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I am interested in finitely generated groups which, endowed with their word metrics, do not admit bilipschitz embeddings into $L_1(0,1)$. I know two classes of such groups:

(1) Heisenberg group $\mathbb{H}(\mathbb{Z})$ (result of Cheeger and Kleiner, Ann. of Math. 171 (2010), 1347-1385).

(2) Gromov's random group (also known as Gromov's monster), Geom. Funct. Anal. 13 (2003), 73-146. These groups do not admit even coarse embeddings into $L_1$ since they contain families of expanders weakly; there are further examples of Osajda arXiv:1406.5015 which do not admit bilipschitz embeddings into $L_1$ for the same reason.

Question: Any other known examples?

Remark. Finitely generated Gromov hyperbolic groups admit bilipschitz embeddings into $L_1$ by the result of Buyalo, Dranishnikov, and Schroeder, Invent. Math. 169 (2007), 153-192. The same is true for finite direct products of such groups.

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  • $\begingroup$ There is not one Gromov random group, there is a recipe providing plenty of groups (and moreover it sometimes means those finitely presented groups containing these infinitely presented groups when they are arranged to be recursively presentable). $\endgroup$
    – YCor
    Commented Aug 21, 2014 at 9:52
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    $\begingroup$ To answer your question, it would be a good PhD work to prove that any non-virtually-abelian nilpotent f.g. group (or, more naturally, any non-abelian simply connected nilpotent Lie group) has no bilipschitz embedding into $L^1$. $\endgroup$
    – YCor
    Commented Aug 21, 2014 at 9:54
  • $\begingroup$ What other classes of f.g. groups do Lipschitz embed into $L_1$? $\endgroup$
    – Bill Johnson
    Commented Aug 21, 2014 at 15:04
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    $\begingroup$ @Bill Free products of groups mentioned in the remark. $\endgroup$
    – Mikhail Ostrovskii
    Commented Aug 21, 2014 at 15:21
  • $\begingroup$ @YCor (Yves) Thank you for the comments. $\endgroup$
    – Mikhail Ostrovskii
    Commented Aug 21, 2014 at 15:21

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