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\section{Introduction}

Analytical methods of integral calculations in the lattice gauge theories (LGTs)
had been an important ingredient since the early days of LGT. Many integrals
appearing in the strong coupling expansion of the LGT can be found in \cite{creutz, lat_rev}.
Important progress relevant for this paper was made in Refs. \cite{weingarten, bars_green, bars, samuel_un,
eriksson, brower, balantekin} for $U(N)$ integrals.
Many $U(N)$ integrals considered here have been studied in Refs. \cite{kontsevich_mod, moerbeke1,
moerbeke2, orlov1, orlov2, orlov3}.
Some integrals for $SU(N)$ group can be found in \cite{creutz, wipf, carlsson}.
Last decade has seen a thorough development of the theory
of the integrals over $U(N)$ group. These integrals are of polynomial type
\begin{equation}
{\cal I}_N(r,s) \ = \  \int_{U(N)}  dU \ \prod_{k=1}^{r} U_{i_k j_k} \prod_{n=1}^{s} U_{m_n l_n}^*
\label{integral_def}
\end{equation}
and can be expressed through summation over permutations of the group indices.
The weight of the permutation is given by the Weingarten function whose theory
has been developed in a series of papers \cite{collins1, collins2, novak, zinn-justin,
novaes1, novaes2, collins3, zuber}.
Precisely, the integral is given by
\begin{equation}
{\cal I}_N(r,s) \ = \ \delta_{r,s} \ {\cal I}_N(r)\ ,
\label{unintg_def}
\end{equation}
\begin{equation}
{\cal I}_N(r) \ = \ \sum_{\tau,\sigma\in S_r}  Wg^N(\tau^{-1}\sigma) \prod_{k=1}^{r}
\delta_{i_k,m_{\sigma(k)}} \delta_{j_k,l_{\tau(k)}} \ .
\label{basic_intg}
\end{equation}
$S_r$ is a group of permutations of $r$ elements and
$Wg^N(\sigma)$ is the Weingarten function which depends only on the length of the cycles
of a permutation $\sigma$. Its explicit form is given by
\begin{equation}
Wg^N(\sigma) = \frac{1}{(r!)^2} \ \sum_{\lambda\vdash r} \ \frac{d^2(\lambda)}{s_{\lambda}(1^N)} \ \chi_{\lambda}(\sigma) \ ,
\label{Wgn_def}
\end{equation}
where $d(\lambda), \chi_{\lambda}(\sigma)$ are the dimension and the character of
the irreducible representation $\lambda$ of $S_r$. The irreducible representations $\lambda$ are enumerated
by partitions $\lambda=(\lambda_1,\lambda_2,\cdots,\lambda_{l(\lambda)})$ of $r$,
{\it i.e.} $\sum_{i=1}^{l(\lambda)} \lambda_i \equiv |\lambda|= r$,
where $l(\lambda)$ is the length of the partition and $\lambda_1\geq\lambda_2\geq \cdots \geq \lambda_{l(\lambda)}> 0$.
The sum in (\ref{Wgn_def}) is taken over all $\lambda$ such that $l(\lambda)\leq N$.
$s_{\lambda}(X)$, $X = (x_1,x_2,\cdots,x_N)$, is the Schur function ($s_{\lambda}(1^N)$ is the dimension
of $U(N)$ representation).
To the best of our knowledge, the character expansion of the form (\ref{Wgn_def}), valid for $r\leq N$, had been derived
for the first time in \cite{samuel_un}, where the recurrence relations for $Wg^N(\sigma)$ were also presented.
This result was rederived in \cite{collins1}, again for the case $r\leq N$. In Ref.\cite{collins2} it was shown that
the result (\ref{basic_intg}) and the expression (\ref{Wgn_def}) remains valid also when $r>N$. In this case the restriction
on the length of the partition $l(\lambda)$ can be omitted. If so, and for $r>N$ the function $Wg^N(\sigma)$, as a function of $N$
will have poles at integers $-r+1,\cdots,r-1$.
These poles will, however cancel after summation over all permutations in (\ref{basic_intg}).

For many applications in the lattice Quantum Chromodynamics (QCD) an extension
of the integral ${\cal I}(r,s)$ to $SU(N)$ group is very important.
Such integrals appear in the strong coupling expansion of the pure gauge action and are useful
in the evaluation of the baryon spectrum. Integrals of this type appear during calculations
of the dual representations of some spin models describing effective interaction between Polyakov loops
\cite{spin_flux1, spin_flux2} and in principal chiral models \cite{su2pcm_dual}.
Many one-link integrals in the strong-coupled LGT can be reduced to the computation of (\ref{integral_def}).
These are one-link integrals with the staggered and the Wilson fermions as well as the one-link integrals
of the scalar QCD. Most importantly, the knowledge of these integrals is necessary for the construction
of the dual formulations of the full lattice QCD \cite{su2_dual, su3_dual, dual1}.

In the paper \cite{dual1} we outlined a certain approach to the duality transformations based on the Taylor
expansion of the Boltzmann weight both for $U(N)$ spin models and for $U(N)$ LGTs. Before presenting the details
of our calculations and their development we think it would be useful to extend the integration methods to
the special unitary group $SU(N)$. It will help in generalizing the approach of Ref. \cite{dual1} to lattice QCD.
Actually, the generating functional for $SU(N)$ integrals ${\cal I}_N(r,s)$ has been calculated long ago (see the Appendix
of the review \cite{lat_rev}). Recently, the case $s-r=N$ has been examined in some details in \cite{zuber}.
In this paper we present our results for $SU(N)$ group integrals of the type (\ref{integral_def}) for arbitrary $r,s$.
In deriving them we use the approach of Ref. \cite{novaes1}.
In Section~2 we obtain and describe in details our results for $SU(N)$ integrals.
As the simplest but important applications we compute various one-link and one-site
integrals relevant for LGT in Section~3. The results and perspectives are discussed in Section~4. In Appendix we collected some
formula used in computation of group integrals.

\section{$SU(N)$ polynomial integrals}

In this section we calculate the following integral over $SU(N)$ group
\begin{equation}
{\cal I}_N(r,s) \ = \  \int_{SU(N)}  dU \ \prod_{k=1}^{r} U_{i_k j_k} \prod_{n=1}^{s} U_{m_n l_n}^* \ .
\label{integral_sundef}
\end{equation}
Similar to $U(N)$ case this integral can be re-written as a sum of the invariant integrals.
The product of the matrix elements in the integrand is presented in the form
\begin{equation}
\prod_{k=1}^{r} U_{i_k j_k} \ = \ \frac{1}{r!} \ \prod_{k=1}^{r} \frac{\partial}{\partial A^*_{i_kj_k}} \
P_{1^r}(A^{\dagger}U) \ ,
\label{prod_matrix}
\end{equation}
where $P_{\tau}(X)$ is the power sum symmetric function (\ref{pwr_sum}). The integral becomes
\begin{equation}
{\cal I}_N(r,s) \ = \  \frac{1}{r! s!} \ \prod_{k=1}^{r} \frac{\partial}{\partial A^*_{i_kj_k}} \
\prod_{n=1}^{s} \frac{\partial}{\partial B_{m_n l_n}} \ {\cal F} \ ,
\label{integral_sun1}
\end{equation}
where
\begin{equation}
{\cal F} \ = \ \int_{SU(N)}  dU \ P_{1^r}(A^{\dagger}U) \ P_{1^s}(B U^{\dagger}) \ .
\label{integral_sun2}
\end{equation}
Using Eq.(\ref{schur_pwrsum_inverse}) from Appendix the last integral can be brought to
\begin{equation}
{\cal F} \ = \ \sum_{\lambda\vdash r} \ \chi_{\lambda}(1^r) \  \sum_{\mu\vdash s} \ \chi_{\mu}(1^s) \
\int_{SU(N)}  dU \ s_{\lambda}(A^{\dagger} U) \ s_{\mu}(B U^{\dagger}) \ .
\label{integral_sun3}
\end{equation}
We assume here the partitions $\lambda$ and $\mu$ are those described after Eq.(\ref{Wgn_def}).
Consider first the following integral
\begin{equation}
{\cal F}_0 \ = \ \int_{SU(N)}  dU \ s_{\lambda}(U) \ s_{\mu}(U^{\dagger}) \ .
\label{integral_sun4}
\end{equation}
Let $s_{\lambda}(U)=s_{\lambda}(u_1,\cdots,u_N)$, where $u_i$ are eigenvalues of $U$. Then
\begin{equation}
s_{\lambda}(U) \ = \ \frac{{\rm det} \ u_i^{\lambda_j+N-j}}{{\rm det} \ u_i^{N-j}} \ .
\label{schur_det}
\end{equation}
Last two integrals can be written as integrals over $U(N)$ measure with additional constraint
\begin{equation}
 {\rm det} \ U \ = \ \prod_{i=1}^N \ u_i \ = \ 1 \ .
\label{sun_det}
\end{equation}
Then, it is straightforward to prove
\begin{equation}
{\cal F}_0 \ = \ \sum_{q=-\infty}^{\infty} \prod_{i=1}^N \ \delta_{\lambda_i-\mu_i+q,0} \ .
\label{integral_sun5}
\end{equation}
Summation over $q$ enforces the constraint (\ref{sun_det}). We thus have
\begin{equation}
\sum_{i=1}^N \ \left ( \mu_i - \lambda_i \right ) \ = \ s -r \ = \ N q \ ,
\label{triality_cond}
\end{equation}
{\it i.e.}, the difference $s-r$ must be multiple of $N$. This is, of course, a consequence
of $Z(N)$ symmetry.
Clearly, the same holds for the integral in (\ref{integral_sun3}), therefore we can write
\begin{equation}
s_{\mu}(B U^{\dagger}) \ = \ s_{\lambda+q^N}(B U^{\dagger}) \ = \ s_{q^N}(B U^{\dagger})\ s_{\lambda}(B U^{\dagger})
\ = \ \left ( {\rm det} \ B U^{\dagger} \right )^q \ s_{\lambda}(B U^{\dagger}) \ ,
\label{schur_factor}
\end{equation}
where $\lambda+q^N=(\lambda_1+q,\cdots,\lambda_N+q)$ and $s_{q^N}(X)=s_{(q,\cdots,q)}(X)$. Last two equalities follow
from the representation (\ref{schur_det}). Suppose that $s\geq r$.
Substituting last expression into (\ref{integral_sun3}) and performing integration we find
\begin{equation}
{\cal F} \ = \ \sum_{q=0}^{\infty} \ \delta_{s-r,Nq} \left ( {\rm det} \ B \right )^q
\ \sum_{\lambda\vdash r} \ d(\lambda) \  d(\lambda+q^N) \
\frac{s_{\lambda}(A^{\dagger} B)}{s_{\lambda}(1^N)} \ ,
\label{integral_sun6}
\end{equation}
which results in the following intermediate result
\begin{eqnarray}
{\cal I}_N(r,s) &=&  \sum_{q=0}^{\infty} \ \delta_{s-r,Nq} \ \frac{1}{r! (r+Nq)!} \ \
\sum_{\lambda\vdash r} \ \frac{d(\lambda) \  d(\lambda+q^N)}{s_{\lambda}(1^N)} \nonumber  \\
&\times& \prod_{k=1}^{r} \frac{\partial}{\partial A^*_{i_kj_k}} \
\prod_{n=1}^{r+Nq} \frac{\partial}{\partial B_{m_n l_n}} \
s_{q^N}(B) \ s_{\lambda}(A^{\dagger} B) \ .
\label{integral_sun7}
\end{eqnarray}
The equality $({\rm det} \ B)^q = s_{q^N}(B)$ was used here.
To take all the derivatives one employs the relation between the Schur functions and the power sum symmetric
functions (\ref{schur_pwrsum}). This gives for the second line of (\ref{integral_sun7})
\begin{eqnarray}
\frac{1}{r! (Nq)!} \  \sum_{\sigma\in S_r} \ \sum_{\tau\in S_{Nq}} \ \chi_{\lambda}(\sigma) \
\chi_{q^N}(\tau) \ \sum_{a_1,\cdots,a_r=1}^N \ \sum_{b_1,\cdots,b_r=1}^N \ \sum_{c_1,\cdots,c_{Nq}=1}^N \nonumber \\
\prod_{k=1}^{r} \frac{\partial}{\partial A^*_{i_kj_k}} \ \prod_{n=1}^{r+Nq} \frac{\partial}{\partial B_{m_n l_n}} \
\prod_{k=1}^{r} \ A^*_{b_k a_k} B_{b_k a_{\sigma (k)}}  \   \prod_{n=1}^{Nq} \ B_{c_n c_{\tau (n)}} \ .
\label{derivativ}
\end{eqnarray}
Taking all derivatives and performing summations over matrix indices $a_i,b_i,c_i$
we finally obtain for $s\geq r$
\begin{eqnarray}
{\cal I}_N(r,s) &=&  \sum_{q=0}^{\infty} \ \delta_{s-r,Nq}  \ \sum_{\sigma\in S_r} \ Wg^{N,q}(\sigma)
\sum_{\rho\in S_{r+Nq}} \ \prod_{k=1}^r \ \delta_{i_k,m_{\rho(k)}} \delta_{j_{\sigma(k)},l_{\rho(k)}}
\nonumber   \\
&\times& \frac{1}{(Nq)!} \ \sum_{\tau\in S_{Nq}} \ \chi_{q^N}(\tau) \
\prod_{k=1}^{Nq} \ \delta_{m_{\rho(\tau(k)+r)},l_{\rho(k+r)}} \ ,
\label{sunintegr_fin}
\end{eqnarray}
where we defined the $SU(N)$ Weingarten function as
\begin{equation}
Wg^{N,q}(\sigma) = \frac{1}{r!(r+Nq)!} \ \sum_{\lambda\vdash\ r} \
\frac{d(\lambda)d(\lambda + q^N)}{s_{\lambda}(1^N)} \ \chi_{\lambda}(\sigma) \ .
\label{Wgn_SUN}
\end{equation}
When $r=s$ one restores the result (\ref{unintg_def}). The extension to $r>s$ is trivial.
The second line in (\ref{sunintegr_fin}) can be re-written in an equivalent form through completely
anti-symmetric tensors with the help of identity
\begin{eqnarray}
\sum_{\tau\in S_{Nq}} \ \chi_{q^N}(\tau) \ \prod_{k=1}^{Nq} \ \delta_{m_{\tau(k)},l_{k}} \ = \
\frac{1}{(N!)^q} \ \sum_{\sigma\in S_{Nq}} \ \epsilon_{m_{\sigma(1)}\cdots m_{\sigma(N)}} \cdots
\nonumber   \\
\epsilon_{m_{\sigma((q-1)N+1)}\cdots m_{\sigma(qN)}} \ \epsilon_{l_{\sigma(1)}\cdots l_{\sigma(N)}} \cdots
\epsilon_{l_{\sigma((q-1)N+1)}\cdots l_{\sigma(qN)}} \ .
\label{epsilon_form}
\end{eqnarray}
While not so compact, this representation is more useful for practical calculations in LGT.
The expressions (\ref{sunintegr_fin}), (\ref{Wgn_SUN}) and (\ref{epsilon_form}) are main result of this paper.
They allow to evaluate many $SU(N)$ integrals and present the result in a compact form.

Similar result for the generating function
\begin{equation}
\int_{SU(N)}  dU \  \left ( \mbox{Tr}J U \right )^r \ \left ( \mbox{Tr}K U^{\dagger} \right )^s
\label{sun_genfunc}
\end{equation}
has recently been derived in \cite{unger_lat18}. In this paper the authors define the $SU(N)$ Weingarten
function as
\begin{equation}
\widetilde{Wg}^{N,q}(\sigma) = \frac{1}{(r!)^2} \ \sum_{\lambda\vdash\ r} \
\frac{d^2(\lambda)}{s_{\lambda}(1^{N+q})} \ \chi_{\lambda}(\sigma) \ .
\label{Wgn_unger}
\end{equation}
The relation between two definitions is given by
\begin{equation}
\prod_{k=0}^{N-1} \ \frac{k!}{(k+q)!}  \ \widetilde{Wg}^{N,q}(\sigma) \ =  \ Wg^{N,q}(\sigma) \ .
\label{Wgn_relation}
\end{equation}
With this relation taken into account our result for the generating functional (\ref{sun_genfunc}) agree
with that of \cite{unger_lat18}.

We finish this section with a conjecture about large-$N$ asymptotic behaviour of the $SU(N)$ Weingarten function.
In case of the $U(N)$ group it is well-known and the leading term of the asymptotic expansion reads \cite{collins2}
\begin{equation}
Wg^N(\sigma) \ = \ \frac{1}{N^{r}} \
\left ( \prod_{k} \ \delta_{\sigma(k),k} +  {\cal{O}} \left ( N^{-1}  \right ) \right ) \ ,
\label{WgUN_asymp}
\end{equation}
if $\sigma\in S_r$. The simple way to obtain the leading term for the $SU(N)$ group is to consider the integral
(\ref{Q_def}) (see next section).  Then, one can readily find
\begin{equation}
(r+Nq)! \ \sum_{\sigma\in S_r} \ Wg^{N,q}(\sigma)  N^{|\sigma|} \ = \ Q_N(r,s) \ .
\label{Wgn_sum}
\end{equation}
Here, $|\sigma|$ is a number of cycles in the permutation $\sigma$ and $Q_N(r,s)$ is given in Eq.(\ref{QSUN}).
Taking asymptotics on the right-hand side of the last expression leads to the equality
\begin{equation}
(r+Nq)! \ \sum_{\sigma\in S_r} \ Wg^{N,q}(\sigma) N^{|\sigma|} \ = \ \frac{(r+Nq)!}{(N!)^q}
\left ( 1 + {\cal{O}} \left ( N^{-1}  \right )  \right) \ .
\label{Wgn_sum_asymp}
\end{equation}
If we suppose that, similar to $U(N)$ group, the leading contribution to the large-$N$ expansion comes from
the identity permutation $\sigma(k)=k$ we obtain
\begin{equation}
Wg^{N,q}(\sigma) \ = \ \frac{1}{N^{r}(N!)^q} \ \left ( \prod_{k} \ \delta_{\sigma(k),k}  +
{\cal{O}} \left ( N^{-1}  \right )  \right ) \ .
\label{WgSUN_asymp}
\end{equation}
Note, that $N^r (N!)^q\sim N^{r+Nq}=N^s$.

\section{Applications}

In this section we apply the formula (\ref{sunintegr_fin}) for the evaluation of various $SU(N)$ integrals
which often appear in some spin models and LGTs.

\subsection{Invariant integrals}

The simplest integrals which can be calculated with the help of (\ref{sunintegr_fin}) are those whose
integrands depend only on the traces of the group elements
\begin{equation}
Q_N(r,s) = \int_G dU \ ({\rm Tr} U)^r \ ({\rm Tr} U^*)^s  \ ,
\label{Q_def}
\end{equation}
where $G=U(N),SU(N)$. Such integrals appear in the effective spin models describing the interaction
of the Polyakov loops in the finite-temperature LGTs. Writing the traces as
\begin{equation}
({\rm Tr} U)^s \ = \ \sum_{i_1=1}^N \ \cdots   \sum_{i_s=1}^N \
\prod_{k=1}^s \ U_{i_ki_k}
\label{race_exp}
\end{equation}
one obtains the integral of the type (\ref{integral_def}) or (\ref{integral_sundef}).
Using (\ref{basic_intg}) and (\ref{sunintegr_fin}) one finds after summation over group indices
\begin{eqnarray}
Q_N(r,s) \ = \
\begin{cases}
\delta_{r,s } \  \sum_{\lambda \vdash s} \ d^2(\lambda) \ ,  \ G=U(N) \ ,  \\
\sum_{q=-\infty}^{\infty} \delta_{s-r,q N} \
\sum_{\lambda \vdash {\rm min}(r,s)} \ d(\lambda) \ d(\lambda + |q|^N) \ , \ G=SU(N) \ .
\end{cases}
\label{QSUN}
\end{eqnarray}
Of course, the simpler way to compute (\ref{Q_def}) is to use the expansion
$({\rm Tr} U)^s=(u_1+u_2+\dots+u_N)^s=\sum_{\lambda \vdash s} d(\lambda)s_{\lambda}(U)$.
Then, the result (\ref{QSUN}) follows from the orthogonality of the Schur functions.

The next example is provided by the coefficients of the character expansion of an invariant function
$H(x,y)$
\begin{equation}
C_{\lambda}(a,b) \ = \ \int_G dU \ s_{\lambda}(U) \
H\left ( \frac{a}{2}{\rm Tr} U, \frac{b}{2} {\rm Tr} U^{\dagger} \right ) \ .
\label{char_coeff_def}
\end{equation}
For the most common case $H(x,y)=e^{x+y}$ the result is well known for many decades and reads for $G=SU(N)$
\begin{equation}
\label{char_coeff_res}
C_{\lambda}(a,b) \ = \ \left ( \frac{b}{a} \right )^{\frac{1}{2}|\lambda|} \
\sum_{m=-\infty}^\infty \ \left ( \frac{b}{a} \right )^{\frac{1}{2}N m}  \
{\rm det} I_{\lambda_i-i+j+m}(\sqrt{ab})_{1\leq i,j \leq N}\ ,
\end{equation}
where $|\lambda|=\sum_i\lambda_i$. Only term with $m=0$ contributes for $U(N)$ group.
An alternative expression for the coefficients $C_{\lambda}$ can be obtained with the help of (\ref{sunintegr_fin}).
First, using the Taylor expansion of the function $H(x,y)$, the Eq.(\ref{char_coeff_def}) is written as
\begin{equation}
\label{char_coeff_res1}
C_{\lambda}(a,b) \ = \ \sum_{r=0}^{\infty} \sum_{s=0}^{\infty} \ h(r,s) \ \frac{(a/2)^{r}}{r!} \ \frac{(b/2)^{s}}{s!}
\int_G dU \ s_{\lambda}(U) \ ({\rm Tr} U)^r \ ({\rm Tr} U^{\dagger})^s \ ,
\end{equation}
where $h(r,s) = {\left. \frac{\partial^{r+s}H(a,b)}{\partial a^{r}\partial b^{s}} \right |}_{a=b=0}$.
The last integral can again be presented as an integral over the power sum symmetric functions following
Eqs.(\ref{prod_matrix})-(\ref{integral_sun2}) and using the relations (\ref{schur_pwrsum}), (\ref{pwr_sum}).
This leads to the computation of the integral of the form
\begin{equation}
\int_G dU \ P_{\sigma}(U) \ P_{1^r}(A^{\dagger}U) \ P_{1^s}(B U^{\dagger}) \ , \
\sigma\in S_{|\lambda|} \ , \ |\lambda| = \sum_i\lambda_i \ .
\label{}
\end{equation}
Repeating all the calculations from the previous section for the last integral we end up with the following result
\begin{equation}
\label{char_coeff_res2}
C_{\lambda}(a,b) \ = \ \sum_{r=0}^{\infty} \sum_{s=0}^{\infty} \ h(r,s)  \ \frac{(a/2)^{r}}{r!} \ \frac{(b/2)^{s}}{s!} \
\sum_{\mu\vdash r} \ \sum_{\nu\vdash s} \ C_{\lambda \ \mu}^{\nu}(G) \ d(\mu) \ d(\nu) \ .
\end{equation}
The Littlewood-Richardson coefficients $C^{\nu}_{\lambda \ \mu}(G)$ appear in the final answer in the form
\begin{equation}
C_{\lambda \ \mu}^{\nu}(U(N)) \ = \ \frac{1}{r!s!} \ \sum_{\sigma\in S_r} \ \sum_{\tau\in S_s} \
\chi_{\lambda}(\sigma) \ \chi_{\mu}(\tau) \ \chi_{\nu}([\sigma : \tau]) \
\label{LR_un}
\end{equation}
for $G=U(N)$ and
\begin{equation}
C_{\lambda \ \mu}^{\nu}(SU(N)) \ = \ \sum_{q=0}^{\infty} \ (Nq)! \ \frac{d(\nu + |q|^N)}{s_{\nu}(1^N)} \
\sum_{\rho\vdash s+Nq} \ \frac{s_{\rho}(1^N)}{d(\rho)} \ C_{\nu \ Nq}^{\rho}(U(N)) \
C_{\lambda \ \mu}^{\rho}(U(N)) \
\label{LR_sun}
\end{equation}
for $G=SU(N)$. The permutation $[\sigma : \tau]\in S_{r+s}$ is defined as
\begin{equation}
[\sigma : \tau](k) \ = \
\begin{cases}
 \sigma(k) \ , \ k\leq r \ , \\
 \tau(k-r)+r \ , \ r < k \leq r+s \ .
\end{cases}
\label{prod_def}
\end{equation}
The constraints on partitions $r+|\lambda|=s$ for $U(N)$ and $r+|\lambda|=s+Nq$ for $SU(N)$ are evident
from the Littlewood-Richardson coefficients. The expression (\ref{char_coeff_res2}) can be further simplified
using the summation formula
\begin{equation}
\sum_{\mu\vdash r} \ C_{\lambda \ \mu}^{\nu}(U(N)) \ d(\mu) \ = \ d\left ( \nu/\lambda  \right ) \ .
\label{Cd_sum}
\end{equation}
Here, $d\left ( \nu/\lambda  \right )$ is the dimension of a skew representation defined by a corresponding
skew Young diagram
\begin{equation}
d\left ( \nu/\lambda  \right ) \ = \ \frac{1}{|\lambda|!} \sum_{\sigma\in S_{|\lambda|}} \ \chi_{\nu}([I : \sigma]) \
\chi_{\lambda}(\sigma) \ .
\label{skew_d}
\end{equation}
Then, {\it e.g.} for $U(N)$ we get a simple expression
\begin{equation}
\label{char_coeff_res3}
C_{\lambda}(a,b) \ = \ \sum_{r=0}^{\infty} \ h(r,r+|\lambda|) \
\frac{(ab/4)^{r}}{r!} \ \frac{(b/2)^{|\lambda|}}{(r+|\lambda|)!} \
\sum_{\nu\vdash r+|\lambda|} \ d\left ( \nu/\lambda  \right ) \ d(\nu)
\end{equation}
and similar formula can be written down for $SU(N)$ if one uses Eqs.(\ref{char_coeff_res2}), (\ref{LR_sun}).

\subsection{One-link integrals}

Consider the following integral over $G=U(N),SU(N)$ with arbitrary $N\times N$ matrices $A,B$
\begin{equation}
Z_G(A,B) \ = \ \int_G dU \ \exp \left [ \frac{\beta}{2} \left ( {\rm Tr} A U
+ {\rm Tr} B U^{\dagger}  \right )   \right  ] \ .
\label{one_linkint_def}
\end{equation}
Expanding the integrand in the Taylor series one gets
\begin{equation}
Z_G(A,B) \ = \ \sum_{r=0}^{\infty}\sum_{s=0}^{\infty} \ \frac{(\beta/2)^{r+s}}{r!s!} \
\int_G \ dU \left ( {\rm Tr} AU \right )^r \ \left ( {\rm Tr} BU^{\dagger} \right )^s \ .
\label{one_linkint_exp1}
\end{equation}
This can be presented as
\begin{equation}
Z_G(A,B) =  \sum_{r=0}^{\infty}\sum_{s=0}^{\infty} \ \frac{(\beta/2)^{r+s}}{r!s!}
\sum_{i_1\cdots i_r} \ \sum_{j_1\cdots j_r} \ \sum_{m_1\cdots m_s} \ \sum_{l_1\cdots l_s}
\prod_{k=1}^r A_{{i_k j_k}} \ \prod_{n=1}^s B_{l_n m_n} \  {\cal I}_N(r,s) \ .
\label{one_linkint_exp2}
\end{equation}
Using the result of the previous section (\ref{sunintegr_fin}) for the group integral
we can calculate all summations over group indices and over all permutations. All necessary formulae
to do this are given in the Appendix. We find for $G=SU(N)$
\begin{eqnarray}
Z_{SU(N)}(A,B)  =  \sum_{s=0}^{\infty} \sum_{q=-\infty}^{\infty} \
\frac{1}{s!} \left ( \frac{\beta}{2} \right )^{2s+N|q|} \ \left ( \mbox{det} C \right )^{|q|}
\sum_{\sigma \in S_s} \  Wg^{N,q}(\sigma) P_{\sigma} (A B) \ .
\label{sun_1link_res_Wgn}
\end{eqnarray}
With the help of (\ref{char_orthog}) and (\ref{schur_pwrsum_inverse}) this can be re-written as
\begin{eqnarray}
Z_{SU(N)}(A,B)  =  \sum_{s=0}^{\infty} \sum_{q=-\infty}^{\infty} \ \left ( \frac{\beta}{2} \right )^{2s+N|q|} \
\sum_{\lambda \vdash s} \ \frac{d(\lambda)d(\lambda + |q|^N)}{s!(s+N|q|)!} \ \frac{s_{\lambda}(A B)}{s_{\lambda}(1^N)}
\left ( \mbox{det} C \right )^{|q|} ,
\label{sun_1link_res}
\end{eqnarray}
where $C=A$ if $q>0$ and $C=B$ if $q<0$. Equivalent form of (\ref{sun_1link_res}) reads
\begin{eqnarray}
Z_{SU(N)}(A,B) &=&  \sum_{q=-\infty}^{\infty} \ \sum_{\lambda_1\geq\lambda_2\geq\cdots\geq\lambda_N\geq 0} \
\left ( \frac{\beta}{2} \right )^{2\sum_i\lambda_i+N|q|} \
\frac{d(\lambda)d(\lambda + |q|^N)}{\left ( \sum_i\lambda_i \right )!\left ( \sum_i\lambda_i+N|q| \right )!} \nonumber   \\
&\times& \ \frac{s_{\lambda}(A B)}{s_{\lambda}(1^N)} \ \left ( \mbox{det} C \right )^{|q|} \ .
\label{sun_1link_fin}
\end{eqnarray}
When $B=0$ the last formula reduces to
\begin{eqnarray}
Z_{SU(N)}(A,0)  =  \sum_{q=0}^{\infty} \ \left ( \frac{\beta}{2} \right )^{Nq} \
\left ( \mbox{det} A \right )^q \
\prod_{k=0}^{N-1} \ \frac{k!}{(q+k)!} \ .
\label{sun_1link_B0}
\end{eqnarray}
This result coincides with \cite{wipf} for $A=a I$, where $I$ is a unit matrix.
For $U(N)$ group (\ref{sun_1link_fin}) simplifies to
\begin{eqnarray}
Z_{U(N)}(A,B) \ = \ \sum_{\lambda_1\geq\lambda_2\geq\cdots\geq\lambda_N\geq 0} \
\left ( \frac{\beta}{2} \right )^{2\sum_i\lambda_i} \
\frac{d^2(\lambda)}{\left ( ( \sum_i\lambda_i )! \right )^2} \
\frac{s_{\lambda}(A B)}{s_{\lambda}(1^N)} \ .
\label{un_1link_res}
\end{eqnarray}
Integrals of the more general form
\begin{equation}
Z_G(A,B,C_k,D_l) \ = \ \int_G dU \ \prod_k \mbox{Tr} C_k U \ \prod_l \mbox{Tr} D_l U^{\dagger}
\exp \left [ \frac{\beta}{2} \left ( {\rm Tr} A U
+ {\rm Tr} B U^{\dagger}  \right )   \right  ]
\label{one_linkint_gen}
\end{equation}
can be evaluated by differentiating the basic integral (\ref{one_linkint_def}), for example
\begin{equation}
Z_G(A,B,C,D) \ = \ {\left. \frac{\partial^{2}}{\partial h_1 \partial h_2} \ Z_G(A+h_1 C , B+h_2 D) \right |}_{h_1=h_2=0} \ .
\label{one_linkint_gen_res}
\end{equation}
The final result is quite cumbersome and will not be given here.

\subsection{Staggered fermions}

Formulae (\ref{sun_1link_res_Wgn})-(\ref{un_1link_res}) can be further specified for many particular cases.
As an important example, let us consider the partition function with staggered fermions.
For $N_f$ flavours of the staggered fermions the matrices $A$ and $B$ from (\ref{one_linkint_def}) read
\begin{equation}
A^{ij} \ = \  \eta_{\nu}(x) \ y_{\nu} \ \sum_{f=1}^{N_f} \
\bar{\psi}^i_f(x) \psi^j_f(x+e_{\nu}) \ ,
\label{A_stagferm}
\end{equation}
\begin{equation}
B^{ij} \ = \ - \eta_{\nu}(x) \ y_{\nu}^{-1} \ \sum_{f=1}^{N_f} \
\bar{\psi}^i_f(x+e_{\nu}) \psi^j_f(x) \ .
\label{B_stagferm}
\end{equation}
$\eta_{\nu}(x)$ is given by
\begin{equation}
\eta_{0}(x) \ = \ 1 \ ; \
\eta_{\nu}(x) \ = \  \xi \ (-1)^{x_0+x_{1}+ \cdots +x_{\nu-1}} \ , \ \nu = 1,\cdots , d \ .
\label{KSferm}
\end{equation}
$\xi=a_t/a_s$ is the lattice anisotropy and the chemical potential $\mu$ is introduced via $y_{\nu}$
\begin{equation}
y_{\nu} \ = \
\begin{cases}
\exp (\mu), \ \nu = 0 \ , \ \mu = a_t \mu_q  \\
1, \  \nu=1,\cdots , d \ .
\end{cases}
\label{chem_pot}
\end{equation}
We define the composite meson fields as
\begin{equation}
\sigma_{f f^{\prime}}(x) \ = \ \sum_{i=1}^N \ \bar{\psi}^{i}_f(x)\psi^i_{f^{\prime}}(x)
\label{meson_comp}
\end{equation}
and the composite (anti-)baryon fields as
\begin{eqnarray}
\label{baryon_comp}
&&B_{f_1...f_{N}}(x) \ = \ \frac{1}{N!} \ \sum_{i_1...i_{N}=1}^{N} \
\epsilon_{i_1...i_{N}} \ \psi^{i_1}_{f_1}(x) \cdots \psi^{i_{N}}_{f_{N}}(x) \ , \\
&&\bar{B}_{f_1...f_{N}}(x) \ = \ \frac{1}{N!} \ \sum_{i_1...i_{N}=1}^{N} \ \epsilon_{i_1...i_{N}} \
\bar{\psi}^{i_{N}}_{f_1}(x) \cdots \bar{\psi}^{i_1}_{f_{N}}(x) \ .
\label{antibaryon_comp}
\end{eqnarray}
Consider first the $U(N)$ model (\ref{un_1link_res}). Using representation (\ref{schur_repr1}) for the Schur functions
and definitions (\ref{A_stagferm}), (\ref{B_stagferm}) it is easy to prove that
\begin{equation}
{\rm Tr} (AB)^j \ = \ \left ( \eta_{\nu}(x) \right )^{2j} \ (-1)^{j+1} \ {\rm Tr} \ \Sigma^j \ ,
\label{AB_trace}
\end{equation}
where we introduced matrix
\begin{equation}
\Sigma_{f_1f_2} \ = \ \sum_{f=1}^{N_f} \sigma_{f_1f}(x)\sigma_{ff_2}(x+e_{\nu}) \ .
\label{sigma_def}
\end{equation}
Substituting this result into (\ref{un_1link_res}) and using $\sum i\tau_i=s$, we obtain ($\beta=1$)
\begin{eqnarray}
Z_{U(N)}(A,B) \ = \ \sum_{s=0}^{N N_f} \ \left ( \frac{\eta_{\nu}(x)}{2} \right )^{2s} \ \sum_{\lambda\vdash\ s} \
\frac{d^2(\lambda)}{(s!)^2} \ \frac{s_{\lambda^{\prime}}(\Sigma)}{s_{\lambda}(1^N)}  \ .
\label{un_1link_stagferm}
\end{eqnarray}
Here, $\lambda^{\prime}$ is a representation dual to $\lambda$. The dual representation is defined by exchanging
raws and columns in the corresponding Young diagram, {\it i.e.} $\lambda^{\prime}_i=\sum_j \mathbf{1}_{\lambda_j\geq i}$.
In order to extend this result to $SU(N)$ we need to calculate determinants of $A$ and $B$ matrices. They are
\begin{eqnarray}
\label{DetA}
\mbox{det} A  = (-1)^{\frac{N(N-1)}{2}} N! (\eta_{\nu}(x) \ y_{\nu})^N \sum_{f_i = 1}^{N_f}
\bar{B}_{f_1...f_{N}}(x) B_{f_1...f_{N}}(x+e_{\nu})  \ , \\
\mbox{det} B  = (-1)^{\frac{N(N-1)}{2}} N! (- \eta_{\nu}(x) \ y_{\nu}^{-1})^N \sum_{f_i = 1}^{N_f}
\bar{B}_{f_1...f_{N}}(x+e_{\nu}) B_{f_1...f_{N}}(x) \ .
\label{DetB}
\end{eqnarray}
The factor $(-1)^{\frac{N(N-1)}{2}}$ appears due to $\sum_{i = 0}^{N-1} i = \frac{N(N-1)}{2}$ commutations
of the fermion fields to gather them into baryons. Combining these expressions with (\ref{un_1link_stagferm}) and
(\ref{sun_1link_res}) we can write down the partition function with an arbitrary number of staggered fermion flavours as
\begin{eqnarray}
Z_{SU(N)}(A,B)  &=&  Z_{U(N)}(A,B)  +
\sum_{q=1}^{N_f} \ (-1)^{\frac{N(N-1)}{2}q} (N!)^q \ \sum_{s=0}^{N (N_f-q)} \ \left ( \frac{\eta_{\nu}(x)}{2} \right )^{2s+Nq}
\nonumber   \\
&\times& \sum_{\lambda \vdash s} \ \frac{d(\lambda)d(\lambda + q^N)}{s!(s+N q)!}
\frac{s_{\lambda^{\prime}}(\Sigma)}{s_{\lambda}(1^N)}  \
\left [ y_{\nu}^{Nq} Q^q +  (-1)^{Nq} y_{\nu}^{-Nq} \bar{Q}^q  \right ] \ ,
\label{PF_stagferm}
\end{eqnarray}
where notations have been used
\begin{eqnarray}
Q \ = \ \sum_{f_i = 1}^{N_f} \ \bar{B}_{f_1...f_{N}}(x) B_{f_1...f_{N}}(x+e_{\nu}) \ , \nonumber \\
\bar{Q} \ = \ \sum_{f_i = 1}^{N_f}  \bar{B}_{f_1...f_{N}}(x+e_{\nu}) B_{f_1...f_{N}}(x) \ .
\label{Qbaryon}
\end{eqnarray}
As the simplest case, consider one-flavour system. Then, the matrix $\Sigma$ becomes a scalar
$\Sigma=\sigma(x)\sigma(x+e_{\nu})$ and the summation over $\lambda$ consists of one term, namely
$\lambda^{\prime}=(s,0,\cdots,0)$ (or $\lambda_i=1, i\in [1,s]$ and zero otherwise).
This, together with (\ref{schur_reduction}), yields
\begin{eqnarray}
Z_{SU(N)}(A,B) \ = \ \sum_{s=0}^{N} \ \left ( \frac{\eta_{\nu}(x)}{2} \right )^{2s} \
\frac{(N-s)!}{s!N!} \ \left ( \sigma(x)\sigma(x+e_{\nu})  \right )^s \nonumber   \\
+ (-1)^{\frac{N(N-1)}{2}}  \left ( \frac{\eta_{\nu}(x)}{2} \right )^N \
\left [ y_{\nu}^N \bar{B}(x) B(x+e_{\nu}) + (-1)^N y_{\nu}^{-N} \bar{B}(x+e_{\nu}) B(x)  \right ] \ .
\label{un_1link_stagferm_nf1}
\end{eqnarray}
This agrees with the well-known result of Ref.\cite{rossi}.
Similar method of computing one-link integral with one flavour of the staggered fermions
has been used in \cite{strcpl_qcd}. Slightly more complicated is the case of two staggered flavours, $N_f=2$.
Parameterizing  $\lambda^\prime$ as $\lambda^\prime = (u_1, u_2)$, $u_1 + u_2 = s$, $N \geqslant u_1 \geqslant u_2 \geqslant 0$
and utilizing the property (\ref{schur_reduction}) of the Schur functions
we present result for the two-flavour partition function as
\begin{equation}
Z_{SU(N)}(A,B) \ = \ Z_0 + Z_1 + Z_2 \ ,
\label{2flavorPF}
\end{equation}
\begin{eqnarray}
Z_0 \ &=& \ \sum_{u_1 \geqslant u_2 \geqslant 0}^N
 \left( \frac{\eta_{\nu}(x)}{2} \right)^{2 (u_1 + u_2)}
 \frac{(u_1 - u_2 + 1) (N - u_1)! (N - u_2 + 1)!}{(u_1 + 1)! u_2! N! (N + 1)!} \nonumber \\
 &\times& \sum_{i = 0}^{\left[ \frac{u_1 - u_2} {2} \right]}
 C_{u_1 - u_2 - i}^{i} (-1)^i (\mbox{det}\Sigma)^{u_2 + i} (\mbox{Tr} \Sigma )^{u_1 - u_2 - 2i} \ ,
\label{2flavorPF0}
\end{eqnarray}
\begin{eqnarray}
\label{2flavorPF1}
&&Z_1  = \sum_{u_1 \geqslant u_2 \geqslant 0}^N
 \left( \frac{\eta_{\nu}(x)}{2} \right)^{2 (u_1 + u_2) +  N}
 \frac{(u_1 - u_2 + 1) (N - u_1 + 1)! (N - u_2 + 2)!}{(u_1 + 1)! u_2! (N + 1)! (N + 2)!}  \\
&&(-1)^{\frac{N(N-1)}{2}} [ y_{\nu}^N Q + (-1)^N y_{\nu}^{-N} \bar{Q} ] \sum_{i = 0}^{\left[ \frac{u_1 - u_2} {2} \right]}
 C_{u_1 - u_2 - i}^{i} (-1)^i (\mbox{det}\Sigma)^{u_2 + i} (\mbox{Tr} \Sigma )^{u_1 - u_2 - 2i} , \nonumber
\end{eqnarray}
\begin{eqnarray}
Z_2 \ = \ \left ( \frac{\eta_{\nu}(x)}{2} \right )^{2 N} \
\frac{1}{(N+1)} \ \left [ y_{\nu}^{2N} Q^2 + y_{\nu}^{-2N} \bar{Q}^2 \right ] \ .
\label{2flavorPF2}
\end{eqnarray}
Here, $C^i_k$ are the binomial coefficients. To calculate the Schur functions we have used their relations
with the complete symmetric functions $h_k$ and the elementary symmetric functions $e_k$,
(\ref{ef_hf})-(\ref{ef_hf_schur}).

\subsection{Reduced principal chiral model}

Our last example is the so-called reduced principal chiral model whose partition function reads
\begin{equation}
Z \ = \ \int_G dU \ \exp \left [ \beta \sum_{\mu=1}^d  {\rm Re} {\rm Tr} U \Gamma_{\mu}
U^{\dagger} \Gamma_{\mu}^{\dagger} \right  ] \ .
\label{pcm_reduced}
\end{equation}
The matrices $\Gamma_{\mu}\in SU(N)$ and satisfy certain commutation relations exact form of which is not
important here. It is expected that the reduced model gives an exact solution for the principal chiral model
in the large-$N$ limit (see, for instance (\cite{vicari}) and refs. therein). Here we discuss the group integration
in (\ref{pcm_reduced}) using the method of the Weingarten functions. For simplicity, we restrict ourself to the two-dimensional
case. Expanding the integrand into two character series and using the relation (\ref{schur_pwrsum}) again one finds
\begin{equation}
Z \ = \ \sum_{\lambda_1} \ \sum_{\lambda_2} \ \frac{C_{\lambda_1}(\beta) \ C_{\lambda_2}(\beta)}{|\lambda_1|!|\lambda_2|!} \
\sum_{\sigma_1\in S_{|\lambda_1|}} \ \sum_{\sigma_2\in S_{|\lambda_2|}} \ \chi_{\lambda_1}(\sigma_1) \ \chi_{\lambda_2}(\sigma_2)
J_{\sigma_1,\sigma_2}(\{ \Gamma_{\mu} \}) \ ,
\label{pcm_reduced_exp}
\end{equation}
where $C_{\lambda_i}(\beta)$ are given by either (\ref{char_coeff_res}) or (\ref{char_coeff_res3}) and the resulting group
integral takes the form
\begin{equation}
J_{\sigma_1,\sigma_2}(\{ \Gamma_{\mu} \}) \ = \ \int_G dU \ P_{\sigma_1}(U\Gamma_1U^{\dagger}\Gamma_1^{\dagger}) \
P_{\sigma_2}(U\Gamma_2U^{\dagger}\Gamma_2^{\dagger}) \ .
\label{Jss_def}
\end{equation}
Applying Eq.(\ref{pwr_sum}) one sees the last integral is of a type (\ref{integral_sundef}), therefore
the equation (\ref{sunintegr_fin}) can be used (it is sufficient to take only the term with $q=0$ in the limit of large $N$)
and gives after partial summation over group indices
\begin{equation}
J_{\sigma_1,\sigma_2}(\{ \Gamma_{\mu} \}) \ = \ \sum_{\tau,\rho\in S_{r_1+r_2}} \
Wg^N(\tau^{-1}\rho) \ \overline{P}_{\tau,\tau}(\{ \Gamma_{\mu} \}) \
\overline{P}_{\rho\sigma_1,\rho\sigma_2}(\{ \Gamma_{\mu}^{\dagger} \}) \ .
\label{Jss_res}
\end{equation}
We introduced here the following invariant function
\begin{equation}
\overline{P}_{\tau,\pi}(\{ \Gamma_{\mu} \})  \ = \ \sum_{i_1,\cdots,i_{r_1+r_2}} \
\prod_{k=1}^{r_1} \Gamma_1^{i_{\tau(k)},i_k} \
\prod_{k=1}^{r_2} \Gamma_2^{i_{\pi(r_1+k)},i_{r_1+k}} \ .
\label{P_inv}
\end{equation}
The right-hand side of the last equation can be expressed in terms of traces of products of matrices $\Gamma_{\mu}$.
Detailed investigation will be published elsewhere. Here we would like to emphasize that we expect certain simplifications
in the large-$N$ limit which is the only one relevant here.
Indeed, taking the asymptotics of the Weingarten function (\ref{WgUN_asymp})
in  Eq.(\ref{Jss_res}) we obtain
\begin{equation}
J_{\sigma_1,\sigma_2}(\{ \Gamma_{\mu} \}) \ = \ \sum_{\rho\in S_{r_1+r_2}} \
\frac{1}{N^{r_1+r_2}} \ \overline{P}_{\rho,\rho}(\{ \Gamma_{\mu} \}) \
\overline{P}_{\rho\sigma_1,\rho\sigma_2}(\{ \Gamma_{\mu}^{\dagger} \}) \ .
\label{Jss_res1}
\end{equation}

\section{Summary and perspectives}

In this article we have calculated certain integrals over $SU(N)$ group.
The basic integral (\ref{integral_sundef}) is of polynomial type whose integrand includes product
of an arbitrary number of the group matrices and their conjugates in the fundamental representation.
The result of the integration is expressed through the summation over permutations of the group indices
which are contracted via products of the Kronecker deltas and the totally anti-symmetric tensors,
Eqs.(\ref{sunintegr_fin}) and (\ref{epsilon_form}). The weight is given by the $SU(N)$ Weingarten
function (\ref{Wgn_SUN}). We have considered several applications of this general result.
In particular, we have calculated the integrals of powers of traces of the group elements,
the coefficients of the character expansion of the invariant function of an arbitrary form and
the general one-link integral appearing in LGT. In the latter case, the result was used to evaluate
one-link integrals with arbitrary number of staggered fermion flavours. Also, we applied our approach
for the computation of the integrals in the expression for the partition function of the reduced principal chiral model.

The method of the Weingarten functions is a powerful and very general tool which can be used both for the
derivation of alternative representations of known integrals and for calculation of new and
more complex ones appearing in lattice QCD and spin models.
Some further applications of this method may include one-link integrals with $N_f$ flavours of the Wilson
fermions, the scalar lattice QCD in the strong coupling region, $SU(N)$ principal chiral model at finite $N$.
Interesting is an extension of 3.3 to the Eguchi-Kawai model and its twisted version.

The main goal of this article was to give a mathematical background to our computations of the dual representations
of 1) the effective Polyakov loop spin models and 2) pure gauge LGT and lattice QCD with the staggered fermions,
as outlined in \cite{dual1}. Details of these calculations will be reported elsewhere.
Also, an interesting and important direction is to explore the large-$N$ expansion of $SU(N)$ models in frameworks of
the present method. This can be done by using the asymptotic expansion of the Weingarten function (\ref{WgSUN_asymp})
and is planned for future investigations.

\section*{Appendix}

Let $\lambda=(\lambda_1,\lambda_2,\cdots,\lambda_N)$ be a partition $\lambda\vdash\ r$,
$\lambda_1\geq\lambda_2\geq \cdots \geq \lambda_N\geq 0$ and $\sum_{i=1}^N\lambda_i=r$.
$\chi_{\lambda}(\sigma)$ denotes a character of $\sigma \in S_r$
in representation $\lambda$. $d(\lambda)=\chi_{\lambda}(1)$ is the dimension of the representation $\lambda$.
The Schur function $s_{\lambda}(X)$ is a character of the unitary group $G$, thus $s_{\lambda}(1^N)$ is
the  dimension of the irreducible representation $\lambda$ of $G$. If $l(\lambda)$ is the length of the partition $\lambda$,
{\it i.e.}, the number of non-vanishing parts $\lambda_i$, then
\begin{equation}
d(\lambda) \ = \ r! \
\frac{\prod_{1\leq i<j \leq l(\lambda)}(\lambda_i-\lambda_j+j-i)}{\prod_{i=1}^{l(\lambda)}(\lambda_i+l(\lambda)-i)!} \ ,
\label{dim_l}
\end{equation}
\begin{equation}
s_{\lambda}(1^N) \ = \ \frac{\prod_{1\leq i<j \leq N}(\lambda_i-\lambda_j+j-i)}{\prod_{i=1}^N( N-i)!} \ .
\label{G_dim}
\end{equation}
Below we list some formulae used in evaluating the group integrals.
Orthogonality relation for the characters of the permutation group is written as
\begin{equation}
\sum_{\tau\in S_r} \ \chi_{\mu}(\tau) \chi_{\lambda}(\tau \sigma) \ = \
r! \ \delta_{\mu,\lambda} \ \frac{\chi_{\lambda}(\sigma)}{d(\lambda)} \ .
\label{char_orthog}
\end{equation}
Relation between Schur functions and power sum symmetric functions of matrix argument is given by
\begin{equation}
s_{\lambda}(X) \ = \ \frac{1}{r!} \ \sum_{\tau\in S_r} \ \chi_{\lambda}(\tau) P_{\tau}(X) \ ,
\label{schur_pwrsum}
\end{equation}
where $X$ has the dimension $N$ and
\begin{equation}
P_{\tau}(X) \ = \ \sum_{i_1i_2\cdots i_r}^N \ \prod_{k=1}^r \ X_{i_k i_{\tau (k)}} \ .
\label{pwr_sum}
\end{equation}
Inverse relation reads
\begin{equation}
P_{\tau}(X) \ = \ \sum_{\mu\vdash r} \ \chi_{\mu}(\tau) s_{\mu}(X) \ .
\label{schur_pwrsum_inverse}
\end{equation}
The power sum symmetric function of the unit matrix equals
\begin{equation}
P_{\tau}(I) \ = \ N^{|\tau|} \ ,
\label{pwrsum_unitmatr}
\end{equation}
where $|\tau|$ is the number of cycles in the permutation $\tau$.
We mention the following formula  ($1 \leq i \leq N$)
\begin{equation}
\sum_{j_1\cdots j_r}^N \ \sum_{m_1\cdots m_r}^N \ \prod_{k=1}^r \
\delta_{j_k,m_{\tau (k)}} \ \delta_{j_k,m_{\sigma (k)}} \ y_{j_k} y_{m_k} \ = \
P_{\tau^{-1}\sigma}(x) \ , \ x_i=y_i^2  \ .
\label{pwr_sum_add}
\end{equation}

Given the complete symmetric functions $h_k$ and the elementary symmetric functions $e_k$
in $m$ variables $x_1,\cdots ,x_m$
\begin{eqnarray}
h_k \ = \ \sum_{1\leq n_1\leq\cdots\leq n_k \leq m} \ x_{n_1} \cdots x_{n_k} \ , \
e_k \ = \ \sum_{1\leq n_1 < \cdots < n_k \leq m} \ x_{n_1} \cdots x_{n_k} \ ,
\label{ef_hf}
\end{eqnarray}
the Schur functions can be computed with the help of identities
\begin{eqnarray}
s_{\lambda}(X) \ = \ s_{(\lambda_1,\cdots,\lambda_m )}(x_1,\cdots,x_m) \ = \
\mbox{det} \left ( h_{\lambda_i-i+j}  \right )_{1\leq i,j \leq m} \ , \nonumber  \\
s_{\lambda}(X) \ = \ s_{(\lambda_1,\cdots,\lambda_m )}(x_1,\cdots,x_m) \ = \
\mbox{det} \left ( e_{\lambda^{\prime}_i-i+j}  \right )_{1\leq i,j \leq m} \ ,
\label{ef_hf_schur}
\end{eqnarray}
where $\lambda^{\prime}$ is a partition dual to $\lambda$ and the following rule is understood
\begin{equation}
s_{(\lambda_1,\cdots,\lambda_n)}(x_1,\cdots,x_{n-1},0) \ = \
\begin{cases}
0, \ {\rm {if}} \ \lambda_n \ne 0  \ \ , \\
s_{(\lambda_1,\cdots,\lambda_{n-1})}(x_1,\cdots,x_{n-1}), \ {\rm {if}} \ \lambda_n=0 \ .
\end{cases}
\label{schur_reduction}
\end{equation}
Another expression for the Schur function used in the text reads
\begin{equation}
s_{\lambda}(X) \ = \ \sum_{\tau_1,...,\tau_s; \sum i\tau_i=s} \ \chi_{\lambda}(\tau)
\prod_{j=1}^s \ \frac{1}{\tau_j! j^{\tau_j}}  \  \left [  {\rm Tr} (X)^j \right ]^{\tau_j}
\label{schur_repr1}
\end{equation}
with $\tau$ being a permutation such that the number of cylcles of length $j$ in $\tau$ is $\tau_j$.

\vspace{0.5cm}

{\bf \large Acknowledgements}

\vspace{0.2cm}

V.C. acknowledges financial support from the INFN HPC{\_}HTC
and NPQCD projects.

\vspace{0.5cm}

redpajama_set_name: RedPajamaArXiv


text:
Column: The Stilllife sickness claims another victim (me)
Photo by Annie Le Baron | The Daily Tar Heel
Student blows nose inside the student union at Alpine Bagel on Oct. 10, 2022.
By Hannah Kaufman
October 13, 2022 | 9:42am EDT
Editor's note: This article is satire.
This past month, I came down with an illness.
It wasn't your average, travel-size Kleenex type of cold. It wasn't COVID-19, strep throat or bronchitis. In fact, it wasn't anything that modern medicine has ever seen — so rare, in fact, that even our esteemed staff at Campus Health was unable to diagnose it.
They're calling it the Stilllife Sickness.
My symptoms began on a Wednesday morning. I woke to the faint sound of EDM, and wondered which of my roommates could possibly be listening to techno so early in the morning. I had slept 12 hours but I was feeling far more tired than usual — so loopy, in fact, it was almost as if I was tipsy.
Wobbling slightly, I made my way to the bathroom and flicked on the switch, closing my eyes so they had a chance to adjust to the light. But that's when things got strange. My eyelids didn't offer their usual cool, dark solace — instead, all I could see in every direction were strobe lights, flashing white and blue across my corneas.
I wrenched my eyes open and jumped back, in shock. How could that be possible? The strobe lights resembled those that would shine at 2 a.m. on a Saturday night, not in my bathroom at 9 a.m. on a Wednesday. My head hurt just thinking about it — but come to think of it, my head had been hurting all morning. If only my roommate would stop playing "Mo Bamba" so I could focus on the medical mystery that was materializing before my strobe-afflicted eyes.
I had gone to bed at 10 p.m. the night before after a night of studying, so why did I feel so… hungover? I hadn't even consumed alcohol since last Friday night at StillLife. I blew my nose, and what came out was not the regular assortment of snot-related gunk, but a brown-tinged liquid that smelled sharply of tequila.
What was happening to me? The EDM music was getting louder and louder, blasting at a volume no JBL is capable of. I slammed open the bathroom door and ran over to my roommate, Abby, who was calmly preparing avocado toast.
"CAN YOU PLEASE TURN THE MUSIC DOWN???" I half-screamed half-shrieked, unable to hide my anger.
"What music?" she replied, startled. "I don't even own a speaker. Do you want a slice of avocado toast?"
And so my Stilllife Sickness began.
Over the course of the next two weeks, every symptom, I realized, was identical to some aspect of the clubbing experience. I heard bass-boosted beat drops everywhere I went. I had sudden urges to dance in the middle of my math lecture. I mistook my friend offering me medicine for a Stilllife promoter trying to give me VIP passes.
I couldn't stop dressing in skirts and crop tops.
Finally, at the end of two weeks, my misery came to an end. I stopped feeling the need to dance up on strangers and, once the deafening sounds of EDM subsided, I was able to listen to my favorite indie folk playlist again. Bon Iver never sounded so good.
But I wasn't alone in what I went through during those brutal weeks. A friend of mine who had attended Might As Well the Thursday before kept feeling the need to yell compliments at random girls in bathrooms, and couldn't understand why Stone & Leaf Cafe wouldn't make her an alcoholic drink named 'The Well.' She was the first to be afflicted by the MAW Mucus.
And still, students dropped like flies. Underclassmen girls and frat boys all but disappeared off campus due to the raging Frat Flu, which resulted in a feeling of full-body stickiness that was impossible to wash off.
If they had gone to Four Corners, it was the Four Corners Fog. TRU, the TRU Tickle-in-my-throat. The Crunkleton, the Crunkleton Cough (there are no known cases of this yet, as no living person has ever been seen stepping foot into The Crunkleton).
Finally, September came to an end, and people began to build up immunity to the countless diseases that were swirling around campus. The first-year girls returned to the frats, the scary bouncers returned to MAW and whoever goes to The Crunkleton returned to The Crunkleton.
And though I feel completely back to normal now, it's still hard for me to imagine entering Still Life again. The strobe lights were too bright, the music too loud, the crop tops too cropped. I'm sure I'll go back there one day, but for now, I'm trying to focus on my health. No more wild Saturday nights, no more weekend benders, no more —
"Abby, can you PLEASE turn down the Lupe Fiasco? Oh, you aren't playing anything right now? Yeah…I'm feeling fine. Don't worry about me."
"Is there any way you could put on some Bon Iver?"
@_hannahkaufman

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text:
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Walmart isolates Black Friday deals into 3 separate events that kick off online
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Matthew Ronald
For customers who can't leave behind Black Friday conventions, Walmart said Wednesday that it actually plans to have in-store occasions highlighting profound limits.
However the occasion deals days will accompany pandemic-related insurances. Stores will open at 5 a.m. neighborhood time. Clients must arrange single-document before they enter. Stores will restrict the quantity of individuals inside. Workers will circulate disinfected shopping baskets. Also, a few, named wellbeing diplomats, will welcome customers and remind them to put on a cover.
Walmart is wanting to find a way to demoralize groups and prod some deal trackers on the web, as well. The organization said it will separate Black Friday into three distinctive occasion deals occasions, stumbled during the time of November. Each will start on its site and hit stores a couple of days after the fact.
Clients can likewise get buys at the store without venturing inside by utilizing curbside pickup.
Scott McCall, leader VP and head marketing official for Walmart U.S., said the organization needs to convey on low costs, comfort and security — regardless of how clients decide to shop.
"We've been very thoughtful as we planned this year's event," he said in a news release. "By spreading deals out across multiple days and making our hottest deals available online, we expect the Black Friday experience in our stores will be safer and more manageable for both our customers and our associates."
Walmart's first Black Friday occasion will begin online Nov. 4 and in stores Nov. 7 and highlight toys, gadgets and home items. The subsequent occasion, zeroed in on hardware, for example, TVs, cell phones, PCs and tablets alongside certain things from other product classes, will start online Nov. 11 and in stores Nov. 14. Also, the third occasion will commence online Nov. 25 and in stores Nov. 27 — the very day as the typical post-Thanksgiving shopping occasion. It will have a scope of things from hardware and toys to clothing and occasional stylistic theme.
Retailers have been mindful about their vacation deals standpoint, as they attempt to anticipate customers' craving for commending the season and trading endowments during a worldwide wellbeing emergency and a downturn. The National Retail Federation, which for the most part puts out its figure toward the beginning of October, still can't seem to say something.
Deloitte and Accenture, notwithstanding, have shared blended pictures of how the regularly bustling shopping time frame may play out. Deloitte anticipated that occasion retail deals will increment somewhere in the range of 1% and 1.5%, yet said that will depend on how much affluent Americans overdo it and how much lower-pay families pull back. Clients studied by Accenture, in any case, said they intend to spend a normal of about $540 — almost $100 not exactly a year ago.
Significant retailers, including Walmart, Target and Best Buy, have attempted to urge clients to begin their vacation shopping sooner than expected by beginning deals occasions in October. Walmart started its "Big Save Event" on Sunday, simply a couple of days before the beginning of Amazon Prime Day on Tuesday. Walmart's occasion goes through Thursday.
Related Topics:3 separate eventsBlack Fridaykick off onlineWalmart
Walmart's plan for three black friday to avoiding crowds this year during the coronavirus
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Matthew Ronald grew up in Chicago. His mother is a preschool teacher, and his father is a cartoonist. After high school Matthew attended college where he majored in early-childhood education and child psychology. After college he worked with special needs children in schools. He then decided to go into publishing, before becoming a writer himself, something he always had an interest in. More than that, he published number of news articles as a freelance author on apstersmedia.com.
Apple's new 11-inch iPad Pro with M1 processor is already $50 off at Walmart
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Ford Motor's market value topped $100 billion for the first time ever as the automaker's stock hit a new 52-week high Thursday.
The organization's shares hopped Thursday by as much as 5.7% to $25.87, hitting another 20 or more year high, prior to shutting at $25.02 a share, up 2.3%. Its fairly estimated worth dropped to $99.99 billion.
The additions have been energized by Ford's plans to build production of electric vehicles, including the Mustang Mach-E crossover and an upcoming electric version of its bestselling F-150 pickup that is expected out this spring. The efforts are part for a Ford+ turnaround plan led by CEO Jim Farley, who assumed control over the helm in October 2020.
Ford's presently worth more than crosstown adversary General Motors, at about $90 billion, as well as electric vehicle start-up Rivian Automotive, at $72 billion, which has failed to support acquires following a blockbuster IPO in November. Ford proceeds to altogether trail Tesla, which has a market cap of more than $1 trillion.
The automaker is evaluated overweight with a value focus of $21.83 an offer, as indicated by a normal of 22 examiners arranged by FactSet. Be that as it may, not all Wall Street examiners haven't totally gotten tied up with Ford's turnaround.
"The stock market's attraction to the Ford EV story continues to take us by surprise," Morgan Stanley analyst Adam Jonas told investors in a Thursday note called "Ford Market Cap Crosses $100bn: What's In the Price?"
Morgan Stanley's value focus for Ford is $12 an offer. Its bull case for the stock is $25 an offer, as indicated by Jonas.
"Ford's share price movement is impressive and management deserve credit for changing the strategic narrative, triggering a re-rating," Jonas said. "However, at this juncture, we believe the risks facing Ford and the sector are rising faster than the opportunity."
Jonas refered to worries including the auto industry's historically cyclical nature returning, challenges in scaling EV production and more competitive and engaging EVs entering the market against Ford.
Brand Buzz
Naseeb Abbas, Also known as Prince Naseeb, a dedicated humanitarian aid worker and a talented faith inspired singer has raised a staggering total of £500,000.00 together on a personal Justgiving page in addition raised an estimate of £3,000,000 collectively in a 3 year period with the generous support of his faithful following gained by the widespread appreciation of his vocal talents.
With a following of nearly 60,000 on all social media platforms collectively, Naseeb has become somewhat of a household named and this together with his transparency on social media has led to thousands of people entrusting him with their charitable donations.
In the past he has worked closely with many reputable charities, including Human Relief Foundation with whom he undertook his first deployment to Jordan in June 2017. Adam Kelwick, International Fundraiser at Human Relief Foundation has previously described Naseeb as being 'extremely likeable and humble, despite his many talents.'
The suffering and anguish of people that he saw in Jordan has led to his deployment to several other countries including Bangladesh, Yemen, Burma and Pakistan to hand deliver aid to thousands of more people.
Naseeb is currently working with Kashmir Orphan Relief Trust (KORT), a charity which provides for the needs of hundreds of orphan children in his hometown of Mirpur, Azad Kashmir as well as serving humanity in and around Pakistan.
He plays a fundamental part in raising funds for the charity through his various social media platforms and hopes to continue working in close association with KORT to serve humanity and help provide a safe and loving environment for the children in their care.
For many of us, life can feel like a never-ending cycle of hard work with little reward. Author-entrepreneur Mark Lachance's new book "The Lucky Formula" offers a way out. In his book, he lays out the formula for attracting good fortune in both business and personal life by harnessing the power of luck. The book includes personal anecdotes and teaches readers how to use a variety of techniques such as visualization and meditation in order to attract success; all with an emphasis on keeping things positive.
With the goal of helping readers create their own luck, this book shares stories on how certain events in his past led him to unlock secrets for bringing success into one's career path – all while revealing the misfortunes and trials he faced along the way.
Mark Lachance has been making waves in the social media marketing world for years now. He's a pioneer of this niche, with a career that would inspire any rookie or contemporary to try their hand at entrepreneurship. His story is one that encompasses both passion and tenacity, as he has been in the trenches but always comes out on top.
His early years as a sports agent served him well, giving him insight into the world of branding and marketing. His time spent working with companies helped shape his ideas for entrepreneurship – launching multiple successful businesses under Mark's belt including Maxy Media the number 1 digital marketing powerhouse in Canada when it comes to monthly spend on Tiktok.
Mark was living the dream. He had it all- until he lost everything. A few bad decisions led him into debt and depression, but this man refused to give up on himself or his dreams. When Mark found out that he could no longer pay his rent, he woke up one morning with a plan to start over from scratch. That day marked the first of many days that would lead him back into success.
For anyone who is down in the dumps, don't give up hope. You are stronger than you think. Keep fighting through life's struggles and see it through.
And if you're interested to see the same success and attract more luck in your life, grab a copy of his upcoming book, The Lucky Formula, and learn how you can stack the odds in your favor and cash in on success. You can start with calculating your Lucky Score. Simply go to TheLuckyFormula.com/Quiz to calculate your ability to attract Luck and Success.
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i watched for years as my dad came home from work exhausted. drained. he hated his job. his job hated him, too, and made sure it sucked as much life from him each day as it could. my dad, when he's been away from work, on a day off, over the weekend, on a vacation...he has this smile, and this laugh, and it brightens your day. he has this energy that, for a guy that's 63 years old, is contagious. sort of comical. but mostly contagious. but not when work was sucking him dry. not when we had financial problems. not when he raised his fist and asked God, "why?"
i said to myself, don't ever let that happen to you. i developed a hard-headed optimism. i opted out of international business and got into teaching so i could have a job that would ensure i made a meaningful impact on people's lives every day. i vowed to live life to its fullest.
after time, one way or another, we learn about life, and living it to its fullest. we learn that sometimes the metaphors we most commonly use for life don't seem to adequately describe what we're going through...we move past life being like a box of chocolates, we're in too much turbulence to imagine taking a leisurely stroll and stumbling upon two roads diverged in a yellow wood. sometimes i imagine life is less like a journey down a highway, and more like a boxing match against a disproportionately large opponent. living life to its fullest sometimes means taking shots from life, ones we expect and ones we don't expect.
life can beat us up. sometimes it gives us a combo of hits that takes us down, and we'd like to just stay on the ground for a while. maybe we feel like we can't get up--the blows have knocked the wind out of us, and we're struggling to pull in breaths of air. maybe we're afraid to get back up, only to take another heavy shot.
for too long i watched life beat up on my dad. that's not even counting what he endured before i was even born.
i don't want to be that, i don't want to fall victim to the same bully that pushed my dad around. but when that left hook comes and rattles my rib cage, that uppercut connects with my jaw, words and thoughts and wishes and plans become dim...and i'm left not being able to see straight, fighting for breath, just trying to remember my name.
all the images evoked from all the fight-centered movies i've seen, the one with the protagonist on his knees, face swollen, blood dripping from the lip come to mind...and i have to say i don't know if my life is like those movies. i don't know if the inspirational music will begin to swell and my legs will find their strength again. i don't know when i'll catch my second wind and get that look of confidence back in my eyes.
my face is on the mat, my eyes are swollen shut, and my hands are crawling, reaching, waiting...waiting for the touch of the only one who can pull me back onto my feet. waiting for the only voice whose words will pierce through all this ringing in my head.


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HomePriyanka Chopra wikiPriyanka Chopra Wiki, Age, Height, Weight, Career, Movie, Husband
Priyanka Chopra Wiki, Age, Height, Weight, Career, Movie, Husband
by - WikiBioOnline on - March 15, 2019
Priyanka Chopra Biography
Priyanka Chopra is the Indian film actress, singer and Miss World title of 2000. Because of his film career, he became Bollywood's most expensive actress. Today Priyanka's name is taken in the well-known artists of India. Along with this, Priyanka Chopra was also the most sexist woman in Asia by the media. In his life till now, he has received many awards and has also been nominated for the award. Which included the National Film Award for Best Actress and five Filmfare Awards and Padma Shri, which the Indian government gave to him in his achievements in 2016.
At the age of 13, Chopra went to United States to study, while she was living with her aunt. In her teenage years Priyanka suffered a lot of problems while living in America. The ethnic problem was mainly. She used to say, "I was a strange girl, she never had any respect, I came from a middle-class family and there were white spots on my feet …. But I used to do hard work. Today my legs are selling 12 brands. "
Along with being a great actress, Priyanka also performs stage shows from Bahut, writes column for the newspaper and also the brand ambassador of the products. His first solo movie "In My City" came in 2012. Their second single came "Exotic" in 2013. This film was also shown in the United States and Canada. Chopra is known for his invaluable work from Bahot, due to which he became UNICEF's Goodwill Ambassador in 2010. Priyanka also works socially from social work and also disseminates the spread of nature, health and education, women education and women's rights and women's protection, Sam sexuality. He never makes his personal life the media's life.
1) Andaz (2003)
2) Mujhse Shaadi Karogi (2004)
3) Atraj (2004)
4) Don (2006)
5) Fashion (2008)
6) Kaminee (2009)
7) whats your Rashi (2009)
8) 7 Khoon Maaf (2011)
9) Don 2 (2011)
10) Barfi (2012)
11) Marry Com (2014)
12) Dil Dharkne Do (2015)
13) Bajirao Mastani (2015)
Priyanka Chopra Awards
All prizes given to Priyanka are given below ….
Priyanka Chopra's Film Award for Fashion (2008),
National Film Award for Best Actress and five Filmfare Awards,
Best Debut Award for Andaz (2003) and
Best Negative Role in Aitbaar (2004),
Best Actress for 7 Khoon Maaf,
Awarded Best Supporting Actor Award for Baji Rao Mastani (2015).
Despite not being a part of the filmmakers' house, Priyanka herself made Bollywood's top actress. All this is the result of their hard work and sincerity in work.
Biography Priyanka Chopra Priyanka Chopra age Priyanka Chopra Biography Priyanka Chopra career Priyanka Chopra height Priyanka Chopra husband Priyanka Chopra movie Priyanka Chopra wiki
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Mahatma Gandhi Biography | Life Story | History & Family
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This is the second year of the first Asia Pacific regional security outlook to be produced by a multinational team of contributors from the region. The outlook contains an overview and chapters on security issues as seen from the perspective of the major countries involved in the intergovernmental ASEAN Regional Forum security consultations. The country chapters deal with three basic issues: changing perceptions of threat, defense doctrines and issues, and national contributions to regional and global security. This year's volume focuses on the developments in these areas during 1997 and projects their implications for the future.
The regional overview aggregates and compares the country perspectives, pointing out differences of emphasis that affect the region's international relations. Last year's Security Outlook placed four broad issues on a "watchlist": the Korean peninsula, territorial disputes, changing large power roles and relations, and arms modernization programs.


text:
Joseph R. Biden
46th President of the United States: 2021 ‐ present
Statement by Vice President Joe Biden on President Trump's Visit to Atlanta
Four years ago, President Trump ran on a platform to make America great again. Yet under President Trump's failed leadership, nearly 6,800 Georgians have died from the coronavirus. Black Georgians have been hit particularly hard by this crisis — 3,000 Black Georgians have died, 430,000 Black Georgians are uninsured, and 11.8% of Black Georgians have been left jobless. And, in the midst of this global health pandemic and economic crisis, President Trump is still working to tear down the Affordable Care Act and take away protections for Georgians with pre-existing conditions.
We are in a battle for the soul of our nation and the stakes of this election could not be higher. Congressman John Lewis understood that part of being an American means doing what we can, with the time we have, to achieve the promise of our nation — that we are all created equal and deserve to be treated equally.
As President, I will work to advance racial equity across the American economy and build back better. I promise to fight for Black working families and direct real investments to advance racial equity as part of our nation's economic recovery.
Joseph R. Biden, Statement by Vice President Joe Biden on President Trump's Visit to Atlanta Online by Gerhard Peters and John T. Woolley, The American Presidency Project https://www.presidency.ucsb.edu/node/345041
Campaign Documents


Just When I Thought I'd Heard it All!
I've got a minor challenge that came up last week I want to share. At this point it's my own internal wrestling match. I got involved defending a completely absurd accusation and just should have walked away. My wife, Sheila, has a favorite line, "Don't should on yourself!"
I was accused on Friday of misleading all of you by calling this a "University". I was told I was dishonest. I was told, "Until they are accredited by the University System or provide proof of accreditation OUS accepts, they should drop the "university" from their name. It's misleading at best."
We set out to have fun with the university metaphor and we're going to continue to do just that. The site is "FREE" and the partners in the project represent some of the best supporters of education in photography.
However, the best part of all was when my accuser let his ego get in the way with the claim, "I guarantee these people I could give them more of an education for no cost than your "university" at any rate." I'm assuming he's about to take on the faculty: three Pulitzer Prize winners and some of the finest artists, photo-educators in the world.
I have no intention of dropping the "U" in SCU or of slowing down in building content on the site to help you build a stronger business model. The SCU site and content is FREE as well as my time whenever you write or call for help. I won't always have the answers, but I've got an amazing network with people who do and I talk to dozens of photographers every week. We're all here to help. In fact, we're going to ramp up in 2014. Stay tuned next week, because we've got some new features and programming we're going to announce for 2014.
Have a great Friday and thanks for listening to my rant! Make it a great day and the doorway to a spectacular weekend. I hope to see many of you in NYC next week. If you're headed to PPE check out ShootNYC. I'm the opening speaker on Thursday morning and the two day event is free, but you have to register. Looking forward to seeing you in NYC!
No doubt this guy is 60+ years old and has been wearing his medals from print competitions from the past 40 years. No doubt he has won awards for decades. Has a "Master of Photography" award plastered all over his website. Not to diss any of those that have all of that fluff as it does take a lot of time and talent to earn those. But a lot of the guys that have those "medals" are mean and nasty men that are against the very thing that SCU is all about. UGH..... Keep on keeping on Skip! (Again, I have respect for the photographers that have earned those titles....but not for the ones who have earned them and feel they are above photographers who do not walk on that path.
Hey Skip, don't forget about iTunes U! "The world's largest online catalog of free education content." I don't think anyone has given Apple a hard time about their choice of the name.
Don't forget about Monsters University.


text:
The Fabulist logo
Heat in Four Movements
By Bernardo Carvalho
Fable●
Translated from the Portuguese by Julia Powers
Dementia caniculae, or heat madness, as I have come to
diagnose it, is a mild, almost imperceptible condition, although
progressive, which is disseminated among humans insidiously
and without mercy until reaching a point of no return.
Dr. Horatio Lummings
The Masochist (102°)
"They were love messages, obviously. She was talking to me when she said, 'The heat wave approaching the interior of the state should reach the city this afternoon with highs of up to 102 degrees,' or, 'For today, a low of 87 and a high of 100.' They were signals, often indirect ones; she meant something else, something, and I could feel that she was delirious with love at times, an encoded language between the two of us, meant for me alone to decipher, things she couldn't say in front of the television viewers or radio listeners, and if she didn't show up later like she'd said she would, always in code, of course, during the TV weather forecast on or on the radio, it was just because she'd been thwarted somehow, and someone, someone other than me, had decoded the messages and prevented her from showing up to meet me that afternoon. They were love messages devised especially for my sake, because she wasn't stupid, not in the least, and she loved me so much, in her encoded way, of course, so much so that it was doubtful anyone else could understand or even conceive of what she truly meant by '104 in the shade'—only I could know. When people would come up to me saying, 'Have you heard? It's going to be a 102 this afternoon,' I could hardly keep from laughing in their faces. It was frightening to think they were that stupid, but all the better, that they didn't understand that she was, deep down, talking to me and no one else. And while the others were dying of heat in the streets, capable of murder even, going mad with the heat in traffic jams, sweating buckets, and running in desperation for the air-conditioning, it was enough for me to simply turn on the radio and listen. On TV they were insisting on the drastic nature of the situation, saying the city had never experienced such a heat wave, and I just laughed to myself thinking of her powers of persuasion, how they fell for it so easily, how they believed and lived, by autosuggestion, what she told them live, and they suffered what she told them to suffer, in a world that didn't exist, when really it was nothing more than a code, between the two of us, just for me. What a remarkable woman."
The Botanist (113°)
As the journalists traveled away from the city, the vegetation thinned out, forest giving way to scrubland, and scrubland to nothing at all, and that was where the botanist lived. He had summoned the press himself, though he tolerated their questions with difficulty. He claimed to have something to say at long last, after years without giving even a sign of life, discredited, as the rumors spread that he no longer dominated his faculties. "For years I have been studying the attraction of certain plants to the heat and I have arrived at a startling conclusion," said the botanist. "Which is what?" asked an impatient journalist. "That above 113 degrees plants emit strange signals, at an amazing frequency. The plants speak more and more with the increase in temperature," replied the botanist. "And if they seek out heat, it is because that is what allows them to speak, and because they have the need to speak, you understand what I'm saying? They too want to say something, just like you and I. The heat arouses their desire to say something," he said. "And what are the consequences of this discovery?" another wanted to know, seeing in it neither reason nor sense. To which the botanist, already losing his grip on himself, in part because of the heat, answered, "Well, how should I know, since, just as in my own case, I don't have the least idea what they're trying to say?"
The Laplandish Student (122°)
With the seasonal shortage of news, the newspaper assigned the columnist to write a weekly summer column narrating his wanderings through the world in search of unknown people, whose names, by some chance, passed through his hands. What he narrated in the first column, and which is here recapitulated, was his trip to Lapland, in search of D.M., a woman who, in 1957, as a prize for placing fourth in the complementary course at the high school for girls in Sodankylä, had received a book of photographs about deserts. The book piqued the columnist's interest (how had it ended up there?) as he rummaged through the secondhand bookshops along the Seine during his vacation. There was a photo of the Sahara printed on the cover and on the title page a stamp in Finnish: "Distribution of Prizes, Complementary Course, Fourth Prize in the Sodankylä High School for Girls, Awarded to D.M., June 2, 1957." In his column, the columnist narrated his adventures through the snow and ice to reach Sodankylä and how, with the help of her surname, he managed to find the house of D.M.'s older sister. Upon opening the door, the woman, rosy and very fat, a bear in her skin coat, informed him that her sister had disappeared years ago, roaming through the desert, and that, after weeks of senseless searches, all they managed to recover was her "sun-tanned skeleton." The columnist asked what D.M. had been doing in the desert, a sort of suicide, after all, and the sister replied that ever since D.M. was a girl, in her high school complementary course, the desert had been an obsession of hers, that D. had tried in vain to quench by it flipping through the pages of a book. "She went to the desert to find what she couldn't see in the photographs." As he stood there shivering and uncomprehending, she had no recourse but to conclude with the obvious: "The heat. She was convinced. She thought she would be able to see it."
The Firefighter (195°)
The moment he enters, he shouts and hears a voice. He needs to know if anyone's still trapped in the debris in the middle of the fire, someone he can still save from the flames. And as he advances into the blaze, on his descent into hell, and the heat mounts to the humanly intolerable, he shouts and hears a voice, and it gets closer and clearer each time. He believes that the voice hears him, too, and calls out to him. To keep it calm in the meantime, he decides to tell it a story, the only one he knows, that of his own life, and it advances in time as he advances through the columns of fire and the smoke that intoxicates, getting closer to the voice, and deeper and deeper into the heat. And as he tells his story, he hears the voice. But then all at once he realizes that, in fact, it's the voice that's telling him the story, that the voice is his, and so the story ends there.
The Memory Colony
I forgot Arabic and French soon after I left Algiers. You could say that I forgot Algiers after I left Algiers. In the past few years, I've started forgetting other things: names, faces, birthdays, anniversaries, appointments, events, words.
By H. M. Naqvi●January 2016
The Caravaggio Affair
Being who I was, a young novelist with the most conventional of middle-class lives, I fell in love completely with the reckless maestro whose sword was engraved with the words nec spe on one side and nec metuon the other: "Without hope, without fear."
By Álvaro Enrigue●February 2016


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What is Dr Disrespect Net Worth? – All About Guy Beahm Internet Personality
By: Author Rabya Atif
Categories Celebrities, Models
Guy Beahm is known worldwide as Dr DisRespect, an online alias that he adopted. He is an award-winning Twitch streamer, however, leveraged from Twitch platform to be known as an internet personality. Dr DisRespect is a perfect real-life example for everyone to believe the fact that everyone should do all that they are good at to attain fame and wealth in this day and age.
Dr DisRespect's net worth and popularity are gained through playing games and efficiently using the streaming platforms available. Thus, he has an impressive fan base translating stardom and huge bucks in this way.
How has Dr Disrespect Taken His Career To The Next Level?
Net Worth How Much Dr DisRespect Earns?
What is the physical appearance?
Who is Dr DisRespect's Wife – Mrs Assassin?
What happened to Dr DisRespect when he cheated on his wife?
Dr DisRespect was born on March 10, 1982, and was originally named Guy Beahm. He is just an ordinary guy before embracing the Dr DisRespect character. For instance, no one knows his birthplace, although all the Dr DisRespect's fans know that he currently lives in Washington. However, Guy Beahm's education has few details that he graduated from California State Polytechnic University in Pomona. Moreover, he also played Division II NCAA basketball during his university time.
His age is 40 years. And reportedly, he holds American nationality, his ethnicity is Caucasian, and his religion is Christianity.
Complete Name Herschel Beahm IV
Nicknames Dr. DisRespect & Guy Beahm
Birthplace California, United States
Occupation Internet Personality, Gaming Streamer
Ethnicity White Caucasian
Spouse Name Assassin (Video Game Streamer)
Children Name Alana Beahm (Daughter)
Started Streaming 2013
Channel Name DrDisRespect
Distinctive Features Mustache
Towering Height
Height 6' 8"
Shoe Size 12.5 (US)
Famous For Playing the multiplayer games Fortnite Battle Royale and H1Z1
Awards 1. Esports Sports Award for the Best Streamer (2017)
2. The Game Award as the Best Gamer (2017)
Current Residence Encinitas, California (United States)
Biceps Size 16.5 inches
Alma Mater California State Polytechnic University in Pomona
Favorite Actor Brad Pitt
Favorite Actress Margot Robbie
Favorite Color Black & Red
Dream Holiday Destination New York
Hobbies Playing video games & Spend time with his family
Favorite Dishes Continental Dishes
Guy Beahm rose as an internet personality when he started playing Halo 2 on the Xbox. While playing on Xbox, he had access to proximity voice chat that engaged other players in real-time streaming.
See Also: Who is Ari Fletcher? - Get To Know Her Age, Net Worth & Personal Life
In these streams, he showcased a character of dominance that helped him be a PRO player of the game. On January 5, 2010, Guy Beahm and his roommates decided to post gaming content on YouTube.
In the early stages of this step, Beham was given the position of community manager of Sledgehamme games. At this time, Beahm managed to rise through the ranks and hit level designer, i.e., how he helped everyone create a multiplayer map for Call of Duty: Advanced Warfare. He left Sledgehammer and joined Justin. tv to manage this entire focus on streaming.
Initially, when Dr Disrespect was struggling to gather followers, he used to play battle royale games like H1Z1. He was reputed to PlayerUnknown's Battlegrounds within no time, and then he moved to Fortnite. Today, he is a famous gaming celebrity with more than 2 million Twitch. tv followers are calling him "Dr DisRespect."
Below are a few best highlights of Dr Disrespect's career:
He has acquired over 3.7 million followers on social platforms.
He won the Game Awards ("Trending Gamer", 2017).
Net Worth: How Much Dr DisRespect Earns?
As of now, the estimated Dr DisRespect net worth is said to be around $6 million, which he earns from his social media profiles.
Dr DisRespect is popular widely for his way of entertaining the streaming industry rather than as a professional gamer. However, Dr Disrespect's character can be defined as similar to a professional wrestler with some cool sunglasses, tactical vests, and a sizeable attractive moustache. With time, Dr DisRespect is ranked in the second position in popularity on gaming platforms like Ninja, yet primarily known as the "Face of Twitch."
Besides prominent one-liners, Dr Disrespect's height is another buzz that constantly made him stand out. He is 6′ 8″ and possesses a 37″ vertical leap.
A post shared by Dr Disrespect (@drdisrespect)
Guy Beahm also keeps his family and personal life matters out of the public discussion. He got married in 2013 when he was working with Sledgehammer Games. Beahm's race to become the most followed streamer constantly drove towards heights. In this race, Dr DisRespect's wife also got an efficient edge. However, due to the massive demand for followers, the couple was occasionally seen in events like the eSports awards in 2017.
Beahm has been a controversial gamer who considered his family to hold a significant position in his life. In December 2017, Dr DisRespect announced that he wanted to focus on his family. Despite staying inactive for a short time, the return of Guy Beahm on the platform after two months clocked more than 388,000 viewers on his return stream. At this moment, when he was at the peak of his entertainment career, Dr DisRespect's wife also joined the streaming platform – Twitch with her own character identity as Mrs Assassin.
See Also: Who is Mor Shapiro? Know Everything About Ben Shapiro's Wife
Dr DisRespect and Mrs Assassin's life has not been a bed of roses. In 2017, there were several allegations on the famous Twitch streamer that he had been cheating on his wife. Moreover, one of Beahm's fans also admitted that he has been dating four women in 2015 and 2017.
Consequently, the couple separated their paths as Guy Beahm himself admitted that he was cheating on his wife. After the mishap, Dr DisRespected delivered a blubbering confession where he admitted his "Stupid f**king mistakes" and was wholeheartedly sorry to his fans, Twitch, and sponsors.
He concluded his confession by saying he will be "taking time off" to put things together. Thus, the couple managed to sort things out and are still together to date. Besides giving him, Mrs Assassin demonstrated that she is quite a supportive and understanding wife. Mrs Assassin came to fame just because of her husband, and more details about her haven't been disclosed yet. Yet, the information about Dr DisRespect's wife clearly states that she is a loving, caring, and supportive wife.
In August 2020, Beahm publicly announced that he was writing a personal memoir called Violence.
Dr Disrespect is pretty close to his mother and often shares snap with his mother on social platforms.
Moreover, the names of Dr Disrespect's parents and siblings are yet unknown.
Dr Disrespect is fond of playing basketball and football as he was a good player at Shasta High School.
Dr Disrespect's hobbies, according to him, include playing guitar and dancing.
Beahm became Dr Disrespect from playing battle royale games, starting with H1Z1 before moving to PlayerUnknown's battleground.
Instagram Follow Dr DisRespect on Instagram
Youtube Watch Dr DisRespect Videos on YouTube
Twitter Follow Dr DisRespect on Twitter
Facebook Follow Dr DisRespect on Facebook
TikTok Follow Dr DisRespect on Tiktok
IMDb Find Dr DisRespect on IMDb
Reddit Find Dr DisRespect on Reddit
Wikipedia Dr DisRespect on Wikipedia
Pinterest Dr DisRespect on Pinterest
Love him or hate him; he is going great with his work. He stands out from all other gaming streamers with his business brand success. Dr Disrespect's fans worldwide admire him for how he has disconnected his personal life from the entertainment industry. And this is something that most famous streamers fail to do. Thus, it gives a clear impression that Guy Beahm has a different personality Dr Disrespect.
What are your thoughts on Dr Disrespect net worth and personal life? Let us know in the comments.
Meet Clary Fisher, Simon Konecki's Ex-Wife
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All About Marzia Kjellberg (PewDiePie's Wife) - News, Age, and More
Wednesday 12th of October 2022
[…] Belonging to the widely viewed YouTube star and Twitch streamer PewDiePie – Marzia Kjellberg has gathered massive fame around the world for her sassy […]
Richard Wineberg
Sunday 31st of July 2022
This man is very talented!!!


Facebook customers have the exceptional capacity to connect with each other, as well as assemble online for all manner of discourse and discussion. Whether you intend to discuss last night's episode of Game of Thrones, or take individuals on in warmed political arguments, chances exist's a Facebook Group for that - Facebook How To Delete Group.
Integral to Facebook's social networking model is the Group, which is a foundation of the popular system. Groups are designed to fit an unfathomably huge range of users, and also their members could be just about any person, from the average joe to a real-life celeb. Whatever the objective of a team, and despite that inhabits them, it's a sure thing to say most Facebook individuals are in arrangement when it pertains to their significance.
Often, nonetheless, Groups shed their popularity. Perhaps just what was once an energetic space has actually shed its core subscription, or a TELEVISION program people once went crazy regarding has shed its charm. Whatever the situation, you might make a decision to delete or leave a group because of this.
If you're not a manager of the Group, and have actually chosen you've lost interest, all you have to do is stop by the Group's homepage, as well as click the switch in the upper-right-hand edge that says "Joined" This will certainly provide you the option to unfollow, or straight-up leave the Group.
If you are an admin or Group developer, there are a couple of unique actions you should comply with. Thankfully, they'll just take a number of mins, depending upon the size of your Group! First, click the switch that says "Members" under the Group's cover photo. Click the drawing of a gear switch next to the name of each participant in the Group, which will certainly let you erase them from said Group. Once all the Group participants are gone, just click the "Leave Group" button alongside your name. As every one of the participants will certainly have currently been deleted, your separation from the Group will efficiently shut it down.
However, it is necessary to note that Facebook uses a little caution about removing Groups. If you're an admin, however didn't produce the Group yourself, the Group's initial creator has to allow themselves be erased before you can totally close the Group down.
No matter your interest, there's a Facebook Group for you. Have a favorite Facebook Group you're currently a part of? Be sure to share it in the comments below.


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Worried Man Blues Ukulele Tutorial With Tab
Firstly, let's learn a three-chord and catchy folk tune with this Worried Man Blues ukulele tutorial complete with lyrics plus chords and tabs on screen! Worried Man Blues is one of the songs in our newest book 21 Easy Ukulele Folk Songs. So don't forget to get a copy when it's out on February 4th.
WORRIED MAN BLUES UKULELE TUTORIAL
Indeed, Worried Man Blues is an easy ukulele song as you'll need just three simple chords to play it. The chords are C, F and G7. For the string and strum version, you can do a D-DU-D-DU (D-down, U-up) pattern as shown by Jenny in the video. Also, for those that love to play melody tabs, Jenny plays the chord melody at the end of the video. So don't miss that part.
By the way, join us on Facebook for a live lesson on Worried Man Blues ukulele tutorial. In case you're not able to join the Facebook Live session for this song, just find the recording on the videos section.
Moreover, subscribe here to get email notifications of our new videos that get released each week. In particular, the notifications include downloadable sheet music with lyrics, chords and tabs that are very useful when learning the ukulele.
FIRST RECORDED BY THE CARTER FAMILY
One feature of most folk songs is that their composers are unknown or unrecorded. Certainly, Worried Man Blues is no different. While A.P. Carter, his wife Sara and sister-in-law Maybelle, collectively known as the Carter Family, first recorded the song in 1930, the writer of the song is not known. Accordingly, A. P. Carter used to travel throughout the United States collecting old songs and Worried Man Blues might be one of the songs he picked up from these travels.
DIFFERENT LYRICS
Another common feature with folk songs is that they have different versions of lyrics. Many other musicians have recorded Worried Man Blues after the Carter Family and most of them have different lyrics. For instance, Woody Guthrie recorded the song in 1940 with lyrics that underscored the number 21. While the lyrics from the Carter Family's recording described shackles with "29 links of chain" and a train which was "16 coaches long," Guthrie changed those numbers to 21. Moreover, Guthrie started with the verse instead of the chorus and he also added some new verses at the end of the song.
OTHER NOTABLE RECORDINGS
Here are other performers from various genres that have recorded Worried Man Blues: Pete Seeger, Cisco Houston, Ramblin' Jack Elliot, Kingston Trio, June Carter with Johnny Cash, The Stanley Brothers, George Jones, Osborne Brothers, Van Morrison and Devo. One notable difference with modern versions of the song is that the singers repeat the line of the verses three times. Similarly, Jenny repeats the line three times in the video tutorial. Whereas in the earlier versions, such as those of the Carter Family and Guthrie, they repeated the lines just twice.
Happy strumming!
Start playing the Ukulele with 21 Songs in 6 Days
You get much more than a book of songs. You'll learn how to play them with complete instruction in the book. Free online videos included with your purchase.
No matter what your learning style, you get to read, listen and watch. This ensures the fastest, easiest learning experience possible.
It's hard to mess this up. You'll learn the easiest songs first. You'll feel your confidence growing and your skills developing in minutes.
Learn directly from a Grammy award nominated music teacher and a former Harvard master class piano teacher.
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