# Expected value of $(1 - X)^{-2}$ over Haar measure of the unitary group, $X \in U(N)$

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?

• are you interested in asymptotics or in exact expressions? Dec 31, 2013 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 $U(N)$. Dec 31, 2013 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 $1$? Dec 31, 2013 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?" Dec 31, 2013 at 23:46
• 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. Jan 1, 2014 at 17:13

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.