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I am trying to find the answer of

$$\int dU \ |Tr(U^m)|^2$$

where $m\in\mathbb{N}$ and $U$'s are unitary matrices in $\textit{U}(n)$ and $dU$ is a normalized Haar measure. In the case $m=1$, the answer seems to be $1$.

I don't know where to start. Does anyone have an idea? Is there a clean answer for $m>1$ ?

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    $\begingroup$ If the map $U\mapsto U^m$ is surjective $U(n)\to U(n)$, then your integral doesn't change if you change $m$ to 1. In this case the pullback of the Haar measure over $p_m$ is the original Haar measure. $\endgroup$
    – Joonas Ilmavirta
    Commented Jan 29, 2015 at 14:25
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    $\begingroup$ I'm sorry but I don't understand the second part of your comment ("The pullback of the Haar measure over $p_m$ is the original Haar measure"). I tried to find the answer for $m=2$ with Mathematica and I found the integral equals 2. $\endgroup$
    – Atnap
    Commented Jan 29, 2015 at 15:17

2 Answers 2

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$$\int_{{\rm U}(n)} dU\,|{\rm Tr}\,(U^m)|^2={\rm min}\,(n,m).$$

see Theorem 2.1.b of Diaconis and Evans (2001). [*]

[*] This 2001 reference corrects an earlier paper by Diaconis and Shahshahani (1994), which would have given as an answer $m$ instead of ${\rm min}\,(n,m)$.

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    $\begingroup$ 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 $m^n$. $\endgroup$
    – David E Speyer
    Commented Jan 29, 2015 at 16:01
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    $\begingroup$ 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 $\endgroup$
    – Carlo Beenakker
    Commented Jan 29, 2015 at 16:04
  • $\begingroup$ 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... $\endgroup$
    – Joonas Ilmavirta
    Commented Jan 29, 2015 at 16:14
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This is an elaboration on my comment.

Let's start in a more general setting since the comment was not about the groups $U(n)$: Let $G$ be a compact group and $H$ a normalized Haar measure on it. Let $p_m:G\to G$ be defined by $p_m(x)=x^m$, $m\in\mathbb Z$. Now for any continuous function $f:G\to\mathbb R$ we have $$ \int_Gf(p_m(x))dH(x) = \int_Gf(x)d(p_m)_*H(x). $$ Since $p_m$ is a homomorphism1, the pushforward $(p_m)_*H$ is a normalized Haar measure on $p_m(G)$. If $p_m$ is surjective, it follows from the uniqueness of Haar measures that $p_{m*}H=H$ and thus $$ \int_Gf(p_m(x))dH(x) = \int_Gf(x)dH(x). $$ This idea made me think that the integral would be independent of $m$.

1 The map is not a homomorphism. The argument works (I think) for homomorphisms, but typically $p_m$ is not a homomorphism in a nonabelian group. If $p_m$ is not a homomorphism, the pushforward doesn't have to be a Haar measure and the proof falls apart.

Maybe I'll let this answer stay here as a warning example...

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    $\begingroup$ 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.) $\endgroup$
    – David E Speyer
    Commented Jan 30, 2015 at 15:56
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    $\begingroup$ 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 $\prod_{j=1}^n \left( \frac{e^{i m \theta_j}-1}{e^{i \theta_j}-1} \right)^n$. $\endgroup$
    – David E Speyer
    Commented Jan 30, 2015 at 15:58
  • $\begingroup$ 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. $\endgroup$
    – David E Speyer
    Commented Jan 30, 2015 at 16:11
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    $\begingroup$ Basically, the problem is that $p_m$ is not a homomorphism: $(xy)^m \neq x^m y^m$ in general for nonabelian groups. $\endgroup$
    – Terry Tao
    Commented Jan 30, 2015 at 18:03
  • $\begingroup$ @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. $\endgroup$
    – Joonas Ilmavirta
    Commented Jan 30, 2015 at 20:10

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