The 3rd de Rham cohomology of a 6th-dimensional oriented compact smooth manifold has even dimension

Question

I'm trying to show that

Given an oriented compact smooth manifold M of dimension 6, $H^3_{dR}(M)$ has even dimension.

I'm not sure where to begin. Any hint or suggestion is greatly appreciated.

Answer 1

The cup product $H^3(M,\Bbb R)\times H^3(M,\Bbb R)\to H^6(M,\Bbb R)\cong\Bbb R$ is a skew-symmetric form. It is also non-degenerate, essentially by Poincare duality. So $H^3(M,\Bbb R)$ is even-dimensional.

Answer 2

Equip $M$ with a Riemannian metric $g$. Then $H^3_{dR}(M;\mathbb{R})$ can be identified with the space $\mathcal{H}^3(M,g)$ of real harmonic $3$-forms on $(M,g)$. But harmonic iff closed and coclosed, so the Hodge star $\star$ restricts to an endomorphism of $\mathcal{H}^3(M,g)$. Now the Hodge star on 3-forms in 6-dimension satisfies $\star^2=(-1)^{3(6-3)}=-1$, so $\dim_\mathbb{R}\mathcal{H}^3(M,g)$ is even.