TITLE: Hodge Theory, intuition?
QUESTION [12 upvotes]: We have the following theorem of Hodge, as follows:
$$\dim \ker \Delta^p = \dim H^p(M) = b_p(M).$$My question is, what is the intuition behind this statement?

REPLY [6 votes]: I would say that the basic intuition is that a Riemannian metric defines an inner product on the space of differential forms, which can be used to define an adjoint operator to the exterior derivative.
Now consider the adjoint $(d_{k-1})^*:\Omega^k\to\Omega^{k-1}$ of $d_{k-1}:\Omega^{k-1}\to\Omega^k$. If the spaces $\Omega^\ell$ were all finite dimensional, then $\ker((d_{k-1})^*)$ would be the orthocomplement of $\DeclareMathOperator{\im}{im}\im(d_{k-1})$. Hence $\ker((d_{k-1})^*)\cap \ker(d_k)$ would be a complement to $\im(d_{k-1})$ in $\ker(d_k)$ and thus project isomorphically onto the cohomology space $H^k$. Hence you see that $\ker((d_{k-1})^*)\cap \ker(d_k)$ would have the same dimension as $H^k$.
Using adjointness again, it is easy to verify that (if things were finite dimensional) defining $\Delta_k=d_{k-1}\circ (d_{k-1})^*+(d_k)^*\circ d_k$ you get $\ker(\Delta_k)+\ker((d_{k-1})^*)\cap \ker(d_k)$. (This is like $\Delta(\phi)=0$ implies $0=d\Delta(\phi)=dd^*d\phi=0$ (since $dd=0$) and using adjointness twice, one gets $d\phi=0$, and likewise for $d^*$.)
Using functional analysis/PDE theory as mentioned in the answer of @GeorgesElencwajg , one proves that this continues to hold on a compact manifold in spite of the fact that one deals with infinite dimensional spaces.