7. the Classical and Exceptional Lie Algebras * Version 1.4 *

Home , Cartan matrix

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

Eâb Eˆcd Eâb Eˆcd = δai δbj δcj δdk δbc Eâd, (7.1.2) · → ij jk → so that [Eâb, Eˆcd] = δbc Eâd 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 Eâb] = λa Eâb λb Eâb, (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 Eâb . 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−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 a6X=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. ρ(3) = α + α + α θ is the highest root. 1 2 3 ≡ The results of the algorithm can be indicated with the branching diagram in Fig. 7.2. Since the roots are non-degenerate [Proposition (VI.), Sec. 5.2.2] and each positive root ρ has a corresponding negative root ρ, we have a complete inventory for the − set of roots ∆ for A3. There are ∆ = 12 roots, so the group dimension d = 12+3 = 15, as expected for su(4). To get a geometrical picture| | of the algebra, we examine the Cartan matrix in Eq. (7.1.23). The angle between simple roots ◦ α1, α2 and between α2, α3 is 120 , while α1 and α3 are orthogonal. Combined with the fact that the algebra is simply laced and the{ inventory} of roots{ (Fig. 7.2),} one ﬁnds that the roots correspond to the twelve vertices of a certain quasiregular polyhedron (a cuboctahedron). See Fig. 7.3.

7.2 sp(2n): Cn

Next we consider Cn, the Lie algebra associated to symplectic Sp(2n) transformations. From module 2A, Eqs. (2A.3.1) and (2A.3.3), the deﬁning conditions for a symplectic transformation Uˆ on a2n-component complex vector are

† Uˆ Uˆ = 1ˆ2n, (7.2.1a) T Uˆ ˆ Uˆ =ˆ, (7.2.1b)

Figure 7.3: Root geometry for su(4) = A3. The twelve roots correspond to the vertices of the green polyhedron (a cuboctahedron). The vertices of the embedded red (blue) tetrahedron correspond to the four weights of the deﬁning “4” (conjugate “4”)¯ representation, as demonstrated module 8. The box arrays on the right are the corresponding Young tableaux, also discussed in module 8.

5 where the 2n 2n matrix ˆ is a block rank-2 Levi-Civita symbol, × 0 1ˆ ˆ n . (7.2.2) → 1ˆn 0 − If we express Uˆ as the exponentiation of an (antihermitian) 2n 2n matrix Yˆ × Uˆ = exp(Yˆ ), then the symplectic condition translates to T Yˆ =ˆ Yˆ ˆ. (7.2.3) As with ˆ, we decompose this into n n blocks, × Yˆ Yˆ Yˆ = 1 2 . (7.2.4) Yˆ Yˆ 3 4 Eq. (7.2.3) then implies that

T Yˆ = Yˆ , 4 − 1 ˆ ˆ T Y2 = Y2 , (7.2.5) ˆ ˆ T Y3 = Y3 .

Let 1 j, k n. We can deﬁne the following basis elements for Yˆ , ≤ ≤ 1 Eˆ Eˆjk Eˆk n,j n, (7.2.6a) jk ≡ − + + 2 2 Eˆ Eˆj,k n + Eˆk,j n = Eˆ , (7.2.6b) jk ≡ + + (jk) 3 3 Eˆ Eˆj n,k + Eˆk n,j = Eˆ . (7.2.6c) jk ≡ + + (jk) The parentheses here denote symmetrization [Eq. (1.4.2)]. The total number of such elements is

n2 + n(n +1) = n(2n + 1), which is the correct group dimension [Eq. (2A.3.5)]. Explicitly for n = 5 [sp(10)], we have (e.g.)

00000 00 0 00 00000 00 0 00  0 0 0 1 0 00 0 00   00000 00 0 00     00000 00 0 00  Eˆ1   , (7.2.7a) 34 →  00000 00 0 00     00000 00 0 00     00000 00 0 00     00000 0 0 1 0 0   −   00000 00 0 00      00000 00000 00000 00000  00000 0 0 0 1 0   00000 0 0 1 0 0     00000 00000  Eˆ2   , (7.2.7b) 34 →  00000 00000     00000 00000     00000 00000     00000 00000     00000 00000     

6 00000 00000 00000 00000  00000 00000   00000 00000     00000 00000  Eˆ3   . (7.2.7c) 25 →  00000 00000     00 00 1 00000     00000 00000     00000 00000     0 1 0 0 0 00000      Using Eq. (7.1.3), one can verify the following Lie brackets implied by the corresponding commutation relations:

1 1 1 1 [e , e ] = δkl e δjp e , (7.2.8a) jk lp jp − lk 1 2 2 2 [ejk, elp] = δkl ejp + δpk ejl, (7.2.8b) 1 3 3 3 [e , e ] = δjp e δjl e , (7.2.8c) jk lp − kl − kp 2 2 [ejk, elp]=0, (7.2.8d) 3 3 [ejk, elp]=0, (7.2.8e) 2 3 1 1 1 1 [ejk, elp] = δkl ejp + δjp ekl + δpk ejl + δjl ekp. (7.2.8f)

Exercise: Verify all of the Lie brackets in Eq. (7.2.8). • 1 A basis for the n-dimensional Cartan subalgebra H is given by the diagonal traceless elements ejj , 1 j n; a generic element h H is { } ≤ ≤ ∈ n 1 h = λj ejj. (7.2.9) j X=1 Then 1 1 [h, e ] = (λj λk) e , (7.2.10a) jk − jk 2 2 [h, ejk] = (λj + λk) ejk, (7.2.10b) 3 3 [h, e ] = (λj + λk) e , (7.2.10c) jk − jk 1,2,3 1 so that the collection of ejk excluding diagonal ejj is the set of root vectors. The simple roots{ can be} associated to the root{ vectors}

root vector root α(h) 1 e12 eα1 α1(h) = λ1 λ2 1 ≡ − e23 eα2 α2(h) = λ2 λ3 . ≡ . − (7.2.11) . . 1 en−1,n eαn−1 αn−1(h) = λn−1 λn 2 ≡ − e eα αn(h)=2λn. n,n ≡ n

1 By deﬁnition, we take α > α >...> αn (lexicographic ordering). Then positive roots are associated to root vectors e with 1 2 { jk} jk) and e3 . The positive root associated to e1 (j

7 The commutation relations between the simple root vectors are given by

1 1 [eα , eα ] = δj,i e δj,i− e , 1 i, j n 1, (7.2.13a) i j +1 i,i+2 − 1 i−1,i+1 ≤ ≤ − 2 [eα , eα ]=2δi,n− e , 1 i n 1, (7.2.13b) i n 1 n−1,n ≤ ≤ − 1 1 [eα , e−α ] = δij e e , 1 i, j n 1, (7.2.13c) i j i,i − i+1,i+1 ≤ ≤ − [eα , e−α ]=0, 1 i n 1, (7.2.13d) i n ≤ ≤ − 1 [eαn , e−αn ]=4 enn. (7.2.13e)

The Killing form between Cartan subalgebra elements

1 0 0 1 h λieii, h λjejj , (7.2.14) ≡ i ≡ j X X is given by

n n n 0 0 1 0 2 0 3 (h,h ) = h, h , epq + h, h , epq + h, h , epq coeff. of e1 coeff. of e2 coeff. of e3 p,q =1 pq p≥q=1 pq p≥q=1 pq p=6Xq X X n n 0 0 0 0 = (λp λq)(λ λ )+2 (λp + λq )(λ + λ ) − p − q p q p,q=1 p≥q X X=1 n n n 0 0 0 0 0 = (λp λq)(λ λ ) + (λp + λq )(λ + λ )+4 λpλ − p − q p q p p,q=1 p,q=1 p=1 X n X X 0 = 4(n + 1) λpλp. (7.2.15) p=1 X Now since 0 0 0 0 λi λi+1, 1 i n 1, (hα ,h ) = αi(h ) = − ≤ ≤ − (7.2.16) i 2λ0 , i = n, n we conclude that

n 1 1 1 (αi) 1 (αi) 1 hα = e e λ e , λ = (δi,j δi ,j) , 1 i n 1, i 4(n + 1) ii − i+1,i+1 ≡ j jj j 4(n + 1) − +1 ≤ ≤ − j=1 n X (7.2.17) 1 1 (αn ) 1 (αn ) 1 hα = e λ e , λ = δj,n. n 2(n + 1) nn ≡ j jj j 2(n + 1) j=1 X Eqs. (7.2.13c), (7.2.13e), (7.2.17) and (7.1.18) therefore give the Killing forms of the simple root vectors,

(eα , e−α ) =4(n + 1), 1 i n 1, i i ≤ ≤ − (7.2.18) (eαn , e−αn ) =8(n + 1).

We can now compute the scalar products between simple roots, using

[h , e ] = αi, αj e αi αj h i αj (αi) (αi) 1 = λ λ eα = (2δi,j δi ,j δi,j ) eα , 1 i, j n 1, j − j+1 j 4(n + 1) − +1 − +1 j ≤ ≤ −

[hα , eα ] = αn, αj eα n j h i j (αn) (αn ) 1 (7.2.19) = λ λ e = δj ,n e , 1 j n 1, j − j+1 αj −2(n + 1) +1 αj ≤ ≤ −

[hα , eα ] = αn, αn eα n n h i n 1 = eα . (n + 1) n

8 Figure 7.4: sp(2n) = Cn.

Therefore 1 αi, αj = (2δi,j δi ,j δi,j ) , 1 i, j n 1, h i 4(n + 1) − +1 − +1 ≤ ≤ − 1 αi, αn = δi,n−1, 1 i n 1, simple root scalar products in C . (7.2.20) h i − 2(n + 1) ≤ ≤ − n 1 αn, αn = , h i (n + 1)

2 2 The last equation implies that αn is a long root, with αn =2 αj , 1 j n 1. Cn is not simply laced. The corresponding Cartan matrix is given by [via| | Eq. (7.1.22| | )] ≤ ≤ −

2 1 0 ... 0 0 1− 2 1 ... 0 0 −0 1− 2 ... 0 0  ˆ − A  . . . . .  , Cartan matrix for Cn. (7.2.21) →  ......     0 0 0 ... 2 1  −   0 0 0 ... 2 2   −   

The Dynkin diagram is shown in Fig. 7.4.

7.3 so(2n): Dn

Next we consider Dn, the Lie algebra associated to special orthogonal SO(2n) transformations on even-dimensional vectors. From Eq. (2A.2.2) in module 2A, the deﬁning condition for an orthogonal transformation Uˆ on a 2n-component vector is

T Uˆ Uˆ = 1ˆ2n. (7.3.1)

If we express Uˆ as the exponentiation of an (antihermitian) 2n 2n matrix Yˆ × Uˆ = exp(Yˆ ), then we have the antisymmetry condition

T Yˆ = Y.ˆ (7.3.2) − Since we would like to identify Cartan subalgebra elements with diagonal matrices, Eq. (7.3.2) is inconvenient. Instead, we make a unitary (but not orthogonal) basis transformation,

0 † † Uˆ Vˆ Uˆ V,ˆ Vˆ Vˆ = 1ˆ n. (7.3.3) ≡ 2 Then T Uˆ 0 Mˆ Uˆ 0 = M,ˆ (7.3.4) T T † Mˆ Vˆ Vˆ = Mˆ , Mˆ Mˆ = 1ˆ n. ≡ 2 For Uˆ 0 = exp(Yˆ0), T Mˆ † Yˆ 0 Mˆ = Yˆ 0. (7.3.5) −

9 We choose Vˆ to have the block form

1 i1ˆn i1ˆn 0 1ˆn Vˆ = − , Mˆ = Mˆ † = . (7.3.6) √2 1ˆn 1ˆn ⇒ 1ˆn 0 − − As with Mˆ , we decompose Yˆ0 into n n blocks, × Yˆ Yˆ Yˆ 0 = 1 2 . (7.3.7) Yˆ Yˆ 3 4 Eq. (7.3.5) then implies that

T Yˆ = Yˆ , 4 − 1 ˆ ˆ T Y2 = Y2 , (7.3.8) − T Yˆ = Yˆ . 3 − 3 This is almost the same as the symplectic case [Eq. (7.2.5)], except the oﬀ-diagonal blocks are now antisymmetric. Let 1 j, k n. We can deﬁne the following basis elements for Yˆ , ≤ ≤ 1 Eˆ Eˆjk Eˆk n,j n, (7.3.9a) jk ≡ − + + 2 2 Eˆ Eˆj,k n Eˆk,j n = Eˆ , (7.3.9b) jk ≡ + − + [jk] 3 3 Eˆ Eˆj n,k Eˆk n,j = Eˆ . (7.3.9c) jk ≡ + − + [jk] The parentheses here denote antisymmetrization [Eq. (1.4.2)]. The total number of such elements is

n2 + n(n 1) = n(2n 1), − − which is the correct group dimension [Eq. (7.3.2)]. Relative to the n = 5 symplectic example in Eq. (7.2.7),

00000 00 0 00 00000 00 0 00  00000 00 0 1 0   00000 0 0 1 0 0   −   00000 00 0 00  Eˆ2   , (7.3.10a) 34 →  00000 00 0 00     00000 00 0 00     00000 00 0 00     00000 00 0 00     00000 00 0 00      0 0 000 00000 0 0 000 00000  0 0 000 00000   0 0 000 00000     0 0 000 00000  Eˆ3   . (7.3.10b) 25 →  0 0 000 00000     0 0 00 1 00000     0 0 000 00000     0 0 000 00000     0 1 0 0 0 00000   −   

10 Using Eq. (7.1.3), the Lie brackets are given by [c.f. Eq. (7.2.8)]

1 1 1 1 [e , e ] = δkl e δjp e , (7.3.11a) jk lp jp − lk 1 2 2 2 [e , e ] = δkl e δpk e , (7.3.11b) jk lp jp − jl 1 3 3 3 [e , e ] = δjp e δjl e , (7.3.11c) jk lp kl − kp 2 2 [ejk, elp]=0, (7.3.11d) 3 3 [ejk, elp]=0, (7.3.11e) 2 3 1 1 1 1 [e , e ] = δkl e + δjp e δpk e δjl e . (7.3.11f) jk lp jp kl − jl − kp Exercise: Verify the Lie brackets in Eq. (7.3.11). • A generic Cartan subalgebra H element is given by n 1 h = λj ejj. (7.3.12) j X=1 Then 1 1 [h, e ] = (λj λk) e , (7.3.13a) jk − jk 2 2 [h, ejk] = (λj + λk) ejk, (7.3.13b) 3 3 [h, e ] = (λj + λk) e . (7.3.13c) jk − jk The simple roots can be associated to the root vectors

root vector root α(h) 1 e12 eα1 α1(h) = λ1 λ2 1 ≡ − e23 eα2 α2(h) = λ2 λ3 . ≡ . − (7.3.14) . . 1 en−1,n eαn−1 αn−1(h) = λn−1 λn 2 ≡ − e e αn(h) = λn− + λn. n−1,n ≡ αn 1 1 By deﬁnition, we take α1 > α2 >...> αn (lexicographic ordering). Then positive roots are associated to root vectors ejk with 2 1 3 1 { } jk) and ejk . The positive root associated to ejk (j i WLoG)

j−1 n−2 λi + λj = (λi λj )+2(λj λn− ) + (λn− λn) + (λn− + λn) = αl +2 αl + αn− + αn. (7.3.15) − − 1 1 − 1 1 l i l j X= X= The commutation relations between the simple root vectors are given by

1 1 [eα , eα ] = δj,i e δj,i− e , 1 i, j n 1, (7.3.16a) i j +1 i,i+2 − 1 i−1,i+1 ≤ ≤ − 2 [eα , eα ] = δi,n− e , 1 i n 1, (7.3.16b) i n 2 n−2,n ≤ ≤ − 1 1 [eα , e−α ] = δij e e , 1 i, j n 1, (7.3.16c) i j i,i − i+1,i+1 ≤ ≤ − [eα , e−α ]=0, 1 i n 1, (7.3.16d) i n ≤ ≤ − 1 1 [eαn , e−αn ] = en−1,n−1 + en,n. (7.3.16e)

The Killing form between Cartan subalgebra elements

1 0 0 1 h λie , h λ e , (7.3.17) ≡ ii ≡ j jj i j X X is given by n 0 0 (h,h ) =4(n 1) λpλ . (7.3.18) − p p=1 X

11 Exercise: Verify Eq. (7.3.18). • Now since 0 0 0 0 λi λi+1, 1 i n 1, (hα ,h ) = αi(h ) = − ≤ ≤ − (7.3.19) i λ0 + λ0 , i = n, n−1 n we conclude that n 1 1 1 (αi) 1 (αi) 1 hα = e e λ e , λ = (δi,j δi ,j ) , 1 i n 1, i 4(n 1) ii − i+1,i+1 ≡ j jj j 4(n 1) − +1 ≤ ≤ − − j=1 − X n (7.3.20) 1 1 1 (αn) 1 (αn) 1 hα = e + e λ e , λ = (δn− ,j + δn,j ) . n 4(n 1) n−1,n−1 n,n ≡ j jj j 4(n 1) 1 j=1 − X − Eqs. (7.3.16c), (7.3.16e), (7.3.20) and (7.1.18) therefore give the Killing forms of the simple root vectors,

(eα , e−α )=4(n 1), 1 i n. (7.3.21) i i − ≤ ≤ We can now compute the scalar products between simple roots, using 1 [hα , eα ] = (2δi,j δi ,j δi,j ) eα , 1 i, j n 1, i j 4(n 1) − +1 − +1 j ≤ ≤ − − 1 [hα , eα ] = δj,n−2 eα , 1 j n 1, (7.3.22) n j − 4(n 1) j ≤ ≤ − − 2 [h , e ] = e . αn αn 4(n 1) αn − Therefore 1 αi, αj = (2δi,j δi ,j δi,j ) , 1 i, j n 1, h i 4(n 1) − +1 − +1 ≤ ≤ − − 1 αi, αn = δi,n−2, 1 i n 1, simple root scalar products in D . (7.3.23) h i − 4(n 1) ≤ ≤ − n − 2 αn, αn = , h i 4(n 1) − Since all roots have the same length, Dn is simply laced. The corresponding Cartan matrix is given by [via Eq. (7.1.22)]

2 1 0 ... 0 0 0 1− 2 1 ... 0 0 0 −0 1− 2 ... 0 0 0  − ˆ  ......  A  ......  , Cartan matrix for Dn. (7.3.24) →    0 0 0 ... 2 1 1  − −   0 0 0 ... 1 2 0   −   0 0 0 ... 1 0 2   −   

The Dynkin diagram is shown in Fig. 7.5.

Figure 7.5: so(2n) = Dn.

12 7.4 so(2n + 1): Bn

Finally we consider Bn so(2n + 1). We can view the Bn generators in the deﬁning representation as equivalent to those for Dn, but with an added “zeroth”⇔ row and “zeroth” column preceding the 2n 2n generators of so(2n); i.e. a vector index i runs from 0 i 2n. × ≤ ≤ T As in Sec. 7.3, we begin by rotating the orthogonal condition Uˆ Uˆ = 1ˆ2n+1 to a diﬀerent basis. Applying the transformation as in Eq. (7.3.3), we choose

√2 0ˆ1,n 0ˆ1,n 1 0ˆ1,n 0ˆ1,n 1 T Vˆ = 0ˆn,1 i1ˆn i1ˆn , Mˆ = Vˆ Vˆ = 0ˆn,1 0 1ˆn . (7.4.1) √2  −  ⇒   0ˆn, 1ˆn 1ˆn 0ˆn, 1ˆn 0 1 − − 1     0 Here 0ˆ1,n (0ˆn,1) denotes an n-fold row (column) of zeroes. The (2n+1) (2n+1) generator Yˆ satisﬁes the condition as in Eq. (7.3.5), which allows the block decomposition [c.f. Eqs. (7.3.7) and (7.3.8)] ×

1 1 2 b cˆ1,n cˆ1,n ˆ 0 ˆ1 ˆ ˆ Y = dn,1 Y1 Y2 . (7.4.2)  ˆ2 ˆ ˆ  dn,1 Y3 Y4   Eq. (7.3.5) then implies that

T Yˆ = Yˆ , 4 − 1 ˆ ˆ T Y2 = Y2 , − T Yˆ = Yˆ , 3 − 3 b1 =0, (7.4.3) T ˆc1 = dˆ2 , 1,n − n,1 h iT ˆc2 = dˆ1 . 1,n − n,1 h i The basis elements for Yˆ include Eˆ1 , Eˆ2 , Eˆ3 (1 j, k n) from Eq. (7.3.9), and { jk jk jk} ≤ ≤ 4 Eˆ Eˆ ,j Eˆj n, , (7.4.4a) j ≡ 0 − + 0 5 Eˆ Eˆj, Eˆ ,j n, (7.4.4b) j ≡ 0 − 0 + both with 1 j n. The total number of such elements is ≤ ≤ n(2n 1)+2n = n(2n + 1). − ˆ1 ˆ2 ˆ3 Relative to the example in Eq. (7.3.10), Ejk, Ejk, Ejk are now block embedded in 11 11 matrices (after one initial row and one initial column of zeroes), while (e.g.) { } ×

0 0 0 0 1 0 00000 0 00000 00000  0 00000 00000   0 00000 00000     0 00000 00000    Eˆ4  0 00000 00000  , (7.4.5a) 4 →    0 00000 00000     0 00000 00000     0 00000 00000     1 00000 00000  −   0 00000 00000     

13 0 00000 0 0 1 0 0 0 00000 00− 0 00  0 00000 00 0 00   1 00000 00 0 00     0 00000 00 0 00    Eˆ5  0 00000 00 0 00  . (7.4.5b) 3 →    0 00000 00 0 00     0 00000 00 0 00     0 00000 00 0 00     0 00000 00 0 00     0 00000 00 0 00      In addition to the Lie brackets in Eq. (7.3.11), we have

1 4 4 [e , e ] = δjl e , (7.4.6a) jk l − k 1 5 5 [ejk, el ] = δkl ej , (7.4.6b) 2 4 5 5 [e , e ] = δjl e δkl e , (7.4.6c) jk l k − j 2 5 [ejk, el ]=0, (7.4.6d) 3 4 [ejk, el ]=0, (7.4.6e) 3 5 4 4 [e , e ] = δjl e δkl e , (7.4.6f) jk l k − j [e4, e4] = e3 , (7.4.6g) j k − jk [e5, e5] = e2 , (7.4.6h) j k − jk [e4, e5] = e1 . (7.4.6i) j k − kj Exercise: Verify the Lie brackets in Eq. (7.4.6). • A generic Cartan subalgebra H element is given by n 1 h = λj ejj. (7.4.7) j X=1 Then 1 1 [h, e ] = (λj λk) e , (7.4.8a) jk − jk 2 2 [h, ejk] = (λj + λk) ejk, (7.4.8b) 3 3 [h, e ] = (λj + λk) e , (7.4.8c) jk − jk 4 4 [h, e ] = λj e , (7.4.8d) j − j 5 5 [h, ej ] = λj ej . (7.4.8e) The simple roots can be associated to the root vectors

root vector root α(h) 1 e12 eα1 α1(h) = λ1 λ2 1 ≡ − e23 eα2 α2(h) = λ2 λ3 . ≡ . − (7.4.9) . . 1 en−1,n eαn−1 αn−1(h) = λn−1 λn 5 ≡ − e eα αn(h) = λn. n ≡ n 1 By deﬁnition, we take α1 > α2 > ...> αn (lexicographic ordering). The positive roots are associated to root vectors ejk with 2 5 1 3 4 1 { } j k), ejk , and ej . The positive root associated to ejk (j i WLoG)

j−1 n−1 λi + λj = (λi λj )+2(λj λn)+2λn = αl +2 αl +2αn. (7.4.10) − − l i l j X= X=

14 Eq. (7.4.10) has almost the same form as the symplectic Cn case, Eq. (7.2.12); the diﬀerence is the prefactor of two for αn in 5 Eq. (7.4.10). In fact we will see that Bn and Cn are closely related (although not identical, except for B2 = C2). For ei , the decomposition is

n−1 λi = (λi λn) + λn = αl + αn. (7.4.11) − Xl=i The commutation relations between the simple root vectors are given by

1 1 [eα , eα ] = δj,i e δj,i− e , 1 i, j n 1, (7.4.12a) i j +1 i,i+2 − 1 i−1,i+1 ≤ ≤ − 5 [eα , eα ] = δi,n− e , 1 i n 1, (7.4.12b) i n 1 n−1 ≤ ≤ − 1 1 [eα , e−α ] = δij e e , 1 i, j n 1, (7.4.12c) i j i,i − i+1,i+1 ≤ ≤ − [eα , e−α ]=0, 1 i n 1, (7.4.12d) i n ≤ ≤ − 1 [eαn , e−αn ] = enn. (7.4.12e)

The Killing form between Cartan subalgebra elements

1 0 0 1 h λie , h λ e , (7.4.13) ≡ ii ≡ j jj i j X X is given by

n 0 1 0 (h,h )=4 n λpλ . (7.4.14) − 2 p p=1 X Exercise: Verify Eq. (7.4.14). • Now since λ0 λ0 , 1 i n 1, (h ,h0) = α (h0) = i i+1 (7.4.15) αi i λ0 ,− i =≤n,≤ − n we conclude that

n 1 1 1 (αi) 1 (αi) 1 hα = e e λ e , λ = (δi,j δi+1,j ) , 1 i n 1, i 4 n 1 ii − i+1,i+1 ≡ j jj j 4 n 1 − ≤ ≤ − 2 j=1 2 − X − (7.4.16) n 1 1 (αn) 1 (αn ) 1 hα = e λ e , λ = δn,j. n 4 n 1 n,n ≡ j jj j 4 n 1 2 j 2 − X=1 − Eqs. (7.4.12c), (7.4.12e), (7.4.16) and (7.1.18) therefore give the Killing forms of the simple root vectors,

1 (e , e− )=4 n , 1 i n. (7.4.17) αi αi − 2 ≤ ≤ We can now compute the scalar products between simple roots, using 1 [hα , eα ] = (2δi,j δi+1,j δi,j+1) eα , 1 i, j n 1, i j 4 n 1 − − j ≤ ≤ − − 2 1 [hαn , eαj ] = δj,n−1 eαj , 1 j n 1, (7.4.18) − 4 n 1 ≤ ≤ − − 2 1 [hα , eα ] = eα . n n 4 n 1 n − 2

15 Figure 7.6: so(2n +1) = Bn.

Therefore 1 αi, αj = (2δi,j δi+1,j δi,j+1) , 1 i, j n 1, h i 4 n 1 − − ≤ ≤ − − 2 1 αi, αn = δi,n−1, 1 i n 1, simple root scalar products in B . (7.4.19) h i − 4 n 1 ≤ ≤ − n − 2 1 αn, αn = . h i 4 n 1 − 2 2 2 The last equation implies that αn is a short root, with αn = αj /2, 1 j n 1. Like Cn, Bn is not simply laced. The corresponding Cartan matrix is given by [via| Eq.| (7.1.22| | )] ≤ ≤ −

2 1 0 ... 0 0 1− 2 1 ... 0 0 −0 1− 2 ... 0 0  ˆ − A  . . . . .  , Cartan matrix for Bn. (7.4.20) →  ......     0 0 0 ... 2 2  −   0 0 0 ... 1 2   −   

The Dynkin diagram is shown in Fig. 7.6. The Cartan matrices/Dynkin diagrams for Cn [Eq. (7.2.21), Fig. 7.4] and Bn [Eq. (7.4.20), Fig. 7.6] are almost identical. In both cases, the only non-vanishing simple root inner products involve “nearest-neighbors” αi and αi ,1 i n 1. The first n 1 +1 ≤ ≤ − − simple roots share the same length, while αn is a long (short) root for Cn (Bn). In fact, the root systems for Bn and Cn have the same geometry, except that long and short roots are interchanged. 1 On the other hand, we will see that Dn and Bn have more in common in terms of the n simplest (so called fundamental ) representations, described in module 8. In particular, both Dn and Bn possess fundamental spinor representations, which are qualitatively distinct and cannot be built by tensoring together defining vector indices. By contrast, every representation of Cn or An can be built by tensoring together indices transforming in the the defining 2n-dimensional [(n + 1)-dimensional] representation.

Exercise: Find the inventory of roots and plot the 3D root geometry for B3 and C3, following the same approach used for A3 in •Sec. 7.1.1.

7.5 Classical Lie algebras to rank 4; equivalent algebras

In Table 1, we show the classical Lie algebras through rank four. Various “accidental” equivalences between diﬀerent algebras of the same rank are indicated, and easily recognized by the identical Dynkin diagrams. All four families are distinct for each rank above three. Of course, equivalence of the Lie algebras does not imply equivalence of the corresponding symmetry groups.

7.6 The exceptional algebras

The four classical families studied above do not exhaust the possibilities for (semi-)simple Lie algebras. It is perhaps surprising that there are only ﬁve additional possibilities corresponding to particular algebras of ﬁxed rank. This can be shown by a careful determination of allowable Dynkin diagrams. The argument is explicated in chapter IX of Cahn [1] and chapter 20 of Georgi [2];

1So far we have used the terms “defining” and “fundamental” representations interchangably, as is common in physics. However, for a given Lie algebra L, there exists of set of “simplest” representations (which always includes the defining one), and mathematicians refer to each member of this set as a fundamental representation. We will define fundamental represenations precisely in module 8, using the fundamental weights.

16 Table 1: Classical Lie algebras to rank 4.

Rank Dynkin Cartan Compact #ofpositiveroots

1 A1 su(2) = sp(2) = so(3) 1

A2 su(3) 3 2 B2 = C2 so(5) = sp(4) 4 D = A A so(4) = su(2) su(2) 2 2 1 × 1 ×

A3 = D3 su(4) = so(6) 6 3 B3 so(7) 9 C3 sp(6) 9

A4 su(5) 10 B4 so(9) 16 4 C4 sp(8) 16

D4 so(8) 12

we won’t belabor it here. Instead, we simply state the Cartan matrices and Dynkin diagrams for the ﬁve exceptional algebras G2, F4, and E6,7,8. Our labeling conventions are the same as chapter 13 of [3].

(7.6.1)

Cartan matrices for the exceptional Lie algebras G2, F4, E6, E7, and E8

2 10 0 0 0 1− 2 10 0 0 2 3 −0 1− 2 1 0 1 AˆG = − , AÊ = − − − , 2 1 2 6  0 0 1 2 1 0  −  − −   0 0 0 1 2 0   −   0 0 10 0 2   −    2 10 0 0 0 0 1− 2 10 0 0 0 2 1 0 0 − −  − 0 1 2 1 0 0 1 1 2 2 0 − − − AˆF = − −  , AÊ =  0 0 1 2 1 0 0  , 4 0 1 2 1 7  − −  (7.6.2) − −  0 0 0 1 2 1 0   0 0 1 2   − −   −   0 0 0 0 1 2 0     −   0 0 10 0 0 2   −    2 1000000 1− 2 10 0 0 0 0 −0 1− 2 10 0 0 0  − −  0 0 1 2 10 0 0  AÊ =  − −  . 8  0 0 0 1 2 1 0 1  − − −   0 0 0 0 1 2 1 0   − −   00 00 0 1 2 0   −   0 0 0 0 10 0 2   −   

17 Figure 7.7: The ﬁve exceptional algebras.

References

[1] Robert N. Cahn, Semi-Simple Lie Algebras and Their Representations (Benjamin/Cummings, Menlo Park, California, 1984). [2] Howard Georgi, Lie Algebras in Particle Physics, 2nd ed. (Westview Press, Boulder, Colorado, 1999). [3] Phillipe Di Francesco, Pierre Mathieu, David S´en´echal, Conformal Field Theory (Springer-Verlag, New York, 1996).