The following came up when reading the definition of the moduli stack of principally polarized abelian varieties in [1].

Let $\pi_1:A_1 \to S_1$ and $\pi_2: A_2 \to S_2$ be abelian schemes over $S_i$, i.e. each $A_i$ is a group scheme over $S_i$, such that $\pi$ is smooth, proper and has geometrically connected fibers. Suppose also that $\lambda_i$ are principal polarizations, i.e. isomorphisms $\lambda_i: A_i \to \hat A_i$.

In [1] Faltings and Chain define a morphism $(A_1, \lambda_1) \to (A_2, \lambda_2)$ to be a homomorphism $$\mu: A_1 \to A_2$$ such that $\mu^*(\lambda_2) = \lambda_1$.

First, I think the authors omitted that we also need a morphism $f: S_1 \to S_2$, such that $\pi_2 \circ \mu = f \circ \pi_1$, right?

But mostly I wonder what exactly $\mu^*(\lambda_2)$ means. The first thing that came to my mind is to take it as the composition $$\mu^*(\lambda_2): A_1 \xrightarrow{\mu} A_2 \xrightarrow{\lambda_2} \hat A_2 \xrightarrow{\hat \mu} \hat A_1,$$ however I'm not sure if $\hat \mu$ is even well-defined, since $A_1$ and $A_2$ do not have a common base. Sure, we could consider $A_1$ as a scheme over $S_2$ via $f$, but depending on the properties of $f$, $A_1$ will not be an abelian scheme over $S_2$. For example, why should it still be smooth?

Any help would be appreciated :)

[1] Faltings, Chai; Degeneration of Abelian Varieties

  • $\begingroup$ I'm not sure if this is what they mean, but you can consider $A_1\times S_2$ and $A_2\times S_1$ over $S_1\times S_2$ and then everything make sense. $\endgroup$
    – ali
    Commented Oct 21, 2021 at 21:09

2 Answers 2


In terms of functors of points you get $\hat\mu$ because line bundles algebraically equivalent to zero pull back to line bundles algebraically equivalent to zero.

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    Commented Sep 21, 2021 at 16:32
  • $\begingroup$ What I find confusing is that by definition the functors are defined over different categories (schemes over $S_1$ and schemes over $S_2$). So we get a natural transformation $\mu^*: Pic^0_{A_2/S_2} \to Pic^0_{A_1/S_2}$, and for $T_1 \to S_1$ we have $Pic^0_{A_1/S_2}(T_1) = Pic^0_{A_1/S_1}(T_1)$. But I'm not sure if $A_1$ even represents $Pic^0_{A_1/S_2}$? $\endgroup$ Commented Sep 22, 2021 at 8:46
  • $\begingroup$ I mean $\hat A_1 \to S_2$ should represent $Pic^0_{A_1/S_2}$. $\endgroup$ Commented Sep 22, 2021 at 8:52

Yes, I think there is implicitly a morphism $f \colon S_1 \rightarrow S_2$ as you say.

When Faltings--Chai write $\mu^*(\lambda_2)$ I believe they mean the following. By pullback along $f$, the map $\lambda_2$ defines a principal polarization of $A_2 \times_{S_2} S_1$, which we call $f^* \lambda_2$. The map $\mu$ induces a homomorphism of abelian schemes $\mu' \colon A_1 \rightarrow A_2 \times_{S_2} S_1$ (over the same base $S_1$). Then $\mu^*(\lambda_2)$ should mean $\mu^*(f^*\lambda_2)$, i.e. the composite

$$A_1 \xrightarrow{\mu'} A_2 \times_{S_2} S_1 \xrightarrow{f^* \lambda_2} \widehat{(A_2 \times_{S_2} S_1)} \xrightarrow{\hat{\mu'}} \hat{A}_1.$$

To summarize, a morphism $\mu \colon (A_1, \lambda_1) \rightarrow (A_2, \lambda_2)$ consists of a morphism of schemes $f \colon S_1 \rightarrow S_2$ and a morphism of principally polarized abelian schemes $(A_1, \lambda_1) \rightarrow f^*(A_2, \lambda_2)$ (which, as Faltings--Chai remarks, is forced to be an isomorphism when $A_1$ and $A_2$ have the same relative dimension over their respective bases).


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