# why Moser theory tells us that tangent space (upto strong isotopy) is $H^2(X,\Bbb{R})$

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 $$H^2(X,\Bbb{R})$$?

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.
• 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. Sep 27, 2022 at 0:41
• @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. Sep 27, 2022 at 13:45