By Skoruppa N.-P.

Those are the notes of a path on Lie algebras which I gave on the college of Bordeaux in spring 1997. The direction used to be a so-called "Cours PostDEA", and as such needed to be held inside of 12 hours. much more hard, no earlier wisdom approximately Lie algebras may be assumed. however, I had the objective to arrive as top of the direction the nature formulation for Kac-Moody algebras, and, even as, to offer entire proofs so far as attainable.

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In fact, each σβ (β ∈ ∆) permutes the set R+ \{β} of all positive roots with ∆ omitted (see step 7 of the proof of the character formula). Hence σβ (ρ) = ρ − β, and, since by definition of σβ we have σβ (ρ) = ρ − ρ(hβ )β, we deduce ρ(hβ ) = 1. Theorem. (Character Formula) Let Λ ∈ H ∗ such that Λ(hα ) ∈ Z≥0 for all α ∈ ∆. Then the character of the simple highest weight module L(Λ) is given by the formula det(w) e(w(Λ + ρ)) chL(Λ) = w∈W e(ρ) α>0 (1 − e(−α))dim Lα Example. Let L = L((2)) = sl2 . Write q = e(− 21 α), α denoting the single element of ∆.

5. CHARACTER FORMULA AND CONSEQUENCES 43 This clearly defines an element of E. As a first application we calculate the character of the Verma module M (Λ). Let R+ be the set of positive roots of L, and, for α ∈ R+ let y(α)1 , . . , y(α)d(α) be a basis of L−α . e. the ideal generated by N+ and h − Λ(h) (h ∈ H). Let α1 , α2 , . . be an enumeration of R+ . Using the PBW theorem we then see that the images of n n n n 2,d(α2 ) 1,d(α1 ) ··· y(α2 )1 2,1 . . y(α2 )d(α y(α1 )1 1,1 · · · y(α1 )d(α 2) 1) under the canonical map U (L) → U (L)/J form a basis of M (L).

Indeed, applying the diagonalization lemma to the root space decomposition of L with respect to H, one finds I ∩ Lα = 0 for some root α of L. Either Lα = H, or else Lα is one dimensional, hence Lα ⊂ I. But in the latter case the one dimensional subspace [Lα , L−α ] ⊂ H is contained in I. It is thus reasonable to remove the ideals I with I ∩ H = 0 from L(A) by setting Definition. L(A) := L(A)/I(A). Note that the kernel of the canonical map π : L(A) → L(A) intersects H trivially. Thus we shall henceforth identify H with its image under π.