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\usepackage{youngtab}
\usepackage{tikz}
%\usepackage[english]{isodate}

\DeclareRobustCommand{\SkipTocEntry}[5]{}

\DeclareMathAlphabet{\mathpzc}{OT1}{pzc}{m}{it}

\newcommand\Myperm[2][^n]{\prescript{#1\mkern-2.5mu}{}P_{#2}}

\newcommand{\ncom}{\newcommand }
\ncom{\ka}{\kappa}
\ncom{\Cbb}{\mathbb{C}}
\ncom{\Z}{\mathbb{Z}}
\ncom{\Grpn}{G(r,p,n)}
\ncom{\Drpn}{\text{D}(r,p,n)}
\ncom{\Lrpn}{L(r,p,n)}
\ncom{\Vk}{V^{\otimes k}}
\ncom{\Vkk}{V^{\otimes (k+1)}}
\ncom{\Vkh}{V^{\otimes (k+\frac{1}{2})}}
\ncom{\vi}{v_{i_1} \otimes v_{i_2} \otimes \cdots \otimes v_{i_k}}
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\ncom{\La}{\lambda}
\ncom{\tla}{\tilde{\La}}
\ncom{\tm}{\tilde{\mu}}
\ncom{\Tab}{\Tabop(r, \La)}
\ncom{\T}{\mathpzc{T}}
\ncom{\Tn}{\mathpzc{T}_k(r,p,n)}
\ncom{\Thn}{\mathpzc{T}_{k+\frac{1}{2}}(r,p,n)}
\ncom{\Tmn}{\mathpzc{T}_{k-\frac{1}{2}}(r,p,n)}
\ncom{\vo}{\varOmega}
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%%%%%%%%% a placer avant \begin{document}


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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

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%\DeclareMathAlphabet{\mathpzc}{OT1}{pzc}{m}{it} %pseudo \mathcal
\newcommand*{\mathpzc}{\mathcal}
\newcommand*{\fold}{\text{-}\textrm{fold}}
\end{DefTralics}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%% Auteur

%%%1
\author{\firstname{Ashish} \lastname{Mishra}}

\address{Instituto de Ci{\^e}ncias Exatas e Naturais\\ 
Universidade Federal do Par{\'a}\\
Bel{\'e}m\\
Par{\'a}\\
Brazil}

\email{ashishmsr84@gmail.com}

%%%2
\author{\firstname{Shraddha} \lastname{Srivastava}}

\address{The Institute of Mathematical Sciences\\
 Chennai\\
 India}
 
\email{maths.shraddha@gmail.com}


\thanks{The first author has been supported by UFPA-CAPES/PNPD fellowship and visiting professorship at UFPA, Brazil. The second author was supported by post-doctoral fellowship NPDF under DST-SERB, India.}


%%%%% Sujet

\keywords{Complex reflection groups, Tanabe algebras, Schur--Weyl duality, Jucys--Murphy elements}
 
\subjclass{05E10, 20F55, 20C15}



%%%%% Gestion

\DOI{10.5802/alco.97}
\datereceived{2018-12-22}
\daterevised{2019-08-11}
\dateaccepted{2019-09-18}


%%%%% Titre et résumé

\title
[On representation theory of Tanabe algebras]
{On representation theory of partition algebras for complex reflection groups}



\begin{abstract}
This paper defines the partition algebra, denoted by $\mathpzc{T}_k(r,p,n)$, for complex reflection group $G(r,p,n)$ acting on $k\fold$ tensor product $(\mathbb{C}^n)^{\otimes k}$, where $\mathbb{C}^n$ is the reflection representation of $G(r,p,n)$. A basis of the centralizer algebra of this action of $G(r,p,n)$ was given by Tanabe and for $p =1$, the corresponding partition algebra was studied by Orellana.
We also define a subalgebra $\mathpzc{T}_{k+\frac{1}{2}}(r,p,n)$ such that $\mathpzc{T}_k(r,p,n) \subseteq \mathpzc{T}_{k+\frac{1}{2}}(r,p,n)\subseteq \mathpzc{T}_{k+1}(r,p,n)$ and establish this subalgebra as partition algebra of a subgroup of $G(r,p,n)$ acting on $(\mathbb{C}^n)^{\otimes k}$. We call the algebras $\mathpzc{T}_k(r,p,n)$ and $\mathpzc{T}_{k+\frac{1}{2}}(r,p,n)$ Tanabe algebras. The aim of this paper is to study representation theory of Tanabe algebras: parametrization of their irreducible modules, and construction of Bratteli diagram for the tower of Tanabe algebras 
\[
\mathpzc{T}_0(r,p,n) \subseteq \mathpzc{T}_{\frac{1}{2}}(r,p,n) \subseteq \mathpzc{T}_1(r,p,n) \subseteq \mathpzc{T}_{\frac{3}{2}}(r,p,n) \subseteq \cdots \subseteq \mathpzc{T}_{\lfloor \frac{n}{2}\rfloor}(r,p,n).
\]
 We conclude the paper by giving Jucys--Murphy elements of Tanabe algebras and their actions on the Gelfand--Tsetlin basis, determined by this multiplicity free tower, of irreducible modules.
\end{abstract}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%




\begin{document}


\maketitle


\tableofcontents

\section{Introduction}\label{int}

The symmetric group $S_{k}$ acts on the $k\fold$ tensor product $\Vk$ of the $n$-dimensional vector space $V = \Cbb^n$ over the field of complex numbers $\Cbb$. The general linear group $GL_{n}(\mathbb{C})$ acts on $\Vk$ diagonally where $V$ is the defining representation of $GL_n(\Cbb)$. These two actions commute; moreover, they generate the centralizers of each other. This is known as the classical Schur--Weyl duality~\cite{Gre07}. 

Jones~\cite{J94} and Martin~\cite{M94}, independently, defined partition algebra $\Cbb A_{k}(q)$, where $q\in \Cbb$, as a generalization of Temperley--Lieb algebras and Potts model in higher dimensional statistical mechanics. The symmetric group $S_{n}$, being the subgroup of permutation matrices in $GL_n(\Cbb)$, acts on $V^{\otimes k}$. Jones~\cite{J94} proved Schur--Weyl duality between the partition algebra $\Cbb A_k(n)$ and the symmetric group $S_n$ acting on $\Vk$. Furthermore, Martin and Saleur studied the structure of partition algebras in~\cite{MS93, MS94} and proved that the partition algebra $\Cbb A_k(n)$ is semisimple unless $n$ is an integer such that $0 \leq n < 2k-1$.


 The subalgebra $\Cbb A_{k+\frac{1}{2}}(n)$ of partition algebra $\Cbb A_{k+1}(n)$ was introduced by Martin and Rollet~\cite{MR98} (See also~\cite{M00}). Halverson and Ram~\cite{HR05} showed Schur--Weyl duality between $\Cbb A_{k+\frac{1}{2}}(n)$ and the subgroup $S_{n-1}$ of $S_n$; and thus established it to be equally important as partition algebra $\Cbb A_k(n)$. It was also shown in~\cite{HR05} that the branching rule is multiplicity free for $\Cbb A_{l-\frac{1}{2}}(n) \subseteq \Cbb A_l(n)$ for $ l \in \frac{1}{2}\Z_{>0}$ whenever both the algebras are semisimple.
Recursively building the Bratteli diagram for the tower of partition algebras 
\[
\Cbb A_0(n) \subseteq \Cbb A_{\frac{1}{2}}(n) \subseteq \Cbb A_1(n) \subseteq \Cbb A_{\frac{3}{2}}(n) \subseteq \cdots,
\] 
the Jucys--Murphy elements of partition algebras were given in~\cite[Theorem~3.37]{HR05}. Later, the seminormal representations of parition algebra were derived by Enyang~\cite{E13}.

The complex reflection group $G(r,p,n)$, where $r,p$ and $n$ are positive integers such that $p$ divides $r$, is a subgroup of $GL_{n}(\Cbb)$. The group $G(r,1,n)$ is the wreath product of the cyclic group $\Z/r\Z$ by the symmetric group $S_n$ and $\Grpn$ is a normal subgroup of index $p$ of $G(r,1,n)$.
Shephard and Todd~\cite{ST54} gave a classification of finite irreducible complex reflection groups. It was shown there that the families of groups $S_n$ for $n>1$, $\Z/r\Z$ for $r >1$, and $\Grpn$ (except when $(r,p,n) = (2,2,2) \text { or } (1,1,1)$) are the only infinite families of finite irreducible complex reflection groups and there are exactly $34$ more finite irreducible complex reflection groups. Also, they characterized the group $\Grpn$, for $n>1$, by showing that these (except the group $G(2,2,2)$) are the only finite irreducible imprimitive complex reflection groups up to isomorphism.

The restriction of the action of $GL_n(\Cbb)$ on $V$ to $\Grpn$ is the reflection representation of $\Grpn$. Tanabe~\cite[Lemma~2.1]{T97} described a basis of the centralizer algebra of the action of $G(r,p,n)$ on the tensor space $V^{\otimes k}$. Orellana~\cite{Ore07} defined a subalgebra $\mathpzc{T}_{k}(n,r)$ of partition algebra $\Cbb A_k(n)$, and proved Schur--Weyl duality between $\T_{k}(n,r)$ and $G(r,1,n)$~\cite[Theorem~5.4]{Ore07}. Also, she recursively constructed the Bratteli diagram for the tower of algebras 
\[
\T_{0}(n,r)\subseteq \T_{1}(n,r)\subseteq \T_{2}(n,r) \subseteq \cdots
\] 
in~\cite[Proposition~5.6]{Ore07}.

In this paper, we define a subalgebra, denoted by $\T_k(r,p,n)$, of partition algebra $\Cbb A_k(n)$ such that there is Schur--Weyl duality between $\T_k(r,p,n)$ and the complex reflection group $\Grpn$. In particular, for $p=1$, the algebra $\T_k(r,1,n)$ is equal to the algebra $\T_k(n,r)$ defined by Orellana. Along the lines of~\cite{HR05}, we introduce a subgroup $\Lrpn$ of $\Grpn$ which plays a role analogous to the subgroup $S_{n-1}$ of $S_n$ in the study of partition algebra. We define a subalgebra, denoted by $\T_{k+\frac{1}{2}}(r,p,n)$, of partition algebra $\Cbb A_{k+\frac{1}{2}}(n)$ and exhibit Schur--Weyl duality between $\T_{k+\frac{1}{2}}(r,p,n)$ and $\Lrpn$. Thus, the algebras $\T_k(r,p,n)$ and $\T_{k+\frac{1}{2}}(r,p,n)$ are partition algebras for the complex reflection group $\Grpn$ and its subgroup $\Lrpn$, respectively. We call the algebras $\Tn$ and $\Thn$ \emph{Tanabe algebras}. 

The main results in this paper are as follows.
\begin{enumerate}[(\alph*)]
 \item\label{intro_a} For Tanabe algebras:
 \begin{enumerate}[(\roman*)]
 \item\label{intro_a_i}  Decomposition of the centralizer algebras $\End _{\Grpn}(\Vk)$ and $\End _{\Lrpn}(\Vk)$ into their irreducible modules which, in particular, gives parametrization of the irreducible modules of Tanabe algebras $\Tn$ and $\Thn$ for $n \geq 2k$ and $n \geq 2k+1$, respectively (Theorem~\ref{irrend}). 
 
 \item\label{intro_a_ii} Construction of Bratteli diagram recursively for the tower 
\begin{center}
\vspace{\abovedisplayskip}
$\displaystyle 
\T_0(r,p,n) \subseteq \T_{\frac{1}{2}}(r,p,n) \subseteq \T_1(r,p,n) \subseteq \T_{\frac{3}{2}}(r,p,n) \subseteq \cdots \subseteq \T_{\lfloor \frac{n}{2}\rfloor}(r,p,n) .$
\vspace{\belowdisplayskip}
\end{center}
 In this case, the Bratteli diagram is a simple graph (Section~\ref{bd}).
 \item\label{intro_a_iii} Description of a specific set of commuting elements, called Jucys--Murphy elements, which act as scalars on the canonical basis, called Gelfand--Tsetlin basis, of each irreducible module of Tanabe algebras (Theorem~\ref{jm}).
 \end{enumerate}
 \item\label{intro_b} For complex reflection groups:
 \begin{enumerate}[(\roman*)]
 \item\label{intro_b_i} Construction of a basis of irreducible $\Grpn$-modules (Theorem~\ref{eb}) using a combination of ideas from Okounkov--Vershik approach to the representation theory of $G(r,1,n)$ in~\cite{MS16}, Clifford theory and higher Specht polynomials in~\cite{MY98}.
 \item\label{intro_b_ii} Branching rule from $\Grpn$ to $\Lrpn$ (Theorem~\ref{brgpl}).
 \item\label{intro_b_iii} Decomposition of $\Vk$ in terms of irreducible $\Grpn$-modules and $\Lrpn$-modules (Theorem~\ref{decvk}).
 \end{enumerate}
\end{enumerate}

Using theory of the basic construction,~\cite[Theorem~3.27]{HR05} shows that the necessary and sufficient condition for the semisimplicity of partition algebra $\Cbb A_l(n)$, for $n \in \Z_{\geq 2}$ and $l \in \frac{1}{2}\Z_{\geq 0}$, is $ l \leq \frac{n+1}{2}$. The corresponding question for Tanabe algebras is open. Also, another important question is to find a presentation by generators and relations for Tanabe algebras. A generating set for $\T_k(r,1,n)$ is given in~\cite[Proposition~5.1]{Ore07}. A set of generators for $\End _{\Grpn}(V^{\otimes k})$ is described in~\cite[Section~3]{T97}.

The inductive approach to the representation theory of symmetric groups was done by Okounkov and Vershik in~\cite{vo1, vo}. This approach considers the chain of symmetric groups 
\[
\{1\} = S_1 \subset S_2 \subset \cdots \subset S_n \subset \cdots 
\]
 to study their representation theory recursively. The advantage over the traditional approach is that the appearance of Young diagrams and standard Young tableaux is given a spectral explanation, and the branching rule is determined simultaneously. The Gelfand--Tsetlin decomposition, the Gelfand--Tsetlin algebra, the canonical Gelfand--Tsetlin basis of the irreducible representations, and the Jucys--Murphy elements, a set of generators of Gelfand--Tsetlin algebra, are fundamental to this approach. The corresponding approach in the case of $G(r,1,n)$ proves fruitful in giving new proofs of some known results and also in establishing new results in this paper.

This paper is organized in the following sections. Section~\ref{pre} gives a brief introduction to partition algebra, Okounkov--Vershik approach to the representation theory of $G(r,1,n)$, and Clifford theory. 
In Section~\ref{ta}, we define Tanabe algebras, $\Tn$ and $\Thn$, as subspaces of partition 
algebras and prove that these subspaces are algebras. 


Section~\ref{crg} contains a description of representation theory of complex reflection group $G(r,p,n)$ and its subgroup $L(r,p,n)$ (Theorems~\ref{irrgp} and~\ref{irrlp}). We review the representation theory of $\Grpn$ using Clifford theory. We parametrize the irreducible $L(r,1,n)$-modules and, then by Clifford theory, determine the representation theory of $L(r,p,n)$. This section concludes with the branching rule from $\Grpn$ to $\Lrpn$. 

\looseness-1
In Section~\ref{swdu}, we demonstrate that $\Tn$ and $\Thn$ are in Schur--Weyl duality with $\Grpn$ and $\Lrpn$, respectively.
Using results from Section~\ref{crg}, Section~\ref{bd} starts with the decomposition of $\Vk$ as $\Grpn$-module and $\Lrpn$-module. Then, we give decomposition of $\Vk$ as $(\Grpn, \Tn)$-bimodule and $(\Lrpn, \Thn)$-bimodule (Theorem~\ref{dvkp}) and use it to construct Bratteli diagram of Tanabe algebras. 
In Section~\ref{jme}, we give Jucys--Murphy elements and their actions on the canonical Gelfand--Tsetlin basis of irreducible modules of Tanabe algebras.

\begin{enonce*}[remark]{Conventions} Throughout this paper, we assume that 
\begin{enumerate}[(\roman*)]
 \item\label{conv_i} $r,p,m$ and $n$ are positive integers such that $p$ divides $r$ and $m = \frac{r}{p}$, and 
 \item\label{conv_ii} we index the components in a $w$-tuple from $1, \ldots, w$, therefore, for a multiple $t$ of $w$, the $t(\modrm w)$-th component means the $w$-th component.
\end{enumerate}
\end{enonce*}

\section{Preliminaries}\label{pre}
In this section, we give an overview of partition algebra, Okounkov--Vershik approach and Clifford theory to set up notations and to state basic definitions and results used in the rest of the paper.

\subsection{Partition algebra}\label{pa}
For $k \in \Z_{\geq 0}$, let $A_k$ be the set of all set partitions of $\{1,2,\ldots,k,1', 2', \ldots, k'\}$. Given an element $d \in A_k$, we say that $i$ and $j$ are in the same block in $d$ if $i$ and $j$ belong to the same set partition in $d$. The elements of $A_k$ can be depicted as graphs, called partition diagrams, with the vertices $\{1,2, \ldots, k\}$ and $\{1',2', \ldots, k'\}$ in the top and bottom rows, respectively and two vertices in the same block are connected by an edge. By $d = \{B_1, B_2, \ldots, B_s\}$, we denote that there are exactly $s$ blocks $B_1, B_2, \ldots, B_s$ in $d$. Also, $|d|$ denotes the number of blocks in $d$. 

The multiplication of two elements $d_1, d_2 \in A_k$, denoted by $d_1 \circ d_2$, is obtained by concatenating the diagrams $d_1$ and $d_2$ in the following way: place $d_1$ above $d_2$, identify the vertices in the bottom row of $d_1$ with the vertices in the top row of $d_2$, then remove all the connected blocks which are entirely in the middle row. The multiplication $\circ$ makes $(A_k, \circ)$ a monoid with the identity element given in Figure~\ref{fig:Id}.

\begin{figure}[htb]\centering
\begin{tikzpicture}[scale=1,mycirc/.style={circle,fill=black, minimum size=0.1mm, inner sep = 1.5pt}]
 \node[mycirc,label=above:{$1$}] (n1) at (0,1) {};
 \node[mycirc,label=above:{$2$}] (n2) at (1,1) {};
 \node[mycirc,label=above:{$3$}] (n3) at (2,1) {};
 \node(a) at (3,0.5) {...};
 \node[mycirc,label=above:{$i$}] (ni) at (4,1) {};
 \node(b) at (5,0.5) {...};
 \node[mycirc,label=above:{$(k-1)$}] (nk-1) at (6,1) {};
 \node[mycirc,label=above:{$k$}] (nk) at (7,1) {};
 \node[mycirc,label=below:{$1'$}] (n1') at (0,0) {};
 \node[mycirc,label=below:{$2'$}] (n2') at (1,0) {};
 \node[mycirc,label=below:{$3'$}] (n3') at (2,0) {};
 \node[mycirc,label=below:{$i'$}] (ni') at (4,0) {};
 \node[mycirc,label=below:{$(k-1)'$}] (nk-1') at (6,0) {};
 \node[mycirc,label=below:{$k'$}] (nk') at (7,0) {};
 \draw (n1) -- (n1');
 \draw (n2) -- (n2');
 \draw (n3) -- (n3');
 \draw (ni) -- (ni');
 \draw (nk-1) -- (nk-1');
 \draw (nk) -- (nk');
\end{tikzpicture}
\caption{Identity element in partition monoid.}
\label{fig:Id}
\end{figure}

Define a subset $A_{k+\frac{1}{2}}$ of $A_{k+1}$ consisting of those elements which have $(k+1)$ and $(k+1)'$ in the same block. It can be easily seen that $A_{k+\frac{1}{2}}$ is a submonoid of $A_{k+1}$. The monoids $A_k$ and $A_{k+\frac{1}{2}}$ are called partition monoids. 

\begin{exam}\label{pd} Taking $k=6$, the elements $d_1$ and $d_2$ in $A_6$ with 
\begin{align*}
d_1 &= \{\{1,2,1'\},\{3,5,3'\},\{4\},\{6,5'\},\{2',4'\},\{6'\}\} \in A_6\\
\intertext{and}
d_2 &= \{\{1,5,2',3'\},\{2,4\},\{3\},\{6,6'\},\{1',4'\},\{5'\}\} \in A_6,
\end{align*}
 can be written in terms of partition diagrams and the multiplication $d_1 \circ d_2$ is illustrated in Figure~\ref{fig:mult}.

\begin{figure}[ht]\centering 
 \begin{tikzpicture}[scale=1,mycirc/.style={circle,fill=black, minimum size=0.1mm, inner sep = 1.5pt}
 ]
 \node (1) at (-1,0.5) {$d_1 =$};
 \node[mycirc,label=above:{$1$}] (n1) at (0,1) {};
 \node[mycirc,label=above:{$2$}] (n2) at (1,1) {};
 \node[mycirc,label=above:{$3$}] (n3) at (2,1) {};
 \node[mycirc,label=above:{$4$}] (n4) at (3,1) {};
 \node[mycirc,label=above:{$5$}] (n5) at (4,1) {};
 \node[mycirc,label=above:{$6$}] (n6) at (5,1) {};
 \node[mycirc,label=below:{$1'$}] (n1') at (0,0) {};
 \node[mycirc,label=below:{$2'$}] (n2') at (1,0) {};
 \node[mycirc,label=below:{$3'$}] (n3') at (2,0) {};
 \node[mycirc,label=below:{$4'$}] (n4') at (3,0) {};
 \node[mycirc,label=below:{$5'$}] (n5') at (4,0) {};
 \node[mycirc,label=below:{$6'$}] (n6') at (5,0) {};
 \draw (n1) -- (n2);
 \draw (n1) -- (n1');
 \draw (n3) .. controls(3,0.7) .. (n5) -- (n3');
 \draw (n2') .. controls(2,0.3) .. (n4');
 \draw (n6) -- (n5');
 

 \node (1) at (-1,-2) {$d_2 =$};
 \node[mycirc,label=above:{$1$}] (n7) at (0,-1.5) {};
 \node[mycirc,label=above:{$2$}] (n8) at (1,-1.5) {};
 \node[mycirc,label=above:{$3$}] (n9) at (2,-1.5) {};
 \node[mycirc,label=above:{$4$}] (n10) at (3,-1.5) {};
 \node[mycirc,label=above:{$5$}] (n11) at (4,-1.5) {};
 \node[mycirc,label=above:{$6$}] (n12) at (5,-1.5) {};
 \node[mycirc,label=below:{$1'$}] (n7') at (0,-2.5) {};
 \node[mycirc,label=below:{$2'$}] (n8') at (1,-2.5) {};
 \node[mycirc,label=below:{$3'$}] (n9') at (2,-2.5) {};
 \node[mycirc,label=below:{$4'$}] (n10') at (3,-2.5) {};
 \node[mycirc,label=below:{$5'$}] (n11') at (4,-2.5) {};
 \node[mycirc,label=below:{$6'$}] (n12') at (5,-2.5) {};
 \draw (n7) -- (n8') -- (n9') -- (n11);
 \draw (n8) .. controls(2, -1.9) .. (n10);
 \draw (n7') .. controls(1.5,-2) .. (n10');
 \draw (n12) -- (n12');
 
 \draw[dashed] (n1') .. controls(-0.5,-1)..(n7);
 \draw[dashed] (n2') .. controls(0.5,-1)..(n8);
 \draw[dashed] (n3') .. controls(1.5,-1)..(n9);
 \draw[dashed] (n4') .. controls(2.5,-1)..(n10);
 \draw[dashed] (n5') .. controls(3.5,-1)..(n11);
 \draw[dashed] (n6') .. controls(4.5,-1)..(n12);
\end{tikzpicture}

\begin{tikzpicture}[scale=1,mycirc/.style={circle,fill=black, minimum size=0.1mm, inner sep = 1.5pt}
 ]
 \node (1) at (-1,0.5) {$d_1 \circ d_2 =$};
 \node[mycirc,label=above:{$1$}] (n1) at (0,1) {};
 \node[mycirc,label=above:{$2$}] (n2) at (1,1) {};
 \node[mycirc,label=above:{$3$}] (n3) at (2,1) {};
 \node[mycirc,label=above:{$4$}] (n4) at (3,1) {};
 \node[mycirc,label=above:{$5$}] (n5) at (4,1) {};
 \node[mycirc,label=above:{$6$}] (n6) at (5,1) {};
 \node[mycirc,label=below:{$1'$}] (n1') at (0,0) {};
 \node[mycirc,label=below:{$2'$}] (n2') at (1,0) {};
 \node[mycirc,label=below:{$3'$}] (n3') at (2,0) {};
 \node[mycirc,label=below:{$4'$}] (n4') at (3,0) {};
 \node[mycirc,label=below:{$5'$}] (n5') at (4,0) {};
 \node[mycirc,label=below:{$6'$}] (n6') at (5,0) {};
 \draw (n1) -- (n2);
 \draw (n1) -- (n2') -- (n3') -- (n6);
 \draw (n3) .. controls(3,0.7) .. (n5);
 \draw (n1') .. controls(1.5,0.3) .. (n4');
 \end{tikzpicture}
\caption{Example of multiplication in partition monoid.}
\label{fig:mult}
\end{figure}
\end{exam}
 
For a complex number $q$, define 
\[
\Cbb A_k(q) := \Cbb\text {-}\spanop \{d \in A_k\}.
\]
The multiplication of basis elements, which when extended linearly makes $\Cbb A_k(q)$ an associative algebra, is defined as: for $d_1, d_2 \in A_k$, define 
\begin{equation*}
 d_1d_2 := q^l d_1\circ d_2
\end{equation*}
where $l$ is the number of blocks removed from the middle row while computing $d_1\circ d_2$. Also, $\Cbb A_{k+\frac{1}{2}}(q)$ is a subalgebra of $\Cbb A_{k+1}(q)$. The algebras $\Cbb A_k(q)$ and $\Cbb A_{k+\frac{1}{2}}(q)$ are called \emph{partition algebras}.

\begin{exam}
In Example~\ref{pd}, the product $d_1d_2$ in $\Cbb A_6(q)$ is illustrated in Figure~\ref{fig:multalg} since one block has been removed from the middle row while computing $d_1\circ d_2$ in Figure~\ref{fig:mult}.

\begin{figure}[ht]
\centering
 \begin{tikzpicture}[scale=0.8,mycirc/.style={circle,fill=black, minimum size=0.1mm, inner sep = 1.5pt}
 ]
 \node (1) at (-1.5,0.5) {$d_1d_2 =$};
 \node (2) at (-0.5,0.5) {$q$};
 \node[mycirc,label=above:{$1$}] (n1) at (0,1) {};
 \node[mycirc,label=above:{$2$}] (n2) at (1,1) {};
 \node[mycirc,label=above:{$3$}] (n3) at (2,1) {};
 \node[mycirc,label=above:{$4$}] (n4) at (3,1) {};
 \node[mycirc,label=above:{$5$}] (n5) at (4,1) {};
 \node[mycirc,label=above:{$6$}] (n6) at (5,1) {};
 \node[mycirc,label=below:{$1'$}] (n1') at (0,0) {};
 \node[mycirc,label=below:{$2'$}] (n2') at (1,0) {};
 \node[mycirc,label=below:{$3'$}] (n3') at (2,0) {};
 \node[mycirc,label=below:{$4'$}] (n4') at (3,0) {};
 \node[mycirc,label=below:{$5'$}] (n5') at (4,0) {};
 \node[mycirc,label=below:{$6'$}] (n6') at (5,0) {};
 \draw (n1) -- (n2);
 \draw (n1) -- (n2') -- (n3') -- (n6);
 \draw (n3) .. controls(3,0.7) .. (n5);
 \draw (n1') .. controls(1.5,0.3) .. (n4');
 \end{tikzpicture}\caption{Example of multiplication in partition algebra.}
 \label{fig:multalg}
\end{figure}
\end{exam}

\subsection{The Okounkov--Vershik approach}\label{ov}
Let $G^n$ denote the direct product of $n$-copies of a finite group $G$ and let $\Cbb[G]$ be the group algebra of $G$.
The action of the symmetric group $S_n$ on $G^n$ by permuting the coordinates defines the semidirect product of $G^n$ by $S_n$. The group $G^n \rtimes S_n$ is also known as wreath product of $G$ by $S_n$. The Okounkov--Vershik approach to the representation theory of $G^n \rtimes S_n$, where $G$ is any finite group, was done in~\cite{MS16}. In this paper, we are interested in the particular case $G = \Z/r\Z$, the cyclic group of order $r$. We use the notation $G(r,1,n) := (\Z/r\Z)^n \rtimes S_n$ where $\Z/r\Z = \langle \zeta \rangle$ with $\zeta$ being a primitive $r$-th root of unity. Thus, 
\[
G(r,1,n) = \{(g_1, g_2, \ldots, g_n, \pi) \mid g_i \in \Z/r\Z \text { for } i = 1, \ldots, n \text { and } \pi \in S_n \}.
\]
In this section, we follow~\cite{MS16} and present here a brief summary of Okounkov--Vershik approach to the representation theory of $G(r,1,n)$.

For $1\leq i \leq n$, let
 $H_{i,n} := \{(g_1,\ldots ,g_n,\pi)\in G(r,1,n) \;|\;
 \pi(j)=j \text { for }i+1\leq j\leq n\}.$
We have the following chain of subgroups of $G(r,1,n)$
\begin{equation} \label{mfic}
H_{1,n}\subseteq H_{2,n} \subseteq \cdots \subseteq H_{n,n} := G(r,1,n).
\end{equation}
The irreducible representations of $H_{1,n} \cong G^n = (\Z/r\Z)^n$ are one-dimensional. 

The following result of Wigner~\cite[p.~677]{wig68b} gives a necessary and sufficient condition for the restriction to be multiplicity free. For a proof using double centralizer theorem, see~\cite[Theorem~4.1]{MS16}. We use this result to prove the multiplicity freeness of chain~~\eqref{mfic}.

\begin{theo} \label{mfc} 
Let $A$ be a complex finite dimensional semisimple algebra and let $B$ be a
semisimple subalgebra. 
Then the relative commutant of the pair $A$ and $B$, denoted by $Z(A,B)$, consisting of all the elements of $A$ that commute with the elements of $B$ is semisimple and the
following conditions are equivalent:
\begin{enumerate}[(\alph*)]
\item\label{theo2.3_a} The restriction of any finite dimensional complex irreducible
representation of $A$ to $B$ is multiplicity free.

\item\label{theo2.3_b} The relative commutant $Z(A,B)$ is commutative.
\end{enumerate}
\end{theo}

Using~\cite[Theorem~4.13]{MS16}, we can conclude that the relative commutant of the pair of group algebras $\Cbb [H_{m,n}]$ and $\Cbb [H_{m-1,n}]$ is commutative for all $2 \leq m \leq n$. Thus, Theorem~\ref{mfc} implies that the chain~\eqref{mfic} is multiplicity free.

For each $i=1,2, \ldots, n$, suppose that $H_{i,n}^{\wedge}$ denotes the indexing set of irreducible $H_{i,n}$-modules and given $\La \in H_{i,n}^{\wedge}$, assume that $V^{\La}$ denotes the corresponding $H_{i,n}$-module. Bratteli diagram of the chain~\eqref{mfic} is a simple graph in which the vertices at $i$-th level are elements of $H_{i,n}^{\wedge}$ and a vertex $\mu\in H_{i-1,n}^{\wedge}$ is joined by an edge with a vertex $\lambda\in H_{i,n}^{\wedge}$ if $V^\mu$ appears in the restriction of $V^\lambda$ to $H_{i-1,n}$. 

For a fixed $1 \leq m \leq n$, consider the $H_{m,n}$-module $V^{\La_m}$, where $\lambda_m\in H_{m,n}^{\wedge}$. The branching rule being multiplicity free implies that the decomposition of $V^{\La_m}$ into irreducible $H_{m-1,n}$-modules is canonical, and the decomposition is 
\[
V^{\lambda_m}=\bigoplus_{\La_{m-1}} V^{\La_{m-1}},
\]
where the sum is over all $\La_{m-1}\in H_{m-1,n}^{\wedge}$ with an edge from $\La_{m-1}$ to $\La_m$ such that $V^{\La_{m-1}}$ is identified with the corresponding submodule of $V^{\lambda_m}$. We iterate the above decomposition for the chain of subgroups of $H_{m,n}$ in chain~\eqref{mfic}. This implies that the following decomposition of $V^{\lambda_m}$ into irreducible $H_{1,n}^{\wedge}$-submodules is canonical:
\begin{equation} \label{decomp}
V^{\lambda_m}=\bigoplus_{T} V_T,
\end{equation}
where the sum is over all possible paths $T=(\lambda_1, \lambda_2, \ldots, \lambda_m)$ from a vertex in $H_{1,n}$ to $\La_m$ in Bratteli diagram with $\lambda_i\in H_{i,n}^{\wedge}$ for $ 1 \leq i \leq m$.

The decomposition given in~\eqref{decomp} is called the \emph{Gelfand--Tsetlin decomposition \textup{(}GZ-decomposition\textup{)}} of $V^{\lambda_m}$ and each $V_T$ in~\eqref{decomp} is called a \emph{Gelfand--Tsetlin subspace \textup{(}GZ-subspace\textup{)}} of $V^{\lambda_m}$. In our case, each GZ-subspace $V_T$ is one-dimensional. Choose a non-zero vector $v_T \in V_T$. The basis 
\[
\{v_T \mid T \text { is a path in the GZ-decomposition of } V^{\La_m}\}
\]
 of $V^{\La_m}$ is called the \emph{Gelfand--Tsetlin basis \textup{(}GZ-basis\textup{)}} of $V^{\La_m}$ and it is unique up to scalars
and
\begin{equation*}
\Cbb[H_{i,n}]\cdot v_T = V^{\lambda_{i}},\;\;\;i=1,2,\ldots ,m.
\end{equation*}
The $GZ$-basis being canonical is significant in the case of $G(r,1,n)$, since, in general, for the wreath product of a finite group by symmetric group, $GZ$-subspaces are canonical, $GZ$-basis may not necessarily be canonical.

Using the algebra isomorphism
\begin{equation}\label{iso}
\Cbb[H_{m,n}] \cong \bigoplus_{\La_m\in H_{m,n}^{\wedge}} \End (V^{\La_m}),
\end{equation}
given by
\[
g \mapsto ( V^{\La_m} \buildrel {g}\over \rightarrow V^{\La_m}\;:\;
\La_m
\in H_{m,n}^{\wedge}),\;\;g\in H_{m,n}, 1 \leq m \leq n, 
\]
we can define the \emph{Gelfand--Tsetlin algebra \textup{(}GZ-algebra\textup{)}}, denoted by $GZ_{m,n}$, a maximal commutative subalgebra of $\Cbb[H_{m,n}]$ based on the GZ-decomposition~\eqref{decomp}:
\begin{multline*}
 GZ_{m,n} = \{ a\in \Cbb[H_{m,n}] \mid a \text { acts diagonally in the GZ-basis of } V^{\lambda_m},\\ 
 \text { for all }\lambda_m\in H_{m,n}^{\wedge}\}.
\end{multline*}

\begin{theo}[{\cite[Theorem~3.1(i)]{MS16}}] Let $Z_{i,n}$ denote the center of $\Cbb[H_{i,n}]$ for each $i =1,2, \ldots,m$. Then, 
\[
GZ_{m,n} = \langle Z_{1,n}, Z_{2,n}, \ldots ,Z_{m,n} \rangle.
\]
\end{theo}

The theorem above implies the following result which is~\cite[Lemma~3.2]{MS16}.

\begin{lemma} \label{sc}\leavevmode 
\begin{enumerate}[(\alph*)]
\item\label{lemma2.5_a} Let $v\in V^{\lambda_m}$ for some $\lambda_m\in H_{m,n}^{\wedge}$ such that $v$ is an eigenvector of every element of $GZ_{m,n}$, then \textup{(}a scalar multiple of\textup{)} $v$ belongs to the $GZ$-basis of $V^{\lambda_m}$.

\item\label{lemma2.5_b} Let $v$ and $u$ be elements in $V^{\La_m}$ and $V^{\mu_m}$, respectively for some $\La_m, \mu_m \in H_{m,n}^{\wedge}$ such that $v$ and $u$ have the same eigenvalues for every element of $GZ_{m,n}$, then $v=u$ and $\La_m = \mu_m$.
\end{enumerate}
\end{lemma}
A $GZ$-subspace of $H_{m,n}$ is a $GZ$-subspace of some irreducible $H_{m,n}$-module $V^{\lambda_m}$, $\lambda_m \in H_{m,n}^{\wedge}$. Thus, using Lemma~\ref{sc}, a $GZ$-subspace of $H_{m,n}$ is $GZ$-subspace of a unique irreducible representation of $H_{m,n}$. 

The Jucys--Murphy elements for the wreath product of a finite group by a symmetric group were given in~\cite{Pu97}. For our particular case $G(r,1,n)$,
the Jucys--Murphy elements can be written as: 
\begin{align} \label{JM}
 X_1 &= 0, \nonumber \\
\intertext{and} 
 X_j &= \sum_{i=1}^{j-1}\sum\limits_{l=0}^{r-1}\zeta_i^{l}\zeta_j^{-l}s_{ij},\;\;2\leq
j\leq n,
\end{align}
where $s_{ij}$ is the transposition $(i,j)$ and 
\[
\zeta_i^{l}\zeta_j^{-l}s_{ij} = (1, \ldots, 1, \g^l, 1 \ldots, 1, \g^{-l}, 1, \ldots, 1, (i,j)) \in G(r,1,n),
\] 
with $\g^l$ and $\g^{-l}$ as $i$-th and $j$-th coordinates, respectively. It is clear that the element $X_j$ can be identified naturally with an element of $H_{j,n}$.

\begin{theo}[{\cite[Theorem~4.4]{MS16}}]\label{yjm3} We have
\[
GZ_{m,n} = \langle Z[\Cbb[G^n]], X_1,X_2,\ldots ,X_m \rangle. 
\]
\end{theo}

A $GZ$-subspace $V$ of $H_{m,n}$ is isomorphic to $\rho_1\otimes \cdots \otimes
\rho_n,\;\rho_i\in G^{\wedge}$ for all $i$. We call $\rho=(\rho_1,\ldots
,\rho_n)$ the \emph{label} of $V$. And define the \emph{weight} $\alpha(V)$ of $V$ by 
\begin{equation}
\alpha(V) = (\rho,\alpha_1,\ldots ,\alpha_m),
\end{equation}
where $\alpha_i = \text {eigenvalue of $X_i$ on $V$}$.
Using Lemma~\ref{sc} and Theorem~\ref{yjm3}, it can be easily shown that a $GZ$-subspace is uniquely determined by its weight.

Let $\Y$ be the set of all Young diagrams. The unique Young diagram with zero boxes is empty Young diagram, denoted by $\emptyset$. For $\lambda \in \Y$, let $|\lambda|$ denote the number of boxes in $\La$. Define 
\[
\Y(r,n) := \{\lambda = (\lambda_1, \lambda_2, \ldots, \lambda_r) \mid \La_i \in \Y \text { for all }i =1,2, \ldots,r \text { and } \sum\limits_{i=1}^{r} |\La_i|=n\}, 
\] 
\ie $\Y(r,n)$ is the set of $r$-tuples of Young diagrams such that the total number of boxes is $n$. 

Let $\mu\in \Y$. A standard Young tableau of shape $\mu$ is obtained by filling the boxes in the Young diagram $\mu$ with the distinct numbers $1,2,\ldots ,| \mu |$ such that the numbers in the boxes strictly increase along each row and each column of $\mu$. For $\La \in \Y(r,n)$, a standard $r$-tuple of Young tableaux of shape $\La$ is obtained by filling the $n$-boxes in the $r$-tuple $\La$ with the distinct numbers $1,2,\ldots ,n$ such that the numbers in the boxes strictly increase along each row and each column of all Young diagrams occurring in $\La$. Define
$\Tab$ as the set of all standard $r$-tuple of Young tableaux and set $\Tabop(r,n) := \cup_{\La \in \Y(r,n)}\Tab$. 

For each $i = 1,2, \ldots,r$, define the irreducible representation $\sigma_i$ of $G$:
\begin{align*}
 \sigma_i:~ G & \rightarrow \Cbb^* \\
 \zeta & \mapsto \zeta^{i-1}.
\end{align*}
The irreducible representations of $G$ are $\sigma_1, \sigma_2, \ldots, \sigma_r$.

We draw Young diagrams by following the convention of writing down matrices with $x$-axis running downwards and $y$-axis running to the right. For a box $b$ of a Young diagram, its content $c(b)$ is its $y$-coordinate $-$ its $x$-coordinate. Given $\La = (\La_1, \ldots, \La_r) \in \Y(r,n)$, $T \in \Tab$ and $1 \leq i \leq n$, the number $i$ resides in exactly one box of one of $\La_1, \ldots, \La_r$, say $\La_{j_i}$, let $b_T(i)$ be this box in $\La_{j_i}$ and let $r_T(i) := \sigma_{j_i}$ . 
 
The following result for $G(r,1,n)$ can be easily seen by~\cite[Theorem~6.5]{MS16}.
\begin{theo} \label{gzt}
Let $\La \in \Y(r,n)$. Then the GZ-subspaces of
$V^{\La}$ can be parametrized by $T \in \Tab$ and the GZ-decomposition of $V^{\La}$ can be written as
\begin{equation} \label{gzd} 
V^{\La} = \bigoplus\limits_{T \in \Tab} V_T,
\end{equation}
where each $V_T$ is closed under the action of $G^n$ and, as a $G^n$-module, is isomorphic to the irreducible
$G^n$-module
\[
r_T(1)\otimes r_T(2)\otimes \cdots \otimes r_T(n)
\] 
For $i=1,\ldots ,n$, the eigenvalue of $X_i$ on $V_T$ is given by
$rc(b_T(i))$. 
\end{theo}

Let $R$ denote the element of $\Tab$ defined as follows: for $\La = (\La_1, \ldots, \La_r)$, we start with $\La_1$ by filling the Young diagram $\La_1$ with the
numbers $1,\ldots ,|\La_1|$ in \emph{row major order}, \ie the first row is filled with $1,2,\ldots ,l_1$ in increasing order where $l_1$ is the length of the first row, the second row is filled with $l_1+1,\ldots ,l_1+l_2$ in increasing order where 
$l_2$ is the length of the second row and so on till the last row of
$\La_1$ has been filled. Then we fill the Young diagram $\La_2$ with $|\La_1|+1,\ldots ,|\La_1|+|\La_2|$ in row major order and so on till the last Young diagram $\La_r$.

The irreducible representations of $G(r,1,n)$ are parametrized by the elements of $\Y(r,n)$ and given $\La \in \Y(r,n)$, the $GZ$-basis elements (and hence, $GZ$-subspaces) of $V^{\La}$ are parametrized by $T \in \Tab$.

\subsection{Clifford Theory}\label{ct}
We give an outline of Clifford theory for a finite group $H$ and its normal subgroup $N$ such that $H/N$ is a cyclic group of order $p$ as done in~\cite{BB07, MY98, S89}. 
The pair $H$ and $N$ on which they have applied Clifford theory is the pair $G(r,1,n)$ and $\Grpn$.
The group $\Grpn$ can be considered as the subgroup of $GL_n(\Cbb)$ consisting of generalized permutation matrices such that the $m$-th power of the product of nonzero entries is one. We discuss the complex reflection group $\Grpn$ and its representation theory in detail in Section~\ref{crg} and review Clifford theory for the rest of this section.

Let $H^{\wedge}$ denote the indexing set of irreducible representations of $H$. 
Identifying $H/N$ with the group $C$ consisting of one-dimensional representations of $H$ which contain $N$ in their kernel, we can define an action of $C$ on the set of irreducible representations of $H$ by
\begin{align*}
V^\rho \mapsto \delta \otimes V^\rho
\end{align*}
where $\delta \in C$ and $V^\rho$ is the irreducible representation of $H$ indexed by $\rho \in H^{\wedge}.$ Denote the orbit of $V^{\rho}$ by $[\rho]$ with respect to the action of $C$. The irreducible representations of $H$ which are in the same orbit are called associates of each other. Assume that $V^\rho$ has $b(\rho)$ associates. Then the stabilizer subgroup of $C$ with respect to $V^\rho$, denoted by $C_{\rho}$, has the order $u(\rho) = \frac{p}{b(\rho)}$. Suppose that $\delta_0$ is a generator of $C_{\rho}$. It is easy to see that there exists a $N$-linear map $A : V^\rho \longrightarrow V^\rho$ such that $A(hv)= \delta_0(h)hA(v)$ for all $h \in H$ and $v \in V^\rho$. Then by Schur's lemma, the $H$-linear map $A^{u(\rho)}$ acts by a nonzero scalar. Normalizing the scalar, we obtain that $A^{u(\rho)}$ is the identity map on $V^\rho$. Such an $A$ is called the \emph{associator} of $V^{\rho}$. Also, if $\mu \in [\rho]$, then the stabilizer subgroup $C_\mu = C_\rho.$ The following theorem gives parametrization of irreducible $N$-modules. 


\begin{theo}\label{cd}\leavevmode 
\begin{enumerate}[(\alph*)]
\item\label{theo2.8_a} The eigenspace decomposition of $V^\rho$ with respect to $A$ is given by
 \begin{equation}\label{esdec}
 V^\rho \cong \bigoplus_{l=0}^{u(\rho)-1} E^{(l)},
 \end{equation}
where $E^{(l)}$ is the eigenspace with eigenvalue $e^{\frac{2\pi i l}{u(\rho)}}$. The group $C_\rho$ can be identified with $\{e^{\frac{2\pi i l}{u(\rho)}} \mid l = 0,1, \ldots, u(\rho)-1\}.$
\item\label{theo2.8_b} The eigenspaces $E^{(l)}$, for $0 \leq l \leq u(\rho)-1$, occuring in~\eqref{esdec} are inequivalent irreducible $N$-modules, each of which is of dimension $\dim(V^\rho)/u(\rho)$. 
\item\label{theo2.8_c} For any $0 \leq l \leq u(\rho)-1$, we have 
\[
\Ind_N^H (E^{(l)}) = \bigoplus\limits_{\mu \in [\rho]} V^\mu.
\] 
\item\label{theo2.8_d} Let $\mathcal{O}$ denote the set of all orbits in $H^{\wedge}$.
 The irreducible $N$-modules are parametrized by the pairs $([\rho], \epsilon)$ where $[\rho] \in \mathcal{O}$ and $\epsilon \in C_\rho$.
\end{enumerate}
\end{theo}

\section{Tanabe algebra}\label{ta}
The partition monoid is a poset with the partial order given as: for $d, d' \in A_k$, $d' \leq d$ if $d'$ is coarser than $d$, \ie if $i$ and $j$ are in the same block of $d$, then $i$ and $j$ are in the same block of $d'$.
For $d \in A_k$, define the unique element $x_d \in \Cbb A_k(n)$ satisfying
\begin{equation}\label{xd}
 d = \sum\limits_{d' \leq d} x_{d'}.
\end{equation}
This partial order on $A_k$ can be extended to a total order on $A_k$. 
It can be easily seen that the transition matrix between $\{d \mid d \in A_k \}$ and $\{x_d \mid d \in A_k \}$ is an upper triangular matrix with $1'$s on the diagonal and thus, $\{x_d \mid d \in A_k \}$ is also a basis of the partition algebra $\Cbb A_k(n)$, see also~\cite[p.~879]{HR05}. 

An \emph{internal block} in $ d_1 \circ d_2$, for $d_1, d_2 \in A_k$, is a block that is entirely in the middle while computing $d_1 \circ d_2$. We say that \emph{the bottom row of $d_1$ matches with the top row of $d_2$} if the following condition is satisfied: $i'$ and $j'$ are in the same block in $d_1$ if and only if $i$ and $j$ are in the same block in $d_2$ for $1 \leq i, j \leq k$. For every $s$ in a block $B$ of $d \in A_k$, if we put $i_s = t$ for some $ 1 \leq t \leq n$, then $t$ is said to be a \emph{mark of the block} $B$. The next lemma and the idea of its proof are from the online notes~\cite{Ram2010S} and it also appears in~\cite[Section~4]{BH19}. It gives the structure constants with respect to the basis $\{x_d \mid d \in A_k\}$ of $\Cbb A_k(n)$. 

\begin{lemma}\label{obprod}
 For $d_1, d_2 \in A_k$, the multiplication of $x_{d_1}$ and $x_{d_2}$ in $\Cbb A_k(n)$ is given by 
\begin{align*}
 x_{d_1}x_{d_2} = 
 \begin{cases}
 \sum\limits_{d \in A_k} c_d x_d, & \text {if the bottom row of } d_1 \text { matches with the top row of } d_2, \\
 0, & \text {otherwise,}
 \end{cases}
\end{align*}
\looseness-1
where the sum is taken over all those $d$ in $A_k$ such that $d$ is coarser than $d_1 \circ d_2$ and the coarsening is done by connecting a block of $d_1$ which is contained entirely in the top row of $d_1$ with a block of $d_2$ which is contained entirely in the bottom row of $d_2$ and 
\begin{equation*}
c_d = (n - |d|)_{[d_1 \circ d_2]}, 
\end{equation*}
where $|d|$ is the number of blocks in $d$, $[d_1 \circ d_2]$ is the number of internal blocks in $d_1 \circ d_2$, and for $a \in \Z$, $b \in \Z_{\geq 0},$
\begin{align*}
 (a)_b := 
 \begin{cases}
 a(a-1) \cdots (a-b+1), & \text {if } b > 0,\\
 1, & \text {if } b = 0, 
 \end{cases}
\end{align*}
such that when $a,b \in \Z_{\geq 0}$ and $a \geq b$, we have $(a)_b = \Myperm[a]{b}$, the number of permutations of $a$ objects taken $b$ at a time.
\end{lemma}

\begin{proof}
 Let $n \geq 2k$. Then $\phi_k: \Cbb A_k(n) \cong \End _{S_n}(\Vk)$ (by Schur--Weyl duality for partition algebras, Theorem~\ref{oswd}). Identifying $x_d$ with $\phi_k(x_d)$, we have 
 \begin{multline*}
 (\vi) (x_{d_1}x_{d_2}) \\
= \sum\limits_{i_{1'}, i_{2'}, \ldots, i_{k'},i_{1''}, i_{2''}, \ldots, i_{k''}}(v_{i_{1''}} \otimes v_{i_{2''}} \otimes \cdots \otimes v_{i_{k''}})(x_{d_1})^{i_1,i_2, \ldots, i_k}_{i_{1'}, i_{2'}, \ldots, i_{k'}}(x_{d_2})^{i_{1'}, i_{2'}, \ldots, i_{k'}}_{i_{1''}, i_{2''}, \ldots, i_{k''}}.
 \end{multline*}
 If the bottom row of $d_1$ does not match with the top row of $d_2$, then using~\eqref{xdcoeff} it can be seen that $x_{d_1}x_{d_2} = 0$. 

 If the bottom row of $d_1$ matches with the top row of $d_2$, then again using~\eqref{xdcoeff} we have 
 \begin{multline*}
 \sum\limits_{i_{1'}, i_{2'}, \ldots, i_{k'},i_{1''}, i_{2''}, \ldots, i_{k''}} (v_{i_{1''}} \otimes v_{i_{2''}} \otimes \cdots \otimes v_{i_{k''}}) (x_{d_1})^{i_1,i_2, \ldots, i_k}_{i_{1'}, i_{2'}, \ldots, i_{k'}}(x_{d_2})^{i_{1'}, i_{2'}, \ldots, i_{k'}}_{i_{1''}, i_{2''}, \ldots, i_{k''}} \\ 
 = \sum\limits_{d}\alpha_d x_d,
 \end{multline*}
where $\alpha_d$ is some positive integer and the sum is over all $d$ obtained by coarsening $d_1 \circ d_2$ which is done by connecting a block of $d_1$ contained entirely in the top row of $d_1$ and a block of $d_2$ contained entirely in the bottom row of $d_2$. So, $ \alpha_d = $ number of ways the internal blocks of $d_1 \circ d_2$ can be marked distinctly after putting distinct marks on the blocks of $d$ $= (n - |d|)_{[d_1 \circ d_2]} = c_d.$

Fix $k$ and vary $n$. For a given $n$, fix $d_{1}, d_{2},d\in A_{k}(n)$. Then the coefficient of $x_{d}$ in the product $x_{d_{1}}x_{d_{2}}$ is a polynomial $f_{d}(n)$ in $n$. Then by above arguments, for $n \geq 2k$, we have $f_{d}(n)=(n-|d|)_{[d_{1}\circ d_{2}]}$. The fundamental theorem of algebra implies that $f_{d}(n)=(n-|d|)_{[d_{1}\circ d_{2}]}$ for all $n$.
\end{proof}

Let $B$ be a block of $d \in A_k$. Suppose that $N(B)$ is the number of elements in $B \cap \{1,2,\ldots, k\}$ and $M(B)$ is the number of elements in $B \cap \{1',2',\ldots, k'\}$. Thus, $N(B)$ and $M(B)$ are the number of elements in the block $B$ in top row and bottom row of $d$, respectively. 

Define the following mutually disjoint subsets of $A_k$:
\begin{align*}
 \Pi_k(r) &\coloneqq \{ d = \{B_1, B_2, \ldots, B_s\} \in A_k \mid s \geq 1 \text { and } \\
 &N(B_i) \equiv M(B_i)~(\modrm r) (1 \leq i \leq s)\}, \\
 \Lambda_k(r,p,n) &\coloneqq \{ d = \{B_1, B_2, \ldots, B_n\} \in A_k \mid 
N(B_i) \equiv M(B_i)~(\modrm m),\\ 
&N(B_i) \not\equiv M(B_i)~(\modrm r), (1 \leq i \leq n), \text { and } \\ 
& N(B_i) - M(B_i) \equiv N(B_j)-M(B_j)~(\modrm r), (1 \leq i, j \leq n)\}, \\
 \Theta_k(r,p,n) &\coloneqq \{ d = \{B_1, B_2, \ldots, B_y\} \in A_k \mid y > n, \\ 
&N(B_i) \equiv M(B_i)~(\modrm m),(1 \leq i \leq y), \\
& \text {and for some } j \in \{1,\ldots,y\}, N(B_j) \not\equiv M(B_j)~(\modrm r) \}.
\end{align*}
See Example~\ref{k2} for the above defined sets in the case $k=2$. Define $A_k(r,p,n)$, a subset of $A_k$, by setting
\begin{equation*}
 A_k(r,p,n) := \Pi_k(r) \cup \Lambda_k(r,p,n) \cup \Theta_k(r,p,n).
\end{equation*}


\begin{defi}
Define $\Tn:= \Cbb$-span$\{x_d \mid d \in A_k(r,p,n)\}$, a subspace of the partition algebra $\Cbb A_k(n)$.
\end{defi}
\begin{defi}
Define $\Cbb A_k(r,p,n):= \Cbb$-span$\{d \mid d \in A_k(r,p,n)\}$, a subspace of the partition algebra $\Cbb A_k(n)$.
\end{defi}
The subspace $\Cbb A_k(r,1,n)$ was defined by Orellana~\cite[p.~610]{Ore07} (she used the notation $\T_k(n,r)$). Also, she proved Schur--Weyl duality between $\Cbb A_k(r,1,n)$ and $G(r,1,n)$~\cite[Theorem~5.4]{Ore07}. In the next lemma, we show that the subspaces $\Tn$ and $\Cbb A_k(r,p,n)$ are equal when $p=1$. For $p \neq 1$, the subspace $\Cbb A_k(r,p,n)$ may not be a subalgebra of $\Cbb A_k(n)$, and $\Cbb A_k(r,p,n)$ may neither contain nor be contained in $\Tn$. See Remark~\ref{pnot1}.
\begin{lemma}\label{p1eq}
 The subspaces $\T_k(r,1,n)$ and $\Cbb A_k(r,1,n)$ are equal. Also, $\T_k(r,1,n)$ is a subalgebra of $\Cbb A_k(n)$.
\end{lemma}

\begin{proof}
 The sets $\Lambda_k(r,1,n)$ and $\Theta_k(r,1,n)$ are empty. So, 
 \[
\T_k(r,1,n) = \Cbb\text{-}\spanop\{x_d \mid d \in \Pi_k(r) \} \text { and }\Cbb A_k(r,1,n) = \Cbb\text {-}\spanop\{d \mid d \in \Pi_k(r) \}.
\]
 Also, for $d \in \Pi_k(r)$, the elements $d' \leq d$ belong to $\Pi_k(r)$ because the difference between the number of elements in top row and bottom row in each block remains $0~(\modrm r)$ even after coarsening. Using the definition of $x_d$ in~\eqref{xd}, we get that $d \in \T_k(r,1,n)$ for all $d \in \Pi_k(r)$. Thus, $\Cbb A_k(r,1,n) \subseteq \T_k(r,1,n)$ and we get the equality because of their dimensions being equal.
 
 The subset $\Pi_k(r)$ is a submonoid of $A_k$. Therefore, $\Cbb A_k(r,1,n)$ is a subalgebra of $\Cbb A_k(n)$.
\end{proof}

The explicit formula for the dimension of $\T_k(r,p,n)$, or even in the case of $\T_k(r,1,n)$, is not known. However, when $m \to \infty$ (recall $m = \frac{r}{p}$), then $\Lambda_k(r,p,n)$ and $\Theta_k(r,p,n)$ are empty sets and thus, 
\[
\displaystyle \lim_{m \to \infty} \dim(\Tn) = \lim_{r \to \infty} \dim(\T_k(r,1,n)).
\]
 Also, $\displaystyle \lim_{r \to \infty} \dim(\T_k(r,1,n))$ is described in~\cite[p.~611]{Ore07}.

Let $V = \Cbb^n$ be the $n$-dimensional vector space on which $GL_n(\Cbb)$ acts naturally. The action of $\Grpn$ on $V$ is given by the restriction of the action of $GL_n(\Cbb)$ on $V$. Also, $\Grpn$ acts on the $k\fold$ tensor product $\Vk$ by the diagonal action.

\begin{rema}\label{remalg}
The proof of Theorem~\ref{vsalg} uses the following two known results (both of which are independent of the algebra structure of $\Tn$):
\begin{enumerate}[(\alph*)]
 \item\label{rema3.5_a} Schur--Weyl duality between the partition algebra $\Cbb A_{k}(n)$ and the symmetric group $S_n$ as stated in Theorem~\ref{oswd}\ref{theo5.1_a} (also see~\cite[Theorem~3.6]{HR05} and~\cite{J94}); and
 \item\label{rema3.5_b} The basis of the centralizer algebra of the action of $\Grpn$ on $\Vk$ as given in Lemma~\ref{ca}\ref{lemma5.2_a} (also see~\cite[Lemma~2.1]{T97}).
\end{enumerate}
\end{rema}

\begin{theo}\label{vsalg}
 The vector space $\Tn$ is a subalgebra of $\Cbb A_{k}(n)$.
\end{theo}
\begin{proof}
Let $d_{1}, d_{2}\in A_{k}(r,p,n)$. It is sufficient to assume that the bottom row of $d_{1}$ matches with the top row of $d_{2}$.
The multiplication $x_{d_{1}}x_{d_{2}}$ is given by 
\begin{equation}\label{stc}
x_{d_{1}}x_{d_{2}}=\sum\limits_{d \in A_k} c_{d} x_{d}
\end{equation} 
\begin{enonce*}[remark]{Case 1}
%\emph{Case 1.} 
If $d_{1}, d_{2}\in \Pi_{k}(r)$, then by Lemma~\ref{p1eq}, we have $x_{d_{1}}x_{d_{2}} \in \T_k(r,1,n) \subseteq \Tn$.
\end{enonce*}
\begin{enonce*}[remark]{Case 2}
%\noindent\emph{Case 2.} 
One of $d_1$ or $d_2$ is in $\Theta_k(r,p,n)$. Without loss of generality, assume that $d_{1}\in\Theta_k(r,p,n)$ and $d_{2}\in A_{k}(r,p,n)$. 

\begin{enonce*}[remark]{Claim}
%\emph{Claim}: 
$c_{d} = 0 \text { for } d \notin A_{k}(r,p,n)$ in~\eqref{stc}. Since $d_{1}$ has more than 
$n$ blocks, therefore using Theorem~\ref{oswd}\ref{theo5.1_a} (see Remark~\ref{remalg}), we get, in~\eqref{stc} 
\[
 \sum_{\substack{d \in A_k, \\ |d| \leq n}} c_d \phi_k(x_d) = 0. 
\]
The linear independence of $\{\phi_k(x_d) \mid |d| \leq n \}$ implies that $c_d = 0$ for $d \in A_k, |d| \leq n.$ Thus, $c_d$ can be nonzero only when $|d| > n$. For such $d$, we show that either $d \in \Pi_{k}(r)$ or $d\in \Theta_{k}(r,p,n)$. 
Suppose $d\notin \Pi_{k}(r)$, then there exists $1 \leq j\leq |d| $ such that $N(B_{j})\not\equiv M(B_{j})({\modrm r)}$.
\end{enonce*}
\begin{enonce*}[remark]{Subcase 2.1}
%\emph{Subcase 2.1.} 
Suppose $d=d_{1}\circ d_{2}$.
If a block $B$ in $d_{1}$ is connected with a block $B'$ in $d_{2}$ then $N(B)\equiv M(B)(\modrm m)$, $N(B')\equiv M(B')(\modrm m)$ and $M(B)=N(B')$. Thus, $N(B)\equiv M(B')(\modrm m)$ and $d \in \Theta_k(r,p,n)$. This also includes the cases when either of $B$ and $B'$ are entirely in the top or bottom row of $d_1$ and $d_2$, respectively.
\end{enonce*}
\begin{enonce*}[remark]{Subcase 2.2}
%\emph{Subcase 2.2.} 
Suppose that $d$ is obtained by coarsening of $d_1 \circ d_2$ as in Lemma~\ref{obprod}. Let the coarsening be done by connecting a block $B$ entirely in the top row of $d_{1}$ with a block $B'$ entirely in the bottom row of $d_{2}$. Then 
\[
N(B)\equiv 0(\modrm m) \text { and } 0 \equiv M(B')(\modrm m).
\]
 Thus, $N(B)\equiv M(B')(\modrm m)$ and $d \in \Theta_k(r,p,n)$.
\end{enonce*}
\end{enonce*}

\begin{enonce*}[remark]{Case 3}
%\noindent\emph{Case 3.} 
One of $d_1$ and $d_2$ is in $\Pi_{k}(r)$ and the other is in $ \Lambda_{k}(r,p,n)$. Without loss of generality, assume that $d_{1}\in \Pi_{k}(r)$ and $d_{2}\in \Lambda_{k}(r,p,n)$. If $|d_1| > n$, then we can argue similar to the case (ii) above. So, assume that $|d_{1}| \leq n$. From~\eqref{stc}, we have 
\[
0 \neq \sum\limits_{d \in A_k} c_{d}\phi_{k}(x_{d}) \in \End _{\Grpn}(\Vk).
\]
 Using the basis of $\End _{\Grpn}(\Vk)$ as given in Lemma~\ref{ca}\ref{lemma5.2_a} (see Remark~\ref{remalg}), it follows that, for $d$ such that $|d| \leq n$, $c_{d}$ can be nonzero only when $d\in \Pi_{k}(r) \cup \Lambda_{k}(r,p,n)$.

If there exists $d$ in~\eqref{stc} with more than $n$ blocks such that $c_d \neq 0$, then by the arguments similar to the case (ii), we get either $d\in \Pi_{k}(r)$ or $d\in\Lambda_{k}(r,p,n)$.\qedhere\end{enonce*}
\end{proof}

Define the following mutually disjoint subsets of $A_{k+\frac{1}{2}}$:
\begin{align*}
 \Pi_{k+\frac{1}{2}}(r) &:= \Pi_{k+1}(r) \cap A_{k+\frac{1}{2}}, \\
 \Lambda_{k+\frac{1}{2}}(r,p,n) &:= \Lambda_{k+1}(r,p,n) \cap A_{k+\frac{1}{2}}, \\
 \Theta_{k+\frac{1}{2}}(r,p,n) &:= \Theta_{k+1}(r,p,n) \cap A_{k+\frac{1}{2}}.
\end{align*}
Also, define $A_{k+\frac{1}{2}}(r,p,n)$, a subset of $A_{k+\frac{1}{2}}$, by setting
\begin{equation*}
A_{k+\frac{1}{2}}(r,p,n) := \Pi_{k+\frac{1}{2}}(r) \cup \Lambda_{k+\frac{1}{2}}(r,p,n) \cup \Theta_{k+\frac{1}{2}}(r,p,n).
\end{equation*}

\begin{defi}
Define $\Thn := \Cbb\text {-}\spanop\{x_d \mid d \in A_{k+\frac{1}{2}}(r,p,n)\}$, a subspace of partition algebra $\Cbb A_{k+\frac{1}{2}}(n)$.
\end{defi}

\begin{theo}
 The vector space $\Thn$ is a subalgebra of 
 $\Cbb A_{k+\frac{1}{2}}(n)$.
\end{theo}

\begin{proof}
Note that $\Thn = \T_{k+1}(r,p,n) \cap \Cbb A_{k+\frac{1}{2}}(n)$, hence $\Thn$ is an algebra.
\end{proof}

\begin{defi}[Tanabe algebra]
We call the algebras $\Tn$ and $\Thn$ Tanabe algebras.
\end{defi}

There is an injective algebra homomorphism (see~\cite[p.~879]{HR05})
\begin{align*}
\Cbb A_{k}(n) & \hookrightarrow \Cbb A_{k+\frac{1}{2}}(n) \\
d & \mapsto d',
\end{align*}
where $d\in A_{k}$ and $d'\in A_{k+\frac{1}{2}}$ has same blocks as $d$ with an additional block $\{(k+1),(k+1)'\}$. It is easy to see that the corresponding element $x_{d}$ is mapped to $(x_{d'}+\sum x_{d''})$, where the sum is over all $d''\in A_{k+\frac{1}{2}} \setminus \{d'\}$ obtained by connecting a block in $d'$ with the block $\{(k+1),(k+1)'\}$. Using the description of the above map in terms of the elements
$x_{d}$, we see that the algebra $\Tn$ can be embedded inside the algebra $\Thn$.

\begin{exam}\label{k2} In this example, we describe the elements of the sets $\Pi_k(r), \Lambda_k(r,p,n)$ and $\Theta_k(r,p,n)$ for various specific values of $r,p$ and $n$ when $k = 2$. The partition monoid $A_2 = \{d_1, d_2, \ldots, d_{15}\}$ with the elements in terms of partition diagrams is given in Figure~\ref{A2}. 

\begin{figure}[htb]
\begin{tikzpicture}[scale=0.9,mycirc/.style={circle,fill=black, minimum size=0.1mm, inner sep = 1.5pt}
 ]
 \node (1) at (-1,0.5) {$d_1 =$};
 \node[mycirc,label=above:{$1$}] (n1) at (0,1) {};
 \node[mycirc,label=above:{$2$}] (n2) at (1,1) {};
 \node[mycirc,label=below:{$1'$}] (n3) at (0,0) {};
 \node[mycirc,label=below:{$2'$}] (n4) at (1,0) {};
 \node (a) at (1.4,0.3) {,};
 
 
 \node (2) at (2.5, 0.5) {$d_2 =$};
 \node[mycirc,label=above:{$1$}] (n5) at (3.5,1) {};
 \node[mycirc,label=above:{$2$}] (n6) at (4.5,1) {};
 \node[mycirc,label=below:{$1'$}] (n7) at (3.5,0) {};
 \node[mycirc,label=below:{$2'$}] (n8) at (4.5,0) {};
 \draw (n5) -- (n6);
 \node (b) at (4.9,0.3) {,};
 
 
 \node (3) at (6.0, 0.5) {$d_3 =$};
 \node[mycirc,label=above:{$1$}] (n9) at (7,1) {};
 \node[mycirc,label=above:{$2$}] (n10) at (8,1) {};
 \node[mycirc,label=below:{$1'$}] (n11) at (7,0) {};
 \node[mycirc,label=below:{$2'$}] (n12) at (8,0) {};
 \draw (n9) -- (n11);
 \node (c) at (8.4,0.3) {,};
 
 
 \node (4) at (9.5, 0.5) {$d_4 =$};
 \node[mycirc,label=above:{$1$}] (n13) at (10.5,1) {};
 \node[mycirc,label=above:{$2$}] (n14) at (11.5,1) {};
 \node[mycirc,label=below:{$1'$}] (n15) at (10.5,0) {};
 \node[mycirc,label=below:{$2'$}] (n16) at (11.5,0) {};
 \draw (n13) -- (n16);
 \node (d) at (11.9,0.3) {,};
 
\end{tikzpicture}

\begin{tikzpicture}[scale=0.9,mycirc/.style={circle,fill=black, minimum size=0.1mm, inner sep = 1.5pt}
 ]
 \node (1) at (-1,0.5) {$d_5 =$};
 \node[mycirc,label=above:{$1$}] (n1) at (0,1) {};
 \node[mycirc,label=above:{$2$}] (n2) at (1,1) {};
 \node[mycirc,label=below:{$1'$}] (n3) at (0,0) {};
 \node[mycirc,label=below:{$2'$}] (n4) at (1,0) {};
 \node (a) at (1.4,0.3) {,};
 \draw (n2) -- (n3);
 
 
 \node (2) at (2.5, 0.5) {$d_6 =$};
 \node[mycirc,label=above:{$1$}] (n5) at (3.5,1) {};
 \node[mycirc,label=above:{$2$}] (n6) at (4.5,1) {};
 \node[mycirc,label=below:{$1'$}] (n7) at (3.5,0) {};
 \node[mycirc,label=below:{$2'$}] (n8) at (4.5,0) {};
 \draw (n6) -- (n8);
 \node (b) at (4.9,0.3) {,};
 
 
 \node (3) at (6.0, 0.5) {$d_7 =$};
 \node[mycirc,label=above:{$1$}] (n9) at (7,1) {};
 \node[mycirc,label=above:{$2$}] (n10) at (8,1) {};
 \node[mycirc,label=below:{$1'$}] (n11) at (7,0) {};
 \node[mycirc,label=below:{$2'$}] (n12) at (8,0) {};
 \draw (n11) -- (n12);
 \node (c) at (8.4,0.3) {,};
 
 
 \node (4) at (9.5, 0.5) {$d_8 =$};
 \node[mycirc,label=above:{$1$}] (n13) at (10.5,1) {};
 \node[mycirc,label=above:{$2$}] (n14) at (11.5,1) {};
 \node[mycirc,label=below:{$1'$}] (n15) at (10.5,0) {};
 \node[mycirc,label=below:{$2'$}] (n16) at (11.5,0) {};
 \draw (n13) -- (n14);
 \draw (n15) -- (n16);
 \node (d) at (11.9,0.3) {,};
 
\end{tikzpicture}

\begin{tikzpicture}[scale=0.9,mycirc/.style={circle,fill=black, minimum size=0.1mm, inner sep = 1.5pt}
 ]
 \node (1) at (-1,0.5) {$d_9 =$};
 \node[mycirc,label=above:{$1$}] (n1) at (0,1) {};
 \node[mycirc,label=above:{$2$}] (n2) at (1,1) {};
 \node[mycirc,label=below:{$1'$}] (n3) at (0,0) {};
 \node[mycirc,label=below:{$2'$}] (n4) at (1,0) {};
 \node (a) at (1.4,0.3) {,};
 \draw (n1) -- (n4);
 \draw (n2) -- (n3);
 
 
 \node (2) at (2.5, 0.5) {$d_{10} =$};
 \node[mycirc,label=above:{$1$}] (n5) at (3.5,1) {};
 \node[mycirc,label=above:{$2$}] (n6) at (4.5,1) {};
 \node[mycirc,label=below:{$1'$}] (n7) at (3.5,0) {};
 \node[mycirc,label=below:{$2'$}] (n8) at (4.5,0) {};
 \draw (n6) -- (n8);
 \draw (n5) -- (n7);
 \node (b) at (4.9,0.3) {,};
 
 
 \node (3) at (6.0, 0.5) {$d_{11} =$};
 \node[mycirc,label=above:{$1$}] (n9) at (7,1) {};
 \node[mycirc,label=above:{$2$}] (n10) at (8,1) {};
 \node[mycirc,label=below:{$1'$}] (n11) at (7,0) {};
 \node[mycirc,label=below:{$2'$}] (n12) at (8,0) {};
 \draw (n11) -- (n12);
 \draw (n11) -- (n9);
 \node (c) at (8.4,0.3) {,};
 
 
 \node (4) at (9.5, 0.5) {$d_{12} =$};
 \node[mycirc,label=above:{$1$}] (n13) at (10.5,1) {};
 \node[mycirc,label=above:{$2$}] (n14) at (11.5,1) {};
 \node[mycirc,label=below:{$1'$}] (n15) at (10.5,0) {};
 \node[mycirc,label=below:{$2'$}] (n16) at (11.5,0) {};
 \draw (n13) -- (n14);
 \draw (n15) -- (n13);
 \node (d) at (11.9,0.3) {,};
 
\end{tikzpicture}

\begin{tikzpicture}[scale=0.9,mycirc/.style={circle,fill=black, minimum size=0.1mm, inner sep = 1.5pt}
 ]
 \node (1) at (-1,0.5) {$d_{13} =$};
 \node[mycirc,label=above:{$1$}] (n1) at (0,1) {};
 \node[mycirc,label=above:{$2$}] (n2) at (1,1) {};
 \node[mycirc,label=below:{$1'$}] (n3) at (0,0) {};
 \node[mycirc,label=below:{$2'$}] (n4) at (1,0) {};
 \node (a) at (1.4,0.3) {,};
 \draw (n1) -- (n2);
 \draw (n2) -- (n4);
 
 
 \node (2) at (2.5, 0.5) {$d_{14} =$};
 \node[mycirc,label=above:{$1$}] (n5) at (3.5,1) {};
 \node[mycirc,label=above:{$2$}] (n6) at (4.5,1) {};
 \node[mycirc,label=below:{$1'$}] (n7) at (3.5,0) {};
 \node[mycirc,label=below:{$2'$}] (n8) at (4.5,0) {};
 \draw (n6) -- (n8);
 \draw (n8) -- (n7);
 \node (b) at (4.9,0.3) {,};
 
 
 \node (3) at (6.0, 0.5) {$d_{15} =$};
 \node[mycirc,label=above:{$1$}] (n9) at (7,1) {};
 \node[mycirc,label=above:{$2$}] (n10) at (8,1) {};
 \node[mycirc,label=below:{$1'$}] (n11) at (7,0) {};
 \node[mycirc,label=below:{$2'$}] (n12) at (8,0) {};
 \draw (n11) -- (n12);
 \draw (n10) -- (n9);
 \draw (n9) -- (n11);
 \draw (n10) -- (n12);
 
\end{tikzpicture}\caption{Elements of $A_2$.}\label{A2}
\end{figure}

\begin{enumerate}[(\alph*)]
 \item\label{exam3.10_a} For $r = 2$, we have $\Pi_2(2) = \{d_8, d_9, d_{10}, d_{15}\}.$ 
 \begin{enumerate}[(\roman*)]
 \item\label{exam3.10_a_i} For $p=2, n=2$, the sets 
\[
\Lambda_2(2,2,2) =\{d_{11}, d_{12}, d_{13}, d_{14}\}
\] 
\[
 \text { and }\Theta_2(2,2,2) = \{d_1, d_2, d_3, d_4, d_5, d_6, d_7 \}.
\]
 Thus, $\T_2(2,2,2)$ is the partition algebra $\Cbb A_2(2)$. 
 \item\label{exam3.10_a_ii} For $p=2, n=3$, $\Lambda_2(2,2,3)$ is an empty set and $\Theta_2(2,2,3) =\{d_1\}.$ 
 \item\label{exam3.10_a_iii} For $p=2, n=4$, we have $\Lambda_2(2,2,4) =\{d_1\}$ and $\Theta_2(2,2,4)$ is an empty set.
 \end{enumerate}
 \item\label{exam3.10_b} For $r \neq 1,2$, we have $\Pi_2(r) = \{d_9, d_{10}, d_{15}\}.$ For $r = 3$, $\Lambda_2(r,p,n)$ is nonempty if and only if $(r,p,n) = (3,3,3)$; and $\Lambda_2(3,3,3) = \{d_2, d_7\}.$ 
 For $r = 4$, $\Lambda_2(r,p,n)$ is nonempty if and only if $(r,p,n) = (4,2,2)$ or $(4,4,2)$; and $\Lambda_2(4,2,2) = \Lambda_2(4,4,2) = \{d_8\}.$ For $r > 4$, $\Lambda_2(r,p,n)$ is empty for all values of $p$ and $n$. In general, for $r > 2k$, $\Lambda_k(r,p,n)$ is empty for all values of $p$ and $n$.
 \end{enumerate}
\end{exam}

\begin{rema}\label{pnot1}
In Example~\ref{k2}\ref{exam3.10_a}, we see that $A_2(2,2,3) = \{d_1, d_8, d_9, d_{10}, d_{15}\}.$ The product $d_8d_1 = d_2$ is not an element of $A_2(2,2,3)$ which implies that $\Cbb A_2(2,2,3)$ is not an algebra. 
Using the definition of $x_d$ in (\ref{xd}), we have (a)~$d_1 = \sum\limits_{i=1}^{15}x_{d_i}$ and (b)~the coefficient of $d_2$ in the expression of $x_{d_1}$ as a linear combination of partition diagrams is $-1$.
So, we can conclude that $\Cbb A_2(2,2,3) \nsubseteq \T_2(2,2,3)$ and $\T_2(2,2,3) \nsubseteq \Cbb A_2(2,2,3)$. 
\end{rema}


\begin{rema}\label{222k}
 For $n \geq 2k$, $\Theta_k(r,p,n)$ is an empty set. Moreover, for $n \geq 2k$, the set $\Lambda_k(r,p,n)$ is nonempty if and only if $(r,p,n) = (2,2,2k)$; $\Lambda_k(2,2,2k) = \{d\}$, where $d$ is a partition diagram with $2k$ blocks, \ie each block consists of a single vertex. Using the multiplication rule in Lemma~\ref{obprod}, it is easy to check that the corresponding $x_d$ is a central element of Tanabe algebra $\T_k(2,2,2k)$. This remark, along with Remark~\ref{cp}, is useful in Section~\ref{jme} for the special case $(r,p,n) = (2,2,2k).$
\end{rema}


\section{Complex reflection groups}\label{crg}
For an $n$-dimensional complex vector space $W$, a linear isomorphism of $W$ of finite order is said to be a \emph{reflection} in $W$ if it has exactly $(n-1)$ eigenvalues equal to $1$. A \emph{complex reflection group} $R$ in $W$ is a group generated by reflections in $W$. The space $W$ is called the \emph{reflection representation} of $R$. We say $R$ is irreducible if the $R$-invariant complement of the subspace $W^R$, which is fixed pointwise by $R$, in $W$ is irreducible (see~\cite[Definitions~(1.1)]{C76}). If there exists a direct sum 
$W=W_{1}\oplus W_{2}\oplus\cdots\oplus W_{t}$, where $W_{i}$ is non-trivial proper subspace of $W$ for each $1 \leq i \leq n$, such that $W_1, W_2, \ldots, W_t$ are permuted among themselves under the action of $R$, then we say that $R$ is \emph{imprimitive}. By Shephard--Todd classification, the groups $\Grpn$, for $n > 1$, (except the group $G(2,2,2)$) are the only finite irreducible imprimitive complex reflection groups~\cite[Section~2]{ST54}. The group $G(2,2,2)$ is imprimitive, but it is not irreducible~\cite[Theorem~(2.4)]{C76}. 

Suppose that $G := \Z/r\Z$ is the cyclic group of order $r$ with $\zeta$, a primitive $r$-th root of unity. Define $\Drpn$ to be the subgroup of $GL_n(\Cbb)$ consisting of diagonal matrices as:
\begin{equation*}
 \Drpn := \left\{ \begin{bmatrix}
 \zeta^{i_1} & 0 & \ldots & 0 \\
 0 & \zeta^{i_2} & \ldots & 0 \\
 \vdots & \vdots & \ddots & \vdots \\
 0 & 0 & \ldots & \zeta^{i_n}\\
 \end{bmatrix} 
 \bigmid i_1 + i_2 + \cdots + i_n \equiv 0 (\modrm p)\right\}.
\end{equation*}

Let $S_n$ be the group of $n \times n$ permutation matrices. Define $\Grpn$ to be the subgroup of $GL(n, \Cbb)$ generated by $\Drpn$ and $S_n$. Since $S_n$ normalizes $\Drpn$ and $\Drpn \cap S_n = \{I_n\}$, where $I_n$ is the identity matrix, the group $\Grpn$ is a semidirect product: 
\[
\Grpn = \Drpn \rtimes S_n.
\]
 Thus, as a subgroup of $GL_n(\Cbb)$, the group $\Grpn$ consists of generalized permutation matrices with nonzero entries being $r$-th roots of unity and the $m$-th power of the product of nonzero entries is one. Also, the elements of $\Grpn$ can be written as $(n+1)\text{-tuple}$:
\[
\Grpn = \{(\zeta^{i_1}, \zeta^{i_2}, \ldots, \zeta^{i_n}, \pi) \mid i_1 + i_2 + \cdots + i_n \equiv 0 (\modrm p), \pi \in S_n\}.
\] 

The particular case when $p=1$ is the group $G(r,1,n)$, the wreath product of the group $G$ by the symmetric group $S_n$, of order $r^nn!$. Taking the exact sequence 
\begin{align*}
 1 \longrightarrow \Grpn \longrightarrow G(r,1,n) & \longrightarrow \Z/p\Z \longrightarrow 1 \\
 (\zeta^{i_1}, \zeta^{i_2}, \ldots, \zeta^{i_n}, \pi) & \mapsto \zeta^{i_1 + i_2 + \cdots + i_n},
\end{align*}
we see that $\Grpn$ is a normal subgroup of the group $G(r,1,n)$ of index $p$. So, the order of the group $\Grpn$ is $(r^nn!)/p$. 

Some families of groups which are special cases of $\Grpn$ are:
\begin{enumerate}[(\alph*)]
 \item\label{sect4_a} cyclic group of order $r$, \ie $ \Z/r\Z = G(r,1,1)$,
 \item\label{sect4_b} dihedral group of order $2r$, $D_{2r} = G(r,r,2)$,
 \item\label{sect4_c} symmetric group on $n$ symbols, $S_n = G(1,1,n)$,
 \item\label{sect4_d} Weyl group of type $B_n$ (also called hyperoctahedral group) is $G(2,1,n)$,
 \item\label{sect4_e} Weyl group of type $D_n$ is $G(2,2,n)$. 
\end{enumerate}

Let $G(n)$ be an isomorphic copy of $G ~ (= \Z/r\Z) $ in $G(r,1,n)$ defined as 
\[
G(n) := \{(1, \ldots, 1,g_n, (1))\mid g_n \in G\}.
\]
 Assume that $S_{n-1}$ is the subgroup of $S_n$ consisting of elements fixing $n$. The groups $G(r,1,n-1) \times G(n)$ and $\Grpn$ are subgroups of $G(r,1,n)$. Let $\Lrpn$ be the subgroup of $\Grpn$ defined as:
\begin{align*}
 \Lrpn := 
 & (G(r,1,n-1) \times G(n))\cap \Grpn \\
 = & ((G^{n-1} \rtimes S_{n-1}) \times G(n))\cap (\Drpn \rtimes S_n) \\ 
 = & (G^n \rtimes S_{n-1})\cap (\Drpn \rtimes S_n) \\ 
 = & \Drpn \rtimes S_{n-1}.
\end{align*}
As a subgroup of $GL_n(\Cbb)$, the group $\Lrpn$ consists of those elements in $\Grpn$ such that the $(n,n)$-th entry is nonzero. 
For $p=1$, we have
\begin{align*}
 L(r,1,n) &= G^n \rtimes S_{n-1} \\ 
 & = \big(G^{n-1} \rtimes S_{n-1}\big) \times G(n) \\ 
 &= G(r,1,n-1) \times G(n).
\end{align*}
The order of $L(r,1,n)$ is $r^n(n-1)!$. Taking the exact sequence 
\begin{align*}
 1 \longrightarrow \Lrpn \longrightarrow L(r,1,n) & \longrightarrow \Z/p\Z \longrightarrow 1 \\
 (g_1, g_2, \ldots, g_n, \pi) & \mapsto g_1g_2 \cdots g_n,
\end{align*}
we see that $\Lrpn$ is a normal subgroup of the group $L(r,1,n)$ of index $p$. 
Thus, the order of $\Lrpn$ is $(r^n(n-1)!)/p$. 

Recall from Section~\ref{ov} that $\Y(r,n)$ is the set of $r$-tuples of Young diagram such that the total number of boxes is $n$. Given an $r$-tuple of Young diagrams $(\lambda_1, \lambda_2, \ldots, \lambda_r) \in \Y(r,n-1)$, choose one $i\in \{1,2, \ldots, r\}$, take $\lambda_i$ ($\lambda_i$ may be empty also), color it by $n$ and denote by $\lambda_i^n$. We note that $\lambda_i^n$ denotes the same Young diagram $\lambda_i$, but it has the color $n$. The \emph{$(n,i)$-colored $r$-tuple of Young diagrams}, denoted by $\La^{(n,i)} := (\lambda_1, \lambda_2, \ldots,\lambda_{i-1}, \lambda_i^n, \lambda_{i+1}, \ldots, \lambda_r)$, consists of the $r$-tuple $(\lambda_1, \lambda_2, \ldots, \lambda_r) \in \Y(r,n-1)$ with $i$-th component $\La_i$ colored by $n$. Let $\Y^n(r,n-1)$ denote the set of all $(n,i)$-colored $r$-tuples of Young diagrams with total $n-1$ boxes for $i = 1,2, \ldots, r$.

\begin{lemma}
\looseness-1
The irreducible $L(r,1,n)$-modules are indexed by the elements of $\Y^n(r,n - 1)$. 
\end{lemma}

\begin{proof}
 The irreducible representations of $G$ are $\sigma_1, \sigma_2, \ldots, \sigma_r$ (defined in Section~\ref{ov}). Suppose that $V^{\lambda}$ is the irreducible representation of $G(r,1,n-1)$ corresponding to $\lambda \in \Y(r,n-1)$. Then,
\[
\{V^{\La} \otimes \sigma_i \mid \La \in \Y(r,n-1) , i = 1, \ldots, r \}
\]
is the set of irreducible representations of $L(r,1,n)$ which is indexed by the elements of the set $\Y^n(r,n-1)$.
\end{proof}

\begin{defi}[Inner corner and outer corner]
 Let $\lambda$ be a Young diagram with $|\lambda|$ boxes. A box, which when removed from $\lambda$, leaves a Young diagram with $|\lambda|-1$ boxes is called an inner corner of $\lambda$. A box, which when added to $\lambda$, gives a Young diagram with $|\lambda|+1$ boxes is called an outer corner of $\lambda$.
\end{defi}


We describe the branching rule from $G(r,1,n)$ to $L(r,1,n)$. For $\mu = (\mu_1,\mu_2, \ldots, \mu_r)$ in $\Y(r,n)$ with $\mu_i \neq \emptyset$, let $\mu \downarrow i$ denote the set of elements $\nu^{(n,i)} \in \Y^n(r,n-1)$ such that $\nu$ is obtained from $\mu$ by removing the box at an inner corner of $\mu_i$ and then coloring the $i$-th component of $\nu$ by $n$ to obtain $(n,i)$-colored $r$-tuple $\nu^{(n,i)}$. Assume that $V^{\mu}$ and $V^{\nu^{(n,i)}}$ are the irreducible $G(r,1,n)$-module and $L(r,1,n)$-module corresponding to $\mu \in \Y(r,n)$ and $\nu^{(n,i)} \in \Y^n(r,n-1)$, respectively.

\begin{theo}[Branching rule from $G(r,1,n)$ to $L(r,1,n)$]\label{brgl1}
We have 
\begin{equation}
 \Res_{L(r,1,n)}^{G(r,1,n)}(V^{\mu}) = \bigoplus_{i=1}^{r}\left(\bigoplus_{\nu \in \mu \downarrow i} V^{\nu^{(n,i)}}\right).
\end{equation}
\end{theo}
\begin{rema}
We take equality in place of isomorphism because the restriction rule is multiplicity free which makes the decomposition canonical and we identify $V^{\nu^{(n,i)}}$ with the corresponding $L(r,1,n)$-submodule of $V^{\mu}$.
\end{rema}
\begin{proof}
 Since $\nu_j = \mu_j$ for $j \neq i$ and $|\nu_i| = |\mu_i| - 1$, therefore given a $GZ$-subspace of $V^{\mu}$, there exists a $GZ$-subspace of $V^{\nu^{(n,i)}} = V^{\nu} \otimes \sigma_i$ with the same label. Also, for $1 \leq i \leq n-1$, the action of $X_i \in GZ_{n-1,n} \subseteq GZ_{n,n}$ on $GZ$-subspace of $V^{\nu^{(n,i)}}$ is same as its action on GZ-subspace of $V^{\mu}$. A GZ-subspace is uniquely determined by its weight and a GZ-subspace uniquely determines the parametrization of irreducible representation. Thus, $V^{\nu^{(n,i)}}$ appears in the restriction of $V^{\mu}$ as a $L(r,1,n)$-module with multiplicity one since the restriction from $G(r,1,n)$ to $L(r,1,n)$ is multiplicity free (follows from chain~\eqref{mfic} since $H_{m-1,n} = L(r,1,n)$).
\end{proof}

The next step is the parametrization of the irreducible representations of $\Grpn$ and $\Lrpn$ using Clifford theory. Consider the one-dimensional representation 
\begin{gather*}
\delta_0 : G(r,1,n) \longrightarrow \Cbb^*
\\
 \delta_0(g_1, g_2, \ldots, g_n, \sigma) = g_1g_2\ldots g_n.
\end{gather*} 
As a $G(r,1,n)$-module, $\delta_0$ is parametrized by $(\emptyset, (n), \emptyset, \ldots, \emptyset)$ and $\Grpn \subseteq \Ker (\delta_0^m)$. We use the same notation $\delta_0$ to denote the restriction of $\delta_0$ to $L(r,1,n)$. It will be clear from the context whether we consider $\delta_0$ as a $G(r,1,n)$-module or as a $L(r,1,n)$-module. As a $L(r,1,n)$-module, $\delta_0$ is parametrized by the $(n,2)$-colored $r$-tuple $(\emptyset, (n-1)^n, \emptyset, \ldots, \emptyset)$ and $\Lrpn \subseteq \Ker (\delta_0^m)$. The cyclic group $C$ generated by $\delta_0^m$ is of order $p$. Thus 
\[
 C \cong G(r,1,n)/\Grpn \cong L(r,1,n)/\Lrpn.
\] 

\begin{defi}
Define the shift map on $\Y(r,n)$ as $\sh : \Y(r,n) \longrightarrow \Y(r,n)$ by 
\[
 (\La_1, \La_2, \ldots, \La_r) \mapsto (\La_r, \La_1, \La_2, \ldots, \La_{r-1}).
\]
Using the same notation, the shift map on $\Y^n(r,n-1)$ is defined as 
\begin{align*}
\sh : \Y^n(r,n-1) & \longrightarrow \Y^n(r,n-1) \text { by } \\ 
(\mu_1, \mu_2, \ldots,\mu_i^n,\ldots,\mu_r) & \mapsto (\mu_r, \mu_1, \mu_2, \ldots,\mu_{i-1}, \mu_i^n,\ldots,\mu_{r-1}),
\end{align*}
where the $r$-tuples on the left hand side and the right hand side are $(n,i)$-colored and $(n,i+1)$-colored, respectively. For the definition of shift map on $\Tab$, where $\La \in \Y(r,n)$, see Definition~\ref{defsh}.
\end{defi}

Suppose that $V^{\La}$ and $V^{\mu^{(n,i)}}$ denote the irreducible representations of $G(r,1,n)$ and $L(r,1,n)$ parametrized by the $r$-tuple $\La = (\La_1, \La_2, \ldots, \La_r) \in \Y(r,n)$ and $(n,i)$-colored $r$-tuple $\mu^{(n,i)} = (\mu_1, \mu_2, \ldots,\mu_i^n,\ldots,\mu_r) \in \Y^n(r,n-1)$ for some $i \in \{1,2,\ldots,r\}$, respectively.

The following lemma is proved using Okounkov--Vershik approach. Part~\ref{lemma4.6_a} is~\cite[Theorem~24]{MRW04} and it was proved there using $*$-rim hook tableaux. 

\begin{lemma}\label{shl}
For $\La \in \Y(r,n)$ and $\mu^{(n,i)} \in \Y^n(r,n-1)$, the following are true:
\begin{enumerate}[(\alph*)]
 \item\label{lemma4.6_a} As $G(r,1,n)$-modules, 
\[
 \delta_0 \otimes V^{\La} \; \cong \; V^{\emph{sh}(\La)}.
\]
 \item\label{lemma4.6_b} As $L(r,1,n)$-modules, 
\[
 \delta_0 \otimes V^{\mu^{(n,i)}} \; \cong \; V^{\emph{sh}(\mu^{(n,i)})}.
\]
\end{enumerate}
\end{lemma}

\begin{proof} \ref{lemma4.6_a}
%\begin{enumerate}[(\alph*)]
% \item\label{proof_lemma4.6_a} 
A $GZ$-subspace of an irreducible representation of $G(r,1,n)$ is uniquely determined by its weight. Also, a $GZ$-subspace uniquely determines the $r$-tuple of Young diagrams in $\Y(r,n)$ which parametrize the irreducible representation of which it is a $GZ$-subspace.

For $\La = (\La_1, \La_2, \ldots, \La_r) \in \Y(r,n)$ with $y_i := |\La_i| $, let $R$ be the standard $r$-tuple of Young tableaux written in row major order. The $GZ$-subspace of type $V_R$ is isomorphic to 
\[
\underbrace{\sigma_1 \otimes \cdots \otimes \sigma_1}_{y_1\fold}\otimes \underbrace{\sigma_2 \otimes \cdots \otimes \sigma_2}_{y_2\fold} \otimes \cdots \otimes \underbrace{\sigma_r \otimes \cdots \otimes \sigma_r}_{y_r\fold}
\]
 as a $G^n$-module. For $i =1,2, \ldots, n$ and $GZ$-basis element 
\[
v_R = \underbrace{v_1 \otimes \cdots \otimes v_1}_{y_1\fold}\otimes \underbrace{v_2 \otimes \cdots \otimes v_2}_{y_2\fold} \otimes \cdots \otimes \underbrace{v_r \otimes \cdots \otimes v_r}_{y_r\fold}
\]
 we have 
$ X_i(v_R) = r c(b_R(i))(v_R)$ (for the definition of Jucys--Murphy elements $X_i$, see~\eqref{JM} in Section~\ref{ov}).

The $GZ$-subspace of $\delta_0$ is given by $n\fold \sigma_2 \otimes \cdots \otimes \sigma_2$ with $GZ$-basis element given by $n\fold v_2 \otimes \cdots \otimes v_2.$ Thus, the $GZ$-subspace of $\delta_0 \otimes V^{\La}$ is 
\begin{align*}
 & (\underbrace{\sigma_1 \otimes \cdots \otimes \sigma_1}_{y_1\fold}\otimes \underbrace{\sigma_2 \otimes \cdots \otimes \sigma_2}_{y_2\fold} \otimes \cdots \otimes \underbrace{\sigma_r \otimes \cdots \otimes \sigma_r}_{y_r\fold}) \otimes (\underbrace{\sigma_2 \otimes \cdots \otimes \sigma_2}_{n \fold}) \\
 & \cong \; \underbrace{\sigma_2 \otimes \cdots \otimes \sigma_2}_{y_1\fold}\otimes \underbrace{\sigma_3 \otimes \cdots \otimes \sigma_3}_{y_2\fold } \otimes \cdots \otimes \underbrace{\sigma_1\otimes \cdots \otimes \sigma_1}_{y_r\fold } \\
 & \cong \; V_{\sh(R)},
\end{align*}
isomorphic as $G^n$-module, with basis element $v'$ being
\[
\underbrace{(v_1 \otimes v_2) \cdots \otimes (v_1 \otimes v_2)}_{y_1\fold }\otimes \underbrace{(v_2 \otimes v_2) \cdots \otimes (v_2 \otimes v_2)}_{y_2\fold } \otimes \cdots \otimes \underbrace{(v_r \otimes v_2) \cdots \otimes (v_r \otimes v_2)}_{y_r\fold }). 
\]
Also, for $1 \leq i \neq j \leq n$, we have 
\[
\g_i^l \g_j^{-l} s_{ij}(v_2 \otimes \cdots \otimes v_2) = v_2 \otimes \cdots \otimes v_2 
\] 
So, for $ 1 \leq i \leq n$,
\begin{align*}
X_i (v') & = (v_2 \otimes \cdots \otimes v_2) \otimes X_i(v_1 \otimes \cdots \otimes v_1\otimes v_2 \otimes \cdots \otimes v_2 \otimes \cdots \otimes v_r \otimes \cdots \otimes v_r) \\
& = r c(b_R(i))(v') \\
& = r c(b_{\sh(R)}(i))(v_{\sh(R)}) = X_i(v_{\sh(R)})
\end{align*}
which implies that $v' = v_{\sh(R)}.$

We have shown that $V_{\sh(R)}$ is a $GZ$-subspace of $\delta_0 \otimes V^{\La}$. Thus, $V^{\sh(\La)}$, corresponding to $r$-tuple $\sh(\La)$, is a $G(r,1,n)$-submodule of $\delta_0 \otimes V^{\La}$. The irreducibility of $\delta_0 \otimes V^{\La}$ implies the result. 

\ref{lemma4.6_b}
%\item\label{proof_lemma4.6_b} 
This part can be proved by arguments similar to those in part~\ref{lemma4.6_a}. To be able to do so, we note that a $GZ$-subspace of an irreducible representation of $L(r,1,n)$ is uniquely determined by its weight, \ie its label and the action of Jucys--Murphy elements $X_1, X_2, \ldots, X_{n-1}$ on it. %\qedhere
%\end{enumerate}
\end{proof}

Lemma~\ref{shl} implies Corollaries~\ref{shl2} and~\ref{shl1}.
Part~\ref{coro4.7_a} of Corollary~\ref{shl2} is~\cite[Corollary~25]{MRW04} and is also stated as Theorem~2.1 in~\cite{Ore07}. 

\begin{coro}\label{shl2}
For $\La \in \Y(r,n)$ and $\mu^{(n,i)} \in \Y^n(r,n-1)$, the following are true for $t \in \Z:$
\begin{enumerate}[(\alph*)]
\item\label{coro4.7_a} As a $G(r,1,n)$-module, 
\[
 \delta_0^t \otimes V^{\La} \cong V^{\emph{sh}^t(\La)}.
\]
\item\label{coro4.7_b} As a $L(r,1,n)$-module, 
\[
 \delta_0^t \otimes V^{\mu^{(n,i)}} \cong V^{\emph{sh}^t(\mu^{(n,i)})}.
\]
\end{enumerate}
\end{coro}



\begin{coro}\label{shl1}
For $t \in \Z:$
\begin{enumerate}[(\alph*)]
\item\label{coro4.8_a} As a $G(r,1,n)$-module, $\delta_0^t$ is parametrized by $(\emptyset, \ldots,\emptyset, (n), \emptyset, \ldots, \emptyset) \in \Y(r,n)$, where $(n)$ occurs at $(t+1)(\modrm r)$-th component. 
\item\label{coro4.8_b} As a $L(r,1,n)$-module, $\delta_0^t$ is parametrized by the $(n,(t+1)(\modrm r))$-colored $r$-tuple $(\emptyset, \ldots,\emptyset, (n-1)^n, \emptyset, \ldots, \emptyset) \in \Y^n(r,n-1)$, where $(n-1)^n$ occurs at $(t+1)(\modrm r)$-th component.
\end{enumerate}
\end{coro}

We define a combinatorial object $(m,p)$-necklace as in~\cite[p.~174]{HR98} which will be useful in parametrization of irreducible $\Grpn$-modules and $\Lrpn$-modules.

Let $\lambda = (\lambda_1, \lambda_2, \ldots, \lambda_r) \in \Y(r,n)$. For each $i$ such that $1 \leq i \leq m$, consider the $p$-tuple 
\[
\tla_{(i)} := (\lambda_{i}, \lambda_{m+i}, \lambda_{2m+i}, \ldots, \lambda_{(p-1)m+i}).
\]
 Depict $\tla_{(i)}$ as a $p$-necklace in the following way: the circular necklace, with centre on the $x$-axis, has $p$ nodes and lies in a vertical $xy$-plane with the first node $\lambda_{i}$ placed at the point, where tangent to the necklace in $(y > 0)$-half plane is parallel to the $x$-axis. The placement of nodes is done in clockwise direction with the $j$-th node being $\lambda_{(j-1)m+i}$ and placed at a clockwise angle of $2\pi/(j-1)$ with $y$-axis for $j =2,\ldots,p$. A \emph{$(m,p)$-necklace} of total $n$ boxes obtained from $\La \in \Y(r,n)$, denoted by $\tla$, is a $m$-tuple 
\[
 \tla = (\tla_{(1)}, \tla_{(2)}, \ldots, \tla_{(m)}),
\]
 where $\tla_{(i)}$ is a $p$-necklace for each $1 \leq i \leq m$. For $ 1 \leq j \leq p$ and $1 \leq i \leq m$, let $\tla_{(i,j)}$ denote the $j$-th node in $\tla_{(i)}$, \ie $\tla_{(i,j)} = \La_{(j-1)m+i}$. Thus, we have 
\begin{equation*}
 \sum\limits_{i=1}^{m}\sum\limits_{j=1}^p\tla_{(i,j)} = n.
\end{equation*}
Two $(m,p)$-necklaces, $\tla$ and $\tm$, both of total boxes $n$, are said to be \emph{equivalent} if for some integer $t$, $\tla_{(i,j)} = \tm_{(i,(j+t)(\modrm p))}$ for all $1 \leq j \leq p$ and $1 \leq i \leq m$. Let $\Y(m,p,n)$ denote the set of inequivalent $(m,p)$-necklaces of total $n$ boxes.

Note that for any element $\mu \in [\La] $, the stabilizer subgroup $C_\mu = C_\La$. So, the stabilizer subgroup of a representative of $[\La]$ can be written as $C_\La$ while considering $(m,p)$-necklace $\tla$.

\begin{exam}\label{exn}
The $(3,4)$-necklace of total $~30$ boxes obtained from 
\[
\La = ((2,1), (3,2), (2,1,1), (1), (1,1), (1,1), (1,1,1), (2,1), (1), (2), (2), (2) ) 
 \] 
 is given in Figure~\ref{fig:nklc}.

\begin{figure}[htb]
\centering
\begin{tikzpicture}[scale=0.8,mycirc/.style={circle,fill=black, minimum size=0.1mm, inner sep = 1.5pt}]

 \node[label=above:{\tiny{$\tla_{(1,1)}$}}] (1) at (0,2) {$\yng(2,1)$};
 \node[label=left:{\tiny{$\tla_{(1,2)}$}}] (2) at (2,0) {$\yng(1)$};
 \node[label=below:{\tiny{$\tla_{(1,3)}$}}] (3) at (0,-2) {$\yng(1,1,1)$};
 \node[label=right:{\tiny{$\tla_{(1,4)}$}}] (4) at (-2,0) {$\yng(2)$};
 \draw (1) to[bend left=30] (2);
 \draw (2) to[bend left=30] (3);
 \draw (3) to[bend left=30] (4);
 \draw (4) to[bend left=30] (1);
 \node(a) at (2.3,-0.7) {,};
 \end{tikzpicture}
 \begin{tikzpicture}[scale=0.8,mycirc/.style={circle,fill=black, minimum size=0.1mm, inner sep = 1.5pt}]
 \node[label=above:{\tiny{$\tla_{(2,1)}$}}] (1) at (0,2) {$\yng(3,2)$};
 \node[label=left:{\tiny{$\tla_{(2,2)}$}}] (2) at (2,0) {$\yng(1,1)$};
 \node[label=below:{\tiny{$\tla_{(2,3)}$}}] (3) at (0,-2) {$\yng(2,1)$};
 \node[label=right:{\tiny{$\tla_{(2,4)}$}}] (4) at (-2,0) {$\yng(2)$};
 \draw (1) to[bend left=30] (2);
 \draw (2) to[bend left=30] (3);
 \draw (3) to[bend left=30] (4);
 \draw (4) to[bend left=30] (1);
 \node(a) at (2.5,-0.4) {,};
 \end{tikzpicture}
 \begin{tikzpicture}[scale=0.8,mycirc/.style={circle,fill=black, minimum size=0.1mm, inner sep = 1.5pt}]
 \node[label=above:{\tiny{$\tla_{(3,1)}$}}] (1) at (0,2) {$\yng(2,1,1)$};
 \node[label=left:{\tiny{$\tla_{(3,2)}$}}] (2) at (2,0) {$\yng(1,1)$};
 \node[label=below:{\tiny{$\tla_{(3,3)}$}}] (3) at (0,-2) {$\yng(1)$};
 \node[label=right:{\tiny{$\tla_{(3,4)}$}}] (4) at (-2,0) {$\yng(2)$};
 \draw (1) to[bend left=30] (2);
 \draw (2) to[bend left=30] (3);
 \draw (3) to[bend left=30] (4);
 \draw (4) to[bend left=30] (1);
 \end{tikzpicture}\caption{The $(3,4)$-necklace obtained from $\lambda$ in Example~\ref{exn}.}
 \label{fig:nklc}
 \end{figure}
 \end{exam}

Theorem~\ref{irrgp} and its proof follow the expositions in~\cite{BB07, MY98, S89}.

\begin{theo}\label{irrgp}
 The irreducible $\Grpn$-modules are parametrized by the ordered pairs $(\tla, \delta)$, where $\tla \in \Y(m,p,n)$ and $\delta \in C_{\lambda}$. Given $\La \in \Y(r,n)$, the restriction of the corresponding $G(r,1,n)$-module $V^{\La}$ to $\Grpn$ has multiplicity free decomposition given as:
 \[
 \Res_{G(r,p,n)}^{G(r,1,n)}(V^{\La}) \; = \; \bigoplus_{\delta \in C_{\La}} V^{(\tla,\delta)}.
\]
 Also, for $\mu \in [\La]$, 
\[
\Res_{G(r,p,n)}^{G(r,1,n)}(V^{\La}) \; \cong \; \Res_{G(r,p,n)}^{G(r,1,n)}(V^{\mu}).
\]
\end{theo}

\begin{proof}
The group $C = \langle \delta_0^m \rangle$ acts on the set of irreducible $G(r,1,n)$-modules. For $\La \in \Y(r,n)$, suppose that $[\La]$ denotes the set of elements in $\Y(r,n)$ which parametrize the irreducible $G(r,1,n)$-modules in the orbit of $V^{\La}$. Using Corollary~\ref{shl2}\ref{coro4.7_a}, we have 
\begin{align*}
 [\La] & = \{{\nu} \mid \nu = \sh^{im}(\La) \text { for some } i=0,1,\ldots,p-1 \}.
 \end{align*}
 Let the order of the orbit $[\La]$ be $b(\La)$. Then, the order of the stabilizer subgroup $C_{\La}$ is $u(\La) := \frac{p}{b(\La)}$. Also, $C_{\La}$ is generated by $\delta_0^{b(\La)m}$. The result follows from Theorem~\ref{cd}.
\end{proof}

Given $\tm \in \Y(m,p,n-1)$, the \emph{$(n,i,j)$-colored $(m,p)$-necklace}, denoted by $\tm^{(n,i,j)}$, is obtained by coloring $\tm_{(i,j)}$ by $n$, for $1 \leq i \leq m$ and $ 1 \leq j \leq p$. The colored $(m,p)$-necklaces, $\tm^{(n,i,j)}$ and $\tilde{\nu}^{(n,s,t)}$, are \emph{equivalent} if and only if 
\begin{enumerate}[(\roman*)]
\item\label{sect4_i} $i=s$, and $j = (t+l) (\modrm p)$ for some $l \in \Z$, and
\item\label{sect4_ii} the corresponding $\tm$ and $\tilde\nu$ are equivalent as $(m,p)$-necklaces using the same~$l$ as in~\ref{sect4_i}, \ie $\tm_{(a,b)} = \tilde\nu_{(a,(b+l)(\modrm p))}$ for all $1 \leq a \leq m$ and $1 \leq b \leq p$.
\end{enumerate}
Let $\Y^n(m,p,n-1)$ be the set of inequivalent $(n,i,j)$-colored $(m,p)$-necklaces of total $n-1$ boxes for all $1 \leq i \leq m$, $1 \leq j \leq p$.



\begin{exam}\label{exncol}
Consider the $(3,4)$-necklace $\tla$ with total $30$ boxes in Example~\ref{exn}. We obtain the $(31,2,3)$-colored $(3,4)$-necklace in Figure~\ref{fig:colnklc}, denoted by $\tla^{(31,2,3)}$, by coloring $\tla_{(2,3)}$ by $31$.

\begin{figure}[ht]
\centering
\begin{tikzpicture}[scale=0.8,mycirc/.style={circle,fill=black, minimum size=0.1mm, inner sep = 1.5pt}]
 \node[label=above:{\tiny{$\tla_{(1,1)}$}}] (1) at (0,2) {$\yng(2,1)$};
 \node[label=left:{\tiny{$\tla_{(1,2)}$}}] (2) at (2,0) {$\yng(1)$};
 \node[label=below:{\tiny{$\tla_{(1,3)}$}}] (3) at (0,-2) {$\yng(1,1,1)$};
 \node[label=right:{\tiny{$\tla_{(1,4)}$}}] (4) at (-2,0) {$\yng(2)$};
 \draw (1) to[bend left=30] (2);
 \draw (2) to[bend left=30] (3);
 \draw (3) to[bend left=30] (4);
 \draw (4) to[bend left=30] (1);
 \node(a) at (2.3,-0.7) {,};
 \end{tikzpicture}
 \begin{tikzpicture}[scale=0.8,mycirc/.style={circle,fill=black, minimum size=0.1mm, inner sep = 1.5pt}]
 \node[label=above:{\tiny{$\tla_{(2,1)}$}}] (1) at (0,2) {$\yng(3,2)$};
 \node[label=left:{\tiny{$\tla_{(2,2)}$}}] (2) at (2,0) {$\yng(1,1)$};
 \node[label=below:{\tiny{$\tla_{(2,3)}$}}] (3) at (0,-2) {${\yng(2,1)}^{\;31}$};
 \node[label=right:{\tiny{$\tla_{(2,4)}$}}] (4) at (-2,0) {$\yng(2)$};
 \draw (1) to[bend left=30] (2);
 \draw (2) to[bend left=30] (3);
 \draw (3) to[bend left=30] (4);
 \draw (4) to[bend left=30] (1);
 \node(a) at (2.5,-0.4) {,};
 \end{tikzpicture}
 \begin{tikzpicture}[scale=0.8,mycirc/.style={circle,fill=black, minimum size=0.1mm, inner sep = 1.5pt}]
 \node[label=above:{\tiny{$\tla_{(3,1)}$}}] (1) at (0,2) {$\yng(2,1,1)$};
 \node[label=left:{\tiny{$\tla_{(3,2)}$}}] (2) at (2,0) {$\yng(1,1)$};
 \node[label=below:{\tiny{$\tla_{(3,3)}$}}] (3) at (0,-2) {$\yng(1)$};
 \node[label=right:{\tiny{$\tla_{(3,4)}$}}] (4) at (-2,0) {$\yng(2)$};
 \draw (1) to[bend left=30] (2);
 \draw (2) to[bend left=30] (3);
 \draw (3) to[bend left=30] (4);
 \draw (4) to[bend left=30] (1);
 \end{tikzpicture}\caption{The $(31,2,3)$-colored $(3,4)$-necklace in Example~\ref{exncol}.}
 \label{fig:colnklc}
 \end{figure}
 \end{exam}

By depicting $\mu = (\mu_1, \mu_2, \ldots,\mu_t^n,\ldots,\mu_r) \in \Y^n(r,n-1)$ as a $(m,p)$-necklace, we get $\tm^{(n,i,j)} \in \Y^n(m,p,n-1)$, where $t = (j-1)m+i$ for a unique pair $(i,j)$ such that $1 \leq i \leq m$ and $1 \leq j \leq p$.
 
\begin{theo}\label{irrlp}
The irreducible $\Lrpn$-modules are parametrized by the elements of $\Y^n(m,p,n-1)$.
For $\mu^{(n,t)} \in \Y^n(r,n-1)$, the restriction of the corresponding irreducible $L(r,1,n)$-module $V^{\mu^{(n,t)}}$ to $\Lrpn$ has multiplicity free decomposition given as:
\[
 \Res^{L(r,1,n)}_{\Lrpn}(V^{\mu^{(n,t)}}) ~ = ~ V^{\tm^{(n,i,j)}},
\]
where $t = (j-1)m+i$ for a unique pair $(i,j)$ such that $1 \leq i \leq m$ and $1 \leq j \leq p$. Also, for any $~\nu^{(n,s)} \in [\mu^{(n,t)}]$, 
\[
 \Res^{L(r,1,n)}_{\Lrpn}(V^{\mu^{(n,t)}}) ~ \cong ~ \Res^{L(r,1,n)}_{\Lrpn}(V^{\nu^{(n,s)}}).
\]
\end{theo}

\begin{proof}
The group $C = \langle \delta_0^m \rangle$ acts on the set of irreducible $L(r,1,n)$-modules. For $\mu^{(n,t)} \in \Y^n(r,n-1)$, suppose that $[\mu^{(n,t)}]$ denotes the set of elements in $\Y^n(r,n-1)$ which parametrize the irreducible $L(r,1,n)$-modules in the orbit of $V^{\mu^{(n,t)}}$. Using Corollary~\ref{shl2}\ref{coro4.7_b}, we have 
\begin{align*}
 [\mu^{(n,t)}] & = \{\omega^{(n,y)} \mid \omega^{(n,y)} = \sh^{zm}(\mu^{(n,t)}) \text { for some } z=0,1,\ldots,p-1 \}.
\end{align*}

Since the color is also shifting, therefore, the number of elements in the orbit is $p$ and thus the stabilizer subgroup consists of identity element only. The results follow from Theorem~\ref{cd}.
\end{proof}

\addtocontents{toc}{\SkipTocEntry}
\subsection*{Branching rule from \texorpdfstring{$G(r,p,n)$}{G(r,p,n)} to \texorpdfstring{$L(r,p,n)$}{L(r,p,n)}}

The construction of higher Specht polynomials for $\Grpn$ from the higher Specht polynomials for $G(r,1,n)$ was described in~\cite{MY98} to decompose a module isomorphic to left regular $\Grpn$-module into its irreducible submodules. Applying a similar (but not identical) construction on the canonical $GZ$-bases of irreducible $G(r,1,n)$-modules obtained in Okounkov--Vershik approach in Section~\ref{pre}, we construct the bases of irreducible $\Grpn$-modules in Theorem~\ref{eb}. We use such constructed basis to show in Theorem~\ref{gl} that the irreducible $\Grpn$-modules $V^{(\tla,\delta_1)}$ and $V^{(\tla,\delta_2)}$, for $\tla \in \Y(m,p,n)$ and $\delta_1, \delta_2 \in C_{\La}$, are isomorphic as $\Lrpn$-modules. Theorem~\ref{gl} is useful in the proof of Theorem~\ref{brgpl} for description of branching rule from $\Grpn$ to $\Lrpn$.

\begin{defi}\label{defsh}
\looseness-1
Fix $\La \in \Y(r,n)$. Define the shift map
$\sh : \Tab \longrightarrow \Tab $ 
by 
\[
(T_1, T_2, \ldots, T_r) \mapsto (T_r, T_1, \ldots, T_{r-1}).
\]
\end{defi}

Since $C_{\La}$ is generated by $\delta_0^{mb(\La)}$, the $G(r,1,n)$-modules $V^{\La}$ and $\delta_0^{mb(\La)} \otimes V^{\La}$ are isomorphic. Suppose that $T \in \Tab$ and $\mathbf{1}_{\delta_0^{mb(\La)}}$ is the basis element of one-dimensional $G(r,1,n)$-module ${\delta_0^{mb(\La)}}$. Using Corollary~\ref{shl2}\ref{coro4.7_a}, define the $G(r,1,n)$-linear isomorphism $\mathpzc{E}: V^{\La} \longrightarrow \delta_0^{mb(\La)} \otimes V^{\La}$ by
\[
 v_T \mapsto \mathbf{1}_{\delta_0^{mb(\La)}} \otimes v_{\sh^{-mb(\La)}(T)}.
\]
Also, the map $\mathpzc{F}: \delta_0^{mb(\La)} \otimes V^{\La} \longrightarrow V^{\La}$ given by 
$ \mathbf{1}_{\delta_0^{mb(\La)}} \otimes v_T \mapsto v_T$
is a $G(r,p,n)$-linear isomorphism.

The associator of $V^{\La}$ is given by 
\begin{align}\label{at}
\mathpzc{A}^{\La} = \mathpzc{F}\mathpzc{E} : V^{\La} & \longrightarrow V^{\La} \nonumber \\
v_T & \mapsto v_{\sh^{-mb(\La)}(T)}.
\end{align}
For $h = 1, 2, \ldots, r$, we define 
\begin{equation*}
\Tab_h = \{T = (T^1, T^2, \ldots, T^r) \in \Tab \mid n \in T^{r-\nu}, 0 \leq \nu < h \}.
\end{equation*}
For $T \in \Tab_{mb(\La)}$, we get the following $u(\La)$ distinct standard $r$-Young tableaux: 
\[
 T, \sh^{mb(\La)}(T), \sh^{2mb(\La)}(T), \ldots, \sh^{(u(\La)-1)mb(\La)}(T). 
\] 
An element $\delta \in C_{\La} = \langle \delta_0^{mb(\La)} \rangle $ can be identified with $\zeta^{lmb(\La)}$ for some $0 \leq l \leq u(\La)-1$.
Fixing $\delta \in C_{\La}$, we define, for each $T \in \Tab_{mb(\La)}$, 
\begin{equation*}
 v^{(\delta)}_T := \sum\limits_{t=0}^{u(\La)-1} \zeta^{tlmb(\La)}v_{\sh^{tmb(\La)}(T)}.
\end{equation*}
The linear independence of $\{v_T \mid T \in \Tab\}$ implies that $\{ v^{(\delta)}_T \mid T \in \Tab_{mb(\La)}\}$, for a fixed $\delta \in C_{\La}$, is linearly independent. 

\begin{theo}\label{eb}
 For $\La = (\La_1, \La_2, \ldots, \La_r) \in \Y(r,n)$, consider $\tla \in \Y(m,p,n)$. For each $\delta \in C_{\La}$, define 
\[
V^{(\tla,\delta)} := \Cbb\text {-}\spanop\{v^{(\delta)}_T \mid T \in \Tab_{mb(\La)}\}.
\]
 The following are true:
 \begin{enumerate}[(\alph*)]
 \item\label{theo4.14_a} The eigenspace decomposition of $V^{\La}$ with respect to the associator $\mathpzc{A}^{\La}$ is:
 \begin{equation}
 V^{\La} = \bigoplus_{\delta \in C_{\La}}V^{(\tla,\delta)}.
 \end{equation}
 \item\label{theo4.14_b} The eigenspace $V^{(\tla,\delta)}$, for $\delta \in C_{\La}$, is an irreducible $\Grpn$-module.
 \item\label{theo4.14_c} The set $\{V^{(\tla,\delta)} \mid \tla \in \Y(m,p,n), \delta \in C_{\La} \}$ is the complete set of irreducible $\Grpn$-modules.
 \end{enumerate}
\end{theo}

\begin{proof}
It can be seen from the definition of the associator $\mathpzc{A}^{\La}$ in~\eqref{at} that 
\[
\mathpzc{A}^{\La}(v^{(\delta)}_T) = \zeta^{lmb(\La)}(v^{(\delta)}_T), \text { for }\delta \in C_{\La}.
\]
 This implies that the subspaces $V^{(\tla,\delta)}$, for $\delta \in C_{\La}$, are contained in the distinct eigenspaces of $\mathpzc{A}^{\La}$. Thus, we have 
 \begin{equation}\label{sub}	
 \bigoplus_{\delta \in C_{\La}}V^{(\tla,\delta)} \subset V^{\La}. 
 \end{equation}
 Also for each $\delta \in C_{\La}$, the dimension of $V^{(\tla,\delta)}$ is equal to the number of elements in $\Tab_{mb(\La)}$, denoted by $\#(\Tab)$. This implies that we have 
\[
\dim V^{(\tla,\delta)} = \frac{1}{u(\La)}\#(\Tab) = \frac{1}{u(\La)}\dim V^{\La}.
\]
 Thus, the dimensions of both sides in~\eqref{sub} are equal which implies equality in~\eqref{sub}. This proves part~\ref{theo4.14_a}. 
 The proofs of parts~\ref{theo4.14_b} and~\ref{theo4.14_c} follow from Clifford theory and part~\ref{theo4.14_a} of this theorem.
\end{proof}

\begin{theo}\label{gl}
 For a fixed $\tla \in \Y(m,p,n)$ and $\delta_1, \delta_2 \in C_{\La}$, we have
 \begin{equation*}
 \Res_{L(r,p,n)}^{G(r,p,n)}(V^{(\tla,\delta_1)}) ~ \cong ~ \Res_{L(r,p,n)}^{G(r,p,n)}(V^{(\tla,\delta_2)}).
 \end{equation*}
\end{theo}

\begin{proof}
 The linear map $ \theta: V^{(\tla,\delta_1)} \longrightarrow V^{(\tla,\delta_2)} $ defined by setting
 \[
 v^{(\delta_1)}_T \mapsto v^{(\delta_2)}_T, ~ \text {for } T \in \Tab_{mb(\La)},
\]
is an $\Lrpn$-module isomorphism.
\end{proof}

Given $\tla \in \Y(m,p,n), 1 \leq i \leq m, 1 \leq j \leq p$, let $\tla \downarrow (i,j)$ denote the set of all elements in $\Y^n(m,p,n-1)$ obtained by deleting a box from an inner corner in $\tla_{(i,j)}$ and then coloring the corresponding node by $n$. For a fixed $1 \leq i \leq m$, define $J(i) \subseteq \{1,2, \ldots,p\} $ such that for $s,t \in J(i), s \neq t$, we have $\tla \downarrow (i,s) \cap \tla \downarrow (i,t) = \emptyset$. 
 If $\tla \downarrow (i,s) \cap \tla \downarrow (i,t) \neq \emptyset$, then $\tla \downarrow (i,s) = \tla \downarrow (i,t)$.

\begin{theo}[Branching rule from $\Grpn$ to $\Lrpn$]\label{brgpl} 
 For $\tla \in \Y(m,p,n)$ and $\delta \in C(\La)$, we have
 \begin{equation*}
\Res_{L(r,p,n)}^{G(r,p,n)}(V^{(\tla, \delta)}) \cong \bigoplus_{i=1}^{m}\bigoplus_{j \in J(i)}\left(\bigoplus_{\tm^{(n,i,j)} \in \tla \downarrow (i,j)} V^{\tm^{(n,i,j)}}\right), 
 \end{equation*}
and the branching rule from $\Grpn$ to $\Lrpn$ is multiplicity free.
\end{theo}

\begin{proof}
We use the transitivity of restriction from $G(r,1,n)$ to $\Lrpn$:
\[
G(r,1,n) \supset L(r,1,n) \supset L(r,p,n) \text { and } G(r,1,n) \supset G(r,p,n) \supset L(r,p,n).
\]
 Given $\tla$, we have $ \La \in \Y(r,n)$. Considering $V^{\La}$ as $L(r,1,n)$-module, Theorem~\ref{brgl1} implies that 
\begin{equation}\label{gl1v}
 \Res_{L(r,1,n)}^{G(r,1,n)}(V^{\La}) \cong \bigoplus_{t=1}^{r}\big(\bigoplus_{\mu \in \La \downarrow t} V^{\mu^{(n,t)}}\big).
\end{equation}
Writing $t = (j-1)m+i$, where $1 \leq i \leq m, 1 \leq j \leq p$, we note that $u(\La)$ distinct elements of $\Y^n(r,n-1)$ 
\begin{equation}\label{shu}
\mu^{(n,t)}, \sh^{mb(\La)}(\mu^{(n,t)}), \ldots, \sh^{(u(\La)-1)mb(\La)}(\mu^{(n,t)})
\end{equation}
give rise to the same $\tm^{(n,i,j)} \in \Y^n(m,p,n-1)$ and $j \in J(i)$. Also, from $\mu^{(n,s)}$ (not in~\eqref{shu}) such that $s= (y-1)m+i$, where $ 1 \leq y \leq p$, we get $\tm^{(n,i,y)} \in \Y^n(m,p,n-1)$, not equivalent to $\tm^{(n,i,j)}$, and thus $y \in J(i)$ and $y \neq j$. Restricting $V^{\La}$ as $\Lrpn$-module in~\eqref{gl1v}, Theorem~\ref{irrlp} implies that 
\begin{equation}\label{glla}
 \Res_{\Lrpn}^{G(r,1,n)}(V^{\La}) \cong \left(\bigoplus_{i=1}^{m}\bigoplus_{j \in J(i)}\left(\bigoplus_{\tm^{(n,i,j)} \in \lambda \downarrow (i,j)} V^{\tm^{(n,i,j)}}\right)\right)^{\oplus u(\La)}.
\end{equation}

Considering $V^{\La}$ as $\Grpn$-module, Theorem~\ref{irrgp} implies that 
\begin{equation}\label{a1}
\Res_{\Grpn}^{G(r,1,n)}(V^{\La}) = \bigoplus_{\delta \in C_{\La}} V^{(\tla,\delta)}.
\end{equation}
Further restricting $V^{\La}$ as $\Lrpn$-module in~\eqref{a1}, using Theorem~\ref{gl} and the order of $C_{\La}$ being $u(\La)$, we get 
\begin{equation}\label{ggl}
\Res_{\Lrpn}^{G(r,1,n)}(V^{\La}) \cong \bigg(\Res_{\Lrpn}^{\Grpn}(V^{(\tla,\delta)})\bigg)^{\oplus u(\La)},
\end{equation}
where $\delta \in C_{\La}.$
The result follows from~\eqref{glla} and~\eqref{ggl}. 
\end{proof}

Next, we illustrate Theorem~\ref{brgpl} with an example.

\eject
\begin{exam}
 For the group $G(4,2,8)$, consider its irreducible representation $V^{((\tla_1, \tla_2), \delta)}$ of $G(4,2,8)$ parametrized by $((\tla_1, \tla_2), \delta)$, where $(\tla_1, \tla_2)$ is the $(2,2)$-necklace obtained from $\La = ((2,1),(1),(2,1),(1))$ and $\delta$ is a fixed element of the stabilizer subgroup $C_\La$. Thus, the $(2,2)$-necklace $(\tla_1, \tla_2)$ is as given in Figure~\ref{fig:22nklc}.

 \begin{figure}[htb]
 \centering
 \begin{tikzpicture}[scale=0.8,mycirc/.style={circle,fill=black, minimum size=0.1mm, inner sep = 1.5pt}]
 \node (1) at (0,1) {$\yng(2,1)$};
 \node (2) at (0,-1) {$\yng(2,1)$};
 \draw (1) to[bend left=60] (2);
 \draw (2) to[bend left=60] (1);
 \node(a) at (1.4,0) {,};
 \end{tikzpicture}
 \begin{tikzpicture}[scale=0.8,mycirc/.style={circle,fill=black, minimum size=0.1mm, inner sep = 1.5pt}]
 \node (1) at (2,1.2) {$\yng(1)$};
 \node (2) at (2,-1.2) {$\yng(1)$};
 \draw (1) to[bend left=60] (2);
 \draw (2) to[bend left=60] (1);
 \end{tikzpicture}
 \caption{The $(2,2)$-necklace obtained from $\La = ((2,1),(1),(2,1),(1))$.}
 \label{fig:22nklc}
 \end{figure}
 
 It can be easily seen that the sets $\tla \downarrow (1,1)$ and $\tla \downarrow (1,2)$ are equal and so, $J(1)$ contains only one element which can be chosen to be either $1$ or $2$. We choose $J(1) = \{2\}$; and $\tla \downarrow (1,2)$ contains two elements written as $(8,1,2)$-colored $(2,2)$-necklaces as given in Figure~\ref{fig:ld12}.
 
 \begin{figure}[ht]
 \centering
 \begin{tikzpicture}[scale=0.8,mycirc/.style={circle,fill=black, minimum size=0.1mm, inner sep = 1.5pt}]
 \node (1) at (0,1) {$\yng(2,1)$};
 \node (2) at (0,-1) {${\yng(2)}^{\;8}$};
 \draw (1) to[bend left=60] (2);
 \draw (2) to[bend left=60] (1);
 \node(a) at (1.4,0) {,};
 \end{tikzpicture}
 \begin{tikzpicture}[scale=0.8,mycirc/.style={circle,fill=black, minimum size=0.1mm, inner sep = 1.5pt}]
 \node (1) at (2,1.2) {$\yng(1)$};
 \node (2) at (2,-1.2) {$\yng(1)$};
 \draw (1) to[bend left=60] (2);
 \draw (2) to[bend left=60] (1);
 \node(a) at (4,0) { and };
 \end{tikzpicture}
 \begin{tikzpicture}[scale=0.8,mycirc/.style={circle,fill=black, minimum size=0.1mm, inner sep = 1.5pt}]
 \node (1) at (0,1) {$\yng(2,1)$};
 \node (2) at (0,-1) {${\yng(1,1)}^{\;8}$};
 \draw (1) to[bend left=60] (2);
 \draw (2) to[bend left=60] (1);
 \node(a) at (1.4,-0.2) {,};
 \end{tikzpicture}
 \begin{tikzpicture}[scale=0.8,mycirc/.style={circle,fill=black, minimum size=0.1mm, inner sep = 1.5pt}]
 \node (1) at (2,1.2) {$\yng(1)$};
 \node (2) at (2,-1.2) {$\yng(1)$};
 \draw (1) to[bend left=60] (2);
 \draw (2) to[bend left=60] (1);
 \end{tikzpicture}
 \caption{Elements of $\tla \downarrow (1,2)$.}
 \label{fig:ld12}
 \end{figure}
 
Also, the sets $\tla \downarrow (2,1)$ and $\tla \downarrow (2,2)$ are equal. We choose $J(2) = \{2\}$; and $\tla \downarrow (2,2)$ contains a single element written as a $(8,2,2)$-colored $(2,2)$-necklace as given in Figure~\ref{fig:ld22}.

 \begin{figure}[ht]
 \centering
 \begin{tikzpicture}[scale=0.8,mycirc/.style={circle,fill=black, minimum size=0.1mm, inner sep = 1.5pt}]
 \node (1) at (0,1) {$\yng(2,1)$};
 \node (2) at (0,-1) {${\yng(2,1)}$};
 \draw (1) to[bend left=60] (2);
 \draw (2) to[bend left=60] (1);
 \node(a) at (1.4,0) {,};
 \end{tikzpicture}
 \begin{tikzpicture}[scale=0.8,mycirc/.style={circle,fill=black, minimum size=0.1mm, inner sep = 1.5pt}]
 \node (1) at (2,1.2) {$\yng(1)$};
 \node (2) at (2,-1.2) {${\emptyset}^{8}$};
 \draw (1) to[bend left=60] (2);
 \draw (2) to[bend left=60] (1);
 \end{tikzpicture}
 \caption{Element of $\tla \downarrow (2,2)$.}
 \label{fig:ld22}
 \end{figure} 
 
 Theorem~\ref{brgpl} implies that as a $L(4,2,8)$-module, $V^{((\tla_1, \tla_2), \delta)}$ has a multiplicity free decomposition into a direct sum of irreducible representations	 parametrized by the elements of $\tla \downarrow (1,2) \cup \tla \downarrow (2,2).$

\end{exam}




\section{Schur--Weyl duality for Tanabe algebras}\label{swdu}
Let $V = \Cbb^n$ be the $n$-dimensional vector space with standard basis $\{v_1, v_2, \ldots, v_n\}$. There is a natural action of $GL_n(\Cbb)$ on $V$. For $k \in \Z_{\geq 0}$, consider the $k$-fold tensor product $\Vk = V \otimes V \otimes \cdots \otimes V$ with the basis 
\[
\{v_{i_1} \otimes v_{i_2} \otimes \cdots \otimes v_{i_k} \mid 1 \leq i_1, i_2, \ldots, i_k \leq n \}.
\]
 With respect to this basis, $F \in \End (\Vk)$ can be written as a matrix $\left(F_{i_{1'}, \cdots, i_{k'}}^{i_1, \cdots, i_k}\right)$ such that 
\[
(\vi)F = \sum\limits_{1 \leq i_{1'}, i_{2'}, \ldots, i_{k'} \leq n} F_{i_{1'}, \cdots, i_{k'}}^{i_1, \cdots, i_k} (v_{i_{1'}} \otimes v_{i_{2'}} \otimes \cdots \otimes v_{i_{k'}}).
\]
The action of $GL_n(\Cbb)$ on $\Vk$ is given by 
\[
g(v_{i_1} \otimes v_{i_2} \otimes \cdots \otimes v_{i_k}) = g v_{i_1} \otimes g v_{i_2} \otimes \cdots \otimes g v_{i_k}
\]
 for $ g \in GL_n(\Cbb)$ and $v_{i_1} \otimes v_{i_2} \otimes \cdots \otimes v_{i_k} \in \Vk$. The symmetric group $S_n$ can be identified with the subgroup of permutation matrices of $GL_n(\Cbb)$. Also, we can identify the subgroup $S_{n-1}$ of $S_n$ fixing $n$ with the subgroup of of permutation matrices having $(n,n)$-th entry as $1$ of $GL_n(\Cbb)$. The action of $S_n$ on $\Vk$ is given by the restriction of the action of $GL_n(\Cbb)$ to $S_n$. Define $\Vkh := V^{\otimes k} \otimes v_n$, a subspace of $\Vkk$, which is isomorphic to $\Vk$ as a $S_{n-1}$-module.

Define a map 
\begin{align*}
\phi_k : ~ \Cbb A_k(n) & \rightarrow \End (V^{\otimes k}) \\
 d & \mapsto \phi_k(d) 
\end{align*}
such that for $d \in A_k$ and for $1 \leq i_1, i_2, \ldots, i_k, i_{1'}, i_{2'}, \ldots, i_{k'} \leq n$, 
\[
(\vi)(\phi_k(d)) = \sum\limits_{1\leq i_{1'}, i_{2'}, \ldots, i_{k'} \leq n}(\phi_k(d))^{i_1,i_2, \ldots, i_k}_{i_{1'}, i_{2'}, \ldots, i_{k'}}(\vip)
\]
where
\begin{equation}\label{dcoeff}
 (\phi_k(d))^{i_1,i_2, \ldots, i_k}_{i_{1'}, i_{2'}, \ldots, i_{k'}} =
 \begin{cases}
	 1, & \text { if } i_r = i_s \text { when } r \text { and } s \text { are} \\ 
	 &\qquad\qquad\text { in the same block of } d, \\ 
 0, & \text { otherwise.} 
 \end{cases}
\end{equation}
This defines a right action of $\Cbb A_k(n)$ on $\Vk$ as:
\[
(\vi)d \; := \; (\vi)(\phi_k(d)).
\] 
It follows from~\eqref{xd} and~\eqref{dcoeff} that for all $d \in A_k$, 
\begin{align}\label{xdcoeff}
 (\phi_k(x_d))^{i_1,i_2, \ldots, i_k}_{i_{1'}, i_{2'}, \ldots, i_{k'}} =
 \begin{cases}
 1, & \text { if } i_r = i_s \text { if and only if } r \text { and } s \\ 
 & \qquad\qquad\text { are in the same block of } d, \\ 
 0, & \text { otherwise.} 
 \end{cases}
\end{align}
The action of the partition algebra $\Cbb A_{k+\frac{1}{2}}(n)$ on $V^{\otimes (k+\frac{1}{2})}$ is 
\[
\phi_{k+\frac{1}{2}}: \Cbb A_{k+\frac{1}{2}}(n) \longrightarrow \End (V^{\otimes (k+\frac{1}{2})})
\]
 given by $\phi_{k+\frac{1}{2}} = {\phi_{k+1}}_{|_{\Cbb A_{k+\frac{1}{2}}(n)}}$.

The following theorem is~\cite[Theorem~3.6]{HR05} which shows that $\Cbb A_k(n)$ and $\Cbb A_{k+\frac{1}{2}}(n)$ are in Schur--Weyl duality with $S_n$ and $S_{n-1}$ acting on $\Vk$ and $\Vkh$, respectively.

\begin{theo}\label{oswd}\leavevmode 
\begin{enumerate}[(\alph*)]
\item\label{theo5.1_a} The image of the map $\phi_k : ~ \Cbb A_k(n) \rightarrow \End (\Vk)$ is given by $\End _{S_n}(\Vk)$ and the kernel is given by 
\[
\Cbb \text {-}\spanop \{x_d \mid d \text { has more than } n \text { blocks}\}.
\]
 Thus, the partition algebra $\Cbb A_k(n)$ is isomorphic to $\End _{S_n}(\Vk)$ if and only if $n \geq 2k$. 
 
\item\label{theo5.1_b} The image of the map $\phi_{k+\frac{1}{2}} : ~ \Cbb A_{k+\frac{1}{2}}(n) \rightarrow \End (\Vkh)$ is given by $\End _{S_{n-1}}(\Vkh)$ and the kernel is given by 
\[
\Cbb\text {-}\spanop \{x_d \mid d \text { has more than } n \text { blocks}\}.
\]
 Thus, the partition algebra $\Cbb A_{k+\frac{1}{2}}(n)$ is isomorphic to $\End _{S_{n-1}}(\Vkh)$ if and only if $n \geq 2k+1$. 
\end{enumerate}
\end{theo}

Let $\Pi_k(r,n)$ and $\Pi_{k+\frac{1}{2}}(r,n)$ be subsets of $\Pi_k(r)$ and $\Pi_{k+\frac{1}{2}}(r)$ (defined in Section~\ref{ta}), respectively, consisting of those elements which have at most $n$ blocks. Define
\begin{align*}
 \Pi_k(r,p,n) &\coloneq \Pi_k(r,n) \cup \Lambda_k(r,p,n), \text { and } \\ 
 \Pi_{k+\frac{1}{2}}(r,p,n) &\coloneq \Pi_{k+\frac{1}{2}}(r,n) \cup \Lambda_{k+\frac{1}{2}}(r,p,n), 
\end{align*}
subsets of $A_k(r,p,n)$ and $A_{k+\frac{1}{2}}(r,p,n)$, respectively.

The actions of $\Grpn$ and $\Lrpn$ on $V$ are given by restrictions of the action of $GL_n(\Cbb)$ on $V$. Also, $V$ is the reflection representation of $G(r,p,n)$. We note that $\Cbb$-span$\{v_n\}$ is a $\Lrpn$-invariant subspace of $V$.
The following lemma gives bases of the centralizer algebras of the diagonal actions of $\Grpn$ and $\Lrpn$ on $\Vk$ and $\Vkh$, respectively. Part~\ref{theo5.1_a} is~\cite[Lemma~2.1]{T97} and we follow the proof there to prove part~\ref{lemma5.2_b} here.

\begin{lemma}\label{ca}\leavevmode 
\begin{enumerate}[(\alph*)]
\item\label{lemma5.2_a} $ \{\phi_k(x_d)) \mid d \in \Pi_k(r,p,n)\}$ is a basis of $\End _{\Grpn}(V^{\otimes k})$.
\item\label{lemma5.2_b} $ \{\phi_{k+\frac{1}{2}}(x_d) \mid d \in \Pi_{k+\frac{1}{2}}(r,p,n)\}$ is a basis of $\End _{\Lrpn}(V^{\otimes ({k+\frac{1}{2}})})$. 
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma5.2_b} An element $d \in \Pi_{k+\frac{1}{2}}(r,p,n)$ has at most $n$ blocks. By part~\ref{theo5.1_b} of Theorem~\ref{oswd}, $\phi_{k+\frac{1}{2}}(x_d) \neq 0$. Also, 
\[
 \{\phi_{k+\frac{1}{2}}(x_d) \mid d \in \Pi_{k+\frac{1}{2}}(r,p,n)\} \subset \{\phi_{k+\frac{1}{2}}(x_d) \mid d \in A_{k+\frac{1}{2}}\}
 \] 
is a linearly independent set.

Since $S_{n-1}$ is a subgroup of $\Lrpn$, thus we have 
\[
\End _{\Lrpn}(V^{\otimes ({k+\frac{1}{2}})}) \subset \End _{S_{n-1}}(V^{\otimes ({k+\frac{1}{2}})}).
\]
Choose $0 \neq F \in \End _{\Lrpn}(V^{\otimes ({k+\frac{1}{2}})})$. Then, $F$ can be written as 
 \begin{align*}
 F & = \sum\limits_{d \in A_{k+\frac{1}{2}}}a_d \phi_{k+\frac{1}{2}}(x_d)\\
 & = \sum\limits_{d \in \Pi_{k+\frac{1}{2}}(r,p,n)}a_d \phi_{k+\frac{1}{2}}(x_d)
 + \sum_{\substack{d \in A_{k+\frac{1}{2}}, \\ d \notin \Pi_{k+\frac{1}{2}}(r,p,n)}}a_d \phi_{k+\frac{1}{2}}(x_d).
 \end{align*}
with $\phi_{k+\frac{1}{2}}(x_d) \neq 0$ and $a_d \neq 0$ for some $d \in A_{k+\frac{1}{2}}.$
Fix such a $d \in A_{k+\frac{1}{2}}$ and let $1 \leq i_1, \ldots,i_k, i_{1'}, \ldots, i_{k'} \leq n$ with $i_{k+1}=i_{(k+1)'}=n$ such that 
\[
(\phi_k(x_d))^{i_1,i_2, \ldots, i_k,n}_{i_{1'}, i_{2'}, \ldots, i_{k'},n} = 1.
\]
For $1 \leq u \leq n$, define 
\[
B_u := \{j \in \{1, \ldots, k+1, 1',\ldots, (k+1)'\} \mid i_j = u \}.
\]
 Note that $d = \{B_1, B_2, \ldots, B_n\}$, where some of the blocks $B_1, B_2, \ldots, B_{n-1}$ may be empty and $\{k+1, (k+1)'\} \subseteq B_n$.

For $ 1 \leq i \leq n$, define 
\[
t_i := (1, \ldots, 1, \zeta^p, 1, \ldots, 1),
\]
 where $\zeta^p$ is $i$-th component, and for $ 1 \leq i \neq j \leq n$, define 
\[
h_{ij} := (1, \ldots, 1, \zeta, 1, \ldots, 1, \zeta^{-1}, 1, \ldots, 1),
\]
 where $\zeta$ and $\zeta^{-1}$ are $i$-th and $j$-th components, respectively. The elements $t_i$, for $1 \leq i \leq n$, and the elements $h_{ij}$, for $1 \leq i \neq j \leq n$, together generate $\Drpn$.

For $1 \leq i \leq n$,
\[
t_i^{-1} F t_i (v_{i_1} \otimes v_{i_2} \otimes \cdots \otimes v_{i_k}\otimes v_{i_{k+1}}) = (v_{i_1} \otimes v_{i_2} \otimes \cdots \otimes v_{i_k}\otimes v_{i_{k+1}})F
\]
implies that 
\begin{align}\label{ti}
 & \sum\limits_{i_{1'}, i_{2'}, \ldots, i_{k'}} \zeta^{p(N(B_i)-M(B_i))} F_{i_{1'}, \cdots, i_{k'},i_{(k+1)'}}^{i_1, \cdots, i_k,i_{k+1}} (v_{i_{1'}} \otimes v_{i_{2'}} \otimes \cdots \otimes v_{i_{k'}}\otimes v_{i_{(k+1)'}}) \nonumber \\ 
 & = \sum\limits_{i_{1'}, i_{2'}, \ldots, i_{k'}} F_{i_{1'}, \cdots, i_{k'},i_{(k+1)'}}^{i_1, \cdots, i_k,i_{k+1}} (v_{i_{1'}} \otimes v_{i_{2'}} \otimes \cdots \otimes v_{i_{k'}}\otimes v_{i_{(k+1)'}}).
\end{align}
For $1 \leq i \neq j \leq n$,
\[
h_{ij}^{-1} F h_{ij} (v_{i_1} \otimes v_{i_2} \otimes \cdots \otimes v_{i_k}\otimes v_{i_{k+1}}) = (v_{i_1} \otimes v_{i_2} \otimes \cdots \otimes v_{i_k}\otimes v_{i_{k+1}})F
\]
implies that 
\begin{align}\label{hij}
 \sum\limits_{i_{1'}, i_{2'}, \ldots, i_{k'}} & \Mk\Mk \zeta^{(N(B_i)-M(B_i))-(N(B_j)-M(B_j))} F_{i_{1'}, \cdots, i_{k'},i_{(k+1)'}}^{i_1, \cdots, i_k,i_{k+1}} (v_{i_{1'}} \Mk\otimes\Mk v_{i_{2'}} \Mk\otimes\Mk \cdots \Mk\otimes\Mk v_{i_{k'}}\Mk\otimes\Mk v_{i_{(k+1)'}}) \nonumber \\ 
 & = \sum\limits_{i_{1'}, i_{2'}, \ldots, i_{k'}} F_{i_{1'}, \cdots, i_{k'},i_{(k+1)'}}^{i_1, \cdots, i_k,i_{k+1}} (v_{i_{1'}} \otimes v_{i_{2'}} \otimes \cdots \otimes v_{i_{k'}}\otimes v_{i_{(k+1)'}}).
\end{align} 
From~\eqref{ti} and~\eqref{hij} we have
\begin{align}
 & N(B_i) \equiv M(B_i)(\modrm m), \text { for } 1 \leq i \leq n, \label{nbmbm} \\
 & N(B_i) - M(B_i) \equiv N(B_j)- M(B_j) (\modrm r), \text { for } 1 \leq i \neq j
 \leq n. \label{nbmbr}
\end{align}

The following two cases arise.
\begin{enumerate}[(\roman*)]
 \item\label{proof_lemma5.2_i} If $N(B_1) \equiv M(B_1) (\modrm r)$, then~\eqref{nbmbr} implies that $N(B_i) \equiv M(B_i) (\modrm r)$ for all $1 \leq i \leq n$. So, we have 
\[
d \in \Pi_{k+\frac{1}{2}}(r,n).
\] 
 \item\label{proof_lemma5.2_ii} If $N(B_1) \not\equiv M(B_1) (\modrm r)$, then~\eqref{nbmbr} implies that $N(B_i) \not\equiv M(B_i)(\modrm r)$ for all $1 \leq i \leq n$. Thus, the number of elements, $ N(B_i) + M(B_i)$, in the block $B_i$ is nonzero for all $1 \leq i \leq n$. So, all the $n$ blocks, $B_1, \ldots, B_n$, in $d$ are nonempty. Along with~\eqref{nbmbm}, we get $d \in \Lambda_{k+\frac{1}{2}}(r,p,n)$. 
\end{enumerate}
Combining both the cases we get that $d \in \Pi_{k+\frac{1}{2}}(r,p,n)$.
\end{proof} 

Recall from Section~\ref{ta} that Tanabe algebras $\Tn$ and $\Thn$ are subalgebras of partition algebras $\Cbb A_k(n)$ and $\Cbb A_{k+\frac{1}{2}}(n)$, respectively. The actions of $\Tn$ and $\Thn$ on $\Vk$ and $\Vkh$, respectively, are given by: 
\begin{align*}
 & \psi_k: ~ \Tn \longrightarrow \End (\Vk) \text { with } \psi_k := {\phi_k}_{|_{\Tn}},\\
 \text {and } & \psi_{k+\frac{1}{2}} : ~ \Thn \longrightarrow \End (\Vkh) \text { with } \psi_{k+\frac{1}{2}} := {\phi}_{{k+\frac{1}{2}}{|_{\Thn}}}.
\end{align*}

Using Theorem~\ref{swd} and Corollary~\ref{corswd}, we get that $\Tn$ and $\Thn$ are in Schur--Weyl duality with $\Grpn$ and $\Lrpn$ acting on $\Vk$ and $\Vkh$, respectively.

\goodbreak
\begin{theo}\label{swd}\leavevmode 
 \begin{enumerate}[(\alph*)]
\item\label{theo5.3_a} The image of the map $ \psi_k\Mk:\Mk \Tn \Mk\rightarrow \Mk\End (\mk V^{\otimes k}\mk) $ is given by $ \End _{\Grpn}(\mk V^{\otimes k}\mk )$ and the kernel is given by 
\[
 \Cbb\text {-}\spanop\{ x_d \mid d \in A_k(r,p,n) \text { has more than } n \text { blocks}\}.
\]
 Thus, Tanabe algebra $\Tn$ is isomorphic to $\End _{\Grpn}(V^{\otimes k})$ if and only if $n \geq 2k$.

\item\label{theo5.3_b} The image of the map $ \psi_{k+\frac{1}{2}}: \Thn \rightarrow \End (V^{\otimes ({k+\frac{1}{2}})})$ is given by $ \End _{\Lrpn}(V^{\otimes ({k+\frac{1}{2}})})$ and the kernel is given by 
\[
 \Cbb\text {-}\spanop\{ x_d \mid d \in A_{k+\frac{1}{2}}(r,p,n) \text { has more than } n \text { blocks}\}.
\]
 Thus, Tanabe algebra $\Thn$ is isomorphic to $\End _{\Lrpn}(V^{\otimes ({k+\frac{1}{2}})})$ if and only if $n \geq 2k+1$.
\end{enumerate}
\end{theo}

\goodbreak
\begin{proof} 
%\leavevmode 
%\begin{enumerate}[(\alph*)]
%\item\label{proof_theo5.3_a} 
\ref{theo5.3_a} For $r = 1$, this is Theorem~\ref{oswd}\ref{theo5.1_a} which is Schur--Weyl duality between $\Cbb A_k(n)$ and $S_n$ acting on $\Vk$. Now consider $r \geq 2$.
Using Lemma~\ref{ca}\ref{lemma5.2_a}, we have 
\[
\End _{\Grpn}(V^{\otimes k}) = \Cbb\text {-}\spanop\{\psi_k(x_d) \mid d \in \Pi_k(r,p,n)\} \subset \psi_k(\Tn).\]
 The element $d \in A_k(r,p,n) \setminus \Pi_k(r,p,n) $ has more than $n$ blocks. So, Theorem~\ref{oswd}\ref{theo5.1_a} implies that 
\[
(\vi)\psi_k(x_d) = 0,
\]
for $\vi \in \Vk$.
Thus, we get the image and kernel as stated in the theorem. 
The kernel of $\psi_k$ is zero if and only if $n \geq 2k$.

\ref{theo5.3_b} 
%\item\label{proof_theo5.3_b} 
The proof of this part is along the similar lines as that of part~\ref{theo5.3_a} using Lemma~\ref{ca}\ref{lemma5.2_b} and Theorem~\ref{oswd}\ref{theo5.1_b}. 
%\qedhere
%\end{enumerate}
\end{proof}

\begin{rema}
Putting $p = 1$ in Theorem~\ref{swd}\ref{theo5.3_a} we recover Schur--Weyl duality between $\T_k(r,1,n)$ and $G(r,1,n)$ as given in~\cite[Theorem~5.4]{Ore07}. 
\end{rema}

Using the diagonal actions of $\Grpn$ and $\Lrpn$ on $\Vk$ and $\Vkh$, respectively, we define the following maps 
\[
\vartheta_k : \Cbb[\Grpn] \rightarrow \End (\Vk) \text { and } \vartheta_{k+\frac{1}{2}} : \Cbb[\Lrpn] \rightarrow \End (\Vkh).
\]

\begin{coro}\label{corswd}
Assume that $n \geq 2k$. Then,
\begin{enumerate}[(\alph*)]
 \item\label{coro5.5_a} The image of the map $\vartheta_k : \Cbb[\Grpn] \rightarrow \End (\Vk)$ is given by $\End _{\Tn}(\Vk)$.
 \item\label{coro5.5_b} The image of the map $\vartheta_{k+\frac{1}{2}} : \Cbb[\Lrpn] \rightarrow \End (\Vkh)$ is given by $\End _{\Thn}(\Vkh)$.
\end{enumerate}
\end{coro}

\begin{proof}
%\leavevmode 
%\begin{enumerate}[(\alph*)]
% \item\label{proof_coro5.5_a} 
\ref{coro5.5_a} 
Theorem~\ref{swd}\ref{theo5.3_a} implies that for $n \geq 2k$, the algebra $\Tn$ is isomorphic to $\End _{\Grpn}(\Vk)$ and the action of $\Tn$ on $\Vk$ commutes with the action of $\Grpn$ on $\Vk$. So, $\vartheta_k(\Cbb[\Grpn]) \subseteq \End _{\Tn}(\Vk).$ Using double centralizer theorem~\cite[Theorem~1.3]{bloss02}, we get that $\vartheta_k(\Cbb[\Grpn]) = \End _{\Tn}(\Vk).$ 

\ref{coro5.5_b} 
% \item\label{proof_coro5.5_b} 
Theorem~\ref{swd}\ref{theo5.3_b} and double centralizer theorem imply part \ref{coro5.5_b}. %\qedhere
%\end{enumerate}
\end{proof}


\section{Bratteli diagram of Tanabe algebras}\label{bd}

Let us first study the decomposition of $\Vk$ and $\Vkh$ as $\Grpn$-module and as $\Lrpn$-module, respectively. For the rest of the paper, we assume that $r \geq 2$. 

It can be easily seen using Okounkov--Vershik approach that the $G(r,1,n)$-module $V$ is an irreducible module parametrized by $((n-1),(1), \emptyset, \ldots, \emptyset) \in \Y(r,n)$. Using the theory of Section~\ref{crg}, we see that for $(r,p,n) \neq (2,2,2)$, the $\Grpn$-module $V$ is an irreducible module parametrized by $(\tla, \delta)$, where $\tla \in \Y(m,p,n)$ and $\delta \in C_{\La}$ are as follows:
\begin{enumerate}[(\roman*)]
 \item\label{sect6_i} If $p \neq r$, then $\tla = (\tla_{(1)}, \ldots, \tla_{(m)})$ with 
 \begin{align*}
 \tla_{(1)} & = ((n-1), \emptyset, \ldots, \emptyset), \\
 \tla_{(2)} & = ((1), \emptyset, \ldots, \emptyset), \\
 \tla_{(i)} & = (\emptyset, \ldots, \emptyset), \; \text { for } i = 3, \ldots, m,
 \end{align*}
and $\delta = 1$ since $C_{\La} = \{1\}.$
 \item\label{sect6_ii} If $p =r$ and $(r,p,n) \neq (2,2,2)$, then $\tla = (\tla_{(1)}) = ((n-1),(1), \emptyset, \ldots, \emptyset)$ and $\delta = 1$ since $C_{\La} = \{1\}.$ 
\end{enumerate}
For $(r,p,n) = (2,2,2)$, $V$ is the direct sum of irreducible $G(2,2,2)$-modules parametrized by $(((1),(1)),1)$ and $(((1),(1)),-1)$.

Suppose that $\mathbf{1}_n$ is the trivial representation of $G(r,1,n)$. Then, $ \sigma = \mathbf{1}_{n-1} \otimes \sigma_2$ is a one-dimensional representation of $L(r,1,n)$ and thus, by restriction, a representation of $L(r,p,n)$. The parametrization of $\sigma$ as a $L(r,1,n)$-module is $\mu = ((n-1), \emptyset^{n}, \emptyset, \ldots, \emptyset) \in \Y^n(r,n-1)$. The parametrization of $\sigma$ as a $\Lrpn$-module is $\tm^{(n,i,j)} \in \Y^n(m,p,n-1)$ given as follows: 
\begin{enumerate}[(\roman*)]
 \item\label{sect6bis_i} If $p \neq r$, then $i=2, j =1$ and $\tm^{(n,2,1)} = (\tm_{(1)}, \tm_{(2)}^{(n,1)}, \ldots, \tm_{(m)})$ with 
 \begin{align*}
 & \tm_{(1)} = ((n-1), \emptyset, \ldots, \emptyset), \\
 & \tm_{(2)}^{(n,1)} = (\emptyset^n, \emptyset, \ldots, \emptyset), \\
 & \tm_{(i)} = (\emptyset, \ldots, \emptyset) \text { for } i = 3, \ldots, m.
 \end{align*}
 \item\label{sect6bis_ii} If $p =r$, then $i=1, j =2$ and $\tm^{(n,1,2)} = (\tm_{(1)}^{(n,2)}) =((n-1),\emptyset^n, \emptyset, \ldots, \emptyset)$.
\end{enumerate}
 Using the above parametrizations of $V$ and $\sigma$, by Frobenius reciprocity and Theorem~\ref{brgpl}, we have (for $(r,p,n) = (2,2,2)$ also) 
\[
 V \cong \Ind_{\Lrpn}^{\Grpn}(\sigma).
\]
 
 Let $M$ be a $\Grpn$-module. Then using the tensor identity, we have 
\begin{align*}
\Ind_{L(r,p,n)}^{G(r,p,n)}\bigg(\Res_{L(r,p,n)}^{G(r,p,n)}(M) \otimes \sigma \bigg) & \cong M \otimes \Ind_{L(r,p,n)}^{G(r,p,n)}(\sigma) \\ 
& \cong M \otimes V.
\end{align*}
Thus, taking $M = V^{\otimes(k-1)}$ for $k \geq 1$, we have 
\begin{align}
 \Ind_{L(r,p,n)}^{G(r,p,n)}\bigg(\Res_{L(r,p,n)}^{G(r,p,n)}(V^{\otimes(k-1)}) \otimes \sigma \bigg) & \cong V^{\otimes k} \label{ind}\\
 \text {and }\Res_{L(r,p,n)}^{G(r,p,n)}\bigg(\Ind_{L(r,p,n)}^{G(r,p,n)}\bigg(\Res_{L(r,p,n)}^{G(r,p,n)}(V^{\otimes(k-1)}) \otimes \sigma \bigg)\bigg) & \cong V^{\otimes k} \label{res}
 \end{align}
 as $\Grpn$-module and $\Lrpn$-module, respectively. 
 
It will be clear from the context whether we consider $\sigma$ as a $L(r,1,n)$-module or as a $\Lrpn$-module.
Given $\La^{(n,t)} \in \Y^n(r,n-1)$, assume that $V^{\La^{(n,t)}}$ is the corresponding irreducible $L(r,1,n)$-module.
\begin{lemma}\label{tp1} 
For $\La^{(n,t)} \in \Y^n(r,n-1)$
 \begin{equation}
 V^{\La^{(n,t)}} \otimes \sigma = V^{\La^{(n,z)}}
 \end{equation}
where $\La^{(n,z)} \in \Y^n(r,n-1)$ and $z = (t+1)(\modrm r).$
\end{lemma}
\begin{proof}
 Noting that the $GZ$-subspace of $\sigma$ is given by $\underbrace{\sigma_1 \otimes \cdots \otimes \sigma_1}_{(n-1) \fold } \otimes \sigma_2$ with $GZ$-basis element given by $\underbrace{v_1 \otimes \cdots \otimes v_1}_{(n-1)\fold } \otimes v_2$, the proof is similar to that of Theorem~\ref{shl}. 
\end{proof}

Given an $(n,i,j)$-colored $(m,p)$-necklace $\tla^{(n,i,j)} \in \Y^n(m,p,n-1)$, suppose that $V^{\tla^{(n,i,j)}}$ is the corresponding irreducible $\Lrpn$-module.

\begin{lemma}\label{tp} 
For $\tla^{(n,i,j)} \in \Y^n(m,p,n-1)$
 \begin{equation}
 V^{\tla^{(n,i,j)}} \otimes \sigma = V^{\tla^{(n,x,y)}}
 \end{equation}
where $\tla^{(n,x,y)} \in \Y^n(m,p,n-1)$ is obtained from $\tla^{(n,i,j)}$ by the following rule: 
\begin{enumerate}[(\alph*)]
 \item\label{lemma6.2_i} If $i < m$, then $x=i+1$ and $y=j$;
 \item\label{lemma6.2_ii} If $i = m$, then $x=1$ and $y=(j+1)(\modrm p)$.
\end{enumerate}
\end{lemma}

\begin{proof}
 The proof follows by using Lemma~\ref{tp1} and Theorem~\ref{irrlp}.
\end{proof}

Define the sets $\vo_k(r,p,n)$ and $\vo_{k+\frac{1}{2}}(r,p,n)$ as follows. 
Let 
\[
{\vo}_0(r,p,n)= \{(\tla,1)\}
\]
 where $\tla = (((n),\emptyset, \ldots, \emptyset), (\emptyset, \ldots, \emptyset), \ldots, (\emptyset, \ldots, \emptyset)) \in \Y(m,p,n)$.
For $k \in \Z_{>0}$ the sets $\vo_k(r,p,n) \subseteq \Y(m,p,n) \times C $ and $\vo_{k+\frac{1}{2}}(r,p,n) \subseteq \Y^n(m,p,n-1)$ are obtained by the following recursive rule. 

\emph{From $\vo_k(r,p,n)$ to $\vo_{k+\frac{1}{2}}(r,p,n)$}:

For $(\tla,\delta) \in \vo_k(r,p,n)$, let $\tla_{(i,j)-} \in \Y(m,p,n-1)$ be the set of $(m,p)$-necklaces obtained by deleting an inner corner from $\tla_{(i,j)}$. 
For $\tm \in \tla_{(i,j)-}$, color $\tm$ by $(n,i,j)$ to obtain $\tm^{(n,i,j)} \in \vo_{k+\frac{1}{2}}(r,p,n)$. 

\emph{From $\vo_{k+\frac{1}{2}}(r,p,n)$ to $\vo_{k+1}(r,p,n)$}:

For $\tm^{(n,i,j)} \in \vo_{k+\frac{1}{2}}(r,p,n)$, remove the color $(n,i,j)$ to get $\tm \in \Y(m,p,n-1)$ and then add a box to an outer corner, either in the $j$-th node of $(i+1)$-th component of $\tm$ if $1 \leq i \leq m-1$ or in the $(j+1)(\modrm p)$-th node of the first component of $\tm$ if $i = m$, to obtain $\tilde{\nu} \in \Y(m,p,n)$. Let $C_{\nu}$ be the corresponding stabilizer subgroup. For $\delta \in C_\nu \subseteq C$, the element $(\tilde{\nu}, \delta) \in \vo_{k+1}(r,p,n)$.

\begin{theo}\label{decvk}
The indexing sets of the irreducible $\Grpn$-modules occuring in $\Vk$ and of the irreducible $\Lrpn$-modules occuring in $\Vkh$ are $\vo_k(r,p,n)$ and $\vo_{k+\frac{1}{2}}(r,p,n)$, respectively.
\end{theo}
\begin{proof}
 The proof follows from~\eqref{ind},~\eqref{res}, Lemma
~\ref{tp}, branching rule from $\Grpn$ to $\Lrpn$ in Theorem~\ref{brgpl}, Frobenius reciprocity and the observation that the spaces $\Vkh$ and $\Vk$ are isomorphic as $\Lrpn$-modules.
\end{proof}

\begin{theo}\label{irrend}
 The indexing sets of the irreducible $\End _{\Grpn}(\Vk)$-modules and of the irreducible $\End _{\Lrpn}(\Vkh)$-modules are $\vo_k(r,p,n)$ and $\vo_{k+\frac{1}{2}}(r,p,n)$, respectively.
\end{theo}

\begin{proof}
 The proof is a consequence of the centralizer theorem (\cite[Theorem~5.4]{HR05}) and Theorem~\ref{decvk}. 
\end{proof}

\begin{theo}\label{dvkp}
 Let $n$ and $k$ be nonnegative integers.
 \begin{enumerate}[(\alph*)]
 \item\label{theo6.5_a} For $n \geq 2k$, as $(\Cbb[G(r,p,n)], \Tn)$-bimodule,
 \[
 \Vk \cong \bigoplus\limits_{(\tla,\delta) \in \vo_k(r,p,n)} \bigg(V^{(\tla, \delta)} \otimes \T_k^{(\tla, \delta)}\bigg),
\]
 where $V^{(\tla, \delta)}$ is the irreducible $G(r,p,n)$-module and $\T_k^{(\tla,\delta)}$ is the irreducible $\Tn$-module parametrized by $(\tla,\delta) \in \vo_k(r,p,n)$. Also
 \begin{align*}
 \dim(\T_k^{(\tla, \delta)}) = & \text { the number of paths from }(((n),\emptyset, \ldots, \emptyset),1) \in \vo_0(r,p,n) \\ 
 &\text { to } (\tla, \delta) \in \vo_k(r,p,n) \text { in the Bratteli diagram }\widehat{\T}(r,p,n).
 \end{align*}
 
 \item\label{theo6.5_b} For $n \geq 2k+1$, as $(\Cbb[L(r,p,n)], \Thn))$-bimodule,
 \[
 \Vkh \cong \bigoplus\limits_{\tm^{(n,i,j)} \in \vo_{k+\frac{1}{2}}(r,p,n)} \bigg(V^{\tm^{(n,i,j)}} \otimes \T_{k+\frac{1}{2}}^{\tm^{(n,i,j)}}\bigg),
\]
 where $V^{\tm^{(n,i,j)}}$ is the irreducible $L(r,p,n)$-module and $\T_{k+\frac{1}{2}}^{\tm^{(n,i,j)}}$ is the irreducible 
 $\Thn$-module parametrized by $\tm^{(n,i,j)} \in \vo_{k+\frac{1}{2}}(r,p,n)$ and 
 \begin{align*}
 \dim(\T_k^{\tm^{(n,i,j)}}) = & \text { the number of paths from }(((n),\emptyset, \ldots, \emptyset),1) \in \vo_0(r,p,n) \\ 
 &\text { to } \tm^{(n,i,j)} \in \vo_{k+\frac{1}{2}}(r,p,n) \text { in the Bratteli diagram }\widehat{\T}(r,p,n).
 \end{align*}
 \end{enumerate}

\end{theo}
 
\begin{proof}
 The proofs of~\ref{theo6.5_a} and~\ref{theo6.5_b} follow from Theorem~\ref{swd}\ref{theo5.3_a} and~\ref{theo5.3_b}, respectively along with the centralizer theorem, Theorem~\ref{decvk} and Theorem~\ref{irrend}.
 \end{proof}
 
For $(\tla, \delta) \in \vo_k(r,p,n)$, define the set $A_{k-\frac{1}{2}}^{(\tla, \delta)}$ as consisting of the elements $\tm^{(n,i,j)} \in \vo_{k-\frac{1}{2}}(r,p,n)$ for some $ 1 \leq i \leq m$ and $1 \leq j \leq p$ such that $(\tla, \delta)$ is obtained from $\tm^{(n,i,j)}$ while constructing $\vo_k(r,p,n)$ from $\vo_{k-\frac{1}{2}}(r,p,n)$.
For $\tm^{(n,i,j)} \in \vo_{k+\frac{1}{2}}(r,p,n)$, define the set $A_{k}^{\tm^{(n,i,j)}}$ as consisting of the elements $(\tla, \delta) \in \vo_{k}(r,p,n)$ such that $\tm^{(n,i,j)}$ is obtained from $(\tla, \delta)$ while constructing $\vo_{k+\frac{1}{2}}(r,p,n)$ from $\vo_k(r,p,n)$. 
 
%\goodbreak
 \begin{coro}\label{brtkh} \leavevmode 
 \begin{enumerate}[(\alph*)]
 \item\label{coro6.6_a} For $n\geq 2k$ and for $(\tla, \delta) \in \vo_k(r,p,n)$, we have 
\[
\Res_{\Tmn}^{\Tn}(\T^{(\tla,\delta)}_{k}) = \bigoplus_{\tm^{(n,i,j)} \in A_{k-\frac{1}{2}}^{(\tla, \delta)}} \T_{k-\frac{1}{2}}^{\tm^{(n,i,j)}}.
\]
 \item\label{coro6.6_b}  For $n\geq 2k+1$ and for $\tm^{(n,i,j)} \in \vo_{k+\frac{1}{2}}(r,p,n)$, we have 
\[
\Res_{\Tn}^{\Thn}(\T^{\tm^{(n,i,j)}}_{k+\frac{1}{2}}) = \bigoplus_{(\tla, \delta) \in A_k^{\tm^{(n,i,j)}}} \T_k^{(\tla, \delta)}.
\]
 \end{enumerate}
 \end{coro}
 
 \begin{proof}
 %\leavevmode 
 %\begin{enumerate}[(\alph*)]
 %\item\label{proof_coro6.6_a} 
 \ref{coro6.6_a} 
Using Theorem~\ref{dvkp}\ref{theo6.5_a} and~\eqref{ind},
 \begin{align*}
 \T^{(\tla,\delta)}_{k}&\cong \Hom_{\Grpn}(V^{\otimes k},V^{(\tla,\delta)})\\
 &\cong \Hom_{\Grpn}({\Ind ^{\Grpn}_{\Lrpn}}((\Res ^{\Grpn}_{\Lrpn}V^{\otimes (k-1)})\otimes \sigma),V^{(\tla,\delta)}).
 \end{align*}
 From the above isomorphism and Frobenius reciprocity, we get
 \begin{multline}\label{restkh}
  \Res _{\Tmn}^{\Tn} \T^{(\tla,\delta)}_{k}  \\
\begin{aligned}
 &\cong \Res _{\Tmn}^{\Tn}\Hom _{\Lrpn}((\Res ^{\Grpn}_{\Lrpn}V^{\otimes (k-1)})\otimes \sigma,
 {\Res ^{\Grpn}_{\Lrpn}}V^{(\tla,\delta)}) \\
 &\cong \Hom _{\Lrpn}(V^{\otimes (k-\frac{1}{2})}, \sigma' \otimes {\Res ^{\Grpn}_{\Lrpn}}V^{(\tla,\delta)}),
\end{aligned}
 \end{multline}
 where $\sigma'$ is the contragredient representation of $\sigma$. As $L(r,1,n)$-representation, $\sigma'= \sigma^{-1} = \underbrace{\sigma \otimes \cdots \otimes \sigma}_{(r-1)\fold }$ is parametrized by $((n-1),\emptyset,\ldots,\emptyset,\emptyset^{n})\in \Y^{n}(r,n)$. First using Theorem~\ref{brgpl} and then by the repeated application of Lemma~\ref{tp}, we compute $\sigma' \otimes {\Res ^{\Grpn}_{\Lrpn}}V^{(\tla,\delta)}$. Then, from~\eqref{restkh} and Theorem~\ref{dvkp}\ref{theo6.5_b}, we have the restriction rule.

 \ref{coro6.6_b} 
%\item\label{proof_coro6.6_b} 
The proof is along the similar lines as that of part~\ref{coro6.6_a} using Theorem~\ref{dvkp}\ref{theo6.5_b},~\eqref{res}, Theorem~\ref{brgpl}, Frobenius reciprocity and Theorem~\ref{dvkp}\ref{theo6.5_a}.
%\qedhere
% \end{enumerate}
\end{proof}

Orellana~\cite[p.~614]{Ore07} describes the rule for recursively constructing Bratteli diagram for the tower of algebras 
\[
 \T_0(r,1,n) \subseteq \T_1(r,1,n) \subseteq \T_2(r,1,n) \subseteq \cdots.
\]
We consider the tower of Tanabe algebras 
\begin{equation}\label{bt} 
\T_0(r,p,n) \subseteq \T_{\frac{1}{2}}(r,p,n) \subseteq \T_1(r,p,n) \subseteq \T_{\frac{3}{2}}(r,p,n) \subseteq \cdots \subseteq \T_{\lfloor \frac{n}{2} \rfloor}(r,p,n)
\end{equation}
and using Theorems~\ref{decvk},~\ref{irrend} and Corollary~\ref{brtkh}, construct the corresponding Bratteli diagram $\widehat{\T}(r,p,n)$ recursively by the following rule: For $l \in \frac{1}{2}\Z_{\geq 0}$, the vertices at $l$-th level of Bratteli diagram are elements of the set $\vo_l(r,p,n)$. A vertex $\mathpzc{v}_l$ at $l$-th level is connected by an edge with a vertex $\mathpzc{v}_{l+\frac{1}{2}}$ at $(l+\frac{1}{2})$-th level if and only if $\mathpzc{v}_{l+\frac{1}{2}}$ is obtained from $\mathpzc{v}_l$ while constructing $\vo_{l+\frac{1}{2}}(r,p,n)$ from $\vo_l(r,p,n)$. The Bratteli diagram of Tanabe algebras is a simple graph.

\begin{rema}\label{cp}
For $t \in \Z_{\geq 0}$, $t \leq \lfloor \frac{n}{2} \rfloor $ and $ (\tla, \delta) \in \vo_t(r,p,n)$, the stabilizer subgroup $C_{\La}$ is non-trivial if and only if $(r,p,n) =(2,2,2k)$ and $t = k$; in this case $C_{\La} = \{1,-1\}$. Thus, for $n \geq 2k$, there is a one-to-one correspondence between the irreducible representations of the same degree occuring at $t$-th level in Bratteli diagrams for $\T_k(r,1,n)$ and $\Tn$ if and only if $(r,p,n, t) \neq (2,2,2k,k)$; the correspondence in terms of parametrization is $\La \mapsto (\tla, 1)$.
\end{rema}

We draw Bratteli diagram, up to level $k=2$, of Tanabe algebras for $(r,p,n) = (2,2,4)$ in Figure~\ref{fig:bd1}. Note that at level $k = 2$, a node parametrized by $(\tla, -1)$ also appears when $\La = ((2),(2))$ because $C_\La$ is nontrivial. 

\begin{figure}[htb]
\centering
\includegraphics[%height=7cm, 
width=\textwidth]{Figures/bd1figure.pdf}
\caption{Bratteli diagram, up to level $k=2$, of Tanabe algebras for $(r,p,n) = (2,2,4)$.}
\label{fig:bd1}
\end{figure}

In Figure~\ref{fig:bd2}, we draw Bratteli diagram, up to level $k = \frac{5}{2}$, of Tanabe algebras for $(r,p,n) = (6,2,6)$. Note that the representation $\T_\frac{5}{2}^{\pv}$ corresponding to $\pv = \left(((4), \, \emptyset), \, ((1)^{6}, \,\emptyset),\,( \emptyset, \, \emptyset))\right)$ is of dimension two.
To accommodate the figure, the last two vertices in the level $k=2$ in Figure~\ref{fig:bd2} have been denoted by $\pv_1$ and $\pv_2$, where $\pv_1 = (((4)^{6}, \, \emptyset), \, (\emptyset, \,\emptyset),\,((1), \, \emptyset)))$ and $\pv_2 = (((5), \, \emptyset), \, (\emptyset, \,\emptyset),\,( \emptyset^{6}, \, \emptyset))).$



 


\begin{figure}[htb]
\centering
\includegraphics[%height=7cm, 
width=\textwidth]{Figures/bd2figure_H.pdf}
\caption{Bratteli diagram, up to level $k = \frac{5}{2}$, of Tanabe algebras for $(r,p,n) = (6,2,6)$.}
\label{fig:bd2}
\end{figure}






 
\section{Jucys--Murphy elements for Tanabe algebras}\label{jme}
 Recall from Section~\ref{pre} that the Jucys--Murphy elements for $G(r,1,n)$ are: 
\begin{align*}
& X_1 = 0, \\
\text { and }& X_j = \sum_{i=1}^{j-1}\sum\limits_{l=0}^{r-1}\zeta_i^{l}\zeta_j^{-l}s_{ij},\;\;2\leq
j\leq n.
\end{align*}
For $T \in \Tab$, we have 
\[
\sum\limits_{b \in \lambda}c(b) := \sum\limits_{j=1}^n c(b_{T}(j))
\]
and it is easily seen that $\sum\limits_{b \in \lambda}c(b)$ is independent of the choice of $T \in \Tab$.

\begin{lemma}\label{ka} \leavevmode 
\begin{enumerate}[(\alph*)]
 \item\label{lemma7.1_a} For $r,n \in \Z_{\geq 0}$, 
\[
 \ka_{r,n} := \frac{1}{r}\sum\limits_{l=0}^{r-1}\sum\limits_{1 \leq i < j \leq n}\zeta_i^{l}\zeta_j^{-l}s_{ij}\]
 is a central element of $\Cbb[G(r,p,n)]$ and $\ka_{r,n} = \sum\limits_{b \in \lambda}c(b)$ as operators on $V^{(\tla,\delta)},$ the irreducible $G(r,p,n)$-module parametrized by $(\tla, \delta) \in \Y(m,p,n) \times C_\La$.
 \item\label{lemma7.1_b} For $r,n \in \Z_{\geq 0}$, 
\[
 \ka_{r,n-1} := \frac{1}{r}\sum\limits_{l=0}^{r-1}\sum\limits_{1 \leq i < j \leq n-1}\zeta_i^{l}\zeta_j^{-l}s_{ij}
\]
 is a central element of $\Cbb[L(r,p,n)]$ and $\ka_{r,n-1} = \sum\limits_{b \in \mu}c(b)$ as operators on $V^{\mu^{(n,i)}}$, the irreducible $L(r,1,n)$-module parametrized by $\mu^{(n,i)} \in \Y^n(r,n-1)$.
 \end{enumerate}
 \end{lemma}
 
 %\goodbreak
\begin{proof}   
%\leavevmode 
%\begin{enumerate}[(\alph*)]
%\item\label{proof_lemma_7.1_a} 
 \ref{lemma7.1_a} 
First, we consider the case $p=1$. Being the sum of elements in the conjugacy class of $(1,1, \ldots, 1, s_{12})$, $\ka_{r,n}$ is a central element of $\Cbb[G(r,1,n)]$ and 
 \begin{equation*}
 \ka_{r,n} = \frac{1}{r}\sum_{j=1}^{n} X_j.
 \end{equation*}
 For the irreducible $G(r,1,n)$-module $V^{\La}$ parametrized by $\La \in \Y(r,n)$, the canonical decomposition of $V^\La$ into $GZ$-subspaces is 
\[
V^{\lambda} = \bigoplus\limits_{T \in \emph{Tab}(r, \La)}V_T.
\] 
Using Theorem~\ref{gzt} for the action of Jucys--Murphy elements of $G(r,1,n)$, we have 
\[
\ka_{r,n}(v_T) = \sum\limits_{j=1}^n c(b_T(j))(v_T) = \bigg(\sum\limits_{b \in \lambda}c(b)\bigg)v_T
\]
 where $v_T$ is $GZ$-basis element corresponding to $T \in \Tab$. Thus, $\ka_{r,n} = \sum\limits_{b \in \lambda}c(b) $ as operators on $V^{\La}$. 
For a divisor $p$ of $r$, note that $\ka_{r,n} \in \Cbb[\Grpn] \subseteq \Cbb[G(r,1,n)]$. Thus, $\ka_{r,n}$ is a central element of $\Cbb[\Grpn]$ also, and its action on the irreducible $\Grpn$-module $V^{(\tla, \delta)}$ follows by restricting the action of $G(r,1,n)$ on $V^\La$.

 \ref{lemma7.1_b} 
%\item\label{proof_lemma_7.1_b} 
The proof is along the similar lines as that of part~\ref{lemma7.1_a}. 
%\qedhere
%\end{enumerate}
\end{proof}

Now, we describe a particular central element in $\Cbb[G(2,2,2k)]$. The conjugacy class $C$ of the element $(1,1,\ldots,1,(1,2)(3,4) \cdots (2k-1, 2k))$ in $G(2,1,2k)$ consists of the elements of the 
form $(a_1, a_2, \ldots, a_{2k}, (i_1, i_2)(i_3,i_4) \cdots (i_{2k-1}, i_{2k}))$ such 
that $(i_1, i_2), (i_3,i_4), \ldots, (i_{2k-1}, i_{2k})$ are mutually disjoint transpositions in $S_{2k}$, and $a_{i_j}a_{i_{j+1}} = 1 $ for all $j = 1, 3, \ldots, 2k-1$ with $a_i \in G = \Z/2\Z = \{1,-1\}$ for all $i=1,\ldots, 2k$. Using~\cite[Theorem~11]{Re77}, the conjugacy class of $(1,1,\ldots,1,(1,2)(3,4) \cdots (2k-1, 2k))$ in $G(2,1,2k)$ decomposes into two conjugacy classes, denoted by $C_1$ and $C_2$, in $G(2,2,2k)$ with representatives 
\begin{align*}c_1 & = (-1,-1,1, \ldots, 1, (1,2)(3,4) \cdots (2k-1, 2k)) \\
\text { and } c_2 & = (1,1, \ldots, 1, (1,2)(3,4) \cdots (2k-1, 2k)),
\end{align*}
respectively. The classes $C_1$ and $C_2$ consist of those elements in $C$ such that the number of pairs $(a_{i_j},a_{i_{j+1}}) = (-1,-1) $, where $j = 1, 3, \ldots, 2k-1$, is odd and even, respectively. Let $z_1$ and $z_2$ be the conjugacy class sums of $C_1$ and $C_2$, respectively. Define $z = z_2 - z_1$ which is a central element in $\Cbb[G(2,2,2k)]$.
 
\begin{lemma}\phantomsection\label{actz}
Let $\La \in \Y(2,2k)$. Then, 
\begin{enumerate}[(\alph*)]
 \item\label{lemma_7.2_a} For $\La \neq ((k),(k))$, $z=0$ as operators on the irreducible $G(2,2,2k)$-module $V^{(\tla, 1)}$. 
 \item\label{lemma_7.2_b} For $\La = ((k),(k))$, $z = 2^k k!$ as operators on the irreducible $G(2,2,2k)$-module $V^{(\tla, 1)}$ and $z = -2^k k!$ as operators on the irreducible $G(2,2,2k)$-module $V^{(\tla, -1)}$.
\end{enumerate}
\end{lemma}

\begin{proof} 
In the following, we use~\cite[Theorem~6.10]{MS16} to describe the action of $z$ on irreducible $G(2,2,2k)$-modules. 
The irreducible $G(2,1,2k)$-module $V^{\La}$ parametrized by $\La = (\La_1, \La_2) \in \Y(2,2k)$ has a $GZ$-basis element $v_R$ where $R = (R_1, R_2)$ is the element of $\Tabop(2,\La)$ written in row major order, \ie the entries in $R_1$ are in from $1 ,\ldots, |\La_1|$ and entries in $R_2$ are from $|\La_1|+1, \ldots, |\La_1|+|\La_2|$, both filled in row major order.
We have the following cases: 

%\begin{enumerate}[(\alph*)]
% \item\label{proof_lemma_7.2_a} 
 \ref{lemma_7.2_a} 
$\La \neq ((k),(k))$: $V^{\La}$ remains irreducible as $G(2,2,2k)$-module and $V^{(\tla, 1)} = V^\La$ with $v_R^{(1)} = v_R$ as one of the basis elements using the parametrization of irreducible $G(2,2,2k)$-module in Theorem~\ref{irrgp} and construction of basis of irreducible $G(2,2,2k)$-modules. 
 
 Let $Y$ be the set of those $\pi \in S_{2k}$ which can be written as a product of disjoint transpositions such that the elements of each transposition are either in $R_1$ or in $R_2$. For a fixed $\pi \in Y$, the action of $\sum(a_1, \ldots, a_{2k}, \pi)$ on $v_R^{(1)}$, where the sum is over all such elements in $C_1$, is equal to the action of $\sum(b_1, \ldots, b_{2k}, \pi)$ on $v_R^{(1)}$, where the sum is over all such elements in $C_2$. 
 
 The coefficient of $v_R$ in $t v_R^{(1)}$ is zero for any $t \in C_1 \cup C_2$ which is of the form $t = (a_1, a_2, \ldots, a_{2k}, (i_1, i_2)(i_3,i_4) \cdots (i_{2k-1}, i_{2k}))$ such that there is at least one transposition $(i_y, i_{y+1})$ with one of $i_y, i_{y+1}$ being from the entries in $R_1$ and the other being from the entries in $R_2$. 
 
 Thus, we have $zv_R^{(1)} = 0.$

 \ref{lemma_7.2_b} 
% \item\label{proof_lemma_7.2_b} 
$\La = ((k),(k))$: $V^{\La}$ decomposes into two irreducible as $G(2,2,2k)$-modules $V^{(\tla, 1)}$ and $V^{(\tla, -1)}$ with $v_R^{(1)} = v_R+v_{\sh(R)}$ and $v_R^{(-1)} = v_R-v_{\sh(R)}$ as one of their basis elements, respectively. Analogous to part~\ref{lemma_7.2_a}, for a fixed $\pi \in Y$, the action of $\sum(a_1, \ldots, a_{2k}, \pi)$ on $v_R$ and $v_{\sh(R)}$, where the sum is over all such elements in $C_1$, is equal to the action of $\sum(b_1, \ldots, b_{2k}, \pi)$ on $v_R$ and $v_{\sh(R)}$, where the sum is over all such elements in $C_2$, respectively.
 
 Let $P$ be the set of those $\beta \in S_{2k}$ which can be written as a product of disjoint transpositions such that one element of each transposition is from $1,\ldots,k$ and the other one is from $k+1, \ldots, 2k$. The order of $P$ is $k!$. For $(a_1, \ldots, a_{2k}, \beta) \in C_1$ and $\beta \in P$, $(a_1, \ldots, a_{2k}, \beta)v_R = - v_{\sh(R)}$ and $(a_1, \ldots, a_{2k}, \beta)v_{\sh(R)} = - v_R$.
 For $(a_1, \ldots, a_{2k}, \beta) \in C_2$ and $\beta \in P$, $(a_1, \ldots, a_{2k}, \beta)v_R = v_{\sh(R)}$ and $(a_1, \ldots, a_{2k}, \beta)v_{\sh(R)} = v_R$.
 
 For those elements $(a_1, \ldots, a_{2k}, \gamma) \in C_1 \cup C_2$ such that $\gamma \notin Y \cup P$, the coefficients of both the elements $v_R$ and $v_{\sh(R)}$ in both $(a_1, \ldots, a_{2k}, \gamma)v_R$ and $(a_1, \ldots, a_{2k}, \gamma)v_{\sh(R)}$ are zero.
 Thus,
 \[
 z v_R^{(1)} = (2^k k!) v_R^{(1)} \text { and } z v_R^{(-1)} = -(2^k k!) v_R^{(-1)}. 
\]
 Thus, we get the scalars as stated in the theorem. 
 %\qedhere
 %\end{enumerate}
\end{proof}

Assume that $S$ is a subset of $\{1, 2, \ldots, k\}$, $I$ is a subset of $S \cup S'$ and $I^c$ denotes the complement of $I$ in $S \cup S'$, where $S'$ is the set of all $j'$ such that $j \in S$. Define the elements $b_S$ and $ d_I$ of the partition monoid $A_k$:
\begin{align*}
 b_S = \{S \cup S', \{l,l'\}_{l \notin S} \} \text { and } d_I = \{I, I^c, \{l,l'\}_{l \notin S} \}.
\end{align*}
Thus, $b_S \in \Pi_k(r)$. Also, it is easy to see that 
\[
d_I = d_{I^c}, \; d_{S \cup S'} = d_{\emptyset} = b_S, \; \text { and } d_{\{l,l'\}} = d_{\{l,l'\}^c} = b_{S\setminus\{l\}}.
\] 

\begin{exam}
For $k=6$, $S=\{1,2,4\}$, and $I=\{1,4'\}\subset S\cup S'$, $b_S$ and $d_I$ are given in Figure~\ref{fig:bsdi}.

\begin{figure}[ht]
\centering
 \begin{tikzpicture}[scale=0.9,
 mycirc/.style={circle,fill=black, minimum size=0.1mm, inner sep = 1.5pt}
 ]
 \node (1) at (-1,0.5) {$b_S =$};
 \node[mycirc,label=above:{$1$}] (n1) at (0,1) {};
 \node[mycirc,label=above:{$2$}] (n2) at (1,1) {};
 \node[mycirc,label=above:{$3$}] (n3) at (2,1) {};
 \node[mycirc,label=above:{$4$}] (n4) at (3,1) {};
 \node[mycirc,label=above:{$5$}] (n5) at (4,1) {};
 \node[mycirc,label=above:{$6$}] (n6) at (5,1) {};
 \node[mycirc,label=below:{$1'$}] (n1') at (0,0) {};
 \node[mycirc,label=below:{$2'$}] (n2') at (1,0) {};
 \node[mycirc,label=below:{$3'$}] (n3') at (2,0) {};
 \node[mycirc,label=below:{$4'$}] (n4') at (3,0) {};
 \node[mycirc,label=below:{$5'$}] (n5') at (4,0) {};
 \node[mycirc,label=below:{$6'$}] (n6') at (5,0) {};
 \draw (n1) -- (n2);
 \draw (n1) -- (n1');
 \draw (n1')-- (n2');
 \draw (n2) .. controls(2,0.7) .. (n4);
 \draw (n2').. controls(2,0.3) .. (n4');
 \draw (n3) -- (n3');
 \draw (n5) -- (n5');
 \draw (n6) -- (n6');
 \node(a) at (5.4,0.3) {,};
 

 \node (1) at (6.3,0.5) {$d_I =$};
 \node[mycirc,label=above:{$1$}] (n7) at (7.3,1) {};
 \node[mycirc,label=above:{$2$}] (n8) at (8.3,1) {};
 \node[mycirc,label=above:{$3$}] (n9) at (9.3,1) {};
 \node[mycirc,label=above:{$4$}] (n10) at (10.3,1) {};
 \node[mycirc,label=above:{$5$}] (n11) at (11.3,1) {};
 \node[mycirc,label=above:{$6$}] (n12) at (12.3,1) {};
 \node[mycirc,label=below:{$1'$}] (n7') at (7.3,0) {};
 \node[mycirc,label=below:{$2'$}] (n8') at (8.3,0) {};
 \node[mycirc,label=below:{$3'$}] (n9') at (9.3,0) {};
 \node[mycirc,label=below:{$4'$}] (n10') at (10.3,0) {};
 \node[mycirc,label=below:{$5'$}] (n11') at (11.3,0) {};
 \node[mycirc,label=below:{$6'$}] (n12') at (12.3,0) {};
 \draw (n7) -- (n10');
 \draw (n8) -- (n7');
 \draw (n8) .. controls(9.5,0.7) .. (n10);
 \draw (n7') .. controls(8,0.3) .. (n8');
 \draw (n9) -- (n9');
 \draw (n11) -- (n11');
 \draw (n12) -- (n12'); 
\end{tikzpicture}\caption{Example of $b_S$ and $d_I$.}
\label{fig:bsdi}
\end{figure}
\end{exam}

Following the notation of Section~\ref{ta}, let $N(I)$ and $M(I)$ denote the number of elements in top row and bottom row of the block $I$, respectively. 
For $k \in \Z_{\geq 0}$, we define an element $Z_{k,r} \in \T_k(r,1,n) \subseteq \Tn$:
\begin{equation*}
 Z_{k,r} = \binom{n}{2}+\sum\limits_{|S| \geq 1}(-1)^{|S|}\bigg((n-1)b_S + \sum\limits_{N(I) \equiv M(I) (\modrm r)}(-1)^{N(I)-M(I)}(d_I - b_S)\bigg),
\end{equation*}
where the outer sum is over all the nonempty subsets $S$ of $\{1,2,\ldots,k\}$ and the inner sum is over $I \subseteq S \cup S'$ such that $d_I \in \Pi_k(r)$ and $d_I \neq b_S$. 

Define an element $Z_{k+\frac{1}{2},r} \in \T_{k+\frac{1}{2}}(r,1,n) \subseteq \Thn$ as follows:
\begin{multline*}
 Z_{k+\frac{1}{2},r}  = \binom{n}{2} + \!\!\sum\limits_{\substack{|S| \geq 1\\ k+1 \notin S}}(-1)^{|S|}\bigg((n-1)b_S + \sum\limits_{N(I) \equiv M(I) (\modrm r)}(-1)^{N(I)-M(I)}(d_I - b_S)\!\!\bigg) \\ 
+ \sum\limits_{\substack{ k+1 \in S \\|S| \geq 1}}(-1)^{|S|}\bigg((n-1)b_S + \sum\limits_{\substack{\{k+1, (k+1)'\} \subset I \text { or } I^c \\ N(I) \equiv M(I) (\modrm r)}}(-1)^{N(I)-M(I)}(d_I - b_S)\bigg),
\end{multline*}
where the first outer sum is over all the nonempty subsets $S$ of $\{1,2,\ldots,k+1\}$ such that $k+1 \notin S$ and the inner sum in that is over $I \subseteq S \cup S'$ such that $d_I \in \Pi_{k+\frac{1}{2}}(r)$ and $d_I \neq b_S$; 
and the second outer sum is over all the nonempty subsets $S$ of $\{1,2,\ldots,k, k+1\}$ such that $k+1 \in S$ and the inner sum in that is over $I \subseteq S \cup S'$ such that 
\begin{center}
$\{k+1, (k+1)'\} \subseteq I \text { or } I^c$, $d_I \in \Pi_{k+\frac{1}{2}}(r)$ and $d_I \neq b_S$.
\end{center}

The elements $Z_{k,r}$ and $Z_{k+\frac{1}{2},r}$ and the idea of the proof of the next theorem are from the online notes~\cite{Ram2010M}.

\begin{theo}%\phantomsection
\label{zkr}\leavevmode 
\begin{enumerate}[(\alph*)]
\item\label{theo7.4_a} Let $k \in \Z_{\geq 0}$. Then,
\begin{equation*}
 \ka_{r,n} = Z_{k,r} \text { and } \ka_{r,n-1} = Z_{k+\frac{1}{2}, r} 
\end{equation*}
as operators on $\Vk$ and $\Vkh$, respectively.

\item\label{theo7.4_b} Let $k \in \Z_{\geq 0}$. Then $Z_{k,r}$ is a central element of $\Tn$. For $n \geq 2k$ 
\[
Z_{k,r} = \sum\limits_{b \in \lambda}c(b) 
\]
 as operators on $\T_k^{(\tla, \delta)}$, the irreducible $\Tn$-module parametrized by 
$(\tla, \delta) \in \vo_k(r,p,n)$. 

Also, $Z_{k+\frac{1}{2}, r}$ is a central element of $\Thn$. For $n \geq 2k+1$, 
\[
Z_{k+\frac{1}{2},r} = \sum\limits_{b \in \mu}c(b) 
\]
 as operators on $\T_{k+\frac{1}{2}}^{\tm^{(n,i,j)}}$, the irreducible $\Thn$-module parametrized by $\tm^{(n,i,j)} \in \vo_{k+\frac{1}{2}}(r,p,n)$.
\end{enumerate}
\end{theo}

\begin{proof}
%\leavevmode 
%\begin{enumerate}[(\alph*)]
%\item\label{proof_theo7.4_a} 
 \ref{theo7.4_a} 
We express the action of $\ka_{r,n}$ in terms of matrices $E_{i,j}$.
\begin{multline}\label{kact}
  2 \ka_{r,n}(\vi) \\ 
\begin{aligned}
 &= \frac{1}{r}\sum\limits_{l=0}^{r-1}\sum\limits_{1 \leq i \neq j \leq n}\g_i^{l}\g_j^{-l}s_{ij} v_{i_1} \otimes \g_i^{l}\g_j^{-l}s_{ij} v_{i_2}
 \otimes \cdots \otimes \g_i^{l}\g_j^{-l}s_{ij} v_{i_k} \\
 &= \frac{1}{r}\sum\limits_{l=0}^{r-1}\sum\limits_{1 \leq i \neq j \leq n} 
 (1-E_{ii}-E_{jj}+\g^{l}E_{ij}+\g^{-l}E_{ji})v_{i_1} \otimes \cdots  
\end{aligned}\\
 \otimes (1-E_{ii}-E_{jj}+\g^{l}E_{ij}+\g^{-l}E_{ji})v_{i_k}.
 \end{multline}

Let $S$ be a subset of $\{1,2, \ldots, k\}$ such that $S^c$ corresponds to the tensor positions where $1$ is acting, $I \subset S \cup S'$ corresponds to the tensor positions that must equal $i$ and $I^c$ corresponds to the tensor positions that must equal $j$. Let 
\begin{multline*}
c_{S,I} :=  \prod_{t \in S^c} (\delta_{i_ti_{t'}})(-1)^{\#(\{t,t'\}\subset I) + \#(\{t,t'\}\subset I^c)} \\ 
 \times \zeta^{l(\#(\{t \in I,t' \in I^c\}) - \#(\{t \in I^c ,t' \in I\}))}\prod_{t \in I}(\delta_{i_ti})\prod_{t \in I^c}(\delta_{i_tj}) 
\end{multline*}
Thus, expanding~\eqref{kact}, we get that $2 \ka_{r,n}(\vi)$ equals 
\begin{equation}\label{exp}
\frac{1}{r}\sum\limits_{S \subset \{1,\ldots, k\}} \sum\limits_{\itp}\sum\limits_{l=0}^{r-1}\sum\limits_{1 \leq i \neq j \leq n}\sum\limits_{I \subset S \cup S'}c_{S,I}(\vip).
\end{equation}

Now, we take various cases of $S$ and $I$ to compute the above expression~\eqref{exp}. Let $|S| = 0$, then $I$ is empty set and 
\[
c_{S,I} = c_{\emptyset, \emptyset} = \prod\limits_{t \in \{1, \ldots,k \}}\delta_{i_ti_{t'}}.
\]
 The corresponding summand in~\eqref{exp} is 
\[
(n^2-n)(\vi).
\]

Assume that $|S| \geq 1$ and we consider various cases of $I \subset S \cup S'$. Since the whole sum is symmetric in $i$ and $j$ and in $I$ and $I^c$, therefore, the sum obtained is same when $I$ is interchanged with $I^c$. If $I = S \cup S'$, then 
\[
c_{S,I} = \prod_{t \in S^c} (\delta_{i_ti_{t'}})(-1)^{|S|}\prod_{t \in S \cup S'}(\delta_{i_ti}),
\]
 and thus the corresponding summand in expression~\eqref{exp} is 
\begin{multline*}
\frac{1}{r} \sum\limits_{\itp}\sum\limits_{l=0}^{r-1}(n-1)\sum\limits_{1 \leq i \leq n}c_{S,I}(\vip) \\ 
=  (n-1)(-1)^{|S|}b_S(\vi).
\end{multline*}
We get an identical summand for the case $I = \emptyset$.

Consider $I \subsetneq S \cup S'$ and $N(I) \not\equiv M(I) (\modrm r)$. Let 
\begin{align*}
T(I) &= \{t \in \{1,2,\ldots, k\} \mid t \in I \},\\
D(I) &= \{t' \in \{1',2',\ldots, k'\} \mid t' \in I \}, 
 \intertext{and}
 B(I) &= \{t \in I \mid t' \in I \}.
 \end{align*}
  Thus, $N(I) = |T(I)|, ~M(I) = |D(I)|$. Also, we can see that 
\begin{align*}
\#(\{t \in I,t' \in I^c\}) &= N(I) - |B(I)| 
 \intertext{and}
\#(\{t \in I^c ,t' \in I\}) &= M(I) - |B(I)|.
 \end{align*} 
 Thus,
\begin{align*}
 \#(\{t \in I,t' \in I^c\})-\#(\{t \in I^c ,t' \in I\}) = N(I)-M(I) \not\equiv 0(\modrm r).
\end{align*}
In this case, since the sum of all the $r$-th roots of unity is zero, the summand for all such $I$ in expression~\eqref{exp} is zero. 

Now, consider those subsets $I \subsetneq S \cup S'$ such that $N(I) \equiv M(I) (\modrm r)$. Define $B(I)' := \{t'\mid t \in B(I)\}$, thus 
\begin{gather*}
 \{t \in \{1,2,\ldots, k\} \mid \{t,t'\}\subset I\} = B(I), 
\\
\text { and } \{ t \in \{1,2,\ldots, k\} \mid \{t,t'\}\subset I^c\} = (S \setminus T(I)) \setminus (D(I) \setminus B(I)').
\end{gather*} 
This implies that 
\begin{align*}
(-1)^{\#(\{t,t'\}\subset I) + \#(\{t,t'\}\subset I^c)} & = (-1)^{|B(I)| + (|S| - N(I)) -(M(I) - |B(I)|)} \\
& = (-1)^{|B(I)| + (|S| - N(I))+(M(I) - |B(I)|)} \\
& = (-1)^{|S| - (N(I)- M(I))} \\
& = (-1)^{|S| + (N(I) - M(I))}.
\end{align*}
Thus, for the subsets $I$ such that $N(I) \equiv M(I) (\modrm r)$, we get 
the summand in expression~\eqref{exp} as:
\begin{align*}
 &\frac{1}{r}(-1)^{\#(\{t,t'\}\subset I) + \#(\{t,t'\}\subset I^c)} \sum\limits_{\itp}\sum\limits_{l=0}^{r-1}\prod_{t \in S^c} (\delta_{i_ti_{t'}})\\ 
 &\hspace*{47mm} \times \sum\limits_{1 \leq i \neq j \leq n} \prod_{t \in I} (\delta_{i_ti})\prod_{t \in I^c}(\delta_{i_tj})(\vip) \\
 & \hspace*{20mm}= (-1)^{|S| + (N(I) - M(I))}\sum\limits_{\itp}\prod_{t \in S^c} (\delta_{i_ti_{t'}}) \left(\sum\limits_{1 \leq i, j \leq n} \prod_{t \in I} (\delta_{i_ti})\prod_{t \in I^c}(\delta_{i_tj})\right.\\ 
 &\hspace*{47mm}\left. -\sum\limits_{1 \leq i = j \leq n} \prod_{t \in I} (\delta_{i_ti})\prod_{t \in I^c}(\delta_{i_tj})\right)(\vip)\\
 & \hspace*{20mm}= (-1)^{|S| + (N(I) - M(I))}(d_I-{b_S})(\vi).
\end{align*}
Also, for the subsets $I$ such that $N(I) \equiv M(I) (\modrm r)$, we also have $N(I^c) \equiv M(I^c) (\modrm r)$ and thus we get an identical summand by interchanging $I$ and $I^c$.

Combining all the above cases together, we get that, as operators on $\Vk$, 
\begin{equation*}
 \ka_{r,n} = Z_{k,r}.
\end{equation*}


Now we prove the second part of~\ref{theo7.4_b}. We have 
\begin{equation*}
 (1-E_{ii}-E_{jj}+E_{ii}E_{jj})(v_n) = 
 \begin{cases}
 0, & \text {if } i = n \text { or } j=n, \\
 v_n, & \text {otherwise.}
 \end{cases}
\end{equation*}
Thus, 
\begin{multline}\label{kact1}
 2 \ka_{r,n-1}(\vi \otimes v_n)  \\
 \begin{aligned} 
 &=  \frac{1}{r}\sum\limits_{l=0}^{r-1}\sum\limits_{1 \leq i \neq j \leq n-1}\g_i^{l}\g_j^{-l}s_{ij}(\vi \otimes v_n)  \\
 &= \frac{1}{r}\sum\limits_{l=0}^{r-1}\sum\limits_{1 \leq i \neq j \leq n}\g_i^{l}\g_j^{-l}s_{ij} v_{i_1} \otimes \g_i^{l}\g_j^{-l}s_{ij} v_{i_2}
 \otimes \\
 & \quad\quad \cdots \otimes \g_i^{l}\g_j^{-l}s_{ij} v_{i_k} \otimes (1-E_{ii}-E_{jj}+E_{ii}E_{jj})(v_n) \\
& =  \frac{1}{r}\sum\limits_{l=0}^{r-1}\sum\limits_{1 \leq i \neq j \leq n}\g_i^{l}\g_j^{-l}s_{ij}(\vi)\otimes v_n  \\
& \quad +  \frac{1}{r}\sum\limits_{l=0}^{r-1}\sum\limits_{1 \leq i \neq j \leq n} 
 \bigg((1-E_{ii}-E_{jj}+\g^{l}E_{ij}+\g^{-l}E_{ji})v_{i_1} \otimes \cdots  \\ 
 & \quad \otimes (1-E_{ii}-E_{jj}+\g^{l}E_{ij}+\g^{-l}E_{ji})v_{i_k}\bigg) \otimes (-E_{ii}-E_{jj})v_n \\
 &\quad +  \frac{1}{r}\sum\limits_{l=0}^{r-1}\sum\limits_{1 \leq i \neq j \leq n}\g_i^{l}\g_j^{-l}s_{ij} v_{i_1} \otimes \g_i^{l}\g_j^{-l}s_{ij} v_{i_2}
 \otimes \cdots \otimes \g_i^{l}\g_j^{-l}s_{ij} v_{i_k} \otimes E_{ii}E_{jj}v_n.
 \end{aligned}
 \end{multline}

In the expression~\eqref{kact1}, the first summand is equal to $2\ka_{r,n}(\vi)$ which has been calculated in the first part of~\ref{theo7.4_b}. Since $i \neq j$, the last summand is zero. Expanding the middle summand gives 
\begin{equation*}
\frac{1}{r}\sum\limits_{\substack{S \subset \{1,\ldots, k+1\} \\ k+1 \in S} } \sum\limits_{\itp}\sum\limits_{l=0}^{r-1}\sum\limits_{1 \leq i \neq j \leq n}\sum\limits_{\substack{I \subset S \cup S' \\ \{k+1, (k+1)'\} \subset I \text{ or } I^c}}c_{S,I}(\vip).
\end{equation*}

The case $|S| = 0$ does not arise because $k+1 \in S$. For $|S| >1$, we consider various cases of $I \subset S \cup S'$ which are: 
\begin{enumerate}[(\roman*)]
\item\label{proof_theo7.4_b_i} $I = S \cup S'$, 
 
\item\label{proof_theo7.4_b_ii} $\{k+1,(k+1)'\} \subset I \subsetneq S \cup S',~N(I) \not\equiv M(I) (\modrm r)$, 
and  
\item\label{proof_theo7.4_b_iii} $ \{k+1,(k+1)'\} \subset I \subsetneq S \cup S',~N(I) \equiv M(I) (\modrm r)$.
\end{enumerate}

\noindent
and identical summands arise when $I$ is interchanged with $I^c$ in the 
cases~\ref{proof_theo7.4_b_i}, \ref{proof_theo7.4_b_ii},  and~\ref{proof_theo7.4_b_iii}. Thus, the middle summand gives us 
\begin{equation*}
 \sum\limits_{\substack{ k+1 \in S \\|S| \geq 1}}(-1)^{|S|}2\bigg((n-1)b_S + \sum\limits_{\substack{\{k+1, (k+1)'\} \subset I \text{ or } I^c \\ N(I) \equiv M(I) (\modrm r)}}(-1)^{N(I)-M(I)}(d_I - b_S)\bigg).
\end{equation*}

So, as operators on $\Vkh$, we have $\ka_{r,n-1} = Z_{k+\frac{1}{2},r}.$



\ref{theo7.4_b}
%\item\label{proof_theo7.4_b} 
Using Theorem~\ref{dvkp}\ref{theo6.5_a} and using Lemma~\ref{ka}\ref{lemma7.1_a}, we get that for $n \geq 2k$, $Z_{k,r}$ acts on $\T_k^{(\tla, \delta)}$ as the constant stated in the theorem. Therefore, $Z_{k,r}$ is a central element of $\Tn$ for $n\geq 2k$. Since the multiplication of elements of $\T_k(r,p,n)$ is a polynomial in $n$, therefore the equality 
\[
Z_{k,r} x_d = x_d Z_{k,r} \text { for all } x_d \in \T_k(r,p,n) \text { and } n \geq 2k 
\]
 implies the equality 
\[
 Z_{k,r} x_d = x_d Z_{k,r} \text { for all } x_d \in \T_k(r,1,n) \text { and for all } n.
\] 

Theorem~\ref{dvkp}\ref{theo6.5_b} and Lemma~\ref{ka}\ref{lemma7.1_b} along with the arguments similar to the above imply the result for $Z_{k+\frac{1}{2},r}.$
%\qedhere
%\end{enumerate}
\end{proof}

In the light of Remarks~\ref{222k} and~\ref{cp}, $(r,p,n) = (2,2,2k)$ is the special case. 
Define $M_{k,2,2} := x_d \in \T_k(2,2,2k)$, where $d$ is the only element in $\Lambda_k(2,2,2k)$ and $d$ consists of $2k$ blocks, each vertex being a block. The element $M_{k,2,2}$ is a central element of $\T_k(2,2,2k)$.

\begin{theo}\label{mk22}\leavevmode 
\begin{enumerate}[(\alph*)]
\item\label{theo7.5_a} Let $k \in \Z_{\geq 0}$. Then,
$M_{k,2,2} = \frac{1}{2^k}z$
as operators on $\Vk$.

\item\label{theo7.5_b} Let $k \in \Z_{\geq 0}$. Then, for $\La \neq ((k),(k))$, $M_{k,2,2} = 0$ as operators on the irreducible $\T_k(2,2,2k)$-module $\T_k^{(\tla, 1)}$. 
For $\La = ((k),(k))$, 
\[
M_{k,2,2} = k!
\] 
as operators on the irreducible $\T_k(2,2,2k)$-module $\T_k^{(\tla, 1)}$ and 
\[
M_{k,2,2} = -k!
\]
 as operators on the irreducible $\T_k(2,2,2k)$-module $\T_k^{(\tla, -1)}$.
\end{enumerate}
\end{theo}

\begin{proof}
%\leavevmode 
% \begin{enumerate}[(\alph*)]
%\item\label{proof_theo7.5_a} 
\ref{theo7.5_a}
The action of $M_{k,2,2}$ on $\Vk$ is:
\begin{align*}
 M_{k,2,2} (v_{i_1} \otimes \cdots \otimes v_{i_k}) = 
 \begin{cases}
 \sum\limits_{\pi} v_{\pi(j_1)} \otimes \cdots \otimes v_{\pi(j_k)}, & \text {if } i_1, i_2, \ldots, i_k \text { are distinct } \\[-6pt]
 & \text {elements of } \{1, \ldots, 2k\}, \\[2pt]
 0, & \text {otherwise,}
 \end{cases}
\end{align*}
where $\pi$ varies over all the permutations of the set 
\[
 \{j_1, \ldots, j_k\} = \{1,\ldots, 2k\} \setminus \{i_1, \ldots, i_k\}.
\]

Now, we discuss the action of $z$ on $\Vk$.
Consider $v_{i_1} \otimes \cdots \otimes v_{i_k} \in \Vk$ such that $i_1, \ldots, i_k$ are distinct elements of $\{1, \ldots, 2k\}$. Then, 
\[
(a_1, \ldots, a_{2k}, (i_1, j_1) \cdots (i_{k},j_{k})) (v_{i_1} \otimes \cdots \otimes v_{i_k}) = -(v_{j_1} \otimes \cdots \otimes v_{j_k}) 
\]
if $(a_1, \ldots, a_{2k}, (i_1, j_1) \cdots (i_{k},j_{k})) \in C_1 $,
and 
\[
(a_1, \ldots, a_{2k}, (i_1, j_1) \cdots (i_{k},j_{k})) (v_{i_1} \otimes \cdots \otimes v_{i_k}) = (v_{j_1} \otimes \cdots \otimes v_{j_k}) 
\]
if $(a_1, \ldots, a_{2k}, (i_1, j_1) \cdots (i_{k},j_{k})) \in C_2$,
where in each case 
\[
 \{j_1, \ldots, j_k\} = \{1,\ldots, 2k\} \setminus \{i_1, \ldots, i_k\}.
\]
For a fixed $(i_1, j_1) \cdots (i_{k},j_{k})$ element in $S_{2k}$, there are $2^{k-1}$ elements of the form $(a_1, \ldots, a_{2k},(i_1, j_1) \cdots (i_{k},j_{k}))$ in each of $C_1$ and $C_2$. 

Consider an element of the form $(a_1, \ldots, a_{2k}, (x_1, y_1) \cdots (x_{k},y_{k})) \in C$ such that at least one pair, say $\{x_1, y_1\} \subset \{i_1, \ldots, i_k\}.$ Then, one of $x_2, \ldots, x_k$, say $x_k$, is different from $i_1, \ldots, i_k$ and one can choose $y_k \in \{1, \ldots, 2k \} \setminus \{i_1, \ldots, i_k, x_k, y_2, \ldots, y_{k-1}\}.$ Now, $(a_{x_k}, a_{y_k}) = (1,1)$ or $(a_{x_k}, a_{y_k}) = (-1,-1)$ keeps the sign of the action of $(a_1, \ldots, a_{2k}, (x_1, y_1) \cdots (x_{k},y_{k}))$ on $(v_{i_1} \otimes \cdots \otimes v_{i_k})$ same. 

Given $(b_1, \ldots, b_{2k}, (x_1, y_1) \cdots (x_{k},y_{k})) \in C_1$ such that $(b_{x_1}, b_{y_1}) = (1,1)$, we have the element $(f_1, \ldots, f_{2k}, (x_1, y_1) \cdots (x_{k},y_{k})) \in C_2$, such that $(f_{x_i}, f_{y_i}) = (b_{x_i}, b_{y_i})$ for $i \neq k$ and $(f_{x_k}, f_{y_k}) = - (b_{x_k}, b_{y_k})$ and 
\begin{multline*}
  (b_1, \ldots, b_{2k}, (x_1, y_1) \cdots (x_{k},y_{k})) (v_{i_1} \otimes \cdots \otimes v_{i_k}) \\
  = (f_1, \ldots, f_{2k}, (x_1, y_1) \cdots (x_{k},y_{k})) (v_{i_1} \otimes \cdots \otimes v_{i_k}).
\end{multline*}
A similar analysis can be done if $(b_{x_1}, b_{y_1}) = (-1,-1)$.

If at least two of $i_1, \ldots, i_k$ are same, say $i_1 = i_2$, then we can find a pair $(a_{x_k}, a_{y_k})$ such that the action of $(a_1, \ldots, a_{2k}, (x_1, y_1) \cdots (x_{k},y_{k}))$ on $(v_{i_1} \otimes \cdots \otimes v_{i_k})$ has the same sign whether $(a_{x_k}, a_{y_k}) = (1,1)$ or $(-1,-1)$. A similar analysis as above shows that corresponding to any element $(b_1, \ldots, b_{2k}, (x_1, y_1) \cdots (x_{k},y_{k})) \in C_1$ such that we can find the element $(f_1, \ldots, f_{2k}, (x_1, y_1) \cdots (x_{k},y_{k})) \in C_2$, such that
\begin{multline*}
  (b_1, \ldots, b_{2k}, (x_1, y_1) \cdots (x_{k},y_{k})) (v_{i_1} \otimes \cdots \otimes v_{i_k}) \\
 = (f_1, \ldots, f_{2k}, (x_1, y_1) \cdots (x_{k},y_{k})) (v_{i_1} \otimes \cdots \otimes v_{i_k}).
\end{multline*}
Collecting all the cases, we have 
\begin{align*}
 z (v_{i_1} \otimes \cdots \otimes v_{i_k}) = 
 \begin{cases}
 (2^k)\sum\limits_{\pi} v_{\pi(j_1)} \otimes \cdots \otimes v_{\pi(j_k)}, & \text {if } i_1, \ldots, i_k \text { are distinct } \\
 & \text {elements of } \{1, \ldots, 2k\}, \\
 0, & \text {otherwise,}
 \end{cases}
\end{align*}
where $\pi$ varies over all the permutations of the set 
\[
 \{j_1, \ldots, j_k\} = \{1,\ldots, 2k\} \setminus \{i_1, \ldots, i_k\}.
\]

%\item\label{proof_theo7.5_b} 
\ref{theo7.5_b}
The proof is clear by using part~\ref{theo7.5_a} of this theorem, Theorem~\ref{dvkp}~\ref{theo6.5_a} and Lemma~\ref{actz}.
%\end{enumerate}
\end{proof}


%For $ l \in \frac{1}{2}\Z_{> 0}$, define the Jucys--Murphy elements of $\T_l(r,p,n)$ as follows:
%\begin{align*}
%M_{\frac{1}{2},r} & = 1, \\
%\text {and }M_{l,r} & = Z_{l,r} - Z_{l-\frac{1}{2},r}.
%\end{align*}
For $ l \in \frac{1}{2}\Z_{> 0}$, define the Jucys--Murphy elements of $\T_l(r,p,n)$ as follows:
\begin{align*}
M_{\frac{1}{2},r} & = 1, \\
\text{and }M_{y,r} & = Z_{y,r} - Z_{y-\frac{1}{2},r}, \text{ for } y \in \frac{1}{2}\Z_{> 0} \text{ and } 1 \leq y \leq l.
\end{align*}
In addition to these elements, $\T_k(2,2,2k)$ has one more Jucys--Murphy element which is $M_{k,2,2}$.

\begin{theo}\label{jm}
 Let $l \in \frac{1}{2}\Z_{\geq 0}$ and let $n$ be a positive integer.
 
 \begin{enumerate}[(\alph*)]
 \item\label{theo7.6_a} The elements $M_{\frac{1}{2},r}, M_{1,r}, \ldots, M_{l-\frac{1}{2},r},M_{l,r}$ commute with each other in Tanabe algebra $\T_l(r,p,n).$
 
 \item\label{theo7.6_b} Assume that $n \geq 2l$. Let $\mathpzc{v}_l \in \vo_l(r,p,n)$ and $\T_{l}^{\pv_l}$ be the irreducible $\T_l(r,p,n)$-module parametrized by $\pv_l$. Then there is a unique, up to scalars, basis 
\[
\{u_\mathpzc{P} \mid \PP \text { is a path in } \widehat{\T}(r,p,n) \text { from } \pv_0 = ((n), \emptyset, \ldots, \emptyset) \text { to } \pv_l\}
\]
 of $\T_{l}^{\pv_l}$ such that, for all $\PP = (\pv_0, \pv_{\frac{1}{2}}, \pv_1, \ldots, \pv_l)$, and for all $k \in \Z_{\geq 0}$, $k \leq l$
 \begin{equation*}
 M_{k,r}(u_\PP) = c(\pv_k / \pv_{k-\frac{1}{2}})u_\PP,
 \end{equation*}
and 
\begin{equation*}
M_{k+\frac{1}{2},r}(u_\PP) = - c(\pv_k / \pv_{k+\frac{1}{2}})u_\PP,
\end{equation*}
where $\pv_k / \pv_{k-\frac{1}{2}} $ and $\pv_k / \pv_{k+\frac{1}{2}}$ denote the box by which $\pv_k$ differs from $\pv_{k-\frac{1}{2}}$ and $\pv_{k+\frac{1}{2}}$ as $r$-tuple of Young diagrams, respectively.
\item\label{theo7.6_c} For $(r,p,n) = (2,2,2k)$, the element $M_{k,2,2}$ commutes with the elements $M_{\frac{1}{2},2},$ $M_{1,2},$ $\ldots, M_{2k-\frac{1}{2},2},M_{2k,2}$. The scalars by which the Jucys--Murphy elements of $\T_k(2,2,2k)$ act on the basis \textup{(}as given by part~\ref{theo7.6_b}\textup{)} of $\T_{k}^{(((k),(k)),1)}$ and $\T_{k}^{(((k),(k)),-1)}$ are same except for $M_{k,2,2}$.
\end{enumerate}
\end{theo}

%\goodbreak
\begin{proof}
%\leavevmode 
% \begin{enumerate}[(\alph*)]
% \item\label{proof_theo7.6_a} 
\ref{theo7.6_a}
For $i, j \in \frac{1}{2}\Z_{\geq 0}$ and $i \leq j \leq l$, we have $Z_{i,r}, Z_{j,r} \in \T_j(r,p,n)$ and $Z_{j,r}$ is a central element of $\T_j(r,p,n) \subseteq \T_l(r,p,n)$, thus $Z_{i,r} Z_{j,r} = Z_{j,r}Z_{i,r}.$ Since $M_{j,r} = Z_{j,r} - Z_{j-\frac{1}{2},r}$, thus Jucys--Murphy elements commute with each other in $\T_l(r,p,n)$.

\ref{theo7.6_b}
% \item\label{proof_theo7.6_b}  
The branching rule from $\T_j(r,p,n)$ to $\T_{j-\frac{1}{2}}(r,p,n)$ is multiplicity free for all $j \in \frac{1}{2}\Z_{\geq 0}$ and $n \geq 2j$. Thus, $\T_{l}^{\pv_l}$ has canonical decomposition as irreducible $\T_{l-\frac{1}{2}}$-module:
 \[
 \T_{l}^{\pv_l} = \bigoplus\limits_{\pv_{l-\frac{1}{2}} \in \vo_{l-\frac{1}{2}}(r,p,n)} \T^{\pv_{l-\frac{1}{2}}}_{l-\frac{1}{2}}
 \]
 such that there is an edge from $\pv_{l-\frac{1}{2}}$ to $\pv_l$ in $\widehat{\T}(r,p,n)$. Further, iterating this decomposition, a canonical decomposition of $\T_{l}^{\pv_l}$ into irreducible $\T_{0}(r,p,n)$-modules is obtained:
 \begin{equation}\label{cdec}
 \T_{l}^{\pv_l} = \bigoplus\limits_{\PP} \T_{\PP},
 \end{equation}
 where $\T_{\PP}$ are one-dimensional $\T_{0}(r,p,n)$-modules and the sum is over all paths $\PP = (\pv_0, \pv_{\frac{1}{2}}, \pv_1, \ldots, \pv_l) $ such that $\pv_j \in \vo_j(r,p,n).$ The basis of $\T_{l}^{\pv_l}$ is obtained by choosing a nonzero vector $u_\PP$ in each $\T_{\PP}$ in the decomposition~\eqref{cdec}. Such a basis is called the \emph{Gelfand--Tsetlin basis} of the corresponding irreducible representation and it is unique, up to scalars. Using the decomposition~\eqref{cdec} and the definition of $u_{\PP}$, we get
 \[
\T_j(r,p,n) u_{\PP} = \T_{j}^{\pv_j}, 
\] 
for all $j \in \frac{1}{2}\Z_{\geq 0}$ and $j \leq l$ ,which implies that $u_{\PP}$ is a basis element of $\T_{j}^{\pv_j}$. Thus, for all $j \in \frac{1}{2}\Z_{\geq 0}$ and $j \leq l$, the action of $Z_{j,r}$ on $u_{\PP}$ is as a scalar given in Theorem~\ref{zkr}\ref{theo7.4_b}. 
 Now, by the definition of Jucys--Murphy elements, we get their actions on $u_{\PP}$. 
 
\ref{theo7.6_c}
% \item\label{proof_theo7.6_c}  
The element $M_{k,2,2}$ is a central element of $\T_k(2,2,2k)$. For $(r,p,n) = (2,2,2k)$ and $\La = ((k),(k))$, let $\pv_k = (\tla, 1) \in \vo_k(2,2,2k)$, $\pv'_k = (\tla, -1) \in \vo_k(2,2,2k)$. Then 
\[
\T_k^{\pv_k} \cong \T_k^{\pv'_k}
\] 
as $\T_{k-\frac{1}{2}}(r,p,n)$-modules. Thus, the part of the paths from $\pv_0$ to $\pv_k$ and $\pv'_k$ are same for $l <k $, $l \in \frac{1}{2}\Z_{\geq 0}$ and so, we have 
\[
M_j(u_\PP) = M_j (u_{\PP'}), \; \; \; j \leq k \text { and } j \in \frac{1}{2}\Z_{\geq 0}
\] 
where $u_\PP$ and $u_{\PP'}$ are Gelfand--Tsetlin basis elements of $\T_k^{\pv_k}$ and $\T_k^{\pv'_k}$, respectively. However, by Theorem~\ref{mk22}\ref{theo7.5_b}, we get that 
\[
M_{k,2,2}u_\PP = (k!)u_\PP \text { and } M_{k,2,2}u_{\PP'} = -(k!) u_{\PP'},
\]
 which proves the result. \qedhere
% \end{enumerate}
\end{proof}
 

\longthanks{The authors thank Arun Ram for suggesting the problem, for valuable insights and for his online notes~\cite{Ram}. The second author gratefully acknowledges the workshop ``Representation Theory of Symmetric Groups and Related Algebras'' at Institute for Mathematical Sciences, NUS, Singapore, where she had the opportunity to have discussions with Arun Ram.}

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