# Difference between stabilizer and automorphism group of subvariety of an abelian variety

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

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

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.