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Engel’s theorem∗ rmilson† 2013-03-21 14:13:46 Before proceeding, it will be useful to recall the definition of a nilpotent Lie algebra. Let g be a Lie algebra. The lower central series of g is defined to be the filtration of ideals D0 g ⊃ D1 g ⊃ D2 g ⊃ . . . , where D0 g = g, Dk+1 g = [g, Dk g], k ∈ N. To say that g is nilpotent is to say that the lower central series has a trivial termination, i.e. that there exists a k such that Dk g = 0, or equivalently, that k nested bracket operations always vanish. Theorem 1 (Engel) Let g ⊂ End V be a Lie algebra of endomorphisms of a finite-dimensional vector space V . Suppose that all elements of g are nilpotent transformations. Then, g is a nilpotent Lie algebra. Lemma 1 Let X : V → V be a nilpotent endomorphism of a vector space V . Then, the adjoint action ad(X) : End V → End V is also a nilpotent endomorphism. Proof. Suppose that Xk = 0 for some k ∈ N. We will show that ad(X)2k−1 = 0. ∗ hEngelsTheoremi created: h2013-03-21i by: hrmilsoni version: h32991i Privacy setting: h1i hTheoremi h17B30i h15A57i † This text is available under the Creative Commons Attribution/Share-Alike License 3.0. You can reuse this document or portions thereof only if you do so under terms that are compatible with the CC-BY-SA license. 1 Note that ad(X) = l(X) − r(X), where l(X), r(X) : End V → End V, are the endomorphisms corresponding, respectively, to left and right multiplication by X. These two endomorphisms commute, and hence we can use the binomial formula to write ad(X)2k−1 = 2k−1 X (−1)i l(X)2k−1−i r(X)i . i=0 Each of terms in the above sum vanishes because l(X)k = r(X)k = 0. QED Lemma 2 Let g be as in the theorem, and suppose, in addition, that g is a nilpotent Lie algebra. Then the joint kernel, \ ker g = ker a, a∈g is non-trivial. Proof. We proceed by induction on the dimension of g. The claim is true for dimension 1, because then g is generated by a single nilpotent transformation, and all nilpotent transformations are singular. Suppose then that the claim is true for all Lie algebras of dimension less than n = dim g. We note that D1 g fits the hypotheses of the lemma, and has dimension less than n, because g is nilpotent. Hence, by the induction hypothesis V0 = ker D1 g is non-trivial. Now, if we restrict all actions to V0 , we obtain a representation of g by abelian transformations. This is because for all a, b ∈ g and v ∈ V0 we have abv − bav = [a, b]v = 0. Now a finite number of mutually commuting linear endomorphisms admits a mutual eigenspace decomposition. In particular, if all of the commuting endomorphisms are singular, their joint kernel will be non-trivial. We apply this result to a basis of g/D1 g acting on V0 , and the desired conclusion follows. QED 2 Proof of the theorem. We proceed by induction on the dimension of g. The theorem is true in dimension 1, because in that circumstance D1 g is trivial. Next, suppose that the theorem holds for all Lie algebras of dimension less than n = dim g. Let h ⊂ g be a properly contained subalgebra of minimum codimension. We claim that there exists an a ∈ g but not in h such that [a, h] ⊂ h. By the induction hypothesis, h is nilpotent. To prove the claim consider the isotropy representation of h on g/h. By Lemma 1, the action of each a ∈ h on g/h is a nilpotent endomorphism. Hence, we can apply Lemma 2 to deduce that the joint kernel of all these actions is non-trivial, i.e. there exists a a ∈ g but not in h such that [b, a] ≡ 0 mod h, for all b ∈ h. Equivalently, [h, a] ⊂ h and the claim is proved. Evidently then, the span of a and h is a subalgebra of g. Since h has minimum codimension, we infer that h and a span all of g, and that D1 g ⊂ h. (1) Next, we claim that all the Dk h are ideals of g. It is enough to show that [a, Dk h] ⊂ Dk h. We argue by induction on k. Suppose the claim is true for some k. Let b ∈ h, c ∈ Dk h be given. By the Jacobi identity [a, [b, c]] = [[a, b], c] + [b, [a, c]]. The first term on the right hand-side in Dk+1 h because [a, b] ∈ h. The second term is in Dk+1 h by the induction hypothesis. In this way the claim is established. Now a is nilpotent, and hence by Lemma 1, ad(a)n = 0 for some n ∈ N. We now claim that Dn+1 g ⊂ D1 h. By (??) it suffices to show that n times z }| { [g, [. . . [g, h] . . .]] ⊂ D1 h. Putting g1 = g/D1 h, h1 = h/D1 h, this is equivalent to n times z }| { [g1 , [. . . [g1 , h1 ] . . .]] = 0. 3 (2) However, h1 is abelian, and hence, the above follows directly from (??). Adapting this argument in the obvious fashion we can show that Dkn+1 g ⊂ Dk h. Since h is nilpotent, g must be nilpotent as well. QED Historical remark. In the traditional formulation of Engel’s theorem, the hypotheses are the same, but the conclusion is that there exists a basis B of V , such that all elements of g are represented by nilpotent matrices relative to B. Let us put this another way. The vector space of nilpotent matrices Nil, is a nilpotent Lie algebra, and indeed all subalgebras of Nil are nilpotent Lie algebras. Engel’s theorem asserts that the converse holds, i.e. if all elements of a Lie algebra g are nilpotent transformations, then g is isomorphic to a subalgebra of Nil. The classical result follows straightforwardly from our version of the Theorem and from Lemma 2. Indeed, let V1 be the joint kernel g. We then let U2 be the joint kernel of g acting on V /V0 , and let V2 ⊂ V be the subspace obtained by pulling U2 x back to V . We do this a finite number of times and obtain a flag of subspaces 0 = V0 ⊂ V1 ⊂ V2 ⊂ . . . ⊂ Vn = V, such that gVk+1 = Vk for all k. The choose an adapted basis relative to this flag, and we’re done. 4