Medical, Pharma, Engineering, Science, Technology and Business

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.

**Visit for more related articles at** Journal of Generalized Lie Theory and Applications

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.

Lie algebra; Subalgebras; Reductive pairs

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 for any such that:

for any a∈F,

and

We call 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:

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) can be computed:

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 where and are 3 linearly independent vectors.

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

**Part I. Maximal subalgebras of Lie algebra g**

Now we start to determine that a 3-dimensional subspace is a subalgebra of Lie algebra *g* when vectors 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 be the basis (1) for *h*. Evauate Lie products

So, and

The system of 3 equations for is: We have two different solutions: and This means that the following 1-parameter set of subalgebras *h _{1}* and one special subalgebra

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

So, we have Consequently, the following system of equations appears for components *a _{3}, b_{3}*:

This system of equations has only one solution *a _{3}*=0,

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

So,

So,

The system of equations has only one solution *a _{2}*=0. The corresponding subalgebra is

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

The vector doesn’t belong to the subspace 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:

**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. Let be a possible 1-dimensional complement. To simplify our evaluation, we consider 2 possible cases for the generating vector

and

**Subalgebra h_{1} Case 1:** Multiply basic vectors

from *h _{1}* by vector We have:

(it’s an identity),

From the vector equalities above, we obtain the following system of conditions for the components *d _{2}, d_{3}, d_{4}* and coefficients

From the equations we receive

*d _{3}*=0, and

If *d _{2}*=0,

**Subalgebra h_{1} Case 2:** Multiply basic vectors from

(an identity),

From the vector equalities above, we obtain the following system of equations for the components *d _{2}, d_{3}, d_{4}* and coefficients

*zd _{4}=d_{2}*. The last system of equations has just the zero solution for

**Subalgebra h_{2}. Case 1:** Multiply basic vectors from

From the last system of vector equalities, we obtain a system of equations for *d _{2}, d_{3}, d_{4}* that has just one solution

**Subalgebra h_{2}. Case 2:** Multiply basic vectors of

Transforming this system of vector equalities into a system of equations for the components *d _{2}, d_{3}, d_{4}* and solving the system, the only zero solution

Subalgebras *h _{3}* and

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,

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

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

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

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

So,

So,

So,

The conditions found above produce the following system of 3 equations for components *d _{2}, d_{3}, d_{4}*:

The unknown variables are components *d _{2}, d_{3}, d_{4}.* All other components are supposed to be done. Solve the system starting with the first equation. We have:

Two solutions are possible:

1) If *a _{4}*=1 then from the second and third equations we obtain The last two equations produce two expressions for

Comparing these two equalities for *d _{4}*, we obtain

These solutions produce the following two reductive pairs:

These two reductive pairs are particular cases of the pairs {*h _{1}, m_{1}*}, {

1. If *b _{4}d_{2}=c_{4}d_{3}* we can suppose that a4 is any real number. Now, like in the previous step 1, we obtain the following system of three equations:

From the 2^{nd} and 3^{rd} equations above, we can find the 4^{th} component *d _{4}*:

Comparing the last two equalities for the same component *d _{4}*,

we have and

The last equality produces two results: or *a _{4}+1+2b_{4}c_{4}*=0. Compute the component

For the condition *a _{4}+1+2b_{4}c_{4}*=0, we have

Consider the special case when *c _{4}*=0. Then we have the following system of equations for If

If *a _{4}*=−1, then

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

Multiply vectors by vector . We have:

So,

So,

So,

In this case, the following system of 3 equations for unknown components *d _{2}, d_{3}, d_{4}* is obtained

Solve the system of equations starting with the first equation. We have the equality which produces *a _{4}*=1 or

1. If *a _{4}*=1, then we obtain from the 2

Comparing two different expressions for the same component *d _{4}*, we find or we obtain

However, this pairs are particular cases of the reductive pair {*h _{6}, m_{6}*} that will be found in the next subcase.

Suppose now that *b _{4}d_{2}*=

This produces the new reductive pair:

If *c _{4}*=0, then the corresponding system of equations for

If *a _{4}*=−1 then

If *d _{2}=d_{4}*=0 then we obtain a nonreductive pair

If *a*_{4}≠−1, then *d _{2}*=0, and a new reductive pair is

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

So,

So,

So,

These conditions generate the following system of equations for unknown components *d _{2}, d_{3}, d_{4}*:

From the last equation, we have *d _{3}=−b_{3}d_{2}*. Substitute this value of

If *a _{3}*≠0,

If *a _{3}*=0,

This pair is not reductive because *h⊂m*.

If *a _{3}*≠0,

If *a _{3}*=0,

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

So,

So,

So,

These conditions generate the next system of equations for unknown components *d _{2}, d_{3}, d_{4}:*

Simplifying this system of equations, we obtain

If *a _{3}*≠0,

If *a _{3}*=0,

If *a _{3}*=0,

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

So,

So,

So, *x _{3}*=0,

These conditions imply *d _{2}*=0, . If

If *a _{2}*=0 then the following pair appears: . This pair is not reductive because

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

So,

So,

So,

These conditions imply . If *a _{2}0* and

If *a _{2}*=0 and

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

So,

So,

So, *x _{3}*=−

These equalities imply *d _{3}*=0,

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

So,

So,

So,

These equalities imply *d _{3}*=0,

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:

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

- Shtukar U (2016) Classification of Canonical Bases for (n-1)-dimensional Subspaces of n-dimensional Vector Space. Journal of Generalized Lie Theory and Applications 10: 245.
- Nomizu K (1954) Invariant Affine Connections on Homogeneous Spaces. American Journal of Mathematics 76: 33-65.
- Patera J, Winternitz P (1977) Subalgebras of real three- and four-dimensional Lie algebras. Journal of Mathematical Physics 18: 1449-1455.
- Sagle A, Winter DJ (1967) On Homogeneous Spaces and Reductive Subalgebras of Simple Lie Algebras. Transactions of the American Mathematics Society 128: 142-147.

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