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I'm not sure whether this is a research-level question, but upon skimming through Mumford book of Abelian Varieties I noticed he gives this definition $$ \begin{equation} \label{eq} \text{Pic}^0(A)=\{\mathcal{L}\in\text{Pic}(A)\text{ such that }t_a^*\mathcal{L}\simeq\mathcal{L}\text{ for all }a\in A\} \end{equation} $$ for an abelian variety $A$. I would really like to find a proof that these are exactly the line bundles algebraically equivalent to zero, which avoids all sorts of representability results. My attemps so far have yielded only meagre results.

  1. Let $\mathcal{L}$ be as above, and $\mathcal{K}$ an ample line bundle. Then Mumford proves that for some $a\in A$ the sheaf $\mathcal{K}^{-1}\otimes t_a^*\mathcal{K}$ is isomorphic to $\mathcal{L}$, so the sheaf $m^*\mathcal{K}\otimes\mathcal{K}^{-1}$ is an algebraic deformation of $\mathcal{L}$ to $\mathcal{O}_A$.
  2. And let us now consider a sheaf $\mathcal{C}$ on a curve $C\times A$ which deforms $\mathcal{L}$ to $\mathcal{O}_A$ (for convenience I also set $\mathcal{C}_{C\times0}=\mathcal{O}_C$). The hope would be to show that every fibre of $\mathcal{C}$ is in Mumford's set, and this is where I've come short of ideas :/ An (infinitesimal) direction I explored was trying to show that $\mathcal{C}_a:=\mathcal{C}^{-1}\otimes (id\times t_a)^*\mathcal{C}$ has trivial fibres; this is true a posteriori and looks kind of cute, also because $\mathcal{C}_a$ has the nice property that, if I'm not mistaken, $$ \mathcal{C}_{a+b}\simeq (\text{id}\times t_b)^*(\mathcal{C}_a) $$ but I haven't managed any progress with it.

I would be equally satisfied of a heuristic argument which refutes the existence of such an elementary proof, actually.

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    $\begingroup$ Welcome new contributor. The operation of $A$ on first-order deformations $\mathcal{L}$ of $\mathcal{O}_A$ sending $\mathcal{L}$ to $\mathcal{L}^\vee\otimes_{\mathcal{O}_A} t_a^*\mathcal{L}$ induces a morphism from $A$ to $\textbf{GL}(H^1(A,\mathcal{O}_A))$. The domain is proper, and the target is affine, therefore the morphism is constant. The same argument applies for higher-order deformations as well. Thus, your sheaf $\mathcal{C}_a$ is isomorphic to $\mathcal{O}_{C\times A}$ to all infinitesimal orders near $\mathcal{O}_A$. Now use the theorem on formal functions. $\endgroup$
    – Jason Starr
    Commented Aug 1, 2023 at 10:14
  • $\begingroup$ . . . Correction: replace $\textbf{GL}$ by $\text{Hom}$. $\endgroup$
    – Jason Starr
    Commented Aug 1, 2023 at 12:04
  • $\begingroup$ Mm I'm not sure I understood everything that's implied in your answer, may I ask some further questions, to steer my doubts away? For one, what kind of space classifies higher-order deformations? Does it represent some functor, so as to guarantee the map from $A$ is a morphism? And secondly I tried writing down the argument with the formal function theorem: it implies that the pushforward of $\mathcal{C}_a$ onto C is locally free, and thus has a global nowhere-vanishing section on an open set $U\times A$. But the set where it has trivial fibres is closed so its fibres must all be free. Thanks! $\endgroup$
    – Basil
    Commented Aug 4, 2023 at 23:43

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