18.745 Introduction to Lie Algebras Fall 2004

Lecture 4: Nilpotent and Solvable Lie Algebras

Prof. Victor Kač Transcribed by: Anthony Tagliaferro

Definition Given a Lie algebra 𝔤, define the following descending sequences of ideals of 𝔤;

(central series) 𝔤 ⊃ [𝔤, 𝔤] = 𝔤^{1} ⊃ [[𝔤, 𝔤], 𝔤] = 𝔤^{3} ⊃ [𝔤^{3}, 𝔤] = 𝔤^{4} ⊃ ⋯ ⊃ [𝔤^{k−1}, 𝔤] = 𝔤^{k} ⊃ … (1)
(derived series) 𝔤 ⊃ [𝔤, 𝔤] = 𝔤′ ⊃ [𝔤′, 𝔤′] = 𝔤′′ ⊃ ⋯ ⊃ [𝔤^{(k−1)}, 𝔤^{(k−1)}] = g^{(k)} ⊃ …  (2)

Exercise 4.1 Show that:

a) All members of the central and derived series are ideals of 𝔤

b) 𝔤^{k+1} ⊃ 𝔤^{(k)}

Solution a) First, let’s show that 𝔤^{k} is an ideal of 𝔤. Notice that:

[𝔤^{k}, 𝔤] = 𝔤^{k+1} ⊂ 𝔤^{k}  (3)

It therefore follows that 𝔤^{k} is an ideal for all k. Now, let’s show that 𝔤^{(k)} is an ideal. To do this, we’ll use induction on k. First notice that:

[𝔤^{(1)}, 𝔤] = [[𝔤, 𝔤], 𝔤] ⊂ [𝔤, 𝔤] = 𝔤^{(1)}  (4)

this is true since we know [𝔤, 𝔤] ⊂ 𝔤. It follows that 𝔤^{(1)} is an ideal of 𝔤. Now all we have to prove is that 𝔤^{(k)} is an ideal if 𝔤^{(k−1)} is one. Therefore, consider an arbitrary element x = [a, b] ∈ [𝔤^{(k)}, 𝔤] where a ∈ 𝔤^{(k)} and b ∈ 𝔤. Then, by definition of 𝔤^{(k)}, a = [c, d] where c, d ∈ 𝔤^{(k−1)}. Using the Jacobi identity, we find that:

[[c, d], b] = [[c, b], d] + [c, [d, b]]  (5)

but, since 𝔤^{(k−1)} is an ideal, [c, b], [d, b] ∈ 𝔤^{(k−1)}. Therefore, it follows that:

[a, b] = [[c, d], b] = [[c, b], d] + [c, [d, b]] ∈ [𝔤^{(k−1)}, 𝔤^{(k−1)}] = 𝔤^{(k)}  (6)

Hence, since the element x = [a, b] was arbitrary, [𝔤^{(k)}, 𝔤] ⊂ 𝔤^{(k)} and therefore 𝔤^{(k)} is an ideal for all k.

b)We need to show that 𝔤^{k+1} ⊃ 𝔤^{(k)}. Again, we’ll use induction on k.

𝔤^{(1)} = [𝔤, 𝔤] = 𝔤^{2} ⇒ 𝔤^{(1)} ⊂ 𝔤^{2}  (7)

Now consider an arbitrary element of 𝔤^{(}k), x = [a, b] where a, b ∈ 𝔤^{(k−1)}, we’re assuming that since a ∈ 𝔤^{(k−1)}, a ∈ 𝔤^{k} and also notice that b ∈ 𝔤. Therefore [a, b] ∈ [𝔤^{k}, 𝔤] = 𝔤^{k+1}. Hence 𝔤^{(k)} ⊂ 𝔤^{k+1} for all k. □

Definition A Lie algebra is called nilpotent (resp. solvable) if 𝔤^{k} = 0 (resp. 𝔤^{(k)} = 0) for k sufficiently large.

Remark Notice that it follows from Exercise 4.1b that a nilpotent Lie algebra is also solvable.

Examples

1) Abelian Lie algebras are nilpotent.

2) The Heisenberg Lie Algebras H_{k} are nilpotent since H_{k}^{(1)} = ℂ and [[H_{k}, H_{k}], H_{k}] = 0

3) 𝔟_{2}= 𝔽a + 𝔽b, [a, b] = 0 is solvable. since 𝔟_{2}^{(1)} = 𝔽b, 𝔟_{2}^{(2)} = 0 but it is not nilpotent since 𝔟_{2}^{n} = 𝔽b for n ≥ 2.

Exercise 4.2 Let 𝔟_{k}(𝔽) be the subalgebra of 𝔤𝔩_{k}(𝔽) consisting of all upper-triangular matrices and 𝔫_{k}(𝔽) be the subalgebra of 𝔤𝔩_{k}(𝔽) consisting of all strictly upper triangular matrices. Show that:

a) 𝔟_{k}(𝔽) is a solvable Lie algebra (but not nilpotent)

b) 𝔫_{k}(𝔽) is a nilpotent Lie algebra (in particular, 𝔫_{3}(𝔽) ≈ H_{1})

Solution The gist of this problem is to show that the Lie bracket of any two upper triangular matrices is even more upper triangular. It will actually prove easier to prove part (b) first.

b) First notice that if e_{ij} is a basis element of 𝔤𝔩_{k}(𝔽), then the quantity j − i tells us which diagonal the nonzero entry of e_{ij} lies on. Let’s compute the Lie bracket [e_{ij}, e_{mn}] = δ_{jm}e_{in} − δ_{ni}e_{mj}. Let’s say that j − i = r and n − m = s. Then, when j = m, n − i = n + r − j = n + r − m = s + r, so n − i = r + s and similarly if n = i, then j − m = s + r. The payoff from this little bit of arithmetic comes from the following observation. First, define:

𝔥_{k,m} = {∑_{i,j} x_{ij}e_{ij}|x_{ij} ∈ 𝔽; j − i ≥ m; i, j ≤ k}  (8)

Then, we’ve shown that [𝔥_{k,m}, 𝔥_{k,n}] ⊂ 𝔥_{k,m+n}. Notice also that 𝔫_{k}(𝔽) = 𝔥_{k,1} and 𝔟_{k}(𝔽) = 𝔥_{k,0}. Also notice from the definition that 𝔥_{k,k+1} = 0. Proving that 𝔫_{k}(𝔽) is nilpotent is now trivial:

𝔫_{k}(𝔽)^{2} = 𝔥_{k,1}^{2} ⊂ 𝔥_{k,2}
𝔫_{k}(𝔽)^{3} = [𝔥_{k,1}^{2}, 𝔥_{k,1}] ⊂ [𝔥_{k,2}, 𝔥_{k,1}] ⊂ 𝔥_{k,3}
⋮
𝔫_{k}(𝔽)^{k} = 𝔥_{k,1}^{k} ⊂ 𝔥_{k,k+1} = 0

Therefore 𝔫_{k}(𝔽) is nilpotent. Now, on to proving 𝔟_{k}(𝔽) is solvable.

a) To prove this, we will use a proposition that will appear later in these notes. The proposition states that if a Lie algebra contains a solvable ideal and if the Lie algebra modulo this ideal is solvable, then the ideal itself is solvable. Notice that 𝔫_{k}(𝔽) ⊂ 𝔟_{k}(𝔽) and that 𝔫_{k}(𝔽) is solvable (since it’s nilpotent). Also notice that 𝔟_{k}(𝔽)/𝔫_{k}(𝔽) = the set of diagonal matrices, which is abelian, hence solvable. Therefore, by the proposition, 𝔟_{k}(𝔽) is solvable. Notice that in particular,

𝔫_{3}(𝔽) = span \{q = (\begin{matrix}0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) , p = (\begin{matrix}0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix}) , c = (\begin{matrix}0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix})\}  (9)

These matrices exactly satisfy the conditions for the Heisenberg algebra H_{1}. Hence 𝔫_{3}(𝔽) ≈ H_{1}.

Also, we have proved that 𝔟_{k}(𝔽) is solvable, but we have yet to show that is not nilpotent. To demonstrate this, consider the commutator: [e_{ii}, e_{ij}] = e_{ij} for i < j. Since e_{ii} is in 𝔟_{k}(𝔽), it follows that e_{ij} ∈ 𝔟_{k}(𝔽)^{2}, and therefore, e_{ij} ∈ [𝔟_{k}(𝔽), 𝔟_{k}(𝔽)^{2}] = 𝔟_{k}(𝔽)^{3}. Hence e_{ij} ∈ 𝔟_{k}(𝔽)^{k} for all k follows by induction. Thus 𝔟_{k}(𝔽) is not nilpotent. □

Remark Notice that any subalgebra and factor (quotient) algebra of a nilpotent (resp. solvable) Lie algebra is a nilpotent (resp. solvable) Lie algebra.

Proposition 1) The center of a nonzero nilpotent Lie algebra 𝔤 is nonzero.

2) If 𝔤/center(𝔤) is a nilpotent Lie algebra, then 𝔤 is a nilpotent Lie algebra.

Proof a) Let k ≥ 2 be the minimal integer such that 𝔤^{k} = 0. Then 𝔤^{k−1} ≠ 0 and [𝔤^{k−1}, 𝔤] = 𝔤^{k} = 0. Hence 𝔤^{k−1} ⊂ center(𝔤)

b) If 𝔤/center(𝔤) is a nilpotent Lie algebra then any commutator in 𝔤/center(𝔤) of length k (k ≫ 0) is zero:

0 = [[… [a_{1} + e, a_{2} + e], a_{3} + e], …, a_{k} + e]  (10)
= [… [a_{1}, a_{2}], a_{3}], …, a_{k}] + e  (11)

Hence in 𝔤, we have: [… [a_{1}, a_{2}], …, a_{k}] ∈ e, where e denotes the center of 𝔤. Hence any commutator of length k + 1 in 𝔤 is zero, in other words, 𝔤^{k+1} = 0.

Theorem (The Engel Characterization of Nilpotent Lie Algebras) A finite dimensional Lie algebra 𝔤 is nilpotent if and only if ad a is a nilpotent operator for any a ∈ 𝔤.

Proof If 𝔤 is nilpotent, then for k ≫ 0, any commutator of length k is zero, in particular, (ad a)^{k}b = 0, being a commutator of length k + 1. Conversely, suppose (ad a)^{k} = 0 for all a ∈ 𝔤, k ≫ 0. Conside the adjoint representation defined by:

𝔤 → 𝔤𝔩_{𝔤}  (12)
a ↦ ad a  (13)

Then ad 𝔤 ⊂ 𝔤𝔩_{𝔤} is isomorphic to 𝔤/center(𝔤) since center(𝔤) = ker ad . So it suffices to prove that ad 𝔤 is a nilpotent Lie algebra. Now, we can apply Engel’s Theorem: all operators from ad 𝔤 ⊂ 𝔤𝔩_{𝔤} are nilpotent, hence ad 𝔤 ⊂ 𝔫_{𝔤} (the Lie algebra of strictly upper-triangular matrices, for some choice of a basis of 𝔤), and so ad 𝔤 is a subalgebra of a nilpotent Lie algebra, hence is a nilpotent Lie algebra itself. In particular, we see that (ad a)^{dim𝔤} = 0.

Discussion What follows is a discussion of the classification of finite dimensional nilpotent Lie algebras.

Consider only the 2-step nilpotent Lie algebras: 𝔤 ⊃ 𝔢 where 𝔢 = center(𝔤) and 𝔤̄ = 𝔤/𝔢 is abelian (if and only if 𝔤^{3} = 0). We have the following invariant of 𝔤: the map φ,

φ : 𝔤̄ ⊗ 𝔤̄ → e  (14)
φ(ā ⊗ b̄) = [a, b]  (15)

Where a and b are the pre-images in 𝔤 of ā and b̄. The only condition on φ is skew-symmetry:

φ(b̄ ⊗ ā) = −φ(ā ⊗ b̄)  (16)

So we have a bijective correspondence between the set of 2-step nilpotent Lie algebras and the set of skew-symmetric maps (φ : 𝔤̄ ⊗ 𝔤̄ → center(𝔤)).

Exercise 4.3 Show that if 𝔤 is 2-step nilpotent and dim center(𝔤) = 1, then 𝔤 ≈ H_{k}.

Solution By definition, 𝔤 is nilpotent if 𝔤/center(𝔤) is abelian. Therefore, [𝔤, 𝔤] ⊂ center(𝔤). If the dimension of the center of 𝔤 is one, the we can conclude that the derived algebra of 𝔤also has dimension one. Therefore, by a Exercise 2.3, we can conclude that: 𝔤 = 𝔟_{2} ⊕ ab_{n−2} or 𝔤 = H_{k} ⊕ ab_{n−2k−1} where ab_{n} denotes the n-dimensional Lie abelian algebra. However, 𝔟_{2} ⊕ ab_{n−2} is not nilpotent. Therefore 𝔤 = H_{k} ⊕ ab_{n−2k−1}. However, we know that dim(center(𝔤))= 1, hence 𝔤 = H_{k}. □

Remark As discussed in lecture, the next case, dim center(𝔤) is a “wild” problem and the general classification of nilpotent Lie algebras is believed to be impossible, if not very difficult.

Proposition Let 𝔤 be a Lie algebra and let 𝔥 be an ideal of 𝔤. Then 𝔤 is solvable if and only if 𝔤/𝔥 is solvable and 𝔥 is solvable.

Remark This is not true if “solvable” is replaced by “nilpotent”, as the case of 𝔟_{2}, where [a, b] = b, demonstrates.

Proof 𝔤 is solvable, therefore 𝔥 is solvable and 𝔤/𝔥 is solvable. Conversely, if 𝔤̄ = 𝔤/𝔥 is solvable, then 𝔤^{(k)} ⊂ 𝔥. However, if in addition, 𝔥 is solvable (i.e. 𝔥^{(l)} = 0 for some l ≥ 1), then 𝔤^{(k+l)} ⊂ 𝔥^{(l)} = 0, so 𝔤^{(k+l)} = 0. Hence 𝔤 is solvable.