# Zhenghe's Blog

## 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. $\square$