When are $k$-sectors of a Lie groupoid a manifold? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T03:45:22Z http://mathoverflow.net/feeds/question/108830 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108830/when-are-k-sectors-of-a-lie-groupoid-a-manifold When are $k$-sectors of a Lie groupoid a manifold? H. Shindoh 2012-10-04T15:52:24Z 2013-01-02T20:02:16Z <p>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 <a href="http://arxiv.org/abs/math/0605534" rel="nofollow">arXiv:math/0605534</a> for more details.)</p> <p>My question is: how do we prove that the space $(\mathcal{G}^k)_0$ is a manifold?</p> <p>(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 $Z=\lbrace(x,\dots,x) \in G_0^{2k}\mid x \in G_0 \rbrace$ ?)</p> <p><strong>[Additional explanation]</strong> </p> <p>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.)</p> <p>I have found a relavant explanation in <a href="http://arxiv.org/abs/math/0203100" rel="nofollow">Moerdijk</a>. 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. </p> <p>Moreover for a proper foliation groupoid, we can also use "local charts" in the sense of <a href="http://www.math.uiuc.edu/K-theory/0446/" rel="nofollow">Crainic-Moerdijk</a> to show that the smoothness of $S_\mathcal{G}$. So I am checking Crainic-Moerdijk.</p> <p>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. </p> http://mathoverflow.net/questions/108830/when-are-k-sectors-of-a-lie-groupoid-a-manifold/109352#109352 Answer by David Roberts for When are $k$-sectors of a Lie groupoid a manifold? David Roberts 2012-10-11T03:12:30Z 2012-11-28T06:25:16Z <p>You don't try to show that the map is transverse to $Z$, but rather take the iterated pullback $G_1\times_{G_0}G_1 \times_{G_0} \cdots \times_{G_0} G_1$.</p> <p>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.</p> <hr> <p>Edit:</p> <p>Ah, I found the result I was looking for in Mackenzie's <em>Lie groupoids and Lie algebroids in differential geometry</em> (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 <em>differentiable</em> 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.</p> <p>However, in the newer <em>General theory of Lie groupoids and Lie algebroids</em> (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.</p>