M fibers over the circle then construct a symplectic form I'm trying to prove that if a 3-manifold $M$ ﬁbers over the circle, then $M\times S^1$ admits a symplectic structure. I know that it is an standard result. 
Probably it is very easy, but I can't see it. First at all what means "fibers over..."?
 I don't know if it is in the homotopic sense. 
I appreciate any idea or hint.
 A: Let $S$ be a surface, and $\varphi: S \to S$ a homeomorphism. The mapping
torus of $S*_{\varphi}$ is the manifold $S*_{\varphi} := S \times [0, 1]/(s, 1)\sim  (\varphi(s),0)$. A 3-manifold is said to fiber
over the circle, to admit the structure of a fibration, or to be a surface bundle over the circle if it is homeomorphic to a manifold of the form $S*_{\varphi}$. The surfaces $S\times  t$ are called fibers.
It is known that if a closed, oriented 3-manifold, $M$, fibers over the circle, then
$S^1\times M$ admits a symplectic form. This was first observed by W. P. Thurston (see
Some Simple Examples of Symplectic Manifolds). 
One way to see this is as follows. Suppose $f : M \to S^1$ is a smooth map with
no critical values. Then, one can find a Riemannian metric on $M$ so that $df := f^*\rm{dt}$ is a harmonic 1-form. Here, dt denotes the the volume form on $S^1$ with respect to the standard Riemannian metric. As a result, the 2-form $\omega_f := {\rm dt} \wedge df + *df$ is a symplectic form on $S^1\times  M$, which is self-dual and harmonic with respect to the product Riemanian metric. 
The converse is a conjecture (often attributed to Taubes in the literature).
