Isotropy subgroupoid of a regular Lie groupoid

Question

Let $(G\rightrightarrows M)$ be a Lie groupoid (i.e. a groupoid with source map $s$ and target map $t$ such that $G,M$ are smooth manifolds and the structural maps are all smooth (and $s$,$t$ are submersions).

Defining $$IG := \{g \in G \mid s(g) = t(g) \}$$ we obtain the so called isotropy subgroupoid $IG \rightrightarrows M$ which is in general only a subgroupoid but not a sub LIE groupoid, in that $IG$ need not be a submanifold of $G$ (for an example consider the action groupoid of the canonical action $\mathbb{S}^1 \times \mathbb{R}^2 \rightarrow \mathbb{R}^2$ of the circle group via rotation. In this case $G=\mathbb{S}^1 \times \mathbb{R}^2$ and thus $IG = \mathbb{S}^1 \sqcup \bigsqcup_{x\in \mathbb{R}^2} \{1\}$, whence the isotropy subgroupoid of the action groupoid can not be a Lie groupoid with respect to the subspace topologies.

Now I have read in several places (cf. e.g. 1) that the isotropy groupoid is an embedded Lie subgroupoid if the Lie groupoid is regular. Recall that a Lie groupoid is regular if its anchor $$(s,t)\colon G \rightarrow M\times M, g \mapsto (s(g),t(g))$$ is a mapping of constant rank (there are some other equivalent formulations, using e.g. the orbit foliation of the Lie groupoid, see e.g. 2). Unfortunately, I was not able to track down a proof of this folklore fact in the literature.

The old book by Mackenzie (Lie Groupoids and Lie Algebroids in Differential Geometry) proves a similar statement using foliation charts for the orbit foliation. Note that the statement in the book claims that it is true for all Lie groupoids (there called differentiable groupoids), but the proof holds only for locally trivial Lie groupoids (and seems to be intended to be only for this class). For this class however, the anchor $(s,t)$ is a submersion (and $IG = (s,t)^{-1} (\Delta M)$, where $\Delta M$ is the diagonal) whence it is easy to see why $IG$ is an (embedded) submanifold of $G$.

I wonder now how one tackles the general case for regular Lie groupoids. Surely there must be a proof somewhere in the literature? Or alternatively an easy argument which eludes me?

Answer 1

This is not exactly an answer to your question, but I hope it helps.

If it would be fine to first pass to the connected components of the identity of each isotropy group, I believe you find the proof of the analogous statement (that you then obtain an embedded subgroupoid) in Proposition 2.5 of

I. Moerdijk, On the Classification of Regular Groupoids

P.S. - See update 1 for a counterexample, and update 2 for an even easier counterexample to the general case.

---

Update 1

After thinking a bit more about this, I believe that in general one really has to first pass to the connected component of the identity in the isotropy Lie groups. One problem that might arise in general is seen in the following example of an action groupoid (which comes from family of actions of the reals on a torus).

Consider the action of the Lie group $(\mathbb{R},+)$ on $T \times (-1,1)$, where $T=\mathbb{R}^2/ \mathbb{Z}^2$ is a 2-torus, given by $$\lambda \cdot ([x,y],\epsilon)= ([x+\lambda, y+\lambda\epsilon],\epsilon), $$ for $\lambda\in \mathbb{R}$, $[x,y]\in T$ and $\epsilon\in (-1,1)$.

The associated action groupoid $G=\mathbb{R}\times T \times (-1,1) \rightrightarrows T \times (-1,1)$ is a regular Lie groupoid, because all the orbits are diffeomorphic to either circles or lines.

The isotropy of a point $([x,y],\epsilon)$ is:

• a copy of $\mathbb{Z}$ inside of $\mathbb{R}\times \{([x,y],\epsilon)\}$ if $\epsilon$ is rational;

• the trivial group $\{ 0\} \times \{([x,y],\epsilon)\}$ if $\epsilon$ is irrational.

Moreover, all points along a same orbit have ''the same'' isotropy in the sense that if $([x,y],\epsilon)$ and $([x',y'],\epsilon)$ are in the same orbit, and $(\lambda , [x,y],\epsilon)$ is in the isotropy of $([x,y],\epsilon)$, then $(\lambda , [x',y'],\epsilon)$ is in the isotropy of $([x',y'],\epsilon)$. This means in particular that the isotropy subgroupoid, if it were an embedded submanifold should have dimension at least one.

So take for example $\epsilon=0$, $\lambda = 1$. Then $(1,[x,y],0)$ is a point in $IG$, but there is no neighborhood of it inside of $G$, which when intersected with $IG$ is diffeomorphic to some euclidean space, because of all the points $([x,y],\epsilon)$ with $\epsilon$ small that have trivial isotropy. So $IG$ is not an embedded submanifold.

I get the impression this would work one dimension lower, taking just a circle instead of a torus, but for me it is easier to visualize the situation when the orbits are one-dimensional.

Update 2 (May 7, 2019)

Thanks to a question of Alexander Schmending, I later realised that a comment I previously made (and now deleted to avoid confusion) was not true. In the comment I suggested that if $G$ is regular and proper, $IG$ should be a Lie subgroupoid. But again, this is not necessarily the case, and the example is way easier than the one above.

This can be seen by considering the action groupoid $G= \{-1,1\}\times \mathbb{R^2}$ associated with the action of $\{-1,1\}$ on the plane $\mathbb{R^2}$ by reflection through the origin. This Lie groupoid is proper and regular. All points except the origin have trivial isotropy, while the origin has $\{-1,1\}$ as isotropy.

The source map $s$ restricted to $IG$ cannot be a submersion, as $s_{|IG}$ has rank 0 at $(-1,0)\in G=\{-1,1\}\times \mathbb{R^2}$.

So things go wrong in a different way than that of the example of update 1: $IG$ is Hausdorff, it is even an embedded submanifold of $G$ (with components of different dimensions), just not a Lie groupoid.