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|<c, \frac{1}{c}\leq\Im E\leq c for some c>0. Let

\mathcal C=\{(a,E)\in\mathbb R\times\mathbb H:|a|\leq c and |\Re E|<c, \frac{1}{c}\leq\Im E\leq c\}.

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 


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

=\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


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

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

Blog at WordPress.com.

%d bloggers like this: