# Poincaré bundle and Weil pairing for Abelian schemes

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?)

• For the perfectness of the Weil pairing, see Oda "The first de Rham cohomology and Dieudonne modules", esp. Thm. 1.1. – Kestutis Cesnavicius Mar 7 '14 at 13:22
• 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. – user76758 Mar 7 '14 at 13:47

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.

• Thank you. Do you have a reference for the existence of the Poincaré bundle? – TKe Mar 7 '14 at 12:50
• I think I've found it in [FGA explained, Kleiman, The Picard scheme], p. 262, Exercise 9.4.3. – TKe Mar 7 '14 at 13:02
• 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). – Kestutis Cesnavicius Mar 7 '14 at 13:20
• 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). – user76758 Mar 7 '14 at 13:50

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?

• 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)? – user76758 Mar 7 '14 at 13:44
• 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. – TKe Mar 7 '14 at 13:47
• 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? – user76758 Mar 7 '14 at 13:55
• 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. – TKe Mar 7 '14 at 14:01
• 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$^0$?). – user76758 Mar 7 '14 at 14:41