Smooth quotients of bi-elliptic surfaces

Abstract

We consider the quotient X of bi-elliptic surface by a finite automorphism group. If X is smooth, then it is a bi-elliptic surface or ruled surface with irregularity one. As a corollary any bi-elliptic surface cannot be Galois covering of projective plane, hence does not have any Galois embedding.

1 Statement of result

We consider a covering of surface, i.e., let $X_1$ and $X_2$ be connected normal complex surfaces and $\pi : X_1 \longrightarrow X_2$ a finite surjective proper holomorphic map. There are lots of studies of the covering. We are interested in the following cases (1) and (2):

• (1) $X_2$ is a rational surface, in particular the projective plane $\mathbb P^2$, for example (Friedman and Leyenson 2010)

• (2) $X_1$ has Kodaira dimension 0, for example (Muammed Uludaǧ 2005; Yoshihara 2007)

In this note we consider the case where $X_1$ is a bi-elliptic surface and $\pi$ is a Galois covering. The definition of bi-elliptic surface is as follows, which has been called a hyperelliptic surface (Suwa 1970).

Definition 1

A bi-elliptic surface is a surface with the geometric genus zero and having an abelian surface as its unramified covering.

First we note the following:

Remark 2

Let S be a bi-elliptic surface. If $\pi : S \longrightarrow X$ is a Galois covering and X is smooth, then X has no curve with negative self-intersection number. In particular, if X is rational, then it is $\mathbb P^2$ or $\mathbb {P}^1 \times \mathbb {P}^1$.

Proof

Suppose the contrary. Then, there exists an irreducible curve C in X such that $C^2 <0$. We have that $(\pi ^*(C))^2=\deg \pi \cdot C^2 <0$. Thus there exists an irreducible component $C'$ in $\pi ^*(C)$ with $C'^2 < 0$. which is a contradiction. Because there exists no curve with negative self-intersection number in bi-elliptic surface. $\square$

Our result is stated as follows:

Theorem 3

Let S be a bi-elliptic surface and G a finite subgroup of $\mathrm {Aut}(S)$, the automorphism group of S. Let $X=S/G$ be the quotient surface. If X is smooth, then it is a bi-elliptic surface or a ruled surface with irregularity one.

As a corollary we have the following.

Corollary 4

Bi-elliptic surface cannot be a Galois covering of any smooth rational surface.

Miranda (2005) presents a construction of bi-elliptic surface as covering of rational elliptic surface. We present the other examples.

Example 5

Let A be the abelian surface with the period matrix

$$\begin{aligned} \left( \begin{array}{l@{\quad }l@{\quad }l@{\quad }l} 1 &{} 0 &{} \lambda &{} 0 \\ 0 &{} 1 &{} 0 &{} e_n \end{array}\right) , \end{aligned}$$

where $\mathfrak {I}\lambda >0$ and $e_{n}=\exp (2\pi \sqrt{-1}/n ), \ (n=4,\ 6)$. Let $\sigma _i \ (i=1,\ 2)$ be the automorphism of A induced by

$$\begin{aligned} \sigma _1z= \left( \begin{array}{l@{\quad }l} 1 &{} 0 \\ 0 &{} e_{n} \end{array}\right) z + \left( \begin{array}{l@{\quad }l} 1/n \\ 0 \end{array}\right) \quad \ \mathrm {and} \quad \ \sigma _2z =\left( \begin{array}{l@{\quad }l} 1 &{} 0 \\ 0 &{} e_{n} \end{array}\right) z, \end{aligned}$$

where $z \in \mathbb C^2$. Then $S=A/\langle \sigma _1^m \rangle \ (n=2m)$ and $X=A/\langle \sigma _1 \rangle$ are bi-elliptic surfaces. Note that $\pi : S \longrightarrow X$ is a double covering. On the other hand, we have $\sigma _1^m \sigma _2=\sigma _2 \sigma _1^m$, hence $\sigma _2$ induces an automorphism $\overline{\sigma }_2$ on S. It is easy to see that $S/\langle \overline{\sigma }_2 \rangle$ is isomorphic to $E \times \mathbb P^1$, where E is an elliptic curve with the period matrix $(1/2, \lambda )$.

Remark 6

[Yoshihara (2000), Example 2.4] Let $A_i \ (i = 1, 2)$ be the abelian surface defined by the following period matrix:

$$\begin{aligned} \begin{array}{l} \Omega _1 \ = \ \left( \begin{array}{l@{\quad }l@{\quad }l@{\quad }l} 1 &{} 0 &{} \omega &{} 0 \\ 0 &{} 1 &{} 0 &{} \omega \end{array} \right) \quad \mathrm{and} \quad \Omega _2 \ = \ \left( \begin{array}{l@{\quad }l@{\quad }l@{\quad }l} 1 &{} 0 &{} (\omega - 1)/3 &{} 0 \\ 0 &{} 1 &{} (\omega - 1)/3 &{} \omega \end{array} \right) , \ \mathrm{respectively}, \end{array} \end{aligned}$$

where $\omega = \exp (2\pi \sqrt{-1}/3)$. Let $\sigma _i$ be the automorphism of $A_i$ defined by

$$\begin{aligned} \sigma _1z= \left( \begin{array}{ll} 1 &{}\quad 0 \\ 0 &{}\quad \omega \end{array}\right) z + \left( \begin{array}{ll} (\omega + 2)/3 \\ 0 \end{array}\right) \ \ \mathrm {and} \ \ \sigma _2z =\left( \begin{array}{ll} 1 &{}\quad 0 \\ 0 &{} \quad \omega \end{array}\right) z, + \left( \begin{array}{ll} 1/3 \\ 0 \end{array}\right) \end{aligned}$$

Then $\sigma _{i}^{3} = id$ on $A_i$ and $S_i = A_i/\sigma _i$ is a bi-elliptic surface. Moreover letting

$$\begin{aligned} \tau _1z= \left( \begin{array}{ll} \omega &{}\quad 0 \\ 0 &{}\quad \omega \end{array}\right) z \ \ \mathrm {and} \ \ \sigma _2z =\left( \begin{array}{ll} \omega &{}\quad 0 \\ 0 &{} \quad \omega \end{array}\right) z + \left( \begin{array}{ll} 2/3 \\ 0 \end{array}\right) \end{aligned}$$

we see that $\tau _i$ defines an automorphism of $S_i$ and $S_i/\tau _i$ turns out to be a rational surface. Note that the rational surfaces in these examples have singular points.

We have defined Galois embedding of algebraic variety and applied it to study several varieties. In particular we have shown that many abelian surfaces have Galois embeddings (Yoshihara 2007). However, as a corollary of Theorem 3, we have the following.

Corollary 7

There does not exist any Galois embedding of bi-elliptic surface.

2 Proof

Let S be a bi-elliptic surface and $\pi _1 : S \longrightarrow X$ a finite Galois covering. Then S can be expressed as $\mathbb {C}^{2}/\Gamma _{1}$, where $\Gamma _{1}$ is the complex crystallographic group given in [Suwa (1970), Theorem]. Let $A=\mathbb C^{2}/\Gamma _{0}$ be the abelian surface, where $\pi _2 : A \longrightarrow S$ is an unramified covering such that $\Gamma _0$ is the normal subgroup of $\Gamma _1$ and $|\Gamma _1 : \Gamma _0|=2, 3, 4 \ \mathrm {or} \ 6$. Put $n=|\Gamma _1 : \Gamma _0 |$. Let G be the finite subgroup of $\mathrm {Aut}(S)$ such that $X=S/G$. For $g \in G$, we let $\widetilde{g}$ be the lift of g on the universal covering $\mathbb C^2$.

Claim 8

The $\widetilde{g}$ turns out to be an affine transformation.

Proof

For each $s \in \Gamma _0^n$, there exists $t \in \Gamma _0$ such that $s=t^n$. Since $\Gamma _0 \subset \Gamma _1$ and $\Gamma _1$ is a discrete group, we have $u \in \Gamma _1$ such that $\widetilde{g}t=u\widetilde{g}$. Hence we get $\widetilde{g}t^n=u^n\widetilde{g}$ and $u^n \in \Gamma _0$. This implies $\widetilde{g}s \equiv \widetilde{g} ( \mod \Gamma _0)$. Let $\mathcal L_0$ be the lattice defining the abelian surface $A=\mathbb C^2/\Gamma _0$. Putting $z={}^t(z_1, z_2) \in \mathbb C^2$, $\widetilde{g}(z_1, z_2)=(\widetilde{g}_1(z_1, z_2), \ \widetilde{g}_2(z_1, z_2))$, we have

$$\begin{aligned} \widetilde{g}_i(z_1+\alpha _1, z_2+\alpha _2)=\widetilde{g}_i(z_1, z_2)+\beta _i, \end{aligned}$$

where $i=1, 2$, ${}^t(\alpha _1, \alpha _2) \in n \mathcal L_0$ and ${}^t(\beta _1, \beta _2) \in \mathcal L_0$. Hence we get

$$\begin{aligned} s^{*} \left( \frac{\partial \widetilde{g_i}}{\partial z_j} \right) = \frac{\partial \widetilde{g_i}}{\partial z_j} \ \ \ (i, j = 1, 2). \end{aligned}$$

This means $\partial \widetilde{g_i}/\partial z_j$ is a holomorphic function on $\mathbb C^2/{\Gamma _0^n}$, which is also an abelian surface. Therefore $\partial \widetilde{g_i}/\partial z_j$ is a constant, i.e., $\widetilde{g}$ is an affine transformation. $\square$

Thus $\widetilde{g}$ has the representation $\widetilde{g}z=M(g)z+v(g)$, where $z \in \mathbb C^2, \ M(g) \in GL(2, \mathbb C)$ and $v(g) \in \mathbb C^2$. We use this expression hereafter. Let $\Gamma _2$ be the affine transformation group generated by $\{ \ \widetilde{g} \ | \ g \in G \}$ and $\Gamma _1$. Then we have $X=\mathbb C^2/{\Gamma _2}$. Since $\pi _1 : S \longrightarrow X$ is a Galois covering, we have $\Gamma _0$ (resp. $\Gamma _1$) is a normal subgroup of $\Gamma _1$ (resp. $\Gamma _2$) and $\Gamma _2/\Gamma _1 \cong G$. Let $\Gamma _1/{\Gamma _0}$ be generated by $\sigma$. Then, we have the expression $\widetilde{\sigma }z=M(\sigma )z+v(\sigma )$, where

Claim 9

The X has two fibrations.

Proof

Since $\Gamma _1$ is a normal subgroup of $\Gamma _2$, there exists an integer r such that $\widetilde{g}\widetilde{\sigma }\widetilde{g}^{-1}=t {\widetilde{\sigma }}^r$, where $g \in G$ and $t \in \Gamma _0$. This means $M(g)M(\sigma )M(g)^{-1}=M(\sigma )^r$, hence $r=1$. Therefore we have $M(g)M(\sigma )=M(\sigma )M(g)$. If we put , it is easy to see that M(g) is a diagonal matrix. Hence the lift of each element $\gamma$ of $\Gamma _2$ can be expressed as

$$\begin{aligned} \widetilde{\gamma }(z)= \left( \begin{array}{ll} \alpha _1 &{}\quad 0 \\ 0 &{}\quad \alpha _2 \end{array}\right) z + \left( \begin{array}{ll} a_1 \\ a_2 \end{array}\right) . \qquad \qquad \qquad \qquad {(*)} \end{aligned}$$

Since $\mathcal L_0$ is the lattice defining the abelian surface A, we have $\Gamma _0=\{ \ \ell \ | \ \ell (z)=z+{}^t(b_1, b_2), \ \mathrm {where} \ {}^t(b_1, b_2) \in \mathcal {L}_0 \}$.

Put $\Gamma _{0i}=\{ \ \ell _i \ | \ \ell _i(z_i)=z_i+b_i, \mathrm {where} \ {}^t(b_1, b_2) \in \mathcal {L}_0 \}$. Referring to the list of abelian surfaces in Suwa (1970), we infer that $\mathbb C/{\Gamma _{0i}}$ is an elliptic curve. Using the above representation $(*)$, we define

$$\begin{aligned} \Gamma _{2i}=\{ \ \gamma _i \ | \ \gamma _i(z_i)=\alpha _i z_i+a_i, \mathrm {where} \ \gamma \in \widetilde{\Gamma _2} \}. \end{aligned}$$

Then $\Gamma _{0i}$ is a subgroup of $\Gamma _{2i}$ with a finite index. Hence $\Gamma _{0i}$ is a discrete subgroup and $\mathbb C/\Gamma _{2i}$ is a smooth curve. Thus we have the following diagram

where $i=1, 2$. Consequently we get two fibrations $\bar{p_i} : X \longrightarrow C_i$. $\square$

Suppose that X is smooth. Then we have

$$\begin{aligned} \dim \mathrm {H}^1(X,\ \mathcal O_X) \le \dim \mathrm {H}^1(S,\ \mathcal O_S) =1, \ \mathrm {and} \end{aligned}$$

$$\begin{aligned} \dim \mathrm {H}^2(X,\ \mathcal O_X) \le \dim \mathrm {H}^2(S,\ \mathcal O_S) =0 \end{aligned}$$

and the Kodaira dimension of X is less than or equal to 0. By Remark 2 and the classification theorem of algebraic surfaces, we infer that X is $\mathbb P^1 \times \mathbb P^1$, a ruled surface with irregularity one or a bi-elliptic surface.

Claim 10

The X cannot be $\mathbb P^1 \times \mathbb P^1$.

Proof

In the representation above $(*)$ there exists i such that $\alpha _i \ne 1$. Then the fiber space $\bar{p_i} : X \longrightarrow C_i$ has a multiple singular fiber. Let $F_i$ be the fiber of the projection $\pi _i : \mathbb P^1 \times \mathbb P^1 \longrightarrow \mathbb P^1$. Then each divisor D is linearly equivalent to $n_1F_1 + n_2F_2$. If $D^2=0$, then we have $n_1=0$ or $n_2=0$. So that there are only two fibrations from $\mathbb P^1 \times \mathbb P^1$ to $\mathbb P^1$, which are the first and second projections. Neither of projections have any multiple fiber. Combining the above assertions we infer readily the conclusion.

$\square$

This completes the proof.