function of the AHR model for the single species cases is identical to the one of the TASEP [87]. The important detail of the AHR model’s wave function lies in the interaction between different types of particles.

1.2 A.2. Two species case

Now consider the case that two types of particles are present. We start with the simplest case with one plus and one minus particles. When $x\ne y\pm 1$, the corresponding eigenvalue equation reads as

$$\begin{aligned} \varLambda _{1,1} \psi (x;y) = \alpha \psi (x;y+1) + \beta \psi (x-1;y) - (\beta +\alpha )\psi (x;y), \end{aligned}$$

(A.7)

while for $x=y+1$ and $y=x+1$,

$$\begin{aligned} \varLambda _{1,1}\psi (y+1;y) =&\psi (y;y+1) - (\beta +\alpha )\psi (y+1;y), \end{aligned}$$

(A.8)

$$\begin{aligned} \varLambda _{1,1}\psi (x;x+1) =&\beta \psi (x-1;x+1) + \alpha \psi (x;x+2) - \psi (x;x+1). \end{aligned}$$

(A.9)

Expression (A.7) is satisfied by the wave function (A.6) and the eigenvalue $\varLambda _{1,1}=\beta (z^{-1}-1) +\alpha (w^{-1}-1)$. In the same way, (A.8) and (A.9) are satisfied by imposing appropriate boundary conditions. Comparing (A.7) and (A.8) gives the first condition,

$$\begin{aligned} \psi (y;y+1) = \alpha \psi (y+1;y+1) + \beta \psi (y;y). \end{aligned}$$

This indicates a piece-wise wave function

$$\begin{aligned} \psi (x;y) = \left\{ \begin{array}{l} \displaystyle C_{+-} z^{x} w^{-y}\quad x<y\\ \displaystyle C_{-+} z^{x} w^{-y}\quad x\ge y \end{array}\right. , \end{aligned}$$

so that (A.8) is satisfied by setting the amplitude ratio to $C_{+-}/C_{-+}=\alpha z+\beta w$. Fortunately, when $\beta +\alpha =1$ equation (A.9) agrees with (A.7) without any extra condition. Incidentally, the condition $\beta +\alpha =1$ leads to a factorised stationary state [82]. The reasoning above suggests to modify (A.6) to

$$\begin{aligned} \psi (\mathbf {x};\mathbf {y})=&{} \sum \limits _{\pi \in S_n}{{\,\mathrm {sign}\,}}(\pi ) \prod \limits _{i=1}^{n} \left( \frac{1}{1-z_{\pi _i}}\right) ^i z_{\pi _i}^{x_i} \sum \limits _{\sigma \in S_m}{{\,\mathrm {sign}\,}}(\sigma ) \prod \limits _{j=1}^{m} \left( \frac{1}{1-w_{\sigma _j}}\right) ^{-j} w_{\sigma _j}^{-y_j}C_{\mathbf {x},\mathbf {y},\pi ,\sigma },\nonumber \\ \end{aligned}$$

(A.10)

where $C_{\mathbf {x},\mathbf {y},\pi ,\sigma }$ depends on the relative position of $\mathbf {x},\mathbf {y}$. For general n, m, we only need to impose the following boundary conditions on (A.10),

$$\begin{aligned}&\psi (\mathbf {x}; y_1,\dots ,y_j=x_i+1,\dots ,y_m) = \beta \psi (\mathbf {x}; y_1,\dots ,y_j=x_i,\dots ,y_m) \nonumber \\ {}&\quad + \alpha \psi (x_1,\dots ,x_{i-1},x_i+1,x_{i+1},\dots ,x_n; y_1,\dots ,y_j=x_i+1,\dots ,y_m). \end{aligned}$$

(A.11)

As in the single species case, there are no extra boundary condition for sectors with more than two particles. Substituting (A.10) into (A.11) gives the ratio

$$\begin{aligned} \frac{C_{y_j>x_i}}{C_{y_j\le x_i}} = \alpha z_{\pi _i}+\beta w_{\sigma _j}. \end{aligned}$$

This suggests a form of the coefficient $C_{\mathbf {x},\mathbf {y},\pi ,\sigma }$ as

$$\begin{aligned} C_{\mathbf {x},\mathbf {y},\pi ,\sigma } = \prod _{j=1}^m \prod _{k=1}^{r_j} \frac{1}{\alpha z_{\pi _{n-k+1}}+\beta w_{\sigma _j}}, \end{aligned}$$

where $r_j$ is the number of plus particles to the right of the $j^\mathrm{th}$ minus particle, i.e., $r_j=\#\{x_i\in \mathbf {x}\mid x_i \ge y_j\}$.

In conclusion, the general Bethe wave function is given by

$$\begin{aligned} \psi (\mathbf {x};\mathbf {y})= \sum _{\pi \in S_n}{{\,\mathrm {sign}\,}}(\pi ) \prod _{i=1}^{n} \left( \frac{1}{1-z_{\pi _i}}\right) ^i z_{\pi _i}^{x_i} \sum _{\sigma \in S_m}{{\,\mathrm {sign}\,}}(\sigma ) \prod _{j=1}^{m} \left( \frac{1}{1-w_{\sigma _j}}\right) ^{-j} w_{\sigma _j}^{-y_j} \prod _{k=1}^{r_j} \frac{1}{\alpha z_{\pi _{n-k+1}}+\beta w_{\sigma _j}},\nonumber \\ \end{aligned}$$

(A.12)

with eigenvalue

$$\begin{aligned} \varLambda _{n,m} = \beta \sum _{j=1}^n (z_j^{-1}-1) + \alpha \sum _{k=1}^m (w_k^{-1}-1). \end{aligned}$$

(A.13)

This result leads to an integral formula for the transition probability, as stated in Sect. 2.

B. Boundary Conditions for the Transition Probability

In this appendix we provide details of the proof that (2.7) satisfies the boundary conditions (2.3)–(2.5). Throughout the entire proof, we shall call the factor $\prod _{k=1}^{m}\prod _{j=1}^{r_k} (\alpha z_{\pi _{n-j+1}}+\beta w_{\sigma _k})^{-1}$ the scattering factor because it corresponds to scatterings of plus and minus particles.

1.1 B.1. Proof of boundary condition (2.3)

On both sides of (2.3), i.e., when $x_{i+1}=x_i$ on the left hand side and $x_{i+1}=x_i+1$ on the right hand side, the scattering factor $\prod _{k=1}^{m}\prod _{j=1}^{r_k}(\alpha z_{\pi _{n-j+1}}+\beta w_{\sigma _k})^{-1}$ remains unchanged within the physical regions $\varOmega ^{n+m}$. Moreover, the scattering factor is symmetric in $z_{\pi _i},z_{\pi _{i+1}}$ since there is no minus particle between the $i^\mathrm{th}$ and the $i+1^\mathrm{st}$ plus particles.

From (2.7) we observe that $x_i$ and $x_{i+1}$ only appear as exponents of $z_{\pi _i}$ and $z_{\pi _{i+1}}$, and as there is no change in the scattering factor we only need to compare the $z_{\pi _i}$ and $z_{\pi _{i+1}}$ factors

$$\begin{aligned} \left( \frac{1}{1-z_{\pi _i}}\right) ^i \left( \frac{1}{1-z_{\pi _{i+1}}}\right) ^{i+1} z_{\pi _i}^{x_i}z_{\pi _{i+1}}^{x_{i+1}} \end{aligned}$$

(B.1)

in the integrand for both sides of (2.3). For the left hand side of (2.3), i.e., when $x_{i+1}=x_{i}$, the factor (B.1) reads as

$$\begin{aligned} LHS = \left( \frac{1}{1-z_{\pi _i}}\right) ^i \left( \frac{1}{1-z_{\pi _{i+1}}}\right) ^{i+1} (z_{\pi _i}z_{\pi _{i+1}})^{x_i}. \end{aligned}$$

When $x_{i+1}=x_i+1$, (B.1) becomes

$$\begin{aligned} RHS = \left( \frac{1}{1-z_{\pi _i}}\right) ^i \left( \frac{1}{1-z_{\pi _{i+1}}}\right) ^{i+1} (z_{\pi _i}z_{\pi _{i+1}})^{x_i}z_{\pi _{i+1}}. \end{aligned}$$

It follows that

$$\begin{aligned} LHS-RHS =&\left[ \frac{1}{(1-z_{\pi _i})(1-z_{\pi _{i+1}})}\right] ^i (z_{\pi _i}z_{\pi _{i+1}})^{x_i} \left( \frac{1}{1-z_{\pi _{i+1}}}- \frac{z_{\pi _{i+1}}}{1-z_{\pi _{i+1}}}\right) \\ =&\left[ \frac{1}{(1-z_{\pi _i})(1-z_{\pi _{i+1}})}\right] ^i (z_{\pi _i}z_{\pi _{i+1}})^{x_i}. \end{aligned}$$

This factor is symmetric in $z_{\pi _i},z_{\pi _{i+1}}$. Moreover, the scattering factor is also symmetric in $z_{\pi _i},z_{\pi _{i+1}}$, and hence summing over $\pi \in S_n$ in (2.7) gives a zero integrand due to the factor ${{\,\mathrm{sign}\,}}(\pi )$.

1.2 B.2. Proof of boundary condition (2.4)

The proof of this boundary condition is very similar to the one above. First we notice that when $y_{i-1}=y_i$ and $y_{i-1}=y_i-1$, the scattering factor is unchanged and symmetric in $w_{\sigma _{i-1}},w_{\sigma _i}$. Hence as before, we only need to consider

$$\begin{aligned} \left( \frac{1}{1-w_{\sigma _{i-1}}}\right) ^{-i+1} \left( \frac{1}{1-w_{\sigma _i}}\right) ^{-i} w_{\sigma _{i-1}}^{-y_{i-1}}w_{\sigma _i}^{-y_i}. \end{aligned}$$

(B.2)

Then difference of this factor between the left hand side and right hand side of (2.4) is given by

$$\begin{aligned} LHS-RHS =&\left[ \frac{1}{(1-w_{\sigma _{i-1}}) (1-w_{\sigma _i})}\right] ^{-i} (w_{\sigma _{i-1}}w_{\sigma _i})^{-y_i} \left( \frac{1}{1-w_{\sigma _{i-1}}}- \frac{w_{\sigma _{i-1}}}{1-w_{\sigma _{i-1}}} \right) \\ =&\left[ \frac{1}{(1-w_{\sigma _{i-1}}) (1-w_{\sigma _i})}\right] ^{-i} (w_{\sigma _{i-1}}w_{\sigma _i})^{-y_i}, \end{aligned}$$

which is symmetric in $w_{\sigma _{i-1}},w_{\sigma _i}$. Thus summing over $\sigma \in S_m$ in (2.7) gives a zero integrand, as required.

1.3 B.3. Proof of boundary condition (2.5)

We consider two cases when $y_j=x_i$ and $y_j=x_i+1$. From the integrand of (2.7), in these two cases, one only needs to check the factor

$$\begin{aligned} \left( \frac{1}{1-z_{\pi _i}}\right) ^i \left( \frac{1}{1-w_{\sigma _j}}\right) ^{-j} z_{\pi _i}^{x_i}w_{\sigma _j}^{-y_j} \prod _{k=1}^{r_j} \frac{1}{\alpha z_{\pi _{n-k+1}} + \beta w_{\sigma _j}}, \end{aligned}$$

for both sides of (2.5). For the left hand side of (2.5), $y_j=x_i+1$, which indicates that the $i+1^\mathrm{st}$ plus particle is sitting at the right hand side of $y_j$. Therefore $r_j=n-i$ and then

$$\begin{aligned} LHS = \left( \frac{1}{1-z_{\pi _i}}\right) ^i \left( \frac{1}{1-w_{\sigma _j}}\right) ^{-j} \left( \frac{z_{\pi _i}}{w_{\sigma _j}}\right) ^{x_i} w_{\sigma _j}^{-1} \prod _{k=1}^{n-i} \frac{1}{\alpha z_{\pi _{i+k}}+ \beta w_{\sigma _j}}. \end{aligned}$$

Now let us consider the right hand side of (2.5) where $y_j=x_i$. In this case $r_j=n-i+1$, since the $i^\mathrm{th}$ plus particle is sitting at $y_j$. It follows that

$$\begin{aligned} RHS =&\beta \left( \frac{1}{1-z_{\pi _i}}\right) ^i \left( \frac{1}{1-w_{\sigma _j}}\right) ^{-j} \left( \frac{z_{\pi _i}}{w_{\sigma _j}}\right) ^{x_i} \prod _{k=0}^{n-i} \frac{1}{\alpha z_{\pi _{i+k}}+ \beta w_{\sigma _j}} \\ {}&+ \alpha \left( \frac{1}{1-z_{\pi _i}}\right) ^i \left( \frac{1}{1-w_{\sigma _j}}\right) ^{-j} \left( \frac{z_{\pi _i}}{w_{\sigma _j}}\right) ^{x_i+1} \prod _{k=0}^{n-i} \frac{1}{\alpha z_{\pi _{i+k}}+\beta w_{\sigma _j}} \\ =&\left( \frac{1}{1-z_{\pi _i}}\right) ^i \left( \frac{1}{1-w_{\sigma _j}}\right) ^{-j} \left( \frac{z_{\pi _i}}{w_{\sigma _j}}\right) ^{x_i} \prod _{k=0}^{n-i} \frac{1}{\alpha z_{\pi _{i+k}}+ \beta w_{\sigma _j}} \left( \beta + \alpha \frac{z_{\pi _i}}{w_{\sigma _j}}\right) \\ =&\left( \frac{1}{1-z_{\pi _i}}\right) ^i \left( \frac{1}{1-w_{\sigma _j}}\right) ^{-j} \left( \frac{z_{\pi _i}}{w_{\sigma _j}}\right) ^{x_i} w_{\sigma _j}^{-1} \prod _{k=1}^{n-i} \frac{1}{\alpha z_{\pi _{i+k}}+\beta w_{\sigma _j}}, \end{aligned}$$

which is exactly the same as LHS.

C. Proof of Symmetrisation Identity Lemma 3.1

In this appendix we provide detail of the proof of Lemma 3.1. In Appendix C.1, we show the direct proof by induction. On the other hand, the identity Lemma 3.1 can also be derived by specialising equation (9) of [95], and we show detail of the proof in Appendix C.2.

1.1 C.1. Direct proof by induction

We first recall the statement of Lemma 3.1,

$$\begin{aligned} \sum _{\pi \in S_n}{{\,\mathrm{sign}\,}}(\pi )\prod _{i=1}^n \left( \frac{z_{\pi _i}-1}{z_{\pi _i}}\right) ^i \frac{1}{(1-(1-\rho )\prod _{j=1}^i z_{\pi _j})} = \frac{\prod _{1 \le i<j \le n}(z_j-z_i) \prod _{i=1}^{n}(z_i-1)}{\prod _{i=1}^{n}z_i^n(1-(1-\rho )z_i)}. \end{aligned}$$

(C.1)

We prove this identity by mathematical induction in n. One can easily see that (C.1) holds for $n=1$. We assume that it holds for $n-1$, and prove it for n. Let us denote the left hand side of the identity by $f_n(z_1,\dots ,z_n)$. To make use of the induction assumption, we need to find the relation between $f_n$ and $f_{n-1}$. The sum over $S_n$ can be split into a sum over $k \in [1, n]$ such that $\pi _n=k$ and then sum over $S_{n-1}$. We observe that ${{\,\mathrm{sign}\,}}(\pi )=(-1)^{n-k}{{\,\mathrm{sign}\,}}(\sigma )$ where $\sigma$ is the permutation $\pi$ restricted on $\{1,2,\dots ,k-1,k+1,\dots ,n\}$, and $(-1)^{n-k}$ is the signature of the permutation $(1,\dots ,k-1,k+1,\dots ,n,k)$. Therefore,

$$\begin{aligned} f_n(z_1,\dots ,z_n) = \sum _{k=1}^n(-1)^{n-k} \left( \frac{z_k-1}{z_k}\right) ^n \frac{1}{1-(1-\rho )\prod _{i=1}^nz_{i}} f_{n-1}(z_1,\dots ,z_{k-1},z_{k+1},\dots ,z_n). \end{aligned}$$

Substituting our induction assumption for $f_{n-1}(z_1,\dots ,z_{k-1},z_{k+1},\dots ,z_n)$ and after some rearrangements, we obtain

$$\begin{aligned} f_n(z_1,\dots ,z_n)&= \frac{\prod _{1 \le i<j \le n}(z_j-z_i)}{1-(1-\rho )\prod _{i=1}^nz_i} \prod _{i=1}^n \frac{z_i-1}{z_i^{n-1}} \sum _{k=1}^n \frac{(z_k-1)^{n-1}}{z_k} \prod _{i = 1, i\ne k}^{n} \frac{1}{(z_k-z_i)(1-(1-\rho )z_i)}. \end{aligned}$$

In order to show the sum over k gives expected result, we consider the function defined by

$$\begin{aligned} F(z) = \frac{(z-1)^{n-1}(1-(1-\rho )z)}{z\prod _{i=1}^n(z-z_i)(1-(1-\rho )z_i)}. \end{aligned}$$

By the residue theorem,

$$\begin{aligned} \sum _{k=1}^{n}{{\,\mathrm{Res}\,}}_{z=z_k}F(z) =&\sum _{k=1}^n \frac{(z_k-1)^{n-1}}{z_k} \prod _{i=1,i\ne k}^{n} \frac{1}{(z_k-z_i)(1-(1-\rho )z_i)} \\ =&-{{\,\mathrm{Res}\,}}_{z=0}F(z) - {{\,\mathrm{Res}\,}}_{z=\infty }F(z) \\ =&\frac{1}{\prod _{i=1}^nz_i(1-(1-\rho )z_i)} - \frac{(1-\rho )}{\prod _{i=1}^n(1-(1-\rho )z_i)}, \end{aligned}$$

which gives the required result to complete the proof.

1.2 C.2. Alternative proof

We start from the identity, which is equation (9) of [95]:

$$\begin{aligned}&\sum _{\sigma \in S_n} {{\,\mathrm{sign}\,}}(\sigma ) \prod _{1 \le i< j \le n} \frac{1}{ p + q \xi _{\sigma _i} \xi _{\sigma _j} - \xi _{\sigma _i} } \prod _{i = 1}^{n} \frac{1}{ ( \prod _{j=i}^{n} \xi _{\sigma _j} ) - (1 - \rho ) - \tau ^{n-i+1} \rho } \\&\quad = q^{n(n-1)/2} \frac{ \prod _{1 \le i < j \le n} (\xi _i - \xi _j) }{ \prod _{i=1}^{n}( \xi _i - (1 - \rho ) - \tau \rho ) \prod _{1 \le i \ne j \le n} ( p + q \xi _i \xi _j - \xi _i) }, \end{aligned}$$

where $0 < \rho \le 1$, $p + q = 1$ and $\tau = p/q$. Setting $p=0, q = 1, \tau = 0$ and simplifying the product yield

$$\begin{aligned}&\sum _{\sigma \in S_n} {{\,\mathrm{sign}\,}}(\sigma ) \prod _{i = 1}^{n} { \left( \frac{1}{\xi _{\sigma _i}} \right) }^{n-i} {\left( \frac{1}{\xi _{\sigma _i} - 1} \right) }^{i-1} \frac{1}{ ( \prod _{j=i}^{n} \xi _{\sigma _j} ) - (1 - \rho ) } \\&\quad = q^{n(n-1)/2} \frac{ \prod _{1 \le i < j \le n} (\xi _i - \xi _j) }{ \prod _{i=1}^{n}( \xi _i - (1 - \rho ) ) \xi _i^{n-1} {(\xi _i - 1)}^{n-1} }. \end{aligned}$$

Changing variables $\xi _i = z_i^{-1}$ for all $i \in [1,n]$, this equation is written as

$$\begin{aligned} \sum _{\sigma \in S_n} {{\,\mathrm{sign}\,}}(\sigma ) \prod _{i=1}^{n}{ z_{\sigma _i}^{n-i} {\left( \frac{z_{\sigma _i}}{1 - z_{\sigma _i}} \right) }^{i-1} \frac{ \prod _{j=i}^{n}{ z_{\sigma _j} } }{ 1 - (1 - \rho ) \prod _{j=i}^{n}{ z_{\sigma _j} } } } = \frac{ \prod _{i=1}^{n}{ z_i^{n} } \prod _{1 \le i < j \le n}{ (z_j - z_i) } }{ \prod _{i=1}^{n}{ {(1 - z_i)}^{n-1} (1 - (1 - \rho ) z_i) } }. \end{aligned}$$

Replacing $\sigma = (\sigma _1, \dots , \sigma _n)$ with its reversal $\pi = (\sigma _n , \dots , \sigma _1)$, the left hand side is written as

$$\begin{aligned} \sum _{\pi \in S_n} {(-1)}^{n(n-1)/2} {{\,\mathrm{sign}\,}}(\pi ) \prod _{i=1}^{n}{ \frac{ z_{\pi _i}^{n-1} {(1 - z_{\pi _i})}^{i-n} \prod _{j=1}^{i}{ z_{\pi _j} } }{ 1 - (1 - \rho ) \prod _{j=1}^{i}{ z_{\pi _j} } } }, \end{aligned}$$

and canceling like terms on both sides yields

$$\begin{aligned} \sum _{\pi \in S_n} {(-1)}^{n(n-1)/2} {{\,\mathrm{sign}\,}}(\pi ) \prod _{i=1}^{n}{ \frac{ {(1 - z_{\pi _i})}^{i} \prod _{j=1}^{i}{ z_{\pi _j} } }{ 1 - (1 - \rho ) \prod _{j=1}^{i}{ z_{\pi _j} } } } = \frac{ \prod _{i=1}^{n}{ z_i ( 1 - z_i ) } \prod _{1 \le i < j \le n}{ (z_j - z_i) } }{ \prod _{i=1}^{n}{ (1 - (1 - \rho ) z_i) } }. \end{aligned}$$

Considering the equality

$$\begin{aligned} \prod _{i=1}^{n}{ \prod _{j=1}^{i}{ z_{\pi _j} } } = \prod _{i=1}^{n}{ z_{\pi _i}^{n+1-i} } = \prod _{i=1}^{n}{ z_i^{n+1} } \prod _{i=1}^{n}{ z_{\pi _i}^{-i} }, \end{aligned}$$

and dividing $\prod _{i=1}^{n}{z_i^{n+1}}$ into the both sides, we obtain

$$\begin{aligned}&\sum _{\pi \in S_n} {(-1)}^{n(n-1)/2} {{\,\mathrm{sign}\,}}(\pi ) \prod _{i=1}^{n}{ {\left( \frac{1 - z_{\pi _i}}{ z_{\pi _i} } \right) }^i } \frac{ 1 }{ 1 - (1 - \rho ) \prod _{j=1}^{i}{ z_{\pi _j} } } \\&\quad = \frac{ \prod _{i=1}^{n}{ ( 1 - z_i ) } \prod _{1 \le i < j \le n}{ (z_j - z_i) } }{ \prod _{i=1}^{n}{ z_i^{n} (1 - (1 - \rho ) z_i) } }, \end{aligned}$$

and then, accounting for ${(-1)}^{n(n-1)/2} {(-1)}^{n(n+1)/2} = {(-1)}^{n^2} = {(-1)}^{n}$, we arrive at equation (C.1).

D. Asymptotic of the Rescaled Kernel

1.1 D.1. Proof of Lemma 6.1

The facts that $w_{*}$ is a double root and $w_{2}$ is a single root can be checked easily. To prove $\varGamma$ gives the steepest descent contour, we first show that ${{\,\mathrm{Re}\,}}(g_1)$ is decreasing along $\varGamma _2\cup \varGamma _3$. On $\varGamma _2$, we find

$$\begin{aligned}&\frac{\mathrm{d}{{\,\mathrm{Re}\,}}(g_1)(s)}{\mathrm{d}s} \\&\quad = \frac{ 2 s^2 \Big [ - 3 (3 - \rho ) (1 - \rho )^2 (1 + \rho ) - 2 s (1 - \rho )(3 + 10 \rho - 5 \rho ^2) - 2 s^2 (11 - 6 \rho + 3 \rho ^2) - 12 s^3 (1 - \rho ) - 8 s^4 \Big ] }{\big (3s^2 + (1 + \rho - s)^2\big ) \big (3s^2 + (1 + s - \rho )^2\big ) \big (3s^2 + (3 + s - \rho )^2\big ) } . \end{aligned}$$

One can verify that $\mathrm{d}{{\,\mathrm{Re}\,}}(g_1)(s)/\mathrm{d}s<0$ for any $s\in [0,2]$ and any $\rho \in (0,1)$, implying that ${{\,\mathrm{Re}\,}}(g_1)$ is decreasing along $\varGamma _2$. It follows by symmetry that, ${{\,\mathrm{Re}\,}}(g_1)$ is increasing along $\varGamma _1$. In fact, ${{\,\mathrm{Re}\,}}(\ln (w))=\ln (|w|)$, so ${{\,\mathrm{Re}\,}}(g_1)$ is symmetric with respect to the horizontal axis.

It remains to check the monotonicity of ${{\,\mathrm{Re}\,}}(g_1)$ along $\varGamma _3$. Again taking the derivative along $\varGamma _3$, we have

$$\begin{aligned} \frac{\mathrm{d}{{\,\mathrm{Re}\,}}(g_1)(\theta )}{\mathrm{d}\theta } = \frac{ 2 (1 + \rho )^2 \sin (\theta ) \cos (\theta /2)^2 [17 - 5 \rho + 3 \rho ^2 + \rho ^3 + 8 (1 + \rho ) \cos (\theta )] }{ [5 + 2 \rho +\rho ^2 + 4 (1 + \rho ) \cos (\theta )] [17 + 2\rho +\rho ^2 + 8 (1 + \rho ) \cos (\theta )] }, \end{aligned}$$

which equals zero only at $\theta =0,\pi$, i.e. ${{\,\mathrm{Re}\,}}(g_1)$ decreases when $s\in [-2\pi /3,0]$, and increases when $s\in [0,2\pi /3]$. In fact $17 - 5 \rho + 3 \rho ^2 + \rho ^3 + 8 (1 + \rho ) \cos (\theta )\ge 9 - 13 \rho + 3 \rho ^2 + \rho ^3$. The derivative of $9 - 13 \rho + 3 \rho ^2 + \rho ^3$ with respect to $\rho$ is $-13+6\rho +3\rho ^2 \le -13+6+3<0$, indicating that $9 - 13 \rho + 3 \rho ^2 + \rho ^3\ge 9-13+3+1=0$. Similarly, one can see that $5 + 2 \rho +\rho ^2 + 4 (1 + \rho ) \cos (\theta )$ and $17 + 2\rho +\rho ^2 + 8 (1 + \rho ) \cos (\theta )$ are non-negative for any $\theta$ and $\rho \in (0,1)$.

Therefore, we conclude that ${{\,\mathrm{Re}\,}}(g_1)$ is strictly monotone along $\varGamma$ except at its minimum point $w=2-\rho '/2$ and maximum point $w_{*}=\rho '/2$.

Similarly, the fact that the contour $\varSigma$ is a steepest descent path for $-g_1(w)$ can be proved by calculating $\tfrac{\mathrm{d}{{\,\mathrm{Re}\,}}(g_1)(\theta )}{\mathrm{d}\theta }$ along $\varSigma$.

1.2 D.2. Proof of uniform convergence of the rescaled kernel, Proposition 6.2

Consider the steepest descent contour $\varGamma \times \varSigma =(\bigcup _{i=1}^3 \varGamma _i) \times (\bigcup _{i=1}^4 \varSigma _i)$ defined in (6.11) and (6.12). We observe that the integrand of the kernel contains the factor $(w-z)^{-1}$, which requires that the z, w-contours do not intersect with each other. As a consequence, we need to deform the contour $\varGamma$ away from the saddle point $w_{*}$, and replaced by a vertical line through $w_{*}(1+\delta )$ (see Fig. 11 and (D.1)). We choose $\delta =t^{-1/3}$. Under such a deformed contour, we now can bound $|w-z|^{-1}$ by $(w_{*} \delta )^{-1}$ along $\varGamma ' \times \varSigma$, where

$$\begin{aligned} \varGamma ' = \left\{ z \in \varGamma \big | | z - w_{*} | > 2 w_{*} \delta \right\} + \left\{ z \in {\mathbb {C}} \big | z = w_{*} + w_{*} \delta (1 - s \mathrm{i}) , s \in \left[ -\sqrt{3} , \sqrt{3} \right] \right\} . \nonumber \\ \end{aligned}$$

(D.1)

Fig. 11

The deformed contour $\varGamma '$ with the blue part $\varGamma _\mathrm{vert}$

To estimate the integrand of ${\mathcal {K}}_t^{c } (\xi , \zeta )$ in $\varGamma ' \times \varSigma$, we separate the contour into two parts: far away from the point $w_{*}$ and the neighbourhood of $w_{*}$, denoted by

$$\begin{aligned} \varGamma ^{\varDelta }&= \left\{ z \in \varGamma \big | | z - w_{*} | \le \varDelta \right\} ,\,\,\,\, \varGamma '^{\varDelta } = \left\{ z \in \varGamma ' \big | | z - w_{*} | \le \varDelta \right\} , \end{aligned}$$

(D.2)

$$\begin{aligned} \varSigma ^{\varDelta }&= \left\{ w \in \varSigma \big | | w - w_{*}| \le \varDelta \right\} , \end{aligned}$$

(D.3)

where we choose⁴$\varDelta =t^{-1/9}$ so that the integrand along the distant parts can be bounded by ${{\,\mathrm{e}\,}}^{-a_1 \varDelta ^3 t} \rightarrow 0$ as t tends to infinity. We will see this later in the proof. The deformed contour is now separated into four parts: $\varGamma '^{\varDelta } \times \varSigma ^{\varDelta }$, $(\varGamma ' \backslash \varGamma '^{\varDelta }) \times \varSigma ^{\varDelta }$, $\varGamma '^{\varDelta } \times (\varSigma \backslash \varSigma ^{\varDelta })$, $(\varGamma ' \backslash \varGamma '^{\varDelta }) \times (\varSigma \backslash \varSigma ^{\varDelta })$. As will be shown later, the main contribution in the long time limit is from the first part $\varGamma '^{\varDelta } \times \varSigma ^{\varDelta }$, i.e., paths of integration passing near a saddle point $w_{*}$. It will be also shown that other parts and the error terms of the first parts converge uniformly to zero on ${[-L , L]}^2$.

The integrand in (6.7) is estimated in three parts: z-integrand $\mathrm {e}^{ f(z, t,\xi ) - f(w_{*} , t, \xi ) + g_{\phi }(z) }$; w-integrand $\mathrm {e}^{ - f(w, t,\zeta ) + f(w_{*} , t, \zeta ) + g_{\psi }(w) }$; and $(w-z)^{-1}$. We have the following bounds of the factor $|w-z|$ along the deformed contour,

$$\begin{aligned} \underset{ (z , w) \in \varGamma '^{\varDelta } \times \varSigma ^{\varDelta } }{\min }{ | w - z | } =&w_{*} \delta , \end{aligned}$$

(D.4a)

$$\begin{aligned} \underset{ (z , w) \in (\varGamma ' \times \varSigma ) \backslash (\varGamma '^{\varDelta } \times \varSigma ^{\varDelta }) }{\min }{ | w - z | } \ge&\frac{\sqrt{3}}{2} \varDelta . \end{aligned}$$

(D.4b)

(i) Main contribution on $\varvec{\varGamma '^{\varDelta } \times \varSigma ^{\varDelta }}$. Let us now give the main contribution on $\varGamma '^{\varDelta } \times \varSigma ^{\varDelta }$. Consider the z-integrand $\mathrm {e}^{ f(z, t,\xi ) - f(w_{*} , t, \xi ) }$ and introduce a new variable: $z = w_{*} + \lambda ^{-1} Z$, where $\lambda$ is some constant which is defined in (6.14). Using the Taylor expansions given in (6.9), the z-integrand $\mathrm {e}^{ f(z, t,\xi ) - f(w_{*} , t, \xi ) + g_{\phi }(z) }$ is rewritten as

$$\begin{aligned} \begin{aligned}&\mathrm {e}^{ f(z, t,\xi ) - f(w_{*} , t, \xi ) + g_{\phi }(z) } \\&\quad = \mathrm {e}^{ \{ g_1(z) - g_1(w_{*}) \} t + \{ g_2(z) - g_2(w_{*}) \} t^{1/2} + \{ g_3(z , \xi ) - g_3(w_{*} , \xi ) \} t^{1/3} + g_{\phi }(z) } \\&\quad = \mathrm {e}^{ 2 a_1 {(\lambda ^{-1} Z)}^3 t + b_{3,\xi } \lambda ^{-1} Z t^{1/3} + g_{\phi }(w_{*}) + {\mathcal {O}}( Z^4 t , Z^2 t^{1/2} , Z^2 t^{1/3} , LZ^2t^{1/3} , Z ) } \\&\quad = \mathrm {e}^{ \frac{1}{3} Z^3 t - (s_2 + \xi ) Z t^{1/3} + {\mathcal {O}}( Z^4 t , Z^2 t^{1/2} , Z^2 t^{1/3} , Z ) }, \end{aligned} \end{aligned}$$

(D.5)

where in the last line, $g_{\phi }(w_{*})=0$, $2 a_1 \lambda ^{-3} = 1/3$ and $b_{3,\xi } \lambda ^{-1} = - (s_2 + \xi )$ with $\lambda$ and $\lambda _c$ defined in (6.14) and (6.5), respectively. Note that L is fixed and hence ${\mathcal {O}}(LZ^2t^{1/3})$ can be absorbed into ${\mathcal {O}}(Z^2t^{1/3})$.

With respect to the new variable Z, let $\gamma ^{\varDelta }$ be the corresponding path of the line integral: $\gamma ^{\varDelta } = \left\{ Z \in {\mathbb {C}} \big | w_{*} + \lambda ^{-1} Z \in \varGamma '^{\varDelta } \right\}$. It is easy to see that any $Z \in \gamma ^{\varDelta }$ satisfy $|Z| \le \lambda \varDelta$.

We then factorise the integrand into two terms: main contribution and the error term $R_z$:

$$\begin{aligned} \mathrm {e}^{ \frac{1}{3} Z^3 t - (s_2 + \xi ) Z t^{1/3} + {\mathcal {O}}( Z^4 t , Z^2 t^{1/2} , Z^2 t^{1/3} , L Z^2 t^{1/3} , Z ) } = \mathrm {e}^{ \frac{1}{3} Z^3 t - (s_2 + \xi ) Z t^{1/3} } + R_z, \end{aligned}$$

where

$$\begin{aligned} R_z = \mathrm {e}^{ \frac{1}{3} Z^3 t - (s_2 + \xi ) Z t^{1/3} } \left( {{\,\mathrm{e}\,}}^{{\mathcal {O}}( Z^4 t , Z^2 t^{1/2} , Z^2 t^{1/3} , Z ) } - 1 \right) . \end{aligned}$$

We claim that the error term $R_z$ gives a zero integral in the long time limit, which will be shown in the second part (ii) of the proof. To get rid of the variable t in the integrand, we change the variable Z to $v = t^{1/3} Z$. The deformed contour now becomes $\gamma ^{\varDelta t^{1/3}} = \left\{ v \in {\mathbb {C}} \mid w_{*} + v /(\lambda t^{1/3}) \in {\varGamma '}^{\varDelta } \right\}$. The z-integrand is then rewritten into

$$\begin{aligned} \mathrm {e}^{ \frac{1}{3} v^3 - (s_2 + \xi ) v } + R_v, \end{aligned}$$

(D.6)

where

$$\begin{aligned} R_v = \mathrm {e}^{ \frac{1}{3} v^3 - (s_2 + \xi ) v } \left( \mathrm {e}^{ {\mathcal {O}}( v^4 t^{-1/3} , v^2 t^{-1/6} , v^2 t^{-1/3} , v t^{-1/3} ) } - 1 \right) . \end{aligned}$$

(D.7)

Let us now consider the w-integrand $\mathrm {e}^{ - f(w, t,\zeta ) + f(w_{*} , t, \zeta ) + g_{\psi }(w) }$. Similarly, we change the variable to $w=w_{*}+\lambda ^{-1}W$ with $W=t^{-1/3}u$, then the contour now becomes $\sigma ^{\varDelta t^{1/3}} = \left\{ u \in {\mathbb {C}} \big | w_{*} + u /(\lambda t^{1/3}) \in \varSigma ^{\varDelta } \right\}$ and the integrand is written as

$$\begin{aligned} \frac{1}{w_{*} + c}\mathrm {e}^{- \frac{1}{3} u^3 + (s_2 + \zeta ) u } + R_u, \end{aligned}$$

(D.8)

where

$$\begin{aligned} R_u = \frac{1}{w_{*} + c} \mathrm {e}^{ -\frac{1}{3} u^3 + (s_2 + \zeta ) u } \left( {{\,\mathrm{e}\,}}^{ {\mathcal {O}}( u^4 t^{-1/3} , u^2 t^{-1/6} , u^2 t^{-1/3} , u t^{-1/3} ) } - 1 \right) . \end{aligned}$$

(D.9)

Note that the coefficient $(w_{*} + c)^{-1}$ comes from the factor ${{\,\mathrm{e}\,}}^{g_{\psi }(w_{*})}=(w_{*} + c)^{-1}$, while ${{\,\mathrm{e}\,}}^{g_{\phi }(w_{*})}=1$.

Here we focus on a part of the integral (6.7) whose integral path is limited to $\varGamma '^\varDelta \times \varSigma ^\varDelta$. By the change of variables $v = t^{1/3} \lambda (z - w_{*})$, $u = t^{1/3} \lambda (w - w_{*})$, and combining (D.6), (D.8) and $(z-w)^{-1}$, a part of such an integral that does not involve error terms $R_v$ and $R_u$ now becomes

$$\begin{aligned} \int _{{\overline{\gamma }}^{\varDelta t^{1/3}} \times \sigma ^{\varDelta t^{1/3}}} \frac{\mathrm {d} v}{2 \pi \mathrm{i}} \frac{\mathrm {d} u}{2 \pi \mathrm{i}} \frac{1}{v - u}\mathrm {e}^{ \frac{1}{3} v^3 - (s_2 + \xi ) v - \frac{1}{3} {u}^3 + (s_2 + \zeta ) u}, \end{aligned}$$

(D.10)

where ${\overline{\gamma }}^{\varDelta t^{1/3}}$ denotes a path obtained by reversing the orientation of $\gamma ^{\varDelta t^{1/3}}$. In the following, let ${\overline{C}}$ denote a path obtained by reversing the orientation of C.

Recall the Airy kernel A(x, y) in (1.11). Considering the definition of the Airy function (1.10) and carrying out $\lambda$-integration by using the identity $\int _{0}^{\infty } \mathrm{d}\lambda {{\,\mathrm{e}\,}}^{- \lambda a} = 1/a$ which holds for $a \in {\mathbb {C}}$ satisfying ${{\,\mathrm{Re}\,}}(a) > 0$, we can obtain another expression of A(x, y) as

$$\begin{aligned} A(x,y) = \int _{{\overline{\gamma }}^\infty \times \sigma ^\infty } \frac{\mathrm {d} v}{2 \pi \mathrm{i}} \frac{\mathrm {d} u}{2 \pi \mathrm{i}} \frac{1}{v-u}\mathrm {e}^{ \frac{1}{3} v^3 - x v - \frac{1}{3} {u}^3 + y u}, \end{aligned}$$

(D.11)

where

$$\begin{aligned}&\gamma ^\infty \,{:=}\, \left\{ v \in {\mathbb {C}} \mid v = - s {{\,\mathrm {e}\,}}^{ \mathrm {i}\pi /3 } , s \in \left( - \infty , -2 w_{*} \delta \lambda t^{1/3} \right) \right\} \cup \left\{ v \in {\mathbb {C}} \mid v = w_{*} \delta \lambda t^{1/3} (1 - s \mathrm {i}), s \in \left[ -\sqrt{3} , \sqrt{3} \right] \right\} \\&\quad \cup \left\{ v \in {\mathbb {C}} \mid v = s {{\,\mathrm {e}\,}}^{- \mathrm {i}\pi / 3} , s \in \left( 2 w_{*} \delta \lambda t^{1/3} , \infty \right) \right\} \\&\sigma ^\infty {:=}\left\{ u \in {\mathbb {C}} \mid u = -s {{\,\mathrm {e}\,}}^{- \mathrm {i}2 \pi / 3} , s \in \left( -\infty , 0 \right] \right\} \cup \left\{ u \in {\mathbb {C}} \mid u = s {{\,\mathrm {e}\,}}^{ \mathrm {i}2 \pi /3 } , s \in \left( 0, \infty \right) \right\} . \end{aligned}$$

We choose such contours because they are convenient for the subsequent calculations.

Using similar arguments to those explained in [16], one can show that, in the long time limit, (D.10) behaves as the kernel of GUE Tracy–Widom distribution, i.e.,

$$\begin{aligned} \left| \int _{{\overline{\gamma }}^{\varDelta t^{1/3}} \times \sigma ^{\varDelta t^{1/3}}} \frac{\mathrm {d} v}{2 \pi \mathrm{i}} \frac{\mathrm {d} u}{2 \pi \mathrm{i}} \frac{1}{v - u} \mathrm {e}^{ \frac{1}{3} v^3 - (s_2 + \xi ) v - \frac{1}{3} {u}^3 + (s_2 + \zeta ) u} -A(s_2+\xi ,s_2+\zeta ) \right| \le {\mathcal {O}}( \mathrm {e}^{ - c_1 \varDelta ^3 t } ) \end{aligned}$$

(D.12)

holds for t large enough where $c_1$ is some positive constant. To see this, we first notice, from the form of the Airy kernel (D.11), that the left hand side of (D.12) is bounded above by

$$\begin{aligned}&\left| \int _{{\overline{\gamma }}^\infty } \frac{\mathrm {d} v}{2 \pi \mathrm{i}} \mathrm {e}^{ \frac{1}{3} v^3 - (s_2 + \xi ) v} \int _{ \sigma ^\infty - \sigma ^{\varDelta t^{1/3}} } \frac{\mathrm {d} u}{2 \pi \mathrm{i}} \mathrm {e}^{ - \frac{1}{3} {u}^3 + (s_2 + \zeta ) u} \frac{1}{v - u} \right| \nonumber \\&\quad + \left| \int _{{\overline{\gamma }}^\infty - {\overline{\gamma }}^{\varDelta t^{1/3}}} \frac{\mathrm {d} v}{2 \pi \mathrm{i}} \mathrm {e}^{ \frac{1}{3} v^3 - (s_2 + \xi ) v} \int _{ \sigma ^{\varDelta t^{1/3}}} \frac{\mathrm {d} u}{2 \pi \mathrm{i}} \mathrm {e}^{ - \frac{1}{3} {u}^3 + (s_2 + \zeta ) u} \frac{1}{v - u} \right| , \end{aligned}$$

(D.13)

where $\gamma _1 - \gamma _2$ denotes the concatenation of the contours $\gamma _1$ and $\overline{\gamma _2}$. The Airy contour $\sigma ^\infty$ differs from the contour $\sigma ^{\varDelta t^{1/3}}$ by

$$\begin{aligned} \sigma ^\infty - \sigma ^{\varDelta t^{1/3}} = \left\{ u\in {\mathbb {C}} \mid u= - s{{\,\mathrm {e}\,}}^{ - \mathrm {i}2 \pi /3} , s \in \left( - \infty , - \lambda \varDelta t^{1/3} \right) \right\} \cup \left\{ u\in {\mathbb {C}} \mid u= s {{\,\mathrm {e}\,}}^{\mathrm {i}2 \pi /3} , s\in \left( \lambda \varDelta t^{1/3}, \infty \right) \right\} .\nonumber \\ \end{aligned}$$

(D.14)

We can also find that, in the integrands of (D.13), $|v - u| \ge D \varDelta t^{1/3}$ holds with some positive constant D. Then, the first term of (D.13) is bounded above by

$$\begin{aligned} \begin{aligned}&\int _{ {\overline{\gamma }}^\infty } \frac{| \mathrm {d} v |}{2 \pi } \left| \mathrm {e}^{ \frac{1}{3} v^3 - (s_2 + \xi ) v} \right| \int _{ \sigma ^\infty - \sigma ^{\varDelta t^{1/3}} } \frac{| \mathrm {d} u |}{2 \pi } \left| \mathrm {e}^{ - \frac{1}{3} {u}^3 + (s_2 + \zeta ) u} \right| \left| \frac{1}{v - u} \right| \\&\quad \le \frac{c'}{\varDelta t^{1/3}} \int _{0}^{\infty } \mathrm{d}r {{\,\mathrm{e}\,}}^{- \frac{r^3}{3} - \frac{1}{2}(s_2 + \xi )r} \int _{\lambda \varDelta t^{1/3}}^{\infty } \mathrm{d}s {{\,\mathrm{e}\,}}^{- \frac{s^3}{3} - \frac{1}{2}(s_2 + \zeta )s}. \end{aligned} \end{aligned}$$

(D.15)

where $c'$ is some positive constant. ${{\,\mathrm{e}\,}}^{-r^3/3}$ and ${{\,\mathrm{e}\,}}^{-s^3/3}$ are dominant in the integrands and it follows from $\int _{\lambda \varDelta t^{1/3}}^{\infty } \mathrm{d}x {{\,\mathrm{e}\,}}^{-x^3/3} \le {{\,\mathrm{e}\,}}^{- c_1 \varDelta ^3 t}$ with some positive constant $c_1$ that the first term of (D.13) is bounded above by ${\mathcal {O}}({{\,\mathrm{e}\,}}^{- c_1 \varDelta ^3 t})$. For the second term of (D.13), ${\overline{\gamma }}^\infty$ also differs from ${\overline{\gamma }}^{\varDelta t^{1/3}}$ by (D.14) (${\overline{\gamma }}^\infty - {\overline{\gamma }}^{\varDelta t^{1/3}}=$ (D.14)). Then, the second term of (D.13) can be evaluated in the same fashion as the first term. Hence it turns out that (D.13) is bounded above by ${\mathcal {O}}({{\,\mathrm{e}\,}}^{- c_1 \varDelta ^3 t})$, and then we arrive at (D.12). Because $\varDelta =t^{-1/9}$, ${\mathcal {O}}( \mathrm {e}^{ - c_1 \varDelta ^3 t } )$ decays exponentially with respect to t. Hence we have shown that without error terms, the integral in $\varGamma '^{\varDelta } \times \varSigma ^{\varDelta }$ tends to an Airy kernel as t goes to ${\infty }$. Next we will prove the integral involving error terms vanishes.

(ii) Estimate of the error term ${\varvec{R}}$ in $\varvec{\varGamma '^{\varDelta } \times \varSigma ^{\varDelta }}$. Rewrite the target integral into the main contribution and the error terms:

$$\begin{aligned}&\lambda _c t^{1/3} \int _{{\varGamma '}^{\varDelta } \times {\varSigma }^{\varDelta }} \frac{\mathrm {d} z}{2 \pi \mathrm{i}} \frac{\mathrm {d} w}{2 \pi \mathrm{i}} \frac{1}{ w-z} \frac{ \mathrm {e}^{ f(z, t,\xi ) - f(w_{*} , t, \xi ) + g_{\phi }(z)} }{ \mathrm {e}^{ f(w, t,\zeta ) - f(w_{*} , t, \zeta ) - g_{\psi }(w)} } - (\mathrm{D.10}) \nonumber \\&\quad = \int _{\gamma ^{\varDelta t^{1/3}}} \frac{\mathrm {d} v}{2 \pi \mathrm{i}} \int _{ -\sigma ^{\varDelta t^{1/3}}} \frac{\mathrm {d} u}{2 \pi \mathrm{i}} \frac{1}{u-v} \left( R_vR_u + R_u {{\,\mathrm{e}\,}}^{\frac{1}{3} v^3 - (s_2 + \xi ) v} + R_v {{\,\mathrm{e}\,}}^{\frac{1}{3} u^3 - (s_2 + \xi ) u} \right) \nonumber \\&\quad = \int _{\gamma ^{\varDelta t^{1/3}}} \frac{\mathrm {d} v}{2 \pi \mathrm{i}} \int _{ -\sigma ^{\varDelta t^{1/3}}} \frac{\mathrm {d} u}{2 \pi \mathrm{i}} \frac{1}{u-v} {{\,\mathrm{e}\,}}^{h(v,\xi ) + h(u,\zeta )} \left[ {{\,\mathrm{e}\,}}^{ O(v) + O(u) } - 1 \right] , \end{aligned}$$

(D.16)

where $h(v,\xi )=\frac{1}{3} v^3 - (s_2 + \xi ) v$, and $O(v)={\mathcal {O}}( v^4 t^{-1/3} , v^2 t^{-1/6} , v^2 t^{-1/3} , v t^{-1/3} )$. Let us now show that the above vanishes in the long time limit. First recall (D.4a), we have the estimate $|u-z|^{-1} \le w_{*}^{-1} \lambda \delta ^{-1} t^{-1/3} = \lambda / w_{*}$, as $\delta = t^{-1/3}$. Then by the inequality $|{{\,\mathrm{e}\,}}^x - 1| \le |x| {{\,\mathrm{e}\,}}^{|x|}$, we can bound (D.16) by the product of two line integrals as

$$\begin{aligned} (\mathrm {D.16}) \le \frac{\lambda }{w_{*}} \int _{\gamma ^{\varDelta t^{1/3}}} \int _{ -\sigma ^{\varDelta t^{1/3}}} \left| \frac{\mathrm {d} v}{2 \pi \mathrm {i}} \right| \left| \frac{\mathrm {d} u}{2 \pi \mathrm {i}} \right| {{\,\mathrm {e}\,}}^{h(v,\xi ) + h(u,\zeta )} {{\,\mathrm {e}\,}}^{ O(v) + O(u) } \left| O(u) + O(v) \right| {:=} \, R.\nonumber \\ \end{aligned}$$

(D.17)

We now only need to show that R goes to zero as t tends to infinity. Let us now consider O(v). Clearly, ${\mathcal {O}}(v^4 t^{-1/3})$ is dominated by ${\mathcal {O}}(v^4 t^{-1/6})$ for large enough t. Likewise ${\mathcal {O}}(v^2 t^{-1/6})$ dominates ${\mathcal {O}}(v^2 t^{-1/3})$, and ${\mathcal {O}}( v t^{-1/6})$ dominates ${\mathcal {O}}(v t^{-1/3} )$ when t is large enough. As a result, ${\mathcal {O}}(v^4 t^{-1/3}, v^2 t^{-1/6}, v^2 t^{-1/3} , v t^{-1/3})$ $\ll t^{-1/6} {\mathcal {O}}(v^4 , v^2, v)$. Then let us consider ${{\,\mathrm{e}\,}}^{O(v)}$. We observe that along $\gamma ^{\varDelta t^{1/3}}$, $\mathrm{max} |v| \le \varDelta t^{1/3} \le t^{1/4}$ as we choose $\varDelta = t^{-1/9}$. When t is large enough, $|v^4 t^{-1/3}| \le |v|^4 t^{-1/3} \le |v|^{3} t^{-1/12} < |v|^3/6$, i.e. ${\mathcal {O}}(v^4 t^{-1/3}) < |v|^3/6$. Similarly, ${\mathcal {O}}(v^2 t^{-1/6}, v^2 t^{-1/3}) < |v|^2/6$ and ${\mathcal {O}}( v t^{-1/3}) < 1$ when t is large enough. Therefore, ${\mathcal {O}}(v^4 t^{-1/3} , v^2 t^{-1/6} , v^2 t^{-1/3} , v t^{-1/3})$ is bounded by $(|v|^3 + |v|^2)/6+1$. Consequently, R is now bounded as

$$\begin{aligned} R < t^{- 1/6} \frac{\lambda }{w_{*} } \int _{\gamma ^{\varDelta t^{1/3}}} \int _{ -\sigma ^{\varDelta t^{1/3}}} \left| \frac{\mathrm {d} v}{2 \pi \mathrm{i}} \right| \left| \frac{\mathrm {d} u}{2 \pi \mathrm{i}} \right| {{\,\mathrm{e}\,}}^{{\tilde{h}}(v,\xi ) + {\tilde{h}}(u,\zeta )} \left| {\tilde{O}}(v) + {\tilde{O}}(u) \right| , \end{aligned}$$

(D.18)

where ${\tilde{O}}(v) = {\mathcal {O}}(v^4 , v^2 , v)$, and ${\tilde{h}}(v,\xi ) = \frac{1}{3} v^3 - (s_2 +\xi ) v + \frac{1}{6} |v|^3 + \frac{1}{6} |v|^2 + 1$. One sees that the dependence of t only appears in the integration boundary $u = v = \lambda \varDelta t^{1/3} e^{\pm \mathrm{i}\pi /3}$. Since $\varDelta = t^{-1/9}$, i.e., $\varDelta t^{1/3} \gg 1$, the integrand is dominated by the term ${{\,\mathrm{e}\,}}^{\frac{1}{3} v^3 + \frac{1}{6} |v|^3}{{\,\mathrm{e}\,}}^{\frac{1}{3} u^3 + \frac{1}{6} |u|^3} = {{\,\mathrm{e}\,}}^{ - \frac{1}{3} \lambda ^3 \varDelta ^3 t }$ at the boundary. This implies that the integral (D.18), without the prefactor $t^{-1/6}$, is bounded in the limit $t\rightarrow \infty$. Namely, for large enough t, there exists a constant $c_3$ such that

$$\begin{aligned} R < c_3 t^{-1/6}. \end{aligned}$$

(iii) Estimate of $\varvec{(\varGamma ' \times \varSigma ) \backslash ( \varGamma '^{\varDelta } \times \varSigma ^{\varDelta }) }$. Next, we evaluate the rest of three terms in $(\varGamma ' \times \varSigma ) \backslash ( \varGamma '^{\varDelta } \times \varSigma ^{\varDelta })$: $(\varGamma ' \backslash \varGamma '^{\varDelta }) \times \varSigma ^{\varDelta }$, $\varGamma '^{\varDelta } \times (\varSigma \backslash \varSigma ^{\varDelta })$, $(\varGamma ' \backslash \varGamma '^{\varDelta }) \times (\varSigma \backslash \varSigma ^{\varDelta })$. Here we only show the proof sketch of $(\varGamma ' \backslash \varGamma '^{\varDelta }) \times \varSigma ^{\varDelta }$, while the proofs of the other two follow exactly in the same way.

Let us consider $(\varGamma ' \backslash \varGamma '^{\varDelta }) \times \varSigma ^{\varDelta }$. By Lemma 6.1, we know $\varGamma$ is the steepest descent contour along which the real part of $g_1(z)$ takes the maximum value at $z = w_{*}$ on $\varGamma$. For t large enough, the real part of $g_1(z)$ takes the maximum value at the points $z = w_{*} + \varDelta \mathrm {e}^{ \pm \mathrm{i}\pi /3 }$ on $\varGamma ' \backslash \varGamma '^{\varDelta }$. By the Taylor expansion (6.8), one can see that along $\varGamma ' \backslash \varGamma '^{\varDelta }$, we have

$$\begin{aligned} {{\,\mathrm {Re}\,}} (g_1(z)-g_1(w_{*})) \le -2 a_1 \varDelta ^3 + \varDelta ^4 {{\,\mathrm {Re}\,}} ({{\,\mathrm{e}\,}}^{\pm 4 \mathrm{i}\pi / 3} h_1(w_{*} + \varDelta \mathrm {e}^{ \pm i \pi /3 })) \le - a_1 \varDelta ^3, \end{aligned}$$

where $a_1>0$ is given in (6.10), and the second inequality holds when t is large enough. Since $\varDelta = t^{-1/9}$ and the function $h_1(z)$ is bounded along $\varGamma '$, then $\varDelta | h_1(z)| \le a_1$ for t large enough. Therefore, the z-integrand except $(w-z)^{-1}$ is bounded by

$$\begin{aligned} |{{\,\mathrm{e}\,}}^{f(z , t , \xi ) - f(w_{*} , t , \xi )}| \le {{\,\mathrm{e}\,}}^{ -a_1 \varDelta ^3 t +{\mathcal {O}}(t^{1/2})}. \end{aligned}$$

Note that the bound is uniform for any fixed L. Then by (D.4b), we have the bound of $|w-z|^{-1} \le 2 / \sqrt{3} \varDelta ^{-1} \ll 2 t^{1/3}$. Since $\varDelta = t^{-1/9}$, then ${{\,\mathrm{e}\,}}^{-\varDelta ^3 t }= {{\,\mathrm{e}\,}}^{-t^{2/3}}$. Obviously ${{\,\mathrm{e}\,}}^{-t^{2/3}} t^{1/3} {{\,\mathrm{e}\,}}^{{\mathcal {O}}(t^{1/2})} \rightarrow 0$ as $t \rightarrow {\infty }$. In conclusion, for large enough t,

$$\begin{aligned} |{{\,\mathrm{e}\,}}^{f(z , t , \xi ) - f(w_{*} , t , \xi )}(w-z)^{-1}| \le {{\,\mathrm{e}\,}}^{ -a_1 \varDelta ^3 t/2 }, \end{aligned}$$

(D.19)

where $a_1>0$.

Now we are left with the w-integrand (except for $|w-z|^{-1}$). Using the change of variable again $u = t^{1/3} \lambda (w - w_{*})$, we have

$$\begin{aligned} t^{1/3} \mathrm {e}^{ - f(w, t,\zeta ) + f(w_{*} , t, \zeta ) + g_{\psi }(w)} =\frac{\lambda ^{-1}}{w_{*} + c}\mathrm {e}^{ h(u, \zeta )} {{\,\mathrm{e}\,}}^{O(u) }, \end{aligned}$$

where $h(u, \zeta ) =\frac{1}{3} u^3 - (s_2 + \zeta ) u$ and $O(u) = {\mathcal {O}}( u^4 t^{-1/3} , u^2 t^{-1/6} , u^2 t^{-1/3} , u t^{-1/3} )$. By the same analysis as in part (ii), one can see that

$$\begin{aligned} \int _{\sigma ^{\varDelta t^{1/3} }} \left| \frac{\mathrm{d}u}{2 \pi \mathrm{i}} \right| \left| \frac{\lambda ^{-1}}{w_{*} + c}\mathrm {e}^{ h(u, \zeta )} {{\,\mathrm{e}\,}}^{O(u) } \right| \le c_4, \end{aligned}$$

(D.20)

where $c_4$ is some positive constant independent of t.

Combining (D.19) and (D.20), we obtain

$$\begin{aligned}&\left| \lambda _c t^{1/3} \int _{( \varGamma ' \backslash {\varGamma '}^{\varDelta } ) \times \varSigma ^{\varDelta } } \frac{\mathrm {d} z}{2 \pi \mathrm {i}} \frac{\mathrm {d} w}{2 \pi \mathrm {i}} \mathrm {e}^{ f(z, t, \xi ) - f(w_{*} , t, \xi ) + g_{\phi }(z) } \mathrm {e}^{ - f(w , t, \zeta ) + f(w_{*} , t, \zeta ) + g_{\psi }(w) } \frac{1}{w-z} \right| \le c_5 \mathrm {e}^{ - \frac{1}{2} a_1 \varDelta ^3 t }, \end{aligned}$$

(D.21)

where $c_5$ is some positive constant, and hence (D.21) goes to zero as t goes to infinity.

1.3 D.3. Proof of (6.18)

Since $g_1(z)$ is analytic along $\varGamma '$ and $\varSigma '$, there exists some positive constant $H_1$ such that $|h_1(z)| < H_1$ for z along $\varGamma '$ and $\varSigma '$. Similarly, we also have $|h_2(z)| < H_2$, $|{\bar{h}}_3(z)| < H_3$, $|h_{\phi }(z)| < H_{\phi }$ and $|h_{\psi }(z)| < H_{\psi }$. Consider $g_1(z)$ near $z = w_{*}$, we have

$$\begin{aligned}&\left| g_1(z)-g_1(w_{*})\right| = \left| 2 a_1 (z-w_{*})^3 + (z-w_{*})^4 h_1(z) \right| \nonumber \\&\quad \le 2 a_1 |z-w_{*}|^3 + H_1 |z-w_{*}|^4 \le 3a_1 |z-w_{*}|^3, \end{aligned}$$

(D.22)

where the last inequality holds under the restriction that $|z-w_{*}| \le H_1 / a_1$. Consequently, if $|z-w_{*}|<A$ where $A\,{:=}\,\max \{H_1 / a_1, H_2 / |b_2|, H_3 / |{\bar{b}}_3|, H_{\phi } / |b_{\phi }| , H_{\psi } / |b_{\psi }| \}$, the following inequalities hold:

$$\begin{aligned} \left| g_1(z)-g_1(w_{*})\right| \le&3a_1 |z-w_{*}|^3, \end{aligned}$$

(D.23a)

$$\begin{aligned} \left| g_2(z)-g_2(w_{*})\right| \le&2|b_2| |z-w_{*}|^2, \end{aligned}$$

(D.23b)

$$\begin{aligned} \left| {\bar{g}}_3(z)-{\bar{g}}_3(w_{*})\right| \le&2|{\bar{b}}_3| |z-w_{*}|, \end{aligned}$$

(D.23c)

$$\begin{aligned} \left| g_{\phi }(z)-g_{\phi }(w_{*})\right| \le&2|b_{\phi }| |z-w_{*}|, \end{aligned}$$

(D.23d)

$$\begin{aligned} \ \left| g_{\psi }(z)-g_{\psi }(w_{*})\right| \le&2|b_{\psi }| |z-w_{*}|. \end{aligned}$$

(D.23e)

Furthermore, if $|z-w_{*}| > 12 |b_2| t^{-1/2} / a_1$, then $2|b_2| |z-w_{*}|^2 t^{1/2} \le \frac{1}{6} a_1 |z-w_{*}|^3 t$. Therefore, if $|z-w_{*}| > \max \{ \frac{12 |b_2|}{a_1} t^{-1/2} ,\sqrt{\frac{12 |{\bar{b}}_3|}{a_1}} t^{-1/3}, \sqrt{\frac{12 |b_{\phi }|}{a_1}} t^{-1/2}, \sqrt{\frac{12 |b_{\psi }|}{a_1}} t^{-1/2}\}$,

$$\begin{aligned} \left| g_2(z)-g_2(w_{*})\right| t^{1/2} \le&2|b_2| |z-w_{*}|^2 t^{1/2} \le \frac{1}{6} a_1 |z-w_{*}|^3 t, \end{aligned}$$

(D.24a)

$$\begin{aligned} \left| {\bar{g}}_3(z)-{\bar{g}}_3(w_{*})\right| t^{1/3} \le&2|{\bar{b}}_3| |z-w_{*}| t^{1/3} \le \frac{1}{6} a_1 |z-w_{*}|^3 t, \end{aligned}$$

(D.24b)

$$\begin{aligned} \left| g_{\phi }(z)-g_{\phi }(w_{*})\right| \le&2|b_{\phi }| |z-w_{*}| \le \frac{1}{6} a_1 |z-w_{*}|^3 t, \end{aligned}$$

(D.24c)

$$\begin{aligned} \left| g_{\psi }(z)-g_{\psi }(w_{*})\right| \le&2|b_{\psi }| |z-w_{*}| \le \frac{1}{6} a_1 |z-w_{*}|^3 t. \end{aligned}$$

(D.24d)

When t is large enough, $t^{-1/3} \gg t^{-1/2}$, i.e., we only require $|z-w_{*}| > \sqrt{\frac{12 |{\bar{b}}_3|}{a_1}} t^{-1/3} \,{:=} c_1 t^{-1/3}$, then the above inequalities hold. Using the above inequalities, we have the following estimations.

$\bullet$ Contribution from $\varGamma _\mathrm{vert}$: (6.18a) For $z \in \varGamma _\mathrm{vert}$, $|z-w_{*}| \le 2 w_{*} \delta \le \varDelta = t^{-1/9}$. Since A is a fixed constant, one can choose t large enough such that $|z-w_{*}| \le \varDelta < A$. Moreover, we restrict $\delta > c_1 t^{-1/3} / w_{*}$ so that $|z - w_{*}| \ge w_{*} \delta > c_1 t^{-1/3}$. Thus, from (D.23) and (D.24) we have,

$$\begin{aligned} \left| {{\,\mathrm{e}\,}}^{{\bar{f}}(z,t) - {\bar{f}}(w_{*}, t) + g_{\phi }(z) } \right| \le {{\,\mathrm{e}\,}}^{(3a_1 +a_1/2)|z-w_{*}|^3 t} {{\,\mathrm{e}\,}}^{g_{\phi }(w_{*})} \le {{\,\mathrm{e}\,}}^{28 a_1 (w_{*} \delta )^3t} {{\,\mathrm{e}\,}}^{g_{\phi }(w_{*})}. \end{aligned}$$

$\bullet$ Contribution from $\varGamma '^{\varDelta } \backslash \varGamma _\mathrm{vert}$: (6.18b)

Note that we can also bound this part by ${{\,\mathrm{e}\,}}^{28 a_1 (w_{*} \delta )^3}$ using the same method as above. However, since the contour now is along the direction ${{\,\mathrm{e}\,}}^{\pm \mathrm{i}\pi /3}$, we could refine the bound of $g_1(z)$. We parameterise z along $\varGamma '^{\varDelta } \backslash \varGamma _\mathrm{vert}$ by $z=w_{*} + v {{\,\mathrm{e}\,}}^{\pm \mathrm{i}\pi /3}$ where $v \in (2 w_{*} \delta , \varDelta ]$. Therefore,

$$\begin{aligned} \left| {{\,\mathrm{e}\,}}^{g_1(z) - g_1(w_{*})} \right| = \left| {{\,\mathrm{e}\,}}^{-2a_1 v^3 + h_1(z) v^4 {{\,\mathrm{e}\,}}^{\pm 4 \mathrm{i}\pi /3}}\right| \le {{\,\mathrm{e}\,}}^{-a_1 v^3}, \end{aligned}$$

when $|z-w_{*}| < A$. In fact, we have $|z-w_{*}| \le \varDelta = t^{-1/9} < A$ when t is large enough. Then we repeat the above estimate (D.24) for $g_2(z) t^{1/2} + {\bar{g}}_3(z) t^{1/3} + g_{\phi }(z)$ and we obtain for $\varGamma '^{\varDelta } \backslash \varGamma _\mathrm{vert}$,

$$\begin{aligned} \left| {{\,\mathrm{e}\,}}^{{\bar{f}}(z,t) - {\bar{f}}(w_{*}, t) + g_{\phi }(z) } \right| \le {{\,\mathrm{e}\,}}^{(-a_1 +a_1/2)|z-w_{*}|^3 t} {{\,\mathrm{e}\,}}^{g_{\phi }(w_{*})} \le {{\,\mathrm{e}\,}}^{-a_1 v^3 t /2} {{\,\mathrm{e}\,}}^{g_{\phi }(w_{*})}. \end{aligned}$$

(D.25)

In fact, when $z \in \varGamma '^{\varDelta } \backslash \varGamma _\mathrm{vert}$, we have $|z-w_{*}| \ge 2 w_{*} \delta> w_{*} \delta > c_1 t^{-1/3}$, which satisfies the restriction of (D.24).

$\bullet$ Contribution from $\varGamma ' \backslash \varGamma '^{\varDelta }$: (6.18c) This estimate is exactly the same as part (iii) in the Appendix D.2. Recall that we have proved in Appendix D.2, along $\varGamma ' \backslash \varGamma '^{\varDelta }$,

$$\begin{aligned} \left| {{\,\mathrm{e}\,}}^{{\bar{f}}(z,t) - {\bar{f}}(w_{*},t) + g_{\phi }(z)} \right| \le {{\,\mathrm{e}\,}}^{- a_1 \varDelta ^3 t + {\mathcal {O}}(t^{1/2})} \le {{\,\mathrm{e}\,}}^{- a_1 \varDelta ^3 t /2}, \end{aligned}$$

(D.26)

where ${{\,\mathrm{e}\,}}^{{\mathcal {O}}(t^{1/2})} < {{\,\mathrm{e}\,}}^{ a_1 \varDelta ^3 t /2}$ for t large enough.

1.4 D.4. Proof of (6.20)

Along $\varGamma '$, the minimum value of $z+c$ is taken at the point $z = w_{*} ( 1+ \delta )$ (See Fig. 6). Therefore,

$$\begin{aligned} \left| \frac{w_{*} + c}{z + c}\right| \le \left| \frac{w_{*} + c}{w_{*}(1 + \delta ) + c}\right| = \left| \frac{1}{ 1 + \frac{w_{*}\delta }{w_{*} + c}}\right| \le {{\,\mathrm{e}\,}}^{-\frac{1}{2}\frac{w_{*}}{w_{*} + c} \delta }, \end{aligned}$$

where the last inequality follows by $1/(1+x) < {{\,\mathrm{e}\,}}^{-x/2}$ when $0<x<2$. This restriction is satisfied if we suppose $\delta < 1$. Similarly, we consider the term involving $\zeta$. Along $\varSigma '$, when $c > 1$, $|w + c|$ takes the maximum value at $w = w_{*} - w_{*} \delta (1 \pm \sqrt{3} \mathrm{i})$ and, as easily seen from Fig. 6, it is given by

$$\begin{aligned} \left| w + c\right| \le \left| \sqrt{(w_{*}+c)^2 + 4 \delta ^2 w_{*}^2 - 2 \delta w_{*}(w_{*}+c)}\right| = \sqrt{w_{*}^2(4\delta ^2 - 2\delta +1) +2w_{*}c(1-\delta ) +c^2}. \end{aligned}$$

This can be seen for the following reason. Actually, for $\delta \in (0, 1/(1-\rho ))$, if c satisfied $0< c < \{ \delta (2 \delta - 1) {( 1- \rho )}^2 - 4(1 + \rho ) \} / \{ 2 \delta (1 - \rho ) - 8 \} =: C(\rho , \delta )$, $|w+c|$ takes the maximum value $(3 + \rho )/2 - c$ at $w = -(3 + \rho )/2$. For fixed $\rho \in (0, 1)$, $C(\rho , \delta )$ is an increasing function of $\delta$ in a region $\delta \in (0,1/4)$ and hence $\max _{\delta \in (0, 1/4)}{C(\rho ,\delta )} = C(\rho , 1/4) = \{ {(15 + \rho )}^2 - 192 \} / \{ 4 (15 + \rho ) \}$. Since $C(\rho ,1/4)$ is also an increasing function of $\rho$ in the region $\rho \in (0,1)$, it turns out that $\max _{\rho \in (0,1)}{C(\rho , 1/4)} = C(1,1/4) = 1$. Therefore, when $c>1$, $|w+c|$ always takes the maximum value at $w = w_{*} - w_{*} \delta (1 \pm \sqrt{3} \mathrm{i})$ on $\varSigma '$ for any $\rho \in (0,1)$ and $\delta \in (0,1/4)$. Then we bound $4\delta ^2-2\delta +1$ by $1-\delta$, which is valid when $0< \delta < 1/4$. It follows that

$$\begin{aligned} |w+c| \le&\sqrt{(w_{*}^2+2w_{*}c+c^2)(1-\delta )+\delta c^2} \\ =&\sqrt{(w_{*} + c)^2 -\delta [(w_{*} + c)^2-c^2]} \\ \le&\sqrt{(w_{*} + c)^2 -\delta w_{*}(w_{*} + c)}. \end{aligned}$$

Therefore,

$$\begin{aligned} \left| \frac{w + c}{w_{*} + c}\right| \le \sqrt{1- \frac{\delta w_{*}}{w_{*} +c}} \le {{\,\mathrm{e}\,}}^{-\frac{1}{2}\frac{w_{*}}{w_{*} + c}\delta }, \end{aligned}$$

where the last line follows by $\sqrt{1-x} \le {{\,\mathrm{e}\,}}^{-x/2}$ when $x>0$.

E. The proofs of Lemmas and Proposition in Sect. 7

In this appendix we prove Lemmas 7.1, 7.5 and Proposition 7.2 in Sect. 7.

1.1 E.1. Proof of Lemma 7.1, on commutativity between the integrals and the sums

In the following, we will prove that the determinant of the matrix $K^c(x_i , x_j , \mathbf {z})$ has the dominant series as

$$\begin{aligned} \left| \det \left[ K^c(x_i , x_j, \mathbf {z}) \right] _{1 \le i , j \le k} \right| \le M \prod _{i=1}^{k}{ \mathrm {e}^{ - \frac{w_{*}}{4(w_{*} + c)} \delta x_i } }, \end{aligned}$$

(E.1)

where M is some constant depending only on $n, m \in {\mathbb {N}}$, $t \in (0,\infty )$ and $k \in [1,m]$. Since the series in the right hand side of (E.1) is a geometric series, we can perform the infinite sums from $x_i = 1$ to $\infty$ for all $i \in [1,k]$ and it turns out to be finite:

$$\begin{aligned} \sum _{x_1 = 1}^{\infty }{ \sum _{x_2 = 1}^{ \infty }{ \cdots \sum _{x_k = 1}^{\infty }{ M \prod _{i=1}^{k}{ {{\,\mathrm{e}\,}}^{ - \frac{w_{*}}{4(w_{*} + c)} \delta x_i } } } } } = M {\left[ \frac{1}{ {{\,\mathrm{e}\,}}^{ (w_{*} \delta )/[4(w_{*} + c)] } - 1 } \right] }^k < \infty . \end{aligned}$$

From the Weierstrass M-test, which is specialisation of the Lebesgue’s dominated convergence theorem, it turns out that we can commute the sums over $x_i \in {\mathbb {N}}$ for all $i \in [1,k]$ and the integrals with respect to $z_j$ for all $j \in [1,n-1]$.

In order to show (E.1), we will prove the inequality

$$\begin{aligned} \left| {(w_{*} + c)}^{x - y} K^c(x , y , \mathbf {z}) \right| \le C {{\,\mathrm{e}\,}}^{ - \frac{w_{*}}{4(w_{*} + c)} \delta (x + y) }, \end{aligned}$$

(E.2)

where C is some constant depending on $n,m \in {\mathbb {N}}$ and $t \in (0,\infty )$. From the formula of $K^c(x,y,\mathbf {z})$, which is given by (7.1), we can find the following bound:

$$\begin{aligned} \begin{aligned}&\left| {(w_{*} + c)}^{x - y} K^c(x,y,\mathbf {z}) \right| \le \oint _{1} \frac{ \left| \mathrm {d}z \right| }{2 \pi } \left| \frac{z + \rho '}{z + 1} \right| {{\,\mathrm {e}\,}}^{ - \frac{t}{2} {{\,\mathrm {Re}\,}}(z) } {\left| \frac{z}{z - 1} \right| }^m {\left| \frac{1 + z}{z} \right| }^n {\left| \frac{w_{*} + c}{z + c} \right| }^x \prod _{j=1}^{n-1}{ \left| \frac{1 + z_j z}{1 + z} \right| } \\ {}&\quad \quad \times \oint _{0 , - \rho ' , \{ - z_j^{-1} \}_{j=1}^{n-1} } \frac{ \left| \mathrm {d}w \right| }{2 \pi } \left| \frac{w + 1}{(w + \rho ')(w + c )} \right| {{\,\mathrm {e}\,}}^{ \frac{t}{2} {{\,\mathrm {Re}\,}}(w) } {\left| \frac{w - 1}{w} \right| }^m {\left| \frac{w}{1 + w} \right| }^n {\left| \frac{w_{*} + c}{w + c} \right| }^{-y} \prod _{j=1}^{n-1}{ \left| \frac{1 + w}{1 + z_j w} \right| } \frac{1}{ \left| w-z \right| }. \end{aligned} \end{aligned}$$

Let us choose the contours of z-integral and w-integral as $\varGamma '$ and $\varSigma '$ defined in the proof of Proposition 6.3, respectively. This implies that, for any z and w on the contours, the parts depending on $x \in {\mathbb {N}}$ and $y \in {\mathbb {N}}$ are bounded above as

$$\begin{aligned} {\left| \frac{w_{*} + c}{z + c} \right| }^x&\le {{\,\mathrm{e}\,}}^{ - \frac{w_{*}}{4(w_{*} + c)} \delta x } , \\ {\left| \frac{w_{*} + c}{w + c} \right| }^{-y}&\le {{\,\mathrm{e}\,}}^{ - \frac{w_{*}}{4(w_{*} + c)} \delta y } . \end{aligned}$$

Since the contour of z-integral does not pass the points at $z=-c$ and $|z|, |w| < \infty$ and $|w - z| > 0$ holds for any $(z,w) \in \varGamma ' \times \varSigma '$, the other parts of integrands are bounded by some constant depending on $n, m \in {\mathbb {N}}$ and $t \in (0,\infty )$. In addition, considering that the lengths of the contours are finite, it turns out that (E.2) holds.

Since a determinant of a matrix whose (i, j) entry is given by $a_{i,j}$ and a matrix whose (i, j) entry is given by $b^{x_i - x_j} a_{i,j}$ are equivalent, we obtain the equality

$$\begin{aligned} \left| \det \left[ K^c(x_i , x_j , \mathbf {z}) \right] _{1 \le i , j \le k } \right| = \left| \det \left[ {(w_{*} + c)}^{x_i - x_j} K^c(x_i , x_j , \mathbf {z}) \right] _{1 \le i , j \le k } \right| . \end{aligned}$$

(E.3)

Application of the Hadamard’s inequality to the right hand side of (E.3) yields

$$\begin{aligned} \left| \det \left[ {(w_{*} + c)}^{x_i - x_j} K^c(x_i , x_j , \mathbf {z}) \right] _{1 \le i , j \le k } \right| \le \prod _{i=1}^{k}{ \sqrt{ \sum _{j=1}^{k}{ {\left| {(w_{*} + c)}^{x_i - x_j} K^c(x_i , x_j , \mathbf {z}) \right| }^2 } } }. \end{aligned}$$

(E.4)

Finally, using the inequality (E.2), we get

$$\begin{aligned} \begin{aligned} \sqrt{ \sum _{j=1}^{k}{ {\left| {(w_{*} + c)}^{x_i - x_j} K^c(x_i , x_j , \mathbf {z}) \right| }^2 } } \le C \sqrt{ \sum _{j=1}^{k}{ \mathrm {e}^{ - \frac{w_{*}}{2(w_{*} + c)} \delta (x_i + x_j) } } } = C \mathrm {e}^{ - \frac{w_{*}}{4(w_{*} + c)} \delta x_i } \sqrt{ \sum _{j=1}^{k}{ \mathrm {e}^{ - \frac{w_{*}}{2(w_{*} + c)} \delta x_j } } } . \end{aligned} \end{aligned}$$

(E.5)

Since $\frac{ w_{*} }{ 2( w_{*} + c ) } \delta > 0$ and $x_j \ge 1$ hold for any $j \in [1,k]$, it turns out that $\sum _{j = 1}^{k}{ {{\,\mathrm{e}\,}}^{- \frac{ w_{*} }{ 2( w_{*} + c ) } \delta x_j } } < k$ holds. Therefore, it follows from (E.5) that we obtain

$$\begin{aligned} \prod _{i=1}^{k}{ \sqrt{ \sum _{j=1}^{k}{ {\left| {(w_{*} + c)}^{x_i - x_j} K^c(x_i , x_j , \mathbf {z}) \right| }^2 } } } \le { k }^{\frac{k}{2}} C^k \prod _{i=1}^{k}{ \mathrm {e}^{ - \frac{w_{*}}{4(w_{*} + c)} \delta x_i } }. \end{aligned}$$

(E.6)

It follows from (E.3), (E.4) and (E.6) that the inequality (E.1) holds with a constant M set as $M = { k }^{\frac{k}{2}} C^k$.

1.2 E.2. Proof of the regularity of the kernel, Lemma 7.5

A complex function $f : {\mathbb {C}}^{n-1} \rightarrow {\mathbb {C}}$ is complex analytic if and only if it is holomorphic. In the following, we will show the kernel is holomorphic by proving it is complex analytic in ${S(1,r)}^{n-1}$ for $r \in (0, 1)$. In other words, we will show that, for any $\mathbf {b} = (b_1,\ldots ,b_{n-1}) \in {S(1,r)}^{n-1}$, there exists an open polydisc $S(b_1, r_1) \times \cdots \times S(b_{n-1}, r_{n-1}) \subset {S(1,r)}^{n-1}$ such that the kernel $K^c(x,y,\mathbf {z})$ has a power series expansion

$$\begin{aligned} K^c(x,y,\mathbf {z}) = \sum _{k_1,\ldots ,k_{n-1} = 0}^{\infty }{ c_{k_1,\ldots ,k_{n-1}} \prod _{j=1}^{n-1}{{(z_j - b_j)}^{k_j}} } , \end{aligned}$$

which converges for any $\mathbf {z} \in S(b_1, r_1) \times \cdots \times S(b_{n-1}, r_{n-1})$, where coefficients $c_{k_1,\ldots ,k_{n-1}}$ are independent of $\mathbf {z}$. In this proof, we start from the form

$$\begin{aligned} K^c(x,y,\mathbf {z}) = \oint _{1} \frac{\mathrm{d}z}{2 \pi \mathrm{i}} F^c(z,x) \prod _{j=1}^{n-1}{ \frac{1 + z_j z}{1 + z} } \oint _{0 , - \rho ' , \{ - z_j^{-1} \}_{j=1}^{n-1} } \frac{\mathrm{d}w}{2 \pi \mathrm{i}} G^c(w,y) \prod _{j=1}^{n-1}{\frac{1 + w}{1 + z_j w}} \frac{1}{w-z}, \nonumber \\ \end{aligned}$$

(E.7)

where the functions $F^c(z,x)$ and $G^c(w,y)$ are defined in (7.5).

As stated in the proof of Lemma 7.3, we can choose the contour such that

$$\begin{aligned} \left| \frac{1 + w}{w} \right|> r > \left| b_j - 1 \right| \end{aligned}$$

(E.8)

hold for any $\mathbf {b} = (b_1, \ldots , b_{n-1}) \in {S(1,r)}^{n-1}$ and w on the contour. Using the reverse triangle inequality and (E.8), we obtain

$$\begin{aligned} \left| \frac{1+w}{w} + (b_j - 1) \right| \ge \left| \frac{1 + w}{w} \right| - |b_j - 1|> r - |b_j - 1| > 0. \end{aligned}$$

(E.9)

Suppose that the open polydisc $S(b_1, r_1) \times \cdots \times S(b_{n-1}, r_{n-1})$ is included in ${S(1,r)}^{n-1}$, $r_j$ and $b_j$ satisfy

$$\begin{aligned} r_j < r - |b_j - 1| \end{aligned}$$

(E.10)

for any $j \in [1,n-1]$. It follows from (E.9) and (E.10) that the equality and inequalities

$$\begin{aligned} \left| \frac{ w(b_j - z_j)}{1 + b_j w} \right| = \left| \frac{b_j - z_j}{ (1 + w)/w + (b_j - 1) } \right| \le \frac{|z_j - b_j|}{ \left| ( 1 + w)/w \right| - \left| b_j - 1 \right| }< \frac{r_j}{r - | b_j - 1 |} < 1 \end{aligned}$$

(E.11)

hold for any $\mathbf {z} \in S(b_1, r_1) \times \cdots \times S(b_{n-1}, r_{n-1}) \subset {S(1,r)}^{n-1}$ and any w on the contour.

The right hand side of (E.7) depends on $z_j$ via the factor $(1+z_j z)/(1 + z_j w)$, and (E.11) guarantees that a Taylor series expansion of it around $z_j = b_j$:

$$\begin{aligned} \frac{1+z_j z}{1 + z_j w} = \sum _{k=0}^{\infty }{ d_k(z,w,b_j) {(z_j - b_j)}^k } \end{aligned}$$

(E.12)

converges, where

$$\begin{aligned} d_k(z,w,b_j) = {\left\{ \begin{array}{ll} \frac{1+b_j z}{1+b_j w} &{} \mathrm {for}\, k=0 \\ \frac{w-z}{w(1 + b_j w)} {\left( \frac{-w}{1+b_j w} \right) }^k &{} \mathrm {for}\, k \ge 1 . \end{array}\right. } \end{aligned}$$

Since a Taylor series (E.12) converges uniformly for any z and w on the contours, we can move the sum over $k \in {\mathbb {N}} \cup \{ 0 \}$ to the outside of z, w-integrals. Therefore, the kernel can be expanded as

$$\begin{aligned} K^c(x,y,\mathbf {z}) = \sum _{k_1,\ldots , k_{n-1} = 0}^{\infty } c_{k_1,\ldots ,k_{n-1}} \prod _{j=1}^{n-1} {(z_j - b_j)}^{k_j}, \end{aligned}$$

(E.13)

where

$$\begin{aligned} c_{k_1,\ldots ,k_{n-1}} = \oint _1 \frac{\mathrm {d}z}{2 \pi \mathrm {i}} \oint _{0, - \rho ', {\{ - b^{-1}_j \}}_{j=1}^{n-1} } \frac{ \mathrm {d}w }{2 \pi \mathrm {i}} \frac{F^c(z,x) G^c(w,y) }{w - z} {\left( \frac{1+w}{1+z} \right) }^{n-1} \prod _{j=1}^{n-1}{d_{k_j}(z,w,b_j)}. \end{aligned}$$

To confirm convergence of a power series in the right hand side of (E.13), we will evaluate an upper bound of $|c_{k_1,\ldots ,k_{n-1}}|$. Using the Cauchy’s integral theorem, we can choose the contours of z, w-integrals such that the absolute value of integrand except for $\prod _{j=1}^{n-1}{d_{k_j}(z,w,b_j)}$ is finite for any $n,m \in {\mathbb {N}}$ and $t \in (0,\infty )$, i.e., the contour of z-integral does not pass the points at $z=-c$ and the lengths of both contours are finite. From (E.9) and (E.10), we also have

$$\begin{aligned} \left| \frac{w}{1 + b_j w} \right| < \frac{1}{r - |b_j - 1|} \le \frac{1}{r_j}, \end{aligned}$$

hence we can find an upper bound of $|d_k(z,w,b_j)|$ as

$$\begin{aligned} \left| d_0(z,w,b_j) \right|&= \left| \frac{1+b_j z}{1 + b_j w} \right| < C_1 , \end{aligned}$$

(E.14a)

$$\begin{aligned} \left| d_k(z,w,b_j) \right|&= \left| \frac{w-z}{w(1 + b_j w)} {\left( \frac{-w}{1+b_j w} \right) }^k \right| < \frac{C_2}{r_j^{k}} , \end{aligned}$$

(E.14b)

where $C_1$ and $C_2$ are some positive constants. Utilising (E.14), it turns out that $| c_{k_1,\ldots ,k_{n-1}} |$ is bounded as

$$\begin{aligned} \begin{aligned} \left| c_{k_1,\ldots ,k_{n-1}} \right|&\le \oint _1 \frac{\left| \mathrm{d}z \right| }{2 \pi } \oint _{0, - \rho ', {\{ - b^{-1}_j \}}_{j=1}^{n-1}} \frac{ \left| \mathrm{d}w \right| }{2 \pi } \left| \frac{F^c(z,x) G^c(w,y) }{w-z} \right| {\left| \frac{1 + w}{1 + z} \right| }^{n-1} \prod _{j=1}^{n-1}{ \left| d_{k_j}(z,w,b_j) \right| } \\&< \frac{C}{r_1^{k_1} \cdots r_{n-1}^{k_{n-1}}} , \end{aligned} \end{aligned}$$

where C is some constant depending on $n, m \in {\mathbb {N}}$ and $t \in (0, \infty )$. This implies that a power series in the right hand side of (E.13) converges.

1.3 E.3. Proof of extension of decoupling to determinant, Proposition 7.2

In order to simplify the notations, we introduce the abbreviations

$$\begin{aligned} K_{ij}(\mathbf {z})&= K^c(x_i , x_j, \mathbf {z}),&{\bar{K}}_{ij}&= {\bar{K}}^c(x_i , x_j),&A_i(\mathbf {z})&= A^c(x_i, \mathbf {z}), \\ {(A_p)}_i&= A^c_p(x_i),&B_i&= B^c(x_i), \end{aligned}$$

and the vectors

$$\begin{aligned} \mathbf {K}^{(k)}_{i}(\mathbf {z})&= \left( \begin{array}{cccc} K_{i1}(\mathbf {z}),&K_{i2}(\mathbf {z}),&\cdots ,&K_{ik}(\mathbf {z}) \end{array} \right) ,\\ \mathbf {\bar{K}}^{(k)}_{i}&= \left( \begin{array}{cccc} {\bar{K}}_{i1},&{\bar{K}}_{i2},&\cdots ,&{\bar{K}}_{ik} \end{array} \right) , \\ \mathbf {B}^{(k)}&= \left( \begin{array}{cccc} B_{1},&B_{2},&\cdots ,&B_{k} \end{array} \right) . \end{aligned}$$

In the following, using k-dimensional row vectors $\mathbf {v}_i = (v_{i1}, \cdots , v_{ik})$, we represent the determinant of a $k \times k$ matrix $\det [v_{ij}]_{1 \le i , j \le k}$ as

$$\begin{aligned} \left| \left[ \begin{array}{c} \mathbf {v}_1 \\ \mathbf {v}_2 \\ \vdots \\ \mathbf {v}_k \end{array} \right] \right| = \left| \begin{array}{cccc} v_{11} &{} v_{12} &{} \cdots &{} v_{1k} \\ v_{21} &{} v_{22} &{} \cdots &{} v_{2k} \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ v_{k1} &{} v_{k2} &{} \cdots &{} v_{kk} \end{array} \right| . \end{aligned}$$

Using these notations, the claim of Proposition 7.2 can be represented as

$$\begin{aligned}&\oint _{1} \prod _{j=1}^{n-1}{ \frac{\mathrm{d}z_j}{2 \pi \mathrm{i}} } \frac{ \prod _{1 \le i< j \le n-1}{(z_j - z_i)} }{ \prod _{j=1}^{n-1}{{(z_j -1)}^{j+1}} } \prod _{j=1}^{n-1}{h(z_j)} \left| \left[ \begin{array}{c} \mathbf {K}^{(k)}_{1}(\mathbf {z}) \\ \vdots \\ \mathbf {K}^{(k)}_{k}(\mathbf {z}) \end{array} \right] \right| \nonumber \\= & {} \oint _{1} \prod _{j=1}^{n-1}{ \frac{\mathrm{d}z_j}{2 \pi \mathrm{i}} } \frac{ \prod _{1 \le i < j \le n-1}{(z_j - z_i)} }{ \prod _{j=1}^{n-1}{{(z_j -1)}^{j+1}} } \prod _{j=1}^{n-1}{h(z_j)} \left| \left[ \begin{array}{c} \mathbf {\bar{K}}^{(k)}_{1} - \sum _{p=1}^{n-1}{ \prod _{q=1}^{p}{ (z_q - 1) {(A_p)}_1 } } \mathbf {B}^{(k)} \\ \vdots \\ \mathbf {\bar{K}}^{(k)}_{k} - \sum _{p=1}^{n-1}{ \prod _{q=1}^{p}{ (z_q - 1) {(A_p)}_k } } \mathbf {B}^{(k)} \end{array} \right] \right| .\qquad \qquad \end{aligned}$$

(E.15)

Since $K_{ij}(\mathbf {z})$ is invariant under exchanges of any two $\mathbf {z}$-variables and is a holomorphic function in the vicinity of $z_j = 1$ for any $j \in [1,n-1]$ as shown in Lemma 7.5, we can apply Lemma 7.4 to one row of $\det [ K_{ij}(\mathbf {z}) ]_{1 \le i , j \le k}$. Applying Lemma 7.3 to the first row and using the multilinearity of determinant, the left hand side of (E.15) is divided into two terms as

$$\begin{aligned}&\oint _{1} \prod _{j=1}^{n-1}{ \frac{\mathrm{d}z_j}{2 \pi \mathrm{i}} } \frac{ \prod _{1 \le i< j \le n-1}{(z_j - z_i)} }{ \prod _{j=1}^{n-1}{{(z_j -1)}^{j+1}} } \prod _{j=1}^{n-1}{h(z_j)} \left| \left[ \begin{array}{c} \mathbf {K}^{(k)}_{1}(\mathbf {z}) \\ \vdots \\ \mathbf {K}^{(k)}_{k}(\mathbf {z}) \end{array} \right] \right| \nonumber \\ =&\oint _{1} \prod _{j=1}^{n-1}{ \frac{\mathrm{d}z_j}{2 \pi \mathrm{i}} } \frac{ \prod _{1 \le i< j \le n-1}{(z_j - z_i)} }{ \prod _{j=1}^{n-1}{{(z_j -1)}^{j+1}} } \prod _{j=1}^{n-1}{h(z_j)} \left| \left[ \begin{array}{c} \mathbf {\bar{K}}^{(k)}_{1} - (z_1 - 1) A_1(\mathbf {z}) \mathbf {B}^{(k)}\\ \mathbf {K}^{(k)}_{2}(\mathbf {z}) \\ \vdots \\ \mathbf {K}^{(k)}_{k}(\mathbf {z}) \end{array} \right] \right| \nonumber \\ =&\oint _{1} \prod _{j=1}^{n-1}{ \frac{\mathrm{d}z_j}{2 \pi \mathrm{i}} } \frac{ \prod _{1 \le i < j \le n-1}{(z_j - z_i)} }{ \prod _{j=2}^{n-1}{{(z_j -1)}^{j+1}} } \prod _{j=1}^{n-1}{h(z_j)} \left\{ \frac{1}{{(z_1 - 1)}^2} \left| \left[ \begin{array}{c} \mathbf {\bar{K}}^{(k)}_{1} \\ \mathbf {K}^{(k)}_{2}(\mathbf {z}) \\ \vdots \\ \mathbf {K}^{(k)}_{k}(\mathbf {z}) \end{array} \right] \right| - \frac{1}{(z_1 - 1)} \left| \left[ \begin{array}{c} A_1(\mathbf {z}) \mathbf {B}^{(k)} \\ \mathbf {K}^{(k)}_{2}(\mathbf {z}) \\ \vdots \\ \mathbf {K}^{(k)}_{k}(\mathbf {z}) \end{array} \right] \right| \right\} . \end{aligned}$$

(E.16)

For the same reason, we can apply Lemma 7.3 to the second row of the first term of the right hand side of (E.16) and divide the term into further two terms. Carrying out the same calculation from the second row to the $k^{\mathrm {th}}$ row in order, the determinant is divided into $k+1$ terms as

$$\begin{aligned} \begin{aligned}&\oint _{1} \prod _{j=1}^{n-1}{ \frac{\mathrm{d}z_j}{2 \pi \mathrm{i}} } \frac{ \prod _{1 \le i< j \le n-1}{(z_j - z_i)} }{ \prod _{j=1}^{n-1}{{(z_j -1)}^{j+1}} } \prod _{j=1}^{n-1}{h(z_j)} \left| \left[ \begin{array}{c} \mathbf {K}^{(k)}_{1}(\mathbf {z}) \\ \vdots \\ \mathbf {K}^{(k)}_{k}(\mathbf {z}) \end{array} \right] \right| \\&\quad = \oint _{1} \prod _{j=1}^{n-1}{ \frac{\mathrm{d}z_j}{2 \pi \mathrm{i}} } \frac{ \prod _{1 \le i < j \le n-1}{(z_j - z_i)} }{ \prod _{j=2}^{n-1}{{(z_j -1)}^{j+1}} } \prod _{j=1}^{n-1}{h(z_j)} \left\{ \frac{1}{{(z_1 - 1)}^2} \left| \left[ \begin{array}{c} \mathbf {\bar{K}}^{(k)}_{1} \\ \vdots \\ \vdots \\ \vdots \\ \vdots \\ \mathbf {\bar{K}}^{(k)}_{k} \end{array} \right] \right| - \frac{1}{(z_1 - 1)} \sum _{\ell =1}^{k}{ \left| \left[ \begin{array}{c} \mathbf {\bar{K}}^{(k)}_{1} \\ \vdots \\ \mathbf {\bar{K}}^{(k)}_{\ell -1} \\ A_\ell (\mathbf {z}) \mathbf {B}^{(k)} \\ \mathbf {K}^{(k)}_{\ell +1}(\mathbf {z}) \\ \vdots \\ \mathbf {K}^{(k)}_{k}(\mathbf {z}) \end{array} \right] \right| } \right\} . \end{aligned} \end{aligned}$$

(E.17)

For convenience, we define $K'^c(x,y, \mathbf {z})$ and $A'^c(x, \mathbf {z})$ as the function which are obtained by substituting 1 into $z_1$ of $K^c(x,y, \mathbf {z})$ and $z_2$ of $A(x, \mathbf {z})$, respectively. In other words, $K'^c(x,y, \mathbf {z})$ and $A'^c(x, \mathbf {z})$ are given by

$$\begin{aligned} \begin{aligned} K'^c(x,y, \mathbf {z}) =&\left. K^c(x,y,\mathbf {z}) \right| _{z_1 = 1}\\ =&\oint _{1} \frac{\mathrm{d}z}{2 \pi \mathrm{i}} F^c(z, x) \prod _{j=2}^{n-1}{ \frac{1 + z_j z}{1 + z} } \oint _{0 , - \rho ' , -1 , \{ - z_j^{-1} \}_{j=2}^{n-1} } \frac{\mathrm{d}w}{2 \pi \mathrm{i}} G^c(w, y)\prod _{j=2}^{n-1}{\frac{1 + w}{1 + z_j w}} \frac{1}{w-z}, \\ A'^c(x, \mathbf {z})=&\left. A^c(x,\mathbf {z}) \right| _{z_2 = 1} = \oint _{1} \frac{ \mathrm{d}z }{2 \pi \mathrm{i}} F^c(z, x) \frac{1}{1 + z} \prod _{j=3}^{n-1}{ \frac{1 + z_j z}{1 + z} }. \end{aligned} \end{aligned}$$

In addition, we introduce the abbreviations

$$\begin{aligned}&K'_{ij}(\mathbf {z}) = K'^c(x_i , x_j, \mathbf {z}), \quad A'_i(\mathbf {z}) = A'^c(x_i, \mathbf {z}) , \end{aligned}$$

and the vector

$$\begin{aligned} \mathbf {K}'^{(k)}_{i}(\mathbf {z})&= \left( \begin{array}{cccc} K'_{i1}(\mathbf {z}),&K'_{i2}(\mathbf {z}),&\cdots ,&K'_{ik}(\mathbf {z}) \end{array} \right) . \end{aligned}$$

We focus on each summand in the sum over $\ell \in [1,k]$ on the right hand side of (E.17). Since $K_{ij}(\mathbf {z})$ and $A_i(\mathbf {z})$ are holomorphic functions in the vicinity of $z_1 = 1$, the $z_1$-integrand is a function with a single pole of order 1 at $z_1 = 1$ and the $z_1$-integration is carried out by substituting 1 into $z_1$. Hence, the $r^{\mathrm {th}}$ summand ($r \in [1,k]$) in the sum over $\ell \in [1,k]$ on the right hand side of (E.17) is written as

$$\begin{aligned} \begin{aligned} h(1) \oint _{1} \prod _{j=2}^{n-1}{ \frac{\mathrm{d}z_j}{2 \pi \mathrm{i}} } \frac{ \prod _{2 \le i < j \le n-1}{(z_j - z_i)} }{ \prod _{j=2}^{n-1}{{(z_j -1)}^{j}} } \prod _{j=2}^{n-1}{h(z_j)} \left| \left[ \begin{array}{c} \mathbf {\bar{K}}^{(k)}_{1} \\ \vdots \\ {\mathbf {\bar{K}}}^{(k)}_{r -1} \\ A_r (\mathbf {z}) \mathbf {B}^{(k)} \\ \mathbf {K}'^{(k)}_{r +1}(\mathbf {z}) \\ \vdots \\ \mathbf {K}'^{(k)}_{k}(\mathbf {z}) \end{array} \right] \right| . \end{aligned} \end{aligned}$$

(E.18)

Removing the variable $z_{n-1}$ and shifting the indices of $\mathbf {z}$-variables by 1 as $\{ z_1,\ldots ,z_{n-2} \} \rightarrow \{ z_2,\ldots ,z_{n-1} \}$ in the claim of Lemma 7.3, we obtain

$$\begin{aligned} \begin{aligned}&\oint _{1} \prod _{j=2}^{n-1}{ \frac{ \mathrm{d}z_j}{2 \pi \mathrm{i}} } \frac{ \prod _{2 \le i< j \le n-1}{(z_j - z_i)} }{ \prod _{j=2}^{n-1}{{(z_j -1)}^{j}} } \prod _{j=2}^{n-1}{h(z_j)} K'^c(x ,y , \mathbf {z}) \\&\quad = \oint _{1} \prod _{j=2}^{n-1}{ \frac{\mathrm{d}z_j}{2 \pi \mathrm{i}} } \frac{ \prod _{2 \le i < j \le n-1}{(z_j - z_i)} }{ \prod _{j=2}^{n-1}{{(z_j -1)}^{j}} } \prod _{j=2}^{n-1}{h(z_j)} \left[ {\bar{K}}^c(x , y) - (z_2 - 1) A'^c(x, \mathbf {z}) B^c(y) \right] . \end{aligned} \end{aligned}$$

(E.19)

Since $K'_{ij}(\mathbf {z})$ and $A_i(\mathbf {z})$ are symmetric under the exchange of any two variables in $\{ z_2,\ldots ,z_{n-1} \}$, we can apply the equality (E.19) to one of the rows from $r +1$ $^{\mathrm {st}}$ to $k^{\mathrm {th}}$ row of the determinant (E.18). Applying the equality (E.19) to $r +1^{\mathrm {st}}$ row of the determinant in (E.18), it is replaced with ${\mathbf {\bar{K}}}^{(k)}_{r +1} - (z_2 - 1) A'_{r +1}(\mathbf {z}) \mathbf {B}^{(k)}$ and (E.18) is divided into two terms as

$$\begin{aligned}&h(1) \oint _{1} \prod _{j=2}^{n-1}{ \frac{\mathrm{d}z_j}{2 \pi \mathrm{i}} } \frac{ \prod _{2 \le i< j \le n-1}{(z_j - z_i)} }{ \prod _{j=2}^{n-1}{{(z_j -1)}^{j}} } \prod _{j=2}^{n-1}{h(z_j)} \left| \left[ \begin{array}{c} {\mathbf {\bar{K}}}^{(k)}_{1} \\ \vdots \\ {\mathbf {\bar{K}}}^{(k)}_{r -1} \\ A_r (\mathbf {z}) \mathbf {B}^{(k)} \\ {\mathbf {\bar{K}}}^{(k)}_{r +1} - (z_2 - 1) A'_{r +1}(\mathbf {z}) \mathbf {B}^{(k)} \\ \mathbf {K}'^{(k)}_{r +2}(\mathbf {z}) \\ \vdots \\ \mathbf {K}'^{(k)}_{k}(\mathbf {z}) \end{array} \right] \right| \nonumber \\&\quad = h(1) \oint _{1} \prod _{j=2}^{n-1}{ \frac{\mathrm{d}z_j}{2 \pi \mathrm{i}} } \frac{ \prod _{2 \le i < j \le n-1}{(z_j - z_i)} }{ \prod _{j=3}^{n-1}{{(z_j -1)}^{j}} } \prod _{j=2}^{n-1}{h(z_j)} \left\{ \frac{1}{{(z_2 - 1)}^2} \left| \left[ \begin{array}{c} {\mathbf {\bar{K}}}^{(k)}_{1} \\ \vdots \\ {\mathbf {\bar{K}}}^{(k)}_{r -1} \\ A_r (\mathbf {z}) \mathbf {B}^{(k)} \\ {\mathbf {\bar{K}}}^{(k)}_{r +1} \\ \mathbf {K}'^{(k)}_{r +2}(\mathbf {z}) \\ \vdots \\ \mathbf {K}'^{(k)}_{k}(\mathbf {z}) \end{array} \right] \right| - \frac{A_r (\mathbf {z}) A'_{r +1}(\mathbf {z})}{(z_2 - 1)} \left| \left[ \begin{array}{c} {\mathbf {\bar{K}}}^{(k)}_{1} \\ \vdots \\ {\mathbf {\bar{K}}}^{(k)}_{r -1} \\ \mathbf {B}^{(k)} \\ \mathbf {B}^{(k)} \\ \mathbf {K}'^{(k)}_{r +2}(\mathbf {z}) \\ \vdots \\ \mathbf {K}'^{(k)}_{k}(\mathbf {z}) \end{array} \right] \right| \right\} . \end{aligned}$$

Obviously, the second term vanishes because the $r$ $^{\mathrm {th}}$ row and the $r+1$ $^{\mathrm {st}}$ row are equivalent. Carrying out the same calculations to the $r+2^{\mathrm{nd}}$ and subsequent rows of the first term, we obtain

$$\begin{aligned} h(1) \oint _{1} \prod _{j=2}^{n-1}{ \frac{\mathrm{d}z_j}{2 \pi \mathrm{i}} } \frac{ \prod _{2 \le i < j \le n-1}{(z_j - z_i)} }{ \prod _{j=2}^{n-1}{{(z_j -1)}^{j}} } \prod _{j=2}^{n-1}{h(z_j)} \left| \left[ \begin{array}{c} {\mathbf {\bar{K}}}^{(k)}_{1} \\ \vdots \\ {\mathbf {\bar{K}}}^{(k)}_{r -1} \\ A_r(\mathbf {z}) \mathbf {B}^{(k)} \\ {\mathbf {\bar{K}}}^{(k)}_{r +1} \\ \vdots \\ {\mathbf {\bar{K}}}^{(k)}_{k} \end{array} \right] \right| . \end{aligned}$$

Since ${\bar{K}}_{ij}$ is independent of $\mathbf {z}$-variables and $A_i(\mathbf {z})$ is independent of $z_1$, it is allowed to revive $z_1$-integral as

$$\begin{aligned} \oint _{1} \prod _{j=1}^{n-1}{ \frac{\mathrm{d}z_j}{2 \pi \mathrm{i}} } \frac{ \prod _{1 \le i < j \le n-1}{(z_j - z_i)} }{ \prod _{j=1}^{n-1}{{(z_j -1)}^{j+1}} } \prod _{j=1}^{n-1}{h(z_j)} \left| \left[ \begin{array}{c} {\mathbf {\bar{K}}}^{(k)}_{1} \\ \vdots \\ {\mathbf {\bar{K}}}^{(k)}_{r-1} \\ (z_1 - 1) A_r(\mathbf {z}) \mathbf {B}^{(k)} \\ {\mathbf {\bar{K}}}^{(k)}_{r +1} \\ \vdots \\ {\mathbf {\bar{K}}}^{(k)}_{k} \end{array} \right] \right| . \end{aligned}$$

In addition, since ${\bar{K}}_{ij}$ is independent of $\mathbf {z}$-variables, we can apply Lemma 7.4 to $r^\mathrm{th}$ row and obtain

$$\begin{aligned} \oint _{1} \prod _{j=1}^{n-1}{ \frac{\mathrm{d}z_j}{2 \pi \mathrm{i}} } \frac{ \prod _{1 \le i < j \le n-1}{(z_j - z_i)} }{ \prod _{j=1}^{n-1}{{(z_j -1)}^{j+1}} } \prod _{j=1}^{n-1}{h(z_j)} \left| \left[ \! \begin{array}{c} {\mathbf {\bar{K}}}^{(k)}_{1} \\ \vdots \\ {\mathbf {\bar{K}}}^{(k)}_{r -1} \\ \sum _{p=1}^{n-1}{ \prod _{q=1}^{p}{ (z_q - 1) {(A_p)}_r } } \mathbf {B}^{(k)} \\ {\mathbf {\bar{K}}}^{(k)}_{r +1} \\ \vdots \\ {\mathbf {\bar{K}}}^{(k)}_{k} \end{array}\! \right] \right| . \end{aligned}$$

(E.20)

This implies that we can rewrite $r^\mathrm{th}$ summand in the sum over $\ell \in [1,k]$ on the right hand side of (E.17) as (E.20) for all $r \in [1,k]$. Therefore, we have

$$\begin{aligned} \begin{aligned}&\oint _{1} \prod _{j=1}^{n-1}{ \frac{\mathrm{d}z_j}{2 \pi \mathrm{i}} } \frac{ \prod _{1 \le i< j \le n-1}{(z_j - z_i)} }{ \prod _{j=1}^{n-1}{{(z_j -1)}^{j+1}} } \prod _{j=1}^{n-1}{h(z_j)} \left| \left[ \begin{array}{c} \mathbf {K}^{(k)}_{1}(\mathbf {z}) \\ \vdots \\ \mathbf {K}^{(k)}_{k}(\mathbf {z}) \end{array} \right] \right| \\&= \oint _{1} \prod _{j=1}^{n-1}{ \frac{\mathrm{d}z_j}{2 \pi \mathrm{i}} } \frac{ \prod _{1 \le i < j \le n-1}{(z_j - z_i)} }{ \prod _{j=1}^{n-1}{{(z_j -1)}^{j+1}} } \prod _{j=1}^{n-1}{h(z_j)} \left\{ \left| \left[ \begin{array}{c} {\mathbf {\bar{K}}}^{(k)}_{1} \\ \vdots \\ \vdots \\ \vdots \\ \vdots \\ {\mathbf {\bar{K}}}^{(k)}_{k} \end{array} \right] \right| - \sum _{\ell =1}^{k}{ \left| \left[ \begin{array}{c} {\mathbf {\bar{K}}}^{(k)}_{1} \\ \vdots \\ {\mathbf {\bar{K}}}^{(k)}_{\ell -1} \\ \sum _{p=1}^{n-1}{ \prod _{q=1}^{p}{ (z_q - 1) {(A_p)}_\ell } } \mathbf {B}^{(k)} \\ {\mathbf {\bar{K}}}^{(k)}_{\ell +1} \\ \vdots \\ {\mathbf {\bar{K}}}^{(k)}_{k} \end{array} \right] \right| } \right\} . \end{aligned} \end{aligned}$$

(E.21)

By the Laplace expansion, we can see that the terms in the curly brackets the right hand side of (E.21) is equivalent to that of (7.25) with $a_{i,j} = {\bar{K}}_{ij}$, $u_i = \sum _{p=1}^{n-1} \prod _{q=1}^{p} (z_q - 1) {(A_p)}_i$ and $v_j = B_j$. Hence we obtain

$$\begin{aligned} \begin{aligned}&\left| \left[ \begin{array}{c} {\mathbf {\bar{K}}}^{(k)}_{1} \\ \vdots \\ \vdots \\ \vdots \\ \vdots \\ {\mathbf {\bar{K}}}^{(k)}_{k} \end{array} \right] \right| - \sum _{\ell =1}^{k}{ \left| \left[ \begin{array}{c} {\mathbf {\bar{K}}}^{(k)}_{1} \\ \vdots \\ {\mathbf {\bar{K}}}^{(k)}_{\ell -1} \\ \sum _{p=1}^{n-1}{ \prod _{q=1}^{p}{ (z_q - 1) {(A_p)}_\ell } } \mathbf {B}^{(k)} \\ {\mathbf {\bar{K}}}^{(k)}_{\ell +1} \\ \vdots \\ {\mathbf {\bar{K}}}^{(k)}_{k} \end{array} \right] \right| } = \left| \left[ \begin{array}{c} {\mathbf {\bar{K}}}^{(k)}_{1} - \sum _{p=1}^{n-1}{ \prod _{q=1}^{p}{ (z_q - 1) {(A_p)}_1 } } \mathbf {B}^{(k)} \\ \vdots \\ \vdots \\ \vdots \\ \vdots \\ {\mathbf {\bar{K}}}^{(k)}_{k} - \sum _{p=1}^{n-1}{ \prod _{q=1}^{p}{ (z_q - 1) {(A_p)}_k } } \mathbf {B}^{(k)} \end{array} \right] \right| \end{aligned} \end{aligned}$$

(E.22)

and, finally, it follows from (E.21) and (E.22) that the equality (E.15) holds.

F. Proof of the Uniform Convergence of ${\varvec{\mathcal {K}}_t}$ on a Bounded Set, Proposition 7.10

A rigorous proof falls into the same pattern as the proof of Proposition 6.2 shown in Appendix D.2. To avoid reiterating ourselves, here we only give a basic idea of the proof. Recall that ${\mathcal {K}}_t(\xi , \zeta )$ is given by

$$\begin{aligned} {\mathcal {K}}_t(\xi ,\zeta ) = - \lambda _2 t^{1/2} \oint _1 \frac{\mathrm{d}w}{2 \pi \mathrm{i}} {{\,\mathrm{e}\,}}^{f(w , t, \xi ) -f(\rho ' , t, \xi ) + g_4(w)} , \end{aligned}$$

where $f(w,t,\xi )=g_1(w)t+g_2(w,\xi )t^{1/2}+g_3(w)t^{1/3}$ with $g_i(w)$ and $g_2(w,\xi )$ given in (7.40). Solving $g_1'(w)=0$ gives us $w_1=\rho '$ and $w_2=-\rho '/2$. One can obtain a steepest descent through $w_1=\rho '$, since the one passing $w_2$ would include extra poles at origin and hence vary the estimate of the integral.

One can see that the contour $\varTheta =\varTheta _1\cup \varTheta _2$ (see Fig. 12), where

$$\begin{aligned} \varTheta _1=&\left\{ w=\rho '-\frac{s}{\sqrt{3}} \mathrm{i}, \,\,\,s\in [-\rho ,\rho ]\right\} , \end{aligned}$$

(F.1a)

$$\begin{aligned} \varTheta _2=&\left\{ w=1+ \frac{2\rho }{\sqrt{3}}{{\,\mathrm{e}\,}}^{\mathrm{i}s} , \,\,\,s\in [-5\pi /6,5\pi /6]\right\} , \end{aligned}$$

(F.1b)

is a steepest descent path of $g_1(w)$ passing through $\rho '$. Namely, $w=\rho '$ is the strict global maximum point of $\mathrm{Re}(g_1)$ along $\varTheta$. This can be proved by calculating $\frac{\mathrm{d}\mathrm{Re}(g_1)(s)}{\mathrm{d}s}$ along $\varTheta$.

Fig. 12

steepest descent contour of integration in ${\mathcal {K}}_t$ passing the saddle point at $w_1=\rho '$

As shown in the proof of Proposition 6.2, we can prove that for large enough t, only the part $\varTheta ^{\delta }\,{:=}\,\{w\in \varTheta \mid |w-w_1|\le \delta \}$, where $\delta =t^{-1/6}$, contributes to the integral. Near $w_1=\rho '$, the Taylor expansion of $g_i(w)$ are given by

$$\begin{aligned} g_1(w)-g_1(w_1) =&\frac{9(1-\rho )(w-\rho ')^2}{16(2-\rho )\rho }+ {\mathcal {O}}[(w-\rho ')^3], \end{aligned}$$

(F.2a)

$$\begin{aligned} g_2(w,\xi )-g_2(w_1,\xi ) =&-\frac{c_\mathrm{g} s_\mathrm{g} + 2 \rho (2-\rho ) \lambda _2 \xi }{2\rho (1-\rho )(2-\rho )}(w-\rho ')+ {\mathcal {O}}[(w-\rho ')^2], \end{aligned}$$

(F.2b)

$$\begin{aligned} g_3(w)-g_3(w_1) =&{\mathcal {O}}[(w-\rho ')^2], \end{aligned}$$

(F.2c)

$$\begin{aligned} g_4(w)-g_4(w_1)=&{\mathcal {O}}[(w-\rho ')]. \end{aligned}$$

(F.2d)

Denote the function $g_i (w)$ without the error terms by ${\bar{g}}_i(w)$. As in Proposition 6.2, we can show that for large t, only the term ${{\,\mathrm{e}\,}}^{{\bar{g}}_1(w)t+{\bar{g}}_2(w,\xi _1,\kappa _1)t^{1/2}+{\bar{g}}_3(w)t^{1/3}+{\bar{g}}_4(w)}$ contributes to the integral. We re-parameterise $\varGamma _{\delta }$ by

$$\begin{aligned} w-\rho '=\mathrm{i}v\frac{2 }{3}\sqrt{\frac{\rho (2-\rho )}{t(1-\rho )}}, \end{aligned}$$

where $-c \delta t^{1/2} \le v \le c \delta t^{1/2}$, and $c=\tfrac{3}{2}\sqrt{\tfrac{(1-\rho )}{\rho (2-\rho )}}$. Thus we are left with

$$\begin{aligned}&\lim _{t\rightarrow \infty } ( - \lambda _2 t^{1/2}) \int _{\varGamma _{\delta }}\frac{\mathrm{d}w}{2\pi \mathrm{i}} {{\,\mathrm{e}\,}}^{{\bar{g}}_1(w)t+{\bar{g}}_2(w, \xi )t^{1/2}+{\bar{g}}_3(w)t^{1/3}+{\bar{g}}_4(w)-f(\rho ',t,\xi _1)}\\&\quad =\lim _{t\rightarrow \infty } ( - \lambda _2 t^{1/2} ) \int _{\varGamma _{\delta }}\frac{\mathrm{d}w}{2\pi \mathrm{i}} \exp \left( \frac{9(1-\rho )(w-\rho ')^2}{16(2-\rho )\rho }t-\frac{c_\mathrm{g} s_\mathrm{g} + 2 \rho (2 - \rho ) \lambda _2 \xi }{2 \rho (1-\rho )(2-\rho )}(w-\rho ')t^{1/2} - \ln (\rho ') \right) \\&\quad =\lim _{t\rightarrow \infty } \frac{- 1}{2 \sqrt{2} \pi } \int _{c\delta t^{1/2}}^{-c\delta t^{1/2}} \mathrm{d}v {{\,\mathrm{e}\,}}^{-v^2/4-v\mathrm{i}(s_\mathrm{g} + \xi )/\sqrt{2}} = \frac{ 1 }{2 \sqrt{2} \pi } \int _{-\infty }^{\infty } \mathrm{d}v {{\,\mathrm{e}\,}}^{-v^2/4-v\mathrm{i}(s_\mathrm{g} + \xi )/\sqrt{2}} = \frac{1}{ \sqrt{2\pi } } {{\,\mathrm{e}\,}}^{-(s_\mathrm{g} + \xi )^2/2}, \end{aligned}$$

where $\lambda _2$ is defined in (7.37).