Interlacing adjacent levels of $\beta$–Jacobi corners processes

Abstract

We study the asymptotics of the global fluctuations for the difference between two adjacent levels in the $\beta$–Jacobi corners process (multilevel and general $\beta$ extension of the classical Jacobi ensemble of random matrices). The limit is identified with the derivative of the 2d Gaussian free field. Our main tools are integral forms for the (Macdonald-type) difference operators originating from the shuffle algebra.

Appendices

Appendices

Dimension reduction identities

In this appendix we discuss integral identities, which are widely used in proofs in the main text. A special case ($m=1$) of the following result was communicated to the authors by Alexei Borodin, and we present our own proof here.

For any positive integer n, let $\sigma _n$ denote the cycle $(12\cdots n)$, and let $S^{cyc}(n)$ denote the n-element subgroup of the symmetric group spanned by $\sigma _n$.

Theorem A.1

Let $n\ge 2$, and $f_1, \ldots , f_n : \mathbb {C} \rightarrow \mathbb {C}$ be meromorphic with possible poles at $\{\mathfrak {p}_1, \ldots , \mathfrak {p}_m \}$. Then we have the identity

$$\begin{aligned}&\sum _{\sigma \in S^{cyc}(n)} \frac{1}{(2\pi {\mathbf {i}})^n} \oint \cdots \oint \frac{f_{\sigma (1)}(u_{1}) \cdots f_{\sigma (n)}(u_{n}) }{(u_2 - u_1) \cdots (u_n - u_{n-1})} du_1\cdots du_n\nonumber \\&\quad \quad = \frac{1}{2\pi {\mathbf {i}}} \oint f_1(u)\cdots f_n(u) du, \end{aligned}$$

(A.1)

where the contours in both sides are positively oriented, enclosing $\{\mathfrak {p}_1, \ldots , \mathfrak {p}_m \}$, and for the left hand side we require $|u_1| \ll \cdots \ll |u_n|$.

Proof

Let $\mathfrak {C}_1, \ldots , \mathfrak {C}_{2n-1}$ be closed paths around $\{\mathfrak {p}_1, \ldots , \mathfrak {p}_m \}$, and each $\mathfrak {C}_i$ is inside $\mathfrak {C}_{i+1}$, $1\le i \le 2n-2$. Also, to simplify notations, set $f_{n+t} = f_t$ and $u_{n+t} = u_t$ for any $1 \le t \le n-1$. Then the left hand side of (A.1) can be written as

$$\begin{aligned} \sum _{t=0}^{n-1} \frac{1}{(2\pi {\mathbf {i}})^n} \oint _{\mathfrak {C}_{1}} \cdots \oint _{\mathfrak {C}_{n}} \frac{f_{1+t}(u_1) \cdots f_{n+t}(u_n) }{(u_2 - u_1) \cdots (u_n- u_{n-1})} du_n \cdots du_1. \end{aligned}$$

(A.2)

When $n=2$, we have

$$\begin{aligned}&\frac{1}{(2\pi {\mathbf {i}})^2} \oint _{\mathfrak {C}_{1}} \oint _{\mathfrak {C}_{2}} \frac{f_1(u_1) f_2(u_2)}{u_2 - u_1} du_2 du_1 + \frac{1}{(2\pi {\mathbf {i}})^2} \oint _{\mathfrak {C}_{1}} \oint _{\mathfrak {C}_{2}} \frac{f_2(u_1) f_1(u_2)}{u_2 - u_1} du_2 du_1 \nonumber \\&\quad = \frac{1}{(2\pi {\mathbf {i}})^2} \oint _{\mathfrak {C}_{1}} \oint _{\mathfrak {C}_{2}} \frac{f_1(u_1) f_2(u_2)}{u_2 - u_1} du_2 du_1 + \frac{1}{(2\pi {\mathbf {i}})^2} \oint _{\mathfrak {C}_{3}} \oint _{\mathfrak {C}_{2}} \frac{f_1(u_1) f_2(u_2)}{u_1 - u_2} du_2 du_1 \nonumber \\&\quad = \frac{1}{(2\pi {\mathbf {i}})^2} \oint _{\mathfrak {C}_{3} - \mathfrak {C}_{1}} \oint _{\mathfrak {C}_{2}} \frac{f_1(u_1) f_2(u_2)}{u_1 - u_2} du_2 du_1, \end{aligned}$$

(A.3)

where $\oint _{\mathfrak {C}_{3} - \mathfrak {C}_{1}}$ is a notation for the difference of integrals over $\mathfrak {C}_3$ and $\mathfrak {C}_1$. Further, (A.3) equals to

$$\begin{aligned} \frac{1}{2\pi {\mathbf {i}}} \oint _{\mathfrak {C}_{2}} f_1(u) f_2(u) du, \end{aligned}$$

(A.4)

since as a function of $u_1$, $\frac{f_1(u_1) f_2(u_2)}{u_1 - u_2}$ has a single pole at $u_2$ between $\mathfrak {C}_{3}$ and $\mathfrak {C}_{1}$; and the residue at this pole equals $f_1(u_2)f_2(u_2)$. This proves the case of $n=2$.

When $n\ge 3$, we argue by induction and assume that Theorem A.1 is true for $n-1$. For any $1 \le t \le n-1$, we have that

$$\begin{aligned}&\frac{1}{(2\pi {\mathbf {i}})^n} \oint _{\mathfrak {C}_{1}} \cdots \oint _{\mathfrak {C}_{n}} \frac{f_{1+t}(u_1) \cdots f_{n+t}(u_n) }{(u_2 - u_1) \cdots (u_n- u_{n-1})} du_n \cdots du_1 \nonumber \\&\quad = \frac{1}{(2\pi {\mathbf {i}})^n} \oint _{\mathfrak {C}_{1+t}} \cdots \oint _{\mathfrak {C}_{n+t}} \frac{f_{1}(u_1) \cdots f_{n}(u_n) }{(u_{2+t} - u_{1+t}) \cdots (u_{n+t}- u_{n-1+t})} du_{n+t} \cdots du_{1+t} \nonumber \\&\quad = \frac{1}{(2\pi {\mathbf {i}})^n} \oint _{\mathfrak {C}_{n+1}} \cdots \oint _{\mathfrak {C}_{n+t}} \oint _{\mathfrak {C}_{t+1}} \cdots \oint _{\mathfrak {C}_{n}}\nonumber \\&\quad \quad \times \frac{f_{1}(u_1) \cdots f_{n}(u_n) }{(u_{2+t} - u_{1+t}) \cdots (u_{n+t}- u_{n-1+t})} du_{n} \cdots du_{1}. \end{aligned}$$

(A.5)

Now we can move the contours of $u_1, \ldots , u_t$ from $\mathfrak {C}_{n+1}, \ldots , \mathfrak {C}_{n+t}$ to $\mathfrak {C}_{1}, \ldots , \mathfrak {C}_{t}$, respectively. We move the contours one by one starting from $u_1$, and each move is across $\mathfrak {C}_{t+1}, \ldots , \mathfrak {C}_{n}$. For $u_1 (= u_{n+1})$, the only pole between $\mathfrak {C}_{n+1}$ and $\mathfrak {C}_{1}$ is $u_n$; for any $u_i$, $1 < i \le t$, there is no pole between $\mathfrak {C}_{n+i}$ and $\mathfrak {C}_{i}$. Thus we have that

$$\begin{aligned}&\frac{1}{(2\pi {\mathbf {i}})^n} \oint _{\mathfrak {C}_{n+1}} \cdots \oint _{\mathfrak {C}_{n+t}} \oint _{\mathfrak {C}_{t+1}} \cdots \oint _{\mathfrak {C}_{n}} \frac{f_{1}(u_1) \cdots f_{n}(u_n) }{(u_{2+t} - u_{1+t}) \cdots (u_{n+t}- u_{n-1+t})} du_{n} \cdots du_{1} \nonumber \\&\quad = \frac{1}{(2\pi {\mathbf {i}})^n} \oint _{\mathfrak {C}_{1}} \cdots \oint _{\mathfrak {C}_{n}} \frac{f_{1}(u_1) \cdots f_{n}(u_n) }{(u_{2+t} - u_{1+t}) \cdots (u_{n+t}- u_{n-1+t})} du_{n} \cdots du_{1} \nonumber \\&\quad \quad + \frac{1}{(2\pi {\mathbf {i}})^{n-1}} \oint _{\mathfrak {C}_{1}} \cdots \oint _{\mathfrak {C}_{n-1}}\nonumber \\&\quad \quad \times \frac{f_{1+t}(u_1) \cdots f_{n}(u_{n-t}) f_{n+1}(u_{n-t}) \cdots f_{n+t}(u_{n-1}) }{(u_2 - u_1) \cdots (u_{n-1}- u_{n-2})} du_{n-1} \cdots du_1. \end{aligned}$$

(A.6)

Notice that (taking into account that $u_{n+t} = u_t$)

$$\begin{aligned} \sum _{t=0}^{n-1} \frac{f_{1}(u_1) \cdots f_{n}(u_n) }{(u_{2+t} - u_{1+t}) \cdots (u_{n+t}- u_{n-1+t})} = 0, \end{aligned}$$

(A.7)

and by the induction assumption (applied to $f_2, \ldots , f_{n-1}, f_n f_1$), we have that

$$\begin{aligned}&\sum _{t=0}^{n-1} \frac{1}{(2\pi {\mathbf {i}})^n} \oint _{\mathfrak {C}_{1}} \cdots \oint _{\mathfrak {C}_{n}} \frac{f_{1+t}(u_1) \cdots f_{n+t}(u_n) }{(u_2 - u_1) \cdots (u_n- u_{n-1})} du_n \cdots du_1 \nonumber \\&\quad = \sum _{t=1}^{n-1} \frac{1}{(2\pi {\mathbf {i}})^{n-1}} \oint _{\mathfrak {C}_{1}} \cdots \oint _{\mathfrak {C}_{n-1}}\nonumber \\&\quad \quad \times \frac{f_{1+t}(u_1) \cdots f_{n}(u_{n-t}) f_{n+1}(u_{n-t}) \cdots f_{n+t}(u_{n-1}) }{(u_2 - u_1) \cdots (u_{n-1}- u_{n-2})} du_{n-1} \cdots du_1 \nonumber \\&\quad = \frac{1}{2\pi {\mathbf {i}}} \oint f_1(u)\cdots f_n(u) du. \end{aligned}$$

(A.8)

$\square$

Corollary A.2

Let s be a positive integer. Let f, $g_1, \ldots , g_s$ be meromorphic functions with possible poles at $\{\mathfrak {p}_1, \ldots , \mathfrak {p}_m \}$. Then for $n\ge 2$,

$$\begin{aligned}&\frac{1}{(2\pi {\mathbf {i}})^n} \oint \cdots \oint \frac{1}{(v_2-v_1)\cdots (v_n-v_{n-1})} \prod _{i=1}^n f(v_i) dv_i \prod _{i=1}^s \left( \sum _{j=1}^n g_i(v_j)\right) \nonumber \\&\quad = \frac{n^{s-1}}{2\pi {\mathbf {i}}}\oint f(v)^n \prod _{i=1}^s g_i(v) dv, \end{aligned}$$

(A.9)

where the contours in both sides are around all of $\{\mathfrak {p}_1, \ldots , \mathfrak {p}_m \}$, and for the left hand side we require $|u_1| \ll \cdots \ll |u_n|$.

Proof

Take disjoint sets $U_1, \ldots , U_n$, with $\bigcup _{i=1}^n U_i = \{1, \ldots , s\}$ (some of which might be empty). In Theorem A.1 we let $f_{i} = f\prod _{j\in U_{i} } g_j$ for each $1 \le i \le n$, and get

$$\begin{aligned}&\sum _{\sigma \in S^{cyc}(n)} \frac{1}{(2\pi {\mathbf {i}})^n} \oint \cdots \oint \frac{1}{(v_2-v_1)\cdots (v_n-v_{n-1})} \prod _{i=1}^n \left( f(v_i) \prod _{j\in U_{\sigma (i)}} g_j(v_i) dv_i \right) \nonumber \\&\quad = \frac{1}{2\pi {\mathbf {i}}}\oint f(v)^n \prod _{i=1}^s g_i(v) dv. \end{aligned}$$

(A.10)

Summing over all $n^s$ partitions $U_1, \ldots , U_n$ of $\{1, \ldots , s\}$ into n disjoint sets, we obtain (A.9). $\square$