cohomological dimension of a group acting on a product

Question

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.

Answer 1

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.

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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.

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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.

Answer 2

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.