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.
My favorite reference for this (Poincare duality via differential forms) is Bott and Tu "Differential Forms in Algebraic Topology", Chapter 1, section 5.