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arXiv math/0601063math.AG
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关于p_g=q=1与曲线乘积同构的一般型曲面

On surfaces of general type with p_g=q=1 isogenous to a product of curves

Francesco Polizzi

摘要:

光滑代数曲面$S$称为\emph{isogenous to a product of unmixed type},如果存在两条光滑曲线$C, F$和一个有限群$G$,忠实地作用在$C$和$F$上,自由地作用在它们的乘积上,使得$S=(C \times F)/G$ .本文在假定群$G$是交换群的条件下,对与非混合积同构的一般型曲面进行了分类,得到了与非混合积同构的$p_g=q=1$曲面.它们属于四个族,我们称之为$I, II, III, IV$型曲面。模空间$\mathfrak{M}_{I}, \mathfrak{M}_{II}, \mathfrak{M}_{IV}$是不可约的,而$\mathfrak{M}_{III}$是两个不可约分量的不交并。在最后一节中,我们通过构造几个例子来分析$G$不是阿贝尔的情况。

英文摘要:

A smooth algebraic surface $S$ is said to be \emph{isogenous to a product of unmixed type} if there exist two smooth curves $C, F$ and a finite group $G$, acting faithfully on both $C$ and $F$ and freely on their product, so that $S=(C \times F)/G$. In this paper we classify the surfaces of general type with $p_g=q=1$ which are isogenous to an unmixed product, assuming that the group $G$ is abelian. It turns out that they belong to four families, that we call surfaces of type $I, II, III, IV$. The moduli spaces $\mathfrak{M}_{I}, \mathfrak{M}_{II}, \mathfrak{M}_{IV}$ are irreducible, whereas $\mathfrak{M}_{III}$ is the disjoint union of two irreducible components. In the last section we start the analysis of the case where $G$ is not abelian, by constructing several examples.

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