TITLE: Nonnegativity of an integral over the unitary group
QUESTION [10 upvotes]: For an $n$-by-$n$ unitary matrix $U$ and a permutation $\sigma\in S_n$, let
$$w_\sigma=(-1)^\sigma\det(U^*)\prod_{i=1}^n U_{i,\sigma(i)}.$$
Is $\int_{U(n)}\mathrm{Re}(w_{\sigma_1})\mathrm{Re}(w_{\sigma_2})dU\ge 0$ for all $\sigma_1,\sigma_2\in S_n$?

The integral can be expanded to
$$\frac{1}{4}\int_{U(n)} (w_{\sigma_1}+\bar w_{\sigma_1})(w_{\sigma_2}+\bar w_{\sigma_2})dU=\frac{1}{2}\int_{U(n)} w_{\sigma_1}w_{\sigma_2}+w_{\sigma_1}\bar w_{\sigma_2}dU.$$
Once the $\det(U^*)$ factors have been expanded as sums over permutations, this can be evaluated using Weingarten functions (see Collins 2003). The latter term is
\begin{align}\int_{U(n)} w_{\sigma_1}\bar w_{\sigma_2}dU&=(-1)^{\sigma_1\sigma_2}\int_{U(n)}U_{1\sigma_1(1)}\cdots U_{n\sigma_1(n)}U^*_{\sigma_2(1)1}\cdots U^*_{\sigma_2(n)n}dU\\
&=(-1)^{\sigma_1\sigma_2}Wg(\sigma_2^{-1}\sigma_1,n).\end{align}
By Novak 2010, the element $\sum_\sigma Wg(\sigma,n)\sigma$ of the group algebra $\mathbb{C}[S_n]$ can be written as a product of elements $(n+J_k)^{-1}$ of the form $\sum_\sigma (-1)^\sigma|a_\sigma|\sigma$, hence $(-1)^\sigma Wg(\sigma,n)\ge 0$. Computer experiments suggest the former term is nonnegative as well, but I haven't been able to prove it.
Note that $\int_{U(n)}\mathrm{Re}(w_\sigma)dU=1/n!$ by a symmetry argument (see here).

I can show that
$$I_{\sigma_1,\sigma_2}=\frac{1}{n!(n+1)!}\sum_{\pi\in C}\sum_{\tau_1,\tau_2\in R}(-1)^{\sigma\pi} [\tau_1\pi\tau_2=\sigma],$$
where $C$ and $R$ are the Young subgroups $\mathrm{Sym}(1,\dots,n)\times\mathrm{Sym}(n+1,\dots,2n)$ and $\mathrm{Sym}(1,n+1)\times\cdots\times\mathrm{Sym}(n,2n)$ respectively and $\sigma=\sigma_1\oplus\sigma_2\in C$. Is there a way to link this with Carlo's formula or otherwise show it's nonnegative?
Proof: By expanding the $\det U^*$ factors and applying Weingarten functions,
\begin{align*}
I_{\sigma_1,\sigma_2}&=(-1)^{\sigma_1\sigma_2}\sum_{\pi_1,\pi_2}(-1)^{\pi_1\pi_2}\int_{U(n)}\prod_{i=1}^n U_{i\sigma_1(i)}U_{i\sigma_2(i)}U^*_{\pi_1(i)i}U^*_{\pi_2(i)i}dU\\
&=(-1)^{\sigma}\sum_{\pi\in C}(-1)^{\pi}\sum_{\tau_1,\tau_2\in R}Wg(\sigma^{-1}\tau_1\pi\tau_2,n)\\
&=(-1)^{\sigma}\sum_{\lambda\vdash 2n}\frac{\chi^\lambda(1)^2}{(2n)!^2s_{\lambda,n}(1)}\sum_{\pi\in C}(-1)^{\pi}\sum_{\tau_1,\tau_2\in R}\chi^\lambda(\sigma^{-1}\tau_1\pi\tau_2).
\end{align*}
For each $\lambda$, $\chi^\lambda$ is being summed over cosets of $R$ and the dual character $\chi^{\lambda'}=\mathrm{sgn}\cdot\chi^\lambda$ is being summed over cosets of $C$. The sum of a character $\chi$ over a coset of a subgroup $H$ is $0$ unless $\mathrm{res}_H\chi$ contains the trivial character with multiplicity at least $1$. By the Frobenius-Young correspondence, the only irreducible contained in both $\mathrm{ind}^{S_{2n}}_R 1$ and $\mathrm{ind}^{S_{2n}}_C \mathrm{sgn}$ is $\chi^\mu$ with multiplicity $1$, where $\mu=(2^n)$. Therefore,
$$c_\lambda:=\sum_{\pi\in C}(-1)^{\pi}\sum_{\tau_1,\tau_2\in R}\chi^\lambda(\sigma^{-1}\tau_1\pi\tau_2)=\begin{cases}0 & \text{if }\lambda\neq\mu\\ c_\mu&\text{if }\lambda=\mu.\end{cases}$$
Evaluating the above formula for the character of the regular representation yields
$$\chi^\mu(1)c_\mu=\sum_{\lambda\vdash 2n}\chi^\lambda(1)c_\lambda=c_{reg}=\sum_{\pi\in C}(-1)^{\pi}\sum_{\tau_1,\tau_2\in R}(2n)![\sigma^{-1}\tau_1\pi\tau_2=1].$$
It's easy to show with the hook-length formula that $\chi^\mu(1)=\frac{(2n)!}{n!(n+1)!}$ and even easier to show that there is only one semistandard Young tableau of shape $\mu$ with entries in $[n]$, so $s_{\mu,n}(1)=1$.

REPLY [4 votes]: I don't understand the matrix integrals, but the sum in the last paragraph is easy to deal with. Here is a story to visualize the permutation groups. Imagine $n$ couples lined up in two lines in a hall, perhaps to dance. A permutation in $C$ permutes each line within itself, breaking up partnerships. A permutation in $R$ makes some couple switch places with each other. Note that every permutation in $R$ is an involution.
Let's first prove positivity. Let $\tau_s$ switch $k_s$ pairs of couples.  If we are to have $\tau_1 \sigma \tau_2 \in C$, then each person who switches under $\tau_1$ must switch again until $\tau_2$ to get back to their original line. So $k_1 = k_2$ and $(-1)^{\tau_1} = (-1)^{\tau_2}$. This implies that $(-1)^{\pi} = (-1)^{\tau_1 \sigma \tau_2}$ and thus $(-1)^{\sigma \pi}=1$.
Now, let's compute the sum. Let $T_s$ be the set of indices $k$ for which $\tau_s(k) = k+n$. In order to have $\tau_1 \sigma \tau_2 \in C$, each person must cross the hall under $\tau_1$ if and only if they do under $\tau_2$. Looking at the people who start on the $\{1,2,\ldots, n \}$ side of the hall, we must have $\sigma_2(T_1) = T_2$. Looking at the people who start on the other side, we must also have $\sigma_1(T_1) = T_2$. Thus $\sigma_1^{-1} \sigma_2 (T_1) = T_1$. So $T_1$ is a union of orbits of $\sigma_1^{-1} \sigma_2$. Each such union of orbits has a unique compatible $T_2$, by the equation $\sigma_1(T_1) = T_2$, and each such $(\tau_1, \tau_2)$ determines a unique $\pi$ by $\pi = \tau_1 \sigma \tau_2$. So the number of nonzero terms is the number of subsets of orbits of $\sigma_1^{-1} \sigma_2$. The number of orbits is $n-\ell_T(\sigma_1^{-1} \sigma_2)$ where $\ell_T$ is the length with respect to the set of all reflections, so the sum is $2^{n-\ell_T(\sigma_1^{-1} \sigma_2)}$ as requested.