Riemannian metrics on homogeneous spaces

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.

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 $G/H$.