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Let $\omega_t$ be closed 2-forms in a $2n$-manifold $M$ with $ t \in [0,1]$, and suppose that in some point $x \in M$ $(\omega_t)_x$ is nondegenerate for all $ t \in [0,1]$. I want to show that there exists a neighbourhood of $x$ such that $\omega_t$ is nondegenerate for all $t$.

The point is: being nondegenerate is an open condition, meaning that for every $t$ one can find a neighbourhood $U_t$ of $x$ such that $\omega_t$ is nondegenerate in $U_t$. But how to find a neighbourhood that fulfils the statement for all $t$? I would think in taking the intersection, but I could obtain the point ... I guess that it has to do with the fact that $[0,1]$ is compact... .

The question arises when considering Moser's proof of the Darboux theorem: one has $\omega_0$ and $\omega_1$ closed 2-forms and define $\omega_t := (1-t) \omega_0 + t \omega_1$, with $(\omega_0)_x$ and $(\omega_1)_x$ nondegenerate.

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$\omega_t$ is not degenerated is equivalent to saying that $\Lambda_t=\wedge^n\omega_t$ is a volume form. Consider a neighborhood $V$ of $x$ diffeomorphic to an open subset of $\mathbb{R}^{2n}$, you can find vector fields $(X_1,...,X_{2n})$ which does not vanish on $V$ such that $\Lambda_t(X_1(x),...,X_{2n}(x))\neq 0$, the function defined on $f:[0,1]\times V\rightarrow \mathbb{R}$ by $f(t,y)=\Lambda_t(X_1(y),...,X_{2n}(y))$ is continuous; this implies that $f^{-1}(\mathbb{R}-\{0\})$ contains a neighborhood of $[0,1]\times \{x\}$, for every $t\in [0,1]$ there exists an open neighborhood $I_t\subset [0,1]$ which contains $t$, and a neighborhood $U_t$ of $x$ such that $I_t\times U_t\subset f^{-1}(\mathbb{R}-\{0\})$, since $[0,1]$ is compact, you can find finitely many $t_0,...,t_n$ such that $\cup_{i=1}^{i=n}I_{t_i}=[0,1]$ $\cap_{i=0}^{i=n}U_{t_i}$ is the requested neighborhood.

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  • $\begingroup$ Sorry for the confusion, but if $f(t,y)=\Lambda_t(X_1(y),...,X_{2n}(y))$ is nonzero in $[0,1] \times \{ x \}$, why do you say that $I_t\times U_t\subset f^{-1}(0)$ ? $\endgroup$
    – Minkowski
    Nov 12, 2017 at 17:17

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