$X$ = 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?
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$\begingroup$ Hi, Tuan, Did you see the springer online reference on <a href="eom.springer.de/a/a011640.htm" > Automorphisms of algebraic surfaces </a>? $\endgroup$– Fei YEApr 7, 2010 at 6:27
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$\begingroup$ yes, Fei, if $X$ is not ruled and an elliptic surface, then $Aut^0(X)$ is an abelian algebraic group of dimension 1. $\endgroup$– TuanApr 7, 2010 at 20:01
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$\begingroup$ hah, I also have a proof that the previous statement holds for elliptic ruled surfaces also! $\endgroup$– TuanApr 8, 2010 at 4:55
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2 Answers
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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.
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$\begingroup$ 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. $\endgroup$– TuanApr 12, 2010 at 2:09
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$\begingroup$ Yes, thanks, I misremembered. Trivial either way. $\endgroup$– VA.Apr 12, 2010 at 2:37
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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.
