1.745 Intoduction to Lie Algebas Novembe 1, 010 Lectue 16 Root Systems and Root Lattices Pof. Victo Kac Scibe: Michael Cossley Recall that a oot system is a pai (V, ), whee V is a finite dimensional Euclidean space ove R with a positive definite bilinea fom (, ) and is a finite subset, such that: 1. 0 / ; R =V ;. If α, then nα if and only if n = ±1; 3. (Sting popety) if α, β, then {β +jα j Z} ( 0) = {β pα,..., β,..., β+qα}whee p q = (α,β) (α,α). (V, ) is called indecomposable if it cannot be decomposed into a non-tivial othogonal diect sum. = dimv is called the ank of (V, ) and elements of ae called oots. Definition 16.1. An isomophism of an indecomposable oot system (V, ) and (V 1, 1 ) is a vecto space isomophism ϕ : V V 1, such that ϕ( ) = 1, and (ϕ(α),ϕ(β)) 1 = c(α, β) fo all α, β, whee c is a positive constant, independent of α and β. In paticula, eplacing (, ) by c(, ), whee c>0, we get, by definition, an isomophic oot system. Example 16.1. Root systems of ank 1: (R, ={α, α}), α = 0, (α, β) =αβ. This oot system is isomophic to that of sl (F), so 3 (F), and sp (F). Poposition 16.1. Let (V, ) be an indecomposable oot system with the bilinea fom (, ). Then 1. Any othe bilinea fom (, ) 1 fo which the sting popety holds is popotional to (, ), i.e. (α, β) 1 = c(α, β) fo some positive c R, independentofα and β.. If (α, α) Q fo some α, then(β,γ) Q fo all β,γ. Poof. Fix α. Since (V, ) is indecomposable, fo any β thee exists a sequence γ 0,γ 1,...,γ k such that α = γ 0, β = γ k, (γ i,γ i+1 ) = 0 fo all i = 0,...,k 1. Define c by (α, α) 1 = c(α, α). By the sting popety p q = (α,γ 1) (α,α) = (α,γ 1) 1 (α,α) 1. Hence (α, γ 1 ) 1 = c(α, γ 1 ). Likewise, by the sting popety, (α,γ 1) (γ 1,γ 1 ) = (α,γ 1) 1 (γ 1,γ 1 ) 1. Hence (γ 1,γ 1 )=c(γ 1,γ 1 ). Continuing this way, we show that (γ,γ ) 1 = c(γ,γ ),...(β,β) 1 = c(β,β). Since (α,β) (α,α) = (α,β) 1 (α,α) 1, we conclude that (α, β) 1 = c(α, β) fo all α, β. Since spans V, we conclude that (1) holds. The same agument poves (). Definition 16.. A lattice in an Euclidean space V is a discete subgoup (Q, +) of V, which spans V ove R, i.e. RQ = V. Fo example, Z n R n. Poposition 16.. If is a finite set in an Euclidean space V,spanningV ove R, suchthat (α, β) Q fo all α, β, thenz is a lattice in V. 1
Poof. The only thing to pove is that Z is a discete set. Choose a basis β 1,...,β of V among the vectos of. Then fo any α, wehaveα = c iβ i, c i R. Hence, (α, β j )= c i(β i,,β j ). But ((β i,β j )) i,j=1 is a Gamm matix of a basis, hence it is non-singula. Hence the c i s can be computed by Came s ule, so all c i Q. So Z Q{β 1,...,β }. But since is finite, we conclude that Z 1 N Z{β 1,...,β } whee N is a positive intege. But 1 N Z{β 1,...,β } is discete, hence Z is discete. Example 16.. {1, } R, then Z{1, }is not a discete set. Coollay 16.3. If (V, ) is a oot system, then Q := Z is a lattice, called the oot lattice. Poof. By the two popositions, the coollay holds if (V, ) is indecomposable, hence holds fo any oot system (V, ). We will list fou seies of oot systems of ank known to us. In all cases ( i, j )=δ ij. Type g V Q +1 +1 +1 +1 A sl +1(F) { a i i a i R, a i =0} { i j 1 i, j +1} { a i i a i Z, a i =0} B so +1(F) { a i i a i R} {± i ± j, ± i 1 i, j, i = j} { a i i a i Z} C sp (F) { a i i a i R} {± i ± j, ± i 1 i, j, i = j} { a i i a i Z, a i Z} D so (F), 3 { a i i a i R} {± i ± j 1 i, j, i = j} { a i i a i Z, a i Z} Remak Explanation: in case A, V is a facto space of V = +1 a i i by a 1-dimensional subspace R( 1 +...+ +1 ). Notice that ( 1 +...+ +1 ) = V in the table, so V = V R( 1 +...+ +1 ) with diect sum. Secondly, why is A = { +1 a i i a i Z, +1 a i =0}. Clealy, Z{ i j i = j} is included in this set. To show the evese inclusion, wite +1 Q a i i = a 1 ( 1 )+(a 1 +a )( 3 )+...+(a 1 +a +...+a )( +1 )+(a 1 +...+a +1 )( +1 ), whee the coefficient of the last tem is zeo. So the evese inclusion is also tue. Fo case B, the fom of the oot lattice is clealy coect. Execise 16.1. Explain the oot lattices in cases C and D. Poof. C. Z C Q C as each membe of C is of the fom a 1 i + a j with a 1 = ±1, a = ±1. So each element of Z C is of the fom a i i whee a i Z, a i Z. Q C Z C as any a i i with a i Z, a i Z can be witten in the fom a i i = a 1 ( 1 )+(a +a 1 )( 3 )+...+(a +a 1 +...+a 1 )( 1 )+(a +...+a 1 ) 1. All these coefficients belong to Z and the final coefficient of 1 belongs to Z so can be witten as a+...+a 1 1, hence a i i Z C.
D. The poof fo Q D is identical except i D, so wite a i i = a 1 ( 1 )+(a + a 1 )( 3 )+...+ a +...+ a 1 as 3, and this belongs to Z D. ( 1 )+ a +...+ a 1 ( )+ a +...+ a 1 ( 1 + ) Definition 16.3. A lattice Q is called integal (espectively even) if (α, β) Z (espectively (α, α) Z) fo all α, β Q. Note that an even lattice is always integal: if α, β Q, Q even, then (α + β,α + β) =(α, α)+(β,β) + (α, β), (α, α), (β,β) Z. Hence (α, β) Z, so (α, β) Z. Example 16.3. Fo a positive intege let E = { a i i eithe all a i Z o all a i Z + 1, and a i Z}, with E R,( i, j )=δ ij. Poposition 16.4. E is an even lattice if and only if is divisible by. Poof. We use that a ± a Z if a Z. Let α = a i i E. Case 1: all a i Z, then (α, α) = a i = a i mod 0 mod by the condition of E,so(α, α) Z. Case : wite α = ρ + β whee ρ =( 1,..., 1 ) and β has intege coefficients b i. Then (α, α) =(ρ, ρ) + (ρ, β)+ (β,β) =(ρ, ρ)+ (b i + b i)= 4 + n, n Z. So (α, α) is even if and only if is a multiple of. Theoem 16.5. Let Q be an even lattice in an Euclidean space V,andassumethesubset = {α Q (α, α) =} spans V ove R. Then(V, )is a oot system. Poof. Axiom (1) of a oot system is clea, as is axiom (): if (α, α) =, then (nα, nα) =iff. n = ±1. It emains to show the sting popety. Revesing the sign of α if necessay, we may assume (α, β) 0. Note that fo α, β we have: 0 (α β,α β) =(α, α) (α, β)+(β,β) =4 (α, β), whee (α, β) Z, (α, β) 0. So the only possibilities ae (α, β) =0, 1 o. In the last case, hence q =0, p =,sop q =(α, β) =and the sting popety is satisfied. Execise 16.. Complete the poof, fo (α, β) =0o 1. Poof. Fo (α, β) =1, α β, α+ β/, sop =1, q =0,andp q =(α, β). Fo (α, β) =0, α + β/, α β/, p =0, q =0,andp q =(α, β). So the sting popety holds geneally, and the theoem holds. 3
The most emakable lattice is E (which is even by poposition.) Execise 16.3. Show that E := {α E (α, α) =} = {± i ± j i = j} { 1 (± 1 ±...± ) even numbe of minus signs}, that E =40,andthatR E =V. Poof. E = { a i i all a i Z, a i Z} { a i i all a i Z + 1, a i Z}. The elements fom the fist set satisfying (α, α) =ae clealy {± i ± j i = j} as ( i, j )=δ ij, and the second set must have all a i = ± 1 else (α, α) >, andas a i Z, thee must be an even numbe of minus signs. E = { i + j i = j} + { i j i = j} + { i j i = j} + {1 (± 1 ±...± ) even numbe of minus signs} =.7 +.7 +.7+7 = 40. Clealy i = i+ j + i j R E,and{ i } fom a basis of V, hence R =V. So (R, E ) is a oot system by the theoem, which is called the oot system of type E. Execise 16.4. Conside the following subsystem of the oot system of type E : take ρ =( 1,..., 1 ) and let E7 = {α E (α, ρ) =0}, Q E7 = {α Q E (α, ρ) =0}, V E7 = {v V E (v, ρ) =0}. Show that (V E7, E7 ) is a oot system of ank 7, and that E7 = 16. Poof. Clealy E7 = {α Q E7 (α, α) =}. Q E7 is an even lattice in V E7 as it is a subgoup of Q E and clealy RQ E7 = V E7,so(V E7, E7 )is a oot system of ank 7 as V E = V E7 R( 1,..., 1 ), V E7 =( 1,..., 1 ),anddimv E =. E7 = { i j i = j} { ± 1 ±...± 4minussigns.} Hence, E7 = 56 + = 16 4 Execise 16.5. Let E6 = {α E7 (α, 7 + ) = 0}, Q E6 = {α Q E7 (α, 7 + ) = 0}, V E6 = {v V E7 (v, 7 + )=0}. Show that (V E6, E6 ) is a oot system of ank 6, and that E6 = 7. 4
Poof. Clealy E6 = {α Q E6 (α, α) =}. Q E6 is an even lattice in V E6 as it is a subgoup of Q E7 and clealy RQ E6 = V E6,so(V E6, E6 ) is a oot system of ank 6 as V E7 = V E6 R( 7 + ), V E6 =( 7 + ), and dimv E7 =7. E6 = { i j if i =7,j =, if i =,j =7} { ± 1 ±...± 4minussignsand 7, have opposite signs} Hence, 6 E6 =6.5++ = 7. 3 5