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Let $M$ be a compact manifold of dimension $2n$. Let $\omega$ be a 2 form on $M$ such that the induced bundle map $\tilde{\omega}: TM \to T^*M$ defined by $\tilde{\omega}(X)(Y) = \omega(X,Y)$ is a bundle isomorphism. Show that if $d \omega = 0$, then $[\omega^n] \in H_{dR}^{2n}(M)$ is nonzero.

I recognize that on oriented compact manifolds top cohomology is nontrivial, but this question doesn't seem to assume that. I tried assuming $\omega^n$ was exact and getting some contradiction but didn't get anywhere. Any advice would be appreciated.

Emilio Minichiello
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    If you write $\omega$ in normal coordinates as $\omega = a_{ij} \mathrm{d} x_i \wedge \mathrm{d} x_j$, then the condition that $\tilde{\omega}$ is an isomorphism is equivalent to saying that $\det(a_{ij}) \neq 0$. But if you compute $\omega^n$ in these same coordinates, you'll find that $\omega^n$ is just the function $\det(a_{ij})$ times the volume form. But then $\omega^n$ is a nowhere vanishing $2n$-form and therefore its pairing with the fundamental class is nonzero, revealing that $[\omega^n] \neq 0$. – Or Eisenberg May 14 '19 at 22:16

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Or Eisenberg's comment goes in the correct direction, but it doesn't appear to quite work, since as written $\omega^n$ cannot possibly be $\det(a_{ij})$ times the volume form, since $\omega^n$ has degree $n$ in the entries of the matrix, but $(a_{ij})$ appears to be $2n\times 2n$.

Instead, let's do the following. Work on a framed open set $U$. Choose $X_1\in TM(U)$, choose $X_2$ such that $\omega(X_1,X_2)=1$ (since $\omega$ is nondegenerate). Then $\ker \omega(X_1,-)\cap \ker \omega(X_2,-)$ is $2n-2$ dimensional, so inductively choose $X_{2i-1}$, $X_{2i}$ such that $\omega(X_{2i-1},X_{2i})=1$ and when $i<j$, $\omega(X_{i},X_{j})=0$ otherwise. We may have to reduce the size of the open subset, so that we can select all of our vector fields.

Then given our basis $X_1,\ldots,x_{2n}$, let $\theta_1,\ldots,\theta_{2n}$ be the dual basis for $TM^*$. Observe that by construction of our basis, $$\omega = \sum_{i=1}^n\theta_{2i-1}\wedge \theta_{2i}.$$ Thus $\omega^n = n! \theta_1\wedge \cdots \wedge \theta_{2n}$. The $n!$ comes from the $n!$ ways to permute the $n$ 2-forms that make up $\omega$, since each permutation arises exactly once as a term in the product. All other terms in the product contain repeated 1-forms, and thus vanish.

Thus locally $\omega^n$ is a nowhere vanishing $2n$-form, so globally $\omega^n$ is a nowhere vanishing $2n$-form, i.e., a volume form.

On any (connected) compact manifold a volume form has nonzero integral, see here for a justification/explanation. Thus $\omega^n$ has nonzero integral, so its cohomology class is nonzero.

You may notice that we haven't used that $d\omega=0$. This is fine. Wikipedia confirms that $\omega^n$ is a volume form on almost symplectic manifolds as well.

jgon
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