Lie subalgebras of $so(3,1)$ up to conjugacy Open Access

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.

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

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

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

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

Then conjugating S by P, we obtain

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

Now the characteristic polynomial of this reduced S is given by

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

Then we find that

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

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

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 $so(3,1)$

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

where

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

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

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

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

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].

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

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

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

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

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

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

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

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

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

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

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:

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

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).

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:

The matrix T engenders the following Lie algebra

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 $so(3,1)$⁠.

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.

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.

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

If we make the following change of basis

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

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

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.