Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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$.