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Question about curve stabilisers acting on annular curve graphs, plus context since I'm interested in being fact-checked.

Definition: let the group $G$ act by isometries on a metric space $(X,d)$. The stable translation length of $g\in G$ is $\tau(g) = \lim_{n\rightarrow\infty} d(x,g^nx)/n$, (for any $x\in X$).

Context: if $S$ is a connected, oriented, hyperbolic surface of sufficiently high finite complexity, and $C(S)$ is its curve graph, then the action of the mapping class group $MCG(S)$ on $C(S)$ is acylindrical, and this implies the existence of $K>0$, depending on $S$, such that $\tau(g)\geq K$ for $g$ pseudo-Anosov (Theorem 1.3 and Lemma 2.2 in Bowditch).

My question is about annular subsurfaces of $S$. Fix an isotopy class $\gamma$ of essential simple closed curves, and let $G < MCG(S)$ be the stabiliser of $\gamma$.

Let $C(\gamma)$ be the curve graph of the annulus in $S$ with core curve $\gamma$, defined in Section 2.4 of Masur and Minsky, where it is explained that $C(\gamma)$ is quasi-isometric to $\mathbb R$, and $G$ acts by isometries, and the Dehn twist $t$ satisfies $\tau(t) =1$ for this action.

Question: Does there exist $K>0$ (allowed to depend on $S$) such that the $G$--action on $C(\gamma)$ has the property that for all $g\in G$, either $\tau(g)=0$ or $\tau(g)>K$?

I tentatively guess "no" but I don't know an argument.

On the other hand, I don't think one can use the above result about acylindricity to say "yes": I believe it's well-known (and seems fairly straightforward) that the image $\bar G$ of $G$ in $Isom(C(\gamma))$ does not in general act acylindrically on $C(\gamma)$. Actually, I also struggle to find a reference for non-acylindricity of $\bar G$ on $C(\gamma)$; is there one or is this MCG folklore?

(My coauthors and I have a result which is "sharp" given a "no" to the question, we're writing the introduction to the paper, and want to comment on MCGs.)

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I think that the answer is “yes, such a constant $K$ exists.”

Suppose that $g$ is the given mapping class in the stabiliser of $\gamma$. Replace $g$ by a large power (bounded by the topology of $S$) so that it fixes the curves of its canonical reducing system (and also their orientations and their sides). In each component of the complement of the reducing system, this new $g$ acts as either the identity, or as a pseudo-Anosov map.

If we only have the former, $g$ is a product of powers of disjoint Dehn twists, and we are done.

If we have some of the latter, then we take a further power (again bounded by the topology of $S$) so that $g$ fixes the switches of its stable train-tracks. In particular, we can control the hyperbolic geometry of a neighbourhood of $\gamma$. The action of $g$ on arcs of $C(\gamma)$ sends endpoints into the “cusps” of the stable track, and otherwise does a power of a Dehn twist. The motion of the endpoints (once close enough to the cusps) is sufficiently slow that it does not contribute to the stable translation length, and we are done.

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    $\begingroup$ Another thing you can do is to estimate the stable translation length by using an element of the arc complex which goes straight to a cusp on both sides of $\gamma$. It follows that (this further power of) $g$ has integer valued stable translation length, just like a Dehn twist. $\endgroup$ Commented Feb 6, 2023 at 6:03

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