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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

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  • $\begingroup$ 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
    Oct 21, 2021 at 21:09

2 Answers 2

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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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    $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    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 $Pic^0_{A_1/S_2}$? $\endgroup$ Sep 22, 2021 at 8:46
  • $\begingroup$ I mean $\hat A_1 \to S_2$ should represent $Pic^0_{A_1/S_2}$. $\endgroup$ Sep 22, 2021 at 8:52
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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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