Automorphism group of bi-elliptic surface - MathOverflow

Automorphism group of bi-elliptic surface

2010-04-07T06:03:35Z

$X$ = bi-elliptic surface (smooth and over $\mathbb{C}$), Aut($X$) = the group of automorphisms of $X$, Aut$^0(X)$ = connected component of the identity in Aut($X$).

Is Aut$^0(X)$ always an affine algebraic group?

Answer by Holger Partsch for Automorphism group of bi-elliptic surface

2010-04-11T08:37:22Z

actually, you can describe the automorphism scheme quite explicitly: Such a surface $X$ has an etale galois covering $E \times F \to X$ where $E$ and $F$ are elliptic curves.

The automorphisms scheme of $E \times F$ is easy to understand and for $X$ you can use descend. An automorphisms of $E \times F$ descends to $X$, if it commutes with the galois action. For example, if $X = (E \times F)/(\mathbb Z/2\mathbb Z)$ where the action is given by $(x, y) \mapsto (-x, y + c)$ for a non trivial two torsion point $c$ of $E$, you will find that $Aut^0(X) = F/\left < c \right >$.

In general, you can prove in that way, that the reduction of $Aut^0(X)$ is just the Albanese of $X$. In characteristic two or three, it can happen that $Aut^0(X)$ is non-reduced.

Answer by VA for Automorphism group of bi-elliptic surface

2010-04-11T14:09:14Z

The answer is always "no". By classification, a bielliptic surface over $\mathbb C$ has the form $(E\times F)/G$ where $E,F$ are elliptic curves, $G=\subset Aut(E,0)$ is an abelian group acting by complex multiplications on $E$ and by translations on $F$. ($G$ is not necessarily cyclic as Tuan correctly points out.)

($X$ maps to an elliptic curve $F/G$ and every fiber is isomorphic to an elliptic curve $E$, hence the name bielliptic.)

Then $F$ acts on $E\times F$ by $(x,y)\mapsto (x,y+f)$, and this action commutes with the $G$-action. Thus, $F\subset Aut^0(X)$. As $F$ is a projective variety, $Aut^0(X)$ is not affine.