Is complex Abelian variety isogenic to its dual? 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?
 A: 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:


*

*
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!

*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.

*
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 $A\to A^\vee$.
