Lecture 10 - Representation Theory III: Theory of Weights

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1 Lecture 10 - Representation Theory III: Theory of Weights February 18, Terminology One assumes a base = {α i } i has been chosen. Then a weight Λ with non-negative integral Dynkin coefficients Λ i = Λ, α i is called a dominant weight. If all coefficients are all positive, it is called strongly dominant. The element δ = 1 α (1) 2 recurs frequently enough that we give it a name, the Weyl vector. Its weight is (1,..., 1) and lies in the fundamental Weyl chamber. If Λ = (Λ 1,..., Λ n ) is a weight, it components Λ i = Λ, α i = 2 (Λ, α i) (α i, α i ) (2) are called its Dynkin coefficients. The vectors λ i = (0,..., 1,..., 0) (3) (a 1 in the i th position) are then dual to the vectors 2α i /(α i, α i ), in the sense that ( ) Λ i 2α j, = δj i (4) (α j, α j ) Weights are Z-linear combinations of the λ i, so we can produce nice visualizations of representations, at least in low-dimensional cases. 1

2 2 Automorphisms Lie algebra automorphisms produce automorphisms of their representations. Recall that the automorphism group of a semisimple algebra g has the form Out Inn, inner times outer automorphisms. The inner automorphisms consist of Weyl transformations (or re-selection of the base) and diagonal transformations, and the outer automorphisms are permutations of a fixed base (and correspond to the Dynkin diagram automorphisms). The diagonal automorphisms have no effect on the weight schemes. That is they don t affect what vectors go to which, they just alter coefficients. Any Weyl transformation re-orders the weights in a given representation. Any weights conjugate to a highest weight under some Weyl transformation is called an extreme weight. Outer automorphism permute representations. 3 Partial Order and Height Given two weights λ, µ, we say that λ > µ if λ µ is not zero and is a non-negative sum of positive weights. That is, λ µ is nonzero, but in the positive cone of. This is not the same as saying λ µ is dominant, which would mean it is a non-negative positive sum of positive fundamental weights. If λ µ is neither in the positive cone nor in the negative cone of, the weights are incomeasurable. Given a representation, it is possible to define the height (or length) of a weight vector. If v 0 V Λ is a highest weight vector and y i1... y im.v 0 = 0 (not necessarily ordered) where m is minimal, we say v 0 has height 1 2 (m 1). The height of any other weight vector v is m n where v = y j1... y jn.v 0 and n is minimal (again, the monomial is not necessarily ordered). Given a representation, heights are not necessarily integral, but are uniquely defined, which is due to the fact that is a basis of h. In the case of the adjoint representation, this corresponds to the previously defined height of a root. 4 Casimir Operator Let = {α i } i be the simple roots of some semi-simple algebra with corresponding {h i, x i, y i }. Define coefficients κ ij = κ(h i, h j ), and define the matrix (κ ij ) to be the inverse of (κ ij ). ( 1 2 Recalling that κ(x i, y j ) = (α i,α i)) δij and κ(x i, x j ) = κ(y i, y j ) = 0, we can write the 2

3 Casimir element c = κ ij h i h j + (x α y α + y α x α ). (5) We know that c = const on V Λ. To compute the value of this constant, let v 0 V Λ be the highest weight. Since x α.v 0 = 0 we have c.v 0 = = κ ij h i h j.v 0 + (x α y α + y α x α ) κ ij (Λ, h i ) (Λ, h j ) v 0 + x α y α.v 0 ([x α, y α ] + y α x α ).v 0 = (Λ, Λ) v 0 + 2h α.v 0 (6) = (Λ, Λ) v 0 + = (Λ, Λ) v (Λ, α) v 0 ( ) = (Λ, Λ) + (Λ, α) v 0 so that c V Λ = (Λ, Λ) + 2 (Λ, δ) (7) 5 The Casimir on sl(2, C) Now on sl 2 we have the adjoint representation, where {h, x, y} is the basis, and ad h x = α(h)x, where α is the root and α(h) = 2. With κ = (8) we have (h, h) = 8, so that with h = is then 2α (α,α) 4 we have 8 = (α,α) so = 1 2. The Casimir c = 1 8 h.h. + 1 (x.y. + y.x.) (9) 4 3

4 Let V Λ be the irreducible weight space of highest weight Λ, which is just a nonnegative integer. Then Λ is characterized by a single Dynkin coefficient Λ 1 = 2(Λ,α) (α,α) Z, so Λ = 1 2 Λ 1α. Then The Weyl symbol δ is 1 2α, so that (Λ, Λ) = 1 4 Λ2 1 = 1 8 Λ2 1. (10) c = (Λ, Λ + α) = 1 8 Λ Λ 1 = 1 8 Λ 1 (Λ 1 + 2). (11) 6 Freudenthal s Dimension Formula Consider the weight space of weight µ in V Λ µ, which we denote V Λ µ. If the dimension of the weight space is N µ, and we trace the Casimir over this weight space we obtain T r c V Λ µ = N µ (Λ, Λ + 2δ) (12) On the other hand we can attempt a direct calculation of the action of c on V Λ µ. With c = κ ij h i h j + (x α y α + y α x α ) (13) The first part is easy: κ ij h i h j.v µ = κ ij h i (µ, h j ).v µ = κ ij (µ, h i ) (µ, h j ).v µ (14) = (µ, µ) v µ (15) Thus T r κ ij h i h j V Λ µ We have to compute the value of the second term. = (µ, µ) N µ (16) Now any vector v µ V Λ µ is in the middle of the α-weight string of highest weight t. Restricting to this sl 2 representation, the Casimir is c = 1 8 h αh α (x αy α + y α x α ) 1 8 t (t + 2) v m = 1 8 h αh α.v m (x αy α + y α x α ).v m = (m, α) h α.v m (x αy α + y α x α ).v m = 1 ( ) 2 2 (m, α) 2 v m (x αy α + y α x α ).v m (17) 4

5 Then (x α y α + y α x α ).v µ = 1 2 t (t + 2) v µ 2 (µ, α)2 2 v µ. (18) It will be convenient to re-express this in a slightly different way. There is a non-negative integer k so that µ + kα is the highest weight. We have ( ) (µ, α) t v µ+kα = h α.v µ+kα = 2 + 2k v µ+kα (19) so that (x α y α + y α x α ).v µ = = 2 ((µ, α) + k) ((µ, α) + (k + 1)) 2 (µ, α)2 2 v µ 2 v µ (20) 2(2k + 1) (µ, α) + 2k(k + 1) v µ On an individual vector we have (x α y α + y α x α ).v µ = ((2k + 1) (µ, α) + k(k + 1)) v µ (21) so that T r (x α y α + y α x α ) = dim Λ µ(k) ((2k + 1) (µ, α) + k(k + 1)) (22) where dim(µ, k) is the dimension of the subspace of Vµ Λ sl 2 -representations of highest weight µ + kα. Thus spanned by those vectors within dim(µ, k) = N µ+kα N µ+(k+1)α (23) 5

6 so that T r (x α y α + y α x α ) ( ) Nµ+kα N µ+(k+1)α ((2k + 1) (µ, α) + k(k + 1)) = k=0 = N µ+kα ((2k + 1) (µ, α) + k(k + 1)) k=0 N µ+(k+1)α ((2k + 1) (µ, α) + k(k + 1)) k=0 = N µ+kα ((2k + 1) (µ, α) + k(k + 1)) k=0 N µ+kα ((2k 1) (µ, α) + (k 1)k) k=1 = N µ (µ, α) + 2 N µ+kα ((µ, α) + k) k=1 = 2N µ (µ, δ) + 2 N µ+kα ((µ, α) + k) k=1 (24) Bringing it together, we have N µ (Λ, Λ + 2δ) = N µ (µ, µ) + 2N µ (µ, δ) + 2 N µ+kα ((µ, α) + k) k=1 N µ ((Λ, Λ + 2δ) (µ, µ + 2δ)) = 2 N µ+kα ((µ, α) + k) N µ (Λ + µ + 2δ, Λ µ) = 2 k=1 N µ+kα ((µ, α) + k) k=1 (25) We arrive at Freudenthal s formula: N µ = 2 k=1 N µ+kα ((µ, α) + k) (Λ + µ + 2δ, Λ µ) (26) 6