TITLE: Codimension one foliations induced by a map to the circle.
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: 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