10 2. THE SERRE RELATIONS Relations: eifj fjei =δijhi, hihj = hjhi, hiej ejhi = ⎪2ej (i = j) −ej (i j = ±1) 0 (otherwise) , hifj fjhi = ⎪−2fj (i = j) fj (i j = ±1) 0 (otherwise) , ei 2 ej 2eiejei + ejei 2 = 0 (i j = ±1), eiej = ejei (otherwise), f 2 i fj 2fifjfi + fjfi2 = 0 (i j = ±1), fifj = fjfi (otherwise). Proof. We define elements Lij U (i = j) by induction on |j i|: Li,i+1 = ei, Lij = [Lik, Lkj] (i k j), Li+1,i = fi, Lij = [Lik, Lkj] (i k j). For a, b U, [a, b] is ab ba by definition. Assertion 1 The Lij are well-defined. We prove this by induction on N = |j −i|. Since the proof for i j is the same as the proof for i j, we prove it only for i j. Assume that it is already proved up to N 1. We shall show for 1 l N 2 that [Li,i+l+1, Li+l+1,i+N] = [Li,i+l, Li+l,i+N] . (2.1) Since we have Li,i+l+1 = [Li,i+l, ei+l] by the induction hypothesis, the left hand side of (2.1) equals [[Li,i+l, ei+l] , Li+l+1,i+N], which is the same as [ei+l, [Li,i+l, Li+l+1,i+N]] + [Li,i+l, [ei+l, Li+l+1,i+N]] . We now notice that Li,i+l and Li+l+1,i+N are non-commutative polynomials in ei,...,ei+l−1 and ei+l+1,...,ei+N−1 respectively. Hence the first term is 0. The second term equals the right hand side of (2.1) by the induction hypothesis. Assertion 2 The Lie algebra g is an irreducible U-module via the following action: ei ad(Ei,i+1), fi ad(Ei+1,i), hi ad(Eii Ei+1,i+1), for 1 i n 1. (Recall that the Eij are matrix units.) To prove that this defines an action of U, we check that ad(Ei,i+1), ad(Ei+1,i) and ad(Eii −Ei+1,i+1), for 1 i n−1, satisfy the defining relations of U. By the Jacobi identity, it is enough to see that Ei,i+1,Ei+1,i and Eii Ei+1,i+1 satisfy the relations, but this is easily checked by direct computation. We remark here that we also have seen that Lij ad(Eij). Next we show that g is irreducible as a U-module. Let a be a non-zero U- submodule of g. It is easy to see that a is not contained in h and that a has a simultaneous eigenspace decomposition with respect to ad(h). So a contains a matrix unit Ekl (k = l). We shall show that a contains all Eij (i = j). If i = k and j = l, then Eij = ad(Eik)Ekl a. If i = k and j = l, then Eij = −ad(Elj)Ekl a. If i = k and j = l, we consider ad(Eik)ad(Elj)Ekl and we have that Eij is in the
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