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$\def\Pic{\mathcal{Pic}}\def\Gm{\mathbb{G}_m}\def\Hom{\mathop{Hom}}\def\HOM{\mathcal{Hom}}$ If $A/S$ is an abelian scheme, the fppf sheaf $\Pic^0_{A/S}$ is representable by an abelian scheme $\hat{A}$. (By general theory, since A/S is cohomologically flat in dimension 0, $\Pic^\tau_{A/S}$ is a proper algebraic space, it is smooth by the same argument as in Mumford's GIT, hence an abelian algebraic space, hence an abelian scheme by Raynaud's theorem. The books by Faltings-Chai and BLR on Néron Models are the standard references on this).

This $\hat{A}$ is called the dual abelian variety. There are several ways to see it as a "dual":

  1. the first is that by the above definition, $\hat{A}=\HOM_S(A, B\Gm)$, where $B\Gm$ is the classifying $1$-stack of $\Gm$. (I think this is correct if we require the morphisms to be group morphism, so that they land in the relative connected component, ie corresponds to elements of $\Pic^0$).
  2. An alternative is to see $\hat{A}$ as $\mathcal{Ext}^1(A,\Gm)$. The proof I know is as follow: to a line bundle $\mathcal{L}$ algebraically equivalent to $0$ on $A$, one associate Mumford's theta group $G(\mathcal{L})$ which in this case is a commutative group, and a central extension of $A$ by $\Gm$.
  3. These induce a canonical duality as follow: if $f:A \to B$ is an isogeny, then $\mathop{Ker} f$, the kernel of $f$ is the Cartier dual of $\mathop{Ker} \hat{f}$, the kernel of the dual isogeny $\hat{f}: \hat{B} \to \hat{A}$. This can be seen from 2) using the long exact sequence induced by applying $\mathop{Hom}(\cdot, \Gm)$ on $0 \to \mathop{Ker} f \to A \to B \to 0$.

I have two questions:

  1. Is there a direct "abstract nonsense" proof of 1. $\Rightarrow$ 2.?
  2. Using that $\Hom_S(A,\Gm)=1$, we can reinterpret 2. as $\hat{A}=(\tau_{\leq 1} R\HOM_S(A,\Gm))[1]$. The fact that there is a $1$-truncation here yields to the following natural question: would it make sense to define $\mathbb{R}\hat{A}$ in derived algebraic geometry as $\mathbb{R}\HOM_S(A, B \Gm)$? Whould we then have an identification between $\mathbb{R}\hat{A}$ and the full $R\HOM_S(A,\Gm)$?
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    $\begingroup$ 1) Under the Dold-Kan correspondence between simplicial abelian groups and chain complexes, B corresponds to [1]. 2) Ext^1 is a quotient of that RHom, not a subspace, so this is an example of a quotient space rather than an underived quotient. I can't see a distinction between $B_{\infty}$ and $B$. $\endgroup$
    – Jon Pridham
    Commented Dec 12, 2020 at 23:19
  • $\begingroup$ Thanks for your comment. You are right that I should have asked about $\mathbb{R}Hom(A,BG_m)$ instead, I have edited (hopefully this make sense even if $BG_m$ is not derived). Perhaps I should have just asked if there was a well defined notion of derived dual abelian scheme. $\endgroup$
    – Damien Robert
    Commented Dec 13, 2020 at 9:47

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