# Zhenghe's Blog

## April 4, 2011

### Notes 11: Renormalization (III)-Convergence of Renormalization

I’ve been back to Evanston from Toronto. But I guess I still have 10 more notes to post. It will take me a very long time to finish.

In this post we will discuss the convergence of renormalization. As pointed out in last post, it’s necessary that we should assume zero Lyapunov to get convergence. In fact, we will assume $L^2$-conjugacy to rotations. This is somehow natural because Kotani theory tells us that in the Schrodinger cocycle case, for almost every energy, zero Lyapunov exponent implies $L^2$-conjugacy. More concretely, we assume for $A\in C^{\omega}(\mathbb R/\mathbb Z, SL(2,\mathbb R)), \alpha\in\mathbb R\setminus\mathbb Q$ that $(\alpha,A)$ is $L^2$-conjugate to $SO(2,\mathbb R)$-valued cocycles. Namely, there is a measurable map $B:\mathbb R/\mathbb Z\rightarrow SO(2,\mathbb R)$ such that $\int_{\mathbb R/\mathbb Z}\|B(x)\|^2dx<\infty$ and $B(x+\alpha)A(x)B(x)^{-1}\in SO(2,\mathbb R)$ for almost every $x\in\mathbb R/\mathbb Z.$  Let $S(x)=\sup_{n\geq 1}\frac{1}{n}\sum^{n-1}_{k=0}\|B(x+k\alpha)\|^2,$ which is finite almost everywhere by the Maximal Ergodic Theorem. WLOG, we can assume that $A$ can be holomorphically extended to $\Omega_{\delta}=\{z\in\mathbb C/\mathbb Z: |\Im z|<\delta\}$ which is also Lipschitz in $\Omega_{\delta}.$ Then we have the following lemma.

Lemma 1: There exists $C>0$ such that for almost every $x_0,$ we have for every $n\geq 1$ and $x\in\Omega_{\delta}$

$\|A_n(x_0)^{-1}(A_n(x)-A_n(x_0))\|\leq Ce^{C\|B(x_0)\|^2S(x_0)n|x-x_0|}.$

Proof: As explained in Notes 2, $A_k(x)$ is bounded in $L^1.$ In fact, $\|A_k(x)\|\leq\|B(x+k\alpha)\|\|B(x)\|$ for almost every $x.$ Let $x_0$ be such a point. Note $A_n(x_0)^{-1}A_n(x)=A(x_0)^{-1}\cdots A(x_0+(n-1)\alpha)^{-1}A(x+(n-1)\alpha)\cdots A(x).$ If we let $A(x_0+k\alpha)^{-1}A(x+k\alpha)=H_k(x)+id,$ then by Lipschitz condition $H_k(x)\leq C|x-x_0|.$ Obviously, $A_n(x_0)^{-1}A_n(x)=A_{n-1}(x_0)^{-1}H_{n-1}(x)A_{n-1}(x)+A_{n-1}(x_0)^{-1}A_{n-1}(x).$ Hence by induction we have

$A_n(x_0)^{-1}A_n(x)=id+\sum^{n-1}_{k=0}A_k(x_0)^{-1}H_k(x)A_k(x).$

This obviously implies that

$\|A_n(x_0)^{-1}A_n(x)-id\|\leq e^{\sum^{n-1}_{k=0}\|A_k(x_0)\|^2\|H_k(x)\|}-1.$

Thus we obtain

$\|A_n(x_0)^{-1}A_n(x)-id\|\leq e^{\sum^{n-1}_{k=0}\|B(x_0+k\alpha)\|^2\|B(x_0)\|^2\|H_k(x)\|}-1.$

Hence,

$\|A_n(x_0)^{-1}A_n(x)-id\|\leq Ce^{S(x_0)n\|B(x_0)\|^2C|x-x_0|},$

which completes the proof. $\square$

Assume further that $x_0$ is a measurable continuity point of $S$ and $B.$ Here for example, $x_0$ is a measurable continuity point of $S,$ if it is a Lebesgue density point of $S^{-1}(S(x_0)-\epsilon, S(x_0)+\epsilon)$ for every $\epsilon.$ It’s standard result that this is a full measure set since $S$ is measurable and almost everywhere finite. Same definition can be applied to $\|B(x)\|.$ By definition, it’s easy to see the portion of $x$ that $S(x)$ is close to $S(x_0)$ is getting larger and larger in smaller and smaller neighborhood of $x_0.$ Let $x_0$ be that $S(x_0)<\infty,$ which is a full measure condition. Then Lemma 1 implies the following estimate

Lemma 2:  Let $x_0$ be as above. Then for for every $d>0,$ there exist a $n_0(d)>0$ such that

$\|A_{(-1)^nq_n}(x)\|\leq \inf\limits_{x'-x_0\in [-\frac{d}{q_n}, \frac{d}{q_n}]}C(x_0)e^{C(x_0)q_n|x-x'|}$

as long as $n\geq n_0(d).$

Proof: If $n$ is sufficiently large, measurable continuity hyperthesis implies that for every $x'\in [x_0-\frac{d}{q_n}, x_0+\frac{d}{q_n}],$ we can find some $x_0'$ with $|x_0'-x'|\leq\frac{1}{q_n}$ and such that $B(x_0'), B(x_0'+\beta_n)$ are close to $B(x_0)$ and $S(x_0'), S(x_0'+\beta_n)$ are close to $S(x_0).$ WLOG, we can assume $n$ is even. Then Lemma 1 together with our choice of $x_0$ implies that

$\|A_{q_n}(x_0')^{-1}(A_{q_n}(x)-A_{q_n}(x_0'))\|\leq Ce^{C\|B(x_0)\|^2S(x_0)n|x-x_0'|}.$

And we can of course assume $x_0'$ such that $\|A_{q_n}(x_0')\|\leq\|B(x_0')\|\|B(x_0'+\beta_{n})\|.$  Since $\|A^{-1}B\|\geq\|B\|\|A\|^{-1}$ and $|x_0'-x'|<\frac{1}{q_n}.$ Combined these together we get the estimate we want. For $n$ odd, we apply the same discussion to $\|A_{-q_n}(x_0')^{-1}(A_{-q_n}(x)^{-1}-A_{-q_n}(x_0')^{-1})\|.$ $\square$

If we renormalize around $x_0,$ we know from last post that

$\mathcal R^n(\Phi)(1,0)=(1,A_{(-1)^{n-1}q_{n-1}}(x_0+\beta_{n-1}(\cdot)))=(1,A^{(n,0)})$ and
$\mathcal R^n(\Phi)(0,1)=(\alpha_n, A_{(-1)^{n}q_{n}}(x_0+\beta_{n-1}(\cdot)))=(\alpha_n, A^{(n,1)}).$

Note $q_{n-1} Then we have the following obvious corollary from Lemma 2

Corollary 3: Let $\Omega_{\delta/\beta_{n-1}}(\mathbb R)=\{z\in\mathbb C: |\Im z|<\delta/\beta_{n-1}\}.$ Then

$\|A^{(n,i)}(x)\|\leq\inf\limits_{x'\in [-d,d]}Ce^{C|x-x'|},$ where $i=0,1$ and $x\in\Omega_{\delta/\beta_{n-1}}(\mathbb R).$

Thus by homorphicity, $A^{(n,i)}$ are precompact in $C^{\omega}.$ So we can take limit along some subsequence. Denote the limit by $\tilde{A}.$ Then by estimate in Corollary 3 we  get $\|\tilde A(z)\|\leq Ce^{|\Im z|}.$

We in fact also have that $B(x_0)\tilde A(x) B(x_0)^{-1}\in SO(2,\mathbb R)$ for $x\in \mathbb R.$ This is given by the following lemma

Lemma 4: Let $A, x_0$ be as above, then for every $d>0$ and every $\epsilon>0,$ there exists a $n_0(d,\epsilon)$ such that if $n>n_0(d, \epsilon)$ and $\|\alpha n\|_{\mathbb R/\mathbb Z}\leq\frac{d}{n},$ then

$B(x_0)A_n(x)B(x_0)^{-1}$ is $\epsilon$ close to $SO(2,\mathbb R)$ for every $x\in[x_0-\frac{d}{n}, x_0+\frac{d}{n}].$

Proof: For $n$ sufficiently large, for every $x\in[x_0-\frac{d}{n}, x_0+\frac{d}{n}],$ as in proof of Lemma 2, we can find some $x'$ $\frac{\epsilon}{n}$ which is close to $x$ and $S(x'), B(x'), B(x'+n\alpha))$ are $\epsilon$ close to $S(x_0), B(x_0)$. Then the same argument of proof of Lemma 2 implies that $A_n(x)$ and $A_n(x')$ are $\epsilon$ close.

Thus we can reduce the proof to the case $B(x_0)A_n(x')B(x_0)^{-1}$ is $\epsilon$ close to $SO(2,\mathbb R).$ But this is clear since we can furthermore choose $x'$ such that $B(x'+\alpha n)A_n(x')B(x')^{-1}\in SO(2,\mathbb R),$ and we’ve already had $B(x'), B(x'+n\alpha)$ are $\epsilon$ close to $B(x_0).$ $\square$

Thus we can write $\tilde A(z)=B(x_0)^{-1}R_{\phi (z)}B(x_0),$ where $\phi:\mathbb C\rightarrow\mathbb C$ satisfying $|\Im{\phi(z)}|\leq C+C|\Im z|$ is an entire function.  Thus $\phi$ must be linear. We’ve basically proved the following theorem

Theorem 5: If the real analytic cocycle dynamics $(\alpha, A)$ is $L^2$ conjugate to rotations, then for almost every $x_0\in\mathbb R/\mathbb Z$ there exists $B(x_0)\in SL(2,\mathbb R),$ and a sequence of affine functions with bounded coefficients $\phi^{(n,0)}, \phi^{(n,1)}:\mathbb R\rightarrow\mathbb R$ such that

$R_{-\phi^{(n,0)}(x)}BA^{(n,0)}(x)B^{-1}\rightarrow id$ and $R_{-\phi^{(n,1)}(x)}BA^{(n,1)}(x)B^{-1}\rightarrow id.$

as $n\rightarrow\infty.$

To conclude, Theorem 5 implies the following final version of renormalization convergence theorem

Theorem 6: Let $(\alpha, A)$ be as in Theorem 5; let deg be the topological degree of map $A.$ Then there exists a sequence of renomalization representatives $(\alpha_n, A^{(n)})$ and $\theta_n\in\mathbb R,$ such that

$R_{-\theta_n-(-1)^ndeg x}A^{(n)}(x)\rightarrow id$ in $C^{\omega}$ as $n\rightarrow\infty$

Proof:  Let $B, \phi^{(n,0)}(x)=a_{n,0}x+b_{n,0}$ and $\phi^{(n,1)}(x)=a_{n,1}x+b_{n,1}$ be as in Theorem 5. Let $n$ be large and let $\tilde B(x)=R_{a_{n,0}\frac{x^2-x}{2}+b_{n,0}x}B.$ Then $\tilde A(x)=\tilde B(x+1)A^{(n,0)}(x)\tilde B(x)^{-1}$ is $C^{\omega}$ close to identity and $\tilde B(x+\alpha_n)A^{(n,1)}(x)\tilde B(x)^{-1}$ is $C^{\omega}$ close to $R_{\psi^n(x)},$ where $\psi^n(x)=(a_{n,0}\alpha_n+a_{n,1})x+\frac{\alpha_{n}^2-\alpha_n}{2}+b_{n,0}+b_{n,1}.$ By Lemma 1 of Notes 10, we know there exists $C\in C^{\omega}(\mathbb R, SL(2,\mathbb R))$ which is $C^{\omega}$ close to identity such that $C(x+1)\tilde A(x)C(x)^{-1}=id.$

Thus $B^{(n)}=C\tilde B$ is a normalizing map for $A^{(n,0)}$ and $A^{(n)}(x)=B^{(n)}(x+\alpha_n)A^{(n,1)}(x)B^{(n)}(x)^{-1}$ is $C^{\omega}$ close to some $R_{\psi_n(x)},$ where $\psi_n$ is linear. Since the way we get renormalization representatives preserving homotopic relation and the degree of n-th renormalization representative of $(\alpha, R_{deg x})$ is $(-1)^n deg,$ we get that degree of $(\alpha_n, A_n)$ is $(-1)^ndeg.$ Thus the linear coefficient of $\psi_n$ must be close to $(-1)^ndeg$ and $A^{(n)}$ must be close to $R_{\theta_n-(-1)^ndeg x}$ for some $\theta_n\in\mathbb R.$ $\square$

I’ve finished the serial posts about renormalization. It’s a powerful technique in the way that we can use it to reduce global problem to local problem and apply perturbation theory like KAM thoerem. More precisely, we can start with $L^2$ conjugacy to rotations and end up as $C^{\omega}$ close to rotations. If degree is zero and $\alpha$ satisfying some arithematic properties, we can then apply standard KAM theorem to get reducibility.

For these posts, I am following Artur‘s course and he and Krikorian‘s papers. Here I only do the $C^{\omega}$ case while they’ve considered smooth cases in there papers.

Next post will be something about distribution of eigenvalues of the our old friend: One dimensional discrete Schrodinger operator:)

## March 27, 2011

### Notes 10: Renormalization (II)-Renormalization Representatives

Filed under: Schrodinger Cocycles — Zhenghe @ 6:01 pm
Tags: ,

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. $\square$

Next post, I will do some computation concerning the limit of renormalization process in Schrodinger case in zero Lyapunov exponents regime.

## March 3, 2011

### Notes 9: Averaging and Renormalization(I)-Defining the Renormalization Operator

This is post will be a breif introduction about some averaging and renormalization procedures in Schrodinger cocycles. Renormalization is everywhere in dynamical systems and it has been used by many people in Schrodinger cocycles for a while. Lot’s of interesting results have been obtained. As for averaging, I guess it has not been very widely used in Schrodinger cocycles yet. Although it has been a powerful technique in Hamiltonian dynamics for a very long time in studying longtime evolution of action variables. Artur has already used them to construct counter-examples to Kotani-Last Conjecture and Schrodinger Conjecture. We’ve introduced Kotani-Last Conjecture in previous post, namely, the counter-example is that the base dynamics is not almost periodic but the corresponding Schrodinger operator admits absolutely continuous spectrum. For Schrodinger conjecture, the counter example is a Schrodinger operator whose absolutely continous spectrum admits unbounded generalized eigenfunctions.

I am not familar with both averaging and renormalization.  So I may need a little bit longer time to finish this post. But I will try my best to give a brief and clear introduction.

(I) Averaging

Let the frequency $\alpha$ be very Louville number. For instance, let $\alpha=\frac{1}{a_1+\frac{1}{a_2+\frac{1}{a_3+\frac{1}{\ddots}}}}$ be the continued fraction expansion. Thus $\frac{p_n}{q_n}=\frac{1}{a_1+\frac{1}{a_2+\frac{1}{a_3+\frac{1}{\ddots+\frac{1}{a_n}}}}}$ be the n’th step approximant. We let $a_n,$ hence $q_n,$ grows sufficiently fast (see my third post for introduction of Liouville number). Let the potential $v$ be a very small real analytic function. Combining averaging procedure with some renormalization and KAM procedure,  Avila, Fayad and Krikorian proved in their paper ‘A KAM scheme for $SL(2,\mathbb R)$ cocycles with Liouvillean frequencies’ the following theorem

Theorem:  Let $v: \mathbb R/\mathbb Z\rightarrow\mathbb R$ be analytic and close to a constant. For every $\alpha\in\mathbb R,$ there exists a positive measure set of $E\in\mathbb R$ such that $( \alpha, A^{(v,E)})$ is conjugate to a cocycle of rotations.

The main difficulty is Liouvillean frequency case, where they used the idea of averaging. We will not prove this theorem. We are going to give an idea how does the averaging procedure look like in Schrodinger cocycle.

Let’s start with the base dynamics $x\mapsto x.$ Namely, we consider one-parameter family of cocycles over fixed point. We have two different ways to consider the spectrum, which is the following

$I=\cup_x\Sigma_x:\{E\in\mathbb R: |tr(A^{(E-v)}(x))|=|E-v(x)|\leq 2$ for at least one $x \},$
$I^0=\cap_x\Sigma_x:\{E\in\mathbb R: |tr(A^{(E-v)}(x))|=|E-v(x)|\leq 2$ for all $x \}.$

Let’s start with $E\in I^0.$ For simplicity, set $A(x)=A^{(E-v)}(x)$ and we omit the dependence on $E.$ Let $u(x)$ be the invariant direction of $A(x).$ Let $B^0:\mathbb R/\mathbb Z\rightarrow SL(2,\mathbb R)$ be that $B^0(x)\cdot u(x)=i.$ Thus $B^0(x)A(x)B^0(x)^{-1}=R_{\psi^0(x)}$  for some $\psi^0:\mathbb R/\mathbb Z\rightarrow\mathbb R$ analytic. Thus there exists a constant $C_0$ such that

$B^0(x+\frac{p}{q})A(x)B^0(x)^{-1}=R_{\psi^0(x)}+\frac{C_0}{q}.$

Then for any closed interval $I^1$ inside the interior of $I^0,$ we can choose large $q$ such that $R_{\psi^0(x)}+\frac{C_0}{q}$ is again elliptic. Hence similarly, there is some $B^1:\mathbb R/\mathbb Z\rightarrow SL(2,\mathbb R)$ which is $\frac{1}{q}$ close to identity and $\psi^1:\mathbb R/\mathbb Z\rightarrow\mathbb R$ such that

$B^1(x)(R_{\psi^0(x)}+\frac{C_0}{q})B^1(x)^{-1}=R_{\psi^1(x)}.$ Hence

$B^1(x+\frac{1}{q})(R_{\psi^0(x)}+\frac{C_0}{q})B^1(x)^{-1}=R_{\psi^1(x)}+\frac{C_1}{q^2}$ for some constant $C_1.$

Thus $(\frac{1}{q},A)$ is conjugate to $(\frac{1}{q}, R_{\psi^1(x)}+\frac{C_1}{q^2}).$ Iterating $q$ times we get $R_{\psi_{q}^1(x)}+\frac{C_1}{q},$ where $\psi_{q}^1(x)$ is the n-th Birkhoff sume $\sum^{q-1}_{j=0}\psi(x+\frac{j}{q}).$ It’s not difficult to see that $\psi^1$ is close to its averaging $q\hat{\psi}(0).$ Thus except those bad $E$ such that $\psi_{q}^{1}(x)=\pm id$ for some $x,$ $(\frac{1}{q}, A^{(E-v)})$ stay ellipitc. By monotonicity of the Schrodinger cocycle with respect to energy and suitable choice of $q,$ these bad $E$ are of arbitrary small measure.

(II) Renormalization

Here we only introduce renormalization in quasiperiodic $SL(2,\mathbb R)$-cocycle where the frequency is one dimensional (i.e. the base dynamics is one dimensional). Let’s denote the renormalization operator by $\mathcal R.$ This will be a operator defined on the space of all cocycle dynamics. More concretely, it’s defined on the space of $\mathbb Z^2$ action on cocycle dynamics. We will continue to use the continued fraction expansion and approximants as in Averaging.

A natural way to see that the cocycle dynamics $(\alpha, A):(\mathbb R/\mathbb Z)\times\mathbb R^2\rightarrow (\mathbb R/\mathbb Z)\times\mathbb R^2$ is to see it as $(\alpha, A):\mathbb R\times\mathbb R^2\rightarrow \mathbb R\times\mathbb R^2,$ which commutes with $(1, id):\mathbb R\times\mathbb R^2\rightarrow \mathbb R\times\mathbb R^2.$ Because it’s easy to check that

$A(x)=A(x+1)\Leftrightarrow (\alpha,A)\cdot(1, id)=(1, id)\cdot(\alpha,A).$

This is similar to the following case. Denote the orientation preserving diffeomorphism on $\mathbb R$ by $Diffeo_+(\mathbb R).$ Then $F:\mathbb R/\mathbb Z\rightarrow\mathbb R/\mathbb Z$ is orientation preserving diffeomorphism if and only if it can be lifted to $Diffeo_+(\mathbb R)$ and commutes with $G:\mathbb R\rightarrow\mathbb R,$ where $G(x)=x+1.$ More generally, consider a commuting pair $(F,G)\in Diffeo_+(\mathbb R)$ where G has no fixed point. Then $F$ define a dynamical systems on $\mathbb R/G.$ Because $F(G(x))=G(F(x))\sim F(x)$ preserving the equivalent relation. Here $\mathbb R/G$ is diffeomorphic to $\mathbb R/\mathbb Z,$ but not in a canonical way.

Thus it’s more natural to view the pair $(F,G)$ as a $\mathbb Z^2$ group action, which is the following group homomorphism

$\Phi:\mathbb Z^2\rightarrow Diffeo_+(\mathbb R)$ with $\Phi(1,0)=G$ and $\Phi(0,1)=F.$

This automatically encodes the commuting relation.

The reason we introduce above procedure is that after renormalization, we will have no canonical way to glue $\mathbb R$ into $\mathbb R/\mathbb Z.$

Now let’s define the renormalization operator $\mathcal R$ step by step. We will renormalize around $0\in\mathbb R/\mathbb Z.$

(1) Define the $\mathbb Z^2$-action.
Let $\Lambda^r=\mathbb R\times C^r(\mathbb R,SL(2,\mathbb R))$ be subgroup of $Diff^r(\mathbb R\times\mathbb R^2).$ Then for $(\alpha,A)\in\Lambda^r$ satisfying $A(x)=A(x+1),$ we can define $\mathbb Z^2$ action $\Phi:\mathbb Z^2\rightarrow \Lambda^r$ such that $\Phi(1,0)=(1,id), \Phi(0,1)=(\alpha,A).$ Let $\Phi(n,m)=(\alpha_{n,m}, A_{n,m}).$

(2) Define two operations on the above $\mathbb Z^2$-action.
The first is the rescaling operator $M_{\lambda},$ which is given by

$M_{\lambda}(\Phi)(n,m)=(\lambda^{-1}\alpha_{n,m}, x\mapsto A_{n,m}(\lambda x)).$

The second is base changing operator $N_{U}.$ For $U\in GL(2,\mathbb Z),$ it’s given by

$N_{U}(\Phi)(n.m)=\Phi(n',m'), \binom{n'}{m'}=U^{-1}\binom{n}{m}.$

Obviously these two operations commutes with each other.

(3) More facts about continued fraction expansion.
Let $\alpha$ be as in averaging. Let $G$ be the Gauss map $G(\alpha)=\{\alpha^{-1}\}.$ Denote $G^n(\alpha)=\alpha_n.$ Thus $a_n=[\alpha_{n-1}^{-1}],$ where $a_n$ comes from the continued fraction expansion. Define a map

$U(x)=\begin{pmatrix}[x^{-1}]&1\\1&0\end{pmatrix}$ for $x\in (0,1).$

Let $\beta_n=\alpha_{n}\cdots\alpha.$ Then it’s easy to see that we also have $\beta_n=(-1)^n(q_n\alpha-p_n)=\frac{1}{q_{n+1}+\alpha_{n+1}q_n}.$

(4) Define the renormalization operator.
$\mathcal R:\Lambda^r\rightarrow\Lambda^r$ is given by

$\mathcal R(\Phi)=M_{\alpha}(N_{U(\alpha)}(\Phi)),$ where $\alpha$ is from $\Phi(0,1)=(\alpha,A).$ $\square$

It easy to see that

$\mathcal R(\Phi)(1,0)=M_{\alpha}(\Phi)(0,1)=(1,A(\alpha(\cdot)))$ and
$\mathcal R(\Phi)(0,1)=M_{\alpha}(\Phi)(1,-a_1)=M_{\alpha}(1-a_1\alpha, A_{-a_1})=(\alpha_1,A_{-a_1}(\alpha(\cdot))).$

Thus geometrically, if we  look at $\mathbb R\times\mathbb R\mathbb P^1,$ it looks like we glue $\{x\}\times\mathbb R\mathbb P^1$ and $\{x+\alpha\}\times\mathbb R\mathbb P^1$ via $(\alpha,A).$ Then by commuting relation $(1-a_1\alpha, A_{-a_1})$ define a dynamical systems on $(\mathbb R\times \mathbb R\mathbb P^1)/(\alpha,A).$ Then we rescale the first coordinate to be $\mathbb R/\mathbb Z$ again. Finally we get the new dynamics $\mathcal R(\Phi)(0,1).$

Let $Q_n=U(\alpha_{n-1})\cdots U(\alpha).$ Then it’s easy to see that

$\mathcal R^n(\Phi)=M_{\alpha_{n-1}}\circ N_{U(\alpha_{n-1})}\circ\cdots\circ M_{\alpha}\circ N_{U_{\alpha}}(\Phi)=M_{\beta_{n-1}}(N_{Q_n}(\Phi)).$

Thus we are looking at smaller and smaller space scale but larger and larger time scale.

Next post we will show how are the dynamics of original cocycle and these of renormalized cocycles related. And we will normalize $\mathcal R^n(\Phi)$ so that $\mathcal R^n(\Phi)(1,0)=(1,id)$ and do some computation concerning the limit of the renormalzation.

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