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Is there any length function on additive group of $\mathbb{Q}$ such that $\mathbb{Q}$ is of polynomial growth WRT this length function? What about the multiplicative group of $\mathbb{Q}$ instead?

edited May 22 2012 at 3:14

asked May 21 2012 at 17:23

Vahid Shirbisheh
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Is $\mathbb{Q} =\mathcal{Q}$ the rationals? – Marc Palm May 21 2012 at 17:34

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Take the length function defined by the generating set which is the entire group. – Lee Mosher May 21 2012 at 17:41

sorry my mistake I meant $\mathbb{Q}$. If I take the length function defined by a generating set its growth would be infinite. I need somehow manipulate the length function coming from a generating set to get a polynomial growth. And if it is impossible I need to prove it. – Vahid Shirbisheh May 21 2012 at 18:01

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@Vahid: What is your definition of a length function? The standard notion of word length is used only for finitely-generated groups. I guess, you want to define $|x|$ axiomatically so that $|x+y|\le |x|+|y|$, $\{x: |x|\le n\}$ is finite for all $n\in {\mathbb N}$, etc. Until you clearly state what axioms you want, your question makes no sense. – Misha May 21 2012 at 18:36

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@Misha. You're right; actually for a f.g. group, the word length achieves the minimal growth rate among all lengths. The interest is to make things meaningful with infinitely generated groups. In this case there is no "largest" length and, in general, no length with "smallest " growth. Lengths are also useful when you want to consider countable groups as metric spaces up to coarse equivalence (extending viewing f.g. groups as metric spaces up to QI). [I refer to Lyndon's lengths as "Lyndon lengths" because it's a much more specific context in my opinion.] – Yves Cornulier May 21 2012 at 21:39

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Yves Cornulier · Accepted Answer · 2012-05-21 22:09:52Z UTC

Yes for $\mathbf{Q}$, no for $\mathbf{Q}^*$.

For $\mathbf{Q}$, write it as the union of an increasing sequence $L_n$ with $L_1=\mathbf{Z}$ and $L_n$ of finite index over $L_1$. Pick a function $F$ with fast growth and define $l'(r)=|r|+F(\sup{n:r\notin L_n})$.

For $\mathbf{Q}^*$, it contains a subgroup isomorphic to $\mathbf{Z}^d$ for every $d$, so for every length the growth is at least polynomial of degree $d$. So it can't be polynomial.

Added-1: for $\mathbf{Q}$ we can arrange so that $[L_n:L_1]\le n$ for all $n$. Then we can pick $F$ to be the identity and then the growth is at most quadratic.

Added-2: the argument extends, showing that an abelian group admits a length with polynomial growth iff it's countable and has finite $\mathbf{Q}$-rank.

growth of infinitely generated groups

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