What happens to inertia groups after blow ups?

Question

Say we have a branched $G$-Galois covering $X \rightarrow Y$ of surfaces, and assume that the branch locus (in $Y$) is a divisor with normal crossings. I will always assume that the ramification is tame, and you can pretend we're doing this with varieties over $\mathbb{C}$ if you prefer. This implies that the inertia groups of any two irreducible components of the ramification divisor commute, because they live in the inertia group of the point which is abelian. I've been told the following fact, and have been spending way too much time trying to understand why it's true:

Say the irreducible components $D_1$ and $D_2$ of the branch locus meet at a node. Let's say we pick a $P_1$ and $P_2$ irreducible preimages of $D_1$ and $D_2$ respectively, that meet.

If we blow up that node of the branch locus in $Y$ (and get the new scheme/variety $Y'$) and then normalize in $\kappa(X)$ (to get $X'$), we get a map $X' \rightarrow Y'$ where over $D_1$, $D_2$ and $E$ (the exceptional divisor) are $P_1$, $P_2$ and an irreducible divisor connecting them $P_3$; such that if $I_1$ was the inertia group of $P_1$ and $I_2$ the inertia group of $P_2$ then $I_1I_2$ (which is defined because $I_1$ and $I_2$ commute) is the inertia group of $P_3$.

Why is this true? I've been blowing things up and normalizing for way too long and it has been thoroughly uninsightful. I must be missing an important heuristic.

Answer 1

I'll do this over $\mathbb{C}$, since you'll let me get away with it.

The geometric fact you need to know is the following: If $D \subset Y$ is a branch divisor for the $G$ covering $X \to Y$, then the inertia group of $D$ is generated by the monodromy around a small loop $\gamma$ encircling $D$. In other words, near any smooth point of $D$, the divisor $D$ locally looks like $\{ z_n=0 \} \subset \mathbb{C}^n$, and you are supposed to take the monodromy around $(0,0,\ldots, 0, r e^{i \theta} )$ for some small $r$. Well, that's not quite right because (1) the inertia group is only defined up to conjugacy and (2) I need to specify a path from my chosen basepoint in $Y$ to the loop I am computing monodromy around. Specifying the path in (2) lets me talk about inertia groups without saying "up to conjugacy". I'll ignore these issues.

So, suppose that $x$ is a point of $Y$ at which two components, $D_1$ and $D_2$, of the branch divisor cross normally. Passing to an analytic neighborhood of $x$, we may assume that $Y$ is the unit ball in $\mathbb{C}^2$, and that $D_i = \{ z_i =0 \}$. I'll use your notations $Y'$ and $E$. Inside $Y'$, consider small loops $\gamma_1$, $\gamma_2$ and $\delta$ around $D_1$, $D_2$ and $E$. The monodromy around these loops will generate the inertia groups at $D_1$, $D_2$ and $E$. The key fact is that,

in $\pi_1(Y' \setminus (D_1 \cup D_2 \cup E))$, we have $\delta= \gamma_1 + \gamma_2$.

Let's see why. $Y' \setminus (D_1 \cup D_2 \cup E) \cong Y \setminus (D_1 \cup D_2) \cong \{ (z_1, z_2) \in (\mathbb{C}^*)^2 : |z_1|^2+|z_2|^2<1 \}$. This last space clearly retracts onto $(S^1)^2$, with fundamental group $\mathbb{Z}^2$. Explicitly, a small loop around $D_1$ is given by $(r e^{i \theta}, s)$, for some small $r$ and $s$, and hence has class $(1,0)$. Similarly, a small loop around $D_2$ has class $(0,1)$. A small loop around $E$ (this is the one you might have to think about) is given by $(r e^{i \theta}, s e^{i \theta})$ for $r$ and $s$ small, and hence has class $(1,1)$. As desired, $(1,1)=(1,0)+(0,1)$.