Universality of the Stochastic Bessel Operator

Abstract

We establish universality at the hard edge for general beta ensembles assuming that: the background potential V is a polynomial such that $x \mapsto V(x^2)$ is strongly convex, $\beta \ge 1$, and the “dimension-difference” parameter $a\ge 0$. The method rests on the corresponding tridiagonal matrix models, showing that their appropriate continuum scaling limit is given by the Stochastic Bessel Operator. As conjectured in Edelman and Sutton (J Stat Phys 127:1121–1165, 2007) and rigorously established in Ramírez and Rider (Commun Math Phys 288:887–906, 2009), the latter characterizes the hard edge in the case of linear potential and all $\beta$ (the classical “beta-Laguerre” ensembles).

Appendix

We include here the derivation that our matrix model $B(X,Y) B(X,Y)^{T}$ [with (X, Y) sampled from the measure P] realizes the joint eigenvalue density (1.1). To simplify notation a bit we take $c \prod _{i < j} |\lambda _i - \lambda _j |^{\beta } \prod _{i=1}^n \lambda _i^{\gamma } e^{-V(\lambda _i)}$ as the target density, with any $\gamma > -1$ and polynomial V.

Also, to make a more direct connection with the derivation of the $\beta$-Laguerre ensemble one finds in the literature (in say [9]) consider first an upper bidiagonal matrix M with coordinates labeled in decreasing order: $M_{i,i} = x_{n-i+1}$ for $i=1, \dots , n$ and $M_{i, i+1} = y_{n-i}$ for $i =1,\dots , n-1$, with all $x_i$, $y_i$ positive. Also introduce the tridiagonal coordinates through a Jacobi matrix $T = T(a, b)$ with $T_{i,i} = a_{n-i+1}$ for $i=1, \dots , n$ and $T_{i, i+1} = T_{i+1, i} = b_{n-i}$ for $i = 1, \dots , n-1$. Here each $a_i \in {\mathbb {R}}$ and each $b_i \in {\mathbb {R}}_+$. We track the calculation from eigenvalue/eigenvector coordinates to (x, y) coordinates via $Q \Lambda Q^{\dagger } = T = M M^T$. Here Q is the eigenvector matrix, of which we only need the first components. These can be chosen to be real positive, and are denoted $(q_1, \dots , q_{n-1})$, noting that $q_n$ is specified by $\sum _{i=1}^n q_i^2 = 1$.

Next, we have that the Jacobians for the maps from $(\lambda , q)$ to (a, b), and then from (a, b) to (x, y) are given by

$$\begin{aligned} J = q_n \frac{\prod _{i=1}^n q_i}{\prod _{i=1}^{n-1} b_i}, \quad J' = 2^n x_1 \prod _{i=2}^{n} x_i^2, \end{aligned}$$

respectively. See [13, Eq. 1.156] for the former. The latter is derived from the identities $a_i = x_i^2 + y_{i}^2$ and $b_i = x_{i+1} y_{i}$ (where $y_n = 0$ is understood). We will also need the well-known relation,

$$\begin{aligned} \prod _{i <j} (\lambda _i -\lambda _j)^2 = \frac{\prod _{i=1}^{n-1} b_i^{2i} }{\prod _{i=1}^{n} q_i^2}, \end{aligned}$$

for which see [13, Eq. 1.148].

Since we obviously have that $\sum _{i=1}^n V(\lambda _i) = \mathrm {tr}V(M M^T)$, the necessary computation is:

$$\begin{aligned}&\left( \prod _{i=1}^n \lambda _i \right) ^{\gamma } \prod _{i < j} |\lambda _i - \lambda _j |^{\beta } \left( q_n^{-1} \prod _{i=1}^n q_i^{\beta -1} \right) \, d q \wedge d \lambda \\&\quad = \left( \prod _{i=1}^n x_i^{2 \gamma } \right) \left( \frac{\prod _{i=1}^{n-1} (x_{i+1} y_i)^{\beta i}}{\prod _{i=1}^{n-1} q_i^{\beta }} \right) J J' \, dx \wedge dy \\&\quad = 2^n \prod _{i=1}^{n} x_i^{2 \gamma + \beta (i-1) +1} \prod _{i=1}^{n-1} y_i^{\beta i -1} dx \wedge dy. \end{aligned}$$

Putting in $\gamma = \frac{\beta }{2}(a+1) - 1$ we recognize the factors in $x_i^{ \beta (a+i) -1}$ and $y_i^{\beta i -1}$ in the claimed bidiagonal matrix density (1.5). Here we have decided to work with $B = S M S^{-1}$ where S is the antidiagonal matrix of alternating signs. This transformation does not effect the joint density of the individual coordinates, and the eigenvalues of $B B^T$ and $M M^T$ agree.