text:
\section{Introduction}\lbl{intro}
\setcounter{equation}{0}
Let $M_n$ be an $n\times n$ Haar-invariant unitary matrix, that is, the entries of unitary matrix $M_n$ are random variables satisfying that the probability distribution of the entries of $M_n$ is the same as that of $UM_n$ and that of $M_nU$ for any $n\times n$ unitary matrix $U.$ Diaconis and Evans (Theorem 2.1 from \cite{Evans}) proved that

{\it (a) Consider $a=(a_1, \cdots, a_k)$ and $b=(b_1, \cdots, b_k)$ with $a_j, b_j\in \{0,1,2,\cdots\}.$ Then for $n\geq \sum_{j=1}^{k}ja_j \vee \sum_{j=1}^kjb_j,$
\begin{eqnarray}\lbl{DiaEvans}
\mathbb{E}\Big[\prod_{j=1}^k(\Tr(M_n^j))^{a_j}\overline{(\Tr(M_n^j))^{b_j}} \,\Big] =\delta_{ab}\prod_{j=1}^kj^{a_j}a_j!
\end{eqnarray}
where $\delta_{ab}$ is Kronecker's delta.

(b) For any positive integers $j$ and $k,$
\begin{eqnarray}\lbl{weapon}
\mathbb{E}\big[ \Tr(M_n^j)\overline{\Tr(M_n^k)} \, \big]=\delta_{jk} \cdot j \wedge n.
\end{eqnarray}}
The idea of the proof is based on the group representation theory of unitary group $U(n).$ Some other derivations for (\ref{DiaEvans}) and (\ref{weapon}) are given in \cite{Shahshahani, Pastur, Rains, Stolz}.

Notice an $n\times n$ Haar-invariant unitary matrix is also called a CUE, which belongs to the Circular Ensembles of three members: the Circular Orthogonal Ensemble (COE), the Circular Unitary Ensemble (CUE) and the Circular Symplectic Ensemble (CSE), see Figure \ref{figure-ensemble} for the relationship, where the left circle consists of matrices which generate the Haar measures on the orthogonal group $O(n),$ Haar measure on the unitary group $U(n)$ and Haar measure on the real symplectic group $Sp(n),$ respectively.

Let $e^{i\theta_1}, \cdots, e^{i\theta_n}$ be the the eigenvalues of an $n\times n$ Haar-invariant unitary matrix, or equivalently, an $n\times n$ CUE, it is known (see, e.g., \cite{Forrester, Mehta}) that the density function of $\theta_1, \cdots \theta_n$ is $f(\theta_1, \cdots, \theta_n|\beta)$ with $\beta=2,$ where
\begin{eqnarray}\lbl{bensemble}
f(\theta_1, \cdots, \theta_n|\beta)=(2\pi)^{-n}\cdot\frac{\Gamma(1+\beta/2)^n}{\Gamma(1+\beta n/2)}\prod_{1\leq j<k \leq n}|e^{i\theta_j}-e^{i\theta_k}|^{\beta}
\end{eqnarray}
with $\beta>0$ and $\theta_i\in [0, 2\pi)$ for $1\leq i \leq n.$ The density function of $\theta_1, \cdots,\theta_n$ for the COE is $f(\theta_1, \cdots, \theta_n|\beta)$ with $\beta=1,$ and that for the CSE is $f(\theta_1, \cdots, \theta_n|\beta)$ with $\beta=4.$

The purpose of this paper is to study the analogues of (\ref{DiaEvans}) and (\ref{weapon}) for the circular $\beta$-ensembles with density function $f(\theta_1, \cdots, \theta_n|\beta)$ in (\ref{bensemble}) for any $\beta>0.$ Before stating the main results, we next introduce some background about the circular $\beta$-ensembles.

The circular ensembles were first introduced by physicist Dyson \cite{Dyson62a, Dyson62b, Dyson62c} for the study of nuclear
scattering data. They are modifications of the Gaussian matrix ensembles. In fact, as studied in \cite{Dyson62a}, Dyson shows that the consideration of time reversal symmetry leading to the three Gaussian ensembles behaves  equally well to unitary matrices. A time reversal symmetry requires that $U = U^T,$ no time reversal symmetry has no constraint, and a time reversal symmetry for a system with an odd number of spin 1/2 particles requires $U = U^D,$ where $D$ denotes the quaternion dual. Choosing such matrices with a uniform probability then gives COE, CUE and CSE, respectively (see, e.g., \cite{Forresterbook, Mehta}). The entries of COE and CUE are asymptotically complex normal random variables when the sizes of the matrices are large \cite{Jiang09a, Jiang09c, Jiang06}.

Let $U$ be an $n\times n$ Haar-invariant unitary matrix. As mentioned earlier, $U$ is also a CUE; the matrix $U^TU$ gives a COE. A similar but a bit more involved construction of CSE can be found in Ch. 9 from \cite{Mehta}. For the relations among the zonal polynomials, the Schur functions, the zonal spherical functions, the Gelfand pairs and the three circular ensembles, see, e.g., Chap. VII in  \cite{Macdonald1998} or Section 2.7 in \cite{Blower} for reference.
\begin{figure}\lbl{figure-ensemble}
\centering
\epsfig{file=figure-ensemble.eps, width=0.50\linewidth,clip=}
\caption{{\it Circular Ensembles and Haar-invariant Matrices from Classical Compact Groups}}
\end{figure}

Now we consider the moments in (\ref{DiaEvans}) and (\ref{weapon}) for the circular $\beta$-ensembles. Taking $\beta=1$ in (\ref{bensemble}), that is, choosing $W_n$ such that it is an $n\times n$ COE, by an elementary check in Lemma \ref{blue}, we have
\begin{eqnarray}\lbl{glue}
\mathbb{E}\big[ |\Tr(W_n)|^2 \big]=\frac{2n}{n+1}
\end{eqnarray}
for all $n\geq 2.$ This suggests that, unlike the right hand sides of  (\ref{DiaEvans}) or (\ref{weapon}) that are free of $n,$ the moments for the general  circular $\beta$-ensemble may depend on $n$ for $\beta\ne 2.$ In fact, by using the Jack functions, we will soon see from (\ref{hehe}) below that the second moment in (\ref{glue}) does depend on $n$ except   $\beta=2,$ in which case $W_n$ is an $n\times n$ CUE.

In this paper, we will prove some inequalities on the moments in (\ref{DiaEvans}) and (\ref{weapon}) for the circular $\beta$-ensembles with arbitrary $\beta>0.$ In particular, some of our inequalities for $\beta=2$ recover the equality in (\ref{DiaEvans}) by Diaconis and Evans \cite{Evans}. Further, we evaluate the limiting behavior by letting $n\to\infty$ for the left hand side in (\ref{DiaEvans}) and $k\to\infty$ for the left hand side in (\ref{weapon}) respectively. Their limits exist and look quite similar to the right hand sides of (\ref{DiaEvans}) and (\ref{weapon}).

Now we state our main results. Let $\lambda=(\lambda_1, \lambda_2, \cdots)$ be a partition, that is, the sequence is in non-increasing order and only finite of $\lambda_i$'s are non-zero. The weight of $\lambda$ is $|\lambda|=\lambda_1+\lambda_2 + \cdots.$ Denote by $m_i(\lambda)$ the multiplicity of $i$ in $(\lambda_1, \lambda_2, \cdots)$ for each $i,$  and $l(\lambda)$ the length of $\lambda$: $l(\lambda)=m_1(\lambda)+ m_2(\lambda) + \cdots.$ Recall the convention $0!=1.$ Set
\begin{eqnarray}\lbl{insect}
 z_{\lambda}=\prod_{i\geq 1}i^{m_i(\lambda)}m_i(\lambda)!.
\end{eqnarray}
Let $\rho=(\rho_1, \rho_2, \cdots)$ be a partition, and
\begin{eqnarray}\lbl{remote}
 p_{\rho}=\prod_{i=1}^{l(\rho)}p_{\rho_i},\ \mbox{where}\ \ p_k(x_1, x_2, \cdots)=x_1^k + x_2^k + \cdots
\end{eqnarray}
for integer $k\geq 1$ and indeterminates $x_i$'s. The function $p_{\rho}$ is called  the power-sum symmetric function. For real number $\alpha>0,$ integers $K\geq 1$ and $n\geq 1,$ define
two constants $A=A(n,K,\alpha)$ and $B=B(n,K,\alpha)$ by
\begin{eqnarray}\lbl{meso}
A=
\Big(1-\frac{|\alpha-1|}{n-K+\alpha}\,\delta(\alpha \geq 1)\Big)^K
\quad \mbox{and}\quad B=\Big(1+\frac{|\alpha-1|}{n-K+\alpha}\,\delta(\alpha < 1)\Big)^K,
\end{eqnarray}
where $\delta(\alpha \geq 1)=1-\delta(\alpha <1)$ is $1$ if $\alpha \geq 1$, or $0$ otherwise.
With these notation, our main result in this paper is stated as follows.

\begin{theorem}\lbl{face} Let $\beta>0$ and $\theta_1, \cdots, \theta_n$ have density $f(\theta_1, \cdots, \theta_n|\beta)$ as in (\ref{bensemble}). Set $Z_n=(e^{i\theta_1}, \cdots, e^{i\theta_n})$ and $\alpha=2/\beta.$ For partitions $\mu$ and $\nu,$ the following hold.
\begin{itemize}
\item[(a)] If $n\geq K=|\mu|$, then
 $$
 A\leq  \frac{\mathbb{E} \big[ |p_{\mu}(Z_n)|^2 \big]}{\alpha^{l(\mu)}z_{\mu}}\leq B.
 $$
\item[(b)]  If $|\mu|\ne |\nu|$, then $\mathbb{E} \Big[p_{\mu}(Z_n)\overline{p_{\nu}(Z_n)} \Big]=0$.
If $\mu\ne \nu$ and $n\geq K=|\mu| \vee |\nu|$, then
$$
 \Bigg|\mathbb{E} \Big[ p_{\mu}(Z_n)\overline{p_{\nu}(Z_n)} \Big] \Bigg|
 \leq  \max\{|A-1|,|B-1|\}\cdot \alpha^{(l(\mu)+l(\nu))/2}(z_{\mu}z_{\nu})^{1/2}.
$$
\item[(c)] There exists a constant $C$ depending only on $\beta$ such that for any $m\geq 1$ and $n\geq 2$, we have
$$
\Big| \mathbb{E}\big[ |p_m(Z_n)|^2 \big]-n \Big|\leq C\frac{n^32^{n\beta}}{m^{1\wedge \beta}}.
$$
\end{itemize}
\end{theorem}

Take $\beta=2$ in (a) and (b) of Theorem \ref{face}, then $A=1$ and $B=1.$ The two results recover the result of Diaconis and Evans  in (\ref{DiaEvans}). Further, letting $n\to\infty$ in (a) and (b) of Theorem \ref{face}, we see that $A$ and $B$ (depending on $n$) converge to $1;$ letting $m\to\infty$ in (c) of the theorem, then the last term in (c) goes to $0.$ So we obviously have the following results.
\begin{coro}\lbl{main} Let the conditions be as in Theorem \ref{face}.  Then, for any $\beta>0,$
\begin{eqnarray*}
& (a) & \lim_{n\to\infty}\mathbb{E} \Big[p_{\mu}(Z_n)\overline{p_{\nu}(Z_n)}\,\Big] = \delta_{\mu \nu}\left(\frac{2}{\beta}\right)^{l(\mu)}z_{\mu};\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\
& (b) & \lim_{m\to\infty}\mathbb{E}\big[ |p_m(Z_n)|^2 \big] = n \quad \mbox{for any $n\geq 2$}.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{eqnarray*}
\end{coro}
Part (b) of the above corollary says that, as $m\to\infty,$ the limit of $\mathbb{E} \big[|p_m(Z_n)|^2 \big] $ does not depend on parameter $\beta,$ which is consistent with (\ref{weapon}). By studying $A$ and $B$ in (\ref{meso}), we have the following corollary from Theorem \ref{face}.
\begin{coro}\lbl{theory} Let $\beta>0$ and $f(\theta_1, \cdots, \theta_n|\beta)$ be as in (\ref{bensemble}). Set $\alpha =2/\beta$ and $Z_n=(e^{i\theta_1}, \cdots, e^{i\theta_n}).$ Let $\mu$ and $\nu$ be partitions with $\mu\ne \nu$ and  $K= |\mu|\vee |\nu|.$ If $n\geq 2K,$ then
\begin{eqnarray*}
& (a) & \Big| \frac{\mathbb{E} [|p_{\mu}(Z_n)|^2 ]}{\alpha^{l(\mu)}z_{\mu}}-1\Big |\leq \frac{6|1-\alpha|K}{n}; \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \\
& (b) &  \Big|\mathbb{E} \Big [p_{\mu}(Z_n)\overline{p_{\nu}(Z_n)}\Big]\, \Big| \leq \frac{6|1-\alpha|K}{n}\cdot \alpha^{(l(\mu)+l(\nu))/2}(z_{\mu}z_{\nu})^{1/2}.
\end{eqnarray*}
\end{coro}
The above results are in the forms of inequalities or limits. We actually derive an exact formula in (\ref{gosh}) to compute $\mathbb{E} \big[ |p_{\mu}(Z_n)|^2 \big]$ for every partition $\mu.$ In general it is not easy to evaluate this quantity for arbitrary $\mu,$ however, we are able to do so when $\mu$ is special. For instance, by using the exact formula we calculate the moment in (\ref{glue})  for any $\beta>0$ as follows.

\noindent \textbf{Example}\ For any $n\geq 1,$
\begin{eqnarray}\lbl{hehe}
\mathbb{E} [|p_1(Z_n)|^2]
=\frac{2}{\beta}\frac{n}{n-1+2\beta^{-1}}=
\begin{cases}
\frac{2n}{n+1},& \text{if $\beta=1$;}\\
1, & \text{if $\beta=2$};\lbl{cry}\\
\frac{n}{2n-1}, & \text{if $\beta=4$.}
\end{cases}
\end{eqnarray}
The verification of this formula through (\ref{gosh}) is provided in Appendix. We also give $\mathbb{E}[|p_1(Z_n)|^4]$, $\mathbb{E}[|p_2(Z_n)|^2]$ and $\mathbb{E}[p_2(Z_n)\overline{p_1(Z_n)^2}]$ in closed forms in Appendix.

The main tool used in our proofs is the Jack functions. Diaconis and Evans \cite{Evans} and Diaconis and Shahshahani \cite{Shahshahani} use the group representation theory to study (\ref{DiaEvans}) and (\ref{weapon}) because  $U(n)$ is a compact Lie group. The situations for the Circular Orthogonal Ensembles ($\beta=1$) and the Circular Symplectic Ensembles ($\beta=4$) are different. In fact, the two ensembles are not groups.

The proofs of (\ref{DiaEvans}) and (\ref{weapon}), however, involve with the Schur functions, which have a strong relationship with the unitary groups. Looking at Figure \ref{figure-ensemble}, an Haar-invariant unitary matrix is also a CUE. From the perspective of symmetric functions, the COE is connected to the zonal polynomials, and the  CSE to zonal spherical functions. The three functions are special cases of the Jack polynomial $J_{\lambda}^{(\alpha)}$ with $\alpha =1, 2$ and $1/2,$ respectively, where $\lambda$ is a partition. See Section \ref{proofs} next for this or \cite{Macdonald1998} for general properties of the Jack polynomials. By using the Jack functions, we are able to prove (a) and (b) in Theorem \ref{face}. Part (c) in the theorem is proved by evaluating the expectation/integral with respect to $f(\theta_1, \cdots, \theta_n|\beta)$ in (\ref{bensemble}) directly.

\noindent\textbf{Remark}\ Treating $n$ as a variable, the bound $n^32^{n\beta}m^{-(1\wedge \beta)}$ in (c) of Theorem \ref{face} seems quite large. It is likely to be improved.

The organization of the rest of the paper is as follows. We provide some necessary backgrounds of the Jack functions, the proofs of (a) and (b) of Theorem \ref{face} and Corollary \ref{theory} in Section \ref{proofs}. The proof of part (c) is given in Section \ref{party}. In Appendix we prove (\ref{glue}) by two ways different from the method of the Jack functions. Some other explicit formulas of moments are also given in the same section.

\section{Proofs of (a) and (b) of Theorem \ref{face} and Corollary \ref{theory} by Using Jack Functions}\lbl{proofs}
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For a partition $\lambda,$ the notation $\lambda'=(\lambda_1', \lambda_2',\cdots)$ represents the conjugate partition of $\lambda,$ whose Young diagram is obtained by transposing the Young diagram of $\lambda.$ Given $\alpha>0$ and partition $\lambda,$ we denote by $J_{\lambda}^{(\alpha)}$ the Jack polynomial, see, e.g., Chapter VI from \cite{Macdonald1998} or \cite{Forresterbook}.

Recall power-symmetric function $p_{\rho}$ in (\ref{remote}). Let $\theta_{\rho}^{\lambda}(\alpha)$ denote the coefficient of $p_{\rho}$ in $J_{\lambda}^{(\alpha)},$ that is,  \begin{eqnarray}\lbl{Christmas}
J_{\lambda}^{(\alpha)}=\sum_{\rho: |\rho|=|\lambda|}\theta_{\rho}^{\lambda}(\alpha)p_{\rho}.
\end{eqnarray}
The $\theta_{\rho}^{\lambda}(\alpha)$ are real numbers. Inversely, let $\Theta_{\rho}^{\lambda}(\alpha)$ be the coefficient of $J_{\lambda}^{(\alpha)}$ in  $p_{\rho},$ that is,
\begin{eqnarray}\lbl{eve}
p_{\rho}=\sum_{\lambda: |\lambda|=|\rho|}\Theta_{\rho}^{\lambda}(\alpha) J_{\lambda}^{(\alpha)}.
\end{eqnarray}
Set
\begin{eqnarray}
C_{\lambda}(\alpha)=\prod_{(i,j)\in \lambda}\Big\{ (\alpha(\lambda_i -j ) + \lambda_j'-i+1)(\alpha(\lambda_i -j ) + \lambda_j'-i+\alpha) \Big\}, \lbl{ski}\ \ \ \ \ \ \ \ \ \ \
\end{eqnarray}
where $(i,j)$ runs over all cells of the Young diagram of $\lambda.$
\begin{lemma}\lbl{summer} Recalling $\theta_{\rho}^{\lambda}(\alpha)$ in (\ref{Christmas}) and $\Theta_{\rho}^{\lambda}(\alpha)$ in (\ref{eve}). Then, for any partitions $\lambda$ and $\rho$ with $|\lambda|=|\rho|,$ we have
\begin{eqnarray}\lbl{winter}
\Theta_{\rho}^{\lambda}(\alpha)= \frac{\alpha^{l(\rho)}z_{\rho}}{C_{\lambda}(\alpha)}\,\theta_{\rho}^{\lambda}(\alpha).
\end{eqnarray}
\end{lemma}
\textbf{Proof}. A scalar product relevant to the Jack functions is defined by
\begin{eqnarray}\lbl{scalar}
\langle p_{\lambda}, p_{\mu}\rangle_{\alpha}=\delta_{\lambda \mu}\,\alpha^{l(\lambda)}z_{\lambda}
\end{eqnarray}
for any partitions $\lambda$ and $\mu,$ where $z_{\lambda}$ is as in (\ref{insect}), see Section 10 of Chapter VI from \cite{Macdonald1998}. Following from (10.22) in \cite{Macdonald1998} and the definition of  Jack functions, we see that
\begin{eqnarray}\lbl{relation}
\langle J_{\lambda}^{(\alpha)}, J_{\mu}^{(\alpha)}\rangle_{\alpha}= \delta_{\lambda \mu}\, C_{\lambda}(\alpha).
\end{eqnarray}
It follows from (\ref{Christmas}) and (\ref{scalar}) that
\begin{eqnarray*}
\langle J_{\lambda}^{(\alpha)}, p_{\rho}\rangle_{\alpha} = \langle \sum_{v}\theta_{v}^{\lambda}(\alpha)p_v, p_{\rho}\rangle_{\alpha} = \sum_{v}  \theta_{v}^{\lambda}(\alpha)\langle p_v, p_{\rho}\rangle_{\alpha}= \theta_{\rho}^{\lambda}(\alpha) \alpha^{l(\rho)}z_{\rho}.
\end{eqnarray*}
Similarly, by (\ref{eve}) and (\ref{relation}),
\begin{eqnarray*}
\langle J_{\lambda}^{(\alpha)}, p_{\rho}\rangle_{\alpha} = \langle J_{\lambda}^{(\alpha)}, \sum_{v}\Theta_{\rho}^{v}(\alpha)J_v^{(\alpha)}\rangle_{\alpha} = \sum_{v}  \Theta_{\rho}^{v}(\alpha)\langle J_{\lambda}^{(\alpha)}, J_{v}^{(\alpha)}\rangle_{\alpha}= \Theta_{\rho}^{\lambda}(\alpha) C_{\lambda}(\alpha).
\end{eqnarray*}
These two equalities lead to (\ref{winter}).\ \ \ \ \ \ \ $\blacksquare$\\

The coefficients $\theta_{\rho}^{\lambda}$'s satisfy the following orthogonality relations ((10.31) and (10.32) from \cite{Macdonald1998}):
\begin{eqnarray}
& & \sum_{\rho}z_{\rho}\alpha^{l(\rho)}\theta_{\rho}^{\lambda}(\alpha)\theta_{\rho}^{\mu}(\alpha)
=\delta_{\lambda\mu}C_{\lambda}(\alpha);\lbl{Florida}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\
& & \sum_{\lambda}\frac{1}{C_{\lambda}(\alpha)}\theta_{\rho}^{\lambda}(\alpha)\theta_{\sigma}^{\lambda}(\alpha)
=\delta_{\rho\sigma}z_{\rho}^{-1}\alpha^{-l(\rho)}.\lbl{Markov}
\end{eqnarray}
In other words, if $a_{\lambda\rho}:=\left(z_{\rho}\alpha^{l(\rho)}/C_{\lambda}(\alpha)\right)^{1/2}\theta_{\rho}^{\lambda}(\alpha)$,
then $A_m=(a_{\lambda\rho})_{|\lambda|=|\rho|=m}$ is an orthogonal matrix of size $p(m)$
for $m\geq 1$. Here $p(m)$ is the number of  partitions  of $m.$  The following are some special cases of the Jack polynomials. \\

\noindent\textbf{Example}\ Let $\alpha=1,\, s_{\lambda}$ be the Schur polynomial and $\chi_{\mu}^{\lambda}$ the character value for the irreducible representation of the symmetric groups. It is well known that $J_{\lambda}^{(1)}=h(\lambda)s_{\lambda}$ with $h(\lambda)=\sqrt{C_{\lambda}(1)}$ as the hook-length product. Further, by (7.8)
from Chapter VI of \cite{Macdonald1998} and (\ref{eve}) that
\begin{eqnarray*}
\theta_{\mu}^{\lambda}(1)=\frac{h(\lambda)\chi_{\mu}^{\lambda}}{z_{\mu}}\ \ \mbox{and}\ \
\Theta_{\mu}^{\lambda}(1)=\frac{\chi_{\mu}^{\lambda}}{h(\lambda)}.
\end{eqnarray*}

\noindent\textbf{Example}\ Let $\alpha=2$. Then $J_\lambda^{(2)}$ coincides with the zonal polynomial $Z_\lambda$.
By (2.13) and (2.16) from Chapter VII of \cite{Macdonald1998}, we have
\begin{eqnarray*}
\theta_{\mu}^{\lambda}(2)=\frac{2^k k!}{2^{l(\mu)} z_\mu}  \omega_{\mu}^{\lambda}\ \ \mbox{and}\ \
\Theta_{\mu}^{\lambda}(2)=\frac{2^k k! }{h(2\lambda)}\omega_{\mu}^{\lambda},
\end{eqnarray*}
with $k=|\lambda|=|\mu|$,
where $h(2\lambda)=C_\lambda(2)$ is the hook-length product of $2\lambda=(2\lambda_1,2\lambda_2,\dots)$ and
$\omega^\lambda_\mu$ is the value of the zonal spherical function of a Gelfand pair
$(\mathfrak{S}_{2k}, \mathfrak{B}_k)$.
Here $\mathfrak{S}_{2k}$ is the symmetric group and $\mathfrak{B}_k$
is the hyperoctahedral group in $\mathfrak{S}_{2k}$.

\noindent\textbf{Example} (Example 1(a) on p. 383 from \cite{Macdonald1998})\ For each partition $\rho$ of $k$, we have
\begin{eqnarray}\lbl{piano}
\theta_{\rho}^{(k)}(\alpha)=\frac{k!}{z_{\rho}}\alpha^{k- l(\rho)}\ \ \mbox{and}\ \ \theta_{\rho}^{(1^k)}(\alpha)=\frac{k!}{z_{\rho}}(-1)^{k-l(\rho)}.
\end{eqnarray}
For each partition $\lambda$ with $l(\lambda) \le n$, we define
\begin{eqnarray*}
\mathcal{N}_{\lambda}^{\alpha}(n)= \prod_{(i,j)\in \lambda}\frac{n+(j-1)\alpha-(i-1)}{n+j\alpha-i},
\end{eqnarray*}
which is a positive real number.

\begin{lemma}\lbl{nice} Let $\lambda$ and $\mu$ be two partitions. Let $\alpha>0$ and $n\geq 2$.
 Then
\begin{eqnarray*}
& & \frac{1}{(2\pi)^n}
\int_{[0,2\pi)^n} J_{\lambda}^{(\alpha)}(e^{i\theta_1},\dots,e^{i\theta_n})
J_{\mu}^{(\alpha)}(e^{-i\theta_1},\dots,e^{-i\theta_n})\prod_{1\leq p<q\leq n}|e^{i \theta_p}-e^{i\theta_q}|^{2/\alpha}\, d\theta_1 \cdots d \theta_n \\
&= & \delta_{\lambda \mu}\cdot \delta(l(\lambda)\leq n)\cdot \frac{\Gamma(\frac{n}{\alpha}+1)}{\Gamma(1+\frac{1}{\alpha})^n}C_\lambda(\alpha) \mathcal{N}_{\lambda}^{\alpha}(n).
\end{eqnarray*}
\end{lemma}
\textbf{Proof}. Since $J_\lambda^{(\alpha)}(x_1,\dots,x_n)=0$ if $l(\lambda) > n,$ we assume $l(\lambda) \leq n$ in the following discussion. It is known (e.g., Theorem 12.1.1 from \cite{Mehta}) that
\begin{eqnarray}\lbl{mehta}
\frac{1}{(2\pi)^n}
\int_{[0,2\pi)^n}\prod_{1 \le p < q \le n}|e^{i \theta_p} -e^{i\theta_q}|^{2/\alpha}
d\theta_1 \cdots d \theta_n
=\frac{\Gamma(\frac{n}{\alpha}+1)}{\Gamma(1+\frac{1}{\alpha})^n}.
\end{eqnarray}
From (10.22), (10.35) and (10.37) in \cite{Macdonald1998}, we see that
\begin{eqnarray*}
& & \frac{1}{(2\pi)^n n! \, C_{\lambda}(\alpha)} \int_{[0,2\pi)^n}J_{\lambda}^{(\alpha)}(e^{i\theta_1},\dots,e^{i\theta_n})J_{\mu}^{(\alpha)}(e^{-i\theta_1},\dots,e^{-i\theta_n})
\prod_{1 \le p < q \le n}|e^{i \theta_p} -e^{i\theta_q}|^{2/\alpha} d\theta_1 \cdots d\theta_n
\\
&= & c_n \mathcal{N}_\lambda^\alpha(n)
\end{eqnarray*}
where
\begin{eqnarray*}
c_n:=\frac{1}{(2\pi)^n n!}\int_{[0,2\pi)^n}
\prod_{1 \le p < q \le n}|e^{i \theta_p} -e^{i\theta_q}|^{2/\alpha} d\theta_1 \cdots d\theta_n =\frac{1}{n!}\cdot\frac{\Gamma(\frac{n}{\alpha}+1)}{\Gamma(1+\frac{1}{\alpha})^n}
\end{eqnarray*}
by (\ref{mehta}).
Hence the desired conclusion follows.\ \ \ \ \ \ $\blacksquare$

\begin{prop}\lbl{exact} Let $\beta>0$ be a constant. Suppose $\theta_1, \cdots, \theta_n$ have a joint density as in (\ref{bensemble}). Let $Z_n=(e^{i\theta_1}, \cdots, e^{i\theta_n}).$ Given partitions $\mu$ and $\nu$ of weight $K,$ then
\begin{eqnarray}\lbl{gosh}
& & \mathbb{E} \Big[p_{\mu}(Z_n)\overline{p_{\nu}(Z_n)}\,\Big]\nonumber\\
&= &
\alpha^{l(\mu)+l(\nu)}z_{\mu}z_{\nu}\sum_{\lambda \vdash K:\, l(\lambda)\leq n}\frac{\theta_{\mu}^{\lambda}(\alpha)\theta_{\nu}^{\lambda}(\alpha)}{C_{\lambda}(\alpha)}
\mathcal{N}_\lambda^\alpha(n).\lbl{yahoo}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{eqnarray}
\end{prop}
\textbf{Proof}. Reviewing (\ref{bensemble}), by (\ref{eve}) and Lemma \ref{nice}, we have
\begin{eqnarray*}
\mathbb{E} \Big[p_{\mu}(Z_n)\overline{p_{\nu}(Z_n)}\,\Big] =\sum_{\lambda \vdash K:\, l(\lambda)\leq n}\Theta_{\mu}^{\lambda}(\alpha)\Theta_{\nu}^{\lambda}(\alpha)C_\lambda(\alpha)\mathcal{N}_{\lambda}^{\alpha}(n)
\end{eqnarray*}
where $\alpha=2/\beta.$ By Lemma \ref{summer}, the above is identical to
\begin{eqnarray*}
& & \alpha^{l(\mu)+l(\nu)}z_{\mu}z_{\nu}\sum_{\lambda \vdash K:\, l(\lambda)\leq n}
\frac{\theta_{\mu}^{\lambda}(\alpha)\theta_{\nu}^{\lambda}(\alpha)}
{C_{\lambda}(\alpha)}\mathcal{N}_{\lambda}^{\alpha}(n).
\end{eqnarray*}
The proof is completed.\ \ \ \ \ \ \ \ $\blacksquare$\\

For positive integers $n$ and $K$ and real number $\alpha >0.$
Define
\begin{eqnarray}
& & \Gamma_{n, K}^{\alpha}
=\max_{\lambda \vdash K:\, l(\lambda)\leq n} \mathcal{N}_\lambda^\alpha(n)
=\max_{\lambda \vdash K:\, l(\lambda)\leq n}\prod_{(i,j)\in \lambda}\frac{n+(j-1)\alpha-(i-1)}{n+j\alpha-i};\lbl{er}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\
& & \gamma_{n, K}^{\alpha}
=\min_{\lambda \vdash K:\, l(\lambda)\leq n} \mathcal{N}_\lambda^\alpha(n)
=\min_{\lambda \vdash K:\, l(\lambda)\leq n}\prod_{(i,j)\in \lambda}\frac{n+(j-1)\alpha-(i-1)}{n+j\alpha-i}.\lbl{three}
\end{eqnarray}

\begin{lemma}\lbl{sheep} Let $\alpha>0,\, K\geq 1$  and $\Gamma_{n, K}^{\alpha}$ be as in (\ref{er}) and $\gamma_{n, K}^{\alpha}$ be as in (\ref{three}).  If $n\geq  K$, then $ A\leq \gamma_{n, K}^{\alpha}\leq  \Gamma_{n, K}^{\alpha} \leq B$
where $A$ and $B$ are as in (\ref{meso}). Further, if $n\geq K,$ then
\begin{eqnarray}\lbl{star}
 \max_{\lambda \vdash K}| \mathcal{N}_\lambda^\alpha(n)-1 | \leq  \max\{|A-1|, |B-1|\}.
\end{eqnarray}
\end{lemma}
\textbf{Proof}. For $\lambda \vdash K$ such that $l(\lambda)\leq n$ and $(i,j)\in \lambda,$ it is easy to check that
\begin{eqnarray}\lbl{silent1}
1\leq i\leq \min\{n, K\}\ \ \mbox{and}\ \ 1\leq j \leq K.
\end{eqnarray}
Thus, $n+(j-1)\alpha-(i-1)\geq n-i+1>0\ \ \mbox{and}\ \ n+j\alpha-i\geq j\alpha>0.$ It follows that
\begin{eqnarray}\lbl{mama1}
b_{i,j}(\alpha):=\frac{n+(j-1)\alpha-(i-1)}{n+j\alpha-i}>0.
\end{eqnarray}
Write
\begin{eqnarray}\lbl{ding1}
b_{i,j}(\alpha)= 1+ \frac{1-\alpha}{n+j\alpha-i}.
\end{eqnarray}

\noindent{\it Case 1:} $\alpha\geq 1.$ By (\ref{mama1}) and (\ref{ding1}), we see that $b_{i,j}(\alpha) \in [0, 1]$ for all $\lambda \vdash K$ such that $l(\lambda)\leq n$ and $(i,j)\in \lambda,$ which concludes $\Gamma_{n, K}^{\alpha}\leq 1$.

 Further, by (\ref{silent1}), $n+j\alpha-i\geq n-K + \alpha>0$ for all $\lambda \vdash K$ such that $l(\lambda)\leq n$ and $(i,j)\in \lambda.$ Thus, noticing $1-\alpha\leq 0,$ we get
\begin{eqnarray*}
b_{i,j}(\alpha)\geq 1+\frac{1-\alpha}{n-K+\alpha}=1-\frac{|1-\alpha|}{n-K+\alpha}>0
\end{eqnarray*}
for all $n\geq K.$ This yields
\begin{eqnarray*}
\gamma_{n, K}^{\alpha}\geq \left(1-\frac{|1-\alpha|}{n-K+\alpha}\right)^K.
\end{eqnarray*}
The above two conclusions lead to that
\begin{eqnarray}\lbl{grass}
\left(1-\frac{|1-\alpha|}{n-K+\alpha}\right)^K \leq \gamma_{n, K}^{\alpha}\leq \Gamma_{n, K}^{\alpha}\leq 1
\end{eqnarray}
for all $n\geq K$ and $\alpha \geq 1.$

\noindent{\it Case 2:}  $\alpha\in (0,1].$ By (\ref{ding1}), $b_{i,j}(\alpha)\geq  1$ for all $\lambda \vdash K$ such that $l(\lambda)\leq n$ and $(i,j)\in \lambda,$ which shows $\gamma_{n, K}^{\alpha}\geq 1.$

Moreover, by (\ref{silent1}) again, $n+j\alpha-i\geq n-K+\alpha$ for all $\lambda \vdash K$ such that $l(\lambda)\leq n$ and $(i,j)\in \lambda.$ Thus, with $1-\alpha> 0,$ we have from (\ref{ding1})  that
\begin{eqnarray*}
b_{i,j}(\alpha)\leq 1+\frac{1-\alpha}{n-K+\alpha}.
\end{eqnarray*}
By the definition of $\Gamma_{n, K}^{\alpha}$ and the earlier conclusion, we get
\begin{eqnarray*}
1\leq \gamma_{n, K}^{\alpha}\leq \Gamma_{n, K}^{\alpha}\leq \left(1+\frac{1-\alpha}{n-K+\alpha}\right)^K
\end{eqnarray*}
 for all $n\geq K$ and $\alpha \in (0,1].$ This and (\ref{grass}) prove the first part of the lemma.

Finally, by the definitions in (\ref{er}) and (\ref{three}),
\begin{eqnarray*}
\gamma_{n, K}^{\alpha} \leq  \mathcal{N}_\lambda^\alpha(n)=\prod_{(i,j)\in \lambda}b_{i,j}(\alpha) \leq \Gamma_{n, K}^{\alpha}
\end{eqnarray*}
for all $\lambda \vdash K$ since $l(\lambda)\leq n$ holds automatically if $n\geq K.$ By the proved conclusion,
\begin{eqnarray*}
A-1 \leq  \mathcal{N}_\lambda^\alpha(n) -1\leq B-1
\end{eqnarray*}
for all $\lambda \vdash K.$ This implies (\ref{star}). \ \ \ \ \ \ \ $\blacksquare$\\

\noindent\textbf{Proof of (a) and (b) of Theorem \ref{face}}. (a) By Proposition \ref{exact},  take $\mu=\nu$ with weight $K$ to have
\begin{eqnarray}
\mathbb{E} \big[ |p_{\mu}(Z_n)|^2 \big]=
\alpha^{2l(\mu)}z_{\mu}^2\sum_{\lambda \vdash K: l(\lambda)\leq n}\frac{\theta_{\mu}^{\lambda}(\alpha)^2}{C_{\lambda}(\alpha)}\mathcal{N}_\lambda^\alpha(n).
\end{eqnarray}
Lemma \ref{sheep} says that $\Gamma_{n, K}^{\alpha}>0$ and $\gamma_{n, K}^{\alpha}>0$ for all $n\geq K.$ By the definitions of  $\Gamma_{n, K}^{\alpha}$ in (\ref{er}) and $\gamma_{n, K}^{\alpha}$ in (\ref{three}), since $C_{\lambda}(\alpha)>0$ for any partition $\lambda$ and $\alpha>0,$
\begin{eqnarray}
\gamma_{n, K}^{\alpha}\cdot \alpha^{2l(\mu)}z_{\mu}^2\sum_{\lambda \vdash K: l(\lambda)\leq n}\frac{\theta_{\mu}^{\lambda}(\alpha)^2}{C_{\lambda}(\alpha)}
& \leq & \mathbb{E} \big[ |p_{\mu}(Z_n)|^2 \big] \nonumber\\
& \leq & \Gamma_{n, K}^{\alpha}\cdot \alpha^{2l(\mu)}z_{\mu}^2\sum_{\lambda \vdash K}\frac{\theta_{\mu}^{\lambda}(\alpha)^2}{C_{\lambda}(\alpha)}.\lbl{west2}
\end{eqnarray}
 From assumption $n\geq K,$ if $\lambda \vdash K$, we know $l(\lambda)\leq n$ automatically. Therefore, from (\ref{Markov}) the two sums in (\ref{west2}) are both equal to $z_{\mu}^{-1}\alpha^{-l(\mu)}.$ Consequently,
\begin{eqnarray*}
\gamma_{n, K}^{\alpha} \leq \frac{\mathbb{E} \big[ |p_{\mu}(Z_n)|^2 \big]}{\alpha^{l(\mu)}z_{\mu}} \leq \Gamma_{n, K}^{\alpha}.
\end{eqnarray*}
The conclusion (a) then follows from the first part of Lemma \ref{sheep}.

(b) First, assume $|\mu|\ne |\nu|.$ Notice
\begin{eqnarray*}
& & \mathbb{E} \Big[p_{\mu}(Z_n)\overline{p_{\nu}(Z_n)}\,\Big]\\
&=&
Const\cdot\int_0^{2\pi}\cdots\int_{0}^{2\pi} p_{\mu}(e^{i\theta_1}, \cdots, e^{i\theta_n})\overline{p_{\nu}(e^{i\theta_1}, \cdots, e^{i\theta_n})}\prod_{1\leq j<k \leq n}|e^{i\theta_j}-e^{i\theta_k}|^{\beta}\,d\theta_1\cdots d\theta_n.
\end{eqnarray*}
For an integrable function $h(x),$ we know $\int_0^{2\pi}h(e^{ix})\,dx=\int_b^{b+2\pi}h(e^{ix})\,dx$ for any $b\in\mathbb{R}.$ Using the induction and the Fubini theorem, we see that
\begin{eqnarray*}
& & \mathbb{E} \Big[p_{\mu}(Z_n)\overline{p_{\nu}(Z_n)}\,\Big]\\
&=&
Const\cdot\int_b^{b+2\pi}\cdots\int_{b}^{b+2\pi} p_{\mu}(e^{i\theta_1}, \cdots, e^{i\theta_n})\overline{p_{\nu}(e^{i\theta_1}, \cdots, e^{i\theta_n})}\prod_{1\leq j<k \leq n}|e^{i\theta_j}-e^{i\theta_k}|^{\beta}\,d\theta_1\cdots d\theta_n.
\end{eqnarray*}
Making transform $\eta_j=\theta_j-b$ for $1\leq j \leq n,$ noting that $p_{\mu}(e^{ib+i\eta_1}, \cdots, e^{ib+i\eta_n})=e^{ib|\mu|}p_{\mu}(e^{i\eta_1}, \cdots, e^{i\eta_n})$ for any $b\in \mathbb{R},$ we obtain that
\begin{eqnarray*}
\mathbb{E} \Big[p_{\mu}(Z_n)\overline{p_{\nu}(Z_n)}\,\Big]=e^{ib(|\mu|-|\nu|)}\mathbb{E} \Big[p_{\mu}(Z_n)\overline{p_{\nu}(Z_n)}\,\Big]
\end{eqnarray*}
for any $b\in \mathbb{R}.$
If $|\mu|\ne |\nu|$, since $b$ is arbitrary,
 we then conclude
\begin{eqnarray*}
\mathbb{E} \Big[p_{\mu}(Z_n)\overline{p_{\nu}(Z_n)}\,\Big]=0
\end{eqnarray*}
for all $n\geq 2$.

To prove the second part of (b), by the first part, it suffices to prove the conclusion for $n\geq |\mu|=|\nu|=K$. Observe that $l(\lambda)\leq n$ if $\lambda \vdash K.$ Thus, it follows from Proposition \ref{exact} that
\begin{eqnarray}
& & \mathbb{E} \Big[p_{\mu}(Z_n)\overline{p_{\nu}(Z_n)}\,\Big]\nonumber\\
&= &
\alpha^{l(\mu)+l(\nu)}z_{\mu}z_{\nu}\sum_{\lambda \vdash K}\frac{\theta_{\mu}^{\lambda}(\alpha)\theta_{\nu}^{\lambda}(\alpha)}{C_{\lambda}(\alpha)}\mathcal{N}_\lambda^\alpha(n)\nonumber\\
& = & \alpha^{l(\mu)+l(\nu)}z_{\mu}z_{\nu}\Bigg[ \sum_{\lambda \vdash K}\frac{\theta_{\mu}^{\lambda}(\alpha)\theta_{\nu}^{\lambda}(\alpha)}{C_{\lambda}(\alpha)} + \sum_{\lambda \vdash K}\frac{\theta_{\mu}^{\lambda}(\alpha)\theta_{\nu}^{\lambda}(\alpha)}{C_{\lambda}(\alpha)}
\Big( \mathcal{N}_\lambda^\alpha(n)-1\Big)\Bigg] \nonumber\\
& = & \alpha^{l(\mu)+l(\nu)}z_{\mu}z_{\nu} \sum_{\lambda \vdash K}\frac{\theta_{\mu}^{\lambda}(\alpha)\theta_{\nu}^{\lambda}(\alpha)}{C_{\lambda}(\alpha)}\Big( \mathcal{N}_\lambda^\alpha(n)-1\Big)
\end{eqnarray}
where the last identity comes from the orthogonal property in (\ref{Markov}). Therefore,
\begin{eqnarray*}
& & \Big|\mathbb{E} \Big [p_{\mu}(Z_n)\overline{p_{\nu}(Z_n)}\,\Big]\Big| \\
&\leq & \max_{\lambda \vdash K} |  \mathcal{N}_\lambda^\alpha(n)-1 |
\cdot \alpha^{l(\mu)+l(\nu)}z_{\mu}z_{\nu} \sum_{\lambda \vdash K}\frac{|\theta_{\mu}^{\lambda}(\alpha)|\cdot|\theta_{\nu}^{\lambda}(\alpha)|}{C_{\lambda}(\alpha)}.
\end{eqnarray*}
Now, by the Cauchy-Schwartz inequality the sum above is bounded by
\begin{eqnarray*}
\Big(\sum_{\lambda \vdash K}\frac{|\theta_{\mu}^{\lambda}(\alpha)|^2}{C_{\lambda}(\alpha)}\Big)^{1/2}\cdot \Big(\sum_{\lambda \vdash K}\frac{|\theta_{\nu}^{\lambda}(\alpha)|^2}{C_{\lambda}(\alpha)}\Big)^{1/2}=z_{\mu}^{-1/2}z_{\nu}^{-1/2}\alpha^{-(l(\mu)+l(\nu))/2}
\end{eqnarray*}
by (\ref{Markov}). The above two inequalities imply
\begin{eqnarray*}
\Big|\mathbb{E}\Big[p_{\mu}(Z_n)\overline{p_{\nu}(Z_n)}\,\Big]\Big| & \leq &  \max_{\lambda \vdash K}|  \mathcal{N}_\lambda^\alpha(n)-1 | \cdot \alpha^{(l(\mu)+l(\nu))/2}(z_{\mu}z_{\nu})^{1/2}\\
& \leq & \max\{|A-1|,|B-1|\}\cdot \alpha^{(l(\mu)+l(\nu))/2}(z_{\mu}z_{\nu})^{1/2}
\end{eqnarray*}
by (\ref{star}).\ \ \ \ \ \ \ \ $\blacksquare$

\begin{lemma}\lbl{vote} Let $A$ and $B$ be as in (\ref{meso}) with $\beta>0.$ Set $\alpha=2/\beta.$ If $n\geq 2K,$ then
\begin{eqnarray*}
\max\{|A-1|, |B-1|\} \leq \frac{6|1-\alpha|K}{n}.
\end{eqnarray*}
\end{lemma}
\textbf{Proof}. By the definitions of $A$ and $B$, it suffices to show that, as $n\geq 2K,$
\begin{eqnarray}
& & 1-\Big(1-\frac{\alpha -1}{n-K+\alpha}\Big)^K \leq \frac{6|1-\alpha|K}{n}\qquad \mbox{for $\alpha\geq 1;$} \lbl{stick}\\
& & \Big(1+\frac{1-\alpha}{n-K+\alpha}\Big)^K-1 \leq \frac{6|1-\alpha|K}{n}\qquad \mbox{for $\alpha\in (0,1)$.} \lbl{pogo}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{eqnarray}
First, if $\alpha\geq 1,$ then $(\alpha-1)/(n-K+\alpha)\in [0, 1).$ Notice $(1+x)^K\geq 1+Kx$ for all $x\geq -1$ (see, e.g., Theorem 42 on p.40 from \cite{Hardy}), we have
\begin{eqnarray*}
1-\Big(1-\frac{\alpha -1}{n-K+\alpha}\Big)^K \leq \frac{K(\alpha-1)}{n-K+\alpha}\leq \frac{2K|1-\alpha|}{n}
\end{eqnarray*}
since $n-K+\alpha\geq n/2$ as $n\geq 2K.$ This proves (\ref{stick}).

Second, for $\alpha\in (0,1),$ it is easy to verify that $(1-\alpha)/(n-K+\alpha) \leq 1/K$ provided $n\geq 2K.$ By the fact that $(1+x)^K \leq 1+ 3Kx$ for all $0\leq x\leq 1/K$, we obtain
\begin{eqnarray*}
\Big(1+\frac{1-\alpha}{n-K+\alpha}\Big)^K-1\leq \frac{3(1-\alpha)K}{n-K+\alpha} \leq \frac{6|1-\alpha|K}{n}
\end{eqnarray*}
since $n-K+\alpha\geq n/2$ if $n\geq 2K$ as used earlier. This concludes (\ref{pogo}).\ \ \ \ \ \ \ \ $\blacksquare$\\

\noindent\textbf{Proof of Corollary \ref{theory}}. (a) By Theorem \ref{face}
\begin{eqnarray*}
A-1\leq  \frac{\mathbb{E} \big[|p_{\mu}(Z_n)|^2\big] }{\alpha^{l(\mu)}z_{\mu}}-1\leq B-1
\end{eqnarray*}
Thus,
\begin{eqnarray*}
\Big| \frac{\mathbb{E} \big[|p_{\mu}(Z_n)|^2 \big]}{\alpha^{l(\mu)}z_{\mu}}-1\Big |\leq \max\{|A-1|, |B-1|\}.
\end{eqnarray*}
The conclusion (a) then follows from Lemma \ref{vote}.

(b) The conclusion obviously holds if $|\mu|\ne |\nu|$ by (b) of Theorem \ref{face}. If $|\mu|=|\nu|=K,$ by (b) of Theorem \ref{face} and Lemma \ref{vote}, we get the desired result.\ \ \ \ \ \ \  $\blacksquare$\\

\section{Proof of (c) of Theorem \ref{face}}\lbl{party}
\setcounter{equation}{0}
We start the proof through a series of lemmas.
\begin{lemma}\lbl{stirling} Let $a$ be a real number. Then
\begin{eqnarray*}
\frac{\Gamma(x+a)}{\Gamma(x)} \sim x^a
\end{eqnarray*}
as $x \to +\infty$, where $\Gamma(x)$ is the gamma function.
\end{lemma}
\textbf{Proof}. Recall the Stirling formula (see, e.g., p.368 from \cite{Gamelin} or (37) on
p.204 from \cite{Ahlfors}):
\begin{eqnarray*}
\log\Gamma(z)=z\log z - z -\frac{1}{2}\log z+ \log \sqrt{2\pi}
+\frac{1}{12z} +O\left(\frac{1}{x^3}\right)
\end{eqnarray*}
as $x=\mbox{Re}\,(z)\to +\infty$. Then
\begin{eqnarray}\lbl{alpha}
\log \frac{\Gamma(x+a)}{\Gamma(x)} =(x+a)\log (x+a) -x\log x-a-\frac{1}{2}\left(\log (x+a)-\log x\right) +O\left(\frac{1}{x}\right)
\end{eqnarray}
as $x\to\infty.$ First, use the fact that $\log(1+t)\sim t+O(t^2)$ as $t\to 0$ to get
\begin{eqnarray*}
(x+a)\log (x+a) -x\log x &= & (x+a)\left(\log x +\log \Big(1+\frac{a}{x}\Big)\right) -x\log x\\
& = & (x+a)\Big(\log x +\frac{a}{x} +O(x^{-2})\Big)-x\log x\\
& = & a\log x + a + O\left(\frac{1}{x}\right)
\end{eqnarray*}
as $x\to\infty.$ Evidently, $\log (x+a)-\log x=\log (1+ax^{-1})=O(1/x)$ as $x\to\infty.$ Plugging  these two assertions into (\ref{alpha}), we have
\begin{eqnarray*}
\log \frac{\Gamma(x+a)}{\Gamma(x)} =a\log x+ O\left(\frac{1}{x}\right)
\end{eqnarray*}
as $x\to \infty.$ The proof is complete.\ \ \ \ \ \ \ \ \ $\blacksquare$

\begin{lemma}\lbl{Jelly} Let $\beta>0.$ For positive integers $m$ and $k$ and real numbers $a_1, \cdots, a_k,$ define
\begin{eqnarray*}
D=\int_{0}^{\pi}\cos (2mt)\Big|\prod_{i=1}^k\sin\big(t+a_i\big)\Big|^{\beta}\,dt.
\end{eqnarray*}
Then $|D|\leq 6(1+\beta)(\frac{k}{m})^{1\wedge \beta }.$
\end{lemma}
\textbf{Proof}. First, since $|D|\leq \int_0^{\pi}1\,dt=\pi,$ the conclusion obviously holds for $m=1.$ Now we assume $m\geq 2.$ Set $s=mt$. Then
\begin{eqnarray}
D
&= & \frac{1}{m}\int_0^{m\pi}\cos (2s)\,\Big|\prod_{i=1}^k\sin\big(\frac{s}{m}+a_i\big)\Big|^{\beta}\,ds\nonumber\\
& = & \frac{1}{m}\sum_{j=0}^{m-1}\int_{j\pi}^{(j+1)\pi}\cos (2s)\,\Big|\prod_{i=1}^k\sin\big(\frac{s}{m}+a_i\big)\Big|^{\beta}\,ds\nonumber\\
& = & \frac{1}{m}\sum_{j=0}^{m-1}\int_{0}^{\pi}\cos (2s)\,\Big|\prod_{i=1}^k\sin\big(\frac{s+j\pi}{m}+a_i\big)\Big|^{\beta}\,ds\nonumber\\
&= &\int_0^{\pi}L_m(s)\cos (2s)\,ds\lbl{identity}
\end{eqnarray}
where we make a transform: $s\to s-j\pi$ in the second identity to get the third one, and
\begin{eqnarray}\lbl{bao}
L_m(s)=\frac{1}{m}\sum_{j=0}^{m-1}\Big|\prod_{i=1}^k\sin\big(b_{ij} + \frac{s}{m}\big)\Big|^{\beta}
\end{eqnarray}
for $0\leq s \leq \pi$ and $b_{ij}=a_i +\frac{j\pi}{m}.$ Note that the inequality $|\,|x|^{\beta} -|y|^{\beta}|\leq (1+\beta)|x-y|^{1 \wedge \beta}$ for any $\beta>0$ and  $x, y \in [-1, 1],$ it follows that
\begin{eqnarray}
& & \Bigg|L_m(s)-\frac{1}{m}\sum_{j=0}^{m-1}\Big|\prod_{i=1}^k\sin b_{ij}\Big|^{\beta}\Bigg|\nonumber\\
& \leq & \frac{1}{m}\sum_{j=0}^{m-1}\Bigg| \,\Big|\prod_{i=1}^k\sin\big(b_{ij} + \frac{s}{m}\big)\Big|^{\beta}-\Big|\prod_{i=1}^k\sin b_{ij}\Big|^{\beta}\,\Bigg|\nonumber\\
& \leq & \frac{1+\beta}{m}\sum_{j=0}^{m-1}\Bigg|\prod_{i=1}^k\sin\big(b_{ij} + \frac{s}{m}\big)-\prod_{i=1}^k\sin b_{ij}\Bigg|^{\beta \wedge 1}.\lbl{sichuan}
\end{eqnarray}
Now, by the chain rule, $\Big(\prod_{i=1}^k\sin(b_{ij} + t)\Big)'=\sum_{l=1}^k\cos (b_{lj}+t)\prod_{1\leq i\leq k,\, i\ne l}\sin(b_{ij} + t)$ for any $t\in\mathbb{R}.$ Thus the absolute value of the derivative is bounded by $k$ for ant $t\in \mathbb{R}.$ By the mean-value theorem,
\begin{eqnarray*}
\Bigg|\prod_{i=1}^k\sin\big(b_{ij} + \frac{s}{m}\big)-\prod_{i=1}^k\sin b_{ij}\Bigg| \leq \frac{ks}{m}.
\end{eqnarray*}
This implies that the last term in (\ref{sichuan}) is controlled by
\begin{eqnarray*}
\frac{1+\beta}{m}\sum_{j=0}^{m-1}\Big(\frac{ks}{m}\Big)^{1\wedge \beta }=(1+\beta)\Big(\frac{ks}{m}\Big)^{1\wedge \beta }.
\end{eqnarray*}
It follows from (\ref{sichuan}) that
\begin{eqnarray}
\Bigg|L_m(s)-\frac{1}{m}\sum_{j=0}^{m-1}\Big|\prod_{i=1}^k\sin b_{ij}\Big|^{\beta}\Bigg| \leq (1+\beta)\Big(\frac{ks}{m}\Big)^{1\wedge \beta }.
\end{eqnarray}
Set $C=\frac{1}{m}\sum_{j=0}^{m-1}\Big|\prod_{i=1}^k\sin b_{ij}\Big|^{\beta}.$ Notice $\int_0^{\pi}\cos (2s)\,ds=0.$ From the above we use the simple fact that $|\cos(2s)|\leq 1$ to have
\begin{eqnarray*}
\Big|\int_0^{\pi}L_m(s)\cos (2s)\,ds\Big|
& = & \Big|\int_0^{\pi}C\cos (2s)\,ds+ \int_0^{\pi}(L_m(s)-C)\cos (2s)\,ds \Big|\\
& \leq & (1+\beta)\Big(\frac{k}{m}\Big)^{1\wedge \beta }\int_0^{\pi}s^{1 \wedge \beta}\,ds.
\end{eqnarray*}
Now the last integral above is bounded by $\int_0^1 1\,ds + \int_1^{\pi}s\,ds=(\pi^2 + 1)/2\leq 6.$ The proof is completedby using (\ref{identity}).\ \ \ \ \ \ \ $\blacksquare$

\begin{lemma}\lbl{button} For $\beta>0$, let $f(\theta_1, \cdots, \theta_n|\beta)$ be as in (\ref{bensemble}). Define
\begin{eqnarray*}
I(m,n)=\int_0^{2\pi}\cdots \int_0^{2\pi}\cos (m(\theta_2 -\theta_1))f(\theta_1, \cdots, \theta_n|\beta)\,d\theta_1\cdots\,d\theta_n,\ \ m\geq 0,\, n\geq 2.
\end{eqnarray*}
Then, for some constant $K=K(\beta)$, we have $|I(m,n)|\leq (K n2^{n\beta})m^{-(1\wedge \beta)}$ for all $m\geq 1$ and $n\geq 2.$
\end{lemma}
\noindent\textbf{Proof}. Evidently, since $f(\theta_1, \cdots, \theta_n|\beta)$ is a probability density function, we know
\begin{eqnarray}\lbl{Fateev}
I(0,n)=1
\end{eqnarray}
for all $n\geq 2.$ Since $|e^{ix} -e^{iy}|^2=|1-e^{i(x-y)}|^2=(1-\cos(x-y))^2+\sin^2(x-y)=2(1-\cos(x-y))=4\sin^2 ((x-y)/2)$ for any $x, y \in \mathbb{R},$ the probability density function in (\ref{bensemble}) becomes
\begin{eqnarray}\lbl{Einstein}
f(\theta_1, \cdots, \theta_n|\beta)=C_n\prod_{1\leq j<k \leq n}\Big|\sin\Big(\frac{\theta_j - \theta_k}{2}\Big)\Big|^{\beta}
\end{eqnarray}
where $\theta_1, \cdots, \theta_n\in [0, 2\pi]$ and
\begin{eqnarray*}
C_n=2^{n(n-1)\beta/2}(2\pi)^{-n}\cdot\frac{\Gamma(1+\beta/2)^n}{\Gamma(1+\beta n/2)}.
\end{eqnarray*}
Now,
\begin{eqnarray}
 I(m,n)&= &\int_0^{2\pi}\cdots \int_0^{2\pi}\cos (m(\theta_2 -\theta_1))f(\theta_1, \cdots, \theta_n|\beta)\,d\theta_1\cdots\,d\theta_n\lbl{Hornets}\\
& = & C_n\int_0^{2\pi}\cdots \int_0^{2\pi}\cos (m(\theta_2 -\theta_1))\cdot \prod_{1\leq j<k \leq n}\Big|\sin\Big(\frac{\theta_j - \theta_k}{2}\Big)\Big|^{\beta}\,d\theta_2\cdots\,d\theta_n\,d\theta_1.\nonumber
\end{eqnarray}
Making transforms $x_i=\theta_i-\theta_1$ for $i=2,3,\cdots,n,$ we obtain that
\begin{eqnarray}
I(m,n)=C_n\int_0^{2\pi}\int_{-\theta_1}^{2\pi-\theta_1}\cdots \int_{-\theta_1}^{2\pi-\theta_1}\cos (mx_2)\cdot G_n(x)\,dx_2\cdots\,dx_n\,d\theta_1
\end{eqnarray}
where
\begin{eqnarray}\lbl{useful}
G_n(x)=\prod_{i=2}^n\Big|\sin\Big(\frac{x_i}{2}\Big)\Big|^{\beta}\cdot\prod_{2\leq j<k \leq n}\Big|\sin\Big(\frac{x_j - x_k}{2}\Big)\Big|^{\beta}
\end{eqnarray}
where the second product is understood to be $1$ if $n=2.$
For a periodic and integrable function $h(x)$ with period $2\pi$, we know that $\int_b^{b+2\pi}h(x)\,dx=\int_0^{2\pi}h(x)\,dx.$ By induction and the Fubini theorem, we have
\begin{eqnarray}
I(m,n)&= & C_n\int_0^{2\pi}\cdots \int_{0}^{2\pi}\cos (mx_2)\cdot G_n(x)\,dx_2\cdots\,dx_n\,d\theta_1\nonumber\\
& = & (2\pi)C_n\int_{0}^{2\pi}\cdots \int_{0}^{2\pi}\cos (mx_2)\cdot G_n(x)\,dx_2\cdots\,dx_n\lbl{Lamar}\\
& = & (2\pi)C_n\int_{0}^{2\pi}\cdots \int_{0}^{2\pi}\cos (mx_2)J_n(x) H_n(x)\,dx_2\cdots\,dx_n\lbl{hotwater}
\end{eqnarray}
where $G_n(x)=J_n(x)H_n(x)$ and
\begin{eqnarray*}
H_n(x)=\begin{cases}
\prod_{i=3}^n\big|\sin\Big(\frac{x_i}{2}\Big)\big|^{\beta}\cdot \prod_{3\leq j<k \leq n}\big|\sin\Big(\frac{x_j - x_k}{2}\Big)\big|^{\beta}, & \text{if $n\geq 4$};\\
\big|\sin\Big(\frac{x_3}{2}\Big)\big|^{\beta}, & \text{if $n= 3$};\\
 1, & \text{if $n=2$}
\end{cases}
\end{eqnarray*}
and
\begin{eqnarray*}
J_n(x)=\begin{cases}
\big|\sin\Big(\frac{x_2}{2}\Big)\big|^{\beta}\prod_{i=3}^n\big|\sin\Big(\frac{x_2 - x_i}{2}\Big)\big|^{\beta}, & \text{if $n\geq 3$};\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\
 \big|\sin\Big(\frac{x_2}{2}\Big)\big|^{\beta}, & \text{if $n=2$.}
\end{cases}
\end{eqnarray*}
In particular,
\begin{eqnarray}\lbl{Santafe}
I(m, 2)=2\pi C_2\int_0^{2\pi}\cos (mx_2)J_2(x)\,dx_2.
\end{eqnarray}
Taking $m=0$ in (\ref{Lamar}), we know from (\ref{Fateev}) that
\begin{eqnarray*}
\int_0^{2\pi}\cdots \int_{0}^{2\pi}\prod_{i=2}^n\Big|\sin\Big(\frac{x_i}{2}\Big)\Big|^{\beta}\cdot \prod_{2\leq j<k \leq n}\Big|\sin\Big(\frac{x_j - x_k}{2}\Big)\Big|^{\beta}\,dx_2\,dx_3\cdots\,dx_n=\frac{1}{2\pi C_n}
\end{eqnarray*}
for all $n\geq 2,$ where the second product above is understood to be $1$ if $n=2.$ This implies
\begin{eqnarray}\lbl{back}
\int_{0}^{2\pi}\cdots\int_0^{2\pi} H_n(x)\,dx_3\cdots\,dx_n=\frac{1}{2\pi C_{n-1}}
\end{eqnarray}
for all $n\geq 3.$ Now, recalling the definition of $J_n(x),$ let $t=x_2/2,$ we have
\begin{eqnarray*}
\int_{0}^{2\pi}\cos (mx_2)J_n(x)\,dx_2=2\int_{0}^{\pi}\cos (2mt)\Big|\prod_{i=1}^{n-1}\sin\big(t+a_i\big)\Big|^{\beta}\,dt
\end{eqnarray*}
for all $n\geq 2,$ where $a_1=0,\, a_i=-x_{i+1}/2$ for $i=2,\cdots,n-1.$ By Lemma \ref{Jelly},
\begin{eqnarray}\lbl{regular}
\Big|\int_{0}^{2\pi}\cos (mx_2)J_n(x)\,dx_2\Big|\leq 12(1+\beta)\left(\frac{n}{m}\right)^{1\wedge \beta }
\end{eqnarray}
for all $n\geq 2.$ Therefore, this and (\ref{Santafe}) imply that for some constant $K_1=K_1(\beta),$
\begin{eqnarray}\lbl{bar}
|I(m,2)| \leq \frac{K_1}{m^{(1\wedge \beta)}}.
\end{eqnarray}
 Now assume $n\geq 3.$ By (\ref{hotwater}) and (\ref{regular}), and then (\ref{back}), we obtain
\begin{eqnarray}
|I(m,n)| &\leq & 24\pi (1+\beta)C_n\Big(\frac{n}{m}\Big)^{1\wedge \beta }\int_{0}^{2\pi}\cdots\int_0^{2\pi} H_n(x)\,dx_3\cdots\,dx_n\nonumber\\
& = & 12(1+\beta)\Big(\frac{n}{m}\Big)^{1\wedge \beta }\frac{C_n}{C_{n-1}}\lbl{dong}
\end{eqnarray}
for all $n\geq 3.$ Now,
\begin{eqnarray}\lbl{li}
\frac{C_n}{C_{n-1}}=\frac{\Gamma(1+\beta/2)}{2\pi}\cdot\frac{\Gamma(1+\beta n/2-\beta/2)}{\Gamma(1+\beta n/2)}\cdot 2^{(n-1)\beta}
\end{eqnarray}
for all $n\geq 3.$ By Lemma \ref{stirling}, there exists a constant $K_2=K_2(\beta)$ such that
\begin{eqnarray*}
\frac{\Gamma(1+\beta n/2-\beta/2)}{\Gamma(1+\beta n/2)} \leq \frac{K_2}{n^{\beta/2}}
\end{eqnarray*}
for all $n\geq 1.$ This, (\ref{dong}) and (\ref{li}) imply that there exists a constant $K=K(\beta)$  such that
\begin{eqnarray*}
|I(m,n)| \leq K\cdot \Big(\frac{n}{m}\Big)^{1\wedge \beta}\cdot \frac{1}{n^{\beta/2}}\cdot 2^{n\beta}= K n^{(1\wedge \beta) -\beta/2}\frac{2^{n\beta}}{m^{1\wedge \beta}}\leq K\frac{n2^{n\beta}}{m^{1\wedge \beta}}
\end{eqnarray*}
for all $n\geq 3.$ This together with (\ref{bar}) proves the lemma. \ \ \ \ \ \ \ \ \ $\blacksquare$\\

\noindent\textbf{Proof of (c) of Theorem \ref{face}}. Observe that, for any real numbers $x_1, \cdots, x_n,$
\begin{eqnarray*}
\Big|\sum_{j=1}^ne^{ix_j}\Big|^2 & = &  \sum_{j=1}^ne^{ix_j}\cdot \sum_{j=1}^ne^{-ix_j}\\
&= & n + \sum_{j\ne k}e^{i(x_j-x_k)}= n + \sum_{1\leq j< k\leq n}\Big(e^{i(x_j-x_k)} + e^{-i(x_j-x_k)}\Big)\\
& = & n+ 2\sum_{1\leq j< k\leq n}\cos(x_j-x_k).
\end{eqnarray*}
Thus, by the symmetry of $f(\theta_1, \cdots, \theta_n|\beta),$
\begin{eqnarray}
\mathbb{E} \big[ |p_m(Z_n)|^2 \big]=\mathbb{E}
\Bigg[\Big|\sum_{j=1}^ne^{im\theta_j}\Big|^2 \Bigg]= n + n(n-1)\cdot \mathbb{E}[\cos \{m(\theta_1-\theta_2)\}].
\end{eqnarray}
The conclusion then follows from Lemma \ref{button}. \ \ \ \ \ \ \ \ \ $\blacksquare$

\section{Appendix}\lbl{appendix1}
\setcounter{equation}{0}

In this section we calculate some moments for the circular $\beta$-ensembles. The first result below is an independent check of the second moment of the trace of a COE given in (\ref{glue}). The derivation does not depend on the Jack function as used in Section \ref{proofs}. It only uses the distribution of the entries of the COE.
\begin{lemma}\lbl{blue} Let $W_n$ be an $n\times n$ circular orthogonal ensemble (COE), that is, $W_n=U_n^T U_n$ for some Haar-invariant unitary matrix $U_n$.
Then $\mathbb{E}\big[|\Tr(W_n)|^2 \big]=2n/(n+1)$ for all $n\geq 2.$
\end{lemma}
\noindent\textbf{First Proof of Lemma \ref{blue}}. We prove the lemma in three steps.

{\it Step 1}. Write $U_n=(u_{rs}).$ First, we claim that
\begin{eqnarray}\lbl{snow}
\mathbb{E}[u^2_{rs}\bar{u}_{pq}^2]=0
\end{eqnarray}
 if $r\ne p$ or $s\ne q.$ In fact, since $U_n$ is Haar-invariant unitary, the distributions of $UU_n$ and $U_nU$ are the same as that of $U_n$ for any unitary matrix $U$. In particular, take $U=\mbox{diag}(e^{i\theta_k})_{1\leq k \leq n}$ to obtain that
\begin{eqnarray}\lbl{grass3}
\mathcal{L}\left(\left(e^{i\theta_r}u_{rs}\right)_{1\leq r, s\leq n}\right)=\mathcal{L}\left(\left(e^{i\theta_s}u_{rs}\right)_{1\leq r, s\leq n}\right)=\mathcal{L}\left(\left(u_{rs}\right)_{1\leq r, s\leq n}\right)
\end{eqnarray}
for any $\theta_1, \cdots, \theta_n \in \mathbb{R},$ where $\mathcal{L}(X)$ is the joint distribution of the entries of random matrix $X.$
 If $r\ne p$, taking $\theta_r-\theta_p=\pi/2$, then by (\ref{grass3}), we have that
\begin{eqnarray*}
\mathbb{E}[u^2_{rs}\bar{u}_{pq}^2]=e^{2i(\theta_r-\theta_p)}\mathbb{E}[u^2_{rs}\bar{u}_{pq}^2]=-\mathbb{E}[u^2_{rs}\bar{u}_{pq}^2]
\end{eqnarray*}
which means (\ref{snow}). The case for $s=q$ can be proved similarly.

{\it Step 2}. Recall notation
$(2m-1)!\,!=(2m-1)(2m-3)\cdots 3\cdot 1$ for any integer $m\geq 1,$
and $(-1)!!=1$ by convention. We have the following fact (Lemma 2.4 from \cite{Jiang09b}):
\begin{eqnarray}\lbl{sushi}
\mathbb{E}[\xi_1^{a_1}\xi_2^{a_2}\cdots
\xi_{n}^{a_{n}}]=\frac{\prod_{i=1}^{n}(2a_i-1)!\,!}{\prod_{i=1}^{a}(n+2i-2)}.
\end{eqnarray}
where  $a_1,\cdots, a_{n}$ are non-negative integers with $a=\sum_{i=1}^{n}a_i$,\ $\xi_i=X_i^2/(X_1^2+\cdots +  X_n^2)$ and $X_1, \cdots, X_n$ are i.i.d.
random variables with $X_1\sim N(0,1).$

{\it Step 3}. Evidently, $\Tr(W_n)=\sum_{1\leq i, j\leq n}u_{ij}^2.$ Notice, from the invariant property, by exchanging some rows and some columns of $U_n$, we see that the distributions of $u_{rs}$ and $u_{11}$ are identical for any $1\leq r, s\leq n$. By (\ref{snow}),

\begin{eqnarray}\lbl{morning}
\mathbb{E}\big[|\Tr(W_n)|^2\big]
 =  \mathbb{E}\Bigg[ (\sum_{r,s}u_{rs}^2)(\sum_{p,q}\bar{u}_{p,q}^2) \Bigg]
 =\mathbb{E}\Bigg[\sum_{r,s}|u_{rs}|^4 \Bigg]=n^2E \big[|u_{11}|^4 \big].
\end{eqnarray}
It is known (e.g., Lemma 2.1 in \cite{Jiang09a, Jiang09b}) that the probability distribution of $|u_{11}|^2$ is the same as that of $(X_1^2 + X_2^2)/\sum_{i=1}^{2n}{X_i}^2$.
 By (\ref{sushi}),
\begin{eqnarray*}
\mathbb{E}[\xi_1^2]=\frac{3}{2n(2n+2)}\ \ \mbox{and}\ \ \ \mathbb{E}[\xi_1\xi_2]=\frac{1}{2n(2n+2)}.
\end{eqnarray*}
Then
\begin{eqnarray*}
\mathbb{E}[|u_{11}|^4]=\mathbb{E}[(\xi_1 + \xi_2)^2]=2\mathbb{E}[\xi_1^2] + 2\mathbb{E}[\xi_1\xi_2]=\frac{2}{n(n+1)}.
\end{eqnarray*}
Substitute this into (\ref{morning}) to see that $\mathbb{E}\big[|\Tr(W_n)|^2\big]=2n/(n+1).$\ \ \ \ \ \ \ \ $\blacksquare$\\

\noindent\textbf{Second Proof of Lemma \ref{blue}}.
We use the following formula due to Collins \cite{Collins} (see also \cite{Matsumoto}):
let $(u_{ij})_{1 \le i,j \le n}$ be an $n \times n$ CUE matrix
(or equivalently, an Haar-distributed unitary matrix) and
let $i_1,\dots,i_k$, $j_1,\dots,j_k$, $i_{1}',\dots,i_k'$, $j_1'\dots,j_k'$
be elements in $\{1,2,\dots,n\}$. Then
\begin{eqnarray}
\mathbb{E} \big[
u_{i_1 j_1} \cdots u_{i_k j_k} \overline{u_{i_1'j_1'} \cdots u_{i_k'j_k'}}
\big]
= \sum_{\sigma, \tau \in \mathfrak{S}_k} \mathrm{Wg}_{n,k}(\sigma^{-1} \tau)
\Big( \prod_{p=1}^k \delta_{i_p, i_{\sigma(p)}'} \Big)
\Big( \prod_{q=1}^k \delta_{j_q, j_{\tau(q)}'} \Big).
\lbl{WgFormula}
\end{eqnarray}
Here $\mathfrak{S}_k$ is the symmetric group and $\mathrm{Wg}_{n,k}$ is a
class function on $\mathfrak{S}_k$,
called the Weingarten function for the unitary group.
For our purpose, we do not need the explicit definition of $\mathrm{Wg}_{n,k}$ but
 use the case for $k=2$. In fact, for $n \ge 2$, we know (see (5.2) of \cite{Collins})
\begin{eqnarray}
\mathrm{Wg}_{n,2}(\mathrm{id}_2) = \frac{1}{n^2-1} \qquad \text{and} \qquad
\mathrm{Wg}_{n,2}((1 \ 2)) = -\frac{1}{n(n^2-1)},
\lbl{Wg}
\end{eqnarray}
where $\mathrm{id}_2$ and $(1 \ 2)$ are the identity permutation and
the transposition on $\{1,2\}$, respectively.

We have $|\Tr(W_n)|^2= \sum_{r,s,p,q} u_{rs}^2 \overline{u}_{pq}^2$.
By \eqref{WgFormula}, $\mathbb{E} [ u_{rs}^2 \overline{u}_{pq}^2]$ is zero unless
$r=p$ and $s=q.$ Moreover, $\mathbb{E} [ u_{rs}^2 \overline{u}_{rs}^2] =
\mathbb{E}[|u_{11}|^4]$ for all $1\leq r,\, s\leq n.$
Therefore, using \eqref{WgFormula} and \eqref{Wg}, we obtain
\begin{eqnarray*}
\mathbb{E} \big[ |\Tr(W_n)|^2 \big] = n^2 \mathbb{E} [|u_{11}|^4]
=2n^2 \Big\{ \mathrm{Wg}_{n,2}( \mathrm{id}_2)+ \mathrm{Wg}_{n,2}( (1\ 2)) \Big\}=
\frac{2n}{n+1}.\ \ \ \ \ \ \ \ \blacksquare\\
\end{eqnarray*}
Lemma \ref{blue} corresponds to the conclusion for $\beta=1$ in (\ref{cry}), which is derived through Proposition \ref{exact} by the Jack functions.
Now we apply the same proposition  to derive some other moments for the circular  $\beta$-ensembles. Let $p_k$ and $Z_n$ be as in Theorem \ref{face}.\\

\noindent \textbf{Example}\ Assume $\alpha=2/\beta>0.$ For $n\geq 2,$
\begin{eqnarray}
\mathbb{E} [|p_1(Z_n)|^4]
&= & \frac{2n\alpha^2(n^2+  2(\alpha-1)n-\alpha)}{(n+\alpha-1)(n+\alpha-2)(n+ 2\alpha-1)}\lbl{right}\\
& = &
\begin{cases}
\frac{8(n^2+2n-2)}{(n+1)(n+3)},& \text{if $\beta=1$;}\nonumber\\
2, & \text{if $\beta=2$};\nonumber\\
\frac{2n^2-2n-1}{(2n-1)(2n-3)}, & \text{if $\beta=4$.}\nonumber
\end{cases}
\end{eqnarray}

\noindent \textbf{Example}\ Assume $\alpha=2/\beta>0.$ For $n\geq 2,$
\begin{eqnarray}
& & \mathbb{E} [|p_2(Z_n)|^2]=\frac{2\alpha n(n^2 + 2(\alpha-1)n + \alpha^2-3\alpha+1)}{(n+\alpha-1)(n+2\alpha-1)(n+\alpha-2)}\lbl{left}\\
&= & \begin{cases}
\frac{4(n^2+2n-1)}{(n+1)(n+3)},& \text{if $\beta=1$;}\nonumber\\
2,& \text{if $\beta=2$;}\\
\frac{4n^2-4n-1}{(2n-1)(2n-3)},& \text{if $\beta=4$.}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{cases}
\end{eqnarray}

\noindent \textbf{Example}\ Assume $\alpha=2/\beta>0.$ For $n\geq 2,$
\begin{eqnarray}
& & \mathbb{E} \Big [p_2(Z_n)\overline{p_1(Z_n)^2}\,\Big]
 =  \mathbb{E} \Big [\,\overline{p_2(Z_n)}p_1(Z_n)^2\,\Big]\nonumber\\
&= &
\frac{2\alpha^2(\alpha-1)n}{(n+\alpha -1)(n+2\alpha -1)(n+\alpha-2)}\lbl{upper}\\
& = &
\begin{cases}
\frac{8}{(n+1)(n+3)},& \text{if $\beta=1$;}\nonumber\\
0,& \text{if $\beta=2$;}\nonumber\\
\frac{-1}{(2n-1)(2n-3)},& \text{if $\beta=4$.}\nonumber
\end{cases}
\end{eqnarray}
In particular, if $\beta\ne 2,$ as $n\to\infty,$
\begin{eqnarray}\lbl{Chinese}
\mathbb{E} \Big [p_2(Z_n)\overline{p_1(Z_n)^2}\,\Big] \sim 2\alpha^2(\alpha-1)n^{-2}.
\end{eqnarray}

\noindent\textbf{Proofs of (\ref{cry}), (\ref{right}), (\ref{left}) and (\ref{upper})}. Let $n\geq 2$, $\mu$ and $\nu$ be partitions of $2.$ Set $\alpha=2/\beta.$ By Proposition \ref{exact}
and (\ref{piano}), we have
\begin{eqnarray}
\mathbb{E}\Big[p_{\mu}(Z_n)\overline{p_{\nu}(Z_n)}\, \Big]=\alpha^{l(\mu)+ l(\nu)}\Big(\frac{4\alpha^{2-l(\mu)}\alpha^{2-l(\nu)}}{2\alpha^2(\alpha +1)}\frac{n(n+\alpha)}{(n+\alpha -1)(n+2\alpha-1)}\nonumber\\
\ \ \ \ \ \ \ \ \ \ \  + \frac{4(-1)^{2-l(\mu)}(-1)^{2-l(\nu)}}{2\alpha (\alpha +1)}\frac{n(n-1)}{(n+\alpha -1)(n+\alpha-2)}
\Big)\nonumber\\
=\frac{2\alpha^{l(\mu)+ l(\nu)}n}{\alpha^2(\alpha +1)(n+\alpha -1)}\Big(\frac{\alpha^{4-l(\mu)-l(\nu)}(n+\alpha)}{n+2\alpha -1} +\frac{(-1)^{l(\mu) + l(\nu)}\alpha (n-1)}{n+\alpha-2}\Big).\lbl{long}\\
\nonumber
\end{eqnarray}
\indent (i) Take $\mu=\nu=(1)$ in Proposition \ref{exact}. Since $\theta_{(1)}^{(1)}(\alpha)=1,\, C_{(1)}(\alpha)=\alpha$ for any $\alpha>0,$ we obtain (\ref{cry}).

(ii) Taking $\mu=\nu=(1,1)$ in (\ref{long}), (\ref{right}) follows.

(iii) Taking $\mu=\nu=(2)$ in (\ref{long}), (\ref{left}) follows.

(iv) Taking $\mu=(2)$ and $\nu=(1,1)$ in (\ref{long}), we get the identity for the first expectation in (\ref{upper}). Since the value of the expectation is real, the identity for the second expectation follows. With the earlier conclusion, (\ref{Chinese}) is obvious.\ \ \ \ \ \ \ \ $\blacksquare$\\

\noindent\textbf{Acknowledgement}\ The first author thanks Professor Persi Diaconis very much for introducing to him the problem studied in this paper; he also thanks the NSF for a partial support for this research. We thank Professor Benoit Collins for a very helpful private communication.

\baselineskip 12pt
\def\ref{\par\noindent\hangindent 25pt}

redpajama_set_name: RedPajamaArXiv


text:
Freight Traffic Between Russia and Poland Faces Suspension
Freight traffic between Russia and Poland may be suspended because Poland refuses to follow new Russian regulations, Russia's Transport Ministry announced on Friday, the Interfax news agency reported.
The new regulations, according to the ministry, are aimed at tighter control of Russian permits issued for motor transportation of goods from other countries, but Polish officials insist the regulations imply a higher level of state control over freight traffic.
The ministry points out that Poland hasn't offered any solution to the problem other than demanding that the federal law be changed, Interfax reported.
In 2014 a law was passed in Russia introducing stricter registration rules for international freight traffic.
Hailed as Hero at EU Summit, Zelensky Urges Faster Arms Supplies
Poland to Summon Russian Envoy as Diplomatic Spat Drags On
Poland said on Saturday it would summon Russia's envoy after he said Warsaw was partly to blame for the outbreak of World War II, continuing a spat...
Poland Has Right to Dismantle Soviet-Era Monuments — Ambassador
Poland has the right to take down Soviet-era monuments, the Polish Ambassador to Russia said after the Russian Foreign Ministry issued a protest against...
Russia's Envoy to Poland Withdraws World War II Remarks
Russia's ambassador to Poland has partly backtracked from an accusation that Poland bears some blame for starting World War II because of its policies...
Ukraine's President Volodymyr Zelensky received a hero's welcome in Brussels on Thursday as he lobbied star-struck EU leaders to speed the delivery of...

redpajama_set_name: RedPajamaCommonCrawl


text:
module Hbc::Verify; end

require "hbc/verify/checksum"
require "hbc/verify/gpg"

module Hbc::Verify
  module_function

  def verifications
    [
      Hbc::Verify::Checksum
      # TODO: Hbc::Verify::Gpg
    ]
  end

  def all(cask, downloaded_path)
    odebug "Verifying download"
    verifications = for_cask(cask)
    odebug "#{verifications.size} verifications defined", verifications
    verifications.each do |verification|
      odebug "Running verification of class #{verification}"
      verification.new(cask, downloaded_path).verify
    end
  end

  def for_cask(cask)
    odebug "Determining which verifications to run for Cask #{cask}"
    verifications.select do |verification|
      odebug "Checking for verification class #{verification}"
      verification.me?(cask)
    end
  end
end

redpajama_set_name: RedPajamaGithub


text:
Wearing a full black mask, Ye to Alex Jones: 'I like Hitler'
Ye continues to find himself in a pickle; either involuntarily or NOT.
Kanye West in a controversial interview
During an interview with conspiracy theorist Alex Jones on his platform Infowars, Ye, wearing a black mask completely covering his face, said he likes Adolf Hitler, leader of Nazi Germany whose regime cost the lives of millions.
"I like Hitler," Ye told Jonse. "I love Jewish people, but I also love Nazis."
Even though West hid his face, there seemed no doubt it was him.
You know Kanye has reached a point of no return when he makes Alex Jones look normal.
— anthony "light 1" fantano (@theneedledrop) December 1, 2022
This comes in light of recent accusations held against the rapper over racist and anti-semitic remarks, which caused him to lose deals with Adidas, Balenciaga, Gap, Footlocker, and other brands.
Ye further said he did not like the adjective "evil" to refer to Nazis, stressing that every human being has something of value, "especially Hitler".
"Every human being has something of value that they brought to the table, especially Hitler" -Ye pic.twitter.com/05dwYsKLXf
— Heisenberg Fan Account (PARODY) (@methalcohol) December 1, 2022
Neo-Nazi Nick Fuentes also appeared on the show and said women shouldn't be allowed to vote.
Ye was recently spotted at Trump's Mar-a-Lago country club and home in the company of self-proclaimed white nationalist Nick Fuentes in November.
Republicans have normalized hate so much in America that Kanye can wildly declare how much he loves Nazis and Hitler, and folks are like ¯\_(ツ)_/¯.
So disgusted right now.
— BrooklynDad_Defiant!☮ (@mmpadellan) December 1, 2022
In the subsequent video, titled "Mar-A-Lago Debrief," Ye said after he asked Trump to run as his vice-president in the 2024 elections, the former President "started basically screaming at me at the table, telling me I'm going to lose. Has that ever worked for anyone in history? I'm like, 'Wait, hold on Trump, you're talking to Ye.'"
Fuentes, on his part, is reputed for his provocative racist, misogynistic, and fascist opinions.
In 2020, he founded a white nationalist organization called America First Foundation.
He has frequently been called a white supremacist by the US Department of Justice and has been banned from multiple social media platforms.
Read more: Arkansas Gov: Trump dinner with Kanye, Nick Fuentes, not accidental
Musk suspends Kanye West's account after swastika tweet


In two years time the technology of making ribbon microphones had been worked out and the result has surpassed all expectations! The ones who know what kind of microphone it is will certainly say that a ribbon microphone is a studio microphone. Even though it seems to be not enough of high frequencies, the reality of the transmission of the timbre can not be compared to any other microphone. Many beginner sound engineers do not even know what it is. Although recently almost all firms have in their line-up a "ribbon" pattern.
I would not insist that this is a vocal microphone, rather than instrumental, but it is perfect for the baritone. Using a pop-filter is obligatory! It also sounds very good on strings, winds, percussion, xylophone and as overheads. And guitar amps sound fantastic!!!
by analogy with the well-known brands such as Samar Audio and Lundahl.
It is also possible to select a stereo couple of RM BIV-1, which allows to use them in MS and Blumleine (XY) systems.
You can also order a microphone with a shok mount.
A stereo version was created because of the great desire to use MS and Blumleine (XY) recording systems. It is also possible to select a stereo pair of microphones of two RM BIV-1, which allows use them in MS and Blumleine (XY) systems.


text:
Home Bollywood Neena Gupta joins Rishi Kapoor in 'Mulk'
Neena Gupta joins Rishi Kapoor in 'Mulk'
Mumbai–Actress Neena Gupta has been roped in for Anubhav Sinha's "Mulk", which also stars veteran actor Rishi Kapoor.
Neena will essay the role of Rishi Kapoor's wife in the film and has begun her training for the role already. This is the first time she and Rishi will be teaming up, said a statement.
"I met Anubhav at a time when I was desperately seeking quality work. It was a really thoughtful gesture on his part to reach out to me at a time when I was undergoing this inner turmoil. 'Mulk' is a phenomenal film and I'm quite excited to be a part of this experimental film," Neena said.
In August, Neena took to social media to request work, posting a picture on Instagram with the caption: "I live in Mumbai and working, am a good actor looking for good parts to play."
Sinha says he has always been a fan of Neena's work.
"She fits the requirement of the role perfectly. After I saw her Instagram post, I didn't hesitate at all to pick the phone and call her because there are some people in the industry who are timeless and she is one of them," he said.
Sinha, who is both director and producer of "Mulk" added: "I couldn't, for the longest time, find anyone who could match the description of the role, maybe she was destined to be a part of it. When I was dabbling in television in my earlier days I distinctly remember how she delivered some path-breaking performances in 'Saans' and 'Buniyaad', a time when Indian television was actually progressive and intelligently entertaining. She is a forerunner of her times and her involvement will enhance the multi dimensional script even further."
A social thriller based on true life events, "Mulk" will go on floors in October. It will also feature Taapsee Pannu, Prateik Babbar and Rajat Kapur. The film will be shot in Varanasi and Lucknow and is scheduled to release early next year. The film is produced by Benaras Media Works.
Screenplay writer-producer Mushtaq Shiekh will be the associate producer of "Mulk". The plot of the film revolves around the story of a joint family who hails from a small town in India, who are out to reclaim their honour after being embroiled in a controversy. (IANS)
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Conman case: Delhi court allows Jacqueline Fernandez to travel abroad


Many questions are still unanswered about what led to the death of 2½-year-old Gabriella Maria Boyd after Mamaroneck village police shot the girl's mother as she stabbed officers at her home on April 28.
Sunday, April 22: The girl's father, New Rochelle resident Stephen Boyd, received a message from her mother, Cynthia Arce, who told Boyd that she had been thrown out of her Chestnut Avenue home by her mother, Gabriella's grandmother, and had nowhere to stay, said Martin Rosen, Boyd's lawyer.
Thursday, April 26: At some point Arce was allowed back into her home, Rosen said. But on Thursday, Boyd received a troublesome phone call from Arce.
"My client was very concerned about it because of the phone call he received, which I heard, which was very disturbing, and sounded that the woman was in terrible distress and was perhaps about to hurt herself," Rosen said.
He and Boyd submitted an application to Westchester Family Court seeking temporary custody of Gabriella.
The order was signed in the afternoon by Judge Hal Greenwald and granted Boyd temporary physical custody of his daughter and barred Arce from contact with Boyd or Gabriella.
Boyd went to police headquarters at 4:45 pm requesting that they serve the order on Arce. At 5:14 pm police arrived at the Chestnut Avenue house. The papers are served but at 5:42 pm one of the officers reports that Arce refused to turn over the toddler and he is returning to headquarters to "investigate further the situation."
"He was really helpless at that time," Rosen said of Boyd. "He just asked the police to implement the court order and gain entry to the house and take the child. They refused. They said they had some doubts whether they had the authority to enter the house to arrest her."
Arce's lawyer, Richard Portale, says that the interaction was "non-adversarial" and that she told police that the matter could be handled the following day when Boyd was scheduled to be with Gabriella.
After discussion with prosecutors, police were instructed that they did not have authority to forcibly remove the girl from the house.
The police union said that officers notified child protective services and were told that CPS was opening an investigation. The union did not provide a timeframe for that communication.
At 7:15 pm, the New York Statewide Central Register of child abuse and maltreatment was notified by police via its hotline about Gabriella. The register notified Westchester child protective services.
Westchester County officials have said that CPS went to the house later that night but could not gain access to the house. Portale said that Arce's mother, who owns the house, observed two people outside the house at 11:35 pm who she believes may have been CPS workers but she was unaware of any effort they made to gain entry.
Shortly before 9 am, Arce informs Boyd that she won't be bringing Gabriella to New Rochelle for his weekly visitation because the girl was sick.
At 9:09 am, Boyd calls village police and says he will come to headquarters "in about an hour" so they can update the police report to reflect the lack of visitation.
Following Boyd's visit, he calls police again at 12:12 pm to let them know Arce is taking Gabriella to the doctor. He figures if she leaves the house with Gabriella, police can take the girl.
The desk officer confers with colleagues and tells Boyd they still can't help him, it's a civil matter.
According to Portale, Arce's lawyer, Solorzano went to work in the morning and returned at about 1:00 pm.
1:30 p.m.: Police responded to a 911 call from Solorzano and found the dying toddler at the house. They started CPR and Arce ended up stabbing Officers Osvaldo Ramos and Joseph Gaglione and was shot by another officer after efforts to subdue her with a stun gun failed. She had also cut her throat at some point. She and Gabriella were taken to the hospital, where the girl was pronounced dead.
A few hours later, a detective calls Boyd and leaves a message. Boyd calls back at 4:12 pm and is asked to come to headquarters. Once there, they talk to him for several minutes about the events of the previous night before telling him for the first time that Gabriella had been taken to the emergency room but had died.
At 6:46 pm: the New York Statewide Central Register gets child fatality notification for Gabriella but it is not transmitted to Westchester County until two hours later.
Rosen details the police "refusal" to enforce the court order.He said an autopsy is being conducted to find out how Gabriella died. If it was of natural causes, he said, the matter is over. But if there was foul play, which he suspects, "that's going to bring some serious consequences," he said.
Arce is charged with two counts of attempted aggravated murder of a police officer. She is arraigned in her hospital bed.
Westchester District Attorney Anthony Scarpino announces that it would take a few more weeks before tests results are available to help the Medical Examiner determine Gabriella's cause of death.
Amid silence from authorities, the Village of Mamaroneck Police Benevolent Association issues a statement detailing police response on April 27. It said officers had no indication Gabriella was in danger and that they were instructed that they had no authority to forcibly remove the girl from the house.
Arce, still hospitalized, does not appear at her first court hearing in Mamaroneck Village Court.
Arce again is not well enough to appear in court. Portale agrees to allow the case to proceed to the grand jury.
Two days after what would have been Gabriella's 3rd birthday, a 5K run/walk was hosted by the Gabriella Boyd Memorial Foundation in New Rochelle.
Westchester prosecutors confirm that the Medical Examiner's Office has ruled Gabriella's death a homicide, though the exact cause of death is undetermined.
Hours later, Portale tells The Journal News/lohud that while Arce caused her daughter's death she was not criminally responsible due to mental disease or defect. He contends that her bipolar disorder and undiagnosed post-partum depression combined to cause a psychotic break April 28 - and that one of the triggers could have been Stephen Boyd's effort to wrest custody of Gabriella from her.


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Protect the use of the Earth's limited water supply and prioritize its use for the purpose of sustaining all life.
End patterns of exploitation and replace our current profit- and growth-based economies with economies that meet the real needs of humans and other life forms and that protect, restore, and sustain our global environment.
Make communities climate resilient—provide everyone, especially frontline and vulnerable populations, the resources needed to adapt to and reduce the impacts of climate change. This could include affordable housing, health care, education and job training, food and clean water.
Support the leadership and sovereignty of indigenous nations and tribal peoples who have tens of thousands of years of experience effectively sustaining their lands and adapting to change. Support the preservation, revitalization, and widespread adoption of Traditional Ecological Knowledge (TEK, also known as indigenous science) practices. Protect those practices from exploitation. Honor the treaties, the leadership and experience of Native people, and their sovereign rights (recognized or not).
Create the conditions where women can control their own bodies and reproduction without compromise, including through voluntary family planning, human rights for women, and education and opportunity for girls and young women. Birth rates decrease under these conditions.
All national, tribal, state, provincial, and local governments enact and enforce policies that are in accordance with this draft program. Require all industries and businesses to make ending climate change part of their goals above being profitable, in line with the Paris Agreement. All national governments immediately strengthen and meet their nationally determined contributions (NDCs) to reduce GHG emissions under the Paris Agreement and the 2018 IPCC Report.
Rapidly reduce fossil fuel use, production (including fracking), exploration, and infrastructure. Remove the subsidies that continue to encourage the use of fossil fuels. Support divestment from fossil fuels.
Refocus tribal, national, and local economies on a fossil-free future, budgeting the necessary resources to meet these goals.
Reduce energy consumption to the level of rational need through conservation and energy efficiency. Improve energy efficiency at all levels of use.
Prioritizing community-based/community-led solutions, develop and transition to renewable energy sources, such as solar, wind, geothermal, small scale and micro-hydro, as well as storage and delivery systems, and convert to all-electric power usage.
Decommission existing nuclear reactors where they can be replaced with renewable energy, not fossil fuels. No new current generation nuclear reactors should be built.
Ensure that rapidly reducing GHG emissions and decommissioning nuclear reactors not be at the expense of workers in the affected industries. Support training (with pay) and resources for workers in the fossil fuel and nuclear industries, the military, and other related institutions and industries that need to be phased out to achieve the emissions reduction goal.
Support the wide adoption of climate friendly farming techniques that reduce emissions from producing crops and livestock and/or remove carbon dioxide from the atmosphere and store it in soil and perennial plants like trees. Ensure that frontline populations are educated in these techniques.
Support methods that make farm and grazing land more productive with agroecological techniques, which can also reduce pressure on forests.
Reduce emissions from deforestation and industrial agriculture by greatly reducing consumption of livestock protein in wealthy nations and putting strong limits on biofuel. End deforestation while backing the leadership of and protecting the sovereignty and stewardship of indigenous peoples.
End agricultural subsidies that support inefficient and polluting practices.
Promote, where possible, local food production and consumption. Support food sovereignty for frontline communities.
Protect and restore natural carbon sinks (such as oceans, forests, peatlands, and wetlands), with full engagement of, inclusion of and leadership by the people who inhabit those places.
Develop new safe, affordable, and effective practices and technologies to capture CO2 from the atmosphere and store it.
End war and support universal demilitarization.
Use military budgets to fund the transition to a sustainable, "clean, renewable energy," future.
Reorganize our societies so people live and work in local communities and reduce reliance on travel and shipping.
Provide widespread and affordable access to public transportation powered by renewable energy.
Provide lowest emissions commercial shipping and air travel, while reducing air travel.
Adopt low and zero emissions construction materials and techniques. Change our industrial economy to one with more efficient use and re-use of materials.
Design and construct new buildings and rehabilitate existing buildings to be energy efficient and environmentally friendly, primarily in wealthy nations, with priority to low-income communities.
Reduce consumption, primarily in wealthy nations, through lifestyle change and zero-waste strategies. Rethink whole industries that generate GHG emissions with limited human value, such as armaments, fashion, advertising, psychopharmacology, and beautification.
Ensure that wealthy nations provide the necessary technological and financial assistance to other nations and frontline populations so they may create and/or implement locally-appropriate initiatives that enable them to adapt to climate change, mitigate its effects on their populations, stay on their own land, and rapidly gain access to renewable energy and carbon farming techniques.
End policies that limit the rights of climate migrants, refugees, and displaced people seeking the right to a safe and dignified life in a new land. Recognize that there can be a conflict that will need to be resolved between safeguarding sovereign native rights to their land and the needs of these populations of migrants.
Communicate to people everywhere about climate change: the causes (including human distress recordings), results, disparate impact, and solutions in a way that will move them to take individual and collective actions in support of this draft program.
Build an international movement committed to ending climate change as part of transforming society to one that ends the exploitation of humans, all forms of oppression, irrational and destructive use of resources, and creates a sustainable, equitable future for all humans, all life on earth, and the planet.
Join organizations focused on climate change and work with these organizations to: (a) back the leadership of indigenous people, people of the global majority, poor and working class people, women and youth, (b) assist people to understand and address the connections between climate change, racism, genocide, and other oppressions, and the economic system that both drives and benefits from them; (c) educate people about and implement elements of the above program as part of the work of the organization, and (d) support the work of the organization by building relationships, supporting good leadership, helping with its work, and introducing the tools and practices of RC.
Join organizations focused on social justice and ending oppression and work with them to (a) understand and address the connections between their issues and climate change, (b) educate people about and implement elements of the above program as part of the work of the organization, and (c) support the work of the organization by building relationships, supporting good leadership, helping with its work, and introducing the tools and practices of RC.
Encourage these groups to build alliances to end climate change, environmental degradation, capitalism, oppression, war, and poverty and create an equitable, sustainable future.
Advocate and agitate for good policy and action wherever you are, through whatever means you have. Where possible, seek to end patterns of exploitation and advocate for transformative change (ending class society) rather than simply reform. When working for reforms, use the opportunity to create the conditions for people to understand the need for transformative change and move toward it. Include nonviolent strategies of protest, resistance, and direct action in support of this program.
While the massive changes that are needed will require action by every level of government, industry, and all institutions; personal changes are also required, and especially by those of us in the Global North.
One way to start would be to develop a personal program for reducing greenhouse gas emissions in your household. Set a goal to reduce your emissions by 10% (and another 10% when the first goal is achieved, and so on) and research how you can accomplish that, addressing each point above that is relevant in your family. Develop a plan that all family members can participate in. Try your plan for a week and see if it reduced emissions. How was it for you and your family? Have a session about it and see if you need to revise your plan. Can you then involve another family in your neighborhood in trying it for a week with you? Could it become a neighborhood project? A town project?
Learn about how you can help preserve carbon sinks in your geographical region and implement practices that support these efforts, for example: gardening that restores soil carbon and planting trees. Is anyone near you involved in these efforts? Would they welcome you and others in their efforts?
We can discharge the worries and fears that would interfere with our ability to think about and act rationally, with integrity and courage, in the potentially widespread social upheavals as climate change progresses.
We can take the RC care of the environment goal into sessions and make it a personal goal.
We can work in our classes and workshops on the RC care of the environment goal, the draft policy, and draft program, and help people make them their own.
We can support the projects of Sustaining All Life, United to End Racism, and No Limits for Women.
We can help organizations and leaders in the climate movement to lead more effectively by sharing the tools, theory and practice of RC with them.
We can organize our neighbors to reduce energy use, plant trees, and so on.
We can reduce green house gas emissions and consumption, especially those of us in the countries contributing the most greenhouse gas emissions.
We can always build unity.
We can work to find a rational economic system that is in everyone's interest.
 More detailed information on all of these issues can be found in the article Why We Prioritize Addressing Climate Change.
 We define vulnerable populations as people with disabilities, children, older people, homeless and displaced people, and people in institutions—populations often not considered in developing climate change policy.
 Upholding treaty obligations between sovereign nations would be a big step toward resolving climate change.
 This could include ending sexism, sexual exploitation, and forced marriage; educating girls and boys about sexual coercion and family planning; and providing free and universal access to rational and safe birth control.
 The wealthy nations must reduce emissions in line with principles of equity (the idea that countries with more responsibility for causing the problem, and more capacity to act, should do more than those who lack capacity and contributed few emissions).
 By the terms of the Paris Agreement, the U.S. remains a party until November 2020.
 Mega-dams should not be supported because they destroy local ecosystems and take land, often from poor and indigenous people.
 Nuclear reactors are unsafe, expensive, and harmful to life and the environment. But we would not decommission them if it requires they be replaced with fossil fuel power sources, which would result in emissions that are more dangerous at this time than the reactors.
 Food sovereignty is the right to healthy and culturally appropriate food produced through ecologically sound and sustainable methods, and their right to define their own food and agriculture systems.
 Food waste in landfills is also a major source of methane, which can be reduced by composting and other techniques.
 When indigenous and tribal people have sovereign rights to their forest land, emissions from deforestation and degradation are greatly reduced.
 Adopting Zero Waste plans and practices will result in significant reductions in consumption and trash disposal in the short-term and eventual complete elimination of waste. This includes increased recycling, reuse and composting as well as reduction in consumption of materials.
 This is to avoid the destructive path taken by the wealthy nations. It is in everyone's interest that developing nations not pursue a fossil fuel-based economy and instead be provided needed resources to build their economies as they determine using renewable energy sources.
 This includes insuring that wealthy nations fulfill their commitment to mobilize US$100 billion per year in climate finance from public and private sources for developing countries by 2020.


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Turbo-Charged Freakin' Turkeysaurus Rex
Turbo-charged freakin' Turkeysaurus Rex!
SALT LAKE CITY - Fossils discovered in southern Utah are from a new species of birdlike dinosaur that resembled a 7-foot-tall brightly colored turkey and could run up to 25 mph, scientists said Tuesday.
Fossils of the meat-eater's hand-like claw and foot were found in the Grand Staircase-Escalante National Monument near the Arizona border, giving paleontologists reason to believe some dinosaurs known as raptors roamed from Canada to northern New Mexico about 75 million years ago.
Much smaller variations of the dinosaur had been found previously in Montana, South Dakota and the Canadian province of Alberta.
The dinosaur had a strong toothless beak, powerful arms and formidable claws that made it capable of eating animals and plants. Large feathers grew on its hind end, giving it a resemblance to a turkey, Zanno said.
Scientists are not sure what purpose the feathers served, but it was not for flying. "It's quite different from modern birds," she said.
"This is the last great, unexplored dinosaur bone yard in the lower 48 states," said Scott Sampson, the museum's chief curator who wrote the journal article with Zanno.
Three other dinosaurs discovered at the monument are expected to be named soon, Sampson said, including a meat-eating tyrannosaur, a horned dinosaur and a duckbilled dinosaur with a 7-foot-long head.
Turbo-charged. Freakin'. Turkeysaurus Rex.
Scientists speculate the species died out due to overhunting by a primative species of mammals resembling John Madden.
I always like it when guys studying bones report that their find was "brightly-colored." Like, ummm, okay. You know that. Because, uhhh, the bones are brightly colored or something, right?
Paleontologists are scientists, but they sure do feel comfortable doing a lot of speculating, don't they?
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South Africa has high data prices compared to other countries globally. The parliamentary portfolio committee on telecommunications and postal services conducted a two day hearing in September (2016) to investigate the high data prices in South Africa. In these hearings the South African mobile network operators (MNOs) submitted that the primary reason for high data costs was the "spectrum crunch" in large urban areas.
Participation by a greater number of firms and individuals in the South African economy is hindered by a range of barriers identified in recent CCRED studies, including access to finance.


text:
BY Elizabeth Callen
From the October 2017 St. Croix Valley Magazine issue
Mom-and-Pop Shop
The small grocery store Len's Family Foods is a hidden gem with deep Stillwater roots.
Tate Carlson
For more than 100 years, a small grocery store has stood on the corner of Myrtle and Owens streets in Stillwater. Although ownership has passed through various hands several times, Len's Family Foods has occupied the corner store for more than two decades—and has no plans to leave any time soon. Open 365 days of the year, the grocery store is a neighborhood gem, well-loved and well-known in the Valley as a warm and welcoming place to shop.
The family in Len's Family Foods is not only a part of the name, it is a part of the store's identity, as well. Named for owner Len Nelson, the beloved corner store is a family-owned and -operated business, which makes it not only one of the best-loved grocery stores in the St. Croix Valley, but one of the more unusual, as well.
This year marks the 22nd anniversary of Len's, but the building itself has been around since 1900 and has always housed a grocery store. Nelson says that storied history is a big part of what makes it so special. "We are, in fact, the oldest business in operation in the same location in Stillwater," he says. He first opened the store back in the early 1990s to avoid growing too bored during retirement. After spending his life in the grocery business—his first job as a teenager was in a grocery store—opening his own seemed like the obvious choice. And, of course, the location was already perfect.
"We are one of the last of what are called mom-and-pop stores around, so we're very proud of that, and we're proud to serve our community," Nelson says. With so many chain grocery stores popping up throughout the Valley, family-owned operations are a rarity. Even in the midst of such changes, however, Len's continues to hold its own.
Throughout the years, Len's Family Store has remained a family affair. Nelson's son Kevin works as the store manager; his daughter-in-law Tammy does the books for the store, and his grandson Jacob runs the meat department. Though the staff is small, they work hard and love what they do, and they make a point to treat their customers like family.
"We pride ourselves on our fast and friendly service," Nelson says. At his store, even running in to pick up a dozen eggs or a loaf of bread is a positive experience, and Nelson works to ensure that it's as easy as possible; you'll never have to wait in line too long. "It's always quick and easy shopping, easy parking," he says. "I sound like a record, don't I? But we really do pride ourselves on that."
As a full-service grocery store, Len's has just about everything that could be on your weekly shopping list, despite its small size. There's fresh produce, a bakery and a dairy department, but the store's meat market is particularly noteworthy. "One of the things that we have that's special is our meat department, which has a sterling reputation in the St. Croix Valley," Nelson says. They use the highest-grade products available, and, of course, the service is second to none. Employees greet customers with friendly hellos and are more than happy to answer questions, make recommendations or make you a delicious sandwich for lunch.
So the next time your fridge gets empty—especially as we enter the holiday season, with the store's 365-days-a-year policy—consider paying Len's Family Foods a visit. The colorful mural on the side of the building makes it easy to find, and its friendly service makes supporting a local business even more rewarding. Nelson says that, more than anything, he and his family are grateful that their work gives them the opportunity to serve people throughout the community they have long called home.
"I consider myself a native of Stillwater, because I used to work at Hooley's in the olden days and the Red Owl when that was still here," he says. "I sure love the Stillwater area. I've been here and been a grocer all my life."
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« Ten points if you catch the movie reference.
Last night I headed into Naxx donning my shiny new 13/53/7 spec.
I'd only tested it out briefly, a few minutes with the Target Dummy in Silvermoon (no one else ever uses the ones in Silvermoon!) and then a short romp through Heroic Violet Hold this weekend.
My overall impression from the raid? MM is fun, but sometimes watching all the cooldowns feels like it's just busywork. But I think I've found my temporary happy medium between the horrible DPS of Beast Mastery and the unbearable clunkiness of Survival, so I'll probably stick it out at least until 3.1 hits.
When the raid started, I was… a bit lost. I am still new to this, and I've had to move my buttons around. "Where's Volley?!" I found myself panicking. I couldn't remember what the icon even looked like, I just knew it was always '6′. Well, now it's 'Alt-2′ apparently.
The first thing I noticed was my Volleys were critting SO much. I check my character sheet… I'm up to nearly 47% crit while raid buffed… holy crap! I'm sitting at 36% unbuffed, which is a lot more than my 30% in BM. Crits are shiny.
The second thing I noticed was on our first boss fight – we started with Military, and we were fighting Instructor Razuvious. About 30 seconds into the fight, I glance at my mana bar. I'd used only a scintilla of mana, sitting at 92%! CRAP! I forgot to take Viper off again! I thought to myself. But I check and double check, and I'm most definitely rocking the Dragonhawk.
Even with just 2 points in Efficiency and 1 point in Rapid Recuperation, I don't think I dropped below 50% mana on any fight.
I also need to pay more attention to when I need to use Aimed and when to use Multi. I try to switch between them depending on how many mobs I'm fighting, but sometimes I just hit one without paying attention.
I'm glad I've decided to try this spec out, though! And I was happy to see myself hit 3rd on DPS for one of the boss fights It's no 1st place ZOMG OP BM, but it'll do!
I'm glad you like it! But um…how on earth were you using less mana as MM than as Surv?? This is confusing me. MM is very mana-intensive.
You weren't having any mana issues as a MM? Were you in Naxx 10 or 25?
No no, I was using less mana as MM than I was as BM I had zero points in Efficiency or anything like that as BM, and I even had Multi-shot as part of my shot rotation. I had to use Viper almost every boss fight. I only raided as Survival for 1 or 2 nights, and I don't remember what my mana consumption was like.
Korixys: I was in Naxx 25. I was having no mana issues at all – not really sure why! I know that MM is usually considered very mana-intensive, but I certainly was happy with it.
Yeah, when I am in 25, I never have issues with mana at all whether I'm MM or SV (I switch back and forth quite a bit). You will notice a huge difference in mana when you run heroics and 10 mans as a MM. I like MM's rotation and its definitely the most enjoyable to play IMO, but I love SV because I'm not worrying about running out of mana.
Ah, I thought you were in a 10-man. That explains it. lol Though compared to SV, MM is a black hole for mana. When I was SV, it's like I didn't even need a mana bar.
Yeah, for some reason as BM I have serious mana issues even in 25s. I don't do 10 mans and rarely heroics, which is why I took few mana talents in the first place, because everyone tells me mana shouldn't be an issue in 25s.
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The "immersive spectacle" could be the very cure for your Glastonbury-based withdrawl symptoms next year.
Arcadia - the section responsible for that huge mesmerising or menacing spider (depending on how well your Glasto experience is going) is hosting its own festival in 2018.
Arcadia is coming to London's Queen Elizabeth Olympic Park on 5-6 May, and it looks like they'll be banging out some tunes too!
The people behind the much-loved area at Glastonbury will celebrate 10 years of insane metal sculptures and high-octane installations with help from Leftfield, who'll play the best of their back catalogue, and the Andy C-created RAM records.
Glastonbury Festival 2017 saw Radiohead, Foo Fighters and Ed Sheeran headline, with the Foos finally making up for their cancellation back in 2015.
Florence + The Machine stood in for the band when Dave Grohl broke his leg that year and covered covered their Times Like These single.