Timeline for Milnor-Wolf theorem for topological groups
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Mar 8, 2017 at 14:20 | vote | accept | Snoop Catt | ||
| Mar 8, 2017 at 14:19 | comment | added | YCor | @iPe Jenkins gives a uniformly discrete free subsemigroup, which is enough for exponential growth, but doesn't claim it to be QI-embedded. About English version: don't know. | |
| Mar 8, 2017 at 14:02 | comment | added | Snoop Catt | @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? | |
| Mar 8, 2017 at 13:53 | comment | added | YCor | @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). | |
| Mar 8, 2017 at 13:33 | comment | added | Snoop Catt | @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) | |
| Mar 8, 2017 at 13:14 | comment | added | YCor | 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. | |
| Mar 8, 2017 at 13:07 | comment | added | YCor | 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. | |
| Mar 8, 2017 at 11:48 | history | edited | Colin Reid | CC BY-SA 3.0 |
added 18 characters in body
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| Mar 8, 2017 at 11:40 | history | answered | Colin Reid | CC BY-SA 3.0 |