de Cornulier, Tessera, and Valette (Geom. Funct. Anal. 17 (2007), 770-792) conjectured that a finitely generated group $G$ with its word metric admits a bilipschitz embedding into a Hilbert space if and only if $G$ contains a subgroup of finite index isomorphic to $\mathbb{Z}^n$ for some $n$. As far as I know this conjecture is still open. I would like to know whether the situation changes if we allow any uniformly convex space as a target space. More precisely:

Question: Does there exist an infinite finitely generated group $G$ such that

(1) $G$ does not contain a subgroup of finite index isomorphic to $\mathbb{Z}^n$; (2) $G$, endowed with its word metric, admits a bilipschitz embedding into some uniformly convex Banach space?


1 Answer 1


A negative answer to (1) sounds like a natural extension of our conjecture.

Some evidence: in the main two cases for which the conjecture is known to hold in the Hilbert case, the same argument also works for arbitrary uniformly convex Banach spaces (UCB): on the one hand non-virtually abelian nilpotent groups (which don't embed bilipschitz into a UCB, by a result of Pauls), and groups with a bilipschitz-embedded 3-regular tree (which don't embed into a UCB by Bourgain, and include non-amenable f.g. groups by Benjamini-Schramm, and also include non-virtually nilpotent solvable f.g. groups).

  • $\begingroup$ (Yves) Thank you very much for the comment. It was very interesting for me to get your opinion on this matter. So it looks like you do not feel that the extension of the class of target spaces to UCB can help to embed some groups. (I cannot accept your answer because it does not answer my `Yes or No' question.) $\endgroup$ Aug 21, 2014 at 15:28

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