Journal of Generalized Lie Theory and Applications Canonical Bases for Subspaces of a Vector Space, and 5-Dimensional Subalgebras of Lie Algebra of Lorentz Group

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.


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  be a general basis for arbitrary 5-dimensional subspace S of a 6-dimensional vector space V with its standard basis 1

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 1 / a a  first, and the operations after the first one. The following basis is obtained 1

Remark
The first components of vectors , , , , a b c d f       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 after the first one. The following new basis is obtained after the first one. The following canonical basis is obtained 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 The following basis is obtained . The following basis is obtained The coefficient f 6 is not zero at the last basis. Perform the operation after the first one. We obtain the new canonical basis If f 5 ≠ 0 at the basis (5a), then perform the operation 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:  (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 after the first one. The following basis is 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 ), so it's not a new one. (8)  .
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 2 / a a  first, and the operations  after the first one. The following new basis appears   (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 after the first one. The following basis is obtained  first, and the operations after the first one. The following basis is obtained 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 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 5 ) 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.
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.