The Moser theorem says If $M$ is a compact manifold with $\omega_0, \omega_1$ are two isotopic symplectic forms, then $\omega_0,\omega_1$ are strong isotopic.

I don't understand why this theorem tells us that tangent space (up to strong isotopy) is $H^2(X,\Bbb{R})$

I know if they are isotopic then $[\omega_t]$ is constant in $H^2(X,\Bbb{R})$, and the strong isotopic will not change the cohomology class, that is :

$$(\{\text{space of symplectic form}\}/\sim )\to H^{2}(X,\Bbb{R})\\ \omega \mapsto [\omega]$$ is well defined where $\sim$ is the equivalent class that are strong isotopic.

Why does this theorem tell us that the tangent space (up to strong isotopy) is $H^2(X,\Bbb{R})$?


1 Answer 1


It is not completely clear from your question which space you are interested in. Anyway, here are some thoughts, tell me what you think:

Let $\mathcal{S}$ be the space of all symplectic forms on $M$. If $\omega\in\mathcal{S}$ and $\eta$ is any closed $2$-form, then for small $t$ also $\omega+t\,\eta$ will be a symplectic form. This implies that $T_\omega\mathcal{S}$ can be identified with $\mathrm{ker}\left(d:\mathcal{A}^2\to\mathcal{A}^3\right)$.

Of course, we are actually interested in the tangent space of $\mathcal{S}/\sim$ at a point $[\omega]_{\sim}$. This can be identified with the quotient of two different tangent spaces: $T_\omega\mathcal{S}$ and $T_\omega[\omega]_{\sim}$. We claim that this last space is the range of $d:\mathcal{A}^1\to\mathcal{A}^2$, so that the quotient will be $$\frac{\mathrm{ker}\left(d:\mathcal{A}^2\to\mathcal{A}^3\right)}{\mathrm{ran}\left(d:\mathcal{A}^1\to\mathcal{A}^2\right)}=H^2(M,\mathbb{R}).$$ This is where the Moser Theorem comes into play: any element of $[\omega]_\sim$, i.e. a symplectic form strongly isotopic to $\omega$, can be written as $\omega+d\,\vartheta$ for some $1$-form $\vartheta$. Then the tangent space of $[\omega]_\sim$, as we wanted.

The whole argument is not particularly rigorous, but I hope it can be useful anyway.

  • $\begingroup$ Hi Lemmon, can you explain a bit why $\omega+t\,\eta $ with small $t$ is symplectic form, and why this implies $\mathrm{ker}\left(d:\mathcal{A}^2\to\mathcal{A}^3\right) $ can be identified with the $T_\omega\mathcal{S} $? I can't figure it out. $\endgroup$
    – yi li
    Sep 27, 2022 at 0:41
  • 1
    $\begingroup$ @yili sure, I was not very precise in my answer. I am happy to clarify. The tangent space at $\omega$ is the set of all possible derivations $\partial_{t=0}\omega_t$, for $\{\omega_t\mid -\varepsilon<t<\varepsilon\}$ a path of symplectic forms in $\mathcal{S}$. The conditions for $\omega_t$ being symplectic are that it should be closed and nondegenerate. Nondegeneracy is an open condition, so it does not impose any condition on $\dot{\omega}_0$, closedness of $\omega_t$ instead tells you that $d\dot{\omega}_0=0$. So the tangent space is indeed the kernel of $d$ on $2$-forms. $\endgroup$ Sep 27, 2022 at 13:45
  • $\begingroup$ that's cool , thank you $\endgroup$
    – yi li
    Sep 27, 2022 at 14:03

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