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 $d$.


1 Answer 1


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 $d$.

  • 2
    $\begingroup$ 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). $\endgroup$
    – YCor
    Feb 3, 2016 at 13:50
  • $\begingroup$ @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... $\endgroup$ Feb 3, 2016 at 18:20
  • $\begingroup$ @Anton: I intend to put this proof in the appendix of a paper I am writing. If you want any credit for it, tell me. (Personally, I feel this turned out to be a trivial question, so I guess you won't bother, but I felt it's safer to ask than be sorry later:). $\endgroup$ Feb 13, 2016 at 15:48
  • $\begingroup$ I agree, it's trivial. $\endgroup$
    – Echo
    Feb 14, 2016 at 17:48

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