# Zhenghe's Blog

## January 26, 2011

### Notes 2: Kotani theory (I)-Zero Lyapunov Exponents and L^2 rotation conjugacy

This and next posts will be about Kotani theory, which is part of my syllabus. Kotani theory is first found by Shinichi Kotani when he studies spectrum theory of $Schr\ddot odinger$ operator with ergodic potentials. It gives a complete decription of  the absolutely continuous part of spectrum of  $Schr\ddot odinger$ operators in terms of Lyapunov exponents of the  corresponding of $Schr\ddot odinger$ cocycles. Thus it builds a deep and beautiful relation between operator theory and dynamical systems.

Later Artur Avila and Raphael Krikorian generalize Kotani theory to more general coycle dynamics setting and it becomes a powerful tool to study Schrodinger operators. What I am going to introduce here are main results and some preliminary stuff. Let’s start with one dimensional (1D) discrete $Schr\ddot odinger$ operators and $Schr\ddot odinger$ cocyles. From now on I set $L(E)=L(f,A^{(E-v)})$.

For simplicity, I will continue to use the triple $(X,f,\mu)$ as in last post. Assume $v\in C(X,R)$ be  a contiuous function. Then we can define a cocycle map via

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

The correspoding cocycle dynamics $(f, A^{(v)})$ is called $Schr\ddot odinger$ cocyles. It arises from the following $Schr\ddot odinger$ operator $H_{f,v,x}$ on $l^2(\mathbb Z)$

$(H_{f,v,x}u)_n=u_{n+1}+u_{n-1}+v(f^n(x))u_n,$

where $u=(u_n)_{n\in\mathbb Z}\in l^2(\mathbb Z).$ Then $u\in\mathbb C^{\mathbb Z}$ solving the eigenfunction equation $H_{f,v,x}u=Eu, E\in\mathbb R$ if and only if

$A^{(E-v)}(f^n(x))\binom{u_n}{u_{n-1}}=\binom{u_{n+1}}{u_n}.$

Thus the growth rate of $|u_n|$ with respect to $n$ characterizes both the spectrum type of the energy $E$ and the dynamics of cocycle $(f,A^{(E-v)})$, which allows one to go back and forth between spectrum theory and dynamical systems. Let $\Sigma_x$ be the spectrum of the bounded linear selfadjoint operator $H_{f,v,x}$, then the first basic fact relate operator and cocycle is for a.e. $x,$

$\Sigma_x=\{E: (f, A^{(E-v)})\notin \mathcal U\mathcal H\}.$

If furthermore $f$ is minimal, then the above relation is in fact true for all $x.$ Thus  for a.e. $x,$ the spectrum $\Sigma_x$ is independent of $x\in X,$ let’s denote it as $\Sigma$. For each $x,$ we can further decompose $\Sigma_x$ as $\Sigma_x=\Sigma_{x,pp}\cup\Sigma_{x,ac}\cup\Sigma_{x,sc}$, which correspond to pure point, absolutely continuous and singular continuous part of the spectrum of the operator $H_{f,v,x}$. These are defined by the spectral measure of the operator, which one can find in any standard functional analysis book.  It turns out that in our case, there exists sets $\Sigma_{\bullet}, \bullet\in\{pp, ac, sc\}$ such that $\Sigma_{x,\bullet}=\Sigma_{\bullet}$ for a.e. $x$ and $\bullet\in\{pp, sc,ac\}.$ (If in addition $f$ is minimal, then in fact $\Sigma_{x,ac}=\Sigma_{ac},$ for all $x,$ which is in general not true for pp and sc.)

Now we denote $\mathcal Z=\{E: L(E)=0\}.$ And for any set $\mathcal S\subset\mathbb R,$ the essential support of the set $\mathcal S$ is given by

$\overline{\mathcal S}^{ess}=\{E\in\mathbb R: Leb(\mathcal S\cap (E-\epsilon, E+\epsilon))>0$ for every $\epsilon>0\}.$

Then the next deep relation between dynamics and spectrum is the following

$\Sigma_{ac}=\overline{\mathcal Z}^{ess}.$

The relation $\Sigma_{ac}\subset\overline{\mathcal Z}^{ess}$ is relatively easy since positive Lyapunov exponents give exponential growth of the solution $u$ to the eigenfunction equation which in some sense contradicts with the absolute continuity of spectrum. The part $\overline{\mathcal Z}^{ess}\subset\Sigma_{ac}$ is a rather deep result called Kotani theory. It in fact  is a theory about that, under the assumption that Lyapunov exponents are zero, when can one in some sense conjugate the $SL(2,\mathbb R)-$valued cocycles to $SO(2,\mathbb R)-$valued cocycles. Obviously, if the cocycle map $A$ takes value in $SO(2,\mathbb R)$, then all orbits $\{A_n(x)w\}_n$ are bounded. While zero Lyapunv exponent in general just means that $\|A_n(x)w\|_n$ grows subexponentially.  What Kotani theory tells us is in fact that the following theorem

Theorem 1: For almost every $E$, if $L(E)=0,$ then there exists a map $B:X\rightarrow SL(2,\mathbb R)$ such that$B(f(x))A^{(E-v)}(x)B(x)^{-1}\in SO(2,\mathbb R)$ and $\int_X\|B(x)\|^2d\mu<\infty.$

Thus the generalized eigenfunctions for absolutely continous spectrum of the operators in question oscillate in some $L^2$ sense. Which are kind of wave like solutions and far from eigenfunctions of real eigenvalues, which decay in $l^2$ sense. As I said this is obviously stronger then zero Lyapunov exponents. Indeed, in this case we have

$\frac{1}{n}\int_X\ln\|A_n(x)\|d\mu\leq\frac{1}{n}\int_X\|A_n(x)\|d\mu\leq\frac{1}{n}\int_X\|B(f^n(x))^{-1}\|\|B(x)\|d\mu$
$\leq\frac{1}{n}\int_X\|B(x)\|^2d\mu\rightarrow 0,$ as $n\rightarrow\infty$

Now let me explain how can one construct the above $B:X\rightarrow SL(2,\mathbb R).$  We first prove the following key lemma

Lemma 2: Assume $E\in\mathbb R$ satisfying $L(E+i\epsilon)=O(\epsilon),$ for $\epsilon>0$ small. Then $(f,A^{(E-v)})$ is $L^2-$conjugate to $SO(2,\mathbb R)-$valued cocycles.
Proof: First we note by the same reason that $(f,R_{i\theta}A)\in\mathcal U\mathcal H$ as in last blog, we have $(f,A^{(E+i\epsilon-v)})\in\mathcal U\mathcal H$. Thus there is invariant section $u^{\epsilon}:X\rightarrow \mathcal H$ which is the unstable direction. There are different ways to calculate Lyapunov exponents of cocylce dynamics via invaiant section of corresponding projective dynamics. We use the following: $-\frac{1}{2}$ of the contraction rate measured in $Poincar\acute e$ metric of mobius transformation at $u^{\epsilon}$. In our case we need to consider the following composition of map. Let $\mathbb H_{\epsilon}=\{E:\Im E>\epsilon\}$ with standard $Poincar\acute e$ metric, then the composition is

$\mathbb H\overset{A^{(E+i\epsilon-v)}}{\longrightarrow}\mathbb H_{\epsilon}\overset{i_{\epsilon}}{\longrightarrow}\mathbb H,$

where the first map is a isometry and the second one (which is the inclusion map) is a contraction. Thus we consider the contraction of the second map at invariant section. Then the Lyapunov exponents is given by

$L(E+i\epsilon)=-\frac{1}{2}\int_X\ln(1-\frac{\epsilon}{\Im u^{\epsilon}})d\mu.$

For simplicity, let me first assume that:

$\lim\limits_{\epsilon\rightarrow 0}u^{\epsilon}(x)$ exists for $a.e. x$ and we denote it by $u(x)$.
(then obviously, $u(x)$ is invariant for a.e. $x$, i.e. $A^{(E-v)}(x)\cdot u(x)=u(f(x))$ for a.e.$x$.)

Assuming this, we have the following straightforward estimate via Fatou’s lemma and our assumption in Lemma :

$\frac{1}{2}\int_X\frac{1}{\Im u}d\mu\leq\liminf\limits_{\epsilon\rightarrow 0}-\frac{1}{2\epsilon}\int_X\ln(1-\frac{\epsilon}{\Im u^{\epsilon}})d\mu=O(1)<\infty.$

Since $u(f(x))=E-v(x)-\frac{1}{u(x)}$, we have

$\frac{1}{2}\int_X\frac{1}{\Im u}d\mu=\frac{1}{2}\int_X\frac{1}{\Im u(f(x)}d\mu=\frac{1}{2}\int_X\frac{|u|^2}{\Im u}d\mu<\infty$.

Hence

$\int_X\frac{1+|u|^2}{\Im u}d\mu<\infty.$

Let $\phi(u)=\frac{1+|u|^2}{\Im u}$. Now we define $B:X\rightarrow SL(2,\mathbb R)$ such that $B(x)\cdot u=i,$ thus $B(f(x))A(x)B(x)^{-1}\cdot i=i$ which implies $B(f(x))A(x)B(x)^{-1}\in SO(2,\mathbb R)$. On the other hand, it’s easy to see for quite general reason $\|B(x)\|_{HS}^2=\phi(u).$ Thus $\int_X\|B(x)\|_{HS}^2d\mu<\infty.$
Here for $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL(2,\mathbb R),$ $\|A\|_{HS}^2$ is the Hilbert-Schmit norm of $A$ which is just $a^2+b^2+c^2+d^2.$ It’s easy to see that if $A\cdot z=i,$ then $\|A\|_{HS}^2=\phi(z)$. Since all norms of $SL(2,R)$ are equivalent, we complete the proof of the Lemma up to the assumption.    $\square$

To get around the assumption, we need conformal barycenter, which is a Borelian function $\mathcal B:\mathcal M\rightarrow\mathbb H,$ where $\mathcal M$ is the space of probability measures on $\mathbb H.$ This function is equivariant with respect to $SL(2,\mathbb R)$ change of coordinates. i. e. for

$A\in SL(2,\mathbb R), \nu\in\mathcal M,$ we have $\mathcal B(A_*\nu)=A\cdot\mathcal B(\nu).$

Now let’s consider the probability measures $\nu_{\epsilon}=\mu\otimes\delta_{u^{\epsilon}}$ on $X\times\overline{\mathbb H}$. Then there is a subsequence converging to some $\nu=\mu\otimes\nu_x$ such that $\int_{X\times\overline{\mathbb H}}\phi d\nu<\infty$. Then apply conformal barycenter to $\nu_x$, we get a point $u(x)=\mathcal B(\nu_x)\in\mathbb H$ such that $u$ is an invariant section and

$\int_X\phi(u(x))d\mu=\int_{X\times\overline{\mathbb H}}\phi d\delta_{u(x)}d\mu\leq\int_{X\times\overline{\mathbb H}}\phi d\nu<\infty.$

Now we can construct the map $B$ as  in the proof of lemma.

Although we introduce conformal barycenter to complete the proof of Lemma, to prove the Theorem we actually only need to apply Fubini and Fatou theorem, to pass the existence of limits from For every $x$, converges for  a.e. $E$  to  For a.e. $E$, converges for a.e. $x$. Then we get the conclusion in assumption for a.e. E.

In the next post, I will prove how can we get the condition in Lemma under the condtion in Theorem, which will complete the proof of theorem.

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