Global Proof of Serre Duality - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T12:29:44Z http://mathoverflow.net/feeds/question/6111 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/6111/global-proof-of-serre-duality Global Proof of Serre Duality John McCarthy 2009-11-19T15:25:00Z 2009-11-19T21:42:31Z <p>Does anyone know of a global proof (involving no local argument) of Serre Duality at the level of varieties or manifolds (as opposed to schemes).</p> http://mathoverflow.net/questions/6111/global-proof-of-serre-duality/6116#6116 Answer by Ben Webster for Global Proof of Serre Duality Ben Webster 2009-11-19T15:45:59Z 2009-11-19T15:45:59Z <p>Have you looked at the one in Griffiths and Harris? That's at least rather different from the general nonsense one in Hartshorne.</p> http://mathoverflow.net/questions/6111/global-proof-of-serre-duality/6120#6120 Answer by David Speyer for Global Proof of Serre Duality David Speyer 2009-11-19T15:56:07Z 2009-11-19T15:56:07Z <p>You might like the proof in section 5.3 of Voisin's book <em>Hodge theory and complex algebraic geometry</em>.</p> http://mathoverflow.net/questions/6111/global-proof-of-serre-duality/6149#6149 Answer by Spinorbundle for Global Proof of Serre Duality Spinorbundle 2009-11-19T18:52:49Z 2009-11-19T18:52:49Z <p>Since a $\bar{\partial}$-Operator is an elliptic Operator, you can use elliptic theory in order to prove the Serre duality. In fact the Serre duality is a kind of corollary of the"fundamental theorem" (as I Know it). In fact, you do not need the Hodge theorem, since the Hodge theorem itself is a corollary of the theorem. </p> <p>For a reference of this "fundamental theorem" (perhaps slightly reformulated) I would refer to one of the following:</p> <ul> <li>Wells, Differential Analysis on Complex Manifolds;</li> <li>Gilkey, Invariance Theory, the Heat Equation and the Atiyah Singer Index Theorem</li> </ul> <p>(a complete proof for pseudodifferential operators)</p> <ul> <li>Warner, Foundations of Differentiable Manifolds</li> </ul> <p>(the theorem is included as an exercise on the last page</p> <ul> <li>Kazdan, Lecture Notes on Applications of Partial Differential Equations to Some Problems in Differential Geometry (online available <a href="http://www.math.upenn.edu/~kazdan/japan/japan.pdf" rel="nofollow"> here</a>) </li> </ul> <p>(Corollary 2.5, for a sketch of the proof)</p> http://mathoverflow.net/questions/6111/global-proof-of-serre-duality/6158#6158 Answer by lemega for Global Proof of Serre Duality lemega 2009-11-19T19:54:32Z 2009-11-19T19:54:32Z <p>I like the presentation from Analytic methods in algebraic geometry by Demailly. Here is the link: <a href="http://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/eem2007.pdf" rel="nofollow">http://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/eem2007.pdf</a>.</p> http://mathoverflow.net/questions/6111/global-proof-of-serre-duality/6178#6178 Answer by Greg Stevenson for Global Proof of Serre Duality Greg Stevenson 2009-11-19T21:42:31Z 2009-11-19T21:42:31Z <p>I thought I'd offer a high-tech alternative for certain varieties. If $X$ is smooth and projective over a field $k$ then <a href="http://front.math.ucdavis.edu/0204.5218" rel="nofollow">Bondal and van den Bergh</a> give a proof here that $D^b(\mathrm{Coh}X)$ is saturated which is a strong representability condition on cohomological/homological functors to the category of $k$ vector spaces. It follows immediately that $D^b(\mathrm{Coh}X)$ has a Serre functor by using the fact that $Hom(A,-)^*$ is representable for every bounded complex of coherent sheaves $A$.</p>