Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I recently came across an interesting result of Kobayashi [Corollary 5.5], a special case of which is the following: Suppose $\Gamma$ is a discrete torsion free subgroup of $SL_n(\mathbb{R})$ which acts properly discontinuously on the homogeneous space $X=SL_n(\mathbb{R})/SL_{n-1}(\mathbb{R})$. Then the cohomological dimension of $\Gamma$ is less than or equal to $n$.

The homogeneous space $X$ above is diffeomorphic to a fiber bundle with base space $S^{n-1}$ and fibers $\mathbb{R}^n$. This motivates my question:

Suppose $G$ is a discrete torsion free group acting properly discontinuously on $M \times \mathbb{R}^n$ where $M$ is a compact manifold. What can be said about the cohomological dimension of $G$? Is it less than or equal to $n$?

Kobayashi's proof uses spectral sequences, a tool which I am not familiar with. So before spending time learning about these objects, I was wondering if there is an obvious obstruction to the generalization of Kobayashi's result mentioned above.

share|improve this question
1  
Consider the case when $n=0$. –  Tom Goodwillie Jul 15 '12 at 23:02
    
Even with $n=1$, for $cd(G)=1$ iff $G$ is free. You shouldn't expect anything unless your space is a $K(G,1)$-complex, for then $cd(G)\le$ geometric dimension of $G$ (minimal dimension of such a complex). –  Chris Gerig Jul 16 '12 at 0:04
    
Thanks for your comments. I have modified the question to give some background. –  Andrew Zimmer Jul 16 '12 at 0:38
1  
@Chris: have you got a nontrivial (= with no finite index cyclic subgroup) example of a discrete group acting properly on $M\times\mathbf{R}$ with $M$ compact manifold? @Tom: if I don't miss anything, for $n=0$ it's clear that the conclusion is that $G$ is finite (properness means that for any compact $K$, the set of $g$ such that $gK\cap K$ is nonempty is finite; then take $K=M$) i.e. trivial if $G$ is assumed torsion-free. –  Yves Cornulier Jul 16 '12 at 9:15
    
Oh, sorry, I think I overlooked the words "torsion free". –  Tom Goodwillie Jul 16 '12 at 17:00
show 1 more comment

2 Answers

up vote 6 down vote accepted

I think Kobayashi's approach also works in the OP's situation:

If $M$ is an oriented compact connected manifold then $cd_\mathbb{R}(G) \le n$.

Proof: As in Kobayashi the base ring is the field of real numbers. Let $X = M \times \mathbb{R}^n$ and let $A$ be an $\mathbb{R}G$-module. Since $G$ is torsion-free and acts properly discontinuously, the action is actually free. Thus by [Cartan-Eilenberg: Homological Algebra, XVI §9] there is a spectral sequence $$E_2^{i,j}=H^i(G;H^j(X;A)) \Rightarrow H^{i+j}(X/G;A)$$ (see also Kobayashi, p. 14). Before diving into technical details let's sketch the basic idea. Suppose $\dim M=m$ and $cd_\mathbb{R}(G) = d$. Since $X \simeq M$, $H^j(X;A)=0$ if $j>m$. Futhermore $H^i(G;-)=0$ if $i > d$. Hence the $E_2$-term looks like $$\begin{array}{lcccr} - & - & - & - & \bullet \newline | & & & & | \newline | & & & & | \newline - & - & - & - & - \newline \end{array}$$ Outside the rectangle all entries are zero and the bullet has coordinate $(d,m)$. For positional reasons $E_2^{d,m}=E_\infty^{d,m}$.

Now suppose $E_2^{d,m} \neq 0$. Hence the abutment $H^{d+m}(X/G;A) \neq 0$. But $X/G$ is a manifold of dimension $m+n$. Hence $H^i(X/G;A)=0$ if $i> m+n$ and consequently $d+m \le n+m$, i.e. $d \le n$ as to be shown.


The details: By $X \simeq M$ and Poincare duality we have isomorphisms $$H^m(X)\cong H^m(M) \cong H_0(M) \cong \mathbb{R}$$ as $\mathbb R$-vector spaces. Under this isomorphism the action of $g \in G$ on $H^m(X)$ corresponds to scalar multiplication on $\mathbb R$ by an $\rho(g) \in \mathbb R^\times$ satisfying $\rho(g)\rho(h)=\rho(gh)$ for $g,h \in G$. Moreover, by universal coefficients $$H^m(X;A) \cong H^m(X) \otimes_\mathbb{R} A \cong A \hspace{110pt}(1)$$ as vector spaces and hence $E_2^{i,m}=H^i(G;A)$.

Assume for the moment we know that the action $\odot$ of $G$ on $A$ in $(1)$ is given by $$g \odot a = \rho(g)(g \cdot a).\hspace{150pt}(2)$$ Let $(B,\ast)$ be an $\mathbb R$-module such that $H^d(G;B) \neq 0$. Define a $G$-action $\cdot$ on $A := B$ by $$g \cdot a := \rho(g^{-1})(g\ast a)$$ (this is indeed an action because multiplication in $\mathbb R$ is commutative). Hence $\odot = \ast$ and $E_2^{d,m}=H^d(G;B) \neq 0$ as desired.


It remains to show $(2)$. This is a straightforward, but tedious calculation using the definition of the various $G$-actions. So I think it's best -- to leave it to the reader ? No, but I think it's best to write it down here only if one is really interested in the demonstation.

share|improve this answer
1  
@Chris, Ralph. I think there's a single definition of "properly discontinuous action" (which is a weird way of saying that a discrete group acts properly): for every compact $K$ in the space, the set of $g$ with $gK\cap K\neq\emptyset$ is finite. And indeed for a torsion-free group this forces the action be free. –  Yves Cornulier Jul 17 '12 at 3:00
    
@Ralph: Of course, good clarification. –  Chris Gerig Jul 17 '12 at 3:33
add comment

As Tom pointed out, the case that $n=0$ is trivial.

For $n=1$, if the group acts properly discontinously and cocompactly, then the group will have two ends, and therefore is isomorphic to $\mathbb{Z}$ by a result of Wall (Lemma 4.1)

One may see that any such properly discontinuous non-trivial torsion-free action must be cocompact. Suppose there is $g\in G$ such that $g(M\times\{0\}) \cap (M\times \{0\})=\emptyset$. Then by homology there is a compact submanifold $K\subset M\times \mathbb{R}$ such that $\partial K= (M\times\{0\}) \cup g(M\times\{0\})$. Then one sees that $K$ forms a fundamental domain for the action of $\langle g\rangle$ on $M\times \mathbb{R}$ by properness, and thus the action of $G$ is cocompact.

Otherwise, one has $g(M\times\{0\})\cap (M\times\{0\})\neq \emptyset$ for all $g\in G$. This violates proper discontinuity, since the preimage of the compact set $(M\times\{0\})\times (M\times\{0\})$ under the map $G\times (M\times\mathbb{R})\to (M\times\mathbb{R})\times (M\times\mathbb{R})$ is not compact.

Another case one can deal with partially is for $M=S^1, n=2$. Then if $G$ acts cocompactly, then $G$ is as surface group by Theorem 1.2 of Maillot (although this theorem is attributed to Geoff Mess in an earlier unpublished preprint). This also follows now from the geometrization theorem now, but note that Maillot also partially treats the non-cocompact action case with some extra hypotheses.

share|improve this answer
    
I think you are right. Thanks for the hint. I would be glad if you could also have a look on what I hope is a complete answer in the oriented case. –  Ralph Jul 16 '12 at 22:43
1  
For $n=1$ this even gives a conclusion without assuming $G$ torsion-free, namely that $G$ has an infinite cyclic subgroup of finite index. –  Yves Cornulier Jul 17 '12 at 3:19
    
You usually don't have to assume your group is torsion-free, because by a lemma by Selberg any discrete subgroup of a matrix group has a torsion-free subgroup of finite index. In this case, however, the usual cohomological dimension is infinite and one sidesteps this by using the virtual cohomological dimension $\mathrm{vcd}(G)$, which is by definition the $\mathrm{cd}$ of any finite-index subgroup. A lemma by Serre ensures that these subgroups all have equal cohomological dimension. –  Earthliŋ Jul 17 '12 at 3:49
    
@ Yves & s.barmeier: yes, I knew the torsion case would be virtually cyclic, but the question was for torsion-free groups, so I stuck to that case. Same for Maillot's theorem. Assuming no torsion is natural to avoid a finite group acting on M. But there may be interesting groups with torsion acting on $M\times R^n$ which have no finite-index torsion-free subgroups. In fact, it would be interesting to find such a group which doesn't act discretely on $R^n$. –  Ian Agol Jul 17 '12 at 4:35
1  
@KBuck: it seems that some discrete subgroups of $SL_2(\mathbf{R})$ can contain elements of all prime order and thus are not virtually torsion-free although these groups have a (faithful) proper action on the plane. –  Yves Cornulier Jul 17 '12 at 6:43
show 5 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.