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},


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.


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.


\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.


\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.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.


%d bloggers like this: