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Let $M$ be a compact manifold, let $\tilde{M}$ be its universal cover, and suppose that the Euler characteristic $\chi(\tilde{M})=0$. My question is: does this imply that $\chi(M)=0$? This is clear if $\pi_1(M)$ is finite, but I am interested in the case $|\pi_1(M)|=\infty$.

It might not feel right, but I can't think of any counterexample, either.

Thank you very much in advance!


EDIT: I was rightfully asked what I mean by Euler characteristic of the (non compact) manifold $\tilde{M}$. My answer right now is: the one you want!

What I am thinking of, is $\chi(\tilde{M})=\sum_i (-1)^i\dim H_i(\tilde{M},k)$, with $k=\mathbb{Q}$ or $\mathbb{R}$, and $H_i$ are either the usual or the compactly supported cohomology groups.

In my case, $\tilde{M}$ retracts to a compact Lie group.

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HW, I am not sure what you are asking. In your example $\chi(\tilde{M})$ is not 0... if $M_1\to M_2$ is a $n$-sheeted covering then $\chi(M_1)=n\cdot\chi(M_2)$, so $\chi(M_1)=0$ iff $\chi(M_2)=0$. Probably I didn't understand your comment though. –  Marco Radeschi Oct 9 '12 at 12:38
    
Sorry, I just realised I miscalculated the Euler characteristic! My mistake. I'm going to delete the comment, since it was silly. –  HJRW Oct 9 '12 at 12:40
    
And, as you say, it is obvious for finite-sheeted covers. –  HJRW Oct 9 '12 at 12:40
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If $\pi_1(M)$ is infinite, then $\widetilde{M}$ is not compact: in what sense of Euler characteristic do you mean in this case? –  Oscar Randal-Williams Oct 9 '12 at 12:53
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If $\tilde{M}$ is a Lie group and the fundamental group $\pi$ a subgroup, then there is a nonvanishing $\pi$-equivariant vector field on $\tilde{M}$ that descends to a nonvanishing vector field on $M$, whence $\chi(M)=0$. What is the case you are interested in? –  Johannes Ebert Oct 14 '12 at 15:46

2 Answers 2

Take a hyperbolic, compact, connected, oriented $3$-manifold $M$. Then The universal cover $\tilde{M}$ is contractible and thus $\chi(\tilde{M})=\dim H_0(\tilde{M})=1$.

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Hi Liviu, thanks for the answer. The thing is, my question is the other way round. I am assuming that $\tilde{M}$ has zero Euler characteristic, and I ask wether the same is true for $M$. –  Marco Radeschi Oct 9 '12 at 13:40
    
Ooops! Missed the $\tilde{}$. –  Liviu Nicolaescu Oct 9 '12 at 14:29

Here is sketch proof that $\chi(\tilde{M})=0$ implies that $\chi(M)=0$ for fundamental groups with certain finiteness properties.

The idea is to adapt the usual spectral sequence proof that the Euler characteristic is multiplicative on fibrations. Let $k$ be a field of characteristic zero, and let $\pi=\pi_1(M)$. The Cartan-Leray spectral sequence of the regular cover $\tilde M\to M$ has $E^2 = H_p(\pi;H_q(\tilde M;k))$ and converges to $H_{p+q}(M;k)$. But we can also go one page back and start at $E^1 =F_p\otimes_{k[\pi]} H_q(\tilde M;k)$, where $F_\bullet$ is a free resolution of $k$ by $k[\pi]$-modules. (Reference: K. Brown, Cohomology of groups, VII.5 and VII.7.) Now in a spectral sequence the Euler characteristic of each page is the Euler characteristic of the previous one, so it suffices to check that $\chi(E^1)=0$ when $\chi(\tilde M)=0$.

If the trivial module $k$ admits a finite free resolution by $k[\pi]$-modules, we can choose each $F_p$ to be a finitely generated free $k[\pi]$ module, so $F_p$ is a direct sum $k[\pi]^{n(p)}$ for some $n(p)$. I think we then get $F_p\otimes_{k[\pi]} H_q(\tilde M;k)\cong k^{n(p)}\otimes_k H_q(\tilde M;k)$ as vector spaces, and the result follows in the usual way since the Euler characteristic of a tensor product of finite dimensional graded vector spaces is the product of their Euler characteristics.

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@Mark: Isn't your argument a variant/generalization of the following: suppose that $\tilde M$ (although not necessarily compact!) has the homotopy type of a finitely dominated space. Consider the fibration $\tilde M \to M \to B\pi$. Then if $B\pi$ is finitely dominated, the Euler characteristic of $M$ is a product of the Euler characteristic of $\tilde M$ with the Euler characteristic of $B\pi$ (because the fibration is orientable). –  John Klein Oct 14 '12 at 16:01
    
@John: Yes, I think that's right. Although I don't really see why your fibration is orientable? (In particular, couldn't $\pi$ act non-trivially on the homology of $\tilde M$?) Meanwhile, it seems orientability is not really necessary for this argument; see mathoverflow.net/questions/80326/… –  Mark Grant Oct 14 '12 at 16:18
    
@Mark: your're right, the fibration doesn't need to be orientable, and in fact is might not be. –  John Klein Oct 14 '12 at 17:54
    
What does this condition mean in terms of ordinary group theory or topology? –  Will Sawin Oct 14 '12 at 22:22
    
@Will: In group theoretical notion, the condition says that $\pi$ is of type $FP$. –  Ralph Oct 14 '12 at 22:42

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