Let $X$ be a smooth closed subvariety of a complex abelian variety $A$. Assume $X$ is of general type and of codimension one with $\omega_X$ ample.

Often, people speak about the stabilizer $\mathrm{Stab}_A(X)$ of $X$ in $A$. This is the group of $a$ in $A$ such that $X+a = X$.

What is the relation of $\mathrm{Stab}_A(X)$ to $\mathrm{Aut}(X)$?

They are both finite. Are they equal? If the stabilizer is trivial, does that imply $\mathrm{Aut}(X)$ is trivial? What about vice versa? Does $\mathrm{Stab}_a(X)$ inject into $\mathrm{Aut}(X)$?

Crossposted from stackexchange, because I didn't get any replies there unfortunately: https://math.stackexchange.com/questions/4446811/difference-between-stabilizer-and-automorphism-group

  • 1
    $\begingroup$ It might be more interesting to ask whether $Aut(X)$ is the subgroup of $Aut(A)$ (automorphisms as a variety) which preserves $X$. $\endgroup$
    – naf
    May 11, 2022 at 5:27
  • $\begingroup$ @naf Yes, that's interesting. I've posted it here now mathoverflow.net/questions/422387/… $\endgroup$
    – Hinter
    May 12, 2022 at 15:28

1 Answer 1


They have absolutely no reason to be equal. Consider the case where $A$ is the Jacobian of a genus 2 curve $C$, and $X=C$ embedded in $A$ by $x\mapsto [x]-[p]$ for some fixed point $p\in C$. Then $X$ is a Theta divisor, so $\operatorname{Stab}_A(X) $ is trivial. But $X$ has always a nontrivial automorphism, the hyperelliptic involution, and it may have more in some cases.

Of course $\operatorname{Stab}_A(X) $ injects into $\operatorname{Aut}(X) $, since a nontrivial translation does not fix any point of $A$. But that is all you can say.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.