Suppose $\Gamma$ is the fundamental group of a closed, oriented surface $S$. Let $B$ be a finitely generated, infinite index subgroup of $\Gamma$, and let $\Gamma_B$ be the compact core of the $B$-covering space of $S$ (well-defined up to isotopy in the $B$-covering space). Lifting hyperbolic structures gives a well-defined continuous function from $T(\Gamma)$, the Teichmuller space of hyperbolic structures $\Gamma$, to $T(\Gamma_B)$, the Teichmuller space of hyperbolic structures with totally geodesic boundary on $\Gamma_B$.

The general question I would like to ask is: How rigid is the map $T(\Gamma) \to T(\Gamma_B)$? However, as it stands that question suffers from the problem that the boundary components of $T(\Gamma_B)$ can be arbitrarily long. So, in order to sweep that problem under the rug, I want to compose with the map from $T(\Gamma_B)$ to the Masur-Minsky marking complex $M(\Gamma_B)$, defined in their paper "Geometry of the curve complex II".

Question 1: How rigid is the composed map $T(\Gamma) \to T(\Gamma_B) \to M(\Gamma_B)$?

EDIT: What I mean by "rigidity" here is, loosely speaking, some kind of measurement of the extent of the image of this map in the marking complex $M(\Gamma_B)$. Here are two examples which exhibit opposite extremes of rigidity.

For example, if $B$ is the fundamental group of an essential subsurface of $\Gamma$ then the composed map is essentially surjective. This is the "least rigid" possibility.

For another example, if $B$ has rank $2$ and $\Gamma_B$ is a pair of pants then the composed map has bounded image in $M(\Gamma_B)$. This is the "most rigid" possibility (although this example is uninteresting because $M(\Gamma_B)$ is itself just a single point).

Question 2: Are there examples of subgroups $B$ where $\Gamma_B$ is not just a pair of pants but the image of the composed map $T(\Gamma) \to M(\Gamma_B)$ has finite diameter?

As a reminder of how the map $T(\Gamma_B) \to M(\Gamma_B)$ is defined, one fixes an appropriate Margulis constant $\epsilon$ for $\Gamma_B$, chosen so that for any hyperbolic structure in $T(\Gamma_B)$ the collection of closed geodesics and closed proper arcs of length $\le \epsilon$ fill $\Gamma_B$, and from that collection one uses surgery to construct a pants decomposition of $\Gamma_B$ and a transverse closed curve for each pants curve.

This question occurred to me while pondering a question of Benjamin Steinberg. I am not aware of any literature on this question.

It would also make sense to ask this question with regard to a natural "lifting" map $M(\Gamma) \to M(\Gamma_B)$, but the image of this map is probably coarsely the same as the image of the map $T(\Gamma) \to M(\Gamma_B)$.

  • $\begingroup$ @Lee: It will never be bounded (except for the pair of pants example): Just apply powers your favorite pseudo-Anosov to a non-peripheral element of $B$. Could you clarify what do you mean by rigidity in Question 1? The setup sounds somewhat similar to McMullen's work on theta-operator (or, even more remotely, skinning map). The restriction map $T(\Gamma)\to T(\Gamma_B)$ is holomorphic, so it contracts Teichmuller metric. It might follow from McMullen's work that the map is a strict contraction. $\endgroup$
    – Misha
    May 3, 2012 at 18:05
  • $\begingroup$ I edited to try to clarify what "rigid" means, although it is still a somewhat vague notion. $\endgroup$
    – Lee Mosher
    May 3, 2012 at 18:11
  • $\begingroup$ @Misha: just because the lengths of all elements of $B$ are uniformly long does not mean that the image is unbounded in the marking complex $M(\Gamma_B)$. For example, $\Gamma_B$ could be a one-holed torus with three very short properly embedded arcs hitting the boundary at right angles, cutting the torus into two right-angled hexagons. On this $\Gamma_B$, one could force the boundary length off to infinity without changing which three arcs are short, and the marking would be unchanged. $\endgroup$
    – Lee Mosher
    May 3, 2012 at 18:20
  • $\begingroup$ added: also, as the boundary length goes off to infinity, all lengths of all closed curves would have a lower bound that goes off to infinity. $\endgroup$
    – Lee Mosher
    May 3, 2012 at 18:21
  • $\begingroup$ @Lee: If $B$ contains a simple non-peripheral loop $\beta$ which projects to a non-simple loop in $S$ then the restriction map $f$ will always miss the thin part of $T(\Gamma_B)$ corresponding to $\beta$. Thus, the map is never coarsely surjective unless $B$ is a subsurface or a pair of pants. Also, for dimension reasons, the image is "too small" if $-\chi(B)> -\chi(S)$. An interesting question then would be if $f$ is (sometimes/always) coarsely onto thick part of $T(\Gamma_B)$ when $-\chi(B)\le -\chi(S)$ and $B$ is not embedded. "Sometimes" is probably true, "always" is probably false. $\endgroup$
    – Misha
    May 3, 2012 at 18:28

1 Answer 1


Regarding Question 2, you get lots of examples that are rigid for the lifting map $M(\Gamma) \to M(\Gamma_B)$.

Let $B$ be finitely generated subgroup of $\Gamma$ (considered a fuchsian group) such that every nontrivial element of $B$ fills $S$ (there are lots of these). Proposition 5.1 of [Kent, Leininger, and Schleimer. Trees and mapping class groups. J. Reine Angew. Math., 637:1–21, 2009] states that if you pull back all the essential simple closed curves on $S$ to the cover corresponding to $B$ and intersect with the convex core $\mathrm{core}(B)$, you get a finite set of arcs up to proper isotopy.

The idea of the proof is this: Suppose that there are arcs obtained by pulling back simple closed curves to $\mathrm{core}(B)$ whose lengths tend to infinity. These arcs limit (after passing to a subsequence) to a lamination $\widetilde \lambda$ on $\mathrm{core}(B)$ that covers a lamination $\lambda$ on $S$. But then there is a loop in $\mathrm{core}(B)$ that is not filling in $S$ (since the boundary of the supporting subsurface of $\widetilde \lambda$ will project to a curve in $S$ that misses $\lambda$). So there is a uniform bound to length of any of the arcs in our collection, and so the collection of isotopy classes is bounded.

So, if you lift any marking, you land in a finite set of markings for $\Gamma_B$.

It seems to me that, with some more work, this argument should give you a finite set of such arcs even if you allow $\Gamma$ to range over all of Teichmuller space (Dowdall and Leininger and I have been considering similar things, so I'll talk to them), which would answer Question 2 in its original form. Though it's probably easier to just say that the lifting map is essentially the same as $T(\Gamma) \to M(\Gamma_B)$, as you suggest.

  • $\begingroup$ Very nice! $\endgroup$
    – Lee Mosher
    May 6, 2012 at 23:24
  • 1
    $\begingroup$ Leininger has pointed out to me that the set of arcs does not depend on $\Gamma$, since they are determined by topological data (the lifts of simple curves on $S$ to $\mathbb{H}^2$ ``link" the limit set, and this linking data, which is really topological, determines the arcs you get). So the map $T(\Gamma) \to M(\Gamma_B)$ will be bounded for these filling groups $B$. $\endgroup$ May 8, 2012 at 20:12
  • 1
    $\begingroup$ Also, you should get other rigid examples by taking subgroups that are purely filling with respect to a subsurface. $\endgroup$ May 8, 2012 at 20:13

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