ERT7: Furstenberg-Weiss Topological Multiple Recurrence Theorem

Posted by: yglima | February 21, 2010

ERT7: Furstenberg-Weiss Topological Multiple Recurrence Theorem

As promissed in ERT6, this post intends to prove

Theorem 1 Let ${T:X\rightarrow X}$ be a homeomorphism of the compact metric space ${(X,d)}$ . Then, for any ${k\in\mathbb{N}}$ and ${\varepsilon>0}$ , there exist ${x\in X}$ and ${n\in\mathbb{N}}$ such that $\displaystyle d\left(T^{in}x,x\right)<\varepsilon\, ,\ i=1,2,\ldots,k.$

Moreover, given any dense subset ${Y\subset X}$ , we can take ${x\in Y}$ .

Actually, we prove the more general version

Theorem 2 Let ${X}$ be a compact metric space and ${T_1,\ldots,T_k}$ commutative homeomorphisms of ${X}$ . Then some ${x\in X}$ is multiply recurrent, that is: there exists a sequence ${n_j\rightarrow\infty}$ such that

$\displaystyle {T_i}^{n_j}x\rightarrow x,\ \ i=1,\ldots,k.$

1. Homogeneity

Consider a homeomorphism ${T:X\rightarrow X}$ .

Definition 3 ${T}$ is homogeneous if there is a group ${G}$ of homeomorphisms of ${X}$ , each element of ${G}$ commuting with ${T}$ , such that the action ${G:X\rightarrow X}$ is minimal.

We remark that the notion of minimality of a group action is the same as that of a single homeomorphism:

$\displaystyle \overline{\{gx\,;\,g\in G\}}=X,\ \forall\,x\in X.$

Definition 4 ${A\subset X}$ is homogeneous if there is a group ${G}$ of homeomorphisms of ${X}$ , each element of ${G}$ commuting with ${T}$ , such that ${GA=A}$ and the action ${G:A\rightarrow A}$ is minimal.

In particular, ${A}$ is closed. Note that ${A}$ does not need to be ${T}$ -invariant. This is the good point of the above definition: we analyse the dynamics of ${A}$ not by the original map ${T}$ , but through the action of ${G}$ . We now proceed to the technical details of the proof.

Proposition 5 Let ${T:X\rightarrow X}$ be a homeomorphism and ${A\subset X}$ homogeneous. Suppose that, for each ${\varepsilon>0}$ , there exist ${x,y\in A}$ and ${n\ge 1}$ such that

$\displaystyle d\left(T^nx,y\right)<\varepsilon.$

Then, for every ${\varepsilon>0}$ , there exist ${z\in X}$ and ${n\ge 1}$ such that

$\displaystyle d\left(T^nz,z\right)<\varepsilon.$

In other words, the property assumed also holds for a pair on the diagonal ${\Delta=\{(z,z)\,;\,z\in A\}}$ .

Proof: Let ${G}$ be the group associated to the homegeneity of ${A}$ and ${\varepsilon>0}$ . The ${G}$ -orbit of every ${x\in A}$ is dense in ${A}$ and then becomes as close as we want to any other ${y\in A}$ .

Let us prove that, in a scale of ${\varepsilon/2}$ , the ${G}$ -orbits in ${A}$ are finite. More specifically, we prove the existence of ${g_1,\ldots,g_k\in G}$ for which

$\min_{1\le i\le k}d\left(g_ix,y\right)<\dfrac{\varepsilon}{2}\,, \forall\,x,y\in A.$

Consider an open cover of ${A}$ by ${V_1,\ldots,V_r}$ , each of them with diameter at most ${\varepsilon/2}$ . For each ${j\in\{1,\ldots,r\}}$ , the family ${\{g^{-1}V_j\,;\,g\in G\}}$ is an open cover of ${A}$ (in fact, if ${x\in A}$ , there exists ${g\in G}$ such that ${gx\in V_j}$ , that is, ${x\in g^{-1}V_j}$ ). Reduce this cover to a finite one:

$\displaystyle A={g_{j1}}^{-1}V_j\cup\cdots\cup {g_{js_j}}^{-1}V_j.$

The finite set ${\{g_{ji}\,;\,1\le j\le r,1\le i\le s_j\}}$ satisfies the required property: if ${x,y\in A}$ , then ${y\in V_j}$ , for some ${j}$ , and ${x\in {g_{ji}}^{-1}V_j}$ , for some ${i\in\{1,\ldots,s_j\}}$ , implying that

$\displaystyle d\left(g_{ji}x,y\right)<\frac{\varepsilon}{2}\,\cdot$

Just rename the ${g_{ji}}$ .

We now prove that, for every ${y\in A}$ , there exist ${x\in A}$ and ${n\ge 1}$ such that

$\displaystyle d\left(T^nx,y\right)<\varepsilon.$

Let ${\delta>0}$ be such that

$\displaystyle d(z,w)<\delta\ \Longrightarrow\ d(g_iz,g_iw)<\frac{\varepsilon}{2}\,,\ i=1,\ldots,k$

and ${x_0,y_0\in A}$ , ${n\ge 1}$ such that

$\displaystyle d(T^nx_0,y_0)<\delta.$

Taking ${i\in\{1,\ldots,r\}}$ for which

$\displaystyle d(g_iy_0,y)<\frac{\varepsilon}{2}$

and ${x=g_ix_0}$ , we conclude the inequalities

$\displaystyle \begin{array}{rcl} d(T^nx,y)&= & d(g_iT^nx_0,y)\\ &\le & d(g_iT^nx_0,g_iy_0)+d(g_iy_0,y)\\ &\le &\frac{\varepsilon}{2}+\frac{\varepsilon}{2}\\ &= &\varepsilon. \end{array}$

Finally, we prove the proposition. Take ${x_0,x_1\in A}$ and ${n_1\ge 1}$ such that

$\displaystyle d\left(T^{n_1}x_1,x_0\right)<\varepsilon_1=\frac{\varepsilon}{2}\,\cdot$

By continuity, there is ${\varepsilon_2<\varepsilon_1}$ for which

$\displaystyle d(x,x_1)<\varepsilon_2\ \Longrightarrow\ d(T^{n_1}x,x_0)<\varepsilon_1.$

Take ${x_2\in A}$ and ${n_2\ge 1}$ satisfying

$\displaystyle d(T^{n_2}x_2,x_1)<\varepsilon_2\,.$

Proceeding by induction, if

$\displaystyle d(T^{n_i}x_i,x_{i-1})<\varepsilon_i\,,$

take ${\varepsilon_{i+1}<\varepsilon_i}$ such that

$\displaystyle d(x,x_i)<\varepsilon_{i+1}\ \Longrightarrow\ d(T^{n_i}x,x_{i-1})<\varepsilon_i$

and ${x_{i+1}\in A}$ , ${n_{i+1}\ge 1}$ for which

$\displaystyle d\left(T^{n_{i+1}}x_{i+1},x_i\right)<\varepsilon_{i+1}\,.$

Then, if ${i<j}$ ,

$d\left(T^{n_j+\cdots+n_{i+1}}x_j,x_i\right)<\varepsilon_{i+1}\,.$

But ${i,j}$ may be chosen such that ${d(x_i,x_j)<\varepsilon_1}$ and then, taking ${n=n_j+\cdots+n_{i+1}}$ , (1) implies the inequality

$\displaystyle d\left(T^nx_j,x_j\right)\le d\left(T^nx_j,x_i\right)+d\left(x_i,x_j\right)<\varepsilon_{i+1}+\varepsilon_1\le 2\varepsilon_1=\varepsilon.$

$\Box$

Proposition 6 Under the same conditions of Proposition 5, there exists ${x\in A}$ recurrent for ${T}$ .

Proof: Define ${f:A\rightarrow\mathbb R}$ by

$\displaystyle f(x)\doteq\inf_{n\in\mathbb N}d\left(T^nx,x\right).$

From Proposition 5, the image of ${f}$ has values arbitrarily close to zero. In addition, ${f}$ is upper semicontinuous. In fact, given ${x\in A}$ and ${\varepsilon>0}$ , let ${n\ge 1}$ such that

$\displaystyle d\left(T^nx,x\right)<f(x)+\varepsilon.$

By continuity of ${T^n}$ , there is ${\delta>0}$ for which

$\displaystyle d(x,y)<\delta\ \ \Longrightarrow\ \ d(T^ny,y)<f(x)+\varepsilon.$

Then

$\displaystyle f(y)<f(x)+\varepsilon,\ \forall\,y\in B(x,\delta)$

and so, as ${\varepsilon>0}$ is arbitrary,

$\displaystyle \limsup_{y\rightarrow x}f(y)\le f(x).$

Then the set ${C_f}$ of continuity points of ${f}$ is residual (in particular, it is non-empty). Let us show that ${f|_{C_f}=0}$ . This will conclude the proof.

By contradiction, suppose that ${f(x_0)>0}$ for some ${x_0\in C_f}$ . Take a neighbourhood ${V}$ of ${x_0}$ and ${\delta>0}$ such that

$\displaystyle f(x)>\delta\,,\ \forall\,x\in V. \ \ \ \ \ (1)$

We now use the homogeneity of ${A}$ to prove that a similar inequality to (1) holds in all of ${A}$ .

Let ${G}$ be the group associated to the homogeneity of ${A}$ . By compactness of ${A}$ and minimality of ${G:A\rightarrow A}$ , there are ${g_1,\ldots,g_k\in G}$ such that

$\displaystyle A=\bigcup_{i=1}^{k}{g_i}^{-1}V.$

Take ${\eta>0}$ for which $\displaystyle d(x,y)<\eta\ \Longrightarrow\ d(g_ix,g_iy)<\delta,\ i=1,\ldots,k. \ \ \ \ \ (2)$

Let us prove that ${f(x)\ge\eta}$ , for every ${x\in A}$ . If ${f(x)<\eta}$ , there exists ${n\ge 1}$ such that

$\displaystyle d\left(T^nx,x\right)<\eta$

and then, taking ${i}$ for which ${y=g_ix\in V}$ , (2) guarantees that

$\displaystyle d\left(T^ny,y\right)=d\left(g_iT^nx,g_ix\right)<\delta,$

which contradicts (1). $\Box$

2. Proof of Theorem 2

We proceed by induction. For ${k=1}$ , it is sufficient to take any point of a minimal set (which exists by Zorn’s Lemma). Suppose the result is true for ${k-1\ge 1}$ and consider ${k}$ commutative homeomorphisms ${T_1,\ldots,T_k}$ of ${X}$ . Let ${G=\langle T_1,\ldots,T_k\rangle}$ be the group generated by ${T_1,\ldots,T_k}$ . We can assume that ${G}$ acts minimaly on ${X}$ . If this is not the case, we restrict ${X}$ to a ${G}$ -invariant closed subset ${A}$ for which ${G:A\rightarrow A}$ is minimal (such minimal set exists again by Zorn’s Lemma).

Let ${\Delta=\{(x,\ldots,x)\,;\,x\in X\}\subset X^k}$ be the diagonal and consider the product transformation ${T=T_1\times\cdots\times T_k}$ . We wish to show that there exists ${x^*\in\Delta}$ recurrent for ${T}$ . To this matter, it suffices to check the hypotheses of Proposition 5:

I. ${\Delta\subset X^k}$ is homogeneous for ${T}$ .

II. For each ${\varepsilon>0}$ , there exist ${x^*,y^*\in\Delta}$ and ${n\ge 1}$ such that

$\displaystyle d\left(T^nx^*,y^*\right)<\varepsilon.$

The action of ${G}$ can be induced in ${X^k}$ in a natural way, associating ${g\in G}$ to the map ${\tilde g=g\times\cdots\times g}$ . In this setting, if ${\tilde G=\{\tilde g\,;\, g\in G\}}$ , the pairs ${(X,G)}$ and ${(\Delta,\tilde G)}$ are isomorphic and then ${\tilde G}$ acts minimaly on ${\Delta}$ . In particular, ${\Delta}$ is homogeneous for ${T}$ , establishing I.

To prove II, define ${R_i=T_i{T_k}^{-1}}$ , ${i=1,\ldots,k-1}$ . By the induction hypothesis, there are ${x\in X}$ and ${n\ge 1}$ such that

$\displaystyle d\left({R_i}^nx,x\right)<\varepsilon,\ i=1,\ldots,k-1.$

Taking

$\displaystyle x^*=({T_k}^{-n}x,\ldots,{T_k}^{-n}x)\ \ \text{ and }\ \ y^*=(x,\ldots,x),$

we get

$\displaystyle \begin{array}{rcl} d\left(T^nx^*,y^*\right)&=&d\left(({T_1}^n{T_k}^{-n}x,\ldots,{T_{k-1}}^n{T_k}^{-n}x,x),(x,\ldots,x,x)\right)\\ &=&d\left(({R_1}^nx,\ldots,{R_{k-1}}^nx,x),(x,\ldots,x,x)\right)\\ &<&\varepsilon\,, \end{array}$

concluding the proof of the theorem.

Previous posts: ERT0, ERT1, ERT2, ERT3, ERT4, ERT5, ERT6.

Like this:

Be the first to like this post.

Posted in expository, guest blog, math.DS, Mathematics | Tags: Furstenberg-Weiss theorem, homogeneous set, homogeneous transformation, multiple recurrence

M	T	W	T	F	S	S
« Jan				Mar »
1	2	3	4	5	6	7
8	9	10	11	12	13	14
15	16	17	18	19	20	21
22	23	24	25	26	27	28

	matheuscmss on Diffusion in Ehrenfest wind-tr…
	yglima on Diffusion in Ehrenfest wind-tr…
	yglima on CF1: uniform distribution and …
	vdelecroix on CF1: uniform distribution and …
	matheuscmss on CF1: uniform distribution and …

Disquisitiones Mathematicae

ERT7: Furstenberg-Weiss Topological Multiple Recurrence Theorem

Like this:

Leave a Reply Cancel reply

Categories

Archives

Mathematics

politics

Recent Posts

Tags

Top Posts

Recent Comments

Follow “Disquisitiones Mathematicae”