Timeline for Poisson kernel for the orthogonal groups

Current License: CC BY-SA 4.0

16 events

when toggle format	what		by	license	comment
May 14 at 18:42	history	bounty ended	thedude
May 14 at 18:42	vote	accept	thedude
May 14 at 6:14	comment	added	Carlo Beenakker		I have restricted the answer to ${\rm SO}(N)={\rm O}_+(N)$; for ${\rm O}_-(N)$ the reproducing property fails even for the Haar measure, so for $Z=0$, since $\int_{{\rm O}_-}X^2 d\mu(X)\neq 0$, while $Z^2=0$.
May 14 at 6:04	history	edited	Carlo Beenakker	CC BY-SA 4.0	restricted to SO(N); the reproducing property does not hold even for scalar Z when det O = -1
May 13 at 20:52	history	edited	Carlo Beenakker	CC BY-SA 4.0	reproducing property only holds for scalar Z
May 13 at 20:46	comment	added	Carlo Beenakker		yes, the orthogonal group is disconnected, you average over either sector; I am unsure about the expansion you mention; one complication is that for the orthogonal group the reproducing property only holds for $Z$ proportional to the unit matrix.
May 13 at 20:44	history	edited	Carlo Beenakker	CC BY-SA 4.0	reproducing property only holds for scalar Z
May 13 at 20:42	comment	added	thedude		Also, have you ever seen a discussion about expanding this kernel as an infinite series using irreducible characters?
May 13 at 20:41	comment	added	thedude		By $O_\pm(N)$ you mean I take either the +, i.e. $SO(N)$, or the $-$?
May 13 at 18:47	history	edited	Carlo Beenakker	CC BY-SA 4.0	checked reproducing property for N=2
May 13 at 18:38	history	edited	Carlo Beenakker	CC BY-SA 4.0	deleted 225 characters in body
May 13 at 18:18	history	edited	Carlo Beenakker	CC BY-SA 4.0	reproducing property
May 13 at 16:40	history	edited	Carlo Beenakker	CC BY-SA 4.0	added 49 characters in body
May 13 at 16:17	history	edited	Carlo Beenakker	CC BY-SA 4.0	added 812 characters in body
May 13 at 12:54	history	edited	Carlo Beenakker	CC BY-SA 4.0	added 4 characters in body
May 13 at 12:49	history	answered	Carlo Beenakker	CC BY-SA 4.0

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