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Suppose $\omega$ is symplectic structure on $\mathbb R^n$. Let $\omega_0:=\omega|_{x=0}$.

Let $\overline{\omega}= \omega_0-\omega$ and for $t\in[0,1]; \omega_t:= \omega+ t\overline{\omega}$.

How to show that there is some neighborhood of origin such that all $\omega_t$ are symplectic.

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  • $\begingroup$ I don't understand what $\omega_0$ is. Restriction of the form to a single point? $\endgroup$
    – user7530
    Jul 17, 2013 at 4:32
  • $\begingroup$ Let us say $\sigma:=\omega_0$ is a two form on $\mathbb R^n$ such that for any $p\in \mathbb R^n$, $\sigma(p):= \omega_0$ with natural parallelism in $\mathbb R^n$. $\endgroup$
    – Junu
    Jul 17, 2013 at 4:51

1 Answer 1

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I presume you are working through Moser's proof of Darboux's theorem.

The family of forms $\omega_t = (1-t) \omega + t\omega_0$ are all closed since both $\omega$ and $\omega_0$ are.

Observe that at the point $x = 0$, $\omega_t = \omega_0$, so the form is non-degenerate on the tangent space at $0$ $T_0 \mathbb{R}^{2n}$. Non-degeneracy is an open property and $[0,1]$ is compact so there is a small neighbourhood of $0$ for which $\omega_t$ is non-degenerate for all $t \in [0,1]$.

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  • $\begingroup$ Thanks for the answer.. Yes I was working on Moser proof. $\endgroup$
    – Junu
    Jul 25, 2013 at 16:18

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