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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 $\Delta=d\delta +\delta d$. (http://en.wikipedia.org/wiki/Hodge_theory)

It isn't clear to me how this relates to the following "Hodge theorem" from http://math.bu.edu/people/sr/articles/book.pdf:

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. All eigenvalues are nonnegative, 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?

This is cross-posted from https://math.stackexchange.com/questions/617557/relation-of-hodge-theorem-to-eigenfunction-basis-of-laplacian. References would be appreciated!

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    $\begingroup$ Sometimes a mathematician will prove more than one theorem... $\endgroup$ Commented Dec 25, 2013 at 5:13
  • $\begingroup$ @QiaochuYuan I do think that these theorems are unrelated, but they seem to be related in some way in the book I cited. Just wanted to check! $\endgroup$
    – user44650
    Commented Dec 25, 2013 at 5:55
  • $\begingroup$ This is a well-formed, interesting question, but M.SE might be a better home. In any case, we don't tend to like cross-posting, and especially over Christmas you should allow questions to linger for a few days before seeking wider audience. Do please correct the version there following your edits here. $\endgroup$ Commented Dec 26, 2013 at 19:05

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In addition to what you said, you need to know that Green's operator $G$ is compact and self adjoint (cf Griffiths & Harris). Therefore you have a spectral decomposition for $G$, where the eigenspaces have finite dimension and are orthogonal to each other. From the equation $Id=\pi+\Delta G$, you can see that eigenforms of $G$ are eigenforms for $\Delta$ (with different eigenvalues). Now do Gram-Schmid on each eigenspace. The non negativity of the eigenvalues of $\Delta$, I leave as an exercise (Hint: if $A$ is a square matrix, what can you say about the eigenvalues of $A^*A$?)

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