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PUMaC 2017 · 加试 · 第 4 题

PUMaC 2017 — Power Round — Problem 4

专题
Discrete Math / 离散数学
难度
L3
来源
PUMaC

题目详情

Problem 4.1.3. (10 points) Show that x ∈ L is semisimple if and only if ad x is semisimple. Problem 4.1.4. ′ ′ (i) (3 points) Let I = { x ∈ L | κ ( x, y ) = 0 ∀ y ∈ I } for I ⊆ L an ideal. Show that I is also an ideal. (ii) (5 points) If L is semisimple, show that there are n simple subalgebras L i (1 ≤ n ≤ dim L ) such that n ⊕ L = L . i i =1 4.2 Root space decomposition (31 points) Definition 4.2.A. The center of a subalgebra S ⊆ L is { x ∈ L | [ x, y ] = 0 ∀ y ∈ S } . Theorem 4.2.1. Any Lie algebra L has a nontrivial subalgebra H of semisimple elements that is equal to its center in L ; H is known as the maximal toral subalgebra . There exists a basis of L such that each basis element is an eigenvector to each element of H . In particular [ h, v ] = λ ( h ) v for some function λ : H → C . For example, for the i i i i sub-basis spanning H consisting of { v , . . . , v } (renumbered as necessary), λ ( h ) = 0 1 dim H i identically for 1 ≤ i ≤ dim H , since H is equal to the center, where the bracket is abelian (but this would not necessarily hold for an element not in H ). ∗ Definition 4.2.B. λ is called a root of L , and H is the vector space with spanned by i { λ | 1 ≤ i ≤ dim L } . i Definition 4.2.C. The root space L corresponding to the root λ consists of the elements λ of L that are λ -eigenvectors of H – i.e., L = { v ∈ L | [ h, v ] = λ ( h ) v ∀ h ∈ H } . λ PUMaC 2017 Power Round Page 16 This yields the root space decomposition of L . From each root λ we then are able to derive a convenient isomorphism: there are three elements e ∈ L , f ∈ L and h ∈ H for which sl is isomorphic to span { e , f , h } via λ λ λ − λ λ 2 λ λ λ ϕ : v 7 → e 1 λ v 7 → f 2 λ v 7 → h 3 λ and the additional condition that λ ( h ) = 2. λ ∗ We are now also ready to talk about H as a vector space having an inner product. ∗ Definition 4.2.D. t ∈ H corresponds to λ ∈ H such that κ ( t , h ) = λ ( h ) for each λ λ h ∈ H . Then, we get a convenient inner product for our space space of functions! We let the ∗ inner product on H be 〈 λ , λ 〉 = κ ( t , t ). 1 2 λ λ 1 2

解析

Problem 4.1.3. (10 points) Show that x ∈ L is semisimple if and only if ad x is semisimple. Solution : We note that ad is injective, because if ad x = ad y then ad ( x − y ) = 0, but then κ ( x − y, z ) = 0 for all z ∈ L , so x − y = 0 by nondegeneracy. Any element of a matrix space can be uniquely written as the sum of a semisimple and nilpotent component (where those components commute, though this won’t be necessary for us right now). Thus if we can show that x semisimple implies ad x semisimple we’ll be done, since the inverse of the PUMaC 2017 Power Round Page 10 ad map will take a semisimple adjoint plus a zero nilpotent part back to some other part plus a zero nilpotent part in L . Suppose x has eigenvectors e with eigenvalue λ . We may i i give ad L ’s matrix space a subbasis of e , the dim L × dim L matrix with 1 at row i and i,j column j . Computation gives that (ad x ) e = ( λ − λ ) e , so the sum of ad x ’s eigenspace’s i,j i j i,j dimensions is maximal, since that is equivalent to being diagonalizable (since the structure of eigenspaces is preserved by similarity). Therefore we see that since x semisimple implies ad x semisimple, we are done.