Proving Hodge decomposition without using the theory of elliptic operators? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T21:27:57Z http://mathoverflow.net/feeds/question/28265 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/28265/proving-hodge-decomposition-without-using-the-theory-of-elliptic-operators Proving Hodge decomposition without using the theory of elliptic operators? Konrad Voelkel 2010-06-15T15:42:12Z 2010-06-24T13:23:59Z <p>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:</p> <blockquote> <p>Let P : E &rarr; 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:</p> <p>smooth sections of E = Ker P &oplus; Im P*</p> <p>(where P* is the formal L²-adjoint of P)</p> </blockquote> <p>In the case of Hodge Theory, we consider the elliptic self-adjoint operator &Delta;, the Laplacian (or: Hodge-Laplacian, or Laplace-Beltrami operator).</p> <p>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!</p> <hr> <p>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), <strong>I wonder if this "functional analysis" is really always necessary</strong>.</p> <p><strong>Do you know of any class of complex manifolds</strong> (most likely some restricted class of complex projective varieties) <strong>where you can get the theorem above without using the theory of elliptic operators</strong> (or at least, where you can simplify the proofs that much that you don't notice you're working with elliptic operators)<strong>?</strong> Maybe the general theorem really requires functional analysis (I think so), but the <strong>Hodge decomposition might follow from other arguments</strong>.</p> <p>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).</p> <p>If this ends up being argumentative and subjective, feel free to use the community wiki hammer.</p> http://mathoverflow.net/questions/28265/proving-hodge-decomposition-without-using-the-theory-of-elliptic-operators/28269#28269 Answer by Kevin Lin for Proving Hodge decomposition without using the theory of elliptic operators? Kevin Lin 2010-06-15T16:10:30Z 2010-06-15T16:10:30Z <p>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.</p> http://mathoverflow.net/questions/28265/proving-hodge-decomposition-without-using-the-theory-of-elliptic-operators/28307#28307 Answer by Andy Putman for Proving Hodge decomposition without using the theory of elliptic operators? Andy Putman 2010-06-15T19:52:50Z 2010-06-15T19:52:50Z <p>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.</p> http://mathoverflow.net/questions/28265/proving-hodge-decomposition-without-using-the-theory-of-elliptic-operators/28549#28549 Answer by Donu Arapura for Proving Hodge decomposition without using the theory of elliptic operators? Donu Arapura 2010-06-17T18:19:49Z 2010-06-24T13:23:59Z <p>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.</p> <p>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 </p> <p>$$closed(X)= exact(X)\oplus exact(X)^\perp$$</p> <p>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:</p> <p>$$\overline{closed(X)}= \overline{exact(X)}\oplus \overline{exact(X)}^\perp$$</p> <p>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.</p> <p>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. </p> <p>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</p> <p><a href="http://www.math.purdue.edu/~dvb/preprints/book.pdf" rel="nofollow">http://www.math.purdue.edu/~dvb/preprints/book.pdf</a></p> <p>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.</p> <p><strong>Addendum</strong> (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 <em>opposed</em> 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 $[\frac{dz}{2\pi iz}]$.</p> http://mathoverflow.net/questions/28265/proving-hodge-decomposition-without-using-the-theory-of-elliptic-operators/28692#28692 Answer by Ari for Proving Hodge decomposition without using the theory of elliptic operators? Ari 2010-06-18T23:39:39Z 2010-06-18T23:39:39Z <p>The Hodge decomposition can be proved, in a very nice, abstract-functional-analysis setting, on so-called <em>Hilbert complexes</em>. Brüning and Lesch wrote an <a href="http://dx.doi.org/10.1016/0022-1236%2892%2990147-B" rel="nofollow">excellent paper</a> on the topic in J. Funct. Anal., first developing the theory on arbitrary Hilbert complexes, and then discussing the application to elliptic complexes.</p>