I am working with Friedland entropy and there is a proof I cannot figure out how to do. Friedland entropy is defined for $\mathbb{Z}^k$ continuos actions $\mathcal{T}$ on a topological metric space $X$. Let $T_i$ for $i=1,\dots,k$ be the generators of the actions, i.e. the images of the vectors of the canonical base $\mathcal{T}(e_i)=T_i$.

The orbit space is the defined $$ \mathcal{X}=\{(x_n)_{n \in \mathbb{N}}\in \prod_{n \in \mathbb{N}}X \ \ | \ \ \forall n \in \mathbb{N},\exists i=1\dots,k:T_i(x_n)=x_{n+1}\}. $$ On this space we have the shift transformation $\sigma$ defined with $\sigma((x_n)_{n \in \mathbb{N}})=(x_{n+1})_{n \in \mathbb{N}}$. We define the entropy of the action with $e(\mathcal{T})=h_{top}(\sigma)$: it is equal to the topological entropy of the shift.

I read in a paper by Pollicott ad Geller that if we have a $\mathbb{Z}^2$ action generated by a homeomorphism on a compact metric space $X$ and by the identity, then the Friedland entropy of the actions is equal to the topological entropuy of the homeomorphism. The article says it is trivial but I cannot figure out how to do that. Could you help me?

  • $\begingroup$ The claim you are quoting sounds wrong: I have not seen what the Geller and Pollicott paper you mention says, but if $X$ is the unit circle and $T$ is an irrational rotation by $\alpha$ say, its topological entropy is 0. But if you look at $\mathcal X$, the orbit space generated by $T_1=T$ and $T_2=\text{id}$, its entropy seems to be $\log 2$. For any $x$ and any two of the $2^n$ sequences of $T_1$'s and $T_2$'s, the $n$-orbit segments are far apart (by at least $\alpha$), so that the topological entropy is at least $\log 2$. $\endgroup$ Nov 20 '20 at 7:22
  • $\begingroup$ Ok, I think I understand what you are saying. The problem with the paper, his title is "An Entropy for Z^2 action with finite entropy generators", is that doesn't justify the claim. Could, in your opinion, change something, if we assume that the topological entropy of $T$ is positive? $\endgroup$
    – user502940
    Nov 20 '20 at 8:16
  • $\begingroup$ Zero entropy counts as finite entropy, so this is not where the difficulty lies. (Also, I think even if T is the full shift on 2 symbols, I think $e(\mathcal T)=\log 3$ from a small calculation I wrote down). I did find the paper, and found the claim you refer to (Proposition 1(2)). However, section 4 makes it clear (taking $\beta=0$) that $e(T,I)=\log 2$ which is not equal to $h(T)$, so unless I'm misunderstanding something, Prop 1(2) is false as stated. $\endgroup$ Nov 20 '20 at 8:47
  • $\begingroup$ Could I ask you to explain me the calculation you wrote down about the example with the full shift on 2 symbols? $\endgroup$
    – user502940
    Nov 20 '20 at 8:48
  • $\begingroup$ Sure. It’s not 100% rigorous, but I believe it can be made rigorous. I want to count the number of words of length $n$ that you see. We’ll split this up by the total number of times, $k$ that you apply shift (with the other times applying the identity). If you shift $k$ times, then you see $k$ symbols of thee was original word.; and there are $\binom nk$ choices for the $k$ times that you apply $T$ not $I$. So the number of words is $\sum_{k=0}^n\binom nk2^k$ words. This is $3^n$ by the binomial theorem. $\endgroup$ Nov 20 '20 at 16:32

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