Lie Algebras

Lie Algebras

Tudor Ciurca

January 19, 2020

Contents

1 Course information 2

2 Introduction 3 2.1 Engel’s theorem ...... 6 2.2 Lie’s theorem ...... 8

3 Classification 10

1 1 Course information

Office Hour: Mondays 12:00 - 13:00 Coursework: 2 tests and 1 assessed coursework

All purple text was added post-lecture.

2 2 Introduction

Definition 2.1. A Lie algebra a over a field K is a K-vector space quipped with a bilinear product [, ] satisfying

1. [a, a] = 0 for all a ∈ a

2. [a, [b, c]] + [b, [c, a]] + [c, [a, b]] = 0 for all a, b, c ∈ a

Definition 2.2. A Lie algebra a is called Abelian if [a, b] = 0 for all a, b ∈ a.

Proposition 2.3. Let a be a Lie algebra. Then [, ] is skew-symmetric so that [a, b] = −[b, a] for all a, b ∈ a.

Proof. The result follows directly from

0 = [a + b, a + b] = [a, a] + [b, b] + [a, b] + [b, a] = [a, b] + [b, a]

Definition 2.4. A morphism of Lie algebras f : a → b is a K-linear map satisfying [f(x), f(y)] = f([x, y]) for all x, y ∈ a.

Proposition 2.5. There is a functor Θ: AlgK → LieK for any field K that sends any K-algebra A to its associated Lie algebra Θ(A) with Lie bracket defined by [a, b] = ab − ba for all a, b ∈ A.A K-algebra map becomes a morphism of Lie algebras.

Proof. One firstly verifies that Θ(A) is indeed a Lie algebra

1.[ a, a] = a2 − a2 = 0 for all a ∈ A

2.[ a, [b, c]] + [b, [c, a]] + [c, [a, b]] = a[b, c] − [b, c]a + b[c, a] − [c, a]b + c[a, b] − [a, b]c = abc − acb − bca + cba + bca − bac − cab + acb + cab − cba − abc + bac = 0 for all a, b, c ∈ A.

Suppose f : A → B is a map of K-algebras. Then it is clearly K-linear, and

f([a, b]) = f(ab − ba) = f(ab) − f(ba) = f(a)f(b) − f(b)f(a) = [f(a), f(b)]

Thus f becomes a morphism of Lie algebras.

Example 2.6. 1. The previous proposition gives us a lot of examples. Notable examples include gln(K) := Θ(Gln(K)) for a field K. Many examples are not of this form, but we can get many more as subalgebras of such Lie algebras, as you can see below.

2. sln(K) ⊂ gln(K) defined as the subspace of zero trace matrices. Closure under Lie bracket is easily verified since T r(ab) = T r(ba) for all matrices a, b ∈ gln(K).

T 3. son(K) ⊂ gln(K) defined as the subspace of skew-symmetric matrices. Verify that (ab−ba) = bT aT − aT bT = ba − ab = −(ab − ba)

4. tn(K) ⊂ gln(K) defined as the subspace of upper triangular matrices. This is in fact a subring so clearly it is also a Lie subalgebra with the inherited Lie bracket.

3 Definition 2.7. Let a be a Lie algebra over K. A Lie subalgebra is a subspace b ⊂ a closed under the Lie bracket of a. Clearly b becomes a Lie algebra.

An ideal in a is a Lie subalgebra b ⊂ a satisfying [a, b] ∈ b for all b ∈ b and a ∈ a. The quotient Lie algebra a/b is the corresponding quotient vector space with Lie bracket inherited from a. To prove this is well defined, suppose a − b ∈ b and x − y ∈ b for some a, b, x, y ∈ a. Then

[a, x] − [b, y] ≡ [a, x] − [b, y] + [b, y − x] + [b − a, x] ≡ 0 (mod b)

Remark 2.8. The kernel of a Lie algebra morphism is an ideal. Let φ : a → b be a Lie algebra morphism, and consider the subspace ker(φ) ⊂ a. We verify that for all x ∈ ker(φ) and y ∈ a we have φ[x, y] = [φ(x), φ(y)] = [0, φ(y)] = 0 and so [x, y] ∈ ker(φ) as required.

Definition 2.9. Let a be a Lie algebra, with ideals m, n. We define the Lie subalgebra [m, n] to be the subspace of a spanned by the set {[m, n]: m ∈ m, n ∈ n}. This inherits the Lie bracket of a. It becomes an ideal of a since for all x ∈ m,a ∈ n and t ∈ a.

[t, [x, a]] = −[x, [a, t]] − [a, [t, x]] = [n, x] + [m, a] ∈ [m, n] for n = [a, t] ∈ n and m = [t, x] ∈ m.

Definition 2.10. Let a be a Lie algebra. We define the derived series of a via the following

• a(0) = a

• a(n+1) = [a(n), a(n)]

(n) for all n ∈ N. Each a is an ideal of a by inductively applying Definition 2.9. (n) Definition 2.11. A Lie algebra a is said to be solvable if a = 0 for some n ∈ N. Every abelian Lie algebra is solvable.

Proposition 2.12. Let a be a Lie algebra and [a, a] its derived subalgebra. Then the Lie algebra a0 = a/[a, a] is Abelian.

Proof. For all x, y ∈ a0 we have [x, y] ≡ 0 by definition and so we are done.

Definition 2.13. Let a be a Lie algebra. We define the (lower) central series series of a via the following

• a0 = a

• an+1 = [an, a]

n for all n ∈ N. Each a is an ideal of a by inductively applying Definition 2.9. n Definition 2.14. A Lie algebra a is said to be nilpotent if a = 0 for some n ∈ N. Every abelian Lie algebra is nilpotent, and every nilpotent Lie algebra is solvable. This is clear since a0 = a(0) and if we assume by induction that an ⊃ a(n) then an+1 = [an, a] ⊃ [a(n), a(n)] = a(n+1).

Definition 2.15. Let a be a Lie algebra over a field K. A representation of a is a Lie algebra morphism φ : a → gl(V ) for some vector space V over K.

Example 2.16. We have the adjoint representation ad : a → gl(a) defined by ad(x) = [x, ].

4 Proof. ad is K-linear since ad(kx + y) = [kx + y, ] = k[x, ] + [y, ] for all k ∈ K and x, y ∈ a, by bilinearity of the Lie bracket. Now for all x, y, z ∈ a we show

ad([x, y])(z) = [[x, y], z] = −[z, [x, y]] = [x, [y, z]] + [y, [z, x]] =

= [x, [y, z] − [y, [x, z]] = ([x, ][y, ] − [y, ][x, ])(z) = [ad(x), ad(y)](z)

Definition 2.17. The center of a Lie algebra a over a K is ker(ad) = {z ∈ a :[z, x] = 0 ∀x ∈ a}

Therefore the adjoint representation induces an injective (or faithful) representation of the Lie algebra a/ker(ad) into gl(a). The adjoint representation will be of great use to us.

Definition 2.18. Suppose φ : a → gl(V ) is a representation of a. Let W ⊂ V be a K-subspace which is φ(a)−invariant so that φ(x)(w) ∈ w for all w ∈ W and x ∈ a. Then we get an induced quotient representation φW : a → gl(V/W ). We will also be using this construction often. It is easy to verify that we do indeed get a mor- phism of Lie algebras φW : a → gl(V/W ).

Proposition 2.19. Let b ⊂ a be a Lie subalgebra. Then the following are true

1. If a is solvable, then b is solvable.

2. If a is nilpotent, then b is nilpotent.

Proof. It suffices to prove by induction that b(n) ⊂ a(n) and bn ⊂ an. The base case is precisely the assumption b ⊂ a. Suppose that bn−1 ⊂ an−1 and b(n−1) ⊂ a(n−1). Then bn = [bn−1, b] ⊂ [an−1, a] = an and b(n) = [b(n−1), b(n−1)] ⊂ [a(n−1), a(n−1)] = an so we are done.

We may refine this result when we are dealing with ideals in the solvable case.

Proposition 2.20. Let a be a Lie algebra with an ideal b ⊂ a. Then a is solvable if and only if b and a/b are solvable.

Proof. The forward direction is clear. Let us assume that b and a/b are solvable. Then there (n) is some n ∈ N so that a ⊂ b under the assumption that a/b is solvable. Now there is also (m) some m ∈ N so that b = 0 under the assumption that b is solvable. Finally we see that a(m+n) = (a(n))(m) = b(m) = 0.

Proposition 2.21. Suppose a is a Lie algebra with an ideal b ⊂ a. Then there is a one-to- one inclusion preserving correspondence between ideals containing b and ideals of the quotient Lie algebra a/b.

Proof. An ideal of a containing b naturally yields an ideal of a/b through the quotient. Now take an ideal c ⊂ a/b and define c0 = b + c as a vector space. Then c0 is an ideal since given b ∈ b, c ∈ c and a ∈ a we have [a, b + c] = [a, b] + [a, c] ∈ c0 since [a, b] ∈ b and [a, c] ∈ c. One can now easily check that this correspondence is inclusion preserving.

5 2.1 Engel’s theorem We will be proving our first important result; a theorem of Engel. For this section we make no assumptions on the field K, but we require that every Lie algebra is finite-dimensional.

Theorem 2.22 (Engel). Let a be a Lie algebra. Then a is nilpotent if and only if ad(x) is nilpotent for all x ∈ a.

We gain some inspiration from the following linear algebra results.

Proposition 2.23. Let V be a finite dimensional vector space over a field K, and L a linear transformation on V . If det(L) = 0, then there is an eigenpair (v, 0) for L, where v is some nonzero element of V .

Proof. Suppose not. Then ker(L) = 0 and so L : V → V is injective. By the first isomorphism theorem we get that L is an isomorphism and so det(L) 6= 0, a contradiction.

Proposition 2.24. A linear transformation L on a K-space V of dimension n is nilpotent if and only if there is a basis of V whereby L becomes a strictly upper triangular matrix.

Proof. The converse is clear, so we prove the forward direction. Given a nilpotent linear trans- formation L on V , we form an increasing sequence of subspaces 0 = V0 ⊂ V1 ⊂ ... ⊂ Vn = V as follows. Given Vm, and assuming that it is stable under L, we get an induced linear map Lm on 0 V/Vm, which is still nilpotent. We have det(Lm) = 0 and so there is some eigenpair (vm, 0) for Lm 0 0 for some nonzero vm ∈ V/Vm. Let vm be some lift of vm in V and define Vm+1 = Vm + hvmi. We have Lvm ∈ Vm and so Vm+1 is again stable under L. We let v0 . . . vn−1 be our basis, and we can see that under this basis L becomes a strictly upper triangular matrix.

Lemma 2.25. Suppose we have a faithful representation ρ : a → gl(V ) whose image consists of nilpotent matrices. Here a,V are vector spaces over K. Then the image of the adjoint representation ad : a → gl(a) also consists of nilpotent matrices.

2ny−1 ny Proof. For all x, y ∈ a one can see that ρ([x, ] (y)) = 0 if ρ(y) = 0 for some ny ∈ N. However this is true by the assumptions of the lemma. By faithfulness, [x, ]2ny−1(y) = 0. Choose n to be 2n−1 the largest of the {ny1 . . . nym } for some basis {y1 . . . ym} of a. Then [x, ] (y) = 0 for all y ∈ a and so we are done because [x, ]2n−1 = 0 and so [x, ] = ad(x) is nilpotent.

The main content of Engel’s theorem is contained in the following technical result.

Theorem 2.26. Let ρ : a → gl(V ) be a Lie algebra representation, where both a and V are vector spaces over a field K. Assume that ρ(x) is nilpotent for all x ∈ a. Then there is a nonzero element v ∈ V so that ρ(x)(v) = 0 for all x ∈ a.

Proof. We prove this by induction on the dimension of a. Let us take care of the base case, where m dim(a) = 1. We have some basis {y} of a. Then ρ(y) is nilpotent so ρ(y) = 0 for some m ∈ N, and so det(ρ(y)) = 0. As a result there is an eigenpair (v, 0) for ρ(y) for some nonzero v ∈ V . Thus ρ(ky)(v) = kρ(y)(v) = 0 for all k ∈ K and we are done since hyi = a.

Suppose dim(a) = n and assume the induction hypothesis. We may assume that the represen- tation ρ is injective, otherwise dim(a/ker(ρ)) < n and so we may apply the induction hypothesis to the induced faithful representation ρ0 : a/ker(ρ) → gl(V ). This gives us a nonzero vector v ∈ V so that ρ0(x)(v) = 0 for all x ∈ a/ker(ρ). It follows that ρ(x)(v) = ρ(q(x))0(v) = 0 for all x ∈ a where

6 q : a → a/ker(ρ) is the quotient Lie algebra map. We would then be done.

Observe that we have a faithful representation ρ : a → gl(V ) whose image consists of nilpotent matrices. It follows by Lemma 2.25 that the image of the adjoint representation ad : a → gl(a) also consists of nilpotent matrices. Let b be a maximal proper Lie subalgebra of a. Consider the restric- tion of the adjoint representation ad|b : b → gl(a). By definition of Lie subalgebra, the subspace b ⊂ a is ad|b(b)−invariant and so we get an induced quotient representation (ad|b)b : b → gl(a/b). Now dim(b) < dim(a) and the image of (ad|b)b consists of nilpotent matrices so by the induction hypothesis there is some z ∈ a/b so that (ad|b)b(x)(z) = 0 for all x ∈ b.

0 0 0 Let z be a lift of z in a. Then ad|b(z )(x) = [z , x] ∈ b for all x ∈ b by the construction above. It follows that b + hz0i is a Lie subalgebra of a properly containing b, and so by assumption we must have a = b + hz0i, since b was chosen to be a maximal proper Lie subalgebra.

Now we draw our attention back to gl(V ). Consider the subspace U ⊂ V defined by

U = {v ∈ V |ρ(x)(v) = 0 ∀x ∈ b}

which is nonzero by the induction hypothesis. We show that U is also a-invariant. Firstly ρ(x)(u) = 0 for all u ∈ U and x ∈ b. Now given u ∈ U, we have ρ(z0)(u) ∈ U as long as ρ(x)ρ(z0)(u) = [ρ(x), ρ(z0)](u) = ρ([x, z0])(u) = 0 for all x ∈ b. But this is clear since [x, z0] ∈ b for all x ∈ b by the construction above.

Finally, since we’ve shown that hz0i acts on U via ρ, and ρ(z0) is nilpotent, there must be some non-zero u ∈ U so that ρ(z0)(u) = 0. Since ρ(x)(u) = 0 for all x ∈ b then it follows that ρ(y)(u) = 0 for all y ∈ b + hz0i = a as required.

We will prove the following version of Engel’s theorem first.

Lemma 2.27. Let a ⊂ gl(V ) be a Lie subalgebra for an m-dimensional K-space V . Suppose a consists of nilpotent matrices. Then there is a strictly decreasing sequence of subspaces V = V m ⊃ V m−1 ⊃ ... ⊃ V 0 = 0 so that for all x ∈ a we have xV i+1 ⊂ V i for all i = 0 . . . m − 1.

Proof. We naturally get a faithful representation ρ : a → gl(V ) whose image consists of nilpo- tent matrices. We construct by induction a sequence of strictly increasing spaces 0 = V 0 ⊂ V 1 ⊂ ... ⊂ V m = V so that ρ(x)V i+1 ⊂ V i for all x ∈ a and i = 0 . . . m−1. The base case V 0 = 0 is done.

Given V n, we know that is it invariant under ρ(a) by assumption and so we get an induced quotient representation ρ0 : a → gl(V/V n). We see that the image of a under ρ0 consists of nilpotent 0 n matrices and so we can apply Theorem 2.26 to get some nonzero vn ∈ V/V which we lift to an n n+1 n n+1 n element vn ∈ V which is not in V . Define V = V +hvni. Then it is clear that ρ(x)V ⊂ V for all x ∈ a as required.

Corollary 2.1. Let ρ : a → gl(V ) be a representation whose image consists of nilpotent matrices. Then there is a basis of V whereby ρ(x) becomes a strictly upper triangular matrix for all x ∈ a.

Proof. Suppose V is m-dimensional over K. We may realize the image of ρ as a/ker(ρ) ⊂ gl(V ), and then apply Lemma 2.27 to get a strictly descending sequence of subspaces V = V m ⊃ V m−1 ⊃ ... ⊃ V 0 = 0 so that for all x ∈ a we have ρ(x)V i+1 ⊂ V i for all i = 0 . . . m−1. By the construction i+1 i i+1 in the lemma we see for all i = 0 . . . m − 1 that V = V + hvii for some vi ∈ V . From this

7 we can deduce that ρ(x) is strictly upper triangular in the basis {v0 . . . vm−1} for all x ∈ a as required.

It is now easy to complete the original statement of Engel’s theorem. Consider the adjoint representation ad : a → gl(a). If its image consists of nilpotent matrices, then there is a basis of a where ad(x) is strictly upper triangular for all x ∈ a. If dim(a) = n then one can easily deduce that an−1 = 0 and so a in nilpotent.

m The forward direction is easier. If a is nilpotent then a = 0 for some m ∈ N and so for all x, y ∈ a we have ([x, ]m+1)(y) = 0 since ([x, ]m+1)(y) ∈ am. Then it follows that ad(x)m+1 = ([x, ]m+1) = 0 for all x ∈ a and so ad(x) is nilpotent for all x ∈ a as required.

2.2 Lie’s theorem

In this section we will work over the field K = C. Definition 2.28. Let λ ∈ V ∗ and ρ : a → gl(V ) a representation. Then define V λ(ρ) := {v ∈ V : ρ(x)v = λ(v)v ∀x ∈ a}. This is called the generalized eigenspace for λ. Example 2.29. A simultaneous nonzero eigenvector v ∈ V for a representation ρ : a → gl(V ) gives rise to a function λ : a → C satisfying ρ(x)v = λ(x)v for all x ∈ a. This function is C-linear since given k ∈ C and x, y ∈ a we get λ(kx + y)v = ρ(kx + y)v = kρ(x)v + ρ(y)v = kλ(x)v + λ(y)v and so λ(kx + y) = kλ(x) + λ(y) since v 6= 0. The main content of Lie’s theorem will be contained in the following technical result. Lemma 2.30 (Dynkin). Suppose ρ : a → gl(V ) is a Lie algebra representation and b ⊂ a an ideal. λ ∗ Then V (ρ|b) is a ρ-invariant subspace for all λ ∈ V .

∗ λ λ Proof. Fix λ ∈ V and v ∈ V (ρ|b). Given t ∈ a we must show tv ∈ V (ρ|b). Equivalently we must show ρ(x)ρ(t)v = λ(x)ρ(t)v for all x ∈ b, but the left hand side simplifies to

ρ(x)ρ(t)v = [ρ(x), ρ(t)]v + ρ(t)ρ(x)v = ρ([x, t])v + ρ(t)λ(x)v

and so it suffices to show that λ([x, t]) = ρ([x, t]) = 0, since b is an ideal and so [x, t] ∈ b. As- n sume v 6= 0, since otherwise the result is clear. Consider the sequence of vectors {ρ(t) v : n ∈ N0} i in V . There is a corresponding increasing sequence of subspaces defined by Ui = hv, ρ(t)v . . . ρ(t) vi m for all i ∈ N0. There is a smallest m so that ρ(t) v ∈ Um−1, because V is finite dimensional. We claim that Um−1 is ρ(t)-invariant and ρ(x)-invariant, for all x ∈ b.

Firstly, Um−1 is clearly ρ(t)-invariant by construction. Take x ∈ b, and note that ρ(x)v = j λ(x)v ∈ Um−1. This is the base case. Now assume that ρ(x)ρ(t) v ∈ Um−1 for all j < n, for some n > 0. We have the key calculation X n ρ(([ , t]n)(x))v = ([ , ρ(t)]n)(ρ(x))v = ρ(x)ρ(t)nv + (−1)n−j ρ(t)n−jρ(x)ρ(t)jv j 0≤j

n n n Now ([ , t] )(x) ∈ b because b is an ideal, and so ρ(([ , t] )(x))v = λ(([ , t] )(x))v ∈ Um−1. It follows n that ρ(x)ρ(t) v ∈ Um−1 by applying the inductive hypothesis to the right hand side.

8 By the assumption that b is an ideal, we have that b + hti is a Lie subalgebra of a and that m−1 Um−1 is invariant under this Lie algebra. Under the basis {v, ρ(t)v . . . ρ(t) v} of Um−1, we see that ρ([x, t]) becomes an upper triangular matrix with diagonal entries all equal to λ([x, t]), since

X n ρ([x, t])ρ(t)nv = ρ(([ , t]n)([x, t]))v − (−1)n−j ρ(t)n−jρ([x, t])ρ(t)jv j 0≤j

n−1 X n X ρ([x, t])ρ(t)nv = λ(([ , t]n)([x, t]))v − ( (−1)n−j ρ(t)n−jλ([x, t])ρ(t)jv) + a ρ(t)iv j i 0≤j

Remark 2.31. Dynkin’s lemma is true for arbitrary fields K as long as dim(V ) < char(K) in positive characteristic. This becomes relevant in the last line of the proof above.

Theorem 2.32. Let a ⊂ gl(V ) be a solvable Lie subalgebra. Then there is a nonzero simultaneous eigenvector v ∈ V so that for all x ∈ a we have xv = λ(x)v, for some λ ∈ V ∗.

Proof. We prove this by induction on dim(a). For the base case suppose dim(a) = 1 so that a = hxi for some x ∈ a. Then by the Jordan normal form theorem x has a nonzero eigenvector v ∈ V , which turns out to be an eigenvector for all y ∈ a as a result, and so we are done.

We’ve shown before that a/[a, a] is abelian, and every subspace of an abelian Lie algebra is an ideal. Since a is solvable, we must have a 6= [a, a] and so a/[a, a] 6= 0. Thus we may select a codimension 1 ideal c ⊂ a/[a, a]. This corresponds by via lifting to a codimension 1 ideal c0 ⊂ a.

By the induction hypothesis, there is a nonzero simultaneous eigenvector v ∈ V so that xv = λ(x)v for all x ∈ c0 and some λ ∈ V ∗. We naturally get a faithful representation ρ : a → gl(V ). Then λ 0 0 by Dynkin’s lemma we have that V (ρ|c0 ) is a-invariant. Writing a = c +hxi for some x ∈ a\c , this λ means that V (ρ|c0 ) is nonempty and x-stable and so again by the Jordan normal form theorem λ there is a nonzero eigenvector v ∈ V (ρ|c0 ) for x. This makes v a simultaneous eigenvector for the entirety of a, and we are done.

Theorem 2.33 (Lie). Let ρ : a → gl(V ) a representation of a solvable Lie algebra. Then there is a basis of V where all matrices become upper triangular matrices.

Proof. This is easy now that we have Theorem 2.32. Supposing that dim(V ) = n, it suffices to construct a basis {v0 . . . vn−1} so that the increasing chain of subspaces Ui := hv0 . . . vii are ρ- invariant for all i = 0 . . . n − 1. Supposing that we have constructed such a space Ui, we may apply 0 Theorem 2.32 on the induced faithful representation ρ : a/ker(ρUi ) → gl(V/Ui) to get a nonzero 0 simultaneous eigenvector vi+1 ∈ V/Ui, which we lift to an element vi+1 ∈ V . It is clear now that Ui+1 := Ui + hvi+1i is ρ-invariant. We are done by induction, since the base case is vacuous.

9 3 Classification

Definition 3.1. The Killing form for a Lie algebra a is a bilinear form on a ⊗C a defined by K(x, y) = T r(ad(x) · ad(y))

Remark 3.2. Bilineary is clear from additivity of trace and linearity of the adjoint representation. The Killing form is also symmetric, since for all x, y ∈ a we have

K(x, y) − K(y, x) = T r(ad(x) · ad(y) − ad(y) · ad(x)) = T r([ad(x), ad(y)]) = 0

10