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It is common to see the following consequence of Cartan-Hadamard stated as a significant result relating curvature and topology:

Let $M$ be a compact Riemannian manifold with non-positive sectional curvature $K \leq 0$. Then the fundamental group of $M$ is infinite.

A common proof is using the Cartan-Hadamard theorem to guarantee that $M$ is a quotient of $\mathbb{R}^n$ and concluding by stating without proof that a finite quotient of $\mathbb{R}^n$ is never compact. I have seen a proof of this consequence using geometrical means (a compact group acting on $M$ by isometries has a fixed point), but I would really like to see a purely topological proof.

Matheus Andrade
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1 Answers1

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Thanks to Deane and Ivo for showing me the way! This answer is based on their comments.

Let $\widetilde{M}$ be the universal cover of $M$. We know that $M =\dfrac{ \widetilde{M}}{\pi_1(M)}$. If $\pi_1(M)$ was finite, then $\widetilde{M}$ would be compact as well (because anytime we have a finitely sheeted covering map $p: X \to Y$ with $Y$ compact, $X$ is compact too - for a proof of this, see this question or this one), but since $\widetilde{M}$ is diffeomorphic to $\mathbb{R}^n$, it is noncompact, so $\pi_1(M)$ cannot be finite.

Alternatively, any covering map $p: \widetilde{M} \to M$ is a fiber bundle over $M$ (with fibers $\{p^{-1}(q) \}_{q \in M}$) with discrete fibers (as seen here or here). If $\pi_1(M)$ is finite, then these fibers are not only discrete but finite (and all of them have the same cardinality, namely $|\pi_1(M) |$) and therefore compact. Now, since the total space (in this case, $\widetilde{M}$) of a fiber bundle with a compact base (in this case, $M$) and compact fibers (in this case, $\{ p^{-1}(q)\}_{q \in M}$) is compact, the finiteness of $\pi_1(M)$ would again imply the compactness of $\widetilde{M}$ (this is a consequence of a more general fact - the total space of a fiber bundle is compact if and only if both its base and fibers are compact, as seen here and here), which we know cannot happen because $\widetilde{M}$ is diffeomorphic to $\mathbb{R}^n$. Hence $\pi_1(M)$ cannot be finite.

One can note that these arguments generalize a bit further: any compact manifold with noncompact universal cover has an infinite fundamental group (so, for example, $\mathbb{S}^2 \times \mathbb{S}^1$ also has an infinite fundamental group, even though it satisfies $K \geq 0$ - although explicitly computing its fundamental group is trivial even without geometry).

UPDATE: For the more general case where the quotient is not a covering space, one can take a look at this blog post.

Matheus Andrade
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  • This solves a certain case of the question you posed, but what if the quotient $\Bbb{R}^n \to \Bbb{R}^n/G$ is not a covering map? For instance, what about when you quotient $\Bbb{R}^2$ by $\Bbb{Z}/k\Bbb{Z}$ acting as roots of unity in $S^1$? You can probably upgrade your proof to this case by proving that points with trivial stabilizer are open and then running your argument or something like that... – Alekos Robotis Nov 02 '22 at 18:53
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    @AlekosRobotis you are correct, thanks for pointing it out. I failed to appropriately title my question, the only case I had in mind were the ones where the quotient is a covering space. I'll change the title. I think this blog post appropriately answer the general question, though, and so I'll link it in my answer as well. – Matheus Andrade Nov 02 '22 at 18:58