# The $Pic ^ 0$ of an abelian variety

dear, I need a precise reference to the fact that the connected component of $Pic (A)$, where $A$ is an abelian variety over a field $k$ consists of the following set $\{L \in Pic (A) : T ^*_x L \cong L \ \textrm{for all} \ x \in X \}$ where $Tx$ is the translation by $x$

In the books of abelian varieties and the Mumford Milne not have that.

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I mean the connected component of identity in $Pic (A)$ – Flávio May 22 '11 at 0:18
Are you sure this isn't in Mumford's Abelian Varieties? I don't have my copy with me, but I'm pretty sure I remember it being in there. – Matt May 22 '11 at 2:54
Actually Mumford defines $\textrm{Pic}^0$ in this way: see Chapter II, Section 8 – Francesco Polizzi May 22 '11 at 8:46
– Qiaochu Yuan May 22 '11 at 10:45
I'm sure this is not the book of Munford. This is here (google.com.br/…), but unfortunately this is not an article. – Flávio May 22 '11 at 12:26

As noted, both sources essentially define $Pic^{0}(A)$ to consist of the $L$ for which $T_{x}^{\ast }L\approx L$, but the questioner asks for a proof that this is the identity component of $Pic(A)$. Both sources show that $Pic^{0}(A)$ is represented by an abelian variety, so it is a question of showing that the quotient $NS(A)=Pic(A)/Pic^{0}(A)$ is discrete. But it is shown that $NS(A)$ is a finitely generated group, and hence must be discrete.

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By definition, $L$ is in $Pic^0$ if the homomorphism $\phi_L$ is identically zero, where $\phi: X \rightarrow Pic(X)$ is defined by sending $x$ to the isomorphism class of $T_x^*L \otimes L^{-1}$.

The map $\phi_L$ being identically zero means precisely that $T_x^*L \otimes L^{-1} \cong \mathcal O_X$ for every $x \in X$, i.e. that $T_x^*L \cong L$.

Read pages 74-75 in Mumford's book very carefully.

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expensive, found a demo for this question.

Set $m: A \times A \rightarrow A$ be the group law over $A$, $p_{12}: \textrm{Pic}^0(A) \times A \times A \rightarrow \textrm{Pic}^0(A) \times A$ be the projection over the first and second factors, $p_{13}:\textrm{Pic}^0(A) \times A \times A \rightarrow \textrm{Pic}^0(A) \times A$ be the projection over the first and third factors and $L$ be the Poincaré bundle over $A \times \textrm{Pic}^0(A)$ trivialized in neutral element $e$ of $A$, this is, $L \mid _{ \{e\} \times \textrm{Pic}^0(A)}$ is trivial.

Now, define $(\textrm{Id},m):\textrm{Pic}^0(A) \times A \times A \rightarrow \textrm{Pic}^0(A) \times A$, where Id is the identity map, and consider $$M:= (\textrm{Id},m)^*L \otimes P_{12}^*L^{-1} \otimes p_{13}^*L^{-1}$$ line bundle over $\textrm{Pic}^0(A) \times A \times A$. Then, since $M \mid_{\{\mathscr{O}_{\textrm{Pic}^0(A)}\} \times A \times A}, \ M \mid _{ \textrm{Pic}^0(A) \times \{e\} \times A} \ \textrm{and} \ M \mid _{ \textrm{Pic}^0(A) \times A \times \{e\}}$ are trivial, follow of the Cube theorem that $M$ is trivial. Hence, $M \mid _{\{\mathscr{N}\} \times \{x\} \times A} \cong t^*_x\mathscr{N} \otimes \mathscr{N}^{-1}$ is trivial for any $\mathscr{N} \in \textrm{Pic}^0(A)$ and $x \in A$, this is, $\textrm{Pic}^0(A) \subseteq \Sigma :=\{ \mathscr{L} \in \textrm{Pic}(A) \mid t_x^* \mathscr{L} \cong \mathscr{L} \forall \ x \in X\}.$ So, since $\Sigma$ is an Abelian variety, see [Mumford section 8], and $\textrm{Pic}^0(A)$ is open and closed in $\textrm{Pic}(A)$, see [FAG proposition 0.5.3], follow that $\textrm{Pic}^0(A)=\Sigma$.

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[FAG proposition 9.5.3] – Flávio May 23 '11 at 18:33