Timeline for Does every CAT(0) space embed in a measurable integral of $\mathbb{R}$-trees?
Current License: CC BY-SA 3.0
13 events
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| May 5, 2015 at 16:59 | comment | added | YCor | Oops, my argument only shows that in an $\ell^1$-product of trees, the square root of the distance is Hilbertian. Right now I'm not sure whether it holds for the $\ell^2$-product. It would work if it were true that the $\ell^2$-product of 2 $\ell^1$-spaces is always $\ell^1$-embeddable. | |
| May 5, 2015 at 14:44 | comment | added | Dylan Thurston | Let us continue this discussion in chat. | |
| May 5, 2015 at 14:04 | comment | added | YCor | No it's certainly not true that for any metric space $(X,d)$, the metric space $(X,\sqrt{d})$ embeds isometrically into a Hilbert space. A counterexample is the graph on 5 vertices that is complete-bipartite 2+3. | |
| May 5, 2015 at 14:02 | comment | added | Dylan Thurston | @YCor, your argument would apply in much greater generality, since, I believe, for any metric space, if you take the square root of the distances you get a space that embeds in Hilbert space. (This amounts to saying that the PSD cone contains a polyhedral cone.) But then your conclusion seems to be too strong, if I'm understanding right. | |
| May 5, 2015 at 13:53 | comment | added | Dylan Thurston | Having read up on spaces with measured walls, I'm now confused how you are getting an action on a space with measured walls from the map to a tree. The map to a tree gives a set of walls, but it won't be group-invariant, and to make it group-invariant you might kill the finiteness properties. Spaces with measured walls were apparently introduced here: Cherix, Martin, and Valette, Spaces with measured walls, the Haagerup property and property (T), Ergod. Thy. & Dynam. Sys., 24 (2004), 1895--1908. | |
| May 5, 2015 at 13:49 | comment | added | Dylan Thurston | @YCor, can you post your proposed answer separately, so that we can discuss it properly rather than in a comment section? | |
| May 5, 2015 at 11:46 | comment | added | YCor | PS: the problem is not to "find a good" extension of the action, because if we have an isometric action on a subset a Hilbert space, the isometric extension to its closed affine hull is unique. All we have to care is how this extension behaves given information about the original action. And for isometric actions (on arbitrary metric spaces!), metric properness is nice because it can be read on any nonempty invariant subspace. | |
| May 5, 2015 at 11:21 | comment | added | YCor | I didn't say this. But by definition the action of $Isom(X)$ on $X$ is proper, and hence the action of $Isom(X)$ on the Hilbert space containing isometrically $X$ will be metrically proper (in particular if the orbits of $Isom(X)$ in $X$ are unbounded, then there will be no fixed point, but it's not the best formulation, properness is more natural here and yields stronger consequences). | |
| May 5, 2015 at 11:16 | comment | added | HJRW | @YCor, to be sure I understand you, when you assert that 'any group of isometries of a subset of a Hilbert space extends to a group of isometries of the whole Hilbert space', you also want that a fixed-point free action extends to a fixed-point free action? | |
| May 5, 2015 at 8:47 | comment | added | YCor | I don't understand Misha's argument and how you get an action fro a bare isometric embedding. Here's a replacement. Any subset of a ($\ell^2$) product of real trees has an isometric embedding into a Hilbert space endowed with the square root of its distance. While any group of isometries of a subset of a Hilbert space extends to a group of isometries of the whole Hilbert space. Hence if $X$ is a proper metric space embedding isometrically into a l2 product of real trees then $\mathrm{Isom}(X)$ has the Haagerup Property (hence is not T unless compact). | |
| May 5, 2015 at 7:53 | comment | added | Dylan Thurston | Also, for completeness, I suppose it's easy to construct a NPC symmetric space whose isometry group has property T? If I understand the wikipedia entry correctly, lattices in a quaternionic hyperbolic space should do the job here. | |
| May 5, 2015 at 2:50 | comment | added | Dylan Thurston | What's a "space with measurable walls", and how do you get such an action from a projection onto an unbounded tree? | |
| May 5, 2015 at 2:41 | history | answered | Misha | CC BY-SA 3.0 |