# Applying the Abramov-Rokhlin skew product entropy formula to a bounded-to-one factor

Let $$(X, \mathcal{B}, \mu, S)$$ and $$(Y, \mathcal{C}, \nu, T)$$ be invertible probability-measure-preserving systems, with a measurable factor map $$\pi: X \to Y$$, i.e. $$\pi \circ S = T \circ \pi$$. Suppose that there is some $$N \in \mathbb{N}$$ such that $$\nu$$-almost every $$y \in Y$$ has at most $$N$$ preimages under $$\pi$$, i.e. $$\# \pi^{-1}(\{y\}) \leq N$$. If necessary, I'm happy to assume that $$X$$ and $$Y$$ are compact metric spaces, $$\mu$$ and $$\nu$$ are Borel, and $$\pi, S, T$$ are continuous.

I'm trying to show that $$\pi$$ preserves measure-theoretic entropy, i.e. $$h(\nu) = h(\mu)$$. I'd like to apply the Abramov-Rokhlin entropy formula, which expresses the entropy of a skew product transformation as the sum of the entropy of the base and the entropy of the fibre. To do this, I'd like to realize $$X$$ as the direct product $$Y \times S$$ (or a subset of this), with $$F = \{ 1, \dots, N \}$$, together some probability vector $$\mathbf{p} \in \mathbb{R}^N$$, such that $$\mu = \nu \times \mathbf{p}$$. I'd then try to realize $$S$$ as $$S(y,k) = (T(y), \alpha(y)(k))$$ where for each $$y$$, $$\alpha(y)$$ is an invertible measure-preserving transformation of $$(F, \mathbf{p})$$.

The problem is that if the $$\alpha$$'s give a transitive action of $$S_N$$, this would require that $$\mathbf{p}$$ be uniform. In particular, it seems that the fiber cardinality would have to be constant, not just bounded. Even forgetting about the measure, this is an issue, because the $$\alpha$$'s would have to be bijections.

Two ideas I've played with:

• The number of preimages $$\# \pi^{-1}(\{y\})$$ is constant along the orbit of $$y$$, so I could try partitioning $$X$$ and $$Y$$ into at most $$N$$ disjoint subsystems on which $$\pi$$ is constant-to-one, then applying the Abramov-Rokhlin formula separately on each one, but I don't know how to show that this decomposition would be measurable.

• Bögenschutz and Crauel (1990) have a generalization of the Abramov-Rokhlin formula for a skew product transformation (with the notation above) in which we no longer require a fixed measure on the fiber, but only that $$\mu$$ be invariant with fixed $$Y$$-marginal $$\nu$$. This might solve the problem by letting me add a null set of fixed points to $$X$$, for instance, to make $$\pi$$ constant-to-one.

I'm also hoping to apply this in the case where $$S$$ and $$T$$ are really actions of a countable amenable group $$G$$, using Ward and Zhang's generalization of the Abramov-Rokhlin formula. I haven't worked through the proofs to see if the Bögenschutz-Crauel and Ward-Zhang generalizations are compatible with each other.

• Jisang Yoo's paper: arxiv.org/abs/1612.08648 has a lemma (Lemma 3.1) showing measurability of the decomposition by number of pre-images. – Anthony Quas Mar 31 '20 at 17:18
• Thanks! It looks like this decomposition may not quite be measurable, but it is universally measurable (in particular, it becomes measurable if you replace $\mathcal{B}$ and $\mathcal{C}$ by their completions with respect to $\mu$ and $\nu$, respectively), which is enough. Universal measurability will actually help elsewhere in this project, too, so I'm glad to have learned about it here. – Sophie MacDonald Mar 31 '20 at 17:53

• Thank you for the suggestion to look at Rokhlin's paper! The idea I was missing was similar to the one in my reply to Anthony Quas's comment on the question: work with a complete $\sigma$-algebra, rather than with Borel sets only. – Sophie MacDonald Apr 1 '20 at 23:24