Existence non-trivial parallel $p$-form implies non-triviality of $p$-th cohomology group using De Rham cohomology

Question

Cross-post from MSE.

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Suppose $(M,g)$ be a closed Riemannian manifold. Because every parallel (nontrivial) $p$-form $\omega$ is harmonic so the $p$-th Betti number should be positive i.e. $b_p\geq 1$. How to see this using De Rham cohomology? (of course without using properties of Hodge theory like $\delta$, 1-1 correspondence between De Rham and Harmonic forms, etc.)

It is well-known that parallel forms are closed. The more involved part is showing that it is not exact.

Such an easy argument based on harmonic forms I think there must be similar easy argument based on De Rham cohomology or it needs somewhat difficult and tricky argument similar to the proof of Poincaré lemma?

Answer 1

Here is an argument in the orientable case: If $\omega$ is a $g$-parallel $p$-form on an orientable Riemannian $n$-manifold $(M^n,g)$, then it is closed since $\mathrm{d}\omega = \pi_{p+1}(\nabla^g\omega) = 0$, where $$\nabla^g:\Omega^p(M)=C^\infty(\Lambda^p(T^*M))\to C^\infty(T^*M\otimes\Lambda^p(T^*M))$$ is the Levi-Civita connection and $\pi_{p+1}:T^*M\otimes\Lambda^p(T^*M)\to\Lambda^{p+1}(T^*M)$ is the obvious skew-symmetrization. Moreover, the $(n{-}p)$-form $\ast_g\omega$ is also $g$-parallel, so it, too, is closed. Now $$ \omega\wedge \ast_g\omega = |\omega|^2_g\,\,\,{\ast_g}1, $$ so, unless $\omega\equiv0$, we will have $$ \int_M \omega\wedge \ast_g\omega >0. $$ Thus, $\omega$ can't be exact, since $\omega = \mathrm{d}\eta$ would give $$ \int_M \omega\wedge \ast_g\omega = \int_M \mathrm{d}(\eta\wedge \ast_g\omega) = 0, $$ by Stokes' Theorem. No Hodge theory required.

Finally, if $M$ is not orientable, pull $\omega$ up to the orientation double cover $\tilde M$, observe that it is not exact there, and conclude it couldn't have been exact on $M$ either.