MATH 223A NOTES 2011 LIE ALGEBRAS 35

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1 MATH 3A NOTES 011 LIE ALGEBRAS Abstract root systems We now attempt to reconstruct the Lie algebra based only on the information given by the set of roots Φ which is embedded in Euclidean space E Definition. Any finite subset Φ of Euclidean space E satisfying the conditions of Theorem 8.5. will be called a root system. To repeat: Definition A root system is defined to be a subset Φ of standard Euclidean space E (i.e., finite dimensional real vector space with a positive definite inner product (, )) satisfying the following. (1) Φ is finite, spans E and 0 / Φ. () If Φ then Φ and not other real multiple of is in Φ. (3) If, β Φ then β, := (β,) (, ) Z (4) If, β Φ then β β, Φ Note that, =. Proposition Let σ : E E be the linear mapping given by σ (x) =x x, Then σ is reflection through the hyperplane perpendicular to. Proof. σ sends to and leaves fixed every vector perpendicular to. Condition (4) says that σ (Φ) = Φ for all Φ. Definition The Weyl group W is defined to be the subgroup of GL(E) generated by the reflections σ for all Φ. β σ (β) =β ( 3) Figure 1. In this example, β, = 3 and σ (β) =β +3 Φ. We will prove later that the vectors in the middle: β +, β + are also in Φ. Example (1) A n 1 = {± i 1 i n} E = R n is a root system. σ i sends i to i and leaves the other unit vectors j fixed. The matrix of this reflection map is the diagonal matrix with 1 in the ith position with other diagonal entries equal to 1. Therefore, the Weyl group is the group of diagonal matrices with ±1 on the diagonal: W = Z n = Z Z Z. (Z = Z/Z)

2 36 MATH 3A NOTES 011 LIE ALGEBRAS () A n 1 = {±( i j ) 1 i<j n} (3) is a root system in E = {x R n x i =0}. For = i j, the reflection σ switches i and j and leaves the other k fixed. This clearly sends the set A n 1 to itself. The reflections are transpositions. So, the Weyl group is the symmetric group S n. D n = {±( i ± j ) 1 i<j n} is a root system. For = i + j, the reflection σ switches i and j (and i j ) and leaves the other k fixed. This clearly sends the set D n to itself. One can show that the Weyl group is the group of signed permutations with an even number of negative signs. Exercise Find an explicit isomorphism D 3 = A Two roots. Lemma Suppose that, β Φ and β = ±. Then, ββ, =0, 1, or 3. Proof. This follows from the Schwartz inequality:, ββ, = where equality holds iff, β are collinear. 4(, β)(β,) (β,β)(, ) = 4(, β) β 4 This is also equal to 4 cos θ where θ is the angle between, β. cos θ = ± cos θ = θ or π θ = π π π π If the product, ββ, is 0 then, β are perpendicular and σ (β) =β. This is illustrated in the first example above. If the product of, β, β, is nonzero, one of them must be ±1. By symmetry assume, β = ±1. Then β, = ±1, ±, ±3. β, = β,, β = (β,)/ (β,)/ β = β =1,, 3 This looks like 6 cases, but this reduces to 3 cases with the following observation. Let γ = σ (β) =β β, Then γ, = β, β, = β, So, by replacing β with γ if necessary, we may assume that β, is negative.

3 MATH 3A NOTES 011 LIE ALGEBRAS root system A. Suppose that β, = 1. Then = β and θ =π/3. Also σ (β) =β β, = β + β β β + β This is the root system of sl(3,f). Also, it is a special case of example above with n = 3. The correspondence is with = β = and = 1, β = 3 ( + β = 1 3 ) β, = (β,) (, ) = = 1 In terms of H, the Cartan subalgebra of L = sl(3,f), i H is defined by i (h) =h i. So, which agrees with the earlier terminology. (h) = 1 (h) (h) =h 1 h Exercise 9... Generalize this correspondenc to show that Example is the root system for sl(n, F ). An animation of A 3 = D 3, the root system of sl(4,f) is on the webpage. In that rotating figure, the green arrow is = 1, the red arrow is β = 3 and the blue arrow is γ = 3 4. The angles are:, β =, β,γ =, θ =π/3 θ =π/3, γ =0, θ = π/ The white arrow is the sum of these three roots: + β + γ = 1 4. Note that all roots have the same length:.

4 38 MATH 3A NOTES 011 LIE ALGEBRAS 9... root system B = C. Suppose that β, =. Then β = and θ = 3π/4. Also σ (β) =β β, = β + β β + β + β β β This root system has short roots of length 1 and long roots of length. Exercise Show that the 6 vectors in R 3 given by ± i (6vectors), ± i ± j (1vectors), ± 1 ± ± 3 (8vectors) do not form a root system. (These vectors form a cube just as the figure above forms a square.) Definition The root system B n in R n is defined to be the union of the set of n short roots ± i and the 4 n long roots ±i ± j. The long roots of B n form a subsystem D n as in Example (3) and the short roots form A n 1 as in Example (1). For example, for n = 3, there are 6 short roots and 1 long roots and these long roots form D 3 = A 3. Definition The root system C n in R n is defined to be the union of the set of n long roots ± i and the 4 n short roots ±i ± j. The short roots of C n form a subsystem D n.forn = 3 there are 1 short roots and 6 long roots. The case B 3,C 3 can be seen in these animations: B 3, C 3. The short roots are red and the long roots are blue in both root system G. Suppose that β, = 3. Then β = 3 and θ =5π/6. Also σ (β) =β β, = β +3

5 MATH 3A NOTES 011 LIE ALGEBRAS 39 β +3 β β + β + β +3 β 3 β β β β root system A 1 A 1. When β, = 0 the two roots are perpendicular. They could have any length. β This is the root system of sl(,f) sl(,f) Irreducible root systems. β Definition We say that a root system Φ decomposes if it is a disjoint union Φ 1 Φ of two nonempty subsets so that every root in Φ 1 is perpendicular to every root of Φ. We say that Φ is irreducible if there is no such decomposition. If Φ decomposes, then E also decomposes as an orthogonal direct sum E = E 1 E where E i is the span of Φ i.eachφ i E i is a root system. Exercise Show that the root system of a product L = L 1 L of two semisimple Lie algebras decomposes as the union of the root systems of L 1,L. Conversely, any decomposition of the root system of L comes from such a factorization of L. Proposition The decomposition of a root system Φ into irreducible components is unique. Proof. If Φ = Φ i and Φ = Ψ j are any two decompositions then so is Φ = Φ i Ψ j.

6 40 MATH 3A NOTES 011 LIE ALGEBRAS 9.4. Bases for root systems. Definition A subset = { 1,,, n } of Φ is called a base for the root system if (1) is a basis for E () Every element β Φ can be written as β = k i i where all k i are integers with the same sign (or zero). Given a choice of base, the root in are called simple roots, roots which are positive linear combinations of simple roots are called positive roots the set of positive roots is denoted Φ +. The set of negative roots (negatives of positive roots) is Φ. Thus Φ=Φ + Φ Example In the root systems A,B,G,A 1 A 1 above, the roots, β were chosen to form a base for the root system. Example In the root system A n 1 = { i j R n } the roots i := i i+1, i =1,,n 1 form a base. In the A 3 -animation, the simple roots are green, red and blue. Example In the second set of animations for the root systems B 3 and C 3 the simple roots have green tips, and the positive roots are in the foreground (with negative roots behind a semitransparent plane). Lemma If, β Φ and (, β) < 0 then + β is a root. Similarly, if (, β) > 0 then β is a root. Proof. By symmetry we may assume, β = ±1 and this has the same sign as (, β). Then + β if (, β) < 0 σ β () = β if (, β) > 0 and σ β () Φ by definition of root system. Theorem Every root system has a base. Proof. First choose a vector x which is not perpendicular to any root. Since there are only finitely many roots and their perpendicular hyperplane have zero measure, their union also has zero measure. Take any point in the complement. Define a root β Φtobepositive if (β,x) > 0. Let S be the set of all real numbers (β,x) > 0forβ Φ. Then S is finite. Define a positive root to be indecomposable if it cannot be written as a sum of other positive roots. Claim 1 Every positive root β can be written as a sum of indecomposable positive roots. Pf: This holds by induction on the number of elements in the set S which are less than (β,x). If this number is zero then β is indecomposable. If β decomposes into a sum β = β 1 +β, we have (β i,x) < (β,x). So, by induction, each β i is a sum of indecomposable positive roots.

7 MATH 3A NOTES 011 LIE ALGEBRAS 41 Claim If, β are indecomposable positive roots then (, β) 0. Pf: If (, β) > 0 then γ = β is a root and therefore either γ or γ is a positive room. In the first case, = β +γ is not indecomposable. In the second case, β = +( γ) is not indecomposable. Claim 3 Let be the set of indecomposable positive roots. Then is a base for Φ. Pf: Since the second condition in the definition of a base is satisfied by construction, it suffices to show that is a basis for E. Since spans Φ it also spans E. So, it suffices to show is linearly independent. Suppose not. Then there is a linear dependence k i i = 0 where i. Separate the positive and negative coefficients κ i to obtain a relation: si i = t j j where s i,t j are all 0 and i = j. Let z denote this sum. Then (z,z) = s i t j ( i, j ) 0 by Claim. So, z = 0 proving Claim 3 and the theorem.