# Is every etale cover a principal bundle?

Let $$f: X\rightarrow Y$$ be proper etale morphism between varieties over the field of complex numbers. Does there exists a finite group $$G$$ such that $$Y$$ is the categorical quotient of $$X$$ under the free action of $$G$$? Or in other words is every etale cover a principal bundle?

We can consider the group scheme $$Aut_Y(X)$$ on $$Y$$, whose $$T$$ valued points are $$Aut_{T}(X\times_Y T)$$ i.e., the group of automorphisms of $$X\times_Y T$$ which commute with the projection to $$T$$. It is clear the $$X$$ is the categorical quotient of $$X$$ by $$Aut_Y(X)$$. But it is not clear why $$Aut_Y(X)$$ should be the group $$Y\times G$$, for some finite group $$G$$.

• Not in general: the function field of $X$ might not be a Galois extension of the function field of $Y$. But there always exists a cover of the type you want which dominates $X$, i.e. a $G$-torsor $Z\to Y$ and a subgroup $H\subseteq G$ such that $X=Z/H$. The problem is that $H$ might not be normal in $G$. The question is probably better for math.SE. – Piotr Achinger Jan 21 at 9:36

Not every etale covering is a principal bundle under a group $$G$$. This is easiest to see using the Galois correspondence for etale coverings: the category of finite etale coverings of $$X$$ is equivalent to the category of finite continuous $$\pi_1^{et}(X,x)$$-sets, for a chosen basepoint $$x\in X$$. Passed across this formalism, you are asking whether for every finite continuous $$\pi_1^{et}(X,x)$$-set $$S$$ (corresponding to the covering $$Y\to X$$), there is a free action of a finite group $$G$$ on $$S$$, commuting with the $$\pi_1^{et}(X,x)$$-action, whose quotient is the single point (corresponding to the trivial covering $$X\to X$$). In other words, the action of $$G$$ on $$S$$ should be freely transitive.
But in general there are plenty of examples where there is no $$\pi_1^{et}(X,x)$$-equivariant freely transitive action of a finite group $$G$$ on $$S$$. For example, suppose that $$H\leq\pi_1^{et}(X,x)$$ is a non-normal open subgroup. Then the coset space $$S:=\pi_1^{et}(X,x)/H$$ with the left-regular action of $$\pi_1^{et}(X,x)$$ provides a counterexample. Indeed, since $$H$$ is non-normal, there is an element $$g\in\pi_1^{et}(X,x)$$ such that $$gHg^{-1}\neq H$$. But the stabiliser of the identity coset $$H\in S$$ is $$H$$, whereas the stabiliser of the coset $$gH\in S$$ is $$gHg^{-1}$$. Hence there is no $$\pi_1^{et}(X,x)$$-equivariant permutation of $$S$$ taking $$H$$ to $$gH$$. So there's no chance of a $$\pi_1^{et}(X,x)$$-equivariant freely transitive action on $$S$$.