# Isotropy group of a Lie groupoid is a Lie group

I am trying to see that Isotropy group/object group/vertex group of a Lie groupoid is a Lie group.

Let $\mathcal{G}$ be a Lie groupoid and $x$ be an object in $\mathcal{G}$ i.e., $x\in \mathcal{G}_0$. By an isotropy group of $\mathcal{G}$ we mean the collection of all arrows from $x$ to itself. Some people write $\mathcal{G}(x,x)$ or $\mathcal{G}_x$ for this set.

This is a group as we are in the case of groupoid where all arrows are invertible and as source and target are same for each map.

If we write down what actually is $\mathcal{G}_x$ is, we have $$\mathcal{G}_x=\{g:x\rightarrow x\}=s^{-1}(x)\cap t^{-1}(x)$$ where $s,t:\mathcal{G}_1\rightarrow \mathcal{G}_0$ are source and target maps. Only extra information given on them is that they are smooth submersions. As $s:\mathcal{G}_1\rightarrow \mathcal{G}_0$ is a submersion, $s^{-1}(x)$ is an embedded submanifold of $\mathcal{G}_1$ and similarly as $t:\mathcal{G}_1\rightarrow \mathcal{G}_0$ is a submersion, $t^{-1}(x)$ is an embedded submanifold of $\mathcal{G}_1$. I am not able to see why would their intersection is an embedded submanifold. I guess it has something to do with transversal intersections.

Two maps $F,G:M\rightarrow N$ are said to intersect transversally if $$F_{*,a}(T_aM)+G_{*,b}(T_bM)= T_p(N)$$ where $a,b\in M$ such that $F(a)=G(b)=p$.

If atleast one of maps $F,G:M\rightarrow N$ is a submersion, then they intersect transversally.

As $F_{*,a}$ or $G_{*,b}$ is surjective, we already have $F_{*,a}(T_aM))=T_pN$ or $G_{*,b}(T_bM))=T_pN$, which means $$F_{*,a}(T_aM)+G_{*,b}(T_bM)= T_p(N).$$

In particular, for source and target maps $s,t$ as above, we have transversal intersection. Thus, by transversal intersection theorem. we see that $$\{(g,h):s(g)=t(h)\}\subset \mathcal{G}_1\times \mathcal{G}_1$$ is a smooth manifold.

But here what I am looking for is little extra.

The set $s^{-1}(x)\cap t^{-1}(x)$ is a subset of $\{(g,h):s(g)=t(h)\}=\mathcal{G}_1\times_{\mathcal{G}_0} \mathcal{G}_1$. How do I see that $\mathcal{G}_x=s^{-1}(x)\cap t^{-1}(x)$ is also a submanifold? How do I see that with this smooth structure, this is a Lie group?

Any suggestions are welcome. I am reading Orbifolds as Groupoids.

Any outline would also be accepted as an answer.

• The result that you want is Theorem III.1.4 on page 86 of the book "Lie Groupoids and Lie Algebroids in Differential Geometry" by Kirill Mackenzie. There is some work involved; I haven't digested it. – Neil Strickland May 31 '18 at 14:08
• @NeilStrickland Thank you :) I would check that right now :) – Praphulla Koushik May 31 '18 at 14:28
• For some one who are not comfortable reading that old type set, you can see the book Introduction to Foliations and Groupoids by Ieke Moerdijk and Janez Mrčun Theorem 5.4 Page 115.. I do not know anything about Foliations so I can not understand that completely... It would be useful if some one can give an out line... Thanks – Praphulla Koushik May 31 '18 at 15:15

There is also a rundown in the more modern book of Mackenzie: General Theory of Lie groupoids and Lie algebroids (p.26 Corollary 1.4.11)

Let me summarise the idea: Fix a Lie groupoid $$G \rightrightarrows M$$ with source map $$s$$ and target map $$t$$.

Idea:

1. Every fibre $$s^{-1} (x)$$ is a closed embedded submanifold of $$G$$ (as $$s$$ is a submersion). Thus it makes sense to consider $$t|_{s^{-1}(x)}$$ (restriction of $$t$$ to the source fibre) as a smooth map.
2. Prove that $$t|_{s^{-1}(x)}$$ is a map of constant rank for each $$x \in G$$. Here Mackenzie uses the existence of enough local bisections (see below).
3. the isotropy group $$G_x^x = (t|_{s^{-1}(x)})^{-1}(x)$$ is a closed embedded submanifold by the constant rank theorem (see your favorite book on differential geometry or the very general version in https://arxiv.org/pdf/1502.05795.pdf Theorem F). Note that this shows that $$G_x^x$$ is a submanifold of $$s^{-1}(x)$$ but since this is an embedded submanifold it is also a submanifold of $$G$$.
4. This submanifold structure turns $$G^x_x$$ into a Lie group. Multiplication and inversion are restrictions of the groupoid operations which are smooth on $$G$$. Now $$G_x^x$$ is a closed embedded submanifold of $$G$$ and thus inherits smooth multiplication and inversion from the groupoid.

Now to carry out 2. One constructs first a local bisection through each arrow, i.e. a smooth local section $$\sigma \colon U \rightarrow G$$ ($$U \subseteq M$$ open) of $$s$$ such that $$t\circ \sigma$$ is a diffeomorphism and $$\sigma(s(g))=g$$. This exists by some linear algebra and an application of the inverse function theorem. (Details: You take a section of the source map at the point, use linear algebra to show that one can arrange the image of the derivative of the section to be a simultaneous complement to the kernels of the derivatives of source and target map. Then the inverse function theorem yields the desired properties).

Now this was the finite-dimensional case (in which you are interested I assume). Since I enjoy infinite-dimensional groupoids, let me hijack your question to push this further:

The above proof works as presented for any infinite-dimensional groupoid (using Bastiani calculus to make sense of differentiability beyond Banach spaces) over a finite-dimensional base $$M$$. See e.g. Appendix A of https://arxiv.org/pdf/1506.05415.pdf for a proof. However, due to the local bisection trick and the constant rank theorem this is as far as one gets.

The nice thing about the Moerdijk, Mrcun approach referenced above is that it can be pushed further. Namely if we are dealing with Banach groupoids (not necessarily finite-dimensional base but now everything is modeled on a Banach space), then this generalised approach yields the Lie group structure of the isotropy groups. This was just recently observed in https://arxiv.org/pdf/1802.09430.pdf (Theorem 3.3 ii)

As far as I know, the question is still open if the isotropy groups of an infinite-dimensional Lie groupoid always inherit a canonical Lie group structure.

• This does give some idea but I need some more time to really digest, so can not accept it now. Thanks for the answer. – Praphulla Koushik Jun 6 '18 at 12:57
• I edited slightly to make it easily readable.. I asked this $8/9$ months ago, I did not come across the need for this result so did not spend time on this... Now, I want to understand this... I will accept in 4/5 hours If I do not have further questions... – Praphulla Koushik Feb 15 at 7:19