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Given a smooth projective variety $X$ of any dimension, is there a canonical way to associate to a (very) ample line bundle $\mathcal{L}$ on $X$ an isogeny $\mathrm{Alb}(X) \to \mathrm{Alb}(X)^\vee$?

EDIT: Another related question: When is the Abel-Jacobi map $X \to \mathrm{Alb}(X)$ a closed immersion?

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    $\begingroup$ Hi Timo, regarding the Albanese, I suggest you look at mathoverflow.net/questions/2548/… $\endgroup$ – Lars Feb 14 '12 at 22:25
  • $\begingroup$ This seems unlikely. A polarization of the albanese canonically gives a line bundle on the Albanese, which one can pull back to a line bundle on the variety. This gives a map from the line bundles of $X$ to the line bundles of $X$. Since not all line bundles always come from the Albanese, this map is not the identity. What could it possibly be? $\endgroup$ – Will Sawin Apr 16 '12 at 23:23
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The following seems (I have to think about it) to work [EDIT: no, it doesn't]: By the universal property of the Albanese, we just have to construct a morphism $X \to \mathrm{Alb}(X)^\vee$:

Given a line bundle $\mathcal{L}$ on $X$, we can pull it back to $\mathcal{L}$ on $X \times \mathrm{Alb}(X)$, and we have to check that it is trivial when pulled back by the zero section and in $\mathrm{Pic}^0(\mathrm{Alb}(X))$ when pulled back by all $x \in X$ [EDIT: This is wrong.]. Then Mumford, Abelian Varieties, p. 125 gives us a unique $\phi: X \to \mathrm{Alb}(X)^\vee$ with $\mathcal{L} = (\phi \times 1_{\mathrm{Alb}(X)})^*\mathcal{P}$.

For the induced map $\psi: \mathrm{Alb}(X) \to \mathrm{Alb}(X)^\vee$ to be a polarisation, we have to check that $\psi^\vee = \psi$ in $\mathrm{Hom}(\mathrm{Alb}(X), \mathrm{Alb}(X)^\vee)$.

Conversely, a polarisation is given by a line bundle on $\mathrm{Alb}(X)$, which we can pull back by the Abel-Jacobi map.

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