2.1. THE SERRE RELATIONS 11 submodule a. Since all off-diagonal matrix units are in the submodule, we have that ad(Ei,i+1)Ei+1,i ∈ a for all i. We have proved that any non-zero submodule must coincide with g. Assertion 3 (1) The algebra U is a U-module via ad(x)(a) = xa − ax (x = ei,fi,hi). (2) The space L = ∑ n−1 i=1 Chi + ∑ i=j CLij is stable under ad(L). In particular, it is a U-submodule. (3) Define a linear map ι : g → L by Eij → Lij (i = j), Eii − Ei+1,i+1 → hi. Then ι is an isomorphism of U-modules. Let a, b, c be elements of U. Then the Jacobi identity [a, [b, c]] + [b, [c, a]] + [c, [a, b]] = 0 holds and thus ad ([a, b]) = [ad(a), ad(b)]. In particular, the defining relations of U imply that ad(ei),ad(fi) and ad(hi) all satisfy the same defining relations. Hence (1) is proved. The Jacobi identity also implies the following Leibniz rule. ad(x)[a, b] = [ad(x)a, b] + [a, ad(x)b] (x, a, b ∈ U) In order to prove (2) and (3), we establish explicit formulas for [x, Ljk] (x = ei,fi,hi) in the following steps. (step 1) [ hi, Ljk ] = (δij − δi+1,j − δik + δi+1,k)Ljk (j = k). Exercise 2.3. Prove this formula. (step 2) [ei, Lik] = 0 (k = i). For k = i ± 1,i + 2, this is a direct consequence of the defining relations. For other values of k, it is proved by induction. (step 3) [ ei, Lj,i+1 ] = 0 (j = i + 1). The proof is the same as (step 2). (step 4) The following hold. In particular, we have [ei, Ljk] ∈ L for any j, k. (i) [ei, Ljk] = 0 (j = i + 1, k = i, j = k). (ii) [ei, Li+1,k] = Lik (k = i, i + 1). (iii) [ei, Lji] = −Lj,i+1 (j = i + 1,i). (iv) [ei, Li+1,i] = hi. We start with (i). It is obvious for the cases j k, i j k and j k i. The cases j = i k and j i = k − 1 k are nothing but (step 2) and (step 3). In the remaining case j i k − 1, we know that [ei, Ljk] = 0 by the following computation. [ei, [Lj,i+1, Li+1,k]] = [Lj,i+1, [ei, Li+1,k]] = [Lj,i+1, Lik] = [[Lji, ei], Lik] = [[Lji, Lik], ei] = −[ei, Ljk] Next we prove (ii). It is obvious for the case k ≥ i + 2. If k ≤ i − 1, it follows from [ei, [fi, Lik]] = [hi, Lik] = Lik. The proof of (iii) is the same as (ii), and (iv) is obvious. Hence we have proved the assertion.

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