# Fibration with total space a symplectic manifold and base space connected such that fibres are symplectic submanifolds

In Dusa McDuff and Dietmar Salamon's book Introduction to Symplectic Topology, lemma 6.1.3(page 253) states that:

Let $$π\colon M\to B$$ be a locally trivial fibration with connected base and $$\omega \in \Omega^2(M)$$ be a symplectic form such that the fibres are all symplectic submanifolds of M. Then $$π\colon M\to B$$ admits the structure of a symplectic fibration which is compatible with $$\omega$$.

Here we assume that $$M$$ is closed and connected, and the fibre $$(F,\sigma)$$ is itself a symplectic manifold. By compatible with $$\omega$$ we mean the restriction of $$\omega$$ to each fibre agrees with the pullback form of $$\sigma$$ along local trivializations. We use $$l_b$$ to denote the inclusion of the fibre $$F_b$$ at $$b\in B$$ into $$M$$.

Now the proof starts from an exercise:

First use Stokes’ theorem to prove that the symplectic forms $$\sigma_b=l_b^*(\sigma)$$ all represent the same cohomology class in $$H^2(F)$$ under the local trivializations $${\phi_{\alpha}}(b)\colon F_b\to F$$.

My question is, how could we do this? I'm a beginner in studying symplectic topology. I feel (vaguely) this should be true because for any two points in $$B$$ we can connect them by a sequence of local charts. But how does Stoke's theorem come into play? Thanks.

Let $$(U_i,f_i)$$ be a trivialization of the fibration, where $$f_i:U_i\times F\rightarrow M$$. Consider $$(u_0,b_0); (u_1,b_1)\in U_i\times F, u_0,u_1\in U_i, b_0,b_1\in F$$. We suppose that there exists a path $$p_t:[0,1]\rightarrow U_i\times F$$ such that $$p_0=(u_0,b_0)$$ and $$p_1=(u_1,b_1)$$. This defines a map $$P:[0,1]\times F\rightarrow M$$ by $$P(t,f)=f_i(p_t,f)$$
Let $$[c]\in H_2(F)$$, it can be represented by a map $$c:S\rightarrow F$$ where $$F$$ is a surface. Consider $$C:[0,1]\times S\rightarrow [0,1]\times F$$ defined by $$C(t,s)=(t,c(s))$$, the $$2$$-form $$\omega_S=C^*P^*\omega$$ is closed, we deduce that $$d\omega_S=0$$, this implies that $$\int_{[0,1]\times S}d\omega_S=\int_{0\times S}i_{u_0}^*\omega-\int_{S_1}i_{u_1}\omega=0$$. This implies that $$[\omega_{u_0}]=[\omega_{u_1}]$$.
• Thanks! Could you elaborate a bit? I think you mean $S$ is a surface? Also, in the last implication, we use the Poincare duality, right? Jul 1 '21 at 12:15