Lie Algebras and Representation Theory Homework 7

Transcription

1 Lie Algebras and Representation Theory Homework 7 Debbie Matthews

2 Problem 10.5 If σ W can be written as a product of t simple reflections, prove that t has the same parity as l(σ). Let = {α 1, α 2,..., α l } be a base for some root system Φ. We write s i = σ αi W. Recall that each simple reflection s i permutes the positive roots except for α i. (Lemma B) This means we can consider each simple reflection as a signed permutation of the simple roots, where s i sends α i to it s negative and permutes all other simple roots without changing their sign. Now suppose σ = s a1 s a2 s at = s b1 s b2 s bp. Then s bp 1 s b2 s b1 s a1 s a2 s at = s bp. On the right side, we have only one negative object in the associated signed permutation. So that must also be the case on the left side. Then t + p 1 is odd by the above realization. Therefore, t + p is even. Hence any expression of σ as a product of simple reflections will have the same parity. In particular, that parity agrees with the length. 1

3 Problem 10.6 Define a function sn : W {±1} by sn(σ) = ( 1) l(σ). Prove that sn is a homomorphism. Given the previous problem, this is straightforward. For two elements σ, γ W, we know sn(σ)sn(γ) = ( 1) l(σ)+l(γ). At the same time, since σ can be written as a product of l(σ) simple reflections and γ can be written as a product of l(gamma) simple reflections, then we know σγ can be written as a product of l(σ) + l(γ) simple reflections. By the previous problem, l(σγ) will have the same parity as l(σ) + l(γ). Therefore, ( 1) (l(σ)+l(γ) = ( 1) l(σγ) = sn(σγ). Hence, sn(σ)sn(γ) = sn(σγ) and sn is a group homomorphism. 2

4 Problem 10.7 Prove that the intersection of positive open half-spaces associated with any basis γ 1,..., γ l of E is nonvoid. Let γ 1,..., γ t be a basis for E. Let δ i be the projection of γ i onto the orthogonal complement of the span of all basis vectors except γ i. Denote the span of all basis vectors except γ i by E i = span{γ 1,..., γ i,..., γ t }. Hence δ i E i. Consider γ = r i δ i for r i > 0. For each k, we have (γ k, γ) = (γ k, r i δ i ). But for i k we know (γ k, δ i ) = 0 since γ k E i and δ i E i. Thus (γ k, γ) = r k γ k > 0 and so the intersection fo the open half-spaces associated to the basis {γ 1,..., γ t } is nonempty. 3

5 Problem 10.8 Let be a base of Φ, α β simple roots, Φ αβ the rank 2 root system in E αβ = Rα + Rβ. The Weyl group W αβ of Φ αβ is generated by the restrictions τ α, τ β to E αβ of σ α, σ β, and W αβ may be viewed as a subgroup of W. Prove that the length of an element of W αβ (relative to τ α, τ β ) coincides with the length of the corresponding element in W. Since we can view W αβ as a subgroup of W is cleat the length of some σ in W αβ is greater than or equal to its length in W since any product of simple reflections that can be written in W αβ can also be written in W. So we need to prove that the additional element in W do not allow a further reduction of σ beyond what can be achieved in W αβ. Denote α = α 1 and β = α 2. Let α 3, α 4,..., α l be the remaining simple roots in. Let s i denote the simple reflection associated with α i. Let t 1 = τ α and t 2 = τ β. Suppose σ W αβ such that σ = s a1 s ap = s b1 s bq where p is minimal in W αβ and q is minimal in W. That is, 1 a i 2 and 1 b i l. From a corollary, we know that since s a1 s ap and s b1 s bq are reduced expressions that σ(α ap ), σ(α bq ) Φ. However, σ can be generated by s 1 and s 2. Furthermore, each s i permutes the positive simple roots except α i. Finally, since the restrictions t 1 and t 2 permute roots in E αβ Φ, we know that an arbitrary product composed of s 1 and s 2 must permute the positve roots outside of E αβ. Therefore, b q is 1 or 2. But then, we can consider s a1 s ap s bq W αβ. Since s b1 s bq 1 must be a reduced expression in W, we conclude similarly that b q 1 is 1 or 2. We can continue and ultimately conclude, each b i is 1 or 2 and therefore, the lengths must agree. 4

6 Problem 10.9 Prove that there is a unique element σ in W sending Φ + to Φ (relative to ). Prove that any reduced expression for σ must involve all σ α (α ). Discuss l(σ). First we deal with uniqueness. Suppose σ, γ W map Φ + Φ. Then σγ maps Φ + Φ + and in particular it sends the base to itself. By Theroem 10.3 part (e), this implies σγ = id and so γ = σ 1. But the same holds for σσ and γγ. Hence, σ = γ. Let σ = s a1 s ap be a reduced expression. Then since σ(α i ) is negative for each i, we have from Lemma C that σs i = s a1 ŝ ak s ap. Hence σ = s a1 ŝ ak s ap s i is another reduced expression for σ. We know the Weyl group is a coxeter group and σ will be the longest element. 5