On a compact, boundless, Riemannian manifold, the dimension of the space of harmonic k-forms is equal to the k-th Betti number. Is this correct (by Hodge theory)? For example, on the surface of a unit sphere S^2 and any equivalent topology, the 0-th Betti number is 1, which is consistent with the fact that any harmonic function on a compact, boundless, Riemannian manifold is a constant. More importantly to me, what about the 1st Betti number? I think the only harmonic 1-form on the topological class of S^2 is zero --- you can't comb every hair on a sphere towards the same direction --- right?

Basically, it is easy to prove both results (1st betti number = 0 and the only harmonic 1-form is zero) on the ideal case of S^2. But if I want to go further, some topological tool has to be involved?

I am a PDE guy, so if there is any PDE-based proof to the above question, it'd be highly appreciated!

On a compact, boundaryless, Riemannian manifold, the dimension of the space of harmonic k-forms is equal to the k-th Betti number. Is this correct (by Hodge theory)? For example, on the surface of a unit sphere $S^2$ and any equivalent topology, the 0-th Betti number is 1, which is consistent with the fact that any harmonic function on a compact, boundless, Riemannian manifold is a constant. More importantly to me, what about the 1st Betti number? I think the only harmonic 1-form on the topological class of $S^2$ is zero --- you can't comb every hair on a sphere towards the same direction --- right?

Basically, it is easy to prove both results (1st betti number = 0 and the only harmonic 1-form is zero) on the ideal case of $S^2$. But if I want to go further, some topological tool has to be involved?

I am a PDE guy, so if there is any PDE-based proof to the above question, it'd be highly appreciated!

Dear Experts,

On a compact, boundless, Riemannian manifold, the dimension of the space of harmonic k-forms is equal to the k-th Betti number. Is this correct (by Hodge theory)? For example, on the surface of a unit sphere S^2 and any equivalent topology, the 0-th Betti number is 1, which is consistent with the fact that any harmonic function on a compact, boundless, Riemannian manifold is a constant. More importantly to me, what about the 1st Betti number? I think the only harmonic 1-form on the topological class of S^2 is zero --- you can't comb every hair on a sphere towards the same direction --- right?

Basically, it is easy to prove both results (1st betti number = 0 and the only harmonic 1-form is zero) on the ideal case of S^2. But if I want to go further, some topological tool has to be involved?

I am a PDE guy, so if there is any PDE-based proof to the above question, it'd be highly appreciated!

Thank you for your valuable time!

Bin Cheng

On a compact, boundless, Riemannian manifold, the dimension of the space of harmonic k-forms is equal to the k-th Betti number. Is this correct (by Hodge theory)? For example, on the surface of a unit sphere S^2 and any equivalent topology, the 0-th Betti number is 1, which is consistent with the fact that any harmonic function on a compact, boundless, Riemannian manifold is a constant. More importantly to me, what about the 1st Betti number? I think the only harmonic 1-form on the topological class of S^2 is zero --- you can't comb every hair on a sphere towards the same direction --- right?

Basically, it is easy to prove both results (1st betti number = 0 and the only harmonic 1-form is zero) on the ideal case of S^2. But if I want to go further, some topological tool has to be involved?

I am a PDE guy, so if there is any PDE-based proof to the above question, it'd be highly appreciated!