Anderson localization for multi-frequency quasi-periodic operators on $\pmb {\mathbb {Z}}^{d}$

Abstract

We establish Anderson localization for general analytic k-frequency quasi-periodic operators on ${\mathbb {Z}}^d$ for arbitraryk, d.

Appendix A

In the following, we will prove the several variables matrix-valued Cartan estimate, i.e., Lemma 3.5. The proof is similar to that in [Bou05, Bou02]. Before going to the details, we recall some useful lemmas. The first result is the standard Schur’s complement theorem. For convenience, we include a proof here.

Lemma A.1

Let T be the matrix

$$\begin{aligned} T=\left( \begin{array}{cc} T_1&{}T_2\\ T_2^t&{}T_3 \end{array} \right) , \end{aligned}$$

where $T_1$ is an invertible $n\times n$ matrix , $T_2$ is an $n\times k$ matrix and $T_3$ is a $k\times k$ matrix. Let

$$\begin{aligned} S=T_3-T_2^tT_1^{-1} T_2. \end{aligned}$$

Then T is invertible if and only if S is invertible, and

$$\begin{aligned} \Vert S^{-1}\Vert \le \Vert T^{-1}\Vert \le C(1+\Vert T_1^{-1}\Vert )^2(1+\Vert S^{-1}\Vert ), \end{aligned}$$

(A.1)

where C depends only on $\Vert T_2\Vert$.

Proof

It is easy to check that

$$\begin{aligned} T=\left( \begin{array}{cc} T_1&{}T_2\\ T_2^t&{}T_3 \end{array} \right) =\left( \begin{array}{cc} I&{}0\\ T_2^{t}T_1^{-1}&{}I \end{array} \right) \left( \begin{array}{cc} I&{}T_2\\ 0&{}S \end{array} \right) \left( \begin{array}{cc} T_1&{}0\\ 0&{}I \end{array} \right) . \end{aligned}$$

(A.2)

It implies T is invertible if and only if S is invertible and also the second inequality of (A.1). By (A.2), one has

$$\begin{aligned} T^{-1}= & {} \left( \begin{array}{cc} T_1&{}0\\ 0&{}I \end{array} \right) ^{-1}\left( \begin{array}{cc} I&{} T_2 \\ 0&{} S \end{array} \right) ^{-1} \left( \begin{array}{cc} I&{}0\\ T_2^{t}T_1^{-1}&{}I \end{array} \right) ^{-1}\\= & {} \left( \begin{array}{cc} T_1^{-1}&{}0\\ 0&{}I \end{array} \right) \left( \begin{array}{cc} I&{}-T_2S^{-1}\\ 0&{} S^{-1} \end{array} \right) \left( \begin{array}{cc} I&{}0\\ -T_2^{t}T_1^{-1}&{}I \end{array} \right) \\= & {} \left( \begin{array}{cc} \star &{}\star \\ \star &{} S^{-1} \end{array} \right) . \end{aligned}$$

implying the first inequality of (A.1). $\square$

We then introduce the higher dimensional Cartan sets Lemma of Goldstein-Schlag [GS08]. We denote by ${\mathcal {D}}(z,r)$ the standard disk on ${\mathbb {C}}$ of center z and radius $r>0$.

Lemma A.2

[GS08, Lemma 2.15] Let $f(z_1,\ldots ,z_J)$ be an analytic function defined in a ploydisk ${\mathcal {P}}=\prod \limits _{1\le i\le J}{\mathcal {D}}(z_{i,0},1/2)$ and $\phi =\log |f|$. Let $\sup \limits _{{\underline{z}}\in {\mathcal {P}}}\phi ({\underline{z}})\le M,m\le \phi ({\underline{z}}_0)$, ${\underline{z}}_0=(z_{1,0},\ldots ,z_{J,0})$. Given $F\gg 1$, there exists a set ${\mathcal {B}}\subset {\mathcal {P}}$ such that

$$\begin{aligned} \phi ({\underline{z}})>M-C(J)F(M-m),\ \mathrm {for}\ \forall \ {\underline{z}}\in \prod \limits _{1\le i\le J}{\mathcal {D}}(z_{i,0},1/4)\setminus {\mathcal {B}}, \end{aligned}$$

(A.3)

and

$$\begin{aligned} \mathrm {mes}({\mathcal {B}}\cap {\mathbb {R}}^J)\le C(J)e^{-F^{1/J}}. \end{aligned}$$

(A.4)

Proof of Lemma 3.5

The proof is similar to that of Proposition 14.1 in [Bou05] in case $J=1$ and Lemma 1.43 in [Bou02] without explicit bounds. In the following proof, $C=C(B_1,J)$ and $c=c(B_1,J)$.

Let

$$\begin{aligned} \mu =10^{-2}{J^{-1}}\delta (1+B_1)^{-1}(1+B_2)^{-1}. \end{aligned}$$

Fix

$$\begin{aligned} x_0\in \left[ -\delta /2, \delta /2\right] ^{J} \end{aligned}$$

and consider T(z) with $|z-x_0|=\sup \limits _{1\le i\le J}|z_i-x_{0,i}|<\mu$. Thanks to Cauchy’s estimate and (3.12), one obtains for $|z-x_0|<\mu$,

$$\begin{aligned} \Vert {\partial _{z_i} T(z)}\Vert \le \frac{4 B_1}{\delta },\quad i=1,2,\ldots , J, \end{aligned}$$

which implies

$$\begin{aligned} \Vert T(z)-T(x_0)\Vert \le \frac{4JB_1\mu }{\delta }\le 25^{-1}(1+B_2)^{-1}. \end{aligned}$$

From the assumption (ii) of Lemma 3.5, we can find $V=V(x_0)$ so that $|V|\le M$ and (3.13) is satisfied. Denote by $V^c=[1,N]\setminus V$. Thus using the standard Neumann series argument and (3.13), one has

$$\begin{aligned} \Vert (R_{V^c}T(z)R_{V^c})^{-1}\Vert \le 2B_2\ \mathrm {for}\ |z-x_0|<\mu .\end{aligned}$$

(A.5)

We define for $|z-x_0|<\mu$ the analytic self-adjoint function

$$\begin{aligned} S(z)=R_{V}T(z)R_{V}-R_{V}T(z)R_{V^c}(R_{V^c}T(z)R_{V^c})^{-1}R_{V^c}T(z)R_{V}. \end{aligned}$$

(A.6)

Then by (A.5) and (A.6), we have

$$\begin{aligned} \Vert S(z)\Vert \le 3B_1^2B_2. \end{aligned}$$

(A.7)

Recalling Lemma A.1, if S(z) is invertible, so is T(z) and by (A.1),

$$\begin{aligned} \Vert S^{-1}(z)\Vert \le C\Vert T^{-1}(z)\Vert \le CB_2^2(1+\Vert S^{-1}(z)\Vert ). \end{aligned}$$

(A.8)

For $x\in {\mathbb {R}}^{J}$, one has

$$\begin{aligned} ||S(x)||^M\ge |\det S(x)|=\prod _{\lambda \in \sigma (S(x))}|\lambda |\ge \Vert S^{-1}(x)\Vert ^{-M}. \end{aligned}$$

(A.9)

By (A.7), one has

$$\begin{aligned} \Vert S^{-1}(x)\Vert \le \frac{\Vert S(x)\Vert ^{M-1}}{|\det S(x)|}\le \frac{(3B_1^2B_2)^M}{|\det S(x)|}. \end{aligned}$$

(A.10)

Let

$$\begin{aligned} \phi (z)=\log |\det S(x_0+\mu z)|,\ |z|<1. \end{aligned}$$

Then by (A.9) and (A.7),

$$\begin{aligned} \sup _{|z|<1}\phi (z)\le C M\log B_2. \end{aligned}$$

(A.11)

By (3.14) and the definition of $\mu$, there is some $x_1$ with $|x_0-x_1|<\mu /10$ such that

$$\begin{aligned} \Vert T^{-1}(x_1)\Vert \le B_3. \end{aligned}$$

(A.12)

Hence by (A.8), $\Vert S^{-1}(x_1)\Vert \le CB_3$, and from (A.9),

$$\begin{aligned} \phi (a)\ge -CM\log B_3, \end{aligned}$$

(A.13)

where $a=\frac{x_1-x_0}{\mu }$, so $|a|<1/10$. Let

$$\begin{aligned} {\mathcal {P}}=\prod _{1\le i\le J}{\mathcal {D}}(a_i,{1}/{2}). \end{aligned}$$

Then one has

$$\begin{aligned} \sup _{z\in {\mathcal {P}}}\phi (z)\le C M\log B_2, \phi (a)\ge - CM\log B_3. \end{aligned}$$

Applying Lemma A.2 and recalling (A.3), (A.4), for any $F\gg 1$, there is some set ${\mathcal {B}}\subset \prod \limits _{1\le i\le J}{\mathcal {D}}(a_i,{1}/{4})$ with

$$\begin{aligned} \phi (z)\ge -CF M\log (B_2+B_3)\ \mathrm {for}\ z\in \prod \limits _{1\le i\le J}{\mathcal {D}}(a_i,{1}/{4})\setminus {\mathcal {B}}, \end{aligned}$$

(A.14)

and

$$\begin{aligned} \mathrm {mes}({\mathcal {B}}\cap {\mathbb {R}}^{J})\le Ce^{-F^{1/J}}. \end{aligned}$$

(A.15)

For $0<\epsilon <1$, let

$$\begin{aligned} F =\frac{-c\log \epsilon }{ M\log (B_2+B_3)}. \end{aligned}$$

Then by (A.14) and (A.15),

$$\begin{aligned}&\mathrm {mes}\left\{ x\in {\mathbb {R}}^{J}: \ |x-x_1|<\mu /4\ \mathrm {and}\ |\det S(x)|\le \epsilon \right\} \\&\quad = \mu ^{{J}}\mathrm {mes}\left\{ x\in {\mathbb {R}}^{J}:\ |x-a|<1/4\ \mathrm {and}\ \phi (x)\le \log \epsilon \right\} \\&\quad \le C\mu ^{{J}} e^{-F^{1/J}}. \end{aligned}$$

Since $|x_0-x_1|<\mu /10$, we have

$$\begin{aligned} \mathrm {mes}\left\{ x\in {\mathbb {R}}^{J}:\ |x-x_0|<\mu /8\ \mathrm {and}\ |\det (S(x))|\le \epsilon \right\} \le C \mu ^{{J}} e^{-c\left( \frac{\log \epsilon ^{-1}}{M\log (B_2+B_3)}\right) ^{1/J}}.\nonumber \\ \end{aligned}$$

(A.16)

Recalling (A.8), (A.10) and (3.15), one has for $|x-x_0|<\mu /8$ and $|\det S(x)|\ge \epsilon$,

$$\begin{aligned} \Vert T^{-1}(x)\Vert \le C(1+B_2^2)(1+\epsilon ^{-1}(3B_1^2{B_2})^M)\le C\epsilon ^{-2}. \end{aligned}$$

(A.17)

Covering $[-\frac{\delta }{2},\frac{\delta }{2}]^{J}$ by cubes of side $\mu /4$, and combining (A.16) and (A.17), one has

$$\begin{aligned} \mathrm {mes}\left\{ x\in \left[ -\delta /2, \delta /2\right] ^{J}:\ \Vert T^{-1}(x)\Vert \ge \epsilon ^{-2}\right\} \le C\delta ^{J}e^{-c\left( \frac{\log \epsilon ^{-1}}{M\log (B_2+B_3)}\right) ^{1/J}}. \end{aligned}$$

$\square$