Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $M=\mathbb{R}^n/G$ be a closed flat manifold, and let $F\to M \to N$ be a locally trivial submersion, where $F$ and $N$ are closed manifolds.

My question is simple: are $F$ and $N$ homeomorphic to flat manifolds?

This question seems quite natural to me, and I would expect the answer to this fact to be well known.

Any reference will be welcome. Thank you in advance.

share|improve this question
    
They are clearly homotopy equivalent to flat manifolds, so they are also homonorphic to flat manifolds by Farrel and Hsiang (or Farrel and Jones if you like). This probably also works smoothly but would require more work. –  Misha Jan 31 '13 at 19:23
    
Misha, why is the base $N$ aspherical, eg why does it have a trivial $\pi_2$? –  Igor Belegradek Jan 31 '13 at 19:33
    
Igor, you are right, it is unclear: I was thinking about fibrations with aspherical fiber/base. On the other hand, I do not know any examples of finite-dimensional bundles with contractible total space ond non-contractible fiber/base. Farrel had some old papers on this which I would have to look up. –  Misha Jan 31 '13 at 20:32
    
Misha, isn't it false is the smooth category? even if F is zero dimensional? –  Ben Wieland Feb 1 '13 at 4:20
add comment

2 Answers

So, the question was answered by Alexander Lytchak. I am writing down the answer for the sake of completeness.

Consider the fibration between the universal covers $F'\to\tilde{M}\to \tilde{N}$. $\tilde{M}$ is contractible and $\tilde{N}$ is simply connected, thus we can apply the Serre spectral sequence with integral coefficients, and from it we obtain that $F'$ and $\tilde{N}$ are contractible. In fact, if $H^*(F')$ has cohomological dimension $a$, and $H^*(\tilde{N})$ has cohomological dimension $b$, then $H^*(\tilde{M})$ would have cohomological dimension $a+b$ and this has to be $0$.

Then $N$ is acyclic, and from the exact sequence in homotopy so is $F$. Moreover we have and exact sequence

$$ 1\to\pi_1(F)\to \pi_1(M)\to \pi_1(N)\to 1$$

where $\pi_1(M)$ is a Bieberbach group, i.e. a torsion free group with a finite index normal abelian subgroup. As for $\pi_1(F)$, a subgroup of a Bieberbach group is again a Bieberbach group and therefore $F$ is homeomorphic to a flat manifold.

As for $N$, one can prove that there exists a normal abelian subgroup of $\pi_1(N)$ of finite index. Moreover, $\pi_1(N)$ is torsion free, since otherwise there would be a finite cyclic subgroup acting on the contractible manifold $\tilde{N}$ without fixed points. This is not possible, and therefore $\pi_1(N)$ is a Bieberbach group. Again, this implies that $N$ is a Bieberbach manifold.

share|improve this answer
    
Marco, do you understand Lytchak's proof? I do not, at several places. Say, why is $\tilde M$ contactible? I assume it is the pullback of $M$ under universal cover of $N$, eg if $M=N\times S^1$, then $\tilde M$ is not contractible. Of course, one can precompose with universal cover but then the fiver $F^\prime$ seems non-connected. I do not understand teh second paragraph either. –  Igor Belegradek Jan 31 '13 at 23:06
    
Igor: As you say, you precompose the pullback with the universal cover. By the exact sequence in homotopy the fiber F′ is connected. What is not working in the second paragraph? –  Marco Radeschi Feb 1 '13 at 8:58
    
Marco: I agree that $F^\prime$ is connected, but am confused on the spectral sequence argument. There exist fiber bundles with contractible total space and simply-connected non-acyclic base, eg the universal $G$-bundle $G\to EG\to BG$ of a compact connected Lie group $G$. The only difference with the example at hand is that $BG$ has infinite cohomological dimension. How would Lytchak's argument fail for this fiber bundle? –  Igor Belegradek Feb 1 '13 at 13:18
1  
The point is: if $H^a(\tilde{N})$ is the highest nonzero cohomology group of $\tilde{N}$, and $H^b(F')$ the highest cohomology group of $F'$, then in the second page of the spectral sequence the term $E_2^{a,b}$ is nonzero, and no nonzero differentials land on it, or leave it. For this fact to be true it is important that both cohomologies of $F'$ and $\tilde{N}$ have only a finite number of nonzero cohomology group, so this passage does not apply in the situations you described. It follows that $E^{a,b}_2$ survives to the infinity page, and in particular $H^{a+b}(\tilde{M})\neq 0$. –  Marco Radeschi Feb 1 '13 at 13:49
add comment

If the submersion is Riemannian, the answer is yes for $N$: Any Riemannian submersion with complete flat total space and compact base must have a flat base space. This was proved by Luis Guijarro and Peter Petersen in Annales Scientifiques de l’École Normale Supérieure Volume 30, Issue 5, 1997, Pages 595–603.

However, this uses more hypotheses than you have (and gets a stronger result than you wish for).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.