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Canonical Bases for Subspaces of a Vector Space, and 5-Dimensional Subalgebras of Lie Algebra of Lorentz Group | OMICS International
ISSN: 1736-4337
Journal of Generalized Lie Theory and Applications
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Canonical Bases for Subspaces of a Vector Space, and 5-Dimensional Subalgebras of Lie Algebra of Lorentz Group

Uladzimir Shtukar*

Math/Physics Department, North Carolina Central University, Durham, NC 27707, USA

*Corresponding Author:
Uladzimir Shtukar
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: August 23, 2016; Accepted Date: September 21, 2016; Published Date: September 30, 2016

Citation: Shtukar U (2016) Canonical Bases for Subspaces of a Vector Space, and 5-Dimensional Subalgebras of Lie Algebra of Lorentz Group. J Generalized Lie Theory Appl 10: 243. doi:10.4172/1736-4337.1000243

Copyright: © 2016 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

Canonical bases for subspaces of a vector space are introduced as a new effective method to analyze subalgebras of Lie algebras. This method generalizes well known Gauss-Jordan elimination method.

Keywords

Vector space; Subspaces; Lie algebras; Subalgebras

Introduction

This article has two parts. In Part I, the canonical bases for 5-dimensional subspaces of a 6-dimensional vector spaces are introduced, and all of them are found. Then the nonequivalent canonical bases are classified in Theorem 1. The corresponding evaluation in Part I has a universal character, and it can be called a generalization of the well-known Gauss - Jordan elimination method [1]. Canonical bases for subspaces of 3-, 4-, and 5-dimensional vector spaces are already found too, and they will be demonstrated in the separate manuscripts. The canonical bases for the (n–1)-dimensional subspaces of vector spaces of dimension n>6 can be constructed in the way similar to this in Part I. This new method of canonical bases helps to study all objects associated with subspaces of vector spaces.

In Part II, this method is applied to study subalgebras of Lie algebra of Lorentz group. It’s a fact that a classification problem of subalgebras of low dimensional real Lie algebras was discussed during 1970-1980 years. That classification of subalgebras of all real Lie algebras of dimension n ≤ 4 only was obtained in the form of representatives for equivalent classes of subalgebras considering under their groups of inner automorphisms [2,3]. The subalgebras of real Lie algebras of dimension n ≥ 5 were not classified before. As a step of the further classification, the 5-dimensional hypothetical subalgebras of 6-dimensional Lie algebra of Lorentz group are investigated in Part II [4]. The corresponding procedure involves nonequivalent canonical bases from Part I. It is proved that Lie algebra of Lorentz group has no subalgebras of the dimension 5. This means also that Lorentz group has no connected 5-dimensional subgroups.

Part I

Canonical bases for 5-dimensional subspaces of a 6-dimensional vector space

Let image

image (I)

image be a general basis for arbitrary 5-dimensional subspace S of a 6-dimensional vector space V with its standard basis image. We associate the next matrix M with the basis (I)

image

Definition 1

The basis (I) is called canonical if its associated matrix M is in reduced row echelon form.

Definition 2

Two bases are called equivalent if they generate the same subspace of a given vector space, and two bases are nonequivalent if they generate different subspaces.

We start our transformation procedure for the basis (I) to find all canonical nonequivalent bases for the subspace S.

Suppose that at least one coefficient from a1, b1, c1, d1, f1 is not zero. Without any loss in the generality, we can propose that a1≠0. Perform the linear operation image first, and the operations first, and the operations image after the first one. The following basis is obtained imageimageimageimage

Remark

The first components of vectors image are changed as the result of the operations performed but all other components of them are saved just for convenience. This idea will be used throughout of Part I.

Suppose now that at least one coefficient from b2, c2, d2, f2 in the basis (a) is not zero. Without any loss in generality, let b2 ≠ 0. Perform the first linear operation image, and the operations imageimage after the first one. The following new basis is obtained

image

image (1)

image

Suppose that at least one coefficient among c3, d3, f3 in the basis (1) is not zero. Again, without any loss in the generality, let c3≠0. Perform the first operation image, and the operations imageimage then. We obtain the following basis

image (2)

Suppose now that at least one coefficient from d4, f4 in the basis (2) is not zero. Let d4≠0. Perform the operation image first, and then the operations image. The new transformed basis is

image (3)

At least one coefficient from f5, f6 is not zero in the basis (3). If f5 ≠ 0, then perform the operation image first, and the operations imageimage after the first one. The following canonical basis is obtained

image

If f6 ≠ 0, then perform operation image first, and the operations image

image after the first one. The new basis is obtained

image

The last basis is equivalent to the basis (a1) if f5 ≠ 0. So, f5=0, and the new canonical basis is obtained

image

1. Suppose that both coefficients d4, f4 at the basis (2) are zero. We have

image

Suppose that at least one coefficient from d5, f5 at (4) is not zero. It’s easy to see that the alternative case with d5=0 and f5=0 is impossible because the corresponding vectors image are linearly dependent. Let d5 ≠ 0. Perform operation image first, and the operations image next. The following basis is obtained

image

image

It’s obvious that f6 ≠ 0 for the vector image at the last basis. Perform the operation image first, and the operations imageimage after the first one. We obtain the new canonical basis

image

If f5 ≠ 0 at the basis (4), perform the operation image first, and then the operations image. The following basis is obtained

image

image

We have d6 ≠ 0 in the last basis. Perform the first operation image and then the operations image. The following canonical basis is obtained

image

This basis is not new, it’s equivalent to the basis (a3).

2. Suppose, in opposition to the step 2, that all coefficients c3, d3, f3 are zero in the basis (1). We have

image image (5)

image

Consider coefficients c4, d4, f4 in the basis (5). Suppose that at least one of them is not zero.

Let c4≠0 (without any loss in generality). Perform the operation image first, and then the operations imageimage. The following basis is obtained

image

At least one coefficient among d5, f5 is not zero in the basis (5a). If both coefficients d5, f5 are zero, then image, and vectors image are linearly dependent but it’s impossible for any basis. Let d50. Perform the operation image first, and then the operations imageimage. The following basis is obtained

image

image

The coefficient f6 is not zero at the last basis. Perform the operation image first, and The operations image after the first one. We obtain the new canonical basis

image (a4)

If f6 ≠ 0 at the basis (5a), then perform the operation image first, and the operations image after the first operation. The following basis is obtained

image

The previous basis generates the next canonical basis

image

This basis is obviously equivalent to the basis (a4), so it’s not a new one.

3. If all coefficients c4, d4, f4 are zero at (5), then the basis has the following structure:

image

It’s obvious that 3 vectors image at (6) are located at the same plane determined by vectors image. So, they are linearly dependent that contradicts the fact that all vectors image are linearly independent. So, the case c4=0, d4=0, f4=0 doesn’t generate any canonical basis.

4. Suppose, in opposition to the Step 1, that the second coefficients b2, c2, d2, f2 at the basis (a) are zero. We obtain the following basis

image

Consider coefficients b3, c3, d3, f3 at the basis (7). Suppose that at least one of them is not zero. Without any loss in the generality, let b3 ≠ 0. Perform the operation image first, and then the operations image image. The following basis is obtained

image

Consider coefficients c4, d4, f4 at the previous basis. Suppose that at least one of them is not zero. Let c4 ≠ 0. Perform the operation image first, and the operations image after the first one. The following basis is obtained

image

At least one coefficient among d5, f5 in the basis (8) is not zero. If both coefficients d5, f5 are zero, then vectors image are linearly dependent but it’s impossible for (8) to be a basis. Let d5 ≠ 0. Perform the operation image first, and the operations image image after the first one at the basis (8). The following new basis is obtained

image

It’s obvious that f6 ≠ 0 at the last basis, and it generates the new canonical basis

image (a5)

If f5 ≠ 0 at the basis (8), perform the operation image first, and then the operations image, image at the basis (8). The following basis obtained

image

It’s obvious that d6 ≠ 0 in the last basis. So, the following canonical basis is generated

image

The last basis is equivalent to the basis (a5/sub>), so it’s not a new one.

5. Suppose that coefficients b3/sub>, c3/sub>, d3/sub>, f3/sub> at (8) are zero. We receive the basis

image

image (9)

image

The four vectors image in the basis (9) are linearly dependent because they are located at the same 3-dimensional subspace image but it contradicts to the fact that vectors image form a basis. So, this case doesn’t generate any canonical basis.

A. Suppose that a1=b1=c1=d1=f1=0 in the basis (I). The following basis (b) is obtained

image

Suppose that at least one coefficient among a2, b2, c2, d2, f2 is not zero in (b). Like before, we can suppose that a2≠0. Perform the operation image first, and the operations image after the first one. The following new basis appears

image (1)

Consider coefficients b3, c3, d3, f3 in the basis (1). Suppose that at least one of them is not zero. We can suggest that b3≠0 (without any loss in the generality). Perform the operation image first, and the operations image after the first one. The following basis is obtained

image (2)

If all coefficients b3, c3, d3, f3 in (1) are zero, then vectors image are linearly dependent but it’s impossible because these vectors form a basis.

1. Consider coefficients c4, d4, f4 in the basis (2). At least one of them is not zero. If all of them are zero, then vectors

image are linearly dependent because they are located at the same 2-dimensional subspace generated by vectors image. This means that the case c4=d4=f4=0 is impossible.

Suppose that c4 ≠ 0. Perform the operation image in the basis (2) first, and the operations image after the first operation. The following basis is obtained

image (3)

Consider coefficients d5, f5 in the basis (3). At least one of them is not zero. Otherwise, the vectors image and image are linearly dependent but it’s impossible. Suppose that d5 ≠ 0. Perform the operation image first, and the operations image image after the first one. The following basis is obtained

image(4)

The coefficient f6 is not zero in the basis (4). Perform the operation image in the basis (4) first, and the linear operations image image after the first one. We obtain the following new canonical basis

image (b1)

If f5 ≠ 0 in the basis (4), we obtain the same canonical basis (b1).

2. Suppose that all coefficients a2, b2, c2, d2, f2 are zero in the basis (b). The following possible basis is obtained

image (5)

These 5 vectors image in (5) are linearly dependent because they are located at the same 4-dimensional subspace generated by vectors image. So, the system (5) of the vectors image doesn’t produce any canonical basis.

The research performed for the set of coefficients a1, b1, c1, d1, f1 in the cases A and B produces 6 canonical bases (a1) – (a5) and (b1). To find other canonical bases, we should repeat the similar evaluations considering the following five sets of coefficients {a2, b2, c2, d2, f2}, {a3, b3, c3, d3, f3}, {a4, b4, c4, d4, f4},{a5, b5, c5, d5, f5}, and {a6, b6, c6, d6, f6} in the basis (I). According the equity principle, we will obtain 6 similar canonical bases for each set of coefficients. Details are very close to those in the cases A, B, and we omit them. The total list of canonical bases is

image

image

image

Analyze all these bases comparing them step by step. The bases (b1), (d1), (f1), (h1), (j1), (l1) are particular cases of (c5), (a5), (c4), (a3), (a2), (a1) respectively. The bases (c1) – (c4) are obviously equivalent to the bases (a1) – (a4). The basis (c5) is equivalent to the basis (a5) if a1≠0, and (c5) is equivalent to the basis (b1) if a1=0. The bases

(e1) – (e3) are equivalent to the bases (a1) – (a3). If a2≠0 then the basis (e4) is equivalent to the basis (a4); if a2=0 then (e4) is equivalent to (a5). The basis (e3) is a particular case of the basis (a4) if a1 ≠ 0, and (e5) is a particular case of (g5) if a1=0. The bases (g1), (g2) are obviously equivalent to the bases (a1), (a2). The basis (g3) is equivalent to the basis (a3) if a3 ≠ 0, and (g3) is equivalent to (a1) if a3=0. The basis (g4) is equivalent to the basis (a3) if a2 ≠ 0, and (g4) is equivalent to (a5) if a2=0. The basis (g5) is a particular case of the basis (a3) if a1 ≠ 0, and (g5) is a particular case of (i5) if a1=0. The basis (i1) is equivalent to the basis (a1). The basis (i2) is equivalent to the basis (a2) if a4 ≠ 0, so consider the basis (i2) if a4=0. The new basis (i2) is a particular case of the basis (a1) if f4=0, and (i2) is equivalent to (a3) if f4=0. The basis (i3) is a particular case of the basis (a2) if a3 ≠ 0, and the basis (i3) is equivalent to the basis (a4) if a3=0. The basis (i4) is a particular case of the basis (a2) if a2 ≠ 0, and (i4) is equivalent to (a5) if a2=0. The basis (i5) is a particular case of the basis (a2) if a1 ≠ 0, and (i5) is a particular case of the basis (k5) if a1=0. The basis (k1) is equivalent to the basis (a2). The basis (k2) is equivalent to the basis (i2). The basis (k3) is a particular case of the basis (a1) if a3 ≠ 0, and (k3) is equivalent to the basis (a4) if a3=0. The basis (k4) is a particular case of the basis (a1) if a2 ≠ 0, and (k4) is equivalent to the basis (a5) if a2=0. The basis (k5) is a particular case of the basis (a1) if a1 ≠ 0, and (k5) is equivalent to the basis (b1) if a1=0.

The analysis performed above implies the following statement.

Theorem 1

Each basis of any 5-dimensional subspace in a 6-dimensional vector space is equivalent to one and only one of the following 6 canonical bases

image

Part II

5-dimensional subalgebras of Lie algebra of Lorentz group

Introduction: Lorentz group is the group of transformations of Minkowski space-time R4. This group is not compact, not abelian, and not connected 6-dimensional real Lie group. The identity component of Lorentz group is the group SO+ (3, 1). This component contains the generators for boots along x-, y- and z-axis, and it contains the generators for rotations in Minkowski space-time [4]. Lie algebra of the group SO+ (3, 1) is 6-dimensional real Lie algebra denoted below by L that has the following standard basis:

image

The Lie product of any two square matrices A, B is defined by [A, B]=AB – BA. For the standard basis of Lie algebra of Lorentz group, the non-zero products are

image

image

To determine which 5-dimensional subspace h of the given Lie algebra L is a subalgebra of L, we will check the condition [h, h]⊂ h applying to the nonequivalent canonical bases that are described in the Theorem 1.

Let the subspace h1 be generated by the canonical basis (a1). Compute all products between vectors image in this basis. Utilizing the table of products (*), we have

image

The equation image has no solution in the set of all real numbers. This means that no 5-dimensional subalgebra of Lie algebra L with the basis (a1) exists.

Let the subspace h2 be generated by the canonical basis (a2). Compute all products between vectors image in this basis. Utilizing the table of products (*), we have

image

image

The last equation d25 = −1 has no solution in the set of all real numbers. This means that no 5-dimensional subalgebra of Lie algebra L with the basis (a2) exists.

Let the subspace h3 be generated by the canonical basis (a3). Consider all products between vectors image of this basis. Utilizing the table of products (*), we have

image

So, n1=0, n2=0, n3=0, n4=0, n5=0, and 0=–1.

The last contradiction 0=–1 proves that Lie algebra L has no 5-dimensional subalgebra generated by the basis (a3).

Let the subspace h4 be generated by the canonical basis (a4). Consider all products between vectors image in this basis. Utilizing the table of products (*), we have

image

So, x1=0, x2=0, x3=1, x4=b3, x5=–a3, and 0=0.

image

image

The last condition a23 = −1 is impossible in the set of all real numbers. This means that the product image doesn’t belong to the 5-dimensional subspace generated by the basis (a4). Thus, this subspace is not subalgebra of Lie algebra L.

Let the subspace h5 be generated by the canonical basis (a5). Consider all products between vectors image in this basis. Utilizing the table of products (*), we have

image

So, x1=0, x2=0, x3=0, x4=1, x5=a2, and 0=0.

image

image

The last condition a22 = −1 is not satisfied in the set of all real numbers. This means that no 5-dimensional subalgebra with the basis (a5) exists in Lie algebra L.

Let the subspace h6 be generated by the canonical basis (b1). Consider all products between vectors image in this basis. Utilizing the table of products (*), we have

image

So, x1=0, x2=0, x3=0, x5=1, and 1=1.

image

So, y1=0, y2=0, y3=0, y4=0, y5=0, and 0=– 1.

The last contradiction 0=– 1 shows that Lie algebra L has no 5-dimensional subalgebra generated by the basis (b1).

The evaluations performed in Part II prove the following statements.

Theorem 2

Lie algebra of the Lorentz group doesn’t contain any 5-dimendional subalgebra.

Corollary

Lorentz group doesn’t contain any connected 5-dimensional subgroup.

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