Let $\Gamma$ be a discrete subgroup of a connected finite dimensional Lie group $G$. Let $K$ be a maximal compact subgroup of $G$ and denote $X=G/K$. It is well-known that $\Gamma$ acts properly on $X$ via multiplication on the left.


Now suppose that a discrete group $\tilde{\Gamma}$ contains $\Gamma$ as a finite index subgroup (I'm not assuming that $\tilde{\Gamma}$ is a subgroup of $G$.)


When can the action of $\Gamma$ on $X$ be extended to a proper action of $\tilde{\Gamma}$ on $X$?
Can you give an example where this is not the case?


**Edit:** Let my emphasize that I only require the action of $\tilde{\Gamma}$ on $X$ to be proper (i.e. any continuous action with discrete orbits and finite point stablizers). I am not asking whether $\tilde{\Gamma}$ is a subgroup of $G$. The action of $\tilde{\Gamma}$ on $X$ is allowed to have a finite kernel.


Thanks.



discrete subgroups of Lie groups and actions on homogeneous spaces
Here is a simple counter-example. Let $p : S\_3 \to S\_2$ be a degree 2 covering map from the closed genus 3 surface to the closed genus 2 surface, inducing an index 2 injection $p\_\* : \pi\_1(S\_3) \to \pi\_1(S\_2)$. Pulling back any hyperbolic structure from $S\_2$ to $S\_3$ via $p$ defines an embedding of Teichmuller spaces $p^\* : T(S\_2) \to T(S\_3)$, the domain having dimension 6 and the range having dimension 12. Pick a point in $T(S\_3)$ which is not in the image of this embedding nor in any translate of the image under the action of the mapping class group $MCG(S\_3)$ on $T(S\_3)$. This point represents a discrete subgroup $\Gamma \approx \pi\_1(S\_3)$ of $Isom(H^2) = PSL(2,R)$ such that, under the index 2 inclusion of $\Gamma \hookrightarrow \tilde\Gamma \approx \pi\_1(S\_2)$, the action of $\Gamma$ does not extend to an action of $\tilde\Gamma$.


ADDED AFTER INITIAL COMMENTS: The OP has clarified that he intends only to ask about the existence of topological actions of $\tilde\Gamma$, in which setting this is not a counter-example as pointed out by Yves.


However, in the other direction, rigidity theorems apply to give many interesting positive examples. For instance, suppose $\Gamma$ is a uniform lattice in $Isom(H^n) = SO(n,1)$, $n \ge 3$, meaning a discrete cocompact subgroup. If $\tilde\Gamma$ is a finite index supergroup of $\Gamma$, and if $\tilde\Gamma$ has no finite normal subgroups, then the given isometric action of $\Gamma$ extends to a faithful isometric action of $\tilde\Gamma$.


To prove it, since $\tilde\Gamma$ is quasi-isometric to $\Gamma$ which is quasi-isometric to $H^n$, applying the Sullivan-Tukia quasi-isometric rigidity theorem for $H^n$ it follows that there is a cocompact discrete action of $\tilde\Gamma$ on $H^n$ with finite kernel, but we have hypothesized the kernel away so this action is faithful. We now have two discrete cocompact actions of $\Gamma$: the given one, and the restriction of the one on $\tilde\Gamma$ that we got by applying Sullivan-Tukia. We can therefore apply Mostow rigidity and conjugate the $\tilde\Gamma$ action by an isometry of $H^n$, so that its restriction to $\Gamma$ agrees with the given $\Gamma$ action, and we are done.


Basically the same argument will work whenever the symmetric space satisfies Mostow rigidity and some reasonably strong form of quasi-isometric rigidity, which is true of just about every symmetric space associated to an irreducible semi-simple Lie group (except for $SL(2,R)$ of course).


And if it is OK for the action to have a finite kernel, then just drop the hypothesis that $\Gamma$ has no finite index normal subgroup, and the same argument still works.


ADDED AFTER FURTHER COMMENTS: See Misha's comment for an improvement on this argument that avoids QI-rigidity altogether, using solely Mostow rigidity.


