From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: polar transformation for integration over R^{n x p} Date: 28 Jul 1998 04:00:58 GMT In article <35B41254.26BBD52E@irl.cri.nz>, Bob Valkenburg wrote: >I am interested in performing an integral over the nxp real matrices >(n>=p). I would like to know if there is a simple analog of the polar >transformation for the volume form. > > >When p=1 then the integral is over R^n. If x = (x1 ... xn)^t then >the polar decomposition is, > >x = t r where t = x/|x| and r = |x| > >and the volume form is given by; > >dx = dx1^dx2^ ... ^dxn > = r^(n-1) dr^dt > >where, >| . | - matrix norm >^ - wedge operator >dt = the volume form on S^(n-1) = O(n,1) the unit sphere in R^n > > >Is there an analog for nxp matrices ?, > >i.e. If X is an nxp matrix (n>=p) and > >X = T R is the polar decomposition >where, > T is nxp orthogonal T'T = Ip > R is pxp symmetric positive semidefinite (s.p.s.d.) > >then is there a simple analog. Something like, > >dX = g dH^dT > >where, >g is some sort of gram determinant (to be specified). >dH is the volume form over the pxp s.p.s.d. matrices. >dT is the volume form over the nxp orthogonal matrices O(n,p) (the n-p >Stiefel manifold). This is not really my baliwick, but as no one else is responding I'll muddle through to arrive at what I'm pretty sure is the right answer: dX = (detR)^((2n-p-1)/2) dH dT. Of course you can change coordinates, at least when integrating over regions where the map (T,R) -> T.R is a local homeomorphism. You just want to know the volume element, that is, the determinant of the appropriate Jacobian to use in the Change-of-Variables theorem (your "g"). Since the measure on M(n,p) (the set of n x p matrices) is invariant under O(n), this g is independent of T. Since the measure is also invariant under O(p), g depends only on the eigenvalues of R -- more precisely it must depend on their symmetric functions, hence the characteristic polynomial of R. Unfortunately, I cannot recall the similar 1-sentence explanation as to why in fact g depends only on det(R), but I'm pretty sure that's true: g(T,R) = g(det R) for some g: R-> R. It's then easy to determine what g must be. Let S1 and S2 be compact subsets of the Stiefel manifold and the symmetric space, respectively. What is the volume of the set { T.(cR) ; T in S1, R in S2 } where c is a fixed real number? Well, on the one hand, this is just a scaling by c, whose effect on M(n,p) is to multiply volume by c^(n*p). On the other hand, this volume should be found by integration: it's int_S1 int_(c S2) g dR dT. Now, a simple change of variables on R allows us to write this as int_S1 int_S2 g( c^p detR ) (c^(p+1 choose 2) dR) (dT) So, whatever the set S2, the value of int_S2 g(c^p detR) c^((p+1 choose 2) - n*p) dR should be independent of c. Taking sets S2 which diminish to a point, we conclude g(c^p detR) c^((p+1 choose 2) - n*p) depends on R but not on c. Writing d = c^p, this says g(d detR) = d^( n - (p+1)/2 ) * g(det R) i.e., g is homogeneous, and in so in fact g(detR) = (detR)^( n - (p+1)/2 ). So I believe the correct volume element is dX = (detR)^((2n-p-1)/2) dR dT = (det R'R)^((2n-p-1)/4) dR dT = (det X'X)^((2n-p-1)/4) dR dT according to taste. Note that when p=1 this gives dX = r^(n-1) dr dT as you noted. Also when p = n, it gives dX = (det X)^((n-1)/2) dR dT which I believe I have seen in Lie group texts as a formula for the measure on GL(n). This post should be sufficient incentive for someone who does this regularly to step in and correct me :-) . dave >P.S. I am new to using news groups so could you please email >responses to me directly (aswell). I hope this was a swell response.