On the properties of Homogeneous Spaces

In summary, in page 181 of Nakahara's Geometry, Topology, and Physics, the author discusses the concept of homogeneous spaces. These are manifolds obtained by coset spaces G/H, where G is a Lie group and H is a subgroup of G. The author notes that the dimension of G/H is equal to the dimension of G minus the dimension of H. Additionally, if G acts transitively on a manifold M and H is the isotropy group of a point p in M, then the coset space G/H is also a homogeneous space. This can be shown to be homeomorphic to M under certain conditions. An example is given where G=SO(3) acts on R^3 and H=SO
  • #1
Redsummers
163
0
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,
 
Last edited:
Physics news on Phys.org
  • #2
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 because [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^{n-1}[/itex] together with a rotation in [itex]SO(n-1)[/itex] "fixing [itex]x[/itex]", that is, [itex]SO(n)[/itex] is an [itex]SO(n-1)[/itex]-bundle over [itex]S^{n-1}[/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(n-1)[/itex] acts on the fibers of [itex]SO(n)/H[/itex], which are all isomorphic to [itex]SO(n-1)[/itex]. Or maybe I'm making something simple into something unnecessarily complicated...sometimes I can't tell. ;-)
 
Last edited:
  • #3
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
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
Michael_1812 said:
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).
 
  • #6
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
 

1. What is a homogeneous space?

A homogeneous space is a mathematical concept that describes a space where every point looks the same in terms of its geometric properties. This means that there exists a symmetry group that acts transitively on the space, meaning any point can be transformed into any other point by a specific transformation.

2. How do homogeneous spaces relate to Lie groups?

Homogeneous spaces are closely related to Lie groups, as they can be defined as quotients of Lie groups by their subgroups. In other words, a homogeneous space is a space that has the same local geometric structure as a Lie group.

3. What are some applications of homogeneous spaces?

Homogeneous spaces have a wide range of applications in mathematics, physics, and engineering. They are used to study symmetry and group actions in geometry, as well as in the theory of Lie groups and their representations. In physics, homogeneous spaces are used to describe the symmetries of physical systems, such as the space-time in general relativity.

4. Can every space be classified as a homogeneous space?

No, not every space can be classified as a homogeneous space. In order for a space to be homogeneous, it must have a symmetry group that acts transitively on the space. This is not the case for all spaces, as some may have different geometric properties at different points.

5. How do you determine the properties of a homogeneous space?

The properties of a homogeneous space can be determined by studying its symmetry group and its orbit structure. By understanding the transformations that map one point to another in the space, we can determine its geometric properties. Additionally, the representations of the symmetry group can provide further insight into the structure of the space.

Similar threads

  • Differential Geometry
Replies
9
Views
422
Replies
4
Views
1K
Replies
4
Views
1K
  • Differential Geometry
Replies
17
Views
3K
Replies
4
Views
2K
  • Differential Geometry
Replies
7
Views
2K
  • Differential Geometry
Replies
1
Views
1K
  • Differential Geometry
Replies
4
Views
2K
  • Differential Geometry
Replies
1
Views
1K
Replies
12
Views
2K
Back
Top