most recent 30 from http://mathoverflow.net 2013-05-26T05:14:10Z http://mathoverflow.net/feeds/question/78188 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/78188/decomposition-of-haar-measure-other-than-hurwitzs John Jiang 2011-10-15T05:02:36Z 2011-10-15T19:41:32Z

<p>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 $R(1,2;\theta_1)R(2,3;\theta_2) \ldots R(n-1,n;\theta_{n-1}) R(2,3;\theta_n) \ldots R(n-1,n;\theta_{\binom{n}{2}})$ 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 Saloff-Coste: www-stat.stanford.edu/~cgates/PERSI/papers/kac10.pdf</p>

http://mathoverflow.net/questions/78188/decomposition-of-haar-measure-other-than-hurwitzs/78227#78227 Carlo Beenakker 2011-10-15T19:41:32Z 2011-10-15T19:41:32Z

<p>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.</p>