# Regarding the dual of an abelian variety

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?

• For the second part, abelian varieties are tori, so in a given dimension they are all diffeomorphic. I don't know a nice example for the first part. Commented Jan 24, 2017 at 9:45
• @Nefertiti thanks for the comment! it is helpful to me Commented Jan 24, 2017 at 14:52

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.

• Maybe you forget to mention that $A$ is principally polarised. Else, how do you get the existence of the integer $n$?
– Doug
Commented Jun 3, 2023 at 11:18
• Yes, nice catch. Commented Jun 3, 2023 at 19:41