Relationship between distances on homogeneous spaces and their Lie groups

Question

Consider the (round) sphere $M=\mathbb{S}^{n-1}$ as a homogeneous $O(n)$-space. Then for $x,y\in\mathbb{S}^{n-1}$ there is $g\in O(n)$ such that $y=g\cdot x$. Denote the Riemannian distance on $\mathbb{S}^{n-1}$ by $d_{\mathbb{S}^{n-1}}$. Intuitively, if $y$ and $x$ are not far apart then $g$ should be almost the identity (because the $O(n)$-action is smooth). I am able to explicitly construct such a rotation $g$ so that \begin{align} \|g-\operatorname{Id_{\mathbb{R}^n}}\| \leq 2\; d_{\mathbb{S}^{n-1}}(y,x) \end{align} where $\|\cdot\|$ is the operator norm for matrices.

Analoguously, if $M=\mathrm{Gr}_m(\mathbb{R}^n)$ is the Grassmannian, then by a similar construction using principle angles I can find a rotation $g$ such that $F=g\cdot E$ for $m$-planes $E,F$ and \begin{align} \|g-\operatorname{Id_{\mathbb{R}^n}}\| \leq 2m\; d_{\mathrm{Gr}_m(\mathbb{R}^n)}(F,E) \end{align} where $d_{\mathrm{Gr}_m(\mathbb{R}^n)}$ is the angle metric on $\mathrm{Gr}_m(\mathbb{R}^n)$ (e.g. here).

However, I find these constructions rather unsatisfying and would like to understand if there is a more abstract underlying principle at play.

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Here is my question: Given a homogeneous $G$-space $M$, are there always ($G$-invariant ?) metrics on $M$ and $G$ such that for all $x,y\in M$ there is $g\in G$ such that $y=g\cdot x$ and which fulfils the quantitative estimate \begin{align} d_G(g,e) \leq C \; d_M(y, x) ? \end{align}

Feel free to add any hypotheses (such as compactness etc) that apply to $\mathbb{S}^{n-1}$, $\mathrm{Gr}_m(\mathbb{R}^n)$ and $O(n)$.

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EDIT: As proposed in the comments by levap and Moishe Kohan, I have edited my question.

Answer 1

With the revised version the answer is positive (with $C=1$) assuming that $G$ acts properly on $M$, which in your setting means that it acts with compact point-stabilizers. This condition is satisfied in your examples since your groups $G$ are compact.

I will also assume that $G$ is connected, although with a tiny bit more work the proof works in general (replace $G$ with the identity component in this group).

Furthermore, I will assume that $M$ is a smooth manifold and $G$ is a Lie group acting smoothly and transitively on $M$.

Pick a point $m\in M$ and let $K$ denote its $G$-stabilizer $G_m$. Then the orbit map $$ g\mapsto g\cdot m $$
defines a homeomorphism (actually, a diffeomorphism) $G/K\to M$. The projection $\pi: G\to G/K=M$ is a principal $K$-bundle. The group $G$ admits a Riemannian metric which is left $G$-invariant (this is true regardless of compactness of $K$) and right $K$-invariant (here we need compactness of $K$). To get such a metric, average a $G$-left invariant Riemannian metric on $G$ via the right $K$-action.

Because the metric is right $K$-invariant it descends to a Riemannian metric on $M=G/K$ so that for every $g\in G$ the orthogonal complement $N_g$ to $T_{g}(gK)$ in $T_gG$ maps isometrically to $T_mM$ ($m=\pi(g)$) via the differential $d\pi_g: T_gG\to T_mM$. In particular, the map $\pi$ is 1-Lipschitz with respect to these two Riemannian metrics on $G$ and $M$. Furthermore, since the metric on $G$ was left $G$-invariant, the Riemannian metric on $M$ will be also $G$-invariant.

Observe also that the orthogonal complements $N_g$ define a connection $\nabla$ (in the sense of Ehresmann) on the bundle $\pi: G\to M$.

Now, define the distance function $d_G$ to be the Riemannian distance function of the chosen metric on $G$. (Here I am using the assumption that $G$ is connected.) Similarly, let $d_M$ denote the Riemannian distance function on $M$. As we noted above, the map $\pi$ is 1-Lipschitz. In particular, $$ d_G(g,h)\ge d_M(\pi(g), \pi(h)). $$ What you want, however, is the opposite Lipschitz inequality. To get this, we do the following. Pick two points $x, y\in M$ and a distance-minimizing geodesic $c$ from $x$ to $y$. (It is a standard fact that homogeneous Riemannian manifolds are complete, hence, such $c$ will always exist.)

For every $g\in \pi^{-1}(x)$, the connection $\nabla$ yields a lift $\tilde{c}$ of $c$, which is a smooth curve in $G$ starting at $g$ and terminating at some $h\in\pi^{-1}(y)$, such that for each $t$, the velocity vector $\tilde{c}'(t)$ belongs to the horizontal subspace of $T_{\tilde{c}(t)}G$ given by the connection $\nabla$ and $$ \pi\circ \tilde{c}= c. $$ The existence of such a lift is the only mildly nontrivial ingredient of the proof and it is a standard fact about Ehresmann connections, which amounts to the short-term existence of solutions of ODEs with the given initial conditions.

In particular, by the construction of the metric on $M$, the norms of $\tilde{c}'(t)$ and ${c}'(t)$ are the same. Thus, the length of $\tilde{c}$ is the same as the length of $c$, i.e. equals $d(x,y)$.

In particular, $$ d_G(g,h)\le d(x,y). $$ The 1-Lipschitz inequality above implies the equality $$ d_G(g,h)= d_M(x,y) $$ Since the metric on $G$ is $G$-left invariant, we obtain $$ d_G(g,h)= d_G(h^{-1}g,e). $$ Taking $f=h^{-1}g$, we then obtain an element $f\in G$ such that $f(x)=y$ and $$d_G(f,e)=d_M(x,y)$$ qed

Lastly, the reference for the theory of Ehresmann connections that I like is:

I.Kollar, P.Michor, J.Slovak, "Natural operators in differential geometry", Springer-Verlag, 1993.