Suppose $M$ is a (closed, connected, oriented, smooth) manifold.

If $M$ is aspherical, i.e., if the inversal covering $\tilde{M}$ is contractible, $M$ is a $B\pi_1(M)$. This is often enforced by geometry, for instance it holds if $M$ admits a metric of non-positive sectional curvature (Cartan--Hadamard).

We deduce that the inequality $$\text{vcd}(\pi_1(M)) \leq \text{dim}(M)$$ holds (it is actually an equality in this case).

I learned recently that for a simply-connected Lie group $G$ and a lattice $\Gamma < G$, we can pick a maximal compact $K < G$, then $K \backslash G$ is contractible, hence $K \backslash G / \Gamma \sim_{\mathbf Q} B\Gamma$. Thus for $M = G/\Gamma$, the above inequality is also satisfied.

Furthermore, and going in the same direction, Mostow proved that fundamental groups of arbitrary homogeneous spaces, if they are solvable, have rank at most the dimension of the space. The above inequality thus also holds in this case.

However, it is (of course) easy to construct manifolds for which the above inequality fails, as any finitely presented group is the fundamental group of some $d$-manifold for all $d \geq 4$.

Is there a common generalization of being aspherical or a homogeneous space of the type described that ensures that the above inequality is satisfied? What are other conditions on $M$ that ensure that it holds?

  • 1
    $\begingroup$ There seems to be something missing in your statement attributed to Mostow. Aspherical is not enough. Indeed the quotient of the universal covering of $SL_2$ by a lattice is homogeneous, aspherical, and its $\pi_1$ is not solvable. $\endgroup$
    – YCor
    Mar 9, 2021 at 13:26
  • $\begingroup$ Mostov's result assumes solvability of the fundamental group. $\endgroup$ Mar 9, 2021 at 13:44
  • $\begingroup$ You are of course totally right, I really misunderstood what Mostow proved. I have edited the question, I hope now it makes more sense. $\endgroup$ Mar 9, 2021 at 14:58
  • 1
    $\begingroup$ This inequality holds for all 3-manifold groups. $\endgroup$ Mar 9, 2021 at 18:08

1 Answer 1


Let $M$ be an orientable manifold. The relation between the cohomological dimension of $\pi_1(M)$ and that of $M$ comes from the following:

Proposition: $M$ has the homotopy type of a fibration over $B\pi_1(M)$ with fiber $\tilde M$.

Proof: Let $B\pi_1(M) = E\pi_1(M)/\pi_1(M) $ be a classifying space for $\Gamma$ and consider the quotient $X = (E\pi_1(M)\times \tilde M )/\pi_1(M)$ where $\pi_1(M)$ acts diagonally on the product. One can view $X$ in two different ways:

  1. As a fibration over $M$ with fibers $E\pi_1(M)$. Since $E\pi_1(M)$ is contractible, this is a homotopy equivalence and we deduce that $H^\bullet(M) = H^\bullet(X)$.

  2. As a fibration over $B\pi_1(M)$ with fibers $\tilde{M}$. QED

Now, the Leray--Serre spectral sequence computes (in theory) the cohomology of $M$ in terms of the cohomology of $\Gamma$ and $\tilde M$.

Let us make the following assumptions:

(i) $\Gamma$ has finite virtual cohomological dimension $d_1$

(ii) if $d_2$ is the cohomological dimension of $\tilde M$ then the cohomology group with twisted coefficients $$H^{d_1}(\pi_1(M), H^{d_2}(\tilde M))$$ is non-zero. (Here, $H^{d_2}(\tilde M)$ is seen as a $\pi_1(M)$-module).

Then the non-zero terms of the second page of the Leray--Serre spectral sequence are all contained in the rectangle $[0,d_1] \times [0,d_2]$ and the top right corner term $E_k^{d_1,d_2}$ stabilizes at page $2$. One deduces that $X$ has cohomological dimension $d_1+d_2$ and $$H^{d_1+d_2}(X) = H^{d_1}(\pi_1(M), H^{d_2}(\tilde M)).$$

On the other side, $X$ is homotopy equivalent to $M$ and thus has cohomological dimension at most $\dim(M)$, with equality if and only if $M$ is closed. We conclude that $$d_1\leq \dim(M) - d_2~,$$ with equality if and only if $M$ is closed.

I might be missing some condition to guaranty the convergence of the Leray--Serre spectral sequence, but I know that the argument works in the following setting: let $G/H$ be a homogeneous space an $\Gamma$ a discrete and torsion-free subgroup of $G$ acting properly discontinuously on $G/H$. Then $G$ acts trivially on $H^\bullet(G/H,\mathbb Z)$ by continuity, hence so does $\Gamma$. Applying the argument above to $M= \Gamma \backslash G/H$, one gets that $$\mathrm{coh.dim}(\Gamma)\leq \dim(G/H) - \mathrm{coh.dim}(G/H)~,$$ with equality if and only if the action of $\Gamma$ is cocompact.

The cohomological dimension of $G/H$ is easily computed because $G/H$ retracts to $K_G/K_H$ where $K_G$ and $K_H$ are maximal compact subgroups of $G$ and $H$ respectively.


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