18.745: Lecture 5

Professor: Victor Kač Scribe: Zhongtao Wu

Definition 1. If 𝔥 is a Lie algebra and π : 𝔥 → gl_{V} is its representation, and λ ∈ 𝔥^{∗}, then the subspace V_{λ}^{𝔥} = {v ∈ V |π(a)v = λ(a)v for all a ∈ 𝔥} is called a weight space of 𝔥 attached to λ.

Lemma 2 (Lie’s lemma). Let 𝔽 be a field of characteristic 0; let π be a representation of 𝔤 in a finite-dimensional vector space V, and let 𝔥 be an ideal of 𝔤. Then for any λ ∈ 𝔥^{∗}, π(𝔤)V_{λ}^{𝔥} ⊂ V_{λ}^{𝔥}. (i.e weight spaces of 𝔥 are 𝔤invariant.)

Proof. We may assume V_{λ}^{𝔥} ≠ 0. We want to prove that if π(g)v ∈ V_{λ}^{𝔥}, then π(g)v ∈ V_{λ}^{𝔥}, namely:

π(h)(π(g)v) = λ(h)π(g)v, ∀h ∈ 𝔥, g ∈ 𝔤, v ∈ V_{λ}^{𝔥}

Using AB = BA + [A, B], we can rewrite the equation as:

π(g)π(h)v + π([h, g])v = λ(h)π(g)v.

Since 𝔥 is an ideal, [h, g] ∈ 𝔥. So we can use the definition of V_{λ}^{𝔥} to further simplify the equation to:

λ(h)π(g)v + λ([h, g])(v) = λ(h)π(g)v

or equivalently λ([h, g])(v) = 0. So we need to show that if V_{λ}^{𝔥} ≠ 0, then λ([h, g]) = 0 for all h ∈ 𝔥, g ∈ 𝔤.

Pick a non-zero vector v ∈ V_{λ}^{𝔥} and take any g ∈ 𝔤. Denote by W_{m} the span of vectors v, π(g)v, …, π(g)^{m}v. We have an increasing sequence of subspaces W_{0} = 𝔽v ⊂ W_{1} = 𝔽v + 𝔽π(g)v ⊂ ⋯ in V. Take the smallest N, such that the vectors v, π(g)v, …, π(g)^{N−1}v are linearly independent, but v, …, π(g)^{N}v are linearly dependent, i.e, π(g)^{N}v = linear combination of v, …, π(g)^{N−1}v; and the same holds for π(g)^{N+1}v, . . .. Hence,

π(g)W_{N−1} ⊂ W_{N−1} = W_{N} = W_{N+1} = ⋯

We shall prove that for any h ∈ 𝔥, π(h)W_{m} ⊂ W_{m} for all m less than N, and moreover, that in the basis v, π(g)v, …, π(g)^{m}v, the operator π(h) has the

matrix form (\begin{matrix}λ(h) & ∗ \\ . & . & . \\ 0 & & λ(h) \end{matrix})

We do it by induction on n: For n = 0, we have matrix (λ(h)). Suppose this is true for < n, then:

π(h)π(g)^{n}v = π(h)π(g)π(g)^{n−1}v = π([h, g])π(g)^{n−1}v + π(g)π(h)π(g)^{n−1}v.

Since [h, g] ∈ 𝔥, the first term π([h, g])π(g)^{n−1} ∈ W_{n−1}. For our second term, we use the induction hypothesis on π(h)π(g)^{n−1}v to get:

π(g)π(h)π(g)^{n−1}v = π(g)(λ(h)π(g)^{n−1}v + c_{n−2}π(g)^{n−2}v + ⋯ + c_{0}v)
= λ(h)π(g)^{n}v + c_{n−2}π(g)^{n−2}v + ⋯ + c_{0}π(g)v

So we proved the inductive step. In particular, both π(h) and π(g) are operators on W_{N−1} and π(h) is upper triangular on W_{N−1} with λ(h) on the diagonal. But π([h, g]) = [π(h), π(g)] on W_{n−1}; if we take tr_{W_{N−1}} of both sides, we get Nλ([h, g]) = 0. So λ([h, g]) = 0 since char 𝔽 = 0.

From the above lemma, we can easily prove the following main theorem in this section.

Theorem 3 (Lie’s theorem). Let 𝔽 be an algebraically closed field of characteristic 0. Let 𝔤 be a solvable Lie algebra and let π : 𝔤 → gl_{V} be a representation of 𝔤 in a finite dimensional vector space V over 𝔽. Then there exists a common eigenvector v ∈ V for all operators π(a), a ∈ 𝔤. i.e, π(a)v = λ(a)v, where λ(a) ∈ 𝔽, v ≠ 0.

Proof. Note that π(g) ⊂ gl_{V} is a finite dimensional subalgebra since dim gl_{V} = (dim V)^{2}, so we may assume that dim 𝔤 < ∞ and prove Lie’s theorem by induction on n = dim 𝔤.

If n = 1, i.e, 𝔤 = 𝔽a, then we take for v any eigenvector of π(a). (We use here that 𝔽 is algebraically closed.)

For any n > 1, since 𝔤 is solvable, [𝔤, 𝔤] ⊊ 𝔤; so we can take a subspace 𝔥 of 𝔤 of codimension 1 containing [𝔤, 𝔤]. Then 𝔥 is an ideal of 𝔤 since [𝔥, 𝔤] ⊂ [𝔤, 𝔤] ⊂ 𝔥. dim 𝔥 = n − 1, so we can apply the inductive assumption and find v ∈ V such that π(h)v = λ(h)v for all h ∈ 𝔥. Obviously λ ∈ 𝔥^{∗}, so V_{λ}^{𝔥} ≠ 0. Apply Lie’s lemma, we get π(a)V_{λ}^{𝔥} ⊂ V_{λ}^{𝔥} for all a ∈ 𝔤. Write 𝔤 = 𝔥 + 𝔽a, we have π(a)V_{λ}^{𝔥} ⊂ V_{λ}^{𝔥}. But V_{λ}^{𝔥} is a finite dimensional vector space over an algebraically closed field, hence the operator π(a) has an eigenvector v ∈ V_{λ}^{𝔥}. This is the desired v.

Exercise 5.1: Lie’s lemma and hence Lie’s theorem hold over an algebraically closed field of char 𝔽 = p > dim V.

Proof. The only place we used char 𝔽 = 0 in our proof of Lie’s lemma is concluding λ([𝔥, 𝔤]) = 0 from the identity Nλ([𝔥, 𝔤]) = 0, where N is the dimension of the subspace W_{N−1}. But if char 𝔽 = p > dim V > dim W_{N−1} = N, such an argument is still valid.

Exercise 5.2: Take 𝔤 = ℌ_{1} acting on 𝔽[x] by P = \frac{d}{dx}, Q = x, c = 1. Suppose char 𝔽 = p > 0, then U = ⊕_{j≥p}𝔽x^{j} is a subrepresentation. Hence V = 𝔽[x]/U is a p-dimensional representation of ℌ_{1}. It can be shown that there is no common eigenvector for P and Q, so Lie’s theorem fails.

Proof. Suppose f ∈ U, then Q(f) ∈ U. If deg f > p, then deg P (f) ≥ p and hence P (f) ∈ U. Otherwise, deg f = p, i.e, f = cx^{p} for some constant c. Then P (f) = cpx^{p−1} = 0 since char 𝔽 = p. Hence, U is a subrepresentation, and the quotient V = 𝔽[x]/U is a p-dimensional representation of 𝔤.

Next, notice that the only nonzero eigenvector for Q is a scalar of x^{p−1}. But x^{p−1} is not an eigenvector for P, since P (x^{p−1}) = (p−1)x^{p−2}. So Lie’s theorem fails.

Exercise 5.3: If 𝔤 is abelian, then λ([h, g]) = 0 for all λ ∈ 𝔥^{∗}, g, h ∈ 𝔤. So Lie’s lemma and hence Lie’s theorem hold over any algebraically closed field, even if char 𝔽 ≠ 0.

Proof. If 𝔤 is abelian, then the commutator [π(g), π(h)] is identically 0. So we have

π(h)(π(g)v) = π(g)π(h)v = λ(h)(π(g)v), ∀h ∈ 𝔥, g ∈ 𝔤.

This means that π(g)V_{λ}^{𝔥} ⊂ V_{λ}^{𝔥}, hence Lie’s lemma holds.

Lie’s theorem implies the following corollary:

Corollary 4. (a) For any representation π of a solvable Lie algebra 𝔤 in a finite dimensional vector space V, there exists a basis of V for which the matrices of all operators π(g), g ∈ 𝔤 are upper triangular.

(b) A subalgebra 𝔤 ⊂ gl_{V}, dim V < ∞, is solvable if and only if in some basis the matrices of all operators from 𝔤 are upper triangular.

(c) If 𝔤 is a finite dimensional solvable Lie algebra, then [𝔤, 𝔤] is a nilpotent Lie algebra.

Proof. (a) By Lie’s theorem, there exists a common eigenvector v for π(𝔤). Let v_{1} = v. The subspace 𝔽v_{1} is π(𝔤) invariant, hence we may consider the representation π_{V/𝔽v_{1}} of 𝔤 in V/𝔽v_{1}. Apply Lie’s theorem, we can find a common eigenvector v′_{2} ∈ V/𝔽v_{1} for π_{V/𝔽v_{1}}(𝔤). This means that if v_{2} ∈ V is a preimage of v′_{2}, then π(𝔤)v_{2} ∈ 𝔽v_{1} + 𝔽v_{2}. Next, consider V/(𝔽v_{1} + 𝔽v_{2}) and construct v_{3} ∈ V such that π(𝔤)v_{3} ∈ 𝔽v_{1} + 𝔽v_{2} + 𝔽v_{3}, etc.

So for any a ∈ 𝔤,

π(a)v_{1} ∈ 𝔽v_{1}
π(a)v_{2} ∈ 𝔽v_{1} + 𝔽v_{2}
π(a)v_{3} ∈ 𝔽v_{1} + 𝔽v_{2} + 𝔽v_{3}
⋯

But this exactly means that π(a) is upper triangular in the basis v_{1}, …, v_{n}.

(b) If the matrices of all operators from 𝔤 are upper triangular in some basis, then it is a subalgebra of the solvable Lie algebra b_{k}(𝔽), which consists of all upper triangular matrices. Hence, 𝔤 is also solvable.

Conversely, if 𝔤 ⊂ gl_{V} is solvable, then we can apply part (a) to find a basis in which all matrices are upper triangular.

(c) Consider ad : 𝔤 → gl_{𝔤}, the adjoint representation of 𝔤. Ker ad = center(𝔤) is abelian, hence is a solvable ideal. Also, ad(𝔤) = 𝔤/ center(𝔤) ⊂ gl_{𝔤} and ad([𝔤, 𝔤]) = [𝔤, 𝔤]/[𝔤, 𝔤] ∩ center(𝔤) ⊂ gl_{𝔤}. In order to prove [𝔤, 𝔤] is a nilpotent Lie algebra, it suffices to prove ad[𝔤, 𝔤] is nilpotent. But ad 𝔤 ⊂ gl_{𝔤} is a solvable subalgebra, hence is an upper triangular subalgebra of gl_{𝔤} in some basis of 𝔤 by part (b). But the commutator of two upper triangular matrices is strictly upper triangular, hence ad[𝔤, 𝔤] ⊂ strictly upper triangular. So ad[𝔤, 𝔤] is a nilpotent Lie algebra; hence [𝔤, 𝔤] is a nilpotent Lie algebra.