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I have just started studying symplectic geometry, and one fact has me completely stumped.

Apparently a symplectic manifold $V$ such that $\mathrm{dim}V = n$ has volume form: $\frac{\omega^n}{n!}$.

Now I recall a volume form is a non-vanishing, top-dimensional form on $V$.

However I have no clue how to show $\frac{\omega^n}{n!}$ is either non-vanishing or a top-dimensional form. This last condition in particular seems like something that will have an 'obvious' explanation, but one that I can't figure out.

Most of these definitions have been floating around in my head barely a day. I'm not far into the topic, and my differential geometry/experience with differential forms is still fairly basic, so the simpler an explanation someone can offer, the better.

leob
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1 Answers1

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Let $V$ be a $2n$-dimensional symplectic manifold.

$\omega$ is a $2$-form, so $\omega^n$ is a $2n$-form. Top-dimensional means that the dimension of the form is equal to the dimension of the space. In this case $2n = 2n$.

Choose a point $p \in V$, and choose a basis $v^1, \ldots, v^{2n}$ of $T_pV$ such that $$\omega = dv_1 \wedge dv_2 + dv_3 \wedge dv_4 + \ldots + dv_{2n-1}\wedge dv_{2n}.$$

Then according to this question, we get $\omega^n(v^1, \ldots, v^{2n}) = n! \cdot (dv_1 \wedge dv_2 \wedge dv_3 \wedge dv_4 \wedge \cdots \wedge dv_{2n-1} \wedge dv_{2n})(v^1, \ldots, v^{2n}) = n! \cdot 1.$

Thus $\omega$ does not vanish at $p$.

Magma
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