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I would like to know examples of $G/K$ where $G$ is a locally compact group and $K$ is a compact subgroup of $G$.

I know about Riemannian symmetric spaces of Euclidean, compact and non-compact type. They can be realized as $G/K$. I am wondering what are other prominent spaces (discrete and continuous) which can be written as $G/K$, where $G$ can be unimodular or non-unimodular. Perhaps homogeneous tree is an example outside the set of Riemannian symmetric spaces.

flag	asked Jul 21 at 9:02 dhurup 11●2

	A homogeneous tree is not topologically homogeneous and thus can't be $G/K$. For $(G,K)=(SL_2(\mathbf{Q}_p),SL_2(\mathbf{Z}_p))$, indeed $G/K$ can be viewed as the vertex set of a tree. – Yves Cornulier Jul 21 at 13:55
	Also note that $G/K$ is canonically a topological space with a transitive $G$-action. In many cases it carries a left-invariant distance but this is not canonical. So maybe you might specify a little what you have in mind. – Yves Cornulier Jul 21 at 13:58
	What is with $K$ trivial? ;) The question should be more precise. Perhaps of interest two you, every connected locally compact group contains a normal subgroup such that the quotient is a Lie group. – Marc Palm Jul 21 at 14:02
	Thanks for the answers and comments and sorry if you find the question not-so-precise. Let me try again. I want examples of pair $(G, K)$ where $G$ is a locally compact group and $K$ a non trivial compact subgroup of $G$ such that $G/K$ is not a Riemannian symmetric space, i.e. $G/K$ is either not a symmetric space or it is not a Riemannian manifold, but for some reason the space $G/K$ is an important object. Robert Bryant has given an example. – dhurup Jul 21 at 15:01
	@dhurup: you comment is still as ambiguous as the question. The statement "$G/K$ is not a symmetric space" is meaningless in two respects: 1/ is can mean "is isometric" or something weaker such as "is homomorphic to"; 2/ if you mean "is isometric to", it depends on the choice of a Riemannian metric. – Yves Cornulier Jul 22 at 2:25

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Robert Bryant · Accepted Answer · 2012-07-21 13:32:14Z UTC

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There are lots of Riemannian homogeneous spaces that are not symmetric. For example, consider the chain $\mathrm{SO}(2)\subset \mathrm{SO}(3)\subset \mathrm{SO}(4)$. The quotient $\mathrm{SO}(4)/\mathrm{SO}(2)$ is not a symmetric space.

answered Jul 21 at 13:32

Robert Bryant
21.8k●3●75●97

Actually in general if $G$ is a Lie group and $K$ is a compact subgroup, there are several $G$-invariant Riemannian metrics on $G/K$ and it might happen that some of them are symmetric. In your example I guess you mean that none is symmetric. – Yves Cornulier Jul 21 at 13:52

@Yves: Actually, no, I didn't mean that. That quotient actually turns out to be $S^2\times S^3$ topologically, so it actually can be written as a symmetric space, just not this way. If you want an example that can't be written as a symmetric space, take $\mathrm{SU(3)}$ instead of $\mathrm{SO(4)}$ in the above sequence and consider $\mathrm{SU}(3)/\mathrm{SO}(2)$. – Robert Bryant Jul 21 at 14:45

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@Robert: I'm now confused: I understand from your comment that $SO(4)/SO(2)$ is homeomorphic to a symmetric space. But is it a symmetric space for some choice of $SO(4)$-invariant metric? – Yves Cornulier Jul 21 at 14:58

@Yves: Well, there is a 4-parameter family of $\mathrm{SO}(4)$-invariant metrics on $\mathrm{SO}(4)/\mathrm{SO}(2)$, and one of them might be isometric to $S^3\times S^2 = \mathrm{SO}(4)/\mathrm{SO}(3)\times\mathrm{SO}(3)/\mathrm{SO}(2)$. So maybe that was not the best example. What I was trying to get at in my answer was that not all homogeneous spaces are symmetric spaces. A better example is $\mathrm{SO}(5)/K$, where $K$ is the subgroup of $\mathrm{SO}(5)$ that is isomorphic to $\mathrm{SO}(3)$ and that acts irreducibly on $\mathbb{R}^5$, as it's not homeomorphic to any symmetric space. – Robert Bryant Jul 22 at 13:48

Examples of locally compact group quotiented by compact subgroup, but not Riemannian symmetric space

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