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! 



Betti number and harmonic forms
Just to add some details to Paul's answer:


It is indeed correct that the $k$-th Betti number of a compact Riemannian manifold $M$ without boundary is equal to the dimension of the space of harmonic $k$-forms. This is the Hodge isomorphism theorem. A good proof of this may be found in Demailly's book (available for free [here](https://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/agbook.pdf)) at the beginning of chapter 6.


Basically what happens is this: we use a Riemannian metric to define an inner product on the space $E^k := C^\infty(M, \bigwedge^k T\_M^\*)$ of smooth $k$-forms on $M$. The Riemannian metric also gives us the Laplacian $\Delta$, which turns out to be an elliptic differential operator on $E^k$. Thus we get an orthogonal direct sum decomposition


$$C^\infty(M, \bigwedge^k T\_M^\*) = \mathcal H^k(M) \oplus Im \Delta = \mathcal H^k(M) \oplus Im d \oplus Im d^{\*}$$


where $\mathcal H^k(M) := Ker \Delta$ is the space of harmonic $k$-forms on $M$, and $d^{\*}$ is the formal adjoint of the exterior derivative $d$. The subspace of $d$-closed forms of $E^k$ is exactly $\mathcal H^k(M) \oplus Im d$, and thus we obtain an isomorphism between $\mathcal H^k(M)$ and the $k$-th De Rham cohomology group $H^k(M,\mathbb{R})$, whose dimension is equal to the $k$-th Betti number of $M$.


Morally speaking this isomorphism is interesting because it gives a link between the topological and geometric structures of the manifold. The Betti numbers only depend on the topology of $M$, while the space of harmonic forms is defined by a Riemannian metric. With this theorem we can interpret the $k$-th Betti number as counting the number of linearly independent harmonic $k$-form on the original manifold. In particular, as you say, if a $k$-th Betti number is zero, then the only harmonic $k$-form on $M$ is the zero form. 


