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](https://mathoverflow.net/questions/95709/are-all-free-factors-of-finitely-generated-subgroups-of-free-groups-geometric/95768#95768). 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 $T(\Gamma) \to M(\Gamma\_B)$.



Does the fundamental group of a surface have rigid subgroups?
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.


