Edge Scaling of the β-Jacobi Ensemble

Abstract

We study the scaling limit of the spectrum of the β-Jacobi ensemble at the soft edge and hard edge for general values of β. We show that the limiting point processes correspond respectively to the stochastic Airy and Bessel point processes introduced in Ramírez et al. (J. Am. Math. Soc. 24(4):919–944, 2011) and Ramírez and Rider (Commun. Math. Phys. 288(3):887–906, 2009).

Appendix: Proof of Theorem 2

We will be working primarily with a lower bidiagonal matrix M, and the symmetric tridiagonal matrix MM ^T. For convenience we will adopt the following notations. Denote the diagonal entries of our symmetric tridiagonal matrix MM ^T by $\underline {a}= \{a_{1},\ldots,a_{n}\}$ and its off-diagonal entries by $\underline {b}=\{b_{1},\ldots, b_{n-1} \}$. For the entries of the bi-diagonal matrix M use $\underline {x}= \{x_{1},\ldots,x_{n}\}$ to denote the diagonal entries and $\underline {y}=\{ y_{1},\ldots,y_{n-1} \}$ for its sub-diagonal entries. We will also use another tridiagonal matrix which arises in the following lemma (see e.g. [7]):

Lemma 23

Let M be an n×n bidiagonal matrix with a ₁,a ₂,…,a _n in the diagonal and b ₁,b ₂,…,b _n−1 in the off-diagonal. Consider the 2n×2n symmetric tridiagonal matrix L which has zeros in the main diagonal and a ₁,b ₁,a ₂,b ₂,…,a _n above and below the diagonal. If the singular values of M are λ ₁,λ ₂,…,λ _n then the eigenvalues of L are ±λ _i ,i=1…n.

Recall the definitions of M _β,n ,C _k and $\tilde{C}_{k}$ from the statement of Theorem 2. It will be convenient write $\{C_{i},S_{i}\}_{i=1}^{n}$ and $\{\tilde{C}_{i}, \tilde{S}_{i}\}_{i=1}^{n-1}$ in terms of a set of random angles $\underline {\alpha }= \{ \alpha_{1}, \ldots, \alpha_{n} \}$ and $\underline {\theta }=\{\theta_{1},\ldots, \theta_{n-1}\}$ where C _k =cos(α _k ) and $\tilde{C}_{k} = \cos(\theta_{k})$. Then, S _k =sin(α _k ) and $\tilde{S}_{k} = \sin(\theta_{k})$, and the densities of the random angles α _k and θ _k are:

with a,b defined as in Theorem 2 and B(x,y) the beta function. By independence, the joint density of $(\underline {\alpha }, \underline {\theta })$ is given by the product of the densities. For convenience we will denote

$$\tilde{Z}_{\beta,n}= \frac{2^{2n-1}}{\prod_{k=1}^nB ( \frac {\beta}{2}(n_1-n+k), \frac{\beta}{2}(n_2-n+k) ) \prod_{k=1}^{n-1}B ( \frac {\beta}{2}k, \frac{\beta}{2}(n_1+n_2-2n+k+1) )} $$

This will be the necessary normalizing constant.

We will now map $(\underline {\alpha }, \underline {\theta })$ to the entries of M and from there to $(\underline {\lambda }, \underline {q})$ where the λ _i are the eigenvalues of MM ^T and the q _i are the positive leading entries of the corresponding normalized eigenvectors. This second map will actually by the composition of several maps.

From the angles we map to the entries of the bidiagonal matrix:

Lemma 24

The Jacobian of the transform $T:(\underline {\alpha }, \underline {\theta }) \to (\underline {x}, \underline {y})$, where

(for notational convenience we take θ _n =π/2) is given by

Proof

The matrix of the partial derivatives with ordering (α ₁,…,α _n ,θ ₁,…,θ _n−1)→(x ₁,…,x _n ,y ₁,…,y _n−1) has diagonal entries

$$- \sin\alpha_1 \sin\theta_1,\ldots.,-\sin \alpha_{n-1}\sin\theta_{n-1}, -\sin\alpha_n, - \sin \alpha_2 \sin\theta_1,\ldots, -\sin\alpha_n \sin \theta_{n-1}. $$

Row and column reduction gives us that the determinant is given by the product of the diagonal, therefore

$$J_T= -\frac{\sin^2(\alpha_n)}{\sin(\alpha_1)} \prod_{k=1}^{n-1} \sin^2(\alpha_k)\sin^2(\theta_k) $$

 □

We finish by mapping from the bidiagonal to the tridiagonal matrix, which in turn is mapped to by the eigenvalues. Denote by q ₁,…,q _n the leading entries of the eigenvectors associated with the ordered eigenvalues λ ₁<λ ₂<⋯<λ _n of the tridiagonal matrix normalized so that q _i >0 and $\sum q_{i}^{2} =1$.

Lemma 25

([6], Lemmas 2.7, 2.9 and 2.11)

For $\underline {x},\underline {y},\underline {a},\underline {b},\underline {\lambda }$ and $\underline {q}$ defined as above we have the following:

• (1) The Jacobian of the map $\psi: (\underline {x}, \underline {y}) \to(\underline {a}, \underline {b})$ can be written as

  $$J_\psi= 2^n x_1 \prod _{i=2}^{n} x_i^2. $$

• (2) The Vandermonde determinant for the ordered eigvenvalues of a symmetric tridiagonal matrix with positive sub-diagonal b=(b _n−1,…,b ₁) is given by

  $$\Delta( \underline {\lambda })= \prod_{i<j}( \lambda_i-\lambda_j)= \frac{\prod_{i=1}^{n-1} b_i^i}{\prod_{i=1}^n q_i}. $$

• (3) The Jacobian of the map $\phi:(\underline {a}, \underline {b}) \to(\underline {\lambda }, \underline {q})$ can be written as

  $$J_\phi= \frac{\prod_{i=1}^{n-1} b_i}{\prod_{i=1}^n q_i}. $$

Written in terms of our $\underline {x}, \underline {y}, \underline {a}$, and $\underline {b}$ with $Q_{n}= \prod_{i=1}^{n} q_{i}$ we get

The remainder of the proof of Theorem 2, is to show that the Jacobian of the transformation gives the desired result.

We substitute in the following pieces: first, notice that

$$(\det M_{n,\beta})^2 = \prod_{i=1}^n \lambda_i = \prod_{k=1}^n \cos^2 \alpha_k \cdot\prod_{k=1}^{n-1} \sin^2 \theta_k. $$

Then, applying Lemma 23, the singular values of M _n,β are the squares of the eigenvalues of the symmetric tridiagonal matrix L with zeroes in the main diagonal and

$$C_n, S_n \tilde{C}_{n-1}, C_{n-1} \tilde{S}_{n-1}, \dots, S_2 \tilde{C}_1, C_1 \tilde{S}_1, $$

in the off-diagonal. We can check that L+I can be written as AA ^T where A is the bidiagonal matrix with

$$1, S_n, \tilde{S}_{n-1}, S_{n-1}, \dots, S_1, \quad\mbox{and}\quad C_n, \tilde{C}_{n-1}, C_{n-1}, \tilde{C}_{n-2}, \dots, C_1 $$

in the diagonal and below the diagonal respectively. Using the characterization from Lemma 23, we find that

$$\det(L+I) = \prod_{{}^{k=1,\ldots,n}_{ \epsilon=+,-}} 1 + \epsilon \sqrt{ \lambda_k} = \prod^n_{k=1} (1- \lambda_k) = (\det A)^2 = \prod _{k=1}^n \sin^2 \alpha_k \prod_{k=1}^{n-1} \sin^2 \theta_k. $$

Remark 26

At this point one could again apply Lemma 23 to A to find a 4n×4n matrix with zeros in the diagonal and

$$1, C_n , S_n, \tilde{C}_{n-1}, \tilde{S}_{n-1}, C_{n-1},\ldots,C_1,S_1 $$

above and below the diagonal. This matrix will have eigenvalue $\pm \sqrt{1\pm\sqrt{\lambda_{k}}}$.

Lastly recall ([6], Lemma 2.7) that we have

$$\Delta(\lambda)= \frac{1}{Q_n} \prod_{k=1}^{n-1} b_k^k= \frac {1}{Q_n}\prod _{k=1}^n \sin^{k-1}\alpha_k \cos^{k-1}\alpha_k \cdot\prod_{k=1}^{n-1} \cos^k\theta_k \cdot\prod_{k=1}^{n-1} \sin^{k-1}\theta_k. $$

Making all the appropriate substitutions this gives us that

From this we can see that the joint density function of $\underline {q}$ and $\underline {\lambda }$ separate. As in the Hermite and Laguerre cases we have that $\underline {q}\sim( \chi_{\beta},\ldots,\chi_{\beta})$ normalized to unit length [6]. This give us that for the unordered eigenvalues

$$f_{\beta,n,n_1,n_2}(\underline {\lambda })= \frac{\tilde{Z}_{\beta,n}}{2^n n!}\frac{ [\varGamma(\frac{\beta}{2}) ]^n}{\varGamma( \frac{\beta n}{2} )}\prod _{i<j}|\lambda_i- \lambda_j|^\beta \cdot\prod_{k=1}^n \lambda_k^a \cdot\prod_{k=1}^n (1- \lambda_k)^b. $$

This completes the proof of Theorem 2.  □

Remark 27

The normalizing constant on the final density can be written as

$$Z_{\beta,n} = \biggl[ \varGamma\biggl(\frac{\beta}{2} \biggr) \biggr]^n \prod_{k=1}^n \frac{\varGamma ( \frac{\beta}{2} (n_1+n_2-n+k) )}{\varGamma( \frac{\beta}{2}(n_1-n+k) )\varGamma( \frac{\beta }{2}(n_2-n+k) )\varGamma( \frac{\beta}{2}k )}. $$

This gives an alternate derivation for the Selberg integral.