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Let ${\mathcal{G} = \lbrace s,t:G_1 \to G_0 \rbrace}$ be a Lie groupoid. Define $$(\mathcal{G}^k)_0:=\lbrace (a_1,\dots,a_k) \in G_1^k\mid s(a_1)=t(a_1)=\dots=s(a_k)=t(a_k) \rbrace$$ (This is the space of objects of $k$-sectors $\mathcal{G}^k$. See Adem-Ruan-Zhang arXiv:math/0605534 for more details.)

My question is: how do we prove that the space $(\mathcal{G}^k)_0$ is a manifold?

(Or how do we see that the map $$ G_1^k \to G_0^{2k}: (a_1,\dots,a_k)\mapsto(s(a_1),t(a_1),\dots,s(a_k),t(a_k)) $$ is transverse to $ Z=\lbrace(x,\dots,x) \in G_0^{2k}\mid x \in G_0 \rbrace $ ?)

[Additional explanation]

Let the circle group $S^1$ act on the unit sphere $S^2 \subset \mathbb{R}^3$ as rotations about the $z$-axis. For the action groupoid $\mathcal{G}$ of the action, the space $(\mathcal{G}^1)_0=\lbrace (t,x) \in S^1 \times S^2 \mid\ t=1 \text{ if } x \ne (0,0,\pm 1) \rbrace$. This is not a manifold. So if the space $(\mathcal{G}^k)_0$ is to be a manifold, then we have to assume some conditions. (This is the reason why I have changed the title of this question.)

I have found a relavant explanation in Moerdijk. In section 6.4, he deals with the inertia orbifold (groupoid). According to the paper, if $\mathcal{G}$ is étale, then we can show by taking local chats that $S_\mathcal{G} (=(\mathcal{G}^1)_0)$ is a manifold.

Moreover for a proper foliation groupoid, we can also use "local charts" in the sense of Crainic-Moerdijk to show that the smoothness of $S_\mathcal{G}$. So I am checking Crainic-Moerdijk.

But the problem still remains even if we understand that $(\mathcal{G}^1)_0$ is a manifold. Let $\pi_k:(\mathcal{G}^k)_0 \to G_0$ is the map sending $(a_1,\dots)$ to $s(a_1)$. Then $(\mathcal{G}^{k+1})_0$ is the fiber product of $\pi_k$ and $\pi_1$ as a topological space. But $\pi_1$ is not a submersion. In fact, let the circle group acts on the unit 3-sphere in $\mathbb{C}^2$ with multiplicity $(1,p)$. (The quotient is the orbifold called a tear drop.) Then $(\mathcal{G}^1)_0$ consists of the original 3-sphere with $p-1$ circles. Therefore $\pi_1$ is not a submersion.

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The map is transverse to $Z$ because it is a submersion; it is a submersion because $s$ and $t$ are submersions (by definition of a Lie groupoid). – Konrad Waldorf Oct 7 '12 at 14:11
Thanks for your comment. But I don't think that the map is a submersion, because the dimension of the target $G_0^{2k}$ can be greater than one of the domain $G_1^k$. (e.g. the action groupoid of an $S^1$-manifold $M$.) Therefore the differential map of the map is not surjective in general. – H. Shindoh Oct 7 '12 at 21:14
Note that Moerdijk claims it to be true for a Lie groupoid with discrete hom-spaces. This rules out your example. – David Roberts Oct 17 '12 at 6:31
Yes. The example which I wrote is not a foliation groupoid. So I think that the space is a manifold for a proper foliation groupoid. – H. Shindoh Oct 17 '12 at 8:57
I have roughly checked that for a locally free $H$-action on $M$ the space $(\mathcal{G}^1)_0$ of the action groupoid $\mathcal{G} = H \ltimes M$ is a manifold. Maybe it is better to see locally the space $(\mathcal{G}^1)_0$ by using slices, instead of the transversality of the above map. Regarding a proper foliation groupoid $\mathcal{G}$, I think that we can show that $(\mathcal{G}^1)_0$ is a manifold if we can take "slices" for a foliation structure on $G_0$ induced by the groupoid $\mathcal{G}$. – H. Shindoh Jan 2 '13 at 20:12

You don't try to show that the map is transverse to $Z$, but rather take the iterated pullback $G_1\times_{G_0}G_1 \times_{G_0} \cdots \times_{G_0} G_1$.

For the case of $G_1\times_{G_0}G_1$ what you have is the pullback of $s$ along $t$, both submersions, so the projections are submersions, and hence the new map you are going to pull back is a submersion (being the composite of one of these projections and a source or target map). You iterate this and at each step the projection is a submersion, so the next step is do-able.


Ah, I found the result I was looking for in Mackenzie's Lie groupoids and Lie algebroids in differential geometry (LMS lecture note series no. 124), namely Proposition III.1.17, on page 92. It says that for any Lie groupoid $\Omega \rightrightarrows B$ (there called a differentiable groupoid - for him Lie groupoids are a specialised notion) the inertia groupoid (there denoted $G\Omega$) is a sub-Lie groupoid, and the arrows of $G\Omega$ form a closed embedded submanifold of $\Omega$. In particular this result implies the source = target map $G\Omega \to B$ is a submersion.

However, in the newer General theory of Lie groupoids and Lie algebroids (LMS lecture note series no. 213), he corrects this, in a comment just after example 1.2.12, and only claims it for locally trivial Lie groupoids, which was probably what he was thinking of in the earlier book.

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Thank you for your answer. Could you explain a relation between $G_1 \times_{G_0} G_1$ and $(\mathcal{G}^1)_0$? I do not think that they agree. (I guess that the fibred products which you mention is the nerve of the Lie groupoid.) – H. Shindoh Oct 12 '12 at 9:26
Ah you are right, I misread the question. But what you want follows from what I wrote and the result that for any Lie groupoid G the inertia groupoid ΛG, defined as the groupoid with same objects and arrows the pullback of (s,t):G_1 -> G_0^2 along the diagonal is also a Lie groupoid. The proof of this is in Mackenzie's first book on Lie groupoids, not too far from the start. – David Roberts Oct 12 '12 at 22:27
I am checking the book which you mentioned. But I have not found helpful description. I was wondering if you could tell me precise pages in the book. – H. Shindoh Oct 16 '12 at 15:34
Thank you very much for your detailed comment and sorry for my late reply. As Mackenzie corrects the his statement (Prop III.1.17 in LMS 124) in the newer book, I doubt his proof of the statement. (The cartesian square is wrong.) Regarding the comments after example 1.2.12 in LMS 213, the condition of locally triviality is too strong, but the discussion relating to the inner subgroupoid can be helpful to solve my question. – H. Shindoh Dec 31 '12 at 19:21

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