0
$\begingroup$

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.

$\endgroup$
8
  • $\begingroup$ 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. $\endgroup$
    – YCor
    Jul 21, 2012 at 13:55
  • $\begingroup$ 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. $\endgroup$
    – YCor
    Jul 21, 2012 at 13:58
  • $\begingroup$ 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. $\endgroup$
    – Marc Palm
    Jul 21, 2012 at 14:02
  • $\begingroup$ 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. $\endgroup$
    – dhurup
    Jul 21, 2012 at 15:01
  • $\begingroup$ @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. $\endgroup$
    – YCor
    Jul 22, 2012 at 2:25

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$
    – YCor
    Jul 21, 2012 at 13:52
  • $\begingroup$ @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 $\mathrm{SU}(3)/\mathrm{SO}(2)$. $\endgroup$ Jul 21, 2012 at 14:45
  • 2
    $\begingroup$ @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? $\endgroup$
    – YCor
    Jul 21, 2012 at 14:58
  • $\begingroup$ @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. $\endgroup$ Jul 22, 2012 at 13:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.