Almost Sure Weyl Law for Quantized Tori

Abstract

We study the eigenvalues of the Toeplitz quantization of complex-valued functions on the torus subject to small random perturbations given by a complex-valued random matrix whose entries are independent copies of a random variable with mean 0, variance 1 and bounded fourth moment. We prove that the eigenvalues of the perturbed operator satisfy a Weyl law with probability close to one, which proves in particular a conjecture by Christiansen and Zworski (Commun Math Phys 299:305–334, 2010).

Appendix A. Estimate on the Smallest Singular Value

We present for the reader’s convenience a complex version of a result due to Sankar, Spielmann and Teng [22, Lemma 3.2], see also [28, Theorem 2.2].

Theorem 23

There exists a constant $C>0$ such that the following holds. Let $N\ge 2$, let $X_0$ be an arbitrary complex $N\times N$ matrix, and let Q be an $N\times N$ complex Gaussian random matrix whose entries are all independent copies of a complex Gaussian random variable $q\sim \mathcal {N}_{\mathbb {C}}(0,1)$. Then, for any $\delta >0$

$$\begin{aligned} \mathbf {P}\left( s_N(X_0 + \delta Q) < \delta t \right) \le C N t^2. \end{aligned}$$

The proof, which is a straightforward modification of the proof of [22, Lemma 3.2], is presented here for the reader’s convenience.

Lemma 24

There exists a $C>0$ such that for any $\nu \in \mathbb {C}^N$ with $\Vert \nu \Vert =1$, we have that for any $\tau >0$

$$\begin{aligned} \mathbf {P}\left( \Vert (X_0 + \delta Q)^{-1}\nu \Vert > \tau \right) \le C \frac{1}{ \tau ^2 \delta ^2}. \end{aligned}$$

Proof

• 1. Since Q is a Gaussian random matrix, it is clear that the zero set of the map $\mathbb {C}^{N\times N} \ni Q\mapsto \det (X_0 + \delta Q)$ has Lebesgue measure 0. Hence, $X:=X_0 + \delta Q$ is almost surely invertible.

  Let U be a $N\times N$ unitary matrix such that $U^*e_1 = \nu$, where $e_1$ is the unit vector in $\mathbb {C}^N$ with 1 in the first entry, and 0 in the other entries. Write $B=UX$ and $B_0 = UX_0$, then, almost surely,

  $$\begin{aligned} \Vert X^{-1} \nu \Vert = \Vert X^{-1} U^*e_1\Vert = \Vert B^{-1} e_1\Vert . \end{aligned}$$

  We denote by $\widehat{b}$ the first column of $B^{-1}$, and by $b_j$, $j=1,\dots ,N$, the rows of B, hence $\widehat{b}\cdot b_1=1$ and $\widehat{b}\cdot b_j=0$, $j=2,\dots , N$. Notice that the $b_i$ are linearly independent and let t denote the unit vector which is orthogonal to the space spanned by the $\overline{b}_j$ for $j=2,\dots , N$. Then,

  $$\begin{aligned} \widehat{b} = ( \widehat{b} | t ) \,t, \quad \text {and } \quad ( \widehat{b} | t ) (t \cdot b_1)=1. \end{aligned}$$

  Hence, $\Vert B^{-1} e_1\Vert = | t \cdot b_1|^{-1}$, and

  $$\begin{aligned} \mathbf {P}\left( \Vert (X_0 + \delta Q)^{-1}\nu \Vert > \tau \right) = \mathbf {P}\left( | t \cdot b_1| < \tau ^{-1} \right) . \end{aligned}$$

  (0.1)

• 2. Since U is unitary it follows that the entries of $\delta UQ$ are independent and identically distributed complex Gaussian random variables $\sim \mathcal {N}_{\mathbb {C}}(0,\delta ^2)$. Since t is a unit vector depending only on $b_i$, $i=2,\dots , N$, it follows that when fixing these row vectors, $t \cdot b_1-\lambda$, with $\lambda = ( B_0 t)_1$, is a complex Gaussian random variables $\sim \mathcal {N}_{\mathbb {C}}(0, \delta ^2)$. Then,

  $$\begin{aligned} \mathbf {P}\left( | t \cdot b_1|< \tau ^{-1} \right) \le \mathbf {P}\left( | t \cdot b_1 - \lambda | < \tau ^{-1} \right) \le \frac{1}{\tau ^2 \delta ^2}, \end{aligned}$$

and we conclude the second statement of the Lemma. Here, the second inequality follows from a straightforward calculation. To see the first inequality, it is enough to show that for a complex Gaussian random variable $u\sim \mathcal {N}_{\mathbb {C}}(0, \delta ^2)$, we have that for any $b>0$, $x \ge 0$,

$$\begin{aligned} \mathbf {P}\left( | x + \mathrm {Re}\,u |< b \right) \le \mathbf {P}\left( | \mathrm {Re}\,u | < b \right) . \end{aligned}$$

(0.2)

The left hand side is equal to the integral

$$\begin{aligned} I(x) = \frac{1}{\delta \sqrt{\pi }} \int _{-(x+b)}^{b-x} \mathrm {e}^{-t^2/\delta ^2} dt. \end{aligned}$$

Since

$$\begin{aligned} \frac{d}{dx}I(x) = \frac{1}{\delta \sqrt{\pi }} \left[ \mathrm {e}^{-(x+b)^2/\delta ^2} - \mathrm {e}^{-(b-x)^2/\delta ^2} \right] \le 0, \end{aligned}$$

the map $x\mapsto I(x)$ is decreasing, so (0.2) holds, as it is trivially true for $x=0$. $\square$

Lemma 25

There exists a constant $C>1$ such that the following holds. Let $N\ge 2$, and let $\nu$ be a uniformly distributed random unit vector in $\mathbb {C}^N$. Then, for any $0 < c \le 1$

$$\begin{aligned} \mathbf {P}\left( |\nu _1| \ge \sqrt{\frac{c}{CN}} \right) \ge (1-\mathrm {e}^{-2}) \mathbf {P}\left( |z| > \sqrt{c}\right) , \end{aligned}$$

where $z\sim \mathcal {N}_{\mathbb {C}}(0,1)$.

Proof

Let $z\in \mathbb {C}^N$ be a random vector whose entries $z_j\sim \mathcal {N}_{\mathbb {C}}(0,1)$, $j=1,\dots ,N$ are independent and identically distributed complex Gaussian random variables. Then,

$$\begin{aligned} \nu _*(d\mathbf {P}) = \left( \frac{z}{\Vert z\Vert }\right) _*(d\mathbf {P}). \end{aligned}$$

Writing $z=(z_1,z')$, we get that

$$\begin{aligned} \mathbf {P}\left( |\nu _1| \ge \sqrt{\frac{c}{C N}} \right)&= \mathbf {P}\left( \frac{|z_1^2|}{\Vert z\Vert ^2} \ge \frac{c}{CN} \right) \\&= \mathbf {P}\left( \frac{(CN-1)|z_1^2|}{\Vert z'\Vert ^2} \ge \frac{(CN-1)c}{CN-c} \right) \\&\ge \mathbf {P}\left( \frac{(CN-1)|z_1^2|}{\Vert z'\Vert ^2} \ge c \right) \\&\ge \mathbf {P}\left( \Vert z'\Vert ^2 \le (CN-1) \right) \mathbf {P}\left( |z_1^2|\ge c \right) . \end{aligned}$$

Since $z'$ is a complex Gaussian random vector in $\mathbb {C}^{N-1}$ with independent and identically distributed entries $\sim \mathcal {N}_{\mathbb {C}}(0,1)$, we get from Markov’s inequality that for $C>1$ large enough

$$\begin{aligned} \mathbf {P}\left( \Vert z'\Vert ^2 \le (CN-1) \right) \ge 1 -\mathrm {e}^{-N}, \end{aligned}$$

and the statement of the Lemma follows. $\square$

Proof of Theorem 23

Let $\nu$ be a uniformly distributed random unit vector in $\mathbb {C}^N$. By Lemma 24 we know that for any $\tau >0$

$$\begin{aligned} \mathbf {P}\left( \Vert (X_0 + \delta Q)^{-1}\nu \Vert > \tau \right) \le C \frac{1}{ \tau ^2 \delta ^2}. \end{aligned}$$

(0.3)

Write $X:=X_0 + \delta Q$, and let u be the unit eigenvector of $XX^*$ corresponding to its smallest eigenvalue $t_1^2 \ge 0$, i.e.

$$\begin{aligned} XX^* u = t_1^2 u. \end{aligned}$$

(0.4)

Then, almost surely,

$$\begin{aligned} \Vert X^{-1} u\Vert = t_1^{-1}\Vert u \Vert = \Vert X^{-1}\Vert . \end{aligned}$$

(0.5)

Writing $\nu = \nu _0 u + \nu ^{\perp }$, with $\nu _0=(\nu | u)$ and $\nu ^{\perp }$ orthogonal to u, we see that, almost surely,

$$\begin{aligned} \Vert X^{-1} \nu \Vert ^2&= ( (XX^*)^{-1} \nu _0 u + \nu ^{\perp }| \nu _0 u + \nu ^{\perp }) \\&= |\nu _0|^2 ( (XX^*)^{-1} u |u ) +( (XX^*)^{-1} \nu ^{\perp } |\nu ^{\perp }) \\&\ge |\nu _0|^2\Vert X^{-1}\Vert ^2. \end{aligned}$$

Let $C>1$ be as in Lemma 25. Then, using the above, we get that for any $0 < c\le 1$ and any $\tau >0$,

$$\begin{aligned} \mathbf {P}\left( \Vert X^{-1} \nu \Vert> \tau \sqrt{\frac{c}{C N}} \right)&\ge \mathbf {P}\left( |\nu _0|\Vert X^{-1}\Vert> \tau \sqrt{\frac{c}{C N}} \right) \nonumber \\&\ge \mathbf {P}\left( \Vert X^{-1}\Vert> \tau \text { and } |\nu _0|> \sqrt{\frac{c}{C N}}\right) \nonumber \\&\ge \mathbf {P}\left( \Vert X^{-1}\Vert> \tau \right) \min \limits _{ Q: \Vert X^{-1}\Vert> \tau } \mathbf {P}\left( |\nu _0| > \sqrt{\frac{c}{C N}} \right) . \end{aligned}$$

(0.6)

Since the distribution of $\nu$ is invariant a under unitary change of variables, we may express $\nu$ in an orthonormal basis of $\mathbb {C}^N$ which has u as its first vector, wherefore the first component of $\nu$ is $(\nu |u)$. Thus, using Lemma 25 we obtain from (0.6) that

$$\begin{aligned} \mathbf {P}\left( \Vert X^{-1} \nu \Vert> \tau \sqrt{\frac{c}{C N}} \right) \ge \mathbf {P}\left( \Vert X^{-1}\Vert> \tau \right) (1-\mathrm {e}^{-2}) \mathbf {P}\left( |z| > \sqrt{c}\right) , \end{aligned}$$

where $z \sim \mathcal {N}_{\mathbb {C}}(0,1)$ is a complex Gaussian random variable. This, together with (0.3), then yields that there exists a constant $C>0$ such that

$$\begin{aligned} \mathbf {P}\left( \Vert X^{-1}\Vert> \tau \right) \le \frac{CN}{c\,\mathbf {P}\left( |z| > \sqrt{c}\right) \tau ^2 \delta ^2}. \end{aligned}$$

Since, we may choose $c \in ]0,1]$, we take $c=1$, which gives that $c\,\mathbf {P}\left( |z| > \sqrt{c}\right) = e^{-1}$. Recall that $\Vert X^{-1} \Vert = s_N(X)^{-1}$, so taking $t =(\tau \delta )^{-1}$, we deduce that there exists a constant $C>0$ such that for any $N\ge 2$

$$\begin{aligned} \mathbf {P}\left( s_N(X) < t\delta \right) \le CNt^2. \end{aligned}$$

$\square$