18.745 Introduction to Lie Algebras Fall 2004 Lecture 4: Nilpotent and Solvable Lie Algebras Prof. Victor Kaˇc Transcribed by: Anthony Tagliaferro
Definition Given a Lie algebra g, define the following descending sequences of ideals of g;
− (central series) g ⊃ [g, g] = g1 ⊃ [[g, g], g] = g3 ⊃ [g3, g] = g4 ⊃ · · · ⊃ [gk 1, g] = gk ⊃ . . . (1) 0 0 0 00 − − (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) gk+1 ⊃ g(k)
Solution a) First, let’s show that gk is an ideal of g. Notice that:
[gk, g] = gk+1 ⊂ gk (3)
It therefore follows that gk 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 gk+1 ⊃ g(k). Again, we’ll use induction on k.
g(1) = [g, g] = g2 ⇒ g(1) ⊂ g2 (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 ∈ gk and also notice that b ∈ g. Therefore [a, b] ∈ [gk, g] = gk+1. Hence g(k) ⊂ gk+1 for all k.
Definition A Lie algebra is called nilpotent (resp. solvable) if gk = 0 (resp. g(k) = 0) for k sufficiently large.
1 Remark Notice that it follows from Exercise 4.1b that a nilpotent Lie algebra is also solvable. Examples 1) Abelian Lie algebras are nilpotent. (1) C 2) The Heisenberg Lie Algebras Hk are nilpotent since Hk = and [[Hk, Hk], Hk] = 0 F F (1) F (2) n F 3) b2= a + b, [a, b] = 0 is solvable. since b2 = b, b2 = 0 but it is not nilpotent since b2 = b for n ≥ 2.
Exercise 4.2 Let bk(F) be the subalgebra of glk(F) consisting of all upper-triangular matrices F F and nk( ) be the subalgebra of glk( ) consisting of all strictly upper triangular matrices. Show that: a) bk(F) is a solvable Lie algebra (but not nilpotent) b) nk(F) is a nilpotent Lie algebra (in particular, n3(F) ≈ H1)
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 eij is a basis element of glk(F), then the quantity j − i tells us which diagonal the nonzero entry of eij lies on. Let’s compute the Lie bracket [eij, emn] = δjmein − δniemj. 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 = x e |x ∈ F; j − i ≥ m; i, j ≤ k (8) k,m nX ij ij ij o i,j
Then, we’ve shown that [hk,m, hk,n] ⊂ hk,m+n. Notice also that nk(F) = hk,1 and bk(F) = hk,0. Also notice from the definition that hk,k+1 = 0. Proving that nk(F) is nilpotent is now trivial: F 2 2 nk( ) = hk,1 ⊂ hk,2 F 3 2 nk( ) = [hk,1, hk,1] ⊂ [hk,2, hk,1] ⊂ hk,3 . . F k k nk( ) = hk,1 ⊂ hk,k+1 = 0
Therefore nk(F) is nilpotent. Now, on to proving bk(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 nk(F) ⊂ bk(F) and that nk(F) is solvable (since it’s nilpotent). Also notice that bk(F)/nk(F) = the set of diagonal matrices, which is abelian, hence solvable. Therefore, by the proposition, bk(F) is solvable. Notice that in particular, 0 1 0 0 0 0 0 0 1 F n3( ) = span q = 0 0 0 , p = 0 0 1 , c = 0 0 0 (9) 0 0 0 0 0 0 0 0 0 These matrices exactly satisfy the conditions for the Heisenberg algebra H1. Hence n3(F) ≈ H1.
2 Also, we have proved that bk(F) is solvable, but we have yet to show that is not nilpotent. To demonstrate this, consider the commutator: [eii, eij] = eij for i < j. Since eii is in bk(F), it follows 2 2 3 k that eij ∈ bk(F) , and therefore, eij ∈ [bk(F), bk(F) ] = bk(F) . Hence eij ∈ bk(F) for all k follows by induction. Thus bk(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 gk = 0. Then gk−1 6= 0 and [gk−1, g] = gk = 0. Hence gk−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 = [[. . . [a1 + e, a2 + e], a3 + e], . . . , ak + e] (10)
= [. . . [a1, a2], a3], . . . , ak] + e (11)
Hence in g, we have: [. . . [a1, a2], . . . , ak] ∈ e, where e denotes the center of g. Hence any commutator of length k + 1 in g is zero, in other words, gk+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)kb = 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 → glg (12) a 7→ ad a (13)
Then ad g ⊂ glg 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 ⊂ glg are nilpotent, hence ad g ⊂ ng (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.
3 Consider only the 2-step nilpotent Lie algebras: g ⊃ e where e = center(g) and g¯ = g/e is abelian (if and only if g3 = 0). We have the following invariant of g: the map ϕ,
ϕ : g¯ ⊗ g¯ → e (14) ϕ(a¯ ⊗ ¯b) = [a, b] (15)
Where a and b are the pre-images in g of a¯ and ¯b. The only condition on ϕ is skew-symmetry:
ϕ(¯b ⊗ a¯) = −ϕ(a¯ ⊗ ¯b) (16)
So we have a bijective correspondence between the set of 2-step nilpotent Lie algebras and the set of skew-symmetric maps (ϕ : g¯ ⊗ g¯ → center(g)).
Exercise 4.3 Show that if g is 2-step nilpotent and dim center(g) = 1, then g ≈ Hk.
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 = b2 ⊕ abn−2 or g = Hk ⊕ abn−2k−1 where abn denotes the n-dimensional Lie abelian algebra. However, b2 ⊕ abn−2 is not nilpotent. Therefore g = Hk ⊕ abn−2k−1. However, we know that dim(center(g))= 1, hence g = Hk.
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 b2, where [a, b] = b, demonstrates.
Proof g is solvable, therefore h is solvable and g/h is solvable. Conversely, if g¯ = 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.
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