Zhenghe's Blog

January 30, 2011

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

Filed under: Schrodinger Cocycles — Zhenghe @ 10:00 pm
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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 Part of the contents of this course are even from Artur’s unpublished work.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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