Notes to Lie Algebras and Representation Theory [version 5 Jun 2019 ed.]

Table of contents :
1. February 26th, Introduction to Lie algebra......Page 2
2. March 5th, Solvable and nilpotent Lie algebras......Page 12
3. March 12th, Lie theorem, Jordan decomposition......Page 26
4. March 19th, Cartan's criterion, Killing form......Page 36
5. March 26th, Semisimple decomposition and Lie modules......Page 49
6. April 2nd, Casimir element, Weyl's theorem......Page 61
7. April 9th, Representation of `39`42`"613A``45`47`"603Asl(2, F) , toral subalgebras......Page 73
8. April 16th, Centralizer of H ; Orthogonal, integral properties......Page 83
9. April 23th, Rationality properties, reflections, root systems......Page 98
10. April 30th, Bases and Weyl chambers......Page 110
11. May 7th, Weyl group and its actions......Page 122
12. May 14th, Irreducible root systems, two root lengths and Cartan matrix......Page 133
13. May 21th, Coxeter graphs, Dynkin diagrams......Page 147
14. May 28th, Classification, irreducible root systems of types A, B and C......Page 157
15. June 4th, Irreducible root systems of types D,E,F,G......Page 169
16. June 11th, Weyl group of each type, Automorphisms of the Dynkin diagram, Weights......Page 185
References......Page 205

Citation preview

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 at Shantou University taught by me. • Targeted audience: Graduate students in pure mathematics.

Contents 1. February 26th, Introduction to Lie algebra 2 2. March 5th, Solvable and nilpotent Lie algebras 12 3. March 12th, Lie theorem, Jordan decomposition 26 4. March 19th, Cartan’s criterion, Killing form 36 5. March 26th, Semisimple decomposition and Lie modules 49 6. April 2nd, Casimir element, Weyl’s theorem 61 7. April 9th, Representation of sl(2, F ), toral subalgebras 73 8. April 16th, Centralizer of H; Orthogonal, integral properties 83 9. April 23th, Rationality properties, reflections, root systems 98 10. April 30th, Bases and Weyl chambers 110 11. May 7th, Weyl group and its actions 122 12. May 14th, Irreducible root systems, two root lengths and Cartan matrix 133 13. May 21th, Coxeter graphs, Dynkin diagrams 147 14. May 28th, Classification, irreducible root systems of types A, B and C 157 15. June 4th, Irreducible root systems of types D,E,F,G 169 16. June 11th, Weyl group of each type, Automorphisms of the Dynkin diagram, Weights 185 References 205

1

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ZHENGYAO WU

1. February 26th, Introduction to Lie algebra Sophus Lie (1842-1899) established the theory in late 1880s in Oslo, Norway. Definition 1.1 Let F be a field. Let L be a F -vector space. We say that L is a F -Lie algebra if there exists a map 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 1.3 Let 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 1.4 Let 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 ] = δjk eil − δli ekj where eij is the n × n matrix whose (i, j)-entry is 1 and all other entries are 0; δij = 1 if i = j, otherwise δij = 0. Subalgebras of gl(V ) ' gl(n, F ) are called linear algebras. Definition 1.5 Let L, L0 be two F -Lie algebras. A map f : L → L0 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-Iwasawa Every F -Lie algebra is isomorphic to a linear F -Lie algebra. Proof. Omit. Definition 1.7 Classical algebras are the following proper subalgebras of gl(V ) of type Al , Bl , Cl , Dl , l ≥ 1.



NOTES TO HUMPHREYS

3

Example 1.8 Type 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 1.9 Type 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 

  . Il    

0

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 = −xt s}

=

     0        t −b  2           −bt1



        

b1

b2  

m

 ∈ gl(2l + 1, F ) : nt = −n, pt = −p n   

p



−mt

 

      

dimF (Bl ) = 2l2 + l since Bl has standard basis {ei+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 1.10 Type Cl (l ≥ 1): symplectic algebras 



0



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

−Il

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 = −xt s}

Il  .  0

4

ZHENGYAO WU

=

   m     

p



n   −mt

   

∈ gl(2l, F ) : nt = n, pt = p .



  

dimF (sp(2l, F )) = 2l2 + l since it has standard basis {ei,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 1.11 Type Dl (l ≥ 1): orthogonal algebras of even degree: 



 0 Il  . Let f be a non-degenerate symmertic bilinear form on V whose matrix is s =    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 = −xt s}

=

   m       

p



n  −mt

 

   

∈ gl(2l, F ) : nt = −n, pt = −p .   

dimF (o(2l, F )) = 2l2 − l since it has standard basis {ei,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 1.12 A derivation of an F -algebra A is an F -linear map δ : A → A such that δ(ab) = aδ(b) + δ(a)b for all a, b ∈ A. Let Der(A) be the set of all derivations of A. We have Der(A) ⊂ gl(A). Definition 1.13 Let L be a F -Lie algebra. The map ad : L → gl(L) such that (ad x)(y) = [x, y] for all x, y ∈ L is the adjoint representation of L. Lemma 1.14 Let L be a F -Lie algebra. Then ad(L) ⊂ Der(L). Proof. We need to show that (ad x)([y, z]) = [(ad x)(y), z] + [y, (ad x)(z)] for all x, y, z ∈ L.

NOTES TO HUMPHREYS

Steps

Statements

Reasons

1.

(ad x)([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.

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

Definition 1.13

5.

(ad x) is a derivation for all x ∈ L

Definition 1.12

5

 Definition 1.15 Elements in ad(L) are called inner derivations, elements in Der(L) − ad(L) are called outer derivations. Example 1.16 Let 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 and adgl(n,F ) (x) 6= 0 for n ≥ 2. Definition 1.17 A F -Lie algebra L is abelian if [x, y] = 0 for all x, y ∈ L. Example 1.18 Every F -Lie algebra L of dimension 1 is abelian by Definition 1.1(L2). Example 1.19 A two dimensional F -Lie algebras is either abelian or isomorphic to F x + F y such that [x, y] = x. Proof. Suppose L is not abelian and L = F a + F b. Steps

Statements

Reasons

1.

[L, L] = F x for x = [a, b].

Definition 1.1(L2)

2.

There exists z ∈ L − F x.

dimF L = 2

3.

[x, z] = cx for some c ∈ F ∗ .

step 1

4.

Let y = c−1 z. Then [x, y] = x

Definition 1.1(L1)

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ZHENGYAO WU

Steps

Statements

Reasons

and L = F x + F y.

step 2

 Definition 1.20 A 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. P (5) If I and J are ideals of L, then [I, J] = { [xi , yi ] : xi ∈ I, yi ∈ J} is an ideal of L. Definition 1.22 We call L simple if . (1) L only has ideals 0 and L; (2) [L, L] 6= 0. Lemma 1.23 If L is simple, then Z(L) = 0. Proof. Steps

Statements

Reasons

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



NOTES TO HUMPHREYS

7

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

Statements

Reasons

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 1.25 If char(F ) 6= 2, then sl(2, F ) is simple. Proof. A standard basis for L = sl(2, F ) is 



0

1

x= 

0 0

, 





0

0

y= 

1 0



, 



1 0  . h=   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: Steps

Statements

Reasons

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.

char F 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.

char F 6= 2

7.

I = L.

x, y, h ∈ I by steps 3,4,6.

8

ZHENGYAO WU

The case b 6= 0: Steps

Statements

Reasons

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.

char F 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.

char F 6= 2

7.

I = L.

x, y, h ∈ I by steps 3,4,6.

The case a = b = 0: Steps

Statements

Reasons

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 1.26 Let 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).

NOTES TO HUMPHREYS

9

Definition 1.27 A linear map φ : L → L0 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) = L0 ; isomorphism if it is both a monomorphism and an epimorphism; automorphism of L0 = L and it is an isomorphism. Let Aut(L) denote the group of automorphisms of L.

Example 1.28 Let φ : L → L0 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 for all x, y ∈ L is a F -Lie algebra, called the quotient algebra. There is a canonical homomorphism π : 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 → L0 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 → L0 such that the following diagram commutes φ

L π

/

LO 0

ψ

L/I (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 1.30 Let 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 1.31 Let 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 [(ad x), (ad y)](z) = ad([x, y])(z) for all x, y, z ∈ L.

10

ZHENGYAO WU

Steps

Statements

Reasons

1.

[(ad x), (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.

[(ad x), (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 (ad x) = 0 iff [x, L] = 0 iff x ∈ Z(L). (3) Steps

Statements

Reasons

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

δn . n=0 n! k−1 P

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

Statements

1.

δ n ([x, y]) =

2.

Reasons n   P n [δ i (x), δ n−i (y)]. i=0 n P

Leibniz rule of Definition 1.12

i

δn δi δ n−i ([x, y]) = (x), (y) n! (n − i)! i=0 i! "

#

!

n n! = and Definition 1.1(L1) i i!(n − i)!

NOTES TO HUMPHREYS

11

Steps

Statements

Reasons

3.

(exp δ)([x, y]) = [(exp δ)(x), (exp δ)(y)].

exp δ =

δn and Definition 1.1(L1) n=0 n! k−1 P

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

k−1 P i=1

δi and η k = 0.  i!

Example 1.33 Suppose char(F ) = 0 and x ∈ L such that (ad x)k = 0 for some k > 0. Then exp(ad x) = 1 + (ad x) +

(ad x)2 (ad x)k−1 + ··· + ∈ Aut(L). 2! (k − 1)!

Definition 1.34 The subgroup of Aut(L) generated by exp(ad x), x ∈ L is denoted by int(L), its elements are called inner automorphisms. Lemma 1.35 int(L) is a normal subgroup of Aut(L). Proof. Steps

Statements

Reasons

1.

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

For all y ∈ L, (φ ◦ (ad x) ◦ 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(ad x) ◦ φ−1 = exp(ad(φ(x))) ∈ int(L) and Definition 1.34

 Example 1.36 Let L ⊂ gl(V ) and x ∈ L. Let λx be the left multiplication by x. Let ρx be the right multiplication −1 −1 −1 by x. Then λ−1 x = left multiplication by x ; ρx = right multiplication by x . Also, λx ◦ ρy = ρy ◦ λx for all x, y ∈ L. (1) ad x = λx − ρx since (ad x)(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 .

12

ZHENGYAO WU

2. March 5th, Solvable and nilpotent Lie algebras Lemma 2.1 Suppose char(F ) = 0. Let L be a F -Lie algebra, L ⊂ gl(V ) for some F -vector space V . If xk = 0 for some integer k > 0 in gl(V ), then exp(ad x) = int(exp x). Proof. exp(ad x) = exp(λx − ρx ) = exp(λx ) ◦ exp(ρx )−1 = λexp(x) ◦ ρ−1 exp(x) = int(exp x).



Example 2.2 



0

1

Let L = sl(2, F ) and x =  

0 0

 





0 0 . Let σ = exp(ad x) exp(ad(−y)) exp(ad x) ∈ and y =    1 0 



 0 1 . int(L). Then σ = int(s), where s = exp(x) exp(−y) exp(x) =    −1 0 







1 1  1 0 ; Since y 2 = 0, exp(−y) =  . Proof. Since x2 = 0, exp(x) =      0 1 −1 1 





1

1  1

Then s =  

0 1

 



0 1 1

−1 1

 

0 1

 





 0 1 . =   −1 0



Definition 2.3 Let 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. Steps

Statements

Reasons

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

NOTES TO HUMPHREYS

13

Steps

Statements

Reasons

5.

L is non-solvable.

Definition 2.3

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

 

 a1  

0



a1 a2 a1 b2 + b1 c2 

0











b1  a2 b2  a1 b1  a2 b2  a2 b2  a1 b1    −   =  ,          0 c1 0 c2 0 c2 0 c1 0 c2 c1



= 



c1 c2





 a1 a2 a2 b 1 + b 2 c 1 

−  

0

 

c1 c2





0 a1 b2 + b1 c2 − a2 b1 − b2 c1 

=  0

0

 

Elements of L(2) have the form 

 







0  

b1  0 b2  0 0 ,  =        0 0 0 0 0 0

So L(2) = 0, L is solvable. Proposition 2.6 If L is a solvable F -Lie algebra and K is a subalgebra of L, then K is solvable. Proof. Steps

Statements

Reasons

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)

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)

⊂

Suppose K (m−1) ⊂ L(m−1) .

14

ZHENGYAO WU

 Proposition 2.7 If L is solvable F -Lie algebra, then its homomorphic images are solvable. Proof. Suppose φ : L → L0 is a homomorphism of F -Lie algebras and M = φ(L). Steps

Statements

Reasons

0.

M is a subalgebra of L0 .

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)

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)

= =

Suppose φ(L(m−1) ) = M (m−1) .

 Proposition 2.8 Referenced 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. Steps

Statements

Reasons

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)

NOTES TO HUMPHREYS

15

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

Statements

Reasons

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 Proposition 1.29(3)

3.

I + J is solvable.

J and (I + J)/J are solvable and Proposition 2.8,

 Lemma 2.10 Let 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. Steps

Statements

Reasons

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

16

ZHENGYAO WU

Steps

Statements

Reasons

8.

Rad(L) = S.

steps 2 and 7

9.

Rad(L) is unique.

(2)

 Definition 2.11 Let 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 2.12 Let L be a finite dimensional simple F -Lie algebra. Then L is semisimple. Proof. Steps

Statements

Reasons

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 2.13 For 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. Let R = π −1 (Rad(L/ Rad(L))). Steps

Statements

Reasons

1.

R is an ideal of L.

π(R) = Rad(L/ Rad(L)) is an ideal of L/ 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)

NOTES TO HUMPHREYS

Steps

Statements

3.

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

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.

17

Reasons =

Lemma 2.10(1)

 Definition 2.14 Let 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. (ad x1 ◦ (ad x2 ) ◦ · · · ◦ (ad xn ))(y) = 0 for all xi , y ∈ L. Example 2.15 Any abelian algebra L is nilpotent, since L1 = 0. For example, d(n, F ), Z(L) are nilpotent. Lemma 2.16 Nilpotent F -Lie algebras are solvable. Proof. Let L be a nilpotent F -Lie algebra. Steps

Statements

Reasons

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)

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ZHENGYAO WU

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

 0  

d

0 0

 

by Example 2.5. Elements of L2 has form



 















b 0 d a b 0 d 0 d a b  −   ,  =           0 c 0 0 0 0 0 c 0 c 0 0

a  













0 ad 0 cd 0 (a − c)d − =  =       0 0 0 0 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 2.18 Let n(n, F ) be the set of strictly upper triangular matrices. Then n(n, F ) is nilpotent. We only 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 2.19 If L is a nilpotent F -Lie algebra and K is a subalgebra of L, then K is nilpotent. Proof. Steps

Statements

Reasons

1.

K m ⊂ Lm for all integers m ≥ 0.

Induction.

1.1

K 0 = K ⊂ L = L0 .

given.

1.2

K m = [K, K m−1 ] ⊂ [L, Lm−1 ] = Lm

Suppose K m−1 ⊂ Lm−1 .

2.

Ln = 0 for some integer n ≥ 0.

L is nilpotent and Definition 2.14(2)

3.

K n = 0.

step 1

4.

K is nilpotent.

Definition 2.14(2)



NOTES TO HUMPHREYS

19

Proposition 2.20 If L is nilpotent F -Lie algebra, then its homomorphic images are nilpotent. Proof. Suppose φ : L → L0 is a homomorphism of F -Lie algebras and M = φ(L). Steps

Statements

Reasons

1.

φ(Lm ) = M m for all integers m ≥ 0.

Induction.

1.1

φ(L0 ) = φ(L) = M = M 0 .

M = φ(L)

1.2

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

2.

Ln = 0 for some integer n ≥ 0.

L is nilpotent and Definition 2.14(2)

3.

M n = φ(Ln ) = φ(0) = 0.

step 1

4.

M is nilpotent.

Definition 2.14(2)

=

Suppose φ(Lm−1 ) = M m−1 .

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

Statements

Reasons

1.

(L/Z(L))n = 0 for some integer n > 0.

L/Z(L) is 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)

nilpotent

and

Defini-

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

20

ZHENGYAO WU

Steps

Statements

Reasons

1.

Suppose n ≥ 0 is the least integer with Ln = 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 2.23 Let L be a F -Lie algebra. For x ∈ L, x is ad-nilpotent if ad x is a nilpotent endomorphism in gl(L). Lemma 2.24 If L is nilpotent, then every element of L is ad-nilpotent. Proof. By Definition 2.14(2), ((ad x1 ) ◦ (ad x2 ) ◦ · · · ◦ (ad xn ))(y) = 0 for all xi , y ∈ L. Let x = x1 = · · · = xn . We have (ad x)n = 0.  Lemma 2.25 Let V be an F -vector space. If x ∈ gl(V ) is a nilpotent endomorphism, then ad x is also nilpotent. Proof. Steps

Statements

Reasons

1.

Suppose xn = 0 for some integer n > 0.

x is nilpotent

2.

λnx = 0 and ρnx = 0.

Example 1.36

3.

(ad x)2n = (λx − ρx )2n .

Example 1.36(1)

=

2n   P 2n i=0

=0

i

(λx )i (−ρx )2n−i

Binomial theorem either i ≥ n or 2n − i ≥ n



NOTES TO HUMPHREYS

21

Example 2.26 In ∈ gl(n, F ) is ad-nilpotent but not nilpotent, since ad(In ) = 0 and Inm = In for all integer m ≥ 1. Lemma 2.27 Let V be an F -vector space. If f ∈ gl(V ) is nilpotent, then there exists 0 6= v ∈ V such that f (v) = 0. Proof. Steps

Statements

Reasons

1.

Suppose n ≥ 0 is the least integer such that f n = 0.

f is nilpotent

2.

If n = 0, there exists 0 6= v ∈ V such that f (v) = 0.

f =0

3.

If n > 0, then f n−1 6= 0.

n is the least.

4.

There exists 0 6= w ∈ V such that f n−1 (w) 6= 0.

5.

There exists 0 6= v ∈ V and f (v) = 0.

v = f n−1 (w) and f n (w) = 0

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

Statements

Reasons

1.

The quotient algebra L/K and the canonical homomorphism π : L → L/K exists.

K is an ideal of L

2.

If dimF (L/K) > 1, then it has a proper 1-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 homomorphism.

22

Steps

ZHENGYAO WU

Statements

Reasons

π −1 (S) is proper.

dimF (π −1 (S)) = 1 + dimF (K) dimF (L/K) + dimF (K) = dimF (L).




5.

A contradiction. Thus dimF (L/K) = 1.

K is a maximal proper subalgebra of L.

6.

L = K + F z for some z ∈ L − K.

Suppose {z + K} is a basis of L/K.

=

1 + dimF (K)

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

Statements

Reasons

1.

Every element of ad(L) is nilpotent.

L is linear, every element of L is nilpotent and 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 such that 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



NOTES TO HUMPHREYS

23

Theorem 2.30 Let L be a subalgebra of gl(V ), where V is finite dimensional, V 6= {0}. If every element of L is nilpotent, 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 ). By Lemma 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. Steps

Statements

Reasons

1.

There exists a maximal proper nonzero subalgebra K of L.

dimF (L) > 1

2.

K is an ideal of L.

Every element of L is nilpotent and Lemma 2.29

3.

L = K + F z 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 inductive hypothesis

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 for all 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 Engel Let L be a subalgebra of gl(V ), where V is finite dimensional, V 6= {0}. If every element of L is ad-nilpotent, then L is nilpotent.

24

ZHENGYAO WU

Proof. Steps

Statements

Reasons

1.

If dimF (L) = 1, then L is nilpotent.

L is abelian, Example 1.18 and Example 2.15

2.

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

Every element of L is ad-nilpotent and Theorem 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 2.32 Let 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 2.33 Let L be a subalgebra of gl(V ), where V is an n-dimensional F -vector space, V 6= {0}. Suppose x ∈ L. If every element of L is nilpotent, then there exists a flag (Vi ) in V , stable under L, with x(Vi ) ⊂ Vi−1 for all i, i.e. there exists a basis of V relative to which the matrices of L are all in n(n, F ). Proof. We prove by induction. For n = 1, take the flag {0} = V0 ⊂ V1 = V . If x = 0, then 0V = {0}. If x 6= 0, then there exists 0 6= v ∈ V such that x(v) = 0 by Lemma 2.27. Since V = F v, x(V ) = {0}. Then v form a basis of V relative to which the matrices of L are all (0) and n(1, F ) = {(0)}. Now we suppose n > 1. Steps

Statements

Reasons

1.

There exists v ∈ V such that x(v) = 0 for all x ∈ L.

Every element of L is nilpotent and Theorem 2.30

NOTES TO HUMPHREYS

25

Steps

Statements

Reasons

2.

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

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

3.

Let π : V → V /(F v) be the quotient map. Let Vi = π −1 (Wi−1 ) for all 1 ≤ i ≤ n. To be precise, V0 = 0, V1 = F v, V2 = π −1 (W1 ), . . . , Vn = π −1 (Wn−1 ) = π −1 (V /(F v)) = 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 /F v has a basis (e1 , . . . , en−1 ) relative to which the matrices of L/(F v) are all in n(n − 1, F ).

Inductive hypothesis

6.

V has a basis (v, e1 , . . . , en−1 ) relative to which 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 3.1 Let L be a subalgebra of gl(V ), where V is finite dimensional, V 6= {0}. Let L be a nilpotent F -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 3.2 Let F be an algebraically closed field of characteristic 0. Let L be a solvable F -Lie algebra of dimension n > 0. Then there exists an ideal K of L of codimension 1. Proof. Steps

Statements

Reasons

1.

[L, L] ( L.

L is solvable, n > 0 and Definition 2.3(2)

2.

L/[L, L] has a codimension 1 subspace K 0 .

dimF (L/[L, L]) ≥ 1

3.

Let π : L → L/[L, L] be the canonical quotient map and K = π −1 (K 0 ).

3.1.

K is an ideal of L.

[L, K] ⊂ [L, L] = π −1 (0) ⊂ π −1 (K 0 ) = K.

3.2.

K is of codimension 1.

dimF (K) = dimF (K 0 ) + dimF ([L, L]) = dimF (L/[L, L])−1+dimF ([L, L]) = n−1.

 Lemma 3.3 Let F be an algebraically closed field of characteristic 0. Let V be a finite dimensional F -vector space, V 6= {0}. Let L be a solvable subalgebra of gl(V ). Let K be an ideal of L of codimension 1. 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 that w, x(w), . . . , xn (w) are linearly dependent. Let W0 = 0; Wi = SpanF {w, x(w), x2 (w) . . . , xi−1 (w)}, 1 ≤ i ≤ n. Let y ∈ K.

NOTES TO HUMPHREYS

Steps

Statements

Reasons

Steps

Statements

Reasons

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 and x(Wn ) ⊂ Wn .

defn of Wi , (0 ≤ i ≤ n)

2.

y(xi (w)) ≡ λ(y)xi (w) mod Wi , 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) + w0 for some w0 ∈ Wi−1 .

Inductive hypothesis

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) + w0 ) − λ([x, y])xi−1 (w)

step 2.2

27

= λ(y)xi (w) + x(w0 ) − λ([x, y])xi−1 (w) where x(w0 ) − 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

x(w) ∈ W .

defn of W

7.

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



Theorem 3.4 Let F be an algebraically closed field of characteristic 0. Let V be a finite dimensional F -vector space, V 6= {0}. Let L be a solvable subalgebra of gl(V ). Then there exists in V a common eigenvector 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 an eigenvector of 0. Now we suppose dimF (L) > 0. Steps

Statements

Reasons

1.

There exists a codimension one ideal K of L.

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

2.

K is solvable.

L is solvable, K is a subalgebra of L and Proposition 2.6

3.

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

Inductive hypothesis

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 with an 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 ∈ K and c ∈ F . Define λ(x) = λ(y) + cλ(z).

L = K + F z 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 theorem Let F be an algebraically closed field of characteristic 0. Let V be an n-dimensional F -vector space, 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. Steps

Statements

Reasons

1.

There exists a common eigenvector v of L.

L is a solvable subalgebra of gl(V ) and Theorem 3.4

NOTES TO HUMPHREYS

29

Steps

Statements

Reasons

2.

L acts on V /(F v).

x(v) ∈ F v for all x ∈ L

3.

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

dimF (V /(F v)) = n − 1 and inductive hypothesis

4.

Let π : V → V /(F v) 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 /F v has a basis (e1 , . . . , en−1 ) relative to which the matrices of L/(F v) are all in t(n − 1, F ).

Inductive hypothesis

7.

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

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

= ⊂

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

Statements

Reasons

1.

ad(L) is solvable.

L is solvable and Proposition 2.7

2.

There exists a flag (Li )0≤i≤n of subspaces of 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 3.7 Let F be an algebraically closed field of characteristic 0. Let L be finite dimensional solvable F -Lie algebra. If x ∈ [L, L] then ad x is nilpotent. Furthermore, [L, L] is nilpotent. Proof. Steps

Statements

Reasons

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 such that Li = SpanF {x1 , . . . , xi } relative to which 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 3.8 Let F be an algebraically closed field (of arbitrary characteristic). Let V be a finite dimensional F -vector space. We call x ∈ EndF (V ) semisimple if the roots of its minimal polynomial over F are distinct. Lemma 3.9 Let F be an algebraically closed field. Let V be a finite dimensional F -vector space. Then x ∈ 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 is the size of the largest Jordan block of a. Each block has the form 



a 1 0 · · · 0 0    0    .. .    0   

a 1 ··· 0 .. .

.. .

..

.

.. .

0 0 ··· a

   0   ..  . .    1   

0 0 0 ··· 0 a

NOTES TO HUMPHREYS

31

Since x is similar to its Jordan canonical form, x is semisimple iff every Jordan block of x has size 1 × 1 iff the Jordan canonical form of x is diagonal.  Lemma 3.10 Let F be an algebraically closed field. Let V be a finite dimensional F -vector space. Let W be a subspace 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 . Steps

Statements

Reasons

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 3.11 Let 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 ) such that 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 for all 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 algebraically closed. Steps

Statements

Reasons

1.

x(Eλ ) ⊂ Eλ for all eigenvalues λ of y.

y(x(v)) = x(y(v)) = x(λv) = λx(v) for all v ∈ Eλ

2.

x is semisimple.

F is algebraically closed, x is diagonalizable and Lemma 3.9

3.

x|Eλ is semisimple.

step 1 and Lemma 3.10

32

ZHENGYAO WU

Steps

Statements

Reasons

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 ) ◦ zλ−1 is diagonal.

6.

zλ ◦ (y|Eλ ) ◦ zλ−1 = λIm , m = dimF (Eλ ).

7.

Let z =

L

row vectors of zλ are eigenvectors.

zλ . Then z◦x◦z −1 and z◦y◦z −1

V =

λ

L

Eλ and steps 5,6 for all λ

λ

are diagonal.

(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, there exists w ∈ GL(V ) such that w ◦ z ◦ xi ◦ z −1 ◦ w−1 is diagonal.  Lemma 3.12 Let 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 Q

(T − ai )mi , where a1 , . . . , am are distinct.

i=1

Then there exists a polynomial p(T ) ∈ F [T ] such that p(T ) ≡ ai mod(T − ai )mi , p(T ) ≡ 0 mod T. 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   1 2   ∈ C2×2 , its characteristic polynomial is (T − 1)(T + 1). We need to find p(T ) For x =    0 −1 such that p(T ) ≡ 1 mod(T − 1), p(T ) ≡ −1 mod(T + 1) and p(T ) ≡ 0 mod T . In fact, 1 T (T + 1)y1 ≡ 1 mod(T − 1) =⇒ y1 ≡ mod(T − 1). 2 1 T (T − 1)y2 ≡ 1 mod(T + 1) =⇒ y2 ≡ mod(T + 1). 2 (T − 1)(T + 1) · y3 ≡ 1 mod T

=⇒ y3 ≡ −1 mod T.

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

NOTES TO HUMPHREYS

33

Proposition 3.14 Jordan-Chevalley decomposition Let 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 is semisimple; xn is nilpotent; and xs xn = xn xs . (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 xs y = yxs and xn y = 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 is Let Vi = ker(x − ai · 1)mi . (a) Existence.

k Q

(T − ai )mi , where a1 , . . . , am are distinct.

i=1

Steps

Statements

Reasons

1.

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

Lemma 3.12

2.

Define xs = p(x), xn = x − p(x). Then x = xs + xn . Thus xs xn = xn xs .

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 P

Vi and Lemma 3.9

i=1

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 P

Vi

i=1

(b) Steps

Statements

Reasons

1.

p(0) = 0.

p(T ) ≡ 0 mod T 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

Steps

Statements

Reasons

4.

xn = q(x).

step 2 of (a) Existence.

5.

If xy = yx, then xs y = yxs and xn y = 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 = x0s + x0n are two decompositions. Steps

Statements

Reasons

6.

xxs = xs x and xxn = xn x.

step 2 of (a)

7.1.

xs x0s = x0s xs .

step 5 of (b) and step 6 here

7.2.

xs − x0s is diagonalizable.

xs , xs0 are diagonalizable and Lemma 3.11

8.1.

xn x0n = x0n xn .

step 5 of (b) and step 6 here

8.2.

x0n − xn is nilpotent.

Binomial theorem

9.

xs − x0s = x0n − xn = 0.

It is both semisimple and nilpotent

 Example 3.15  1

In C2×2 , x =  



2  

0 −1





1

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



0  

0 −1







0 2 . and n =    0 0 

2 0 −2 , ns =  , sn 6= ns.    0 0 0 0

0

Although x = s + n, s is semisimple and n is nilpotent, sn =   So t = s + n is not the Jordan decomposition of t. Example 3.16 

1 2  ∈ C2 , its characteristic polynomial is (T − 1)2 . We need to find p(T ) such that For x =    0 1

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

NOTES TO HUMPHREYS

35

In fact, T y1 ≡ 1 mod(T − 1)2 =⇒ (2T − 1)y1 ≡ T mod(T − 1)2 (T − 1)2 y2 ≡ 1 mod T

=⇒ T 2 y1 ≡ T mod(T − 1)2 =⇒ y1 ≡ 2 − T mod(T − 1)2 . =⇒ y2 ≡ 1 mod T.

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





1 2 1 −2

 xs = x(2 − x) =    0 0 1

1

 





1 0





0 2

.  , x n = x − xs =  =     0 0 0 1

36

ZHENGYAO WU

4. March 19th, Cartan’s criterion, Killing form Lemma 4.1 Let F be an algebraically closed field. Let V be a finite dimensional F -vector space. If x ∈ gl(V ) is semisimple, then ad x ∈ gl(gl(V )) is semisimple. (In Lemma 2.25, we proved that if x ∈ gl(V ) is nilpotent, then ad x ∈ gl(gl(V )) is nilpotent.) Proof. Steps

Statements

Reasons

1.

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

F is algebraically closed, x is semisimple and Lemma 3.9

2.

[x, eij ] = (ai − aj )eij .

[x, eij ] = [ n P

n P k=1

ak ekk , eij ] =

n P

ak [ekk , eij ] =

k=1

ak (δki ekj − δjk eik ) = (ai − aj )eij

k=1

3.

ad x is diagonalizable.

The matrix of (ad x) is diagonal relative to the basis {eij : 1 ≤ i, j ≤ n} of gl(V ).

4.

ad x is semisimple.

F is algebraically closed and Lemma 3.9

 Lemma 4.2 Let 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 ad x = ad(xs )+ad(xn ) is the Jordan decomposition of ad x in gl(gl(V )). Proof. Steps

Statements

Reasons

1.

ad x = ad(xs ) + ad(xn ).

x = xs + xn and ad is linear

2.

ad(xs ) is semisimple.

xs is semisimple and Lemma 4.1

3.

(ad xn ) is nilpotent.

xn is nilpotent and Lemma 2.25

4.

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

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

5.

ad(xs ) = (ad x)s and (ad xn ) = (ad x)n .

Uniqueness of the Jordan decomposition of ad x Proposition 3.14(a)

NOTES TO HUMPHREYS

37

 Lemma 4.3 Let 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 is

k Q

(T − ai )mi , where a1 , . . . , am are distinct.

i=1

Let Vi = ker(x − ai · 1)mi . Steps

Statements

Reasons

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) = ai yz + yaj z = (ai + aj )yz.

4.

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

5.

(x − (ai + aj ) · 1)mi +mj (yz) =

miP +mj

(x −

x ∈ Der(A)

Leibniz Rule

k=0

ai · 1)mi +mj −k (y) · (x − aj · 1)k (z). 6.

yz ∈ ker(x − (ai + aj ) · 1)mi +mj .

7.

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

7.1.

If ker(x − (ai + aj ) · 1)mi +mj = {0}, then yz = 0.

7.2.

If ker(x − (ai + aj ) · 1)mi +mj 6= {0}, then ai + aj is an eigenvalue with eigenvector yz.

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

8.

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

steps 3, 7

9.

xs ∈ Der(A).

A=

either k ≥ mi or mi + mj − k ≥ mj

L

Vi .

i

10.

xn ∈ Der(A).

x n = x − xs

38

ZHENGYAO WU

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

Statements

Reasons

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 Interpolation Let F be a field. Suppose a1 , . . . , an , b1 , . . . , bn ∈ F . If ai 6= aj for all 1 ≤ i < j ≤ n, then there exists a polynomial r(T ) ∈ F [T ] such that r(ai ) = bi . Proof. Lagrange: n X



n Y



x − aj   bi . r(T ) = i=1 j=1,j6=i ai − aj  Lemma 4.6 Let F be an algebraically closed field of characteristic 0. Let V be a finite dimensional F -vector space. Let A ⊂ B be two subspaces of gl(V ). Let M = {y ∈ gl(V ) : [y, B] ⊂ A}. If x ∈ M such that Tr(xy) = 0 for all y ∈ M , then x is nilpotent. Proof. Steps

Statements

Reasons

1.

Let x = xs + xn be the Jordan decomposition.

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 = SpanQ {ai : 1 ≤ i ≤ n}. Then any linear function f : E → Q is 0.

See the next table

NOTES TO HUMPHREYS

39

Steps

Statements

Reasons

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

Steps

Statements

Reasons

3.1.

[xs , eij ] = (ai − aj )eij .

xs =

n P

ak ekk

k=1

3.2.

Let y = diag(f (a1 ), . . . , f (an )). [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(ad x), p(T ) = 0.

Proposition 3.14(b)

3.7

(ad y) = r(p(ad x)), r(p(0)) = 0.

composition of step 3.3, 3.6

3.8

(ad x)(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(xs y) =

n P

ai f (ai ).

Then

Tr(xn y) = 0

i=1

3.11

n P

f (ai )2 = 0.

Apply f on both sides

i=1

3.12

f (ai ) = 0 for all 1 ≤ i ≤ n.

im(f ) ⊂ Q

3.13

f = 0.

defn of E

40

ZHENGYAO WU

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

Statements

Reasons

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 criterion Let F be an algebraically closed field of characteristic 0. Let V be a finite dimensional F -vector space. Let L be a subalgebra of gl(V ). Suppose that Tr(xy) = 0 for all x ∈ [L, L], y ∈ L, then L is solvable. Proof. Steps

Statements

Reasons

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.

ad x 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

NOTES TO HUMPHREYS

41

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

Statements

Reasons

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 4.10 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional F -Lie algebra. The Killing form of L is κ(x, y) = Tr((ad x) ◦ (ad y)), ∀x, y ∈ L. Lemma 4.11 Let 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 κ is an associative, symmetric bilinear form over L. Proof. Steps

Statements

Reasons

1.

κ is associative.

Lemma 4.7

2.

κ is symmetric.

Tr(A ◦ B) = Tr(B ◦ A) for all A, B ∈ gl(L)

3.

κ is bilinear.

Tr and ad are linear



42

ZHENGYAO WU

Lemma 4.12 Let F be an algebraically closed field of characteristic 0. Let V be a finite dimensional F -vector space. 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. Steps

Statements

Reasons

1.

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

Let W be a sub-vector space of V .

2.

φ(ei ) = n.

3.

m P

aij ej , aij = 0 for m + 1 ≤ j ≤

φ(V ) ⊂ W

j=1

Tr(φ) = Tr(φ|W ).

n P

aii =

i=1

m P

aii

i=1

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

Statements

Reasons

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

κ|I×I is the Killing form of I.

Definition 4.10

3.

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

NOTES TO HUMPHREYS

43

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

Statements

Reasons

1.

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

Definition 4.14(1)

2.

β is non-degenerate iff βb is injective

Definition 4.14(2)

3.

iff βb is invertible

dimF (V ) = n < ∞

4.

iff the matrix (β(xi , xj ))1≤i,j≤n of βb is nonsingular.

 Example 4.17 Let F be an algebraically closed field of characteristic 0. The Killing form of sl(2, F ) is nondegenerate. Proof. Take the standard basis {x, h, y}. We have   x       (ad x)   h    

y



=

0    0   

0

 

 



 

x x 2 0 −2 0                   , (ad h)   = 0 0 0 1 h h 

0

0

   

y

   

y

 

 



 

x x  0 0 0  x  0                         , (ad y)   =   . 0  h h −1 0 0 h    

0 0 −2

y

   

y

 

0

2 0

   

y

44

ZHENGYAO WU

(ad x) ◦ (ad x) =

0    0   

0 −2  0

0

0 0

0

0    0   

0

(ad h) ◦ (ad x) =

0

0 0

2    0   

  , −2   

4    0   

(ad h) ◦ (ad h) =

0

  , 0   

(ad y) ◦ (ad h) =

(ad h) ◦ (ad y) =

0 0

(ad y) ◦ (ad y) =

0 −4 0

0 0 0

 0    −2   

0

0 

  , 0   



0 0   

, 0 0   

0 0







0    0   

2

  , 0   



0 0 4

0 0 

0 0 

0 0 2

0

0 0  0

0    0   

(ad y) ◦ (ad x) =







2

  , 0   

0

0

0

  , 0   

−4 0 



0 



(ad x) ◦ (ad y) =

0    0   





(ad x) ◦ (ad h) =

  ,   













       

0 0

0 0  0

  , 0   

−2 0 0

Take traces, κ has matrix  0    0   



0 4  8

  0   

4 0 0

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



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

Statements

Reasons

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.

NOTES TO HUMPHREYS

Steps

Statements

Reasons

1.

Rad(L) 6= 0.

Definition 2.11

2.

There exists a least integer n ≥ 1 such that Rad(L)(n) = {0}.

Definition 2.3

3.

Rad(L)(n−1) is a nonzero abelian subalgebra of L. Rad(L)(n−1) is an ideal of L.

45

[Hum78, p.14, Exercise 1]

 Lemma 4.19 Let 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. Steps

Statements

Reasons

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 4.20 Let 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. If I is an abelian ideal I of L, then I ⊂ Rad(κ). Proof.

46

ZHENGYAO WU

Steps

Statements

Reasons

1.

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

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

2.

((ad x) ◦ (ad y))2 (L) = 0.

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

3.

κ(x, y) = Tr((ad x) ◦ (ad y)) = 0.

(ad x) ◦ (ad y) is nilpotent

4.

I ⊂ Rad(κ).

Definition 4.14(1)

 Theorem 4.21 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional F -Lie algebra. Then L is semisimple iff its Killing form κ is non-degenerate. (See [Hum78, p.24, Exercise 6] for the case char F = p > 0.) Proof. Suppose L is semisimple. Steps

Statements

Reasons

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. Steps

Statements

Reasons

1.

Rad(κ) = {0}.

Definition 4.14(2)

2.

{0} is the only abelian ideal of L.

Every abelian ideal is in Rad(κ) by Lemma 4.20

3.

L is semisimple.

Lemma 4.18

NOTES TO HUMPHREYS

47

 Definition 4.22 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional F -Lie algebra. (1) L is the direct sum of ideals if there exist ideals I1 , . . . , It such that L = I1 + · · · + It as direct sum 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 4.23 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional F -Lie algebra. If I is an ideal of L, then so is I ⊥ . Proof. Suppose x ∈ I ⊥ and y ∈ L, Steps

Statements

Reasons

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)

[x, y] ∈ I ⊥ . Thus I ⊥ is an ideal of L.

Definition 4.22(2)

3.

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

Statements

Reasons

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 dimF (I ∗ )

→

I ∗ ))

=

dimF (L) −

48

ZHENGYAO WU

Steps

Statements

Reasons

4.

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



NOTES TO HUMPHREYS

49

5. March 26th, Semisimple decomposition and Lie modules Theorem 5.1 Let 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. Steps

Statements

Reasons

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 Definition 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 Definition 2.11

6.

⊥ L⊥ 1 is a direct sum of simple ideals of L1 (and of L).

dimF (L⊥ 1 ) < dimF (L) and inductive hypothesis

7.

L = L1 ⊕ L2 ⊕ · · · ⊕ Lt .

Suppose L⊥ 1 = L2 ⊕ · · · ⊕ Lt .

 Theorem 5.2 Let 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. Steps

Statements

Reasons

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

Steps

Statements

Reasons

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 ] = 0 for 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 tion 1.22(1)

6=

{0} and Defini-

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

Statements

Reasons

1.

Suppose L = L1 ⊕ · · · ⊕ Lt where Li are simple 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.

4.

[L, L] =

 t L

Li ,

i=1

= =

t L

[Li , Li ]

i=1 t L

Li = L.

t L

Definition 4.22



Li

step 1

i=1

Definition 1.1(L1) step 2

i=1



NOTES TO HUMPHREYS

51

Corollary 5.4 All ideals of a semisimple F -Lie algebra L are semisimple. Proof. Let I be an ideal of L. Steps

Statements

Reasons

1.

L = I ⊕ I ⊥.

Lemma 4.24

2.

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

Lemma 4.13



3.



κ|I

κ= 

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 5.5 Each 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 5.6 All homomorphic images of a semisimple F -Lie algebra L are semisimple. Proof. Let φ : L → L0 be a homomorphism of Lie algebras and M = φ(L). Steps

Statements

Reasons

1.

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

φ is a homomorphism.



52

ZHENGYAO WU

Steps

Statements

Reasons

2.

If φ(I) has a nonzero proper ideal J, then I 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 a simple ideal of M .

4.

L = L1 ⊕ · · · ⊕ Lt where Li are simple ideals of L.

Theorem 5.1

5.

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

steps 3,4

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

Statements

Reasons

1.

[δ, ad x](y) = δ((ad x)(y)) − (ad x)(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 5.8 If L is a semisimple Lie algebra, then L is linear. Proof. Steps

Statements

Reasons

1.

Rad(L) = {0}

L is semisimple and Definition 2.11

NOTES TO HUMPHREYS

Steps

Statements

Reasons

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)

53

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

Statements

Reasons

1.

ad(L) ' L.

Proposition 5.8

2.

ad(L) is semisimple.

L is semisimple

3.

ad(L) has a non-degenerate Killing form κ.

Theorem 4.21

4.

ad(L) ⊂ Der(L).

Lemma 1.14

5.

[Der(L), ad(L)] ⊂ ad(L), i.e. ad(L) is an ideal of Der(L).

Lemma 5.7

6.

κ = κ0 |ad(L) where κ0 is the Killing form of Der(L).

Lemma 4.13

7.

The complement ad(L)⊥ of ad(L) in Der(L) relative to κDer(L) is an ideal of Der(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 steps 5,7

54

ZHENGYAO WU

Steps

Statements

10.

ad(L)⊥ = 0.

11.

Reasons

If δ ∈ ad(L)⊥ , then ad(δ(x)) = [δ, ad x] = 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

Der(L) = ad(L).

 Proposition 5.10 If 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. Steps

Statements

Reasons

1.

Let ad x = (ad x)s + (ad x)n be the Jordan decomposition of ad x in gl(L).

Proposition 3.14(a)

2.

ad x ∈ Der(L).

Lemma 1.14

3.

(ad x)s , (ad x)n ∈ Der(L).

Lemma 4.3

4.

ad(L) = Der(L). Thus (ad x)s , (ad x)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) such that ad(xs ) = (ad x)s and (ad xn ) = (ad x)n .

7.

x = xs + xn .

ad x = (ad x)s + (ad x)n , ad is linear and step 5

NOTES TO HUMPHREYS

Steps

55

Statements

Reasons

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.

(ad x)s ◦ (ad x)n = (ad x)n ◦ (ad x)s and step 5

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

Statements

Reasons

1.

Let x = xs + xn be the usual Jordan decomposition 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 decomposition of x.

Proposition 5.10



56

ZHENGYAO WU

Definition 5.12 Let 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 that φ(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) is not satisfied. Lemma 5.14 x · 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 5.15 A homomorphism of L-modules φ : V → W is an isomorphism if it is an isomorphism of F -vector spaces, i.e. V and W afford equivalent representations of L. Definition 5.16 An 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 = F v 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 5.18 An 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 0 , i.e. an L-submodule such that V = W ⊕ W 0 (p.30, exercise 2).

NOTES TO HUMPHREYS

57

Example 5.19 A semisimple algebra L is a completely reducible L-module by Theorem 5.1. Lemma 5.20 Schur Let 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. Steps

Statements

Reasons

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 5.21 Let 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) Steps

Statements

1.

((ax + by) · f )(v)

Reasons

= (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

Steps

Statements

2.

(ax + by) · f = a(x · f ) + b(y · f )

Reasons

(M2) Steps

Statements

1.

(x · (af + bg))(v)

Reasons

= 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). 2.

defn of ·

x · (af + bg) = a(x · f ) + b(x · g)

(M3) Steps

Statements

1.

([x, y] · f )(v)

2.

Reasons

= [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

(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 ·

NOTES TO HUMPHREYS

Steps

Statements

59

Reasons

= 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 5.22 Let V and W 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 of L-modules. Steps

Statements

1.

φ(x · (f ⊗ w))(v)

2.

Reasons

= 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 .

(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 generators f ⊗ w.

steps 1,2

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ZHENGYAO WU

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

Statements

1.1.

Suppose g ∈ HomF (V, W ) such that P g(vi ) = gij wj .

Reasons

j

1.2.

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

1.3.

φ is surjective.

g(vi ) =

P j

2.

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

3.

φ is an isomorphism of vector spaces.

Therefore φ is an isomorphism of L-modules.

P

fj (vi )wj = φ(

fj ⊗ w j )

j

steps 1,2



Definition 5.23 Let 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 5.24 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimple F -Lie algebra. The Killing form κ of L is the trace form of the adjoint representation ad. Lemma 5.25 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimple F -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. 

NOTES TO HUMPHREYS

61

6. April 2nd, Casimir element, Weyl’s theorem (3) Let S = Rad(β). Steps

Statements

Reasons

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)

5.

Rad(L) = 0

L is semisimple and Definition 2.11

6.

β is non-degenerate.

S = 0 and Definition 4.14(2)

Lemma 6.1 Let 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 ∈ L such that [x, xi ] =

n P

aij xj and [x, yi ] =

j=1

n P

bij yj . Then aik = −bki .

j=1

Proof. Steps

Statements

Reasons

1.

β([xi , x], yk ) = β(xi , [x, yk ])

associativity by Lemma 5.25(1)

2.

β(−

n P

aij xj , yk ) = β(xi ,

n P

bkj yj )

j=1

j=1

n P

Remark 1.2(L2’), [x, xi ] =

aij xj and

j=1

[x, yi ] =

n P

bij yj

j=1

3.

−

n P

aij β(xj , yk ) =

j=1

4.

−aik = bki .

n P

bkj β(xi , yj ).

Definition 1.1(L1)

j=1

β(xi , yj ) = δij =

   

1, i = j;

  

0, i 6= j.



62

ZHENGYAO WU

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

Statements

Reasons

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 6.3 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimple F -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 P

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

i=1

for all x ∈ L. Proof. Steps

Statements

1.

[φ(x), cφ (β)] = [φ(x),

Reasons n P

φ(xi ) ◦ φ(yi )]

defn of cφ (β)

i=1

= = = =

n P

[φ(x), φ(xi ) ◦ φ(yi )]

Definition 1.1(L1)

i=1 n P

[φ(x), φ(xi )] ◦ φ(yi ) + φ(xi ) ◦ [φ(x), φ(yi )]

i=1 n P i=1 n P i=1

φ([x, xi ]) ◦ φ(yi ) +

n P

φ(xi ) ◦ φ([x, yi ])

Lemma 6.2 φ is a homomorphism

i=1

φ(

n P

j=1

aij xj ) ◦ φ(yi ) +

n P

φ(xi ) ◦ φ(

i=1

n P

bij yj )

Suppose [x, xi ] =

j=1

j=1

and [x, yi ] =

n P j=1

=

n P n P

(aji + bij )φ(xi ) ◦ φ(yj ).

i=1 j=1

n P

φ is linear

bij yj .

aij xj

NOTES TO HUMPHREYS

Steps

63

Statements

Reasons

= 0.

Lemma 6.1

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

Statements

Reasons

1.

Tr(cφ ) = Tr(

n P

φ(xi ) ◦ φ(yi ))

Lemma 6.3

i=1

= = =

n P i=1 n P i=1 n P

Tr(φ(xi ) ◦ φ(yi ))

Tr is linear

β(xi , yi )

Definition 5.23

1 = n = dimF (L)

β(xi , yj ) = δij

i=1

 Lemma 6.5

dimF (L) · 1V . dimF (V ) In particular, cφ is independent of the choice of basis. Proof. If V is irreducible, then cφ =

Steps

Statements

Reasons

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).

5.

a=

dimF (L) . dimF (V )

Lemma 6.4 steps 3,4

64

ZHENGYAO WU

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

3 · 1V . 2

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 











1 0 0 −1 0 0 ,  , xy =   , xh =  x2 =        0 0 0 0 0 0 











0 1 1 0   0 0  , h2 =   , hy =  , hx =        0 0 0 1 −1 0 



0

0

yx =  

0 1

Then





(Tr(ei ej ))1≤i,j≤3 =





0 0 0 0   , y2 =  . yh =      1 0 0 0

, 

0    0   







0 1  2

  , 0   

(Tr(ei ej ))−1 1≤i,j≤3 =

1 0 0



0    0   

0 1  





1 2



0 

1 0 0

 

1 Hence the dual basis is (x, h, y)(Tr(ei ej ))−1 1≤i,j≤3 = (y, 2 h, x). Hence





2







0 0  0 1  0 1 0 0 1 1 1 0   +    +  cφ = xy + h2 + yx =           2 2 0 0 1 0 0 −1 1 0 0 0 



1

0

=  where

0 0



1 2 

+  

0



0 1 2



0 +  



0

0 1

 



3 2 

=

0



0 3 2

. 

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



Definition 6.7 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimple F -Lie algebra. Let φ : L → gl(V ) be a finite dimensional representation of L (not necessarily faithful). By Lemma 4.24, φ0 = φ|ker(φ)⊥ is faithful. We call cφ = cφ0 the Casimir element of φ. Remark 6.8 (1) Since φ(L) = φ0 (ker(φ)⊥ ), by Lemma 6.3, cφ = a · 1 for some a ∈ F .

NOTES TO HUMPHREYS

65

(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 = F v 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}. If ker(φ) = {0}, then φ is faithful. Otherwise ker(φ) = L and hence x · v = 0 for all v ∈ V . Since V is irreducible, dimF V ≤ 1, we are in case (2a)(2b). Lemma 6.9 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimple F -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. Steps

Statements

Reasons

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

Statements

Reasons

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 =

3.2.

ker(cφ ) ∩ W 6= W .

dimF (L) · 1 6= 0. dimF (W )

W is irreducible and Lemma 6.5

66

ZHENGYAO WU

Steps

Statements

Reasons

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 P

φ(xi ) ◦ φ(yi )

i=1

6.

dimF (ker(cφ )) = 1.

steps 4,5

7.

ker(cφ ) = W 0 .

steps 3,6

 Lemma 6.11 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimple F -Lie algebra. Let φ : L → gl(V ) be a finite dimensional representation of L. If V has a Lsubmodule 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. Steps

Statements

Reasons

1.

W has a nonzero proper submodule W 0 .

W is reducible

2.

0 → W/W 0 → V /W 0 → F → 0 is exact.

V /W 0 ' V /W ' F W/W 0

3.

dimF (V /W 0 ) < dimF (V ).

W 0 6= {0}.

4.

f, W f /W 0 ⊕ W/W 0 = V /W 0 . There exists W

Inductive hypothesis

5.

f ) = dim (W 0 ) + 1 dimF (W F

f /W 0 ) dimF (W = dimF (V /W 0 ) − dimF (W/W 0 ) = dimF (V ) − dimF (W ) = 1

≤ dimF (W )

W0 ( W

< dimF (V )

dimF (W ) = dimF (V ) − 1

NOTES TO HUMPHREYS

Steps

Statements

Reasons

6.

f. There exists X, X ⊕ W 0 = W

Inductive hypothesis

7.

f ∩W X ∩W ⊂W

f X⊂W

⊂ X ∩ W0

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

67

 Lemma 6.12 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimple F -Lie algebra. Let φ : L → gl(V ) be a finite dimensional representation of L. Let W be a nonzero proper L-submodule of V . Let V = {f ∈ Hom(V, W ) : f |W = a · 1W , ∃a ∈ F }. Let W = {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 , Steps

Statements

Reasons

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)

x·f ∈W.

defn of W

3.

e

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



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

68

ZHENGYAO WU

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

Statements

Reasons

1.

dimF (W ) = dimF (V ) − 1.

Lemma 6.12

2.

W has a complement W 0 in V .

Lemma 6.11

3.

Let f : V → W be the generator of W 0 such that f |W = 1W .

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

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 )).

9.

W ⊕ ker(f ) = V .

=

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

steps 4,8

 Lemma 6.14 Let F be an algebraically closed field of characteristic 0. Let L ⊂ gl(V ) be a finite dimensional semisimple 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 ). Hence 1V ∈ N − L. (2) Let ad = adgl(V ) . Steps

Statements

Reasons

1.

Let x = xs + xn be the Jordan decomposition of x in gl(V ).

Proposition 3.14(a)

2.

(ad x)(L) ⊂ L.

x ∈ L ( Ngl(V ) (L)

3.

(ad x)s (L) ⊂ L, (ad x)n (L) ⊂ L.

Proposition 3.14(c)

NOTES TO HUMPHREYS

Steps

Statements

Reasons

4.

ad(xs )(L) ⊂ L, ad(xn )(L) ⊂ L.

Lemma 4.2

5.

xs , xn ∈ Ngl(V ) (L).

Example 1.26(1)

69

 Lemma 6.15 Let F be an algebraically closed field of characteristic 0. Let V be a finite dimensional F -vector space. 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, Steps

Statements

Reasons

1.

y(W ) ⊂ W .

W is an L-module

2.

L = [L, L].

L is semisimple and Corollary 5.3

3.

y=

4.

Tr(y|W ) = Tr( [ai , bi ]|W ) P Tr([ai |W , bi |W ]) = 0.

5.

L ⊂ LW .

P

[ai , bi ] for some ai , bi ∈ L. P

=

Tr is linear and commutative

(3) Steps

Statements

Reasons

1.

Suppose x ∈ L. Then x(W ) ⊂ W and Tr(x|W ) = 0.

(2)

2.

xs (W ) ⊂ W , xn (W ) ⊂ W .

Proposition 3.14(c)

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ZHENGYAO WU

Steps

Statements

Reasons

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 6.16 Let F be an algebraically closed field of characteristic 0. Let V be a finite dimensional F -vector space. Let L ⊂ gl(V ) be a semisimple F -Lie algebra. If x ∈ L, then xs , xn ∈ L. Proof. Let L0 = Ngl(V ) (L) ∩

T W ⊂V

LW . Suppose x ∈ L.

Steps

Statements

Reasons

1.

Let x = xs + xn be the Jordan decomposition 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 ∈ L0 .

defn of L0

4.

x ∈ L0 and hence L ⊂ L0 .

step 1

5.

L0 is a subalgebra of Ngl(V ) (L).

Example 1.26(1) and Lemma 6.15(1)

6.

L0 = L ⊕ M for a L-submodule M of L0 .

L is semisimple and Weyl’s Theorem 6.13

7.

L acts on M trivially.

[L, L0 ] ⊂ [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 ∈ L0 ⊂ LW

11.

y|W = 0.

a=0

NOTES TO HUMPHREYS

71

Steps

Statements

Reasons

12.

y = 0.

V is a direct sum of irreducible Lsubmodules by Weyl’s Theorem 6.13

13.

M = 0 and hence xn , xs ∈ L0 = L.

step 3

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

Statements

Reasons

1.

Let x = xs + xn be the usual Jordan decomposition of x in gl(V ).

Proposition 3.14(a)

2.

xs , xn ∈ L.

Theorem 6.16

3.

ad x = ad(xs )+ad(xn ) is the usual Jordan decomposition of ad x in gl(L).

Lemma 4.2

4.

x = xs + xn is the abstract Jordan decomposition of x in L.

Proposition 5.10

 Corollary 6.18 Let 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. If x = 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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ZHENGYAO WU

Steps

Statements

Reasons

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 of adφ(L) (φ(xs )).

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.

10.

φ(x) = φ(xs ) + φ(xn ) is the abstract Jordan decomposition.

steps 5,8,9 and Proposition 5.10

11.

φ(x) = φ(xs ) + φ(xn ) is the usual Jordan decomposition.

Theorem 6.17

= =

φ is linear

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



NOTES TO HUMPHREYS

73

7. April 9th, Representation of sl(2, F ), toral subalgebras Review 7.1 In sl(2, F ), 











0 1 0 0 1 0  , y =  , h =   , [h, x] = 2x, [h, y] = −2y, [x, y] = h. x=       0 0 1 0 0 −1

Definition 7.2 Let 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 and Vλ is called a weight space. Lemma 7.3 Let F be an algebraically closed field of characteristic 0. Let φ : sl(2, F ) → gl(V ) be a finite ` dimensional 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 7.4 Let 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 7.5 Let 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 a maximal vector. Lemma 7.6 Let F be an algebraically closed field of characteristic 0. Let V be a finite dimensional irreducible 1 sl(2, F )-module. Let v0 ∈ Vλ be a maximal vector; Define v−1 = 0, vi = y i · v0 . Then (1) i! h · vi = (λ − 2i)vi ; (2) y · vi = (i + 1)vi+1 ; (3) x · vi = (λ − i + 1)vi−1 . 1 i 1 y · v0 = (i + 1) y i+1 · v0 = (i + 1)vi+1 . i! (i + 1)! (1) Since v0 ∈ Vλ , h · v0 = λv0 . Suppose h · vi = (λ − 2i)vi . Then Proof. (2) y · vi = y ·

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ZHENGYAO WU

h · vi+1 = h · (

1 y · vi ), i+1

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

by (2) by Definition 5.12(M1) 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 7.7 Let F be an algebraically closed field of characteristic 0. Let V be a finite dimensional irreducible sl(2, F )-module. Then V '

a

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

µ

where dimF (V ) = m + 1 and dimF (Vµ ) = 1 for all µ. Proof. Let λ be the highest weight with weight space Vλ . Steps

Statements

1.

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

1.1.

Suppose

1.2.

=

n P

n P

ai vi = 0 for ai ∈ F .

i=0 n P

n P

i=0

i=0

0=h·( =

Reasons

ai vi ) =

ai (h · vi )

Definition 5.12(M2)

(λ − 2i)ai vi

Lemma 7.6(1)

i=0 n P

n P

i=0

i=0

(λ − 2i)ai vi − (λ − 2n)

ai vi

step 1.1

NOTES TO HUMPHREYS

Steps

Statements =2

75

Reasons

n−1 P

(n − i)ai vi

i=0

1.3.

a0 = · · · = an−1 = 0.

Inductive hypothesis: (v0 , . . . , vn−1 ) are linearly 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.

Span{v0 , . . . , vm } is a nonzero submodule of V with vi ∈ Vλ−2i .

Lemma 7.6

4.

V = Span{v0 , . . . , vm }.

V is irreducible

5.

Vλ−2i = F vi for all i.

V ⊂

L-

m `

Vλ−2i = V,

i=0

 Remark 7.8 Let 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 maximal vector 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 7.9 Let F be an algebraically closed field of characteristic 0. Let V be an sl(2, F )-module. Then eigenvalues 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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ZHENGYAO WU

Steps

Statements

Reasons

Steps

Statements

Reasons

1.

V '

t `

V (mi ), where V (mi ) are irre-

Theorem 6.13 and Remark 7.8

i=1

ducible submodules. 2.

Eigenvalues

of

are

V

t S

{mi , mi −

Theorem 7.7

i=1

2, . . . , −mi } 3.

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

 Corollary 7.10 Let F be an algebraically closed field of characteristic 0. Let V be an sl(2, F )-module. In any decomposition of V into direct sum of irreducible submodules, the number of summands is dimF (V0 ) + dimF (V1 ). Proof. Suppose V 6= 0. Steps

Statements

1.

V '

t `

Reasons

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

i=1

ules. 2.

t=

P

1+

mi even

=

P mi even

P

Theorem mark 7.8

6.13

and

Re-

1

mi odd

dimF ((V (mi ))0 ) +

P

dimF ((V (mi ))1 ).

Theorem 7.7

mi odd

= dimF (V0 ) + dimF (V1 ).

 Example 7.11 The 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).

NOTES TO HUMPHREYS

77

Lemma 7.12 Let F be an algebraically closed field of characteristic 0. Let φ : sl(2, F ) → gl(V (m)) be a representation. 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. Steps

Statements

Reasons

1.

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

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))

'

◦

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

φ is a homomorphism   0 



1

= φ(shs−1 ) = φ(−h) = −φ(h)

 as in Example 2.2 s=  −1 0

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 7.13 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzero semisimple F -Lie algebra. Let T be a nonzero subalgebra of L. We say that T is toral if it consists of ad-semisimple elements. Lemma 7.14 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzero semisimple F -Lie algebra. If T is a toral subalgebra of L, then T is abelian. Proof. Steps

Statements

Reasons

1.

adT (x) is semisimple for all x ∈ T .

Definition 7.13

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ZHENGYAO WU

Steps

Statements

Reasons

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 for some a ∈ F ∗ , 0 6= y ∈ T .

3.2.

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

step 2

3.3.

−ay = [y, x] = a1 b1 z1 + · · · + an bn zn where x = a1 z1 + · · · + an zn , ai ∈ F .

steps 3.1, 3.2, and Remark 1.2(L2’)

3.4.

0 = [y, −ay] = a1 b21 z1 + · · · + an b2n zn .

steps 3.1, 3.2, and Definition 1.1(L2)

3.5.

ai b2i = 0 for all 1 ≤ i ≤ n.

step 3.2

3.6.

ai bi = 0 for all 1 ≤ i ≤ n.

(ai bi )2 = ai (ai b2i ) = ai · 0 = 0

3.7.

ay = a1 b1 z1 + · · · + an bn zn = 0, a contradiction.

to step 3.1

4.

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

step 4

 Lemma 7.15 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzero semisimple 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}. Steps

Statements

Reasons

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

NOTES TO HUMPHREYS

79

Steps

Statements

Reasons

4.

Span{xs : x ∈ S} is a toral subalgebra of L.

S 6= ∅, Lemma 7.14 and Lemma 3.11.

 Definition 7.16 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzero semisimple F -Lie algebra. Let H be a toral subalgebra of L. We call H a maximal if for all toral subalgbra H 0 of L such that H ⊂ H 0 , we have H = H 0 . Example 7.17 Let F be an algebraically closed field of characteristic 0. Let L = sl(n, F ). The set H of diagonal n × n matrices of trace 0 is a maximal toral subalgebra of L. We only verify that H = F h is a maximal toral subalgebra of sl(2, F ). If H 0 is another toral subalgebra of sl(2, F ) such that H ⊂ H 0 , then [H 0 , H 0 ] = 0 by Lemma 7.14. Then [H 0 , H] = 0 ⊂ H and hence H 0 ⊂ Nsl(2,F ) (H). Since [h, x] = 2x 6∈ H and [h, y] = −2y 6∈ H, Nsl(2,F ) (H) = H. Therefore H 0 = H. Definition 7.18 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzero semisimple 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 7.19 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzero semisimple 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 Decomposition Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzero F semisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. Then L = L0 ⊕ Lα . α∈Φ

Proof. Steps

Statements

Reasons

1.

H is abelian.

H is toral and Lemma 7.14

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ZHENGYAO WU

Steps

Statements

Reasons

2.

Elements of adL (H) are simultaneously diagonalizable.

Lemma 3.11

3.

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

 Lemma 7.21 Φ is finite, since dimF (L) < ∞ and Lemma 7.20. Example 7.22 Let F be an algebraically closed field of characteristic 0. The Cartan decomposition of sl(2, F ). The maximal toral subalgebra is F h. 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 = F h. • If a 6= 0, then α(h) = 2 and b = c = 0, we have L2 = F x. • If b = 6 0, then α(h) = −2 and a = c = 0, we have L−2 = F y. Therefore sl(2, F ) = F0 ⊕ F2 ⊕ F−2 = F h ⊕ F x ⊕ F y is the Cartan decomposition. Proposition 7.23 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzero semisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. [Lα , Lβ ] ⊂ Lα+β for all α, β ∈ H ∗ . Proof. Suppose x ∈ Lα and y ∈ Lβ . Steps

Statements

Reasons

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)

NOTES TO HUMPHREYS

Steps

Statements

Reasons

3.

[x, y] ∈ Lα+β .

Definition 7.18

81

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

Statements

Reasons

1.

If y ∈ Lβ , then (ad x)n (y) ∈ Lnα+β .

Proposition 7.23

2.

{n ∈ N : Lnα+β 6= {0}} is finite.

Lemma 7.21

3.

(ad x)|Lβ is nilpotent for all β ∈ Φ.

4.

ad x is nilpotent.

Lemma 7.20

 Proposition 7.25 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzero semisimple 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 , Steps

Statements

Reasons

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

82

ZHENGYAO WU

Steps

Statements

Reasons

5.

κ(x, y) = 0.

step 1



NOTES TO HUMPHREYS

83

8. April 16th, Centralizer of H; Orthogonal, integral properties Corollary 8.1 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzero semisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. κ|L0 is non-degenerate. Proof. Suppose z ∈ L0 . Steps

Statements

Reasons

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 ⊕

F

Lα by Lemma 7.20

α∈Φ

4.

κ is non-degenerate.

L is semisimple and Theorem 4.21

5.

z = 0 and κ|L0 is non-degenerate.

Definition 4.14(2)

 Lemma 8.2 Let F be a field. Let V be a finite dimensional F -vector space. Suppose x, y ∈ End(V ) such that x ◦ y = y ◦ x and y is nilpotent. Then x ◦ y is nilpotent and Tr(x ◦ y) = 0. Proof. Suppose y n = 0. Since x ◦ y = y ◦ x, (x ◦ y)n = xn ◦ y n = 0. Since xy is nilpotent, its eigenvalues are all 0 and hence their sum Tr(xy) = 0.



Lemma 8.3 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzero semisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. If x ∈ CL (H) and x is ad-semisimple, then x ∈ H. Proof. Steps

Statements

Reasons

1.

Every element of H +F x is ad-semisimple.

x ∈ CL (H) and Lemma 3.11

2.

H + F x is a toral subalgebra of L.

Definition 7.13

3.

H + F x = H and hence x ∈ H.

H is maximal and Definition 7.16

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ZHENGYAO WU

 Lemma 8.4 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzero semisimple 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. Steps

Statements

Reasons

1.

Let x = xs + xn be the abstract Jordan decomposition of x. Then ad x = ad(xs )+ ad(xn ) is the usual Jordan decomposition of ad x.

Proposition 5.10

2.

For all x ∈ CL (H), xn ∈ CL (H).

Proposition 3.14(c)

3.

[ad xn , ad h] = ad([xn , h]) = 0.

ad is a homomorphism

4.

κ(h, xn ) = Tr(ad(h)(ad xn )) = 0.

(ad xn ) 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 all x ∈ CL (H).

steps 3,6

9.

h = 0.

Corollary 8.1

10.

κ|H is non-degenerate.

Definition 4.14(2)

 Lemma 8.5 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzero semisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. Then CL (H) is nilpotent. Proof.

NOTES TO HUMPHREYS

Steps

Statements

Reasons

1.

Let x = xs + xn be the abstract Jordan decomposition of x. Then ad x = ad(xs )+ ad(xn ) is the usual Jordan decomposition of ad x.

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

85

 Lemma 8.6 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzero semisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. Then H ∩[CL (H), CL (H)] = 0. Proof. Steps

Statements

1.

κ(H, H ∩ [CL (H), CL (H)]) κ(H, [CL (H), CL (H)])

2.

Reasons ⊂

= κ([H, CL (H)], CL (H))

κ is associative by Lemma 4.11

= κ(0, CL (H)) = 0.

defn of CL (H)

H ∩ [CL (H), CL (H)] = 0.

κ|H is non-degenerate by Lemma 8.4



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Lemma 8.7 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzero semisimple 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. Steps

Statements

Reasons

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 [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} by Lemma 8.6

4.3.

x is not ad-semisimple.

5.

Let x = xs + xn be the abstract Jordan decomposition of x. Then ad x = ad(xs )+ ad(xn ) is the usual Jordan decomposition of ad x.

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 all y ∈ CL (H).

x ∈ Z(CL (H)) and Proposition 3.14(c)

9.

[ad xn , ad y] = 0 for all y ∈ CL (H).

ad is a homomorphism.

10.

κ(xn , y) = 0 for all y ∈ CL (H).

ad xn is nilpotent and Lemma 8.2

11.

xn = 0, a contradiction to step 6.

κ|CL (H) is non-degenerate by Corollary 8.1 and Lemma 7.19

∈

Z(CL (H)) ∩



NOTES TO HUMPHREYS

87

Proposition 8.8 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzero semisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. Then CL (H) = H. Proof. Suppose the opposite H ( CL (H). Steps

Statements

Reasons

1.

There exists x ∈ CL (H) − H and x is not ad-semisimple.

Lemma 8.3

2.

Let x = xs + xn be the abstract Jordan decomposition of x. Then ad x = ad(xs )+ ad(xn ) is the usual Jordan decomposition of ad x.

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.

[ad xn , ad y] = 0 for all y ∈ CL (H).

ad is a homomorphism.

6.

κ(xn , y) = 0 for all y ∈ CL (H).

ad xn 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 8.9 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzero semisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. There exists a bijection t : 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 is surjective by definition.  Proposition 8.10 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzero semisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. Then Let Φ be the set of roots of L relative to H. Φ spans H ∗ . Proof. Suppose Span Φ ( H ∗ .

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ZHENGYAO WU

Steps

Statements

Reasons

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 in Corollary 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 decomposition Lemma 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 8.11 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzero semisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. Let Φ be the set of roots of L relative to H. If α ∈ Φ, then −α ∈ Φ. Proof. Suppose the opposite −α 6∈ Φ. Steps

Statements

Reasons

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

4.

κ is non-degenerate.

L is semisimple and Theorem 4.21

5.

Lα = 0, a contradiction.

to α ∈ Φ

NOTES TO HUMPHREYS

89

 Proposition 8.12 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzero semisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. Let Φ be the set of roots of L relative to H. Let α ∈ Φ, x ∈ Lα and y ∈ L−α . Then [x, y] = κ(x, y)tα . Proof. Steps

Statements

Reasons

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

[x, y] = κ(x, y)tα

κ|H is non-degenerate by Lemma 8.4

2.

 Proposition 8.13 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzero semisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. Let Φ be the set of roots of L relative to H. If α ∈ Φ, then [Lα , L−α ] = F tα . Proof. Steps

Statements

Reasons

1.

[Lα , L−α ] ⊂ F tα .

Proposition 8.12

2.1.

Suppose κ(Lα , L−α ) = 0. κ(Lα , Lβ ) = 0 for all β ∈ Φ.

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.

Then

Proposition 7.25 for β 6= −α

90

ZHENGYAO WU

Steps

Statements

Reasons

4.

[Lα , L−α ] 6= 0.

Proposition 8.12

5.

[Lα , L−α ] = F tα .

steps 1, 4

 Proposition 8.14 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzero semisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. Let Φ be the set of roots of L relative to H. Then α(tα ) = κ(tα , tα ) 6= 0 for all α ∈ Φ. Proof. Suppose the opposite α(tα ) = 0. Steps

Statements

Reasons

1.

There exists 0 6= x ∈ Lα and 0 6= y ∈ L−α such that κ(x, y) 6= 0.

Proposition 8.13

2.

Let x0 =

3.

Consider the F -Lie Span{x0 , y, tα }.

3.1.

[tα , x0 ] = α(tα )x0 = 0.

x, x0 ∈ 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 whose eigenvalues are all 0

10.

tα ∈ Z(L) ⊂ Rad(L) = 0.

L is semisimple

11.

tα = 0, a contradiction.

to step 12

x . Then [x0 , y] = tα . κ(x, y) algebra

S

Proposition 8.12 =

NOTES TO HUMPHREYS

91

Steps

Statements

Reasons

12.

tα 6= 0.

α 6= 0 (since α ∈ Φ) and Corollary 8.9

 Proposition 8.15 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzero semisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. Let Φ be the set of roots of L relative to H. For all α ∈ Φ and 0 6= xα ∈ Lα , there exists yα ∈ L−α such that Sα = Span{xα , yα , hα = [xα , yα ]} ' sl(2, F ) 



0

1

xα 7→ x =  

0 0

 , yα 









0 0 1 0   , hα 7→ h =   7→ y =      1 0 0 −1

Proof. Steps

Statements

Reasons

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 =

4.

5.

2 y ∈ L−α . Then κ(tα , tα )κ(xα , y) 2 . κ(xα , yα ) = κ(tα , tα ) 2tα Let hα = . Then [xα , yα ] = κ(tα , tα ) 2tα κ(xα , yα )tα = = hα . κ(tα , tα ) 2tα 2[tα , xα ] [hα , xα ] = [ , xα ] = κ(tα , tα ) κ(tα , tα ) 2α(tα )xα = κ(tα , tα ) = 2xα

6.

2tα 2[tα , yα ] , yα ] = κ(tα , tα ) κ(tα , tα ) −2α(tα )yα = κ(tα , tα ) [hα , yα ] = [

= −2yα

κ is bilinear

Proposition 8.12

κ is bilinear x α ∈ Lα α(tα ) = κ(tα , tα ) by Corollary 8.9 κ is bilinear x α ∈ Lα α(tα ) = κ(tα , tα ) by Corollary 8.9

92

ZHENGYAO WU

Steps

Statements

Reasons

7.

S ' sl(2, F ).

steps 4,5,6 and [x, y] = h, [h, x] = 2x, [h, y] = −2y in sl(2, F )

 Proposition 8.16 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimple F -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. 2tα Suppose α ∈ Φ and Sα ' sl(2, F ) with (xα , yα , hα ) 7→ (x, y, h). Then (1) hα = and (2) κ(tα , tα ) α(hα ) = 2. Proof. We provide a new proof besides step 4 of Proposition 8.15. Steps

Statements

Reasons

1.

hα = [xα , yα ]

S ' sl(2, F )

= κ(xα , yα )tα .

Proposition 8.12

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 ,

x α ∈ Lα

2.

3. 4. 5.

2 2 = . α(tα ) κ(tα , tα ) 2tα 2tα hα = = . α(tα ) κ(tα , tα ) 2α(tα ) α(hα ) = = 2. α(tα ) κ(xα , yα ) =

defn of t in Corollary 8.9 steps 1,3 α is linear and steps 4

 Proposition 8.17 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimple F -Lie algebra over F . Let H be a maximal toral subalgebra of L. Let Φ be the set of roots of L

NOTES TO HUMPHREYS

93

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) t−α = −tα and (2) h−α = −hα . Proof. Steps

Statements

Reasons

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−α κ(t−α , t−α ) −2tα 2tα = =− κ(−tα , −tα ) κ(tα , tα ) = −hα

Proposition 8.16(1) step 3 and κ is bilinear Proposition 8.16(1)

 Proposition 8.18 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimple F -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. Suppose α ∈ Φ and Sα ' sl(2, F ) with (xα , yα , hα ) 7→ (x, y, h). The only scalar multiples of α in Φ are ±α. Proof. Let M = L0 ⊕

F c∈F ∗

Lcα , where L0 = H by Lemma 7.19.

Steps

Statements

Reasons

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 = Span{xα , yα , hα } and below

2.1

[z, xα ] = α(z)xα = 0.

x α ∈ 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) ⊕ F hα .

below

94

ZHENGYAO WU

Steps

Statements

Reasons

3.1.

{0} = 6 H/ ker(α) ' F

α : H → F is nonzero and Proposition 1.29(1)

3.2.

dimF (ker(α)) = dimF (H) − 1.

1 ≤ dimF (H/ ker(α)) ≤ dimF (F ) = 1.

3.3.

ker(α) ∩ F hα = 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 Proposition 8.16(2)

where c =

n for n ∈ Z. 2

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’s theorem Theorem 6.13

M ' ker(α) ⊕ Sα ⊕ W ,

step 3

where ker(α) is the direct sum of onedimensional trivial sl(2, F )-modules.

step 2

and W is the direct sum of irreducible sl(2, F )-modules. 7.

Every weight appears at most once in M .

defn of M

8.

Weights of irreducible submodules of W are not even.

Weight 0 already appears in Sα and Theorem 7.7

9.

For α ∈ Φ, ±2α 6∈ Φ.

steps 4,8

10.

Weights of irreducible submodules of W are not odd.

below

10.1

Otherwise weight 2c = 1 appears in W . α ∈ Φ. 2

Theorem 7.7

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 . The only scalar multiples of α in Φ are ±α.

steps 6,11

10.2

α (hα ) 2

= 1 by Proposition 8.16(2)

NOTES TO HUMPHREYS

95

 Proposition 8.19 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimple F -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. If α ∈ Φ, then dimF (Lα ) = 1. In particular, Sα = Hα ⊕ Lα ⊕ L−α , where Hα = [Lα , L−α ]. Proof. Suppose α ∈ Φ. Steps

Statements

Reasons

1.

dimF (Lα ) = dimF (L−α ) = 1

Theorem 7.7

2.

Sα = Lα ⊕ L−α ⊕ Hα , where Lα = F xα , L−α = F yα and Hα = F hα .

Sα ' sl(2, F ) by Proposition 8.15, and Example 7.22

 Proposition 8.20 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimple F -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 α, β ∈ Φ, β 6= ±α. Let r be the largest integer such that β − rα ∈ Φ. Let q be 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 =

`

Lβ+iα .

i∈Z

Steps

Statements

Reasons

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 irreducible component.

step 2,3 and Corollary 7.10

5.

V is irreducible, we write V ' V (m).

Remark 7.8

96

ZHENGYAO WU

Steps

Statements

Reasons

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 such that β + qα ∈ Φ.

6.2.

−m = β(hα ) − 2r.

step 1 and r is the largest integer such that β − 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 8.21 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimple F -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. 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 8.22 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimple F -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. If α, β, α + β ∈ Φ, then [Lα , Lβ ] = Lα+β . Proof. Steps

Statements

1.

V =

q `

Lβ+iα is an irreducible Sα '

Reasons Proposition 8.20

i=−r

sl(2, F )-module. 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

NOTES TO HUMPHREYS

97

Steps

Statements

Reasons

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 8.23 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimple F -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. Then L is generated by {Lα : α ∈ Φ} as F -Lie algebra. Proof. Steps

Statements

Reasons

1.

Φ spans H ∗ .

Proposition 8.10

2.

t(Φ) spans H.

Corollary 8.9

3.

{hα : α ∈ Φ} span H.

F tα = F hα by Proposition 8.16(1)

4.

L is generated by {Lα : α ∈ Φ}.

L=H⊕

`

Lα by Lemma 7.20

α∈Φ



98

ZHENGYAO WU

9. April 23th, Rationality properties, reflections, root systems Definition 9.1 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimple F -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. 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 9.2 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimple F -Lie algebra over F . Let H be a maximal toral subalgebra of L. Let Φ be the set of roots of L 2(β, α) relative to H. Suppose α, β ∈ Φ. Then β(hα ) = . (α, α) Proof. Steps

Statements

Reasons

1.

β(hα ) = κ(tβ , hα ) ! 2tα 2κ(tβ , tα ) = κ tβ , = κ(tα , tα ) κ(tα , tα ) 2(β, α) = . (α, α)

defn of t in Corollary 8.9 Proposition 8.16(1) and κ is bilinear Definition 9.1

 Lemma 9.3 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimple F -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. By Proposition 8.10, we may choose a basis {α1 , . . . , αl } of H ∗ in Φ. If β=

l P i=1

ci αi ∈ Φ, then ci ∈ Q.

Proof. Steps

Statements

1.

(β, αj ) =

l P

Reasons ci (αi , αj ).

i=1

2.

We have l equations with l unknowns ci whose coefficients and constants are intel 2(αi , αj ) 2(β, αj ) P gers = ci , 1 ≤ j ≤ l. (αj , αj ) i=1 (αj , αj )

β=

l P

ci αi ∈ Φ and Definition 9.1

i=1

Lemma 9.2 and Corollary 8.21

NOTES TO HUMPHREYS

99

Steps

Statements

Reasons

3.

det((αi , αj ))1≤i,j≤l 6= 0.

{αi : 1 ≤ i ≤ l} form a basis of H ∗ and κ is non-degenerate since L is semisimple.

!

4.

2(αi , αj ) det (αj , αj ) 1≤i,j≤l l 2 det((αi , αj )1≤i,j≤l ) 6= 0 l Q (αj , αj )

=

j=1

5.

The linear system has a unique solution in Q. Thus, ci ∈ Q for all 1 ≤ i ≤ l.

steps 2,4 and Cramer’s Rule

 Lemma 9.4 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimple F -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. 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. Steps

Statements

Reasons

1.

Take a basis of L consisting of some elements of H and xα ∈ Lα for α ∈ Φ.

Lemma 7.20

2.

For all λ ∈ H ∗ , [tλ , xα ] = α(tλ )xα .

x α ∈ Lα

[tλ , H] = 0.

H is abelian by Lemma 7.14

For all λ, µ ∈ H ∗ , (λ, µ) = κ(tλ , tµ )

Definition 9.1

= Tr(ad(tλ ) ◦ ad(tµ ))

Definition 4.10

3.

=

P

α(tλ )α(tµ )

step 1,2

(α, λ)(α, µ).

Definition 9.1

α∈Φ

=

P α∈Φ

4. 5.

1 P (α, β)2 = . (β, β) α∈Φ (β, β)2 2(β, α) = α(hβ ) ∈ Z. (α, α)

For β ∈ Φ, (β, β) =

P

(α, β)2 .

α∈Φ

Corollary 8.21 and Lemma 9.2

100

ZHENGYAO WU

Steps

Statements

Reasons

6.

(β, β) ∈ Q for all β ∈ Φ.

7.

(α, β) ∈ Q for all α, β ∈ Φ.

8.

(•, •) extends to a non-degenerate symmetric bilinear form EQ × EQ → Q.

linear extension

9.

(•, •) is positive definite on EQ .

For all λ ∈ EQ , (λ, λ) =

4 ∈ Z by steps 4,5. (β, β) 1 2(α, β) (α, β) = · · (β, β) ∈ Q. 2 (β, β)

P

(α, λ)2 ≥ 0.

α∈Φ

 Theorem 9.5 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional semisimple F -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 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. 2(β, α) α ∈ Φ. (c) If α, β ∈ Φ, then β − (α, α) 2(β, α) (d) If α, β ∈ Φ, then ∈ Z. (α, α) 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 9.6 Let E be a finite dimensional vector space over R. Let (•, •) : E × E → R be a positive definite symmetric bilinear form. We call (•, •) an inner product of E and (E, (•, •)) a Euclidean space. An R-linear transformation f : E → E is orthogonal if (f (x), f (y)) = (x, y) for all x, y ∈ E. Definition 9.7 A 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.

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101

Lemma 9.8 Let 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 + ai x ∈ E, pi ∈ P , ai ∈ R, i ∈ {1, 2}, we have (p1 + a1 x, p2 + a2 x) = (p1 , p2 ) + a1 a2 (x, x), and (σ(p1 + a1 x), σ(p2 + a2 x)) = (p1 − a1 x, p2 − a2 x) = (p1 , p2 ) + a1 a2 (x, x).  Example 9.9 Let 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(β, α) (2) σα (β) = β − hβ, αiα where hβ, αi = for all β ∈ E. (α, α) Proof. (1) β ∈ Pcα iff (β, cα) = c(β, α) = 0 iff β ∈ Pα . (2) If β ∈ Pα , then (β, α) = 0. Then hβ, αi = 0 and hence σα (β) = β. Also, if (β, Pα ) = 0, then x k α, β = cα for some c ∈ R∗ . Then hβ, αi = 2c and σα (β) = β − 2cα = −β. It follows from Definition 9.7 that σα is a reflection.  Lemma 9.10 Let E be a Euclidean space. Let α be a nonzero vector in E. (1) hα, αi = 2. (2) h•, •i is linear on the first coordinate. (3) σα2 = 1E . 2(α, α) = 2. (α, α) (2) Because (•, •) is bilinear. But h•, •i is not linear on the second coordinate because of the 1 denominator. For example, hβ, 2αi = hβ, αi for β ∈ E. 2 (3) Proof. (1) hα, αi =

Steps

Statements

Reasons

1.

σα2 (β) = σ(β) − hσ(β), αiα

Example 9.9(2)

2.

= β − hβ, αiα − hβ − hβ, αiα, αiα

Example 9.9(2)

3.

= β − hβ, αiα − hβ, αiα + hβ, αihα, αiα

Lemma 9.10(2)

4.

= β − 2hβ, αiα + 2hβ, αiα = β

Lemma 9.10(1)



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Lemma 9.11 Let E be a Euclidean space. Suppose Φ is a finite set, spans E and 0 6∈ Φ. Suppose all reflections {σα : α ∈ Φ} leave Φ invariant. If σ ∈ GL(E) leaves Φ invariant, fixes pointwise a hyperplane P of E and σ(α) = −α for some α ∈ Φ, then σ = σα for some α ∈ Φ. Proof. Let τ = σ ◦ σα . Steps

Statements

Reasons

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 not all distinct in Φ. Thus there exists kβ > 0 in Z such that τ kβ = 1.

Φ is finite

7.

Let k = max{kβ : β ∈ Φ}. Then τ k = 1. Then the minimal polynomial of τ divides T 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 9.12 A 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 hα, βi ∈ Z.

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103

Lemma 9.13 Let E be a Euclidean space. Let Φ be a root system of E. Then Φ = −Φ by Definition 9.12(R2). Definition 9.14 Let 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 9.15 Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. Let SΦ 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 9.16 Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. If σ ∈ GL(E) such that σ(Φ) = Φ, then σ ◦ σα ◦ σ −1 = σσ(α) . Proof. Steps

Statements

Reasons

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 9.17 Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. If σ ∈ GL(E) such that σ(Φ) = Φ, then hβ, αi = hσ(β), σ(α)i for all α, β ∈ Φ. Proof. Steps

Statements

Reasons

1.

σσ(α) = σ ◦ σα ◦ σ −1 .

Lemma 9.16

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Statements

Reasons

σσ(α) (σ(β)) = σ ◦ σα (β). 2.

σ(β) − hσ(β), σ(α)iσ(α) = hβ, αiα) = σ(β) − hβ, αiσ(α)

3.

hσ(β), σ(α)i = hβ, αi.

σ(β −

Example 9.9(2)

 Definition 9.18 Let E, E 0 be a Euclidean spaces. Let Φ, Φ0 be root systems of E, E 0 respectively. An isomorphism φ : (Φ, E) → (Φ0 , E 0 ) is a R-vector space isomorphism φ : E → E 0 such that (1) φ(Φ) = Φ0 and (2) hφ(β), φ(α)i = hβ, αi. Remark 9.19 Let φ : (Φ, E) → (Φ0 , E 0 ) 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 from Lemma 9.17. Lemma 9.20 Let φ : (Φ, E) → (Φ0 , E 0 ) be an isomorphism of root systems. Then σφ(α) (φ(β)) = φ(σα (β)) for all α, β ∈ Φ. Proof. Steps

Statements

Reasons

1.

σφ(α) (φ(β)) = φ(β) − hφ(β), φ(α)iφ(α)

Example 9.9(2)

= φ(β) − hβ, αiφ(α)

Definition 9.18(2)

= φ(β − hβ, αiα)

φ is F -linear

= φ(σα (β))

Definition 9.18(2)



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Lemma 9.21 Let E, E 0 be a Euclidean spaces. Let Φ, Φ0 be root systems of E, E 0 respectively. Let W , W 0 be Weyl groups of Φ, Φ0 , respectively. Let φ : (Φ, E) → (Φ0 , E 0 ) be an isomorphism of root systems. Then int(φ) : W 0 → 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 0 , σ 0 7→ φ−1 ◦ σ −1 ◦ φ. Definition 9.22 The dual of α ∈ Φ is α∨ =



2α . The dual of Φ is Φ∨ = {α∨ : α ∈ Φ}. (α, α)

Lemma 9.23 Let F be an algebraically closed field of characteristic 0. Let L be a finite dimensional nonzero semisimple F -Lie algebra. Let H be a maximal toral subalgebra of L. Let Φ be the set of roots of L relative to H and α ∈ Φ. Let tα satisfy α(h) = κ(tα , h) for all h ∈ H (as in Corollary 8.9). Let 2tα hα satisfy hα = (as in Proposition 8.16). Then tα∨ = hα . κ(tα , tα ) Proof. For all h ∈ H, we have Steps

Statements

Reasons

1.

κ(tα∨ , h) = α∨ (h) 2α(h) = (α, α) 2κ(tα , h) = . κ(tα , tα ) ! 2tα ,h κ(hα , h) = κ κ(tα , tα ) 2κ(tα , h) . = κ(tα , tα )

Corollary 8.9

tα∨ = hα .

κ|H is non-degenerate by Lemma 8.4

2.

3.

Definition 9.22 Corollary 8.9 and Definition 9.1 Proposition 8.16 κ is bilinear

 Definition 9.24 Let E be a Euclidean space. Let Φ be a root system of E. We call l = dimR (E) the rank of Φ. Example 9.25 The only rank one root system is (A1 ). −α

α

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Example 9.26 The following root system has rank two. We say that it has type (A1 × A1 ). β

−α

α

−β Example 9.27 The following root system has rank two. We say that it has type (A2 ). α+β

β

−α

α

−β

−(α + β)

Example 9.28 The following root system has rank two. We say that it has type (B2 ). β

α+β

−α

−(2α + β)

2α + β

α

−(α + β)

−β

Example 9.29 The following root system has rank two. We say that it has type (G2 ).

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107

3α + 2β

α+β

β

2α + β

3α + β

−α

−(3α + β)

α

−(2α + β) −(α + β)

−β

−(3α + 2β) Lemma 9.30 Let E be a Euclidean space. Let Φ be a root system of E. Let α, β ∈ Φ such that α 6= ±β and kβk ≥ kαk > 0. Let 0 ≤ θ ≤ π be the angle between α and β. Then the following are the only possibilities. hα, βi

hβ, αi

θ

kβk2 /kαk2

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

3

Proof. Steps

Statements

Reasons

1.

(α, β) = kαkkβk cos(θ).

defn of (•, •)

2.1.

hβ, αi = 2

2.2.

kβk cos(θ), kαk kαk hα, βi = 2 cos(θ). kβk

Example 9.9(2)

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Steps

Statements

Reasons

3.

hβ, αi, hα, βi ∈ Z.

Definition 9.12(R4)

4.

hβ, αihα, βi = 4 cos2 (θ) ∈ Z.

5.

4 cos2 (θ) ∈ {0, 1, 2, 3}

5.0.

4 cos2 (θ) = 0.

| cos θ| ≤ 1

θ = π/2.

0≤θ≤π

hβ, αi = hα, βi = 0

step 2.

kβk2 /kαk2 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. kαk = 1, kβk2 /kαk2 = 1. hα, βi = kβk

0≤θ≤π

hβ, αi = 1.

step 4, 5.1

If cos(θ) = −1/2, then θ = 2π/3. kαk hα, βi = − = −1, kβk2 /kαk2 = 1. kβk

0≤θ≤π

hβ, αi = −1.

steps 4, 5.1

4 cos2 (θ) = 2. √ If cos(θ) = 2/2, then θ = π/4. √ kαk √ hα, βi = 2 ∈ [1, 2]. kβk

0≤θ≤π

5.1.2

5.2. 5.2.1

steps 2.2, 3 and kβk ≥ kαk > 0

steps 2.2, 3 and kβk ≥ kαk > 0

steps 2.2, 3 and kβk ≥ kαk > 0

Thus hα, βi = 1, kβk2 /kαk2 = 2. hβ, αi = 2. 5.2.2

√ If cos(θ) = − 2/2, then θ = 3π/4. √ kαk √ hα, βi = 2 ∈ [− 2, −1]. kβk

step 4, 5.2 0≤θ≤π steps 2.2, 3 and kβk ≥ kαk > 0

Thus hα, βi = −1, kβk2 /kαk2 = 2. hβ, αi = 2. 5.3.

4 cos2 (θ) = 3.

step 4, 5.2

NOTES TO HUMPHREYS

Steps 5.3.1

Statements √ If cos(θ) = 3/2, then θ = π/6. √ √ kαk ∈ [1, 3]. hα, βi = 3 kβk

109

Reasons 0≤θ≤π steps 2.2, 3 and kβk ≥ kαk > 0

Thus hα, βi = 1, kβk2 /kαk2 = 3. hβ, αi = 3. 5.3.2

√ If cos(θ) = − 3/2, then θ = 5π/6. √ kαk √ hα, βi = 3 ∈ [− 3, −1]. kβk

step 4, 5.2 0≤θ≤π steps 2.2, 3 and kβk ≥ kαk > 0

Thus hα, βi = −1, kβk2 /kαk2 = 3. hβ, αi = −3.

step 4, 5.2

 Lemma 9.31 Let 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) Steps

Statements

Reasons

1.

If kβk ≥ kαk > 0, then hα, βi = 1.

(α, β) > 0 and Lemma 9.30

2.

α − β = α − hα, βiβ = σβ (α) is a root.

Example 9.9(2) and Definition 9.12(R3)

3.

If kαk ≥ kβk > 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 10.1 Let E be a Euclidean space. Let Φ be a root system of E. Let α, β ∈ Φ such that α 6= ±β. Then {i ∈ Z : β + iα ∈ Φ} = [−r, q] ∩ Z for some integers r, q ≥ 0. Proof. Steps

Statements

Reasons

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, then there exist −r < s ≤ p < q in Z such that β + (s − 1)α ∈ Φ, β + sα 6∈ Φ, β + pα 6∈ Φ and 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 and Lemma 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, by Lemma 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 10.2 Let E be a Euclidean space. Let Φ be a root system of E. Let α, β ∈ Φ such that α 6= ±β. We call {β + iα ∈ Φ : −r ≤ i ≤ q} the α-string through β. Lemma 10.3 Let E be a Euclidean space. Let Φ be a root system of E. Let α, β ∈ Φ such that α 6= ±β. The α-string through β is of length r + q + 1 ≤ 4. Proof.

NOTES TO HUMPHREYS

Steps

Statements

Reasons

1.

σα (β − rα) = σα (β) + rα

σα is R-linear by Definition 9.7

= β + (r − hβ, αi)α

Example 9.9(2)

∈Φ

β − rα ∈ Φ and Definition 9.12 (R3)

2.

r − hβ, αi ≤ q.

Lemma 10.1

3.

σα (β + qα) = σα (β) − qα

σα is R-linear by Definition 9.7

= β − (q + hβ, αi)α

Example 9.9(2)

∈Φ

β + qα ∈ Φ and Definition 9.12 (R3)

4.

−r ≤ −(q + hβ, αi).

Lemma 10.1

5.

hβ, αi = 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 hβ − rα, αi

7.

= hβ, αi − rhα, αi

Lemma 9.10(2)

= (r − q) − 2r = −(r + q) ≤ 0

step 5 and hα, αi = 2 by Lemma 9.10(1)

hβ − rα, αi ∈ {0, −1, −2, −3}

Lemma 9.30

The length is r+q+1 = −hβ −rα, αi+1 ∈ {1, 2, 3, 4}.

Lemma 10.1 and step 6.2

111

 Definition 10.4 Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. We call ∆ ⊂ Φ a base and we call elements of ∆ simple roots if (B1) ∆ is a R-basis of the R-vector space E; P (B2) For all β ∈ Φ, β = kα α, kα ∈ Z such that kα ≥ 0 for all α ∈ ∆ or kα ≤ 0 for all α ∈ ∆. 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 if P α − β ∈ Φ+ . The height of β ∈ Φ relative to ∆ is ht(β) = kα . α∈∆

Remark 10.5 (1) Card(∆) = dimR E by (B1). P kα α is unique since ∆ is R-linearly independent by (B1). (2) The expression β = α∈∆

(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 10.6 Let 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. Steps

Statements

Reasons

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 10.7 Let 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. Steps

Statements

Reasons

1.

There are infinitely many translates of P1 .

R is infinite

2.

Then there exists a translate P of P1 such that P 6= Pi for all 1 ≤ i ≤ n.

n < ∞.

3.

{P ∩ Pi , 2 ≤ i ≤ n} are codimension ≤ 1 subspaces of P .

dimR P − dimR (P ∩ Pi ) ≤ dimR E − dimR (Pi ) = 1

NOTES TO HUMPHREYS

Steps

Statements

4.

There exits a point Q ∈ P −

113

Reasons n S

Pi .

Inductive hypothesis

i=2

5.

Q∈E−

n S

Q ∈ P ⊂ E − P1 and step 4

Pi .

i=1

 Another proof for the case n > 1. Deleting duplicate ones, we may assume that Pi are distinct. Steps

Statements

Reasons

1.

Suppose y ∈ E − P1 .

dimF (P1 ) < dimF (E)

2.

For all x ∈ P1 , x + ay ∈ E − P1 for all a ∈ R∗ .

P1 is a linear subspace of E.

3.

If P1 ∪ · · · ∪ Pn = E, then there exists Pj , 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 + a0 y ∈ Pj and a 6= a0 . (x + ay) − (x + a0 y) ∈ Pj . Then y = a − a0

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 = dimR E − 1.

7.

A contradiction.

to our assumption that Pi are distinct.

 Definition 10.8 Let E be a Euclidean space. Let Φ be a root system of E. For each α ∈ Φ, let Pα = {β ∈ E : (β, α) = 0}. S (1) We call γ ∈ E regular if γ ∈ E − Pα ; otherwise we call γ singular. α∈Φ

(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 10.9 Suppose 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 10.10 Let E be a Euclidean space. Let Φ be a root system of E. Suppose γ ∈ E is regular. Then Φ+ (γ) ⊂ SpanZ (∆(γ)). Similarly, Φ− (γ) ⊂ SpanZ (∆(γ)). Proof. Suppose the opposite: Let S = Φ+ (γ) − SpanZ (∆(γ)) 6= ∅. Steps

Statements

Reasons

1.

Let α ∈ S such that (α, γ) min{(β, γ) : β ∈ S}.

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

=

Φ is finite, then S is finite

 Lemma 10.11 Let 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. Steps

Statements

Reasons

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)

NOTES TO HUMPHREYS

Steps

Statements

Reasons

3.1.

If β − α ∈ Φ+ (γ), then β = β − α + α.

Remark 10.9(2)

3.2.

A contradiction.

to β ∈ ∆(γ)

115

(2) Suppose the opposite (α, β) > 0. Steps

Statements

Reasons

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 10.12 Let E be a Euclidean space and γ ∈ E. Let ∆ be a subset of {α ∈ E : (α, γ) > 0} such that for each pair of vectors α, β ∈ ∆, (α, β) ≤ 0. Then ∆ is linearly independent. Proof. Suppose

P α∈∆

rα α = 0 for rα ∈ R. We need to show that rα = 0 for all α ∈ ∆. We write    

sα ,



−tβ , rβ ≤ 0.

rα =  

Steps

Statements

1.

P

sα α =

P

rα ≥ 0,

Reasons P

tβ β.

rα α = 0.

α∈∆

we denote ε = 2.1.

P

0 ≤ (ε, ε) P

=( =

P

sα α,

sα α =

P

tβ β. (•, •) is positive definite by Definition 9.6

P

tβ β)

sα tβ (α, β)

step 1 (•, •) is bilinear by Definition 9.6

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Steps

ZHENGYAO WU

Statements

Reasons

≤ 0.

The angle between each pair α, β ∈ ∆ is obtuse.

2.2.

ε = 0.

3.1.

0 = (γ, ε) =

P

sα (γ, α)

≥ 0 for all rα ≥ 0. 3.2.

sα = 0 for all rα ≥ 0.

4.1.

0 = (γ, ε) =

P

tβ (γ, β)

≥ 0 for all rβ ≤ 0. 4.2.

tβ = 0 for all rβ ≤ 0.

5.

rα = 0 for all α ∈ ∆.

step 1 and (•, •) is bilinear ∆ ⊂ {α ∈ E : (α, γ) > 0} and sα ≥ 0

step 1 and (•, •) is bilinear ∆ ⊂ {α ∈ E : (α, γ) > 0} and tβ ≥ 0

steps 3.2, 4.2

 Theorem 10.13 Let 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. Steps

Statements

Reasons

1.

Φ ⊂ SpanZ (∆(γ)).

Lemma 10.10

Thus Definition 10.4(B2) holds. 2.

E = SpanR (∆(γ)).

step 1 and Φ spans E by Definition 9.12(R1)

3.

∆(γ) is linearly independent.

Lemma 10.12

4.

Definition 10.4(B1) holds.

steps 3,4



NOTES TO HUMPHREYS

117

Theorem 10.14 Let 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. Steps

Statements

Reasons

1.

There exists γ ∈ E such that (γ, α) > 0 for 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(∆) = dimR E = Card(∆(γ)).

Definition 10.4(B1) and Theorem 10.13

9.

∆ = ∆(γ).

steps 7,8

 Definition 10.15 S Let E be a Euclidean space. Let Φ be a root system of E. The connected components of E − Pα α∈Φ

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 10.16 Let E be a Euclidean space. Let Φ be a root system of E. Let γ, γ 0 be regular elements of E. The following are equivalent: (1) C(γ) = C(γ 0 ). (2) γ, γ 0 lie on the same side of each Pα (α ∈ Φ).

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(3) Φ+ (γ) = Φ+ (γ 0 ). (4) ∆(γ) = ∆(γ 0 ). In particular, (1) iff (4) provides a bijection between the set of all Weyl chambers of Φ and the set of all bases of Φ given by C(γ) ←→ ∆(γ). Proof. (1) implies (2). Steps

Statements

Reasons

1.

Let c : [0, 1] → E be a path from γ to γ 0 , i.e. a continuous map such that c(0) = γ, c(1) = γ 0 and c([0, 1]) ⊂ C(γ).

(1)

2.

We may assume that there exists α ∈ Φ such that (γ, α) > 0 and (γ 0 , α) < 0.

Suppose (2) does not hold, i.e. γ and γ 0 are on different sides of Pα

3.

Define f (x) = (c(x), α), x ∈ [0, 1]. Then there exists 0 < ξ < 1 such that f (ξ) = 0.

f is continuous, f (0) > 0, f (1) < 0 and intermediate 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 0 ), there exists Pα between γ and γ 0 , a contradiction to (2). (2) implies (3) Steps

Statements

Reasons

1.

α ∈ Φ+ (γ) iff (α, γ) > 0

Definition 10.8(2)

iff (α, γ 0 ) > 0

(2)

iff α ∈ Φ+ (γ 0 ) .

Definition 10.8(2)

(3) implies (2) Suppose α ∈ Φ. Steps

Statements

Reasons

1.

(α, γ) > 0 iff α ∈ Φ+ (γ)

Definition 10.8(2)

iff α ∈ Φ+ (γ 0 )

(3)

iff (α, γ 0 ) > 0

Definition 10.8(2)

NOTES TO HUMPHREYS

Steps

Statements

Reasons

2.

(α, γ) < 0 iff (α, γ 0 ) < 0

γ is regular and step 1

119

(3) implies (4) Steps

Statements

Reasons

1.

∆(γ) (resp. ∆(γ 0 )) is the set of indecomposable roots of Φ+ (γ) (resp. Φ+ (γ 0 )).

Definition 10.8(3)

2.

∆(γ) = ∆(γ 0 ).

(3)

(4) implies (3). Steps

Statements

Reasons

1.

Φ+ (γ) = Φ ∩ SpanN (∆(γ))

Definition 10.4(B2)

= Φ ∩ SpanN (∆(γ 0 ))

(4)

= Φ+ (γ 0 )

Definition 10.4(B2)

 Definition 10.17 Let E be a Euclidean space. Let Φ be a root system of E. Let ∆ be a base of Φ. If ∆ = ∆(γ) for a regular element γ ∈ E, then we call C(∆) = C(γ) the fundamental Weyl chamber relative to ∆. Remark 10.18 The chamber C(∆) is open and convex since it is the intersection of finitely many open half-spaces. Example 10.19 The fundamental Weyl chamber of A1 × A1 is the yellow-green area.

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ZHENGYAO WU

β

−α

α

−β Example 10.20 The fundamental Weyl chamber of A2 is the yellow-green area.

α+β

β

−α

α

−β

−(α + β)

Example 10.21 The fundamental Weyl chamber of B2 is the yellow-green area.

β

α+β

−α

−(2α + β)

2α + β

α

−(α + β)

−β

Example 10.22 The fundamental Weyl chamber of G2 is the yellow-green area.

NOTES TO HUMPHREYS

121

3α + 2β

α+β

β

2α + β

−α

−(3α + β)

3α + β

α

−(2α + β) −(α + β)

−β

−(3α + 2β) Lemma 10.23 Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. If σ ∈ 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(ha, bi) for all a, b ∈ E. Steps

Statements

1.

x ∈ σ(C(γ)) iff σ −1 (x) ∈ C(γ)

Reasons

iff sgn((σ −1 (x), α)) = sgn((γ, α)) for all α∈Φ

Lemma 10.16 (1) iff (2)

iff sgn((x, σ(α)) = sgn((σ(γ), σ(α))) for all α ∈ Φ

σ ∈ W preserves signs of inner products by Lemma 9.17

iff x ∈ C(σ(γ))

σ(α) runs through tion 9.12(R3)

Φ

by

Defini-



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11. May 7th, Weyl group and its actions Lemma 11.1 Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. If σ ∈ W and γ ∈ E is regular, then σ(∆(γ)) = ∆(σ(γ)). Proof. Steps

Statements

Reasons

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 11.2 Let 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 ∈ ∆. Steps

Statements

Reasons

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 =

P γ∈∆

kγ γ, kγ ∈ N.

α ∈ Φ+

4.

kγ 0 > 0 for some γ 0 6= β.

otherwise α = kβ β = β ∈ ∆ by Definition 9.12(R2), a contradiction to α 6∈ ∆.

5.

kγ 0 γ 0 is a summand of α − β.

γ 0 6= β.

6.

α − β ∈ Φ+ .

step 2 and kγ 0 > 0.

NOTES TO HUMPHREYS

123

Case 2. (α, β) ≤ 0 for all β ∈ ∆. Steps

Statements

Reasons

1.

∆ ∪ {α} is linearly independent.

Lemma 10.12

2.

A contradiction.

to Definition 9.12(B1).

 Corollary 11.3 Let 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 sum α1 + · · · + αi ∈ Φ+ , 1 ≤ i ≤ k. Proof. Use induction on ht(β) = k. When k = 1, β = α1 ∈ ∆. Suppose k > 1, Steps

Statements

Reasons

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 hypothesis

4.

β = α1 + · · · + αk ∈ Φ+ .

given

 Lemma 11.4 Let 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}, Steps

Statements

1.

Suppose β =

Reasons P γ∈∆

kγ γ.

Definition 10.4(B2)

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ZHENGYAO WU

Steps

Statements

Reasons

2.

There exists γ 6= α, kγ > 0.

otherwise β = kα α = α by Definition 9.12(R2), a contradiction to β 6= α

3.

kγ γ is a summand of σα (β).

γ 6= α and σα (β) = β − hβ, αiα.

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 σα2 = 1 by Lemma 9.10(3)

 Corollary 11.5 Let E be a Euclidean space. Let Φ be a root system of E. Let ∆ be a base of Φ. Set δ =

1 P β. 2 β>0

Then σα (δ) = δ − α for all α ∈ ∆. Proof. Steps

Statements P

1.

Reasons P

β=

β∈Φ+ −{α}

2.

σα (β).

Lemma 11.4

β∈Φ+ −{α}

2δ − α =

P

β

defn of δ

β∈Φ+ −{α}

=

P

σα (β)

step 1

β∈Φ+ −{α}

= σα (

P β∈Φ+ −{α}

3.

β)

σα is R-linear

= σα (2δ − α)

defn of δ

= 2σα (δ) + α.

σα is R-linear and σα = −α

σα (δ) = δ − α.



NOTES TO HUMPHREYS

125

Lemma 11.6 Let E be a Euclidean space. Let Φ be a root system of E. Let ∆ be a base of Φ. Suppose α1 , . . . , 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.

Steps

Statements

Reasons

1.

β0 < 0

given σ1 ◦ · · · ◦ σt−1 (αt ) < 0

βt−1 > 0

given αt ∈ ∆ ⊂ Φ+

2.

There exists 1 ≤ s < t such that βs > 0 and s = min{i : βi > 0}.

3.

σs (βs ) = βs−1 < 0.

defn of βs and defn of s

4.

βs = α s .

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−1 ◦σs+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 11.7 Let E be a Euclidean space. Let Φ be a root system of E. Let ∆ be a base of Φ. Suppose α1 , . . . , αt ∈ ∆ (not necessarily distinct). If σ = σ1 ◦ · · · ◦ σt ∈ W , σi = σαi and t is the smallest possible 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 11.8 Let 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 0 be the subgroup of W generated by reflections σα , α ∈ ∆. If γ ∈ E is regular, then there exists σ ∈ W 0 such that (σ(γ), α) > 0 for all α ∈ ∆, i.e. σ(γ) ∈ C(∆). 1 P Proof. Let δ = α. 2 α∈Φ+ Steps

Statements

Reasons

1.

Choose σ ∈ W 0 such that (σ(γ), δ) = max{(τ (γ), δ) : τ ∈ W 0 }.

W 0 is finite since W is finite Lemma 9.15

2.

σα ◦ σ ∈ W 0 for all α ∈ ∆.

σα , σ ∈ W 0

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 11.9 Let 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 0 be the subgroup of W generated by reflections σα , α ∈ ∆. Then W 0 acts transitively on Weyl chambers, i.e. for all Weyl chambers C(γ), C(γ 0 ), there exists τ ∈ W 0 such that τ (C(γ)) = C(γ 0 ).

NOTES TO HUMPHREYS

127

Proof. Let C(γ) and C(γ 0 ) be Weyl chambers. Steps

Statements

Reasons

1.1.

There exists σ ∈ W 0 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 0 ∈ W 0 s.t. σ 0 (γ 0 ) ∈ C(∆).

Theorem 11.8

2.2.

σ 0 (C(γ 0 )) = C(σ 0 (γ 0 )) = C(∆).

Lemma 10.23

3.

Let τ = σ 0 −1 ◦ σ ∈ W 0 . Then τ (C(γ)) = σ 0 −1 (σ(C(γ))) = σ 0 −1 (C(∆)) = C(γ 0 ).

steps 1.2, 2.2

 Theorem 11.10 Let 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 0 be the subgroup of W generated by reflections σα , α ∈ ∆. Then W 0 acts transitively on bases, i.e. if ∆0 is another base of Φ, then σ(∆) = ∆0 for some σ ∈ W 0 . Proof. Steps

Statements

Reasons

1.

Suppose ∆ = ∆(γ) and ∆0 = ∆(γ 0 ).

Theorem 10.13

2.

There exists σ ∈ W 0 s.t. σ(C(γ)) = C(γ 0 ).

Theorem 11.9

3.

C(σ(γ)) = C(γ 0 ).

σ(C(γ)) = C(σ(γ)) by Lemma 10.23

4.

∆(σ(γ)) = ∆(γ 0 ).

Lemma 10.16 (1) iff (4)

5.

σ(∆) = σ(∆(γ)) = ∆(σ(γ))

Lemma 11.1

= ∆(γ 0 ) = ∆0 .

step 4

 Lemma 11.11 Let E be a Euclidean space. Let Φ be a root system of E. For all α ∈ Φ, there exists a base ∆ of Φ such that α ∈ ∆.

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ZHENGYAO WU

Proof. If Φ = A1 , then {α} is a base. If Φ 6= A1 , then there exists β ∈ Φ − {±α}. Steps

Statements

1.

There exists γ ∈ Pα −

Reasons S

Pβ .

β∈Φ−{±α}

2.

3.

4.

(γ, α) = 0. Let m = min{|(γ, β)|, β ∈ Φ − {±α}} > 0.

S β∈Φ−{±α}

(Pβ ∩ Pα ) ( Pα by Lemma 10.7

Φ is finite

m > (γ 0 , α) > 0 There exists γ 0 such that 2 m for all β ∈ Φ − {±α}. and |(γ 0 , β)| > 2

The continuity of the first coordinate of (•, •)

α ∈ ∆(γ 0 ).

below

α ∈ Φ+ (γ 0 )

(γ 0 , α) > 0

α is indecomposable.

|(γ 0 , β)| > (γ 0 , α) for all β ∈ Φ − {±α}

 Theorem 11.12 Let 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 0 be the subgroup of W generated by reflections σα , α ∈ ∆. For all α ∈ Φ, there exists σ ∈ W 0 such that σ(α) ∈ ∆. Proof. Steps

Statements

Reasons

1.

There exists a base ∆0 of Φ such that α ∈ ∆0 .

Lemma 11.11

2.

There exists σ ∈ W 0 such that σ(∆0 ) = ∆.

Theorem 11.10

3.

σ(α) ∈ σ(∆0 ) = ∆.

 Theorem 11.13 Let 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 0 be the subgroup of W generated by reflections σα , α ∈ ∆. Then W = W 0 .

NOTES TO HUMPHREYS

129

Proof. By definition W 0 ⊂ W . Conversely, it suffices to show that σα ∈ W 0 for all α ∈ Φ. Steps

Statements

Reasons

1.

There exists σ ∈ W 0 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 0 .

σβ , σ ∈ W 0 .

 Corollary 11.14 Let 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 11.15 Let 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. Steps

Statements

Reasons

1.

σ = σ1 ◦ · · · ◦ σt , σi = σαi , αi ∈ ∆ with the smallest length.

Theorem 11.13

2.

σ(αt ) < 0.

Corollary 11.7

3.

A contradiction.

to αt ∈ ∆ and σ(∆) = ∆ ⊂ Φ+ .

130

ZHENGYAO WU

 Definition 11.16 Let 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 the decomposition 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 11.17 Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ and σ ∈ 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 reduced decomposition. Steps

Statements

Reasons

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(σ◦σt ) → 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.

σt−1 = σt by Lemma 9.10(3)

6.

n(σ ◦ σt ) = Card(S(σ ◦ σt ))

Definition 11.16(2)

= Card(S(σ) − {αt })

NOTES TO HUMPHREYS

Steps

Statements

Reasons

= 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

131

 Definition 11.18 Let E be a Euclidean space. Let Φ be a root system of E. Let ∆ be a base of Φ. A simple root is an element of ∆. A simple reflection has the form σα , α ∈ ∆. Definition 11.19 Let X be a topological space and D ⊂ X. Let G be a group with an action on X. We call D a fundamental domain for this action if for all x ∈ X, there exists a unique g ∈ G such that gx ∈ D. Example 11.20 Let 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 fundamental domain for the action of W on E, by [Hum78, p.55, Exercise 14]. Lemma 11.21 Let 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 simple reflections 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. Steps

Statements

Reasons

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)

132

ZHENGYAO WU

Steps

Statements

Reasons

4.

Suppose σ(α) =

P

kβ β, where kβ ≤ 0

Definition 10.4(B2)

β∈∆

for all β ∈ ∆. 5.1.

(µ, σ(α)) =

P

kβ (µ, β) ≤ 0.

(µ, β) ≥ 0 for all β ∈ ∆

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(σ ◦σα ) = n(σ ◦σα ) = n(σ)−1 = l(σ)−1.

Lemma 11.17

9.

σ ◦ σα is the product of simple reflections which fix λ.

σ ◦ σα (λ) = µ and Induction

10.

σ is the product of simple reflections which fix λ.

σα2 = 1 by Lemma 9.10(3)



NOTES TO HUMPHREYS

133

12. May 14th, Irreducible root systems, two root lengths and Cartan matrix Definition 12.1 Let 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 12.2 Root systems of types A1 , A2 , B2 , G2 are irreducible. The root system of type A1 × A1 is reducible. Lemma 12.3 Let 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, Steps

Statements

Reasons

1.

σα (σβ (γ)) = σα (γ − hγ, βiβ)

Example 9.9(2)

= σα (γ) − hγ, βiσα (β)

σα is R-linear

= γ − hγ, αiα − hγ, βiβ.

Example 9.9(2) and σα (β) = β

2.

σβ (σα (γ)) = γ − hγ, αiα − hγ, βiβ.

 Lemma 12.4 Let E be a Euclidean space. Let Φ be a root system of E. Let ∆ be a base of Φ. Then Φ is reducible iff ∆ is reducible. Or equivalently, Φ is irreducible iff ∆ is irreducible. Proof. Suppose Φ is reducible. Steps

Statements

Reasons

1.

Suppose Φ = Φ1 tΦ2 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)

134

ZHENGYAO WU

Steps

Statements

Reasons

2.3.

Φ2 = {0}.

(•, •) is non-degenerate since it is positive definite

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. Steps

Statements

Reasons

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 for some 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 ) Lemma 12.3

4.3.

σβj (σ(α)) = σ(α).

(σ(α), βj ) = 0 since (∆1 , ∆2 ) = {0}

=

0 and

NOTES TO HUMPHREYS

135

Steps

Statements

Reasons

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 12.5 Let 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 have β < 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 by Definition 10.4(B1). Since (•, •) is non-degenerate, β = 0, a contradiction to Definition 9.12(R1).  Lemma 12.6 Let E be a Euclidean space. Let Φ be an irreducible root system of E. Let ∆ be a base of Φ. P Suppose Φ has a maximal root β. If β = kγ γ, then kγ ≥ 1 for all γ ∈ ∆. γ∈∆

Proof. Let ∆1 = {γ ∈ ∆ : kγ > 0} = 6 ∅ and ∆2 = {γ ∈ ∆ : kγ = 0}. Then ∆ = ∆1 t ∆2 by Definition 10.4(2). Steps

Statements

Reasons

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) =

P

kγ (α, γ)

γ∈∆1

4.

∆ is irreducible.

Φ is irreducible and Lemma 12.4

5.

There exists γ 0 ∈ ∆1 such that (α, γ 0 ) 6= 0.

Otherwise (∆1 , ∆2 ) = 0, a contradiction to Definition 12.1

136

ZHENGYAO WU

Steps

Statements

Reasons

6.

(α, γ 0 ) < 0.

step 3

7.

(α, β) < 0.

steps 4,7

8.

A contradiction.

to Lemma 12.5(1).

 Lemma 12.7 Let E be a Euclidean space. Let Φ be an irreducible root system of E. There exists a unique maximal root β relative to 0.

Lemma 12.5(2)

2.

(β 0 , β) ≥ (α, β).

β 0 ≥ α and Lemma 12.6

3.

(β 0 , β) > 0 and β 6= −β 0 .

steps 1,2

4.

β − β 0 ∈ Φ.

Lemma 9.31

5.

If β − β 0 ∈ Φ+ , then a contradiction.

to the maximality of β 0

If β 0 − 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 12.8 The maximal root of A2 is the black vector.

NOTES TO HUMPHREYS

137

α+β

β

−α

α

−β

−(α + β) Example 12.9 The maximal root of B2 is the black vector. α+β

β

−α

2α + β

α

−(2α + β)

−(α + β)

−β

Example 12.10 The maximal root of G2 is the black vector. 3α + 2β

α+β

β

2α + β

−α

−(3α + β)

3α + β

α

−(2α + β) −(α + β)

−(3α + 2β)

−β

138

ZHENGYAO WU

Lemma 12.11 Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. If Φ is irreducible, 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 0 be a subspace of E. If σα (E 0 ) = E 0 , then α ∈ E 0 or E 0 ⊂ Pα . ” (1) Suppose the opposite that E = E 0 + E 00 with (E 0 , E 00 ) = 0, E 0 is W -invariant and E 00 is W -invariant. Steps

Statements

Reasons

1.

σα (E 0 ) = E 0 for all α ∈ Φ.

E 0 is W -invariant

2.

Either α ∈ E 0 or E 0 ⊂ Pα .

[Hum78, Sec.9, Exercise 1]

3.

Φ = (Φ ∩ E 0 ) t (Φ ∩ E 00 ).

Either α ∈ E 0 or α ∈ E 00 , but not both.

4.

Φ = Φ ∩ E 0 or Φ = Φ ∩ E 00 .

Φ is irreducible

Φ ⊂ E 0 or Φ ⊂ E 00 . 6.

E = E 0 or E = E 00 .

Φ spans E by Definition 9.12(R1)

(2) Suppose the opposite that E 0 = SpanR (W β) for some β ∈ Φ and E 0 ( E. Let E 00 be the orthogonal complement of E 0 . Steps

Statements

1.

For all α ∈ Φ, (E 0 , σα (E 00 ))

Reasons

= (σα (E 0 ), σα (E 00 ))

E 0 is W -invariant.

= (E 0 , E 00 ) = 0.

Lemma 9.8

2.

E 00 is W -invariant.

σα (E 00 ) ⊂ E 00 for all α ∈ Φ and σα2 = 1

3.

A contradiction.

to (1)



NOTES TO HUMPHREYS

139

Lemma 12.12 Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. If Φ is irreducible, then at most two root lengths occur in Φ. Proof. Suppose α, β ∈ Φ. Steps

Statements

Reasons

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 kβk ≥ kαk.

kαk = kσ(α)k for all simple reflections σ and Theorem 11.13

4.

kβk2 /kαk2 ∈ {1, 2, 3}.

Lemma 9.30

5.

If there are ≥ 3 root lengths squares a < √ √ b < c, then ac = 3, ab = cb = 2. √ √ c = cb · ab = 2 · 2 = 2, a contradiction. a

c a

6.

>

to

c a

b a

> 1 and

=

√

c a

>

c b

> 1.

3

 Definition 12.13 If an irreducible root system has two root length, then we call them long or short. Lemma 12.14 Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. If Φ is irreducible, then W acts transitively on all roots of a given length. Proof. Suppose α, β ∈ Φ with kαk = kβk. Steps

Statements

Reasons

1.

The W -orbit of α spans E.

Lemma 12.11

140

ZHENGYAO WU

Steps

Statements

Reasons

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). kβk = kσ(α)k.

kαk = kσ(α)k as in step 3 of Lemma 12.12

5.

hβ, σ(α)i = ±1.

Lemma 9.30

6.

If hβ, σ(α)i = −1, then hσβ (β), σ(α)i = h−β, σ(α)i = 1.

Lemma 9.11 and Lemma 9.10(2)

7.

We may assume that hβ, αi = 1.

σβ , σ ∈ W

8.

hα, βi = 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 12.15 Let E be a Euclidean space. Let Φ be a root system of E. Let W be the Weyl group of Φ. If Φ is irreducible with two root lengths, then its maximal root β is long. Proof. Steps

Statements

Reasons

1.

For each α ∈ Φ there exists σ ∈ W such that 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

NOTES TO HUMPHREYS

Steps

Statements

Reasons

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

141

 Definition 12.16 Let 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 (hαi , αj i)1≤i,j≤l . Its entries are called Cartan integers. Example 12.17 The Cartan matrix of A1 × A1 is 



2  

0

0 2

, 

relative to α1 = (x, 0), x > 0, α2 = (0, y), y > 0. Example 12.18 The Cartan matrix of A2 is    



2

−1

−1 2

, 

√

1 3 ). relative to α1 = (1, 0), α2 = (− , 2 2 Example 12.19 The Cartan matrix of B2 is    



2

−1

relative to α1 = (−1, 1), α2 = (1, 0).

−2 2

, 

142

ZHENGYAO WU

Example 12.20 The Cartan matrix of G2 is 

   

−1

2

−3

2

, 

√ 3 3 ). relative to α1 = (1, 0), α2 = (− , 2 2 Lemma 12.21 The 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 reflections σ and α, β ∈ E. We have hαi , αj i = hσ(αi ), σ(αj )i for all σ ∈ W and 1 ≤ i, j ≤ l.  Lemma 12.22 The Cartan matrix is nonsingular. Proof. Steps 1.

Statements

Reasons

det (hαi , αj i)1≤i,j≤l = det 2l

=

l Q

2(αi , αj ) (αj , αj )

!

det((αi , αj )1≤i,j≤l ) 6= 0

1≤i,j≤l

(•, •) is non-degenerate

(αj , αj )

j=1

 Lemma 12.23 Let E be a Euclidean space. Let Φ, Φ0 be root systems of E with bases ∆, ∆0 , respectively. Every bijection ∆ → ∆0 , αi 7→ αi0 such that hαi , αj i = hαi0 , αj0 i for all 1 ≤ i, j ≤ l extends uniquely to an isomorphism φ : E → E 0 such that (1) φ(Φ) = Φ0 ; (2) hφ(α), φ(β)i = hα, βi 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 0 , φ(

l P

i=1

(1)

xi αi ) =

l P i=1

xi αi0 , xi ∈ R, 1 ≤ i ≤ l.

NOTES TO HUMPHREYS

Steps

Statements

1.

For all β 0 ∈ Φ0 , β 0 =

143

Reasons l P i=1

xi αi0 for xi ∈ Z with

Definition 10.4(B2)

xi ≥ 0 for all i or xi ≤ 0 for all i. 2.

Let β =

l P

xi αi . Then φ(β) = β 0 .

defn of φ

i=1

(2) Steps

Statements

Reasons

1.

hφ(α), φ(β)i = hα, βi for all α ∈ Φ and β ∈ ∆.

h•, •i is linear on the first coordinate by Lemma 9.10(2)

2.

For all β ∈ Φ, there exists σ ∈ W such that 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.

hφ(σγ (α)), φ(σγ (β))i

8.

= hσφ(γ) (φ(α)), σφ(γ) (φ(β))i

step 5

= hφ(α), φ(β)i

Lemma 9.17

hφ(σγ (α)), φ(σγ (β))i = hσγ (α), σγ (β)i

σγ (β) ∈ ∆ and step 1

= hα, βi

Lemma 9.17

9.

hφ(α), φ(β)i = hα, βi.

steps 7,8

10.

Suppose l(σ) > 1 and σ = τ ◦ σγ for some γ ∈ ∆ and τ ∈ W with l(τ ) = l(σ) − 1.

Definition 11.16

11.

hφ(σγ (α)), φ(σγ (β))i = hσγ (α)), σγ (β)i

τ (σγ (β)) ∈ ∆ and inductive hypothesis

= hα, βi

Lemma 9.17

144

ZHENGYAO WU

Steps

Statements

Reasons

12.

hφ(α), φ(β)i = hα, βi.

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 and q = −hαi , αj i, 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 = hγ, αj i by step 5 of the proof of 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 



2 0 . Determine A1 × A1 . The Cartan matrix of A1 × A1 is    0 2

• • • •

The roots of height one are α1 , α2 . Since hα1 , α2 i = 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 hα1 , α2 i = −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 hα1 + α2 , α2 i = −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 no roots of height three (and above). • All of positive roots are α1 , α2 , α1 + α2 . • All of negative roots are −α1 , −α2 , −α1 − α2 .

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145

Example 12.26 



 2 −2 . Determine B2 . The Cartan matrix of B2 is    −1 2

• The roots of height one are α1 , α2 . • The only root of height two is α1 + α2 . – Since hα1 , α2 i = −2 = r − q and r = 0, q = 2. The α2 -string through α1 is {α1 , α1 + α2 , α1 + 2α2 }. – Since hα1 , α2 i = −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 hα1 + α2 , α2 i = −2 + 2 = 0 = r − q and r = 1, q = 1. The α2 -string through α1 + α2 is {α1 , α1 + α2 , α1 + 2α2 }. – Since hα1 + α2 , α1 i = 2 − 1 = 1 = r − q and r = 1, q = 0. The α1 -string through α1 + α2 is {α2 , α1 + α2 }. • There is no root of height four and above. – Since hα1 + 2α2 , α2 i = −2 + 2 ∗ 2 = 2 = r − q and r = 2, q = 0. The α2 -string through α1 + 2α2 is {α1 , α1 + α2 , α1 + 2α2 }. – Since hα1 + 2α2 , α1 i = 2 + 2 ∗ (−1) = 0 = r − q and r = 0, q = 0. The α1 -string through 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 −3

−1 2

, 

Determine G2 .

• The roots of height one are α1 , α2 . • The only root of height two is α1 + α2 . – Since hα1 , α2 i = −1 = r−q and r = 0, q = 1. The α2 -string through α1 is {α1 , α1 +α2 }. – Since hα1 , α2 i = −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 hα1 + α2 , α2 i = −1 + 2 = 1 = r − q and r = 1, q = 0. The α2 -string through α1 + α2 is {α1 , α1 + α2 }. – Since hα1 + α2 , α1 i = 2 − 3 = −1 = r − q and r = 1, q = 2. The α1 -string through α1 + α2 is {α2 , α1 + α2 , 2α1 + α2 , 3α1 + α2 }. • The only root of height four is 3α1 + α2 . – Since h2α1 + α2 , α2 i = 2 ∗ (−1) + 2 = 0 = r − q and r = 0, q = 0. The α2 -string through 2α1 + α2 is {2α1 + α2 }.

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•

•

• •

– Since h2α1 + α2 , α1 i = 2 ∗ 2 − 3 = 1 = r − q and r = 2, q = 1. The α1 -string through 2α1 + α2 is {α2 , α1 + α2 , 2α1 + α2 , 3α1 + α2 }. The only root of height five is 3α1 + 2α2 . – Since h3α1 + α2 , α2 i = 3 ∗ (−1) + 2 = −1 = r − q and r = 0, q = 1. The α2 -string through 3α1 + α2 is {3α1 + α2 , 3α1 + 2α2 }. – Since h3α1 + α2 , α1 i = 3 ∗ 2 − 3 = 3 = r − q and r = 3, q = 0. The α1 -string through 3α1 + α2 is {α2 , α1 + α2 , 2α1 + α2 , 3α1 + α2 }. There are no root of height six (and above). – Since h3α1 + 2α2 , α2 i = 3 ∗ (−1) + 2 ∗ 2 = 1 = r − q and r = 1, q = 0. The α2 -string through 3α1 + 2α2 is {3α1 + α2 , 3α1 + 2α2 }. – Since h3α1 + 2α2 , α1 i = 3 ∗ 2 + 2 ∗ (−3) = 0 = r − q and r = 0, q = 0. The α1 -string through 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 .

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147

13. May 21th, Coxeter graphs, Dynkin diagrams Definition 13.1 Let E be a Euclidean space. Let Φ be a root system of E with base ∆ = {α1 , . . . , αl }. The Coxeter graph of Φ has • l vertices, the i-th vertex corresponds αi . • i-th and j-th vertices are joined by hαi , αj ihαj , αi i ∈ {0, 1, 2, 3} edges (see Lemma 9.30). Example 13.2 The Coxeter graph of A1 × A1 is

Example 13.3 The Coxeter graph of A2 is

Example 13.4 The Coxeter graph of B2 is

Example 13.5 The 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 hαi , αj i = hαj , αi i = 0; If there is one edge, then hαi , αj i = hαj , αi i = −1 by Lemma 9.30. (2) If there are n ≥ 2 edges, then (hαi , αj i = −1 and hαj , αi i = −n) or (hαi , αj i = −n and hαj , αi i = −1) by Lemma 9.30.  Definition 13.7 Let E be a Euclidean space. Let Φ be a root system of E. On the Coxeter graph of Φ. On each double or triple edge, we add an arrow from the long root to the short root, then we obtain the Dynkin diagram of Φ. Example 13.8

√ 1 3 The Dynkin diagram of B2 with base α1 = (− , ), α2 = (1, 0) is 2 2 1

2

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ZHENGYAO WU

Example 13.9

√ 3 3 The Dynkin diagram of G2 with base α1 = (1, 0), α2 = (− , ) is 2 2 1

2

Lemma 13.10 Let E be a Euclidean space. Let Φ be a root system of E. The Dynkin diagram of Φ determines its Cartan matrix. Proof. Suppose kαi k ≥ kαj k. Between the i-th and j-th vertex, by Lemma 9.30, • There is no edge iff hαi , αj i = hαj , αi i = 0; • There is one edge iff hαi , αj i = hαj , αi i = −1; • If there are n ∈ {2, 3} edges, then hαi , αj i = −n, hαj , αi i = −1.  Example 13.11 The 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

0

2

−2

−1

2

0

−1

0

   0    −1   

2

Proposition 13.12 Let 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 hαi , αj i = hαj , αi i = 0. Let ∆ = ∆1 t · · · t ∆t , (∆i , ∆j ) = 0 for all i 6= j, where the Coxeter graphs of ∆i are connected components of the Coxeter graph of Φ. Let Ei = SpanR (∆i ). Let Φi = SpanZ (∆i ) ∩ Φ. By Definition 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.

NOTES TO HUMPHREYS

149

(2) Steps

Statements

Reasons

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 13.13 Let E be a Euclidean space. Let A = {ε1 , . . . , εn } be a subset of unit vectors of E. We call A admissible 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 = Proof. Let εi =

αl α1 ,..., kα1 k kα1 k

)

is admissible.

αi . Then kεi k = 1 for all 1 ≤ i ≤ l. kαi k

Steps

Statements

Reasons

1.

∆ is linearly independent. Then A is linearly independent.

Definition 10.4(B1)

2.

(αi , αj ) ≤ 0 for all i 6= j. Then (εi , εj ) ≤ 0 for all i 6= j.

Lemma 10.6

3.

4(εi , εj )2 = hαi , αj ihαj , αi i

hαi , αj i

2(αi , αj ) (αj , αj ) 2(εi kαi k, εj kαj k) kαi k =2 (εi , εj ) 2 kαj k kαj k =

=

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ZHENGYAO WU

Steps

4.

Statements

Reasons

∈ {0, 1, 2, 3} for all i 6= j.

Lemma 9.30

A is admissible.

steps 1,2,3 and Definition 13.13

 Definition 13.15 Let 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 and all incident edges. Lemma 13.17 Let E be a Euclidean space. Let A = {ε1 , . . . , εn } be an admissible set. Let Γ be the Coxeter graph of A. The number of edges of Γ (without multiplicity) is N ≤ n − 1. n P

Proof. Let ε =

εi .

i=1

Steps

Statements

Reasons

1.

ε 6= 0.

A is linearly independent by Definition 13.13(1)

2.

For i 6= j, if there exists at least one edge between the i-th and the j-th vertex, then √ √ 2(εi , εj ) ∈ {−1, − 2, − 3}.

Definition 13.13(2)(3)

2(εi , εj ) ≤ −1. 3.

(•, •) is positive definite

0 < (ε, ε) =( =

n P

i=1 n P

εi ,

n P

εi )

(εi , εi ) + 2

i=1

defn of ε

i=1

P 1≤i k

2.

(ε, η0 ) 6= 0.

{ε, η1 , . . . , ηk } is linearly independent as a subset of A

3.

ε=

k P

{η0 , η1 , . . . , ηk } is SpanR {ε, η1 , . . . , ηk }

(ε, ηi )ηi .

i=0

4.

(ηi , ηj ) = 0 for all i 6= j.

Lemma 13.18

5.

1 = (ε, ε)

ε∈A

k P

= (ε,

(ε, ηi )ηi )

step 3

i=0

=

k P

(ε, ηi )2 >

i=0

k P

(ε, ηi )2 .

i=1

(•, •) is bilinear

a

basis

of

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Steps

Statements

Reasons

6.

The number of edges connected to ε is

Definition 13.15

k P

4(ε, ηi )2 < 4.

i=1

 Corollary 13.20 The only connected Coxeter graph with a triple edge is

Proof. Since Lemma 13.19.



Lemma 13.21 Let E be a Euclidean space. Let A = {ε1 , . . . , εn } be an admissible set. Let Γ be the Coxeter graph of A. Suppose {ε1 , . . . , εk } ⊂ A has Coxeter graph ··· Then A0 = (A − {ε1 , . . . , εk }) ∪ {ε} is admissible for ε =

k P

εi .

i=1

Proof. Steps

Statements

Reasons

1.

A0 is linearly independent.

ε=

k P

εi and A is linearly independent by

i=1

Definition 13.13(1) 2.

(ε, ε) = (

k P i=1

=

k P

εi ,

k P

(εi , εi ) + 2

i=1

εi )

defn of ε

i=1

P

(εi , εj )

= k − (k − 1) = 1.

3.

(•, •) is bilinear

1≤i 1.

and

2(εi , εj )

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)

=

NOTES TO HUMPHREYS

153

Steps

Statements

Reasons

5.

4(η, ε)2 ∈ {0, 1, 2, 3}.

step 3 and Definition 13.13(3)

6.

A0 is admissible.

steps 1,4,5 and Definition 13.13

 Lemma 13.22 Let E be a Euclidean space. Let A = {ε1 , . . . , εn } be an admissible set. Let Γ be the Coxeter graph 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 13.23 Let E be a Euclidean space. Let A = {ε1 , . . . , εn } be an admissible set. Let Γ be the Coxeter graph 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.

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ZHENGYAO WU

• Suppose Γ has no triple edges. For a vertex is connected to exactly three other distinct vertices, 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 13.24 Let E be a Euclidean space. Let A be an admissible set. Let Γ be the Coxeter graph of A. If Γ has type ··· ··· ε1 ε2 εp−1 εp ηq ηq−1 η2 η1 Then it is one of the following ··· Proof. Let ε =

p P

iεi and η =

i=1

Statements

1.

(ε, ε) = (

=

p P

p P i2 (ε

jηj .

iεi ,

p P

iεi )

defn of ε

i=1

i , εi )

P

+2

ij(εi , εj )

(•, •) is symmetric bilinear

1≤i (ε, η)2

ε, η are linearly independent 2 2

5.

p(p + 1) q(q + 1) pq · > 2 2 2 (p + 1)(q + 1) > 2pq.

canceling

pq 4

NOTES TO HUMPHREYS

Steps

6.

Statements

Reasons

(p − 1)(q − 1) ≤ 1.

pq − p − q < 1

155

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 13.25 1 1 1 Find positive integer solutions to + + > 1, p ≥ q ≥ r ≥ 1. . p q r Proof. Steps

Statements

Reasons

1.

If r = 1, then (p, q, 1) is a solution.

1 1 1 1 1 + + = + +1>1 p q r p q

2.1

If r ≥ 2, then

2.2.

r ≤ 2.

1
. p q 2

4.

If q = 2, then (p, 2, 2) is a solution.

1 1 1 + + 1 p q r p

5.1.

If q ≥ 3, then

5.2.

q ≤ 3.

1 1 1 2 < + ≤ 2 p q q

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.

6.3

We have solutions (3, 3, 2), (4, 3, 2) and (5, 3, 2).

7.

Solutions are (p, q, r) ∈ steps 1, 4, 6.3 {(p, q, 1), (p, 2, 2), (3, 3, 2), (4, 3, 2), (5, 3, 2)}.

1 1 1 1 ≤ ≤ ≤ . p q r 2 1 1 1 3 + + ≤ . p q r r

1 1 1 ≤ ≤ . p q 3

1 1 1 < + by step 3 2 p 3 If r ≥ 2 and q ≥ 3, then p ∈ {3, 4, 5}

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 Lemma 13.26 Let E be a Euclidean space. Let A be an admissible set. Let Γ be the Coxeter graph of A. If Γ has type ζ1 ζ2

··· ε1

ε2

···

ψ

ζr−1

εp−1

ηq−1

··· η2 η1

for p, q, r ≥ 2,

then it is one of the following ε1

ε2

···

ψ

ζ1

εp−1

η1 ψ ε1

where p ≥ 2,

ζ1

ε2

η1 η2 ψ ε1

ε2

,

ζ1

ε3

η1 η2 ψ ε1

ε2

ε3

ε4

,

ζ1 η1 η2

.

NOTES TO HUMPHREYS

157

14. May 28th, Classification, irreducible root systems of types A, B and C p−1 P

Proof. Suppose ε =

iεi , η =

i=1

q−1 P

iηi and ζ =

i=1

r−1 P

iζi . Let θ1 , θ2 , θ3 be the angle between ψ and

i=1

ε, η, ζ, respectively. Steps

Statements

Reasons

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 0 SpanR {ε, η, ζ, ψ} such that (ψ 0 , ε) (ψ 0 , η) = (ψ 0 , ζ) = 0 and (ψ, ψ 0 ) 6= 0.

4.

ψ = (ψ, ε)

∈ =

ε η ζ +(ψ, η) +(ψ, ζ) + 2 2 kεk kηk kζk2

(ψ, ψ 0 )ψ 0 . 5.

1 = (ψ, ψ) (ψ, ε)2 = + kεk2 (ψ, ε)2 > + kεk2

2

(ψ, η) (ψ, ζ) + + (ψ, ψ 0 )2 2 kηk kζk2 (ψ, η)2 (ψ, ζ)2 + kηk2 kζk2

7.

8. 9. 10.

p(p − 1) , (η, η) 2 q(q − 1) r(r − 1) , (ζ, ζ) = . 2 2 2 (ε, ψ) cos2 (θ1 ) = (ε, ε)(ψ, ψ) ((p − 1)εp−1 , ψ)2 = (ε, ε)(ψ, ψ) (p − 1)2 (− 12 )2 1 1 = = (1 − ). p(p−1) 2 p ·1 2 1 1 1 cos2 (θ2 ) = (1 − ), cos2 (θ3 ) = (1 − 2 q 2 1 1 1 1 1 1 (1 − ) + (1 − ) + (1 − ) < 1. 2 p 2 q 2 r (ε, ε)

ε η ζ , , , ψ 0 } is orthonormal by 2 2 kεk kηk kζk2 steps 1,3 {

ψ is a unit 2

(•, •) is bilinear and step 4 (ψ, ψ 0 ) 6= 0 by steps 2,3

= cos2 (θ1 ) + cos2 (θ2 ) + cos2 (θ3 ). 6.

Similar to steps 1,2 of the proof of Lemma 13.19

ψ is a unit and defn of (•, •)

=

=

similar to step 1 of the proof of Lemma 13.24 defn of (•, •) (εi , ψ) = 0 for all 1 ≤ i ≤ p − 2. 2(εp−1 , ψ) = −1, ψ is a unit and step 6

1 ). r

similar to step 6 steps 5,7,8

The following are the only possibilities.

Lemma 13.25

(p, q, 1) is not a solution

it does not have a node.

158

Steps

ZHENGYAO WU

Statements

Reasons

Solutions (p, 2, 2), p ≥ 2 has the first graph; Solutions (3, 3, 2), (4, 3, 2) and (5, 3, 2) has the second, third and fourth graph, respectively.

 Theorem 14.1 Let E be a Euclidean space. Let Φ be an irreducible root system of rank l. Let Γ be the Coxeter graph of Φ. The Dynkin diagram of Γ is one of the following A` (` ≥ 1) B` (` ≥ 2) C` (` ≥ 3) D` (` ≥ 4)

α1

α2

α3

α1

α2

α3

α1

α2

α3

α1

α2

α3

··· ··· ··· ···

α`−1

α`

α`−1

α`

α`−1

α` α`−1

α`−2 α`

α2 E6

α1

α3

α4

α5

α6

α2 E7

α1

α3

α4

α5

α6

α7

α2 E8

α1

α3 F4

α4 α1 G2

α5

α6

α2

α3

α1

α2

Proof. Let Dyn(Φ) be the Dynkin diagram of Φ.

α7 α4

α8

NOTES TO HUMPHREYS

159

Steps

Statements

Reasons

1.

Dyn(Φ) is one of Al (l ≥ 1) or G2 , or has a 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 14.2 Let E be a Euclidean space. Let Φ be an irreducible root system of rank l. Then the Cartan matrix of Φ is one of the following:

Al , (l ≥ 1) :

 2    −1     0          0   

0

−1 2 −1

0

···

0

−1 · · ·

0

···

0

2

... 0

0

···

0

0

· · · −1

2

0 

  0    0  ,       −1   









Bl , (l ≥ 2) :

 2    −1          0     0   

0

2

−1 · · · 2

··· ..

0

0

0

0

2

−1

.

0

···

0

· · · −1

0

···

0

2 −1

0 

  0         0    −2   

2

 

Cl , (l ≥ 3) :



 2    −1          0     0   

0

−1 · · · 2

···

0 0

0 0

... 0

···

2

0

· · · −1

0

···

0

−1 2 −2

0 

  0      ,   0    −1   

2

Dl , (l ≥ 4) :



 2    −1          0     0     0   

0

−1 · · ·

0

0

0

0 

···

0

0

0

 0 

2

... 0

···

−1

0

0

· · · −1

2

−1

0

···

0

−1

2

0

···

0

−1

0

2



      0     −1    0   

2

160

ZHENGYAO WU



  2     0    −1      0     0   

E6 :

 2     0    −1      0     0     0   





0

0

−1

0

0

2

0

−1

0

0

2

−1

0

−1 −1

2

−1

0

0

−1

2

0

0

0

−1

0 

  0    0  ,   0    −1   

E7 :

2

0

−1

0

0

0

2

0

−1

0

0

0

2

−1

0

0

2

−1

0

−1 −1 0

0

−1

2

−1

0

0

0

−1

2

0

0

0

0

−1

0

   0    0    0     0    −1   

2





E8 :

0

 2     0    −1     0      0     0     0   

0

0 2 0

−1 0 2

−1 −1

0 −1 −1 2

0 0 0 −1

0 0 0 0

0 0 0 0

0

0

−1

2

−1

0

0

0

0

−1

2

−1

0

0

0

0

−1

2

0

0

0

0

0

−1

0

   0    0    0  ,   0    0    −1   



F4 :



 2    −1     0   

0

−1 2

0 −2

−1

2

0

−1

0

   0  ,  −1   





G2 :

  

2

−3

2

2

Proof. It follows from Lemma 13.6 and Theorem 14.1.



Definition 14.3 A lattice in Rn is a free Z-module of rank n. Example 14.4 Let ε1 , . . . , εn be an orthonormal basis of Rn . Let I = SpanZ {εi : 1 ≤ i ≤ n}. Then I is a lattice. Lemma 14.5 Let ε1 , . . . , εn be an orthonormal basis of Rn . Let I = SpanZ {εi : 1 ≤ i ≤ n}. If x ∈ I and (x, x) = 1, then x = ±εi for some 1 ≤ i ≤ n. Proof. Steps

Statements

Reasons

1.

Suppose x = x1 ε1 + · · · + xn εn , xi ∈ Z for all 1 ≤ i ≤ n.

x∈I

−1 2

. 

NOTES TO HUMPHREYS

Steps

Statements

Reasons

2.

x21 + · · · + x2n = 1.

(x, x) = 1

3.

x2i = 1 for some i and x2j = 0 for all j 6= i.

x2i ∈ Z≥0 for all i.

161

x = ±εi .

 Lemma 14.6 Let ε1 , . . . , εn be an orthonormal basis of Rn . Let I = SpanZ {εi : 1 ≤ i ≤ n}. If x ∈ I and (x, x) = 2, then x = ±(εi ± εj ), i 6= j. Proof. Steps

Statements

Reasons

1.

Suppose x = x1 ε1 + · · · + xn εn , xi ∈ Z for all 1 ≤ i ≤ n.

x∈I

2.

x21 + · · · + x2n = 2.

(x, x) = 2

3.

x2i = x2j = 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 14.7 Let n ∈ Z. Then n2 + n = 0 or n2 + n ≥ 2. Proof. Steps

Statements

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.

Reasons

n2 + n = |n|(|n| ± 1) ≥ 2 · (2 − 1) = 2.

162

ZHENGYAO WU

 Lemma 14.8 Let ε1 , . . . , εn be an orthonormal basis of Rn . Let I = SpanZ {εi : 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) Steps

Statements

Reasons

1.

Φ is finite.

Lemma 14.5 and Lemma 14.6

2.

Φ spans E.

Given

3.

0 6∈ Φ.

0 6∈ A

Steps

Statements

Reasons

1.

Suppose x, cx ∈ Φ, c ∈ R.

2.1.

If A = {1} or {2}, then kcxk = kxk ∈ √ {1, 2}.

Card(A) = 1

2.2.

|c| = 1, c ∈ {±1}.

kcxk = |c|kxk

3.1.

If A = {1, 2} and kcxk = kxk, then c = ±1.

similar to step 2

3.2.

If A = {1, 2} and kcxk 6= kxk, then |c|2 = √ √ kcxk2 /kxk2 ∈ {2, 21 }, c = ± 2 or ± 22 .

{kcxk2 , kxk2 } = {1, 2}

3.3.

c ∈ Q.

x, cx ∈ I

3.4

a contradiction.

steps 3.2, 3.3

(R2)

(R4)

NOTES TO HUMPHREYS

Steps

Statements

1.

For all x ∈ Φ,

2.

For all x, y ∈ I, (x, y) ∈ Z.

3.

For all x, y ∈ Φ, hy, xi =

163

Reasons 2 (x,x)

∈ Z.

defn of Φ defn of I and (•, •)

2 (x, y) ∈ Z. (x, x)

Example 9.9

(R3) Steps

Statements

Reasons

1.

For all x, y ∈ I, σx (y) = y − hy, xix ∈ I.

hy, xi ∈ 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 14.9 There is an irreducible root system of type Al , (l ≥ 1). Proof. Construction: • Let E = {x ∈ Rl+1 :

l+1 P

xi = 0}.

i=1

• Let I = SpanZ {εi : 1 ≤ i ≤ l + 1}, I 0 = I ∩ E. • Let Φ = {α ∈ I : (α, α) = 2} and Φ0 = Φ ∩ E. Since E is the hyperplane of Rl+1 orthogonal to 0

l+1 P

εi , it is a Euclidean space of dimension l under

i=1

(•, •). Next, we show that Φ is a root system, using Definition 9.12. Steps

Statements

Reasons

0.

Φ is a root system.

Lemma 14.8

1.

(R1) of Φ0 is true.

(R1) of Φ is true and below.

Φ0 is finite.

Φ is finite and Φ0 ⊂ Φ

0 6∈ Φ0 .

0 6∈ Φ and Φ0 ⊂ Φ

164

ZHENGYAO WU

Steps

2.

3.

Statements

Reasons

Φ0 spans E.

Φ spans Rl+1 and Φ0 = Φ ∩ E.

(R2) of Φ0 is true.

below

α ∈ Φ0 iff −α ∈ Φ0 .

α ∈ I iff −α ∈ I, α ∈ E iff −α ∈ E and (−α, −α) = (α, α)

If α ∈ Φ0 , then Rα ∩ Φ0 = {±α}.

Rα ∩ Φ = {±α} by (R2) of Φ

(R3) of Φ0 is true.

below

For all α ∈ Φ0 , σα (Φ0 ) ⊂ Φ.

Φ0 ⊂ Φ and (R3) of Φ

If (β,

l+1 P

εi ) = 0, then (σα (β),

i=1

l+1 P

εi ) = 0.

Lemma 9.8, and σα (

l+1 P

εi ) =

i=1

i=1

(α,

l+1 P

l+1 P

εi since

i=1

εi ) = 0.

i=1

4.

σα (Φ0 ) ⊂ Φ0 .

Φ0 = Φ ∩ E

(R4) of Φ0 is true.

below

For all α, β ∈ Φ0 , hα, βi ∈ Z.

(R4) of Φ and Φ0 ⊂ Φ

Description: • Φ0 = {εi − εj : 1 ≤ i 6= j ≤ l + 1}; Card(Φ0 + ) = l2 + l. • Φ0 has base ∆ = {αi = εi − εi+1 : 1 ≤ i ≤ l}. The description of Φ0 follows from Lemma 14.6. Card(Φ0 + ) = (B2) of Definition 10.4, ∆ is a base of Φ0 . Steps

Statements

Reasons

5.

(B1) is true.

below

5.1.

∆ is linearly independent.

5.2.





l+1 2

= l2 + l. Verifying (B1) and

If c1 α1 + · · · + cl αl = 0, then c1 ε1 + (c2 − c1 )ε2 + · · · + (cl − cl1 )εl − cl εl+1 = 0.

αi = εi − εi+1

Then c1 = c2 −c1 = · · · = cl −cl−1 = cl = 0 and hence ci = 0 for all 1 ≤ i ≤ l.

{εi : 1 ≤ i ≤ l + 1} are linearly independent

∆ is a basis of E.

step 1.1 and Card(∆) = l.

NOTES TO HUMPHREYS

165

Steps

Statements

Reasons

6.

(B2) is true.

below

6.1.

εi − εj = αi + αi+1 + · · · + αj−1 for all 1 ≤ i < j ≤ l.

αk = εk − εk+1 , 1 ≤ i ≤ k ≤ j − 1 ≤ l − 1

6.2.

εi − εj = −αj − αj+1 − · · · − αi−1 for all 1 ≤ j < i ≤ l.

αk = εk − εk+1 , 1 ≤ j ≤ k ≤ i − 1 ≤ l − 1

We proceed to show that Φ0 , ∆ has Cartan matrix 

  2    −1     0          0   

0

−1 2 −1

···

0

0 

−1 · · ·

0

0 

0

2

··· ..

0

.

0

0

···

0

0

· · · −1

2

 

  0         −1   

2

It follows by Theorem 14.2 that Φ0 is irreducible of type Al , (l ≥ 1). Steps

Statements

Reasons

7.1.

If i < j, then (αi , αj ) = (εi − εi+1 , εj − εj+1 ) = −δi+1,j .

{εi : 1 ≤ i ≤ l + 1} is an orthonormal basis of Rn+1

7.2.

(αj , αj ) = (εj − εj+1 , εj − εj+1 ) = 2 for all 1 ≤ j ≤ l + 1.

{εi : 1 ≤ i ≤ l + 1} is an orthonormal basis of Rn+1

7.3.

hαi , αj i =

8.

If i > j, then hαi , αj i = −δi,j+1 .

2(αi , αj ) = −δi+1,j . (αj , αj )

steps 1,2 similar to step 7

 Proposition 14.10 There is an irreducible root system of type Bl , (l ≥ 2).

166

ZHENGYAO WU

Proof. Construction: • Let E = Rl and I = SpanZ {εi : 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 Verifying (B1) and (B2) of Definition 10.4, ∆ is a base of Φ. Steps

Statements

Reasons

1.

(B1) is true.

below

1.1.

∆ is linearly independent.

l 2

= l2 .

If c1 α1 + · · · + cl αl = 0, then c1 ε1 + (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 = 0 and hence ci = 0 for all 1 ≤ i ≤ l.

{εi : 1 ≤ i ≤ l + 1} are linearly independent

1.2.

∆ is a basis of E.

step 1.1 and Card(∆) = l.

2.

(B2) is true.

below

2.1.

εi − εj = αi + αi+1 + · · · + αj−1 for all 1 ≤ i < j ≤ l.

αk = εk − εk+1 , 1 ≤ i ≤ k ≤ j − 1 ≤ l − 1

2.2.

εi = εi − εl + εl = αi + · · · + αl−1 + αl for all 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 all 1 ≤ j < i ≤ l.

step 2.1

2.5.

−εi = −(εi −εl )−εl = −αi −· · ·−αl−1 −αl for all 1 ≤ i ≤ l

step 2.2.

NOTES TO HUMPHREYS

167

Steps

Statements

Reasons

2.6.

−εi − εj = −(εi − εl ) − (εj − εl ) − 2εl = −αi − · · · − αj−1 − 2αj − · · · − 2αl for all 1 ≤ i < j ≤ l.

step 2.3

We proceed to show that Φ, ∆ has Cartan matrix 

  2    −1          0     0   

−1 · · ·

0

2

··· ..

0

0

0

0

2

−1

.

0

···

0

· · · −1

0

···

0

2 −1

0 

  0         0    −2   

2

It follows by Theorem 14.2 that Φ is irreducible of type Bl , (l ≥ 2). Steps

Statements

Reasons

3.

If 1 ≤ i < j ≤ l except when i = l − 1 and j = l, then hαi , αj i = −δi+1,j .

similar to step 7 of Proposition 14.9

If 1 ≤ j < i ≤ l, then hαi , αj i = −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 basis of Rn

4.2.

(αl , αl ) = (εl , εl ) = 1

{εi : 1 ≤ i ≤ l} is an orthonormal basis of Rn

4.3.

hαl−1 , αl i =

2(αl−1 , αl ) = −2. (αl , αl )

steps 4.1, 4.2

 Proposition 14.11 There is an irreducible root system of type Cl , (l ≥ 3). Proof. Construction:

168

ZHENGYAO WU

• Let E = Rl and Φ the root system of type Bl as in Proposition 14.10. 2α • Let Φ∨ = {α∨ = : α ∈ Φ}. (α, α) 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 }. Steps

Statements

Reasons

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 · · ·

0

0

0 

···

0

0

0 

2

... 0

···

0

· · · −1

0

···

2

0

−1 2 −2

 

       0    −1   

2

It follows by Theorem 14.2 that Φ is irreducible of type Cl , (l ≥ 3). Steps

Statements

Reasons

7.

hβ ∨ , α∨ i = hα, βi 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 transpose of that of Φ of type Bl



NOTES TO HUMPHREYS

169

15. June 4th, Irreducible root systems of types D,E,F,G Proposition 15.1 There 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 (B2) of Definition 10.4, ∆ is a base of Φ. Steps

Statements

Reasons

1.

(B1) is true.

below

1.1.

∆ is linearly independent.

l 2

= l2 − l. Verifying (B1) and

If c1 α1 + · · · + cl αl = 0, then c1 ε1 + (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 hence ci = 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

1.2.

∆ is a basis of E.

step 1.1 and Card(∆) = l.

2.

(B2) is true.

below

2.1.

εi − εj = αi + αi+1 + · · · + αj−1 for all 1 ≤ i < j ≤ l.

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 +αl for all 1 ≤ i < j ≤ l.

step 2.1

2.3.

εj − εi = −αj − αj+1 − · · · − αi−1 for all 1 ≤ j < i ≤ l.

step 2.1

170

ZHENGYAO WU

Steps

Statements

Reasons

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

We proceed to show that Φ, ∆ has Cartan matrix 

  2    −1          0     0     0   

0

−1 · · · 2

··· ..

0

0

0

0

0

0

2

−1

0

2

−1

.

0

···

0

· · · −1

0

···

0

−1

2

0

···

0

−1

0

0

   0        0     −1    0   

2

It follows by Theorem 14.2 that Φ is irreducible of type Dl , (l ≥ 4). Steps

Statements

Reasons

3.

If 1 ≤ i < j ≤ l except when i = l − 1 and j = l, then hαi , αj i = −δi+1,j .

similar to step 7 of Proposition 14.9

If 1 ≤ j < i ≤ l, then hαi , αj i = −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 basis of Rn

4.2.

(αl−1 , αl−1 ) = (εl−1 − εl , εl−1 − εl ) = 2.

{εi : 1 ≤ i ≤ l} is an orthonormal basis of Rn

4.3.

hαl−2 , αl−1 i =

5.1.

(αl−2 , αl ) = (εl−2 − εl−1 , εl−1 + εl ) = −1.

{εi : 1 ≤ i ≤ l} is an orthonormal basis of Rn

5.2.

(αl , αl ) = (εl−1 + εl , εl−1 + εl ) = 2.

{εi : 1 ≤ i ≤ l} is an orthonormal basis of Rn

2(αl−2 , αl−1 ) = −1. (αl−1 , αl−1 )

steps 4.1, 4.2 and Example 9.9

NOTES TO HUMPHREYS

Steps

Statements

5.3.

hαl−2 , αl i =

6.1.

(αl−1 , αl ) = (εl−1 − εl , εl−1 + εl ) = 0.

6.2.

hαl−1 , αl i =

171

Reasons 2(αl−2 , αl ) = −1. (αl , αl )

steps 5.1, 5.2 and Example 9.9 {εi : 1 ≤ i ≤ l} is an orthonormal basis of Rn

2(αl−1 , αl ) = 0. (αl , αl )

steps 6.1, and Example 9.9

 Proposition 15.2 There is an irreducible root system of type E8 . Proof. Construction: 1 • Let E = R8 with orthonormal basis {εi : 1 ≤ i ≤ 8}. Define ε9 = (ε1 + · · · + ε8 ) 2 8 9 P P 0 00 ci is even.}. • Let I = SpanZ {εi : 1 ≤ i ≤ 8}, I = I + Zε9 and I = { ci εi ∈ I 0 : i=1

i=1

• Let Φ = {α ∈ I 00 : (α, α) = 2}. Description: • Φ = {±(εi ± εj ), 1 ≤ i < j ≤ 8} ∪ { Card(Φ+ ) = 120.

8 8 1P P (−1)k(i) εi : k(i) ∈ {0, 1}, k(i) is even.}; 2 i=1 i=1

1 (ε1 + ε8 − (ε2 + · · · + ε7 )), α2 = ε1 + ε2 , α3 = ε2 − ε1 , α4 = 2 ε3 − ε2 , α5 = ε4 − ε3 , α6 = ε5 − ε4 , α7 = ε6 − ε5 , α8 = ε7 − ε6 , }.

• Φ has base ∆ = {α1 = We first describe Φ. Steps

Statements

Reasons

1.

Suppose α ∈ Φ. Then α =

9 P

ci εi for

α ∈ I 00

i=1

ci ∈ Z where 2.

8 P

ci is even.

i=1

We assume c9 ∈ {0, 1}. If c9 = 2q + r, 0 ≤ r < 2, then α =

8 P

1 ε9 = (ε1 + · · · + ε8 ) 2

(ci + q)εi + rε9 .

i=1

3.

(α, α) =

8 P

(ci +

i=1

8 8 c9 2 P P ) = c2i + c9 ci + 2 i=1 i=1

8 P

α=

(ci +

i=1

c9 )εi 2

2c29 = 2. 4.

If c9 = 0, then {±(εi ± εj ), 1 ≤ i < j ≤ 8} ⊂ Φ.

8 P c2 i=1

i

= 2 and Lemma 14.6

172

ZHENGYAO WU

Steps

Statements

5.

If c9 = 1, then {

Reasons 8 1P (−1)k(i) εi : k(i) ∈ 2 i=1

8 P

Lemma 14.7

{0, 1}} ⊂ Φ. 5.1.

(c2i + ci ) = 0 and ci ∈ {−1, 0} by

i=1

8 P

Card{1 ≤ i ≤ 8 : ci = −1} is even.

ci is even.

i=1

5.2.

8 P

k(i) = 0 iff ci = 0; k(i) = 1 iff ci = −1

k(i) is even in step 5

i=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. 

1 2

     1    −1     0      0     0     0   

0



− 12

− 21

− 12

− 21

− 12

− 12

1 2 

1

0

0

0

0

0

0 

1

0

0

0

0

0

0 

−1

1

0

0

0

0

0

−1

1

0

0

0

0

0

−1

1

0

0

0

0

0

−1

1

0

0

0

0

0

−1

1

   

  0     0    0    0   

0

Steps

Statements

Reasons

6.

∆ is linearly independent over R.

above

7.

(R1) is true.

below

7.1.

Φ is finite.

Card(Φ+ ) = 2 82 + 12 ( 12 28 ) = 56 + 64 = 120, Card(Φ) = 240

7.2.

Φ spans E.

∆ is linearly independent, Card(∆) = 8 and ∆ ⊂ Φ

7.3.

0 6∈ Φ.

(0, 0) = 0 6= 2

8.

(R2) is true. If α, cα ∈ Φ and c ∈ R, then c = ±1.

|c| = 1 since |c|kαk = kcαk = 2 and kαk = 2

 

NOTES TO HUMPHREYS

Steps

Statements

Reasons

9.

(R4) is true.

below

9.1.

For all α, β ∈ Φ, hα, βi =

9.2.

If α = ±(εs ± εt ) and β = ±(εi ± εj ) then (α, β) ∈ Z.

9.3.

If α = ±(εs ± εt ) and β = 8 8 1P P (−1)k(i) εi : k(i) ∈ {0, 1} with k(i) 2 i=1 i=1 even, then (α, β) ∈ Z.

9.4.

If α = with

8 1P (−1)k(i) εi : 2 i=1

8 P

2(α, β) = (α, β) (β, β)

k(i) ∈ {0, 1}

k(i) = 2k even, and β

=

i=1 8 1P (−1)h(i) εi : 2 i=1 8 P i=1

h(i) ∈ {0, 1} with

173

(β, β) = 2 for all β ∈ Φ {εi : 1 ≤ i ≤ 8} is an orthonormal basis of R8 1 1 (α, β) = ± ± ∈ {1, 0, −1} 2 2

Suppose Card({1 ≤ i ≤ 8 : k(i) = h(i) = 1}) = t. Then (α, β) = 8 1 1P (−1)k(i)+h(i) = ((8 − 2k − 2h + t) − 4 i=1 4 (2h − t) − (2k − t) + t) = 2 − h − k + t ∈ Z

h(i) = 2h even, then (α, β) ∈ Z.

10.

(R3) is true.

below

10.1.

If β ∈ I 00 , then σα (β) = β − hβ, αiα ∈ I 00

(R4) and defn of I 00

10.2.

If β ∈ Φ, then (σα (β), σα (β)) = (β, β) = 2, σα (β) ∈ Φ.

Lemma 9.8 and defn of Φ

Verifying (B1) and (B2) of Definition 10.4, ∆ is a base of Φ. Steps

Statements

Reasons

11.

(B1) is true, i.e. ∆ is a basis of E

Card(∆) = 8 and ∆ is linearly independent by the discussion just before step 6

12.

(B2) is true.

below

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 ) = ∓ i 1 and (a + 1, b) is generated by {(i, i + 1) : 1 ≤ i ≤ l}.

1.3.

(a, b) = (a, a + 1)(a + 1, b)(a, a + 1) is generated by {(i, i + 1) : 1 ≤ i ≤ l}.

step 1.2.

2.

Transpositions (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 generated by transpositions.

2.3.

(a1 , a2 , . . . , an ) = (a1 , . . . , an−1 )(a1 , an ) is generated by transpositions.

3.

Cycles generate permutations.

generate

cycles

If a permutation σ does not have cycles, then a, σ(a), σ(σ(a)), . . . are distinct, a contradiction.

step 2.2

to the fact that {1, 2, . . . , l + 1} is finite.

 Proposition 16.2 Let E be the Eulidean space {x ∈ Rl+1 :

l+1 P i=1

xi = 0}. Let Φ(Al ) be the irreducible root system

of type Al , (l ≥ 1) in E with base ∆(Al ) = {αi = εi − εi+1 : 1 ≤ i ≤ l}. Let W (Al ) be the

186

ZHENGYAO WU

Weyl group of Φ(Al ). Let Sl+1 be the permutation group of {1, 2, . . . , l + 1}. Then we have an ∼ isomorphism W (Al ) − → Sl+1 ; σαi 7→ (i, i + 1) and the order of W (Al ) is (l + 1)!. Proof. Steps

0.

Statements

σεi −εi+1 (εj ) =

Reasons                 

εj ,

j 6∈ {i, i + 1}

εi ,

hεj , εi −εi+1 i = (εj , εi −εi+1 ) since (εj , εi − εi+1 , εj , εi −εi+1 ) = 2; and Example 9.9(2)

.

εi+1 , j = i j =i+1

1.

W (Al ) is generated by {σαi : αi ∈ ∆}, with the following relations.

Theorem 11.13

1.1.

For j > i + 1 or j + 1 < σαj ◦ σαi (εk ) = σαi ◦ σαj (εk )

i, =

αi = εi − εi+1 and step 0

=

αi = εi − εi+1 and step 0

                                

1.2.

εk ,

εi+1 , k = i εi ,

.

k =i+1

εj+1 , k = j εj ,

σ

αi+1    εj ,          εi+2 ,            

k 6∈ {i, i + 1, j, j + 1}

k =j+1 ◦

σαi (εj )

j 6∈ {i, i + 1, i + 2} j=i .

εi ,

j =i+1

εi+1

j =i+2

1.3.

σα2 i = 1.

Lemma 9.10(3)

2.

Sl+1 is generated by {(i, i+1) : 1 ≤ i ≤ l} with 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) fixes i, i + 1

2.2.

(i, i + 1)(i + 1, i + 2) = (i + 1, i, i + 2).

2.3.

(i, i + 1)2 = 1.

NOTES TO HUMPHREYS

187

Steps

Statements

Reasons

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 16.3 Let G be a group. Let N be a normal subgroup of G. Let H be a subgroup of G. We call G an inner semi-direct product of N and H and we write G = N o H if G = N H and N ∩ H = {e}. Proposition 16.4 Let E ' Rl be the Eulidean space. Let Φ(Bl ) be the irreducible root system of type Bl , (l ≥ 2) in E with base ∆(Bl ) = {αi = εi − εi+1 : 1 ≤ i ≤ l − 1} ∪ {αl = εl }. Let W (Bl ) be the Weyl group of Φ(Bl ). Let Sl be the permutation group of {1, 2, . . . , l}. Then W (Bl ) ' (Z/2Z)l o Sl and it has order 2l l!. 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 a subgroup N of W . Let ei ∈ (Z/2Z)n whose i-th entry is 1 and all other entries are 0. Steps 0.

1.1.

Statements σεi (εj ) =

Reasons

   

εj ,

  

−εi , j = i

j 6= i .

hεi , εj i =

   

0, j 6= i

  

2, j = i

For all j 6= i, σεj ◦ σεi (εk ) = σεi ◦ σεj (εk ) =                 

εk ,

k 6∈ {i, j}

−εi , k = i −εj , k = j

1.2.

σε2i = 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 ≤ l} with 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 .

188

ZHENGYAO WU

Steps

Statements

Reasons

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 ◦ σ αi =

               

4.1.

For j 6∈ {i, i + 1}, σαi ◦ σεj ◦ σαi (εk ) =       

4.2.

−εj , k = j

  

0,

◦

σ

αi    ε   

4.3.

k 6= j

   

k,

σεi

σεj ,

j 6∈ {i, i + 1};

σεi+1 , j = i; σεi ,

, see below

j = i + 1.

step 0; and step 0 of Proposition 16.2

= σεj (εk ). ◦

k 6= i + 1

−εi+1 , k = i + 1

σαi (εk )    

=

step 0; and step 0 of Proposition 16.2

=

step 0; and step 0 of Proposition 16.2

= σεi+1 (εk ).

      

σαi ◦ σεi+1 ◦ σαi (εk ) =

  

k 6= i

   

−εi , k = i

  

εk ,

σεi (εk ). 5.

W (Bl ) = N H.

below

5.1.

N H is a subgroup of W (Bl ).

step 4

5.2.

Elements of W (Bl ) are products of reflections of the form σαi , 1 ≤ i ≤ l.

Theorem 11.13

5.3.

W (Bl ) is a subgroup of N H.

Use step 4 to move σαl to the left.

6.

H ∩ N = {1}.

below

6.1.

Otherwise there exists τ =

Q i∈I

Q j∈J

σαi =

N ∩ H 6= {1}.

σεj 6= 1 in H ∩N , where I ⊂ {1, . . . , l−

1}, J ⊂ {1, . . . , l} and the left hand side is the reduced decomposition. 6.2.

There exists εk such that τ (εk ) Q σαi (εk ) = εt , t 6= k. i∈I

=

τ 6= 1 and step 0 of Proposition 16.2

NOTES TO HUMPHREYS

Steps

Statements

6.3.

τ (εk ) =

Q j∈J

189

Reasons σεj (εk ) = ±εk .

step 0

6.4.

a contradiction.

steps 6.2 and 6.3.

7.

W (Bl ) = N oH ' (Z/2Z)l oSl and hence Card(W (Bl )) = 2l l!.

steps 4,5,6 and Definition 16.3

 Proposition 16.5 Let E ' Rl be the Eulidean space. Let Φ(Cl ) be the irreducible root system of type Cl , (l ≥ 3) in E with base ∆(Cl ) = {αi = εi − εi+1 : 1 ≤ i ≤ l − 1} ∪ {αl = 2εl }. Let W (Cl ) be the Weyl group of Φ(Cl ). Let Sl be the permutation group of {1, 2, . . . , l}. Then W (Cl ) ' (Z/2Z)l o Sl and it has order 2l l!. Proof. Steps

Statements

Reasons

1.

Φ(Cl ) = Φ(Bl )∨ .

Proposition 14.11

2.

σ α = σ α∨ .

α∨ =

3.

W (Cl ) = W (Bl ).

Theorem 11.13

2 α and Example 9.9(1) (α, α)

 Proposition 16.6 Let E ' Rl be the Eulidean space. Let Φ(Dl ) be the irreducible root system of type Dl , (l ≥ 4) in E with base ∆(Dl ) = {αi = εi − εi+1 : 1 ≤ i ≤ l − 1} ∪ {αl = εl−1 + εl }. Let W (Dl ) be the Weyl group 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−1 l!. Proof. Since W is generated by σα , α ∈ Φ(Dl ) = {±(εi ± εj ) : 1 ≤ i < j ≤ l}, we suppose {σαi : 1 ≤ i ≤ l − 1} generates a subgroup H of W and {σε1 +εj ◦ σε1 −εj : 1 ≤ j ≤ l} generates a subgroup N of W . Let ei ∈ (Z/2Z)n whose i-th entry is 1 and all other entries are 0.

190

Steps

0.

ZHENGYAO WU

Statements

Reasons         

εk ,

σε1 +εj ◦ σε1 −εj (εk ) =  −ε1 , k = 1,

        

σε1 ±εj (εk ) =  ∓εj , k = 1,

−εj , k = j,

      

∓ε1 , k = j,

      

εk ,

k 6∈ {1, j},

k 6∈ {1, 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 =

step 0; and step 0 of Proposition 16.2

                                

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 .

step 0; and step 0 of Proposition 16.2

5.

W (Dl ) = N H.

below

5.1.

N H is a subgroup of W (Dl ).

step 4

5.2.

Elements of W (Dl ) are products of reflections of the form σαi , 1 ≤ i ≤ l.

Theorem 11.13

5.3.

W (Dl ) is a subgroup of N H.

Use step 4 to move σαl to the left.

NOTES TO HUMPHREYS

Steps

Statements

Reasons

6.

H ∩ N = {1}.

below

6.1.

Otherwise there exists τ =

Q i∈I

Q j∈J

σαi =

191

N ∩ H 6= {1}.

(σε1 +εj ◦ σε1 −εj ) 6= 1 in H ∩ N , where

I ⊂ {1, . . . , l − 1}, J ⊂ {1, . . . , l} and the left hand side is the reduced decomposition. 6.2.

There exists εk such that τ (εk ) Q σαi (εk ) = εt , t 6= k.

=

τ 6= 1 and step 0 of Proposition 16.2

i∈I

6.3.

τ (εk ) =

Q 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 and hence Card(W (Dl )) = 2l−1 l!.

steps 4,5,6 and Definition 16.3

 Theorem 16.7 Let 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

2l l!

Cl , (l ≥ 3)

(Z/2Z)l o Sl

2l l!

Dl , (l ≥ 4)

(Z/2Z)l−1 o Sl

2l−1 l!

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 · Ω8 : 2, see [CCN 85], pp.[85]-[87]

214 ∗ 35 ∗ 52 ∗ 7 = 696729600

F4

1+4 GO+ : S3 × S3 , see [Wil09, p.103] 4 (3) ' 2

27 ∗ 32 = 1152

192

ZHENGYAO WU

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 16.8 Let 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 Aut(Φ) = W o Γ. Proof. Steps

Statements

Reasons

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 ∈ ∆} Theorem 11.13

3.

Γ is a subgroup of Aut(Φ).

defn of Γ

4.

Γ ∩ W = {1}.

W acts simply transitively on bases of Φ by Theorem 11.15

5.

For all τ ∈ Aut(Φ), τ (∆) is a base of Φ.

∆ is a base of Φ = τ −1 (Φ) and Definition 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 16.9 Let E be a Euclidean space. Let Φ be a root system with a base ∆. Let W be the Weyl group of

NOTES TO HUMPHREYS

193

Φ. Let Γ = {σ ∈ Aut(Φ) : σ(∆) = ∆}. Then Γ is identified with the group A of Dynkin diagram automorphisms (if there is only one root length, it is also called a graph automorphism). Proof. Define φ : Γ → Aut(Dyn(∆)). Steps

Statements

Reasons

1.

φ(σ) sends the i-th vertex to the τ (i)-th vertex for some τ ∈ Sl .

There exists τ ∈ Sl such that σ(αi ) = ατ (i) .

2.

For all σ ∈ Γ, hσ(αi ), σ(αj )i = hαi , αj i for all 1 ≤ i, j ≤ l.

Lemma 9.17

3.

The multiplicity and possible arrow between the i-th and the j-th vertices is the same as those between the τ (i)-th and the τ (j)-th vertices.

Definition 13.1 and Lemma 9.30

4.

φ(σ) is a diagram automorphism.

Definition 13.7

 Example 16.10 If 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 16.11 Let E be an Euclidean space. Let Φ be a root system in E. Let Λ = {λ ∈ E : hλ, αi ∈ Z, ∀α ∈ Φ}. We call λ ∈ Λ a weight. Lemma 16.12 Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let Λ be the set of weights of Φ. Then (1) (Λ, +) is a subgroup of (E, +). (2) Φ ⊂ Λ. (3) λ ∈ Λ iff hλ, αi ∈ Z for all α ∈ ∆. Proof. (1)(2) follow from Lemma 9.10(2), Definition 9.12(R4), respectively. P For (3), suppose β ∈ Φ. hλ, βi = hβ ∨ , λ∨ i suppose β = kα α. α∈∆

194

ZHENGYAO WU

Steps

Statements

Reasons

1.

hλ, βi = hβ ∨ , λ∨ i

[Hum78, p.46, Exercise 2]

=h

kα α∨ , λ∨ i

P

α∈∆

[Hum78, p.54, Exercise 1]; and β ∨ = P kα α∨ by Definition 10.4(B2) α∈∆

=

P

kα hα∨ , λ∨ i

Lemma 9.10(2)

kα hλ, αi

[Hum78, p.46, Exercise 2]

α∈∆

=

P α∈∆

∈ Z.

λ ∈ Λ; and kα ∈ Z by Definition 10.4(B2)

 Definition 16.13 Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let Λ be the group of weights of Φ. The subgroup Λr of Λ generated by Φ is called the root lattice. Example 16.14 Let 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 16.15 Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let Λ be the group of weights of Φ and λ ∈ Λ. (1) We call λ ∈ Λ strongly dominant if hλ, αi > 0, for all α ∈ ∆. (2) We call λ dominant if hλ, αi ≥ 0 for all α ∈ ∆. Let Λ+ be the set of dominant weights. Lemma 16.16 Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let Λ be the group of weights 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 16.17 Let E be an Euclidean space. Let Φ be a root system in E with a base ∆ = {α1 , . . . , αl }. Suppose

NOTES TO HUMPHREYS

λ1 , . . . , λl ∈ E such that hλi , αj i = δij =

   

0, j 6= i

  

1, j = i

195

. We call {λi : 1 ≤ i ≤ l} fundamental

dominant weights relative to ∆. Lemma 16.18 Let E be an Euclidean space. Let Φ be a root system in E with a base ∆ = {α1 , . . . , αl }. Let Λ be the group of weights of Φ. Then Λ is a lattice. We call Λ the weight lattice of Φ relative to ∆. Proof. Suppose mi = hλ, αi i. Steps

Statements

Reasons

1.

Λ = SpanZ {λi : 1 ≤ i ≤ l}.

λi ∈ Λ for all 1 ≤ i ≤ l and below

1.1.

For all 1 ≤ j ≤ l, hλ −

l P

mi λi , αj i =

Lemma 9.10(2)

i=1

hλ, αj i − = mj −

l P

mi hλi , αj i

i=1 l P

mi δij = mj − mj = 0.

Definition 16.17

i=1

1.2.

For all 1 ≤ j ≤ l, (λ −

l P

mi λi , αj ) = 0.

Example 9.9(2)

i=1

1.3.

λ=

l P

(•, •) is non-degenerate and ∆ is a basis of E

mi λi .

i=1

2.

{λi : 1 ≤ i ≤ l} are linearly independent.

If c1 λ1 + · · · + cl λl = 0, then 0 = hc1 λ1 + · · · + cl λl , αj i = cj for all 1 ≤ j ≤ l.

3.

Λ is a free Z-module with basis {λi : 1 ≤ i ≤ l}.

steps 1,2

 Lemma 16.19 Let E be an Euclidean space. Let Φ be a root system in E with a base ∆ = {α1 , . . . , αl }. Let Λr be the root lattice of Φ. Let Λ be the weight lattice of Φ relative to ∆. Then Λ/Λr is a finite group, called the fundamental group of Φ. Proof. Let C be the Cartan matrix. Suppose αi =

l P j=1

mij λj .

196

ZHENGYAO WU

Steps

Statements

1.

For all 1 h

l P

Reasons ≤

mij λj , αk i =

j=1 l P

=

i, k l P

≤

l, hαi , αk i

=

Lemma 9.10(2)

mij hλj , αk i

j=1

mij δjk = mik .

Definition 16.17

j=1

2.

C −1 sends ∆ to {λj : 1 ≤ j ≤ l}.

C is the change of bases of E from {λj : 1 ≤ j ≤ l} to ∆.

3.

Denominators of C −1 are factors of det(C).

C −1 = C ∗ / det(C) where C ∗ is the matrix of cofactors of C

4.

Λ/Λr is a finite group with order dividing det(C).

λi + Λr has order dividing det(C) for all 1≤i≤l

 Example 16.20 Let E be an Euclidean space. Let Φ be a root system in E of type A1 with a base ∆ = {α1 }. Let Λr be the root lattice of Φ. Let Λ be the weight lattice of Φ relative to ∆. Then Λ/Λr ' Z/2Z. Proof. Steps

Statements

Reasons

1.

λ1 = 12 α1 .

Lemma 9.10(1)(2)

2.

Λ = Zλ1 .

Lemma 16.18

Λr = Zα1 .

Definition 16.13

Λ/Λr ' Z/2Z.

step 1

 Lemma 16.21 Let E be an Euclidean space. Let Φ be an irreducible root system in E with a base ∆. If there exists 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 entry unless for A1 . 

NOTES TO HUMPHREYS

197

Example 16.22 Let 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 . Steps

Statements

Reasons 



1.



C= 

2 −1

−1 2

. 

Theorem 14.2 



1 2 1 . =   3 1 2

2.

det(C) = 3 and C −1

3.

λ1 = 23 α1 + 13 α2 , λ2 = 31 α1 + 23 α2 .

step 2 of Lemma 16.19

4.

Λ/Λr ' Z/3Z.

Λr ( Λ and step 4 of Lemma 16.19

Linear algebra

α2 λ2 λ1 α1

 Example 16.23 Fundamental 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

Steps

Statements

Reasons

1.

The order of Λ/Λr divides the determinant of 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 16.24 Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let Λ be the weight lattice of Φ relative to ∆. For λ, µ ∈ Λ, we say that λ > µ if λ − µ is a sum of positive roots (it is possible that λ − µ 6∈ Φ+ ). Lemma 16.25 Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let W be the Weyl group of Φ. Let Λ be the weight lattice of Φ relative to ∆. For all λ ∈ Λ, there exists σ ∈ W such that 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(∆). Since τ ◦ σ −1 (σ(λ)) = τ (λ), by Lemma 11.21, σ(λ) = τ (λ).  Lemma 16.26 Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let W be the Weyl group 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. Steps

Statements

Reasons

1.

For all α, β ∈ ∆, hα, βi ∈ Z

Definition 9.12(R4)

NOTES TO HUMPHREYS

Steps

Statements

Reasons

2.

For all µ ∈ Λ, hµ, αi, hµ, βi ∈ Z

Definition 16.11

3.

hσα (µ), βi = hµ − hµ, αiα, βi

Example 9.9(2)

= hµ, βi − hµ, αihα, βi

Lemma 9.10(2)

∈ Z.

steps 1,2

4.

For all µ ∈ Λ and reflection σα for some α ∈ ∆, σα (µ) ∈ Λ.

Definition 16.11

5.

Suppose σ = σ1 ◦ · · · σn is a reduced decomposition for αi1 , . . . , αin ∈ ∆ and σ i = σ αi .

Theorem 11.13

6.

We induct on l(σ) = n.

7.

For n = 1, σα (λ) = λ − hλ, αiα ≤ λ for all α ∈ ∆.

hλ, αi ≥ 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 Steps

Statements

Reasons

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 < hλ, αi

λ ∈ Λ+ and Definition 16.15(1)

= hσ(λ), σ(α)i

Lemma 9.8 and Theorem 11.13

= hλ, σ(α)i

σ(λ) = λ

≤ 0, a contradiction.

λ ∈ Λ+ and step 3

199

200

ZHENGYAO WU

 Lemma 16.27 Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let W be the Weyl group of Φ. Let Λ be the weight lattice of Φ relative to ∆ and λ ∈ Λ. If λ ∈ Λ+ , then {µ ∈ Λ+ : µ ≤ λ} is finite. Proof. Steps

Statements

Reasons

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 : (λ, λ)} is compact.

(x, x) ≤

 Example 16.28 P Let δ = 21 α. It is possible that δ 6∈ Λr . For A1 , δ = 21 α1 . It is also possible that δ ∈ Λr . For A2 , δ =

α∈Φ+ 1 (α1 + 2

α2 + (α1 + α2 )) = α1 + α2 .

Lemma 16.29 Let E be an Euclidean space. Let Φ be a root system in E with a base ∆ = {αi : 1 ≤ i ≤ l}. Let P α. {λi : 1 ≤ i ≤ l} be the set of fundamental dominant weights relative to ∆. Let δ = 21 α∈Φ+

Then (1) δ =

l P

λi . (2) δ ∈ Λ+ .

i=1

Proof. Steps

Statements

1.

δ=

l P

hδ, αi iλi .

Reasons step 1 of Lemma 16.19

i=1

2.

δ − hδ, αi iαi = σi (δ) = δ − αi .

Example 9.9(2) and Corollary 11.5

3.

hδ, αi i = 1, i.e. (2) is true.

Definition 16.15(1)

4.

(1) is true.

steps 1, 3

NOTES TO HUMPHREYS

201

 Lemma 16.30 Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let W be the Weyl group of 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. Steps

Statements

Reasons

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

with equality iff (µ − ν, δ) = 0 iff µ = ν.

Otherwise (µ − ν, δ) > 0 by steps 1,7

8.

202

ZHENGYAO WU

 Definition 16.31 Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let W be the Weyl group of Φ. Let Λ be the weight lattice of Φ relative to ∆. A subset Π of Λ is saturated if for all λ ∈ Π, α ∈ Φ, λ − iα ∈ Π for all i between 0 and hλ, αi. Lemma 16.32 Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let W be the Weyl group of Φ. Let Λ be the weight lattice of Φ relative to ∆. Let Π be a saturated subset of Λ. Then σ(Π) ⊂ Π for all σ ∈ W . Proof. If λ ∈ Π, then σα (λ) = λ − hλ, αiα ∈ Π by Definition 16.31. The assertion follows since every σ ∈ W is a product of simple reflections by Theorem 11.13.  Definition 16.33 Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let W be the Weyl group of Φ. Let Λ be the weight lattice of Φ relative to ∆. Let Π be a saturated subset of Λ. We say 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 16.35 Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let W be the Weyl group of Φ. Let Λ be the weight lattice of Φ relative to ∆. Let Π be a saturated subset of Λ. If Π has a highest weight, then it is finite. Proof. Let λ be the highest weight of Π. Steps

Statements

Reasons

1.

For all µ ∈ Π, there exists σ ∈ W such that 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

NOTES TO HUMPHREYS

203

Steps

Statements

Reasons

5.

W is finite.

Lemma 9.15

6.

Card(Π) is finite.

≤ Card(W ) · Card({ν ∈ Λ+ : ν ≤ λ})

 Lemma 16.36 Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let W be the Weyl group of Φ. Let Λ be the weight lattice of Φ relative to ∆. Let Π be a saturated subset of Λ with highest weight λ. If µ ∈ Λ+ and µ ≤ λ, then µ ∈ Π. Proof. Suppose µ0 = µ +

P α∈∆

kα α, kα ≥ 0, kα ∈ Z and ht(µ0 − µ) > 0. We show that if µ0 ∈ Π,

then there exists β ∈ ∆ such that for some β ∈ ∆, kβ > 0 and µ0 − β ∈ Π. Since λ ≥ µ, we start from the given λ ∈ Π and obtain µ ∈ Π by induction. Steps

Statements

1.

(

P

kα α,

α∈∆

2.

P

(

P

Reasons kα α) > 0.

ht(µ0 − µ) > 0.

α∈∆

kα α, β) > 0 for some β ∈ ∆ with

α∈∆

kβ > 0. 3.

h

P

kα α, βi > 0.

Example 9.9 and

α∈∆

4.

If µ0 ∈ Π, then µ0 − β ∈ Π.

2 >0 (β, β)

Definition 16.31

 Corollary 16.37 Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let W be the Weyl group of Φ. Let Λ be the weight lattice of Φ relative to ∆. Let Π be a saturated subset of Λ with highest weight λ. Then Π = {µ ∈ Λ : ∃σ ∈ W , σ(µ) ∈ Λ+ , σ(µ) ≤ λ}. Proof.

204

ZHENGYAO WU

Steps

Statements

Reasons

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 16.38 Let E be an Euclidean space. Let Φ be a root system in E with a base ∆. Let W be the Weyl group of Φ. Let Λ be the weight lattice of Φ relative to ∆. Let Π be a saturated subset of Λ with highest weight λ. If µ ∈ Π, then (µ + δ, µ + δ) ≤ (λ + δ, λ + δ). The equality holds iff µ = λ. Proof. Steps

Statements

Reasons

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

NOTES TO HUMPHREYS

Steps

Statements

Reasons

π = 0, i.e. µ = λ.

λ + δ is strongly dominant and step 2

205

 References [CCN+ 85] J. H. Conway, R. T. Curtis, S. P. Norton, R. A. Parker, and R. A. Wilson, Atlas of finite groups, Oxford University Press, Eynsham, 1985, Maximal subgroups and ordinary characters for simple groups, With computational assistance from J. G. Thackray. MR 827219 →191 [Hum78] James E. Humphreys, Introduction to Lie algebras and representation theory, Graduate Texts in Mathematics, 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-Verlag London, Ltd., London, 2009. MR 2562037 →191 Department of Mathematics, Shantou University, 243 Daxue Road, Shantou, Guangdong, China 515063 E-mail address: [email protected]