7. the classical and exceptional lie algebras * version 1.4mf23.web.rice.edu/la_7_v1.4 classical and...

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7. The classical and exceptional Lie algebras * version 1.4 * Matthew Foster November 14, 2016 Contents 7.1 su(n): A n-1 1 7.1.1 Example: su(4) = A 3 . ................................................... 4 7.2 sp(2n): C n 5 7.3 so(2n): D n 9 7.4 so(2n +1): B n 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 A n , B n , C n , and D n . 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 G 2 , F 4 , and E 6,7,8 . 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 B n [so(2n + 1)] and D n [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): A n-1 Consider the space of all n × n matrices with real or complex elements. A basis for this space can be defined as follows: ˆ E ab ij = δ ai δ bj ˆ E ab ··· b-1 b b+1 ··· . . . a-1 a 1 a+1 . . . . (7.1.1) 1

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Page 1: 7. The classical and exceptional Lie algebras * version 1.4mf23.web.rice.edu/LA_7_v1.4 Classical and exceptional Lie algebras.pdf[so(2n+ 1)] and Dn [so(2n)] turn out to give a useful

7. The classical and exceptional Lie algebras

* version 1.4 *

Matthew Foster

November 14, 2016

Contents

7.1 su(n): An−1 17.1.1 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 fromthe Cartan matrix or Dynkin diagram. The latter encodes the inner products and norm ratios of the simple roots. In this module, weassemble the Cartan matrix from the simple roots for the four classical Lie algebra families An, Bn, Cn, and Dn. The needed datais extracted from the defining representation for each algebra. We will also state (but not derive) the results for the five exceptionalalgebras 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 (finitedimensional) 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 exploitthis 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:

(Eab

)

ij= δaiδbj ⇒ Eab →

··· b−1 b b+1 ···

...a−1

a 1a+1

...

. (7.1.1)

1

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All unmarked entries are zero. These elements satisfy the multiplication rule

Eab · Ecd →(Eab

)

ij

(Ecd

)

jk= δai δbj δcj δdk → δbc Ead, (7.1.2)

so that

[Eab, Ecd] = δbc Ead − δda Ecb. (7.1.3)

Note that∑

c

[Eab, Ecc] = 0, (7.1.4)

which implies that the identity

1n =∑

c

Ecc (7.1.5)

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 elementof the Cartan subalgebra can be written as

H =

n∑

i=1

λi Eii,

n∑

i=1

λi = 0. (7.1.6)

Now

adh (eab) → [H, Eab] = λa Eab − λb Eab, (7.1.7)

which implies that eab is a root vector:

[h, eab] = αab(h) eab, αab

(∑

i

λieii

)= λa − λb. (7.1.8)

We identify the positive root vectors with set of upper triangular {Eab}. For a generic Cartan subalgebra element h ∈ Hdefined 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)e12 ≡ eα1

α1(h) = λ1 − λ2

e23 ≡ eα2 α2(h) = λ2 − λ3

......

en−1,n ≡ eαn−1αn−1(h) = λn−1 − λn

(7.1.9)

Proof: A generic positive root is

αij(h) = λi − λj , i < j. (7.1.10)

This can be written as

αij(h) = (λi − λi+1) + (λi+1 − λi+2) + (λi+2 − λi+3) + . . . + (λj−1 − λj) =

j−1∑

l=i

αl. (7.1.11)

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 ispositive, but not simple. Then

αi(h) = λi − λi+1 =∑

j 6=i

κ(i)

j αj(h) = . . . + κ(i)

i−1(λi−1 − λi) + κ(i)

i+1(λi+1 − λi+2) + . . . (7.1.12)

Since all the κ(i)

j are positive, there is no way to get λi+1 with a negative coefficient (contradiction).

2

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For the simple roots, we therefore have the associated Lie brackets

[eαi, eαj

] = δj,i+1 ei,i+2 − δj,i−1 ei−1,i+1,

[eαi, e−αj

] = δij (eii − ei+1,i+1) ,

[h, eαi] = (λi − λi+1) eαi

, h defined as in Eq. (7.1.6).

(7.1.13a)

(7.1.13b)

(7.1.13c)

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 thesimple roots. Consider the Killing form between generic elements of H . We use the algorithm described in module 4 [Eq. (4.2.2)].For

h ≡∑

i

λieii, h′ ≡∑

j

λ′jejj , (7.1.14)

we have the Killing form

(h, h′) =

n∑

a,ba6=b

=1

([h, [h′, eab]]

)

coeff. of eab

=

n∑

a,b=1

(λa − λb)(λ′a − λ′

b) = 2n

n∑

a=1

λaλ′a, (7.1.15)

since∑

b λb = 0. Therefore, since

(hαi, h′) = αi(h

′) = λ′i − λ′

i+1, (7.1.16)

we conclude that

hαi=

eii − ei+1,i+1

2n≡

n∑

j=1

λ(αi)

j ejj, λ(αi)

j =1

2n(δi,j − δi+1,j) . (7.1.17)

Recall that [Eq. (4.4.10)]

hαi=

[eαi, e−αi

]

(eαi, e−αi

). (7.1.18)

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

=(λ

(αi)

j − λ(αi)

j+1

)eαj

=1

2n(δi,j − δi+1,j − δi,j+1 + δi+1,j+1) eαj

.(7.1.20)

Therefore

〈αi, αj〉 =1

2n(2δi,j − δi+1,j − δi,j+1) , simple root scalar products in An−1. (7.1.21)

The Cartan matrix is defined as [Eqs. (5.3.8) and (5.3.9)]

Aij =2〈αi, αj〉〈αj, αj〉

= −pαi;αj(i 6= j), (7.1.22)

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

Aij = 2δij − δi+1,j − δi,j+1 ⇒ A →

2 −1 0 . . . 0 0−1 2 −1 . . . 0 00 −1 2 . . . 0 0...

...... . . .

......

0 0 0 . . . 2 −10 0 0 . . . −1 2

, Cartan matrix for An−1. (7.1.23)

3

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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

]

(α1) (α2) (α3)(7.1.24a)

2. At k = 2, the sum ρ(2)

1 ≡ α1 + α2 is a valid root since [Eq. (7.1.22)]

pα1 ;α2 = 1,

pα2 ;α1 = 1.(7.1.24b)

The sum ρ(2)

2 ≡ α2 + α3 is a valid root since

pα2 ;α3 = 1,

pα3 ;α2 = 1.(7.1.24c)

The Dynkin labels for these level 2 roots are

[1 1 −1

] [−1 1 1

]

ρ(2)1 ρ

(2)2

(7.1.24d)

The non-zero depths are

mρ(2)1 ;α1

= mρ(2)1 ;α2

= 1, mρ(2)2 ;α2

= mρ(2)1 ;α3

= 1. (7.1.24e)

3. At k = 3, we compute

pρ(2)1 ;α1

=mρ(2)1 ;α1

− 1 = 0,

pρ(2)1 ;α2

=mρ(2)1 ;α2

− 1 = 0,

pρ(2)1 ;α2

=mρ(2)1 ;α2

+ 1 = 1,

pρ(2)2 ;α1

=mρ(2)1 ;α1

+ 1 = 1,

pρ(2)2 ;α2

=mρ(2)1 ;α2

− 1 = 0,

pρ(2)2 ;α2

=mρ(2)1 ;α2

− 1 = 0.

(7.1.24f)

Thus ρ(3) ≡ ρ(2)

1 + α3 = ρ(2)

2 + α1 = α1 + α2 + α3 is the only k = 3 root,

[1 0 1

]

ρ(3)(7.1.24g)

The depths are

mρ(3);α1= 1, mρ(3) ;α2

= 0, mρ(3) ;α3= 1. (7.1.24h)

4

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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);α1=mρ(3) ;α1

− 1 = 0,

pρ(3);α2=mρ(3) ;α2

− 0 = 0, pρ(3);α3= mρ(3) ;α3

− 1 = 0.(7.1.24i)

We conclude that there are no level k = 4 roots. ρ(3) = α1 + α2 + α3 ≡ θ is the highest root.

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 theset 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 andthe inventory of roots (Fig. 7.2), one finds that the roots correspond to the twelve vertices of a certain quasiregular polyhedron (acuboctahedron). 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 U on a 2n-component complex vector are

U †U = 12n, (7.2.1a)

UT ε U = ε, (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

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where the 2n× 2n matrix ε is a block rank-2 Levi-Civita symbol,

ε →[

0 1n

−1n 0

]. (7.2.2)

If we express U as the exponentiation of an (antihermitian) 2n × 2n matrix Y

U = exp(Y ),

then the symplectic condition translates to

Y T = ε Y ε. (7.2.3)

As with ε, we decompose this into n × n blocks,

Y =

[Y1 Y2

Y3 Y4

]. (7.2.4)

Eq. (7.2.3) then implies that

Y4 = − Y T

1 ,

Y2 = Y T

2 ,

Y3 = Y T

3 .

(7.2.5)

Let 1 ≤ j, k ≤ n. We can define the following basis elements for Y ,

E1jk ≡ Ejk − Ek+n,j+n, (7.2.6a)

E2jk ≡ Ej,k+n + Ek,j+n = E2

(jk), (7.2.6b)

E3jk ≡ Ej+n,k + Ek+n,j = E3

(jk). (7.2.6c)

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.)

E134 →

0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 00 0 0 1 0 0 0 0 0 00 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 −1 0 00 0 0 0 0 0 0 0 0 0

, (7.2.7a)

E234 →

0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 1 00 0 0 0 0 0 0 1 0 00 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0

, (7.2.7b)

6

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E325 →

0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 00 0 0 0 1 0 0 0 0 00 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 00 1 0 0 0 0 0 0 0 0

. (7.2.7c)

Using Eq. (7.1.3), one can verify the following Lie brackets implied by the corresponding commutation relations:

[e1jk, e1

lp] = δkl e1jp − δjp e1

lk, (7.2.8a)

[e1jk, e2

lp] = δkl e2jp + δpk e2

jl, (7.2.8b)

[e1jk, e3

lp] = − δjp e3kl − δjl e

3kp, (7.2.8c)

[e2jk, e2

lp] = 0, (7.2.8d)

[e3jk, e3

lp] = 0, (7.2.8e)

[e2jk, e3

lp] = δkl e1jp + δjp e1

kl + δpk e1jl + δjl e1

kp. (7.2.8f)

• Exercise: Verify all of the Lie brackets in Eq. (7.2.8).

A basis for the n-dimensional Cartan subalgebra H is given by the diagonal traceless elements {e1jj}, 1 ≤ j ≤ n; a generic

element h ∈ H is

h =

n∑

j=1

λj e1jj. (7.2.9)

Then

[h, e1jk] = (λj − λk) e1

jk,

[h, e2jk] = (λj + λk) e2

jk,

[h, e3jk] = − (λj + λk) e3

jk,

(7.2.10a)

(7.2.10b)

(7.2.10c)

so that the collection of {e1,2,3jk } excluding diagonal {e1

jj} is the set of root vectors.The simple roots can be associated to the root vectors

root vector root α(h)e112 ≡ eα1

α1(h) = λ1 − λ2

e123 ≡ eα2 α2(h) = λ2 − λ3

......

e1n−1,n ≡ eαn−1

αn−1(h) = λn−1 − λn

e2n,n ≡ eαn

αn(h) = 2λn.

(7.2.11)

By definition, we take α1 > α2 > . . . > αn (lexicographic ordering). Then positive roots are associated to root vectors {e1jk} with

j < k, and to {e2jk}. Negative root vectors are {e1

jk} (j > k) and {e3jk}. The positive root associated to e1

jk (j < k) is expressed in

terms of simple roots identically as in Eq. (7.1.11). For e2ij , the decomposition is (taking j ≥ i WLoG)

λi + λj = (λi − λj) + 2(λj − λn) + 2λn =

j−1∑

l=i

αl + 2

n−1∑

l=j

αl + αn. (7.2.12)

7

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The commutation relations between the simple root vectors are given by

[eαi, eαj

] = δj,i+1 e1i,i+2 − δj,i−1 e1

i−1,i+1, 1 ≤ i, j ≤ n − 1,

[eαi, eαn

] = 2δi,n−1 e2n−1,n, 1 ≤ i ≤ n − 1,

[eαi, e−αj

] = δij

(e1i,i − e1

i+1,i+1

), 1 ≤ i, j ≤ n − 1,

[eαi, e−αn

] = 0, 1 ≤ i ≤ n − 1,

[eαn, e−αn

] = 4 e1nn.

(7.2.13a)

(7.2.13b)

(7.2.13c)

(7.2.13d)

(7.2.13e)

The Killing form between Cartan subalgebra elements

h ≡∑

i

λie1ii, h′ ≡

j

λ′je

1jj , (7.2.14)

is given by

(h, h′) =

n∑

p,qp6=q

=1

( [h,[h′, e1

pq

]] )

coeff. of e1pq

+

n∑

p≥q=1

( [h,[h′, e2

pq

]] )

coeff. of e2pq

+

n∑

p≥q=1

([h,[h′, e3

pq

]] )

coeff. of e3pq

=n∑

p,q=1

(λp − λq)(λ′p − λ′

q) + 2n∑

p≥q=1

(λp + λq)(λ′p + λ′

q)

=

n∑

p,q=1

(λp − λq)(λ′p − λ′

q) +

n∑

p,q=1

(λp + λq)(λ′p + λ′

q) + 4

n∑

p=1

λpλ′p

=4(n + 1)

n∑

p=1

λpλ′p. (7.2.15)

Now since

(hαi, h′) = αi(h

′) =

{λ′

i − λ′i+1, 1 ≤ i ≤ n − 1,

2λ′n, i = n,

(7.2.16)

we conclude that

hαi=

1

4(n + 1)

(e1ii − e1

i+1,i+1

)≡

n∑

j=1

λ(αi)

j e1jj, λ

(αi)

j =1

4(n + 1)(δi,j − δi+1,j) , 1 ≤ i ≤ n − 1,

hαn=

1

2(n + 1)e1nn ≡

n∑

j=1

λ(αn)

j e1jj, λ(αn)

j =1

2(n + 1)δj,n.

(7.2.17)

Eqs. (7.2.13c), (7.2.13e), (7.2.17) and (7.1.18) therefore give the Killing forms of the simple root vectors,

(eαi, e−αi

) =4(n + 1), 1 ≤ i ≤ n − 1,

(eαn, e−αn

) =8(n + 1).(7.2.18)

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

[hαi, eαj

] = 〈αi, αj〉eαj

=(λ

(αi)

j − λ(αi)

j+1

)eαj

=1

4(n + 1)(2δi,j − δi+1,j − δi,j+1) eαj

, 1 ≤ i, j ≤ n − 1,

[hαn, eαj

] = 〈αn, αj〉eαj

=(λ(αn)

j − λ(αn)

j+1

)eαj

= − 1

2(n + 1)δj+1,n eαj

, 1 ≤ j ≤ n − 1,

[hαn, eαn

] = 〈αn, αn〉eαn

=1

(n + 1)eαn

.

(7.2.19)

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Figure 7.4: sp(2n) = Cn.

Therefore

〈αi, αj〉 =1

4(n + 1)(2δi,j − δi+1,j − δi,j+1) , 1 ≤ i, j ≤ n − 1,

〈αi, αn〉 = − 1

2(n + 1)δi,n−1, 1 ≤ i ≤ n − 1,

〈αn, αn〉 =1

(n + 1),

simple root scalar products in Cn. (7.2.20)

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

A →

2 −1 0 . . . 0 0−1 2 −1 . . . 0 00 −1 2 . . . 0 0...

...... . . .

......

0 0 0 . . . 2 −10 0 0 . . . −2 2

, Cartan matrix for Cn. (7.2.21)

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. FromEq. (2A.2.2) in module 2A, the defining condition for an orthogonal transformation U on a 2n-component vector is

UTU = 12n. (7.3.1)

If we express U as the exponentiation of an (antihermitian) 2n × 2n matrix Y

U = exp(Y ),

then we have the antisymmetry condition

−Y T = Y. (7.3.2)

Since we would like to identify Cartan subalgebra elements with diagonal matrices, Eq. (7.3.2) is inconvenient. Instead, we make aunitary (but not orthogonal) basis transformation,

U ′ ≡ V † U V, V †V = 12n. (7.3.3)

Then

U ′TMU ′ = M,

M ≡ V TV = MT, M †M = 12n.(7.3.4)

For U ′ = exp(Y ′),

−M † Y ′T M = Y ′. (7.3.5)

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We choose V to have the block form

V =1√2

[i1n −i1n

−1n −1n

], ⇒ M = M † =

[0 1n

1n 0

]. (7.3.6)

As with M , we decompose Y ′ into n × n blocks,

Y ′ =

[Y1 Y2

Y3 Y4

]. (7.3.7)

Eq. (7.3.5) then implies that

Y4 = − Y T

1 ,

Y2 = − Y T

2 ,

Y3 = − Y T

3 .

(7.3.8)

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 Y ,

E1jk ≡ Ejk − Ek+n,j+n, (7.3.9a)

E2jk ≡ Ej,k+n − Ek,j+n = E2

[jk], (7.3.9b)

E3jk ≡ Ej+n,k − Ek+n,j = E3

[jk]. (7.3.9c)

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),

E234 →

0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 1 00 0 0 0 0 0 0 −1 0 00 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0

, (7.3.10a)

E325 →

0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 00 0 0 0 1 0 0 0 0 00 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 00 −1 0 0 0 0 0 0 0 0

. (7.3.10b)

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

[e1jk, e1

lp] = δkl e1jp − δjp e1

lk, (7.3.11a)

[e1jk, e2

lp] = δkl e2jp − δpk e2

jl, (7.3.11b)

[e1jk, e3

lp] = δjp e3kl − δjl e

3kp, (7.3.11c)

[e2jk, e2

lp] = 0, (7.3.11d)

[e3jk, e3

lp] = 0, (7.3.11e)

[e2jk, e3

lp] = δkl e1jp + δjp e1

kl − δpk e1jl − δjl e1

kp. (7.3.11f)

• Exercise: Verify the Lie brackets in Eq. (7.3.11).

A generic Cartan subalgebra H element is given by

h =

n∑

j=1

λj e1jj. (7.3.12)

Then

[h, e1jk] = (λj − λk) e1

jk,

[h, e2jk] = (λj + λk) e2

jk,

[h, e3jk] = − (λj + λk) e3

jk.

(7.3.13a)

(7.3.13b)

(7.3.13c)

The simple roots can be associated to the root vectors

root vector root α(h)

e112 ≡ eα1 α1(h) = λ1 − λ2

e123 ≡ eα2

α2(h) = λ2 − λ3

......

e1n−1,n ≡ eαn−1 αn−1(h) = λn−1 − λn

e2n−1,n ≡ eαn

αn(h) = λn−1 + λn.

(7.3.14)

By definition, we take α1 > α2 > . . . > αn (lexicographic ordering). Then positive roots are associated to root vectors {e1jk} with

j < k, and to {e2jk}. Negative root vectors are {e1

jk} (j > k) and {e3jk}. The positive root associated to e1

jk (j < k) is expressed in

terms of simple roots identically as in Eq. (7.1.11). For e2ij , the decomposition is (taking j > i WLoG)

λi + λj = (λi − λj) + 2(λj − λn−1) + (λn−1 − λn) + (λn−1 + λn) =

j−1∑

l=i

αl + 2

n−2∑

l=j

αl + αn−1 + αn. (7.3.15)

The commutation relations between the simple root vectors are given by

[eαi, eαj

] = δj,i+1 e1i,i+2 − δj,i−1 e1

i−1,i+1, 1 ≤ i, j ≤ n − 1,

[eαi, eαn

] = δi,n−2 e2n−2,n, 1 ≤ i ≤ n − 1,

[eαi, e−αj

] = δij

(e1i,i − e1

i+1,i+1

), 1 ≤ i, j ≤ n − 1,

[eαi, e−αn

] = 0, 1 ≤ i ≤ n − 1,

[eαn, e−αn

] = e1n−1,n−1 + e1

n,n.

(7.3.16a)

(7.3.16b)

(7.3.16c)

(7.3.16d)

(7.3.16e)

The Killing form between Cartan subalgebra elements

h ≡∑

i

λie1ii, h′ ≡

j

λ′je

1jj , (7.3.17)

is given by

(h, h′) = 4(n − 1)

n∑

p=1

λpλ′p. (7.3.18)

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• Exercise: Verify Eq. (7.3.18).

Now since

(hαi, h′) = αi(h

′) =

{λ′

i − λ′i+1, 1 ≤ i ≤ n − 1,

λ′n−1 + λ′

n, i = n,(7.3.19)

we conclude that

hαi=

1

4(n − 1)

(e1ii − e1

i+1,i+1

)≡

n∑

j=1

λ(αi)

j e1jj, λ

(αi)

j =1

4(n − 1)(δi,j − δi+1,j) , 1 ≤ i ≤ n − 1,

hαn=

1

4(n − 1)

(e1n−1,n−1 + e1

n,n

)≡

n∑

j=1

λ(αn)

j e1jj, λ(αn)

j =1

4(n − 1)(δn−1,j + δn,j) .

(7.3.20)

Eqs. (7.3.16c), (7.3.16e), (7.3.20) and (7.1.18) therefore give the Killing forms of the simple root vectors,

(eαi, e−αi

) = 4(n − 1), 1 ≤ i ≤ n. (7.3.21)

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

[hαi, eαj

] =1

4(n − 1)(2δi,j − δi+1,j − δi,j+1) eαj

, 1 ≤ i, j ≤ n − 1,

[hαn, eαj

] = − 1

4(n − 1)δj,n−2 eαj

, 1 ≤ j ≤ n − 1,

[hαn, eαn

] =2

4(n − 1)eαn

.

(7.3.22)

Therefore

〈αi, αj〉 =1

4(n − 1)(2δi,j − δi+1,j − δi,j+1) , 1 ≤ i, j ≤ n − 1,

〈αi, αn〉 = − 1

4(n − 1)δi,n−2, 1 ≤ i ≤ n − 1,

〈αn, αn〉 =2

4(n − 1),

simple root scalar products in Dn. (7.3.23)

Since all roots have the same length, Dn is simply laced.The corresponding Cartan matrix is given by [via Eq. (7.1.22)]

A →

2 −1 0 . . . 0 0 0−1 2 −1 . . . 0 0 00 −1 2 . . . 0 0 0...

...... . . .

......

...0 0 0 . . . 2 −1 −10 0 0 . . . −1 2 00 0 0 . . . −1 0 2

, Cartan matrix for Dn. (7.3.24)

The Dynkin diagram is shown in Fig. 7.5.

Figure 7.5: so(2n) = Dn.

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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 from0 ≤ i ≤ 2n.

As in Sec. 7.3, we begin by rotating the orthogonal condition UTU = 12n+1 to a different basis. Applying the transformationas in Eq. (7.3.3), we choose

V =1√2

√2 01,n 01,n

0n,1 i1n −i1n

0n,1 −1n −1n

, ⇒ M = V TV =

1 01,n 01,n

0n,1 0 1n

0n,1 1n 0

. (7.4.1)

Here 01,n (0n,1) denotes an n-fold row (column) of zeroes. The (2n+1)×(2n+1) generator Y ′ satisfies the condition as in Eq. (7.3.5),which allows the block decomposition [c.f. Eqs. (7.3.7) and (7.3.8)]

Y ′ =

b1 c1

1,n c21,n

d1n,1 Y1 Y2

d2n,1 Y3 Y4

. (7.4.2)

Eq. (7.3.5) then implies that

Y4 = − Y T

1 ,

Y2 = − Y T

2 ,

Y3 = − Y T

3 ,

b1 =0,

c11,n = −

[d2

n,1

]T,

c21,n = −

[d1

n,1

]T.

(7.4.3)

The basis elements for Y include {E1jk, E

2jk, E

3jk} (1 ≤ j, k ≤ n) from Eq. (7.3.9), and

E4j ≡ E0,j − Ej+n,0, (7.4.4a)

E5j ≡ Ej,0 − E0,j+n, (7.4.4b)

both with 1 ≤ j ≤ n. The total number of such elements is

n(2n − 1) + 2n = n(2n + 1).

Relative to the example in Eq. (7.3.10), {E1jk, E

2jk, E

3jk} are now block embedded in 11 × 11 matrices (after one initial row and one

initial column of zeroes), while (e.g.)

E44 →

0 0 0 0 1 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0−1 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0

, (7.4.5a)

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E53 →

0 0 0 0 0 0 0 0 −1 0 00 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 01 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0

. (7.4.5b)

In addition to the Lie brackets in Eq. (7.3.11), we have

[e1jk, e4

l ] = − δjl e4k, (7.4.6a)

[e1jk, e5

l ] = δkl e5j , (7.4.6b)

[e2jk, e4

l ] = δjl e5k − δkl e

5j , (7.4.6c)

[e2jk, e5

l ] = 0, (7.4.6d)

[e3jk, e4

l ] = 0, (7.4.6e)

[e3jk, e5

l ] = δjl e4k − δkl e

4j , (7.4.6f)

[e4j , e

4k] = − e3

jk, (7.4.6g)

[e5j , e

5k] = − e2

jk, (7.4.6h)

[e4j , e

5k] = − e1

kj . (7.4.6i)

• Exercise: Verify the Lie brackets in Eq. (7.4.6).

A generic Cartan subalgebra H element is given by

h =

n∑

j=1

λj e1jj. (7.4.7)

Then

[h, e1jk] = (λj − λk) e1

jk,

[h, e2jk] = (λj + λk) e2

jk,

[h, e3jk] = − (λj + λk) e3

jk,

[h, e4j ] = − λj e4

j ,

[h, e5j ] =λje

5j .

(7.4.8a)

(7.4.8b)

(7.4.8c)

(7.4.8d)

(7.4.8e)

The simple roots can be associated to the root vectors

root vector root α(h)

e112 ≡ eα1 α1(h) = λ1 − λ2

e123 ≡ eα2 α2(h) = λ2 − λ3

......

e1n−1,n ≡ eαn−1 αn−1(h) = λn−1 − λn

e5n ≡ eαn

αn(h) = λn.

(7.4.9)

By definition, we take α1 > α2 > . . . > αn (lexicographic ordering). The positive roots are associated to root vectors {e1jk} with

j < k, {e2jk}, and {e5

j}. Negative root vectors are {e1jk} (j > k), {e3

jk}, and {e4j}. The positive root associated to e1

jk (j < k) is

expressed in terms of simple roots identically as in Eq. (7.1.11). For e2ij, the decomposition is (taking j > i WLoG)

λi + λj = (λi − λj) + 2(λj − λn) + 2λn =

j−1∑

l=i

αl + 2

n−1∑

l=j

αl + 2αn. (7.4.10)

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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 inEq. (7.4.10). In fact we will see that Bn and Cn are closely related (although not identical, except for B2 = C2). For e5

i , thedecomposition is

λi = (λi − λn) + λn =

n−1∑

l=i

αl + αn. (7.4.11)

The commutation relations between the simple root vectors are given by

[eαi, eαj

] = δj,i+1 e1i,i+2 − δj,i−1 e1

i−1,i+1, 1 ≤ i, j ≤ n − 1,

[eαi, eαn

] = δi,n−1 e5n−1, 1 ≤ i ≤ n − 1,

[eαi, e−αj

] = δij

(e1i,i − e1

i+1,i+1

), 1 ≤ i, j ≤ n − 1,

[eαi, e−αn

] = 0, 1 ≤ i ≤ n − 1,

[eαn, e−αn

] = e1nn.

(7.4.12a)

(7.4.12b)

(7.4.12c)

(7.4.12d)

(7.4.12e)

The Killing form between Cartan subalgebra elements

h ≡∑

i

λie1ii, h′ ≡

j

λ′je

1jj , (7.4.13)

is given by

(h, h′) =4

(n − 1

2

) n∑

p=1

λpλ′p. (7.4.14)

• Exercise: Verify Eq. (7.4.14).

Now since

(hαi, h′) = αi(h

′) =

{λ′

i − λ′i+1, 1 ≤ i ≤ n − 1,

λ′n, i = n,

(7.4.15)

we conclude that

hαi=

1

4(n − 1

2

)(e1ii − e1

i+1,i+1

)≡

n∑

j=1

λ(αi)

j e1jj, λ

(αi)

j =1

4(n − 1

2

) (δi,j − δi+1,j) , 1 ≤ i ≤ n − 1,

hαn=

1

4(n − 1

2

)e1n,n ≡

n∑

j=1

λ(αn)

j e1jj, λ(αn)

j =1

4(n − 1

2

)δn,j.

(7.4.16)

Eqs. (7.4.12c), (7.4.12e), (7.4.16) and (7.1.18) therefore give the Killing forms of the simple root vectors,

(eαi, e−αi

) = 4

(n − 1

2

), 1 ≤ i ≤ n. (7.4.17)

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

[hαi, eαj

] =1

4(n − 1

2

) (2δi,j − δi+1,j − δi,j+1) eαj, 1 ≤ i, j ≤ n − 1,

[hαn, eαj

] = − 1

4(n − 1

2

) δj,n−1 eαj, 1 ≤ j ≤ n − 1,

[hαn, eαn

] =1

4(n − 1

2

)eαn.

(7.4.18)

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

Therefore

〈αi, αj〉 =1

4(n − 1

2

) (2δi,j − δi+1,j − δi,j+1) , 1 ≤ i, j ≤ n − 1,

〈αi, αn〉 = − 1

4(n − 1

2

)δi,n−1, 1 ≤ i ≤ n − 1,

〈αn, αn〉 =1

4(n − 1

2

) .

simple root scalar products in Bn. (7.4.19)

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

A →

2 −1 0 . . . 0 0−1 2 −1 . . . 0 00 −1 2 . . . 0 0...

...... . . .

......

0 0 0 . . . 2 −20 0 0 . . . −1 2

, Cartan matrix for Bn. (7.4.20)

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, 1 ≤ i ≤ n− 1. The first n− 1simple 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 thesame geometry, except that long and short roots are interchanged.

On the other hand, we will see that Dn and Bn have more in common in terms of the n simplest (so called fundamental1)representations, described in module 8. In particular, both Dn and Bn possess fundamental spinor representations, which arequalitatively distinct and cannot be built by tensoring together defining vector indices. By contrast, every representation of Cn orAn 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 inSec. 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 thesame rank are indicated, and easily recognized by the identical Dynkin diagrams. All four families are distinct for each rank abovethree. 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 surprisingthat there are only five additional possibilities corresponding to particular algebras of fixed rank. This can be shown by a carefuldetermination 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 algebraL, there exists of set of “simplest” representations (which always includes the defining one), and mathematicians refer to each member of this set as afundamental representation. We will define fundamental represenations precisely in module 8, using the fundamental weights.

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Table 1: Classical Lie algebras to rank 4.

Rank Dynkin Cartan Compact # of positive roots

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

A2 su(3) 32 B2 = C2 so(5) = sp(4) 4

D2 = A1 × A1 so(4) = su(2) × su(2) 2

A3 = D3 su(4) = so(6) 63 B3 so(7) 9

C3 sp(6) 9

A4 su(5) 10B4 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].

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

AG2 =

[2 −3−1 2

], AE6 =

2 −1 0 0 0 0−1 2 −1 0 0 00 −1 2 −1 0 −10 0 −1 2 −1 00 0 0 −1 2 00 0 −1 0 0 2

,

AF4 =

2 −1 0 0−1 2 −2 00 −1 2 −10 0 −1 2

, AE7 =

2 −1 0 0 0 0 0−1 2 −1 0 0 0 00 −1 2 −1 0 0 −10 0 −1 2 −1 0 00 0 0 −1 2 −1 00 0 0 0 −1 2 00 0 −1 0 0 0 2

,

AE8 =

2 −1 0 0 0 0 0 0−1 2 −1 0 0 0 0 00 −1 2 −1 0 0 0 00 0 −1 2 −1 0 0 00 0 0 −1 2 −1 0 −10 0 0 0 −1 2 −1 00 0 0 0 0 −1 2 00 0 0 0 −1 0 0 2

.

(7.6.1)

(7.6.2)

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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 Senechal, Conformal Field Theory (Springer-Verlag, New York, 1996).

18