Let $G$ be a locally compact Hausdorff (second countable) groupoid with Hausdorff (second countable) unit space $X$. Assume $G$ is étale, i.e., the source and range maps of $G$ are local homeomorphisms. We say that $G$ is proper if the map $(s,r)\colon G \to X\times X$ is proper, i.e., the preimage of a compact subset is compact.

I think the following theorem is true, but I can't find a proof. I'd greatly appreciate any pointer.

if $G$ is proper then each $x\in X$ admits an open neighborhood $U\subseteq X$ equipped with an action of $G_x^x$ (the automorphisms group at $x$) such that the restriction $G|_U$ is isomorphic to the action groupoid $U\rtimes G_x^x$ and the map (whenever it is defined) $$ G\times_{G_x^x}U \to X$$ sending $[g,x]$ to $gx$ is a $G$-equivariant homeomorphism onto an open neighborhood of (the orbit of) $x\in X$.

EDIT: I would be interested to know if anything is known in the non-Lie non-étale case. When $G$ is a locally compact group (acting properly), a result of Abels says that $X$ is locally compactly induced, but the compact subgroup cannot (in general) be chosen to be a stabilizer. By analogy, for a locally compact proper groupoid $G$ it seems reasonable to prove that

$G$ is locally Morita equivalent to an action groupoid by a compact group. More precisely, $X$ can be covered by opens $V$ which such that $G|_V$ is Morita equivalent to $U\rtimes H$ for $H$ a compact group acting on a subset $U\subseteq V$. Note that $U$ (the slice) need not be open.

  • 1
    $\begingroup$ mathoverflow.net/questions/303183/… ??? $\endgroup$ Jul 30, 2018 at 11:01
  • $\begingroup$ WHat does Principal bundles tag has anything to do with this question? $\endgroup$ Jul 30, 2018 at 13:41
  • 2
    $\begingroup$ Thanks for your link. I somehow missed it. I'll think about it for a bit because it doesn't directly answer my question (or at least I don't see it yet). When the action is free, the theorem should say that you get a principal bundle, moreover the local model will always involve a principal bundle of some sort. That's why I used that tag. $\endgroup$
    – vap
    Jul 30, 2018 at 13:43
  • $\begingroup$ May be you should specify what principal bundle you are looking for... $\endgroup$ Jul 30, 2018 at 16:08
  • 1
    $\begingroup$ Perhaps the literature on orbispaces will have some hints... $\endgroup$
    – David Roberts
    Jul 31, 2018 at 0:18

1 Answer 1


It looks like the result suggested by Praphulla Koushik here is enough in the étale case. Indeed, the image of the map in my question can be rewritten as the image of $$r\colon s^{-1}(U)\to X.$$ Since $r$ is a local homeomorphism, its image is open and obviously $G$-invariant. It is easy to see that by modding out $G|_U$ we get a bijection.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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