Seminar Sophus Lie 2 (1992) 57–58 Version of July 20, 1995

An example of a solvable Lie algebra

Michael Wüstner

We shall consider a solvable real Lie algebra 𝔤 which is linearly spanned by the set comp 𝔤 of its compact elements and use the notation of [1]. In particular, the diagram in [1] below Proposition 2.10 gives a good orientation on the subalgebras involved in our discussion. It was shown in [1] that a Cartan algebra in a real Lie algebra spanned by its compact elements has nilpotent class ≤ 2. It was left open whether there are algebras in which the Cartan algebras are nonabelian. In the following we present an example of dimension 13 showing that this can happen.

Let 𝔤 be a real vector space with a basis

{u, x, y, z, t, e_{1}, e_{2}, f_{1}, f_{2}, g_{1}, g_{2}, h_{1}, h_{2}}

and define the following relations on the basis elements:

[u, a_{1}] = a_{2},  [u, a_{2}] = −a_{1} for a = e, f, g, h,
[e_{1}, e_{2}] = x,  [f_{1}, f_{2}] = y,
[e_{1}, f_{2}] = −t, [e_{2}, f_{1}] = t,
[x, f_{i}] = h_{i}, [y, e_{i}] = g_{i} for i = 1, 2, [x, y] = z,
[t, e_{i}] = −h_{i}, [t, f_{i}] = −g_{i} for i = 1, 2,
[e_{1}, g_{2}] = −\frac{1}{2}z, [e_{2}, g_{1}] = \frac{1}{2}z, [f_{1}, h_{2}] = \frac{1}{2}z,  [f_{2}, h_{1}] = −\frac{1}{2}z.

All other brackets between basis elements are set equal to 0. Bilinear extension yield a bilinear product [·, ·]: 𝔤 × 𝔤 → 𝔤.

We define on 𝔤 the structure of a graded vector space by setting 𝔤^{0} = ℝ·u, 𝔤^{1} = span{e_{1}, e_{2}, f_{1}, f_{2}}, 𝔤^{2} = span{x, y, t}, 𝔤^{3} = span{g_{1}, g_{2}, h_{1}, h_{2}}, 𝔤^{4} = ℝ·z. It is readily verified that [𝔤^{m}, 𝔤^{n}] ⊆ 𝔤^{m+n}. Thus 𝔤 is a graded algebra with respect to the bracket. Lemma. 𝔤 is a solvable Lie algebra. Proof. We verify the validity of the Jacobi identity by straightforward, but tedious calculation. I have written a program which executes the necessary verifications. The gradation permits us a quick inspection of the solvability of 𝔤.

We verify quickly that the subspace 𝔥 = span{u, x, y, z, t} is a Cartan subalgebra and 𝔱 =^{def} 𝔥 ∩ comp 𝔤 = span{u, z}. Also, the sum of the weight spaces with respect to 𝔱 is 𝔱^{+} = span{e_{1}, e_{2}, f_{1}, f_{2}, g_{1}, g_{2}, h_{1}, h_{2}} (cf. [1]), and

𝔤′ = C^{∞} = ⟨𝔱^{+}⟩ = span{e_{1}, e_{2}, f_{1}, f_{2}, g_{1}, g_{2}, h_{1}, h_{2}, t, x, y, z}. It follows that 𝔢 =^{def}

𝔥∩𝔤′ = span{x, y, t, z}. Further, in the notation of [1], we have 𝔷 = ℝ·z = 𝔧 and may take 𝔢_{1} = span{x, y, t} as a complement of 𝔧 in 𝔢, and 𝔱_{1} = ℝ·u as a complement of 𝔷 in 𝔱. With these choices we obtain 𝔤′ = 𝔢_{1} ⊕ 𝔱^{+}. Therefore, we have dim 𝔱 = 2, dim 𝔥 = 5, dim 𝔢 = 4, dim 𝔢_{1} = 3, dim 𝔱^{+} = 8, dim 𝔤′ = dim⟨𝔱^{+}⟩ = 12.

The following diagram records the gradation and the relevant subalgebras:

degree  𝔥  𝔱^{+}
𝔱  𝔢_{1}
0  u
1  e_{1} e_{2} f_{1} f_{2}
2  x y t
3  g_{1} g_{2} h_{1} h_{2}
4  z

The following equations will show that 𝔤 = span(comp(𝔤)). From [1], Lemma 2.2(iii) we know that comp = Γ·𝔱. We will be finished if we can show that 𝔤 is spanned by elements of the form e^{ad w}(u) with w ∈ C^{∞}. Now we compute:

e^{ad(e_{1}+e_{2})}(u) = u − e_{2} + e_{1} − 2x,
e^{ad(e_{1}−e_{2})}(u) = u − e_{2} − e_{1} − 2x,
e^{ad(e_{2}−e_{1})}(u) = u + e_{1} + e_{2} − 2x,
e^{ad(f_{1}+f_{2})}(u) = u − f_{2} + f_{1} − 2y,
e^{ad(f_{1}−f_{2})}(u) = u − f_{2} − f_{1} − 2y,
e^{ad(f_{2}−f_{1})}(u) = u + f_{1} + f_{2} − 2y,
e^{ad(e_{1}+f_{1})}(u) = u − e_{2} − f_{2} − x − y + 3g_{1} + 3h_{1} + 2t,
e^{ad(a_{1})}(u)  = u − a_{2},
e^{ad(a_{2})}(u)  = u + a_{1}  for a = g, h.

We see that the span of these elements is 𝔤 and this shows that 𝔤 = span comp.

References

[1] Hofmann, K.,H., Compact elements in solvable real Lie algebras, Seminar Sophus Lie 2 (1992), this volume.

Fachbereich Mathematik Technische Hochschule Schloßgartenstr. 7 D-6100 Darmstadt

Received May 12, 1992