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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?

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1 Answer 1

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.

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Thanks Dan! This is interesting. –  LMN Mar 27 '13 at 22:46
    
More generally, any non-simply-connected graph of simply connected varieties joined transversely has a universal cover given in a combinatorial fashion. –  S. Carnahan Mar 31 '13 at 3:44

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