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**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.

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

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.

Vector space; Subspaces; Lie algebras; Subalgebras

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

(I)

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

**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 a_{1}, b_{1}, c_{1}, d_{1}, f_{1} is not zero. Without any loss in the generality, we can propose that a_{1}≠0. Perform the linear operation first, and the operations first, and the operations after the first one. The following basis is obtained

**Remark**

The first components of vectors 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 b_{2}, c_{2}, d_{2}, f_{2} in the basis (a) is not zero. Without any loss in generality, let b_{2} ≠ 0. Perform the first linear operation , and the operations after the first one. The following new basis is obtained

(1)

Suppose that at least one coefficient among c_{3}, d_{3}, f_{3} in the basis (1) is not zero. Again, without any loss in the generality, let c_{3}≠0. Perform the first operation , and the operations then. We obtain the following basis

(2)

Suppose now that at least one coefficient from d_{4}, f_{4} in the basis (2) is not zero. Let d_{4}≠0. Perform the operation first, and then the operations . The new transformed basis is

(3)

At least one coefficient from f_{5}, f_{6} is not zero in the basis (3). If f_{5} ≠ 0, then perform the operation first, and the operations after the first one. The following canonical basis is obtained

If f_{6} ≠ 0, then perform operation first, and the operations

after the first one. The new basis is obtained

The last basis is equivalent to the basis (a_{1}) if f_{5} ≠ 0. So, f_{5}=0, and the new canonical basis is obtained

1. Suppose that both coefficients d_{4}, f_{4} at the basis (2) are zero. We have

Suppose that at least one coefficient from d_{5}, f_{5} at (4) is not zero. It’s easy to see that the alternative case with d_{5}=0 and f_{5}=0 is impossible because the corresponding vectors are linearly dependent. Let d_{5} ≠ 0. Perform operation first, and the operations next. The following basis is obtained

It’s obvious that f_{6} ≠ 0 for the vector at the last basis. Perform the operation first, and the operations after the first one. We obtain the new canonical basis

If f_{5} ≠ 0 at the basis (4), perform the operation first, and then the operations . The following basis is obtained

We have d_{6} ≠ 0 in the last basis. Perform the first operation and then the operations . The following canonical basis is obtained

This basis is not new, it’s equivalent to the basis (a_{3}).

2. Suppose, in opposition to the step 2, that all coefficients c_{3}, d_{3}, f_{3} are zero in the basis (1). We have

(5)

Consider coefficients c_{4}, d_{4}, f_{4} in the basis (5). Suppose that at least one of them is not zero.

Let c_{4}≠0 (without any loss in generality). Perform the operation first, and then the operations . The following basis is obtained

At least one coefficient among d_{5}, f_{5} is not zero in the basis (5a). If both coefficients d_{5}, f_{5} are zero, then , and vectors are linearly dependent but it’s impossible for any basis. Let d_{5}0. Perform the operation first, and then the operations . The following basis is obtained

The coefficient f_{6} is not zero at the last basis. Perform the operation first, and The operations after the first one. We obtain the new canonical basis

(a_{4})

If f_{6} ≠ 0 at the basis (5a), then perform the operation first, and the operations after the first operation. The following basis is obtained

The previous basis generates the next canonical basis

This basis is obviously equivalent to the basis (a_{4}), so it’s not a new one.

3. If all coefficients c_{4}, d_{4}, f_{4} are zero at (5), then the basis has the following structure:

It’s obvious that 3 vectors at (6) are located at the same plane determined by vectors . So, they are linearly dependent that contradicts the fact that all vectors are linearly independent. So, the case c_{4}=0, d_{4}=0, f_{4}=0 doesn’t generate any canonical basis.

4. Suppose, in opposition to the Step 1, that the second coefficients b_{2}, c_{2}, d_{2}, f_{2} at the basis (a) are zero. We obtain the following basis

Consider coefficients b_{3}, c_{3}, d_{3}, f_{3} at the basis (7). Suppose that at least one of them is not zero. Without any loss in the generality, let b_{3} ≠ 0. Perform the operation first, and then the operations . The following basis is obtained

Consider coefficients c_{4}, d_{4}, f_{4} at the previous basis. Suppose that at least one of them is not zero. Let c_{4} ≠ 0. Perform the operation first, and the operations after the first one. The following basis is obtained

At least one coefficient among d_{5}, f_{5} in the basis (8) is not zero. If both coefficients d_{5}, f_{5} are zero, then vectors are linearly dependent but it’s impossible for (8) to be a basis. Let d_{5} ≠ 0. Perform the operation first, and the operations after the first one at the basis (8). The following new basis is obtained

It’s obvious that f_{6} ≠ 0 at the last basis, and it generates the new canonical basis

(a_{5})

If f_{5} ≠ 0 at the basis (8), perform the operation first, and then the operations , at the basis (8). The following basis obtained

It’s obvious that d_{6} ≠ 0 in the last basis. So, the following canonical basis is generated

The last basis is equivalent to the basis (a_{5/sub>), so it’s not a new one.}

5. Suppose that coefficients b_{3/sub>, c3/sub>, d3/sub>, f3/sub> at (8) are zero. We receive the basis}

(9)

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

A. Suppose that a_{1}=b_{1}=c_{1}=d_{1}=f_{1}=0 in the basis (I). The following basis (b) is obtained

Suppose that at least one coefficient among a_{2}, b_{2}, c_{2}, d_{2}, f_{2} is not zero in (b). Like before, we can suppose that a_{2}≠0. Perform the operation first, and the operations after the first one. The following new basis appears

(1)

Consider coefficients b_{3}, c_{3}, d_{3}, f_{3} in the basis (1). Suppose that at least one of them is not zero. We can suggest that b_{3}≠0 (without any loss in the generality). Perform the operation first, and the operations after the first one. The following basis is obtained

(2)

If all coefficients b_{3}, c_{3}, d_{3}, f_{3} in (1) are zero, then vectors are linearly dependent but it’s impossible because these vectors form a basis.

1. Consider coefficients c_{4}, d_{4}, f_{4} in the basis (2). At least one of them is not zero. If all of them are zero, then vectors

are linearly dependent because they are located at the same 2-dimensional subspace generated by vectors . This means that the case c_{4}=d_{4}=f_{4}=0 is impossible.

Suppose that c_{4} ≠ 0. Perform the operation in the basis (2) first, and the operations after the first operation. The following basis is obtained

(3)

Consider coefficients d_{5}, f_{5} in the basis (3). At least one of them is not zero. Otherwise, the vectors and are linearly dependent but it’s impossible. Suppose that d_{5} ≠ 0. Perform the operation first, and the operations after the first one. The following basis is obtained

(4)

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

(b_{1})

If f_{5} ≠ 0 in the basis (4), we obtain the same canonical basis (b_{1}).

2. Suppose that all coefficients a_{2}, b_{2}, c_{2}, d_{2}, f_{2} are zero in the basis (b). The following possible basis is obtained

(5)

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

The research performed for the set of coefficients a_{1}, b_{1}, c_{1}, d_{1}, f_{1} in the cases A and B produces 6 canonical bases (a_{1}) – (a_{5}) and (b_{1}). To find other canonical bases, we should repeat the similar evaluations considering the following five sets of coefficients {a_{2}, b_{2}, c_{2}, d_{2}, f_{2}}, {a_{3}, b_{3}, c_{3}, d_{3}, f_{3}}, {a_{4}, b_{4}, c_{4}, d_{4}, f_{4}},{a_{5}, b_{5}, c_{5}, d_{5}, f_{5}}, and {a_{6}, b_{6}, c_{6}, d_{6}, f_{6}} 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

Analyze all these bases comparing them step by step. The bases (b_{1}), (d_{1}), (f_{1}), (h_{1}), (j_{1}), (l_{1}) are particular cases of (c_{5}), (a_{5}), (c_{4}), (a_{3}), (a_{2}), (a_{1}) respectively. The bases (c_{1}) – (c_{4}) are obviously equivalent to the bases (a_{1}) – (a_{4}). The basis (c_{5}) is equivalent to the basis (a_{5}) if a_{1}≠0, and (c_{5}) is equivalent to the basis (b_{1}) if a_{1}=0. The bases

(e_{1}) – (e_{3}) are equivalent to the bases (a_{1}) – (a_{3}). If a_{2}≠0 then the basis (e_{4}) is equivalent to the basis (a_{4}); if a_{2}=0 then (e_{4}) is equivalent to (a_{5}). The basis (e_{3}) is a particular case of the basis (a_{4}) if a_{1} ≠ 0, and (e_{5}) is a particular case of (g_{5}) if a_{1}=0. The bases (g_{1}), (g_{2}) are obviously equivalent to the bases (a_{1}), (a_{2}). The basis (g_{3}) is equivalent to the basis (a_{3}) if a_{3} ≠ 0, and (g_{3}) is equivalent to (a_{1}) if a_{3}=0. The basis (g_{4}) is equivalent to the basis (a_{3}) if a_{2} ≠ 0, and (g_{4}) is equivalent to (a_{5}) if a_{2}=0. The basis (g_{5}) is a particular case of the basis (a_{3}) if a_{1} ≠ 0, and (g_{5}) is a particular case of (i_{5}) if a_{1}=0. The basis (i_{1}) is equivalent to the basis (a_{1}). The basis (i_{2}) is equivalent to the basis (a_{2}) if a_{4} ≠ 0, so consider the basis (i_{2}) if a_{4}=0. The new basis (i_{2}) is a particular case of the basis (a_{1}) if f_{4}=0, and (i_{2}) is equivalent to (a_{3}) if f_{4}=0. The basis (i_{3}) is a particular case of the basis (a_{2}) if a_{3} ≠ 0, and the basis (i_{3}) is equivalent to the basis (a_{4}) if a_{3}=0. The basis (i_{4}) is a particular case of the basis (a_{2}) if a_{2} ≠ 0, and (i_{4}) is equivalent to (a_{5}) if a_{2}=0. The basis (i_{5}) is a particular case of the basis (a_{2}) if a_{1} ≠ 0, and (i_{5}) is a particular case of the basis (k_{5}) if a_{1}=0. The basis (k_{1}) is equivalent to the basis (a_{2}). The basis (k_{2}) is equivalent to the basis (i_{2}). The basis (k_{3}) is a particular case of the basis (a_{1}) if a_{3} ≠ 0, and (k_{3}) is equivalent to the basis (a_{4}) if a_{3}=0. The basis (k_{4}) is a particular case of the basis (a_{1}) if a_{2} ≠ 0, and (k_{4}) is equivalent to the basis (a_{5}) if a_{2}=0. The basis (k_{5}) is a particular case of the basis (a_{1}) if a_{1} ≠ 0, and (k_{5}) is equivalent to the basis (b_{1}) if a_{1}=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

**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:

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

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 (a_{1}). Compute all products between vectors in this basis. Utilizing the table of products (*), we have

The equation has no solution in the set of all real numbers. This means that no 5-dimensional subalgebra of Lie algebra L with the basis (a_{1}) exists.

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

The last equation d^{2}_{5} = −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 (a_{2}) exists.

Let the subspace h_{3} be generated by the canonical basis (a_{3}). Consider all products between vectors of this basis. Utilizing the table of products (*), we have

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 (a_{3}).

Let the subspace h_{4} be generated by the canonical basis (a_{4}). Consider all products between vectors in this basis. Utilizing the table of products (*), we have

So, x_{1}=0, x_{2}=0, x_{3}=1, x_{4}=b_{3}, x_{5}=–a_{3}, and 0=0.

The last condition a^{2}_{3} = −1 is impossible in the set of all real numbers. This means that the product doesn’t belong to the 5-dimensional subspace generated by the basis (a_{4}). 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 in this basis. Utilizing the table of products (*), we have

So, x_{1}=0, x_{2}=0, x_{3}=0, x_{4}=1, x_{5}=a_{2}, and 0=0.

The last condition a^{2}_{2} = −1 is not satisfied in the set of all real numbers. This means that no 5-dimensional subalgebra with the basis (a_{5}) exists in Lie algebra L.

Let the subspace h_{6} be generated by the canonical basis (b_{1}). Consider all products between vectors in this basis. Utilizing the table of products (*), we have

So, x_{1}=0, x_{2}=0, x_{3}=0, x_{5}=1, and 1=1.

So, y_{1}=0, y_{2}=0, y_{3}=0, y_{4}=0, y_{5}=0, and 0=– 1.

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

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.

- Holt J (2013) Linear Algebra with Applications. W.H. Freeman and Company, New York.
- Patera J, Sharp RT, Winternitz P, Zassenhaus H (1976) Invariants of Real Low Dimension Lie Algebras. Journal of Mathematical Physics 17: 986-994.
- Patera J, Winternitz P (1977) Subalgebras of real three- and four-dimensional Lie algebras. Journal of Mathematical Physics 18: 1449-1455.
- Bincer AM (2012) Lie Groups and Lie Algebras âï¿½ï¿½ A Physicistâï¿½ï¿½s Perspective. Oxford University Press.

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