Q. HO-KIM--Group Theory: A Problem Book

Group TheoryA Problem Book

Q. Ho-Kim

c© Q. Ho-Kim (Hồ Kim Quang). Group Theory: A Problem Book 2015-12.Cover illustration from image by Jo Edkins, used with permission.

1

Group Theory: A Problem Book

This book contains suggested solutions to the end-of-chapter problemsin Group Theory: A Physicist’s Primer by the same author, to bereferred to as GTAPP, which is freely available athttp://www.scribd.com/doc/207786199/Group-Theory-A-Physicist-s-Primer.Additional information on notations and definitions can be found inthe corresponding chapters of GTAPP.

Problems from Chapter 1 1–10Problems from Chapter 2 11–27Problems from Chapter 3 27–48Problems from Chapter 4 48–70Problems from Chapter 5 71–79Problems from Chapter 6 79–86Problems from Chapter 7 87-99Subject Index 100–101

Chapter 1. Group Structure

1.1 Show that the set of all rotations Rz(α) around the z axis by an angle αacting on the coordinates (x, y) for all values of α forms a group.

SOLUTION 1.1 In an active transformation, r is rotated by an angle α aboutthe coordinate origin to become r′ = (x′, y′) in the original frame ex, ey:

r 7→ r′ = Rz(α)r :

{

x′ = x cos α − y sin αy′ = x sin α + y cos α

Making use of the identities cos(a + b) = cos a cos b− sin a sin b and sin(a +b) = sin a cos b + cos a sin b, we obtain Rz(α)Rz(β) = Rz(α + β)by ordinarymatrix multiplication. It follows that: (i) Rz(α + β)Rz(γ) = Rz(α)Rz(β + γ);(ii) Rz(α)Rz(0) = Rz(α), which means Rz(0) = 1 is the identity element;(iii) Rz(α)Rz(−α) = Rz(0), which means Rz(−α) is the inverse of Rz(α).Since Rz(α)TRz(α) = Rz(−α)Rz(α) = 1, with T denoting transposition ofmatrix, we have a continuous abelian group formed by a set of orthogonal2 × 2 real matrices with matrix product as the group binary operation. �

2 CHAPTER 1

1.2 Prove that a group is abelian if the order of any of its elements, exceptthe identity, is 2.

SOLUTION 1.2 Let a, b 6= e be any two elements of a group G; then ab = c isalso in G. We want to prove that ab = ba. By assumption, a2 = b2 = c2 = e.First, we have c · b = ab · b = ab2 = a → c · a = c · cb = c2b = b. And soba = ca · a = c. Since ab = c, we have proved that ab = ba, for any a, b ∈ G.Thus G is abelian. �

1.3 Find the structure of the groups of order 4, 6, 8, and 10 with elementsall of order 2.

SOLUTION 1.3 We know from the preceding problem that any group withall elements, other than the identity, of order 2 is an abelian group. So thegroups to be found must be abelian.

|G| = 4: G = {e, a, b, c}, with a2 = b2 = c2 = e. The product ab isin G, but it cannot be e (because then a = b−1 = b), nor can it be a or b(because then b = e or a = e); so it must be c. We have then the four-groupV ∼= C2 × C2.

|G| = 6: G = {e, a, . . . , c, d, . . .}. Consider the product of any two ele-ments cd ∈ G. It cannot be e, c, or d because that would lead to contradic-tions (just as in the previous case), so it must be some distinct element, sayt. Then e, c, d, t form the group V. By Lagrange’s theorem (GTAPP § 1.3), 4must divide 6, but it does not. So an order-6 group with all order-2 elementscannot exist.

|G| = 8: An argument similar to the case of order 4 leads to a group ofthe form G = {e, a, b, ab, σ, σa, σb, σab} which is abelian and isomorphic tothe direct product V × C2

∼= C2 × C2 × C2.|G| = 10: An argument similar to the case of order 6 shows that this

group cannot exist, because 4 is not a divisor of 10. �

1.4 Prove: the intersection of two subgroups of a group is also a subgroup.

SOLUTION 1.4 Let H1 and H2 be two subgroups of a group G, and H3 be theset of common elements a, b, . . . (the intersection) of H1 and H2. We wantto verify that H3 satisfies the group axioms. (i) Since H3 is a set in G, itselements satisfy G’s multiplication law and associativity. (ii) The productab of two common elements of H1 and H2 belongs to both subgroups, andtherefore must belong to H3. (iii) The unique identity element of G is com-mon to H1 and H2, therefore is in H3. (iv) The inverse a−1 of any elementa ∈ H3 must be in both H1 and H2, and therefore also in H3. The intersec-tion of any subgroups of a group contains at least one element in common,the identity. �

1.5 Use Cayley’s theorem to find all distinct groups of order 6.

SOLUTION 1.5 It follows from Cayley’s theorem that all groups of order nare isomorphic to the regular-permutations subgroups of Sn. Since regular

PROBLEMS & SOLUTIONS 3

permutations, when resolved into independent cycles, must have all cyclesequal in length `, ` must be a divisor of the order n. In the present case,n = 6 has divisors 6, 3, 2, 1.

First, suppose the group has a 6-cycle permutation p = (123456). Tak-ing successive powers, we have p2 = (135)(246), p3 = (14)(25)(36), p4 =(153)(264), p5 = (654321), and p6 = (1)(2)(3)(4)(5)(6) = e, which com-plete a period. So we have a structure isomorphic to C6.

Next, suppose the group has no 6-cycles, then we need two 3-cycles orthree 2-cycles for each permutation. Let us start with pa = (123)(654),which leads to p2

a = (321)(456) and p3a = e. We need another generator,

which must have order 2. We take pb = (14)(25)(36), which has p2b =

e. By direct calculation, we prove that pbpa = (15)(26)(34), and pb pa =p−1

a pb. Altogether, these relations suffice to define the group structure. It isisomorphic to D3.

Starting with other permutations leads to no new structures. So, up toisomorphisms, there are just two groups of order 6, namely C6 and D3. �

1.6 Let G be a group of even order |G| that contains a cyclic subgroup H =〈a〉 of order |H|, such that |H| = |G|/2. Find the coset of H in G, and,using Lagrange’s theorem (GTAPP § 1.3), identify the structure of G for|G| = 4, 6, 8.

SOLUTION 1.6 Since the subgroup H has order |H| = |G|/2, hence an indexkH = 2, it has only a single coset, which together with H itself gives backG. As its left coset equals its right coset, the subgroup H is invariant in G.We have the elements of H and its coset bH given by

H = {e = ap, a, . . . , ap−1}; p = |H|,bH = {b, ba . . . , bap−1} def

= {b0, b1, . . . , bp−1}

Later we will use the notations bj+p = bj, bj+m = bjam (with p = |H| and

j, m any integer). To solve this problem, we have to answer two questions:What are the orders of bj for j = 0, . . . , p − 1? What are the commutationrelations between the elements of G? We know, by Lagrange’s theorem(GTAPP § 1.3), that the order of an element is one of the divisors of |G|.Whereas order 1 corresponds to the identity element, an order equal to |G|is also ruled out in our case because of the presence of an element, a, oforder |G|/2. So we need to consider only divisors d such that 1 < d < |G|.

• Order |G| = 4 and divisor 2. With p = 2, we simply have subgroupH = {e = a2, a}, coset bH = {b, ba}, and group G = {e = a2, a, b, ba}.

Order of the elements: b2 must be one of the elements of G. It cannotbe b, because it would mean b = e. It cannot be ba, because it would meanb = a. It cannot be a, because that would imply b4 = a2 = e, i.e. b wouldhave order 4. Finally, b2 = e leads to no contradictions, and this gives 2 as

4 CHAPTER 1

the order of b. By the same reasoning we also can show that ba also hasorder 2. So we have a2 = b2 = (ba)2 = e.

Commutation relation: (ba)2 = e implies ba = a−1b−1 = ab.The resulting group G is isomorphic to V ∼= D2.• Order |G| = 6 and divisors 3, 2. Subgroup H = {e = a3, a, a2}, coset

bH = {b, ba, ba2}, and group G = {e, a, a2, b, ba, ba2}.Order of the elements: For any j, we cannot have b2

j = a or b2j = a2,

because in both cases, bj would have order 6. We cannot have b2j = bk (any

given j, some k), because it would mean bj would be one of the elementse, a, a2. We are left with the only possibility, b2

j = e for any j, which leads tono contradictions.

Commutation relations: First, (ba)2 = e means ba = a−1b−1 = a2b.Second, (ba2)2 = e means ba2 = a−2b−1 = ab.

G is isomorphic to D3.• Order |G| = 8 and divisors 4, 2. Subgroup H = {e = a4, a, a2, a3},

coset bH = {b, ba, ba2, ba3} = {b0, b1, b2, b3}, and group G = H + bH.Order of the elements: For any j, we cannot have b2

j = bk (any given j,

some k) because it would mean bj would be one of the elements e, a, a2, a3. Ifb2

j = a, then b8j = a4 = e and bj has order 8. If b2

j = a3, then b8j = a12 = e and

bj has order 8. All these possibilities are ruled out. Of the two remaining,the first b2

j = a2 implies b4j = e, i.e. 4th order elements; while the second,

b2j = e, or second-order elements.

− If b2j = a2 for all j, we can derive from the general relation bj+m = bja

m

several equivalent results:bjbj+m = am+2 , then bja

2 = am+2bj+m , a2bj+m = bjam+2. (Note: These

relations should be used only when all bj have order 4.)− If b2

j = e for all j, we have bjbj+m = am , then bjam = a−mbj . (Note:

These relations should be used only when all bj have order 2.)Therefore, there are three cases to be considered: All elements bj of

order 4; some bj of order 4, and some of order 2; and lastly, all bj of order 2.. Case 1: b2

j = a2, j = 0, 1, 2, 3. The canonical commutation relation is

bja2 = a2bj. This group is isomorphic to the quaternion group Q8 by the

correspondence a 7→ k and b 7→ j where j and k (together with i) are unitsof the quaternion algebra (see GTAPP §1.1 and 1.4).

. Case 2: b20 = b2

2 = e, b21 = b2

3 = a2. We can verify that bja = abj, fromwhich bja

m = ambj for j = 0, 1, 2, 3. Hence {e = a4, a, a2, a3, b, ba, ba2, ba3} isisomorphic to the direct product C4 × C2.

. Case 3: b2j = e, for j = 0, 1, 2, 3. The canonical commutation relation is

bja = a3bj. The group G is isomorphic to D4. �


1.7 Show that there are five distinct groups of order 8, namely, the cyclicgroup C8, the group of symmetries of the square D4, the quaternion groupQ8, and two product groups, C4 × C2 and D2 × C2.

SOLUTION 1.7 A group G of order 8 contains elements of order 1, 2, 4, 8.The only element of order 1 is the identity, now identify elements of order8, 4, 2.

If there is at least one element of order-8, we have the cyclic group C8.Now assume there is no order-8 elements, but there is at least one ele-

ment a of order 4. Then G contains the subgroup H = {a, a2, a3, a4 = e},which is invariant in G with index kH = 2. The problem reduces to deter-mining its coset bH = {b, ba, ba2, ba3}, already considered in Problem 1.6.We have learned then that G must be Q8, or D4, or C4 × C2.

Lastly, if there are no elements of order 8 or 4, then all elements, exceptthe identity, must have order 2. From the answer to Problem 1.3, we havelearned that G is an abelian group isomorphic to D2 × C2. �

1.8 Enumerate the conjugacy classes, subgroups and normal subgroups, aswell as the factor groups, of the symmetric groups S2, S3, and S4.

SOLUTION 1.8(a) Group S2 has two elements, e = (1)(2)and (12), forming two classes,

[e] and [(12)]. It has no proper subgroups, and is isomorphic to C2: S2∼= C2.

(b) Groups S3 has six elements divided into 3 conjugate classes, [e], [(12),(23), (31)], and [(123), (321)]. It contains four subgroups, three of whichhave order 2, and one of order 3, which are listed below together with theirrespective left cosets:

• H12 = {e, (12)}; (23)H1

2 = (321)H12 = {(23), (321)}

(31)H12 = (123)H1

2 = {(31), (123)}• H2

2 = {e, (23)}; (12)H22 = (123)H2

2 = {(12), (123)}(31)H2

2 = (321)H22 = {(31), (321)}

• H32 = {e, (31)}; (12)H3

2 = (321)H32 = {(12), (321)}

(23)H32 = (123)H3

2 = {(23), (123)}• H3 = {e, (123), (321)};

(12)H3 = (23)H3 = (31)H3 = {(12), (23), (31)} .

Of these subgroups of S3, there is just one invariant, H3∼= A3 (alternating

group). The corresponding factor group S3/H3 = S3/A3 is isomorphic toC2, or to any of the order-2 subgroups Hi

2.(c) Group S4, the symmetric group of degree 4, is of order 4! = 24.

Its 24 elements are divided into 5 classes, each with a characteristic cyclestructure:

6 CHAPTER 1

[e];[(12), (13), (14), (23), (24), (34)];[(12)(34), (13)(24), (14)(23)];[(123), (132), (124), (142), (134), (143), (234), (243)];[(1234), (1243), (1324), (1342), (1423), (1432) ].We can check the number of classes obtained by counting the ways in

which the degree n = 4 is partitioned into sums of integers:4 = 1 + 1 + 1 + 1 = 2 + 1 + 1 = 2 + 2 = 3 + 1 = 4 + 0.The order of a proper subgroup of S4 must be one of the divisors of 24,

that is 12, 8, 6, 4, 3, 2. We give below a partial list of the S4 subgroups:• Order 12: A4, isomorphic to the tetrahedral group T and clearly in-

variant in S4 since it contains elements in complete classes:T ∼= {e, (12), (13), (14), (23), (24), (34),

(123), (132), (124), (142), (134), (143), (234), (243)}.• Order 8: A subgroup isomorphic to D4

D4∼= {e, (13), (24), (13)(24), (12)(34), (14)(23), (1234), (1432)}.

• Order 6: Four subgroups, all isomorphic to S3 and composed of per-mutations of any three of the four objects, leaving the fourth unchanged.

• Order 4: Three subgroups isomorphic to C4, of the form {e, p, p2, p3},where p is (1234), or (1243), or (1324) with e = p4; plus one invariantsubgroup isomorphic to Klein’s four-group:

V ∼= {e, (12)(34), (13)(24), (14)(23)}.• Order 3: Four of the form {e = p3, p, p2}, with p any 3-cycle.• Order 2: Three subgroups with two 2-cycles of the form {e, (ij)(kl)},

plus six with one 2-cycle of the form {e, (ij)}.From the two invariant subgroups, T ∼= A4 and V ∼= D2, we construct

their cosets and the factor groups of S4, which leads to• S4/V = {V, (12)V, (23)V, (13)V, (123)V, (321)V}, isomorphic to S3.• S4/T = {T, (ij)T} with (ij) any 2-cycle. S4/T is isomorphic to C2. �

1.9 The dihedral group D4 is the group of the symmetry transformationsof a square, generated by a rotation b through an angle π about a diag-onal, and a counterclockwise rotation a through an angle π/2 about theaxis perpendicular to the square plane and passing through its center. Thegenerators satisfy the relations a4 = e, b2 = e, ba = a−1b.

(a) Give the multiplication table.(b) Identify the subgroup of S8 isomorphic to D4.(c) Enumerate all the conjugacy classes.(d) Enumerate all the subgroups, identifying the normal subgroups.(e) Give the cosets of the subgroups and the factor groups associated

with the normal subgroups.


SOLUTION 1.9 (a) D4 has 8 elements, e, b, ba, ba2, ba3, a, a2, a3, of respectiveorders 1, 2, 2, 2, 2, 4, 2, 4. Their pairwise products can be derived from thebasic relations between the generators of the group: a4 = e, b2 = e, ba =a−1b, from which we also have ab = ba−1. The group table is as follows:

D4 e b ba ba2 ba3 a a2 a3

e e b ba ba2 ba3 a a2 a3

b b e a a2 a3 ba ba2 ba3

ba ba a3 e a a2 ba2 ba3 bba2 ba2 a2 a3 e a ba3 b baba3 ba3 a a2 a3 e b ba ba2

a a ba3 b ba ba2 a2 a3 ea2 a2 ba2 ba3 b ba a3 e aa3 a3 ba ba2 ba3 b e a a2

(b) By Cayley’s theorem, the order-8 group D4 must be isomorphic toa subgroup of S8, formed by the regular permutations of 8 symbols. Eachelement g of D4 is mapped to such a permutation pg, as shown below (seeGTAPP § 1.6 for additional details):

initially → 1 2 3 4 5 6 7 8

e 7→ pe 1 2 3 4 5 6 7 8b 7→ pb 2 1 6 7 8 3 4 5

ba 7→ pba 3 8 1 6 7 4 5 2ba2 7→ pba2 4 7 8 1 6 5 2 3ba3 7→ pba3 5 6 7 8 1 2 3 4

a 7→ pa 6 5 2 3 4 7 8 1a2 7→ pa2 7 4 5 2 3 8 1 6a3 7→ pa3 8 3 4 5 2 1 6 7

The eight permutations are

pe = (1)(2) . . . (7)(8) ,

pb = (12)(36)(47)(58) ,

pba = (13)(28)(46)(57) ,

pba2 = (14)(27)(38)(56) ,

pba3 = (15)(26)(37)(48) ,

pa = (1678)(5432) ,

pa2 = (17)(24)(35)(68) ,

pa3 = (2345)(8761) .

8 CHAPTER 1

By the rules of permutation multiplication, we obtain the following rela-tions:

p4a = p2

b = pe; and pb pa = p−1a pb;

p2a = pa2 , p3

a = pa3 ;pb pa = pba, pb pa2 = pba2 , pb pa3 = pba3 .

They show that the eight permutations pi form a group isomorphic to D4.(c) One quick way of identifying conjugate elements is to place the g-

row and g-column, one above the other. The boxes that have the same en-tries in the row and the corresponding column indicate that the columnlabel and the row label are possible conjugates. Take for example the twopairs of rows and columns corresponding to elements b and ba, we see thate and a2 are self-conjugate, and a, a3; b, ba2; ba, ba3 are pairs of conjugates.

labels e b ba ba2 ba3 a a2 a3

row b b e a a2 a3 ba ba2 ba3

column b b e a3 a2 a ba3 ba2 ba

row ba ba a3 e a a2 ba2 ba3 bcolumn ba ba a e a3 a2 b ba3 ba2

Proceeding in this way for all elements g will produce the conjugate classes,• Order-1 element : [e] ;• Order-2 elements: [a2], [b, ba2], [ba, ba3] ;• Order-4 elements: [a, a3] .(d) Examining the multiplication table, we can identify the following

subgroups of D4:• Order-2 subgroups: H1

2 = {e, a2}, {e, b}, {e, ba}, {e, ba2}, {e, ba3} ;• Order-4: H1

4 = {e, a, a2, a3}, H24 = {e, b, ba2, a2}, H3

4 = {e, ba, ba3, a2} .The subgroups H1

2, Hi4 (i = 1, 2, 3), which contain complete classes, are

the invariant subgroups in D4.(e) Cosets of invariant subgroups and factor groups:− Cosets of H1

2 = {e, a2}: This subgroup has an index equal to 4, and sohas 3 cosets: bH1

2 = {b, ba2}, aH12 = {a, a3}, baH1

2 = {ba, ba3}. The squareof each coset is the subgroup itself, which means we can map H1

2 7→ e′,aH1

2 7→ a′, bH12 7→ b′ , baH1

2 7→ c′, with a′2 = b′2 = c′2 = e′. The factorgroup D4/H1

2 = {e′, a′, b′, c′} is isomorphic to V.H1

4 , H24, H3

4 have an index equal to 2, and so have one coset each:− Coset of H1

4: bH14 = {b, ba, ba2, ba3}, with (bH1

4)2 = H1

4.− Coset of H2

4: aH24 = {a, a3, ba, ba3}, with (aH2

4)2 = H2

4.− Coset of H3

4: aH3

4= {a, b, ba2, a3}, with (aH3

4)2 = H3

4.

In these cases the factor groups D4/Hi4 are isomorphic to C2. �

1.10 Let Dn be the dihedral group generated by a of order n and b of order2. Show that the subgroup 〈a〉 is normal in Dn.


SOLUTION 1.10 Let A = 〈a〉 = {ak; k = 0, . . . , n − 1}. Each element of Dn

is of the form g = aibj, with i = 0, . . . , n − 1 and j = 0, 1, defined so that(ab)2 = e. So we want to prove that gak g−1 ∈ A. The case of j = 0 is trivial,so we just consider j = 1.

From (ab)2 = e, we have bab−1 = bab = a−1 ∈ A. We also have a−k =an−k , where k = 0, . . . , n − 1. So that:

(i) bakb−1 = (bab)k ∈ A.(ii) (ab)ak(ab)−1 = a(bakb)a−1 ∈ A.(iii) (aib)ak(aib)−1 = ai(bakb)a−i ∈ A.

This proves that gak g−1 ∈ A for every g ∈ Dn, and so A = 〈a〉 is a normalsubgroup in Dn . �

1.11 The quaternion group Q8 is generated by two order-4 elements a andb, such that (bam)2 = a2, with m = 0, 1, 2, 3.

(a) Give the multiplication table.(b) The name of the group arises from the fact that its elements can be

represented by 4 objects, e.g. 1, x, y, z, and its negatives; or four indepen-dent 2 × 2 matrices, such as the identity I matrix and the Pauli matrices:

σ1 =

[

0 11 0

]

σ2 =

[

0 −ii 0

]

σ3 =

[

1 00 −1

]

Making use of the relations σiσj = I + iεijk σk, establish the correspondencebetween the three representations.

(c) Identify the conjugacy classes and subgroups of Q8. For each normalsubgroup, give the factor group of Q8.

SOLUTION 1.11(a) Let us define bj = baj, where j = 0, 1, 2, 3 and bj + 4 = bj. By

assumption b2j = a2 for all j. From the general relation bj+m = bja

m, we canderive several equivalent relations:

bjbj+m = am+2 , then bja2 = am+2bj+m , a2bj+m = bja

m+2.With these relations, the group table can be constructed:

Q8 e b0 b1 b2 b3 a a2 a3

e e b0 b1 b2 b3 a a2 a3

b0 b0 a2 a3 e a b1 b2 b3

b1 b1 a a2 a3 e b2 b3 b0

b2 b2 e a a2 a3 b3 b0 b1

b3 b3 a3 e a a2 b0 b1 b2

a a b3 b0 b1 b2 a2 a3 ea2 a2 b2 b3 b0 b1 a3 e aa3 a3 b1 b2 b3 b0 e a a2

10 CHAPTER 1

(b) We note in particular the relations:b2

0 = b23 = a2

b3b0 = a2 · a, b0a = a2b3, ab3 = a2b0.Given the properties of the Pauli matrices, the results suggest the mapping:

e 7→ I, b0 7→ iσ2, b3 7→ iσ1, a 7→ iσ3.Alternatively, we can also map

e 7→ 1, a2 7→ −1;b0 7→ y, b3 7→ x, a 7→ z;

where the new symbols satisfy the relationsx2 = y2 = z2 = −1, xy = −z, yz = −x, zx = −y.With the order of the elements slightly changed to make the symmetries

hidden in the table more apparent, the multiplication table of Q8 with thenew symbols becomes

Q8 1 x y z −1 −x −y −z

1 1 x y z −1 −x −y −zx x −1 −z y −x 1 z −yy y z −1 −x −y −z 1 xz z −y x −1 −z y −x 1

−1 −1 −x −y −z 1 x y z−x −x 1 z −y x −1 −z y−y −y −z 1 x y z −1 −x−z −z y −x 1 z −y x −1

(c) The group Q8 has the following characteristics:. Order of elements: The orders of 1,−1,±x,±y,±z are respectively 1,

2, 4, 4, 4.. Five conjugate classes: [1], [−1], [x,−x], [y,−y], [z,−z].. Four invariant subgroups: H2 = {1,−1}, H1

4 = {1,−1, x,−x}, H24 =

{1,−1, y,−y}, H34 = {1,−1, z,−z}.

. Cosets and factor groups: The invariant group H2 is isomorphic tothe second-order inversion group V2. Its cosets are {±x}, {±y}, {±z}, thesquare of each of which is H2, so the factor group Q8/H2, consisting of H2

and its three cosets, is isomorphic to Klein’s group V. Each of the order-4invariant subgroups Hi

4 has an index equal to 2, and a single coset, whosesquare is the invariant subgroup itself. Its factor group, formed by Hi

4 andits coset, is isomorphic to the inversion group V2 ≈ C2. �


Chapter 2. Group Representations

2.1 (Representations in function space) In a two-dimensional Euclideanspace, define a transformation R on the Cartesian coordinates (x, y) by

[

x′

y′

]

= D(R)

[

xy

]

.

The set of all homogeneous scalar functions of degree 2 in x, y invariantunder R form a linear vector space V3 of dimension 3 spanned by the basisvectors ψ1(x, y) = x2, ψ2(x, y) =

√2xy and ψ3(x, y) = y2. Given D(R) as

one of the following matrices:

(a)

[

−1 00 1

]

, (b)

[

cos φ − sin φsin φ cos φ

]

, (c)

[

−1/2 −√

3/2√3/2 −1/2

]

,

find the corresponding matrix representation of R in V3 = {ψ1, ψ2, ψ3}.

SOLUTION 2.1 First recall that a representation matrix M in representationspace Vn is defined by the action of an operator π(g) on a basis {ψi} of Vn

such that

π(R)ψj =n

∑i=1

ψiMij(R) .

In addition, if ψ is a scalar function of a variable x, then it also satisfies therelation π(R)ψ(D(R)x) = ψ(x) or, equivalently, π(R)ψ(x) = ψ(D(R−1)x).Applying it to the definition of M, we have for π(R)ψj(x) = ψj′(x) therelation

ψj′(x) = ψj(x′′) =n

∑i=1

ψi(x)Mij(R) where x′′ = D(R−1)x .

Keep in mind two points: (i) x′′ = D(R−1)x, and not x′, is present; (ii) eachcalculation of ψj gives the entries of a column (not a row).

(a) In this case we have x′′ = −x, y′′ = y, so thatψ1′(x, y) = x

′′2 = x2 = ψ1(x, y),ψ2′(x, y) =

√2x

′′y′′

= −√

2xy = −ψ2(x, y), andψ3′(x, y) = y

′′2 = y2 = ψ3(x, y).It follows that the representation matrix is

Ma(R) =

1 0 00 −1 00 0 1

.

12 CHAPTER 2

(b) In this case, D(R−1(φ)) = D(R(−φ)) and so we have x′′ = cx + sy,y′′ = −sx + cy, where c = cos φ, s = sin φ. To obtain the representationmatrix M, we calculate

π(R)ψ1(x, y) = x′′2 = x2c2 + 2csxy + y2s2 = ψ1c2 +

√2csψ2 + ψ3s2 ,

π(R)ψ2(x, y) =√

2x′′y′′ = −ψ1√

2cs + ψ2(c2 − s2) + ψ3√

2cs ,

π(R)ψ3(x, y) = y′′2 = s2x2 − 2scxy + c2y2 = ψ1s2 −

√2scψ2 + ψ3c2 .

The representation matrix in V3 is therefore

Mb(R) =

c2 −√

2cs s2

√2cs c2 − s2 −

√2cs

s2√

2cs c2

.

(c) x′′ and y′′ are given by

[

x′′

y′′

]

= D(R−1)

[

xy

]

=

[ −1/2√

3/2

−√

3/2 −1/2

] [

xy

]

.

The transformed basis functions are then calculated to be

π(R)ψ1(x, y) = x′′2 =

1

4(ψ1 −

√6ψ2 + 3ψ3)

π(R)ψ2(x, y) =√

2x′′y′′ =1

4

[√6ψ1 − 2ψ2 −

√6ψ3

]

π(R)ψ3(x, y) = y′′2 =

1

4

[

3ψ1 +√

6ψ2 + ψ3]

.

From this follows the representation matrix

Mc(R) =1

4

1√

6 3

−√

6 −2√

6

3 −√

6 1

.

It can be checked that in every case the matrix Mi is orthogonal, MTi Mi = 1.

The reason comes from our choice of the basis functions, and in particularof the numerical factor in ψ2 =

√2xy. It makes the transformed vectors in

the given basis {ψ1, ψ2, ψ3} orthogonal. For example, in case (c) the threevectors (1,−

√6, 3), (

√6,−2,−

√6) and (3,

√6, 1) are mutually orthogonal.

2.2 (Block diagonalization) Find the basis that reduces the representationmatrix M(R) of Problem 2.1 Case (c) to a block-diagonal form.

SOLUTION 2.2 The basis used in Problem 2.1 was formed by the functionsψ1 = x2, ψ2 =

√2xy and ψ3 = y2. We could try a basis in which one


vector is quadratic and completely symmetric in x, y, that is x2 + y2. So wetake as a new basis the functions φ1 = (ψ1 + ψ3)/

√2, φ2 = ψ2, and φ3 =

(ψ1 − ψ3)/√

2. From the transformation rules for ψi under M(R) alreadyderived (Case (c) of Problem 2.1), we can check that φ1 transforms into itselfand so is the basis of a one-dimensional subspace invariant under M(R),whereas φ2 and φ3 transform among themselves, and therefore constitutethe basis of a two-dimensional subspace invariant under M(R). So M(R) isreduced to a block diagonal form by the similarity transformation S definedby the new basis:

S = S−1 =

1√2

1 0 1

0√

2 01 0 −1

.

By matrix multiplication, we can check that

SM(R)S−1 =1

4S

1√

6 3

−√

6 −2√

6

3 −√

6 1

S−1 =

1 0 0

0 −1/2 −√

3/2

0√

3/2 −1/2

.

2.3 (Regular representations) The regular representation of a group G isdefined by the left action of its group elements (called g, h, q, . . .):

qg = ∑h

hDhg(q) , Dhg(q) =

{

1 : qg = h0 : qg 6= h

(where we have used group elements as indices to label matrix elements).But it can also be defined by the right action

gq = ∑h

Dgh(q)h , Dgh(q) =

{

1 : gq = h0 : gq 6= h

(a) How are D and D related when G is abelian?(b) Find their relationship when G is a finite group.(c) Find D, D, and X for the groups C3 and D3.

SOLUTION 2.3 (a) If G is abelian, then qg = gq and Dgh(q) = Dhg(q) =

DTgh(q), that is D is the transpose of D.

(b) In general, if we can find a matrix X independent of group elementssuch that D = XDX

−1, then we will have shown that D and D are equivalent.Therefore, if

∑p

Dhp(q)Xpg = ∑p

Xhp Dpg(q) ,

then X must have elements such that Xhq,g = Xh,qg . This equation says thattwo matrix entries of X are equal if the products of their row index and their

14 CHAPTER 2

column index (as group elements) are equal: (hq)g = h(qg) = g′ , where g′

is the resulting group element. This suggests that we may build X by fixingg′ (e.g. g′ = e) in the following way: Xpq = 1 if pq = e, and Xpq = 0 ifpq 6= e. By letting g′ take all allowed values, we will have |G| possible X.

(c) To find D, D, and X, it is simplest to start from the group table anduse its labels for all the matrices in an identical manner.

Group C3. This abelian group has three elements: e, a, b = a2, corre-sponding to indices 1, 2, 3. As usual, the rows and columns of the multipli-cation table are both labeled in that order:

1 2 3

e 1 2 3a 2 3 1b 3 1 2

From this table, we obtain the (left-action) representation matrices Dij(g)

D(e) =

1 0 00 1 00 0 1

, D(a) =

0 0 11 0 00 1 0

, D(b) =

0 1 00 0 11 0 0

.

The (right-action) representation matrices Dij (g) are simply the transposesof Dij(g). A similarity transformation is given by the matrix

X =

1 0 00 0 10 1 0

.

Group D3. The multiplication table of D3 is shown below

1 2 3 4 5 6

e 1 2 3 4 5 6b 2 1 5 6 3 4

ba 3 6 1 5 4 2ba2 4 5 6 1 2 3

a 5 4 2 3 6 1a2 6 3 4 2 1 5

The nonzero matrix elements of Dhg(q) are found by using the grouptable: Look at row q of the table, which has entries h corresponding to thecolumn labels g (being given in the first row), then Dhg(q) = 1. Similarly,for the right action regular representation Dhg(q): At column q, the entriesare h all the way down the column, which correspond to the row labels


g, then Dhg(q) = 1. In this way, we obtain the following nonzero matrixelements (all = 1), which we give in the order they appear in the multipli-cation table:

Dii(e) : i = 1, . . . , 6.

Dij(b) : (ij) = (21), (12), (53), (64), (35), (46).

Dij(ba) : (ij) = (31), (62), (13), (54), (45), (26).

Dij(ba2) : (ij) = (41), (52), (63), (14), (25), (36).

Dij(a) : (ij) = (51), (42), (23), (34), (65), (16).

Dij(a2) : (ij) = (61), (32), (43), (24), (15), (56).

Dii(e) : i = 1, . . . , 6.

Dij(b) : (ij) = (12), (21), (36), (45), (54), (63).

Dij(ba) : (ij) = (13), (25), (31), (46), (52), (64).

Dij(ba2) : (ij) = (14), (26), (35), (41), (53), (62).

Dij(a) : (ij) = (15), (23), (34), (42), (56), (61).

Dij(a2) : (ij) = (16), (24), (32), (43), (51), (61).

A matrix X that makes the transformation D = XDX−1 is given by

X =

1 0 0 0 0 00 1 0 0 0 00 0 1 0 0 00 0 0 1 0 00 0 0 0 0 10 0 0 0 1 0

.

As X is independent of g, it defines the equivalence of the representationsD(g) and D(g) for every g ∈ G.

2.4 (Inner products) In GTAPP §. 2.4 we have defined an inner product {, }by the relation

{x, y} def= |G|−1 ∑

h∈G

〈D(h)x|D(h)y〉, x, y ∈ V ,

where D(G) is a representation of a group G of order |G| in a linear spaceV , and 〈|〉 is a complex scalar product. Show that {, } satisfies the scalar-product axioms, i.e., linearity, complex conjugation, and positivity of norm.

SOLUTION 2.4 To simplify notations, we drop the overall factor |G|−1.

16 CHAPTER 2

(a) Complex conjugate:

{x, y} = ∑h∈G

〈D(h)x|D(h)y〉

= ∑h∈G

〈D(h)y|D(h)x〉∗ = {y, x}∗ .

(b) Linearity:

{x, ay + bz} = ∑h∈G〈D(h)x|D(h)[ay + bz]〉

= ∑h∈G[a 〈D(h)x|D(h)y〉+ b 〈D(h)x|D(h)z〉]

= a ∑h∈G

〈D(h)x|D(h)y〉+ b ∑h∈G

〈D(h)x|D(h)z〉

= a {x, y}+ b {x, z} .

However, note that just as for the product 〈x|y〉, the product {x, y} is anti-linear in x: {ax + bz, y} = a∗{x, y}+ b∗{z, y}.

(c) Positivity of the norm:

{x, x} = ∑h∈G

〈D(h)x|D(h)x〉 ≥ 0 ,

since each term satisfies 〈D(h)x|D(h)x〉 ≥ 0. Also {x, x} = 0 if and only ifall 〈D(h)x|D(h)x〉 = 0, which in turn is satisfied iff x = 0.

2.5 (D3 two-dimensional representation) Group D3 can be realized in atwo-dimensional Euclidean space by the symmetry transformations thatleave an equilateral triangle invariant. Let us label the vertices of such atriangle by 1, 2, 3 counterclockwise.

(a) To begin, choose a Cartesian basis e1, e2 in which the origin coincideswith the triangle center, the axis y = e2 passes through a vertex , and theaxis x = e1, perpendicular to y in the conventional sense. Find the matrixrepresentation of the six elements of D3. Is it unitary, is it reducible?

(b) Now take a new basis defined by a reference frame with axes u1

and u2 each passing through a vertex. Find the similarity transformationthat takes one basis to the other, Sim = 〈ei|um〉. Find the equivalent grouprepresentation in this new basis. Is it unitary, is it reducible?

SOLUTION 2.5 (a) In the Cartesian basis {e1, e2}, the matrix representa-tion is defined by π(g)ej = eiDij(g). A clockwise rotation of 2π/3 of the

reference frame produces x′ = π(a)e1 = − 12e1 +

√3

2e2, y′ = π(a)e2 =

−√

32

e1 − 12e2. This gives Dij(a). On the other hand a reflection about the y

axis produces x′ = −x and y′ = +y. We identify this transformation withgroup element ba, and so we get Dij(ba). Then we have three matrices


D(e) =

[

1 00 1

]

D(a) =

[

−1/2 −√

3/2√3/2 −1/2

]

D(ba) =

[

−1 00 1

]

.

The other three representation matrices are obtained from matrix mul-tiplication and group properties: D(a2) = D(a)D(a), D(b) = D(ba)D(a2),and D(ba2) = D(ba)D(a), yielding the result

D(a2) = 12

[

−1√

3

−√

3 −1

]

D(b) = 12

[

1 −√

3

−√

3 −1

]

D(ba2) = 12

[

1√

3√3 −1

]

.

All six matrices are unitary, as seen by matrix multiplication: D†D =1. We have a unitary representation because the representation basis isorthonormal. To check whether the representation is irreducible or not, weuse the relation

∑µ

cµχ∗µχµ ≥ |G| ,

where the equality sign is satisfied if and only if the representation is irre-ducible. We calculate χµ from the above matrices

[1] = [e]: c1 = 1, χ1 = 2;[2] = [b]: c2 = 3, χ3 = 0;[3] = [a]: c3 = 2, χ2 = −1;

so that ∑µ cµχ∗µχµ = 1.22 + 3.02 + 2(−1)2 = 6 = |G|, which implies irre-

ducibility. (If y = e2 passes through vertex 1 and the vertices are labeled123 counterclockwise, we can identify D3 elements with the permutationsof the labels: a = (123), a2 = a−1 = (321), b = (12), ba = (23), ba2 = (13)).

(b) Let u2 coincide with e2, passing through vertex 1, and u1 a unit vectorthrough vertex 3. The bases are related by u1 = 1

2(√

3e1 − e2) and u2 = e1.So that we have the similarity transformation matrix S and its inverse

S =

[√3/2 0

−1/2 1

]

S−1 =

[

2/√

3 0

1/√

3 1

]

The representation in the ui basis is given by D = S−1DS. Explicitly,

D(e) =

[

1 00 1

]

D(a) =

[

0 −11 −1

]

D(a2) =

[

−1 1−1 0

]

D(b) =

[

1 −10 −1

]

D(ba) =

[

−1 0−1 1

]

D(ba2) =

[

0 11 0

]

.

This representation is not unitary because the basis being used is not or-thogonal. However, it is still irreducible because the characters remain un-changed in similarity transformations. If we want to check it explicitly, wemust use the sum ∑µ cµχµ′χµ, where [µ′] designates the class of the inverseelements of [µ]. In the present case [µ′] = [µ] for all µ, and so χµ′ = χµ.

2.6 (Representation of sum of class elements) Let [µ] be a nontrivial conju-gacy class of a finite group G. Prove that an irreducible representation of

18 CHAPTER 2

the sum of the group elements in [µ] is a multiple of the identity, λI, anddetermine the constant factor λ.

SOLUTION 2.6 Define q = ∑h∈[µ] h. In any irreducible representation α ofG, q is represented by Dα(q) = ∑h∈[µ] Dα(h). Taking its trace, we get

Tr Dα(q) = ∑

h∈[µ]

Tr Dα(h) = cµχα

µ ,

where cµ is the number of elements in [µ]. For any g ∈ G and any h ∈ [µ],we have ghg−1 ∈ [µ]. So the operator q commutes with every g ∈ G, gq =qg. It follows

Dα(g)Dα(q) = D

α(q)Dα(g) for all g ∈ G .

By Schur’s Lemma Dα(q) must be a multiple of the identity: Dα(q) = λI.Taking the trace of both sides leads to cµχα

µ = λdα, where dα = TrI is thedimension of the representation. This gives us the constant λ = cµχα

µ/dα

and the representation Dα(q) = (cµχα

µ/dα) I.

2.7 (Group algebra & completeness relation) (a) Consider a finite group Gand define the set C(G) = {∑g∈G agg} with complex numbers ag. Showthat C(G) is closed under addition and multiplication of any two members.A linear vector space that is closed under some multiplication law (of twovectors) is called an algebra.

(b) Let Kµ = ∑cµ

i gµi be the sum of all elements of the conjugacy class [µ]

of size cµ ≥ 1, where µ = 1, . . . , nc. Show that KµKν = ∑γ cλµνKλ, where cλ

µν

are positive integers. In the following µ = 1 indicates the identity.(c) Consider any irreducible representation Dα(G), and define, for any

conjugacy class [µ], the sum of matrices Dµ = ∑g∈[µ] Dα(g). Show that Dµ

commutes with Dα(h) for every h ∈ G, and therefore (why?) Dµ = λµ I.Calculate λ.

(d) Show that χαµχα

ν = ∑γ Nγµνχα

γχα1, where χα

1 is the character of theidentity e, and N

γµν is a number that depends only on the group, not its

representations. Give an interpretation of this result.(e) Show that the inverses of the elements of any class [µ] form another

complete class [µ′], and prove that the coefficients defined in (b) are suchthat c1

µν = cµ if ν ∈ [µ′]; otherwise, it is zero. From this and (b), show that

cν

κ

∑α

χαµχα

ν = |G|δνµ′ ,

where κ is the number of distinct simple representations. If Dα is unitary,Dα(g−1) = (Dα(g))†, which implies χα

µ′ = χα∗µ . What does this result mean


about χαµ; what does it say about the values of κ (the number of representa-

tions) and nc (the number of classes)?

SOLUTION 2.7 (a) Let A and B be any two members of C(G). Then

A + B = ∑g

(ag + bg)g = ∑g

cgg = C ,

AB = ∑gh

agbhgh

= ∑g′

(∑h

ag′h−1 bh)g′ = D .

We have clearly, by definition, C and D in C(G).(b) Kµ is by definition invariant under h ∈ G, that is hKµh−1 = Kµ. So is

the product KµKν for any classes µ, ν, that is KµKν = h KµKν h−1. But in thisequation, h permutes the elements in the classes, and so KµKν must containcomplete classes, which means KµKν = ∑γ c

γµν Kγ, where c

γµν are positive

integers, characteristic of the group structure.(c) For the same reason that we have hKµh−1 = Kµ for any h ∈ G, we

also have, for any irreducible representation,

Dα(h)DαµDα(h−1) = ∑

g∈[µ]

Dα(h)Dα(g)Dα(h−1)

= ∑g′∈[µ]

Dα(g′) = Dαµ .

So Dαµ commutes with every Dα(h). Schur’s Lemma tells us that Dα

µ = λµ I.To find the constant λµ take the trace on both sides: the left side gives cµχα

µ,whereas the right side gives λµdα = λµχα

1, resulting in λµ = cµχαµ/χα

1

(d) Corresponding to KµKν = ∑γ cγµνKγ we have the matrix representa-

tion DµDν = ∑γ cγµνDγ. From the result obtained in (c) we have λµλν =

∑γ cγµνλγ. Substituting the values for λ, we have

cµcνχαµχα

ν = χα1 ∑

γ

cγµνcγχα

γ .

Thus, the result is of the form χαµχα

ν = ∑γ Nγµνχα

γχα1. The product of two

primitive characters of different classes in the same representation is givenby a linear combination of the characters of all classes of that representa-tion, the coefficients being independent of the representation, apart froman overall factor, dα = χα

1.(e) Let g, g′ be conjugates under G, i.e. hgh−1 = g′ for all h ∈ G. If we

take the inverse of both sides of this equation, hg−1h−1 = g′−1 , we see thatg−1 and g′−1 are also conjugates of each other. If [µ] is a conjugacy class,

20 CHAPTER 2

the inverses of its elements also form a complete class, called [µ′], with thesame size, cµ′ = cµ. The product [µ][µ′] contains the identity cµ times, butthe product [µ][ν] of any two classes such that ν 6= µ′ has no element e. Weencapsulate this result in the equation c1

µν = cµδνµ′ . Now take the equationgiven in (d) and sum both sides over α = 1, 2, . . . , dα:

cµcν ∑α

χαµχα

ν = ∑γ

cγµνcγ ∑

α

χα1χα

γ = |G|c1c1µν ,

where the second = is based on a result obtained in GTAPP § 2.7. Now, wehave c1 = 1 and c1

µν = cµδνµ′ , and hence the relation

cν

κ

∑α

χαµχα

ν = |G|δνµ′ .

(Recall that µ′ on the right-hand side indicates the class of the inverses ofthe elements of the class µ present on the left-hand side.) This equationholds for both unitary and non-unitary representations. If Dα is unitary,we have Dα(g−1) = (Dα(g))†, which implies χα

µ′ = χα∗µ , and the above

equation becomes

cν

κ

∑α

χα∗µ χα

ν = |G|δµν (for unitary representations).

This equation can be interpreted as a statement of the orthogonality of thenc vectors χµ, labeled by classes, in a κ-dimensional space. Since there canbe no more linear independent vectors than the dimension, we have nc ≤ κ.Taking this result with the inequality κ ≤ nc derived in GTAPP §2.7, weconclude that κ = nc. This proof of completeness differs from that given inthe Chapter in that it relies only on the characters; it does not depend onthe completeness of the representation matrices. �

2.8 (Sums of characters) Show that the sum of the simple characters of theelements of a finite group vanishes in any irreducible representation exceptthe trivial one-dimensional representation. Show also that the sum over thesimple representations, ∑α dαχα

µ with µ 6= [e] and dα being the representa-tion dimensions, also vanishes.

SOLUTION 2.8 The stated results follow from the orthogonality between thefirst row in the character table and every other row, and the orthogonalitybetween the first column and every other column. Orthogonality between

representation α = 1 (χ1µ = 1) and representation β 6= 1 gives: ∑µ cµχ

βµ =

∑g∈G χβ(g) = 0, β 6= 1. Orthogonality between class ν = 1 (χα1

= dα) andany class µ 6= 1 gives ∑α dαχα

µ = 0, (µ 6= 1). These two linear (rather


than quadratic) relations are useful in finding a character if all the othersare known.

2.9 (D3 representation in V4) Let V4 be the four-dimensional vector spaceof the functions of two real variables f (x, y) = c1x3 + c2x2y + c3xy2 + c4y3,with a basis consisting of ψ1 = x2, ψ2 = x2y, ψ3 = xy2, and ψ4 = y3.It is assumed that the variables x, y are Cartesian coordinates that linearlytransform according to x′i = Dij(g)xj (x1 = x, x2 = y, similarly for x′1, x′2)

[

x′

y′

]

= D(g)

[

xy

]

,

where D(g), with g = e, b, ba, ba2, a, a2, are the two-dimensional represen-tation matrices of the group G = D3 (given in the SOLUTION 2.5).

(a) Find the representation matrices M(G) in the basis {ψi} of V4.(b) Calculate the characters in this representation.(c) Find the transformation that reduces M(G) to direct sums of the ir-

reducible representations of D3.(d) Find the bases of the fully reduced representation in terms of {ψi}.

SOLUTION 2.9 (a) Recall that representation matrix M(g) is defined by

π(g)ψj(x) = ψj(x′′, y′′) = ∑i

ψi(x)Mij(g) , x′′ = D(g−1)x .

We need consider just the group generators, since the other group elementscan be obtained by combinations. We take a and ba as the group generators,and so we need the matrices

D(a−1) = D(a2) =1

2

[

−1√

3

−√

3 −1

]

, D((ba)−1) = D(ba) =

[

−1 00 −1

]

.

First consider π(a), where x′′ = (1/2)(−x+√

3y), y′′ = (−1/2)(√

3x + y).

π(a)ψ1 = (x′′)3

=1

8(−ψ1 + 3

√3ψ2 − 9ψ3 + 3

√3ψ4) ,

π(a)ψ2 = (x′′)2y′′

=1

8(−

√3ψ1 + 5ψ2 −

√3ψ3 − 3ψ4) ,

π(a)ψ3 = x′′(y′′)2

=1

8(−3ψ1 +

√3ψ2 + 5ψ3 +

√3ψ4) ,

π(a)ψ4 = (x′′)3

=1

8(−3

√3ψ1 − 9ψ2 − 3

√3ψ3 − ψ4) .

22 CHAPTER 2

Next, for π(ba), we have x′′ = −x and y′′ = y, and

π(a)ψ1 = (x′′)3 = −ψ1

π(a)ψ2 = (x′′)2y′′ = ψ2

π(a)ψ3 = x′′(y′′)2 = −ψ3

π(a)ψ4 = (x′′)3 = ψ4 .

The representation matrices can be read off from these results:

M(a) =1

8

−1 −√

3 −3 −3√

3

3√

3 5√

3 −9

−9 −√

3 5 −3√

3

3√

3 −3√

3 −1

M(ba) =

−1 0 0 00 1 0 00 0 −1 00 0 0 1

Of course M(e) = diag [1, 1, 1, 1] is the usual diagonal matrix.(b) As we know, D3 has three conjugacy classes, namely, [1] = {e}, [2] =

{b, ba, ba2}, and [3] = {a, a2}, with respective element numbers cµ = 1, 3, 2.Note that every class coincides with the class of the inverses of its elements,i.e. [µ′] = [µ]. Their characters in this representation are given by thetraces of the matrices M(g): χ = (4, 0, 1) on the conjugacy classes. To checkwhether this representation is irreducible or not we calculate the sum

∑µ

cµχµ′ χµ = 1 · 42 + 3 · 0 + 2 · 12 = 18 .

This sum being greater than |G| = 6, the representation is reducible.We further know (GTAPP § 2.3) that D3 has three irreducible represen-

tations of dimensions (1,1,2) and characters χ1µ = (1, 1, 1), χ2

µ = (1,−1, 1),and χ3

µ = (2, 0,−1). We can then apply the formula

aα = |G|−1∑µ

cµχαµχµ

to obtain the multiplicities for the representation M(G), which are aα = 1for every α. Therefore, M can be reduced to the direct sum ⊕αDα(G).

(c) The question is to find a matrix S independent of g such that

D(ba) = SD(ba)S−1 = ⊕αDα(ba) =

−1 0 0 00 1 0 00 0 −1 00 0 0 1

and

D(a) = SD(a)S−1 = ⊕αDα(a) =

1 0 0 00 1 0 0

0 0 −1/2 −√

3/2

0 0√

3/2 −1/2

.


With these transformation equations written as SD(g) = D(g)S, the equa-tion for g = ba requires S to be of the form

S =

a1 0 c1 00 b1 0 d1

a2 0 c2 00 b2 0 d2

.

The equation for g = a imposes further restrictions on Sij , leaving three ele-ments undetermined, which can be arbitrarily chosen, leading to the matrixS that we give below together with its inverse:

S =1

2

−1 0 1 00 −1 0 13 0 1 00 1 0 3

S−1 =

1

2

−1 0 1 00 −3 0 13 0 1 00 1 0 1

.

(d) The reduced representation matrices D(g) = SD(g)S−1 are expressedin the transformed basis functions ψj = ψi(S−1)ij , which are

ψ1 = −ψ1 + 3ψ3 = −x(x2 − 3y2) ,

ψ2 = −3ψ2 + ψ4 = −y(3x2 − y2) ,

ψ3 = ψ1 + ψ3 = x(x2 + y2) ,

ψ4 = ψ2 + ψ4 = y(x2 + y2) .

Using the transformation rules derived in (a), one can prove that ψ1, ψ2,and {ψ3, ψ4} span invariant subspaces, two of dimension 1 and one of di-mension 2, which are the spaces of the irreducible representations of D3.

Remarks: It is useful to note the method applied in (c) to find a simi-larity transformation matrix: One should start with an element g whoserepresentation D(g) is diagonal, then proceed to the next simplest matrixD(g′). Proceeding in this order simplifies your work. In addition, if S

−1

is eventually needed, but not S itself, it would be wiser to determine theinverse (rather than S) directly from the calculation DS

−1 = S−1

D, savingyou from extra work.

2.10 (Character table for D4) Construct the character table for D4 (the groupof symmetries for the square) studied in Problem 1.10.

SOLUTION 2.10 The dihedral group D4 has 8 elements divided into 5 con-jugacy classes: [1]1 = {e}, [2]1 = {a2}, [3]2 = {a, a3}, [4]2 = {b, ba2}, and[5]2 = {ba, ba3} (with sizes indicated as subscripts). As nc = 5, there are5 distinct irreducible representations of dimensions dα that satisfy the con-dition ∑α d2

α = 8, which means dα = (1, 1, 1, 1, 2) on the simple represen-tations. Besides an invariant subgroup of order 2, which does not concern

24 CHAPTER 2

us, D4 has three invariant subgroups of order 4. The factor groups derivedfrom them are:

D4/H1 = {H1, bH1} = {[1], [2], [3]; [4], [5]} ,D4/H2 = {H2, aH2} = {[1], [2], [4]; [3], [5]} ,

D4/H3 = {H3, aH3} = {[1], [2], [5]; [3], [4]} ,all of which are isomorphic to C2. Knowing the characters of the cyclicgroup, we can now determine the characters of the one-dimensional repre-sentations of D4. For example, the mapping H1 7→ e ∈ C2 and bH1 7→ c ∈C2 shows that χ2

1 = χ22 = χ2

3 = 1 and χ24 = χ2

5 = −1. Finally, the charactersof the two-dimensional representation, calculated with the orthogonalityand completeness relations, allow us to complete the character table.

D4 [1]1 [2]1 [3]2 [4]2 [5]2

1 1 1 1 1 1

2 1 1 1 −1 −1

3 1 1 −1 1 −1

4 1 1 −1 −1 1

5 2 −2 0 0 0

2.11 (Character table of T ∼= A4) T is the group of rotations that leave aregular tetrahedron invariant (this four-sided pyramid has 4 three-fold axesand 3 two-fold axes, found in CH4 for example). Using the group structureand the orthogonality and completeness relations, construct the charactertable for T. Note that T is a subgroup of S4, isomorphic to the alternatinggroup A4 (see Problem 1.8).

SOLUTION 2.11 T is a subgroup of S4 with 12 elements, divided into 4classes:

[1] = {e} (c1 = 1),[2] = {(12)(34), (13)(24), (14)(23)} (c2 = 3),[3] = {(123), (142), (134), (243)} (c3 = 4)[4] = {(132), (124), (143), (234)} (c4 = 4)It has an invariant subgroup: N = {e, (12)(34), 13)(24), (14)(23)} con-

sisting of all elements of classes [1] and [2]. The corresponding factor groupof T is T/N = {N, (123)N, (321)N}, isomorphic to C3.

Since there are 4 classes, there are also 4 inequivalent simple represen-tations, and the equation ∑α d2

α = 12 admits the unique solution dα =(1, 1, 1, 3): there are 3 representations of one-dimension, and 1 rep of threedimensions. To obtain the characters for the mentioned one-dimensionalirreps (α = 1, 2, 3), we may use T

hom7→ T/Niso7→ C3, that is, e, g ∈ [2] 7→ N;

g ∈ [3] 7→ (123)N; g ∈ [4] 7→ (321)N; and known results for C3.


The partially filled table is (ω = exp(i2π/3))

↓ α [1]1 [2]3 [3]4 [4]4

1 1 1 1 1

2 1 1 ω ω2

3 1 1 ω2 ω

4 3 χ42? χ4

3? χ44?

The characters for α = 4 are then determined by ∑α dαχαµ = 0 with µ =

2, 3, 4. This completes the table. Alternatively, the permutational represen-tation πP on 4 letters for T has the character χP = (4, 0, 1, 1); since πP − α1

has |χ|2 = 1, it is simple, and so must be ∼= α4.

e (12)(34) (123) (132)

T [1]1 [2]3 [3]4 [4]4α1 : 1 1 1 1 1

α2 : 2 1 1 ω ω2

α3 : 3 1 1 ω2 ω

α4 : 4 3 −1 0 0

2.12 (Character table for S4) Find the characters for the group S4.

SOLUTION 2.12 The 24 elements of S4 are divided into 5 classes:[1] = {e} (c1 = 1),[2] = {(12), (13), (14), (23), (24), (34)} (c2 = 6),[3] = {(12)(34), 13)(24), (14)(23)} (c3 = 3),[4] = {(123), (132), (124), (142), (134), (143), (234), (243)} (c4 = 8),[5] = {(1234), (1243), (1324), (1342), (1423), (1432)} (c5 = 6).Since there are 5 classes, there are also 5 irreducible representations,

whose dimensions must satisfy ∑α d2α = 24. This equation admits the

unique solution dα = (1, 1, 2, 3, 3).S4 has two invariant subgroups, T (order 12) and V (order 4).T contains the complete classes [1], [3], [4], and its coset (12)T contains

all the elements of [2], [5]. The factor group S4/T = {T, (12)T} is isomor-phic to C2. This allows us to determine the characters of the 1-dimensionalrepresentations

χ1µ = (1, 1, 1, 1, 1)

χ2µ = (1,−1, 1, 1,−1) .

The second invariant subgroup of S4 is V ≈ {[1], [3]}. The associatedfactor group of S4 is S4/V = {V, (12)V, (23)V, (13)V, (123)V, (321)V}, iso-

26 CHAPTER 2

morphic to D3. Recalling that D3 has 3 classes, which we call [e]D, [b]D and[a]D, we have the isomorphic mapping S4/V 7→ D3:

[1] ∪ [3] 7→ [e]D, [2]∪ [5] 7→ [b]D, [4] 7→ [a]D .Since the 2-dimensional representation of D3 has the character (2, 0,−1)

on the conjugacy classes [e]D, [b]D and [a]D, we have for symmetric groupS4 the character χ3

µ = (2, 0, 2,−1, 0).The remaining characters, in representations α = 4, 5, are then deter-

mined from ∑α dαχαµ = 0 (µ 6= 0) and ∑α |χα

µ|2 = 24/cµ. The results aretabulated below, with a representative of each class given on the top line.

Table 2.1: Character Table of S4

e (12) (12)(34) (123) (1234)

S4 [1]1 [2]6 [3]3 [4]8 [5]6α1 : 1 1 1 1 1 1

α2 : 2 1 −1 1 1 −1

α3 : 3 2 0 2 −1 0

α4 : 4 3 1 −1 0 −1

α5 : 5 3 −1 −1 0 1

2.13 Find the permutational representation on 4 letters for S4; find its de-composition. What is the product α2 × α4 (where α2 is a nontrivial one-dimensional representation and α4 a three-dimensional simple representa-tion of S4)? On the basis of these results, show a way to obtain the charac-ters for S4.

SOLUTION 2.13 The character of πP is χP = (4, 2, 0, 1, 0) on the conjugacyclasses. As χP = χ1 + χ4 (from the character table for S4), we have πP ∼=α1 + α4. On the other hand we have χ2χ4 = (3,−1,−1, 0, 1), of norm 1, andso α2 × α4 is simple and has dimension 3. It is equivalent to α5.

These results suggest another way of obtaining the character table forS4. Once the two one-dimensional representations α1 and α2 are known,we obtain α4 = πP − α1, then α5 = α2 × α4. The remaining α3 is determinedby orthogonality.

2.14 (Matrix representation of product group) Suppose that a finite groupG is the product group of its two subgroups, G = H × K. Using the matri-ces and not the characters, show that the direct product Dα(H)× Dβ(K) oftwo simple representations of H and K is a simple representation of G.

SOLUTION 2.14 By definition, hk = kh for any h ∈ H any k ∈ K. Let L1

be the representation space of H in which the irreducible representation


Dα(H) is defined, and L2 the representation space of K for the irreduciblerepresentation Dα(K). We assume that L1 and L2 do not have non-trivialinvariant subspaces for H and K. Let us proceed in three steps.

(i) First we want to prove that Dα(H)× Dβ(K) is a representation of G.Let h, h′ ∈ H and k, k′ ∈ K. If we have the correspondence from hk, h′k′ ∈ Gto reps D(hk) = Dα(h)× Dβ(k) and D(h′k′) = Dα(h′) × Dβ(k′); then wehave the correspondence from hkh′k′ = hh′kk′ ∈ G to reps

D(hk)D(h′k′) = Dα(h)× Dβ(k) · Dα(h′) × Dβ(k′)

= Dα(h)Dα(h′)× Dβ(k)Dβ(k′)

= Dα(hh′) × Dβ(kk′)

= D(hh′ · kk′).

So the set D(hk) forms a representation of G, to be denoted by D(G) =Dα(H) × Dβ(K), in the representation space L = L1 × L2. If x ∈ L1 andy ∈ L2, then x × y is in L.

(ii) We next want to show that D(G) is irreducible, which is equivalentto show that the only possible invariant subspace in L is a trivial subspace.

If there is a nonzero invariant subspace L′ in L with respect to D(G),then L′ = {x × y} must contain all vectors of L1 × y because Dα(H) ×Dβ(e) ⊂ D(G). Furthermore L′ must also contain all vectors in L1 × L2

because Dα(e)× Dβ(K) ⊂ D(G). So L′ and L1 × L2 coincide, which impliesL′ = L.

(iii) If either Dα or Dβ is replaced by a different inequivalent irreduciblerepresentation, then the irreducible representation of the direct product isalso changed to another distinct irreducible representation. The numberof classes of the direct product G is given by the product of the numbersof classes of the subgroups H and K, and is equal to the product of thenumbers of distinct irreducible representations of the subgroups. So eachirreducible representation of the direct product H × K is a direct productof irreducible representations of the two subgroups, Dα(H)× Dβ(K). �

Chapter 3. Lie Groups & Lie Algebras

3.1 (Rotations, O(2), SO(2)) A rotation R(θ) in the plane R2 about a refer-ence origin through an angle θ, is defined by a real matrix Rij, such that

R(θ)ei = ejRji(θ) , R(θ) =

[

cos θ − sin θsin θ cos θ

]

where e1, e2 are the reference orthonormal unit vectors in R2.(a) Show that the set 〈R(θ)〉 form a Lie group: specify its product law,

identity element, and inverse element. What is the range of values of θ?

28 CHAPTER 3

(b) The group O(2) is defined as the set of 2 × 2 real nonsingular ma-trices A obeying the condition ATA = I. What physical situations do thetwo possibilities, det A = ±1, correspond to? Define SO(2) = {A ∈ O(2) :det A = 1}. Show that there is a one-to-one correspondence between rota-tions in a plane and SO(2) matrices.

SOLUTION 3.1 (a) With usual matrix multiplication, we verify R(θ′)R(θ) =R(θ′ + θ), R−1(θ) = R(−θ) and R(0) = I (the 2 × 2 identity matrix), so thatthe continuous set {R(θ)} satisfies the group axioms for any real value of θ.In addition, although θ and θ + 2πn with any integer n, correspond to dif-ferent group elements, they represent the same transformation; thereforewe require on geometrical and physical grounds that R(θ + 2πn) = R(θ).Then, in this parameterization, the underlying one-dimensional manifoldis defined by the real variable θ in the interval 0 ≤ θ < 2π. The composi-tion function φ(θ′, θ) = θ′ + θ is analytic in both arguments. So we have aone-dimensional abelian Lie group. The range of values of θ is determinedby the product rule: if the range is smaller, there would be rotations whoseproduct is outside the range; if it is greater, there would be rotations givenmore than once. The group parameter space is the entire unit circle.

(b) From the condition AT A = I that the 2 × 2 real matrices of O(2)must satisfy, we have (det A)2 = 1, which implies det A = 1 or det A =−1. When A acts on a two-dimensional coordinate space, it represents alltransformations that conserve the scalar product of any two real vectors:a pure rotation if det A = 1; and a rotation times a mirror reflection ifdet A = −1. Group O(2) is a mixed (continuous and discrete) group.

An arbitrary matrix A of O(2) has the form:

A =

[

a bc d

]

a, b, c, d ∈ R.

The condition AT A = I yields: (i) a2 + c2 = 1, (ii) b2 + d2 = 1, (iii) ab + cd =0. From (i) we have a = cos α and c = sin α; from (ii), d = cos β andb = sin β; and finally from (iii), sin(α + β) = 0 or β = nπ − α, for anyinteger n = 0, 1, . . .. So that, in general, A is one of the two forms

A =

[

cos α − sin αsin α cos α

]

, A =

[

cos α sin αsin α − cos α

]

.

So O(2) is specified by a continuous parameter, 0 ≤ α < 2π, and a discretenumber, taking two values ±1. It represents rotations in the plane adjoinedto mirror reflections. The special orthogonal group SO(2) contains onlymatrices A with det A = 1, parameterized by α, and represents properrotations in the plane. SO(2) is connected to the identity, A 7→ 1 as α 7→ 0,and is infinitely connected since there are an infinite number of operators


obeying the identity R(2πn) = exp(i2πn) = 1 for all integer n—there areclosed paths on the unit circle which wind around it n times, for any integern, and which cannot be continuously deformed into each other. �

3.2 (O(2), SO(2), and U(1)) (a) Find the conjugacy classes of O(2). (b) Provethat the transformations that leave the magnitude of complex numbers in-variant form a one-parameter Lie group (called U(1)) isomorphic to SO(2).

SOLUTION 3.2 (a) O(2) contains a Lie subgroup, namely SO(2). In addition,I = diag[1, 1], I1 = diag[−1, 1], I2 = diag[1,−1], Ir = diag[−1,−1] are allelements of O(2), but I1 and I2 are not in SO(2). If R is in SO(2), then, forany g ∈ O(2), so is gRg−1 because det gRg−1 = det R = 1, so SO(2) is aninvariant subgroup. Take any A ∈ O(2) with det A = −1 (and so clearlynot in SO(2)), then det(I2A) = 1 which means I2A = R for some R ∈ SO(2).It follows A = I2R. In conclusion any A ∈O(2) is either R ∈ SO(2) or I2R,and O(2) is a disjoint union of {R : R ∈ SO(2)} and {I2R : R ∈ SO(2)}.These are the cosets making up the quotient group O(2)/SO(2) isomorphicto C2

∼= 〈I, I2〉.(b) Consider the transformation z 7→ z′ = αz, where z, z′ are complex

variables and α a complex-valued parameter. The invariance |z′| = |z|is satisfied if and only if |α| = 1. This condition is equivalent to α =exp(iθ), for real θ. These transformations form an abelian group because wehave the multiplication rule U(θ1)U(θ2) = U(θ1 + θ2), the identity elementU(0) = 1, and the inverse element U(θ)−1 = U(−θ). To each value of theparameter, there must correspond a single group element, which impliesexp[i(θ + 2πn)] = exp(iθ) for any integer n. It follows that 0 ≤ θ < 2π. Ifthe range were smaller, there would be group transformations whose prod-uct is outside the range; if it were greater, there would be transformationsgiven by more than one θ value within the range.

Since this group, called U(1), has the same number of parameters de-fined on the same range of values as SO(2) it must be isomorphic to it. �

3.3 (U(2) and SU(2)) (a) Find the general parameterization of the U(2) andSU(2) matrices. (b) In a complex vector space C2, the inner product is de-fined by 〈x, y〉 = ∑

2i=1 x∗i yi. Show that the group of linear transformations

in C2 that preserve the inner product is an U(2) group.

SOLUTION 3.3 (a) An arbitrary complex matrix A of U(2),

A =

[

a bc d

]

a, b, c, d ∈ C,

satisfies the unitarity condition A† A = I, or explicitly : (i) |a|2 + |c|2 = 1,(ii) |b|2 + |d|2 = 1, (iii) a∗b + c∗d = 0. Eq. (i) has the solution

a = cos θ eiα, c = − sin θ eiβ

30 CHAPTER 3

where 0 ≤ θ ≤ π/2 and 0 ≤ α, β < 2π. From (ii) we have

b = sin φ eiγ, d = cos φ eiδ.

With these expressions, (iii) becomes

cos θ sin φ ei(γ−α) = sin θ cos φ ei(δ−β) .

Equating the magnitudes of the two sides, we obtain sin(θ − φ) = 0, whichhas the only solution θ = φ in the allowed ranges of values of θ, φ. Equatingthe phases of the two sides, we have γ − α = δ − β or α + δ = β + γ =2λ(modulo 2π). This leads to the restrictions on the parameters:

α = λ + ζ , β = λ + η,

γ = λ − η, δ = λ − ζ ,

(all modulo 2π), where λ, ζ , η are arbitrary real phases. Therefore, the gen-eral unitary matrix A ∈ U(2) has the form

A = eiλ

[

cos θ eiζ − sin θ eiη

sin θ e−iη cos θ e−iζ

]

.

The four essential parameters have the ranges of values: 0 ≤ θ ≤ π/2,0 ≤ λ < π, and 0 ≤ ζ , η < 2π.

SU(2) is the subgroup of U(2), with det A = 1. This condition is satisfiedif and only if eiλ = 1, or λ = 0. A general matrix U ∈ SU(2) is specified bythree parameters, as in the expression

U =

[

cos θ eiζ − sin θ eiη

sin θ e−iη cos θ e−iζ

]

.

(b) Let x, y be vectors in a two-dimensional complex vector space C2. Alinear operator T in C2 acts on x, y to give x′ = Tx, y′ = Ty. Then thecondition that the inner product be preserved under T means 〈x′|y′〉 =〈x|T†T|y〉 = 〈x|y〉. It is satisfied if and only if T† T = I. That is, thegroup of linear transformations that preserve the inner products in C2 isa U(2) group. The special unitary group SU(2) arises when we requirethat the volume element in C2 be also preserved: dx′1 dx′2 = dx1 dx2; asdx′1 dx′2 = |∂(x′1x′2)/∂(x1, x2)|dx1 dx2, this implies det T = 1 (where xi arethe coordinates of x, and Tij = ∂x′i/∂xj). �

3.4 (O(1;1) and SO(1;1)) Write down the general matrices in O(1; 1) andSO(1; 1).

SOLUTION 3.4 A matrix A in O(1; 1) on a space with metric g = diag[1,−1]is real, and satisfies the relation AT gA = g. Writing A =

[

a bc d

]

, we get


a2 − c2 = 1, d2 − b2 = 1, and ab − cd = 0. The solution to each of thefirst two equations is a hyperbola in the plane (a, c) and (d, b) respectively:a = α cosh t, c = γ sinh t, d = δ cosh s, b = β sinh s, where α, β, γ, δ are the± signs. The third equation ab − cd = 0 requires s = ±t, and determinesthe possible signs. It turns out that A may be in one of the four forms:

[

ch t sh tsh t ch t

]

,

[

−ch t sh tsh t −ch t

]

,

[

ch t −sh tsh t −ch t

]

,

[

−ch t −sh tsh t ch t

]

,

where ch = cosh, sh = sinh, and t is a real number in (−∞, ∞). Ma-trices of the first two forms have det A = +1; those of the last two formshave det A = −1; they compose the four disjoint components of the groupO(1; 1). Matrices A with determinant equal to one form the group SO(1; 1)which consists of two disjoint components: one with c = cosh t is connectedto the identity (this identity component is associated with proper Lorentztransformations – rotations plus boosts), and the other with c = − cosh t (itis associated with time-reversal transformations).

3.5 (Dimension of orthogonal groups) Find the dimension of the orthogo-nal Lie groups O(n, R) and SO(n, R).

SOLUTION 3.5 O(n) = O(n, R) is the group of all real n × n matrices Asatisfying the orthogonality condition AT A = I, which can be written interms of the matrix elements as ∑k akiakj = δij, where i, j = 1, 2, . . . , n.Taking i = j these relations give n distinct conditions; and with i 6= jthey give (n2 − n)/2 additional conditions. So we have n + (n2 − n)/2 =n(n + 1)/2 distinct conditions on the n2 matrix elements. There remainn2 − n(n + 1)/2 = n(n − 1)/2 independent matrix elements. So the di-mension of the Lie group O(n) is n(n− 1)/2.

AT A = I implies det A = ±1. To select either possibility does notimpose a new condition. So SO(n) = SO(n, R), with det A = 1, has thesame dimension as O(n), which is n(n− 1)/2. �

3.6 (Dimension of unitary groups) Find the dimension of the unitary Liegroups U(n, C) and SU(n, C).

SOLUTION 3.6 U(n) = U(n, C) is the group of all n× n complex matrices Asatisfying the unitarity condition A† A = I, which can be written in termsof the matrix elements as ∑k a∗kiakj = δij , where i, j = 1, 2, . . . , n. Settingi = j, we have n real relations; requiring i 6= j we have (n2 − n)/2 complexrelations or equivalently 2(n2 − n)/2 real conditions. So we have a total ofn2 real functional relations among the 2n2 matrix elements. The number ofindependent parameters is 2n2 − n2 = n2: this gives the dimension of thegroup U(n). Note that for any A ∈ U(n), we have det A = eiλ, where λ is areal phase.

32 CHAPTER 3

SU(n) is a subgroup of U(n) restricted by det A = 1. This additionalconstraint eliminates one parameter from U(n) by fixing the phase param-eter, e.g. λ = 0. So SU(n) is of dimension (n2 − 1). �

3.7 (Dimension of symplectic groups) (a) Show that Sp(n, F) contains alsothe inverse and the transpose of every one of its elements. Find the di-mension of Sp(n, R). (b) Same questions for the compact symplectic groupSp(n). Assume F = R or C.

SOLUTION 3.7 (a) The metric J for symplectic groups is such that J2 = −I2n

and JT J = I2n, where I` = diag[1, 1, . . . , 1] with ` entries.If S ∈ Sp(n, F), then ST JS = J, which may be rewritten as JS−1 = ST J,

or still (ST)−1 JS−1 = J → (S−1)T JS−1 = J (where (AT)−1 = (A−1)T forany invertible matrix). This shows if S ∈ Sp(n, F), then S−1 ∈ Sp(n, F).

Noting S−1 = − JST J (using J2 = −I), we rewrite (S−1)T JS−1 = J as

J = (− JST J)T J(− JST J) = JTSJST J

=⇒ SJST = J =⇒ (ST)T JST = J .

So if S is in Sp(n, F), then so is ST.Contrary to the convention used in GTAPP, we take the m = 2n indices

in the following order: a = 1, 1, 2, 2, . . . , n, n (a may be either µ or µ with1 ≤ µ ≤ n) and define J as follows:

Jab =

1 : a = µ, b = µ−1 : a = µ, b = µ

0 : otherwiseJ =

0 1 0 0 0 0 ..−1 0 0 0 0 0 ..0 0 0 1 0 0 ..0 0 −1 0 0 0 ..0 0 0 0 .. .. ..

(J has the same properties as before: J2 = −I and JT J = I.) Then writingout ST JS = J in terms of the components S = [sab], we get

ST JS = J =⇒ ∑µµ

[sµasµb − sµa sµb] = Jab .

Now assume F = R. When a = b, Jaa = 0, we have an identity leadingto no conditions. But when a 6= b, there are (m2 − m)/2 real relationalfunctions among the m2 real matrix elements. Therefore, the number ofindependent parameters are m2 − (m2 − m)/2 = m(m + 1)/2, where m =2n. So the dimension of Sp(n, R) is n(2n + 1).

(b) We can prove as in (a) that (for U† = U−1) U−1 and UT satisfyUT JU = J if U does. Note also: U† = U−1 = − JUT J, from which U∗ =− JUJ.


From U∗ = − JUJ, we have the following relations:

u∗µν = uµν , u∗

µν = −uµν . (#)

Each gives 2n2 real constraints; so together they give 4n2 real constraints.Next, unitarity, ∑c u∗

caucb = δab , imposes the following:(i) a = b: ∑ν u∗

νauνa + ∑ν u∗νauνa = 1. Whether a = µ or a = µ, we have

the same set of n equations (to see this, we use Eqs. (#) So this case yields nindependent real constraints.

(ii) a = µ, b = ν 6= µ: ∑τ u∗τµuτν + u∗

τµuτν = 0. These are n(n − 1)/2complex constraints. They are the same as the relations obtained with a = νand b = µ 6= ν as well as the relations obtained by exchanging a and b(again use Eqs. (#)).

(iii) a = µ, b = ν 6= µ: ∑τ u∗τµuτν + u∗

τµuτν = 0. Again, n(n − 1)/2complex constraints. The indices a = ν, b = µ 6= ν yield no new conditions.

Recalling that one complex condition is equivalent to two real condi-tions, we have the total number of real constraints given by

2 × 2n2 + n + 2 × 2 × 12

n(n− 1) = 6n2 − n .

As the number of real parameters is 2(2n)2 = 8n2, the number of inde-pendent real parameters in Sp(n) is 8n2 − (6n2 − n) = n(2n + 1). So thedimension of the Lie group Sp(n) is n(2n + 1).

3.8 (Product law) A Lie group is defined in terms of the complete set of itsgenerators Xµ, obeying the brackets [Xµ, Xν ] = cµν

λXλ . Find the productlaw for its elements exp(αµXµ) and exp(βνXν) up to the total third order ofα, β.

SOLUTION 3.8 The product of two elements defined by exponential mapsmust be an exponential map given by exp(αµXµ) exp(βνXν) = exp(γλXλ).Using the CH formula given GTAPP § 3.5 and the bracket relation betweenthe generators, we obtain for the product element exp(γλXλ):

γλ = αλ + βλ +1

2αµβνcµν

λ +1

12αµβνcµν

αcκαλ(ακ − βκ) + . . . . �

3.9 (Identities with matrix exponentials) (a) Prove eXeY = eX+Y for arbi-trary n × n commuting matrices. (b) Prove exp(SXS−1) = S exp(X)S−1 forarbitrary X ∈ M(n, F) and some invertible matrix S ∈ M(n, F).

SOLUTION 3.9 (a) Use the power series for eX and eY , regroup terms ofequal total powers in X, Y, and use the binomial formula (you may assume

34 CHAPTER 3

XY ≡ YX):

eXeY = I + (X + Y) +( X2

2!+ XY +

Y2

2!

)

+ . . .

=∞

∑m=0

m

∑k=0

Xm−k Yk

(m− k)!k!=

∞

∑m=0

(X + Y)m

m!= eX+Y .

(b) We have S(1 + X + 12

X2 + . . .)S−1 = 1 + SXS−1 + 12

SX2S−1 + . . .,and this equals exp(SXS−1) because SXmS−1 = (SXS−1)m. �

3.10 (Polar decomposition of a matrix) (a) Prove that for any A ∈ SL(n, R)there is a unique decomposition A = RS, where R ∈ SO(n), and S a sym-metric positive definite matrix of determinant one. (b) Given A ∈ SL(2, R),determine R and S.

SOLUTION 3.10 (a) If A = RS with R ∈ SO(n), then AT A = ST RT RS = STSsince RT R = I. Requiring ST = S, we have AT A = S2. If S is positivedefinite (〈x, Sx〉 > 0 for x 6= 0), then S may be regarded as the ‘length’of A. The condition det S = 1 follows automatically from det A = 1 anddet R = 1.

(b) Let A =[ α β

γ δ

]

where det A = αδ − βγ = 1, and R =[

c −ss c

]

where

det R = c2 + s2 = 1. Then define S = R−1A = RT A, or explicitly S =[

cα + sγ cβ + sδcγ − sα cδ − sβ

]

. So that we get det S = (c2 + s2)(α − βγ) = 1, while

ST = S produces c(β −γ)+ s(α + δ) = 0, which yields s/c = (γ− β)/(δ+α). If we let c = cos θ and s = sin θ, then θ = tan−1(γ − β)/(δ + α). Thus,R and S are completely determined in terms of A. �

3.11 (Nilpotent matrices) Find the lowest power for which the followingmatrices A, B, C nullify; and compute eA, eB and eC.

A =

[

0 a0 0

]

, B =

0 a b0 0 c0 0 0

, C =

0 a b c0 0 d e0 0 0 f0 0 0 0

.

SOLUTION 3.11 (a) A2 = 0. B2 =

0 0 ac0 0 00 0 0

, and B3 = 0.

C2 =

0 0 ad ae + b f0 0 0 d f0 0 0 00 0 0 0

, C3 =

0 0 0 ad f0 0 0 00 0 0 00 0 0 0

, and C4 = 0


(b) eA = I + A =

[

1 a0 1

]

.

eB = I + B + 12

B2 =

1 a b + 12

ac0 1 c0 0 1

.

eC = I + C + 12

C2 + 16

C3 =

1 a b + 12

ad c + 12(ae + b f ) + 1

6ad f

0 1 d e + 12

d f0 0 1 f0 0 0 1

.

�

3.12 (Computing eX) Compute eX for the following real matrices X:[

0 aa 0

]

,

[

0 −aa 0

]

,

[

a b0 a

]

,

[

a b−b a

]

,

[

a 0b a

]

,

[

1 10 0

]

,

SOLUTION 3.12 We shall use matrices obeying J2 = I, K2 = −I, N2 = 0:

I =

[

1 00 1

]

, J =

[

0 11 0

]

, K =

[

0 −11 0

]

, N =

[

0 10 0

]

.

(a) X =

[

0 aa 0

]

= aJ. Summing separately even and odd powers of X,

and using J2 = I, we get:

eX = I(1 + 12

a2 + · · · ) + J(a +1

3!a3 + . . .) = I cosh a + J sinh a.

(b) X =

[

0 −aa 0

]

= aK. Summing separately even and odd powers of

X, and using K2 = −I, we get:

eX = I(1− 12

a2 + . . .) + K(a− 1

3!a3 + · · · ) = I cos a + K sin a.

(c) X =

[

a b0 a

]

= aI + bN. Remarking that NI = I N and N2 = 0, we

have

eX = eaI+bN = eaI ebN =

[

ea 00 ea

] [

1 b0 1

]

=

[

ea bea

0 ea

]

= ea(I + bN).

(d) X =

[

a b−b a

]

= aI − bK. Using IK = KI and results from (b):

eX = eaI−bK = ea

[

cos b sin b− sin b cos b

]

= ea(I cos b − K sin b)

36 CHAPTER 3

(e) X =

[

a 0b a

]

= (aI + bN)T. Using results from (c), we get

eX = exp(aI + bN)T = (exp(aI + bN))T =

[

ea 0bea ea

]

= ea(I + bNT).

(f) X =

[

1 10 0

]

. As X is idempotent (X2 = X), we have

eX = I + X(1 +1

2+

1

3!+ · · · ) = I + X(e − 1) =

[

e e − 10 1

]

.

�

3.13 (Is given Y equal eX with real X?) Are the following (real) matrices ofthe form eX? If so, compute X.

[

1 00 0

]

,

[

0 11 0

]

,

[

a 00 b

]

,

[

0 −11 0

]

,

[

1 10 1

]

,

[

1 1−1 1

]

,

SOLUTION 3.13 Recalling that det exp(X) = exp(TrX), we know we musthave det eX > 0 for real X.

(a) Y =

[

1 00 0

]

. Since det Y = 0, Y cannot be of the form eX .

(b) Y =

[

0 11 0

]

. We have det Y = −1, and so Y 6= eX .

(c) Y =

[

a 00 b

]

. If a, b are positive real then λ = log a, µ = log b, and

Y = eΛ where Λ = diag[λ, µ].

(d) Y =

[

0 −11 0

]

. We have seen in Problem 12 (b) that, with X = aK,

eX = I cos a + K sin a. In order to have Y = K, we fix a = π/2, which givesus X = (π/2)K.

(e) Y =

[

1 10 1

]

= I + N. But I + N = eN , since N2 = 0; so Y = eN .

(f) Y =

[

1 1−1 1

]

= I − K. Referring to Problem 12 (d), we know that if

b = π/4, then

X =

[

a π/4−π/4 a

]

=⇒ eX =ea

√2

[

1 1−1 1

]

=ea

√2

Y.

So we have Y =√

2eX−a = eZ , where Z =

[

12

log 2 π/4

−π/4 12

log 2

]

. �

3.14 (Dimension-three algebras) Find the generators of rotation acting on(a) three-dimensional Euclidean space of metric g = diag[1, 1, 1]; and (b)three-dimensional space of metric g = diag[1, 1,−1].


SOLUTION 3.14 To determine the generators of a Lie group, we need to con-sider its properties near the identity I, so that we write A ≈ I + M for anygroup element A, and define the generators Xα by M = + ∑α αXα, whereα are the independent essential parameters in the group space. (NB: Thisdefinition of Xα differs in sign, for all α, from the definition in GTAPP Chap-ter 3; the algebras for the two options are isomorphically equivalent underXα 7→ −Xa.) Under the action of A, the Cartesian coordinates transformas x′i = Aij xj = xi + dxi, where dxi = Mij xj = Mxi. This is equivalentto M = ∑i dxi(∂/∂xi) (where M is now an operator on the coordinates).Writing in terms of Xα , we have

∑α

αXα = ∑i

dxi∂

∂xi=⇒ Xα = + ∑

i

∂(dxi)

∂α

∂

∂xi.

(Here again, there is a sign difference with GTAPP, as explained above.)(a) Let A ∈ SO(3), so that AT A = I and det A = 1. Writing A ≈ I + M,

we have MT = −M (antisymmetric matrix). Following are the generalexpression for M and the variations dxi:

M =

0 c −b−c 0 ab −a 0

dx = cy − bzdy = −cx + azdz = bx − ay

Using the formula for Xα given above, we get

Xa = z∂

∂y− y

∂

∂z, Xb = x

∂

∂z− z

∂

∂x, Xc = y

∂

∂x− x

∂

∂y,

which bracket according to [Xa, Xb] = Xc, [Xb, Xc] = Xa, [Xc, Xa ] = Xb .(b) Any B ∈ SO(2; 1) obey BT gB = g, where g = diag[1, 1,−1]. With

B ≈ I + L, the matrix L satisfies LT g + gL = 0. So gL is antisymmetric, andwe can take L = gM, with M ∈ SO(3). Therefore,

L =

0 c −b−c 0 a−b a 0

dx = cy − bzdy = −cx + azdz = −bx + ay

So that the generators of SO(2; 1) are:

Xa = z∂

∂y+ y

∂

∂z, Xb = −x

∂

∂z− z

∂

∂x, Xc = y

∂

∂x− x

∂

∂y,

which obey [Xa, Xb] = −Xc, [Xb, Xc] = Xa, [Xc, Xa ] = Xb . �

3.15 (Exponential of rotation) (a) Referring to Problem 3.1, write the rota-tion matrix for a very small angle δθ as R(δθ) = I + δθR: sum the expan-sion series that defines exp(θR). (b) When the rotation is applied on the

38 CHAPTER 3

Cartesian coordinates x1, x2 of a two-dimensional Euclidean space, we candefine the corresponding Lie group of rotations, generated by a differen-tial operator X. Find X and calculate ϕX(θ) = exp(−θX) (the minus signin exp(−θX) arises because the operator is applied on the coordinates, notthe basis vectors)

SOLUTION 3.15 (a) For a very small angle δθ, we have from Problem 3.1

R(δθ) ≈[

1 −δθδθ 1

]

= I + δθR =⇒ R =

[

0 −11 0

]

.

Now the power series that defines exp(θR) for arbitrary real θ is

exp(θR) =∞

∑n=0

1

(2n)!(θR)2n +

∞

∑n=0

1

(2n + 1)!(θR)2n+1

Noting that R2 = −I, the series can be summed to give

exp(θR) =∞

∑n=0

θ2n

(2n)!(−1)n + R

∞

∑n=0

θ2n+1

(2n + 1)!(−1)n

= cos θ + R sin θ ,

which reproduces the original rotation matrix R(θ).(b) The transformation functions are given by

f1(θ; x1, x2) = x1 cos θ − x2 sin θ,

f2(θ; x1, x2) = x1 sin θ + x2 cos θ.

From the general formula, the generator is defined as

X = − ∂ f1

∂θ

∣

∣

∣

∣

θ=0

∂

∂x1

− ∂ f2

∂θ

∣

∣

∣

∣

θ=0

∂

∂x2

= x2

∂

∂x1

− x1

∂

∂x2

,

so that Xx1 = x2, Xx2 = −x1. The exponential map ϕX(θ) = e−θX , definedby the usual series of powers of X, is applied on x1, x2:

ϕX(θ)x1 = x1 − θXx1 +1

2!θ2X2x1 −

1

3!θ3X3 x1 · · ·

= x1 cos θ − x2 sin θ = f1(θ; x1, x2) ;

ϕX(θ)x2 = x2 − θXx2 +1

2!θ2X2x2 −

1

3!θ3X3 x2 · · ·

= x1 sin θ + x2 cos θ = f2(θ; x1, x2) .

In the proof, we have used the following identities

X2n+1 x1 = (−1)nx2, X2n+2 x1 = (−1)n+1x1,

X2n+1 x2 = (−1)n+1x1, X2n+2 x2 = (−1)n+1x2.


So the exponential map ϕX(θ) reproduces ( f1, f2) as it should. �

3.16 (Lorentz boost) (a) Referring to the relevant discussion in the Chapter,show that for any one-dimensional Lie group, with the composition func-tion φ(β, α), one can find a new parameter α′ such that the compositionfunction is additive, i.e. φ(β′, α′) = β′ + α′ (one-parameter Lie groups areabelian).

(b) In a two-dimensional space with metric g = diag[1,−1], define theLorentz boost L(v) along the x axis by the matrix

L(v) =

[

γ γvγv γ

]

where 0 ≤ v < 1, γ = 1/√

1 − v2.

(i) Show that L(v2)L(v1) = L(φ(v2, v1)), with φ(v2, v1) to be found. (ii)Find the parameter ω(v), such that L(ω2)L(ω1) = L(ω2 + ω1). (iii) Findthe group generator X such that the exponential map exp(ωX) reproducesL(ω).

SOLUTION 3.16 (a) Assuming one dimensional group space, we use the fol-lowing formula found in GTAPP § 3.3

dx(α)

dα= −ψ(α)Xx(α), where ψ(α) =

[

∂φ(β, α)

∂β

∣

∣

∣

∣

β=0

]−1

.

If ρ(α) is a representation of α acting on x so that x(α) = ρ(α)x(0), we have

dρ

dα= −ψ(α)Xρ(α) , ρ(0) = 1,

dρ

dα

∣

∣

∣

∣

0

= ρ′(0) = −X.

This differential equation can be solved to give

ρ(α) = exp[−ω(α)X] , ω(α) =∫ α

0ψ(s)ds .

Since for β · α = β + α, we have the representation ρ(β · α) = ρ(β) + ρ(α),the corresponding exponential map exp[−ω(α)X] implies the compositionrule ω(β · α) = ω(β)+ ω(α).

(b) By matrix multiplication we can check that

L(v2)L(v1) =

[

λ λφλφ λ

]

= L(

φ(v2, v1))

,

where λ = 1/√

1 − φ2, which implies φ(v2, v1) = v1v2/(1 + v2v1), and so

∂φ(v2, v1)

∂v2

∣

∣

∣

∣

v2=0

= 1 − v21 =⇒ ψ(v) =

1

1 − v2.

40 CHAPTER 3

Hence the new parameter is (with v < 1)

ω(v) =∫ v

0

1

1 − s2ds = 1

2log

1 + v

1 − v= tanh−1(v),

or equivalently v = tanh ω, γ = cosh ω, and γv = sinh ω. With this newvariable, we have

L(v) ≡ L(ω) =

[

cosh ω sinh ωsinh ω cosh ω

]

,

and check directly that this special Lorentz group is abelian: L(ω2)L(ω1) =

L(ω2 + ω1) = L(ω1)L(ω2). Expanding to first order L(ω) = 1 + ω dLdω , we

define the generator X (again, with a + sign) by

X =dL

dω

∣

∣

∣

∣

0

=

[

0 11 0

]

Now, we sum the power series of the exponential, noting X2 = I:

exp(ωX) = I +ω2

2!X2 +

ω4

4!X4 + · · ·+ ωX +

ω3

3!X3 + · · ·

= X2

(

1 +ω2

2!+ · · ·

)

+ X

(

ω +ω3

3!+ · · ·

)

= I cosh ω + X sinh ω ,

and recover the matrix L(ω) obtained before. Note that, as the L(ω) matrixelements have no upper limits, the Lorentz group is non compact. �

3.17 (Heisenberg group) Define the sets of matrices:

H =

1 a b0 1 c0 0 1

; a, b, c ∈ R

, h =

0 α β0 0 γ0 0 0

; α, β, γ ∈ R

.

(a) Show that H is a Lie group, and find its center (invariant abeliansubgroup). (b) Show that for any X ∈ h, the exponential eX ∈ H, andconversely if X is any matrix such that etX ∈ H, then X = detX /dt|t=0 is inh. Define a basis for h, together with its algebra. (c) Find the center C(h) ofh. Using the CH formula, calculate exp X for an arbitrary X ∈ h.

SOLUTION 3.17 (a) It is evident that I = diag[1, 1, 1] ∈ H. If A, A′ ∈ H,matrix arithmetic shows that AA′ and the inverse of A are in H:

AA′ =

1 a + a′ b + b′ + ac′

0 1 c + c′

0 0 1

, A−1 =

1 −a ac − b0 1 −c0 0 1

.


The limit of matrices of the form in H is also of that form, so H is a closedsubgroup of GL(3, R)and a (connected, simply connected) matrix Lie group.

From the product law, we see that matrices of the form

1 0 d0 1 00 0 1

commute with each other and with every A ∈ H. So they form the centerC(H) of H, defined as C(H) = {A ∈ H; AB = BA for all B ∈ H}.

(b) Every X ∈ h is nilpotent: X2 =

0 0 αγ0 0 00 0 0

, X3 = 0. So we have:

etX = I + tX + 12

t2X2 =

0 tα tβ + 12t2αγ

0 1 tγ0 0 1

.

For every X ∈ h, eX is in H. On the other hand, if A(t) ∈ H such thata(t) = ta′(0) + · · · , b(t) = tb′(0) + · · · , and c(t) = tc′(0) + · · · , thenlimt→0 [(A(t)− I)/t] is strictly upper triangular and so belongs to h. Thus,the Lie algebra of H (a space of upper triangular matrices) is h (a space ofstrictly upper triangular matrices); conversely the Lie algebra h is exponen-tially mapped in a unique way to the Lie group H.

(c) The center C(h) = {X ∈ h; [X, Y] = 0, ∀Y ∈ h} consists of matrices

of the form

0 0 β0 0 00 0 0

with β ∈ R. If X, Y ∈ h, then [X, Y] ∈ C(h), and so

[[X, Y], T] = 0 for all T ∈ h. In particular, [[X, Y], X] = 0 and [X, Y], Y] = 0.Choose for basis in h

P =

0 1 00 0 00 0 0

, Q =

0 0 00 0 10 0 0

, E =

0 0 10 0 00 0 0

,

with the commutation relations [P, Q] = E, [E, P] = 0, [E, Q] = 0, andthe center C(h) = 〈E〉. The only non-zero bilinear product is PQ = E.Any X ∈ h can be written as X = αP + βE + γQ, with real α, β, γ. As wehave seen eX can be calculated using the fact that X is nilpotent. Now werecalculate it making use of the CH formula. First note that as E ∈ C(h),exp X = exp βE · exp(αP + γQ). Further, as [P, Q] ∈ C(h) the CH formulasimplifies so that exp αP · exp γQ = exp(αP + γQ + 1

2αγ[P, Q]).

42 CHAPTER 3

We have now

eX = eαP+βE+γQ = e(β− 12

αγ)E eαP eγQ

= [I + (β − 12αγ)E][I + αP][I + γQ]

= I + (β − 12

αγ)E + αP + γQ + αγPQ

= I + (β + 12

αγ)E + αP + γQ.

This agrees with result obtained in (b). �

3.18 (Euclidean group E(2)) In classical physics, space is assumed to be ho-mogeneous and isotropic, so that physical phenomena should not dependon the specific location or orientation of the physical system. In mathe-matical terms, this means a physical Euclidean space, with its symmetriesdescribed by the Euclidean group, which consists of uniform translationsand uniform rotations. Take, as an example, the Euclidean group E(2) de-fined by the following transformations of the Cartesian coordinates x1, x2

in R2:

x′1 = x1 cos θ − x2 sin θ + α1

x′2 = x1 sin θ + x2 cos θ + α2 ,

in terms of a displacement distance (−∞ < αi < ∞; i = 1, 2) and a rotationangle (0 ≤ θ ≤ 2π). We call A(θ, α) the operator effecting this transforma-tion: A(θ, α)xi = x′i = Rij(θ)xj + αi.

(a) Verify that A(θ, α) form a Lie group, and that A(θ, 0) and A(0, α)are its two subgroups. Calculate the transformation functions as well asthe composition functions of E(2).

(b) Calculate the generators (X0, X1, X2) of E(2) and their brackets.

SOLUTION 3.18 We shall call the rotation angle θ ≡ α0, and use the notationα ≡ (α0, α1, α2) ≡ (α0, α). Similarly for β or γ.

(a) Apply A(β) on x′ = A(α)x, we have

x′′i = Rij(β0)x′j + βi

= Rij(β0)[Rjk(α0)xk + αj] + βi

= Rik(β0 + α0)xk + Rij(β0)αj + βi ,

which implies the product rule: A(β)A(α) = A(

β0 + α0, R(β0)α + β)

. Withthe identity A(0, 0) = e, the inverse follows: A(α)−1 = A(−α0,−R(−α0)α).


Let us also check associativity:

A(γ) · (A(β)A(α)) = A(γ) · A[

β0 + α0, R(β0)α + β]

= A[

γ0 + β0 + α0, R(γ0 + β0)α + R(γ0)β + γ]

;

(A(γ)A(β)) · A(α) = A[

γ0 + β0, R(γ0)β + γ]

· A(α)

= A[

γ0 + β0 + α0, R(γ0 + β0)α + R(γ0)β + γ]

.

Hence A(γ) ·(

A(β)A(α))

=(

A(γ)A(β))

· A(α).The transformation functions are given by

fi(α; x) = Rij(α0)xj + αi; i = 1, 2 ,

and the composition functions

φ0(β, α) = β0 + α0; φi(β, α) = Rij(β0)αj + βi, i = 1, 2 .

Closure and associativity can be readily verified for these functions.Since the composition functions are analytic in its arguments, they define aLie group. We remark that

A(ϕ, 0)A(θ, 0) = A(ϕ + θ, 0) ,

A(0, β)A(0, α) = A(0, β + α) ,

which imply closure for the sets {A(θ, 0)} and {A(0, α)}. Since the gen-eral group properties are also satisfied, they form subgroups of E(2): thesubgroup of rotation R = {A(θ, 0)} and the subgroup of translation T ={A(0, α)}.

(b) The generators are obtained from the transformation functions bythe formula (with a conventional sign):

Xµ(x) = − ∂ f j(α; x)

∂αµ

∣

∣

∣

∣

α=0

∂

∂xjµ = 0, 1, 2 .

The nonzero derivatives are ∂ f1/∂θ|0 = −x2, ∂ f1/∂α1 = 1, ∂ f2/∂θ|0 = x1,∂ f2/∂α2 = 1. So we have the generators in geometric space

X0(x) = −x1

∂

∂x2

+ x2

∂

∂x1

,

X1(x) = − ∂

∂x1

, X2(x) = − ∂

∂x2

,

which bracket according to [X0, X1] = X2, [X0, X2] = −X1, [X1, X2] =0. These operators are anti-hermitian in an inner-product complex vector

44 CHAPTER 3

space. In physics, it is customary to define hermitian operators instead:

Jdef= iX0(x) = −i

(

x1

∂

∂x2

− x2

∂

∂x1

)

,

Pidef= iXi(x) = −i

∂

∂xi, (i = 1, 2) ,

together with the brackets [ J, P1] = iP2, [ J, P2] = −iP1, [P1, P2] = 0 �

3.19 (Linearization of E(2)) Let L be the vector space spanned by vectors

v =

x1

x2

1

, where[

x1

x2

]

is a vector in two-dimensional Euclidean space. The

object of this problem is to show that E(2) is isomorphic to a matrix Liegroup, by showing that E(2) group and algebra elements (Problem 3.18)are expressible as 3 × 3 matrices in space L, denoted M(3, R).

(a) Find the generators of E(2) and their brackets as elements of M(3, R).(b) Prove in M(3, R) that A(θ, α) = A(0, α)A(θ, 0).(c) Show that the exponential series of exp(θX0) and exp(α1X1 + α2X2)

sum to A(θ, 0) and A(0, α) respectively.(d) What is A(θ, 0)A(0, α)? Compare it to A(0, α)A(θ, 0).(e) In question (b), we have proved that exp(α1X1 + α2X2) exp(θX0) is

identical to the group element A(θ, α). But the theory tells us that, in gen-eral, it is the map exp(θX0 + α1X1 + α2X2) that should reproduce the groupelements. How do you explain the difference?

SOLUTION 3.19 We will use the same symbols as in Problem 3.18 althoughnow they refer to their matrix forms. Thus, the general transformation ofE(2) acting on R3 vectors is given by the 3 × 3 real matrix

A(θ, α1, α2) =

cos θ − sin θ α1

sin θ cos θ α2

0 0 1

≡[

R2(θ) α

0 1

]

,

and the elements of the (abelian) rotation group R = {A(θ, 0)} and abeliantranslation group T = {A(0, α)} are

A(θ, 0) =

cos θ − sin θ 0sin θ cos θ 0

0 0 1

, A(0, α) =

1 0 α1

0 1 α2

0 0 1

.

(a) The group generators are identified with the first-order terms in theexpansion series of the group element, A(θ, α1, α2) = I + θX0 + α1X1 +α2X2 + · · · ,

X0 =

[

0 −1 01 0 00 0 0

]

, X1 =

[

0 0 10 0 00 0 0

]

, X2 =

[

0 0 00 0 10 0 0

]

.


With matrix arithmetic, we find XiX0 = 0, X0X1 = X2, and X0X2 = −X1.Hence the brackets: [X0, X1] = X2, [X0, X2] = −X1, and[X1, X2] = 0.

(b) Rewrite R(θ) = A(θ, 0) and T(α) = A(0, α) in the form

R(θ) =

[

R2(θ) 0

0 1

]

T(α) =

[

I2 α

0 1

]

,

where R2(θ) and I2 are 2 × 2 matrices. Now take the product

T(α)R(θ) =

[

I2 α0 1

] [

R2(θ) 0

0 1

]

=

[

R2(θ) α0 1

]

= A(θ, α),

Another way to obtain the same result is to apply the group product rule:

A(θ, α)A(−θ, 0) = A(θ − θ, α) ≡ T(α)

7→ A(θ, α) = T(α)A(θ, 0) = T(α)R(θ).

(c) To sum the series for exp(θX0), note first the identities (simple toverify): X2n

0 = (−)n+1X20 and X2n+1

0= (−)nX0, and also

I = E − X20 , E =

[

0 0 00 0 00 0 1

]

, X20 =

[−1 0 00 −1 00 0 0

]

.

Then, the power series in question is

exp(θX0) =∞

∑n=0

1

(2n)!(θX0)

2n +∞

∑n=0

1

(2n + 1)!(θX0)

2n+1

= E − X20

∞

∑n=0

(−1)n

(2n)!(θ)2n + X0

∞

∑n=0

(−1)n

(2n + 1)!(θ)2n+1

= E − X20 cos θ + X0 sin θ

= A(θ, 0) = R(θ).

As for exp(α1X1 + α2X2), it suffices to notice that X21 = X2

2 = 0 andX1X2 = X2X1 = 0, so that

exp(α1X1 + α2X2) = (1 + α1X1)(1 + α2X2)

= I + α1X1 + α2X2

= A(0, α) = T(α).

So exp(θX0) generates the subgroup of rotation R, and exp(α1X1 +α2X2) the subgroup of translation T .

(d) Applying group multiplication, we get A(θ, 0)A(0, α) = A[θ, R2(θ)α].So we have here R(θ)T(α) = A[θ, R2(θ)α], whereas in (b) we obtained

46 CHAPTER 3

T(α)R(θ) = A(θ, α). The matrices R(θ) and T(α) are both elements of E(2),and so their products must also be elements of the group. As they do notcommute, their products depend on the order of the factors.

(e) In Question (b), we have proved that exp(α1X1 + α2X2) exp(θX0) isidentical to the group element A(θ, α). But the theory tells us that, in gen-eral, it is exp(θX0 + α1X1 + α2X2) that should reproduce the group ele-ments. The first few powers of the power series of this exponential are:

M = θX0 + α1X1 + α2X2,

M2 = θ2X20 − θα2X1 + θα1X2,

M3 = −θ3X0 − θ2α1X1 − θ2α2X2,

M4 = −θ4X20 + θ3α2X1 − θ3α1X2,

M5 = θ5X0 + θ4α1X1 + θ4α2X2.

It follows the exponential series

exp M = I + M +1

2!M2 + · · ·

= (E − X20) + X0

(

θ − θ3

3!+ · · ·

)

− X20

(

−θ2 +θ4

4!+ · · ·

)

+ X1

[

α1

θ

(

θ − θ3

3!+ · · ·

)

− α2

θ

(θ2

2!− θ4

4!+ · · ·

)

]

+ X2

[

α2

θ

(

θ − θ3

3!+ · · ·

)

+α1

θ

(θ2

2!− θ4

4!+ · · ·

)

]

= E + X0 sin θ − X20 cos θ + X1α1 + X2α2.

where

α1 =

[

α1

sin θ

θ− α2

1 − cos θ

θ

]

≡ S1j(θ)αj,

α2 =

[

α1

1 − cos θ

θ+ α2

sin θ

θ

]

≡ S2j(θ)αj,

Thus we have shown that the exponential map of the algebra 〈X0, X1, X2〉gives

exp(θX0 + α1X1 + α2X2) = E + X0 sin θ − X20 cos θ + X1α1 + X2α2

= A(θ, α1, α2) = A[θ, S(θ)α].

That is, exp M generates the same Lie group, E(2), but the elements areparameterized differently from the original definition, A(θ, α). The twoparameterizations are equivalent, being related by a nonsingular similarity


transformation. (Let s = sin(θ/2)/(θ/2) and Σ = S/s, then ΣTΣ = I, andso Σ−1 = ΣT .) �

3.20 (Invariant subgroup and factor group of E(2)) With the notations andresults in Problem 3.19, answer the following questions:

(a) Prove that eθX0T(α)e−θX0 = T[R(θ)α]. From this result, discuss themeaning of the commutation relations between X0 and X1, X2.

(b) Prove that T = {A(0, α)} is the invariant subgroup of E(2), and thatthe factor group E2/T is isomorphic to SO(2).

(c) Is E(2) simple, semi-simple, compact?

SOLUTION 3.20(a) We use associativity of group product to calculate

eθX0T(α)e−θX0 = A(θ, 0)A(0, α)A(−θ, 0)

= A(θ, 0)A(−θ, α)

= A[0, R2(θ)α] ≡ T[R2(θ)α].

From this result and T(α) = I + α1X1 + α2X2, we have

eθX0 (2

∑j=1

αjXj)e−θX0 =2

∑j,k=1

Xk Rkj(θ)αj.

Since αj are arbitrary, it follows eθX0 Xje−θX0 = ∑

2k=1 Xk Rkj(θ) which is equiv-

alent to the bracket relations

[X0, Xj ] = εjk Xk, (ε12 = −ε21 = 1).

These commutation relations between X0 and X1, X2 say that (X1, X2) trans-forms as a vector under rotation.

(b) First, use the group product rule

A(θ, β)T(α)A(−θ, β) = A(0, β)A(θ, 0) ·T(α) · A(θ, 0)−1 A(0, β)−1

= A(0, β)T[R2(θ)α]A(0, β)−1

= T(β)T[R2(θ)α]T(β)−1 = T[R2(θ)α].

The last line arises from the commutativity of translations. Hence the abeliansubgroup T = {A(0, α)} is the center of E(2).

(b) As T is an invariant subgroup, we have E(2) = T + T A(θ, 0) (withθ 6= 0 in the second term). Letting θ range over all allowed values, wehave a continuous family of cosets in one-to-one correspondence with theelements of {A(θ, 0)} ∼= SO(2). So the factor group E(2)/T is isomorphicwith SO(2).

48 CHAPTER 4

(c) As E(2) has a normal subgroup, T , it is not simple; and since T isabelian, E(2) is not semi-simple. As the range of values of the translationparameters α1, α2 is unbounded, the Lie group is not compact. �

3.21 (The adjoint representation) Consider a Lie algebra g of dimension η.When it has a p-dimensional subalgebra, say h, we denote the first η − pbasis vectors by Xα , · · · and the last p vectors by Xi, · · · , which span h. Givethe structure of the matrices R(Xi) and R(Xα) in the adjoint representationin the following situations: (a) g is abelian. (b) g has a subalgebra h. (c) g

has an invariant subalgebra h. (d) g is semisimple. Note: This problem is notin GTAPP.

SOLUTION 3.21 In general the matrices are of the form

R(Xα) =

γ k

β cγαβ ck

αβ

j cγαj ck

αj

=

[

R11(Xα) R12(Xα)R21(Xα) R22(Xα)

]

R(Xi) =

γ k

β cγiβ ck

iβ

j cγij ck

ij

=

[

R11(Xi) R12(Xi)R21(Xi) R22(Xi)

]

(a) g is abelian: R(Xα) = 0, R(Xi) = 0.(b) h is a subalgebra: [h, h] ⊆ h, and so R21(Xi) = 0, and

R(Xα) =

[

R11(Xα) R12(Xα)R21(Xα) R22(Xα)

]

, R(Xi) =

[

R11(Xi) R12(Xi)0 R22(Xi)

]

(c) h is an invariant subalgebra: [g, h] ⊆ h, and so R21(Xα) = 0 andR11(Xi) = R21(Xi) = 0.

(d) g is semisimple:

R(Xα) =

[

R11(Xα) R12(Xα)∗ R22(Xα)

]

, R(Xi) =

[

∗ R12(Xi)∗ R22(Xi)

]

,

where ∗ indicates that not all the elements in this block can be zero for anychoice of the basis. �

Chapter 4. sl(2, C)

4.1 (Casimir invariant) Let h, e, f be the canonical basis of sl(2, C). Findthe quadratic Casimir invariant in any representation, and its value in anirreducible representation. Hint: Use the angular momentum of su(2)C.


SOLUTION 4.1 In any representation of sl(2, C), we have πz 7→ Z, withZ related to the angular momentum operators Ji of su(2)C by J3 = 1/2H,J1 = 1/2(E + F), J2 = − i/2(E − F). Now, Ji satisfies [ Ji, Jj] = iεijk Jk, whereεijk is the usual antisymmetric unit tensor. Define J2 = Ji Ji, then for anyj = 1, 2, 3, we have [J2, Jj] = Ji[ Ji, Jj] + [ Ji, Jj] Ji = iεijk( Ji Jk + Jk Ji) = 0. Sofor any X = C J1 + C J2 + C J3, we also have [J2, X] = 0. Conclusion: J2 is aquadratic Casimir operator in su(2)C. Going to sl(2, C), it has the form J2 =1/4H2 + 1/2(EF + FE), which satisfies [J2, Z] = 0 for any Z ∈ π(sl(2, C)).In the simple representation of highest weight n (wrt H), sl(2, C) acts on thehighest weight vector v according to Hv = nv, Ev = 0, and EFv = [E, F]v =nv, and so J2v = (1/4n2 + 1/2n)v, which gives the invariant value of J2 inthe simple representation of highest weight n as 1/4n(n + 2) = j(j + 1),with n = 2j. �

4.2 (Brackets) Let h, e, f be the canonical basis of sl(2, C). Calculate thebrackets [h, em], [h, f m], [ f , em], and [e, f m].

SOLUTION 4.2

(a) [h, em] = 2mem

(b) [h, f m] = −2m f m

(c) [ f , em] = −mem−1h − m(m − 1)em−1

(d) [e, f m] = m f m−1h − m(m − 1) f m−1

4.3 (Spin-one representation) Let ξ , η ∈ C2 transform into ξ ′ = aξ + cη,η′ = bξ + dη under SU(2) (so that d = a∗, c = −b∗, and aa∗ + bb∗ = 1).Let ψ+ = ξ2, ψ0 =

√2ξη, and ψ− = η2. Show that the ψµ’s transform as a

spin-one representation; find the representation matrix D1 in terms of theEuler angles.

SOLUTION 4.3 With the ψµ’s and the transformation of ξ , η as given, wehave

ψ′+ = ξ ′2 = a2ψ+ −

√2ab∗ψ0 + b∗2ψ−

ψ′0 =

√2ξ ′η′ =

√2ab ψ+ + (aa∗ − bb∗)ψ0 −

√2a∗b∗ψ−

ψ′− = η′2 = b2ψ+ +

√2a∗b ψ0 + a∗2ψ−

The first line gives:

D1,1 = a2 = e−iαc2e−iγ; D0,1 = −√

2ab∗ =√

2cs e−iγ;

D−1,1 = b∗2 = e+iαs2e−iγ ;

50 CHAPTER 4

where we have used a = e−i(α+γ)/2 c, b = −e−i(α−γ)/2 s, c = cos(β/2), ands = sin(β/2). The last two lines give:

D1,0 =√

2ab = −√

2cs e−iα; D0,0 = aa∗ − bb∗ = c2 − s2

D−1,0 = −√

2a∗b∗ =√

2cseiα

D1,−1 = b2 = e−iαs2eiγ; D0,−1 =√

2a∗b = −√

2cs eiγ

D−1,−1 = a∗2 = eiαc2eiγ.

With D1mm′ (α, β, γ) = exp

(

−i(mα + m′γ))

d1mm′ (β), we have

d1mm′ (β) =

c2 −√

2cs s2

√2cs c2 − s2 −

√2cs

s2√

2cs c2

.

4.4 (Quaternion groups) A quaternion q is a number that can be writtenas q = ∑

3µ=0 xµqµ where xµ are real numbers, and qµ = {q0, qi} (with µ =

0, . . . , 3 and i = 1, 2, 3) are defined such that

q0q0 = q0, q0qi = qiq0 = qi, qiqj = −δijq0 + εijkqk ;

and q∗0 = q0, q∗i = −qi under complex conjugation.(a) Let Q be a one-dimensional quaternion (q)-valued vector space, with

typical vector η. The set of all non-zero mappings η′ = Qη form a Liegroup, called GL(1,Q). Find its generators and its Lie algebra.

(b) Show that {Q ∈ GL(1, Q)|Q∗Q = 1} is a Lie group. Find its associ-ated Lie algebra.

(c) Evaluate the exponential mapping of the above Lie algebras.

SOLUTION 4.4 (a) Let Q be the mapping Q 7→ Q: η 7→ Qη, where η is a qnumber (one-dimensional vector) in Q, and Q = ∑

3µ=0 ωµqµ a q-valued op-

erator, depending on the real parameters ωµ. Clearly QQ′ = Q′′ (closure),and I = 1 q0 is the identity operator. So the set of all non-zero Q forms afour-dimensional non compact Lie group, called GL(1,Q). Near the iden-tity, Q ≈ I + δQ ≡ 1 q0 − 1/2δωµ qµ, so that by definition Xµ = − 1/2qµ, withµ = 0, . . . , 3, are the generators of GL(1,Q). From the defining properties ofqµ, the Xµ’s obey the bracket rules

[X0, Xi] = 0, [Xi, Xj] = −εijk Xk ,

and so define a Lie algebra gl(1, Q) which is manifestly a direct sum of twocommuting subalgebras: gl(1, Q) = 〈X0〉 ⊕ 〈X1, X2, X3〉.

(b) {Q ∈ GL(1, Q)|Q∗Q = 1} is a closed and bounded subgroup ofGL(1, Q), and so is a Lie group (called SL(1, Q)). If Q is required to beunimodular, Q∗Q = 1, we have near the identity

Q∗Q ≈ (1 + δQ)∗(1 + δQ) = (1 − δω0)q0 + O(δω2).


It follows that δω0 = 0, i.e. ω0 is fixed, leaving ω1, ω2, ω3 as the remainingfree parameters. The Lie algebra (gl(1, Q)) spanned by X1, X2, X3 has theproduct rules [Xi, Xj] = −εijk Xk.

(c) First, define ω2 = ∑i ω2i and ω = ω/ω, then we have

(

∑i

ωiXi

)2= ∑

i

ω2i X2

i + 1/2 ∑ij

ωiωj(XiXj + XjXi) = −1

4ω2 q0,

which means that ∑i ωiXi is cyclic, so that the mapping exp sl(1, Q), whichgives SL(1, Q), has a closed form:

exp(

∑i

ωiXi

)

= ∑n

1

n!

(

∑i

ωiXi

)n= q0 cos

ω

2− ω · q sin

ω

2.

On the other hand, when mapping gl(1, Q) onto GL(1, Q), we may usethe identity exp(X0 + Xi) = exp X0 exp Xi since [X0, Xi] = 0, and so wehave

exp(

∑µ

ωµXµ

)

= exp(ω0X0) exp(∑i

ωiXi) = e−ω0 /2(

q0 cosω

2− ω · q sin

ω

2

)

.

GL(1, Q) is non compact because ω0 has unlimited range, but in con-trast SL(1, Q) is compact, because exp(∑ ωiXi) involves periodic functionsof ω, and its elements are such that (ω, ω) = (ω + 4π, ω) = −(ω + 2π, ω).We may choose the parameter space to be all points ω such that 0 ≤ ω ≤2π, so that the group elements are in one-to-one correspondence with el-ements of the algebra within a radius of 2π of the origin. Since further(2π, ω) = −(0, ω) = −q0, all points on the surface of this sphere are identi-fied with a single group operation, −q0. Note the isomorphism of GL(1, Q)and U(2) on the one hand, and SL(1, Q) and SU(2) on the other hand. �

4.5 (det R = 1) Show that the invariance of the tensor εijk under rotationsis equivalent to det R = 1 for any orthogonal real 3 × 3 matrix R.

SOLUTION 4.5 Invariance of εijk under rotation means that for any rotationR, we have

Ri`Rjm Rknε`mn = εijk .

First, let (ijk) = (123), then R1`R2mR3nε`mn = 1. The LHS is just det R, andso the equation is: det R = 1. Secondly, if (ijk) = (11k) for any k, then theRHS= 0, and the LHS is

Rk1(R12R13 − R13R12) + Rk2(R11R13 − R13R11) + Rk3(R11R12 − R12R11)

which is identically equal to zero, no new information. And so the invari-ance relation for εijk is just equivalent to the condition det R = 1.

52 CHAPTER 4

4.6 (Invariant axis) Show that a proper rotation in an odd-dimensional spacepossesses an invariant axis.

SOLUTION 4.6 Let R be an n × n real orthogonal matrix with det R = 1.Its eigenvalue problem is Ru = λu, yielding the characteristic equationdet(R − λI) = 0. It is a polynomial equation in λ of order n, with real-valued coefficients (given by the real values entries of R); if λ is a root, sois λ∗. The roots must be either real or come in pairs of complex conjugates.Since R is unitary (real orthogonal),

〈u, u〉 = 〈Ru, Ru〉 = |λ|2〈u, u〉,

so that |λ| = 1 is a condition that all the roots must satisfy. It follows thatthe root λ = 1 can occur only when n is odd. Then the correspondingeigenvector u(1) satisfies the equation Ru(1) = u(1) and the eigenvector u(1)

defines the direction of the axis of rotation, which is of course the invariantdirection. �

4.7 (Similarity transformation) Consider the group of pure rotation in R3:{

R ∈ M(3, R3)|RTR = I; det R = 1}

.(a) Find the direction of the axis and the angle of rotation in terms of

the matrix elements Rij .(b) Find the transformation U that diagonalizes R, so that U−1 RU = Λ

is a diagonal matrix.(c) Find the transformation that gives Λ in a real orthogonal basis.

SOLUTION 4.7 From the preceding problem, we know that R must havethe eigenvalues λ1 = eiω, λ2 = e−iω (0 ≤ ω ≤ π), and λ3 = 1 (so that thecorresponding eigenvector u(3) is the direction of the axis of rotation).

(a) Consider Ru(3) = u(3). Since RT R = I, we also have RTu(3) = u(3) . Itfollows that (R− RT)u(3) = 0. Writing this out in components, we have

(R12 − R21)u(3)2 +(R13 − R31)u

(3)3 = 0

(R21 − R12)u(3)1

+(R23 − R32)u(3)2

= 0

(R31 − R13)u(3)1 +(R32 − R23)u

(3)2 = 0.

These equations give the components of u(3):

u(3)1

: u(3)2

: u(3)3

= (R23 − R32) : (R31 − R13) : (R12 − R21) ,

which specify the direction of the axis of rotation. As for the rotation angleω, we take the trace of R, which is also equal to its diagonalized form:

Tr R = R11 + R22 + R33 = eiω + e−iω + 1 = 1 + 2 cos ω .

This gives ω in terms of matrix elements.


(b) To find the two other eigenvectors, first note that u(1) = u(2)∗ fromthe fact that R = R∗ and λ1 = λ∗

2, and so we only need to calculate thevector u(1); it can be found from the condition that it is orthogonal to u(3) .The evaluation of the eigenvectors is completed by normalizing them to 1.

We define a matrix U such that the entries of its columns are given by

the components of the eigenvectors: Ujk = u(k)j . Its unitarity follows from

the orthonormality of the eigenvectors. The eigenequations then appear as

Riju(k)j = λku

(k)i = u

(j)i δjkλk ,

or in matrix form, RU = UΛ, where Λ = diag [λ1, λ2, λ3] expressed in thecomplex basis {u(i)}.

(c) Noting that u(1) = u(2)∗ and u(3) = u(3)∗ , we now define a real basiswith e1 = (u1 + u2)/

√2, e2 = i(u1 − u2)/

√2, and e3 = u3. The matrix,

V = [Vma ], defined by ea = ∑m umVma , connecting the 2 bases is given by

V =

1/√

2 i/√

2 0

1/√

2 −i/√

2 00 0 1

.

V transforms the diagonal matrix Λ in the complex basis {u} into a non-diagonal matrix R in the real basis {e}:

V−1ΛV =

cos ω − sin ω 0sin ω cos ω 0

0 0 1

≡ R(ω, e3) .

Putting this together with the result in (b), we have

V−1U−1R(ω, u3)UV = R(ω, e3)

Thus, the original R parameterized in angle ω and axis u3 is first diagonal-ized by U in the {u} basis, then transformed by V into a matrix in a realbasis. Put it another way, S = UV converts a rotation R(ω, u3) about theaxis u3 into a rotation about the axis e3 through the same angle ω. �

4.8 (Angle-axis rotation matrix) Write out the 3× 3 matrix of rotation R(ω, n)in terms of the rotation angle ω and the spherical coordinates θ, ϕ of the axisn; or, alternatively in terms of ω and the Cartesian components n1, n2, n3 ofn in an orthonormal basis {ei}, with n2 = 1.

SOLUTION 4.8 We use the formula R(ω, n) = S(θ, ϕ)R(ω, e3)S−1(θ, ϕ),where

n = sθcϕe1 + sθsϕe2 + cθ e3 = eini ,

S(θ, ϕ) = R(ϕ, e3)R(θ, e2)

cθ = cos θ; sθ = sin θ; etc.

54 CHAPTER 4

The matrix S and its inverse (in the Cartesian basis {ei}) are:

S =

cθ cϕ −sϕ sθcϕ

cθsϕ cϕ sθsϕ

−sθ 0 cθ

, S−1 =

cθ cϕ cθsϕ −sθ

−sϕ cϕ 0sθcϕ sθsϕ cθ

.

In addition, we know the matrix for R(ω, e3) from GTAPP § 4.4. So we canevaluate, by matrix multiplication, the elements Rij(ω, n) ≡ Tij :

T11 = s2θc2

ϕ(1 − cω) + cω T12 = s2θcϕsϕ(1 − cω) − cθ sω

T13 = sθcθcϕ(1 − cω) + sθsϕsω T21 = s2θsϕcϕ(1 − cω) + cθ sω

T22 = cω + s2θs2

ϕ(1 − cω) T23 = sθcθsϕ(1 − cω)− sθcϕsω

T31 = sθcθcϕ(1 − cω)− sθsϕsω T32 = sθcθsϕ(1 − cω) + sθcϕsω

T33 = 1 − s2θ(1 − cω).

In terms of the Cartesian components ni, expressed in spherical coordi-nates, n1 = sθcϕ, n2 = sθsϕ, and n3 = cθ, we have in the Cartesian basis

T =

n21(1 − cω) + cω n1n2(1 − cω) − n3sω n1n3(1 − cω) + n2sω

n2n1(1 − cω) + n3sω n22(1 − cω) + cω n2n3(1 − cω)− n1sω

n3n1(1 − cω) − n2sω n3n2(1 − cω) + n1sω n23(1 − cω) + cω

We can do some checking here. First, calculate the trace of T, which isT11 + T22 + T33 = 1 + 2cω = 1 + 2 cos ω, to be compared with

Tr R(ω, n) = Tr (SR(ω, e3)S−1) = Tr R(ω, e3) = 1 + 2 cos ω.

Secondly, we can use the invariance property R(ω, n)n = n to verify thatTijnj coincides with ni as it should. �

4.9 (Euler angles rotation matrix) Write out the rotation matrix R(α, β, γ)in a Cartesian basis in terms of the Euler angles.

SOLUTION 4.9 In terms of the Euler angles α, β, γ, the rotation matrix canbe written as

R(α, β, γ) = R(α, e3)R(β, e2)R(γ, e3).

Making use of the expressions for R(ω, ei) given in § 4.4, we can calculatethe rotation matrix by simple matrix multiplications, leading to the result:

R(α, β, γ) =

cαcβcγ − sαsγ −cαcβsγ − sαcγ cαsβ

sαcβcγ + cαsγ −sαcβsγ + cαcγ sαsβ

−sβcγ sβsγ cβ

,

where cα = cos α, sα = sin α, etc.. �


4.10 (Relations between parameters) Find the relations between the angle-axis parameters and the Euler angles for the same rotation.

SOLUTION 4.10 This proof relies on the results given in the solutions 4.8and 4.9. The two matrices T ≡ R(ω, n) and R ≡ R(α, β, γ) are equal, com-ponent by component. We need three independent relations between thetwo sets of parameters. One such relates the traces of T and R: We alreadyhave Tr T = 1 + 2cω. In addition,

Tr R(α, β, γ) = R11 + R22 + R33 = −1 + 4 cos2

(

α + γ

2

)

cos2 β

2.

This determines ω via cω = cos ω:

cω = −1 + 2 cos2

(

α + γ

2

)

cos2 β

2. (4.1)

Next we have (by trigonometry)

R11 = cαcβcγ − sαsγ = cα+γ cos2(β/2)− cα−γ sin2(β/2),

and with (4.1)’s help, we get

T11 = −1 + 2c2ϕ sin2(β/2)+ 2 cos2(β/2) cos2((α + γ)/2).

Then 2c2ϕ = 1 + cos(2ϕ) and T11 = R11 together yield

cos(2ϕ) + cos(α − γ) = 0 =⇒ ϕ =π + α − γ

2. (4.2)

Finally, consider T33 = R33, which gives 1 − s2θ(1 − cω) = cβ, leading to

tan2 θ = s2θ/(1− s2

θ) = (1 − cβ)/[(1− cω)− (1 − cβ)]. With the expression(4.1) for cω, we get

tan θ =tan(β/2)

sin((α + γ)/2). (4.3)

The equations (4.1)–(4.3) relate the two sets of rotation parameters.Another tactic is to first derive (4.1) as above, then use the invariance of

n, that is R(α, β, γ)n = n, with the components expressed in θ, ϕ variables,to obtain (4.2) and (4.3). This approach requires knowledge of the functionsRij(α, β, γ) only, and not of Tij (ω, θ, ϕ). Conversely, one can evaluate theEuler angles from the angle and axis parameters:

α − γ = 2ϕ − π

cos β = 1 − (1 − cos ω) sin2 θ

cos(α + γ) =1 − cos β − 2 cos ω

1 + cos β.

56 CHAPTER 4

4.11 (R(α, β, γ)at β = 0, π) Show that R(α, 0, γ) is function of the sum α + γonly, and R(α, π, γ) of the difference α − γ only.

SOLUTION 4.11(i) R(α, 0, γ) = R(α, e3)R(0, e2)R(γ, e3) = R(α, e3)R(γ, e3)

= R(α + γ, e3).(ii) We evaluate by matrix multiplication

R(α, π, γ) = R(α, e3)R(π, e2)R(γ, e3)

=

cα −sα 0sα cα 00 0 1

−1 0 00 1 00 0 −1

cγ −sγ 0sγ cγ 00 0 1

= −

cos(α − γ) sin(α − γ) 0sin(α − γ) − cos(α − γ) 0

0 0 1

.

4.12 (To rotate a vector) Show that the rotation R(α, n) of a position vectorx in R3 produces the following vector (with α ≡ αn, n2 = 1):

x′ = x +sin α

α(α × x) +

1 − cos α

α2[α × (α × x)]. (4.4)

SOLUTION 4.12 Let x = x/|x|, and define an orthonormal basis formedby the unit vectors n × x/ sin θ, u ≡ n × (n × x)/ sin θ, and n, where θ isthe angle between n and x. In this basis, x = − sin θ u + cos θ n, and therotation in question has the expression

R(α, n) =

cos α − sin α 0sin α cos α 0

0 0 1

.

The rotated vector x′ is given by

x′ = |x|R(α, n)x

= (sin α) n × x − cos α sin θ |x| u + cos θ |x| n

= cos α x + sin α (n× x) + (1 − cos α)(n · x)n

or equivalently, with (α · x)α = α × (α × x) + α2x,

x′ = x +sin α

α(α × x) +

1 − cos α

α2[α × (α × x)].

Therefore, to the lowest order of the rotation angle, the rotation by the di-rected angle α applied on x produces the vector x′ = x + (α × x) + O(α2).


4.13 (Commutation of angular momentum) (a) Show that the sequence ofinfinitesimal rotations by the directed angles α, β, −α, and −β on the posi-tion vector x is equivalent to the rotation by the directed angle β × α on x.(b) In Hilbert space, a rotation by α is represented by the unitary operatorU(α) = eA, where A = α · X with Xi denoting the generators of rotation.Following the argument used in (a), derive the commutation relations forXi (see also § 3.3 GTAPP).

SOLUTION 4.13 (a) The four successive infinitesimal rotations in questionapplied on x give

A : x 7→ x′ = x + α × x

B : x′ 7→ x′′ = x′ + β × x′ = x + α × x + β × (α × x) + β × x

−A : x′′ 7→ x′′′ = x′′ − α × x′′ = x + β × x + β × (α × x)− α × (β × x)

−B : x′′′ 7→ x′′′′ = x′′′ − β × x′′′ = x + (β × α) × x .

So the sequence A, B, −A, −B is effectively equivalent to a rotation by thedirected angle β × α, transforming x into x + (β × α)× x.

(b) A rotation R(α) is represented by the unitary operator U(α) = eA

in Hilbert space, where A = α · X with Xi denoting the (anti-Hermitian)generators of rotation, related to the Hermitian operators by Xi = −iJi,with i = 1, 2, 3.

The sequence A, B, −A, −B considered in (a) is therefore representedby the operator U(−β) U(−α) U(β) U(α) in Hilbert space. To the first fewterms, U(α) = eA ≈ 1 + A, and this product of exponentials is given by

U(−β) U(−α) U(β) U(α) ≈ I + [β · X, α · X] .

On the other hand, as we have seen in (a), this sequence of rotations takentogether is equivalent to a rotation by the directed angle β × α, representedby the operator U(β × α) ≈ I + (β × α) · X. Equating the two results, wehave

[β · X, α · X ] = (β × α) · X + iζ(β, α)I ,

where the extra additive constant term, ζ(β, α)I = −ζ(α, β)I, is necessaryfor generality. In Cartesian components it follows

[Xi, Xj ] = εijk(Xk + iξk I) (summing over repeated index).

Now we can redefine Xk + iξk I → Xk for k = 1, 2, 3, then we obtain thecommutation relations for the rotation generators

[Xi, Xj] = εijk Xk .

58 CHAPTER 4

If we now introduce the Hermitian operators Ji via Xi = −iJi, we obtain

[ Ji, Jj] = iεijk Jk .

These are the bracket relations for the angular momentum operators. �

4.14 (Exponential mapping) Re-express R(ω, n) = exp(−iωnkLk) in a non-transcendental form by summing the power series of the exponential. Lk

are 3× 3 matrices of the angular momentum operators in a Cartesian basis.

SOLUTION 4.14 The 3 × 3 matrices Lk, with k = 1, 2, 3, that define so(3) aregiven by

L1 =

0 0 00 0 −i0 i 0

L2 =

0 0 i0 0 0−i 0 0

L3 =

0 −i 0i 0 00 0 0

.

With (n1, n2, n3) denoting the Cartesian components of a unit vector n, or-dinary matrix operations produce

n · L =

0 −in3 in2

in3 0 −in1

−in2 in1 0

(n · L)2 =

1 − n21 −n1n2 −n1n3

−n1n2 1 − n22 −n2n3

−n1n3 −n2n3 1 − n23

,

and (n · L)3 = (n · L). So n · L is a cyclic matrix, and the power series forexp(−iωn · L) divides into even and odd powers of n · L:

exp(−iωn · L) = 1 + ∑m=0

(−iωn · L)2m+1

(2n + 1)!+ ∑

m=0

(−iωn · L)2m+2

(2n + 2)!

= 1 − in · L sin ω − (n · L)2(1 − cos ω),

which is the result we want. An easier way is to use R(ω, n) = SR(ω, e3)S−1,where

R(ω, e3) = exp(−iωL3) = 1 − iL3 sin ω − L23(1 − cos ω),

S = R(ϕ, e3)R(θ, e2) =⇒ SL3S−1 = n · L,

(the vector n is in direction (θ, ϕ)). Then, we get

R(ω, n) = SR(ω, e3)S−1

= 1 − i SL3S−1 sin ω − SL23S−1(1 − cos ω),

= 1 − i n · L sin ω − (n · L)2(1 − cos ω).

This result is in agreement with the summation of the power series ob-tained above, and with with the solution to Problem 4.8. �


4.15 (From Cartesian to spherical basis) Show that the matrices Jk ≡ D1( Jk)in the spin j representation (in which D1( J3) is diagonal) are equivalent bya similarity transformation to the defining matrices Lk of the algebra so(3)given in the preceding problem.

SOLUTION 4.15 The matrices Jk ≡ D1( Jk) in the j = 1 angular-momentumrepresentation are given by

J1 =1√2

[

0 1 01 0 10 1 0

]

J=i√2

[

0 −1 01 0 −10 1 0

]

J3 =

[

1 0 00 0 00 0 −1

]

.

We want to calculate the matrix U that transforms L3 into the diagonalmatrix J3. This is equivalent to solving the eignevalue equation L3uλ = λuλ.There are three roots λ = +1, 0,−1 with corresponding normalized vectors

u+ =1√2

−1−i0

u0 =

001

u− =1√2

1−i0

.

The entries in each column of the transformation matrix {Uiµ} ≡ {u(µ)i } are

given by the components of each eigenvector (+0− for both rows, columns):

U =1√2

−1 0 1−i 0 −i

0√

2 0

U−1 =i√2

−1 i 0

0 0√

21 i 0

.

Then U−1 LkU = Jk for k = 1, 2, 3, as we can check. U converts a Cartesianbasis into the angular-momentum canonical basis in three dimensions. An-other way of finding U, given Li and Ji, is to use the equations

L3U = UJ3 and L1U = UJ1 ,

to determine all the elements of U up to an overall constant. Requiring thatU be unitary, we recover the matrix obtained by the first method.

As a corollary, we can relate the defining matrix R(α, β, γ) of SO(3) (cf.Problem 9) to the irreducible representation D1[R(α, β, γ)] as follows:

U−1 R(α, β, γ)U = D1(α, β, γ) .

Of course, this is true for any parameterization; e.g. in (ω, n), knowingR(ω, n) = exp(−iωnkLk) from Problem 14, we can get exp(−iωnkJk):

D1[R(ω, n)] = U−1 R(ω, n)U

= 1 − i n · U−1LU sin ω − (n · U−1 LU)2(1 − cos ω)

= 1 − i n · J sin ω − (n · J)2(1 − cos ω)

60 CHAPTER 4

gives the j = 1 representation of rotation R(ω, n) in the angle-axis parame-terization. �

4.16 (Non standard parameterization) In GTAPP we emphasize the stan-dard parameterization (ω1, ω2, ω3) of SU(2). But there are times when it ismore advantageous to use a non-standard parameterization (e.g. to makethe exponential mapping simpler). In SU(2) the use of a parameteriza-tion associated with the operators J± = J1 ± iJ2 has a further advantagewhen one has to study transitions between states belonging to the same ir-reducible representation. In this problem, we examine just such mappingsinvolving the elements J±, J3.

(a) Evaluate Uo(ω+, ω−, ω3) = exp[−i(ω+J+ + ω− J− + ω3 J3)] and re-late the parameters ω+, ω−, ω3 to the standard parameters ω1, ω2, ω3.

(b) Evaluate Ua(α+, α−, α3) = exp[−i(α+J+ + α− J−)] exp[−iα3J3], anddiscuss the characteristics of the parameters.

(c) Repeat with Ub(β+, β−, β3) = exp[−iβ+J+] exp[−iβ3J3] exp[−iβ−J−].

SOLUTION 4.16 We can rewrite an element of su(2)C in the form

ω1J1 + ω2 J2 + ω3J3 = ω+ J+ + ω− J− + ω3J3

J± = J1 ± iJ2, ω± =1

2(ω1 ∓ iω2)

ω1, ω2, ω3 real ω+ = ω∗−

ω =√

ω21+ ω2

2 + ω23 =

√

4ω+ω− + ω23 .

In the standard parameterization and labeling, each element ∑ ωi Ji ismapped onto

U(ω, n) = exp[−i(ω1J1 + ω2J2 + ω3 J3)]

=

cos ω2− iω3

sin(ω/2)ω −i(ω1 − iω2)

sin ω/2ω

−i(ω1 + iω2)sin ω/2

ω cos ω2

+ iω3sin(ω/2)

ω

.

(a) The exponential mapping of ω+ J+ + ω− J− + ω3 J3 leads to

Uo(ω+, ω−, ω3) = exp[−i(ω+J+ + ω− J− + ω3J3)]

=

cos ω2− iω3

sin(ω/2)ω −2iω+

sin ω/2ω

−2iω− sin ω/2ω cos ω

2+ iω3

sin(ω/2)ω

.


(b) We have in this case

Ua(α+, α−, α3) = exp[−i(α+J+ + α− J−)] exp[−iα3J3]

=

cos α2

−2iα+sin α/2

α

−2iα− sin α/2α cos α

2

e−iα3/2 0

0 e+iα3/2

=

cos α2

e−iα3/2 −2iα+sin α/2

α e+iα3/2

−2iα− sin α/2α e−iα3/2 cos α

2e+iα3/2

.

Here α = 2√

α+α−, and the parameters α3 = α∗3 and α+ = α∗

− are related toω± and ω3 (and hence to the Cartesian parameters ωi) for the same trans-formation (group element) through the Cayley-Klein parameters.

(c) In this case, we have

Ub(β+, β−, β3) = exp(−iβ+J+) exp(−iβ3J3) exp(−iβ−J−)

=

[

1 −iβ+

0 1

] [

e−iβ3/2 0

0 e+iβ3/2

] [

1 0−iβ− 1

]

=

[

e−iβ3/2 − β+β−e+iβ3/2 −iβ+e+iβ3/2

−iβ−e+iβ3/2 e+iβ3/2

]

.

Here all three variables βµ are complex, constrained by the conditions

d = a∗ : e+iβ3/2 = eiβ∗3/2 − β∗

+β∗−e−iβ∗

3/2 ,

c = −b∗ : −iβ−e+iβ3/2 = +iβ∗+e−iβ∗

3/2 .

�

4.17 (Matrices in non standard parameters) Find Djm′m [U] for the different

U found in Problem 4.16.

SOLUTION 4.17 The most direct way is to apply the formula Djm,m(a, b)

given in GTAPP § 4.5, with the appropriate a, b found in Problem 4.16. Inthe cases where U is expressed in terms of the standard parameters ωi, orin terms of ω± and ω3, the coefficients a and b consist each of two terms,which make their binomial expansions rather messy. That’s why it is usefulto consider alternatives. For Ua and Ub, the elements a (or a∗) and b consisteach of a single term, and the general formula applies without difficulty. �

4.18 (j = 1/2 rotation matrices) (a) Write down the matrices for the samegeneral rotation in the Euler angles, R(α, β, γ), and the angle-axis parame-ters, R(ω, n), both in the 2-dimensional j = 1/2 angular-momentum basis.

(b) By comparing the expressions for D1/2[R(α, β, γ)] and D1/2[R(ω, n)],derive the relationship between the two sets of parameters.

62 CHAPTER 4

SOLUTION 4.18 (a) For the Euler angles, we will use the matrix given inGTAPP § 4.5 (both rows and columns are labeled in the order (+,−)):

D1/2m′m [R(α, β, γ)] = e−iαm′

d1/2m′m(β)e−iγm

= e−iαm′ 〈 12m′|e−iβσ2/2 | 1

2m〉e−iγm

=

(

e−i(α+γ)/2 cos(β/2) −e−i(α−γ)/2 sin(β/2)e+i(α−γ)/2 sin(β/2) ei(α+γ)/2 cos(β/2)

)

.

For the ω, n parameters, we will use the matrix found in Solution 4.11,in which σ3 is diagonal:

D1/2m′m [R(ω, n)] =

[

cos(ω/2)− i sin(ω/2) cos θ −i sin(ω/2) sin θe−iϕ

−i sin(ω/2) sin θeiϕ cos(ω/2)+ i sin(ω/2) cos θ

]

.

(b) These two matrices are directly comparable. If they represent thesame rotation, they are identically equal, and we can then derive a relation-ship between their respective parameters. We shall call R ≡ D1/2[R(α, β, γ)]and T ≡ D1/2[R(ω, n)].

(i) TrR = TrT yields

cosω

2= cos

α + γ

2cos

β

2.

From this equation we obtain a result to be used in the next step:

sin2(ω/2)− sin2(β/2) = sin2 α + γ

2cos2 β

2.

(ii) R+−R−+ = T+−T−+ yields

sin2 θ =sin2(β/2)

sin2(ω/2)

from which we have

tan2 θ =sin2 θ

1 − sin2 θ=

sin2 θ

sin2(ω/2)− sin2(β/2)

=tan2(β/2)

sin2((α + γ)/2).

(iii) R+− + R−+ = T+− + T−+ yields

−2i sinω

2sin θ cos ϕ = −2i sin

β

2cos

α − γ + π

2,

which implies ϕ = α − γ + π/2. The results found here agree with thoseobtained in Problem 4.10. �


4.19 (Schwinger’s model) J. Schwinger gave a model of the algebra of an-gular momentum based on the algebra of two independent harmonic oscil-lators represented by (a+, a†

+) and (a−, a†−), which satisfy the commutation

relations typical of uncoupled harmonic oscillators:[

a+, a†+

]

= 1,[

a−, a†−]

= 1, (4.5)[

a+, a†−]

= 0,[

a−, a†+

]

= 0. (4.6)

(a) Define the number operators N+ = a†+a+ and N− = a†

−a−. Showthat they commute with one another, and thus have common eigenkets|n+, n−〉 of respective eigenvalues n+, n− where the vacuum ket |0, 0〉 ≡ |0〉is defined by a+|0〉 = 0, and a−|0〉 = 0. Construct the normalized ket|n+, n−〉 from the normalized vacuum state |0, 0〉.

(b) Define the operators J+ = a†+a−, J− = a†

−a+, J3 = 12(a†

+a+ − a†−a−),

J2 = 12( J+ J− + J− J+) + J2

3. Derive their bracket rules.(c) Evaluate J2|n+, n−〉, J3|n+, n−〉, J+|n+, n−〉, and J−|n+, n−〉. Inter-

pret |n+, n−〉 as a vector | jm〉 describing a system of spin j in state m.(d) From the rotation property of the angular momentum kets | jm〉, ob-

tain the explicit formula for the rotation functions djm′m(β).

SOLUTION 4.19 (a) N+ and N− are clearly Hermitian, and further commuteby (4.6). They have common eigenkets, and their eigenvalues are real:

N+|n+, n−〉 = n+ |n+, n−〉, N−|n+, n−〉 = n− |n+, n−〉 .

These kets are orthogonal to one another and assumed normalized to 1. Inaddition, since the norm of a±|n±〉 is positive and n± = 〈n±|N±|n±〉, thenumbers n± must be both non negative: n± ≥ 0. From (4.5), N± also satisfy[N±, a±] = −a±, and

[

N±, a†±]

= a†±. Consider first the (+) operators:

N+a†+|n+〉 = (

[

N+, a†+

]

+ a†+N+)|n+〉 = (n+ + 1)a†

+|n+〉.

And similarly, N+a+|n+〉 = (n+ − 1)a+|n+〉. Therefore, a+|n+〉 is an eigen-ket of N+ of eigenvalue n+ − 1, and so must be proportional to |n+ − 1〉,up to a normalization factor. Similarly a†

+|n+〉 is proportional to |n+ + 1〉.To find the normalization factor, first notice that 〈n+|a†

+a+|n+〉 = n+. Sothat we have a+|n+〉 =

√n+|n+ − 1〉 If we keep on applying a+ on the ket

|n+〉, we will get a+|n+〉, a2+|n+〉, a3

+|n+〉, . . . , with successively smallercoefficients

√n+,

√

n+(n+ − 1),√

n+(n+ − 1)(n+ − 2), . . ., until the se-quence terminates with a

n++ |n+〉 =

√n+! |0〉. If the process is reversed with

a†+|n+〉 =

√n+ + 1|n+ + 1〉, then starting from the vacuum, we will get

(a†+)n+ |0〉 =

√n+! |n+〉. And so, assuming |0〉 is normalized, we have the

eigenkets of N+, all normalized to one,

|n+〉 =1√n+!

(a†+)n+ |0〉 .

64 CHAPTER 4

Obviously the same reasoning applies to the (−) type oscillators as well,and we have the simultaneous normalized eigenkets for N+ and N− givenby

|n+, n−〉 =1√

n+!n−!(a+)n+ (a−)n− |0〉.

(b) Given (4.5)–(4.6), we can prove [ J3, J±] = ± J± and [ J+, J−] = 2J3,which are the usual angular-momentum commutation relations. In addi-tion, defining N = N+ + N−, we have the quadratic Casimir operator:

J2 =1

2( J+ J− + J− J+) + J2

3 =N

2

(

N

2+ 1

)

.

Let n = n+ + n−, we can easily obtain

J2|n+, n−〉 =n

2

(n

2+ 1

)

|n+, n−〉,

J3|n+, n−〉 =1

2(N+ − N−)|n+, n−〉 =

1

2(n+ − n−)|n+, n−〉,

J+|n+, n−〉 = a†+a−|n+, n−〉 =

√

n−(n+ + 1)|n+ + 1, n− − 1〉,

J−|n+, n−〉 = a†−a+|n+, n−〉 =

√

n+(n− + 1)|n+ − 1, n− + 1〉.

Notice that these operations leave the sum n = n+ + n− unchanged.We can interpret the quantum oscillators as ‘particles’ of spin 1/2: states

of spin up (+ 1/2) are associated with type (+) operators, and states of spindown (− 1/2) with type (−) operators; thus |n+, n−〉 represents a state withn+ spin up particles and n− spin down particles. This is expressed by thefact that the eigenvalue of J3 in this state is 1/2(n+ − n−) ≡ m and theeigenvalue of J2 is n/2(n/2 + 1) ≡ j(j + 1). The operator J+ destroys oneunit of spin down and creates one unit of spin up, thus increasing the zcomponent of angular momentum by one unit. The operator J− destroysone particle of spin up and creates one particle of spin down, which hasthe effect of reducing the z component of the angular momentum by oneunit.

Now making the correspondence n/2 7→ j, (n+ − n−)/2 7→ m, we write

| jm〉 = Njm(a†+)j+m(a†

−)j−m |0〉,(

Njm =1

√

(j + m)!(j− m)!

)

(4.7)

as the eigenket of J2 and J3, with eigenvalues j(j + 1) and m.From the theory of angular momentum we know that when we add,

for example, the spins of two spin 1/2 particles, we obtain states with spinj = 1 and 0; if we have three such particles, we have states with spin j = 3/2


and 1/2. Generally, adding the spins of 2j spin 1/2 particles, we obtain stateswith angular momentum j, j − 1, j − 2, . . . , 0 (or 1/2). In Schwinger’s modelof angular momentum, we have a construct of 2j spin 1/2 particles, andobtain only states of angular momentum j. The reason is that the oper-ators a†

µ, aµ that create or destroy quanta of spin 1/2 are bosonic operators,satisfying commutation (not anticommutation) relations, and the n particlequantum states (4.7) are totally symmetric under permutation, with totalangular momentum j = n × 1/2 = n/2.

(d) As we are concerned only with the rotation function djm′m(β), we

need to consider rotations through the angle β about the y-axis:

U[R(0, β, 0)] = exp(−iJ2β).

We want to examine how (4.7) transforms under U[R]:

U[R]| jm〉 = Njm

(

U[R]a†+U−1 [R]

)j+m (

U[R]a†−U−1[R]

)j−mU[R]|0〉 .

(4.8)(Notice that on the LHS, U[R] should be regarded as a rank 2j tensor op-erator applied on a direct-product vector.) Examine now each of the fac-tors on the RHS. First, U[R]|0〉 = |0〉, because the vacuum is invariant (i.e.Ji|0〉 = 0). Next evaluate various commutators of a†

+ with J2 = (a†+a− −

a†−a+)/(2i):

[

− J2, a†+

]

=1

2i

[

a†−a+, a†

+

]

=1

2ia†− ,

[

− J2,[

− J2, a†+

]

]

=1

2i

[

− J2, a†−]

=1

4a†+ .

Keeping on in this way bracketing more factors of J2, we obtain either a†+ or

a†−. Summing terms in the exponential series of U[R]a†

+U−1[R], we obtain(and in a similar way for a†

−):

U[R]a†+U−1[R] = cos(β/2) a†

+ + sin(β/2) a†− ,

U[R]a†−U−1[R] = cos(β/2) a†

−− sin(β/2) a†+ .

These results are equivalent to giving the rules for rotating spin-up andspin-down vectors ( a†

±|0〉 ) through the angle β about the y axis. Substitut-ing them into (4.8) and using the binomial expansion twice, we get

U[R]|jm〉 = ∑µ

∑ν

√

(j + m)!(j− m)!

(j + m − µ)!µ!(j− m − ν)!ν![a†

+ cos(β/2)]j+m−µ

× [a†− sin(β/2)]µ[−a†

+ sin(β/2)]j−m−ν [a†− cos(β/2)]ν |0〉 .

66 CHAPTER 4

Let us replace ν by m′, defined as m′ = j − µ − ν, then the above be-comes

U[R]|jm〉 = ∑m′

(a†+)j+m′

(a†−)j−m′

√

(j + m′)!(j− m′)!

× ∑ν

(−)m′−m+µ

√

(j + m′)!(j− m′)!(j + m)!(j− m)!

µ!(j− m′ − µ)!(j + m − µ)!(m′ − m + µ)!

× (cos(β/2))2j+m−m′−2µ (sin(β/2))2µ+m′−m |0〉 .

By definition, the matrix representation of U[R] in the angular momentumbasis is given by the usual equation:

U[R]| jm〉 = ∑m′

| jm′〉 Djm′m [R] .

In our case, R = R(0, β, 0), and so Djm′m[R(0, β, 0)] = d

jm′m(β), and hence

the above result should be compared with

U[R(0, β, 0)]|jm〉 = ∑m′

| jm′〉 djm′m(β) ,

which yields Wigner’s formula for the rotation functions:

djm′m(β) = ∑

ν

(−)m′−m+µ

√

(j + m′)!(j− m′)!(j + m)!(j− m)!

µ!(j− m′ − µ)!(j + m − µ)!(m′− m + µ)!

× [cos(β/2)]2j+m−m′−2µ [sin(β/2)]2µ+m′−m .

Its transpose is given by

djmm′ (β) = (−)m−m′

djm′m(β)

= ∑ν

(−)µ

√

(j + m′)!(j− m′)!(j + m)!(j− m)!

µ!(j− m′ − µ)!(j + m − µ)!(m′− m + µ)!

× [cos(β/2)]2j+m−m′−2µ [sin(β/2)]2µ+m′−m .

This agrees with the formula obtained by other means in the Chapter. �

4.20 (Weight diagrams) We have studied in this Chapter how to decom-pose tensor products of the kind SymaV (1) × SymbV (1), where V (1) ∼= C2

is the standard (two-dimensional) representation of sl(2, C). In this prob-lem we want to apply the same approach to find the decomposition ofsymmetric powers of V (2), i.e. SymaV (2). Here, V (n) = SymnV (1) is the(n + 1)-dimensional irreducible representation of sl(2, C). In order to do


this, first find the decomposition of Sym2V (2) and Sym3V (2); then general-ize to SymaV (2).

SOLUTION 4.20 (a) Although Sym2V (1) is irreducible, Sym2V (2) is not, aswe will see now. Take as a basis for V (2) the three objects x, z, y, thenSym2V (2) consists of the six symmetric objects xx, xz, xy, zz, zy, yy (‘sym-metric’ means xy = yx, xz = zx, etc.). So we have in all six eigenvaluesfor J3, namely, −2,−1, 0 (twice), 1, 2. The presence of a multiplicity of twotells us that this representation is not irreducible. The quintet −2,−1, 0, 1, 2composes V (4), while the remaining 0 is V (0). In terms of spherical harmon-ics, the first set corresponds to D2

m and the second to D00. The decomposi-

tion is Sym2V (2) = V (4) + V (0):

s

0es

−1s

1s

−2s

2

(b) As for Sym3V (2) , it consists of the symmetric triple products of x, z, y,namely, x3, x2z, x2y, xz2, xzy, xy2, z3, z2y, zy2 and y3, corresponding to theeigenvalues −3,−2,−1 (twice), 0 (twice), 1 (twice), 2, 3. It is most likely(dimensional considerations) that they belong to the irreducible represen-tations V (6) and V (2). So we have the relation Sym3V (2) = V (6) + V (2):

s

0es

−1e s

1es

−2s

2s

−3s

3

(c) Finally, SymnV (2), for some positive integer n, consists of all the ho-mogeneous n order polynomials in x, y, z. The number of symmetric mono-mials of the form xaybzc with a + b + c = n is d(n) = 1

2(n + 1)(n + 2), for

any integer n. In other words, there are d(n) eigenspaces νm, with m =−n,−n+ 1, . . . , n − 1, n, with various multiplicities.

The maximally symmetric space contained in SymnV (2) is V (2n) , havingthe eigenvalues −n,−n + 1, . . . , n − 1, n. Its complement must have thedimension

d(n) − (2n + 1) =1

2(n− 1)n = d(n−2) ,

corresponding to the number of the remaining eigenvalues. The maximallysymmetric subspace of V (2n) is therefore V (2(n−2)) ; it has dimension 2n − 3.The number of remaining eigenvalues is given by

d(n−2) − (2n− 3) =1

2(n − 3)(n− 2) = d(n−4) .

The representation of dimension 2(n − 4) + 1 is V (2(n−4)) . The sequenceterminates either with V (2) or V (0) . So the symmetric n-powers of V (2) isfully reducible, with the decomposition into a direct product given by

SymnV (2) = V (2n) ⊕V (2n−4) ⊕ · · ·⊕ V (2)( or V (0)) .

68 CHAPTER 4

We can check the result by calculating the dimensions. As we have seen,SymnV (2) has dimension d(n) = 1

2(n + 1)(n + 2). It must match the sum of

the dimension on the RHS. When n is even, that sum is

n/2

∑k=0

(2n − 4k + 1) = (2n + 1)n/2

∑k=0

1 − 4n/2

∑k=1

k =1

2(n + 1)(n + 2) .

When n is odd, the sum has the upper limit k = (n − 1)/2, and we stillobtain

(n−1)/2

∑k=0

(2n− 4k + 1) = (2n + 1)((n−1)/2

∑k=0

1)− 4((n−1)/2

∑k=1

k) =1

2(n + 1)(n + 2) .

The two sides agree in dimensions. �

4.21 (Reduction of direct product) Calculate the Clebsch–Gordan (CG) co-efficients for the direct product of irreducible representations of SO(3): (a)D1/2 × D1, and (b) D1 × D1. A good way is to make use of the propertiesof states of highest weight and the raising/lowering operators.

SOLUTION 4.21 Consider the addition rule J = j× I + I × j′ of angular mo-mentums, corresponding to the reduction of the product Dj × Dj′ = ⊕DJ

(J being summed over | j − j′ |, | j − j′ | + 1, . . . , j + j′). In the representationspace of Dj × Dj′ , the basis state is | jm〉| j′m′〉 ≡ | jm, j′m′〉, while in the ‘cou-pled’ representation DJ , the basis state |(jj′) J M〉; both are eigenstates of J3,respectively, of eigenvalue m + m′ and M. The state of the highest weight inDj × Dj′ is | jj, j′ j′〉; it is also the state of highest weight in the representationspace of Dj+j′ , so that

|(jj′) j + j′ , j + j′〉 = | jj, j′ j′〉 .

Applying powers of J− on this state, we can obtain all vectors | j+ j′, M〉. Weproceed similarly in the subspace orthogonal to the representation spaceof Dj+j′ , seeking the highest weight state belonging to a different DJ andusing J− to generate |(jj′) J M〉 for M = J, J − 1, . . . ,− J. As j, j′ are fixed,we can use the simplified notations: | jm, j′m′〉 ≡ |m : m′〉 and |(jj′) J M〉 ≡| J, M〉, suppressing explicit reference to j, j′ . (Note that the two bases aredifferentiated by the distinctive signs (: and ,).) The following formula will


be useful:

J−| J, M〉 = [( J + M)( J − M + 1)]1/2| J, M− 1〉

= (j− × I + I × j′−)j

∑m=−j

j′

∑m′=−j′

|m : m′〉〈jm, j′m′| J M〉

= ∑mm′

{

[(j + m)(j− m + 1)]1/2|m − 1 : m′〉

+ [(j′ + m′)(j′ − m′ + 1)]1/2|m : m′ − 1〉}

〈jm, j′m′| J M〉

(a) Product D1/2 × D1 = D3/2 ⊕ D1/2.. Representation J = 3/2:

| 3/2, 3/2〉 = | 1/2 : 1〉,| 3/2, 1/2〉 = (1/

√3) J−| 3/2, 3/2〉 = (1/

√3)[

√2| 1/2 : 0〉+ | − 1/2 : 1〉]

| 3/2,− 1/2〉 = (1/2) J−| 3/2, 1/2〉= (1/

√12)[2| 1/2 : −1〉+

√2| − 1/2 : 0〉+

√2| − 1/2 : 0〉]

= (1/√

3)[| 1/2 : −1〉+√

2| − 1/2 : 0〉],| 3/2,− 3/2〉 = | − 1/2 : −1〉 .

. Representation J = 1/2: The highest weight state is | 1/2, 1/2〉, whichmust be orthogonal to | 3/2, 1/2〉. Its component | 1/2 : 0〉 has a positive co-efficient, consistent with the phase convention adopted for the CG coeffi-cients.

| 1/2, 1/2〉 = (1/√

3)[| 1/2 : 0〉 −√

2| − 1/2 : 1〉| 1/2,− 1/2〉 = J−| 1/2, 1/2〉

= (1/√

3)[√

2| 1/2 : −1〉+ | − 1/2 : 0〉 − 2| − 1/2 : 0〉]= (1/

√3)[

√2| 1/2 : −1〉 − | − 1/2 : 0〉] .

(b) Product D1 × D1 = D2 ⊕ D1 ⊕ D0.. Representation J = 2:

|2, 2〉 = |1 : 1〉,|2, 1〉 = (1/2)J−|2, 2〉 = (1/

√2)[|1 : 0〉 + |0 : 1〉]

|2, 0〉 = (1/√

6)|2, 1〉= (1/

√6)|1 : −1〉+ |0 : 0〉+ |0 : 0〉+ | − 1 : 1〉]

= (1/√

6)|1 : −1〉+ 2|0 : 0〉+ | − 1 : 1〉]|2,−1〉 = (1/

√2)[|0 : −1〉+ | − 1 : 0〉]

|2,−2〉 = | − 1 : −1〉 .

70 CHAPTER 4

. Representation J = 1: The state of highest weight |1, 1〉 can be foundfrom its orthogonality to |2, 1〉 found above, and the phase convention ofthe C-G coefficients. Then, apply J− on |1, 1〉.

|1, 1〉 = (1/√

2)[|1 : 0〉 − |0 : 1〉]|1, 0〉 = (1/

√2) J−|1, 1〉

= (1/√

2)[|1 : −1〉+ |0 : 0〉 − |0 : 0〉 − | − 1 : 1〉]= (1/

√2)[|1 : −1〉 − | − 1 : 1〉]

|1,−1〉 = (1/√

2) J−|1, 0〉 = (1/√

2)[|0 : −1〉 − | − 1 : 0〉]. Representation J = 0: This representation has just one basis state

|0, 0〉; it is a linear combination of three |m : m′〉 vectors:

|0, 0〉 = a|1 : −1〉+ b|0 : 0〉+ c| − 1 : 1〉.J+ annihilates this state:

(1/√

2) J+|0, 0〉 = a|1 : 0〉+ b|1 : 0〉+ b|0 : 1〉+ c|0 : 1〉 = 0,

which yields a = −b = c. The normalized state is

|0, 0〉 = (1/√

3)|1 : −1〉 − |0 : 0〉+ | − 1 : 1〉.It is orthogonal to both |2, 0〉 and |1, 0〉 as it should. �

4.22 (Rotation of tensor product) Show that Tjm = ∑µν〈`µ, sν|jm〉X`

µYsν trans-

forms as a rank j tensor if X`µ and Ys

ν transform as tensors of rank ` and s.

SOLUTION 4.22 “ X`µ and Ys

ν transform as tensors of rank ` and s” means

U[R] X`µ U[R−1] = ∑µ′ X`

µ′ D`µ′µ[R], U[R] Ys

ν U[R−1] = ∑ν′ Ysν′D

sν′ν [R].

We have to show that Tj transforms like a tensor of rank j. Making use ofthe CG series for Dj and the orthogonality of the CG, we obtain under anarbitrary rotation R:

UTjmU−1 = ∑

µν

〈`µ, sν|jm〉UX`µUU−1Ys

νU−1

= ∑µνµ′ ν′

〈`µ, sν|jm〉D`µ′µDs

ν′ν X`µ′Ys

ν′

= ∑µνµ′ ν′

〈`µ, sν|jm〉 ∑JMM′

〈`µ′, sν′| J M′〉〈`µ, sν|JM〉DJM′M X`

µ′Ysν′

= ∑µ′ν′M′

〈`µ′, sν′| J M′〉DjM′m X`

µ′Ysν′ = ∑

M′T

jM′ D

jM′m.

This completes our proof. �


Chapter 5. sl(3, C)

5.1 (Matrices Eij) The matrices Eij have a single non-zero entry, equal to 1,at row i, column j, i.e. (Eij )ab = δiaδjb . Rewrite the basic elements of sl(3, C)in terms of 3 × 3 matrices Eij , and check their commutation relations.

SOLUTION 5.1 The basic eight elements of sl(3, C), in terms of the matricesEij , are: T1 = E11 − E22, T2 = E22 − E33, E1 = E12, E2 = E23, E3 = E13,F1 = E21, F2 = E32, and F3 = E31.

From their definition, Eij satisfy Eij Eab = Eib δja . In particular, with nosummation over repeated indices, Eaa Eab = Eab . These relations can beused to check the commutation relations among the Ti , Ei , and Fi. �

5.2 (Spin and hypercharge) In physics, the CSA of su(3)C is often definedin a different basis: tz = 1/2t1 and y = 1/3t1 + 2/3t2. The other elements ofthe algebra remain the same as we have in GTAPP, with new notations:t+ = e1, t− = f1, u+ = e2, u− = f2, v+ = e3, and v− = f3. In thestandard representation, where πz 7→ Z, they are related to the 3 × 3 Gell-Mann matrices λi by Tz = 1/2λ3 and Y = 1/

√3λ8, T± = 1/2(λ1 ± iλ2),

U± = 1/2(λ6 ± iλ7), and V± = 1/2(λ4 ± iλ5).(a) Calculate the quadratic traces Tr(hihj) in the two bases of the CSA.(b) Calculate the roots in the physical basis h = {tz, y}.(c) Calculate the Killing forms (y : y), (tz : tz), and (y : tz).(d) Calculate hαi from its definition, and then the inner products 〈αi, αj〉,

with i, j = 1, 2, 3.

SOLUTION 5.2(a) In our basis Tr T1

2 = Tr T22 = 2, Tr(T1T2) = −1. In the physical

basis: Tr T2z = 1/2, Tr Y2 = 2/3, Tr (YTz) = 0.

(b) We display the commutation relations of tz and of y with variouselements of the algebra in the following table

tz y t+ t− u+ u− v+ v−tz 0 0 t+ −t− − 1/2u+

1/2u− 1/2v+ − 1/2v−y 0 0 0 0 u+ −u− v+ −v−

α (00) (0

0) (1

0) (−1

0) (−

1/21

) (1/2−1

) (1/21

) (−1/2−1

)

t±, u±, v± are simultaneous eigenvectors of ad tz and ad y with nonzeroeigenvalues, and so are the root elements of the algebra. From the first tworows, the roots are obtained and given in the last row. The positive roots,relative to tz, y, are α1 = (1, 0), α2 = (− 1/2, 1), and α3 = (1/2, 1) = α1 + α2;the other three are their negatives.

(Note: H. Georgi Ch. VII uses H1 = tz and H2 =√

3y/2 for the CSA. Soour roots αi are written in his basis as α1 7→ (1, 0), α2 7→ (− 1/2,

√3/2), and

72 CHAPTER 5

α3 7→ (1/2,√

3/2). Further, he chooses α ≡ α3 and β ≡ −α2 for the simpleroots, so that α1 = α + β. His H1 axis is along α1 and H2 perpendicular toit, and so is perpendicular to the line α2–α3.)

(c) We calculate the Killing form directly from its definition, (x : y) =Tr (adx ady), with adx ady = [x, [y, ]]. For (tz : tz), we have

[tz, [tz, tz]] = 0, [tz, [tz, y]] = 0, [tz, [tz, t±]] = t±, [tz, [tz, u±]] = 1/4u±,[tz, [tz, v±]] = 1/4v±. So that

(tz : tz) = 0 + 0 + 1 + 1 + 1/4 + 1/4 + 1/4 + 1/4 = 3. Similarly from thecommutation relations, and in the same order, we have

(y : y) = 0 + 0 + 0 + 0 + 1 + 1 + 1 + 1 = 4,(tz : y) = 0 + 0 + 0 + 0 − 1

2− 1

2+ 1

2+ 1

2= 0. If the bilinear forms among

the t1 and t2 are assumed known, then the above results are arrived at morequickly. Note that (tz : y) = 0 tells us that tz ⊥ y.

(d) Let hαi = citz + diy with i = 1, 2, 3. We know, on one hand, (hαi :k) = αi(k), and, on the other hand, (citz + diy : tz) = 3ci and (citz + diy :y) = 4di. With the roots known from (b), we obtain ci and di, leading tohα1 = 1/3tz, hα2 = − 1/6tz + 1/4y, hα3 = 1/6tz + 1/4y. The inner productson h∗0 are obtained via the Killing forms, 〈αi, αj〉 = (hαi : hαj), leading to〈αi, αi〉 = 1/3 (with i = 1, 2, 3), 〈α1, α3〉 = 〈α2, α3〉 = −〈α1, α2〉 = 1/6. Clearlythese properties of the roots are independent of the CSA basis used to dothe calculations. �

5.3 (Casimir operator) Find the quadratic Casimir operator in sl(3, C), andcalculate its invariant value in the irreducible representation of highest weight(n, m). Hint: Use the properties of the Gell-Mann matrices λi.

SOLUTION 5.3 We have sl(3, C) = 〈h1, h2, e1, f1, e2, f2, e3, f3〉 in GTAPP’s no-tations. In the standard representation πz 7→ Z, they are related to theGell-Mann matrices in su(3)C by λ3 = H1, λ8 = 1/

√3(H1 + 2H2), λ1 =

E1 + F1, λ2 = −i(E1 − F1), λ4 = E3 + F3, λ5 = −i(E3 − F3), λ6 = E2 + F2,λ7 = −i(E2 − F2). Define Ji = 1/2λi, with i = 1, . . . , 8, then they have theproperties: (i) [ Ji, Jj] = i fijk Jk, and (ii) Tr ( Ji Jj) = Nπδij . Remark: These tworelations are true in any representation, only the value of Nπ varies with the rep-resentation π. For π = (1, 0), with the Gell-Mann matrices, N1,0 = 1/2. Forπ = (1, 1), with ( Ji)jk = −i fijk, one has N1,1 = 3. It follows that fijk is anti-symmetric wrt all three indices: fijk = − f jik because of (i); and fijk = − fikj

because i/2 fijk = Tr ( Ji Jj Jk − Jj Ji Jk) = Tr ( Jk Ji Jj − Ji Jk Jj) by the trace cyclicity.

Define J2 = Ji Ji. Then [J2, Jj] = Ji[ Ji, Jj] + [ Ji, Jj] Ji = i fijk( Ji Jk + Jk Ji) =

0 for all j = 1, . . . , 8. So J2 is a quadratic Casimir operator of su(3)C.To rewrite it in terms of the canonical basis in sl(3, C), we use the corre-spondence written above, and obtain J2 = 1/3(H2

1 + H22) + 1/6(H1H2 +

H2H1) + 1/2(E1F1 + F1E1 + E2 F2 + F2E2 + E3F3 + F3E3). The Casimir in-variant for the irreducible representation of highest weight (n, m) (relative


to H1, H2) can be calculate by applying J2 on the highest weight vectorv, making use of (H1, H2)v = (n, m)v, H3v = (n + m)v, and Eiv = 0,EiFiv = [Ei, Fi]v = Hiv (i = 1, 2, 3). This gives us J2v = Cπv, whereCn,m = 1/3[(n + m)(n + m + 3)− nm]. �

5.4 The structure constants Nαβ of a simple Lie algebra are defined in theequation [eα, eβ] = Nαβ eα+β if α + β is a non-zero root. Assuming this is thecase and going into a finite representation, prove the symmetry relations:Nαβ = −Nβα = N−β,−α = Nβ,−α−β = N−α−β,α .

SOLUTION 5.4 Let Eα = π(eα) be defined in a finite representation π cho-sen such that E†

α = E−α and N∗αβ = Nαβ. The relation Nαβ = −Nβα follows

trivially from the defining equation. Applying Hermitian conjugation onboth sides of the same equation leads to [E−β, E−α ] = N∗

αβE−α−β , whichyields: Nαβ = N−β,−α . The invariance of the Killing form – expressed inthe identity

(

[a, b] : c) = (a : [b, c])

, or in our case,(

[Eα, Eβ ] : Eγ

)

=(

Eα :

[Eβ, Eγ ])

– gives us Nαβδα+β+γ,0 = Nβγδα+β+γ,0 , or simply Nαβ = Nβ,−α−β ,and interchanging the indices, it is equivalent to Nβα = Nα,−α−β . This com-pletes the proof. �

5.5 (Weight vectors) Let s = 〈h, e, f 〉 be an sl(2, C) subalgebra of sl(3, C),and let π(h) = H, π(e) = E, and π( f ) = F. Further, let [H, E] = α(h)E,[H, F] = −α(h)E, and [E, F] = H; and finally Hv = µ(h)v. (Hint: ReviewGTAPP §5.2 Weights and Weight Vectors) Prove for any integer k ≥ 0:

(a) [H, Ek] = kα(h)Ek and [H, Fk] = −kα(h)Fk;(b) HEkv = [µ(h)+ kα(h)]Ekv and HFkv = [µ(h)− kα(h)]Fkv;(c) [E, Fk ] = kFk−1 H − xkα(h)Fk−1; find xk.

SOLUTION 5.5 (a) By definition, we have [H, E] = α(h)E. We assert that[

H, Ek]

= kα(h)Ek, which can be proved by induction:[

H, Ek+1]

= E[

H, Ek]

+[

H, E]

Ek = E(kα(h)Ek) + α(h)EEk

= (k + 1)α(h)Ek+1 .

Similarly, starting from [H, F] = −α(h)F, we have [H, Fk] = −kα(h)Fk.(b) HEkv = Ek Hv +

[

H, Ek]

v =[

µ(h)+ kα(h)]

Ekv . This equation tellsus that Ekv is a weight vector with weight µ(h)+ kα(h).

(c) With the given result, we compute [E, Fk+1 ] = (k + 1)Fk H − (xk +k)αFk. For the given relation to hold we must have xk+1 = xk + k. Because[E, F] = H, we have x1 = 0, and so xk = k(k − 1)/2. �

5.6 (π2,0 is group invariant) Consider the irreducible representation of high-est weight (2, 0) of sl(3, C), to be called π2,0. Enumerate its weights andweight vectors, and prove that the rep space is group invariant.

SOLUTION 5.6 Let b = {H1, H2, E1, E2, E3, F1, F2, F3} be the basis of sl(3, C),satisfying the standard bracket relations. We call s1 = {H1, E1, F1} and

74 CHAPTER 5

s2 = {H2, E2, F2}. From GTAPP § 5.2, the representation ((2, 0),V ) hasweights (2, 0), (0, 1), (1,−1), (−2, 2), (−1, 0), and (0,−2), relative to H1

and H2. They correspond, respectively, to the weight vectors v, F1v, F2F1v,F2

1 v, F2F21 v, and F2

2 F21 v. We call this set W , and V = 0 ∪W . By definition

Eiv = 0 for i = 1, 2, 3.To prove that V is group invariant, it suffices to prove that it is invariant

under each of the basis elements b. Each of the vectors in W is a jointeigenvectors of H1 and H2, and so W is invariant under these two elements.

(i) F2 and Ei nullify v, and we have F1v ∈ W and F3v = [F2, F1]v =F2F1v ∈ W . So the action of b leaves v in V .

(ii) Consider now w2 = F1v. Clearly, Fiw2 ∈ W for i = 1, 2. Now,F3w2 can be calculated in two ways. First, F3F1v = F1F2F1v making use of[F3, F1] = 0. Second, F3F1v = (F2F2

1− F1F2F1)v making use of F3 = [F2, F1].

Equating the two results gives F2F21 v = 2F1F2F1v. The end result is F3w2 ∈

W . As for the action of Ei , we can use the bracket relations [E1, F1] = H1,[E2, F1] = 0, and [E3, F1] = −E2 to show that E1w2 = µ?v, whereas E2 andE3 nullify w2. This takes care of F1v of W .

(iii) As for the remaining vectors wi of W , one argues by induction.Assuming the assertion is true for wi, then using the bracket relations, oneproves that the action of b on wi+1 either gives 0 or a vector in W . �

5.7 (Representation of highest weight (2, 0)) Construct the irreducible rep-resentation of highest weight (2, 0) of sl(3, C) (also called a2). Give the ma-trix elements of the lowering operators F1 and F2 in a normalized basis ofthe representation.

SOLUTION 5.7 The construction is based on the notion of string of weights,Sαi(µ) = {µ + pαi, . . . , µ, . . . , µ − qαi}, where αi is a simple root, and p, qare non-negative integers related by q − p = µ(hi). The algebra a2 has twosimple roots, α1 = (2,−1) and α2 = (−1, 2); and every weight is of theform µ = (µ1, µ2) where µi = µ(hi). As λ = (2, 0) is the highest weight(corresponding to the weight vector v) of the representation πλ, all otherweights of πλ must be of the form λ − αi − αj − · · · .

(i) The strings Sαi(λ) must have q = 0 since λ + αi is not a weight forboth i = 1, 2. The formula p = q + λi produces p = 0 + 2 = 2 for α1 andp = 0 + 0 for α2, and we obtain two weights µ1 = λ − α1 = (0, 1) (weightvector F1v) and µ2 = λ − 2α1 = (−2, 2) (weight vector F2

1 v). On the otherhand, λ − α2 and λ − 3α1 are not weights of πλ.

(ii) Consider Sα2(µ1): here p = 0 because µ1 + α2 = λ − α1 + α2 is not aweight, hence q = p + µ1

2 = 1. So µ3 = µ1 − α2 = λ − α1 − α2 = (1,−1) isa weight (for the vector F2F1v).

(iii) Consider Sα2(µ2): we have here q = p + µ22 = 0 + 2 = 2, and so two

new weights, µ4 = µ2 − α2 = (−1, 0) (vector F2F21 v) and µ5 = µ2 − 2α2 =

(0,−2) (vector F22 F2

1v).


(iv) From µ3 = (1,−1), we can build two strings. For Sα1(µ3), we haveq = p + µ3

1= 0 + 1 = 1 (p = 0 because λ− α2 is not a weight), but µ3 − α1 =

(−1, 0) = µ4 is not new. For Sα2(µ3) we have q = p + µ32 = 1 − 1 = 0, and

so µ3 − α2 = λ − α1 − 2α2 is not a weight.(v) From µ4 = (−1, 0), we can check that the string Sα1(µ4) has q =

p + µ41 = 1 − 1 = 0, and Sα2 µ4) has q = p + µ4

2 = 1 + 0 = 1, but µ4 − α2 =(0,−2) is not new.

(vi) And finally, from µ5 = (0,−2), we can check that the string Sα1(µ5)has q = p + µ5

1= 0, and Sα2(µ5) has q = p + µ5

2 = 2 − 2 = 0.This completes the description of π2,0. The results are summarized in

the following table (λ = (2, 0) is the highest weight defining the irrep π2,0):

(2, 0) v0 |λ, (2, 0)〉(0, 1) F1v0 |λ, (0, 1)〉(−2, 2), (1,−1) F2

1 v0, F2F1v0 |λ, (−2, 2)〉, |λ, (1,−1)〉(−1, 0) F2F2

1 v0 |λ, (−1, 0)〉(0,−2) F2

2 F21 v0 |λ, (0,−2)〉

The last column gives the normalized kets. (−2, 2), (0, 1), (2, 0) form a s1

triplet, whereas (0, 1), (1,−1)a doublet and (−2, 2), (−1, 0), (0,−2)a tripletof s2. To find the matrix elements of Fi we generalize a formula in Prob-lem 1. For a string of weights µ, µ− αi, . . . , µ − qαi giving a complete mul-tiplet of si, we have

Fi |µ− sαi〉 = |µ − (s + 1)αi〉√

(q − s)(s + 1).

This formula is valid for one-dimensional weight spaces. In our case, itgives

F1|λ, (2, 0)〉 = |λ, (0, 1)〉√

2, F1|λ, (0, 1)〉 = |λ, (−2, 2)〉√

2,F2|λ, (0, 1)〉 = |λ, (1,−1)〉,F2|λ, (−2, 2)〉 = |λ, (−1, 0)〉

√2, F2|λ, (−1, 0)〉 = |λ, (0,−2)〉

√2. �

5.8 (The adjoint representation (1, 1)) Construct the irrep of highest weightλ = (1, 1) of sl(3, C). Find the matrix elements of F1 and F2.

SOLUTION 5.8 Proceeding as in the preceding problem, we find

(1, 1) v0 |λ, (1, 1)〉(−1, 2), (2,−1) F1v0, F2v0 |λ, (−1, 2)〉, |λ, (2,−1)〉(0, 0) F1F2v0, F2F1v0 |λ, (0, 0)1〉, |λ, (0, 0)0〉(1,−2), (−2, 1) F2

2 F1v0, F21 F2v0 |λ, (1,−2)〉, |λ, (−2, 1)〉

(−1,−1) F1F22 F1v0 |λ, (0,−2)〉 .

Here (−1,−1) is one-dimensional because F1F22 F1 = F2F2

1 F2, and so thematrix elements of F1 and F2 can be found from formula given in the pre-ceding problem. The difficulty lies in the (0, 0) states. As F1F2 6= F2F1,

76 CHAPTER 5

(0, 0) is a two-dimensional weight space spanned by independent combi-nations of F1F2v0 and F2F1v0. Choose the combinations such that one, called(0, 0)0, is a singlet of s1, and the other, called (0, 0)1, is part of the triplet(−2, 1), (0, 0)1, (2,−1) of s1. Then the (0, 0) state that is part of the triplet(−1, 2), (0, 0), (1,−2) of s2 must be a linear combination of the eigen statesof s1 (0, 0)0 and (0, 0)1:

F2|λ, (−1, 2)〉 = a|λ, (0, 0)1〉+ b|λ, (0, 0)0〉.

Using the results found in Problem 1, we have a2 + b2 = 2. As (0, 0)0

and (0, 0)1 belong to the singlet and triplet of s1, we have

E1F2|λ, (−1, 2)〉 = aE1|λ, (0, 0)1〉 + bE1|λ, (0, 0)0〉 = a√

2|λ, (2,−1〉.

On the other hand, since E1F2 = F2E1, we also have E1F2|λ, (−1, 2)〉 =F2E1|λ, (−1, 2)〉 = F2|(1, 1)〉 = |(2,−1)〉. Hence a = 1/

√2, and, picking a

phase, b =√

3/2.State (0, 0) can also be reached via E2|λ, (1,−2)〉 = c|λ, (00)1〉+ d|λ, (00)0〉,

where c2 + d2 = 2. Proceeding in the same way, we obtain c = 1/√

2 andd =

√3/2. Now, as F2E2 = E2F2 − H2, we have the equation

F2|λ, (0, 0)1〉+√

3F2|λ, (0, 0)0〉 = 2√

2|λ, (1,−2)〉 ,

leading to√

2F2|λ, (0, 0)1〉 = |λ, (1,−2)〉,√

2F2|λ, (0, 0)0〉 =√

3|λ, (1,−2)〉.�

5.9 (Dual representation) Let G be a Lie group, and g its Lie algebra.(a) Consider the dual space V defined wrt some bilinear form 〈u, v〉 for

all v ∈ V and u ∈ V. Let Π be a representation, Π : G → GL(V), and thedual representation Π : G → GL(V). Show that Π(g) = Π(g−1)T for allg ∈ G, where T indicates transposition.

(b) Let π be the representation of g on V (i.e. dual to π). What is π(X)for X ∈ g? Check that π preserves the Lie algebra products.

(c) Now take the case of sl(3, C). How are the weights of π and π re-lated? If (n, m) is the highest weight of the irreducible representation π,what is the highest weight of its dual π?

SOLUTION 5.9 (a) The bilinear form 〈u, v〉 must be invariant for any v ∈ V,

u ∈ V and g ∈ G: 〈u, v〉 = 〈Π(g)u, Π(g)v〉 = 〈u, ΠT(g)Π(g)v〉. It follows

that ΠT(g)Π(g) = I, or Π(g) = ΠT(g−1) for any g ∈ G.

(b) Let g = ex, with x ∈ g; then Π(g) = eπ(x) . Near the identity, therelation 〈u, v〉 = 〈Π(g)u, Π(g)v〉 becomes 〈π(x)u, v〉+ 〈u, π(x)v〉. andso π(x) = −π(x)T. Let g be defined by the products [xi, xj] = cijk xk, or[π(xi), π(xj)] = cijk π(xk) in any π. Making the transposition the product


law becomes [π(xi)T , π(xj)

T ] = −cijk π(xk)T . But this is [π(xi), π(xj)] =

cijk π(xk), and so the product law is preserved (meaning that π is indeed arepresentation).

(c) Let π be an arbitrary representation. Then in the weight vector basis,H is diagonal, that is, H = Λ = diag [µ1, µ2, . . .] for H = π(h) with h ∈ h.Since HT = ΛT = Λ, H = π(h) = −HT has the same set of eigenvalues,or π has the same set of weights as π, only with a sign change for all theweights. On the other hand E = −ET = −F and F = −FT = −E, thehighest weight vector in π becomes the lowest weight vector in π; and viceversa. Since to every vector in π corresponds a vector in π, each invariantsubspace in V corresponds to an invariant subspace in V . Therefore if π isirreducible, so is π.

As in GTAPP, call the fundamental representations V (1,0) = 〈ξ1, ξ2, ξ3〉and V (0,1) = 〈ξ1, ξ2, ξ3〉, where ξ i have the weights κ1 = (1, 0), κ2 = (−1, 1),and κ3 = (0,−1), and ξ i have the weights −κi for i = 1, 2, 3 (all with re-spect to the canonical basis 〈h1, h2〉 for the CSA). Think of the simple rep-resentation V (n,m) as being built up from the tensor product SymnV (1,0) ⊗SymmV (0,1), then the highest weight vector in V (n,m) is ξ1nξ3

m having theweight nκ1 − mκ3 = (n, m). The vector ξ3nξ1

m is the lowest weight vectorhaving the weight nκ3 − mκ1 = (−m,−n). On the other hand, in V (m,n) ,the vector ξ1mξ3

n has the highest weight given by (m, n). But V (m,n) is the

dual to V (n, m), i.e. V(n,m)= V (m,n) , and so we see that the highest weight

of V(n,m)is the negative of the lowest weight of V (n,m) .

5.10 (Mesons and baryons) Let ξ i correspond to the quarks u, d, s, and ξ i tothe anti-quarks u, d, s. As in Problem 2, the isospin is Tz = 1/2λ3, and thehypercharge Y = 1/

√3λ8; in addition, the (electric) charge is Q = Tz + 1

2Y.

Given that mesons are pairs of quark-antiquark, and baryons are made upof three quarks, find the quark content of the meson and of the baryonoctets, and give the values of Tz, Y, and Q for their members.SOLUTION 5.10 (a) Any tensor product ξ iξ j in 3 ⊗ 3 may be split into twoparts of definite symmetry of the form: ξ iξ j = Sδi

j + Mij, where S = 1/3ξ iξ i

and Mij = ξ iξ j − 1/3δi

jξiξ i. In terms of the quark flavors, (ξ1, ξ2, ξ3) =

(u, d, s) and (ξ1, ξ2, ξ3) = (u, d, s), the singlet S ∈ 1 and the octet Mij ∈ 8

take the form

S = 1/3(uu + dd + ss)

M =

1/3(2uu− dd − ss) ud usdu 1/3(−uu + 2dd − ss) dssu sd 1/3(−uu− dd + 2ss)

.

The quarks u, d, s have, respectively, isospin Tz = (1/2,− 1/2, 0), hyper-charge Y = (1/3, 1/3,− 2/3), and charge Q = (2/3,− 1/3,− 1/3), whereas the

78 CHAPTER 5

antiquarks u, d, s just have the same values with opposite signs in all cases.The singlet S has zero quantum numbers; the different entries of M havethe values for Tz, Y, Q as shown:

Tz, Y, Q =

0, 0, 0 1, 0, 1 1/2, 1, 1−1, 0,−1 0, 0, 0 − 1/2, 1, 0

− 1/2,−1,−1 1/2,−1, 0 0, 0, 0

.

The values of Tz indicate that there are 4 isospin subrepresentations: T = 0,T = 1/2 (2 times), and T = 1. The two isospin doublets can be assignedunambiguously to two pairs of mesons (provided their spin and parity arealso known), and similarly for the two charged isospin vectors (1, 0, 1 and−1, 0,−1). As for its neutral component, it is admixed with the isospinsinglet in the three diagonal entries (there are only two independent fieldsbecause Tr M = 0). For example, if M is to describe the pseudo-scalarmesons, then we will have the following assignment for the entries of thematrix:

M =

1/√

2 π0 + 1/√

6 η π+ K+

π− − 1/√

2 π0 + 1/√

6 η K0

K− K0 − 2/√

6 η

.

(b) 3 ⊗ 3 ⊗ 3 can be reduced in two steps. In the first step the tensorproduct ξ iξ j in 3 ⊗ 3 is written as ξ iξ j = 1

2εijk ηk + 1

2Sij , which combines a

3 tensor, ηk = εkmnξmξn, with a 6 tensor, Sij = ξ iξ j + ξ jξ i. In the secondstep, the product 3 ⊗ 3 and 6 ⊗ 3 are decomposed into 1 ⊕ 8 and 8 ⊕ 10respectively:

ηkξc =1

3δc

k(ηmξm) + [ηkξc − 1

3δc

k(ηmξm)]

Sabξc =1

3(εackSbm + εbckSam)ξnεmnk +

1

3(Sabξc + Sbcξa + Scaξb).

Putting altogether, we have the product

ξ iξ jξk =1√6

εijk S +1√2

εijm Bkm +

1√6(εikm N j

m + εjkm Nim) + Dijk .

The four terms correspond to the representations 1, 8, 8, 10, respectively.In particular, the Ba

b octet is given by:√

2Bab = ηbξa − 1

3δa

b(ηmξm), orexplicitly in terms of u, d, s:√

2B11 = −uds + usd + dus− sud

√2B1

2 = (su− us)u√

2B13 = (ud− du)u

√2B2

1 = (ds− sd)d√

2B22 = −uds− dsu + dus + sdu

√2B2

3 = (ud − du)d√

2B33 = usd− dsu− sud + sdu

√2B3

2 = (su− us)s


The table of the values of Tz, Y, Q for M is valid for B as well, and so is ourremark about the mixing of the neutral fields. The following table showsexplicitly the different subrepresentations in the octet of the spin 1

2baryons:

Y ↓ T ↓ Tz → −1 − 1/2 0 1/2 1

0 1 Σ− Σ0 Σ+

1 1/2 n p−1 1/2 Ξ− Ξ0

0 0 Λ0

Chapter 6. Simple Lie Algebras: Structure

6.1 (Find ∆ for b2) Find all the roots of Lie algebra b2.

SOLUTION 6.1 The associated Dynkin diagram B2 indicates there are twosimple roots, α1 (short) and α2 (long), producing the Cartan integers a12 =a21 = −1. The simple roots α1 and α2 are the only roots at level one. Fromthese, we construct two chains: (i) α1 + pα2, with p = −a21 = 1, which tellsus that α1 + α2 is a root, but not α1 + 2α2; (ii) α2 + pα1, with p = −a12 = 2,which says that 2α1 + α2 is a root, but not 3α1 + α2. So α1 + α2 is the only(positive) root at level 2, and 2α1 + α2 is the only (positive) root at level 3.At the next level, (2α1 + α2) + α1 = 3α1 + α2 is not a root (as we alreadyknow); and (2α1 + α2) + α2 = 2(α1 + α2) is not a root. So there are no rootsat level 4 and higher. So the root system of so(5, C) ∼= sp(2, C) is:

Σ = {α1, α2}, ∆ = {±α1, ±α2, ±(α1 + α2), ±(2α1 + α2)}. �

6.2 (Find ∆+ for d4) We have seen in GTAPP § 6.5 that the set of roots of asimple Lie algebra g can be found from its given fundamental system Σ andCartan matrix A. Use this recursive method to find all positive roots of thed4 Lie algebra.

SOLUTION 6.2 We recall that any positive root β can be written β = ∑i αiki

where the ki are non-negative integers; the sum ∑i ki = n is called the levelof β. The method is based on the fact that if all the roots up to level n areknown, then the roots at higher levels can be found by examining strings ofroots of the type Sαj(β) = {β + kαj| − q ≤ k ≤ p}, where p = q − ∑i ajiki,with known q and ki. There are new (positive) roots if and only if p > 0;they are β + αj, . . . , β + pαj.

The Lie algebra d4 has four simple roots, αi; i = 1, 2, 3, 4, such that the

80 CHAPTER 6

Cartan matrix is

A =

2 −1 0 0−1 2 −1 −1

0 −1 2 00 −1 0 2

.

. Level 1: α1, α2, α3, α4.

. Level 2: All roots are of the form αi + pαj. Since q = 0 and ki = 0, 1,we will have the possibilities:

α1 + pα2: p = −a21 = 1 → β1 ≡ α1 + α2,α1 + pα3: p = −a31 = 0 ,α1 + pα4: p = −a41 = 0,α2 + pα3: p = −a32 = 1 → β2 ≡ α2 + α3,α2 + pα4: p = −a42 = 1 → β3 ≡ α2 + α4,α3 + pα4: p = −a43 = 0 .. Level 3: Roots at this level are of the form βi + pαj. Note that α1 + 2α2

is not a root since we already know that p = 1 for the string α1 + pα2; andsimilarly α2 + 2α3 and α2 + 2α4 are not roots. Hence we need to consideronly the following cases:

β1 + pα1: p = 1 − a11 − a12 = 0 ,β1 + pα3: p = −a31 − a32 = 1 → β1 + α3 = α1 + α2 + α3,β1 + pα4: p = −a41 − a42 = 1 → β1 + α4 = α1 + α2 + α4,β2 + pα1: p = −a12 − a13 = 1 → β2 + α1 = α2 + α3 + α1,β2 + pα2: p = 1 − a22 − a23 = 0 ,β2 + pα4: p = −a42 − a43 = 1 →β2 + α4 = α2 + α3 + α4,β3 + pα1: p = −a12 − a14 = 1 → β3 + α1 = α2 + α4 + α1,β3 + pα2: p = 1 − a22 − a24 = 0 ,β3 + pα3: p = −a32 − a34 = 1 → β3 + α3 = α2 + α4 + α3.There are three roots at level 3: γ1 = α1 + α2 + α3, γ2 = α1 + α2 + α4,

and γ3 = α2 + α3 + α4.. Level 4: Roots at this level are of the form γi + pαj.γ1 + pα2: p = −a21 − a22 − a23 = 0 ,γ1 + pα4: p = −a41 − a42 − a43 = 1 → γ1 + α4 = α1 + α2 + α3 + α4,γ2 + pα2: p = −a21 − a22 − a24 = 0 ,γ2 + pα3: p = −a31 − a32 − a34 = 1 → γ2 + α3 = α1 + α2 + α4 + α3,γ3 + pα1: p = −a12 − a13 − a14 = 1 → γ3 + α1 = α2 + α3 + α4 + α1,γ3 + pα2: p = −a22 − a23 − a24 = 0 .There is a single root at level 4: δ = α1 + α2 + α3 + α4,. Level 5: As 2α1 + α2 + α3 + α4, α1 + α2 + 2α3 + α4 and α1 + α2 + α3 +

2α4 are not roots, we need to consider only α1 + α2 + α3 + α4 + pα2. Herethe integer p = −a21 − a22 − a23 − a24 = 1, so that the only root at level 5 isε = α1 + 2α2 + 2α3 + α4, which also implies that α1 + 3α2 + 2α3 + α4 is nota root. So that at the next level, we need to consider only 3 cases.


. Level 6:ε + pα1: p = −a11 − 2a12 − a13 − a14 = 0 ,ε + pα3: p = −a31 − 2a32 − a33 − a34 = 0 ,ε + pα4: p = −a41 − 2a42 − a43 − a44 = 0 .There are no roots at level 6, or higher.In conclusion, d4 has 12 positive roots: α1, α2, α3, α4; α1 + α2, α2 + α3,

α2 + α4; α1 + α2 + α3, α1 + α2 + α4, α2 + α3 + α4; α1 + α2 + α3 + α4; andα1 + 2α2 + 2α3 + α4. Thus, d4 has 24 roots, is of rank 4, and order 28. This isthe root system of so(8, C). �

6.3 (Rules for Dynkin diagrams) Identification of the allowed Dynkin dia-grams (those that are consistent with the definition of a fundamental sys-tem of roots (FSR)) is based on a set of general rules derived from the defi-nition of FSR of semisimple Lie algebras. Prove:

(a) Rule 1: A Dynkin diagram contains more vertices than joined pairs.It cannot have closed polygons. (A closed polygon, or loop, is a set of pointssequentially connected and the last point is also connected to the first.)

(b) Rule 2: The maximum number of lines issuing from a vertex is 3.(c) Rule 3: If an allowed Dynkin diagram Γ contains an A2 simple chain

as a sub diagram, then the diagram obtained from Γ by replacing the simplechain with a vertex point is also an allowed Dynkin diagram.

SOLUTION 6.3 Let Σ = {α1, α2, . . . , α`} be an FSR, and Γ the correspondingDynkin diagram. Further let ui = αi/|αi|, so that 〈ui, ui〉 = 1 and 2〈ui, uj〉 =0,−1,−

√2, or −

√3 for i 6= j. Note that 4〈ui, uj〉2 is the number of lines

joining i and j; if it is equal to 0 the two vertices i, j are not connected.(a) Let u = ∑

`i=1 ui. Since 〈u, u〉 > 0, we have ` > −2 ∑i<j 〈ui, uj〉 ≥ n,

where n is the number of joined vertices. Hence n < ` or Rule 1.If a diagram contains a subdiagram (1, 2, . . . , k) that is a loop, e.g. ui is

linked with ui+1, and uk linked with u1, then in this subdiagram the numberof vertices is equal to the number of joined pairs, contradicting Rule 1.

(b) Given Rule 1 assume you have a diagram in which normalized ver-tex u is linked with normalized vertices v1, . . . , vk. No two vi are connected(otherwise there would be loops); this means 〈vi, vj〉 = 0 with i 6= j. Sinceu, v1, . . . , vk are linear independent, they span a space in which there is avector v0 normalized to 1 and orthogonal to all other vi. So {v0, . . . , vk}forms an orthonormal basis, and u may be written as u = ∑

ki=0 vi〈vi, u〉.

Further 〈u, v0〉 6= 0, otherwise one could write u in terms of the vi, contraryto assumption. We have then

1 = 〈u, u〉 = 〈u, v0〉2 + 〈u, v1〉2 + . . . + 〈u, vk〉2.

So 〈u, v1〉2 + . . . + 〈u, vk〉2 < 1, or ∑ki=1 4〈u, vi〉2 < 4, which tells us that the

number of lines joining vertex u to all vertices vi must be less than 4.

82 CHAPTER 6

(c) Let diagram Γ contain a simple chain, consisting of (normalized)vertices v1, . . . , vk linked pairwise by basic A2 links, such that 2〈vi, vi+1〉 =−1, with i = 1, . . . , k − 1. Let v = ∑

ki=1 vi; then 〈v, v〉 = k + 2 ∑i<j 〈vi, vj〉 =

k − (k − 1) = 1. Let u be a vertex of Γ, but not in the chain, connectedto any one of the vi, say vj (but to no other, by Rule 1). Then we have〈u, v〉 = 〈u, ∑i vi〉 = 〈u, vj〉, so that 4〈u, v〉2 = 4〈u, vj〉2 = 0, 1, 2, or 3. Thesetwo results, 〈v, v〉 = 1 and 4〈u, v〉2 ≤ 3, mean that v behaves like any othervertex. The simple chain v1, . . . , vk is shrunk to vertex v, and a vertex u of Γ

connected to a vj by n lines is now connected to v by n lines. �

6.4 (Non allowed diagrams) Show that configurations Fig.6.7(c.1), Fig.6.8(d.1)–(d.2) in GTAPP are not allowed by showing that the bilinear form 〈, 〉defined in the respective spaces h∗0 is not positive definite.

SOLUTION 6.4 By definition an FSR is a basis of an inner-product Eu-clidean space V = h∗0, and so any non-zero vector v ∈ V must be suchthat 〈v, v〉 > 0.

Configuration Fig.6.7(c.1): Let u1, . . . , u5 the normalized vectors (roots)corresponding to the vertices of the diagram such that 〈u1, u2〉 = 〈u3, u4〉 =〈u4, u5〉 = − 1/2, and 〈u2, u3〉 = −1/

√2. Consider the vector v =

√2u1 +

2√

2u2 + 3u3 + 2u4 + u5, which is nonzero. But 〈v, v〉 = 0, contrary to thedefinition of an FSR.

Configuration Fig.6.8(d.1): We label the vertices such that 〈ui, ui+1〉 =− 1/2 for i = 1, 2, 3, 4, 6, and 〈u3, u6〉 = − 1/2. Now, take v = 3u3 + 2(u2 +u4 + u6) + u1 + u5 + u7. But 〈v, v〉 = 0, in contradiction.

Configuration Fig.6.8(d.2): We label the vertices such that 〈ui, ui+1〉 =− 1/2 for i = 1, 2, . . . , 7 and 〈u6, u9〉 = − 1/2, and consider the nonzero vectorv = u1 + 2u2 + 3u3 + 4u4 + 5u5 + 6u6 + 4u7 + 2u8 + 3u9. It turns out that〈v, v〉 = 0, in contradiction. �

6.5 (Generators of g2) Given that the Lie algebra g2 is defined by a 2 × 2Cartan matrix, with elements a11 = a22 = 2, a12 = −3, and a21 = −1,identify its generators.

SOLUTION 6.5 We know that from the Cartan matrix of g2 we can deter-mine its simple (positive) roots α1 and α2, as well as its other four posi-tive roots: α3 = α1 + α2, α4 = α1 + α3 = α1 + α1 + α2, α5 = α1 + α4 =α1 + α1 + α1 + α2, and α6 = α2 + α5 = α2 + α1 + α1 + α1 + α2. Note that eachis of the form αi1 + . . . + αik such that the partial sums αi1 + αi2 + . . . + αir

are also roots for each r ≤ k.Applying the structure theory, we know that associated with the simple

roots, α1 and α2, we may define the subalgebras si = 〈hi, ei, fi〉 ∼= sl(2, C)with i = 1, 2, such that [ei, f j] = δijhi, [hi, ej] = aijej, and [hi, f j] = −aijej, sothat the simple roots relative to hi are αj(hi) = aij.

We need 8 more generators associated with the remaining roots in ∆.


We can define them as e3 = [e1, e2], e4 = [e1, e3], e5 = [e1, e4], e6 = [e2, e5],and similarly f3, f4, f5, f6. According to the general recipe, h1, h2, e1, . . . , e6,f1, . . . , f6 form a basis for the 14-dimensional algebra g2, with h1 and h2 abasis for h, ei a generator of the root space gαi , and fi a generator of the rootspace g−αi , for i = 1, 2, . . . , 6. �

6.6 (Multiplication table for g2) With the same definitions as in Problem 5,establish the multiplication table for the generators of the Lie algebra g2.

SOLUTION 6.6 Using a simplified notation: [ab] ≡ [a, b], we first recall theJacobi identity, written in equivalent forms:

[

x[ab]]

=[

[xa]b]

+[

a[xb]]

and[

[ab]x]

=[

a[bx]]

−[

b[ax]]

= ada adb x − adb ada x.We begin by calculating the products of h1, h2 with the other generators,

with the result that ei, fi are eigen vectors with eigenvalues ±αi in the rootsystem ∆.

For i, j = 1 or 2, we know that [hi, ej] = aijej, and [hi, f j] = −aijej, withthe simple roots relative to hi given by αj(hi) = aij , where aii = 2, a12 = −3,a21 = −1. For the remaining root vectors, we have for any h = ah1 + bh2:

[he3] =[

[he1]e2

]

+[

e1[he2]]

= (α1(h) + α2(h))[e1e2] = α3(h)e3

[he4] =[

[he1]e3

]

+[

e1[he3]]

= (α1(h) + α3(h))[e1e3] = α4(h)e4

[he5] =[

[he1]e4

]

+[

e1[he4]]

= (α1(h) + α4(h))[e1e4] = α5(h)e5

[he6] =[

[he2]e5

]

+[

e2[he5]]

= (α2(h) + α5(h))[e2e5] = α6(h)e6

And similarly for [h f j] = −αj(h) fj. Since all roots αj are known in terms ofαj(hi) = aij, with i, j = 1 or 2, we obtain the products listed in the followingtable:

e1 e2 e3 e4 e5 e6 f1 f2 f3 f4 f5 f6

h1 2e1 −3e2 −e3 e4 3e5 0 −2 f1 3 f2 f3 − f4 −3 f5 0h2 −e1 2e2 e3 0 −e5 e6 f1 −2 f2 − f3 0 f5 − f6

Since each ei is in root space gαi , its adjoint action carries ej to 0 or gαi+αj ,such that [eiej] = 0 if αi + αj 6∈ ∆. Similarly, we must have [ fi f j] = 0 if−(αi + αj) 6∈ ∆ and [ei f j] = 0 if αi − αj 6∈ ∆. The following tables recordthese results, with * denoting the products still to be determined.

84 CHAPTER 6

e1 e2 e3 e4 e5 e6

e1 0 e3 e4 e5 0 0e2 0 0 0 e6 0e3 0 ∗ 0 0e4 0 0 0e5 0 0e6 0

f1 f2 f3 f4 f5 f6

f1 0 f3 f4 f5 0 0f2 0 0 0 f6 0f3 0 ∗ 0 0f4 0 0 0f5 0 0f6 0

f1 f2 f3 f4 f5 f6

e1 h1 0 ∗ ∗ ∗ 0e2 0 h2 ∗ 0 0 ∗e3 ∗ ∗ ∗ ∗ 0 ∗e4 ∗ 0 ∗ ∗ ∗ ∗e5 ∗ 0 0 ∗ ∗ ∗e6 0 ∗ ∗ ∗ ∗ ∗

g2 decomposes under s1 and s2, and so the adjoint action of e1, f1 ande2, f2 gives direct information on the elements that span the subrepresenta-tions. Therefore, we start the calculation with ade2 and ad f2 acting on theother basis vectors of s2 subrepresentations:

[e2 f3] = [[e2 f1] f2] + [ f1[e2 f2]] = 0 + [ f1h2] = − f1

[e2 f6] = [[e2 f2] f5] + [ f2[e2 f5]] = [h2 f5] + 0 = f5

[ f2e3] = [[ f2e1]e2] + [e1[ f2e2]] = 0 − [e1h2] = −e1

[ f2e6] = [[ f2e2]e5] + [e2[ f2e5]] = −[h2e5] + 0 = e5

In a similar way, we consider ade1 and ad f1 acting on the other basisvectors of s1 subrepresentations:

[e1 f3] = [[e1 f1] f2] + [ f1[e1 f2]] = [h1 f2] = 3 f2

[e1 f4] = [[e1 f1] f3] + [ f1[e1 f3]] = [h1 f3] + 3[ f1 f2] = 4 f3

[e1 f5] = [[e1 f1] f4] + [ f1[e1 f4]] = [h1 f4] + 4[ f1 f3] = 3 f4

[ f1e3] = 3e2

[ f1e4] = 4e3

[ f1e5] = 3e4

As for the remaining entries, we calculate in the same way, proceed-ing from the simple to the more complicated cases. For example, for theproducts involving e3, f3, we have

[e3 f3] = ade1 ade2 f3 − ade2 ade1 f3 = −h1 − 3h2

[e3 f4] = ade1 ade2 f4 − ade2 ade1 f4 = 0 − 4ade2 f3 = 4 f1


[e3 f6] = ade1 ade2 f6 − ade2 ade1 f6 = ade1 f5 − 0 = 3 f4

[e3e4] = ade1 ade2 e4 − ade2 ade1 e4 = 0 − 4ade2 e5 = −e6

[ f3 f4] = ad f1 ad f2 f4 − ad f2 ad f1 f4 = 0 − ad f2 f5 = − f6

[ f3e4] = ad f1 ad f2 e4 − ad f2 ad f1 e4 = 0 − ad f2 e3 = 4e1

[ f3e6] = ad f1 ad f2 e6 − ad f2 ad f1 e6 = ad f1 e5 − 0 = 3e4

We calculate another one:[e4 f4] = ade1 ade3 f4 − ade3 ade1 f4 = 4h1 + 4(h1 + 3h2) = 8h1 + 12h2.The completed results are given in the following table.

e1 e2 e3 e4 e5 e6 f1 f2 f3 f4 f5 f6

e1 0 e3 e4 e5 0 0 h1 0 3 f2 4 f3 3 f4 0e2 0 0 0 e6 0 0 h2 − f1 0 0 f5

e3 0 −e6 0 0 −3e2 e1 −h3 4 f1 0 3 f4

e4 0 0 0 −4e3 0 −4e1 h4 −12 f1 12 f3

e5 0 0 −3e4 0 0 12e1 −h5 36 f2

e6 0 0 −e5 −3e4 −12e3 −36e2 h6

f1 0 f3 f4 f5 0 0f2 0 0 0 f6 0f3 0 − f6 0 0f4 0 0 0f5 0 0f6 0

where h3 = h1 + 3h2, h4 = 8h1 + 12h2, h5 = 36(h1 + h2), h6 = 36(h1 + 2h2).

6.7 (Positivity of norms in classical series) Check that vectors in the rootspaces associated with Dynkin diagrams A`, B`, C`, and D` have positivedefinite norm squares, as expected of Euclidean spaces.

SOLUTION 6.7 Let αi, with i = 1, . . . , ` be the FSR corresponding to a givenallowed diagram, ui = αi/|αi| be the normalized simple roots, and φ anyvector in h∗0, written as φ = ∑i xiα

i = ∑i xi|αi|ui, where xi are arbitrary realnumbers. We have 2〈ui, uj〉 = −1,−

√2,−

√3 , with i 6= j, respectively for

the A2, B2 and G2 basic links. We will show that in each case 〈φ, φ〉may bewritten as a sum of squares of reals.

Define ak = 1/2[(x1 − x2)2 + (x2 − x3)

2 + · · ·+ (xk−1 − xk)2].

. A`: 〈φ, φ〉 = ∑i x2i − (x1x2 + · · ·+ x`−1x`)

= 1/2x21+ a` + 1/2x2

`.

. B`: All vertices have weights 2, except α` of weight 1.〈φ, φ〉 = 2(x2

1 + · · ·+ x2`−1

) + x2`− ∑

`−1i=1

xixi+1 = x21 + 2a`.

. C`: All vertices have weights 1, except α` of weight 2.〈φ, φ〉 = x2

1 + · · ·+ x2`−1

+ 2x2`− ∑

`−2i=1 xixi+1 − 2x`−1x`

= 1/2x21 + 2a`−1 + 1/2(x`−1 − 2x`)

2.

86 CHAPTER 6

. D`: The branch point α`−2 is linked with the endpoints α`−1 and α`.〈φ, φ〉 = ∑i x2

i − ∑`−2i=1

xixi+1 − x`−2x`

= a`−2 + 1/2x21 + (1/2x`−2 − x`−1)

2 + (1/2x`−2 − x`)2. �

6.8 (Positivity of norms in the exceptional algebras) Same question as in thepreceding problem for the five exceptional algebras.

SOLUTION 6.8. G2: 〈φ, φ〉 = x2

1 + 3x22 − 3x1x2 = (x1 − 3/2x2)2 + 3/4x2

2.. F4: 〈φ, φ〉 = x2

1 + x22 + 2x2

3 + 2x24 − x1x2 − 2x2x3 − 2x3x4 = 1/4[x2

1 +(x1 − 2x4)

2 + (x1 − 2x3 + 2x4)2 + (x1 − 2x2 + 2x3)2].

. E6: 〈φ, φ〉 = ∑6i=1 x2

i − x1x3 − x2x4 − x3x4 − x4x5 − x5x6 = 1/2[(1/2x1 +x2 − x3)2 + (− 1/2x1 + x2 + x3− x4)

2 + (− 1/2x1 + x4− x5)2 + (− 1/2x1 + x5−x6)

2 + (− 1/2x1 + x6)2 + 3/4x2

1].. E7: 〈φ, φ〉 = ∑

7i=1 x2

i − x1x3 − x2x4 − x3x4 − x4x5 − x5x6 − x6x7 =1/2[(1/2x1 + x2 − x3)

2 + (− 1/2x1 + x2 + x3 − x4)2 + (− 1/2x1 + x4 − x5)

2 +(− 1/2x1 + x5 − x6)2 + (− 1/2x1 + x6 − x7)2 + (− 1/2x1 + x7)2 + 1/2x2

1].. E8: 〈φ, φ〉 = ∑

8i=1 x2

i − x1x3 − x2x4 − x3x4 − x4x5 − x5x6 − x6x7 −x7x8 = 1/2[(1/2x1 + x2 − x3)2 + (− 1/2x1 + x2 + x3 − x4)

2 + (− 1/2x1 + x4 −x5)

2 + (− 1/2x1 + x5 − x6)2 + (− 1/2x1 + x6 − x7)

2 + (− 1/2x1 + x7 − x8)2 +

(− 1/2x1 + x8)2 + 1/4x21]. �

6.9 (Maximal roots) Find the maximal roots of e6, e7, and e8.

SOLUTION 6.9e6: α1 + 2α2 + 2α3 + 3α4 + 2α5 + α6 = 1/2(ε1 + ε2 + ε3 + ε4 + ε5 +

√3ε6).

e7: 2α1 + 2α2 + 3α3 + 4α4 + 3α5 + 2α6 + α7 =√

2ε7.e8: 2α1 + 3α2 + 4α3 + 6α4 + 5α5 + 4α6 + 3α7 + 2α8 = ε7 + ε8. �

6.10 (Data on g2) Rank-two Lie algebra g2 is defined by the simple roots α1

and α2 related by |α2| =√

3|α1| and separated by an angle of 150◦. Intro-duce the basis ei (i = 1, 2, 3) for h0 and the dual basis εi (i = 1, 2, 3) for h∗0obeying εi(ej) = δij and ε1 + ε2 + ε3 = 0. Express the fundamental co-rootshi and the fundamental roots and weights αi and ωi, where i = 1, 2 andωi(hj) = δij , in the appropriate basis.

SOLUTION 6.10 Choose α1 = ε2, then α2 = ε1 − ε2. As h1 and h2 must beof the form ae1 + be2 + ce3, we can determine the coefficients a, b, c fromαj(hi) = aij (the Cartan integers) and (ε1 + ε2 + ε3)(h) = 0, leading to h1 =

−e1 + 2e2 − e3 and h2 = e1 − e2. Since ω1 and ω2 in h∗0 may be writtenmε1 + nε2 + qε3 = (m− q)ε1 + (n− q)ε2, we determine the coefficients fromωi(hj) = δij , leading to ω1 = ε1 + ε2 and ω2 = 2ε1 + ε2 = ε1 − ε3. �


Chapter 7. Simple Lie Algebras: Representations

7.1 Weyl group for a2. Consider the Weyl group for a2 (see GTAPP § 7.1 Ex-ample 1). (a) Give the transformations of the positive roots and an arbitraryweight under the Weyl group. (b) Give the orbits of the dominant weights(0, 0), (1,0), (0,1), (1,1), (2,0), (0,2), (3,0) and (0,3).

SOLUTION 7.1 (a) Let µ(h) = (m1, m2) where mi = µ(hαi) be an arbitraryweight on h∗0. Roots –weights of the adjoint representation– can be denotedin the same way. From the general definition wα · µ = µ− µ(hα)α, it followsfor the simple roots α1 = (2,−1) and α2 = (−1, 2) the basic relations

w1(m1, m2) = (m1, m2)− m1(2,−1) = (−m1, m1 + m2)

w2(m1, m2) = (m1, m2)− m2(−1, 2) = (m1 + m2,−m2).

Hence the results for (a):

wi wi · (m1, m2) wi · α1 wi · α2 wi · α3

w0 (m1, m2) α1 α2 α3

w1 (−m1, m1 + m2) −α1 α3 α2

w2 (m1 + m2,−m2) α3 −α2 α1

w3 (−m2,−m1) −α2 −α1 −α3

w4 (−m1 − m2, m1) α2 −α3 −α1

w5 (m2,−m1 − m2) −α3 α1 −α2

(b) The Weyl orbit of a weight consists of the distinct equivalent weightsof the given weight. From Weyl reflections w · (m1, m2) given above weobtain the following results (s indicates the orbital size):

(0, 0) →(0,0) [s = 1];(1, 0) →(1,0) (-1,1) (0,-1) [s = 3]; (0, 1) → (0,1) (1,-1) (-1,0) [s = 3];(1, 1) →(1,1) (-1,2) (2,-1) (-2,1) (1,-2) (-1,-1) [s = 6];(2, 0) →(2,0) (-2,2) (0,-2) [s = 3]; (0, 2) →(0,2) (2,-2) (-2,0) [s = 3];(3, 0) →(3,0) (-3,3) (0,-3) [s = 3]; (0, 3) →(0,3) (3,-3) (-3,0) [s = 3]. �

7.2 Weyl group for b2. (a) Find the Weyl group for Lie algebra b2. (b) Givethe orbits of the dominant weights (0,0), (1,0), (0,1), (1,1), (2,0), (0,2).

SOLUTION 7.2 The Lie algebra b2 is defined by the Cartan matrix A (or itsinverse A−1)

A =

[

2 −1−2 2

]

, A−1 =

[

1 1/2

1 1

]

.

Its two simple roots are α1 = (2,−2) and α2 = (−1, 2) (or equivalently α1 =2ω1 − 2ω2 = ε1 − ε2 and α2 = −ω1 + 2ω2 = ε2). The system of positiveroots ∆+ consists of α1, α2, α3 = α1 + α2 = (1, 0), and α4 = α1 + 2α2 = (0, 2),the latter being the maximal root.

88 CHAPTER 7

(a) Let µ = (m1, m2) be an arbitrary weight, w1 = wα1 , and w2 = wα2

the Weyl reflections with respect to the simple roots, defined, as usual, bywi · (m1, m2) = (m1, m2)− miα

i (no summation intended). Explicitly

w1(m1, m2) = (−m1, 2m1 + m2)

w2(m1, m2) = (m1 + m2,−m2)

w1w2(m1, m2) = (−m1 − m2, 2m1 + m2)

w2w1(m1, m2) = (m1 + m2,−2m1 − m2)

w21(m1, m2) = (m1, m2) = w2

2(m1, m2).

Thus we have in general w21 = w2

2 = 1, w1w2 = −w2w1. We can pick w1

and w2 to generate the elements of the Weyl group W for b2: ±1, ±w1, ±w2,and ±w1w2. The element 1 has order 1; whereas −1,±w1,±w2 have order2; and ±w1w2 have order 4. Element w1 can be regarded as reflection inthe horizontal of the square (root diagram), and w1w2 four-fold rotation(through an angle of π/2) about the (vertical) z-axis. W is of order 8, and isisomorphic to the symmetry group D4 of the square.

(b) Weyl orbits and their sizes s:(0, 0) →(0,0) [s = 1];(1, 0) →(1,0) (-1,2) (1,-2) (-1,0) [s = 4](0, 1) →(0,1) (1,-1) (-1,1) (0,-1) [s = 4](1, 1) →(1,1) (-1,3) (2,-1) (-2,3) (2,-3) (1,-3) (-2,1) (-1,-1) [s = 8](2, 0) →(2,0) (-2,4) (2,-4) (-2,0) [s = 4](0, 2) →(0,2) (2,-2) (-2,2) (0,-2) [s = 4].The orbits of the dominant weights (1,0), (0,1), (2,0) and (0,2) are square,

while the orbit of (1,1) is octagonal. �

7.3 Properties of Weyl reflections. Given a Lie algebra g, prove the followingproperties of its Weyl reflections:

(a) The Weyl reflections are isometries on h∗, i.e transformations thatpreserve the scalar product.

(b) The reflection wα on the hyperplane perpendicular to a simple rootα permutes all other positive roots.

(c) Let δ = 1/2 ∑β∈∆+ β, i.e. half the sum of the positive roots, and w 6= 1be an arbitrary Weyl reflection. δ − w · δ is a sum of distinct positive roots.

(d) δ defined in (c) is the sum of the fundamental weights, and so is aweight.

SOLUTION 7.3 (a) For a Weyl reflection wα in any root α, direct applicationof the definition wαµ = µ − 2α〈µ, α〉/〈α, α〉 produces 〈wαµ, wαλ〉 = 〈µ, λ〉for arbitrary µ, λ ∈ h∗.


(b) Recall first that any root β has a unique integral linear expansion interms of the simple roots, β = ∑i kiα

i with all ki ≥ 0 (if β is positive) or allki ≤ 0 (if β is negative). Assume now that β � 0 (if β ≺ 0 consider instead−β), and let αi be a given simple root.

If β = αi, then wαi β = −β. So assume positive β 6= αi from now on,then

wαi β = (ki − β(hαi))αi + ∑j6=i

k jαj .

Since β 6= αi, there must be some k j > 0, with j 6= i; since wαi β is a root, allk j ≥ 0, and so wαi · β is a positive root.

(c) Consider first a reflection wαi in a simple root, then from the defini-tion of δ it follows wαi · δ = δ− αi, or δ− wαi · δ = αi. More generally, for anyw ∈ W different from the identity element, $ = δ − w · δ obeys w$ = −$and is given by

$ = 1/2 ∑β�0

(β − w · β) = 1/2 ∑β�0

(1 ± 1)β ,

and so δ − w · δ is the sum of the positive roots β that obey wβ = −β.(d) Applying the isometry property in h∗ on 〈δ, αi〉 with the reflection

wαi for any simple root αi, we have 〈wαiδ, wαiαi〉 = 〈δ, αi〉, or 2〈δ, αi〉 =〈αi, αi〉. But this implies δ(hαi) = 1 for any and therefore every αi. Thismeans δ = 1/2 ∑β∈∆+ β = ∑i ωi is half the sum of positive roots or, equiva-lently, the sum of the fundamental weights. It follows that δ is a (dominant)weight, written in the basis {ωi} as (1, 1, . . . , 1, 1). �

7.4 Dual representations. We know that the necessary and sufficient condi-tion for an irrep πλ with highest weight λ to be self-dual is that its lowestweight is −λ. Determine the self-duality of the irreps of a2, b2, and g2.

SOLUTION 7.4 a2: Irrep π10 has highest weight ω1 = (1, 0) and lowestweight −ω2 = (0,−1), whereas π01 has highest weight ω2 = (0, 1) andlowest weight −ω1 = (−1, 0). Irrep πnm has highest weight nω1 + mω2,that is n times highest weight of π10 plus m times highest weight of π01.Similarly its lowest weight is equal to n times lowest weight of π10 plus mtimes lowest weight of π01, that is n(−ω2) + m(−ω1) = −mω1 − nω2, or(−m,−n). For self-duality, (n, m) = −(−m− n), which is satisfied iff n =m. If wα3 is Weyl reflection wrt positive root α3, then wα3 · (n, m) = −(m, n).So wα3 = −1 when restricted to weight subspace (n, n).

b2: Irrep π10 has lowest weight −(1, 0), and so is self-dual (or, equiva-lently, self-conjugate). Irrep π01 has lowest weight −(0, 1), and so is self-dual. The lowest weight of irrep πnm is n(−1, 0) + m(0,−1) = −(n, m),and so πnm is self dual for every dominant weight. If we look at the Weylgroup W of b2, we see that there is an element w ∈ W such that w2 = −1.

90 CHAPTER 7

Calling wα1 and wα2 the Weyl reflections wrt the fundamental roots, thenw = wα1 wα2. Since w2(n, m) = −(n, m) for every dominant weight (n, m),every irrep of b2 is self-dual.

g2: The same reasoning applies. Irrep πnm has lowest weight −(n, m)and so is self-dual. Alternatively there is an element y ∈ W of g2 equal to−1, so that y(n, m) = −(n, m) for any non negative integers n, m, and soevery irrep of g2 is self-dual. That element y is (w1w2)3, where w1 = wα1

and w2 = wα2 are the Weyl reflections wrt the simple roots α1 and α2. �

7.5 Inner products of weights. Inner products of weights can be calculated inthe same way as for roots, since the two entities live on the same space, h∗0.We take sl(3, C) to illustrate, and set µ = (m1, m2) for any weight, wheremi = µ(hi) and hi are the fundamental coroots. (a) Calculate hµ defined by(hµ : h) = µ(h), where h is any vector of h0. (b) Calculate 〈µ, µ′〉 for anyweights µ, µ′ ∈ h∗0.

SOLUTION 7.5 Let ∆+ = {α1, α2, α3} be the set of positive roots of sl(3, C),in which α3 = α1 + α2, the sum of the two simple roots. Let hi be thefundamental coroots so that αi(hi) = 2 with i = 1, 2. Let h = ah1 + bh2

be any vector on h0. We know that α1(h) = 2a − b and α2(h)− a + 2b. Inaddition from (h : h′) = 2 ∑α∈∆+ α(h)α(h′), we deduce (h1 : h1) = (h2 :h2) = 12 and (h1 : h2) = −6, or more generally (h : h′) = 12(aa′ + bb′) −6(ab′ + ba′).

(a) From this expression for (h : h′) and the defining relation (hµ : h) =µ(h), we obtain a system of two linear equations which yield the desiredsolution: hµ = ch1 + dh2 = 1/18(2m1 + m2)h1 + 1/18(m1 + 2m2)h2.

(b) The inner product 〈µ, µ′〉 for any weights µ, µ′ ∈ h∗0 can be calculatedfrom 〈µ, µ′〉 = µ′(hµ) = cµ′(h1) + dµ′(h2) = cm′

1 + dm′2, with c, d found in

(a). Hence 〈µ, µ′〉 = 1/18(2m1m′1 + 2m2m′

2 + m1m′2 + m2m′

1). Note that theweight (m1, m2) has norm square 〈µ, µ〉 = 1/9(m2

1+ m2

2 + m1m2), which ismaximal for a highest weight (with both m1, m2 ≥ 0) or a lowest weight(with both m1, m2 ≤ 0). �

7.6 Fundamental weights. Find the expansion coefficients of ωi in {εj} forthe algebras b3, d4, g2, f4, e6, e7, and e8.

SOLUTION 7.6 For each algebra, let us define the ` × ` matrices A = {aij},B = A−1, C = {cij}, and D = CB = {dij}. Given that αi = ∑j ω jaji andαi = ∑i εjcji (where aij and cij are known), we must have ωi = ∑j εjdji,where D is to be found. In the following we list for each algebra C, B andD = CB.. b3:

C =

1 0 0−1 1 00 −1 1

B =

1 1 1/2

1 2 11 2 3/2

D =

1 1 1/2

0 1 1/2

0 0 1/2

.


The entries of each column of C give the expansion coefficients in α1 =ε1 − ε2, α2 = ε2 − ε3, α3 = ε3. The entries of each column of B give theexpansion coefficients in ω1 = α1 + α2 + α3, ω2 = α1 + 2(α2 + α3), ω3 =1/2α1 + α2 + 3/2α3. And finally, the entries of each column of D give theexpansion coefficients in ω1 = ε1, ω2 = ε1 + ε2, ω3 = 1/2(ε1 + ε2 + ε3).. d4:

C =

1 0 0 0−1 1 0 00 −1 1 10 0 −1 1

B =

1 1 1/21/2

1 2 1 11/2 1 1 1/2

1/2 1 1/2 1

D =

1 1 1/21/2

0 1 1/21/2

0 0 1/21/2

0 0 − 1/21/2

.

. g2:

C =

[

1 − 3/2

0√

3/2

]

, B =

[

2 31 2

]

, D =

[

1/2 0√3/2

√3

]

.

. f4:

C =

1/2 0 0 0− 1/2 0 0 1− 1/2 0 1 −1− 1/2 1 −1 0

, B =

2 3 4 22 6 8 42 4 6 31 2 3 2

D =

1 3/2 2 10 1/2 1 10 1/2 1 00 1/2 0 0

.

. e6:

C =

1/2 1 −1 0 0 0−1/2 1 1 −1 0 0−1/2 0 0 1 −1 0−1/2 0 0 0 1 −1−1/2 0 0 0 0 1√

3/2 0 0 0 0 0

, B =1

3

4 3 5 6 4 23 6 6 9 6 35 6 10 12 8 46 9 12 18 12 64 6 8 12 10 52 3 4 6 5 4

D =1

2

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

4/√

3

√3 5/

√3 2

√3 4/

√3

2/√

3

.

92 CHAPTER 7

. e7:

C = B =

1/2 1 −1 0 0 0 0−1/2 1 1 −1 0 0 0−1/2 0 0 1 −1 0 0−1/2 0 0 0 1 −1 0−1/2 0 0 0 0 1 −1−1/2 0 0 0 0 0 11/√

2 0 0 0 0 0 0

,

2 2 3 4 3 2 12 7/2 4 6 9/2 3 3/2

3 4 6 8 6 4 24 6 8 12 9 6 33 9/2 6 9 15/2 5 5/2

2 3 4 6 5 4 21 3/2 2 3 5/2 2 3/2

D =1

2

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

2√

2 2√

2 3√

2 4√

2 3√

2 2√

2√

2

.

. e8:

C = B =

1/2 1 −1 0 0 0 0 0−1/2 1 1 −1 0 0 0 0−1/2 0 0 1 −1 0 0 0−1/2 0 0 0 1 −1 0 0−1/2 0 0 0 0 1 −1 0−1/2 0 0 0 0 0 1 −1−1/2 0 0 0 0 0 0 1

1/2 0 0 0 0 0 0 0

,

4 5 7 10 8 6 4 25 8 10 15 12 9 6 37 10 14 20 16 12 8 4

10 15 20 30 24 18 12 68 12 16 24 20 15 10 56 9 12 18 15 12 8 44 6 8 12 10 8 6 32 3 4 6 5 4 3 2

D =1

2

0 1 −1 0 0 0 0 00 1 1 0 0 0 0 00 1 1 2 0 0 0 00 1 1 2 2 0 0 00 1 1 2 2 2 0 00 1 1 2 2 2 2 00 1 1 2 2 2 2 24 5 7 10 8 6 4 2

.

7.7 Representations of b2. Find the irreducible representations of b2 withhighest weights (0,1), (1,0),(0,2) and (2,0). Discuss the representations ofthe tensor products π0,1 ⊕ π0,1, π1,0 ⊕ π1,0, and π0,1 ⊕ π1,0.

SOLUTION 7.7 As in similar problems for a2 (Chapter 4), a solution maymake use of the notion of string of weights spaced by simple weights andthe formula q = p + µ(hi). As usual, call si the subalgebra {Hi, Ei , Fi}, withi = 1, 2, and the simple roots α1 = (2,−2) and α2 = (−1, 2).


(a) Representation of highest weight λ = (0, 1). We have successively thestrings of weights:

Sα2 · (0, 1) : (0, 1)(1,−1) (µ(h2) = 1: s2 doublet);Sα1 · (1,−1) : (1,−1)(−1, 1) (µ(h1) = 1: s1 doublet);Sα2 · (−1, 1) : (−1, 1)(0,−1) (s2 doublet).The process ends here because the last weight, (0,-1) has no positive

components. So π0,1 has dimension 4, a fact confirmed by the multiplicityand the W-orbital size of the only dominant weight (0,1). Note that π0,1 isself-conjugate because its lowest weight is the negative of its own highestweight; (0,−1) = −(0, 1).

The matrix elements of Fi in normalized kets can be calculated from theformula Fi|µ − kαi〉 = |µ − (k + 1)αi〉

√

(q − k)(k + 1), with 0 ≤ k ≤ q,obtained in GTAPP Chapter 7 Sec. 7.2. In our case, we have

F2|λ, (0, 1)〉 = |λ, (1,−1)〉, F1|λ, (1,−1)〉 = |λ, (−1, 1)〉 ,F2|λ, (−1, 1)〉 = |λ, (0,−1)〉.(b) Representation of highest weight λ = (1, 0).

Strings of weights:Sα1 · ( 1, 0) : (1, 0)(−1, 2) (s1 doublet);Sα2 · (−1, 2) : (−1, 2)(0, 0)(1,−2) (µ(h2)=2: s2 triplet);Sα1 · (1,−2) : (1,−2)(−1, 0) (s1 doublet);The process ends with the weight (−1, 0), which has no positive com-

ponent. The weight (0,0) can be reached from (1,0) only through one path,and so must have multiplicity 1. There are two dominant weights, (1,0) and(0,0), each of multiplicity one, and W-orbits of size 4 for one and 1 for theother. Hence the dimension of π1,0 is 5.

The matrix elements of Fi are given byF1|λ, (1, 0)〉 = |λ, (−1, 2)〉,F2|λ, (−1, 2)〉 =

√2|λ, (0, 0)〉, F2|λ, (0, 0)〉 =

√2|λ, (1,−2)〉,

F1|λ, (1,−2)〉 = |λ, (−1, 0)〉.(c) Representation of highest weight λ = (0, 2).

Sα2 · ( 0, 2) : (0, 2)(1, 0)(2,−2) (s2 triplet);Sα1 · ( 1, 0) : (1, 0)(−1, 2) (s1 doublet);Sα2 · (−1, 2) : (−1, 2)(0, 0)(1,−2) (s2 triplet);Sα1 · (2,−2) : (2,−2)(0, 0)(−2, 2) (s1 triplet);Sα1 · (1,−2) : (1,−2)(−1, 0) (s1 doublet);Sα2 · (−2, 2) : (−2, 2)(−1, 0)(0,−2) (s2 triplet).We have reached the weights (−1, 0) and (0,−2), which signal the end

of the process. There are three dominant weights (0,2), (1,0) and (0,0). Theweight (0,2) being the highest must be single, and so are the equivalents onits W-orbit. (1,0) can be reached from (0,2) by one path only, and so mustalso be single, as are its Weyl equivalents. The weight (0,0) has multiplicity

94 CHAPTER 7

2, because it can be reached via two independent paths. Hence, we havedim π0,2 = 4 × 1 + 4 × 1 + 1 × 2 = 10. It contains, apart from (0,0), theweights (2,−2) = α1, (−1, 2) = α2, (1, 0) = α1 + α2, (0, 2) = α1 + 2α2 andtheir negatives, i.e. all of ∆, and so it is the adjoint representation of b2. As(0,−2) = −(0, 2), it is self-conjugate.

The calculation of the matrix elements of Fi has only one difficulty inthe double multiplicity of (0,0) as a weight of πλ, but it is a problem wehave met in GTAPP Chapter 7 Sec. 7.2 with the adjoint representation of a2.We just follow the same approach. The results (for λ = (0, 2)) are:

F2|λ, (0, 2)〉 =√

2|λ, (1, 0)〉, F2|λ, (1, 0)〉 =√

2|λ, (2,−2)〉,F1|λ, (1, 0)〉 = |λ, (−1, 2)〉,F2|λ, (−1, 2)〉 =

√2|λ, (00)1〉,

F1|λ, (2,−2)〉 = |λ, (00)1〉+ |λ, (00)0〉,F1|λ, (00)1〉 = |λ, (−2, 2)〉, F1|λ, (00)0〉 = |λ, (−2, 2)〉,F2|λ, (00)1〉 =

√2|λ, (1,−2)〉,

F2|λ, (−2, 2)〉 =√

2|λ, (−1, 0)〉,F2|λ, (1,−2)〉 = |λ, (−1, 0)〉, F2|λ, (−1, 0)〉 = |λ, (0,−2)〉.It is useful here to have a graphical aid: One may keep track of the

progress of the calculation with, for example, a diagram on the weight lat-tice, as shown below for π0,2.

@@

@@

@@

@

@@

@@

@@

@

@@

@@

@@

@

−2, 2s

−1, 2s

−1, 0s

0, 2c

0,−2s

0, 0cf

1, 0c

1,−2s

2,−2sπ0,2[b2]

(d) Representation of highest weight λ = (2, 0). From explicit construc-tion, we see that π20 has four dominant weights: (2,0), (0,2) and (1,0) withmultiplicities equal to one and W-orbital sizes equal to four; and (0,0) withmultiplicity two and W-orbital size one. So it is a 14-dimensional represen-tation, as shown in the accompanying figure.


s

s

s

s s

s s

qd s

s s

s s

π2,0[b2]

(e) Tensor products π0,1 ⊕ π0,1, π1,0 ⊕ π1,0, and π0,1 ⊕ π1,0. Weights in theproduct representations are obtained as pairwise sums of weights of theproduct factors. The resulting weights are partitioned into the weights ofirreducible representations, and in this way a likely decomposition is sur-mised. Confirmation is made by the usual Cartan–Weyl construction withthe ladder operators Ei , Fi applied on the maximal dominant weight. Fromthe weights contained in the products, we infer the following decomposi-tions:

π0,1 ⊕ π0,1 = ∧2π0,1 ⊕ Sym2π0,1,

∧2π0,1∼= C ⊕ π0,1,

Sym2π0,1∼= π0,2;

π1,0 ⊕ π1,0 = ∧2π1,0 ⊕ Sym2π1,0,

∧2π1,0∼= π0,2

∼= Sym2π0,1,

Sym2π1,0∼= C ⊕ π2,0.

The Sym2π1,0 weight diagram is presented in the figure (see next page).As for π0,1 ⊕ π1,0, it contains (1, 1) as the maximal dominant weight, hencethe tensor product must be reducible with the 16-dimensional π1,1 in itsdecomposition, the remaining weights belong to π0,1, according to

π0,1 ⊕ π1,0∼= π0,1 ⊕ π1,1 .

7.8 Representations of g2. (a) Find the Weyl group for the Lie algebra g2.(b) Give the orbits of the dominant weights (1,0), (0,1), (1,1), (2,0), (0,2).

96 CHAPTER 7

s

s

s

s s

s s

qce s

s s

s s

Sym2π1,0[b2]

(c) Give the weights of the irreducible representation having (0,1) as thehighest weight.

SOLUTION 7.8 The Lie algebra g2 has the following characteristics:Σ: α1 = (2,−1), α2 = (−3, 2) in the basis of the fundamental weights

ω1 = 1/2ε1 +√

3/2ε2, ω2 =√

3ε2.∆+: α1, α2, α1 + α2, 2α1 + α2, 3α1 + α2, 3α1 + 2α2.(a) Call w1 = wα1 , w2 = wα2 :

w1(m1, m2) = (−m1, m1 + m2),

w2(m1, m2) = (m1 + 3m2,−m2).

Both w1 and w2 are of order 2, whereas w1w2 is of order 6. Now leta = w1, b = w2 and c = w1w2. Then the Weyl group for g2 has the elements:1, a, c, c2 = −ba, c3 = −1, c4 = −c, c5 = ba, ac = b, ac2 = bc, ac3 = −a,ac4 = −b, and ac5 = −bc. The order of the elements are as follows: order1:1; order 2: −1,±a,±b,±bc; order 6: ±c,±ba. The multiplication table,with columns and rows labeled in order 1, a, b, c, d = ba, f = bc, −1, −a,−b,−c, −d,− f , is divided into quarters (++, +−,−+,−−), each obtainedfrom any other by changes of signs. The ++ quarter is given by:

1 a b c d f

1 1 a b c d fa a 1 c b − f −db b d 1 f a cc c − f a −d 1 bd d b f 1 −c −af f −c d −a b 1

The action of the Weyl group on the weights (m1, m2) of g2 is shown inthe accompanying table (e.g. w1(m1, m2) = (−m1, m1 + m2); w1(m, 0) =(−m, m), etc.).


1 m1, m2 m, 0 0, m m, ma = w1 −m1, m1 + m2 −m, m 0, m −m, 2mb = w2 m1 + 3m2,−m2 m, 0 3m,−m 4m,−m

c = w1w2 −m1 − 3m2, m1 + 2m2 −m, m −3m, 2m −4m, 3md = w2w1 2m1 + 3m2,−m1 − m2 2m,−m 3m,−m 5m,−2m

f = w2w1w2 2m1 + 3m2,−m1 − 2m2 2m,−m 3m,−2m 5m,−3m

The table must be completed with the results of the action of the six nega-tive elements −1, −w1, −w2, −w1w2, −w2w1, and −w2w1w2.

(b) From the results of (a), the W-orbits of the given dominant weightsare as follows (µ

a→ µ′ means a · µ → µ′):

(1, 0)a→ (−1, 1)

b→ (2,−1)a→ (−2, 1)

b→ (1,−1)a→ (−1, 0)

(0, 1)b→ (3,−1)

a→ (−3, 2)b→ (3,−2)

a→ (−3, 1)b→ (0,−1)

(2, 0)a→ (−2, 2)

b→ (4,−2)a→ (−4, 2)

b→ (2,−2)a→ (−2, 0)

(0, 2)b→ (6,−2)

a→ (−6, 4)b→ (6,−4)

a→ (−6, 2)b→ (0,−2)

(1, 1)a→ (−1, 2)

b→ (5,−2)a→ (−5, 3)

b→ (4,−3)a→ (−4, 1)

b→ (−1,−1)

(1, 1)b→ (4,−1)

a→ (−4, 3)b→ (5,−3)

a→ (−5, 2)b→ (1,−2)

a→ (−1,−1).The W-orbital size is 12 for (1, 1), and 6 for the others.

(c) Representation π0,1: Starting from weight (0,1), we have the strings:Sα2 : (0, 1), (3,−1); s2 doublet;Sα1 : (3,−1)), (1, 0), (−1, 1), (−3, 2); s1 quadruplet;Sα2 : (−1, 1), (2,−1); s2 doublet;Sα1 : (2,−1), (0, 0), (−2, 1); s1 triplet;Sα2 : (−3, 2), (0, 0), (3,−2); s2 triplet;Sα1 : (3,−2), (1,−1), (−1, 0), (−3, 1); s1 quadruplet;Sα2 : (−3, 1), (0,−1); s2 doublet.The dominant weights (0, 1) and (1, 0) have multiplicity 1 and W-orbits

of size 6; whereas (0, 0) as a weight of π0,1 has multipliciy two and orbitalsize equal to one. Hence the dimension of π0,1 is 14. All non-zero weightsare also the roots of the algebra, and so π0,1 is the adjoint representation ofg2, illustrated in the accompanying diagram (see next page).

7.9 Find the integral expansion ∑i kiαi (where αi are simple roots) for the

positive roots of the classical algebras a`, b`, c`, and d`. (An error in GTAPPVer. 2014-12 has been corrected: positive weights → positive roots.)

SOLUTION 7.9. a` = sl(` + 1, C)

Σ: αi ≡ εi − εi+1 (i = 1, . . . , `)∆+ (`(` + 1)/2):

98 CHAPTER 7

s

s s s s

s sf s

s s s s

s

π0,1[g2]

αij ≡ εi − εj (1 ≤ i < j ≤ ` + 1).=⇒ αij = αi + αi+1 + · · ·+ αj−1 (1 ≤ i < j ≤ ` + 1).

. b` = so(2` + 1, C)Σ: αi = εi − εi+1, (i = 1, . . . , ` − 1), α` = ε`.∆+ (`2):α

ij− ≡ εi − εj; α

ij+ ≡ εi + εj (1 ≤ i < j ≤ `); αi

0 ≡ εi (1 ≤ i ≤ `).

=⇒ αij− = αi + αi+1 + · · ·+ αj−1 (1 ≤ i < j ≤ `).

αij+ = αi + · · ·+ αj−1 + 2αj + · · ·+ 2α` (1 ≤ i < j ≤ `).

αi0 = αi + αi+1 + · · ·+ α` (i = 1, . . . , ` − 1), α`

0 = α`.

. c` = sp(`, C)Σ: αi ≡ εi − εi+1, (i = 1, . . . , ` − 1), α` ≡ 2ε`.∆+ (`2):α


ij+ ≡ εi + εj (1 ≤ i < j ≤ `); αi

0 ≡ 2εi (1 ≤ i ≤ `).

=⇒ αij− = αi + αi+1 + · · ·+ αj−1 (1 ≤ i < j ≤ `).

αij+ = αi + · · ·+ αj−1 + 2αj + · · ·+ 2α`−1 + α` (1 ≤ i < j ≤ ` − 1).

αi`+ = αi + · · ·+ α`−1 + α` (1 ≤ i ≤ ` − 1).

αi0 = 2αi + · · ·+ 2α`−1 + α` (i = 1, . . . , `− 1), α`

0 = α`.. d` = so(2`, C)

Σ: αi ≡ εi − εi+1, (i = 1, . . . , ` − 1), α` ≡ ε`−1 + ε`.∆+ (`(`− 1)):α


ij+ ≡ εi + εj (1 ≤ i < j ≤ `);

=⇒ αij− = αi + αi+1 + · · ·+ αj−1 (1 ≤ i < j ≤ `).

αij+ = αi + · · ·+ αj−1 + 2αj + · · ·+ 2α`−2 + α`−1 + α` (1 ≤ i < j ≤ `− 2);

αi,`−1+ = αi + · · ·+ α`−1 + α` (i = 1, . . . , ` − 2);

αi,`+ = αi + · · ·+ α`−2 + α` (i = 1, . . . , ` − 2);

α`−1,`+ = α`. �


7.10 Dimension of the irreps of a`. Find the dimension of the irreps of a` forarbitrary `.

SOLUTION 7.10 All positive roots of a` are of the form αij = εi − εj, with1 ≤ i < j ≤ ` + 1. As the simple roots are defined to be αi = αi,i+1 , withi = 1, . . . , `, all the positive roots may be written

αi,j = αi + αi+1 + · · ·+ αj−1, 1 ≤ i < j ≤ ` + 1.

The coefficients kαs in the expansion α = αi,j = ∑s kα

s αs (in terms of simpleroots αs) are given by ks = 1 if i ≤ s ≤ j − 1, and ks = 0 otherwise. So thatthe dimension formula for an irrep πλ of a` takes the form

dim πλ[a`] =`

∏i=1

`+1

∏j=i+1

∑s kijs (ms + 1)

∑s kijs

=`

∏i=1

`+1

∏j=i+1

∑j−1

s=i (ms + 1)

j − i.

It is the product of factors listed below, with the corresponding (ij) for eachfactor shown in the first column:

(12)(23) . . . (`, ` + 1) : m1+11

m2+11

· · · m`+11

(13)(24) . . . (`− 1, ` + 1) : m1+m2+22

m2+m3+22

· · · m`−1+m`+2

2

· · · : · · · · · · · · ·· · · : · · · · · · · · ·

(1, `)(2, `+ 1) :m1+m2+···+m`−1 +`−1

`−1

m2+m3+···+m`+`−1

`−1

(1, ` + 1) : m1+m2+···+m`+`

`

We can check that this general formula reproduces the values for ` =1, 2 given in GTAPP Sec. 7.3. Let us take two further examples:

In sl(4, C), irrep πm1,m2 ,m3has the dimension given by

112

(m1 + 1)(m2 + 1)(m3 + 1)(m1 + m2 + 2)(m2 + m3 + 2)(m1 + m2 + m3 + 3),

whereas in sl(5, C), irrep πm1,m2 ,m3 ,m4has the dimension

1288

(m1 + 1)(m2 + 1)(m3 + 1)(m4 + 1)(m1 + m2 + 2)(m2 + m3 + 2)(m3 + m4 + 2)

× (m1 + m2 + m3 + 3)(m2 + m3 + m4 + 3)(m1 + m2 + m3 + m4 + 4).

In particular, dimπ111 = 26 and dimπ1111 = 210, of the general form 2p

where p = `(` + 1)/ is the number of positive roots of the algebra a`. �

Index

a1, sl(2, C), 48–70a2, sl(3, C), 71–79

representation, 73–79, 87a`, sl(` + 1, C), 97, 99A4, 6

character table, 24abelian group, 2angular momentum, 57

b2, so(5, C), 79, 87representation, 92–96

b` , so(2` + 1, C), 98baryon, 77block diagonalization, 12

c` , sp(`, C), 98C2 × C2, 2C2 × C2 × C2, 2C3, 14C4 × C2, 5C6, 3C8, 5Cartan matrix, 80Casimir operator, 49, 72Cayley theorem, 2, 7CH, Campbell–Hausdorff, 33, 40, 42character of representation, 17Clebsch–Gordan coefficient, 68

d4, so(8, C), 79, 81d`, so(2`, C), 98D2, 4D2 × C2, 5D3, 3, 4

multiplication table, 14representation, 16, 21

D4, 5, 6character table, 24conjugacy class, 6

multiplication table, 6normal subgroup, 7

Dn, 8dimension of

orthogonal group, 31irreducible representation, 99symplectic group, 32unitary group, 31

dual (of representation), 76, 89Dynkin diagram, 81

e6–e8, 86Euclidean group, 42–48exponential map, 33–36, 38, 44, 58

f4, 90

g2, 82–86, 88, 96Gell-Mann matrices, 71group algebra, 18

Heisenberg group, 40hypercharge, 71

inner product, 15, 90

Lagrange theorem, 2, 3level (root), 79Lie bracket, 33, 49Lorentz group, 39

matrix representation, 27maximal root, 86meson, 77

nilpotent matrix, 34

O(1;1), SO(1;1), 31O(2), SO(2), 27, 29O(3), SO(3), 59

100

INDEX 101

Pauli matrices, 9permutation, 3positivity of norm, 15

in classical algebras, 85in exceptional algebras, 86

product of representations, 27

Q8, 4, 5, 9multiplication table, 9normal subgroup, 9

quaternion (Lie) group, 50

representation, 11adjoint, 48dual, 76, 77, 89fundamental, 77irreducible, 17, 73regular, 13

root, 71, 74, 79rotation, 1, 27, 38, 51–62

S2, S3, S4, 5character table, 26permutational representation, 26

Schur Lemma, 18, 19Schwinger model, 63

self duality, 89SL(1, Q), 51SL(n, R), 34SO(2;1), 37structure constant, 73SU(2), 29, 51subgroup, 2

normal, 7, 9symmetric power of representation, 67

T, see A4

tensor product of representation, 67, 77

U(1), 29U(2), 29, 51

V, 2, 4

weight, 49, 73diagram, 67fundamental, 90

Weyl group fora2, 87b2, 87g2, 88, 96

102 INDEX

c© Q. Ho-Kim (Hồ Kim Quang). Group Theory: A Problem Book Ver. 2015-12.Cover illustration from image by Jo Edkins, used with permission.This book contains suggested solutions to the end-of-chapter problems inGroup Theory: A Physicist’s Primer by the same author, available athttp://www.scribd.com/doc/207786199/Group-Theory-A-Physicist-s-Primer.

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Q. HO-KIM--Group Theory: A Problem Book