18.745 Introduction to Lie Algebras Fall 2004

Lecture 19 — November 16, 2004

Prof. Victor Kač Scribe: Patrick Lam

We have been proving the following,

Theorem 1 (Classification theorem) Let 𝔤 be a finite-dimensional semisimple Lie algebra over an algebraically closed field of characteristic 0. Then 𝔤 is isomorphic to a direct sum of simple Lie algebras, and a complete and non-redundant list of the latter is as follows:

𝔰𝔩_{n}(𝔽) (n ≥ 2), 𝔰𝔬_{n}(𝔽) (n ≥ 7), 𝔰𝔭_{n}(𝔽) (n ≥ 4, even), E_{6}, E_{7}, E_{8}, F_{4}, G_{2}.

The strategy of the proof is given in the following diagram.

𝔤 ⟶^{choose 𝔥} Δ_{root system} = abstract root system ⟶^{choose f ∈ V^{∗}} Π ⟶ A_{Cartan matrix} = abstract
↑  Cartan
𝔤(A) ⟵ matrix A

First construct the following set of generators E_{i}, F_{i}, H_{i} (i = 1, …, r = dim 𝔥 = rank (𝔤)) as follows: let α_{i} be a simple root, and choose E_{i} ∈ 𝔤_{α_{i}}, F_{i} ∈ 𝔤_{−α_{i}} such that [E_{i}, F_{i}] = \frac{2ν^{−1}(α_{i})}{K(α_{i},α_{j})} (recall that [𝔤_{α_{i}}, 𝔤_{−α_{i}}] = 𝔽ν^{−1}(α_{i}), 𝔤_{α_{i}} = 𝔽E_{i}, 𝔤_{−α_{i}} = 𝔽F_{i}.) Then we have:

[H_{i}, H_{j}] = 0, [H_{i}, E_{j}] = a_{ij}E_{j}, [H_{i}, F_{j}] = −a_{ij}F_{j}, [E_{i}, F_{j}] = δ_{ij}H_{i}  (∗)

where a_{ij} := \frac{2K(α_{i},α_{j})}{K(α_{i},α_{i})} are the entries of the Cartan matrix A = (a_{ij}).

To check these: [H_{i}, E_{j}] = α_{j}(H_{i})E_{j} = \frac{2α_{j}(ν^{−1}(α_{i}))}{K(α_{i},α_{i})} = \frac{2K(α_{j},α_{i})}{K(α_{i},α_{i})} = a_{ij}; [H_{i}, F_{j}] = −a_{ij}F_{j} is clear; [E_{i}, F_{i}] has been checked. [E_{i}, F_{j}] = 0 because it belongs to 𝔤_{α_{i}−α_{j}}, so it is not a root by part (a) of the Theorem in Lecture 17.

Next, we denote by 𝔫_{+} (respectively 𝔫_{−}) the subalgebra of 𝔤 generated by all E_{i}s (respectively F_{i}s). Then

𝔫_{+} = ⨁_{α∈Δ_{+}} 𝔤_{α}, 𝔫_{−} = ⨁_{α∈−Δ_{+}} 𝔤_{α}.

Indeed, let α ∈ Δ_{+} \ Π. Then by part (c) of the Theorem from Lecture 17, α − α_{i} ∈ Δ_{+} for some simple root α_{i}, hence 𝔤_{α} = [𝔤_{α−α_{i}}, E_{i}] since dim 𝔤_{α} = 1. [𝔤_{α}, 𝔤_{β}] = 𝔤_{α+β} if α, β, α + β ∈ Δ by part (d) of Theorem 3 from Lecture 13.

This proves that all 𝔤_{α}, with α ∈ Δ_{+}, are in 𝔫_{+}. Likewise all 𝔤_{−α}(α ∈ Δ_{+}) are in 𝔫_{−}. Note that we have the so-called triangular decomposition 𝔤 = 𝔫_{−} ⊕ 𝔥 ⊕ 𝔫_{+} (a direct sum as vector spaces).

Exercise 19.1. (a) Show that for 𝔰𝔩_{n}, 𝔫_{±} = strictly upper lower triangular matrices, if 𝔥 = diagonal matrices, and if f chosen as we did. (b) Find the triangular decompositions for 𝔰𝔬_{n} and 𝔰𝔭_{n}.

Solution. (a) In Lecture 14, we noted that

𝔰𝔩_{n}(𝔽) = 𝔥 ⊕ ( ⨁_{\begin{matrix}i, j = 1 \\ i ≠ j \end{matrix}}^{n} 𝔽E_{ij}) ,

where 𝔥 was the diagonal matrices, and 𝔽E_{ij} is the root space attached to the root ε_{i} −ε_{j}. We need to distinguish the positive roots from the negative roots; the f we have chosen gives positive roots for each E_{ij} where i > j. The direct sum of the spaces 𝔽E_{ij} where i > j is indeed the desired space 𝔫_{+} of strictly upper triangular matrices, and 𝔫_{−} is the space of strictly lower triangular matrices, showing the triangular decomposition.

(b) For 𝔰𝔬_{n}(𝔽), we have

𝔥 = \{( \begin{matrix}a_{1} & & & & & & 0 \\ & a_{2} \\ & & . & . & . \\ & & & & & −a_{2} \\ 0 & & & & & & −a_{1} \end{matrix} )\} .

and 𝔰𝔬_{n}(𝔽) = 𝔥 ⊕ (⨁_{i,j} 𝔽F_{ij}), where F_{ij} = E_{ij} − E_{n+1−j,n+1−i}. Then the positive roots are ε_{i} ± ε_{j}(i < j), ε_{i}. Hence the space 𝔫_{+} is the set of matrices A such that A′ = −A which are strictly upper-triangular, while the 𝔫_{−} is the set of matrices A such that A′ = −A which are strictly lower-triangular.

For 𝔰𝔭_{n}(𝔽), we have the same Cartan subalgebra 𝔥; this time, the F_{ij} = E_{ij} − E_{n+1−j,n+1−i} for 1 ≤ i, j ≤ r(i ≠ j) and F_{ij} = E_{ij} + E_{n+1−j,n+1−i} (1 ≤ i ≤ r, r + 1 ≤ j ≤ n. The corresponding root vectors are Δ_{𝔰𝔭_{2r}} = {ε_{i} −ε_{j}(i, j = 1, …, r, i ≠ j), ε_{i} +ε_{j}, −ε_{i} −ε_{j}(i, j = 1, …, r)}; the positive roots are now ε_{i} ± ε_{j}(i < j) and 2ε_{i} if n is even. Finally, 𝔫_{+} is the set of upper triangular matrices (in

block form) ( \begin{matrix}a & b \\ c & d \end{matrix} ) such that d = −a′, b = b′, c = c′, where ′ denotes transposition with respect

to the antidiagonal, and 𝔫_{−} is the set of lower triangular matrices which satisfy this condition.

Remark 1 Any non-zero ideal of 𝔤 has a non-zero intersection with 𝔥.

Lemma 1 Let 𝔥 be a finite-dimensional abelian Lie algebra, and let π be a diagonalizable representation of 𝔥 in a vector space V (not necessarily finite-dimensional), i.e. V = ⨁_{λ∈𝔥^{∗}} V_{α}, where V_{λ} = {v ∈ V | π(h)v = λ(h)v, h ∈ 𝔥}. Then for any π(𝔥)-invariant subspace U ⊂ V, we have U = ⨁_{λ∈𝔥^{∗}}(U ∩ V_{λ}).

Proof. Take u ∈ U and write u = ∑_{λ} v_{λ}, where v_{λ} ∈ V_{λ}. Take h ∈ 𝔥 such that λ_{1}(h) ≠ λ_{2}(h) (scale so that λ_{1}(h) = 1.) Then U ∋ π(h) = λ_{1}(h)v_{λ_{1}} + λ_{2}(h)v_{λ_{2}} + · · ·. Hence

U − π(h)u = (1 − λ_{2}(h))v_{λ_{2}} + (1 − λ_{3}(h))v_{λ_{3}} + ⋯,

and we may apply induction on the number of summands in u. □

Proof of the Remark. If I is a non-zero ideal of 𝔤 which intersects 𝔥 trivially, then by the lemma, 𝔤_{α} ⊂ I for some root α. But then [𝔤_{α}, g_{−α}] = 𝔽ν^{−1}(α) ⊂ I, a contradiction. □

Let 𝔤̃(A) be the Lie algebra on generators E_{i}, F_{i}, H_{i} (i = 1, …, r) subject to relations (*).

Exercise 19.2. Show that 𝔤̃((2)) = 𝔰𝔩_{2}(𝔽) but 𝔤̃ (( \begin{matrix}2 & −1 \\ −1 & 2 \end{matrix} )) is infinite dimensional. Find the

elements of J ( \begin{matrix}2 & −1 \\ −1 & 2 \end{matrix} ).

Solution. For the case 𝔤̃((2)) = 𝔰𝔩_{2}(𝔽), we have r = 1, giving the generators E_{1}, F_{1}, H_{1} with the relationships [E_{1}, F_{1}] = H_{1}; [H_{1}, H_{1}] = 0; [H_{1}, E_{1}] = 2E_{1}; [H_{1}, F_{1}] = −2F_{1}. Let

H_{1} = ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ); E_{1} = ( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ); F_{1} = ( \begin{matrix}0 & 0 \\ 1 & 0 \end{matrix} );

simple calculations show that these relationships are satisfied, and these matrices generate 𝔰𝔩_{2}(𝔽).

To show that 𝔤̃ (( \begin{matrix}2 & −1 \\ −1 & 2 \end{matrix} )) is infinite-dimensional, we note that 𝔫̃_{+} and 𝔫̃_{−} are freely generated

over the generators E_{1}, E_{2}, which is an infinite-dimensional space. Since 𝔤̃ = 𝔫̃_{+} ⊕ 𝔥̃ ⊕ 𝔫̃_{−}, it is also infinite-dimensional.

The maximal ideal J ( \begin{matrix}2 & −1 \\ −1 & 2 \end{matrix} ) is the ideal containing the inverse images of the triple commu-

tators E_{1}E_{2}E_{1} and E_{2}E_{1}E_{2}, as well as F_{1}F_{2}F_{1} and F_{2}F_{1}F_{2}. □

Lemma 2 (a) 𝔤̃(A) = 𝔫̃_{−} ⊕ 𝔥 ⊕ 𝔫̃_{+}, where 𝔫̃_{+} (resp. 𝔫̃_{−}) is generated by the E_{i}s (resp. F_{i}s) and 𝔥 = span of H_{i}s. (b) 𝔤̃(A) has a maximal ideal J (A) among ideals intersecting 𝔥 trivially.

Proof. (Exercise 19.3.) First we show by induction on n that any commutator of length n:

[[b_{i_{1}}, b_{i_{2}}], …, b_{i_{n}}] (where b_{i_{k}} ∈ {E_{i}, F_{i}, H_{i}}),

lie in either 𝔫_{+} or in 𝔫_{−} or in 𝔥.

Solution. The base case is obvious. Now assume b = [[b_{i_{1}}, b_{i_{2}}], …, b_{i_{n}}] is in 𝔫_{+}, 𝔫_{−} or 𝔥; we show that b′ = [[[b_{i_{1}}, b_{i_{2}}], …, b_{i_{n}}], b_{i_{n+1}}] is in 𝔫_{+}, 𝔫_{−} or 𝔥. The following table summarizes the possible values of the n + 1-length commutator b′, ignoring irrelevant constant factors in front of the b.

\begin{matrix}[figure: b; b_{i_{n+1}}] & E_{i} & F_{i} & H_{i} \\ E_{j} & 0 & δ_{ji}H_{j} & −a_{ij}E_{j} \\ F_{j} & −δ_{ij}H_{i} & 0 & −a_{ij}F_{j} \\ H_{j} & a_{ji}E_{i} & −a_{ji}F_{i} & 0 \end{matrix}

Recalling that 𝔫_{+} is defined to be the subalgebra generated by the E_{i}, that 𝔫_{−} is generated by the F_{i} and that 𝔥 is generated by the H_{i}, we observe that b′, the n + 1-length commutator, always belongs to one of 𝔫_{+}, 𝔫_{−} or 𝔥 if all the b_{i_{k}}s are E_{i}, F_{i} or H_{i}. □

Hence 𝔤̃(A) = 𝔫̃_{−} + 𝔥 + 𝔫̃_{+}. This sum is direct since 𝔫̃_{+} = ⨁ 𝔤̃(A)_{α}, α = ∑_{i=1}^{n} k_{i}α_{ij}, k_{i} ∈ ℤ_{i}, not all 0, and similarly for 𝔫̃_{−}, but 𝔥 = 𝔤̃(A)_{0}. (This proves (a).)

But by Lemma 1, any ideal I of 𝔤̃(A) is of the form I = I_{−} ⊕ I_{0} ⊕ I_{+}, where I_{±} ⊂ 𝔫̃_{±}, I_{0} ⊂ 𝔥. So if I ∩ 𝔥 = 0, then I = I_{−} ⊕ I_{+}, I_{±} ⊂ 𝔫̃_{±}. Hence (b) follows, since J (A), as a sum of all ideals intersecting 𝔥 trivally, is a proper maximal ideal. We let 𝔤(A) = 𝔤̃(A)/J (A). Now consider the surjective homomorphism 𝔤̃(A) →^{φ} 𝔤 defined by φ(E_{i}) = E_{i}, φ(F_{i}) = F_{i}, φ(H_{i}) = H_{i} (which generate 𝔤). Moreover, J (A) ⊂ ker φ, otherwise φ(J (A)) contradicts Remark 1, and furthermore, J (A) = ker φ since J (A) is maximal. Hence we have an isomorphism 𝔤(A) ˜→𝔤. (Note that φ is mapping the infinite-dimensional 𝔫_{+} and 𝔫_{−} to finite-dimensional 𝔫̃_{+} and 𝔫̃_{−}). □

So, if we know that 𝔤 with a given root system exists, like 𝔤 = 𝔰𝔩_{n}, 𝔰𝔬_{n}, or 𝔰𝔭_{n}, then 𝔤(A) = 𝔤 is the Lie algebra with this root system. We have proved that any simple Lie algebra over 𝔽 is isomorphic to 𝔰𝔩_{n}(𝔽), 𝔰𝔬_{n}(𝔽), 𝔰𝔭_{n}(𝔽), or possibly the simple Lie algebras whose root systems are Δ_{E_{i}}, i = 6, 7, 8, Δ_{F_{4}}, Δ_{G_{2}}, called the exceptional Lie algebras. So it remains to prove the existence of the latter. Note that we have the same Cartan matrices in the following cases:

A_{1} = B_{1} = C_{1} (1-d root systems are identical)
B_{2} = C_{2}  [figure: ◯; ◯; ◯; ◯]
D_{3} = A_{3}  ◯  ◯  ◯
D_{2} = A_{1} ⊕ A_{1}

or, on the level of Lie algebras, 𝔰𝔩_{2}(𝔽) ≃ 𝔰𝔬_{3}(𝔽) ≃ 𝔰𝔭_{2}(𝔽); 𝔰𝔬_{5}(𝔽) ≃ 𝔰𝔭_{4}(𝔽); 𝔰𝔬_{6}(𝔽) ≃ 𝔰𝔩_{4}(𝔽); 𝔰𝔬_{4} ≃ 𝔰𝔩_{2}(𝔽) ⊕ 𝔰𝔩_{2}(𝔽).

Now we want to construct the simple Lie algebras from their root systems. (We will carry out an explicit construction; this construction doesn’t show uniqueness.)

First we consider the simply-laced case A = A^{T}, i.e. all roots have the same length. Let (V, Δ) be a simply-laced root system, with (·, ·) such that (α, α) = 2 for any root α, and let Q = ℤΔ be the root lattice. Then Δ = {α ∈ Q | (α, α) = 2}, in all examples A_{r}, D_{r}, E_{6}, E_{7}, E_{8}; the first two are known to exist, whereas the last three are new.

Consider the following space over 𝔽: 𝔤 = 𝔥 ⊕ (⨁_{α∈Δ} 𝔽E_{α}), where 𝔥 = 𝔽 ⊗_{ℚ} (ℚΔ) and E_{α} are some non-zero vectors. Extend the bilinear form from (·, ·) on V to 𝔥 by bilinearity. Define brackets on 𝔤 as follows:

(i) [h, h′] = 0 for h, h′ ∈ 𝔥.

(ii) [h, E_{α}] = (h, α)E_{α}

(iii) [E_{α}, E_{−α}] = −α (for convenience)

(iv) [E_{α}, E_{β}] = 0 if α + β ∉ Δ ∪ {0}

(v) [E_{α}, E_{β}] = ε(α, β)E_{α+β} if α + β ∈ Δ.

Next time, we will study how to construct ε so that the Jacobi identity holds. To do this, we have to check the Jacobi identities of any triple of distinct basis elements of 𝔤:

(i) If this triple is h, h′, h′′ ∈ 𝔥, the identity is obvious. (ii) If h, h′, E_{α} or h, E_{α}, E_{β}, the Jacobi identity holds for any choice of ε(α, β) (Exercise 19.4). (iii) (next time) If E_{α}, E_{β}, E_{γ}, we will have to choose ε appropriately.

Solution. A straightforward calculation shows the Jacobi identity:

[h, [h′, E_{α}]] + [h′, [E_{α}, h]] + [E_{α}, [h, h′]] = [h, [h′, E_{α}]] + [h′, [E_{α}, h]]
= [h, (h′, α)E_{α}] + [h′, −(h, α)E_{α}]
= (h′, α)[h, E_{α}] − (h, α)[h′, E_{α}]
= (h′, α)(h, α)E_{α} − (h, α)(h′, α)E_{α}
= 0

and

[h, [E_{α}, E_{β}]] + [E_{α}, [E_{β}, h]] + [E_{β}, [h, E_{α}]]
= (α + β, h)[E_{α}, E_{β}] − (β, h)[E_{α}, E_{β}] + (α, h)[E_{β}, E_{α}]
= (α + β, h)[E_{α}, E_{β}] − (β, h)[E_{α}, E_{β}] − (α, h)[E_{α}, E_{β}]
= 0.

□