# symplectic strucutre

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.

• I don't understand what $\omega_0$ is. Restriction of the form to a single point? Jul 17, 2013 at 4:32
• 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 $\mathbb R^n$.
– Junu
Jul 17, 2013 at 4:51

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 $t \in [0,1]$.