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ISSN: 1736-4337
Journal of Generalized Lie Theory and Applications
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Classification of Maximal Subalgebras and Corresponding Reductive Pairs of Lie Algebra of All 2 × 2 Real Matrices

Shtukar U*

Math/Physics Department, North Carolina Central University, Durham, USA

*Corresponding Author:
Shtukar U
Associate Professor, Math/Physics Department
North Carolina Central University
1801 Fayetteville Street, Durham, NC 27707, USA
Tel: 919-597-0375
E-mail: [email protected]

Received date: February 19, 2017; Accepted date: April 10, 2017; Published date: April 17, 2017

Citation: Shtukar U (2017) Classification of Maximal Subalgebras and Corresponding Reductive Pairs of Lie Algebra of All 2 × 2 Real Matrices. J Generalized Lie Theory Appl 11: 263. doi: 10.4172/1736-4337.1000263

Copyright: © 2017 Shtukar U. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.

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Abstract

The purpose of the article is to describe all 3-dimensional subalgebras and all corresponding reductive pairs of Lie algebra of all 2 × 2 real matrices. This Lie algebra is 4-dimensional as a vector space, it’s not simple, and it’s not solvable. The evaluation procedure utilizes the canonical bases for subspaces that were introduced. In Part I of this article, all 3-dimensional subalgebras of the given Lie algebra g are classified. All reductive pairs {h, m} with 3-dimensional subalgebras h are found in Part II. Surprisingly, there is only one reductive pair {h, m} with special 3-dimensional subalgebra h and 1-dimensional complement m. Finally, all reductive pairs {h, m} with 1-dimensional subalgebras h of algebra g are classified in Part III of the article.

Keywords

Lie algebra; Subalgebras; Reductive pairs

Introduction

Reductive homogeneous spaces appeared for the first time in the fundamental manuscript [1,2] of Katsumi Nomizu, in which the author investigated invariant affine connections and Riemannian metrics on them. Sagle and Winter in their article [3] analyzed algebraic structures generated by reductive pairs of simple Lie algebras. The next problem studied by some authors was classification of subalgebras of some Lie algebras. For example, Patera and Winternitz have classified all subalgebras of real Lie algebras of dimensions d=3 and d=4 in their manuscript [4]. This classification of subalgebras of low dimensional real Lie algebras was done by a representative of each conjugacy class where the conjugacy was considered under the group of inner automorphisms of Lie algebras. The articles mentioned above have stimulated this research for all subalgebras and all reductive pairs at Lie algebra g of all real 2 × 2 matrices. In contrast to the article [4], this research is utilized a different method. Our method involves canonical bases for subspaces [1] that allow us to find all 3-dimensional subalgebras and the corresponding reductive pairs of the given Lie algebra g. Our classification of reductive pairs is done here for the first time. The classification of 2-dimensional subalgebras with its reductive pairs of the same Lie algebra will be done at the separate article. New knowledge concerning the structure of this Lie algebra is important for Algebra, Geometry, and Physics.

We start with standard definitions for the readers’ convenience.

Definition 1

Let g be a vector space over a field F. Then g is called a Lie algebra over F if there exists a Lie bracket operation equation for anyequation such that:

equation for any a∈F,

equation and

equation

We call equation a Lie product.

Definition 2

Let g be a Lie algebra. A subspace h⊂g is called a (Lie) subalgebra of g, if [h, h]⊂h.

Definition 3

Let g be a Lie algebra, h be subalgebra of g. If there exists a subspace m of g such that h ⊕ m= g and [h, m]⊂m, then {h, m} is called a reductive pair of g, and {g,h, m} is called a reductive triple. We say also that subspace m is a reductive complement for h.

Lie algebra g and its standard basis:

This Lie algebra contains all 2 × 2 matrices over the field of all real numbers. The standard basis of this algebra consists of the next four matrices:

equation

It is well known that the Lie multiplication operation [A, B] for any two square matrices A and B of the same size is defined to be [A, B] = ABBA. According this rule, the fundamental products of the basic vectors (matrices) equation can be computed:

equation

All other products of basic vectors are zeros.

Let h be any 3-dimensional subspace of Lie algebra g. We can describe subspace h as equation whereequation andequation are 3 linearly independent vectors.

According to the article [1], all canonical bases for 3-dimensional subspaces of 4-dimensional vector space are:

equation

Part I. Maximal subalgebras of Lie algebra g

Now we start to determine that a 3-dimensional subspace equation is a subalgebra of Lie algebra g when vectorsequation form one of the canonical bases (1), (2), (3), or (4) listed above. We have to check that the condition [h, h]⊂h is true for each of these bases. The necessary evaluation procedure follows.

Let equation be the basis (1) for h. Evauate Lie products equation

equation

So, equation and equation

The system of 3 equations for equation is:equationequation We have two different solutions:equation and equation This means that the following 1-parameter set of subalgebras h1 and one special subalgebra h2 exist for this case:

equation

2. Let equation be the basis (2) for a possible subalgebra h. Evaluate Lie products equation

equation

So, we have equationequation Consequently, the following system of equations appears for components a3, b3:

equation

This system of equations has only one solution a3=0, b3=0. This solution produces the following subalgebra:

equation

Let equation be the basis (3) for h. Evaluate Lie products equation. for this case. We have:

equation So,equationequation

equation So,equationequation

equation

The system of equations has only one solution a2=0. The corresponding subalgebra is equation

Consider the last possible basis equation Evaluate Lie products equation for this basis. We obtain:

equation

The vector equation doesn’t belong to the subspace equation at this case. So, this subspace is not subalgebra of algebra g.

The next statement describes all 3-dimensional subalgebras of Lie algebra g.

Theorem 1: All different 3-dimensional subalgebras of Lie algebra of all 2 × 2 real matrices are listed here:

equation

equation

Corollary: The subalgebras above are maximal for the given Lie algebra.

Part II. Reductive pairs with 3-dimensional subalgebras h of Lie algebra g

How many of 3-dimensional subalgebras h form reductive pairs {h, m} in this Lie algebra? To answer this question, we will use the conditions from the Definition 3, i.e. [h, m]⊂m, g= h⊕ m where m is an appropriate 1-dimensional reductive complement for a given 3-dimensional subalgebra h. The list of all 3-dimensional subalgebras from Theorem 2 will be used to find all possible reductive complements. Letequation be a possible 1-dimensional complement. To simplify our evaluation, we consider 2 possible cases for the generating vector equation

equation and equationequation

Subalgebra h1 Case 1: Multiply basic vectors

equation from h1 by vector equation We have:

equation (it’s an identity),

equation

equationequation

From the vector equalities above, we obtain the following system of conditions for the components d2, d3, d4 and coefficients y, z: y=d3,equation zd4=d2. These equalities produce the system of 6 equations for d2, d3, d4:

equation

From the equations equation we receiveequationequation

d3=0, and d4=1. If d4=−1 the equation and d3=b4. The first solution is the vector equation and the subspace equation is a possible reductive complement for h1. Unfortunately, the condition g= h1⊕ m is not satisfied because equation and m is a subspace of h1, m⊂h1. So, m is not reductive +e complement for h1.

If d2=0, d3=0, and d4=1 then vector equation is the consequential result. In this case, m⊂h1 again. This means that a reductive complement for subalgebra h1 doesn’t exist for this case.

Subalgebra h1 Case 2: Multiply basic vectors from h1 by vector equation

equation (an identity),

equation

equation

equation

From the vector equalities above, we obtain the following system of equations for the components d2, d3, d4 and coefficients y, z:

equationzd4=d2. The last system of equations has just the zero solution for d2, d3, d4:d3=0, d2=0, d4=0. The zero vectors equation is the only solution for the system. This means that a nonzero reductive complement for h1 doesn’t exist.

Subalgebra h2. Case 1: Multiply basic vectors equation from h2 by vector equation We have:

equation

equation

equation

From the last system of vector equalities, we obtain a system of equations for d2, d3, d4 that has just one solution d2=0, d3=0, d4=1. The subspace equation generated by vectorequation satisfies the conditions equation is a reductive pair for this case where equation

Subalgebra h2. Case 2: Multiply basic vectors of h2 by vector equation We have:

equation

equation

equation

Transforming this system of vector equalities into a system of equations for the components d2, d3, d4 and solving the system, the only zero solution d2=0, d3=0, d4=0 is obtained. So, a nonzero reductive complement for h2 doesn’t exist for this case.

Subalgebras h3 and h4 don’t produce any reductive pair. The details are similar for the cases of subalgebras h1 and h2, therefore they are omitted.

The total analysis conducted in this Part III establishes the following statement.

Theorem 2: The only one reductive pair with 3-dimensional subalgebra exists for Lie algebra g of all real 2 × 2 matrices; it is {h, m} where, equation

Corollary: The subspace equation is a 1-dimensional ideal of Lie algebra g. Moreover, the subalgebra equation is a 3-dimensional ideal of Lie algebra g.

Part III. Reductive pairs {h, m} with 1-dimensional subalgebras h of Lie algebra g

It is well known that each 1-dimensional subspace h of algebra g is an abelian subalgebra of g. The corresponding reductive complements m for each h should be 3-dimensional subspaces m such that equation and g= h⊕m. Therefore, the canonical bases for 3-dimensional subspaces that are found in the Part I can be utilized for m. The list of all canonical bases contains the next 4 bases:

equation

equation

equation

equation

Let equation be 1-dimensional subalgebra in algebra g. We will consider two cases for the generating vector equation

equation

To determine if a subalgebra h forms a reductive pair with some complement m, we will check that the conditions equation are satisfied.

1a. Consider the basis (1) for a complement m and case (a) for equation Multiply vectorsequation by vector equation We have:

equation

equation

So, equation

equation

So, equation

equation

So, equation

The conditions found above produce the following system of 3 equations for components d2, d3, d4:

equationequation

The unknown variables are components d2, d3, d4. All other components are supposed to be done. Solve the system starting with the first equation. We have:

equation

Two solutions are possible: equation

1) If a4=1 then from the second and third equations we obtain equation The last two equations produce two expressions for d4:

equation

Comparing these two equalities for d4, we obtain b4c4=−1 or c4d3=b4d2. Using these conditions, we obtain for this case a4=1 the following solutions:

equation

equation

These solutions produce the following two reductive pairs:

equation

equation

equation

equation

These two reductive pairs are particular cases of the pairs {h1, m1}, {h2, m2} obtained in the next step 2.

1. If b4d2=c4d3 we can suppose that a4 is any real number. Now, like in the previous step 1, we obtain the following system of three equations:

equation

From the 2nd and 3rd equations above, we can find the 4th component d4:

equation

Comparing the last two equalities for the same component d4,

we have equation andequation

The last equality produces two results: equation or a4+1+2b4c4=0. Compute the component d4 in terms of a4, b4, c4, d2 when equation. We have equation . So, we obtain the following reductive pair:

equation

For the condition a4+1+2b4c4=0, we have d4=1+b4d2+c4d3, and the corresponding reductive pair is:

equation

Consider the special case when c4=0. Then we have the following system of equations for equationequation If a4−1, then d2=0, and equation with any b4. As the result, the following new reductive pair is obtained:

equation

If a4=−1, then b4d2=0, b4(d4−1)=0, and we obtain the following new reductive pairs:

equation

equation

1b. Consider the basis (1) for a complement m and the case (b) for vector equation

Multiply vectors equation by vector equation. We have:

equation

So, equation

equation

So, equation

equation

So, equation

In this case, the following system of 3 equations for unknown components d2, d3, d4 is obtained equation equation

Solve the system of equations starting with the first equation. We have the equality equation which produces a4=1 or equation

1. If a4=1, then we obtain equationequation from the 2nd and 3rd equations. Find d4 from these two equalities:

equation

Comparing two different expressions for the same component d4, we find equation or we obtain b4c4=−1, and equation this means that two new reductive pairs are found:

equation

equation

However, this pairs are particular cases of the reductive pair {h6, m6} that will be found in the next subcase.

Suppose now that b4d2=c4d3, and equivalently equation Utilizing this equality at the second and third equalities, we have the same result for d4 from both of them:

equation

This produces the new reductive pair:

equation

If c4=0, then the corresponding system of equations for d2, d3, d4 is: equation

If a4=−1 then b4d2=0, b4d4=0, and b4=0 or d2=d4=0. In the 1st case we obtain a reductive pair:

equation

If d2=d4=0 then we obtain a nonreductive pair equation

If a4≠−1, then d2=0, equation and a new reductive pair is equation

2a. Consider the basis (2) for the complement m and the case (a) for vector equation . Multiply vectorsequation by vectorequation We have:

equation

equation

So, equation

equation

So, equation

equation

So, equation

These conditions generate the following system of equations for unknown components d2, d3, d4:

equation

From the last equation, we have d3=−b3d2. Substitute this value of d3 into the first and the second equations: 0

equation

If a3≠0, b3≠0 then d4=1, d2=0, d3=0, and the following reductive pair is obtained equation

If a3=0, b3=0 then d3=0, and the following pair is obtained equation

This pair is not reductive because h⊂m.

If a3≠0, b3=0 then d3=0, d4=1−a3d2, and the following reductive pair is obtained equation equation

If a3=0, b3≠0 then d3=−b3d2, d4=1, and the following reductive pair is obtained equation

2b. Consider the basis (2) for the complement m and the case (b) for vector equation Multiply vectorsequation by vectorequation We have:

equation

So, equation

equation

So, equation

equation

So, equation

These conditions generate the next system of equations for unknown components d2, d3, d4:

equation

Simplifying this system of equations, we obtain equation equation

If a3≠0, b3≠0 then d3=−b3d2, d4=a3d2, and the following reductive pair is obtained equation equation

If a3=0, b3=0 then a nonreductive pair {h, m} is obtained because h⊂m. Similarly, if a3≠0, b3=0 then a nonreductive pair {h, m} is obtained because h⊂m.

If a3=0, b3≠0 then the following reductive pair is obtained equation but this pair is a particular case of the pair {h12, m12}.

3a. Consider the basis (3) and the case (a) for the vector equation. Multiply vectors equation by vector equation We have:

equation

So, equation

equation

So, equation

equation

So, x3=0, y3=d3, z3=0 and d2=0.

These conditions imply d2=0, equation. If a20 then d4=1+a2d3, and the following reductive pair is obtained equation

If a2=0 then the following pair appears: equationequation. This pair is not reductive because h⊂m.

3b. Consider the basis (3) and the case (b) for the vector equation. Multiply vectors equation by vector equation We have

equation

So, equation

equation

So, equation

equation

So, equation

These conditions imply equation . If a20 and d2=0 then the pair equation equation is obtained but it is not reductive because h⊂m.

If a2=0 and d2=0, then we have a nonreductive pair equation

4a. Consider the basis (4) and the case (a) for vector equation. Multiply vectors equation by vector equation We have:

equation

So, equation

equation

So, equation

equation

So, x3=−d2, y3=d3, z3=0, and 0=0 (it’s an identity).

These equalities imply d3=0, d2=0. So, the following reductive pair is obtained equation

4b. Consider the basis (4) and the case (b) for vector equation . Multiply vectors equation by vector equation We have:

equation

So, equation

equation

So, equation

equation

So, equation

These equalities imply d3=0, d2=0. So, the following pair is obtained equation

But this pair is not reductive because h⊂m.

Next statement describes all results that are found in the Part III.

Theorem 3: All different reductive pairs {h, m} with 1-dimensional sub algebras h and 3-dimensiomal complements m of Lie algebra of all 2 × 2 real matrices are:

equation

Remark

It’s unknown what reductive pairs among equation are equivalent with respect to inner automorphisms of the given Lie algebra.

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