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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

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    $\begingroup$ 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

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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.

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