# 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.

## April 4, 2011

### Notes 11: Renormalization (III)-Convergence of Renormalization

I’ve been back to Evanston from Toronto. But I guess I still have 10 more notes to post. It will take me a very long time to finish.

In this post we will discuss the convergence of renormalization. As pointed out in last post, it’s necessary that we should assume zero Lyapunov to get convergence. In fact, we will assume $L^2$-conjugacy to rotations. This is somehow natural because Kotani theory tells us that in the Schrodinger cocycle case, for almost every energy, zero Lyapunov exponent implies $L^2$-conjugacy. More concretely, we assume for $A\in C^{\omega}(\mathbb R/\mathbb Z, SL(2,\mathbb R)), \alpha\in\mathbb R\setminus\mathbb Q$ that $(\alpha,A)$ is $L^2$-conjugate to $SO(2,\mathbb R)$-valued cocycles. Namely, there is a measurable map $B:\mathbb R/\mathbb Z\rightarrow SO(2,\mathbb R)$ such that $\int_{\mathbb R/\mathbb Z}\|B(x)\|^2dx<\infty$ and $B(x+\alpha)A(x)B(x)^{-1}\in SO(2,\mathbb R)$ for almost every $x\in\mathbb R/\mathbb Z.$  Let $S(x)=\sup_{n\geq 1}\frac{1}{n}\sum^{n-1}_{k=0}\|B(x+k\alpha)\|^2,$ which is finite almost everywhere by the Maximal Ergodic Theorem. WLOG, we can assume that $A$ can be holomorphically extended to $\Omega_{\delta}=\{z\in\mathbb C/\mathbb Z: |\Im z|<\delta\}$ which is also Lipschitz in $\Omega_{\delta}.$ Then we have the following lemma.

Lemma 1: There exists $C>0$ such that for almost every $x_0,$ we have for every $n\geq 1$ and $x\in\Omega_{\delta}$

$\|A_n(x_0)^{-1}(A_n(x)-A_n(x_0))\|\leq Ce^{C\|B(x_0)\|^2S(x_0)n|x-x_0|}.$

Proof: As explained in Notes 2, $A_k(x)$ is bounded in $L^1.$ In fact, $\|A_k(x)\|\leq\|B(x+k\alpha)\|\|B(x)\|$ for almost every $x.$ Let $x_0$ be such a point. Note $A_n(x_0)^{-1}A_n(x)=A(x_0)^{-1}\cdots A(x_0+(n-1)\alpha)^{-1}A(x+(n-1)\alpha)\cdots A(x).$ If we let $A(x_0+k\alpha)^{-1}A(x+k\alpha)=H_k(x)+id,$ then by Lipschitz condition $H_k(x)\leq C|x-x_0|.$ Obviously, $A_n(x_0)^{-1}A_n(x)=A_{n-1}(x_0)^{-1}H_{n-1}(x)A_{n-1}(x)+A_{n-1}(x_0)^{-1}A_{n-1}(x).$ Hence by induction we have

$A_n(x_0)^{-1}A_n(x)=id+\sum^{n-1}_{k=0}A_k(x_0)^{-1}H_k(x)A_k(x).$

This obviously implies that

$\|A_n(x_0)^{-1}A_n(x)-id\|\leq e^{\sum^{n-1}_{k=0}\|A_k(x_0)\|^2\|H_k(x)\|}-1.$

Thus we obtain

$\|A_n(x_0)^{-1}A_n(x)-id\|\leq e^{\sum^{n-1}_{k=0}\|B(x_0+k\alpha)\|^2\|B(x_0)\|^2\|H_k(x)\|}-1.$

Hence,

$\|A_n(x_0)^{-1}A_n(x)-id\|\leq Ce^{S(x_0)n\|B(x_0)\|^2C|x-x_0|},$

which completes the proof. $\square$

Assume further that $x_0$ is a measurable continuity point of $S$ and $B.$ Here for example, $x_0$ is a measurable continuity point of $S,$ if it is a Lebesgue density point of $S^{-1}(S(x_0)-\epsilon, S(x_0)+\epsilon)$ for every $\epsilon.$ It’s standard result that this is a full measure set since $S$ is measurable and almost everywhere finite. Same definition can be applied to $\|B(x)\|.$ By definition, it’s easy to see the portion of $x$ that $S(x)$ is close to $S(x_0)$ is getting larger and larger in smaller and smaller neighborhood of $x_0.$ Let $x_0$ be that $S(x_0)<\infty,$ which is a full measure condition. Then Lemma 1 implies the following estimate

Lemma 2:  Let $x_0$ be as above. Then for for every $d>0,$ there exist a $n_0(d)>0$ such that

$\|A_{(-1)^nq_n}(x)\|\leq \inf\limits_{x'-x_0\in [-\frac{d}{q_n}, \frac{d}{q_n}]}C(x_0)e^{C(x_0)q_n|x-x'|}$

as long as $n\geq n_0(d).$

Proof: If $n$ is sufficiently large, measurable continuity hyperthesis implies that for every $x'\in [x_0-\frac{d}{q_n}, x_0+\frac{d}{q_n}],$ we can find some $x_0'$ with $|x_0'-x'|\leq\frac{1}{q_n}$ and such that $B(x_0'), B(x_0'+\beta_n)$ are close to $B(x_0)$ and $S(x_0'), S(x_0'+\beta_n)$ are close to $S(x_0).$ WLOG, we can assume $n$ is even. Then Lemma 1 together with our choice of $x_0$ implies that

$\|A_{q_n}(x_0')^{-1}(A_{q_n}(x)-A_{q_n}(x_0'))\|\leq Ce^{C\|B(x_0)\|^2S(x_0)n|x-x_0'|}.$

And we can of course assume $x_0'$ such that $\|A_{q_n}(x_0')\|\leq\|B(x_0')\|\|B(x_0'+\beta_{n})\|.$  Since $\|A^{-1}B\|\geq\|B\|\|A\|^{-1}$ and $|x_0'-x'|<\frac{1}{q_n}.$ Combined these together we get the estimate we want. For $n$ odd, we apply the same discussion to $\|A_{-q_n}(x_0')^{-1}(A_{-q_n}(x)^{-1}-A_{-q_n}(x_0')^{-1})\|.$ $\square$

If we renormalize around $x_0,$ we know from last post that

$\mathcal R^n(\Phi)(1,0)=(1,A_{(-1)^{n-1}q_{n-1}}(x_0+\beta_{n-1}(\cdot)))=(1,A^{(n,0)})$ and
$\mathcal R^n(\Phi)(0,1)=(\alpha_n, A_{(-1)^{n}q_{n}}(x_0+\beta_{n-1}(\cdot)))=(\alpha_n, A^{(n,1)}).$

Note $q_{n-1} Then we have the following obvious corollary from Lemma 2

Corollary 3: Let $\Omega_{\delta/\beta_{n-1}}(\mathbb R)=\{z\in\mathbb C: |\Im z|<\delta/\beta_{n-1}\}.$ Then

$\|A^{(n,i)}(x)\|\leq\inf\limits_{x'\in [-d,d]}Ce^{C|x-x'|},$ where $i=0,1$ and $x\in\Omega_{\delta/\beta_{n-1}}(\mathbb R).$

Thus by homorphicity, $A^{(n,i)}$ are precompact in $C^{\omega}.$ So we can take limit along some subsequence. Denote the limit by $\tilde{A}.$ Then by estimate in Corollary 3 we  get $\|\tilde A(z)\|\leq Ce^{|\Im z|}.$

We in fact also have that $B(x_0)\tilde A(x) B(x_0)^{-1}\in SO(2,\mathbb R)$ for $x\in \mathbb R.$ This is given by the following lemma

Lemma 4: Let $A, x_0$ be as above, then for every $d>0$ and every $\epsilon>0,$ there exists a $n_0(d,\epsilon)$ such that if $n>n_0(d, \epsilon)$ and $\|\alpha n\|_{\mathbb R/\mathbb Z}\leq\frac{d}{n},$ then

$B(x_0)A_n(x)B(x_0)^{-1}$ is $\epsilon$ close to $SO(2,\mathbb R)$ for every $x\in[x_0-\frac{d}{n}, x_0+\frac{d}{n}].$

Proof: For $n$ sufficiently large, for every $x\in[x_0-\frac{d}{n}, x_0+\frac{d}{n}],$ as in proof of Lemma 2, we can find some $x'$ $\frac{\epsilon}{n}$ which is close to $x$ and $S(x'), B(x'), B(x'+n\alpha))$ are $\epsilon$ close to $S(x_0), B(x_0)$. Then the same argument of proof of Lemma 2 implies that $A_n(x)$ and $A_n(x')$ are $\epsilon$ close.

Thus we can reduce the proof to the case $B(x_0)A_n(x')B(x_0)^{-1}$ is $\epsilon$ close to $SO(2,\mathbb R).$ But this is clear since we can furthermore choose $x'$ such that $B(x'+\alpha n)A_n(x')B(x')^{-1}\in SO(2,\mathbb R),$ and we’ve already had $B(x'), B(x'+n\alpha)$ are $\epsilon$ close to $B(x_0).$ $\square$

Thus we can write $\tilde A(z)=B(x_0)^{-1}R_{\phi (z)}B(x_0),$ where $\phi:\mathbb C\rightarrow\mathbb C$ satisfying $|\Im{\phi(z)}|\leq C+C|\Im z|$ is an entire function.  Thus $\phi$ must be linear. We’ve basically proved the following theorem

Theorem 5: If the real analytic cocycle dynamics $(\alpha, A)$ is $L^2$ conjugate to rotations, then for almost every $x_0\in\mathbb R/\mathbb Z$ there exists $B(x_0)\in SL(2,\mathbb R),$ and a sequence of affine functions with bounded coefficients $\phi^{(n,0)}, \phi^{(n,1)}:\mathbb R\rightarrow\mathbb R$ such that

$R_{-\phi^{(n,0)}(x)}BA^{(n,0)}(x)B^{-1}\rightarrow id$ and $R_{-\phi^{(n,1)}(x)}BA^{(n,1)}(x)B^{-1}\rightarrow id.$

as $n\rightarrow\infty.$

To conclude, Theorem 5 implies the following final version of renormalization convergence theorem

Theorem 6: Let $(\alpha, A)$ be as in Theorem 5; let deg be the topological degree of map $A.$ Then there exists a sequence of renomalization representatives $(\alpha_n, A^{(n)})$ and $\theta_n\in\mathbb R,$ such that

$R_{-\theta_n-(-1)^ndeg x}A^{(n)}(x)\rightarrow id$ in $C^{\omega}$ as $n\rightarrow\infty$

Proof:  Let $B, \phi^{(n,0)}(x)=a_{n,0}x+b_{n,0}$ and $\phi^{(n,1)}(x)=a_{n,1}x+b_{n,1}$ be as in Theorem 5. Let $n$ be large and let $\tilde B(x)=R_{a_{n,0}\frac{x^2-x}{2}+b_{n,0}x}B.$ Then $\tilde A(x)=\tilde B(x+1)A^{(n,0)}(x)\tilde B(x)^{-1}$ is $C^{\omega}$ close to identity and $\tilde B(x+\alpha_n)A^{(n,1)}(x)\tilde B(x)^{-1}$ is $C^{\omega}$ close to $R_{\psi^n(x)},$ where $\psi^n(x)=(a_{n,0}\alpha_n+a_{n,1})x+\frac{\alpha_{n}^2-\alpha_n}{2}+b_{n,0}+b_{n,1}.$ By Lemma 1 of Notes 10, we know there exists $C\in C^{\omega}(\mathbb R, SL(2,\mathbb R))$ which is $C^{\omega}$ close to identity such that $C(x+1)\tilde A(x)C(x)^{-1}=id.$

Thus $B^{(n)}=C\tilde B$ is a normalizing map for $A^{(n,0)}$ and $A^{(n)}(x)=B^{(n)}(x+\alpha_n)A^{(n,1)}(x)B^{(n)}(x)^{-1}$ is $C^{\omega}$ close to some $R_{\psi_n(x)},$ where $\psi_n$ is linear. Since the way we get renormalization representatives preserving homotopic relation and the degree of n-th renormalization representative of $(\alpha, R_{deg x})$ is $(-1)^n deg,$ we get that degree of $(\alpha_n, A_n)$ is $(-1)^ndeg.$ Thus the linear coefficient of $\psi_n$ must be close to $(-1)^ndeg$ and $A^{(n)}$ must be close to $R_{\theta_n-(-1)^ndeg x}$ for some $\theta_n\in\mathbb R.$ $\square$

I’ve finished the serial posts about renormalization. It’s a powerful technique in the way that we can use it to reduce global problem to local problem and apply perturbation theory like KAM thoerem. More precisely, we can start with $L^2$ conjugacy to rotations and end up as $C^{\omega}$ close to rotations. If degree is zero and $\alpha$ satisfying some arithematic properties, we can then apply standard KAM theorem to get reducibility.

For these posts, I am following Artur‘s course and he and Krikorian‘s papers. Here I only do the $C^{\omega}$ case while they’ve considered smooth cases in there papers.

Next post will be something about distribution of eigenvalues of the our old friend: One dimensional discrete Schrodinger operator:)

## March 27, 2011

### Notes 10: Renormalization (II)-Renormalization Representatives

Filed under: Schrodinger Cocycles — Zhenghe @ 6:01 pm
Tags: ,

Recall in the last post we define $\Lambda^r=\mathbb R\times C^r(\mathbb R, SL(2,\mathbb R)).$ Then given a SL(2,R) cocycle dynamics $(\alpha, A):\mathbb R/\mathbb Z\times\mathbb R^2\rightarrow \mathbb R/\mathbb Z\times\mathbb R^2$ is equivalent to a $\mathbb Z^2$ action $\Phi:\mathbb Z^2\rightarrow \Lambda^r$ such that $\Phi(1,0)=(1,id), \Phi(0,1)=(\alpha,A).$

We’ve also defined the renormalization operator $\mathcal R$ around $0\in\mathbb R/\mathbb Z$ such that $\mathcal R(\Phi)=M_{\alpha}N_{U(\alpha)}(\Phi),$ where $M_{\alpha}$ is rescaling operator and $N_{U(\alpha)}$ is the basing change operator. By definition we have

$\mathcal R^n(\Phi)(1,0)=(1,A_{(-1)^{n-1}q_{n-1}}(\beta_{n-1}(\cdot)))$ and $\mathcal R^n(\Phi)(0,1)=(\alpha_n, A_{(-1)^{n}q_{n}}(\beta_{n-1}(\cdot))).$

Obviously, we are looking at smaller and smaller space scale but larger and larger time scale. Thus if we want to study the limit of renormalization process, it is necessary to assume zero Lyapunov exponent to start with. See last post for all the details.

In this post we will first analytically normalize $\mathcal R^n(\Phi)(1,0)$ to be $(1,id)$ for $r=\omega.$ Because then the correponding $\mathcal R^n(\Phi)(0,1)$ is well-defined on $\mathbb R/\mathbb Z\times\mathbb R^2$ by commuting relation. Secondly, we will explain the relation between the dynamics of original system and these of renormalized systems.

(I) Normalizing $\mathcal R^n(\Phi)(1,0)$ to be $(1,id)$

We state the normalizing result in the following Lemma

Lemma 1: For any $\mathbb Z^2$ action $\Phi:\mathbb Z^2\rightarrow\Lambda^r$ with $\pi_1(\Phi(1,0))=1,$ we can $C^r$ conjugate it to be that $\Phi(1,0)=(1,id).$ Here $r\in\mathbb N\cup\{\infty,\omega\}.$ Here $\pi_1$ is the projection to the first coordinate.

Proof: Denote $\Phi(1,0)=(1,C), \Phi(0,1)=(\alpha, B).$ Then it suffices to find some $C^r D:\mathbb R\rightarrow SL(2,\mathbb R)$ such that $D(x+1)C(x)D(x)^{-1}=id.$ Then automatically, $(\alpha, D(x+\alpha)B(x)D(x)^{-1})$ is well-defined on $\mathbb R/\mathbb Z\times\mathbb R^2.$

First let’s consider the case $r\leq\infty.$ Obviously, $D(x+1)C(x)D(x)^{-1}=id$ is equivalent to $D(x+1)=D(x)C(x)^{-1}.$ Thus it’s sufficient to define $D(x)$ on an arbitrary interval with length biger than 1.  One way to do it is to let $D(x)=id$ around a small interval around 0, hence $D(x)=C(x-1)^{-1}$ around 1, then $C^r$ extends it to an open connected interval containing $[0,1].$ It’s easy to see that if $C(x)$ is $C^r$ close to identity on $(-\epsilon_1, 1+\epsilon_2),$ then we can choose $D$ such that so is $D$ on $(-\epsilon_1, 1+\epsilon_2)$ for some $\epsilon_1, \epsilon_2>0.$

The case $r=\omega$ needs a little bit more computation. We will first deal with case $C$ is close to identity near a neighborhood of $\mathbb R$ in $\mathbb C.$ It’s enough to find some $D'=ED,$ where $D$ is from above and $E$ is something satisfying

$E(x+\alpha)E(x)^{-1}=id$ and $\bar{\partial}(ED)=(\bar{\partial}E)D+E(\bar{\partial}D)=0$ (this obviously imlies that D’ is holomorphic).

Thus it’s necessary $E^{-1}\bar{\partial}E=-D^{-1}\bar{\partial}D.$

Let’s first note that the analyticity of  $C$ implies the periodicity of $D^{-1}\bar{\partial}D.$ Indeed, we have $D(x+1)C(x)=D(x).$ By analyticity, we get $\bar{\partial}D(x+1)C(x)=\bar{\partial}D(x).$ Combined with $C(x)=D(x+1)^{-1}D(x),$ we get

$\bar{\partial}D(x+1)D(x+1)^{-1}=\bar{\partial}D(x)D(x)^{-1}\in sl(2,\mathbb R).$

Thus we can define $\phi=E^{-1}\bar{\partial}E=-\bar{\partial}DD^{-1}:\mathbb R/\mathbb Z\rightarrow sl(2,\mathbb R).$ And we can of course $C^r$ extend $\phi$ to $\Omega_{\epsilon}=\{z:|\Im z|<\epsilon\}$ for some $\epsilon>0.$ It’s easy to see that $\phi$ is sufficiently close to zero if $C$ is suffciently close to identity. Let’s introduce the following operator

$T=e^{-P(\cdot)}\bar{\partial}e^{P(\cdot)},$ where $P(\phi)(z)=\frac{-1}{\pi}\int_{\bar{\Omega}_{\epsilon}}\frac{\phi(w)}{z-w}dxdy, w=x+iy$ is the Cauchy transform.

It’s a standard result that  $P$ inverts $\bar{\partial}.$ A direct computation show that $T(0)=0,$ $DT(0)=id$ and  Thus $T$ is invertible near zero function. Thus if $\phi$ is sufficiently small, we can find some $\psi$ such that $T(\psi)=\phi.$ Then $E=e^{P(\psi)}$ can be our choice and $E$ is near identity. This addresses the local case. Note we in this argument we only need the $C^{\infty}$ smallness of $C-id.$

For the global case, we can first find some $C^{\infty}$ B such that $A(x)=B(x+1)B(x)^{-1},$  and then approximate $B$ in $C^{\infty}$ topology by some $C^{\omega}$ $B'.$ Then $B'(x+1)A(x)B(x)'$ is $C^{\infty}$ close to identity and we can apply the local argument.  $\square$

(I omited some details, one can find the detailed proof in Avila and Krikorian‘s paper ‘Reducibility or nonuniform hyperbolicity for quasiperiodic Schrodinger cocylces‘)

Back to the case $\Phi(1,0)=(1,id), \Phi(0,1)=(\alpha,A).$ Make the notation $R^n(\Phi)(1,0)=(1,A^{(n,0)})$ and $R^n(\Phi)(0,1)=(\alpha_n, A^{(n,1)}).$ Let $B^{(n)}$ be the normalizing map such that $B^{(n)}(x+1)A^{(n,0)}(x)B^{(n)}(x)^{-1}=id.$ Let $A^{(n)}(x)=B^{(n)}(x+\alpha_n)A^{(n,1)}(x)B^{(n)}(x)^{-1}.$ Then

$(\alpha_n, A^{(n)})$ is called a representative of the n-th renomalization of $(\alpha, A).$

By the proof of Lemma 1, it’s not difficult to see that representative is not unique but all of them are conjugate. In fact, any element of conjugacy classs of $A^{(n)}.$

(II) One of the relation between the dynamics of $(\alpha,A)$ and the $(\alpha_n,A^{(n)})$.

It’s given by the next lemma

Lemma 2: If a $C^r$-renormalization representative $(\alpha_n, A^{(n)})$ is $C^r$ conjugate to rotations, then $(\alpha, A)$ is $C^{r}$ conjugate to rotations. Here again $r\in\mathbb N\cup\{\infty,\omega\}.$

Proof:  Recall

$\mathcal R^n(\Phi)(1,0)=(1,A_{(-1)^{n-1}q_{n-1}}(\beta_{n-1}(\cdot)))=(1,A^{(n,0)})$ and
$\mathcal R^n(\Phi)(0,1)=(\alpha_n, A_{(-1)^{n}q_{n}}(\beta_{n-1}(\cdot)))=(\alpha_n, A^{(n,1)}).$

And $B^{(n)}$ is the normalizing map. By assumption and dicussion following the proof of Lemma 1, we can assume that $B^{(n)}(x+\alpha_n)A^{(n)}(x)B^{(n)}(x)^{-1}\in SO(2,\mathbb R)$ for every $x.$ If we set $B'(\beta_{n-1}x)=B^{(n)}(x),$ then it’s not difficult to see that the above choice of $B^{(n)}$ implies that

$\tilde{A}^{(1)}(x)=B'(x+\alpha)A(x)B'(x)^{-1}\in SO(2,\mathbb R)$ for every $x,$ which in turn implies that $\tilde{A}^{(0)}(x)=B'(x+1)A(x)B'(x)^{-1}\in SO(2,\mathbb R)$ for every $x.$

Consider the commuting pair $(1,\tilde{A}^{(0)}(x))$ and $(\alpha, \tilde{A}^{(1)}(x)).$ A simpler version of proof of Lemma 1 implies that we can choose some $C^r$ normalizaing map $\tilde B\in SO(2,\mathbb R)$ for $(1,\tilde{A}^{(0)}(x)).$ Thus if we set $B(x)=\tilde B(x)B'(x),$ we have $B(x+\alpha)A(x)B(x)^{-1}\in SO(2,\mathbb R)$ and it is 1-periodic. This completes the proof. $\square$

Next post, I will do some computation concerning the limit of renormalization process in Schrodinger case in zero Lyapunov exponents regime.

## March 3, 2011

### Notes 9: Averaging and Renormalization(I)-Defining the Renormalization Operator

This is post will be a breif introduction about some averaging and renormalization procedures in Schrodinger cocycles. Renormalization is everywhere in dynamical systems and it has been used by many people in Schrodinger cocycles for a while. Lot’s of interesting results have been obtained. As for averaging, I guess it has not been very widely used in Schrodinger cocycles yet. Although it has been a powerful technique in Hamiltonian dynamics for a very long time in studying longtime evolution of action variables. Artur has already used them to construct counter-examples to Kotani-Last Conjecture and Schrodinger Conjecture. We’ve introduced Kotani-Last Conjecture in previous post, namely, the counter-example is that the base dynamics is not almost periodic but the corresponding Schrodinger operator admits absolutely continuous spectrum. For Schrodinger conjecture, the counter example is a Schrodinger operator whose absolutely continous spectrum admits unbounded generalized eigenfunctions.

I am not familar with both averaging and renormalization.  So I may need a little bit longer time to finish this post. But I will try my best to give a brief and clear introduction.

(I) Averaging

Let the frequency $\alpha$ be very Louville number. For instance, let $\alpha=\frac{1}{a_1+\frac{1}{a_2+\frac{1}{a_3+\frac{1}{\ddots}}}}$ be the continued fraction expansion. Thus $\frac{p_n}{q_n}=\frac{1}{a_1+\frac{1}{a_2+\frac{1}{a_3+\frac{1}{\ddots+\frac{1}{a_n}}}}}$ be the n’th step approximant. We let $a_n,$ hence $q_n,$ grows sufficiently fast (see my third post for introduction of Liouville number). Let the potential $v$ be a very small real analytic function. Combining averaging procedure with some renormalization and KAM procedure,  Avila, Fayad and Krikorian proved in their paper ‘A KAM scheme for $SL(2,\mathbb R)$ cocycles with Liouvillean frequencies’ the following theorem

Theorem:  Let $v: \mathbb R/\mathbb Z\rightarrow\mathbb R$ be analytic and close to a constant. For every $\alpha\in\mathbb R,$ there exists a positive measure set of $E\in\mathbb R$ such that $( \alpha, A^{(v,E)})$ is conjugate to a cocycle of rotations.

The main difficulty is Liouvillean frequency case, where they used the idea of averaging. We will not prove this theorem. We are going to give an idea how does the averaging procedure look like in Schrodinger cocycle.

Let’s start with the base dynamics $x\mapsto x.$ Namely, we consider one-parameter family of cocycles over fixed point. We have two different ways to consider the spectrum, which is the following

$I=\cup_x\Sigma_x:\{E\in\mathbb R: |tr(A^{(E-v)}(x))|=|E-v(x)|\leq 2$ for at least one $x \},$
$I^0=\cap_x\Sigma_x:\{E\in\mathbb R: |tr(A^{(E-v)}(x))|=|E-v(x)|\leq 2$ for all $x \}.$

Let’s start with $E\in I^0.$ For simplicity, set $A(x)=A^{(E-v)}(x)$ and we omit the dependence on $E.$ Let $u(x)$ be the invariant direction of $A(x).$ Let $B^0:\mathbb R/\mathbb Z\rightarrow SL(2,\mathbb R)$ be that $B^0(x)\cdot u(x)=i.$ Thus $B^0(x)A(x)B^0(x)^{-1}=R_{\psi^0(x)}$  for some $\psi^0:\mathbb R/\mathbb Z\rightarrow\mathbb R$ analytic. Thus there exists a constant $C_0$ such that

$B^0(x+\frac{p}{q})A(x)B^0(x)^{-1}=R_{\psi^0(x)}+\frac{C_0}{q}.$

Then for any closed interval $I^1$ inside the interior of $I^0,$ we can choose large $q$ such that $R_{\psi^0(x)}+\frac{C_0}{q}$ is again elliptic. Hence similarly, there is some $B^1:\mathbb R/\mathbb Z\rightarrow SL(2,\mathbb R)$ which is $\frac{1}{q}$ close to identity and $\psi^1:\mathbb R/\mathbb Z\rightarrow\mathbb R$ such that

$B^1(x)(R_{\psi^0(x)}+\frac{C_0}{q})B^1(x)^{-1}=R_{\psi^1(x)}.$ Hence

$B^1(x+\frac{1}{q})(R_{\psi^0(x)}+\frac{C_0}{q})B^1(x)^{-1}=R_{\psi^1(x)}+\frac{C_1}{q^2}$ for some constant $C_1.$

Thus $(\frac{1}{q},A)$ is conjugate to $(\frac{1}{q}, R_{\psi^1(x)}+\frac{C_1}{q^2}).$ Iterating $q$ times we get $R_{\psi_{q}^1(x)}+\frac{C_1}{q},$ where $\psi_{q}^1(x)$ is the n-th Birkhoff sume $\sum^{q-1}_{j=0}\psi(x+\frac{j}{q}).$ It’s not difficult to see that $\psi^1$ is close to its averaging $q\hat{\psi}(0).$ Thus except those bad $E$ such that $\psi_{q}^{1}(x)=\pm id$ for some $x,$ $(\frac{1}{q}, A^{(E-v)})$ stay ellipitc. By monotonicity of the Schrodinger cocycle with respect to energy and suitable choice of $q,$ these bad $E$ are of arbitrary small measure.

(II) Renormalization

Here we only introduce renormalization in quasiperiodic $SL(2,\mathbb R)$-cocycle where the frequency is one dimensional (i.e. the base dynamics is one dimensional). Let’s denote the renormalization operator by $\mathcal R.$ This will be a operator defined on the space of all cocycle dynamics. More concretely, it’s defined on the space of $\mathbb Z^2$ action on cocycle dynamics. We will continue to use the continued fraction expansion and approximants as in Averaging.

A natural way to see that the cocycle dynamics $(\alpha, A):(\mathbb R/\mathbb Z)\times\mathbb R^2\rightarrow (\mathbb R/\mathbb Z)\times\mathbb R^2$ is to see it as $(\alpha, A):\mathbb R\times\mathbb R^2\rightarrow \mathbb R\times\mathbb R^2,$ which commutes with $(1, id):\mathbb R\times\mathbb R^2\rightarrow \mathbb R\times\mathbb R^2.$ Because it’s easy to check that

$A(x)=A(x+1)\Leftrightarrow (\alpha,A)\cdot(1, id)=(1, id)\cdot(\alpha,A).$

This is similar to the following case. Denote the orientation preserving diffeomorphism on $\mathbb R$ by $Diffeo_+(\mathbb R).$ Then $F:\mathbb R/\mathbb Z\rightarrow\mathbb R/\mathbb Z$ is orientation preserving diffeomorphism if and only if it can be lifted to $Diffeo_+(\mathbb R)$ and commutes with $G:\mathbb R\rightarrow\mathbb R,$ where $G(x)=x+1.$ More generally, consider a commuting pair $(F,G)\in Diffeo_+(\mathbb R)$ where G has no fixed point. Then $F$ define a dynamical systems on $\mathbb R/G.$ Because $F(G(x))=G(F(x))\sim F(x)$ preserving the equivalent relation. Here $\mathbb R/G$ is diffeomorphic to $\mathbb R/\mathbb Z,$ but not in a canonical way.

Thus it’s more natural to view the pair $(F,G)$ as a $\mathbb Z^2$ group action, which is the following group homomorphism

$\Phi:\mathbb Z^2\rightarrow Diffeo_+(\mathbb R)$ with $\Phi(1,0)=G$ and $\Phi(0,1)=F.$

This automatically encodes the commuting relation.

The reason we introduce above procedure is that after renormalization, we will have no canonical way to glue $\mathbb R$ into $\mathbb R/\mathbb Z.$

Now let’s define the renormalization operator $\mathcal R$ step by step. We will renormalize around $0\in\mathbb R/\mathbb Z.$

(1) Define the $\mathbb Z^2$-action.
Let $\Lambda^r=\mathbb R\times C^r(\mathbb R,SL(2,\mathbb R))$ be subgroup of $Diff^r(\mathbb R\times\mathbb R^2).$ Then for $(\alpha,A)\in\Lambda^r$ satisfying $A(x)=A(x+1),$ we can define $\mathbb Z^2$ action $\Phi:\mathbb Z^2\rightarrow \Lambda^r$ such that $\Phi(1,0)=(1,id), \Phi(0,1)=(\alpha,A).$ Let $\Phi(n,m)=(\alpha_{n,m}, A_{n,m}).$

(2) Define two operations on the above $\mathbb Z^2$-action.
The first is the rescaling operator $M_{\lambda},$ which is given by

$M_{\lambda}(\Phi)(n,m)=(\lambda^{-1}\alpha_{n,m}, x\mapsto A_{n,m}(\lambda x)).$

The second is base changing operator $N_{U}.$ For $U\in GL(2,\mathbb Z),$ it’s given by

$N_{U}(\Phi)(n.m)=\Phi(n',m'), \binom{n'}{m'}=U^{-1}\binom{n}{m}.$

Obviously these two operations commutes with each other.

(3) More facts about continued fraction expansion.
Let $\alpha$ be as in averaging. Let $G$ be the Gauss map $G(\alpha)=\{\alpha^{-1}\}.$ Denote $G^n(\alpha)=\alpha_n.$ Thus $a_n=[\alpha_{n-1}^{-1}],$ where $a_n$ comes from the continued fraction expansion. Define a map

$U(x)=\begin{pmatrix}[x^{-1}]&1\\1&0\end{pmatrix}$ for $x\in (0,1).$

Let $\beta_n=\alpha_{n}\cdots\alpha.$ Then it’s easy to see that we also have $\beta_n=(-1)^n(q_n\alpha-p_n)=\frac{1}{q_{n+1}+\alpha_{n+1}q_n}.$

(4) Define the renormalization operator.
$\mathcal R:\Lambda^r\rightarrow\Lambda^r$ is given by

$\mathcal R(\Phi)=M_{\alpha}(N_{U(\alpha)}(\Phi)),$ where $\alpha$ is from $\Phi(0,1)=(\alpha,A).$ $\square$

It easy to see that

$\mathcal R(\Phi)(1,0)=M_{\alpha}(\Phi)(0,1)=(1,A(\alpha(\cdot)))$ and
$\mathcal R(\Phi)(0,1)=M_{\alpha}(\Phi)(1,-a_1)=M_{\alpha}(1-a_1\alpha, A_{-a_1})=(\alpha_1,A_{-a_1}(\alpha(\cdot))).$

Thus geometrically, if we  look at $\mathbb R\times\mathbb R\mathbb P^1,$ it looks like we glue $\{x\}\times\mathbb R\mathbb P^1$ and $\{x+\alpha\}\times\mathbb R\mathbb P^1$ via $(\alpha,A).$ Then by commuting relation $(1-a_1\alpha, A_{-a_1})$ define a dynamical systems on $(\mathbb R\times \mathbb R\mathbb P^1)/(\alpha,A).$ Then we rescale the first coordinate to be $\mathbb R/\mathbb Z$ again. Finally we get the new dynamics $\mathcal R(\Phi)(0,1).$

Let $Q_n=U(\alpha_{n-1})\cdots U(\alpha).$ Then it’s easy to see that

$\mathcal R^n(\Phi)=M_{\alpha_{n-1}}\circ N_{U(\alpha_{n-1})}\circ\cdots\circ M_{\alpha}\circ N_{U_{\alpha}}(\Phi)=M_{\beta_{n-1}}(N_{Q_n}(\Phi)).$

Thus we are looking at smaller and smaller space scale but larger and larger time scale.

Next post we will show how are the dynamics of original cocycle and these of renormalized cocycles related. And we will normalize $\mathcal R^n(\Phi)$ so that $\mathcal R^n(\Phi)(1,0)=(1,id)$ and do some computation concerning the limit of the renormalzation.

## February 24, 2011

### Notes 8:Examples(II)-Limit Periodic Potentials and Almost Mathieu Operator

This notes will be some further examples which are some natural generalization of periodic potentials. Example 3 will be limit periodic potential, where we will construct a potential with positive measure set of absolutely continuous spectrum. Example 4 will be quasi-periodic potential, in fact, the Almost Mathieu operator. I will only state some main results for the quasi-periodic example.

Example 3: Limit periodic potentials

There are two different equivalent ways to define limit periodic potentials. The first is that consider the $(X,f,\mu),$ where $X$ is a compact Cantor group, $f$ is a minimal translation and $\mu$ is Haar measure. Let $v:X\rightarrow\mathbb R$ be continuous. Then this potential $v$ is limit periodic. The equivalent say it’s limit periodic is that, start with any triple and consider the sequence $(v(f^n(x)))_{n\in\mathbb Z}.$ It’s limit periodic if it can be approximated in $l^{\infty}(\mathbb Z)$ by periodic sequence. In fact if so , the hull of  $(v(f^n(x)))_{n\in\mathbb Z}$ in $l^{\infty}(\mathbb Z)$ will be compact cantor group. Thus it’s not very difficult to see the equivalence between these two descriptions.

For simplicity, let’s consider the the Cantor group of 2-adic integers $X=\mathbb Z_2=\varprojlim\mathbb Z/(2^n\mathbb Z).$ Where the topology is induced from product topology. Taking minimal translation $f(x)=x+1.$ Note in this topology $\mathbb Z$ is dense and $2^n\rightarrow 0$ as $n\rightarrow\infty.$ Then potential $v$ on this space can be approximated by sequence of potentials $v^{(i)}$ of period $2^i.$ Let’s start with some $v^{(i)}.$ Define $v^{(i+1)}$ by induction as follows

$v^{(i+1)}(j)=v^{(i)}(j), 0\leq j\leq 2^i-1$ and
$v^{(i+1)}(j)=v^{(i)}(j-2^i)+\epsilon_i, 2^i\leq j\leq 2^{i+1}-1,$

where $\epsilon_i>0$ is sufficiently small. Thus

$A^{(i+1)}_{2^{i+1}}(0)=\tilde{A}^{(i)}_{2^{i}}(0)A^{(i)}_{2^{i}}(0)=A^{(i)}_{2^{i}}(0)^2+O(\epsilon_i).$

So if $E\in\mathcal K\subset \mathcal U\mathcal H,$ then $A^{(i+1)}_{2^{i+1}}(0)$ remains $\mathcal U\mathcal H$ for suitable $\epsilon_i.$ Here $\mathcal K$ can be any compact set. On the other hand, in the interior of each band we have no control for two types of points: boundary points, where the matrices is parabolic; these $E$ such that $tr(A^{(i)}_{2^i})=0.$ Because then $A^{(i)}_{2^{i}}(0)^2=-id$ and small perturbation may lead to the appearance of new gaps. But we can always ignore a small interval around these $E.$ Thus a large part of each band persists (note each band can be  broken into two bands).

Each time we choose a suitable smaller perturbation $\epsilon_i$ so that $\epsilon\rightarrow 0$ as $i\rightarrow\infty.$ Eventually, we can get spectrum $\Sigma$ such that $Leb(\Sigma)>0,$ which may also be a Cantor set.

On the other hand, for each $i,$ let $u^{(i)}(j)\in\mathbb H$ be the invariant direction of $A^{(i)}_{2^{i}}(j).$ Then it’s easy to see  $u^{(i)}(j)$ are invariant section of cocylce $A^{(i)}(j).$ Let $P^{i}:\mathbb Z_2\rightarrow \mathbb Z/(2^i\mathbb Z)$ be the projection. By above induction procedure, it’s not difficult to see that for each $l\in\mathbb Z_2$ and $E\in\Sigma,$ there are some $u(l,E)\in\mathbb H$ such that

$\lim\limits_{i\rightarrow\infty} u^{(i)}(P^{i}(l),E)=u(l,E).$ Obviously, $\lim\limits_{i\rightarrow\infty} A^{(i)}(P^{i}(l))=A^{(E-v)}(l).$

These imply that $u(l,E)$ is an invariant section of $A^{(E-v)}$ which takes values in $\mathbb H.$ Thus Lyapunov exponents stay zero through $\Sigma.$ Thus by Kotani Theory we get absolutely continous spetrum.

Example 4: Quasiperiodic Potentials-Amost Mathieu Operator

This type of potentials are of most interest. For simplicity, let’s focus on one dimensional frequency case, where the triple is $(\mathbb R/\mathbb Z, R_{\alpha}, Leb).$ As in the Corollary of Notes 6, $\alpha\in\mathbb R\setminus\mathbb Q$ is the frequency. Let $v\in C^{r}(\mathbb R/\mathbb Z,\mathbb R), r\in\mathbb N\cup\{\infty,\omega\}$ which is the quasiperiodic potential. Let’s recall the operator is for $u\in l^2(\mathbb Z)$

$(H_{x,\alpha}u)_n=u_{n+1}+u_{n-1}+v(x+n\alpha)u_n.$

The cocycle associated with the family of spectral equation $H_{x,\alpha}u=Eu,x\in\mathbb R/\mathbb Z$ is

$(\alpha, A^{(E-v)}),$ where $A^{(E-v)}\in C^r(\mathbb R/\mathbb Z,SL(2,\mathbb R))$ is defined as $A^{(E-v)}(x)=\begin{pmatrix}E-v(x)&-1\\1&0\end{pmatrix}.$

One of the mostly studied model is the so-called Almost Mathieu Operator, where $v(x)=2\lambda\cos(2\pi x)$ and $\lambda$ is the coupling constant. Here are some of the Theorems concerning this model

Theorem 1 (Bourgain-Jitomirskaya 2002): $L(E)=\max\{0,\ln|\lambda|\}$ for all $\lambda\in\mathbb R$ and $\alpha\in\mathbb R\setminus\mathbb Q.$

Theorem 2 (Avila-Jitmirskaya 2009): $\Sigma$ is a Cantor set for all $\lambda\neq0\in\mathbb R$ and $\alpha\in\mathbb R\setminus\mathbb Q.$

Theorem 3 (Avila-Krikorian 2006): $Leb(\Sigma)=|4-4|\lambda||$ for al $\lambda\in\mathbb R$ and $\alpha\in\mathbb R\setminus\mathbb Q$

From now on we will mostly focus on quasiperiodic potential case. Next posts I will do some averaging and renormalization procedures.

## February 21, 2011

### Notes 7: Examples (I)-Periodic Potentials

Filed under: Schrodinger Cocycles — Zhenghe @ 7:23 pm
Tags: , ,

From now on, I will mainly focus on the Schrodinger cocyles case.

Example 1: Constant Potentials

Let’s start with the simplest example: potential over fixed point. Equivalently, we consider constant $SL(2,\mathbb R)$– matrix $A$.

Since eigenvalues are invariant under conjugacy, we can up to a conjugacy assume $A$ is one of  the following

$\begin{pmatrix}\lambda&0\\ 0&\lambda^{-1}\end{pmatrix}, \begin{pmatrix}1&a\\ 0&1\end{pmatrix}, \begin{pmatrix}\cos2\pi\theta&-\sin2\pi\theta\\\sin2\pi\theta&\cos2\pi\theta\end{pmatrix}.$

Under mobius transformation, they have invariant directions in $\mathbb C\mathbb P^1$ as $\infty,\infty,i.$ Recall in Notes 3 we define functon

$\tau(A,m)$ such that $A\binom{m}{1}=\tau(A,m)\binom{A\cdot m}{1}.$

In particular if $u$ is the unstable invariant direction of $A,$ then $\ln\tau(A,u)=L(A)+i2\pi\rho(A).$ As the above cases, $u$ can be $\infty.$ But $\tau(A,u)$ (eigenvalue) is invariant under conjugacy. Thus we can always move the computation to unit disk to avoid $\infty.$ For simplicity let’s always denote it as $\tau(A,u).$ Then for above matrices, we get

$L(A)+i2\pi\rho(A)=\ln\tau(A,u)=\ln\lambda, 0, i2\pi\theta$ (note for each fixed $A, \rho(A)$ is well-defined but not uniquely determined).

Consider $A^{(E)}=\begin{pmatrix}E&-1\\1&0\end{pmatrix}$ and the function $\rho:\mathbb R\rightarrow\mathbb R, E\rightarrow\rho(E).$ Let’s use $Q=\frac{-1}{1+i}\begin{pmatrix}1&-i\\1&i\end{pmatrix}$ to move to unit disk $\mathbb D.$ Then $\mathbb R$ becomes the unit circle $\mathbb S.$ Let’s consider the function $N(E)$ (see notes 3) instead of $\rho(E).$ It’s easy to see that for nonelliptic matrices, $N$ is integer valued. By monotonicity of Notes 3, $N:\mathbb S\rightarrow\mathbb R$ is not well-defined. But it’s well-defined as $N:\mathbb S\rightarrow\mathbb R/\mathbb Z.$  By the proof of Lemma 2 of Notes 3, we know when $E$ changes from $-\infty$ to $\infty,$ so is some $\cot(\theta(E))=A^{(E)}\cdot y.$  Thus the change of correponding angle of the point in $\mathbb S$ is exactly $2\pi.$ This implies that $\deg N=1.$ WLOG, we can set $N(-\infty)=-1, N(\infty)=0.$ Thus $\rho(-\infty)=\frac{1}{2}, \rho(\infty)=0.$

The above argument can be applied to any $A^{(E-v)}=\begin{pmatrix}E-v&-1\\1&0\end{pmatrix},v\in R.$ Combining the equivalent description of spectrum in Notes 2 give the following description of the corresponding operator:

– the spectrum $\Sigma$ is $[v-2,v+2];$
$\rho|_{(-\infty,v-2]}=\frac{1}{2},$  $\rho|_{[v+2,\infty)}=0;$
$\rho$ is analytic and strictly decreasing in the interior of the spectrum $(v-2, v+2).$

Example 2: Periodic Potentials

Next let’s consider periodic potential case. Namely, $(X,f,\mu)=(\mathbb Z/(q\mathbb Z), f, \mu),$ where $f: \mathbb Z/(q\mathbb Z)\rightarrow \mathbb Z/(q\mathbb Z), x\mapsto x+1$ and $\mu$ is the averaged counting measure. Let $v:\mathbb Z/(q\mathbb Z)\rightarrow \mathbb R,$ hence $A^{(v)}:\mathbb Z/(q\mathbb Z)\rightarrow SL(2,\mathbb R).$

For any $A:\mathbb Z/(q\mathbb Z)\rightarrow SL(2,\mathbb R),$ we have for any $j, A_q(j+1)A(j)=A(j)A_q(j).$ This implies that

-the eigenvalue $\tau(A_q(j), u_q(j))$ of $A_q(j)$ is independent of $j.$ Denote it by $\tau(A_q, u_q).$
$L(A)=\frac{1}{q}\ln\delta(A_q).$

Here for any bounded linear operator $A$ on any Banach space, $\delta(A)=\lim\limits_{n\rightarrow\infty}\|A^n\|^{\frac{1}{n}}=\inf_{n\leq1}\|A^n\|^{\frac{1}{n}}$ is the spectral radius of $A.$  From the above facts we get

$L(A)+i2\pi\rho(A)=\frac{1}{q}\ln\tau(A_q,u_q).$

We also have the following obvious equivalent relations

$L(A)>0\Leftrightarrow |trA_q|>2\Leftrightarrow (f,A)\in\mathcal U\mathcal H$ and  $|trA_q|<2\Leftrightarrow$ Elliptic,

where $tr$ stands for trace. For the second case we also have $tr(A_q)=2\cos(2\pi \rho(A)q).$

Back to Schrodinger case, there is a nice graph of the function $\psi(E)=trA^{(E-v)}_q.$ (which is also $2\cos(2\pi\rho(E)q)$ for elliptic case). It’s obviously a polynomial in $E$ of degree $q.$ It has exactly $q-1$ critical points with critical values $y$ satisfying $|y|\geq2.$ We have in fact the following description of the spectrum $\Sigma$ of the correponding operator $H_v$

Theorem: $\Sigma$ consists of $q$ bands and there are $q-1$ spectral gaps ( bands may touch at the boundary points, or equivalently, gap may collapse).

Proof: By the same argument for Schrodinger cocycle over fixed point above and the fact $\rho(AB)=\rho(A)+\rho(B)$, we have the following easy facts for $\rho(E)$ as a function on real line:

-it’s continuous on $\mathbb R$ and is analytic in the interior of spectrum;
-it’s $\frac{k}{2q}$-valued outside the interior of spectrum thus constant in each connected components of the resolvent set;
$\rho(-\infty)=\frac{1}{2}$ and $\rho(\infty)=0$ and nonincreasing.

These together implies that:

$\psi^{-1}[-2,2]$ consists of $q$ compact intervals. Some of them may touch at the boundary points. These are spectrum bands. $\rho'(E)<0$ on $\psi^{-1}(-2,2);$
-between each two bands is the so-called spectrum gap and there are $q-1$ of them (some of them may collasped). On each of them $\rho(E)=\frac{k}{2q},k=q-1,\cdots,1.$ These are labelings of spectral gaps. $\square$

The proof obviously implies the properties of the polynomial $\psi(E).$

A natural question is that when are all gaps open (not collapsed). Obviously by the proof of Theorem, all gaps are open if and only if all the roots of polynomials $\psi(E)-2, \psi(E)+2$ are simple. Let’s give another description of these polynomials. Let’s restrict the operator $H_v$ to $l^2(\mathbb Z/(q\mathbb Z))$ with three different type of boundary conditions. Let $u\in l^2(\mathbb Z/(q\mathbb Z))$

1. First let’s restrict to any subinterval $[j,j+n-1]\subset[0,q-1]$ and consider Dirichlet boundary condition, i.e. $u(j-1)=u(j+n)=0.$ Denote it by $H^{n}_{v,j}$. In this case it can be represented as a $n\times n$ symmetric matrix

$H^{n}_{v,j}=\begin{pmatrix}v(j)&1&0&0\\1&v(j+1)&1&\vdots\\ 0&1&v(j+2)&\vdots\\ 0&0&1&\vdots\\\vdots&\vdots&0&1\\\vdots&\vdots&\vdots&v(j+n-1)\end{pmatrix}$

Let $\det(H^{n}_{v,j}-E)=P_n(j,E).$ Then by induction it’s easy to see that

$A^{(v-E)}_n(j)=\begin{pmatrix}P_n(j,E)&-P_{n-1}(j+1,E)\\P_{n-1}(j,E)&-P_{n-2}(j+1,E)\end{pmatrix}.$

Hence $\psi(E)=P_{q}(0,E)-P_{q-2}(1,E).$ In case 2 and 3 we will only restrict to $[0,q-1].$

2. Periodic boundary condition, i.e. $u(-1)=u(q-1), u(q)=u(0).$ Denote it by $H^{p}_v,$ then it is

$H^{p}_v=\begin{pmatrix}v(0)&1&0&1\\1&v(1)&1&0\\ 0&1&v(2)&\vdots\\ 0&0&1&0\\ 0&\vdots&0&1\\1&0&\vdots&v(q-1)\end{pmatrix}$

3. Antipeiodic boundary condition, i.e. $u(-1)=-u(q-1), u(q)=-u(0).$ Denote it by $H^{a}_v,$ then it is

$H^{a}_v=\begin{pmatrix}v(0)&1&0&-1\\1&v(1)&1&0\\ 0&1&v(2)&\vdots\\ 0&0&1&0\\ 0&\vdots&0&1\\-1&0&\vdots&v(q-1)\end{pmatrix}$

Then by case 1 it’s not difficult to see that for some integer $l,$

$\psi(E)-2=\begin{cases}\det(H^{p}_v-E)&\text{ if }q=2l\\\det(H^{a}_v-E)&\text{ if }q=2l-1\end{cases}$ and $\psi(E)+2=\begin{cases}\det(H^{a}_v-E)&\text{ if } q=2l\\\det(H^{p}_v-E)&\text{ if } q=2l-1\end{cases}.$

Thus all the spectral gaps are open if and only if all eigenvalues of the operators $H^{p}_v$ and $H^{a}_v$ are simple. An easy case is that assume $v(j), j=0,\cdots,q-1$ are sufficiently large and distinct. Then after scaling, all nondigonal coefficients of $H^{p}_v$ and $H^{a}_v$ are sufficiently small. Namely, $H^{p}_v$ and $H^{a}_v$ are small perturbation of digonal matrices with distinct eigenvalues. Then so are $H^{p}_v$ and $H^{a}_v$ themself. Which implies that all gaps are open.

For simplicity, consider $q$ even. Denote eigenvalues of $H^{p}_v$ by $\mu_1<\cdots<\mu_q$ and $H^{a}_v$ by $\lambda_1<\cdots<\lambda_q.$ Then spectrum bands are

$[\mu_j,\lambda_j], j$ odd; $[\lambda_j,\mu_j], j$ even.

And spectral gaps are

$(\mu_j,\mu_{j+1}), j$ even; $(\lambda_j,\lambda_{j+1}), j$ odd.

Finally let’s consider $L(E)+i2\pi\rho(E):\mathbb H\rightarrow\mathbb C.$ Then it’s easy to see that it can be analytically extends through interior of each bands and through gaps. But they cannot be globally defined, since there are nontrivial winding of the invariant direction $u_q(E)$ around each $\mu_j, \lambda_j.$ This winding comes from the parabolicity. Indeed, if we instead consider the eigenvalue $\lambda(E)$ of $A^{(E-v)}_q,$ then

$\frac{1}{q}\ln\lambda(E)=L(E)+i2\pi\rho(E)$ and $\lambda(E)=\frac{1}{2}(\psi(E)\pm\sqrt{\psi(E)^2-4}),$

of which the derivative has singularity at parabolicity. This also explain why $L(E)$ and $\rho(E)$ as functions on the whole real line can at most be $\frac{1}{2}-H\ddot older$ continous.

Note for periodic potential and in spectrum bands, it’s always $L(E)=0$ since they are just parabolic and elliptic matrices. Thus by Theorem 2 of Notes 4, all the spectrum are purely absolutely continous.

Assume potentials are uniformly bounded. Then it’s interesting to note that as $q\rightarrow\infty,$ there are more and more bands which are also thiner and thiner. Thus in quasiperiodic potential case, it’s natural to expect the spectrum is Cantor set under some assumption. We will give an example in next post.

## February 16, 2011

### Notes 6: Density of Positive Lyapunov Exponents for SL(2,R)-cocycles in any Regularity Class

This post is about the density of positive Lyapunov exponents for $SL(2,\mathbb R)$ cocylces in any regular class. It’s again from Artur Avilas course here in Fields institute, Toronto and from his paper Density of Positive Lyapunov Exponents for $SL(2,\mathbb R)$ cocylces.’ Since there are very detailed descriptions and proofs in the paper, I will only state one of the main theorems and give the idea of proof and point out how are they related to previous posts.

Recall that the Corollary of last post is a stronger result, but only in $C^0$ class. Obviously, it’s more difficult to obtain density results in higher regularity class.

Again I will use the base space $(X,f,\mu);$ assume $X=supp(\mu)$ and $f$ is not periodic. I will use $L(A)$ to denote the Lyapunov exponent of the corresponding cocyle map $A:X\rightarrow SL(2,\mathbb R).$ Let’s first introduce a concept to state the main theorem.

Definition: A topological space $\mathfrak B\subset C(X,SL(2,\mathbb R))$ is ample if there exists some dense vector space $\mathfrak b\subset C(X,sl(2,\mathbb R))$, endowed with some finer (than uniform) topological vector space structure, such that for every $A\in\mathfrak B, e^{b}A\in\mathfrak B$ for every $b\in\mathfrak b,$ and the map $b\mapsto e^{b}A$ from $\mathfrak b$ to $\mathfrak B$ is continous.

Remark: Note that if $X$ is a $C^r$ manifold, $r\in\mathbb N\cup\{\infty,\omega\},$ then $C^r(X,SL(2,\mathbb R))$ is ample. Namely we can take $\mathfrak b=C^r(X,sl(2,\mathbb R)).$

The main theorem is the following

Theorem 1: Let $\mathfrak B\subset C(X, SL(2,\mathbb R))$ be ample. Then the Lyapunov exponent is positive for a dense subset of $\mathfrak B.$

Remark: This is basically an optimal result for density of positive Lyapunov exponents for $SL(2,\mathbb R)$ cocycles.

The key theorem lead to Theorem 1 is the next theorem. Let $\|\cdot\|_*$ denote the sup norm in $C(X,sl(2,\mathbb R))$ and $C(X,sl(2,\mathbb C)),$ and for $r>0$ let $\mathcal B_*(r)$ and $\mathcal B_*^{\mathbb C}(r)$ be the correponding $r$-balls. Then

Theorem 2: There exists $\eta>0$ such that if $b\in C(X,sl(2,\mathbb R))$ is $\eta$-close to $\begin{pmatrix}0&1\\-1&0\end{pmatrix},$ then for $\epsilon>0$ and every $A\in C(X,SL(2,\mathbb R)),$ the map

$\Phi(\cdot; A):\mathcal B_*(\eta)\rightarrow\mathbb R, a\mapsto\int_{-1}^{1}\frac{1-t^2}{|t^2+2it+1|}L(e^{\epsilon(tb+(1-t^2)a)}A)dt$

is an analytic function, which depends contiously (as an analytic function) on $A.$

Remark: It will be clear later why this leads to Theorem 1. All the main ingredients for proving Theorem 2 have in fact already been included in previous posts.

Idea of Proof of Theorem 2: The key point is to find $\eta>0$ such that we can check:
1. For $z\in\partial{\mathbb D}\cap\mathbb H$ and for $z=(\sqrt2-1)i, e^{\epsilon(zb+(1-z^2)a)}A$  is $\mathcal U\mathcal H,$ provided $a\in\mathfrak B_*^{\mathbb C}(\eta),$
2. For $z\in\mathbb D\cap\mathbb H, e^{\epsilon(zb+(1-z^2)a)}A$ is $\mathcal U\mathcal H,$ provided $a\in\mathfrak B_*(\eta).$

On the other hand, we can write down the explicit conformal transformation $\psi:\mathbb D\rightarrow\mathbb D\cap\mathbb H$ such that $\psi(z)=\phi^{-1}(\sqrt{\phi(z)}),$ where $\phi:\mathbb D\rightarrow\mathbb H, z\mapsto i\frac{1-z}{1+z}.$ Notice that $\psi(0)=(\sqrt2-1)i.$ Let’s denote $\rho(z,a)=L(e^{\epsilon(zb+(1-z^2)a)}A).$ Once we have these facts, by pluriharmonic theorem in the postProof of HAB formula‘ and mean value formula for harmonic functions, we have

a.$\rho(\psi(0),a)=\int_0^1\rho(\psi(e^{2\pi i\theta}),a)d\theta$ for $a\in\mathfrak B_*(\eta)$ by fact 2; thus
b.$\int_0^{1/2}\rho(\psi(e^{2\pi i\theta}),a)d\theta=\rho(\psi(0),a)-\int_{1/2}^{1}\rho(\psi(e^{2\pi i\theta}),a)d\theta;$ but
c.$\rho(\psi(0),a)-\int_{1/2}^{1}\rho(\psi(e^{2\pi i\theta}),a)d\theta$ is pluriharmonic for $a\in\mathfrak B_*(\eta)$ by fact 1,

from which plus some additional direct computation will establish the result of Theorem 2. This argument is similar to Lemma 4 of last post and the proof of HAB formula.

The idea to obtain facts 1 and 2 is to check that for $m\in\mathbb R,$ we define function $m(\epsilon)=e^{\epsilon(zb+(1-z^2)a)}\cdot m\in\mathbb P\mathbb C^1$ and check that $\frac{d}{d\epsilon}m(\epsilon)$ at $\epsilon=0$ points inside $\mathbb H$ for $z, a$ in facts 1 or 2. Thus $\mathbb H$ will be an invariant conefield for $e^{\epsilon(zb+(1-z^2)a)}A$ for any $A\in C(X,SL(2,\mathbb R))$ and $\epsilon>0$ small, which implies $\mathcal U\mathcal H.$ This is similar to the cases in Kotani theory or HAB formula, where when we complexify $E$ or $\theta$ we get $\mathcal U\mathcal H.$ For detailed proof see Artur’s paper. $\square$

Proof of Theorem 1: We must show that for every $A_1\in\mathfrak B,$  there exists a $A_2\in\mathfrak B$ sufficiently close to $A_1$ in $\mathfrak B$ and $L(A_2)>0.$

For any $\epsilon>0.$ Let $\gamma(t)=L(e^{\epsilon(tb+(1-t^2)a)}A_1).$ Then by subharmonicity of $\gamma$, more concretely, by upper semicontinuity and sub-mean value property, we can choose suitable closed path to see that if $\gamma(0)>0,$ then $\Phi(a;A_1)>0.$

Since $\mathfrak B$ is ample, we can  choose suffciently small $\epsilon>0$ and some $b$ as in Theorem 2 such that $e^{b}A_1\in\mathfrak B$ and $e^{\epsilon tb}A_1$ is sufficient close to $A_1$ in $\mathfrak B$ for every $t\in [-1,1].$ Then by above observation and Corollary of last post we can find some $a\in\mathcal B_*(\eta)$ such that $\Phi(a;A_1)>0.$

Again by the assumption that $\mathcal B$ is ample and the analyticity of the map $\Phi(\cdot;A_1),$ we can assume $a$ such that  $e^{\epsilon(tb+(1-t^2)a)}A_1\in\mathfrak B.$

By Theorem 2, the function $\phi(s)=\Phi(sa;A_1)$ is analyic in $s\in [-1,1].$ Since $\phi(1)>0$, we have for every sufficiently small s>0, $\phi(s)>0.$ Thus we can choose sufficiently small $s_0$ and some $t_0\in[-1,1]$ such that $A_2=e^{\epsilon(t_0b+(1-t_0^2)s_0a)}A_1$. $\square$

For me it’s very interesting to see how Kotani Theory, Uniform Hyperbolicity and Mean value formula lie at the bottom of this density result.

Let me mention an interesting application of Theorem 1. Consider the case $(X,f,\mu)=(\mathbb R/\mathbb Z,R_{\alpha}, Leb),$ where $R_{\alpha}:\mathbb R/\mathbb Z\rightarrow\mathbb R/\mathbb Z, x\mapsto x+\alpha$ and $\alpha$ is irrational.

Let’s consider the cocylce space $C^{\omega}(\mathbb R/\mathbb Z, SL(2,\mathbb R))$ endowed with some inductive limit topology via subspace $C^{\omega}_{\delta}(\mathbb R/\mathbb Z, SL(2,\mathbb R)).$  Here $\delta>0$ and $C^{\omega}_{\delta}(\mathbb R/\mathbb Z, SL(2,\mathbb R))$ is the space of  real analytic cocycle maps which can be extended to $\{z\in\mathbb C/\mathbb Z: |\Im z|<\delta\}.$

Then there is a theorem started with the Schrodinger cocycles in the regime of positive Lyapunov exponents and Diophantine frequencies in Goldstein and Schlag‘s paper ‘Holder Continuity of the IDS for quasi-periodic Schrodinger equations and averages of shifts of subharmonic functions’ , continued as all irrational frequencies and all Lyapunov exponents Schrodinger case in Bourgain and Jitomirskaya‘s paper ‘Continuity of the Lyapunov Exponent for Quasiperiodic Operators with Analytic Potential‘ and ended up as the general real analytic $SL(2,\mathbb R)$ cocyle case in Jitomirkaya, Koslover and Schulteispaper ‘Continuity of the Lyapunov Exponent for analytic quasiperiodic cocycles’ such that

Theorem: The Lyapunov exponent $L:(\mathbb R\setminus\mathbb Q)\times C^{\omega}(\mathbb R/\mathbb Z, SL(2,\mathbb R)), (\alpha, A)\mapsto L(\alpha,A)$ is jointly continuous.

(Artur will talk about the proof of this theorem in future classes, so maybe I will post the idea of proof in the future ). Combining with Theorem 1 we obviously have the following Corollary

Corollary: For any fixed irrational frequency $\alpha,$ Lyapunov exponent is positive for an open and dense subset of $C^{\omega}(\mathbb R/\mathbb Z, SL(2,\mathbb R)).$

To finish topics closely related to Kotani Theory, let’s mention the follow Kotani-Last Conjecture:

In Schrodinger cocycle case: $Leb\{E:L(E)=0\}>0\Rightarrow$ Almost Periodicity of the base dynamics.

Recall that by Notes 4, we know $Leb\{E:L(E)=0\}>0\Rightarrow$ determinism. On the other hand almost periodicity is stronger then determinsim, which has the following equivalent description:

For any $\forall\epsilon>0, \exists\delta>0$ and $N\in\mathbb Z^+$ such that
$|v(f^n(x))-v(f^n(y))|<\delta,\forall -N\leq n\leq 0\Rightarrow |v(f^n(x))-v(f^n(y))|<\epsilon, \forall n\in\mathbb Z.$

It means that sufficiently precise finite information determines the whole potential to specified precision. It is obviously stronger than determinism. Thus it seems natural to pose above conjecture. Unfortunately, it turns out this is not true. Artur already has a counter example.

## February 11, 2011

### Notes 5: Application of Kotani Theory (II)-Genericity of Singular Spectrum

Filed under: Schrodinger Cocycles — Zhenghe @ 12:29 am

This post is about the main theorem of Artur Avila and David Damaniks paper Generic singular spectrum for ergodic $Schr\ddot odinger$ operators’. It is another application of Kotani Theory, which also has its own interests. I will break the proof of main theorem down to several lemmas and point out the key ideas. I will also try to carry out all the details. I am always grateful to Artur who is always willing to explain to me ideas and details whenever I want.

In this post, I will use $L(E,v)$ to denote the Lyapunov exponents of the system $(f,A^{(E-v)}).$ The main theorem is the following

Theorem: Assume $f$ is not periodic ($f^k\neq Id$ for any $k\geq 1$). Then for generic $v\in C(X,\mathbb R)$ (generic means residual), $L(E,v)>0$ for a.e. $E.$

Remark: By Kotani theory this Theorem implies that for generic continuous $v,$ $\Sigma_{ac}(x)=\varnothing$ for a.e. $x.$ Thus for generic continuous potentials, the spectrum is of singular type, i.e. singular continuous and pure points.

Let’s consider the unstable direction $u$ as a function $l^{\infty}(\mathbb Z^-)\times\mathbb H\rightarrow\mathbb H,$ where $\mathbb Z^-$ are nonpositive integers and $l^{\infty}(\mathbb Z^-)$ stands for bounded real-valued sequence on $\mathbb Z^-$.  Let

$B_{1,r}=(L^1(X,d\mu)\cap B_r(L^{\infty}(X,d\mu)), \|\cdot\|_1),$

where $r>0$ and $B_r(L^{\infty}(X,d\mu))$ is the ball around zero function in $L^{\infty}(X,d\mu)$ with radius $r.$ Now let’s state and prove our lemmas.

Lemma 1: For any bounded subset $\mathcal K\subset l^{\infty}(\mathbb Z^-)\times\mathbb H,$ $u(\mathcal K)\subset\mathbb H$ is bounded. Here boundedness in $\mathbb H$ is with respect to hyperbolic metric.

Proof: WLOG, I can assume for any $(v,E)\in\mathcal K,$ $\|v\|_{\infty}\leq c$ and $|\Re E| for some $c>0.$ Let

$\mathcal C=\{(a,E)\in\mathbb R\times\mathbb H:|a|\leq c$ and $|\Re E|

Then we show that $u(\mathcal K)\subset A^{(\mathcal C)}A^{(\mathcal C)}\cdot\mathbb H,$ of which the latter is a bounded set in $\mathbb H.$ Here I use $A^{(a,E)}$ to denote the map $(a,E)\mapsto A^{(E-a)}.$

Indeed, since $u(v,E)=\lim\limits_{n\rightarrow\infty}A^{(E-v_{-1})}\cdots A^{(E-v_{-n})}\cdot i=\lim\limits_{n\rightarrow\infty}s_n(v,E).$ We can of course take limit along even integers. Thus the inclusion $u(\mathcal K)\subset A^{\mathcal (C)}A^{\mathcal (C)}\cdot\mathbb H$ is obvious. For boundedness it’s easy to see that for any $(a,E)\in\mathcal C,$

$A^{(E-a)}\cdot\mathbb H\subset\mathbb H_{\frac{1}{c}}$ and $A^{(E-a)}\cdot\mathbb H_{\frac{1}{c}}\subset \{y\in\mathbb H: |\Re y|\leq 3c, |\Im y|\leq 2c\}.$ $\square$

Lemma 2: Assume $v^{(l)},v\in [-c,c]^{\mathbb Z^-}$ for some $c>0$. Let $v^{(l)}\rightarrow v$ as $l\rightarrow\infty$ pointwisely. Then $\lim\limits_{l\rightarrow\infty}u(v^{(l)},\cdot)=u(v,\cdot)$ in compact open topology as functions on $\mathbb H$ (i.e. uniform convergence in any compacta in $\mathbb H$)

Proof:  By the proof of Theorem 1, we actually see that $s_n(v,E)$ converge to $u(v,E)$ in compact open topology  and the convergence is independent of $v\in [-c,c]^{\mathbb Z^-}$. This is due to the same reason of the proof of Theorem 2 of last post. Namely, $s_n(v,E), n\geq 1$ are holomorphic functions takes value in $\mathbb H$, thus they are normal family. The independence with respect to $v$ is due to the uniform shrink rate of invariant cone field under projectivized action.

Now we have

$|u(v^{(l)},E)-u(v,E)|\leq|u(v^{(l)},E)-s_n(v^{(l)},E)|$
$+|s_n(v^{(l)},E)-s_n(v,E)|+|s_n(v,E)-u(v,E)|.$

Thus for any $\epsilon>0,$ we can choose $N$ large such the first and third terms in the summation above are both less than $\frac{\epsilon}{3}$ on any compacta in $\mathbb H$. For this fixed $N,$ $|s_N(v^{(l)},E)- s_N(v,E)|<\frac{\epsilon}{3}$ for any $l$ large enough and $E$ in any compacta. $\square$

Lemma 3: For any fixed $E\in\mathbb H,$ the function $L(E,\cdot): B_{1,r}\rightarrow \mathbb R$ is continuous.

Proof: By passing to subsequence we can assume $v^{(l)}(x), v(x)\in B_{1,r}$ such that $v^{(l)}(x)\overset{\|\cdot\|_1}{\rightarrow}v(x)$ and pointwisely as $n\rightarrow\infty.$

Thus for a.e. $x\in X$ we have the $(v^{(l)}(f^n(x)))_{n\in\mathbb Z^-}$ converges to $(v(f^n(x))_{n\in\mathbb Z^-}$ pointwisely as $l\rightarrow\infty.$ By Lemma 1 this implies that $u(x,E,v^{(l)})$ converges to $u(x,E,v)$ for a.e. $x$. And they lie in a compact set in $\mathbb H$ by our choice of the ball $B_{1,r}.$ Hence by Bounded Convergence Theorem, we have

$\lim\limits_{l\rightarrow\infty}L(E,v^{(l)})=\lim\limits_{l\rightarrow\infty}\int_X\ln|u(x,E,v^{(l)})|d\mu$
$=\int_X\ln|u(x,E,v)|d\mu=L(E,v).$ $\square$

Remark: Note what we need for $v^{(l)}$ in Lemma 3 are just uniform boundedness and pointwise convergence.

Lemma 4: Fix any interval $[-N, N]\subset\mathbb R.$ Then the function $\int_{-N}^{N}L(E,\cdot)dE: B_{1,r}\rightarrow \mathbb R$ is continuous.

Remark: As a function, $L(E,v), E\in\mathbb R$ behave badly when $(f,A^{(E-v)})\notin\mathcal U\mathcal H$. The only thing that is always true is that it’s upper semicontinuous and plurisubharmonic. This Lemma shows that after taking an appropriate integral, it behaves nicely in $L^1$ sense.

Proof: Consider the semicircle $\gamma$ such that $\gamma=[-N,N]\cup S^+(N),$ where $S^+(N)$ is upperhalf part of  the circle centered at origin with radius $N.$ Pick a point $E_0$ inside $\gamma.$ Then there is a harmonic measure $\nu_{E_0}$ on $\gamma$ such that

$\int_{\gamma}L(E,v)d\nu_{E_0}=L(E_0,v).$

By our assumption, Lemma 3 and bounded convergence theorem, it’s easy to see that $\int_{S^+(N)}L(E,\cdot)d\nu_{E_0}$ and $L(E_0,\cdot)$ are continuous. Thus $\int_{[-N,N]}L(E,v)d\nu_{E_0}$ is also continuous. Via the conformal transformation which transform the unit disk  to the region inside $\gamma,$  it’s not difficult to see that $\int_{[-N,N]}L(E,v)dE$ is also continuous. $\square$

Recall that $\mathcal Z(v)=\{E:L(E,v)=0\}.$ We define a function $\mathcal M: B_{1,r}\rightarrow \mathbb R$ such that $\mathcal M(v)=Leb(\mathcal Z(v))$. Then

Lemma 5: $\mathcal M: B_{1,r}\rightarrow \mathbb R$ is upper semicontinuous.

Proof: We only need to restrict our self on $E\in[-2-r,2+r]=I_r.$ Because outside this interval $E$ is always in resolvent set hence the systems are always in $\mathcal U\mathcal H$ and Lyapunov exponents are positive.

It’s obvious that it suffices to show for any $v\in B_{1,r}$ such that $\mathcal M(v)<4r+4$ and any $0<\epsilon<4r+4-\mathcal M(v),$ there exists a $\delta>0$ such that whenever $v'\in B_{1,r}$ and $\|v'-v\|_1<\delta$ we have $\mathcal M(v')<\mathcal M(v)+\epsilon.$

Let’s show how to choose such a $\delta>0.$ By definition of $\mathcal M(v)$ we can choose a $\alpha>0$ such that $Leb\{\mathcal Z_{\alpha}(v)\}<\mathcal M(v)+\epsilon,$ where $\mathcal Z_{\alpha}(v)=\{E: L(E,v)<\alpha\}.$ Then by Lemma 4 we can choose $\delta$ such that $\int_{I_r}|L(E,v)-L(E,v')|dE<\beta$ with $0<\beta<\frac{\alpha}{2}(4r+4-\mathcal M(v)-\epsilon).$ Indeed we then have

$Leb(I_r\setminus\mathcal Z_{\alpha}(v))\times(\inf_{E\in I_r\setminus\mathcal Z_{\alpha}(v)}\{|L(E,v)-L(E,v')|\})$
$\leq\int_{I_r\setminus\mathcal Z_{\alpha}(v)}|L(E,v)-L(E,v')|dE<\beta<\frac{\alpha}{2}(4r+4-\mathcal M(v)-\epsilon).$

Which implies that $\inf_{E\in I_r\setminus\mathcal Z_{\alpha}(v)}\{|L(E,v)-L(E,v')|\}<\frac{\alpha(4r+4-\mathcal M(v)-\epsilon)}{2Leb(I_r\setminus\mathcal Z_{\alpha}(v))}<\frac{\alpha}{2}.$

Which implies that $L(E,v')>\frac{\alpha}{2}>0$ on $I_r\setminus\mathcal Z_{\alpha}(v),$ thus $\mathcal Z(v')\subset\mathcal Z_{\alpha}(v).$ This obviously implies what we want to show. $\square$

Lemma 6: There exists a dense set $\mathfrak S\subset L^{\infty}(X,d\mu)$ such that if $s\in\mathfrak S,$ then
(1) $s(x), x\in X,$ takes finitely many values;
(2) $(s(f^n(x)))_{n\in\mathbb Z}$ is not periodic for a.e. $x\in X.$

Proof: (This proof is due to Artur. It’s simpler than the one in their paper) For a simple function $s,$ let’s define $\Phi(s,x)\in \mathbb Z^+\cup\{\infty\}$ to be the period of the sequence $(s(f^n(x)))_{n\in\mathbb Z}.$ Then by ergodicity for a.e. $x,$ $\Phi(s,x)=\Phi(s).$ Again by ergodicity it’s not difficult to see that if $\Phi(s)<\infty,$ then $\mu(s^{-1}(a))\in\mathbb Q.$ Thus by our assumption on the triple $(X,f,\mu),$ if $\Phi(s)<\infty,$ we can always change the value on a small measure set to produce a new simple function $\hat s$ sufficiently close to $s$ in $L^{\infty}$ such that $\Phi(\hat s)=\infty.$ $\square$

Lemma 7: For any $\delta>0, \{v\in C(X,\mathbb R), \mathcal M(v)<\delta\}$ is dense in $C(X,\mathbb R)$ in $L^{\infty}$ topology.

Proof: Fix abitrary $v\in C(X,\mathbb R).$ By Lemma 6, we can choose $s\in\mathfrak S$ such that $s$ is sufficiently close to $f$ in $L^{\infty}.$ By Theorem 4 of last post, we have $\mathcal M(s)=0.$ By standard real analysis theorem and by Lemma 5, we can choose $v'\in C(X,\mathbb R)$ such that $v'$ is sufficiently close to $s$ both in $L^{\infty}$ and $L^1$ such that $\mathcal M(v')<\delta.$ $\square$

Now we are ready to prove our main Theorem

Proof of Theorem: For any $n\in\mathbb Z^+,$ we have by Lemma 5 and Lemma 7 $M_{n}=\{v\in C(X,\mathbb R):\mathcal M(v)<\frac{1}{n}\}$ is both open and dense. Thus $\cap_{n\in\mathbb Z^+}M_n$ is residual, which completes the proof. $\square$

Corollary: There is generic set $\mathcal C\subset C(X,SL(2,\mathbb R))$ such that for each $A\in\mathcal C,$
$Leb\{\theta\in\mathbb R/\mathbb Z: L(f,R_{\theta}A)=0\}=0.$
Where $R_{\theta}$ is the rotation matrix with rotation angle $2\pi\theta.$

Sketch of Proof: Use HAB formula we can get the corresponding Lemma 4 in terms of the function $A\mapsto\int_{\mathbb R/\mathbb Z}L(f,R_{\theta}A)d\theta=\int_X \ln\frac{\|A(x)\|+\|A(x)\|^{-1}}{2}d\mu,$ hence Lemma 5 in terms of the map $\hat{\mathcal M}(A)=Leb\{\theta\in\mathbb R/\mathbb Z: L(f,R_{\theta}A)=0\}.$ On the other hand, Kotani theory can be carried over to this one parameter family $(f, R_{\theta}A),$ thus we can have similar Theorem 4 of last post, hence similar result of Lemma 7 for $\hat{\mathcal M}.$ Then the conclusion follows easily. $\square$

## February 9, 2011

### Notes 4: Application of Kotani Theory (I)-Determinism

Filed under: Schrodinger Cocycles — Zhenghe @ 4:34 am
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In this post, I will give several theorems which are applications of Kotani Theory. But I will not give complete proof of all of them. These theorems are basically all Kotani’s work, among which Theorem 1-3 are in his original paper of Kotani Theory Ljapunov indices determine absolutely continuous spectra of stationary random one-dimensional $Schr\ddot odinger$ operators.’  And theorem 4 is in his paper `Jacobi matrices with random potentials taking finitely many values.’

I will use the notations of  previous blogs. E.g. the triple $(X, f,\mu)$, $v\in C(X,\mathbb R)$ a continuous and real valued function, $A^{(E-v)}$ the $Schr\ddot odinger$ cocycle map,  $L(E)$ the Lyapunov exponent  and $u (s)$ the unstable (stable) direction.  We further assume that $supp({\mu})=X.$ Then the first theorem is that

Theorem 1: For a.e. $E$, $L(E)=0\Rightarrow u(E,x)=\bar s(E,x)$ for a.e. $x$.

Remark: Unstable direction comes from the past and stable direction comes from the future (one can take a look at the post ‘A simple proof of HAB formula’ : how to use invariant cone field to obtain invariant direction). Thus this theorem tells us that zero Lyapunov exponents implies that past determines the future, which is kind of determinism.

Idea of Proof: From the proof of Main theorem of Kotani theory we know that for a.e. $E\in\mathbb R,$ $L(E)=0$ implies there exists invariant sections $u(E,x)\in\mathbb H$ and $s(E,x)\in\mathbb H_-$ which are measurable and integrable in some sense.

We didn’t prove the existence of $s$, but it follows from the same argument of the existence of $u.$  Indeed, we only need to replace $A^{(E+i\epsilon-v)}$ by $(A^{(E+\epsilon-v)})^{-1}.$ Then it’s easy to see that $\overline{(A^{(E-v)})^{-1}(x)\cdot\mathbb H_-}\subset\mathbb H_-$ and everything follows from the same argument  of the existence of $u(E,x)$.

Since $A^{(E-v)}, E\in\mathbb R$ is real, we see obviously that $\bar u$ and $\bar s$ are also invariant sections. If $u(x)\neq\bar s(x)$ for a positive measure of  $x$, then by invariance of $u$ and $\bar s$ and ergodicity we have $u(x)\neq\bar s(x)$ for a.e. $x.$ Then there exists a $B:X\rightarrow PSL(2,\mathbb R)$ send $u, \bar s$ to $i, ti$ for some positive $t\neq 1.$  This implies that $B(f(x))A(x)B(x)^{-1}$ fix both $i$ and $ti$ under mobius transformation. This happens only when $B(f(x))A(x)B(x)^{-1}=Id$ in $PSL(2,\mathbb R).$

On the other hand, from last post, we know $A^{(E-v)}$ is monotonic in $E$ in the sense that for any fixed $x\in X$ and $u\in\mathbb R,$ $A^{(E-v)}(x)\cdot u$ is monotonic in $E.$ Thus if $B(E,f(x))A^{(E-v)}(x)B(E,x)^{-1}=Id$ for a positive measure of $E$, then it contradicts with monotonicity under some microscopic computation. $\square$

The second theorem is

Theorem 2: Let $I\in\mathbb R$ be an open interval (then it’s also bounded by boundedness of  the $Schr\ddot odinger$ operator). If $L(E)=0$ on $I,$ then there exists a map $B:I\times X\rightarrow SL(2,\mathbb R)$ such that it’s contiuous on both variable, holomophic in $I$ and $B(E,f(x))A^{(E-v)}(x)B(E,x)^{-1}\in SO(2,\mathbb R).$

Remark: This theorem in particular implies that the spectrum of the corresponding $Schr\ddot odinger$ operator are purely absolutely continuous on the interval $I.$ The generalized eigenfunctions are all wave like (not just in some $L^2$ sense.)

Proof: Define funcition $m:(\mathbb H\cup\mathbb H_-)\times X\rightarrow \mathbb H$ such that

$m(E,x)=u(E,x), for\ E\in\mathbb H$ and
$m(E,x)=\bar s(E,x), for\ E\in\mathbb H_-.$

Then by the proof of Theorem 1, we have

$\lim\limits_{\epsilon\rightarrow 0}m(E+i\epsilon,x)=\lim\limits_{\epsilon\rightarrow 0}m(E-i\epsilon,x)$ for a.e. $x.$ and a.e. $E\in I$.

Thus by Morera’s theorem, for a.e.$x,$ $m(\cdot,x)$ extends through $I$ to be a holomorphic function on $\mathbb C\setminus (\mathbb R\setminus I).$ On the other hand, if we conformal transform $\mathbb H$ to $\mathbb D,$ we see that $m(E,x)_{x}$ is a normal family. Thus for any compacta $\mathcal K$ in $\mathbb C\setminus (\mathbb R\setminus I),$ $m(\cdot,x)$ is uniformly Lipschitz in $E.$  Namely, there exists a constant $c=c(\mathcal K)$ depends only on the compacts such that

$|m(E_1,x)-m(E_2,x)|\leq c|E_1-E_2|$ for any $E_1, E_2\in\mathcal K.$

Since $supp(\mu)=X,$ we can for any $x\in X$ pick a sequence $x_n\in X$ converging to $x.$ Then we get a holomophic function $m(\cdot,x)=\lim\limits_{x_n\rightarrow x}m(\cdot,x_n).$ On the other hand, for $E\notin I,$ we know that $m(E,\cdot)$ is continous in $X$ since they are invariant sections of $\mathcal U\mathcal H$ systems. Thus by holomophicity, $m(\cdot,x)$ is uniquely defined for all $x$ as holomorphic function of $E.$ Since $m(E,\cdot)$ is obtained by taking limits, it automatically continous in $x.$ $\square$

Now let’s state a determinism theorem. First we define a map $\phi:X\rightarrow l^{\infty}(\mathbb Z)$ such that $\phi(x)=(v(f^{n}(x))).$ Thus $\tilde{\mu}=\phi_*\mu$ is a probability measure on $l^{\infty}(\mathbb Z).$ $supp(\tilde{\mu})$ is the hull in the product topology. For example,

if $(v(f^n(x)))_n$ is periodic then the hull is finite;
if $(v(f^n(x)))_n$ is limit periodic and not periodic, then the hull is a compact cantor group;
if the $(v(f^n(x)))_n$ is quasiperiodic with $d$ dimensional rational independent frequency, then hull the $d$ dimensional torus $\mathbb T^{d}$.

Thus up to a homeomorphism, we can actually have that $( X,\mu,f)=(supp(\tilde{\mu}), \tilde{\mu},\sigma),$ where $\sigma$ is the left shift. Namely $(\sigma(x))_n=x_{n+1}.$ Then, we have the following theorem

Theorem 3: Assume $Leb\{E:L(E)=0\}>0.$ Then for all $y, \hat y$ in hull, if $y_n=\hat y_n$ for all $n\leq 0,$ then $y_n=\hat y_n$ for all $n\in\mathbb Z.$

Proof: Here we use the following facts:

(a) for each $x\in X,$ $\lim\limits_{\epsilon\rightarrow 0^+}m(E+i\epsilon,x)=\lim\limits_{\epsilon\rightarrow 0^+}m(E-i\epsilon,x)$ for a.e. $E,$ where $m$ is as in Theorem 2. This follows from Theorem 2 and Fatou’s theorem via a contradiction argument.

(b) $u(E) (s(E)), E\in\mathbb H$ determines $(v(n))_{n\leq0} ((v(n))_{n\geq1}).$ This is not difficult to see by a contradiction argument.

(c) $u(E) (s(E))$ are uniquely determined by their limits on a positive measure set of $E\in\mathbb R.$

Now for each $x\in X,$  $v(f^n(x)), n\leq 0$ obviously determines $u(E,x), E\in\mathbb H.$ But $u(\cdot, x):\mathbb H\rightarrow\mathbb H$ determines $s(\cdot, x):\mathbb H\rightarrow\mathbb H_-$ by fact (a) and (c). Finally, $s(E,x):\mathbb H\rightarrow\mathbb H_-$ determines $v(f^n(x)), n>0$ by fact (b). $\square$

Remark: The conclusion of this theorem is exactly the definition of the deterministic process. Namely a stochastic process $\{v_n(x)\}_{n\in\mathbb Z}$ is said to be deterministic if for all $y\in supp(\tilde{\mu})$ $(y_n)_{n\in\mathbb Z}$ is determined by $\{y_n\}_{n\leq 0}.$  Otherwise it is nondeterministic. Obviously, I.I.d. process are far away from deterministic. In particular, if the process $v_n(x)$ is i.i.d, then for a.e. $E.$ $L(E)>0.$

Next Theorem is that

Theorem 4: If $v$ takes finitely many values and  $Leb\{E:L(E)=0\}>0,$ then $v$ is periodic (or equivalently, hull is finite).

Proof: This is an immediate consequence of Theorem 3. Indeed if $v$ is not periodic then $v$ is nondeterministic. But this cannot happen by our assumption and Theorem 3. $\square$

## January 30, 2011

### Notes 3: Kotani Theory (II)-Fibred Rotation Number (IDS)

Filed under: Schrodinger Cocycles — Zhenghe @ 10:00 pm
Tags: , ,

Up to now, all my posts started from Jan.22.2011 are based on Artur‘s course here in Fields institute, Toronto. Part of the contents of this course are even from Artur’s unpublished work.

This time let me prove the following theorem

Theorem 1: Let $(f, A^{(E-v)})$ as last time, then for a.e. $E,$ $L(E)=o$ implies that $L(E+i\epsilon)=O(\epsilon),$ for small $\epsilon>0$.

As I mentioned last, this Theorem together with the Lemma of last post imply the main theorem of kotani theory.

The proof this theorem make use of the harmonicity of $L(E)$ in upperhalf plane. In particular there is a harmonic function $\rho(E)$ such that $L+i\rho:\mathbb H\rightarrow \mathbb C$ is holomophic. We are going to show this $\rho$ is in fact the fibred rotation number of the correspoding $Schr\ddot odinger$ cocyles and which is also basically the IDS (integrated density of states) of the $Schr\ddot odinger$ operators. This will be the key object of this post. We will prove the following facts about $\rho$:

1. $\rho(E)$ well-defined for all $E\in \overline{\mathbb H}$ and is continuous up to $\partial{\mathbb H}=\mathbb R$.
2. $(f, A^{E-v})$ is monotonic in $E\in\mathbb R$ in some sense which implies the monotonicity of $\rho$ in $E\in\mathbb R$.
3. Thus $\rho(E)$ is differentiable for a.e. $E\in R$ and Cauchy-Riemann equation will imply the conclusion of our theorem.

Let me carry out all the details.

I have to say, for me the fibred rotation number is always a subtle concept. This time I am going to expore as detailed information about it as I can.

From last post we know there exist invariant section $u(E,x)$ for projective dynamics $(f, A^{(E-v)}), \Im E>0$. Thus

$A^{(E-v)}(x)\binom{u(E,x)}{1}=u(E,x)\binom{u(E,f(x))}{1},$

from which it’s easy to see that another way to calculate Lyapunov Exponents via invariant section is

$L(E)=\int_X\ln |u(x)|d\mu,$ thus

$(L+i\rho)(E)=\int_X\ln u(E,x)d\mu:\mathbb H\rightarrow \mathbb C$ is holomorphic.

From which we see that $\rho(E)=\int_X\arg u(E,x)d\mu$ (Here $\arg u(E,x)$ is well-defined. Because by last post, more concretely proof of Lemma 2, it’s easy to see that $u:\mathbb H\times X\rightarrow\mathbb H$. Thus there is no nontrivial loop around origin). For obvious reason, it’s convenient to instead consider $\rho(E)=\frac{1}{2\pi}\int_X\arg u(E,x)d\mu$ (so $L+2\pi \rho i$ is holomorphic functon). By Birkhoff Ergodic Theorem, we have for a.e. x

$\lim\limits_{n\rightarrow\infty}\frac{1}{2\pi n}\sum^{n-1}_{j=0}(\arg u(E,f^{j+1}(x))\cdots u(E,x)-\arg u(E,f^j(x))$ $\cdots u(E,x))$
$=\frac{1}{2\pi n}\sum^{n-1}_{j=0}\arg u(E,f^{j+1}(x))$
$=\rho(E),$

which implies that $\rho$ is some sort of averaged rotation, i.e. a rotation number.

Before proving the next Lemma, I need to do some preparation. To consider rotation number in more general setting, we need go from the $Poincar\acute e$ upperhalf plane $\mathbb H$ to the $Poincar\acute e$ disk $\mathbb D$ via the following matrix

$Q=\frac{-1}{1+i}\begin{pmatrix}1& -i\\1& i\end{pmatrix}\in \mathbb U(2).$

It’s easy to see that $Q\cdot\mathbb H=\mathbb D.$ And $QSL(2,\mathbb R)Q^*=SU(1,1)$, where $SU(1, 1)$ is the subgroup of $SL(2,\mathbb C)$ preserving the unit disk in $\mathbb C\mathbb P^1=\mathbb C\cup\{\infty\}$ under Mobius transformation. For

$A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL(2, \mathbb C),$ let

$\hat A=QAQ^*=\begin{pmatrix}\frac{1}{2}[(a+d)+(b-c)i],&\frac{1}{2}[(a-d)-(b+c)i]\\\frac{1}{2}[(a-d)+(b+c)i],&\frac{1}{2}[(a+d)-(b-c)i]\end{pmatrix}=\begin{pmatrix}\hat u&\hat v\\v&u\end{pmatrix}.$

Then it’s easy to see

for $A\in SL(2,\mathbb R),$ i.e. $A\cdot\mathbb H=\mathbb H,$ then $\hat A\cdot\mathbb D=\mathbb D,$
$\hat u=\bar u,\hat v=\bar v,$ and $|u|^2-|v|^2=1;$
for $\overline{A\cdot\mathbb H}\subset\mathbb H,$ then $\overline{\hat A\cdot\mathbb D}\subset\mathbb D,$
$|u|^2-|v|^2>1.$ Let’s denote this class by $\mathfrak{B}$
for $\overline{A\cdot\mathbb H_-}\subset\mathbb H_-,$ then $\overline{\hat A\cdot\overline{\mathbb D}^c}\subset\overline{\mathbb D}^c,$
$|u|^2-|v|^2<1.$

And all these sets of $A$, or equivalently of $\hat A$ are multiplicative.

In the following Lemma, I always consider the equivalent dynamics $(f, \hat A^{(E-v)}).$ The Lemma is

Lemma 2: $\rho(E)$ is well-defined for all $E\in\overline{\mathbb H}$ and is continuous on $\overline{\mathbb H}$.

Proof: First let’s show that, as long as the cocycle map $\hat A^{(E-v)}(x)\in\mathfrak B,$ or equivalently, $\Im E >0$, we can define $\rho(E)$ via any continuous section $m:X\rightarrow \mathbb D$ (not necessary invariant).

Let’s define $m_n, \tau_n(E,x,m)$ be that

$\hat A^{(E-v)}_n (x)\binom{m}{1}=\tau_n(E,x,m)\binom{m_n}{1}.$

Then obviously $\tilde u(E,x)=Q\cdot u(E,x)$ is the unstable invariant section of $(f, \hat A^{(E-v)})$, thus $\rho(E)$ can be defined as

$\rho(E)=\frac{1}{2\pi}\int_X\arg \tau_1(E,x,\tilde u(E,x))d\mu$
$=\lim\limits_{n\rightarrow\infty}\frac{1}{2\pi n}\sum^{n-1}_{j=1}\arg \tau_1(E,f^{j+1}(x),\tilde u(E,f^{j+1}(x)))$

for a.e. $x.$

Then we show that $\tilde u$ can be replaced by any continuous section $m$ and the convergence is independent of the choice of such $m$. Indeed, we always have that for any $m, m'\in \mathbb D,$

$|\arg\tau_n(E,x,m)-\arg\tau_n(E,x,m')|<\pi$

(hence $|\arg\tau_n(E,x,m)-\arg\tau_n(E,x,m')|\leq\pi,$ for all $E\in\overline{\mathbb H}$).

In fact, by our choice of cocyle map, if we denote $\hat A^{(E-v)}_n(x)=\begin{pmatrix}\hat u_n&\hat v_n\\v_n&u_n\end{pmatrix},$  then

$\tau_n(E,x,\mathbb D)=v_n\mathbb D+u_n,$

which is a disk stay away from $0$ with distance at least $1.$ Thus the above estimate follows easily (Note this is not true for Lyapunov exponents, i.e. $\left |\ln|\tau_n(E,x,m)|-\ln|\tau_n(E,x,m')|\right|$ cannot necessary be bounded). Now we may fix constant section $m(x)\equiv m\in\mathbb D$ to do the remaing computation.

It’s easy to see that $\tau_{n+l}(E,x,m)=\tau_{n}(E,x,m)\tau_{l}(E,f^n(x),m_n)$, so

$\int_X\arg\tau_{n+l}(E,x,m)d\mu=\int_X\arg\tau_{n}(E,x,m)d\mu+\int_X\arg\tau_{l}(E,f^n(x),m_n)d\mu.$

For simplicity let $a_n(E,m)=\frac{1}{2\pi}\int_X\arg\tau_{n}(E,x,m)d\mu.$ Then the above formula implies that

$a_{n+l}(E,m)=a_n(E,m)+a_l(E,m_n),$ and obviously $\lim\limits_{n\rightarrow\infty}\frac{a_n(E,m)}{n}=\rho(E).$

We then have

$|\frac{a_n(E,m)}{n}-\frac{a_l(E,m)}{l}|$
$\leq\frac{1}{nl}\sum_{j=1}^{l-1}|a_n(E,m)-a_n(E,m_{jn})|+\frac{1}{nl}\sum_{p=1}^{n-1}|a_l(E,m)-a_n(E,m_{pl})|$
$\leq \pi(\frac{1}{n}+\frac{1}{l})\rightarrow 0,$ as $n, l\rightarrow\infty.$

Hence, the convergence is uniform.  In the similar way, we can show that $\rho(E)$ is uniform contious in $\mathbb H$. Indeed, it’s easy to see for any fixed $n,$ $a_n(E,m)$ is unform continuous on $\overline{\mathbb H}.$ So we can choose $n_0$ such that $a_{l}(E,m), l=0,\cdots, n_0$ are equi-uniform continuous. Now for any $\epsilon>0,$ we can choose sufficiently small $\delta>0$ such that for $|E-E'|<\delta, E, E'\in\overline{\mathbb H},$

$|a_l(E,m)-a_l(E',m)|<\epsilon,$ for all $l=0,\cdots,n_0.$

Now for arbitrary $n\geq0,$ we have $n=kn_0+l, 0\leq l\leq n_0-1.$ Thus
$|\frac{1}{n}(a_n(E,m)-a_n(E',m))|$
$\leq\frac{1}{kn_0}\sum_{j=0}^{k-1}(|a_{n_0}(E,m_{l+(j-1)n_0})-a_{n_0}(E',m_{l+(j-1)n_0})|+|a_l(E,m)-a_l(E',m)|)$
$\leq\frac{\pi}{k}+\frac{1}{kn_0}|a_l(E,m)-a_l(E',m)|+\frac{1}{n_0}|a_{n_0}(E,m)-a_{n_0}(E',m)|<\epsilon,$ for $k$ large.  Since $n$ is arbitrary, we see $|\rho(E)-\rho(E')|\leq\epsilon,$ for $E, E'\in\mathbb H$.

Thus we can extend $\rho(E)$ to $\overline{\mathbb H}$ which is continous up to $\partial{\mathbb H}=\mathbb R.$ Again we denote it by $\rho(E).$  The above computation actually shows that for $E\in\mathbb R,$

$\rho(E)=\lim\limits_{n\rightarrow\infty}\frac{1}{n}a_n(E,m).$

Indeed, if not we may without loss of generality assume

$\lim\limits_{n\rightarrow\infty}\frac{1}{n}{a_n(E,m)}=a\neq \rho(E).$

Then we can choose $E'\in\mathbb H$ sufficient close to $E$ and $n$ sufficiently large such that all the following terms are less than $\frac{1}{4}|a_0-\rho(E)|$:

$|a_0-\frac{1}{n}a_n(E,m)|, |\frac{1}{n}(a_n(E,m)-a_n(E',m))|,$
$|\frac{1}{n}a_n(E',m)-\rho(E')|, |\rho(E')-\rho(E)|,$

which is obvious a contradiction. $\square$

Our next lemma is

Lemma 3: $\rho(E), E\in\mathbb R$ is nonincreasing.

Proof: This in fact follows from the monotonicity of the following function.  Fix arbitrary $u\in R, x\in\mathbb R/\mathbb Z$ consider the function in $E\in\mathbb R$

$g(E)=A^{(E-v)}(x)\cdot u=E-v(x)-\frac{1}{u}.$

To make everything clear, let’s introduce another way to study the fibred rotation number. Fix $m\in \partial{\mathbb D}$, consider

$N(E)=\lim\limits_{n\rightarrow\infty}\frac{1}{n}\sum_{j=1}^{n}(\arg m_j(E)-\arg m_{j-1}(E)).$

A direct computation shows that $\arg m_j(E)-\arg m_{j-1}(E)=-2\arg\tau_1(E,f^{j}(x),m_{j-1}).$ Indeed,

$m_j=\begin{pmatrix}u(f^jx)&v(f^jx)\\\bar v(f^jx)&\bar u(f^jx)\end{pmatrix}\cdot m_{j-1}=\frac{u(f^jx)m_{j-1}+v(f^jx)}{\bar v(f^jx)m_{j-1}+\bar u(f^jx)}$. Thus

$\frac{u(f^jx)m_{j-1}+v(f^jx)}{\bar v(f^jx)m_{j-1}+\bar u(f^jx)}/m_{j-1}=\frac{v(f^jx)\bar m_{j-1}+u(f^jx)}{\bar v(f^jx)m_{j-1}+\bar u(f^jx)}(m_j\in\partial{\mathbb D},\forall j\geq 0)$. So

$\arg\frac{v(f^jx)\bar m_{j-1}+u(f^jx)}{\bar v(f^jx)m_{j-1}+\bar u(f^jx)}=-2(\bar v(f^jx)m_{j-1}+\bar u(f^jx))=-2\arg\tau_1(E,f^{j}(x),m_{j-1})$ and $N(E)=-2\rho(E).$

Consider $u=Q^*\cdot m\in\mathbb R.$ Then the relation between $u$ and $m$ are $\cot\theta$ and $e^{-2i\theta}.$ Thus it’s not difficult to see that

$E'>E\Rightarrow g(E')>g(E)\Rightarrow \arg m_1(E')>\arg m_1(E).$

Now since we start with the same $m,$ we obviously have $\arg m_1(E')>\arg m_1(E)$ and

$\arg m_2(E')=\arg (\hat A^{(E'-v)}(x)\cdot m_1(E'))$
$>\arg (\hat A^{(E'-v)}(x)\cdot m_1(E))$
$>\arg (\hat A^{(E-v)}(x)\cdot m_1(E))=\arg m_2(E),$

where the first inequality follows from the fact that $\hat A$ preserves order and the second one follows from monotonicity. So by induction we have

$\arg m_n(E)>\arg m_n(E'),$ for all $n>0.$

Thus  $N(E')\geq N(E)$ and  $\rho(E')\leq\rho(E),$ which completes the proof. $\square$

Now we are ready to prove the theorem of this post.

Proof of Theorem 1: By standard harmonic theorem it’s easy to see in our case, $\rho(E+i\epsilon)$ is Poisson integral of $\rho(E), E\in\mathbb R$. Obviously, $\partial_E{\rho}$ is again harmonic. Since $\rho(E), E\in\mathbb R$ is monotonic, $\partial_E{\rho}$ is in fact the Poisson integral of $d\rho(E), E\in\mathbb R.$ Then Fatou’s Theorem tells us for Leb a.e. $E,$ $\lim\limits_{\epsilon\rightarrow 0}\partial_E\rho(E+i\epsilon)=\frac{\partial\rho}{\partial E}(E).$

Now by Cauchy-Riemann equation we have that

$L(E+i\epsilon)=L(E+i0^+)+\int^{\epsilon}_{0}\partial_s(L(E+is))ds$
$=L(E+i0^+)-\int^{\epsilon}_{0}\partial_E(\rho(E+is))ds.$

Now since Lyapunov exponents is a nonnegative upper semicontinuous  function, it’s continuous at $E$, where $L(E)=0.$  Thus the above discussion shows that for a.e. E with $L(E)=0,$ we have

$\lim\limits_{\epsilon\rightarrow 0}\frac{L(E+i\epsilon)}{\epsilon}=-\partial_E\rho(E),$

which completes the proof the Theorem. $\square$

Now I’ve already finished the proof, but probabily I will show that $\Sigma_{ac}=\overline{\mathcal Z}^{ess}$ in the future. As I said in the last post, $\Sigma_{ac}\subset\overline{\mathcal Z}^{ess}$ is relatively easy. It lies in the fact that the generalized eigenfunctions of absolutely continuous spectrum grow at most polynomially fast, which obviously contradicts with positive Lyapunov exponents. And the other part due to Kotani theory has already been contained in these two posts. Let me go back to this in the future.

Next post I will give some application of Kotani theory on deterministic potential and problems concerning density of positive Lyapunov exponents.

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