Finite dimensional irreducible representations of the nullity 2 centreless core $\mathfrak{g}_{2n,\rho}(\mathbb{C}_q)$

TL;DR

This paper classifies finite-dimensional irreducible representations of a specific nullity 2 centreless core Lie algebra by analyzing its structure as a BC_n-graded Lie algebra over an involutory associative algebra.

Contribution

It provides a detailed study of the structure and representation theory of the nullity 2 centreless core Lie algebra $rak{g}_{2n, ho}(bC_q)$, extending understanding of its finite-dimensional irreducible modules.

Findings

Classification of finite-dimensional irreducible representations.

Structural insights into BC_n-graded Lie algebras.

Connection between algebra structure and representation theory.

Abstract

We study the finite-dimensional irreducible representations of the nullity 2 centreless core $\mathfrak{g}_{2n,\rho}(\mathbb{C}_q)$ by investigating the structure of the $\mathrm{BC}_n$-graded Lie algebra $\mathfrak{g}_{2n,\rho}(R)$, where $R$ is a unital involutory associative algebra over a field $k$ of characteristic zero.

Equations

$$\begin{array}{rl}{f}_{ij}(a)& =e_{ij}(a)-e_{n+j,n+i}(\bar{a}),1\le i,j\le n\\ g_{ij}(a)& =e_{i,n+j}(a)-\rho e_{j,n+i}(\bar{a}),1\le i,j\le n\\ h_{ij}(a)& =e_{n+i,j}(a)-\rho e_{n+j,i}(\bar{a}),1\le i,j\le n.\end{array}$$

$$\begin{array}{rl}[f_{ij}(a),f_{kl}(b)]& ={\delta }_{jk}{f}_{il}(ab)-{\delta }_{il}{f}_{kj}(ba),\\ [f_{ij}(a),g_{kl}(b)]& ={\delta }_{jk}{g}_{il}(ab)+{\delta }_{jl}{g}_{ki}(b\bar{a}),\\ [f_{ij}(a),h_{kl}(b)]& =-{\delta }_{il}{h}_{kj}(ba)-{\delta }_{ik}{h}_{jl}(\bar{a}b),\\ [g_{ij}(a),g_{kl}(b)]& =0=[h_{ij}(a),h_{kl}(b)],\\ [g_{ij}(a),h_{kl}(b)]& ={\delta }_{jk}{f}_{il}(ab)+{\delta }_{li}{f}_{jk}(\bar{a}\bar{b})-\rho {\delta }_{jl}{f}_{ik}(a\bar{b})-\rho {\delta }_{ki}{f}_{jl}(\bar{a}b).\end{array}$$

$$g_{ij}(a)=-\rho g_{ji}(\bar{a}),h_{ij}(a)=-\rho h_{ji}(\bar{a}).$$

$$[{\mathfrak{g}}_{2n,\rho }(R),{\mathfrak{g}}_{2n,\rho }(R)]={\mathfrak{g}}_{2n,\rho }(R).$$

$$\left[f_{ik}(a),f_{kj}(1)\right]=f_{ij}(a).$$

$$\left[f_{ik}(a),g_{kj}(1)\right]=g_{ij}(a),$$

$$\left[f_{kj}(-1),h_{ik}(a)\right]=h_{ij}(a).\text{qed}$$

$$R=R_{+}\oplus R_{-}\text{ and }{R}_{+}+[R,R]=R_{+}\oplus (R_{-}\cap [R,R]).$$

$$L=\{a\in \mathcal{A}|\phi (a)=-a\}$$

$$M=\left(\begin{array}[]{cc}0&I_{n}\\ \rho I_{n}&0\\ \end{array}\right)\in M_{2n}(R).$$

$${}^{\ast }:M_{2n}(R)\to M_{2n}(R)\text{ by }{A}^{\ast }=M^{-1}{\overline{A}}^{t}M.$$

$$\begin{array}{rl}(A^{\ast }{)}^{\ast }=& (M^{-1}{\overline{A}}^{t}M{)}^{\ast }=M^{-1}(\overline{M^{-1}{\overline{A}}^{t}M}{)}^{t}M=M^{-1}{\overline{M}}^{t}A({\overline{M}}^t{)}^{-1}M\\ =& M^{-1}\rho MA{\rho }^{-1}{M}^{-1}M=A,\end{array}$$

$$(AB{)}^{\ast }=M^{-1}{\overline{AB}}^{t}M=M^{-1}{\overline{B}}^t{\overline{A}}^{t}M=B^{\ast }{A}^{\ast }.$$

$${\mathcal{L}}_{\rho }=\{A\in M_{2n}(R)|A^{\ast }=-A\}.$$

$$\left(\begin{array}[]{cc}P&S\\ T&-\bar{P}^{t}\\ \end{array}\right)\quad\text{with }\overline{S}^{t}=-\rho S\quad\text{and}\quad\overline{T}^{t}=-\rho T,$$

$$\begin{split}\big{\{}f_{11}(a)|a\in R\big{\}}\cup&\Big{(}\bigcup_{i=2}^{n}\left\{f_{ii}(a)-f_{11}(a)|a\in R\right\}\Big{)}\cup\Big{(}\bigcup_{1\leq i\neq j\leq n}\left\{f_{ij}(a)|a\in R\right\}\Big{)}\\ \cup&\Big{(}\bigcup_{1\leq i<j\leq n}\left\{g_{ij}(a)|a\in R\right\}\Big{)}\cup\Big{(}\bigcup_{1\leq i\leq n}\left\{g_{ii}(a)|a\in R\right\}\Big{)}\\ \cup&\Big{(}\bigcup_{1\leq i<j\leq n}\left\{h_{ij}(a)|a\in R\right\}\Big{)}\cup\Big{(}\bigcup_{1\leq i\leq n}\left\{h_{ii}(a)|a\in R\right\}\Big{)}.\end{split}$$

$$\mathcal{G}_{\rho}=\bigl{\{}Y\in\mathcal{L}_{\rho}\,|\,\mathrm{tr}\,(Y)\equiv 0\bmod[R,R]\bigr{\}}.$$

$${\mathcal{G}}_{\rho }\subset \left\{Y\in {\mathcal{L}}_{\rho }|\mathrm{tr}(Y)\equiv 0\mathrm{mod}[R,R]\right\}.$$

$$\begin{array}{rl}Y=& f_{11}(p)+\sum _{i=2}^n(f_{ii}(q_i)-f_{11}(q_i))+\sum _{1\le i\ne j\le n}{f}_{ij}(r_{ij})+\sum _{1\le i<j\le n}{g}_{ij}(s_{ij})\\ & +\sum _{1\le i\le n}{g}_{ii}(t_i)+\sum _{1\le i<j\le n}{h}_{ij}(u_{ij})+\sum _{1\le i\le n}{h}_{ii}(v_i),\end{array}$$

$$\mathrm{tr}\,(Y)=\mathrm{tr}\,(f_{11}(p))+\sum_{i=2}^{n}\Big{(}\mathrm{tr}\,(f_{ii}(q_{i}))-\mathrm{tr}\,(f_{11}(q_{i}))\Big{)}=p-\overline{p}=\mathrm{tr}\,(f_{11}(p)).$$

$$\begin{array}{rl}{f}_{11}(p)=& f_{11}(a)+\sum _k{f}_{11}\left([b_k,c_k]\right)\\ =& [f_{12}(a),f_{21}(1/2)]+[g_{12}(1/2),h_{21}(a)]+\sum _k\left[f_{11}(b_k),f_{11}(c_k)\right]\end{array}$$

$$\left\{Y\in {\mathcal{L}}_{\rho }|\mathrm{tr}(Y)\equiv 0\mathrm{mod}[R,R]\right\}\subset {\mathcal{G}}_{\rho }.\text{qed}$$

$${\mathcal{G}}_{\rho }={\mathcal{G}}_0\oplus \sum _{i\ne j}{\mathcal{G}}_{{ϵ}_i-{ϵ}_j}\oplus \sum _{i<j}({\mathcal{G}}_{{ϵ}_i+{ϵ}_j}\oplus {\mathcal{G}}_{-{ϵ}_i-{ϵ}_j})\oplus \sum _i({\mathcal{G}}_{2{ϵ}_i}\oplus {\mathcal{G}}_{-2{ϵ}_i})$$

$${\mathcal{G}}_{{ϵ}_i-{ϵ}_j}=f_{ij}(R)(i\ne j),{\mathcal{G}}_{{ϵ}_i+{ϵ}_j}=g_{ij}(R),{\mathcal{G}}_{-{ϵ}_i-{ϵ}_j}=h_{ij}(R),$$

$${\mathcal{G}}_0=f_{11}(R_{+}\oplus (R_{-}\cap [R,R]))⨁\sum _i(f_{ii}-f_{11})(R),$$

$${\mathcal{G}}_0=\sum _{\mu \in {\Delta }_C}[{\mathcal{G}}_{\mu },{\mathcal{G}}_{-\mu }],$$

$$f_{11}([c,d])=[f_{21}(c),f_{12}(d)]-[f_{21}(cd),f_{12}(1)]\in \left[{\mathcal{G}}_{{ϵ}_2-{ϵ}_1},{\mathcal{G}}_{-\left({ϵ}_2-{ϵ}_1\right)}\right],\forall c,d\in R.\text{qed}$$

$${\mathfrak{g}}_{2n,\rho }(R)=f_{11}(R_{+}+[R,R])\oplus \sum _{i=2}^n(f_{ii}-f_{11})(R)\oplus \sum _{i\ne j}{f}_{ij}(R)\oplus \sum _{i\le j}{g}_{ij}(R)\oplus \sum _{i\le j}{h}_{ij}(R).$$

$$f_{ij}(E_{kl})\mapsto f_{(i-1)m+k,(j-1)m+l}(1),\text{ for all }1\le i\ne j\le n,1\le k,l\le m;$$

$$g_{ij}(E_{kl})\mapsto g_{(i-1)m+k,(j-1)m+l}(1),h_{ij}(E_{kl})\mapsto h_{(i-1)m+k,(j-1)m+l}(1);$$

Full text

Keywords: $\mathrm{BC}_{n}$-graded Lie algebra, centreless core, extended affine Lie algebra, involutive ideal, finite-dimensional irreducible representation.

**AMS classification: ** 17B05, 17B10, 17B22, 17B65, 17B67, 17B70.

*Corresponding author: H. Chen, hjchen@ustc.edu.cn

1. Introduction

Extended affine Lie algebras, or EALAs for short, were first introduced by physicists Høegh-Krohn and Torresani in [14] (under the name of quasi-simple Lie algebras), as a generalization of finite-dimensional simple Lie algebras and affine Kac-Moody Lie algebras over the complex numbers $\mathbb{C}$. The structure theory of EALAs has been extensively studied for decades (See [1], [2], [4], [5], [19], [20] and [21]). In particular, Allison, Azam, Berman, Gao and Pianzola in [1] proved the Kac conjecture which implies that the root systems of EALAs are examples of extended affine root systems which were previously introduced by Saito in [22].

The representations of EALAs are much less well understood. Different with the finite and affine setting, there is no successful highest weight theory of EALAs since the lack of a triangular decomposition. And earlier work considered only the untwisted toroidal Lie algebras and a few other isolated examples, such as [11], [12], [24], [25]. Recently, Billig and Lau in [8] constructed irreducible modules for twisted toroidal Lie algebras and extended affine Lie algebras by combining the representation theory of untwisted toroidal algebras [6] with the technique of thin coverings introduced in their paper [7].

As a result of [2] and [19], it is clear that, all but one family of centreless Lie tori can be constructed as an extension of a twisted multiloop algebra. Moreover, Lau gave the classification of the finite-dimensional simple modules of multiloop algebras in [17]. But in general, from an extend affine Lie algebra (or the centreless Lie tori) it is not easy to give the corresponding multiloop algebra structure (see Remark 5.4). In the present paper, we will study the finite-dimensional irreducible representations over the centreless core $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})$ by investigating the structure of the $\mathrm{BC}_{n}$-graded Lie algebra $\mathfrak{g}_{2n,\rho}(R)$. From any ideal $\mathcal{J}$ of $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})$ such that $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})/\mathcal{J}$ is finite-dimensional and semisimple, we relate it to an involutive ideal $J$ of $\mathbb{C}_{q}$. The corresponding quotient $\mathbb{C}_{q}/J$ is just copies of an involutive algebra $R$, where $R$ could be $\mathbb{C}$, $\mathbb{C}\oplus\mathbb{C}^{\mathrm{op}}$, the matrix algebra of order two with different involutions, or the group algebra of the Klein four group, and so on (for more details see Section 3). In any of these cases, the Lie algebra $\mathfrak{g}_{2n,\rho}(R)$ is a finite-dimensional simple Lie algebra for which the finite-dimensional simple modules are well known. Hence we obtain a complete result of the finite-dimensional representations of $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})$.

The paper is organized as follows. In Section 2, we investigate the structure of the root graded Lie algebra $\mathfrak{g}_{2n,\rho}(R)$. Moreover, we discuss the relation between the involutive ideals (invariant under involution) of $R$ and ideals of $\mathfrak{g}_{2n,\rho}(R)$. In Section 3, we focus on the elementary quantum torus in two variables with the natural involution. We define some involutive ideals and study the quotients. In the Section 4 and Section 5, we talk about our main results about the finite-dimensional semisimple quotients and finite-dimensional irreducible modules of $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})$.

Throughout this paper, denote by $\mathbb{Z}$, $\mathbb{N}$, $\mathbb{C}$ and $\mathbb{C}^{*}$ the sets of integers, positive integers, complex numbers and nonzero complex numbers respectively.

2. $\mathfrak{g}_{2n,\rho}(R)$ is a $\mathrm{BC}_{n}$-graded Lie algebra

Let us give the definition of $\mathfrak{g}_{2n,\rho}(R)$ first. Let $k$ be a field of characteristic [math] and $R$ a unital associative $k$-algebra with involution. An involution $\,\bar{\cdot}\,$ in this paper means an anti-involution as defined in [16], that is, a $k-$linear map from $R$ to $R$ satisfying $\overline{ab}=\,\overline{b}\,\,\overline{a}\,$ and $\bar{\bar{a}}=a$ for all $a,b\in R$. Given integer $n\geq 1$ and $\rho\in\{1,-1\}$, let $\mathfrak{g}_{2n,\rho}(R)$ be the subalgebra of the Lie algebra $\mathfrak{gl}_{2n}(R)$ generated by $\{f_{ij}(a),1\leq i\neq j\leq n\}$, $\{g_{ij}(a),1\leq i\leq j\leq n\}$ and $\{h_{ij}(a),1\leq i\leq j\leq n\}$, where $f_{ij}(a)$, $g_{ij}(a)$ and $h_{ij}(a)$ are defined:

[TABLE]

We will frequently need to know how the generators of $\mathfrak{g}_{2n,\rho}(R)$ bracket with one another:

Proposition 2.1**.**

Given $1\leq i,j,k,l\leq n$ and $a,b\in R$,

[TABLE]

Also note that

[TABLE]

Proof.

Direct calculation yields this proposition. ∎

Proposition 2.2**.**

If $n\geq 3$, then the Lie algebra $\mathfrak{g}_{2n,\rho}(R)$ is perfect, i.e.,

[TABLE]

Proof.

Since $\mathfrak{g}_{2n,\rho}(R)$ is a subalgebra, $\left[\mathfrak{g}_{2n,\rho}(R),\ \mathfrak{g}_{2n,\rho}(R)\right]\subset\mathfrak{g}_{2n,\rho}(R)$. To show the reverse inclusion, we show that the generators of $\mathfrak{g}_{2n,\rho}(R)$ can be expressed as elements of $\left[\mathfrak{g}_{2n,\rho}(R),\ \mathfrak{g}_{2n,\rho}(R)\right]$. Indeed let $a\in R$ and $1\leq i,j\leq n$ with $i\neq j$. Since $n\geq 3$, we may choose an index $k$ such that $k\neq i$ and $k\neq j$. But then

[TABLE]

Next take $1\leq i,j\leq n$, allowing $i=j$. Again choose $k$ such that $k\neq i$ and $k\neq j$. Then

[TABLE]

and

[TABLE]

We begin by examining the structure of $\mathfrak{g}_{2n,\rho}(R)$. Let $R_{\pm}=\{a\in R\,|\,\bar{a}=\pm a\}$. Then

[TABLE]

We will make use of the following lemma:

Lemma 2.3**.**

Let $\mathcal{A}$ be an associative $k$-algebra and $\varphi$ an involution on $\mathcal{A}$, then the subspace

[TABLE]

of $\mathcal{A}$ is a Lie algebra under the operation $[a,b]=ab-ba$.

Let

[TABLE]

Then $M$ is an invertible $2n$ by $2n$ matrix and $\overline{M}^{t}=\rho M$, where $A^{t}$ means the transpose of a matrix $A$. Using the matrix $M$, define a map

[TABLE]

Since $\overline{M}^{t}=\rho M$,

[TABLE]

and

[TABLE]

That is, ∗ is an involution on the associative algebra $M_{2n}(R)$. By Lemma 2.3,

[TABLE]

is a Lie subalgebra of $\mathfrak{gl}_{2n}(R)$. The general form of a matrix in $\mathcal{L}_{\rho}$ is

[TABLE]

where $P,S,T\in M_{n}(R)$.

For $i=2,\ldots,n$ and $a\in R$, observe that $f_{ii}(a)=\left(f_{ii}(a)-f_{11}(a)\right)+f_{11}(a)$. So, using the description in (2.2), a spanning set for $\mathcal{L}_{\rho}$ is

[TABLE]

Let $\mathcal{G}_{\rho}$ denote the commutator algebra of $\mathcal{L}_{\rho}$, that is, $\mathcal{G}_{\rho}=[\mathcal{L}_{\rho},\mathcal{L}_{\rho}]$. For given $Y=(y_{ij})\in M_{2n}(R)$, the trace of $Y$ is defined to be $\mathrm{tr}\,Y=\sum\limits_{i=1}^{2n}y_{ii}\in R$. Then we have the following Lemma:

Lemma 2.4**.**

For $n\geq 2$, the elements of $\mathcal{G}_{\rho}$ are precisely the elements of $\mathcal{L}_{\rho}$ whose trace lies in $[R,R]$, i.e.,

[TABLE]

Proof.

For any $A,B\in\mathfrak{gl}_{2n}\,(R)$, $\mathrm{tr}\,([A,B])\in[R,R]$ implies that

[TABLE]

To show the reverse inclusion, we begin by observing that a typical element $Y$ in $\mathcal{L}_{\rho}$ can be described as

[TABLE]

where $p$, $q_{i}$, $r_{ij}$, $s_{ij}$, $u_{ij}$, $t_{i}$ and $v_{i}$ are elements of $R$. Also note that

[TABLE]

So let us take such an element $Y$ in $\mathcal{L}_{\rho}$ such that its trace lies in $[R,R]$, that is, $p-\overline{p}\in[R,R]$. We will show that each term in the above sum lies in $\mathcal{G}_{\rho}=[\mathcal{L}_{\rho},\,\mathcal{L}_{\rho}]$. Indeed, observe that

1. (1)

for $2\leq i\leq n$, $f_{ii}(q_{i})-f_{11}(q_{i})=\left[f_{i1}(q_{i}),\ f_{1i}(1)\right]$, 2. (2)

for $1\leq i\neq j\leq n$, $f_{ij}(r_{ij})=\left[f_{ii}(r_{ij}),f_{ij}(1)\right]$, 3. (3)

for $1\leq i<j\leq n$, $g_{ij}(s_{ij})=\left[f_{ii}(s_{ij}),g_{ij}(1)\right]$, $h_{ij}(u_{ij})=\left[f_{jj}(-1),h_{ij}(u_{ij})\right]$, 4. (4)

for $1\leq i\leq n$, $g_{ii}(t_{i})=\left[f_{ii}(\frac{1}{2}),g_{ii}(t_{i})\right]$ and $h_{ii}(v_{i})=\left[f_{ii}(-\frac{1}{2}),h_{ii}(v_{i})\right]$.

A slightly fancier argument is needed to show that $f_{11}(p)\in[\mathcal{L}_{\rho},\,\mathcal{L}_{\rho}]$. Since $p\in R$, there exist unique $x\in R_{+}$ and $y\in R_{-}$ such that $p=x+y$. So $p-\overline{p}=2y\in R_{-}\cap[R,R]$. Since $R_{+}\oplus\left(R_{-}\cap[R,R]\right)=R_{+}+[R,R]$, we can rewrite $p$ as $p=a+\sum_{k}[b_{k},\,c_{k}]$ for some $a\in R_{+}$ and $b_{k},c_{k}\in R$. Since $a\in R_{+}$, we have now

[TABLE]

is also a member of $[\mathcal{L}_{\rho},\,\mathcal{L}_{\rho}]=\mathcal{G}_{\rho}$ like the other summands of $Y$. Hence $Y\in[\mathcal{L}_{\rho},\,\mathcal{L}_{\rho}]$ and

[TABLE]

$\mathcal{G}_{\rho}$ has an abelian subalgebra \mathcal{H}=\Bigl{\{}\sum\limits_{i=1}^{n}a_{i}(e_{ii}-e_{n+i,n+i})|a_{i}\in k\Bigr{\}} of dimension $n$. The linear functions $\epsilon_{i}$ in the dual space $\mathcal{H}^{*}$, defined by \epsilon_{i}\biggl{(}\sum\limits_{j=1}^{n}a_{j}(e_{jj}-e_{n+j,n+j})\biggr{)}=a_{i} for $i=1,\ldots,n$, permit us to recognize $\mathcal{G}_{\rho}$’s structure as that of a graded Lie algebra:

[TABLE]

where $\mathcal{G}_{\alpha}=\{x\in\mathcal{G}_{\rho}|[h,x]=\alpha(h)x,\text{ for all }h\in\mathcal{H}\}$ for a given $\alpha\in\mathcal{H}^{*}$. Moreover,

[TABLE]

and

[TABLE]

where the right hand sides of (2.4) and (2.5) are defined in a natural way.

Now we need

Lemma 2.5**.**

For $n\geq 2$,

[TABLE]

where $\Delta_{C}$ denotes the root system of type $C$.

Proof.

By the definition of $\mathcal{G}_{\alpha}$, we have that $\sum\limits_{\mu\in\Delta_{C}}[\mathcal{G}_{\mu},\mathcal{G}_{-\mu}]\subset\mathcal{G}_{0}$. The reverse inclusion follows from the proof of Lemma 2.4 and

[TABLE]

Proposition 2.6**.**

If $n\geq 2$, then $\mathcal{G}_{\rho}$ is a $\mathrm{BC}_{n}$-graded Lie algebra with grading subalgebra of type $C_{n}$ if $\rho=-1$ and type $D_{n}$ if $\rho=1$. Furthermore, $\mathcal{G}_{\rho}=\mathfrak{g}_{2n,\rho}(R)$.

Proof.

The first part of the theorem comes from (2.3)-(2.5), Lemma 2.5 and Example 1.16 in [3]. For more information on $\mathrm{BC}_{n}$-graded Lie algebras, see [3]. Now, by Lemma 2.5 again, we get that $\mathcal{G}_{\rho}$ and $\mathfrak{g}_{2n,\rho}(R)$ have the same generators. ∎

Remark 2.7**.**

We gave two definitions of the Lie algebra $\mathfrak{g}_{2n,\rho}$: It is easy to show the perfectness from the definition given by generators. The second one can be used for writing the elements of $\mathfrak{g}_{2n,\rho}$ explicitly and for presenting the $\mathrm{BC}$-graded structure of $\mathfrak{g}_{2n,\rho}$.

Thus,

Corollary 2.8**.**

If $n\geq 2$, then

[TABLE]

We get some familiar Lie algebras in $\mathfrak{g}_{2n,\rho}(R)$ for different choices of $R$.

Example 2.9**.**

1. (1)

Let $R=\mathbb{C}$ with the identity map as its involution. Then $\mathfrak{g}_{2n,\rho}(\mathbb{C})$ is a semisimple Lie algebra for $n\geq\frac{3+\rho}{2}$ in [13]. More explicitly, $\mathfrak{g}_{2n,-1}(\mathbb{C})$ is the simple Lie algebra of type $C_{n}$ for $n\geq 1$ and $\mathfrak{g}_{2n,1}(\mathbb{C})$ is the simple Lie algebra of type $D_{n}$ for $n\geq 3$. For $n=2$, $\mathfrak{g}_{4,1}(\mathbb{C})$ is the semisimple Lie algebra of type $D_{2}$, which is two copies of $\mathfrak{sl}_{2}(\mathbb{C})$. 2. (2)

Let $R=\mathbb{C}[t^{\pm 1}]$ be the Laurent polynomial and the involution is also the identity map. Then $\mathfrak{g}_{2n,\rho}(\mathbb{C}[t^{\pm 1}])\cong\mathfrak{g}_{2n,\rho}(\mathbb{C})\otimes\mathbb{C}[t^{\pm 1}]$ is the corresponding centerless core of an untwisted affine Kac-Moody algebra in [15].

Example 2.10**.**

Let $R=M_{m}(k)$ be the $m$ by $m$ matrix algebra with involution $\bar{\;\;}$ defined by $\overline{A}=A^{t}$. Then $\mathfrak{g}_{2n,\rho}(M_{m}(k))$ is isomorphic to $\mathfrak{g}_{2nm,\rho}(k)$ for $n\geq 2$.

Proof.

The isomorphism $\varphi$ is given by

[TABLE]

and for all $1\leq i<j\leq n,\;1\leq k,l\leq m\,$

[TABLE]

if $\rho=-1$, for all $1\leq i\leq n,\;1\leq k\leq l\leq m$,

[TABLE]

if $\rho=1$, for all $1\leq i\leq n,\;1\leq k<l\leq m$,

[TABLE]

where $E_{ij}$ is the $m\times m$ matrix with $1$ at $(i,j)$ position and [math] otherwise. ∎

Remark 2.11**.**

For $n\geq 2$ and $R=M_{m}(k)$, from Corollary 2.8 we have

[TABLE]

and

[TABLE]

Example 2.12**.**

Let $S$ be a unital associative $\mathbb{C}$-algebra and $R=S\oplus S^{op}$ with involution $\overline{(a,b)}=(b,a)$ for $a,b\in S$. Then $\mathfrak{g}_{2n,\rho}(R)$ is isomorphic to $\mathfrak{sl}_{2n}(S)$ for all $n\geq 2$. In particular, if $S=M_{m}(\mathbb{C})$, where $m\geq 1$, then $\mathfrak{g}_{2n,\rho}(M_{m}(\mathbb{C})\oplus M_{m}(\mathbb{C})^{op})\cong\mathfrak{sl}_{2nm}(\mathbb{C})$.

Proof.

Let $\varphi$ : $\mathfrak{g}_{2n,\rho}(R)\rightarrow\mathfrak{sl}_{2n}(S)$ be defined by

[TABLE]

[TABLE]

for all $1\leq i\neq j\leq n$ and

[TABLE]

[TABLE]

for all $1\leq i,j\leq n$.

In the reverse direction, we can define $\psi$ : $\mathfrak{sl}_{2n}(S)\rightarrow\mathfrak{g}_{2n,\rho}(R)$ by

[TABLE]

[TABLE]

[TABLE]

and

[TABLE]

We get that $\varphi\circ\psi=\mathrm{id}_{\mathfrak{sl}_{2n}(S)}$ and $\psi\circ\varphi=\mathrm{id}_{\mathfrak{g}_{2n,\rho}(R)}$. Therefore $\varphi$ is an isomorphism of Lie algebras. ∎

Remark 2.13**.**

If $R$ is an associative algebra over $\mathbb{C}$, then for $n\geq 2$ we have

[TABLE]

where $\mathfrak{eu}_{2n}(R,\bar{\;\;\;})$ is the elementary unitary Lie algebra studied by Zheng, Chang and Gao in [28]. Moreover, finite-dimensional irreducible representations of the elementary unitary Lie algebras $\mathfrak{eu}_{2n}(\mathbb{C}_{q},\bar{\;\;\;})$ have been studied in [10].

Proof.

The following map is an isomorphism between the two algebras:

[TABLE]

where

[TABLE]

Given an involutive ideal $J$ of $R$, that is, an ideal closed under involution, and $n\geq 2$, consider the three subspaces of $\mathfrak{g}_{2n,\rho}(R)$ defined as follows:

[TABLE]

[TABLE]

[TABLE]

where

[TABLE]

$\hat{J}$ is the ideal $\{a\in R\,|\,[a,R]\subset J\}$ of Lie algebra $R$, and $\widehat{J_{+}}=\{a\in R\,|\,a+\bar{a}\in J\}$.

Remark 2.14**.**

If $J$ is an involutive ideal of $R$ then $\underline{\mathfrak{g}_{2n,\rho}}(J)\subset\overline{\mathfrak{g}_{2n,\rho}}(J)\subset\widetilde{\mathfrak{g}_{2n,\rho}}(J)$. Moreover, if

[TABLE]

that is, $[R,R]\cap J_{-}\subset[J,R]$, then

[TABLE]

Lemma 2.15**.**

For any involutive ideal $J$ of $R$, $\widetilde{\mathfrak{g}_{2n,\rho}}(J)$, $\overline{\mathfrak{g}_{2n,\rho}}(J)$ and $\underline{\mathfrak{g}_{2n,\rho}}(J)$ are all ideals of $\mathfrak{g}_{2n,\rho}(R)$. ∎

We study the quotient $\mathfrak{g}_{2n,\rho}(R)/\overline{\mathfrak{g}_{2n,\rho}}(J)$ in the following theorem:

Proposition 2.16**.**

If $J$ is an involutive ideal of $R$, then the involution $\,\bar{\;}\,$ on $R$ induces an involution on the quotient $R/J$ in a natural way, which we again denote $\,\bar{\;}\,$. For $n\geq 2$, we have $\mathfrak{g}_{2n,\rho}(R)/\overline{\mathfrak{g}_{2n,\rho}}(J)\cong\mathfrak{g}_{2n,\rho}(R/J)$.

Proof.

Let $\pi:\mathfrak{g}_{2n,\rho}(R)\to\mathfrak{g}_{2n,\rho}(R/J)$ be given by $(a_{ij})\mapsto(a_{ij}+J)$. $\pi$ is a surjective homomorphism. Furthermore, $(a_{ij}+J)=(0+J)$ if and only if $a_{ij}\in J$ for all $i,j$. That is, $\ker\pi=\mathfrak{gl}_{2n}(J)\cap\mathfrak{g}_{2n,\rho}(R)$. By Corollary 2.8 and the definition of $\overline{\mathfrak{g}_{2n,\rho}}(J)$, this means that $\ker\pi=\overline{\mathfrak{g}_{2n,\rho}}(J)$. ∎

Conversely, given an ideal $\mathcal{J}$ in $\mathfrak{g}_{2n,\rho}(R)$, the following theorem gives us an involutive ideal $J$ of $R$ corresponding to $\mathcal{J}$.

Theorem 2.17**.**

Let $n\geq 3$ and $\mathcal{J}$ be an ideal of $\mathfrak{g}_{2n,\rho}(R)$.

1. (1)

The subsets

[TABLE]

are all involutive ideals of R and independent of the choice of $(i,j)$. We denote them $J_{1}$, $J_{2}$ and $J_{3}$, respectively. Moreover, we have

[TABLE] 2. (2)

Let

[TABLE]

Then

[TABLE] 3. (3)

Let

[TABLE]

Then they are independent of $i$ and we denote the common set by $I$. Furthermore,

[TABLE] 4. (4)

Let

[TABLE]

Then

[TABLE]

and

[TABLE]

Furthermore,

[TABLE] 5. (5)

$\mathcal{J}$ has the following vector space decomposition

[TABLE]

where

[TABLE]

and

[TABLE]

Moreover,

[TABLE]

Proof.

(1). Fix $1\leq i\neq j\leq n$ and $1\leq k\neq l\leq n$. Take $a\in J_{ij}$.

1. (i)

Because $f_{ij}(a)\in\mathcal{J}$, then

[TABLE]

So $a\in J_{ji}$ and $J_{ij}\subset J_{ji}$. 2. (ii)

Suppose $l\notin\{i,j\}$. Then we have $[f_{ij}(a),f_{jl}(1)]=f_{il}(a)\in\mathcal{J}$. So $a\in J_{il}$ and $J_{ij}\subset J_{il}$.

The following table illustrates how $J_{ij}\subset J_{kl}$ across the other various possibilities for $k$ and $l$.

[TABLE]

Thus, $J_{ij}=J_{kl}$ for all $1\leq i\neq j\leq n$ and $1\leq k\neq l\leq n$. Let $J_{1}$ denote this common set $J_{ij}$.

Similarly, we have $J_{i,n+j}=J_{k,n+l}$ and $J_{n+i,j}=J_{n+k,l}$ for all $1\leq i\neq j\leq n$, $1\leq k\neq l\leq n$ respectively. Let $J_{2}$ $J_{3}$ denote these common set $J_{i,n+j}$ and $J_{n+i,j}$ respectively.

Note that, for all $a\in R$, we have

[TABLE]

and

[TABLE]

Hence $J_{1}\subset J_{2}\subset J_{3}\subset J_{1}$, i.e., $J_{1}=J_{2}=J_{3}$. We denote this common set by $J$.

$J$ is, in fact, an involutive ideal of $R$. Indeed, let $a,b\in J$, $\alpha\in k$, and $r\in R$. Then $f_{12}(a)+f_{12}(b)=f_{12}(a+b)\in\mathcal{J}$ and $\alpha f_{12}(a)=f_{12}(\alpha a)\in\mathcal{J}$. That is, $a+b$ and $\alpha a$ are members of $J$. Since $[f_{12}(a),f_{23}(r)]=f_{13}(ar)$ and $[f_{31}(r),f_{12}(a)]=f_{32}(ra)$. That is, $ar\in J$ and $ra\in J$. Thus $J$ is an ideal of $R$. Furthermore, we have that $[f_{12}(a),g_{32}(1)]=g_{31}(\,\overline{a}\,)\in\mathcal{J}$. That is, $\,\overline{a}\,\in J$. So $J$ is closed under involution.

(2) Fix $1\leq i\neq j\leq n$. Take $a\in I_{ij}$ and choose $p\notin\{i,j\}$, then

[TABLE]

which implies that $a\in J_{i,n+p}=J$. So $I_{ij}\subset J$.

For the reverse inclusion, take $a\in J$, then

[TABLE]

which implies that $a\in I_{ij}$. That is, $J\subset I_{ij}$.

(3) Let $1\leq i\leq n$ and $a\in I_{i}$. Then $f_{ii}(a)\in\mathcal{J}$. Take any $1\leq j\leq n$ with $j\neq i$, then

[TABLE]

But then $f_{jj}(a)=\left(f_{jj}(a)-f_{ii}(a)\right)+f_{ii}(a)\in\mathcal{J}$, implying that $a\in I_{j}$. That is, $I_{i}=I_{j}$. We denote this common set by $I$.

Next, let us show that $J_{+}+[J,R]\subset I$. Take $a\in J_{+}$. Then $\,\overline{a}\,=a$ and

[TABLE]

Hence $a\in I_{i}=I$ and $J_{+}\subset I$. If $a\in J$ and $b\in R$, then

[TABLE]

So $[a,b]\in I_{i}=I$ and $[J,R]\subset I$. Thus $J_{+}+[J,R]\subset I$.

Next, we show that $I\subset\left(R_{+}+[R,R]\right)\cap J$. Take $a\in I$. Since $f_{11}(a)\in\mathcal{J}\subset\mathfrak{g}_{2n,\rho}(R)$. But then $a$ must be a member of $R_{+}+[R,R]$. Hence $I\subset R_{+}+[R,R]$. Moreover, $f_{11}(a)\in\mathcal{J}$ implies that

[TABLE]

That is, $a\in J_{1,n+2}=J$. So $I\subset J$. Thus, $I\subset\left(R_{+}+[R,R]\right)\cap J$.

(4) Fix $1\leq i\leq n$. Since $g_{ii}(a)=-\rho g_{ii}(\,\overline{a}\,)$, thus $J_{i,n+i}=\overline{\overline{J_{i,n+i}}}\subset\overline{J_{i,n+i}}\subset J_{i,n+i}$. That is $\overline{J_{i,n+i}}=J_{i,n+i}$. The proof that $\overline{J_{n+i,i}}=J_{n+i,i}$ is similar.

Take $a\in J$. Choose $1\leq j\leq n$ such that $j\neq i$. Since

[TABLE]

So $a\in J_{i,n+i}$ and $J\subset J_{i,n+i}$.

Next, we show that $J_{i,n+i}\subset J_{-\rho}\oplus R_{\rho}$. Take $a\in\left(J_{i,n+i}\right)_{-\rho}$. Then $g_{ii}(a)\in\mathcal{J}$ and $\,\overline{a}\,=-\rho a$. Choose an index $j$ between $1$ and $n$ such that $j\neq i$. Then

[TABLE]

So $2a$ and, hence, $a\in J_{i,n+j}=J$. That is, $\left(J_{i,n+i}\right)_{-\rho}\subset J=J_{-\rho}\oplus J_{\rho}\subset J_{i,n+i}$, it follows $\left(J_{i,n+i}\right)_{-\rho}=J_{-\rho}$. But then $J_{i,n+i}=\left(J_{i,n+i}\right)_{-\rho}\oplus\left(J_{i,n+i}\right)_{\rho}=J_{-\rho}\oplus\left(J_{i,n+i}\right)_{\rho}\subset J_{-\rho}\oplus R_{\rho}$.

The proof that $J\subset J_{n+i,i}\subset J_{-\rho}\oplus R_{\rho}$ is similar.

Next, we show that $g_{ii}(J_{i,n+i})\subset g_{ii}(J)$. Take $g_{ii}(a)\in g_{ii}(J_{i,n+i})$. So $a\in J_{i,n+i}$. Since $J_{i,n+i}\subset J_{-\rho}\oplus R_{\rho}$, we may write $a=a_{-\rho}+a_{\rho}$ for some $a_{-\rho}\in J_{-\rho}$ and $a_{\rho}\in R_{\rho}$. But then

[TABLE]

Thus $g_{ii}(J_{i,n+i})=g_{ii}(J)$. The proof that $h_{ii}(J_{n+i,i})=h_{ii}(J)$ is similar.

(5). Using Proposition 1.5 in Kac [K], it is enough to show (2.7). Assume $u=\sum\limits_{i=1}^{n}f_{ii}(a_{i})\in\mathcal{J}$. Then

[TABLE]

By part (1), we have $a_{i}-a_{j}\in J$. Set $a=\frac{\sum\limits_{i=1}^{n}a_{i}}{n}$. Then $a_{i}-a=\frac{\sum\limits_{j=1}^{n}(a_{i}-a_{j})}{n}\in J$. Now, rewrite $u$ as follows

[TABLE]

Hence $(\sum\limits_{i=1}^{n}f_{ii})(a)\in\mathcal{J}$, and (2.7) holds.

From parts (1)-(4), we first have $\mathcal{J}\supset\underline{\mathfrak{g}_{2n,\rho}}(J)$. In order to prove (2.8), it is enough to show

[TABLE]

Let $u=(\sum\limits_{i=1}^{n}f_{ii})(a)\in\mathcal{J}$. For $b\in R$, we have $[u,f_{12}(b)]=f_{12}([a,b])\in\mathcal{J}$, which implies $[a,b]\in J$ for all $b\in R$, i.e., $a\in\hat{J}$. On the other hand, we have $[u,g_{12}(b)]=g_{12}(a+\bar{a})\in\mathcal{J}$, i.e., $a+\bar{a}\in J$ and $a\in\widehat{J_{+}}$. ∎

Corollary 2.18**.**

Let $R$ be an involutive simple associative algebra $($has no nontrivial involutive ideal$)$ and $[R,R]\cap Z(R)\cap R_{-}=0$, then $\mathfrak{g}_{2n,\rho}(R)$ is a simple Lie algebra for any $n\geq 3$.

Proof.

Let $\mathcal{J}$ be an ideal of $\mathfrak{g}_{2n,\rho}(R)$. $J$ is the ideal of $R$ corresponding to $\mathcal{J}$ . Since $R$ is involutive simple, we have $J=0$ or $R$. If $J=0$, then $\hat{J}=Z(R)$ and $\widehat{J_{+}}=R_{-}$, it follows from (2.8) that $\mathcal{J}=0$. If $J=R$, we also have $\mathcal{J}=\mathfrak{g}_{2n,\rho}(R)$. ∎

Remark 2.19**.**

In Example 2.10, since

[TABLE]

and $M_{m}(k)$ is simple, hence without proof we can claim that $\mathfrak{g}_{2n,\rho}(M_{m}(k))$ are simple for $n\geq 3$.

In Example 2.12, since $M_{m}(\mathbb{C})\oplus M_{m}(\mathbb{C})^{\rm{op}}$ is involutive simple and $Z(M_{m}(\mathbb{C})\oplus M_{m}(\mathbb{C})^{\rm{op}})\cap[M_{m}(\mathbb{C})\oplus M_{m}(\mathbb{C})^{\rm{op}},M_{m}(\mathbb{C})\oplus M_{m}(\mathbb{C})^{\rm{op}}]=0$. Then we have that $\mathfrak{g}_{2n,\rho}(M_{m}(\mathbb{C})\oplus M_{m}(\mathbb{C})^{\rm{op}})$ are simple for $n\geq 3$.

We end the section by giving more examples:

Example 2.20**.**

Let $\mathcal{K}_{4}=\{1,\tau,\gamma,\tau\gamma\}$ be the Klein four group and we denote it simply by $\mathcal{K}$. Let $\mathbb{C}\mathcal{K}$ be the group algebra of $\mathcal{K}$ over $\mathbb{C}$ and the involution given by

[TABLE]

Then we have

[TABLE]

Proof.

Firstly, we have $\overline{\tau\gamma}=\bar{\gamma}\bar{\tau}=\tau\gamma$ and

[TABLE]

Then all the involutive ideals of $\mathbb{C}\mathcal{K}$ are trivial ideals [math], $\mathbb{C}\mathcal{K}$ and

[TABLE]

It is obvious that

[TABLE]

where we consider $J_{1}$ and $J_{2}$ as associative algebras with identity $\frac{1+\tau\gamma}{2}$ and $\frac{1-\tau\gamma}{2}$ respectively and the involution induced by the involution on $\mathbb{C}\mathcal{K}$. Moreover, by Lemma 4.2 in what follows we have

[TABLE]

Now by Example 2.12, we just need show

[TABLE]

In fact, the isomorphisms are given by

[TABLE]

Example 2.21**.**

Define a new involution $\,\widetilde{\;}\,$ on $M_{2}(\mathbb{C})$ given by

[TABLE]

Now we denote $M_{2}(\mathbb{C})$ with this involution by $\widetilde{M}_{2}(\mathbb{C})$. Then we have

[TABLE]

Proof.

By Example 2.10, we just need prove

[TABLE]

At first, we have $\widetilde{M}_{2}(\mathbb{C})$ is a simple associative algebra, $Z(\widetilde{M}_{2}(\mathbb{C}))=\mathbb{C}I_{2}$ and $\widetilde{E_{ab}}=(-1)^{a+b}E_{3-b,3-a}$, therefore we have

[TABLE]

and $\mathfrak{g}_{2n,\rho}(\widetilde{M}_{2}(\mathbb{C}))$ is a simple Lie algebra. Computing the dimension of these Lie algebras we have

[TABLE]

Now we give an isomorphism $\varphi$ from $\mathfrak{g}_{2n,-\rho}(M_{2}(\mathbb{C}))$ to $\mathfrak{g}_{2n,\rho}(\widetilde{M}_{2}(\mathbb{C}))$:

[TABLE]

and for all $1\leq i,j\leq n$

[TABLE]

3. The quantum torus $\mathbb{C}_{q}$

In this section, we discuss the properties of the quantum torus $\mathbb{C}_{q}[t_{1}^{\pm 1},t_{2}^{\pm 1}]$ as an associative algebra with involution. The main goal of this section is the involutive ideals of $\mathbb{C}_{q}$.

The quantum torus $\mathbb{C}_{q}[t_{1}^{\pm 1},t_{2}^{\pm 1}]$ is the associative algebra over $\mathbb{C}$ generated by $t_{1}^{\pm 1}$, $t_{2}^{\pm 1}$ and subject to the relations $t_{1}t_{1}^{-1}=t_{1}^{-1}t_{1}=t_{2}t_{2}^{-1}=t_{2}^{-1}t_{2}=1$, $t_{2}t_{1}=qt_{1}t_{2}$ (see [18]). We denote it simply by $\mathbb{C}_{q}$. It is easy to see that $t_{1}^{i}t_{2}^{j}$, $i,j\in\mathbb{Z}$ form a basis of $\mathbb{C}_{q}$ and $\mathbb{C}_{q}$ is a $\mathbb{Z}\times\mathbb{Z}$-graded algebra. If $q=\pm 1$, then $\mathbb{C}_{q}$ is called an elementary quantum torus (cf. [27]). Yoshii in his PhD thesis proved that

Proposition 3.1** (Proposition 3.2.7 in [26]).**

There exists a graded involution on $\mathbb{C}_{q}$ if and only if the quantum torus $\mathbb{C}_{q}$ is elementary. In this case, $\overline{t_{1}}=\theta_{1}t_{1}$ and $\overline{t_{2}}=\theta_{2}t_{2}$ with $\theta_{i}=\pm 1$ for $i=1,2$.

In particular, $\mathbb{C}_{q}$ is symmetric about $t_{1}$ and $t_{2}$. So, instead of looking at all eight possible triples $(q,\theta_{1},\theta_{2})$, we can, without loss of generality, restrict our attention to the parameters in the following six cases:

[TABLE]

We denote the set on the right by $\mathcal{P}$.

Lemma 3.2** (Proposition 2.44 in [9]).**

Let $q$ be a primitive root of unit of order $m$. Then we have

1. (1)

The center $Z(\mathbb{C}_{q})$ of $\mathbb{C}_{q}$ has a basis consisting of monomials $t_{1}^{k}t_{2}^{l}$ for $m|k$ and $m|l$. 2. (2)

The Lie algebra $[\mathbb{C}_{q},\mathbb{C}_{q}]$ has a basis consisting of monomials $t_{1}^{k}t_{2}^{l}$ for $m\nmid k$ or $m\nmid l$. 3. (3)

$\mathbb{C}_{q}=[\mathbb{C}_{q},\mathbb{C}_{q}]\oplus Z(\mathbb{C}_{q})$.

In the rest part of this paper, we always assume that $q=\pm 1$ and $m$ is the order of $q$. Note that $m=1$ or $2$.

Definition 3.3**.**

For $\alpha,\beta\in\mathbb{C}^{*}$ and $(q,\theta_{1},\theta_{2})\in\mathcal{P}$, we define the ideal $J(\alpha,\beta)$ of $\mathbb{C}_{q}$ as follows:

1. (1)

If $q=-1$ or $(q,\theta_{1},\theta_{2})=(1,-1,-1)$, then $J(\alpha,\beta)$ is generated by $(t_{1}-\alpha)(t_{1}+\alpha)$, $(t_{2}-\beta)(t_{2}+\beta)$. 2. (2)

If $(q,\theta_{1},\theta_{2})=(1,1,1)$, then $J(\alpha,\beta)$ is generated by $t_{1}-\alpha$, $t_{2}-\beta$. 3. (3)

If $(q,\theta_{1},\theta_{2})=(1,1,-1)$, then $J(\alpha,\beta)$ is generated by $t_{1}-\alpha$, $(t_{2}-\beta)(t_{2}+\beta)$.

Remark 3.4**.**

By Lemma 3.2, for $\alpha,\beta\in\mathbb{C}^{*}$, we know that the generators of the ideal $J(\alpha,\beta)$ always lie in the center of $\mathbb{C}_{q}$. So each element of $J(\alpha,\beta)$ can be written as a $\mathbb{C}_{q}-$linear combination of the two generators.

Lemma 3.5**.**

Let $\alpha,\beta\in\mathbb{C}^{*}$, then $J(\alpha,\beta)$ is an involutive ideal.

Proof.

Let $x=(t_{1}-\alpha)(t_{1}+\alpha)$, $y=(t_{2}-\beta)(t_{2}+\beta)$, $u=t_{1}-\alpha$, $v=t_{2}-\beta$, and $(q,\theta_{1},\theta_{2})\in\mathcal{P}$. Note that

[TABLE]

Similarly, $\overline{y}=y$. Also, $\overline{u}=\theta_{1}t_{1}-\alpha$ and $\overline{v}=\theta_{2}t_{2}-\beta$.

1. (i)

Suppose $q=-1$ or $(q,\theta_{1},\theta_{2})=(1,-1,-1)$. Since a typical element of $J(\alpha,\beta)$ is of the form $ax+by$ for some $a,b\in\mathbb{C}_{q}$. But then

[TABLE] 2. (ii)

Suppose $(q,\theta_{1},\theta_{2})=(1,1,1)$. Take a typical element $au+bv$ in $J(\alpha,\beta)$, where $a,b\in\mathbb{C}_{q}$. Note that $\overline{u}=u$ and $\overline{v}=v$ since $\theta_{1}=\theta_{2}=1$. Then

[TABLE] 3. (iii)

Suppose $(q,\theta_{1},\theta_{2})=(1,1,-1)$. Take a typical element $au+by$ in $J(\alpha,\beta)$, where $a,b\in\mathbb{C}_{q}$. Note that $\overline{u}=u$ since $\theta_{1}=1$. Then

[TABLE]

So, across all the possibilities for $(q,\theta_{1},\theta_{2})$, $J(\alpha,\beta)$ is closed under involution. ∎

Lemma 3.6**.**

For $\alpha,\beta\in\mathbb{C}^{*}$ and $(q,\theta_{1},\theta_{2})\in\mathcal{P}$, we have

1. (1)

If $(q,\theta_{1},\theta_{2})=(-1,1,1)$ or $(-1,1,-1)$, there is an isomorphism

[TABLE]

preserving the involutions. 2. (2)

If $(q,\theta_{1},\theta_{2})=(-1,-1,-1)$, there is an isomorphism

[TABLE]

preserving the involutions, where the involution of $\widetilde{M}_{2}(\mathbb{C})$ is defined as in Example 2.5. 3. (3)

If $(q,\theta_{1},\theta_{2})=(1,1,1)$, there is an isomorphism

[TABLE]

preserving the involutions, where the involution of $\mathbb{C}$ is the identity map. 4. (4)

If $(q,\theta_{1},\theta_{2})=(1,1,-1)$, there is an isomorphism

[TABLE]

preserving the involutions, where the involution of $\mathbb{C}\oplus\mathbb{C}^{\rm{op}}$ is defined as in Example 2.12. 5. (5)

If $(q,\theta_{1},\theta_{2})=(1,-1,-1)$, there is an isomorphism

[TABLE]

preserving the involutions, where the involution of $\mathbb{C}\mathcal{K}$ is defined as in Example 2.20.

Proof.

1. (1)

Define an algebra homomorphism $\varphi:\mathbb{C}_{q}\to M_{2}(\mathbb{C})$ by \varphi(t_{1})=\left[\begin{array}[]{rr}\alpha&0\\ 0&-\alpha\end{array}\right] and \varphi(t_{2})=\left[\begin{array}[]{rr}0&\sqrt{\theta_{2}}\beta\\ \frac{1}{\sqrt{\theta_{2}}}\beta&0\end{array}\right]. Since

[TABLE]

we see that $\varphi$ respects the relation $t_{2}t_{1}=qt_{1}t_{2}$. $\varphi$ also preserves the involution since

[TABLE]

and

[TABLE]

Since $\varphi(t_{1}^{2}-\alpha^{2})=0$ and $\varphi(t_{2}^{2}-\beta^{2})=0$, $J(\alpha,\beta)\subset\ker\varphi$. So $\varphi$ induces an algebra homomorphism $\widetilde{\varphi}:\mathbb{C}_{q}/J(\alpha,\beta)\to M_{2}(\mathbb{C})$. Since $\widetilde{\varphi}(1+J(\alpha,\beta))$, $\widetilde{\varphi}(t_{1}+J(\alpha,\beta))$, $\widetilde{\varphi}(t_{2}+J(\alpha,\beta))$ and $\widetilde{\varphi}(t_{1}t_{2}+J(\alpha,\beta))$ are linearly independent in $M_{2}(\mathbb{C})$, $\text{rank}\>\widetilde{\varphi}=4$. That is, $\widetilde{\varphi}$ is surjective. Since $\dim\left(\mathbb{C}_{q}/J(\alpha,\beta)\right)=4$, $\widetilde{\varphi}$ is also injective. Hence, $\widetilde{\varphi}$ is an isomorphism and $\mathbb{C}_{q}/J(\alpha,\beta)\cong M_{2}(\mathbb{C})$.

Similarly we can prove (2)-(5) with the following homomorphisms: 2. (2)

$\varphi:\mathbb{C}_{q}\to\widetilde{M_{2}}(\mathbb{C})$ by t_{1}\mapsto\left[\begin{array}[]{rr}\alpha&0\\ 0&-\alpha\end{array}\right], t_{2}\mapsto\left[\begin{array}[]{rr}0&\beta\\ \beta&0\end{array}\right]. 3. (3)

$\varphi:\mathbb{C}_{q}\to\mathbb{C}$ by $t_{1}\mapsto\alpha$ and $t_{2}\mapsto\beta$. 4. (4)

$\varphi:\mathbb{C}_{q}\to\mathbb{C}\oplus\mathbb{C}^{\text{op}}$ by $t_{1}\mapsto(\alpha,\alpha)$ and $t_{2}\mapsto(\beta,-\beta)$. 5. (5)

$\varphi:\mathbb{C}_{q}\to\mathbb{C}\mathcal{K}$ by $t_{1}\mapsto\alpha\tau$ and $t_{2}\mapsto\beta\gamma$. ∎

Definition 3.7**.**

Let $(q,\theta_{1},\theta_{2})\in\mathcal{P}$, and $f(T),g(T)\in\mathbb{C}[T]$ with $f(0)g(0)\neq 0$ and $\deg f\deg g>0$. Then $J(f,g)$ is the ideal of $\mathbb{C}_{q}$ generated by $f(t_{1}^{m})$ and $g(t_{2}^{m})$, where $m$ is the order of $q$.

Lemma 3.8**.**

Let $(q,\theta_{1},\theta_{2})\in\mathcal{P}$ and $f(T),g(T)\in\mathbb{C}[T]$ with $f(0)g(0)\neq 0$ and $\deg f\deg g>0$. If $f(\theta_{1}^{m}T)=f(T)$ and $g(\theta_{2}^{m}T)=g(T)$, then $J(f,g)$ is invariant under the involution $\bar{\,\;}$.

Proof.

First we have

[TABLE]

where the last equality follows by hypothesis. Similarly $\overline{g(t_{2}^{m})}=g(t_{2}^{m})$. Since $f(t_{1}^{m})$ and $g(t_{2}^{m})$ are in the center of $\mathbb{C}_{q}$, a typical element of $J(f,g)$ is of the form $af(t_{1}^{m})+bg(t_{2}^{m})$ for some $a,b\in\mathbb{C}_{q}$. But then

[TABLE]

and $\,\overline{a}\,f(t_{1}^{m})+\,\overline{b}\,g(t_{2}^{m})$ lies in $J(f,g)$. That is, $J(f,g)$ is closed under involution. ∎

Remark 3.9**.**

For a polynomial $f(T)\in\mathbb{C}[T]$ with $\deg f>0$ and $f(0)\neq 0$, the roots of $f$ are all nonzero. The condition $f(\eta^{m}T)=f(T)$ always holds if $\eta=1$ or $\eta=-1$, $m=2$. For $\eta=-1$ and $m=1$ the condition is equivalent to that if $\alpha$ is a root of $f$, then $-\alpha$ is also a root of $f$ with the same multiplicity, i.e., $f(T)$ can be written as

[TABLE]

where $\lambda\neq 0$, $s$ is a positive integer, $\pm\alpha_{1},\cdots,\pm\alpha_{s}\neq 0$ and distinct, $k_{i}>0$, $i=1,\cdots,s$.

Conversely, any polynomial $f$ of form (3.1) satisfies $f(-T)=f(T)$.

Remark 3.10**.**

Let $\mathcal{A}$ and $\mathcal{B}$ be two associative algebras with involutions. The direct sum $\mathcal{A}\oplus B$ is an associative algebra. A new involution $\,\bar{\;}\,$ can be defined by $\overline{(a,b)}=(\bar{a},\bar{b})$, for $a\in\mathcal{A}$, $b\in\mathcal{B}$. $\mathcal{A}\oplus\mathcal{B}$ becomes an associative algebra with involution. The involutions in the following lemma always have this meaning.

In the rest part of the paper, if $f,g\in\mathbb{C}[T]$, the greatest common divisor of $f$ and $g$ is denoted by $(f,g)$ and the formal derivative of $f$ is denoted by $f^{\prime}$. $f|g$ means that $f$ is a divisor of $g$.

Lemma 3.11**.**

Let $(q,\theta_{1},\theta_{2})\in\mathcal{P}$ and $f_{1},f_{2},g_{1},g_{2}\in\mathbb{C}[T]$ with $(f_{1},f_{2})=1$, $(g_{1},g_{2})=1$, $f_{1}\cdot f_{2}\cdot g_{1}\cdot g_{2}(0)\neq 0$, $\deg f_{1}\deg f_{2}\deg g_{1}\deg g_{2}>0$ and

[TABLE]

for $i=1,2$. Then there are isomorphisms of associative algebras preserving the involutions

[TABLE]

and

[TABLE]

Proof.

Let $f=f_{1}f_{2}.$ The hypotheses, in combination with Lemma 3.8, tell us that $J(f,g_{1})$, $J(f_{1},g_{1})$, and $J(f_{2},g_{1})$ are involutive ideals of $\mathbb{C}_{q}$. The involution on $\mathbb{C}_{q}$ induces involutions on the quotient algebras $\mathbb{C}_{q}/J(f,g_{1})$, $\mathbb{C}_{q}/J(f_{1},g_{1})$, and $\mathbb{C}_{q}/J(f_{2},g_{1})$. The map $\varphi:\mathbb{C}_{q}\to\mathbb{C}_{q}/J(f_{1},g_{1})\oplus\mathbb{C}_{q}/J(f_{2},g_{1})$ given by

[TABLE]

is an associative algebra homomorphism that preserves the involution on $\mathbb{C}_{q}$. Moreover, it is surjective. Indeed take any $(a+J(f_{1},g_{1}),\ b+J(f_{2},g_{1}))\in\mathbb{C}_{q}/J(f_{1},g_{1})\oplus\mathbb{C}_{q}/J(f_{2},g_{1})$, where $a,b\in\mathbb{C}_{q}$. Since $(f_{1},f_{2})=1$, there exist $h_{1},h_{2}\in\mathbb{C}[T]$ such that

[TABLE]

Let

[TABLE]

and

[TABLE]

Then

[TABLE]

and

[TABLE]

Since the kernel of this homomorphism $\varphi$ is equal to $J(f_{1},g_{1})\cap J(f_{2},g_{1})$,

[TABLE]

We now show that $J(f_{1},g_{1})\cap J(f_{2},g_{1})=J(f,g_{1})$. Since $f=f_{1}f_{2}$ and $f_{1}(t_{1}^{m})$, $f_{2}(t_{1}^{m})$ lie in the centre of $\mathbb{C}_{q}$, it follows that $J(f,g_{1})\subset J(f_{1},g_{1})\cap J(f_{2},g_{1})$. To show the reverse inclusion, let $a\in J(f_{1},g_{1})\cap J(f_{2},g_{1})$. So there exist $b_{1}$, $b_{2}$, $c_{1}$, and $c_{2}\in\mathbb{C}_{q}$ such that

[TABLE]

But then, by (3.2),

[TABLE]

which is an element of $J(f,g_{1})$. Hence $J(f_{1},g_{1})\cap J(f_{2},g_{1})=J(f,g_{1})$ and, by (3.3),

[TABLE]

The proof that $\mathbb{C}_{q}/J(f_{1},g_{1}g_{2})\ \cong\ \mathbb{C}_{q}/J(f_{1},g_{1})\oplus\mathbb{C}_{q}/J(f_{1},g_{2})$ follows similarly. ∎

4. Finite dimensional quotients of $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})$

Suppose $\mathcal{J}$ is an ideal of $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})$ such that $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})/\mathcal{J}$ is finite-dimensional and semisimple. $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})/\mathcal{J}$ are determined in this section. The classification of finite-dimensional quotients of $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})$ is the premise of the classification of finite dimensional irreducible representations of $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})$.

Lemma 4.1**.**

Let $f(T),g(T)\in\mathbb{C}[T]$ such that $J(f,g)$ is involutive ideal of $\mathbb{C}_{q}$, then for $n\geq 2$ we have

[TABLE]

Proof.

We take advantage of Remark 2.14 which says that it suffices to prove the inclusion relation $[\mathbb{C}_{q},\mathbb{C}_{q}]\cap J(f,g)_{-}\subset[\mathbb{C}_{q},J(f,g)]$. Let $u$ be an arbitrary element of $[\mathbb{C}_{q},\mathbb{C}_{q}]\cap J(f,g)_{-}$. We may write it as

[TABLE]

for some $a,b\in\mathbb{C}_{q}$. Since $\mathbb{C}_{q}=[\mathbb{C}_{q},\mathbb{C}_{q}]\oplus Z(\mathbb{C}_{q})$, we can express $a$ as $a=\tilde{a}+c$ and $b$ as $b=\tilde{b}+d$ uniquely for some $\tilde{a}=\sum\limits_{i=1}^{k}[a_{i},a_{i}^{\prime}]$ and $\tilde{b}=\sum\limits_{j=1}^{l}[b_{j},b_{j}^{\prime}]$ in $[\mathbb{C}_{q},\mathbb{C}_{q}]$, with $a_{i},a_{i}^{\prime},b_{j},b_{j}^{\prime}\in\mathbb{C}_{q}$, and $c,d\in Z(\mathbb{C}_{q})$. But then

[TABLE]

with the last equality following from $f(t_{1}^{m})$ and $g(t_{2}^{m})$ being members of $Z(\mathbb{C}_{q})$. Since $u\in[\mathbb{C}_{q},\mathbb{C}_{q}]$ and $\mathbb{C}_{q}=[\mathbb{C}_{q},\mathbb{C}_{q}]\oplus Z(\mathbb{C}_{q})$, the $Z(\mathbb{C}_{q})$-component of $u$, that is, $cf(t_{1}^{m})+dg(t_{2}^{m})$, must equal [math]. So

[TABLE]

Lemma 4.2**.**

Let $\mathcal{A}$ and $\mathcal{B}$ be two associative algebras with involutions and $\mathcal{A}\oplus\mathcal{B}$ be the associative algebra with involution introduced in Remark 3.10. Then we have

[TABLE]

Proof.

The map $\varphi:\mathfrak{g}_{2n,\rho}(\mathcal{A}\oplus\mathcal{B})\to\mathfrak{g}_{2n,\rho}(\mathcal{A})\oplus\mathfrak{g}_{2n,\rho}(\mathcal{B})$ given by

[TABLE]

is an isomorphism. ∎

Lemma 4.3**.**

Let $(q,\theta_{1},\theta_{2})\in\mathcal{P}$ and $f(T),g(T)\in\mathbb{C}[T]$ with $\deg f\deg g>0$, $f(0)g(0)\neq 0$, $(f,f^{\prime})=1$, $(g,g^{\prime})=1$ and $f,g$ satisfy

[TABLE]

Then for $n\geq 2$ we have

1. (1)

If $(q,\theta_{1},\theta_{2})=(-1,1,1)$ or $(-1,1,-1)$, there is an isomorphism

[TABLE]

where $r$ and $s$ are the degree of $f$ and $g$, respectively. 2. (2)

If $(q,\theta_{1},\theta_{2})=(-1,-1,-1)$, there is an isomorphism

[TABLE]

where $r$ and $s$ are the degree of $f$ and $g$, respectively. 3. (3)

If $(q,\theta_{1},\theta_{2})=(1,1,1)$, there is an isomorphism

[TABLE]

where $r$ and $s$ are the degree of $f$ and $g$, respectively. 4. (4)

If $(q,\theta_{1},\theta_{2})=(1,1,-1)$, there is an isomorphism

[TABLE]

where $r$ and $2s$ are the degree of $f$ and $g$, respectively. 5. (5)

If $(q,\theta_{1},\theta_{2})=(1,-1,-1)$, there is an isomorphism

[TABLE]

where $2r$ and $2s$ are the degree of $f$ and $g$, respectively.

Proof.

• (1)

Let $(q,\theta_{1},\theta_{2})=(-1,1,1)\text{ or }(-1,1,-1)$. The conditions $f(0)g(0)\neq 0$, $\deg f\deg g>0$, and $(f,f^{\prime})=(g,g^{\prime})=1$ imply that $f(T)=\lambda_{1}(T-\alpha_{1})\cdots(T-\alpha_{r})$ and $g(T)=\lambda_{2}(T-\beta_{1})\cdots(T-\beta_{s})$ for some $\lambda_{1},\lambda_{2}\in\mathbb{C}^{*}$, $r,s\geq 1$, distinct $\alpha_{1},\ldots,\alpha_{r}\in\mathbb{C}^{*}$, and distinct $\beta_{1},\ldots,\beta_{s}\in\mathbb{C}^{*}$.

Recall that $J(f,g)$ is generated by

[TABLE]

and

[TABLE]

So, by Lemma 3.11 and Lemma 3.6(1), we have

[TABLE]

Hence Lemma 4.2 and Example 2.10 imply

[TABLE]

• (2)

Let $(q,\theta_{1},\theta_{2})=(-1,-1,-1)$. The argument follows exactly the same as in part (1) except that now $\mathbb{C}_{q}/J(f,g)\cong\widetilde{M_{2}}(\mathbb{C})$, by Lemma 3.6 (2). So

[TABLE]

and by Lemma 4.2 and Example 2.21

[TABLE]

• (3)

Let $(q,\theta_{1},\theta_{2})=(1,1,1)$. The conditions $\deg f\deg g>0$, $f(0)g(0)\neq 0$, and $(f,f^{\prime})=(g,g^{\prime})=1$ imply that $f(T)=\lambda_{1}(T-\alpha_{1})\cdots(T-\alpha_{r})$ and $g(T)=\lambda_{2}(T-\beta_{1})\cdots(T-\beta_{s})$ for some $\lambda_{1},\lambda_{2}\in\mathbb{C}^{*}$, $r,s\geq 1$, distinct $\alpha_{1},\ldots,\alpha_{r}\in\mathbb{C}^{*}$, and distinct $\beta_{1},\ldots,\beta_{s}\in\mathbb{C}^{*}$. The ideal $J(f,g)$ is generated by $f(t_{1})=\lambda_{1}(t_{1}-\alpha_{1})\cdots(t_{1}-\alpha_{r})$ and $g(t_{2})=\lambda_{2}(t_{2}-\beta_{1})\cdots(t_{2}-\beta_{s})$. Hence by Lemma 3.11 we have

[TABLE]

and by Lemma 4.2

[TABLE]

• (4)

Let $(q,\theta_{1},\theta_{2})=(1,1,-1)$. We begin as in part (3) except the condition $g(\theta_{2}^{m}T)=g\left((-1)^{1}(T)\right)=g(-T)$ imposes the constraint that if $\beta_{j}$ is a root of $g$, then so is $-\beta_{j}$. So $f(T)=\lambda_{1}(T-\alpha_{1})\cdots(T-\alpha_{r})$ and $g(T)=\lambda_{2}(T-\beta_{1})(T+\beta_{1})\cdots(T-\beta_{s})(T+\beta_{s})$ for some $\lambda_{1},\lambda_{2}\in\mathbb{C}^{*}$, $r,s\geq 1$, distinct $\alpha_{1},\ldots,\alpha_{r}\in\mathbb{C}^{*}$, and distinct $\beta_{1},\ldots,\beta_{s}\in\mathbb{C}^{*}$.

Since $J(f,g)$ is generated by $f(t_{1})$ and $g(t_{2})$,

[TABLE]

and by Lemma 4.2 and Example 2.12 we arrive

[TABLE]

• (5)

Let $(q,\theta_{1},\theta_{2})=(1,-1,-1)$. Now, as in part (4), $f(T)=\lambda_{1}(T-\alpha_{1})(T+\alpha_{1})\cdots(T-\alpha_{r})(T+\alpha_{r})$ and $g(T)=\lambda_{2}(T-\beta_{1})(T+\beta_{1})\cdots(T-\beta_{s})(T+\beta_{s})$ for some $\lambda_{1},\lambda_{2}\in\mathbb{C}^{*}$, $r,s\geq 1$, distinct $\alpha_{1},\ldots,\alpha_{r}\in\mathbb{C}^{*}$, and distinct $\beta_{1},\ldots,\beta_{s}\in\mathbb{C}^{*}$. So,

[TABLE]

and by Lemma 4.2 and Example 2.20 we get

[TABLE]

Remark 4.4**.**

If we denote the elementary quantum torus with above involution by $\mathbb{C}_{q,\theta_{1},\theta_{2}}$, then Lemma 3.2.12 in [26] implies that $\mathbb{C}_{1,-1,1}\cong\mathbb{C}_{1,1,-1}\cong\mathbb{C}_{1,-1,-1}$ and $\mathbb{C}_{-1,1,1}\cong\mathbb{C}_{-1,1,-1}\cong\mathbb{C}_{-1,-1,1}$.

Lemma 4.5**.**

Suppose $n\geq 3$. Let $(q,\theta_{1},\theta_{2})\in\mathcal{P}$ and $f(T),g(T)\in\mathbb{C}[T]$ with $\deg f\deg g>0$, $f(0)g(0)\neq 0$ and $f,g$ satisfy

[TABLE]

Let $f_{0}=f/(f,f^{\prime})$ and $g_{0}=g/(g,g^{\prime})$. Then

[TABLE]

and $\overline{\mathfrak{g}_{2n,\rho}}(J(f_{0},g_{0}))/\overline{\mathfrak{g}_{2n,\rho}}(J(f,g))$ is the radical of $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})/\overline{\mathfrak{g}_{2n,\rho}}(J(f,g))$.

Proof.

The assumptions $f(T)=f(\theta_{1}^{m}T)$, $g(T)=g(\theta_{2}^{m}T)$, $f_{0}=f/(f,f^{\prime})$, and $g_{0}=g/(g,g^{\prime})$ imply that $f_{0}(\theta_{1}^{m}T)=f_{0}(T)$ and $g_{0}(\theta_{2}^{m}T)=g_{0}(T)$ (see Remark 3.9). So, by Lemma 3.8 and Lemma 4.1, we have $\overline{\mathfrak{g}_{2n,\rho}}(J(f,g))=\underline{\mathfrak{g}_{2n,\rho}}(J(f,g))$ and $\overline{\mathfrak{g}_{2n,\rho}}(J(f_{0},g_{0}))=\underline{\mathfrak{g}_{2n,\rho}}(J(f_{0},g_{0}))$. Moreover, the assumption implies $J(f,g)\subset J(f_{0},g_{0})$. So we have

[TABLE]

Hence, by Proposition 2.16,

[TABLE]

We worked out the various possibilities for $\mathfrak{g}_{2n,\rho}\left(\mathbb{C}_{q}/J(f_{0},g_{0})\right)$ in Lemma 4.3, all of which are semisimple. So $\overline{\mathfrak{g}_{2n,\rho}}(J(f_{0},g_{0}))/\overline{\mathfrak{g}_{2n,\rho}}(J(f,g))\supset\mathcal{R}$, where $\mathcal{R}$ denotes the radical of $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})/\overline{\mathfrak{g}_{2n,\rho}}(J(f,g))$.

Let us show that $\overline{\mathfrak{g}_{2n,\rho}}(J(f_{0},g_{0}))/\overline{\mathfrak{g}_{2n,\rho}}(J(f,g))$ is, in fact, equal to $\mathcal{R}$. We do this by showing that $\overline{\mathfrak{g}_{2n,\rho}}(J(f_{0},g_{0}))/\overline{\mathfrak{g}_{2n,\rho}}(J(f,g))$ is nilpotent and, hence, solvable. Let

[TABLE]

It is easy to check that

[TABLE]

where $J(f_{0},g_{0})^{1}=J(f_{0},g_{0})$ and $J(f_{0},g_{0})^{k}=J(f_{0},g_{0})^{k-1}\cdot J(f_{0},g_{0})$ for $k\geq 2$. By induction, the elements of $J(f_{0},g_{0})^{k}$ can be written as $\sum\limits_{i=0}^{k}a_{i}f_{0}(t_{1}^{m})^{i}g_{0}(t_{2}^{m})^{k-i}$, for $a_{0},a_{1},\ldots,a_{k}\in\mathbb{C}_{q}$. By the definition of $f_{0}$ and $g_{0}$, there exist positive integers $k_{1}$ and $k_{2}$ such that $f|f_{0}^{k_{1}}$ and $g|g_{0}^{k_{2}}$. Take $k>k_{1}+k_{2}$ we have

[TABLE]

It follows that for $k$ large enough,

[TABLE]

Hence $\overline{\mathfrak{g}_{2n,\rho}}(J(f_{0},g_{0}))/\overline{\mathfrak{g}_{2n,\rho}}(J(f,g))$ is nilpotent. ∎

Theorem 4.6**.**

Suppose $n\geq 3$ and $(q,\theta_{1},\theta_{2})\in\mathcal{P}$. Let $\mathcal{J}$ be an ideal of $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})$ such that $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})/\mathcal{J}$ is finite-dimensional and semisimple. Then there exist $f,g\in\mathbb{C}[T]$ with $f(0)g(0)\neq 0$, $\deg f\deg g>0$, $(f,f^{\prime})=1$, $(g,g^{\prime})=1$ and

[TABLE]

such that

[TABLE]

Proof.

Let $J$ be the involutive ideal of $\mathbb{C}_{q}$ corresponding to $\mathcal{J}$ as in Theorem 2.17. Since $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})/\mathcal{J}$ is finite-dimensional and semisimple, $J\neq\mathbb{C}_{q}$, and $\mathbb{C}_{q}/J$ is finite-dimensional. Since $1\notin J$, there exists an integer $r\geq 1$ such that

[TABLE]

is linearly independent in $\mathbb{C}_{q}/J$ but

[TABLE]

is linearly dependent. That is

[TABLE]

for some $\lambda_{0},\lambda_{1},\ldots,\lambda_{r}\in\mathbb{C}$ with $\lambda_{r}\neq 0$. We must have that $\lambda_{0}\neq 0$ for if it were then (4.2) would imply that $\lambda_{1}t_{1}^{m}+\cdots+\lambda_{r}t_{1}^{rm}\in J$ and hence

[TABLE]

lies in $J$. Contradiction. Similarly there is a positive integer $s$ such that $1+J,\ t_{2}^{m}+J,\ t_{2}^{2m}+J,\ \ldots,\ t_{2}^{(s-1)m}+J$ are linearly independent, and there exist $\mu_{0},\mu_{1},\ldots,\mu_{s}$ with $\mu_{0}\mu_{s}\neq 0$ such that

[TABLE]

Let

[TABLE]

Since $\widetilde{f}(t_{1}^{m})$ and $\overline{\widetilde{f}(t_{1}^{m})}=\widetilde{f}\left(\overline{t_{1}}^{\,m}\right)=\widetilde{f}\left(\theta_{1}^{m}t_{1}^{m}\right)$ both lie in $J$, so does their difference

[TABLE]

This implies that $\theta_{1}^{im}=1$ whenever $\lambda_{i}\neq 0$. Hence $\widetilde{f}\left(\theta_{1}^{m}T\right)=\widetilde{f}(T)$. Similarly $\widetilde{g}\left(\theta_{2}^{m}T\right)=\widetilde{g}(T)$. By Theorem 2.17, Lemma 3.8, and Lemma 4.1, we get

[TABLE]

Let $f=\widetilde{f}/(\widetilde{f},\widetilde{f}^{\prime})$ and $g=\widetilde{g}/(\widetilde{g},\widetilde{g}^{\prime})$. Then $f(0)g(0)\neq 0$, $\deg(f)\deg(g)>0$, $(f,f^{\prime})=(g,g^{\prime})=1$, $f(\theta_{1}^{m}T)=f(T)$, and $g(\theta_{2}^{m}T)=g(T)$. By Lemma 4.5, $\overline{\mathfrak{g}_{2n,\rho}}(J(f,g))/\overline{\mathfrak{g}_{2n,\rho}}(J(\widetilde{f},\widetilde{g}))$ is the radical of $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})/\overline{\mathfrak{g}_{2n,\rho}}(J(\widetilde{f},\widetilde{g}))$. On the other hand, the semisimplicity of

[TABLE]

implies that

[TABLE]

That is, $\mathcal{J}\supset\overline{\mathfrak{g}_{2n,\rho}}(J(f,g))$. ∎

5. Finite dimensional irreducible representations of $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})$

Let $V$ be a finite-dimensional irreducible module over $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})$. Then ${\rm Ann}(V)=\{x\in\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})|x.V=0\}$ is an ideal of $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})$ and $V$ can be regarded as a faithful irreducible representation of $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})/{\rm Ann}(V)$. And there is a monomorphism of Lie algebras $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})/{\rm Ann}(V)\hookrightarrow\mathfrak{gl}(V)$, i.e., $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})/{\rm Ann}(V)$ can be regarded as a subalgebra of $\mathfrak{gl}(V)$. Moreover, by Proposition 2.2, we have that

[TABLE]

Theorem 5.1**.**

Let $n\geq 3$. If $V$ is an irreducible representation of $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})$ and $\dim V\leq 1$, then $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})$ acts trivially on $V$.

Proof.

Since $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})$ is perfect, the Lie algebra homomorphism corresponding to the given representation is into $\mathfrak{sl}(V)$. But $\mathfrak{sl}(V)=\{0\}$ when $\dim V\leq 1$. Hence $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})$ acts trivially on $V$. ∎

Lemma 5.2** (Proposition 19.1 in [13]).**

Let $\mathfrak{g}\subset\mathfrak{sl}(V)$ ($V$ finite-dimensional) be a nonzero Lie algebra acting irreducibly on $V$. Then $\mathfrak{g}$ is semisimple. ∎

Theorem 5.3**.**

Let $n\geq 3$. $V$ is a finite-dimensional irreducible representation of $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})$ and $\dim V\geq 2$.

1. (1)

If $(q,\theta_{1},\theta_{2})\in\mathcal{P}$ with $q=-1$ or $(q,\theta_{1},\theta_{2})=(1,1,1)$, then there exists $k\geq 0$ such that

[TABLE]

where $m$ is the order of $q$ as a primitive root of unit. 2. (2)

If $(q,\theta_{1},\theta_{2})=(1,1,-1)$ or $(1,-1,-1)$, then there exists $k\geq 0$ such that

[TABLE]

Proof.

Since $2\leq\dim V\leq+\infty$ and $V$ is irreducible, $0\neq\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})/{\rm Ann}(V)\hookrightarrow\mathfrak{sl}(V)$. By Lemma 5.2, $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})/{\rm Ann}(V)$ is semisimple. So, by Theorem 4.6, there exist $f,g\in\mathbb{C}[T]$ with

[TABLE]

and

[TABLE]

Hence $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})/{\rm Ann}(V)$ is isomorphic to a quotient of $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})/\overline{\mathfrak{g}_{2n,\rho}}(J(f,g))$ which, by Proposition 2.16, is isomorphic to $\mathfrak{g}_{2n,\rho}\left(\mathbb{C}_{q}/J(f,g)\right)$. Lemma 4.3, thus, tells us that $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q})/{\rm Ann}(V)$ is, depending on $(q,\theta_{1},\theta_{2})$, isomorphic to $\mathfrak{g}_{2mn,\theta_{1}\rho}(\mathbb{C})^{k}$ or $\mathfrak{sl}_{2n}(\mathbb{C})^{k}$, for some $k\geq 0$. ∎

Remark 5.4**.**

By Section 3 and Section 4 in [23], we know that $\mathfrak{g}_{2n,\rho}(\mathbb{C}_{q,\theta_{1},\theta_{2}})$ is isomorphic to some double (twisted) loop algebra. For example, we have

1. (1)

$\mathfrak{g}_{2n,\rho}(\mathbb{C}_{1,1,1})$ is isomorphic to $\mathfrak{g}_{2n,\rho}(\mathbb{C})\otimes\mathbb{C}[t_{1}^{\pm 1},t_{2}^{\pm 1}]$, which is the untwisted double loop algebra of $\mathfrak{g}_{2n,\rho}(\mathbb{C})$ of type $X_{n}^{(1,1)}$ for $X_{n}$ equal to $C_{n}$ or $D_{n}$; 2. (2)

$\mathfrak{g}_{2n,\rho}(\mathbb{C}_{1,1,-1})$ is isomorphic to the twisted loop algebra of bi-affine algebra of type $A_{2n-1}^{(1,2)}$ which is a subalgebra of the double loop algebra $\mathfrak{sl}_{2n}(\mathbb{C})\otimes\mathbb{C}[t_{1}^{\pm 1},t_{2}^{\pm 1}]$; 3. (3)

$\mathfrak{g}_{2n,-1}(\mathbb{C}_{-1,1,1})$ is isomorphic to the twisted loop algebra of bi-affine algebra of type $C_{n}^{(1,1)*}$ which is a subalgebra of the double loop algebra $\mathfrak{g}_{4n,-1}(\mathbb{C})\otimes\mathbb{C}[t_{1}^{\pm 1},t_{2}^{\pm 1}]$, and so on.

Moreover, the isomorphisms make our results coincide with the results obtained in [17].

Acknowledgments

Main part of the research was carried out during the visit of the second author at York University in 2009. The hospitality and financial support of York University are gratefully acknowledged. The second author is partially supported by the Recruitment Program of Global Youth Experts of China, by the start-up funding from University of Science and Technology of China, by the Fundamental Research Funds for the Central Universities, by NSF of China (Grant 11401551, 11471294, 11771410). The third author is partially supported by NSERC of Canada.