2
$\begingroup$

In this post, it is said that a functor from the fundamental groupoid of a space $X$ (denoted by $\Pi(X)$) to the category $\mathrm{Vect}$ of vector spaces gives a flat vector bundle over $X$. But I feel this picture might be generalized. For example,

(1) Let $X$ be a smooth manifold, and assume there is a functor from $\Pi(X)$ to a Lie group $G$ (as a Lie groupoid with one object). Under certain assumptions, does this give us a principle $G$-bundle? Or, even more generally, what if we replace $G$ by any arbitrary fiber manifold $F$?

(2) Let $X$ be a scheme, and assume a functor from $\Pi(X)$ to the category of schemes, say $x\mapsto Y_x$. Then, under certain nice conditions, what can we say about the union $Y:=\cup_{x\in X} Y_x$ and the natural map $Y\to X$? For example, is $Y$ still a scheme(probably no)?

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ This paper of Barwick, Glasman and Haine may be relevant arxiv.org/abs/1807.03281 $\endgroup$
    – Gregory Arone
    Commented Sep 7, 2018 at 14:25
  • 6
    $\begingroup$ (1) This is true but not quite what you actually want to say. It gives you a flat principal $G$-bundle. The point is that a principal $G$-bundle with connection over a connected manifold is (essentially) the same thing as a continuous functor $\Omega M \to G$, giving parallel transport maps. But the assumption that parallel transport only depends on the underlying homotopy class is the same as saying that the bundle has no curvature by the Ambrose-Singer theorem. If you write this as a representation $\rho: \pi_1 M \to G$, then your bundle is $\widetilde M \times_\rho G$. $\endgroup$
    – mme
    Commented Sep 7, 2018 at 14:29
  • 1
    $\begingroup$ Replacing $G$ with $\text{Diff}(F)$ you obtain the notion of foliated $F$-bundle, which is the appropriate notion of flat connection for fiber bundles. $\endgroup$
    – mme
    Commented Sep 7, 2018 at 14:31
  • $\begingroup$ @MikeMiller Thanks for a great comment. Here by $\Omega M$ you mean the space of loops or the space of paths? $\endgroup$
    – Hang
    Commented Sep 8, 2018 at 16:41
  • 1
    $\begingroup$ @Hang Loops, but only to avoid changing the codomain. If you want to use the path-space you should start by fixing a principal $G$-bundle to begin with, and the category in the codomain should be the topological category whose space of objects is $X$ and whose space of morphisms $x \to y$ is $\text{Isom}(P_x, P_y) \cong G$. I do not think this is avoidable. In the discrete/flat case, you do not need to pay much attention to the topology, and so you should be able to replace the latter category (which is a groupoid with automorphism group $G$) with the one-object category with morphism group G. $\endgroup$
    – mme
    Commented Sep 8, 2018 at 16:58

0

Reset to default

You must log in to answer this question.