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NOTES TO LIE ALGEBRAS AND REPRESENTATION THEORY

ZHENGYAO WU

Abstract. • Main reference: [Hum78, Parts I, II, III].• Lecture notes to the graduate course “Finite dimensional algebra” during Spring 2019 atShantou University taught by me.

• Targeted audience: Graduate students in pure mathematics.

Contents

1. February 26th, Introduction to Lie algebra 22. March 5th, Solvable and nilpotent Lie algebras 123. March 12th, Lie theorem, Jordan decomposition 264. March 19th, Cartan’s criterion, Killing form 365. March 26th, Semisimple decomposition and Lie modules 496. April 2nd, Casimir element, Weyl’s theorem 617. April 9th, Representation of sl(2, F ), toral subalgebras 738. April 16th, Centralizer of H; Orthogonal, integral properties 839. April 23th, Rationality properties, reflections, root systems 9810. April 30th, Bases and Weyl chambers 11011. May 7th, Weyl group and its actions 12212. May 14th, Irreducible root systems, two root lengths and Cartan matrix 13313. May 21th, Coxeter graphs, Dynkin diagrams 14714. May 28th, Classification, irreducible root systems of types A, B and C 15715. June 4th, Irreducible root systems of types D,E,F,G 16916. June 11th, Weyl group of each type, Automorphisms of the Dynkin diagram, Weights 185References 205

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1. February 26th, Introduction to Lie algebra

Sophus Lie (1842-1899) established the theory in late 1880s in Oslo, Norway.

Definition 1.1Let F be a field. Let L be a F -vector space. We say that L is a F -Lie algebra if there exists amap L× L→ L, (x, y) 7→ [x, y] such that(L1) [x, y] is bilinear over F . (L2) [x, x] = 0 for all x ∈ L.(L3) Jacobi identity [x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0 for all x, y, z ∈ L.

Remark 1.2(L2’) [x, y] = −[y, x] for all x, y ∈ L.(1) If L is a F -Lie algebra, then (L2’) holds.(2) If char(F ) 6= 2, then (L2’) implies (L2).(3) Let H,K be subsets of L. We write [H,K] = SpanF[x, y], x ∈ H, y ∈ K. Then [H,K] =[K,H].

Definition 1.3Let K be a subset of L. We call K a subalgebra of L if(1) If K is a sub-F -vector space of L, and(2) [K,K] ⊂ K, then K is a F -Lie algebra.

Example 1.4Let V be a F -vector space. Let EndF (V ) be the space of endomorphisms (linear transformations)of V . For all x, y ∈ EndF (V ), define [x, y] = x y− y x, where means the composition of maps.Show that gl(V ) = (EndF (V ), [•, •]) is a F -Lie algebra, called the general linear algebra.If dimF (V ) = n, then we write gl(n, F ) the F -Lie algebra of all n×n matrices such that [eij, ekl] =δjkeil− δliekj where eij is the n×n matrix whose (i, j)-entry is 1 and all other entries are 0; δij = 1if i = j, otherwise δij = 0.Subalgebras of gl(V ) ' gl(n, F ) are called linear algebras.

Definition 1.5Let L,L′ be two F -Lie algebras. A map f : L→ L′ is an isomorphism if(1) f is a linear isomorphism of vector F -spaces, and(2) f([x, y]) = [f(x), f(y)] for all x, y ∈ L.

Theorem 1.6 Ado-IwasawaEvery F -Lie algebra is isomorphic to a linear F -Lie algebra.

Proof. Omit.

Definition 1.7Classical algebras are the following proper subalgebras of gl(V ) of type Al, Bl, Cl, Dl , l ≥ 1.

NOTES TO HUMPHREYS 3

Example 1.8Type Al (l ≥ 1): special linear algebras

sl(V ) = x ∈ gl(V ) : Tr(x) = 0, dimF (V ) = l + 1.

sl(l + 1, F ) = x ∈ gl(l + 1, F ) : Tr(x) = 0.

dimF (sl(l + 1, F )) = l2 + 2l since it has standard basis

eii − ei+1,i+1 : 1 ≤ i ≤ l ∪ eij : 1 ≤ i 6= j ≤ l + 1.

Example 1.9Type Bl (l ≥ 1): orthogonal algebras of odd degree.

Let f be a non-degenerate symmertic bilinear form on V whose matrix is s =

1 0 0

0 0 Il

0 Il 0

.

o(V ) = x ∈ gl(V ) : f(x(v), w) = −f(v, x(w)), ∀v, w ∈ V , dimF (V ) = 2l + 1,

o(2l + 1, F ) = x ∈ gl(2l + 1, F ) : sx = −xts

=

0 b1 b2

−bt2 m n

−bt1 p −mt

∈ gl(2l + 1, F ) : nt = −n, pt = −p

dimF (Bl) = 2l2 + l since Bl has standard basisei+1,i+1 − el+i+1,l+i+1 : 1 ≤ i ≤ l ∪ ei+1,j+1 − el+j+1,l+i+1 : 1 ≤ i 6= j ≤ l∪e1,l+i+1 − ei+1,1 : 1 ≤ i ≤ l ∪ e1,i+1 − el+i+1,1 : 1 ≤ i ≤ l∪ei+1,l+j+1 − ej+1,l+i+1 : 1 ≤ i < j ≤ l ∪ el+i+1,j+1 − el+j+1,i+1 : 1 ≤ j < i ≤ l.

Example 1.10Type Cl (l ≥ 1): symplectic algebras

Let f be a non-degenerate skew-symmertic bilinear form on V whose matrix is s =

0 Il

−Il 0

.sp(V ) = x ∈ gl(V ) : f(x(v), w) = −f(v, x(w)), ∀v, w ∈ V , dimF (V ) = 2l,

sp(2l, F ) = x ∈ gl(2l, F ) : sx = −xts

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=

m n

p −mt

∈ gl(2l, F ) : nt = n, pt = p

.dimF (sp(2l, F )) = 2l2 + l since it has standard basisei,i − el+i,l+i : 1 ≤ i ≤ l ∪ ei,j − el+j,l+i : 1 ≤ i 6= j ≤ l∪ei,l+i : 1 ≤ i ≤ l ∪ ei,l+j + ej,l+i : 1 ≤ i < j ≤ l∪el+i,i : 1 ≤ i ≤ l ∪ el+i,j + el+j,i : 1 ≤ i < j ≤ l.

Example 1.11Type Dl (l ≥ 1): orthogonal algebras of even degree:

Let f be a non-degenerate symmertic bilinear form on V whose matrix is s =

0 Il

Il 0

.o(V ) = x ∈ gl(V ) : f(x(v), w) = −f(v, x(w)), ∀v, w ∈ V , dimF (V ) = 2l.

o(2l, F ) = x ∈ gl(2l, F ) : sx = −xts

=

m n

p −mt

∈ gl(2l, F ) : nt = −n, pt = −p

.dimF (o(2l, F )) = 2l2 − l since it has standard basisei,i − el+i,l+i : 1 ≤ i ≤ l ∪ ei,j − el+j,l+i : 1 ≤ i 6= j ≤ l∪ei,l+j − ej,l+i : 1 ≤ i < j ≤ l ∪ el+i,j − el+j,i : 1 ≤ j < i ≤ l.

Definition 1.12A derivation of an F -algebra A is an F -linear map δ : A→ A such that δ(ab) = aδ(b) + δ(a)b forall a, b ∈ A. Let Der(A) be the set of all derivations of A. We have Der(A) ⊂ gl(A).

Definition 1.13Let L be a F -Lie algebra. The map ad : L→ gl(L) such that (adx)(y) = [x, y] for all x, y ∈ L isthe adjoint representation of L.

Lemma 1.14Let L be a F -Lie algebra. Then ad(L) ⊂ Der(L).

Proof. We need to show that (adx)([y, z]) = [(adx)(y), z] + [y, (adx)(z)] for all x, y, z ∈ L.


Steps Statements Reasons

1. (adx)([y, z]) = [x, [y, z]] Definition 1.13

2. = −[z, [x, y]]− [y, [z, x]] Definition 1.1(L3)

3. = [[x, y], z] + [y, [x, z]] Definition 1.1(L2)

4. = [(adx)(y), z] + [y, (adx)(z)] Definition 1.13

5. (adx) is a derivation for all x ∈ L Definition 1.12

Definition 1.15Elements in ad(L) are called inner derivations, elements in Der(L) − ad(L) are called outerderivations.

Example 1.16Let K be a subalgebra of L. Then adK(x) and adL(x) are different in general.Let d(n, F ) be the set of diagonal matrices. For x ∈ d(n, F ) ⊂ gl(n, F ), add(n,F )(x) = 0 andadgl(n,F )(x) 6= 0 for n ≥ 2.

Definition 1.17A F -Lie algebra L is abelian if [x, y] = 0 for all x, y ∈ L.

Example 1.18Every F -Lie algebra L of dimension 1 is abelian by Definition 1.1(L2).

Example 1.19A two dimensional F -Lie algebras is either abelian or isomorphic to Fx+Fy such that [x, y] = x.

Proof. Suppose L is not abelian and L = Fa+ Fb.


1. [L,L] = Fx for x = [a, b]. Definition 1.1(L2)

2. There exists z ∈ L− Fx. dimF L = 2

3. [x, z] = cx for some c ∈ F ∗. step 1

4. Let y = c−1z. Then [x, y] = x Definition 1.1(L1)

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and L = Fx+ Fy. step 2

Definition 1.20A subset I of a F -Lie algebra L is an ideal if(1) I is a sub-F -vector space of L. (2) [I, L] ⊂ I.By (2), I is a subalgebra of L.

Example 1.21 Ideals(1) 0 and L are ideals of L, called trivial ideals of L.(2) The center Z(L) = z ∈ L : [z, L] = 0 of L is an ideal of L.(3) The derived algebra [L,L] = Span[x, y] : x, y ∈ L of L is an ideal of L. In particular,[L,L] = 0 iff L is abelian. If L is a classical algebra, then L = [L,L].(4) If I and J are ideals of L, then I + J = x+ y : x ∈ I, y ∈ J is an ideal of L.(5) If I and J are ideals of L, then [I, J ] = ∑[xi, yi] : xi ∈ I, yi ∈ J is an ideal of L.

Definition 1.22We call L simple if .(1) L only has ideals 0 and L;(2) [L,L] 6= 0.

Lemma 1.23If L is simple, then Z(L) = 0.Proof.


1. Z(L) = 0 or L. Definition 1.22(1)

2. Z(L) 6= L. Definition 1.22(2)

3. Z(L) = 0. steps 1,2


Lemma 1.24If L is simple, then L = [L,L].Proof.


1. [L,L] = 0 or L. Definition 1.22(1)

2. [L,L] 6= 0. Definition 1.22(2)

3. [L,L] = L. steps 1,2

Example 1.25If char(F ) 6= 2, then sl(2, F ) is simple.

Proof. A standard basis for L = sl(2, F ) is

x =

0 1

0 0

, y =

0 0

1 0

, h =

1 0

0 −1

.such that [x, y] = h, [h, x] = 2x, [h, y] = −2y.Let I 6= 0 be an ideal of L. We need to show that I = L. Suppose 0 6= ax+ by + ch ∈ I.The case a 6= 0:


1. [ax+ by + ch, y] = ah− 2cy ∈ I. ax+ by + ch ∈ I and y ∈ L

2. [ah− 2cy, y] = −2ay ∈ I. ah− 2cy ∈ I and y ∈ L

3. y ∈ I. charF 6= 2 and a 6= 0

4. [x, y] = h ∈ I. x ∈ L and y ∈ I

5. [h, x] = 2x ∈ I. h ∈ I and x ∈ L

6. x ∈ I. charF 6= 2

7. I = L. x, y, h ∈ I by steps 3,4,6.

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The case b 6= 0:


1. [ax+ by + ch, x] = −bh+ 2cx ∈ I. ax+ by + ch ∈ I and x ∈ L

2. [−bh+ 2cx, x] = −2bx ∈ I. −bh+ 2cx ∈ I and x ∈ L

3. x ∈ I. charF 6= 2 and b 6= 0

4. [x, y] = h ∈ I. x ∈ I and y ∈ L

5. [h, y] = −2y ∈ I. h ∈ I and y ∈ L

6. y ∈ I. charF 6= 2

7. I = L. x, y, h ∈ I by steps 3,4,6.

The case a = b = 0:


1. c 6= 0. ax+ by + ch 6= 0

2. h ∈ I. ax+ by + ch = ch ∈ I

3.1 x ∈ I. steps 5,6 of case a 6= 0

3.2 y ∈ I. steps 5,6 of case b 6= 0

4. I = L. x, y, h ∈ I by steps 3,4.

Hence L is simple.

Example 1.26Let L be a F -Lie algebra.(1) Let K be a sub-F -vector space of L. Its normalizer is NL(K) = x ∈ L : [x,K] ⊂ K

• NL(K) is a subalgebra of L.• K is an ideal of NL(K).• If A is subalgebra of L and K is an ideal of A, then A ⊂ NL(K).

(2) Let X be a subset of L. Its centralizer CL(X) = x ∈ L : [x,X] = 0 is a subalgebra of L.In particular, CL(L) = Z(L).


Definition 1.27A linear map φ : L→ L′ between F -Lie algebras over F is a

• homomorphism if φ([x, y]) = [φ(x), φ(y)] for all x, y ∈ L;• monomorphism if it is a homomorphism and ker(φ) = 0;• epimorphism if it is a homomorphism and im(φ) = L′;• isomorphism if it is both a monomorphism and an epimorphism;• automorphism of L′ = L and it is an isomorphism. Let Aut(L) denote the group ofautomorphisms of L.

Example 1.28Let φ : L→ L′ be a homomorphism between F -Lie algebras over F . Then(1) ker(φ) is an ideal of L. (2) φ(L) is a subalgebra of L.(3) If I is an ideal of L, then the quotient vector F -space L/I with [x + I, y + I] = [x, y] + I forall x, y ∈ L is a F -Lie algebra, called the quotient algebra. There is a canonical homomorphismπ : L→ L/I such that π(x) = x+ I for all x ∈ L. We have ker(π) = I and im(π) = L/I.

Proposition 1.29(1) Let φ : L→ L′ be a homomorphism between F -Lie algebras over F . Then L/ ker(φ) ' im(φ).(1’) If I ⊂ ker(φ) is an ideal of L, then there exists a unique homomorphism ψ : L/I → L′ suchthat the following diagram commutes

Lφ//

π

L′

L/I

ψ

OO

(2) If I ⊂ J are ideals of L, then J/I is an ideal of L/I and (L/I)/(J/I) ' L/J .(3) If I and J are ideals of L, then (I + J)/J ' I/(I ∩ J).

Definition 1.30Let L be a F -Lie algebra.(1) A representation of L is a homomorphism φ : L→ gl(V ) for some vector space V over F .(2) We call φ faithful if it is a monomorphism.

Lemma 1.31Let L be a F -Lie algebra.(1) ad: L→ gl(L) is a representation. (2) ker(ad) = Z(L).(3) If L is simple, then ad is faithful. Thus L is isomorphic to a linear F -Lie algebra.

Proof. (1) We need to show that [(adx), (ad y)](z) = ad([x, y])(z) for all x, y, z ∈ L.

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1. [(adx), (ad y)](z) = [x, [y, z]]− [y, [x, z]]. Definition 1.13

2. = [x, [y, z]] + [y, [z, x]] Definition 1.1(L2)

3. = −[z, [x, y]] Definition 1.1(L3)

4. = [[x, y], z] Definition 1.1(L2)

5. = ad([x, y])(z) Definition 1.13

6. [(adx), (ad y)] = ad([x, y]) for all x, y ∈ L. steps 1-5

7. ad: L→ gl(L) is a representation. Definition 1.30(1)

(2) x ∈ ker(ad) iff (adx) = 0 iff [x, L] = 0 iff x ∈ Z(L).(3)


1. Z(L) = 0. L is simple and Lemma 1.23

2. ker(ad) = 0. (2)

3. ad is faithful. Definition 1.30(2)

4. L ' ad(L) ⊂ gl(L). Proposition 1.29(1)

Example 1.32Suppose char(F ) = 0 and δ ∈ Der(L) such that δk = 0 for some k > 0. Define exp δ =

k−1∑n=0

δn

n! .Then (1)(exp δ)([x, y]) = [(exp δ)(x), (exp δ)(y)]. (2) exp δ ∈ Aut(L).

Proof. (1)


1. δn([x, y]) =n∑i=0

(ni

)[δi(x), δn−i(y)]. Leibniz rule of Definition 1.12

2. δn

n! ([x, y]) =n∑i=0

[δi

i! (x), δn−i

(n− i)!(y)] (

n

i

)= n!i!(n− i)! and Definition 1.1(L1)



3. (exp δ)([x, y]) = [(exp δ)(x), (exp δ)(y)]. exp δ =k−1∑n=0

δn

n! and Definition 1.1(L1)

(2) In fact, exp(δ) = 1− η has inverse 1 + η + η2 + · · ·+ ηk−1, where η = −k−1∑i=1

δi

i! and ηk = 0.

Example 1.33Suppose char(F ) = 0 and x ∈ L such that (adx)k = 0 for some k > 0. Then

exp(adx) = 1 + (adx) + (adx)2

2! + · · ·+ (adx)k−1

(k − 1)! ∈ Aut(L).

Definition 1.34The subgroup of Aut(L) generated by exp(adx), x ∈ L is denoted by int(L), its elements are calledinner automorphisms.

Lemma 1.35int(L) is a normal subgroup of Aut(L).Proof.


1. For all φ ∈ Aut(L) and x ∈ L, φ (adx) φ−1 = ad(φ(x)).

For all y ∈ L, (φ (adx) φ−1)(y) =φ([x, φ−1(y)]) = [φ(x), y] = ad(φ(x))(y)

2. int(L) is a normal subgroup of Aut(L). For all x ∈ L, φ exp(adx) φ−1 =exp(ad(φ(x))) ∈ int(L) and Defini-tion 1.34

Example 1.36Let L ⊂ gl(V ) and x ∈ L. Let λx be the left multiplication by x. Let ρx be the right multiplicationby x. Then λ−1

x = left multiplication by x−1; ρ−1x = right multiplication by x−1. Also, λx ρy =

ρy λx for all x, y ∈ L.(1) adx = λx − ρx since (adx)(y) = xy − yx.(2) If g ∈ GL(V ) and int(g)(x) = gxg−1 for all x ∈ L, then int(g) = λg ρ−1

g .

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2. March 5th, Solvable and nilpotent Lie algebras

Lemma 2.1Suppose char(F ) = 0. Let L be a F -Lie algebra, L ⊂ gl(V ) for some F -vector space V . If xk = 0for some integer k > 0 in gl(V ), then exp(adx) = int(expx).

Proof. exp(adx) = exp(λx − ρx) = exp(λx) exp(ρx)−1 = λexp(x) ρ−1exp(x) = int(exp x).

Example 2.2

Let L = sl(2, F ) and x =

0 1

0 0

and y =

0 0

1 0

. Let σ = exp(adx) exp(ad(−y)) exp(ad x) ∈

int(L). Then σ = int(s), where s = exp(x) exp(−y) exp(x) =

0 1

−1 0

.

Proof. Since x2 = 0, exp(x) =

1 1

0 1

; Since y2 = 0, exp(−y) =

1 0

−1 1

.

Then s =

1 1

0 1

1 0

−1 1

1 1

0 1

=

0 1

−1 0

.

Definition 2.3Let L be a F -Lie algebra.(1) The derived series of ideals of L is

L(0) = L, L(1) = [L,L], · · · , L(i) = [L(i−1), L(i−1)], · · ·

(2) L is solvable if L(n) = 0 for some n.

Example 2.4(1) An abelian F -Lie algebra L is solvable. In fact, L(1) = 0 by Definition 1.17.(2) A simple F -Lie algebra L is non-solvable.


1. L = L(1). L is simple and ??

2. L = L(m) for all integers m ≥ 0. Induction

3. L(1) 6= 0. L is simple and Definition 1.22(2)

4. L(m) 6= 0 for all m. steps 1-3



5. L is non-solvable. Definition 2.3

Example 2.5Let t(n, F ) be the F -Lie algebra of upper triangular matrices in gl(n, F ). Then t(n, F ) issolvable.We only verify that L = t(2, F ) is solvable. Elements of L(1) have the form

a1 b1

0 c1

,a2 b2

0 c2

=

a1 b1

0 c1

a2 b2

0 c2

−a2 b2

0 c2

a1 b1

0 c1

=

a1a2 a1b2 + b1c2

0 c1c2

−a1a2 a2b1 + b2c1

0 c1c2

=

0 a1b2 + b1c2 − a2b1 − b2c1

0 0

Elements of L(2) have the form

0 b1

0 0

,0 b2

0 0

=

0 0

0 0

So L(2) = 0, L is solvable.

Proposition 2.6If L is a solvable F -Lie algebra and K is a subalgebra of L, then K is solvable.Proof.


1. K(m) ⊂ L(m) for all integers m ≥ 0. Induction.

1.1 K(0) = K ⊂ L = L(0). given.

1.2 K(m) = [K(m−1), K(m−1)] ⊂[L(m−1), L(m−1)] = L(m)

Suppose K(m−1) ⊂ L(m−1).

2. L(n) = 0 for some integer n ≥ 0. L is solvable and Definition 2.3(2)

3. K(n) = 0. step 1

4. K is solvable. Definition 2.3(2)

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Proposition 2.7If L is solvable F -Lie algebra, then its homomorphic images are solvable.

Proof. Suppose φ : L→ L′ is a homomorphism of F -Lie algebras and M = φ(L).


0. M is a subalgebra of L′. Example 1.28(2)

1. φ(L(m)) = M (m) for all integers m ≥ 0. Induction.

1.1 φ(L(0)) = φ(L) = M = M (0). M = φ(L)

1.2 φ(L(m)) = φ([L(m−1), L(m−1)]) =[φ(L(m−1)), φ(L(m−1))]) =[M (m−1),M (m−1)] = M (m)

Suppose φ(L(m−1)) = M (m−1).

2. L(n) = 0 for some integer n ≥ 0. L is solvable and Definition 2.3(2)

3. M (n) = φ(L(n)) = φ(0) = 0. step 1

4. M is solvable. Definition 2.3(2)

Proposition 2.8Referenced on pages 15, 16, 38 and 41.Let L be a F -Lie algebra. Let I be an ideal of L. If I and L/I are solvable, then L is solvable.Proof.


1. (L/I)(n) = 0 for some integer n ≥ 0. L/I is solvable and Definition 2.3(2)

2. L(n) is a subalgebra of I. L(n+1) ⊂ L(n) ⊂ I.

3. L(n) is solvable. I is solvable and Proposition 2.6

4. L(n+m) = 0 for some integer m ≥ 0. Definition 2.3(2)

5. L is solvable. Definition 2.3(2)


Proposition 2.9Let L be a F -Lie algebra. If I, J are solvable ideals of L, then I + J is solvable.Proof.


1. I/(I ∩ J) is solvable. I is solvable and Proposition 2.7

2. (I + J)/J is solvable. (I + J)/J ' I/(I ∩ J) by Proposi-tion 1.29(3)

3. I + J is solvable. J and (I + J)/J are solvable and Propo-sition 2.8,

Lemma 2.10Let L be a finite dimensional F -Lie algebra. There exists a unique ideal Rad(L) of L such that(1) Rad(L) is a solvable.(2) If I is a solvable ideal of L, then I ⊂ Rad(L).

Proof. Let A be the set of all solvable ideals of L.


1. A 6= ∅. 0 is a solvable ideal of L

2. There exists S ∈ A such that dimF (I) ≤dimF (S) for all I ∈ A.

dimF (L) is finite

3. For all I ∈ A, S + I ∈ A . Proposition 2.9

4. dimF (S) ≤ dimF (S + I) S ⊂ S + I

5. dimF (S) = dimF (S + I) step 2

6. S = S + I. S ⊂ S + I and dimF (L) is finite

7. I ⊂ S for all I ∈ A

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8. Rad(L) = S. steps 2 and 7

9. Rad(L) is unique. (2)

Definition 2.11Let L be a finite dimensional F -Lie algebra. We call Rad(L) the radical of L. If Rad(L) = 0,then L is called semisimple.

Example 2.12Let L be a finite dimensional simple F -Lie algebra. Then L is semisimple.Proof.


1. Rad(L) exists. L is finite dimensional and Lemma 2.10

2. L is non-solvable. L is simple and Example 2.4(2)

3. Rad(L) 6= L Lemma 2.10(1).

4. Rad(L) = 0. L is simple and Definition 1.22(1)

5. L is semisimple. Definition 2.11.

Lemma 2.13For all finite dimensional F -Lie algebra L, L/Rad(L) is semisimple.

Proof. Let π : L→ L/Rad(L) be the canonical surjection of F -Lie algebras. LetR = π−1(Rad(L/Rad(L))).


1. R is an ideal of L. π(R) = Rad(L/Rad(L)) is an ideal ofL/Rad(L) and π is a homomorphism.

2. R ⊃ π−1(0) = Rad(L) in L. Rad(L/Rad(L)) ⊃ 0 in L/Rad(L)



3. Rad(L) and R/Rad(L) =Rad(L/Rad(L)) are solvable.

Lemma 2.10(1)

4. R is solvable. Proposition 2.8

5. R = Rad(L). step 2 and Lemma 2.10(2)

6. Rad(L/Rad(L)) = 0. Rad(L/Rad(L)) = R/Rad(L)

7. L/Rad(L) is semisimple. Definition 2.11.

Definition 2.14Let L be a F -Lie algebra.(1) The descending central series or lower central series of ideals of L is

L0 = L, L1 = [L,L], · · · , Li = [L,Li−1], · · ·

(2) L is nilpotent if Ln = 0 for some n, i.e. (adx1 (adx2) · · · (adxn))(y) = 0 for all xi, y ∈ L.

Example 2.15Any abelian algebra L is nilpotent, since L1 = 0. For example, d(n, F ), Z(L) are nilpotent.

Lemma 2.16Nilpotent F -Lie algebras are solvable.

Proof. Let L be a nilpotent F -Lie algebra.


1. L(m) ⊂ Lm for all integer m ≥ 0. Induction

1.1. L(0) = L = L0, L(1) = [L,L] = L1. Definition 2.3(1) and Definition 2.14(1)

1.2. If L(n) ⊂ Ln, then L(n+1) = [L(n), L(n)] ⊂[L,L(n)] ⊂ [L,Ln] = Ln+1.

Definition 2.3(1) and Definition 2.14(1)

2. Ln = 0 for some n. L is nilpotent and Definition 2.14(2)

3. L(n) = 0 for some n. step 1

4. L is solvable. Definition 2.3(2)

18 ZHENGYAO WU

Example 2.17t(n, F ) is not nilpotent. We only verify that L = t(2, F ) is not nilpotent. Elements of L1 has form0 d

0 0

by Example 2.5. Elements of L2 has form

a b

0 c

,0 d

0 0

=

a b

0 c

0 d

0 0

−0 d

0 0

a b

0 c

=

0 ad

0 0

−0 cd

0 0

=

0 (a− c)d

0 0

By taking a = 1 and c = 0, we have that L2 = L1. Therefore Ln = L1 6= 0 for all integer n ≥ 1.

Example 2.18Let n(n, F ) be the set of strictly upper triangular matrices. Then n(n, F ) is nilpotent. Weonly verify that n(2, F ) is nilpotent. In fact, elements of n(2, F )1 = t(2, F )(2) = 0 by Example 2.5.

Proposition 2.19If L is a nilpotent F -Lie algebra and K is a subalgebra of L, then K is nilpotent.Proof.


1. Km ⊂ Lm for all integers m ≥ 0. Induction.

1.1 K0 = K ⊂ L = L0. given.

1.2 Km = [K,Km−1] ⊂ [L,Lm−1] = Lm Suppose Km−1 ⊂ Lm−1.

2. Ln = 0 for some integer n ≥ 0. L is nilpotent and Definition 2.14(2)

3. Kn = 0. step 1

4. K is nilpotent. Definition 2.14(2)


Proposition 2.20If L is nilpotent F -Lie algebra, then its homomorphic images are nilpotent.

Proof. Suppose φ : L→ L′ is a homomorphism of F -Lie algebras and M = φ(L).


1. φ(Lm) = Mm for all integers m ≥ 0. Induction.

1.1 φ(L0) = φ(L) = M = M0. M = φ(L)

1.2 φ(Lm) = φ([L,Lm−1]) =[φ(L, φ(Lm−1)]) = [M,Mm−1] = Mm

Suppose φ(Lm−1) = Mm−1.

2. Ln = 0 for some integer n ≥ 0. L is nilpotent and Definition 2.14(2)

3. Mn = φ(Ln) = φ(0) = 0. step 1

4. M is nilpotent. Definition 2.14(2)

Proposition 2.21If L/Z(L) is nilpotent, then so is L.Proof.


1. (L/Z(L))n = 0 for some integer n > 0. L/Z(L) is nilpotent and Defini-tion 2.14(2).

2. Ln ⊂ Z(L).

3. Ln+1 = [L,Ln] ⊂ [L,Z(L)] = 0. Definition 2.14(1) and defn of Z(L)

4. L is nilpotent. Definition 2.14(2)

Proposition 2.22If L is nilpotent and L 6= 0, then Z(L) 6= 0.Proof.

20 ZHENGYAO WU


1. Suppose n ≥ 0 is the least integer withLn = 0.

L is nilpotent and Definition 2.14(2)

2. n ≥ 1. L0 = L 6= 0

3. Ln−1 6= 0. n is the least

4. Ln−1 ⊂ Z(L). [L,Ln−1] = Ln = 0

5. Z(L) 6= 0. steps 3,4

Definition 2.23Let L be a F -Lie algebra. For x ∈ L, x is ad-nilpotent if adx is a nilpotent endomorphism ingl(L).

Lemma 2.24If L is nilpotent, then every element of L is ad-nilpotent.

Proof. By Definition 2.14(2), ((adx1) (adx2) · · · (adxn))(y) = 0 for all xi, y ∈ L. Let x = x1 =· · · = xn. We have (adx)n = 0.

Lemma 2.25Let V be an F -vector space. If x ∈ gl(V ) is a nilpotent endomorphism, then adx is also nilpotent.Proof.


1. Suppose xn = 0 for some integer n > 0. x is nilpotent

2. λnx = 0 and ρnx = 0. Example 1.36

3. (adx)2n = (λx − ρx)2n. Example 1.36(1)

=2n∑i=0

(2ni

)(λx)i(−ρx)2n−i Binomial theorem

= 0 either i ≥ n or 2n− i ≥ n


Example 2.26In ∈ gl(n, F ) is ad-nilpotent but not nilpotent, since ad(In) = 0 and Imn = In for all integer m ≥ 1.

Lemma 2.27Let V be an F -vector space. If f ∈ gl(V ) is nilpotent, then there exists 0 6= v ∈ V such thatf(v) = 0.Proof.


1. Suppose n ≥ 0 is the least integer suchthat fn = 0.

f is nilpotent

2. If n = 0, there exists 0 6= v ∈ V such thatf(v) = 0.

f = 0

3. If n > 0, then fn−1 6= 0. n is the least.

4. There exists 0 6= w ∈ V such thatfn−1(w) 6= 0.

5. There exists 0 6= v ∈ V and f(v) = 0. v = fn−1(w) and fn(w) = 0

Lemma 2.28Let L be a F -Lie algebra. Let K be a maximal proper subalgebra of L and an ideal of L. ThenL = K + Fz for some z ∈ L−K.Proof.


1. The quotient algebra L/K and the canon-ical homomorphism π : L→ L/K exists.

K is an ideal of L

2. If dimF (L/K) > 1, then it has a proper1-dimensional subalgebra S.

Example 1.18

3. π−1(S) is a subalgebra of L. S is a subalgebra of L/K and π is a ho-momorphism.

22 ZHENGYAO WU


π−1(S) is proper. dimF (π−1(S)) = 1 + dimF (K) <

dimF (L/K) + dimF (K) = dimF (L).

4. π−1(S) ) K. dimF (π−1(S)) = 1 + dimF (K) >

dimF (K).

5. A contradiction. Thus dimF (L/K) = 1. K is a maximal proper subalgebra of L.

6. L = K + Fz for some z ∈ L−K. Suppose z +K is a basis of L/K.

Lemma 2.29Let L be a Lie subalgebra of gl(V ), where V is a finite dimensional F -vector space, V 6= 0. LetK be a maximal proper nonzero subalgebra of L. If every element of L is nilpotent, then K is anideal of L.Proof.


1. Every element of ad(L) is nilpotent. L is linear, every element of L is nilpotentand Lemma 2.25

2. ad ∈ gl(L/K). [K,L] ⊂ [L,L] ⊂ L and [K,K] ⊂ K

3. ad is nilpotent. dimF (L) <∞ and step 1

4. There exists 0 + K 6= x + K ∈ L/K suchthat ad(x+K) = 0.

Lemma 2.27

5. x ∈ L−K and [x, L] ⊂ K.

6. K ( NL(K). x ∈ NL(K)−K

7. NL(K) is a subalgebra of L. Example 1.26(1)

8. K is an ideal of L. NL(K) = L by the maximality of K


Theorem 2.30Let L be a subalgebra of gl(V ), where V is finite dimensional, V 6= 0. If every element of L isnilpotent, then there exists nonzero v ∈ V such that x(v) = 0 for all x ∈ L.

Proof. We prove by induction. If dimF (L) = 1, then L is generated by some f ∈ gl(V ). ByLemma 2.27, there exists 0 6= v ∈ V such that f(v) = 0. Then x(v) = 0 for all x ∈ L.Now we suppose dimF (L) > 1.


1. There exists a maximal proper nonzerosubalgebra K of L.

dimF (L) > 1

2. K is an ideal of L. Every element of L is nilpotent andLemma 2.29

3. L = K + Fz for some z ∈ L−K. Lemma 2.28

4. Let W = w ∈ V : y(w) = 0, ∀y ∈ K.Then W 6= 0.

dimF (K) = dimF (L) − 1 and inductivehypothesis

5. For all w ∈ W and y ∈ K, we have [x, y] ∈K.

step 2

6. y(x(w)) = x(y(w))−[x, y](w) = x(0)−0 =0 for all y ∈ K.

step 4

7. x(W ) ⊂ W for all x ∈ L. In particular,z ∈ gl(W ).

defn of W

8. There exists 0 6= v ∈ W , z(v) = 0. z is nilpotent and Lemma 2.27

9. x(v) = (y + cz)(v) = y(v) + cz(v) = 0 forall x ∈ L.

where x = y + cz, y ∈ K, c ∈ F , step 3

and y(v) = 0. y ∈ K, v ∈ W and defn of W

Theorem 2.31 EngelLet L be a subalgebra of gl(V ), where V is finite dimensional, V 6= 0. If every element of L isad-nilpotent, then L is nilpotent.

24 ZHENGYAO WU

Proof.


1. If dimF (L) = 1, then L is nilpotent. L is abelian, Example 1.18 and Exam-ple 2.15

2. There exists 0 6= v ∈ L such that [L, v] =0.

Every element of L is ad-nilpotent andTheorem 2.30

3. dimF (L/Z(L)) < dimF (L). Z(L) 6= 0

4. Every element of L/Z(L) is ad-nilpotent. Every element of L is ad-nilpotent

5. L/Z(L) is nilpotent. step 3 and inductive hypothesis

6. L is nilpotent. Proposition 2.21

Definition 2.32Let V be a finite dimensional F -vector space.(1) A flag in V is a chain of subspaces

0 = V0 ⊂ V1 ⊂ · · · ⊂ Vn = V, dimF (Vi) = i

(2) If x ∈ gl(V ), we say that x stablizes (or leaves invariant) this flag if x(Vi) ⊂ Vi for all i.

Corollary 2.33Let L be a subalgebra of gl(V ), where V is an n-dimensional F -vector space, V 6= 0. Supposex ∈ L. If every element of L is nilpotent, then there exists a flag (Vi) in V , stable under L, withx(Vi) ⊂ Vi−1 for all i, i.e. there exists a basis of V relative to which the matrices of L are all inn(n, F ).

Proof. We prove by induction. For n = 1, take the flag 0 = V0 ⊂ V1 = V . If x = 0, then0V = 0. If x 6= 0, then there exists 0 6= v ∈ V such that x(v) = 0 by Lemma 2.27. SinceV = Fv, x(V ) = 0. Then v form a basis of V relative to which the matrices of L are all (0) andn(1, F ) = (0). Now we suppose n > 1.


1. There exists v ∈ V such that x(v) = 0 forall x ∈ L.

Every element of L is nilpotent and The-orem 2.30



2. There exists a flag (Wi)0≤i≤n−1 in V/(Fv)such that x(Wi) ⊂ Wi−1 for all 1 ≤ i ≤n− 1.

L acts on V/(Fv) and dimF (V/(Fv)) =n− 1 and inductive hypothesis.

3. Let π : V → V/(Fv) be the quo-tient map. Let Vi = π−1(Wi−1) for all1 ≤ i ≤ n. To be precise, V0 =0, V1 = Fv, V2 = π−1(W1), . . . , Vn =π−1(Wn−1) = π−1(V/(Fv)) = V .

4. x(V1) = 0 and x(Vi) = π−1(x(Wi−1)) ⊂π−1(Wi−2) ⊂ Vi−1 for all 2 ≤ i ≤ n.

steps 1, 2

5. Suppose V/Fv has a basis (e1, . . . , en−1)relative to which the matrices of L/(Fv)are all in n(n− 1, F ).

Inductive hypothesis

6. V has a basis (v, e1, . . . , en−1) relative towhich the matrices of L are all in n(n, F ).

the first column is zero by step 1

26 ZHENGYAO WU

3. March 12th, Lie theorem, Jordan decomposition

Lemma 3.1Let L be a subalgebra of gl(V ), where V is finite dimensional, V 6= 0. Let L be a nilpotentF -Lie algebra. Let K be an ideal of L. If K 6= 0, then K ∩ Z(L) 6= 0. In particular, Z(L) 6= 0.

Proof. Since L is nilpotent, every element of ad(L) is nilpotent by Lemma 2.24. Since [L,K] ⊂ K,by Theorem 2.30, there exists 0 6= v ∈ K such that [L, v] = 0. Hence v ∈ K ∩ Z(L).

Lemma 3.2Let F be an algebraically closed field of characteristic 0. Let L be a solvable F -Lie algebra ofdimension n > 0. Then there exists an ideal K of L of codimension 1.Proof.


1. [L,L] ( L. L is solvable, n > 0 and Definition 2.3(2)

2. L/[L,L] has a codimension 1 subspaceK ′. dimF (L/[L,L]) ≥ 1

3. Let π : L → L/[L,L] be the canonicalquotient map and K = π−1(K ′).

3.1. K is an ideal of L. [L,K] ⊂ [L,L] = π−1(0) ⊂ π−1(K ′) = K.

3.2. K is of codimension 1. dimF (K) = dimF (K ′) + dimF ([L,L]) =dimF (L/[L,L])−1+dimF ([L,L]) = n−1.

Lemma 3.3Let F be an algebraically closed field of characteristic 0. Let V be a finite dimensional F -vectorspace, V 6= 0. Let L be a solvable subalgebra of gl(V ). Let K be an ideal of L of codimension1. Let λ : K → F be a linear function. Let W = w ∈ V : y(w) = λ(y)w, ∀y ∈ K. If W 6= 0,then x(W ) ⊂ W for all x ∈ L.

Proof. Suppose w ∈ W . We want to show that x(w) ∈ W . Let n be the smallest integer such thatw, x(w), . . . , xn(w) are linearly dependent. LetW0 = 0; Wi = SpanFw, x(w), x2(w) . . . , xi−1(w), 1 ≤i ≤ n. Let y ∈ K.




0. K is an ideal of L and [x, y] ∈ K. K exists by Lemma 3.2

1. x(Wi−1) ⊂ Wi for all 1 ≤ i ≤ n andx(Wn) ⊂ Wn.

defn of Wi, (0 ≤ i ≤ n)

2. y(xi(w)) ≡ λ(y)xi(w) modWi, i ≥ 0. Induction

2.1. If i = 0, then y(w) = λ(y)w. w ∈ W

2.2. Suppose i ≥ 1. Then y(xi−1(w)) =λ(y)xi−1(w) + w′ for some w′ ∈ Wi−1.


2.3. y(xi(w)) = x(y(xi−1(w)))− [x, y](xi−1(w)) defn of [x, y]

= x(λ(y)xi−1(w) + w′)− λ([x, y])xi−1(w) step 2.2

= λ(y)xi(w) + x(w′)− λ([x, y])xi−1(w)

where x(w′)− λ([x, y])xi−1(w) ∈ Wi. w ∈ Wi−1 and step 1

3. y(Wi) ⊂ Wi and TrWn(y) = nλ(y). step 2

4. nλ([x, y]) = TrWn([x, y]) = TrWn(x y) −TrWn(y x) = 0.

step 3 and property of trace

5. λ([x, y]) = 0. char(F ) = 0

6. y(x(w)) = x(y(w))− [x, y](w) defn of [x, y]

= x(λ(y)w)− λ([x, y])(w) w ∈ W and defn of W

= λ(y)x(w)− λ([x, y])(w) = λ(y)x(w). step 6

7 . x(w) ∈ W . defn of W

We did not use dimF (K) = dimF (L)− 1 in this proof.

Theorem 3.4Let F be an algebraically closed field of characteristic 0. Let V be a finite dimensional F -vectorspace, V 6= 0. Let L be a solvable subalgebra of gl(V ). Then there exists in V a commoneigenvector for all elements of L.

28 ZHENGYAO WU

Proof. Induct on dimF (L). If dimF (L) = 0, then L = 0. Every nonzero vector of V is aneigenvector of 0. Now we suppose dimF (L) > 0.


1. There exists a codimension one ideal K ofL.

L is solvable, dimF (L) > 0 and Lemma 3.2

2. K is solvable. L is solvable, K is a subalgebra of L andProposition 2.6

3. There exists a common eigenvector 0 6=v ∈ V for all elements of K.


4. Let W = w ∈ V : y(w) = λ(y)w, ∀y ∈K and W 6= 0.

v as in step 3 belongs to W

5. z(W ) ⊂ W . Lemma 3.3

6. There exists an eigenvalue λ(z) of z withan eigenvector v0 ∈ W , i.e. z(v0) = λ(z)v0

F is algebraically closed.

7. For all x ∈ L, x = y + cz for some y ∈ Kand c ∈ F . Define λ(x) = λ(y) + cλ(z).

L = K + Fz by Lemma 2.28

8. v0 is a common eigenvector of L. x(v0) = y(v0) + cz(v0) = λ(y)v0 +cλ(z)v0 = λ(x)v0

Corollary 3.5 Lie’s theoremLet F be an algebraically closed field of characteristic 0. Let V be an n-dimensional F -vectorspace, V 6= 0. Let L be a solvable subalgebra of gl(V ). Then L stabilizes some flag in V ,i.e. there exists a basis of V relative to which the matrices of L are all in t(n, F ).

Proof. We prove by induction. For n = 1, take the flag 0 = V0 ⊂ V1 = V . Also, t(1, F ) =gl(1, F ). Now we suppose n > 1.


1. There exists a common eigenvector v of L. L is a solvable subalgebra of gl(V ) andTheorem 3.4



2. L acts on V/(Fv). x(v) ∈ Fv for all x ∈ L

3. There exists a flag (Wi)0≤i≤n−1 in V/(Fv)such that x(Wi) ⊂ Wi for all x ∈ L.

dimF (V/(Fv)) = n − 1 and inductive hy-pothesis

4. Let π : V → V/(Fv) be the quotient map.Let V0 = 0, Vi = π−1(Wi−1) for all 1 ≤i ≤ n. Then (Vi)1≤i≤n is a flag in V .

dimF (Vi) = dimF (Wi−1)+1 = i−1+1 = i.

5. x(V0) = 0 = V0 and x(Vi) =x(π−1(Wi−1)) = π−1(x(Wi−1)) ⊂π−1(Wi−1) = Vi for all 1 ≤ i ≤ n.

6. Suppose V/Fv has a basis (e1, . . . , en−1)relative to which the matrices of L/(Fv)are all in t(n− 1, F ).


7. V has a basis (v, e1, . . . , en−1) relative towhich the matrices of L are all in t(n, F ).

entries of the first column are all zero ex-cept at the (1, 1) entry by step 2

Corollary 3.6Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional solvableF -Lie algebra. Then there exists a flag of ideals of L.Proof.


1. ad(L) is solvable. L is solvable and Proposition 2.7

2. There exists a flag (Li)0≤i≤n of subspacesof L such that [L,Li] ⊂ Li for all i.

Corollary 3.5

3. Li is an ideal of L for all 0 ≤ i ≤ n.

30 ZHENGYAO WU

Corollary 3.7Let F be an algebraically closed field of characteristic 0. Let L be finite dimensional solvable F -Liealgebra. If x ∈ [L,L] then adx is nilpotent. Furthermore, [L,L] is nilpotent.Proof.


1. There exists a flag of ideals (Li)0≤i≤n. L is solvable and Corollary 3.6

2. Let (x1, x2, . . . , xn) be a basis of L suchthat Li = SpanFx1, . . . , xi relative towhich the matrices in ad(L) are in t(n, F ).

3. The matrices of ad([L,L]) are in n(n, F ). ad([L,L]) = [ad(L), ad(L)],[t(n, F ), t(n, F )] = n(n, F )

4. Every element of [L,L] is ad-nilpotent. Example 2.18

5. [L,L] is nilpotent. Engel’s theorem Theorem 2.31

Definition 3.8Let F be an algebraically closed field (of arbitrary characteristic). Let V be a finite dimensionalF -vector space. We call x ∈ EndF (V ) semisimple if the roots of its minimal polynomial over Fare distinct.

Lemma 3.9Let F be an algebraically closed field. Let V be a finite dimensional F -vector space. Thenx ∈ EndF (V ) is semisimple iff x is diagonalizable.

Proof. The minimal polynomial of the Jordan block diagonal matrix of a is (X − a)n where n isthe size of the largest Jordan block of a. Each block has the form

a 1 0 · · · 0 0

0 a 1 · · · 0 0... ... ... . . . ... ...

0 0 0 · · · a 1

0 0 0 · · · 0 a

.


Since x is similar to its Jordan canonical form, x is semisimple iff every Jordan block of x has size1× 1 iff the Jordan canonical form of x is diagonal.

Lemma 3.10Let F be an algebraically closed field. Let V be a finite dimensional F -vector space. Let W be asubspace of V . If x ∈ EndF (V ) is semisimple and x(W ) ⊂ W , then x|W is semisimple.

Proof. Let f(X) be the minimal polynomial of x. Let g(X) be the minimal polynomial of x|W .


1. f(x|W ) = f(x)|W = 0. Cayley-Hamilton theorem

2. g(X)|f(X). g(X) is the minimal polynomial of x|W

3. Roots of f(X) are distinct. x is semisimple and Definition 3.8

4. Roots of g(X) are distinct. step 2

5. x|W is semisimple. Definition 3.8

Lemma 3.11Let F be an algebraically closed field. Let V be a finite dimensional F -vector space.(1) Suppose x, y ∈ End(V ) are diagonalizable and x y = y x. Then there exists z ∈ GL(V ) suchthat z x z−1 and z y z−1 are diagonal.(2) Suppose n ≥ 2 is an integer, x1, x2, . . . , xn ∈ End(V ) are diagonalizable and xi xj = xj xi forall 1 ≤ i, j ≤ n. Then there exists z ∈ GL(V ) such that z xi z−1 are diagonal for all 1 ≤ i ≤ n.

Proof. (1) Let Eλ = v ∈ V : y(v) = λv be eigenspaces of y. They exist since F is algebraicallyclosed.


1. x(Eλ) ⊂ Eλ for all eigenvalues λ of y. y(x(v)) = x(y(v)) = x(λv) = λx(v) for allv ∈ Eλ

2. x is semisimple. F is algebraically closed, x is diagonaliz-able and Lemma 3.9

3. x|Eλ is semisimple. step 1 and Lemma 3.10

32 ZHENGYAO WU


4. x|Eλ is diagonalizable. F is algebraically closed and Lemma 3.9

5. There exists a basis of Eλ and zλ ∈GL(Eλ) such that zλ (x|Eλ) z−1

λ is diag-onal.

6. zλ (y|Eλ) z−1λ = λIm, m = dimF (Eλ). row vectors of zλ are eigenvectors.

7. Let z = ⊕λzλ. Then zxz−1 and zyz−1

are diagonal.V = ⊕

λEλ and steps 5,6 for all λ

(2) By (1), there exists z ∈ GL(V ) such that z xi z−1 are diagonal for all i = 1, 2. Since(z xi z−1) (z xj z−1) = (z xj z−1) (z xi z−1) for all 2 ≤ i, j ≤ n, by (1) again, thereexists w ∈ GL(V ) such that w z xi z−1 w−1 is diagonal.

Lemma 3.12Let F be an algebraically closed field. Let V be a finite dimensional F -vector space. For t ∈EndF (V ), suppose the characteristic polynomial of t is

k∏i=1

(T −ai)mi , where a1, . . . , am are distinct.Then there exists a polynomial p(T ) ∈ F [T ] such that

p(T ) ≡ ai mod(T − ai)mi , p(T ) ≡ 0 modT.

Proof. Case 1: There exists i such that ai = 0. Then (T − ai)mi , 1 ≤ i ≤ k are coprime.Case 2: For all 1 ≤ i ≤ k, ai 6= 0. Then (T − ai)mi , 1 ≤ i ≤ k and T are coprime.In either case, p(T ) exists by Chinese Remainder Theorem.

Example 3.13

For x =

1 2

0 −1

∈ C2×2, its characteristic polynomial is (T − 1)(T + 1). We need to find p(T )

such that p(T ) ≡ 1 mod(T − 1), p(T ) ≡ −1 mod(T + 1) and p(T ) ≡ 0 modT .In fact,

T (T + 1)y1 ≡ 1 mod(T − 1) =⇒ y1 ≡12 mod(T − 1).

T (T − 1)y2 ≡ 1 mod(T + 1) =⇒ y2 ≡12 mod(T + 1).

(T − 1)(T + 1) · y3 ≡ 1 modT =⇒ y3 ≡ −1 modT.Therefore

p(T ) ≡ 1∗T (T +1)∗ 12 +(−1)∗T (T −1)∗ (1

2)+0∗ (T −1)(T +1)∗ (−1) ≡ T modT (T −1)(T +1).


Proposition 3.14 Jordan-Chevalley decompositionLet F be an algebraically closed field. Let V be a finite dimensional F -vector space.(a) For all x ∈ EndF (V ), there exists unique xs, xn ∈ EndF (V ) such that x = xs + sn; xs issemisimple; xn is nilpotent; and xsxn = xnxs.(b) There exists p(T ), q(T ) ∈ F [T ] such that p(0) = 0, q(0) = 0; xs = p(x), xn = q(x); If xy = yx,then xsy = yxs and xny = yxn.(c) If A ⊂ B ⊂ V and x(B) ⊂ A, then xs(B) ⊂ A and xn(B) ⊂ A.We call xs the semisimple part of x; xn the nilpotent part of x.

Proof. Suppose the characteristic polynomial of x isk∏i=1

(T − ai)mi , where a1, . . . , am are distinct.Let Vi = ker(x− ai · 1)mi .(a) Existence.


1. There exists p(T ) ∈ F [T ] such thatp(T ) ≡ ai mod(T − ai)mi and p(T ) ≡0 modT .

Lemma 3.12

2. Define xs = p(x), xn = x − p(x). Thenx = xs + xn. Thus xsxn = xnxs.

F [T ] is commutative

3.1. (xs)|Vi = ai · 1 is diagonalizable. p(T ) ≡ ai mod(T −ai)mi and Vi = ker(x−ai · 1)mi .

3.2. xs is semisimple. V =k∑i=1

Vi and Lemma 3.9

4.1. (T − p(T ))mi ≡ 0 mod(T − ai)mi . p(T ) ≡ ai mod(T − ai)mi

4.2. ((xn)|Vi)mi = 0. Vi = ker(x− ai · 1)mi .

4.3. xn is nilpotent. V =k∑i=1

Vi

(b)


1. p(0) = 0. p(T ) ≡ 0 modT in step 1 of (a) Existence.

2. xs = p(x). step 2 of (a) Existence.

3. Let q(T ) = T − p(T ). Then q(0) = 0. step 1

34 ZHENGYAO WU


4. xn = q(x). step 2 of (a) Existence.

5. If xy = yx, then xsy = yxs and xny =yxn.

step 2,4 and F [T ] is commutative.

(c) follows from p(0) = 0, q(0) = 0; xs = p(x), xn = q(x) in (b) and A ⊂ B.(a) Uniqueness. Suppose x = xs + xn = x′s + x′n are two decompositions.


6. xxs = xsx and xxn = xnx. step 2 of (a)

7.1. xsx′s = x′sxs. step 5 of (b) and step 6 here

7.2. xs − x′s is diagonalizable. xs, xs′ are diagonalizable and Lemma 3.11

8.1. xnx′n = x′nxn. step 5 of (b) and step 6 here

8.2. x′n − xn is nilpotent. Binomial theorem

9. xs − x′s = x′n − xn = 0. It is both semisimple and nilpotent

Example 3.15

In C2×2, x =

1 2

0 −1

is semisimple, so xs = x and xn = 0. Let s =

1 0

0 −1

and n =

0 2

0 0

.

Although x = s + n, s is semisimple and n is nilpotent, sn =

0 2

0 0

, ns =

0 −2

0 0

, sn 6= ns.

So t = s+ n is not the Jordan decomposition of t.

Example 3.16

For x =

1 2

0 1

∈ C2, its characteristic polynomial is (T − 1)2. We need to find p(T ) such that

p(T ) ≡ 1 mod(T − 1)2 and p(T ) ≡ 0 modT .


In fact,

Ty1 ≡ 1 mod(T − 1)2 =⇒ T 2y1 ≡ T mod(T − 1)2

=⇒ (2T − 1)y1 ≡ T mod(T − 1)2 =⇒ y1 ≡ 2− T mod(T − 1)2.

(T − 1)2y2 ≡ 1 modT =⇒ y2 ≡ 1 modT.

Therefore p(T ) ≡ 1 ∗ T ∗ (2− T ) + 0 ∗ (T − 1)2 ∗ 1 ≡ T (2− T ) modT (T − 1)2.

xs = x(2− x) =

1 2

0 1

1 −2

0 1

=

1 0

0 1

, xn = x− xs =

0 2

0 0

.

36 ZHENGYAO WU

4. March 19th, Cartan’s criterion, Killing form

Lemma 4.1Let F be an algebraically closed field. Let V be a finite dimensional F -vector space. If x ∈ gl(V )is semisimple, then adx ∈ gl(gl(V )) is semisimple.(In Lemma 2.25, we proved that if x ∈ gl(V ) is nilpotent, then adx ∈ gl(gl(V )) is nilpotent.)Proof.


1. x is diagonalizable. Suppose that relativeto a basis of V , x = diag(a1, . . . , an).

F is algebraically closed, x is semisimpleand Lemma 3.9

2. [x, eij] = (ai − aj)eij. [x, eij] = [n∑k=1

akekk, eij] =n∑k=1

ak[ekk, eij] =n∑k=1

ak(δkiekj − δjkeik) = (ai − aj)eij

3. adx is diagonalizable. The matrix of (adx) is diagonal relativeto the basis eij : 1 ≤ i, j ≤ n of gl(V ).

4. adx is semisimple. F is algebraically closed and Lemma 3.9

Lemma 4.2Let F be an algebraically closed field. Let V be a finite dimensional F -vector space. Let x ∈ gl(V ).If x = xs+xn is the Jordan decomposition, then adx = ad(xs)+ad(xn) is the Jordan decompositionof adx in gl(gl(V )).Proof.


1. adx = ad(xs) + ad(xn). x = xs + xn and ad is linear

2. ad(xs) is semisimple. xs is semisimple and Lemma 4.1

3. (adxn) is nilpotent. xn is nilpotent and Lemma 2.25

4. [ad(xs), (adxn)] = 0. [ad(xs), (adxn)] = ad([xs, xn]) = ad(0) =0.

5. ad(xs) = (ad x)s and (adxn) = (ad x)n. Uniqueness of the Jordan decompositionof adx Proposition 3.14(a)


Lemma 4.3Let F be an algebraically closed field. Let A be a finite dimensional F -algebra. If x ∈ Der(A),then xs, xn ∈ Der(A).

Proof. Suppose the characteristic polynomial of x isk∏i=1

(T − ai)mi , where a1, . . . , am are distinct.Let Vi = ker(x− ai · 1)mi .


1. xs = p(x). Proposition 3.14(b)

2. xs|Vi = ai · 1. p(T ) ≡ ai mod(T − ai)mi

3. For all y ∈ Vi and z ∈ Vj, xs(y)z +yxs(z) = aiyz + yajz = (ai + aj)yz.

4. (x− (ai + aj) · 1)(yz) = (x− ai · 1)(y)z +y(x− aj · 1)(z).

x ∈ Der(A)

5. (x − (ai + aj) · 1)mi+mj(yz) =mi+mj∑k=0

(x −

ai · 1)mi+mj−k(y) · (x− aj · 1)k(z).

Leibniz Rule

6. yz ∈ ker(x− (ai + aj) · 1)mi+mj . either k ≥ mi or mi +mj − k ≥ mj

7. xs(yz) = (ai + aj)yz.

7.1. If ker(x − (ai + aj) · 1)mi+mj = 0, thenyz = 0.

7.2. If ker(x − (ai + aj) · 1)mi+mj 6= 0, thenai + aj is an eigenvalue with eigenvectoryz.

p(T ) ≡ ai+aj mod(T −ai−aj)m for somem ≥ mi +mj

8. xs(yz) = xs(y)z+yxs(z) for all y ∈ Vi andz ∈ Vj.

steps 3, 7

9. xs ∈ Der(A). A = ⊕iVi.

10. xn ∈ Der(A). xn = x− xs

38 ZHENGYAO WU

Lemma 4.4Let F be an algebraically closed field of characteristic 0. If [L,L] is nilpotent, then L is solvable.Proof.


1. [L,L] is solvable. [L,L] is nilpotent and Lemma 2.16

2. L/[L,L] is solvable. L/[L,L] is abelian and Example 2.4(1)

3. L is solvable. Proposition 2.8

Lemma 4.5 InterpolationLet F be a field. Suppose a1, . . . , an, b1, . . . , bn ∈ F . If ai 6= aj for all 1 ≤ i < j ≤ n, then thereexists a polynomial r(T ) ∈ F [T ] such that r(ai) = bi.

Proof. Lagrange:

r(T ) =n∑i=1

n∏j=1,j 6=i

x− ajai − aj

bi.

Lemma 4.6Let F be an algebraically closed field of characteristic 0. Let V be a finite dimensional F -vectorspace. Let A ⊂ B be two subspaces of gl(V ). Let M = y ∈ gl(V ) : [y,B] ⊂ A. If x ∈ M suchthat Tr(xy) = 0 for all y ∈M , then x is nilpotent.Proof.


1. Let x = xs + xn be the Jordan decompo-sition.

Proposition 3.14(a)

2. There exists a basis of V such that xs =diag(a1, . . . , an) ∈ gl(V ).

xs is semisimple and Lemma 3.9

3. Let E = SpanQai : 1 ≤ i ≤ n. Thenany linear function f : E → Q is 0.

See the next table



4. E∗ = HomQ(E,Q) = 0. step 3

5. E = 0. dimQ(E) = dimQ(E∗) = 0

6. xs = 0. step 2

7. x is nilpotent. x = xn


3.1. [xs, eij] = (ai − aj)eij. xs =n∑k=1

akekk

3.2. Let y = diag(f(a1), . . . , f(an)). Then[y, eij] = (f(ai)− f(aj))eij.

3.3. There exists a polynomial r(T ) ∈ F [T ]such that r(0) = 0 and r(ai−aj) = f(ai)−f(aj) and r is well-defined.

Lemma 4.5 and f is linear.

3.4 r(ad(xs)) = ad y. r(ad(xs))(eij) = r(ai − aj)eij = (f(ai) −f(aj))eij = (ad y)(eij).

3.5 ad(xs) = (ad x)s. Lemma 4.2

3.6 ad(xs) = p(adx), p(T ) = 0. Proposition 3.14(b)

3.7 (ad y) = r(p(adx)), r(p(0)) = 0. composition of step 3.3, 3.6

3.8 (adx)(B) = [x,B] ⊂ A. x ∈M

3.9 (ad y)(B) ⊂ A, i.e. y ∈M . step 3.7

3.10 0 = Tr(xy). Given

= Tr(xsy) =n∑i=1

aif(ai). Tr(xny) = 0

3.11n∑i=1

f(ai)2 = 0. Apply f on both sides

3.12 f(ai) = 0 for all 1 ≤ i ≤ n. im(f) ⊂ Q

3.13 f = 0. defn of E

40 ZHENGYAO WU

Lemma 4.7Let F be an algebraically closed field of characteristic 0. Let V be a finite dimensional F -vectorspace. Tr([x, y]z) = Tr(x[y, z]) for all x, y, z ∈ gl(V ).Proof.


1. Tr([x, y]z)−Tr(x[y, z]) = Tr(xyz−yxz)−Tr(xyz − xzy)

defn of [, ]

2. = −Tr(xzy) + Tr(yxz) Tr is additive

3. = 0 Tr(AB) = Tr(BA) for all A,B ∈ gl(n, F )

Theorem 4.8 Cartan’s criterionLet F be an algebraically closed field of characteristic 0. Let V be a finite dimensional F -vectorspace. Let L be a subalgebra of gl(V ). Suppose that Tr(xy) = 0 for all x ∈ [L,L], y ∈ L, then Lis solvable.Proof.


1. Let M = c ∈ gl(V ) : [c, L] ⊂ [L,L].Then L ⊂M .

2. Tr([a, b]c) = Tr(a[b, c]) Lemma 4.7

= 0 for all a, b ∈ L. Given

3. Tr(xc) = 0 for all x ∈ [L,L], c ∈M . [a, b] : a, b ∈ L generates [L,L]

4. x is nilpotent for all x ∈ [L,L]. Lemma 4.6

5. adx is nilpotent for all x ∈ [L,L]. Lemma 2.25

6. [L,L] is nilpotent. Theorem 2.31

7. L is solvable. Lemma 4.4


Corollary 4.9Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional F -Liealgebra such that Tr((ad x) (ad y)) = 0 for all x ∈ [L,L], y ∈ L. Then L is solvable.Proof.


1. ad(L) is solvable. Theorem 4.8

2. ker(ad) = Z(L). Lemma 1.31(2)

3. L/Z(L) = L/ ker(ad) ' ad(L) is solvable. Proposition 1.29(1)

4. Z(L) is solvable. Z(L) is abelian and Example 2.4(1)

5. L is solvable. Proposition 2.8

Definition 4.10Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional F -Liealgebra. The Killing form of L is κ(x, y) = Tr((adx) (ad y)), ∀x, y ∈ L.

Lemma 4.11Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional F -Liealgebra. Let κ be the Killing form of L. Then κ is an associative, symmetric bilinear form over L.Proof.


1. κ is associative. Lemma 4.7

2. κ is symmetric. Tr(AB) = Tr(B A) for all A,B ∈ gl(L)

3. κ is bilinear. Tr and ad are linear

42 ZHENGYAO WU

Lemma 4.12Let F be an algebraically closed field of characteristic 0. Let V be a finite dimensional F -vectorspace. Let W be a sub-vector space of V . If φ ∈ EndF (V ) such that φ(V ) ⊂ W , then Tr(φ) =Tr(φ|W ).Proof.


1. Suppose e1, . . . , em is a basis of W ande1, . . . , em, em+1, . . . , en is a basis of V .

Let W be a sub-vector space of V .

2. φ(ei) =m∑j=1

aijej, aij = 0 for m + 1 ≤ j ≤n.

φ(V ) ⊂ W

3. Tr(φ) = Tr(φ|W ).n∑i=1

aii =m∑i=1

aii

Lemma 4.13Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional F -Liealgebra. Let I be an ideal of L. Then κ|I×I is the Killing form of I.Proof.


1. For all x ∈ I, (adL x)|I = adI x. I is an ideal of L

2. κ(x, y) = Tr((adL x) (adL y)) Definition 4.10

= Tr((adL x)|I (adL y)|I) x, y ∈ I and ad(I) ⊂ I

= Tr((adI x) (adI y)) for all x, y ∈ I. step 1

3. κ|I×I is the Killing form of I. Definition 4.10

Definition 4.14Let F be a field. Let V be a finite dimensional F -vector space. Let β : V ×V → F be a symmetricbilinear form.


(1) The radical of β is Rad(β) = x ∈ V : β(x, y) = 0,∀y ∈ V .(2) We call β non-degenerate if Rad(β) = 0.

Lemma 4.15Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional F -Liealgebra. Let κ be the Killing form of L. Rad(κ) is an ideal of L.

Proof. For all x ∈ Rad(κ) and y ∈ L, κ([x, y], z) = κ(x, [y, z]) = 0, ∀z ∈ L by Lemma 4.11. Then[x, y] ∈ Rad(κ).

Lemma 4.16Let F be a field. Let V be a vector space with basis x1, . . . , xn. Then a symmetric bilinear formβ is non-degenerate iff det((β(xi, xj))1≤i,j≤n) 6= 0.Proof.


1. Rad(κ) = ker(β) where β : V → V ∗,β(x)(y) = β(x, y).

Definition 4.14(1)

2. β is non-degenerate iff β is injective Definition 4.14(2)

3. iff β is invertible dimF (V ) = n <∞

4. iff the matrix (β(xi, xj))1≤i,j≤n of β is non-singular.

Example 4.17Let F be an algebraically closed field of characteristic 0. The Killing form of sl(2, F ) is non-degenerate.

Proof. Take the standard basis x, h, y. We have

(adx)

x

h

y

=

0 −2 0

0 0 1

0 0 0

x

h

y

, (adh)

x

h

y

=

2 0 0

0 0 0

0 0 −2

x

h

y

, (ad y)

x

h

y

=

0 0 0

−1 0 0

0 2 0

x

h

y

.

44 ZHENGYAO WU

(adx) (adx) =

0 0 −2

0 0 0

0 0 0

, (adh) (adx) =

0 −4 0

0 0 0

0 0 0

, (ad y) (adx) =

0 0 0

0 2 0

0 0 2

,

(adx) (adh) =

0 0 0

0 0 −2

0 0 0

, (adh) (adh) =

4 0 0

0 0 0

0 0 4

, (ad y) (adh) =

0 0 0

−2 0 0

0 0 0

,

(adx) (ad y) =

2 0 0

0 2 0

0 0 0

, (adh) (ad y) =

0 0 0

0 0 0

0 −4 0

, (ad y) (ad y) =

0 0 0

0 0 0

−2 0 0

,

Take traces, κ has matrix 0 0 4

0 8 0

4 0 0

whose determinant is −4 ∗ 8 ∗ 4 = −128 6= 0. By Lemma 4.16, κ is non-degenerate.

Lemma 4.18Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional F -Liealgebra. Then L is not semisimple iff it has a nonzero abelian ideal.

Proof. Suppose L has a nonzero abelian ideal I.


1. I is solvable. I is abelian and Example 2.4(1)

2. 0 6= I ⊂ Rad(L) Lemma 2.10(2)

3. Rad(L) 6= 0. I 6= 0

4. L is not semisimple. Definition 2.11

Suppose L is not semisimple.



1. Rad(L) 6= 0. Definition 2.11

2. There exists a least integer n ≥ 1 suchthat Rad(L)(n) = 0.

Definition 2.3

3. Rad(L)(n−1) is a nonzero abelian subalge-bra of L.

Rad(L)(n−1) is an ideal of L. [Hum78, p.14, Exercise 1]

Lemma 4.19Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional F -Lie algebra.Let κ be the Killing form of L. Then Rad(κ) ⊂ Rad(L). (It is possible that Rad(κ) 6= Rad(L),see [Hum78, p.24, Exercise 4].)Proof.


1. Suppose x ∈ Rad(κ). Then 0 = κ(x, y) =Tr((ad x) (ad y)) for all y ∈ L.

Definition 4.14(1)

2.1. Rad(κ) is solvable. Cartan’s criterion Corollary 4.9

2.2. Rad(κ) is an ideal of L. Lemma 4.15

3. Rad(κ) ⊂ Rad(L). Lemma 2.10(2)

Lemma 4.20Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional F -Liealgebra. Let κ be the Killing form of L. If I is an abelian ideal I of L, then I ⊂ Rad(κ).Proof.

46 ZHENGYAO WU


1. Suppose x ∈ I. For all y ∈ L, ((adx) (ad y))(L) ⊂ I.

((adx) (ad y))(L) = [x, [y, L]] ⊂ [x, L] ⊂I since I is an ideal

2. ((adx) (ad y))2(L) = 0. ((adx) (ad y))2(L) ⊂ ((adx) (ad y))(I) = [x, [y, I]] ⊂ [I, I] = 0since I is an abelian ideal

3. κ(x, y) = Tr((adx) (ad y)) = 0. (adx) (ad y) is nilpotent

4. I ⊂ Rad(κ). Definition 4.14(1)

Theorem 4.21Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional F -Liealgebra. Then L is semisimple iff its Killing form κ is non-degenerate. (See [Hum78, p.24, Exercise6] for the case charF = p > 0.)

Proof. Suppose L is semisimple.


1. Rad(L) = 0. Definition 2.11

2. Rad(κ) = 0. Rad(κ) ⊂ Rad(L) by Lemma 4.19

3. κ is non-degenerate. Definition 4.14(2)

Suppose κ is non-degenerate.


1. Rad(κ) = 0. Definition 4.14(2)

2. 0 is the only abelian ideal of L. Every abelian ideal is in Rad(κ) byLemma 4.20

3. L is semisimple. Lemma 4.18


Definition 4.22Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional F -Liealgebra.(1) L is the direct sum of ideals if there exist ideals I1, . . . , It such that L = I1 + · · ·+It as directsum of sub-vector spaces. We write L = I1 ⊕ · · · ⊕ It. (For all 1 ≤ i 6= j ≤ t [Ii, Ij] ⊂ Ii ∩ Ij = 0.)(2) Let I be an ideal of L. Define I⊥ = x ∈ L : κ(x, I) = 0.

Lemma 4.23Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional F -Liealgebra. If I is an ideal of L, then so is I⊥.

Proof. Suppose x ∈ I⊥ and y ∈ L,


1. For all z ∈ I, [y, z] ∈ I. I is an ideal of L

2. κ([x, y], z) = κ(x, [y, z]) Lemma 4.11

= 0. x ∈ I⊥ and Definition 4.22(2)

3. [x, y] ∈ I⊥. Thus I⊥ is an ideal of L. Definition 4.22(2)

Lemma 4.24Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional Lie F -algebrawith an ideal I. If L is semisimple, then L = I ⊕ I⊥.Proof.


1. I → L∗, x 7→ κ(x, •) is injective. L is semisimple and Theorem 4.21

2. L ' L∗∗ → I∗, x 7→ κ(•, x) is surjective. dimF (L) and duality

3. dimF (I) + dimF (I⊥) = dimF (L) dimF (ker(L → I∗)) = dimF (L) −dimF (I∗)

48 ZHENGYAO WU


4. Let J = I ∩ I⊥. Then κ(x, y) = 0 for allx ∈ [J, J ], y ∈ J .

Definition 4.22(2)

5. J is a solvable ideal of L. Cartan’s criterion Corollary 4.9

6. J ⊂ Rad(L). Lemma 2.10(2)

7. Rad(L) = 0 and thus J = 0. L is semisimple and Definition 2.11(2)

8. L = I ⊕ I⊥. steps 3,7


5. March 26th, Semisimple decomposition and Lie modules

Theorem 5.1Let L be a semisimple Lie algebra. There exist simple ideals L1, · · · , Lt of L such that L =L1 ⊕ · · · ⊕ Lt.

Proof. If L is simple, then it is a simple ideal of itself.Now suppose L is not simple.


1. L has a minimal nonzero proper ideal L1. L is not simple and Definition 1.22

2. L1 is semisimple. Rad(L1) ⊂ Rad(L) = 0 and Defini-tion 2.11

3. L1 is simple. Minimality of L1.

4. L = L1 ⊕ L⊥1 . Lemma 4.24.

5. L⊥1 is semisimple. Rad(L⊥1 ) ⊂ Rad(L) = 0 and Defini-tion 2.11

6. L⊥1 is a direct sum of simple ideals of L⊥1(and of L).

dimF (L⊥1 ) < dimF (L) and inductive hy-pothesis

7. L = L1 ⊕ L2 ⊕ · · · ⊕ Lt. Suppose L⊥1 = L2 ⊕ · · · ⊕ Lt.

Theorem 5.2Let L be a semisimple Lie algebra and L = L1 ⊕ · · · ⊕ Lt is the decomposition into simple ideals.Every nonzero proper simple ideal of L coincides one of Li.

Proof. Let I be a nonzero simple ideal of L.


1. Rad(L) = 0 L is semisimple and Definition 2.11

2. Z(L) is a solvable ideal of L. Example 1.21(2) and Example 2.4(1)

3. Z(L) ⊂ Rad(L) Lemma 2.10(2)

4. Z(L) = 0. steps 1,2

50 ZHENGYAO WU


5. [I, L] is an ideal of I. I is an ideal of L and Definition 1.1(L3)

6. [I, L] = 0 or [I, L] = I. I is simple and Definition 1.22(1)

7. [I, L] = I. [I, L] 6= 0 by step 4

8. I = [I, L1]⊕ · · · ⊕ [I, Lt]. L = L1 ⊕ · · · ⊕ Lt exists by Theorem 5.1

9. So I = [I, Li] for some i and [I, Lj] = 0for all j 6= i.

I is simple and Definition 1.22(1)

10. I = [I, Li] is an ideal of Li. I, Li are ideals of L and Definition 1.1(L3)

11. I = Li. Li is simple, I 6= 0 and Defini-tion 1.22(1)

Corollary 5.3If L is semisimple Lie algebra, then L = [L,L].Proof.


1. Suppose L = L1 ⊕ · · · ⊕ Lt where Li aresimple ideals of L.

Theorem 5.1

2. [Li, Li] = Li. Li are simple and Lemma 1.24

3. [Li, Lj] = 0 for all i 6= j. Definition 4.22

4. [L,L] =[t⊕i=1

Li,t⊕i=1

Li

]step 1

=t⊕i=1

[Li, Li] Definition 1.1(L1)

=t⊕i=1

Li = L. step 2


Corollary 5.4All ideals of a semisimple F -Lie algebra L are semisimple.

Proof. Let I be an ideal of L.


1. L = I ⊕ I⊥. Lemma 4.24

2. Let κ be the Killing form of L. Then κ|Iis the Killing form of I and κ|I⊥ is theKilling form of I⊥.

Lemma 4.13

3. κ =

κ|I 0

0 κ|I⊥

. steps 1,2

4. det(κ) = det(κ|I) det(κ|I⊥).

5. κ is non-degenerate. L is semisimple and Theorem 4.21

6. det(κ) 6= 0. Definition 4.14

7. det(κ|I) 6= 0. steps 4,6

8. κ|I is non-degenerate. Definition 4.14

9. I is semisimple. Theorem 4.21

Corollary 5.5Each ideal I of L is a direct sum of certain simple ideals of L.

Proof. By Corollary 5.4, I is semisimple. Then Theorem 5.1 provides a decomposition.

Corollary 5.6All homomorphic images of a semisimple F -Lie algebra L are semisimple.

Proof. Let φ : L→ L′ be a homomorphism of Lie algebras and M = φ(L).


1. If I is an ideal of L, then φ(I) is an idealof M .

φ is a homomorphism.

52 ZHENGYAO WU


2. If φ(I) has a nonzero proper ideal J , thenI has a nonzero proper ideal φ−1(J).

φ is a homomorphism.

3. If I is a simple ideal of L, then φ(I) is asimple ideal of M .

4. L = L1 ⊕ · · · ⊕ Lt where Li are simpleideals of L.

Theorem 5.1

5. M = φ(L) = φ(L1) ⊕ · · · ⊕ φ(Lt) whereφ(Li) are simple ideals of M .

steps 3,4

Lemma 5.7Let L be a Lie algebra. [δ, adx] = ad(δ(x)) for all x ∈ L, δ ∈ Der(L).

Proof. For all y ∈ L,


1. [δ, adx](y) = δ((adx)(y))− (adx)(δ(y)) Example 1.4

= δ([x, y])− [x, δ(y)] Definition 1.13

= [δ(x), y] Definition 1.12

= ad(δ(x))(y) Definition 1.13

Proposition 5.8If L is a semisimple Lie algebra, then L is linear.Proof.





2. Z(L) is a solvable ideal of L. Example 1.21(2) and Example 2.4(1)

3. Z(L) ⊂ Rad(L) Lemma 2.10(2)

4. Z(L) = 0. steps 1,2

5. ker(ad) = Z(L) = 0. Lemma 1.31(2)

6. L = L/ ker(ad) ' ad(L) ⊂ gl(L). Proposition 1.29(1)

Theorem 5.9If L is a semisimple Lie algebra, then ad(L) = Der(L) i.e. every derivation of L is inner.Proof.


1. ad(L) ' L. Proposition 5.8

2. ad(L) is semisimple. L is semisimple

3. ad(L) has a non-degenerate Killing formκ.

Theorem 4.21

4. ad(L) ⊂ Der(L). Lemma 1.14

5. [Der(L), ad(L)] ⊂ ad(L), i.e. ad(L) is anideal of Der(L).

Lemma 5.7

6. κ = κ′|ad(L) where κ′ is the Killing form ofDer(L).

Lemma 4.13

7. The complement ad(L)⊥ of ad(L) inDer(L) relative to κDer(L) is an ideal ofDer(L).

Lemma 4.23

8. ad(L) ∩ ad(L)⊥ = 0. κ is non-degenerate by step 3

9. [ad(L), ad(L)⊥] = 0. [ad(L), ad(L)⊥] ⊂ ad(L)∩ad(L)⊥ by steps5,7

54 ZHENGYAO WU


10. ad(L)⊥ = 0.

If δ ∈ ad(L)⊥, then ad(δ(x)) = [δ, adx] =0 for all x ∈ L.

Lemma 5.7

δ(x) = 0 for all x ∈ L. Hence δ = 0. ad is injective by Proposition 5.8

11. Der(L) = ad(L).

Proposition 5.10If L is semisimple, then for all x ∈ L, there exists unique xs, xn ∈ L such that

• x = xs + xn

• xs is ad-semisimple (i.e. ad(xs) is semisimple).• xn is ad-nilpotent (i.e. ad(xn) is nilpotent).• [xs, xn] = 0

We call it the abstract Jordan decomposition of x in L.Proof.


1. Let adx = (adx)s+(adx)n be the Jordandecomposition of adx in gl(L).

Proposition 3.14(a)

2. adx ∈ Der(L). Lemma 1.14

3. (adx)s, (adx)n ∈ Der(L). Lemma 4.3

4. ad(L) = Der(L). Thus (adx)s, (adx)n ∈ad(L).

L is semisimple and Theorem 5.9

5. L ' ad(L). L is semisimple and Proposition 5.8

6. There exists unique xs, xn ∈ gl(L) suchthat ad(xs) = (adx)s and (adxn) =(adx)n.

7. x = xs + xn. adx = (ad x)s + (ad x)n, ad is linear andstep 5



xs is ad-semisimple. ad(xs) is semisimple by step 1

xn is ad-nilpotent. ad(xn) is nilpotent by step 1

[xs, xn] = 0. (adx)s (adx)n = (adx)n (adx)s andstep 5

Remark 5.11If L = sl(V ), then the abstract and the usual Jordan decomposition coincide.Proof.


1. Let x = xs + xn be the usual Jordan de-composition in gl(V ).

Proposition 3.14(a)

2.1. Tr(xn) = 0. xn is nilpotent

2.2. xn ∈ L. defn of sl(V )

3.1. Tr(xs) = Tr(x)− Tr(xn) = 0− 0 = 0. xs = x− xn and Tr is linear

3.2. xs ∈ L. defn of sl(V )

4.1. adgl(V )(xs) is semisimple. xs is semisimple and Lemma 4.1

4.2. adL(xs) is semisimple. adL(xs)(L) ⊂ [L,L] ⊂ L and Lemma 3.10

5.1. adgl(V )(xn) is nilpotent. xn is nilpotent and Lemma 2.25

5.2. adL(xn) is nilpotent. adL = adgl(V ) |L

6. [xs, xn] = 0. [xs, xn] = xs xn − xn xs = 0

7. x = xs+xn is the abstract Jordan decom-position of x.

Proposition 5.10

56 ZHENGYAO WU

Definition 5.12Let L be a F -Lie algebra.(1) Let V be a vector space with an operation L× V → V , (x, v) 7→ x · v,(M1) (ax+ by) · v = a(x · v) + b(y · v),(M2) x · (av + bw) = a(x · v) + b(x · w),(M3) [x, y] · v = x · (y · v)− y · (x · v),for all x, y ∈ L, v, w ∈ V and a, b ∈ F . We call V an L-module.(2) Let V,W be L-modules. A homomorphism of L-modules is a map φ : V → W such thatφ(x · v) = x · φ(v) for all x ∈ L and v ∈ V .

Example 5.13(1) Let φ : V → W be a homomorphism of L-modules. Then ker(φ) is an L-module.(2) L is an L-module with x · y = [x, y].(3) Suppose x · v = v for all x ∈ L and v ∈ V 6= 0. Then V is not an L-module since (M3) isnot satisfied.

Lemma 5.14x · v = ϕ(x)(v), x ∈ L, v ∈ V .(1) Given a representation ϕ : L→ gl(V ), this defines an L-module.(2) Given an L-module V , this defines a representation ϕ : L→ gl(V ).

Proof. (M1) iff ϕ is F -linear.(M2) iff ϕ(x) : V → V is F -linear for all x ∈ L.(M3) iff ϕ is a homomorphism of F -Lie algebras.

Definition 5.15A homomorphism of L-modules φ : V → W is an isomorphism if it is an isomorphism of F -vectorspaces, i.e. V and W afford equivalent representations of L.

Definition 5.16An L-module V is irreducible if L has precisely two L-submodules 0 and V .

Example 5.17(1) 0 is not an irreducible L-module.(2) Suppose V = Fv such that x · v = 0 for all x ∈ L. Then V is an irreducible L-module.(3) A simple algebra L is an irreducible L-module.

Definition 5.18An L-module is called completely reducible if(1) V is a direct sum of irreducible L-submodules. Or, equivalently,(2) Each L-submodule W of V has a complement W ′, i.e. an L-submodule such that V = W ⊕W ′

(p.30, exercise 2).


Example 5.19A semisimple algebra L is a completely reducible L-module by Theorem 5.1.

Lemma 5.20 SchurLet φ : L→ gl(V ) be irreducible. Then ψ ∈ EndF (V ) : φ(x) ψ = ψ φ(x) = F · 1.

Proof. Suppose ψ ∈ gl(V ) such that φ(x) ψ = ψ φ(x) for all x ∈ L.


1. ψ has eigenvalues. F is algebraically closed

2. ker(ψ−a ·1) 6= 0 where a is an eigenvalue. Eigenvectors of ψ are nonzero.

3. ker(ψ − a · 1) = V . V is irreducible and Definition 5.16

4. ψ = a · 1. ψ − a · 1 = 0.

Example 5.21Let V and W be L-modules. Suppose x ∈ L, v ∈ V and w ∈ W .(1) V ⊕W is an L-module with x · (v, w) = (x · v, x · w)(2) V ∗ is an L-module by (x · f)(v) = −f(x · v), f ∈ V ∗.(3) V ⊗F W is a L-module by x · (v ⊗ w) = x · v ⊗ w + v ⊗ x · w.(4) HomF (V,W ) is a L-module by (x · f)(v) = x · f(v)− f(x · v), f ∈ HomF (V,W ).

Proof. We only prove (4). Suppose a, b ∈ F , v, w ∈ V , f, g ∈ HomF (V,W )(M1)


1. ((ax+ by) · f)(v)

= (ax+ by) · f(v)− f((ax+ by) · v) defn of ·

= ax · f(v) + by · f(v)− f(ax · v + by · v) (M1) of V,W

= ax ·f(v)+ by ·f(v)−af(x ·v)− bf(y ·v) f is F -linear

= a(x ·f(v)−f(x ·v))+b(y ·f(v)−f(y ·v))

= a(x · f)(v) + b(y · f)(v). defn of ·

58 ZHENGYAO WU


2. (ax+ by) · f = a(x · f) + b(y · f)

(M2)


1. (x · (af + bg))(v)

= x · (af + bg)(v)− (af + bg)(x · v) defn of ·

= ax ·f(v) + bx ·g(v)−af(x ·v)− bg(x ·v) (M2) of V,W

= a(x ·f(v)−f(x ·v))+b(x ·g(v)−g(x ·v))

= a(x · f)(v) + b(x · g)(v). defn of ·

2. x · (af + bg) = a(x · f) + b(x · g)

(M3)


1. ([x, y] · f)(v)

= [x, y] · f(v)− f([x, y] · v) defn of ·

= x · (y · f(v)) − y · (x · f(v)) − f(x · (y ·v)− y · (x · v))

(M3) of V,W

= x · (y · f(v)) − y · (x · f(v)) − f(x · (y ·v)) + f(y · (x · v))

f is F -linear

2. (x · (y · f))(v)− (y · (x · f))(v)

= x · ((y · f)(v)) − (y · f)(x · v) − y · ((x ·f)(v)) + (x · f)(y · v)

defn of ·

= x · (y · f(v) − f(y · v)) − (y · f(x · v) −f(y · (x · v)))− y · (x · f(v)− f(x · v)) + x ·f(y · v)− f(x · (y · v))

defn of ·



= x · (y · f(v))− x · f(y · v)− y · f(x · v) +f(y · (x · v))− y · (x · f(v)) + y · f(x · v) +x · f(y · v)− f(x · (y · v))

= x · (y · f(v)) − y · (x · f(v)) − f(x · (y ·v)− y · (x · v))

3. [x, y] · f = x · (y · f)− y · (x · f). steps 1,2

Lemma 5.22Let V andW be L-modules. There exists an isomorphism of L-modules φ : V ∗⊗W → HomF (V,W )given by φ(f ⊗ w)(v) = f(v)w. In particular, V ∗ ⊗ V ' EndF (V ).

Proof. Suppose v ∈ V , w ∈ W and f ∈ V ∗. We first showt that φ is a homomorphism ofL-modules.


1. φ(x · (f ⊗ w))(v)

= φ((x · f)⊗ w + f ⊗ (x · w))(v) defn of · of V ∗ ⊗W

= φ((x · f)⊗ w)(v) + φ(f ⊗ (x · w))(v) φ is linear.

= (x · f)(v)w + f(v)(x · w) defn of φ.

= −f(x · v)w + f(v)(x · w) defn of · of V .

2. (x · φ(f ⊗ w))(v)

= x · φ(f ⊗ w)(v)− φ(f ⊗ w)(x · v) defn of · of HomF (V,W )

= x · (f(v)w)− f(x · v)w defn of φ.

= −f(x · v)w + f(v)(x · w)

3. φ(x · (f ⊗w)) = x · φ(f ⊗w) for all gener-ators f ⊗ w.

steps 1,2

60 ZHENGYAO WU

Next, we show that φ is an isomorphism of vector spaces. Let (vi) be a basis of V and (wj) be abasis of W .


1.1. Suppose g ∈ HomF (V,W ) such thatg(vi) = ∑

jgijwj.

1.2. Define fj ∈ V ∗ such that fj(vi) = gij.

1.3. φ is surjective. g(vi) = ∑jfj(vi)wj = φ(∑

jfj ⊗ wj)

2. dimF (V ∗ ⊗ W ) = dimF (V ) dimF (W ) =dimF (Hom(V,W )).

3. φ is an isomorphism of vector spaces. steps 1,2

Therefore φ is an isomorphism of L-modules.

Definition 5.23Let F be an algebraically closed field of characteristic 0. Let L be a semisimple F -Lie algebra.Let φ : L → gl(V ) be a faithful representation. Define a symmetric bilinear form β(x, y) =Tr(φ(x) φ(y)), x, y ∈ L. We call it the trace form of φ.

Example 5.24Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimpleF -Lie algebra. The Killing form κ of L is the trace form of the adjoint representation ad.

Lemma 5.25Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimpleF -Lie algebra. Let φ : L→ gl(V ) be a faithful representation. Let β be the trace form of φ. Then(1) β is an associative, symmetric bilinear form over L.(2) Rad(β) is an ideal of L.(3) β is non-degenerate.

Proof. The proof of (1) is similar to Lemma 4.11.The proof of (2) is similar to Lemma 4.15.


6. April 2nd, Casimir element, Weyl’s theorem

(3) Let S = Rad(β).


1. For all x ∈ [S, S], y ∈ S, Tr(φ(x)φ(y)) =β(x, y) = 0.

defn of S

2. φ(S) is solvable. Cartan’s criterion Theorem 4.8

3. S is solvable. S ' φ(S) since φ is faithful

4. S ⊂ Rad(L). Lemma 2.10(2)


6. β is non-degenerate. S = 0 and Definition 4.14(2)

Lemma 6.1Let L be a semisimple F -Lie algebra. Let β be any non-degenerate symmetric bilinear form on L.Let (x1, . . . , xn) be a basis of L and (y1, . . . , yn) its dual basis with respect to β. Suppose x ∈ Lsuch that [x, xi] =

n∑j=1

aijxj and [x, yi] =n∑j=1

bijyj. Then aik = −bki.Proof.


1. β([xi, x], yk) = β(xi, [x, yk]) associativity by Lemma 5.25(1)

2. β(−n∑j=1

aijxj, yk) = β(xi,n∑j=1

bkjyj) Remark 1.2(L2’), [x, xi] =n∑j=1

aijxj and

[x, yi] =n∑j=1

bijyj

3. −n∑j=1

aijβ(xj, yk) =n∑j=1

bkjβ(xi, yj). Definition 1.1(L1)

4. −aik = bki. β(xi, yj) = δij =

1, i = j;

0, i 6= j.

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Lemma 6.2In gl(V ), [x, y z] = [x, y] z + y [x, z].Proof.


1. [x, y z] = x y z − y z x. Example 1.4

2. [x, y] z − y [x, z] = (x y − y x) z +y (x z − z x)

Example 1.4

= x y z− y x z + y x z− y z x

= x y z − y z x.

3. [x, y z] = [x, y] z + y [x, z]. steps 1,2

Lemma 6.3Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimpleF -Lie algebra. Let φ : L → gl(V ) be a faithful representation. Let β be the trace form of φ.Let (x1, . . . , xn) be a basis of L and (y1, . . . , yn) its dual basis with respect to β. Let cφ(β) =n∑i=1

φ(xi) φ(yi). We call cφ = cφ(β) ∈ gl(V ) the Casimir element of φ. Then [φ(x), cφ(β)] = 0for all x ∈ L.Proof.


1. [φ(x), cφ(β)] = [φ(x),n∑i=1

φ(xi) φ(yi)] defn of cφ(β)

=n∑i=1

[φ(x), φ(xi) φ(yi)] Definition 1.1(L1)

=n∑i=1

[φ(x), φ(xi)] φ(yi) + φ(xi) [φ(x), φ(yi)] Lemma 6.2

=n∑i=1

φ([x, xi]) φ(yi) +n∑i=1

φ(xi) φ([x, yi]) φ is a homomorphism

=n∑i=1

φ(n∑j=1

aijxj) φ(yi) +n∑i=1

φ(xi) φ(n∑j=1

bijyj) Suppose [x, xi] =n∑j=1

aijxj

and [x, yi] =n∑j=1

bijyj.

=n∑i=1

n∑j=1

(aji + bij)φ(xi) φ(yj). φ is linear



= 0. Lemma 6.1

Lemma 6.4Let L be a finite dimensional semisimple F -Lie algebra. Let φ : L→ gl(V ) be a faithful represen-tation. Then Tr(cφ) = dimF (L).Proof.


1. Tr(cφ) = Tr(n∑i=1

φ(xi) φ(yi)) Lemma 6.3

=n∑i=1

Tr(φ(xi) φ(yi)) Tr is linear

=n∑i=1

β(xi, yi) Definition 5.23

=n∑i=1

1 = n = dimF (L) β(xi, yj) = δij

Lemma 6.5If V is irreducible, then cφ = dimF (L)

dimF (V ) · 1V .In particular, cφ is independent of the choice of basis.Proof.


1. [φ(x), cφ] = 0 for all x ∈ L. Lemma 6.3

2. cφ = a · 1V for some a ∈ F . Lemma 5.20

3. Tr(cφ) = a dimF (V ).

4. Tr(cφ) = dimF (L). Lemma 6.4

5. a = dimF (L)dimF (V ) . steps 3,4

64 ZHENGYAO WU

Example 6.6Let L = sl(2, F ), V = F 2 and φ : L→ gl(V ) is the identity. Then cφ = 3

2 · 1V .

Proof. The trace form of φ is β(s, t) = Tr(st). Let (e1, e2, e3) = (x, h, y) be a basis of L. Then

x2 =

0 0

0 0

, xh =

0 −1

0 0

, xy =

1 0

0 0

,

hx =

0 1

0 0

, h2 =

1 0

0 1

, hy =

0 0

−1 0

,

yx =

0 0

0 1

, yh =

0 0

1 0

, y2 =

0 0

0 0

.Then

(Tr(eiej))1≤i,j≤3 =

0 0 1

0 2 0

1 0 0

, (Tr(eiej))−11≤i,j≤3 =

0 0 1

0 12 0

1 0 0

Hence the dual basis is (x, h, y)(Tr(eiej))−1

1≤i,j≤3 = (y, 12h, x). Hence

cφ = xy + 12h

2 + yx =

0 1

0 0

0 0

1 0

+ 12

1 0

0 −1

2

+

0 0

1 0

0 1

0 0

=

1 0

0 0

+

12 0

0 12

+

0 0

0 1

=

32 0

0 32

.where 3

2 = dimF (L)dimF (V ) . Or use the fact that sl(2, F ) is irreducible with Lemma 6.5.

Definition 6.7Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimpleF -Lie algebra. Let φ : L → gl(V ) be a finite dimensional representation of L (not necessarilyfaithful). By Lemma 4.24, φ′ = φ|ker(φ)⊥ is faithful. We call cφ = cφ′ the Casimir element of φ.

Remark 6.8(1) Since φ(L) = φ′(ker(φ)⊥), by Lemma 6.3, cφ = a · 1 for some a ∈ F .


(2) If L is simple, then the only non-faithful irreducible representations are(2a) The 0-dimensional module 0 such that x · 0 = 0.(2b) The 1-dimensional module V = Fv such that x · v = 0 for all x ∈ L.(2c) All other representations are faithful. In fact, since L is simple, ker(φ) ∈ 0, L. Ifker(φ) = 0, then φ is faithful. Otherwise ker(φ) = L and hence x · v = 0 for all v ∈ V . Since Vis irreducible, dimF V ≤ 1, we are in case (2a)(2b).

Lemma 6.9Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimpleF -Lie algebra. Let φ : L→ gl(V ) be a finite dimensional representation of L. Then φ(L) ⊂ sl(V ).In particular, L acts trivially on all one-dimensional L-modules.

Proof. Suppose dimF V = n.


1. L = [L,L]. L is semisimple and Corollary 5.3

2. φ(L) = φ([L,L]) = [φ(L), φ(L)]. φ is a homomorphism

⊂ [gl(V ), gl(V )] φ(L) ⊂ gl(V )

⊂ sl(V ). Tr(AB −BA) = 0 for all AB ∈ gl(n, F )

If n = 1, then sl(V ) = 0. Thus x · v = 0 for all v ∈ V .

Lemma 6.10Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimpleF -Lie algebra. Let φ : L → gl(V ) be a finite dimensional representation of L. If V has anirreducible L-submodule W of codimension 1, then W ⊕ ker(cφ) = V .Proof.


1. cφ is an endomorphism of L-module V . Lemma 6.3

2. ker(cφ) is an L-submodule of V . Example 5.13(1)

3.1. cφ|W = dimF (L)dimF (W ) · 1 6= 0. W is irreducible and Lemma 6.5

3.2. ker(cφ) ∩W 6= W .

66 ZHENGYAO WU


3.3. ker(cφ) ∩W = 0. W is irreducible and Definition 5.16

4. dimF (ker(cφ)) ≤ dimF (V ) − dimF (W ) +dimF (ker(cφ) ∩W )

W ⊕ ker(cφ) ⊂ V .

= n− (n− 1) + 0 = 1 codimF W = 1 and step 3.3

5.1. φ(x)(V ) ⊂ W for all x ∈ L. dimF (V/W ) = 1 and Lemma 6.9

5.2. Tr(cφ) = Tr(cφ|W ) 6= 0. Tr(cφ|V/W ) = 0 and Lemma 6.5

5.3. dimF (ker(cφ)) ≥ 1. cφ =n∑i=1

φ(xi) φ(yi)

6. dimF (ker(cφ)) = 1. steps 4,5

7. ker(cφ) = W ′. steps 3,6

Lemma 6.11Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimpleF -Lie algebra. Let φ : L → gl(V ) be a finite dimensional representation of L. If V has a L-submodule W of codimension 1, then its complement X exists.

Proof. If W is irreducible, then X = ker(cφ) by Lemma 6.10. Suppose W is reducible.


1. W has a nonzero proper submodule W ′. W is reducible

2. 0→ W/W ′ → V/W ′ → F → 0 is exact. V/W ′

W/W ′ ' V/W ' F

3. dimF (V/W ′) < dimF (V ). W ′ 6= 0.

4. There exists W , W/W ′⊕W/W ′ = V/W ′. Inductive hypothesis

5. dimF (W ) = dimF (W ′) + 1 dimF (W/W ′) = dimF (V/W ′) −dimF (W/W ′) = dimF (V )− dimF (W ) = 1

≤ dimF (W ) W ′ ( W

< dimF (V ) dimF (W ) = dimF (V )− 1



6. There exists X, X ⊕W ′ = W . Inductive hypothesis

7. X ∩W ⊂ W ∩W X ⊂ W

⊂ X ∩W ′ step 4

= 0 step 6

8. dimF (X) = 1 steps 5,6

9. W ⊕X = V . steps 7,8 and dimF (W ) = dimF (V )− 1

Lemma 6.12Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimpleF -Lie algebra. Let φ : L → gl(V ) be a finite dimensional representation of L. Let W be anonzero proper L-submodule of V . Let V = f ∈ Hom(V,W ) : f |W = a · 1W ,∃a ∈ F. LetW = f ∈ V : f |W = 0.(1) L(V ) ⊂ W . W is an L-submodule of V .(2) dimF (W ) = dimF (V )− 1.

Proof. (1) For all x ∈ L and f ∈ V ,


1. If w ∈ W then x · w ∈ W . W is a submodule of V

2. (x · f)(w) = x · f(w)− f(x · w) Example 5.21(4)

= (x · aw)− a(x · w) = 0. Definition 5.12(M1)

3. x · f ∈ W . defn of W

(2) 0→ W → Ve−→ F → 0 is exact, where e sends f ∈ V to the eigenvalue of f |W .

Theorem 6.13 WeylLet F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimpleF -Lie algebra. Let φ : L→ gl(V ) be a finite dimensional representation of L. Then V is completelyreducible.

68 ZHENGYAO WU

Proof. Let W be a nonzero proper L-submodule of V . We need to find its complement. LetV = f ∈ Hom(V,W ) : f |W = a · 1W ,∃a ∈ F. Let W = f ∈ V : f |W = 0.


1. dimF (W ) = dimF (V )− 1. Lemma 6.12

2. W has a complement W ′ in V . Lemma 6.11

3. Let f : V → W be the generator of W ′

such that f |W = 1W .If f |W = a · 1W , a ∈ F ∗, then replace fwith a−1f .

4. ker(f) ∩W = 0. step 3

5. 0 = (x ·f)(v) = x ·f(v)−f(x ·v), ∀x ∈ L. step 1 and Lemma 6.9

6. ker(f) is a L-submodule of V . If v ∈ ker(f), then x · v ∈ ker(f).

7. im(f) = W . step 3.

8. dimF (W ) = dimF (V/ ker(f)) =dimF (V )− dimF (ker(f)).

V/ ker(f) ' im(f) by Proposition 1.29(1)

9. W ⊕ ker(f) = V . steps 4,8

Lemma 6.14Let F be an algebraically closed field of characteristic 0. Let L ⊂ gl(V ) be a finite dimensionalsemisimple F -Lie algebra. (1) L ( Ngl(V )(L); (2) If x ∈ L, then xs, xn ∈ Ngl(V )(L).

Proof. (1) L ( Ngl(V )(L) since L is a subalgebra of gl(V ). By Lemma 6.9, L ⊂ sl(V ). Hence1V ∈ N − L.(2) Let ad = adgl(V ).


1. Let x = xs + xn be the Jordan decompo-sition of x in gl(V ).

Proposition 3.14(a)

2. (adx)(L) ⊂ L. x ∈ L ( Ngl(V )(L)

3. (adx)s(L) ⊂ L, (adx)n(L) ⊂ L. Proposition 3.14(c)



4. ad(xs)(L) ⊂ L, ad(xn)(L) ⊂ L. Lemma 4.2

5. xs, xn ∈ Ngl(V )(L). Example 1.26(1)

Lemma 6.15Let F be an algebraically closed field of characteristic 0. Let V be a finite dimensional F -vectorspace. Let L ⊂ gl(V ) be a semisimple F -Lie algebra. Let W be any L-submodule of V .

LW = y ∈ gl(V ) : y(W ) ⊂ W, Tr(y|W ) = 0.

(1) LW is a subalgebra of gl(V ); (2) L ⊂ LW ; (3) If x ∈ L, then xs, xn ∈ LW ; (4) LV = sl(V ).

Proof. (1) For all y, z ∈ LW and w ∈ W , [y, z](w) = y(z(w))− z(y(w)) ∈ W and Tr([y, z]) = 0. So[y, z] ∈ LW and hence [LW , LW ] ⊂ LW .(2) For all y ∈ L,


1. y(W ) ⊂ W . W is an L-module

2. L = [L,L]. L is semisimple and Corollary 5.3

3. y = ∑[ai, bi] for some ai, bi ∈ L.

4. Tr(y|W ) = Tr(∑[ai, bi]|W ) =∑Tr([ai|W , bi|W ]) = 0.Tr is linear and commutative

5. L ⊂ LW .

(3)


1. Suppose x ∈ L. Then x(W ) ⊂ W andTr(x|W ) = 0.

(2)

2. xs(W ) ⊂ W , xn(W ) ⊂ W . Proposition 3.14(c)

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3. Tr(xn|W ) = 0. xn|W is nilpotent since xn is nilpotent

4. Tr(xs|W ) = Tr(x|W − xn|W ) = Tr(x|W ) −Tr(xn|W ) = 0.

Tr is linear and steps 1,3

5. xs, xn ∈ LW . defn of LW

(4) follows from definitions of LW and sl(V ).

Theorem 6.16Let F be an algebraically closed field of characteristic 0. Let V be a finite dimensional F -vectorspace. Let L ⊂ gl(V ) be a semisimple F -Lie algebra. If x ∈ L, then xs, xn ∈ L.

Proof. Let L′ = Ngl(V )(L) ∩ ⋂W⊂V

LW . Suppose x ∈ L.


1. Let x = xs + xn be the Jordan decompo-sition of x in gl(V ).

Proposition 3.14(a)

2.1. xs, xn ∈ Ngl(V )(L). Lemma 6.14(2)

2.2. xs, xn ∈ LW . Lemma 6.15(3)

3. xs, xn ∈ L′. defn of L′

4. x ∈ L′ and hence L ⊂ L′. step 1

5. L′ is a subalgebra of Ngl(V )(L). Example 1.26(1) and Lemma 6.15(1)

6. L′ = L⊕M for a L-submodule M of L′. L is semisimple and Weyl’s Theorem 6.13

7. L acts on M trivially. [L,L′] ⊂ [L,N ] ⊂ L

8. If y ∈M then [L, y] = 0.

9. For all irreducible submodule W ⊂ V ,y|W = a · 1W for some a ∈ F .

Schur’s Lemma 5.20

10. a dimF (W ) = Tr(y|W ) = 0. y ∈ L′ ⊂ LW

11. y|W = 0. a = 0



12. y = 0. V is a direct sum of irreducible L-submodules by Weyl’s Theorem 6.13

13. M = 0 and hence xn, xs ∈ L′ = L. step 3

Theorem 6.17Let F be an algebraically closed field of characteristic 0. Let V be a finite dimensional F -vectorspace. Let L ⊂ gl(V ) be a semisimple F -Lie algebra. The abstract and usual Jordan decompositionof L coincide.Proof.


1. Let x = xs + xn be the usual Jordan de-composition of x in gl(V ).

Proposition 3.14(a)

2. xs, xn ∈ L. Theorem 6.16

3. adx = ad(xs)+ad(xn) is the usual Jordandecomposition of adx in gl(L).

Lemma 4.2

4. x = xs+xn is the abstract Jordan decom-position of x in L.

Proposition 5.10

Corollary 6.18Let F be an algebraically closed field of characteristic 0. Let L be a semisimple F -Lie algebra.Let V be a finite dimensional F -vector space. Let φ : L → gl(V ) be a representation of L. Ifx = xs + xn is the abstract Jordan decomposition of x ∈ L, then φ(x) = φ(xs) + φ(xn) is the(abstract and usual) Jordan decomposition of φ(x), i.e. φ(x)s = φ(xs), φ(x)n = φ(xn).Proof.

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1. adL(xs) is semisimple. Proposition 5.10

2. adL(xs) is diagonalizable. Lemma 3.9

3. L is spanned by eigenvectors of adL(xs).

4. φ(L) is spanned by eigenvectors ofadφ(L)(φ(xs)).

φ is linear

5. adφ(L)(φ(xs)) is diagonalizable.

6. adφ(L)(φ(xs)) is semisimple. Lemma 3.9

7. adL(xn) is nilpotent. Proposition 5.10

8. adφ(L)(φ(xn)) is nilpotent. φ is a homomorphism

9. [adφ(L)(φ(xs)), adφ(L)(φ(xn))] =adφ(L)([φ(xs), φ(xn)]) =adφ(L)(φ([xs, xn])) = 0.

adφ(L) and φ are homomorphisms and[xs, xn] = 0.

10. φ(x) = φ(xs) + φ(xn) is the abstract Jor-dan decomposition.

steps 5,8,9 and Proposition 5.10

11. φ(x) = φ(xs) + φ(xn) is the usual Jordandecomposition.

Theorem 6.17


7. April 9th, Representation of sl(2, F ), toral subalgebras

Review 7.1In sl(2, F ),

x =

0 1

0 0

, y =

0 0

1 0

, h =

1 0

0 −1

, [h, x] = 2x, [h, y] = −2y, [x, y] = h.

Definition 7.2Let F be an algebraically closed field of characteristic 0. Let V be a finite dimensional sl(2, F )-module. Let Vλ = v ∈ V : h · v = λv, λ ∈ F . If Vλ 6= 0, then λ is called a weight of h in V andVλ is called a weight space.

Lemma 7.3Let F be an algebraically closed field of characteristic 0. Let φ : sl(2, F ) → gl(V ) be a finitedimensional representation. Then V = ∐

Vλ.

Proof. Since F is algebraically closed and h is semisimple, by Lemma 3.9, φ(h) is diagonalizable.Therefore its eigenvectors span V .

Lemma 7.4Let F be an algebraically closed field of characteristic 0. Let V be a finite dimensional sl(2, F )-module. If v ∈ Vλ, then x · v ∈ Vλ+2 and y · v ∈ Vλ−2.

Proof. h · (x · v) = [h, x] · v + x · (h · v) = 2x · v + x · (λv) = (λ+ 2)x · v.h · (y · v) = [h, y] · v + y · (h · v) = −2y · v + y · (λv) = (λ− 2)y · v.

Definition 7.5Let F be an algebraically closed field of characteristic 0. Let V be a finite dimensional sl(2, F )-module. There exists Vλ 6= 0 and Vλ+2 = 0. We call λ the highest weight and 0 6= v0 ∈ Vλ amaximal vector.

Lemma 7.6Let F be an algebraically closed field of characteristic 0. Let V be a finite dimensional irreduciblesl(2, F )-module. Let v0 ∈ Vλ be a maximal vector; Define v−1 = 0, vi = 1

i!yi · v0. Then (1)

h · vi = (λ− 2i)vi; (2) y · vi = (i+ 1)vi+1; (3) x · vi = (λ− i+ 1)vi−1.

Proof. (2) y · vi = y · 1i!y

i · v0 = (i+ 1) 1(i+ 1)!y

i+1 · v0 = (i+ 1)vi+1.

(1) Since v0 ∈ Vλ, h · v0 = λv0. Suppose h · vi = (λ− 2i)vi. Then

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h · vi+1 = h · ( 1i+ 1y · vi), by (2)

= 1i+ 1(h · (y · vi)), by Definition 5.12(M1)

= λ− 2i− 2i+ 1 y · vi by Lemma 7.4

= (λ− 2(i+ 1))vi+1 by (2)

(3) x · v0 ∈ Vλ+2 = 0. Suppose x · vi−1 = (λ− i+ 2)vi−2. Then

ix · vi

= x · (y · vi−1), by (2)

= [x, y] · vi−1 + y · (x · vi−1), defn of [, ]

= h · vi−1 + y · ((λ− i+ 2)vi−2), x · vi−1 = (λ− i+ 2)vi−2

= (λ− 2(i− 1))vi−1 + (λ− i+ 2)y · vi−2, by (1)

= (λ− 2i+ 2)vi−1 + (λ− i+ 2)(i− 1)vi−1, by (2)

= i(λ− i+ 1)vi−1.

Hence x · vi = (λ− i+ 1)vi−1.

Theorem 7.7Let F be an algebraically closed field of characteristic 0. Let V be a finite dimensional irreduciblesl(2, F )-module. Then

V '∐µ

Vµ, weights µ = m,m− 2, . . . ,−(m− 2),−m

where dimF (V ) = m+ 1 and dimF (Vµ) = 1 for all µ.

Proof. Let λ be the highest weight with weight space Vλ.


1. (vi)0≤i≤n is linearly independent for all n.

1.1. Supposen∑i=0

aivi = 0 for ai ∈ F .

1.2. 0 = h · (n∑i=0

aivi) =n∑i=0

ai(h · vi) Definition 5.12(M2)

=n∑i=0

(λ− 2i)aivi Lemma 7.6(1)

=n∑i=0

(λ− 2i)aivi − (λ− 2n)n∑i=0

aivi step 1.1



= 2n−1∑i=0

(n− i)aivi

1.3. a0 = · · · = an−1 = 0. Inductive hypothesis: (v0, . . . , vn−1) arelinearly independent.

1.4. an = 0. step 1.1

2. There exists an integer m such that vm 6=0 and vm+1 = 0.

dimF (V ) <∞

3. Spanv0, . . . , vm is a nonzero L-submodule of V with vi ∈ Vλ−2i.

Lemma 7.6

4. V = Spanv0, . . . , vm. V is irreducible

5. Vλ−2i = Fvi for all i. V ⊂m∐i=0

Vλ−2i = V,

Remark 7.8Let V be an irreducible sl(2, F )-module of dimension m+ 1. We write V = V (m).(1) The highest weight λ = m.In fact, since vm+1 = 0, 0 = x ·vm+1 = (λ−m)vm by Lemma 7.4. Since vm 6= 0, we have λ−m = 0.(2) The maximal vector v0 ∈ Vm is unique up to scalar multiples, by step 5 of Theorem 7.7.(3) V has basis v0, . . . , vm by steps 1,4 of Theorem 7.7.(4) Up to isomorphism, there exists at most one irreducible sl(2, F )-module of each dimension ≥ 1.In fact, suppose V and W are two irreducible sl(2, F )-modules of dimension m+ 1 with maximalvector v0 and w0, respectively. Define φ(v0) = w0 and let φ be a homomorphism. Then y · φ(v) =φ(y · v) for all v ∈ V . Hence φ(vi) = wi, φ is an isomorphism.

Corollary 7.9Let F be an algebraically closed field of characteristic 0. Let V be an sl(2, F )-module. Theneigenvalues of h on V are all integers, and n is an eigenvalue of h iff −n is so.

Proof. Suppose V 6= 0.

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1. V 't∐i=1

V (mi), where V (mi) are irre-ducible submodules.

Theorem 6.13 and Remark 7.8

2. Eigenvalues of V aret⋃i=1mi,mi −

2, . . . ,−miTheorem 7.7

3. mi − 2j is an eigenvalue of h iff −(mi −2j) = mi − 2(mi − j) is so.

Corollary 7.10Let F be an algebraically closed field of characteristic 0. Let V be an sl(2, F )-module. Inany decomposition of V into direct sum of irreducible submodules, the number of summandsis dimF (V0) + dimF (V1).

Proof. Suppose V 6= 0.


1. V 't∐i=1

V (mi), where V (mi) are irreducible submod-ules.

Theorem 6.13 and Re-mark 7.8

2. t = ∑mi even

1 + ∑mi odd

1

= ∑mi even

dimF ((V (mi))0) + ∑mi odd

dimF ((V (mi))1). Theorem 7.7

= dimF (V0) + dimF (V1).

Example 7.11The trivial representation of sl(2, F ) of dimension one is isomorphic to V (0).The natural representation of sl(2, F ) of dimension two is isomorphic to V (1).The adjoint representation of sl(2, F ) of dimension three is isomorphic to V (2).


Lemma 7.12Let F be an algebraically closed field of characteristic 0. Let φ : sl(2, F ) → gl(V (m)) be a rep-resentation. Let τ = exp(φ(x)) exp(φ(−y)) exp(φ(x)) be an automorphism of V (m), m ≥ 1.Then τ(Vm−2i) = V−(m−2i) for all 0 ≤ i ≤ m.Proof.


1. φ is faithful and hence sl(2, F ) 'φ(sl(2, F )).

sl(2, F ) is semisimple and Remark 6.8(2)

2. int(exp(φ(x))) = exp(ad(φ(x))) Lemma 2.1

3. int(τ) = exp(ad(φ(x)))exp(ad(φ(−y)))exp(ad(φ(x)))

defn of τ

4. int(τ)(φ(h)) = φ(exp(ad(x)) exp(ad(−y)) exp(ad(x))(h))

φ is a homomorphism

= φ(shs−1) = φ(−h) = −φ(h) s =

0 1

−1 0

as in Example 2.2

5. h · τ(vi) = −τ(h · vi) = −(m− 2i)τ(vi). vi ∈ Vm−2i

6. τ(Vm−2i) = V−(m−2i). τ(vi) ∈ V−(m−2i)

Definition 7.13Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzerosemisimple F -Lie algebra. Let T be a nonzero subalgebra of L. We say that T is toral if it consistsof ad-semisimple elements.

Lemma 7.14Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzerosemisimple F -Lie algebra. If T is a toral subalgebra of L, then T is abelian.Proof.


1. adT (x) is semisimple for all x ∈ T . Definition 7.13

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2. adT (x) is diagonalizable for all x ∈ T . F is algebraically closed and Lemma 3.9

3. All eigenvalues of adT (x) are 0. below

3.1. Suppose that adT (x)(y) = [x, y] = ay forsome a ∈ F ∗, 0 6= y ∈ T .

3.2. Linearly independent eigenvectors zi ofadT (y) span T where adT (y)(zi) =[y, zi] = bizi for some bi ∈ F .

step 2

3.3. −ay = [y, x] = a1b1z1 + · · ·+anbnzn wherex = a1z1 + · · ·+ anzn, ai ∈ F .

steps 3.1, 3.2, and Remark 1.2(L2’)

3.4. 0 = [y,−ay] = a1b21z1 + · · ·+ anb

2nzn. steps 3.1, 3.2, and Definition 1.1(L2)

3.5. aib2i = 0 for all 1 ≤ i ≤ n. step 3.2

3.6. aibi = 0 for all 1 ≤ i ≤ n. (aibi)2 = ai(aib2i ) = ai · 0 = 0

3.7. ay = a1b1z1 + · · · + anbnzn = 0, a contra-diction.

to step 3.1

4. adT (x) = 0 for all x ∈ T and hence[T, T ] = 0.

step 4

Lemma 7.15Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzerosemisimple F -Lie algebra. If L is not nilpotent, then L contains a toral subalgebra.

Proof. Let S = x ∈ L : x = xs + xn, xs 6= 0.


1. If S = ∅, then x = xn for all x ∈ L. defn of S

2. x is ad-nilpotent for all x ∈ L. Proposition 5.10

3. L is nilpotent, a contradiction. Engel’s theorem Theorem 2.31



4. Spanxs : x ∈ S is a toral subalgebra ofL.

S 6= ∅, Lemma 7.14 and Lemma 3.11.

Definition 7.16Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzerosemisimple F -Lie algebra. Let H be a toral subalgebra of L. We call H a maximal if for all toralsubalgbra H ′ of L such that H ⊂ H ′, we have H = H ′.

Example 7.17Let F be an algebraically closed field of characteristic 0. Let L = sl(n, F ). The set H of diagonaln × n matrices of trace 0 is a maximal toral subalgebra of L. We only verify that H = Fh isa maximal toral subalgebra of sl(2, F ). If H ′ is another toral subalgebra of sl(2, F ) such thatH ⊂ H ′, then [H ′, H ′] = 0 by Lemma 7.14. Then [H ′, H] = 0 ⊂ H and hence H ′ ⊂ Nsl(2,F )(H).Since [h, x] = 2x 6∈ H and [h, y] = −2y 6∈ H, Nsl(2,F )(H) = H. Therefore H ′ = H.

Definition 7.18Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzerosemisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. Let α ∈ H∗ and Lα = x ∈L : [h, x] = α(h)x, ∀h ∈ H. If α 6= 0 and Lα 6= 0, then we call α a root and Lα a root space.We write Φ = α ∈ H∗ : α 6= 0, Lα 6= 0.

Lemma 7.19Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzerosemisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. Then L0 = CL(H).

Proof. By Definition 7.18, L0 = x ∈ L : [h, x] = 0,∀h ∈ H = CL(H).

Lemma 7.20 Cartan DecompositionLet F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzerosemisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. Then L = L0 ⊕

⊔α∈Φ

Lα.Proof.


1. H is abelian. H is toral and Lemma 7.14

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2. Elements of adL(H) are simultaneously di-agonalizable.

Lemma 3.11

3. For all h ∈ H, (simultaneous) eigenspacesof h are of the form Lα and they span L.

Lemma 7.21Φ is finite, since dimF (L) <∞ and Lemma 7.20.

Example 7.22Let F be an algebraically closed field of characteristic 0. The Cartan decomposition of sl(2, F ).The maximal toral subalgebra is Fh. If 0 6= ax+by+ch ∈ Lα, then [h, ax+by+ch] = 2ax−2by =α(h)(ax+ by + ch). So a(α(h)− 2) = b(α(h) + 2) = cα(h) = 0.

• If a = b = 0, then c 6= 0 and α(h) = 0, we have L0 = Fh.• If a 6= 0, then α(h) = 2 and b = c = 0, we have L2 = Fx.• If b 6= 0, then α(h) = −2 and a = c = 0, we have L−2 = Fy.

Therefore sl(2, F ) = F0 ⊕ F2 ⊕ F−2 = Fh⊕ Fx⊕ Fy is the Cartan decomposition.

Proposition 7.23Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzerosemisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. [Lα, Lβ] ⊂ Lα+β for allα, β ∈ H∗.

Proof. Suppose x ∈ Lα and y ∈ Lβ.


1.1. [h, x] = α(h)x for all h ∈ H. x ∈ Lα

1.2. [h, y] = β(h)y for all h ∈ H. y ∈ Lβ

2. [h, [x, y]] = −[x, [y, h]]− [y, [h, x]] Definition 1.1(L3)

= [x, [h, y]] + [[h, x], y] Remark 1.2(L2’)

= [x, β(h)y] + [α(h)x, y] step 1

= (β(h) + α(h))[x, y] = (α + β)(h)[x, y]. Definition 1.1(L1)



3. [x, y] ∈ Lα+β. Definition 7.18

Proposition 7.24Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzerosemisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. If x ∈ Lα, 0 6= α ∈ H∗,then adx is nilpotent.Proof.


1. If y ∈ Lβ, then (adx)n(y) ∈ Lnα+β. Proposition 7.23

2. n ∈ N : Lnα+β 6= 0 is finite. Lemma 7.21

3. (adx)|Lβ is nilpotent for all β ∈ Φ.

4. adx is nilpotent. Lemma 7.20

Proposition 7.25Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzerosemisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. If α, β ∈ H∗ and α+β 6= 0,then Lα is orthogonal to Lβ relative to the Killing form κ of L.

Proof. For all x ∈ Lα and y ∈ Lβ,


1. There exists h ∈ H s.t. α(h) + β(h) 6= 0. α + β 6= 0

2. κ([x, h], y) = κ(x, [h, y]) Lemma 4.11

3. κ(−α(h)x, y) = κ(x, β(h)y) x ∈ Lα, y ∈ Lβ and Definition 1.1(L2)

4. −α(h)κ(x, y) = β(h)κ(x, y) κ is bilinear

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5. κ(x, y) = 0. step 1


8. April 16th, Centralizer of H; Orthogonal, integral properties

Corollary 8.1Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzerosemisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. κ|L0 is non-degenerate.

Proof. Suppose z ∈ L0.


1. κ(z, Lα) = 0 for all α 6= 0. Proposition 7.25

2. If z ∈ Rad(κ|L0), then κ(z, L0) = 0. Definition 4.14(1)

3. κ(z, L) = 0. L = L0 ⊕⊔α∈Φ

Lα by Lemma 7.20


5. z = 0 and κ|L0 is non-degenerate. Definition 4.14(2)

Lemma 8.2Let F be a field. Let V be a finite dimensional F -vector space. Suppose x, y ∈ End(V ) such thatx y = y x and y is nilpotent. Then x y is nilpotent and Tr(x y) = 0.

Proof. Suppose yn = 0. Since x y = y x, (x y)n = xn yn = 0.Since xy is nilpotent, its eigenvalues are all 0 and hence their sum Tr(xy) = 0.

Lemma 8.3Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzerosemisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. If x ∈ CL(H) and x isad-semisimple, then x ∈ H.Proof.


1. Every element ofH+Fx is ad-semisimple. x ∈ CL(H) and Lemma 3.11

2. H + Fx is a toral subalgebra of L. Definition 7.13

3. H + Fx = H and hence x ∈ H. H is maximal and Definition 7.16

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Lemma 8.4Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzerosemisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. Then κ|H is non-degenerate.

Proof. Suppose h ∈ H such that κ(h,H) = 0.


1. Let x = xs + xn be the abstract Jordandecomposition of x. Then adx = ad(xs)+ad(xn) is the usual Jordan decompositionof adx.

Proposition 5.10

2. For all x ∈ CL(H), xn ∈ CL(H). Proposition 3.14(c)

3. [adxn, adh] = ad([xn, h]) = 0. ad is a homomorphism

4. κ(h, xn) = Tr(ad(h)(adxn)) = 0. (adxn) is nilpotent and Lemma 8.2

5. xs ∈ CL(H). Proposition 3.14(c)

6. xs ∈ H. Lemma 8.3

7. κ(h, xs) = 0. κ(h,H) = 0.

8. κ(h, x) = κ(h, xs) + κ(h, xn) = 0 for allx ∈ CL(H).

steps 3,6

9. h = 0. Corollary 8.1

10. κ|H is non-degenerate. Definition 4.14(2)

Lemma 8.5Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzerosemisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. Then CL(H) is nilpotent.Proof.




Proposition 5.10

2. If x ∈ CL(H), then xs ∈ CL(H). Proposition 3.14(c)

3. xs ∈ H and hence adCL(H)(xs) = 0. Lemma 8.3

4. adCL(H)(xn) is nilpotent. xn is ad-nilpotent by Proposition 5.10

5. adCL(H)(x) = adCL(H)(xs)+adCL(H)(xn) =adCL(H)(xn) is nilpotent for all x ∈ CL(H).

steps 3,4

6. CL(H) is nilpotent. Engel’s theorem Theorem 2.31

Lemma 8.6Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzerosemisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. Then H∩ [CL(H), CL(H)] =0.Proof.


1. κ(H,H ∩ [CL(H), CL(H)]) ⊂κ(H, [CL(H), CL(H)])

= κ([H,CL(H)], CL(H)) κ is associative by Lemma 4.11

= κ(0, CL(H)) = 0. defn of CL(H)

2. H ∩ [CL(H), CL(H)] = 0. κ|H is non-degenerate by Lemma 8.4

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Lemma 8.7Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzerosemisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. Then CL(H) is abelian.

Proof. Suppose the opposite [CL(H), CL(H)] 6= 0.


1. CL(H) is nilpotent. Lemma 8.5

2. Z(CL(H)) ∩ [CL(H), CL(H)] 6= 0. Lemma 3.1

3. Take 0 6= x ∈ Z(CL(H)) ∩[CL(H), CL(H)].

4.1. If x is ad-semisimple, then x ∈ H. x ∈ CL(H) and Lemma 8.3,

4.2. x = 0, a contradiaction. x ∈ H ∩ [CL(H), CL(H)] = 0 byLemma 8.6

4.3. x is not ad-semisimple.


Proposition 5.10

6. xn 6= 0. step 4.3, 5

7. xn ∈ CL(H). x ∈ CL(H) and Proposition 3.14(c)

8. xn ∈ Z(CL(H)), i.e. [xn, y] = 0 for ally ∈ CL(H).

x ∈ Z(CL(H)) and Proposition 3.14(c)

9. [adxn, ad y] = 0 for all y ∈ CL(H). ad is a homomorphism.

10. κ(xn, y) = 0 for all y ∈ CL(H). adxn is nilpotent and Lemma 8.2

11. xn = 0, a contradiction to step 6. κ|CL(H) is non-degenerate by Corollary 8.1and Lemma 7.19


Proposition 8.8Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzerosemisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. Then CL(H) = H.

Proof. Suppose the opposite H ( CL(H).


1. There exists x ∈ CL(H)−H and x is notad-semisimple.

Lemma 8.3


Proposition 5.10

3. xn ∈ CL(H). x ∈ CL(H) and Proposition 3.14(c)

4. [xn, y] = 0 for all y ∈ CL(H). CL(H) is abelian by Lemma 8.7

5. [adxn, ad y] = 0 for all y ∈ CL(H). ad is a homomorphism.

6. κ(xn, y) = 0 for all y ∈ CL(H). adxn is nilpotent and Lemma 8.2

7. xn = 0. κ|CL(H) is non-degenerate by Corollary 8.1

8. x = xs is ad-semisimple, a contradiction. to step 1

Corollary 8.9Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzerosemisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. There exists a bijectiont : H∗ → H, φ 7→ tφ such that φ(h) = κ(tφ, h) for all h ∈ H.

Proof. The map t well-defined and injective since κ|H is non-degenerate by Lemma 8.4. It issurjective by definition.

Proposition 8.10Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzerosemisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. Then Let Φ be the set ofroots of L relative to H. Φ spans H∗.

Proof. Suppose Span Φ ( H∗.

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1. t(Span Φ) ( H. Corollary 8.9

2. t(Span Φ)⊥ 6= 0.

3. Take 0 6= h ∈ t(Span Φ)⊥, then α(h) =κ(tα, h) = 0 for all α ∈ Φ.

Definition 4.22(2) and the defn of t inCorollary 8.9

4. [h, Lα] = 0 for all α ∈ Φ. Definition 7.18

5. [h, L0] = [h,H]. L0 = H by Proposition 8.8

= 0. H is abelian by Lemma 7.14

6. [h, L] = 0. steps 4,5 and the Cartan decompositionLemma 7.20

7. h ∈ Z(L) ⊂ Rad(L) = 0. L is semisimple and Definition 2.11

8. h = 0, a contradiction. to step 3

Proposition 8.11Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzerosemisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. Let Φ be the set of roots ofL relative to H. If α ∈ Φ, then −α ∈ Φ.

Proof. Suppose the opposite −α 6∈ Φ.


1. α + β 6= 0 for all β ∈ Φ. −α 6∈ Φ.

2. κ(Lα, Lβ) = 0 for all β ∈ Φ. Proposition 7.25

3. κ(Lα, L) = 0. Proposition 8.10 and Lemma 7.20


5. Lα = 0, a contradiction. to α ∈ Φ


Proposition 8.12Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzerosemisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. Let Φ be the set of roots ofL relative to H. Let α ∈ Φ, x ∈ Lα and y ∈ L−α. Then [x, y] = κ(x, y)tα.Proof.


1. For all h ∈ H, κ(h, [x, y]) = κ([h, x], y) Lemma 4.11

= α(h)κ(x, y) x ∈ Lα

= κ(tα, h)κ(x, y) the defn of t in Corollary 8.9

= κ(h, κ(x, y)tα) κ is symmetric and bilinear

2. [x, y] = κ(x, y)tα κ|H is non-degenerate by Lemma 8.4

Proposition 8.13Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzerosemisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. Let Φ be the set of roots ofL relative to H. If α ∈ Φ, then [Lα, L−α] = Ftα.Proof.


1. [Lα, L−α] ⊂ Ftα. Proposition 8.12

2.1. Suppose κ(Lα, L−α) = 0. Thenκ(Lα, Lβ) = 0 for all β ∈ Φ.

Proposition 7.25 for β 6= −α

2.2. κ(Lα, L) = 0. the Cartan decomposition Lemma 7.20

2.3. κ is non-degenerate. L is semisimple and Theorem 4.21

2.4. Lα = 0, a contradiction. to α ∈ Φ

3. κ(Lα, L−α) 6= 0, i.e. there exists 0 6= x ∈L−α and 0 6= y ∈ L−α s.t. κ(x, y) 6= 0.

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4. [Lα, L−α] 6= 0. Proposition 8.12

5. [Lα, L−α] = Ftα. steps 1, 4

Proposition 8.14Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzerosemisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. Let Φ be the set of roots ofL relative to H. Then α(tα) = κ(tα, tα) 6= 0 for all α ∈ Φ.

Proof. Suppose the opposite α(tα) = 0.


1. There exists 0 6= x ∈ Lα and 0 6= y ∈ L−αsuch that κ(x, y) 6= 0.

Proposition 8.13

2. Let x′ = x

κ(x, y) . Then [x′, y] = tα. Proposition 8.12

3. Consider the F -Lie algebra S =Spanx′, y, tα.

3.1. [tα, x′] = α(tα)x′ = 0. x, x′ ∈ Lα

3.2. [tα, y] = −α(tα)y = 0. y ∈ L−α

4. S is solvable. dimF [S, S] = 1 and Example 1.18

5. adL(S) is solvable. Proposition 2.7

6. [adL(S), adL(S)] is nilpotent. Corollary 3.7

7. adL(tα) is nilpotent. tα ∈ [S, S] by step 2

8. adL(tα) is semisimple. H is toral and tα ∈ H

9. adL(tα) = 0. by steps 7,8, it is diagonalizable whoseeigenvalues are all 0

10. tα ∈ Z(L) ⊂ Rad(L) = 0. L is semisimple

11. tα = 0, a contradiction. to step 12



12. tα 6= 0. α 6= 0 (since α ∈ Φ) and Corollary 8.9

Proposition 8.15Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzerosemisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. Let Φ be the set of roots ofL relative to H. For all α ∈ Φ and 0 6= xα ∈ Lα, there exists yα ∈ L−α such that

Sα = Spanxα, yα, hα = [xα, yα] ' sl(2, F )

xα 7→ x =

0 1

0 0

, yα 7→ y =

0 0

1 0

, hα 7→ h =

1 0

0 −1

Proof.


1. There exists y ∈ L−α s.t. κ(xα, y) 6= 0. Proposition 8.13

2. κ(tα, tα) 6= 0. Proposition 8.14

3. Let yα = 2κ(tα, tα)κ(xα, y)y ∈ L−α. Then

κ(xα, yα) = 2κ(tα, tα) .

κ is bilinear

4. Let hα = 2tακ(tα, tα) . Then [xα, yα] =

κ(xα, yα)tα = 2tακ(tα, tα) = hα.

Proposition 8.12

5. [hα, xα] = [ 2tακ(tα, tα) , xα] = 2[tα, xα]

κ(tα, tα) κ is bilinear

= 2α(tα)xακ(tα, tα) xα ∈ Lα

= 2xα α(tα) = κ(tα, tα) by Corollary 8.9

6. [hα, yα] = [ 2tακ(tα, tα) , yα] = 2[tα, yα]

κ(tα, tα) κ is bilinear

= −2α(tα)yακ(tα, tα) xα ∈ Lα

= −2yα α(tα) = κ(tα, tα) by Corollary 8.9

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7. S ' sl(2, F ). steps 4,5,6 and [x, y] = h, [h, x] =2x, [h, y] = −2y in sl(2, F )

Proposition 8.16Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimpleF -Lie algebra over F . Let H be a maximal toral subalgebra of L. Let Φ be the set of roots ofL relative to H. Let t : H∗ → H be the map α 7→ tα such that α(h) = κ(tα, h) for all h ∈ H.Suppose α ∈ Φ and Sα ' sl(2, F ) with (xα, yα, hα) 7→ (x, y, h). Then (1) hα = 2tα

κ(tα, tα) and (2)

α(hα) = 2.

Proof. We provide a new proof besides step 4 of Proposition 8.15.


1. hα = [xα, yα] S ' sl(2, F )

= κ(xα, yα)tα. Proposition 8.12

2. 2xα = [hα, xα] S ' sl(2, F )

= [κ(xα, yα)tα, xα] = κ(xα, yα)[tα, xα] κ is bilinear

= κ(xα, yα)α(tα)xα, xα ∈ Lα

3. κ(xα, yα) = 2α(tα) = 2

κ(tα, tα) . defn of t in Corollary 8.9

4. hα = 2tαα(tα) = 2tα

κ(tα, tα) . steps 1,3

5. α(hα) = 2α(tα)α(tα) = 2. α is linear and steps 4

Proposition 8.17Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimpleF -Lie algebra over F . Let H be a maximal toral subalgebra of L. Let Φ be the set of roots of L


relative to H. Let t : H∗ → H be the map α 7→ tα such that α(h) = κ(tα, h) for all h ∈ H. Supposeα ∈ Φ and Sα ' sl(2, F ) with (xα, yα, hα) 7→ (x, y, h). Then (1) t−α = −tα and (2) h−α = −hα.Proof.


1. (−α)(h) = −α(h) defn of −α

2. κ(t−α, h) = −κ(tα, h) = κ(−tα, h) defn of t in Corollary 8.9 and κ is bilinear

3. t−α = −tα. κ|H is non-degenerate by Lemma 8.4

4. h−α = 2t−ακ(t−α, t−α) Proposition 8.16(1)

= −2tακ(−tα,−tα) = − 2tα

κ(tα, tα) step 3 and κ is bilinear

= −hα Proposition 8.16(1)

Proposition 8.18Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimpleF -Lie algebra over F . Let H be a maximal toral subalgebra of L. Let Φ be the set of roots ofL relative to H. Let t : H∗ → H be the map α 7→ tα such that α(h) = κ(tα, h) for all h ∈ H.Suppose α ∈ Φ and Sα ' sl(2, F ) with (xα, yα, hα) 7→ (x, y, h). The only scalar multiples of α inΦ are ±α.

Proof. Let M = L0 ⊕⊔

c∈F ∗Lcα, where L0 = H by Lemma 7.19.


1. M is an Sα-module by ad : Sα → gl(M). Proposition 7.23

2. [ker(α), Sα] = 0. For all z ∈ ker(α) Sα = Spanxα, yα, hα and below

2.1 [z, xα] = α(z)xα = 0. xα ∈ Lα

2.2 [z, yα] = −α(z)yα = 0. yα ∈ L−α

2.3 [z, hα] = 0. H is abelian by Lemma 7.14

3. H = ker(α)⊕ Fhα. below

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3.1. 0 6= H/ ker(α) ' F α : H → F is nonzero and Proposi-tion 1.29(1)

3.2. dimF (ker(α)) = dimF (H)− 1. 1 ≤ dimF (H/ ker(α)) ≤ dimF (F ) = 1.

3.3. ker(α) ∩ Fhα = 0. α(hα) = 2 by Proposition 8.16(2)

4. Weights of hα on M are 0 and 2c, If Lcα 6= 0, then cα(hα) = 2c by Proposi-tion 8.16(2)

where c = n

2 for n ∈ Z. 2c ∈ Z by Corollary 7.9

5. M is an Sα-module. Proposition 7.23

6. M is a direct sum of irreducible Sα-submodules.

Sα ' sl(2, F ) is semisimple and Weyl’stheorem Theorem 6.13

M ' ker(α)⊕ Sα ⊕W , step 3

where ker(α) is the direct sum of one-dimensional trivial sl(2, F )-modules.

step 2

and W is the direct sum of irreduciblesl(2, F )-modules.

7. Every weight appears at most once in M . defn of M

8. Weights of irreducible submodules of Ware not even.

Weight 0 already appears in Sα and The-orem 7.7

9. For α ∈ Φ, ±2α 6∈ Φ. steps 4,8

10. Weights of irreducible submodules of Ware not odd.

below

10.1 Otherwise weight 2c = 1 appears in W . Theorem 7.7

10.2 α

2 ∈ Φ. α2 (hα) = 1 by Proposition 8.16(2)

10.3 A contradiction. to step 9 and α ∈ Φ

11. W = 0. steps 8,10

12. M ' ker(α) ⊕ Sα = L0 ⊕ Lα ⊕ L−α. Theonly scalar multiples of α in Φ are ±α.

steps 6,11


Proposition 8.19Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimpleF -Lie algebra over F . Let H be a maximal toral subalgebra of L. Let Φ be the set of rootsof L relative to H. If α ∈ Φ, then dimF (Lα) = 1. In particular, Sα = Hα ⊕ Lα ⊕ L−α, whereHα = [Lα, L−α].

Proof. Suppose α ∈ Φ.


1. dimF (Lα) = dimF (L−α) = 1 Theorem 7.7

2. Sα = Lα ⊕ L−α ⊕ Hα, where Lα = Fxα,L−α = Fyα and Hα = Fhα.

Sα ' sl(2, F ) by Proposition 8.15, andExample 7.22

Proposition 8.20Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimpleF -Lie algebra over F . Let H be a maximal toral subalgebra of L. Let Φ be the set of roots of Lrelative to H. Let α, β ∈ Φ, β 6= ±α. Let r be the largest integer such that β − rα ∈ Φ. Let qbe the largest integer such that β + qα ∈ Φ. Then (1) β + iα ∈ Φ for all −r ≤ i ≤ q, and (2)β(hα) = r − q.

Proof. Let V = ∐i∈Z

Lβ+iα.


1. β + iα ∈ Φ iff its weight is (β + iα)(hα) =β(hα) + 2i.

Proposition 8.16(2)

2. Only one of 0 and 1 can be weight.

3. If β + iα ∈ Φ, then dimF (Lβ+iα) = 1. Proposition 8.19

4. V has dimF (V0) + dimF (V1) = 1 irre-ducible component.

step 2,3 and Corollary 7.10

5. V is irreducible, we write V ' V (m). Remark 7.8

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6. Weights of V are m,m− 2, . . . ,−m. Theorem 7.7

6.1. m = β(hα) + 2q. step 1 and q be the largest integer suchthat β + qα ∈ Φ.

6.2. −m = β(hα)− 2r. step 1 and r is the largest integer such thatβ − rα ∈ Φ.

7. β(hα) = r − q and m = r + q. steps 6.1, 6.2

8. β + iα ∈ Φ for all −r ≤ i ≤ q. Its weight β(hα) + 2i = m − 2(q − i) ∈[−m,m].

Corollary 8.21Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimpleF -Lie algebra over F . Let H be a maximal toral subalgebra of L. Let Φ be the set of roots of Lrelative to H. If α, β ∈ Φ, then β(hα) ∈ Z and β − β(hα)α ∈ Φ. We call β(hα) Cartan integers.

Proof. By Proposition 8.20(2), β(hα) = r − q ∈ Z.Since −r ≤ −β(hα) = q − r ≤ q, by Proposition 8.20(1), β − β(hα)α ∈ Φ.

Proposition 8.22Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimpleF -Lie algebra over F . Let H be a maximal toral subalgebra of L. Let Φ be the set of roots of Lrelative to H. If α, β, α + β ∈ Φ, then [Lα, Lβ] = Lα+β.Proof.


1. V =q∐

i=−rLβ+iα is an irreducible Sα '

sl(2, F )-module.Proposition 8.20

2. V has basis v0, . . . , vm. Proposition 8.20

3. [Lα, Lβ] 6= 0. For xα ∈ Lα and vq ∈ Lβ, [xα, vq] = (m−q + 1)vq−1 = (r + 1)vq−1 6= 0.

4. [Lα, Lβ] ⊂ Lα+β. Proposition 7.23



5. 0 < dimF ([Lα, Lβ]) ≤ dimF (Lα+β). steps 3,4

6. dimF ([Lα, Lβ]) = dimF (Lα+β) = 1. Proposition 8.19(1)

7. [Lα, Lβ] = Lα+β. steps 4,6

Proposition 8.23Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimpleF -Lie algebra over F . Let H be a maximal toral subalgebra of L. Let Φ be the set of roots of Lrelative to H. Then L is generated by Lα : α ∈ Φ as F -Lie algebra.Proof.


1. Φ spans H∗. Proposition 8.10

2. t(Φ) spans H. Corollary 8.9

3. hα : α ∈ Φ span H. Ftα = Fhα by Proposition 8.16(1)

4. L is generated by Lα : α ∈ Φ. L = H ⊕ ∐α∈Φ

Lα by Lemma 7.20

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9. April 23th, Rationality properties, reflections, root systems

Definition 9.1Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimpleF -Lie algebra over F . Let H be a maximal toral subalgebra of L. Let Φ be the set of roots ofL relative to H. For λ, µ ∈ H∗, define (λ, µ) = κ(tλ, tµ) = λ(tµ) = µ(tλ). It is an associative,symmetric bilinear form by Lemma 4.11.

Lemma 9.2Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimpleF -Lie algebra over F . Let H be a maximal toral subalgebra of L. Let Φ be the set of roots of Lrelative to H. Suppose α, β ∈ Φ. Then β(hα) = 2(β, α)

(α, α) .Proof.


1. β(hα) = κ(tβ, hα) defn of t in Corollary 8.9

= κ

(tβ,

2tακ(tα, tα)

)= 2κ(tβ, tα)

κ(tα, tα) Proposition 8.16(1) and κ is bilinear

= 2(β, α)(α, α) . Definition 9.1

Lemma 9.3Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimpleF -Lie algebra over F . Let H be a maximal toral subalgebra of L. Let Φ be the set of rootsof L relative to H. By Proposition 8.10, we may choose a basis α1, . . . , αl of H∗ in Φ. Ifβ =

l∑i=1

ciαi ∈ Φ, then ci ∈ Q.Proof.


1. (β, αj) =l∑

i=1ci(αi, αj). β =

l∑i=1

ciαi ∈ Φ and Definition 9.1

2. We have l equations with l unknowns ciwhose coefficients and constants are inte-gers 2(β, αj)

(αj, αj)=

l∑i=1

2(αi, αj)(αj, αj)

ci, 1 ≤ j ≤ l.

Lemma 9.2 and Corollary 8.21



3. det((αi, αj))1≤i,j≤l 6= 0. αi : 1 ≤ i ≤ l form a basis of H∗ and κis non-degenerate since L is semisimple.

4. det(


)1≤i,j≤l

=

2ll∏

j=1(αj, αj)

det((αi, αj)1≤i,j≤l) 6= 0

5. The linear system has a unique solution inQ. Thus, ci ∈ Q for all 1 ≤ i ≤ l.

steps 2,4 and Cramer’s Rule

Lemma 9.4Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimpleF -Lie algebra over F . Let H be a maximal toral subalgebra of L. Let Φ be the set of roots of Lrelative to H. The form (•, •) as in Definition 9.1 is a positive definite form on EQ = SpanQ(Φ).(Remark that EQ ⊗ F ' H∗ by Proposition 8.10. )Proof.


1. Take a basis of L consisting of some ele-ments of H and xα ∈ Lα for α ∈ Φ.

Lemma 7.20

2. For all λ ∈ H∗, [tλ, xα] = α(tλ)xα. xα ∈ Lα

[tλ, H] = 0. H is abelian by Lemma 7.14

3. For all λ, µ ∈ H∗, (λ, µ) = κ(tλ, tµ) Definition 9.1

= Tr(ad(tλ) ad(tµ)) Definition 4.10

= ∑α∈Φ

α(tλ)α(tµ) step 1,2

= ∑α∈Φ

(α, λ)(α, µ). Definition 9.1

4. 1(β, β) = ∑

α∈Φ

(α, β)2

(β, β)2 . For β ∈ Φ, (β, β) = ∑α∈Φ

(α, β)2.

5. 2(β, α)(α, α) = α(hβ) ∈ Z. Corollary 8.21 and Lemma 9.2

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6. (β, β) ∈ Q for all β ∈ Φ. 4(β, β) ∈ Z by steps 4,5.

7. (α, β) ∈ Q for all α, β ∈ Φ. (α, β) = 12 ·

2(α, β)(β, β) · (β, β) ∈ Q.

8. (•, •) extends to a non-degenerate sym-metric bilinear form EQ × EQ → Q.

linear extension

9. (•, •) is positive definite on EQ. For all λ ∈ EQ, (λ, λ) = ∑α∈Φ

(α, λ)2 ≥ 0.

Theorem 9.5Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimpleF -Lie algebra over F . Let H be a maximal toral subalgebra of L. Let Φ be the set of roots of Lrelative to H. Let EQ = SpanQ(Φ). Let E = R ⊗Q EQ. Extend (•, •) to an inner product of E.Then E is a Euclidean space and(a) 0 6∈ Φ and Φ spans E.(b) If α ∈ Φ, then −α ∈ Φ and the only multiples of α in Φ are ±α.

(c) If α, β ∈ Φ, then β − 2(β, α)(α, α) α ∈ Φ.

(d) If α, β ∈ Φ, then 2(β, α)(α, α) ∈ Z.

Proof. The space E with (•, •) is Euclidean by Lemma 9.4.(a) follows from Definition 7.18 and Proposition 8.10;(b) follows from Proposition 8.11 and Proposition 8.18;(c) and (d) are Corollary 8.21 .

Definition 9.6Let E be a finite dimensional vector space over R. Let (•, •) : E × E → R be a positive definitesymmetric bilinear form. We call (•, •) an inner product of E and (E, (•, •)) a Euclideanspace.An R-linear transformation f : E → E is orthogonal if (f(x), f(y)) = (x, y) for all x, y ∈ E.

Definition 9.7A hyperplane of E is a subspace of codimension one.A reflection σ in E is an invertible R-linear transformation with(1) σ|P = 1P for some hyperplane P of E. We call P the reflecting hyperplane of σ.(2) If (x, P ) = 0, then σ(x) = −x.


Lemma 9.8Let E be a Euclidean space. All reflections of E are orthogonal.

Proof. Let σ be a reflection of E. Suppose E = P ⊕ Rx where P is the reflecting hyperplane and(x, P ) = 0. For all pi + aix ∈ E, pi ∈ P , ai ∈ R, i ∈ 1, 2, we have

(p1 + a1x, p2 + a2x) = (p1, p2) + a1a2(x, x), and

(σ(p1 + a1x), σ(p2 + a2x)) = (p1 − a1x, p2 − a2x) = (p1, p2) + a1a2(x, x).

Example 9.9Let E be a Euclidean space. Let α be a nonzero vector in E. Let Pα = β ∈ E : (β, α) = 0.There exists a reflection σα with reflecting hyperplane Pα. Then(1) Pα = Pcα for all c ∈ R∗.

(2) σα(β) = β − 〈β, α〉α where 〈β, α〉 = 2(β, α)(α, α) for all β ∈ E.

Proof. (1) β ∈ Pcα iff (β, cα) = c(β, α) = 0 iff β ∈ Pα.(2) If β ∈ Pα, then (β, α) = 0. Then 〈β, α〉 = 0 and hence σα(β) = β. Also, if (β, Pα) = 0, thenx ‖ α, β = cα for some c ∈ R∗. Then 〈β, α〉 = 2c and σα(β) = β − 2cα = −β. It follows fromDefinition 9.7 that σα is a reflection.

Lemma 9.10Let E be a Euclidean space. Let α be a nonzero vector in E.(1) 〈α, α〉 = 2. (2) 〈•, •〉 is linear on the first coordinate. (3) σ2

α = 1E.

Proof. (1) 〈α, α〉 = 2(α, α)(α, α) = 2.

(2) Because (•, •) is bilinear. But 〈•, •〉 is not linear on the second coordinate because of thedenominator. For example, 〈β, 2α〉 = 1

2〈β, α〉 for β ∈ E.(3)


1. σ2α(β) = σ(β)− 〈σ(β), α〉α Example 9.9(2)

2. = β − 〈β, α〉α− 〈β − 〈β, α〉α, α〉α Example 9.9(2)

3. = β − 〈β, α〉α− 〈β, α〉α + 〈β, α〉〈α, α〉α Lemma 9.10(2)

4. = β − 2〈β, α〉α + 2〈β, α〉α = β Lemma 9.10(1)

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Lemma 9.11Let E be a Euclidean space. Suppose Φ is a finite set, spans E and 0 6∈ Φ. Suppose all reflectionsσα : α ∈ Φ leave Φ invariant. If σ ∈ GL(E) leaves Φ invariant, fixes pointwise a hyperplane Pof E and σ(α) = −α for some α ∈ Φ, then σ = σα for some α ∈ Φ.

Proof. Let τ = σ σα.


1. τ(Φ) = Φ. σα and σ leaves Φ invariant

2. τ(α) = α. σα(α) = −α and σ(α) = −α

3. τ acts on P as identity. Given

4. All eigenvalues of τ are 1 steps 2,3

5. The minimal polynomial of τ divides (T −1)l, l = dimR(E).

defn of min poly

6. For each β ∈ Φ, β, τ(β), τ 2(β), . . . are notall distinct in Φ. Thus there exists kβ > 0in Z such that τ kβ = 1.

Φ is finite

7. Let k = maxkβ : β ∈ Φ. Then τ k = 1.Then the minimal polynomial of τ dividesT k − 1.

Φ is finite

8. The minimal polynomial of τ divides T k−1.

defn of min poly

9. The minimal polynomial of τ divides T−1. steps 5,8 and gcd(T k−1, (T −1)l) = T −1

10. σ = σα. τ = 1 and Lemma 9.10(3)

Definition 9.12A subset Φ ⊂ E is a root system if(R1) Φ is a finite set, spans E and 0 6∈ Φ.(R2) If α ∈ Φ, then the only multiples of α in Φ are ±α.(R3) If α ∈ Φ, then σα(Φ) ⊂ Φ.(R4) If α, β ∈ Φ, then 〈α, β〉 ∈ Z.


Lemma 9.13Let E be a Euclidean space. Let Φ be a root system of E. Then Φ = −Φ by Definition 9.12(R2).

Definition 9.14Let E be a Euclidean space. Let Φ be a root system of E. Let W be the subgroup of GL(E)generated by σα : α ∈ Φ. We call W the Weyl group of Φ.

Lemma 9.15Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. LetSΦ be the permutation group of Φ. Then (1) W ⊂ SΦ and (2) W is finite.

Proof. (1) By Definition 9.12(R3), W permutes Φ.(2) By Definition 9.12(R1), Φ is finite. Then Card(W ) ≤ Card(SΦ) = Card(Φ)!, W is finite.

Lemma 9.16Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. Ifσ ∈ GL(E) such that σ(Φ) = Φ, then σ σα σ−1 = σσ(α).Proof.


1. σ σα σ−1(σ(α)) = σ σα(α) = σ(−α) Definition 9.7(2)

= −σ(α). σ is R-linear.

2. For all σ(β) ∈ Pσ(α), β ∈ Pα. (β, α) = (σ(β), σ(α)) = 0 by Lemma 9.8

3. σ σα σ−1(σ(β)) = σ σα(β) = σ(β). Definition 9.7(1)

4. σ σα σ−1 = σσ(α). Lemma 9.11

Lemma 9.17Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. Ifσ ∈ GL(E) such that σ(Φ) = Φ, then 〈β, α〉 = 〈σ(β), σ(α)〉 for all α, β ∈ Φ.Proof.


1. σσ(α) = σ σα σ−1. Lemma 9.16

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σσ(α)(σ(β)) = σ σα(β).

2. σ(β) − 〈σ(β), σ(α)〉σ(α) = σ(β −〈β, α〉α) = σ(β)− 〈β, α〉σ(α)

Example 9.9(2)

3. 〈σ(β), σ(α)〉 = 〈β, α〉.

Definition 9.18Let E,E ′ be a Euclidean spaces. Let Φ,Φ′ be root systems of E,E ′ respectively. An isomorphismφ : (Φ, E)→ (Φ′, E ′) is a R-vector space isomorphism φ : E → E ′ such that(1) φ(Φ) = Φ′ and (2) 〈φ(β), φ(α)〉 = 〈β, α〉.

Remark 9.19Let φ : (Φ, E)→ (Φ′, E ′) be an isomorphism of root systems.(1) φ is not necessarily an isometry with respect to (•, •).(2) φ : (Φ, E) → (Φ, E) is an automorphism iff φ(Φ) = Φ. In fact, Definition 9.18(2) follows fromLemma 9.17.

Lemma 9.20Let φ : (Φ, E)→ (Φ′, E ′) be an isomorphism of root systems. Then σφ(α)(φ(β)) = φ(σα(β)) for allα, β ∈ Φ.Proof.


1. σφ(α)(φ(β)) = φ(β)− 〈φ(β), φ(α)〉φ(α) Example 9.9(2)

= φ(β)− 〈β, α〉φ(α) Definition 9.18(2)

= φ(β − 〈β, α〉α) φ is F -linear

= φ(σα(β)) Definition 9.18(2)


Lemma 9.21Let E,E ′ be a Euclidean spaces. Let Φ,Φ′ be root systems of E,E ′ respectively. Let W ,W ′ beWeyl groups of Φ,Φ′, respectively. Let φ : (Φ, E) → (Φ′, E ′) be an isomorphism of root systems.Then int(φ) : W ′ → W , σ 7→ φ σ φ−1 is an isomorphism of Weyl groups.

Proof. It is a homomorphism since φ (σ τ) φ−1 = (φ σ φ−1) (φ τ φ−1).Its inverse is int(φ−1) : W → W ′, σ′ 7→ φ−1 σ−1 φ.

Definition 9.22The dual of α ∈ Φ is α∨ = 2α

(α, α) . The dual of Φ is Φ∨ = α∨ : α ∈ Φ.

Lemma 9.23Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzerosemisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. Let Φ be the set of roots ofL relative to H and α ∈ Φ. Let tα satisfy α(h) = κ(tα, h) for all h ∈ H (as in Corollary 8.9). Lethα satisfy hα = 2tα

κ(tα, tα) (as in Proposition 8.16). Then tα∨ = hα.

Proof. For all h ∈ H, we have


1. κ(tα∨ , h) = α∨(h) Corollary 8.9

= 2α(h)(α, α) Definition 9.22

= 2κ(tα, h)κ(tα, tα) . Corollary 8.9 and Definition 9.1

2. κ(hα, h) = κ

(2tα

κ(tα, tα) , h)

Proposition 8.16

= 2κ(tα, h)κ(tα, tα) . κ is bilinear

3. tα∨ = hα. κ|H is non-degenerate by Lemma 8.4

Definition 9.24Let E be a Euclidean space. Let Φ be a root system of E. We call l = dimR(E) the rank of Φ.

Example 9.25The only rank one root system is (A1).

α−α

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Example 9.26The following root system has rank two. We say that it has type (A1 × A1).

α−α

β

−β

Example 9.27The following root system has rank two. We say that it has type (A2).

α−α

β

−β

α + β

−(α + β)

Example 9.28The following root system has rank two. We say that it has type (B2).

α−α

α + β

−(α + β)

β

−β

2α + β

−(2α + β)

Example 9.29The following root system has rank two. We say that it has type (G2).


α−α

3α + 2β

−(3α + 2β)

β

−β

3α + β

−(3α + β)

2α + β

−(2α + β)

α + β

−(α + β)

Lemma 9.30Let E be a Euclidean space. Let Φ be a root system of E. Let α, β ∈ Φ such that α 6= ±β and‖β‖ ≥ ‖α‖ > 0. Let 0 ≤ θ ≤ π be the angle between α and β. Then the following are the onlypossibilities.

〈α, β〉〈β, α〉 θ ‖β‖2/‖α‖2

0 0 π/2 undetermined

1 1 π/3 1

−1 −1 2π/3 1

1 2 π/4 2

−1 −2 3π/4 2

1 3 π/6 3

−1 −3 5π/6 3Proof.


1. (α, β) = ‖α‖‖β‖ cos(θ). defn of (•, •)

2.1. 〈β, α〉 = 2‖β‖‖α‖

cos(θ), Example 9.9(2)

2.2. 〈α, β〉 = 2‖α‖‖β‖

cos(θ).

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3. 〈β, α〉, 〈α, β〉 ∈ Z. Definition 9.12(R4)

4. 〈β, α〉〈α, β〉 = 4 cos2(θ) ∈ Z.

5. 4 cos2(θ) ∈ 0, 1, 2, 3 | cos θ| ≤ 1

5.0. 4 cos2(θ) = 0.

θ = π/2. 0 ≤ θ ≤ π

〈β, α〉 = 〈α, β〉 = 0 step 2.

‖β‖2/‖α‖2 is un-dertermined. any real number in (0,+∞)

5.1. 4 cos2(θ) = 1.

5.1.1 If cos(θ) = 1/2, then θ = π/3. 0 ≤ θ ≤ π

〈α, β〉 = ‖α‖‖β‖

= 1, ‖β‖2/‖α‖2 = 1. steps 2.2, 3 and ‖β‖ ≥ ‖α‖ > 0

〈β, α〉 = 1. step 4, 5.1

5.1.2 If cos(θ) = −1/2, then θ = 2π/3. 0 ≤ θ ≤ π

〈α, β〉 = −‖α‖‖β‖

= −1, ‖β‖2/‖α‖2 = 1. steps 2.2, 3 and ‖β‖ ≥ ‖α‖ > 0

〈β, α〉 = −1. steps 4, 5.1

5.2. 4 cos2(θ) = 2.

5.2.1 If cos(θ) =√

2/2, then θ = π/4. 0 ≤ θ ≤ π

〈α, β〉 =√

2‖α‖‖β‖∈ [1,

√2]. steps 2.2, 3 and ‖β‖ ≥ ‖α‖ > 0

Thus 〈α, β〉 = 1, ‖β‖2/‖α‖2 = 2.

〈β, α〉 = 2. step 4, 5.2

5.2.2 If cos(θ) = −√

2/2, then θ = 3π/4. 0 ≤ θ ≤ π

〈α, β〉 =√

2‖α‖‖β‖∈ [−

√2,−1]. steps 2.2, 3 and ‖β‖ ≥ ‖α‖ > 0

Thus 〈α, β〉 = −1, ‖β‖2/‖α‖2 = 2.

〈β, α〉 = 2. step 4, 5.2

5.3. 4 cos2(θ) = 3.



5.3.1 If cos(θ) =√

3/2, then θ = π/6. 0 ≤ θ ≤ π

〈α, β〉 =√

3‖α‖‖β‖∈ [1,

√3]. steps 2.2, 3 and ‖β‖ ≥ ‖α‖ > 0

Thus 〈α, β〉 = 1, ‖β‖2/‖α‖2 = 3.

〈β, α〉 = 3. step 4, 5.2

5.3.2 If cos(θ) = −√

3/2, then θ = 5π/6. 0 ≤ θ ≤ π

〈α, β〉 =√

3‖α‖‖β‖∈ [−

√3,−1]. steps 2.2, 3 and ‖β‖ ≥ ‖α‖ > 0

Thus 〈α, β〉 = −1, ‖β‖2/‖α‖2 = 3.

〈β, α〉 = −3. step 4, 5.2

Lemma 9.31Let E be a Euclidean space. Let Φ be a root system of E. Let α, β ∈ Φ such that α 6= ±β.(1) If (α, β) > 0, then α− β is a root. (2) If (α, β) < 0, then α + β is a root.

Proof. (1)


1. If ‖β‖ ≥ ‖α‖ > 0, then 〈α, β〉 = 1. (α, β) > 0 and Lemma 9.30

2. α− β = α− 〈α, β〉β = σβ(α) is a root. Example 9.9(2) and Definition 9.12(R3)

3. If ‖α‖ ≥ ‖β‖ > 0, then β − α = −(α− β)is a root.

Lemma 9.13

(2) If (α, β) < 0, then (α,−β) > 0. By (1), α + β = α− (−β) is a root.

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10. April 30th, Bases and Weyl chambers

Lemma 10.1Let E be a Euclidean space. Let Φ be a root system of E. Let α, β ∈ Φ such that α 6= ±β. Theni ∈ Z : β + iα ∈ Φ = [−r, q] ∩ Z for some integers r, q ≥ 0.Proof.


1. There exist integers r, q ≥ 0 such that β+qα ∈ Φ, β−rα ∈ Φ and i ∈ Z : β+ iα ∈Φ ⊂ [−r, q] ∩ Z.

Φ is finite by Definition 9.12 (R1)

2. If i ∈ Z : β+ iα ∈ Φ ( [−r, q]∩Z, thenthere exist −r < s ≤ p < q in Z such thatβ + (s− 1)α ∈ Φ, β + sα 6∈ Φ, β + pα 6∈ Φand β + (p+ 1)α ∈ Φ.

Intermediate value theorem

3. (β + (s− 1)α, α) ≥ 0. β + (s − 1)α ∈ Φ and β + sα 6∈ Φ andLemma 9.31(2)

(s− 1)(α, α) ≥ −(β, α).

4. (β + (p+ 1)α, α) ≤ 0 β + pα 6∈ Φ and β + (p + 1)α ∈ Φ, byLemma 9.31(1),

(p+ 1)(α, α) ≤ −(β, α).

5. p+ 1 ≤ s− 1 (p+ 1)(α, α) ≤ −(β, α) ≤ (s− 1)(α, α).

A contradiction. to s ≤ p

Definition 10.2Let E be a Euclidean space. Let Φ be a root system of E. Let α, β ∈ Φ such that α 6= ±β. Wecall β + iα ∈ Φ : −r ≤ i ≤ q the α-string through β.

Lemma 10.3Let E be a Euclidean space. Let Φ be a root system of E. Let α, β ∈ Φ such that α 6= ±β. Theα-string through β is of length r + q + 1 ≤ 4.Proof.



1. σα(β − rα) = σα(β) + rα σα is R-linear by Definition 9.7

= β + (r − 〈β, α〉)α Example 9.9(2)

∈ Φ β − rα ∈ Φ and Definition 9.12 (R3)

2. r − 〈β, α〉 ≤ q. Lemma 10.1

3. σα(β + qα) = σα(β)− qα σα is R-linear by Definition 9.7

= β − (q + 〈β, α〉)α Example 9.9(2)

∈ Φ β + qα ∈ Φ and Definition 9.12 (R3)

4. −r ≤ −(q + 〈β, α〉). Lemma 10.1

5. 〈β, α〉 = r − q. steps 2,4

6.1. If β − rα = ±α, then β = (r ± 1)α ∈α,−α, a contradiction to β 6= ±α.

Definition 9.12(R2)

6.2. So β − rα 6= ±α. Then 〈β − rα, α〉

= 〈β, α〉 − r〈α, α〉 Lemma 9.10(2)

= (r − q)− 2r = −(r + q) ≤ 0 step 5 and 〈α, α〉 = 2 by Lemma 9.10(1)

〈β − rα, α〉 ∈ 0,−1,−2,−3 Lemma 9.30

7. The length is r+q+1 = −〈β−rα, α〉+1 ∈1, 2, 3, 4.

Lemma 10.1 and step 6.2

Definition 10.4Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. Wecall ∆ ⊂ Φ a base and we call elements of ∆ simple roots if(B1) ∆ is a R-basis of the R-vector space E;(B2) For all β ∈ Φ, β = ∑

α∈∆kαα, kα ∈ Z such that kα ≥ 0 for all α ∈ ∆ or kα ≤ 0 for all α ∈ ∆.

• When kα ≥ 0 for all α ∈ ∆, we call β is positive and write β > 0.• When kα ≤ 0 for all α ∈ ∆, we call β is negative and write β < 0.

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We write Φ+ = β ∈ Φ : β > 0, Φ− = β ∈ Φ : β < 0. For α, β ∈ Φ, we say that β < α ifα− β ∈ Φ+. The height of β ∈ Φ relative to ∆ is ht(β) = ∑

α∈∆kα.

Remark 10.5(1) Card(∆) = dimRE by (B1).(2) The expression β = ∑

α∈∆kαα is unique since ∆ is R-linearly independent by (B1).

(3) If α, β ∈ Φ+ and α + β ∈ Φ, then α + β ∈ Φ+ by (B2).(4) ∆ ⊂ Φ+, Φ− = −Φ+, Φ+ ∩ Φ− = ∅, and Φ = Φ+ ∪ Φ−.(5) The relation ≤ is a partial order in Φ. We have α ≤ β in Φ implies that ht(α) ≤ ht(β).

Lemma 10.6Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. Let ∆be a base of Φ. If α, β ∈ ∆ and α 6= β, then (1) α− β 6∈ Φ and (2) (α, β) ≤ 0.

Proof. (1) α− β 6∈ Φ by Definition 10.4(B2).(2) Suppose (α, β) > 0.


1. α 6= −β. Thus α 6= ±β. α, β ∈ ∆ and Definition 10.4(B1)

2. α− β ∈ Φ. (α, β) > 0 and Lemma 9.31(1)

3. A contradiction. to (1).

Lemma 10.7Let E be a Euclidean space. Let P1, . . . , Pn be hyperplanes of E. Then P1 ∪ · · · ∪ Pn ( E.

Proof. If n = 1, then E − P1 6= ∅ since dimF (P1) < dimF (E).Now we suppose n > 1.


1. There are infinitely many translates of P1. R is infinite

2. Then there exists a translate P of P1 suchthat P 6= Pi for all 1 ≤ i ≤ n.

n <∞.

3. P ∩ Pi, 2 ≤ i ≤ n are codimension ≤ 1subspaces of P .

dimR P − dimR(P ∩ Pi) ≤ dimRE −dimR(Pi) = 1



4. There exits a point Q ∈ P −n⋃i=2

Pi. Inductive hypothesis

5. Q ∈ E −n⋃i=1

Pi. Q ∈ P ⊂ E − P1 and step 4

Another proof for the case n > 1. Deleting duplicate ones, we may assume that Pi are distinct.


1. Suppose y ∈ E − P1. dimF (P1) < dimF (E)

2. For all x ∈ P1, x + ay ∈ E − P1 for alla ∈ R∗.

P1 is a linear subspace of E.

3. If P1 ∪ · · · ∪ Pn = E, then there existsPj, j ≥ 2 such that Pj ∩x+ay : a ∈ R∗is an infinite set.

n is finite and the Pigeonhole principle

4. Suppose x + ay, x + a′y ∈ Pj and a 6= a′.

Then y = (x+ ay)− (x+ a′y)a− a′

∈ Pj.

Pj is a linear subspace of E

5. x = (x+ ay)− ay ∈ Pj. Pj is a linear subspace of E

6. P1 ⊂ Pj. Hence P1 = Pj. dimR P1 = dimR Pj = dimRE − 1.

7. A contradiction. to our assumption that Pi are distinct.

Definition 10.8Let E be a Euclidean space. Let Φ be a root system of E. For each α ∈ Φ, let Pα = β ∈E : (β, α) = 0.(1) We call γ ∈ E regular if γ ∈ E − ⋃

α∈ΦPα; otherwise we call γ singular.

(2) For each regular γ ∈ E, let Φ+(γ) = α ∈ Φ : (α, γ) > 0, Φ−(γ) = α ∈ Φ : (α, γ) < 0.(3) We call α ∈ Φ+(γ) decomposable if α = β1 + β2 for some β1 ∈ Φ+(γ) and β2 ∈ Φ+(γ);otherwise we call α indecomposable. Let ∆(γ) be the set of indecomposable roots of Φ+(γ).

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Remark 10.9Suppose γ ∈ E is regular.(1) Φ+(γ) ∩ Φ−(γ) = ∅ and Φ = Φ+(γ) ∪ Φ−(γ).(2) If α, β, α + β ∈ Φ+(γ), then α + β ∈ Φ+(γ); If α, β ∈ Φ−(γ), then α + β ∈ Φ−(γ).

Lemma 10.10Let E be a Euclidean space. Let Φ be a root system of E. Suppose γ ∈ E is regular. ThenΦ+(γ) ⊂ SpanZ(∆(γ)). Similarly, Φ−(γ) ⊂ SpanZ(∆(γ)).

Proof. Suppose the opposite: Let S = Φ+(γ)− SpanZ(∆(γ)) 6= ∅.


1. Let α ∈ S such that (α, γ) =min(β, γ) : β ∈ S.

Φ is finite, then S is finite

2. α = β1 + β2 for some βi ∈ Φ+(γ). α 6∈ ∆(γ)

3. (βi, γ) > 0 for i ∈ 1, 2. βi ∈ Φ+(γ)

4. (βi, γ) < (α, γ) for i ∈ 1, 2. (β1, γ) + (β2, γ) = (α, γ)

5. βi ∈ SpanZ(∆(γ)) for i ∈ 1, 2. βi 6∈ S by step 1

6. α = β1 + β2 ∈ SpanZ(∆(γ)) (SpanZ(∆(γ)),+) is a subgroup of (E,+)

7. A contradiction. to α ∈ S

Lemma 10.11Let E be a Euclidean space. Let Φ be a root system of E. Suppose γ ∈ E is regular. If α, β ∈ ∆(γ)and α 6= β, then (1) α− β 6∈ Φ and (2) (α, β) ≤ 0.This lemma is analogous to Lemma 10.6.

Proof. (1) Suppose α− β ∈ Φ.


1. Either α− β ∈ Φ+(γ) or β − α ∈ Φ+(γ). Remark 10.9(1)

2.1 If α− β ∈ Φ+(γ), then α = α− β + β. Remark 10.9(2)

2.2. A contradiction. to α ∈ ∆(γ)



3.1. If β − α ∈ Φ+(γ), then β = β − α + α. Remark 10.9(2)

3.2. A contradiction. to β ∈ ∆(γ)

(2) Suppose the opposite (α, β) > 0.


1. β 6= −α. Thus β 6= ±α. α, β ∈ ∆(γ) and Definition 10.4(B1)

2. α− β ∈ Φ. (α, β) > 0 and Lemma 9.31(1)

3. A contradiction. to (1)

Lemma 10.12Let E be a Euclidean space and γ ∈ E. Let ∆ be a subset of α ∈ E : (α, γ) > 0 such that foreach pair of vectors α, β ∈ ∆, (α, β) ≤ 0. Then ∆ is linearly independent.

Proof. Suppose ∑α∈∆

rαα = 0 for rα ∈ R. We need to show that rα = 0 for all α ∈ ∆. We write

rα =

sα, rα ≥ 0,

−tβ, rβ ≤ 0.


1. ∑sαα = ∑

tββ.∑α∈∆

rαα = 0.

we denote ε = ∑sαα = ∑

tββ.

2.1. 0 ≤ (ε, ε) (•, •) is positive definite by Definition 9.6

= (∑ sαα,∑tββ) step 1

= ∑sαtβ(α, β) (•, •) is bilinear by Definition 9.6

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≤ 0. The angle between each pair α, β ∈ ∆ isobtuse.

2.2. ε = 0.

3.1. 0 = (γ, ε) = ∑sα(γ, α) step 1 and (•, •) is bilinear

≥ 0 for all rα ≥ 0. ∆ ⊂ α ∈ E : (α, γ) > 0 and sα ≥ 0

3.2. sα = 0 for all rα ≥ 0.

4.1. 0 = (γ, ε) = ∑tβ(γ, β) step 1 and (•, •) is bilinear

≥ 0 for all rβ ≤ 0. ∆ ⊂ α ∈ E : (α, γ) > 0 and tβ ≥ 0

4.2. tβ = 0 for all rβ ≤ 0.

5. rα = 0 for all α ∈ ∆. steps 3.2, 4.2

Theorem 10.13Let E be a Euclidean space. Let Φ be a root system of E. Suppose γ ∈ E is regular. Then ∆(γ)is a base for Φ.Proof.


1. Φ ⊂ SpanZ(∆(γ)). Lemma 10.10

Thus Definition 10.4(B2) holds.

2. E = SpanR(∆(γ)). step 1 and Φ spans E by Defini-tion 9.12(R1)

3. ∆(γ) is linearly independent. Lemma 10.12

4. Definition 10.4(B1) holds. steps 3,4


Theorem 10.14Let E be a Euclidean space. Let Φ be a root system of E. Every base ∆ of Φ is of the form ∆(γ)for some regular γ ∈ E.Proof.


1. There exists γ ∈ E such that (γ, α) > 0for all α ∈ ∆.

[Hum78, p.54, Exercise 7]

2. (γ, α) > 0 for all α ∈ Φ+. Definition 10.4

(γ, α) < 0 for all α ∈ Φ−.

3. γ is regular. Definition 10.8(1)

4. Φ+ ⊂ Φ+(γ) and Φ− ⊂ Φ−(γ). Definition 10.8(2) and step 2

5. Φ = Φ+ ∪ Φ− ⊂ Φ+(γ) ∪ Φ−(γ) = Φ. Remark 10.5(4) and Remark 10.9(1)

Φ+ ∩ Φ− = ∅ and Φ+(γ) ∩ Φ−(γ) = ∅.

6. Φ+ = Φ+(γ) and Φ− = −Φ+(γ). steps 4,5

7. Elements of ∆ are indecomposable. Uniqueness of Definition 10.4(B2)

Then ∆ ⊂ ∆(γ). Definition 10.8(3)

8. Card(∆) = dimRE = Card(∆(γ)). Definition 10.4(B1) and Theorem 10.13

9. ∆ = ∆(γ). steps 7,8

Definition 10.15Let E be a Euclidean space. Let Φ be a root system of E. The connected components of E− ⋃

α∈ΦPα

are called Weyl chambers.Each regular γ ∈ E belongs to precisely one Weyl chamber, denoted C(γ).Recall that a subset of a Euclidean space is connected iff it is path-connected.

Lemma 10.16Let E be a Euclidean space. Let Φ be a root system of E. Let γ, γ′ be regular elements of E. Thefollowing are equivalent:(1) C(γ) = C(γ′). (2) γ, γ′ lie on the same side of each Pα(α ∈ Φ).

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(3) Φ+(γ) = Φ+(γ′). (4) ∆(γ) = ∆(γ′).In particular, (1) iff (4) provides a bijection between the set of all Weyl chambers of Φ and the setof all bases of Φ given by C(γ)←→ ∆(γ).

Proof. (1) implies (2).


1. Let c : [0, 1] → E be a path from γ to γ′,i.e. a continuous map such that c(0) = γ,c(1) = γ′ and c([0, 1]) ⊂ C(γ).

(1)

2. We may assume that there exists α ∈ Φsuch that (γ, α) > 0 and (γ′, α) < 0.

Suppose (2) does not hold, i.e. γ and γ′

are on different sides of Pα

3. Define f(x) = (c(x), α), x ∈ [0, 1]. Thenthere exists 0 < ξ < 1 such that f(ξ) = 0.

f is continuous, f(0) > 0, f(1) < 0 andintermediate value theorem

4. c(ξ) ∈ Pα, a contradiction. to c(ξ) ∈ C(γ) by step 1

(2) implies (1). If C(γ) 6= C(γ′), there exists Pα between γ and γ′, a contradiction to (2).(2) implies (3)


1. α ∈ Φ+(γ) iff (α, γ) > 0 Definition 10.8(2)

iff (α, γ′) > 0 (2)

iff α ∈ Φ+(γ′) . Definition 10.8(2)

(3) implies (2) Suppose α ∈ Φ.


1. (α, γ) > 0 iff α ∈ Φ+(γ) Definition 10.8(2)

iff α ∈ Φ+(γ′) (3)

iff (α, γ′) > 0 Definition 10.8(2)



2. (α, γ) < 0 iff (α, γ′) < 0 γ is regular and step 1

(3) implies (4)


1. ∆(γ) (resp. ∆(γ′)) is the set of indecom-posable roots of Φ+(γ) (resp. Φ+(γ′)).

Definition 10.8(3)

2. ∆(γ) = ∆(γ′). (3)

(4) implies (3).


1. Φ+(γ) = Φ ∩ SpanN(∆(γ)) Definition 10.4(B2)

= Φ ∩ SpanN(∆(γ′)) (4)

= Φ+(γ′) Definition 10.4(B2)

Definition 10.17Let E be a Euclidean space. Let Φ be a root system of E. Let ∆ be a base of Φ. If ∆ = ∆(γ) fora regular element γ ∈ E, then we call C(∆) = C(γ) the fundamental Weyl chamber relativeto ∆.

Remark 10.18The chamber C(∆) is open and convex since it is the intersection of finitely many open half-spaces.

Example 10.19The fundamental Weyl chamber of A1 × A1 is the yellow-green area.

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α−α

β

−β

Example 10.20The fundamental Weyl chamber of A2 is the yellow-green area.

α−α

β

−β

α + β

−(α + β)

Example 10.21The fundamental Weyl chamber of B2 is the yellow-green area.

α−α

α + β

−(α + β)

β

−β

2α + β

−(2α + β)

Example 10.22The fundamental Weyl chamber of G2 is the yellow-green area.


α−α

3α + 2β

−(3α + 2β)

β

−β

3α + β

−(3α + β)

2α + β

−(2α + β)

α + β

−(α + β)

Lemma 10.23Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. Ifσ ∈ W and γ ∈ E is regular, then σ(C(γ)) = C(σ(γ)).

Proof. Let sgn : R→ 1, 0,−1 be the sign function. By Example 9.9(2), sgn((a, b)) = sgn(〈a, b〉)for all a, b ∈ E.


1. x ∈ σ(C(γ)) iff σ−1(x) ∈ C(γ)

iff sgn((σ−1(x), α)) = sgn((γ, α)) for allα ∈ Φ

Lemma 10.16 (1) iff (2)

iff sgn((x, σ(α)) = sgn((σ(γ), σ(α))) forall α ∈ Φ

σ ∈ W preserves signs of inner productsby Lemma 9.17

iff x ∈ C(σ(γ)) σ(α) runs through Φ by Defini-tion 9.12(R3)

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11. May 7th, Weyl group and its actions

Lemma 11.1Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. Ifσ ∈ W and γ ∈ E is regular, then σ(∆(γ)) = ∆(σ(γ)).Proof.


1. x ∈ σ(∆(γ)) iff σ−1(x) ∈ ∆(γ)

iff (σ−1(x), γ) > 0 and Definition 10.8(3)

σ−1(x) is indecomposable

iff (x, σ(γ)) > 0 and Lemma 9.17

x is indecomposable σ is linear

iff x ∈ ∆(σ(γ)). Definition 10.8(3)

Lemma 11.2Let E be a Euclidean space. Let Φ be a root system of E. Let ∆ be a base of Φ. If α ∈ Φ+ −∆,then there exists β ∈ ∆ such that α− β ∈ Φ+.

Proof. Case 1. (α, β) > 0 for some β ∈ ∆.


1.1. β 6= α. β ∈ ∆ and α 6∈ ∆

1.2. β 6= −α. (α, β) > 0.

2. α− β ∈ Φ. Lemma 9.31(1)

3. Suppose α = ∑γ∈∆

kγγ, kγ ∈ N. α ∈ Φ+

4. kγ′ > 0 for some γ′ 6= β. otherwise α = kββ = β ∈ ∆ by Defini-tion 9.12(R2), a contradiction to α 6∈ ∆.

5. kγ′γ′ is a summand of α− β. γ′ 6= β.

6. α− β ∈ Φ+. step 2 and kγ′ > 0.


Case 2. (α, β) ≤ 0 for all β ∈ ∆.


1. ∆ ∪ α is linearly independent. Lemma 10.12

2. A contradiction. to Definition 9.12(B1).

Corollary 11.3Let E be a Euclidean space. Let Φ be a root system of E. Let ∆ be a base of Φ. Each β ∈ Φ+

can be written as β = α1 + · · ·+ αk, (αi ∈ ∆) not necessarily distinct such that each partial sumα1 + · · ·+ αi ∈ Φ+, 1 ≤ i ≤ k.

Proof. Use induction on ht(β) = k. When k = 1, β = α1 ∈ ∆. Suppose k > 1,


1. β ∈ Φ+ −∆. ht(β) > 1

2. There exists αk ∈ ∆ s.t. β − αk ∈ Φ+. Lemma 11.2

3. β−αk = α1 + · · ·+αk−1 such that αi ∈ ∆and α1+· · ·+αi ∈ Φ+ for all 1 ≤ i ≤ k−1.

ht(β − αk) = k− 1 and inductive hypoth-esis

4. β = α1 + · · ·+ αk ∈ Φ+. given

Lemma 11.4Let E be a Euclidean space. Let Φ be a root system of E. Let ∆ be a base of Φ. Suppose α ∈ ∆.Then σα permutes Φ+ − α.

Proof. For all β ∈ Φ+ − α,


1. Suppose β = ∑γ∈∆

kγγ. Definition 10.4(B2)

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2. There exists γ 6= α, kγ > 0. otherwise β = kαα = α by Defini-tion 9.12(R2), a contradiction to β 6= α

3. kγγ is a summand of σα(β). γ 6= α and σα(β) = β − 〈β, α〉α.

4. σα(β) ∈ Φ+ kγ > 0.

5. β 6= −α. β ∈ Φ+ and −α 6∈ Φ+

6. σα(β) 6= α.

7. σα(Φ+ − α) = Φ+ − α. steps 4,6 and σ2α = 1 by Lemma 9.10(3)

Corollary 11.5Let E be a Euclidean space. Let Φ be a root system of E. Let ∆ be a base of Φ. Set δ = 1

2∑β>0

β.

Then σα(δ) = δ − α for all α ∈ ∆.Proof.


1. ∑β∈Φ+−α

β = ∑β∈Φ+−α

σα(β). Lemma 11.4

2. 2δ − α = ∑β∈Φ+−α

β defn of δ

= ∑β∈Φ+−α

σα(β) step 1

= σα( ∑β∈Φ+−α

β) σα is R-linear

= σα(2δ − α) defn of δ

= 2σα(δ) + α. σα is R-linear and σα = −α

3. σα(δ) = δ − α.


Lemma 11.6Let E be a Euclidean space. Let Φ be a root system of E. Let ∆ be a base of Φ. Supposeα1, . . . , αt ∈ ∆ (not necessarily distinct). Write σi = σαi to avoid sub-subscript. If σ1 · · · σt−1(αt) < 0, then there exists s ∈ N, 1 ≤ s < t such that

σ1 · · · σt = σ1 · · · σs−1 σs+1 · · · σt−1.

Proof. Write

βi =

σi+1 · · ·σt−1(αt), 0 ≤ i ≤ t− 2,

αt, i = t− 1.


1. β0 < 0 given σ1 · · · σt−1(αt) < 0

βt−1 > 0 given αt ∈ ∆ ⊂ Φ+

2. There exists 1 ≤ s < t such that βs > 0and s = mini : βi > 0.

3. σs(βs) = βs−1 < 0. defn of βs and defn of s

4. βs = αs. below

Otherwise βs ∈ Φ+ − αs. step 2

βs−1 ∈ Φ+ − αs. Lemma 11.4

A contradiction. to step 3

5. Suppose 1 ≤ s ≤ t− 2.

5.1. σs = σαs = σβs = σσs+1···σt−1(αt) step 4 and defn of βs

= (σs+1· · ·σt−1)σt(σs+1· · ·σt−1)−1 Lemma 9.16

5.2. σs σs+1 · · · σt = σs+1 · · · σt−1. right compose σs+1 · · · σt on both sides

5.3. σ1· · ·σt = σ1· · ·σs−1σs+1· · ·σt−1. left compose σ1 · · · σs−1 on both sides

6. Suppose s = t− 1.

6.1. σs = σαs = σβs = σαt = σt. step 4 and defn of βs

6.2. σ1 · · · σt = σ1 · · · σt−2 σt−1 = σs = σt.

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Corollary 11.7Let E be a Euclidean space. Let Φ be a root system of E. Let ∆ be a base of Φ. Supposeα1, . . . , αt ∈ ∆ (not necessarily distinct). If σ = σ1 · · · σt ∈ W , σi = σαi and t is the smallestpossible length of the decomposition, then σ(αt) < 0.

Proof. If σ(αt) = −σ1 · · · σt−1(αt) > 0, then σ1 · · · σt−1(αt) < 0. By Lemma 11.6, σ =σ1 · · · σs−1 σs+1 · · · σt−1 has length t− 2 < t, a contradiction to the minimality of t.

Theorem 11.8Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. Let∆ be a base of Φ. Let W ′ be the subgroup of W generated by reflections σα, α ∈ ∆. If γ ∈ E isregular, then there exists σ ∈ W ′ such that (σ(γ), α) > 0 for all α ∈ ∆, i.e. σ(γ) ∈ C(∆).

Proof. Let δ = 12∑

α∈Φ+α.


1. Choose σ ∈ W ′ such that (σ(γ), δ) =max(τ(γ), δ) : τ ∈ W ′.

W ′ is finite since W is finite Lemma 9.15

2. σα σ ∈ W ′ for all α ∈ ∆. σα, σ ∈ W ′

3. (σ(γ), δ) ≥ (σα(σ(γ)), δ) steps 1,2

= (σ(γ), σα(δ)) σα is orthogonal by Lemma 9.8

= (σ(γ), δ − α) σα(δ) = δ − α by Corollary 11.5

= (σ(γ), δ)− (σ(δ), α) (•, •) is bilinear

4. (σ(γ), α) ≥ 0 for all α ∈ ∆.

5. (σ(γ), α) > 0 for all α ∈ ∆. γ is regular, (σ(γ), α) = (γ, σ−1(α)) 6= 0

Theorem 11.9Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. Let ∆be a base of Φ. Let W ′ be the subgroup of W generated by reflections σα, α ∈ ∆. Then W ′ actstransitively on Weyl chambers, i.e. for all Weyl chambers C(γ),C(γ′), there exists τ ∈ W ′ suchthat τ(C(γ)) = C(γ′).


Proof. Let C(γ) and C(γ′) be Weyl chambers.


1.1. There exists σ ∈ W ′ s.t. σ(γ) ∈ C(∆). Theorem 11.8

1.2. σ(C(γ)) = C(σ(γ)) = C(∆). Lemma 10.23

2.1. There exists σ′ ∈ W ′ s.t. σ′(γ′) ∈ C(∆). Theorem 11.8

2.2. σ′(C(γ′)) = C(σ′(γ′)) = C(∆). Lemma 10.23

3. Let τ = σ′−1 σ ∈ W ′. Then τ(C(γ)) =σ′−1(σ(C(γ))) = σ′−1(C(∆)) = C(γ′).

steps 1.2, 2.2

Theorem 11.10Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. Let ∆be a base of Φ. Let W ′ be the subgroup of W generated by reflections σα, α ∈ ∆. Then W ′ actstransitively on bases, i.e. if ∆′ is another base of Φ, then σ(∆) = ∆′ for some σ ∈ W ′.Proof.


1. Suppose ∆ = ∆(γ) and ∆′ = ∆(γ′). Theorem 10.13

2. There exists σ ∈ W ′ s.t. σ(C(γ)) = C(γ′). Theorem 11.9

3. C(σ(γ)) = C(γ′). σ(C(γ)) = C(σ(γ)) by Lemma 10.23

4. ∆(σ(γ)) = ∆(γ′). Lemma 10.16 (1) iff (4)

5. σ(∆) = σ(∆(γ)) = ∆(σ(γ)) Lemma 11.1

= ∆(γ′) = ∆′. step 4

Lemma 11.11Let E be a Euclidean space. Let Φ be a root system of E. For all α ∈ Φ, there exists a base ∆ ofΦ such that α ∈ ∆.

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Proof. If Φ = A1, then α is a base. If Φ 6= A1, then there exists β ∈ Φ− ±α.


1. There exists γ ∈ Pα −⋃

β∈Φ−±αPβ.

⋃β∈Φ−±α

(Pβ ∩ Pα) ( Pα by Lemma 10.7

2. (γ, α) = 0. Let m = min|(γ, β)|, β ∈Φ− ±α > 0.

Φ is finite

3. There exists γ′ such that m2 > (γ′, α) > 0

and |(γ′, β)| > m

2 for all β ∈ Φ− ±α.

The continuity of the first coordinate of(•, •)

4. α ∈ ∆(γ′). below

α ∈ Φ+(γ′) (γ′, α) > 0

α is indecomposable. |(γ′, β)| > (γ′, α) for all β ∈ Φ− ±α

Theorem 11.12Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. Let ∆be a base of Φ. Let W ′ be the subgroup of W generated by reflections σα, α ∈ ∆. For all α ∈ Φ,there exists σ ∈ W ′ such that σ(α) ∈ ∆.Proof.


1. There exists a base ∆′ of Φ such that α ∈∆′.

Lemma 11.11

2. There exists σ ∈ W ′ such that σ(∆′) = ∆. Theorem 11.10

3. σ(α) ∈ σ(∆′) = ∆.

Theorem 11.13Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. Let ∆be a base of Φ. Let W ′ be the subgroup of W generated by reflections σα, α ∈ ∆. Then W = W ′.


Proof. By definition W ′ ⊂ W . Conversely, it suffices to show that σα ∈ W ′ for all α ∈ Φ.


1. There exists σ ∈ W ′ s.t. β = σ(α) ∈ ∆. Theorem 11.12

2. σβ = σσ(α) = σ σα σ−1. Lemma 9.16

3. σα = σ−1 σβ σ ∈ W ′. σβ, σ ∈ W ′.

Corollary 11.14Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. Let ∆be a base of Φ.(1) For all γ ∈ E regular, there exists σ ∈ W such that σ(γ) ∈ C(∆).(2) W acts transitively on Weyl chambers.(3) W acts transitively on bases.(4) For all α ∈ Φ, there exists σ ∈ W such that σ(α) ∈ ∆.

Proof. (1) follows from Theorem 11.8 and Theorem 11.13.(2) follows from Theorem 11.9 and Theorem 11.13.(3) follows from Theorem 11.10 and Theorem 11.13.(4) follows from Theorem 11.12 and Theorem 11.13.

Theorem 11.15Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. Let ∆be a base of Φ. Then W acts simply transitively on bases of Φ, i.e. if σ ∈ W such that σ(∆) = ∆,then σ = 1.

Proof. Suppose the opposite σ 6= 1.


1. σ = σ1 · · · σt, σi = σαi , αi ∈ ∆ with thesmallest length.

Theorem 11.13

2. σ(αt) < 0. Corollary 11.7

3. A contradiction. to αt ∈ ∆ and σ(∆) = ∆ ⊂ Φ+.

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Definition 11.16Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. Let ∆be a base of Φ.(1) Suppose σ = σ1 · · · σt ∈ W where σi = σαi αi ∈ ∆ and t is minimal. We call it thedecomposition reduced and t = l(σ) the length of σ. We have l(σ) = 0 iff σ = 1.(2) Let S(σ) = α > 0 : σ(α) < 0, n(σ) = Card(S(σ)).

Lemma 11.17Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ andσ ∈ W . Let ∆ be a base of Φ. Then l(σ) = n(σ).

Proof. First of all, l(σ) = 0 iff σ = 1 iff S(σ) = ∅ iff n(σ) = 0.Suppose l(τ) = n(τ) for all τ ∈ W such that l(τ) < l(σ). Let σ = σ1 · · · σt be a reduceddecomposition.


1. l(σ σt) = l(σ)− 1 = t− 1 below

1.1. l(σ σt) ≤ t− 1. σ σt = σ1 · · · σt−1

1.2. Suppose l(σ σt) = s < t− 1 and σ σt =σβ1 · · · σβs is a reduced expression.

l(σ) ≤ s+ 1 < t, a contradiction. σ = σβ1 · · · σβs σt

2. σ(αt) < 0. Corollary 11.7

3. αt ∈ S(σ). Definition 11.16(1)

4. σt : S(σσt)→ S(σ)−αt is well-defined. below

4.1. αt 6∈ S(σ σt). step 2

4.2. For all β ∈ S(σ σt), σt(β) 6∈ αt. Lemma 11.4

4.3. σ(σt(β)) < 0. β ∈ S(σ σt)

σt(β) ∈ S(σ). Definition 11.16(2)

5. σt : S(σ σt)→ S(σ)−αt is a bijection. σ−1t = σt by Lemma 9.10(3)

6. n(σ σt) = Card(S(σ σt)) Definition 11.16(2)

= Card(S(σ)− αt)



= Card(S(σ))− 1 step 3

= n(σ)− 1. Definition 11.16(2)

7. l(σ σt) = n(σ σt). step 1 and inductive hypothesis

8. l(σ) = l(σ σt) + 1 = n(σ σt) + 1 = n(σ). steps 1,5,6

Definition 11.18Let E be a Euclidean space. Let Φ be a root system of E. Let ∆ be a base of Φ. A simple rootis an element of ∆. A simple reflection has the form σα, α ∈ ∆.

Definition 11.19Let X be a topological space and D ⊂ X. Let G be a group with an action on X. We call Da fundamental domain for this action if for all x ∈ X, there exists a unique g ∈ G such thatgx ∈ D.

Example 11.20Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. Let ∆be a base of Φ. The closure C(∆) of the fundamental Weyl chamber C(∆) in E is a fundamentaldomain for the action of W on E, by [Hum78, p.55, Exercise 14].

Lemma 11.21Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. Let∆ be a base of Φ. Let λ, µ ∈ C(∆). If σ(λ) = µ for some σ ∈ W , then σ is the product of simplereflections which fix λ. In particular, λ = µ.

Proof. If l(σ) = 0, then σ = 1 and hence λ = µ. Now we suppose l(σ) > 0.


1. n(σ) = l(σ) > 0. Lemma 11.17

2. There exists α > 0 such that σ(α) < 0. S(σ) 6= ∅ by Definition 11.16(2)

3. There exists α ∈ ∆ such that σ(α) < 0. by Definition 10.4(B2)

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4. Suppose σ(α) = ∑β∈∆

kββ, where kβ ≤ 0

for all β ∈ ∆.

Definition 10.4(B2)

5.1. (µ, σ(α)) = ∑β∈∆

kβ(µ, β) ≤ 0. (µ, β) ≥ 0 for all β ∈ ∆

5.2. (µ, σ(α)) = (σ−1(µ), α) = (λ, α) ≥ 0. σ(λ) = µ, Theorem 11.13 and Lemma 9.8

6. (λ, α) = 0. step 5

7. σα(λ) = λ. Definition 9.7(1)

8.1. Bijection σα : S(σ σα) ' S(σ)− α. below

α ∈ S(σ) step 2

For all β ∈ S(σ σα), σα(β) ∈ S(σ). Definition 11.16(2)

For all β ∈ S(σ σα), σα(β) 6∈ α. Lemma 11.4

8.2. l(σσα) = n(σσα) = n(σ)−1 = l(σ)−1. Lemma 11.17

9. σ σα is the product of simple reflectionswhich fix λ.

σ σα(λ) = µ and Induction

10. σ is the product of simple reflections whichfix λ.

σ2α = 1 by Lemma 9.10(3)


12. May 14th, Irreducible root systems, two root lengths and Cartan matrix

Definition 12.1Let E be a Euclidean space. Let Φ be a root system of E. A subset S of Φ is reducible if(1) S = S1 ∪ S2 such that S1 ∩ S2 = ∅ and Si 6= ∅ for i ∈ 1, 2. We write S = S1 t S2.(2) (α, β) = 0 for all α ∈ S1 and β ∈ S2. In other words, (S1, S2) = 0.Otherwise it is called irreducible.

Example 12.2Root systems of types A1, A2, B2, G2 are irreducible. The root system of type A1×A1 is reducible.

Lemma 12.3Let E be a Euclidean space. For all 0 6= α, β ∈ E, if (α, β) = 0, then σα σβ = σβ σα.

Proof. Since (α, β) = 0, we have σα(β) = β and σβ(α) = α. For all γ ∈ E,


1. σα(σβ(γ)) = σα(γ − 〈γ, β〉β) Example 9.9(2)

= σα(γ)− 〈γ, β〉σα(β) σα is R-linear

= γ − 〈γ, α〉α− 〈γ, β〉β. Example 9.9(2) and σα(β) = β

2. σβ(σα(γ)) = γ − 〈γ, α〉α− 〈γ, β〉β.

Lemma 12.4Let E be a Euclidean space. Let Φ be a root system of E. Let ∆ be a base of Φ. Then Φ isreducible iff ∆ is reducible. Or equivalently, Φ is irreducible iff ∆ is irreducible.

Proof. Suppose Φ is reducible.


1. Suppose Φ = Φ1tΦ2 with (Φ1,Φ2) = 0.Let ∆i = ∆ ∩ Φi.

Definition 12.1

2. ∆2 = ∆ ∩ Φ2 6= ∅. below

2.1. If ∆ ⊂ Φ1, then (∆,Φ2) = 0. (Φ1,Φ2) = 0

2.2. (E,Φ2) = 0. Definition 10.4(B1)

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2.3. Φ2 = 0. (•, •) is non-degenerate since it is positivedefinite

2.4. A contradiction. to Definition 9.12(R1)

3. ∆1 = ∆ ∩ Φ1 6= ∅. Similar to step 2

4.1. ∆ = ∆ ∩ Φ ∆ ⊂ Φ

= ∆ ∩ (Φ1 ∪ Φ2) = (∆ ∩ Φ1) ∪ (∆ ∩ Φ2) Φ = Φ1 ∪ Φ2

= ∆1 ∪∆2. ∆i = ∆ ∩ Φi

4.2. ∆1 ∩∆2 = ∅ ∆1 ∩∆2 ⊂ Φ1 ∩ Φ2 = ∅

4.3. (∆1,∆2) = 0. (∆1,∆2) ⊂ (Φ1,Φ2) = 0

5. ∆ is reducible. step 4

Conversely, suppose ∆ is reducible.


1. Let ∆ = ∆1 t∆2 with (∆1,∆2) = 0. Definition 12.1

2. Φi = α ∈ Φ : ∃σ ∈ W , σ(α) ∈ ∆i, i ∈1, 2.

3.1. Φ = Φ1 ∪ Φ2. Theorem 11.12 and ∆ = ∆1 ∪∆2

3.2. Φ1 ∩ Φ2 = ∅. ∆1 ∩∆2 = ∅ and step 2

3.3. Φi 6= ∅ for i ∈ 1, 2. ∆i ⊂ Φi and ∆i 6= ∅

4.1. For all α ∈ Φ1, suppose σ(α) ∈ ∆1 forsome σ ∈ W .

step 2

4.2. Suppose σ−1 = σα1 · · ·σαs σβ1 · · ·σβt , αi ∈ ∆1, 1 ≤ i ≤ s, βj ∈ ∆2, 1 ≤j ≤ t.

Theorem 11.13, (∆1,∆2) = 0 andLemma 12.3

4.3. σβj(σ(α)) = σ(α). (σ(α), βj) = 0 since (∆1,∆2) = 0



4.4. Φ1 ⊂ SpanZ(∆1). α = σαs · · · σα1(σ(α)) ∈ SpanZ(∆1)

5. Φ2 ⊂ SpanR(∆2). Similar to step 4

6. (Φ1,Φ2) = 0. steps 4,5 and (∆1,∆2) = 0.

7. Φ is reducible. steps 3,6

Lemma 12.5Let E be a Euclidean space. Let Φ be an irreducible root system of E. Let ∆ be a base of Φ.Suppose Φ has a maximal root β. (1) β ∈ C(∆). (2) There exists α ∈ ∆ such that (α, β) > 0.

Proof. (1) If there exists α ∈ ∆ such that (α, β) < 0, then β +α ∈ Φ by Lemma 9.31(2). We haveβ < β + α, a contradiction to the maximality of β.(2) From (1), (α, β) ≥ 0 for all α ∈ ∆. If (α, β) = 0 for all α ∈ ∆, then (E, β) = 0 byDefinition 10.4(B1). Since (•, •) is non-degenerate, β = 0, a contradiction to Definition 9.12(R1).

Lemma 12.6Let E be a Euclidean space. Let Φ be an irreducible root system of E. Let ∆ be a base of Φ.Suppose Φ has a maximal root β. If β = ∑

γ∈∆kγγ, then kγ ≥ 1 for all γ ∈ ∆.

Proof. Let ∆1 = γ ∈ ∆ : kγ > 0 6= ∅ and ∆2 = γ ∈ ∆ : kγ = 0. Then ∆ = ∆1 t ∆2 byDefinition 10.4(2).


1. If ∆2 = ∅, then we are done.

2. Suppose ∆2 6= ∅. For all α ∈ ∆2, (α, γ) ≤0 for all γ ∈ ∆1.

Lemma 10.6(2)

3. (α, β) ≤ 0. (α, β) = ∑γ∈∆1

kγ(α, γ)

4. ∆ is irreducible. Φ is irreducible and Lemma 12.4

5. There exists γ′ ∈ ∆1 such that (α, γ′) 6= 0. Otherwise (∆1,∆2) = 0, a contradictionto Definition 12.1

136 ZHENGYAO WU


6. (α, γ′) < 0. step 3

7. (α, β) < 0. steps 4,7

8. A contradiction. to Lemma 12.5(1).

Lemma 12.7Let E be a Euclidean space. Let Φ be an irreducible root system of E. There exists a uniquemaximal root β relative to <. In particular, if α ∈ Φ+ and α 6= β, then ht(α) < ht(β).

Proof. Existence since Φ is finite by Definition 9.12(R1).Uniqueness. Let β′ be another maximal root.


1. There exists α ∈ ∆ such that (α, β) > 0. Lemma 12.5(2)

2. (β′, β) ≥ (α, β). β′ ≥ α and Lemma 12.6

3. (β′, β) > 0 and β 6= −β′. steps 1,2

4. β − β′ ∈ Φ. Lemma 9.31

5. If β − β′ ∈ Φ+, then a contradiction. to the maximality of β′

If β′ − β ∈ Φ+, then a contradiction. to the maximality of β

If α ∈ Φ+ and α 6= β, by the uniqueness of β, α < β. Then β − α > 0. Therefore ht(β − α) > 0,ht(α) < ht(β).

Example 12.8The maximal root of A2 is the black vector.


α−α

β

−β

α + β

−(α + β)

Example 12.9The maximal root of B2 is the black vector.

α−α

α + β

−(α + β)

β

−β

2α + β

−(2α + β)

Example 12.10The maximal root of G2 is the black vector.

α−α

3α + 2β

−(3α + 2β)

β

−β

3α + β

−(3α + β)

2α + β

−(2α + β)

α + β

−(α + β)

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Lemma 12.11Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. If Φ isirreducible, then(1) E cannot be decomposable into the orthogonal sum of two nonzero W -invariant subspaces.(2) The W -orbit of each root spans E.

Proof. We review [Hum78, Sec.9, Exercise 1]:“Let E ′ be a subspace of E. If σα(E ′) = E ′, then α ∈ E ′ or E ′ ⊂ Pα. ”(1) Suppose the opposite that E = E ′ + E ′′ with (E ′, E ′′) = 0, E ′ is W -invariant and E ′′ isW -invariant.


1. σα(E ′) = E ′ for all α ∈ Φ. E ′ is W -invariant

2. Either α ∈ E ′ or E ′ ⊂ Pα. [Hum78, Sec.9, Exercise 1]

3. Φ = (Φ ∩ E ′) t (Φ ∩ E ′′). Either α ∈ E ′ or α ∈ E ′′, but not both.

4. Φ = Φ ∩ E ′ or Φ = Φ ∩ E ′′. Φ is irreducible

Φ ⊂ E ′ or Φ ⊂ E ′′.

6. E = E ′ or E = E ′′. Φ spans E by Definition 9.12(R1)

(2) Suppose the opposite that E ′ = SpanR(W β) for some β ∈ Φ and E ′ ( E. Let E ′′ be theorthogonal complement of E ′.


1. For all α ∈ Φ, (E ′, σα(E ′′))

= (σα(E ′), σα(E ′′)) E ′ is W -invariant.

= (E ′, E ′′) = 0. Lemma 9.8

2. E ′′ is W -invariant. σα(E ′′) ⊂ E ′′ for all α ∈ Φ and σ2α = 1

3. A contradiction. to (1)


Lemma 12.12Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. If Φ isirreducible, then at most two root lengths occur in Φ.

Proof. Suppose α, β ∈ Φ.


1. The W -orbit of α spans E. Lemma 12.11

2. (σ(α), β) 6= 0 for some σ ∈ W . below

2.1. Otherwise (σ(α), β) = 0 for all σ ∈ W ,(E, β) = 0.

step 1

2.2. β = 0. (•, •) is non-degenerate

2.3. A contradiction. to Definition 9.12(R1)

3. We may assume that (α, β) 6= 0 and‖β‖ ≥ ‖α‖.

‖α‖ = ‖σ(α)‖ for all simple reflections σand Theorem 11.13

4. ‖β‖2/‖α‖2 ∈ 1, 2, 3. Lemma 9.30

5. If there are ≥ 3 root lengths squares a <b < c, then c

a=√

3, ba

= cb

=√

2.

ca> b

a> 1 and c

a> c

b> 1.

6. ca

= cb· ba

=√

2 ·√

2 = 2, a contradiction. to ca

=√

3

Definition 12.13If an irreducible root system has two root length, then we call them long or short.

Lemma 12.14Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. If Φ isirreducible, then W acts transitively on all roots of a given length.

Proof. Suppose α, β ∈ Φ with ‖α‖ = ‖β‖.


1. The W -orbit of α spans E. Lemma 12.11

140 ZHENGYAO WU


2. (σ(α), β) 6= 0 for some σ ∈ W . as in step 2 of Lemma 12.12

3. If σ(α) = β, then we are done.

4. Suppose β 6= σ(α).

‖β‖ = ‖σ(α)‖. ‖α‖ = ‖σ(α)‖ as in step 3 of Lemma 12.12

5. 〈β, σ(α)〉 = ±1. Lemma 9.30

6. If 〈β, σ(α)〉 = −1, then 〈σβ(β), σ(α)〉 =〈−β, σ(α)〉 = 1.

Lemma 9.11 and Lemma 9.10(2)

7. We may assume that 〈β, α〉 = 1. σβ, σ ∈ W

8. 〈α, β〉 = 1. Lemma 9.30

9. σα(β) = β − α and σβ(α) = α− β. Example 9.9

10. (σα σβ σα)(β) = (σα σβ)(β − α) step 9

= σα(−β − (α− β)) Lemma 9.11 and step 9

= σα(−α) = α. Lemma 9.11

Lemma 12.15Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. If Φ isirreducible with two root lengths, then its maximal root β is long.Proof.


1. For each α ∈ Φ there exists σ ∈ W suchthat σ(α) ∈ C(∆).

Example 11.20

2. β − σ(α) ∈ Φ+. Lemma 12.7

3. (γ, β − σ(α)) ≥ 0 for all γ ∈ C(∆). defn of C(∆)

4. β ∈ C(∆). Lemma 12.5



5. (β, β) > (β, σ(α)). Take γ = β in step 3, and step 4

6. (σ(α), β) > (σ(α), σ(α)) = (α, α). Take γ = σ(α) in step 3, and step 4

7. (β, β) ≥ (α, α) and β is long. steps 5,6

Definition 12.16Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. Let∆ = α1, . . . , αl be a base of Φ. The Cartan matrix of Φ is (〈αi, αj〉)1≤i,j≤l. Its entries arecalled Cartan integers.

Example 12.17The Cartan matrix of A1 × A1 is 2 0

0 2

,relative to α1 = (x, 0), x > 0, α2 = (0, y), y > 0.

Example 12.18The Cartan matrix of A2 is 2 −1

−1 2

,

relative to α1 = (1, 0), α2 = (−12 ,√

32 ).

Example 12.19The Cartan matrix of B2 is 2 −2

−1 2

,relative to α1 = (−1, 1), α2 = (1, 0).

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Example 12.20The Cartan matrix of G2 is 2 −1

−3 2

,

relative to α1 = (1, 0), α2 = (−32 ,√

32 ).

Lemma 12.21The Cartan matrix is independent of the choice of ∆ (but depend on the chosen ordering).

Proof. By Lemma 11.1, W acts transitively on bases. Since (σ(α), σ(β)) = (α, β) for all reflectionsσ and α, β ∈ E. We have 〈αi, αj〉 = 〈σ(αi), σ(αj)〉 for all σ ∈ W and 1 ≤ i, j ≤ l.

Lemma 12.22The Cartan matrix is nonsingular.Proof.


1. det (〈αi, αj〉)1≤i,j≤l = det(


)1≤i,j≤l

= 2ll∏

j=1(αj, αj)

det((αi, αj)1≤i,j≤l) 6= 0 (•, •) is non-degenerate

Lemma 12.23Let E be a Euclidean space. Let Φ,Φ′ be root systems of E with bases ∆,∆′, respectively.Every bijection ∆→ ∆′, αi 7→ α′i such that 〈αi, αj〉 = 〈α′i, α′j〉 for all 1 ≤ i, j ≤ l extends uniquelyto an isomorphism φ : E → E ′ such that(1) φ(Φ) = Φ′;(2) 〈φ(α), φ(β)〉 = 〈α, β〉 for all α, β ∈ Φ;(3) The Cartan matrix determines Φ up to isomorphism.

Proof. By Definition 10.4(B1), there exists a unique vector space isomorphism φ : E → E ′,φ(

l∑i=1

xiαi) =l∑

i=1xiα

′i, xi ∈ R, 1 ≤ i ≤ l.

(1)



1. For all β′ ∈ Φ′, β′ =l∑

i=1xiα

′i for xi ∈ Z with

xi ≥ 0 for all i or xi ≤ 0 for all i.Definition 10.4(B2)

2. Let β =l∑

i=1xiαi. Then φ(β) = β′. defn of φ

(2)


1. 〈φ(α), φ(β)〉 = 〈α, β〉 for all α ∈ Φ andβ ∈ ∆.

〈•, •〉 is linear on the first coordinate byLemma 9.10(2)

2. For all β ∈ Φ, there exists σ ∈ W suchthat σ(β) ∈ ∆.

Theorem 11.12

3. We induct on l(σ). Theorem 11.13

4. When l(σ) = 0, we are done. σ = 1, β ∈ ∆ and step 1

5. When l(σ) = 1, σ = σγ for some γ ∈ ∆. defn of l(σ)

6. φ σγ = σφ(γ) φ step 1

7. 〈φ(σγ(α)), φ(σγ(β))〉

= 〈σφ(γ)(φ(α)), σφ(γ)(φ(β))〉 step 5

= 〈φ(α), φ(β)〉 Lemma 9.17

8. 〈φ(σγ(α)), φ(σγ(β))〉

= 〈σγ(α), σγ(β)〉 σγ(β) ∈ ∆ and step 1

= 〈α, β〉 Lemma 9.17

9. 〈φ(α), φ(β)〉 = 〈α, β〉. steps 7,8

10. Suppose l(σ) > 1 and σ = τ σγ for someγ ∈ ∆ and τ ∈ W with l(τ) = l(σ)− 1.

Definition 11.16

11. 〈φ(σγ(α)), φ(σγ(β))〉 = 〈σγ(α)), σγ(β)〉 τ(σγ(β)) ∈ ∆ and inductive hypothesis

= 〈α, β〉 Lemma 9.17

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12. 〈φ(α), φ(β)〉 = 〈α, β〉. steps 7,11

(3) Uniqueness: by (1)(2) and Definition 9.18.Existence: Given ∆, we construct Φ.

• Roots of height 1 are precisely elements of ∆.• For αi 6= αj in ∆, αi − αj 6∈ Φ. By step 5 of the proof of Lemma 10.3, r = 0 andq = −〈αi, αj〉, we have the αj-string through αi. Repeat this for all pairs of simple roots,we obtain all roots of height 2, and more.• Let γ be a root of height 2 and αj ∈ ∆. Suppose i ∈ Z : γ + iαj ∈ Φ = [−r, q] ∩ Z.Then r ∈ 0, 1 (since ht(γ) = 2 and 2αj 6∈ Φ) and r − q = 〈γ, αj〉 by step 5 of the proofof Lemma 10.3. We obtain all roots of height 3.• By Theorem 11.13, we obtain all positive roots inductively.• Use symmetry to get all negative roots.

Example 12.24

The Cartan matrix of A1 × A1 is

2 0

0 2

. Determine A1 × A1.

• The roots of height one are α1, α2.• Since 〈α1, α2〉 = 0 = r− q and r = 0, q = 0. There are no roots of height two (and above).• All of positive roots are α1, α2.• All of negative roots are −α1,−α2.

Example 12.25

The Cartan matrix of A2 is

2 −1

−1 2

,. Determine A2.

• The roots of height one are α1, α2.• Since 〈α1, α2〉 = −1 = r − q and r = 0, q = 1. The α2-string through α1 is α1, α1 + α2.The only root of height two is α1 + α2.• Since 〈α1 + α2, α2〉 = −1 + 2 = 1 = r− q and r = 1, q = 0. The α2-string through α1 + α2

is α1, α1 + α2. Similarly, the α1-string through α1 + α2 is α2, α1 + α2. There are noroots of height three (and above).• All of positive roots are α1, α2, α1 + α2.• All of negative roots are −α1,−α2,−α1 − α2.


Example 12.26

The Cartan matrix of B2 is

2 −2

−1 2

. Determine B2.

• The roots of height one are α1, α2.• The only root of height two is α1 + α2.

– Since 〈α1, α2〉 = −2 = r − q and r = 0, q = 2. The α2-string through α1 is α1, α1 +α2, α1 + 2α2.

– Since 〈α1, α2〉 = −1 = r−q and r = 0, q = 1. The α1-string through α2 is α2, α1+α2.• The only root of height three is α1 + 2α2.

– Since 〈α1 + α2, α2〉 = −2 + 2 = 0 = r − q and r = 1, q = 1. The α2-string throughα1 + α2 is α1, α1 + α2, α1 + 2α2.

– Since 〈α1 + α2, α1〉 = 2 − 1 = 1 = r − q and r = 1, q = 0. The α1-string throughα1 + α2 is α2, α1 + α2.

• There is no root of height four and above.– Since 〈α1 + 2α2, α2〉 = −2 + 2 ∗ 2 = 2 = r− q and r = 2, q = 0. The α2-string throughα1 + 2α2 is α1, α1 + α2, α1 + 2α2.

– Since 〈α1 + 2α2, α1〉 = 2 + 2 ∗ (−1) = 0 = r − q and r = 0, q = 0. The α1-stringthrough α1 + 2α2 is α1 + 2α2.

• All of positive roots are α1, α2, α1 + α2, α1 + 2α2.• All of negative roots are −α1,−α2,−α1 − α2,−α1 − 2α2.

Example 12.27

The Cartan matrix of G2 is

2 −1

−3 2

, Determine G2.

• The roots of height one are α1, α2.• The only root of height two is α1 + α2.

– Since 〈α1, α2〉 = −1 = r−q and r = 0, q = 1. The α2-string through α1 is α1, α1+α2.– Since 〈α1, α2〉 = −3 = r − q and r = 0, q = 3. The α1-string through α2 is α2, α1 +α2, 2α1 + α2, 3α1 + α2.

• The only root of height three is 2α1 + α2.– Since 〈α1 + α2, α2〉 = −1 + 2 = 1 = r − q and r = 1, q = 0. The α2-string throughα1 + α2 is α1, α1 + α2.

– Since 〈α1 + α2, α1〉 = 2 − 3 = −1 = r − q and r = 1, q = 2. The α1-string throughα1 + α2 is α2, α1 + α2, 2α1 + α2, 3α1 + α2.

• The only root of height four is 3α1 + α2.– Since 〈2α1 + α2, α2〉 = 2 ∗ (−1) + 2 = 0 = r − q and r = 0, q = 0. The α2-stringthrough 2α1 + α2 is 2α1 + α2.

146 ZHENGYAO WU

– Since 〈2α1 + α2, α1〉 = 2 ∗ 2− 3 = 1 = r − q and r = 2, q = 1. The α1-string through2α1 + α2 is α2, α1 + α2, 2α1 + α2, 3α1 + α2.

• The only root of height five is 3α1 + 2α2.– Since 〈3α1 + α2, α2〉 = 3 ∗ (−1) + 2 = −1 = r − q and r = 0, q = 1. The α2-stringthrough 3α1 + α2 is 3α1 + α2, 3α1 + 2α2.

– Since 〈3α1 + α2, α1〉 = 3 ∗ 2− 3 = 3 = r − q and r = 3, q = 0. The α1-string through3α1 + α2 is α2, α1 + α2, 2α1 + α2, 3α1 + α2.

• There are no root of height six (and above).– Since 〈3α1 + 2α2, α2〉 = 3 ∗ (−1) + 2 ∗ 2 = 1 = r − q and r = 1, q = 0. The α2-stringthrough 3α1 + 2α2 is 3α1 + α2, 3α1 + 2α2.

– Since 〈3α1 + 2α2, α1〉 = 3 ∗ 2 + 2 ∗ (−3) = 0 = r − q and r = 0, q = 0. The α1-stringthrough 3α1 + 2α2 is 3α1 + 2α2.

• All of positive roots are α1, α2, α1 + α2, 2α1 + α2, 3α1 + α2, 3α1 + 2α2.• All of negative roots are −α1,−α2,−α1 − α2,−2α1 − α2,−3α1 − α2,−3α1 − 2α2.


13. May 21th, Coxeter graphs, Dynkin diagrams

Definition 13.1Let E be a Euclidean space. Let Φ be a root system of E with base ∆ = α1, . . . , αl. TheCoxeter graph of Φ has

• l vertices, the i-th vertex corresponds αi.• i-th and j-th vertices are joined by 〈αi, αj〉〈αj, αi〉 ∈ 0, 1, 2, 3 edges (see Lemma 9.30).

Example 13.2The Coxeter graph of A1 × A1 is

Example 13.3The Coxeter graph of A2 is

Example 13.4The Coxeter graph of B2 is

Example 13.5The Coxeter graph of G2 is

Lemma 13.6(1) If all roots have the same length, then the Coxeter graph determines the Cartan matrix.(2) If there are two root lengths, then the Coxeter graph cannot determine the Cartan matrix.

Proof. (1) Between the i-th and j-th vertex, if there are no edges, then 〈αi, αj〉 = 〈αj, αi〉 = 0; Ifthere is one edge, then 〈αi, αj〉 = 〈αj, αi〉 = −1 by Lemma 9.30.(2) If there are n ≥ 2 edges, then (〈αi, αj〉 = −1 and 〈αj, αi〉 = −n) or (〈αi, αj〉 = −n and〈αj, αi〉 = −1) by Lemma 9.30.

Definition 13.7Let E be a Euclidean space. Let Φ be a root system of E. On the Coxeter graph of Φ. On eachdouble or triple edge, we add an arrow from the long root to the short root, then we obtain theDynkin diagram of Φ.

Example 13.8

The Dynkin diagram of B2 with base α1 = (−12 ,√

32 ), α2 = (1, 0) is

1 2

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Example 13.9

The Dynkin diagram of G2 with base α1 = (1, 0), α2 = (−32 ,√

32 ) is

1 2

Lemma 13.10Let E be a Euclidean space. Let Φ be a root system of E. The Dynkin diagram of Φ determinesits Cartan matrix.

Proof. Suppose ‖αi‖ ≥ ‖αj‖. Between the i-th and j-th vertex, by Lemma 9.30,• There is no edge iff 〈αi, αj〉 = 〈αj, αi〉 = 0;• There is one edge iff 〈αi, αj〉 = 〈αj, αi〉 = −1;• If there are n ∈ 2, 3 edges, then 〈αi, αj〉 = −n, 〈αj, αi〉 = −1.

Example 13.11The Dynkin diagram of F4 is defined to be

1 2 3 4

By Lemma 13.10, we have the Cartan matrix

2 −1 0 0

−1 2 −2 0

0 −1 2 −1

0 0 −1 2

Proposition 13.12Let E be a Euclidean space. Let Φ be a root system of E. Let ∆ = α1, . . . , αl be a base of Φ.Let W be the Weyl group of Φ. There exist unique Φi, Ei such that(1) Φi is an irreducible root system in Ei(2) E ' E1 ⊕ · · · ⊕ Et is an orthogonal sum where each Ei is W -invariant.(3) Φ = Φ1 t · · · t Φt is a partition.

Proof. Existence: There is no edge between the i-th and the j-th vertices iff 〈αi, αj〉 = 〈αj, αi〉 = 0.Let ∆ = ∆1 t · · · t∆t, (∆i,∆j) = 0 for all i 6= j, where the Coxeter graphs of ∆i are connectedcomponents of the Coxeter graph of Φ. Let Ei = SpanR(∆i). Let Φi = SpanZ(∆i) ∩ Φ. ByDefinition 9.12, Φi is a root system in Ei.Uniqueness follows from uniqueness of connected components.(1) A root system is irreducible iff its Coxeter graph is connected. Since the Coxeter graph of each∆i is connected, ∆i is irreducible. By Lemma 12.4, Φi is irreducible.


(2)


1. E ' E1 ⊕ · · · ⊕ Et. (∆i,∆j) = 0

2. W is generated by σα(α ∈ ∆). Theorem 11.13

3. If β ∈ ∆i, then σβ(Ei) = Ei. Definition 9.12(R1)(R3)

4. If β ∈ ∆−∆i, then Ei ⊂ Pβ. [Hum78, p.45, Exercise 1]

σβ|Ei = 1.

(3) We have Φ = Φ1 ∪ · · · ∪ Φt by Φ ⊂ SpanZ(∆) Lemma 10.10 and defn of Φi. Also, Φi ∩ Φj = ∅for all i 6= j by (2).

Definition 13.13Let E be a Euclidean space. Let A = ε1, . . . , εn be a subset of unit vectors of E. We call Aadmissible if

(1) A is linearly independent.(2) (εi, εj) ≤ 0 for all i 6= j.(3) 4(εi, εj)2 ∈ 0, 1, 2, 3 for all i 6= j.

Example 13.14

If ∆ = α1, . . . , αl is a base of a root system, then A =α1

‖α1‖, . . . ,

αl‖α1‖

is admissible.

Proof. Let εi = αi‖αi‖

. Then ‖εi‖ = 1 for all 1 ≤ i ≤ l.


1. ∆ is linearly independent. Then A is lin-early independent.

Definition 10.4(B1)

2. (αi, αj) ≤ 0 for all i 6= j. Then (εi, εj) ≤ 0for all i 6= j.

Lemma 10.6

3. 4(εi, εj)2 = 〈αi, αj〉〈αj, αi〉〈αi, αj〉 = 2(αi, αj)(αj, αj)

=

2(εi‖αi‖, εj‖αj‖)‖αj‖2 = 2‖αi‖

‖αj‖(εi, εj)

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∈ 0, 1, 2, 3 for all i 6= j. Lemma 9.30

4. A is admissible. steps 1,2,3 and Definition 13.13

Definition 13.15Let E be a Euclidean space. Let A = ε1, . . . , εn be an admissible set. Its Coxeter graph Γ has

• n vertices, the i-th vertex corresponds εi,• the i-th vertex and the j-th vertex are joined by 4(εi, εj)2 ∈ 0, 1, 2, 3 edges.

Remark 13.16(1) Any nonempty subset of an admissible set is admissible.(2) The Coxeter graph of an admissible set generalizes the Coxeter graph of a root system.(3) The Coxeter graph of an admissible subset is obtained by deleting corresponding vertices andall incident edges.

Lemma 13.17Let E be a Euclidean space. Let A = ε1, . . . , εn be an admissible set. Let Γ be the Coxetergraph of A. The number of edges of Γ (without multiplicity) is N ≤ n− 1.

Proof. Let ε =n∑i=1

εi.


1. ε 6= 0. A is linearly independent by Defini-tion 13.13(1)

2. For i 6= j, if there exists at least one edgebetween the i-th and the j-th vertex, then2(εi, εj) ∈ −1,−

√2,−√

3.

Definition 13.13(2)(3)

2(εi, εj) ≤ −1.

3. 0 < (ε, ε) (•, •) is positive definite

= (n∑i=1

εi,n∑i=1

εi) defn of ε

=n∑i=1

(εi, εi) + 2 ∑1≤i<j≤n

(εi, εj) (•, •) is bilinear



≤ n−N εi are units and step 2

4. N ≤ n− 1.

Lemma 13.18Let E be a Euclidean space. Let A = ε1, . . . , εn be an admissible set. Let Γ be the Coxetergraph of A. Then Γ does not contain cycles.

Proof. If Γ has a cycle with n vertices and n edges for some n ≥ 2, then it contradicts Lemma 13.17,because the circle is the Coxeter graph of a subset B ⊂ A, where B is admissible by Re-mark 13.16(1).

Lemma 13.19Let E be a Euclidean space. Let A be an admissible set. Let Γ be the Coxeter graph of A. Letε ∈ A. Let η1, . . . , ηk correspond vertices of A connected to the vertex of ε. Then there are nomore than 3 edges (an edge with multiplicity n is counted as n edges) can originate at ε.Proof.


1. There exists a unit vector η0 ∈SpanRε, η1, . . . , ηk such that (η0, ηi) = 0for all 1 ≤ i ≤ k.

dimR SpanRε, η1, . . . , ηk = k + 1 > k

2. (ε, η0) 6= 0. ε, η1, . . . , ηk is linearly independent as asubset of A

3. ε =k∑i=0

(ε, ηi)ηi. η0, η1, . . . , ηk is a basis ofSpanRε, η1, . . . , ηk

4. (ηi, ηj) = 0 for all i 6= j. Lemma 13.18

5. 1 = (ε, ε) ε ∈ A

= (ε,k∑i=0

(ε, ηi)ηi) step 3

=k∑i=0

(ε, ηi)2 >k∑i=1

(ε, ηi)2. (•, •) is bilinear

152 ZHENGYAO WU


6. The number of edges connected to ε isk∑i=1

4(ε, ηi)2 < 4.Definition 13.15

Corollary 13.20The only connected Coxeter graph with a triple edge is

Proof. Since Lemma 13.19.

Lemma 13.21Let E be a Euclidean space. Let A = ε1, . . . , εn be an admissible set. Let Γ be the Coxetergraph of A. Suppose ε1, . . . , εk ⊂ A has Coxeter graph

· · ·

Then A′ = (A− ε1, . . . , εk) ∪ ε is admissible for ε =k∑i=1

εi.Proof.


1. A′ is linearly independent. ε =k∑i=1

εi and A is linearly independent byDefinition 13.13(1)

2. (ε, ε) = (k∑i=1

εi,k∑i=1

εi) defn of ε

=k∑i=1

(εi, εi) + 2 ∑1≤i<j≤k

(εi, εj) (•, •) is bilinear

= k − (k − 1) = 1. εi are units and 2(εi, εj) =1, |i− j| = 1;

0, |i− j| > 1.

3. Any η ∈ A−ε1, . . . , εl can be connectedto at most one of ε1, . . . , εl.

Lemma 13.18

(η, ε) ∈ 0, (η, εi) for some 1 ≤ i ≤ k.

4. (η, ε) ≤ 0 for all η ∈ A− ε1, . . . , εl. step 3 and Definition 13.13(2)



5. 4(η, ε)2 ∈ 0, 1, 2, 3. step 3 and Definition 13.13(3)

6. A′ is admissible. steps 1,4,5 and Definition 13.13

Lemma 13.22Let E be a Euclidean space. Let A = ε1, . . . , εn be an admissible set. Let Γ be the Coxetergraph of A. Then Γ does not contain subgraph of the following forms

· · ·

· · ·

· · ·

Proof. Suppose the opposite. By Lemma 13.21, Γ contains one of the following subgraph

, ,

a contradiction to Lemma 13.19.

Lemma 13.23Let E be a Euclidean space. Let A = ε1, . . . , εn be an admissible set. Let Γ be the Coxetergraph of A. If Γ is connected and admissible, then it is one of the following

,· · · ,

· · ·

···

···

.· · · · · · ,

Proof. The Dynkin diagram is a tree by Lemma 13.18.• If Γ contains a triple edge, then it is the first graph by Corollary 13.20.

154 ZHENGYAO WU

• Suppose Γ has no triple edges. For a vertex is connected to exactly three other distinctvertices, we call the vertex a lem-node. A node is either

, or .– If Γ has no nodes, then it is of the type of the second graph.– If Γ has only one node, then it is of the type of the third or the fourth graph.– If Γ has at least two nodes, then it is of the type of the fourth graph by Lemma 13.22.

Lemma 13.24Let E be a Euclidean space. Let A be an admissible set. Let Γ be the Coxeter graph of A. If Γhas type

ε1 ε2· · ·

εp−1 εp ηq ηq−1· · ·

η2 η1

Then it is one of the following· · · , or .

Proof. Let ε =p∑i=1

iεi and η =q∑j=1

jηj.


1. (ε, ε) = (p∑i=1

iεi,p∑i=1

iεi) defn of ε

=p∑i=1

i2(εi, εi) + 2 ∑1≤i<j≤p

ij(εi, εj) (•, •) is symmetric bilinear

=p∑i=1

i2 −p−1∑i=1

i(i+ 1) εi are units and (εi, εj) = −1 iff |i−j| = 1

= p2 +p−1∑i=1

i = p2 − p(p− 1)2 = p(p+ 1)

2 .

2. (η, η) = q(q + 1)2 . similar to step 1

3. (ε, η) = (p∑i=1

iεi, η =q∑j=1

jηj) defn of ε and η

= (pεp, qηq) (εi, ηj) = 0 for i < p or j < q

= pq(εp, ηq) = −√

22 pq. 4(εp, ηq)2 = 2 and Definition 13.13(2)

4. (ε, ε)(η, η) ≥ (ε, η)2 The Cauchy-Schwartz inequality

(ε, ε)(η, η) > (ε, η)2 ε, η are linearly independent

5. p(p+ 1)2 · q(q + 1)

2 >p2q2

2(p+ 1)(q + 1) > 2pq. canceling pq4



(p− 1)(q − 1) ≤ 1. pq − p− q < 1

6. If p = 1 or q = 1, then the first graph.

If p ≥ 2 and q ≥ 2, then the second graph. p = q = 2 by step 5

Lemma 13.25Find positive integer solutions to 1

p+ 1q

+ 1r> 1, p ≥ q ≥ r ≥ 1. .

Proof.


1. If r = 1, then (p, q, 1) is a solution. 1p

+ 1q

+ 1r

= 1p

+ 1q

+ 1 > 1

2.1 If r ≥ 2, then 1p≤ 1q≤ 1r≤ 1

2 .

2.2. r ≤ 2. 1 < 1p

+ 1q

+ 1r≤ 3r.

2.3. If r ≥ 2, then r = 2. steps 2.1, 2.2

3. 1p

+ 1q>

12 .

1p

+ 1q

+ 12 < 1

4. If q = 2, then (p, 2, 2) is a solution. 1p

+ 1q

+ 1r

= 1p

+ 1 > 1

5.1. If q ≥ 3, then 1p≤ 1q≤ 1

3 .

5.2. q ≤ 3. 12 <

1p

+ 1q≤ 2q

5.3. If q ≥ 3, then q = 3. steps 5.1, 5.2

6.1 If r ≥ 2 and q ≥ 3, then p ≥ 3.

6.2 If r ≥ 2, then p < 6. 12 <

1p

+ 13 by step 3

6.3 We have solutions (3, 3, 2), (4, 3, 2) and(5, 3, 2).

If r ≥ 2 and q ≥ 3, then p ∈ 3, 4, 5

7. Solutions are (p, q, r) ∈(p, q, 1), (p, 2, 2), (3, 3, 2), (4, 3, 2), (5, 3, 2).

steps 1, 4, 6.3

156 ZHENGYAO WU

Lemma 13.26Let E be a Euclidean space. Let A be an admissible set. Let Γ be the Coxeter graph of A. If Γhas type

ε1 ε2· · ·

εp−1

···

···

ψ ζr−1

ηq−1

ζ2

η2

ζ1

η1 for p, q, r ≥ 2,then it is one of the following

ε1 ε2· · ·

εp−1

ψ ζ1

η1 where p ≥ 2,

ε1 ε2

ψ ζ1

η1η2 ,

ε1 ε2 ε3

ψ ζ1

η1η2 ,

ε1 ε2 ε3 ε4

ψ ζ1

η1η2 .


14. May 28th, Classification, irreducible root systems of types A, B and C

Proof. Suppose ε =p−1∑i=1

iεi, η =q−1∑i=1

iηi and ζ =r−1∑i=1

iζi. Let θ1, θ2, θ3 be the angle between ψ andε, η, ζ, respectively.


1. ε, η, ζ are pariwise orthogonal. εi, ηj, ζk are pariwise orthogonal for all 1 ≤i ≤ p−1, 1 ≤ j ≤ q−1 and 1 ≤ k ≤ r−1.

2. There exists a unit vector ψ′ ∈SpanRε, η, ζ, ψ such that (ψ′, ε) =(ψ′, η) = (ψ′, ζ) = 0 and (ψ, ψ′) 6= 0.

Similar to steps 1,2 of the proof ofLemma 13.19

4. ψ = (ψ, ε) ε

‖ε‖2 +(ψ, η) η

‖η‖2 +(ψ, ζ) ζ

‖ζ‖2 +

(ψ, ψ′)ψ′.

ε

‖ε‖2 ,η

‖η‖2 ,ζ

‖ζ‖2 , ψ′ is orthonormal by

steps 1,3

5. 1 = (ψ, ψ) ψ is a unit

= (ψ, ε)2

‖ε‖2 + (ψ, η)2

‖η‖2 + (ψ, ζ)2

‖ζ‖2 + (ψ, ψ′)2 (•, •) is bilinear and step 4

>(ψ, ε)2

‖ε‖2 + (ψ, η)2

‖η‖2 + (ψ, ζ)2

‖ζ‖2 (ψ, ψ′) 6= 0 by steps 2,3

= cos2(θ1) + cos2(θ2) + cos2(θ3). ψ is a unit and defn of (•, •)

6. (ε, ε) = p(p− 1)2 , (η, η) =

q(q − 1)2 , (ζ, ζ) = r(r − 1)

2 .

similar to step 1 of the proof ofLemma 13.24

7. cos2(θ1) = (ε, ψ)2

(ε, ε)(ψ, ψ) defn of (•, •)

= ((p− 1)εp−1, ψ)2

(ε, ε)(ψ, ψ) (εi, ψ) = 0 for all 1 ≤ i ≤ p− 2.

=(p− 1)2(−1

2)2

p(p−1)2 · 1

= 12(1− 1

p). 2(εp−1, ψ) = −1, ψ is a unit and step 6

8. cos2(θ2) = 12(1− 1

q), cos2(θ3) = 1

2(1− 1r

). similar to step 6

9. 12(1− 1

p) + 1

2(1− 1q

) + 12(1− 1

r) < 1. steps 5,7,8

10. The following are the only possibilities. Lemma 13.25

(p, q, 1) is not a solution it does not have a node.

158 ZHENGYAO WU


Solutions (p, 2, 2), p ≥ 2 has the firstgraph; Solutions (3, 3, 2), (4, 3, 2) and(5, 3, 2) has the second, third and fourthgraph, respectively.

Theorem 14.1Let E be a Euclidean space. Let Φ be an irreducible root system of rank l. Let Γ be the Coxetergraph of Φ. The Dynkin diagram of Γ is one of the following

α1 α2 α3 α`−1 α`· · ·A`(` ≥ 1)

α1 α2 α3 α`−1 α`· · ·B`(` ≥ 2)

α1 α2 α3 α`−1 α`· · ·C`(` ≥ 3)

α1 α2 α3 α`−2α`

α`−1· · ·D`(` ≥ 4)

α1 α3 α4

α2

α5 α6E6

α1 α3 α4

α2

α5 α6 α7E7

α1 α3 α4

α2

α5 α6 α7 α8E8

α1 α2 α3 α4F4

α1 α2G2

Proof. Let Dyn(Φ) be the Dynkin diagram of Φ.



1. Dyn(Φ) is one of Al(l ≥ 1) or G2, or hasa double edge or a node.

Lemma 13.23

2. If has a double edge, then it is F4 or Bl(l ≥2) or Cl(l ≥ 3).

Lemma 13.24

3. If Dyn(Φ) has a node, then it is Dl(l ≥ 4)or E6 or E7 or E8.

Lemma 13.26

Theorem 14.2Let E be a Euclidean space. Let Φ be an irreducible root system of rank l. Then the Cartanmatrix of Φ is one of the following:

Al, (l ≥ 1) :

2 −1 0 · · · 0 0

−1 2 −1 · · · 0 0

0 −1 2 · · · 0 0. . .

0 0 0 · · · 2 −1

0 0 0 · · · −1 2

, Bl, (l ≥ 2) :

2 −1 · · · 0 0 0

−1 2 · · · 0 0 0. . .

0 0 · · · 2 −1 0

0 0 · · · −1 2 −2

0 0 · · · 0 −1 2

Cl, (l ≥ 3) :

2 −1 · · · 0 0 0

−1 2 · · · 0 0 0. . .

0 0 · · · 2 −1 0

0 0 · · · −1 2 −1

0 0 · · · 0 −2 2

, Dl, (l ≥ 4) :

2 −1 · · · 0 0 0 0

−1 2 · · · 0 0 0 0. . .

0 0 · · · 2 −1 0 0

0 0 · · · −1 2 −1 −1

0 0 · · · 0 −1 2 0

0 0 · · · 0 −1 0 2

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E6 :

2 0 −1 0 0 0

0 2 0 −1 0 0

−1 0 2 −1 0 0

0 −1 −1 2 −1 0

0 0 0 −1 2 −1

0 0 0 0 −1 2

, E7 :

2 0 −1 0 0 0 0

0 2 0 −1 0 0 0

−1 0 2 −1 0 0 0

0 −1 −1 2 −1 0 0

0 0 0 −1 2 −1 0

0 0 0 0 −1 2 −1

0 0 0 0 0 −1 2

E8 :

2 0 −1 0 0 0 0 0

0 2 0 −1 0 0 0 0

−1 0 2 −1 0 0 0 0

0 −1 −1 2 −1 0 0 0

0 0 0 −1 2 −1 0 0

0 0 0 0 −1 2 −1 0

0 0 0 0 0 −1 2 −1

0 0 0 0 0 0 −1 2

, F4 :

2 −1 0 0

−1 2 −2 0

0 −1 2 −1

0 0 −1 2

, G2 :

2 −1

−3 2

.

Proof. It follows from Lemma 13.6 and Theorem 14.1.

Definition 14.3A lattice in Rn is a free Z-module of rank n.

Example 14.4Let ε1, . . . , εn be an orthonormal basis of Rn. Let I = SpanZεi : 1 ≤ i ≤ n. Then I is a lattice.

Lemma 14.5Let ε1, . . . , εn be an orthonormal basis of Rn. Let I = SpanZεi : 1 ≤ i ≤ n. If x ∈ I and(x, x) = 1, then x = ±εi for some 1 ≤ i ≤ n.Proof.


1. Suppose x = x1ε1 + · · ·+ xnεn, xi ∈ Z forall 1 ≤ i ≤ n.

x ∈ I



2. x21 + · · ·+ x2

n = 1. (x, x) = 1

3. x2i = 1 for some i and x2

j = 0 for all j 6= i. x2i ∈ Z≥0 for all i.

x = ±εi.

Lemma 14.6Let ε1, . . . , εn be an orthonormal basis of Rn. Let I = SpanZεi : 1 ≤ i ≤ n. If x ∈ I and(x, x) = 2, then x = ±(εi ± εj), i 6= j.Proof.


1. Suppose x = x1ε1 + · · ·+ xnεn, xi ∈ Z forall 1 ≤ i ≤ n.

x ∈ I

2. x21 + · · ·+ x2

n = 2. (x, x) = 2

3. x2i = x2

j = 1 for some i 6= j and x2k = 0 for

all k 6∈ i, j.x2i ∈ Z≥0 for all i.

x = ±(εi ± εj), i 6= j.

Lemma 14.7Let n ∈ Z. Then n2 + n = 0 or n2 + n ≥ 2.Proof.


1. If n ∈ −1, 0, then n2 + n = 0.

2. If n = 1, then n2 + n = 2.

3. If |n| ≥ 2, then n2 + n ≥ 2. n2 + n = |n|(|n| ± 1) ≥ 2 · (2− 1) = 2.

162 ZHENGYAO WU

Lemma 14.8Let ε1, . . . , εn be an orthonormal basis of Rn. Let I = SpanZεi : 1 ≤ i ≤ n. Let Φ = x ∈I : (x, x) ∈ A for A = 1 or 2 or 1, 2. If Φ spans E, then Φ is a root system.

Proof. (R1)


1. Φ is finite. Lemma 14.5 and Lemma 14.6

2. Φ spans E. Given

3. 0 6∈ Φ. 0 6∈ A

(R2)


1. Suppose x, cx ∈ Φ, c ∈ R.

2.1. If A = 1 or 2, then ‖cx‖ = ‖x‖ ∈1,√

2.Card(A) = 1

2.2. |c| = 1, c ∈ ±1. ‖cx‖ = |c|‖x‖

3.1. If A = 1, 2 and ‖cx‖ = ‖x‖, then c =±1.

similar to step 2

3.2. If A = 1, 2 and ‖cx‖ 6= ‖x‖, then |c|2 =‖cx‖2/‖x‖2 ∈ 2, 1

2, c = ±√

2 or ±√

22 .

‖cx‖2, ‖x‖2 = 1, 2

3.3. c ∈ Q. x, cx ∈ I

3.4 a contradiction. steps 3.2, 3.3

(R4)



1. For all x ∈ Φ, 2(x,x) ∈ Z. defn of Φ

2. For all x, y ∈ I, (x, y) ∈ Z. defn of I and (•, •)

3. For all x, y ∈ Φ, 〈y, x〉 = 2(x, x)(x, y) ∈ Z. Example 9.9

(R3)


1. For all x, y ∈ I, σx(y) = y − 〈y, x〉x ∈ I. 〈y, x〉 ∈ Z by (R4)

2. For all x, y ∈ Φ, (σx(y), σx(y)) = (y, y) ∈A, Thus σx(y) ∈ Φ.

Lemma 9.8

σx(Φ) ⊂ Φ.

Since (R1)(R2)(R3)(R4) are verified, it follows from Definition 9.12 that Φ is a root system.

Proposition 14.9There is an irreducible root system of type Al, (l ≥ 1).

Proof. Construction:

• Let E = x ∈ Rl+1 :l+1∑i=1

xi = 0.• Let I = SpanZεi : 1 ≤ i ≤ l + 1, I ′ = I ∩ E.• Let Φ = α ∈ I : (α, α) = 2 and Φ′ = Φ ∩ E.

Since E is the hyperplane of Rl+1 orthogonal tol+1∑i=1

εi, it is a Euclidean space of dimension l under(•, •). Next, we show that Φ′ is a root system, using Definition 9.12.


0. Φ is a root system. Lemma 14.8

1. (R1) of Φ′ is true. (R1) of Φ is true and below.

Φ′ is finite. Φ is finite and Φ′ ⊂ Φ

0 6∈ Φ′. 0 6∈ Φ and Φ′ ⊂ Φ

164 ZHENGYAO WU


Φ′ spans E. Φ spans Rl+1 and Φ′ = Φ ∩ E.

2. (R2) of Φ′ is true. below

α ∈ Φ′ iff −α ∈ Φ′. α ∈ I iff −α ∈ I, α ∈ E iff −α ∈ E and(−α,−α) = (α, α)

If α ∈ Φ′, then Rα ∩ Φ′ = ±α. Rα ∩ Φ = ±α by (R2) of Φ


For all α ∈ Φ′, σα(Φ′) ⊂ Φ. Φ′ ⊂ Φ and (R3) of Φ

If (β,l+1∑i=1

εi) = 0, then (σα(β),l+1∑i=1

εi) = 0. Lemma 9.8, and σα(l+1∑i=1

εi) =l+1∑i=1

εi since

(α,l+1∑i=1

εi) = 0.

σα(Φ′) ⊂ Φ′. Φ′ = Φ ∩ E


For all α, β ∈ Φ′, 〈α, β〉 ∈ Z. (R4) of Φ and Φ′ ⊂ Φ

Description:• Φ′ = εi − εj : 1 ≤ i 6= j ≤ l + 1; Card(Φ′+) = l2 + l.• Φ′ has base ∆ = αi = εi − εi+1 : 1 ≤ i ≤ l.

The description of Φ′ follows from Lemma 14.6. Card(Φ′+) =(l+12

)= l2 + l. Verifying (B1) and

(B2) of Definition 10.4, ∆ is a base of Φ′.


5. (B1) is true. below

5.1. ∆ is linearly independent.

If c1α1 + · · · + clαl = 0, then c1ε1 + (c2 −c1)ε2 + · · ·+ (cl − cl1)εl − clεl+1 = 0.

αi = εi − εi+1

Then c1 = c2−c1 = · · · = cl−cl−1 = cl = 0and hence ci = 0 for all 1 ≤ i ≤ l.

εi : 1 ≤ i ≤ l + 1 are linearly indepen-dent

5.2. ∆ is a basis of E. step 1.1 and Card(∆) = l.




6.1. εi − εj = αi + αi+1 + · · · + αj−1 for all1 ≤ i < j ≤ l.

αk = εk − εk+1, 1 ≤ i ≤ k ≤ j − 1 ≤ l − 1

6.2. εi − εj = −αj − αj+1 − · · · − αi−1 for all1 ≤ j < i ≤ l.

αk = εk − εk+1, 1 ≤ j ≤ k ≤ i− 1 ≤ l − 1

We proceed to show that Φ′,∆ has Cartan matrix

2 −1 0 · · · 0 0

−1 2 −1 · · · 0 0

0 −1 2 · · · 0 0. . .

0 0 0 · · · 2 −1

0 0 0 · · · −1 2

It follows by Theorem 14.2 that Φ′ is irreducible of type Al, (l ≥ 1).


7.1. If i < j, then (αi, αj) = (εi − εi+1, εj −εj+1) = −δi+1,j.

εi : 1 ≤ i ≤ l + 1 is an orthonormalbasis of Rn+1

7.2. (αj, αj) = (εj − εj+1, εj − εj+1) = 2 for all1 ≤ j ≤ l + 1.

εi : 1 ≤ i ≤ l + 1 is an orthonormalbasis of Rn+1

7.3. 〈αi, αj〉 = 2(αi, αj)(αj, αj)

= −δi+1,j. steps 1,2

8. If i > j, then 〈αi, αj〉 = −δi,j+1. similar to step 7

Proposition 14.10There is an irreducible root system of type Bl, (l ≥ 2).

166 ZHENGYAO WU

Proof. Construction:• Let E = Rl and I = SpanZεi : 1 ≤ i ≤ l + 1.• Let Φ = α ∈ I : (α, α) ∈ 1, 2.

That Φ is a root system follows from Lemma 14.8.Description:

• Φ = ±εi : 1 ≤ i ≤ l ∪ ±(εi ± εj), 1 ≤ i < j ≤ l; Card(Φ+) = l2.• Φ has base ∆ = αi = εi − εi+1 : 1 ≤ i ≤ l − 1 ∪ αl = εl.

The description of Φ follows from Lemma 14.5 and Lemma 14.6, Card(Φ+) = l + 2(l2

)= l2.

Verifying (B1) and (B2) of Definition 10.4, ∆ is a base of Φ.




If c1α1 + · · · + clαl = 0, then c1ε1 + (c2 −c1)ε2+· · ·+(cl−1−cl−2)εl−1+(cl−cl−1)εl =0.

αi = εi−εi+1 for 1 ≤ i ≤ l−1 and αl = εl

Then c1 = c2 − c1 = · · · = cl − cl−1 = 0and hence ci = 0 for all 1 ≤ i ≤ l.

εi : 1 ≤ i ≤ l + 1 are linearly indepen-dent




αk = εk − εk+1, 1 ≤ i ≤ k ≤ j − 1 ≤ l − 1

2.2. εi = εi − εl + εl = αi + · · ·+ αl−1 + αl forall 1 ≤ i ≤ l

step 2.1

2.3. εi+εj = (εi−εl)+(εj−εl)+2εl = αi+· · ·+αj−1 +2αj + · · ·+2αl for all 1 ≤ i < j ≤ l.

step 2.1

2.4. εj − εi = −αj − αj+1 − · · · − αi−1 for all1 ≤ j < i ≤ l.

step 2.1

2.5. −εi = −(εi−εl)−εl = −αi−· · ·−αl−1−αlfor all 1 ≤ i ≤ l

step 2.2.



2.6. −εi − εj = −(εi − εl) − (εj − εl) − 2εl =−αi − · · · − αj−1 − 2αj − · · · − 2αl for all1 ≤ i < j ≤ l.

step 2.3

We proceed to show that Φ,∆ has Cartan matrix

2 −1 · · · 0 0 0

−1 2 · · · 0 0 0. . .

0 0 · · · 2 −1 0

0 0 · · · −1 2 −2

0 0 · · · 0 −1 2

It follows by Theorem 14.2 that Φ is irreducible of type Bl, (l ≥ 2).


3. If 1 ≤ i < j ≤ l except when i = l− 1 andj = l, then 〈αi, αj〉 = −δi+1,j.

similar to step 7 of Proposition 14.9

If 1 ≤ j < i ≤ l, then 〈αi, αj〉 = −δi,j+1. similar to step 8 of Proposition 14.9

4.1. (αl−1, αl) = (εl−1 − εl, εl) = −1 εi : 1 ≤ i ≤ l is an orthonormal basisof Rn

4.2. (αl, αl) = (εl, εl) = 1 εi : 1 ≤ i ≤ l is an orthonormal basisof Rn

4.3. 〈αl−1, αl〉 = 2(αl−1, αl)(αl, αl)

= −2. steps 4.1, 4.2

Proposition 14.11There is an irreducible root system of type Cl, (l ≥ 3).


168 ZHENGYAO WU

• Let E = Rl and Φ the root system of type Bl as in Proposition 14.10.• Let Φ∨ = α∨ = 2α

(α, α) : α ∈ Φ.

That Φ∨ is a root system follows from [Hum78, p.46, Exercise 2].Description:

• Φ∨ = ±2εi : 1 ≤ i ≤ l ∪ ±(εi ± εj), 1 ≤ i < j ≤ l; Card(Φ∨+) = l2.• Φ∨ has base ∆∨ = αi = εi − εi+1 : 1 ≤ i ≤ l − 1 ∪ αl = 2εl.


1. ε∨i = 2εi. (εi, εi) = 1 and Definition 9.22

2. (εi ± εj)∨ = εi ± εj. (εi ± εj, εi ± εj) = 2 and Definition 9.22

3. Φ∨, ∆∨ are what we desired. steps 1,2 and (−α)∨ = −α∨

4. ∆∨ is a base of Φ∨. [Hum78, p.54, Exercise 1]

We proceed to show that Φ∨,∆∨ has Cartan matrix

2 −1 · · · 0 0 0

−1 2 · · · 0 0 0. . .

0 0 · · · 2 −1 0

0 0 · · · −1 2 −1

0 0 · · · 0 −2 2

It follows by Theorem 14.2 that Φ is irreducible of type Cl, (l ≥ 3).


7. 〈β∨, α∨〉 = 〈α, β〉 for all α, β ∈ Φ. [Hum78, p.46, Exercise 2]

6. The Cartan matrix of Φ∨ is of type Cl. The Cartan matrix of Φ∨ is the transposeof that of Φ of type Bl


15. June 4th, Irreducible root systems of types D,E,F,G

Proposition 15.1There is an irreducible root system of type Dl, (l ≥ 4).

Proof. Construction:• Let E = Rl.• Let Φ = α ∈ I : (α, α) = 2.

That Φ is a root system follows from Lemma 14.8.Description:

• Φ = ±(εi ± εj), 1 ≤ i 6= j ≤ l; Card(Φ+) = l2 − l.• Φ has base ∆ = αi = εi − εi+1 : 1 ≤ i ≤ l − 1 ∪ αl = εl−1 + εl.

The description of Φ follows from Lemma 14.6, Card(Φ+) = 2(l2

)= l2 − l. Verifying (B1) and

(B2) of Definition 10.4, ∆ is a base of Φ.




If c1α1 + · · · + clαl = 0, then c1ε1 + (c2 −c1)ε2 +· · ·+(cl−2−cl−3)εl−2 +(cl−1−cl−2 +cl)εl−1 + (cl − cl−1)εl = 0.

αi = εi − εi+1 for 1 ≤ i ≤ l − 1 and αl =εl−1 + εl

Then c1 = c2 − c1 = · · · = cl−2 − cl−3 =cl−1 − cl−2 + cl = cl − cl−1 = 0 and henceci = 0 for all 1 ≤ i ≤ l−2 and cl±cl−1 = 0.Hence also cl−1 = cl = 0.

εi : 1 ≤ i ≤ l are linearly independent




αk = εk − εk+1, 1 ≤ i ≤ k ≤ j − 1 ≤ l − 1

2.2. εi+εj = (εi−εl−1)+(εj−εl)+(εl−1+εl) =αi+· · ·+αj−1+2αj+· · ·+2αl−2+αl−1+αlfor all 1 ≤ i < j ≤ l.

step 2.1

2.3. εj − εi = −αj − αj+1 − · · · − αi−1 for all1 ≤ j < i ≤ l.

step 2.1

170 ZHENGYAO WU


2.4. −εi − εj = −(εi − εl−1) − (εj − εl−1) −2εl−1 = −αi − · · · − αj−1 − 2αj − · · · −2αl−2 − αl−1 − αl for all 1 ≤ i < j ≤ l.

step 2.2


2 −1 · · · 0 0 0 0

−1 2 · · · 0 0 0 0. . .

0 0 · · · 2 −1 0 0

0 0 · · · −1 2 −1 −1

0 0 · · · 0 −1 2 0

0 0 · · · 0 −1 0 2

It follows by Theorem 14.2 that Φ is irreducible of type Dl, (l ≥ 4).


3. If 1 ≤ i < j ≤ l except when i = l− 1 andj = l, then 〈αi, αj〉 = −δi+1,j.

similar to step 7 of Proposition 14.9

If 1 ≤ j < i ≤ l, then 〈αi, αj〉 = −δi,j+1. similar to step 8 of Proposition 14.9

4.1. (αl−2, αl−1) = (εl−2−εl−1, εl−1−εl) = −1. εi : 1 ≤ i ≤ l is an orthonormal basisof Rn

4.2. (αl−1, αl−1) = (εl−1 − εl, εl−1 − εl) = 2. εi : 1 ≤ i ≤ l is an orthonormal basisof Rn

4.3. 〈αl−2, αl−1〉 = 2(αl−2, αl−1)(αl−1, αl−1) = −1. steps 4.1, 4.2 and Example 9.9

5.1. (αl−2, αl) = (εl−2 − εl−1, εl−1 + εl) = −1. εi : 1 ≤ i ≤ l is an orthonormal basisof Rn

5.2. (αl, αl) = (εl−1 + εl, εl−1 + εl) = 2. εi : 1 ≤ i ≤ l is an orthonormal basisof Rn



5.3. 〈αl−2, αl〉 = 2(αl−2, αl)(αl, αl)

= −1. steps 5.1, 5.2 and Example 9.9

6.1. (αl−1, αl) = (εl−1 − εl, εl−1 + εl) = 0. εi : 1 ≤ i ≤ l is an orthonormal basisof Rn

6.2. 〈αl−1, αl〉 = 2(αl−1, αl)(αl, αl)

= 0. steps 6.1, and Example 9.9

Proposition 15.2There is an irreducible root system of type E8.


• Let E = R8 with orthonormal basis εi : 1 ≤ i ≤ 8. Define ε9 = 12(ε1 + · · ·+ ε8)

• Let I = SpanZεi : 1 ≤ i ≤ 8, I ′ = I + Zε9 and I ′′ = 9∑i=1

ciεi ∈ I ′ :8∑i=1

ci is even..• Let Φ = α ∈ I ′′ : (α, α) = 2.

Description:

• Φ = ±(εi ± εj), 1 ≤ i < j ≤ 8 ∪ 12

8∑i=1

(−1)k(i)εi : k(i) ∈ 0, 1,8∑i=1

k(i) is even.;Card(Φ+) = 120.• Φ has base ∆ = α1 = 1

2(ε1 + ε8 − (ε2 + · · · + ε7)), α2 = ε1 + ε2, α3 = ε2 − ε1, α4 =ε3 − ε2, α5 = ε4 − ε3, α6 = ε5 − ε4, α7 = ε6 − ε5, α8 = ε7 − ε6, .

We first describe Φ.


1. Suppose α ∈ Φ. Then α =9∑i=1

ciεi for

ci ∈ Z where8∑i=1

ci is even.

α ∈ I ′′

2. We assume c9 ∈ 0, 1. If c9 = 2q + r,0 ≤ r < 2, then α =

8∑i=1

(ci + q)εi + rε9.ε9 = 1

2(ε1 + · · ·+ ε8)

3. (α, α) =8∑i=1

(ci + c9

2 )2 =8∑i=1

c2i + c9

8∑i=1

ci +2c2

9 = 2.α =

8∑i=1

(ci + c9

2 )εi

4. If c9 = 0, then ±(εi ± εj), 1 ≤ i < j ≤8 ⊂ Φ.

8∑i=1

c2i = 2 and Lemma 14.6

172 ZHENGYAO WU


5. If c9 = 1, then 12

8∑i=1

(−1)k(i)εi : k(i) ∈0, 1 ⊂ Φ.

8∑i=1

(c2i + ci) = 0 and ci ∈ −1, 0 by

Lemma 14.7

5.1. Card1 ≤ i ≤ 8 : ci = −1 is even.8∑i=1

ci is even.

5.2.8∑i=1

k(i) is even in step 5 k(i) = 0 iff ci = 0; k(i) = 1 iff ci = −1

Next, we show that Φ is a root system, using Definition 9.12. The matrix of ∆ relative to εi : 1 ≤i ≤ 8 is invertible since its determinant is 1.

12 −1

2 −12 −

12 −

12 −

12 −

12

12

1 1 0 0 0 0 0 0

−1 1 0 0 0 0 0 0

0 −1 1 0 0 0 0 0

0 0 −1 1 0 0 0 0

0 0 0 −1 1 0 0 0

0 0 0 0 −1 1 0 0

0 0 0 0 0 −1 1 0


6. ∆ is linearly independent over R. above

7. (R1) is true. below

7.1. Φ is finite. Card(Φ+) = 2(

82

)+ 1

2(1228) = 56 + 64 =

120, Card(Φ) = 240

7.2. Φ spans E. ∆ is linearly independent, Card(∆) = 8and ∆ ⊂ Φ

7.3. 0 6∈ Φ. (0, 0) = 0 6= 2

8. (R2) is true. If α, cα ∈ Φ and c ∈ R, thenc = ±1.

|c| = 1 since |c|‖α‖ = ‖cα‖ = 2 and ‖α‖ =2




9.1. For all α, β ∈ Φ, 〈α, β〉 = 2(α, β)(β, β) = (α, β) (β, β) = 2 for all β ∈ Φ

9.2. If α = ±(εs± εt) and β = ±(εi± εj) then(α, β) ∈ Z.

εi : 1 ≤ i ≤ 8 is an orthonormal basisof R8

9.3. If α = ±(εs ± εt) and β =12

8∑i=1

(−1)k(i)εi : k(i) ∈ 0, 1 with8∑i=1

k(i)even, then (α, β) ∈ Z.

(α, β) = ±12 ±

12 ∈ 1, 0,−1

9.4. If α = 12

8∑i=1

(−1)k(i)εi : k(i) ∈ 0, 1

with8∑i=1

k(i) = 2k even, and β =12

8∑i=1

(−1)h(i)εi : h(i) ∈ 0, 1 with8∑i=1

h(i) = 2h even, then (α, β) ∈ Z.

Suppose Card(1 ≤ i ≤ 8 : k(i) =h(i) = 1) = t. Then (α, β) =14

8∑i=1

(−1)k(i)+h(i) = 14((8− 2k − 2h+ t)−

(2h− t)− (2k− t) + t) = 2−h− k+ t ∈ Z


10.1. If β ∈ I ′′, then σα(β) = β − 〈β, α〉α ∈ I ′′ (R4) and defn of I ′′

10.2. If β ∈ Φ, then (σα(β), σα(β)) = (β, β) =2, σα(β) ∈ Φ.

Lemma 9.8 and defn of Φ



11. (B1) is true, i.e. ∆ is a basis of E Card(∆) = 8 and ∆ is linearly indepen-dent by the discussion just before step 6


12.1. ε8 − ε7 = 2α1 + 2α2 + 3α3 + 4α4 + 3α5 +2α6 + α7.

Right to left calculation by assuming ε8−ε7 = 2α1+xα2+yα3+4α4+3α5+2α6+α7.

12.2. ±(εi − εj) = ∓j−1∑k=i

(εk+1 − εk) for all 1 ≤i < j ≤ 8

step 2.1 and defn of αi, (3 ≤ i ≤ 8)

174 ZHENGYAO WU


12.3. ±(εi + εj) = ±((εi − ε1) + (εj − ε2) + α2)for all 1 ≤ i < j ≤ 8

step 2.2 and defn of α2

12.4. Endow lexicographic order to the collec-tion of α = 1

28∑i=1

(−1)k(i)εi : k(i) ∈ 0, 1

with8∑i=1

k(i) even, positive iff k(8) = 0.

Totally 28

2 = 128 roots.

12.5. From α1, by adding consecutively αi, (3 ≤i ≤ 8) or εi + εj, (1 ≤ i < j ≤ 7), weobtain 6 positive roots as in step 2.4.

12.6. Every root with 4,6,8 positive coefficentsincluding that of ε8 is what we desired.

Totally(

77

)+(

75

)+(

73

)= 1 + 21 + 35 = 57

roots

12.7. α1 and roots of steps 2.5, 2.6 consists ofall positive roots as in step 2.4.

1+6+57 = 64 positive roots, exactly halfof 128.


2 0 −1 0 0 0 0 0

0 2 0 −1 0 0 0 0

−1 0 2 −1 0 0 0 0

0 −1 −1 2 −1 0 0 0

0 0 0 −1 2 −1 0 0

0 0 0 0 −1 2 −1 0

0 0 0 0 0 −1 2 −1

0 0 0 0 0 0 −1 2

It follows by Theorem 14.2 that Φ is irreducible of type E8.


13. Define βi = α8−i for 1 ≤ i ≤ 7. The Cartan matrix of βi : 1 ≤ i ≤ 7 isD7



14. It suffices to determine the first row andthe first column.

We obtain the lower-right 7 × 7 block,which is the Cartan matrix of D7 in re-verse order.

15. (α1, α1) = 8 ∗ (12)2 = 2. The coefficients of εi in α1 are ±1

2(αj, αj) = 2 ∗ 12 = 2 for all 2 ≤ j ≤ 8. The coefficients of εi in αj are ±1

16. 〈αi, αj〉 = 2(αi, αj)(αj, αj)

= (αi, αj) for all 1 ≤i, j ≤ 8.

step 3

17. 〈α1, α2〉 = 〈α2, α1〉 = 0. (α1, α2) = (12(ε1 + ε8− (ε2 + · · ·+ ε7), ε1 +

ε2) = 12(1 ∗ 1 + (−1) ∗ 1) = 0.

〈α1, α3〉 = 〈α3, α1〉 = −1 (α1, α3) = (12(ε1 + ε8− (ε2 + · · ·+ ε7), ε2−

ε1) = 12((−1) ∗ 1 + 1 ∗ (−1)) = −1

〈α1, αi〉 = 〈αi, α1〉 = 0 for all 4 ≤ i ≤ 8. (α1, αi) = (12(ε1 +ε8−(ε2 + · · ·+ε7), εi−1−

εi−2) = 12((−1) ∗ 1 + (−1) ∗ (−1)) = 0


Proof. Construction:• Let E = R8 have orthonormal basis εi : 1 ≤ i ≤ 8. Let Φ = ±(εi ± εj), 1 ≤ i 6= j ≤

8∪12

8∑i=1

(−1)k(i)εi : k(i) ∈ 0, 1,8∑i=1

k(i) is even. be the irreducible root system of type

E8 with base ∆ = α1 = 12(ε1 + ε8 − (ε2 + · · · + ε7)), α2 = ε1 + ε2, α3 = ε2 − ε1, α4 =

ε3 − ε2, α5 = ε4 − ε3, α6 = ε5 − ε4, α7 = ε6 − ε5, α8 = ε7 − ε6, .• Let ∆′ = αi : 1 ≤ i ≤ 7 and Φ′ = Φ ∩ SpanZ(∆′).

Description:• E ′ = SpanR(∆′) is the hyperplane in E orthogonal to ε7 + ε8.• Φ′ = ±(εi ± εj), 1 ≤ i < j ≤ 6 ∪ ±(ε7 − ε8) ∪ ±1

2(ε7 − ε8 +6∑i=1

(−1)k(i)εi) : k(i) ∈

0, 1,6∑i=1

k(i) is odd.. Card(Φ′+) = 63.• Φ′ has base ∆′ follows from construction.

Firstly, E ′ is the hyperplance since (αi, ε7 + ε8) = 0 for all 1 ≤ i ≤ 7.

176 ZHENGYAO WU


1. Φ′ ⊂ Φ. defn of Φ′

2.1. ±(εi − εj), 1 ≤ i < j ≤ 8 has a summand∓α8 = ε7 − ε6 iff j = 7 or “i ≤ 6 andj = 8”.

steps 12.1, 12.2 of Proposition 15.2

2.2. ±(εi + εj), 1 ≤ i < j ≤ 8 has a summand±α8 = ε7 − ε6 iff i, j ∩ 7, 8 6= ∅.

step 12.3 of Proposition 15.2 and step 2.1

2.3 We obtain ±(εi ± εj), 1 ≤ i < j ≤ 6 ∪±(ε7 − ε8) with 2

(62

)+ 1 = 31 positive

roots.

deleting roots in steps 2.1, 2.2 from±(εi − εj), 1 ≤ i < j ≤ 8

3.1. If α = 12

8∑i=1

(−1)k(i)εi ∈ Φ+, k(i) ∈ 0, 1

with8∑i=1

k(i) even has a summand α8 =ε7 − ε6 iff k(7) = 1 and k(8) = 0.

k(8) = 0 since α must have summand 1α1;

Also, 0 = (α, ε7 + ε8) = (−1)k(7) + 12 .

3.2. α = 12(−ε7 + ε8 +

6∑i=1

(−1)k(i)εi), k(i) ∈

0, 1 with6∑i=1

k(i) odd.

k(7) = 1 and k(8) = 0

3.3 We obtain ±12(−ε7 + ε8 +

6∑i=1

(−1)k(i)εi), k(i) ∈ 0, 1,6∑i=1

k(i)

odd; with(

61

)+(

63

)+(

65

)= 32 positive

roots.

deleting ± roots in steps 3.2from ±1

28∑i=1

(−1)k(i)εi, k(i) ∈

0, 1,8∑i=1

k(i) even

4. Totally 31 + 32 = 63 positive roots. steps 2.3, 3.3

Finally, deleting the last row and the last column of the Cartan matrix of type E8, we obtain theCartan matrix of Φ′,∆′:


2 0 −1 0 0 0 0

0 2 0 −1 0 0 0

−1 0 2 −1 0 0 0

0 −1 −1 2 −1 0 0

0 0 0 −1 2 −1 0

0 0 0 0 −1 2 −1

0 0 0 0 0 −1 2

which is of type E7.


Proof. Construction:• Let E = R8 have orthonormal basis εi : 1 ≤ i ≤ 8. Let E ′ = SpanR(∆′) be thehyperplane in E orthogonal to ε7 + ε8. Let Φ′ = ±(εi ± εj), 1 ≤ i < j ≤ 6 ∪ ±(ε7 −

ε8) ∪ ±12(ε7 − ε8 +

6∑i=1

(−1)k(i)εi) : k(i) ∈ 0, 1,6∑i=1

k(i) is odd. be the irreducible root

system of type E7 with base ∆′′ = α1 = 12(ε1 + ε8 − (ε2 + · · · + ε7)), α2 = ε1 + ε2, α3 =

ε2 − ε1, α4 = ε3 − ε2, α5 = ε4 − ε3, α6 = ε5 − ε4, α7 = ε6 − ε5.• Let ∆′′ = αi : 1 ≤ i ≤ 6 and Φ′′ = Φ′ ∩ SpanZ(∆′′).

Description:• E ′′ = SpanR(∆′′) is the hyperplane in E ′ orthogonal to ε6 + ε7 + 2ε8.• Φ′′ = ±(εi ± εj), 1 ≤ i < j ≤ 5 ∪ ±1

2(ε8 − ε7 − ε6 +5∑i=1

(−1)k(i)εi) : k(i) ∈

0, 1,5∑i=1

k(i) is even.. Card(Φ+) = 36.• Φ′′ has base ∆′′ follows from construction.

Firstly, E ′′ is the hyperplane since (αi, ε6 + ε7 + 2ε8) = 0 for all 1 ≤ i ≤ 6.


1. Φ′′ ⊂ Φ′. defn of Φ′′

2.1. ±(εi − εj), 1 ≤ i < j ≤ 6 has a summand∓α7 = ε6 − ε5 iff j = 6.

steps 12.1, 12.2 of Proposition 15.2

178 ZHENGYAO WU


2.2. ±(εi + εj), 1 ≤ i < j ≤ 6 has a summand±α7 = ε6 − ε5 iff i = 6 or j = 6.

step 12.3 of Proposition 15.2 and step 2.1

2.3. ±(ε7 − ε8) has a summand ±α7. ε8 − ε7 = 2α1 + 2α2 + 3α3 + 4α4 + 3α5 +2α6 + α7

2.4 We obtain ±(εi ± εj), 1 ≤ i < j ≤ 5with 2

(52

)= 20 positive roots.

deleting roots in steps 2.1, 2.2, 2.3 from±(εi − εj), 1 ≤ i < j ≤ 6

3.1. If α = 12(−ε7 + ε8 +

6∑i=1

(−1)k(i)εi) : k(i) ∈

0, 1,6∑i=1

k(i) is odd has a summandα7 = ε6 − ε5 iff k(6) = 1.

step 3.1 of Proposition 15.3 and 0 =

(α, ε6 + ε7 + 2ε8) = (−1)k(6) − 1 + 22 .

3.2. α = 12(−ε6−ε7+ε8+

5∑i=1

(−1)k(i)εi), k(i) ∈

0, 1 with6∑i=1

k(i) even.

k(6) = k(7) = 1 and k(8) = 0

3.3 We obtain ±12(−ε6 − ε7 + ε8 +

6∑i=1

(−1)k(i)εi), k(i) ∈ 0, 1,6∑i=1

k(i)

even; with(

50

)+(

52

)+(

54

)= 16 positive

roots.

deleting ± roots in steps 3.2 from±1

2 − ε7 + ε8 +6∑i=1

(−1)k(i)εi, k(i) ∈

0, 1,8∑i=1

k(i) even

4. Totally 20 + 16 = 36 positive roots. steps 2.4, 3.3

Finally, deleting the last row and the last column of the Cartan matrix of type E7, we obtain theCartan matrix of Φ′′,∆′′:

2 0 −1 0 0 0

0 2 0 −1 0 0

−1 0 2 −1 0 0

0 −1 −1 2 −1 0

0 0 0 −1 2 −1

0 0 0 0 −1 2

which is of type E6.


Proposition 15.5There is an irreducible root system of type F4.


• Let E = R4 with orthonormal basis εi : 1 ≤ i ≤ 4. Let ε5 = 12(ε1 + ε2 + ε3 + ε4).

• Let I = SpanZεi : 1 ≤ i ≤ 4 and I ′ = I + Zε5.• Let Φ = α ∈ I ′ : (α, α) ∈ 1, 2.

Description:• Φ = ±εi : 1 ≤ i ≤ 4 ∪ ±(εi ± εj) : 1 ≤ i < j ≤ 4 ∪ ±1

2(ε1 ± ε2 ± ε3 ± ε4);Card(Φ+) = 24.• Φ has base ∆ = α1 = ε2 − ε3, α2 = ε3 − ε4, α3 = ε4, α4 = 1

2(ε1 − ε2 − ε3 − ε4).We first describe Φ.


1. Suppose α ∈ Φ. Then α =5∑i=1

ciεi forci ∈ Z, c5 ∈ 0, 1.

α ∈ I ′ and step 2 of Proposition 15.2

2. c5 = 0 iff α ∈ ±εi : 1 ≤ i ≤ 4∪±(εi±εj) : 1 ≤ i < j ≤ 4.

c21 + c2

2 + c23 + c2

4 ∈ 1, 2, Lemma 14.5 andLemma 14.6

3. c5 = 1 iff ±12(ε1 ± ε2 ± ε3 ± ε4).

4∑i=1

(c2i + ci) ∈ 0, 1 =⇒ ci ∈ −1, 0 for

all 1 ≤ i ≤ 4 by Lemma 14.7

Next, we show that Φ is a root system, using Definition 9.12. The matrix of ∆ relative to εi : 1 ≤i ≤ 4 is invertible since its determinant is −1

2 .

0 1 −1 0

0 0 1 −1

0 0 0 1

12 −

12 −

12 −

12


4. ∆ is linearly independent over R. above


180 ZHENGYAO WU


5.1. Φ is finite. Card(Φ+) = 4 + 2(

42

)+ 24

2 = 24.

5.2. Φ spans E. ∆ is linearly independent, Card(∆) = 4 =dimRE and ∆ ⊂ Φ

5.3. 0 6∈ Φ. (0, 0) = 0 6∈ 1, 2

6. (R2) is true. similar to (R2) of Lemma 14.8 and x, cx ∈I ′ implies that c ∈ Q


7.1. For all α, β ∈ ±εi : 1 ≤ i ≤ 4, 〈α, β〉 =2(α, β)(β, β) ∈ 0,±2 ⊂ Z.

(α, β) ∈ 0,±1 and (β, β) = 1

7.2. For all α ∈ ±εi : 1 ≤ i ≤ 4 and β ∈±(εi ± εj) : 1 ≤ i < j ≤ 4, 〈α, β〉 ∈0,±1, 〈β, α〉 ∈ 0,±2.

(α, β) ∈ 0,±1, (α, α) = 1 and (β, β) =2

7.3. For all α ∈ ±εi : 1 ≤ i ≤ 4 and β ∈±1

2(ε1 ± ε2 ± ε3 ± ε4), 〈α, β〉 = 〈β, α〉 ∈±1.

(α, β) ∈ ±12, (α, α) = (β, β) = 1

7.4. For all α, β ∈ ±(εi±εj) : 1 ≤ i < j ≤ 4,

〈α, β〉 = 2(α, β)(β, β) ∈ 0,±1,±2 ⊂ Z

(α, β) ∈ 0,±1,±2 and (β, β) = 2

7.5. For all α ∈ ±(εi ± εj) : 1 ≤ i < j ≤ 4and β ∈ ±1

2(ε1 ± ε2 ± ε3 ± ε4), 〈β, α〉 ∈0,±1 and 〈α, β〉 ∈ 0,±2.

(α, β) ∈ 0,±1, (α, α) = 2 and (β, β) =1

7.6. For all α, β ∈ ±12(ε1 ± ε2 ± ε3 ± ε4),

〈α, β〉 ∈ 0,±1,±2.(α, β) ∈ 0,±1

2 ,±1 and (β, β) = 1


8.1. If α, β ∈ I ′, then σα(β) = β−〈β, α〉α ∈ I ′ (R4) and defn of I ′

8.2. If α, β ∈ Φ, then (σα(β), σα(β)) =(β, β) ∈ 1, 2, σα(β) ∈ Φ.





9. (B1) is true, i.e. ∆ is a basis of E Card(∆) = 4 and ∆ is linearly indepen-dent by the discussion just before step 4

10. (B2) is true. below by exhaustion

10.1. ε1 − ε2 = α2 + 2α3 + 2α4

10.2. We obtain εi − εj for all 1 ≤ i < j ≤ 4, 6positive roots.

step 12.1, defn of α1, α2and εi − εj =j−1∑k=i

(εk − εk+1)

10.3. We obtain εi for all 1 ≤ i ≤ 4, 4 positiveroots.

step 12.2, defn of α3 and εi = εi − ε4 + ε4

10.4. We obtain εi + εj for all 1 ≤ i < j ≤ 4, 6positive roots.

step 12.3

10.5. We obtain 12(ε1 ± ε2 ± ε3 ± ε4), 8 positive

roots.

12(ε1 ± ε2 ± ε3 ± ε4) − α1 = aε2 + bε3 +cε4, a, b, c ∈ 0, 1 is a sum of positiveroots by step 12.3, 12.4

10.6 Totally 6 + 4 + 6 + 8 = 24 positive roots.

Add negative signs on each of them, weobtain all negative roots.


2 −1 0 0

−1 2 −2 0

0 −1 2 −1

0 0 −1 2

It follows by Theorem 14.2 that Φ is irreducible of type F4.


11. α1, α2, α3 spans a root system of type B3. Proposition 14.10

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12. 〈α1, α4〉 = 0. Similarly, (α1, α4) = (ε2− ε3,12(ε1− ε2− ε3− ε4)) =

1 ∗ (−12) + (−1) ∗ (−1

2) = 0

〈α4, α1〉 = 〈α2, α4〉 = 〈α4, α2〉 = 0.

13. 〈α3, α4〉 = 2(α3, α4)(α4, α4) = −1. (α3, α4) = (ε4,

12(ε1 − ε2 − ε3 − ε4)) = −1

2and (α4, α4) = 4 ∗ (1

2)2 = 1

Similarly, 〈α3, α4〉 = −1.

Proposition 15.6There is an irreducible root system of type G2.

Proof. Construction:• Let R3 have orthonormal basis ε1, ε2, ε3. Let E = x ∈ R3 : x1 + x2 + x3 = 0.• Let I = SpanZε1, ε2, ε3 and I ′ = I ∩ E.• Let Φ = α ∈ I ′ : (α, α) ∈ 2, 6.

Description:• E is the subspace of R3 orthogonal to ε1 + ε2 + ε3.• Φ = ±ε1 − ε2, ε2 − ε3, ε1 − ε3, 2ε1 − ε2 − ε3, 2ε2 − ε1 − ε3, 2ε3 − ε1 − ε2; Card(Φ+) = 6.• Φ has base ∆ = α1 = ε1 − ε2, α2 = −2ε1 + ε2 + ε3.

We first describe Φ. Suppose α ∈ Φ.


1. (α, α) = 2 iff α ∈ ±ε1−ε2, ε2−ε3, ε1−ε3. Lemma 14.6

2. Suppose α = c1ε1 + c2ε1 + c3ε1, c1, c2, c3 ∈Z and c1 + c2 + c3 = 0.

3. (α, α) = 6 iff α ∈ ±2ε1 − ε2 − ε3, 2ε2 −ε1 − ε3, 2ε3 − ε1 − ε2.

c21 + c2

2 + c23 = 6 iff ci = ±2 for one i and

cj = ∓1 for j 6= i.

4. Card(Φ+) = 6. steps 1,3


Next, we show that Φ is a root system, using Definition 9.12.


5. ∆ is linearly independent over R. α1 = ε1−ε2 is not parallel to α2 = −2ε1 +ε2 + ε3


6.1. Φ is finite. Card(Φ+) = 6.

6.2. Φ spans E. ∆ is linearly independent, Card(∆) = 2 =dimRE and ∆ ⊂ Φ

6.3. 0 6∈ Φ. (0, 0) = 0 6∈ 2, 6

7. (R2) is true. Description of Φ


8.1. For all α ∈ Φ and β ∈ ±(εi − εj) : 1 ≤i < j ≤ 3, 〈α, β〉 ∈ Z

(α, β) ∈ Z since α, β ∈ I and (β, β) = 2

8.2. For all α ∈ ±(εi − εj) : 1 ≤ i < j ≤ 3and β ∈ ±(2εi − εj − εk) : i, j, k =1, 2, 3, 〈α, β〉 ∈ 0,±1 ⊂ Z,

(α, β) ∈ 0,±3; and (β, β) = 6

8.2. For all α, β ∈ ±(2εi−εj−εk) : i, j, k =1, 2, 3, 〈α, β〉 ∈ ±1 ⊂ Z,

(α, β) ∈ ±3 and (β, β) = 6


9.1. If α, β ∈ I ′, then σα(β) = β−〈β, α〉α ∈ I ′ (R4) and defn of I ′

9.2. If α, β ∈ Φ, then (σα(β), σα(β)) =(β, β) ∈ 2, 6, σα(β) ∈ Φ.




10. (B1) is true, i.e. ∆ is a basis of E Card(∆) = 2 and ∆ is linearly indepen-dent by step 5

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11. (B2) is true. exhaustion of positive roots below

11.1. ε1 − ε2 = α1, −2ε1 + ε2 + ε3 = α2 given

11.2. −ε1 + ε3 = α1 + α2, ε3 − ε2 = 2α1 + α2

ε1 − 2ε2 + ε3 = 3α1 + α2

−ε1 − ε2 + 2ε3 = 3α1 + 2α2

12. Add negative signs on each of them, weobtain all negative roots.

Theorem 15.7For each connected Dynkin diagram or Cartan matrix, there exists an irreducible root systemhaving the given Dynkin diagram.

Proof. We have a classification of connected Dynkin diagrams by Theorem 14.1.Type Al, (l ≥ 1): Proposition 14.9; Type Bl, (l ≥ 2): Proposition 14.10;Type Cl, (l ≥ 3): Proposition 14.11; Type Dl, (l ≥ 4): Proposition 15.1;Type E6: Proposition 15.4; Type E7: Proposition 15.3; Type E8: Proposition 15.2.Type F4: Proposition 15.5; Type G2: Proposition 15.6.


16. June 11th, Weyl group of each type, Automorphisms of the Dynkin diagram,Weights

Lemma 16.1(i, i+ 1) : 1 ≤ i ≤ l generates the permutation group Sl+1.

Proof. We assume that composition of cycles are from left to right.


1. (i, i + 1) : 1 ≤ i ≤ l generate all trans-positions (a, b), a < b.

below

1.1. If b− a = 1, then (a, b) = (a, a+ 1).

1.2. Suppose b− a > 1 and (a+ 1, b) is gener-ated by (i, i+ 1) : 1 ≤ i ≤ l.

1.3. (a, b) = (a, a+ 1)(a+ 1, b)(a, a+ 1) is gen-erated by (i, i+ 1) : 1 ≤ i ≤ l.

step 1.2.

2. Transpositions generate cycles(a1, a2, . . . , an).

below

2.1. If n = 2, then (a1, a2) is a transposition.

2.2. Suppose n > 2 and (a1, . . . , an−1) is gen-erated by transpositions.

2.3. (a1, a2, . . . , an) = (a1, . . . , an−1)(a1, an) isgenerated by transpositions.

step 2.2

3. Cycles generate permutations.

If a permutation σ does not have cy-cles, then a, σ(a), σ(σ(a)), . . . are distinct,a contradiction.

to the fact that 1, 2, . . . , l + 1 is finite.

Proposition 16.2Let E be the Eulidean space x ∈ Rl+1 :

l+1∑i=1

xi = 0. Let Φ(Al) be the irreducible root systemof type Al, (l ≥ 1) in E with base ∆(Al) = αi = εi − εi+1 : 1 ≤ i ≤ l. Let W (Al) be the

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Weyl group of Φ(Al). Let Sl+1 be the permutation group of 1, 2, . . . , l + 1. Then we have anisomorphism W (Al) ∼−→Sl+1; σαi 7→ (i, i+ 1) and the order of W (Al) is (l + 1)!.Proof.


0. σεi−εi+1(εj) =

εj, j 6∈ i, i+ 1

εi+1, j = i

εi, j = i+ 1

. 〈εj, εi−εi+1〉 = (εj, εi−εi+1) since (εj, εi−εi+1, εj, εi−εi+1) = 2; and Example 9.9(2)

1. W (Al) is generated by σαi : αi ∈ ∆,with the following relations.

Theorem 11.13

1.1. For j > i + 1 or j + 1 < i,σαj σαi(εk) = σαi σαj(εk) =

εk, k 6∈ i, i+ 1, j, j + 1

εi+1, k = i

εi, k = i+ 1

εj+1, k = j

εj, k = j + 1

.

αi = εi − εi+1 and step 0

1.2. σαi+1 σαi(εj) =

εj, j 6∈ i, i+ 1, i+ 2

εi+2, j = i

εi, j = i+ 1

εi+1 j = i+ 2

.

αi = εi − εi+1 and step 0

1.3. σ2αi

= 1. Lemma 9.10(3)

2. Sl+1 is generated by (i, i+1) : 1 ≤ i ≤ lwith the following relations.

Lemma 16.1

2.1. For j > i+1 or j+1 < i, (i, i+1)(j, j+1) =(j, j + 1)(i, i+ 1).

(i, i + 1) fixes j, j + 1 and (j, j + 1) fixesi, i+ 1

2.2. (i, i+ 1)(i+ 1, i+ 2) = (i+ 1, i, i+ 2).

2.3. (i, i+ 1)2 = 1.



3. σαi 7→ (i, i+ 1) induces a homomorphism. steps 1,2

σαi 7→ (i, i+ 1) induces an isomorphism. Its inverse is induced by (i, i+ 1) 7→ σαi

Definition 16.3Let G be a group. Let N be a normal subgroup of G. Let H be a subgroup of G. We call G aninner semi-direct product of N and H and we write G = N oH if G = NH and N ∩H = e.

Proposition 16.4Let E ' Rl be the Eulidean space. Let Φ(Bl) be the irreducible root system of type Bl, (l ≥ 2) inE with base ∆(Bl) = αi = εi − εi+1 : 1 ≤ i ≤ l − 1 ∪ αl = εl. Let W (Bl) be the Weyl groupof Φ(Bl). Let Sl be the permutation group of 1, 2, . . . , l. Then W (Bl) ' (Z/2Z)l o Sl and ithas order 2ll!.

Proof. Since W is generated by σα, α ∈ Φ(Bl) = ±εi : 1 ≤ i ≤ l∪ ±(εi± εj) : 1 ≤ i < j ≤ l,we suppose σαi : 1 ≤ i ≤ l − 1 generates a subgroup H of W and σεi : 1 ≤ i ≤ l generates asubgroup N of W . Let ei ∈ (Z/2Z)n whose i-th entry is 1 and all other entries are 0.


0. σεi(εj) =

εj, j 6= i

−εi, j = i

. 〈εi, εj〉 =

0, j 6= i

2, j = i

1.1. For all j 6= i, σεj σεi(εk) = σεi σεj(εk) =

εk, k 6∈ i, j

−εi, k = i

−εj, k = j

1.2. σ2εi

= 1. Lemma 9.10(3)

2. σεi 7→ ei induces a homomorphism N →(Z/2Z)l.

(Z/2Z)l is generated by ei : 1 ≤ i ≤ lwith relations ei+ej = ej+ei and 2ei = 0.

σεi 7→ ei induces an isomorphism N →(Z/2Z)l.

Its inverse is induced by ei 7→ σεi .

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3. H ' Sl. similar to Proposition 16.2

4. N is a normal subgroup of W (Bl). W (Bl) acts on N is given by σαi σεj

σαi =

σεj , j 6∈ i, i+ 1;

σεi+1 , j = i;

σεi , j = i+ 1.

, see below

4.1. For j 6∈ i, i + 1, σαi σεj σαi(εk) =0, k 6= j

−εj, k = j

= σεj(εk).

step 0; and step 0 of Proposition 16.2

4.2. σαi σεi σαi(εk) =εk, k 6= i+ 1

−εi+1, k = i+ 1

= σεi+1(εk).


4.3. σαi σεi+1 σαi(εk) =

εk, k 6= i

−εi, k = i

=

σεi(εk).


5. W (Bl) = NH. below

5.1. NH is a subgroup of W (Bl). step 4

5.2. Elements of W (Bl) are products of reflec-tions of the form σαi , 1 ≤ i ≤ l.

Theorem 11.13

5.3. W (Bl) is a subgroup of NH. Use step 4 to move σαl to the left.

6. H ∩N = 1. below

6.1. Otherwise there exists τ = ∏i∈Iσαi =∏

j∈Jσεj 6= 1 in H∩N , where I ⊂ 1, . . . , l−

1, J ⊂ 1, . . . , l and the left hand sideis the reduced decomposition.

N ∩H 6= 1.

6.2. There exists εk such that τ(εk) =∏i∈Iσαi(εk) = εt, t 6= k.

τ 6= 1 and step 0 of Proposition 16.2



6.3. τ(εk) = ∏j∈J

σεj(εk) = ±εk. step 0

6.4. a contradiction. steps 6.2 and 6.3.

7. W (Bl) = NoH ' (Z/2Z)loSl and henceCard(W (Bl)) = 2ll!.

steps 4,5,6 and Definition 16.3

Proposition 16.5Let E ' Rl be the Eulidean space. Let Φ(Cl) be the irreducible root system of type Cl, (l ≥ 3) inE with base ∆(Cl) = αi = εi − εi+1 : 1 ≤ i ≤ l− 1 ∪ αl = 2εl. Let W (Cl) be the Weyl groupof Φ(Cl). Let Sl be the permutation group of 1, 2, . . . , l. Then W (Cl) ' (Z/2Z)l o Sl and ithas order 2ll!.Proof.


1. Φ(Cl) = Φ(Bl)∨. Proposition 14.11

2. σα = σα∨ . α∨ = 2(α, α)α and Example 9.9(1)

3. W (Cl) = W (Bl). Theorem 11.13

Proposition 16.6Let E ' Rl be the Eulidean space. Let Φ(Dl) be the irreducible root system of type Dl, (l ≥ 4) inE with base ∆(Dl) = αi = εi − εi+1 : 1 ≤ i ≤ l − 1 ∪ αl = εl−1 + εl. Let W (Dl) be the Weylgroup of Φ(Dl). Let Sl be the permutation group of 1, 2, . . . , l. Then W (Dl) ' (Z/2Z)l−1 o Sl

and W (Dl) has order 2l−1l!.

Proof. Since W is generated by σα, α ∈ Φ(Dl) = ±(εi ± εj) : 1 ≤ i < j ≤ l, we supposeσαi : 1 ≤ i ≤ l − 1 generates a subgroup H of W and σε1+εj σε1−εj : 1 ≤ j ≤ l generates asubgroup N of W . Let ei ∈ (Z/2Z)n whose i-th entry is 1 and all other entries are 0.

190 ZHENGYAO WU


0. σε1+εj σε1−εj(εk) =

εk, k 6∈ 1, j,

−ε1, k = 1,

−εj, k = j,

σε1±εj(εk) =

εk, k 6∈ 1, j,

∓εj, k = 1,

∓ε1, k = j,

1.1. For all j 6= i, (σε1+εj σε1−εj) (σε1+εi σε1−εi) = σεi+εj σεi−εj = (σε1+εi σε1−εi)(σε1+εj σε1−εj).

step 0

1.2. For all 1 ≤ i ≤ l, (σε1+εj σε1−εj)2 = 1. step 0

2. N → (Z/2Z)l−1; σε1+εj σε1−εj 7→ ej−1

induces a homomorphism.steps 1,2

N → (Z/2Z)l−1; σε1+εj σε1−εj 7→ ej−1

induces an isomorphism.Its inverse is induced by ej 7→ σε1+εj+1 σε1−εj+1

3. H ' Sl. similar to Proposition 16.2

4. N is a normal subgroup of W (Dl). W (Dl) acts on N , see below

4.1. For 1 ≤ i ≤ l − 1 and 2 ≤j ≤ l, σαi (σε1+εj σε1−εj) σαi =

σε1+εj σε1−εj , 1, j 6∈ i, i+ 1;

σε2+εj σε2−εj , i = 1, j > 2

σε1+ε2 σε1−ε2 , i = 1, j = 2

σε1+εi+1 σε1−εi+1 , j = i

σε1+εi σε1−εi , j = i+ 1


4.2. For i = l and 2 ≤ j ≤ l, σαl (σε1+εj σε1−εj) σαl = σε1+εj σε1−εj .


5. W (Dl) = NH. below

5.1. NH is a subgroup of W (Dl). step 4

5.2. Elements of W (Dl) are products of reflec-tions of the form σαi , 1 ≤ i ≤ l.

Theorem 11.13

5.3. W (Dl) is a subgroup of NH. Use step 4 to move σαl to the left.



6. H ∩N = 1. below

6.1. Otherwise there exists τ = ∏i∈Iσαi =∏

j∈J(σε1+εj σε1−εj) 6= 1 in H ∩ N , where

I ⊂ 1, . . . , l − 1, J ⊂ 1, . . . , l and theleft hand side is the reduced decomposi-tion.

N ∩H 6= 1.

6.2. There exists εk such that τ(εk) =∏i∈Iσαi(εk) = εt, t 6= k.

τ 6= 1 and step 0 of Proposition 16.2

6.3. τ(εk) = ∏j∈J

(σε1+εj σε1−εj)(εk) = ±εk. step 0

6.4. a contradiction. steps 6.2 and 6.3.

7. W (Dl) = N o H ' (Z/2Z)l−1 o Sl andhence Card(W (Dl)) = 2l−1l!.

steps 4,5,6 and Definition 16.3

Theorem 16.7Let E be a Eulidean space. Let Φ(Xl) be an irreducible root system of type Xl in E. Let W (Xl)be the Weyl group of Φ(Xl). Then its structure is

Types W Card(W )

Al, (l ≥ 1) Sl+1 (l + 1)!

Bl, (l ≥ 2) (Z/2Z)l o Sl 2ll!

Cl, (l ≥ 3) (Z/2Z)l o Sl 2ll!

Dl, (l ≥ 4) (Z/2Z)l−1 o Sl 2l−1l!

E6 GO−6 (2)× 2 ' U4(2) : 2, see [CCN+85], pp.[26]-[27] 27 ∗ 34 ∗ 5 = 51840

E7 2×GO7(2) ' 2× Sp6(2), see [CCN+85], pp.[46]-[47] 210 ∗ 34 ∗ 5 ∗ 7 = 2903040

E8 2 ·GO+8 (2) ' 2 · Ω+

8 : 2, see [CCN+85], pp.[85]-[87] 214 ∗ 35 ∗ 52 ∗ 7 = 696729600

F4 GO+4 (3) ' 21+4 : S3 × S3, see [Wil09, p.103] 27 ∗ 32 = 1152

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Types W Card(W )

G2 The dihedral group D6 (or D12 by some others) 22 ∗ 3 = 12

Proof. We have a classification of connected Dynkin diagrams by Theorem 14.1.Type Al, (l ≥ 1): Proposition 16.2; Type Bl, (l ≥ 2): Proposition 16.4;Type Cl, (l ≥ 3): Proposition 16.5; Type Dl, (l ≥ 4): Proposition 16.6;Type Xl ∈ E6, E7, E8, F4, see citations. Type G2: [Hum78, p.46, Exercises 3,4].

Lemma 16.8Let E be a Euclidean space. Let Φ be a root system with a base ∆. Let W be the Weyl group ofΦ. Let Γ = σ ∈ Aut(Φ) : σ(∆) = ∆. Then Aut(Φ) = W o Γ.Proof.


1. σ σα σ−1 = σσ(α) for all α ∈ ∆. Lemma 9.16

2. W is a normal subgroup of Aut(Φ). W is generated by σα : α ∈ ∆ Theo-rem 11.13

3. Γ is a subgroup of Aut(Φ). defn of Γ

4. Γ ∩W = 1. W acts simply transitively on bases of Φby Theorem 11.15

5. For all τ ∈ Aut(Φ), τ(∆) is a base of Φ. ∆ is a base of Φ = τ−1(Φ) and Defini-tion 10.4

6. There exists σ ∈ W such that σ(τ(∆)) =∆. i.e. σ τ ∈ Γ.

Theorem 11.10

7. Aut(Φ) = W Γ. τ = σ−1 (σ τ)

8. Aut(Φ) = W o Γ. steps 4, 7 and Definition 16.3

Lemma 16.9Let E be a Euclidean space. Let Φ be a root system with a base ∆. Let W be the Weyl group of


Φ. Let Γ = σ ∈ Aut(Φ) : σ(∆) = ∆. Then Γ is identified with the group A of Dynkin diagramautomorphisms (if there is only one root length, it is also called a graph automorphism).

Proof. Define φ : Γ→ Aut(Dyn(∆)).


1. φ(σ) sends the i-th vertex to the τ(i)-thvertex for some τ ∈ Sl.

There exists τ ∈ Sl such that σ(αi) =ατ(i).

2. For all σ ∈ Γ, 〈σ(αi), σ(αj)〉 = 〈αi, αj〉 forall 1 ≤ i, j ≤ l.

Lemma 9.17

3. The multiplicity and possible arrow be-tween the i-th and the j-th vertices is thesame as those between the τ(i)-th and theτ(j)-th vertices.

Definition 13.1 and Lemma 9.30

4. φ(σ) is a diagram automorphism. Definition 13.7

Example 16.10If Φ is irreducible, then the diagram automorphism group Γ satisfies:

Type A1 Al, (l ≥ 2) Bl, (l ≥ 2) Cl, (l ≥ 3) D4 Dl, (l ≥ 5) E6 E7 E8 F4 G2

Γ 1 Z/2Z 1 1 S3 Z/2Z Z/2Z 1 1 1 1

Definition 16.11Let E be an Euclidean space. Let Φ be a root system in E. Let Λ = λ ∈ E : 〈λ, α〉 ∈ Z, ∀α ∈ Φ.We call λ ∈ Λ a weight.

Lemma 16.12Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let Λ be the set ofweights of Φ. Then(1) (Λ,+) is a subgroup of (E,+). (2) Φ ⊂ Λ. (3) λ ∈ Λ iff 〈λ, α〉 ∈ Z for all α ∈ ∆.

Proof. (1)(2) follow from Lemma 9.10(2), Definition 9.12(R4), respectively.For (3), suppose β ∈ Φ. 〈λ, β〉 = 〈β∨, λ∨〉 suppose β = ∑

α∈∆kαα.

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1. 〈λ, β〉 = 〈β∨, λ∨〉 [Hum78, p.46, Exercise 2]

= 〈 ∑α∈∆

kαα∨, λ∨〉 [Hum78, p.54, Exercise 1]; and β∨ =∑

α∈∆kαα

∨ by Definition 10.4(B2)

= ∑α∈∆

kα〈α∨, λ∨〉 Lemma 9.10(2)

= ∑α∈∆

kα〈λ, α〉 [Hum78, p.46, Exercise 2]

∈ Z. λ ∈ Λ; and kα ∈ Z by Definition 10.4(B2)

Definition 16.13Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let Λ be the group ofweights of Φ. The subgroup Λr of Λ generated by Φ is called the root lattice.

Example 16.14Let E be an Euclidean space. Let Φ be a root system in E. Let Λ be the group of weights of Φ.Then the root lattice Λr is a lattice in E.

Proof. Since Λr = SpanZ(∆) ' Zl, it is a lattice by Definition 14.3.

Definition 16.15Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let Λ be the group ofweights of Φ and λ ∈ Λ.

(1) We call λ ∈ Λ strongly dominant if 〈λ, α〉 > 0, for all α ∈ ∆.(2) We call λ dominant if 〈λ, α〉 ≥ 0 for all α ∈ ∆. Let Λ+ be the set of dominant weights.

Lemma 16.16Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let Λ be the group ofweights of Φ.(1) Λ ∩ C(∆) is the set of strongly dominant weights; (2) Λ+ = Λ ∩ C(∆).

Proof. (1) follows from C(∆) = x ∈ E : (x, α) > 0, ∀α ∈ ∆ and Definition 16.15(1).(2) follows from C(∆) = x ∈ E : (x, α) ≥ 0, ∀α ∈ ∆ and Definition 16.15(2).

Definition 16.17Let E be an Euclidean space. Let Φ be a root system in E with a base ∆ = α1, . . . , αl. Suppose


λ1, . . . , λl ∈ E such that 〈λi, αj〉 = δij =

0, j 6= i

1, j = i

. We call λi : 1 ≤ i ≤ l fundamental

dominant weights relative to ∆.

Lemma 16.18Let E be an Euclidean space. Let Φ be a root system in E with a base ∆ = α1, . . . , αl. Let Λbe the group of weights of Φ. Then Λ is a lattice. We call Λ the weight lattice of Φ relative to∆.

Proof. Suppose mi = 〈λ, αi〉.


1. Λ = SpanZλi : 1 ≤ i ≤ l. λi ∈ Λ for all 1 ≤ i ≤ l and below

1.1. For all 1 ≤ j ≤ l, 〈λ −l∑

i=1miλi, αj〉 =

〈λ, αj〉 −l∑

i=1mi〈λi, αj〉

Lemma 9.10(2)

= mj −l∑

i=1miδij = mj −mj = 0. Definition 16.17

1.2. For all 1 ≤ j ≤ l, (λ−l∑

i=1miλi, αj) = 0. Example 9.9(2)

1.3. λ =l∑

i=1miλi. (•, •) is non-degenerate and ∆ is a basis

of E

2. λi : 1 ≤ i ≤ l are linearly independent. If c1λ1 + · · · + clλl = 0, then 0 = 〈c1λ1 +· · ·+ clλl, αj〉 = cj for all 1 ≤ j ≤ l.

3. Λ is a free Z-module with basis λi : 1 ≤i ≤ l.

steps 1,2

Lemma 16.19Let E be an Euclidean space. Let Φ be a root system in E with a base ∆ = α1, . . . , αl. LetΛr be the root lattice of Φ. Let Λ be the weight lattice of Φ relative to ∆. Then Λ/Λr is a finitegroup, called the fundamental group of Φ.

Proof. Let C be the Cartan matrix. Suppose αi =l∑

j=1mijλj.

196 ZHENGYAO WU


1. For all 1 ≤ i, k ≤ l, 〈αi, αk〉 =〈l∑

j=1mijλj, αk〉 =

l∑j=1

mij〈λj, αk〉

Lemma 9.10(2)

=l∑

j=1mijδjk = mik. Definition 16.17

2. C−1 sends ∆ to λj : 1 ≤ j ≤ l. C is the change of bases of E fromλj : 1 ≤ j ≤ l to ∆.

3. Denominators of C−1 are factors ofdet(C).

C−1 = C∗/ det(C) where C∗ is the matrixof cofactors of C

4. Λ/Λr is a finite group with order dividingdet(C).

λi + Λr has order dividing det(C) for all1 ≤ i ≤ l

Example 16.20Let E be an Euclidean space. Let Φ be a root system in E of type A1 with a base ∆ = α1. LetΛr be the root lattice of Φ. Let Λ be the weight lattice of Φ relative to ∆. Then Λ/Λr ' Z/2Z.Proof.


1. λ1 = 12α1. Lemma 9.10(1)(2)

2. Λ = Zλ1. Lemma 16.18

Λr = Zα1. Definition 16.13

Λ/Λr ' Z/2Z. step 1

Lemma 16.21Let E be an Euclidean space. Let Φ be an irreducible root system in E with a base ∆. If thereexists α ∈ ∆ such that α is a dominant weight, then Φ has type A1.

Proof. By Theorem 14.2, every row of the Cartan matrix of an irreducible has a negative entryunless for A1.


Example 16.22Let E be an Euclidean space. Let Φ be a root system in E of type A2 with a base ∆ = α1, α2.Let Λr be the root lattice of Φ. Let Λ be the weight lattice of Φ relative to ∆. Then Λ/Λr ' Z/3Z.

Proof. Let C be the Cartan matrix of A2.


1. C =

2 −1

−1 2

. Theorem 14.2

2. det(C) = 3 and C−1 = 13

2 1

1 2

. Linear algebra

3. λ1 = 23α1 + 1

3α2, λ2 = 13α1 + 2

3α2. step 2 of Lemma 16.19

4. Λ/Λr ' Z/3Z. Λr ( Λ and step 4 of Lemma 16.19

α1

α2

λ1

λ2

Example 16.23Fundamental groups of irreducible root systems are

Type Al, (l ≥ 1) Bl, (l ≥ 2) Cl, (l ≥ 3) Dl, (l ≥ 4 odd) Dl, (l ≥ 4 even)

Λ/Λr Z/(l + 1)Z Z/2Z Z/2Z Z/4Z Z/2Z× Z/2Z

Type E6 E7 E8 F4 G2

Λ/Λr Z/3Z Z/2Z 1 1 1

Proof. By [Hum78, p.63, Exercise 2],

198 ZHENGYAO WU

Type Al, (l ≥ 1) Bl, (l ≥ 2) Cl, (l ≥ 3) Dl, (l ≥ 4) E6 E7 E8 F4 G2

det(C) l + 1 2 2 4 3 2 1 1 1


1. The order of Λ/Λr divides the determinantof the Cartan matrix C.

step 4 of Lemma 16.19

2. For types A and D, Λ/Λr are as desired. [Hum78, p.71, Exercise 4]

3. For each other type, Λ/Λr is cyclic. Λr ( Λ by Lemma 16.21; and each det(C)is prime

Definition 16.24Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let Λ be the weightlattice of Φ relative to ∆. For λ, µ ∈ Λ, we say that λ > µ if λ− µ is a sum of positive roots (it ispossible that λ− µ 6∈ Φ+).

Lemma 16.25Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let W be the Weylgroup of Φ. Let Λ be the weight lattice of Φ relative to ∆. For all λ ∈ Λ, there exists σ ∈ W suchthat σ(λ) is dominant and σ(λ) is unique.

Proof. Existence. [Hum78, p.55, Exercise 14].Uniqueness. Suppose σ, τ ∈ W such that σ(λ), τ(λ) are dominant, i.e. σ(λ), τ(λ) ∈ C(∆). Sinceτ σ−1(σ(λ)) = τ(λ), by Lemma 11.21, σ(λ) = τ(λ).

Lemma 16.26Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let W be the Weylgroup of Φ. Let Λ be the weight lattice of Φ relative to ∆ and λ ∈ Λ.(1) If λ is dominant, then σ(λ) ≤ λ for all σ ∈ W .(2) If λ is strongly dominant, then σ(λ) = λ iff σ = 1.

Proof. (1) Suppose σ ∈ W and λ ∈ Λ.


1. For all α, β ∈ ∆, 〈α, β〉 ∈ Z Definition 9.12(R4)



2. For all µ ∈ Λ, 〈µ, α〉, 〈µ, β〉 ∈ Z Definition 16.11

3. 〈σα(µ), β〉 = 〈µ− 〈µ, α〉α, β〉 Example 9.9(2)

= 〈µ, β〉 − 〈µ, α〉〈α, β〉 Lemma 9.10(2)

∈ Z. steps 1,2

4 . For all µ ∈ Λ and reflection σα for someα ∈ ∆, σα(µ) ∈ Λ.

Definition 16.11

5. Suppose σ = σ1 · · · σn is a reduceddecomposition for αi1 , . . . , αin ∈ ∆ andσi = σαi .

Theorem 11.13

6. We induct on l(σ) = n.

7. For n = 1, σα(λ) = λ−〈λ, α〉α ≤ λ for allα ∈ ∆.

〈λ, α〉 ≥ 0 since λ ∈ Λ+

8. σ(λ) = σ1 · · · σn−1(σn(λ)) ≤ σn(λ) l(σ1 · · · σn−1) = n− 1 case

≤ λ. the n = 1 case

(2) If σ = 1, then σ(λ) = λ. Conversely, suppose σ(λ) = λ


1. If σ 6= 1, then l(σ) > 0. σ = 1 iff l(σ) = 0

2. n(σ) = l(σ) > 0. Lemma 11.17

3. σ(α) < 0 for some α > 0. Definition 11.16

4. 0 < 〈λ, α〉 λ ∈ Λ+ and Definition 16.15(1)

= 〈σ(λ), σ(α)〉 Lemma 9.8 and Theorem 11.13

= 〈λ, σ(α)〉 σ(λ) = λ

≤ 0, a contradiction. λ ∈ Λ+ and step 3

200 ZHENGYAO WU

Lemma 16.27Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let W be the Weyl groupof Φ. Let Λ be the weight lattice of Φ relative to ∆ and λ ∈ Λ. If λ ∈ Λ+, then µ ∈ Λ+ : µ ≤ λis finite.Proof.


1. λ− µ is a sum of positive roots. µ ≤ λ and Definition 16.24

2. (λ, λ)− (µ, µ) = (λ+ µ, λ− µ) ≥ 0. λ+ µ ∈ Λ+

3. µ ∈ Λ+ : µ ≤ λ = Λ+ ∩ x ∈E : (x, x) ≤ (λ, λ) is finite.

Λ+ is discrete and x ∈ E : (x, x) ≤(λ, λ) is compact.

Example 16.28Let δ = 1

2∑

α∈Φ+α. It is possible that δ 6∈ Λr. For A1, δ = 1

2α1. It is also possible that δ ∈ Λr. For

A2, δ = 12(α1 + α2 + (α1 + α2)) = α1 + α2.

Lemma 16.29Let E be an Euclidean space. Let Φ be a root system in E with a base ∆ = αi : 1 ≤ i ≤ l. Letλi : 1 ≤ i ≤ l be the set of fundamental dominant weights relative to ∆. Let δ = 1

2∑

α∈Φ+α.

Then (1) δ =l∑

i=1λi. (2) δ ∈ Λ+.

Proof.


1. δ =l∑

i=1〈δ, αi〉λi. step 1 of Lemma 16.19

2. δ − 〈δ, αi〉αi = σi(δ) = δ − αi. Example 9.9(2) and Corollary 11.5

3. 〈δ, αi〉 = 1, i.e. (2) is true. Definition 16.15(1)

4. (1) is true. steps 1, 3


Lemma 16.30Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let W be the Weyl groupof Φ. Let Λ be the weight lattice of Φ relative to ∆. Suppose µ ∈ Λ+, σ ∈ W and ν = σ−1(µ).Then (ν + δ, ν + δ) ≤ (µ+ δ, µ+ δ). The equality holds iff ν = µ.Proof.


1. δ ∈ Λ+. Lemma 16.29(2)

2. σ(δ) ≤ δ. Lemma 16.26(1)

3. (ν + δ, ν + δ) = (σ(ν + δ), σ(ν + δ)) Lemma 9.8

= (µ+ σ(δ), µ+ σ(δ)) σ is linear

= (µ, µ) + 2(µ, σ(δ)) + (σ(δ), σ(δ)) (•, •) is bilinear

= (µ, µ) + 2(µ, σ(δ)) + (δ, δ) Lemma 9.8

4. (µ+ δ, µ+ δ)− 2(µ, δ − σ(δ))

= (µ, µ) + 2(µ, δ) + (δ, δ) − 2(µ, δ) +2(µ, σ(δ))

(•, •) is bilinear

= (µ, µ) + 2(µ, σ(δ)) + (δ, δ)

5. (ν+δ, ν+δ) = (µ+δ, µ+δ)−2(µ, δ−σ(δ))

≤ (µ+ δ, µ+ δ). step 2 and µ ∈ Λ+

6. ν = σ−1(µ) ≤ µ. µ ∈ Λ+ and Lemma 16.26

7. (µ, δ − σ(δ)) = (µ, δ)− (µ, σ(δ)) (•, •) is bilinear

= (µ, δ)− (σ−1(µ), δ) Lemma 9.8

= (µ, δ)− (ν, δ) defn of ν

= (µ− ν, δ) (•, •) is bilinear

≥ 0. step 1,6

8. with equality iff (µ− ν, δ) = 0 iff µ = ν. Otherwise (µ− ν, δ) > 0 by steps 1,7

202 ZHENGYAO WU

Definition 16.31Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let W be the Weylgroup of Φ. Let Λ be the weight lattice of Φ relative to ∆. A subset Π of Λ is saturated if for allλ ∈ Π, α ∈ Φ, λ− iα ∈ Π for all i between 0 and 〈λ, α〉.

Lemma 16.32Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let W be the Weylgroup of Φ. Let Λ be the weight lattice of Φ relative to ∆. Let Π be a saturated subset of Λ. Thenσ(Π) ⊂ Π for all σ ∈ W .

Proof. If λ ∈ Π, then σα(λ) = λ − 〈λ, α〉α ∈ Π by Definition 16.31. The assertion follows sinceevery σ ∈ W is a product of simple reflections by Theorem 11.13.

Definition 16.33Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let W be the Weylgroup of Φ. Let Λ be the weight lattice of Φ relative to ∆. Let Π be a saturated subset of Λ. Wesay that Π has highest weight λ if λ ∈ Π and for all µ ∈ Π, µ ≤ λ.

Example 16.34(1) 0 is saturated.(2) Let Φ be a root system. Then Φ ∪ 0 is saturated.(3) Let Φ be an irreducible root system. If λ is the maximal root relative to ∆ (see Lemma 12.7),then it is also the highest weight in Φ ∪ 0.

Lemma 16.35Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let W be the Weylgroup of Φ. Let Λ be the weight lattice of Φ relative to ∆. Let Π be a saturated subset of Λ. If Πhas a highest weight, then it is finite.

Proof. Let λ be the highest weight of Π.


1. For all µ ∈ Π, there exists σ ∈ W suchthat σ(µ) ∈ Λ+.

Lemma 16.25

2. σ(µ) ∈ Π. Π is saturated and Lemma 16.32

3. σ(µ) ≤ λ. λ is the highest weight of Π

4. ν ∈ Λ+ : ν ≤ λ is finite. Lemma 16.27



5. W is finite. Lemma 9.15

6. Card(Π) is finite. ≤ Card(W ) · Card(ν ∈ Λ+ : ν ≤ λ)

Lemma 16.36Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let W be the Weylgroup of Φ. Let Λ be the weight lattice of Φ relative to ∆. Let Π be a saturated subset of Λ withhighest weight λ. If µ ∈ Λ+ and µ ≤ λ, then µ ∈ Π.

Proof. Suppose µ′ = µ + ∑α∈∆

kαα, kα ≥ 0, kα ∈ Z and ht(µ′ − µ) > 0. We show that if µ′ ∈ Π,then there exists β ∈ ∆ such that for some β ∈ ∆, kβ > 0 and µ′ − β ∈ Π. Since λ ≥ µ, we startfrom the given λ ∈ Π and obtain µ ∈ Π by induction.


1. ( ∑α∈∆

kαα,∑α∈∆

kαα) > 0. ht(µ′ − µ) > 0.

2. ( ∑α∈∆

kαα, β) > 0 for some β ∈ ∆ withkβ > 0.

3. 〈 ∑α∈∆

kαα, β〉 > 0. Example 9.9 and 2(β, β) > 0

4. If µ′ ∈ Π, then µ′ − β ∈ Π. Definition 16.31

Corollary 16.37Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let W be the Weylgroup of Φ. Let Λ be the weight lattice of Φ relative to ∆. Let Π be a saturated subset of Λ withhighest weight λ. Then

Π = µ ∈ Λ : ∃σ ∈ W , σ(µ) ∈ Λ+, σ(µ) ≤ λ.Proof.

204 ZHENGYAO WU


1. Suppose µ ∈ Λ such that σ(µ) ∈Λ+, σ(µ) ≤ λ for some σ ∈ W .

1.1. σ(µ) ∈ Π. Lemma 16.36

1.2. µ ∈ σ−1(Π) ⊂ Π. Π is saturated and Lemma 16.32

2. Conversely, suppose µ ∈ Π.

2.1. There exists σ ∈ W s.t. σ(µ) ∈ Λ+. Lemma 16.25

2.2. σ(µ) ≤ λ. Definition 16.33

Lemma 16.38Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let W be the Weylgroup of Φ. Let Λ be the weight lattice of Φ relative to ∆. Let Π be a saturated subset of Λ withhighest weight λ. If µ ∈ Π, then (µ+ δ, µ+ δ) ≤ (λ+ δ, λ+ δ). The equality holds iff µ = λ.Proof.


1. There exists σ ∈ W such that σ(µ) ∈ Λ+

and σ(µ) ≤ λ.Corollary 16.37

2. λ− σ(µ) = π is a sum of positive roots. Definition 16.24

3. (µ+ δ, µ+ δ) ≤ (σ(µ) + δ, σ(µ) + δ). Lemma 16.30

≤ (π, σ(µ) + δ) + (σ(µ) + δ, σ(µ) + δ) σ(µ) + δ ∈ Λ+

= (λ+ δ, σ(µ) + δ) step 2

≤ (λ+ δ, π) + (λ+ δ, σ(µ) + δ) λ+ δ ∈ Λ+

= (λ+ δ, λ+ δ)

4.1. If µ = λ, then the equality holds.

4.2. Conversely, if the equality holds, then (λ+δ, π) = (π, σ(µ) + δ) = 0 and σ(µ) = µ.

(•, •) is non-degenerate



π = 0, i.e. µ = λ. λ+ δ is strongly dominant and step 2

References

[CCN+85] J. H. Conway, R. T. Curtis, S. P. Norton, R. A. Parker, and R. A. Wilson, Atlas of finite groups, OxfordUniversity Press, Eynsham, 1985, Maximal subgroups and ordinary characters for simple groups, Withcomputational assistance from J. G. Thackray. MR 827219 →191

[Hum78] James E. Humphreys, Introduction to Lie algebras and representation theory, Graduate Texts in Mathe-matics, vol. 9, Springer-Verlag, New York-Berlin, 1978, Second printing, revised. MR 499562 →1, →45,→46, →117, →131, →138, →149, →168, →192, →194, →197, →198

[Wil09] Robert A. Wilson, The finite simple groups, Graduate Texts in Mathematics, vol. 251, Springer-VerlagLondon, Ltd., London, 2009. MR 2562037 →191

Department of Mathematics, Shantou University, 243 Daxue Road, Shantou, Guangdong, China515063E-mail address: [email protected]

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