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I'm reading about the Flux homomorphism in Symplectic Topology and I'm trying to show that it is surjective.

I know that if $\psi_{t}$ is the flow of a symplectic vector field $X$, then Flux({$\psi_{t}$}) = [$i(X) \omega$] (here $\omega$ is the symplectic form, so it must be nondegenerate) I'm trying to use this to construct a right inverse, but I don't really know how to proceed.

Steppewolf
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Fix a class $[\alpha]\in H^{1}(M)$. Since $\omega$ is non-degenerate, there exists a unique vector field $X\in\mathfrak{X}(M)$ such that $$ \iota_{X}\omega=\alpha. $$ Since $\alpha$ is closed, the vector field $X$ is symplectic: $$ L_{X}\omega=d\iota_{X}\omega+\iota_{X}d\omega=d\alpha=0. $$ Let $\psi_{t}$ denote the flow of $X$, and denote by $\{\psi_t\}$ its homotopy class. Then $$ Flux\{\psi_{t}\}=[\iota_{X}\omega]=[\alpha]. $$

studiosus
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  • Thanks a lot! This is what I was missing! – Steppewolf Dec 22 '20 at 16:12
  • Don't you need to check that this is well defined ? If you take another representative of the cohomology class and take the different flows generated by the vector fields that these are going to be homotopic ? @studiosus – Someone Apr 03 '21 at 09:34