Question on a stable geodesic lamination on a closed hyperbolic surface

Question

Let me first state a theorem in Casson-Bleiler Automorphisms of Surfaces after Nielsen and Thurston.

Theorem 5.5: Let $h:F\to F$ be a non-periodic irreducible automorphism of a closed orientable hyperbolic surface. Any lift of a strictly positive power of $h$ has finitely many fixed points on $S^1_\infty$, alternating contracting and expanding. There is a unique perfect lamination $L^S$, invariant under $\hat{h}$, such that $\tilde{L}^S$ contains the geodesics joining consecutive contracting fixed points of any lift of a strictly positive power of $h$. Every leaf of $L^S$ is dense in $L^S$.

What I'm wondering is that each positive power of $h$ has finitely many fixed points on $S^1_\infty$ so each associates an ideal polygon formed by connecting contracting fixed points. Since $L^S$ has no closed leaves, each such ideal polygon isometrically embedded on $F$ and there are finitely many ideal polygons as principal regions. Suppose there are only two $A$ and $B$. Then the ideal vertices of $A$ on the cover are all contracting fixed points and so are $B$. Here's where something contradictory thing happens: If I iterate $h$, then all the ideal vertices of $A$ and $B$ are fixed but they share only one ideal vertex as all leaves in $L^S$ are not isolated. This means that the ideal vertices of $A$ and $B$ are no longer fixed as I iterate $h$. I'm misunderstanding something here. Can anyone explain why this is wrong?

Answer 1

It looks like you are confusing ideal polygons on $\mathbb H^2$ with ideal polygons on $F$.

To explain this, let me use the standard terminology of principle regions for the ideal polygons that arise in this manner, but I'm going to be careful and speak separately about principle regions (of $h$) in $\mathbb H^2$ versus principle regions (of $h$) in $F$. Let me also introduce notation for the universal covering map $p : \mathbb H^2 \to F$.

So, while it is true that there are only finitely many principle regions in $F$, in fact there are infinitely many different principle regions in $\mathbb H^2$. Namely, if $\alpha \subset F$ is a principle region of $h$ in $F$, then:

1. $p^{-1}(\alpha) \subset \mathbb H^2$ has a countable infinity of connected components, permuted around freely and transitively by the deck transformation action of $\pi_1 F$ on $\mathbb H^2$;
2. For each connected component $A$ of $p^{-1}(\alpha)$, $A$ is a principle region of $h$ in $\mathbb H^2$, and the restriction map $p : A \to \alpha$ is a homeomorphism.

These facts are all established in the midst of the proof of the theorem you have cited.

So the statement that there are only finitely many principle regions of $h$ in $F$ is equivalent to the statement that the total collection of principle regions of $h$ in $\mathbb H^2$ is countably infinite, and it follows furthermore that any two principle regions of $h$ in $\mathbb H^2$ are disjoint from each other.

What might be confusing you is the following.

Let's suppose $H : \mathbb H^2 \to \mathbb H^2$ is a lift of a positive power of $h$, and that its set $\text{Fix}_+(H)$ of attracting fixed points has $\ge 3$ points. It is indeed true in this situation that the ideal polygon $A$ with endpoints on $\text{Fix}_+(H)$ is a principle region of $h$ in $\mathbb H^2$, and that $\alpha = p(A)$ is a principle region $\alpha$ of $h$ in $F$.

And while (yes) there do exist other principle regions $B$ of $h$ in $\mathbb H^2$, and while (yes) they are disjoint from $A$, and while (no) they do not share any vertices with $A$, in fact the vertex sets of $A$ and $B$ are disjoint from each other, no vertex of $B$ is fixed by $H$. Nonetheless, the vertices of $B$ will be the set of attracting fixed points of some lift of some power of $h$, just not of $H$ itself.