18.745 Introduction to Lie Algebras Fall 2004

Lecture 14

Prof. Victor Kač Scribe: Maksim Lipyanskiy

Recall that g = h ⊕ (⊕_{α∈Δ}g_{α}), dim g_{α} = 1 and h is abelian. Let us compute the Killing form on h:

K(h_{1}, h_{2}) = tr_{g}(ad(h_{1})ad(h_{2})) = tr_{h}(ad(h_{1})ad(h_{2})) + ∑ tr_{g_{α}}(ad(h_{1})ad(h_{2}))

Thus, K(h_{1}, h_{2}) = ∑_{α} α(h_{1})α(h_{2}). On h^{∗}, K(λ, μ) = ∑_{α} K(λ, α)K(μ, α).

Theorem 4. (a) Δ span h^{∗} over 𝔽.

(b) K(α, β) ∈ ℚ if α, β ∈ Δ

(c) K on h^{∗}ℚ is a positive-definite, ℚ-valued, symmetric bilinear form.

Proof. (a) In the contrary case, there exists a nonzero h_{1} ∈ h such that α(h_{1}) = 0 for all α ∈ Δ. Hence [h_{1}, g_{α}] = 0, and [h_{1}, h] = 0 hence h_{1} is in the center which is not possible.

(b) Recall that 2K(α, β)K(α, α)^{−1} = p − q ∈ ℤ. But by the formula above K(α, α) = ∑ K(α, γ)^{2} thus, 4K(α, α)^{−1} = ∑_{γ} 2K(α, γ)K(α, α)^{−1} ∈ ℤ. The result follows from polarization identity.

(c) K(λ, λ), (λ ∈ h^{∗}) = ∑_{α} K(α, λ)^{2} ≥ 0 with equality iff all K(α, λ) = 0. Hence λ = 0 since K on h is nondegenerate.

Proposition. If g is semisimple and a ⊂ g is an ideal then K restricted to a is nondegenerate and g = a ⊕ a^{⊥}. Thus, a is semisimple.

Proof. K(a∩a^{⊥}, a∩a^{⊥}) = 0, hence by Cartan’s criterion a∩a^{⊥} is a solvable ideal. This contradicts the semisimplicity of g. Since dim(g) ≤ dim(a)+dim(a^{⊥}) the result follows.

Definition. A Lie algebra is simple if it is not abelian and has no proper nonzero ideals.

Corollary. g is semisimple iff it is a direct sum of simple Lie algebras.

Exercise 14.1 Show that decomposition is unique up to permutation and any ideal is a subsum of the ideals in decomposition. Solution. It suffices to prove the second statement. So let g = ⨁_{i} g_{i} where the g_{i} are simple. If a is an ideal a ∩ g_{i} is either zero or g_{i}. We need only establish that a is homogeneous, i.e. a = ⨁_{i} a ∩ g_{i}. If a has a nontrivial projection to g_{i} we have [g_{i}, a] = g_{i} since g_{i} has trivial center. But [g_{i}, a] ⊂ a.

Let g = g_{1} ⊕ g_{2} be a direct sum of semisimple Lie algebras. If h_{1}, h_{2} are Cartan subalgebras then h_{1} ⊕ h_{2} is a Cartan subalgebra. If g_{1} = h_{1} ⊕ ⨁_{α∈Δ_{1}} g_{α}, g_{2} = h_{2} ⊕ ⨁_{β∈Δ_{2}} g_{β} then g has root space Δ = Δ_{1} ⊔ Δ_{2} where we extend Δ_{i} by zero to the other Cartan subalgebra.

This decomposition has the property that: (*) If α ∈ Δ_{1}, β ∈ Δ_{2} then β + α ∉ Δ ∪ 0.

Definition. A set of roots is indecomposable if there is no nontrivial decomposition such that (*) holds. Clearly an indecomposable semisimple Lie algebra must be simple. Δ is indecomposable iff for any α, β ∈ Δ there exists a sequence α = γ_{1}, ..., γ_{n} = β such that γ_{i} + γ_{i+1} ∈ Δ or 0.

Example 1. g = sl_{n}(n ≥ 2), h = diagonal matrices with trace zero. Let ϵ_{i}(a_{jj}) = δ_{ij} and restrict ϵ_{i} to h. The eigenvectors for the Cartan subalgebra are E_{ij}. Note that the corresponding root is ϵ_{i} −ϵ_{j}. Hence we have root space decomposition sl_{n}𝔽 = h⊕⨁_{i,j} 𝔽E_{ij}. The root space ϵ_{i} −ϵ_{j} where i ≠ j is indecomposable. Indeed, (ϵ_{i} − ϵ_{j}) + (ϵ_{l} − ϵ_{k}) is a root or zero if j = l and (ϵ_{i} − ϵ_{j}) + (ϵ_{j} − ϵ_{l}), (ϵ_{j} − ϵ_{l}) + (ϵ_{l} − ϵ_{k}) are roots or zero otherwise.

Example 2. g = so_{n}(𝔽) = A ∈ gl_{n}(𝔽) such that A^{T} B+BA = 0. The best choice of B is B_{i,n+1−i} = 1 and zero otherwise.

Exercise 14.2 A ∈ so_{n}(𝔽) iff A′ = −A where A′_{ij} = A_{n−j+1,n−i+1}. Solution. We have (AB)_{ij} = A_{i,n+1−j} and (BA)_{ij} = A_{n+1−i,j}. From this it is easy to see that A^{T} B + BA = 0 iff A_{n+1−j,i} + A_{n+1−i,j} = 0. Make the substitution i′ = n + 1 − i. We have: A_{n+1−j,n+1−i′} + A_{i′,j} = 0 which is what we wanted.

In this case the Cartan subalgebra h is diag(a_{1}, ..., a_{r}, −a_{r}, ..., −a_{1}). The roots are F_{ij} = E_{ij} − E_{n+1−j,n+1−i} where the i, j are determined below. Observe that ϵ_{i} = −ϵ_{n+1−i} when restricted to h. F_{ij} is a root vector with root ϵ_{i} − ϵ_{j}. Note that a basis for the root system is formed by the vectors e_{1}, ..., e_{r} where either n = 2r or n = 2r + 1. In the case n = 2r + 1 the roots are {ϵ_{i} − ϵ_{j}, ϵ_{i}, −ϵ_{i}, ϵ_{i} + ϵ_{j}, −ϵ_{i} − ϵ_{j}} where i ≠ j and i, j ≤ r. This follows from the fact that e_{r+1} = 0 and that ϵ_{i} = −ϵ_{n+1−i}. The case n = 2r is almost identical. Here the roots are {ϵ_{i} − ϵ_{j}, ϵ_{i} + ϵ_{j}, −ϵ_{i} − ϵ_{j}} where i ≠ j and i, j ≤ r.

Exercise 14.3 Do the case r = 2n. Solution. See above.

Exercise 14.4 (a) Δ_{so_{2r+1}} is indecomposable iff r ≥ 1. Thus so_{2r+1} is simple iff r ≥ 1.

(b) Δ_{so_{2r}} is indecomposable iff r ≥ 3. Δ_{so_{4}} = {±(ϵ_{1} − ϵ_{2})} ∪ {±(ϵ_{1} + ϵ_{2})}, Thus, it is not simple. (so_{2} is abelian). Solution. (a) Notice that each ϵ_{i} is connected to ϵ_{j}. Note the obvious fact that roots are connected to their inverses. Furthermore, each ±ϵ_{i} ± ϵ_{j} is connected to an ±ϵ_{i} by choosing the sign of this last root judiciously.

(b) Note that ±ϵ_{i} + μϵ_{j} is connected to −μϵ_{j} ± ϵ_{k} where i, j, k are distinct and μ = ±1 (r ≥ 3). We have shown that one can change one index of any root without leaving the component. Thus we can connect all roots by changing one index at a time.