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It is a well-known theorem first proved by Bourgain that any map $\varphi:T_n\to H$ from the binary tree of height $n$ to a Hilbert space has distortion at least $C \sqrt{\ln n}$ where $C$ is a universal constant. Distortion is the least $D$ such that for some $s$ and all $x,y\in T_n$, we have $$ s d(x,y) \leq d(\varphi(x),\varphi(y)) \leq sD d(x,y)$$ (This also generalizes in some way to many more Banach spaces).

Some of the proofs use the linear structure of $H$, but others are purely metric. I therefore wonder:

Has this distortion estimate been generalized to spaces of non-negative curvature (Alexandrov spaces)?

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  • $\begingroup$ You might want a narrower focus which rules out negatively curved spaces in which the infinite binary tree embeds with uniform value of $D$. $\endgroup$
    – Lee Mosher
    Commented Jun 12, 2013 at 15:39
  • $\begingroup$ @Lee Mosher: I did, by asking the question for spaces of non-negative curvature. $\endgroup$
    – Benoît Kloeckner
    Commented Jun 12, 2013 at 19:22
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    $\begingroup$ Oops, sorry, my eyes saw "non-negative" and my brain saw "non-positive". $\endgroup$
    – Lee Mosher
    Commented Jun 12, 2013 at 20:34

2 Answers 2

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See the direct proof of markov convexity in proposition 2.1 of the following paper

Mendel, Manor; Naor, Assaf Markov convexity and local rigidity of distorted metrics. J. Eur. Math. Soc. (JEMS) 15 (2013), no. 1, 287–337.

the only thing that is used there is lemma 2.3 of the above references, which holds with p=2 for alexandrov spaces of nonnegative curvature. Therefore the answer to your question is known to be positive. The observation that the proof of the above reference works in this setting has been shown in talks that I heard over the past few years, and also stated explicitly in the following paper

Sean Li, Coarse differentiation and quantitative nonembeddability for Carnot groups

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This distortion estimate has been generalized to what are called Markov p-convex metric spaces (Lee, James R.; Naor, Assaf; Peres, Yuval; Trees and Markov convexity).

Alexandrov spaces are known to have Markov type ( Ohta, Shin-Ichi; Markov type of Alexandrov spaces of non-negative curvature), which is philosophically opposite to convexity since it's closely related to smoothness (Naor, Assaf; Peres, Yuval; Schramm, Oded; Sheffield, Scott; Markov chains in smooth Banach spaces and Gromov-hyperbolic metric spaces).

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  • $\begingroup$ It seems to me that "opposite" is the wrong word: did you mean "dual"? As both Hilbert space and trees have Markov type 2, there is little chance that Ohta's result is relevant to my question. However you are right that I should check in the direction of Markov convexity. $\endgroup$
    – Benoît Kloeckner
    Commented Jun 12, 2013 at 19:25

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