# A compact Lie group $G$ acting on a compact Lie group $K$ transitively. Is there a $C$ such that $d(gx,gy)\leq Cd(x,y)$?

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.

• 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.
– gpr1
Commented Jul 26, 2023 at 18:34
• It's still not clear for me. Commented Jul 27, 2023 at 20:30
• 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? Commented Jul 31, 2023 at 16:26
• The action $\Psi: G\times K\to K$ given by $\Psi(g,k)=g\cdot k$ is $C^\infty$ Commented Jul 31, 2023 at 16:28
• Then gpr1's comment suffices. Commented Jul 31, 2023 at 16:29