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  <b.valkenburg@irl.cri.nz> 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.