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In other words jy j this implies that j and y span a

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Unformatted text preview: tient L/J . By induction on dimension, there is a nonzero vector y + J ∈ L/J so that adJ kills this vector. In other words, [Jy ] ⊆ J . This implies that J and y span a subalgebra of L with dimension one more than the dimension of J . By maximality of J , this subalgebra is equal to L. So, J has codimension 1. Also, J is an ideal since [LJ ] ⊆ [JJ ] + [yJ ] ⊆ J . Since J is smaller than L, there is a nonzero vector v ∈ V so that J (v ) = 0. Let W be the set of all v ∈ V so that J (v ) = 0. Then W is a nonzero vector subspace of V . Take y ∈ L, y ∈ J . / Claim 2: y (W ) ⊆ W . Pf: We need to show that, for any x ∈ J and w ∈ W , xy (w) = 0. This follows from the following calculation: xy (w) = [xy ](w) + yx(w) = 0 + 0 = 0 since [xy ] ∈ J and J (w) = 0. Since y is nilpotent and sends W into W , there is some nonzero w ∈ W so that y (w) = 0. Since J (w) = 0, we conclude that L(w) = 0 proving the lemma. ￿ Exercise 3.1.4. (1) Show that n(n, F ) is nilpotent of class n − 1. Hint: there is a filtration of V = F n by vector subspaces 0 = V0 ⊂ V1 ⊂ V2 ⊂ · · · Vn so that x(Vi ) ⊆ Vi−1 for all x ∈ n(n, F ). Such a sequence of subspaces of V is called a flag. (2) Show that, for any linear Lie algebra L ⊆ gl(V ), the existence of a flag in V with the property that x(Vi ) ⊆ Vi−1 implies t...
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