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 = ( [[Lα i , Lα] , Lβ] ι ( [[Eα i , Eα] , Eβ] )) + ( [ Lα, [ i , ] ] ι ( [ Eα, [ i , ] ] )) . 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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