Regarding the dual of an abelian variety

Question

I know that if $X$ is an abelian variety, then it need not be isomorphic to its dual. But I don't know any example: Jacobians, elliptic curves etc are all isomorphic to their duals. Does anyone know an example. Also is $X$ homeomorphic to its dual?

Answer 1

Let $k$ be an algebraically closed field of characteristic zero and let $A$ be a principally polarized abelian variety over $k$ with $\mathrm{End}(A)= \mathbb{Z}$ and $\dim A =2$. Let $P$ be a point of order $2$, and let $G$ be the subgroup generated by $P$. Let $B= A/G$.

Claim. $B$ is not isomorphic to its dual $B^t$.

Proof. By contradiction. Suppose that $f:B\to B^t$ is an isomorphism. Let $q:A\to B$ be the quotient map. Consider the endomorphism $g:=q \circ f\circ q^{\vee}$, i.e., the composed homomorphism $$ A\to B \cong B^t \to A^t $$ Note that $$\deg(g) = \deg(q)\deg(f) \deg(q^\vee) = \deg(q)^2 = 4.$$ On the other hand, since $\mathrm{End}(A) = \mathbb{Z}$ and $A^t\cong A$, there is an integer $n$ such that $\deg g= \deg [n]$, so that $4 = \deg(g) = \deg([n]) = n^4$. Since $4$ is not a fourth power, we get a contradiction. This proves the claim. QED

This argument also shows that there exist non-principally polarizable abelian varieties; see

https://mathoverflow.net/questions/16992/non-principally-polarized-complex-abelian-varieties

Remark. In an earlier version of this answer I made the inaccurate statement that an abelian variety isomorphic to its dual has a principal polarization.