7. The classical and exceptional Lie algebras * version 1.4 * Matthew Foster November 14, 2016 Contents 7.1 su(n): An−1 1 7.1.1 Example: su(4) = A3................................................... 4 7.2 sp(2n): Cn 5 7.3 so(2n): Dn 9 7.4 so(2n +1): Bn 13 7.5 Classical Lie algebras to rank 4; equivalent algebras 16 7.6 The exceptional algebras 16 In Sec. 5.3, we demonstrated how the root system (adjoint representation) for a generic Lie algebra can be constructed from the Cartan matrix or Dynkin diagram. The latter encodes the inner products and norm ratios of the simple roots. In this module, we assemble the Cartan matrix from the simple roots for the four classical Lie algebra families An, Bn, Cn, and Dn. The needed data is extracted from the defining representation for each algebra. We will also state (but not derive) the results for the five exceptional algebras G2, F4, and E6,7,8. In the next module 8, we will show how the Cartan matrix can be used to obtain the full set of weights in any (finite dimensional) irreducible representation. In addition, the weights of the defining representation for the orthogonal algebras Bn ∗ [so(2n + 1)] and Dn [so(2n)] turn out to give a useful alternative basis for H , since these prove to be orthonormal. We will exploit this fact later in module 10 to construct spinor representations. The discussion here largely follows chapter VIII of [1]. 7.1 su(n): An 1 − Consider the space of all n n matrices with real or complex elements. A basis for this space can be defined as follows: × ··· b−1 b b+1 ··· . a−1 a 1 Eˆab = δaiδbj Eˆab a+1 . (7.1.1) ij ⇒ → . . 1 All unmarked entries are zero. These elements satisfy the multiplication rule Eˆab Eˆcd Eˆab Eˆcd = δai δbj δcj δdk δbc Eˆad, (7.1.2) · → ij jk → so that [Eˆab, Eˆcd] = δbc Eˆad δda Eˆcb. (7.1.3) − Note that [Eˆab, Eˆcc]=0, (7.1.4) c X which implies that the identity 1ˆn = Eˆcc (7.1.5) c X generates a one-dimensional abelian ideal. We want to construct root systems for generic semi-simple Lie algebras (no abelian ideals); we should therefore restrict ourselves to traceless n n matrices, i.e. the su(n)= An−1 Lie algebra. In this case, an arbitrary element of the Cartan subalgebra can be written as × n n Hˆ = λi Eˆii, λi =0. (7.1.6) i i X=1 X=1 Now adh (eab) [H,ˆ Eˆab] = λa Eˆab λb Eˆab, (7.1.7) → − which implies that eab is a root vector: b [h, eab] = αab(h) eab, αab λieii = λa λb. (7.1.8) − i ! X We identify the positive root vectors with set of upper triangular Eˆab . For a generic Cartan subalgebra element h H defined via Eq. (7.1.6), we claim that the set of simple roots Π can be associated{ } to the first diagonal root vectors: ∈ root vector root α(h) e e α (h) = λ λ 12 ≡ α1 1 1 − 2 e23 eα2 α2(h) = λ2 λ3 (7.1.9) . ≡ . − . en− ,n e αn− (h) = λn− λn 1 ≡ αn−1 1 1 − Proof: A generic positive root is αij(h) = λi λj , i<j. (7.1.10) − This can be written as j−1 αij(h) = (λi λi ) + (λi λi ) + (λi λi ) + ... + (λj− λj ) = αl. (7.1.11) − +1 +1 − +2 +2 − +3 1 − l i X= Therefore every positive root can be expressed as a linear combination of the αi with non-negative integral coefficients. Moreover, { } αi cannot be expressed as a linear combination of the other proposed simple roots. To see this, assume the converse, i.e. that αi is positive, but not simple. Then (i) (i) (i) αi(h) = λi λi = κ αj (h) = ... + κ (λi− λi) + κ (λi λi ) + ... (7.1.12) − +1 j i−1 1 − i+1 +1 − +2 j6 i X= (i) Since all the κj are positive, there is no way to get λi+1 with a negative coefficient (contradiction). 2 For the simple roots, we therefore have the associated Lie brackets [eα , eα ] = δj,i ei,i δj,i− ei− ,i , (7.1.13a) i j +1 +2 − 1 1 +1 [eα , e−α ] = δij (eii ei ,i ) , (7.1.13b) i j − +1 +1 [h, eα ] = (λi λi ) eα , h defined as in Eq. (7.1.6). (7.1.13c) i − +1 i ˆ Here the root vector associated to αj is e−αj Ej+1,j. Eq. (7.1.13b) is consistent with the fact that simple roots are lowest weight states in mutually generated strings− of the adjoint→ representation [Eq. (5.3.3)]. It is useful to construct the hαi (i 1, 2,...,n 1 ), i.e. the elements of the Cartan subalgebra H that represent the simple roots. Consider the Killing form{ between} ∈ { generic ele−ments} of H. We use the algorithm described in module 4 [Eq. (4.2.2)]. For 0 0 h λieii, h λ ejj , (7.1.14) ≡ ≡ j i j X X we have the Killing form n n n 0 0 0 0 0 (h,h ) = [h, [h , eab]] = (λa λb)(λa λb)=2n λaλa, (7.1.15) coeff. of e − − a,b =1 ab a,b=1 a=1 aX=6 b X X since b λb = 0. Therefore, since 0 0 0 0 P (hαi ,h ) = αi(h ) = λi λi+1, (7.1.16) − we conclude that n eii ei+1,i+1 (αi ) (αi) 1 hα = − λ ejj, λ = (δi,j δi ,j) . (7.1.17) i 2n ≡ j j 2n − +1 j=1 X Recall that [Eq. (4.4.10)] [eαi , e−αi ] hαi = . (7.1.18) (eαi , e−αi ) Eqs. (7.1.13b), (7.1.17) and (7.1.18) therefore give the Killing form of the simple root vectors, (eαi , e−αi )=2n. (7.1.19) We can now compute the scalar products between simple roots, using [hαi , eαj ] = αi, αj eαj h i (7.1.20) (αi) (αi ) 1 = λ λ eα = (δi,j δi ,j δi,j + δi ,j ) eα . j − j+1 j 2n − +1 − +1 +1 +1 j Therefore 1 αi, αj = (2δi,j δi ,j δi,j ) , simple root scalar products in An− . (7.1.21) h i 2n − +1 − +1 1 The Cartan matrix is defined as [Eqs. (5.3.8) and (5.3.9)] 2 αi, αj Aij = h i = pαi;αj (i = j), (7.1.22) αj, αj − 6 h i where pαi;αj 0, 1, 2, 3 determines the number of states above αi in the root string generated by eαj . For An−1, Eq. (7.1.21) implies that ∈ { } 2 1 0 ... 0 0 1− 2 1 ... 0 0 −0 1− 2 ... 0 0 ˆ − Aij =2δij δi+1,j δi,j+1 A . , Cartan matrix for An−1. (7.1.23) − − ⇒ → . ... 0 0 0 ... 2 1 − 0 0 0 ... 1 2 − 3 The corresponding Dynkin diagram is shown in Fig. 7.1. Because all roots have the same length, su(n) = An−1 is simply laced. Figure 7.1: su(n) = An−1. 7.1.1 Example: su(4) = A3. Using the root-building algorithm from Sec. 5.3.2, we construct the roots for A3. 1. At k = 1, we have the simple roots 2 1 0 1 2 1 0 1 2 − − − − (7.1.24a) (α ) (α ) (α ) 1 2 3 2. At k = 2, the sum ρ(2) α + α is a valid root since [Eq. (7.1.22)] 1 ≡ 1 2 pα ;α =1, 1 2 (7.1.24b) pα2 ;α1 =1. The sum ρ(2) α + α is a valid root since 2 ≡ 2 3 pα ;α =1, 2 3 (7.1.24c) pα3 ;α2 =1. The Dynkin labels for these level 2 roots are 1 1 1 1 1 1 (2) − − (2) (7.1.24d) ρ ρ 1 2 The non-zero depths are m (2) = m (2) =1, m (2) = m (2) =1. (7.1.24e) ρ1 ;α1 ρ1 ;α2 ρ2 ;α2 ρ1 ;α3 3. At k = 3, we compute pρ(2);α = mρ(2) ;α 1=0, 1 1 1 1 − pρ(2);α = mρ(2) ;α 1=0, 1 2 1 2 − pρ(2);α = mρ(2) ;α +1=1, 1 2 1 2 (7.1.24f) p (2) = m (2) +1=1, ρ2 ;α1 ρ1 ;α1 pρ(2);α = mρ(2) ;α 1=0, 2 2 1 2 − pρ(2);α = mρ(2) ;α 1=0. 2 2 1 2 − Thus ρ(3) ρ(2) + α = ρ(2) + α = α + α + α is the only k = 3 root, ≡ 1 3 2 1 1 2 3 1 0 1 (7.1.24g) ρ(3) The depths are (3) (3) (3) mρ ;α1 =1, mρ ;α2 =0, mρ ;α3 =1. (7.1.24h) 4 Figure 7.2: Root tree for su(4) = A3. Roots are specified by their Dynkin labels. The level k = 1 simple roots appear at the top; linear combinations of these give the other positive roots. The labels on diagonal lines indicate the simple root that gives the level (k + 1)th root from the kth one. 4. At k = 4, pρ(3);α = mρ(3) ;α 1=0, 1 1 − (7.1.24i) p (3) = m (3) 0=0,p (3) = m (3) 1=0. ρ ;α2 ρ ;α2 − ρ ;α3 ρ ;α3 − We conclude that there are no level k = 4 roots.