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C.F.G
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Cross-post from MSE.

Suppose $(M,g)$ be a (for simplicity consider 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?

Cross-post from MSE.

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?

Cross-post from MSE.

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?

Source Link
C.F.G
C.F.G
  • 4.2k
  • 6
  • 31
  • 65

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

Cross-post from MSE.

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?