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  "body": "\n## Definition\n\nLet $A$ be an [[abelian variety|R7W03TXRS9]] over a field $k$, with unit $e \\in A(k)$.\nFor a point $x$ of $A$, let $t_x \\colon A \\to A$ be translation by $x$.\nLet $\\bar{k}$ be an algebraic closure of $k$.\nLet $\\operatorname{Pic}^0(A)$ be the subgroup of the [[Picard group|3Z9DCWYXQZ]] of classes of [[line bundles|7JCF2TE3V5]] $L$ with $t_x^{*} L \\cong L$ for all $x \\in A(\\bar{k})$.\nThe *dual abelian variety* of $A$ is an abelian variety $A^\\vee$ over $k$ with a line bundle $P$ on $A \\times_k A^\\vee$, the Poincaré bundle, such that:\n\n1. $P$ restricted to $\\set{e} \\times A^\\vee$ is trivial;\n2. $P$ restricted to $A \\times \\set{y}$ lies in $\\operatorname{Pic}^0$ for every point $y$ of $A^\\vee$;\n3. $(A^\\vee, P)$ is universal, in the following sense.\n   Let $T$ be a $k$-scheme and $L$ a line bundle on $A \\times_k T$.\n   Assume $L$ is trivial on $\\set{e} \\times T$ and in $\\operatorname{Pic}^0$ on each fibre $A \\times \\set{t}$.\n   Then there is a unique morphism $f \\colon T \\to A^\\vee$ with $L \\cong (\\id_A \\times f)^{*} P$.\n\n## Remarks\n\nThe pair $(A^\\vee, P)$ exists and is unique up to unique isomorphism.\nWhen $k$ is algebraically closed, clause 3 with $T = \\operatorname{Spec} k$ gives a bijection $A^\\vee(k) \\cong \\operatorname{Pic}^0(A)$.\nThe dimension of $A^\\vee$ equals the dimension of $A$, and $(A^\\vee)^\\vee \\cong A$.\n\nExample: let $E$ be an elliptic curve over an algebraically closed field with origin $O$.\nThe map $P \\mapsto \\mathcal{O}_E(P - O)$ is a bijection from $E(k)$ onto the line bundles of degree $0$, and these are exactly the translation-invariant ones.\nSo $E^\\vee \\cong E$.\n\nNon-example: the full Picard group of $E$ is not the dual.\nThe line bundle $\\mathcal{O}_E(O)$ has degree $1$, and $t_x^{*} \\mathcal{O}_E(O) \\cong \\mathcal{O}_E(O)$ fails for every $x \\ne O$, so it is not in $\\operatorname{Pic}^0(E)$.\n\nOver $\\C$, if $A = V / \\Lambda$, then $A^\\vee$ is the quotient of the space of $\\C$-antilinear forms $f$ on $V$ by the lattice of those $f$ with $\\operatorname{Im} f(\\Lambda) \\subseteq \\Z$.\n",
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