most recent 30 from http://mathoverflow.net 2013-05-24T08:25:02Z http://mathoverflow.net/feeds/question/1765 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/1765/kodaira-spencer-theory-and-moduli-of-curves David Zureick-Brown 2009-10-22T00:20:42Z 2009-10-22T01:45:55Z

<p>I was looking at a <a href="http://www.google.com/url?sa=t&amp;source=web&amp;ct=res&amp;cd=1&amp;ved=0CA4QFjAA&amp;url=http%3A%2F%2Fwww.mathematik.hu-berlin.de%2F~farkas%2Fsdg-2.pdf&amp;ei=3KTfSoXmFo-0sgPoydHhCA&amp;usg=AFQjCNHJWSHKXWGpaxtDsY6CxCJEDGYIBg&amp;sig2=0Yq2ygZ4ikWs5tBgAakIeA" rel="nofollow">paper</a> of Farkas and the following confusing point came up.</p> <p>Let M_g be the moduli stack of smooth genus g curves and let pi:C \to M_g be the universal curve. Let F be \Omega^1_\pi \otimes \Omega^1_\pi, where \Omega^1_\pi is the sheaf of relative differentials of \pi. Then the pushforward \pi_* F is isomorphic \Omega^1_{M_g}.</p> <p>Why is this true? Farkas says this follows from Kodaira-Spencer theory. I googled for a while and asked a few students, but couldn't figure this out.</p>

http://mathoverflow.net/questions/1765/kodaira-spencer-theory-and-moduli-of-curves/1779#1779 S. Carnahan 2009-10-22T01:45:55Z 2009-10-22T01:45:55Z

<p>By standard deformation theory (see e.g., Hartshorne III Ex 4.10, but there are probably better references), the tangent sheaf of M_g is R¹pi_{<code>*</code>}(C, T_C/Mg), which is Serre dual to pi_{<code>*</code>}F. The tangent sheaf is dual to what you wanted.</p>