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Let $\lambda_1, \dots, \lambda_n$ be the eigenvalues of a random Unitary matrix. I am interested in the expected value:

$$\mathbb{E}_{X \in U(N)}\left[ \prod_{i=1}^n \frac{1}{(1 - \lambda_i)^2}\right]$$

Symmetric functions of eigenvalues can be thought of as characters as some representation, $\rho$ and we are counting how many copies of the identity representation in there:

$$ \rho = 1 \oplus \dots $$

It looks like my product is the resolvent of the regular defining representation.

$$ \det_R \ (1-X)^{-2} $$

Can we evaluate this matrix integral directly using properties of the regular representation or is it easier to use the Weyl integration formula?

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john mangual
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  • are you interested in asymptotics or in exact expressions? – ofer zeitouni Dec 31 '13 at 21:42
  • @oferzeitouni asymptotics are okay. i thought it could be tractable since I am only counting the copies of the trivial representation $\mathbf{1}$. The original question was phrased in terms of generating functions with no mention of representation theory representation theory. $$ \int_{\mathbb{T}n} \prod{i=1}^n \frac{dx_i}{2\pi i x_i} \frac{1}{(1 - x_i)^2} \prod_{1 \leq i < j \leq n} \frac{1}{ x_i - x_j} $$ Clearly they are integrating along the maximal torus of $U(N)$. – john mangual Dec 31 '13 at 23:17
  • The weights of the powers of the defining representation (which is not the regular representation in any sense!) are all nontrivial, so isn't the answer just $1$? – Qiaochu Yuan Dec 31 '13 at 23:25
  • @QiaochuYuan Maybe I got it wrong? I said it was the resolvent of the regular representation. And then you take the determinant of that - which is the product of the eigenvalues. If we set $x = e^{2\pi i t}$ we get the Weyl integration formula, where $$ f(x) = \prod_{i=1}^n \frac{1}{(1 - x_i)^2}$$ So then I asked myself "Which representation is this?" – john mangual Dec 31 '13 at 23:46
  • I don't understand what you mean by the resolvent of a representation. – Qiaochu Yuan Jan 01 '14 at 00:24
  • @QiaochuYuan By resolvent I just mean $(1 - X)^{-1}=1 + X + X^2 + \dots$, let $X = \rho(G)$ where $\rho:G \to \mathrm{GL}(V)$ is a representation. Definitely mis-using the word resolvent here, but can't think of a better word for it. – john mangual Jan 01 '14 at 00:39
  • Where $X$ is the defining representation, right? The regular representation of a group $G$ is a representation on $L^2(G)$ which is not the representation you consider here. – Qiaochu Yuan Jan 01 '14 at 01:17
  • I suspect the integral you wrote diverges. That is, the expectation of $\prod_i 1/|1-\lambda_i|^2$ diverges. You can check it already for $n=2$. Did you mean to take principal values, or somehow use cancellations to define the integral? – ofer zeitouni Jan 01 '14 at 06:25
  • Also, isn't there a discrepancy between the power in the title and what you write in the question? – ofer zeitouni Jan 01 '14 at 06:27
  • @oferzeitouni Sharp eye! I took the integral from a paper by D. Zeilberger it says $$ \int_{\mathbb{T}n} \prod{i=1}^n \frac{dx_i}{2\pi i x_i} \frac{1}{(1 - x_i)^2} \prod_{1 \leq i < j \leq n} \frac{1}{ x_i - x_j} = \prod_{i=1}^n \frac{1}{i+1} \binom{2i}{i}$$ I am learning if we integrate with respect to Haar measure there would be an extra factor of the Vandermonde determinant. $$\int_{\mathbb{T}n} \prod{i=1}^n \frac{dx_i}{2\pi i x_i} \frac{1}{(1 - x_i)^2} \prod_{1 \leq i < j \leq n} \frac{1}{ (x_i - x_j)^{\color{red} 2}} $$ – john mangual Jan 01 '14 at 14:35
  • @oferzeitouni Browsing through random matrix notes here and here, this might be related to orthogonal matrices in $SO(n)$ rather than unitary $SU(n)$. – john mangual Jan 01 '14 at 15:25
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    Well, if you mean SO(n), please state so in the question, then asymptotics can be computed; of course, if n is odd in the SO(n) case then you have an eigenvalue at $1$ and the integral is meaningless. In any case, your computation above is wrong - you should not divide by vandermonde, you should multiply by vandermonde (Hint: Zelberger never integrates in his note). But this Vandermonde correction does not resolve the singularity in any case. So I am still puzzled as to what is the precise, correct question. – ofer zeitouni Jan 01 '14 at 17:13
  • @oferzeitouni Oops... $SU(n)$ Haar measure is $\Delta^2$ while Zeilberger has $\Delta^{-1}$. I have written an interesting quantity different from Zeilberger's. He does not integrate, he cites Morris identity, basically Selberg integral.

    What have I written then? Does $\displaystyle \prod_{1 \leq i < j \leq n} \frac{1}{(x_i - x_j)^2}$ have a matrix meaning, or is it easier to use residue calculus directly at this point?

     – john mangual Jan 01 '14 at 17:47
  • The inverse of vandermonde does not have an RMT interpretation as far as I know. Analytically, it is of course not integrable. In Selberg's integral and it's variants, the negative power of the Vandermonde is strictly smaller than 1. – ofer zeitouni Jan 01 '14 at 17:52

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The expectation of $\det(1-A)^s$ for Haar-distributed $A$ on $U(N)$ has been computed exactly, as a holomorphic function of $s$, first by Keating and Snaith (Comm. Math. Phys. 214, 2000) using the Selberg integral (there have been a number of different proofs since then, with a representation-theoretic one due to Bump and Gamburd, although it considers $s=2k$ and then performs analytic continuation). The result is in terms of products and ratios of gamma functions. There are poles at $-1$, $-2$, $\ldots$ (and the integral stated does indeed diverge), and the asymptotic behavior with respect to $N$ is relatively well-understood in terms of the Barnes $G$-function.

I think if you look at direct sums of powers of the standard representation to determine (heuristically at least...) this integral, it should be relatively easy to see that it contains infinitely many copies of the trivial representation.

Denis Chaperon de Lauzières
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