# Given an object in a Lie groupoid, does there exist a subgroupoid for which the object has no automorphisms but retains its equivalence class?

For this question, I am in the smooth finite-dimensional category. All objects are $\mathcal C^\infty$ manifolds and all maps are smooth. Recall the notion of groupoid, and let's restrict our attention to those for which the source map (and hence also the target map) is a (necessarily surjective) submersion. If $G\rightrightarrows X$ is a groupoid, me define a subgroupoid over $X$ to be an immersed submanifold $H\hookrightarrow G$ so that $H\rightrightarrows X$ is a groupoid. (If this is not the standard definition, let me know.)

Pick $x\in X$. Does there exist a subgroupoid $H\hookrightarrow G$ so that: (1) $x$ has no automorphisms in $H$, and (2) if $y\in X$ with $x\cong y$ in $G$, then $x\cong y$ in $H$? I.e.: the groupoid $H$ should be as small as possible but still preserve the isomorphism class of $x$. I don't mind if other isomorphism classes shrink in size.

Consider the case of a group $K$ acting on $X$. Restricting to the orbit of the point $x$, by the orbit-stabilizer theorem $X$ is identified with $K/\text{Stab}(x)$. Your question seems to amount to asking whether the projection $K\times K/\text{Stab}(x)\to K/\text{Stab}(x)\times K/\text{Stab}(x)$ given by $(\gamma,p)\mapsto (\gamma\cdot p,p)$ admits an immersed section. Restricting such a section to a fiber $K/\text{Stab}(x)\times \text{pt}$ would give an section of the map $K\to K/\text{Stab}(x)$, which will not exist in general. (Edit: rewritten. Thanks to David Carchedi for pointing out the mistake in my notation.)

A concrete counterexample: consider the action of $K=\mathbb{R}$ on $X=S^1$ by translation. Then $G$ is diffeomorphic to $\mathbb{R}\times S^1$, and $H$ would have to be diffeomorphic to $S^1\times S^1$, but $S^1\times S^1$ does not immerse in $\mathbb{R}\times S^1$. Moreover the restriction to a fiber $S^1\times \text{pt}$ would be an immersion of $S^1$ into $\mathbb{R}$, which also cannot exist.

But this is not a phenomenon just of fundamental group or of equi-dimensional immersions. For example, consider $K=\text{Isom}(\mathbb{H}^2)=\text{PSL}_2\mathbb{R}$ acting on a hyperbolic surface $\Sigma$. A section of $\text{PSL}_2\mathbb{R}\times \Sigma\to \Sigma\times \Sigma$ restricts to an immersed section of $\text{PSL}_2\mathbb{R}\to \Sigma$. Of course $\Sigma$ does immerse, in fact embed, in $\text{PSL}_2\mathbb{R}$ (it's a 3-manifold). But if we had a continuous section $\varphi:\Sigma\to \text{PSL}_2\mathbb{R}$, we could translate a nonzero vector $v$ to each point $p$ by the isometry $\varphi(p)$. This would yield a nonzero vector field on $\Sigma$, contradicting the Gauss-Bonnet theorem. The same works for $\text{SO}(3)$ acting on $S^2$ or $\mathbb{R}P^2$, where the fundamental group is not the issue.

• (I removed my previous comment because this answer has been rewritten) – David Carchedi Jun 1 '10 at 17:21

My gut would be no. Let us analyze what happens if this were to be true. Then, between any $z$,$w$ in the orbit of $x$, there is a unique arrow (since any two guys in the same orbit have isomorphic isotropy groups). On the other hand, $H$ is transitive when restricted to the orbit of $x$. Call $O$ the orbit of $x$. Then $H$ restricted to $O$ is isomorphic to the pair groupoid on $O$. So, in particular, restricting the immersion $H \hookrightarrow G$ to $H|_{O}$, one obtains an immersion $O \times O \hookrightarrow G$. However, this immersion needs to be a Lie groupoid homomorphism (I presume). This says that there is a smooth way of picking for any two $z$ and $w$ in $O$ a morphism in $G$ between them. Clearly, such a set-theoretical choice exists, but, it seems very unlikely that it can be done smoothly in general.

• What is your motivation by the way? Perhaps there's more to be said, depending on what you would like to accomplish. – David Carchedi Jun 1 '10 at 1:29

Edit: cross posted with David Carchedi.

Let $(x) \subset X$ be those objects which are isomorphic to $x$. If this was a problem in ordinary groupoids, then you could let $H = [(x)\times (x)] \cup_{(x)} X$ where the pushout is taken of the diagonal and the inclusion. The inclusion then requires a choice of section $(x)\times (x) \to G$ of the map $(s,t):G \to X\times X$. On $X \subset H$ just take the inclusion of the objects via the unit map $X \to G$.

Going back to the smooth case, this construction requires the section as given. This is just the simplest construction I could think of, but it seems to me to be minimal, in that any construction as you want will contain it. Since you require $x$ to have no automorphisms in $H$, but still connected to the rest of $(x)$, none of the objects $y\in (x)$ can have automorphisms either, and so the restriction of $H$ to the objects in $(x)$ is necessarily $codisc(x)$ - the codiscrete (=chaotic etc) groupoid with object manifold $(x)$ (I'm assuming this is a manifold, or I don't think the question makes sense). Now if your groupoid is locally trivial (a la Ehresmann), $(x)$ is a sum of connected components of $X$, and then you can take the 'rest' of $H$ to be the restriction of $G$ to the complement of $(x)$ in $X$. This is the maximal such construction. But I can't see how you get away from needing a section $(x)\times (x) \to G$.