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Solutions to Problem Set 2 Problem 1. Let I be an ideal of L. (1) Show I i is a ideal of L. For i = 0, trivial. Assume I i−1 is an ideal. I i is spanned by [y, z], y ∈ I, z ∈ I i−1 . Then [x, [y, z]] = −[y, [z, x]] − [z, [x, y]] ∈ [I, I i−1 ] + [I i−1 , I] = I i . (2) Show I (i) is an ideal of L. i = 0 is trivial, assume L(i−1) is an ideal of L. I (i) is spanned by elements of the form [y, z], y, z ∈ I (i−1) . Then [x, [y, z]] = −[y, [z, x]] − [z, [x, y]] ∈ [I (i−1) , I (i−1) ] + [I (i−1) , I (i−1) ] = I (n) . Problem 2. (Only if) Take Li = L(i) . (If) Assume there exists a finite sequence L = L0 ⊃ L1 . . . ⊃ Lk = 0 of subalgebras and Li−1 /Li is abelian. Then for each i, we have Li ⊃ [Li−1 , Li−1 ] since [Li−1 , Li−1 ] is in the kernel of the quotient Li−1 → Li−1 /Li . Now L0 = L = L(0) , by induction we have L(i) = [L(i−1) , L(i−1) ] ⊂ [Li−1 , Li−1 ] ⊂ Li , hence the derived series of L stops. Problem 7. L is nilpotent, so the lower central series Li descends to 0. For i large enough, Li ⊂ K. There exists a minimal i such that Li ⊂ K. Then ∀x ∈ Li−1 \K, we have x ∈ NL (K)\K. Problem 8. Since L is nilpotent, [L, L] 6= L. The quotient L/[L, L] is an abelian Lie algebra; thus every vector subspace of L/[L, L] is a Lie subalgebra. Let V be a codimension one dimensional subspace of L/[L, L] and K be the preimage of V in L. Then K is a codimension one Lie subalgebra of L. Problem 9. By problem 8, there is an ideal K of codimension 1 and an element x such that L = K + Cx. L is nilpotent and non-zero, the lower central series satisfies Li = 0, Li−1 6= 0. Then [Li−1 , K] ⊂ Li = {0} hence CL (K) ⊃ Li−1 6= 0. Let n be the maximal integer that CL (K) ⊂ Ln and CL (K) * Ln+1 . Let z ∈ CL (K) − Ln+1 be an arbitrary element. Define δ : K 7→ 0, x 7→ z, then δ is linear. 1 Easy to check δ([a, b]) = 0 = [δ(a), b] + [a, δ(b)], ∀a, b ∈ L. Finally, we claim that δ is outer. Indeed, assume the contrary: δ = ad(y), then δ(K) = [y, K] = 0 and y ∈ CL (K) ⊂ Ln and [y, x] = z ∈ Ln+1 , a contradiction. Problem 10. Only need to show ad(x) is nilpotent for all x ∈ L. L/K is nilpotent, hence ad(x)|L/K is nilpotent, i.e. when m large enough, (ad(x))m : L → K, now ad(x)|K is also nilpotent, when n large enough (ad(x)|K )n = 0, hence (ad(x))m+n = 0 on L. Problem A. Take b to be the upper triangular Borel subalgebra of gl(n, C), n is the ideal of b with diagonal entries zero. n is nilpotent, b/n is the diagonal algebra, which is also nilpotent. But b is not nilpotent. Problem B(b) Assume U is a nonzero common invariant subspace of all B ∈ L. Choose v1 v = v2 ∈ U , where at least one vi ∈ Cn is nonzero. v3 Case 1: v3 6= 0. Then ∀u ∈ Cn there exists a A ∈ gl(n, C) such that Av3 = u. Then n 0 In 0 v1 C 0 A 0 u 0 0 A 0 0 In v2 = 0 ∴ 0 ⊆ U. 0 0 0 0 0 0 0 0 v3 0 0 u 0 0 0 In 0 = u 0 0 ∴ Cn ⊆ U. 0 0 0 −In 0 0 0 0 0 0 0 0 −In 0 0 u = 0 ∴ 0 ⊆ U. 0 In 0 0 u Cn Hence U = C3n . The only nonzero common invariant subspace is C3n . Case 2: v1 6= 0. Then 2 0 0 0 v1 0 In 0 0 v2 = 0 , 0 −In 0 v3 −v1 and Case 1 applies. 2 Case 3: v2 6= 0 but v1 = v3 0 0 In 0 0 −In = 0 (otherwise reduce to case 1 or 2). Then 0 0 0 0 v2 = 0 , 0 0 −v2 and, once again, Case 1 applies. 3