Suppose $G$ is a compact Lie group, $K$ a closed Lie subgroup and $\langle\cdot,\cdot\rangle$ is a Riemannian metric on $G$ that is either
- $G$-bi-invariant
or - $G$-left-invariant and $K$-right-invariant.
Note that since $G$ and thus $K$ is compact, we can find such metrics by averaging left-invariant metrics by right-translations by elements of either $G$ or $K$.
Let $M=G/K$ (this is a homogeneous $G$-space) and let $\pi\colon G\to M$ be the projection which sends $g$ to the coset $gK$.
The subgroup $K$ acts on $G$ from the right by right multiplication: $G\times K\to G, (g,k)\mapsto R_k(g)=g\cdot k=gk$.
In the language of e.g. [Lee, p.23] this action is
- vertical, i.e. $\pi(R_k(g))=\pi(g)$ for all $g\in G$ and $k\in K$.
- transitive on fibers, i.e. for every $g,h\in G$ with $\pi(g)=\pi(h)$ there is $k$ (here $=g^{-1}h)\in K$ such that $R_k(g)=h$.
- and isometric, i.e. the map $G\to G, g\mapsto R_k(g)$ is an isometry for every $k\in K$.
The first two properties hold by definition of $\pi$ and the last one holds because the metric on $G$ is supposed to be at least $K$-right-invariant.
Then [Lee, Thm. 2.28] states that there is a unique Riemannian metric on $M$ such that $\pi$ turns into a smooth Riemannian submersion.
Furthermore, one easily sees that this quotient metric is $G$-invariant.
Question 1:
If $G=O(n)$ and $K=O(n-1)$ then $M=\mathbb{S}^{n-1}$. If I'm not mistaken than the quotient metric resulting from property (2) is exactly the usual round metric on the sphere.
What is the quotient metric resulting from property (1)?
Question 2:
"How different" are the quotient metrics resulting from property (1) and (2) in general? Can this be quantified?
Somehow it feels like property (1) should lead to something "worse" because the isometry group of $G$ is bigger than the one for property (2) and thus taking the quotient by $K$ should lead to more elements in the resulting isometry group.
The questions arose because in my application I have two subgroups of $G=O(n)$, namely $K=O(n-1)$ and $L=O(k)\times O(n-k)$. Hence to get metrics on both quotients I would like to choose one Riemannian metric on $G$, i.e. fulfilling property (1) rather than choosing two different metrics on $G$ which fulfil property (2), respectively.
So, hopefully, the answer to Q2 is "not very different" or so?
[Lee] Lee, Introduction to Riemannian manifolds, 2nd ed.
Best Answer
Regarding Q1: (1) implies (2): Each biinvariant metric is also $G$-left invariant and right $K$-invariant for each $K< G$. In the example of $G=O(n), K=O(n-1)$, the resulting metric is isometric to the induced metric on a round sphere in $E^n$ of some radius.
I am not sure what exactly you mean by Q2: This class of metrics given by (2) is, in general, much larger than by (1). A good example to consider is $G=SU(2)$ and $K=\{1\}$. Then $G/K=G$ is diffeomorphic to $S^3$. However, most metrics arising from (2) will not have constant curvature. Interesting examples of metrics on $S^3$ obtained this way (assuming that the metrics are right $SO(2)$-invariant) are "Berger spheres". This is a 1-parameter family $g_t$ of homogeneous Riemannian metrics on $S^3$ (with $g_1$ having constant curvature) admitting foliations by closed geodesics (fibers of a Hopf fibration), the common lengths these geodesics is $L_t$, and $\lim_{t\to 0} L_t=0$.