How do branched coverings of complex surfaces "fit" with branched coverings of curves?

2010-07-31T15:26:49Z

Since I'm used to working with algebraic $\pi_1$'s, which don't work well with surfaces, I find myself lacking geometric intuition when I attempt to do these types of purely geometric arguments. I'm hoping someone can point out a fact or two that will help me gain some insight.

Let's say we have a complex surface, $X$, and we are given a map $X \rightarrow \mathbb{A}^1_{\mathbb{C}}$ such that the fibers are all $\mathbb{P}^1_{\mathbb{C}}$'s. Let's say we are given a prescribed branch divisor, $B$, made up of the prime divisors $B_1, ..., B_r$ (I want them to be horizontal - meaning that each $B_i$ considered over any $a$ in $\mathbb{A}^1_{\mathbb{C}}$ is just a point). Assume for the sake of simplicity that $B_1$ meets $B_2$ once and no two other branch points meet.

Let's fix a complex $t$ (over which $B_1$ and $B_2$ don't meet), and pick our basepoint to be some point over $t$ that doesn't meet the branch points. A $G$-Galois cover $Y \rightarrow X$ branched at most at $B$ gives a $G$-Galois cover of $t \times_{\mathbb{A}^1_{\mathbb{C}}} X \cong \mathbb{P}^1_{\mathbb{C}}$. This cover can now be described as $\gamma_i \mapsto g_i$ where $\gamma_i$ is a loop around $B_i \times_{\mathbb{A}^1_{\mathbb{C}}} t$ (meaning the fiber of $B_i$ over $t$). This would imply in the cover of the curve $\langle g_1 \rangle$ is the inertia of some point over $B_1 \times_{\mathbb{A}^1_{\mathbb{C}}} t$ and $\langle g_2 \rangle$ is the inertia of some point over $B_2 \times_{\mathbb{A}^1_{\mathbb{C}}} t$. In fact, we know what point it is, if you fix a basepoint above. Fix such a basepoint.

It seems that this data alone determines the cover over the entire surface. Is that right? By this I mean: take $\pi_1(X \setminus B, basept)$ and map $\gamma_i$, which is a loop in our curve which lies in our surface, $X$, to $g_i$. And if it is true -- can we somehow use this to know what the inertia groups are for the prime divisors over $B_1$ and $B_2$ that meet? (after all, they don't all meet. all we know is there is some prime divisor over $B_1$ that meets some prime divisor over $B_2$.) Finding the inertia groups of those divisors in terms of whatever I can get my hands on through curves (which means including the cover over $t$) is really my ultimate goal.

Answer by Francesco Polizzi for How do branched coverings of complex surfaces "fit" with branched coverings of curves?

2010-08-01T22:51:01Z

My answer does not intend to completely solve your problem, but I wish to write down a couple of examples that (I hope) could help you gain some geometrical insight.

I will take $G$ abelian, since in this case the map $\pi_1(X \setminus B) \to G$ factors through $H_1(X \setminus B, \mathbb{Z}) \to G$. Moreover I will take $X=\mathbb{P}^2$ (if you want a map with fibre $\mathbb{P}^1$, just blow-up a point).

EXAMPLE 1.

$G$=$\mathbb{Z}_2=\langle g | g^2=1 \rangle$,

$B_1$ and $B_2$ two distinct lines intersecting in a point $p$,

$B=B_1 + B_2$.

Then $H_1(X \setminus B, \mathbb{Z})$ is generated by the two loops $\gamma_1$ and $\gamma_2$ (around $B_1$ and $B_2$, respectively) with the relation $\gamma_1+\gamma_2=0$, hence it is isomorphic to $\mathbb{Z}$. The unique $\mathbb{Z}_2$-cover $Y \to X$ branched on $B$ is obtained by sending $\gamma_1$, and consequently $ \gamma_2$, to the generator $g$. The inertia group (stabilizer) over $B_1 + B_2$ is of course isomorphic to $\mathbb{Z}_2$, but notice that the preimage of $p$ in $Y$ is $singular$. In fact, $Y$ is isomorphic to a quadric cone in $\mathbb{P}^3$.

EXAMPLE 2.

$G$=$\mathbb{Z}_2 \times \mathbb{Z}_2=\langle g_1, g_2, g_3 | g_i^2=1, g_1g_2g_3=1, [g_i, g_j]=1 \rangle$,

$B_1$, $B_2$, $B_3$ three distinct lines intersecting pairwise in three distinct points; set $p_{ij}:=B_i \cap B_j$.

$B=B_1+B_2+B_3$.

Then $H_1(X \setminus B, \mathbb{Z})$ is generated by the three loops $\gamma_1$, $\gamma_2$ and $\gamma_3$ (around $B_1$, $B_2$ and $B_3$, respectively) with the relation $\gamma_1+\gamma_2+\gamma_3=0$, hence it is isomorphic to $\mathbb{Z}^2$. Up to permutation of the indices, the unique $\mathbb{Z}_2 \times \mathbb{Z}_2$-cover $Y \to X$ branched on $B$ is obtained by sending $\gamma_i$ to $g_i$, for all $i=1,2,3$.

The inertia group of a point over $B_1$, distinct from $p_{12}$ and $p_{13}$, is isomorphic to $\langle g_1 \rangle$ and similarly for the other two lines. But the inertia group over the three points $p_{ij}$ is the whole group $\mathbb{Z}_2 \times \mathbb{Z}_2$.

In this case $Y$ is smooth, in fact it is isomorphic to $\mathbb{P}^2$. However, there are three "intermediate covers" $Y_1$, $Y_2$, $Y_3$, corresponding to the three non-trivial subgroups of $G$; each $Y_i$ is then isomorphic to a quadric cone.

A good reference for these topics is Pardini's paper "Abelian covers of algebraic varieties". If you like a more topological approach, you can read Catanese's article "On the moduli space of surfaces of general type", where the case $G=\mathbb{Z}_2 \times \mathbb{Z}_2$ is developed in full details.

How do branched coverings of complex surfaces "fit" with branched coverings of curves? - MathOverflow

How do branched coverings of complex surfaces "fit" with branched coverings of curves?

Answer by Francesco Polizzi for How do branched coverings of complex surfaces "fit" with branched coverings of curves?