Finite Dimensional Simple Modules over Some GIM Lie Algebras

Abstract

GIM Lie algebras are the generalizations of Kac–Moody Lie algebras. However, the structures of GIM Lie algebras are more complex than the latter, so they have encountered many new difficulties to study their representation theory. In this paper, we classify all finite dimensional simple modules over the GIM Lie algebra ${\mathcal{Q}}_{n+1}(2,1)$ as well as those over ${\Theta }_{2n+1}$.

1. Introduction

A generalized intersection matrix (GIM) is a matrix $\left(m_{i,j}\right)$ with integer entries, 2’s along the diagonal, and either $m_{i,j}{m}_{j,i}>0$ or $m_{i,j}=m_{j,i}=0$ if $i\ne j$. Slodowy [1] used these matrices to define the GIM Lie algebras that encompassed all Kac–Moody Lie algebras.

Definition 1

([1,2]). Given a GIM $M={\left(m_{i,j}\right)}_{l\times l}$. The GIM Lie algebra $\mathrm{gim}\left(M\right)$ is a Lie algebra over $\mathbb{C}$ generated by the elements

$$\begin{array}{c}\hfill \{e_i,f_i,h_i|1\le i\le l\}\end{array}$$

subject to the generating relations

$$\begin{array}{ccc}& & [h_i,e_j]=m_{i,j}{e}_j,[h_i,f_j]=-m_{i,j}{f}_j,[e_i,f_i]=h_i,\hfill \\ & & [e_i,f_j]=[f_i,e_j]=0,{\left(\mathrm{ad}{e}_i\right)}^{-m_{i,j}+1}{e}_j={\left(\mathrm{ad}{f}_i\right)}^{-m_{i,j}+1}{f}_j=0,\mathrm{if}{m}_{i,j}\le 0,\hfill \\ & & [e_i,e_j]=[f_i,f_j]=0,{\left(\mathrm{ad}{e}_i\right)}^{m_{i,j}+1}{f}_j={\left(\mathrm{ad}{f}_i\right)}^{m_{i,j}+1}{e}_j=0,\mathrm{if}{m}_{i,j}>0.\hfill \end{array}$$

Slodowy [1] and, later, Berman [3] showed that every GIM Lie algebra is isomorphic to the fixed point subalgebra of involution on some larger Kac–Moody Lie algebra.

The GIM Lie algebras play important roles in the classification of the root-graded Lie algebras (see [4,5,6,7]) and are closely related to the representations of tilted algebras via Ringel–Hall algebras (see [8,9]). Their quantized analogues are isomorphic to the Hopf co-ideals of some Kac–Moody quantum groups (see [10,11,12,13]).

For any $n\ge 3$, let $O_n={\left(o_{i,j}\right)}_{n\times n}$ be the GIM whose entries are given by

$$\begin{array}{c}\hfill o_{i,j}=\left\{\begin{array}{cc}2\hfill & i=j,\hfill \\ -1\hfill & |i-j|=1,\hfill \\ 1\hfill & \{i,j\}=\{1,n\},\hfill \\ 0\hfill & \mathrm{otherwise},\hfill \end{array}\right.\end{array}$$

which corresponds to the Dynkin diagram:

For convenience, we denote the GIM Lie algebra $\mathrm{gim}\left(O_n\right)$ by ${\mathcal{O}}_n$, and we say that it is of type $O_n$. ${\mathcal{O}}_n$ should be the simplest oriented GIM Lie algebra because it is the fixed point subalgebra of affine Kac–Moody Lie algebra ${\widehat{\mathfrak{sl}}}_{2n}$. Roughly, a GIM Lie algebra is called oriented if it is not a Kac–Moody Lie algebra. In [14], all finite dimensional simple modules over ${\mathcal{O}}_n$ are constructed and classified.

For $a,b\in {\mathbb{Z}}_{+}$, let ${\mathcal{Q}}_{n+1}(a,b)$ be the GIM Lie algebra defined via the structural matrix

$$\begin{array}{c}\hfill Q_{n+1}(a,b)=\left(q_{i,j}\right)=\left\{\begin{array}{cc}2\hfill & i=j,\hfill \\ -1\hfill & |i-j|=1,i,j\le n,\hfill \\ 1\hfill & \{i,j\}=\{1,n\},\hfill \\ -a\hfill & i=n,j=n+1,\hfill \\ -b\hfill & i=n+1,j=n,\hfill \\ 0\hfill & \mathrm{otherwise}.\hfill \end{array}\right.\end{array}$$

We say that ${\mathcal{Q}}_{n+1}(a,b)$ is of type $Q_{n+1}(a,b)$. As in the definition above, we assume that ${\mathcal{Q}}_{n+1}(2,1)$ is generated by the $e_i$’s, $f_i$’s, $h_i$’s. Then, ${\mathcal{O}}_n$ is identified as the Lie subalgebra generated by $\{e_i,f_i,h_i|1\le i\le n\}$. In particular, the Dynkin diagram of ${\mathcal{Q}}_{n+1}(2,1)$ is

In this paper, our main purpose is to classify all finite dimensional simple modules over ${\mathcal{Q}}_{n+1}(2,1)$ for all $n\ge 3$. As a surprised byproduct, we also obtain the classification of all finite dimensional simple modules over a class of ‘larger’ oriented GIM Lie algebras ${\Theta }_{2n+1}$ (see Definition 2 below).

We use the word larger with quotation marks, because ${\mathcal{Q}}_{n+1}(2,1)$ seems to be isomorphic to some fixed-point subalgebra ${\Theta }_{2n+1}^{\sigma }$ of ${\Theta }_{2n+1}$. However, we can only give an epimorphism $\xi$ from ${\mathcal{Q}}_{n+1}(2,1)$ to ${\Theta }_{2n+1}^{\sigma }$. Quite fortunately, this epimorphism is sufficient for our purpose.

Throughout the paper, we denote the set of complex numbers, the set of integers, the set of positive integers and the set of non-negative integers using $\mathbb{C},\mathbb{Z},{\mathbb{Z}}_{+},\mathbb{N}$, respectively.

2. GIM Lie Algebras

2.1. Lie Algebra ${\Theta }_{2n+1}$

For any $n\ge 2$, let $R={\left(r_{i,j}\right)}_{(2n+1)\times (2n+1)}$ be the GIM such that

$$\begin{array}{c}\hfill r_{i,j}=\left\{\begin{array}{cc}2\hfill & i=j,\hfill \\ -1\hfill & |i-j|=1,i,j\le 2n\hfill \\ -1\hfill & \{i,j\}=\{1,2n+1\}\hfill \\ 1\hfill & \{i,j\}=\{n,2n+1\},\hfill \\ 0\hfill & \mathrm{otherwise}.\hfill \end{array}\right.\end{array}$$

Definition 2.

The Lie algebra ${\Theta }_{2n+1}$ is the GIM Lie algebra over $\mathbb{C}$ generated by the elements

$$x_i,y_i,{\kappa }_i,1\le i\le 2n+1$$

subject to the generating relations

$$\begin{array}{ccc}& & [{\kappa }_i,x_j]=r_{i,j}{x}_j,[{\kappa }_i,y_j]=-r_{i,j}{y}_j,[x_i,y_i]={\kappa }_i,\hfill \\ & & [x_i,y_j]=[y_i,x_j]=0,{\left(\mathrm{ad}{x}_i\right)}^2{x}_j={\left(\mathrm{ad}{y}_i\right)}^2{y}_j=0,\mathrm{if}{r}_{i,j}=-1,\hfill \\ & & [x_i,y_j]=[x_i,x_j]=[y_i,y_j]=[y_i,x_j]=0,\mathrm{if}{r}_{i,j}=0,\hfill \\ & & [x_i,x_j]=[y_i,y_j]=0,{\left(\mathrm{ad}{x}_i\right)}^2{y}_j={\left(\mathrm{ad}{y}_i\right)}^2{x}_j=0,\mathrm{if}{r}_{i,j}=1.\hfill \end{array}$$

The affine Lie algebra ${\widehat{\mathfrak{sl}}}_{2n}$ can be identified to be the Lie subalgebra generated by the elements $x_i,y_i,{\kappa }_i,1\le i\le 2n$. Moreover, ${\Theta }_{2n+1}$ also is simply-laced and its Dynkin diagram is

2.2. An Involution of ${\Theta }_{2n+1}$

Define a map $\sigma$ such that

$$\begin{array}{ccc}& & x_i\mapsto -y_{i+n},y_i\mapsto -x_{i+n},{\kappa }_i\mapsto -{\kappa }_{i+n},i\le n,\hfill \\ & & x_i\mapsto -y_{i-n},y_i\mapsto -x_{i-n},{\kappa }_i\mapsto -{\kappa }_{i-n},n+1\le i\le 2n,\hfill \\ & & x_{2n+1}\mapsto x_{2n+1},y_{2n+1}\mapsto y_{2n+1},{\kappa }_{2n+1}\mapsto {\kappa }_{2n+1}.\hfill \end{array}$$

Lemma 1.

The map σ defines an automorphism of ${\Theta }_{2n+1}$.

Proof. 

It is straightforward. □

There is no confusion to denote this automorphism by $\sigma$. In particular, $\sigma$ is an involution as well as its restriction ${\sigma |}_{{\widehat{\mathfrak{sl}}}_{2n}}$.

Lemma 2

(see [3]). The fixed-point subalgebra of ${\sigma |}_{{\widehat{\mathfrak{sl}}}_{2n}}$ is isomorphic to ${\mathcal{O}}_n$.

2.3. The Fixed-Point Subalgebra

Let ${\Theta }_{2n+1}^{\sigma }$ be the fixed-point subalgebra of the involution $\sigma$. Then, it can be generated by the elements

$$\begin{array}{c}\hfill x_i-y_{n+i},y_i-x_{n+i},{\kappa }_i-{\kappa }_{n+i},x_{2n+1},y_{2n+1},{\kappa }_{2n+1},1\le i\le n.\end{array}$$

Lemma 3.

For $n\ge 3$, the map ξ

$$\begin{array}{ccc}& & e_i\mapsto x_i-y_{n+i},f_i\mapsto y_i-x_{n+i},h_i\mapsto {\kappa }_i-{\kappa }_{n+i},1\le i\le n,\hfill \\ & & e_{n+1}\mapsto x_{2n+1},f_{n+1}\mapsto y_{2n+1},h_{n+1}\mapsto {\kappa }_{2n+1},\hfill \end{array}$$

defines a Lie algebra epimorphism from ${\mathcal{Q}}_{n+1}(2,1)$ to ${\Theta }_{2n+1}^{\sigma }$.

Proof. 

By Lemma 2, it is sufficient to check the generating relationships for $i=n,n+1$. Notice that the elements $x_i,y_i,{\kappa }_i,i=1,n+1,2n+1$ generate a Lie algebra of type $A_3$. In fact, the restriction ${\sigma |}_{A_3}$ is a diagram automorphism. Then there is a fixed-point subalgebra of type $C_2$ (or $B_2$), and the lemma follows. □

3. Basics and Main Results

Lemma 4

(see [14]). Let V be a finite dimensional vector space and $\mathfrak{g}$ a GIM Lie algebra. Then, V is a simple $\mathfrak{g}$-module if, and only if, there exists an ideal I of $\mathfrak{g}$ such that $\mathfrak{g}/I$ is finite dimensional semi-simple, and V is a simple $\mathfrak{g}/I$-module.

Note that, for a general Lie algebra, the quotient $\mathfrak{g}/I$ perhaps contains a one dimensional direct summand. However, this situation doesn’t happen for GIM Lie algebras (of course, for Kac–Moody Lie algebras).

Let ${\mathcal{Y}}_{2n}$ denote the Lie subalgebra of ${\Theta }_{2n+1}$ generated by the elements

$$\begin{array}{c}\hfill x_i,y_i,{\kappa }_i,i\ne n+1.\end{array}$$

Similar to the above, we say that ${\mathcal{Y}}_{2n}$ is of type $Y_{2n}$. In particular, $Y_4=D_4,Y_6=E_6,Y_8=E_7^{\left(1\right)}$.

Lemma 5.

For $n\ge 5$, the Lie algebra ${\mathcal{Y}}_{2n}$ has no nontrivial finite dimensional module.

Proof. 

The Dynkin diagram of ${\mathcal{Y}}_{2n}$ is a subdiagram of ${\mathcal{Y}}_{2n+2}$. It is sufficient to prove it for $n=5$.

Note that the subalgebra $\mathcal{L}$ generated by the elements $x_i,y_i,{\kappa }_i,i=1,2,3,4,8,9,10,11$, is of type $E_7^{\left(1\right)}$. So

$$\begin{array}{c}\hfill c=4{\kappa }_1+3{\kappa }_2+2{\kappa }_3+{\kappa }_4+{\kappa }_8+2{\kappa }_9+3{\kappa }_{10}+2{\kappa }_{11}\end{array}$$

is the fundamental central element of $\mathcal{L}$ (see Table Aff1 and §6.2 in [15]). Let V be a finite dimensional module over ${\mathcal{Y}}_{10}$, then it is a module over $\mathcal{L}$. This implies $c·V=0$. However, $[c,x_5]=-x_5,[c,y_5]=y_5$. It forces $x_5·V=y_5·V={\kappa }_5·V=0$. Then ${\mathcal{Y}}_{10}·V=0$. □

Define the loop Lie algebra

$$\begin{array}{c}\hfill L\left({\mathcal{Y}}_{2n}\right)=\mathbb{C}[z,z^{-1}]\otimes {\mathcal{Y}}_{2n},\end{array}$$

which has the bracket

$$\begin{array}{c}\hfill [z^i\otimes x,z^j\otimes y]=z^{i+j}\otimes [x,y].\end{array}$$

The following two lemmas are well-known for all loop Lie algebras.

Lemma 6.

If ${(z-a)}^2|g\left(z\right)$ for some $a\in \mathbb{C}$, the quotient $L\left({\mathcal{Y}}_{2n}\right)/g\left(z\right)L\left({\mathcal{Y}}_{2n}\right)$ has a nontrivial nilpotent ideal.

Proof. 

It is straightforward. For example, let $I=\frac{g\left(z\right)}{z-a}L\left({\mathcal{Y}}_{2n}\right)$, it is clearly an ideal, and $I⊈g\left(z\right)L\left({\mathcal{Y}}_{2n}\right)$. However, $[I,I]\subseteq \frac{g\left(z\right)}{{(z-a)}^2}g\left(z\right)L\left({\mathcal{Y}}_{2n}\right)\subseteq g\left(z\right)L\left({\mathcal{Y}}_{2n}\right)$. □

Lemma 7.

Suppose $g\left(z\right)=(z-a_1)\cdots (z-a_t)$ for pairwise different constants $a_1,...,a_t\in {\mathbb{C}}^{*}$, we have

$$\begin{array}{c}\hfill L\left({\mathcal{Y}}_{2n}\right)/g\left(z\right)L\left({\mathcal{Y}}_{2n}\right)\cong {\mathcal{Y}}_{2n}^{\oplus t}.\end{array}$$

Proof. 

Since $g\left(z\right)$ has no multiple root, any common divisor of all $\frac{g\left(z\right)}{z-a_i},1\le i,\le t$, must be a constant. Then there exist some constants $u_i\in {\mathbb{C}}^{*}$ such that

$$\begin{array}{c}\hfill 1=\sum _{i=1}^t{u}_i\frac{g\left(z\right)}{z-a_i}.\end{array}$$

Due to the fact that $\frac{g\left(z\right)}{z-a_i}L\left({\mathcal{Y}}_{2n}\right)/g\left(z\right)L\left({\mathcal{Y}}_{2n}\right)\cong {\mathcal{Y}}_{2n}$ for each i, we have

$$\begin{array}{c}\hfill L\left({\mathcal{Y}}_{2n}\right)/g\left(z\right)L\left({\mathcal{Y}}_{2n}\right)=\underset{i=1}{\overset{t}{⨁}}\frac{g\left(z\right)}{z-a_i}L\left({\mathcal{Y}}_{2n}\right)/g\left(z\right)L\left({\mathcal{Y}}_{2n}\right)\cong {\mathcal{Y}}_{2n}^{\oplus t}.\end{array}$$

□

Lemma 8.

Let V be a finite dimensional vector space.

(1) For $n\ge 5$, the Lie algebra $L\left({\mathcal{Y}}_{2n}\right)$ has no nontrivial finite dimensional simple module.

(2) V is a simple module over $L\left({\mathcal{Y}}_4\right)$ if, and only if, it is a module over $D_4^{\oplus t}$ for some $t\in {\mathbb{Z}}_{+}$.

(3) V is a simple module over $L\left({\mathcal{Y}}_6\right)$ if, and only if, it is a module over $E_6^{\oplus t}$ for some $t\in {\mathbb{Z}}_{+}$.

(4) V is a simple module over $L\left({\mathcal{Y}}_8\right)$ if, and only if, it is a module over $E_7^{\oplus t}$ for some $t\in {\mathbb{Z}}_{+}$.

Proof. 

By Lemma 5, $L\left({\mathcal{Y}}_{2n}\right)$ has no nontrivial finite dimensional simple module for $n\ge 5$.

For $n=2,3,4$, let I be the ideal such that $L\left({\mathcal{Y}}_{2n}\right)/I$ is finite dimensional semi-simple. It is easy to find some monic polynomial $g\left(z\right)\in \mathbb{C}\left[z\right]$ with $g\left(0\right)\ne 0$ such that $g\left(z\right)L\left({\mathcal{Y}}_{2n}\right)\subseteq I$. By Lemma 6, we may assume

$$\begin{array}{c}\hfill g\left(z\right)=\prod _{i=1}^t(z-a_i),\end{array}$$

such that $a_i\ne a_j$ for all $i\ne j$.

For $n=2$, we have $L\left({\mathcal{Y}}_{2n}\right)/g\left(z\right)L\left({\mathcal{Y}}_{2n}\right)\cong D_4^{\oplus t}$ by Lemma 7.

For $n=3$, we have $L\left({\mathcal{Y}}_{2n}\right)/g\left(z\right)L\left({\mathcal{Y}}_{2n}\right)\cong E_6^{\oplus t}$ by Lemma 7.

For $n=4$, we have $L\left({\mathcal{Y}}_{2n}\right)/g\left(z\right)L\left({\mathcal{Y}}_{2n}\right)\cong {E_7^{\left(1\right)}}^{\oplus t}$ by Lemma 7. Moreover, each summand is the untwisted affine Lie algebra. It is known that its semi-simple quotient is a multiple of $E_7$.

The lemma follows. □

Next, we list our main results as follows.

Theorem 1.

(1) 

For $n\ge 5$, ${\Theta }_{2n+1}$ has some nontrivial finite dimensional simple modules.

(2) 

A finite dimensional vector space V is a simple ${\Theta }_5$-module if, and only if, it is a simple $D_4^{\oplus t}$-module for some $t\in {\mathbb{Z}}_{+}$.

(3) 

A finite dimensional vector space V is a simple ${\Theta }_7$-module if, and only if, it is a simple $E_6^{\oplus t}$-module for some $t\in {\mathbb{Z}}_{+}$.

(4) 

A finite dimensional vector space V is a simple ${\Theta }_9$-module if, and only if, it is a simple $E_7^{\oplus t}$-module for some $t\in {\mathbb{Z}}_{+}$.

Theorem 2.

(1) 

For $n\ge 5$, ${\mathcal{Q}}_{n+1}(2,1)$ has some nontrivial finite dimensional simple modules.

(2) 

A finite dimensional vector space V is a simple module over ${\mathcal{Q}}_4(2,1)$ if, and only if, it is a simple module over $C_4^{\oplus r}\oplus F_4^{\oplus s}\oplus E_6^{\oplus t}$ for some $t\in \mathbb{N}$ and $r,s\in \{0,1\}$.

(3) 

A finite dimensional vector space V is a simple module over ${\mathcal{Q}}_5(2,1)$ if, and only if, it is a simple module over $A_7^{\oplus r}\oplus E_6^{\oplus s}\oplus E_7^{\oplus t}$ for some $r,s,t\in \mathbb{N}$.

Immediately, from above theorems we can get the following result directly.

Corollary 1.

(1) 

Let I be a maximal ideal of the finite co-dimension of ${\Theta }_{2n+1}$. Then

$$\begin{array}{ccc}\hfill {\Theta }_{2n+1}/I& \cong & \left\{\begin{array}{cc}{D}_4,\hfill & n=2,\hfill \\ E_6,\hfill & n=3,\hfill \\ E_7,\hfill & n=4,\hfill \\ 0,\hfill & n>4.\hfill \end{array}\right.\hfill \end{array}$$

(2) 

Let I be a maximal ideal of finite co-dimension of ${\mathcal{Q}}_{n+1}(2,1)$. Then

$$\begin{array}{ccc}\hfill {\mathcal{Q}}_{n+1}(2,1)/I& \cong & \left\{\begin{array}{cc}{E}_6\mathrm{or}{F}_4\mathrm{or}{C}_4,\hfill & n=3,\hfill \\ E_7\mathrm{or}{E}_6\mathrm{or}{A}_7,\hfill & n=4,\hfill \\ 0,\hfill & n>4.\hfill \end{array}\right.\hfill \end{array}$$

4. Proof of Theorem 1

Let $y_0=\mathrm{ad}{x}_{n+2}\cdots \mathrm{ad}{x}_{2n}\mathrm{ad}{x}_n\cdots \mathrm{ad}{x}_2\left(x_1\right)$, $x_0=\mathrm{ad}{y}_{n+2}\cdots \mathrm{ad}{y}_{2n}\mathrm{ad}{y}_n\cdots \mathrm{ad}{y}_2\left(y_1\right)$ and ${\kappa }_0={\kappa }_{n+1}-{\sum }_{i=1}^{2n}{\kappa }_i$. It is easy to check that

$$\begin{array}{c}\hfill [{\kappa }_0,x_0]=2x_0,[{\kappa }_0,y_0]=-2y_0,[x_0,y_0]={\kappa }_0.\end{array}$$

Lemma 9.

There exists an epimorphism of Lie algebras

$$\begin{array}{c}\hfill \rho :{\Theta }_{2n+1}\to L\left({\mathcal{Y}}_{2n}\right)\end{array}$$

defined by

$$\begin{array}{ccc}& & x_i\mapsto 1\otimes x_i,y_i\mapsto 1\otimes y_i,{\kappa }_i\mapsto 1\otimes {\kappa }_i,i\ne n+1,\hfill \\ & & x_{n+1}\mapsto z\otimes x_0,y_{n+1}\mapsto z^{-1}\otimes y_0,{\kappa }_{n+1}\mapsto 1\otimes {\kappa }_0.\hfill \end{array}$$

Proof. 

First, the elements

$$1\otimes x_i,1\otimes y_i,1\otimes {\kappa }_i,z\otimes x_0,z^{-1}\otimes y_0,1\otimes {\kappa }_0,1\le i\le 2n+1,i\ne n+1$$

generate the loop algebra $L\left({\mathcal{Y}}_{2n}\right)$.

It is sufficient to check the serre relations involving $x_{n+1},y_{n+1}$. Explicitly, we have

$$\begin{array}{ccc}\hfill [1\otimes x_n,z\otimes x_0]& =& -z\otimes \mathrm{ad}{y}_{n+2}\cdots \mathrm{ad}{y}_{2n}\mathrm{ad}{y}_{n-1}\cdots \mathrm{ad}{y}_2\left(y_1\right),\hfill \\ \hfill [1\otimes x_{n+2},z\otimes x_0]& =& -z\otimes \mathrm{ad}{y}_{n+3}\cdots \mathrm{ad}{y}_{2n}\mathrm{ad}{y}_n\cdots \mathrm{ad}{y}_2\left(y_1\right),\hfill \\ \hfill [1\otimes y_{2n+1},z\otimes x_0]& =& z\otimes \mathrm{ad}{y}_{n+2}\cdots \mathrm{ad}{y}_{2n}\mathrm{ad}{y}_n\cdots \mathrm{ad}{y}_2\left([y_{2n+1},y_1]\right),\hfill \\ \hfill [1\otimes y_n,z^{-1}\otimes y_0]& =& z^{-1}\otimes \mathrm{ad}{x}_{n+2}\cdots \mathrm{ad}{x}_{2n}\mathrm{ad}{x}_{n-1}\cdots \mathrm{ad}{x}_2\left(x_1\right),\hfill \\ \hfill [1\otimes y_{n+2},z^{-1}\otimes y_0]& =& z^{-1}\otimes \mathrm{ad}{x}_{n+3}\cdots \mathrm{ad}{x}_{2n}\mathrm{ad}{x}_n\cdots \mathrm{ad}{x}_2\left(x_1\right),\hfill \\ \hfill [1\otimes x_{2n+1},z^{-1}\otimes y_0]& =& z^{-1}\otimes \mathrm{ad}{x}_{n+2}\cdots \mathrm{ad}{x}_{2n}\mathrm{ad}{x}_n\cdots \mathrm{ad}{x}_2\left([x_{2n+1},x_1]\right),\hfill \end{array}$$

and

$$\begin{array}{ccc}& & [1\otimes y_n,z\otimes x_0]=[1\otimes y_{n+2},z\otimes x_0]=[1\otimes x_{2n+1},z\otimes x_0]=0,\hfill \\ & & [1\otimes x_n,z^{-1}\otimes y_0]=[1\otimes x_{n+2},z^{-1}\otimes y_0]=[1\otimes y_{2n+1},z^{-1}\otimes y_0]=0,\hfill \\ & & [1\otimes x_n,[1\otimes x_n,z\otimes x_0]]=[1\otimes x_{n+2},[1\otimes x_{n+2},z\otimes x_0]]=0,\hfill \\ & & [1\otimes y_{2n+1},[1\otimes y_{2n+1},z\otimes x_0]]=[1\otimes y_n,[1\otimes y_n,z^{-1}\otimes y_0]]=0,\hfill \\ & & [1\otimes y_{n+2},[1\otimes y_{n+2},z^{-1}\otimes y_0]]=[1\otimes x_{2n+1},[1\otimes x_{2n+1},z^{-1}\otimes y_0]]=0.\hfill \end{array}$$

□

Corollary 2.

Any simple $L\left({\mathcal{Y}}_{2n}\right)$-module is a simple ${\Theta }_{2n+1}$-module.

Lemma 10.

Any finite dimensional simple ${\Theta }_{2n+1}$-module must be a simple $L\left({\mathcal{Y}}_{2n}\right)$-module.

Proof. 

Assume that I is an ideal of ${\Theta }_{2n+1}$ such that $S={\Theta }_{2n+1}/I$ is finite dimensional and semi-simple. Let $J=I\cap {\widehat{\mathfrak{sl}}}_{2n}$, then $c_1={\sum }_{i=1}^{2n}{\kappa }_i\in J$. In particular, $c_1$ is a central element, and $\rho \left(c_1\right)=0$. For convenience, we can regard $\rho$ as the map defined over ${\Theta }_{2n+1}/\mathbb{C}{c}_1$. For this situation, ${\rho |}_{{\widehat{\mathfrak{sl}}}_{2n}}$ is injective.

Due to the the fact that ${\widehat{\mathfrak{sl}}}_{2n}/J$ is finite dimensional, there exists a polynomial $g\left(z\right)$ such that

$$\begin{array}{c}\hfill \rho \left(J\right)=g\left(z\right)\mathbb{C}[z,z^{-1}]\otimes {\widehat{\mathfrak{sl}}}_{2n}.\end{array}$$

Let $I_1={\rho }^{-1}(g\left(z\right)\mathbb{C}[z,z^{-1}]\otimes {\Theta }_{2n+1})$, then $S_1=(I_1+I)/I$ is an ideal of S. In addition, $S_1$ is the direct summand of S, since S is semi-simple. Let $\pi :S\to S_1$ be the canonical projection. Then $J=I_1\cap {\widehat{\mathfrak{sl}}}_{2n}\subseteq I$ implies $\pi ({\widehat{\mathfrak{sl}}}_{2n}/J)=0$. In other words, ${\Theta }_{2n+1}$ has an ideal $I_2$ such that ${\widehat{\mathfrak{sl}}}_{2n}\subseteq I_2$ and ${\Theta }_{2n+1}/I_2=S_1$. It is easy to get ${\Theta }_{2n+1}\subseteq I_2$ and $S_1=0$. So $I_1\subset I$. Moreover, $L\left({\mathcal{Y}}_{2n}\right)/\rho \left(I\right)\cong S_1$.

By Lemma 4, we have finished the proof. □

Construction 1.

Let J be the ideal of$L\left({\mathcal{Y}}_{2n}\right)$generated by a polynomial$g\left(z\right)=(z-a_1)\cdots (z-a_p)$with pairwise distinct constants$a_1,\cdots ,a_p\in {\mathbb{C}}^{*}$. Then, by Lemma 8,

$$\begin{array}{ccc}\hfill L\left({\mathcal{Y}}_{2n}\right)/J& \cong & \left\{\begin{array}{cc}{D}_4^{\oplus p},\hfill & n=2,\hfill \\ E_6^{\oplus p},\hfill & n=3,\hfill \\ {\left(E_7^{\left(1\right)}\right)}^{\oplus p},\hfill & n=4.\hfill \end{array}\right.\hfill \end{array}$$

Let$\pi :L\left({\mathcal{Y}}_{2n}\right)\to L\left({\mathcal{Y}}_{2n}\right)/J$be the natural projection and let$\psi :L\left({\mathcal{Y}}_{2n}\right)/J\to \mathrm{End}\left(V\right)$be a finite dimensional simple representation. Then$\rho \circ \pi \circ \psi$defines a simple finite dimensional representation. In particular, the image of$\rho \circ \pi \circ \psi$is as described in Theorem 1.

Proof of Theorem 1.

Corollary 2 and Lemma 10 imply that a finite vector space V is a simple module over ${\Theta }_{2n+1}$ if, and only if, it is a simple module over $L\left({\mathcal{Y}}_{2n}\right)$. Then Theorem 1 holds by Lemma 8 and Construction 1. □

5. Proof of Theorem 2

Similar to the previous section, we regard $\rho$ as the map defined over ${\Theta }_{2n+1}/\mathbb{C}{c}_1$. It is easy to see that $c_1\notin {\Theta }_{2n+1}^{\sigma }$, so ${\xi |}_{{\mathcal{O}}_n}$ is still injective.

Lemma 11.

Any finite dimensional simple module over ${\mathcal{Q}}_{n+1}(2,1)$ is a simple module over $\rho \left({\Theta }_{2n+1}^{\sigma }\right)$.

Proof. 

Suppose that I is an ideal, such that $S={\mathcal{Q}}_{n+1}(2,1)/I$ is finite dimensional and semi-simple. Let $J=I\cap {\mathcal{O}}_n$. Then $(\rho \circ \xi )\left(J\right)$ is an ideal of $(\rho \circ \xi )\left({\mathcal{O}}_n\right)$. Moreover, there exists $\theta \left(z\right)$ such that

$$\begin{array}{c}\hfill \left(\theta \left(z\right)L\left({\mathcal{Y}}_{2n}\right)\right)\cap (\rho \circ \xi )\left({\mathcal{O}}_n\right)\subseteq (\rho \circ \xi )\left(J\right).\end{array}$$

In fact, it holds that

$$\begin{array}{c}\hfill \left(\theta \left(z\right)L\left({\mathcal{Y}}_{2n}\right)\right)\cap (\rho \circ \xi )\left({\mathcal{O}}_n\right)=(\theta \left(z\right)\rho (\widehat{{\mathfrak{sl}}_{2n}}))\cap (\rho \circ \xi )\left({\mathcal{O}}_n\right)=\theta \left(z\right)(\rho \circ \xi )\left({\mathcal{O}}_n\right).\end{array}$$

For details, one is referred to page 4806 of [14].

Similar to the proof of Lemma 10, we have

$$\begin{array}{c}\hfill {(\rho \circ \xi )}^{-1}\left(\theta \left(z\right)\rho \left({\Theta }_{2n+1}^{\sigma }\right)\right)\subseteq I.\end{array}$$

Moreover, S is isomorphic to some quotient of $\rho \left({\Theta }_{2n+1}^{\sigma }\right)/\theta \left(z\right)\rho \left({\Theta }_{2n+1}^{\sigma }\right)$, and hence some quotient of ${\Theta }_{2n+1}^{\sigma }$. □

Corollary 3.

Let V be a finite dimensional vector space. Then V is a simple ${\mathcal{Q}}_{n+1}(2,1)$-module if, and only if, it is a simple $\rho \left({\Theta }_{2n+1}^{\sigma }\right)$-module.

Corollary 4.

Suppose that I is an ideal, such that $S={\mathcal{Q}}_{n+1}(2,1)/I$ is finite dimensional and semi-simple. Then S is a subquotient of $L\left({\mathcal{Y}}_{2n}\right)$.

Proof. 

It follows from the fact that $\rho \left({\Theta }_{2n+1}^{\sigma }\right)/\theta \left(z\right)\rho \left({\Theta }_{2n+1}^{\sigma }\right)$ is a subalgebra of $L\left({\mathcal{Y}}_{2n}\right)/\theta \left(z\right)L\left({\mathcal{Y}}_{2n}\right)$. □

Lemma 12.

If ${(z-a)}^2|\theta \left(z\right)$ for some $a\in \mathbb{C}$, then the quotient $L\left({\mathcal{Y}}_{2n}\right)/\theta \left(z\right)L\left({\mathcal{Y}}_{2n}\right)$ has a nontrivial nilpotent ideal.

Lemma 13.

Suppose $\theta \left(z\right)=(z-a_1)\cdots (z-a_p)$ for some positive integer p and pairwise distinct constants $a_1,...,a_p\in {\mathbb{C}}^{*}$, we have

$$\begin{array}{c}\hfill \rho \left({\Theta }_7^{\sigma }\right)/(\theta \left(z\right)L\left({\mathcal{Y}}_6\right)\cap \rho \left({\Theta }_7^{\sigma }\right))\cong C_4^{\oplus r}\oplus F_4^{\oplus s}\oplus E_6^{\oplus t},\\ \hfill \rho \left({\Theta }_9^{\sigma }\right)/(\theta \left(z\right)L\left({\mathcal{Y}}_8\right)\cap \rho \left({\Theta }_9^{\sigma }\right))\cong {A_7^{\left(2\right)}}^{\oplus r}\oplus {E_6^{\left(2\right)}}^{\oplus s}\oplus {E_7^{\left(1\right)}}^{\oplus t},\end{array}$$

for some $r,s\in \{0,1\}$ and $t\in \mathbb{N}$.

Proof .

(i) $n=3$.

By the main result of [14], we have

$$\begin{array}{c}\hfill \rho \left(\xi \left({\mathcal{O}}_3\right)\right)/(\theta \left(z\right)L\left({\mathcal{Y}}_6\right)\cap \rho \left(\xi \left({\mathcal{O}}_3\right)\right))\cong A_3^{\oplus r}\oplus C_3^{\oplus s}\oplus A_5^{\oplus t},\end{array}$$

for some $r,s\in \{0,1\}$ and $t\in \mathbb{N}$.

Moreover, if $\rho \left(\xi \left({\mathcal{O}}_3\right)\right)/(z-a)L\left({\mathcal{Y}}_6\right)\rho \left(\xi \left({\mathcal{O}}_3\right)\right)\cong A_3$, we infer that

$$\begin{array}{c}\hfill \rho \circ \xi \left([[e_1,e_2],e_3]\right),\rho \circ \xi \left([[e_1,e_2],f_3]\right),\rho \circ \xi \left([[f_1,f_2],e_3]\right),\rho \circ \xi \left([[f_1,f_2],f_3]\right)\in (z-a)L\left({\mathcal{Y}}_6\right),\end{array}$$

the quotient Lie algebra $\rho \left({\Theta }_7^{\sigma }\right)/(z-a)L\left({\mathcal{Y}}_6\right)\rho \left({\Theta }_7^{\sigma }\right)\cong C_4$.

If $\rho \left(\xi \left({\mathcal{O}}_3\right)\right)/(z-a)L\left({\mathcal{Y}}_6\right)\rho \left(\xi \left({\mathcal{O}}_3\right)\right)\cong C_3$, the quotient Lie algebra

$$\rho \left({\Theta }_7^{\sigma }\right)/(z-a)L\left({\mathcal{Y}}_6\right)\rho \left({\Theta }_7^{\sigma }\right)\cong F_4.$$

If $\rho \left(\xi \left({\mathcal{O}}_3\right)\right)/(z-a)L\left({\mathcal{Y}}_6\right)\rho \left(\xi \left({\mathcal{O}}_3\right)\right)\cong A_5$, the quotient Lie algebra

$$\rho \left({\Theta }_7^{\sigma }\right)/(z-a)L\left({\mathcal{Y}}_6\right)\rho \left({\Theta }_7^{\sigma }\right)\cong E_6.$$

(ii) $n=4$.

By the main result of [14], we have

$$\begin{array}{c}\hfill \rho \left(\xi \left({\mathcal{O}}_4\right)\right)/(\theta \left(z\right)L\left({\mathcal{Y}}_8\right)\cap \rho \left(\xi \left({\mathcal{O}}_4\right)\right))\cong D_4^{\oplus r}\oplus C_4^{\oplus s}\oplus A_7^{\oplus t},\end{array}$$

for some $r,s\in \{0,1\}$ and $t\in \mathbb{N}$.

If $\rho \left(\xi \left({\mathcal{O}}_4\right)\right)/(z-a)L\left({\mathcal{Y}}_8\right)\rho \left(\xi \left({\mathcal{O}}_4\right)\right)\cong D_4$, the quotient Lie alpha

$$\rho \left({\Theta }_9^{\sigma }\right)/(z-a)L\left({\mathcal{Y}}_8\right)\rho \left({\Theta }_9^{\sigma }\right)\cong A_7^{\left(2\right)}.$$

If $\rho \left(\xi \left({\mathcal{O}}_4\right)\right)/(z-a)L\left({\mathcal{Y}}_8\right)\rho \left(\xi \left({\mathcal{O}}_4\right)\right)\cong C_4$, the quotient Lie algebra

$$\rho \left({\Theta }_9^{\sigma }\right)/(z-a)L\left({\mathcal{Y}}_8\right)\rho \left({\Theta }_9^{\sigma }\right)\cong E_6^{\left(2\right)}.$$

If $\rho \left(\xi \left({\mathcal{O}}_4\right)\right)/(z-a)L\left({\mathcal{Y}}_8\right)\rho \left(\xi \left({\mathcal{O}}_4\right)\right)\cong A_7$, the quotient Lie algebra

$$\rho \left({\Theta }_9^{\sigma }\right)/(z-a)L\left({\mathcal{Y}}_6\right)\rho \left({\Theta }_9^{\sigma }\right)\cong E_7^{\left(1\right)}.$$

□

Lemma 14.

(1) 

A finite dimensional vector space V is a simple module over $\rho \left({\Theta }_7^{\sigma }\right)$ if, and only if, it is a simple module over $C_4^{\oplus r}\oplus F_4^{\oplus s}\oplus E_6^{\oplus t}$ for some $t\in \mathbb{N}$ and $r,s\in \{0,1\}$.

(2) 

A finite dimensional vector space V is a simple module over $\rho \left({\Theta }_9^{\sigma }\right)$ if, and only if, it is a simple module over $A_7^{\oplus r}\oplus E_6^{\oplus s}\oplus E_7^{\oplus t}$ for some $r,s,t\in \mathbb{N}$.

Proof .

(1) holds by Lemma 13.

Notice that any semi-simple quotient of $A_7^{\left(2\right)}$ is a multiple of $A_7$, any semi-simple quotient of $E_6^{\left(2\right)}$ is a multiple of $E_6$, and the semi-simple quotient of $E_7^{\left(1\right)}$ is a multiple of $E_7$. Thus (2) also holds by Lemma 13. □

Construction 2.

Consider the map in Construction 1, let I be the kernel of the map

$$\xi \circ \rho \circ \pi \circ \psi ,$$

${\mathcal{Q}}_{n+1}(2,1)/I$as in Theorem 2 for$n=3,4$. Then any finite dimensional simple modules over${\mathcal{Q}}_{n+1}(2,1)/I$can be constructed directly. Those are simple${\mathcal{Q}}_{n+1}(2,1)$-modules in the natural way.

Proof of Theorem 2.

For $n\ge 5$, ${\mathcal{Q}}_{n+1}(2,1)$ contains a Lie subalgebra $\mathcal{L}$ generated by the elements

$$\begin{array}{c}\hfill e_i,f_i,h_i,i=1,n-1,n,n+1,\end{array}$$

which is of type $A_5^{\left(2\right)}$. In particular, $h=-h_1+h_{n-1}+2h_n+2h_{n+1}$ is a central element. If V is a finite dimensional module over ${\mathcal{Q}}_{n+1}(2,1)$, we have $h·V=0$. Due to the fact that $n\ge 5$, we have $[h,e_2]=e_2,[h,f_2]=-f_2$, and $e_2·V=f_2·V=h_2·V=0$. It is easy to obtain ${\mathcal{Q}}_{n+1}(2,1)·V=0$.

Assume $n\le 4$. Then it holds by Corollary 3, Lemma 14, and Construction 2. The theorem is proved. □

6. Conclusions

In conclusion, all finite dimensional simple modules over the GIM Lie algebra ${\mathcal{Q}}_{n+1}(2,1)$ as well as those over ${\Theta }_{2n+1}$ are classified in this paper (Theorems 1 and 2). Meanwhile, such research for ${\mathcal{O}}_n$ was completed in [14]. Certainly, the study of finite dimensional modules over GIM Lie algebras is still in its infancy.