2
$\begingroup$

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?

Improve this question
$\endgroup$
14
  • $\begingroup$ are you interested in asymptotics or in exact expressions? $\endgroup$
    – ofer zeitouni
    Commented Dec 31, 2013 at 21:42
  • $\begingroup$ @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 $U(N)$. $\endgroup$
    – john mangual
    Commented Dec 31, 2013 at 23:17
  • $\begingroup$ 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 $1$? $\endgroup$
    – Qiaochu Yuan
    Commented Dec 31, 2013 at 23:25
  • $\begingroup$ @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?" $\endgroup$
    – john mangual
    Commented Dec 31, 2013 at 23:46
  • 1
    $\begingroup$ 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. $\endgroup$
    – ofer zeitouni
    Commented Jan 1, 2014 at 17:13

1 Answer 1

Reset to default
2
$\begingroup$

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.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .