Conference rusure::math

>	The special orthogonal group, SO(2n), of Euclidean 2n 
>	space, R^(2n), is homeomorphic to product of n copies of the 
>	(2n-1)-sphere, S^(2n-1).

	Irrespective of the connectivity issues which Franklin correctly 
raises, the conjuecture didn't seem plausible to me, although I haven't got the
Topological Group Theory to prove it.   Intuitively, SO(2n) seems much
"richer" than n copies of S^(2n-1).


	Here's another conjecture:

	SO(n) / S^(n-1) is homeomorphic to SO(n-1)

	I've a feeling that this is false, that the special orthogonal groups
have fundamental groups which are alternately trivial and C2, as n is 
alternately odd and even.   So maybe you need to modify this to say that we're 
just considering the components which contain the identity in trying to 
establish this result.   [This is essentially an analogous modification to the 
one that Frnaklin suggested to Jim's original allegation.]

	Alternatively, we could conjecture:

	{SO(n) / S^(n-1)} / S^(n-2) is homeomorphic to SO(n-2)

as something which could hold for all n.   Here the topological dimensions and
the fundamental groups will match.


	I don't know *anything* about this area, I'm just trying to guess
intelligently :-).

Andrew.

           <<< Note 1478.1 by HERON::BUCHANAN "object occidented" >>>

>>	The special orthogonal group, SO(2n), of Euclidean 2n 
>>	space, R^(2n), is homeomorphic to product of n copies of the 
>>	(2n-1)-sphere, S^(2n-1).

>	Irrespective of the connectivity issues which Franklin correctly 
> raises, the conjuecture didn't seem plausible to me, although I haven't got the
> Topological Group Theory to prove it.   Intuitively, SO(2n) seems much
> "richer" than n copies of S^(2n-1).

    You're intuition is better than mine, I guess :-(

>	Here's another conjecture:
>
>	SO(n) / S^(n-1) is homeomorphic to SO(n-1)

    Actually, this is close but doesn't make real sense - an n-sphere isn't
    a "group".

    What is true is that the manifold of SO(n)/SO(n-1) can be identified with
    the n-1 sphere.  This leads to a very nice way of choosing a random
    N-dimensional orthogonal rotation matrix uniformly with respect to Haar
    measure:  Select a set of random k dimensional unit vectors (using, say,
    the usual trick of normalizing k-dimensional normal variates) and then
    use Householder reflections to successively build up SO(k) from
    SO(k-1) and your unit k-vector.

    More generally, SO(n)/SO(n-k) is known as a Steiffel manifold -
    the manifold of orthonormal k-frames in n-space.

    Even more generally, the space of k-dimensional subspaces (relaxing
    the orthogonality of the basis vectors) is known as a Grassman
    manifold and is SO(n)/(SO(n-k)*SO(k)).

    The best reference I know of as introduction to this is, oddly enough,
    a paper by a statistician: A.T. James, "The Normal Distribution and
    the Special Orthogonal Group".  It appeared in 1949 in the American
    Journal of Statistics.

    A great book about this stuff is "Spaces of Constant Curvature" by
    Joseph Wolf.  J. F. Adams little book "Lecturs on Lie Groups"
    is another really nice introduction...

    There's a beautiful blend of analysis, geometry, topology as well
    as some combinatorics in this area.  I wish I had more time and
    talent to understand more than I do...

    - Jim

>    You're intuition is better than mine, I guess :-(

	You flatter me.   The intuitions were larded with misunderstandings
in .1, which I would replace if you hadn't already replied to it.


>    Actually, this is close but doesn't make real sense - an n-sphere isn't
>    a "group".

	Yup.   Franklin gave group action to S^0, S^1, S^3 & I unthinkingly
extended it to all S^n.   Thanks for re-expressing the statement in a correct
way.


Thanks,
Andrew.