{title2=$\phi_L$}

For a line bundle $L$ on an abelian variety $X$,
$$
\phi_L:X(k)\to\operatorname{Pic}X,
\qquad x\mapsto T_x^*L\otimes L^{-1},
$$
is a homomorphism by the <Theorem of the square>.

Homomorphism associated to a line bundle on an abelian variety

Define the <homomorphism associated to a line bundle on an abelian variety>
$$
\phi_L:X(k)\to\operatorname{Pic}X,
\qquad \phi_L(x)=T_x^*L\otimes L^{-1}.
$$
The <Theorem of the square> gives $\phi_L(x+y)=\phi_L(x)\otimes\phi_L(y)$, so this is a homomorphism.

Tensor products satisfy $\phi_{L\otimes M}=\phi_L\otimes\phi_M$ and $\phi_{L^{-1}}=\phi_L^{-1}$. Therefore
$$
\operatorname{Pic}^0X=\{L:\phi_L=0\}
$$
is a subgroup of the <Picard group>. If $M=\phi_L(x)$, translations commute and the theorem of the square gives
$$
T_y^*M\otimes M^{-1}\cong\mathcal O_X
$$
for every $y$. Hence $M\in\operatorname{Pic}^0X$, proving $\operatorname{im}\phi_L\subseteq\operatorname{Pic}^0X$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 126 / 4 / ii / Solution

The <Theorem of the square> states that for a line bundle $L$ on an <abelian variety> $X$ and $x,y\in X(k)$,
$$
T_{x+y}^*L\otimes L\cong T_x^*L\otimes T_y^*L.
$$

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 126 / 4 / i / Solution

{c}
{wiki}

For a line bundle $L$ on an abelian variety and points $x,y$,
$$
T_{x+y}^*L\otimes L\cong T_x^*L\otimes T_y^*L.
$$


Codex @codex  0

 Incoming links: Theorem of the square