My question concerns the Picard-Functor for abelian schemes, as it is for example introduced in the book of Birkenhake and Lange "Complex Abelian Varieties" in Appendix F.

So let $X\rightarrow S$ be an abelian scheme over a locally noetherian basis $S$. Let $\mathcal L$ be a line bundle on $X$. It is called rigidified along the zero section $e$ of $X$ if one has an isomorphism $e^*L \simeq \mathcal O_S$.

For each locally noetherian $T\rightarrow S$ they now define the following abelian group

$Pic_{X/S}(T)$:= Isomorphism classes of line bundles on $X\times_S T$ rigidified along the zero section of $X\times_S T$.

Then they say that the resulting functor is representable by the Picard scheme etc. Similar discussion for $Pic^0$.

Now my question:

What do they exactly mean? Do they mean:

(1) the group of iso classes of line bundles, which additionally have the property of being rigidified; an iso here just an iso of line bundles (though both have the addition of being rigidified)


(2) the group of iso classes of pairs $(\mathcal L, \phi)$, where $\mathcal L$ is a line bundle on $X\times_S T$, and $\phi$ is a rigidification of it. Here an isomorphism of such a pair is defined in the obvious way.

Addendum: Even with the answers below one point still isn't clear to me:

Take an abelian variety $X$ over a field $k$. Then Mumford defines the dual $\hat{X}$ of it by saying that $Hom_k(S,\hat{X}) \simeq$ Isoclasses of line bundles on $X\times S$ which (i) are in $Pic^0(X)$ fibrewise with respect to fibers of $S$ and (!) (ii) $\mathcal L$ is trivial on $0\times S$. This definitely doesn't look like (2), but rather (1). My problem is that if he meant (2), then for the same bundle satisfying (i) and (ii) one would get different morphisms $S \rightarrow \hat{X}$, as one can change the rigidification always by a nonzero scalar in $k$. But he always uses the dual of $X$ as if just the data (i) and (ii) would induce a unique morphism $S\rightarrow \hat{X}$.

To sum it up: my problem is that in the case (2) one gets much more morphisms to the representing object than in the other case (1), and this can't be true I think. So where is my fault?

  • 1
    $\begingroup$ They probably mean (2). First of all, these are essentially the same. If $L$ is "rigidifiable" by $\phi$, and if $\psi$ is a second rigidification, then there exists a global section $u$ of $\mathcal{O}^*$ on $S$ such that $\psi$ equals $u\phi$. Thus scaling $L$ by $\pi^* u$ gives an isomorphism from $(L,\phi)$ to $(L,\psi)$. But the formulation in (2) more obviously satisfies glueing. $\endgroup$ Nov 15, 2011 at 14:06
  • $\begingroup$ @Jason: Why not posting this as an answer? $\endgroup$ Nov 15, 2011 at 14:18
  • $\begingroup$ They mean (2). See "Néron models", chap. 8, 8.2, Prop. 4, p. 211. The point is that the rigidification prevents a (rigidified) line bundle from having a non-trivial automorphism. Otherwise the moduli problem is not representable by a scheme (algebraic stacks are introduced in part to deal with such issues - this is also explained in Jason Starr's comment). $\endgroup$ Nov 15, 2011 at 15:46
  • $\begingroup$ But in the one case, namely (2), I get much more morphisms to the representing object. This seems strange... I have added an addendum which explains the problem in the case of an abelian variety and it's dual. $\endgroup$
    – Veen
    Nov 15, 2011 at 17:05
  • $\begingroup$ If you change the rigidification by an invertible scalar you get an isomorphic rigidification by letting the scalar act on $L$. $\endgroup$ Nov 15, 2011 at 17:23

1 Answer 1


They mean (1) (sort of).

There is a canonical choice of the rigidification $\phi$, so being rigidified doesn't mean there is some map which takes it to the structure sheaf, it is predetermined. Explicitly, the rigidification along the zero section means that $(\varepsilon \circ f, 1_T)^*(L) \cong \mathcal{O}_T$, where $f : T \rightarrow S$ defines $X \times_S T$.

See Mumford's Geometric Invariant Theory, section (d) of 0.5 (pg. 22) for a reference.

  • $\begingroup$ Why is the isomorphism predetermined? One could easily compose a given isomorphism with any isomorphism $\mathcal O_T \to \mathcal O_T$. $\endgroup$ Sep 6, 2021 at 11:09

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