Cross-post from [MSE][1].


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Suppose $(M,g)$ be a (for simplicity consider closed) Riemannian manifold. Because every parallel $p$-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][2]?


  [1]: https://math.stackexchange.com/q/4064240/272127
  [2]: https://en.wikipedia.org/wiki/Closed_and_exact_differential_forms#Poincar%C3%A9_lemma