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.

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.

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.

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.

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 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.

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.