Hurwitz defined a decomposition of the Haar measure on $SO(n)$ based on Given's rotation. So by left multiplication of Givens rotation one can always bring an orthogonal matrix into the identity. The order of multiplication is $R(1,2;\theta_1)R(2,3;\theta_2) \ldots R(n1,n;\theta_{n1}) R(2,3;\theta_n) \ldots R(n1,n;\theta_{\binom{n}{2}})$ where the $\theta$'s are independent and have certain beta distributions (not identically distributed). My question is whether this decomposition holds for other product of Givens rotations. Also if we take uniform iid $\theta_j$'s, is the singularity always on the set where some $\theta_j = 0 $ or $\pi$? This is the case for the Hurwitz decomposition. If not please describe the singular set for iid uniform $\theta_j$'s. For a reference of Hurwitz decomposition, see the paper by Diaconis and SaloffCoste: wwwstat.stanford.edu/~cgates/PERSI/papers/kac10.pdf

$\begingroup$ Some similar sounding question (but may be not that much related): Consider any matrix M and its SVD made with the help of the Jacobi algorithm. We get infinite products of Givens rotations: M = \prod R(a_i, b_i, \Theta_i ) Diag \prod R(a_j, b_j, \Theta_j ). Is it known how Theta_i, Theta_j are distributed depending on distribution on space of matrices M ? This would be helpful for understanding the convergence properties of the Jacobi algorithm... $\endgroup$– Alexander ChervovOct 15, 2011 at 6:40
1 Answer
An alternative to the Hurwitz decomposition of the Haar measure on $SO(n)$ has been developed by Julie Mitchell, "Sampling rotation groups by successive orthogonal images", SIAM J. Sci. Comput. 30, 525–547 (2007). The planar rotations used in this work are similar, but not identical to Given's rotations, which may or may not be what you want.

$\begingroup$ Thank you for the reference. I read the whole thing and I think the things they proved are more or less expected. Although it would be nice to prove similar global covering and separation results for discretization of the Kac random walk generators, i.e., Givens rotations. $\endgroup$ Oct 23, 2011 at 5:56