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}<q_{n}<\beta_{n-1}^{-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:)

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