Let $G$ be a group equipped with a left-invariant metric $d$: that is, $(G,d)$ is a metric space and $d(xy,xz) = d(y,z)$ for all $x,y,z \in G$. Suppose further that $(G,d)$ is connected, locally compact, and bounded. Must $(G,d)$ be compact?

I can show from the local compactness that $(G,d)$ is complete. I don't see how to get it to be totally bounded, but I also can't think of a counterexample.

If it helps, you can assume that $(G,d)$ is a topological group (i.e. right translation and inversion are homeomorphisms; we already know that left translation is an isometry). I would even be interested in the case where $G$ is a finite-dimensional Lie group and $d$ induces the manifold topology, but I do not want to assume that $d$ is induced by a left-invariant Riemannian metric.

  • $\begingroup$ This question is interesting. By the way, do you know if existence of a bounded left invariant Riemannian metric implies compactness of the Lie group? $\endgroup$ – Asaf Shachar Mar 12 '17 at 10:12
  • $\begingroup$ @AsafShachar: I believe that follows from the metric completeness thanks to the Hopf-Rinow theorem. $\endgroup$ – Nate Eldredge Mar 13 '17 at 16:30
  • $\begingroup$ Thanks! you are right, of course. $\endgroup$ – Asaf Shachar Mar 13 '17 at 16:47

The answer is no. Consider $\mathbb R$ with the bounded left invariant metric $$d(x,y)=\min(|x-y|,1)$$

  • $\begingroup$ Why is this left invariant? Maybe it's obvious but I'm not seeing it. $\endgroup$ – T. Eskin Jun 2 '15 at 4:53
  • $\begingroup$ @ThomasE. $|(x+a)-(y+a)|=|x-y|$. $\endgroup$ – Matt Samuel Jun 2 '15 at 10:38
  • $\begingroup$ right! Because it's an additive group. Thanks. $\endgroup$ – T. Eskin Jun 2 '15 at 13:15

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