Let $A$ be an abelian variety over a field $k$ with group operation $m\colon A\times A\to A$, and let $A'$ be the dual abelian variety. I know that $A'(k)$ is isomorphic to the subgroup $\operatorname{Pic}^0(A)$ of $\operatorname{Pic}(A)$ composed of the line bundles $\mathscr{L}$ satisfying $m^\*\mathscr{L}\simeq \mathscr{L}\boxtimes \mathscr{L}$.


Now, let $S$ be a $k$-scheme. Can we give a similar description to $A'(S)$ using the group operation $m\_S\colon A\_S\times\_S A\_S\to A\_S$ of $A\_S = A\times S$? For example, is it true that $A'(S)$ is isomorphic to the subgroup of $\operatorname{Pic}(A\_S)$ composed of the line bundles $\mathscr{L}$ satisfying $m\_S^\*\mathscr{L}\otimes p\_S^\*e\_S^\*\mathscr{L}\simeq \mathscr{L}\boxtimes \mathscr{L}$, where $e\_S\colon S\to A\_S$ is the identity section and $p\_S\colon A\_S\times\_S A\_S\to S$ is the structure map?


(I know that $A'(S)$ is the group of isomorphism classes of invertible sheaves on $A\_S$ with rigidification along $e\_S$ whose fibers are on $\operatorname{Pic}^0$. But I'm looking for a "global", not fiber-wise, description more or less on the same lines as what I propose above.)



If $A$ is an abelian variety over $k$ and $S$ is a $k$-scheme, what's $A'(S)$ geometrically?
Let $f:A \to \operatorname{Spec}k$ be your abelian variety and let $P:=\operatorname{Pic}\_{A/k}$ denote the relative Picard functor of $A/k$, i.e. the functor $(Sch/k)^{op} \to Sets$ taking $S$ to $\operatorname{Pic}(A\_S)/\operatorname{Pic}(S)$. This functor is an fppf sheaf and coincides with the rigidified Picard functor along the unit section. Write $P^0:=\operatorname{Pic}^0\_{A/k}$ for the fiberwise-connected component of identity of $P$. As you mentioned, $A'$ represents $P^0$.


Now, let $\mathcal L$ be a $S$-section of $P$. I claim that $\mathcal L$ is in $P^0(S)$ if and only if $m\_S^\*\mathcal L = \mathcal L \boxtimes \mathcal L$ modulo $(f\times f)^\*\operatorname{Pic}(S)$. Using the fact that $\mathcal L$ is in $P^0$ if and only if all its fibres are, the claim reduces to proving that $m\_S^\*\mathcal L = \mathcal L \boxtimes \mathcal L$ $\operatorname{mod} (f\times f)^\*\operatorname{Pic}(S)$ holds if and only if it holds on fibres. The "only if" is clear so let's prove the "if": from now on we assume $(\mathcal L \boxtimes \mathcal L) \otimes m\_S^\*\mathcal L^\vee$ has trivial fibres.


Let $e,\sigma: S \to \operatorname{Pic\_{A\_S\times\_S A\_S/S}}$ be respectively the unit section and the section corresponding to the line bundle $(\mathcal L \boxtimes \mathcal L) \otimes m\_S^\*\mathcal L^\vee$. The base change $e^\*\sigma\colon T \to S$ is a closed immersion since $e$ is (because $\operatorname{Pic\_{A\_S\times\_S A\_S/S}}$ is separated over $S$). By hypothesis the fibres of this closed immersion are isomorphisms, so $e^\*\sigma$ itself is an isomorphism and we are done.


