Riemannian metrics on homogeneous spaces

Question

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

Answer 1

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:

1. If $H$ is trivial, $\mathcal{X}_{G,H}$ is thus quite big (a non-empty open cone in dimension $n(n+1)/2$ where $n=\dim(G)$). More generally, if $H$ is normal, this still holds (with $n=\dim(G/H)$).
2. But there are cases where $\dim(G/H)>1$ and it's a half-line, i.e., the $G$-invariant Riemannian metric on $G/H$ is unique up to rescaling. This holds, in particular, when $G$ is a simple Lie group and $H$ is a maximal compact subgroup. It also holds when $G$ is the isometry group of some irreducible symmetric space of compact type, which includes the case of $\mathrm{SO}(3)/\mathrm{SO}(2)$.
3. If $H$ is compact, then $\mathcal{X}_{G,H}$ is not empty.
4. If $H$ is non-compact, then $\mathcal{X}_{G,H}$ can be non-empty (as when $H$ is normal, see above), or can be empty, for instance when $\dim(G)=2$ and $H$ is a non-normal 1-parameter subgroup.
5. If $G=\mathrm{SL}_2(\mathbf{R})$ (or more generally any simple Lie group) and $H$ is an arbitrary non-compact closed subgroup, then $\mathcal{X}_{G,H}$ is empty ($\star$).

($\star$): 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.

Answer 2

The answer to your question is no for several reasons.

First, just because you can write $M = G/H$ does not mean that any of the natural metrics on $M$ have isometry group $G$ - it is sometimes larger.

For example, consider the homogeneous space $S^6 = G_2/SU(3)$. As YCor states, the set of $G_2$ invariant metrics on $G_2/SU(3)$ is in bijection with $SU(3)$ invariant inner products on $\mathfrak{g}/\mathfrak{h}$ under the adjoint action. Well, in this case, the adjoint action of $SU(3)$ on the $6$-d real vector space $\mathfrak{g}/\mathfrak{h}$ is nothing but the usual $SU(3)$ action on $\mathbb{C}^3\cong \mathbb{R}^6$. In particular, the action is transitive, so irreducible.

It follows that there is, up to scale, a unique invariant metric on $S^6 = G_2/SU(3)$. But the usual round metric is one example of such an invariant metric, so it must be the only example (up to scaling) - and it has isometry group $O(7)$, strictly containing $G_2$.

Another issue is that there may be multiple metrics which are $G$-invariant. This is already hiding in point 1 of YCor's post, but I wanted to make it more explicit. For example, consider $S^3 = SU(2)$. How many metrics (not including scaling) are $SU(2)$ invariant? Well, of course the usual round metric is, but here is another example. Thinking of $S^3\subseteq \mathbb{C}^2$, at each point $p\in S^3$, we can conisder the tangent vector $ip$. Define a new Riemannian metric where the vector $ip$ has length $\lambda \neq 1$, but where the metric on the orthogonal complement is the usual round metric. This new metric is still $SU(2)$ invariant for any choice of $\lambda > 0$, but is only round if $\lambda = 1$. (This new metric gives a so called Berger sphere.)

Answer 3

If you suppose that $G$ is also compact then you have a left invariant metric on $G$. Let $p:G\rightarrow G/H$ the projection and $x$ be a point of $G$, you can identify the tangent space of $p(x)$ to the orthogonal $N_x$ of the subspace of $T_xG$ tangent to the orbit $H(x)$. The restriction of the invariant metric on $N_x$ defines a metric on $G/H$.