# On the properties of Homogeneous Spaces

1. Feb 27, 2012

### 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:

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:

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,

Last edited: Feb 27, 2012
2. Feb 27, 2012

### 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 beacause $H$ 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. ;-)

Last edited: Feb 27, 2012
3. Feb 27, 2012

### 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,

4. Mar 30, 2012

### 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!!

5. Mar 30, 2012

### Tinyboss

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

6. Mar 31, 2012

### Office_Shredder

Staff Emeritus
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