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Let $X$ and $Y$ be compact metric spaces with Borel $\sigma$-algebras $\mathcal{B}$ and $\mathcal{C}$. Let $S: X \to X$ and $T: Y \to Y$ be homeomorphisms. Let $\pi: X \to Y$ be a bounded-to-one continuous factor map, i.e. $\pi$ is surjective, $\pi \circ S = T \circ \pi$, and for some $1 \leq M < \infty$ and all $y \in Y$, $\mathrm{card} \,\pi^{-1}(\{ y \}) \leq M$. Let $\mu$ be an $S$-invariant Borel probability measure on $(X, \mathcal{B})$, and equip $(Y, \mathcal{C})$ with the $T$-invariant measure $\nu = \pi_* \mu$.

Is it the case that $h_{\nu}(T) = h_{\mu}(S)$? If so, how much can my hypotheses be weakened?

I have tried to apply the Abramov-Rokhlin formula, which states that $$ h_{\mu}(S) = h_{\nu}(T) + h_{\mu}(S \, | \, T) $$ where $h_{\mu}(S \, | \, T) = h_{\mu}( S \, | \, \pi^{-1} \mathcal{C} )$ is the relative entropy. I think I should be able to use the bounded-to-one hypothesis to show that this relative entropy is $0$, but I don't know how to do this.

Alp Uzman
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Sophie M
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    why can't you take $Y=X, \pi = id$ and then just $\mu,\nu$ different measures? like $X$ is a Bernoulli shift on two symbols and $\mu = \frac{1}{2}-\frac{1}{2}, \nu = \frac{1}{3}-\frac{2}{3}$; the entropies are different. maybe im missing something – mathworker21 Apr 26 '20 at 06:11
  • Oops! You're very right. Forgot the extremely point that $\nu = \pi_* \mu$. Will edit the question. – Sophie M Apr 26 '20 at 06:33
  • Have you looked at the case of symbolic spaces, let's say with 4 and 2 symbols with the map 0,2 -> 0 and 1,3 -> 1, and see what happens with the uniform measure in this setting? – Felipe Pérez Apr 27 '20 at 01:58
  • @FelipePérez If you mean the full $4$-shift mapping to the full $2$-shift, this is not finite-to-one. If for the domain you mean a pair of disjoint copies of the full $2$-shift (one with alphabet ${0,2}$, one with alphabet ${ 1,3 }$) each carrying a scaled uniform measure, I've tried playing with generators there (tried to show that a generator for $T$ lifts to one for $S$) but haven't actually done the entropy calculation! I'll try that. – Sophie M Apr 27 '20 at 02:36
  • @SophieMacDonald what did u mean "if for the domain you mean..."? can u describe that setup for me plz – mathworker21 Apr 27 '20 at 15:23
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    @mathworker21 I just mean if $X$ (the domain of $\pi$) is the disjoint union of two full $2$-shifts, i.e. $X = { 0,1 }^{\mathbb{Z}} \cup { 2,3 }^{\mathbb{Z}}$. – Sophie M Apr 27 '20 at 15:39
  • Also in my reply to @FelipePérez, it should have been ${0,1}$ and ${2,3}$, not ${0,2}$ and ${1,3}$. – Sophie M Apr 27 '20 at 15:41
  • @SophieMacDonald entropy is obv log2 in either case – mathworker21 Apr 27 '20 at 15:42
  • @mathworker21 yes, but with the block map proposed ($0,2 \mapsto 0$; $1,3 \mapsto 1$), my original statement would have sent the pair of $2$-shifts to a pair of points. – Sophie M Apr 27 '20 at 15:46
  • @SophieMacDonald idk what u mean – mathworker21 Apr 27 '20 at 16:00

1 Answers1

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The answer is yes, by an Abramov-Rohlin type formula of Ledrappier and Walters ("A relativised variational principle for continuous transformations", p.569, Thm.2.1; they prove the statement for pressure).

Theorem (Ledrappier-Walters): Let $X,Y$ be compact metric, $S:X\to X$, $T:Y\to Y$ be continuous, $\pi:X\to Y$ be a continuous surjection such that $\pi\circ S=T\circ \pi$ (so that $T$ is a topological factor of $S$). Then for any $\nu\in\operatorname{Prob}(Y,T)$:

\begin{align*} \operatorname{ent}_{\nu}(T) &\leq \inf_{\substack{\mu\in\operatorname{Prob(X,S)}\\\pi_\ast(\mu)=\nu}} \operatorname{ent}_{\mu}(S) \\ &\leq \sup_{\substack{\mu\in\operatorname{Prob(X,S)}\\\pi_\ast(\mu)=\nu}} \operatorname{ent}_{\mu}(S) = \operatorname{ent}_{\nu}(T) + \int_Y \operatorname{topent}(S;\pi^{-1}(y))\, d\nu(y). \end{align*}

Here $\operatorname{topent}(S;\pi^{-1}(y))$ is defined using $(S,n,\epsilon)$-separated subsets of $\pi^{-1}(y)$ (or sets that $(S,n,\epsilon)$-span $\pi^{-1}(y)$), and $y\mapsto \operatorname{topent}(S;\pi^{-1}(y))$ is measurable. When a fiber $\pi^{-1}(y)$ is finite it's straighforward that $\operatorname{topent}(S;\pi^{-1}(y))=0$; when each fiber is finite (not necessarily uniformly so) the integral on the RHS vanishes.

Let me also mention that this theorem builds on an analogous theorem on topological entropy of Bowen:

Theorem (Bowen): With $X,Y,S,T,\pi$ as in the previous theorem,

\begin{align*} \operatorname{topent}(T) &\leq \operatorname{topent}(S) \\ &\leq \operatorname{topent}(T) + \sup_{y\in Y} \operatorname{topent}(S;\pi^{-1}(y)). \end{align*}

Bowen's proof of this is combinatorial, and the theorem gives the topological version of what you are asking. Of course via the variational principle the Ledrappier-Walters formula also gives it. Robinson's Dynamical Systems: Stability, Symbolic Dynamics, and Chaos has a proof of the uniformly finite fibers case for the topological entropy (p.340,Thm.1.8 of the first edition). I haven't read it but one might be able to adapt it to bypass Ledrappier-Walters.

Alp Uzman
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