# On bi-invariant metrics on groups

I'm looking for a quick, snap-your-fingers proof of the following result:

A continuous length metric on $\mathbb{R}^n$ that is invariant under translations comes from a norm.

To be clear about the terminology: I'll say that a metric $d$ comes from a norm if $d(x,y) = \|x - y \|$.

I'm also looking for a reference where the following question is addressed:

Given a continuous bi-invariant length metric on a Lie group, is it true that one-parameter subgroups are geodesics and/or that the metric comes from a bi-invariant continuous Finsler metric?

Remark 1. Please pay attention to the level of generality. These things are really easy if you assume a bit more regularity than one should.

Edit: May 12, 2014. I had written earlier that I had a proof which used convolution and the action of the affine or linear group on translationn-invariant distances to regularize the metric. In fact, this approach is wrong or incomplete. It is not clear (or not true?) that the regularized metric is a length metric.

On the other hand, the OP does follow from the much, more general results of Berestovskii cited by Yves in his comment.

• Berestovskii gave a general description of left-invariant length metrics on connected Lie groups. Link: link.springer.com/article/10.1007%2FBF00972413
– YCor
May 5 '14 at 10:39
• Thanks. I guess this raises the question of whether there are bi-invariant sub-Finslerian metrics or maximally non-integrable bi-invariant distributions in Lie groups. If there are none, his theorem should yield a positive answer to the second problem. May 5 '14 at 10:49
• In most Lie groups there's no continuous bi-invariant distance at all. For instance, there's none in $SL_2(\mathbf{R})$ (and more generally all connected Lie groups of exponential growth) because the unit is in the closure of some conjugacy class. Also there's none in most groups of polynomial growth: for instance in the Heisenberg group, because all nontrivial cosets of the center are conjugacy classes, the length has to be constant on each nontrivial coset, hence is constant on the center and hence can't be a distance.
– YCor
May 5 '14 at 11:16
• Does not the following work: Let $B=\lbrace x\in\mathbb R^n: d(0,x)\le 1\rbrace$. You have to show that $B$ is absolutely convex, bounded (should follow from complete), and closed (because $d$ is continuous). One has to use that $d$ is a length metric to show that $B$ is absolutely convex, and to show that $d(0,\lambda.x)=\lambda.d(0,x)$ for each $\lambda\ge0$, I guess. May 5 '14 at 12:45
• @YvesCornulier: the existence of a bi-invariant metric is part of the hypotheses. I know they are very rare at least if they are Finsler or Riemannian (I think the group must be the product of a compact group and $\mathbb{R}^n$). May 5 '14 at 13:26

Here is a nice quick proof if one assumes that every two points in $(\mathbb{R}^n,d)$ have a unique metric midpoint.

A continuous length metric on $\mathbb{R}^n$ that is invariant under translations and with unique midpoints comes from a norm.

What follows is a simplified presentation of ideas found in Busemann's Geometry of Geodesics (Sections 17 and 50).

Sketch of the proof. The point of departure will be (Busemann's?) characterization of metrics on $\mathbb{R}^n$ that come from norms:

A complete, continuous metric on $\mathbb{R}^n$ comes from a norm if and only if the affine midpoint of any two points is also a metric midpoint.

Recall that a point $b$ is a metric midpoint of two points $a$ and $c$ if $d(a,b) = d(b,c) = d(a,c)/2$.

It is a standard fact that any two points in a complete length metric space have at least one metric midpoint. Also notice that under the hypothesis of translation invariance completeness comes for free from the fact that $\mathbb{R}^n$ is locally compact.

If we assume that the metric midpoint is unique, then the proof of theorem is very simple: Let $a$ and $c$ be any two distinct points in $\mathbb{R}$ and let $b$ be their metric midpoint. Note that by translation invariance $$d(a, a + (c-b)) = d(a + (b-a), a + (c-b) + (b-a)) = d(b,c) = d(a,c)/2$$ and $$d(a + (c-b), c) = d(a + (c-b), a + (b-a) + (c-b)) = d(a, a + (b-a)) = d(a,b),$$ which is also $d(a,c)/2$. It follows that the point $a + (c - b)$ is the metric centre of $a$. The upshot: the affine midpoint of two arbitrary points $a$ and $c$, $(a + c)/2$ equals their metric center and the metric must come from a norm.

Remark. The proof shows that the map $x \mapsto a + c - x$ maps the set of metric midpoints of the pair $(a,c)$ into itself. It is also easy to see that this map is an isometric involution. If we could prove that this involution has a fixed point, we would have the general result (i.e., without assuming the uniqueness of metric midpoints).