0
$\begingroup$

In which situations is there a Poincaré bundle for Abelian schemes? In [Mumford, Abelian varieties] only the case of Abelian varieties is treated.

The same question for the Weil pairing $\mathscr{A}[n] \times \mathscr{A}^\vee[n] \to \mu_n$. (Why is it a perfect pairing?)

$\endgroup$
  • 1
    $\begingroup$ For the perfectness of the Weil pairing, see Oda "The first de Rham cohomology and Dieudonne modules", esp. Thm. 1.1. $\endgroup$ – Kestutis Cesnavicius Mar 7 '14 at 13:22
  • 1
    $\begingroup$ Oort's book "Commutative group schemes" has a very nice discussion of both the representability of functor $T \mapsto {\rm{Ext}}^1_T(A_T, {\mathbf{G}}_m)$ by the dual abelian scheme when the latter exists (which is always the case, by the result of Raynaud) and not only the relation of its $n$-torsion with Cartier dual of that of $A$ but also the more subtle issue of relating double-duality on both sides. Oda's paper addresses the double-duality aspect (and much more) in its first section. $\endgroup$ – user76758 Mar 7 '14 at 13:47
3
$\begingroup$

Always, because the dual abelian scheme/space can be defined as the connected component of the (fine) moduli space of invertible sheaves trivialized at $0$. The poincare bundle is the universal object.

PS: Defined as above it is clear from general theory that the dual abelian something is an algebraic space. It was shown I think by Raynaud that it is a scheme in most cases of interest (for example if the abelian scheme is projective over the base, so for example over a normal base scheme), but I think later that fact was established in general (not 100% sure). This is related to the question of whether any abelian algebraic space over a base is representable.

$\endgroup$
  • $\begingroup$ Thank you. Do you have a reference for the existence of the Poincaré bundle? $\endgroup$ – TKe Mar 7 '14 at 12:50
  • $\begingroup$ I think I've found it in [FGA explained, Kleiman, The Picard scheme], p. 262, Exercise 9.4.3. $\endgroup$ – TKe Mar 7 '14 at 13:02
  • 1
    $\begingroup$ The representability of the dual abelian scheme (by a scheme) is discussed in Chai, Faltings "Degeneration of abelian varieties", section I.1 (esp. Thm. 1.9). $\endgroup$ – Kestutis Cesnavicius Mar 7 '14 at 13:20
  • 1
    $\begingroup$ The Poincar\'e bundle is tautologically part of the very meaning of "dual abelian scheme" or representability of the (rigidified) Picard functor (say as an algebraic space, which in turn is a special case of Artin's theorem on Picard functors). $\endgroup$ – user76758 Mar 7 '14 at 13:50
0
$\begingroup$

I've found it in [FGA explained, Kleiman, The Picard scheme], p. 262, Exercise 9.4.3.

A universal sheaf/Poincaré sheaf exists iff $\mathbf{Pic}_{X/S}$ represents $\mathrm{Pic}_{X/S}$ or if $f: \mathscr{A} \to S$ has a section.

Edit: This gives us a Poincaré bundle on $\mathscr{A} \times \mathbf{Pic}_{\mathscr{A}/S}$, but I need it on $\mathscr{A} \times \mathbf{Pic}^0_{\mathscr{A}/S}$! Perhaps [FGA explained, Kleiman, The Picard scheme], p. 289, Remark 9.5.24 does help?

$\endgroup$
  • 1
    $\begingroup$ The functor ${\rm{Pic}}^0_{A/S}$ is a subfunctor of ${\rm{Pic}}_{A/S}$ (defined by a condition on geometric fibers), so what is the meaning of the question in the "Edit" that isn't a tautology (via pullback)? $\endgroup$ – user76758 Mar 7 '14 at 13:44
  • $\begingroup$ The question is if the universal property still holds (for the modified Poincaré bundle as in [FGA explained, Kleiman, The Picard scheme], p. 289, Remark 9.5.24. $\endgroup$ – TKe Mar 7 '14 at 13:47
  • 1
    $\begingroup$ If you have a given abelian scheme $B$ with line bundle $L$ on $A \times B$ equipped with trivialization $i$ of its pullback to $A \times \{0\}$ then to check if the resulting map $B \rightarrow A^{\vee}$ is an isomorphism (thereby giving the universal property to $(B, L, i)$ it suffices to check on geometric fibers, where various results in Mumford's book are applicable. I don't know what "modified Poincar\'e bundle" means (FGA Explained not nearby at the moment), but would that address whatever is concerning you? $\endgroup$ – user76758 Mar 7 '14 at 13:55
  • $\begingroup$ The normalised Poincaré bundle is $\mathscr{P} \otimes f_{A^\vee}^*g_{A^\vee}^*\mathscr{P}$ with $f: A \to X$ and $g$ the zero section. $\endgroup$ – TKe Mar 7 '14 at 14:01
  • 1
    $\begingroup$ OK, this is what I think is usually called the "rigidified" Poincar\'e bundle (typo: you missed an inversion on the 2nd tensor factor). But encoding such rigdification is part of the very content of building a Poincar\'e sheaf on the entire Picard scheme (or algebraic space), so I remain puzzled as to where the point of confusion is arising for Pic$^0$ versus Pic in terms of Poincar\'e bundles (i.e., if you are happy for Pic then why not for Pic$^0$?). $\endgroup$ – user76758 Mar 7 '14 at 14:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.