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Let $A$ be an Abelian variety (over an algebraically closed field of characteristic zero), and denote by $A^*$ its dual, which parameterizes degree zero line bundles on $A$. If $Q$ is a point of $A^*$, the corresponding degree zero line bundle on $A$ arises via pull-back of the Poincaré bundle $\mathcal{P} \to A \times A^*$ along $A \xrightarrow{\sim} A \times \{Q\} \xrightarrow{(\mathrm{Id}_A,Q)} A \times A^*$.

Question: If I have a morphism $\lambda: A \to A^*$, I can pull-back $\mathcal{P}$ along the graph morphism $\Gamma_\lambda :A \xrightarrow{(\mathrm{Id}_A,\lambda)} A \times A^*$. Is there any way to know what line bundle on $A$ I obtain, i.e., what point of $A^*$ corresponds to it?

A similar question, possibly of secondary importance: the Poincaré bundle is itself a degree zero line bundle on $A \times A^*$, so it should correspond to a point on $(A \times A^*)^* \cong A \times A^*$. But which one?

57Jimmy
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    I thought about the case of an elliptic curve. In that case I think you get the sum of the fixed points of $\lambda$ (in the group law, here thinking of $E^* \cong E$). In general I think you get, well, the divisor obtained as the intersection product of the graph of $\lambda$ with the divisor class of the Poincaré bundle. But the latter is not an effective divisor for $g>1$, so this is necessarily a little tricky to calculate. – Jake Levinson Nov 27 '21 at 21:03

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