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The Milnor-Wolf theorem states that any finitely generated solvable group has either polynomial or exponential growth.

Is there an analogous result for locally compact compactly generated groups? (or rather, for some smaller class of group? connected ones? connected Lie?)

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I found the following paper:

Yves Guivarc'h, Croissance polynomiale et périodes des fonctions harmoniques, Bulletin de la Société Mathématique de France (1973) Volume: 101, page 333-379

The analogous result (see Corollaire III.3) is proved for all compactly generated soluble locally compact groups as well as some other cases. Here 'growth' is measured in terms of Haar measure, i.e. the asymptotic growth rate of $\mu(U^n)$ where $U$ is some compact generating set with nonempty interior.

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    $\begingroup$ Great! I think one can expect, like in the discrete case, a strengthening saying that in the case of exponential growth, there is a quasi-isometrically embedded free sub-semigroup on 2 generators. $\endgroup$
    – YCor
    Commented Mar 8, 2017 at 13:07
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    $\begingroup$ Also the type of growth can be characterized with no reference to the Haar measure. Let $K$ be a compact symmetric generating neighborhood of 1 in $G$. Let $X$ be a maximal subset for the property that $x,y\in X$, $x\neq y$ implies $x^{-1}y\notin K$. Make $X$ a graph saying $x-y$ if $x^{-1}y\in K^3$. Then this is a connected graph of bounded degree, and its type of growth (up to the usual asymptotic equivalence) does not depend on the choices, and indeed is the same as that obtained with the Haar measure. $\endgroup$
    – YCor
    Commented Mar 8, 2017 at 13:14
  • $\begingroup$ @YCor How does the above result by Guivarch relate to this result by Jenkins: sciencedirect.com/science/article/pii/002212367390092X Namely, a connected separable locally compact group is either of polynomial or exponential growth. (By the way - indeed in the exponential case there is a quasi-isometrically embedded free sub-semigroup on 2 generators) $\endgroup$
    – Snoop Catt
    Commented Mar 8, 2017 at 13:33
  • $\begingroup$ @iPe the relation is quite clear: the connected case is a particular case of the general case. You claim a positive answer to my question on QI-embedded free semigroups, but do you refer to the connected case or in general (I knew for the connected case). $\endgroup$
    – YCor
    Commented Mar 8, 2017 at 13:53
  • $\begingroup$ @YCor Oh, sorry, I meant for the connected case, I was merely detailing the content of Jenkins' paper. Do you by any chance know of anywhere that I can read about the above result by Guivarc'h in English? $\endgroup$
    – Snoop Catt
    Commented Mar 8, 2017 at 14:02

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