Timeline for Expectation of trace of nth power of unitary matrices
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jan 30, 2015 at 11:30 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
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| Jan 29, 2015 at 16:14 | comment | added | Joonas Ilmavirta | I elaborated on my comment in the form of an answer (it was too long for a comment). The argument itself was ok, but all the assumptions do no hold in the present situation... | |
| Jan 29, 2015 at 16:11 | vote | accept | Atnap | ||
| Jan 29, 2015 at 16:04 | comment | added | Carlo Beenakker | thanks; I just realized that the eigenvalues of $U^m$ become independent for $m\geq n$, so this also indicates that taking the power of a unitary matrix changes the distribution. arxiv.org/abs/math/0008079 | |
| Jan 29, 2015 at 16:01 | comment | added | David E Speyer | The $m$-th power map does not pull back Haar measure to Haar measure on $U(n)$. For example, on $U(2)$, all matrices with eigenvalues $(1,-1)$ has square equal to $\mathrm{Id}$. So the squaring map collapses all these matrices, which form a surface, down to a point. We see that the differential of the squaring map has a kernel on these points, so the pull back of Haar measure along squaring is zero at these points. Also, even if Haar measure pulled back to Haar measure, you would have to multiply by the topological degree of the $m$-th power map, which I believe is $m^n$. | |
| Jan 29, 2015 at 15:40 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
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| Jan 29, 2015 at 15:32 | history | answered | Carlo Beenakker | CC BY-SA 3.0 |