---
abstract: 'In this paper, we establish Schur-Weyl reciprocity for the $q$-analogue of the alternating group.  We analyze the sign $q$-permutation representation of the Hecke algebra $\mathcal{H}_{\mathbb{Q}(q),r}(q)$ on the $r$th tensor product of $\mathbb{Z}_2$-graded $\mathbb{Q}$-vector space $V=V_{\overline{0}}{\otimes}V_{\overline{1}}$ in detail, and examine its restriction to the $q$-analogue of the alternating group $\mathcal{H}_{\mathbb{Q}(q),r}^1(q)$. In consequence, we find out that if ${\dim}V_{\overline{0}}={\dim}V_{\overline{1}}$, then the centralizer of $\mathcal{H}_{\mathbb{Q}(q),r}^1(q)$ is a $\mathbb{Z}_2$-crossed product of the centralizer of $\mathcal{H}_{\mathbb{Q}(q),r}(q)$ and obtain Schur-Weyl reciprocity between $\mathcal{H}_{\mathbb{Q}(q),r}^1(q)$ and its centralizer.  Though the structure of the centralizer is more complicated for the case ${\dim}V_{\overline{0}}{\neq}{\dim}V_{\overline{1}}$, we obtain some results about the case.  When $q=1$, Regev has proved Schur-Weyl reciprocity for alternating groups in [@Regev].  Therefore, our result can be regarded as an extension of Regev’s work.' 
author:
- Hideo Mitsuhashi
title: 'Schur-Weyl reciprocity for the $q$-analogue of the alternating group'
---

Department of Information Technology\
Kanagawa Prefectural Junior College for Industrial Technology\
2–4–1 Nakao, Asahi–ku, Yokohama–shi, Kanagawa–ken 241–0815, Japan

Introduction
============

The purpose of this study is to research Schur-Weyl reciprocity for the $q$-analogue of the alternating group.  In our previous paper [@Mit2], we established Schur-Weyl reciprocity between the Hecke algebra $\mathcal{H}_{\mathbb{Q}(q),r}(q)$ and the quantum super Lie algebra $U^\sigma_q\big{(}{\mathfrak{g}\mathfrak{l}}(m,n)\big{)}$. In that paper, we defined the $q$-permutation representation of $\mathcal{H}_{\mathbb{Q}(q),r}(q)$, and showed that the image of the $q$-permutation representation is the centralizer of the image of the vector representation of the quantum super Lie algebra $U^\sigma_q\big{(}{\mathfrak{g}\mathfrak{l}}(m,n)\big{)}$ on the $r$th tensor product of a $\mathbb{Z}_2$-graded $(m+n)$-dimensional $\mathbb{Q}(q)$-vector space $V=V_{\overline{0}}{\oplus}V_{\overline{1}}$. In this paper, we find out the centralizer of the $q$-analogue of the alternating group as the restriction of the $q$-permutation representation.  When $q=1$, Regev has already shown Schur-Weyl reciprocity for the alternating group in [@Regev].  Hence our result is regarded as an extension of Regev’s work. 

Let $q$ be an indeterminate and $K=\mathbb{Q}(q)$. Let $(\pi_r,V^{{\otimes}r})$ be the $q$-permutation representation of $\mathcal{H}_{K,r}(q)$ (definition of the $q$-permutation representation is at (4.1)) and $(\rho_r,V^{{\otimes}r})$ the vector representation of $U^\sigma_q\big{(}{\mathfrak{g}\mathfrak{l}}(m,n)\big{)}$ (definition of the vector representation is at (4.4)).  We have proved in [@Mit2] that $\mathcal{A}_q=\pi_r\big{(}\mathcal{H}_{K,r}(q)\big{)}$ and $\mathcal{B}_q=\rho_r\big{(}U^\sigma_q\big{(}{\mathfrak{g}\mathfrak{l}}(m,n)\big{)}\big{)}$ are full centralizers of each other, namely: $$\mathcal{B}_q=\operatorname{End}_{\mathcal{A}_q}V^{{\otimes}r}{\quad}{\text{and}}{\quad}
\mathcal{A}_q=\operatorname{End}_{\mathcal{B}_q}V^{{\otimes}r}.$$ Let $R_0$ be a commutative domain which includes an invertible element $q$. We further assume that $2$ and $q+q^{-1}$ are invertible elements of $R_0$. Then we can define the $q$-analogue of the alternating group $\mathcal{H}_{R_0,r}^1(q)$ (see Definition 3.1 and Proposition 3.6) in $\mathcal{H}_{R_0,r}(q)$. In [@Mit], we defined the $q$-analogue of the alternating group as a subalgebra of Iwahori-Hecke algebra of type $A$ and obtained defining relations (see Proposition 3.7) for the first time.  In this paper, we show that $\mathcal{H}_{R_0,r}(q)$ is isomorphic to the $\mathbb{Z}_2$-crossed product which is obtained from the crossed system $(\mathcal{H}_{R_0,r}^1(q),\mathbb{Z}_2,\psi_0,\alpha_0)$ (definition of $\psi_0$ and $\alpha_0$ are at (3.2) and (3.3) respectively). 

$\mathcal{H}_{R_0,r}(q)$ is isomorphic to the $\mathbb{Z}_2$-crossed product $\mathcal{H}_{R_0,r}^1(q)_{\alpha_0}^{\psi_0}[\mathbb{Z}_2]$ as $R_0$-algebras. 

Let $\mathcal{C}_q=\pi_r\big{(}\mathcal{H}_{K,r}^1(q)\big{)}$ and $\mathcal{D}_q=\operatorname{End}_{\mathcal{C}_q}V^{{\otimes}r}$. Our main subject is to solve the relation between $\mathcal{D}_q$ and $\mathcal{B}_q$. From (1.1), one can immediately see that $\mathcal{B}_q{\subseteq}\mathcal{D}_q$. But the structure of $\mathcal{D}_q$ is not trivial.  Indeed, the structure of $\mathcal{D}_q$ depends on the dimensions $m=\dim_{K}V_{\overline{0}}$ and $n=\dim_{K}V_{\overline{1}}$. In this paper, we show that if $m=n$, then $\mathcal{D}_q$ is isomorphic to $\mathbb{Z}_2$-crossed product which is obtained from the crossed system $(\mathcal{B}_q,\mathbb{Z}_2,\psi_1,\alpha_1)$ (definition of $\psi_0$ and $\alpha_0$ are at (5.4) and (5.5) respectively). 

If $m=n$, then $\mathcal{D}_q$ is isomorphic to the $\mathbb{Z}_2$-crossed product ${\mathcal{B}_q}_{\alpha_1}^{\psi_1}[\mathbb{Z}_2]$ as $K$-algebras. 

From this theorem, we immediately obtain that $\dim_{K}\mathcal{D}_q=2\dim_{K}\mathcal{B}_q$. Moreover, we show Schur-Weyl reciprocity for $\mathcal{H}_{K,r}^1(q)$.

$\operatorname{End}_{\mathcal{C}_q}V^{{\otimes}r}
=\mathcal{D}_q$ and $\operatorname{End}_{\mathcal{D}_q}V^{{\otimes}r}
=\mathcal{C}_q$ hold. 

In the general case, the matter is more complicated, but we can find out to some extent if we exchange the base field from $K$ to its algebraic closure $\bar{K}$. Let $\bar{U}^\sigma_q\big{(}{\mathfrak{g}\mathfrak{l}}(m,n)\big{)}=U^\sigma_q\big{(}{\mathfrak{g}\mathfrak{l}}(m,n)\big{)}
{\otimes_{K}}\bar{K}$, $\bar{\mathcal{A}}_q=\mathcal{A}_q{\otimes_{K}}\bar{K}$, $\bar{\mathcal{B}}_q=\mathcal{B}_q{\otimes_{K}}\bar{K}$ and $\bar{\mathcal{C}}_q=\mathcal{C}_q{\otimes_{K}}\bar{K}$. Then we have the following theorem by the theory of semisimple algebras. 

$\bar{\mathcal{A}}_q$ and $\bar{\mathcal{C}}_q$ have direct sum decompositions $\bar{\mathcal{A}}_q=\bar{\mathcal{A}}_q^0{\oplus}\bar{\mathcal{A}}_q^1$ and $\bar{\mathcal{C}}_q=\bar{\mathcal{C}}_q^0{\oplus}\bar{\mathcal{C}}_q^1$ respectively, which are satisfy the following relations. 

1.   $\bar{\mathcal{C}}_q^0{\subseteq}\bar{\mathcal{A}}_q^0$ and $\dim_{\bar{K}}\bar{\mathcal{A}}_q^0
    =2\dim_{\bar{K}}\bar{\mathcal{C}}_q^0$

2.   $\bar{\mathcal{C}}_q^1=\bar{\mathcal{A}}_q^1$. Especially $\dim_{\bar{K}}\bar{\mathcal{A}}_q^1
    =\dim_{\bar{K}}\bar{\mathcal{C}}_q^1$.

As a corollary, we obtain an anomalous phenomenon for non-super case as follows. 

Let $n=0$. If $m^2<r$, then $\bar{\mathcal{A}}_q=\bar{\mathcal{C}}_q$ and $\operatorname{End}_{\bar{\mathcal{C}}_q}\bar{V}^{{\otimes}r}=
\operatorname{End}_{\bar{\mathcal{A}}_q}\bar{V}^{{\otimes}r}$.

The details of the decompositions of $\bar{\mathcal{A}}_q$ and $\bar{\mathcal{C}}_q$ are described in section 6.  We also obtain the similar result to Theorem 6.1 about the endomorphism algebras $\operatorname{End}_{\bar{\mathcal{A}}_q}\bar{V}^{{\otimes}r}$ and $\operatorname{End}_{\bar{\mathcal{C}}_q}\bar{V}^{{\otimes}r}$; there exist two $\mathcal{H}_{\bar{K},r}(q)$-submodules $W_0$ and $W_1$ which satisfy the following properties. 

$\operatorname{End}_{\bar{\mathcal{C}}_q}W_0{\supseteq}\operatorname{End}_{\bar{\mathcal{A}}_q}W_0$ and $\operatorname{End}_{\bar{\mathcal{C}}_q}W_1=\operatorname{End}_{\bar{\mathcal{A}}_q}W_1$.

Although the relation between $\bar{\mathcal{A}}_q$ and $\bar{\mathcal{C}}_q$ is made clear by (6.4)-(6.7), that between $\operatorname{End}_{\bar{\mathcal{A}}_q}W$ and $\operatorname{End}_{\bar{\mathcal{C}}_q}W$ is not clear except for Corollary 6.2 at this point. 

Preliminaries
=============

Let $R$ be a commutative ring with $1$ and $G$ a group.  In this section, we shall review the definition and some properties about $G$-crossed products.  A full account about $G$-graded algebras and $G$-crossed products is given in [@N-O]. 

An $R$-algebra $A$ is said to be *$G$-graded if there exist a family of $R$-submodules $\{A_{\sigma}|\sigma{\in}G\}$ of $A$ indexed by elements of $G$ which satisfies the following two conditions:*

1.   $A=\bigoplus_{\sigma{\in}G}A_{\sigma}$,

2.   $A_{\sigma}A_{\tau}{\subseteq}A_{\sigma\tau}$ for $\sigma,\tau{\in}G$.

Moreover, $A$ is said to be *strongly $G$-graded when (G2) is replaced by the following condition:*

1.   $A_{\sigma}A_{\tau}=A_{\sigma\tau}$ for $\sigma,\tau{\in}G$.

We notice that if $A=\bigoplus_{\sigma{\in}G}A_{\sigma}$ is a $G$-graded algebra, then $A_{1_G}$ ($1_G$ means the identity element of $G$) is a subalgebra of $A$ and $1_A{\in}A_{1_G}$.

A $G$-graded $R$-algebra $A=\oplus_{\sigma{\in}G}A_{\sigma}$ is said to be a $G$-crossed product if each $A_{\sigma}$ has an invertible element. 

We notice that a $G$-crossed product is a strongly $G$-graded algebra.  Indeed, if $A$ is a $G$-crossed product, then for an invertible element $u_{\sigma}{\in}A_{\sigma}$, $u_{\sigma}^{-1}{\in}A_{\sigma^{-1}}$ and $1_A=u_{\sigma}u_{\sigma}^{-1}{\in}A_{\sigma}A_{\sigma^{-1}}$. So, we have $A_{\sigma\tau}=1_AA_{\sigma\tau}{\subseteq}(A_{\sigma}A_{\sigma^{-1}})A_{\sigma\tau}
=A_{\sigma}(A_{\sigma^{-1}}A_{\sigma\tau}){\subseteq}A_{\sigma}A_{\tau}$.

Let $A$ be an algebra and $G$ a group.  Suppose that there exist two maps $$\psi:G{\longrightarrow}\operatorname{Aut}(A),$$ (we denote $\psi(\sigma)(a)$ by ${^{\sigma}a}$ for brevity), and $$\alpha:G{\times}G{\longrightarrow}A^{\times},$$ where $A^{\times}$ is the multiplicative group of units of $A_1$, which satisfy the relations: $$\begin{aligned}
{^\sigma}({^\tau}a)
&=&\alpha(\sigma,\tau){^{({\sigma}{\tau})}}a\alpha(\sigma,\tau)^{-1} \\
{^{\sigma_1}}\alpha(\sigma_2,\sigma_3)\alpha(\sigma_1,\sigma_2\sigma_3)
&=&\alpha(\sigma_1,\sigma_2)\alpha(\sigma_1\sigma_2,\sigma_3) \\
\alpha(\sigma,1)
&=&\alpha(1,\sigma)=1, \end{aligned}$$ for $\sigma,\tau,\sigma_1,\sigma_2,\sigma_3{\in}G,a{\in}A$. $(A,G,\psi,\alpha)$ is said to be a *crossed system.  $\psi$ is called a *weak action of $G$ on $A$, and $\alpha$ is called a $\psi$-*cocycle.*** 

We denote by $A_{\alpha}^{\psi}[G]$ the free left $A$-module with the basis $\{u_{\sigma}|\sigma{\in}G\}$ and the following multiplication: $$(a_1u_{\sigma})(a_2u_{\tau})=a_1{^\sigma}a_2\alpha(\sigma,\tau)u_{\sigma\tau},$$ for $a_1,a_2{\in}A,\sigma,\tau{\in}G$.

$A_{\alpha}^{\psi}[G]$ is a $G$-crossed product. 

Every $G$-crossed product is of the form $A_{\alpha}^{\psi}[G]$ for some algebra $A$, some weak action $\psi$ and some $\psi$-cocycle $\alpha$.

When $G$ is finite and a strongly $G$-graded algebra $A=\bigoplus_{\sigma{\in}G}A_{\sigma}$ is finitely generated over $R$ as modules, $A$ is said to have a *$G$-graded Clifford system $\{A_{\sigma}|\sigma{\in}G\}$ if $A$ satisfies (C1).* 

1.   For each $\sigma{\in}G$, there exists an invertible element $a_{\sigma}{\in}A$ such that $A_{\sigma}=a_{\sigma}A_{1_G}=A_{1_G}a_{\sigma}$.

It is clear that such $a_{\sigma}$ is in $A_{\sigma}$. An exposition about group graded Clifford systems can be found in [@C-R1], section 11C.

The $q$-analogue of the alternating group and its representation
================================================================

Let $(W,S=\{s_1,\hdots,s_r\})$ be a Coxeter system of rank $r$. Let $R_0$ be a commutative domain with $1$, and let $q_i(i=1,\hdots,r)$ be any invertible elements of $R_0$ such that $q_i=q_j$ if $s_i$ is conjugate to $s_j$ in $W$. Further we assume that $2$ and $q_i+q_i^{-1}$($i=1,2,\hdots,r$) are invertible elements of $R_0$. The Iwahori-Hecke algebra $\mathcal{H}_{R_0}(W,S)$ is an $R_0$-algebra generated by $\{T_{s_i}|s_i{\in}S\}$ with the defining relations:

1.   $T_{s_i}^2 = (q_i-q_i^{-1})T_{s_i}+1$ if $i=1,2,\hdots,r$,

2.   $(T_{s_i}T_{s_j})^{k_{ij}}=(T_{s_j}T_{s_i})^{k_{ij}}$ if $m_{ij}=2k_{ij}$,

3.   $(T_{s_i}T_{s_j})^{k_{ij}}T_{s_i}=(T_{s_j}T_{s_i})^{k_{ij}}T_{s_j}$ if $m_{ij}=2k_{ij}+1$,

where $m_{ij}$ is the order of $s_is_j$ in $W$. We write $T_i=T_{s_i}$ for brevity. 

If $(W,S)$ is of type $A$ and of rank $r-1$, then $W$ is isomorphic to the symmetric group $\mathfrak{S}_r$. Furthermore, all the elements of $S$ are conjugate to each other, hence we may assume $q_1=\cdots=q_{r-1}=q$. The Iwahori-Hecke algebra $\mathcal{H}_{R_0,r}(q)=\mathcal{H}_{R_0}(W,S)$ of type $A$ has the defining relations:

1.   $T_i^2 = (q-q^{-1})T_i+1$ if $i=1,2,\hdots,r-1$,

2.   $T_iT_{i+1}T_i = T_{i+1}T_iT_{i+1}$ if $i=1,2,\hdots,r-2$,

3.   $T_iT_j = T_jT_i$ if $|i-j|>1$.

Let $\hat{}$ be the Goldman involution.  This is an involution on $\mathcal{H}_{R_0,r}(q)$ defined by $$\hat{T_i}=(q-q^{-1})-T_i.$$

We define $\mathcal{H}_{R_0,r}^{\pm1}(q)$ to be the eigenspaces of $\mathcal{H}_{R_0,r}(q)$ corresponding to the eigenvalues $\pm1$ of $\hat{}$ respectively. 

We notice that $\mathcal{H}_{R_0,r}^1(q)$ is a subalgebra of $\mathcal{H}_{R_0,r}(q)$. Let $T'_i$ ($i=1,2,\hdots,r-1$) be the elements of $\mathcal{H}_{R_0,r}(q)$ defined by $$T'_i=\dfrac{T_i-\hat{T}_i}{q+q^{-1}}= \dfrac{2T_i-(q-q^{-1})}{q+q^{-1}} 
\quad \text{for $i=1,2,\hdots,r-1$}.$$ Then one can immediately check $\hat{T}'_i=-T'_i$.

$T'_i$($i=1,2,\hdots,r$) generate $\mathcal{H}_{R_0,r}(q)$ and satisfy the following defining relations:

1.   ${T'_i}^2 = 1$ if $i=1,2,\hdots,r-1$,

2.   $T'_iT'_{i+1}T'_i = T'_{i+1}T'_iT'_{i+1}-\Big{(}\dfrac{q-q^{-1}}{q+q^{-1}}\Big{)}^2
    (T'_i-T'_{i+1})$ if $i=1,2,\hdots,r-2$,

3.   $T'_iT'_j = T'_jT'_i$ if $|i-j|>1$.

From the equations $$T_i=\dfrac{1}{2}\{(q+q^{-1})T'_i+(q-q^{-1})\} \quad \text{for $i=1,2,\hdots,r-1$},$$ which are obtained from the definition of $T'_i$, we see that $T'_i$($i=1,2,\hdots,r$) generate $\mathcal{H}_{R_0,r}(q)$. The defining relations are obtained from a direct computation. 

Consider the following sets of monomials: $$\begin{split}
\mathcal{C}_1&=\{1,T_1\} \\
\mathcal{C}_2&=\{1,T_2,T_2T_1\} \\
\mathcal{C}_3&=\{1,T_3,T_3T_2,T_3T_2T_1\} \\
&: \\
\mathcal{C}_{r-1}&=\{1,T_{r-1},T_{r-1}T_{r-2},\hdots,T_{r-1}T_{r-2}{\cdots}T_1\}
\end{split}$$ We shall say that $M_1M_2{\cdots}M_{r-1}$ is a *monomial in $T_i$-normal form in $\mathcal{H}_{R_0,r}(q)$ if $M_i{\in}\mathcal{C}_i$ for $i=1,2,\hdots,r-1$. The following fact is well-known in the theory of the Iwahori-Hecke algebra.* 

$\operatorname{rank}_{R_0}\mathcal{H}_{R_0,r}(q)=r!$ and $$\mathcal{H}_{R_0,r}(q)=\bigoplus_{M_i{\in}\mathcal{C}_i}R_0M_1M_2{\cdots}M_{r-1}$$

We derive from this fact that all monomials in $T'_i$-normal form also constitute a basis of $\mathcal{H}_{R_0,r}(q)$.

Let $\mathcal{C}'_i=\{1,T'_i,T'_iT'_{i-1},\hdots,T'_iT'_{i-1}{\cdots}T'_1\}$ for $i=1,2,\hdots,r-1$. Then we have $$\mathcal{H}_{R_0,r}(q)=\bigoplus_{M'_i{\in}\mathcal{C}'_i}R_0M'_1M'_2{\cdots}M'_{r-1}.$$

Consider the map $f$ from $\{T_i\}_{i=1,2,\hdots,r-1}$ to $\{T'_i\}_{i=1,2,\hdots,r-1}$ which is defined by $f(T_i)=T'_i$. This map induces the $R_0$-endomorphism $\bar{f}$ of $\mathcal{H}_{R_0,r}(q)$. $\bar{f}$ is an $R_0$-isomorphism because the inverse $\bar{g}$ which is induced from the map $g$ from $\{T'_i\}_{i=1,2,\hdots,r-1}$ to $\{T_i\}_{i=1,2,\hdots,r-1}$ defined by $$g(T'_i)=\dfrac{1}{2}\{(q+q^{-1})T_i+(q-q^{-1})\}$$ exists.  Hence we conclude that $M'_1M'_2{\cdots}M'_{r-1}$($M'_i{\in}\mathcal{C}'_i$), which are images of $M_1M_2{\cdots}M_{r-1}$($M_i{\in}\mathcal{C}_i$), are linearly independent and constitute a basis of $\mathcal{H}_{R_0,r}(q)$.

Let $\mathcal{E}_r$ (respectively $\mathcal{O}_r$) be the set of all monomials in $T'_i$-normal form in $\mathcal{H}_{R_0,r}(q)$ which are products of even (respectively odd) numbers of $T'_i$’s.  Then the following holds. 

$|\mathcal{E}_r|=|\mathcal{O}_r|=2^{-1}r!$ for $r>1$.

The proof is done by induction on $r$. It is trivial for $r=2$. Let $M'_1M'_2{\cdots}M'_{r-1}{\in}\mathcal{E}_r$ with $r>2$ and $M'_i{\in}\mathcal{C}'_i$. Then $M'_1M'_2{\cdots}M'_{r-2}$ is considered as a monomial in $T'_i$-normal form in $\mathcal{H}_{R_0,r-1}(q)$. Let $(\mathcal{C}'_{r-1})^e$ (respectively $(\mathcal{C}'_{r-1})^o$) be the subset of $\mathcal{C}'_{r-1}$ which consists of products of even (respectively odd) numbers of $T'_i$’s.  We can readily see that $M'_1M'_2{\cdots}M'_{r-2}{\in}\mathcal{E}_{r-1}$ if and only if $M'_{r-1}{\in}(\mathcal{C}'_{r-1})^e$. By induction, $|\mathcal{E}_{r-1}|=|\mathcal{O}_{r-1}|=2^{-1}(r-1)!$ and hence we obtain the following $$|\mathcal{E}_r|=|(\mathcal{C}'_{r-1})^e||\mathcal{E}_{r-1}|
+|(\mathcal{C}'_{r-1})^o||\mathcal{O}_{r-1}| 
=r{\times}2^{-1}(r-1)!=2^{-1}r!$$ as desired. 

Now we characterize $\mathcal{H}_{R_0,r}^1(q)$ as a $q$-analogue of the alternating group. 

$\operatorname{rank}_{R_0}\mathcal{H}_{R_0,r}(q)=2\operatorname{rank}_{R_0}\mathcal{H}_{R_0,r}^1(q)$. Moreover $\mathcal{H}_{R_0,r}^1(q)$ is the subalgebra which consists of all the products of even numbers of $T'_i$’s. 

Let $\mathcal{H}_{R_0,r}^e(q)=\oplus_{M{\in}\mathcal{E}_r}R_0M$ and $\mathcal{H}_{R_0,r}^o(q)=\oplus_{M{\in}\mathcal{O}_r}R_0M$. Then we can see immediately that $\mathcal{H}_{R_0,r}(q)=\mathcal{H}_{R_0,r}^e(q){\oplus}\mathcal{H}_{R_0,r}^o(q)$. Furthermore we obtain $\mathcal{H}_{R_0,r}^e(q)=\mathcal{H}_{R_0,r}^1(q)$ and $\mathcal{H}_{R_0,r}^o(q)=\mathcal{H}_{R_0,r}^{-1}(q)$ from the property $\hat{T}'_i=-T'_i$ .  Combined with Lemma 3.5 we have $\operatorname{rank}_{R_0}\mathcal{H}_{R_0,r}(q)=2\operatorname{rank}_{R_0}\mathcal{H}_{R_0,r}^1(q)$. Let $\bar{\mathcal{H}}_{R_0,r}^1(q)$ be the set of all the linear combinations of products of even numbers of $T'_i$’s.  Obviously $\mathcal{H}_{R_0,r}^1(q){\subseteq}\bar{\mathcal{H}}_{R_0,r}^1(q)$. From the defining relations (A’1)-(A’3), one can see that if a monomial in $T'_i$-normal form which consists of even (respectively odd) number of $T'_i$’s is expressed in a linear combination of other expressions, then each term consists of even (respectively odd) number of $T'_i$’s.  Hence if we express an element of $\bar{\mathcal{H}}_{R_0,r}^1(q)$ by a linear combination of monomials in $T'_i$-normal form, each term is in $\mathcal{H}_{R_0,r}^1(q)$. Consequently $\bar{\mathcal{H}}_{R_0,r}^1(q)=\mathcal{H}_{R_0,r}^1(q)$.

When we suppose that $R_0=\mathbb{C}$ and take a limit $q{\rightarrow}1$, $\mathcal{H}_{\mathbb{C},r}^1(1)$ is isomorphic to the group algebra $\mathbb{C}[\mathfrak{A}_n]$ of the alternating group $\mathfrak{A}_n$.

$\mathcal{H}_{R_0,r}^1(q)$ is isomorphic to the $R_0$-algebra which is generated by $r-2$ elements $X_1,X_2,\hdots,X_{r-2}$ with the defining relations:

1.   $X_1^3=-\Big{(}\dfrac{q-q^{-1}}{q+q^{-1}}\Big{)}^2(X_1^2-X_1)+1$,

2.   $X_i^2=1 \qquad \text{for $i>1$}$,

3.   $(X_{i-1}X_i)^3=-\Big{(}\dfrac{q-q^{-1}}{q+q^{-1}}\Big{)}^2\Big{\{}(X_{i-1}X_i)^2-X_{i-1}X_i\Big{\}}+1 
    \qquad \text{for $i=2,3,\hdots ,r-2$}$,

4.   $(X_iX_j)^2=1 \qquad \text{whenever $|i-j|>1$}$.

An isomorphism is given by $X_i{\longrightarrow}T'_1T'_{i+1}$.

Next, we shall show that $\mathcal{H}_{R_0,r}(q)$ is a $\mathbb{Z}_2$-crossed product.  Since $T'_i$ has an inverse as itself, we can readily see from Proposition 3.6 that $\mathcal{H}_{R_0,r}^1(q)T'_1$ is an $R_0$-submodule of $\mathcal{H}_{R_0,r}(q)$ which consists of all linear combinations of products of odd numbers of $T'_i$’s.  Therefore, we obtain a direct sum decomposition of left $\mathcal{H}_{R_0,r}^1(q)$-modules: $$\mathcal{H}_{R_0,r}(q)=\mathcal{H}_{R_0,r}^1(q){\oplus}\mathcal{H}_{R_0,r}^1(q)T'_1.$$ Let $\mathbb{Z}_2={\langle}1,-1{\rangle}$ be a multiplicative group.  We define two maps $\psi_0$ and $\alpha_0$ to be $$\psi_0:\mathbb{Z}_2{\longrightarrow}\operatorname{Aut}(\mathcal{H}_{R_0,r}^1(q)),\quad 
\psi_0(1)(T)=T,\psi_0(-1)(T)=T'_1TT'_1,\quad T{\in}\mathcal{H}_{R_0,r}^1(q).$$ and $$\alpha_0:\mathbb{Z}_2{\times}\mathbb{Z}_2{\longrightarrow}
(\mathcal{H}_{R_0,r}^1(q))^{\times},\quad 
\alpha_0(\sigma,\tau)=1 \quad\text{for all $\sigma,\tau{\in}\mathbb{Z}_2$}.$$ Then we have the following immediately. 

$\psi_0$ and $\alpha_0$ satisfy (2.1)–(2.3). 

Thus, we obtain a $\mathbb{Z}_2$-crossed product $\mathcal{H}_{R_0,r}^1(q)_{\alpha_0}^{\psi_0}[\mathbb{Z}_2]$ from the crossed system $(\mathcal{H}_{R_0,r}^1(q),\mathbb{Z}_2,\psi_0,\alpha_0)$, where $\psi_0$ and $\alpha_0$ are given by (3.2) and (3.3) respectively. 

$\mathcal{H}_{R_0,r}(q)$ is isomorphic to $\mathcal{H}_{R_0,r}^1(q)_{\alpha_0}^{\psi_0}[\mathbb{Z}_2]$ as $R_0$-algebras. 

Since both $\mathcal{H}_{R_0,r}(q)$ and $\mathcal{H}_{R_0,r}^1(q)_{\alpha_0}^{\psi_0}[\mathbb{Z}_2]$ are free left $\mathcal{H}_{R_0,r}^1(q)$-modules, we may define an isomorphism of $\mathcal{H}_{R_0,r}^1(q)$-modules $${\iota_0}:\mathcal{H}_{R_0,r}^1(q)_{\alpha_0}^{\psi_0}[\mathbb{Z}_2]{\longrightarrow}
\mathcal{H}_{R_0,r}(q),\quad 
{\iota_0}(u_1)=1,{\iota_0}(u_{-1})=T'_1$$ From (2.4) and (3.2) and (3.3), we can determine the multiplication law as follows.  $$(a_1u_{\sigma})(a_2u_{\tau})=
\begin{cases}
a_1a_2u_{\sigma\tau}& \text{if $\sigma=1$,}\\
a_1T'_1a_2T'_1u_{\sigma\tau}& \text{if $\sigma=-1$,}
\end{cases}$$ where $a_1,a_2{\in}\mathcal{H}_{R_0,r}^1(q)$ and $\sigma,\tau{\in}\mathbb{Z}_2$. Therefore, we get four formulas in $\mathcal{H}_{R_0,r}(q)$ $$(a_11)(a_21)=a_1a_21,{\quad}(a_11)(a_2T'_1)=a_1a_2T'_1,{\quad}
(a_1T'_1)(a_21)=a_1T'_1a_2T'_1T'_1,{\quad}(a_1T'_1)(a_2T'_1)=a_1T'_1a_2T'_11,$$ which derive the conclusion that ${\iota_0}$ is an isomorphism of $R_0$-algebras. 

We denote by $\bar{K}$ an algebraic closure of a field $K$. Let $q$ be an indeterminate and $K=\mathbb{Q}(q)$. We shall show (split) semisimplicity of $\mathcal{H}_{\bar{K},r}^1(q)$ and the branching rule from $\mathcal{H}_{\bar{K},r}(q)$ to $\mathcal{H}_{\bar{K},r}^1(q)$. The manner of proof given here is credited to K.Uno, who sent me a letter enclosing the outline of this proof.  It is well-known that $\mathcal{H}_{\bar{K},r}(q)$ is split semisimple and that isomorphism classes of simple left $\mathcal{H}_{\bar{K},r}(q)$-modules are parametrized by Young diagrams of total size $n$. Let $\Lambda_r$ be the set of all Young diagrams of total size $r$. Then, provided that $\{M_{q,\lambda}|\lambda\in\Lambda_r\}$ is a set of all isomorphism classes of simple left $\mathcal{H}_{\bar{K},r}(q)$-modules and that $d_{\lambda}={\deg}M_{q,\lambda}$, we may write $$\mathcal{H}_{\bar{K},r}(q)=\bigoplus_{\lambda{\in}\Lambda_r}I_{q,\lambda}{\quad}
\big{(}I_{q,\lambda}{\cong}\operatorname{Mat}_{d_{\lambda}}(\bar{K})\big{)},$$ where each $I_{q,\lambda}$ is the homogeneous component corresponding to $\lambda$. $I_{q,\lambda}$ is isomorphic to a $d_{\lambda}{\times}d_{\lambda}$ matrix algebra $\operatorname{Mat}_{d_{\lambda}}(\bar{K})$ whose entries lie in $\bar{K}$. Since $\hat{}$ is an involution of $\mathcal{H}_{\bar{K},r}(q)$, for each $\lambda{\in}\Lambda_r$ there exists $\mu{\in}\Lambda_n$ such that $\hat{I}_{q,\lambda}=I_{q,\mu}$. Especially, $d_{\lambda}=d_{\mu}$ follows.  Dipper and James defined Specht modules $S^{\lambda}_K$ for Hecke algebras as irreducible submodules of regular modules in [@D-J], and improved the theory of representations of Hecke algebras in the series of articles such as [@D-J; @D-J2; @D-J3].  In particular, they showed in [@D-J3] that if $K$ is a field and $\mathcal{H}_{K,r}(q)$ is semisimple, then $\hat{S}^{\lambda}_K{\cong_{\mathcal{H}_{K,r}(q)}}S^{\lambda'}_K$ where $\lambda'$ denotes the transpose of $\lambda$. In this case, $M_{q,\lambda}$ is equivalent to $S^{\lambda}_{\bar{K}}$, thus $\mu=\lambda'$ follows.  We divide into two cases depending on whether $\lambda$ is self-conjugate or not. 

(Case1) $\lambda{\neq}\lambda'$:\
In this case, $\hat{}$ induces an involution on $I_{q,\lambda}{\oplus}I_{q,\lambda'}$. Let $\tilde{I}_{q,\lambda}=\big{\{}X+Y{\in}I_{q,\lambda}{\oplus}I_{q,\lambda'}|
(X+Y){\sphat}=X+Y\big{\}}$ for ${\lambda}\neq{\lambda'}$. Then we have $$\tilde{I}_{q,\lambda}=\big{\{}X+\hat{X}{\in}I_{q,\lambda}{\oplus}I_{q,\lambda'}|
X{\in}I_{q,\lambda}\big{\}}
{\cong}I_{q,\lambda}\big{(}{\cong}\operatorname{Mat}_{d_{\lambda}}(\bar{K})\big{)},$$ Thereby, the image of the regular representation of $\mathcal{H}^1_{\bar{K},r}(q)$ on $S^{\lambda}_{\bar{K}}$ is isomorphic to $I_{q,\lambda}$, so $\operatorname{res}^{\mathcal{H}_{\bar{K},r}(q)}_{\mathcal{H}^1_{\bar{K},r}(q)}
S^{\lambda}_{\bar{K}}$ is a simple left $\mathcal{H}^1_{\bar{K},r}(q)$-module.  If $g{\in}\mathcal{H}_{\bar{K},r}(q)$ satisfies $\hat{g}=g$, then the matrix coefficients with respect to the basis $x_1,x_2,\hdots,x_{d_{\lambda}}$ of $S^{\lambda}_{\bar{K}}$ is the same as those with respect to the basis $\hat{x}_1,\hat{x}_2,\hdots,\hat{x}_{d_{\lambda}}$ of $\hat{S}^{\lambda}_{\bar{K}}{\cong}
S^{\lambda'}_{\bar{K}}$. Hence we have the isomorphism of simple left $\mathcal{H}^1_{\bar{K},r}(q)$-modules as follows.  $$\operatorname{res}^{\mathcal{H}_{\bar{K},r}(q)}_{\mathcal{H}^1_{\bar{K},r}(q)}S^{\lambda}_{\bar{K}}{\cong}
\operatorname{res}^{\mathcal{H}_{\bar{K},r}(q)}_{\mathcal{H}^1_{\bar{K},r}(q)}S^{\lambda'}_{\bar{K}}.$$ (Case2) $\lambda=\lambda'$:

In this case, $\hat{}$ induces an involution on $I_{q,\lambda}{\cong}\operatorname{Mat}_{d_{\lambda}}(\bar{K})$. By Skolem-Noether Theorem, there exists an invertible element $P$ of $\operatorname{Mat}_{d_{\lambda}}(\bar{K})$ such that $\hat{X}=PXP^{-1}$ for all $X{\in}\operatorname{Mat}_{d_{\lambda}}(\bar{K})$. Since eigenvalues of $P$ are ${\pm}1$, We may also assume that $$P=
\begin{bmatrix}
1 &   &   &   &   & \\
  & \ddots &   &   &   & \\
  &   & 1 &   &   & \\
  &   &   & -1 &   & \\
  &   &   &    & \ddots  & \\
  &   &   &    &   & -1
\end{bmatrix}.$$ Then we may write $$P
\begin{bmatrix}
X_1 & X_2 \\
X_3  & X_4
\end{bmatrix}
P^{-1}=
\begin{bmatrix}
X_1 & -X_2 \\
-X_3  & X_4
\end{bmatrix},$$ for some submatrices $X_1,X_2,X_3,X_4$. Therefore, we obtain $$\big{\{}X{\in}\operatorname{Mat}_{d_{\lambda}}(\bar{K})|\hat{X}=X\big{\}}
=\bigg{\{}X{\in}\operatorname{Mat}_{d_{\lambda}}(\bar{K})|X=
\begin{bmatrix}
X_1 & 0 \\
0  & X_4
\end{bmatrix}\bigg{\}}.$$ Assuming that $1$ appears $m$ times in $P$. Then $\dim_{\bar{K}}\big{\{}X{\in}\operatorname{Mat}_{d_{\lambda}}(\bar{K})|\hat{X}=X\big{\}}
=m^2+(d_{\lambda}-m)^2$. We easily see that $$m^2+(d_{\lambda}-m)^2{\geq}\dfrac{d_{\lambda}^2}{2} \quad \text{equality holds iff $m=\dfrac{d_{\lambda}}{2}$}.$$ Combining this with (3.5) and the fact that $\dim_{\bar{K}}\mathcal{H}_{\bar{K},r}^1(q)=\dfrac{1}{2}\dim_{\bar{K}}\mathcal{H}_{\bar{K},r}(q)$ (Proposition 3.6), we deduce that $m=d_{\lambda}/2$. Thereby, $$\big{\{}X{\in}\operatorname{Mat}_{d_{\lambda}}(\bar{K})|\hat{X}=X\big{\}}
{\cong}\operatorname{Mat}_{d_{\lambda}/2}(\bar{K}){\oplus}\operatorname{Mat}_{d_{\lambda}/2}(\bar{K})$$ holds.  This means that $\operatorname{res}^{\mathcal{H}_{\bar{K},r}(q)}_{\mathcal{H}_{\bar{K},r}^1(q)}S^{\lambda}_{\bar{K}}$ decomposes into two simple left $\mathcal{H}^1_{\bar{K},r}(q)$-modules $\bar{S}^{\lambda+}_{\bar{K}}$ and $\bar{S}^{\lambda-}_{\bar{K}}$ which are mutually non-isomorphic.  Let $\tilde{I}_{q,\lambda}=\big{\{}X{\in}I_{q,\lambda}|\hat{X}=X\big{\}}$ and $\tilde{I}_{q,\lambda}^+$ (resp.  $\tilde{I}_{q,\lambda}^-$) be the homogeneous component corresponding to $\bar{S}^{\lambda+}_{\bar{K}}$ (resp.  $\bar{S}^{\lambda-}_{\bar{K}}$) for ${\lambda}\neq{\lambda'}$. Then (3.6) implies that $$\tilde{I}_{q,\lambda}=\tilde{I}_{q,\lambda}^+{\oplus}\tilde{I}_{q,\lambda}^-,$$ Summarizing our argument, we conclude that $\mathcal{H}_{\bar{K},r}^1(q)$ is isomorphic to the direct sum of minimal two-sided ideals as follows, $$\mathcal{H}_{\bar{K},r}^1(q)=
\big{\{}\bigoplus_{\lambda{\in}\Lambda_r,\lambda>\lambda'}\tilde{I}_{q,\lambda}\big{\}}
\bigoplus\Big{[}\bigoplus_{\lambda{\in}\Lambda_r,\lambda=\lambda'}
\big{\{}\tilde{I}_{q,\lambda}^+{\oplus}\tilde{I}_{q,\lambda}^-\big{\}}\Big{]},$$ where $<$ denotes the lexicographic order on $\Lambda_r$; $\lambda=(\lambda_1,\lambda_2,\hdots)<
\mu=(\mu_1,\mu_2,\hdots)$ iff $\lambda_k<\mu_k$ for the smallest $k$ such that $\lambda_k{\neq}\mu_k$. Consequently we have proved the following result. 

Let $q$ be an indeterminate and $K=\mathbb{Q}(q)$.\
If $\lambda'{\neq}\lambda$, then $\bar{S}^{\lambda}_{\bar{K}}$ is simple and $\bar{S}^{\lambda}_{\bar{K}}{\cong}\bar{S}^{\lambda'}_{\bar{K}}$ where $\bar{S}^{\lambda}_{\bar{K}}$ denotes $\operatorname{res}^{\mathcal{H}_{\bar{K},r}(q)}_{\mathcal{H}_{\bar{K},r}^1(q)}S^{\lambda}_{\bar{K}}$.\
If $\lambda'=\lambda$, then $\bar{S}^{\lambda}_{\bar{K}}$ decomposes into mutually non-isomorphic two simple left $\mathcal{H}^1_{\bar{K},r}(q)$-modules $\bar{S}^{\lambda+}_{\bar{K}}$ and $\bar{S}^{\lambda-}_{\bar{K}}$.\
The simple left $\mathcal{H}^1_{\bar{K},r}(q)$-modules $\bar{S}^{\lambda}_{\bar{K}}, \bar{S}^{\mu+}_{\bar{K}}, \bar{S}^{\mu-}_{\bar{K}} 
(\lambda,\mu{\in}\Lambda_r,\lambda>\lambda',\mu=\mu')$ constitute a basic set of simple left $\mathcal{H}^1_{\bar{K},r}(q)$-modules.  Moreover, $\mathcal{H}_{\bar{K},r}^1(q)$ is a split semisimple $\bar{K}$-algebra and its homogeneous decomposition is as follows: $$\mathcal{H}_{\bar{K},r}^1(q)=
\big{\{}\bigoplus_{\lambda{\in}\Lambda_r,\lambda>\lambda'}\tilde{I}_{q,\lambda}\big{\}}
\bigoplus\Big{[}\bigoplus_{\lambda{\in}\Lambda_r,\lambda=\lambda'}
\big{\{}\tilde{I}_{q,\lambda}^+{\oplus}\tilde{I}_{q,\lambda}^-\big{\}}\Big{]},$$ where each $\tilde{I}_{q,\lambda}$ is the homogeneous component corresponding to $\lambda$ which is isomorphic to $\operatorname{Mat}_{d_{\lambda}}(\bar{K})$ and each direct sum $\tilde{I}_{q,\lambda}^+{\oplus}\tilde{I}_{q,\lambda}^-$ consists of two homogeneous components $\tilde{I}_{q,\lambda}^+$ and $\tilde{I}_{q,\lambda}^-$ both corresponding to $\lambda$ which are isomorphic to $\operatorname{Mat}_{d_{\lambda}/2}(\bar{K})$.

Schur-Weyl reciprocity between $U^\sigma_q\big{(}{\mathfrak{g}\mathfrak{l}}(m,n)\big{)}$ and $\mathcal{H}_{K,r}(q)$
===================================================================================================================

In our previous paper [@Mit2], we defined the $q$-permutation representation and established Schur-Weyl reciprocity between the quantum superalgebra $U^\sigma_q\big{(}{\mathfrak{g}\mathfrak{l}}(m,n)\big{)}$ and $\mathcal{H}_{K,r}(q)$. Schur-Weyl reciprocity between the general Lie superalgebra ${\mathfrak{g}\mathfrak{l}}(m,n)$ and the symmetric group $\mathfrak{S}_r$ was established in, for example, [@B-R; @Ser].  In this section, we shall review the $q$-permutation representation and the vector representation of $U^\sigma_q\big{(}{\mathfrak{g}\mathfrak{l}}(m,n)\big{)}$.

Let $V={\oplus}_{k=1}^{m+n}Kv_k$ be a $\mathbb{Z}_2$-graded $K$-module of rank $m+n$. By $\mathbb{Z}_2$-graded, we mean that $V$ is a direct sum of two submodule $V_{\overline{0}}=\oplus_{k=1}^{m}Kv_k$ and $V_{\overline{1}}=\oplus_{k=m+1}^{m+n}Kv_k$, and that for each homogeneous element the degree map $|{\cdot}|$ $$|v|=
\begin{cases}
0& \text{if $v{\in}V_{\overline{0}}$},\\
1& \text{if $v{\in}V_{\overline{1}}$,}
\end{cases}$$ is given. 

Let $\pi_r$ be the $q$-permutation representation of $\mathcal{H}_{K,r}(q)$ on the tensor space $V^{{\otimes}r}$. $\pi_r$ is given by $\pi_r(T_i)=\operatorname{Id}^{{\otimes}i-1}{\otimes}T{\otimes}\operatorname{Id}^{{\otimes}r-i-1}$ ($i=1,2,\hdots,r-1$) where $T$ is the operator on $V{\otimes}V$ defined by $$Tv_k{\otimes}v_l=
\begin{cases}
\dfrac{(-1)^{|v_k|}(q+q^{-1})+q-q^{-1}}{2}v_k{\otimes}v_l& \text{if $k=l$,}\\
(-1)^{|v_k||v_l|}v_l{\otimes}v_k+(q-q^{-1})v_k{\otimes}v_l& \text{if $k<l$,}\\
(-1)^{|v_k||v_l|}v_l{\otimes}v_k& \text{if $k>l$.}
\end{cases}$$ and $\operatorname{Id}$ is the identity operator on $V$. This representation $\pi_r$ is reduced to the (normal) $q$-permutation representation of $\mathcal{H}_{K,r}(q)$ with $n=0$ and to the sign permutation representation of $\mathfrak{S}_r$ with $q{\rightarrow}1$. Let $T'$ be the operator defined by $$T'=\dfrac{2T-(q-q^{-1})}{q+q^{-1}}$$ Then $T'$ is determined by $$T'v_k{\otimes}v_l=
\begin{cases}
(-1)^{|v_k|}v_k{\otimes}v_l& \text{if $k=l$,}\\
\dfrac{2(-1)^{|v_k||v_l|}}{q+q^{-1}}v_l{\otimes}v_k+\dfrac{q-q^{-1}}{q+q^{-1}}v_k{\otimes}v_l& 
\text{if $k<l$,}\\
\dfrac{2(-1)^{|v_k||v_l|}}{q+q^{-1}}v_l{\otimes}v_k-\dfrac{q-q^{-1}}{q+q^{-1}}v_k{\otimes}v_l& 
\text{if $k>l$.}
\end{cases}$$ $\pi_r$ is also given by $\pi_r(T'_i)=\operatorname{Id}^{{\otimes}i-1}{\otimes}T'{\otimes}\operatorname{Id}^{{\otimes}r-i-1}$ ($i=1,2,\hdots,r-1$)

Next we shall review quantum superalgebras and their vector representations.  Several definitions of quantum superalgebras appear in, for example, [@B-K-K; @K-T; @Yamane].  The much complete definition and detailed observations of quantum superalgebras can be found in [@Yamane].  In this paper, we obey the manner of definition of $U^\sigma_q\big{(}{\mathfrak{g}\mathfrak{l}}(m,n)\big{)}$ in [@B-K-K].  The method of construction of superalgebra depends on [@Kac] basically.  Let $\Pi=\{\alpha_i\}_{i{\in}I}$ be a set of simple roots with the index set $I=\{1,\hdots,r\}$. We assume that $I$ is a disjoint union of two subsets $I_{\rm even}$ and $I_{\rm odd}$. We define a map $p:I{\longrightarrow}\{0,1\}$ to be such that $$p(i)=
\begin{cases}
0& \text{if $i{\in}I_{\rm even}$},\\
1& \text{if $i{\in}I_{\rm odd}$.}
\end{cases}$$ Let $P$ be a free $\mathbb{Z}$-module which includes all $\alpha_i{\in}P$($i{\in}I$).  We assume that a $\mathbb{Q}$-valued symmetric bilinear form on $P$ $(\cdot,\cdot) : P{\times}P{\longrightarrow}{\mathbb{Q}}$ is defined and that the simple coroots $h_i{\in}P^*$($i{\in}I$) are given as data.  The natural pairing ${\langle}\cdot,\cdot{\rangle} : P^*{\times}P{\longrightarrow}{\mathbb{Z}}$ between $P$ and $P^*$ is assumed to satisfy $${\langle}h_i,\alpha_j{\rangle}=
\begin{cases}
2& \text{if $i=j$ and $i{\in}I_{\rm even}$,}\\
0\quad\text{or}\quad2& \text{if $i=j$ and $i{\in}I_{\rm odd}$,}\\
{\leq}0& \text{if $i{\neq}j$.}
\end{cases}$$ We denote by $\Pi^{\vee}=\{h_i|i{\in}I\}$ the set of all coroots.  Furthermore, for each $i{\in}I$ we assume that there exists a nonzero integer $\ell_i$ such that $\ell_i{\langle}h_i,\lambda{\rangle}=(\alpha_i,\lambda)$ for every $\lambda{\in}P$. Then we immediately have the Cartan matrix $A=[{\langle}h_i,\alpha_j{\rangle}]_{ij}$ is symmetrizable because $\ell_i{\langle}h_i,\alpha_j{\rangle}=(\alpha_i,\alpha_j)
=(\alpha_j,\alpha_i)=\ell_j{\langle}h_j,\alpha_i{\rangle}$. We mention that the symmetrized matrix is $A^{\rm sym}=\operatorname{diag}(\ell_1,\hdots,\ell_r)A=
[(\alpha_i,\alpha_j)]_{ij}$. Let $\mathfrak{h}=P^*{\otimes}_{\mathbb{Z}}\mathbb{Q}$. Then $\Phi=(\mathfrak{h},\Pi^{\vee},\Pi)$ is said to be a fundamental root data associated to $A$.  Let ${\mathfrak{g}}={\mathfrak{g}}(\Phi)$ be the contragredient Lie superalgebra obtained from $\Phi$ and $p$. The quantized enveloping algebra $U_q^\sigma({\mathfrak{g}})$ is the unital associative algebra over $K=\mathbb{Q}(q)$ with generators $q^h (h{\in}P^*),e_i,f_i (i{\in}I)$ and an additional element $\sigma$ which satisfy the following defining relations:

1.   $q^h=1$ for $h=0$,

2.   $q^{h_1}q^{h_2}=q^{h_1+h_2}$ for $h_1,h_2{\in}P^*$,

3.   $q^he_i=q^{{\langle}h,\alpha_j{\rangle}}e_iq^h$ for $h{\in}P^*$ and $i{\in}I$,

4.   $q^hf_i=q^{-{\langle}h,\alpha_j{\rangle}}f_iq^h$ for $h{\in}P^*$ and $i{\in}I$,

5.   $[e_i,f_j]=\delta_{ij}\dfrac{q^{\ell_ih_i}-q^{-\ell_ih_i}}{q^{\ell_i}-q^{-\ell_i}}$ for $i,j{\in}I$,

6.   $\sigma^2=1$,

7.   $q^h{\sigma}={\sigma}q^h$ for $h{\in}P^*$,

8.   $e_i{\sigma}=(-1)^{p(i)}{\sigma}e_i$ for $i{\in}I$,

9.   $f_i{\sigma}=(-1)^{p(i)}{\sigma}f_i$ for $i{\in}I$,

where $[e_i,f_j]$ means the supercommutator $$[e_i,f_j]=e_if_j-(-1)^{p(i)p(j)}f_je_i.$$ We assume further conditions:

1.   If $a{\in}\sum_{i{\in}I}U_q(\mathfrak{n}_+)e_iU_q(\mathfrak{n}_+)$ satisfies $f_ia{\in}U_q(\mathfrak{n}_+)f_i$ for all $i{\in}I$, then $a=0$,

2.   If $a{\in}\sum_{i{\in}I}U_q(\mathfrak{n}_-)f_iU_q(\mathfrak{n}_-)$ satisfies $e_ia{\in}U_q(\mathfrak{n}_-)e_i$ for all $i{\in}I$, then $a=0$,

where $U_q(\mathfrak{n}_+)$ (respectively $U_q(\mathfrak{n}_-)$) is the subalgebra of $U_q^\sigma({\mathfrak{g}})$ generated by $\{e_i|i{\in}I\}$ (respectively $\{f_i|i{\in}I\}$).  $U_q^\sigma({\mathfrak{g}})$ is a Hopf algebra whose comultiplication $\triangle_{\sigma}$, counit $\varepsilon_{\sigma}$, antipode $S_{\sigma}$ are as follows.  $$\begin{split}
&\triangle_{\sigma}(\sigma)=\sigma{\otimes}\sigma, \\
&\triangle_{\sigma}(q^h)=q^h{\otimes}q^h \quad \text{for $h{\in}P^*$}, \\
&\triangle_{\sigma}(e_i)=e_i{\otimes}q^{-\ell_ih_i}+\sigma^{p(i)}{\otimes}e_i \quad \text{for $i{\in}I$}, \\
&\triangle_{\sigma}(f_i)=f_i{\otimes}1+\sigma^{p(i)}q^{\ell_ih_i}{\otimes}f_i \quad \text{for $i{\in}I$}, \\
&\varepsilon_{\sigma}(\sigma)=\varepsilon_{\sigma}(q^h)=1 
\quad \text{for $h{\in}P^*$}, \quad 
\varepsilon_{\sigma}(e_i)=\varepsilon_{\sigma}(f_i)=0 \quad \text{for $i{\in}I$}, \\
&S_{\sigma}(\sigma)=\sigma, \quad S_{\sigma}(q^{{\pm}h})=q^{{\mp}h} 
\quad \text{for $h{\in}P^*$}, \\
&S_{\sigma}(e_i)=-\sigma^{p(i)}e_iq^{\ell_ih_i},{\quad}S_{\sigma}(f_i)=-\sigma^{p(i)}q^{-\ell_ih_i}f_i 
\quad \text{for $i{\in}I$}.  
\end{split}$$ The quantized enveloping algebra $U^\sigma_q\big{(}{\mathfrak{g}\mathfrak{l}}(m,n)\big{)}$ is obtained from the fundamental root data as follows. 

-   $I=I_{\rm even}{\cup}I_{\rm odd}$ is defined by $I_{\rm even}=\{1,2,\hdots,m-1,m+1,\hdots,m+n-1\}$ and $I_{\rm odd}=\{m\}$,

-   $P=\oplus_{b{\in}B}\mathbb{Z}\epsilon_b$, where $B=B_+{\cup}B_-$ with $B_+=\{1,\hdots,m\}$ and $B_-=\{m+1,\hdots,m+n\}$,

-   $(\cdot,\cdot):P{\times}P{\longrightarrow}\mathbb{Q}$ is the symmetric bilinear form on $P$ defined by $$(\epsilon_a,\epsilon_{a'})=
    \begin{cases}
    1& \text{if $a=a'{\in}B_+$,}\\
    -1& \text{if $a=a'{\in}B_-$,}\\
    0& \text{otherwise,}
    \end{cases}$$

-   $\Pi=\{\alpha_i|i{\in}I\}$ is defined by $\alpha_i=\epsilon_i-\epsilon_{i+1}$,

-   $\Pi^{\vee}=\{h_i|i{\in}I\}$ is uniquely determined by the formula $\ell_i{\langle}h_i,\lambda{\rangle}=(\alpha_i,\lambda)$ for any $\lambda{\in}P$,\
    where $$\ell_i=
    \begin{cases}
    1& \text{if $i=1,\hdots,m$,}\\
    -1& \text{if $i=m+1,\hdots,m+n-1$.}
    \end{cases}$$

The vector representation ($\rho,V$) of $U^\sigma_q\big{(}{\mathfrak{g}\mathfrak{l}}(m,n)\big{)}$ on the $\mathbb{Z}_2$-graded vector space $V=V_{\overline{0}}{\oplus}V_{\overline{1}}$ (recall that $V_{\overline{0}}=\oplus_{i=1}^{m}Kv_i,V_{\overline{1}}=\oplus_{i=m+1}^{m+n}Kv_i$) is defined by (see [@B-K-K]) $$\begin{split}
&\rho(\sigma)v_j=(-1)^{|v_j|}v_j \quad \text{for $j=1,\hdots,m+n$}, \\
&\rho(q^h)v_j=q^{\epsilon_j(h)}v_j \quad \text{for $h{\in}P^*,j=1,\hdots,m+n$}, \\
&\rho(e_j)v_{j+1}=v_j \quad \text{for $j=1,\hdots,m+n-1$}, \\
&\rho(f_j)v_j=v_{j+1} \quad \text{for $j=1,\hdots,m+n-1$}, \\
&\text{otherwise $0$.}
\end{split}$$ The vector representation ${\rho}_r$ on $V^{{\otimes}r}$ is given by ${\rho}_r(x)=\rho^{{\otimes}r}\circ\triangle^{(r-1)}(x)$ where $\triangle^{(1)}=\triangle_\sigma$ and $\triangle^{(k)}=(\triangle_\sigma{\otimes}\operatorname{Id}^{{\otimes}k-1})\triangle^{(k-1)}$ inductively.  One can readily see that $\rho_r$ is of the following form: $$\begin{split}
&{\rho}_r(\sigma)=\rho(\sigma)^{{\otimes}r}, \\
&{\rho}_r(q^h)=\rho(q^h)^{{\otimes}r} \quad \text{for $h{\in}P^*$}, \\
&{\rho}_r(e_i)=\sum_{k=1}^N\rho(\sigma^{p(i)})^{{\otimes}k-1}{\otimes}\rho(e_i) 
{\otimes}\rho(q^{-\ell_ih_i})^{{\otimes}r-k} \quad 
\text{for $i{\in}I$}, \\
&{\rho}_r(f_i)=\sum_{k=1}^r\rho(\sigma^{p(i)}q^{\ell_ih_i})^{{\otimes}k-1}{\otimes}\rho(f_i)
{\otimes}\operatorname{Id}^{{\otimes}r-k} \quad 
\text{for $i{\in}I$}. 
\end{split}$$ Our precedent works in [@Mit2] are Theorem 4.1 and Theorem 4.2. 

Let $\mathcal{A}_q=\pi_r\big{(}\mathcal{H}_{K,r}(q)\big{)}$ and $\mathcal{B}_q=\rho_r\big{(}U^\sigma_q\big{(}{\mathfrak{g}\mathfrak{l}}(m,n)\big{)}\big{)}$. Then we have $\operatorname{End}_{\mathcal{B}_q}V^{{\otimes}r}
=\mathcal{A}_q$ and $\operatorname{End}_{\mathcal{A}_q}V^{{\otimes}r}
=\mathcal{B}_q$.

Let $\bar{U}^\sigma_q\big{(}{\mathfrak{g}\mathfrak{l}}(m,n)\big{)}=U^\sigma_q\big{(}{\mathfrak{g}\mathfrak{l}}(m,n)\big{)}
{\otimes_{K}}\bar{K}$, $\bar{\mathcal{A}}_q=\mathcal{A}_q{\otimes_{K}}\bar{K}$ and $\bar{\mathcal{B}}_q=\mathcal{B}_q{\otimes_{K}}\bar{K}$. Then, $\pi_r\big{(}\mathcal{H}_{\bar{K},r}(q)\big{)}=\bar{\mathcal{A}}_q$ and $\rho_r\big{(}\bar{U}^\sigma_q\big{(}{\mathfrak{g}\mathfrak{l}}(m,n)\big{)}\big{)}=\bar{\mathcal{B}}_q$ as $\bar{K}$-algebras of operators on $\bar{V}^{{\otimes}r}=(V{\otimes_{K}}\bar{K})^{{\otimes}r}$. We notice that $\operatorname{End}_{\bar{\mathcal{B}}_q}\bar{V}^{{\otimes}r}
=\bar{\mathcal{A}}_q$ and $\operatorname{End}_{\bar{\mathcal{A}}_q}\bar{V}^{{\otimes}r}
=\bar{\mathcal{B}}_q$ hold. 

Let $H(m,n;r)=\{\lambda=(\lambda_1,\lambda_2,\hdots){\in}\Lambda_r|\lambda_j{\leq}n 
\text{ if } j>m\}$. Diagrams of elements of $H(m,n;r)$ are exactly those contained in the $(m,n)$-hooks.  Then the following holds. 

$\bar{\mathcal{A}}_q=\bigoplus_{\lambda{\in}H(m,n;r)}\bar{\mathcal{A}}_{q,\lambda}$ where each $\bar{\mathcal{A}}_{q,\lambda}=\pi_r(I_{q,\lambda})({\cong}I_{q,\lambda})$ is the image of the homogeneous component $I_{q,\lambda}$ corresponding to $\lambda{\in}H(m,n;r)$ as in (3.4).  Moreover, we have $\bar{V}^{{\otimes}r}=\bigoplus_{\lambda{\in}H(m,n;r)}H_{\lambda}{\otimes}V_{\lambda}$ where $H_{\lambda}$’s are mutually non-isomorphic simple left $\mathcal{H}_{\bar{K},r}(q)$-modules indexed by the elements of $H(m,n;r)$, and $V_{\lambda}$’s are mutually non-isomorphic simple left $\bar{U}^\sigma_q\big{(}{\mathfrak{g}\mathfrak{l}}(m,n)\big{)}$-modules indexed by the elements of $H(m,n;r)$.

Schur-Weyl reciprocity for $\mathcal{H}_{K,r}^1(q)$ in case of $m=n$
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Schur-Weyl reciprocity for the alternating group $\mathfrak{A}_r$ has been researched by Regev in [@Regev].  The paper [@Regev] showed that if $\dim V_{\overline{0}}=\dim V_{\overline{1}}$ under the base field $\mathbb{C}$, then the centralizer algebra $\operatorname{End}_{\mathfrak{A}_r}V^{{\otimes}r}$ has remarkable property; $\mathbb{Z}_2$-crossed product for $\operatorname{End}_{\mathfrak{S}_r}V^{{\otimes}r}$. In this paper, we establish a $q$-analogue extension of Regev’s result.  We also show that Schur-Weyl reciprocity is valid even if the base field is $\mathbb{Q}(q)$.

We set $K=\mathbb{Q}(q)$ in succession, and denote $\mathcal{C}_q=\pi_r\big{(}\mathcal{H}_{K,r}^1(q)\big{)}$. Let us consider the relation between $\operatorname{End}_{\mathcal{A}_q}V^{{\otimes}r}$ and $\operatorname{End}_{\mathcal{C}_q}V^{{\otimes}r}$. Since $\mathcal{H}_{K,r}^1(q){\subsetneq}\mathcal{H}_{K,r}(q)$, we immediately have $$\mathcal{C}_q{\subseteq}\mathcal{A}_q \quad 
\text{and} \quad 
\operatorname{End}_{\mathcal{A}_q}V^{{\otimes}r}{\subseteq}
\operatorname{End}_{\mathcal{C}_q}V^{{\otimes}r}.$$ Recall that $V$ is an $m+n$-dimensional $\mathbb{Z}_2$-graded vector space over $K$. In this section, we analyze the structure of $\operatorname{End}_{\mathcal{C}_q}V^{{\otimes}r}$ in case of $m=n$. We will consider the general case in the next section. 

Assume that $m=n$. Recall that $$\mathcal{B}_q=\{f{\in}\operatorname{End}_{K}V^{{\otimes}r}
|\pi_r(T'_i)f=f\pi_r(T'_i) \text{ for $i=1,2,\hdots,r-1$}\} 
=\operatorname{End}_{\mathcal{A}_q}V^{{\otimes}r}.$$ We define $\mathcal{B}_q^{\dagger},\mathcal{C}_q$ to be the subspaces of $\operatorname{End}_{K}V^{{\otimes}r}$ as follows.  $$\begin{split}
\mathcal{B}_q^{\dagger}&=\{f{\in}\operatorname{End}_{K}V^{{\otimes}r}
|\pi_r(T'_i)f=-f\pi_r(T'_i) \text{ for $i=1,2,\hdots,r-1$}\}, \\
\mathcal{D}_q&=\{f{\in}\operatorname{End}_{K}V^{{\otimes}r}
|\pi_r(T'_1T'_{i+1})f=f\pi_r(T'_1T'_{i+1}) \text{ for $i=1,2,\hdots,r-2$}\} 
=\operatorname{End}_{\mathcal{C}_q}V^{{\otimes}r}. 
\end{split}$$

$\mathcal{D}_q=\mathcal{B}_q{\bigoplus}\mathcal{B}_q^{\dagger}$.

It is clear that the sum is direct.  $\supseteq$ is also obvious.  We notice that if $f{\in}\mathcal{D}_q$, then $\pi_r(T'_j)f\pi_r(T'_j)=\pi_r(T'_1)f\pi_r(T'_1)$ for $j=2,3,\hdots,r-1$ from the definition of $\mathcal{D}_q$.

In general, we may write $$f=\dfrac{1}{2}\big{(}f+\pi_r(T'_1)f\pi_r(T'_1)\big{)}+\dfrac{1}{2}\big{(}f-\pi_r(T'_1)f\pi_r(T'_1)\big{)}.$$ If $f{\in}\mathcal{D}_q$, then one can readily see $$\begin{split}
\pi_r(T'_i)\dfrac{1}{2}\big{(}f+\pi_r(T'_1)f\pi_r(T'_1)\big{)}
&=\dfrac{1}{2}\big{(}\pi_r(T'_i)f\pi_r(T'_i)^2+\pi_r(T'_i)\pi_r(T'_1)f\pi_r(T'_1)\big{)}\\
&=\dfrac{1}{2}\big{(}\pi_r(T'_1)f\pi_r(T'_1)\pi_r(T'_i)+f\pi_r(T'_i)\pi_r(T'_1)^2\big{)}\\
&=\dfrac{1}{2}\big{(}\pi_r(T'_1)f\pi_r(T'_1)+f\big{)}\pi_r(T'_i). 
\end{split}$$ Hence we have $2^{-1}\big{(}f+\pi_r(T'_1)f\pi_r(T'_1)\big{)}{\in}\mathcal{B}_q$. In the same fashion, we also get $2^{-1}\big{(}f-\pi_r(T'_1)f\pi_r(T'_1)\big{)}{\in}\mathcal{B}_q^{\dagger}$. Thus we have proved the reverse inclusion $\subseteq$.

We notice that if $V$ is $\mathbb{Z}_2$-graded, then $\operatorname{End}_{K}V$ is also $\mathbb{Z}_2$-graded.  Namely, $$\operatorname{End}_{K}V=(\operatorname{End}_{K}V)_{\overline{0}}
{\oplus}(\operatorname{End}_{K}V)_{\overline{1}},$$ where $$\begin{split}
(\operatorname{End}_{K}V)_{\overline{0}}
&=\big{\{}f{\in}\operatorname{End}_{K}V|f(V_{\overline{i}}){\subseteq}V_{\overline{i}},
i{\in}\{0,1\}\big{\}}, \\
(\operatorname{End}_{K}V)_{\overline{1}}
&=\big{\{}f{\in}\operatorname{End}_{K}V|f(V_{\overline{i}}){\subseteq}V_{\overline{i+1}},
i{\in}\{0,1\}\big{\}}. 
\end{split}$$ Let $\varphi{\in}\operatorname{End}_{K}V$ be given by $\varphi(v_i)=v_{2m-i+1}$. Obviously, $\varphi{\in}(\operatorname{End}_{K}V)_{\overline{1}}$. Let $\varphi^{{\otimes}r}$ be the tensor product of $\varphi$. In general, for homogeneous elements $f_1,f_2,\hdots,f_r$ of $\operatorname{End}_{K}V$, $f_1{\otimes}f_2{\otimes}\cdots{\otimes}f_r{\in}\operatorname{End}_{K}V^{{\otimes}r}$ is given by $$f_1{\otimes}\cdots{\otimes}f_r(u_1{\otimes}\cdots{\otimes}u_r)
=(-1)^{\sum_{i=2}^r(|f_i|\sum_{j=1}^{i-1}|u_j|)}
f_1(u_1){\otimes}\cdots{\otimes}f_r(u_r),$$ where $u_1,\hdots,u_r$ are homogeneous elements of $V$. Hence we have $$\varphi^{{\otimes}r}(u_1{\otimes}\cdots{\otimes}u_r)
=(-1)^{\sum_{i=2}^r(\sum_{j=1}^{i-1}|u_j|)}
\varphi(u_1){\otimes}\cdots{\otimes}\varphi(u_r).$$

$\varphi^{{\otimes}r}$ is an isomorphism of $K$-vector space $V^{{\otimes}r}$ which satisfies following properties: $$(\varphi^{{\otimes}r})^2=(-1)^{r(r-1)/2}\operatorname{I}$$ and $$\pi_r(T'_i)\varphi^{{\otimes}r}=-\varphi^{{\otimes}r}\pi_r(T'_i)\quad\text{for $i=1,2,\hdots,r-1$},$$ where $\operatorname{I}$ is the identity operator on $V^{{\otimes}r}$.

For homogeneous elements $u_1,\hdots,u_r$ of $V$, one can readily see that $$\begin{split}
{(\varphi^{{\otimes}r})^2}(u_1{\otimes}\cdots{\otimes}u_r)
&=(-1)^{\sum_{i=2}^r(\sum_{j=1}^{i-1}|u_j|)}
\varphi(u_1){\otimes}\cdots{\otimes}\varphi(u_r) \\
&=(-1)^{\sum_{i=2}^r(\sum_{j=1}^{i-1}(|u_j|+|\varphi(u_j)|)}
u_1{\otimes}\cdots{\otimes}u_r \\
&=(-1)^{\sum_{i=2}^r\sum_{j=1}^{i-1}1}
u_1{\otimes}\cdots{\otimes}u_r \\
&=(-1)^{(r-1)r/2}
u_1{\otimes}\cdots{\otimes}u_r. 
\end{split}$$ Hence $(\varphi^{{\otimes}r})^2=(-1)^{r(r-1)/2}\operatorname{I}$, and $\varphi^{{\otimes}r}$ has the inverse $(\varphi^{{\otimes}r})^{-1}=(-1)^{(r-1)r/2}\varphi^{{\otimes}r}$.

To prove the last statement, we check three cases of the definition of $\pi_r$ which has appeared in (4.2).  We notice that it suffice to prove only the case $r=2$.\
Case1 : $k=l$ $$\begin{split}
{\varphi^{{\otimes}2}}T'(v_k{\otimes}v_l)&=(-1)^{|v_k|}\varphi^{{\otimes}2}(v_k{\otimes}v_k) \\
&=v_{2m-k+1}{\otimes}v_{2m-k+1}
\end{split}$$ $$\begin{split}
T'{\varphi^{{\otimes}2}}(v_k{\otimes}v_l)&=(-1)^{|v_k|}T'(v_{2m-k+1}{\otimes}v_{2m-k+1}) \\
&=(-1)^{|v_k|}(-1)^{|v_{2m-k+1}|}v_{2m-k+1}{\otimes}v_{2m-k+1} \\
&=-v_{2m-k+1}{\otimes}v_{2m-k+1}
\end{split}$$ Case2 : $k<l$ $$\begin{split}
{\varphi^{{\otimes}2}}T'(v_k{\otimes}v_l)&=\dfrac{2(-1)^{|v_k||v_l|}}{q+q^{-1}}{\varphi^{{\otimes}2}}(v_l{\otimes}v_k)
+\dfrac{q-q^{-1}}{q+q^{-1}}{\varphi^{{\otimes}2}}(v_k{\otimes}v_l) \\
&=\dfrac{2(-1)^{|v_k||v_l|+|v_l|}}{q+q^{-1}}(v_{2m-l+1}{\otimes}v_{2m-k+1})
+\dfrac{q-q^{-1}}{q+q^{-1}}(-1)^{|v_k|}(v_{2m-k+1}{\otimes}v_{2m-l+1})
\end{split}$$ $$\begin{split}
T'{\varphi^{{\otimes}2}}(v_k{\otimes}v_l)&=(-1)^{|v_k|}T'(v_{2m-k+1}{\otimes}v_{2m-l+1}) \\
&=\dfrac{2(-1)^{|v_k|+|v_{2m-k+1}||v_{2m-l+1}|}}{q+q^{-1}}(v_{2m-l+1}{\otimes}v_{2m-k+1})
-\dfrac{q-q^{-1}}{q+q^{-1}}(-1)^{|v_k|}(v_{2m-k+1}{\otimes}v_{2m-l+1})
\end{split}$$ If $|v_k|=|v_l|=0$, then $(-1)^{|v_k||v_l|+|v_l|}=1$ and $(-1)^{|v_k|+|v_{2m-k+1}||v_{2m-l+1}|}=-1$.\
If $|v_k|=0,|v_l|=1$, then $(-1)^{|v_k||v_l|+|v_l|}=-1$ and $(-1)^{|v_k|+|v_{2m-k+1}||v_{2m-l+1}|}=1$.\
If $|v_k|=1,|v_l|=0$, then $(-1)^{|v_k||v_l|+|v_l|}=1$ and $(-1)^{|v_k|+|v_{2m-k+1}||v_{2m-l+1}|}=-1$.\
If $|v_k|=|v_l|=1$, then $(-1)^{|v_k||v_l|+|v_l|}=1$ and $(-1)^{|v_k|+|v_{2m-k+1}||v_{2m-l+1}|}=-1$.\
After all, ${\varphi^{{\otimes}2}}T'(v_k{\otimes}v_l)=-T'{\varphi^{{\otimes}2}}(v_k{\otimes}v_l)$ holds.\
Case3 : $k>l$ $$\begin{split}
{\varphi^{{\otimes}2}}T'(v_k{\otimes}v_l)&=\dfrac{2(-1)^{|v_k||v_l|}}{q+q^{-1}}{\varphi^{{\otimes}2}}(v_l{\otimes}v_k)
-\dfrac{q-q^{-1}}{q+q^{-1}}{\varphi^{{\otimes}2}}(v_k{\otimes}v_l) \\
&=\dfrac{2(-1)^{|v_k||v_l|+|v_l|}}{q+q^{-1}}(v_{2m-l+1}{\otimes}v_{2m-k+1})
-\dfrac{q-q^{-1}}{q+q^{-1}}(-1)^{|v_k|}(v_{2m-k+1}{\otimes}v_{2m-l+1})
\end{split}$$ $$\begin{split}
T'{\varphi^{{\otimes}2}}(v_k{\otimes}v_l)&=(-1)^{|v_k|}T'(v_{2m-k+1}{\otimes}v_{2m-l+1}) \\
&=\dfrac{2(-1)^{|v_k|+|v_{2m-k+1}||v_{2m-l+1}|}}{q+q^{-1}}(v_{2m-l+1}{\otimes}v_{2m-k+1})
+\dfrac{q-q^{-1}}{q+q^{-1}}(-1)^{|v_k|}(v_{2m-k+1}{\otimes}v_{2m-l+1})
\end{split}$$ In the same manner as case2, we have ${\varphi^{{\otimes}2}}T'(v_k{\otimes}v_l)=-T'{\varphi^{{\otimes}2}}(v_k{\otimes}v_l)$.

Let ${\Phi}$ be the endomorphism of the vector space $\operatorname{End}_{K}V^{{\otimes}r}$ which is given by: $${\Phi}(f)={\varphi^{{\otimes}r}}f \quad 
\text{for $f{\in}\operatorname{End}_{K}V^{{\otimes}r}$}.$$ Then we have the following. 

$\Phi$ is an automorphism of the vector space $\operatorname{End}_{K}V^{{\otimes}r}$. The restriction of $\Phi$ to $\mathcal{D}_q$ gives an automorphism of $\mathcal{D}_q$ which satisfies the following properties: $${\Phi}(\mathcal{B}_q)=\mathcal{B}_q^{\dagger}, \quad {\Phi}(\mathcal{B}_q^{\dagger})=\mathcal{B}_q.$$ Especially, we have $\dim_{K}\mathcal{B}_q=\dim_{K}\mathcal{B}_q^{\dagger}$ and $\mathcal{D}_q=\mathcal{B}_q{\oplus}\Phi(\mathcal{B}_q)=\mathcal{B}_q{\oplus}\varphi^{{\otimes}r}\mathcal{B}_q$.

From (5.2), it immediately follows that ${\Phi^2}f=(-1)^{r(r-1)/2}f$ for $f{\in}\operatorname{End}_{K}V^{{\otimes}r}$. Hence $\Phi$ is an isomorphism of $\operatorname{End}_{K}V^{{\otimes}r}$. Let $f{\in}\mathcal{B}_q$. From (5.3), it follows that $$\begin{split}
{\pi_r}(T'_i){\Phi}(f)&={\pi_r}(T'_i){\varphi^{{\otimes}r}}f \\
&=-{\varphi^{{\otimes}r}}{\pi_r}(T'_i)f \\
&=-{\varphi^{{\otimes}r}}f{\pi_r}(T'_i) \\
&=-{\Phi}(f){\pi_r}(T'_i). 
\end{split}$$ Hence ${\Phi}(f){\in}\mathcal{B}_q^{\dagger}$. In the same way, we also obtain ${\Phi}(f){\in}\mathcal{B}_q$ for $f{\in}\mathcal{B}_q^{\dagger}$. Combining with Lemma 5.1, we can conclude that $\Phi$ defines an automorphism of $\mathcal{D}_q=\mathcal{B}_q{\oplus}\mathcal{B}_q^{\dagger}$ satisfying ${\Phi}(\mathcal{B}_q)=\mathcal{B}_q^{\dagger}$ and ${\Phi}(\mathcal{B}_q^{\dagger})=\mathcal{B}_q$.

Let $\omega$ be the endomorphism of the algebra $\operatorname{End}_{K}V^{{\otimes}r}$ which is given by: $${\omega}(f)=(-1)^{r(r-1)/2}{\varphi^{{\otimes}r}}f{\varphi^{{\otimes}r}} \quad 
\text{for $f{\in}\operatorname{End}_{K}V^{{\otimes}r}$}.$$ Indeed, one can easily see that the following.  $$\begin{split}
{\omega}(fg)&=(-1)^{r(r-1)/2}{\varphi^{{\otimes}r}}fg{\varphi^{{\otimes}r}} \\
&={\varphi^{{\otimes}r}}f{\varphi^{{\otimes}r}}{\varphi^{{\otimes}r}}g{\varphi^{{\otimes}r}} \\
&=\big{(}(-1)^{r(r-1)/2}{\varphi^{{\otimes}r}}f{\varphi^{{\otimes}r}}\big{)}
\big{(}(-1)^{r(r-1)/2}{\varphi^{{\otimes}r}}g{\varphi^{{\otimes}r}}\big{)} \\
&={\omega}(f){\omega}(g). 
\end{split}$$ So, $\omega$ is algebraic. 

$\omega$ is an automorphism of order $2$ of the algebra $\mathcal{B}_q$.

If $f{\in}\mathcal{B}_q$, then it follows from (5.3) that $$\begin{split}
{\pi_r(T'_i)}{\omega}(f)&=(-1)^{r(r-1)/2}{\pi_r(T'_i)}{\varphi^{{\otimes}r}}f{\varphi^{{\otimes}r}} \\
&=(-1)^{r(r-1)/2}{\pi_r(T'_i)}{\varphi^{{\otimes}r}}f{\varphi^{{\otimes}r}} \\
&=(-1)^{r(r-1)/2}{\varphi^{{\otimes}r}}f{\varphi^{{\otimes}r}}{\pi_r(T'_i)} \\
&={\omega}(f){\pi_r(T'_i)}
\end{split}$$ Therefore ${\omega}(f){\in}\mathcal{B}_q$. From (5.2), it follows that $$\begin{split}
{\omega^2}(f)&=(-1)^{r(r-1)/2}{\omega}({\varphi^{{\otimes}r}}f{\varphi^{{\otimes}r}}) \\
&=(-1)^{r(r-1)}{(\varphi^{{\otimes}r})^2}f{(\varphi^{{\otimes}r})^2} \\
&=f
\end{split}$$ Therefore $\omega$ is an automorphism of $\mathcal{B}_q$ of order $2$.

Let $H={\langle}1,\omega{\rangle}$. Then $H$ is a subgroup of $\operatorname{Aut}(\mathcal{B}_q)$. $H$ is naturally isomorphic to $\mathbb{Z}_2={\langle}1,-1{\rangle}$ as (multiplicative) groups.  We define two maps $$\psi_1:\mathbb{Z}_2{\longrightarrow}\operatorname{Aut}(\mathcal{B}_q),\quad 
\psi_1(1)=1,\psi_1(-1)=\omega,$$ and $$\alpha_1:\mathbb{Z}_2{\times}\mathbb{Z}_2{\longrightarrow}(\mathcal{B}_q)^{\times},\quad 
\alpha_1(\sigma,\tau)=
\begin{cases}
1& \text{if $\sigma=1$ or $\tau=1$,}\\
(-1)^{r(r-1)/2}& \text{otherwise.} 
\end{cases}$$ Then we have the following lemma. 

$\psi_1$ and $\alpha_1$ satisfy (2.1)–(2.3). 

(2.3) is trivial.  Since $\alpha(\sigma,\tau)={\pm}1$, (2.1) may be reduced to ${^\sigma}({^\tau}a)={^{({\sigma}{\tau})}}a$, so holds obviously.  If $\sigma_1=1$, then both sides of (2.2) equal $\alpha(\sigma_2,\sigma_3)$. In the same manner, if $\sigma_2=1$ (respectively $\sigma_3=1$), then those equal $\alpha(\sigma_1,\sigma_3)$ (respectively $\alpha(\sigma_1,\sigma_2)$).  If $\sigma_1=\sigma_2=\sigma_3=-1$, then both sides of (2.2) become $-1$. Thus (2.2) holds. 

Hence, we obtain a crossed product ${\mathcal{B}_q}_{\alpha_1}^{\psi_1}[\mathbb{Z}_2]$ from the crossed system $(\mathcal{B}_q,\mathbb{Z}_2,\psi_1,\alpha_1)$, where $\psi_1$ and $\alpha_1$ are given in (5.4) and (5.5) respectively. 

If $m=n$, then $\mathcal{D}_q$ is isomorphic to the $\mathbb{Z}_2$-crossed product ${\mathcal{B}_q}_{\alpha_1}^{\psi_1}[\mathbb{Z}_2]$ as $K$-algebras. 

Because $\mathcal{B}_q^{\dagger}=\varphi^{{\otimes}r}\mathcal{B}_q=\mathcal{B}_q\varphi^{{\otimes}r}$ holds from (5.3) and Proposition 5.3, $$\mathcal{D}_q=\mathcal{B}_q1{\oplus}\mathcal{B}_q\varphi^{{\otimes}r}$$ follows.  Since both ${\mathcal{B}_q}_{\alpha_1}^{\psi_1}[\mathbb{Z}_2]$ and $\mathcal{D}_q$ are free left $\mathcal{B}_q$-modules, we may define an isomorphism of $\mathcal{B}_q$-modules: $${\iota_1}:{\mathcal{B}_q}_{\alpha_1}^{\psi_1}[\mathbb{Z}_2]{\longrightarrow}\mathcal{D}_q,\quad 
{\iota_1}(u_1)=1,{\iota_1}(u_{-1})=\varphi^{{\otimes}r}.$$ From (2.4) and (5.4) and (5.5), one can deduce the multiplication law for ${\mathcal{B}_q}_{\alpha_1}^{\psi_1}[\mathbb{Z}_2]$ as follows.  $$(a_1u_{\sigma})(a_2u_{\tau})=
\begin{cases}
a_1a_2u_1& \text{if $\sigma=1$ and $\tau=1$,}\\
a_1a_2u_{-1}& \text{if $\sigma=1$ and $\tau=-1$,}\\
(-1)^{r(r-1)/2}a_1\varphi^{{\otimes}r}a_2\varphi^{{\otimes}r}u_{-1}& \text{if $\sigma=-1$ and $\tau=1$,}\\
a_1\varphi^{{\otimes}r}a_2\varphi^{{\otimes}r}u_1& \text{if $\sigma=-1$ and $\tau=-1$,}\\
\end{cases}$$ where $a_1,a_2{\in}\mathcal{B}_q$. Therefore, we get four formulas in $\mathcal{D}_q$, $$\begin{aligned}
(a_11)(a_21)&=&a_1a_21,\\
(a_11)(a_2\varphi^{{\otimes}r})&=&a_1a_2\varphi^{{\otimes}r},\\
(a_1\varphi^{{\otimes}r})(a_21)&=&(-1)^{r(r-1)/2}a_1\varphi^{{\otimes}r}a_2
\varphi^{{\otimes}r}\varphi^{{\otimes}r},\\
(a_1\varphi^{{\otimes}r})(a_2\varphi^{{\otimes}r})&=&a_1\varphi^{{\otimes}r}a_2
\varphi^{{\otimes}r}1,\end{aligned}$$ which derive the conclusion that $f$ is an isomorphism of $K$-algebras. 

For a commutative domain $R$ and a subalgebra $A$ of $\operatorname{Mat}(m,R)$, we set $\Tilde{A}=\{X{\in}\operatorname{Mat}(m,R)|XY=YX \text{ for all }Y{\in}A\}$. Let $R_1=\mathbb{Q}[q,q^{-1}]$ be the algebra of Laurent polynomials.  Since matrix elements of $\rho_r$ are in $R_1$ by (4.3) and (4.4), we may define $\mathcal{D}'_q$ to be the subalgebra of $\operatorname{Mat}\big{(}(2m)^r,{R_1}\big{)}$ generated by the set $\{\rho_r(\sigma),\rho_r(q^h),\rho_r(e_i),\rho_r(f_i),\varphi^{{\otimes}r}|h{\in}P^*,i{\in}I\}$. Similarly by (4.2), we may also define $\mathcal{C}'_q$ to be the one generated by $\{I_{(2m)^r},(q+q^{-1})^2\pi_r(T'_1T'_i)|i=2,\hdots,r-1\}$ where $I_{(2m)^r}$ is the identity matrix.  Now we shall complete Schur-Weyl reciprocity for the $q$-analogue of the alternating group, namely, $\mathcal{C}_q=\operatorname{End}_{\mathcal{D}_q}V^{{\otimes}r}$.

$\mathcal{D}_q=\Tilde{\mathcal{C}}_q$ is by definition.  We shall show $\mathcal{C}_q=\Tilde{\mathcal{D}}_q$. The specialization to a nonzero complex number $t$ is a ring homomorphism $\varphi_t : R_1{\longrightarrow}\mathbb{C}$ with the condition $\varphi_t(q)=t$. $\mathbb{C}$ becomes $(\mathbb{C},R_1)$-bimodule, with $R_1$ acting from the right via $\varphi_t$. If $t$ is a transcendental number, we can extend the specialization from $R_1$ to its quotient field $K$, namely $\varphi_t : K{\longrightarrow}\mathbb{C}$. Applying the specialization $\varphi_t$, we obtain the specialized algebras $\mathcal{C}_t=\mathbb{C}{\otimes_{R_1}}\mathcal{C}'_q$ and $\mathcal{D}_t=\mathbb{C}{\otimes_{R_1}}\mathcal{D}_q$ which are subalgebras of $\operatorname{Mat}\big{(}(2m)^r,\mathbb{C}\big{)}$. $\mathcal{C}_t$ and $\mathcal{D}_t$ act on the specialized vector space $\mathbb{C}{\otimes}_KV^{{\otimes}r}$ in obvious ways. 

$\mathcal{C}_q=\Tilde{\mathcal{D}}_q$.

Obviously $\mathcal{C}'_q{\subseteq}\Tilde{\mathcal{D}}'_q$, so ${\operatorname{rank}}_{R_1}\mathcal{C}'_q{\leq}{\operatorname{rank}}_{R_1}\Tilde{\mathcal{D}}'_q$ holds. 

Since $R_1$ is a principal ideal domain, the submodules $\mathcal{C}'_q$ and $\mathcal{D}'_q$ of the free $R_1$-module $\operatorname{Mat}\big{(}(2m)^r,R_1\big{)}$ are also free.  Assume $N={\operatorname{rank}}_{R_1}\mathcal{C}'_q$. Let $X_i(q)$($i=1,\hdots,N$) be a basis of $\mathcal{C}'_q$ and $x_i^{k,l}(q){\in}R_1$ the $(k,l)$-entry of $X_i(q)$. Then we immediately have that the specialized elements $X_i(t)=\big{(}x_i^{k,l}(t)\big{)}$($i=1,\hdots,N$) generate $\mathcal{C}_t$ and $\dim_\mathbb{C}\mathcal{C}_t{\leq}{\operatorname{rank}}_{R_1}\mathcal{C}'_q$. Because $X_i(q)$ are linearly independent, $\sum_{i=1}^N\alpha_i(q)X_i(q)=0$ for $\alpha_1(q),\hdots,\alpha_N(q){\in}R_1$ implies $\alpha_1(q)=\hdots=\alpha_N(q)=0$. But the equation $\sum_{i=1}^N\alpha_iX_i(t)=0$ for $\alpha_1,\hdots,\alpha_N{\in}\mathbb{C}$ may admit a nonzero $\alpha_i$. Consider the system of linear equations: $$\begin{bmatrix}
x_1^{1,1}(t) & x_2^{1,1}(t) & \cdots & x_N^{1,1}(t) \\
x_1^{1,2}(t) & x_2^{1,2}(t) & \cdots & x_N^{1,2}(t) \\
&\cdots\cdots\cdots \\
x_1^{(2m)^r,(2m)^r-1}(t) & x_2^{(2m)^r,(2m)^r-1}(t) & \cdots & x_N^{(2m)^r,(2m)^r-1}(t) \\
x_1^{(2m)^r,(2m)^r}(t) & x_2^{(2m)^r,(2m)^r}(t) & \cdots & x_N^{(2m)^r,(2m)^r}(t)
\end{bmatrix}
\times
\begin{bmatrix}
\alpha_1 \\
\alpha_2 \\
\cdots \\
\alpha_N
\end{bmatrix}
=
\begin{bmatrix}
0 \\
0 \\
\cdots \\
0 \\
0
\end{bmatrix}.$$ Reducing to a common denominator for each row, we may assume that each $x_i^{k,l}(t){\in}\mathbb{Q}[t]$. Suppose there exists a non-trivial solution.  Then the rank of above matrix is less than $N$, which implies that every $N$th minor determinant equals $0$. This is impossible if $t$ is transcendental.  Hence $\dim_\mathbb{C}\mathcal{C}_t={\operatorname{rank}}_{R_1}\mathcal{C}'_q$ for a transcendental $t$. In the same manner, $\dim_\mathbb{C}\mathcal{D}_t={\operatorname{rank}}_{R_1}\mathcal{D}'_q$ if $t$ is a transcendental number.  This argument is valid even if $R_1$ is replaced by $K$, thus we also obtain $\dim_\mathbb{C}\mathcal{C}_t=\dim_K\mathcal{C}_q$, $\dim_\mathbb{C}\mathcal{D}_t=\dim_K\mathcal{D}_q$ for a transcendental number $t$.

Next we shall show that ${\operatorname{rank}}_{R_1}\Tilde{\mathcal{D}}'_q=
\dim_\mathbb{C}\Tilde{\mathcal{D}}_t$. Assume that $Y(q)=\big{(}y^{k,l}(q)\big{)}{\in}\Tilde{\mathcal{D}}'_q$. Then $Y(q)$ commutes with all $\rho_r(\sigma),\rho_r(q^h),\rho_r(e_i),\rho_r(f_i),\varphi^{{\otimes}r}(h{\in}P^*,i{\in}I)$. Hence the matrix elements $y^{k,l}(q)(k,l=1,\hdots,(2m)^r$ determine polynomial equations with coefficients in $R_1$. We notice that ${\operatorname{rank}}_{R_1}\Tilde{\mathcal{D}}'_q{\leq}\dim_\mathbb{C}\Tilde{\mathcal{D}}_t$ in general; besides the elements of specialized algebra $\mathbb{C}{\otimes_{R_1}}\Tilde{\mathcal{D}}'_q$, possibly $\Tilde{\mathcal{D}}_t$ contains another element which commutes with specializations of $\rho_r(\sigma),\rho_r(q^h),\rho_r(e_i),\rho_r(f_i),\varphi^{{\otimes}r}(h{\in}P^*,i{\in}I)$. Assume the case.  Reducing to a common denominator, we may deduce that $t$ is a common solution of certain polynomial equations with coefficients in $\mathbb{Q}$. This is impossible if $t$ is transcendental.  Thus ${\operatorname{rank}}_{R_1}\Tilde{\mathcal{C}}'_q=\dim_\mathbb{C}\Tilde{\mathcal{C}}_t$ for a transcendental $t$. In the same manner, we also have ${\operatorname{rank}}_{R_1}\Tilde{\mathcal{D}}'_q=\dim_\mathbb{C}\Tilde{\mathcal{D}}_t$ for a transcendental $t$. Replacing $R_1$ by $K$, we obtain $\dim_K\Tilde{\mathcal{C}}_q=\dim_\mathbb{C}\Tilde{\mathcal{C}}_t$ and $\dim_K\Tilde{\mathcal{D}}_q=\dim_\mathbb{C}\Tilde{\mathcal{D}}_t$. Considering the specialization $\varphi_1$, we obtain the following: $$\dim_{\mathbb{C}}\mathcal{C}_1{\leq}
\dim_{\mathbb{C}}\mathcal{C}_t=
{\operatorname{rank}}_{R_1}\mathcal{C}'_q{\leq}
{\operatorname{rank}}_{R_1}\Tilde{\mathcal{D}}'_q=
\dim_{\mathbb{C}}\Tilde{\mathcal{D}}_t{\leq}
\dim_{\mathbb{C}}\Tilde{\mathcal{D}}_1.$$ In [@Regev], Regev has shown $\mathcal{D}_1=\Tilde{\mathcal{C}}_1$. Applying double centralizer theorem, we have $\mathcal{C}_1=\Tilde{\mathcal{D}}_1$. Therefore ${\operatorname{rank}}_{R_1}\mathcal{C}'_q={\operatorname{rank}}_{R_1}\Tilde{\mathcal{D}}'_q$ and $\dim_K\mathcal{C}_q=\dim_K\Tilde{\mathcal{D}}_q$, which imply $\mathcal{C}_q=\Tilde{\mathcal{D}}_q$.

From Proposition 5.7 we immediately obtain Schur-Weyl reciprocity for $\mathcal{H}_{K,r}^1(q)$ as follows. 

$\operatorname{End}_{\mathcal{C}_q}V^{{\otimes}r}
=\mathcal{D}_q$ and $\operatorname{End}_{\mathcal{D}_q}V^{{\otimes}r}
=\mathcal{C}_q$ hold. 

Schur-Weyl reciprocity for $\mathcal{H}_{\bar{K},r}^1(q)$ in the general case
=============================================================================

Henceforth, we do not assume $m{\neq}n$ and consider the general case.  In this case, results can be obtained in the same way as $q=1$ case which is used in [@Regev].  Recall Theorem 4.2: $$\bar{\mathcal{A}}_q=\bigoplus_{\lambda{\in}H(m,n;r)}\bar{\mathcal{A}}_{q,\lambda},$$ where each $\bar{\mathcal{A}}_{q,\lambda}=\pi_r(I_{q,\lambda})({\cong}I_{q,\lambda})$ is the image of the homogeneous component $I_{q,\lambda}$ of $\mathcal{H}_{\bar{K},r}(q)$ corresponding to $\lambda{\in}H(m,n;r)$. In the same manner as [@Regev], we define $H_0(m,n;r)=\{\lambda{\in}\Lambda_r|\lambda,\lambda'{\in}H(m,n;r)\}$ and $H_1(m,n;r)=H(m,n;r){\backslash}H_0(m,n;r)$. Then we obtain the following from (6.1).  $$\bar{\mathcal{A}}_q=\Big{[}\bigoplus_{\lambda{\in}H_0(m,n;r),\lambda>\lambda'}
\big{\{}\bar{\mathcal{A}}_{q,\lambda}{\oplus}\bar{\mathcal{A}}_{q,\lambda'}\big{\}}\Big{]}
\bigoplus\big{\{}\bigoplus_{\lambda{\in}H_0(m,n;r),\lambda=\lambda'}\bar{\mathcal{A}}_{q,\lambda}\big{\}}
\bigoplus\big{\{}\bigoplus_{\lambda{\in}H_1(m,n;r)}\bar{\mathcal{A}}_{q,\lambda}\big{\}}.$$ Let $\bar{\mathcal{C}}_q=\mathcal{C}_q{\otimes_{\mathbb{Q}(q)}}\bar{K}
=\pi_r\big{(}\mathcal{H}_{\bar{K},r}^1(q)\big{)}$. From Theorem 3.10 we have: $$\mathcal{H}_{\bar{K},r}^1(q)=
\big{\{}\bigoplus_{\lambda{\in}\Lambda_r,\lambda>\lambda'}\tilde{I}_{q,\lambda}\big{\}}
\bigoplus\Big{[}\bigoplus_{\lambda{\in}\Lambda_r,\lambda=\lambda'}
\big{\{}\tilde{I}_{q,\lambda}^+{\oplus}\tilde{I}_{q,\lambda}^-\big{\}}\Big{]}.$$ So we immediately obtain: $$\bar{\mathcal{C}}_q=
\big{\{}\bigoplus_{\lambda{\in}H_0(m,n;r),\lambda>\lambda'}\pi_r(\tilde{I}_{q,\lambda})\big{\}}
\bigoplus\Big{[}\bigoplus_{\lambda{\in}H_0(m,n;r),\lambda=\lambda'}
\big{\{}\pi_r(\tilde{I}_{q,\lambda}^+){\oplus}\pi_r(\tilde{I}_{q,\lambda}^-)\big{\}}\Big{]}
\bigoplus\big{\{}\bigoplus_{\lambda{\in}H_1(m,n;r)}\pi_r(\tilde{I}_{q,\lambda})\big{\}}.$$ If $\lambda{\in}H_0(m,n;r)$ and $\lambda>\lambda'$, then $\pi_r(I_{q,\lambda}){\cong}I_{q,\lambda}$ and $\pi_r(I_{q,\lambda'}){\cong}I_{q,\lambda'}$. Therefore $\pi_r(I_{q,\lambda}{\oplus}I_{q,\lambda'}){\cong}I_{q,\lambda}{\oplus}I_{q,\lambda'}$, which implies $\pi_r(\tilde{I}_{q,\lambda}){\cong}\pi_r(I_{q,\lambda}){\cong}I_{q,\lambda}$ because $I_{q,\lambda}{\cong}\tilde{I}_{q,\lambda}{\subseteq}I_{q,\lambda}{\oplus}I_{q,\lambda'}$.

If $\lambda{\in}H_0(m,n;r)$ and $\lambda=\lambda'$, then $\pi_r(\tilde{I}_{q,\lambda}^+){\cong}\tilde{I}_{q,\lambda}^+$ and $\pi_r(\tilde{I}_{q,\lambda}^-){\cong}\tilde{I}_{q,\lambda}^-$ hold because $\pi_r(\tilde{I}_{q,\lambda}){\cong}\tilde{I}_{q,\lambda}$ and $\tilde{I}_{q,\lambda}^{\pm}{\subseteq}\tilde{I}_{q,\lambda}$.

If $\lambda{\in}H_1(m,n;r)$ then $\pi_r(I_{q,\lambda}){\cong}I_{q,\lambda}$ and $\pi_r(I_{q,\lambda'})=\{0\}$. Since $\tilde{I}_{q,\lambda}=\big{\{}X+\hat{X}{\in}I_{q,\lambda}{\oplus}I_{q,\lambda'}|
X{\in}I_{q,\lambda}\big{\}}$, we have $\pi_r(X+\hat{X})=\pi_r(X)$ for $X+\hat{X}{\in}\tilde{I}_{q,\lambda}$. Thereby $\pi_r(\tilde{I}_{q,\lambda})=\pi_r(I_{q,\lambda})
{\cong}I_{q,\lambda}$ follows. 

Let us define two-sided ideals of $\bar{\mathcal{A}}_q$ and those of $\bar{\mathcal{C}}_q$ as follows.  $$\begin{aligned}
\bar{\mathcal{A}}_q{\supseteq}\bar{\mathcal{A}}_q^0&=&
\Big{[}\bigoplus_{\lambda{\in}H_0(m,n;r),\lambda>\lambda'}
\big{\{}\bar{\mathcal{A}}_{q,\lambda}{\oplus}\bar{\mathcal{A}}_{q,\lambda'}\big{\}}\Big{]}
{\bigoplus}\big{\{}\bigoplus_{\lambda{\in}H_0(m,n;r),\lambda=\lambda'}
\bar{\mathcal{A}}_{q,\lambda}\big{\}}, \\
\bar{\mathcal{A}}_q{\supseteq}\bar{\mathcal{A}}_q^1&=&\bigoplus_{\lambda{\in}H_1(m,n;r)}
\bar{\mathcal{A}}_{q,\lambda}, \\
\bar{\mathcal{C}}_q{\supseteq}\bar{\mathcal{C}}_q^0&=&
\big{\{}\bigoplus_{\lambda{\in}H_0(m,n;r),\lambda>\lambda'}
\pi_r(\tilde{I}_{q,\lambda})\big{\}}{\bigoplus}\Big{[}\bigoplus_{\lambda{\in}H_0(m,n;r),\lambda=\lambda'}
\big{\{}\pi_r(\tilde{I}_{q,\lambda}^+){\oplus}\pi_r(\tilde{I}_{q,\lambda}^-)\big{\}}\Big{]} \\
\bar{\mathcal{C}}_q{\supseteq}\bar{\mathcal{C}}_q^1&=&\bigoplus_{\lambda{\in}H_1(m,n;r)}
\pi_r(\tilde{I}_{q,\lambda})\end{aligned}$$ Comparing dimensions of components between $\bar{\mathcal{A}}_q$ and $\bar{\mathcal{C}}_q$, we obtain the following theorem. 

$\bar{\mathcal{A}}_q$ and $\bar{\mathcal{C}}_q$ have direct sum decompositions $\bar{\mathcal{A}}_q=\bar{\mathcal{A}}_q^0{\oplus}\bar{\mathcal{A}}_q^1$ and $\bar{\mathcal{C}}_q=\bar{\mathcal{C}}_q^0{\oplus}\bar{\mathcal{C}}_q^1$ respectively, which are satisfy the following relations. 

1.   $\bar{\mathcal{C}}_q^0{\subseteq}\bar{\mathcal{A}}_q^0$ and $\dim_{\bar{K}}\bar{\mathcal{A}}_q^0
    =2\dim_{\bar{K}}\bar{\mathcal{C}}_q^0$

2.   $\bar{\mathcal{C}}_q^1=\bar{\mathcal{A}}_q^1$. Especially $\dim_{\bar{K}}\bar{\mathcal{A}}_q^1
    =\dim_{\bar{K}}\bar{\mathcal{C}}_q^1$.

As a special case, when $m=n$, we readily see that $H_1(m,n;r)=\emptyset$. Hence both $\bar{\mathcal{A}}_q^1$ and $\bar{\mathcal{C}}_q^1$ are zero and $\dim_{\bar{K}}\bar{\mathcal{A}}_q=2\dim_{\bar{K}}\bar{\mathcal{C}}_q$. Moreover, if $n=0$, namely non-super case, then we readily see that if $m^2<r$, then all $\lambda{\in}\Lambda_r$ are contained in $H_1(m,0;r)$ and $\bar{\mathcal{A}}_q^0=\bar{\mathcal{C}}_q^0=0$. The Corollary below is immediately obtained from this argument and (6.2), (6.3). 

Let $n=0$. If $m^2<r$, then $\bar{\mathcal{A}}_q=\bar{\mathcal{C}}_q$ and $\operatorname{End}_{\bar{\mathcal{C}}_q}\bar{V}^{{\otimes}r}=
\operatorname{End}_{\bar{\mathcal{A}}_q}\bar{V}^{{\otimes}r}$.

We denote by $W_0$ and $W_1$ direct summands of $\bar{V}^{{\otimes}r}$ which are defined as follows.  $$\begin{aligned}
W_0&=&\Big{[}\bigoplus_{\lambda{\in}H_0(m,n;r),\lambda>\lambda'}
\big{\{}(H_{\lambda}{\otimes}V_{\lambda}){\oplus}(H_{\lambda'}{\otimes}V_{\lambda'})\big{\}}\Big{]}
{\bigoplus}\big{\{}\bigoplus_{\lambda{\in}H_0(m,n;r),\lambda=\lambda'}
(H_{\lambda}{\otimes}V_{\lambda})\big{\}} \\
W_1&=&\bigoplus_{\lambda{\in}H_1(m,n;r)}(H_{\lambda}{\otimes}V_{\lambda})\end{aligned}$$ We notice that $\operatorname{End}_{\bar{\mathcal{A}}_q}\bar{V}^{{\otimes}r}=
\operatorname{End}_{\bar{\mathcal{A}}_q}W_0{\oplus}\operatorname{End}_{\bar{\mathcal{A}}_q}W_1$ and that $\operatorname{End}_{\bar{\mathcal{C}}_q}\bar{V}^{{\otimes}r}=
\operatorname{End}_{\bar{\mathcal{C}}_q}W_0{\oplus}\operatorname{End}_{\bar{\mathcal{C}}_q}W_1$. From Theorem 6.1, we immediately have the following. 

$\operatorname{End}_{\bar{\mathcal{C}}_q}W_0{\supseteq}\operatorname{End}_{\bar{\mathcal{A}}_q}W_0$ and $\operatorname{End}_{\bar{\mathcal{C}}_q}W_1=\operatorname{End}_{\bar{\mathcal{A}}_q}W_1$.

Since $\bar{\mathcal{A}}_q^1W_0=0$ and $\bar{\mathcal{A}}_q^0W_1=0$, $\operatorname{End}_{\bar{\mathcal{A}}_q}W_0=
\operatorname{End}_{\bar{\mathcal{A}}_q^0}W_0$ and $\operatorname{End}_{\bar{\mathcal{A}}_q}W_1=
\operatorname{End}_{\bar{\mathcal{A}}_q^1}W_1$ hold.  Similarly, we have $\operatorname{End}_{\bar{\mathcal{C}}_q}W_0=
\operatorname{End}_{\bar{\mathcal{C}}_q^1}W_0$ and $\operatorname{End}_{\bar{\mathcal{C}}_q}W_1=
\operatorname{End}_{\bar{\mathcal{C}}_q^1}W_1$. Hence we immediately obtain from Theorem 6.1(1) that $\operatorname{End}_{\bar{\mathcal{C}}_q}W_0=\operatorname{End}_{\bar{\mathcal{C}}_q^0}W_0
{\supseteq}\operatorname{End}_{\bar{\mathcal{A}}_q^0}W_0=\operatorname{End}_{\bar{\mathcal{A}}_q}W_0$ and from Theorem 6.1(2) that $\operatorname{End}_{\bar{\mathcal{C}}_q}W_1=\operatorname{End}_{\bar{\mathcal{C}}_q^1}W_1=
\operatorname{End}_{\bar{\mathcal{A}}_q^1}W_1=\operatorname{End}_{\bar{\mathcal{A}}_q}W_1$ also. 

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