## Proving Hodge decomposition without using the theory of elliptic operators?

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In the common Hodge theory books, the authors usually cite other sources for the theory of elliptic operators, like in the Book about Hodge Theory of Claire Voisin, where you find on page 128 the Theorem 5.22:

Let P : E → F be an elliptic differential operator on a compact manifold. Assume that E and F are of the same rank, and are equipped with metrics. Then Ker P is a finite-dimensional subspace of the smooth sections of E and Im P is a closed subspace of finite codimension of the smooth sections of F and we have an L²-orthogonal direct sum decomposition:

smooth sections of E = Ker P ⊕ Im P*

(where P* is the formal L²-adjoint of P)

In the case of Hodge Theory, we consider the elliptic self-adjoint operator Δ, the Laplacian (or: Hodge-Laplacian, or Laplace-Beltrami operator).

A proof for this theorem is in Wells' "Differential Analysis on Complex Manifolds", Chapter IV - but it takes about 40 pages, which is quite some effort!

Now that I'm learning the theory of elliptic operators (in part, because I want to patch this gap in my understanding of Hodge Theory), I wonder if this "functional analysis" is really always necessary.

Do you know of any class of complex manifolds (most likely some restricted class of complex projective varieties) where you can get the theorem above without using the theory of elliptic operators (or at least, where you can simplify the proofs that much that you don't notice you're working with elliptic operators)? Maybe the general theorem really requires functional analysis (I think so), but the Hodge decomposition might follow from other arguments.

I would be very happy to see some arguments proving special cases of Hodge decomposition on, say, Riemann surfaces. I would be even happier to hear why this is implausible (this would motivate me to learn more about these fascinating elliptic differential operators).

If this ends up being argumentative and subjective, feel free to use the community wiki hammer.

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## 4 Answers

The hard part of the proof of the Hodge decomposition (which is where the serious functional analysis is used) is the construction of the Green's operator. In Section 1.4 of Lange and Birkenhake's "Complex Abelian Varieties", they prove the Hodge decomposition for complex tori using an easy Fourier series argument to construct the Green's operator.

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Moreover, studying tori isn't a diversion from the general case; the Sobolev theory can be done neatly on tori using Fourier series, and then transplanted to other manifolds. This is the line taken by Griffiths & Harris, who I think give a very good account of the Hodge theorem. Wells sets things up in the generality needed by Atiyah-Singer - for Hodge theory, this is overkill. – Tim Perutz Jun 15 2010 at 22:15
Thank you for the reference to Lange and Birkenhake, I didn't know this book. It is exactly what I was looking for: a class of Kähler manifolds where we are able to prove Hodge decomposition more easily. This example might be useful to see more directly what's going on (in the proof). – Konrad Voelkel Jun 16 2010 at 17:19

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This is the final revision and probably my final post for a long time (deadlines). I wanted to write a more definitive answer, about how to approach the Hodge theorem, since I've thinking about it for a while. This a bit long. So the short summary is that there is no really easy path, but each is beautiful in it's own way.

Method 1 (orthogonal projections): This is the standard proof, although there are many variations which can be found in Griffiths-Harris, Warner, Wells,... Here's a very rough idea. The basic problem is to show that the space $closed(X)$ of closed $C^\infty$ forms on a compact Riemannian manifold $X$ is a direct sum of the space of exact forms $exact(X)$ and the space of harmonic forms $harm(X)$. A closed form is harmonic if and only if it is orthogonal to closed form (easy), so one might try to first prove that

$$closed(X)= exact(X)\oplus exact(X)^\perp$$

and then identify the latter with $harm(X)$. Since these are infinite dimensional, the decomposition isn't automatic. However, one can make it work by using $L^2$ forms and applying Hilbert space methods to obtain:

$$\overline{closed(X)}= \overline{exact(X)}\oplus \overline{exact(X)}^\perp$$

But at end one wants to come back to $C^\infty$ forms, and here is where the magic of elliptic operators comes in. The basic result which makes this work is the regularity theorem: a weak solution of elliptic equation, e.g Laplace's equation, is in fact a true $C^\infty$ solution. The space on the right of the second decomposition is therefore $harm(X)$, and this is all one needs.

As you said, the full details entail quite a bit of work, but one can look at standard books on Riemann surfaces (Farkas-Kra, Forster, Narasimhan...) for some instructive special cases.

Method 2 (Heat Equation): The heuristic is that if you think of the closed form as an initial temperature, governed by the heat equation, it should evolve toward a harmonic steady state. The nice thing is that this can solved explicity on Euclidean space, and this gives a good (short time) approximation for the general case. If I may make a shameless plug, I wrote up an outine in chapter 8 of my notes

http://www.math.purdue.edu/~dvb/preprints/book.pdf

Method 3 (Deligne-Illusie): This is really an amplification of Kevin Lin's answer. One important consequence of the Hodge theory is the degeneration of the Hodge to De Rham spectral sequence $$H^j(X,\Omega_X^i) \Rightarrow H^{i+j}(X,\Omega_X^\bullet) = H^{i+j}(X,\mathbb{C})$$ when $X$ is smooth and projective along with Kodaira vanishing. The first algebraic proof of this was due to Falting (on the way to Hodge-Tate). Deligne and Illusie gave a comparatively elementary proof. Although as Ravi Vakil commented, it is not the best way to first learn this stuff. Nor does it give the full Hodge decomposition. However, for people who want to go this route, an introduction aimed at students can be found in the book by Hélène Esnault and the late Eckart Viehweg.

Addendum (added June 24): I wanted to briefly address the question of how much of the Hodge decomposition can be understood algebraically. One can define algebraic de Rham cohomology with its Hodge filtration coming from the spectral sequence above. What is missing is a purely algebraic description of the "Betti lattice" $image [H^i(X,\mathbb{Z})\to H^i(X,\mathbb{C})]$ and the fact that the conjugate filtration $\overline{F}$ and $F$ are opposed in Deligne's sense. The first issue already seems serious. Even if $X$ is defined over $\mathbb{Q}$, a basis for the lattice typically involves transcendentals. This is already clear in the simplest example $H^1(\mathbb{A}^1-{0})$, the lattice is spanned by $[\frac{dz}{2\pi iz}]$.

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@Donu: so does your final paragraph mean that getting decomposition from degeneration isn't evident to you either (as it is not to me)? – Boyarsky Jun 17 2010 at 18:42

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@Boyarsky: I think you are correct. I think Hodge decomposition does not follow from degeneration. I think there is no purely algebraic proof of Hodge decomposition... – Kevin Lin Jun 17 2010 at 20:43
You can, however, use Hironaka and Hodge theory for compact complex algebraic varieties to get a mixed hodge structure on the cohomology of noncompact singular complex varieties :-) – Konrad Voelkel Jun 18 2010 at 7:17
I guess Kevin Lin answered the question, but the existence of a (mixed) Hodge structure on cohomology cannot be proved by pure algebra at present. – Donu Arapura Jun 18 2010 at 19:06
I agree with the suggestion of starting with Riemann surfaces. The higher dimensional proof is more technical and involved, but involves no essential new ideas. – Deane Yang Jun 19 2010 at 1:39
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Deligne-Illusie prove degeneration of the Hodge-de Rham spectral sequence for smooth projective varieties using purely algebraic methods. Then probably the corresponding result in the analytic category follows by GAGA or some such thing.

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But one has to prove that complex conjugation splits the Hodge filtration. Is that obvious just from knowing the dimensions add up correctly (which is what we'd get from GAGA)? – Boyarsky Jun 15 2010 at 16:21

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This is great, but hard, and unfortunately seems not an ideal alternate first route into Hodge theory. – Ravi Vakil Jun 15 2010 at 16:52

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Plus, to prove GAGA you need to know that the cohomology groups of a sheaf over a compact manifold are finite-dimensional. I've only seen this proved as an application of some hardcore functional analysis. – Gunnar Magnusson Jun 16 2010 at 8:08

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Just to point out the reference: Voisin mentions this in her Hodge Theory book, too, on page 206. She also mentions GAGA there but the complex conjugation property doesn't seem to follow from this (otherwise she would have mentioned that). – Konrad Voelkel Jun 16 2010 at 16:54

The Hodge decomposition can be proved, in a very nice, abstract-functional-analysis setting, on so-called Hilbert complexes. Brüning and Lesch wrote an excellent paper on the topic in J. Funct. Anal., first developing the theory on arbitrary Hilbert complexes, and then discussing the application to elliptic complexes.

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