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S Nov 10, 2013 at 14:03 history bounty ended IMeasy
S Nov 10, 2013 at 14:03 history notice removed IMeasy
Nov 9, 2013 at 18:26 vote accept IMeasy
Nov 7, 2013 at 11:34 comment added naf In the tame case, my claim follows from Lemma 2.3 of the Fulton--Olsson paper.
Nov 7, 2013 at 4:13 answer added Jonathan Wise timeline score: 7
Nov 6, 2013 at 15:51 vote accept IMeasy
Nov 6, 2013 at 15:51
Nov 5, 2013 at 0:33 answer added Puzzled timeline score: 2
Nov 4, 2013 at 20:34 comment added IMeasy I really would love to see a proof. Do you have a reference?
Nov 4, 2013 at 15:17 comment added Lennart Meier The general statement is, I think, the following: Let $f:\mathcal{X} \to X$ be the map from a DM-stack to its coarse moduli stack, $L$ a line bundle on $\mathcal{X}$ and $n$ the lowest common multiple of the orders of all automorphism groups. Then $f_*L^{\otimes n}$ is a line bundle and $L^{\otimes n} \to f^*f_*L^{\otimes n}$ an isomorphism. At least, I know a proof in the case that $\mathcal{X}$ is separated and of finite type over a noetherian base scheme.
S Nov 4, 2013 at 12:42 history bounty started IMeasy
S Nov 4, 2013 at 12:42 history notice added IMeasy Draw attention
Nov 2, 2013 at 11:09 comment added naf It is easy to see that the 12'th tensor power of the Hodge bundle--and no smaller tensor power--descends to the coarse moduli space: consider the possible automorphism groups of elliptic curves and their actions on the tangent space at the identity.
Nov 2, 2013 at 0:30 comment added S. Carnahan You can also consider the generator $\Delta$ of Pic on the coarse space, and lift to the degree 24 cover $X(3) \to \overline{\mathcal{M}_{1,1}}$. I think degree of $\omega_P$ for a representable elliptic moduli problem $P$ is computed somewhere in Katz-Mazur.
Nov 1, 2013 at 23:14 comment added Lennart Meier If I remember correctly, Mumford only does it over a field not of characteristic 2 or 3. The general reference is Fulton and Olsson: arxiv.org/abs/0704.2214
Nov 1, 2013 at 20:36 comment added IMeasy $Pic(\mathcal{M}_{1,1})$ is $\mathbb{Z}/12\mathbb{Z}$ (see Mumford). It is $Pic(\overline{\mathcal{M}}_{1,1})$ that is $\mathbb{Z}$. This is pretty standard, see for instance sect.6 of arxiv.org/abs/0812.1803v2
Nov 1, 2013 at 18:11 history edited IMeasy CC BY-SA 3.0
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Nov 1, 2013 at 17:06 comment added Joël That seems reasonable. But how do you know that $Pic \mathcal M_{1,1} = \mathbb Z$? You have a proof or reference?
Nov 1, 2013 at 14:15 history asked IMeasy CC BY-SA 3.0