12 2. THE SERRE RELATIONS

Exercise 2.4. Find formulas for [fi, Ljk] and prove them by using the auto-

morphism of U defined by ei → fi,hi → −hi,fi → ei.

(step 5) The space L is stable under ad(hi) and ad(Lij). In other words, L is a

Lie algebra.

By (step 1), L is stable under ad(hi). We shall show by induction on |j − i|

that it is stable under

ad(Lij). If i j for example, we use (step 4) to conclude

[Lij, L] = [[ei, Li+1,j] , L] = [ei, [Li+1,j, L]] + [Li+1,j, [ei, L]] ⊂ L.

The argument for the case i j is similar.

(step 6) We have an isomorphism of U-modules, ι : g L.

In fact, the above formulas imply that ι is a surjective homomorphism of U-

modules. Since g is an irreducible U-module by Assertion 2, it is enough to show

that L = 0. But if L = 0, we are forced to have ei = 0, fi = 0, hi = 0 for all i,

which contradicts the existence of the non-zero irreducible module of Assertion 2.

Assertion 4 We have [ι(X), ι(Y )] = ι ([X, Y ]) (X, Y ∈ g).

If one of X, Y is a diagonal matrix, the assertion follows from [hi, hj] = 0

and (step 1). To treat the other cases, we consider the root space decomposition

g = h ⊕ (⊕i=jCEij) and the root system Φ = {xi − xj}i=j. Using the simple roots

αi = xi − xi+1, we can describe Φ as follows.

Φ = ±(αi + · · · + αj−1)

ij

For the root α = αi + · · · + αj−1 (resp. α = −(αi + · · · + αj−1)), we denote its

simultaneous eigenvector Eij (resp. Eji) by Eα.

For α, β ∈ Φ, we set

vα,β = [Lα, Lβ] − ι ([Eα, Eβ]) .

By Assertion 3, we have vα,β ∈ L.

In the following, assume by way of contradiction that the set

A = { γ ∈

n−1

⊕

i=1

Zαi | γ = α + β, vα,β = 0 (∃ α, β ∈ Φ)}

is non-empty. Introduce a partial order on A as follows.

γ1 γ2 ⇔ γ1 − γ2 ∈

n−1

⊕

i=1

Z≥0 αi

We take a maximal element γ with respect to the order and suppose that γ = α+β

and vα,β = 0.

By Assertion 3(3), ι is an isomorphism of U-modules. Thus

eivα,β =

(

[[Lαi , Lα] , Lβ] − ι

(

[[Eαi , Eα] , Eβ]

))

+

(

[ Lα, [ Lαi , Lβ ] ] − ι

(

[ Eα, [ Eαi , Eβ ] ]

))

.

Note that if α = −αi and β = αi, then vα,β = 0, which is a contradiction. If

α = −αi and β = αi, then we again have vα,β = 0 by the following computation

for j = i or k = i + 1.

[fi, Ljk] − ι([Ei+1,i, Ejk]) = [fi, Ljk] − δijLi+1,k + δi+1,kLji = 0.

We similarly have vα,β = 0 for the case β = −αi. Hence both α = −αi and β = −αi

are impossible.