How do subgroups of fundamental groups relate to torsors

Question

Fix a nice complex algebraic variety $X$ with base point $x$. (We will work in the analytic topology.)

Let $G$ be a finite group. Then the set of normal subgroups of $\pi_1(X,x)$ with quotient $G$ corresponds to the set of $X$-torsors for $G$. (We write $G$ for the constant sheaf on $X$ associated to $G$.)

This is just saying that Galois covers of $X$ correspond to torsors for $G$.

Now, not all covers of $X$ are Galois. In general, finite covers of $X$ correspond (in some sense) to finite index subgroups of $\pi_1(X,x)$ and the degree is given by the index.

I'd like to know if we can replace the set $H^1(X,G)$ of torsors for $G$ by something "torsorish" and still have a correspondence with the set of finite index subgroups of $\pi(X,x)$ of some fixed degree.

Answer 1

Yes, absolutely. There is an equivalence of categories between étale covers of $X$ of degree $n$, and $S_n$-torsors over $X$.

The general principle here is that $H^1(X,\mathscr G)$, where $X$ is some site and $\mathscr G$ is a sheaf of groups, classifies locally trivial things over $X$ whose sheaf of automorphisms is given by $\mathscr G$. This includes facts like that $H^1(X,\mathcal O_X^\ast)$ classifies line bundles (since an automorphism of a line bundle is given by multiplication by a nonzero scalar on each fibre), or that $H^1(X,\mathcal T_X)$ classifies first order deformations when $X$ is smooth (since deformations of smooth $X$ are locally trivial, and an automorphism of a first order deformation is an infinitesimal automorphism which is a vector field). In your case, the statement follows because both étale covers of degree $n$ and $S_n$-torsors have $\underline{S_n}$ as sheaf of automorphisms.

A more concrete way of seeing this particular case is the following. Given a finite étale cover $Y \to X$, form the $n$-fold fibered product $Y \times_X \cdots \times_X Y \to X$ and remove the "big diagonal". Given an $S_n$-torsor $P \to X$, form the quotient $P/S_{n-1} \to X$. These constructions are inverse to each other.