Minimal representations of Lie algebras with non-trivial Levi decomposition

We obtain minimal dimension matrix representations for each of the Lie algebras of dimensions five, six, seven and eight obtained by Turkowski that have a non-trivial Levi decomposition. The key technique involves using the invariant subspaces associated to a particular representation of a semi-simple Lie algebra to help in the construction of the radical in the putative Levi decomposition.


Introduction
Given a real Lie algebra g of dimension n, a well-known theorem due to Ado asserts that g has a faithful representation as a subalgebra of gl( p, R) for some p. In several recently published papers the authors and others have investigated the problem of finding minimal dimensional representations of indecomposable dimensional Lie algebras of dimension five and less [2,3,8,12]. In two recent papers [2,8], the problem of finding such a minimal representation is considered for the four-dimensional Lie algebras. In [8], the main technique is somewhat indirect and depends on a construction known as a left symmetric structure. In [2,3], the minimal representations have been calculated directly without the need for considering left symmetric structures.
Burde [1] has defined an invariant μ(g) of a Lie algebra g to be the dimension of its minimal faithful representation. So far, minimal representations have been found for all the indecomposable and decomposable algebras of dimension five and less [3,5]. In this paper, we investigate the issue of minimal representations for Lie algebras that have a non-trivial Levi decomposition. Such algebras have been classified in dimensions five to eight by Turkowski [13]. Of course it is interesting to ascertain the value of μ from a theoretical point of view. However, an important practical reason is that calculations involving symbolic programs such as Maple and Mathematica use up lots of memory when storing matrices; accordingly, calculations are likely to be faster if one can represent matrix Lie algebras using matrices of a small size. Besides the value of having explicit representations of low-dimensional Lie algebras, they also add to the growing body of results that seek to provide alternatives to Ado's Theorem for the construction of representations [1].
In this paper, we shall use the classification of the low-dimensional indecomposable Lie algebras found in [11], which is in turn taken from [9]. Such algebras are denoted as A m.n , where m denotes the dimension of the algebra and n the nth one in the list. In dimension 2, the algebra A 2.1 is the unique non-abelian algebra. The same notation is adopted in [10], which concerns a class of six-dimensional algebras, and is used in [13], which is the most important reference for the current paper. We shall also use R n to denote the n × n irreducible representation of so (3) and D j for the irreducible representation of order (2 j + 1) × (2 j + 1) of sl(2, R), where j is either an integer or half-integer. Furthermore, k D j denotes k copies of D j and D 0 the trivial one-dimensional representation.
The outline of this paper is as follows. In Sect. 2, we consider the construction of abstract Lie algebras that have a Levi decomposition. In Sect. 3, we present several results about representations that will be needed in the sequel. In Sect. 4, we carry out the construction of the minimal representations proper. A key element here is the use of the invariant subspaces that come from the representation of the semi-simple factor, in all cases here either so(3) or sl(2, R) in a given representation, that acts by commutator on gl(N , R) for various small values of N , to construct a representation of the radical ideal of the algebra. It is argued in every case that the representations thus constructed are of minimal dimension. Finally, in Sect. 5, we give a list of minimal dimension representations for each of the algebras in Turkowski's list [13]. Occasionally, we give more than one such representation, sometimes depending on parameters λ and μ if it seems to be of particular interest although we do not consider the difficult issue of the inequivalence of different representations. Finally, the reader may object that we have not supplied details of the Turkowski Lie algebras, but to do so would involve repeating large sections of [13]. In any case, the Lie brackets can be readily computed from the representations given in Sect. 5.

Constructing Levi algebras in general
Let us consider the problem of constructing a Lie algebra g that has a Levi decomposition σ ρ in general. We have the following structure equations: where 1 ≤ a, b, c, d ≤ s and s + 1 ≤ i, j, k, l ≤ n and {e a } is a basis for the semi-simple subalgebra σ and {e i } is a basis for the radical ρ. Calculation shows that the Jacobi identity is equivalent to the following conditions: We interpret these conditions as follows: we start with a semi-simple algebra so that the first set of conditions above is satisfied. Then the second set says that the matrices C k al make ρ (merely as a vector space) into a σ -module. The third set says that the C k al are derivations of the Lie algebra ρ and the fourth of course that ρ is a Lie algebra. Therefore to find all possible Lie algebras of dimension n that have a Levi decomposition σ ρ, we can proceed as follows: choose a semi-simple algebra σ of dimension r . Then pick any solvable algebra ρ of dimension n − r and consider a representation of σ in ρ considered simply as the vector space R n−r , all of which are known and are completely reducible [6] since σ is semi-simple. Of course it is a difficult question as to whether a representation of a particular semi-simple algebra exists in a certain dimension: see [6] for more details. Finally, it only remains to check that the matrices representing σ act as derivations of the Lie algebra ρ. In the affirmative case we have our sought after Lie algebra; in the negative case there is no such algebra and we have to choose a different representation of σ in ρ. If all such representations lead to a null result then there can be no non-trivial Levi decomposition involving σ and ρ.

Representation results
We quote the following result from Humphreys [6].

Proposition 3.1 Let L be a finite dimensional complex Lie algebra acting irreducibly on the vector space V .
Then L is reductive and the center is of dimension one or zero. If in addition L ⊂ sl(V ) then L is semi-simple.
Notice that the condition of semi-simplicity is a conclusion not an assumption. In practice we apply this result to semi-simple algebras to deduce that for a particular representation there is only a trivial one-dimensional extension to a Levi decomposition or more generally to an irreducible block in a reducible representation.
The following two results lie at opposite extremes.
Theorem 3.2 Suppose that the Lie algebra g is a semi-direct product of a semi-simple subalgebra σ and an r -dimensional abelian ideal ρ in the Levi decomposition. Then g has a (faithful) representation as a subalgebra of gl(r + 1, R).
Conversely, a subalgebra of gl(r + 1, R) with the upper r × r block giving a representation of semi-simple Lie algebra and the first r entries in the last column being arbitrary and the (r + 1)th zero, gives a Levi subalgebra of gl(r + 1, R) whose radical is abelian.
Proof We quote the structure equations in Eq. 2.1, where now we assume that C k i j = 0. To obtain the required representation consider ad(σ ) restricted to ρ, in other words the matrices C k ai . Define E a to be the same matrix as C k ai , but now augmented by an extra bottom row and last column of zeroes. Now define E k to be the (r + 1) × (r + 1) matrix whose only non-zero is 1 in the (k, r + 1)th position.
now E i E a = 0. As regards E a E i it is a matrix that has only non-zero entries in the last column. That last column is C k ai with a and k fixed, augmented by a zero in the (r + 1, r + 1)th position which means that

Theorem 3.3
Suppose that ρ is the radical of a Lie algebra g and that ρ has trivial center. Then g has a faithful representation of g in gl(r, R), where ρ is of dimension r .
Proof Define E a to be the matrix C k ai and E j to be the matrix C k i j . Then in order to have a representation of g we require that However, these conditions are identical with the last three in Eq. 2.2. If the center of ρ is trivial, then we will have a faithful representation of g.
The previous theorem admits the following variation, having been already noted in the solvable case [3].

Theorem 3.4
Suppose that ρ is the radical of a Lie algebra g and that ρ has an abelian nilradical of dimension r − 1, where ρ is of dimension r . Then g has a faithful representation in gl(r, R).
Proof Define E a to be the matrix C k ai and E r to be the matrix C k ri . For 1 ≤ k ≤ r − 1 define E k to be the r × r matrix whose only non-zero entry is a 1 in the (k, r )th position. Then proceed as in Theorem 3.3. The first algebra that has a non-trivial Levi decomposition is denoted either by A 5.40 in [11] or L 5.1 in [13]. We obtain a representation for it in gl(3, R) by using Theorem 3.2.

Algebras of dimension 6
In dimension six, there are four algebras that have a non-trivial Levi decomposition denoted by L 6.1−6.4 in [13]. According to [12] only one of them, L 6.3 , has a representation in gl(3, R) which is supplied by Theorem 3.3. Furthermore, Theorem 3.2 provides representations of minimal dimension for L 6.1 and L 6.4 in gl(4, R). Furthermore, L 6.2 has a representation in gl(4, R) which is also of minimal dimension.

Algebras of dimension 7
According to [12] the only seven-dimensional algebra that has a representation in gl(3, R) is decomposable. There are seven algebras that have a non-trivial Levi decomposition denoted by L 7.1−7.7 in [13]. Theorem 3.3 provides representations for L 7.1 , L 7.3 , L 7.4 , L 7.5 in gl(4, R), whereas Theorem 3.2 yields representations for L 7.2 , L 7.6 and L 7.7 in gl(5, R). In fact L 7.7 also has a representation in gl(4, R).
It remains to show that the representations for L 7.2 , L 7.6 coming from Theorem (3.2) are minimal. We have said already that they cannot be subalgebras of gl(3, R). If they had representations in gl(4, R) there would be a basis in which the radical R 4 would correspond to the upper right 2 × 2 block according to a Theorem of Schur-Jacobson [7]. As such the form of the representation would have to be A C 0 B . The occurrence of the lower left zero 2 × 2 block is a consequence of the fact that the radical is an ideal. In the case of L 7.2 there is then no room to accommodate a representation of so (3); for L 7.6 we could only take either the standard 2 × 2 representation of sl(2, R) or the 4 × 4 "diagonal" representation of sl(2, R). In the former case, we obtain algebra L 7.7 and in the latter a decomposable subalgebra but certainly not L 7.6 .

Algebras of dimension 8
Now we consider Turkowski's algebras of dimension 8. At the outset it is clear that for such algebras μ ≥ 4, because the only eight-dimensional subalgebra of gl(3, R) is sl(3, R), which is simple [12]. We shall find the algebras for which μ = 4. . However, the complement will be solvable only if either w = x = y = 0 or p = q = r = 0 in which case it gives an eight-dimensional decomposable algebra. So the only possibilities, up to isomorphism, for the simple factor is either of the representations In the first case the invariant subspaces may be described as , where u, v, x, y ∈ R 2 and each of them engenders a two-dimensional invariant subspace, together with a kernel cannot consist entirely of the kernel or else we will obtain a decomposable algebra, so one of u, v, x, y must occur. Suppose that y occurs. We will suppose also that x occurs. We will need to add in a one-dimensional subspace of the kernel. It can be one of four kinds: ⎡ It is not our concern here to look at all possible subalgebras of sl(4, R) (see [4]); however, if we put ν = λ + p, μ = 1 + λ in the first matrix in 5.1, we obtain precisely L 8.17 . In the second case, we obtain L 8.16 , in the third L 8.18 and in the fourth L 8.14 . Now suppose that x = 0 but that y = 0. Then another way to obtain an algebra is to take u = 0 and independent of y. We will need to have e = 0 in the kernel. We obtain thereby a radical that is isomorphic to the five-dimensional Heisenberg algebra: it is L 8.13 =−1 in [13]. Other choices of invariant subspaces will lead to different representations, but not to different underlying algebras.
Finally, we come back to the other representation To avoid having a decomposable algebra, we include B and we need to take a two-dimensional kernel. Again to avoid decomposability, we shall have to take, up to equivalence, e = 0 and g = (λ + 1)d. We obtain L 8.12 p=1 .
We know now that for all the remaining algebras μ > 4. Next we observe that Theorem 3.3 is applicable to the following algebras:  8.22 . We note that Theorem 3.2 implies for L 8.5 , L 8.21 , L 8.22 that μ ≤ 6. In fact for L 8.22 we have μ = 5. The cases of L 8.5 and L 8.21 will be discussed below. As regards L 8.8 it depends on a parameter p; for p = 0 the radical has trivial center but for p = 0 there is a one-dimensional center. However, in the case p = 0 we can find a representation by appealing to Theorem 3.4. This representation, unlike the adjoint representation of ρ, extends uniformly for all values of p. Hence for L 8.8 we have μ = 5. The case L 8.15 will be discussed below. We shall also consider L 8.2 , L 8.6 , L 8.13 =1 , L 8.19 , where the radical is the five-dimensional Heisenberg algebra.
5.2 L 8.5 and L 8.21 As stated above, Theorem 3.2 furnishes representations for L 8.5 and L 8.21 in dimension 6. The question is whether representations exist in lower dimensions. We know that the only possibility is a representation in dimension 5. We examine L 8.5 first. We know that so(3) has a unique irreducible representation in every dimension starting at 3. To find a representation in gl(5, R) we must consider the irreducible representations in dimensions n = 3, n = 4 and n = 5. The case n = 5 is impossible according to Proposition 3.1. For the case n = 3, we argue as follows. Without loss of generality the representation for so(3) may be assumed to be of the form A 0 3×2 0 2×3 0 2×2 , where A is the standard irreducible 3 × 3 representation of so(3). Now consider the action of this representation of so(3) on gl(5, R). We write such a matrix in gl(5, R) as where u, v, x, y ∈ R 3 and a, b, c, d, α ∈ R. Now u, v, x, y engender invariant subspaces and so too do the one-dimensional subspaces corresponding to a, b, c, d, α; the latter five span the kernel of the representation. In order to have a 5-dimensional radical we must choose one of u, v, x, y and a 2-dimensional subspace of the kernel. However, we will never obtain as the representation R 5 of so(3) on the radical, the irreducible 5 × 5 representation of so(3). The argument for n = 4 is similar if even easier. The preceding argument lends itself to immediate generalization as follows. Whenever a k × k irreducible representation of a semi-simple algebra is augmented by adding a certain number of zero rows and columns and acts by commutator on gl(N , R) for some N > k, the corresponding invariant subspaces give a number of copies of the associated standard or definition representation, that is, a matrix acting by matrix multiplication on a vector, together with the kernel and a complement to the k × k irreducible representation inside of gl(k, R). In particular, it is impossible to obtain an N × N irreducible representation unless k = N or from the complementary subspace; of course for k = N this subspace is not available as it would change the given representation into gl(N , R). Now we examine L 8.21 and apply the discussion of the last paragraph. The representation of sl(2, R) on the radical is supposed to give D 2 . As such we can exclude the irreducible 4 × 4 representation of sizes 2 × 2, 3 × 3 and 4 × 4. The 5 × 5 representation is also not allowed because of Proposition 3.1. There remain two more representations to consider: In the first case the invariant subspaces can be described as: We see that none of these invariant subspaces are five-dimensional and hence we cannot obtain the representation D 2 .
In the second case the invariant subspaces are given by: ⎡ Proof Each of the three algebras has a one-dimensional center spanned by e 8 . In each case e 8 also spans the center of the Heisenberg algebra of dimension five which is the radical in these cases. In any matrix representation of Heisenberg, a non-zero center must be represented by a nilpotent matrix. Up to isomorphism, in gl(4, R) there are precisely three non-zero nilpotent matrices of rank three, two and one: . This latter space of matrices comprises a decomposable ten-dimensional Levi algebra. In fact it is isomorphic to R ⊕ (sl(2, R) A 6.82δ=0a=b=1 ) [10]. Since the derived algebra of Heisenberg is spanned by the matrix of type (iii), the only way to obtain it as a subalgebra is in the obvious way, that is, by taking a = 0 and i = −e: it gives the algebra L 8.13 =−1 . Now we shall investigate whether algebras that have a 5-dimensional Heisenberg radical can be represented in gl(2, R). Again the central element e 8 must be represented by a nilpotent matrix E 8 . The rank of E 8 cannot be 4 otherwise its centralizer is abelian; nor can it be 3 or else the centralizer is solvable. Suppose then that the rank of E 8 is 2 and we may assume that E 8 = 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 . Its centralizer may be written as where A, B are 2 × 2, x, y ∈ R 2 and α ∈ R. We deduce first of all that this case cannot correspond to L 8.2 since we have to have a representation of so (3).
Now we may assume that A is the standard 2 × 2 representation of sl(2, R) together with a multiple of the identity. In fact we have the "2D 1 2 " representation of the form and a putative representation of 5-dimensional Heisenberg in the form . Now in the case of algebra L 8.6 the kernel of the representation is "3D 0 ", that is, three-dimensional; however, this circumstance could arise only from having λ and α both non-zero and independent in which case we would obtain a decomposable algebra. Hence L 8.6 is excluded. In the cases L 8.13 and L 8.19 the kernel is one-dimensional and is spanned by E 8 . As such we may assume that λ = α = 0. Now consider invariant subspaces. Both x and y give invariant subspaces and so does the trace-free part of B. However, the only way to have an invariant subspace of dimension five, since the element E 8 must be included, is to take B as multiples of the identity, which, however does not produce a subalgebra. The conclusion is that if there is a 5 × 5 representation of an algebra that has the 5-dimensional Heisenberg as its radical, then the element E 8 must have rank one. Now we assume that E 8 has rank one and is of the form E 8 = x, y ∈ R 2 and α ∈ R. In order to obtain L 8.2 we would have to take the three-dimensional representation of so(3) in the upper left 3 × 3 block. However, in that case because (x, y, z) and (u, v, w) span invariant subspaces the only possibility is to obtain the seven-dimensional Heisenberg algebra.
Considering the cases L 8.6 , L 8.13 , L 8.19 we see that the only possibility for the 3 × 3 block is to have either the three-dimensional irreducible representation of or two-dimensional irreducible representation augmented by an extra row and column of zeros of sl(2, R). In the former case we will obtain again the sevendimensional Heisenberg algebra. In the latter case we have to consider the algebras that we are trying to represent: L 8.6 , L 8.13 , L 8.19 . For L 8.13 and L 8.19 the kernel of the representation is one-dimensional, which leads to the self-same representation that we found above for L 8.13 =−1 but now augmented by an extra middle row and column of zeroes. On the other hand, for L 8.6 the representation is given by "D 1 2 ⊕ 3D 0 ", whereas (x, y) and (u, v) will each give D 1 2 unless they are correlated, that is, u = y, v = −x which gives 2D 1 2 . Thus μ > 5 in the cases L 8.6 , L 8.13 =1 , L 8.19 and in fact for these three algebras μ = 6.

L 8.15
Now let us look at L 8.15 . First of all note that unlike the algebras just discussed, where the radical is fivedimensional Heisenberg, although the radical is again nilpotent, A 5.3 , the algebra as a whole has trivial center and so the adjoint representation is faithful. Thus μ ≤ 8. Moreover, it was shown in [3] that μ = 5 for A 5.3 alone. Thus μ ≥ 5 for L 8.15 . In fact in Sect. 5 the reader will see a 6 × 6 representation. Thus all that remains is to show that μ > 5 for L 8.15 .
As in the preceding subsection, we can argue that E 8 must be nilpotent. Again we only have to consider the case where E 8 has rank one or two.

E 8 of rank 2
If it has rank two, as above, its centralizer may be written as and α ∈ R. Although e 8 is not a central element, its centralizer is spanned by {e 1 , e 2 , e 3 , e 4 , e 5 , e 8 }. Hence the simple factor is of the form However, now we find that E 4 = [E 6 , E 8 ] and E 5 = [E 7 , E 8 ] implies that E 4 and E 5 are zero. Hence no such representation can exist.

E 8 of rank 1
As above we assume that E 8 has rank one and is of the form E 8 = Assume first that d = 0 then c = 0, e = 0. At this point we obtain a Lie algebra whose radical is A 5.1 and not A 5.3 . If instead we assume x = 0 then we have k = r = 0. Again we obtain a Lie algebra whose radical is A 5.1 and not A 5.3 .