Does the cup product on de Rham cohomology induce a nondegenerate bilinear form?

I have small issue I came across in the following.

Suppose $M$ is a compact, oriented manifold of dimension $4n+2$. I want to prove that the de Rham cohomology group $H^{2n+1}(M)$ are even dimensional.

As real vector spaces, write $H^{2n+1}(M)=\mathbb{R}^k$, for some $k$. The cup product is alternating, for if $\omega$ is a $2n+1$ form, $[\omega]\cup[\omega]=[\omega\wedge\omega]=[0]$, since $\omega\wedge\omega=(-1)^{(2n+1)(2n+1)}\omega\wedge\omega=-\omega\wedge\omega$. Therefore it is also skew-symmetric. Finally, as a compact oriented manifold, the top-degree cohomology $H^{4n+2}\simeq\mathbb{R}$ as it is $1$-dimensional.

Under these isomorphisms, we get an alternating, skew-symmetric bilinear form $F\colon\mathbb{R}^k\times\mathbb{R}^k\to\mathbb{R}$. We can represent it as a $k\times k$ matrix $B$ such that $B=-B^T$. Then $$\det(B)=\det(-B^T)=(-1)^k\det(B^T)$$ which implies $k$ is even if $\det(B)\neq 0$.

A bilinear form has invertible matrix iff it is nondegenerate, which translates to if $\omega\wedge\eta$ is exact for every closed $(2n+1)$-form $\eta$ for a given closed $(2n+1)$-form $\omega$, then $\omega$ is actually exact.

Is this true? If so, what is the reason? Thanks.

• see Poincaré duality – user8268 Sep 12 '14 at 8:19
• Take $B(\omega, *\omega)$ for the Hodge dual $*\omega$ of $\omega\in \Omega^{2n+1}(M)$. – anomaly Sep 12 '14 at 8:19