18.745 Introduction to Lie Algebras Fall 2004

Lecture 17 — November 5, 2004

Prof. Victor Kač Scribe: Karola Mészáros

Definition 1 Let (V, △) be an abstract root system. Let f be a linear function on V such that f(α) ≠ 0 for all α ∈ △. The set △_{+} (respectively △_{−}) is equal to {α ∈ △|f(α) > 0 (respectively, f(α) < 0)} is called the subset of positive (respectively, negative) roots.

Note that △_{−} = −△_{+} and △ = △_{+} ⊔ △_{−}.

A positive root is simple if it cannot be represented as a sum of two positive roots. Let Π ⊂ △_{+} denote the set of simple roots.

A short philosophical digression. It is often very important to break the symmetry of systems in order to get more information about them; although, ancient Greeks believed one should just observe the system one studies, and so they did not interfere with the systems they were studying, therefore, they also did not break their symmetry, which explains their slow progress in natural sciences. However, we used this symmetry breaking when classifying Lie algebras by choosing a Cartan subalgebra. Here again we are breaking the symmetry of △ by considering △_{+} and △_{−}.

The set Π is called indecomposable if it cannot be decomposed into a union of orthogonal non-empty subsets: Π = Π′ ∪ Π′′, (Π′, Π′′) = 0. A highest root is a root θ with maximal value of f(θ).

Exercise 17.1. Show that θ is unique in an indecomposable root system.

We shall show this after the following theorem.

Theorem 1 (a) If α, β ∈ Π, α ≠ β, then α−β ∉ △ and (α, β) ≤ 0. Also, (θ, α) ≥ 0 and θ+α ∉ △ for any α ∈ △_{+}.

(b) Any positive root is a linear combination of simple roots with nonnegative integer coefficients.

(c) If α ∈ △_{+} \ Π, then α − γ is a root for some γ ∈ Π. Moreover, α − γ ∈ △_{+}.

(d) Π is a basis of V over ℝ, and a basis of the root lattice Q over ℤ. Also, △_{+} ⊂ ℤ_{+}Π, and △_{−} ⊂ −ℤ_{+}Π.

(e) △ is indecomposable if and only if Π is indecomposable.

Proof. (a) Suppose the contrary, α − β = γ ∈ △. If γ ∈ △_{+}, then α = β + γ contradicts the simplicity of α. Similary, if −γ ∈ △_{+}, then β = α + (−γ) contradicts the simplicity of β. Next, by the string property we have that p − q = \frac{2(α,β)}{(β,β)}, but as α − β ∉ △ as we have just seen, it follows that p = 0, and thus (α, β) ≤ 0. Furthermore, it is clear that θ + α ∉ △ since f(θ) + f(α) = f(θ + α) > f(θ), and θ is the highest root. Finally, by the string property (θ, α) ≥ 0. This concludes the proof of (a).

(b) If α is a positive root that is simple, there is nothing to prove. If α ∈ △_{+} and α is not simple, then α = β + γ for some β, γ ∈ △_{+}. But, f(α) = f(β) + f(γ), hence, f(α) > f(β), f(γ) > 0, and applying induction on the finite ordered set {f(α)}_{α∈△_{+}} shows the statement of (b). This shows that Π spans △_{+} over ℤ_{+}.

(c) Supposing the opposite, we obtain (α, γ) ≤ 0 for all γ ∈ Π by the string property. Hence (α, α) = (α, ∑_{γ∈Π} k_{γ}γ) ≤ 0, as k_{γ} ∈ ℤ_{+} for all γ ∈ Π by (b) and (α, γ) ≤ 0 by our assumption. However, (α, α) ≤ 0 is impossible for α ∈ △_{+} \ Π, thus, α − γ ∈ △ ∪ {0}. Since α ∉ △ and γ ∈ Π, it follows that α − γ ∈ △. All we need to show now is that α − γ ∈ △_{+}. Indeed, if α − γ = β < 0, then γ = α + (−β), which contradicts the simplicity of γ. This shows (c).

(d) We already know by (b) that Π spans △_{+} over ℤ_{+}, and thus Π spans Q over ℤ, hence spans V over ℝ. Thus, it suffices to prove linear independence over ℝ (and then it will follow for ℤ). In the contrary case, there is a linear dependence between simple roots: a := ∑_{α∈Π′} k_{α}α = ∑_{β∈Π′′} k_{β}β =: b, k_{α}, k_{β} > 0, and Π′ ∩ Π′′ = ∅, Π′, Π′′ < Π and (Π′, Π′′) ≤ 0. Taking the inner product of both sides with a, we obtain: (a, a) = (a, b) ≤ 0 and since (a, a) ≥ 0 it follows that (a, a) = 0, thus, a = 0. However, then 0 = f(a) = ∑_{α∈Π′} k_{α}f(α) > 0 as k_{α} > 0 for all α ∈ Π′, therefore, Π′ = ∅. Similarly, Π′′ = ∅. Therefore there is no notrivial linear independence of simple roots.

(e) If △ is decomposable, i.e. △ = △′ ⊔ △′′, (△′, △′′) = 0, let Π′ (respectively Π′′) be simple roots in △′ (respectively △′′), so, Π = Π′ ∪ Π′′, (Π′, Π′′) = 0. Using (d) we conclude that Π′ and Π′′ are non-empty.

Conversely, if Π = Π′ ∪ Π′′ and (Π′, Π′′) = 0, we have to show that any α ∈ △_{+} is a linear combination of elements either all in Π′ or all in Π′′. Suppose the contrary: α ∈ △_{+} and α = α′+α′′, where α′ ∈ ℤ_{+}Π′ and α′′ ∈ ℤ_{+}Π′′. Using (c) we may substract simple roots until, say, α′ is simple. So we assume that α′ ∈ Π and α′′ ∈ ℤ_{+}Π′′. But, (α, α′) = (α′, α′) > 0, hence we can use the string property to conclude that α′′ ∈ △_{+}. Using the string property again, we get that α′ − α′′ ∈ △, which is impossible by (d). This conludes the proof of (e).

Proof of Ex.17.1. By Theorem 1 (b) we know that a highest root θ is of form θ = ∑_{α_{i}∈Π} k_{i}α_{i}, such that k_{i} ≥ 0. We now show that if the root system is indecomposable then k_{i} ≥ 1 for all simple roots. Suppose this is not the case, i.e. there exist k_{i}’s such that k_{i} = 0. Since θ is the highest root it is clear that there must also be k_{i}’s such that k_{i} ≠ 0. Let S be the span of all the α_{i}’s in Π such that k_{i} ≠ 0. Then, we show that for every β ∈ S, (β, α_{j}) = 0 for all α_{j} such that k_{j} = 0. Indeed, by Theorem 1 (a) we know that θ + α_{j} ∉ △, and by (b) we know that θ − α_{j} ∉ △ if k_{j} = 0 (since we would get mixed signs in front of the simple roots). Thus, by the string property, (θ, α_{j}) = 0, and on the other hand, (θ, α_{j}) = ∑ k_{i}(α_{i}, α_{j}), and by Theorem 1 (a), (α_{i}, α_{j}) ≤ 0. Therefore, the only way (θ, α_{j}) = 0 is if (α_{i}, α_{j}) = 0 for all α_{i} such that k_{i} ≠ 0 and α_{j} such that k_{j} = 0. Thus, for every β ∈ S, (β, α_{j}) = 0 for all α_{j} such that k_{j} = 0. Indeed, if S′ is the span of all α_{j} such that k_{j} = 0, then we can conclude that (S, S′) = 0, by assumption S and S′ are nonempty. Thus, by the second proposition in lecture 15 we conclude that the root system is decomposable, contrary to assumption. Therefore, in an indecomposable root system if a highest root θ = ∑_{α_{i}∈Π} k_{i}α_{i}, then, k_{i} ≥ 1.

Now, suppose that there are two different highest roots θ_{1} and θ_{2} in an indecomposable root system. It follows immediately that that θ_{1} + θ_{2} and θ_{1} − θ_{2} are not roots, since f(θ_{1} + θ_{2}) > f(θ_{1}), and f(θ_{1} − θ_{2}) = 0. Thus, by the string property we have 0 = (θ_{1}, θ_{2}). However, by the above we know that θ_{1} = ∑_{α_{i}∈Π} k_{i}α_{i}, with k_{i} ≥ 1, and θ_{2} = ∑_{α_{i}∈Π} l_{i}α_{i}, with l_{i} ≥ 1. Also, by Theorem 1 (a), (θ_{1}, α_{i}) ≥ 0 for all α_{i} ∈ Π, and, by (c) we can find an α_{n} such that θ_{1} − α_{n} ∈ △_{+} and since θ_{1} + α_{n} ∉ △_{+} as θ_{1} is a highest root, it follows by the string property that (θ_{1}, α_{n}) > 0. However, expanding (θ_{1}, θ_{2}) by expanding θ_{2} = ∑_{α_{i}∈Π} l_{i}α_{i}, with l_{i} ≥ 1, we obtain that (θ_{1}, θ_{2}) ≠ 0, thus obtaining that in an indecomposable root system there is a unique highest root.

Definition 2 Let (V, △) be an abstract root system, Π = {α_{1}, α_{2}, …, α_{r}} (r = dimV) a set of simple roots. Then, the matrix A = (\frac{2(α_{i},α_{j})}{(α_{i},α_{i})})_{i,j=1}^{r} is called the Cartan matrix of (V, △).

We shall later show that A = (\frac{2(α_{i},α_{j})}{(α_{i},α_{i})})_{i,j=1}^{r} is independent of the choice of f up to the reordering of the index set. Properties of Cartan matrices A = (a_{ij})_{i,j=1}^{r}, where a_{ij} = \frac{2(α_{i},α_{j})}{(α_{i},α_{i})}.

(1) a_{ii} = 2

(2) a_{ij} is nonpositive integer if i ≠ j. If a_{ij} ≠ 0 then a_{ji} ≠ 0.

(3) All principal minors of A are positive, in particular, detA > 0. Recall that a principal minor is a determinant of a submatrix of A obtained from A by deleting the rows indexed by i_{1}, i_{2}, …, i_{t} and deleting the columns indexed by i_{1}, i_{2}, …, i_{t}.

Proof. The statement of (1) is trivial, and the statement of (2) follows from part (a) of Theorem 1. To prove (3), note that A can be written as a product of a diagonal matrix with the (j, j)^{th} entry equal to \frac{2}{(α_{j},α_{j})} and a matrix B = ((α_{i}, α_{j}))_{i,j=1}^{r}, and B is a symmetric positive definite matrix. By Sylvester’s criterion all principal minors of B are positive, hence the same holds for A.

Definition 3 Let α_{0} = θ, where θ is the highest root, and let Ã = (\frac{2(α_{i},α_{j})}{(α_{i},α_{i})})_{i,j=0}^{r}.

Then, Ã satisfies (1) and (2) above, and detÃ = 0, since ((α_{i}, α_{i}))_{i,j=0}^{r} is the Gramm matrix of the linearly dependent set of vectors α_{0}, α_{1}, …, α_{r}. We wish to classify matrices with properties (1), (2), (3), called abstract Cartan matrices. For r = 1, A = (2), r = 2, A = ( \begin{matrix}2 & −a \\ −b & 2 \end{matrix} ), where either a = b = 0 or both a and b are non-negative integers. Since detA = 4 − ab > 0, it follows that up to the reordering of the index set A is one of the following matrices: ( \begin{matrix}2 & 0 \\ 0 & 2 \end{matrix} ) , ( \begin{matrix}2 & −1 \\ −1 & 2 \end{matrix} ) , ( \begin{matrix}2 & −1 \\ −2 & 2 \end{matrix} ) , ( \begin{matrix}2 & −1 \\ −3 & 2 \end{matrix} ). An abstract Cartan matrix A is depicted by a Dynkin diagram D(A) as follows. D(A) is a graph with r vertices such that the connection of the i^{th} vertex and the j^{th} vertex depends on the matrix ( \begin{matrix}a_{ii} & a_{ij} \\ a_{ji} & a_{jj} \end{matrix} ). For the four martices above, we use the following diagrams, respectively:

Obviously, D(A) is a connected graph if and only if Π is indecomposable.