I am reading Orbifolds as Groupoids: an Introduction by Ieke Moerdijk.

In page $8$ when explaining local charts, it says the following :

Let $\mathcal{G}$ be a Lie groupoid. For an open set $U\subseteq \mathcal{G}_0$, we write $\mathcal{G}|_U$ for the full subgroupoid of G with $U$ as a space of objects. In other words, $(\mathcal{G}|_U)_0=U$ and $(\mathcal{G}|_U)_1=\{x\rightarrow y : x,y\in U\}$. If $\mathcal{G}$ is proper and etale, then for each $x\in \mathcal{G}_0$ there exist arbitrary small neighborhoods $U$ of $x$ for which $\mathcal{G}|_U$ is isomorphic to $\mathcal{G}_x\ltimes U$ for an action of the isotropy group $\mathcal{G}_x$ on $U$.

Reference given is Orbifolds, Sheaves and Groupoids by Ieke Moerdijk and D. A. Pronk.

In page $12$, discussing Characterization of Orbifolds, they gave a proof of this which I am not able to understand completely.

I will write down the proof that I understood.

Let $x\in \mathcal{G}_0$. We are expected to find an open set containing $x$ on which $\mathcal{G}_x$ acts.

Let $\gamma\in \mathcal{G}_x$. As $s,t:\mathcal{G}_1\rightarrow \mathcal{G}_0$ are local diffeomorphisms, we can choose a common open set $W_\gamma$ of $\gamma$ such that $s|_{W_\gamma}:W_\gamma\rightarrow s(W_\gamma)$ and $t|_{W\gamma}:W_\gamma\rightarrow t(W_\gamma)$ are diffeomorphisms.

Consider the composition $s(W_\gamma)\xrightarrow{s^{-1}}W_\gamma\xrightarrow{t} t(W_\gamma)$. As $\gamma\in \mathcal{G}_x$, we have $x\in s(W_\gamma)\bigcap t(W_\gamma)$. So, given $\gamma\in \mathcal{G}_x$, we have open sets $s(W_\gamma),t(W_\gamma)$ containing $x$ and a diffeo. $s(W_\gamma)\rightarrow t(W_\gamma)$. Here, we have two possibly different open sets containing $x$ and these open sets as well as the map $s(W_\gamma)\rightarrow t(W_\gamma)$ vary with $\gamma$. What we want is a single open set $V$ containing $x$ for each $\gamma$ giving the action $\mathcal{G}_x\times V\rightarrow V$.

As first step, we can consider $U_x$ to be $\bigcap s(W_\gamma)$. We have for each $\gamma\in \mathcal{G}_x$ map $s(W_\gamma)\rightarrow t(W\gamma)$. So, if we take intersection, we get one single open set, that works (not completely in the sense we want) and we have $\mathcal{G}_x\times U_x\rightarrow \bigcup t(W_\gamma)$. There is no obvious why $U_x=\bigcup t(W_\gamma)$. So, we consider a smaller subspace $V_x$ of $U_x$ containing $x$ so that we get $\mathcal{G}_x\times V_x\rightarrow V_x$.

Choose $V_x\subset U_x$ such that $$(V_x\times V_x)\bigcap (s,t)\left( \mathcal{G}_1\setminus \bigcup W_\gamma\right)=\emptyset.$$

This is still not a good choice of open sets that gives action of $\mathcal{G}_x$.

Denote by $\tilde{\gamma}$ the map $s(W_\gamma)\rightarrow t(W_\gamma)$ defined above. As $V_x\subseteq s(W_\gamma)$, this gives a map $\tilde{\gamma}:V_x\rightarrow t(W_\gamma)$.

We want to produce an open subset $N_x\subseteq V_x$ containing $x$ such that $\mathcal{G}_x$ acts on $N_x$ i.e., for each $\delta\in \mathcal{G}_x$ and $y\in N_x$ we should have $\delta.y=\tilde{\delta}(y)\in N_x$. This suggests a choice for $N_x$, namely $$N_x=\{y\in V_x: \tilde{\delta}(y)\in V_x \forall \delta\in \mathcal{G}_x\}.$$ We see that $\mathcal{G}_x$ acts on $N_x$. Let $\gamma\in \mathcal{G}_x$ and $y\in N_x$. We see that $\tilde{\gamma}(y)\in N_x$ i.e., $\tilde{\delta}(\tilde{\gamma}(y))\in V_x$ for each $\delta\in \mathcal{G}_x$.

As $y\in N_x$, we have $\tilde{\delta'}(y)\in V_x$ for each $\delta'\in\mathcal{G}_x$, in particular $\tilde{\delta\gamma}(y)=\tilde{\delta}(\tilde{\gamma}(y))\in V_x$ for each $\delta\in \mathcal{G}_x$. Thus, $\tilde{\gamma}(y)\in N_x$. Thus, we have an action of $\mathcal{G}_x$ on $N_x$.

It then says,

The restriction of the groupoid $\mathcal{G}|_{N_x}$ is exactly the translation groupoid of the action of $\mathcal{G}_x$ on $N_x$.

I am not able to see how does that follow. Any hints would be useful.

  • $\begingroup$ There is a more general version of this for proper Lie groupoids, and it is a generalisation of the slice theorem for proper group actions. There are various versions, and incremental results due to Weinstein, Zung, Pflaum-Postuma-Tang, Crainic-Struchiner $\endgroup$
    – David Roberts
    Jun 19, 2018 at 23:11
  • $\begingroup$ @DavidRoberts Can you give some reference please $\endgroup$ Jun 20, 2018 at 1:44
  • $\begingroup$ arxiv.org/abs/1103.5245 arxiv.org/abs/1101.0180 $\endgroup$
    – David Roberts
    Jun 20, 2018 at 3:02
  • $\begingroup$ @DavidRoberts I just saw those papers.. they seem to be so many terms that I do not know... that might take 3/4 days to read them, I am afraid it would be a diversion...can you think of some hint that might be helpful in understanding this result.. $\endgroup$ Jun 20, 2018 at 3:19
  • $\begingroup$ @DavidRoberts this is off topic but, Is there any current PhD student you are aware of who is working in Geometry of Orbifolds/Lie groupoids in the set up of Moerdijk or his students.. $\endgroup$ Jun 20, 2018 at 3:20

1 Answer 1


In the following I will use the notation you cited in the problem statement.

The point of the proof should be that you want to construct an open neighborhood $N_x$ which is stable under the $\delta$-action of the stabiliser group $G_x$ and (here comes the crucical property!) that $$ s^{-1} (N_x) = \coprod_{\gamma \in G_x} \underbrace{W_\gamma \cap s^{-1} (N_x)}_{=:O_\gamma} = \coprod_{\gamma \in G_x} W_\gamma \cap (s,t)^{-1}(N_x\times N_x) \quad \text{ (disjoint union!)}. $$ Here the last equality follows from the definition of $N_x$.

We consider now the open subgroupoid (whence itself a Lie groupoid) $s^{-1} (N_x) \rightrightarrows N_x$. Note that the arrow space is the disjoint union of the $O_\gamma$, Hence we can define the following diffeomorphism $$ \Phi \colon s^{-1} (N_x) \rightarrow G_x \times N_x , z \mapsto (\gamma, s(x)), \text{ if } z \in O_\gamma. $$ Here we use that $s$ restricts on each $W_\gamma$ to a diffeomorphism onto an open neighborhood of $N_x$ (whence the restriction to $O_\gamma$ is again a diffeomorphism). Looking at the definition of the $\delta$-action of $G_x$, we see that the diffeomorphism obtained in this way actually becomes a Lie groupoid isomorphism over the identity: $\Phi \colon s^{-1} (N_x) \rightarrow G_x \ltimes N_x$ (where the left denotes the action groupoid with respect to the $\delta$-action). This should answer your question.

Let me mention one point which bothered me personally by the writeup presented in your question (and in both (!) of the sources cited by you). It is exactly the line (more details by me following the sources):

We see that $G_x$ acts on $N_x$. Let $\gamma \in G_x$ and $y=\tilde{\eta}(z)\in N_x$. We see that $\tilde{\gamma}(y) = \tilde{\gamma} \circ \tilde{\eta} (z) = \tilde{\gamma \eta} (z) \in N_x$.

Here the last equality is not clear at all to me. Moreover, both original sources state it to be obvious. From a proof perspective and for the sake of the orbifold community, the equality better holds as we would not get a group action otherwise. However, I do not see why this is true without further arguments (in case one of the original authors is reading this, I would love to hear the reasoning! As it is presented in the original sources I personally think this is a gap). This bothered me enough to redo the whole proof and check that this property can be enforced by some additional technical steps. The details are recorded in the proof of Proposition 3.13 in

Habib Amiri, Helge Glockner, Alexander Schmeding, Lie groupoids of mappings taking values in a Lie groupoid (2018) arXiv:1811.02888.

Before you go and check it out: The argument in 3.13 of loc.cit. is written without using any finite-dimensionality (as in the paper infinite-dimensional Lie groupoids are treated) but instead a mild topological requirement. Hence this should be the most general version one can establish of the Theorem: proper + étale Lie groupoid $\Rightarrow$ locally isomorphic to action groupoid

  • $\begingroup$ I updated to reference so as not to refer to an anonymous arXiv pdf :-) $\endgroup$
    – David Roberts
    Nov 18, 2018 at 22:49
  • $\begingroup$ I thank you for your time. I was little busy in something else. I will read this in detail and respond as soon as possible. Thank you :) $\endgroup$ Nov 22, 2018 at 11:14

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