PUMaC 2017 · 加试 · 第 3 题
PUMaC 2017 — Power Round — Problem 3
题目详情
- 2 ∈ Z 〈 v,v 〉 Elements of the root system are known as roots . (This is related to the definition in Section 4.3.) Note that a root system is not guaranteed to exist. Definition 5.1.B. A reducible root system may be decomposed into into nonempty subsets R ∪ R for which 〈 v , v 〉 = 0 for v ∈ R , v ∈ R , where R and R are also root systems. 1 2 1 2 1 1 2 2 1 2 An irreducible root system is one that is not reducible. 2 Figure 1: Two root systems of R , (a) A × A and (b) A . 1 1 2 2 Figure 1(a) depicts the root system of R {[ ] [ ] [ ] [ ]} R = 2 0 , − 2 0 , 0 1 , 0 − 1 . 1 a {[ ] [ ]} {[ ] [ ]} Note that R = 2 0 , − 2 0 ∪ 0 1 , 0 − 1 and that 1 a {[ ] [ ]} {[ ] [ ]} 2 span R = R = span 2 0 , − 2 0 ⊕ span 0 1 , 0 − 1 . 1 a We see that A × A is a reducible root system. Meanwhile, Figure 1(b) depicts the irre- 1 1 2 ducible root system of R { [ ] [ ] [ ] [ ]} √ √ √ √ [ ] [ ] 1 3 1 3 1 3 1 3 R = 1 0 , − 1 0 , , , , ; − − − − 1 b 2 2 2 2 2 2 2 2 note that only trivially can R be broken up into smaller root spaces (i.e. via { 0 } and R 1 b 1 b itself). Definition 5.1.C. R has base B ⊂ R if B is a basis of V and, for each w ∈ R , there exists ∑ c : B → N ∪ { 0 } such that w = ± c ( v ) v. v ∈ B 3 This represents reflecting w across the plane in V perpendicular to v and passing through 0. PUMaC 2017 Power Round Page 18 {[ ] [ ]} Going back to our examples, A × A can have the base 2 0 , 0 , 1 and A can 1 1 2 { [ ]} √ [ ] 1 3 have the base 1 0 , . − 2 2
解析
Problem 3.1.5. Fix n . Suppose there is some subalgebra S ⊆ gl for which φ : sl → S is 2 n a bijective Lie algebra homomorphism (where gl has the bracket as in Problem 3.1.1). If n [ ] [ ] [ ] 0 1 0 0 1 0 ′ v = , v = , and v = , let V = φ ( v ). Let λ and λ be the greatest 1 2 3 i i 3 3 0 0 1 0 0 − 1 ′ n and least eigenvalue, respectively, of V and let v, v ∈ C be vectors such that V v = λ v 3 3 3 ′ ′ ′ and V v = λ v . 3 3 ′ (i) (5 points) Find, with proof, V v and V v . 1 2 ′ (ii) (7 points) Find, with proof, λ and λ . 3 3 Solution : (i) We first compute [ v , v ] = 2 v and [ v , v ] = − 2 v . φ ([ v , v ]) v = [ φ ( v ) , φ ( v )] v 3 1 1 3 2 2 3 1 3 1 and expanding gets V V v = ( λ + 2) V v . But we assumed λ was the maximal real 3 1 1 eigenvalue – this is a contradiction if V v 6 = 0. Thus we conclude that it is indeed 1 ′ ′ ′ 0. By analogous computation, V V v = ( λ − 2) V v , and V V v = ( λ − 2) V v , which 3 2 2 3 2 2 ′ forces V v = 0. 2 (ii) We repeat the process of multiplying V V v to get a string of eigenvectors ( v = 3 2 ) v , v , . . . , v , v . We know that the string must stop here because each is λ λ − 2 − λ +2 − λ an eigenvector of V and tr V = tr φ ( v ) = tr( φ ([ v , v ])) = tr( V V − V V ) = 0. 3 3 3 1 2 1 2 2 1 n ∑ ( λ − 2 k ) = 0 k =0 ( n + 1) λ − n ( n + 1) = λ = n ′ and so the minimum must be λ = − n .