We present Serre™s theorem in this chapter. At the end of the chapter we show

that each of the exceptional root systems exists. This then proves the existence

of the exceptional simple Lie algebras.

8.1 The Serre relations.

Recall that if ± and β are roots,

(β, ±)

β, ± := 2

(±, ±)

and the string of roots of the form β + j± is unbroken and extends from

β ’ r± to β + q± where r ’ q = β, ± .

In particular, if ±, β ∈ ∆ so that β ’ ± is not a root, the string is

β, β + ±, . . . , β + q±

where

q = ’ β, ± .

Thus

(ade± )’ β,± +1

eβ = 0,

137

138 CHAPTER 8. SERRE™S THEOREM.

for e± ∈ g± , eβ ∈ gβ but

(ad e± )k eβ = 0 for 0 ¤ k ¤ ’ β, ± ,

if e± = 0, eβ = 0. So if ∆ = {±1 , . . . , ± } we may choose

ei ∈ g±i , fi ∈ g’±i

so that

e1 , . . . , , e , f1 , . . . , f

generate the algebra and

[hi , hj ] = 0, 1 ¤ i, j, ¤ (8.1)

[ei , fi ] = hi (8.2)

[ei , fj ] = 0 i = j (8.3)

[hi , ej ] = ±j , ±i ej (8.4)

[hi , fj ] = ’ ±j , ±i fi (8.5)

(ad ei )’ ±j ,±i +1

ej = 0 i=j (8.6)

(ad fi )’ ±j ,±i +1

fj = 0 i = j. (8.7)

Serre™s theorem says that this is a presentation of a (semi-)simple Lie algebra.

In particular, the Cartan matrix gives a presentation of a simple Lie algebra,

showing that for every Dynkin diagram there exists a unique simple Lie algebra.

8.2 The ¬rst ¬ve relations.

Let f be the free Lie algebra on 3 generators, X1 , . . . , X , Y1 , . . . , Y , Z1 , . . . , Z .

If g is a semi-simple Lie algebra with generators and relations (8.1)“(8.7), we

have a unique homomorphism f ’ g where Xi ’ ei , Yi ’ fi , Zi ’ hi . We

want to consider an intermediate algebra, m, where we make use of all but the

last two sets of relations. So let I be the ideal in f generated by the elements

[Zi , Zj ], [Xi , Yj ] ’ δij Zi , [Zi , Xj ] ’ ±j , ±i Xj , [Zi , Yj ] + ±j , ±i Yj .

We let m := f /I and denote the image of Xi in m by xi etc.

We will ¬rst exhibit m as Lie subalgebra of the algebra of endomorphisms of

a vector space. This will allow us to conclude that the xi , yj and zk are linearly

independent and from this deduce the structure of m. We will then ¬nd that

there is a homomorphism of m onto our desired semi-simple Lie algebra sending

x ’ e, y ’ f, z ’ h.

So consider a vector space with basis v1 , . . . , v and let A be the tensor

algebra over this vector space. We drop the tensor product signs in the algebra,

so write

vi1 vi2 · · · vit := vi1 — · · · vit

8.2. THE FIRST FIVE RELATIONS. 139

for any ¬nite sequence of integers with values from 1 to . We make A into an

f module as follows: We let the Zi act as derivations of A, determined by its

actions on generators by

Zj vi = ’ ±i , ±j vj .

Zi 1 = 0,

So if we de¬ne

cij := ±i , ±j

we have

Zj (vi1 · · · vit ) = ’(ci1 j + · · · + cit j )(vi1 · · · vit ).

The action of the Zi is diagonal in this basis, so their actions commute. We let

the Yi act by left multiplication by vi . So

Yj vi1 · · · vit := vj vi1 · · · vit

and hence

[Zi , Yj ] = ’cji Yj = ’ ±j , ±i Yj

as desired. We now want to de¬ne the action of the Xi so that the relations

analogous to (8.2) and (8.3) hold. Since Zi 1 = 0 these relations will hold when

applied to the element 1 if we set

∀j

Xj 1 = 0

and

Xj vi = 0 ∀i, j.

Suppose we de¬ne

Xj (vp vq ) = ’δjp cqj vq .

Then

Zi Xj (vp vq ) = δjp cqj cqi vq = ’cqi Xj (vp vq )

while

Xj Zi (vp vq ) = δjp cqj (cpi + cqi )vq = ’(cpi + cqi )Xj (vj vq ).

Thus

[Zi , Xj ](vp vq ) = cji Xj (vp vq )

as desired.

In general, de¬ne

Xj (vp1 · · · vpt ) := vp1 (Xj (vp2 · · · vpt )) ’ δp1 j (cp2 j + · · · + cpt j )(vp2 · · · vpt ) (8.8)

for t ≥ 2. We claim that

Zi Xj (vp1 · · · vpt ) = ’(cp1 i + · · · + cpt i ’ cji )Xj (vp1 · · · vpt ).

Indeed, we have veri¬ed this for the case t = 2. By induction, we may assume

that Xj (vp2 · · · vpt ) is an eigenvector of Zi with eigenvalue cp2 i + · · · + cpt i ’

140 CHAPTER 8. SERRE™S THEOREM.

cji . Multiplying this on the left by vp1 produces the ¬rst term on the right of

(8.8). On the other hand, this multiplication produces an eigenvector of Zi with

eigenvalue cp1 i + · · · + cpt i ’ cji . As for the second term on the right of (8.8), if

j = p1 it does not appear. If j = p1 then cp1 i +· · ·+cpt i ’cji = cp2 i +· · ·+cpt i . So

in either case, the right hand side of (8.8) is an eigenvector of Zi with eigenvalue

cp1 i + · · · + cpt i ’ cji . But then

[Zi , Xj ] = ±j , ±i Xj

as desired. We have de¬ned an action of f on A whose kernel contains I, hence

descends to an action of m on A.

Let φ : m ’ End A denote this action. Suppose that z := a1 z1 + · · · + a z

for some complex numbers a1 , . . . , a and that φ(z) = 0. The operator φ(z) has

eigenvalues

’ aj cij

when acting on the subspace V of A. All of these must be zero. But the Cartan

matrix is non-singular. Hence all the ai = 0. This shows that the space spanned

by the zi is in fact -dimensional and spans an -dimensional abelian subalgebra

of m. Call this subalgebra z.

Now consider the 3 -dimensional subspace of f spanned by the Xi , Yi and

Zi , i = 1, . . . , . We wish to show that it projects onto a 3 dimensional subspace

of m under the natural passage to the quotient f ’ m = f /i. The image of this

subspace is spanned by xi , yi and zi . Since φ(xi ) = 0 and φ(yi ) = 0 we know

that xi = 0 and yi = 0. Suppose we had a linear relation of the form

ai xi + bi yi + z = 0.