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Let $M$ be $2n$-dimensional manifold, $\omega$ be symplectic 2-form (i.e. closed, nondegenerate). How does one argue that the Stokes' theorem implies $$\int_M \omega^n\neq 0$$ which is the statement that $\omega^n$ is a volume form? I wanted to argue from nondegeneracy but cannot seem to make it work. Partly it is because I hardly work with coordinate-dependent situations. I tried the following: suppose it is zero. Then $$\int_M\omega^n=0\Longrightarrow\omega^n=\omega\wedge...\wedge\omega=0\,,$$ which seems to imply that for any $2n$ vector fields $\{X_j\}$ we have $$\omega^n(X_1,...,X_{2n})\equiv 0\,.$$ and hence when we put each vector field into each $\omega$, it implies that $\omega$ is degenerate, contradicting the nondegeneracy assumption.

Would appreciate if the argument can be sharpened.

Evangeline A. K. McDowell
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  • The usual notion of 'volume form' is somewhat narrower: A volume form on an $n$-manifold $M$ is nowhere-vanishing section $\Omega$ of $\bigwedge^n T^*M$; for any such section, $\int_M \Omega > 0$ (where the orientation of $M$ is taken to be the one defined by $\Omega$). – Travis Willse Jan 30 '17 at 13:56
  • Stoke's theorem has nothing to do with this problem. As $\omega^n$ is a volume form and $M$ is a manifold, its total volume (that's the integration) is non-zero. See http://math.stackexchange.com/questions/417365/how-to-show-omegan-is-a-volume-form-in-a-symplectic-manifold-m-omega?noredirect=1&lq=1 – tessellation Jan 30 '17 at 18:09
  • Be careful with your implication. $\int \omega^n = 0$ only implies that there exists a point where $\omega^n$ vanishes. Thus, for any collection of vector fields, the function $\omega^n(X_1, \dots, X_{2n})$ vanishes at this point. It's now an exercise in linear algebra to show that this forces $\omega$ to be degenerate at this point. – Sam Lisi Feb 09 '17 at 19:31

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