Let $Y_{n}$ be what I'm calling the ordered configuration space, the topological space of all ordered subsets of $\mathbb{C}$ of cardinality $n$. This can be viewed as $$\mathrm{Spec}(\mathbb{C}[t_{1}, ... , t_{n}, \{(t_{i} - t_{j})^{-1}\}_{1 \leq i < j \leq n}]).$$

The (topological) fundamental group of this is the pure braid group $P_{n}$. So by a general version of the Riemann Existence Theorem, the elements of $P_{n}$ must be Galois automorphisms of the maximal Galois extension of the function field $\mathbb{C}(t_{1}, ... , t_{n})$ which is unramified except at the primes $(t_{i} - t_{j})$. How may I describe algebraically the Galois automorphisms corresponding to, say, the generators of $P_{n}$?

I have a guess (something along the lines of each generator $A_{i,j}$ of $P_{n}$ sending $(t_{i} - t_{j})^{1/n} \mapsto \zeta_{n}(t_{i} - t_{j})^{1/n}$, except that seems to make the Galois group abelian) but am not sure how to prove anything systematically, and I can't find this explicitly described in any source. Thank you for any help you can give.

  • $\begingroup$ I've merged your two accounts, but you should register to keep yourself from making new ones. $\endgroup$
    – S. Carnahan
    Jun 28, 2012 at 5:54

1 Answer 1


Your confusion comes from the fact that your description of the Galois action is, I think, correct, but you are not giving the entire Galois action because you are not having it act on the correct field. The covering generated by $(t_i-t_j)^{1/n}$ is an abelian covering (and by Kummer theory is the maximal abelian covering), so of course the Galois action on it is abelian. But there are other unramified covers whose function fields are not generated by elements of the form $(t_i-t_j)^{1/n}$).

To describe the full Galois action, you need to consider all of these covers. However, as far as I know, there is no well-behaved way to do so. For instance, each such scheme fibers over $\mathbb P^1$ minus three points: take the map $(t_1,..,t_n) \to (2t_1-t_2-t_3:-t_1+2t_2-t_3)$. It is well-defined since if both of those are $0$ then $t_1-t_2$ is zero as well, and the inequalities $t_1\neq t_2$, $t_1\neq t_3$, $t_2\neq t_3$ give three points that are not in the image. So take any unramified cover of $\mathbb P^1$ minus three points and take the fibered product with this map. You get an unramified covering of this scheme.

Writing down equations for all the umramified covers of $\mathbb P^1-\{0,1,\infty\}$ is no easy task, and those aren't even all the covers of this scheme! Thus finding an explicit representation for the Galois action is difficult.

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    $\begingroup$ Another maybe more conceptual way to see this: given a configuration of $n$ distinct points on $\mathbf C$, we get a configuration of $n+1$ distinct points on $\mathbf P^1$ by adding the point at infinity. Hence $Y_n$ maps to $M_{0,n+1}$, which when $n \geq 3$ fibers over $M_{0,4} \cong \mathbf P^1 \setminus \{0,1,\infty\}$. $\endgroup$ Jun 22, 2012 at 17:18
  • $\begingroup$ Okay, I see this, and thank you for the answers. I may never be able to describe the maximal extension field, but fortunately, I probably don't need that. Does anyone have any idea how I may compute the Galois action on particular elements of field extensions? For instance, I'm interested in knowing how the generators of $P_{3}$ act on such elements as $\sqrt{\sqrt{t_{1} - t_{3}} - \sqrt{t_{2} - t_{3}}}$. I'm not sure I have the topological intuition to be able to deduce this, especially since we're looking at dimension > 1 spaces. It would be very helpful in general to know how $P_{n}$ acts o $\endgroup$ Jun 27, 2012 at 22:35
  • $\begingroup$ It's not a higher-dimensional space, it's a bunch of points in the plane. The braid tells you how they move through time. The square root function follows them. There probably isn't a general method like you describe, unless it's purely group-theoretic: you know a map from the braid group to a galois group $G$, you have an extension of $G$ to $G'$, and you count the lifts. It seems like it would require luck for there to only be one. $\endgroup$
    – Will Sawin
    Jun 28, 2012 at 2:08

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