Correction To: Jacobi Ensemble, Hurwitz Numbers and Wilson Polynomials

Correction to: Letters in Mathematical Physics (2021) 111:67 https://doi.org/10.1007/s11005-021-01396-z

1 Introduction

Denote ${\mathcal {H}}_N(I)$ the set of Hermitian matrices of size $N=1,2,\dots$ with eigenvalues in the (closed) interval $I\subseteq {\mathbb {R}}$, endowed with the probability measure

$$\begin{aligned} \mathrm dm_N(X)=\frac{1}{C_N}\exp \bigl (\textrm{tr}\,(V(X))\bigr )\,\mathrm dX,\qquad \mathrm dX=\prod _{i=1}^N\mathrm dX_{ii}\prod _{1\le i<j\le N}\mathrm d\textrm{Re}\,X_{ij}\,\mathrm d\textrm{Im}\,X_{ij}.\nonumber \\ \end{aligned}$$

(1.1)

Here $C_N:=\int _{{\mathcal {H}}_N(I)}\exp \textrm{tr}\,V(X)\mathrm dX$ and $V=V(x)$ is a smooth function of $x\in I^\circ$ (the interior of I) so that V(X) is defined for all $X\in {\mathcal {H}}_N(I)$ by the spectral theorem. We assume that V satisfies the following decay assumptions: there exists $\varepsilon >0$ such that $\exp V(x)={\mathcal {O}}\left( |x-x_0|^{-1+\varepsilon }\right)$ as $x\in I^\circ$ approaches a finite endpoint $x_0$ of I; if I extends to $\pm \infty$ we assume that $V(x)\rightarrow -\infty$ fast enough as $x\rightarrow \pm \infty$ in order for the measure (1.1) to have finite moments of all orders. In particular, the family $\bigl (P_n(x)\bigr )_{n\ge 0}$ of orthogonal polynomials associated with the measure $\exp (V(x))\mathrm dx$ on I exists (unique):

$$\begin{aligned} \int _{I}P_m(x)P_n(x)\exp (V(x))\mathrm dx=h_m\delta _{m,n}. \end{aligned}$$

(1.2)

The generating functions of moments

$$\begin{aligned} {\mathscr {C}}_{\ell }(z_1,\dots ,z_\ell ):=\int _{{\mathcal {H}}_N(I)}\prod _{i=1}^\ell \textrm{tr}\,\left[ \left( z_i-X\right) ^{-1}\right] \mathrm dm_N(X),\qquad \ell \ge 1, \end{aligned}$$

(1.3)

are analytic functions of $z_1,\dots ,z_\ell \in {\mathbb {C}}\setminus I$, symmetric in the variables $z_1,\dots ,z_\ell$. The generating functions for cumulants (or, connected moments) are

$$\begin{aligned} {\mathscr {C}}_{\ell }^\textsf{c}(z_1,\dots ,z_\ell ):= \sum _{{\mathcal {P}}\text { partition of }\{1,\dots ,\ell \}}(-1)^{|{\mathcal {P}}|-1}(|{\mathcal {P}}|-1)! \prod _{A\in {\mathcal {P}}}{\mathscr {C}}_{|A|}(\{z_a\}_{a\in A}). \end{aligned}$$

(1.4)

Introduce the $2\times 2$ matrix

$$\begin{aligned} Y_N(z):= \left( \begin{array}{cc} P_N(z) &{} \frac{1}{2\pi \textrm{i}}\int _IP_N(x)\textrm{e}^{V(x)}\frac{\mathrm dx}{x-z}\\ -\frac{2\pi \textrm{i}}{h_{N-1}}P_{N-1}(z) &{} -\frac{1}{h_{N-1}}\int _IP_{N-1}(x)\textrm{e}^{V(x)}\frac{\mathrm dx}{x-z} \end{array}\right) , \end{aligned}$$

(1.5)

which is the well-known solution to the Riemann–Hilbert problem for orthogonal polynomials [4]; it is an analytic function of $z\in {\mathbb {C}}\setminus I$.

Theorem 1.1

(Theorem 1.5 in [6]) Let

$$\begin{aligned} R(z):=Y_N(z)\begin{pmatrix}1 &{} 0\\ 0 &{}0 \end{pmatrix}Y_N^{-1}(z), \end{aligned}$$

(1.6)

with $Y_N(z)$ as in (1.5). Then the cumulant generating functions (1.4) are given by

$$\begin{aligned} {\mathscr {C}}_1^\textsf{c}(z)&=\bigl (Y^{-1}_N(z)Y_N'(z)\bigr )_{1,1}, \end{aligned}$$

(1.7)

$$\begin{aligned} {\mathscr {C}}_2^\textsf{c}(z_1,z_2)&=\frac{\textrm{tr}\,\bigl (R(z_1)R(z_2)\bigr )-1}{(z_1-z_2)^2}, \end{aligned}$$

(1.8)

$$\begin{aligned} {\mathscr {C}}_\ell ^\textsf{c}(z_1,\dots ,z_\ell )&=-\sum _{(i_1,\dots ,i_\ell )\in \textrm{cyc}( (\ell ) )}\frac{\textrm{tr}\,\bigl (R(z_{i_1})\dots R(z_{i_\ell })\bigr )}{(z_{i_1}-z_{i_2})(z_{i_2}-z_{i_3})\cdots (z_{i_\ell }-z_{i_1})},\quad \ell \ge 3, \end{aligned}$$

(1.9)

where $'$ is the derivative with respect to z and $\textrm{cyc}((\ell ))$ is the set of $\ell$-cycles in the symmetric group ${\mathfrak {S}}_\ell$.

The proof originally presented in [6] is correct only for $\ell =1$. This is because the proof is based on formula (3.5) in [6] for the correlators that holds true only when $\ell =1$; the correct formula is

$$\begin{aligned} {\mathscr {C}}_\ell (z_1,\dots ,z_\ell )=\sum _{k=1}^\ell \sum _{\begin{array}{c} \{\Pi _1,\dots ,\Pi _k\}\\ \bigcup _{i=1}^k\Pi _i=\lbrace 1,\dots ,\ell \rbrace \\ \Pi _i\not =\emptyset \not =\Pi _i\cap \Pi _j \end{array}}\int _{I^k}\frac{\rho _k(x_1,\dots ,x_k)\mathrm dx_1\cdots \mathrm dx_k}{\prod _{i=1}^k\prod _{j\in \Pi _i}(z_j-x_i)}, \end{aligned}$$

(1.10)

the sum on the right-hand side running over all distinct (unordered) set partitions of $\lbrace 1,\dots ,\ell \rbrace$, i.e. unordered collections of non-empty pairwise disjoint subsets of $\lbrace 1,\dots ,\ell \rbrace$ whose union is $\lbrace 1,\dots ,\ell \rbrace$. For example,

$$\begin{aligned} {\mathscr {C}}_2(z_1,z_2)&=\int _{I}\frac{\rho _1(x_1)\mathrm dx_1}{(z_1-x_1)(z_2-x_1)}+\int _{I^2}\frac{\rho _2(x_1,x_2)\mathrm dx_1\mathrm dx_2}{(z_1-x_1)(z_2-x_2)},\nonumber \\ {\mathscr {C}}_3(z_1,z_2,z_3)&=\int _{I}\frac{\rho _1(x_1)\mathrm dx_1}{(z_1-x_1)(z_2-x_1)(z_3-x_1)}+\int _{I^2}\frac{\rho _2(x_1,x_2)\mathrm dx_1\mathrm dx_2}{(z_1-x_1)(z_2-x_1)(z_3-x_2)}\nonumber \\&\quad +\int _{I^2}\frac{\rho _2(x_1,x_2)\mathrm dx_1\mathrm dx_2}{(z_1-x_1)(z_2-x_2)(z_3-x_1)}+\int _{I^2}\frac{\rho _2(x_1,x_2)\mathrm dx_1\mathrm dx_2}{(z_1-x_2)(z_2-x_1)(z_3-x_1)}\nonumber \\&\quad +\int _{I^3}\frac{\rho _3(x_1,x_2,x_3)\mathrm dx_1\mathrm dx_2\mathrm dx_3}{(z_1-x_1)(z_2-x_2)(z_3-x_3)}\,. \end{aligned}$$

(1.11)

Here, $\rho _k(x_1,\dots ,x_k)$ is the k-point correlation function of eigenvalues, given by the well-known determinantal formula [3]

$$\begin{aligned} \rho _k(x_1,\dots ,x_k)=\det \bigl (K_N(x_i,x_j)\bigr )_{i,j=1}^k \end{aligned}$$

(1.12)

in terms of the Christoffel–Darboux kernel $K_N(x,y)$, cf. (2.7) below. Therefore, the analysis at [6, page 24] (for $\ell =2$) and at [6, page 27] (for $\ell >2$) is incorrect. For this reason, here we provide a proof (for all $\ell \ge 1$) which follows a different strategy, closer in spirit to an analogous proof in [1, 5].

2 Proof of Theorem 1.1

The strategy of the proof is based on the observation that setting

$$\begin{aligned} {\mathscr {Z}}_N({\textbf{t}},{\textbf{z}}):=\int _{{\mathcal {H}}_N(I)}\exp \biggl (\textrm{tr}\,(V(X)+\sum _{i=1}^\ell t_i(z_i-X)^{-1})\biggr )\mathrm dX, \quad \nonumber \\ {\textbf{t}}=(t_1,\dots ,t_\ell ),\ {\textbf{z}}=(z_1,\dots ,z_\ell ), \end{aligned}$$

(2.1)

we have

$$\begin{aligned} {\mathscr {C}}^\textsf{c}({\textbf{z}})=\frac{\partial ^\ell }{\partial t_1\cdots \partial t_\ell }\log {\mathscr {Z}}_N({\textbf{t}},{\textbf{z}})\biggr |_{t_i=0}. \end{aligned}$$

(2.2)

Here and below it is assumed that $z_i\not \in I$.

2.1 Orthogonal polynomials on the real line and unitary-invariant ensembles

We denote $P_\ell$ the monic orthogonal polynomials, $h_\ell =\int _I P^2_\ell (x)\textrm{e}^{V(x)}\mathrm dx$, see (1.2), and

$$\begin{aligned} \widehat{P}_\ell (\zeta ):= \frac{1}{2 \pi \textrm{i}}\int _IP_\ell (x)\textrm{e}^{V(x)}\frac{\mathrm dx}{x-\zeta },\qquad \zeta \in {\mathbb {C}}\setminus I, \end{aligned}$$

(2.3)

their Cauchy transforms. The matrix

$$\begin{aligned} Y_N(\zeta ):= \left( \begin{array}{cc} P_N(\zeta ) &{} \widehat{P}_N(\zeta )\\ -\frac{2\pi \textrm{i}}{h_{N-1}} P_{N-1}(\zeta ) &{} -\frac{2\pi \textrm{i}}{h_{N-1}}\widehat{P}_{N-1}(\zeta ) \end{array}\right) , \end{aligned}$$

(2.4)

introduced in (1.5), is an analytic function of $\zeta \in {\mathbb {C}}\setminus I$. It satisfies the jump condition

$$\begin{aligned} Y_{N,+}(x)=Y_{N,-}(x)\begin{pmatrix} 1 &{} \textrm{e}^{V(x)} \\ 0 &{} 1 \end{pmatrix}, \qquad x\in I^\circ , \end{aligned}$$

(2.5)

where $Y_{N,\pm }(x):=\lim _{\epsilon \rightarrow 0_+}Y_N(x\pm \textrm{i}\epsilon )$, $x\in I^\circ$ ($I^\circ$ is the interior of the interval I). As $\zeta \rightarrow \infty$, we have

$$\begin{aligned} Y_{N}(\zeta )=\left( {\textbf{1}}+{\mathcal {O}}(\zeta ^{-1})\right) \zeta ^{N\sigma _3}, \end{aligned}$$

(2.6)

where we denote ${\textbf{1}}=\begin{pmatrix} 1&{} 0 \\ 0 &{} 1 \end{pmatrix}$ and $\sigma _3=\begin{pmatrix} 1&{} 0 \\ 0 &{} -1 \end{pmatrix}$. Lastly, we recall the Christoffel–Darboux kernel

$$\begin{aligned} K_N(x,y):= & {} \textrm{e}^{\frac{V(x)+V(y)}{2}}\sum _{i=0}^{N-1}\frac{P_i(x)P_i(y)}{h_i}\nonumber \\= & {} \frac{\textrm{e}^{\frac{V(x)+V(y)}{2}}}{h_{N-1}}\frac{P_N(x)P_{N-1}(y)-P_{N-1}(x)P_{N}(y)}{x-y}. \end{aligned}$$

(2.7)

The last identity is known as Christoffel–Darboux identity and allows to rewrite the Christoffel–Darboux kernel in terms of the matrix $Y_N$ in (2.4) as

$$\begin{aligned} K_N(x,y)= -\frac{\textrm{e}^{\frac{V(x)+V(y)}{2}}}{2\pi \textrm{i}(x-y)}\begin{pmatrix}0&1 \end{pmatrix}Y^{-1}_N(x)Y_N(y)\begin{pmatrix}1 \\ 0\end{pmatrix}, \end{aligned}$$

(2.8)

which is independent of the choice of boundary value of $Y_N$ because of (2.5).

Next, we need to recall the connection of orthogonal polynomials to the theory of unitary-invariant ensembles of random matrices. The main point which is relevant for our present purposes is that [3]

$$\begin{aligned} \int _{{\mathcal {H}}_N(I)}\exp \bigl (\textrm{tr}\,(V(X))\bigr )\mathrm dX = N!h_0^V\cdots h_{N-1}^V, \end{aligned}$$

(2.9)

where it is convenient to explicitly express the dependence of $P_\ell =P_\ell ^V$ and $h_\ell =h_\ell ^V$ on the potential V. Therefore, introducing the modified potential

$$\begin{aligned} V_{{\textbf{t}},{\textbf{z}}}(x):=V(x)+\sum _{i=1}^\ell \frac{t_i}{z_i-x},\qquad {\textbf{t}}=(t_1,\dots ,t_\ell ),\ \ {\textbf{z}}=(z_1,\dots ,z_\ell ), \end{aligned}$$

(2.10)

we have

$$\begin{aligned} {\mathscr {Z}}_N({\textbf{t}},{\textbf{z}})=N!h_0^{V_{{\textbf{t}},{\textbf{z}}}}\cdots h_{N-1}^{V_{{\textbf{t}},{\textbf{z}}}}. \end{aligned}$$

(2.11)

Lemma 2.1

We have

$$\begin{aligned} \partial _{t_j}h_i^{V_{{\textbf{t}},{\textbf{z}}}}=\int _{I}\bigl (P_i^{V_{{\textbf{t}},{\textbf{z}}}}(x)\bigr )^2\textrm{e}^{V_{{\textbf{t}},{\textbf{z}}}(x)}\frac{\mathrm dx}{z_j-x}. \end{aligned}$$

(2.12)

Proof

We have $h_i^{V_{{\textbf{t}},{\textbf{z}}}}=\int _{I}\bigl (P_i^{V_{{\textbf{t}},{\textbf{z}}}}(x)\bigr )^2\textrm{e}^{V_{{\textbf{t}},{\textbf{z}}}(x)}\mathrm dx$ hence

$$\begin{aligned} \partial _{t_j}h_i^{V_{{\textbf{t}},{\textbf{z}}}}= & {} 2\int _{I}P_i^{V_{{\textbf{t}},{\textbf{z}}}}(x)\left( \partial _{t_j}P_i^{V_{{\textbf{t}},{\textbf{z}}}}(x)\right) \textrm{e}^{V_{{\textbf{t}},{\textbf{z}}}(x)}\mathrm dx\nonumber \\{} & {} \quad +\int _{I}\bigl (P_i^{V_{{\textbf{t}},{\textbf{z}}}}(x)\bigr )^2\textrm{e}^{V_{{\textbf{t}},{\textbf{z}}}(x)}\left( \partial _{t_j}V_{{\textbf{t}},{\textbf{z}}}(x)\right) \mathrm dx, \end{aligned}$$

(2.13)

but the first term vanishes by orthogonality because $P_i^{V_{{\textbf{t}},{\textbf{z}}}}(x)$ are normalized to be monic and, therefore, $\partial _{t_j}P_i^{V_{{\textbf{t}},{\textbf{z}}}}(x)$ is a polynomial of degree strictly less than i. $\square$

Remark 2.2

We allow for the potential V to be complex-valued and, accordingly, slightly abuse the standard terminology and still refer to (1.2) as an orthogonality condition on the real line, even though one usually considers only real-valued potentials in such a context. However, in order to be able to consider the analytic generating function ${\mathscr {Z}}_N({\textbf{t}},{\textbf{z}})$, this caveat plays a role.

More importantly, existence of the monic “orthogonal” polynomial as in (1.2) with respect to a complex-valued potential is not ensured for all values of the parameters ${\textbf{t}},{\textbf{z}}$. However, the condition for existence (non-vanishing of the Hankel determinants of the moments) is open in the space of parameters ${\textbf{t}},{\textbf{z}}$, and contains the subspace ${\textbf{t}}=0$ by standard arguments pertaining to the classical real-valued case (in which the Hankel matrices are positive-definite). Therefore, for sufficiently small ${\textbf{t}}$, the existence of $P_i^{V_{{\textbf{t}},{\textbf{z}}}}(x)$ is not an issue and we will restrict to sufficiently small ${\textbf{t}}$ without further mention in what follows, as in the end we are interested in the quantities (2.2).

2.2 Case $\ell =1$

It follows from (2.8) that

$$\begin{aligned} K_N(x,x)=\lim _{y\rightarrow x}K_N(x,y)=\frac{\textrm{e}^{V(x)}}{2\pi \textrm{i}}\begin{pmatrix}0&1 \end{pmatrix}Y^{-1}_N(x)Y_N'(x)\begin{pmatrix}1 \\ 0\end{pmatrix}. \end{aligned}$$

(2.14)

In the following we shall use the notation

$$\begin{aligned} \Delta f(x):=f_+(x)-f_-(x),\qquad x\in I^\circ , \end{aligned}$$

(2.15)

for the jump of a function f across I, namely $f_\pm (x):=\lim _{\epsilon \rightarrow 0_+}f(x\pm \textrm{i}\epsilon )$. The next lemma is well known, see e.g. [2], and we re-prove it here for the reader’s convenience.

Lemma 2.3

We have

$$\begin{aligned} K_N(x,x)=-\frac{1}{2\pi \textrm{i}}\Delta \left[ \textrm{tr}\,\left( Y^{-1}_N(x)\frac{\partial Y_N(x)}{\partial x}\textrm{E}_{1,1}\right) \right] ,\qquad \textrm{E}_{1,1}:=\begin{pmatrix} 1 &{} 0 \\ 0 &{} 0 \end{pmatrix}. \end{aligned}$$

(2.16)

Proof

Let us denote $':=\partial _x$. It follows from the jump condition (2.5) for $Y_N$ that

$$\begin{aligned} Y'_{N,+}(x)=Y'_{N,-}(x)\begin{pmatrix} 1 &{} \textrm{e}^{V(x)} \\ 0 &{} 1 \end{pmatrix}+Y_{N,-}(x)\begin{pmatrix} 0 &{} V'(x)\textrm{e}^{V(x)} \\ 0 &{} 0 \end{pmatrix}, \qquad x\in I^\circ .\nonumber \\ \end{aligned}$$

(2.17)

Therefore we compute

$$\begin{aligned}&\Delta \left[ \textrm{tr}\,\left( Y^{-1}_N(x)Y_N'(x)\textrm{E}_{1,1}\right) \right] =\textrm{tr}\,\left( Y^{-1}_{N,+}(x)Y_{N,+}'(x)\textrm{E}_{1,1}\right) -\textrm{tr}\,\left( Y^{-1}_{N,-}(x)Y_{N,-}'(x)\textrm{E}_{1,1}\right) \nonumber \\&\quad =\textrm{tr}\,\left( \begin{pmatrix} 1 &{} -\textrm{e}^{V(x)} \\ 0 &{} 1 \end{pmatrix} Y^{-1}_{N,-}(x)Y_{N,-}'(x) \begin{pmatrix} 1 &{} \textrm{e}^{V(x)} \\ 0 &{} 1 \end{pmatrix}\textrm{E}_{1,1}\right) - \textrm{tr}\,\left( Y^{-1}_{N,-}(x)Y_{N,-}'(x)\textrm{E}_{1,1}\right) \nonumber \\&\qquad + \textrm{tr}\,\left( \begin{pmatrix} 1 &{} -\textrm{e}^{V(x)} \\ 0 &{} 1 \end{pmatrix} \begin{pmatrix} 0 &{} V'(x)\textrm{e}^{V(x)} \\ 0 &{} 0 \end{pmatrix}\textrm{E}_{1,1}\right) \end{aligned}$$

(2.18)

The last term vanishes and so, by the cyclic property of the trace, we have

$$\begin{aligned} \Delta \left[ \textrm{tr}\,\left( Y^{-1}_N(x)Y_N'(x)\textrm{E}_{1,1}\right) \right] =\textrm{tr}\,\left[ Y^{-1}_{N,-}(x)Y_{N,-}'(x)\left( \begin{pmatrix} 1 &{} \textrm{e}^{V(x)} \\ 0 &{} 1 \end{pmatrix}\textrm{E}_{1,1}\begin{pmatrix} 1 &{} -\textrm{e}^{V(x)} \\ 0 &{} 1 \end{pmatrix}-\textrm{E}_{1,1}\right) \right] \nonumber \\ \end{aligned}$$

(2.19)

which is easily seen to be equivalent to (2.14). $\square$

We are ready for the proof of the case $\ell =1$. In such case ${\textbf{t}}=t$, ${\textbf{z}}=z$, and $V_{{\textbf{t}},{\textbf{z}}}(x)=V(x)+t/(z-x)$. By (2.11), Lemma 2.1, and (2.7), we have

$$\begin{aligned} \partial _{t}\log {\mathscr {Z}}_N({\textbf{t}},{\textbf{z}})= & {} \sum _{i=0}^{N-1}\frac{1}{h_i^{V_{{\textbf{t}},{\textbf{z}}}}}\partial _{t} h_{i}^{V_{{\textbf{t}},{\textbf{z}}}} =\sum _{i=0}^{N-1}\frac{1}{h_i^{V_{{\textbf{t}},{\textbf{z}}}}}\int _{I}\bigl (P_i^{V_{{\textbf{t}},{\textbf{z}}}}(x)\bigr )^2\textrm{e}^{V_{{\textbf{t}},{\textbf{z}}}(x)}\frac{\mathrm dx}{z-x} \nonumber \\= & {} \int _I K_N^{V_{{\textbf{t}},{\textbf{z}}}}(x,x)\frac{\mathrm dx}{z-x}, \end{aligned}$$

(2.20)

where we denote explicitly the dependence of the Christoffel–Darboux kernel on the potential. Let $\Gamma$ be an oriented contour in the complex plane which surrounds I in counterclockwise sense (i.e., I lies on the left of $\Gamma$) and leaves z outside (i.e., z lies to the right of $\Gamma$). Then, using Lemma 2.3, we get

$$\begin{aligned} \partial _{t}\log {\mathscr {Z}}_N({\textbf{t}},{\textbf{z}})&=-\int _I \Delta \left[ \textrm{tr}\,\left( Y^{-1}_N(x;{\textbf{t}},{\textbf{z}})\frac{\partial Y_N(x;{\textbf{t}},{\textbf{z}})}{\partial x}\textrm{E}_{1,1}\right) \right] \frac{\mathrm dx}{2\pi \textrm{i}(z-x)}\nonumber \\&= \int _\Gamma \textrm{tr}\,\left( Y^{-1}_N(\zeta ;{\textbf{t}},{\textbf{z}})\frac{\partial Y_N(\zeta ;{\textbf{t}},{\textbf{z}})}{\partial \zeta }\textrm{E}_{1,1}\right) \frac{\mathrm d\zeta }{2\pi \textrm{i}(z-\zeta )}, \end{aligned}$$

(2.21)

where $Y_N(\cdot ;{\textbf{t}},{\textbf{z}})$ is the matrix (2.4) for the potential $V_{{\textbf{t}},{\textbf{z}}}$. The last contour integral can be evaluated by a residue computation as

$$\begin{aligned} \partial _{t}\log {\mathscr {Z}}_N({\textbf{t}},{\textbf{z}})&={}-\mathop {\textrm{res}}\limits _{\zeta =z}\left[ \textrm{tr}\,\left( Y^{-1}_N(\zeta ;{\textbf{t}},{\textbf{z}})\frac{\partial Y_N(\zeta ;{\textbf{t}},{\textbf{z}})}{\partial \zeta }\textrm{E}_{1,1}\right) \frac{\mathrm d\zeta }{z-\zeta }\right] \nonumber \\&\quad -\mathop {\textrm{res}}\limits _{\zeta =\infty }\left[ \textrm{tr}\,\left( Y^{-1}_N(\zeta ;{\textbf{t}},{\textbf{z}})\frac{\partial Y_N(\zeta ;{\textbf{t}},{\textbf{z}})}{\partial \zeta }\textrm{E}_{1,1}\right) \frac{\mathrm d\zeta }{z-\zeta }\right] . \end{aligned}$$

(2.22)

Remark 2.4

In the last expression (and similarly below), the residue at infinity is a formal residue, namely the limit of integrals on the upper and lower semicircles $|\zeta |=R$, $\pm \textrm{Im}\,\zeta > 0$, as $R\rightarrow +\infty$. Even if the integrand is discontinuous across the real axis, the limit is nevertheless given by (minus) the coefficient of the term $\zeta ^{-1}$ in the asymptotic expansion at $\zeta =\infty$, as the integrand has the same asymptotic expansion as $\zeta \rightarrow \infty$ in both sectors.

It can be checked from (2.6) that the residue at $\zeta =\infty$ vanishes. Therefore

$$\begin{aligned} \partial _{t}\log {\mathscr {Z}}_N({\textbf{t}},{\textbf{z}})=\textrm{tr}\,\left( Y^{-1}_N(z;{\textbf{t}},{\textbf{z}})\left. \frac{\partial Y_N(\zeta ;{\textbf{t}},{\textbf{z}})}{\partial \zeta }\right| _{\zeta =z}\textrm{E}_{1,1}\right) . \end{aligned}$$

(2.23)

Evaluating this identity at $t=0$, taking into account (2.2), we obtain exactly (1.7).

2.3 Case $\ell =2$

Let us first formulate a result that will be needed for all $\ell \ge 2$. Let

$$\begin{aligned} R(\zeta ;z,t):=Y_N(\zeta ;{\textbf{t}},{\textbf{z}})\textrm{E}_{1,1}Y_N^{-1}(\zeta ;{\textbf{t}},{\textbf{z}}),\qquad \zeta \in {\mathbb {C}}\setminus I, \end{aligned}$$

(2.24)

where, again, $Y_N(\cdot ;{\textbf{t}},{\textbf{z}})$ is the matrix (2.4) for the potential $V_{{\textbf{t}},{\textbf{z}}}$.

Lemma 2.5

Let $\ell \ge 2$, ${\textbf{t}}=(t_1,\dots ,t_\ell )$, ${\textbf{z}}=(z_1,\dots ,z_\ell )$, and $V_{{\textbf{t}},{\textbf{z}}}(x)=V(x)+\sum _{i=1}^\ell \tfrac{t_i}{z_i-x}$. For all $1\le j\le \ell$, we have

$$\begin{aligned} \frac{1}{z_j-\zeta }R(\zeta ;{\textbf{t}},{\textbf{z}})+\left( \frac{\partial Y_N}{\partial t_j}(\zeta ;{\textbf{t}},{\textbf{z}})\right) Y_N^{-1}(\zeta ;{\textbf{t}},{\textbf{z}})=\frac{1}{z_j-\zeta }R(z_j;{\textbf{t}},{\textbf{z}}). \end{aligned}$$

(2.25)

Proof

Let us denote by $\Omega _j(\zeta ;{\textbf{t}},{\textbf{z}})$ the left-hand side of (2.25). Using (2.5) we get the identities, for $x\in I^\circ$,

$$\begin{aligned} Y_{N,+}(x;{{\textbf {t}}},{{\textbf {z}}})&=Y_{N,-}(x ;{{\textbf {t}}},{{\textbf {z}}})\begin{pmatrix} 1 &{}{} \text {e}^{V_{{{\textbf {t}}},{{\textbf {z}}}}(x)} \\ 0 &{}{} 1 \end{pmatrix}\,,\end{aligned}$$

(2.26)

$$\begin{aligned} \frac{\partial Y_{N,+}}{\partial t_j}(x;{{\textbf {t}}},{{\textbf {z}}})&=\frac{\partial Y_{N,-}}{\partial t_j}(x;{{\textbf {t}}},{{\textbf {z}}})\begin{pmatrix} 1 &{}{} \text {e}^{V_{{{\textbf {t}}},{{\textbf {z}}}}(x)} \\ 0 &{}{} 1 \end{pmatrix}+Y_{N,-}(x ;{{\textbf {t}}},{{\textbf {z}}})\begin{pmatrix} 0 &{}{} \frac{1}{z_j- x}\text {e}^{V_{{{\textbf {t}}},{{\textbf {z}}}}(x)} \\ 0 &{}{} 0 \end{pmatrix}\,\end{aligned}$$

(2.27)

from which we readily ascertain that $\Delta \Omega _j(x;{\textbf{t}},{\textbf{z}})=0$ for all $x\in I^\circ$. Hence, $\Omega _j(\zeta ;{\textbf{t}},{\textbf{z}})$ is a meromorphic function of $\zeta$ with a single simple pole at $\zeta =z_j$ and which vanishes at $\zeta =\infty$, because of (2.6), and so the statement follows. (Singularities at the endpoints of I are ruled out by our assumptions on V.) $\square$

Let us consider the case $\ell =2$, in which ${\textbf{t}}=(t_1,t_2)$, ${\textbf{z}}=(z_1,z_2)$, and $V_{{\textbf{t}},{\textbf{z}}}(x)=V(x)+\tfrac{t_1}{z_1-x}+\tfrac{t_2}{z_2-x}$. By the argument used for $\ell =1$, cf. (2.23), we obtain

$$\begin{aligned} \partial _{t_1}\log {\mathscr {Z}}_N({\textbf{t}},{\textbf{z}})=\textrm{tr}\,\left( Y^{-1}_N(z_1;{\textbf{t}},{\textbf{z}})\left. \frac{\partial Y_N(\zeta ;{\textbf{t}},{\textbf{z}})}{\partial \zeta }\right| _{\zeta =z_1}\textrm{E}_{1,1}\right) . \end{aligned}$$

(2.28)

Next we have to take a derivative in $t_2$: omitting the explicit dependence on ${\textbf{t}},{\textbf{z}}$, we have

$$\begin{aligned}{} & {} \partial _{t_2}\partial _{t_1}\log {\mathscr {Z}}_N({\textbf{t}},{\textbf{z}})\nonumber \\{} & {} \quad = \textrm{tr}\,\biggl (-Y^{-1}_N(z_1)\frac{\partial Y_N}{\partial t_2}(z_1)Y_N^{-1}(z_1)\left. \frac{\partial Y_N(\zeta )}{\partial \zeta }\right| _{\zeta =z_1}\textrm{E}_{1,1} +Y^{-1}_N(z_1)\left. \frac{\partial ^2 Y_N(\zeta )}{\partial t_2\partial \zeta }\right| _{\zeta =z_1}\textrm{E}_{1,1}\biggr ).\nonumber \\ \end{aligned}$$

(2.29)

We use (2.25) to rewrite the first term inside the trace in the right-hand side as

$$\begin{aligned} -Y^{-1}_N(z_1)\frac{R(z_2)-R(z_1)}{z_2-z_1}\left. \frac{\partial Y_N(\zeta )}{\partial \zeta }\right| _{\zeta =z_1}\textrm{E}_{1,1} \end{aligned}$$

(2.30)

and the second term as

$$\begin{aligned} Y^{-1}_N(z_1)\frac{\partial ^2 Y_N(\zeta )}{\partial \zeta \partial t_2}\bigg |_{\zeta =z_1}\textrm{E}_{1,1}&= Y^{-1}_N(z_1)\frac{\partial }{\partial \zeta }\biggl (\frac{R(z_2)-R(\zeta )}{z_2-\zeta }Y_N(\zeta )\biggr )\bigg |_{\zeta =z_1}\textrm{E}_{1,1}\nonumber \\&=Y^{-1}_N(z_1)\biggl ( \frac{R(z_2)-R(z_1)}{(z_2-z_1)^2}Y_N(z_1) \nonumber \\&\quad -\frac{\biggl [\frac{\partial Y_N(\zeta )}{\partial \zeta }\bigg |_{\zeta =z_1}Y^{-1}_N(z_1),R(z_1)\biggr ]}{z_2-z_1}Y_N(z_1) \nonumber \\&\quad +\frac{R(z_2)-R(z_1)}{z_2-z_1}\frac{\partial Y_N(\zeta )}{\partial \zeta }\bigg |_{\zeta =z_1}\biggr )\textrm{E}_{1,1}, \end{aligned}$$

(2.31)

where $[A,B]:=AB-BA$ is the commutator. The term in the last row exactly cancels with (2.30), and so, rearranging terms,

$$\begin{aligned} \partial _{t_2}\partial _{t_1}\log {\mathscr {Z}}_N({\textbf{t}},{\textbf{z}})=\frac{\textrm{tr}\,\bigl (R(z_1)R(z_2)\bigr )-1}{(z_2-z_1)^2}+ \frac{\textrm{tr}\,\left( \biggl [Y^{-1}_N(z_1)\displaystyle {\frac{\partial Y_N(\zeta )}{\partial \zeta }}\bigg |_{\zeta =z_1},\textrm{E}_{1,1}\biggr ]\textrm{E}_{1,1}\right) }{z_2-z_1}.\nonumber \\ \end{aligned}$$

(2.32)

Since $\textrm{tr}\,([A,B]B)=\textrm{tr}\,([AB,B])=0$, the proof of the case $\ell =2$ is completed by setting $t_1=t_2=0$.

2.4 Case $\ell \ge 3$

As usual, let ${\textbf{t}}=(t_1,\dots ,t_\ell )$ and ${\textbf{z}}=(z_1,\dots ,z_\ell )$. Let us denote, for $\ell \ge 2$,

$$\begin{aligned} S_\ell (z_1,\dots ,z_\ell ;{\textbf{t}}):=-\sum _{(i_1,\dots ,i_\ell )\in \textrm{cyc}((\ell ))}\frac{\textrm{tr}\,\left( R(z_{i_1};{\textbf{t}},{\textbf{z}})\cdots R(z_{i_\ell };{\textbf{t}},{\textbf{z}})\right) }{(z_{i_1}-z_{i_2})\cdots (z_{i_{\ell -1}}-z_{i_\ell })(z_{i_\ell }-z_{i_1})}-\frac{\delta _{\ell ,2}}{(z_1-z_2)^2},\nonumber \\ \end{aligned}$$

(2.33)

where the sum extends over cyclic permutations of $\lbrace 1,\dots ,\ell \rbrace$. We aim at proving that

$$\begin{aligned} \frac{\partial ^\ell \log {\mathscr {Z}}_N({\textbf{t}},{\textbf{z}})}{\partial t_{\ell }\cdots \partial t_{1}}=S_\ell (z_1,\dots ,z_\ell ;{\textbf{t}}), \end{aligned}$$

(2.34)

where $Y_N(x;{\textbf{t}},{\textbf{z}})$, and so $R(x;{\textbf{t}},{\textbf{z}})$, are computed for the potential $V_{{\textbf{t}},{\textbf{z}}}(x)=V(x)+\sum _{i=1}^\ell \frac{t_i}{z_i-x}$. Then, (1.9) follows by taking $t_i=0$.

The proof of (2.34) is by induction on $\ell \ge 2$ and it is similar in spirit to that in [1]. Hence, let us assume (2.34) for some $\ell \ge 2$ and let us prove it for $\ell +1$. Since the potential V is arbitrary, by replacing V with $V+t_{\ell +1}/(z_{\ell +1}-x)$ we can assume (2.34) holds true for $V_{{\textbf{t}},{\textbf{z}}}(x)=V(x)+\sum _{j=1}^{\ell +1}\frac{t_{j}}{z_{j}-x}$, and so we just have to show that $\partial _{t_{\ell +1}}S_\ell (z_1,\dots ,z_\ell ;{\textbf{t}})$ is equal to $S_{\ell +1}(z_1,\dots ,z_\ell ,z_{\ell +1};{\textbf{t}})$. To this end we first observe that by (2.25), we have

$$\begin{aligned} \frac{\partial R(\zeta ;{\textbf{t}},{\textbf{z}})}{\partial t_j}=\biggl [\frac{R(z_j;{\textbf{t}},{\textbf{z}})-R(\zeta ;{\textbf{t}},{\textbf{z}})}{z_j-\zeta },R(\zeta ;{\textbf{t}},{\textbf{z}})\biggr ]=\frac{[R(z_j;{\textbf{t}},{\textbf{z}}),R(\zeta ;{\textbf{t}},{\textbf{z}})]}{z_j-\zeta }.\nonumber \\ \end{aligned}$$

(2.35)

Therefore,

$$\begin{aligned}{} & {} \frac{\partial S_\ell (z_1,\dots ,z_\ell ;{\textbf{t}})}{\partial t_{\ell +1}}\nonumber \\{} & {} \quad =-\sum _{(i_1,\dots ,i_\ell )\in \textrm{cyc}((\ell ))}\sum _{j=1}^\ell \frac{\textrm{tr}\,\left( R(z_{i_1};{\textbf{t}},{\textbf{z}})\cdots [R(z_{\ell +1};{\textbf{t}},{\textbf{z}}),R(z_{i_j};{\textbf{t}},{\textbf{z}})]\cdots R(z_{i_\ell };{\textbf{t}},{\textbf{z}})\right) }{(z_{i_1}-z_{i_2})\cdots (z_{i_\ell }-z_{i_1})(z_{\ell +1}-z_{i_j})}.\nonumber \\ \end{aligned}$$

(2.36)

Expanding $[R(z_{\ell +1};{\textbf{t}},{\textbf{z}}),R(z_{i_j};{\textbf{t}},{\textbf{z}})]=R(z_{\ell +1};{\textbf{t}},{\textbf{z}})R(z_{i_j};{\textbf{t}},{\textbf{z}})-R(z_{i_j};{\textbf{t}},{\textbf{z}}) R(z_{\ell +1};{\textbf{t}},{\textbf{z}})$, we note that in the previous sum, each term involving the expression

$$\begin{aligned} \textrm{tr}\,\left( R(z_{i_1};{\textbf{t}},{\textbf{z}})\cdots R(z_{\ell +1};{\textbf{t}},{\textbf{z}})R(z_{i_j};{\textbf{t}},{\textbf{z}})\cdots R(z_{i_\ell };{\textbf{t}},{\textbf{z}})\right) \end{aligned}$$

(2.37)

appears twice, but with different denominators. Collecting such terms yields

$$\begin{aligned}&\sum _{(i_1,\dots ,i_\ell )\in \textrm{cyc}((\ell ))}\sum _{j=1}^\ell \frac{\textrm{tr}\,\bigl (R(z_{i_1};{\textbf{t}},{\textbf{z}})\cdots R(z_{\ell +1};{\textbf{t}},{\textbf{z}})R(z_{i_j};{\textbf{t}},{\textbf{z}})\cdots R(z_{i_\ell };{\textbf{t}},{\textbf{z}})\bigr )}{(z_{i_1}-z_{i_2})\cdots (z_{i_\ell }-z_{i_1})}\nonumber \\&\qquad \times \biggl (\frac{1}{z_{i_j}-z_{\ell +1}}-\frac{1}{z_{i_{j-1}}-z_{\ell +1}}\biggr ) \nonumber \\&\quad =-\sum _{(i_1,\dots ,i_\ell )\in \textrm{cyc}((\ell ))}\!\sum _{j=1}^\ell \frac{\textrm{tr}\,\!\left( R(z_{i_1};{\textbf{t}},{\textbf{z}})\cdots R(z_{\ell +1};{\textbf{t}},{\textbf{z}})R(z_{i_j};{\textbf{t}},{\textbf{z}})\cdots R(z_{i_\ell };{\textbf{t}},{\textbf{z}})\right) }{(z_{i_1}-z_{i_2})\cdots (z_{i_{j-1}}-z_{\ell +1})(z_{\ell +1}-z_{i_j}) \cdots (z_{i_\ell }-z_{i_1})}\nonumber \\&\quad =S_{\ell +1}(z_1,\dots ,z_\ell ,z_{\ell +1};{\textbf{t}}), \end{aligned}$$

(2.38)

where we set $i_0:=i_\ell$ in the internal summation. The proof is complete.