Lie subalgebras of $\mathfrak{s}\mathfrak{o}\left(\mathrm{3,1}\right)$ up to conjugacy

Abstract

Purpose

This study aims to find all subalgebras up to conjugacy in the real simple Lie algebra so(3,1).

Design/methodology/approach

The authors use Lie Algebra techniques to find all inequivalent subalgebras of so(3,1) in all dimensions.

Findings

The authors find all subalgebras up to conjugacy in the real simple Lie algebra so(3,1).

Originality/value

This paper is an original research idea. It will be a main reference for many applications such as solving partial differential equations. If so(3,1) is part of the symmetry Lie algebra, then the subalgebras listed in this paper will be used to reduce the order of the partial differential equation (PDE) and produce non-equivalent solutions.

Citation

Ghanam, R., Thompson, G. and Bandara, N. (2022), "Lie subalgebras of $\mathfrak{s}\mathfrak{o}\left(\mathrm{3,1}\right)$ up to conjugacy", Arab Journal of Mathematical Sciences, Vol. 28 No. 2, pp. 253-261. https://doi.org/10.1108/AJMS-01-2022-0007

1. Introduction

In the classification of real simple Lie algebras, so(3,1) is the unique simple six-dimensional Lie algebra. The Lie algebra so(3,1) and its associated Lie group SO(3, 1) are of fundamental importance in the theory of relativity, as is very well known. However, in terms of finding representations of so(3,1), the situation is apt to become confusing because the usual approach is to complexify and so(3,1)⊗ℂ≈so(3,ℂ)⊕so(3,ℂ). A closely related idea is to use Weyl’s unitarian trick. In this regard, we refer to [1] where an apparently non-standard representation of so(3,1) is given. We do not know at this time if it is of physical significance.

In [2] Dynkin studied the problem of finding maximal dimension subgroups of a simple Lie group and by extension, maximal dimension subalgebras of its Lie algebra. In [3], the subalgebras of gl(3,R) were classified. In [4], subalgebras of sl(4,R) were studied that are not solvable. In [5], a slightly different direction provides minimal dimension representations of Levi decomposition Lie algebras up to and including dimension eight.

Our goal in this note is to find all Lie subalgebras of so(3,1) up to conjugacy. Most of the Lie subalgebras concerned can be found from consideration of the Cartan subalgebras, so(3,1) being a rank two algebra. Of course it is important to understand that when we say “conjugate,” we mean equivalent under a change of basis that belongs to SO(3, 1). We study the case of one-dimensional subalgebras in Section 3, two-dimensional subalgebras in Section 4, three-dimensional subalgebras in Section 5, show that there are no five-dimensional subalgebras in Section 6 and consider subalgebras of dimension four in Section 7. In Section 8, we give a different representation of so(3,1) and argue that it is not conjugate to the standard representation. Finally, in Section 9, we provide a table of proper subalgebras of so(3,1) up to conjugacy.

2. The Lie algebra so(3,1)

The real simple Lie algebra so(3,1) is defined by the following space of matrices:

(1)S=0−s6−s5s1s60s4s2s5−s40s3s1s2s30.

From equation (1), the Lie brackets of so(3,1) are

(2)[e1,e2]=−e6,[e1,e3]=−e5,[e1,e5]=−e3,[e1,e6]=−e2,[e2,e3]=e4,[e2,e4]=e3,[e2,e6]=e1,[e3,e4]=−e2,[e3,e5]=e1,[e4,e5]=e6,[e4,e6]=−e5,[e5,e6]=e4.

Our goal in this note is to find all Lie subalgebras of so(3,1) up to conjugacy.

3. One-dimensional Lie subalgebras

Starting from (1), there is a transformation in SO(3, 1) of the form A001, where A ∈ SO(3) such that we can reduce s₄ and s₅ to zero. Now consider the matrix

(3)P=cos⁡θsin⁡θ00−sin⁡θcos⁡θ0000100001.

Then conjugating S by P, we obtain

(4)P−1SP=0−s60s1⁡cos⁡θ−s2⁡sin⁡θs600s1⁡sin⁡θ+s2⁡cos⁡θs20000s1⁡cos⁡θ−s2⁡sin⁡θs1⁡sin⁡θ+s2⁡cos⁡θ00.

Note that P∈so(3,1). As such, we can choose θ so that s₂ = 0. The matrix S has been reduced to

(5)S=0−s60s1s6000000s3s10s30.

Now the characteristic polynomial of this reduced S is given by

(6)λ4+(s62−s12−s32)λ2−s32s62=0.

3.1 Zero eigenvalues

If the four roots of (6) are all zero, we must have in the first instance, s₃s₆ = 0. However, if s₆ = 0, then looking at the λ² term, we would have s₁ = s₃ = 0 and S = 0. Hence, for non-zero S, we must have s₃ = 0 and s₆ = ±s₁. It appears as though we have two cases to consider now, but there is just one case as we shall now explain.

Conjugate S by the matrix Q∈so(3,1), where

(7)Q=0100100000100001.

Then we find that

(8)Q−1SQ=0s600−s600s100000s100,

but we may conjugate again by P from (3) with θ=3π2, so as to restore s₁ to the (1, 4)-entry, without disturbing s₆ and arrive finally at

(9)S=0s10s1−s10000000s1000.

Since we require only a generator for a one-dimensional Lie subalgebra, we may further suppose that s₁ = 1 in (9).

3.2 Eigenvalues not all zero

From now on, we shall assume that the eigenvalues of S are not all zero. In this case, we introduce the matrix R that belongs to so(3,1)

(10)R=cos⁡θ0−sin⁡θ00cosh⁡ψ0sinh⁡ψsin⁡θ0cos⁡θ00sinh⁡ψ0cosh⁡ψ.

In this case, matrix (5) may be conjugated to

(11)R−1SR=0−t60t1t60t400−t40t3t10t30

where

(12)t1=(b⁡cos⁡θ+c⁡sin⁡θ)cosh⁡ψ−a⁡sinh⁡ψ⁡cos⁡θ

(13)t3=a⁡sin⁡θ⁡sinh⁡ψ+(c⁡cos⁡θ−b⁡sin⁡θ)cosh⁡ψ

(14)t4=(b⁡sin⁡θ−c⁡cos⁡θ)sinh⁡ψ−a⁡sin⁡θ⁡cosh⁡ψ

(15)t6=a⁡cos⁡θ⁡cosh⁡ψ−(b⁡cos⁡θ+c⁡sin⁡θ)sinh⁡ψ.

It is always possible to choose θ and ψ such that t₁ = 0 and t₄ = 0. Indeed (12) and (14) imply that

(16)tanh⁡2⁡ψ=2aba2+b2+c2,tan⁡2⁡θ=2bcb2−a2−c2.

If b² − a² − c² = 0, we choose θ=π4. The conclusion is that if the eigenvalues of S are not all zero, then S may always be conjugated to the form

(17)S=0−s600s6000000s300s30.

In terms of a one-dimensional Lie subalgebra, we may further suppose that either s₃ = 1 or s₆ = 1.

4. Two-dimensional Lie subalgebras

4.1 Two-dimensional abelian Lie subalgebras

Now we proceed to examine the two-dimensional Lie subalgebras of so(3,1). First of all, it is easy to check that, starting from matrix (9), a matrix in so(3,1) that commutes with (9) other than (9) itself, must be of the form

(18)B=000000−s200s20s200s20.

Putting the matrices (9) and (18) together gives a two-dimensional abelian subalgebra.

Secondly, the only two-dimensional abelian Lie subalgebra to which the matrix (17) belongs is the Cartan subalgebra obtained by taking the span of the matrices s₃ = 1, s₆ = 0 and s₃ = 0, s₆ = 1 in (17). Hence, any two-dimensional abelian Lie subalgebra of so(3,1) is a Cartan subalgebra, and all of them are conjugate: see [6, 7].

4.2 Two-dimensional non-abelian Lie subalgebras

4.2.1 One generator of type (9)

Now we attempt to find two-dimensional non-abelian Lie subalgebras. We shall assume that one generator A is given by (9) and we take a second B in the form (1). In B, by subtracting a multiple of A from B, we may assume that s₆ = 0. Now we find that

(19)[A,B]−μA−νB=0s2−μνs5+s3+s4s2−νs1−μμ−s20−νs4+s5−νs2−s1−(νs5+s3+s4)νs4−s50−νs3−s5s2−νs1−μ−νs2−s1−νs3−s50.

We begin to solve the conditions arising from setting to zero all entries in the matrix that appear on the right hand side of (19). We find

(20)s4=ν2s3−s3,μ=s2,s1=−νs2−s6,s5=−νs3.

At this point, we see that if ν ≠ 0, then B = 0. However, if ν = 0, then (19) is now satisfied. Furthermore, we have now that

(21)B=000000−s3s20s30s30s2s30.

If we assume that s₂ = 0, then we find that [A, B] = 0, whereas we are assuming that our two-dimensional subalgebra is non-abelian. Thus, we may suppose that s₂ ≠ 0, and we find P⁻¹BP where

(22)P=100001−s322s22−s3s2−s322s220s3s21s3s20s322s22s3s21+s322s22.

We have chosen P so that it belongs to so(3,1) and commutes with A. We find that

(23)P−1BP=0000000s200000s200

and hence we may assume s₂ = 1. We now have our two-dimensional non-abelian Lie subalgebra with generators A, B

(24)A=0101−100000001000,B=0000000100000100

and Lie bracket [A, B] = A. This subalgebra is unique up to conjugacy.

4.2.2 One generator of type (17)

Now we shall show that there can be no two-dimensional non-abelian Lie subalgebra when one generator is of type (17). Thus, we assume that

(25)A=0−s600s6000000s300s30,B=0−t6−t5t1t60t4t2t5−t40t3t1t2t30.

Now supposing there exist μ, ν such that [A, B] − μA − νB = 0, leads to the following system of equations:

μs6+νt6=0νt5−s3t1−s6t4=0νt1−s3t5+s6t2=0νt4+s3t2+s6t5=0νt2+s3t4−s6t1=0μs3+νt3=0.

However, it is easy to see that solving this system leads to an abelian subalgebra.

5. Three-dimensional Lie subalgebras

There are, depending how one counts, perhaps six classes of real, solvable, three-dimensional Lie algebras. In this context, we are referring to abstract Lie algebras, and not at the moment necessarily subalgebras of so(3,1). They are comprised of the algebras A_3.1, …A_3.7 and A_2.1⊕ in [8], as well as the abelian three-dimensional Lie subalgebra. Each of these algebras has a two-dimensional abelian ideal. We saw in the previous Section that two-dimensional abelian subalgebras can occur in just two ways, up to isomorphism. One such way is as a Cartan subalgebra. However, we know that Cartan subalgebras are self-normalizing [7]. Therefore, the only possibility for a three-dimensional solvable subalgebra of so(3,1) to have a two-dimensional abelian ideal is if it the subalgebra spanned by the matrices (9) and (18), up to isomorphism.

Next we take a matrix of the form (1) that we call C, and find the conditions on C such that [A, C] and [B, C] are linear combinations of A and B, where A is a matrix of the form (9) and B of the form (18). We may ease the working by assuming that s₁ = 0 and s₆ = 0 in P. A straightforward calculation reveals that in P we must have s₃ = s₄ = 0. If we set A, B, C equal to e₁, e₂, e₃ and s₅ = a and s₂ = b, respectively, we obtain the non-zero Lie brackets:

(26)[e1,e3]=ae1−be2,[e2,e3]=be1+ae2.

Assuming that a² + b² ≠ 0 so that the matrix C does not vanish, we may scale C by a non-zero factor, so we can suppose that either b = 1 or a = 1, b = 0. As abstract Lie algebras, they are A_3.3 and A_3.6/7 in [8].

It remains only to discuss the cases of subalgebras that are isomorphic to sl(2,R) and so(3). Concerning sl(2,R), we see from (2), that we can take the brackets in the form

(27)[e2+e6,e1]=e2+e6,[e1,e2−e6]=e2−e6,[e2−e6,e2+e6]=2e1.

Accordingly, following the discussion at the end of the previous Section, we may put e₂ + e₆ = A and e₁ = B from (25) so that the bracket [e₂ + e₆, e₁] = e₂ + e₆ is satisfied. We will use the remaining brackets to determine e₂ − e₆ and hence e₂ and e₆ separately. However, it is quite straightforward to check that we obtain precisely the span of the three matrices obtained from (2) by putting in turn s₁ = 1, s₂ = s₃ = s₄ = s₅ = s₆ = 0, s₂ = 1, s₁ = s₃ = s₄ = s₅ = s₆ = 0, s₁ = s₂ = s₃ = s₄ = s₅ = 0, s₆ = 1. In particular, all subalgebras of so(3,1) that are isomorphic to sl(2,R) are conjugate. It is interesting to note that the representation of sl(2,R) appearing in so(3,1) is conjugate via a transformation of gl(4,R) (not so(3,1)!) to the direct sum of the adjoint and a one-dimensional trivial representation, as we invite the reader to show: see also the end of Section 8 below.

As regards so(3), there are only two possible representations in gl(4,R), coming from the irreducible 4 × 4 and standard 3 × 3 representations. However, the former is by 4 × 4 skew-symmetric matrices and so cannot be found in (1). Thus, the only possibility of obtaining so(3) at all in (1), is the obvious one, that is, the upper left 3 × 3 block using s₄, s₅, s₆ in (1).

6. Four-dimensional Lie subalgebras

A Borel subalgebra in a semi-simple Lie algebra is a solvable subalgebra of maximal dimension. We may construct a Borel subalgebra by using the positive roots in a Cartan decomposition. Referring to (1), we use the Cartan subalgebra that corresponds to s₃ and s₆. Then we use the positive simple roots 1±i with root vectors e₁ + ∓ie₂ + ±ie₄ + e₅.

We can obtain the Borel subalgebra from the following set of matrices:

(28)T=0−t4t1t1t40t2t2−t1−t20t3t1t2t300.

The matrix T engenders the following Lie algebra

(29)[e1,e3]=e1,[e1,e4]=−e2,[e2,e3]=e2,[e2,e4]=e1,

which is precisely algebra A_4.12 in [8]. We could also arrive at the same conclusion by revisiting the calculation of the previous Section and allowing the parameters s₂ and s₅ to generate independent matrices. It is known [7] that all such Borel subalgebras are conjugate.

There can be no four-dimensional Lie subalgebras of so(3,1) that have a necessarily trivial Levi decomposition, that is sl(2,R)⊕R or so(3)⊕R, for in both cases the centralizers consist of diagonal matrices and do not belong to so(3,1).

7. Five-dimensional Lie subalgebras

Finally, we shall show that so(3,1) does not possess any five-dimensional Lie subalgebras. Since the Borel subalgebras are four-dimensional, there can be no five-dimensional solvable subalgebras. For the same reason as in dimension four, there can be no Levi decomposition subalgebras that have a trivial Levi decomposition. Thus, we have only to show that we cannot obtain the five-dimensional indecomposable Lie algebra, denoted by A_5.40 in [8], which is a semi-direct product of sl(2,R) and R2. The R2 factor here is the radical, which is an ideal. Now according to Section 5, we may assume that the Levi factor sl(2,R) is determined by s₁, s₂, s₆ in (1). However, as such, we have a representation of sl(2,R) that reduces as an irreducible three-dimensional representation and a trivial one-dimensional representation. Hence, there can be no two-dimensional invariant subspace that would be needed to accommodate the radical of the Lie subalgebra A_5.40.

8. Another representation of so(3,1)

In equation (1), we have given the definition of the Lie algebra so(3,1). We now wish to exhibit another 4 × 4 representation of so(3,1), which is not conjugate to the standard representation. Thus, we introduce the following matrix U.

(30)U=s1s2s3s4−s2s1s4−s3s5s6−s1s2s6−s5−s2−s1.

In the same way as in (1), we obtain the following Lie brackets:

(31)[e1,e3]=2e3,[e1,e4]=2e4,[e1,e5]=−2e5,[e1,e6]=−2e6,[e2,e3]=−2e4,[e2,e4]=2e3,[e2,e5]=−2e6,[e2,e6]=2e5,[e3,e5]=e1,[e3,e6]=e2,[e4,e5]=−e2,[e4,e6]=e1.

If we make the following change of basis

(32)−e12,−(e4+e6)2,−(e3+e5)2,−e22,(e5−e3)2,(e6−e4)2

then we will obtain precisely the same Lie brackets as in (1), and so we know that (30) is a representation of so(3,1). The subalgebra of (30) given by putting s₂ = s₄ = s₆ = 0 is isomorphic to sl(2,R). It appears in the “diagonal” representation of sl(2,R).

Referring to (1), the subalgebra given by putting s₃ = s₄ = s₅ = 0 is isomorphic to sl(2,R). Clearly this representation is equivalent to

(33)S=0−s6s10s60s20s1s2000000.

It may be shown that (33) is equivalent to the direct sum of the adjoint representation and a one-dimensional trivial representation, that is,

(34)S=2s62s100s20s1002s2−2s600000.

Begin by finding a linear combination of the matrices (33) that are nilpotent, which inevitably necessitates the introduction of some 2’s. Thus, the representations (1) and (30) are not conjugate.

9. Table of proper subalgebras of so(3,1) up to conjugacy

9.1 One-dimensional Lie subalgebras

0s10s1−s10000000s1000,0−as100as1000000bs100bs10(a=1orb=1).

9.2 Two-dimensional Lie subalgebras

0s10s1−s10−s200s20s2s10s20,0−s100s1000000s200s20,0s10s1−s100s20000s1s200.

9.3 Three-dimensional Lie subalgebras

0s3−s20−s30s10s2−s1000000,0−s3−0s1s300s20−000s1s200,0s1−as3s1−s10−s2bs3as3s20s2s1bs3s20(a=1,b=0orb=1).

9.4 Four-dimensional Lie subalgebras

0−s4s1s1s40s2s2−s1−s20s3s1s2s30.