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.

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 $O(n)$.

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

  • 1
    $\begingroup$ In the first two paragraphs, you talk about the existence of an isometry while in your final question you ask about an estimate that is relevant for all $g$. Note that if $g \in G$ fixes some $x \in M$ then your estimate would imply that $g = e$ which is not true even in your examples (think of a rotation which fixes the north pole). Also, you haven't made explicit what is the relation between the $G$ action and the metrics. Do you want the action to be by isometries? $\endgroup$
    – levap
    Commented May 22, 2019 at 14:17
  • $\begingroup$ @levap I'm not sure what the right question to ask is really. I suppose I should fix my final paragraph and ask whether such an estimate holds for all $x\in M$ and all $g\in G\setminus \operatorname{Stab}(x)$. Does that make more sense? By your last question you mean that the metrics $d_G$ and $d_M$ should be invariant under the group actions of $G$ on itself and on $M$? Sure, if that helps to formulate a "good question". $\endgroup$ Commented May 22, 2019 at 14:50
  • $\begingroup$ This still does not work. Instead you should add a condition on the orthonormal frame bundle at x. $\endgroup$ Commented May 22, 2019 at 15:40

1 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.

  • $\begingroup$ This is a very elegant answer. I am learning many new things, hence my delayed comments :) (1) I believe the orbit map $\theta^{(m)}$ descends to a diffeomorphism $G/K\to M$ because it's surjective and constant on the fibres of $\pi$. (2) If someone else reads the above answer: For the construction of the metric on $M$ it helped me a lot to consider $T_g(gK)$ as $\operatorname{Ker}(d(\theta^{(m)})_g)$ (use $\exp^G$ and the group law for one direction and $\exp^K$ and the definition of $K$ for the other). Also $N_g$ maps isometrically to $T_{g\cdot m} M$ (as $m$ was fixed before). $\endgroup$ Commented Aug 22, 2019 at 15:42
  • $\begingroup$ (3)You mention twice that $\pi$ is $1$-Lipschitz. The first time without introducing the distances. Do you mean that $d\pi_g$ is $1$-Lipschitz as a map of normed spaces $T_gG$ and $T_{g\cdot m} M$? (4)How do the orthogonal complements define an Ehresmann connection? Can you define the orthogonal complements to be the horizontal subspaces of $T_gG$? I am still learning from the referenced book. (5)I think it should be $f=hg^{-1}$ because $x=g\cdot m$ and $y=h \cdot m$ for $m$ as fixed above. This would require $d_G$ to be $G$-right invariant but isn't it only $K$-right invariant? Am I mistaken? $\endgroup$ Commented Aug 22, 2019 at 16:18
  • $\begingroup$ (6) Is the Ehresmann connection of the principal $K$-bundle $\pi\colon G \to G/K$ the same as the Levi-Civita connection of the $G$-invariant Riemannian metric on $M$? Or at least: Can they somehow be related to each other? $\endgroup$ Commented Aug 22, 2019 at 16:53
  • $\begingroup$ @SvenPistre: (1) One also uses injectivity and smoothness of the descended map as well as the fact that its derivative is injective at every point. Then diffeomorphism of $G/K\to M$ follows from the inverse function theorem. (3) Once you have Riemannian metrics the 1-Lipschitz condition follows from the weak contraction property of maps of tangent spaces (this is quite immediate from the definition of the Riemannian distance function). (4) Once you know what an Ehresmann connection is, it is immediate that orthogonal complements to tangent spaces of fibers define an Ehresmann connection. $\endgroup$ Commented Aug 22, 2019 at 18:12
  • $\begingroup$ @SvenPistre: I will answer the rest of your questions later when I have more time. $\endgroup$ Commented Aug 22, 2019 at 18:13

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