From: Robin Chapman <rjc@maths.ex.ac.uk>
Newsgroups: sci.math.research
Subject: Re: topology of a manifold
Date: 20 Nov 1998 03:30:02 -0600

In article <3654DF1F.4CF4@worldnet.att.net>,
  Marjorie Piette & Steve Ellis <mapiette@worldnet.att.net> wrote:
> I'd like to know about the topology (fundamental group, in particular)
> of the space of all unordered triples of orthogonal lines (thru the
> origin) in 3-space.  Any leads?  Thanks.

Each of these configurations is the image of the coordinate frame under a
matrix in SO(3). However each such configuration comes from more than one, in
fact 24, such matrices. These form a subgroup of G SO(3) (the octahedral
subgroup) of order 24. Your manifold M is just a coset space of SO(3).

Alas SO(3) is not simply connected (otherwise G would be the fundamental
group of M) but SO(3) has a two-fold cover by SU(2) the group
of unit quaternions which is simply connected. Then we can represent
M as a coset space of SU(2) coming from the subgroup H, the inverse image
of G in SU(2). This group H has order 48, is called the binary octahedral
group and is a non-split extension of Z_2 by S_4. This group H
is the fundamental group of M.

Robin Chapman                           + "They did not have proper
SCHOOL OF MATHEMATICal Sciences         -  palms at home in Exeter."
University of Exeter, EX4 4QE, UK       +
rjc@maths.exeter.ac.uk                  -  Peter Carey,
http://www.maths.ex.ac.uk/~rjc/rjc.html +    Oscar and Lucinda, chapter 20

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