# What is the pull-back of a polarization of abelian schemes over different bases?

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

• 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.
– ali
Oct 21, 2021 at 21:09

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.

• 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 $Pic^0_{A_1/S_2}$? Sep 22, 2021 at 8:46
• I mean $\hat A_1 \to S_2$ should represent $Pic^0_{A_1/S_2}$. 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).