From: Boudewijn Moonen <Boudewijn.Moonen@ipb.uni-bonn.de>
Subject: Re: Stiefel Manifold is not a symmetric space
Date: Tue, 16 Mar 1999 10:50:57 +0100
Newsgroups: sci.math.research

Alan Edelman wrote:

> I would like to clearly understand why the real Stiefel manifold
> O(n)/O(n-p)  2<p<n-1 is not a symmetric space.

First, you should specify whether you want to consider it as an
affine or Riemannian symmetric space. So you have to specify with
regard to which affine connection or which Riemannian metric you
want to consider the question. Let us suppose you mean Riemannian
symmetric with respect to the natural metric induced from the negative
Killing form on O(n).

> I *don't* want to hear anything about curvature or Lie brackets
> of [H,V] and [V,V]. I consider that cheating!  :-)

You are free to do so, but you're wrong. A symmetric space, or more
general a homogeneous space with an invariant metric, has a transitive
Lie group of of isometries. So it is natural, and unavoidable,
to use this structure in the description and investigation of the
space. And a Lie group is, inevitably, almost completeley encoded
in Lie algebra terms. So it is natural, and unavoidable, to use
Lie brackets in the description and investigation of the space.
So no cheat whatsoever. And, since in the situation in question
being a symmetric space or not means to have a parallel transported
curvature tensor or not, it is a cheat not wishing to talk about
curvature (as it is, by the way, to try to understand any given
Riemannian geometry without talking about curvature. Look at Riemann's
habilitation lecture).

>I want to understand the map that sends exp(X)I into exp(-X)I,
> where X=[ A  B]  and I is the nxp identity.
>                 [-B' 0]
> Actually I have little clue if given a set of p orthogonal vectors, how
> I can numerically calculate this involution.

I do not completely understand the notation, but I guess you want to
understand the geodesic reflections in the points of the Stiefel
manifold. For this, you will have to describe the geodesics. I think
with respect to this you will be going to have a hard time if you do
not want to make use of those techniques that you consider to be a
cheat and which would tell you that, as I presume (terminology
according to Kobayashi-Nomizu, Vol. II)

    -- the canonical connection has torsion (whence O(n)/O(n-p) cannot
       be symmetric)

    -- the natural torsion-free connection coincides with the Levi-Civita
       connection, whence the geodesics are induced by the exponential map

       of O(n)

I could try to make this more precise, but since you said you didn't want
to hear those things I will abstain from doing so.

> Question 1: Does this map preserve geodesics?

Presumably only those passing through the point of reflection, the radial
ones. Otherwise the map would be an isometry and the Stiefel a symmetric
space

> Question 2: If the answer to question 1 is yes, does the map preserve
>             parallel transport?

The answer is no, but even if it were yes preservation of parallel
transport
presumably implies the Stiefel would be symmetric


> Question 3: If the answer to question 2 is yes, what is the
>             simplest quantity that is not preserved thereby
>             keeping it from being a symmetric space?

The answer is no, nevertheless there may be quantities not being
"preserved", but what do you mean by "preserved"? If this is going
to mean "parallel transported", the simplest quantity in my eyes
not being preserved is an object you don't want to see -- the
curvature tensor. If this is going to mean "invariant under
the geodesic symmetries", the simplest quantity in my eyes is
the metric itself, since these symmetries are not isometric. But
to prove this might confront you e.g. with the Jacobi equation,
and then you are going to meet unpleasant stuff again, the curvature
tensor, the bracket and all that.


> The way I figure it, this is the geometrical symmetry that
> has been abstracted over the years -- I want to get in touch
> directly with that geometrical symmetry not the abstractions that
> are in the modern texts.

It has not been abstracted over the years, these "abstractions" are
an adequate description.  In Riemannian geometry there is as much
"geometry" (whatever this means)  as there is in the description ]
of the fundamental object of the theory,  the  Riemannian metric
itself. This is an analytic or algebraic object, and so you have
to make contact with analysis or algebra when trying to understand
the geometry. In particular, in the case of symmetric spaces geometry
IS Lie group theory, in the spirit of Klein. And even Euclidean
geometry is the study of the orthogonal group, maybe in disguise.

> Ultimately what makes a symmetric space "symmetric"?

The fact that the geodesic symmetries are isometries. Or, equivalently,
(in the simply connected case) the torsion tensor vanishes and the
curvature tensor is parallel transported. This equivalence rests
upon the Jacobi equation.

--
    Boudewijn Moonen
    Institut fuer Photogrammetrie der Universitaet Bonn
    Nussallee 15

    D-53115 Bonn

    GERMANY

    e-mail: Boudewijn.Moonen@ipb.uni-bonn.de
    Tel.:   GERMANY +49-228-732910
    Fax.:   GERMANY +49-228-732712