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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.

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Do you want $\tilde\Gamma$ act effectively? Say, if $\tilde\Gamma$ is the product of $\Gamma$ and a finite group, then $\tilde\Gamma$ acts properly via the projection onto $\Gamma$. On the other hand, if you want the action to be effective, there is an easy counterexample: take $X=\mathbb R$, $\Gamma=\mathbb Z$, and $\tilde\Gamma=\mathbb Z\times\mathbb Z_2$. The factors commute and $\mathbb Z_2$ has a fixed point from which one can show that $\mathbb Z_2$ must act trivially. –  Igor Belegradek Mar 2 '13 at 0:14
    
@Igor: what's the point in adding extra unnatural requirements just to give trivial counterexamples? –  YCor Mar 2 '13 at 4:21
    
Igor does have a point. One instance of the original version of OP's question, before the edit, is whether every finite group acts on $S^2$. The answer is yes, trivially. But the question becomes interesting if you restrict to faithful actions. –  Lee Mosher Mar 2 '13 at 13:31

2 Answers 2

up vote 8 down vote accepted

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.

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@Lee: your argument shows that the action does not extend to an action of $\pi_1(S_2)$ by isometries (i.e. thru the ambient Lie group). But it certainly extends to an action by homeomorphisms. I understand the question as whether the action extends to an action by homeomorphisms. –  YCor Mar 2 '13 at 4:28
    
I understood the OP as Lee (nice answer, BTW!), and Yves'comment made me realize that the question is ambiguous... –  Alain Valette Mar 2 '13 at 11:47
    
Sorry for the ambiguity. I have edited the question to make it more clear. –  Dieter Mar 2 '13 at 12:05
    
@unknown: So, as far as I understand, the question is still open (at the MathOF level). It's a bit weird to tick an answer as correct (esp. after less than 1 day) because if somebody finally answers your question, what can you do? do you still expect people to think about it? (of course this answer is interesting and provides relevant info!) –  YCor Mar 2 '13 at 14:34
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@Lee: In fact, you do not have to invoke qi rigidity in the lattice case: Just note that you can always assume that $\Gamma$ is normal in $\tilde\Gamma$ (by replacing $\Gamma$ with its finite index subgroup). Now, it follows from Mostow rigidity that (apart from the $PSL(2,R)$-case) $Aut(\Gamma)$ embeds isometrically in the Lie group $G$. Therefore, modulo the (irrelevant) finite kernel of the homomorphism $\tilde\Gamma\to Aut(\Gamma)$, you have the required embedding. –  Misha Mar 2 '13 at 22:55

Take $\tilde{\Gamma}$ to be a higher genus Baumslag-Solitar group, see section 9 of M. Kapovich, B. Kleiner, Coarse Alexander duality and duality groups, Journal of Differential Geometry, Vol. 69, (2005) Number 2, p. 279-352. Group $\tilde\Gamma$ cannot act properly discontinuously on any contractible 3-manifold, but it contains a finite index subgroup $\Gamma$ which is is Gromov-hyperbolic and isomorphic to the fundamental group of a compact 3-manifold with boundary. Hence, by a corollary of Thurston's geometrization theorem, $\Gamma$ acts isometrically and discretely on hyperbolic 3-space, i.e., is a discrete subgroup of $G=PO(3,1)$.

Edit: Here is an easier example. Note first that every finite group admitting an effective topological action on $R^2$ is either cyclic or dihedral. (B. von Kérékjártó, Über die endlichen topologischen Gruppen der Kugelfläche, Nederl. Akad. Wetensch. Proc. Ser. A vol. 22 (1919) pp. 568-569: There is probably a more accessible reference for this result, but I do not have one.) Now, take, say, a simple noncyclic finite group $F$ and set $\tilde\Gamma= F * F$. Then $\tilde\Gamma$ is virtually free, so it contains a free finite index subgroup $\Gamma$, which embeds in $PSL(2, R)$. On the other hand, every topological $\tilde\Gamma$-action on the plane has to be trivial since each free factor has to act trivially.

A side note: If you are willing to replace the symmetric space $X$ with the $n$-fold product $X^n=X\times ... \times X$, then every isometric action of $\Gamma$ extends to an isometric action of $\tilde\Gamma$ on $X^n$ (where $|\tilde\Gamma:\Gamma|=n$): This is the induced representation construction. Incidentally, the following is an open problem:

Let $\Gamma$ be a discrete isometry group of hyperbolic $n$-space $H^n$ and $\Gamma \subset \tilde\Gamma$ is a finite-index extension. Is it true that $\tilde\Gamma$ admits a properly discontinuous isometric action on $H^m$ for some $m$?

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Misha, where do you prove that "$\tilde\Gamma$ cannot act properly discontinuously on any contractible 3-manifold"? I only found a proof that $\tilde\Gamma$ cannot act on a coarse PD(3) space (such as the universal cover of a compact aspherical 3-manifold). Of course, for your application you just need to show that $\tilde\Gamma$ cannot act properly discontinuously on the hyperbolic $3$-space, so for the example at hand everything works. –  Igor Belegradek Mar 2 '13 at 20:18
    
In fact, the reference points to "Geometry of quasi-planes". Could you give a page reference? –  Igor Belegradek Mar 2 '13 at 20:36
    
Igor, pages in pdf file for some reason do not show up, but it is p. 37-38, after Lemma 3.9. The link is correct. If you prefer archive version arxiv.org/pdf/math/9911003.pdf, it is on pages 41-43. Concerning your 1st question: The result is indeed stated for coarse PD(3) spaces, which does not include contractible 3-manifolds which are not uniformly acyclic. However, if you go through the proof(s), you realize that the proof takes place in a certain metric neighborhood of the group orbit, so the uniform acyclicity assumption can be weakened. –  Misha Mar 2 '13 at 21:13
    
Oh, the reference is to "Hyperbolic groups with low-dimensional boundary" not to "Geometry of quasi-planes"; sorry I got confused –  Igor Belegradek Mar 2 '13 at 21:25
    
Thanks, this is a very nice answer. Together with Lee Mosher's answer it provides a full answer to my question, i.e. both positive and negative examples. I would like to accept them both, but since this is impossible I will accept Lee Mosher's answer because it was first. –  Dieter Mar 4 '13 at 8:41

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