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Suppose $A$ is a complex Abelian variety and let $A^V$ be the dual Abelian variety. $A^V$ is defined as the moduli space of all line bundles on $A$ with $c_1=0$.

It seems to me that $A$ is isogenic to $A^{V}$. But I am not sure that this is correct. So I would like to give the following "proof".

Consider the space of all line bundles on $A$ with a fixed ample class $c$ (which exists since $A$ is ableian), let us call this space $A_c$. It is clear that $A_c$ isomorphic to $A^V$. Now, $A$ is acting on intself by translations and so it is acting on $A_c$. I think that the orbit of the action covers whole $A_c$ if $c$ is ample. So we get a finite covering map $A\to A_c\cong A^{V}$.

Question. Is the above reasoning correct? Are $A$ and $A^V$ isogenious?

agleaner
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1 Answers1

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I think your reasoning is correct.

The way I learnt about this isogeny issue is the following. Work over any algebraically closed field. Every invertible sheaf $\mathscr L$ on $A$ induces a homomorphism of Abelian varieties $$\phi_\mathscr L:A\to \textrm{Pic }A$$ sending $a\mapsto \tau_a^\ast\mathscr L\otimes\mathscr L^\vee$, where $\tau_a$ is translation by $a$. Now, if $m:A\times A\to A$ and $q:A\times A\to A$ denote the multiplication and the second projection respectively, then for each $a\in A$ we have an identification in $\textrm{Pic }A$: $$(m^\ast\mathscr L\otimes q^\ast \mathscr L^\vee)|_{a\times A}\cong \tau_a^\ast\mathscr L\otimes\mathscr L^\vee.$$ There are several things happening:

  1. The locus of $a\in A$ such that the LHS is trivial (by the displayed isomorphism, this is the kernel of $\phi_\mathscr L$) is closed in $A$. This is a good starting point for its possible finiteness!
  2. For any $\mathscr L$, the image of $\phi_\mathscr L$ lands in $A^\vee$. The image is exactly $A^\vee$ when $\mathscr L$ is ample.
  3. If $h^0(A,\mathscr L)>0$, then $\ker\phi_\mathscr L$ is finite if and only if $\mathscr L$ is ample.

So an isogeny $A\to A^\vee$ (called in this case a polarization) is provided by any ample line bundle. Moreover, those $\mathscr L$ with $h^0(A,\mathscr L)=1$ induce principal polarizations, i.e. polarizations of degree $1$, i.e. isomorphisms $A\to A^\vee$.

Brenin
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  • Dear Brenin, did you mean $\chi(L)$ instead of $h^0(L)$ ? – Cantlog Nov 13 '13 at 19:10
  • @Cantlog: I had in mind the following: a principal polarization "is" the choice of a unimodular skew symmetric bilinear form $\xi\in H^2(A,\mathbb Z)$. There is a matrix representing this form, and $h^0(A,L)$ is the square root of the determinant of this matrix. But this det must be $1$ because $\xi$ is unimodular. Nevertheless, I checked on Arithmetic Geometry (Cornell Silverman ed.) and I only found out that $\deg\phi_L=\chi(L)^2$. Does this imply $h^0(A,L)=1$ when $\deg\phi_L=1$? (If you have that book it is Thm. 13.3, p. 127) – Brenin Nov 13 '13 at 19:55
  • We have $\chi(L)=h^0(L)$ if $h^0(L)\ne 0$. So what you say is true under the hypothesis of 3. But in general, $L$ ample does not imply $h^0(L)\ne 0$. – Cantlog Nov 13 '13 at 20:12