From: "j.e.mebius" Subject: Re: Haar measure and rotation group - reply Date: Mon, 06 Dec 1999 13:48:52 +0100 Newsgroups: sci.math Delft - December 6th 1999 Dear ms Van Lith, mr Rubin and other interested persons, No final solution, but some hints on the Haar measure of SO(n, R): (1) Read Euler's article "Problema algebraicum (...) memorabile" (1770; L.Euleri Opera Omnia 1st series, vol 6, p287-315). Euler parametrises SO(3) by means of 3 angles and explains how to parametrise SO(n) for n > 3. (2) Then read the book "Representations of the rotation and Lorenz groups (...)" by I.M.Gel'fand (1963), in particular the explicit computation of the Haar measure of the entire SO(3). (3) In 4D we have SO(4) ~= S3 x S3 / (I, -I); the Haar measure of S3 is known to be 2*pi^2*r^3. Conjecture: is vol (SO(4)) = (2*pi^2*r^3)^2 / 2? Question: How does one attribute a radius to an SO(n) manifold? I do not know the volume of SO(3) by heart, but would it be pi^2*r^3? (because of SO(3) ~= S3 / (I, -I)) All of these things only valid if one can tell what the radius of an SO(n) manifold is. Good luck: Johan E. Mebius (mailto:j.e.mebius@twi.tudelft.nl) ========= reply to ======== Janneke wrote: > Herman Rubin wrote in message > news:80uq2u$41ks@odds.stat.purdue.edu... > > In article <80u4lb$mur$1@pukkie.phys.uu.nl>, > > Janneke wrote: > > >Hi, > > > > >Can anyone help me with the following: > > > > >Any rotation in 3D can be characterized by (\theta,\phi,l): a rotation by > l > > >around the axis (\theta,\phi) in polar coordinates. > > >The Haar measure on the rotation group in 3D can then be written as > > >dm_H=dcos(\theta) d\phi dl > > >where dcos(\theta) d\phi equals the Lebesgue measure on the surface of a > > >sphere and dl is the Lebesgue measure on the interval (0,pi). > > > > >I am interested in the n dimensional case. What is the Haar measure on > the > > >nD rotation group? Is there a connection with Lebesgue measure on the > > >surface of the hypersphere? Can a general nD rotation be written as a > > >product of rotations around fixed axes (like the Euler angles in 3D)? > > > > The confusion arises from the special properties of three > > dimensions. It is not the axis of rotation which is the > > important item, but the plane in which the rotation occurs > > which is the important item. > > > > The only things needed to handle the problem are the > > reduction to appropriate diagonal form of orthogonal > > matrices and the invariance of Haar measure. Any > > orthogonal matrix of determinant 1 can be reduced by an > > orthogonal transformation to a matrix in which there are > > either 1's or 2x2 rotation matrices on the diagonal. For > > almost all matrices, these rotation matrices will be > > non-trivial and distinct, and there will be a 1 only > > if the dimension is odd. > > > > So one has to select n/2 or (n-1)/2 orthogonal planes; > > this can be done by selecting points at random on the > > sphere, each orthogonal to the previous ones, and using > > pairs to define the planes. In 3 dimensions, the odd > > point can be selected first, and when there are only > > two left, the plane is determined. > > > > Again, by the invariance of Haar measure, the angles > > of rotation in the planes must have a uniform distribution. > > > > So, just by using the definitions, we have the > > representation. > > I agree that any rotation can be transformed to the block-diagonal form you > described. But the basis will be different for different rotations. Thus, I > don't think the random selection of points will do. > > We could restrict attention to the subgroup of rotations that are in > block-diagonal form w.r.t. a fixed basis. Then a rotation is completely > specified by the (n/2) or (n-1)/2 angles, and the Haar measure on this > subgroup is just the product measure of uniformly distributed angles. But > all subgroups of this form are negligable in terms of the Haar measure on > the full rotation group. I still don't see how to represent this full Haar > measure. > > Janneke [HTML deleted --djr]