# Talk:Properly discontinuous action

## is there an isotropy subgroup for a properly discontinuous action?

The article on Orbifold says: The main example of underlying space is a quotient space of a manifold under the properly discontinuous action of a possibly infinite group of diffeomorphisms with finite isotropy subgroups. In particular this applies to any action of a finite group;

I want to agree. But I'm confused. If the action of the group were free + properly discontinuous, we'd get a manifold as a quotient, wouldn't we? But how does this make sense? How can a properly discontinuous action be non free? If there is a point that is fixed by some group element, we'll always have that group element in the set ${\displaystyle \{g\in G|gU\cap U\not =\emptyset \}}$. How can that be the case?

And why is freely discontinuous a redirect to this article? What does freely discontinuous mean?

This is probably terrible obvious, but I don't see it right now. Maybe tomorrow. -- JanCK (talk) 22:16, 19 December 2007 (UTC)

A manifold without boundary is a special case of an orbifold, where all groups used to describe the orbifold charts are trivial. I don't know who made the redirect for freely discontinuous. This should be linked more properly to principal fiber bundle (with discrete structure group). The ideas used to understand principal fiber bundles should help you understand this case: the key idea is the local triviality of the bundle, which locally expresses it as a product U x Γ, where U is an open subset of the base space. For a properly discontinuous action which is non-free, think about the cyclic group of order n acting by rotations about the N-S axis on the 2 sphere. Mathsci (talk) 22:38, 19 December 2007 (UTC)
Thanks, for your reply. Searching for freely discontinuous I found a definition at Free regular set, so I'll put the redirect there. I still don't see how to choose the open set U around the northpole N in your example. Isn't it that for every U (${\displaystyle N\in U}$) and for every element g from your cyclic group there is ${\displaystyle N\in gU\cap U}$, so ${\displaystyle \{g\in G|gU\cap U\not =\emptyset \}=G\not =\{e\}}$? But you are right, this isn't the right place to explain this to me, it's the place to discuss the article. I guess, I misunderstand something, I just wanted to point this out, so someone can thing about it, and see whether the definition given is really correct. If you are the one, who tells me, that it's all correct the way it is, I'll think about it a little longer, and I'll see whether I can have it explained to me here(here meaning physically, here at my university.) -- JanCK (talk) 00:30, 20 December 2007 (UTC)
I noticed you seemed hesitant to work through the example on the talk page. If you want to talk more about it on wikipedia, you might consider the refdesk. Talk pages are just for discussing how to improve the articles. Since the articles confused you and you two found ways to improve them, clearly there was reason to ask on the talk pages, but MathSci's answer should be sufficient for the article (he says it is fine, so odds are pretty high it is just dandy). On the off chance you want a place where you can have discussions about the material (not just the article) while still keeping close ties to the wikipedia community, then the reference desk is a nice place to ask. Certainly if you think an article is wrong, ask on the talk page, but if you want to learn the material (and THEN improve the article), the reference desk is better suited. Hopefully, your physical university is even *better* suited, but talking about it on wikipedia might even educate ignorant authors like me. Several people here work in equivariant (co)homotopy theory, and it is occasionally distressing to have no idea what they are talking about. JackSchmidt (talk) 04:52, 20 December 2007 (UTC)
Equally well references like the online Thurston notes or Peter Scott's BLMS article are put there so that people can check with the quoted sources. For example, having worked fairly thoroughly through all of Haefliger's contributions, it seems clear that his definitions were not quite right (this is explained in footnotes in his book with Bridson). Thurston's definition was adopted because (a) it's currently the only one in use in top journals such as Annals of Mathematics, asterique, etc, and (b) it is applicable to Haefliger's case and removes the potential problems in his definition. For an orbifold chart round any point in the two-sphere example, take small geodesic discs about each point. No group is necessary for non-polar points and the cyclic group of rotations is the group for each pole. I don't quite understand your notation JanCK; but please look in the very user friendly Chapter XIII of Thurston wikilinked on the orbifold page. Good luck, Mathsci (talk) 11:50, 20 December 2007 (UTC)
Actually I didn't want to reply that early, and I wanted to reply on WP:RD/MA, but now that I just have one quote to give, I thought I'll add it here, and do some more thinking afterwards. sry.
Thurston p.302: Proposition 13.2.1. If M is a manifold and ${\displaystyle \Gamma }$ is a group acting properly discontinuously on M, then ${\displaystyle M/\Gamma }$ has the structure of an orbifold. -- Proof. For any point ${\displaystyle x\in M/\Gamma }$, choose ${\displaystyle {\tilde {x}}\in M}$ projecting to x. Let I_x be the isotropy group of ${\displaystyle {\tilde {x}}}$ (I_x depends of course on the particular choice ${\displaystyle {\tilde {x}}}$.) There is a neighborhood ${\displaystyle {\tilde {U}}_{x}}$of${\displaystyle {\tilde {x}}}$ invariant by I_x and disjoint from its translates by elements of ${\displaystyle \Gamma }$ not in Ix.
I think forall x, There is a neighborhood ${\displaystyle {\tilde {U}}_{x}}$ of ${\displaystyle {\tilde {x}}}$ invariant by I_x and disjoint from its translates by elements of ${\displaystyle \Gamma }$ not in isotropygroup would make for a good definition. (probably adding finite to isotropygroup)
Oh, okay, maybe guessing isn't the best way to go about it. So inbetween I had a look at Thurston's definition and here it comes:
p174: Definition 8.2.1. If ${\displaystyle \Gamma }$ is a group acting on a locally compact space X, the action is properly discontinuous if for every compact set ${\displaystyle K\subset X}$, there are only finitely many ${\displaystyle \gamma \in \Gamma }$ such that ${\displaystyle \gamma K\cap K\not =\emptyset }$.
So is his definition of properly discontinuous equal ours for discontinuity? Does it matter that he takes compact sets at the space is locally compact in his case? At least it still seems to me that Thurston's definition allows for a non-trivial isotropy group and ours doesn't. -- JanCK (talk) 12:48, 20 December 2007 (UTC)
Oh, sry, sry, sry, my bad. I'm real sry for complaining so long. I see it now. I'm sry. I read the article several times, but for some reason I kept on thinking that all the definitions gotta be equivalent. But the article clearly states that that's not the case. sry. I kept on thinking about the first definition and how it can't be that that's the one for non-trivial stabilizer. I should have listened to myself and thought about it for long in a more awake state. Please excuse the ado I produced. -- JanCK (talk) 12:57, 20 December 2007 (UTC)
I should really stop. But just one more question. As I'm now thinking that the definition for non-trivial stabilizer fits what I have learned once. Is there really someone who uses the first given definition. Isn't the first definition equivalent to the second definition + free? Okay, I don't see that for sure right now, but I guess that it's got to. Could the article maybe make it clearer that there are several non-equivalent definitions in use? -- JanCK (talk) 13:10, 20 December 2007 (UTC)

You seem to be confused. I'm sorry but to me you are not making much sense and I am unable to respond to such poorly articulated comments. Why are you talking about free actions, when isotropy/stabiliser subgroups are precisely the point? Look at a book on compact transformation groups, such as Bredon, to find out how actions of finite groups on manifolds can be locally linearised (the slice theorem). In this particular case it follows from the uniqueness properties of the exponential map on Riemannian manifolds. Do you know about geodesics on a Riemannian manifold? This is not a forum as I've said before. I made a summary which has been looked at by several experts in the subject (i.e. professional researchers in geometric group theory). I'm not sure that it's a very good idea to make suggestions about things which you don't seem to have quite mastered yet and see no purpose therefore in continuing this dialogue. Mathsci (talk) 18:36, 20 December 2007 (UTC)

## further definition : This one allowing: free + properly discontinous yields quotient being manifold

I think this article lacks the following definition as an alternative to all the others. You can find it for example in "Riccardo Benedetti and Carlo Petronio, Lectures on hyperbolic geometry, Universitext, Springer-Verlag, 1992."

${\displaystyle T}$ locally compact, Hausdorff, topological space, ${\displaystyle \Gamma }$ a group of homeomorphisms: ${\displaystyle T}$ operates properly discontinuously if for any pair ${\displaystyle H,K}$ of compact subsets of ${\displaystyle T}$ the set ${\displaystyle \Gamma (H,K)=\{\gamma \in \Gamma :\gamma (K)\cap H\not =\emptyset \}}$ is finite.

Together with operating freely (${\displaystyle \Leftrightarrow \left(y\in \Gamma ,x\in T,\gamma (x)=x\Rightarrow \gamma ={\text{id}}\right)}$ ) one gets that ${\displaystyle T/\Gamma }$ is Hausdorff (B.1.6 in Benedetti's book). [prop B.1.6: free + properly discontinuous <=> ${\displaystyle T/\Gamma }$ Hausdorff and ${\displaystyle T\to T/\Gamma }$ covering mapping <=> ${\displaystyle T/\Gamma }$ Hausdorff and ${\displaystyle \forall x\in T:\quad \exists U\quad {\text{open}}:\quad \forall \gamma \in \Gamma \setminus \{{\text{id}}\}:x\in U\land \gamma U\cap U=\emptyset }$ ]

If one wants ${\displaystyle T/\Gamma }$ to be a manifold, one needs Hausdorff, so it does seem a mayor application to me.

The definitions in the article don't give that the quotient is Hausdorff. See following example. Let ${\displaystyle T}$ be ${\displaystyle \mathbb {R} ^{2}\setminus \{(0,0)\}}$. Let ${\displaystyle \Gamma =\mathbb {Z} }$. The operation of ${\displaystyle \Gamma }$ on ${\displaystyle T}$ is given by ${\displaystyle \mathbb {Z} \to {\text{Homeo}}(T):z\mapsto \left(T\to T:x\mapsto {\begin{pmatrix}{\frac {1}{2^{z}}}&0\\0&2^{z}\end{pmatrix}}x\right)}$ .

The quotient ${\displaystyle T/\Gamma }$ is not Hausdorff. Take ${\displaystyle (0,1)}$ and ${\displaystyle (1,0)}$ in ${\displaystyle T}$ . Take arbitrary neighbourhoods of ${\displaystyle (0,1)}$ and ${\displaystyle (1,0)}$ in ${\displaystyle T/\Gamma }$. Take the preimage of those sets in ${\displaystyle T}$. This is the same as taking open neighbourhoods ${\displaystyle U_{(0,1)}}$, ${\displaystyle U_{(0,1)}}$ in ${\displaystyle T}$ and construct ${\displaystyle \Gamma U_{(0,1)}}$ and ${\displaystyle \Gamma U_{(1,0)}}$ from them. If you look at ${\displaystyle z_{1}(U_{(0,1)})}$ and ${\displaystyle z_{2}(U_{(0,1)})}$ for ${\displaystyle z_{1}\to \infty }$ and ${\displaystyle z_{2}\to -\infty }$. You realize that ${\displaystyle \Gamma U_{(0,1)}}$ and ${\displaystyle \Gamma U_{(1,0)}}$ intersect. So the neighbourhoods in ${\displaystyle T/\Gamma }$ intersect too. They were arbitrary, so the space isn't Hausdorff.

I think, it's worth taking up the definition of properly discontinuous, I quoted above, because it gets rid of this "problem".

At least, I think we should mention in the section Properly discontinuous action that ${\displaystyle X/G}$ is not Hausdorff, even though ${\displaystyle X}$ is. This might make people aware of the complexity of the situation, as I heard there are several sources that claim Hausdorffness for the quotient even though they use a different definition of Properly discontinous.

Thanks for reading -- JanCK (talk) 14:45, 7 November 2009 (UTC)