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 , ] ] ι
(
[ Eα, [ Eαi , ] ]
))
.
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.
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