Let $X$ be a (reasonable) scheme. I'm curious about constructing the constructing the covering space of a scheme algebraically. The covering space functor $F$ (below) can be represented by a projective family of schemes, and I want to know if there are situations where it can be represented by an algebraic spaces.

Specifically, for a geometric point $z \rightarrow X$, the functor $F$ (on FEt/X) given by $Y \mapsto \textrm{Hom}_X(z, Y) = F(Y)$ is, what Milne calls "strictly pro-representable". That is, there is

(1.) a projective system (= inverse limit system) of schemes in FEt/X $X_i$ and epimorphisms $\phi_{ij}: X_j \rightarrow X_i$

(2.) Elements $f_i \in F(X_i)$ such that $f_i = \phi_{ji} f_j$

(3.) The natural map $\lim_{\rightarrow} \textrm{Hom}(X_i, Z) \rightarrow F(Z)$ induced by the $f_i$ is an isomorphism of sets.

$\textbf{Question: }$ Are there general situations where $F$ is representable by an algebraic space which is not a priori a scheme?

Side note: I suspect that the functor $F$ is not ever representable by a scheme locally of finite type over $X$ (but not finite over $X$), but I don't have a proof. Is this true?


One nontrivial example where the universal cover is actually a scheme (locally of finite type) is when $X$ is the nodal cubic curve over a field $k$, i.e. $\mathbf P^1$ with two points identified. Note that when $k = \mathbf C$, we have $\pi_1(X(\mathbf C)) \cong \mathbf Z$. The universal cover $\widetilde X$ is an infinite chain of $\mathbf P^1$, where the point $0$ on the $i$th curve has been glued to the point $\infty$ on the $(i+1)$st curve, for all $i \in \mathbf Z$; the action of $\mathbf Z$ by deck transformation is translating along this chain.

  • $\begingroup$ More generally, any non-simply-connected graph of simply connected varieties joined transversely has a universal cover given in a combinatorial fashion. $\endgroup$ – S. Carnahan Mar 31 '13 at 3:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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