Timeline for complex deformations of abelian varieties
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Jan 19, 2017 at 15:14 | history | edited | Francesco Polizzi | CC BY-SA 3.0 |
added 12 characters in body
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| Apr 18, 2012 at 2:22 | comment | added | Hugo Chapdelaine | Dear Matthew, so when you say that abelian varieties are rigid do you mean that you are looking for complex deformations that preserve their endomorphism ring? If yes, then how does this fit into classical deformation theory "a la Kodaira-Spencer"? | |
| Apr 17, 2012 at 2:36 | comment | added | Emerton | Dear Hugo, Regarding your question about CM abelian varieties: yes, one can see that they are rigid, and hence defined over number fields. (One way to see it is just in terms of the period lattice.) This is a standard fact and standard argument; it is ubiquitous in the Shimura variety literature, but I don't know where else it is discussed. (Perhaps in Shimura and Taniyama's book?) Best wishes, Matthew | |
| Apr 16, 2012 at 13:09 | vote | accept | Hugo Chapdelaine | ||
| Apr 16, 2012 at 12:44 | answer | added | Donu Arapura | timeline score: 7 | |
| Apr 16, 2012 at 4:26 | comment | added | Emerton | Typo: "$H^1(A,\mathcal A)$" should read "$H^1(A, \mathcal O_A)$". | |
| Apr 16, 2012 at 4:26 | comment | added | Emerton | The tangent bundle $\Theta_A$ is trivial of rank $g$, as Donu notes, and so $\dim H^1(A,\Theta_A) = g H^1(A,\mathcal O_A)$. The fact that $H^1(A,\mathcal A)$ has dimension $g$ is a standard fact. One way to prove it is by Hodge symmetry: it has the same dimension as $H^0(A,\Omega^1_A)$, and the latter has dimension $g$ because an every holomorphic one-form on an abelian variety is necessarily invariant, and any $g$-dimensional Lie group has a $g$-dimensional space of invariant one-forms. Regards, | |
| Apr 16, 2012 at 0:39 | comment | added | Hugo Chapdelaine | Of course intuitively I see why we should have $g^2$ since one needs to choose $2g$ $R$-linearly vectors in $C^g$ so we see that the moduli space of $g$-dimensional complex tori should be something like $GL_{2g}(Z)\backspace M_{2g\times 2g}(R)/GL_g(C)$ so the complex dimension is $g^2$ but this is just a heuristic. But at the end one needs to invoke something in order to compute the dimension of this $H^1$ since it boils down to solve some system of differential equations. | |
| Apr 15, 2012 at 21:58 | comment | added | B R | This is also in Kodaira's "Complex Manifolds and Deformation of Complex Structures" for general complex tori (not necessarily algebraic). See pages 216-218 for the calculation. | |
| Apr 15, 2012 at 21:27 | comment | added | Donu Arapura | Hugo, sorry I have to run. The computation of the last thing should be in Mumford's abelian varieties for example. | |
| Apr 15, 2012 at 21:21 | comment | added | Hugo Chapdelaine | So with what you said you need to compute $H^1(A,\mathcal{O}_A)$ | |
| Apr 15, 2012 at 21:18 | comment | added | Hugo Chapdelaine | Thanks Donu for the quick answer. So could you give me more details on how you get the $g^2$? | |
| Apr 15, 2012 at 21:11 | comment | added | Donu Arapura | $\Theta_A$ is trivial of rank $g=\dim A$. So it's $H^1$ has dimensional $g^2$. Note that when $g>1$, this is bigger than the dimension of the moduli space of abelian varieties, in case you were wondering that. | |
| Apr 15, 2012 at 21:04 | history | asked | Hugo Chapdelaine | CC BY-SA 3.0 |