Timeline for A question on deformations of Theta divisor in the Jacobian of a complex curve
Current License: CC BY-SA 3.0
7 events
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| Apr 28, 2012 at 15:48 | comment | added | roy smith | I am glad this helped. Note also that a similar argument, taking the g fold pontryagin product of the class of the curve, and dividing by g!, shows that the class of the image of the abel map from C^(g) is the fundamental class of the jacobian. Thus the g fold abel map has degree one. This implies "Jacobi inversion" by topological methods as well. By the easy connectedness theorem for fibers of this map to a smooth target, and the weak direction of abel's theorem, and Riemann Roch, the fibers are finite unions of projective spaces, so the converse of abel follows too. | |
| Apr 28, 2012 at 13:47 | comment | added | aglearner | Dear Roy, thanks a lot for the details! This makes a perfect sense for me. | |
| Apr 28, 2012 at 3:30 | comment | added | roy smith | A reference should be chapter 11.2, p. 328, of Birkenhake and Lange's Complex abelian varieties. | |
| Apr 28, 2012 at 3:20 | comment | added | roy smith | I was afraid you would ask that. Ok, first the curve embeds in its jacobian with homology class A1xB1+...AgxBg, where Ai,Bj is a symplectic homology basis of H^1 of the curve. Then the theta divisor is the image of the symmetric product of the curve g-1 times, which is covered by a degree (g-1)! map by the cartesian product of the curve. I claim it follows that the theta divisor has homology class = to 1/(g-1)! times the (g-1) fold pontyragin product of the class of the curve, i.e. A1xB1x...xAg-1xBg-1 + .... +A2xB2x...AgxBg. Then you have to check this is Poincare dual to what I said. | |
| Apr 28, 2012 at 0:28 | comment | added | aglearner | Dear Roy, I have to say, I spent a lot of time today trying to understand how one could prove topologically that $\Theta^g=g!$, and I failed... Why is the Poincare dual given by the form you wrote down? | |
| Apr 27, 2012 at 19:30 | comment | added | roy smith | Since the fact in the corollary is purely topological it also has a topological proof. From the definition of Theta given above, its homology class is Poincare dual to the deRham class dX1^dY1 +...+dXg^dYg, where the one - forms dXi and dYj are Poincare dual to a symplectic homology basis on the curve. Hence the self intersection number of Theta is the coefficient of the g fold wedge product of this two form, i.e. g! | |
| Apr 27, 2012 at 10:17 | history | answered | inkspot | CC BY-SA 3.0 |