Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

$X$ = 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?

share|improve this question
    
Hi, Tuan, Did you see the springer online reference on <a href="eom.springer.de/a/a011640.htm"; > Automorphisms of algebraic surfaces </a>? –  Fei YE Apr 7 '10 at 6:27
    
thanks. then I guess it's never affine! –  Tuan Apr 7 '10 at 13:57
    
yes, Fei, if $X$ is not ruled and an elliptic surface, then $Aut^0(X)$ is an abelian algebraic group of dimension 1. –  Tuan Apr 7 '10 at 20:01
    
hah, I also have a proof that the previous statement holds for elliptic ruled surfaces also! –  Tuan Apr 8 '10 at 4:55
add comment

2 Answers 2

up vote 4 down vote accepted

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.

share|improve this answer
    
I also used the same trick that F acts on X to conclude that $\text{Aut}^0(X)$ is not affine. Just a minor point though, the group $G$ doesn't have to be cyclic. –  Tuan Apr 12 '10 at 2:09
    
Yes, thanks, I misremembered. Trivial either way. –  VA. Apr 12 '10 at 2:37
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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