# Zhenghe's Blog

## May 18, 2011

### Notes 12: Distributions of Eigenvalues of Ergodic 1D discrete Schrodinger operators in a.c. spectrum region

Filed under: Schrodinger Cocycles — Zhenghe @ 11:45 pm
Tags: ,

This post will be something about the distribution of eigenvalues of one dimensional discrete Schrodinger operators in absolutely continuous spectrum region. Namely, given a triple $(X, f, \mu)$ and a potential function $v:X\rightarrow\mathbb R.$ Then these together generates a family of operators $H_{v,x}: l^2(\mathbb Z)\rightarrow l^2(\mathbb Z),$ which is given by

$(H_{v,x}u)_n=u_{n+1}+u_{n-1}+v(f^n(x))u_n,$ for $u\in l^2(\mathbb Z).$

For eigenfunction equations $H_{v,x}u=Eu, x\in X,$ there is an associated Schrodinger cocycle dynamics $(f, A^{E-v}):X\times\mathbb R^2\rightarrow X\times\mathbb R^2.$ The cocycle map $A:X\rightarrow SL(2,\mathbb R)$ is given by

$A^{(E-v)}(x)=\begin{pmatrix}E-v(x)&-1\\ 1&0\end{pmatrix}.$

Let $L(E)$ be it’s Lyapunov exponent (see notes 1 for definition). One way to consider the distribution of eigenvalues is to consider the finite approximation. Namely, as in Notes 7, we will restrict the operator to subinterval $[0, n-1]$ and consider Dirichlet boundary condition, i.e.  $u_{-1}=u_{n}=0.$ Denote it by $H^{n}_{v,x}$.  Let $\lambda_1(x)<\cdots<\lambda_n(x)$ be the eigenvalues of $H^{n}_{v,x}.$ Then we consider the sequence of measure $S_n(x)=\frac{1}{n}\Sigma^{n}_{j=1}\delta_{\lambda_j(x)}.$ It’s standard result that as $n\rightarrow\infty,$ $S_n(x)\rightarrow dN$ in measure  for $\mu-a.e. x.$ Here as a function of energy, $N(E)=\int^{E}_{-\infty}dN:\mathbb R\rightarrow [0,1]$ is the so-called integrated density of states (IDS). Obviously $N$ is nondecreasing. In fact, the relation between IDS and fibered rotation number $\rho$ is that

Lemma 1$N(E)=1-2\rho(E).$

Thus by notes 3 we know, $N$ is flat on resolvent set. Namely, the support of the measure $dN$ are precisely the spectrum of $H_{v,x}$ for $\mu-a.e. x$.

Proof: To see why $N(E)=1-2\rho(E),$ we let $\det(H^{j}_{v,x}-E)=P_j(x,E),$ which a polynomial in $E$ of degree $n.$ We set $P_{-1}(x, E)=0$ and $P_{0}(x, E)=1.$ Then by induction it’s easy to see that

$A^{(E-v)}_j(x)=\begin{pmatrix}P_j(x,E)&-P_{j-1}(fx,E)\\P_{j-1}(x,E)&-P_{j-2}(fx,E)\end{pmatrix},$  and

$P_{j+1}(x,E)+P_{j-1}(x,E)=(v(f^j(x))-E)P_j(x,E), j\geq 0.$

Thus $((-1)^jP_j(x,E))_{0\leq j\leq n-1}$ is an eigenvector of $H^{n}_{v,x}$ if and only $P_{n}(x,E)=0.$ Hence, if and only if $A^{(E-v)}_n(x)e_1\perp e_1,$ where $e_1=\binom{1}{0}.$

Then we note the following facts:

(1)  For $E$ near $-\infty$, it’s easy to see there is constant invariant cone field. Thus there is no rotation for  $A^{(E-v)}_n(x)$ for any $x$ and for any $n$.

(2) $N(E)=0$ for $E$ near $-\infty$. Because $(-\infty,\inf\limits_{x\in X}v(x)-2)$ lies in resolvent set.

(3) $A^{(E-v)}_n(x)$ is monotonic in $E$.

Thus as $E$ goes from $-\infty$ to $E$, $S_n(x)(-\infty, E)$ measures the averaged times of the vector $A^{(E-v)}_n(x)e_1$ passing through $\frac{\pi}{2}\in \mathbb R\mathbb P^1=\mathbb R/(\pi\mathbb Z).$ So as $n\rightarrow \infty,$ $N(E)=\int^{E}_{-\infty}dN=\lim\limits_{n\rightarrow\infty}S_n(x)(-\infty, E)$ is some doubled fibered rotation number. Since $N$ is nondecreasing and $N(E)=0$ for $E$ near $-\infty,$ it’s necessary that $N(E)=1-2\rho(E)$ by Notes 3. $\square$

Now I would like to Sketch a rough proof the following main theorem of this post, which is contained in the paper ‘Bulk Universality and Clock Spacing of Zeros for Ergodic Jacobi Matrices with A.C. Spectrum‘ by, Yoram Last and Barry Simon.

Theorem 2: Assume $Leb(\{E: L(E)=0\})>0$. Then for Leb a.e. $E_0$ with $L(E_0)=0$, we have that $\{\delta_{N(\lambda_j-E_0)}, j=1,\ldots,N\}$ tends to be an arithematic progression. Namely, it’s going to be something like:

$\ldots,\delta_{a-2b},\delta_{a-b},\delta_{a},\delta_{a+b},\delta_{a+2b},\ldots$

And $b$ is nothing other than $\frac{1}{dN/dE(E_0)}$. The convergence is independent of $x$.

Remark: This theorem tells that if we look at a window around $E_0$ with size $O(\frac{1}{N})$, where we expect finitely many eigenvalues. Then if we rescale it by $N$, we will find arithematic progression as $N$ goes to infinity. This is nothing other than the local distribution of eigenvalues in absolutely continuous spectrum region. Now let’s sketch a proof.

Proof: Note that by the proof of Lemma 1, we know that $E$ is an eigenvalue of  $H^{n}_{v,x}$ if and only if $=0$. By kotani theory (see Notes 2 and 3), for a.e. E with $L(E)=0$, we have a $L^2$ map $B: X\rightarrow SL(2,R)$ such that

$B^E(f(x))A^{(E-v)}(x)B^E(x)^{-1}=R_{\phi^E(x)}=D^E(x)\in SO(2,R)$.

Thus

$=\\==0,$

where $\phi^{E}_N(x)=\sum^{N-1}_{j=0}\phi^E(x+j\alpha), D^{E}_N=R_{\phi_{N}^E}$. In fact, for $N$ sufficiently large, what we need to find are these $E$ such that

$=0$.

Assume that $E_0$ is a measurable continuity point for both maps $E\mapsto B^E(x)$ and $E\mapsto B^E(f^N(x))$ (see Notes 11 for definition and properties of measurable continuity points). Then it suffices to show that in some sense $D^{E}_N(x)\rightarrow R_{a+\pi b^{-1}E}$ as $N\rightarrow\infty$.

In fact, replacing $x, x_0$ by $E, E_0$,  the same arguments of Lemma1-Lemma4 give the following results:

(1) $\|A^{(E-v)}_N(x)\|\leq ce^{c|E-E_0|N}$.
(2) Complexifying $E$ at $E_0$ with magnitude $\frac{1}{N}$, we get $A^{(E+\frac{y}{N}-v)}_N(x)$.
(3) Then $A^{(E+\frac{y}{N}-v)}_N(x)$ converges to some entire function $\tilde A(y)$ with $\|\tilde A(y)\|\leq ce^{c|y|}$.
(4) $B^{E_0}(f^N(x))A^{(E_0-v)}(x)B^{E_0}(x)^{-1}\in SO(2,R)$ implies that $B^{E_0}(f^N(x))\tilde A(y)B^{E_0}(x)^{-1}$ sufficiently close to $SO(2,R)$ for large $N$.
(5) By passing to a suitable subsequence, we get that
$B^{E_0}(f^N(x))A^{(E_0+\frac{y}{N}-v)}_N(x)B^{E_0}(x)^{-1}\rightarrow B^{E_0}(x)\tilde A(y)B^{E_0}(x)^{-1}\in SO(2,R)$
for all $y\in R$ as $N\rightarrow\infty$.
(6) Thus we can write it as $B^{E_0}(x)\tilde A(y)B^{E_0}(x)^{-1}=R_{\psi(y)}$. By (3) we have $|\Im\psi(y)|\leq cy$, which implies that $\psi$ is affine.

Finally, we want to show that $\frac{d\psi}{dy}=b^{-1}$.  We need to compute the following

$B^{E_0}(x)A^{(E_0-v)}_N(x)^{-1}B^{E_0}(f^N(x))^{-1}\frac{d}{dE}|_{E_0}(B^{E_0}(f^N(x))A^{(E-v)}_N(x)B^{E_0}(x)^{-1})$
$=\sum^{N-1}_{k=0}(B^{E_0}(x)A^{(E_0-v)}_k(x)^{-1}B^{E_0}(f^k(x))^{-1}B^{E_0}(f^k(x))\begin{pmatrix}0&0\\-1&0\end{pmatrix}\cdot$
$\cdot B^{E_0}(f^k(x))^{-1}B^{E_0}(f^k(x))A^{(E_0-v)}_k(x)B^{E_0}(x)^{-1})$
$=\sum^{N-1}_{k=0}R_{-\phi^{E}_k(x)}B^{E_0}(f^k(x))\begin{pmatrix}0&0\\-1&0\end{pmatrix}B^{E_0}(f^k(x))^{-1}R_{\phi^{E}_k(x)}$

The above formula lies in $sl(2,R)=<\begin{pmatrix}x&0\\ 0&-x\end{pmatrix},\begin{pmatrix}0&y\\y&0\end{pmatrix},\begin{pmatrix}0&-z\\z&0\end{pmatrix}>$

Since $B^{E_0}(f^N(x))A^{(E_0+\frac{y}{N}-v)}_N(x)B^{E_0}(x)^{-1}$ converges to $SO(2,R)$, we only need to take care the $\begin{pmatrix}0&-z\\z&0\end{pmatrix}$ part. Then we have

$Proj_z(\sum^{N-1}_{k=0}R_{-\phi^{E}_k(x)}B^{E_0}(f^k(x))\begin{pmatrix}0&0\\-1&0\end{pmatrix}B^{E_0}(f^k(x))^{-1}R_{\phi^{E}_k(x)})$
$=\sum^{N-1}_{k=0}Proj_z(B^{E_0}(f^k(x))\begin{pmatrix}0&0\\-1&0\end{pmatrix}B^{E_0}(f^k(x))^{-1})$

Let $\begin{pmatrix}a(x)&b(x)\\c(x)&d(x)\end{pmatrix}=B^{E_0}(x)$. Then it’s easy to see that the above formula becomes

$\sum^{N-1}_{k=0}b^2(f^k(x))+d^2(f^k(x))$.

Since for $\tilde A(y)$, we’ve scaled by $N$, we get

$\frac{d\psi}{dy}=\lim\limits_{N\rightarrow\infty}\frac{1}{N}\sum^{N-1}_{k=0}b^2(f^k(x))+d^2(f^k(x))=\int_{X}(b^2(x)+d^2(x))d\mu$
$=\frac{1}{2}\int_{X}\|B(x)\|_{HS}^2d\mu=\pi\frac{dN}{dE}(E_0),$

where $\|\cdot\|_{HS}$ is the Hilbert-Schmit norm (see notes 2). This completes the proof.  $\square$

Here we used the fact that $\frac{dN}{dE}(E_0)=\frac{1}{2\pi}\int_{X}\|B(x)\|_{HS}^2d\mu$. See Theorem 4 of Artur Avila and David Damanik‘s paper ‘Absolute continuity of the IDS for the almost Mathieu operator with non-critical coupling‘ for detailed information.