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On the properties of Homogeneous Spaces 
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#1
Feb2712, 02:34 PM

P: 163

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: Anyway, here comes the example that he gives, which even complicates more my understanding: 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 superdumb... but I better ask it here so that I can sleep with my mind in peace. Thank you in advance, 


#2
Feb2712, 04:08 PM

P: 234

I think in this case we have [itex]M=S^2[/itex], because [itex]G=SO(3)[/itex] is supposed to act transitively, which it does not do on [itex]\mathbb{R}^3[/itex]. The isotropy group [itex]H<SO(3)[/itex] of a point on [itex]S^2[/itex] is isomorphic to [itex]SO(2)[/itex] (which is homeomorphic to [itex]S^1[/itex]) and, as you said, the coset space [itex]SO(3)/H[/itex] doesn't have the quotient group structure beacause [itex]H[/itex] is not normal in [itex]SO(3)[/itex]. But it does have the differentiable structure of [itex]S^2[/itex], which makes sense because, up to something in the isotropy subgroup of [itex]x[/itex], any rotation in [itex]SO(3)[/itex] is determined by where it sends [itex]x[/itex].
In general, a rotation in [itex]SO(n)[/itex] is determined by a point [itex]x\in S^{n1}[/itex] together with a rotation in [itex]SO(n1)[/itex] "fixing [itex]x[/itex]", that is, [itex]SO(n)[/itex] is an [itex]SO(n1)[/itex]bundle over [itex]S^{n1}[/itex]. 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 [itex]SO(n1)[/itex] acts on the fibers of [itex]SO(n)/H[/itex], which are all isomorphic to [itex]SO(n1)[/itex]. Or maybe I'm making something simple into something unnecessarily complicated...sometimes I can't tell. ;) 


#3
Feb2712, 05:08 PM

P: 163

OH!
Thank you a lot for your response Tinyboss! –That's a nice result once we generalize it for the nth 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
Mar3012, 01:55 PM

P: 21

On the properties of Homogeneous Spaces
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
Mar3012, 02:23 PM

P: 234




#6
Mar3112, 12:33 PM

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PF Gold
P: 4,500

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



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