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Hi all

I have a question from a statement made in van der Geer's paper "Cycles on the Moduli Space of Abelian Varieties".

The statement is as follows. In the paper we are looking at the Hodge bundle $\mathbb{E}$, which as I'm sure you know is the pushforward of the sheaf of relative differentials on the universal abelian variety $X_g$ over $A_g$ (viewed as a stack over $\mathbb{Z}$ or with as a scheme with sufficiently high level structure). The statement is then "Since the cohomology of an abelian variety is the exterior algebra on $H^1$, the derived sheaf $\pi_!(\mathcal{O}_{X_g}) = \sum_{i=1}^g (-1)^i\wedge^i \mathbb{E}^\vee$."

I think the answer will be obvious to someone more experienced with abelian schemes (certainly van der Geer must think it obvious since he makes no remarks about it) as it seems to just be a generalization of facts about abelian varieties to the more general scheme case (namely that for an abelian variety $X$ over a field $k$ we have a perfect pairing $H^1(X,\mathcal{O}_X) \times H^0(X, \Omega^1) \rightarrow k$ and that $H^*(X,\mathcal{O}_X) = \wedge H^1(X,\mathcal{O}_X)$). However not having any experience with stacks or abelian schemes I have spent a good week searching through the literature trying to find any reference that would give me the desired generalizations but to no avail. If anyone knows where I could find results that would explain van der Geer's statement that would be greatly appreciated.

Thank you.

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  • $\begingroup$ For a reference, look into the first chapter of Berthelot-Breen-Messing, "Théorie de Dieudonné...", Lecture Notes in Math. 930. Mind also that the result as you state it is true if the abelian scheme is principally polarized (or endowed with an étale polarization) but probably not otherwise. The idea is that the result for abelian varieties implies that the cohomology of an abelian scheme is locally free and compatible with base-change by the semi-continuity theorem. Then you consider the morphism of sheaves $\Lambda^i(R^1\pi_*{\cal O}_{A_g})\to R^i\pi_*{\cal O}_{A_g}$ given by the cup product $\endgroup$
    – Damian Rössler
    Commented Sep 13, 2012 at 8:08
  • $\begingroup$ ...and you conclude that it is an isomorphism, because its base-change to any fibre is an isomorphism (use Nakayama). To conclude, notice that the polarization gives an isomorphism $E^\vee\simeq R^1\pi_*{\cal O}_{A_g}$. $\endgroup$
    – Damian Rössler
    Commented Sep 13, 2012 at 8:08
  • $\begingroup$ Thank you very much. That makes sense and the reference was useful. $\endgroup$
    – user18844
    Commented Sep 14, 2012 at 16:44

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