Entropy of first return map and suspension flows

Question

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~

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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~

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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 $h(\tilde{f}_t,\tilde{\mu})=h(f,\mu)\cdot\frac{t}{\int rd\mu}$.

Answer 1

What may seems intuitive to you depends on your mental model[s] for entropy. This is not static, since we change our mental models and learn new models with time and experience. One good intuitive way to think about entropy is in terms of information theory, as developed by Claude Shannon, where he showed how entropy can be interepreted as the lower bound number of the number of bits required to encode a process, as the encoding becomes more efficient with larger and larger batches. These days it is commonplace to talk about units such as Mbps to measure digital data connections, or channel capacity, without all the complications of noise that were the big concern in Shannon's day. The entropy of a measure-preserving dynamical system is the number of bits per unit time needed to describe an orbit of an average point using a codec (coding-decoding algorithm) of limiting high efficiency. More precisely, it is the infimum of the mean number of bits per unit time per point to describe the orbits of a large batch of uniformly-distributed randomly-chosen points for a long time.

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.