Let G be a Lie group and H be a compact subgroup. The (left) coset space G/H is, up to an isomorphism, equivalent to the smooth homogeneous manifold M. My question is, is it possible to impose an explicit Riemannian structure on this manifold by specifying G as an isometry group with corresponding isotropy group H? For example, consider the group SO(3). The coset space SO(3)/SO(2) is then isomorphic to the 2 sphere. If one specifies SO(3) as the group of isometries, is it possible to recover an explicit expression for the metric? Would this necessarily be the "traditional" metric inherited from embedding in Euclidean 3D space? Thanks
[Math] Riemannian metrics on homogeneous spaces
differential-geometryhomogeneous-spaceslie-algebraslie-groupsriemannian-geometry
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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 $\lim_{t\to 0} L_t=0$.
Claim: Let $M^n$ be an aspherical closed Riemannian homogeneous manifold. Then $M$ is a flat torus $T^n$.
Let's write $M=G/H$ where $G=Isom_0(M)$ is the connected component of the isometry group of $M$ and $H$ is the isotropy group of a point. Both $G$ and $H$ are compact. $G$ admits a left invariant metric which is biinvariant under $H$ and which induces the original Riemannian metric on $M$ (this is true for all Riemannian homogeneous spaces even if $G$ is not compact).
Let $\tilde G$ be the universal cover of $G$ and $\tilde H$ be the preimage of $H$ under the projection $\pi: \tilde G\to G$. Then $M=\tilde G/\tilde H$.
By the theory of compact Lie groups we have that $\tilde G=\mathbb R^k\times \hat G$ where $G$ is compact semisimple (it's a product of several simple factors).
Since all Lie groups have trivial $\pi_2$ it must hold that the identity component $\tilde H_0$ is simply connected since otherwise $M$ would have nontrivial $\pi_2$. So $ \tilde H_0$ is also isomorphic to $\mathbb R^l\times \hat H$ where $\hat H$ is semisimple and simply connected. Then $\hat H\subset \hat G$ is a closed subgroup. The manifold $\hat G/\hat H$ is closed and simply connected. If it's not a point it has nontrivial top homology and hence has a nontrivial $\pi_k$ for some $k>1$. But that would imply that $M$ also has nontrivial $\pi_k$. Therefore $\hat H=\hat G$. Therefore we can "cancel" $\hat G$ in the homogeneous space $M=\tilde G/\tilde H$. This means that the $\mathbb R^k$ factor in $\tilde G$ already acts transitively on $M$.
Now, the punchline is that any left invariant Riemannian metric on $\mathbb R^k$ is flat and any closed subgroup of $\mathbb R^k$ is flat too. This immediately implies that $M=\tilde G/\tilde H$ is flat. It's well known that the only closed Riemannian homogeneous flat manifolds are flat tori so $M$ is a flat $T^n$.
Best Answer
Let $G$ be a connected Lie group and $H$ a closed subgroup (not necessarily compact).
The set $\mathcal{X}_{G,H}$ of $G$-invariant Riemannian metrics on $G/H$ is in canonical bijection with the set of $H$-invariant scalar products on $\mathfrak{g}/\mathfrak{h}$ under the adjoint representation of $G$. If $H$ is compact, the latter is non-empty.
Examples:
($\star$): assume that $H$ is non-compact (so $G$ is non-compact) and that $G/H$ carries an invariant Riemannian metric. Since $H$ is non-compact, the action of $G$ on $G/H$ is non-proper. Hence it has bounded orbits (reference), so $G/H$ is compact. Since moreover it is Riemannian, it has finite volume and it follows that $H$ is a cocompact lattice; in particular it is Zariski-dense. Since the stabilizer of a Riemannian metric of $\mathfrak{g}/\mathfrak{h}=\mathfrak{g}$ is Zariski-closed, it follows that the metric on $\mathfrak{g}$ is $G$-invariant for the adjoint representation. Since $G$ is non-compact, there is no such metric (the set of invariant quadratic forms being 1-dimensional, generated by the Killing form) and we get a contradiction.