Near-Extreme Eigenvalues and the First Gap of Hermitian Random Matrices

Abstract

We study the phenomenon of “crowding” near the largest eigenvalue $\lambda _\mathrm{max}$ of random $N \times N$ matrices belonging to the Gaussian Unitary Ensemble of random matrix theory. We focus on two distinct quantities: (i) the density of states (DOS) near $\lambda _\mathrm{max}$, $\rho _\mathrm{DOS}(r,N)$, which is the average density of eigenvalues located at a distance $r$ from $\lambda _\mathrm{max}$ and (ii) the probability density function of the gap between the first two largest eigenvalues, $p_\mathrm{GAP}(r,N)$. In the edge scaling limit where $r = \mathcal{O}(N^{-1/6})$, which is described by a double scaling limit of a system of unconventional orthogonal polynomials, we show that $\rho _\mathrm{DOS}(r,N)$ and $p_\mathrm{GAP}(r,N)$ are characterized by scaling functions which can be expressed in terms of the solution of a Lax pair associated to the Painlevé XXXIV equation. This provides an alternative and simpler expression for the gap distribution, which was recently studied by Witte et al. in Nonlinearity 26:1799, 2013. Our expressions allow to obtain precise asymptotic behaviors of these scaling functions both for small and large arguments.

Appendices

Appendix 1: Rewriting the Lax System in Terms of Rescaled Variables

In this appendix, we derive two identities relating the Hastings–McLeod solution of Painlevé II with $\alpha = 0$, denoted by $q(x)$ (17), and the one with $\alpha =1/2$, denoted by $q_{1/2}$ (62). We start with the relation between $q(x)$ and $q_{1/2}(x)$ [28, 50]

$$\begin{aligned} q_{1/2}(x) = -2^{-1/3} \frac{q'(-2^{-1/3}x)}{q(-2^{-1/3}x)}. \end{aligned}$$

(123)

By taking the derivative with respect to $x$ on both sides of Eq. (123), one obtains

$$\begin{aligned} q_{1/2}'(x) = 2^{-2/3} \frac{q''(-2^{-1/3}x)}{q(-2^{-1/3}x)} - 2^{-2/3} \left( \frac{q'(-2^{-1/3}x)}{q(-2^{-1/3}x)}\right) ^2. \end{aligned}$$

(124)

By combining Eqs. (123) and (124), one obtains

$$\begin{aligned} x + 2 q_{1/2}^2(x) + 2 q_{1/2}'(x) = 2^{1/3} \left( 2^{-1/3}x + \frac{q''(-2^{-1/3}x)}{q(-2^{-1/3}x)} \right) . \end{aligned}$$

(125)

Finally, using that $q(x)$ is solution of the Painlevé II equation (17) one obtains

$$\begin{aligned} q_{1/2}^2(x) + q_{1/2}'(x) + \frac{x}{2} = 2^{1/3} q^2(-2^{-1/3}x). \end{aligned}$$

(126)

We now derive a second identity by considering the following function

$$\begin{aligned} J(x) = - x - 2 q_{1/2}^2(x) + 2 q_{1/2}'(x) . \end{aligned}$$

(127)

It is straightforward to check that $J(x)$ satisfies

$$\begin{aligned} J'(x) = - 2 q_{1/2}(x) J(x) - 2. \end{aligned}$$

(128)

The solution of the homogenous equation, $J'(x) = - 2 q_{1/2}(x) J(x)$ is, using (123), $J(x) = A/q^2(-2^{-1/3}x)$. By varying the constant, one finds the solution of (128) under the form:

$$\begin{aligned} J(x) = \frac{A(x)}{q^{2}(-2^{-1/3})x}, \; A(x) = 2^{4/3} \int _{-2^{-1/3}x}^\infty q^2(u) du \; + \; a, \end{aligned}$$

(129)

where $a$ is a constant, independent of $x$. On the other hand, from the large $x$ behavior $q_{1/2}(x) \sim 1/(2x)$, one sees that $J(x)$ in (127) behaves like $J(x) \sim -x$, when $x \rightarrow \infty$. This implies that the constant $a$ in Eq. (129) is $a=0$. Hence, one obtains a second identity:

$$\begin{aligned}&- q_{1/2}^2(x) + q_{1/2}'(x) -\frac{x}{2} = -2^{1/3} \frac{\int _{-2^{-1/3}x}^\infty q^2(u) du}{q^2(-2^{-1/3}x)}. \end{aligned}$$

(130)

One can then use these identities (126) and (130) to write the matrix elements of the matrices $\mathbf{A}$ and $\mathbf{B}$ in Eq. (68) in terms of $q(x)$ only—and not $q_{1/2}(x)$. This yields ultimately, with an appropriate change of variable, the expression of the matrices ${\tilde{\mathbf{A}}}$ and ${\tilde{\mathbf{B}}}$ in Eq. (79).

Appendix 2: Expansion of the Solution of the Lax System for Small $\tilde{r}$

In this appendix, we give some details concerning the expansion of $\tilde{\rho }_\mathrm{edge}(\tilde{r})$ [which is actually similar to the one of $\tilde{p}_\mathrm{typ}(\tilde{r})$, see Eq. (23)] for small argument $\tilde{r}$.

1.1 General Structure of the Psi-Functions $\tilde{f}(\tilde{r},x)$ and $\tilde{g}(\tilde{r},x)$ at Small $\tilde{r}$

The small $\tilde{r}$ expansion of $\tilde{\rho }_\mathrm{edge}(\tilde{r})$ necessitates the expansion of the solution of the Lax pair $\tilde{f}(\tilde{r},x)$ and $\tilde{g}(\tilde{r},x)$, which we recall are solutions of the system of differential equations

$$\begin{aligned} \frac{\partial }{\partial \tilde{r}} \left( \begin{array}{c} \tilde{f}(\tilde{r},s) \\ \tilde{g}(\tilde{r},s) \end{array} \right) = {\tilde{\mathbf{A}}}\left( \begin{array}{c} f(\tilde{r},s) \\ g(\tilde{r},s) \end{array} \right) , \; \frac{\partial }{\partial s} \left( \begin{array}{c} \tilde{f}(\tilde{r},s) \\ \tilde{g}(\tilde{r},s) \end{array} \right) = {\tilde{\mathbf{B}}}\left( \begin{array}{c} \tilde{f}(\tilde{r},s) \\ \tilde{g}(\tilde{r},s) \end{array} \right) , \; \end{aligned}$$

(131)

where ${\tilde{\mathbf{A}}}$ and ${\tilde{\mathbf{B}}}$ are $2 \times 2$ matrices given by

$$\begin{aligned} {\tilde{\mathbf{A}}} = \left( \begin{array}{cc} -\frac{q'(s)}{q(s)} &{} 1+q^2(s)/\tilde{r} \\ -\tilde{r}-\frac{\int _s^\infty q^2(u)du}{q^2(s)} &{} \frac{q'(s)}{q(s)} \end{array} \right) , \; {\tilde{\mathbf{B}}} = \left( \begin{array}{cc} \frac{q'(s)}{q(s)} &{} -1 \\ \tilde{r} &{} - \frac{q'(s)}{q(s)} \end{array} \right) ,\; \end{aligned}$$

(132)

where the solutions $\tilde{f}(r,s)$ and $\tilde{g}(r,s)$ are characterized by the asymptotic behaviors given in Eq. (80) in the text. We have already shown that $\tilde{f}(\tilde{r} =0,x)$ exists, and in fact $\tilde{f}(\tilde{r} =0,x) = 2^{-1/6} \sqrt{\pi } q(x)$ and given the $\tilde{r}$-dependence of the matrices ${\tilde{\mathbf{A}}}$ and ${\tilde{\mathbf{B}}}$ one expects that $\tilde{f}(\tilde{r},x)$ admits the following expansion

$$\begin{aligned} \tilde{f}(\tilde{r},x) = 2^{-1/6} \sqrt{\pi } q(x) + \sum _{n = 1}^\infty \tilde{r}^n \tilde{f}_n(x). \end{aligned}$$

(133)

To obtain the small $\tilde{r}$ expansion of $\tilde{g}(\tilde{r},x)$, we show that it can be actually expressed in terms of $\tilde{f}(\tilde{r},x)$. Using the relation (93) shown in the text, one obtains from (133) that $\tilde{g}(\tilde{r},x)$ admits the following expansion

$$\begin{aligned} \tilde{g}(\tilde{r},x) = \sum _{n=1}^\infty \tilde{r}^n \tilde{g}_n(x), \; \tilde{g}_n(x) = -\frac{1}{q(x)} \int _x^\infty q(u) \tilde{f}_{n-1}(u) \; du, \;\; n \ge 1. \end{aligned}$$

(134)

By injecting the expansion of $\tilde{f}(\tilde{r},x)$ (133) into the ‘$\tilde{\mathbf B}$-equation’ satisfied by $\tilde{f}(\tilde{r},x)$ [see Eqs. (131), (132)] one finds the following equations

$$\begin{aligned}&\partial _x \tilde{f}_k(x) = \frac{q'(x)}{q(x)} \tilde{f}_k(x) - \tilde{g}_k(x), \end{aligned}$$

(135)

where $\tilde{g}_k(x)$ defined in Eq. (134) can be expressed in terms of $\tilde{f}_{k-1}(x)$. Hence this set of equations (135) can be solved iteratively for successive values of $k$, starting from $k=1$, to yield the first functions given in Eq. (100) in the text.

1.2 Lowest Order Expansion: Calculation of the Integral $a_2$

In this section of Appendix, we show that the amplitude $a_2$ defined through the rather complicated integral [see Eq. (103) in the text]

$$\begin{aligned} a_2 = \int _{-\infty }^\infty \left[ (q' + q R)^2 - \frac{1}{4}(q^2- R^2)^2 \right] \mathcal{F}_2 dx, \end{aligned}$$

(136)

has actually a very simple expression, namely $a_2 = 1/2$. First we recall that

$$\begin{aligned} R(x) = \int _{x}^\infty q^2(u) \;du = \frac{\mathcal{F}'_2(x)}{\mathcal{F}_2(x)}. \end{aligned}$$

(137)

This identity (137) is the crucial one as it allows us to compute this integral in (136), by using successive integration by parts. To this purpose, we first expand the squares in the integrand in (136) and decompose it as

$$\begin{aligned}&a_2 = \int _{-\infty }^\infty \left( q'^2 - \frac{1}{4}q^4 \right) \mathcal{F}_2 \;dx + J_1 + J_2 + J_3, \end{aligned}$$

(138a)

$$\begin{aligned}&J_1 = 2 \int _{-\infty }^\infty q q' R \mathcal{F}_2 \;dx = 2 \int _{-\infty }^\infty q q' \mathcal{F}'_2 \;dx, \end{aligned}$$

(138b)

$$\begin{aligned}&J_2 = \frac{3}{2} \int _{-\infty }^\infty q^2 R^2 \mathcal{F}_2 \;dx = \frac{3}{2} \int _{-\infty }^\infty q^2 R \mathcal{F}'_2 \;dx, \end{aligned}$$

(138c)

$$\begin{aligned}&J_3 = - \frac{1}{4} \int _{-\infty }^\infty R^4 \mathcal{F}_2 \;dx = - \frac{1}{4} \int _{-\infty }^\infty R^3 \mathcal{F}'_2 \;dx, \end{aligned}$$

(138d)

where we used the shorthand notations $q\equiv q(x), R\equiv R(x)$ and $\mathcal{F}_2 \equiv \mathcal{F}_2(x)$. We compute the integral $J_1$ in (138b) by using an integration by part [integrating $\mathcal{F}_2'(x)$], which yields

$$\begin{aligned} J_1 = - 2 \int _{-\infty }^\infty \left( q'^2 + 2 q^4 + x q^2\right) \mathcal{F}_2 \;dx, \end{aligned}$$

(139)

where we have used that $q(x)$ is solution of the Painlevé II equation (17) as well as the asymptotic behavior of $\mathcal{F}_2(x)$ for large negative argument (116). Similarly, the integral $J_2$ in (138c) can be transformed by using a similar integration by part [again integrating $\mathcal{F}_2'(x)$]. This yields

$$\begin{aligned} J_2 = \frac{3}{2} \int _{-\infty }^\infty q^4 \mathcal{F}_2 \;dx - \frac{3}{2} J_1 = \int _{-\infty }^\infty \left( 3q'^2 + \frac{15}{2} q^4 + 3 x q^2 \right) \mathcal{F}_2 \;dx. \end{aligned}$$

(140)

The integral $J_3$ in (138d) can be transformed using the same procedure as

$$\begin{aligned} J_3 = - \frac{J_2}{2} = -\frac{1}{2} \int _{-\infty }^\infty \left( 3q'^2 + \frac{15}{2} q^4 + 3 x q^2 \right) \mathcal{F}_2 \;dx. \end{aligned}$$

(141)

Combining Eqs. (138a)–(141) one obtains

$$\begin{aligned} a_2 = \frac{1}{2} \int _{-\infty }^\infty \left( q'^2 -q^4 - xq^2 \right) \mathcal{F}_2 \;dx. \end{aligned}$$

(142)

Finally, this last integral can be computed exactly by using the identity:

$$\begin{aligned} R(x) = \int _x^\infty q^2(u) \;du = q'^2 - q^4 - xq^2, \end{aligned}$$

(143)

which can be checked easily by taking derivative with respect to $x$ on both sides, and using that $q(x)$ is solution of the Painlevé II equation (17). Hence using this identity (143), $a_2$ in (142) can be computed as

$$\begin{aligned} a_2 = \frac{1}{2} \int _{-\infty }^\infty R(x) \mathcal{F}_2(x) \;dx = \frac{1}{2} \int _{-\infty }^\infty \mathcal{F}'_2(x) \;dx = \frac{1}{2}, \end{aligned}$$

(144)

as given in Eq. (103) in the text.

Appendix 3: Large $\tilde{r}$ Expansion of $\tilde{p}_\mathrm{typ}(\tilde{r})$: Beyond the Leading Order

In this appendix, we analyze in detail the large $\tilde{r}$ asymptotic of $\tilde{p}_\mathrm{typ}(\tilde{r})$. We obtain in particular the first sub-leading corrections to the leading term obtained in the text, in Eq. (118), yielding the rather precise asymptotics for $\tilde{p}_\mathrm{typ}(\tilde{r})$ given in Eq. (119).

This expansion, beyond leading order, requires the determination of the first correction, of order $\mathcal{O}(\tilde{r}^{-1/2})$ to the asymptotic behavior of $\tilde{f}(-\tilde{r},x)$, for large $\tilde{r}$, given in Eq. (110), which we first focus on. We expect, from (110), the following asymptotic behavior valid for large $\tilde{r}$:

$$\begin{aligned} \tilde{f}(-\tilde{r}, x) = \frac{1}{2^{7/6}} \tilde{r}^{-1/4} e^{-\frac{2}{3}\tilde{r}^{2/3} - x \sqrt{\tilde{r}}} \left( 1 + \frac{1}{\sqrt{\tilde{r}}} F_1(x) + o(\tilde{r}^{-1/2}) \right) . \end{aligned}$$

(145)

To compute the function $F_1(x)$, we use that $\tilde{f}(-\tilde{r},x)$ satisfies the following Schrödinger equation

$$\begin{aligned} \partial _x^2 \tilde{f}(-\tilde{r},x) - (x+2q^2(x))\tilde{f}(-\tilde{r},x) = \tilde{r} \tilde{f}(-\tilde{r},x). \end{aligned}$$

(146)

By injecting the asymptotic expansion (145) into Eq. (146), one obtains $F_1(x)$ as

$$\begin{aligned} F_1(x) = -\frac{1}{2} \int _{-\infty }^x \left[ u + 2q^2(u)\right] \;du, \end{aligned}$$

(147)

where we have used that $\lim _{x \rightarrow -\infty }F_1(x) = 0$. Although this is a reasonable assumption we have not been able to establish it rigorously. In the following, we will need the behavior of $F_1(x)$ for large negative argument. It can be obtained from the large negative argument of $q(s)$:

$$\begin{aligned} q(s) = \sqrt{-\frac{s}{2}} \left( 1 + \frac{1}{8s^3} + \mathcal{O}(s^{-6})\right) , \; \mathrm{when \;} s \rightarrow -\infty , \end{aligned}$$

(148)

which yields, for $F_1(x)$ in (147)

$$\begin{aligned} F_1(x) = -\frac{1}{8x} + o(|x|^{-1}), \; \mathrm{when \;} x \rightarrow -\infty . \end{aligned}$$

(149)

More generally, the analysis of Eq. (146) suggests that $\tilde{f}(-\tilde{r}, x)$ admits an expansion of the form

$$\begin{aligned} \tilde{f}(-\tilde{r}, x) = \frac{1}{2^{7/6}} \tilde{r}^{-1/4} e^{-\frac{2}{3}\tilde{r}^{2/3} - x \sqrt{\tilde{r}}} \left( 1 + \sum _{n=1}^\infty \frac{1}{\tilde{r}^{n/2}} F_n(x) \right) , \; F_n(x) \underset{x \rightarrow -\infty }{\sim } \alpha _n |x|^{-n}.\nonumber \\ \end{aligned}$$

(150)

Equipped with this asymptotic expansion (150) we can now compute the large $\tilde{r}$ asymptotics of $\tilde{p}_\mathrm{typ}(\tilde{r})$ beyond leading order. As this was done in the main text, the starting point of our analysis is the following expression

$$\begin{aligned} \tilde{p}_\mathrm{typ}(\tilde{r}) = \frac{2^{1/3}}{\pi } \int _{-\infty }^\infty \left[ \tilde{f}^2(-\tilde{r},x) -\frac{1}{r^2} q^2(x) \tilde{g}^2(-\tilde{r},x) \right] \mathcal{F}_2(x) \;dx. \end{aligned}$$

(151)

We analyze the first term in (151)—the integral involving $\tilde{f}^2(-\tilde{r},x)$—by injecting the large $\tilde{r}$ expansion obtained above (150). It yields

$$\begin{aligned}&\frac{2^{1/3}}{\pi } \int _{-\infty }^\infty \tilde{f}^2(-\tilde{r},x) \;dx = \frac{1}{4\pi } \frac{1}{\tilde{r}^{1/2}} e^{-\frac{4}{3} \tilde{r}^{3/2}}\nonumber \\&\times \int _{-\infty }^\infty e^{-2 x \sqrt{\tilde{r}}}\left( 1 + \frac{2}{\sqrt{r}} F_1(x) + o(\tilde{r}^{-1/2})\right) \mathcal{F}_2(x) \;dx. \end{aligned}$$

(152)

As we have seen before [see Eq. (117) in the text], the integral over $x$ in (153) is dominated by the region where $x < 0$ where, as shown in the main text, one can use the saddle point method. Indeed, as shown previously in (117), the saddle point is reached for $x^*= 2\sqrt{2} \, \tilde{r}^{1/4}$. It is thus natural to perform the change of variable $x = u \tilde{r}^{1/4}$ and use the asymptotic behavior of $F_1(x)$ for large negative argument given in (149) as well as the one for $\mathcal{F}_2(x)$ [55]:

$$\begin{aligned} \mathcal{F}_2(x) = \tau _2 |x|^{-1/8} e^{-\frac{1}{12} |x|^3} \left( 1 + \frac{3}{2^6 |x|^3} + \mathcal{O}(|x|^{-6}) \right) , \; \tau _2 = 2^{1/24} e^{\zeta '(-1)}, \end{aligned}$$

(153)

to obtain after straightforward (though tedious) manipulations the following expansion:

$$\begin{aligned}&\frac{2^{1/3}}{\pi } \int _{-\infty }^\infty \tilde{f}^2(-\tilde{r},x) \;dx = A \; \exp {\left( -\dfrac{4}{3}\tilde{r}^{3/2} + \dfrac{8}{3}\sqrt{2}\tilde{r}^{3/4}\right) }{\tilde{r}^{-{21}/{32}}}\nonumber \\&\qquad \times \,\left( 1 + \dfrac{131 \sqrt{2}}{1536} \tilde{r}^{-3/4} + \mathcal{O}(\tilde{r}^{-3/2})\right) , \end{aligned}$$

(154)

with $A = 2^{-91/48} e^{\zeta '(-1)}/\sqrt{\pi }$.

We now analyze the second term in Eq. (151), which involves $\tilde{g}(-\tilde{r},x)$ that does not contribute to leading order when $\tilde{r} \rightarrow \infty$. To analyze this term, it is sufficient to expand $\tilde{g}(-\tilde{r},x)$ using Eq. (110) as well as $\mathcal{F}_2(x)$ using Eq. (153) to leading order. One can then perform a large $\tilde{r}$ expansion of this term using again the saddle point method, as shown in the main text (117). One obtains, after some manipulations:

$$\begin{aligned} -\frac{2^{1/3}}{\pi \tilde{r}^2} \int _{-\infty }^\infty \left[ q^2(x) \tilde{g}^2(-\tilde{r},x) \right] \mathcal{F}_2(x) \;dx&= A \; \exp {\left( -\dfrac{4}{3}\tilde{r}^{3/2} + \dfrac{8}{3}\sqrt{2}\tilde{r}^{3/4}\right) }{\tilde{r}^{-{21}/{32}}}\nonumber \\&\times \left( -\sqrt{2} \tilde{r}^{-3/4} + \mathcal{O}(\tilde{r}^{-3/2})\right) .\qquad \end{aligned}$$

(155)

Finally, combining these asymptotic expansions (154) and (155) one obtains from (151)

$$\begin{aligned} \tilde{p}_\mathrm{typ}(\tilde{r}) = A \; \exp {\left( -\dfrac{4}{3}\tilde{r}^{3/2} + \dfrac{8}{3}\sqrt{2}\tilde{r}^{3/4}\right) }{\tilde{r}^{-{21}/{32}}} \left( 1 - \dfrac{1405 \sqrt{2}}{1536} \tilde{r}^{-3/4} + \mathcal{O}(\tilde{r}^{-3/2})\right) ,\nonumber \\ \end{aligned}$$

(156)

as announced in the text in Eq. (119).