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
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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 1N(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 Artur Avila, 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 <A^{(E-v)}_n(x)e_1, e_1>=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

<A^{(E-v)}_N(x)e_1, e_1>=<R_{\phi_{N}^E(x)}B^E(x)^{-1}e_1,B^E(f^N(x))^*e_1>\\=<D^{E}_N(x)B^E(x)^{-1}e_1,B^E(f^N(x))^*e_1>=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

<D^{E}_N(x)B^{E_0}(x)^{-1}e_1,B^{E_0}(f^N(x))^*e_1>=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.

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