# Show me a book about hyper-sphere (n-sphere)

1. Mar 21, 2012

### thanhsonsply

I find that group structure of n-sphere is SO(n+1)/SO(n) (at http://en.wikipedia.org/wiki/N_sphere). So I want to find a book show that.

Please, show me a title of book and author. Thanks!

2. Mar 21, 2012

### mathwonk

many spaces can be realized as a group acting on a set, modulo a subgroup fixing certain points. I.e. if a group acts on a set transitively, and a point has a certain subgroup fixing it, then the set is the group mod the subgroup in some sense. probably this is what is going on. rather than read some book, think about this phenomenon.

e.g. is the sphere in three space somehow equal to the rotation group on three space, modulo the subgroup of rotations fixing one point of the sphere? i.e. is it SO(3)/SO(2)?

I'm not saying it is, just that you will learn more by deciding it for yourself.

3. Mar 22, 2012

### quasar987

John Lee's Introduction the Smooth manifold. Look for one of the examples in chapter 9.

4. Mar 22, 2012

### homeomorphic

Most spheres are not Lie groups. You could always turn the underlying set into a group, but then there's no point in it being a sphere.

I think this has something to do with division algebras, maybe. I think if it were a Lie group, maybe you could take the group algebra and get a division algebra. Division algebras only exist in certain dimensions: 1, 2, 4, and 8. So you get spheres of dimension 1 less: 1, 3, and 7. And 7 doesn't work because the octonians, the dimension 8 division algebra, are non-associative, so you can't get a group by taking the unit octonians.

So the only spheres that are groups are S^1 and S^3.

Don't quote me on this, though, since I haven't quite thought it through. Just came up with it off the top of my head. I think it's right, though.

The end result is correct, anyway:

http://en.wikipedia.org/wiki/3-sphere

5. Mar 22, 2012

### morphism

You kind of have it the other way around. If we can endow $\mathbb R^n$ with some "sufficiently nice" multiplication operation, then $S^{n-1}$ will be a group. But the converse isn't clear - why can't we have a group operation on $S^{n-1}$ that doesn't necessarily extend to a nice multiplication on $\mathbb R^n$?

Anyway, the OP should really try to understand mathwonk's post, which is on the money.

6. Mar 22, 2012

### homeomorphic

It isn't clear, but according to wikipedia and probably some other sources I remember, it is true that only S^1 and S^3 are Lie groups. So, my speculation was that given a group operation on S^n-1, you can construct a division algebra of dimension n somehow. But it's not clear how. The wikipedia article also seems to indicate that the proof should proceed along these lines.

7. Mar 22, 2012

### morphism

Yes, that result is true: S^1 and S^3 (and S^0) are the only spheres that are Lie groups.

Most proofs I know don't mention real division algebras; it typically goes the other way around: once you have this result, you get a restriction on the possible division algebras.

8. Mar 22, 2012

### homeomorphic

That only works for associative division algebras, though, which is a bit unfair to the octonians, but I guess the result is really that those are the only spheres that are H-spaces, which have some kind of product that's not necessarily associative.

It should also be mentioned the they can't even be topological groups (with the usual topology).

This might also be of interest:

http://www.unizar.es/acz/05Publicaciones/Revistas/Revista62/p075.pdf