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Question: Is there a metabelian group, so that for some $\epsilon >0$ and all $p \in [1, \infty[$ the [equivariant] compression exponent in [any] $L^p$-space is bounded by $1-\epsilon$ (bound does not depend on $p$).

A rough formulation of the contrapositive negation would be, is it true that for any metabelian group there is a $C>0$ and a sequence of $C$-Lipschitz embeddings $\phi_n$ in the $L^{p_n}$-spaces $X_n$, so that these embeddings become "arbitrarily close" to being bi-Lipschitz.(see YCor's comment) so that there is a sequence $\alpha_n$ with $\alpha_n \to 1$ and for every $x,y \in G$, $\|\phi_n(x)-\phi_n(y)\|_{X_n} \geq d_G(x,y)^{\alpha_n}$.

Tessera showed it holds for polycyclic groups and the metabelianisation of Baumslag-Solitar groups (see Theorems 9 and 10 in "Isoperimetric profile and random walks on locally compact solvable groups", keeping $p=2$. Naor & Peres showed (Lemma 7.8 in "Embeddings of discrete groups and the speed of random walks" and Theorem 6.1 in "Lp -compression, travelling salesmen, and stable walks") showed that $\mathbb{Z} \wr \mathbb{Z}$ has compression exponent $\max\{ \frac{p}{2p-1}, \frac{2}{3} \}$, as $p \to 1$, it tends to 1. When the base group has polynomial growth, then they also obtain a bound of $\max\{\tfrac{1}{p}, \tfrac{1}{2}\}$ (again take $p \to 1$). Sale (in "Metric Behaviour of the Magnus Embedding") shows that the free solvable group of rank 2 has $L^p$-compression $1/p$ for $p \in [1,2]$.

If instead of looking at metabelian one considers groups of solvable rank 3, there are positive answers to the question (see Brieussel & Zheng "Speed of random walks, isoperimetry and compression of finitely generated groups")

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  • $\begingroup$ I think that the "rough formulation" is deceptive. For instance, the "Hilbert compression exponent" of the Heisenberg group is 1. But this does not mean that it has a sequence of embeddings that is "arbitrary close" to be bilipschitz. The right question is "what is the possible distortion for a metabelian group into $L^p$-spaces". The exponent is a rough, non-subtle way to quantify this. One cannot reduce the growth of a function to its degree of growth: especially there are fine phenomena near exponent 1 (little distortion) and 0 (superpolynmial distortion). $\endgroup$
    – YCor
    Commented Jan 16 at 13:21
  • $\begingroup$ Absolutely correct (in fact when the compression exponent is 1 [and $p>1$] it is exceptional to have a bi-Lipshitz map), I wanted to add something like "in the sense that exponents à la Hölder approaches to one". But then I realised that either the reader knows what compression is, and then this is kind of obvious, or I might as well write down the rigourous definition. I might do the latter, since I should also have specified $p=1$ is also ok $\endgroup$
    – ARG
    Commented Jan 16 at 14:36
  • $\begingroup$ By "contrapositive", do you mean "negation"? $\endgroup$ Commented Jan 16 at 14:54
  • $\begingroup$ @mathworker21 yes, correct. I actually believe the question to have an affirmative answer, but could not find an example. $\endgroup$
    – ARG
    Commented Jan 16 at 15:00

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