The classification of surfaces with p_g=q=1 isogenous to a product of curves

TL;DR

This paper classifies all algebraic surfaces with geometric genus and irregularity equal to one that are constructed as quotients of products of two curves by a finite group, enriching the understanding of their structure.

Contribution

It provides a complete classification of surfaces with p_g=q=1 that are isogenous to a product, a previously unresolved problem in algebraic geometry.

Findings

Complete classification of surfaces with p_g=q=1 that are isogenous to a product.

Identification of all possible group actions on product of curves.

Structural insights into the geometry of these surfaces.

Abstract

A projective surface S is said to be isogenous to a product if there exist two smooth curves C, F and a finite group G acting freely on C \times F so that S=(C \times F)/G. In this paper we classify all surfaces with p_g=q=1 which are isogenous to a product.

Tables

			IdSmall
$g\left(F\right)=g_{\text{<em>alb</em>}}$	$g\left(C\right)$	$G$	Group$\left(G\right)$	Type	$D$	$N$
$3$	$3$	${\left({\mathbb{Z}}_2\right)}^2$	$G(4,2)$	unmixed $(\ast )$	$5$	$1$
$3$	$5$	${\left({\mathbb{Z}}_2\right)}^3$	$G(8,5)$	unmixed $(\ast )$	$4$	$1$
$3$	$5$	${\mathbb{Z}}_2\times {\mathbb{Z}}_4$	$G(8,2)$	unmixed $(\ast )$	$3$	$2$
$3$	$9$	${\mathbb{Z}}_2\times {\mathbb{Z}}_8$	$G(16,5)$	unmixed $(\ast )$	$2$	$1$
$3$	$5$	$D_4$	$G(8,3)$	unmixed	$3$	$1$
$3$	$7$	$D_6$	$G(12,4)$	unmixed $(\ast \ast )$	$3$	$1$
$3$	$9$	${\mathbb{Z}}_2\times D_4$	$G(16,11)$	unmixed	$3$	$1$
$3$	$13$	$D_{2,12,5}$	$G(24,5)$	unmixed	$2$	$1$
$3$	$13$	${\mathbb{Z}}_2\times A_4$	$G(24,13)$	unmixed	$2$	$1$
$3$	$13$	$S_4$	$G(24,12)$	unmixed	$2$	$1$
$3$	$17$	${\mathbb{Z}}_2⋉\left({\mathbb{Z}}_2\times {\mathbb{Z}}_8\right)$	$G(32,9)$	unmixed	$2$	$1$
$3$	$25$	${\mathbb{Z}}_2\times S_4$	$G(48,48)$	unmixed	$2$	$1$
$4$	$3$	$S_3$	$G(6,1)$	unmixed $(\ast \ast )$	$4$	$1$
$4$	$5$	$D_6$	$G(12,4)$	unmixed	$3$	$1$
$4$	$7$	${\mathbb{Z}}_3\times S_3$	$G(18,3)$	unmixed	$2$	$2$
$4$	$7$	${\mathbb{Z}}_3\times S_3$	$G(18,3)$	unmixed	$1$	$1$
$4$	$9$	$S_4$	$G(24,12)$	unmixed $(\ast \ast )$	$2$	$1$
$4$	$13$	$S_3\times S_3$	$G(36,10)$	unmixed	$1$	$1$
$4$	$13$	${\mathbb{Z}}_6\times S_3$	$G(36,12)$	unmixed	$1$	$1$
$4$	$13$	${\mathbb{Z}}_4⋉{\left({\mathbb{Z}}_3\right)}^2$	$G(36,9)$	unmixed	$1$	$2$
$4$	$21$	$A_5$	$G(60,5)$	unmixed $(\ast \ast )$	$1$	$1$
$4$	$25$	${\mathbb{Z}}_3\times S_4$	$G(72,42)$	unmixed	$1$	$1$
$4$	$41$	$S_5$	$G(120,34)$	unmixed	$1$	$1$
$5$	$3$	$D_4$	$G(8,3)$	unmixed $(\ast \ast )$	$4$	$1$
$5$	$4$	$A_4$	$G(12,3)$	unmixed $(\ast \ast )$	$2$	$2$
$5$	$5$	${\mathbb{Z}}_4⋉{\left({\mathbb{Z}}_2\right)}^2$	$G(16,3)$	unmixed	$2$	$3$
$5$	$7$	${\mathbb{Z}}_2\times A_4$	$G(24,13)$	unmixed	$2$	$2$
$5$	$7$	${\mathbb{Z}}_2\times A_4$	$G(24,13)$	unmixed	$1$	$1$
$5$	$9$	${\mathbb{Z}}_8⋉{\left({\mathbb{Z}}_2\right)}^2$	$G(32,5)$	unmixed	$1$	$1$
$5$	$9$	${\mathbb{Z}}_2⋉D_{2,8,5}$	$G(32,7)$	unmixed	$1$	$1$
$5$	$9$	${\mathbb{Z}}_4⋉\left({\mathbb{Z}}_4\times {\mathbb{Z}}_2\right)$	$G(32,2)$	unmixed	$1$	$1$
$5$	$9$	${\mathbb{Z}}_4⋉{\left({\mathbb{Z}}_2\right)}^3$	$G(32,6)$	unmixed	$1$	$1$
$5$	$13$	${\left({\mathbb{Z}}_2\right)}^2\times A_4$	$G(48,49)$	unmixed	$1$	$1$
$5$	$17$	${\mathbb{Z}}_4⋉{\left({\mathbb{Z}}_2\right)}^4$	$G(64,32)$	unmixed	$1$	$2$
$5$	$21$	${\mathbb{Z}}_5⋉{\left({\mathbb{Z}}_2\right)}^4$	$G(80,49)$	unmixed	$1$	$2$
$5$	$5$	$D_{2,8,3}$	$G(16,8)$	mixed	$2$	$1$
$5$	$5$	$D_{2,8,5}$	$G(16,6)$	mixed	$2$	$3$
$5$	$5$	${\mathbb{Z}}_4⋉{\left({\mathbb{Z}}_2\right)}^2$	$G(16,3)$	mixed	$2$	$1$

	IdSmall
$G$	Group$\left(G\right)$	$𝐦$
$D_4$	$G(8,3)$	$(2^2,4^2)$
$D_6$	$G(12,4)$	$(2^3,6)$
${\mathbb{Z}}_2\times D_4$	$G(16,11)$	$(2^3,4)$
$D_{2,12,5}$	$G(24,5)$	$(2,4,12)$
${\mathbb{Z}}_2\times A_4$	$G(24,13)$	$(2,6^2)$
$S_4$	$G(24,12)$	$(3,4^2)$
${\mathbb{Z}}_2⋉\left({\mathbb{Z}}_2\times {\mathbb{Z}}_8\right)$	$G(32,9)$	$(2,4,8)$
${\mathbb{Z}}_2\times S_4$	$G(48,48)$	$(2,4,6)$

	IdSmall
$G$	Group$\left(G\right)$	$𝐦$
$S_3$	$G(6,1)$	$\left(2^6\right)$
$D_6$	$G(12,4)$	$\left(2^5\right)$
${\mathbb{Z}}_3\times S_3$	$G(18,3)$	$(2^2,3^2)$
${\mathbb{Z}}_3\times S_3$	$G(18,3)$	$(3,6^2)$
$S_4$	$G(24,12)$	$(2^3,4)$
$S_3\times S_3$	$G(36,10)$	$(2,6^2)$
${\mathbb{Z}}_6\times S_3$	$G(36,12)$	$(2,6^2)$
${\mathbb{Z}}_4⋉{\left({\mathbb{Z}}_3\right)}^2$	$G(36,9)$	$(3,4^2)$
$A_5$	$G(60,5)$	$(2,5^2)$
${\mathbb{Z}}_3\times S_4$	$G(72,42)$	$(2,3,12)$
$S_5$	$G(120,34)$	$(2,4,5)$

	IdSmall
$G$	Group$\left(G\right)$	$𝐦$
$D_4$	$G(8,3)$	$\left(2^6\right)$
$A_4$	$G(12,3)$	$\left(3^4\right)$
${\mathbb{Z}}_4⋉{\left({\mathbb{Z}}_2\right)}^2$	$G(16,3)$	$(2^2,4^2)$
${\mathbb{Z}}_2\times A_4$	$G(24,13)$	$(2^2,3^2)$
${\mathbb{Z}}_2\times A_4$	$G(24,13)$	$(3,6^2)$
${\mathbb{Z}}_8⋉{\left({\mathbb{Z}}_2\right)}^2$	$G(32,5)$	$(2,8^2)$
${\mathbb{Z}}_2⋉D_{2,8,5}$	$G(32,7)$	$(2,8^2)$
${\mathbb{Z}}_4⋉\left({\mathbb{Z}}_4\times {\mathbb{Z}}_2\right)$	$G(32,2)$	$\left(4^3\right)$
${\mathbb{Z}}_4⋉{\left({\mathbb{Z}}_2\right)}^3$	$G(32,6)$	$\left(4^3\right)$
${\left({\mathbb{Z}}_2\right)}^2\times A_4$	$G(48,49)$	$(2,6^2)$
${\mathbb{Z}}_4⋉{\left({\mathbb{Z}}_2\right)}^4$	$G(64,32)$	$(2,4,8)$
${\mathbb{Z}}_5⋉{\left({\mathbb{Z}}_2\right)}^4$	$G(80,49)$	$(2,5^2)$

	IdSmall		IdSmall
$G^{\circ }$	Group$\left(G^{\circ }\right)$	$G$	Group$\left(G\right)$
$D_4$	$G(8,3)$	$D_{2,8,3}$	$G(16,8)$
${\mathbb{Z}}_2\times {\mathbb{Z}}_4$	$G(8,2)$	$D_{2,8,5}$	$G(16,6)$
${\left({\mathbb{Z}}_2\right)}^3$	$G(8,5)$	${\mathbb{Z}}_4⋉{\left({\mathbb{Z}}_2\right)}^2$	$G(16,3)$

Equations

$$\begin{array}{rl}{K}_S^2& =\frac{K_{C\times F}^2}{|G|}=\frac{8(g(C)-1)(g(F)-1)}{|G|}\\ \chi ({\mathcal{O}}_S)& =\frac{\chi ({\mathcal{O}}_{C\times F})}{|G|}=\frac{(g(C)-1)(g(F)-1)}{|G|},\end{array}$$

$$\mathcal{V}=\{g_1,\dots ,g_r;h_1,\dots ,h_{2{\mathfrak{g}}^{\prime }}\}$$

$$\begin{split}2g(F)-2&=|G|\bigg{(}-2+\sum_{i=1}^{r}\bigg{(}1-\frac{1}{\;m_{i}}\bigg{)}\bigg{)},\qquad\\ 2g(C)-2&=|G|\sum_{j=1}^{s}\bigg{(}1-\frac{1}{\;n_{j}}\bigg{)}.\end{split}$$

$$(U)\left(\bigcup _{\sigma \in G}\bigcup _{i=1}^r⟨\sigma g_i{\sigma }^{-1}⟩\right)\cap \left(\bigcup _{\sigma \in G}\bigcup _{j=1}^s⟨\sigma {\ell }_j{\sigma }^{-1}⟩\right)=\{1_G\}.$$

$$\begin{array}{rl}\mathbb{C}=H^0({\Omega }_S^1)& =(H^0({\Omega }_C^1)\oplus H^0({\Omega }_C^1){)}^G=(H^0({\Omega }_C^1{)}^{G^{\circ }}\oplus H^0({\Omega }_C^1{)}^{G^{\circ }}{)}^{G/G^{\circ }}\\ & =(H^0({\Omega }_E^1)\oplus H^0({\Omega }_E^1){)}^{G/G^{\circ }}.\end{array}$$

$$C\times C\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$1⟶G^{\circ }⟶G⟶{\mathbb{Z}}_2⟶1$$

$$\{{\ell }_1,\dots ,{\ell }_s\}\cap \{g{\ell }_1{g}^{-1},\dots ,g{\ell }_s{g}^{-1}\}=\varnothing ;$$

$$g^2\notin \bigcup _{j=1}^s\bigcup _{\sigma \in G^{\circ }}⟨\sigma {\ell }_j{\sigma }^{-1}⟩.$$

$$2g(F^{\ast })-2=|G^{\circ }{|}^2\cdot \sum _{j=1}^{s}2\left(1-\frac{1}{n_j}\right).$$

$$\begin{array}{rl}2g(F_{\text{alb}})-2=\frac{1}{|G|}(2g(F^{\ast })-2)& =|G^{\circ }|\sum _{j=1}^s\left(1-\frac{1}{n_j}\right)\\ & =2g(C)-2,\end{array}$$

$$\Theta (\mathbf{m}):=-2+\sum _{i=1}^r\left(1-\frac{1}{m_i}\right),\alpha (\mathbf{m}):=\frac{2}{\Theta (\mathbf{m})}.$$

$$\mathcal{T}=\left\{\begin{array}{cccccc}(2,3,7{)}_{84},& (2,3,8{)}_{48},& (2,4,5{)}_{40},& (2,3,9{)}_{36},& (2,3,10{)}_{30},& (2,3,12{)}_{24},\\ (2,4,6{)}_{24},& (3^2,4{)}_{24},& (2,5^2{)}_{20},& (2,3,18{)}_{18},& (2,4,8{)}_{16},& (3^2,5{)}_{15},\\ (2,4,12{)}_{12},& (2,6^2{)}_{12},& (3^2,6{)}_{12},& (3,4^2{)}_{12},& (2,5,10{)}_{10},& (3^2,9{)}_9,\\ (2,8^2{)}_8,& (4^3{)}_8,& (3,6^2{)}_6,& (5^3{)}_5,& (2^3,3{)}_{12},& (2^3,4{)}_8,\\ (2^3,6{)}_6,& (2^2,3^2{)}_6,& (2^2,4^2{)}_4,& (3^4{)}_3,& (2^5{)}_4,& (2^6{)}_2\end{array}\right\}$$

$$\begin{array}[]{c|c}G^{\circ}&G\\ \hline\cr D_{4}&D_{2,8,3}\\ \mathbb{Z}_{2}\times\mathbb{Z}_{4}&D_{2,8,5}\\ \mathbb{Z}_{2}\times\mathbb{Z}_{4}&\mathbb{Z}_{2}\times Q_{8}\\ \mathbb{Z}_{2}\times\mathbb{Z}_{4}&D_{4,4,-1}\\ (\mathbb{Z}_{2})^{3}&\mathbb{Z}_{4}\ltimes(\mathbb{Z}_{2})^{2}\\ \end{array}$$

$$1⟶{\mathbb{Z}}_3\times S_3⟶G⟶{\mathbb{Z}}_2⟶1.$$

$$\begin{array}[]{c|c|c|c|c|c|c|c|c}t&2&4&5&6&7&8&12&17\\ \hline\cr n_{2}(G(32,t))&7&3&7&11&11&3&3&3\\ \end{array}$$

$$1⟶G(32,t)⟶G(64,s)⟶{\mathbb{Z}}_2⟶1$$

$$\begin{array}[]{c|c}G^{\circ}&G\\ \hline\cr G(32,5)&G(64,5)\\ G(32,6)&G(64,33)\\ G(32,7)&G(64,33)\\ G(32,6)&G(64,35)\\ G(32,8)&G(64,37)\end{array}$$

$$G^{\circ }=⟨x,y,z|x^8=y^2=z^2=1,[y,z]=[x,z]=1,[x,y]=z⟩.$$

$$x:=\text{f1},y:=\text{f2},z:=\text{f3},w:=\text{f4},v:=\text{f5},u:=\text{f6}$$

$$\begin{array}{rl}G(64,33)=⟨x,y,z,w,v,u|& z^2=w^2=v^2=u^2=1,x^2=w,y^2=u,\\ & [x,zy]=z,[x,vz]=v,[x,vu]=u,\\ & [y,z]=[y,v]=[z,v]=[w,v]=[x,u]=1⟩\end{array}$$

$$\begin{array}{rl}G(64,35)=⟨x,y,z,w,v,u|& w^2=v^2=u^2=1,z^2=y^2=u,x^2=w,\\ & [y,z]=[z,w]=u,[x,yz]=z,[x,z]=uv,\\ & [y,v]=[z,v]=[w,v]=[x,u]=1⟩\end{array}$$

$$\begin{array}{rl}G(64,37)=⟨x,y,z,w,v,u|& v^2=u^2=1,w^2=z^2=y^2=u,x^2=w,\\ & [y,z]=[z,w]=u,[x,yz]=z,[x,z]=uv,\\ & [y,v]=[z,v]=[w,v]=1⟩.\end{array}$$

$$N_1:=⟨x,z,w,v,u⟩,N_2:=⟨xy,z,w,v,u⟩.$$

$$N_3:=⟨x,z,w,v,u⟩,N_4:=⟨xy,z,w,v,u⟩.$$

$$N_5:=⟨x,z,w,v,u⟩,N_6:=⟨xy,z,w,v,u⟩.$$

$$\text{dim }\mathfrak{M}(G,\mathbf{m},\mathbf{n})=r+s-3,\text{dim }\mathfrak{M}(G,\mathbf{n})=s.$$

$$\begin{array}{rl}{\text{Mod}}_{0,[r]}:=⟨{\sigma }_1,\dots ,{\sigma }_r|& {\sigma }_i{\sigma }_{i+1}{\sigma }_i={\sigma }_{i+1}{\sigma }_i{\sigma }_{i+1},\\ & {\sigma }_i{\sigma }_j={\sigma }_j{\sigma }_i\text{if}|i-j|\ge 2,\\ & {\sigma }_{r-1}{\sigma }_{r-2}\cdots {\sigma }_1^2\cdots {\sigma }_{r-2}{\sigma }_{r-1}=1⟩,\end{array}$$

$${\text{Mod}}_{1,1}:=⟨t_{\alpha },t_{\beta },t_{\gamma }|t_{\alpha }{t}_{\beta }{t}_{\alpha }=t_{\beta }{t}_{\alpha }{t}_{\beta },(t_{\alpha }{t}_{\beta }{)}^3=1⟩,$$

$$\begin{array}{rl}{\text{Mod}}_{1,[2]}:=⟨t_{\alpha },t_{\beta },t_{\gamma },\rho |t_{\alpha }{t}_{\beta }{t}_{\alpha }& =t_{\beta }{t}_{\alpha }{t}_{\beta },t_{\alpha }{t}_{\gamma }{t}_{\alpha }=t_{\gamma }{t}_{\alpha }{t}_{\gamma },\\ t_{\beta }{t}_{\gamma }& =t_{\gamma }{t}_{\beta },(t_{\alpha }{t}_{\beta }{t}_{\gamma }{)}^4=1,\\ t_{\alpha }\rho & =\rho t_{\alpha },t_{\beta }\rho =\rho t_{\beta },t_{\gamma }\rho =\rho t_{\gamma }⟩.\end{array}$$

Full text

0. Introduction

The classification of smooth, complex surfaces $S$ of general type with small birational invariants is quite a natural problem in the framework of algebraic geometry. For instance, one may want to understand the case where the Euler characteristic $\chi(\mathcal{O}_{S})$ is $1$, that is, when the geometric genus $p_{g}(S)$ is equal to the irregularity $q(S)$. All surfaces of general type with these invariants satisfy $p_{g}\leq 4$. In addition, if $p_{g}=q=4$ then the self-intersection $K_{S}^{2}$ of the canonical class of $S$ is equal to $8$ and $S$ is the product of two genus $2$ curves, whereas if $p_{g}=q=3$ then $K_{S}^{2}=6$ or $8$ and both cases are completely described ([CCML98], [HP02], [Pir02]). On the other hand, surfaces of general type with $p_{g}=q=0,1,2$ are still far from being classified. We refer the reader to the survey paper [BaCaPi06] for a recent account on this topic and a comprehensive list of references.

A natural way of producing interesting examples of algebraic surfaces is to construct them as quotients of known ones by the action of a finite group. For instance Godeaux constructed in [Go31] the first example of surface of general type with vanishing geometric genus taking the quotient of a general quintic surface of $\mathbb{P}^{3}$ by a free action of $\mathbb{Z}_{5}$. In line with this Beauville proposed in [Be96, p. 118] the construction of a surface of general type with $p_{g}=q=0$, $K_{S}^{2}=8$ as the quotient of a product of two curves $C$ and $F$ by the free action of a finite group $G$ whose order is related to the genera $g(C)$ and $g(F)$ by the equality $|G|=(g(C)-1)(g(F)-1)$. Generalizing Beauville’s example we say that a surface $S$ is isogenous to a product if $S=(C\times F)/G$, for $C$ and $F$ smooth curves and $G$ a finite group acting freely on $C\times F$. A systematic study of these surfaces has been carried out in [Ca00]. They are of general type if and only if both $g(C)$ and $g(F)$ are greater than or equal to $2$ and in this case $S$ admits a unique minimal realization where they are as small as possible. From now on, we tacitly assume that such a realization is chosen, so that the genera of the curves and the group $G$ are invariants of $S$. The action of $G$ can be seen to respect the product structure on $C\times F$. This means that such actions fall in two cases: the mixed one, where there exists some element in $G$ exchanging the two factors (in this situation $C$ and $F$ must be isomorphic) and the unmixed one, where $G$ acts faithfully on both $C$ and $F$ and diagonally on their product.

After [Be96], examples of surfaces isogenous to a product with $p_{g}=q=0$ appeared in [Par03] and [BaCa03], and their complete classification was obtained in [BaCaGr06].

The next natural step is therefore the analysis of the case $p_{g}=q=1$. Surfaces of general type with these invariants are the irregular ones with the lowest geometric genus and for this reason it would be important to provide their complete description. So far, this has been obtained only in the cases $K_{S}^{2}=2,3$ ([Ca81], [CaCi91], [CaCi93], [Pol05], [CaPi06]).

The goal of the present paper is to give the full list of surfaces with $p_{g}=q=1$ that are isogenous to a product. Our work has to be seen as the sequel to the article [Pol07], which describe all unmixed cases with $G$ abelian and some unmixed examples with $G$ nonabelian. Apart from the complete list of the genera and groups occurring, our paper contains the first examples of surfaces of mixed type with $q=1$. The mixed cases turn out to be much less frequent than the unmixed ones and, as when $p_{g}=q=0$, they occur for only one value of the order of $G$. However, in contrast with what happens when $p_{g}=q=0$, the mixed cases do not correspond to the maximum value of $|G|$ but appear for a rather small order, namely $|G|=16$.

Our classification procedure involves arguments from both geometry and computational group theory. We will give here a brief account on how the result is achieved.

If $S$ is any surface isogenous to a product and satisfying $p_{g}=q$ then $|G|$, $g(C)$, $g(F)$ are related as in Beauville’s example and we have $K^{2}_{S}=8$. Besides, if $p_{g}=q=1$ such surfaces are necessarily minimal and of general type (Lemma 2.1).

If $S=(C\times F)/G$ is of unmixed type then the two projections $\pi_{C}\colon C\times F\longrightarrow C$, $\pi_{F}\colon C\times F\longrightarrow F$ induce two morphisms $\alpha\colon S\longrightarrow C/G$, $\beta\colon S\longrightarrow F/G$, whose smooth fibres are isomorphic to $F$ and $C$, respectively. Moreover, the geometry of $S$ is encoded in the geometry of the two coverings $h\colon C\longrightarrow C/G$, $f\colon F\longrightarrow F/G$ and the invariants of $S$ impose strong restrictions on $g(C)$, $g(F)$ and $|G|$. Indeed we have $1=q(S)=g(C/G)+g(F/G)$ so we may assume that $E:=C/G$ is an elliptic curve and $F/G\cong\mathbb{P}^{1}$. Then $\alpha\colon S\longrightarrow E$ is the Albanese morphism of $S$ and the genus $g_{\textrm{alb}}$ of the general Albanese fibre equals $g(F)$. It is proven in [Pol07, Proposition 2.3] that $3\leq g(F)\leq 5$; in particular this allows us to control $|G|$. The covers $f$ and $h$ are determined by two suitable systems of generators for $G$, that we call $\mathcal{V}$ and $\mathcal{W}$, respectively. Besides, in order to obtain a free action of $G$ on $C\times F$ and a quotient $S$ with the desired invariants, $\mathcal{V}$ and $\mathcal{W}$ are subject to strict conditions of combinatorial nature (Proposition 2.2). The geometry imposes also strong restrictions on the possible $\mathcal{W}$ and the genus of $C$, so the existence of $\mathcal{V}$ and $\mathcal{W}$ and the compatibility conditions can be verified through a computer search. It is worth mentioning that the classification of finite groups of automorphisms acting on curves of genus lesser than or equal to $5$ could have also been retrieved from the existing literature ([Br90], [Ki03], [KuKi90], [KuKu90]).

If $S=(C\times C)/G$ is of mixed type then the index two subgroup $G^{\circ}$ of $G$ corresponding to transformations that do not exchange the coordinates in $C\times C$ acts faithfully on $C$. The quotient $E=C/G^{\circ}$ is isomorphic to the Albanese variety of $S$ and $g_{\textrm{alb}}=g(C)$ (Proposition 2.5). Moreover $g(C)$ may only be $5$, $7$ or $9$, hence $|G|$ is at most $64$ (Proposition 2.10). The cover $h\colon C\longrightarrow E$ is determined by a suitable system of generators $\mathcal{V}$ for $G^{\circ}$ and since the action of $G$ on $C\times C$ is required to be free, combinatorial restrictions involving the elements of $\mathcal{V}$ and those of $G\setminus G^{\circ}$ have to be imposed (Proposition 2.6). Our classification is obtained by first listing those groups $G^{\circ}$ for which $\mathcal{V}$ exists and then by looking at the admissible extensions $G$ of $G^{\circ}$. We find that the only possibility occurring is for $g(C)=5$ so that $|G|$ is necessarily $16$ (Propositions 4.1, 4.2, 4.3).

In the last part of the paper we examine the structure of the subset of the moduli space corresponding to surfaces isogenous to a product with $p_{g}=q=1$. It can be explicitly described by calculating the number of orbits of the direct product of certain mapping class groups with $\textrm{Aut}(G)$ acting on the set (of pairs) of systems of generators (Proposition 5.1). In particular it is possible to determine the number of irreducible connected components and their respective dimensions, see the forthcoming article [Pe08].

Our computations were carried out by using the computer algebra program GAP4, whose database includes all groups of order less than $2000$, with the exception of $1024$ (see [GAP4]). For the reader’s convenience we included the scripts in the Appendix.

Now let us state the main result of this paper.

Main Theorem**.**

Let $S=(C\times F)/G$ be a surface with $p_{g}=q=1$, isogenous to a product of curves. Then $S$ is minimal of general type and the occurrences for $g(F)$, $g(C)$, $G$, the dimension $D$ of the moduli space and the number $N$ of its connected components are precisely those in the table below.

Here IdSmallGroup$(G)$ denotes the label of the group $G$ in the GAP4 database of small groups. The calculation of $N$ is due to Penegini and Rollenske, see [Pe08], except for the cases marked with $(*)$, which were already studied in [Pol07]. The cases marked with $(**)$ also appeared in [Pol07], but the computation of $N$ was missing.

This work is organized as follows.

In Section 1 we collect the basic facts about surfaces isogenous to a product, following the treatment given by Catanese in [Ca00] and we fix the algebraic setup.

In Section 2 we apply the structure theorems of Catanese to the case $p_{g}=q=1$ and this leads to Propositions 2.2 and 2.6, that provide the translation of our classification problem from geometry to algebra. All these results are used in Sections 3 and 4, which are the core of the paper and give the complete lists of the occurring groups and genera in the unmixed and mixed cases, respectively.

Finally, Section 5 is devoted to the description of the moduli spaces.

$\mathbf{Notations\;and\;conventions}$. All varieties, morphisms, etc. in this article are defined over $\mathbb{C}$. By “surface” we mean a projective, non-singular surface $S$, and for such a surface $K_{S}$ denotes the canonical class, $p_{g}(S)=h^{0}(S,\;K_{S})$ is the geometric genus, $q(S)=h^{1}(S,\;K_{S})$ is the irregularity and $\chi(\mathcal{O}_{S})=1-q(S)+p_{g}(S)$ is the Euler characteristic. Throughout the paper we use the following notation for groups:

• •

$\mathbb{Z}_{n}$: cyclic group of order $n$.

• •

$D_{p,q,r}=\mathbb{Z}_{p}\ltimes\mathbb{Z}_{q}=\langle x,y\;|\;x^{p}=y^{q}=1,\;xyx^{-1}=y^{r}\rangle$: split metacyclic group of order $pq$. The group $D_{2,n,-1}$ is the dihedral group of order $2n$ and it will be denoted by $D_{n}$.

• •

$S_{n},\;A_{n}$: symmetric, alternating group on $n$ symbols.

• •

If $x,\,y\in G$, their commutator is defined as $[x,y]=xyx^{-1}y^{-1}$.

• •

If $x\in G$ we denote by $\textrm{Int}_{x}$ the inner automorphism of $G$ defined as $\textrm{Int}_{x}(g)=xgx^{-1}$.

• •

IdSmallGroup$(G)$ indicates the label of the group $G$ in the GAP4 database of small groups. For instance IdSmallGroup$(D_{4})=G(8,3)$ and this means that $D_{4}$ is the third in the list of groups of order $8$.

$\mathbf{Acknowledgements}.$ The authors wish to thank M. Penegini and S. Rollenske for giving them a preliminary version of [Pe08] and for kindly allowing them to include their results in the Main Theorem. Moreover they are indebted with the referee for several valuable comments and suggestions to improve this article.

1. Basic on surfaces isogenous to a product

In this section we collect for the reader’s convenience some basic results on groups acting on curves and surfaces isogenous to a product, referring to [Ca00] for further details.

Definition 1.1**.**

A complex surface $S$ of general type is said to be isogenous to a product if there exist two smooth curves $C$, $F$ and a finite group $G$ acting freely on $C\times F$ so that $S=(C\times F)/G$.

There are two cases: the unmixed one, where $G$ acts diagonally, and the mixed one, where there exist elements of $G$ exchanging the two factors $($and then $C$, $F$ are isomorphic$)$.

In both cases, since the action of $G$ on $C\times F$ is free, we have

[TABLE]

hence $K_{S}^{2}=8\chi(\mathcal{O}_{S})$.

Let $C$, $F$ be curves of genus $\geq 2$. Then the inclusion $\textrm{Aut}(C\times F)\supset\textrm{Aut}(C)\times\textrm{Aut}(F)$ is an equality if $C$ and $F$ are not isomorphic, whereas $\textrm{Aut}(C\times C)=\mathbb{Z}_{2}\ltimes(\textrm{Aut}(C)\times\textrm{Aut}(C))$, the $\mathbb{Z}_{2}$ being generated by the involution exchanging the two coordinates. If $S=(C\times F)/G$ is a surface isogenous to a product, we will always consider its unique minimal realization. This means that

• •

in the unmixed case, we have $G\subset\textrm{Aut}(C)$ and $G\subset\textrm{Aut}(F)$ (i.e. $G$ acts faithfully on both $C$ and $F$);

• •

in the mixed case, where $C\cong F$, we have $G^{\circ}\subset\textrm{Aut}(C)$, for $G^{\circ}:=G\cap(\textrm{Aut}(C)\times\textrm{Aut}(C))$.

(See [Ca00, Corollary 3.9 and Remark 3.10]).

Definition 1.2**.**

Let $G$ be a finite group and let $\mathfrak{g}^{\prime}\geq 0$, and $m_{r}\geq m_{r-1}\geq\ldots\geq m_{1}\geq 2$ be integers. A generating vector for $G$ of type $(\mathfrak{g}^{\prime}\;|\;m_{1},\ldots,m_{r})$ is a $(2\mathfrak{g}^{\prime}+r)$-ple of elements

[TABLE]

such that: the set $\mathcal{V}$ generates $G$;  $|g_{i}|=m_{i}$ and $g_{1}g_{2}\cdots g_{r}\Pi_{i=1}^{\mathfrak{g}^{\prime}}[h_{i},h_{i+\mathfrak{g}^{\prime}}]=1$. If such a $\mathcal{V}$ exists, then $G$ is said to be $(\mathfrak{g}^{\prime}\;|\;m_{1},\ldots,m_{r})$-generated.

For convenience we make abbreviations such as $(4\;|\;2^{3},3^{2})$ for $(4\;|\;2,2,2,3,3)$ when we write down the type of the generating vector $\mathcal{V}$.

By Riemann’s existence theorem a finite group $G$ acts as a group of automorphisms of some compact Riemann surface $X$ of genus $\mathfrak{g}$ with quotient a Riemann surface $Y$ of genus $\mathfrak{g}^{\prime}$ if and only if there exist integers $m_{r}\geq m_{r-1}\geq\ldots\geq m_{1}\geq 2$ such that $G$ is $(\mathfrak{g}^{\prime}\;|\;m_{1},\ldots,m_{r})$-generated and $\mathfrak{g}$, $\mathfrak{g}^{\prime}$, $|G|$ and the $m_{i}$ are related by the Riemann-Hurwitz formula. Moreover, if $\mathcal{V}=\{g_{1},\ldots,g_{r};\;h_{1},\ldots,h_{2\mathfrak{g}^{\prime}}\}$ is a generating vector for $G$, the subgroups $\langle g_{i}\rangle$ and their conjugates are precisely the nontrivial stabilizers of the $G$-action ([Br90, Section 2], [Bre00, Chapter 3], [H71]). The description of surfaces isogenous to a product can be therefore reduced to finding suitable generating vectors. Requiring that $S$ has given invariants $p_{g}$ and $q$ imposes numerical restrictions on the order of the group $G$ and the genus of the curves $C$ and $F$. Our goal is to classify all surfaces with $p_{g}=q=1$ isogenous to a product. The aim of the next section is to translate this classification problem from geometry to algebra.

2. The case $p_{g}=q=1$. Building data

Lemma 2.1**.**

Let $S=(C\times F)/G$ be a surface isogenous to a product with $p_{g}=q=1$. Then

• $(i)$

$K_{S}^{2}=8$.

• $(ii)$

$|G|=(g(C)-1)(g(F)-1)$.

• $(iii)$

$S$* is a minimal surface of general type.*

Proof.

Claims $(i)$ and $(ii)$ follow from (1). Now let us consider $(iii)$. Since $C\times F$ is minimal and the cover $C\times F\longrightarrow S$ is étale, $S$ is minimal as well. Moreover $(ii)$ implies either $g(C)=g(F)=0$ or $g(C)\geq 2$, $g(F)\geq 2$. The first case is impossible otherwise $S=\mathbb{P}^{1}\times\mathbb{P}^{1}$ and $p_{g}=q=0$; thus the second case occurs, hence $S$ is of general type. ∎

2.1. Unmixed case

If $S=(C\times F)/G$ is a surface with $p_{g}=q=1$, isogenous to an unmixed product, then $g(C)\geq 3$, $g(F)\geq 3$ and up to exchanging $F$ and $C$ one may assume $F/G\cong\mathbb{P}^{1}$ and $C/G\cong E$, where $E$ is an elliptic curve. Moreover $\alpha\colon S\longrightarrow C/G$ is the Albanese morphism of $S$ and $g_{\emph{alb}}=g(F)$, see [Pol07, Proposition 2.2]. This leads to

Proposition 2.2**.**

$($[Pol07, Proposition 3.1]$)$* Let $G$ be a finite group which is both $(0\;|\;m_{1},\ldots,m_{r})$ and $(1\;|\;n_{1},\ldots,n_{s})$-generated, with generating vectors $\mathcal{V}=\{g_{1},\ldots,g_{r}\}$ and $\mathcal{W}=\{\ell_{1},\ldots,\ell_{s};\;h_{1},h_{2}\}$, respectively. Let $g(F),\;g(C)$ be the positive integers defined by the Riemann-Hurwitz relations*

[TABLE]

Assume moreover that $g(C)\geq 3$, $g(F)\geq 3$, $|G|=(g(C)-1)(g(F)-1)$ and

[TABLE]

Then there is a free, diagonal action of $G$ on $C\times F$ such that the quotient $S=(C\times F)/G$ is a minimal surface of general type with $p_{g}=q=1$, $K_{S}^{2}=8$. Conversely, every surface with $p_{g}=q=1$, isogenous to an unmixed product, arises in this way.

Here, condition $(U)$ ensures that the $G$-action on $C\times F$ is free.

Set $\mathbf{m}:=(m_{1},\ldots,m_{r})$ and $\mathbf{n}:=(n_{1},\ldots,n_{s})$; if $S=(C\times F)/G$ is a surface with $p_{g}=q=1$ which is constructed by using the recipe in Proposition 2.2, it will be called an unmixed surface of type $(G,\,\mathbf{m},\,\mathbf{n})$.

Proposition 2.3**.**

$($[Pol07, Proposition 2.3]$)$* Let $S=(C\times F)/G$ be an unmixed surface of type $(G,\,\mathbf{m},\,\mathbf{n})$. Then there are exactly the following possibilities:*

• $(a)$

$g(F)=3,\quad\mathbf{n}=(2^{2})$**

• $(b)$

$g(F)=4,\quad\mathbf{n}=(3)$**

• $(c)$

$g(F)=5,\quad\mathbf{n}=(2)$.

The following lemma gives a restriction on $\mathbf{m}$ instead.

Lemma 2.4**.**

Let $S=(C\times F)/G$ be an unmixed surface of type $(G,\,\mathbf{m},\,\mathbf{n})$. Then every $m_{i}$ divides $\frac{|G|}{(g(F)-1)}$.

Proof.

Since $\langle g_{i}\rangle$ is a stabilizer for the $G$-action on $F$ and since $G$ acts freely on $(C\times F)$, the subgroup $\langle g_{i}\rangle\cong\mathbb{Z}_{m_{i}}$ acts freely on $C$. By Riemann-Hurwitz formula applied to the cover $C\longrightarrow C/\langle g_{i}\rangle$ we have $g(C)-1=m_{i}(g(C/\langle g_{i}\rangle)-1)$. Thus $m_{i}$ divides $g(C)-1=\frac{|G|}{(g(F)-1)}$. ∎

2.2. Mixed case

Proposition 2.5**.**

Let $S=(C\times C)/G$ be a surface with $p_{g}=q=1$ isogenous to a mixed product. Then $E:=C/G^{\circ}$ is an elliptic curve isomorphic to the Albanese variety of $S$.

Proof.

We have (see [Ca00, Proposition 3.15])

[TABLE]

Since $S$ is of mixed type, the quotient $\mathbb{Z}_{2}=G/G^{\circ}$ exchanges the last two summands, whence $h^{0}(\Omega_{E}^{1})=1$. Thus $E$ is an elliptic curve and there is a commutative diagram

[TABLE]

showing that the Albanese morphism $\alpha$ of $S$ factors through the Abel-Jacobi map $\hat{\alpha}$ of the double symmetric product $E^{(2)}$ of $E$. ∎

By Lemma 2.1 we have $|G|=(g(C)-1)^{2}$. In this case [Ca00, Proposition 3.16] becomes

Proposition 2.6**.**

Assume that $G^{\circ}$ is a $(1\;|\;n_{1},\ldots,n_{s})$-generated finite group with generating vector $\mathcal{V}=\{\ell_{1},\ldots,\ell_{s};\;h_{1},h_{2}\}$ and that there is a nonsplit extension

[TABLE]

which gives an involution $[\varphi]$ in Out$(G^{\circ})$. Let $g(C)\in\mathbb{N}$ be defined by the Riemann-Hurwitz relation $2g(C)-2=|G^{\circ}|\sum_{j=1}^{s}\left(1-\frac{1}{\;n_{j}}\right)$. Assume, in addition, that $|G|=(g(C)-1)^{2}$ and that

• $(M1)$

for all $g\in G\setminus G^{\circ}$ we have

[TABLE]

• $(M2)$

for all $g\in G\setminus G^{\circ}$ we have

[TABLE]

*Then there is a free, mixed action of $G$ on $C\times C$ such that the quotient $S=(C\times C)/G$ is a minimal surface of general type with $p_{g}=q=1$, $K_{S}^{2}=8$.

Conversely, every surface $S$ with $p_{g}=q=1$, isogenous to a mixed product, arises in this way.*

Here, conditions $(M1)$ and $(M2)$ ensure that the $G$-action on $C\times C$ is free.

Remark 2.7**.**

The surface $S$ is not covered by elliptic curves because it is of general type (Lemma 2.1), so the map $C\longrightarrow C/G^{\circ}=E$ is ramified. Therefore condition $(M1)$ implies that $G$ is not abelian.

Remark 2.8**.**

The exact sequence (4) is non split if and only if the number of elements of order $2$ in $G$ equals the number of elements of order $2$ in $G^{\circ}$.

Proposition 2.9**.**

Let $S=(C\times C)/G$ be a surface with $p_{g}=q=1$, isogenous to a mixed product. Then $g_{\emph{alb}}=g(C)$.

Proof.

Let us look at diagram (3). The Abel-Jacobi map $\hat{\alpha}$ gives to $E^{(2)}$ the structure of a $\mathbb{P}^{1}$-bundle over $E$ ([CaCi93]); let $\mathfrak{f}$ be the generic fibre of this bundle and $F^{*}:=\rho^{*}\varepsilon^{*}(\mathfrak{f})$. If $F_{\textrm{alb}}$ is the generic Albanese fibre of $S$ we have $F_{\textrm{alb}}=\pi(F^{*})$. Let $\mathbf{n}=(n_{1},\ldots,n_{s})$ be such that $G^{\circ}$ is $(1\,|\,n_{1},\ldots n_{s})$-generated and $2g(C)-2=|G^{\circ}|\sum_{j=1}^{s}\left(1-\frac{1}{\;n_{j}}\right)$. The $(G^{\circ}\times G^{\circ})$-cover $\rho$ is branched exactly along the union of $s$ “horizontal” copies of $E$ and $s$ “vertical” copies of $E$; moreover for each $i$ there are one horizontal copy and one vertical copy whose branching number is $n_{i}$. Since $\varepsilon^{*}(\mathfrak{f})$ is an elliptic curve that intersects all these copies of $E$ transversally in one point, by Riemann-Hurwitz formula applied to $F^{*}\longrightarrow\varepsilon^{*}(\mathfrak{f})$ we obtain

[TABLE]

On the other hand the $G$-cover $\pi$ is étale, so we have

[TABLE]

whence $g_{\textrm{alb}}=g(C)$. ∎

If $S=(C\times C)/G$ is a surface with $p_{g}=q=1$ which is constructed by using the recipe of Proposition 2.6, it will be called a mixed surface of type $(G,\,\mathbf{n})$. The analogue of Proposition 2.3 in the mixed case is

Proposition 2.10**.**

Let $S=(C\times C)/G$ be a mixed surface of type $(G,\,\mathbf{n})$. Then there are at most the following possibilities:

• •

$g(C)=5,\quad\mathbf{n}=(2^{2}),\quad|G|=16$;

• •

$g(C)=7,\quad\mathbf{n}=(3),\quad\;\,|G|=36$;

• •

$g(C)=9,\quad\mathbf{n}=(2),\quad\;\,|G|=64$.

Proof.

By Proposition 2.6 we have $2g(C)-2=|G^{\circ}|\sum_{j=1}^{s}\left(1-\frac{1}{\;n_{j}}\right)$ and $|G^{\circ}|=\frac{1}{2}(g(C)-1)^{2}$, so $g(C)$ must be odd and we obtain $4=(g(C)-1)\sum_{j=1}^{s}\left(1-\frac{1}{\;n_{j}}\right)$. Therefore $4\geq\frac{1}{2}(g(C)-1)$ and the only possibilities are $g(C)=3,5,7,9$.

The case $g(C)=3$ is ruled out because $G$ cannot be abelian by Remark 2.7.

If $g(C)=5$ then $\sum_{j=1}^{s}\left(1-\frac{1}{\;n_{j}}\right)=1$, so $\mathbf{n}=(2^{2})$ and $|G|=16$.

If $g(C)=7$ then $\sum_{j=1}^{s}\left(1-\frac{1}{\;n_{j}}\right)=\frac{2}{3}$, so $\mathbf{n}=(3)$ and $|G|=36$.

If $g(C)=9$ then $\sum_{j=1}^{s}\left(1-\frac{1}{\;n_{j}}\right)=\frac{1}{2}$, so $\mathbf{n}=(2)$ and $|G|=64$. ∎

We will see in Section 2.10 that only the case $g(C)=5$ actually occurs.

3. The unmixed case

The classification of surfaces of general type with $p_{g}=q=1$ isogenous to an unmixed product is carried out in [Pol07] when the group $G$ is abelian. Therefore in this section we assume that $G$ is nonabelian.

Following [BaCaGr06, Section 1.2], for an $r$-ple $\mathbf{m}=(m_{1},\ldots,m_{r})\in\mathbb{N}^{r}$ we set

[TABLE]

If $S$ is an unmixed surface of type $(G,\,\mathbf{m},\,\mathbf{n})$ then we necessarily have $2\leq m_{1}\leq\ldots\leq m_{r}$ and $\Theta(\mathbf{m})>0$. Besides, by Proposition 2.2 we have $\alpha(\mathbf{m})=\frac{|G|}{g(F)-1}=g(C)-1\in\mathbb{N}$ and by Lemma 2.4 each integer $m_{i}$ divides $\alpha(\mathbf{m})$. Then we get

Proposition 3.1**.**

Let $S=(C\times F)/G$ be a surface with $p_{g}=q=1$ isogenous to an unmixed product of type $(G,\,\mathbf{m},\,\mathbf{n})$. Then the possibilities for $\mathbf{m}$ and $\alpha(\mathbf{m})$, written in the format $\mathbf{m}_{\alpha(\mathbf{m})}$, lie in the set $\mathcal{T}$ below:

[TABLE]

Proof.

This follows combining [BaCaGr06, Proposition 1.4] with Lemma 2.4.∎

By abuse of notation, we write $\mathbf{m}\in\mathcal{T}$ instead of $\mathbf{m}_{\alpha(\mathbf{m})}\in\mathcal{T}$.

Now we analyze the three cases in Proposition 2.3 separately, according to the value of $g(F)$. Note that if $g(F)=3$, $4$, $5$ then $|\textrm{Aut}(F)|\leq 168$, $120$, $192$, respectively ([Bre00, p. 91]).

Proposition 3.2**.**

If $g(F)=3$ we have precisely the following possibilities.

Proof.

Since $\mathbf{n}=(2^{2})$ it follows that $G$ is $(1\;|\;2^{2})$-generated and by the second relation in $(\ref{generi1})$ we have $|G|=2(g(C)-1)$. So we must describe all unmixed surfaces of type $(G,\mathbf{m},\mathbf{n})$ with $\mathbf{m}\in\mathcal{T}$, $\mathbf{n}=(2^{2})$ and $|G|=2\alpha(\mathbf{m})$. By a computer search through the $r$-tuples in Proposition 3.1 we can therefore list all possibilities, proving our statement. See the GAP4 script $1$ in the Appendix to see how this procedure applies to an explicit example.

∎

Proposition 3.3**.**

If $g(F)=4$ we have precisely the following possibilities.

Proof.

Since $\mathbf{n}=(3)$ it follows that $G$ is $(1\;|\;3)$-generated and by the second relation in (2) we have $|G|=3(g(C)-1)$. Therefore our statement can be proven searching by computer calculation all unmixed surfaces of type $(G,\mathbf{m},\mathbf{n})$ with $\mathbf{m}\in\mathcal{T}$, $\mathbf{n}=(3)$, $|G|=3\alpha(\mathbf{m})$ and $\alpha(\mathbf{m})\leq 40$. ∎

Proposition 3.4**.**

If $g(F)=5$ we have precisely the following possibilities.

Proof.

Since $\mathbf{n}=(2)$, it follows that $G$ is $(1\;|\;2)$-generated and by the second relation in (2) we have $|G|=4(g(C)-1)$. Therefore our statement can be proven searching by computer calculation all unmixed surfaces of type $(G,\mathbf{m},\mathbf{n})$ with $\mathbf{m}\in\mathcal{T}$, $\mathbf{n}=(2)$, $|G|=4\alpha(\mathbf{m})$ and $\alpha(\mathbf{m})\leq 48$. ∎

4. The mixed case

In this section we use Proposition 2.6 in order to classify the surfaces with $p_{g}=q=1$ isogenous to a mixed product. By Proposition 2.10 we have $g(C)=5$, $7$ or $9$. Let us consider the three cases separately.

4.1. The case $g(C)=5,\;|G|=16$

Proposition 4.1**.**

If $g(C)=5,\;|G|=16$ we have precisely the following possibilities.

Proof.

In this case $\mathbf{n}=(2^{2})$, so our first task is to find all nonsplit sequences of type (4) for which $G^{\circ}$ is a $(1\;|\;2^{2})$-generated group of order $8$. The three abelian groups of order $8$ and $D_{4}$ are $(1\;|\;2^{2})$-generated whereas the quaternion group $Q_{8}$ is not.

Since $\mathbb{Z}_{8}$ has only one element $\ell$ of order $2$, condition $(M1)$ in Proposition 2.6 cannot be satisfied for any choice of $\mathcal{V}$. By Remark 2.7 we are left to analyze the possible embeddings of $\mathbb{Z}_{2}\times\mathbb{Z}_{4}$, $D_{4}$ and $(\mathbb{Z}_{2})^{3}$ in nonabelian groups of order $16$. The groups $\mathbb{Z}_{2}\times\mathbb{Z}_{4}$, $D_{4}$ and $(\mathbb{Z}_{2})^{3}$ have $3$, $5$ and $7$ elements of order $2$, respectively. Therefore if $n_{2}$ denotes the number of elements of order $2$ in $G$, by Remark 2.8 we must consider only those groups $G$ of order $16$ with $n_{2}\in\{3,\,5,\,7\}$. The nonabelian groups of order $16$ with $n_{2}=3$ are $D_{2,8,5},\,\mathbb{Z}_{2}\times Q_{8}$ and $D_{4,4,-1}$ and they all contain a copy of $\mathbb{Z}_{2}\times\mathbb{Z}_{4}$. The only nonabelian group of order $16$ with $n_{2}=5$ is $D_{2,8,3}$ and it contains a subgroup isomorphic to $D_{4}$. The nonabelian groups of order $16$ with $n_{2}=7$ are $\mathbb{Z}_{4}\ltimes(\mathbb{Z}_{2})^{2}=G(16,3)$ and $\mathbb{Z}_{2}\ltimes Q_{8}$, and only the former contains a subgroup isomorphic to $(\mathbb{Z}_{2})^{3}$ (cfr. [Wi05]).

Summarizing, we are left with the following cases:

[TABLE]

Let us analyze them separately.

$\bullet$ $G^{\circ}=D_{4},\;\;G=D_{2,8,3}=\langle x,y\;|\;x^{2}=y^{8}=1,\;xyx^{-1}=y^{3}\;\rangle$

We consider the subgroup $G^{\circ}:=\langle x,\,y^{2}\rangle\cong D_{4}$. Set $\ell_{1}=\ell_{2}=x$ and $h_{1}=h_{2}=y^{2}$. Condition $(M1)$ holds because $C_{G}(x)=\langle x,\,y^{4}\rangle\subset G^{\circ}$. Condition $(M2)$ is satisfied because the conjugacy class of $x$ in $G^{\circ}$ is contained in the coset $x\langle y^{2}\rangle$ while for every $g\in yG^{\circ}$ we have $g^{2}\in\langle y\rangle$. Therefore this case occurs by Proposition 2.6.

$\bullet$ $G^{\circ}=\mathbb{Z}_{2}\times\mathbb{Z}_{4},\;\;G=D_{2,8,5}=\langle x,y\;|\;x^{2}=y^{8}=1,\;xyx^{-1}=y^{5}\rangle$

We consider the subgroup $G^{\circ}:=\langle x,\,y^{2}\rangle\cong\mathbb{Z}_{2}\times\mathbb{Z}_{4}$. Set $\ell_{1}=\ell_{2}=x$ and $h_{1}=h_{2}=y^{2}$. Conditions $(M1)$ and $(M2)$ are verified as in the previous case, so this possibility occurs.

$\bullet$ $G^{\circ}=\mathbb{Z}_{2}\times\mathbb{Z}_{4},\;\;G=\mathbb{Z}_{2}\times Q_{8}$ and $G^{\circ}=\mathbb{Z}_{2}\times\mathbb{Z}_{4},\;\;G=D_{4,4,-1}$.

All elements of order $2$ in $G$ are central so condition $(M1)$ cannot be satisfied and these cases do not occur.

$\bullet$ $G^{\circ}=(\mathbb{Z}_{2})^{3},\;\;G=\mathbb{Z}_{4}\ltimes(\mathbb{Z}_{2})^{2}=\langle x,y,z\;|\;x^{4}=y^{2}=z^{2}=1,xyx^{-1}=yz,\;[x,z]=[y,z]=1\rangle$

We consider the subgroup $G^{\circ}:=\langle y,\,z,\,x^{2}\rangle\cong(\mathbb{Z}_{2})^{3}$. Set $\ell_{1}=\ell_{2}=y$ and $h_{1}=z,\;\;h_{2}=x^{2}$. Condition $(M1)$ holds because $G^{\circ}$ is abelian and $[x,y]\neq 1$. Condition $(M2)$ is satisfied because if $g\in xG^{\circ}$ then $g^{2}\in\langle z,\,x^{2}\rangle$. Therefore this case occurs. ∎

4.2. The case $g(C)=7,\;|G|=36$

Proposition 4.2**.**

The case $g(C)=7,\;|G|=36$ does not occur.

Proof.

In this case $\mathbf{n}=(3)$, so $G^{\circ}$ is a group of order $18$ which is $(1\;|\;3)$-generated. There are five groups of order $18$ up to isomorphism. By computer search or direct calculation we see that the only one which is $(1\;|\;3)$-generated is $\mathbb{Z}_{3}\times S_{3}=G(18,3)$. Thus $G$ would fit into a short exact sequence

[TABLE]

A computer search shows that the only groups of order $36$ containing a subgroup isomorphic to $\mathbb{Z}_{3}\times S_{3}$ are $G(36,10)=S_{3}\times S_{3}$ and $G(36,12)=\mathbb{Z}_{6}\times S_{3}$ (see GAP4 script $2$ in the Appendix). They contain $15$ and $7$ elements of order $2$, respectively. On the other hand $\mathbb{Z}_{3}\times S_{3}$ contains $3$ elements of order $2$, so by Remark 2.8 all possible extensions of the form (5) are split and this case cannot occur. ∎

4.3. The case $g(C)=9,\;|G|=64$

Proposition 4.3**.**

The case $g(C)=9,\;|G|=64$ does not occur.

The proof will be the consequence of the results below. First notice that, since $\mathbf{n}=(2)$, the group $G^{\circ}$ must be $(1\,|\,2)$-generated.

Computational Fact 4.4**.**

There exist precisely $8$ groups of order $32$ which are $(1\;|\;2)$-generated, namely $G(32,t)$ for $t\in\{2,4,5,6,7,8,12,17\}$. The number $n_{2}$ of their elements of order $2$ is given in the table below:

[TABLE]

Proof..

Slightly modifying the first part of GAP4 script $1$ in the Appendix we easily find that the groups of order $32$ which are $(1\;|\;2)$-generated are exactly those in the statement. The number of elements of order $2$ in each case are found by a quick computer search: see again the Appendix, GAP4 script $3$. ∎

Computational Fact 4.5**.**

Let $t\in\{2,4,5,6,7,8,12,17\}$. A nonsplit extension of the form

[TABLE]

*exists if and only if the pair $(t,s)$ is one of the following:

$(2,9)$, $(2,57)$, $(2,59)$, $(2,63)$, $(2,64)$, $(2,68)$, $(2,70)$, $(2,72)$, $(2,76)$, $(2,79)$, $(2,81)$, $(2,82)$,

$(4,11)$, $(4,28)$, $(4,122)$, $(4,127)$, $(4,172)$, $(4,182)$,

$(5,5)$, $(5,9)$, $(5,112)$, $(5,113)$, $(5,114)$, $(5,132)$, $(5,164)$, $(5,165)$, $(5,166)$,

$(6,33)$, $(6,35)$,

$(7,33)$,

$(8,37)$,

$(12,7)$, $(12,13)$, $(12,14)$, $(12,15)$, $(12,16)$, $(12,126)$, $(12,127)$, $(12,143)$, $(12,156)$,

$(12,158)$, $(12,160)$,

$(17,28)$, $(17,43)$, $(17,45)$, $(17,46)$.*

Proof..

Assume $t=2$. Using the GAP4 script $4$ in the Appendix we find that the groups of order $64$ containing a subgroup isomorphic to $G(32,2)$ are $G(64,s)$ for $s\in\{8,\,9,\,56,\,57,\,58,\,59,\,61,\,62,\,63$, $64,\,66,\,67,\,68,\,69,\,70,\,72,\,73,\,74,\,75,\,76,\,77,\,78,\,79,\,80,\,81,\,82\}$. By Remark 2.8 and Computational Fact 4.4, in order to detect all the groups $G(64,s)$ fitting in some nonsplit extension of type (6) with $t=2$, it is sufficient to select from the previous list the groups containing exactly $n_{2}=7$ elements of order $2$. This can be done with the GAP4 script $5$ in the Appendix, proving the claim in the case $t=2$. The proof for the other values of $t$ may be carried out exactly in the same way. ∎

Let us denote by $[G,\,G]_{2}$ and $[G^{\circ},\,G^{\circ}]_{2}$ the subsets of elements of order $2$ in $[G,\,G]$ and $[G^{\circ},G^{\circ}]$, respectively.

Lemma 4.6**.**

Assume $g(C)=9$ and that one of the following situations occur:

• •

$[G,G]_{2}\subseteq Z(G)$;

• •

there exists some element $y\in G\setminus G^{\circ}$ commuting with all elements in $[G^{\circ},\,G^{\circ}]_{2}$.

Then given any generating vector $\mathcal{V}=\{\ell_{1};\;h_{1},h_{2}\}$ of type $(1\,|\,2)$ for $G^{\circ}$, condition $(M1)$ in Proposition 2.6 cannot be satisfied.

Proof.

Since $\ell_{1}\in[G^{\circ},\,G^{\circ}]_{2}\subseteq[G,\,G]_{2}$, in any of the above situations $C_{G}(\ell_{1})$ is not contained in $G^{\circ}$, so $(M1)$ cannot hold. ∎

Computational Fact 4.7**.**

Let $G=G(64,s)$ be one of the groups appearing in the list of Computational Fact 4.5. Then $[G,G]_{2}$ is not contained in $Z(G)$ if and only if $s=5$, $33$, $35$, $37$.

Proof.

See the GAP4 script $6$ in the Appendix. ∎

Computational facts 4.5, 4.7 and Lemma 4.6 imply that we only need to analyze the following pairs $(G^{\circ},\,G)$:

[TABLE]

Proposition 4.8**.**

The case $G^{\circ}=G(32,5)$ does not occur.

Proof.

A presentation for the group $G^{\circ}$ is

[TABLE]

Its derived subgroup contains exactly one element of order $2$, namely $z$. It follows that if $\{\ell_{1};\,h_{1},h_{2}\}$ is any generating vector of type $(1\,|\,2)$ for $G^{\circ}$, then $\ell_{1}=z$. Since $[G^{\circ},\,G^{\circ}]$ is characteristic in $G^{\circ}$, condition $(M1)$ cannot be satisfied for any embedding of $G^{\circ}$ into $G$. ∎

By using the two instructions P:=PresentationViaCosetTable(G) and TzPrintRelators(P) and setting in the output

[TABLE]

one obtains the following presentations for $G(64,33)$, $G(64,35)$ and $G(64,37)$.

[TABLE]

[TABLE]

[TABLE]

Computational Fact 4.9**.**

Referring to presentations (7), (8) and (9), we have the following facts.

• •

The group $G(64,33)$ contains exactly one subgroup $N_{1}$ isomorphic to $G(32,6)$ and one subgroup $N_{2}$ isomorphic to $G(32,7)$, namely

[TABLE]

• •

The group $G(64,35)$ contains exactly two subgroups $N_{3}$, $N_{4}$ isomorphic to $G(32,6)$, namely

[TABLE]

• •

The group $G(64,37)$ contains exactly two subgroups $N_{5}$, $N_{6}$ isomorphic to $G(32,8)$, namely

[TABLE]

In addition, for every $i\in\{1,\ldots,6\}$ we have

• $(a)$

$[N_{i},\,N_{i}]=\langle v,u\rangle\cong\mathbb{Z}_{2}\times\mathbb{Z}_{2}$.

• $(b)$

$y\notin N_{i}$* and $y$ commutes with all elements in $[N_{i},\,N_{i}]$.*

Proof.

See the GAP4 script $7$ in the Appendix. ∎

Proposition 4.10**.**

The cases $G^{\circ}=G(32,6),\,G(32,7),\,G(32,8)$ do not occur.

Proof.

By Lemma 4.6 and Computational Fact 4.9 it follows that, given any nonsplit extension of type (6) with $G^{\circ}$ as above, condition $(M1)$ in Proposition 2.6 cannot be satisfied. ∎

Summing up, we finally obtain

*Proof of Proposition *4.3. It follows from Propositions 4.8 and 4.10.

$\Box$

5. Moduli spaces

Let $\mathfrak{M}_{a,b}$ be the moduli space of smooth minimal surfaces of general type with $\chi(\mathcal{O}_{S})=a,\;K_{S}^{2}=b$; by an important result of Gieseker, $\mathfrak{M}_{a,b}$ is a quasiprojective variety for all $a,b\in\mathbb{N}$ (see [Gie77]). Obviously, our surfaces are contained in $\mathfrak{M}_{1,8}$ and we want to describe their locus there. We denote by $\mathfrak{M}(G,\mathbf{m},\mathbf{n})$ the moduli space of unmixed surfaces of type $(G,\mathbf{m},\mathbf{n})$ and by $\mathfrak{M}(G,\mathbf{n})$ the moduli space of mixed surfaces of type $(G,\mathbf{n})$. We know that $\mathbf{n}=(2^{2})$, $(3)$ or $(2)$ in the unmixed case, whereas $\mathbf{n}=(2^{2})$ in the mixed one. By a general result of Catanese ([Ca00]), both $\mathfrak{M}(G,\mathbf{m},\mathbf{n})$ and $\mathfrak{M}(G,\mathbf{n})$ consist of finitely many irreducible connected components of $\mathfrak{M}_{1,8}$, all of the same dimension. More precisely, we have

[TABLE]

Consider the mapping class groups in genus zero and one:

[TABLE]

[TABLE]

[TABLE]

One can prove that

[TABLE]

where $\Sigma_{1}$ is the torus $S^{1}\times S^{1}$ ([Schn03], [CattMu04]). This implies that we can define actions of these groups on the set of generating vectors for $G$ of type $(0\;|\;m_{1},\ldots,m_{r})$, $(1\;|\;n)$ and $(1\;|\;n^{2})$, respectively.

If $\mathcal{V}:=\{g_{1},\ldots,g_{r}\}$ is of type $(0\;|\;m_{1},\ldots,m_{r})$ then the action is given by

[TABLE]

If $\mathcal{W}:=\{\ell_{1};\,h_{1},h_{2}\}$ is of type $(1\;|\;n)$ then

[TABLE]

If $\mathcal{W}:=\{\ell_{1},\ell_{2};\,h_{1},h_{2}\}$ is of type $(1\;|\;n^{2})$ then

[TABLE]

These are called Hurwitz moves and the induced equivalence relation on generating vectors is said Hurwitz equivalence (see [BaCa03], [BaCaGr06], [Pol07]).

Now let $\mathfrak{B}(G,\,\mathbf{m},\,\mathbf{n})$ be the set of pairs of generating vectors $(\mathcal{V},\;\mathcal{W})$ such that the assumptions in Proposition 2.2 are satisfied; then we denote by $\mathfrak{R}$ the equivalence relation on $\mathfrak{B}(G,\,\mathbf{m},\,\mathbf{n})$ generated by Hurwitz moves on $\mathcal{V}$, Hurwitz moves on $\mathcal{W}$ and the simultaneous action of $\textrm{Aut}(G)$ on $\mathcal{V}$ and $\mathcal{W}$. Similarly, let $\mathfrak{B}(G,\,\mathbf{n})$ be the set of generating vectors $\mathcal{V}$ such that the assumptions of Proposition 2.6 are satisfied; then we denote by $\mathfrak{R}$ the equivalence relation on $\mathfrak{B}(G,\,\mathbf{n})$ generated by the Hurwitz moves and the action of $\textrm{Aut}(G)$ on $\mathcal{V}$.

Proposition 5.1**.**

The number of irreducible components in $\mathfrak{M}(G,\,\mathbf{m},\,\mathbf{n})$ equals the number of $\mathfrak{R}$-classes in $\mathfrak{B}(G,\,\mathbf{m},\,\mathbf{n})$. Analogously, the number of irreducible components in $\mathfrak{M}(G,\,\mathbf{n})$ equals the number of $\mathfrak{R}$-classes in $\mathfrak{B}(G,\,\mathbf{n})$.

Proof.

We can repeat exactly the same argument used in [BaCaGr06, Propositions 5.2 and 5.5]; we must just replace, where it is necessary, the mapping class group of $\mathbb{P}^{1}$ with the mapping class group of the elliptic curve $E$. ∎

Proposition 5.1 in principle allows us to compute the number of connected components of the moduli space in each case. In practice, this task may be too hard to be achieved by hand, but it is not out of reach if one uses the computer. Recently, M. Penegini and S. Rollenske developed a GAP4 script that solves this problem in a rather short time. We put the result of their calculations in the Main Theorem (see Introduction), referring the reader to the forthcoming paper [Pe08] for further details.

6. Appendix

In this Appendix we include, for the reader’s convenience, some of the GAP4 scripts that we have used in our computations; all the others are similar and can be easily obtained modifying the ones below.

Let us show how the procedure in the proof of Proposition 3.2 applies to an explicit example, namely $\mathbf{m}_{\alpha(\mathbf{m})}=(2,4,12)_{12}$. First we find all the nonabelian groups of order $24$ that are $(0\,|\,2,4,12)$-generated. This is done using GAP4 as below; the output tells us that there is only one such a group, namely $G=G(24,5)$.

gap> # -------------- SCRIPT 1 ------------------ gap> s:=NumberSmallGroups(24);; set:=[1..s]; [1..15] gap> for t in set do

c:=0; G:= SmallGroup(24,t); Ab:=IsAbelian(G); for g1 in G do for g2 in G do g3:=(g1*g2)^-1; H:= Subgroup(G, [g1,g2]); if Order(g1)=2 and Order(g2)=4 and Order(g3)=12 and Order(H)=Order(G) and Ab=false then c:=c+1; fi; if Order(g1)=2 and Order(g2)=4 and Order(g3)=12 and Order(H)=Order(G) and Ab=false and c=1 then Print(IdSmallGroup(G)," "); fi; od; od; od; Print("\n"); [24,5]

By using the two instructions P:=PresentationViaCosetTable(G) and TzPrintRelators(P) we see that $G$ has the presentation $\langle x,y\;|\;x^{2}=y^{12}=1,\;xyx^{-1}=y^{5}\rangle$, hence it is isomorphic to the metacyclic group $D_{2,12,5}$.

In order to speed up further computations, we define the sets $G2$, $G4$ given by the elements of $G$ having order $2$ and $4$, respectively.

gap> G:=SmallGroup(24,5);; gap> G2:=[];; G4:=[];; gap> for g in G do

if Order(g)=2 then Add(G2,g); fi; if Order(g)=4 then Add(G4,g); fi; od;

Then we check whether $G$ is actually $(1\,|\,2^{2})$-generated; if not, it should be excluded.

gap> c:=0;; gap> for l2 in G2 do

for h1 in G do for h2 in G do l1:=(l2h1h2h1^-1h2^-1)^-1; K:=Subgroup(G, [l2, h1, h2]); if Order(l1)=2 and Order(K)=Order(G) then Print(IdSmallGroup(G), " is (1 | 2,2)-generated", "\n"); c:=1; fi; if c=1 then break; fi; od; if c=1 then break; fi; od; if c=1 then break; fi; od; [24,5] is (1 | 2,2)-generated

We finish the proof by checking whether the surface $S$ actually exists; the procedure is to look for a pair $(\mathcal{V},\,\mathcal{W})$ of generating vectors for $G$ satisfying the assumptions of Proposition 2.2.

gap> c:=0;; gap> for g1 in G2 do

for g2 in G4 do g3:=(g1g2)^-1; H:=Subgroup(G, [g1, g2]); for l2 in G2 do for h1 in G do for h2 in G do l1:=(l2h1h2h1^-1*h2^-1)^-1; K:=Subgroup(G, [l2, h1, h2]); Boole1:=l1 in ConjugacyClass(G, g1); Boole2:=l1 in ConjugacyClass(G, g2^2); Boole3:=l1 in ConjugacyClass(G, g3^6); Boole4:=l2 in ConjugacyClass(G, g1); Boole5:=l2 in ConjugacyClass(G, g2^2); Boole6:=l2 in ConjugacyClass(G, g3^6); if Order(g3)=12 and Order(l1)=2 and Order(H)=Order(G) and Order(K)=Order(G) and Boole1=false and Boole2=false and Boole3=false and Boole4=false and Boole5=false and Boole6=false then Print("The surface exists "); c:=1; fi; if c=1 then break; fi; od; if c=1 then break; fi; od; if c=1 then break; fi; od; if c=1 then break; fi; od; if c=1 then break; fi; od; Print("\n"); The surface exists

The script above can be easily modified in order to obtain the list of all admissible pairs $(\mathcal{V},\,\mathcal{W})$; for instance, one of such pairs is given by

[TABLE]

Finally, here are the GAP4 scripts used in Section 4.

gap> # -------------- SCRIPT 2 ------------------ gap> s:=NumberSmallGroups(36);; set:=[1..s]; [1..14] gap> for t in set do

c:=0; G:=SmallGroup(36,t); N:=NormalSubgroups(G); for G0 in N do if IdSmallGroup(G0)=[18,3] then c:=c+1; fi; if IdSmallGroup(G0)=[18,3] and c=1 then Print(IdSmallGroup(G), " "); fi; od; od; Print("\n"); [36,10] [36,12]

gap> # -------------- SCRIPT 3 ------------------ gap> set:=[2,4,5,6,7,8,12,17];; gap> for t in set do

n2:=0; G0:=SmallGroup(32,t); for g in G0 do if Order(g)=2 then n2:=n2+1; fi; od; Print(IdSmallGroup(G0), " "); Print(n2, " "); od; Print("\n"); [32,2] 7 [32,4] 3 [32,5] 7 [32,6] 11 [32,7] 11 [32,8] 3 [32,12] 3 [32,17] 3

gap> # -------------- SCRIPT 4 ------------------ gap> s:=NumberSmallGroups(64);; set:=[1..s]; [1..267] gap> for t in set do

c:=0; G:=SmallGroup(64,t); N:=NormalSubgroups(G); for G0 in N do if IdSmallGroup(G0)=[32,2] then c:=c+1; fi; if IdSmallGroup(G0)=[32,2] and c=1 then Print(IdSmallGroup(G), " "); fi; od; od; Print("\n"); [64,8] [64,9] [64,56] [64,57] [64,58] [64,59] [64,61] [64,62] [64,63] [64,64] [64,66] [64,67] [64,68] [64,69] [64,70] [64,72] [64,73] [64,74] [64,75] [64,76] [64,77] [64,78] [64,79] [64,80] [64,81] [64,82]

gap> # -------------- SCRIPT 5 ------------------ gap> set:=[8,9,56,57,58,59,61,62,63,64,66,67,68,69,70,

72,73,74,75,76,77,78,79,80,81,82];; gap> for t in set do n2:=0; G:=SmallGroup(64,t); for g in G do if Order(g)=2 then n2:=n2+1; fi; od; if n2=7 then Print(IdSmallGroup(G), " "); fi; od; Print("\n"); [64,9] [64,57] [64,59] [64,63] [64,64] [64,68] [64,70] [64,72] [64,76] [64,79] [64,81] [64,82]

gap> # -------------- SCRIPT 6 ------------------ gap> set:=[5,7,9,11,13,,14,15,16,28,33,35,37,43,45,46,

57,59,63,64,68,70,72,76,79,81,82,112,113,114, 122,126, 127,132,143,156,158,160,164,165,166,172,182];; gap> for t in set do c:=0; G:=SmallGroup(64,t); D:=DerivedSubgroup(G); for d in D do B:=d in Center(G); if Order(d)=2 and B=false then c:=c+1; fi; if Order(d)=2 and B=false and c=1 then Print(IdSmallGroup(G), " "); fi; od; od; Print("\n"); [64,5] [64,33] [64,35] [64,37]

gap> # -------------- SCRIPT 7 ------------------ gap> s:=[33, 35, 37];; I:=[1, 2, 3];; gap> r:=[ [[32,6], [32,7]], [[32,6]], [[32,8]] ];;

for i in I do G:=SmallGroup(64, s[i]); Print(IdSmallGroup(G), "\n"); for N in NormalSubgroups(G) do if IdSmallGroup(N) in r[i] then Print(N, "="); Print(IdSmallGroup(N), " "); Print(DerivedSubgroup(N), "\n"); fi; od; Print("\n"); od; [64,33] Group( [ f1*f2, f3, f4, f5, f6 ] )=[32,7] Group( [ f5, f6 ] ) Group( [ f1, f3, f4, f5, f6 ] )=[32,6] Group( [ f5, f6 ] )

[64,35] Group( [ f1*f2, f3, f4, f5, f6 ] )=[32,6] Group( [ f5, f6 ] ) Group( [ f1, f3, f4, f5, f6 ] )=[32,6] Group( [ f5, f6 ] )

[64,37] Group( [ f1*f2, f3, f4, f5, f6 ] )=[32,8] Group( [ f5, f6 ] ) Group( [ f1, f3, f4, f5, f6 ] )=[32,8] Group( [ f5, f6 ] )