Infinite-dimensional necklace Lie algebras and some finite-dimensional important subalgebras

Abstract

In this paper, a new infinite-dimensional necklace Lie algebra is studied and the left and right index arrays of a necklace word in necklace Lie algebra is first defined. Using the left and right index arrays, we divide the necklace words into 5 classes. We discuss finite-dimensional Lie subalgebras of necklace Lie algebras intensively and prove that some subalgebras are isomorphism to simple Lie algebra sl\left(n\right).

Graphical abstract

Keywords

Necklace Lie algebra / left and right index arrays / subalgebra

1 Introduction

Necklace Lie algebra is a relatively new class of infinite-dimensional Lie algebras characterized by 0. In the study of noncommutative geometry, when considering the role of \mathbb{C}-algebraic automorphism group \mathrm{A}\mathrm{u}\mathrm{t}{A}_1(\mathbb{C}) of Weyl algebra A_1(\mathbb{C}) on orbital {\text{Weyl}}_n (in this case \mathbb{C} is the field of complex numbers), we get an \mathrm{A}\mathrm{u}\mathrm{t}{A}_1(\mathbb{C}) on n\times n matrix orbital space {\text{Calo}}_n on migration. However, this effect is non-differentiable and therefore non-algebraic. Berest and Wilson propose whether {\text{Calo}}_n can be equated with some kind of coadjoint orbit of an infinite-dimensional Lie algebra. Both Lothaire and Reutenauer [4, 8] have failed to use necklace words to characterize the basis of free Lie algebras in an attempt to solve the problem posed by Berest and Wilson. Ginzburg [2] and Bocklandt [1] introduced necklace Lie algebra, respectively, to solve this problem.

A necklace Lie algebra is an infinite-dimensional vector space generated by all necklace characters in a repetitive arrow diagram \overline{Q} induced by an arrow diagram Q, and the special Lie operations are defined on this vector space. The application of the arrow diagram [3, 6] in algebraic representation theory has become more and more widespread, and has become the basic concept of algebraic representation theory. Necklace Lie algebras have important applications in noncommutative geometry and singularity theory, quantum groups, and other fields. We have studied the isomorphisms and homomorphisms of some necklace Lie algebras [11, 12]. In [12], we have studied the isomorphisms of some special arrow diagrams, including unidirectional cyclic arrow diagrams, mixed cyclic arrow diagrams, and the isomorphisms of vertically superimposed arrow diagrams and horizontally superimposed arrow diagrams. We have also studied the homomorphisms of some special arrow diagrams in [11], including homomorphisms with only 2 vertices, 2 arrow diagrams in different directions, only 3 vertices, 3 arrow diagrams in the same direction, and n vertices, n arrow diagrams in different directions. Mei [5] studied some special necklaces Lie algebra structures.

Post [7] studied the structure of transitive Lie algebras, focusing on finite-dimensional Lie subalgebras of infinite-dimensional transitive Lie algebras, and proved that finite-dimensional maximal Lie subalgebras contain semi-simple Lie algebras, and finite-dimensional maximum Lie subalgebras have nontrivial modules. We have also considered some special infinite-dimensional Lie algebras [9, 10]. This paper focuses on finite-dimensional subalgebras of infinite-dimensional necklace Lie algebras, and proves that some of these subalgebras are isomorphic to the canonical simple Lie algebra \text{sl}(n). Follow-up research will explore some important properties of necklace Lie algebra, such as semi-singleness, solvability, and decomposability.

2 Definition of necklace Lie algebra and index arrays of necklace Lie algebra

Let Q=\left(Q_0,Q_1,s,t\right) be any connected directed diagram with two or more vertices, where Q_0=\left\{v_1,v_2,\dots ,v_n\right\} is the set of vertices of Q. Directed edges in Q are called arrows, Q_1 is the set of all arrows in Q, Q_1=\left\{{\alpha }_1,{\alpha }_2,\dots ,{\alpha }_m\right\}, s,t is a mapping from Q_1 to Q_0 such that for any \alpha \in Q_1,s(\alpha )=v is the starting point of \alpha, t(\alpha )=v^{\mathrm{\prime }} is the end of \alpha. Denoted \alpha :v⟶v^{\mathrm{\prime }}, and s(\alpha )\ne t(\alpha ), and Q is called an arrow diagram. There are arrow sequences {\alpha }_{i_1},{\alpha }_{i_2},\dots ,{\alpha }_{i_u},u⩾1 in Q, here i_1,i_2,\dots ,i_u\in \{1,2,\dots ,m\}. If t\left({\alpha }_{i_j}\right)=s\left({\alpha }_{i_{j+1}}\right), then c={\alpha }_{i_1}{\alpha }_{i_2}\cdots {\alpha }_{i_u} is the road in Q. If there is also t\left({\alpha }_{i_u}\right)=s\left({\alpha }_{i_1}\right), then c is said to be a cycle, and u is called the length of the cycle. Define the relationship \sim on the set of all loops in the arrow diagram Q as follows. Let c be a loop in Q. If c^{\mathrm{\prime }} is the loop obtained by rotating the arrows in c sequentially, define c^{\mathrm{\prime }}\sim c. Obviously \sim is an equivalence relation. A circular equivalent class of Q is called a necklace word of Q. If c={\alpha }_{i_1}{\alpha }_{i_2}\cdots {\alpha }_{i_u} is a loop, then the corresponding necklace word is represented by a diagram (see Fig.1).

Represented by \omega ={\alpha }_{i_1}{\alpha }_{i_2}\cdots {\alpha }_{i_u}, c is a representative of the \omega cyclic equivalence class. Each loop in the loop equivalence class has the same length, which is called the length of the necklace word, denoted by |\omega |.

For each arrow \alpha :v⟶v^{\mathrm{\prime }} in Q, add \alpha about v,v^{\mathrm{\prime }} of symmetric arrow {\alpha }^{\ast }, i.e., {\alpha }^{\ast }:v^{\mathrm{\prime }}⟶v.{\alpha }^{\ast } is called the starization of \alpha, and the repetitive arrow diagram \overline{Q} is induced by Q. Note Q_1^{\ast }=\left\{{\alpha }^{\ast }:v_j⟶v_i\mid \mathrm{\forall }\alpha :v_j⟶v_i\in Q_1\right\}, then {\overline{Q}}_0=Q_0, and {\overline{Q}}_1=Q_1\cup Q_1^{\ast }. Similarly, it can be defined as: \mathrm{\forall }\beta \in Q_1^{\ast },\beta :v_1⟶v_1^{\mathrm{\prime }},{\beta }^{\ast }:v_1^{\mathrm{\prime }}⟶v_1,{\beta }^{\ast } is also known as the starification of \beta. Obviously, from the definition, we know \mathrm{\forall }\beta \in {\overline{Q}}_1,{\left({\beta }^{\ast }\right)}^{\ast }=\beta. All cyclic equivalence classes in \overline{Q} are all necklace words form the set {\text{NW}}_{\overline{Q}}, based on all elements in this set on \mathbb{C} into vector space N_Q.

For any {\omega }_1={\alpha }_1{\alpha }_2\cdots {\alpha }_r and {\omega }_2={\beta }_1{\beta }_2\cdots {\beta }_s\in {\text{NW}}_{\overline{Q}} (once the subscript order of the arrows {\omega }_1 and {\omega }_2 is determined, it cannot be replaced), and any i\in \{1,2,\dots ,r\},j\in \{1,2,\dots ,s\},\mathrm{\forall }\alpha \in Q_1, binary operations are defined in set {\text{NW}}_{\overline{Q}} as follows [3, 12]:

\begin{array}{rl}& {\sigma }_{ij}^{\alpha }\left({\omega }_1,{\omega }_2\right)\\ & =\{\begin{array}{ll}{\alpha }_1{\alpha }_2\cdots {\alpha }_{i-1}{\beta }_{j+1}\cdots {\beta }_s{\beta }_1{\beta }_2\cdots {\beta }_{j-1}{\alpha }_{i+1}alph{a}_{i+2}\cdots {\alpha }_r,& \text{if}{\alpha }_i=\alpha \text{and}{\beta }_j={\alpha }^{\ast };\\ 0,& \text{other.}\end{array}\end{array}

The new necklace word {\sigma }_{ij}^{\alpha }\left({\omega }_1,{\omega }_2\right) is defined as a first-order conection of {\omega }_1,{\omega }_2 at position i,j with respect to \alpha. The Lie operation that defines the necklace word is as follows:

(1)\begin{array}{r}\left[{\omega }_1,{\omega }_2\right]=\sum _{\alpha \in Q_1}\sum _{i=1}^r\sum _{j=1}^s{\sigma }_{ij}^{\alpha }\left({\omega }_1,{\omega }_2\right)-\sum _{\alpha \in Q_1}\sum _{i=1}^r\sum _{j=1}^s{\sigma }_{ij}^{\alpha }\left({\omega }_2,{\omega }_1\right).\end{array}

Extend the above Lie operation linearly to N_Q,\mathrm{\forall }{a}_i,b_j\in \mathbb{C},x,y\in N_Q, let x=\sum _{i=1}^r{a}_i{\omega }_i,y=\sum _{j=1}^s{b}_j{\omega }_j, definition

[x,y]=\left[\sum _{i=1}^r{a}_i{\omega }_i,\sum _{j=1}^s{b}_j{\omega }_j\right]=\sum _{i=1}^r\sum _{j=1}^s{a}_i{b}_j\left[{\omega }_i,{\omega }_j\right].

It can be verified that the definition of the Lie operation \left[{\omega }_1,{\omega }_2\right] is independent of {\omega }_2, {\omega }_1 the selected loop representation, and the order in which the arrows are arranged.

According to the illustration of equation (1) and the necklace word, we can visualize the Lie operation: \mathrm{\forall }{\omega }_1,{\omega }_2\in {\text{NW}}_{\overline{Q}}, see Fig.2, i.e., \mathrm{\forall }\alpha \in Q_1, if \alpha appears in {\omega }_1, and then look for {\alpha }^{\ast } in {\omega }_2. If so, delete both \alpha and {\alpha }^{\ast } to open the two necklaces of {\omega }_1 and {\omega }_2 and connect the two paths after opening (the same vertex is connected together), to form a new necklace word. If \alpha appears n_1 times at {\omega }_1, and if \alpha appears n_2 times at {\omega }_2, the process needs to be repeated n_1{n}_2 times, and the combination of n_1{n}_2 necklace words and the sum of the necklace words constitutes a new necklace word. If {\alpha }^{\ast } is not in {\omega }_2, the new necklace word is treated as 0. Then look for the above \alpha in {\omega }_2, look for {\alpha }^{\ast } in {\omega }_1, repeat the above operation, get another new necklace word combination. Subtract the necklace word combination obtained by subtracting the new necklace word combination obtained first, and finally iterate \alpha \in Q_1, add up all the necklace words obtained after subtraction, and the sum is \left[{\omega }_1,{\omega }_2\right].

Since

\left[{\omega }_2,{\omega }_1\right]=\sum _{\alpha \in Q_1}\sum _{i=1}^r\sum _{j=1}^s{\sigma }_{ij}^{\alpha }\left({\omega }_2,{\omega }_1\right)-\sum _{\alpha \in Q_1}\sum _{i=1}^r\sum _{j=1}^s{\sigma }_{ij}^{\alpha }\left({\omega }_1,{\omega }_2\right),

we have \left[{\omega }_1,{\omega }_2\right]=-\left[{\omega }_2,{\omega }_1\right]. Then this Lie operation satisfies the antisymmetry on the vector space N_Q, and it can also be verified that this Lie operation satisfies the bilinearity, closure, and the Jacobi identity.

For a given arrow diagram Q, note Q_1=\left\{{\alpha }_1,{\alpha }_2,\dots ,{\alpha }_m\right\}, then Q_1^{\ast }=\left\{{\alpha }_1^{\ast },{\alpha }_2^{\ast },\dots ,{\alpha }_m^{\ast }\right\}, {\alpha }_i^{\ast } is \alpha of starvation. Take any \omega \in {\text{NW}}_{\overline{Q}}, let c be a loop in the class of circular equivalence represented by \omega. From apex v_{i_1}, the arrows in Q_1 that c passes through in turn is {\alpha }_{i_1},{\alpha }_{i_2},\dots ,{\alpha }_{i_l}, and let L_{\omega }=\left\{{\alpha }_{i_1},{\alpha }_{i_2},\dots ,{\alpha }_{i_l}\right\}; the arrow in the Q_1^{\ast } is {\alpha }_{j_1}^{\ast },{\alpha }_{j_2}^{\ast },\dots ,{\alpha }_{j_r}^{\ast }, and let R_{\omega }=\left\{{\alpha }_{j_1}^{\ast },{\alpha }_{j_2}^{\ast },\dots ,{\alpha }_{j_r}^{\ast }\right\}. Now take out all the subscripts of all elements in L_{\omega } and denote L_{\omega }^0, that is, L_{\omega }^0=\left(i_1,i_2,\dots ,i_l\right), and we can also get R_{\omega }^0=\left(j_1,j_2,\dots ,j_r\right), which are called the left index array and the right index array of \omega, respectively. We stipulate that such an index array is unordered. Since the arrows in the c can be repeated, the elements in the index array are repeatable. If an index array is represented by I, then when i_k is an element in I, denote i_k\in I. The number of elements in I is denoted by |I|. If there are no elements in I, remember I=\mathrm{\varnothing }.

Let I_1=\left(i_1,i_2,\dots ,i_l\right),I_2=\left(j_1,j_2,\dots ,j_r\right) be any two index arrays, and we define the relationship and movement between index arrays as follows:

(i) I_1=I_2 if and only if \mathrm{\forall }{i}_k\in I_1, the number of occurrences of i_k in I_1 is equal to the number of times it occurs in I_2, \mathrm{\forall }{j}_h\in I_2, the number of occurrences of j_h in I_2 is equal to the number of times it occurs in I_1. Otherwise, I_1 is not equal to I_2 and denoted as I_1\ne I_2.

(ii) I_1\subseteq I_2 if and only if \mathrm{\forall }{i}_k\in I_1, there must be i_k\in I_2, and i_k does not appear more times in I_1 than it appears in I_2.

(iii) I_1\subset I_2 if and only if I_1\subseteq I_2 and at least one i_k\in I_2 does not appear in I_1 or appears in I_2 more times than it appears in I_1.

(iv) I_1\cup I_2=\{t_k\mid t_k\in I_1 or t_k\in I_2, and the number of occurrences of t_k is equal to the number of occurrences of t_k in I_1 plus the number of occurrences in I_2\}.

(v) I_1\cap I_2=\{t_k\mid t_k\in I_1,t_k\in I_2, and the number of occurrences of t_k is equal to the lesser of the number of occurrences of t_k in I_1 and the number of occurrences in I_2\}. When there are no common elements in I_1 and I_2, define I_1\cap I_2=\mathrm{\varnothing },\mathrm{\varnothing } index array without any elements.

(vi) I_1-I_2=\{i_k\mid i_k\in I_1, and the number of occurrences of i_k is equal to the number of occurrences of i_k in I_1 minus the number of occurrences of i_k in I_2\}.

By definition, the relationship between the two index arrays and the operation and the set are both related to and different from the operation. For example, if I_1=(1,2,3,1),I_2=(2,1,3,1), then I_1=I_2; if I_1=(1,1),I_2=(1), then I_1\ne I_2,I_2\subset I_1, I_1\cap I_2=(1),I_1\cup I_2=(1,1,1),I_1-I_2=(1),I_2-I_1=\mathrm{\varnothing }.

Example 1　Set Q as shown in Fig.3.

In \overline{Q}, take \omega ={\alpha }_1{\alpha }_1^{\ast }{\alpha }_1{\alpha }_2{\alpha }_2^{\ast }{\alpha }_2{\alpha }_3, then \omega \in {\text{NW}}_{\overline{Q}},L_{\omega }^0=(1,1,2,2,3),R_{\omega }^0=(1,2). There is R_{\omega }^0\subset L_{\omega }^0,R_{\omega }^0\cap L_{\omega }^0=(1,2).

\mathrm{\forall }\omega \in {\text{NW}}_{\overline{Q}}, the relationship between the index arrays is as follows:

A) R_{\omega }^0\subset L_{\omega }^0, contains R_{\omega }^0=\mathrm{\varnothing }. In Example 1, take \omega ={\alpha }_1{\alpha }_1^{\ast }{\alpha }_1{\alpha }_2{\alpha }_2^{\ast }{\alpha }_2{\alpha }_3, then R_{\omega }^0=(1,2),L_{\omega }^0= (1,1,2,2,3), there is R_{\omega }^0\subset L_{\omega }^0.

B) L_{\omega }^0\subset R_{\omega }^0, contains L_{\omega }^0=\mathrm{\varnothing }. In Example 1, take \omega ={\alpha }_1{\alpha }_1^{\ast }{\alpha }_3^{\ast }{\alpha }_2^{\ast }{\alpha }_2{\alpha }_2^{\ast }{\alpha }_1^{\ast }, then L_{\omega }^0=(1,2),L_{\omega }^0= (1,1,3,2,2), there is L_{\omega }^0\subset R_{\omega }^0.

C) L_{\omega }^0=R_{\omega }^0, contains R_{\omega }^0=\mathrm{\varnothing }. In Example 1, take \omega ={\alpha }_1{\alpha }_4{\alpha }_4^{\ast }{\alpha }_1^{\ast }{\alpha }_3^{\ast }{\alpha }_3, then L_{\omega }^0=(1,4,3),R_{\omega }^0=(4,1,3), with L_{\omega }^0=R_{\omega }^0.

D) L_{\omega }^0\ne \mathrm{\varnothing },R_{\omega }^0\ne \mathrm{\varnothing }, and L_{\omega }^0\cap R_{\omega }^0=\mathrm{\varnothing }. As shown in Fig.4.

Take \omega ={\alpha }_4{\alpha }_5{\alpha }_6{\alpha }_3^{\ast }{\alpha }_2^{\ast }{\alpha }_1^{\ast }, then L_{\omega }^0=(4,5,6),R_{\omega }^0=(1,2,3),L_{\omega }^0\cap R_{\omega }^0=\mathrm{\varnothing }.

E) L_{\omega }^0\cap R_{\omega }^0\ne L_{\omega }^0,L_{\omega }^0\cap R_{\omega }^0\ne R_{\omega }^0, and L_{\omega }^0\cap R_{\omega }^0\ne \mathrm{\varnothing }. As shown in Fig.5.

Take \omega ={\alpha }_2{\alpha }_3{\alpha }_4{\alpha }_5{\alpha }_1^{\ast }{\alpha }_3^{\ast }{\alpha }_2^{\ast }, then L_{\omega }^0=(2,3,4,5),R_{\omega }^0=(1,2,3),L_{\omega }^0\cap R_{\omega }^0=(2,3).

As the definitions of L_{\omega }^0 and R_{\omega }^0, for any definite \omega \in {\text{NW}}_{\overline{Q}}, L_{\omega }^0 and R_{\omega }^0 are determined. Since the above categories A)–E) already contain all the relations between any two index arrays, the elements in {\text{NW}}_{\overline{Q}} can be divided into the above 5 categories.

By equation (1), \mathrm{\forall }{\omega }_1,{\omega }_2\in {\text{NW}}_{\overline{Q}},\left[{\omega }_1,{\omega }_2\right]=\sum _{\alpha }\sum _{i,j}{\sigma }_{ij}^{\alpha }\left({\omega }_1,{\omega }_2\right)-\sum _{\alpha }\sum _{i,j}{\sigma }_{ij}^{\alpha }\left({\omega }_2,{\omega }_1\right). If {\sigma }_{ij}^{\alpha }\left({\omega }_1,{\omega }_2\right)\ne 0, record {\sigma }_{ij}^{\alpha }\left({\omega }_1,{\omega }_2\right)={\omega }_{ij}, corresponding record {\sigma }_{ij}^{\alpha }\left({\omega }_2,{\omega }_1\right)={\omega }_{ij}^{\mathrm{\prime }}. Using the definitions of the index array L_{\omega }^0 and R_{\omega }^0, it is known that \mathrm{\exists }{\alpha }_k\in Q_1 such that

\begin{array}{ll}{L}_{{\omega }_{ij}}^0=\left(L_{{\omega }_1}^0-(k)\right)\cup L_{{\omega }_2}^0,& R_{{\omega }_{ij}}^0=\left(R_{{\omega }_2}^0-(k)\right)\cup R_{{\omega }_1}^0,\\ L_{{\omega }_{ij}^{\mathrm{\prime }}}^0=\left(L_{{\omega }_2}^0-(k)\right)\cup L_{{\omega }_1}^0,& R_{{\omega }_{ij}^{\mathrm{\prime }}}^0=\left(R_{{\omega }_1}^0-(k)\right)\cup R_{{\omega }_2}^0.\end{array}

3 Main results

Set the arrow diagram Q_1 as shown in Fig.6:

All necklace words of length 2 in Fig.6 are h_1={\alpha }_1{\alpha }_1^{\ast },h_2={\alpha }_2{\alpha }_2^{\ast },{\alpha }_1{\alpha }_2^{\ast } and {\alpha }_1^{\ast }{\alpha }_2. The order g_1 is a linear space generated by h_1,h_2,{\alpha }_1{\alpha }_2^{\ast } and {\alpha }_1^{\ast }{\alpha }_2.

Proposition 1　 g_1 is a subalgebra of N_Q, and the derivative algebra g_1^{(1)} of g_1 is a subspace generated by {\alpha }_1{\alpha }_2^{\ast },{\alpha }_1^{\ast }{\alpha }_2 and h_2-h_1.

Since g_1 is the linear space generated by all necklace words of length 2, \mathrm{\forall }{\omega }_1,{\omega }_2\in g_1, then any non-zero necklace word in \left[{\omega }_1,{\omega }_2\right] has a length of 2, and \left[{\omega }_1,{\omega }_2\right]\in g_1,g_1 is a subalgebra of N_Q. Because

\left[{\alpha }_1{\alpha }_2^{\ast },h_1\right]={\alpha }_1{\alpha }_2^{\ast },\left[{\alpha }_1^{\ast }{\alpha }_2,h_1\right]=-{\alpha }_1^{\ast }{\alpha }_2,\left[{\alpha }_1{\alpha }_2^{\ast },{\alpha }_1^{\ast }{\alpha }_2\right]=h_2-h_1,

the subspaces generated by {\alpha }_1{\alpha }_2^{\ast },{\alpha }_1^{\ast }{\alpha }_2 and h_2-h_1 must contain the derivative algebra g_1^{(1)}=\left[g_1,g_1\right] of g_1. And it can be concluded that h_1\notin g_1^{(1)}. Conversely, if h_1\in g_1^{(1)}, then there must be the necklace word x,y such that h_1 appears in the expression of [x,y]. Since the length of h_1 is 2, there must be necklace words x,y with length of 2, and x={\alpha }_1{\alpha }_2^{\ast },y={\alpha }_1^{\ast }{\alpha }_2 or x={\alpha }_1^{\ast }{\alpha }_2,y={\alpha }_1{\alpha }_2^{\ast }. As

\left[{\alpha }_1{\alpha }_2^{\ast },{\alpha }_1^{\ast }{\alpha }_2\right]=h_2-h_1,\left[{\alpha }_1^{\ast }{\alpha }_2,{\alpha }_1{\alpha }_2^{\ast }\right]=h_1-h_2,

h_1 can never be solved. Thus, Proposition 1 holds.

Proposition 2　 g_1^{(1)} is a three-dimensional simple Lie algebra.

Proof　The three-dimensional simple Lie algebra \text{sl}(2) exists based on e,f, and h, such that

[h,e]=2e,[h,f]=-2f,[e,f]=h.

g_1^{(1)} is a subspace generated by {\alpha }_1{\alpha }_2^{\ast },{\alpha }_1^{\ast }{\alpha }_2 and h_2-h_1, so

\left[h_2-h_1,{\alpha }_1{\alpha }_2^{\ast }\right]=2{\alpha }_1{\alpha }_2^{\ast },\left[h_2-h_1,{\alpha }_1^{\ast }{\alpha }_2\right]=-2{\alpha }_1^{\ast }{\alpha }_2,\left[{\alpha }_1{\alpha }_2^{\ast },{\alpha }_1^{\ast }{\alpha }_2\right]=h_2-h_1.

The isomorphic map between g_1^{(1)} and \text{sl}(2) is constructed as follows:

g_1^{(1)}\to \text{sl}(2),h_2-h_1\mapsto h,{\alpha }_1{\alpha }_2^{\ast }\mapsto e,{\alpha }_1^{\ast }{\alpha }_2\mapsto f.

The basis vectors expand linearly, then g_1^{(1)}\cong \text{sl}(2). The original proposition holds.

Proposition 3　 The center c\left(g_1\right)=b_1\left(h_1+h_2\right) of g_1, \mathrm{\forall }{b}_1\in \mathbb{C}.

Proof　Let x=b_1{h}_1+b_2{h}_2+b_3{\alpha }_1{\alpha }_2^{\ast }+b_4{\alpha }_1^{\ast }{\alpha }_2\in c\left(g_1\right). Thus,

\left[x,h_1\right]=b_3{\alpha }_1{\alpha }_2^{\ast }-b_4{\alpha }_1^{\ast }{\alpha }_2=0.

And {\alpha }_1{\alpha }_2^{\ast } and {\alpha }_1^{\ast }{\alpha }_2 are linear independent, b_3=b_4=0,

x\in c\left(g_1\right),\left[x,{\alpha }_1{\alpha }_2^{\ast }\right]=\left(-b_1+b_2\right){\alpha }_1{\alpha }_2^{\ast }=0.

Thus, b_1=b_2. Thereby, Proposition 3 holds.

Proposition 4　 g_1 only has two proper ideals, c\left(g_1\right) and g_1^{(1)}, and g_1 can be decomposed into the direct sum of two proper ideals g_1=c\left(g_1\right)\oplus g_1^{(1)}.

Proof　 g_1^{(1)} is a three-dimensional subspace generated by {\alpha }_1{\alpha }_2^{\ast },{\alpha }_1^{\ast }{\alpha }_2, and h_2-h_1, while c(g) is a subspace generated by h_1+h_2, and g_1=c\left(g_1\right)\cap g_1^{(1)}=\{0\}. As the direct sum of subspaces, then c\left(g_1\right)+g_1^{(1)},c\left(g_1\right) and g_1^{(1)} are ideal. So, there is g_1=c\left(g_1\right)\oplus g_1^{(1)}.

Assuming that u is ideal for g_1, then u\cap g_1^{(1)} is also ideal for g_1^{(1)}, and g_1^{(1)} is a simple Lie algebra, thus u\cap g_1^{(1)}=0 or u\cap g_1^{(1)}=g_1^{(1)}.

If u\cap g_1^{(1)}=0, then u=0 or u=c\left(g_1\right).

If u\cap g_1^{(1)}=g_1^{(1)}, then g_1^{(1)}\subseteq u, while g_1^{(1)}\subseteq u\subseteq g_1,\dim \left(g_1\right)-\dim \left(g_1^{(1)}\right)=1, thus u=g_1^{(1)} or u=g_1. In summary, Proposition 4 holds.

Set the arrow diagram Q_2 as shown in Fig.7.

All necklace words of length 2 are {\overline{h}}_1={\alpha }_1{\alpha }_1^{\ast },{\overline{h}}_2={\alpha }_2{\alpha }_2^{\ast },{\alpha }_1{\alpha }_2 and {\alpha }_1^{\ast }{\alpha }_2^{\ast }. The order g_2 is a linear space generated by {\overline{h}}_1,{\overline{h}}_2,{\alpha }_1{\alpha }_2 and {\alpha }_1^{\ast }{\alpha }_2^{\ast }.

Proposition 5　 g_2 is a subalgebra of N_{Q_2}, and the derivative algebra g_2^{(1)} of g_2 is a linear space generated by {\alpha }_1{\alpha }_2,{\alpha }_1^{\ast }{\alpha }_2^{\ast }, and {\overline{h}}_2+{\overline{h}}_1.

Proof　Similar as the proof of Proposition 1.

Proposition 6　 g_2^{(1)} is a three-dimensional simple Lie algebra.

Proof　The three-dimensional simple Lie algebra \text{sl}(2) exists based on e,f, and h, such that

[h,e]=2e,[h,f]=-2f,[e,f]=h.

g_2^{(1)} is a subspace generated by {\alpha }_1{\alpha }_2,{\alpha }_2^{\ast }{\alpha }_1^{\ast }, and {\overline{h}}_2+{\overline{h}}_1. So

\left[{\overline{h}}_2+{\overline{h}}_1,-{\alpha }_1^{\ast }{\alpha }_2^{\ast }\right]=-2{\alpha }_1^{\ast }{\alpha }_2^{\ast },\left[{\overline{h}}_2+{\overline{h}}_1,{\alpha }_1{\alpha }_2\right]=-2{\alpha }_1{\alpha }_2,\left[-{\alpha }_1^{\ast }{\alpha }_2^{\ast },{\alpha }_1{\alpha }_2\right]={\overline{h}}_2+{\overline{h}}_1.

The isomorphism map between g_2^{(1)} and \text{sl}(2) is constructed as follows:

g_2^{(1)}\to \mathrm{sl}(2),{\overline{h}}_2+{\overline{h}}_1\mapsto h,-{\alpha }_1^{\ast }{\alpha }_2^{\ast }\mapsto e,{\alpha }_1{\alpha }_2\mapsto f.

Then the question of the basis vector is linearly expanded, and g_2^{(1)}\cong \text{sl}(2). Proposition 6 holds.

From Proposition 2 and Proposition 6, the derivative algebra g_1^{(1)} of g_1 and the derivative algebra g_2^{(1)} of g_2 are isomorphic to \text{sl}(2), so we find the implementation of two different arrow diagrams \text{sl}(2) when n=2. Since g_1^{(1)}\cong g_2^{(1)}, is g_1 isomorphic with g_2? Now, we extend from the isomorphism of g_1^{(1)} to g_2^{(1)} to the isomorphism between g_1 and g_2.

Proposition 7　 g_2 is isomorphic to g_1.

Proof　Construct the isomorphic map \sigma as follows:

\sigma :g_2\to g_1,\sigma \left({\overline{h}}_2\right)=h_2,\sigma \left({\overline{h}}_1\right)=-h_1,\sigma \left({\alpha }_1^{\ast }{\alpha }_2^{\ast }\right)=-{\alpha }_1{\alpha }_2^{\ast },\sigma \left({\alpha }_1{\alpha }_2\right)={\alpha }_1^{\ast }{\alpha }_2.

\sigma expands linearly between the basis vectors of g_2. g_2^{(1)}\cong \text{sl}(2), and the correspondence of the basis vector is

{\overline{h}}_2+{\overline{h}}_1\mapsto h,-{\alpha }_1^{\ast }{\alpha }_2^{\ast }\mapsto e,{\alpha }_1{\alpha }_2\mapsto f.

g_1^{(1)}\cong \text{sl}(2), and the correspondence of the basis vector is

h_2-h_1\mapsto h,{\alpha }_1{\alpha }_2^{\ast }\mapsto e,{\alpha }_1^{\ast }{\alpha }_2\mapsto f.

The construction process from \sigma has

\sigma \left({\overline{h}}_2+{\overline{h}}_1\right)=\sigma \left({\overline{h}}_2\right)+\sigma \left({\overline{h}}_1\right)=h_2-h_1,\sigma \left(-{\alpha }_1^{\ast }{\alpha }_2^{\ast }\right)={\alpha }_1{\alpha }_2^{\ast },\sigma \left({\alpha }_1{\alpha }_2\right)={\alpha }_1^{\ast }{\alpha }_2.

So, \sigma is an isomorphism of g_2^{(1)} to g_1^{(1)}.

Further,

\begin{array}{ll}& \left[\sigma \left({\overline{h}}_2\right),\sigma \left(-{\alpha }_1^{\ast }{\alpha }_2^{\ast }\right)\right]=\left[h_2,{\alpha }_1{\alpha }_2^{\ast }\right]={\alpha }_1{\alpha }_2^{\ast }=\sigma \left(-{\alpha }_1^{\ast }{\alpha }_2^{\ast }\right)=\sigma \left(\left[{\overline{h}}_2,-{\alpha }_1^{\ast }{\alpha }_2^{\ast }\right]\right),\\ & \left[\sigma \left({\overline{h}}_2\right),\sigma \left({\alpha }_1{\alpha }_2\right)\right]=\sigma \left(\left[{\overline{h}}_2,{\alpha }_1{\alpha }_2\right]\right),\\ & \left[\sigma \left({\overline{h}}_1\right),\sigma \left({\alpha }_1^{\ast }{\alpha }_2^{\ast }\right)\right]=\sigma \left(\left[{\overline{h}}_1,{\alpha }_1^{\ast }{\alpha }_2^{\ast }\right]\right),\\ & \left[\sigma \left({\overline{h}}_1\right),\sigma \left({\alpha }_1{\alpha }_2\right)\right]=\sigma \left(\left[{\overline{h}}_1,{\alpha }_1{\alpha }_2\right]\right).\end{array}

Thus, \sigma is an isomorphism of g_2 to g_1.

We can actually construct an isomorphic map {\sigma }_1 from g_2 to g_1 as follows:

{\sigma }_1\left({\overline{h}}_2\right)=-h_1,{\sigma }_1\left({\overline{h}}_1\right)=h_2,{\sigma }_1\left({\alpha }_1^{\ast }{\alpha }_2^{\ast }\right)=-{\alpha }_1{\alpha }_2^{\ast },{\sigma }_1\left({\alpha }_1{\alpha }_2\right)={\alpha }_1^{\ast }{\alpha }_2.

{\sigma }_1 also maintains the isomorphism of g_1^{(1)} to g_2^{(1)}.

Set the arrow diagram Q_3 as shown in Fig.8.

All necklace words of length 2 are {\alpha }_1{\alpha }_2^{\ast },{\alpha }_1^{\ast }{\alpha }_2,{\alpha }_1{\alpha }_3^{\ast },{\alpha }_1^{\ast }{\alpha }_3,{\alpha }_2{\alpha }_3^{\ast },{\alpha }_2^{\ast }{\alpha }_3 and h_i={\alpha }_i{\alpha }_i^{\ast }(i=1,2,3). Let g_3 be the linear space generated by all necklace words of length 2. g_3 is a subalgebra of N_{Q_3}. The derivative algebra g_3^{(1)} of g_3 is the linear space generated by {\alpha }_1{\alpha }_2^{\ast },{\alpha }_1^{\ast }{\alpha }_2,{\alpha }_1{\alpha }_3^{\ast },{\alpha }_1^{\ast }{\alpha }_3,{\alpha }_2{\alpha }_3^{\ast },{\alpha }_2^{\ast }{\alpha }_3,h_1-h_2, and h_1-h_3.

Proposition 8　 The center c\left(g_3\right)=k_7\left(h_1+h_2+h_3\right) of g_3, \mathrm{\forall }{k}_7\in \mathbb{C}.

Proof　Let

x=k_1{\alpha }_1{\alpha }_2^{\ast }+k_2{\alpha }_1^{\ast }{\alpha }_2+k_3{\alpha }_1{\alpha }_3^{\ast }+k_4{\alpha }_1^{\ast }{\alpha }_3+k_5{\alpha }_2{\alpha }_3^{\ast }+k_6{\alpha }_2^{\ast }{\alpha }_3+k_7{h}_1+k_8{h}_2+k_9{h}_3\in c\left(g_3\right).

Then

\left[x,h_1\right]=k_1{\alpha }_1{\alpha }_2^{\ast }-k_2{\alpha }_1^{\ast }{\alpha }_2+k_3{\alpha }_1{\alpha }_3^{\ast }-k_4{\alpha }_1^{\ast }{\alpha }_3=0.

{\alpha }_1{\alpha }_2^{\ast },{\alpha }_1^{\ast }{\alpha }_2,{\alpha }_1{\alpha }_3^{\ast }, and {\alpha }_1^{\ast }{\alpha }_3 are linear independent. So,

\begin{array}{ll}& k_1=k_2=k_3=k_4=0,\\ & x=k_5{\alpha }_2{\alpha }_3^{\ast }+k_6{\alpha }_2^{\ast }{\alpha }_3+k_7{h}_1+k_8{h}_2+k_9{h}_3\in c\left(g_3\right),\\ & \left[x,h_2\right]=k_5{\alpha }_2{\alpha }_3^{\ast }-k_6{\alpha }_2^{\ast }{\alpha }_3.\end{array}

{\alpha }_2{\alpha }_3^{\ast } and {\alpha }_2^{\ast }{\alpha }_3 are linearly independent, thus k_5=k_6. Therefore,

\begin{array}{ll}& x=k_7{h}_1+k_8{h}_2+k_9{h}_3\in c\left(g_3\right),\\ & \left[x,{\alpha }_1{\alpha }_2^{\ast }\right]=-k_7+k_8=0\Rightarrow k_7=k_8,\\ & \left[x,{\alpha }_2{\alpha }_3^{\ast }\right]=-k_8+k_9=0\Rightarrow k_8=k_9,\end{array}

Thus, there must be x=k_7\left(h_1+h_2+h_3\right)\left(\mathrm{\forall }{k}_7\in \mathbb{C}\right). Conversely, when x=k_7\left(h_1+h_2+h_3\right)\left(\mathrm{\forall }{k}_7\in \mathbb{C}\right), it can be verified that there are [x,y]=0 for \mathrm{\forall }y\in g_3. In fact, for any necklace word \omega of length 2, [x,\omega ]=0, and g_3 is all necklace words of length 2. Thus, Proposition 8 holds.

Proposition 9　 c\left(g_3\right)\cap g_3^{(1)}=\{0\}.

Proof　Let y\in c\left(g_3\right)\cap g_3^{(1)}. Then there exist a,a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8\in \mathbb{C}, such that

\begin{array}{rl}y& =a\left(h_1+h_2+h_3\right)\\ & =a_1\left({\alpha }_1{\alpha }_2^{\ast }\right)+a_2\left({\alpha }_1^{\ast }{\alpha }_2\right)+a_3\left({\alpha }_1{\alpha }_3^{\ast }\right)+a_4\left({\alpha }_1^{\ast }{\alpha }_3\right)\\ & +a_5\left({\alpha }_2{\alpha }_3^{\ast }\right)+a_6\left({\alpha }_2^{\ast }{\alpha }_3\right)+a_7\left(h_1-h_2\right)+a_8\left(h_1-h_3\right).\end{array}

Then

a_1=a_2=a_3=a_4=a_5=a_6=0,

and

a=a_7+a_8,a=-a_7,a=-a_8\Rightarrow a=a_7=a_8=0.

Therefore, y=0. Proposition 9 holds.

According to Proposition 2, the derivative algebra g_3^{(1)} contains at least 3 three-dimensional simple Lie algebras, namely g_{12},g_{13}, and g_{23}. g_{12} is a linear space generated by {\alpha }_1{\alpha }_2^{\ast }, {\alpha }_1^{\ast }{\alpha }_2, and h_1-h_2, g_{13} is a linear space generated by {\alpha }_1{\alpha }_3^{\ast },{\alpha }_1^{\ast }{\alpha }_3, and h_1-h_3, g_{23} is a linear space generated by {\alpha }_2{\alpha }_3^{\ast },{\alpha }_3^{\ast }{\alpha }_2, and h_2-h_3. However, it can be verified that g_{12},g_{13}, and g_{23} are not ideal for g_3^{(1)}. For linear subspaces g_{12},g_{13}, and g_{23}, any two of them can be the direct sum of linear subspaces, but g_{12},g_{13}, and g_{23} cannot be the direct sum of linear subspaces, otherwise the dimension of g_3^{(1)} is 9. While \dim \left(g_3^{(1)}\right)=8, actually because h_2-h_3=\left(h_1-h_2\right)-\left(h_1-h_3\right), and h_2-h_3\in g_{23},h_1-h_2\in g_{12},h_1-h_3\in g_{13}.

\text{sl}(3) is a typical simple-Lie algebra on \mathbb{C}, with a set of bases of E_{12},E_{21},E_{13}, E_{31}, E_{23},E_{32},E_{11}-E_{22}, and E_{11}-E_{33} (E_{ij} denotes the matrix with the elements of the i\mathrm{t}\mathrm{h} row and j\mathrm{t}\mathrm{h} column as 1 and the remaining elements as 0).

Proposition 10　 g_3^{(1)} is a simple Lie algebra.

Proof　A typical simple Lie algebra \text{sl}(3) has a set of bases of E_{12},E_{21},E_{13},E_{31},E_{23},E_{32},E_{11}-E_{22}, and E_{11}-E_{33}, while g_3^{(1)} has a set of bases of {\alpha }_1{\alpha }_2^{\ast },{\alpha }_1^{\ast }{\alpha }_2,{\alpha }_1{\alpha }_3^{\ast },{\alpha }_1^{\ast }{\alpha }_3,{\alpha }_2{\alpha }_3^{\ast },{\alpha }_2^{\ast }{\alpha }_3,h_1-h_2, and h_1-h_3, construct isomorphic maps as follows:

\begin{array}{ll}{\sigma }_2\left({\alpha }_1^{\ast }{\alpha }_2\right)=E_{12},& {\sigma }_2\left({\alpha }_2^{\ast }{\alpha }_1\right)=E_{21},\\ {\sigma }_2\left({\alpha }_1^{\ast }{\alpha }_3\right)=E_{13},& {\sigma }_2\left({\alpha }_3^{\ast }{\alpha }_1\right)=E_{31},\\ {\sigma }_2\left({\alpha }_2^{\ast }{\alpha }_3\right)=E_{23},& {\sigma }_2\left({\alpha }_3^{\ast }{\alpha }_2\right)=E_{23},\\ {\sigma }_2\left(h_1-h_2\right)=E_{11}-E_{22},\\ {\sigma }_2\left(h_1-h_3\right)=E_{11}-E_{33}.\end{array}

Then expand linearly between the basis vectors of g_3^{(1)}. \mathrm{\forall }i,j,k,l\in \{1,2,3\},

\begin{array}{rl}& \left[{\alpha }_j{\alpha }_i^{\ast },{\alpha }_l{\alpha }_k^{\ast }\right]=\{\begin{array}{ll}{\alpha }_i{\alpha }_i^{\ast }-{\alpha }_j{\alpha }_j^{\ast },& j=k,\text{and}i=l,\\ {\alpha }_l{\alpha }_i^{\ast },& j=k,\text{and}i\ne l,\\ -{\alpha }_j{\alpha }_k^{\ast },& j\ne k,\text{and}i=l,\\ 0,& j\ne k,\text{and}i\ne k;\end{array}\\ & \left[E_{ij},E_{kl}\right]=E_{ij}{E}_{kl}-E_{kl}{E}_{ij}=\{\begin{array}{ll}{E}_{ii}-E_{jj},& j=k,\text{and}i=l,\\ E_{il},& j=k,\text{and}i\ne l,\\ -E_{kj},& j\ne k,\text{and}i=l,\\ 0,& j\ne k,\text{and}i\ne k.\end{array}\end{array}

In the above discussion, when i=j, {\alpha }_i^{\ast }{\alpha }_j=h_i; when l=k, {\alpha }_l^{\ast }{\alpha }_k=h_k, i.e., g_3^{(1)} and \text{sl}(3) have exactly the same operation rules. For any necklace word {\omega }_1,{\omega }_2\in g_3^{(1)},{\sigma }_2\left[{\omega }_1,{\omega }_2\right]=\left[{\sigma }_2\left({\omega }_1\right),{\sigma }_2\left({\omega }_2\right)\right],g_3^{(1)}\cong \text{sl}(3), and \text{sl}(3) is a simple Lie algebra. Thus, g_3^{(1)} is a simple Lie algebra.

Set the arrow diagram Q_4 as shown in Fig.9.

All necklace words of length 2 are {\alpha }_i{\alpha }_j^{\ast }(\mathrm{\forall }i\ne j,i,j\in \{1,2,\dots ,n\}),h_i={\alpha }_i{\alpha }_i^{\ast }(i\in \{1,2,\dots ,n\}). Let g_4 be the linear space generated by all necklace words of length 2, and it can be similarly shown that g_4 is a subalgebra of N_Q, while the derivative algebra g_4^{(1)} of g_4 is a linear subspace generated by {\alpha }_i{\alpha }_j^{\ast }(\mathrm{\forall }i\ne j,i,j\in \{1,2,\dots ,n\}) and h_i-h_{i+1}(i\in \{1,2,\dots ,n-1\}).

Proposition 11　 g_4^{(1)} isomorphic to \text{sl}(n), g_4^{(1)} is a simple Lie algebra.

Proof　Typical simple Lie algebra \text{sl}(n) has a set base of E_{ij}(\mathrm{\forall }i\ne j,i,j\in \{1,2,\dots ,n\}) and E_{ii}-E_{(i+1)(i+1)}, and the set base for g_4^{(1)} is {\alpha }_i{\alpha }_j^{\ast }(\mathrm{\forall }i\ne j,i,j\in \{1,2,\dots ,n\}) and h_i-h_{i+1}(i\in \{1,2,\dots ,n-1\}), construct the isomorphic map {\sigma }_3 as follows:

\begin{array}{rl}{\sigma }_3:g^{(1)}\to & \mathrm{sl}(n),{\sigma }_3\left({\alpha }_i{\alpha }_j^{\ast }\right)=E_{ji}(\mathrm{\forall }i\ne j,i,j\in \{1,2,\dots n\}),\\ & {\sigma }_3\left(h_i-h_{i+1}\right)=E_{ii}-E_{(i+1)(i+1)}.\end{array}

Then expand linearly between the basis vectors of g_4^{(1)}. \mathrm{\forall }i,j,k,l\in \{1,2,\dots ,n\},

\begin{array}{rl}& \left[{\alpha }_j{\alpha }_i^{\ast },{\alpha }_l{\alpha }_k^{\ast }\right]=\{\begin{array}{ll}{\alpha }_i{\alpha }_i^{\ast }-{\alpha }_j{\alpha }_j^{\ast },& j=k,\text{and}i=l,\\ {\alpha }_l{\alpha }_i^{\ast },& j=k,\text{and}i\ne l,\\ -{\alpha }_j{\alpha }_k^{\ast },& j\ne k,\text{and}i=l,\\ 0,& j\ne k,\text{and}i\ne k;\end{array}\\ & \left[E_{ij},E_{kl}\right]=E_{ij}{E}_{kl}-E_{kl}{E}_{ij}=\{\begin{array}{ll}{E}_{ii}-E_{jj},& j=k,\text{and}i=l,\\ E_{il},& j=k,\text{and}i\ne l,\\ -E_{kj},& j\ne k,\text{and}i=l,\\ 0,& j\ne k,\text{and}i\ne k.\end{array}\end{array}

In the above discussion, when i=j, {\alpha }_i^{\ast }{\alpha }_j=h_i; when l=k, {\alpha }_l^{\ast }{\alpha }_k=h_k, that is, g_4^{(1)} has exactly the same operation rules as \text{sl}(n). For any necklace word {\omega }_1,{\omega }_2\in g_4^{(1)},{\sigma }_2\left[{\omega }_1,{\omega }_2\right]=\left[{\sigma }_2\left({\omega }_1\right),{\sigma }_2\left({\omega }_2\right)\right],g_4^{(1)}\cong \text{sl}(n), and \text{sl}(n) is a simple Lie algebra, so g_4^{(1)} is a simple Lie algebra.

Set the arrow diagram Q_5 as shown in Fig.10.

h_i={\alpha }_i{\alpha }_i^{\ast }(i=1,2,3),{\omega }_1={\alpha }_1{\alpha }_2{\alpha }_3. Let g_5 be the linear space generated by h_i={\alpha }_i{\alpha }_i^{\ast }(i=1,2,3) and {\omega }_1={\alpha }_1{\alpha }_2{\alpha }_3. g_{51} is the linear space generated by {\omega }_1={\alpha }_1{\alpha }_2{\alpha }_3, g_{52} is the linear space generated by {\omega }_1 and h_1, g_{53} is the linear space generated by {\omega }_1,h_1, and h_2.

Proposition 12　 g_5 is the solvable subalgebra of N_Q. g_{51},g_{52},g_{53} are solvable ideals, and \{0\}\subset g_{51}\subset g_{52}\subset g_{53}\subset g_5.

Proof　Since \left[{\omega }_1,h_i\right]={\omega }_1,\left[h_j,h_i\right]=0,\mathrm{\forall }i,j\in \{1,2,3\}, thus \left[g_5,g_5\right]=g_{51}. While g_{51} is a one-dimensional commutative Lie algebra, g_{51},g_{52},g_{53} are solvable ideals, and \{0\}\subset g_{51}\subset g_{52}\subset g_{53}\subset g_5, thus Proposition 12 holds.

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