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Assume $G$ is a compact Lie group of finite dimension with bi-invariant Riemannian metric. Is the diameter of $G$ bounded above by some constant that can be determined? Alternatively, does there exists a bi-invariant metric on $G$ such that its diameter is bounded above by the diameter of some orthogonal group?

Thoughts:

$G$ being compact Lie group means that $G$ can be embedded as a subgroup in $O(k)$ for some $k$.

We can give $O(k)$ a bi-invariant Riemannian metric and that restricts to bi-invariant metric on $G$, provided the metric restricts to a bi-invariant one on G.

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    For every compact Riemannian manifold $M$, the diameter is finite. Furthermore, by simply scaling the Riemannian metric on $M$ by an appropriately chosen constant, you can make the diameter be any positive number whatsoever. – Lee Mosher Jun 06 '23 at 17:29
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    If $g$ is bi-invariant, so is $\lambda^2 g$ for all $\lambda >0$. Note that $\mathrm{diam}{\lambda^2 g}(G) = \frac{1}{\lambda}\mathrm{diam}{g}(G)$. – Didier Jun 06 '23 at 17:30
  • @Didier Thanks so much! Out of curiosity. If we give $G$ a biinvariant riemannian metric and then scale the metric, is the volume wrt Haar measure affected? or is the volume wrt Haar measure independent of Riemannian structure on G? –  Jun 19 '23 at 16:56

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