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EDIT: In the book "Principles of Algebraic Geometry" by Griffiths and Harris the authors prove the Hodge decomposition for the Dolbeault operator $\bar\partial$ on differential forms on a compact complex manifold (see Ch. 0). Then they mention without proof (see p. 89) that similar decomposition holds for the de Rham operator $d$ on forms on any compact Riemannian manifold, and the proof is similar (they actually use this fact on p. 116).

I am teaching a course following mostly this book. I prefer to prove both cases of the Hodge decomposition. I am wondering if there is a unified approach of doing that.

For example de Rham and Dolbeault complexes are examples of elliptic complexes of differential operators between vector bundles.

Is there a Hodge decomposition theorem for elliptic complexes on a compact manifold in the class of infinitely smooth sections?

A reference would be very helpful, especially if it is of introductory level, i.e. appropriate for the course.

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Yes, there is a Hodge decomposition for elliptic complexes on compact oriented Riemannian manifolds.

$L^2$-version. Let $(M,g)$ be a compact oriented Riemannian manifold and let $$ 0 \to \Gamma(E_0) \to \Gamma(E_1) \to \dots \to \Gamma(E_m) \to 0 $$ be an elliptic differential complex with the arrows given by the differential operators $D_0,\dotsc,D_{m-1}$, where each $E_i$ is equipped with a metric and a compatible connection. Thus we have formal adjoints and can form the Laplacians: $$ \Delta_i := D_i^* \circ D_i + D_{i-1} \circ D_{i-1}^*: \Gamma(E_i) \to \Gamma(E_i). $$ Then we get the Hodge decomposition $$ L^2(E_i) = \ker(\Delta_i) \oplus \operatorname{ran}(D_{i-1}) \oplus \operatorname{ran}(D_i^*). $$

I learnt this from our own Liviu Nicolaescu's Lectures on the Geometry of Manifolds. More precisely, this is Chapter 10.4.3 in the notes (my version is from 9. Sep 2018).

EDIT: As per OP's comment, I am adding the smooth version I know of.

$C^{\infty}$-version. Consider the total vector bundle $E := \bigoplus_{i=0}^{m} E_i$ and the Laplacian $\Delta = D \circ D^* + D^* \circ D$, i.e. $\Delta$ is just a tuple of Laplacians $\Delta_i: \Gamma(E_i) \to \Gamma(E_i)$. The following statements are equivalent:

(1) $\Delta$ is Fredholm and $\Gamma(E) = \ker(\Delta) + \operatorname{ran}(\Delta)$ (note, this is just a sum, not a direct sum);

(2) The complex itself is Fredholm and satisfies the Hodge decomposition: $$ \Gamma(E) = \ker(\Delta) \oplus \operatorname{ran}(D) \oplus \operatorname{ran}(D^*) $$ This can be found in Chapter 1 of van den Ban and Crainic's lectures on Analysis on Manifolds, more precisely exercise 1.3.15, if you don't mind following the exercises and working it out from the preceding theory, but I don't know a more self-contained source.

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    $\begingroup$ @LSpice: The notes are also freely available, the last version is here: www3.nd.edu/~lnicolae/Lectures.pdf $\endgroup$
    – M.G.
    Commented Mar 14, 2023 at 19:06
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    $\begingroup$ Thank you. I would prefer to have Hodge decomposition for infinitely smooth sections rather than $L^2$. (Another less important remark is that in the mentioned lecture notes the differential operators have the first order.) $\endgroup$
    – asv
    Commented Mar 14, 2023 at 19:07
  • $\begingroup$ @asv: I believe I have seen the Hodge decomposition stated for $\Gamma(E)$, where $E$ is the total vector bundle. Would that work for you? In fact, I think I even remember where. $\endgroup$
    – M.G.
    Commented Mar 14, 2023 at 19:15
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    $\begingroup$ @LSpice: I think, that's fine. That way we have both links. $\endgroup$
    – M.G.
    Commented Mar 14, 2023 at 19:26
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    $\begingroup$ @asv: Ah, I see now what you meant. Indeed one has to get the ingredients to establish (1) for concrete cases from elsewhere, so it's not self-contained in that regard. I'm sorry I couldn't be of more help. $\endgroup$
    – M.G.
    Commented Mar 15, 2023 at 15:07

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