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.

  • $\begingroup$ Dear Makhalan, could you please explain what "inertia of some point" means? $\endgroup$ Aug 1, 2010 at 9:20
  • $\begingroup$ Sure. I'm not sure exactly what you're asking, so I'll answer two questions: 1. in general when I say inertia group, it is in the context of either a specific point in Y, or a specific prime divisor in Y. Then inertia means the subgroup of G that fixes that point, or that fixes that divisor pointwise. 2. If you're talking about that exact quote, then I meant that <g_1> is the inertia group of some point in Y above the point in X which is the intersection of the fiber over t and B_1. There are several points in Y that map to the point in X. This would be the inertia of one of those. $\endgroup$ Aug 1, 2010 at 15:48

1 Answer 1


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).


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

$B_1$ 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 $\mathbb{P}^3$.


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

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


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.


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