4
$\begingroup$

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

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 $d$.

$\endgroup$

1 Answer 1

6
$\begingroup$

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 $d$.

$\endgroup$
4
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.