Let $G$ be a compact connected Lie group acting transitively and smoothly on another compact Lie group $K$. Let $d$ be the distance in $K$ that is not $G$-invariant. Is there a constant $C$ such that $d(gx,gy)\leq Cd(x,y)$ for every $g\in G$ and $x,y\in K$? Moreover, the action is not isometric.

The path I tried by MVT is leading me to bound $ \lVert dg_{x}\rVert_{\operatorname{op}}$ for $g\in G$ and for $x\in K$:

Let $\gamma$ be a geodesic connecting $x$ to $y$. For each $g\in G$, $g\gamma$ is a path connecting $gx$ to $gy$, so $$d(gx,gy)\leq \int_0^1|(g\gamma)'(t)|dt=\int_0^1|dg_{\gamma(t)}\gamma'(t)|dt\leq\int_0^1\lVert dg_{\gamma(t)}\rVert_{\operatorname{op}}|\gamma'(t)|dt$$

It would suffice to show that $(g,x)\mapsto dg_{x}$ is continuous in the topology of the operator norm. But it is proving to be harder than I thought.

  • 4
    $\begingroup$ You are implicitly considering a Riemannian metric on $M$, so you can instead write $|\mathrm{d}g_{\gamma(t)}\gamma'(t)|\leq |X^g_{\gamma(t)}||\gamma'(t)|$, where $X^g$ is the vector field evaluated at $x\in M$ as the Riemannian type change of $T_xM\ni v\mapsto dg_xv$. The result then follows from the fact that $G\ni g\mapsto X^g\in\mathfrak{X}(M)$ and $|\cdot|$ induced by the Riemmanian metric are continuous. $\endgroup$
    – gpr1
    Commented Jul 26, 2023 at 18:34
  • $\begingroup$ It's still not clear for me. $\endgroup$
    – Gomes93
    Commented Jul 27, 2023 at 20:30
  • 2
    $\begingroup$ You should clarify your definition of "acting smoothly", I do not think it is the standard one. It is especially unclear since $K$ is also a Lie group: Do you assume an action by group automorphisms? $\endgroup$ Commented Jul 31, 2023 at 16:26
  • $\begingroup$ The action $\Psi: G\times K\to K$ given by $\Psi(g,k)=g\cdot k$ is $C^\infty$ $\endgroup$
    – Gomes93
    Commented Jul 31, 2023 at 16:28
  • 1
    $\begingroup$ Then gpr1's comment suffices. $\endgroup$ Commented Jul 31, 2023 at 16:29


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