By Andreescu T., Feng Z.

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2 that L = L and therefore L(m) = L for all m ≥ 1, so sl(2, C) is not solvable. If L is solvable, then the derived series of L provides us with an “approximation” of L by a ﬁnite series of ideals with abelian quotients. This also works the other way around. 3 If L is a Lie algebra with ideals L = I0 ⊇ I1 ⊇ . . ⊇ Im−1 ⊇ Im = 0 such that Ik−1 /Ik is abelian for 1 ≤ k ≤ m, then L is solvable. 1 Solvable Lie Algebras 29 Proof We shall show that L(k) is contained in Ik for k between 1 and m. Putting k = m will then give L(m) = 0.

Let a be the diagonal matrix with entries (α1 , . . , αn ). The eigenvalue of a on ei is αi , but this varies as a runs through the elements of A, so we must be prepared to let diﬀerent elements of A act with diﬀerent eigenvalues. We can specify the eigenvalues of elements of A by giving a function λ : A → F . The corresponding eigenspace is then Vλ := {v ∈ V : a(v) = λ(a)v for all a ∈ A} . 1 (i) Check that the eigenspaces Vλ deﬁned above are vector subspaces of V . 3 The Invariance Lemma 39 (ii) Using the notation above, deﬁne εi : A → F by εi (a) = αi .

Perhaps surprisingly, there are no other Lie algebras with this property. We shall now prove the following theorem. 2 Let F be any ﬁeld. There is a unique 3-dimensional Lie algebra over F such that L is 1-dimensional and L is not contained in Z(L). This Lie algebra is the direct sum of the 2-dimensional non-abelian Lie algebra with the 1-dimensional Lie algebra. 22 3. Low-Dimensional Lie Algebras Proof We start by picking some non-zero element x ∈ L . Since x is not central, there must be some y ∈ L with [x, y] = 0.