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Let $M$ be a closed connected $n$-manifold. If there exists a submersion $p:M\to S^1$, then $p$ is both proper and onto (since $M$ is compact and $S^1$ is connected). Therefore, by Ehresmann's theorem, $p$ is a fiber bundle and $M$ admits a codimension one foliation. In particular, the "easy" Thurston's theorem (the easy implication I mean) implies that $M$ has Euler characteristic zero (see here for example).

I am interested in the converse of this proposition:

If $M$ be a closed connected $n$-manifold with Euler characteristic $0$, is there always a submersion $$M\longrightarrow S^1~?$$

If $M$ has Euler characteristic $0$ the hard implication of Thurston theorem says that there is a codimension 1 foliation on $M$. Of course this foliation doesn't have to be induced by a submersion to the circle (the leaves could be non compact), but maybe there is another foliation on $M$ coming from a submersion to the circle.

There should be some obvious counter examples but I can't find one. If there are indeed counter examples to this statement, I'd be happy to know if they are counter examples where $M$ admits a flat metric, that is

Is there a closed connected $n$-manifold $M$ which admits a flat metric (and hence $\chi(M)=0$ by Chern-Gauss-Bonnet theorem) such that there is no submersion $M\to S^1$?

Remark: In dimension $1$ and $2$, the only closed connected manifolds with Euler characteristic $0$ are the circle, the Klein bottle and the torus, and they all admit a flat metric and a submersion to the circle.

Thanks in advance.

Adam Chalumeau
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2 Answers2

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There are many counterexamples to your first question.

Suppose $M$ is a closed manifold which admits a subersion $f:M\rightarrow S^1$. Then $f_\ast:\pi_1(M)\rightarrow \mathbb{Z}$ must be non-trivial.

So, for example, no odd dimensional sphere has a submersion to $S^1$; more generally, no closed manifold with finite fundamental group has such a submersion.

The proof is not too bad: if $f_\ast$ is the trivial map, then covering space theory implies $f$ lifts to a map $\tilde{f}:M\rightarrow \mathbb{R}$. Since being a submersion is a local property, $\tilde{f}$ is a submersion. But then $\tilde{f}$ is an open map. This means $\tilde{f}(M)$ is open in $\mathbb{R}$, but it is also closed in $\mathbb{R}$ because $M$ is compact. But that means $\tilde{f}(M) = \mathbb{R}$, so $\mathbb{R}$ is compact, a contradiction.

I don't know the answer to the second question, but it seems harder. If $M$ admits a flat metric, it is finitely covered by a torus, and of course, a torus admits such a map.

Jason DeVito - on hiatus
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There exists a flat riemannian manifold $M$ with no submersion $M\rightarrow S^1$. An example of such a manifold is the Hantzsche Wendt 3 dimensional manifold. You may read the book of Charlap to have have a full description of this manifold.

The first homology of this manifold is finite, and a surjection $M\rightarrow S^1$ induces a surjective map $\pi_1(M)\rightarrow\mathbb{Z}$ (Serre exact sequence associated to a fibration induces the exact sequence $\pi_1(M)\rightarrow\pi_1(S^1)=\mathbb{Z}\rightarrow\pi_0(F)\rightarrow 1$ where $F$ is the fibre of $f$. Since $F$ has a finite number of connected component, we deduce that the image of $\pi_1(M)\rightarrow\pi_1(S^1)=\mathbb{Z}$ is not trivial. This implies that $H_1(M,\mathbb{Z})$ the abelianization of $\pi_1(M)$ has a rank greater than $1$. This is not true for the HW manifold.

Charlap. Bieberbach groups and flat manifold

Tsemo Aristide
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  • Nice. I was searching for a list of flat $3-$manifolds to try to do something like this, but I failed. Incidentally, I don't understand your argument that the map on $\pi_1$ surjective. I understand the long exact sequence, but why is the fiber connected? – Jason DeVito - on hiatus Aug 02 '19 at 06:25