TITLE: Betti number and harmonic forms
QUESTION [11 upvotes]: 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!

REPLY [8 votes]: If I understood your question correctly, it is: how can I compute the dimension of the space of harmonic forms? 
There is one class of Riemann manifolds where it is possible to write down the harmonic forms explcitly, without topology. These are the symmetric spaces (for example spheres, projective spaces, compact Lie groups, tori, Grassmann manifolds, flag manifolds). The space of harmonic forms on the symmetric space $M=G/H$ is the same as the invariant forms, which is $\Lambda^* (\mathfrak{g}/\mathfrak{h})^G$. And you can compute this algebraically. I guess that each explicit computation of harmonic forms, including your example on $S^2$, is based on that principle (maybe for Riemann surfaces and nonsymmetric homogeneous spaces there are exceptions to this rule).
As pointed out by Gunnar, the dimension of the space of harmonic forms on $M$ does not depend on the metric. So you can conclude that there is no harmonic 1-form on $S^2$, in any wild metric.
Of course, most manifolds do not carry a symmetric metric. Here, if you want to compute the dimension of the space of harmonic forms, it might be easiest to compute first the cohomology and then appeal to the Hodge theorem.