Timeline for Expectation of trace of nth power of unitary matrices
Current License: CC BY-SA 3.0
8 events
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| Jan 30, 2015 at 21:10 | comment | added | David E Speyer | @JoonasIlmavirta You're welcome! And I note also that what I wrote in the comment above isn't quite right either: $(S \exp(iH))^m = S^m \exp(i H) \exp(i S^{-1} H S) \cdots \exp i S^{-(m-1)} H S^{m-1})$, but this is only equal to $S^m \exp(i (H + S^{-1} H S + \cdots + S^{-(m-1)} H S^{(m-1)})$ up to $O(H^2)$. Fortunately, that's good enough for the first order computation that we care about. | |
| Jan 30, 2015 at 20:14 | history | edited | Joonas Ilmavirta | CC BY-SA 3.0 |
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| Jan 30, 2015 at 20:10 | comment | added | Joonas Ilmavirta | @DavidSpeyer, ah, thanks! I had left my original comment too hastily but it's good I wrote this answer to set things right. (Attempts to save bad ideas can teach, but sometimes in a humiliating way...) The argument seems ok if $p_m$ is replaced with a homomorphism, but this fails. | |
| Jan 30, 2015 at 18:03 | comment | added | Terry Tao | Basically, the problem is that $p_m$ is not a homomorphism: $(xy)^m \neq x^m y^m$ in general for nonabelian groups. | |
| Jan 30, 2015 at 16:11 | comment | added | David E Speyer | Basic computation: Let the source matrix be $S$. We can write matrices near $S$ as $S \exp(i H)$, where $H$ is a small Hermitian matrix. Then $(S \exp(i H))^m = S^m \exp(i \left( H + S^{-1} H S + S^{-2} H S^2 + \cdots + S^{-(m-1)} H S^{m-1} \right)$. We just have to find the determinant of the linear map $H \mapsto H + S^{-1} H S + S^{-2} H S^2 + \cdots + S^{-(m-1)} H S^{m-1}$, from Hermitian matrices to Hermitian matrices. | |
| Jan 30, 2015 at 15:58 | comment | added | David E Speyer | Letting the source matrix have eigenvalues $e^{i \theta_1}$, $e^{i \theta_2}$, ..., $e^{i \theta_n}$. I compute that the $m$-th power map multiplies Haar measure by $\prod_{j=1}^n \left( \frac{e^{i m \theta_j}-1}{e^{i \theta_j}-1} \right)^n$. | |
| Jan 30, 2015 at 15:56 | comment | added | David E Speyer | I don't think this is right. Taking $m$-th powers IS surjecive from $U(n)$ to $U(n)$ because any unitary can be diagonalized and any diagonal unitary has a diagonal unitary $m$-th root. It just isn't true that the squaring map pulls back Haar measure to Haar measure. (I notice that I am saying pull back and you are saying pushforward; I'm not sure if this is relevant. I am saying pullback because I am thinking of Haar measure as a volume form.) | |
| Jan 29, 2015 at 16:12 | history | answered | Joonas Ilmavirta | CC BY-SA 3.0 |