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?

  • 2
    $\begingroup$ 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. $\endgroup$
    – Nefertiti
    Jan 24, 2017 at 9:45
  • $\begingroup$ @Nefertiti thanks for the comment! it is helpful to me $\endgroup$
    – Rüdiger
    Jan 24, 2017 at 14:52

1 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


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.

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

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .