Timeline for Is there a simple proof that a group of linear growth is quasi-isometric to Z?
Current License: CC BY-SA 2.5
9 events
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| Mar 3, 2012 at 20:01 | comment | added | Alain Valette | There is a proof of quasi-isometric rigidity of $\mathbb{Z}^n$ which avoids Gromov's polynomial growth theorem, in my paper with Cornulier and Tessera: Isometric group actions on Hilbert spaces: growth of cocycles. Geom. Funct. Anal. 17 (2007), 770-792. | |
| Sep 17, 2010 at 15:46 | comment | added | Sam Nead | @Mark - thank you for the reference. | |
| Sep 17, 2010 at 3:16 | comment | added | user6976 | @Sam Nead: The fact that if $G$ is q.i. to ${\mathbb Z}^n$, then $G$ is virtually ${\mathbb Z}^n$ is an easy part of Gromov's argument. The main difficulty in Gromov's paper is the use of Montgomery-Zippin to deduce that the asymptotic cone is a manifold, and its isometry group is a Lie group with finitely many components. If $G$ is q.i. to ${\mathbb Z}^n$, then the asymptotic cones are ${\mathbb R}^n$ and Montgomery-Zippin is not needed. The rest of Gromov's proof is elementary mod Tits alternative. Shalom's paper is in Acta Math. 192 (2004), no. 2, 119--185. | |
| Apr 16, 2010 at 22:46 | comment | added | Sam Nead | I should have pointed out (ie realized) that that there are pointed graphs which are one-ended and have linear growth and are not quasi-isometric to a ray. For example let $C_{2n}$ (for $n \in N$) be the circle of perimeter $2n$ and glue each of these to the next along antipodal points... So the group structure appears to be needed in a more serious fashion. | |
| Apr 16, 2010 at 20:44 | comment | added | Tom Church | @Henry: that's not true in Z, so it can't be as straightforward as you suggest -- you'll need to use the one-ended assumption. | |
| Apr 16, 2010 at 19:20 | comment | added | Sam Nead | I agree that linear growth should imply that, but I don't see a slick proof (or any proof at the moment). I will think about it -- it would be nice to finish this proof. The Wilkie-van den Dries proof looks very intricate. | |
| Apr 16, 2010 at 17:36 | comment | added | HJRW | Surely linear growth implies that every element of your group is a bounded distance from the ray? That should fill in your 'vague bit'. Or am I missing something? | |
| Apr 16, 2010 at 16:20 | history | edited | Sam Nead | CC BY-SA 2.5 |
Added context.
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| Apr 16, 2010 at 16:07 | history | answered | Sam Nead | CC BY-SA 2.5 |