Posted by: matheuscmss | April 24, 2008

What’s the size of certain unbalanced products of SL(2,R) matrices? (part II)

As I promised in my previous post, I’ll discuss a recent result of Fayad and Krikorian about the exponential growth of the norm of certain products of two SL(2,\mathbb{R}) matrices A and B (where A is hyperbolic and B is arbitrary) under very restrictive conditions on the relative number of appearences of the matrix B inside these products. Let me start with the precise statement of Fayad and Krikorian’s theorem:

Theorem 1 (Fayad and Krikorian, 2008). Let A_0\in SL(2,\mathbb{R}) be a hyperbolic matrix (i.e., |tr(A_0)|>2). Fix 1<\gamma<|tr(A_0)|/2 and 0<\alpha<1/2. Then, for almost every A_1\in SL(2,\mathbb{R}), we can find a constant C>0 such that

\|A_{w_N}\dots A_{w_1}\|\geq C\gamma^N

for any binary word W=(w_1,\dots,w_N)\in\{0,1\}^N verifying the following frequency condition:

(1) \#\{j\in\{1,\dots,N\}: w_j=1\}\leq N^{\alpha}.

We note that the frequency condition (1) above is more restrictive than the frequency condition from Bochi and Fayad’s problem (compare with the equation (1) of my previous post).

Let us present the proof of Fayad and Krikiorian’s result. Note that up to a linear change of coordinates one can assume that

(2) A_0 = \left( \begin{array}{cc} a & 1\\ -1 & 0 \end{array}\right) where a=tr(A_0)>2.

Also, the reader can easily check that one can write every matrix A_1\in SL(2,\mathbb{R}), A_1=\left(\begin{array}{cc} r & s \\ t & u\end{array}\right) with u\neq 0 as a product

(3) A_1 = A_1(b_1,b_2,b_3) = \left(\begin{array}{cc} b_1 & 1 \\ -1 & 0\end{array}\right) \left(\begin{array}{cc} b_2 & 1 \\ -1 & 0\end{array}\right) \left(\begin{array}{cc} b_3 & 1 \\ -1 & 0\end{array}\right), where

(4) b_1=-(s+1)/u, \quad b_2=-u, \quad b_3=(t-1)/u.

Remark 1. The set of matrices A_1\in SL(2,\mathbb{R}) of the form A_1=\left(\begin{array}{cc} r & s \\ t & 0\end{array}\right) (i.e., the matrices of SL(2,\mathbb{R}) which can’t be represented as (3) above) has zero Haar measure. Nevertheless, among the set of matrices represented by (3), it is not hard to see that zero Lebesgue measure sets of (b_1,b_2,b_3)\in\mathbb{R}^3 corresponds to zero Haar measure sets of SL(2,\mathbb{R}) matrices. Indeed, this last fact follows directly from (4).

Using this remark, we see that the proof of theorem 1 is complete once we show the following proposition:

Proposition 1. Let A_0\in SL(2,\mathbb{R}) be a hyperbolic matrix of the form (2) (where a = tr(A_0)>2). Fix 1<\gamma<a/2 and 0<\alpha<1/2. Then, for any given (\theta,\theta')\in\mathbb{R}^2, there exists a set of full Lebesgue measure of b\in\mathbb{R} so that we can find a constant C>0 with

\|A_{w_N}\dots A_{w_1}\|\geq C\gamma^N

for any word W=(w_1,\dots,w_N)\in\{0,1\}^N verifying the frequency condition

\#\{j\in\{1,\dots,N\}: w_j=1\}\leq N^{\alpha},

where A_1=A_1(b,b+\theta,b+\theta') is defined by (3).

Before starting the proof of this proposition, let us introduce some notation.

Definition 1. Given a binary word W=(w_1,\dots,w_N)\in\{0,1\}^N and two matrices A, B\in SL(2,\mathbb{R}), we define W(A,B) = A_{w_N}\dots A_{w_1} where A_{w_j}=A if w_j=0 and A_j=B if w_j=1.

Definition 2. We denote by \mathcal{C}_n(\alpha) the set of binary words W=(w_1,\dots,w_n)\in\{0,1\}^n verifying the frequency condition \#\{j\in\{1,\dots,n\}: w_j=1\}\leq n^{\alpha}.

Next, we recall three useful facts (for later use). Firstly, we begin with the following simple formula:

(5) \left(\begin{array}{cc}a_n & 1\\ -1 & 0\end{array}\right) \dots \left(\begin{array}{cc}a_1 & 1\\ -1 & 0\end{array}\right) = \left(\begin{array}{cc}\mathcal{O}_n & \star \\ -\mathcal{O}_{n-1} & \star\end{array}\right),

where \mathcal{O}_n:=\det\left(\begin{array}{cccccc}a_1 & 1 & 0 & 0 & \dots & 0\\ 1 & a_2 & 1 & 0 &\dots&0\\ 0&1&\ddots& &\dots&0\\ & & &&a_{n-1}&1\\ & & &0&1&a_n\end{array}\right). This classical elementary formula follows easily by induction as the reader can check.

Secondly, we note an easy estimate for the determinants \mathcal{O}_n considered above:

Lemma 1. Let k\geq 1 and \Delta_k:=\det\left(\begin{array}{cccccc}d_1 & 1/a & 0 & 0 & \dots & 0\\ 1/a & d_2 & 1/a & 0 &\dots&0\\ 0&1/a&\ddots& &\dots&0\\ & & &&d_{k-1}&1/a\\ & & &0&1/a&d_k\end{array}\right), where a\geq 2 and d_1,\dots,d_k\geq 1. Then, it holds

\Delta_k\geq 1/2^k.

Proof of lemma 1. Putting \Delta_0:=1, a direct computation reveals that \Delta_k = d_k\Delta_{k-1} - (1/a)^2\Delta_{k-2}. Hence, using that a\geq 2 and d_i\geq 1, we see that \mu_i:=\Delta_i/\Delta_{i-1} satisfies \mu_i\geq 1 - \frac{1}{4\mu_{i-1}}. Since \mu_1\geq 1/2, we obtain by induction that \mu_i\geq 1/2 for every 1\leq i\leq k. Thus, \Delta_k = \mu_k\dots\mu_1\geq 1/2^k. \square

Thirdly, we’ll need the following fact:

Lemma 2. Let P(x) be a polynomial of degree \leq n. Given a closed interval I\subset\mathbb{R}, denote by \|P\|_{I} := \max\limits_{x\in I} |P(x)|. Then, for any \epsilon>0, it holds

\lambda(\{x\in I: |P(x)|\leq\epsilon\}) \leq 2n(n+1)^{1/n} \left(\frac{\epsilon}{\|P\|_I}\right)^{1/n} \lambda(I).

Proof of lemma 2. For the short proof of this lemma, we follow the presentation of Kleinbock and Margulis (proposition 3.2, pages 345 and 346). Denote by \ell = \lambda(I) and \sigma = \frac{1}{\ell}\lambda(\{x\in I: |P(x)|\leq\epsilon\}). Since B:=\{x\in I: |P(x)|\leq\epsilon\} has Lebesgue measure \sigma\ell, it follows that there are k+1 points x_1,\dots,x_{k+1}\in B such that |x_i-x_j|\geq\sigma\ell/2k for all 1\leq i<j\leq k+1. Using Lagrange’s interpolation formula, we know that the polynomial P of degree \leq k is completely determined by its values on the points x_1,\dots,x_{k+1}:

P(x) = \sum\limits_{i=1}^{k+1} f(x_i) \prod\limits_{\substack{1\leq j\leq k+1 \\ j\neq i}}\frac{(x-x_j)}{(x_j-x_i)}.

Because |x_j-x_i|\geq\sigma\ell/2k for every 1\leq i<j\leq k+1, we obtain

\|P\|_I\leq (k+1)\epsilon \frac{\ell^k}{(\sigma\ell/2k)^k}.

Since \ell:=\lambda(I) and \lambda(\{x\in I: |P(x)|\leq\epsilon\}):=\sigma\ell, the proof of the lemma 2 is complete. \square

After these preliminaries, we are ready to show the desired proposition.

Proof of proposition 1. Fix (\theta,\theta')\in\mathbb{R}^2. Given M a large number and W\in\mathcal{C}_n(\alpha), we define

\mathcal{E}(W):=\{|b|\leq M: \|W(A_0, A_1(b,b+\theta,b+\theta')\|\leq\gamma^n\} and

\mathcal{E}_n:=\bigcup\limits_{W\in\mathcal{C}_n(\alpha)}\mathcal{E}(W).

From Borel-Cantelli’s lemma, our task is reduced to show that \sum\limits_{n\in\mathbb{N}} \lambda(\mathcal{E}_n)<\infty, where \lambda stands for the Lebesgue measure on \mathbb{R}.

Since it is fairly standard to check that \#\mathcal{C}_n(\alpha)\leq n^{\alpha} n^{n^{\alpha}}, it suffices to prove the following affirmative:

Claim. It holds

(6) \sum\limits_{n\in\mathbb{N}} n^{\alpha}n^{n^{\alpha}}\max\limits_{W\in\mathcal{C}_n(\alpha)} \lambda(\mathcal{E}(W))<\infty.

In order to show this claim, we apply (5) so that

W(A_0, A_1(b,b+\theta,b+\theta') = \left(\begin{array}{cc}\Delta (W,b,\theta,\theta') &\star \\ \star & \star\end{array}\right)

with \Delta (W,b,\theta,\theta'):=\det\left(\begin{array}{cccccc}z_1 & 1 & 0 & 0 & \dots & 0\\ 1 & z_2 & 1 & 0 &\dots&0\\ 0&1&\ddots& &\dots&0\\ & & &&z_{m-1}&1\\ & & &0&1&z_m\end{array}\right),

where z_i=a corresponds to the appearance of a letter A_0 in W (i.e., w_i=0), a string z_i=b, z_{i+1}=b+\theta, z_{i+2}=b+\theta' corresponds to the appearance of a letter A_1 in W (i.e., w_i=1) and m := \#\{ j: w_j=0\}+3\#\{ j: w_j=1\}\geq n. In particular,

(7) \|W(A_0, A_1(b,b+\theta,b+\theta')\|\geq |\Delta (W,b,\theta,\theta')|.

At this point we define

( 8 ) P_W(b) := a^{-m}\Delta(W,b,\theta,\theta') (where a:=|tr A_0|>2).

Note that P_W is a polynomial of degree 3\#\{1\leq j\leq n : w_j=1\}\leq 3n^{\alpha} when W\in\mathcal{C}_n(\alpha). Also, a direct application of lemma 1 implies that P_W(a+|\theta|+|\theta'|)\geq 1/2^m. Assuming M large so that M\geq a+|\theta|+|\theta'| (without loss of generality), we can combine these two previous facts with the estimate of lemma 2 to obtain that

(9) \lambda(\{|b|\leq M : |P_W(b)|<\epsilon\})\leq 24 M n^{\alpha} (2^m\epsilon)^{1/3n^{\alpha}}.

Therefore, by taking \epsilon = (\gamma/a)^m in (9) and using that m\geq n, we deduce that

\lambda(\mathcal{E}(W))\leq 24 M n^{\alpha}(2\gamma/a)^{m/3n^{\alpha}}\leq 24 M n^{\alpha} \tau^{n^{1-\alpha}},

where \tau=(2\gamma/a)^{1/3}<1. Using this last estimate, one obtains

(10) \sum\limits_{n\in\mathbb{N}} n^{\alpha}n^{n^{\alpha}}\max\limits_{W\in\mathcal{C}_n(\alpha)} \lambda(\mathcal{E}(W))\leq 24 M\sum\limits_{n\in\mathbb{N}} n^{2\alpha} n^{n^{\alpha}} \tau^{n^{1-\alpha}}.

Because \alpha<1/2, we know that \sum\limits_{n\in\mathbb{N}} n^{2\alpha}n^{n^{\alpha}} \tau^{n^{1-\alpha}}<\infty, so that (10) implies the desired estimate (6) of our claim. This ends the proof of proposition 1. \square

Closing this post, we would like to point out that Fayad and Krikorian believe that the requirement \alpha<1/2 in the frequency condition (1) is not optimal although their method uses this restriction in an essential way (as we saw above). Basically, they suspect that if we do some combinatorics on words verifying the frequency condition (1) (perhaps using some multiscale analysis), one should be able to answer positively the following conjecture:

Conjecture (Fayad and Krikorian, 2008). The statement of theorem 1 is true for any 0\leq\alpha<1.

Posted in Mathematics, expository, math.DS | Tags: , ,

« What’s the size of certain unbalanced products of SL(2,R) matrices?
The differentiable sphere theorem of Brendle and Schoen »

Responses

  1. [...] O leitor curioso por mais detalhes pode ver a (curta) demonstração do lemma 2 do meu post em ingles sobre o teorema de Fayad e Krikorian (cuja prova é, por sua vez, baseada nestes argumentos de [...]

    By: Algumas ideias da prova do teorema de Kaloshin e Rodnianski « Disquisitiones Mathematicae (versão portuguesa) on August 7, 2008
    at 6:36 pm

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