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Do the additive group or the multiplicative group of $\mathbb{Q}$ have property (RD) (Rapid Decay)?

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    $\begingroup$ And what's property (RD)? $\endgroup$ May 21, 2012 at 18:12
  • $\begingroup$ Surely yes for any abelian group, by Cauchy-Schwarz? Unless there is some subtlety in the definition of length function that I have missed... $\endgroup$
    – Yemon Choi
    May 21, 2012 at 18:13
  • $\begingroup$ @Felix: there is a nice exposition at the start of I. Chaterji's thesis math.ethz.ch/u/burger/chaterji.pdf $\endgroup$
    – Yemon Choi
    May 21, 2012 at 18:14
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    $\begingroup$ Property (RD) depends on the metric. For example, a cyclic group with exponentially distorted metric does not have property (RD) (as any amenable group of exponential growth). So the answer depends on the metric you choose. For a finitely generated group, the metric is usually the standard word metric, but in your case the groups are not finitely generated. $\endgroup$
    – user6976
    May 21, 2012 at 19:30
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    $\begingroup$ Please use the "edit" link to refine your question. $\endgroup$
    – S. Carnahan
    May 22, 2012 at 1:05

1 Answer 1

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Thanks to 'Yves Cornulier's answer to my other question about the growth of $\mathbb{Q}$, we now know (1) there is a length function on the additive group of $\mathbb{Q}$ which makes $\mathbb{Q}$ of polynomial growth. (2) there is no length function on $\mathbb{Q}^\times$ making it of polynomial growth.

We can modify a theorem by Jolissaint which says: if $G$ is an amenable (finitely generated) group, then $G$ has (RD) if and only if $G$ is of polynomial growth. To generalize this theorem to infinitely generated groups one only needs to show that if $G$ has (RD) w.r.t. some length function $L$ then $\{ g\in G; L(g)\leq r\}$ is finite for all $r\geq 0$. This is easily done by introducing to sequence of functions in $\mathbb{C}G$ (I will give details in the next few days).

Now, since $\mathbb{Q}$ and $\mathbb{Q}^\times$ are both amenable, $\mathbb{Q}^\times$ does note have (RD) and $\mathbb{Q}$ has (RD).

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