Let $A$ be a (semi-)abelian variety over an algebraically closed field $K$, and $X$ be a closed irreducible subvariety. Can $X$ have a non-trivial finite stabilizer ? By stabilizer, I mean the closed subgroup $S\subset A$ such that $X$ is stable by translation by any point in $S$. This question is equivalent to the one in the title since you can always take the quotient by the neutral component of $S$ to reduce to a finite group.
$\begingroup$
$\endgroup$
1
-
1$\begingroup$ No. Let $f:A \to B$ be a separable isogeny and $D$ a smooth ample divisor in $B$. Then $f^{−1}(D)$ is irreducible if $\dim(X)>1$ (since it is smooth and connected) and is stabilized by the kernel of $f$. $\endgroup$– nafCommented May 15, 2013 at 13:12
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
No. Take a curve in its Jacobian and pull it back by multiplication by some n. The resulting pullback is a curve invariant by the n torsion.