2.4 Root space decomposition

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1 34 CHAPTER 2. SEMISIMPLE LIE ALGEBRAS 2.4 Root space decomposition Let L denote a semisimple Lie algebra in this section. We will study the detailed structure of L through its adjoint representation Maximal toral subalgebra and root space decomposition The semisimple Lie algera L contains the semisimple part and nilpotent part of all its elements. Since L is not nilpotent, Engel s theorem ensures that there is a non-nilpotent x L, which has a nonzero semisimple part x s L. Hence F x s is a nonzero subalgebra consisting of semisimple elements, called a toral subalgebra. So there exists a nonzero maximal toral subalgebra in L. The following result is a rough analogy to Engel s theorem. Lem A toral subalgebra of L is abelian. Proof. Suppose T is a toral subalgebra of L. We need ad T x = 0 for all x T. The semisimplicity of x implies that ad T x is diagonalizable. If ad T x has a nonzero eigenvalue α, then there is nonzero y T such that [x, y] = αy. Then (ad T y) 2 (x) = [y, [y, x]] = 0. However, y is also semisimple, and ad T y is diagonalizable. We can express x as a linear combination of linear independent eigenvectors {x i } of ad T y: x = a i x i, a i 0, ad T y(x i ) = λ i x i. i Then 0 = (ad T y) 2 (x) = a i λ 2 i x i, which implies that all λ i = 0. Therefore, αy = (ad T y)(x) = i a i λ i x i = 0, which is a contradiction. So T must be abelian. i Now fix a maximal toral subalgebra H of L. Then ad L H is a commuting family of semsimple elements. So ad L H is simiultaneously diagonalizable (exercise). L is the direct sum of some common eigenspaces of ad L H: In particular, L α = {x L [h, x] = α(h)x}, α H. (2.4) H L 0 = C L (H), the centralizer of H in L. (2.5) Each nonzero α H for which L α 0 is called a root; the set of all roots is denoted by Φ. Then we get the root space decomposition: L = C L (H) α Φ L α. (2.6) Ex. Suppose L = sl(l + 1, F). Let h i := e i,i e i+1,i+1 for i [l]. Then H := i [l] Fh i is a maximal torus subalgebra of L, and C L (H) = H. Denote by ɛ i H the linear functional that takes the i-th diagonal entry of elements of H. Then each e ij (i, j [l + 1], i j) is a common eigenvector for elements of H, such that for h H, (ad L h)(e ij ) = [h, e ij ] = (ɛ i ɛ j )(h)e ij. Therefore, sl(l + 1, F) has the root space decomposition (exercise) sl(l + 1, F) = H i j i,j [l+1] Fe ij, where the set of roots Φ := {ɛ i ɛ j i, j [l + 1], i j}.

2 2.4. ROOT SPACE DECOMPOSITION 35 Thm For α, β H, [L α, L β ] L α+β. In particular, if α 0 and x L α, then ad x is nilpotent. 2. If α, β H and α + β 0, then L α L β relative to the Killing form κ of L. Proof. Let α, β H, h H, x L α, y L β be arbitrary. 1. ad h([x, y]) = [[h, x], y] + [x, [h, y]] = α(h)[x, y] + β(h)[x, y] = (α + β)(h)[x, y]. So [L α, L β ] L α+β. If α 0 and x L α, then ad x is nilpotent according to the root space decomposition of L. 2. If α + β 0, then κ([h, x], y) = κ([x, h], y) = κ(x, [h, y]) α(h)κ(x, y) = β(h)κ(x, y) Hence L α L β relative to the Killing form κ of L. (α + β)(h)κ(x, y) = 0 for any h H κ(x, y) = 0. Theorem 2.17 and the nondegeneracy of κ in L imply κ(l α, L α ) 0 and the following result. Cor The restriction of the Killing form to L 0 = C L (H) is nondegenerate. Proof. In the root space decomposition L = C L (H) α Φ L α, C L (H) = L 0 is orthogonal to all L α for α Φ relative to κ. Since κ is nondegenerate on L, it must be nondegenerate on C L (H) as well. Remark. The restriction of κ to C is κ(x, y) = Tr (ad x ad y) for x, y C, instead of κ (x, y) = Tr (ad C x ad C y). In fact, H is in the radical of κ, so that κ is dengerate. Ex. We verify Theorem 2.17 for L = sl(l + 1, F) with the root space decomposition: sl(l + 1, F) = H Fe ij i j i,j [l+1] 1. The root space Fe ij = L ɛi ɛ j. We have [L ɛi ɛ j, L ɛp ɛ q ] = [Fe ij, Fe pq ] = δ jp Fe iq δ qi Fe pj = 0 if p j, q i Fe iq if p = j, q i L ɛi ɛ j +ɛ p ɛ q. Fe pj if p j, q = i F(e ii e jj ) if p = j, q = i 2. In sl(l + 1, F), it is known that κ(x, y) = 2(l + 1)Tr(xy). We can examine the orthogonal relationship of the root spaces, using the basis B := {h i i [l]} {e ij i, j [l + 1], i j}.

3 36 CHAPTER 2. SEMISIMPLE LIE ALGEBRAS (a) κ(h i, h i ) = 4(l + 1), κ(h i, h i+1 ) = 2(l + 1), κ(h i, h j ) = 0 for j i > 1. In particular, a basis dual to {h i i [l]} consists of g i := l + 1 i 2(l + 1) 2 We can verify that κ(h i, g j ) = δ ij. i i e ss 2(l + 1) 2 s=1 l+1 t=i+1 e tt. (2.7) (b) κ(h i, e pq ) = 2(l + 1)Tr(h i e pq ) = 2(l + 1)(ɛ p ɛ q )(h i )Tr(e pq ) = 0. So H Fe pq. { 2(l + 1), (p, q) = (j, i), (c) κ(e ij, e pq ) = 2(l + 1)Tr(e ij e pq ) = 0, otherwise. So Fe ij Fe pq whenever (p, q) (j, i). The above case study shows that L α L β whenever α + β 0. Thm Any maximal toral subalgebra H of L satisfies that H = C L (H). Proof. The proof is proceeded in several steps: (1) C contains the semisimple and nilpotent parts of its elements. If x C, then ad x(h) H, so that ad x s (H) = (ad x) s (H) H, ad x n (H) = (ad x) n (H) H. Hence x s, x n C. (2) All semisimple elements of C lie in H. If x C is semisimple, then H + F x is an abelian subalgebra consisting of semisimple elements. Since H is a maximal toral subalgbra, we have H + F x = H and thus x H. (3) C is nilpotent. For any x C, x = x s + x n. The semisimple part x s lies in H, so that ad C x s = 0, and thus ad C x = ad C x n is nilpotent. By Engel s Theorem, C is nilpotent. (4) The restriction of κ to H is nondegenerate. Let κ(h, H) = 0 for some h H; we show that h = 0. If x C is nilpotent, then ad x is nilpotent and commutes with ad h (since [x, h] = 0). So ad x ad h is nilpotent and κ(x, h) = Tr (ad x ad h) = 0. By (1) and (2), we see that κ(h, C) = 0, which forces h = 0 by the nondegeneracy of κ on C. (5) C is abelian. Otherwise [C, C] 0, C being nilpotent by (3). Then Z(C) [C, C] 0, since the nilpotent ad C-action on [C, C] annihilates a nonzero element z, which must be in Z(C) [C, C]. The associativity of κ implies that κ(h, [C, C]) = 0. The nondegeneracy of κ on H by (4) implies that H [C, C] = 0. So z H, and its nilpotent part z n C {0}. Then ad z n is nilpotent and commutes with ad C, so that κ(z n, C) = 0, contarary to the nondegeneracy of κ on C (Corollary 2.18). (6) C = H. Otherwise C contains a nonzero nilpotent element x by (1), (2). Then ad x is nilpotent, and [x, C] = 0 by (5) implies that κ(x, C) = Tr (ad x ad C) = 0, contrary to the nondegeneracy of κ on C. Now the root space decomposition of L becomes L = H α Φ L α (2.8) where L 0 = H = C L (H); every h H is semisimple; for α Φ, every x L α is nilpotent; relative to the Killing form κ, L α L β for any α + β 0.

4 2.4. ROOT SPACE DECOMPOSITION 37 Cor The restriction of κ to H = C L (H) is nondegenerate. The corollary implies that there is a bijection H H such that every α H corresponds to the unique t α H that satisfies α(h) = κ(t α, h) for all h H. (2.9) Ex. The Lie algebra L = sl(l + 1, F) has the root system Φ = {ɛ i ɛ j i, j [l + 1], i j}. Let us compute t α for α = ɛ i ɛ j. Recall that we have κ(x, y) = 2(l + 1)Tr(xy) for L = sl(l + 1, F). Suppose t α = l+1 k=1 t ke kk H, then for any h = l+1 k=1 s ke kk H, we have l+1 κ(t α, h) = 2(l + 1) t k s k = α(h) = (ɛ i ɛ j )(h) = s i s j. k=1 Therefore, t ɛi ɛ j = 1 2(l + 1) (e ii e jj ) Properties of the Root Space Decomposition The root space decomposition and the Killing form imply a rich structure of L. Thm The Φ, H, and L α have the following properties: 1. Φ spans H. 2. If α Φ, then α Φ; both L α and L α have dimension 1; for x L α, y L α, we have [x, y] = κ(x, y)t α. (2.10) In particular, [L α, L α ] = Ft α. The t α satisfies that α(t α ) = κ(t α, t α ) 0. For any nonzero x α L α, there exists unique y α L α such that S α := span (x α, y α, h α := [x α, y α ]) is a simple subalgebra of L isomorphic to sl(2, F) via Moreover, h α = x α 3. If α Φ, then Fα Φ = {±α}. [ ] 0 1, y 0 0 α [ ] [ ] , h 1 0 α. (2.11) 0 1 2t α κ(t α, t α ) = h α is independent of the choice of x α. 4. If α, β Φ, then β(h α ) Z, and β β(h α )α Φ. (The numbers β(h α ) are called Cartan integers.) 5. Let α, β Φ, β ±α. Let r, q be (resp.) the largest integers for which β rα, β + qα are roots. Then all β + iα Φ for r i q, and the Cartan integer β(h α ) = r q. 6. If α, β, α + β Φ, then [L α, L β ] = L α+β. Proof. 1. If Φ does not span H, there exists a nonzero h H s.t. α(h) = 0 α Φ. So [h, L α ] = 0 α Φ. Moreover, [h, H] = 0. Hence h Z( L) = 0, a contradiction.

5 38 CHAPTER 2. SEMISIMPLE LIE ALGEBRAS 2. If α Φ but α Φ, then α + β 0 for any β = 0 or β Φ, so that κ(l α, L β ) = 0. Thus κ(l α, L) = 0, contradicting the nondegeneracy of κ. So α Φ. Suppose x L α, y L α are nonzero. Then for any h H, κ(h, [x, y] κ(x, y)t α ) = κ(h, [x, y]) κ(h, κ(x, y)t α ) = κ([h, x], y) κ(x, y)κ(h, t α ) = α(h)κ(x, y) α(h)κ(x, y) = 0. Therefore, [x, y] = κ(x, y)t α. The relation also holds if one of x and y is zero. Now κ(l α, L α ) 0 since otherwise κ(l α, L) = 0, a contradiction to nondegeneracy of κ. So [L α, L α ] = Ft α. If 0 = κ(t α, t α ) = α(t α ), then [t α, x] = 0 = [t α, y]. The Lie subalgebra S = span(x, y, t α ) is solvable. So ad L S is solvable, and Lie s theorem implies that ad L t α ad L [S, S] is nilpotent. Then t α is nilpotent. However, t α H is semisimple. We have t α = 0, a contradiction. Hence κ(t α, t α ) 0. 2 For any nonzero x α L α, let y α L α such that κ(x α, y α ) = κ(t α, t α ) ; let h α := [x α, y α ]. Clearly h α = 2t α κ(t α, t α ) = h α, [h α, x α ] = α(h α )x α = 2x α, [h α, y α ] = 2y α. So the Lie subalgebra S α := span(x α, y α, h α ) is isomorphic to sl(2, F) via (2.11). Consider the subspace M of L spanned by H along with all root spaces of the form L cα (c F ). M is a S α -submodule so that it is a direct sum of irreducible S α -modules V (m) for m Z +. The Ker α is a codimension 1 subspace in H complement to Fh α, that is, Ker α Fh α = H. Since S α acts trivially on Ker α, the V (m) components of M with even integers m is S α Ker α. In particular, 2α Φ. Then 1 2α Φ as well (otherwise, replacing α by 1 2α and we will get contradiction). So M doesn t have V (m) summands for odd integers m. Hence M = S α Ker α. (2.12) In particular, dim L α = dim L α = It is obvious by (2.12). 4. For β Φ, we consider the S α -action on L β. The subspace M β := i Z L β+iα (2.13) is a S α -module of L. In particular, L β is a weight space of h α with the weight β(h α ), which must be in Z. Then β(h α ) Z is also a weight of M β. Since β(h α ) = (β β(h α )α)(h α ), we have β β(h α )α Φ. 5. Now assume β ±α in the above paragraph. Then no β +iα can be 0. Each root space is one dimensional, and the integral weights appearing in M β has the form (β+iα)(h α ) = β(h α )+2i, which covers 0 or 1 exactly once. Therefore, M β is an irreducible S α -module. Let r, q be (resp.) the largest integers for which β rα, β + qα are roots. Then all β + iα Φ for r i q. The highest (resp. lowest) weight is (β +qα)(h α ) = β(h α )+2q (resp. β(h α ) 2r.) The symmetry of weights implies that β(h α ) + 2q + β(h α ) 2r = 0, that is, β(h α ) = r q. We call β rα,, β,, β + qα the α-string through β.

6 2.4. ROOT SPACE DECOMPOSITION In the S α -module M β, obviously [L α, L β ] = L α+β. Consider F = C case. Let E := α Φ Rα be the real span of Φ. Then E H with dim R E = dim F H since Φ spans H. We may tranfer the nondegenerate Killing form from H to H by Some properties of Φ can be rephrased as follow. (γ, δ) := κ(t γ, t δ ) for all γ, δ H. (2.14) Thm Let (Φ, E) be defined from (L, H) as above. Then: 1. Φ spans E, and 0 Φ. 2. If α Φ, then α Φ, and no other scalar multiple of α is in Φ. 3. If α, β Φ, then β 2(β, α) (α, α) Φ, where 2(β, α) (α, α) = β(h α) Z is a Cartan integer. The set Φ is called a root system in E. We will completely classify all (Φ, E) for simple Lie algebras in the next Chapter.