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