# Any bi-invariant distance on a group is inverse-invariant?

$\newcommand{\inv}{\text{inv}}$

Let $G$ be a Lie group. Assume $d$ is a metric on $G$ (in the sense of metric spaces) which is bi-invariant. Is it true that the inverse automorphism must be an isometry of $(G,d)$?

I do not assume $d$ is induced by a Riemannian metric*.

*If $d$ is induced by a Riemannian metric $g$, than the answer is positive:

By the Myers–Steenrod theorem, $g$ is also bi-invariant.

Note that $\inv = R_{s^{-1}}\circ \inv \circ L_{s^{-1}}$, so $$(d\inv)_s = (dR_{s^{-1}})_e \circ (d\inv)_e \circ (dL_{s^{-1}})_s$$

Since $(d\inv)_e:T_eG \to T_eG$ is the minus operation $(v \mapsto -v)$, we get:

$$(d\inv)_s = - (dR_{s^{-1}})_e \circ (dL_{s^{-1}})_s$$

So, bi-invariance of the metric $g$ implies inverse-invariance of $g$, which implies invers-invariance of $d$.

I understand the bi-invariance as $$d(ax,ay)=d(x,y)=d(xa,ya)$$ for any $a,x,y\in G$. Then $$d(x,y)=d(1,x^{-1}y)=d(y^{-1},x^{-1})=d(x^{-1},y^{-1}).$$ The last step is the symmetry of $d$.
• note that this has nothing to do with Lie groups and distances, it holds for arbitrary bi-invariant symmetric kernels on groups (here kernel means any function from $G\times G$ to any set).
• @Anton: Thanks. So this turned out to be quite trivial... I guess I was blinded since the original question I thought of was only in the Riemannian setting, and there I had to use the fact $(d\text{inv})_e$ was an isometry, and lift it to other points... Feb 3, 2016 at 18:20