Correction To: Jacobi Ensemble, Hurwitz Numbers and Wilson Polynomials

(1.1) Here CN := ∫ HN (I ) exp tr V (X)dX and V = V (x) is a smooth function of x ∈ I ◦ (the interior of I ) so that V (X) is defined for all X ∈ HN (I ) by the spectral theorem. We assume that V satisfies the following decay assumptions: there exists ε > 0 such that exp V (x) = O |x − x0|−1+ε ) as x ∈ I ◦ approaches a finite endpoint x0 of I ; if I extends to ±∞ we assume that V (x) → −∞ fast enough as x → ±∞ in order for the measure (1.1) to have finite moments of all orders. In particular, the family


Introduction
Denote H N (I ) the set of Hermitian matrices of size N = 1, 2, . . . with eigenvalues in the (closed) interval I ⊆ R, endowed with the probability measure (1. 2) The generating functions of moments are analytic functions of z 1 , . . . , z ∈ C \ I , symmetric in the variables z 1 , . . . , z . The generating functions for cumulants (or, connected moments) are which is the well-known solution to the Riemann-Hilbert problem for orthogonal polynomials [4]; it is an analytic function of z ∈ C \ I .
with Y N (z) as in (1.5). Then the cumulant generating functions (1.4) are given by where is the derivative with respect to z and cyc(( )) is the set of -cycles in the symmetric group S .

Proof of Theorem 1.1
The strategy of the proof is based on the observation that setting Here and below it is assumed that z i / ∈ I .

Orthogonal polynomials on the real line and unitary-invariant ensembles
We denote P the monic orthogonal polynomials, h = I P 2 (x)e V (x) dx, see (1.2), and their Cauchy transforms. The matrix , (2.4) introduced in (1.5), is an analytic function of ζ ∈ C \ I . It satisfies the jump condition where we denote 1 = 1 0 0 1 and σ 3 = 1 0 0 −1 . Lastly, we recall the Christoffel-Darboux kernel 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 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] where it is convenient to explicitly express the dependence of P = P V and h = h V on the potential V . Therefore, introducing the modified potential but the first term vanishes by orthogonality because P V t,z i (x) are normalized to be monic and, therefore, is a polynomial of degree strictly less than i.

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 Z N (t, 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 t, z. However, the condition for existence (non-vanishing of the Hankel determinants of the moments) is open in the space of parameters t, z, and contains the subspace t = 0 by standard arguments pertaining to the classical real-valued case (in which the Hankel matrices are positive-definite). Therefore, for sufficiently small t, the existence of P V t,z i (x) is not an issue and we will restrict to sufficiently small t without further mention in what follows, as in the end we are interested in the quantities (2.2).

Case = 1
It follows from (2.8) that (2.14) In the following we shall use the notation for the jump of a function f across I , namely f ± (x) := lim →0 + f (x ± i ). 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
Proof Let us denote := ∂ x . It follows from the jump condition (2.5) for Y N that (2.17) Therefore we compute The last term vanishes and so, by the cyclic property of the trace, we have which is easily seen to be equivalent to (2.14).
We are ready for the proof of the case = 1. In such case t = t, z = z, and V t,z (x) = V (x) + t/(z − x). By (2.11), Lemma 2.1, and (2.7), we have where we denote explicitly the dependence of the Christoffel-Darboux kernel on the potential. Let be an oriented contour in the complex plane which surrounds I in counterclockwise sense (i.e., I lies on the left of ) and leaves z outside (i.e., z lies to the right of ). Then, using Lemma 2.3, we get , (2.21) where Y N (·; t, z) is the matrix (2.4) for the potential V t,z . The last contour integral can be evaluated by a residue computation as

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 |ζ | = R, ±Im ζ > 0, as R → +∞. Even if the integrand is discontinuous across the real axis, the limit is nevertheless given by (minus) the coefficient of the term ζ −1 in the asymptotic expansion at ζ = ∞, as the integrand has the same asymptotic expansion as ζ → ∞ in both sectors.
It can be checked from (2.6) that the residue at ζ = ∞ vanishes. Therefore Evaluating this identity at t = 0, taking into account (2.2), we obtain exactly (1.7).

Lemma 2.5 Let
Proof Let us denote by j (ζ ; t, z) the left-hand side of (2.25). Using (2.5) we get the identities, for x ∈ I • , from which we readily ascertain that j (x; t, z) = 0 for all x ∈ I • . Hence, j (ζ ; t, z) is a meromorphic function of ζ with a single simple pole at ζ = z j and which vanishes at ζ = ∞, because of (2.6), and so the statement follows. (Singularities at the endpoints of I are ruled out by our assumptions on V .) Let us consider the case = 2, in which t = (t 1 , t 2 ), z = (z 1 , z 2 ), and V t,z (x) = V (x) + t 1 z 1 −x + t 2 z 2 −x . By the argument used for = 1, cf. (2.23), we obtain Next we have to take a derivative in t 2 : omitting the explicit dependence on t, z, we have (2.29) We use (2.25) to rewrite the first term inside the trace in the right-hand side as and the second term as

Case ≥ 3
As usual, let t = (t 1 , . . . , t ) and z = (z 1 , . . . , z ). Let us denote, for ≥ 2, tr R(z i 1 ; t, z) · · · R(z i ; t, z) where the sum extends over cyclic permutations of {1, . . . , }. We aim at proving that ∂t · · · ∂t 1 = S (z 1 , . . . , z ; t), (2.34) where Y N (x; t, z), and so R(x; t, z), are computed for the potential Then, (1.9) follows by taking t i = 0. The proof of (2.34) is by induction on ≥ 2 and it is similar in spirit to that in [1]. Hence, let us assume (2.34) for some ≥ 2 and let us prove it for + 1. Since the potential V is arbitrary, by replacing V with V + t +1 /(z +1 − x) we can assume (2.34) holds true for V t,z (x) = V (x) + +1 j=1 t j z j −x , and so we just have to show that ∂ t +1 S (z 1 , . . . , z ; t) is equal to S +1 (z 1 , . . . , z , z +1 ; t). To this end we first observe that by (2.25), we have (2.35) Therefore, tr R(z i 1 ; t, z) · · · [R(z +1 ; t, z), R(z i j ; t, z)] · · · R(z i ; t, z) (2.36) Expanding [R(z +1 ; t, z), R(z i j ; t, z)] = R(z +1 ; t, z)R(z i j ; t, z) − R(z i j ; t, z)R(z +1 ; t, z), we note that in the previous sum, each term involving the expression tr R(z i 1 ; t, z) · · · R(z +1 ; t, z)R(z i j ; t, z) · · · R(z i ; t, z) (2.37) appears twice, but with different denominators. Collecting such terms yields tr R(z i 1 ; t, z) · · · R(z +1 ; t, z)R(z i j ; t, z) · · · R(z i ; t, z) Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.