representation theory of finite dimensional lie algebrastorres/media/lie algebras notes.pdf ·...

Representation Theory of Finite dimensionalLie algebras

July 21, 2015

Contents

1 Introduction to Lie Algebras 51.1 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Definition of the classical simple Lie algebras . . . . . . . . . . 101.3 Nilpotent and solvable Lie algebras . . . . . . . . . . . . . . . 111.4 g-modules, basic definitions . . . . . . . . . . . . . . . . . . . 141.5 Testing solvability and semisimplicity . . . . . . . . . . . . . . 171.6 Jordan Decomposition and

Proof of Cartan’s Criterion . . . . . . . . . . . . . . . . . . . . 201.6.1 Properties . . . . . . . . . . . . . . . . . . . . . . . . . 22

1.7 Theorems of Levi and Malcev . . . . . . . . . . . . . . . . . . 231.7.1 Weyl’s complete irreducibility theorem. . . . . . . . . . 261.7.2 Classification of irreducible finite dimensional sl(2,C)

modules . . . . . . . . . . . . . . . . . . . . . . . . . . 311.8 Universal enveloping algebras . . . . . . . . . . . . . . . . . . 33

2 Representations of Lie algebras 452.1 Constructing new representations . . . . . . . . . . . . . . . . 45

2.1.1 Pull-back and restriction . . . . . . . . . . . . . . . . . 452.1.2 Induction . . . . . . . . . . . . . . . . . . . . . . . . . 46

2.2 Verma Modules . . . . . . . . . . . . . . . . . . . . . . . . . . 492.3 Abstract Jordan Decomposition . . . . . . . . . . . . . . . . . 52

3 Structure Theory of Semisimple Complex Lie Algebras 573.1 Root Space Decomposition . . . . . . . . . . . . . . . . . . . . 583.2 Root Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

3.2.1 Changing scalar . . . . . . . . . . . . . . . . . . . . . . 693.2.2 Bases of root system . . . . . . . . . . . . . . . . . . . 753.2.3 Weyl Chambers . . . . . . . . . . . . . . . . . . . . . . 77

3

4 CONTENTS

3.2.4 Subsets of roots . . . . . . . . . . . . . . . . . . . . . . 803.2.5 Classification of a parabolic subset over a fixed R+(B) . 82

3.3 Borel and Parabolic subalgebras of a complex semi simple Liealgebra. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

4 Highest Weight Theory 854.1 Construction of highest weight modules . . . . . . . . . . . . . 914.2 Character formula . . . . . . . . . . . . . . . . . . . . . . . . . 964.3 Category O . . . . . . . . . . . . . . . . . . . . . . . . . . . . 974.4 Cartan matrices and Dynkin diagrams . . . . . . . . . . . . . 100

4.4.1 Classification of irreducible, reduced root systems/ DynkinDiagrams . . . . . . . . . . . . . . . . . . . . . . . . . 105

Literature

Humphreys:

• Introduction to Lie algebras and Representation Theory

• Complex Reflection Groups

• Representations of semi simple Lie Algebras

• Knapp: Lie groups: beyond an introduction

• V.S. Varadarajan: Lie Groups, Lie Algebras and their Representations

Chapter 1

Introduction to Lie Algebras

(Lecture 1)

1.1 Basic Definitions

Definition 1.1.1. A Lie algebra is a vector space g over some field k, togetherwith a bilinear map [−,−] : g× g→ g such that:

[x, x] = 0∀x ∈ g(anti-symmetry) (1.1)

[x, [y, z]] + [y, [z, x]] + [z, [x, y]](Jacobi identity) (1.2)

Remarks.

• [−,−]is called a Lie bracket.

• ∀x, y ∈ g, [x, y] = −[y, x](because 0 = [x + y, x + y] = [x, x] + [x, y] +[y, x] + [y, y])

• Often: Say a k-Lie algebra if we have a Lie algebra over k.

Examples.

V a k-vector space, [v, w] = 0∀v, w ∈ V defines a k-Lie algebra.

Any associative k-algebra is naturally a k-Lie algebra by

[a, b] := ab− ba∀a, b ∈ A

. Excercise: check 1.2.

5

6 CHAPTER 1. INTRODUCTION TO LIE ALGEBRAS

In particular: The vector space

gl(n, k) = {A ∈ Mn×n(k)}

is a lie algebra via [A,B] = AB−BA (usual commutator of matrices). Moregenerally:

gl(V) = {ρ : V→ Vk − linear};

V a k-vector space, is a lie algebra via [f, g] = f ◦ g − g ◦ f ∀f, g ∈ gl(V).These are the general linear Lie algebras.

Let g1, g2 be Lie algebras, then g1 ⊕ g2 is a Lie algebra via

[(x, y), (x′, y′)] = [[x, x′], [y, y′]](take the Lie algebra component wise)

. The Lie algebra g1 ⊕ g2 is called the direct sum of g1 and g2.

Definition 1.1.2. Given g1, g2 k-Lie algebras, a morphism f : g1 → g2 ofk-Lie algebras is a k-linear map such that f([x, y]) = [f(x), f(y)].

Remarks. • id : g→ g is a Lie algebra homomorphism.

• f : g1 → g2, g : g2 → g3 Lie algebra homomorphisms, then g ◦ f : g1 →g2 is a Lie algebra homomorphism. (because

g ◦ f([x, y]) = g([f(x), f(y)]) = [g ◦ f(x), g ◦ f(y)]).

Hence: k-Lie algebras with Lie algebra homomorphisms form a cate-gory.

Example 1.1.3. Let g be a Lie algebra.

ad :g→ gl(g)

x 7→ ad(x)

where ad(x)(y) = [x, y] ∀x, y ∈ g is a lie algebra homomorphism. It is calledthe adjoint representation. (because: linear clear, since [−,−] is bilinear, and

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

= [x, [y, z]]− [y, [x, z]]Jacobi−id

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

1.1. BASIC DEFINITIONS 7

Definition 1.1.4. Let g be a k-Lie algebra, l ⊂ g a vector subspace.

• l is called a sub-lie algebra if [x, y] ∈ l for all x, y ∈ l.

• l is called an ideal if [x, y] ∈ l for all x ∈ l, y ∈ g.(denoted l C g(⇐⇒ [x, y] ∈ l for all x ∈ g, y ∈ l)).

Given a Lie algebra g, I C g, then the vector space g/I becomes a Liealgebra via [x+ I, y + I] = [x, y] + I.

To check this is well defined:

Assume x+I = x′+I, y+I = y′+I =⇒ x′ = x+u, y′ = y+v, u, v ∈ I =⇒

[x′ + I, y′ + I] = [u+ x+ I, v + y + I] =

[u+ x, v + y] + I = [u, v] + [u, y] + [x, v] + [x, y] + I = [x, y] + I =⇒

well defined.

Proposition 1.1.5. Let f : g1 → g2 be a Lie algebra homomorphism. Then

1 ker(f)C g1.

2 im(f) is a Lie subalgebra of g2.

3 Universal Property: If I C g1, ker(f) ⊂ I, the following diagram com-mutes:

x g1 g2

x+ I ∈ g1/I

@@@@R

-f

?

In particular: g1/ ker(f) ∼= im(f) is an isomorphism of Lie algebras.

Proof. : Standard.

Remark 1.1.6. There are the usual isomorphism Theorems:

a. I, JC g, I ⊆ J, then J/IC g/I and

(g/I)/(J/I) ∼= g/J


b. I, JC g. Then I + JC g; I ∩ JC g, and

I/I ∩ J ∼= I + J/J

Definition 1.1.7. A Lie algebra g is called simple if g contains no idealsIC g except for I = 0 or I = g, and if [g, g] 6= 0.

Remark 1.1.8. [g, g] = 0Def.⇐⇒ [x, y] = 0∀x, y ∈ g (i.e g is abelian).

If g is simple, then:

a. Z(g) = {x ∈ g : [x, y] = 0∀y ∈ g}

b. [g, g] = g(because: Z(g)Cg and g not abelian, hence Z(g) = 0); Z(g)Cg because∀x, y ∈ g, z ∈ Z(g):

[[x, z], y] = [x, [z, y]] + [z, [y, x]]

[g, g] is obviously an ideal, because for x, y, z ∈ g, [x, [y, z]] ∈ [g, g].[g, g] is called the derived lie algebra of g. Also, [g, g] is the smallestideal of g such that g/[g, g] is abelian.

Example 1.1.9.

sl(2, k) := {A ∈ Mat(2× 2, k)|Tr(A) = 0}

Is a lie subalgebra of gl(2, k), becauseTr([A,B]) = Tr([A,B]) = Tr(AB− BA) = 0 ∀A,B ∈ sl(2, k).

Fact 1.1.10. sl(2, k) is simple ⇐⇒ char(k) 6= 2.

Proof. Choose a standard basis (sl2 ”triple”) as follows:

e =(

0 10 0

)f =

(0 01 0

)h =

(1 00 −1

)Then we have (excercise!):

[h, e] = 2e

[h, f ] = −2f

[e, f ] = h

These relations define the lie bracket.If char(k) = 2, then the vector subspace h spanned by h is a non-trivialideal because [e, h] = 0, [f, h] = 0. Since [h, h] = 0, then g = sl(2, k) is notsimple.

1.1. BASIC DEFINITIONS 9

Remark 1.1.11. gl(n, k) is not simple, because 1 spans a non-trivial ideal(Note Z(gl(n, k)) 6= 0).

Lemma 1.1.12. ker(ad(g)→ gl(g)) = {x ∈ g| ad(x) = 0} = {x ∈ g|[x, y] =0∀x, y ∈ g} = Z(g).

Proof. Just definitions.

Corollary 1.1.13. Let g be a simple Lie algebra, then g is a linear Lie algebra(i.e. g is a Lie subalgebra of some Lie algebra of matrices, i.e., of gl(V) forsome vector space V).

Proof. Since g is simple, then Z(g) = 0, which implies ad is injective, henceg ∼= ad(g) is an isomorphism of Lie algebras, and ad(g) is a Linear lie algebra.

Theorem 1.1.14. (Theorem od Ado) Every finite dimensional Lie algebrais linear.

Proof. Later.

Theorem 1.1.15. (Cartan-Killing Classification) Every complex finite-dimensionalsimple Lie algebra is isomorphic to exactly one of the following list:

• Classical Lie Algebras:

An sl(n+ 1,C);n ≥ 1

Bn so(2n+ 1,C);n ≥ 2

Cn sp(2n,C);n ≥ 3

Dn so(2n,C);n ≥ 4

• Exceptional Lie Algebras:

E6

E7

E8

F4


G2

Remark 1.1.16. Every finite dimensional complex Lie algebra which is adirect sum of simple Lie algebras is called semi-simple.

Remark 1.1.17. (Without further explanations){connected, compact Lie group with trivial center

} 1:1←→{

s.s. complex f.d. Lie alg}

G 7→ C⊗R Lie(G)

Where Lie(G) is the tangent space of G at the origin.

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

(Lecture 2- 6th April 2011)

1.2 Definition of the classical simple Lie al-

gebras

• sl(n+ 1,C) = {A ∈ gl(n+ 1,C) : Tr(A) = 0}

• sp(2n,C) = {A ∈ gl(2n,C) : f(Ax, y) = −f(x,Ay)∀x, y ∈ C2n} wheref is the bilinear form given by

f(x, y) := xtMy

where M =(

0 In−In 0

).

Explicitly: (A BC D

)t(0 In−In 0

)+

(0 In−In 0

)(A BC D

)= 0

⇐⇒(−Ct At

−Dt Bt

)=

(C D−A −B

)Thus

sp(2n,C) ={( A skew

skew At

)}=⇒ dim(sp(2n,C)) = n2 + n(n− 1) = 2n2 − n.

1.3. NILPOTENT AND SOLVABLE LIE ALGEBRAS 11

• so(2n+ 1,C) ={x ∈ gl(2n+ 1,C) : XtM + MX = 0

}where

M =

1 0 00 0 In0 −In 0

• so(2n,C) = {X ∈ gl(2n+ 1,C) : XtM + MX = 0} where

M =

(0 InIn 0

)

1.3 Nilpotent and solvable Lie algebras

Definition 1.3.1. • Let A be a k-algebra, then x ∈ A is said to benilpotent if xn = 0 for some n ∈ N.

• Let g be a k-Lie algebra, then x ∈ g is said to be ad-nilpotent ifad(x) ∈ gl(g) = Endk(g) is nilpotent.

Definition 1.3.2. Let g be a Lie algebra. Defineg0 := g, g1 := [g, g], gi := [g, gi−1], andg(0) := g, g(1) := [g, g], g(i) := [g(i−1), g(i−1)].For any Lie algebra g, · · · g1 ⊂ g0 is called the central series and · · · g(1) ⊂ g(0)

the derived series.The Lie algebra g is said to be nilpotent if gi = 0 for some i > 0. It is said

to be solvable if g(i) = 0 for some i > 0.

Remark 1.3.3. g nilpotent =⇒ g solvable (ab−ba = 0∀a ∈ g, b ∈ gi−1 ⊂ g).

Examples. • [gl(n,C), gl(n,C)] = sl(n,C) (Property of trace.)

• [sl(n,C), sl(n,C)] = sl(n,C) =⇒ gl(n,C) and sl(n,C) are neithernilpotent nor solvable.

• N+ := { strictly upper triangular matrices } are nilpotent (the samefor strictly lower ones).

• Diagonal matrices are abelian hence nilpotent and solvable.

• Heisenberg Lie algebra K3(R)


• sl(2, k), char(k) = 2 is solvable.

Proposition 1.3.4. Let g be a Lie algebra. The following statements hold.

1. g nilpotent =⇒ any Lie subalgebra or quotient by an ideal is nilpotent.

2. IC g, g/I nilpotent, I ⊂ Z(g) =⇒ g nilpotent.

3. g 6= {0} nilpotent =⇒ Z(g) 6= {0}.

4. g nilpotent, x ∈ g =⇒ x is ad−nilpotent.

5. IC g =⇒ I(i) C g∀i.

Proof.

1. l ⊂ g Lie subalgebra =⇒ [l, l] ⊂ [g, g] =⇒ ∀i, li ⊂ li =⇒ l isnilpotent. Now, let f : g1 → g2 be a lie algebra homomorphism.

Claim. If g1 is nilpotent then so is f(g1):

We have f(gi1) = f(g1)i; which is clear for i = 0, and for i > 0 we have

f(gi1) = f [g1, gi−11 ] = [f(g1), f(gi−1

1 )] = [f(g1), f(g1)i−1] =induction

f(g1)i.

Hence any homomorphic image of gi is nilpotent, so any quotient is.

2. (g/I) nilpotent =⇒ (g/I)i = 0 for some i ∈ N. Let can : g → g/I bethe canonical projection. Then

(g/I)i = can(g)i = can(gi) =⇒ gi ⊂ I ⊂ Z(g) =⇒ [g, gi] = gi+1 = 0.

3. g nilpotent =⇒ gi = {0}, for some i ∈ N which we choose minimal.Then, [g, gi−1] = 0 so the center is not trivial.

4. gi+1 = 0 for some i ∈ N. Let x ∈ g =⇒ ad(x)i(g) ⊂ gi+1 = {0} =⇒ad(x) is nilpotent.

• I, J C g ideals. Then by the Jacobi identity, [I, J] C g. Now applyinduction on i.

1.3. NILPOTENT AND SOLVABLE LIE ALGEBRAS 13

Proposition 1.3.5. Let V 6= 0 be a finite dimensional k-vector space, andg ⊂ gl(V) a Lie subalgebra, such that g contains only nilpotent elements ofEndk(V). Then:

1. There exists v ∈ V, v 6= 0, such that gv = {0}.

2. There is a chain of k-vector spaces {0} ⊂ V1 ⊂ · · · ⊂ Vd = V wheredim(Vi) = i, such that g(Vi) ⊂ Vi−1.

3. There is a basis of V such that g ⊂ {

0 ∗. . .

0 0

}, where elements of

g are viewed as matrices in this basis. (In particular g is nilpotent).

Proof. We know that if x ∈ g is nilpotent, then ad(x) ∈ Endk(gl(V)) isnilpotent because

ad(x)n(y) =n∑j=0

(nj

)(−1)n−j[xj, y]xn−j

1. We proceed by induction on dim(g). If dim(g) = 0, it’s clear. Choosel ⊂ g a maximal Lie subalgebra. Consider ad : g → g, ad : g/l → g/l.Since ad(l) consists of nilpotent elements, there exists x 6= 0 ∈ g/l suchthat ad(l)(x) = 0, by induction. Hence, there exists x 6= 0 ∈ g− l suchthat [l, x] ⊂ l, thus k · x + l is a Lie subalgebra of g, hence is equal tog by maximality of l. Now consider

W = {w ∈ V : l · v = 0}.

By induction hypothesis, W 6= {0}.Claim. xW ⊂W because if y ∈ l, v ∈W, yxv = xyv + [y, x]v = 0.

Since x is nilpotent, there exists v ∈W such that xv = 0, hence g·v = 0because g = k · x+ l.

2. Consider the inclusion p : g → gl(V). The image consists of nilpotentendomorphisms.

Claim. There exists a flag

V0 ⊂ V1 ⊂ · · · ⊂ Vd = V,

where dim(Vi) = i, such that p(g)(Vi) ⊂ Vi−1.


Induction on dim(V):If dim(V) = 1, this is clear. Now let dim(V) > 1. There exists v ∈V, v 6= 0 such that p(g)v = 0. Let V1 := k · v,V′ = V/V1, considercan : V→ V/V1. Let Vi := can−1(Vi−1) and proceed by induction.

1.4 g-modules, basic definitions

Theorem 1.4.1. (Engel) Let g be a finite dimensionalLie algebra. Then, gis nilpotent if and only if all its elements are ad-nilpotent.

Definition 1.4.2. A representation of a k-Lie algebra g is a k-vector spaceV with a Lie algebra homomorphism ρ : g → gl(V). A representation iscalled irreducible or simple if the only sub vector spaces W of V such thatρ(W) ⊂ V are {0} or V.

Remark 1.4.3. A representation ρ : g → gl(V) is also called a g-module:Alternativel, a g-module is a k-linear map

g× V→ V

(x, v) 7→ x · v

such that x · (y · v)− y · (x · v) = [x, y] · v.

Remark 1.4.4. The representation is called indecomposable if there are nosub vector spaces W,W′ such that V = W⊗W′, ρ(g). Clear: For g-modules,irreducible implies indecomposable, however the other implication doesn’thold!

Theorem 1.4.5. (Lie, abstract form) Every irreducible representation of afinite dimensional complex solvable Lie algebra is 1-dimensional.

Theorem 1.4.6. (Lie, concrete form) Let g ⊂ gl(V) be a solvable linearLie algebra, dim(V) < ∞, V complex vector space. Then, there exists v ∈V, v 6= 0, such that g · v = Cv(i.e. v is a common eigen vector for all theelements in g).

Remark 1.4.7. This is wrong for a general field k. For example, sl(2, k) forchar(k) = 2 is solvable, the 2-dimensional standard representation/vector representationsl(2, k)→ gl(k2) is irreducible but not one dimensional.

1.4. G-MODULES, BASIC DEFINITIONS 15

Proof of Engel. The first implication holds by Proposition 1.3.4. If, for allx ∈ g, ad(x) is nilpotent, then, again by Proposition 1.3.4, g is nilpotentbecause ad(g) ∼= g/Z(g) by isomorphism theorem.

———————————————————————————————Lecture 3, 11/04/2011.

Theorem 1.4.8. (Lie) Let g be a finite dimensional sub Lie algebra of gl(V)for V some finite dimensional vector space. Then there exists v ∈ V, v 6= 0such that g · v ⊂ Cv.

Definition 1.4.9. Let V be a k-vector space and and g ⊂ gl(V) a Liesubalgebra. A linear map λ : g→ k is called a weight for g if

Vλ := {v ∈ V : xv = λ(x)v∀x ∈ g} 6= {0}

and then Vλ is called a weight space.

Lemma 1.4.10. (Invariance Lemma) If char(k) = 0, (e.g k = C as in theTheorem) and g ⊂ gl(V) is a Lie subalgebra, where V is a finite dimensionalvector space, ICg, λ : I→ k a weight for I. Then Vλ is g-stable, i.e. gVλ ⊂ Vλ

(i.e. xv ∈ Vλ∀x ∈ g, v ∈ Vλ).

(Proof: Excercise)

Proof of Lie’s Theorem by induction on dim(g).

dim(g) = 1 Clear!

dim(g) > 1

If g is solvable, then [g, g] ⊂ g. Choose a vector subspace U od codimension1 in g, so g = U⊕ Cz for some z ∈ g (as vector space).

So, U is an ideal of g. Since Lie subalgebras and ideals inherit solvabilitywe get that if U is solvable, then by induction hypothesis, there exists w ∈ Vsuch that Uw ⊂ Cw. Let λ : U → C be the corresponding weight. By theInvariance Lemma (1.4.10), Vλ 6= 0 is g-stable, in particular z-stable, i.e.zVλ ⊂ Vλ. Hence, there exists an eigenvector 0 6= v ∈ Vλ for z. Let µ be thecorresponding eigenvalue. Then v is the vector we are looking for:

x ∈ g = U⊕ Cz, x = u+ βz, u ∈ U, β ∈ C.

Then, xv = (u+ βz)v = λ(u)v + βµv = (λ(u) + βµ)v


Remark 1.4.11. One could weaken the assumption for the field to be C bypassing to a subfield which contains at least all of the occurring eigenvalues.However, char(k) = 0 is required.

Remark 1.4.12. Let ϕ : g1 → g2 a Lie algebra morphism. Then g is solvableif and only if ker(ϕ) and im(ϕ) are both solvable.In particular, for any Lie algebra g:

a. I, JC g solvable ideals, then I + J is again a solvable ideal because

I + J/J ∼= I/I ∩ J

and it follows since then I+J/J and J are solvable. In particular, thereis always a unique maximal proper solvable ideal.

b. Let ICg. Take a 2-dimensional Lie algebra with basis x, y and [x, y] = xsolvable, not nilpotent. Then I := span(x) is a 1-dimensional ideal(in particular nilpotent), and the quotient g/I is also 1-dimensionalnilpotent but g is not.

Corollary 1.4.13. A Lie algebra g is solvable if and only if [g, g] is nilpotent.

Proof. The quotient g/[g, g] is abelian, hence solvable, and [g, g] is nilpotent,hence solvable. Thus g is solvable.If g is solvable, then ad(g) ⊂ gl(g) is solvable. By Lie’s Theorem,

ad(g) ⊂

∗ · · · ∗...

. . ....

0 · · · ∗

After some good choice of basis in g, hence

[ad(g), ad(g)] ⊆

0 · · · ∗

.... . .

...0 · · · 0

and therefore a nilpotent Lie subalgebra. Now, ker(ad|[g,g]) = Z([g, g]), hence,we have a short exact sequence

0→ ker(ad)→ [g, g]→ ad([g, g])→ 0

1.5. TESTING SOLVABILITY AND SEMISIMPLICITY 17

Since ker(ad) is an ideal contained in Z(g), by Proposition 1.3.4, [g, g] isnilpotent.

Definition 1.4.14. Let g be a k-Lie algebra. The maximal solvable ideal(which exists by the previous Remark) is called the radical of g and denotedrad(g). The Lie algebra g is called semi-simple if rad(g) = {0}.

1.5 Testing solvability and semisimplicity

Aim. ”Explicit” criterion for solvability/ semi-simplicity.Motivation: V finite dimensional complex vector space, g ⊂ gl(V) a solvableLie subalgebra. Then,

g ⊆

∗ · · · ∗. . .

0 ∗

(Lie’s Theorem)

So: tr(xy) = 0∀x ∈ g, y ∈ [g, g]. We develop traces to find the criterion wewant.

Definition 1.5.1. Let g be a finite dimensional k-Lie algebra. The Killing formis the bilinear form

K(x, y) := Tr(ad(x) ad(y))

Properties of Killing form:

symmetric K(x, y) = K(y, x) (clear!) (1.3)

invariance K([x, y], z) = K(x, [y, z])(”associativity”) (1.4)

To see this, note that for matrices X,Y,Z we have:

[X,Y]Z = XYZ− Y(XZ)

X[Y,Z] = XYZ− (XZ)Y

In particular:


a. rad(g) = {x ∈ g : K(x, y) = 0∀y ∈ g}C g (use invariance)

b. I C g, I⊥ := {x ∈ g : K(x, y) = 0∀y ∈ I} is again an ideal (use invari-ance).

Remark 1.5.2. The condition 1.4 is called ”invariance” because K corre-sponds to an element K ∈ (g ⊗ g)∗ (= dual vector space of g ⊗ g) which isinvariant under the natural action of g. Here invariant means sent to zerounder the action of g. This makes sense by the following:

Let G be an affine algebraic group, and V a representation. Then it isalso a representation of g = Lie(G). Now, v ∈ V is said to be G-invariant inthe sense that gv = v for all g ∈ G =⇒ g-invariant in the sense that xv = 0for all x ∈ g.

Theorem 1.5.3 (Cartan’s irreducibility criterion). Let k be a field, char(k) =0. Let g be a finite dimensional k-Lie algebra.

g solvable ⇐⇒ K(X,Y) = 0∀x ∈ g.y ∈ [g, g]

Proof. First we need a Lemma:

Lemma 1.5.4. (Cartan criterion gl(V)) Let V be a complex vector space,finite dimensional, and g ⊂ gl(V) Lie subalgebra. Then

g solvable ⇐⇒ Tr(xy) = 0∀x ∈ g, y ∈ [g, g].

Assuming the lemma, we can deduce the theorem (k = C):

K(x, y) = 0∀x ∈ g, y ∈ [g, g] ⇐⇒ Tr(ad(x) ad(y)) ⇐⇒ ad(g) solvable.

Now, we have a short exact sequence of Lie algebras:

0→ Z(g)→ g→ ad g→ 0

So, g is solvable by the above.

Theorem 1.5.5 (Characterization of semi-simple Lie algebras). For g a finitedimensional, complex Lie algebra.Then the following are equivalent:

1 g is isomorphic to a direct sum of simple Lie algebras.

1.5. TESTING SOLVABILITY AND SEMISIMPLICITY 19

2 g is a direct sum of its simple ideals (viewed as Lie algebras.)

3 The Killing form K is non-degenerate.

4 g has no non-zero abelian ideals.

5 g has no non-zero solvable ideals.

———————————————————————————————Lecture 4, 13/04/2011.

Proof of Theorem 1.5.5. First note that the Cartan criterion says that ifKg = 0, then g is solvable.

4 ⇐⇒ 5 Every abelian ideal is solvable. On the other hand, if IC g is solvable,then the last non-zero step in the derived series is an abelian ideal.

Claim. Let g be a finite dimensional Lie algebra (over arbitrary k), andIC g an ideal. Then KI = Kg|I×I where KI,Kg are the killing forms ofI, g respectively.

Proof. Assume U ⊂ g is a vector subspace, and ϕ : g → g is a linearmap. Then Tr(ϕ) = Tr(ϕ|U). Now take U = I and ϕ = ad(x) ad(y) forsome x, y ∈ I. Then ad(x) ad(y)(z) ∈ I.

5 =⇒ 3 First, rad(K) = {x ∈ g : K(x, y) = 0∀x ∈ g}. Now, since K is invariant,then rad(K) C g(x ∈ rad(K), z ∈ g =⇒ K([x, z], y) = K([x, [z, y]] =0)), hence by the Claim, Krad(K) = Krad(K)×rad(K) = 0. By Cartan’scriterion, rad(K) is solvable. Since by assumption there are no non-zero solvable ideals, then rad(K) = 0, so K is non-degenerate.

3 =⇒ 4 Assume {0} 6= I C g is an abelian ideal, then (ad(x) ad(y))2 = 0∀x ∈g, y ∈ I, i.e. ad(y) ad(x) ad(y)(z) = [y, [x, [y, z]]] ∈ [I, I] = 0. Thismeans ad(x) ad(y) is nilpotent, so Tr(ad(x) ad(y)) = 0 for x ∈ g, y ∈ I.Hence K is degenerate, a contradiction!

2 =⇒ 1 =⇒ 4 Clear!

5 =⇒ 2 Let I C g. Then I⊥ := {x ∈ g : K(x, y) = 0∀y ∈ I} is an ideal ofg.(x ∈ I⊥, z ∈ g, =⇒ K([x, z], y) = K(x, [z, y]) = 0) By Cartan’scriterion, I ∩ I⊥ is a solvable ideal, hence zero. On the other hand, Kis non-degenerate (because 5 =⇒ 3), so g = I ⊕ I⊥, and hence every


ideal of I or I⊥ is an ideal of g. By induction on dimension, I, I⊥ satisfy2. Hence g is a direct sum of simple ideals. It is left to show that everysimple ideal occurs:Assume that J ⊂ g is a simple ideal. Then,

J = [J, J] = [J, g] =r⊕i=1

[J, Ii]

where g =⊕r

i=1 Ii is our decomposition into a sum of simple ide-als. This means that, since the Ii are simple, and J is also simple,J = [J, Ii] = Ii for some i. Hence every simple ideal occurs in thedecomposition.

Example 1.5.6. The Killing form for sl(2, k) is given by the following ma-

trix

0 0 40 8 04 0 0

, in the basis e =

(0 10 0

), h =

(1 00 −1

), f =

(0 01 0

).

So, sl(2, k) is semisimple if and only if char(k) 6= 2.

Remark 1.5.7. One can show that, for sl(n,C),

K(x, y) = 2n tr(xy)

and for gl(n,C):

K(a, b) = 2n tr(ab)− 2 tr(a) tr(b)

1.6 Jordan Decomposition and

Proof of Cartan’s Criterion

Lemma 1.6.1. Let V be a finite dimensional C-vector space, and x : V→ Va C-linear endomorphism. Then there exist unique xs, xnC-linear endomor-phisms such that xs is diagonalizable and xn is nilpotent, x = xs + xn andxsxn = xnxs. Moreover, every subspace of V stabilized by x is stabilized byxs and xn.

1.6. JORDANDECOMPOSITION ANDPROOFOF CARTAN’S CRITERION21

Example 1.6.2. i.(λ 10 λ

)=

(λ 00 λ

)+

(0 10 0

)x =xs+ xn

ii. (1 20 3

)=

(1 00 3

)+

(0 20 0

)x =xs+ xn

The latter is not a Jordan decomposition because xn and xs don’tcommute.

Proof. Let a1, · · · , ak be the distinct eigenvalues of x, so x(t) =∏k

i=1(t−ai)mifor some mi ∈ N. Let V(x, ai) := ker(x− ai)mi be the generalized eigenspaceof x corresponding to ai. Then,

V =k⊕i=1

V(x, ai). (1.5)

Hence, by the Chinese remainder Theorem, there exists P(t) ∈ C[t] a poly-nomial satisfying:

P(t) = ai(mod(t− ai)mi); 1 ≤ i ≤ r

P(t) = 0 mod(t)

Define

xs :=P(x)

xn :=Q(x)

where Q(t) = t− P(t) ∈ C[t].

Now, xs and xn commute: in fact they commute with whatever x com-mutes with. By construction, xs − ai|V(x,ai) = 0, hence xs is diagonalizablewith basis given by the decomposition 1.5 as above, so xn = x− xs must benilpotent. This proves existence.


To prove uniqueness, assume there exist two such decompositions of x, namelyx = xs + xn and x = x′s + xn′ . We then have:

xs − x′s = x′n − xn (1.6)

The right hand side of (1.6) is nilpotent and the left hand side is diagonaliz-able, hence both sides must be zero, and uniqueness is proven.

1.6.1 Properties

1) Functoriality: Assume the following diagram commutes in the categoryof finite dimensional C-vector spaces.

V W

V W?

x

-f

?

y

-f

Then the two following diagrams commute:

V W

V W?

xs/xn

-f

?

ys/yn

-f

Proof. f ◦x = y◦f implies that f(V(x, λ)) ⊂ V(y, λ). Then go throughthe definition of xs, xn.

2) Im(xs) ⊆ im(x). (This holds since xs = P(x), and P(0) = 0. Thismeans xs(v) = x ◦ r(x)(v) for some r(t) ∈ C[t]).

Lemma 1.6.3. If x = xs+xn is the Jordan decomposition of x, then ad(xs)+ad(xn) is the Jordan decomposition of ad(x).

Remark 1.6.4. This lemma is crucial for defining the ”abstract” Jordandecomposition in a complex finite dimensional Lie algebra g.

1.7. THEOREMS OF LEVI AND MALCEV 23

Proof. We have [ad(xs), ad(xn)] = ad([xn, xs]) = 0, so ad(xs), ad(xn) com-mute. Further, ad(xn) is nilpotent because xn is nilpotent. To see that ad(xn)is diagonalizable, choose an eigenbasis v1, · · · , vk of V for xs, with eigenvaluesxs(vi) = λivi for all vi ∈ Vi. Then Eij is an eigenvector for ad(xs) with eigen-value λi − λj. (Notation: Eij is the ij-matrix unit , i.e. (Eij)kl = δikδjl)

Lemma 1.6.5. Let V be a finite dimensional C-vector space. Let A ⊂ B ⊂End(V) be vector subspaces, and let

T := {z ∈ End(V) : ad(z)(B) ⊂ A}.

Then, if Tr(xz) = 0 for all z ∈ T, then x is nilpotent.

(Proof later!)

Proof of Cartan’s Criterion (Lemma 1.5.4). Assume Tr(xy) = 0 for all y ∈g ⊂ gl(V), x ∈ [g, g]. It is enough to show that [g, g] is nilpotent. We useLemma 1.6.5 in the following way: Take x ∈ [g, g]. We show that (Tr)(xy) =0 for all y ∈ gl(V) such that [y, g] ⊂ [g, g]. Write x =

∑ni=1[xi, yi];xi, yi ∈ g.

Take y ∈ gl(V) as above. Then:

Tr(xy) =n∑i=1

Tr([xi, yi]y) =n∑i=1

Tr(xi[yi, y]) = 0 (by hypothesis.)

1.7 Theorems of Levi and Malcev

Aim. Let g be a finite dimensional Lie algebra. Then rad(g) ⊕ l = g asvector spaces for some semisimple Lie algebra l, and g = rad(g) n l as Liealgebras. Then l is called the Levi compliment of g. It is unique up to someinner automorphism.

Definition 1.7.1. Let g be a k-Lie algebra. We define

Der(g) := {Derivations of g} = {δ ∈ Endk(g) : g→ g : δ([a, b]) = [δ(a), b] + [a, δ(b)]}

It is a Lie subalgebra of Endk(V).

Example 1.7.2. Let x ∈ g. Then ad(x) : g → g is a derivation. (Jacobiidentity!)


Remark 1.7.3. ad(g) := {ad(x) : x ∈ g}CDer(g)

Definition 1.7.4. Assume x ∈ g such that ad(x) is nilpotent, and char(k) =0. Then

exp(ad(x)) :=∑k≥0

(adx)k

k!∈ Aut(g) := {f ∈ Endk(g); f invertible }

An automorphism f ∈ Aut(g) is called inner, if it is contained in the subgroupgenerated by the exp(adx), x ∈ g as above.

Proposition 1.7.5. Let g be a finite dimensional semi-simple Lie algebraover C. Then Der(g) = ad(g)

Proof. Since g is semi-simple, Z(g) = 0. Thus g = ad(g) ⊕ Z(g) ∼= ad(g) asLie algebras. Let I C Der(g) an ideal of Lie algebras. Consider the Killingform KI = KI×I. In particular, I ∩ I⊥ = 0 so [I, I⊥] = 0. Now, let δ ∈ I⊥,hence we have for x, y ∈ g:

0 = [δ, ad(x)](y) = δ([x, y])− [x, δ(y)] = [δx, y]

Hence ad(δ(x)) = 0 for all x ∈ g, hence δ = 0. This means there are noproper ideals, hence ad(g) ∼= Der(g) as Lie algebras.

Definition 1.7.6. Let g be a finite dimensional k-Lie algebra. If

g = I⊕ l

as vector spaces, where I is an ideal and l is a Lie subalgebra, then g is calledthe semi direct product of I and l, and is denoted by

g = In l

If such a decomposition exists, then

α :l→ Der(I)

x 7→ ad(x)|I

(check!)


Conversely, assume that we are given α : l → Der(h) a Lie algebra ho-momorphism for some k-Lie algebras h, l. Then g := h ⊕ l becomes a Liealgebra, via:

[(x, y), (x′, y′)] = (α(y)(x′)− α(y′)x+ [x, x′], [y, y′])

For α = 0 this is just the direct sum of Lie algebras.

Lemma 1.7.7. Let g be a finite dimensional Lie algebra. Then g/ rad(g) issemi-simple.

Proof. We show that rad(g/ rad(g)) is zero. Let I C g/ rad(g) be a solvableideal. Consider

can :g→ g/ rad(g).

Set J := can−1(I) C g, an ideal containing rad(g). We have the followingshort exact sequence:

0 rad(g) J J/ rad(g) 0- - - -

Hence, J must be solvable, hence J ⊂ rad(g), so J = rad(g), so there is nonon-trivial solvable ideal.

Theorem of Levi 1.7.8. Let g be a finite dimensional Lie algebra. If α :g� l is a Lie algebra homomorphism, surjective, and l is semi-simple, thenthere exists β : l→ g, a Lie algebra homomorphism such that α ◦ β = Id |l.

Corollary 1.7.9. Any finite dimensional Lie algebra g is a semi-direct prod-uct of it’s radical and a semi-simple Lie algebra; more precisely,

g = rad(g)n g/ rad(g)

where can = α : g→ g/ rad(g).

Proof of Corollary 1.7.9. The map can : g � g/ rad(g) is a surjective Liealgebra homomorphism. Therefore there exists a Lie algebra homomorphismβ : g/ rad(g) → g as in Levi’s Theorem, hence β(l) ⊂ g is a Lie subalgebra,and

rad(g)⊕ β(l) = g

as vector spaces.


Theorem of Malcev 1.7.10. For two Levi complements l and l′ of g, thereexists some x ∈ [g, rad(g)], ad(x) nilpotent, such that exp(ad(x))(l) = l′.

Proof: Omitted.

Remark 1.7.11. Lie complements are precisely the maximal semi-simpleLie subalgebras (Exercise).

1.7.1 Weyl’s complete irreducibility theorem.

Let g be a k-Lie algebra. Let X,W be k-vector spaces, and let

ϕ : g→ gl(V)

ψ : g→ gl(W)

be two representations of Lie algebras.

Constructions of ’new’ representations

• Homk(V,W) becomes a g-module via:

x · f(v) = x(fv)− f(xv)

(Check: [x, y] · f = x · y · f − y · x · f) In particular, if W = k, x · λ = 0for all λ ∈ k, x ∈ g. Then, Homk(V, k) = V∗ is a g-module via:

x · f(v) = −f(x · v)

• V ⊗k W is a g-module via

x · v ⊗ w = x(v ⊗ w) = x · v ⊗ w + v ⊗ x · w

For x ∈ g, v ∈ V, w ∈W.

Remark 1.7.12.

Homg(V,W) = {f ∈ Homk(V,W) : f(x · v) = x · f(v)}Homk(V,W)g = {f ∈ Homk(V,W) : x · f = 0∀x ∈ g}

The elements in Homg(V,W) are called g-module homomorphisms ormorphisms of representations from V to W.


Schur’s Lemma 1.7.13. Let V be a finite dimensional complex vector spaceand g a finite dimensional complex Lie algebra. Assume V is a g-module,irreducible (i.e. there is no vector subspace stable under the action of g).Then, Endg(V) := Homg(V,V) = C Id.

Proof: excercise

Remark 1.7.14. It also holds in case dim(V) is countable.

Remark 1.7.15. Schur’s Lemma doesn’t hold for k = R. Let g = R (withzero Lie bracket), and V = C: Consider the representation

R = g→ gl(V) = EndR(C)

λ 7→ multiplication by λi

This representation is irreducible (check!). However,

EndR(V) = C id 6= R id

Hence in this case Schur’s Lemma doesn’t hold.

Definition 1.7.16. Let g be a k-Lie algebra, and V a g-module. Then,V is completely reducible if it is isomorphic to a direct sum of irreducibleg-modules.

———————————————————————————————Lecture 5, 20/04/2011.

Today: Proof of Weyl’s Theorem (For k = C) only.

Theorem 1.7.17 (Weyl’s Theorem). Let g be a semisimple complex finitedimensional Lie Algebra. Then every finite dimensional representation of gis completely reducible.

Lemma 1.7.18. For a finite dimensional Lie algebra and V a finite dimen-sional representation, the following are equivalent:

• Every subrepresentation W of V has a complement.

• The representation V is completely reducible.

• V is a sum of irreducible representations.


Proof: Omitted.

Example 1.7.19. Consider g = C, with [x, y] = 0 for all x, y ∈ C. Therepresentation

g = C→ gl(C2)

λ 7→(

0 λ0 0

)is not completely reducible. In general, the representation

k → gl(V)(dimk(V) <∞)

1 7→ a

is completely reducible if and only if a is diagonalizable.

Definition 1.7.20. A finite dimensional Lie algebra is called reductive if theadjoint representation is completely reducible.

By Weyl’s Theorem, if a finite dimensional Lie algebra g is semisimple,then it is also reductive. The other direction doesn’t hold:

Example 1.7.21. The Lie algebra g = gl(n,C) is reductive, but not semisim-ple.

The Casimir Operator

Definition 1.7.22. Let g be a finite dimensional Lie algebra. Let β : g ×g → k be a non-degenerate invariant symmetric bilinear form. Let V bea representation of g, and ϕ : g → gl(V). Let {x1, · · ·xn} be a basis ofg. Let {x1, · · · , xn} be the ’dual’ basis with respect to β, i.e. such thatβ(xi, x

j) = δij.

Then define Cβ ∈ Endk(V) by

Cβ : V→ V

v 7→n∑i=1

xixiv

Lemma 1.7.23. Cβ ∈ Endg(V)


Proof. Let

aij(y) = β([xi, y], xj);

[xi, y] =n∑j=1

aij(y)xj

bji(y) = β(xi, [y, xj]);

[y, xi] =n∑i=1

bji(y)xi

Then,

yCβ(v)− Cβ(yv) =n∑i=1

(yxixi − xixiy)v;

(yxixi − xixiy) = [y, xi]x

iv − xi[xi, y]v−

Remark 1.7.24. If β is the Killing form, then Cβ is called the Casimir operator.

Lemma 1.7.25. Let V be a finite dimensional vector space, g a finite di-mensional Lie algebra, g ⊆ gl(V) a Lie subalgebra, char(k) = 0. If g issemisimple, then

1 The form β defined by (x, y) 7→ tr(x, y) is a non-degenerate symmetricbilinear form (a ’trace form’).

2 Let C = Cβ. Then Tr(C) = dim(g)

Proof. 1 The form β is clearly symmetric and invariant. Then rad(β) isa solvable ideal in g. Since g is semisimple, rad(β) = {0}. Hence, β isnon-degenerate.

2 This is the definition of Cβ.

Example 1.7.26. Let g = sl(2,C), and take the standard basis {e, h, f}.Then the dual basis with respect to the trace form is:

(f,h

2, e) =⇒ Ctrace form = ef +

1

2h2 + fe =

(32

00 3

2

)∈ Endg(C2)


In contrast:

CK =1

4(ef + fe) +

h2

8=

1

4(Ctrace form)

Lemma 1.7.27. Let g be a semisimple finite dimensional Lie algebra overC, and let V be a finite dimensional representation. Then,

V = Vg ⊕ gV

where Vg = {x|gx = 0∀g ∈ g}Proof. By induction on dim(V). Without loss of generality, assume V 6=Vg, where ϕ : g → gl(V) is our representation. Now, ϕ(g) ⊂ gl(V) isa Lie subalgebra; it is semisimple since it is the image under a Lie algebrahomomorphism of a semisimple Lie algebra. Consider C = Ctrace form (as inLemma 1.7.25 above). Then V decomposes into a direct sum of generalizedeigenspaces for C (here k = C is being used). Since C ∈ Endg(V), thesegeneralized eigenspaces are g subrepresentations. Hence, if there is morethan one, we can decompose V = V1 ⊕ V2 and the statement follows byinduction. Hence we may assume there is only one generalized eigenspace.Since Tr(C) = dim(g) 6= 0, the eigenvalue is not zero. Hence, V = CV andVg = {0}. Hence, V = gV = gV ⊕ Vg and we are done.

Proof of Weyl’s Theorem (Theorem 1.7.17). Let U ⊆ V be a subrepresenta-tion for a given finite dimensional representation V of g.

to show: U has a complement which is again a representation.

Remark 1.7.28. Weyl’s Theorem doesn’t hold for infinite dimensional rep-resentations.

Example 1.7.29. Let g = sl(2,C), V = C[x].

e 7→ d

dx

h 7→ −2xd

dx

f 7→ −x2 d

dxCheck: This defnes a homomorphism of Lie algebras. This is an irreducibleinfinite dimensional representation.


1.7.2 Classification of irreducible finite dimensional sl(2,C)modules

Theorem 1.7.30.

1. Finite dimensionalirreducible sl(2,C)

modules

/ ∼= 1:1←→ N

V 7−→ dim(V)− 1

V(n)←− [ n

2. With respect to the standard basis {e, f, h}, V(n) decomposes into onedimensional eigenspaces.

3.

V(n)⊗ V(m) ∼= V(n+m)⊕ V(n+m− 2)⊕ · · · ⊕ V(m− n)

e.g. V(n)⊗ V(1) ∼= V(n+ 1) where V(1)is the two dimensional vectorrepresentation.

Remark 1.7.31. The multiplicities [V(n)⊗V(m) : V(j)], which denote thenumber of times that V(j) appears as a summand in V(n)⊗V(m), are calledthe Clebsch-Gordon coefficients.

Example 1.7.32.

V(0) = trivial representation, = C, with x · λ = 0

V(1) = vector representation

V(2) = adjoint representation

Proof of Theorem 1.7.30.

Claim. Two irreducible sl(2,C)-modules of the same dimension are isomor-phic.

Let V be a finite dimensional sl(2,C)-module. For µ ∈ C, let

Vµ := {v ∈ V : h · v = µv}


be the µ- eigenspace for the fixed basis element h ∈ sl(2,C).Then we have

eVµ ⊆ Vµ+2

fVµ ⊆ Vµ−2

Choose µ such that Vµ 6= {0}, and Vµ+2 = 0. Fix v ∈ Vµ − {0}. Then:

hf iv = (µ− 2i)f iv∀i ≥ 0

ef iv = i(µ− i+ 1)f i−1v∀i ≥ 0

(this can be checked by induction on i)Hence, the span of {v, fv, · · · f rv · · · } is a subrepresentation of V, so equalto it, by irreducibility of V.

Choose d ∈ N minimal such that fdv = 0 (Vis finite dimensional). Then,

{f iv : 1 ≤ i ≤ d}

is a basis of V. Further,

0 = efdv = d(µ− d+ 1)fd−1v =⇒ µ = d+ 1

Hence, µ is determined by the dimension, hence uniqueness.

For existence: Let V = k[x, y]. Consider the representation given by

e 7→ xd

dy

h 7→ xdx− ydy

f 7→ yd

dx

Check: this defines an infinite dimensional representation.Consider Ud = subspace spanned by all monomials of degree d; i.e.

Ud := {xαiyωi0 ≤ i ≤ d;αi + ωi = d}

1.8. UNIVERSAL ENVELOPING ALGEBRAS 33

Then Ud := V(d) is invariant under sl(2,C) action, and dim(Ud) = d + 1.We further have the formulas:

eωi = iωi−1

fωi = (d− i)ωi+1

hωi = (d− 2i)ωi

For 3., look at the dimension of the h-eigenspaces.

———————————————————————————————Lecture 6, 27/04/2011.

Theorem 1.7.33 (Levi). Let α : g � l a surjective Lie algebra homomor-phism over C, where l is semisimple. Then there is a split

β : l→ g

i.e. a Lie algebra homomorphism such that α ◦ β = idl

1.8 Universal enveloping algebras

Let g be any k-Lie algebra.

Definition 1.8.1. A universal enveloping algebra of g is a pair (U(g), can),where U(g) is a k-algebra, associative, with 1, and can : g → U(g) is a Liealgebra homomorphism such that it satisfies the following universal property:Given a commutative diagram

g A

U(g)?

can

-ϕ

p p p p pp p�∃! unitary alg. homo

Examples. • g = {0}, then U(g) = k

{0} A

k?

-

p p p p pp p p�1 7→1


• g = k, with trivial/ zero Lie bracket. Then U(g) = k[x]:

g A

k[x]?

17→x

-ϕ

p p p p pp p�x 7→ϕ(1)

Remark 1.8.2. Universal enveloping algebras are unique up to unique iso-morphism of algebras. In fact. If g is any k-Lie algebra, and U(g),U′(g) areboth universal enveloping algebras, then we have a commutative diagram:

U(g)

g U′(g)

U(g)

pppppppp?∃!α

��

can

-can′

@@@@R

can

pppppppp?∃!β

Then β ◦ α = id; α ◦ β = id. Hence α, β are isomorphisms.

Aim. g-modules are the same as U(g)-modules

Proposition 1.8.3. Let g be a k-Lie algebra. Let V be a k-vector space.Then {

U(g)-modulestructures onV

}1:1←→

{g-module

structures onV

}Proof. Given a U(g)-module structure on V, i.e. an algebra homomorphism

ϕ : U(g)→ Endk(V),

Then ϕ is in particular a Lie algebra homomorphism. We get thus an inducedmap

−ϕ : g

can→ U(g)ϕ→ Endk(V) = gl(V)


which is a composition of Lie algebra homomorphisms, hence a Lie alge-bra homomorphism; and this defines a g-module structure on V.Now, given a g-module structure φ : g → gl(V) on V, we get, by the uni-versal property of the universal enveloping algebra, a unique unitary algebra

homomorphism∼φ : U(g)→ gl(V ) such that the following diagram commutes:

g gl(V)

U(g)?

can

-φ

p p p p pp p p�∃!∼φ

In particular∼φ defines a U(g)-module structure on V. Ir follows from the

universal property that∼−ϕ = ϕ and

−∼φ = φ.

Corollary 1.8.4. There is an equivalence (even isomorphism) of categories

{U(g)-modules} ∼= {g-modules}

Proof. The corollary is clear on objects by Proposition 1.8.3. On morphisms,it’s just the identity.

Remark 1.8.5. Let g be any k-Lie algebra. Then k is a 1-dimensional(trivial) representation, via

x · λ = 0 ∀x ∈ g, λ ∈ k

This gives a U(g) module, i.e. a unitary algebra homomorphism

U(g)→ Endk(k)

This map is called augmentation map or counit. In fact, U(g) is a Hopfalgebra (non-commutative) in general.

———————————————————————————————Lecture 7, 04/05/2011.

Theorem 1.8.6. Every k-Lie algebra g (over any field k) has a universalenveloping algebra U(g).


Def/Prop 1.8.7. Let V be a k-vector space. Then the tensor algebra TkVis the k-algebra defined as the k-vector space:

TkV := k ⊕ V ⊕ (V ⊗ V)⊕ (V ⊗ V ⊗ V) · · · =⊕r≥0

V⊗r

with V⊗0 := k and multiplication given by

(v1 ⊗ · · · ⊗ vr)(w1 ⊗ · · · ⊗ zs) := v1 ⊗ · · · ⊗ vr ⊗ w1 ⊗ ws ∈ V⊗(r+s)

Extending this linearly turns TkV into an associative, unitary k- algebra.

Proof. Multiplication is well defined. In fact, if λ ∈ k = V⊗0, then define

λ · (w1 ⊗ · · ·ws) = λw1 ⊗ · · ·ws = (w1 ⊗ · · · ⊗ ws)λ,

and for λ, µ ∈ k,

λ · µ = λµ = µ · λ

hence multiplication is well defined.To see that TkV is in fact unitary, let 1 ∈ V⊗0 = k. This is the unit, by theabove multiplication rules.

Proposition 1.8.8 (Universal property of TkV).Let V be a k-vector space. Let TkV be the tensor algebra. Let A be anassociative unitary k-algebra, ϕ : V → A a k-linear map, and j : V → TkVthe embedding into the 2nd summand. Then there exists a unique homo-morphism of unitary algebras

∼ϕ : TkV → A making the following diagram

commute:

V TkV

A?

ϕ

-j ppppppppp ∼ϕ

Proof. The algebra TkV is generated as unitary algebra by elements v ∈V, 1 ∈ k, hence uniqueness (once existence).To show existence, define

∼ϕ(v1 ⊗ · · · ⊗ vr) = ϕ(v1) · · ·ϕ(vr)

and extend linearly. This is the required homomorphism.


Let now V := g be the k-vector space underlying a k-Lie algebra g.Consider Tk(g). Let I be the ideal generated by all the elements of the formx⊗ y − y ⊗ x− [x, y] ∈ Tk(g) for all x, y ∈ g. Then define

U(g) = Tk(g)/I

and

can : gj→ Tk(g)

π→ Tk(g)/I = U(g)

where π is the canonical projection.

Theorem 1.8.9. Given g a k-Lie algebra, (U(g), can) is it’s universal en-veloping algebra.

Proof. Let A be a k-algebra, associative and unitary. Let ϕ : g → A be aLie algebra homomorphism. We have the following maps which make thefollowing diagram commute:

g U(g) = Tk(g)/I

A Tk(g)?

ϕ

-can

QQQQQQs

j

6p

pppppppppppp�∃!∼ϕ

Claim. The unitary algebra homomorphism∼ϕ factors through U(g) (i.e.

maps I to zero).

Note that once the claim is shown we have the desired induced morphism−ϕ : U(g)→ A which marked the diagram above still commute. Now, we havefor x, y ∈ g,

∼ϕ(x⊗ y − y ⊗ x− [x, y]) = ϕ(x)ϕ(y)− ϕ(y)ϕ(x)− ϕ([x, y]) = 0.

Hence the claim is shown. We still have to check that can is a Lie algebrahomomorphism. Let x, y ∈ g, then


can([x, y]) = p ◦ j([x, y]) = p([x, y]) = p(x⊗ y − y ⊗ x)

= p(x)⊗ p(y)− p(y)⊗ p(x) = [can(x), can(y)]

Poincare-Birkhoff-Witt Theorem 1.8.10 (concrete version). Let g bea k-Lie algebra, Let {xi : i ∈ I} be an ordered basis indexed by a totally

ordered set (I, <). Then the monomials−xi1 · · ·

−xir for r ∈ N with

−xi =

can(xi) ∈ V⊗1 = g where i1 ≤ · · · ≤ ir (For r = 0 there is a unique monomialdenoted by ∅ or 1) form a k-basis of U(g)

Before the proof, we will state the Theorem in a more ’elegant’ way. Forthis, we need the following:

Definition 1.8.11. Let A k-algebra. A filtration of A is a sequence A≤0 ⊆· · · ⊆ A of vector spaces such that:

1. A =⋃i≥0

A≤i

2. A≤iA≤j ⊆ A≤i+j

If A has a filtration, then it is called afiltered algebra.

Definition 1.8.12. Let A be a filtered algebra. Then the multiplication inA induces a map

A≤i/A≤i−1 ⊗ A≤j/Aj−1 → A≤(i+j)/A≤(I+j−1)

Hence, the associated graded algebra defined as

gr(A) = ⊕1≥0A≤i/A≤(i−1)

becomes an associative k- algebra with multiplication by the above inducedmap.

Remark 1.8.13. The associated graded algebra gr(A) = ⊕i≥0Ai, (Ai =A≤i/A≤(i−1)) is a graded algebra.


Example 1.8.14. Consider A= k[X], the polynomial algebra in one variable.Define A≤i to be the span of all homogeneous polynomials of degree less orequal that i, i.e.

A≤i = {1, x, · · ·xi}

Then A≤iA≤j ⊆ A≤(i+j) and A =⋃i≥0

A≤i and this defines a filtration of A.

Note: In the remark above, Ai denotes the polynomials of degree i, andA = Ai

i≥0, AiAj ⊆ A(i+j)

Remark 1.8.15. Given a graded algebra, i.e. A = Aii≥0

; Ai vector subspaces

such that AiAj ⊆ A(i+j), then A≤j := Ai0≤i≤j

defines a filtration on A.

Definition 1.8.16. Let V be a k-vector space, TkV it’s tensor algebra. De-fine Sk(V) := TkV/J, where J is the ideal generated by x ⊗ y − y ⊗ x, forx.y ∈ V. This is called the symmetric algebra associated to V (It’s commu-tative!).

Exercise 1.8.17. If dim(V) = n < ∞, then S(V) ∼= k[x1, · · · , xn] as k-algebra. In particular, it has a natural filtration as in Example 1.8.14.

Now, the more ’elegant’ statement.

Theorem 1.8.18 (Poincare-Birkhoff-Witt ’abstract version’). If Tk(g) isfiltered by T≤i = span{x1 ⊗ · · · ⊗ xj : j ≤ i} and U(g) is filtered byp(Tk(g)≤i) = U(g)≤i then the following two maps have the same kernel:

Sk(g) = Tk(g)proj←− Tk(g) = gr(Tk)(g)→ gr(U(g))

In particular, gr(U(g)) ∼= Sk(g).

Proof of concrete version.

1. The monomials−xi1 · · ·

−xir ; i1 ≤ · · · ≤ ir span U(g) = T(g)/I.

It is hence generated as an algebra by elements from g, hence as vec-

tor space by elements−xi1 · · ·

−xir ; i1 ≤ · · · ≤ ir (with not necessarily


increasing indices) where we write short xy for x⊗ y. Using this nota-tion, since, in U(g), xy − yx = [x, y] for all x, y ∈ p(g) ⊂ U(g). Thismeans that for a permutation σ ∈ Sr such that σ(i1) ≤ · · · ≤ σ(ir), wehave

xi1 · · ·x1r = xσ(i1) · · · xσ(ir) + rest;

where rest ∈ U(g)≤(r−1). Now, for r = 0 we have ∅ = 1 is in the basisand every element in U(g)≤0 is of the form λ1 for λ ∈ k. We are thendone by induction.

2. The monomials−xi1 · · ·

−xir ; i1 ≤ · · · ≤ ir are linearly independent.

Main idea: Construct a representation U(g)ϕ−→ Endk(S) where S =

k[xi : i ∈ I] such that ϕ(−xi1 · · ·

−xir) i1 ≤ · · · ≤ ir for i1 ≤ · · · ≤ ir are

linearly independent endomorphisms. (Proof of this next time!)

Interesting consequence of PBW

Corollary 1.8.19. Assume a k-Lie algebra g may be written as g = n⊕ b,where the sum is as vector spaces, and assume n, b are both Lie subalgebrasof g. Then, the multiplication

U(n)⊗ U(b) −→ U(g)

p(n1 ⊗ · · · ⊗ nr)⊗ p(b1 ⊗ · · · ⊗ bs) 7−→ p(n1 ⊗ · · · ⊗ nr ⊗ b1 ⊗ · · · ⊗ bs)

defines an isomorphism of U(n),U(b)-bimodules.

———————————————————————————————Lecture 9, 05/05/2011

Recall Corollary 1.8.19. The isomorphism given is, in general, not anisomorphism of algebras.

Example 1.8.20. Let g = sl(2,C); n =

{(0 ∗0 0

)}= Ce, h =

{(x 00 y

)}=

Ch, n− =

{(0 0∗ 0

)}= Cf . We have

g = n− ⊕ h⊕ n


However, multiplication cannot be an algebra homomorphism, because, onone side U(n−) ⊗ U(h) ⊗ U(n) = S(n−) ⊗ S(h) ⊗ S(n) is commutative, andU(g) is not.

Proof of Corollary 1.8.19. To see that it’s an isomorphism of vector spaces,take a basis of n, {xi : i ∈ I}, and a basis of b, {xj : j ∈ J} so that {xk : k ∈ I∪J} is a basis of g. Choose an ordering on I∪J such that ik < jm, and the resultis a direct consequence of the PBW Theorem. To see that multiplication isan isomorphism of (U(n),U(b))-bimodules, let x ∈ n, can(x) ∈ U(n) (theseelements generate U(n) as an algebra). Let

can(x)−xi1 · · ·

−xir =

∑b∈PBW basis

αbb

Then,

m(can(x)−xi1 · · ·

−xir ⊗

−xj1 · · ·

−xjs) = m(

∑b

αbb⊗−xj1 · · ·

−xjs)

=∑b

αbb−xj1 · · ·

−xjs

= can(x)(−xi1 · · ·

−xir

−xj1 · · ·

−xjs)

Hence, multiplication is an isomorphism of left U(n)-modules. Multiplicationis analogously an isomorphism of right U(b)-modules.

Remark 1.8.21. In particular, if M is a U(b) module, then U(g) ⊗M is aleft U(n)-module, even a U(g)-module.

Corollary 1.8.22. The canonical map can : g → U(g) from the definitionof U(g) is injective, hence g is a vector subspace of U(g).

Proof. The elements−xi = can(xi) where {xi : i ∈ I} is a basis of g are part

of a PBW basis, so a basis of g is mapped to linearly independent vectors inU(g).

Rest of proof of PBW ’concrete version’.

To show: monomials−xi1 · · ·

−xir are linearly independent!

Fix a basis {xi : i ∈ I} for g. Let S = k[xi] for i ∈ I. For λ = (λ1, · · ·λr) ∈Ir, denote Zλ := xλ1···λr ∈ S; r is called the length of λ, denoted |λ| =


r, (λ1 · · ·λr) increasing if λ1 ≤ · · ·λr. Now, we know that {Zλ : λ increasing}is a basis of S, and S is filtered by s≤r = span(zλ : |λ| ≤ r). For λ ∈ Ir, wesay j ≤ λ if j ≤ λi for all i ≤ |λ|.

Lemma 1.8.23. For each m ∈ N, there exists a linear map

ϕm : g⊗ S≤m −→ S≤(m+1)

x⊗ s 7−→ x · s = ϕm(x⊗ s)

such that the following holds:

1. ϕm−1 is the restriction of ϕm to g⊗ S≤(m− 1)

2. xi · Zλ = xixλ1 · · ·xλr = ZiZλ; λ ∈ Ir, i ∈ I; i ≤ λ

3. xi · Zλ − ZiZλ ∈ S≤m for λ ∈ Im, i ∈ I.

4. xi · xj · y − xj · xy · y = [xi, xj] · y, ∀y ∈ S≤(m−1)

This lemma implies the Theorem: it gives a representation ( follows from4.) ϕ : g → Endk(S) which is well defined by 1., and by the universal

property of U(g), we get an algebra homomorphism∼ϕ : U(g) → Endk(S),

and we get∼ϕ(∼xi1 · · ·

∼xir)(1) = zi1 · · · zir ; i1 ≤ · · · ir; the latter are linearly

independent, hence∼xi1 · · ·

∼xir must be linearly independent because

∼ϕ is an

algebra homomorphism.

———————————————————————————————Lecture 10, 9.05.2011

Lemma 1.8.24. Let A,B be filtered k-algebras; denote by {A≤i}, {B≤j} thecorresponding filtrations. If f : A → B is an algebra homomorphism suchthat f(A≤r) ⊆ B≤r, then we get an induced algebra homomorphism

gr(f) : gr(A)→ gr(B)

Proof. We get induced maps fi : Ai → Bi.

In particular, we get a map gr(p) : Tk(g)→ gr(U(g)) for p : Tk(g)→ U(g)the canonical projection (gr(Tk) ∼= Tk(g)).


Proof of Theorem 1.8.18. Let J =< x ⊗ y − y ⊗ x : x, y ∈ g >= ker(α). Toshow: J ⊂ ker(gr(p)). Then get induced algebra homomorphism

∼p : S(g)→ gr(U(g))

such that∼p ◦ α = gr(p).

We have, for x, y ∈ g,

gr(o)(x⊗ y − y ⊗ x) = [x, y] = 0 ∈ gr(U(g))2

Hence, J ⊂ ker(gr(p)), so∼p exists. Besides,

∼p is an isomorphism because it

maps a basis given by ordered monomials in an ordered basis of g viewed aselements in S(g) to the corresponding PBW basis of ordered monomials in abasis of g.

Chapter 2

Representations of Lie algebras

Aim. Understand representations of g or U(g)-modules.In general: Hopeless! There is not even a classification of irreducible modules(up to isomorphism) for semi-simple complex Lie algebras (apart from g =sl(2,C), sl(3,C))

Reference for sl(2,C): V. Marzorchuk: Lectures on sl(2,C)-modules.

However: it is possible to classify finite dimensional irreducible repre-sentation and so called irreducible highest weight representations. To set thisup we need Verma Modules. We will start constructing representations of gfrom representations of Lie subalgebras of g and vice versa.

2.1 Constructing new representations

2.1.1 Pull-back and restriction

Let g, b be Lie algebras (over some field k). Let α : b → g be a Lie algebrahomomorphism. Let ρ : g→ gl(V) for some k-vector space V be a represen-tation of g. Then ρ ◦α : b→ gl(V) is a representation of b, by pulling ρ backto b.

Special case: If b is a Lie subalgebra of g, and α = incl is the inclusion,then pulling pack is nothing else than restricting. Denote

resgb V = ρ ◦ incl

45

46 CHAPTER 2. REPRESENTATIONS OF LIE ALGEBRAS

Note: The map incl : b → g induces, by universal property, a map−

incl : U(b)→ U(g). By PBW Theorem, this is an inclusion (take a basis of band extend to a basis of g). Then resgb V corresponds to the usual restrictionof the (usual) U(g) module V to the U(b) module V.

2.1.2 Induction

Let g be a k-Lie algebra. Let b ⊂ g be a Lie subalgebra. Let ρ : b → gl(V)be a representation, V is the corresponding U(b) module. Then

indU(g)U(b) := U(g)⊗U(b) V

is a U(g)-module, it is called the induced module from U(b) to U(g), and itis a (U(b),U(g))- bimodule by multiplying with U(g) from the left and withU(b) on the right. In particular U(g)⊗U(b) V is a left U(g)-module.

Proposition 2.1.1 (Universal property of Induction). Let g be a k- Liealgebra. Let b ⊂ g be a Lie subalgebra. Let M,N be U(b) modules, andf : M → N be a U(b)-module homomorphism. Then there exists a unique−f : U(g)⊗U(b) M→ N such that the following diagram commutes:

U(g)⊗U(b) M

M

p p p p p p p p p pR−f

6ϕ

-f

where ϕ(m) = 1⊗m.

Proof. The map f should satisfy

−f(u⊗m) =

−f(u · 1⊗m) = u

−f(1⊗m) = uf(m)

We can thus define it in this way, extending linearly. The map is hencealso automatically unique.

To see that−f is well defined, we must check:

−f(ub⊗m) =

−f(u⊗ bm) for all

2.1. CONSTRUCTING NEW REPRESENTATIONS 47

b ∈ b (since b generates U(b) as an algebra it is enough to take b ∈ b insteadof b ∈ U(b)). By definition we have

−f(ub⊗m) = (ub)

−f(m) = u(bf(m)) =

−f(u⊗ bm)

Proposition 2.1.2 (Adjointness of ⊗ and Hom). There are natural isomor-phisms of vector spaces for M ∈ U(b)−mod and N ∈ U(g)-mod as follows:

HomU(g)(U(g)⊗U(b) M,N) ∼= Hom(M, resU(g)U(b))

∼= HomU(b)(M,HomU(g)(U(g),N))

where

HomU(g)(U(g)⊗U(b) M,N) ∼= HomU(b)(M,HomU(g)(U(g),N))

fΦ7−→∼f

◦g

Φ′←− [ g

where

∼f(m)(u) := f(u⊗m) for u ∈ U(g),m ∈ M◦g(u⊗m) := g(m)(u)

Remark 2.1.3. The isomorphism given by Φ, Φ′ holds in general. Let R, Sbe rings, and let X be a (R, S)-bimodule, M and S-module, N an R-module.Then,

HomR(X⊗S M,N) ∼= HomS(M,HomR(X,N))

and the maps and proof are exactly the same as in Proposition 2.1.2.

Proof. We have:

The map∼f is a U(g)-module homomorphism:

∼f(m)(u1u2) =

∼f(u1u2 ⊗m)

The map∼f is a U(b)-module homomorphism:

∼f(bm)(u) = f(u⊗bm) = f(ub⊗m) =

∼f(m)(ub) = (b

∼f(m))(u) = u1f(u2⊗m) = u1

∼f(m)(u2)


The maps Φ,Φ′ are inverse to each other:

◦∼f(u⊗m) =

∼f(m)(u) = f(u⊗m)

∼◦g(m)(u) =

◦g(u⊗m) = g(m)(u)

The map◦g is well defined:

◦g(ub⊗m) = g(m)(ub) = bg(m)(u) = g(bm)(u) =

◦g(u⊗ bm)

The map◦g is a U(g)-module homomorphism:

◦g(u1u2 ⊗m) = g(m)(u1u2) = u1g(m)(u2)

Hence we have an isomorphism

HomU(g)(U(g)⊗U(b) M,N) ∼= HomU(b)(M,HomU(g)(U(g),N))

Now, we still need to check:

Claim.

HomU(g)(U(g),N) ∼= resU(g)U(b) N

f 7−→ f(1)

as U(b)-modules.

Let f1, f2 ∈ HomU(b)(U(g),N), such that f1(1) = f2(1), then f1 = f2

because f1(u) = f1(u1) = uf1(1) = uf2(1) = f2(u) since they are bothhomomorphisms of U(g)-modules. Hence it is injective, and it is clearlysurjective.

Example 2.1.4. Let g = gl(n, k) or g = sl(n, k). Then there is a decompo-sition into Lie subalgebras:

g = n⊕ h⊕ n+

Where h consists of all the diagonal matrices, n of strictly lower diagonalmatrices, and n+ of strictly upper diagonal ones.

2.2. VERMA MODULES 49

Lemma 2.1.5. If h is an abelian (commutative) Lie algebra over C, thenevery irreducible U(h)-module is one dimensional and the action is given byh·v = λ(h)·v for h ∈ h, v 6= 0 ∈ V for some λ ∈ h∗. Denote the correspondingmodule by Cλ

Next, we will study the (Verma) modules :

M(λ) := U(g)⊗U(b) Cλ = IndgbCλ

Note that the above construction means extending the one dimensional rep-resentation Cλ to U(n+) by n · v = 0 for all n ∈ n+.

———————————————————————————————Lecture 11, 11.05.2011

2.2 Verma Modules

Remark 2.2.1.

• Good universal properties

• Easy to compare dimensions of h-eigenspaces (character formulas)

• Every finite dimensional irreducible g- module for g a semisimple orreductive Lie algebra is isomorphic to a unique quotient of some Vermamodule.

Definition 2.2.2. Let g = gl(n, k) or g = sl(n, k), k any field. Let h ={ diagonal matrices} ⊆ g. Let λ ∈ h∗. Let kλ be the one dimensional(irreducible) h-module given by h · v = λ(h)v for h ∈ h, v ∈ kλ. Extend thisto a module for b = { upper triangular matrices} ⊂ g by n · v = 0 whenevern is strictly upper triangular. Then

M(λ) = Indgb = U(g)⊗U(b) kλ

is called the Verma module with highest weight λ.


Theorem 2.2.3. (Properties of Verma Modules)

1. Any Verma module M(λ) is infinite dimensional; in fact M(λ) ∼= U(n−),where n− = { strictly lower triangular matrices} as vector spaces; evenas left U(n−)-modules.

2. M(λ) = ⊕µ∈h∗

M(λ)µ, where M(λ)µ is the weight space of M(λ) for the

weight µ, i.e.

M(λ)µ := {u ∈ M(λ) : h · u = µ(h)u ∀h ∈ h}

3. M(λ) has a unique maximal proper (by inclusion) submodule N. Inparticular, M(λ)/N is an irreducible U(g)-module denoted L(λ).

Example 2.2.4. Let g = sl(2,C), where h = span

{(1 00 −1

)}, so h∗ is one

dimensional. Identify h∗ ∼= C via λ 7→ λ(h). Consider λ = 0. Then we havethat

M(0) ∼= U(g)⊗U(b) C ∼= C[f ]

f ihjek ⊗ v 7→{

0 if k > 0 or j > 0f i if k = j = 0

is an isomorphism of left U(n−)-modules since

f ihjek ⊗ v = f i ⊗ hjekv =

{0 if k > 0 since e ∈ n+

0 if k = 0 since λ = 0

Hence, by the PBW Theorem, the monomials f i⊗1 form a basis of M(0).We calculate the action of e, h:

e(f ⊗ 1) = ef ⊗ 1 = fe⊗ 1 + h⊗ 1 = 1⊗ h · 1 = λ(h) = 0;

e(1⊗ 1) = e⊗ 1 = 1⊗ e · 1 = 0

e(f 2 ⊗ 1) = ef 2 ⊗ 1 = fef ⊗ 1 + hf ⊗ 1 = fh⊗ 1− 2f ⊗ 1 = −2f ⊗ 1

Proceeding similarly for further powers of f and for the action of h, we getthe following picture:

1⊗ 1 // f ⊗ 1 // f 2 ⊗ 1 · · ·

2.2. VERMA MODULES 51

Note. f 2 ⊗ 1, i ≥ 0 for an eigenbasis for the action of h.

Note. M(0)/N ∼= Trivial one dimensional U(g)-module, i.e., xv = 0 ∀x ∈g, ∀v ∈ M(0)/N.

Fact 2.2.5. N is also irreducible! Actually, N ∼= M(−2)

Note. M(0) � N⊗M(0)/N because the surjection M(0)� M(0)/N doesn’tsplit. Alternatively, EndU(g)(M(0)) ∼= C because every endomorphism ϕ isdetermined by ϕ(1 ⊗ 1) as ϕ(u ⊗ 1) = u · ϕ(1 ⊗ 1) and ϕ(1 ⊗ 1) belongs tothe 0-eigenspace for h which is one dimensional, and spanned by 1⊗1, henceϕ(1⊗ 1) = λ(1⊗ 1). In particular, Weyl’s Theorem doesn’t hold for infinitedimensional modules.

Proof of Theorem 2.2.3. 1. To see this recall U(g) ∼= U(n−) ⊗ U(b) is anisomorphism of (U(n−),U(b))-bimodules. We further have:

M(λ) = U(g)⊗U(b) kλ ∼= (U(n−)⊗ U(b))⊗U(b) kλ ∼= U(n−)⊗ kλ ∼= U(n−)

all as U(n−)-modules.

2. From 1., we know M(λ) ∼= U(n−)⊗ kλ as a vector space.

Claim. Take a PBW basis of U(n−) given by monomials∼xı1 · · ·

∼xir for

i1 ≤ · · · ir where {xi : i ∈ I} is a basis of n−. This basis forms aneigenbasis of M(λ) for the action of h.

For the “empty monomial” 1, we have 1⊗ 1 and h · (1⊗ 1) = 1⊗ h1 =1 ⊗ λ(h)1 ∈ M(λ)λ, so it is a simultaneous eigenvector for h. Nowconsider the particular basis {Eij : i ≤ j} of n− where Eij denotes thematrix with (Eij)ij = 1 and zero otherwise. Now assume g = gl(n, k)(for sl(n, k), excercise!). Then a basis of h is given by {Eii}. Compute

Eii ·∼xı1 · · ·

∼xir recursively by using the formulas:

Eii(EabY)⊗ 1 = EabEiiY ⊗ 1 + [Eii,Eab]Y ⊗ 1

[Eii,Eab] = δaiEab − δibEab

Hence, we get

Eii(EabY)⊗ 1 = EabEiiY ⊗ 1 + scalars(EabY ⊗ 1)

Hence by induction we get a weight/eigenspace decomposition as wanted.


3.

Lemma 2.2.6. Let M be a U(g)-module (g) as above and M = ⊕µ∈h∗

Mµ

be a weight space decomposition. Let N be a U(g)-submodule. ThenN also has a weight space decomposition N = ⊕

µ∈h∗Mµ.

Proof. Let x ∈ N, x 6= 0. Then x =∑n

i=1 αµiuµi where uµi ∈ Mµi . Weclaim that all of the uµi ’s are contained in N. Choose x ∈ N such thatn in the last expression is minimal with respect to x belonging to N.Let h ∈ h. Then clearly µ1(h)x, h · x ∈ N. Hence,

(hx− µ1x) =n∑i=2

u′µi ∈ N

with u′µi = (µ1(h)− µi(h))uµi ∈ Mµi .

3. Let U be a submodule of M(λ). Then U contains M(λ)λ if and onlyif U = M(λ). Hence the sum of all proper submodules is a propersubmodule, hence the existence of one unique maximal submodule I.

2.3 Abstract Jordan Decomposition

Aim. Construct a triangular decomposition

g = n− ⊕ h⊕ n+

for semisimple/ reductive Lie algebras.

Theorem 2.3.1. Let g be a semisimple complex Lie algebra, x ∈ g. Then:

1. There exists a unique decomposition x = s + n, where ad(s) is diago-nalizable/semisimple, ad(n) is nilpotent and [s, n] = 0.

2. If ρ : g → gl(V) is a finite dimensional representation, then ρ(s) =ρ(x)s, ρ(n) = ρ(x)n (i.e. ρ(x) = ρ(x)s + ρ(x)n) is it’s concrete Jordandecomposition.

2.3. ABSTRACT JORDAN DECOMPOSITION 53

3. Let g, g′ be semisimple complex Lie algebras, and ϕ : g → g′ be aLie algebra homomorphism. Let x ∈ g, x = s + n as in 1. Then,ρ(x) = ρ(s) + ρ(n) satisfies the conditions in 1. and hence is theabstract Jordan decomposition of ϕ(x).

Lemma 2.3.2. Let V be a finite dimensional vector space over C, andg ⊆ gl(V) a semisimple Lie subalgebra. If x = xs + xn is the Jordan de-composition, then xs, xn ∈ g

Proof : later.

Proof. 1. Let x ∈ g. Consider the concrete Jordan decomposition

ad(x) = ad(x)s + ad(x)n.

Since g is semisimple, ad is injective. Hence ad(g) ⊂ gl(g) is a semisim-ple Lie subalgebra. By the above Lemma, ad(x)s, ad(x)n ∈ ad(g).Hence there exist elements s, n ∈ g such that ad(s) = ad(x)s, ad(n) =ad(x)n. By definition, x = s + n and by definition of the concreteJordan decomposition, s is diagonalizable and n is nilpotent. Also bydefinition of the concrete Jordan decomposition we have:

ad([s, n]) = [ad(x)s, ad(x)n] = 0

Hence, since ad is injective we have [s, n] = 0.

———————————————————————————————

Lecture 11, 16.05.2011

2. Consider the following diagram

g ρ(g) gl(V)

g ρ(g) gl(V)?

(ad(x))s ad(s)

-α

?

ad(ρ(s)) (ad(ρ))s

-incl

?

ad(ρ(s)) (ad(ρ))s

-ρ

-incl

We know that (ad(x))s = ad(s). Since the leftmost diagram commutesfor the boldface maps and the other maps as well, and since α is sur-jective, (it is ρ with restricted target), we have ρ(s) = ρ(x)s. Similarly,ρ(n) = ρ(x)n.


3. exercise

Proof of Lemma 2.3.2. Let V be a finite dimensional complex vector space,g ⊆ gl(V) a semisimple Lie subalgebra. Let x ∈ g, and x = xs + xn itsconcrete Jordan decomposition. To show: xs, xn ∈ g. Consider the followingset:

D :=

y ∈ gl(V) :a) [y, g] ⊆ gb) yW ⊆Wc) tr(y)|W = 0

∀g− invariantW ⊆ V

Claim 1. The set D is a Lie subalgebra of gl(V).

This is clear. Just take y, y′ ∈ D, and it is clear (using Jacobi identity)that [y, y′] ∈ D.

Claim 2. If y ∈ D then ys, yn ∈ D.

a) Let y ∈ D, then ad(y) ∈ gl(g), hence, (ad(y))s = ad(ys) ∈ gl(g) and(ad(y))n = ad(yn) ∈ gl(g).

b) The maps ys, yn preserve every subspace W ⊂ V preserved by y.

c) Follows since yn is nilpotent and hence tr(yn) = 0.

Claim 3. g ⊆ D

Let y ∈ g then:

a) [y, g] ⊆ g

b) yW ⊂W If W ⊂ V is g-invariant.

c) pendiente

Claim 4. gCD

This is a consequence of a).

Hence, D = g⊕ I for some ideal I (Excercise sheet). In particular: [g, I] ={0} (follows from a) and the fact that I is an ideal.) To show: I = {0}. Let

z ∈ I. Take a g-invariant subspace W ⊂ V. Then, wϕ7−→ zw is a g-module

homomorphism since for x ∈ g, [x, z] = 0 (direct sum), hence

ϕ(xw)− xϕ(w) = zxw − xzw = 0

2.3. ABSTRACT JORDAN DECOMPOSITION 55

Now, decompose V into a direct sum of irreducible g-modules (Weyl’s The-orem):

V =r⊕i=1

Wi

By Schur’s Lemma, the action of z on Wi is given by a scalar λi ∈ C.

Terminology:

x ∈ g ⊆ gl(V), x = xs + xn

The element x is called nilpotent or semisimple if xs = 0, respectively xn = 0.

Note. {semisimple elements}∩{nilpotent elements} = 0

Chapter 3

Structure Theory of SemisimpleComplex Lie Algebras

Definition 3.0.3. Let g be a complex semisimple finite dimensional Liealgebra. A Lie subalgebra h ⊆ g is called a Cartan subalgebra if:

1. The Lie algebra h is abelian and consists only of semisimple elements.

2. It is maximal, with respect to inclusion, with this property.

Remark 3.0.4. In a more general setting, i.e. if h ⊆ g are both finitedimensional Lie algebras over any field k, then h is called a Cartan subalgebraif it is nilpotent.

Remark 3.0.5. In the setting of Definition 3.0.3, the always exists a Cartansubalgebra.

Proof. Assume that g contains only nilpotent elements. Then, by Engel’sTheorem, g is nilpotent. Since g is semisimple, then it must contain at leastone semisimple element different from zero.

Claim. If h ⊂ g consists of only ad-semisimple elements, then h is abelian.

Assume it is not abelian, then there exists x ∈ h such that ad(x)(y) 6= 0for some y ∈ h. Hence, for some z ∈ h, ad(x)(z) = λz for some λ ∈ C, λ 6=0. However, ad2(z)(x) = [z, [z, x]] = [z, λz] = 0 which is a contradiction,because if ad2(x) = 0, then so is ad(y)(x).

Example 3.0.6. For g = sl(n,C), then h = { diagonal matrices with trace zero}is a Cartan subalgebra (Excercise!).

57

58CHAPTER 3. STRUCTURE THEORYOF SEMISIMPLE COMPLEX LIE ALGEBRAS

3.1 Root Space Decomposition

Definition 3.1.1. Let g be a semisimple complex finite dimensional Liealgebra, and let h ⊂ g be a Cartan subalgebra. Then, since h is abelian andcontains only diagonalizable elements, there exists a simultaneous eigenspacedecomposition of g with respect to the action of h via ad :

g =⊕λ∈h∗

gλ

where gλ := {x ∈ g : ad(h)(x) = λ(h)x ∀h ∈ h}. Now, if λ 6= 0 and gλ 6= {0},then λ ∈ h∗ is called a root of g. The set R of roots is called a root system.The eigenspaces gλ for λ ∈ R are called root spaces.

Hence we have

g = g0

⊕λ∈R

gλ

where g0 = {x ∈ g : [x, h] = 0} is the centralizer of h ∈ g.

Example 3.1.2. Let g = sl(n,C), where h ⊂ g is the “standard” Cartansubalgebra of all diagonal matrices. Denote by εi ∈ h∗ the linear map defined

by εi

a1 · · · 0...

0 · · · an

= ai. Then,

g = h⊕⊕λ∈R

gλ

R = {εi − εj : i 6= j}

A basis for the rootspace gεi−εj is given by the matrix Eij, since for h =a1 · · · 0...

0 · · · an

∈ h,

[h,Eij] = aiEij − ajEij = (εi − εj)(h)Eij.

Further, {Eii − Ei+1,i+1} is a basis of h.

3.1. ROOT SPACE DECOMPOSITION 59

Note. g0 = h

Note. Root spaces are one dimensional!

Picture of root systems MISSING

Theorem 3.1.3. Let g be a semisimple complex finite dimensional Lie alge-bra, h ⊂ g a Cartan subalgebra, and R ⊂ h∗ the corresponding root system.Then,

1. g0 = h

2. dim(gλ) = 1 ∀λ ∈ R

3. For every α ∈ R, there exist xα ∈ gα, yα ∈ g−α, hα ∈ [gα, g−α]

4. For α ∈ R, if µλ ∈ R for some µ ∈ C, then µ ∈ {±1} such that themap

sl(2,C) −→ g

e 7−→ xα

f 7−→ yα

h 7−→ hα

is an injective Lie algebra homomorphism.

5. [gα, gβ] = gα+β for any α, β ∈ R such that α + β ∈ R.

Remark 3.1.4. The triples (xα, hα, yα) for α ∈ R are called “sl(2,C) triples”of g.

Theorem 3.1.5 (Jacobson Morosov).

To prepare for the proof of Theorem 3.1.3 we begin with a

Lemma 3.1.6.

1. [gλ, gµ] ⊆ gλ+µ

2. K(gλ, gµ) = 0 unless λ = −µ

3. K|g0×g0 is non degenerate.


Proof. Let x ∈ gλy ∈ gµ.

1. For h ∈ h,

[h, [x, y]] = −[x, [y, h]]−[y, [h, x]] = µ(h)[x, y]+λ(h)[x, y] = (µ+λ)(h)[x, y].

2. We havead(x) ad(y)(gν) ⊂ gλ+µ+ν

Hence, since R is finite, ad(x) ad(y) is nilpotent if λ + µ 6= 0, sotr(ad(x) ad(y)) = 0 unless λ = −µ.

3. K(gα, g0) = 0 by 2. Hence, since g is semisimple, K|g0×g0 is non degen-erate.

———————————————————————————————Lecture 13, 18.05.2011


Proof of 1. Since h is abelian then h ⊆ g0. Let x ∈ g0, and let x = s+n be it’sabstract Jordan Decomposition. By the properties of Jordan Decomposition,we have tha since ad(x)(h) ⊂ {0}, then also ad(s)(h) = ad(x)s(h) ⊂ {0} andad(n)(h) = ad(x)n(h) ⊂ {0}. Hence, if x = s + n ∈ g0, then s, n ∈ g0. Bymaximality of h, this means that s ∈ h. In particular ad(s) = 0, and hencead(x) = ad(n). By Engel’s Theorem, we get

Claim. The Lie algebra g0 is nilpotent.

Hence, by Lie’s Theorem,

ad(g0) ⊆

0 · · · ∗

. . .

0 · · · 0

Assume there is z ∈ g0 which is ad-nilpotent, then K(z, g0) = 0. By Lemma3.1.6, we know that K|g0×g0 is non degenerate. Hence z = 0. Hence, g0 isnilpotent, and thus x = s ∈ h.

Remark 3.1.7. For g a semisimple finite dimensional complex Lie algebra,a Cartan subalgebra is a Cartan subalgebra in the general definition.


Lemma 3.1.8. Set up as above. Let α ∈ R. Then:

a. −α ∈ R

b. dim([gα, g−α]) = 1

c. α doesn’t vanish on the line [gα, g−α] ⊆ g0 = h

Proof. If −α /∈ R, then g−α = {0}; hence K(gα, g−α) = 0, but, since g issemisimple, K is a non-degenerate form on g. By Lemma 3.1.6, K(gα, gβ) = 0except if α = −β. This is a contradiction, hence −α ∈ R, and a. above holds.Now, let x ∈ gα, x 6= 0, y ∈ g−α, y 6= 0. Invariance of the Killing form gives

K(h, [x, y]) = K([h, x], y) = α(h)K(x, y).

Hence, ker(α) ⊆ [gα, g−α]⊥, the orthogonal complement of [gα, g−α] ⊆ g0 = h,and so

dim([gα, g−α]⊥) ≥ dim(h)− 1,

hencedim([gα, gα]) ≤ 1

Now let x 6= 0, x ∈ gα, y 6= 0, y ∈ g−α, h = [x, y]. If we assume that α(h) = 0,then x, y, h span a nilpotent Lie subalgebra, hence a solvable one, thus byLie’s Theorem, ad(x), ad(y), ad(h) are strictly upper triangular matrices forsome basis, so ad(h) = ad([x, y]) is a nilpotent endomorphism. hence, bydefinition of Cartan subalgebra, we get that ad(h) = ad([x, y]) = 0, and bysemisimplicty of g this means that h = 0. Hence c. holds. Now, we knowthat K is not zero on gα × g−α, so there exist non zero x ∈ gα, y ∈ g−α suchthat K(x, y) 6= 0. Now choose h ∈ h such that α(h) 6= 0, then

K(h, [x, y]) = α(h)K(x, y) 6= 0.

In particular, [x, y] 6= 0, so [gα, g−α] 6= 0, hence dim([gα, g−α]) = 1.

Definition 3.1.9. Let g be a semisimple complex finite dimensional Lie

algebra, and let h ⊆ g be a Cartan subalgebra, α ∈ R a root. Define∨αıh by

the properties:

• ∨α ∈ [gα, g−α]

• < α,∨α >= α(

∨α) = 2


This∨α ∈ h is called the dual root to α or the coroot of α.

Remark 3.1.10. Since α does not vanish on the line [gα, g−α], there exists∨α ∈ [gα, g−α] with the desired properties.

We restate 3. in the following

Claim. There is an embedding

sl(2,C)→ gα ⊕ [gα, g−α]⊕ gα

of Lie algebras.

Proof of Claim (3.) Let α ∈ R. Then find x ∈ gα, y ∈ g−α such that h =[x, y] 6= 0, by previous Lemma. Then

[∨α, x] = α(

∨α)x = 2x

[∨α, y] = −α(

∨α)y = −2y

Hence (x,∨α, y) define an sl(2,C) triple for α. Hence

e 7−→ x

h 7−→ ∨α

f 7−→ y

defines the desired embedding.

Example 3.1.11. Let g = sl(3,C). Then

(x = Eij,∨α = Eii − Ejj, y = Eji)

form an sl(2)-triple for the root α = (εi = εj)

Exercise 3.1.12. Find more “nonstandard” sl(2,C)- triples.

Main Idea. For α ∈ R, we have an embedding

sl(2,C)→ gα ⊕ [gα, g−α]⊕ gα

We can then study g as an sl(2,C) module by letting sl(2,C) act via ad. Manystructural problems of g can thus be solved using representation theory offinite dimensional sl(2,C)-modules.


Claim. Let α ∈ R, then Cα ∩ R = {±1}

Proof. Let α ∈ R. Find x ∈ gα, y ∈ g−α,∨α ∈ [gα, g−α]. Consider the sl(2,C)-

module

M =⊕t∈C

gtα ⊕ C∨α

Decompose it into irreducible finite dimensional sl(2,C)-modules.

As possible weights (eigenvalues of ad(∨α)) we have, 0, with one dimen-

sional weight space gα, or 2t, in which case t ∈ 12Z. By sl(2,C)- theory, it

follows that the only even weights are 0, 2,−2 because gα ⊕ C∨α ⊕ g−α is a

summand and any other weight would give a higher dimensional zero weightspace. By sl(2,C)-theory, we get that if βt is a weight for an sl(2,C)-module,then 2β is a root. Hence, 1

2α is not a root, but then α is not a root, and this

means 1 is not a weight. Hence,

M = g−α ⊕ [gα, g−α]⊕ g−α.

So there are only even weights and they are only {2, 0,−2}. Hence, t ∈ {±1}.In particular dim(gα) = 1. We have now finished proving 2. and 5. ofTheorem 3.1.3 and hence all of it.

Remark 3.1.13. The fact that root spaces are one dimensional is a verystrong property and special of finite dimensional Lie algebras. (...Kac-MoodyLie Algebras!)

Proposition 3.1.14. Let g be a semisimple complex finite dimensional Liealgebra, and h ⊂ g a Cartan subalgebra, R the associated set of roots. Then

1. For α, β ∈ R, < α,∨β >:= α(

∨β) ∈ Z.

2. β− < β,∨α > α ∈ R.

3. R spans h∗.

Note. The root system R is not a basis of h∗.

(Motivation, sl3, reflection of a root is again a root)


Proof. Assume β = α, then < α,∨α >= 2 ∈ Z. If α 6= β, Then consider the

sl(2,C) representation

M =⊕i∈Z

gβ+iα.

Each weight space is one dimensional, and∨α acts on gβ+iα with eigenvalue

(β + iα)(∨α) =< β,

∨α > +2i. Hence by sl(2,C) theory, we get < β,

∨α >∈ Z.

Also by sl(2,C) theory, − < β,∨α > is a weight. Hence gβiα 6= 0 for i = − <

β,∨α >, so β− < β,

∨α >∈ R. To show 3., it is enough to show⋂

α∈R

ker(α) = {0}.

Let h ∈ h such that α(h) = 0 for all α ∈ R. Then, [h, gα] = 0 for all α ∈ R,so h ∈ Z(g) (the center of g). But g is semisimple, so Z(g) = {0}, andh = 0.

———————————————————————————————Lecture 14, 23.05.2011

3.2 Root Systems

In this section, V will denote a finite dimensional k-vector space, where k isany field with char(k) = 0.

Definition 3.2.1. An endomorphism s ∈ Endk(V) is called a reflection ifs2 = Id and rank(IdV−s) = 1.

Remark 3.2.2. If dimk(V) = n, and s ∈ Endk(V) is a reflection, then

dimk(Hs := {v ∈ V : s(v) = v}) = n− 1.

This will be the reflection hyperplane of s.

Lemma 3.2.3. An endomorphism s ∈ Endk(V) is a reflection if and only ifthere exist 0 6= a ∈ V, 0 6= a∗ ∈ V∗ such that a∗(a) = 2 and

s(x) = x− a∗(x)a

for all x ∈ V.

3.2. ROOT SYSTEMS 65

Proof. Let s ∈ Endk(V), and assume there exist a ∈ V, a∗ ∈ V∗ as in thestatement of Lemma 3.2.3. Let x ∈ V. Then,

s2(x) = s(x)− a∗(x)s(a) = x− a∗(x)a− a∗(a− a∗a(a)a)

= x− 2a∗(x)a+ a∗(x)a∗(a)a = x

and(Id−s)(x) = a∗(x)a

Since a∗ 6≡ 0, this implies rank(s− Id) = 1.Now let s ∈ Endk(V) be a reflection, so dim(Im(Id−s)) = 1. Hence, thereexist 0 6= a ∈ V, 0 6= a∗ ∈ V∗ such that for all x ∈ V, x − s(x) = a∗(x)a.Since s2 = Id,

−a∗(x)a = s(x)− x = s(x− s(x)) = a∗(x)s(a) = a∗(x)(a− a∗(a)a);

hence a∗(a) = 2.

Example 3.2.4. Let g be a semisimple finite dimensional complex Lie alge-bra, h ⊆ g a Cartan subalgebra, and α ∈ R a root. Then

sα,∨α

: h∗ → h∗

x 7−→ x− < x,∨α > α

and correspondingly s∨α,α

: h→ h are reflections.

Remark 3.2.5. Let s ∈ Endk(V) be a reflection. Then V = V+s ⊕V−s where

V±s is the ±1- eigenspace for s, respectively.

Note. The eigenspace V−s is a one dimensional subspace of V spanned bya ∈ V given in Lemma 3.2.3.

Lemma 3.2.6. Let (− , −) be a bilinear, non-degenerate symmetric formon V, and let s ∈ Endk(V) be a reflection. Then:

1. The reflection s is orthogonal (i.e. (s(x), s(y)) = (x, y) for all x, y ∈ V)if and only if V+

s ∩ (V+s )⊥ = 0 and V−s ∩ (V−s )⊥ = 0

2. If (− , −)|H is non-degenerate for some hyperplane H, then there existsa unique reflection s = sH ∈ Endk(V) such that s(h) = h for all h ∈ Hand if α ∈ H⊥/{0}, then H⊥ = h · α, and

s(x) = x− 2(x, α)

(α, α)α.


In this case, sH is called the reflection in the hyperplane/ orthogonalto the hyperplane H.

Proof. Excercise.

Remark 3.2.7. If s ∈ Endk(V) is a reflection, then s ◦ sH ◦ s−1 = ss(H).

Definition 3.2.8. A subset R ⊂ V is called a root system if the followingholds:

R1 The set R is finite, 0 /∈ V, and R spans V.

R2 ∀α ∈ R, ∃∨α ∈ V∗ such that∨α(α) = 2 and s

α,∨α(R) ⊆ R

R3 ∀α, β ∈ R,∨α(β) ∈ Z.

The root system R is called reduced if kα ∩ R = {±1} for all α ∈ R. Therank of the root system R ⊂ V is the dimension of V.

We have already proven the following:

Theorem 3.2.9. Let g be a semisimple complex finite dimensional Lie alge-bra, and h ⊆ g a Cartan subalgebra. Then R subseth∗, the corresponding setof roots of g, is a reduced root system in the sense of the above definition.

Remark 3.2.10. The∨α in Definition 3.2.8 is uniquely determined: Suppose

∨α,∨β satisfy R2 for α. Then, s

α,∨α(α) = α = s

α,∨β(α), and s(R) = s

α,∨α(R) =

sα,∨β(R). Consider

(s ◦ t)n(x) = x− n(∨β(x)− ∨α(x))α.

Since s, t generate a finite subgroup of GL(V), the order of s ◦ t is finite.Substitute this in place of n above.

Remark 3.2.11. The reflections sα,∨α

generate a finite subgroup W of GL(V),

since they are all contained in the permutation group of the finite set R (R1,R2 i Definition 3.2.8). Let

G(R) := {w ∈ GL(V) : w(R) ⊂ R}.

Then W(R) ⊆ G(R).


Definition 3.2.12. The group W = WR is called the Weyl group associatedwith R ⊂ V.

Example 3.2.13. Let g = sl(3,C), h ⊆ g Cartan subalgebra. Then R ={εi − εj : 1 ≤ i, j ≤ 3}, and WR

∼= S3, the symmetric group on three letters.

We will denote in the future sα = sα,∨α, and W = WR when there is no

risk of confusion.

Remark 3.2.14. Setting as before. Since sα(α) = −α, then −R = R. Theidentity map idV ∈ G(R), but it doesn’t need to be contained in WR(R)

Remark 3.2.15. If V = V1 ⊕ V2, R1 ⊆ V1,R2 ⊆ V2 are root systems, thenR1 ∪ R2 ⊂ V is also a root system (excercise). This is called the direct sumof the root systems R1 and R2.

Definition 3.2.16. The root system R is called irreducible if R is not thedirect sum of two non empty root systems.

If R = R1 ∪R2 is the direct sum of two root systems, then WR = WR1 ×WR2 (excercise) and w|Vi = idVi for each w ∈ Rj, for j 6= i.

Fact 3.2.17. If R ⊂ V is an irreducible root system, then the action of WR

on V defines an irreducible WR-module, i.e. a representation of the groupWR.

Proposition 3.2.18. Let R ⊂ V be a root system. Then∨R = {∨α : α ∈

R} ⊆ V∗ is a root system. It is called the dual root system of R.

Proof. Without loss of generality, may assume R is an irreducible root sys-

tem. Now,∨R is a finite set by definition, 0 /∈

∨R because

∨α(α) = 2. Now we

show that∨R spans V∗. Assume there exists 0 6= x ∈ V such that

∨α(x) = 0

for all α ∈ R. Then k · x is stable under the action of WR which contradictsV being an irreducible WR-module. Now let α, β ∈ R, and let γ = sα(β); θ =

s∨α(β). Then, θ(γ) =

∨β(β− < β,

∨α > α) −

∨β(α)

∨α(β− < β,

∨α > α). Since

∨β(β) = 2, then θ(γ) = 2 and hence sγ,θ = sα ◦ sβ ◦ sα. Using the formulas ◦ sH ◦ s−1 = ss(H), we get that sγ,θ(R) ⊂ R, using. By uniqueness of coroot,

we get θ − ∨γ = s∨α(∨β). Hence s∨

α(∨R) =

∨R and

∨∨α = α.


Remark 3.2.19. We have a bijection

R→∨R

α 7→ ∨α

It induces an isomorphism

WR →W∨R

;

w 7→ (tw)−1

even an isomorphism G(R) ∼= G(∨R).

Why is W(R) interesting?

It finally gives a classification of semisimple finite dimensional complex Lie

algebras, up to isomorphism (< α,∨β >∈ Z).

Theorem 3.2.20 (Harish-Chandra). Let U(g) be the universal envelop-ing algebra of a semisimple complex finite dimensional Lie algebra. LetZ(U(g)) ⊆ U(g) be it’s center. There is an isomorphism

Z(U(g)) ∼= S(h)WR

Where the action of WR on h∗ induces an action on S(h).

Remark 3.2.21. The space S(h)WR is the ring of regular functions on thespace of orbits h∗/WR of the WR action on h∗.

Main Point

Theorem 3.2.22 (Sheppard-Todd). The algebra S(h)WR is again a polyno-mial ring, and h∗/WR is an affine algebraic variety.


3.2.1 Changing scalar

Let R ⊆ V be a root system, over a field k. Let VQ denote the Q span of the

roots α ∈ R, and VQ∗ the Q-span of the coroots∨α ∈

∨R.

Note. For k ⊂ L a field extension, we have that L⊗k R := {1⊗ α : α ∈ R}forms a root system for L⊗k V, as L-vector space, with dual roots 1⊗ ∨α, forα ∈ R.

Proposition 3.2.23. Let R ⊂ V be a root system, notation as above. (V isa k-vectors space, and Q ⊆ k is a field extension.)Then

1. R is a root system in VQ.

2.

k ⊗Q VQ → V

λ⊗ v 7→ λv

is an isomorphism, similarly for VQ

3. VQ∗ = (VQ)∗

Proof. 1. By definition, R ⊆ VQ. BY property R3 of root systems, if

α, β ∈ R then∨α(β) ∈ Z. Hence,

∨α(VQ) ⊆ Q, so that

∨α indeed defines

a linear functional over Q, so this defines a coroot. The properties of aroot system follow from the definition.

2. The map

ki−→ V

λ⊗ v 7−→ λv

is surjective since R spans V over k. Also,

V∗i∗−→ k ⊗Q V∗

∨α 7−→ 1⊗ ∨α

is also surjective, so i must be an isomorphism. Hence 2. and 3. hold.


Corollary 3.2.24. Let R ⊂ V be a root system. This defines a root systemRQ ⊆ V(Q) and then RR := {1 ⊗ α : α ∈ R} ⊂ R ⊗Q VQ a root system overR.

Remark 3.2.25. The Weyl groups WR,WRQ and WRR are all isomorphic.

Recall the bilinear form

γ(x, y) =∑α∈R

∨α(x)

∨α(y).

If x, y ∈ VQ, then γ(x, y) ∈ Q, hence we get bilinear forms

VQ × VQ −→ Q(x, y) 7−→ γ(x, y)

and hence

γR : R⊗Q VQ × R⊗Q VQ −→ R(λ⊗ x, µ⊗ y) 7−→ λµ⊗ γ(x, y)

Then,

γR(x, x) =∑α∈R

(∨α(x))2 ∈ R

and γR(x, x) = 0 if and only if∨α(x) = 0 for all α ∈ R, which happens if

and only if x = 0. We thus get an inner product on the real vector spaceR ⊗Q VQ := VR. Conversely, if (− , −) is an inner product on VR which isWR-invariant, then we can measure angles between roots.Given α, β ∈ R ⊆ VR, let θ = ](α, β) be the angle between α and β. Then

cos(θ) =(α, β)

‖α‖‖β‖

Note. This doesn’t depend on the choice of an inner product. If R is anirreducible root system and α, β ∈ R, then ‖α‖

‖β‖ doesn’t depend on the choiceof the inner form.


Note. We have < β,∨α >= 2(α,β)

(α,α)= 2‖β‖‖α‖ cos(θ), hence < α,

∨β >< β,

∨α >=

4 cos2(θ) ≥ 0. We get the identity

0 ≤< α,∨β >< β,

∨α >≤ 4 (3.1)

By the properties of root systems we conclude that there are only finitely many

possibilities for the value of < α,∨β > for α, β ∈ R.

Note. For α, β ∈ R, (α, β) 6= 0 =⇒ <β,∨α>

<α,∨β>

= (β,β)(α,α)

and

< β,∨α >= 0 ⇐⇒ < α,

∨β >= 0 ⇐⇒ (α, β) = 0

List of all possibilities for < α,∨β >,< β,

∨α > with assumption ‖α‖ ≤ ‖β‖.

< α,∨β > < β,

∨α > θ ‖β‖2

‖α‖2

I 0 0 π2

undeterminedII 1 1 π

31

II −1 −1 2π3

1III 1 2 π

42

III −1 −2 3π4

2IV 1 3 π

63

IV −1 −3 5π6

3don’t occur if R is reduced 1 4 0 4

”” −1 −4 π 4don’t occur if α 6= ±β 2 2 0 1

−2 −2 π 1


Rank two root systems

I

II Type A2, Lie algebra: sl(3,C)

ε1

ε2

ε3 ε1 − ε2 = α

ε1 − ε3 = α + β

ε2 − ε3 = β

III Type B2, Lie algebra:

ε1

ε2 ε1 + ε2

ε1 − ε2

IV Type G2

ε1

ε2

ε3 ε1 − ε2 = α

2ε1 + ε2

ε1 + 2ε2

Proposition 3.2.26. Let R ⊆ V be a root system. Then

1. (x, y) =∑

α∈R

∨α(x)

∨α(y) defines a G(R) (hence WR)-invariant symmet-

ric non degenerate bilinear form on V.


2. If R = R1∪ · · · ∪Rs is a direct sum of irreducible root systems Ri ⊆ Vi

and (− , −) is a symmetric G(R) or WR-invariant bilinear form on

V =n

Vii=1

, then (− , −)|V1×V1 is, up to scalar, the form defined in 1.,

and the spaces Vi are pairwise orthogonal.

Remark 3.2.27. Given g a semisimple complex finite dimensional Lie alge-bra, and h ⊆ g a Cartan subalgebra, sometimes it’s useful to identify h withh∗ via the Killing form:

h −→ h∗

h 7−→ K(h, −).

then, the Killing form induces, via this identification, a symmetric, nondegenerate bilinear form on h∗. One can show that under this isomorphism,we get

∨α =

2α

(α, α)

———————————————————————————————Lecture 15, 30/05/2011

Recall. Let g be a semisimple complex finite dimensional Lie algebra, h ⊆ g

a Cartan subalgebra, R the corresponding root system,∨R ⊆ h∗∗ the dual

root system. We have a bijection

R→∨R

α 7−→ ∨α.

and for each α ∈ R, reflections

sα,∨α

: h→ h

s∨α,α

: h∗ → h∗.

Claim 5. Setting as above. We have s∨α,α

=∗ (s∨α,α

) = (∗(s∨α,α

))−1 where for a

linear map f : W1 →W2, the map

∗f : W∗2 −→W∗

1

ϕ 7−→ (w1 7→ ϕ(f(w1)))

is the dual map.


Proof. Its clear that (∗sα,∨α)2 = id because (s

α,∨α)2 = Idh∗ . We have for

α ∈ h∗, h ∈ h,

(∗sα,∨α)(α)(h) = α(s

α,∨α(h)) = α(h− < h, α >

∨α) = α(h)− < h,

∨α > α(

∨α) = −α(h).

Similarly, ∗sα,∨α(β) = β if β(

∨α) = 0 hence rank(∗s

α,∨α− Id) = 1.

Remark 3.2.28. The map w 7→ (∗w)−1 defines an isomorphism of groupsWR∼= W∨

Rbecause WR is generated by reflections and ∗s1

∗s2 =∗ (s1s2).

Remark 3.2.29. The same holds for G(R) instead of WR.

Proof of Proposition 3.2.26. 1. That γ is bilinear and symmetric is clearfrom its definition. Take g ∈WR. We have,

γ(g(x), g(y)) =∑α∈R

∨α(g(x))

∨α(g(x))

=∑α∈R

∗g(∨α(x))∗g(

∨α(y))

∑β∈R

∨β(x)

∨β(y)

because W∨R

permutes β ∈∨R. Hence γ is WR-invariant. To see it is non

degenerate, let α, β ∈ R. Then,∨α(β) ∈ Z. Hence γ(x, x) ∈ Z≥0. Also

since∨α(α) = 2, we have γ(α, α) ≥ 4. In particular, γ 6= 0. Now let

x ∈ Vi, y ∈ Vj, i 6= j. Then∨α(x)

∨α(y) = 0 because

∨α(x) = 0 for every

α ∈ Ra, β ∈ Rb, a 6= b. Hence we may assume R ⊆ V is irreducible.By Excercise 25, we are done.

2. (The reason one actually needs a proof of this is that the form (−,−)only coincides with γ on the irreducible components a priori.) Define

Ui := {w(x)− x : x ∈ Vi;w ∈WRi}.Then Ui is a WRi-submodule of Vi. Since sα(α) − α = −2α, we getUi 6= 0, so Ui = Vi since Vi is irreducible. Now let x ∈ Vi, y ∈ Vj fori 6= j. Then since w|Vi = IdVj forw ∈WRj , we get

(w(x), y) = (w(x), w(y))

so that w(x)− x is orthogonal to y. Remaining part: Exercise.


3.2.2 Bases of root system

Definition 3.2.30. Let R ⊂ V be a root system, where V is a finite di-mensional R-vector space. Choose a basis {e1, · · · , en} of V. Define thelexicographic order on elements of V by

λ =n∑i=1

λiei �n∑i=1

µiei

⇐⇒∃1 ≤ s ≤ n s. th. λi = µi for 1 ≤ i ≤ s and λs+1 > µs+1

Note. This order, of course, depends on the choice of basis!

Remark 3.2.31. The lexicographic order � defines a total order on elementsof V.

Definition 3.2.32. Let R ⊆ V as above, � a lexicographic order on V.Consider

R+ : = {α ∈ R : α � 0} := positive roots

R− : = {α ∈ R : 0 � α} := negative roots

It follows from the definition of these two sets that R = R+ ∪ R− andR+ ∩ R−. Let

B := {α ∈ R+ : α is not the sum of two positive roots } (3.2)

The set B is called the base or basis of the root system R. Elements of B arecalled simple roots.

Note. Everything depends on the choice of �!

Remark 3.2.33. The set B is compatible with addition, i.e. λ � µ, λ′ �µ′ =⇒ λ+ λ′ � µ+ µ′

Corollary/Definition 3.2.34. Given g, a semisimple complex finite di-mensional Lie algebra and h ⊂ g a Cartan subalgebra, we get a triangulardecomposition:

g = n− ⊕ h⊕ n+

where all the summands are Lie subalgebras, and

n± = ⊕α∈R±

gα.


Proof. As a vector space the sum comes from the definition of the roots. Also,by definition, h is a subalgebra. Now, let α, β ∈ R+, and xα ∈ gα, xβ ∈ gβ.Then [xα, xβ] ⊆ gα+β; by remark above α + β ∈ R+. SImilarly, n− is asubalgebra.

Definition 3.2.35. Let g = n−⊕h⊕n+ as above. Let λ ∈ h∗ and define a onedimensional h-module C by h·v := λ(h)v for h ∈ h, v ∈ C, and call it Cλ. Ex-tend to n+ by x ·v := 0 for all x ∈ n+. The Verma module of highest weightλ is

M(λ) = U(g)⊗U(n+) (C)λ

Theorem 3.2.36. Let R ⊂ V be a root system, V f.d. R-vector space, � alexicographic order, B a basis with respect to �, R+ the corresponding setof positive roots. Then

1 Any α ∈ R+ is a sum of elements in B, i.e. α =∑nibi;ni∈Z, bi ∈ B.

missing example sl3

2. For α, β ∈ B, if α 6= β then < α,∨β >≤ 0 andα− β /∈ R.

3. Elements of B form a basis of V.

4. For α ∈ R+/B, there exists β ∈ B such that < α,∨β > 0 and α−β ∈ R+.

5. Let α ∈ R+, β ∈ B, α 6= β. Then sβ(α) ∈ R+. In particular sβ permutesthe set R+/{β}.

Remark 3.2.37. Let w ∈ WR. We denote by l(w) the minimal number ofreflections sα, α ∈ B needed to write w as a product of such. For sl(3,C),WR∼= S3. We can choose B such that {sα : α ∈ B} corresponds to the set of

simple transpositions in S3. Then l(w) is the usual length of an element in S3

(this holds for any n), and l(w) = #|WR(R)+ ∩ R−|. The correct definitionof VQ is the span of the elements in B, for char(k) 6= 0.

Proof. 1 Let αm � · · · � α1 be the ordered elements of R+. Now, α1 ∈B, otherwise, by definition of B, α = α + β for some α, β ∈ R+, soα1 � α, β which is a contradiction. We proceed by induction. Assumethe statement holds for αj, for 1 ≤ j ≤ i. Assume αi+1 /∈ B. Thenαi+1 = α + β for some α, β ∈ R+, and we are then done by induction.


2 If α−β ∈ R, then −(α−β) = β−α ∈ R, so either α−β or β−α ∈ R+

since α = (β − α) + α and β = (α − β) + β, this means neither αorβ

con belong to B. By “Observations”, < α,∨β >≤ 0.

3 Since R spans V, any α ∈ R+ is a Z-linear combination of elementsin B, and further for any α ∈ R−,−α ∈ R+, we conclude B spansV. To see that the elements if B are linearly independent, assumefor λi ∈ R, that v =

∑si=1 λibi =

∑mi=s+1 λibi, so by 2, < v, v >=<∑s

i=1 λibi,∑m

i=s+1 λibi >< 0, so λj = 0 since bi � 0.

4 Let α ∈ R+ − B, α =∑nibi, with ni ∈ Z≥0 and bi ∈ B. Then,

0 < (α, α) = (α,∑

nibi)

so < α,∨bi >> 0 for some i. By observations, α− bi, bi − α ∈ R, so one

must belong to R+, but it cannot be bi − α because otherwise bi is asum of two positive roots, so α− bi ∈ R+.

5 Write α =∑nibi, ni ∈ Z≥0 and bi ∈ B. Since α 6= bi, there exists j

such that nj 6= 0 and bj 6= β. Hence, sβ(α) = α− < α,∨β >=

∑mibi

where mj = nj for our fixed j. Hence, since by 1, the coefficients ofa root are either all positive or all negative, and mj > 0, we concludesβ(α) ∈ R+.

3.2.3 Weyl Chambers

Let R ⊆ V be a root system, V a f.d. R-vector space and (−,−) the WR-invariant scalar product. Let {e1, · · · en} be an R-basis of V. Let

C := {x =n∑i=1

λi(x)ei ∈ V : λi(x) > 0∀i}

Given α ∈ R, then the hyperplane associated with the reflection sα is

Hα := {x ∈ V :< x,∨α >= 0}


Consider

H :=⋃α∈R

Hα =⋃α∈R+

Hα

Consider V−H. The connected components are called Weyl Chambers.

Remark 3.2.38. The connected components don’t depend on (−,−). Also,WR acts on the set of weyl chambers, since sα is continuous, bijective with

continuous inverse. Also WR(H) ⊂ H because < x,∨α >= 0 ⇐⇒ <

w(x), w(∨α) >= 0 so x ∈ Hα ⇐⇒ sβ(x) ∈ Hsβ(α). Moreover, w ∈ WR

permutes Weyl Chambers.

Lemma 3.2.39. The Weyl group WR acts transitively on Weyl chambers.

Proof. Tauvel

———————————————————————————————Lecture 17, 30/05/2011

Let V be a finite dimensional real vector space, R ⊂ V reduced rootsystem, (−,−) a WR-invariant scalar product.

Proposition 3.2.40. Let B = {β1, · · · , βl} be a base of R. Let {β′1, · · · β′n}be the dual base with respect to (−,−), i.e. (βi, βj) = δij. Then,

C(B) = {x ∈ V : (x, β) > 0∀β ∈ B}= {x ∈ V : (x, βi) > 0∀i ∈ 1, l}

= {∑

xiβi : xi ∈ R≥0}

=l

R≥0β′i

i=1

Is a Weyl chamber, in particular an open simplicial cone.

Proof.

Claim. C(B) ⊆ V/H.Let x ∈ C(B), α ∈ R. Since Hα = H−α, assume w.l.o.g. that α ∈ R+, α =

n1β1 + · · ·+ nlβl. Then,

(x, α) =∑

ni(x, βi) > 0

so x /∈ Hα.


Claim. C(B) is connected.

This is true if and only if C(B) ⊂ C for a Weyl chamber C. Assume thenthat C(B) * C. Let x ∈ C− C(B), then there exists i such that (x, βi) < 0.Let

C+ = {y ∈ C : (y, βi) > 0}C− = {y ∈ C : (y, βj) < 0};

both open in C. Then C− is not empty (x ∈ C−), and C(B) ⊆ C+, so C isn’tconnected, a contradiction.

Theorem 3.2.41. Let R ⊂ V as before. Then

1

{bases of R} ∼−→ {Weyl chambers}B 7→ C(B)

is WR-equivariant.

2 W(R) acts simply transitively on {bases of R} and on {Weyl chambers}.

Definition 3.2.42. If a basis B of R is fixed, then C(B) is a fundamental Weyl chamber.


1 Let B be a base of R. Let C′ be any Weyl chamber, then by Lemmathere exists w ∈WR such that w(C(B)) = C′. Note that w(B) is a baseof R, and w(C(B) = C(w(B))), so the map is surjective and equivariant.Now we check injectivity: Let B = {β1, · · · βl},B′ = {α1, · · ·αl} bebases of R, and let {β′1, · · · , β′l}, {α′1, · · · , α′l} be the dual bases withrespect to (−,−), respectively. Assume that

l⊕i=1R>0β

′i = C(B) = C(B′) =

l⊕i=1R>0α

′i.

In particular, the set of edge rays, i.e. {R>0β′i}, {R>0α

′i} coincide.

2 Let B be a base of R, and w ∈ WR. To show: w(B) = B impliesw = id. Let w = s1 · · · sr be a reduced decomposition of w, wheresi = sβi , βi ∈ B. By problem 29, if r ≥ 1, then w(βr) ∈ R−, whichcontradicts our assumption w(B) = B, so r = 0 and thus w = id.


Notation. We will denote by R+(B) = R+(C) the sums of elements of Bthat are in R.

The set R+(C) determines C:

C = C(B) = {x ∈ V : (x, β) > 0∀β ∈ B}= {x ∈ V : (x, β) > 0∀β ∈ R+(C)}

3.2.4 Subsets of roots

Let R be a reduced root system in a f.d. R-vector space V.

Definition 3.2.43. A subset P ⊂ R is closed if ∀α, β ∈ P if α+β ∈ R, thenα + β ∈ P. It is parabolic if it is closed and R = P ∪ (−P).

Proposition 3.2.44. Let P ⊂ R be closed and such that P∪(−P) = ∅, thenthere exists a Weyl Chamber C of R such that P ⊂ R+(C).

Proof. Assume w.l.o.g. that P 6= ∅.Claim. If α1, · · ·αn ∈ P and n ≥ 1, then α = α1 + · · ·+ αn 6= 0.

We proceed by induction on n. For n = 1, theres nothing to do. If n ≥ 2,then assume α = 0. Thus,

(α1, α2 + · · ·+ αn) = (α1,−α1) < 0

Hence, there exists j ≥ 2 such that (α1, αj) < 0. Since α1 6= −αj becauseP ∩ (−P) = ∅ we have, by ’Observations’, that α1 + αj ∈ R, and since Pis closed, α1 + αj ∈ P. Then α cannot be zero since α1 + αj 6= 0 and byinduction

∑i 6=j,1 αi 6= 0.

Claim. There exists α ∈ P,such that∀β ∈ P, (α, β) ≥ 0.

Otherwise, there exist α1, α2 ∈ P such that (α1, α2) < 0, so α1 + α2 ∈ P.Then there exists α3 ∈ P such that (α1 + α2, α3) < 0, so α1 + α2 + α3 ∈ P.Hence for all i we may find a sequence α1, · · · , αi such that α1 + · · ·+αi ∈ P.Since P is finite there exist αj+1, · · · , αk ∈ P with αj+1 + · · · + αk = 0,contradicting the previous claim.

Claim. There exists an ordered basis of V such that all elements of P arepositive with respect to the lexicographic order. This means, with respect tothis basis, P ⊂ R+ = R+(B) = R+(C).


We proceed by induction on l = dim(V). For l = 1, just choose anelement of P. Now assume l ≥ 2. Then, by the previous claim, there existsv1 ∈ V − {0} such that (v1, β) ≥ 0∀β ∈ P. Consider the hyperplane

H := (Rv1)⊥ = {x ∈ V : (x, v1) = 0}.

We now have that R∩H ⊂ span(R∩H) is a root system, and P∩H ⊂ R∩His closed and (P ∩H) ∩ (−P ∩H) = ∅. By induction, there exists an orderedbasis (v1, · · · , vr) of spanR(R ∩ H) such that elements of P ∩ H are positivewith respect to the lexicographic order. The claim now follows.

Corollary 3.2.45. Let P ⊂ R be a subset. The following are equivalent:

1 There exists a Weyl Chamber C of R such that P = R+(C). If so, thesuch a C is unique.

2 P is parabolic and P ∩ (−P) = ∅

3 P is closed and R = P ∩ (−P).

Proposition 3.2.46. Let P ⊆ R, then the following are equivalent:

1 P is parabolic.

2 P is closed and there exists a Weyl chamber such that R+(C) ⊆ P.

Proof. If R+(C) ⊂ P for some Weyl chamber C, then R = P∪ (−P). Assumenow that P is parabolic, and let C be a Weyl chamber such that |P∩R+(P)|is maximal.

Claim. B ⊂ P

Otherwise, let β ∈ B − P. Since R = P ∪ (−P), then −β ∈ P. LetC′ := sβ(C). Then, since sβ permutes R+(C)− {β}, we have

R+(C′) = sβ(R+(C)) = (R+(C)− {β}) ∪ {−β};

P ∩ R+(C) = (P ∩ R+(C)− {β})◦∪ P ∩ {−β}

Since the latter is impossible, then B ⊆ P.


———————————————————————————————Lecture 18, 08/06/2011

3.2.5 Classification of a parabolic subset over a fixedR+(B)

Lemma 3.2.47. Let B be a base of R, and R+ := R+(C) = R+(B). Let Pbe closed, and assume that R+ ⊆ P (so P is parabolic). Let Σ := B ∩ (−P),and let Q be the set of roots that are sums of elements in −Σ = (−B) ∩ P.Then P = R+ ∪Q.

Lemma 3.2.48. Let B be a base of R, let Σ ⊆ B be a subset. and letQ = Q(Σ) as above. Then P := R+ ∪Q is closed in R (and parabolic).

Corollary 3.2.49. Let B be a base of R, notation as above. Then,{parabolic subsets of R

containing R+

}∼←→ { subsets of B} = P(B)

P 7−→ B ∩ (−P)

R+ ∪Q(Σ)←− [ Σ

In particular, the set on the left has 2|B| elements.

3.3 Borel and Parabolic subalgebras of a com-

plex semi simple Lie algebra.

Recall.

g− complex semisimple Lie algebra,

h ⊆ g Cartan subalgebra

R = R(g, h) ⊂ h∗ root system

g = h⊕⊕α∈R

gα, h = g0.

V ∈ h−mod, λ ∈ h∗,

Vλ := {v ∈ V : hv = λ(h)v∀h ∈ h}P(V) := set of weights of V = {λ ∈ h∗ : Vλ 6= {0}}

3.3. BOREL AND PARABOLIC SUBALGEBRAS OF A COMPLEX SEMI SIMPLE LIE ALGEBRA. 83

Definition 3.3.1. A Borel subalgebra of (g, h) is a Lie subalgebra of g ofthe form

b = h⊕α∈R+

gα

For some choice of B a basis of R. A parabolic subalgebra of (g, h) is a Liesubalgebra Psubseteqg that contains a Borel subalgebra of (g, h).

Proposition 3.3.2. We have an inclusion preserving bijection:

{closed subsets of R} 1:1←→{

Lie subalgebras of gthat contain h

}P 7→ a(P) := h⊕

⊕α∈P

gα{non-zero weights of a

w.r.t.h− action

}P(a)− {0} ←− [ a

Moreover, if P ⊆ R is closed, then:

1 a(P) is a Borel subalgebra of (g, h) ⇐⇒ R = P◦∪ (−P)

2 a(P) is a parabolic subalgebra of (g, h) ⇐⇒ P is parabolic.

Corollary 3.3.3. There are bijections

{Weyl chambers of R} ←→ {bases of R} ←→ {Borel subalgebras of (g, h)}

Proposition 3.3.4. Let B be a base of R, and b the corresponding Borelsubalgebra of (g, h). Then,

{subsets ofB} = P(B)∼−→{

parabolic subalgebras of(g, h) containing b

}σ 7→ h⊕ b

⊕α∈R− a sum of elements in −Σ

gα

———————————————————————————————Lecture 19, 20/06/2011

Chapter 4

Highest Weight Theory

Definition 4.0.5. Let g be a s.s. f.d. complex Lie algebra, and h ⊆ g Cartansubalgebra, V a representation of g, and λ ∈ h∗. Then

V(λ) = Vλ := {v ∈ V : hv = λ(h)v∀h ∈ h}

is the λ-weight space of V. The element λ is called a weight of V if Vλ 6= {0}.Let P(V) be the set of all weights of V. Often elements in h∗ are just called(possible) weights.

Example 4.0.6. Let V = g, the adjoint representation. Then P(V) =R ∪ {0}.

Remark 4.0.7. If Cλ denotes the one dimensional representation, wherehv := λ(h)v, then for any h-module V,

HomU(h)(Cλ,V) ∼= Vλ

f 7→ f(1)

Definition 4.0.8. Let R+ ⊆ R be a choice of positive roots. The dominant integral weightsis defined as

X+ = X+(R+) = {λ ∈ h∗ :< λ,∨α >∈ Z≥0∀α ∈ R+}

and

R+ ⊆ X+ ⊆ X = X(R) := {λ ∈ h∗ :< λ,∨α >∈ Z∀α ∈ R}

85

86 CHAPTER 4. HIGHEST WEIGHT THEORY

is called the set of integral weights. We call the set

Q := {∑

hαα : hα ∈ Z}

the root lattice, and

Q+ := {∑

hαα : hα ∈ Z≥0}

the positive root lattice.

Example 4.0.9. Let g = sl(2,C), then R+ = {α}, h∗ ∼= C, X(R) = Zω1,

where ω1 = α2, since < α,

∨α >= 2.

Remark 4.0.10. In general, the set {∨αi : αi ∈ B, a basis of R, i = 1, · · · , n},forms a basis of h. Choose ωi ∈ h∗ such that < ωi,

∨αj >= δij for all 0 <

i, j ≤ n. Then

X(R) = Zω1 + · · ·+ Zωn ⊂ h∗

It is actually a free Z-module with basis the ω′is. Each ωi is called thei-th fundamental dominant integral weight, and

X+(R) = Nω1 + · · ·+ Nωn ⊂ h∗

Aim. finite dimensional

irreducible representationsof g up to isomorphism

1:1←→ X+(R+)

V 7→ maximal element in P(V)

Recall. For sl(2,C):finite dimensional

irreducible representationsof sl(2,C) up to isomorphism

1:1←→ Z≥01:1←→ X+(R)

V 7→ dim(V)− 1

Definition 4.0.11. Define a partial ordering on X+(R),X(R),Q,Q+ by

λ ≥ µ ⇐⇒ λ− µ ∈ Q+

87

Example 4.0.12. Let g = sl(2,C). Let V be an irreducible f.d. representa-tion of dimension n + 1. Then P(V) = {nω1, (n − 2)ω2, · · · − nω1}, and themaximal element is nω1, the minimal element is −nω1.

Remark 4.0.13. If V is not necessarily finite dimensional, or irreducible,then

• maximal/minimal element doesn’t necessarily exists (example: the Vermamodule M(0))

• P(V) could be empty (U(g) as left g-module)

• maximal/minimal elements need not be unique (take V ⊕ V for V afinite dimensional irreducible sl(2,C) module).

However it is all true for f.d. irreducible g-modules

Lemma 4.0.14. Let V be a representation of g, setup as above. Let 0 6=v ∈ V. The following are equivalent:

1 v¯⊆ Cv

2 hv ⊂ Cv; n+v = {0} where n+ =⊕α∈R+

gα

3 hv ⊂ Cv and gαv = {0}∀α ∈ B, where B is a basis of R with positiveroots R+.

If this holds, then v ∈ V is called a primitive vector of V.

Lemma 4.0.15. Let V be a representation of g, and v ∈ V a primitivevector. Let U be the n−-submodule of V generated by v, where n− =

⊕α∈R−

gα.

Assume that V is generated by v as a U(g)-module. Then:

1 U = V

2 dim(Vµ) <∞, dim(Vλ) = 1∀µ ∈ P(V)

3 V =⊕

µ∈P(V)

Vµ

4 ∀µ ∈ P(V), µ ≤ λ


5 EndU(g)(V) = C idV

Example 4.0.16. Let V = M(0) for g = sl(2,C), and h the standard Cartan

subalgebra. Action of f =

(0 01 0

):

•

��

(weight λ = 0)

•

��•

...

Proof. The space U is generated as a vector space by elements of the formxα1 · · ·xαnv with αi ∈ gi for some αi ∈ R−, and not necessarily αi 6= αj.Note that

x · (xα1 · · ·xαnv) =n∑j=0

xα1 · · ·xαj−1[x, xαj ]xαj+1

· · ·xαn · v + xα1 · · ·xαnxv

(To see this note that x ·xα1xα2 · v = [x, xα1 ]xα2v+xα1xα2xv and then doinduction.) So, if h ∈ h then the above implies that

h · (xα1 · · ·xαnv) = (α1(h) + · · ·+ αn(h) + λ(h))v,

in particular, xα1 · · ·xαnv is a weight vector of weight λ + α1 + · · ·αn. Notethat α1 + · · ·αn is a sum of negative roots. Now take x ∈ n+, then xv = 0because v is primitive and

[x, xαj ] ∈ h⊕⊕β>αj

gβ

so by induction on n, x · (xα1 · · ·xαnv) ∈ U. Hence U is a U(g) submoduleof V containing v ∈ V, hence, since V is generated by v as a U(g)-module,U = V, hence (1). For (2), we have

dim(Vµ) ≤ |{(p1, · · · pn) ∈ Nn|λ−r∑i=1

piβi = µ}| = P(λ− µ)

for B = {β1, · · · βr}.

89

Remark 4.0.17. P(λ−µ) = the Constant’s partition function counting thepossibilities of writing λ − µ as a non negative linear combination of basiselements βi ∈ B. By definition P(λ−λ) = dim(Vλ) = 1, (3) is now clear. For(5), let f ∈ EndU(g)(V), then f(v) ∈ Vλ = Cv, and f(uv) = uf(v). Hence fis determined by f(v), and

EndU(g)(V) ∼= Cvv 7→ f(v)


———————————————————————————————Lecture 20, 22/06/2011

Lemma 4.0.18. Let V be a simple representation of g, and let λ ∈ P(V).The following are equivalent:

1 ∀µ ∈ P(V), µ ≤ λ

2 λ is a maximal weight

3 If α ∈ B then α + λ /∈ P(V)

In this situation, the U(g)-module is called a highest weight module.

Definition 4.0.19. A U(g)-module V (not necessarily simple) is called ahighest weight module if there exists a primitive vector v ∈ Vλ for someλ ∈ P(V) and V is generated as a U(g)-module by v.

Proof. It’s clear that 1 =⇒ 2, and, if λ is maximal, α + λ > λ, so λ + α /∈P(V), so 2 =⇒ 3. Now, let 0 6= v ∈ Vλ, and assume 3. Then, hv ⊆ Cv andxαv = 0 because λ+ α /∈ P(V) for all α ∈ B, xα ∈ gα, hence, v is a primitivevector. So, 3 =⇒ 4. Now assume 4. SInce V is simple, it is generatedby the given primitive vector v. By Lemma 4.0.15, we have µ ≤ λ for allµ ∈ P(V).

Let V be a simple representation of V such that P(V) has a maximalelement λ. Then:

1 dim(Vµ) <∞ for all µ, and dim(Vλ) = 1.

2 Vλ = { primitive vectors }

3 Fro all µ ∈ P(V), µ ≤ λ.

4.1. CONSTRUCTION OF HIGHEST WEIGHT MODULES 91

4.1 Construction of highest weight modules

Properties and Universality of Verma Modules

Setup. g is a complex s.s. Lie algebra, h ⊆ g a Cartan subalgebra, R+ ⊂ Ra choice of positive roots.

g = n− ⊕ h⊕ n+; b the corresponding Borel subalgebra.

Senote by ≤ the partial ordering on h∗.

Theorem 4.1.1. 1

M(λ) = U(g)⊗U(b) Cλ

is the Verma module of highest weight λ, and, as U(n−)-modules,

U(n−) ∼= M(λ)

u 7→ u(1⊗ 1)

as U(n−)-module.

2 M(λ) = M(λ)µµ∈h∗

, ( dim)(M(λ)λ) = 1.

3 dim(M(λ)µ) = P(λ− µ), where P is Konstant’s partition function.

In particular M(λ) is a highest weight module of highest weight λ. It isuniversal in the sense that if µ is another highest weight module of highestweight λ, then there exists a surjection

M(λ)� M

Proof. By the PBW Theorem, we have

M(λ) = U(g)⊗U(b) Cλ ∼= U(n−)⊗ U(b)⊗U(b) Cλ

as vector space, and as U(n−)-left module, it is isomorphic to U(n−) ⊗ Cλ.Now, for each α ∈ R+, choose x−α ∈ g−α; the PBW Theorem says thatU(n−) has as a basis

{xm := xm1−α1· · ·xmnαn } for mi ∈ Z≥0.


Then, xm ·vλ is a weight vector of weight λ−∑n

i=1miαi, and by 1, the xm ·vλform a basis of M(λ). It is now clear that in 2 we even have

M(λ) =⊕

µ∈h∗,µ≤λ

M(λ)µ

Now, by definition of M(λ), we have that vλ = 1 ⊗ 1 generates M(λ) as aU(g)-module, so M(λ) is in fact a highest weight module.Let M be any U(g) module, and let λ ∈ P(V) be maximal.

Claim. HomU(g)(M(λ),M) 6= 0

Indeed, the adjunction of ⊗ and Hom gives

HomU(g)(M(λ),M) = HomU(g)(U(g)⊗U(b) Cλ,M)

= HomU(b)(Cλ,HomU(g)(U(g),M))

= HomU(b)(Cλ,M)

= HomU(h)(Cλ,M) 6= 0

Given by 1 7→ v, v ∈ Mλ. If M is a highest weight module generated byv ∈ Mλ, then fv : M(λ)� M is surjective.

Classification of irreducible/ simple highest weight modules

Theorem 4.1.2. 1 M(λ), for λ ∈ h∗, has a unique proper maximal sub-module rad(M(λ))

2 Let L(λ) = M(λ)/ rad(M(λ)). Then there is a bijection

h∗1:1←→

{irreducible highest weight

U(g)-modules up to isomorphism

}λ 7→ L(λ)

3 if a simple representation has a highest weight, then it is already ahighest weight module.

Proof. 1 Let U ⊆ M(λ) be a U(g)-submodule. Then U =⊕µ∈h∗

Uµ because,

since M(λ) is generated by v ∈ M(λ)λ which has dimension one andso any proper submodule is contained in M

µ 6=λ(λ)µ. In particular, the

sum of all proper submodules must be a proper submodule, equal torad(M(λ)).


2 L(λ) is simple by construction.

Claim. L(λ) is a highest weight module and L(λ) = L(µ) =⇒ λ = µ.

Let can : M(λ) � L(λ) be the canonical surjection, then can(1⊗ 1) ∈L(λ)λ is the highest weight vector, so since L(λ) is simple it is generatedby can(1⊗ 1), so it is a highest weight module.Now assume that M is another simple highest weight module , suchthat f : M(λ) � M. Then ker(f) ⊂ rad(M(λ)), so by isomorphismtheorem we have an induced map L(λ) → M which is non-zero, hencean isomorphism, since M,L(λ) are both simple. We conclude that M(λ)has a unique simple quotient.

Aim. dim(L(λ)) <∞ ⇐⇒ λ ∈ X+(R+)

Lemma 4.1.3. For α ∈ B,

M(sα · λ) ⊆ M(λ)

if sα ◦ λ ≤ λ where

Definition 4.1.4.

sα ◦ λ := sα(λ+ ρ)− ρ;

ρ =1

2

∑α∈R+

α

is called the dot action of WR on h∗.

Remark 4.1.5. < ρ,∨α >= 1∀α ∈ B because, for α ∈ B,

sα(ρ) = ρ− < ρ,∨α > α =

1

2sα(∑β 6=α

β) +1

2sα(α) =

1

2β − 1

2α = ρ− α

Proof of Lemma 4.1.3. Let sα · λ < λ, α ∈ B ⇐⇒ sα(λ + ρ) − ρ < λ ⇐⇒sα(λ) + sα(ρ)− ρ < λ ⇐⇒ sα(λ) + ρ− < ρ,

∨α > α < λ+ ρ =⇒ sα(λ) < λ

in our partial ordering. Hence < λ,∨α >∈ Z≥0.

Claim. Let x−α ∈ g−α. Then ω = xn+1−α vλ 6= 0, where vλ = 1⊗ 1 ∈ M(λ), but

xβxn+1−α vλ = 0∀β ∈ B, in other words, ω is a highest weight vector.


Proof of Claim. We have that xβω = 0 because [xβ, x−α] = 0 and so xβxn+1−α vλ =

xn+1−α xβvλ = 0. Further, xαx

n+1−α vλ = 0 because xαx

j−αvλ0 = j(n−j+1)xj+1

−α vλby sl(2,C)-theory, considering sl(2,C) ∼= g−α ⊕ h ⊕ gα and the sl(2,C)-modules gen by vλ, x−αvλ, · · · .

By universal property of Verma modules, we have

M(sαλ)→ M(λ)

1⊗ 1 7→ ω

because ω has weight sα · λ. It remains to show that the above map isinjective. But for u ∈ U(n−), we have that u(1 ⊗ 1) 7→ uω, the modulesM(sα·λ) and M(λ) are both free U(n−) - modules of rank 1, and the latterhas no zero divisors by the PBW theorem, so the map must be injective.

Theorem 4.1.6. Let g be a s.s. complex Lie algebra. There is a bijection

X+(R+)1:1←→

{irreducible finite dimensional

representations of g up to isomorphism

}λ 7→ L(λ)

highest weight of V←[ v


Question. What is dim(L(λ))? What is dim(L(λ))µ?

Weyl’s dimension formula

Theorem 4.1.7. Let g be a s.s. complex finite dimensional Lie algebra,λ ∈ X+(R). Then,

dim(L(λ)) =

∏α∈R+

< λ+ ρ,∨α >∏

α∈R+

< ρ,∨α >

Example 4.1.8. Let g = sl(2,C), h ⊂ g standard Cartan subalgebra. Then

X+(R+)1:1←→ N

nω1 ← [ n

L(nω1) is by definition the n+1 dimensional irreducible sl(2,C)-module. Ap-plying the dimension formula we get

dim(L(nω1)) =< λ+ ρ,

∨α >

< ρ,∨α >

=< λ,

∨α > +1

1= n+ 1

To express dim(L(λ))µ one usually uses character theory.


4.2 Character formula

Let mλ(µ) = dim(L(λ)µ), and let P be the weight lattice.

Definition 4.2.1. Let f ∈ Maps(P,Z) be a function. Define

Supp(f) := {λ ∈ P : f(λ) 6= 0}

Let

H := {f ∈ Maps(P,Z) : ∃S ⊆ P finite ∀λ ∈ Supp(f), λ ≤ s for some s ∈ S}

We define a ring structure on H by (f + g)(λ) = f(λ) + g(λ) and (fg)(λ) =∑µ∈P f(µ)g(λ−µ), which is well defined because of the definition of H. For

any λ ∈ P, define

eλ(µ) = δλ,µ.

Proposition 4.2.2. We have

1. eλ ∈ H

2. eλeµ = eλ+µ

3. e0 is a 1 for H

Definition 4.2.3. For each finite dimensional U(g) module V, we define

ch(V) :=∑µ∈h∗

dim(Vµ)eµ ∈ Zh∗

This is called the (formal) character of V.

Proposition 4.2.4. Let ρ = 12

∑α∈R+ α. Then

Ch(M(λ))∏α∈R+

(eα/2 − e−α/2) = Ch(M(λ))∑w∈W

(−1)l(w)ew(ρ) = eλ+ρ

Theorem 4.2.5 (Weyl Character formula).

Ch(L(λ))∑w∈W

(−1)l(w)ew(ρ) =∑w∈W

(−1)l(w)ew(λ+ρ)

4.3. CATEGORY O 97

——————————————————————————————Lecture 04/07/2011

4.3 Category ODefinition 4.3.1. Let g be a semi-simple complex finite dimensional Liealgebra with a fixed Cartan subalgebra h ⊆ g and R+ ⊆ R be a systemof positive roots, hence a Borel subalgebra b = h

⊕α∈R+

gα. Then the BGG

category O(g) = O(g, h,R+) is the full subcategory of U(g)-modules givenby all the U(g)-modules M which satisfy

O1 M is finitely generated as a U(g)-module.

O2 There is a weight space decomposition M =⊕λ∈h∗

Mλ.

O3 M is locally U(b)-finite, i.e. for every m ∈ M there exists m ∈ N ⊆ Ma finite dimensional U(b) invariant vector subspace.

Remark 4.3.2. O3 implies that P(M), the set of weights of M, is containedin the set

{finite union of sets of the form λ−Q+}

Example 4.3.3.

• Every finite dimensional U(g)-module is an object in O(g).

• Every Verma module M(λ) is an object of O(g).

Proposition 4.3.4.

1. The category O(g) is noetherian, i.e. every object is a noetherian U(g)-module.

2. O(g) is closed under taking finite direct sums of objects, submodulesand quotients.

3. Given an object M in O(g), and E a finite dimensional U(g)-module,then M⊗ E is an object in O(g).


4. Given an object M in O(g), then it is finitely generated as an U(n−)-module.

Remark 4.3.5. Since M(λ) is in O(g), Proposition 4.3.4 (2) implies thatL(λ) is also an object in O(g).

Proof. 1. Recall that by the PBW theorem, we have that gr(U(g)) ∼= §(g).In particular, U(g) is noetherian. Since every finitely generated moduleover a noetherian ring is noetherian, O1 implies 1.

2. Taking finite sums is ok.Submodules of finitely generated modules over a noetherian ring areagain finitely generated, hence O1 holds, we already proved that O2holds, and O3 id direct.

3. M,E as in the statement.Let M =⊕µ∈h∗

Mµ, E =⊕µ∈h∗

Eν where (M⊗E)λ

is spanned by all vectors m⊗ e where m ∈ Mµ, e ∈ Eν and λ = µ+ ν.Let m1, · · ·mn be generators of M as a U(g)-module and let v1, · · · vrbe a basis of E.

Claim. The set {mi ⊗ vj : 1 ≤ i ≤ n; 1 ≤ j ≤ r} generate M ⊗ E as aU(g)-module.

Since M is already finitely generated as a U(g)-module, it follows.

Remark 4.3.6. Category O(g, b) is an abelian category, i.e. it satisfies:

A0 There is a zero object (namely the module {0}).

A1 For every pair M,N of objects, the direct sum and product is again inO(g, b).

A2 Every morphism between objects has kernel and cokernel.

A3 Every monomorphism is part of a kernel. Every epimorphism is partof a cokernel.

Theorem 4.3.7. Every object M ∈ O(g, b) has a finite filtration

{0} ⊆ M0 ⊆ M1 ⊆ · · · ⊆ Mr = M

of U(g)-modules such that Mi/Mi−1∼= L(µi) for some µi ∈ h∗. This is called

the Jordan Holder series.

4.3. CATEGORY O 99

Remark 4.3.8. This filtration is not unique: Say M = L(λ) ⊕ L(µ), then{0} ⊆ L(λ) ⊆ M, {0} ⊆ L(µ) ⊆ M are both Jordan Holder series but ofcourse different. However, the sub quotients in a given filtration do notdepend on the filtration.

Question. What is [M(λ),L(µ)] = aλµ?

If we know aλ, µ, then, via the following

Fact 4.3.9. If [M(λ) : L(µ)] 6= 0, then µ ∈WR·λ and aλµ 6= 0 for only finitelymany µ′s and in particular, M(λ) = L(λ) if λ is minimal in it’s WR-orbitunder dot-action.

Example 4.3.10. Let g = sl(2,C).

•

��

λ = 0

•

��

−2

•

��

−4

...

We get a short exact sequence

M(sα · 0) ↪→ M(0)� L(0)

and M(sα · 0) = L(sα · 0).

Theorem 4.3.11 (Conjectured by Kazhdan-Lusztig, proven by several peo-ple in the 1980’s).

[M(y · λ) : L(x · λ)] = px,y(1)

Where px,y is a polynomial defined recursively, in the ring Z[q, q−1]. It iscalled a Kazdhan-Lusztig polynomial.


Warning. In general almost nothing is known about irreducible U(g)-modulesnot of the form L(λ). Many statements can be reformulated aas statementsabout objects in O(g, b).

Example 4.3.12. Let M be an irreducible U(g)-module (g a finite dimen-sional f.d. Lie algebra). Almost nothing known!

——————————————————————————————Lecture 06/07/2011

4.4 Cartan matrices and Dynkin diagrams

Let (V,R) be a reduced root system in an n-dimensional vector space V. LetB = {α1, · · · , αn} be a chosen basis with a fixed ordering.

Definition 4.4.1. The Cartan matrix associated to (V,R,B) is the n × nmatrix A = (aij)1≤i,j≤n, where aij :=< αi,

∨αj >=

2(αi,αj)

(αi,αi)

Remark 4.4.2. The Cartan matrix depends on the choice of B and the orderon the simple roots, however, different orderings just gives a Cartan Matrixwhich differs from the original by conjugation with a permutation matrix;and similarly if we choose a different basis.

Proposition 4.4.3. The Cartan matrix A of a root system (V,R) satisfies:

1. Aij ∈ Z.

2. Aii = 2.

3. Aij ≤ 0 if i 6= j.

4. Aij = 0 ⇐⇒ Aji = 0.

5. There exists a diagonal matrix D with positive diagonal entries suchthat S = DA is symmetric.

Definition 4.4.4. A n×nmatrix satisfying the above is called an abstract Cartan Matrix.Two Cartan abstract matrices are isomorphic if they differ by conjugationwith a permutation matrix.

4.4. CARTAN MATRICES AND DYNKIN DIAGRAMS 101

Proposition 4.4.5. A root system (V,R) is reducible if and only if thecorresponding Cartan matrix is block diagonal with more than one block.

Definition 4.4.6. An abstract Cartan matrix is reducible if it is block di-agonal with more than one block after maybe renumbering simultaneouslyrows and columns.

Definition 4.4.7. Let A be a Cartan matrix associated to a reduced rootsystem (V,R,B) as above. Then we associate to A a finite graph which iscalled a Dynkin diagram as follows:

• to each simple root (i.e. to each element of B) we associate a vertex ofthe graph.

• to vertices corresponding to αi ∈ B and αj ∈ B we put AijAji numberof lines connecting the two vertices.

• If αi, αj ∈ B, with i 6= j with vertices connected by at least one linesegment, and ||αi|| > ||αj|| then we indicate this by putting >, someans ||αi|| > ||αj||, if i < j and the nodes are ordered accordingly.

Examples.

g = sl(2,C)⊕ sl(2,C) A =

(2 00 2

)

g = sl(3,C) A =

(2 −1−1 2

)


ε1

ε2

ε3 ε1 − ε2 = α

ε1 − ε3 = α + β

ε2 − ε3 = β

g of type G2; A =

(2 −3−1 2

)

ε1

ε2

ε3 ε1 − ε2 = α

2ε1 + ε2

ε1 + 2ε2

Recall. Given B,B′ two bases of a root system (V,R), then there exists aunique w ∈WR such that w(B) = B′.

Proposition 4.4.8. Let (V,R) be a root system of some finite dimensionalsemi-simple complex Lie algebra, and B ⊂ R+,B′ ⊂ R+ two choices of posi-tive roots. Then the corresponding Cartan matrices are isomorphic.

Proof. Let B = {α1, · · ·αn},B′ = {α′1, · · ·α′n} be ordered in a way such thatw(αi) = α′i. Then:

< α′i,∨α′j > =

2(α′i, α′j)

(α′i, α′i)

=2(w(αi), w(αj))

(w(αi), w(αi))

=2(αi, αj)

(αi, αi)=< αi,

∨αj >


Hence, the corresponding Cartan Matrices agree up to simultaneous permu-tation of rows and columns.

Proposition 4.4.9. Let R,R′ be non-isomorphic reduced root systems withbases B,B′ respectively. Then the Cartan Matrices are not isomorphic.

Proof. Excercise;

Recall. Let (V,R), (V′,R′) be root systems. They are isomorphic if thereexists an isomorphism F of vector spaces F : V → V′ such that F(R) = R′,

and < F(α),F(∨β) >=< α,

∨β >.

Proposition 4.4.10. Let (V,R,B), (V′,R′,B′) de root systems with corre-sponding bases. Assume there exists a bijection f : B → B′ transformingthe Cartan matrix for (V,R,B) into the Cartan matrix for (V′,R′,B′). Then(V,R) is isomorphic to (V′,R′) via an isomorphism F : V → V′ which ex-tends f . In particular, the Cartan matrix defines the root system up toisomorphism.

Proof. Let B = {α1, · · · , αn},B′ = {α1, · · · , αn} be bases of R resp R′, so inparticular they are bases of V, resp. V′. Then we have

sF(α)(F(β)) = F(β)− < F(β),∨

F(α) > F(α) = F(sα(β))

hence

sF(α) ◦ F = F ◦ sα

and so

WR →WR′

w 7→ F ◦ w ◦ F−1

is an isomorphism of the Weyl groups. Further, since F(β) = sF(α)(F(sα(β))) ∈

R′, then F(R) ⊂ R′, and < α,∨β >=< F(α),F(

∨β) > is an exercise.

Proposition 4.4.11. Let (V,R) be a reduced root system, and let Γ be it’sDynkin diagram.Then


1. (V,R) is irreducible ⇐⇒ Γ is connected.

2. Γ has no cycles, i.e. it is a forest. By 1., if it is irreducible, then Γ is atree.

Proof. 1. If R = R1∪R2, and some bases B1,B2 then the Dynkin diagramsΓ(Ri) are disjoint. Conversely, assuming that Γ(R) is not connected,then we have by definition a basis of R, B = B1∪B2 where there are noconnections between the vertices of B1 and the vertices from B2. LetVi denote the span of Bi inside of V. Then, since the Bi are orthogonalwith respect to (−,−), then both Vi are WR-stable. Then Ri = R∩Vi

give decomposition R = R1 ∪ R2.

2. Assume β1, · · · , βn ∈ B correspond to vertices of a cycle, and let γi =βi||βi|| . Now, if there is an edge between vertex βi and vertex βj, then wehave

||n∑i=1

γi||2 = n+ 2∑i<j

(γi, γj)

≤ n+ 2[(γ1, γ2), · · · (γn−1, γn)]

≤ n− n = 0

Contradiction!


Lecture 11/07/2011

4.4.1 Classification of irreducible, reduced root sys-tems/ Dynkin Diagrams

Theorem 4.4.12. Let (V,R) be a reduced irreducible root system, and letΓ = Γ(R) be the corresponding Dynkin Diagram. Then Γ is isomorphic toexactly one diagram from the following list.

An

Bn

Cn

Dn

G2

F4

E6

E7

E8

Remark 4.4.13. It’s easy to see by hand that the Dynkin diagrams listedabove are pairwise non-isomorphic.

Proof of Theorem 4.4.12. Recall that if βi, βj, i < j are simple roots, and if

we denote f(i, j) = ||βi||2||βj ||2 , then we have:

I aij = 0 = aji; θ(βi, βj) = π2

II aij = 1 = aji; θ(βj, βi) = 2π3

III f(j, i) = 2; f(i, j) = 12; aij = −2; aji = −1; ||βj|| =

√2||βi||; θ(βi, βj).

IV f(j, i) = 3, f(i, j) = 13, aij = −3, aji = −1; θ(βi, βj) = 5π

6.


Now: Classify the Dynkin diagrams.

1. Assume Γ has a triple edge. Then there are i, j with f(j, i) = 3 and

(γi, γj) =√

32

. Assume we have an extra vertex k connected to i, w.l.o.g.Then we know (γi, γk) ≤ −1

2; (γj, γk) ≤ 0, hence,

||√

3γj + 2γi + 2γk||2 ≤ 3 + 4 + 1− 2√

3 · 2√

3

2− 4

2= 0

And similarly consider the cases when

1. Assume Γ has two double edges, get a contradiction.

2. Assume Γ has a double edge and a branching point, get a contradiction.

3. Assume Γ has a double edge, get Bn,Cn,F4.

4. Assume Γ has no ramification point, get type An.

5. Assume Γ has more than one branching point, get contradiction.

6. Assume Γ has only one branching point, get Dn,E6,E7,E8

Problem. Find a simple Lie algebra for each Dynkin Diagram from Theorem4.4.12.

Answer. Existence and Uniqueness Theorems from Serre.

Theorem 4.4.14 (Serre). Assume g is a semisimple complex finite dimen-sional Lie algebra with Cartan matrix A ∈ Matn×n(Z). Then g is isomorphicas a Lie algebra to the Lie algebra with generators {ei, fi, hi : 1 ≤ i ≤ n}and relations:

“sl(2,C)-type relations”

1. [hi, hj] = 0

2. [hi, ej] = aijej

3. [hi, fj] = −aijfj


4. [ei, fj] = δijhi

“Serre relations”

5. (ad ei)−aij+1ej = 0, for i 6= j.

6. (ad fi)−aij+1fj = 0, for i = ±j.

Example 4.4.15. For sl(n,C), we have that the Cartan matrix is given byaii+1 = −1 = aii−1. Then relation 5 becomes eiej = ejei if |i − j| > 1 ande2i ej − 2eiej + eie

2j = 0 if |i− j| = 1, and similarly relation 6 for the f ′is.

To define a Lie algebra with generators and relations we need a free Liealgebra:

Definition 4.4.16. A free Lie algebra on a set X (of so called) generatorsis a pair (L, i), where L is a Lie algebra and i is a map i : X→ L, such thatthe following universal property holds:For any Lie algebra g (over the same field as L), and function ϕ : X → g,there is a unique map making the following diagram commute:

X i //

ϕ

��

L

��g

Remark 4.4.17. Free Lie algebra on X is unique up to isomorphism, andexistence is similar to the construction of a free group.

A Lie algebra generated by X with relations R is then the quotient of thefree Lie algebra L by the ideal generated by the relations.

Theorem 4.4.18 (Serre). Given an irreducible Cartan matrix, the Lie alge-bra with generators and relations as in Theorem is a simple complex finitedimensional Lie algebra. Two such algebras are isomorphic if and only iftheir Cartan matrices are isomorphic.

Proof: Humphreys.


Last Lecture!

Last time:

Classified irreducible root systems via their Dynkin Diagram, hence obtaineda classification of complex finite dimensional simple Lie algebras. On theother hand,

Question. Can we classify the Weyl group? Does WR distinguish the rootsystem?

Answer. No: WBn∼= WCn .

Weyl Groups are special examples of Coxeter groups.

Goal. Classify those.

Definition 4.4.19. A Coxeter system is a pair (W, §), where

• W is a group.

• S ⊆ W is a set of generators of W, subject only to the following rela-tions:

(ss′)m(s,s′) = e

for every s, s′ ∈ S, and some m(s, s′) ∈ N ∪∞, such that m(s, s) = 1,and m(s, s′) ≥ 2 for s 6= s′, and m(s, s′) = 0 if there is no relationbetween s, s′.

To a Coxeter system, we may associate a Coxeter graph by putting avertex for each generator s ∈ §, and an edge from s to s′ labelled by m(s, s′).Often one omits the edges labeled by 2 and omits the labels 3.

Examples. 1. Let W = S2, and S = {(12)} ⊂ S2 generates S2 withthe single relation (12)2 = e. More generally, let W = Sn, and S ={(ii+ 1) = si : 1 ≤ i ≤ n−1} generates the symmetric group. One canshow that the relations generating it are

s2i = e(i.e.m(si, si) = 2)

(sisj) = sisjif |i− j| > 1(i.e.m(si, sj) = 2)

sisjsi = sjsisj ⇐⇒ (sisj)3 = e; ( so m(i, j) = 3) if |i− j| = 1.

So that (Sn, {si : 1 ≤ i ≤ n− 1}) is a Coxeter system.


2. The Universal Coxeter System (W, S) is given by generators from Sand relations m(s, s) = 1,m(s, s′) = ∞ if s 6= s′. If |S| = 2, then theUniversal Coxeter system has W = infinite Dihedral group, D∞.

Remark 4.4.20. This group appears as a “Weyl Group” for an infinitedimensional Lie algebra, g = sl(2,C). This Lie algebra is the Liealgebra corresponding to the Dynkin Diagram via generatorsand relations as in Serre’s theorem.

More concrete description: Start with g = sl(2,C). Then consider

gloop := g⊗ C[t, t−1]

which is a Lie algebra via [a⊗ tn, b⊗ tm] := [a, b]⊗ tn+m. Check: Thissatisfies the axioms of a Lie bracket. Now add extra central element c:

gloop ⊕ Cc : = g⊗ C[t, t−1]⊕ Cc[a⊗ tn + αc, b⊗ tm + βc] = [a, b]⊗ tn+m + K(a, b)δm+n,0c

Now add an extra derivation: δ(a ⊗ tm + αc) := t ddt

(a ⊗ tm), whichindeed defines a derivation on gloop ⊕ C. Now, as a vector space,

∧sl2 = sl2 ⊗ C[t, t−1]⊕ Cc⊕ Cd

Where [d,A] = δ(A) makes it, together with the earlier Lie bracket,into a Lie algebra. Then, get a Weyl group via root theory, and get:

Waff = WA1 o ZR+

3. (W, S = {s1, s2, s3}) with m(s1, s2) = 3,m(s1, s3) = 2,m(s2, s3) = ∞.Then W ∼= PSL(2,Z) := SL(2,Z)/(±1), namely, define Φ : W →PSL(2,Z) as:

s1 7→(

0 11 0

)s2 7→

(−1 10 1

)s3 7→

(−1 00 1

)


Well defined:

Φ(s1)2 =

(0 11 0

)(0 11 0

)=

(1 00 1

)Φ(s2)2 =

(−1 10 1

)(−1 10 1

)=

(1 00 1

)Φ(s3)2 =

(−1 00 1

)(−1 00 1

)=

(1 00 1

)Φ(s1s3)2 =

(0 1−1 0

)2

=

(−1 00 −1

)=

(1 00 1

)Φ(s1s2)3 =

(−1 00 −1

)Φ(s2s3)a =

(1 a0 1

)(infinite order!)

Note:

(0 1−1 0

),

(0 11 1

)generate SL(2,Z) as a group, hence, Φ is sur-

jective. To show it is injective, prove that PGL(2,Z) is a free productof subgroups of order 2 and 3 generated by the images above.

Definition 4.4.21. A group W is a Coxeter group, if there exists S ⊂Wsuch that (W, S) is a Coxeter system.

Two Coxeter groups are isomorphic as Coxeter groups if there existS ⊂ W, S′ ⊂ W′ such that (W, S) and (W′, S′) are Coxeter systems andan isomoprhism ϕ : W→W′ of groups such that ϕ(S) = S′.

Theorem 4.4.22. There exists a classification of finite Coxeter groups upto isomorphism. More precisely, they are classified by the following Coxeterdiagrams:

An(n ≥ 1)

Bn4

Dn


F44

E6

E7

E8

H34

H45

I2(m)m

Main idea. Realize W as a reflection group. Given V a vector space onbasis {αs : s ∈ S}, define a bilinear form B on V by:

B(αs, αs′) :=

{− cos( π

m(s,s′)) if m(s, s′) <∞

−1 if m(s, s′) =∞

In particular, B(αs, αs) = 1 and if s 6= s′, then B(αs, αs′) ≤ 0. We get areflection defined by rs(v) = v−B(v, αs)αs, and then a faithful representationcalled the geometric representation defined by s 7→ rs, hence realize Was a finite subgroup of GL(V) generated by reflections.

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