On the properties of Homogeneous Spaces

[Redsummers]

Hello,

I am currently going over Nakahara's Geometry, Topology, and Physics and even though I have bumped into some typos/mistakes, there's something that I am sure is not a mistake but rather a misunderstanding I have of the basic concepts.

Namely, in page 181, he describes the notion of homogeneous space:

Let G be a Lie group and H any subgroup of G. The coset space G/H admits a differentiable structure and G/H becomes a manifold, called a homogeneous space. Note that dim G/H = dimG - dimH. let G be a Lie group which acts on a manifold M transitively and let H(p) be an isotropy group of p in M. [the term 'isotropy group' may be known to others by 'stabiliser'... just saying.] H(p) is a Lie subgroup and the coset space G/H(p) is a homogeneous space. In fact, if G, H(p) and M satisfy technical requirements (e.g. G/H(p) be compact) it can be shown that G/H(p) is homeomorphic to M. See example below...

Thus far, the notion of such space makes total sense to me... however that last statement of homeomorphism is not clear at all... If somebody can provide proofs or some related theorem, I would appreciate it.

Anyway, here comes the example that he gives, which even complicates more my understanding:

Let G = SO(3) be a group acting on R^3 [So I suppose M = R^3...] and H = SO(2) be the isotropy group of x element R^3.

Okay, from this, it's clear that SO(3) acts on S^2 transitively and hence we have that SO(3)/SO(2) is isomorphic to S^2. I.e... G/H = S^2. (However, since SO(2) is not a normal subgroup of SO(3), S^2 does not admit a group structure.)

That said, it is clear to me that G/H(p) is compact (as the requirement above)... but I don't see how S^2 is homeomorphic to R^3. Can anybody explain this?

I mean, I see how –for example– S^2 - {p} is homeomorphic to R^2... but S^2 to R^3??

Maybe it's late and the question is just super-dumb... but I better ask it here so that I can sleep with my mind in peace.Thank you in advance,

 

[Tinyboss]

I think in this case we have M=S^2, because G=SO(3) is supposed to act transitively, which it does not do on \mathbb{R}^3. The isotropy group H<SO(3) of a point on S^2 is isomorphic to SO(2) (which is homeomorphic to S^1) and, as you said, the coset space SO(3)/H doesn't have the quotient group structure because H is not normal in SO(3). But it does have the differentiable structure of S^2, which makes sense because, up to something in the isotropy subgroup of x, any rotation in SO(3) is determined by where it sends x.

In general, a rotation in SO(n) is determined by a point x\in S^{n-1} together with a rotation in SO(n-1) "fixing x", that is, SO(n) is an SO(n-1)-bundle over S^{n-1}.

The part in quotes above is sloppy language, but I hope it suggests the right idea. I think maybe the right way to say it is that the rotation in SO(n-1) acts on the fibers of SO(n)/H, which are all isomorphic to SO(n-1). Or maybe I'm making something simple into something unnecessarily complicated...sometimes I can't tell. ;-)

 

[Redsummers]

OH!

Thank you a lot for your response Tinyboss! –That's a nice result once we generalize it for the n-th orthogonal group ^^
I guess my mistake was on assuming that M=R^3, but I suppose I passed over the actual meaning of 'acting transitively'.

Cheers,

 

[Michael_1812]

Guys,
May I now ask you a (presumably, silly) question. Why is SO(2) not a normal (invariant) subgroup of SO(3) ?
Many thanks!

 

[Tinyboss]

Guys,
May I now ask you a (presumably, silly) question. Why is SO(2) not a normal (invariant) subgroup of SO(3) ?
Many thanks!

It's not closed under conjugation by arbitrary elements of SO(3).

 

[Office_Shredder]

Note the homeomorphis is quite simple - if g is in G and g(p) =q, then g gets mapped to q. modding out by the isotropy group is required to make this map 1 to 1, and the transitive action is what you need for surjectivity