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.


1 Answer 1


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 $[\omega_{u_0}]=[\omega_{u_1}]$.

  • $\begingroup$ 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? $\endgroup$
    – W NIU
    Commented Jul 1, 2021 at 12:15
  • $\begingroup$ A surface here is a 2-dimensional manifold. $\endgroup$ Commented Jul 1, 2021 at 12:16

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