Stationary Half-Space Last Passage Percolation

Abstract

In this paper we study stationary last passage percolation with exponential weights and in half-space geometry. We determine the limiting distribution of the last passage time in a critical window close to the origin. The result is a new two-parameter family of distributions: one parameter for the strength of the diagonal bounding the half-space (strength of the source at the origin in the equivalent TASEP language) and the other for the distance of the point of observation from the origin. It should be compared with the one-parameter family giving the Baik–Rains distributions for full-space geometry. We finally show that far enough away from the characteristic line, our distributions indeed converge to the Baik–Rains family. We derive our results using a related inhomogeneous integrable model having Pfaffian correlations, together with careful analytic continuation, and steepest descent analysis.

Appendices

On Pfaffians and Point Processes

Given an anti-symmetric $2n \times 2n$ matrix $(a_{i,j})$, its Pfaffian is defined as:

$$\begin{aligned} \mathrm {pf}[a_{i,j}]_{1\le i < j\le 2n} = \frac{1}{2^{n} n!}\sum _{\sigma \in S_{2n}}{{\,\mathrm{sgn}\,}}(\sigma )a_{\sigma (1),\sigma (2)} a_{\sigma (3),\sigma (4)} \cdots a_{\sigma (2n-1),\sigma (2n)} \end{aligned}$$

(A.1)

where $S_{2n}$ is the permutation group on 2n letters. Remark that the Pfaffian is determined entirely by the upper triangular part. One can show that

$$\begin{aligned} \left( \mathrm {pf}[a_{i,j}]_{1\le i < j\le 2n}\right) ^2=\det [a_{i,j}]_{1\le i,j\le 2n}. \end{aligned}$$

(A.2)

Suppose that one has a $2 \times 2$ anti-symmetric matrix kernel K(x, y), i.e. K is a $2 \times 2$ matrix function of (x, y) satisfying $K(x, y) = -K^t(y, x)$ with t denoting transposition. Notice the interchange of x and y. Given such a kernel and points (variables or numbers) $x_1, \dots , x_n$, one can define a $2n \times 2n$ anti-symmetric matrix $K^{(n)}$ block-wise as follows: its $2 \times 2$ block (i, j) for $1 \le i, j \le n$ is given by the matrix $K(x_i, x_j)$. Because $K(x, y) = -K^t(y, x)$, $K^{(n)}$ thus defined is even-dimensional anti-symmetric and its Pfaffian well-defined.

Throughout we are interested in probability distributions on configuration spaces $\varLambda$ equipped with a measure $d \mu$ as follows: either $\varLambda = {\mathbb {Z}}$ (or a semi-infinite interval in ${\mathbb {Z}}$) is discrete and $d \mu$ is counting measure, or $\varLambda = {\mathbb {R}}$ (or a semi-infinite interval in ${\mathbb {R}}$) is continuous and $d \mu = d x$ is Lebesgue measure. A point process (measure)¹⁰ on such a configuration space $\varLambda$ is called Pfaffian with $2 \times 2$ matrix correlation kernel K if there exists a $2 \times 2$ matrix kernel K satisfying $K(x, y) = -K^t(y, x)$ such that the n-point correlation functions $\rho _n(x_1, x_2, \dots , x_n)$ of the process, for all $n \ge 1$, are Pfaffians of the associated $2n \times 2n$ matrix $K^{(n)}$:

$$\begin{aligned} \rho _n(x_1, x_2, \dots , x_n) = \mathrm {pf}[K^{(n)} (x_i, x_j)]_{1\le i<j\le n}. \end{aligned}$$

(A.3)

We recall that if $\varLambda$ is discrete, $\rho _n(x_1, x_2, \dots , x_n) = {\mathbb {P}}(S : x_i \in S,\ \forall \,1 \le i \le n)$ is the probability of S containing all the x’s; if $\varLambda$ is continuous, $\rho _n(x_1, x_2, \dots , x_n) d x_1 \dots d x_n$ is the probability of sets S having non-empty intersection with all $[x_i, x_i + d x_i]$. A simple observation states that the one-point function is the $K_{12}$ entry: ${\mathbb {P}}(S:x \in S) = \rho _1(x) = K_{12}(x, x)$ (in the discrete setting; the obvious modification is needed in the continuous setting).

Given a $2 \times 2$ anti-symmetric matrix kernel K defined on a configuration space $\varLambda$ equipped with a measure $d \mu$ (either counting measure or Lebesgue measure depending on the underlying space), the Fredholm Pfaffian of K restricted to the subspace $U \subset \varLambda$ is defined as

$$\begin{aligned} \mathrm {pf}(J + \lambda K)_{L^2(U) \times L^2(U)} = \sum _{n=0}^{\infty }\frac{\lambda ^n}{n!} \int _{U^n} \mathrm {pf}[ K^{(n)} (x_i, x_j)]_{1\le i<j\le n} \prod _{i=1}^n d \mu (x_i) \end{aligned}$$

(A.4)

assuming the sum is finite. Here J is the anti-symmetric matrix kernel $J(x,y)=\delta _{x,y} \left( {\begin{matrix} 0 &{} 1 \\ -1 &{} 0 \end{matrix}} \right)$, but as is oftentimes the case in the literature, this technicality is overlooked and we think of J just as the corresponding $2 \times 2$ matrix. We note that oftentimes a sufficient condition for the Fredholm Pfaffian $\mathrm {pf}(J+\lambda K)_{L^2(U) \times L^2(U)}$ to be finite finite is that K is a trace class operator on $L^2(U) \times L^2(U)$ (or rather it has trace class entries). Moreover, Fredholm Pfaffians are defined up to conjugation, in the following sense. Suppose ${\tilde{K}}$ is the anti-symmetric matrix kernel

$$\begin{aligned} {\tilde{K}}(x,y)= \left( \begin{matrix} e^{f(x)} &{} \quad 0 \\ 0 &{}\quad e^{-f(x)} \end{matrix}\right) K(x,y) \left( \begin{matrix} e^{f(y)} &{}\quad 0 \\ 0 &{} \quad e^{-f(y)} \end{matrix}\right) \end{aligned}$$

(A.5)

for a $d\mu$-measurable function f. Then it is not hard to check that $\mathrm {pf}[K^{(n)} (x_i, x_j)]_{1\le i<j\le 2n} = \mathrm {pf}[{\tilde{K}}^{(n)} (x_i, x_j)]_{1\le i<j\le 2n}$ and so $\mathrm {pf}(J+\lambda K) = \mathrm {pf}(J+\lambda {\tilde{K}})$ whenever both are defined. Importantly, we can use this to define $\mathrm {pf}(J+\lambda K)$ even if K is not a trace class operator provided we find an appropriate f which makes ${\tilde{K}}$ trace class.

Owing to identity (A.2), we have the following relation between Fredholm Pfaffians with $2 \times 2$ matrix kernels K and block Fredholm determinants with related kernel $J^{-1}K$:

$$\begin{aligned} \mathrm {pf}(J + \lambda K)^2_{L^2(U) \times L^2(U)} = \det (\mathbb {1}+ \lambda J^{-1} K)_{L^2(U) \times L^2(U)} \end{aligned}$$

(A.6)

where we remark the Fredholm determinant on the right hand side is defined as in (A.4) with pf replaced by det and $K^{(n)}$ by $(J^{-1}K)^{(n)}$.

Finally, for any two bounded $2 \times 2$ matrix operators $A:L^2(U) \times L^2(U) \rightarrow L^2(V) \times L^2(V)$, $B:L^2(V) \times L^2(V) \rightarrow L^2(U) \times L^2(U)$ for which both sides below are defined, we have

$$\begin{aligned} \mathrm {pf}(J+\lambda JAB)_{L^2(V) \times L^2(V)}=\mathrm {pf}(J+\lambda JBA)_{L^2(U) \times L^2(U)}. \end{aligned}$$

(A.7)

For example, this could happen if the entries of either A or B are trace class (as such operators form an ideal inside the bounded ones), or if the entries of both A and B are Hilbert–Schmidt (as products of two Hilbert–Schmidt operators are trace class), but it could happen in more generality too. We note that oftentimes AB is infinite dimensional while BA is finite dimensional. This fact is immediately implied by the corresponding Fredholm determinant identity $\det (\mathbb {1}+ \lambda AB) =\det (\mathbb {1}+ \lambda BA)$ and (A.6).

For more on Pfaffians, the reader is referred to the the Appendix of [62] for the analytic side and to [75] for the algebraic and combinatorial ones.

Some Determinantal Computations

In this section we prove (3.23). For this we need to show that

$$\begin{aligned} \left\langle Y_1 \left| Z_2 \right\rangle \right. = \left\langle Y_2 \left| Z_1 \right\rangle \right. =0 \text { and } \left\langle Y_1 \left| Z_1 \right\rangle \right. = \left\langle Y_2 \left| Z_2 \right\rangle \right. \end{aligned}$$

(B.1)

where $Z_i, Y_i, i=1,2$ are defined in (3.18). For brevity and consistency, let us rename

$$\begin{aligned} A=(\mathbb {1}-{{\overline{G}}})^{-1}, \quad h_1= g_1, \quad h_2=g_2, \quad f_1 = f_{+}^\beta . \end{aligned}$$

(B.2)

We have

$$\begin{aligned} \left\langle Y_1 \left| Z_2 \right\rangle \right. = \left\langle f_1 \left| A_{12} f_1 \right\rangle \right. \end{aligned}$$

(B.3)

and

$$\begin{aligned} \left\langle Y_2 \left| Z_1 \right\rangle \right. = \left\langle -h_1 \quad h_2 \left| \begin{array}{c} A_{11}h_2+A_{12}h_1 \\ A_{21}h_2+A_{22} h_1 \end{array} \right\rangle \right. . \end{aligned}$$

(B.4)

To show that both are zero, we need the following result.

Proposition 39

We have:

$$\begin{aligned} \begin{aligned} A_{12}(x,y)=-A_{12}(y,x),\quad A_{21}(x,y)=-A_{21}(y,x),\quad A_{11}(x,y)=A_{22}(y,x). \end{aligned} \end{aligned}$$

(B.5)

Proof

Assume for now that we know the norm of ${\overline{G}}$ is less than 1. Then we can expand A as a Neumann series $(\mathbb {1}-\overline{G})^{-1}=\mathbb {1}+{{\overline{G}}}+{{\overline{G}}}^2+{{\overline{G}}}^3+\cdots$. We will prove by induction that, for any $n\ge 1$

$$\begin{aligned} \begin{aligned} {{\overline{G}}}^n_{12}(x,y)=-{{\overline{G}}}^n_{12}(y,x),\quad \overline{G}^n_{21}(x,y)=-{{\overline{G}}}^n_{21}(y,x),\quad \overline{G}^n_{11}(x,y)={{\overline{G}}}^n_{22}(y, x). \end{aligned} \end{aligned}$$

(B.6)

The base case is true by the definition of ${{\overline{G}}}=J^{-1}{{\overline{K}}}$ and by the fact that

$$\begin{aligned} \begin{aligned} {{\overline{K}}}_{12}(x,y)=-{{\overline{K}}}_{21}(y,x),\quad {{\overline{K}}}_{11}(x,y) =-{{\overline{K}}}_{11}(y,x),\quad {{\overline{K}}}_{22}(x,y)=-{{\overline{K}}}_{22}(y,x). \end{aligned} \end{aligned}$$

(B.7)

Now we proceed with the inductive step: assuming that (B.6) holds for $B={{\overline{G}}}^n$, for $C={{\overline{G}}}^{n+1}$ we have

$$\begin{aligned} \begin{aligned} C_{11}(x,y)&=\left( B_{11}{{\overline{G}}}_{11}+B_{12}{{\overline{G}}}_{21}\right) (x,y)\\&=\int _s^{\infty } dz B_{11}(x,z)\overline{G}_{11}(z,y)+\int _s^{\infty } dz B_{12}(x,z){{\overline{G}}}_{21}(z,y)\\&=\int _s^{\infty } dz B_{22}(z,x)\overline{G}_{22}(y,z)+\int _s^{\infty } dz B_{12}(z,x){{\overline{G}}}_{21}(y,z)\\&=C_{22}(y,x). \end{aligned} \end{aligned}$$

(B.8)

Moreover,

$$\begin{aligned} \begin{aligned} C_{12}(x,y)&=\left( B_{11}{{\overline{G}}}_{12}+B_{12}{{\overline{G}}}_{22}\right) (x,y)\\&=\int _s^{\infty } dz B_{11}(x,z)\overline{G}_{12}(z,y)+\int _s^{\infty } dz B_{12}(x,z){{\overline{G}}}_{22}(z,y)\\&=-\int _s^{\infty } dz B_{22}(z,x)\overline{G}_{12}(y,z)-\int _s^{\infty } dz B_{12}(z,x){{\overline{G}}}_{11}(y,z)\\&=-C_{12}(y,x) \end{aligned} \end{aligned}$$

(B.9)

where we have used that $C = B {\overline{G}} = {\overline{G}} B$. An analogous computation holds for $C_{21}$.

Now it may happen that the norm of ${\overline{G}}$, of which we remain ignorant, is $\ge 1$. We then replace ${\overline{G}}$ by $\omega {\overline{G}}$ for $\omega >0$ small enough to make $\omega {\overline{G}}$ of subunit norm. The argument above applies to the corresponding $A(\omega ) = (\mathbb {1}-\omega {\overline{G}})^{-1}$. We then analytically continue in $\omega$, as the spectrum of any trace class operator (and in particular of ${\overline{G}}$) is discrete without non-zero accumulation points. $\quad \square$

Using Proposition 39 in Eq. (B.3), we have

$$\begin{aligned} \begin{aligned} \left\langle Y_1 \left| Z_2 \right\rangle \right.&=\int _s^{\infty } \int _s^{\infty } dx dy f_1(x)A_{12}(x,y)f_1(y)\\&=-\int _s^{\infty } \int _s^{\infty } dx dy f_1(x)A_{12}(y,x)f_1(y) \\&=0. \end{aligned} \end{aligned}$$

(B.10)

For the same reason, (B.4) becomes:

$$\begin{aligned} \left\langle Y_1 \left| Z_2 \right\rangle \right. = -\left\langle h_1 \left| A_{11}h_2 \right\rangle \right. - \left\langle h_1 \left| A_{12}h_1 \right\rangle \right. + \left\langle h_2 \left| A_{22}h_1 \right\rangle \right. + \left\langle h_2 \left| A_{21}h_2 \right\rangle \right. = 0 \end{aligned}$$

(B.11)

since the last two terms are zero, and the first two are equal by Proposition 39:

$$\begin{aligned} \begin{aligned} \left\langle h_2 \left| A_{22}h_1 \right\rangle \right.&= \int _s^{\infty } \int _s^{\infty } dx dy h_2(x)A_{22}(x,y)h_1(y)\\&=\int _s^{\infty } \int _s^{\infty } dx dy h_2(x)A_{11}(y,x)h_1(y)\\&=\left\langle h_1 \left| A_{11}h_2 \right\rangle \right. . \end{aligned} \end{aligned}$$

(B.12)

Now, we show that $\left\langle Y_1 \left| Z_1 \right\rangle \right. = \left\langle Y_2 \left| Z_2 \right\rangle \right.$. The two are explicitly given by

$$\begin{aligned} \left\langle Y_1 \left| Z_1 \right\rangle \right. = \left\langle f_1 \left| A_{11} h_2 \right\rangle \right. + \left\langle f_1 \left| A_{12}h_1 \right\rangle \right. , \quad \left\langle Y_2 \left| Z_2 \right\rangle \right. = -\left\langle h_1 \left| A_{12} f_1 \right\rangle \right. + \left\langle h_2 \left| A_{22}f_1 \right\rangle \right. . \end{aligned}$$

(B.13)

By Proposition 39, as $A_{11}(x, y) = A_{22}(y, x)$ and $A_{12}(x, y) = - A_{12}(y, x)$, it follows that $\left\langle f_1 \left| A_{11} h_2 \right\rangle \right. = \left\langle h_2 \left| A_{22} f_1 \right\rangle \right.$ and that $\left\langle f_1 \left| A_{12} h_1 \right\rangle \right. = -\left\langle h_1 \left| A_{12} f_1 \right\rangle \right.$ proving the desired equality.

Correlations for Geometric Weights

1.1 Correlation kernel for generic geometric weights

In this section we briefly review the main result on integrable last passage percolation with i.i.d. geometric random variables. It is by passing to the exponential limit in this result that we obtain Theorem 13. Consider half-space last passage percolation with i.i.d.  geometric random variables $(W_{i,j})_{1 \le j \le i \le N}$ distributed as follows:

$$\begin{aligned} W_{i,j} = {\left\{ \begin{array}{ll} \mathrm {Geom}(a x_i), &{} i = j, \\ \mathrm {Geom}(x_i x_j), &{} i > j \end{array}\right. } \end{aligned}$$

(C.1)

where $a, x_1, \dots , x_N$ are real parameters satisfying

$$\begin{aligned} 0 \le a< \min _i \tfrac{1}{x_i}, \quad 0< x_1, \dots , x_N < 1 \end{aligned}$$

(C.2)

and a random variable X is geometric$\mathrm {Geom}(q)$ if ${\mathbb {P}}(X = k) = (1-q)q^k, \forall k \in {\mathbb {N}}$. We depict this in Fig. 4. It is helpful to visualize the x’s as indexing the rows and columns of the half-space.

Fig. 4

A possible LPP path (polymer) from (1, 1) to $(N, N-n)$ for $(N, n) = (8, 2)$. The non-diagonal dots are independent geometric random variables with parameters assigned by multiplying the row and column x parameters; the diagonal has an extra parameter a

We have the following result for the last passage percolation time $L_{N,N-n}$ from (1, 1) to $(N,N-n)$ (for fixed $0 \le n \le N-1$). The precise statement as stated below is first rigorously proven and explicitly stated in [4]. Previously the kernel was derived in [72], but not in a completely rigorous way, specifically for the $n=0$ case. The Pfaffian structure for $n=0$ was first derived by Rains [70], and subsequently extended for generic n and/or as multi-point joint distribution in [4, 16, 27, 45, 72]. See also the pioneering algebraic work of Baik–Rains [8, 10] for an alternative but equivalent approach via Toeplitz+Hankel determinants and matrix integrals, as well as the more combinatorial approach of Forrester–Rains [43]. We follow [16] for the exposition hereinafter.

Theorem 40

The distribution of $L_{N,N-n}$ is a Fredholm Pfaffian

$$\begin{aligned} {\mathbb {P}}(L_{N,N-n} \le s) = \mathrm {pf}(J-K^{\mathrm{geo}})_{\ell ^2 (\{s+1, s+2,\dots \})} \end{aligned}$$

(C.3)

with $2 \times 2$ matrix correlation kernel $K^{\mathrm{geo}}: {\mathbb {Z}}^2 \rightarrow \mathrm {Mat}_2({\mathbb {R}})$ given by

$$\begin{aligned} \begin{aligned} K^{\mathrm{geo}}_{11}(k, \ell )&= \frac{1}{(2\pi \mathrm{i})^2}\int \frac{dz}{z^{k}} \int \frac{dw}{w^{\ell }} F(z) F(w) \frac{(z-w) (z-a) (w-a)}{(z^2-1) (w^2-1) (zw-1)}, \\ K^{\mathrm{geo}}_{12}(k, \ell ) = -K^{\mathrm{geo}}_{21}(\ell , k)&= \frac{1}{(2\pi \mathrm{i})^2}\int \frac{dz}{z^{k}} \int \frac{dw}{w^{-\ell +1}} \frac{F(z)}{F(w)} \frac{(zw-1) (z-a)}{(z-w) (z^2-1) (w-a)}, \\ K^{\mathrm{geo}}_{22}(k, \ell )&= \frac{1}{(2\pi \mathrm{i})^2}\int \frac{dz}{z^{-k+1}} \int \frac{dw}{w^{-\ell +1}} \frac{1}{F(z) F(w)} \frac{(z-w)}{ (zw-1) (z-a) (w-a)} \end{aligned} \end{aligned}$$

(C.4)

where

$$\begin{aligned} F(z)= \frac{\prod _{k=1}^N (1-x_k/z)}{\prod _{k=1}^{N-n} (1-x_k z)} \end{aligned}$$

(C.5)

and where the contours are positively oriented circles centered around the origin satisfying the following conditions:

• for $K^{\mathrm{geo}}_{11}$, $1< |z|, |w| < \min _i \frac{1}{x_i}$;

• for $K^{\mathrm{geo}}_{12}$, $\max \{ \max _i x_i, a\}< |w| < |z|$ and $1< |z| < \min _i \frac{1}{x_i}$;

• for $K^{\mathrm{geo}}_{22}$, $\max \{ \max _i x_i, a\} < |w|, |z|$ and $1 < |zw|$.

1.2 Reformulation of the correlation kernel for our model

Now consider the case $x_1=b\in (0,1)$, $x_2=x_3=\ldots =x_N=\sqrt{q}$ for some $q\in (0,1)$ and $a>0$ such that $a\sqrt{q}<1$ and $ab<1$. We will manipulate the kernel so that the limit to the exponential case becomes straightforward. In particular, it is not necessary to do any steepest descent analysis (Laplace method) as was done in [4].

Let us define the following functions:

$$\begin{aligned} H(z)=\frac{(1-\sqrt{q}/z)^{N-1}}{(1-\sqrt{q}z)^{N-1}},\quad B(z)= \frac{1-b/z}{1-bz} \end{aligned}$$

(C.6)

which when combined yield

$$\begin{aligned} F(z)=H(z) B(z) (1-\sqrt{q} z)^n. \end{aligned}$$

(C.7)

Notice that

$$\begin{aligned} H(z^{-1}) = H(z)^{-1},\quad B(z^{-1})=B(z)^{-1},\quad F(z^{-1})=H(z)^{-1} B(z)^{-1} (1-\sqrt{q}/z)^n. \end{aligned}$$

(C.8)

Since in the Fredholm Pfaffian (C.3) the smallest value to be considered is $s=0$, the $k,\ell$ entries of the kernel will be in $\{1,2,3,\ldots \}$ only. Below we only consider the case $b\ne \sqrt{q}$, since the $b=\sqrt{q}$ case is simpler: just set $B(z)=0$ and replace $N-1$ with N in H(z). To see this, take for instance the $b\rightarrow \sqrt{q}$ limit in (C.13).

Lemma 41

Let $b\ne \sqrt{q}$ and $k,\ell \ge 1$. The kernel entry $K^\mathrm{geo}_{11}$ can be expressed as

$$\begin{aligned} \begin{aligned} K^{\mathrm{geo}}_{11}(k,\ell ) =\,&\frac{-1}{(2\pi \mathrm{i})^2}\oint \limits _{\varGamma _{\sqrt{q}}} dw\!\!\! \oint \limits _{\varGamma _{1/\sqrt{q}}} dz \frac{w^{\ell -1}}{z^{k}}\frac{h^{\mathrm{geo}}_{11}(z,w)}{z-w} \\&+ \frac{-1}{(2\pi \mathrm{i})^2}\oint \limits _{\varGamma _{b}} dw \oint \limits _{\varGamma _{1/\sqrt{q}}} dz \frac{w^{\ell -1}}{z^{k}}\frac{h^{\mathrm{geo}}_{11}(z,w)}{z-w}\\&+\frac{-1}{(2\pi \mathrm{i})^2}\oint \limits _{\varGamma _{\sqrt{q}}} dw \oint \limits _{\varGamma _{1/b}} dz \frac{w^{\ell -1}}{z^{k}}\frac{h^\mathrm{geo}_{11}(z,w)}{z-w} \end{aligned} \end{aligned}$$

(C.9)

with

$$\begin{aligned} h^{\mathrm{geo}}_{11}(z,w)=\frac{H(z)B(z)}{H(w)B(w)}(1-\sqrt{q} z)^n(1-\sqrt{q}/w)^n \frac{(zw-1)(z-a)(1-w a)}{(z^2-1)(1-w^2)}. \end{aligned}$$

(C.10)

Proof

Let us start with

$$\begin{aligned} \begin{aligned} K^{\mathrm{geo}}_{11}(k,\ell )&= \oint \frac{dw}{2\pi \mathrm{i}} \oint \frac{dz}{2\pi \mathrm{i}} \frac{H(z)(1-\sqrt{q} z)^nB(z) \, H(w) (1-\sqrt{q}w)^n B(w)}{z^{k}w^{\ell }}\\&\quad \times \frac{(z-w)(z-a)(w-a)}{(z^2-1)(w^2-1)(zw-1)} \end{aligned} \end{aligned}$$

(C.11)

where the integrals can be taken over circles with radii larger than 1 and smaller than $\min \{1/b,1/\sqrt{q}\}$. The change of variable $w\rightarrow w^{-1}$ leads to

$$\begin{aligned} K^{\mathrm{geo}}_{11}(k,\ell ) =\frac{1}{(2\pi \mathrm{i})^2}\oint dw \oint dz \frac{w^{\ell -1}}{z^{k}}\frac{h^{\mathrm{geo}}_{11}(z,w)}{z-w} \end{aligned}$$

(C.12)

where we can take $|w|\in (\max \{\sqrt{q},b\},1)$ and $|z|\in (1,\min \{1/b,1/\sqrt{q}\})$.

The integrand behaves like $w^{N-n+\ell -1}$ as $w\rightarrow 0$ and thus there is no pole at $w=0$ for any $\ell \ge 1$ and $n\le N$. Also, it behaves like $1/(z^{N-n+k+1})\lesssim z^{-2}$ as $z\rightarrow \infty$ for all $n\le N$ and $k\ge 1$. Thus there is no pole in z at $\infty$. This means that we can deform the integration contours to include only the poles at b and $\sqrt{q}$. Moreover, we can open up the z contour and close it down around the poles at 1/b and $1/\sqrt{q}$. By doing this we change of orientation of the contour, which then gives a minus sign as all the closed contours we consider as anticlockwise oriented. Thus we get

$$\begin{aligned} K^{\mathrm{geo}}_{11}(k,\ell ) =\frac{-1}{(2\pi \mathrm{i})^2}\oint \limits _{\varGamma _{b,\sqrt{q}}} dw \oint \limits _{\varGamma _{1/b,1/\sqrt{q}}} dz \frac{w^{\ell -1}}{z^{k}}\frac{h^{\mathrm{geo}}_{11}(z,w)}{z-w} \end{aligned}$$

(C.13)

The claimed result follows from the observation that

$$\begin{aligned} \oint \limits _{\varGamma _b} dw \oint \limits _{\varGamma _{1/b}}dz \frac{w^{\ell -1}}{z^{k}}\frac{h^{\mathrm{geo}}_{11}(z,w)}{z-w} =0. \end{aligned}$$

(C.14)

$\square$

Lemma 42

Let $a,b\sqrt{q}$ be all different and $k,\ell \ge 1$. The kernel entry $K^{\mathrm{geo}}_{12}$ can be expressed as

$$\begin{aligned} \begin{aligned} K^{\mathrm{geo}}_{12}(k,\ell )&=\frac{-1}{(2\pi \mathrm{i})^2}\oint \limits _{\varGamma _{\sqrt{q},a,b}} dw\oint \limits _{\varGamma _{1/\sqrt{q}}} dz \frac{w^{\ell -1}}{z^{k}}\frac{h^{\mathrm{geo}}_{12}(z,w)}{z-w} \\&\quad + \frac{-1}{(2\pi \mathrm{i})^2}\oint \limits _{\varGamma _{\sqrt{q},a}} dw \oint \limits _{\varGamma _{1/b}} dz \frac{w^{\ell -1}}{z^{k}}\frac{h^\mathrm{geo}_{12}(z,w)}{z-w} \end{aligned} \end{aligned}$$

(C.15)

where

$$\begin{aligned} h^{\mathrm{geo}}_{12}(z,w)=\frac{H(z)B(z)}{H(w)B(w)}\frac{(1-\sqrt{q} z)^n}{(1-\sqrt{q}/w)^n} \frac{(zw-1)(z-a)}{(z^2-1)(w-a)}. \end{aligned}$$

(C.16)

Proof

This time the change of variables is not necessary. Initially we have integration paths such that $\max \{\sqrt{q},a,b\}<|w|<|z|$ and $1<|z|<\min \{1/b,1/\sqrt{q}\}$. As before, 0 is not a pole for w and $\infty$ is not a pole for z. Thus we can deform the w and z contours to include only $\sqrt{q},a,b$ and $1/b,1/\sqrt{q}$ respectively. Also in this case the contribution of the poles at $(w=b,z=1/b)$ is equal to zero. We then get the claimed result. $\square$

Lemma 43

Let $a,b,\sqrt{q}$ be all different and $k,\ell \ge 1$. The kernel entry $K^{\mathrm{geo}}_{22}$ can be expressed as

$$\begin{aligned} \begin{aligned} K^{\mathrm{geo}}_{22}(k,\ell )&= E(k,\ell )+\frac{-1}{(2\pi \mathrm{i})^2}\oint \limits _{\varGamma _{\sqrt{q}}}\ \,dw \oint \limits _{\varGamma _{1/\sqrt{q},1/a,1/b}} dz \frac{w^{\ell -1}}{z^{k}}\frac{h^{\mathrm{geo}}_{22}(z,w)}{z-w} \\&\quad + \frac{-1}{(2\pi \mathrm{i})^2}\oint \limits _{\varGamma _{a}}\, dw\oint \limits _{\varGamma _{1/\sqrt{q},1/b}} dz \frac{w^{\ell -1}}{z^{k}}\frac{h^{\mathrm{geo}}_{22}(z,w)}{z-w}\\&\quad +\frac{-1}{(2\pi \mathrm{i})^2}\oint \limits _{\varGamma _{b}}\, dw \oint \limits _{\varGamma _{1/\sqrt{q},1/a}} dz \frac{w^{\ell -1}}{z^{k}}\frac{h^{\mathrm{geo}}_{22}(z,w)}{z-w} \end{aligned} \end{aligned}$$

(C.17)

where

$$\begin{aligned} h^\mathrm{geo}_{22}(z,w)=\frac{H(z)B(z)}{H(w)B(w)}\frac{1}{(1-\sqrt{q}/z)^n(1-\sqrt{q}/w)^n} \frac{(zw-1)}{(w-a)(1-az)} \end{aligned}$$

(C.18)

and

$$\begin{aligned} E(k,\ell )=-{{\,\mathrm{sgn}\,}}(k-\ell ) \oint \limits _{\varGamma _{1/a,1/\sqrt{q}}} \frac{dz}{2\pi \mathrm{i}} z^{\ell -k} \frac{z^{-1}-z}{(1-az)(z-a)}\frac{1}{(1-\sqrt{q}/z)^n (1-\sqrt{q} z)^n}. \end{aligned}$$

(C.19)

Remark 44

The condition that $a,b,\sqrt{q}$ are all different is not really a restriction. If two or more of them are equal, one just has to use a different contour decomposition starting from the integration contours for (w, z) as $\varGamma _{\sqrt{q},a,b}\times \varGamma _{1/\sqrt{q},1/a,1/b}$. For example, if $a=b$ and recalling that $b\in (0,1)$, the given contours are already fine. The decomposition we mention is useful since we want to study the case $b\rightarrow 1/a$ later.

Proof

We start with

$$\begin{aligned} K^{\mathrm{geo}}_{22}(k,\ell )&=\oint \frac{dw}{2\pi \mathrm{i}}\oint \frac{dz}{2\pi \mathrm{i}} \frac{z^{k-1} w^{\ell -1}}{H(z)B(z)H(w)B(w)}\frac{1}{(1-\sqrt{q}z)^n(1-\sqrt{q}w)^n}\nonumber \\&\quad \times \frac{z-w}{(zw-1)(z-a)(w-a)} \end{aligned}$$

(C.20)

where the integration contours can be taken as circles with $|z|=|w|>\max \{\sqrt{q},a,b,1\}$ (the value $1/\sqrt{q}$ is not a pole). The change of variable $z\rightarrow z^{-1}$ leads to

$$\begin{aligned} K^{\mathrm{geo}}_{22}(k,\ell )=\frac{1}{(2\pi \mathrm{i})^2} \oint dw \oint dz \frac{w^{\ell -1}}{z^{k}} \frac{h^{\mathrm{geo}}_{22}(z,w)}{z-w} \end{aligned}$$

(C.21)

where the integration contours need to satisfy $\max \{\sqrt{q},a,b\}<|z|<|w|<\min \{1/b,1/a,1/\sqrt{q}\}$. Notice however that this time we cannot directly shrink the w contour around $\sqrt{q},a,b$ since z is also a pole for w and lies inside the w contour. We can exchange the contours, that is get w inside z, by correcting with the residue at $w=z$. We get

$$\begin{aligned} K^{\mathrm{geo}}_{22}(k,\ell )=E(k,\ell )+\frac{1}{(2\pi \mathrm{i})^2} \oint dw \oint dz \frac{w^{\ell -1}}{z^{k}} \frac{h^{\mathrm{geo}}_{22}(z,w)}{z-w} \end{aligned}$$

(C.22)

where the integration contours now satisfy $\max \{\sqrt{q},a,b\}<|w|<|z|<\min \{1/b,1/a,1/\sqrt{q}\}$, and where

$$\begin{aligned} E(k,\ell ) = \oint \limits _{\varGamma _{a, \sqrt{q}}} \frac{dz}{2\pi \mathrm{i}} \frac{z^{\ell -k-1}(1-z^2)}{(1-\sqrt{q}/z)^n(1-\sqrt{q} z)^n}\frac{1}{(1-az)(z-a)}. \end{aligned}$$

(C.23)

For the double integral, we verify that $w=0$ and $z=\infty$ are not poles and then deform them to include only the poles at $\sqrt{q},a,b$ for w and $1/\sqrt{q},1/a,1/b$ for z (the minus sign comes from the exchange of the orientation in the z path). Then we notice that the contributions of the poles at $(w=a,z=1/a)$ and at $(w=b,z=1/b)$ are 0. We are left with $9-2=7$ contributions which we put together as three integrals in such a way that the paths can be taken as single contours and do not cross (recall the conditions $ab<1$, $a\sqrt{q}<1$ and that $a,b,\sqrt{q}$ are all different).

Concerning $E(k,\ell )$, it is an easy computation to verify that $E(k,k)=0$ (the $n=0$ and $n \ge 1$ are two separate computations), which is actually a consequence of the anti-symmetry of the entry $K_{22}(k,l)$. Furthermore, for $k>\ell$ we have

$$\begin{aligned} E(k,\ell )=-\oint \limits _{\varGamma _{1/a,1/\sqrt{q}}} \frac{dz}{2\pi \mathrm{i}} z^{\ell -k} \frac{z^{-1}-z}{(1-az)(z-a)}\frac{1}{(1-\sqrt{q}/z)^n (1-\sqrt{q} z)^n} \end{aligned}$$

(C.24)

and for $k<\ell$ we have

$$\begin{aligned} E(k,\ell )=\oint \limits _{\varGamma _{a,\sqrt{q}}} \frac{dz}{2\pi \mathrm{i}} z^{\ell -k} \frac{z^{-1}-z}{(1-az)(z-a)}\frac{1}{(1-\sqrt{q}/z)^n (1-\sqrt{q} z)^n} = -E(\ell ,k). \end{aligned}$$

(C.25)

$\square$

1.3 From the geometric to the exponential model: proof of Theorem 13

The exponential model is obtained from the geometric one by taking the $\varepsilon \rightarrow 0$ limit with the following variables:

$$\begin{aligned} (a,b)=(1-\varepsilon \alpha ,1-\varepsilon \beta ),\quad q=1-\varepsilon ,\quad (k,\ell )=\varepsilon ^{-1}(x,y). \end{aligned}$$

(C.26)

Let us consider the accordingly rescaled kernel and conjugated kernel

$$\begin{aligned} K^\varepsilon _{11}(x,y)&=\varepsilon ^{-2-2n}K^{\mathrm{geo}}_{11}(k,\ell ), \quad K^\varepsilon _{12}(x,y) = \varepsilon ^{-1}K^{\mathrm{geo}}_{12}(k,\ell ), \nonumber \\ K^\varepsilon _{21}(x,y)&=\varepsilon ^{-1}K^{\mathrm{geo}}_{21}(k,\ell ), \quad K^\varepsilon _{22}(x,y) = \varepsilon ^{2n}K^{\mathrm{geo}}_{22}(k,\ell ). \end{aligned}$$

(C.27)

Then the following result is straightforward and does not require complicated asymptotic analysis.

Proposition 45

Uniformly for x, y in a bounded set of ${\mathbb {R}}_+$, we have

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0} K^\varepsilon _{ij}(x,y)= K_{ij}(x,y) \end{aligned}$$

(C.28)

where the kernel K is the one of Theorem 13 for the exponential model.

Proof

Let us also change the integration variables as $z=1+Z \varepsilon$, $w=1+W \varepsilon$. Notice that all the variables are in an $\varepsilon$-neighborhood of 1 and thus the contours for Z and W can be fixed independently of $\varepsilon$; they just need to pass/be in the correct position with respect to the poles.

Therefore we can easily get the following point-wise limits:

$$\begin{aligned} \begin{aligned} \lim _{\varepsilon \rightarrow 0} H(z) z^{-k}&=\left[ \frac{\tfrac{1}{2}+z}{\tfrac{1}{2}-z} \right] ^{N-1} e^{-x Z}=\varPhi (x,Z),\\ \lim _{\varepsilon \rightarrow 0} (1-\sqrt{q} z)^n \varepsilon ^{-n}&= (\tfrac{1}{2}-Z)^n,\\ \lim _{\varepsilon \rightarrow 0} (1-\sqrt{q} /w)^n \varepsilon ^{-n}&= (\tfrac{1}{2}+W)^n \end{aligned} \end{aligned}$$

(C.29)

and

$$\begin{aligned} \begin{aligned}&\lim _{\varepsilon \rightarrow 0} \frac{1-b/z}{1-b z}\frac{1-b w}{1-b/w}\frac{(z-a)(1-w a)(zw-1)}{(z^2-1)(1-w^2)(z-w)}\\&\quad =\frac{(Z+\alpha )(Z+\beta )(W-\alpha )(W-\beta )(Z+W)}{4Z W (Z-W)(W+\beta )(Z-\beta )},\\&\lim _{\varepsilon \rightarrow 0} \frac{1-b/z}{1-b z}\frac{1-b w}{1-b/w}\frac{(z-a)(zw-1)}{(z^2-1)(w-a)(z-w)} \varepsilon \\&\quad =\frac{(Z+\alpha )(Z+\beta )(W-\beta )(Z+W)}{2Z (Z-W)(W+\alpha )(W+\beta )(Z-\beta )},\\&\lim _{\varepsilon \rightarrow 0} \frac{1-b/z}{1-b z}\frac{1-b w}{1-b/w}\frac{zw-1}{(1-a z)(w-a)(z-w)} \varepsilon ^{2}\\&\quad =\frac{-(Z+\beta )(W-\beta )(Z+W)}{(Z-W)(W+\alpha )(W+\beta )(Z-\alpha )(Z-\beta )},\\&\lim _{\varepsilon \rightarrow 0} z^{\ell -k} \frac{z^{-1}-z}{(1-a z)(z-a)} \varepsilon =e^{-Z(x-y)}\frac{2Z}{Z^2-\alpha ^2}. \end{aligned} \end{aligned}$$

(C.30)

By choosing the paths for Z, W so that their distances to any of the poles are bounded away from 0 uniformly for all $0<\varepsilon \ll 1$, we have that the integrands (appropriately multiplied by some powers of $\varepsilon$) are uniformly bounded. Thus we can use dominated convergence to take the $\varepsilon \rightarrow 0$ limit inside the integration. $\quad \square$

It remains for us to find appropriate bounds in order to prove the convergence of the Fredholm Pfaffian to its corresponding limit. We do this next.

Lemma 46

Let $\beta \in (0,1/2)$, $\alpha \in (-1/2,1/2)$ with $\alpha +\beta >0$. Let us set

$$\begin{aligned} \delta =(\beta -\max \{0,-\alpha \})/4,\text { and }\mu =\max \{0,-\alpha \}+2\delta . \end{aligned}$$

(C.31)

Then the following bounds hold uniformly in $x,y\in {\mathbb {R}}_+$,

$$\begin{aligned} |K^\varepsilon _{11}(x,y) e^{\mu (x+y)}|&\le C e^{-\delta (x+y)}, \quad |K^\varepsilon _{12}(x,y) e^{\mu x} e^{-\mu y}| \le C e^{-\delta (x+y)}, \nonumber \\ |K^\varepsilon _{21}(x,y) e^{-\mu x} e^{\mu y}|&\le C e^{-\delta (x+y)}, \quad |K^\varepsilon _{22}(x,y)e^{-\mu (x+y)}| \le C e^{-\delta (x+y)}. \end{aligned}$$

(C.32)

Proof

The proof is quite simple, since the $k,\ell$ dependence is only through the term $w^\ell /z^k$ and the rest of the integrand remains nicely bounded and converges as we saw in Proposition 45.

For $K^\varepsilon _{11}(k,l)$, choose contours satisfying $|z|\ge 1+(\mu +\delta )\varepsilon$ and $|w|\le 1-(\mu +\delta )\varepsilon$. This gives $|w^\ell /z^k|\le \frac{(1-(\mu +\delta )\varepsilon )^{y/\varepsilon }}{(1+(\mu +\delta )\varepsilon )^{x/\varepsilon }}\simeq e^{-\mu (x+y)} e^{-\delta (x+y)}$.

For $K^\varepsilon _{12}(k,l)$, choose the contours such that $|z|\ge 1+(\mu +\delta )\varepsilon$ and $|w|\le 1+(\mu -\delta )\varepsilon$. This gives $|w^\ell /z^k|\lesssim e^{-\mu (x-y)} e^{-\delta (x+y)}$.

For the double integrand in $K^\varepsilon _{22}(k,l)$ there are several terms. In the first and third one, we choose contours satisfying $|z|\ge 1-(\mu -\delta )\varepsilon$ and $|w|\le 1-(\mu +\delta )\varepsilon$, while in the second one $|z|\ge 1+(\mu +\delta ) \varepsilon$ and $|w|\le 1+(\mu -\delta )\varepsilon$. Combining the cases, we have $|w^\ell /z^k|\lesssim e^{\mu (x+y)} e^{-\delta (x+y)}$. Finally, for the term coming from $E(k,\ell )$, for $k>\ell$ we take a contour with $|z|\ge 1-(\mu -\delta )\varepsilon$, which gives $|z^{\ell -k}|\lesssim e^{(\mu -\delta )(x-y)}\le e^{\mu x} e^{-\delta (x+y)}$ by using $\mu -\delta \ge \delta$. For $k<\ell$ the bound follows from the anti-symmetry of $E(k,\ell )$. $\square$

Proof of Theorem 13

With Proposition 45 and Lemma 46 in hand, the proof of Theorem 13 is standard. One writes the Fredholm Pfaffian as a series expansion and then uses the Hadamard bound. Exactly the same steps are made in the proof of the convergence of the large time limit of Proposition 20 and thus we omit them here for brevity.