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# Whenare $k$-sectors of a Lie groupoid amanifold?

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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