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Question 1. Does every CAT(0) space embed isometrically inside an integral of $\mathbb{R}$-trees?

Here an integral of $\mathbb{R}$ trees means the set of functions from a measure space $\mathcal{F}$ to a measurable field of based $\mathbb{R}$-trees over $\mathcal{F}$, so that the squared distance from the basepoint is integrable, as described in Proposition 44 and Remark 45 of https://dx.doi.org/10.1090/S0894-0347-06-00525-X .

(Originally the question referred to $\ell^2$ products of $\mathbb{R}$-trees, but Nicolas Monod below gave an argument that these do not suffice, and suggested the generalization.)

Really I'm most interested in the finite version of this question (where we also don't need to talk about $\mathbb{R}$-trees):

Question 2. Does every $k$-element subset of a CAT(0) space embed isometrically in a (finite) product of metric trees?

To go from Question 2 to Question 1, one would presumably attempt a limiting approximation argument, although it's a little complicated, since an approximation for $k$ points doesn't necessarily have an obvious relation to the approximation for $k+1$ points.

If you consider the set of possible squared-distances between $k$ points in any CAT(0) space, you get a subset $MC0_k$ of $\mathbb{R}^{\binom{k}{2}}$. This is clearly closed under scaling, and the fact that a product of CAT(0) spaces is still CAT(0) implies that $MC0_k$ is a convex subset. On the other hand, you could look at $MT_k$, the set of possible squared-distances between $k$ points in a tre. Question 2 is then asking whether the convex hull of $MT_k$ is $MC0_k$. This reformulation makes it clear that, if the answer to Question 2 is affirmative, you would need at most $\binom{k}{2}$ different trees to realize any $k$-element subset of a CAT(0) space.

For $k=4$, the answer to Question 2 appears to be affirmative. Petrunin gives an elegant characterization of which squared-distances between 4 points can be realized in a CAT(0) space: https://arxiv.org/abs/1411.5329 Let me sketch the argument.

If the four distances are realized in $\mathbb{E}^3$, we are done. Otherwise, the four distances can (generically) be realized by a space $X_0$ obtained by gluing together three triangles around a vertex, so that the angle sum around the vertex is greater than $2\pi$. Then consider the space $X_1$ obtained by deleting one of these triangles. This is again CAT(0), and all distances are the same except between one pair of points. The distances in $X_0$ can be realized as a convex combination of he distances in $X_1$ and the distances in another space $X_3$ where that changed length is shortened until the three triangles embed in $\mathbb{E}^2$.

You can proceed in a similar way, deleting the triangles one at a time, until you reduce $X_0$ to a convex combination of a tripod and spaces that embed in $\mathcal{E}^2$.

It looks like one could also push this through for $k=5$, although there are many more cases to consider.

(Moved the question about extreme points here: What are the extremal CAT(0) metrics? )

This is a version of this old question: Length inequalities in trees and CAT(0) spaces However, I started off asking the earlier question in a slightly confusing dual form, so I wanted to restate it with a more easily-understood question first.

(Edited to refer to $\mathbb{R}$-trees)

Nicolas Monod gave an elegant argument against Question 1 using products. I accepted his answer, but edited the question to refer to integrals instead, which is the natural generalization.

(Or, if you prefer finitary questions, Question 2 is still open, and Nicolas' argument against embedding in a product of trees doesn't really give any evidence against it.)

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    $\begingroup$ Maybe split this question into at least two separate questions? $\endgroup$
    – Suvrit
    Commented May 5, 2015 at 1:30
  • $\begingroup$ Thanks for the suggestion, I split off the question about extreme points. $\endgroup$
    – Dylan Thurston
    Commented May 5, 2015 at 3:01
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    $\begingroup$ You might be interested in a result of Buyalo and Schroeder, who showed any hyperbolic group quasi-isometrically embeds into a product of trees: arxiv.org/abs/math/0509355 . $\endgroup$
    – HJRW
    Commented May 5, 2015 at 8:15
  • $\begingroup$ @HJRW: Thanks, although it's not exactly what I'm asking. $\endgroup$
    – Dylan Thurston
    Commented May 5, 2015 at 8:22
  • $\begingroup$ @DylanThurston, that's why I posted a comment, and not an answer! $\endgroup$
    – HJRW
    Commented May 5, 2015 at 8:25

3 Answers 3

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This is not possible; sorry for just posting a sketch but I am not on MO :-(

For instance a generic geodesic triangle in the hyperbolic plane does not embed in a finite product of trees.

Suppose you have a finite product $X$ of uniquely geodesic metric spaces $X_n$ with the "pythagorean" distance on the product. Then it is uniquely geodesic and any geodesic segment $\sigma\colon[a,b]\to X$ is of the form $\sigma(t)=(\sigma_n(a_n t))_n$ where $a_n\geq 0$ is a sequence with $\sum a_n^2 = 1$ and each $\sigma_n$ is a geodesic in $X_n$. This characterisation is not entirely trivial, but very well-known (see e.g. Bridson--Haefliger).

Now in the case at hand you would deduce that the distance between two geodesics issuing from a common point would be a piecewise affine function, which it definitely isn't for hyperbolic geometry.

The fact about geodesics extends to infinite sums, and even to integrals (see e.g. Proposition 44 in https://dx.doi.org/10.1090/S0894-0347-06-00525-X )

The latter part of the argument above (piecewise affine) does not extend as is to this "integral" setting though.

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    $\begingroup$ I don't think you use "generic" anywhere, the argument works for any subset of $\mathbb{H}^2$ containing a pair of non-collinear segments. Also the argument shows that the distance $f(t)=d(u_t,v_t)$ between two speed-one geodesics $(u_t),(v_t)$ will be of the form $f(t)=\sqrt{\sum_i h_i(t)^2}$ with $h_i$ piecewise affine, which is not necessarily piecewise affine but is enough for the argument. $\endgroup$
    – YCor
    Commented May 6, 2015 at 9:16
  • $\begingroup$ @Yves: (1) yes that's what I meant by generic (I'd have said scalene but then I'd have gotten picked on as it's too strong) and (2) yes you're right I guess "non-analytic" would be the lazy overkill! $\endgroup$
    – Nicolas Monod
    Commented May 6, 2015 at 11:10
  • $\begingroup$ Thanks for the nice argument! But you're addressing a question I didn't ask. In Question 1, I'm asking about embedding in an infinite product. In Question 2, I'm asking about finite products, but only about embedding finitely many points in a distance-preserving way, not their convex hull. The three corners of the triangle can trivially be embedded in a product of trees, and your argument doesn't rule out embedding a finite subset of the triangle (although it might well be impossible). As you note, it also doesn't rule out embedding inside infinite constructions. $\endgroup$
    – Dylan Thurston
    Commented May 6, 2015 at 11:21
  • $\begingroup$ Also, welcome to MO! $\endgroup$
    – Dylan Thurston
    Commented May 6, 2015 at 11:23
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    $\begingroup$ @DylanThurston: I understand the argument to work for infinite products. In this case the distance function has the form $u(t)=\sqrt{\sum_i{h_i(a_it)}}$, where $h_i$ is a (convex nonnegative) piecewise affine function with slopes in $\{-2,0,2\}$, $a_i\ge 0$ and $\sum a_i^2=1$, and some additional summability condition ensuring that $u<\infty$. Then a little argument is needed to ensure that $u$ cannot be analytic unless it's affine. $\endgroup$
    – YCor
    Commented May 6, 2015 at 12:56
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Here is an answer to the 1st question using group theory. Any NPC symmetric space whose isometrically group has property T does not embed isometrically into a product of real trees. The reason is that if you have a group acting isometrically on an unbounded subset of a product of trees, projection of the subset to at least one factor is unbounded, hence you can use this factor to construct a space with measurable walls on which the group acts. But such action cannot exist for a group with property T.

However, if you asked me about a product of rank 2 irreducible affine buildings, I would not know how to rule this out.

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  • $\begingroup$ What's a "space with measurable walls", and how do you get such an action from a projection onto an unbounded tree? $\endgroup$
    – Dylan Thurston
    Commented May 5, 2015 at 2:50
  • $\begingroup$ 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. $\endgroup$
    – Dylan Thurston
    Commented May 5, 2015 at 7:53
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    $\begingroup$ 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). $\endgroup$
    – YCor
    Commented May 5, 2015 at 8:47
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    $\begingroup$ @YCor, can you post your proposed answer separately, so that we can discuss it properly rather than in a comment section? $\endgroup$
    – Dylan Thurston
    Commented May 5, 2015 at 13:49
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    $\begingroup$ 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. $\endgroup$
    – YCor
    Commented May 5, 2015 at 14:04
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It meant to be an answer to another question, https://mathoverflow.net/a/406833/, but it is an answer to this question as well.

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