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Let $M$ be $n$ dimensional compact connected smooth manifold without boundary whose universal cover is diffeomorphic to $\mathbb{R}^n$, must the Euler characteristic of $M$ vanish?

This is true in the simplest case, when $M= \mathbb R^n/\Lambda$ is a torus.

Michael Albanese
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Amr
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  • What would all of its homotopy groups be? – Randall Aug 09 '21 at 11:56
  • @Randfall I read somewhere that all of its higher homotopy groups would be zero . However it's fundamental group could be non trivial. I also don't know what will happen to it's betti numbers – Amr Aug 09 '21 at 11:58
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    Let $g\geq 1$. The Euler characteristic of the $g$-genus surface $\Sigma_g:=#_g \Bbb T$ is $2-2g$. Now, the universal cover of $\Sigma_g$ is diffeomorphic to $\Bbb R^n$. Here, $\Bbb T\cong \Bbb S^1\times \Bbb S^1$. – Sumanta Aug 09 '21 at 12:28
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    A different note: If your universal cover $p\colon \widetilde M\to M$ is $m$-fold for some positive integer $m$, then $\widetilde M$ is closed manifold and $\chi(\widetilde M)=m\cdot \chi(M)$ as there exists a CW-structure on $\widetilde M$ such that for each $k$-cell of $M$ we have $m$-many $k$-cell in $\widetilde M$. Now, if $\chi(\widetilde M)$ vanishes then so is $\chi(M)$, and conversely. – Sumanta Aug 09 '21 at 12:37
  • If $n=\dim M$ is odd then $\chi(M)=0$, irrespective of the universal cover. See here – Sumanta Aug 09 '21 at 12:50
  • @Sumanta Can you give me some reference for the result $\chi(\tilde{M})=m \cdot \chi(M)$? – Sebathon Aug 09 '21 at 12:55
  • Read Theorem $10.43.$ and Corollary $10.45.$ on pages $302$ and $305$ of Joseph J. Rotman's An Introduction to Algebraic Topology. Or see here – Sumanta Aug 09 '21 at 12:58

2 Answers2

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There is a difficult open problem along the lines of your question, known as the Hopf-Thurston Conjecture:

If $M$ is a closed aspherical manifold of dimension $2k$, then $(-1)^k\chi(M)\ge 0$.

Moishe Kohan
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This was already answered in the comments but I couldn't resist to give some further examples and my comment ended up being to long for a comment.

Firstly from the case of dimension $2$, one may think that the Euler characteristic of such manifolds is even, but this is not the case. In algebraic geometry there is an interesting class of surfaces called ball quotients. Quotients of a unit open ball in $\mathbb{C}^n$ acting freely and properly discontinuously.

There have been ball quotients of complex dimension two (hence real dimension $4$), which have the same Betti numbers as $\mathbb{CP}^2$ hence Euler characteristic $3$. These are usually called fake projective planes.

At this point, I would like to mention the closely related notion of an aspherical manifold, a manifold whose universal cover is contractible. But note that in dimension $n \geq 4$ this notion differs from being covered by $\mathbb{R}^n$, since there are contractible manifolds not homeomorphic to $\mathbb{R}^n$.

A consequence of the geometrization conjecture (proved by Perelman) says that a closed aspherical $3$-manifold has universal cover homeomorphic to $\mathbb{R}^3$

Luo constructed a closed $4$-manifold, whose universal cover is contractible, which has the same Betti numbers as $S^{4}$ and hence $\chi=2$ (note that taking products of $2$-manifolds can't give such a value). However in his constructions the universal cover is not homeomorphic to $\mathbb{R}^4$.

I think the last noteworthy thing to mention if $M$ is a quotient of a group acting on $\mathbb{R}^n$ freely and properly discontinuously by isometries of the Euclidean metric then the statement is true. This follows from the theorem of Bieberbach that every flat manifold is finitely covered by a torus of the same dimension.

Nick L
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  • +! Thanks for the detailed answer. Your last paragraph says "if M is a quotient of a group acting on ℝRn freely and properly discontinuously by isometries of the Euclidean metric then the statement is true". What if $M$ is the quotient of $\mathbb{R}^n$ by a group of affine isomorphisms of $\mathbb{R}^n$ acting freely and properly discontinuously ? – Amr Aug 09 '21 at 22:53
  • https://math.stackexchange.com/questions/4220890/euler-characteristic-of-a-quotient-of-mathbbrn-by-a-group-of-affine-transf – Amr Aug 09 '21 at 23:59
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    Your paragraph on the geometrization conjecture seems false: the Whitehead manifold is a contractible $3$-manifold which is not homeomorphic to $\mathbb{R}^3$. – Jason DeVito - on hiatus Aug 23 '21 at 15:44
  • You are right, i missed the assumption that the manifold is closed. For the statement for closed manifolds see section 3.2 of http://www.map.mpim-bonn.mpg.de/Aspherical_manifolds. – Nick L Aug 23 '21 at 18:10
  • Cool! I wasn't aware of that result. – Jason DeVito - on hiatus Aug 23 '21 at 20:13