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I'm somewhat confused about the definitions of Betti numbers for Riemannian manifolds. Working with the first Betti number as an example, I have usually taken the definition to be the rank of the homology group $H_1(M)$, where $M$ is the manifold in question. I'm also aware that through a Hodge-theoretic argument, we have that the first Betti number is equal to the dimension of the space of harmonic 1-forms on $M$, and that in fact this space is isomorphic to $H^1(M; \mathbb{R})$.

So my question is essentially: How do we get an isomorphism $H_1(M)\cong H^1(M; \mathbb{R})$?

I know that through Poincaré duality we have the isomorphism $H_1(M) \cong H^{n-1}(M; \mathbb{Z})$ but I can't see how this helps.

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It's not true that $H_1(M)\cong H^1(M;\mathbb R)$, but the ranks are the same. $H_1(M)\otimes\mathbb R\cong H_1(M;\mathbb R),$ and $H^1(M;\mathbb R)\cong H_1(M;\mathbb R)$ by a non-canonical isomorphism. –  Jim Conant Dec 16 '10 at 16:17
They are not naturally isomorphic, but rather are naturally dual, via universal coefficients. (Of course, since any finite dimensional vector space is non-naturally isomorphic to its dual, it follows that they are non-naturally isomorphic, as Jim Conant notes in his comment above, and Steve Gubkin notes in his answer below.) –  Emerton Dec 16 '10 at 16:22

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$H_1(M,\mathbb{R}) \cong H^1(M, \mathbb{R})$ follows from the universal coefficients theorem for cohomology.

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This is only true under some finiteness assumptions on $H_*(M)$, which hold e.g. if $M$ is a compact manifold, or if more generally, $H_*(M)$ is finitely generated. In general, $H^p(M;\mathbb R)$ is isomorphic to $Hom(H_p(M), \mathbb R)$, and $Hom$ turns sums into products, so e.g. if $M$ is the connected sum of infinitely many real projective spaces of dimension $>2$, the $H_1(M;\mathbb R)$ is a direct sum, and $H^1(M;\mathbb R)$ is the direct product of infinitely many copies of $R$. –  Igor Belegradek Dec 16 '10 at 23:43
I should have used tori instead of projective spaces. –  Igor Belegradek Dec 17 '10 at 13:07

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