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Suppose I have a (finite-dimensional) Lie group $(G,\circ)$ with identity element $e\in G$. Then I can always construct a left-invariant metric $$ g_q\colon T_qG\times T_gG \to \mathbb [0,\infty),\qquad (x,y)\mapsto g_q(x,y) = \langle dL_{q^{-1}}(q)\;x, dL_{q^{-1}}(q)\;y\rangle, $$ where $$L_q\colon G \to G, \quad p\mapsto L_q(p) = q\circ p$$ is the left-translation (and $dL_q(p)\colon T_pG \to T_{L_q(p)}G = T_{q\circ p}G$ is its derivative) and $\langle\bullet,\bullet\rangle\colon T_eG\times T_eG \to [0,\infty)$ is a scalar product on the Lie algebra $T_eG$, which is a linear space.

From a metric $g_p$ we can construct a distance function $dst$ on $G$, which makes $(G,dst)$ a metric space: $$ dst\colon G\times G\to [0,\infty),\qquad (q,p)\mapsto dst(q,p) = \inf_{\gamma\in\Gamma(q,p)} L(\gamma),$$ where $\Gamma(q,p)\subseteq G$ is the set of all differentiable curves with $\gamma(0)=q$, $\gamma(1) = p$ and $L(\gamma)$ gives the length of a curve by $$ L(\gamma) = \int_0^1 \sqrt{g_{\gamma(s)}(\gamma'(s),\gamma'(s))}\;ds. $$

If $g_p$ is left-invariant, then $dst$ is also left-invariant in the sense that $$\begin{align}dst(q\circ a, q\circ b) = dst(a,b). \tag{9.1}\end{align}$$

I know that not every Lie group admits a bi-invariant metric (for example $SE(3)$ does not, since it is not the direct product of linear and compact Lie groups). Therefore, not every Lie group is a metric space, where the distance is bi-invariant.

I just read "Lie Group Methods" from Iserles, Munthe-Kaas, Nørsett and Zanna and there they state

"[A]ccording to the Birkhoff–Kakutani theorem (Birkhoff 1936), every Lie group $G$ admits a left-invariant, al- most right-invariant metric which, in addition to (9.1), obeys $$ dst(X\circ Z, Y\circ Z) \leq \rho(Z) dst(X, Y),$$ where the function $\rho$ is finite." (Note that I altered the name of the distance function and used $\circ$ for the Lie group product)

Unfortunately, I don't really understand the paper "A note on topological groups" from Birkhoff (Compositio Mathematica, Volume 3 (1936), p. 427-430) and the Birkhoff-Kakutani seems to be a theorem about whether a toplogical group (or Hausdorff group) is metricizable. I know that a Lie group is a special case of a topological group, but the theorem or the proof do not seem to be concerned with the invariance of the metric.

Can somebody explain to me or point me to a resource, where the existence of a left-invariant and almost right-invariant distance function is discussed? Also, does "$\rho$ is finite" mean that there is a constant $C$ such that $\rho(p)\leq C$ for all $p\in G$?

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  1. Birkhoff's paper is indeed irrelevant for your purposes.

  2. I think, by "finite" they simply mean that $\rho$ takes values in ${\mathbb R}$.

  3. The fact that the usual construction of a left-invariant Riemannian metric gives an almost right-invariant metric can be found in

J. Schiff and S. Shnider, Lie Groups and Error Analysis, Journal of Lie Theory, 11 (2001) 231-254.

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  • $\begingroup$ $SE(3)$ is one of the examples, where there is no biinvariant Riemannian metric, check for example Belta,Kumar's 2002 Corollary 3.7 or Computational Kinematics, p. 303 $\endgroup$ Jan 17 '20 at 7:21
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    $\begingroup$ @Wauzl: Oh, you are right, of course. $\endgroup$ Jan 17 '20 at 9:11
  • $\begingroup$ Note also that the version of the Schiff and Shnider paper you linked is some preprint, not the version in the Journal of Lie theory. In the Journal of Lie Theory the paper has an appendix written by the editor Karl Hoffmann, where he proves the fact in a different (and to me more clear) way. $\endgroup$ Jan 24 '20 at 11:58
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It's easy, namely due to the fact that every operator in finite dimension has a finite norm.

Indeed, fix a Euclidean structure on $T_1G$. For $g\in G$, the conjugation map $h\mapsto ghg^{-1}$ induces an operator on $T_1G$, with some norm $C_g$ with respect to the Euclidean distance. Then it follows the right translation by $g$ is $C_g$-Lipschitz on $G$.

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  • $\begingroup$ Could you please elaborate on this? $\endgroup$ Jan 23 '20 at 14:14

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