Notes 10: Renormalization (II)-Renormalization Representatives

March 27, 2011

Notes 10: Renormalization (II)-Renormalization Representatives

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. $\square$