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Before you jump on the "duplicate" buttom, let me say that I do not want to hear about Weingarten calculus and I do not want to see a character of the symmetric group.

What I would like is a formula for the (normalized) Haar measure integral $$ \int_{U(N)} g_{i_1 j_1}\cdots g_{i_n j_n} {\bar{g}}_{k_1 l_1}\cdots {\bar{g}}_{k_n l_n}\ d\mu(g) $$ of the form $$ \left.\mathcal{D}\ g_{i_1 j_1}\cdots g_{i_n j_n} {\bar{g}}_{k_1 l_1}\cdots {\bar{g}}_{k_n l_n}\right|_{g=\bar{g}=0} $$ where $\mathcal{D}$ is an explicit constant coefficient differential operator of infinite order. Of course, in this formula the $g$'s and $\bar{g}$'s are treated are $2N^2$ completely unrelated formal variables.

As an example of what I would like, in the case of $SU(N)$ and $\bar{g}$-free monomials $$ \mathcal{D}=\sum_{n=0}^{\infty} \frac{0!1!\cdots (N-1)!}{n!(n+1)!\cdots(n+N-1)!}\ ({\rm det}(\partial g))^n $$ works.

As per the "Additional remark" in my second answer to this MO question, I had a vague recollection of seeing a math-physics paper with such a formula, but maybe my memory is faulty. So I think it's better to ask the experts.

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    $\begingroup$ I remember this arxiv.org/pdf/hep-th/9209083v2.pdf but the difference is that instead of the Haar measure there is an Itsykson-Zuber exponent, too. Then however you can apply differential operator to M or N to get "correlation functions" and then set M=N=0. Is that helpful? $\endgroup$
    – Leonid Petrov
    Commented Dec 1, 2016 at 0:31
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    $\begingroup$ :)- why ask on MO when I could have walked down the hallway and knocked on your door...Thanks this looks really interesting. $\endgroup$
    – Abdelmalek Abdesselam
    Commented Dec 1, 2016 at 0:46
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    $\begingroup$ :) the paper does not use differential operators but with them the statement hopefully can be obtained easier. I have not checked details though. $\endgroup$
    – Leonid Petrov
    Commented Dec 1, 2016 at 0:50
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    $\begingroup$ and here arxiv.org/pdf/hep-th/0502041.pdf in introduction there is some discussion 10 years later $\endgroup$
    – Leonid Petrov
    Commented Dec 1, 2016 at 0:51
  • $\begingroup$ Still worth adding a formal answer to this question, even if it's been sorted! $\endgroup$
    – David Roberts ♦
    Commented Dec 1, 2016 at 5:46

1 Answer 1

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To expand on my comments, this paper https://arxiv.org/pdf/hep-th/9209083v2.pdf by Shatashvili deals with ``correlation functions'' of Haar unitary matrices of the form $$ \int_{U(N)}^{} d\mu(U) e_{}^{tr(UAU_{}^{-1}B)} U_{i_1j_1}^{}\bar U_{k_1\ell_1}^{}\ldots U_{i_mj_m}^{}\bar U_{k_m\ell_m}, $$ and provides a certain combinatorial formula for these. Then setting $A=0$ would probably recover what you're asking about.

The same correlation functions (and an alternative formula for them) are also discussed in https://arxiv.org/pdf/hep-th/0502041.pdf

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