When are $k$-sectors of a Lie groupoid a manifold?

2012-10-04T15:52:24Z

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.

Answer by David Roberts for When are $k$-sectors of a Lie groupoid a manifold?

2012-10-11T03:12:30Z

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.

Edit:

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.

When are $k$-sectors of a Lie groupoid a manifold? - MathOverflow

When are $k$-sectors of a Lie groupoid a manifold?

Answer by David Roberts for When are $k$-sectors of a Lie groupoid a manifold?