# Quotient of an abelian surface by an antisymplectic involution

What can we say about the quotient of an abelian surface by an antisymplectic involution?

• I would conjecture that it is always $CP^1 \times T^2$, because I see no other examples Sep 25, 2013 at 14:10

Denote by $\sigma$ the involution on the abelian surface $A$ and set $X:=A/\sigma$. The eigenvalues of the action of $\sigma$ on $H^0(\Omega^1_A)$ are $+1$ and $-1$. So $\sigma$ has no isolated fixed points and it follows that $X$ is smooth with $h^1(\mathcal O_X)=1$. There are two possibilities:

1) $\sigma$ has no fixed point. Then the quotient map $A\to X$ is \'etale and $X$ is a bielliptic surface almost by definition.

2) $\sigma$ has a fixed point. Then, up to a translation, we may assume that $\Sigma$ is an endomorphism. The fixed locus is a union of elliptic curves and $X$ is ruled. This can be seen either by classification of surfaces or observing that $A$ contains a family of translates of elliptic curves on which $\sigma$ acts as multiplcation by $-1$.

• In case 1), if $X=E_1 \times E_2/ G$, how can we relate $A$ with $E_1 \times E_2$? Jun 2, 2014 at 12:20

Averaging a Kaehler class over the involution, and taking the corresponding Ricci-flat metric, we may assume that the involution preserves a flat metric on a torus. At each fixed point, the eigenvalues of the involution are +1, -1, and the fixed point set is a subtorus $T_0\subset T$. This means that the involution acts as -1 on $T_1 :=T/T_0$, hence the quotient is a product $T_0\times T_1/\{\pm 1\}$, that is, $T_0 \times {\Bbb C} P^1$.

• What about the map $T \times T \to T \times T$ that switches the two factors? One can't always write the original torus as a product of the subtorus and the quotient torus. However you can do this up to $2$-torsion so $Sym^2(T)$ is the only counterexample. Sep 25, 2013 at 21:38
• the map which switches two factors is symplectic, and we need antisymplectic Sep 26, 2013 at 6:40
• @Misha: the map that switches factors is antisymplectic, since $dx\wedge dy$ goes to $dy\wedge dx$.
– rita
Sep 26, 2013 at 12:02