2.5 Semisimple Lie Algebras
2.5.3 Geometry of Root Systems
thenα could not be a root, by the argument just given, which is a contradiction.
Thus we conclude that Mα=Chα+Ceα+Cfα. Hence dimgα=1.
(4): The notion ofα root string throughβ from Section 2.4.2 carries over verba- tim, as does Lemma 2.4.3. Hence the argument in Corollary 2.4.5 applies.
(5): This follows from the same argument as Corollary 2.4.4. ut
2.5 Semisimple Lie Algebras 119
By Theorem 2.5.20 (3) we have
tr(ad(Hi)ad(Hj)) =
∑
α∈Φ
hα,Hiihα,Hji.
This is an integer by (2.48), soB(a,b)∈Q. Likewise, B(a,a) =tr(ad(a)2) =
∑
α∈Φ
hα,ai2,
and we have just proved thathα,ai ∈Q. Ifa6=0 then there existsα∈Φ such that hα,ai 6=0, because the center ofgis trivial. HenceB(a,a)>0. ut Corollary 2.5.22.LethRbe the real span of{hα : α∈Φ}and leth∗
Rbe the real span of the roots. Then the Killing form is real-valued and positive definite onhR. Furthermore,hR∼=h∗
Runder the Killing-form duality.
Proof. This follows immediately from (2.49) and Lemma 2.5.21. ut LetE=h∗
Rwith the bilinear form(·,·)defined by the dual of the Killing form.
By Corollary 2.5.22,E is anl-dimensional real Euclidean vector space. We have Φ⊂E, and the coroots are related to the roots by
αˇ = 2
(α,α)α forα∈Φ
by (2.49). Let ˇΦ ={αˇ : α ∈Φ} be the set of coroots. Then(β,αˇ)∈Z for all α,β∈Φ by (2.48).
An elementh∈Eis calledregularif(α,h)6=0 for allα∈Φ. Since the set [
α∈Φ
{h∈E :(α,h) =0}
is a finite union of hyperplanes, regular elements exist. Fix a regular elementh0and define
Φ+={α∈Φ : (α,h0)>0}.
ThenΦ=Φ+∪(−Φ+). We call the elements ofΦ+thepositive roots. A positive rootαis calledindecomposableif there do not existβ,γ∈Φ+such thatα=β+γ (these definitions depend on the choice ofh0, of course).
Proposition 2.5.23.Let∆ be the set of indecomposable positive roots.
1.∆ is a basis for the vector space E.
2. Every positive root is a linear combination of the elements of∆with nonnegative integer coefficients.
3. Ifβ∈Φ+\∆ then there existsα∈∆ such thatβ−α∈Φ+. 4. Ifα,β ∈∆ then theα root string throughβis
β,β+α, . . . ,β+pα, where p=−(β,αˇ)≥0. (2.50)
Proof. The key to the proof is the following property of root systems:
(?) Ifα,β∈Φ and(α,β)>0 thenβ−α∈Φ .
This property holds by Theorem 2.5.20 (4):β−(β,α)αˇ ∈Φ and(β,αˇ)≥1, since (α,β)>0; henceβ−α∈Φ. It follows from(?)that
(α,β)≤0 for allα,β ∈∆withα6=β . (2.51) Indeed, if(α,β)>0 then (?) would imply thatβ−α ∈Φ. Ifβ−α ∈Φ+then α=β+ (β−α), contradicting the indecomposability ofα. Likewise,α−β∈Φ+ would contradict the indecomposability ofβ. We now use these results to prove the assertions of the proposition.
(1): Any real linear relation among the elements of∆ can be written as
∑
α∈∆1
cαα=
∑
β∈∆2
dββ, (2.52)
where∆1and∆2are disjoint subsets of∆ and the coefficientscα anddβ are non- negative. Denote the sum in (2.52) byγ. Then by (2.51) we have
0≤(γ,γ) =
∑
α∈∆1
∑
β∈∆2
cαdβ(α,β)≤0. Henceγ=0, and so we have
0= (γ,h0) =
∑
α∈∆1
cα(α,h0) =
∑
β∈∆2
dβ(β,h0).
Since(α,h0)>0 and(β,h0)>0, it follows thatcα=dβ =0 for allα,β.
(2): The setM={(α,h0):α∈Φ+}of positive real numbers is finite and totally ordered. If m0is the smallest number inM, then anyα ∈Φ+with(α,h0) =m0
is indecomposable; henceα ∈∆. Givenβ ∈Φ+\∆, thenm= (β,h0)>m0and β =γ+δ for someγ,δ ∈Φ+. Since(γ,h0)<mand(δ,h0)<m, we may assume by induction onmthatγ andδ are nonnegative integral combinations of elements of∆, and hence so isβ.
(3): Letβ∈Φ+\∆. There must existα∈∆such that(α,β)>0, since otherwise the set∆∪ {β}would be linearly independent by the argument at the beginning of the proof. This is impossible, since∆ is a basis forE by (1) and (2). Thusγ = β−α ∈Φ by (?). Sinceβ 6=α, there is some δ ∈∆ that occurs with positive coefficient inγ. Henceγ∈Φ+.
(4): Sinceβ−αcannot be a root, theα-string throughβ begins atβ. Now apply
Theorem 2.5.20 (4). ut
We call the elements of∆ thesimple roots(relative to the choice ofΦ+). Fix an enumeration α1, . . . ,αl of ∆ and write Ei=eαi, Fi = fαi, and Hi=hαi
for the elements of the TDS triple associated withαi. Define theCartan integers
2.5 Semisimple Lie Algebras 121
Ci j=hαj,Hiiand thel×l Cartan matrix C= [Ci j]as in Section 2.4.3. Note that Cii=2 andCi j≤0 fori6= j.
Theorem 2.5.24.The simple root vectors {E1, . . . ,El,F1, . . . ,Fl} generateg. They satisfy the relations[Ei,Fj] =0for i6=j and[Hi,Hj] =0, where Hi= [Ei,Fi]. They also satisfy the following relations determined by the Cartan matrix:
[Hi,Ej] =Ci jEj, [Hi,Fj] =−Ci jFj; (2.53) ad(Ei)−Ci j+1Ej=0 for i6= j ; (2.54) ad(Fi)−Ci j+1Fj=0 for i6= j . (2.55) Proof. Letg0be the Lie subalgebra generated by theEiandFj. Since{H1, . . . ,Hl} is a basis forh, we haveh⊂g0. We show thatgβ ∈g0for allβ ∈Φ+by induction on the height ofβ, exactly as in the proof of Theorem 2.4.11. The same argument withβ replaced by−β shows thatg−β⊂g0. Henceg0=g.
The commutation relations in the theorem follow from the definition of the Car-
tan integers and Proposition 2.5.23 (4). ut
The proof of Theorem 2.5.24 also gives the following generalization of Theorem 2.4.11:
Corollary 2.5.25.Definen+=∑α∈Φ+gα andn−=∑α∈Φ+g−α. Thenn+andn− are Lie subalgebras of g that are invariant under adh, and g=n−+h+n+. Furthermore,n+is generated by{E1, . . . ,El}andn−is generated by{F1, . . . ,Fl}. Remark 2.5.26.We define theheight of a root (relative to the system of positive roots) just as for the Lie algebras of the classical groups: ht ∑iciαi=∑ici (the coefficientsciare integers all of the same sign). Then
n−=
∑
ht(α)<0
gα and n+=
∑
ht(α)>0
gα.
Letb=h+n+. Thenbis a maximal solvable subalgebra ofgthat we call aBorel subalgebra.
We call the set ∆ of simple roots decomposable if it can be partitioned into nonempty disjoint subsets∆1∪∆2, with∆1⊥∆2 relative to the inner product on E. Otherwise, we call∆ indecomposable.
Theorem 2.5.27.The semisimple Lie algebragis simple if and only if∆ is inde- composable.
Proof. Assume that∆=∆1∪∆2is decomposable. Letα∈∆1 andβ ∈∆2. Then p=0 in (2.50), since(α,β) =0. Henceβ+α is not a root, and we already know thatβ−α is not a root. Thus
[g±α,g±β] =0 for allα∈∆1andβ ∈∆2. (2.56)
Letmbe the Lie subalgebra ofggenerated by the root spacesg±αwithα ranging over∆1. It is clear from (2.56) and Theorem 2.5.24 thatmis a proper ideal ing.
Hencegis not simple.
Conversely, supposegis not simple. Theng=g1⊕ ··· ⊕gr, where eachgi is a simple Lie algebra. The Cartan subalgebrahmust decompose ash=h1⊕ ··· ⊕hr, and by maximality ofhwe see thathiis a Cartan subalgebra ingi. It is clear from the definition of the Killing form that the roots ofgiare orthogonal to the roots of gj fori6=j. Since∆ is a basis forh∗, it must contain a basis for eachh∗i. Hence∆
is decomposable. ut