13 Roots of a semisimple Lie algebra

Recall that the Killing form K gives a non-degenerate bilinear pairing 𝔤_{α} × 𝔤_{−α} → 𝔽. In addition, [e, f] = K(e, f)ν^{−1}(α) for any e ∈ 𝔤_{α}, f ∈ 𝔤_{−α}. Also, K(α, α) ≠ 0 for any α ∈ Δ.

For each α ∈ Δ, choose a non-zero E ∈ 𝔤_{α} and an F ∈ 𝔤_{−α} such that K(E, F) = \frac{2}{K(α,α)}. Let H = \frac{2ν^{−1}(α)}{K(α,α)}. Then [E, F] = H, [E, H] = −α(H)E = −2E and [F, H] = α(H)F = 2F (as α(ν^{−1}(α)) = K(α, α)). So the span of E, F and H is isomorphic to sl_{2}(𝔽).

Lemma 13.1 (Basic lemma for sl_{2}) Let π be a representation of sl_{2}(𝔽) in a vector space V, and v ∈ V be such that π(E)v = 0 and π(H)v = λv, λ ∈ 𝔽. Then

1. π(H)π(F)^{n}v = (λ − 2n)π(F)^{n}v

2. π(E)π(F)^{n}v = n(λ − n + 1)π(F)^{(n−1)}v

3. If V is finite-dimensional, then λ ∈ ℤ_{+}, the vectors π(F)^{j}v (0 ≤ j ≤ λ) are linearly independent, and π(F)^{(λ+1)}v=0.

Proof. (1) and (2) are proved by induction on n; we leave the details to the reader. For (3), notice that if λ − n + 1 were non-zero for all integer n ≥ 1, then (2) would imply, by induction, that π(F)^{n}v ≠ 0 for any n ∈ ℤ_{+}. In this case (1) would mean that π(H) has infinitely many eigenvalues, which is impossible in a finite-dimensional space. Therefore, λ is a non-negative integer.

Next, it follows from (2) by induction that π(F)^{j}v ≠ 0 for j ≤ λ. Then (1) says that π(F)^{j}v is an eigenvector of π(H) that corresponds to the eigenvalue λ − 2j; therefore v, π(F)v, π(F)^{2}v, … π(F)^{λ}(v) must be linearly independent. Finally, π(F)^{λ+1}(j) must be zero, because otherwise (2) would imply, by induction, that π(F)^{n}v is non-zero for infinitely many values of n, and π(H) would have infinitely many eigenvalues, which cannot possibly happen. □

Exercise 13.1 Prove parts (1) and (2) in the basic lemma for sl_{2}(𝔽).

Solution. Not surprisingly, we use induction on n. Let’s start with part (1). The case n = 0 is given. Assume this is true for some n. Then for n + 1,

π(H)π(F)^{n+1}v = π(F)π(H)π(F)^{n}v − π([F, H])π(F)^{n}v =

= π(F)((λ − 2n)π(F)^{n}v) − 2π(F)π(F)^{n}v = (λ − 2(n + 1))π(F)^{n+1}v

so, by induction, the statement is true for all n.

In part (2), as much as we’d like to, we cannot begin with n = 0, because π(F) may not be invertible. So let’s use n = 1 as the inductive base:

π(E)π(F)v = π(F)π(E)v − π([F, E])v = π(H)v = λv

because π(E)v = 0, π(H)v = λv and [E, F] = H.

So for n = 1, this is true. If it is true for some n, then, for n + 1, we write

π(E)π(F)^{n+1}v = π(F)π(E)π(F)^{n}(v) − π([F, E])π(F)^{n}(v) =
= π(F)(n(λ − n + 1)π(F^{n−1})v) + π(H)π(F)^{n}(v) =

= (n(λ − n + 1) + (λ − 2n))π(F)^{n}v = (n + 1)(λ − (n + 1) + 1)π(F)^{n}v

so, by induction, this statement is also true for all n.

Exercise 13.2 Assume π is a representation of sl_{2}(𝔽) in a vector space V, and v ∈ V is such that π(F)v = 0 and π(H)v = λv. Prove that π(H)π(E)^{n}v = (λ + 2n)π(E)^{n}v, that π(F)π(E)^{n}v = −n(λ + n − 1)π(E)^{(n−1)}v, and that if V is given to be finite-dimensional, then −λ ∈ ℤ_{+}, the vectors π(E)^{j}v, 0 ≤ j ≤ −λ, are linearly independent, and π(E)^{−λ+1}v = 0.

Solution. Rather than rewrite the proof of the basic lemma for sl_{2} with minor changes (which is possible), we’ll reduce this fact to the basic lemma. Namely, consider the isomorphism φ : sl_{2} → sl_{2} for which

φ(E) = F, φ(F) = E, φ(H) = −H

One checks easily that φ is a Lie algebra isomorphism. Therefore, the representation π′ of sl_{2} in 𝔤 defined by π′A(u) = π(φ(A))u is indeed a representation. Obviously, π′ satisfies the assumptions of the basic lemma, except that π′(H)v = −λv. Replacing all λ-s by −λ-s in the conclusions of the basic lemma, we see that, first, π′(H)π′(F)^{n}v = (−λ − 2n)π′(F)^{n}v. Second, π′(E)π′(F)^{n}v = n(−λ−n+1)π′(F)^{(n−1)}v. And third, if V is finite-dimensional, then −λ ∈ ℤ_{+}, the vectors π′(F)^{j}v (0 ≤ j ≤ −λ) are linearly independent, and π′(F)^{(−λ+1)}v=0. All that’s left is to replace π′(E) by π(F), π′(F) by π(E), and π′(H) by −π(H). Once we do that, the last three statemenbs become the three statements we need.

Theorem 13.2 In the above notation,

1. dim(𝔤_{α}) = 1 if α ∈ Δ.

2. If α, β ∈ Δ, then {β + nα : n ∈ ℤ} ∩ (Δ ∪ {0}) is a finite connected string. I.e. it is {β − pα, β − (p − 1)α, … β + (q − 1)α, β + qα}, where p, q ∈ ℤ_{+} and p − q = \frac{2K(α,β)}{K(α,α)}. In particular, \frac{2K(α,β)}{K(α,α)} ∈ ℤ.

3. If α, β, (α + β) ∈ Δ, then [𝔤_{α}, 𝔤_{β}] = 𝔤_{α+β}

4. If α ∈ Δ, then nα ∈ Δ iff n = ±1.

Proof. Part (1) is proved by contradiction. Assume dim 𝔤_{α} > 1. Consider the subalgebra 𝔞_{α} = 𝔽E ⊕𝔽F ⊕𝔽H constructed above, where E ∈ 𝔤_{α}, F ∈ 𝔤_{−α}, H = \frac{2ν^{−1}(α)}{K(α,α)} ∈ 𝔥 and [E, F] = H, [E, H] = −2E, [F, H] = 2F. Since 𝔤_{α} and 𝔤_{−α} are non-degenerately paired by K, dim 𝔤_{−α} > 1. Therefore, there is a nonzero v ∈ 𝔤_{−α} for which K(E, v) = 0. For that v, [E, v] = K(E, v)ν^{−1}(α) = 0, and [H, v] = −α(H)v = −2v. Observe that 𝔞_{α} is isomorphic to sl_{2}(𝔽), and it is represented in 𝔤 by the adjoint representation. Further, 𝔤 is finite-dimensional and contains a non-zero v in 𝔤 for which ad E(v) = 0 and ad H(v) = −2v. This contradicts part 3 of the basic lemma for sl_{2}, therefore dim 𝔤_{α} = 1.

We proceed to part (2). Let q be the largest non-negative integer for which β + qα ∈ Δ ∪ {0}; it must exist, because Δ is a finite set. Pick a non-zero v ∈ 𝔤_{β+qα}, and again consider the adjoint representation of 𝔞_{α} in 𝔤. Then ad E(v) ∈ 𝔤_{β+(q+1)α} = {0}, i.e. ad E(v) = 0. Also,

ad H(v) = [H, v] = ((β + qα)(H))v = (\frac{2K(β, α)}{K(α, α)} + 2q) v

so, once again, the basic lemma for sl_{2} applies. It tells us that \frac{2K(β,α)}{K(α,α)} + 2q

lies in ℤ_{+}, and moreover, β + qα, β + (q − 1)α, … β + qα − 2 (\frac{K(β,α)}{K(α,α)} + q) all

lie in Δ ∪ \{0\}, because \frac{2K(β,α)}{K(α,α)} + 2q, \frac{2K(β,α)}{K(α,α)} + 2q − 2, … − \frac{2K(β,α)}{K(α,α)} − 2q are eigenvalues of ad H.

Denote \frac{2K(β,α)}{K(α,α)} + q by p, and let p′ be the largest non-negative integer for which β − pα ∈ Δ ∪ {0}. Choose a non-zero v′ ∈ 𝔤_{β−p′α}. Then ad F (v′) = 0 (because ad F (v′) ∈ 𝔤_{β−(p+1)α} = \{0\}), and ad H(v′) = (\frac{2K(β,α)}{K(α,α)} − 2p′) v′. Applying the second version of the basic lemma (Exercise 13.2), we conclude that 2p′ −2\frac{K(β,α)}{K(α,α)} ∈ ℤ_{+}, and β −p′α, β −(p′ −1)α, … β −p′α+2 (−\frac{K(β,α)}{K(α,α)} + p′) all

lie in Δ∪\{0\}. Denote −\frac{2K(β,α)}{K(α,α)} +p′ by q′. Since p′ and q are the largest possible,

we have p′ ≥ p and q ≥ q′. But on the other hand, p′ − q′ = p − q = \frac{2K(α,β)}{K(α,α)}. So both inequalities must be equalities, i.e. p′ = p and q′ = q. Therefore, {β + nα : n ∈ ℤ_{+}} ∩ (Δ ∪ {0}) = {β − pα, β − (p − 1)α, … β + qα} as claimed.

Part (3) follows from this easily. Let α, β ∈ Δ, and let p be the maximum non-negative integer for which β − pα ∈ Δ. As before, if v ≠ 0 is an element of 𝔤_{α−pβ}, then ad(F)v = 0 and ad(H)v = (\frac{2K(β,α)}{K(α,α)} − 2p) v. So by the second

version of the basic lemma (Exercise 13.2), ad(E)^{j}v ≠ 0 if 0 ≤ j ≤ 2p − \frac{2K(β,α)}{K(α,α)} = p + q. Also, q ≥ 1, because α + β ∈ Δ. Therefore 0 ≠ ad(E)^{p}v ∈ 𝔤_{β}, 0 ≠ ad(E)^{p+1}v ∈ 𝔤_{α+β}, and of course, 0 ≠ E ∈ 𝔤_{α}. Hence [𝔤_{α}, 𝔤_{β}] = 𝔤_{α+β}, because each of these subspaces is one-dimensional.

Finally, (4) is a consequence of the above. Let β = nα ∈ Δ. Then, by (1), \frac{2K(β,α)}{K(β,β)} ∈ ℤ, i.e. \frac{2}{n} ∈ ℤ. So all we need to show is that n can’t be 2 (the same result for −α will then imply that n can’t be −2). However, 𝔤_{2α} = [𝔤_{α}, 𝔤_{α}] by (3) and [𝔤_{α}, 𝔤_{α}] = 0 by (1). Thus 2α is not in Δ, and n can’t be 2. □