Timeline for Computing Haar measure of matrices sampled from SO(n)
Current License: CC BY-SA 4.0
13 events
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| Aug 21 at 6:04 | comment | added | Carlo Beenakker | a small point: you remark that "Restricting to 𝑆𝑂(𝑛) removes the peaks at −1 and +1"; that is true for even $n$; for odd $n$ a peak a +1 remains. | |
| Aug 20 at 21:52 | comment | added | Yaroslav Bulatov | @CarloBeenakker I see, that gives me uniform eigenphase distribution with complex-valued matrices. It seems there's a way to get uniform eigenphase dist with real-valued matrices by combining $n/2$ 2D rotations using Jordan Normal form, was curious if that density has a name | |
| Aug 20 at 20:34 | comment | added | Carlo Beenakker | circular unitary ensemble? [Haar measure on $U(n)$] | |
| Aug 20 at 19:12 | comment | added | Yaroslav Bulatov | @CarloBeenakker btw, is there a name of the density where the angles are actually uniform? Posted question with simulations here | |
| Aug 20 at 19:00 | history | edited | Yaroslav Bulatov | CC BY-SA 4.0 |
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| Aug 20 at 17:17 | comment | added | Yaroslav Bulatov | @CarloBeenakker thanks for the catch, updated the code which now agrees with the formula | |
| Aug 20 at 17:16 | history | edited | Yaroslav Bulatov | CC BY-SA 4.0 |
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| Aug 20 at 17:02 | history | edited | Yaroslav Bulatov | CC BY-SA 4.0 |
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| Aug 20 at 5:43 | comment | added | Carlo Beenakker | there is a typo in your code; for SO(n) you are still sampling from $n=2$; if you correct that line (replace SO@2 with SO@n) you will find the double-peaked distribution $$p(\theta)=\frac{\cos 2 \theta+2}{4 \pi }.$$ | |
| Aug 19 at 21:29 | comment | added | Yaroslav Bulatov | @CarloBeenakker I've updated code/diagram for $n=4$, eig dist appears uniform with 100k samples of SO(4) | |
| Aug 19 at 21:29 | history | edited | Yaroslav Bulatov | CC BY-SA 4.0 |
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| Aug 19 at 19:34 | comment | added | Carlo Beenakker | do you agree that the uniformity of the eigenvalues of SO($n$) only holds for $n=2$? for $n=4$ you get the double peaked function in my answer, right? | |
| Aug 19 at 17:46 | history | answered | Yaroslav Bulatov | CC BY-SA 4.0 |