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

The Theorem of the square states that for a line bundle $L$ on an abelian variety $X$ and $x,y\in X(k)$,

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Define the homomorphism associated to a line bundle on an abelian varietyThe Theorem of the square gives ${\varphi }_L(x+y)={\varphi }_L(x)\otimes {\varphi }_L(y)$, so this is a homomorphism.

Tensor products satisfy ${\varphi }_{L\otimes M}={\varphi }_L\otimes {\varphi }_M$ and ${\varphi }_{L^{-1}}={\varphi }_L^{-1}$. Thereforeis a subgroup of the Picard group. If $M={\varphi }_L(x)$, translations commute and the theorem of the square givesfor every $y$. Hence $M\in {\mathrm{Pic}}^{0}X$, proving $\mathrm{im}{\varphi }_L\subseteq {\mathrm{Pic}}^{0}X$.

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The mapis a homomorphism. Pullback along $i_1(x)=(x,e_2)$ recovers $L_1$, up to tensoring with a fixed one-dimensional vector space, which is a trivial line bundle; similarly $i_2^{\ast }$ recovers $L_2$. Thus $\tau$ is injective.

It need not be surjective. For an elliptic curve $E$, the line bundle ${\mathcal{O}}_{E\times E}(\Delta )$ of the diagonal restricts to $E\times \{y\}$ as ${\mathcal{O}}_E(y)$, whose class varies with $y$. A line bundle pulled back separately from the two factors has constant class on these fibers. Hence $\mathcal{O}(\Delta )$ is not in the image of $\tau$.

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Part iii already proves injectivity. Pullback along $i_j$ sends translation-invariant line bundles to translation-invariant line bundles, so it definesClearly $\rho \tau =1$.

For $L\in {\mathrm{Pic}}^0(X_1\times X_2)$, put $L_j=i_j^{\ast }L$ andThen $M$ is trivial on both coordinate axes. Since $L$ and the two correcting factors lie in ${\mathrm{Pic}}^0$, translation by $(x_1,e_2)$ leaves $M$ invariant. Hence every restriction $M{|}_{\{x_1\}\times X_2}$ is trivial. The Seesaw theorem and triviality on $X_1\times \{e_2\}$ imply $M\cong {\mathcal{O}}_X$. Thus $L=\tau (L_1,L_2)$, so

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