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The classical Hodge theorem I know of relates the de Rham cohomology groups isomorphically to the space of harmonic forms and shows that $Id=\pi+\Delta G$, where $\pi$ is the harmonic projection of $k$-forms and $G$ is the Green's Operator for the Laplacian $\Delta=d\delta +\delta d$. (

It isn't clear to me how this relates to the following "Hodge theorem" from

Let $(M,g)$ be a compact, connected oriented Riemannian manifold. Then there exist an orthonormal basis of eigenfunctions (eigenforms) for $L^2(M,g)$ (or $L^2\Lambda^k(M,g)$) of the Laplacian. $0$ is an eigenvalue with multiplicity 1, and all other eigenvalues are strictly positive, accumulate at infinity, and have finite multiplicities.

In particular, I don't see the connection to the orthonormal basis of eigenfunctions part: how does this follow from the classical version?

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In general, 0 is not a simple eigenvalue when k>0. – timur Dec 27 '13 at 3:59

The first theorem you mention follows from the second, as I will explain. But more importantly, the theorems together give a rich picture of the de Rham complex on a compact Riemannian metric, and so deserve to be understood in conjunction:

Let $M$ be a compact Riemannian manifold with Laplacian $\Delta$. Note that one trivially has $[\mathrm d,\Delta]=0$. It follows that if $f$ is an eigen-$k$-form for $\Delta$ with eigenvalue $\lambda$, then $\Delta(\mathrm d f) = \lambda \mathrm d f$, and so $\mathrm d f$ is either $0$ or an eigen-$(k+1)$-form. By the second Hodge theorem, the de Rham complex then breaks up as a direct sum of finite-dimensional complexes:

$$ \Omega^\bullet_{\mathrm{dR}}(M) = \bigoplus_{\lambda} \Omega^\bullet_{\mathrm{dR}}(M)_\lambda $$

where $\Omega^k_{\mathrm{dR}}(M)_\lambda$ is the $\lambda$th eigenspace for the action of $\Delta$ on $\Omega^k_{\mathrm{dR}}(M)$. One concludes that*:

$$ \mathrm H^\bullet_{\mathrm{dR}}(M) = \bigoplus_{\lambda} \mathrm H^\bullet \bigl(\Omega_{\mathrm{dR}}(M)_\lambda,\mathrm d\bigr) $$

The operator $\delta$ also commutes with $\Delta$, and so descends to $\Omega^\bullet_{\mathrm{dR}}(M)_\lambda$. Moreover, on $\Omega^\bullet_{\mathrm{dR}}(M)_\lambda$, $\Delta = [\mathrm d,\delta] = \lambda$, and so if $\delta \neq 0$, $\eta = \lambda^{-1}\delta$ is a contracting homotopy for the complex $\Omega^\bullet_{\mathrm{dR}}(M)_\lambda$, i.e. $[\mathrm d,\eta] = \mathrm d \eta + \eta \mathrm d = \mathrm{id}$. It follows that $\Omega^\bullet_{\mathrm{dR}}(M)_\lambda$ is exact, and so:

$$ \mathrm H^\bullet_{\mathrm{dR}}(M) = \mathrm H^\bullet \bigl(\Omega_{\mathrm{dR}}(M)_0,\mathrm d\bigr) $$

Now the only question is to see that $\Omega^\bullet_{\mathrm{dR}}(M)_0$ has trivial differential. But $\Delta = (\mathrm d+\delta)(\delta + \mathrm d) = (\mathrm d+\delta)(\mathrm d+\delta)^*$ is nonnegative; in particular, if $\Delta x = 0$, then $(\mathrm d + \delta)x = 0$. On the other hand, for $x$ a $k$-form, $\mathrm d x$ and $\delta x$ have different degrees, and so if their sum is zero, both are zero.

This completes the proof of the first theorem you mention.

In summary, the de Rham complex looks like: the harmonic forms, with no differentials between them, living at the bottom; exact complexes, with distinguished antiderivative operators, at non-zero eigenvalues.

*Actually, I've lied a bit here, since I'm ignoring issues of completions and topological vector spaces. The direct sum in the first line is not an algebraic direct sum, but a Hilbert space direct sum. The correct proof is: let $\iota : \Omega^\bullet_{\mathrm{dR}}(M)_0 \to \Omega^\bullet_{\mathrm{dR}}(M)$ denote the inclusion of the $0$th eigenspace into the whole complex, and $\pi$ the projection; letting $G$ denote the Green's operator for $\Delta$, the operator $\eta = G\delta = \delta G$ is a homotopy between $\mathrm{id}$ and $\iota\pi$, i.e. $[\mathrm d,\eta] = \mathrm d \eta + \eta \mathrm d = \pm(\mathrm{id} - \iota \pi)$; it follows that $\iota$ and $\pi$ are quasiisomorphisms, and in fact inverse isomorphisms on homology.

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