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Given a discrete group G. Is there a nice criterion to decide, whether there is a compact Hausdorff $G$- space X, that contains the discrete space $G$ as a subspace, such that the stabilizer of every point in $X$ is (virtually) cyclic ?

For example the free group admits such a compactification (As well as any hyperbolic group I think). Is it possible to decide, whether $\mathbb{Z}^2$ admits such a compactification? .

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CAT(0) groups (such as $\mathbb{Z}^2$) admit a `visual boundary'. I don't have a reference to hand, but look at Metric spaces of non-positive curvature by Bridson and Haefliger for details. –  HJRW Sep 22 '10 at 15:44
    
Browsing google hits for 'CAT(0) boundary', I'm reminded that Croke--Kleiner showed that the boundary is not an invariant of the group $G$, so my previous comment was, strictly speaking, slightly inaccurate. But it is well defined for the space $X$, which I think is what you wanted. –  HJRW Sep 22 '10 at 15:48
    
I think, that the CAT(0) boundary has too big stabilizers. In the example of $\mathbb{Z}^2$ the whole group acts trivially on the boundary. However for any CAT(0) group $G$ the the group $G\times \mathbb{Z}$ fixes a point at its boundary. So I think CAT(0) groups don't fit in this scheme. –  HenrikRüping Sep 22 '10 at 15:53
    
Ah yes, good point. On the other hand, the construction is very natural, which suggests that it's the `right' one. Why do you want such small stabilisers? –  HJRW Sep 22 '10 at 15:58
    
well the motivation was the free group acting on a tree. If I drop that condition, I could always take the one point compactification. Furthermore I am wondering, whether the CAT(0) boundary works, if the space doesn't contain $\mathbb{R}^2$ as a subspace. –  HenrikRüping Sep 22 '10 at 16:14
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2 Answers

up vote 11 down vote accepted

There are many compactifications of particular groups. For your example of $\mathbb Z^2$: one construction for a compactification is to first embed it as a subgroup of $S^1 = \mathbb R / \mathbb Z$ by picking two rationally independent numbers for the images of the generators. Now compactify $\mathbb Z^2$ by making large elements connverge toward their image points in $S^1$. The stabilizer of any point is trivial.

The same method works to get a compactification associated with any action of $G$ on a compact space $X$. Just pick a point $x \in X$, and adjoin the closure of the orbit of $X$ at infinity in $G$. If the action has no fixed points in the cloure of the orbit, then stabilizers are trivial. It's easier to avoid all but cyclic stabilizers. To make actions with small stabilizers, you can take products of examples; point stabilizes in the product become intersections of stabilizers in the factors. There are many tricks, some of them useful, for making compactifications that are Hausdorff metric spaces.

There's an ultimate (but non-constructive and of large cardinality) compactification, the Stone-Cech compactification, which has trivial point stabilizers for any group,

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Regarding the question of `whether the CAT(0) boundary works, if the space doesn't contain $\mathbb{R}^2$ as a subspace' (see comments above), the Flat Plane Theorem asserts that any CAT(0) group that acts on a CAT(0) space without an isometrically embedded copy of $\mathbb{R}^2$ is word-hyperbolic. So in this case you can use the usual hyperbolic boundary. See Bridson & Haefliger for details.

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