# Entropy of first return map and suspension flows

There are some well know formulas of Abramov about derived systems.

Firstly let $(X,\mu,f)$ be a probability preserving system and $A\subset X$ is measurable such that $\bigcup_{n\ge0}f^nA=X$. Let $\mu_A$ be conditional probability of $(\mu,A)$ and $f_A:A\to A$ be the first return map with respect to $A$, that is, $f_A(x)=f^{k(x)}x$ where $k(x)=\inf[k\ge1: f^kx\in A]$ (well defined up to a null set). Then Abramov proved that
$h(f_A,\mu_A)\cdot\mu(A)=h(f,\mu)$.

Secondly let $(X,\mu,f)$ be a probability preserving system and $r:X\to (c,\infty)$ be a roof function with $c>0$ and $\int rd\mu<\infty$. Then consider the suspension space $X_r=[(x,y):x\in X,0\le y\le r(x)]/\sim$ where $(x,r(x))\sim(fx,0)$, the suspension measure $\tilde{\mu}$ given by $\tilde{\mu}(A)=\int_X |A_x|d\mu(x)/\int_X rd\mu$, and the suspension flow $\tilde{f}_t:X_r\to X_r,[x,y]\mapsto[x,y+t]$. Abramov also proved that
$h(\tilde{f},\tilde{\mu})\cdot\int_X rd\mu=h(f,\mu)$.

I think there are direct/intuitive proofs of these two entropy formulas. Any explanation will be great.

Thanks~

For example the first can be derived from the discrete version of the second one:

The first return time $k:A\to\mathbb{N}$ can be viewed as a discrete roof function on $A$. Then suspension is $A_k=[(x,k):x\in A,k=0,1,\cdots, k(x)]/\sim$, which is isomorphic to $X$. And the map $A_k\to A_k,[x,k]\to[x,k+1]$ is isomorphic to $T$ on $X$ (note that $\int_A k(x)d\mu(x)=1$, or equally $\int k(x)d\mu_A(x)=1/\mu(A)$). By identifying $A$ with $A\times[0]\subset A_k$, we see that
$h(f,\mu)/\mu(A)=h(f_A,\mu_A)$.

Still I have no idea about the proof about the entropy of suspension flows~

Finally I understood Abramov's proof (indeed his proof is very clear). Pick $t\in(0,c)$ (fixed from now on) and consider the subset $A=[(x,s):0\le s < t]\subset X_r$. Then the first return map $\tilde{f}_A$ of $\tilde{f}_t$ with respect to $A$ is given by $(x,s)\mapsto(fx,s-r(x))$, where $s-r(x)\in\mathbb{T}_t$ (Note that we can view $A=X\times\mathbb{T}_t$ and $\tilde{f}_A$ as a fiber extension of $f$).

1. He showed that $h(\tilde{f}_A,\tilde{\mu}_A)=h(f,\mu)$ (since the extension is isometric on the fiber).

2. As the first return map of $(X_r,\tilde{f}_t)$, $h(\tilde{f}_A,\tilde{\mu}_A)\cdot\tilde{\mu}(A)=h(\tilde{f}_t,\tilde{\mu})$.

Plugging in $\tilde{\mu}(A)=\frac{t}{\int rd\mu}$, he got the desire formula $h(\tilde{f}_t,\tilde{\mu})=h(f,\mu)\cdot\frac{t}{\int rd\mu}$.

-

The theorems of Abramov, thought about in this way become obvious. They're generalizations of the idea that the number of megabits per minute is 60 times the number of megabits-per-second. The orbits contain the same information. [It's important here that the invariant measure is preserved, up to a constant, by the modification of the measurable dynamical system]. The only thing that has changed is the measurement of space$\times$time; the formula gives ratio of volumes of space-time.
By the way, the corresponding formulas fail for topological entropy in place of measure-theoretic entropy. (One characterization of topological entropy is the supremum of measure-theoretic entropy over all possible invariant measures). For example, the shift S on two symbols (i.e., arbitrary two-sided-infinite sequences of 0's and 1's) has topological and measure-theoretic entropy 1 bit per unit of time. Let $X$ be the union of $S$ with a cycle of length $n$, i.e., addition mod n acting on integers mod n. The topological entropy of X is still 1 bit per unit time, but the measure-theoretic entropy is the ratio of the volume of S to the volume of $X$ per unit time. If you slow the two parts down by taking a section $A$, the measure-theoretic entropy changes as per the formulas you've given, but the topological entropy depends only on the intersection of $A$ with the shift portion. In general, the change in topological entropy coming from a change in the time parameter will be quite tricky.
Thanks! Your last paragraph is impressive~ I think this is sufficient to understand another formula of Abramov, which states that the entropy of a flow $\phi_t$ on $(X,\mu)$ satisfies $h(\phi_t,\mu)=|t|\cdot h(\phi_1,\mu)$. – Pengfei Apr 18 '11 at 4:14