Timeline for Growth function of locally compact groups
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Feb 16, 2017 at 20:41 | comment | added | LSpice | Thanks to @YCor for pointing out two problems with my example: the metric I had in mind isn't left-invariant (which can be fixed by working with the additive rather than the multiplicative group), and closures of open balls need not be closed balls of the same radius (which cannot be so fixed). Where should I turn in my p-adic analyst's badge? | |
| Feb 16, 2017 at 20:26 | comment | added | YCor | The edited question looks like "I'd like a locally compact, non-discrete group, with a proper left-invariant metric". This is a trivial question... | |
| Feb 16, 2017 at 19:54 | comment | added | YCor | For the metric on $\mathbf{R}$ of my previous comment: define $L(x)=n+x$ for $n\ge 0$, $x\in [2n,2n+1]$ and $L(x)=n+1$ for $x\in [2n+1,2n+2]$. This is well-defined and subadditive. So the distance $N(x,y)=L(|x-y|)$ is the proper left-invariant distance I referred to. | |
| Feb 16, 2017 at 19:49 | comment | added | YCor | @LSpice The multiplicative group of nonzero $p$-adics is direct product of a compact group $K$ by $\mathbf{Z}$. So it's somewhat a decorated version of a discrete group. In addition, you should specify the metric, because the standard $p$-adic norm is not left-invariant by multiplication. | |
| Feb 16, 2017 at 19:46 | answer | added | LSpice | timeline score: 1 | |
| Feb 16, 2017 at 18:26 | comment | added | Alessandro Carderi | @LSpice Yes, you are right, this is a good example. Probably any totally disconnect group carry a metric with similar properties | |
| Feb 16, 2017 at 18:18 | comment | added | LSpice | If $G$ is the multiplicative group of the $p$-adic numbers, then the metric takes values in the cyclic subgroup of $\mathbb Q^\times$ generated by $p$. It satisfies your condition on ball closures. I don't remember the definition of a proper metric, so I'm not sure if it qualifies. | |
| Feb 16, 2017 at 18:18 | comment | added | Alessandro Carderi | @YCor Can you please explain to me why (and if) you can get a proper metric with your construction? | |
| Feb 16, 2017 at 18:16 | history | edited | Alessandro Carderi | CC BY-SA 3.0 |
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| Feb 16, 2017 at 17:19 | comment | added | Alessandro Carderi | @NateEldredge Yes, sorry, I want to assume that $G$ is not discrete. | |
| Feb 16, 2017 at 16:03 | history | edited | YCor |
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| Feb 16, 2017 at 15:35 | comment | added | YCor | There's a largest left-invariant metric on $\mathbf{R}$ such that $d(x,0)\le x$ for all $x\ge 0$ and $d(x,0)\le 1$ for all $x\in [1,2]$. This metric is compatible and indeed $d(x,0)=\min(x,1)$ for all $x\in [0,2]$. The 1-sphere equals all $\pm[1,2]$ and has positive measure, so we have discontinuity then. | |
| Feb 16, 2017 at 15:32 | comment | added | YCor | $Gr_d$ is continuous at every $r<0$, and is continuous at zero iff $G$ is non-discrete. | |
| Feb 16, 2017 at 15:16 | comment | added | Nate Eldredge | Do you want to assume $G$ is connected? Otherwise any discrete group is a counterexample... | |
| Feb 16, 2017 at 13:23 | history | asked | Alessandro Carderi | CC BY-SA 3.0 |