18.745 Introduction to Lie Algebras  Fall 2004

Lecture 12 - October 19th, 2004

Professor Victor Kac̆  Scribe: David Meyer

Definition. An Abstract Jordan Decomposition of an element of a Lie Algebra 𝔤 is a decomposition of the form a = a_{s} + a_{n} where ad(a_{s}) is a semisimple operator and ad(a_{n}) is a nilpotent operator (on 𝔤), and [a_{s}, a_{n}] = 0.

Example. If 𝔤 = gl_{N}(𝔽), where 𝔽 is algebraically closed, then A = A_{s} + A_{n} ∈ 𝔤 (the concrete Jordan decomposition) is also an abstract Jordan decomposition, since ad(A_{s}) is semisimple, ad(A_{n}) is nilpotent, and [ad(A_{s}), ad(A_{n})] = ad[A_{s}, A_{n}] = 0.

Remark. Note that A′_{s} = A_{s} + λI, A′_{n} = A_{n} − λI is another abstract Jordan decomposition, for any λ ∈ 𝔽. Thus we see that the abstract Jordan decomposition is not unique itself. The uniqueness fails in this case because I is a central element.

Claim. An abstract Jordan decomposition is unique (if it exists) if center(𝔤) = 0. For if a = a_{s} + a_{n} = a′_{s} + a′_{n} are two abstract Jordan decompositions, then:

ad(a) = ad(a_{s}) + ad(a_{n}) = ad(a′_{s}) + ad(a′_{n}),

both usual Jordan decompositions of ad(a). Hence by the uniqueness of the usual Jordan decomposition, ad(a_{s}) = ad(a′_{s}) and ad(a_{n}) = ad(a′_{n}), or ad(a_{s} − a′_{s}) = ad(a_{n} − a′_{n}) = 0. Then since center(𝔤) = 0, we conclude that a_{s} = a′_{s} and a_{n} = a′_{n}.

Remark. In some situations, an abstract Jordan decomposition may not exist, but it is difficult to construct examples of this.

Exercise 12.1. If 𝔤 is the Lie Algebra of an algebraic group over an algebraically closed field 𝔽, then any a ∈ 𝔤 has an abstract Jordan decomposition.

Proof. Beyond the scope of this write up. □

Exercise 12.2. Using Levi’s Theorem, show that any 4-dimensional Lie Algebra over an algebraically closed field of characteristic 0 is solvable or sl_{2}(𝔽)⊕{one dimensional abelian}.

Proof. Let 𝔤 be a 4-dimensional Lie Algebra over an algebraically closed field of characteristic 0. By Levi’s Theorem, there exists a semisimple subalgebra S of 𝔤 such that 𝔤 = S ⊕ R(𝔤), where R(𝔤) is the radical of 𝔤, and S ∩ R(𝔤) = 0. If 𝔤 is solvable, we are done, so let us consider the case where 𝔤 is not solvable. In this case, dim(S) > 0. The cases of dim(S) = 1 and dim(S) = 2 are not possible, since then S would be solvable. Suppose dim(S) = 3. By exercises 8.2 and 8.3, we note that the only semisimple 3-dimensional Lie Algebra (over an algebraically closed field of characteristic 0) is sl_{2}(𝔽). So S = sl_{2}(𝔽), and we have a homomorphism sl_{2}(𝔽) ⟶ Der(R(𝔤)). However, Der(R(𝔤)) is one dimensional, and so the kernel of this map will be an ideal of sl_{2}(𝔽) of dimension 2 or 3. If of dimension 2, the kernel would be solvable, violating the semisimplicity of sl_{2}(𝔽). So the kernel must have dimension 3, and so sl_{2}(𝔽) commutes with R(𝔤). Thus 𝔤 = sl_{2}(𝔽)⊕{one dimensional abelian}.

All that remains is to handle the case where dim(S) = 4. We wish to show there are no semisimple 4-dimensional Lie algebras.

Assume 𝔤 is semisimple. Choose a Cartan subalgebra 𝔥 of 𝔤 (note that 𝔥 must be non-zero), and write the generalized root space decomposition. Note that by Theorem 1(b) in Lecture 12, roots come in pairs, and so dim(𝔥) is even.

Case dim(𝔥) = 2: Then 𝔤 = 𝔥 + 𝔤_{α} + 𝔤_{−α}. Choose nonzero E ∈ 𝔤_{α} and nonzero F ∈ 𝔤_{−α}. Let H = [E, F] ∈ 𝔥. Then, the Jacobi identity,

[H, [E, F]] + [E, [F, H]] + [F, [H, E]] = 0,

implies: [E, [F, H]] = [[H, E], F], with both sides in 𝔥. Since ad(H) cannot be 0, we must have one (and thus both) of [F, H] and [H, E] nonzero.

Write [H, E] = cE, for nonzero c ∈ 𝔽. This forces [F, H] = cF. Now we can choose a final basis vector B for 𝔥 and subtract away a multiple of H such that [B, B] = 0, [B, H] = 0, and [B, E] = 0. But by the Jacobi identity, we have [B,[E,F]] + [E,[F,B]] + [F,[B,E]] = 0 which implies [F, B] = 0, and thus ad(B) = 0 which contradicts semisimplicity.

Case dim(𝔥) = 4: Not possible, since 𝔤 is not nilpotent. □

Proposition. Let 𝔤 be a Lie Algebra (over an algebraically closed field 𝔽) with center 0, such that all derivations of 𝔤 are inner. (In particular, 𝔤 is semisimple). Then any element of 𝔤 admits a (unique) abstract Jordan decomposition.

Proof. Take a ∈ 𝔤. Then A = ad(a) = A_{s} + A_{n} (the usual Jordan decomposition) where A_{s}, A_{n} ∈ gl_{𝔤}, A_{s} semisimple, A_{n} nilpotent, and A_{s}A_{n} = A_{n}A_{s}. (In End(𝔤)). Let 𝔤 = ⨁ 𝔤_{λ}, λ taken over the eigenvalues of A_{s}, be the eigenspace decomposition of 𝔤 with respect to A_{s}.

Let us prove that A_{s} = ad(a_{s}) for some element a_{s} ∈ 𝔤. To do this, we must check that A_{s} is a derivation of 𝔤, ie, that A_{s}([x, y]) = [A_{s}x, y] + [x, A_{s}y]. Luckily, it suffices to check this for a basis of eigenvectors of A_{s}. Take x ∈ 𝔤_{λ} and y ∈ 𝔤_{μ}. Recall that [x, y] ∈ 𝔤_{λ+μ}, since 𝔤_{λ} and 𝔤_{μ} are generalized eigenspaces of ad(a). Now it is easy to see:

LHS = A_{s}([x, y]) = (λ + μ)[x, y]
RHS = [A_{s}x, y] + [x, A_{s}y] = λ[x, y] + μ[x, y] = (λ + μ)[x, y].

So A_{s} is a derivation of 𝔤. Since all derivations of 𝔤 are inner, we have that A_{s} = ad(a_{s}) for some a_{s} ∈ 𝔤. Hence, letting a_{n} = a − a_{s}, we see that ad(a_{n}) = ad(a − a_{s}) = A − A_{s} = A_{n} is nilpotent, and [ad(a_{s}), ad(a_{n})] = 0 = ad[a_{s}, a_{n}]. Since center(𝔤) = 0, it follows that [a_{s}, a_{n}] = 0. Thus a = a_{s} + a_{n} is an abstract Jordan decomposition of a. Uniqueness follows from center(𝔤) = 0. □

From now on, we will assume that 𝔽 is an algebraically closed field of characteristic 0, and that 𝔤 is a finite-dimensional semisimple Lie Algebra over 𝔽.

Let 𝔥 be a Cartan subalgebra and let 𝔤 = ⨁_{α∈𝔥^{∗}} 𝔤_{α} be the generalized root space decomposition, where:

𝔤_{α} = {a ∈ 𝔤|(ad(h) − α(h)I)^{N}a = 0 for some N > 0, for all h ∈ 𝔥}.

Recall that [𝔤_{α}, 𝔤_{β}] ⊂ 𝔤_{α+β}, and 𝔤_{0} = 𝔥. For semisimple algebras, we can say much more however.

Theorem 1.

(a) With respect to the Killing Form, 𝔤_{α}, 𝔤_{β} are orthogonal (ie, K(𝔤_{α}, 𝔤_{β}) = 0), if α + β ≠ 0.

(b) K |_{𝔤_{−α}+𝔤_{α}} is a non-degenerate bilinear form. In particular, K |_{𝔥} is non-degenerate, and K defines a non-degenerate pairing of 𝔤_{α} and 𝔤_{−α}.

(c) 𝔥 is a maximal abelian subalgebra of 𝔤.

(d) 𝔥 consists of semisimple elements. (ie, ad(h) is semisimple for all h ∈ 𝔥).

Proof.

(a) holds in any finite-dimensional Lie Algebra. Let a ∈ 𝔤_{α}, b ∈ 𝔤_{β}. Then for arbitrary γ:

(ad(a))(ad(b))(𝔤_{γ}) ⊂ 𝔤_{γ+α+β}.

Hence, ((ad(a))(ad(b)))^{N}(𝔤_{γ}) ⊂ 𝔤_{γ+N(α+β)} = 0 for N sufficiently large since α + β ≠ 0 and since there are only finitely many γ for which 𝔤_{γ} ≠ 0. So (ad(a))(ad(b)) is a nilpotent operator on 𝔤. Hence tr(ad(a)ad(b)) = 0. Thus, K(𝔤_{α}, 𝔤_{β}) = 0.

(b) follows from (a) and the fact that K is non-degenerate on 𝔤, since by (a) the kernel of K |_{𝔤_{−α}+𝔤_{α}} lies in the kernel of K |_{𝔤}.

(c) By the easy part of Cartan’s Criterion, K(𝔥, [𝔥, 𝔥]) = 0, since 𝔥 is solvable. But by (b), [𝔥, 𝔥] = 0 since K |_{𝔥} is non-degenerate. So 𝔥 is abelian. Also, 𝔥 is maximal abelian since it is maximal among nilpotent subalgebras (being a Cartan subalgebra).

(d) Take h ∈ 𝔥, and write the abstract Jordan decomposition h = h_{s} + h_{n}. For each h′ ∈ 𝔥, [ad(h), ad(h′)] = ad([h, h′]) = 0, hence ad(h_{s}) and ad(h_{n}) commute with ad(h′) (see remark below). Hence since center(𝔤) = 0, [h_{s}, h′] = 0 and [h_{n}, h′] = 0. Therefore, h_{s} ∈ 𝔥 and h_{n} ∈ 𝔥, since 𝔥 is maximal abelian. To show h_{n} = 0, we compute:

K(h_{n}, h′) = tr(ad(h_{n})ad(h′)) = 0.

(since ad(h_{n}) is nilpotent, and ad(h′) commutes with ad(h_{n}), their composition is nilpotent). Hence K(h_{n}, 𝔥) = 0. Hence by (b), h_{n} = 0, and h = h_{s} is semisimple.

□

Remark. We used the following fact from Linear Algebra:

If A and B are commuting operators on a finite-dimensional vector space V, then A_{s} and B are also commuting. Indeed, consider the generalized eigenspace decomposition for A, V = ⨁_{λ} V_{λ}. Each V_{λ} is B-invariant. But A_{s} on V_{λ} is just λI, hence A_{s}B = BA_{s} on V_{λ} for each λ. Hence A_{s}B = BA_{s} on V.

Theorem 1 says that the generalized root space decompositions are just regular root spaces:

𝔤_{α} = {a ∈ 𝔤|[h, a] = α(h)a for all h ∈ 𝔥}

since all ad(h) are semisimple operators.

Let Δ = {α ∈ 𝔥^{∗}|α ≠ 0, 𝔤_{α} ≠ 0}. An element of Δ is called a root of 𝔤, and 𝔤_{α} the attached root space. (α(h) is the eigenvalue of ad(h), hence the word ”root”)

The root space decomposition now becomes:

𝔤 = 𝔥 ⊕ ( ⨁_{α∈Δ} 𝔤_{α}), where
𝔤_{α} = {a ∈ 𝔤|[h, a] = α(h)a for all h ∈ 𝔥},

and 𝔥 is the maximal abelian subalgebra of 𝔤.

Remark. We have a linear map ν : 𝔥 ⟶ 𝔥^{∗} defined by (ν(h))(h′) = K(h, h′). But K is non-degenerate, hence ν is an isomorphism. This gives us a bilinear form on 𝔥^{∗}:

K(ν(h), ν(h′)) = K(h, h′) = ν(h)(h′) = ν(h′)(h).

Theorem 2. (a) If α ∈ Δ, e ∈ 𝔤_{α}, f ∈ 𝔤_{−α}, then [e, f] = K(e, f)ν^{−1}(α) ∈ 𝔥.

(b) If α ∈ Δ, then K(α, α) ≠ 0.

Proof.

(a) [e, f] ∈ [𝔤_{α}, 𝔤_{−α}] ⊂ 𝔤_{0} = 𝔥, so [e, f] − K(e, f)ν^{−1}(α) ∈ 𝔥. To prove that it is 0, we need to check that

K([e, f] − K(e, f)ν^{−1}(α), h′) = 0, ∀h′ ∈ 𝔥.

But this is just a computation:

K([e, f]−K(e, f)ν^{−1}(α), h′) = K([e, f], h′)−K(e, f)K(ν^{−1}(α), h′) = K(e, [f, h′]) − K(e, f)α(h′) = α(h′)K(e, f) − K(e, f)α(h′) = 0.

(b) Assume the contrary, that K(α, α) = 0, ie, α(ν^{−1}(α)) = 0. Consider the following 3-dimensional subalgebra of 𝔤:

𝔽e + 𝔽f + 𝔽ν^{−1}(α),

where e ∈ 𝔤_{α}, f ∈ 𝔤_{−α}, and K(e, f) = 1. (we can choose such elements by Theorem 1(b)).

By (a), [e, f] = ν^{−1}(α). Also, we have:

[ν^{−1}(α), e] = α(ν^{−1}(α))e = 0,
[ν^{−1}(α), f] = −α(ν^{−1}(α))f = 0.

Hence this 3-dimensional subalgebra is solvable (even nilpotent). Hence by Lie’s Theorem, in its adjoint representation, it can be represented by upper triangular matrices in some basis. Hence its derived algebra, 𝔽ν^{−1}(α) can be represented by strictly upper-triangular matrices, which is impossible since ν^{−1}(α) ∈ 𝔥 is a semisimple element.

□