MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
I feel like you're looking for something like the following. Galois covers of a topological space $X$ of degree $n$ correspond to homomorphisms $\pi_1X\to S_n$ where the image is transitive and has trivial point stabilizers. Arbitrary covers of degree $n$ correspond to homomorphisms $\pi_1X\to S_n$ where the image is just transitive. No doubt this can be phrased in terms of something torsorish --- the answer should be approximately '$H^1(X,Y)$ where $Y$ is a transitive-but-not-necessarily-simply-transitive $G$-set'. – HJRW Mar 18 '12 at 19:03
That should have been 'Arbitrary connected covers of degree $n$...' If you don't require connectedness, you can also drop the transitivity assumption. – HJRW Mar 18 '12 at 19:31
There is a mistake in your statement. The group $H^1(X,G)$ is in bijective correspondence with homomorphisms $\pi_1(X,x)\to G$ up to conjugacy. The kernel is a normal subgroup, but the quotient need not equal $G$ -- only a subgroup of $G$. – Jason Starr Mar 18 '12 at 21:11
up vote 6 down vote accepted

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.

share|cite|improve this answer
@Dan Petersen:What you are saying is exactly what I'm puzzled in. Can you provide any reference? Thank you in advance. – Xiaobo Zhuang Jan 4 '13 at 7:10
I mean the finite etale cover part. – Xiaobo Zhuang Jan 4 '13 at 7:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.