6
$\begingroup$

Every locally compact second countable group $G$ has a regular left-invariant measure $h$, the Haar measure. On the other hand the Birkhoff–Kakutani Theorem asserts that such groups also admit a compatible left-invariant metric $d$. Let us denote by $B(id_G,r)$ the $d$-open ball around the identity of radius $r$.

I'm interested in the relations between $h$ and $d$. More precisely I'm curious about the function $$Gr_d:\mathbb R\rightarrow \mathbb R,\quad r\mapsto h(B(id_G,r)).$$

Clearly this function is monotone increasing right-continuous but in general (?) not continuous. The function $Gr_d$ is not continuous at $r\in\mathbb R$ if and only if the boundary of the $r$-ball has positive measure. However since it is monotone there are only countably many point of discontinuity.

My first question is: given a l.c.s.c. group $G$ is there a left-invariant metric $d$ such that $Gr_d$ is continuous?

In most of the groups I know, there exists a canonical metric for which the growth $Gr_d$ is continuous. But in general I have no idea how to construct such a metric.

The second question is about how bad things can go: does anyone have an example of a l.c.s.c. group G and a compatible left-invariant metric $d$ for which $Gr_d$ is not continuous? Can $\mathbb R^2$ have such metrics? Given an at most countable subset $C\subset \mathbb R$ can I find a metric on $\mathbb R^2$ (or your favorite l.c.s.c. group) for which the set of discontinuity points of $Gr_d$ is exactly $C$?

Edit: Here it is what I meant but didn't write (thanks for the comments!):

  • The group $G$ is non-discrete
  • The metric $d$ is proper

And this third point would make me even happier

  • The closure of any open ball is the closed ball of the same radius
Improve this question
$\endgroup$
11
  • $\begingroup$ Do you want to assume $G$ is connected? Otherwise any discrete group is a counterexample... $\endgroup$
    – Nate Eldredge
    Commented Feb 16, 2017 at 15:16
  • $\begingroup$ $Gr_d$ is continuous at every $r<0$, and is continuous at zero iff $G$ is non-discrete. $\endgroup$
    – YCor
    Commented Feb 16, 2017 at 15:32
  • $\begingroup$ 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. $\endgroup$
    – YCor
    Commented Feb 16, 2017 at 15:35
  • $\begingroup$ @NateEldredge Yes, sorry, I want to assume that $G$ is not discrete. $\endgroup$
    – Alessandro Carderi
    Commented Feb 16, 2017 at 17:19
  • 1
    $\begingroup$ @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. $\endgroup$
    – YCor
    Commented Feb 16, 2017 at 19:49

1 Answer 1

Reset to default
1
$\begingroup$

Since Wikipedia tells me that a proper metric space is one in which every closed ball is compact, I think that a modification of my comment will work: namely, the additive group of a $p$-adic field (i.e., complete, non-Archimedean, non-trivially discretely valued field with finite residue field), with its usual metric. (Maybe it violates some spirit of the question, though. For example, it is essentially maximally non-Euclidean; an $(n + 1)$-point subset of $\mathbb Q_p$ embeds isometrically in $\mathbb R^n$, but in no smaller Euclidean space.)

EDIT: I originally suggested the multiplicative group, but @YCor pointed out that doesn't work.

Improve this answer
$\endgroup$
6
  • 1
    $\begingroup$ You need to specify the metric on the multiplicative group. $\endgroup$
    – YCor
    Commented Feb 16, 2017 at 19:54
  • 1
    $\begingroup$ And clearly $\mathbf{Q}_p^*$ admits no proper compatible left-invariant ultrametric. $\endgroup$
    – YCor
    Commented Feb 16, 2017 at 19:56
  • $\begingroup$ @YCor, oh, right, I ignored the crucial 'left-invariant' part. I guess the additive group (with its usual metric) should still work, though. $\endgroup$
    – LSpice
    Commented Feb 16, 2017 at 20:14
  • 1
    $\begingroup$ But in the additive group, the open 1-ball is closed, i.e. equal to its closure, but not equal to the "closed 1-ball". $\endgroup$
    – YCor
    Commented Feb 16, 2017 at 20:22
  • $\begingroup$ I actually don't know which question you claim to answer. $\endgroup$
    – YCor
    Commented Feb 16, 2017 at 20:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .