175
Group Theory Part I, Discrete Groups Michiel Snoek
| Date post: | 25-Oct-2015 |
| Category: |
Documents |
| Upload: | lunarcausticac |
| View: | 48 times |
| Download: | 6 times |
Group Theory
Part I, Discrete Groups
Michiel Snoek
Group Theory
Part I, Discrete Groups
Authors: Ben Bakker and Daniel BoerDepartment of Physics and Astronomy
Vrije Universiteit, Amsterdam
Ninth Edition, 2010
Cover illustration: M.C. Escher, Cirkellimiet III, 1959Weblocatie: www.worldofescher.com
i
Contents
1 Introduction 1
2 Structure of Groups 32.1 Abstract Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.2 Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.3 Conjugacy Classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.4 Invariant Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.5 Cosets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.6 Factor Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.7 Direct and Semi-Direct Products of Groups . . . . . . . . . . . . . . . . . 102.8 Examples of some common groups . . . . . . . . . . . . . . . . . . . . . . . 122.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
3 Representations 153.1 Homomorphic and Isomorphic Mappings . . . . . . . . . . . . . . . . . . . 153.2 Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.3 Reducible and Irreducible Representations . . . . . . . . . . . . . . . . . . 193.4 Characters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.5 The Group Algebra and the Regular Representation . . . . . . . . . . . . 243.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
4 Representations: Function Spaces and Operators 294.1 Projection Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294.2 Tensor Products and the Clebsch-Gordan Series . . . . . . . . . . . . . . . 314.3 Irreducible Tensorial Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . 334.4 The Wigner-Eckart Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 374.5 Representations of Direct Product Groups . . . . . . . . . . . . . . . . . . 404.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
5 Symmetries of Molecules and Solids 435.1 O(3) and SO(3) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435.2 The Euclidian Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445.3 Symmetries of Molecules . . . . . . . . . . . . . . . . . . . . . . . . . . . . 455.4 Example: Vibrations of Molecules . . . . . . . . . . . . . . . . . . . . . . 455.5 Translation Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 565.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
ii
6 The Symmetric Groups Sn 606.1 The Structure of Sn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 606.2 The role of Sn in physics . . . . . . . . . . . . . . . . . . . . . . . . . . . 626.3 The irreps of Sn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
6.3.1 Partitions of n and frames . . . . . . . . . . . . . . . . . . . . . . . 636.3.2 Young Tableaux and Young operators . . . . . . . . . . . . . . . . 646.3.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
6.4 Products of wave functions . . . . . . . . . . . . . . . . . . . . . . . . . . 696.4.1 Outer product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 696.4.2 Inner product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
6.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
A Mathematical Preliminaries 73A.1 Set Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
A.1.1 Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73A.1.2 Mappings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73A.1.3 Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
A.2 Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74A.3 Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
B Bibliography 77
iii
Chapter 1
Introduction
Mathematicians and physicists alike have been interested and sometimes fascinated bythe symmetries in nature, for example the symmetries of forms in nature. In Euclidiangeometry symmetry is invariance under transformations of solid figures, e.g., translations,reflections and rotations.
In a more abstract sense: symmetry is an invariance under a set of operations. Anexample is the symmetry of the coefficients of polynomials as functions of their zeros.Take for instance the quadratic form
x2 + bx+ c = (x− x1)(x− x2) (1.1)
with zeros x1 and x2. Then the coefficients b and c are symmetric polynomials of x1 andx2:
b = −(x1 + x2), c = x1x2. (1.2)
It is easy to convince oneself that this simple pattern can be generalized to polynomialsof any degree normalized such that the term of highest degree has coefficient unity: thecoefficient of the term of degree k in a polynomial of degree n is a symmetric polynomialof order n− k in its zeros.
Algebraic considerations of this sort have led to the idea of a group of substitutions,nowadays better known as permutations. Later on this was generalized to other trans-formations. The early history of group theory is associated with the names of Gauss,Lagrange, Abel, Galois, and Cauchy, to mention the greatest masters of that period.
As early as 1849 Cayley proposed the idea of an abstract group, i.e., an algebraic
structure, independent of the meaning of its elements. Five years later he proved theimportant theorem that any finite group can be understood as a group of permutations.At the time these ideas were not fully appreciated. So it happened that Dedekind couldbecome the effective founder of abstract group theory. Based on the idea of groups ofpermutations he also gave a definition of an abstract group. When Cayley wrote in 1878again on group theory, these papers had more influence.
All these considerations were mostly connected to discrete groups. Lie took up thenotion of continuous transformation groups in the context of the solution of differentialequations. These groups are now named after him and play an enormously important rolein physics.
At the time, the nineteenth century, such considerations were foreign to physicists.Nowadays, however, group theory plays a very important role in as widely different partsof physics as crystallography and particle physics.
1
In physics we are interested in the invariances of the laws of nature under a set oftransformations. For instance
space-time invariancesGalileo relativity ↔ conservation of energy and momentumEinstein relativity ↔ conservation of energy-momentum
internal symmetriesgauge invariance ↔ charge conservation in QED and QCDisospin symmetry ↔ charge symmetry of nuclear forcesPauli principle ↔ permutation symmetry
discrete symmetriesP ≡ space inversionC ≡ charge conjugationT ≡ time reversal
In addition physicists want to understand how the different forms in nature, like crystals,can be explained using the known simple laws of physics.
It is useful to give here a rough classification of groups.
Discrete groups
Discrete groups are sets of elements that can be counted. The number of elements maybe finite or infinite. In the latter case the elements are denumerable.
finite permutation groups Sn, rotations of moleculesinfinite translations in crystals
Continuous groups
The elements of continuous groups depend continuously on some parameters. Examplesare:
O(3), rotations and reflections in three dimensionsSU(n), symmetries of quark systemsLorentz group, Poincare group, Galilei group
In the lectures we shall be concerned with those topics that are most interesting tophysicists. These are not necessarily the simplest notions. In order to define the conceptsof group theory, we shall begin by discussing finite groups. There we can explain in arelatively simple way the most important concepts and illustrate them easily. Next weshall discuss more advanced subjects.
2
Chapter 2
Structure of Groups
The power of abstract algebra is its ability to see general structures in concrete casesand generalize them. In this chapter we shall discuss the most important concepts ofabstract group theory, necessary to understand the theorems that are important for ananalysis of concrete examples. It is of course our aim to apply group theory to situationsrelevant to physics. Many results found in a physical context can be understood usingelementary methods only, but group theory can unify the different results and put them ina proper perspective. So we shall concentrate on the abstract concepts here, occasionallyillustrating them with simple, concrete examples.
2.1 Abstract Groups
Definition group
A group G is the pair (G; )1 where G is a set of elements and is a composition law,such that
(i) T1, T2 ∈ G⇒ T1 T2 ∈ G(ii) T1, T2, T3 ∈ G⇒ T1 (T2 T3) = (T1 T2) T3(iii) ∃E ∈ G ∀T ∈ G : E T = T E = T(iv) ∀T ∈ G ∃T−1 ∈ G : T T−1 = T−1 T = E
Property (i) says that the group is closed under the composition law. Property (ii) is theassociative law. The element E is the neutral element or the identity. T−1 is the elementinverse to T .
Definition order
The number of elements of a group is called the order of the group. If the order of a groupis finite, the group is said to be finite, else infinite.
Example 1G = (Z; +) This is the group of the integers with addition as the composition law.
Example 2G = (R\0;×) This is the group of real numbers except 0 with multiplication as thecomposition law.
1For simplicity we use the same symbol for the set of elements G and the pair (G; ).
3
Example 3Galilei group in 1+1 dimension. Consider the transformation
x′ = x+ vt+ a
t′ = t+ b (2.1)
This transformation can be written in matrix form
x′
t′
1
= T (a, b, v)
xt1
, T (a, b, v) ≡
1 v a0 1 b0 0 1
. (2.2)
The composition law is matrix multiplication. It gives
T (a1, b1, v1) T (a2, b2, v2) = T (a1 + a2 + b2v1, b1 + b2, v1 + v2), (2.3)
the inverse is given byT−1(a, b, v) = T (−a+ bv,−b,−v). (2.4)
Definition Abelian
A group G is said to be Abelian if the composition law is commutative:T1 T2 = T2 T1, ∀T1, T2 ∈ G.
Definition order, period
Let n be the smallest integer such that Xn = E for an element X of the group G withidentity E, then n is said to be the order of X in G.The set of elements X,X2, . . . , Xn = E is called the period of X .
Remark
The period of an element of G is a group. It is a cyclic group, which means that anyelement can be obtained by taking a suitable power of a specific element different fromthe identity.
Clearly, there may be parts of a group that are themselves groups with the samecomposition law as the original group. Such groups are called subgroups and they are thesubject of the next section.
2.2 Subgroups
Definition subgroup, proper subgroup
Let (G; ) be a group. If H ⊂ G and (H ; ) is also a group, then H is said to be asubgroup of G. If H 6= G and H 6= E, then the subgroup H is said to be a propersubgroup of G.
Example 4Let n be an integer, then we denote by nZ the set of all multiples of n. Then (nZ; +) isa subgroup of (Z; +).
Example 5For any group G with identity E, (E; ) is a subgroup of G.
4
Theorem 1Let T0 ∈ G. Consider the set GT0 = TT0|T ∈ G. Then GT0 = G.Proof
(i) G ⊂ GT0. Take T ∈ G then TT−10 ∈ G, so T = TT−1
0 T0 ∈ GT0.(ii) GT0 ⊂ G. This is evident from the definition of a group. As G ⊂ GT0 and
GT0 ⊂ G, GT0 = G.2
Remark
If T1T0 = T2T0 then T1 = T2 ∀T0, T1, T2 ∈ G.
Example 6Let p be an integer and Z+p the set of all integers of the form n+p, n ∈ Z, then Z+p = Z.
Example 7Let z1, . . . , zn be the n distinct roots of the equation zn = 1, so zm = exp(2mπi/n).Then G = (z1, . . . , zn;×) is a group. Let zmG be the set zmz1, . . . , zmzn, then(zmG;×) = G. G has the same structure as the group of the integers under additionmodulo n, which is usually denoted by Zn. G forms a finite subgroup of (C\0;×).
2.3 Conjugacy Classes
If we consider different symmetry transformations, we sense intuitively that some of themare related. For instance, reflections share the property that they change the ”handedness”of a motion or a geometric figure from left- to right-handed and vice versa, whereasrotations do not. This intuition can be made into a precise notion, that of conjugacy ofoperations. This leads to the definition of classes.
Definition conjugate
Two elements T1, T2 ∈ G are called conjugate if there exists an element X ∈ G such thatT1 = XT2X
−1.
Remark
The property of being conjugate is an equivalence relation.
Definition class
A class K of a group G is a subset of G such that all elements of K are conjugate to eachother and no elements of G \K are conjugate to any element of K.
To find a class of G, start with any element and find all the elements conjugate to it.
Theorem 2Let G be a group, Ki the set of its classes. Then the following properties hold
(i) Any element of G belongs to exactly one class.(ii) E is a class.(iii) If G is Abelian than all classes consist of single elements of G.
The proof of this theorem is left as an exercise.
Example 8 The group S3
Consider the equilateral triangle ABC. Its vertices have coordinates a(1, 0, 0), b(0, 1, 0)and c(0, 0, 1). As a plane triangle it has the following symmetries
5
Aa
A
Bb
Cc
S
R
Cb
Aa Bc
Cb
Ac Ba
Figure 2.1: Symmetries of the plane equilateral triangle.
(i) rotations R over an angle of 120 about an axis through its center of gravity andperpendicular to the plane of the triangle.
(ii) reflections in a bisectrix of any of the vertices, denoted by SA, SB, SC .
We take the passive point of view, i.e., the triangle is kept fixed, the frame of reference istransformed. Denote the original triangle as (Aa,Bb, Cc), then the effect of the operationsSA and R is depicted in Fig. 2.1. We find E = R3, S2
α = E, (α = A,B,C). This can besummarized in the following group table:
R R2 SA SB SC
R R2 E SC SA SB
R2 E R SB SC SA
SA SB SC E R R2
SB SC SA R2 E RSC SA SB R R2 E
, where ST TS
In order to analyze the structure of S3 we determine its proper subgroups and conju-gacy classes.
Proper subgroupsR = E,R,R2,SA = E, SA, SB = E, SB, SC = E, SC. Although S3 is not Abelian, these
subgroups are.
We clearly see that for any reflection one has S−1α = Sα and for the rotations R−1 = R2.
ClassesThe classes are [E] = E, [R] = R,R2 and [S] = SA, SB, Sc.Remarks
(i) Any transformation T ∈ S3 permutes the coordinates of the vertices. So we canconsider S3 as the group of permutations of three objects as well. It must have 3!=6elements.
6
R R2
CES
S
SA
B
E
SB
2
SE E
A SC
E
RR
Figure 2.2: The group S3 and its proper subgroups.
E R R2
CS
S
SA
B
Figure 2.3: Classes of the group S3.
(ii) Using the coordinates a, b and c we can map the elements of S3 on 3× 3 matrices.They are
SA =
1 0 00 0 10 1 0
, SB =
0 0 10 1 01 0 0
, SC =
0 1 01 0 00 0 1
,
R =
0 0 11 0 00 1 0
, R2 = R−1 = RT =
0 1 00 0 11 0 0
. (2.5)
(iii) We can also use an embedding of the triangle ABC in R2. The coordinates of A,
B and C in the coordinate frame (ex, ey) are A: (−1
2
√3,−1
2) ≡ a, B: ( 1
2
√3,−1
2) ≡ b, C:
(0, 1) ≡ c. A rotation about the origin over an angle φ takes ex and ey to
Rφex ≡ e′x = cosφ ex + sinφ ey,
Rφey ≡ e′y = − sinφ ex + cosφ ey. (2.6)
The rotations of the basis vectors are given by Eq. (2.6). The components of vectorstransform according to the inverse transformations. We find thus for the matrices
R = R−1120 =
[
−1
2−1
2
√3
1
2
√3 −1
2
]
, R2 = R−1240 = R−1 =
[
−1
2
1
2
√3
−1
2
√3 −1
2
]
. (2.7)
A reflection in the y-axis, SC , takes ex to −ex, but ey is not changed. The correspondingcoordinate transformation matrix is
SC =
[
−1 00 1
]
. (2.8)
7
The other reflections, SA and SB, are easily deduced if one rotates by 30, performs areflection in the new x-axis and rotates back (for SA) or rotates first over −30, performsthe reflection and rotates back again (SB). Then one finds for the transformation matrices
SA =
[
1
2
1
2
√3
1
2
√3 −1
2
]
, SB =
[
1
2−1
2
√3
−1
2
√3 −1
2
]
. (2.9)
Clearly, one must have SAa = a etc. and SAb = c, SAc = b etc. These relations are easyto check.
(iv) If we consider a group G and a subgroup H , then the classes of H are definedrelative to H . This means that they do not necessarily coincide with the intersections ofthe classes of G with H .
2.4 Invariant Subgroups
Definition invariant or normal subgroup
Consider a group G and one of its subgroups H . H is said to be an invariant or normalsubgroup of G if for all S ∈ H and X ∈ G: XSX−1 ∈ H .
Remark
An invariant subgroup H in G consists of complete classes of G.
Example 9Consider again the group S3. The following subgroups are invariant: E and E,R,R2 ≡R. R = [E] ∪ [R].
Definition simple group
A group G is said to be simple if it has no proper invariant subgroups.
2.5 Cosets
If a group G has a proper subgroup H , it can be split into pieces, called cosets. The setof cosets is again a group if H is invariant and a proper composition law is defined. Thisgroup is called the factor group.
Definition left, right coset
Consider a group G with subgroup H . Let T be a fixed element of G. The set ST |S ∈H ⊂ G is called a right coset of H in G, denoted as HT . Similarly, one defines the leftcoset TH = TS|S ∈ H.Remark
If T ∈ H then TH = HT = H .
Example 10Group G = S3
Subgroup H1 = R.Cosets H1SA = H1SB = H1SC = SAH1 = . . . = [S].
Subgroup H2 = E, SA.Cosets H2SA = SAH2 = H2, H2SB = R, SB, H2SC = R2, SC, H2R = R, SB,
8
H2R2 = R2, SC; SBH2 = R2, SB = R2H2, SCH2 = R, SC = RH2.
Remarks
(i) In general the cosets are not subgroups.
(ii) If T ′ ∈ HT then HT ′ = HT .
(iii) Lagrange’s theorem Every element of G belongs to exactly one right and one leftcoset of H . This means that two (left/right) cosets K1 and K2 either coincide orare disjoint.
(iv) If G is a finite group of order g, H a subgroup of order h, then the number of leftand right cosets is g/h. This number is called the index of H in G. Corollary: theorder of any subgroup of a finite group G is a divisor of the order of G.
Theorem 3If H is an invariant subgroup of the group G, then the right and left cosets of H coincide.Proof
H is invariant, so ∀S ∈ H, ∀T ∈ G : TST−1 ∈ H . Consider the coset TH = TS|S ∈H. Let S1 = TST−1 ∈ H . Then it follows that TS = S1T ∈ HT . So, any element ofTH belongs also to HT . In the same way one proves that any element of HT belongs toTH .2
2.6 Factor Groups
Definition product of cosets
Let G be a group and H an invariant subgroup of G. The product ∗ of right cosets isdefined as: HT1 ∗HT2 = H(T1 T2), where T1 and T2 are elements of G.
Remark
If T ′1 ∈ HT1 and T ′
2 ∈ HT2 then HT1 ∗HT2 = HT ′1 ∗HT ′
2. We prove this as follows.
HT ′1 ∗HT ′
2 = H(T ′1 T ′
2) = S T ′1 T ′
2|S ∈ H
If T ′i ∈ HTi then there exists Si ∈ H such that T ′
i = Si Ti (i = 1, 2). This implies
S T ′1 T ′
2 = S S1 T1 S2 T2.
As H is invariant, i.e., T1H = HT1, there is an element S ′2 in H such that T1S2 = S ′
2T1.Hence
S T ′1 T ′
2 = (S S1 S ′2) (T1 T2) ∈ H(T1 T2).
So, H(T ′1 T ′
2) ⊂ H(T1 T2). In the same way one can prove that H(T1 T2) is a subsetof H(T ′
1 T ′2) which completes the proof.
Theorem 4The set of all right(left) cosets of an invariant subgroup H of the group G form a groupwith the composition law ∗.Proof
9
(i) Because for any T1, T2 ∈ G the product T1 T2 also belongs to G, the productHT1 ∗HT2 = H(T1 T2) is a right coset of H .
(ii) Associativity of the composition law.
(HT1 ∗HT2) ∗HT3 = H(T1 T2) ∗HT3= H((T1 T2) T3) = H(T1 (T2 T3))= HT1 ∗H(T2 T3)= HT1 ∗ (HT2 ∗HT3)
(iii) The coset HE = H is the identity, because H ∗HT = H(E T ) = HT .(iv) (HT )−1 = HT−1, for (HT ) ∗ (HT−1) = H(T T−1) = H
2
Definition factor group or quotient group
The set of right cosets HT of the invariant subgroup H of the group G with the compo-sition law ∗ is called the factor group and denoted by G/H .
Example 11Take G = S3, H = R. Then G/H = S3/R = (RE,RSA; ∗). The structure ofS3/R is the same as the structure of (1,−1;×) or Z2 = (0, 1; +(mod2)) and also ofS2 = (E, SA; ).
2.7 Direct and Semi-Direct Products of Groups
The previous section was concerned with the idea to combine sets of elements of a groupinto a coset and treat this as one entity. It may be considered as a movement to simpli-fication. In the present section we combine two groups into one, a movement to richerstructure.
Although we shall discuss representations later, we mention here that direct productgroups and semi-direct product groups share the property that their irreducible represen-tations (see Chapter 3) are intimately related to the irreducible representations of the twofactors producing the product group (see Cornwell 1984).
Consider two groups G1 and G2. Let T1 and T ′1 belong to G1 and T2 and T ′
2 to G2.Then one defines the product ∗ of ordered pairs (T1, T2) and (T ′
1, T′2) as follows
(T1, T2) ∗ (T ′1, T
′2) = (T1 T ′
1, T2 T ′2). (2.10)
The set (T1, T2)|T1 ∈ G1, T2 ∈ G2 with the composition law ∗ forms a group. Theidentity of this group is (E1, E2) if E1 is the identity in G1 and E2 in G2.
Definition direct product
The direct product G1 ⊗G2 of two groups G1 and G2 is (T1, T2)|T1 ∈ G1, T2 ∈ G2 withthe composition law ∗ in Eq. (2.10).
Definition direct-product group
A group G is a direct-product group if it is isomorphic to a direct product of groups.(We shall give a formal definition of “isomorphic” later. Here we say that two groups areisomorphic if there exists a 1-1 mapping of one onto the other that preserves the groupstructure.)
10
It is clear from this definition that any direct product of groups is a direct-productgroup, but beware: the converse is not true, see the next example.
Example 12The group O(3) is defined as the set of all orthogonal 3×3 matrices with the usual matrixmultiplication as the composition law. This group has the same structure as the directproduct SO(3)⊗Z2, where SO(3) is the subgroup of O(3) of all orthogonal 3×3 matriceswith determinant equal to 1. The relation is for instance established via RE ↔ (R, 1),RP ↔ (R,−1) for all R ∈ SO(3) and for E = diag(1, 1, 1) and P = diag(−1,−1,−1).
Remark
The direct product group G1 ⊗G2 has an invariant subgroup (G1, E2) that is isomorphicto G1 and one that is isomorphic to G2, (E1, G2). This is what is meant when below wediscuss for instance the subgroup G1 of G1 ⊗G2.
The following characterization holds true: G is a direct-product group if there existtwo subgroups G1 and G2 of G such that
(i) G1 ∩G2 = E, E is the identity of G(ii) all elements of G1 commute with all elements of G2
(iii) ∀T ∈ G, ∃T1 ∈ G1 ∧ ∃T2 ∈ G2 : T = T1 T2The structure of G is in this case the same as the structure of G1 ⊗G2.
Definition semi-direct product
A group G is called a semi-direct product of the two subgroups G1 and G2, denoted asG = G1 ⊙G2, if the following properties hold
(i) G1 is an invariant subgroup while G2 is any subgroup of G(ii) G1 ∩G2 = E, E is the identity of G(iii) ∀T ∈ G, ∃T1 ∈ G1 ∧ ∃T2 ∈ G2 : T = T1 T2
The differences between G = G1 ⊙G2 and G1 ⊗G2 are(a) G1 is required to be an invariant subgroup of G.(b) In G1 ⊙G2 commutativity of T1 ∈ G1 and T2 ∈ G2 is not required.
Remark
In physics one encounters several examples that seem to contradict the definitions givenabove by containing the direct product of a group with itself. For example, consider thesymmetry SU(N)R⊗SU(N)L. In this case the two SU(N) groups are the same, but theyact on right-handed and left-handed particles respectively. The copies of the same groupact on different objects and hence commute even if the group itself is not Abelian.
Example 13S3 can be considered as a semi-direct product: S3 = R⊙S2. Note that the direct productof the two Abelian groups R and S2 is Abelian, as opposed to S3.
Example 14 The Poincare group (inhomogeneous Lorentz group)If xµ is a four-vector (coordinate vector in Minkowski space) then an element of thePoincare group P is a pair (Λ|t), where Λ is represented in Minkowski space by a real4 × 4 matrix, with matrix elements Lµ
ν , that leaves the metric g = diag(1,−1,−1,−1)invariant,
LTgL = g ↔ (LT)µα(g)αβ(L)βν = (g)µν ,
with (LT)µα = Lαµ, (L)βν = Lβ
ν ,
11
↔ gαβLαµL
βν = gµν , (2.11)
and tµ is an arbitrary four-vector. The action of an element of P on an arbitrary fourvector is (Λ|t)x = Λx+ t, in components ((Λ|t)x)µ = Lµ
νxν + tµ.
The composition law of elements of the Poincare group is
(Λ1|t1)(Λ2|t2) = (Λ1Λ2|Λ1t2 + t1). (2.12)
The Lorentz group L = Λ can be identified with the subgroup of elements (Λ|0) of P .The translations form an invariant subgroup, T , of P . The full Poincare group can beconsidered as the semi-direct product P = T ⊙ L. Every element can be written as
(Λ|t) = (1|t)(Λ|0). (2.13)
2.8 Examples of some common groups
Frequently one encounters the notations Cn, Dn, Sn, and An. These are the names ofcommon groups, which we shall introduce here briefly.
C 4
C 1 C 2 C 3
C 6
D 1 D 2 D 3
D 4 D 6
Figure 2.4: Examples of figures with symmetry Cn and Dn, n = 1, . . . 6.
The cyclic groups Cn
Consider a plane rotation over an angle 2π/n, say R, Then R, R2, . . . , Rn−1 are all differentrotations. However, Rn is equal to the identity, as a rotation over 2π is equivalent to norotation at all. If we now define Rp Rq = R(p+q)modn then the rotations R, R2, . . . ,Rn = I form a group. This group is called the cyclic group of order n. It is said to begenerated by the element R. An application of this group is the symmetry of a directed
n-sided polygon. With directed we mean that the sides have a direction associated withit.
The dihedral groups Dn
Consider again a plane regular n-sided polygon. If the sides of this polygon are notdirected and we consider it embedded in three-dimensional space, we can in addition to
12
the rotation in the plane of the polygon consider also rotations about a line in the planeover an angle π. If we combine those with the rotations belonging to Cn we get the groupDn. Note that this group is a perfectly well realizable symmetry of a rigid body. If wewould consider reflections, we realize that we can not perform in nature the reflection of arigid body physically. That is the deep reason why molecules that are each others mirrorimage may have different properties.
The symmetric groups Sn
Consider ordered n-tuples, say (a1, . . . an). A permutation of this n-tuple changes it intoanother n-tuple. If we consider the permutation σ1 followed by the permutation σ2 byσ = σ1 σ2, we see immediately that σ is also a permutation of this n-tuple. It is easy tosee that all permutations have an inverse and that all the axioms of a group are fulfilledby the permutations of (a1, . . . an). The group is called the symmetric group.
The alternating groups An
Consider the group Sn. One can separate the permutations into two categories, said tobe even and odd, respectively. A permutation σ is said to be even if it can be written as aproduct of an even number of permutations that interchange two elements of the n-tuple.(The latter are called transpositions.) A permutation that is not even is said to be odd.One can immediately draw the conclusion that the product of two even permutations iseven. As the identity can be considered as a permutation that contains no transpositionsat all, it is also even. A little more thought is needed to discover that the inverses of evenpermutations are also even. One may verify that the even permutations themselves forma group that has received the name alternating group and is denoted by An. It is clearlya subgroup of Sn.
2.9 Exercises
Exercise 2.1
Check the relations in Eqs. (2.3) and (2.4).
Exercise 2.2
Prove that the property of being conjugate is an equivalence relation.
Exercise 2.3
Prove Theorem 2.
Exercise 2.4
Show that an invariant subgroup H in G consists of complete classes of G.
Exercise 2.5
Prove the four remarks below Example 10.
Exercise 2.6
Explain why the factor group G/H is only a group if H is an invariant subgroup.
13
Exercise 2.7
Show that the direct product of two groups with the composition law in Eq. (2.10) formsa group.
Exercise 2.8
Show that S3 = R⊙ S2, and similarly, for the Poincare group that P = T ⊙ L.
Exercise 2.9
Prove that the property of four vectors xµ that their length squared, x2, is invariant underLorentz transformations entails that LTgL = g.
Exercise 2.10
Show that the cyclic group Cn is isomorphic to Zn.
Exercise 2.11
Show that D3 is isomorphic to S3 and that D4 is not isomorphic to S4.
Exercise 2.12
Show that the inverse of an even(odd) permutations is even(odd).
Exercise 2.13
Show that Sn and its subgroup An fulfill all the axioms of a group.
14
Chapter 3
Representations
Till now, we have discussed groups rather abstractly. However, in order to be able to applya theory of algebraic structures, one must be able to represent them in a way that lendsitself to practical calculations. A method that is universally accepted for this purposeis the mapping of abstract groups onto groups of matrices. Such a mapping is called arepresentation (rep) of the group. The nature of such maps is the subject of the presentchapter.
Before we give a formal definition of a rep, we introduce special types of mappings,homomorphisms and isomorphisms, which are tools to compare groups. One wants to beable to decide whether two groups are the same.
3.1 Homomorphic and Isomorphic Mappings
Definition homomorphism, isosmorphism
A mapping φ : G → G′ of a group G with composition law onto a group G′ withcomposition law · is said to be homomorphic (a homomorphism), iff ∀T1, T2 ∈ G : φ(T1 T2) = φ(T1) · φ(T2).
The mapping φ is said to be isosmorphic (an isosmorphism) if besides being a homo-morphism it is also bijective (1-1, onto).
Remark
In general, a homomorphism is a many-to-one mapping.
Notation
If G and G′ are isomorphic, we shall write G ∼= G′; if G is mapped onto G′ by a homo-morphism, we shall write G ≃ G′.
Definition kernel
Let G and G′ be groups, φ a homomorphism φ : G → G′. The set K = T ∈ G|φ(T ) =E ′, with E ′ the identity in G′, is said to be the kernel (ker) of φ.
The next theorem tells us what the general situation is: the possible reps map a groupG onto groups that are isomorphic with a factor group of G. This is a powerful theorem,as it connects the different reps to the invariant subgroups of G.
Theorem 5 “First Homomorphism Theorem”Let G and G′ be groups and φ : G→ G′ be a homomorphism with kernel K. Then
(i) K is an invariant subgroup of G.
15
(ii) Let T be an element of G, KT a right coset, then ∀T1, T2 ∈ KT one findsφ(T1) = φ(T2).
(iii) The mapping θ(KT ) = φ(T ) is an isomorphism from G/K onto G′.Proof
(i) K is a subgroup. If T1, T2 ∈ K = KE then, because φ is a homomorphismφ(T1 T2) = φ(T1) · φ(T2) = E ′ · E ′ = E ′. Hence φ(T1 T2) = E ′ so T1 T2 ∈ K. ClearlyK is closed under the composition .
The associative law holds in G and G′, so it also holds in K.K contains the identity E:
∀T ∈ G, φ(T ) = φ(T E) = φ(T ) · φ(E) ⇒ φ(E) = E ′ ⇒ E ∈ K
The inverse of any element of K also belongs to K because
∀T ∈ K, E ′ = φ(E) = φ(T T−1) = φ(T ) · φ(T−1)
⇒ E ′ = E ′ · φ(T−1) ⇒ φ(T−1) = E ′ ⇒ T−1 ∈ K
Next we check that K is invariant
∀X ∈ G, ∀T ∈ K, φ(X T X−1) = φ(X) · φ(T ) · φ(X−1)
= φ(X) ·E ′ · φ(X−1)
= φ(X) · φ(X−1)
= E ′
⇒ X T X−1 ∈ K
(ii) Let T ∈ G and T1, T2 ∈ KT , then ∃K1, K2 ∈ K such that Ti = KiT . Then itfollows that
φ(Ti) = φ(Ki T ) = φ(Ki) · φ(T ) = E ′ · φ(T ) = φ(T ), (i = 1, 2)
So T1 and T2 are mapped onto the same element of G′.(iii) Define the mapping θ from G/K to G′ by θ(KT ) = φ(T ). This mapping is 1-1
from G/K onto G′. First, if θ(KT1) = θ(KT2) then by definition φ(T1) = φ(T2). Weprove that the converse is also true. If T1 = T2 this is trivial. Consider now the caseT1 6= T2. Then there exists a T3 6= E, namely T3 = T1 T−1
2 , such that T1 = T3 T2 soφ(T1) = φ(T3) · φ(T2). Now φ(T1) = φ(T2) so φ(T3) = E ′ hence T3 ∈ K so KT3 = K.But this means that KT1 = K(T3 T2) = (KT3)T2 = KT2, so φ(T1) = φ(T2) impliesKT1 = KT2: the mapping is 1-1.
We check that θ is a homomorphism
θ(KT1) · θ(KT2) = φ(T1) · φ(T2) = φ(T1 T2) = θ(K(T1 T2)) = θ(KT1 ∗KT2)
2
Example 15Before, we have stated that the group (z ∈ C|zn = 1;×) consisting of the n distinct rootsof unity has the same structure as the group Zn of the integers under addition modulo n.We mean here that the two groups are isomorphic. The isomorphism is provided by the
16
mapping m 7→ exp(2πim/n). Using the above theorem we can also demonstrate thesegroups to be isomorphic to the factor group Z/nZ.
Let G = (Z; +) and φ(m) = m(modn), for some fixed positive integer n. Then φ isa homomorphism from G to the group G′ = (0, 1, 2, . . . , n− 1,+(modn)), which is Zn.The kernel of this mapping is K = 0,±n,±2n, . . ., which is the group nZ, an invariantsubgroup of G. Hence, we observe that G/K = Z/nZ ∼= Zn.
Example 16Let G = S3. Every element of S3 can be uniquely written as RE or RS, with R ∈ R =E,R,R2 and S ∈ S = SA, SB, SC. Define φ(T ) as follows
φ(T ) =
1, T ∈ R−1, T ∈ S (3.1)
One can check easily that φ is a homomorphism from S3 onto (−1, 1;×). Its kernel isK = R. The factor group G/K is S3/R = E, SA ∼= (−1, 1;×). The isomorphism θis
θ(R) = 1, θ(RSA) = θ(S) = −1. (3.2)
3.2 Representations
Let GL(d,K) be the “general linear” group of invertible d × d matrices with elementstaken from the field K. The compositon law is ordinary matrix multiplication. In mostcases of interest to physicists K is either C or R.
Definition representation
A homomorphism D from the group G onto a subgroup D(G) of GL(d,K) is said to bea d-dimensional representation (rep) of the group G.
Remark
We shall, somewhat carelessly, write D : G → G′, even if we understand that the rep Dmaps G onto a subgroup of G′.
One may generalize the definition of a rep to cases where GL(d,K) is replaced by agroup GL(V ) of invertible linear operators on some vector space V . This is quite typicalfor Quantum Mechanics or Quantum Field Theory. Upon a choice of basis for V , oneestablishes an isomorphism between GL(V ) and GL(n,K) (i.e., GL(Rn) or GL(Cn)). Inthis way the group multiplication becomes matrix multiplication.
Definition faithful
If a rep D is an isomorphism, it is said to be faithful.
Example 17(Trivial) Let G be any group, G′ = (1;×). The mapping φ(T ) = 1, ∀T ∈ G is a rep.
Example 18Let G = GL(d,K), the group of all d × d matrices with determinant different from 0.Define the mapping φ : G→ (R\0;×) by φ(T ) = det(T ) 6= 0. Then φ is a rep.
In the remainder of this section we consider the rep D : G → GL(d,K). Let Vbe a linear vector space with a basis e1, . . . , ed. It is supposed to be orthonormal:e†men = 〈em|en〉 = δmn. The last expression denotes the inner product in V .
17
Define for all T ∈ G an operator Φ(T ) on V as follows (the order of the subscripts
matters)
Φ(T )en =
d∑
m=1
emDmn(T ) (3.3)
The operator Φ is clearly linear
Φ(T )∑
j
bj ej =∑
j
bjΦ(T )ej. (3.4)
We can express the matrix elements of D(T ) in terms of those of Φ(T )
Dmn(T ) = e†mΦ(T )en = 〈em|Φ(T )en〉. (3.5)
Definition equivalent reps
Let D be a d-dim rep of a group G and let S be a non-singular d×d matrix. The mappingD′ : G→ GL(d,K) with D′(T ) = S−1D(T )S is said to be equivalent to D.
Remark
Note that D′ is also a rep. A similarity transformation S−1D(T )S is merely a change ofbasis, because if
Φ(T )en =∑
m
emDmn(T ), (3.6)
thenΦ(T )e′n =
∑
m
e′mD′mn(T ), (3.7)
provided D′mn(T ) =
∑
k,l(S−1)mkDkl(T )Sln and em = e′k(S
−1)km.
Definition unitary rep
A rep U is said to be unitary if U(T ) is a unitary matrix for all T ∈ G, i.e., U †(T )U(T ) =U(T )U †(T ) = I, the identity in GL(d,K).
Theorem 6If G is a group of finite order g, then every rep of G is equivalent to a unitary rep.Proof
We use the operator Φ defined before. Let ~u and ~v be two vectors in the linear space V .Define an inner product (·, ·) as follows
(~u,~v) ≡ 1
g
∑
T∈G〈Φ(T )~u|Φ(T )~v〉. (3.8)
Then we find for any X ∈ G:
(Φ(X)~u,Φ(X)~v) =1
g
∑
T∈G〈Φ(XT )~u|Φ(XT )~v〉
=1
g
∑
T ′∈G〈Φ(T ′)~u|Φ(T ′)~v〉
= (~u,~v). (3.9)
Clearly, the operator Φ(X) is unitary with respect to the inner product (·, ·).
18
Let ui be an orthonormal basis of V with respect to this inner product and viwith respect to the original inner product 〈·|·〉. So, (ui, uj) = δij and 〈vi|vj〉 = δij . Thenthere exists an operator S : V → V, Sui = vi. Let ~y =
∑
i αiui and ~x =∑
i βiui. Thenwe find
〈S~y|S~x〉 =∑
i,j
α∗iβj〈Sui|Suj〉 =
∑
i,j
α∗iβj〈vi|vj〉 =
∑
i
α∗i βi = (~y, ~x). (3.10)
So we find
〈SΦ(T )S−1~y|SΦ(T )S−1~x〉 = (Φ(T )S−1~y,Φ(T )S−1~x)
= (S−1~y, S−1~x)
= 〈~y|~x〉. (3.11)
Thus, SΦ(T )S−1 is unitary with respect to the inner product 〈·|·〉. Consequently, thetransformed matrices SDS−1 are unitary matrices.2
3.3 Reducible and Irreducible Representations
Definition reducible, irreducible
A rep D of a group G is said to be reducible if it is equivalent to a rep D′ for which everymatrix has the form
D′(T ) =
[
D1(T ) B(T )∅ D2(T )
]
. (3.12)
If D is not reducible it is said to be irreducible (an irrep).
Remarks
(i) The matrices Di, (i = 1, 2) form reps. If Di are not irreps, then they can bebrought into the same form as D′ itself, until the reps Di on the diagonal are irreps.
(ii) The space V has an invariant subspace V1 under the group G with dimension equalto d1, the dimension of D1. In order to see this, let en be a basis in V . Then
Φ(T )en =∑
emDmn(T )
=
d1∑
m=1
emDmn(T ) +d∑
m=d1+1
emDmn(T ). (3.13)
Now take n between 1 and d1, then we have
Φ(T )en =
d1∑
m=1
emDmn(T ) =
d1∑
m=1
em [D1(T )]mn . (3.14)
Hence D1 is a rep. The invariant subspace is V1 = Spane1, . . . , ed1. Its orthonormalcomplement Spaned1+1, . . . , ed1+d2 is, however, not invariant under G.
(iii) If D′ would be unitary, then reducibility is equivalent to full reducibility or de-
composability:
D′ =
[
D1 ∅∅ D2
]
. (3.15)
19
Again, if Di are not irreps, then one can decompose them as well, until all reps on thediagonal are irreps. Since all reps of a finite group are equivalent to unitary reps, theyare always fully reducible.
The next two theorems, Schur’s lemma’s, which state interesting properties of matricesthat connect different irreps or commute with all the matrices of a rep, also provide animportant practical tool to determine whether a rep is reducible or not (See Eq. (3.34).)
Theorem 7 (Schur’s first lemma)Let D and D′ be any two irreps of the group G with dimensions d and d′ resp. Supposethere exists a matrix A, d × d′, such that ∀T ∈ G : D(T )A = AD′(T ). Then eitherA = [∅], the null matrix, or d = d′ and detA 6= 0. In the latter case D and D′ areequivalent.Proof
Let V and V ′ be the two linear spaces associated with the reps D and D′. A maps V ′
into V . Consider the ker of A: kerA = ~v ′ ∈ V ′ : A~v ′ = ~0 and the range RA = ~v ∈ V :~v = A~v ′.
(i) kerA is an invariant subspace. Suppose ~v ′ ∈ kerA, then AD′(T )~v ′ = D(T )A~v ′ = ~0
so also D′(T )~v′ ∈ kerA. As D′ is an irrep, V ′ has no proper invariant subspaces, so either
kerA = V ′, or kerA = ~0. kerA = V ′ implies that A is the null operator. If kerA = ~0,A is 1-1.
(ii) RA is invariant in V . Indeed, take any vector from RA, say ~v = A~v′, thenD(T )A~v ′ = AD′(T )~v ′ This relation tells us that every element ~v from RA, when actedupon with a D(T ) is again an element of RA, because it can be written as a vector ofV ′, say ~v ′′ = D′(T )~v ′ acted upon by A. Now we use that D is an irrep, so RA is eitherall of V or the null vector only. In the latter case we can conclude again that A is thenull operator. In the former case we see that kerA = ~0 and the range of A is V , so Ais surjective. From this fact, taken together with the fact that A is 1-1, we can concludethat A is bijective. So, D and D′ are equivalent. If A is 1-1, it must be a square matrix,which means that d′ = d.2
Theorem 8 (Schur’s second lemma)Let D be a d-dim irrep of the group G, B a d× d-matrix such that ∀T ∈ G : D(T )B =BD(T ), then the matrix B is proportional to the identity matrix I = diag(1, . . . , 1).Proof
Define A = B − λI, λ such that detA = 0. Here, λ can be one of the eigenvaluesof B, which must have at least one (complex) eigenvalue. Then DB = BD impliesD(A+ λI) = (A + λI)D hence DA = AD. According to Schur’s first lemma A = [∅], asdetA = 0. Consequently, B = λI, with λ ∈ C.2
Using these lemma’s we can prove the orthogonality theorem for finite-dimensionalirreps.
Theorem 9 (Orthogonality)Let Dp and Dq be two unitary irreps of the group G with dimensions dp and dq. If p 6= qthese irreps are supposed to be inequivalent. The order of G is g. Then the following
20
relation holds1
g
∑
T∈GDp(T )jkD
q(T )∗ts =1
dpδpqδjtδks. (3.16)
Proof
Let B be an arbitrary dp × dq matrix. Construct the matrix A as follows:
A =1
g
∑
T∈GDp(T )BDq(T−1) (3.17)
then for all S ∈ G
Dp(S)A = Dp(S)1
g
∑
T∈GDp(T )BDq(T−1)Dq(S−1)Dq(S)
=1
g
∑
T∈GDp(ST )BDq([ST ]−1)Dq(S)
= ADq(S). (3.18)
Schur’s first and second lemma imply that A = λIδpq, because Dp and Dq are not equiv-
alent if p 6= q. The constant λ depends on p and B.Now consider all matrices B of the form
B(l, m)jk = δljδmk, 1 ≤ l, j ≤ dp, 1 ≤ m, k ≤ dq. (3.19)
Upon substitution of B in the definition of A and using the expression for A just found,we obtain
1
g
∑
T∈GDp(T )ilD
q(T−1)ms = λδpqδis, 1 ≤ i, l ≤ dp, 1 ≤ m, s ≤ dq. (3.20)
The constant λ may depend on p, l and m. Take now p = q and i = s, and sum over s,then it follows
λdp =1
g
∑
T∈G
dp∑
s=1
Dp(T )slDp(T−1)ms
=1
g
∑
T∈GDp(TT−1)ml = δml (3.21)
so
λ =1
dpδml. (3.22)
Consequently,1
g
∑
T∈GDp(T )jkD
q(T−1)st =1
dpδpqδksδjt. (3.23)
The irreps Dp and Dq are chosen to be unitary, as can be done for a finite group G, so
Dq(T−1)st = Dq(T )∗ts, (3.24)
consequently,1
g
∑
T∈GDp(T )jkD
q(T )∗ts =1
dpδpqδksδjt. (3.25)
2
21
3.4 Characters
We are interested in irreps up to equivalence, i.e., we do not count two irreps as differentif they are equivalent. Therefore we look for a property that is invariant under similaritytransformations. The character is such a function.
Definition character
Let D be a d-dim rep of G. Then the mapping χD : G→ C defined for all elements T ofG by
χD(T ) = TrD(T ) =
d∑
j=1
D(T )jj (3.26)
is said to be the character of the rep D.
Remark
(i) χD(E) = d(ii) Because TrA is invariant under similarity transformations, χD = χD′
for any twoequivalent reps. For the same reason a character is constant on a full class of G.
Theorem 10 (First Orthogonality Theorem of Characters)If χD and χD′
are characters of the irreps D and D′ of the group G of finite order g, andif D is not equivalent to D′, then
1
g
∑
T∈GχD(T )∗χD′
(T ) = δDD′. (3.27)
Proof
(i) D and D′ may be taken unitary.(ii) From the orthogonality theorem we infer
1
g
∑
T∈GD(T )∗jkD
′(T )st =1
dδDD′δjsδkt (3.28)
Take traces, i.e., take j = k and s = t and sum over j and s:
1
g
∑
T∈GχD(T )∗χD′
(T ) =1
g
∑
i
NiχD(Ki)
∗χD′
(Ki) = δDD′ (3.29)
2
Example 19 S3
In the three-dimensional rep given by Eq. (2.5) we find χc(E) = 3, χc(R) = χc(R2) = 0
and χc(SA) = χc(SB) = χc(SC) = 1. So the classes [E], [R] and [S] have the characters3, 0 and 1, respectively. In the two-dimensional rep given by Eqs. (2.7-2.9) we findχb([E]) = 2, χb([R]) = −1 and χb([S]) = 0. If we consider the characters as vectorsin class space, then we have χc = (3, 0, 1) and χb = (2,−1, 0). A third rep is the onedescribed in Eq. (3.1): χa([E]) = χa([R]) = 1, χa([S]) = −1, so χa = (1, 1,−1). We seethat
∑
T χa(T )χb(T ) = 1(1×2)+2(1×(−1))+3(−1×0) = 0 and also the other combination∑
T χa(T )χc(T ) = 0. However,∑
T χb(T )χc(T ) does not vanish. Consequently, not allthree reps can be irreps.
Theorem 11If G is a finite group, then the two irreps D and D′ are equivalent iff χD = χD′
.
22
Proof
If D and D′ are equivalent, then χD = χD′
. Suppose that χD = χD′
, but that D and D′
are not equivalent. From the orthogonality theorem it follows that
1
g
∑
T∈GχD(T )∗χD′
(T ) = 0. (3.30)
We may replace χD′
by χD in this expression, but then the sum is a sum over positivenumbers and must be different from zero; in fact, it must be 1, as follows from the firstorthogonality theorem. This is absurd, so the hypothesis that D and D′ are not equivalentmust be false.2
A theorem that can be used to analyze reducible reps is the following.
Theorem 12Let G be a finite group of order g and D any rep of G. If D is equivalent to the direct
sum of the irreps Dp
D′ =⊕
p
npDp (3.31)
then the numbers np can be computed from
np =1
g
∑
T∈GχD(T )χDp
(T )∗ =1
g
∑
T∈GχD(T )∗χDp
(T ). (3.32)
Proof
If D ∼= D′ =⊕
p npDp, then χD =
∑
p npχDp
. Use the orthogonality theorem:
1
g
∑
T∈G
∑
p′
np′χDp′
(T )
χDp
(T )∗ =∑
p′
np′1
g
∑
T∈GχDp′
(T )χDp
(T )∗
= np (3.33)
2
Corollary
A rep D of a group G of finite order g is an irrep iff
1
g
∑
T∈G
[
χD(T )]2
= 1. (3.34)
One can check easily that this identity is fulfilled by χa and χb of the group S3, butnot by χc. So χa and χb are irreps, χc is not.
There exists a second orthogonality theorem on group characters. In order to stateand prove that theorem, we need the concept of the regular representation. This is thesubject of the next section.
23
3.5 The Group Algebra and the Regular Represen-
tation
The concept of the regular representation is connected to the idea of the group algebra,which we introduce first.
Let G be a group of finite order g. We consider its elements T1, . . . , Tg as basis vectorsin a g-dim space. So, an arbitrary vector has the form
∑
i xiTi. The scalars xi are elementsof some field K. This space is said to be the group algebra, if we define the linear structure
∑
i
xiTi +∑
j
yjTj =∑
l
(xl + yl)Tl (3.35)
and the product
∑
i
xiTi ·∑
k
ykTk =∑
j
[(
∑
i
xijTi Tj)
· yj−1T−1j
]
=∑
i,j
(xijyj−1) Ti. (3.36)
(We use here the indices ij for x and j−1 for y which have of course a unique relationwith the indices of the elements Tn of G.)
The scalar xij is the coordinate of TiTj in the vector∑
l xlTl and yj−1 is the coordinateof T−1
j in the vector∑
l ylTl.If we multiply an arbitrary vector
∑
i xiTi by a group element S, we find a new vector∑
i xi(STi) =∑
i x′iTi, where the x′i form a permutation of the xi, because, if Ti runs
over G, STi runs also over G, although the order may be different.We use this property to define a rep of G, the regular representation, as follows:
D(Ts)kj =
1 if Ts Tj = Tk0 if Ts Tj 6= Tk
(3.37)
(As a mnemonic device you may write D(Ts)kj in bra and ket notation as 〈Tk|Ts|Tj〉.)Example 20Consider the Abelian group R = E,R,R2.
~x = xE + yR+ zR2 ↔ (x, y, z)
R~x = xR + yR2 + zE ↔ (z, x, y)
R2~x = xR2 + yE + zR ↔ (y, z, x) (3.38)
So we find the matrix elements in the regular rep
D(E) =
1 0 00 1 00 0 1
, D(R) =
0 0 11 0 00 1 0
. D(R2) =
0 1 00 0 11 0 0
. (3.39)
It is clear from Eq. (3.38) that the elements of the group act as permutations of(x, y, z). All matrices of this rep have in any row or column precisely one 1, the otherentries being 0. Matrices with this property are called permutation matrices.
We shall show that D is indeed a rep. First, we see that the following relation holds
Ts Tj =∑
k
TkD(Ts)kj. (3.40)
24
Then we have for the product of two elements
(Ts Tt) Tj =∑
k
TkD(Ts Tt)kj. (3.41)
Upon applying the associative law we obtain
(Ts Tt) Tj = Ts (Tt Tj)= Ts
∑
m
TmD(Tt)mj
=∑
k
Tk∑
m
D(Ts)kmD(Tt)mj . (3.42)
Combining the two results we find that
D(Ts Tt)kj =∑
m
D(Ts)kmD(Tt)mj (3.43)
which proves that the matrix representing a product is given by the matrix product ofthe matrices representing the factors in the product.
Let χ(T ) be the character of the regular rep. Then it is obvious that χ(T ) = 0 if T 6= Eand that χ(E) = g. We expand the character into the characters of the non-equivalentirreps Dp of G
χ(T ) =∑
p
npχDp
(T ). (3.44)
For T = E we have χ(E) = g, χDp
(E) = dp. The numbers np are provided by the firstorthogonality theorem:
np =1
g
∑
T∈Gχ(T )χ∗Dp
(T ) =1
gχ(E)χ∗Dp
(E) = dp. (3.45)
So we find the important result that the number of times an irrep is contained in the
regular rep equals its dimension. (Contained is meant here in the sense of Eq. (3.31).)The formula χ(T ) =
∑
p npχDp
(T ) can be rewritten as
χ(T ) =∑
Dp
dpχDp
(T ) =∑
Dp
χDp
(E)χDp
(T ) = gδE T . (3.46)
This result can be generalized and cast into a theorem.
Theorem 13 Second Orthogonality TheoremLet G be a finite group of order g, Ki the set of its conjugacy classes, Dp the set ofits irreps up to equivalence, Nj is the number of elements in the class Kj , then
1
g
∑
Dp
χDp
(Ki)χ∗Dp
(Kj)Nj = δij (3.47)
Proof
Let T(i)1 , . . . , T
(i)Ni
be the elements of Ki and define for any class Ki the element of the
group algebra Ci =∑Ni
l=1 T(i)l . Then we can multiply these operators:
CiCj =∑
l,m
T(i)l T (j)
m =∑
k
cijkCk. (3.48)
25
This follows because the operators Ci are invariant: XCiX−1 = Ci, so the product of
any two is also invariant and must therefore be a linear combination of such operators.The numbers cijk must be nonegative integers. They do not depend on the rep, but arecharacteristics of the group G.
Consider any irrep D of G. The matrix obtained by summing all matrices correspond-ing to the class Ki is denoted by Di:
Di =
Ni∑
j
D(T(i)j ). (3.49)
So Di is the matrix image of Ci. Because Ki is a class we have
∑
j
D(XT(i)j X−1) = D(X)
∑
j
D(T(i)j )D(X−1) = Di, (3.50)
as XT (i)j X−1Ni
j=1 is simply a permutation of the class Ki.Because we have assumed thatD is an irrep andDi commutes according to the formula
we just derived with any D(X), we obtain Di = λiI (Schur’s second lemma). This entailsfor the characters
TrDi = Niχ(Ki) = λiTr I = λiχ(E) ⇒ λi =Niχ(Ki)
χ(E). (3.51)
In this irrep we find the matrix equivalent of the expansion of CiCj:
DiDj =∑
k
cijkDk ⇒ λiλj =∑
k
cijkλk. (3.52)
Upon substitution of the expressions for the lambda’s we obtain
Niχ(Ki)
χ(E)
Njχ(Kj)
χ(E)=∑
k
cijkNkχ(Kk)
χ(E)(3.53)
Consider now the case k = 1 corresponding to the class K1 = [E]. For any i thereexists a unique index ı such that cij1 = Niδjı. This ı is the index of the class that consistsof the inverses of the elements of Ki.
Now we rewrite our last formula for a specific irrep Dp:
NiNjχDp
(Ki)χDp
(Kj) =∑
k
cijkNkχDp
(E)χDp
(Kk) (3.54)
Next sum over Dp and use Eq. (3.46)
NiNj
∑
Dp
χDp
(Ki)χDp
(Kj) =∑
k
cijkNk
∑
Dp
χDp
(E)χDp
(Kk)
=∑
k
cijkNk gδE,Kkfrom Eq. (3.46)
= cij1g
= Niδjıg. (3.55)
26
For a finite group any irrep is equivalent to a unitary irrep. If Dp is unitary, Dpij(T
−1) =Dp∗
ji (T ) so χDp
(T−1) = χDp
(T )∗. Consequently, χDp
(Ki) = χ∗Dp
(Kı). So we find, aftersome redefinition of indices
1
g
∑
Dp
χDp
(Ki)χ∗Dp
(Kj)Nj = δij (3.56)
2
Corollary
Consider the numbers√Niχ
Dp
(Ki) as vectors in the space of irreps. The second orthog-onality theorem states that these vectors are orthogonal. Thus, they must be linearlyindependent. Therefore, the number of classes is not greater than the number of non-equivalent irreps. The first orthogonality theorem can be understood as expressing theorthogonality of the vectors
√Niχ
Dp
(Ki) in the space of classes. Then it follows that thenumber of irreps is not greater than the number of classes. So the final result is
For a finite group the number of irreps up to equivalence, is equal to thenumber of classes.
3.6 Exercises
Exercise 3.1
Prove that φ of Example 18 is indeed a representation.
Exercise 3.2
Show that for Abelian groups all (complex) irreps are one-dimensional.
Exercise 3.3
Show that the matrices D1 and D2 given in Eq. (3.12) indeed form reps.
Exercise 3.4
Let G be a group and φ a representation of G. Consider the maps φ∗, (φT)−1 and (φ†)−1
defined by
φ∗(g) = (φ(g))∗, (3.57)
(φT)−1(g) = ((φ(g))T)−1, (3.58)
(φ†)−1(g) = ((φ(g))†)−1, (3.59)
for all g ∈ G. Here ∗,T, †,−1 denote complex conjugation, transposition, Hermitianconjugation and the inverse, respectively.
(a) Prove that the three maps φ∗, (φT)−1 and (φ†)−1 are also representations.(b) Show that these maps are irreducible representations if φ is an irrep.
Exercise 3.5
27
Let G be a group, φ an irreducible matrix representation of G, and K a class of G. Define
A =∑
T∈Kφ(T ). (3.60)
Show that A is proportional to the identity matrix.
28
Chapter 4
Representations: Function Spacesand Operators
Applications of group theory use operator language. In classical mechanics one is con-cerned with operators in space and time, i.e., differential operators on functions of thecoordinates and the time. In quantum mechanics operators in Hilbert space are the onesused throughout.
Here we shall discuss the connection between transformations in R3 and operators on
functions φ(~r), ~r ∈ R3. Let G be a group of invertible coordinate transformations in R
3.Define the mapping P : G → G′, where G′ is a group of operators on functions of ~r, asfollows
∀T ∈ G, ∀φ(~r) : (P (T )φ)(T~r) = φ(~r). (4.1)
The composition law in G′, say ·, is defined by
∀φ : (P (S) · P (T ))φ = P (S T )φ. (4.2)
Once the connection between transformations in R3 and operations on functions is estab-
lished, we may answer the question how to construct functions with a definite symmetry.(In the sequel it will be clear from the context which composition law is used in the
product of operators. Therefore we shall no longer write or · explicitly.)
4.1 Projection Operators
Let φ(~r) be a square-summable function on R3:∫
d3r|φ(~r)|2 < ∞. Then obviously∫
d2rφ(~r) ≡ φ0(r) is a square-summable function that is spherically symmetric. Ap-parently, the integration over the angular part of ~r projects out a function with rotationalsymmetry. This idea is extended in the following theorem. (L2(R3) is the Hilbert spaceof square-summable functions on R
3).
Theorem 14Any function φ(~r) ∈ L
2(R3) can be expanded in the basis functions of the unitary irrepsof a finite group G of coordinate transformations T in R
3 such that T preserves the lengthof vectors in R
3.Proof
Construct the functions φT (~r) = P (T )φ(~r). If P (T ) is defined as above, then it is unitaryfor all T ∈ G.
29
Suppose the set φT |T ∈ G is d-dimensional. By Schmidt orthogonalization one canconstruct d orthonormal functions ψk, k = 1, . . . , d, that form a basis for this subspace.So, any element of this set can be expanded in this basis.
P (T )φ(~r) =
d∑
k=1
ck(T )ψk(~r) (4.3)
Furthermore, for any S, P (S)ψk(~r) is also a linear combination of the basis functionsψn, n = 1, . . . , d. We can write
P (S)ψk(~r) =
d∑
n=1
ψn(~r)D(S)nk, k = 1, . . . , d. (4.4)
This means that D is a d-dim rep of G. We say that ψ1(~r), . . . , ψd(~r) form a basis of
the rep D. Take for instance T = E, then the corresponding P (T ) = P (E) is the identityoperator and we find, using Eq. (4.3):
P (E)φ(~r) = φ(~r) =
d∑
k=1
ck(E)ψk(~r), (4.5)
so any φ ∈ L2 can be expanded in basis functions of a rep D of G.
2
The rep D may be reducible. Now suppose that D is equivalent to a decomposablerep D′, then there exists a unitary matrix S, such that
D′(T ) = S−1D(T )S ⇒ ψ′n(~r) =
∑
m
Smnψm(~r) (4.6)
are basis functions for D′ and therefore can be grouped together in sets of basis functionsfor the irreps contained in D′:
D′ =⊕
p
Dp ⇒ ψ′n = ψα
1 , . . . , ψαnα, . . . , ψω
1 , . . . , ψωnω (4.7)
Definition projection operator
Let Dp be a unitary irrep of dimension dp of the group G, which is of order g. Then theoperators
Ppmn = dp
1
g
∑
T∈GDp(T )∗mnP (T ) (4.8)
are said to be projection operators on the basis of Dp.In the next theorem we shall use the inner product in the Hilbert space L2. We denote
it by (·, ·).Theorem 15The projection operators Pp
mn have the properties(i) hermiticity (self-adjointness)
∀φ, ψ ∈ L2, (Pp
mnφ, ψ) = (φ,Ppnmψ) (4.9)
30
(ii) orthogonality∀Pp
mn, Pqst : Pp
mnPqst = δpqδnsPq
mt. (4.10)
(iii) If ψpk
dpk=1 forms the basis for the irrep Dp, then
Ppmnψ
ql = δpqδnlψ
pm. (4.11)
(iv) ∀φ ∈ L2:
Ppnnφ = apnφ
pn (4.12)
andφ =
∑
n,p
apnφpn. (4.13)
We leave the proof of this theorem as an exercise.
Remark
If Ppnnφ is not the null element of L2, then it can be used as a basis function for the rep Dp.
The other basis functions can be obtained by acting with Ppmn on φp
n. One may considerthe coefficients apn as elements of a column vector, the index n being the row index. Thisexplains the usual terminology where the function φp
n is said to belong to the n-th row of
the irrep Dp and denote also its label n as its row index.
4.2 Tensor Products and the Clebsch-Gordan Series
A vector is defined by its behaviour under transformations, say R : ~x 7→ ~x′ correspondsto the component relation x′n = Rnmxm. Once a vector is defined this way, tensors aredefined as objects that transform as products of vectors, e.g., Tn1,...,nM
is a tensor elementiff
T ′n′
1,...,n′
M= Rn′
1n1· · ·Rn′
MnMTn1,...,nM
. (4.14)
The object Rn′
1n1· · ·Rn′
MnM
represents the transformation of the tensor. It is called thetensor product of its components R. We can use this idea for any rep of any group G. Inthe following theorem we state that the tensor product of two reps is again a rep, alsoreferred to as a direct product rep.
Theorem 16If Dp and Dq are two unitary irreps of a group G, with dimensions dp and dq resp., thenthe matrices with elements
Dmr,ns(T ) = Dpmn(T )D
qrs(T ) (4.15)
denoted byD(T ) = Dp(T )⊗Dq(T ) (4.16)
form a unitary rep of G of dimension dpdq. The character χ of D is given by χ(T ) =χp(T )χq(T ).Proof
(i) ∀T1, T2 ∈ G:
D(T1)D(T2) = (Dp(T1)⊗Dq(T1)) (Dp(T2)⊗Dq(T2))
= Dp(T1T2)⊗Dq(T1T2)
= D(T1T2) (4.17)
31
so D is a rep.(ii) If Dp and Dq are unitary, then so is D.(iii) Take traces to obtain the characters
χ(T ) =
dp∑
j=1
dq∑
s=1
(Dp(T )⊗Dq(T ))js,js
=∑
j
∑
s
Dp(T )jjDq(T )ss
= χp(T )χq(T ). (4.18)
2
Remark
(i) In general, Dp ⊗Dq is reducible.(ii) If G is of finite order, then Dp ⊗Dq is equivalent to a decomposable rep
Dp ⊗Dq = C
(
⊕
r
nrpqD
r
)
C−1, (4.19)
with Dr irreps. The decomposition⊕
r nrpqD
r is called the Clebsch-Gordan series. Itscoefficients nr
pq indicate how many times a specific irrep appears, using the notation thatDp ⊕Dp = 2Dp. They can be computed using a theorem we proved before. We obtain
nrpq =
1
g
∑
T∈Gχ(T )χr(T )∗
=1
g
∑
T∈Gχp(T )χq(T )χr(T )∗ (4.20)
Example 21 S3
The character table is
χ1 χ2 χ3 Nj
K1 = [E] 1 1 2 1K2 = [R] 1 1 -1 2K3 = [S] 1 -1 0 3
The order of S3 is g = 6, d1 = d2 = 1, d3 = 2. Because the character determines theirrep uniquely, we find immediately
(i) D1 ⊗Dp ≃ Dp ⇒ np1p = np
p1 = 1(ii) D2 ⊗D2 ≃ D1 ⇒ np
22 = δ1p(iii) D2 ⊗D3 ≃ D3 ⇒ np
23 = np32 = δ3p
The only interesting case is D3 ⊗ D3. From χD3⊗D3
(K1) = 4 we conclude that itsdimension is 4, as it must be because D3 is two-dimensional. The other characters areχD3⊗D3
(K2) = 1 and χD3⊗D3
(K3) = 0. The formula D =⊕
p npDp implies
1
g
∑
T∈G
∣
∣χD(T )∣
∣
2=∑
p
n2p (4.21)
32
so we find in our case1
6
∑
T∈S3
∣
∣
∣χD3⊗D3
(T )∣
∣
∣
2
= 3. (4.22)
The only solution is∑
p n2p = 3 = 1 + 1 + 1 so n1 = n2 = n3 = 1. An elementary
computation gives indeed
n133 =
1
6(4 + 2 + 0) = 1, n2
33 =1
6(4 + 2 + 0) = 1, n3
33 =1
6(8− 2 + 0) = 1. (4.23)
Finally:D3 ⊗D3 ≃ D1 ⊕D2 ⊕D3. (4.24)
4.3 Irreducible Tensorial Sets
We introduced the idea of a tensor by means of its behaviour under transformations. Wecan define irreducible tensorial sets following the same logic. The transformations weconsider in the latter case behave like irreps of a group G. Let us illustrate this conceptwith an example of transformations in L
2.Let G be a group of coordinate transformations, G′ its counterpart for L2 and P the
familiar mapping P : G → G′. Let ψpk(~r) and ψq
k(~r) be bases for the irreps Dp andDq respectively. It means that, e.g.,
P (T )ψpk =
dp∑
j=1
ψpjD(T )jk (4.25)
Now form the second-rank tensor ψpkψ
qs . It transforms as
P (T ) [ψpkψ
qs ] = P (T )ψp
k P (T )ψqs
=∑
j,r
ψpjD
p(T )jk ψqrD
q(T )rs
=∑
j,r
[Dp(T )⊗Dq(T )]jr,ksψpjψ
qr . (4.26)
Apparently, Dp⊗Dq has ψpkψ
qs as a basis if they are independent. We call this basis a
tensorial set. Now, in general, Dp⊗Dq is not an irrep, so the tensorial set is not irreducible.In other words, there are subspaces in the space spanned by the basis ψp
kψqs that are
invariant under the transformations of the group G. Suppose Dp ⊗Dq =⊕
r nrpqD
r, thenfor all r such that nr
pq 6= 0 there are among the ψpkψ
qs nr
pq basis sets for Dr, composedof linear combinations of the products of functions ψp
kψqs. We shall denote these basis
functions by θr,αl , r indicates the irrep Dr, l the row index of the basis function and α isan index running from 1 to nr
pq. We write
θr,αl =∑
k,s
(
pq rαks l
)
ψpkψ
qs (4.27)
and find
P (T )θr,αl =
dr∑
t=1
θr,αt Dr(T )tl. (4.28)
33
-0.5 0.0 0.5
-0.4
-0.2
0.0
0.2
0.4
0.6
0.8
1.0
Figure 4.1: A contourplot of the function ψ21 overlayed on the equilateral triangle. The
lines intersecting at the the center of the triangle are the nodal lines of ψ21 , which are also
the lines that remain invariant under the reflections Sα. A rotation over 120 does notchange the sign of this function, a reflection in either of the nodal lines does.
The coefficients
(
pq rαks l
)
are called Clebsch-Gordan (CG) coefficients. These coeffi-
cients define the basis transformation matrix C of Eq. (4.19).
For a finite group G we may stick to unitary irreps. Then the matrix of CG coefficientsturns out to be unitary too, if the right phase convention is adopted. (If θr,αl (~r) is abasis, so is exp(iωr,α
l )θr,αl (~r) when ωr,αl is a real number.) Then we have
(
pq rαks l
)
=
(
rα pql ks
)∗. (4.29)
Example 22 Basis functions for the irreps of S3
We consider again the group S3 of symmetries of the equilateral triangle introduced insect. 2.3. The equations (2.7) - (2.9) define a two-dimensional rep of S3, D
3, with basisfunctions that can be written as ψ3
1 = x exp(−(x2 + y2)) and ψ32 = y exp(−(x2 + y2)).
These functions of x and y are normalizable and acting with any of the 2 × 2 matricesdefined in sect. 2.3 we find
P (T )ψ3k =
∑
j
ψ3jD
3(T )jk. (4.30)
The function exp(−(x2 + y2)) is fully symmetric and can be used, besides providing thenormalizability of ψ3
1 and ψ32, as the basis function ψ1
1 of the fully symmetric rep D1.A function that is antisymmetric under reflections and symmetric under rotations, i.e.,
the basis function ψ21 of D1, can be easily constructed using the basic geometry of the
34
equilateral triangle. The lines x =√3y, x = −√
3y, and x = 0 are the lines that remaininvariant under the reflections SA, SB, and SC , respectively, while they are permutedamong each other by the rotations R and R2. A straightforward computation shows thatthe function
ψ21 = x(x+
√3y)((x−√
3y) exp(−(x2 + y2)) (4.31)
changes sign upon reflections and remains invariant under rotations and thus is the bas-isfunction we were looking for. By direct computation one can easily verify that
E ψ21 = Rψ2
1 = R2 ψ21 = ψ2
1 , SA ψ21 = SB ψ
21 = SC ψ
21 = −ψ2
1 . (4.32)
It is clear from this contruction that we cannot use ψ31 and ψ3
2 in a straightforward wayto construct the four basis functions for D3⊗D3, for the simple reason that the functionsx2, xy, yx, and y2 are not independent.
In order to overcome this difficulty we use two vectors ~r1 = (x1, y1) and ~r2 = (x2, y2).The tensor product D3 ⊗ D3 has the basis x1x2, x1y2, y1x2, y1y2. The basis functionsfor D1, D2 and D3 are
θ11 = (x1x2 + y1y2), symmetric
θ21 = (x1y2 − y1x2), anti-symmetric
θ31 = (x1y2 + y1x2)θ32 = (x1x2 − y1y2)
, mixed symmetric, (4.33)
which can be checked easily by direct computation. Note that we can easily add the factorexp(−(x21 + y21 + x22 + y22)) to ensure that these basis functions belong to L
2(r1)⊗ L2(r2).
These basis functions are not normalized. The CG coefficients that we can read off dotherefore not form a unitary matrix. This can be remedied easily, but we do not want tobother. We have
(
33 111 1
)
=
(
33 122 1
)
= 1,
(
33 212 1
)
= −(
33 221 1
)
= 1,
(
33 312 1
)
=
(
33 321 1
)
= 1,
(
33 311 2
)
= −(
33 322 2
)
= 1. (4.34)
All other combinations
(
33 pkl r
)
are zero.
In quantum mechanics one encounters the CG decomposition when adding angularmomenta. For orbital angular momenta this concerns irreps of the group of rotationsin three dimensions, SO(3), and for intrinsic angular momenta or spin it applies to thespecial unitary group SU(2) of unitary 2 × 2 matrices with determinant 1. The irrepsof these groups are usually labelled by quantum numbers l and j, respectively. A tensorproduct of two states |l, m〉 is decomposed (without the need for an additional label α)as follows:
|l1, m1〉 ⊗ |l2, m2〉 =l1+l2∑
l3=|l1−l2|
l3∑
m3=−l3
(
l1 l2 l3m1 m2 m3
)
|l3, m3〉, (4.35)
35
although it is more common to write the CG coefficients as (l1m1l2m2|l3m3) or〈l1m1l2m2|l3m3〉. 1
Let us consider SO(3) (or finite subgroups of it) in more detail. Usually its elementsare represented as 3 × 3 matrices R that act on the components of a three-vector ~x, i.e.V = R
3. When considering the tensor elements Tn1,...,nMconsisting of tensor products of
a three-vector ~x, i.e. Tn1,...,nM= xn1
. . . xnM, then V = R
3 ⊗ . . . ⊗ R3 where R
3 appearsM times. In general, V has invariant subspaces under the group transformations. Theseinvariant subspaces determine the irreps of SO(3) and the basis functions for those irreps
are the tensor elements Y[l]n1,...,nl which are obtained from Tn1,...,nM
by symmetrizing over allthe indices and subtracting traces (the reason will be addressed later on in these lectures).
This means that Y[l]n1,...,nl is invariant under interchange of any two indices and for which
the contraction with any δnink(with i and k between 1 and l) vanishes. For l = 0 one
simply has Y [0] = 1; for l = 1 the tensor Y[1]n is the vector xn itself; and for l = 2 one
has the traceless tensor Y[2]n1n2 = xn1
xn2− δn1n2
|~x|2/3, which has 5 linearly independentcomponents. In general, one has 2l + 1 linearly independent tensors (labelled by thequantum number m).
In this way the tensor Tn1n2= xn1
xn2can for example be decomposed into a linear
combination of l = 0 and l = 2 Y [l]-tensors having dimensions 1 and 5 respectively:
Tij ≡ xixj =1
3|~x|2Y [0]δij + Y
[2]ij . (4.36)
Under rotations the components of Y[2]ij transform amongst themselves, i.e. they span an
invariant subspace, while the term with Y [0]δij remains invariant.
In quantum mechanics it is common to write the tensors Y[l]n1,...,nl in a complex basis
in which the vector ~x = (x, y, z) is transformed to (−(x+ iy)/√2, z, (x− iy)/
√2). In this
way one obtains, but for a normalization factor, the standard spherical harmonics Ylm.These tensors therefore transform according to the irreps of the rotation group SO(3),meaning that l does not change under rotations, only the quantum number m of Lz does.
If one considers the decomposition of the tensor xiyj consisting of a product of twothree-dimensional vectors ~x and ~y, then there is also a three-dimensional (l = 1) compo-nent present, which is absent in the case of ~x = ~y. One arrives at
Tij ≡ xiyj =1
3δijTkk +
1
2ǫijkǫklmTlm +
1
2
(
Tij + Tji −2
3δijTkk
)
, (4.37)
where the second (antisymmetric) term can also be written as 1
2(Tij−Tji) using ǫijk ǫklm =
δil δjm−δim δjl, and clearly has three linearly independent components. In general, invari-ant subspaces of tensor elements can be obtained by contractions with the two invarianttensors δij and ǫijk.
Eq. (4.37) reflects the general CG decomposition:
3⊗ 3 = 1⊕ 3⊕ 5, (4.38)
1In the classical literature about quantum mechanics there is some confusion about the notation ofthe CG coefficient. Moreover, Wigner has introduced a symbol that is more symmetric that the CGcoefficients. This 3 − j symbol is written in a way very similar to the notation we use here for the CG
symbols, namely
(
l1 l2 l3m1 m2 m3
)
and is identical with (−1)l1−l2−m3(2l3+1)−1/2(l1m1l2m2|l3−m3). A
detailed discussion of these coefficients and the different conventions used may be found in A.R. Edmonds,Angular Momentum in Quantum Mechanics.
36
where the irreps are now labelled by their dimension d = 2l + 1. The 3 rep on the r.h.s.is formed with help of the ǫijk tensor, which allows one to obtain vectors through thevector product ~x × ~y. When considering only rotations, there is no difference betweensuch pseudovectors and ordinary vectors, but if one considers also reflections, then theyare not equivalent.
In general, one has the CG decomposition:
(2l1 + 1)⊗ (2l2 + 1) =
l1+l2⊕
l3=|l1−l2|(2l3 + 1), (4.39)
which is established via the basis transformation in Eq. (4.35). This result is not onlyvalid for SO(3), i.e. integer l, but also for SU(2) if one includes half-integer values j.
The idea of an irreducible tensorial set of operators is formed in analogy with the basistensors.
Definition irreducible tensorial set of operators
The set Qpk
dpk=1 of operators in a Hilbert space is said to be an irreducible tensorial set
iff there exists an irrep Dp of G with dimension dp, such that for all T ∈ G
P (T )QpjP (T )
−1 =
dp∑
k=1
QpkD
p(T )kj. (4.40)
Example 23Let φ(~r) be an arbitrary element of L2 and ψp
k a basis for the irrep Dp of G. Define theoperator Qp
k byQp
kφ(~r) = ψpk(~r)φ(~r), (k = 1, . . . , dp). (4.41)
Then we obtain for all elements T of G
P (T )QpkP (T )
−1φ(~r) = P (T )[
ψpk(~r)P (T )
−1φ(~r)]
= P (T )ψpk(~r) · P (T )
[
P (T )−1φ(~r)]
=∑
l
ψpl (~r)D
p(T )lkφ(~r)
=∑
l
QplD
p(T )lkφ(~r). (4.42)
(Note that the action of P (T ) on a product ψφ is the product P (T )ψP (T )φ.) As φ(~r)was arbitrary, we find that multiplication of an element of L2 by a basis element for Dp
is an irreducible tensor operator.
The next theorem states an important result for the matrix elements of irreducibletensor operators: their dependence on the row index of the basis elements is given by theCG coefficients.
4.4 The Wigner-Eckart Theorem
The matrix elements of irreducible operators have an interesting property: They can besplit in a factor that is independent of the dynamics of the system at hand and one that
37
contains the dynamics. The former is a CG coefficient of the irreps involved and thelatter is the so-called reduced matrix element. This factorization is very useful for theestablishment of selection rules. We formulate this property in the form of a famoustheorem and give a simple example later.
Theorem 17 (Wigner-Eckart)Let G be a group of coordinate transformations (of finite order), Dp, Dq and Dr unitary
irreps of G, with dimensions dp, dq and dr respectively. Let ψpj
dpj=1 and ψr
kdrk=1 be bases
for Dp and Dr and let Qql
dql=1 be an irreducible tensorial set for Dq. Then
(ψrk, Q
qlψ
pj ) =
nrpq∑
α=1
(
qp rαlj k
)∗(r‖Qq‖p)α (4.43)
where (r‖Qq‖p)α is a set of reduced matrix elements that are independent of the indices
j, l, and k and
(
qp rαlj k
)
are the CG coefficients in a unitary rep.
Proof
For all T ∈ G there is a unitary operator P (T ), so
(ψrk, Q
qlψ
pj ) = (P (T )ψr
k, P (T )[
Qqlψ
pj
]
)
= (P (T )ψrk, P (T )Q
qlP (T )
−1P (T )ψpj ) (4.44)
The following three equations
P (T )ψrk =
∑
k′
ψrk′D
r(T )k′k,
P (T )ψpj =
∑
j′
ψpj′D
p(T )j′j,
P (T )QqlP (T
−1) =∑
l′
Qql′D
q(T )l′l, (4.45)
imply
(ψrk, Q
qlψ
pj ) =
∑
k′
∑
l′
∑
j′
Dr(T )∗k′kDq(T )l′lD
p(T )j′j(ψrk′ , Q
ql′ψ
pj′). (4.46)
The l.h.s. is independent of T , so we may sum over T running over all of G and divideby g:
(ψrk, Q
qlψ
pj ) =
1
g
∑
k′
∑
l′
∑
j′
(
∑
T∈GDr(T )∗k′kD
q(T )l′lDp(T )j′j
)
(ψrk′, Q
ql′ψ
pj′). (4.47)
The sum over T can be expressed in terms of CG coefficients. We demonstrate this asfollows. As the CG coefficients form a unitary matrix, the product ψq
l ψpj can be expanded
as follows
ψql ψ
pj =
∑
r
nrqp∑
α=1
dr∑
s=1
(
rα qps lj
)
θr,αs . (4.48)
Acting on this product with the projector Prk′k, defined in the usual way
Prk′k =
drg
∑
T∈GDr(T )∗k′kP (T ) (4.49)
38
we find
Prk′k
∑
α,s
(
rα qps lj
)
θr,αs =∑
α
(
rα qpk lj
)
θr,αk′
=∑
α
(
rα qpk lj
)
∑
l′,j′
(
qp rαl′j′ k′
)
ψql′ψ
pj′. (4.50)
and also
Prk′k
[
ψql ψ
pj
]
=drg
∑
T
Dr(T )∗k′k [P (T )ψql ][
P (T )ψpj
]
=drg
∑
T
Dr(T )∗k′k∑
l′
∑
j′
ψql′D
q(T )l′lψpj′D
p(T )j′j . (4.51)
Equating Eq. (4.50) and Eq. (4.51) we find, using the fact that the basis functions areindependent,
∑
T
Dr(T )∗k′kDq(T )l′lD
p(T )j′j =g
dr
∑
α
(
qp rαl′j′ k′
)(
rα qpk lj
)
. (4.52)
To obtain the desired form, we replace T by T−1, then
Dp(T−1)j′j =[
Dp(T )−1]
j′j= Dp(T )∗jj′ (4.53)
for the unitary irrep Dp, so∑
T∈GDr(T−1)∗k′kD
q(T−1)l′lDp(T−1)j′j =
∑
T∈GDr(T )kk′D
q(T )∗ll′Dp(T )∗jj′
=g
dr
∑
α
(
qp rαlj k
)∗(rα qpk′ l′j′
)∗
=g
dr
∑
α
(
qp rαlj k
)∗(qp rαl′j′ k′
)
. (4.54)
as the CG coefficient matrix was supposed to be unitary. So, finally
(
ψrk, Q
qlψ
pj
)
=1
dr
∑
k′
∑
l′
∑
j′
∑
α
(
qp rαlj k
)∗(qp rαl′j′ k′
)
(
ψrk′ , Q
ql′ψ
pj′
)
. (4.55)
If we define the reduced matrix element as
(r‖Qq‖p)α =1
dr
∑
k′
∑
l′
∑
j′
(
qp rαl′j′ k′
)
(
ψrk′, Q
ql′ψ
pj′
)
(4.56)
we see that it is independent of the indices k′, l′ and j′, so the stated result is obtained.2
Example 24 The Wigner-Eckart theorem is used ubiquitously in applications of rota-tional symmetry. Take as an example the integral over three spherical harmonics
〈l′m′|Yλµ|lm〉 =∫ 2π
0
dφ
∫ π
0
dθ sin θY ∗l′m′(θ, φ)Yλµ(θ, φ)Ylm(θ, φ) = (l′m′|λµ, lm)(l′‖Yλ‖l).
(4.57)
39
Remark: Although the spherical harmonics Ylm are complex functions, the CG-coefficientsturn out to be real.
Here we use the notation for the CG-coefficients that is commonly used in the literatureon the reps of the rotation group:
(l′m′|λµlm) =
(
lλ l′
mµ m′
)
. (4.58)
The CG coefficient (l′m′|λµ lm) vanishes if λ > |l′ − l| or m′ 6= m + µ. In this case thereduced matrix element is
(l′‖Yλ‖l) =√
(2l + 1)(2λ+ 1)
4π(2l′ + 1)(l′0|λ0, l0). (4.59)
Another application is the matrix element of a vector operator, say the magnetic dipolemoment ~d. A vector operator transforms under rotations as a spherical harmonic withangular momentum λ = 1. in this case one can use the same formula as the one above,but the reduced matrix element must be determined separately. The selection rule nowgives that any matrix element of a magnetic dipole vanishes unless the angular momentaof the initial and final states, j and j′, differ by at most 1. A special case is j = j′ = 0:The magnetic moment of a state (particle) with spin 0 must vanish.
4.5 Representations of Direct Product Groups
As a final topic in this chapter we discuss reps of direct product groups, which should notbe confused with direct product reps discussed before.
If D1 and D2 are two reps of groups G1 and G2 resp., then the matrices D((T1, T2))defined for all T1 ∈ G1 and T2 ∈ G2 by
D((T1, T2)) = D1(T1)⊗D2(T2) (4.60)
form a rep of G1 ⊗ G2. It is not hard to show that if D1 and D2 are unitary, then D isunitary. Furthermore, if D1 and D2 are irreps, then D is also an irrep. In this sense Eq.(4.60) is quite different from Eq. (4.16).
If G1 and G2 are both isomorphic to the same group G, then D1 and D2 are two irrepsof G, say Dp and Dq respectively. In that case D((T1, T2)) = Dp(T1)⊗Dq(T2) is an irrepof G ⊗ G. The so-called “diagonal subgroup” formed by elements (T, T ) is isomorphicwith G. In that case D((T, T )) = Dp(T )⊗Dq(T ), which was called D(T ) before in Eq.(4.16), forms a reducible rep of G.
In particle physics one encounters direct product groups of the form G ⊗ G, such asone of the symmetry groups related to the strong interactions: SU(n) ⊗ SU(n). Onecan encounter this product for n = 2 where one SU(2) refers to the quantum numberspin and the other to isospin. It is clear that making a Clebsch-Gordan decomposition ofthis direct product does not make sense. Another case is when the SU(n)’s refer to thechirality of fermions (left- or righthanded) and where n refers to the number of fermion“flavors”. Subscripts L and R are usually attached, i.e. SU(n)L ⊗ SU(n)R, to indicate
40
the distinction between the two SU(n)’s. The diagonal subgroup belonging to this directproduct group is usually called SU(n)V .
This concludes the part of the formal theory of groups of finite order that we wantedto discuss. In the next two chapters we shall give some examples where the formal theoryis applied.
4.6 Exercises
Exercise 4.1
The order of the indices in Eq. (4.4) is essential. Show that in case we would write D(S)knin Eq. (4.4), then D would not be a rep.
Exercise 4.2
Prove Theorem 15.
Exercise 4.3
Show that if D1 and D2 are irreps of the groups G1 and G2 respectively, that then D asdefined in Eq. (4.60) is an irrep of the direct product group G1 ⊗G2.
Exercise 4.4
Write down a direct product rep of a direct product group.
41
42
Chapter 5
Symmetries of Molecules and Solids
In this chapter we discuss the application of the theory of group reps to systems consistingof sets of identical entities–say atoms–placed in a certain ordered configuration in three-dimensional space. The symmetries we shall discuss are translations or reflections androtations in R
3. The latter belong to the group O(3), which we discuss first.
5.1 O(3) and SO(3)
A vector in R3 is an object ~x =
∑3i=1 xiei, where the set e1, e2, e3, is an orthonormal
basis of R3 and the coefficients, or vector components xi are real numbers. The groupO(3) is the set of all transformations of the basis vectors that leave their length invariant.The length of a vector ~x, denoted by ‖~x‖, is defined by
‖~x‖ =
3∑
i=1
x2i . (5.1)
Apparently, the vectors ei have unit length.In the vector space R
3 basis transformations T take the form
e′j = T ej =3∑
i=1
eiTij (5.2)
Upon a basis transformation the components of the vector ~x with respect to the new basisare also transformed. We find
~x =3∑
i=1
xiei =3∑
j=1
x′j e′j =
3∑
i=1
3∑
j=1
x′j eiTij . (5.3)
So we see that
xi =3∑
j=1
Tijx′j . (5.4)
T belongs to O(3) if the length of ~x is invariant, i.e.,∑
i x2i =
∑
j x′2j , from which we find
3∑
i=1
TijTik = δjk, (j, k = 1, 2, 3). (5.5)
43
Matrices that have this property are said to be orthogonal. As a consequence, det T = ±1.If det T = 1, T is said to belong to SO(3).
(This is a convention used generally. Suppose a group of invertible matrices is denotedby “G”, then the subgroup of matrices with determinant equal to unity is denoted by“SG”.)
LemmaIf T ∈ SO(3) then there exists a vector ~f with length ‖~f‖ = 1, such that T ~f = ~f .
Proof
T ~f = ~f means: the matrix T has an eigenvalue equal to 1. This means in its turn thatdet[T − I] = 0, where I = diag(1, 1, 1) is the identity matrix in three dimensions. Now
det[T − I] = det[TT − IT] = det[T−1 − I]
= det T−1 det(−I) det[T − I]
= − det[T − I],
where TT is the transpose of the matrix T , being equal to T−1 and having a determinantequal to 1 as T ∈ SO(3). So we can conclude that det[T − I] = 0: there is an eigenvalueequal to 1.2
The eigenvector may not be unique. This is easy to see: if we take for T the identityI, than any vector different from the null vector is an eigenvector with eigenvalue 1. For agiven ~f of length 1, we may find two normalized vectors ~g and ~h such that ~f,~g,~h forman orthogonal basis for R3. In this basis the matrix T has the form
T =
1 0 00 cos θ sin θ0 − sin θ cos θ
. (5.6)
for some angle θ.Clearly, Tr T = 1 + 2 cos θ. As an application of the foregoing we can conclude that
all rotations about any axis over an angle θ are equivalent: they belong to the same classof SO(3).
Definition Space inversion P
The space inversion P is the transformation that reverses the direction of all basis vectors
P ei = −ei, (so, x′i = −xi). (5.7)
This operation has the property P 2 = I so P−1 = P .Let T ′ ∈ O(3) with determinant det T ′ = −1. Then T = T ′P ∈ SO(3). Consequently,
T ′ = TP , so we can conclude that the elements of O(3) can be written as either a rotation,in which case they belong to SO(3) or as a rotation times the inversion.
5.2 The Euclidian Group
The Euclidian group consists of the transformations of solid bodies in space. Only trans-formations that leave distances and angles unchanged are included.
44
Definition isometry
A transformation T is said to be an isometry if for any two vectors ~x, ~y ∈ R3, T preserves
the length of their difference: ‖T~x− T~y‖ = ‖~x− ~y‖.Definition Euclidian Group
The set of all isometries of R3 is said to be the Euclidian Group in 3 dimensions, denotedby E(3).
Definition translation
The operator T~a is said to be a translation, iff
T~a~x = ~x+ ~a, (~x,~a ∈ R3). (5.8)
Remark
Rotations are linear transformations, translations are not. The translations form anAbelian subgroup of E(3).
In what follows we shall denote the neutral element of R3 by ~0.Every T ∈ E(3) can be uniquely written in the form
T = T~aO ≡ (~a|O), O ∈ O(3). (5.9)
The elements of O(3) are uniquely defined by the property that they leave ~0 invariant.Consider the mapping φ : (~a|O) 7→ detO; this map is a homomorphism from E(3) to
(1,−1;×). The ker of this homomorphism is called E+(3). It is the proper Euclidian
group or the group of rigid motions. E+(3) is an invariant subgroup of E(3).
5.3 Symmetries of Molecules
The term “molecule” is used here in a geometrical sense: it is an ordered set M of pointsin R
3. These points may not all be equivalent: they may be occupied by different “atoms”.
Definition symmetry group
The symmetry group of a molecule M is the subset of E(3) that leaves the moleculeinvariant: G = T ∈ E(3)|TM =M.Definition G-orbit
The set T~x : T ∈ G is said to be the G-orbit of the vector ~x.
There are two different types of symmetry groups: those containing only rotations,i.e., subgroups of SO(3), and those containing reflections as well: G ⊂ O(3), G 6⊂ SO(3).
(For a detailed classification of finite symmetry groups one may consult the literature,e.g. Cornwell 1984.)
5.4 Example: Vibrations of Molecules
The physics of vibrations of molecules can be studied in the classical as well as thequantum domain. In this section we shall limit ourselves to the classical case.
Suppose the molecule M contains N atoms with (generalized) coordinate vectors~q1, . . . , ~qN . It is sometimes more convenient to renumber the coordinates. In three dimen-sions:
q1x, q1y, q1z, . . . , qNx, qNy, qNz → q1, q2, q3, . . . , q3N. (5.10)
45
It is a natural choice for the generalized coordinates to take them proportional to thedifferences of the actual and the equilibrium positions of the atoms. If mk is the mass ofthe atom Ak and ~r its equilibrium position, then we define the displacements
q3k−2 =√mk(xk − xk), q3k−1 =
√mk(yk − yk), q3k =
√mk(zk − zk). (5.11)
The factor√mk is introduced to reduce T to the simple form
T = 1
2
D∑
m=1
q2m, D = 3N. (5.12)
If the qm are indeed chosen such that qm = 0, ∀m corresponds to the situationwhere all atoms are in their equilibrium position, then the approximation of small vibra-tions amounts to truncating the Taylor series for the potential
V (q1, . . . , qD) = V (~0) + 1
2
∂2
∂qi∂qjV (q1, . . . , qD)
∣
∣
∣
∣
~q=~0
qiqj + . . . (5.13)
after the quadratic terms. (The first order terms vanish because ~0 is taken to be the pointof equilibrium.) By a shift in the energy scale we can make V (~0) to vanish. Then we findin the quadratic approximation
V (q1, . . . , qD) = 1
2
D∑
i=1
D∑
j=1
Fijqiqj . (5.14)
The matrix F is called the force matrix. (Actually, the force is the first derivative of thepotential, so this is a misnomer.)
If we collect the coordinates in a column vector q we can write the kinetic and potentialenergies in this approximation as quadratic forms
T = 1
2qTq, V = 1
2qTFq. (5.15)
The object that defines the dynamics is the Lagrangean L = T − V . The equations ofmotion are, of course, the Euler-Lagrange (EL) equations
d
dt
(
∂L
∂qm
)
=∂L
∂qm, (5.16)
which become in the case considered
qm +D∑
n=1
Fmnqn =D∑
n=1
(δmnqn + Fmnqn) = 0. (5.17)
If F would be diagonal, then a solution would be that every qm behaves as a harmonicoscillator. Our strategy here is to use symmetry arguments as much as possible to performthe diagonalization. We substitute tentatively the form
qm = cm cos(ωt+ α). (5.18)
46
Upon substitution in the EL equations we find
Fc = ω2c. (5.19)
The physics of our problem limits F to a real, symmetric matrix. So F is hermitian andcan be diagonalized by an orthogonal matrix
S−1FS = diag(λ1, . . . , λD). (5.20)
It is well known that the columns of S are the normalized eigenvectors c(m) of F , corre-sponding to the eigenvalues λm = ω2
m
S = ( c(1) | c(2) | . . . | c(D) ). (5.21)
A solution of the EL equations qm = c(n)m cos(ωnt+ αn) is said to be a normal mode. The
general solution is
qm =
D∑
n=1
αnc(n)m cos(ωnt+ δn). (5.22)
The sets αnDn=1 and δnDn=1 are determined by the initial conditions qm(0) = am,qm(0) = bm, m = 1, . . . , D.
Definition normal coordinates
The coordinates Qm =∑
n(ST)mnqn =
∑
n qnSnm are said to be the normal coordinatesof the molecule M .
Remark
The Hamiltonian H = T +V becomes the sum of harmonic oscillator hamiltonians in thenormal coordinates:
H =∑
m
(
1
2P 2m + 1
2ω2mQ
2m
)
(5.23)
where
Pm =∂L
∂Qm
= Qm. (5.24)
The coordinates qm correspond to any position of the atoms A1, . . . , AN . However, thecase we consider, small oscillations, is characterized by equilibrium positions that definethe symmetry of the molecule. We anticipate that this symmetry influences the possiblevibrations.
So far we have not used the symmetry of the molecule. Now we discuss the limitationsthe symmetry group G imposes on the vibrations of M . Let T ∈ G be a symmetryoperation, then we can write M ′ = TM , i.e., ~r
′k = R(T )~rj, which expresses the fact that
an element of G permutes the particle labels. (We introduce the notation R(T ) for thetransformation in R
3 in order to distinguish the element of the abstract group G fromthe corresponding transformation in coordinate space.) If the same R(T ) is acting on theactual coordinates ~riNi=1, then the kinetic and potential energies must be invariant.
Till now we have considered the effects of symmetry operations on the coordinates ofthe atoms. The solutions were, however, written in terms of the generalized coordinates,so we shall now see what the effect of an element of G is on the column vectors q. Indeed,
47
the effect of T ∈ G on the displacements is twofold: the atoms are permuted and thedisplacements may be rotated:
Ddisp(T ) = p(T )⊗ R(T ), (5.25)
where Ddisp is aD×D matrix, p(T ) an N×N matrix and R(T ) a 3×3 matrix. If the atomsare in their equilibrium positions then the rotation has no effect as the displacements arezero in that case.
The character of Ddisp, say χDdisp
, can be computed easily. Let u(T ) be the numberof atoms not moved by T and θ be the angle of rotation, then
χDdisp
(T ) = u(T )(1 + 2 cos θ) detR(T ). (5.26)
Ddisp is unitary as it is the direct product of permutation matrices and orthogonal3× 3 matrices. We can find a unitary matrix U that decomposes Ddisp:
U−1DdispU ≡ D′(T ) =⊕
p
npDp(T ). (5.27)
The invariance of the potential energy V entails the invariance of the force matrix Fbecause we have for the coordinates q′ = Ddisp(T )q and V (q′1, . . . , q
′D) = V (q1, . . . , qD),
from which we derive
V = 1
2qTFq = 1
2q′TFq′ = 1
2qTDdispT(T )FDdisp(T )q. (5.28)
As this identity must be valid for any small displacement vector q, we must have
DdispT(T )FDdisp(T ) = F. (5.29)
This means that F is form invariant, because the dependence of V on any qi is the sameas its dependence on q′i after the transformation Ddisp is made. Ddisp is real and unitary,so DdispT(T ) = Ddisp(T )−1, hence
FDdisp(T ) = Ddisp(T )F. (5.30)
Now we perform on F the unitary transformation that reduced Ddisp to D′
F ′ ≡ U−1FU. (5.31)
Then, if we write the block structure of D′ and F ′ as follows
D′(T ) =
D1 ∅. . .
D1
D2
. . .
D2
. . .
Dω
. . .
∅ Dω
(5.32)
48
and
F ′ =
F ′11αβ F ′12
αβ . . . F ′1ωαβ
F ′21αβ F ′22
αβ...
. . .
F ′ω1αβ F ′ωω
αβ
, (5.33)
then we obtain from the identity F ′D′ = D′F ′
Dp(T )F ′pqαβ = F ′pq
αβDq(T ), (5.34)
where F ′pqαβ is a submatrix of dimension dp×dq, α = 1, . . . , np, β = 1, . . . , nq. As the irreps
Dp and Dq could be taken unitary, Schur’s lemma’s tell us
F ′pqαβ =
[∅] p 6= qf pαβIdp p = q
(5.35)
where Idp is the dp × dp identity matrix. For fixed p the f pαβ are n2
p complex numbersthat form a Hermitian matrix.
A further similarity transformation with an orthogonal matrix W will bring F ′ into ablock-diagonal form
F ′′ ≡W−1F ′W =
f 111 . . . f 1
1n1∅
......
f 1n11 . . . f 1
n1n1
. . .
fω11 . . . fω
1nω
......
∅ fωnω1 . . . fω
nωnω
=
f 1 ∅f 2
. . .
∅ fω
,
(5.36)where the block f p is np dimensional. Finally, the matrixWU brings F to this form, so wemay summarize the procedure followed by saying that we have succeeded to diagonalizeF partially by group-theoretic considerations.
The eigenvalue equation det |F − λI| = 0 splits into equations det |f p − λpI| = 0. Asevery matrix f p is occurring dp times, we see that every eigenvalue λpα, α = 1, . . . np, is dpfold degenerate.
In order to illustrate these ideas in a few simple cases, we discuss in some detail thediatomic molecule and a triatomic one.
The diatomic molecule
Consider a diatomic molecule, consisting of two identical atoms, say H2, N2 or O2. Theequilibrium positions are as shown in Fig. 5.1. This system is considered in one dimension.Then the two atoms have the coordinates x1 and x2. Our coordinate vector has the formx = (x1, x2).
We suppose that the forces between the atoms follow Hooke’s law with respect to anequilibrium position. Let the equilibrium positions of the two atoms be x = −d/2 andx = d/2, respectively and their displacements x1 and x2. Then the distance between theatoms is
∆ = |(−d/2 + x1)− (d/2 + x2)| = | − d+ x1 − x2|. (5.37)
49
For small oscillations, |x1 − x2| ≪ d we find
V = 1
2k(∆− d)2 = 1
2k(x1 − x2)
2 = 1
2k(x21 − 2x1x2 + x22) ≡ 1
2Fijxixj , (5.38)
so the matrix F is
F = k
[
1 −1−1 1
]
. (5.39)
It is immediately clear that the symmetry group of the diatomic molecule is S2, thegroup of permutations interchanging the atoms. This group is very simple. It consists ofthe identity E and the reflection S, with E(x1, x2) = (x1, x2) and S(x1, x2) = (−x2,−x1).
A B
x0
Figure 5.1: Geometrical representation of a diatomic molecule.
Now we consider the displacement rep. We have indicated the displacements in Fig. 5.1with the fat arrows. In this rep the potential has the same form as in Eq. (5.38). (Weshall see later that this may become different in other cases.)
In the space of the particle labels the permutations are given by
D(E) =
[
1 00 1
]
, D(S) =
[
0 −1−1 0
]
. (5.40)
The group S2 is Abelian, so its classes consist of single elements: [E] = E, [S] = S .The characters are given by
χ([E]) = 2, χ([S]) = 0. (5.41)
The two different irreps of S2 are one-dimensional, as S2 is Abelian. They are givenin Tab. 5.1
Table 5.1: Character table for the irreps of S2
χ1 χ2
E 1 1S 1 -1
The characters of the rep D are given in Tab. 5.2.The matrices Ddisp(T ), T ∈ G form a rep of G, but not an irrep, as is clear from the
character table above: dimDdisp = dimD = 2, whereas the irreps of S2 have dimension 1.Using the orthogonality theorem for characters we can calculate the Clebsch-Gordan
series D = n1D1 ⊕ n2D
2. We find n1 = n2 = 1.Now we shall show that one can determine the eigenvalues and their degeneracy with-
out diagonalizing the matrix F defined by Eq. (5.38). The technique we use is to take
50
Table 5.2: Character table for the rep D
χE 2S 0
traces of products of elements of the symmetry groups and F in the displacement rep.Suppose F has been diagonalized to
F ′ =
[
λ1 00 λ2
]
. (5.42)
We find for the action of an element of S2 in the rep D′, where F is diagonalized
D′(X)F ′ =
[
λ1D1(X) 00 λ2D
2(X)
]
. (5.43)
Take traces:Tr(D′(X)F ′) = Tr(D(X)F ) = λ1χ
1(X) + λ2χ2(X), (5.44)
from which we find
Tr(D(E)F ) = λ1 + λ2 = 2k,
Tr(D(S)F ) = λ1 − λ2 = 2k. (5.45)
Because both irreps are contained just once in the rep D, the eigenvalues are nondegen-erate, except for accidental degeneracy. We find
λ1 = 2k, λ2 = 0. (5.46)
In order to understand the meaning of the eigenvalue 0 we use elementary methods. Thetransformation
x =1√2(x1 − x2), X =
1√2(x1 + x2) (5.47)
separates the center of mass X from the relative coordinate. The Hamiltonian H = T +Vcan easily be written in terms of x and X :
H = 1
2X2 + 1
2x2 + 1
2(2k)x2. (5.48)
This corresponds to the diagonalized matrix F ′:
F ′ = 2k
[
1 00 0
]
. (5.49)
The motion of the center of mass is linear, so the eigenvalues we found correspond to thetranslation of the center of mass, frequency 0, and the vibration in the relative motion,frequency
√2k.
The vibrations of the diatomic molecule are of a one-dimensional nature. Still, wecould discuss them in a two-dimensional space, where the two dimensions are associated
51
y
A B
x
Figure 5.2: Geometrical representation of a diatomic molecule in two dimensions.
with the coordinates of the two particles. In reality the diatomic molecule is embeddedin a higher dimensional space. We shall now place the molecule in two-dimensional spaceand discuss the consequences for the results. The situation is depicted in Fig. 5.2. Thetwo equilibrium positions are (−d/2, 0) and (d/2, 0). The displacements are (x1, y1) and(x2, y2). The potential energy is in the case Hooke’s law applies
V = 1
2k(∆− d)2. (5.50)
We compute the distance d12 in a way similar to the one used for the one-dimensionalcase. If we write
~ξ1 = (−d/2 + x1, y1), ~ξ2 = (−d/2 + x2, y2), (5.51)
then we find for the distance between the two atoms
∆2 = (~ξ1 − ~ξ2)2 = (d− x1 + x2)
2 + (y1 − y2)2 ≈ d2 − 2d(x1 − x2) +O(x2i , y
2i ), (5.52)
where again the approximation of small oscillations is made. Taking the square root andapproximating again gives
|∆− d| ≈ |x1 − x2|. (5.53)
So we find the same formula as before, Eq. (5.38), as expected, but now this expressionneeds to be used in a four-dimensional context. The group of symmetries is larger. Besidesthe reflection S we had before, we have also a rotation about the z-axis. We define thematrices
r =
[
−1 00 −1
]
, s =
[
−1 00 1
]
. (5.54)
These transformations affect the individual displacements vectors. Furthermore there isthe interchange of the atoms by the rotation and the reflection that must be taken intoaccount. The full 4× 4 matrices are then
D(R) =
[
0 rr 0
]
, D(S) =
[
0 ss 0
]
. (5.55)
The full group contains also the product of a rotation and a reflection. It has the repre-sentation
D(RS) =
[
rs 00 rs
]
. (5.56)
The group table is shown below.
52
Table 5.3: Group table for the symmetry of the diatomic molecule in two dimensions.
R S RSR E RS SS RS E RRS S R E
Table 5.4: Characters for the symmetry of the diatomic molecule in two dimensions.
E R S RSD1 1 1 1 1D2 1 1 −1 −1D3 1 −1 1 −1D4 1 −1 −1 1
The symmetry group is Abelian: SR = RS. For that reason all irreps are one-dimensional. There are four of them. Two are easy to find: the trivial one D1(X) = [1]and the antisymmetric one D2(X) = [1] for X = E,R, and D2(X) = [−1] for X = S,RS.As X2 = E for all X we find that the matrix elements must be 1 or −1. Then theorthogonality theorem of irreps leaves us no choice for the two other characters than theone shown in Tab. 5.4. If we denote the character of the displacement rep simply by χ,then we find
χ(E) = 4, χ(R) = 0, χ(S) = 0, χ(RS) = 0. (5.57)
A straighforward computation gives then
D′ = D1 ⊕D2 ⊕D3 ⊕D4. (5.58)
Using the same trick as before, we calculate the eigenvalues λ1, . . . , λ4 of F . In the two-dimensional case the matrix F is
F =
1 0 −1 00 0 0 0
−1 0 1 00 0 0 0
. (5.59)
Taking traces we find
λ1 + λ2 + λ3 + λ4 = k
λ1 + λ2 − λ3 − λ4 = k
λ1 − λ2 + λ3 − λ4 = k
λ1 − λ2 − λ3 + λ4 = k (5.60)
The solution isλ1 = 2k, λ2 = 0, λ3 = 0, λ4 = 0. (5.61)
The interpretation is now clear. Besides the true vibration, corresponding to λ1 = k, wefind a rotation and two translations, one in the x- and one in the y-direction, all of whichcorrespond to zero frequency.
53
C
BA
y
x
Figure 5.3: Geometrical representation of a triatomic molecule in two dimensions.
Plane triangle
As a next level of complication we consider a triatomic molecule with as an equilibriumconfiguration an equilateral triangle. It is depicted in Fig. 5.3. The equilibrium positionsof the three atoms are
a0 = (−1
2
√3,−1
2) , b0 = ( 1
2
√3,−1
2) , c0 = (0, 1). (5.62)
Following the procedure used before, we find for the expression of the potential in termsof the displacement vectors
V = 1
2k
(xa − xb)2 + [ 1
2(xb − xc)− 1
2
√3(yb − yc)]
2 + [ 12(xa − xc) + 1
2
√3(ya − yc)]
2
(5.63)We can easily reconstruct the matrices of the symmetries of this triangle in the displace-ment rep, if we use the results of Sect. 2.3. For the permutations of the atoms we havefor instance
R =
0 0 11 0 00 1 0
, Sc =
0 1 01 0 00 0 1
, (5.64)
and for the displacements
r =
[
−1
2−1
2
√3
1
2
√3 −1
2
]
, sc =
[
−1 00 1
]
. (5.65)
Using these results together we find for the displacements rep the 6× 6 matrices
D(R) =
0 0 rr 0 00 r 0
, D(Sc) =
0 sc 0sc 0 00 0 sc
. (5.66)
The characters areχ(E) = 6, χ(R) = 0, χ(Sc) = 0. (5.67)
The orthogonality theorem gives for the CG series
D′ = D1 ⊕D2 ⊕ 2D3. (5.68)
54
A
A
x
A
A
2
3
1
4
z
y
Figure 5.4: Geometrical representation of the ammonia molecule.
Taking traces of products of the potential matrix F with the different matrices of D wefind again a set of linear equations for the eigenvalues λi:
λ1 + λ2 + 2(λ31 + λ32) = 6k
λ1 + λ2 − (λ31 + λ32) = 3k/2
λ1 − λ2 = 3k (5.69)
The two eigenvalues λ31 and λ32 correspond to the two mixed-symmetric irreps in the CGseries. The solution is
λ1 = 3k, λ2 = 0, λ31 + λ32 = 3k/2. (5.70)
The physical argument that the plane triangle has three symmetries that are not vibra-tional, like the diatomic molecule, tells us that we must find λ31 = 0, λ32 = 3/2k. Soagain, using some physical arguments and purely algebraic methods we have succeeded toanalyze the spectrum of vibrations: there are two vibrational modes, one with eigenvalue3k, the other with 3/2k. The latter has degeneracy 2.
An example worked out in some detail by Cornwell is the ammonia molecule.
The NH3 molecule
Consider the ammonia molecule, consisting of three hydrogen atoms and one nitrogenatom. We assign coordinates for the equilibrium positions as shown in Fig. 5.4:
~r1 =(
−a2
√3,−a
2, 0)
, ~r2 =(a
2
√3,−a
2, 0)
, ~r3 = (0, a, 0) ,
~r4 = (0, 0, c) .
The atoms A1, A2, and A3 are hydrogen atoms, forming an equilateral triangle and A4 isthe nitrogen atom.
It is immediately clear that the symmetry group of the NH3 molecule is S3, thesymmetry group of the equilateral triangle of the H atoms. The permutations are givenin the space of the four particle labels by
p(T ) =
T ′ 000
0 0 0 1
, (5.71)
55
where T ′ is one of the 3×3 matrices given in sect. 2.3 and also used above for the triatomicmolecule. As before, the classes are [E] = E, [R] = R,R2 and [S] = SA, SB, SC.Again, Ddisp is given by Eq. (5.25). The matrices R(T ) have dimension 3. The charactersare computed using Eq. (5.26) and are found to be
u([E]) = 4, θ = 0, 1 + 2 cos θ = 3, χ([E]) = 12,u([R]) = 1, θ = 120, cos θ = −1
2, χ([R]) = 0 ,
u([S]) = 2, χ(T ′) = 1, χ([S]) = 2.
The matrices Ddisp(T ), T ∈ G form a rep of G, but not an irrep, as is clear from thecharacter table above and application of the criterion for irreducibilty given by Eq. (3.34).(Further details can be found in Cornwell.)
5.5 Translation Symmetry
If atoms are ordered in a lattice, we have, besides the symmetry group that is necessarilya subgroup of O(3), also translation symmetry.
An infinite three-dimensional lattice is defined by(i) basic lattice vectors ~a1,~a2,~a3,(ii) lattice coordinate vectors ~n = (n1, n2, n3) with integer components.
The vectors ~ai define the orientations of the lattice. Any element (atom) of the latticeis characterized by a coordinate vector ~n such that its position is given by
~r = ~r0 + ~tn = ~r0 + n1~a1 + n2~a2 + n3~a3 (5.72)
for some ~n and ~r0. The latter is connected to the choice of the origin of the coordinatesystem, so it can be set to ~0 by a translation of the origin.
The subgroup T∞ of E(3), given by
T∞ = (~tn|E), (5.73)
is said to be the group of primitive translations of the lattice. (E is the identity in O(3).)The set of all pure rotations ~0|R, R ∈ SO(3), such that for all ~tn the vector R~tn isagain a lattice vector, is said to be the maximal point group of the lattice.
The 14 different types of lattices–the Bravais lattices–can be found in Cornwell 1984.We shall not discuss them here in detail.
The fact that T∞ is an infinite Abelian group, which means that it has an infinity ofone-dimensional irreps, makes it difficult to work with. For that reason alone it would beadvantageous to make an approximation that limits the translations taken into accountto a finite, possibly very large number. Fortunately, real crystals are necessarily finite.Therefore translational symmetry is also broken in nature, but only to the extent thatsurface effects are important. For crystals having sizes much larger than the lattice spacingwe may ignore these effects. (Lattice defects like vacancies or impurities also break thesymmetry.)
A simplification is achieved if the group T∞ is replaced by a group corresponding to acyclic repetition of the lattice: the group T∞ is replaced by a torus T , if for some values(N1, N2, N3) the translations (Ni~ai|E) are identified with the identity (~0|E). Physically
56
it means that periodic boundary conditions are imposed. The mapping of T∞ onto thetorus T is a homomorphism. Its ker is characterized by (
∑
i liNi~ai|E), li ∈ Z.T is of finite order g = N1N2N3 and Abelian, so it possesses g inequivalent irreps of
dimension 1. As (~ai|E)ni = (ni~ai|E), we have T = T1 ⊗ T2 ⊗ T3, where Ti is the cyclicgroup of order Ni. All irreps have the form
(~aj |E) 7→ exp(2πipj/Nj), (j = 1, 2, 3), pj ∈ Z. (5.74)
Consequently, the irreps are given by
D~p((∑
j
nj~aj |E)) = exp(−2πi∑
j
njpj/Nj). (5.75)
As (Nj~aj |E) = (~0|E), one must find exp(−2πi(pj + Nj)/Nj) = exp(−2πipj/Nj) so onlyNj different values of pj are allowed, say 0,1, . . . , Nj − 1.
Definition reciprocal lattice
The reciprocal lattice is defined by the basis vectors ~b1,~b2,~b3 such that
~bi = 2πǫijk~aj × ~ak/~ai · (~aj × ~ak). (5.76)
“Allowed” k vectors are~k =
∑
i
ki~bi, ki = pi/Ni, (5.77)
whence~k · ~tn = 2π
∑
j
(pjnj/Nj), (5.78)
orD
~k((∑
j
nj~aj|E)) = exp(−i~k · ~tn). (5.79)
The basis vectors transforming according to D~k must have the property
P (T )ψ~k(~r) = D
~k(T )ψ~k(~r), (5.80)
which meansψ~k(~r − ~tn) = exp(−i~k · ~tn)ψ~k(~r). (5.81)
From this relation it follows that if we write
ψ~k(~r) = ei
~k·~ru~k(~r), (5.82)
then u~k(~r) is a periodic function such that
u~k(~r − ~tn) = u~k(~r), ∀~tn. (5.83)
This is the Bloch condition.
57
5.6 Exercises
Exercise 5.1
If T is an element of O(3), prove that the determinant of T is 1 or −1.
Exercise 5.2
The matrix A is defined as
A =
1 0 00 −1 00 0 1
. (5.84)
(a) Show that A ∈ O(3)(b) A can be written as the product of a rotation R and the space inversion P : A = RP .Determine the rotation axis and the rotation angle.
Exercise 5.3
The lemma in Sect. 5.1 states that any matrix T that belongs to SO(3), has at least oneeigenvalue equal to 1. Formulate a similar lemma about the occurrence of the eigenvalues1 and −1 of the matrices that belong to O(3), and prove it.
Discuss in particular the situation that T has two eigenvalues equal to 1 or equal to−1.
58
59
Chapter 6
The Symmetric Groups Sn
The set of all permutations of n objects is called Sn. In this chapter we shall study thestructure of Sn, it irreps and its role in physics. The irreps are conveniently character-ized by the so-called Young tableaux. We shall discuss the properties of such tableaux,especially concerning product representations.
6.1 The Structure of Sn
A permutation p of n objects (digits, letters, particle labels) is a mapping p : i 7→ p(i), i =1, . . . , n, such that p is invertible. An explicit matrix notation is often used:
p =
(
1 2 . . . np(1) p(2) . . . p(n)
)
. (6.1)
In this notation it is comparatively easy to express the inverse and the product of per-mutations. First, notice that the permutation is defined by the relation between the twoentries in the same column of the 2× n matrix. Therefore, the order of the columns doesnot matter. We use this to write the product of two permutations as follows
p2 p1 =
(
1 . . . np2(1) . . . p2(n)
)(
1 . . . np1(1) . . . p1(n)
)
=
(
p1(1) . . . p1(n)p2(p1(1)) . . . p2(p1(n))
)(
1 . . . np1(1) . . . p1(n)
)
=
(
1 . . . np2(p1(1)) . . . p2(p1(n))
)
(6.2)
where the columns of the first permutation are put in such an order that the second row ofthe first permutation is identical to the first row of the second permutation. The identityis the permutation where both rows are identical. It can be written in n! ways, e.g.,
E =
(
1 . . . n1 . . . n
)
=
(
p(1) . . . p(n)p(1) . . . p(n)
)
, (6.3)
where p is an arbitrary permutation. Now it is clear that the inverse to a permutation pcan be written as
p−1 =
(
p(1) . . . p(n)1 . . . n
)
(6.4)
60
Definition Sn
The set of all permutations of n objects is called Sn.
Several obvious properties of Sn can be noted.(i) As we noticed above, if we define the composition law by
p2 p1 : i 7→ p2(p1(i)) (6.5)
then we see that p2 p1 is also a permutation.(ii) The composition law is associative. We can see this as follows.
(p2 p1) : k 7→ p2(p1(k))
p3 : k 7→ p3(k)
p3 : p2(p1(k)) 7→ p3(p2(p1(k))) (6.6)
so we findp3 (p2 p1) : k 7→ p3(p2(p1(k))). (6.7)
On the other hand
p3 p2 : k 7→ p3(p2(k))
p1 : k 7→ p1(k)
p3 p2 : p1(k) 7→ p3(p2(p1(k))) (6.8)
so(p3 p2) p1 : k 7→ p3(p2(p1(k))). (6.9)
We see that p3 (p2 p1) = (p3 p2) p1, so the composition law is indeed associative.(iii) The identity of Sn is the permutation p(i) = i.(iv) There corresponds to every permutation an inverse, as we have noticed above.
Conclusion: Sn is a group
In order to elucidate the structure of Sn, we introduce the concept of cycles and statesome of their properties.
(i) A transposition or 2-cycle is a permutation p such that p(i1) = i2, p(i2) = i1 andp(i) = i, i 6∈ i1, i2.
(ii) An m-cycle is a permutation such that p(i1) = i2, p(i2) = i3, . . . , p(im−1) = im,p(im) = i1 and p(i) = i, i 6∈ i1, . . . , im. This m-cycle will be denoted by (i1, . . . , im).
(iii) If the two cycles (i1, . . . , im1) and (j1, . . . , jm2
) have no elements in common, thenthey commute.
(iv) Every cycle can be written as a product of transpositions, e.g.,
pm = (i1, . . . , im) = (i1, i2)(i2, i3) . . . (im−1, im) = (i1, im)(i1, im−1) . . . (i1, i2) (6.10)
The sign δm of a permutation pm is given by δm = (−1)m−1, i.e., the parity of the numberof cycles.
(v) Every permutation p can be factorized uniquely into disjoint cycles:p = (i1, . . . , im1
) . . . (k1, . . . , kmp), where the elements not permuted are written as 1-cycles,
e.g., (l1) . . . (lω), or not written at all. So the identity can be written in cycle notation as(1) . . . (n) or simply as (1).
61
(vi) If p is an m-cycle, then for any q ∈ Sn, qpq−1 is also an m-cycle. We prove this
statement as follows. Let q : i 7→ q(i), then
(qpq−1) : q(i) 7→ q(p(i)) (6.11)
so we can write the permutation (qpq−1) in matrix notation as
(qpq−1) =
(
q(1) . . . q(n)q(p(1)) . . . q(p(n))
)
=
(
q (1 . . . n)q (p(1) . . . p(n))
)
(6.12)
Thus the permutation qpq−1 can be viewed as applying q to the rows of p. Since q isbijective, the cycles that are obtained will stay disjoint. Therefore, qpq−1 has the samecycle structure as p.
Corollary
The conjugacy classes are characterized by the cycle structure of their elements.
6.2 The role of Sn in physics
The symmetric groups Sn are groups of permutations of n objects. Permutations play animportant role in physics:
- through application of the (anti-)symmetrization postulates for systems of indistin-guishable bosons (fermions),
- through the symmetry types of tensors.Although we may consider these two as disctinct motivations, we actually combine themwhen constructing n-particle wave functions, as is illustrated by the following example.
Example 25 Spin-1/2 systemsLet u, d be a basis for spin-1/2 space, 〈u|u〉 = 〈d|d〉 = 1, 〈u|d〉 = 〈d|u〉 = 0. Then ann-particle state can be written as
Φ(1, . . . , n) =∑
αi
vα1,...,αnwα1(1) . . . wαn
(n), (6.13)
where wαi∈ u, d and vα1,...,αn is a complex number for any n-tuple (α1, . . . , αn). The
coefficients vα1,...,αn are called tensor coefficients.We consider some examples.
(i) n = 1Φ(1) = vuu(1) + vdd(1). (6.14)
(ii) n = 2
Φ(1, 2) = vuuu(1)u(2) + vudu(1)d(2) + vdud(1)u(2) + vddd(1)d(2). (6.15)
States of definite total spin S, S = 0, 1, can be formed by taking suitable linear combina-tions of products of the basis functions. This is the same as taking particular values forthe coefficients vα1α2 . For instance,
S = 1,
Φ11 = u(1)u(2), vuu = 1
Φ10 = 1√
2(u(1)d(2) + d(1)u(2)), vud = vdu = 1√
2
Φ1−1 = d(1)d(2), vdd = 1
(6.16)
62
and
S = 0, Φ00 =
1√2(u(1)d(2)− d(1)u(2)), vud = −vdu =
1√2. (6.17)
The components vα1α2 not mentioned explicitly in these two formulas vanish. The statesare labelled as ΦS
Sz.
Apparently, the spin-1 state corresponds to symmetric coefficients v and the S = 0state to anti-symmetric ones. Note that the states ΦS
Szdo not mix under S2 transforma-
tions.(iii) n = 3
There are four linearly independent symmetric states, corresponding to symmetric tensorcomponents. An anti-symmetric state cannot be constructed in this case. (It wouldcorrespond to a Slater determinant. A 3× 3 Slater determinant must vanish if there areonly two independent basis states.) In addition, there are four states that are neithersymmetric nor anti-symmetric. Again one notes that the states of different symmetry donot mix under S3. These states are said to transform under irreps of S3, as they spaninvariant subspaces.
Next we will study the irreps of Sn in a general and systematic way.
6.3 The irreps of Sn
In order to elucidate the structure of Sn, we have introduced cycles. Every permutationp can be factorized uniquely into disjoint cycles: p = (i1, . . . , im1
) . . . (k1, . . . , kmp). The
conjugacy classes are characterized by this cycle structure and as we will see each classwill determine an irrep of Sn. Recall that for a finite group the number of irreps up toequivalence, is equal to the number of classes.
6.3.1 Partitions of n and frames
In order to go from classes to irreps of Sn, we will consider so-called partitions of n. Weshall denote the cycle structure by a set of non-negative numbers ν1, . . . , νn such that
n = ν1 + 2ν2 + . . .+ nνn, (6.18)
where νi is the number of i-cycles in a given permutation. In order to facilitate thecounting of irreps, the set ν1, . . . , νn is mapped onto the set λ1, . . . , λn as follows
λ1 = ν1 + . . .+ νn, λ2 = ν2 + . . .+ νn, . . . , λn = νn. (6.19)
Then we have
λm − λm+1 = νm, λ1 + λ2 + . . .+ λn = n, λ1 ≥ λ2 ≥ . . . ≥ λn ≥ 0. (6.20)
The set λ1, . . . , λn is said to form a partition of n. It is common usage to writeλ1, . . . , λp, 0, . . . , 0 as λn1
i1, λn2
i2, . . . if λ1 = λ2 = . . . λn1
≡ λi1 , λn1+1 = . . . = λn1+n2≡
λi2 etc. and drop the zeros. For n = 5 we have e.g.,
22100, 21110, 31100 → 221, 213, 312. (6.21)
Consider a partition λ1, . . . , λn of n. To it we associate a frame as in the figurebelow, with λ1 boxes in the first row, λ2 in the second row, etc. For example for n = 8:
63
λ1 = 3 ν1 = 1
λ2 = 2 ν2 = 0
λ3 = 2 ν3 = 1
λ4 = 1 ν4 = 1
λ5, . . . , λ8 = 0 ν5, . . . , ν8 = 0
The frame shown above corresponds to the class of permutations with one 1-cycle,one 3-cycle and one 4-cycle (which corresponds to the number of boxes in the separatecolumns: one column of 1 box, one of 3 boxes and one of 4 boxes; this is in fact a generalfeature of frames).
Because of the property λ1 ≥ λ2 ≥ . . . ≥ λn ≥ 0, the frames one will encounter areall such that the boxes in the rows are not increasing when going down and idem in thecolumns when moving right. Frames satisfying this property are called legitimate frames.
6.3.2 Young Tableaux and Young operators
A Young tableau is obtained from a frame by filling it with the numbers 1, . . . , n in anyorder, each number being used exactly once. For any tableau T one defines two sets ofpermutations
- R(T) is the set of permutations that permute the numbers in each row of T amongthemselves, but does not mix rows.
- C(T) is a similar set, now for the columns.
Example 27 S6
Consider the tableau
1 4 62 53
The row operations are the permutationsR(T ) = (1), (14), (16), (46), (25), (14)(25),
(16)(25), (46)(25), (146), (164), (146)(25), (164)(25) The column operations are
C(T ) = (1), (12), (13), (23), (45), (123), (132)(12)(45), (13)(45), (23)(45), (123)(45), (132)(45)
(In these formulae the identity is denoted by (1).)
The usefulness of permutation symmetry is based on symmetrizing and anti-symmetrizingobjects (functions, tensors, etc.) Then the group algebra is involved naturally. It is theset Rn defined by
Rn = x|x =∑
p∈Sn
xpp, xp ∈ K, (6.22)
64
where K is the field (R or C) to which the numbers xp belong. So, any element of Rn isan n! dimensional vector. Two special elements of Rn are constructed with the help ofthe tableau T :
P =∑
p∈R(T )
p, Q =∑
q∈C(T )
δqq, (6.23)
where δq is the sign of the permutation q. It is +1 if q can be written as the product ofan even number of transpositions, otherwise it is −1.Definition Young operator
For a given Young tableau T the corresponding Young operator is defined as Y = QP .
For the Young tableau consisting of only one row, the Young operator1 YS =∑
p∈Snp
projects out a one-dimensional invariant subspace of Rn as follows. First one notes thatany permutation p′ ∈ Sn, acts like a 1×1 dimensional matrix on this particular Y : p′YS =YS, such that RnYS = cYS|c ∈ K. The same holds for the Young tableau consisting ofonly one column: YA =
∑
q∈Snδqq, p
′YA = δp′YA, such that again RnYA = cYA|c ∈ K isa one-dimensional invariant subspace of Rn.
The other invariant subspaces of Rn can be obtained by considering the Young opera-tors Yi corresponding to a particular subset of Young tableaux Ti, which will be introducedbelow. Once the space Rn is split into invariant subspaces RnYi, such that Rn =
⊕
iRnYi,all irreps can be found.
Next we will list without proof a number of important results that establish this con-nection between Young tableaux, Young operators, and irreps. Proofs of the statementscan be found in W.-K. Tung (2003) and also (to slightly lesser extent) in Hamermesh(1962).
Theorem 18Any Young operator Y = QP of Rn has the property Y 2 = cY, c 6= 0. It generates aninvariant subspace RnY of Rn. This invariant subspace provides an irrep of Sn. Repsdetermined by different tableaux corresponding to the same frame are equivalent, thosecorresponding to different frames are inequivalent.
Recall that a representation of Sn is a homomorphism from Sn to a subgroup of GL(V )for some finite-dimensional vector space V . Here we are considering the case of V = Rn,such that the n!× n! regular representation is obtained. This representation is now splitinto irreps, since we have split V =
⊕
iRnYi into its invariant subspaces. But often oneis not interested in taking V to be Rn, but rather V = RnΦ for some function Φ of nelements (particle positions or momenta, indices, etc). This latter space is usually muchsmaller than Rn (because some Y ′s “annihilate” Φ), but the action of Sn on this smallerspace can also be split into irrep, since V =
⊕
iRnYiΦ. This clearly works for any Φ.The irrep corresponding to the invariant subspace RnY Φ can be found explicitly by
letting all permutations p act on states Y Φ, and do this for all Y , that is, for all Youngtableaux, belonging to a frame. But it turns out that it is only necessary to consider theso-called standard tableaux of a frame. A tableau is said to be standard if in any row andcolumn the numbers are placed in increasing order.
1The subscripts S and A on the two specific Young tableaux n and 1n reflect the fact that onedeals with fully symmetric and antisymmetric combinations of all permutations. These Young operatorsare proportional to the symmetrizer and antisymmetrizer that one sometimes encounters in textbooks onQuantum Mechanics.
65
Theorem 19The Young operators Y1, . . . , Yd corresponding to the d standard tableaux of a framedetermine a d-dimensional irrep.
6.3.3 Examples
Example 28 S4
Table 6.1: Frames and standard tableaux for S4
4 1234 d = 1
3, 1 1234
1342
1243
d = 3
22 1234
1324
d=2
2, 12 1234
1324
1423
d = 3
14 1234
d = 1
The frames and standard tableaux for S4 are depicted in table 6.1. The operatorsY = QP can be used to construct wave functions or tensors with the desired symmetry.For T corresponding to 1n we obtain the fully anti-symmetric state, for T correspondingto n the fully symmetric one. The other symmetry types are called mixed symmetric.
In order to illustrate the use of Y in a setting where all calculations can be performedeasily, we discuss again S2 and S3.
Example 29 S2
1 2 YS = E + (12) Y 2S = (E + (12))(E + (12)) = 2YS
66
To connect to the spin-1/2 example of section 6.2, we let YS act on all products wi(1)wj(2),where i, j = 1 corresponds to spin up (u) and i, j = 2 to spin down (d):
YSu(1)u(2) = 2u(1)u(2)
YSu(1)d(2) = u(1)d(2) + u(2)d(1) (6.24)
YSd(1)d(2) = 2d(1)d(2)
One observes that the S = 1 states are reproduced, which means that under transforma-tions in spin-1/2 space (i.e., SU(2) transformations) these three states transform amongeach other. Any of these spin-1 states is a basis function for the 1-dimensional irrep 2of S2. Very explicitly, this means that for any permutation p, p acts as the 1 × 1 unitmatrix on the state YSwi(1)wj(2), i.e. pYSwi(1)wj(2) = (1)YSwi(1)wj(2).
Similarly, for the other Young tableau:
12
YA = E − (12) Y 2A = (E − (12))(E − (12)) = 2YA
YAu(1)u(2) = 0
YAu(1)d(2) = u(1)d(2)− u(2)d(1) (6.25)
YAd(1)d(2) = 0
The spin-0 state YAu(1)d(2) is the basis function for the irrep 12 of S2. Hence, for anypermutation p, p acts on the state YAwi(1)wj(2) as pYAwi(1)wj(2) = (δp)YAwi(1)wj(2),i.e. the 1-dimensional irrep given by the sign of a permutation.
Remark
In these examples of spin-1/2 systems we see that the constructed states transform underirreps of Sn (particle interchange), but at the same time we know that under rotations aspin-1/2 system transforms under SU(2). Hence, we want spin wave functions (and wavefunctions for multi-particle states in general), to transform under irreps of Sn and othersymmetry groups, such as the rotations or Lorentz transformations, but also symmetrygroups of internal quantum numbers such as isospin and color. The reason is that wewant physical results to be invariant under these transformations, hence one has to dealwith states that transform under irreps of the symmetry groups. In the example above wehave seen that by considering the wave functions transforming under irreps of Sn, we havealso arrived at a decomposition of products of single particle wave functions into statesthat transform unders irreps of SU(2). This is a very general feature and will becomemore transparent later on in these lectures (see also [Hassani, 1999], pages 856-857).
Example 30 S3
Permutations of S3 can have three distinct cycle structures: 3, 2, 1 or 13. Firstwe consider the fully symmetric irrep corresponding to the frame 3. Its standard tableauis
1 2 3 YS = E + (12) + (13) + (23) + (123) + (132)
Acting with YS on any of the states u(1)u(2)u(3), u(1)u(2)d(3), . . . , d(1)d(2)d(3) givesa fully symmetric state that corresponds to spin S = 3/2. Of all the eight states projectedout by YS only four are independent. They correspond to the four values of Sz.
Similarly, for the fully antisymmetric irrep 13:
67
123
YA = E − (12)− (13)− (23) + (123) + (132)
YA annihilates any product wi(1)wj(2)wk(3) because there are only two different basisstates, so of the three state labels i, j, k two must be the same.
Finally, we consider the mixed symmetric irrep 2, 1. The frame 2, 1 has twostandard tableaux:
1 23
Y1 = (E − (13))(E + 12) = E + (12)− (13)− (123)
1 32
Y2 = (E − (12))(E + (13)) = E − (12) + (13)− (132)
One can verify that indeed
R3 = R3YS ⊕R3YA ⊕ R3Y1 ⊕ R3Y2 (6.26)
The two Young operators of the standard tableaux determine the two dimensionalirrep 2, 1 of S3. Note that this irrep appears twice in the decomposition (6.26) of theregular rep into irreps, as it should.
If Y1 and Y2 act on the states wi(1)wj(2)wk(3) they annihilate many of them, notablythose that are symmetric in 1 and 3 (Y1) or 1 and 2 (Y2). The states that survive are
Y1
Φ11 = u(2)[u(1)d(3)− d(1)u(3)],
Φ12 = d(2)[u(1)d(3)− d(1)u(3)], (6.27)
Y2
Φ21 = u(3)[u(1)d(2)− d(1)u(2)],
Φ22 = d(3)[u(1)d(2)− d(1)u(2)]. (6.28)
Remark
The states obtained this way are linearly independent, but not necessarily orthogonal(despite the fact that Y1Y2 = 0) or normalized. Indeed, Φ11 ⊥ Φ12 and Φ21 ⊥ Φ22, butΦ11 6⊥ Φ21, etc.
Φ11 and Φ21 form a non-orthogonal basis for the 2-dim irrep of S3. Upon orthog-onalization and normalization we find the basis functions
Ψ+1 =
1√2u(3)[u(1)d(2)− d(1)u(2)],
Ψ+2 =
1√6[u(3)[u(1)d(2) + d(1)u(2)]− 2d(3)u(1)u(2)] (6.29)
This is a set of spin-1/2 states with Sz = 1/2. The states with Sz = −1/2 can be obtainedby orthogonalizing the states Φ12 and Φ22.
The above means that any permutation p acts on the two dimensional subspacespanned by Y1wi(1)wj(2)wk(3) and Y2wi(1)wj(2)wk(3) as a 2× 2 matrix. One can verify
68
that the characters of the 2-dimensional irrep obtained this way are indeed the same asthose of the irrep D3 given in Eqs. (2.7)-(2.9) which means that D3 is thus equivalent toit.
Remark
The dimension of an irrep of Sn can also be obtained by computing the hooks factor H .The hook length of a box a in a Young tableau is the number of boxes directly below,plus directly to the right of the box a, and counting the box a itself once. In the followingtableau of S6 we have indicated the hook lengths per box:
5 3 13 11
The hooks factor H is the product of the hook lenghts of all boxes of the tableau. Thedimension of an irrep of Sn is then given by n!/H . In the example shown above H = 45and the dimension of the irrep is therefore 6!/45 = 16. It is straightforward to verify thatin this way the dimensions of the irreps in the examples of S2, S3 and S4 are reproduced.
We recall the general result that holds for finite groups
∑
i
d2i = g, (6.30)
where the sum is over all irreps i, di is the dimension of an irrep and g is the order of thegroup. This formula serves as a useful check on the calculation of the dimensions.
6.4 Products of wave functions
When constructing wave functions one encounters two types of products of representa-tions, which require different decomposition methods. These two types of products arecalled inner and outer products [Hamermesh, 1962], page 249. They correspond to thedirect-product representations of a group and the representations of direct products ofgroups discussed chapter 4. We start with a discussion of the outer product.
6.4.1 Outer product
In constructing an n-particle wave function, one can consider it as a product of 1-particlewave functions, but one can also consider it as a product of several n-particle wave func-tions, each describing a different feature of the total n-particle wave function, e.g. orbitalangular momentum, spin, isospin, color. An example of the first (outer product) case iswhen one restricts oneself to the spin part of the wave function and construct the totalspin wave function from the product of 1-particle spin wave functions wi(j). In that caseone will have to deal with irreps of Sn and in the previous section we have studied allthese irreps. In this sense one is constructing the (outer) product ⊗ ...⊗ of n boxes,according to certain multiplication rules that will be given below. For n = 2 this leads to
⊗ = ⊕ (6.31)
69
In a similar way one can consider the tensor product of irreps of Sn and of Sm, whichyields a irrep of Sn ⊗ Sm, that can be decomposed into irreps of the larger group Sn+m.The multiplication rules are as follows.
When “multiplying” two frames A and B, one starts by labeling the boxes in the toprow of B with a’s, the second row with b’s, etc. Then one starts by taking the boxes fromthe top row of B and attaching them to A, building to the right and/or down. One hasto retain only the legal frames, which are the ones for which the boxes in the rows are notincreasing when going down and idem in the columns when moving right. Also, not morethan one a can appear in each column. Then follows the second row (with b’s), and so on.After all letters have been distributed, one keeps only those constructed frames for whichthe following holds. If one reads the letters along the rows from right to left from the toprow down to the bottom row, then the sequence must satisfy the requirement that to theleft of any symbol there are no fewer a’s than b’s, no fewer b’s than c’s, etc. For example,baabc is not allowed, but ababc is.
In this way one can find for example (exercise 3.3)
⊗ = ⊕ ⊕ ⊕ (6.32)
For the case of n = m = 3 and the irreps 2, 1 ⊗ 2, 1 the above procedure is explicitlyshown in [Hassani, 1999], pages 722-723.
One can verify the correctness of the end result as follows: the sum of the dimensions
of the r.h.s. should equal the product of the dimensions on the l.h.s. times
(
n +mn
)
.
6.4.2 Inner product
The other product (the inner product) one encounters when one wants to consider theproduct of the total n-particle spin wave function and the total n-particle isospin wavefunction. This product is part of the total n-particle wave function. It is therefore aproduct of several n-particle wave functions, so one has to consider the CG decompositionof the direct-product representation for Sn.
Example 31 An application is the construction in SU(2) of fully anti-symmetric wavefunctions of three particles with spin S = 1/2 and isospin I = 1/2. The classic example isthe proton state which consists of two up quarks and one down quark. For the up quarkone has Sz = ±1/2 and Iz = 1/2; for the down quark one has Sz = ±1/2 and Iz = −1/2.One can write isospin wave functions exactly like the spin wave function given in theprevious section, by interpreting u as isospin ‘up’ and d as isospin ‘down’. We denotethe isospin wave functions by X±
i , where X+i is relevant for the proton and X−
i for theneutron. The spin wave functions were denoted by Ψ±
i , see Eq. (6.29) of the previoussection. Both the spin and the isospin wave functions are mixed symmetric.
In this case one needs the CG series for S3 applied to the direct product of mixedirreps: 2, 1 ⊗ 2, 1
⊗ = ⊕ ⊕ (6.33)
70
This corresponds to the decomposition obtained before in Example 21, Eq. (4.24), forwhich the CG coefficients were discussed in Example 22. For general construction rulesfor the CG decomposition in terms of Young tableaux we refer to the literature.
For the product wave function (spin wave function times isospin wave function) wecan construct (suppressing superscripts ±)
ψS =1√2(X1Ψ1 +X2Ψ2)
ψA =1√2(X1Ψ2 −X2Ψ1)
ψMS =1√2(X1Ψ2 +X2Ψ1)
ψMA =1√2(X1Ψ1 −X2Ψ2) (6.34)
The state ψS is fully symmetric, ψA fully anti-symmetric and ψMS and ψMA form the basisof the mixed-symmetric irrep. The proton and neutron spin-isospin states correspond tothe fully symmetric state ψS.
Summarizing: the direct product of Xi and Ψj is expressed in terms of the above fourψ states via the CG decomposition, pictorially represented in Eq. (6.33). It is the CGdecomposition for S3, not of SU(2)spin ⊗ SU(2)isospin, as should be clear by now. Theusefulness of Sn in the construction of n-particle wave functions that are transformingunder irreps of continuous groups such as SU(2) has hopefully also become apparent.Further details will be addressed later. Finally, care must be taken to properly interpretthe symbol ⊗, either as yielding a product rep of a group or a rep of a direct product ofgroups.
6.5 Exercises
Exercise 6.1
Check the correctness of Eq. (6.10).
Exercise 6.2 The irreps of S3
a) Write down the frames, partitions and standard tableaux of S3.b) What are the dimensions of the irreps of S3?c) Write down the elements of S3 in the standard 2 × n matrix notation and in cyclenotation.d) Divide these elements into conjugacy classes. How many classes are there? Is this inaccordance with the answer of question b?e) Verify that the following equation holds for S3 using the answer of question b:
∑
i
d2i = g,
where the sum is over all irreps i, di is the dimension of an irrep and g is the order of thegroup.
71
f) Write down the parity (also called sign) of the permutations of S3. Does this furnish aone-dimensional irrep?
Exercise 6.3
Decompose ⊗ into irreps of S5 and count the dimensions of the irreps in the two
different ways explained in the text.
Exercise 6.4
Decompose ⊗ into irreps of S6 and count the dimensions of the irreps. Verify
the result using the rule given in the text.
72
Appendix A
Mathematical Preliminaries
In this appendix we mention some mathematical concepts used in the lecture notes. Fora more extended introduction the reader may consult e.g. the book by Y. Choquet-Bruhat, C. De Witt-Morette and M. Dillard-Bleick, Analysis, Manifolds and Physics,North-Holland, 1977, in particular Chapter I.
A.1 Set Theory
We describe here sets, mappings and relations.
A.1.1 Sets
Empty setThe empty, or void, or null set ∅ is a set that contains no members.
InclusionWe write a ∈ A to indicate that a belongs to the set A.
SubsetIf all members of a set A are also members of a set B then we say that A is a subset ofB, written as A ⊂ B.
Direct ProductThe direct product of two sets A and B, denoted by A ⊗ B, is the set of ordered pairs(a, b), where a ∈ A and b ∈ B.
A.1.2 Mappings
MappingA mapping f : A→ B associates in a unique manner to every element a of A an elementb of B, written as f : a 7→ b, also written as b = f(a).
RangeThe subset of B that consists of all the images of elements of A is called the range R(f)of f .
Injective, Surjective, BijectiveIf for all b ∈ R(f) there is only one a ∈ A such that f(a) = b then f has an inverse
73
mapping, denoted by f−1 and is said to be 1− 1 or injective.If R(f) = B, the mapping f is said to be onto or surjective. If f is both injective and
surjective it is called bijective.
A.1.3 Relations
RelationA relation between A and B is a subset R of A⊗B. The elements a of A and b of B aresaid to be R-related if (a, b) ∈ R, also written as aRb.
Equivalence RelationA relation R ⊂ A⊗ A is an equivalence relation if it is
reflexive (a, a) ∈ R, ∀a ∈ A,symmetric (a, b) ∈ R implies (b, a) ∈ R, ∀a, b ∈ A,transitive if (a, b) ∈ R and (b, c) ∈ R then (a, c) ∈ R, ∀a, b, c ∈ A.
A.2 Algebra
We shall not repeat the definition of a group here. Other structures that are importantare defined below.
RingA ring is a set X together with two operations, multiplication : (x, y) 7→ x y and addi-tion + : (x, y) 7→ x+ y. The ring is an Abelian group under addition, and multiplicationis associative and distributive with respect to addition.
(x y) z = x (y z)x (y + z) = x y + x z(y + z) x = y x+ z x, ∀x, y, z ∈ X
Zero, Identity, InverseThe neutral element for addition is called the zero of the ring. If there exists also a neutralelement e for multiplication, say ex = xe = x for all x ∈ X , then this element is calledthe identity. If a ring has an identity one can define the inverse of any element x, denotedby x−1, as the element that fulfills xx−1 = x−1x = e, if such an element exists.
FieldA ring with an identity is a field iff all elements different from zero have an inverse. Anexample is the field of the real numbers R.
ModuleA module over the ring R is an Abelian group G together with an operation, called scalarmultiplication, R⊗G→ G, (α, x) 7→ αx, such that
α(x+ y) = αx+ αy
(α+ β)x = αx+ βx
(αβ)x = α(βx), ∀α, β ∈ R, ∀x, y ∈ G.
74
If the ring has an identity e, then
ex = xe = x, ∀x ∈ G.
AlgebraAn algebra A is a module over a ring R with an identity and an associative operation(x, y) 7→ xy, usually also called multiplication, such that
A is a ring and the operations (α, x) 7→ αx and (x, y) 7→ xy have the propertyα(xy) = (αx)y = x(αy).
For simplicity we have not written explicitly the symbol for multiplication either inR itself or for the multiplication of an element of X with an element of R. Usually thelatter are called scalars.
A.3 Symbols
∀x for every x∃x there exists an x∧ and; a ∧ b: proposition a and proposition biff if and only f; a iff b: a⇒ b and b⇒ a⇒ implication a⇒ b: proposition a implies proposition ba, b, c, . . . the set with elements a, b, c, . . .∈ included in⊂ subset of⊗ direct product⊙ semi-direct product⊕ direct sumA→ B A is mapped into Ba 7→ b a is mapped onto bR the set of the real numbersC the set of the complex numbersZ the set of the integersRn n-dimensional Euclidian space
L2 The Hilbert space of quadratically summable functionsA ≃ B A and B are homomorphicA ∼= B A and B are isomorphic∝ proportional todiag(a, b, . . .) the matrix with diagonal elements a, b, . . . and all off-diagonal
elements being zero.Spane1, . . . , en the linear space consisting of the linear combinations of the
vectors e1, . . . , en.Orthogonal complement the set of vectors perpendicular to all vectors belonging to a
given linear space.
75
76
Appendix B
Bibliography
These lecture notes are mainly based on the book by Jones:
Jones 1998 H.F. Jones, Groups, Representations and Physics, Second ed.,Institute of Physics Publishing, 1998
This book is very well suited for self-study. It covers almost all of the subjects treated inthe lectures.
A recent, insightful textbook, with a slightly different approach is
Ramond 2010 Pierre Ramond, Group Theory: A Physicist’s Survey,Cambridge University Press, 2010
A rather comprehensive text on group theory is
Cornwell 1984 J.S. Cornwell, Group Theory in Physics, vols I and II,Academic Press, 1984
Cornwell 1997 J.S. Cornwell, Group Theory in Physics, An Introduction,Academic Press, 1997
The set of two volumes contains much more than the subject matter of the lecture notes.The second reference is a condensed and abridged version of these volumes.
Hamermesh 1962 M. Hamermesh, Group Theory and its Application to Physical Prob-
lems, Addison Wesley, 1962
This work is more easy going than Cornwell, and is particularly recommended for thetreatment of the symmetric groups. It is, however, much more limited in scope thanCornwell. One should keep in mind that at the time Hamermesh published his book thequark model did not exist.
77
Tung 2003 W.-K. Tung, Group theory in physics: an introduction to symmetry
principles, group representations, and special functions in classical
and quantum physics, World Scientific, 2003Hassani 1999 S. Hassani, Mathematical Physics: a Modern Introduction to Its
Foundations, Springer Verlag, 1999
Classic textbooks on group theory and its applications in quantum mechanics are:
Wigner 1931 E.P. Wigner, Gruppentheorie un ihre Anwendung auf die Quanten-
mechanik der Atomspektren, Friedr. Vieweg & Sohn, 1931Weyl 1928 H. Weyl, The Theory of Groups and Quantum Mechanics,
Dover Publications
Special topics are treated in e.g.
Lichtenberg 1970 D.B. Lichtenberg, Unitary Symmetry and Elementary Particles,Academic Press, 1970
Georgi 1982 H. Georgi, Lie Algebras in Particle Physics,Benjamin/Cummings, 1982
78
A. Durer, Melencolia I, 1514
Group Theory in Physics
Part II, Lie Groups & Lie Algebras
Michiel Snoek
Tenth Edition, 2010
Group Theory in Physics
Part II, Lie Groups & Lie Algebras
Authors: Ben Bakker and Daniel BoerDepartment of Physics and Astronomy
Vrije Universiteit, Amsterdam
Tenth Edition, 2010
Cover illustration: M.C. Escher, Cirkellimiet III, 1959Weblocatie: www.worldofescher.com
i
Contents
7 Lie Groups and Lie Algebras 707.1 Linear Lie Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 707.2 Topological properties of linear Lie Groups . . . . . . . . . . . . . . . . . 737.3 Lie Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 747.4 The Exponential Map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 767.5 Comparisons of Lie Algebras . . . . . . . . . . . . . . . . . . . . . . . . . 797.6 Representations of Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . 807.7 Universal Covering Group . . . . . . . . . . . . . . . . . . . . . . . . . . . 827.8 Rotations in Euclidian space: SO(3) and SU(2) . . . . . . . . . . . . . . . 837.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
8 Irreps of su(n) 858.1 Irreps of su(2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 858.2 Irreps of su(3) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 888.3 Irreps of su(n) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 898.4 Application: SU(6) Proton Wave Functions . . . . . . . . . . . . . . . . . 928.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94
9 The Lorentz group 959.1 Lorentz transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
9.1.1 Physical Interpretation . . . . . . . . . . . . . . . . . . . . . . . . 979.1.2 The Lorentz algebra . . . . . . . . . . . . . . . . . . . . . . . . . . 98
9.2 The Group SL(2,C) and its Connection to the Lorentz Group . . . . . . . 1009.2.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
9.3 Representations of the Lorentz group . . . . . . . . . . . . . . . . . . . . . 1059.4 Casimir operators of the Lorentz group . . . . . . . . . . . . . . . . . . . . 1079.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
10 The Poincare group 10910.1 The Poincare Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10910.2 Invariants: Mass and Spin . . . . . . . . . . . . . . . . . . . . . . . . . . . 11010.3 Representations of the Poincare group . . . . . . . . . . . . . . . . . . . . 112
10.3.1 Translations and the little group . . . . . . . . . . . . . . . . . . . 11210.3.2 Spin states . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
10.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
A Bibliography 119
ii
Chapter 7
Lie Groups and Lie Algebras
In Part I of these lectures we have encountered two kinds of groups:(i) discrete groups, which are groups consisting of a discrete set of elements (which canbe of finite or infinite order), such as (Z; +) or Zn = (z ∈ C|zn = 1;×);(ii) continuous groups, where the group elements depend continuously on a set of param-eters, such as O(3) or GL(d,K).
The theory of continuous groups, i.e., continuous sets of elements with a compositionlaw that fulfills the axioms of a group, is much more complicated than the theory ofdiscrete groups. In the latter the elements can be enumerated, while in continuous groupsthe elements are given as functions of some variable or variables. In addition to algebraicconcepts, topological notions like continuity, connectedness and compactness are involved.
In this section we will focus on a particular kind of continuous groups, namely Liegroups. Loosely speaking, a Lie group is a group whose elements depend analytically ona finite number of parameters (the finite number being the dimension of the Lie group).Here we shall not discuss the general definition of a Lie group, which requires the languageof differential geometry, but following Cornwell (1984) we limit our discussion to linear
Lie groups, i.e. Lie groups that can be represented as a group of matrices. All Lie groupsof relevance to physics belong to this category. Our goal here is thus not to give a fullmathematical treatment of Lie groups and their associated Lie algebras, but rather toconvey those results that are of most use to physicists. Therefore, we will mention severaltheorems without proofs, just to inform the reader about their existence. We shall focusmainly on important examples and refer to theorems to indicate that the results areusually more general. At the end of these lecture notes a list of books is given where mostof these theorems can be found including proofs. For most of these theorems the relevantbooks are the two volumes by Cornwell (1984).
7.1 Linear Lie Groups
We shall treat a linear Lie group as a group endowed with certain topological and ana-lytical properties.
Definition Linear Lie Group
A group G is said to be a linear Lie group iff the conditions (i)-(iv) are satisfied:(i) G has a faithful, finite-dimensional rep D : G→ D(G).
70
We can then define a metric in G by introducing a distance function
d(T1, T2) =
√∑
ij
|Dij(T1)−Dij(T2)|2. (7.1)
(ii) There is a sphere Mδ with radius δ > 0 around the identity E of G, such thatall elements T of G inside that sphere (i.e. d(T,E) < δ) can be parametrized with n realparameters x1, . . . , xn. No two sets of parameters x1, . . . , xn correspond to the sameelement in G. The parametrization is usually chosen in such a way that E = T (0, . . . , 0).
(iii) There exists a real number η > 0, such that for any real n-tuple (x1, . . . , xn)with property
∑i x
2i < η2, which together define the sphere Sη in R
n, there is an elementT = T (x1, . . . , xn) ∈Mδ.
(iv) For (x1, . . . , xn) ∈ Sη, the matrix elements Dij(x1, . . . , xn) ≡ Dij(T (x1, . . . , xn))are analytic functions of the parameters.
Remark
The number n of independent parameters is called the dimension of the Lie group. Itmust not be confused with the dimension of the matrices D.
Property (ii) leads to the following theorem:
Theorem 20If the matrices apnp=1 are defined by
[ap]ij =∂
∂xpDij(x1, . . . , xn)
∣∣∣∣xi=0
(7.2)
then they are linearly independent.
The analyticity property (iv) gives rise to another theorem:
Theorem 21In a sufficiently small neighbourhood of the identity the parameters x′′i of T (x′′i ) =T (xi)T (x
′i) and xi of T−1(xi) = T (xi) are analytic functions of x1, . . . , xn, x′1, . . . x′n
and x1, . . . , xn respectively.Example 32The (generalized) torus R
n/Zn = (R/Z)n ∼= (S1)n is a linear Lie group (here S1 denotesthe unit circle, S1 = z ∈ C| |z| = 1, viewed as a multiplicative subgroup of C\0). Theisomorphism1
R/Z ∼= S1 is provided by the mapping t 7→ exp(2πit) (note the similarityto Z/nZ ∼= Zn = (z ∈ C|zn = 1;×)).Example 33 SU(2)The group SU(2) is defined as the set of unitary, 2×2, complex matrices with determinantequal to 1. A real parametrization would be
U =
[α1 + iα2 β1 + iβ2−β1 + iβ2 α1 − iα2
], α2
1 + α22 + β2
1 + β22 = 1. (7.3)
The number of independent parameters is three: the relation between the four real onesdefines the surface of a three-dimensional sphere embedded in four dimensional Euclidianspace.
1Lie group homomorphisms and isomorphisms are group homomorphisms and isomorphisms with theadditional property that the mappings are analytic.
71
If we define the independent parameters
x1 = 2β2, x2 = 2β1, x3 = 2α2 (7.4)
and let α1 > 0 be determined by the relation (7.3), then x1 = x2 = x3 = 0 correspondsto the identity and
d(U,E) = 2
[1−
1− 1
4
(x21 + x22 + x23
) 12
]12
< δ, (7.5)
implies 0 ≤ x21+x22+x
23 < 2δ2− δ4/4 ≡ η2, which also requires δ < 2
√2. This shows that
SU(2) is a linear Lie group of dimension 3.Using this parametrization (and Eq. 7.2) we find the matrices a1, a2 and a3:
a1 =i
2
[0 11 0
], a2 =
i
2
[0 −ii 0
], a3 =
i
2
[1 00 −1
]. (7.6)
These matrices are proportional to the well-known Pauli matrices, ak = (i/2)σk. Thesethree matrices are linearly independent over the real numbers.
Just as in the case of finite groups we can find subgroups of a Lie group. The differenceis that a subgroup need not be a Lie group itself. Otherwise the following definition isappropriate:
Definition Lie-subgroup
If H is a subgroup of the linear Lie group G and H is itself a linear Lie group, then it iscalled a Lie-subgroup of G.
Example 34The general linear Lie group GL(n,R) has a Lie-subgroup called the special linear group,defined as,
SL(n,R) = T ∈ GL(n,R)| det(T ) = 1. (7.7)
Another Lie-subgroup of GL(n,R) is formed by the orthogonal group
O(n) = T ∈ GL(n,R)|TTT = E, (7.8)
where T denotes transposition. O(n) itself has as a Lie-subgroup the special orthogonalgroup
SO(n) = T ∈ O(n)| det(T ) = 1, (7.9)
which is also a Lie-subgroup of SL(n,R). All of these Lie groups have finite subgroupstoo, for instance, the rotational symmetry groups of the (generalized) platonic solids inn-dimensions (called regular polytopes).
If in the above we replace R by C, then we have that GL(n,C) has as Lie subgroupSL(n,C) and the unitary group
U(n) = T ∈ GL(n,C)|T †T = E, (7.10)
where † denotes Hermitian conjugation (i.e. complex conjugation and transposition com-bined). The special unitary group
SU(n) = T ∈ U(n)| det(T ) = 1, (7.11)
is a Lie-subgroup of all of these Lie groups.
72
7.2 Topological properties of linear Lie Groups
By referring to its elements as T (x1, . . . , xn) we can consider the linear Lie group G asa topological space. The distance function allows us to define continuity. Then we canspeak of a continuous curve in this space. If xi(t) is a continuous function of t for all iin a certain domain of t-values, then the elements with parameter values xi(t) are said todefine a continuous curve in G. Two elements of G are said to be (arcwise) connected ifthere is a continuous curve on which both are lying.
Definition connected component
A connected component of the linear Lie group G is the maximal subset of G that can beobtained by continuous variation of the parameters of one element of G.
The importance of this notion becomes clear if we formulate the following theorem(exercise 7.1).
Theorem 22The connected component of a linear Lie group that contains the identity E is an invariant(or normal) Lie-subgroup of G.
We have seen before that in order to find the irreps we need to know the invariantsubgroups of a group.
Another important global property that a Lie group may have, is simple connectedness.It is a topological property defined in terms of continuous curves on the group. We shall asa matter of convention take the parameter space such that the curve T (t) ∈ G is definedfor t ∈ [0, 1]. T (0) and T (1) are its begin and end points.
A very simple, yet important example of a curve is a point: T (t) = T0, ∀t ∈ [0, 1]. Byconsidering different curves, also called paths, with the same begin and end points, wecan form the concept of paths continuously deformable into each other.
Definition continuously deformable
Let T s(t) be a path for all s ∈ [0, 1]. If T s(t) is a continuous function of s for all t, wesay that for all s1 and s2 in [0, 1] the paths T s1(t) and T s2(t) are continuously deformableinto each other (or are homotopic). If a curve belongs to a set of continuously deformablepaths that contains as one of the curves a point, then we say that this curve is contractibleto a point (or is null-homotopic).
Definition simply connected
A linear connected Lie group is said to be simply connected, if every closed path, i.e. apath T (t) such that T (0) = T (1), is contractible to a point (i.e., all closed paths are nullhomotopic).
The third and last global property of Lie groups we will discuss is compactness.
Definition compact Lie group
A linear Lie group of dimension n with a finite number of connected components is saidto be compact if the parameters range over a compact set, i.e. a closed and bounded set.
Example 35The torus is an example of a compact, connected Lie group, which is not simply connected.
Example 36 SU(2)The elements of SU(2) were expressed in Eq. (7.3) in terms of the coordinates of a four-
73
dimensional Euclidian space, restricted to the surface of the unit sphere. This surface issimply connected. Clearly, the group SU(2) is compact.
Example 37 SO(3)SO(3) is the connected component of O(3) that contains the identity. The other connectedcomponent of O(3) does not form a group. Any element of SO(3), a three-dimensionalrotation, can be described by a vector ~α = αn, where α is the angle of rotation, limitedto −π ≤ α ≤ π, and n is the direction about which the rotation takes place. Clearly,the parameter space of SO(3) is a sphere with radius π and hence compact. The interiorof this sphere is simply connected. However, the two points πn and −πn, lying on thesurface, parametrize the same rotation (for each n). Therefore they must be identified.So any line connecting these two pionts, e.g., T (t) = (2t− 1)πn, t ∈ [0, 1], is closed. Sucha curve cannot be contracted to a point, so SO(3) is not simply connected.
Example 38 SL(2,R) and SL(2,C)The group SL(2,R) is not compact, as its parameters range over a non-compact set.Without proof we state that it is connected, but not simply connected, essentially forthe same reason why its subgroup SO(2) ∼= R/Z is not simply connected. In contrast,SL(2,C) is simply connected. It is related to the Lorentz group, which has four connectedcomponents (see Chap. 9.)
In the following table these observations are summarized.
compact connected simply connected
Rn/Zn
GL(n,R)
SL(n,R)
O(n)
SO(n)
GL(n,C)
SL(n,C)
U(n)
SU(n)
7.3 Lie Algebras
Linear Lie groups can be analyzed and characterized locally by their associated Lie alge-bras. We start by giving the definition of a Lie algebra.
Definition Linear Lie algebra
A real linear Lie algebra L of dimension n, (n ≥ 1), is a real linear vector space ofdimension n, endowed with a product [a, b], called the Lie product, defined for all elementsa and b of L, with the properties
(i) [a, b] ∈ L, ∀a, b ∈ L(ii) [αa+ βb, c] = α[a, c] + β[b, c], ∀a, b, c ∈ L, ∀α, β ∈ R
(iii) [a, b] = −[b, a](iv) [a, [b, c]] + [b, [c, a]] + [c, [a, b]] = 0, ∀a, b, c ∈ L. The last property is the so-called
Jacobi identity.
74
We shall mostly encounter matrix representations of Lie algebras, in which case theLie product is the usual commutator: [A,B] = AB − BA.
Because L has finite dimension n, there exists a basis, say a1, . . . , an for L. Considerall products [ap, aq]. They belong to L, so they can be written as linear combinations ofbasis elements:
[ap, aq] =n∑
r=1
f rpqar. (7.12)
The constants f rpq are called the structure constants.
As we will see, the tangent space of a Lie group G at the point E (the identity)is a vector space with the algebraic structure of a Lie algebra. The importance of Liealgebras for the study of Lie groups lies in the fact that the structure of the latter in aneighbourhood of E is determined entirely by the structure constants of the Lie algebracorresponding to it.
Suppose G is a linear Lie group of dimension n. Then we define an analytic curve
in G similar to a continuous curve. Let (x1(t), . . . , xn(t)) be an n-tuple of real analyticfunctions of t, defined on the subset [0, t0) of R, such that t0 > 0 and xi(0) = 0. If theelements of G are given by T (x1(t), . . . , xn(t)), then the set T (x1(t), . . . , xn(t))|t ∈ [0, t0)is said to be an analytic curve T (t) in G.
Definition tangent vector
The tangent vector a to an analytic curve T (t) in G is the m×m matrix defined by
a =dT (t)
dt
∣∣∣∣t=0
. (7.13)
Remark
For the sake of brevity we have identified here T and D(T ), the m×m matrix represen-tation of G. The matrices [ap] defined by Eq. (7.2) are the tangent vectors of the curvesxp(t) = t, xq = 0, ∀q 6= p.
Theorem 23 Group-theoretical commutatorLet A(t) and B(t) be two analytic curves in the Lie group G, with tangent vectors a andb respectively, then [a, b] = ab− ba is the tangent vector of the curve C(t) given by
C(t) = A(√t)B(√t)A(√t)−1B(
√t)−1. (7.14)
From this theorem (exercise 7.3) it follows that we can associate a Lie algebra L withany Lie group G by identifying L with the vector space of tangent vectors of analyticcurves in G, with the Lie product [a, b]. The basis of L is given by the special elementsdefined by Eq. (7.2).
Example 39 su(2)The matrices ai in Eq. (7.6) of example 33 form a basis of the Lie algebra L ≡ su(2) ofthe Lie group SU(2). The elements ai are related to the Pauli matrices via ai = (i/2)σi.From quantum mechanics one recalls that the Pauli matrices satisfy:
[σi2,σj2] = i
3∑
k=1
ǫijkσk2. (7.15)
This implies that for su(2) the structure constants are f rpq = −ǫpqr.
75
7.4 The Exponential Map
We have seen that one can go from the Lie group to the Lie algebra by considering tangentvectors at the identity, one may wonder whether one can also take the opposite route.Here the key element is the exponential map. Since we are mainly interested in matrixreps, we have to define the matrix exponential function. Let A be an m×m matrix. Theexponential of this matrix is defined by the usual power series:
eA =∞∑
n=0
An
n!. (7.16)
If two matrices A and B commute, the exponential of their sum is just what we expect:exp(A+B) = exp(A) exp(B), but otherwise the product can be written as the exponentialof a more complicated argument, given by the Campbell-Baker-Hausdorff formula:
eAeB = eC , C = A+B + 12 [A,B] +
1
12[A, [A,B]] + [[A,B], B]+ . . . (7.17)
The exponential has the following properties:
[eA]−1 = e−A, det[eA] = eTrA. (7.18)
If we introduce a parameter t we can construct a one-parameter group of elementsT (t) defined by
T (t) = exp(tA). (7.19)
The composition law is the usual matrix multiplication. It is not difficult to show thatthe set of these elements exp(tA)|t ∈ R is an Abelian Lie group of m ×m matrices ifA is an m×m matrix. Upon using the definition of the exponential function we find thesingle matrix as defined in Eq. (7.2):
dT (t)
dt= AetA
t→0= A. (7.20)
Example 40
A = Az =
0 1 0−1 0 00 0 0
, T (t) = exp(tA) =
cos t sin t 0− sin t cos t 0
0 0 1
. (7.21)
These matrices are the well-known three-dimensional matrices that describe rotationsabout the z-axis in R
3. Differentiation gives, as expected
dT (t)
dt
∣∣∣∣t=0
= Az. (7.22)
We say that the matrix Az generates rotations about the z-axis.
We have seen that one could very easily construct an Abelian Lie group by expo-nentiating a single matrix times a real parameter. The Lie group obtained this way isone-dimensional. Here we shall use them as subgroups of a multi-dimensional Lie group.
76
Theorem 24For all a ∈ L there exists a one-parameter subgroup of G that can be defined by T (t) =exp(ta), (t ∈ R).
For compact Lie groups one can go one step further. The connected subgroup is closelyconnected to the Lie algebra:
Theorem 25If G is a compact Lie group then every element of the connected subgroup can be expressedas exp a, for some a ∈ L.
In other words, on a compact, connected Lie group G the exponential map (exp : L→G) is always surjective. However, in general it is not injective, since exp(a) = exp(b) doesnot necessarily imply a = b.
Example 41 SU(N)Consider the Lie group SU(N) and its Lie algebra su(N). For all U ∈ SU(N) the matrixU is a unitary N × N matrix with determinant equal to 1. If U is close enough to theidentity, it can be written as U = exp(ta), a ∈ L, t ∈ R. Because the matrices U areunitary, the matrices a must be anti-hermitian:
U−1 = U †, a† = −a. (7.23)
The determinant being unity implies
detU = 1, Tr a = 0. (7.24)
So, su(N) consists of traceless, anti-hermitian N ×N matrices. The number of indepen-dent parameters is then n = dimL = N2 − 1.
Note that in this example we did not need to use the fact that SU(N) is compactand connected, and that therefore all its group elements can always be written in singleexponential form. The reason is that the properties of compactness and connectedness areglobal properties of the group, whereas the properties of the elements of the Lie algebraconcern local properties of the group only.
Example 42 SU(2)As we have seen, the elements U of the unitary group SU(2) are generated by the 3 = 22−1traceless, anti-hermitian 2× 2 matrices ai = (i/2)σi.
As SU(2) is compact and connected, every element U ∈ SU(2) can be written in singleexponential form: U = exp(
∑3i=1 θiai). In terms of the Pauli matrices this leads to
U = exp(i3∑
i=1
θi2σi) = cos
θ
21+ i sin
θ
2n · ~σ, (7.25)
with θi = θni, n = (n1, n2, n3). This is a frequently used parametrization of elements ofSU(2).
Theorem 26Every element of the connected subgroup of any linear Lie group G can be expressed asa finite product of exponentials of its real linear Lie algebra L.
Example 43 SL(2,R)The Lie group SL(2,R) consists of all real 2 × 2 matrices with determinant 1. The
77
corresponding Lie algebra sl(2,R) consists of traceless real 2 × 2 matrices. Consider thefollowing elements of SL(2,R):
E =
[1 00 1
], S =
[−1 00 −1
]. (7.26)
They are connected by an analytic curve that lies completely within SL(2,R):
C(t) =
[cos t sin t− sin t cos t
](t ∈ [0, π]). (7.27)
Consider also the following elements of SL(2,R):
A(r) =
[r 00 r−1
], r ∈ R\0. (7.28)
If for given r we would express the diagonal matrix A(r) in terms of an element of sl(2,R)as A = exp a, then a must also be diagonal. Suppose
a =
[λ1 00 λ2
], (7.29)
then r = exp λ1, r−1 = exp λ2, hence λ1 + λ2 = 0, which relation just says that a is
traceless.For complex r we would be able to write any r as an exponential, but for real r we
can only do that if r > 0. Therefore we see that there are elements of G that cannot beobtained by exponentiation of a single matrix. If we would allow several matrices to beinvolved, then we can proceed as follows. Suppose r = − exp(−λ). We can write A(r) asa product of two exponentials
A(λ) = exp(−λa1) exp(πa2) (7.30)
where the matrices a1 and a2 are given by
a1 =
[1 00 −1
], a2 =
[0 1−1 0
]. (7.31)
One then findseta2 = cos t E + sin t a2, (7.32)
so exp(πa2) = S, and
e−λa1 =
[e−λ 00 eλ
]. (7.33)
Consequently we find
A(λ) =
[−e−λ 0
0 −eλ]. (7.34)
We see that although A(r) with r > 0 and A(r) with r < 0 are connected (through Eand S for example) and so belong to the connected subgroup of SL(2,R) (which in factis the whole group itself), the non-compactness of this group is an impediment to writingall its elements as a single exponential.
In conclusion, it is straightforward to obtain the Lie algebra from the Lie group, butnot vice versa. To characterize the global structure of a Lie group one needs additionaltools, such as homotopy groups, but these will not be discussed here.
78
7.5 Comparisons of Lie Algebras
The notion of a subgroup can be extended, mutatis mutandis, to an algebra. Also theidea of invariance can be used in this context. The relevant concepts are defined in thefollowing two definitions.
Definition subalgebra
The set L′ is said to be a subalgebra of the Lie algebra L iff(i) L′ ⊂ L(ii) L′ endowed with the same Lie product as L is a Lie algebra(iii) If a′, b′ ∈ L′ then [a′, b′] ∈ L′
Definition invariant subalgebra
The set L′ is an invariant subalgebra of the Lie algebra L, if it is a subalgebra and if forall a′ ∈ L′ and all b ∈ L, [a′, b] ∈ L′.
One may wonder if the Lie algebra corresponding to an invariant subgroup of a Liegroup with a corresponding Lie algebra, is an invariant subalgebra. The affirmative answeris given in the next theorem.
Theorem 27If to the Lie groups G and G′ correspond the Lie algebras L and L′, and if G′ ⊂ G,then L′ ⊂ L. Furthermore, if G′ is an invariant subgroup of G, then L′ is an invariantsubalgebra of L.
In order to be able to compare two Lie algebras, we need the concepts of homo- andisomorphisms of Lie algebras.
Definition homomorphism of Lie algebras
A mapping ψ : L → L′ of the Lie algebra L onto the Lie algebra L′ is said to be ahomomorphism iff
(i) ψ is linear, i.e., ψ(αa+ βb) = αψ(a) + βψ(b).(ii) ψ([a, b]) = [ψ(a), ψ(b)].
The homomorphism ψ is said to be an isomorphism if it is 1-1. Because of these propertiesthe structure constants of L and L′ coincide, whenever the two Lie algebras are isomorphic.
Definition discrete subgroup
The subset K ⊂ G of a Lie group G is said to be a discrete subgroup of G if either K is afinite subgroup, or if K is countably infinite and such that there exists a neighbourhoodof the identity E that contains no element of K, except E itself.
Clearly, discrete subgroups are not Lie groups. Nevertheless, they are important here,as they play a role in the mapping of different Lie groups onto each other.
Theorem 28If the kernel K of an analytic homomorphism φ : G → G′ is discrete, then the corre-sponding mapping ψ : L→ L′ is an isomorphism.
Remark
This theorem tells us that different Lie groups may have the same Lie algebra. Thisis intuitively plausible. The Lie algebra is concerned with the different analytic curvesthrough the identity and one can restrict to a small neighbourhood of the identity, suchthat φ and hence ψ are 1-1.
79
Example 44 G = SU(2), G′ = SO(3)We can define a homomorphism φ : U 7→ O from SU(2) onto SO(3) in the following way
Ojk = [φ(U)]jk = 12Tr σjUσkU−1. (7.35)
The indices j and k take three different values, so this mapping takes a two-dimensionalunitary matrix into a three-dimensional orthogonal matrix. The mapping φ is analytic.
The matrices U and −U = SU = US are both mapped into the same matrix O. Thekernel of φ is K = E, S. We see that the kernel of the homomorphism is discrete, sothe Lie algebras of SU(2) and SO(3) must be isomorphic. Indeed, the elements ai, foundto be the basis for su(2), see Eq. (7.6), are mapped onto
ψ(a1) =
0 0 00 0 10 −1 0
, ψ(a2) =
0 0 −10 0 01 0 0
, ψ(a3) =
0 1 0−1 0 00 0 0
. (7.36)
which is a basis for so(3), i.e. they are the generators of rotations in three dimensions. Itis easy to verify that from the commutation relations of these matrices ψ(ai) it followsthat the so(3) structure constants are f r
pq = −ǫpqr, indeed the same ones as for su(2).
7.6 Representations of Lie algebras
Now we would like to know whether one can derive a rep of the Lie algebra from a rep ofthe Lie group. The answer is given in the next theorem.
Theorem 29Let G be a linear Lie group with Lie algebra L, DG an analytic d-dimensional rep of G.Then
(i) There exists a d-dim rep DL of L defined by
DL(a) =d
dtDG(e
ta)
∣∣∣∣t=0
. (7.37)
(ii) For all a ∈ L and for all t ∈ R we have
exp(tDL(a)) = DG(exp(ta)). (7.38)
(iii) If two reps of G are equivalent, then the corresponding reps of L are equivalenttoo. The converse also holds if G is connected.
(iv) DG is irreducible if DL is. The converse also holds if G is connected.
Result (iv) is important because it implies that in order to classify the irreps of a Liegroup, it is sufficient to classify the irreps of its Lie algebra, which is usually much easier.It is, however, important to note that the above theorem presupposes that a rep of theLie group is given. In general, the Lie algebra may have more irreps than the Lie group.As example 44 of the previous section shows different Lie groups may have the same Liealgebra. Therefore, not all reps of a Lie algebra become upon exponentiation reps of allLie groups with that Lie algebra. We will return to this issue at the end of this chapter.
Next we discuss a frequently encountered rep of Lie algebras. The structure constantscan be used to define a rep themselves, the so-called adjoint representation. It is importantfor the classification of Lie algebras
80
Definition adjoint representation
Let L be an n-dimensional Lie algebra with basis a1, . . . , an. The adjoint rep ad isdefined by
[a, aj ] =
n∑
k=1
ak[ad(a)]kj . (7.39)
One can verify easily that the matrix elements of ad(a) are related to the structureconstants:
[ad(ap)]kj = fkpj. (7.40)
In order to check that the structure constants themselves indeed furnish a representationof the Lie algebra, one can apply Eq. (7.40) to Eq. (7.12) to obtain:
f rqmf
spr + f r
mpfsqr + f r
pqfsmr = 0, (7.41)
which is a correct relation due to the Jacobi identity. To verify this insert Eq. (7.12) into
[ap, [aq, am]] + [aq, [am, ap]] + [am, [ap, aq]] = 0. (7.42)
For the so(3) structure constants f rpq = −ǫpqr, Eq. (7.41) also follows from the identity
ǫijk ǫklm = δil δjm − δim δjl. (7.43)
The adjoint rep of a Lie algebra has some resemblance with the regular rep of a finitegroup, discussed in Chap. 3. For the latter the group elements themselves were used asa basis for the rep, forming the so-called carrier space of the rep. For the adjoint repthe basis elements of the Lie algebra form the carrier space. The adjoint rep maps Lonto itself, in a linear and invertible way, it is therefore in GL(L). The adjoint rep is ananalytic homomorphism from L to GL(L).
As before one is interested in finding all irreducible irreps. For finite groups thereare a host of tools to determine whether one is dealing with an irrep and whether onehas found all irreps. For Lie groups the situation is more complicated as the number ofelements is infinite and instead of summing over elements one has to define (invariant)integration over the group. This is done via the Haar measure, but we will not discussthis topic here.
For Lie groups the best tool to determine whether one is dealing with an irrep is stillSchur’s lemma. For compact linear Lie groups every rep is equivalent to a unitary repand for unitary reps of Lie groups Schur’s lemma applies in the following strong form.
Theorem 30A unitary rep D : G → GL(V ) of a Lie group G is an irrep iff the only operatorscommuting with D(T ), ∀T ∈ G, are scalar multiples of the unit operator.
We will discuss two important physics applications of this theorem. The first is thedetermination of the center of SU(n) and the second concerns the use of so-called Casimiroperators.
Definition center of a group
Let G be a group, then the subgroup Z(G) of all its elements that commute with allelements of the group, is called its center.
81
This applies to both discrete and Lie groups. Correspondingly, the center of a Liealgebra L is defined as X ∈ L|[X, Y ] = 0, ∀Y ∈ L.Example 45The center of the Lie group SU(n) is obtained as follows. Consider a d-dimensional irrepD of SU(n) and elements T of its center Z(SU(n)). It follows from TT ′ = T ′T thatD(T )D(T ′) = D(T ′)D(T ), ∀T ′ ∈ SU(n). From Schur’s lemma it follows that D(T ) = λI,where λ ∈ C. In order for D(T ) to be an element of SU(n), det(D(T )) = λn has to beequal to 1.
The defining rep of SU(n) (the n×n unitary matrices with determinant one) is an irrepand by definition faithful, hence, taking for D the defining rep implies that D(T ) = λIis a 1-1 mapping from the center to the n distinct roots of unity. Hence, we concludethat Z(SU(n)) is isomorphic to Zn. It is straightforward to apply similar steps for othergroups, as long as there exists a faithful irrep.
From this example it can also be seen that restricting an irrep of a group G to asubgroup H does not necessarily yield an irrep of H . The center is an Abelian subgroup(in fact the maximal Abelian invariant subgroup) and therefore only has 1-dimensionalirreps.
7.7 Universal Covering Group
The center of a linear Lie group also plays an important role in characterizing the globalproperties of that and related groups. If the discrete kernel of a homomorphism φ : G→ G′
corresponds to a central subgroup, i.e. a subgroup of the center, of the group G, then onecan infer global properties about G and G′.
Theorem 31If G is a connected linear Lie group, then there exists a simply connected Lie group Gsuch that
(i) G is analytically isomorphic to a factor group G/K, where K is a discrete Abelianinvariant subgroup of G.
(ii) If G is simply connected, then G ∼= G.(iii) The real Lie algebras L and L corresponding to G and G respectively, are isomor-
phic.(iv) Every rep of L is obtained by differentiation from a rep of G:
DL(a) =d
dtDG(e
ta)
∣∣∣∣t=0
. (7.44)
Definition universal covering group
If G is associated with G as described in the theorem above, it is said to be the universalcovering group of G.
Example 46SU(2) is the universal covering group of SO(3). The kernel K = E, S of the homo-morphism φ of Eq. (7.35) is the center of SU(2), which is isomorphic to Z2, such that onefinds that SU(2)/Z2
∼= SO(3) and su(2) ∼= so(3).
82
In order to make more tangible what this means physically, we explicitly considerrotations in three dimensional Euclidian space. The connection to quantum mechanicswill become evident.
7.8 Rotations in Euclidian space: SO(3) and SU(2)
An element T of SO(3), a rotation, can be represented by a 3 × 3 matrix acting onthree-dimensional vectors. If θ is the angle of rotation and n its axis, then one finds
R(T )ij = cos θδij + (1− cos θ)ninj + sin θǫijknk, −π ≤ θ ≤ π. (7.45)
The parameter space is a filled sphere of radius π with antipodes identified. This identi-fication led us earlier to the conclusion that SO(3) is not simply connected.
Now we discuss the same rotation in SU(2). It is represented by the 2 × 2 unitarymatrix
U(T ) = cosθ
21+ i sin
θ
2n · ~σ. (7.46)
The matrices σk are the familiar Pauli matrices. The angle θ is now allowed to range overthe interval [−2π, 2π] (in this way the parameter space can be viewed as a sphere of radius2π, with all points on the surface identified with each other) and we see that the anglesθ and θ+ 2π correspond to the same rotation, but the corresponding matrices are U and−U . Recall that a spin-1/2 state (e.g. spin ‘up’) goes over into minus that state after arotation over 2π (around any axis). This two-to-one relation between SU(2) and SO(3)was explicitly seen from the kernel of the homomorphism φ between these two groups inSect. 7.5.
Now we fix n. Then the matrices given in Eq. (7.45) form a one-parameter subgroup.We can easily determine its generator
a = limθ→0
R(T )− 1
θ=
0 n3 −n2
−n3 0 n1
n2 −n1 0
=
∑
i
niψ(ai), (7.47)
where the matrices ψ(ai) are defined by Eq. (7.36).We can do the same for U(T ). Then we find
a′ = limθ→0
U(T )− 1
θ=i
2n · ~σ =
∑
i
niai. (7.48)
As we have seen, the Lie algebras so(3) and su(2) are isomorphic: the matrices ψ(ak) and(i/2)σk satisfy the same commutation relations. They are two different reps of the sameLie algebra generators. Moreover, they are irreps as we will see in the next chapter. Herewe note that ψ(ak) is in fact the Lie algebra basis element ak in the adjoint rep, e.g.
ψ(a1)jk = −ǫ1kj = f j1k = [ad(a1)]jk. (7.49)
In conclusion, we have seen explicitly in this section that not every rep of su(2) yieldsa rep of SO(3) upon exponentiation: exp(
∑3i=1 θiai) does not form a 2-dimensional rep
of elements in SO(3). It forms what is called a projective representation:
D(T1)D(T2) = exp(iφ(T1, T2))D(T1T2), (7.50)
where φ is a real phase that depends on the elements T1 and T2. Spin states of spin-1/2particles thus transform under a projective rep of SO(3).
83
7.9 Exercises
Exercise 7.1
Show that the connected component of a Lie group that contains the identity E is aLie-subgroup of G (hint: first show that a subset H of a group G is a subgroup of G, ifS1S
−12 ∈ H , ∀S1, S2 ∈ H) and moreover, is an invariant Lie-subgroup of G.
Exercise 7.2
Show that SO(3)/SO(2) ∼= S2 and that it is not a Lie group.
Exercise 7.3
Prove Theorem 23.
Exercise 7.4
Demonstrate for the matrices given in Eq. (7.21) that T (t) follows from exponentiatingA.
Exercise 7.5
Verify the correctness of Eq. (7.25).
Exercise 7.6
Derive the matrices in Eq. (7.36) from Eq. (7.35).
Exercise 7.7
Verify that the adjoint rep is a rep.
84
Chapter 8
Irreps of su(n)
In physics the groups SU(2) and SO(3) are very important, especially in quantum me-chanics. Since we can obtain their irreps from the irreps of their Lie algebra su(2) ∼= so(3),we will study the irreps of su(2) in further detail in this chapter. For explicit irreps ofSO(3) and SU(2) we refer to textbooks on quantum mechanics. In this chapter we willalso address the irreps of su(n), and hence of SU(n), for arbitrary n using Young tableaux.
8.1 Irreps of su(2)
As SO(3) and SU(2) are compact, we can take each irrep to be unitary. Let φkdk=1 bea basis for a d-dim irrep of su(2). The operators Φ(a) act on this basis as follows
Φ(a)φk =d∑
j=1
φjD(a)jk. (8.1)
The matrices D must be anti-hermitian, therefore one defines the matrices Jk = −iD(ak),which are hermitian, traceless d× d matrices. They fulfil the familiar commutation rela-tions of the angular momentum operators
[Jj , Jk] = iǫjklJl. (8.2)
Now define the ladder operators J± by
J± = J1 ± iJ2 (8.3)
and the Casimir operator1
J2 = J21 + J2
2 + J23 , (8.4)
then J2 commutes with every Jk. Schur’s lemma then implies that for an irrep D, J2 isa multiple of the unit matrix. Then all basis vectors φk are eigenvectors of J2 with thesame eigenvalue.
What are the eigenvalues? The commutation relations Eq. (8.2) imply
[J2, J±] = 0, [J3, J±] = ±J±, [J+, J−] = 2J3. (8.5)
1A Casimir operator of second order is a quadratic operator (i.e. a linear combination of quadraticproducts of generators), with the property that it commutes with all elements of the Lie algebra. Casimir(1931) published a general procedure of how to construct such operators.
85
Also, J2 and J3 commute, such that we can simultaneously diagonalize them. We denotetheir eigenvalues by λj and m, respectively. The normalized eigenvectors are written as|j,m〉
J2|j,m〉 = λj|j,m〉 J3|j,m〉 = m|j,m〉. (8.6)
Now consider J+|j,m〉. It is an eigenvector of J2, as J2 commutes with J+. The relation
J3J+ = J+J3 + J+ = J+(J3 + 1) (8.7)
tells us thatJ3(J+|j,m〉) = J+(J3 + 1)|j,m〉 = (m+ 1)J+|j,m〉 (8.8)
So J+|j,m〉 is proportional to the eigenvector |j,m+ 1〉How to determine λj and m? We use the relations
J+J− = J2 − J23 + J3, J−J+ = J2 − J2
3 − J3. (8.9)
As J†+ = J− we find for any vector φ
〈φ|J+J−|φ〉 = 〈J−φ|J−|φ〉 ≥ 0, 〈φ|J−J+|φ〉 = 〈J+φ|J+|φ〉 ≥ 0. (8.10)
Apply these two inequalities to |j,m〉
λj −m2 ±m ≥ 0. (8.11)
The definition Eq. (8.4) implies that λj ≥ 0. These two inequalities are satisfied simulta-neously if
|m| ≤ −1 +√1 + 4λj
2. (8.12)
From here we infer that there is for any given value of λj a maximum value for |m|. Callthis value m = j. Then we must have
J+|j,m = j〉 = 0, (8.13)
otherwise there would exist an eigenvector with eigenvalue m = j + 1.Now we prove that λj = j(j + 1), with j ∈ 0, 1/2, 1, . . .. Take an eigenvector |j,m〉
then m ≤ j. Eq. (8.10) with |φ〉 = |j, j〉 tells us that the largest value of m and λj arerelated by
λj = j(j + 1) (8.14)
There exists an integer p ≥ 0 such that j − 1 < m+ p ≤ j. Consider now the vectors
J+|j,m〉, (J+)2|j,m〉, . . . , (J+)p|j,m〉. (8.15)
They are all eigenvectors of J2 and J3. Consider the vector J+(J+)p|j,m〉. As m + p ≤
j, |j,m + p〉 is not the null vector, but J+|j,m + p〉 is an eigenvector with eigenvaluem+ p+1 > j. As such eigenvectors cannot exist, we conclude that the integer p must besuch that m + p = j. Conclusion: j −m = p is an integer. In a completely similar way,now using J− instead of J+, one derives that j +m = q is also an integer. If these twoconditions are taken together one finds that j +m+ j −m = 2j = p+ q is an integer. Soj takes on the values 0, 1/2, 1, . . .
86
If we label the irreps with the value of j, say D[j], then we can easily determine thesimplest cases. Below we will give the irreps for j = 0, 1/2, 1 and 3/2 explicitly. Usingthe general formula
D[j](Jk)φjm =
j∑
m′=−j
φjm′D
[j](Jk)m′m (8.16)
we findD[j](Jk)m′m = (φj
m′, Jkφjm). (8.17)
For j = 0, D[0](Jk) = 0 and for j = 1/2 one obtains
D[1/2](Jk) =σk2. (8.18)
Note that in quantum mechanics the φ1/2m correspond to the two-component spinors used
to characterize a spin-1/2 state. We call the half-integer j reps “spinor reps”.For j = 1
D[1](J1) =1
2
0√2 0√
2 0√2
0√2 0
,
D[1](J2) = − i2
0√2 0
−√2 0√2
0 −√2 0
,
D[1](J3) =
1 0 00 0 00 0 −1
, (8.19)
which is equivalent, but not equal to, Eq. (7.36). The difference is that now the matrixD(J3) is diagonal, which is conventionally done when considering angular momentumstates |j,m〉. The φ1
m could correspond for example to the three m components of a spin-1 state (j = 1) or the components of a vector ~x in R
3 written as (−(x+iy),√2z, x−iy)/
√2
instead of (x, y, z).For j = 3/2 the irreps with J3 diagonal is
D[3/2](J1) =1
2
0√3 0 0√
3 0 2 00 2 0
√3
0 0√3 0
,
D[3/2](J2) = − i2
0√3 0 0
−√3 0 2 00 −2 0
√3
0 0 −√3 0
,
D[3/2](J3) =
32
0 0 00 1
20 0
0 0 −12
00 0 0 −3
2
. (8.20)
87
Remark on isospin
These representations are also frequently used in nuclear and particle physics. One well-known example is related to isospin symmetry. Subatomic particles can be labelled byisospin quantum numbers I, I3 in complete analogy to the quantum numbers j,m. Theapproximate symmetry of Nature under isospin “rotations” (rotations in some abstract“internal” vector space) allows one to classify particles by these quantum numbers. Theisospin rotations form a group (SU(2)) and the particles transform according to the irrepsof this group. For instance, apart from a small mass difference, the proton and the neutrononly differ by their electric charge. Therefore, one can view them as the m = 1/2 and
m = −1/2 components of the set φi=1/2m (usually referred to as a doublet). Under an
isospin rotation the proton and the neutron only transform among themselves via thegroup elements generated by the D[1/2](Ji) given above. Similarly, the φi=1
m (forming
a triplet) and φi=3/2m (forming a quadruplet) are identified with sets of particles (the
π’s and ∆’s, respectively) that only differ by their electric charge (ignoring again smallmass differences). Apart from being a way to classify subatomic particles, one can alsouse these isospin notions (which are due to Heisenberg) to make predictions for ratiosof decay rates and of cross sections. This is another example of the usefulness of grouptheory in physics.
8.2 Irreps of su(3)
The irreps of su(2), and hence of SU(2), are labelled by the eigenvalues of the Casimiroperator J2. It is also common practice to label them simply by the dimension of theirrep: d = 2j + 1. Taking direct products of irreps is a way to generate other irreps, viathe Clebsch-Gordan decomposition which decomposes a direct product of two irreps interms of irreps again. For SU(2) we recall (see Chap. 4) that this is
(2j1 + 1)⊗ (2j2 + 1) =
j1+j2⊕
j3=|j1−j2|
(2j3 + 1). (8.21)
This is a result specific to SU(2) (and SO(3) for integer j). For j1 = j2 = 1 one obtains
3⊗ 3 = 1⊕ 3⊕ 5. (8.22)
If one were to consider SU(3) instead, then the decompositions of direct product reps arein general different, e.g.
3⊗ 3 = 3∗ ⊕ 6, (8.23)
where 3∗ denotes an inequivalent rep of dimension three, the complex conjugate rep.Clearly, if D(T ) forms a rep for elements T , then D∗(T ) also forms a rep, since
D(T1)D(T2) = D(T1T2) implies D∗(T1)D∗(T2) = D∗(T1T2). Sometimes the complex con-
jugate rep can be equivalent to the rep itself, as is for instance the case for SU(2) forwhich the 3 and 3∗ are equivalent, but for SU(3) they are not.
Also, for SU(3) one has3⊗ 3∗ = 1⊕ 8. (8.24)
These decompositions are most easily obtained by using Young tableaux (next section).
88
Although we have labelled the irreps of SU(3) by their dimension, one sees that theremay be inequivalent irreps of the same dimension. The irreps of SU(n) for n ≥ 3 are notuniquely labelled by just one number, but rather by n−1 numbers corresponding to n−1distinct Casimir operators.
8.3 Irreps of su(n)
In order to find the irreps of su(n) and the decomposition of products of its irreps intoirreps, it is convenient to use Young tableaux. It is based on the determination of allirreps of the symmetric group Sm, the (finite) group of all permutations of m objects.
The group SU(n) consists of n×n unitary matrices with determinant equal to 1. Thematrices U ∈ SU(n) act on the complex vector basis ψi with i = 1, . . . , n as lineartransformations. The set ψi forms the basis of the so-called defining rep n. As shouldbe clear by now, tensor products of these ψi are not irreducible in general. The irreducibledecomposition consists of tensor products of specific symmetry properties, just as it wasthe case for SO(3), cf. Eq. (4.37). These symmetry properties are determined by theirreps of Sm, where m is the number of indices of the tensor product ψi1 · · ·ψim . SU(n)transformations preserve the symmetry of the indices under permutations. Invariant sub-spaces and hence irreps are thus classified by these symmetries under Sm. The symmetryunder permutations of the m indices is expressed by a Young tableau and a correspondingoperator, as explained in Chap. 6.
The Young tableaux for a tensor product of m vectors (ψi1 · · ·ψim) are given by aframe with m boxes as illustrated for the case of m = 8 below.
To each standard tableau of this irrep of Sm corresponds a Young operator, which actingon the tensor product ψi1 · · ·ψim will yield an irreducible tensor. For example, i j
corresponds to the symmetric combination ψiψj+ψjψi. Hence, in a tensor product of twovectors from C
n, the symmetric combination forms an irreducible combination: it staysinvariant under the action of S2 (it is a singlet) and the symmetric combinations for alli, j = 1 . . . n transform among themselves under SU(n) transformations. For n = 2 thereare three such combinations (ψ1ψ1, ψ1ψ2+ψ2ψ1, ψ2ψ2) and one antisymmetric combination(ψ1ψ2 − ψ2ψ1). This represents the product decomposition:
⊗ = ⊕ (8.25)
The box represents one ψi, the l.h.s. represents therefore ψiψj and the r.h.s. represents
the symmetric product and the antisymmetric product . For n = 2 this expresses
that 2⊗ 2 = 3⊕ 1.Each frame thus corresponds to an irrep of both Sm and of SU(n). Different frames
correspond to inequivalent irreps, even if the dimension of the irreps turn out to be the
89
same. In Chap. 6 we explained how to calculate the dimension of the irreps of Sm usingstandard tableaux (m!/H , where H is the hooks factor) and how to decompose a productof two irreps. For the irreps of SU(n) similar rules can be given.
We will first explain how to determine the dimension of the irreps of SU(n) from aframe and then how to multiply irreps and decompose them.
Dimension of the irreps
To obtain the dimension of an irrep corresponding to a certain frame we need to countthe number of standard tableaux. For SU(n) these are obtained by assigning a specificindex value (i1, . . . , im ∈ 1, . . . , n) to each box in the frame. The resulting tableau iscalled a standard tableau when the index numbers do not decrease when going from left
to right in a row and always increase from top to bottom. For example,1 12
is a standard
tableau, whereas1 21
and2 11
are not. Such non-standard tableaux either vanish or
are not independent of the standard ones.The dimensionality of an irrep is equal to the number of standard tableaux, which is
given by the number
d(ν1, ν2, . . . , νn−1) = (1 + ν1)(1 + ν2) · · · (1 + νn−1)
×(1 +
ν1 + ν22
)· · ·
(1 +
νn−2 + νn−1
2
)
×(1 +
ν1 + ν2 + ν33
)· · ·
(1 +
νn−3 + νn−2 + νn−1
3
)
×(1 +
ν1 + . . .+ νn−1
n− 1
), (8.26)
where the numbers νi are determined via the number of boxes λi in row i, such thatνi = λi−λi+1. Here i takes at most the value n−1, because there can be no more than nrows in a frame for symmetry reasons. A frame with more rows leads to a Young operatorthat yields a tensor which is antisymmetric in more indices than there are different valuesfor the indices. This always leads to a vanishing tensor combination.
Another way of calculating the dimensionality is to use the hooks factor H . Recallthat the hook length of a box a in a Young tableau is the number of boxes directly below,plus directly to the right of the box a, and counting the box a itself once. In the followingtableau for m = 6 the hook lengths are indicated for each box:
5 3 13 11
The hooks factor H is the product of the hook lenghts of all boxes of the tableau. Inthe example H = 45. The dimension of the irrep of SU(n) corresponding to the frameis equal to the ratio F/H , where F is determined as follows. Assign a factor n to theupper left hand corner of the tableau, then add 1 each time a step to the right is takenand subtract 1 each time a step downward is made. For the example above:
90
n n+1 n+2
n−1 n
n−2
The factor F is determined by multiplying the factors in each box: n2(n2 − 1)(n2 − 4).The dimension of the irrep corresponding to this example for n = 3 is therefore 8, whichindeed coincides with d(1, 1), for n = 4 it is 64, which coincides with d(1, 1, 1).
Multiplication and decomposition of irreps
Next we explain the methodology of CG decomposing a product of irreps of SU(n). Itfollows the rules of the outer product of the symmetric groups explained in Chap. 6, withsome small differences. This is understandable since when multiplying frames of m andk boxes, corresponding to a product of tensors with m and k indices, frames with m+ kboxes result and hence a tensor with m+ k indices.
When “multiplying” two Young tableaux A and B, one starts by labeling the boxesin the top row of B with a’s, the second row with b’s, etc. Then one starts by takingthe boxes from the top row of B and attaching them to A, building to the right and/ordown. One has to retain only the legitimate frames, which are the ones for which theboxes in the rows are not increasing when going down and idem in the columns whenmoving right. Also, not more than one a can appear in each column. Then follows thesecond row (with b’s), and so on. Columns with n boxes do not have to be drawn, theycan be “crossed out”. Frames with columns with more than n boxes are zero, so thoseframes can be left out completely. After all letters have been distributed, one keeps onlythose constructed tableaux for which the following holds. If one reads the letters alongthe rows from right to left from the top row down to the bottom row, then the sequencemust satisfy the requirement that to the left of any symbol there are no fewer a’s thanb’s, no fewer b’s than c’s, etc. For example,
a bb ac (8.27)
yields baabc and is not allowed, as opposed to for instance ababc.The simplest example is the product of two single boxes:
⊗ = ⊕ (8.28)
For n = 3 the l.h.s. corresponds to 3 ⊗ 3 and the r.h.s. to 6 ⊕ 3∗. As explained before,the star indicates that one is dealing with the complex conjugate irrep, more about thisbelow. Thus, Eq. (8.28) and Eq. (8.23) express the same result for SU(3).
A more complicated example is:
⊗ = ⊕ ⊕ ⊕ (8.29)
91
The complex conjugate irrep
To obtain the Young tableaux for conjugate reps the procedure is as follows. Take theYoung tableau for an irrep k and append boxes to each column until all columns have kboxes. In this way one obtains a Young tableau of several columns that can all be crossedout, which yields the 1 irrep. The added boxes together also form a Young tableau whenrotated by 180. This is the Young tableau corresponding to the conjugate rep k∗. For
example, in SU(3) the rep 3 is , hence 3∗ is given by the Young tableau .
With the above rules we can obtain that 3∗ ⊗ 3 = 8⊕ 1:
⊗ a = a ⊕a
(8.30)
Note that the calculation is simpler in this way (3∗⊗3) than calculating it the other wayaround: 3⊗ 3∗. Also, if one is interested in for instance 3∗ ⊗ 3∗ it is simpler to take theresult 3⊗ 3 = 6⊕ 3∗ and taking its complex conjugate: 3∗ ⊗ 3∗ = 6∗ ⊕ 3.
8.4 Application: SU(6) Proton Wave Functions
In this section we return to the study of the wave function of a system of three spin-1/2particles, more specifically, the proton wave function in terms of quark states. The protonwave function must transform under irreps of several groups. The proton consists of threequarks: two up quarks and one down quark. The up and down quark states can be viewedas basis states of the i = 1/2 irrep of the isospin symmetry group SU(2)isospin, or as twoout of the three “flavor” states of SU(3)flavor. In the latter case one takes into accountthe strange quark as well. Each of the quarks has spin 1/2, therefore their spin states arein an s = 1/2 irrep (the defining rep) of the spin symmetry group SU(2)spin. We assumethat the spatial part of the wave function is in an S-wave (symmetric). Quarks also carryanother internal quantum number called color, which transforms according to the definingrep of SU(3)color. For reasons that are beyond the scope of these lectures the color part ofthe wave function must be fully anti-symmetric under the interchange of any two of thethree quarks. This implies that the spin times isospin part of the wave function has to besymmetric, the state ψS of Chap. 6, Example 31. This is from the perspective of S3. Thespin wave function that transforms under an irrep of S3 was multiplied with the isospinwave function that also transforms under an irrep of S3. The product wave function alsoneeds to transform under an irrep of S3, which required performing the Clebsch-Gordandecomposition (the inner product). Now we are going to discuss the spin-isospin/flavorwave function from the perspective of SU(2)spin⊗SU(2)isospin and SU(2)spin⊗SU(3)flavor,viewed as a subgroup of SU(6).
The spin wave function of the proton is itself a product of three spin wave functions ofthe quarks. The latter are given in the defining representation of SU(2). The proton spinwave function is then obtained using the product rules explained in the previous section.Each quark spin state is represented by a box, therefore, we have to calculate the productof three such boxes. This yields:
⊗ ⊗ = ⊕ ⊕ ⊕ (8.31)
92
The Young tableaux on the r.h.s. all correspond to irreps of S3 and the amount of timeseach irrep appears equals the dimension of that S3 irrep. For isospin wave functions thisis completely analogous.
For SU(2) Eq. (8.31) corresponds to
2⊗ 2⊗ 2 = 4⊕ 2⊕ 2, (8.32)
and for SU(3) it corresponds to
3⊗ 3⊗ 3 = 10⊕ 8⊕ 8⊕ 1. (8.33)
The spin-isospin/flavor part of the proton’s wave function is in an irrep of the productgroup SU(2)spin ⊗ SU(2)isospin or SU(2)spin ⊗ SU(3)flavor. From the fact that the protonhas spin-1/2 and isospin-1/2 one can conclude that it is actually in the 2⊗2 or 2⊗8 rep.
The spin wave functions were discussed before in Chap. 6 and were denoted by Ψ±i .
The isospin wave functions were denoted by X±i , where X
+i is relevant for the proton and
X−i for the neutron. Both the spin and the isospin wave functions are mixed symmetric,
but their product should be symmetric under the interchange of any two of the threequarks, as mentioned, which is (suppressing superscripts ±):
ψS =1√2(X1Ψ1 +X2Ψ2) (8.34)
This is an example of an irrep of a direct product group (SU(2)spin ⊗ SU(2)isospin). Fromthe perspective of S3 it is however a direct product irrep. Both Ψ and X transform underan irrep of S3, their direct product is again a rep of S3 which can be CG decomposed intoirreps. One of those irreps is fully symmetric and corresponds to ψS.
Sometimes people consider the product group SU(2)spin ⊗ SU(3)flavor as part of thelarger group SU(6). For SU(6) Eq. (8.31) corresponds to
6⊗ 6⊗ 6 = 56⊕ 70⊕ 70⊕ 20. (8.35)
The proton belongs to the symmetric 56 irrep. Similarly one can use Young tableaux tofind out which representations arise when embedding a direct product of groups (e.g. theStandard Model SU(3)⊗ SU(2)⊗ U(1)) into a larger group (e.g. SU(5) or other “grandunified theory” groups). For a discussion of this we refer to Georgi (1982).
All baryons consisting of the three light quarks fit into this scheme of irreps of SU(2)⊗SU(3) or SU(6), although the strange quark mass breaks the SU(3) symmetry to a largerextent. Similarly, the mesons (consisting of a quark and an antiquark) can be classifiedaccording to irreps of these SU(n) groups. For hadrons involving the heavier quarks (forinstance charm) the flavor symmetry is broken too much due to the large mass differencesto be of much use, other than to classify all possible mesons.
93
8.5 Exercises
Exercise 8.1
Explicit construct the relationship between the matrices ψ(ai) given in Eq. (7.36) andD[1](J1) in Eq. (8.19).
Exercise 8.2
Verify explicitly the correctness of Eq. (8.20).
Exercise 8.3
Check explicitly that D[3/2](J2) is what one expects for j = 3/2.
Exercise 8.4
Show that a rep of SU(2) and its complex conjugate are equivalent.
Exercise 8.5
Decompose ⊗ into irreps for su(n) and count the dimensions of the irreps for
su(3) by using the hooks factors and verify some of these dimensions by using Eq. (8.26).
94
Chapter 9
The Lorentz group
In this chapter we discuss the Lorentz group and its representations. After a brief sectionon the Lorentz group, the connection of this group with the group SL(2,C) of linear trans-formations in a two-dimensional complex space is discussed. Then the representations ofSL(2,C) are given and in addition those of the Lorentz group.
9.1 Lorentz transformations
The mathematical theory of Lorentz transformations can be considered as the geometryof Minkowski space. The basic elements of this space are the four-vectors x. These vectorshave contravariant (denoted by upper indices) components xµ and covariant components(lower indices) xµ. The relation betweem them is given by the metric tensor g in thefollowing way:
xµ = gµνxν , xµ = gµνxν , µ, ν ∈ 0, 1, 2, 3. (9.1)
The geometry of Minkowski space is fully determined by specifying the metric tensor. Ithas the form:
gµν = gµν = diag (1,−1,−1,−1). (9.2)
Actually, by postulating that the relation between contravariant and covariant compo-nents of any tensor is given by a relation of the type Eq. (9.1), one finds that the followingrelation is consistent:
gµν = gµαgαν = δµν = g µν . (9.3)
The symbol δµν is just Kronecker’s delta, and not viewed as a tensor. It is given by:
δµν = δµν = δµν = diag (1, 1, 1, 1). (9.4)
The line-element ds is given by the square root of the expression:
ds2 = gµνdxµdxν = dxµdxµ = dxνdx
ν . (9.5)
Finally, in order to give meaning to the concepts of vector and tensor, we specify theadmissible coordinate transformations: any transformation Λ that leaves the elements ofthe metric tensor invariant is a bona fide transformation in Minkowski space. The matrixelements of such a transformation (a rep of Λ) shall be denoted by Lµ
ν . (It is importantto realize that these numbers are not the elements of a mixed tensor!)
95
So we shall write:x′µ = Lµ
νxν , (9.6)
and require the identity:gµν = Lα
µLβνgαβ. (9.7)
It is useful to interpret Eq. (9.7) as a relation between matrices. For the objects Lµν
we interpret the upper index as a row index, the lower one as a column index, whereasfor g the left index is the row index and the other the column index. Then Eq. (9.7) istranslated into the matrix form:
g = LTgL, (9.8)
where LT is the transposed matrix with elements:
[LT ]νµ = Lµν . (9.9)
Because, as a matrix, g2 = 1, Eq. (9.8) tells us that the inverse of Λ is represented bythe matrix gLTg, which can be translated into:
[L−1
]µν= gµα
[LT
]αβgβν = gµαLβ
αgβν . (9.10)
(If x and y are two four-vectors connected by a Lorentz transformation as follows: yµ =Lµ
νxν , then we find that xν = Lµ
νyµ, which is equivalent to: xν =(gναLβ
αgµβ)yµ.)
A matrix L that satisfies Eq. (9.10) is called a pseudo-orthogonal matrix, in this caseof the type O(3, 1), where the ‘3’ and the ‘1’ reflect the number of minus and plus signsin the metric tensor g. If in addition detL = 1 holds, then L ∈ SO(3, 1).
Working out Eq. (9.10) for some matrix elements we find:
(L−1)00 = L00, (L
−1)0i = −Li0, (L
−1)i j = Lji. (9.11)
Equation (9.7) has the immediate consequence that the line element Eq. (9.5) is in-variant. Actually, if x and y are any two four-vectors, then the dot-product x · y, definedby:
x · y = gµνxµyν = xµyµ = xνy
ν, (9.12)
is invariant: if x′ = Lx and y′ = Ly then x′ · y′ = x · y. An identity that shall prove usefulis:
(Lx) · y = x ·(L−1y
). (9.13)
A consequence of Eq. (9.8) is that (detL)2 = 1, therefore it follows that detL = 1 ordetL = −1. Upon substituting µ = ν = 0 in Eq. (9.7) one finds:
1 = gµνLµ0L
ν0 = (L0
0)2 −
∑
i
(Li0)
2. (9.14)
Thus we find:(L0
0)2 = 1 +
∑
i
(Li0)
2 ≥ 1. (9.15)
One can check (exercise 9.1) that if two Lorentz transformations Λ1 and Λ2 have theproperty L0
0(Λ1) ≥ 1 resp. L00(Λ2) ≥ 1, then their product Λ3 has the same property:
L00(Λ3) ≥ 1.Thus we have found that the Lorentz group has four components:
96
(i) L↑+ : The set of Lorentz transformations with detL=1 and L0
0 ≥ 1, proper or-
thochronous;(ii) L↑
− : detL = −1, L00 ≥ 1, non-proper orthochronous;
(iii) L↓+ : detL = +1, L0
0 ≤ −1, proper non-orthochronous;
(iv) L↓− : detL = −1, L0
0 ≤ −1, non-proper non-orthochronous.Because detL is a continuous function of its matrix elements, these four components
are mutually distinct. Moreover, since det I = 1 and I00=1, only elements of L↑+ (also
called SO+(3, 1)) are continuously connected to the identity of the Lorentz group, I.
9.1.1 Physical Interpretation
Minkowski space is relevant for physics if the four-vector x is interpreted as:
xµ = (x0, x1, x2, x3) = (x0, ~x) = (ct, ~r). (9.16)
We shall study some concrete examples. It is important to stress that we will inter-pret the transformations as active transformations: Eq. (9.6) is a relation between thecomponents of two different vectors x and x′, with respect to one system of coordinateaxes. We may write:
x = xµe[µ], x′ = x′µe[µ]. (9.17)
The unit vectors e[µ] are examples of vectors which can be moved actively to e′[µ] by aLorentz transformation Λ. The covariant components of the unit vectors are given by:
e[µ]ν = δµν . (9.18)
Now we give the detailed form of the matrices representing two particular Lorentztransformations, rotations and pure Lorentz transformations (also called boosts). Theseare important since any Lorentz transformation in L↑
+ can be uniquely written as a rota-tion followed by a boost.
(i) RotationsAny transformation R that leaves x0 unaltered and rotates ~x is a Lorentz transforma-
tion. The associated matrix has the form
Lµν =
1 0 0 000 Ri
j
0
. (9.19)
The matrix R is orthogonal and has determinant 1:
detR = 1, R−1 = RT . (9.20)
A convenient parametrization of rotations is given by specifying the axis of rotation nand the rotation angle θ. Then one finds the explicit formula:
~x ′ = ~x cos θ + (n× ~x) sin θ + n(n · ~x)(1− cos θ). (9.21)
Accordingly, the matrix elements of R are:
Rij = δij cos θ + ninj(1− cos θ) + ǫikjn
k sin θ. (9.22)
97
For a rotation about the z-axis, n = (0, 0, 1), the matrix R has the well-known form
[Ri
j
]=
cos θ − sin θ 0sin θ cos θ 00 0 1
. (9.23)
(ii) Pure Lorentz transformations
The general case of a pure Lorentz transformation is
x′ 0 = γx0 + γ ~β · ~x
~x ′ = γ ~βx0 +γ2
1 + γ~β (~β · ~x) + ~x (9.24)
with~β =
~v
c, γ =
1√1− ~β2
. (9.25)
Because β = |~β| ≤ 1, we can parametrize the pure Lorentz transformations alternatively
in terms of an angle χ. Using the definitions γ = coshχ, γβ = sinhχ and β = ~β/β, oneobtains the representation
x′ 0 = x0 coshχ+ β · ~x sinhχ~x ′ = βx0 sinhχ+ β (β · ~x)(coshχ− 1) + ~x. (9.26)
The case β = e[z] gives
(Lµν) =
γ 0 0 γβ0 1 0 00 0 1 0γβ 0 0 γ
=
coshχ 0 0 sinhχ0 1 0 00 0 1 0
sinhχ 0 0 coshχ
. (9.27)
9.1.2 The Lorentz algebra
In order to find the generators of the Lie algebra related to the Lorentz group we studythe infinitesimal transformations. The Lie algebra has six basis elements.
In the case of a rotation we have:
~x ′ ≈ ~x+ (n× ~x)θ, (9.28)
so for the corresponding matrix we find:
Rij ≈ δij + ǫikjn
kθ ≡(I − iθn · ~J
)i
j. (9.29)
The notation ~θ = θn is useful and shall be employed in Sect. 9.2.1.The generators ~J are a subset of the six generators denoted Mµν of the Lorentz group.
The relation is:Jk = ǫklmM
lm, (no summation over l and m). (9.30)
98
Explicitly:
J1 =M23 =
0 0 0 00 0 0 00 0 0 −i0 0 i 0
,
J2 =M31 =
0 0 0 00 0 0 i0 0 0 00 −i 0 0
,
J3 =M12 =
0 0 0 00 0 −i 00 i 0 00 0 0 0
. (9.31)
A general expression for the matrix elements of the generators M lm can be writtendown easily:
(M lm)jk = i(gljgmk − gmjglk). (9.32)
For pure Lorentz transformations the infinitesimal form is
x′ 0 ≈ x0 + ~x · ~χ~x ′ ≈ x0~χ+ ~x. (9.33)
Here we introduce the vector ~χ = χn, n denoting the axis along which the pure Lorentztransformation is taken. The corresponding generators have the matrix representations
K1 =M01 =
0 i 0 0i 0 0 00 0 0 00 0 0 0
,
K2 =M02 =
0 0 i 00 0 0 0i 0 0 00 0 0 0
,
K3 =M03 =
0 0 0 i0 0 0 00 0 0 0i 0 0 0
. (9.34)
Note that these are anti-hermitian matrices as opposed to the hermitian matrices J i. Itimplies that the finite dimensional reps of the Lorentz group are not unitary (as it wouldbe for a compact group).
These formulae can be taken together to obtain a general form for the matrix elementsof the generators:
(Mµν)αβ = i(gµαgνβ − gναgµβ). (9.35)
99
Summarizing, we have established the following scheme:
M00 M01 M02 M03
M10 M11 M12 M13
M20 M21 M22 M23
M30 M31 M32 M33
=
0 K1 K2 K3
−K1 0 J3 −J2
−K2 −J3 0 J1
−K3 J2 −J1 0
(9.36)
The generators satisfy a set of commutation relations which can be checked easily bysubstitution of Eq. (9.35) for the elements of M . They are:
[Mµν ,Mρσ] = −i(Mµρgνσ +Mνσgµρ −Mνρgµσ −Mµσgνρ). (9.37)
These relations can be written in terms of the more familiar operators ~J and ~K as follows:
[J j, Jk
]= iǫjklJ
l,[J j , Kk
]= iǫjklK
l,[Kj , Kk
]= −iǫjklJ l. (9.38)
We end this section with the exponential form of a general Lorentz transformation.Now that we have the generators of the Lorentz group, we can easily express any Lorentztransformation in terms of them and a set of parameters.
Suppose the set of real numbers ωµν are such that ωµν = −ωνµ. Then we can write
any infinitesimal Lorentz transformation in L↑+ as:
L ≈ 1− i
2ωµνM
µν , (9.39)
with ωµν determined by Eqs. (9.29) and (9.33).Since the Lorentz group is non-compact, it is to be expected that not all elements can
be written as an exponential
L = exp
(− i2ωµνM
µν
). (9.40)
However, since any Lorentz transformation in L↑+ can be uniquely written as a rotation
followed by a boost, a general element of L↑+ is written as:
L = exp(−i ~χ · ~K
)exp
(−i ~θ · ~J
). (9.41)
In the specific 4×4 matrix rep employed above, elements in L↑−, L
↓+, L
↓− can be obtained
from those in L↑+ by multiplication by the matrices P µ
ν = diag (1,−1,−1,−1), T µν =
diag (−1, 1, 1, 1) and P µρT
ρν , respectively.
9.2 The Group SL(2,C) and its Connection to the
Lorentz Group
The group SL(2,C) is the group of complex linear transformations in two dimensions,such that the determinant of a matrix A ∈ SL(2,C) is unity. So, we can immediatelyconclude that SL(2,C) is a six (real) parameter group.
100
Now consider the space of two-dimensional hermitian matrices. This space is inti-mately linked to Minkowski space, as we see when we write an element X of it in thefollowing way:
X =
[x0 − x3 −x1 + ix2
−x1 − ix2 x0 + x3
]. (9.42)
The determinant of X is equal to the length of the four-vector x:
detX = (x0 − x3)(x0 + x3)− (x1 − ix2)(x1 + ix2) = xµxµ. (9.43)
But we can write X in another way, viz. by expanding it in the basis σµ = 1, σi,consisting of the 2× 2 identity matrix and the three Pauli matrices:
X = x01− x1σ1 − x2σ2 − x3σ3 = xµσµ. (9.44)
This expression can be read as an expansion of X in unit vectors. The basis vectorsare:
E[0] = 1, E[i] = −σi. (9.45)
So, it is useful to define(σµ) = (σ0,−~σ). (9.46)
Then we can write for the Pauli matrices the well-known trace identities in the follow-ing way:
Tr(σµσν) = 2gµν . (9.47)
From this identity one derives that the four-vector components xµ are obtained bytaking traces over X times the appropriate Pauli matrix:
xµ =1
2Tr(Xσµ) =
1
2Tr(xνσ
ν σµ) = xνgνµ. (9.48)
Next, consider the transformation
X ′ = AXA†, (9.49)
with A an element of GL(2,C), such that X ′ is also hermitian. In order for this transfor-mation to preserve detX , and, consequently, to be related to a Lorentz transformation,it is required that | detA|2 = 1. The relation to Lorentz transformations is most easilyestablished by a comparison of the two equations:
x′µ = Lµνx
ν (9.50)
and
x′µ =1
2Tr(X ′σµ). (9.51)
Upon substitution of Eq. (9.49) into Eq. (9.51), and using the relation (9.50) we obtain:
Lµνx
ν = x′µ =1
2Tr(X ′σµ) =
1
2Tr(Aσνx
νA†σµ) =1
2Tr(AσνA
†σµ)xν (9.52)
As x is an arbitrary four-vector, we find:
Lµν =
1
2Tr(σµAσνA
†). (9.53)
101
Clearly, any phase θ of detA = eiθ leads to the same Lorentz tranformation. Therefore,one can restrict oneself to detA = 1, which means A ∈ SL(2,C).
From Eq. (9.53) we calculate L00 = 1
2Tr(σ0Aσ0A
†) = 12Tr(AA†) > 0. Apparently,
L00 is positive, so the Lorentz transformations connected to SL(2,C) belong to the or-
thochronous part L↑. As the identity of the Lorentz group is obtained by taking for A the2×2 identity matrix, and the elements Lµ
ν are continuous functions of the matrix elementsof A, one sees that the Lorentz transformations determined by Eq. (9.53) belong to L↑
+.It is clear that A and −A determine the same Lorentz transformation Λ: Λ(A) = Λ(−A).So the map SL(2,C)→ L↑
+ is a homomorphism, but not an isomorphism. In other words,
L↑+ is isomorphic to SL(2,C)/Z2, which is not simply connected. Just like SU(2) was the
universal covering group of SO(3), SL(2,C) is the covering group of L↑+.
9.2.1 Examples
We discuss here the same examples as before: rotations and pure Lorentz transformations.
(i) RotationsThe subgroup SU(2) of SL(2,C) consists of unitary matrices U , which can be written
as:U(n, θ) = exp
(−i~θ · ~σ/2
)= σ0 cos(θ/2)− in · ~σ sin(θ/2). (9.54)
taking appropriate traces. The angle θ follows from:
Tr U(n, θ) = 2 cos(θ/2) (9.55)
and the components of the unit vector n are computed from:
Tr(U(n, θ)σj
)= −2inj sin(θ/2). (9.56)
Now, substitute U for A in Eq. (9.53). Then one obtains:
Lµν =
1
2Tr(σµUσνU
†). (9.57)
For µ = ν = 0 we find:
L00 =
1
2Tr(UU †) = 1, (9.58)
as U is unitary. In view of the relation (9.15) we can conclude that L0i = Li
0 = 0, (i =1, 2, 3). Next we derive the matrix elements Li
j. To do so, we need the following identityfor the σ-matrices:
σnσm = δnm + iǫnmlσl. (9.59)
Then we find, using σi = −σi and σi = −σi, the formula:
Lij =
1
2Tr
(σi
[σ0 cos(θ/2)− inkσk sin(θ/2)
]σj
[σ0 cos(θ/2) + inlσl sin(θ/2)
])
= cos2(θ/2)1
2Tr(σiσj) + sin2(θ/2)nink 1
2Tr(σiσkσjσl)
+ sin(θ/2) cos(θ/2)nl 1
2Tr(σi[σj , σl]). (9.60)
102
The first trace gives δij ; the second can be worked out using the identity (9.59). It gives:
1
2Tr(σiσkσjσl) = δikδjl − δijδkl + δilδjk. (9.61)
The final result is:
Lij = δij cos θ + ninj(1− cos θ) + ǫiljn
l sin θ, (9.62)
which is identical to Eq. (9.22).So, the subgroup SU(2) of SL(2,C) is mapped on the rotations in Minkowski space.
Because SU(2) is a group and the map from SL(2,C) to L↑+ is a homomorphism, the
rotations form a subgroup of L↑+.
(ii) Pure Lorentz Transformations
Consider the hermitian matrices A. The product of two hermitian matrices is notalways hermitian, so this subset of SL(2,C) is not a group. We can expand A in the sameway as X . The analogue of Eq. (9.44) is:
A = Aµσµ, (9.63)
where the four parameters Aµ must be real. We have seen that detA = AµAµ, so in order
to belong to SL(2,C) the matrix A must be such that AµAµ = (A0)2 − ( ~A)2 = 1. This
equation is satisfied by a four-vector with components:
Aµ = (cosh(χ/2), n sinh(χ/2)). (9.64)
Then the matrix A has the form
A = exp(−~χ · ~σ/2) = cosh(χ/2)− n · ~σ sinh(χ/2). (9.65)
The similarity with the expression (9.54) for U is immediately clear, and so is theprocedure one can follow to calculate Lµ
ν(A). We find for the 00-element:
L00(A) =
1
2Tr(AA†) =
1
2Tr(A2) =
1
2Tr(coshχ− sinhχ n · ~σ) = coshχ. (9.66)
A calculation that is almost as simple as the one before gives:
Li0(A) = L0
i(A) =1
2Tr(AσiA) = −2AiA0 = ni sinhχ. (9.67)
The calculation of Lij follows the same pattern as in the previous case. The result is
of course:Li
j = δij + ninj(coshχ− 1), (9.68)
which corresponds to a pure Lorentz transformation along the unit vector n, with velocityc tanhχ.
(iii) Combinations of Pure Lorentz Transformations and Rotations
Because the hermitian matrices do not form a group, the product of two pure Lorentztransformations is not necessarily a pure Lorentz transformation. But one can split any
103
Lorentz transformation into a rotation and a pure Lorentz transformation. We shall showhow this is done on the level of SL(2,C).
Let A belong to SL(2,C). We write A = HU , where U is unitary and H hermitian.This can always be done in a unique way (modulo a sign as shall be shown shortly). Theproduct AA† = HUU †H† = H2 is just the hermitian operator that occurs in Eq. (9.66).So, if we write H = cosh(χ/2)− n · ~σ sinh(χ/2), then we find
AA† = H2 = coshχ− n · ~σ sinh(χ). (9.69)
The hyperbolic angle χ is determined by taking the trace, as in Eq. (9.66), and thecomponents of the vector n sinhχ are computed by taking the following traces:
coshχ =1
2Tr(AA†), ni sinhχ =
1
2Tr(AA†σi). (9.70)
It is clear that H is not uniquely determined: its sign is arbitrary.The matrix U is now calculated as H−1A, with H−1 = cosh(χ/2) + n · ~σ sinh(χ/2). If
we write U = cos(θ/2)− iu · ~σ sin(θ/2) then, by taking again the appropriate traces, wedetermine θ and u:
cos(θ/2) =1
2Tr U, ui sin(θ/2) =
i
2Tr(Uσi). (9.71)
(iv) Boosts to a specific Four-Momentum p
Consider the case of a particle with mass m. In the rest frame the four-momentum of
this particle, denoted byp, is of course
pµ = (m,~0). We want to determine the Lorentz
transformation to a reference frame where pµ = (p0, ~p). If no rotations are included, wecan write immediately:
coshχ = p0/m, sinhχ = |~p |/m, n = p = ~p/|~p |. (9.72)
The boost λ(p) in SL(2,C) is then
(p← p) ≡ λ(p) = cosh(χ/2)− p · ~σ sinh(χ/2). (9.73)
Upon using the identities for hyperbolic functions
cosh(χ/2) =√
(cosh(χ) + 1) /2,
sinh(χ/2) = sinh(χ)/ (2 cosh(χ/2)) ,
we obtain:
λ(p) =m+ pµσ
µ
√2m(p0 +m)
≡ (E +m)− ~p · ~σ√2m(E +m)
. (9.74)
Here we put the velocity of light c = 1 and used the notation E = p0. The corresponding4× 4-matrix is (Kµ
ν = Tr(σµλ(p)σνλ(p))/2):
K(p) =
Em
pxm
pym
pzm
pxm
1 + p2xm(E+m)
pxpym(E+m)
pxpzm(E+m)
pym
pypxm(E+m)
1 +p2y
m(E+m)pypz
m(E+m)pzm
pzpxm(E+m)
pzpym(E+m)
1 + p2zm(E+m)
. (9.75)
The matrix λ(p) transforms the hermitian matrix
P =pµσ
µ = mσ0 into P = pµσµ
according to the rule: P = λ(p)
Pλ(p) (recall X ′ = AXA†). The correctness of this
relation can be checked easily; this is also true for the relation pµ = Kµν(p)
pν .
104
9.3 Representations of the Lorentz group
The Lorentz group is not compact, as the hyperbolical angle χ ranges over the whole realaxis. For finite or compact groups G every rep D is equivalent to a unitary rep. As theLorentz group (and the Poincare group) are non-compact, we can expect some of theirreps to be non-unitary.
Finite-dimensional representations of L↑+
In Sect. 9.1.2 we have written down the commutation relations for the generators ~J and ~Kof the rotations and the pure Lorentz transformations. These relations can be uncoupledby taking the following complex, linear combinations:
~C =1
2( ~J + i ~K), ~D =
1
2( ~J − i ~K), (9.76)
which are hermitian, due to the fact that ~K is anti-hermitian.Straightforward application of the commutation relations eq. (9.38) leads to:
[Cj, Ck] = iǫjklCl,
[Dj, Dk] = iǫjklDl,
[Cj, Dk] = 0. (9.77)
Apparently, ~C and ~D are the generators of the group SU(2). Therefore, the possibleirreducible finite-dimensional reps of L↑
+ are of the form1 D[j1,j2] = D[j1]⊗D[j2], 2j1, 2j2 =0, 1, 2, · · ·. A canonical basis for this rep has dimension (2j1+1)(2j2+1) and is such that
C3f j1,j2m1,m2
= m1fj1,j2m1,m2
,
D3f j1,j2m1,m2
= m2fj1,j2m1,m2
,
C±f j1,j2m1,m2
= (C1 ± iC2)f j1,j2m1,m2
=√j1(j1 + 1)−m1(m1 ± 1)f j1,j2
m1±1,m2,
D±f j1,j2m1,m2
= (D1 ± iD2)f j1,j2m1,m2
=√j2(j2 + 1)−m2(m2 ± 1)f j1,j2
m1,m2±1. (9.78)
The operators ~C2 and ~D2 commute with any of the generators C i and Di and so areCasimir-operators (see next section). They are multiples of the identity in any irrep ofSU(2)⊗ SU(2)2.
Both ~C and ~D may be extended to the whole space by writing for a specific rep [j] ofSU(2):
~C = ~S [j1] ⊗ I [j2], ~D = I [j1] ⊗ ~S [j2]. (9.79)
The operators ~S [j] are the generators of SU(2) and I [j] is the identity-operator in the
carrier space of D[j]. This means that ~S [j] is the (2j+1)× (2j+1) dimensional matrix forthe total angular momentum, whereas I [j] is the (2j + 1)× (2j + 1) dimensional identity
1Note that performing a Clebsch-Gordan decomposition of the tensor product D[j1] ⊗D[j2] does notmake sense for applications in physics. It would require that the states can be labelled by the eigenvaluesof (~C + ~D)2 and C3 +D3, but since situations in physics can depend on (~C − ~D) combinations as well,
these do not form a complete set of commuting variables. E.g., [ ~J2, ~K] = −i[(~C + ~D)2, (~C − ~D)] 6= 0.2Note that su(2)⊕ su(2) ∼= so(3)⊕ so(3) ∼= so(4) 6∼= so(3, 1).
105
matrix. As ~J = ~C + ~D and ~K = −i( ~C − ~D), we find for any element A of SL(2,C),
characterized by ~θ and ~χ the following SU(2)⊗ SU(2) rep:
D[j1,j2](A) = exp(−i~χ · ~K) exp(−i~θ · ~J)= exp(−i~χ · ( ~C − ~D)/i) exp(−i~θ · ( ~C + ~D)) (9.80)
= exp(−~χ · (~S [j1] ⊗ I [j2] − I [j1] ⊗ ~S [j2])) exp(−i~θ · (~S [j1] ⊗ I [j2] + I [j1] ⊗ ~S [j2]))
Clearly these matrices transform states with label (j1, j2) into states with (j1, j2). If thestates could be labelled by one label j, then there would be no mixing among differentvalues of j under arbitrary Lorentz transformations. But they only stay invariant underrotations, since ~J = ~C + ~D. Boosts generated by ~K = −i( ~C − ~D) transform states ofdifferent j into each other. Hence j does not label irreducible subspaces. States labelledby j1, j2 do form irreducible subspaces.
Two questions arise: D[j1,j2] is a rep of SL(2,C), is it a rep of L↑+ too? And, what is the
connection between D[j,0] and D[0,j] ? In order to answer the first question, we consider−I. As A and −A ∈ SL(2,C) give rise to the same Λ ∈ L↑
+, −I must be mapped on the
identity transformation in L↑+. From Eq. (9.54) we see immediately that −I ∈ SL(2,C)
corresponds to a rotation over 2π about any axis. Such a rotation gives a phase factor(−1)2j in D[j] (see the literature). We thus find:
D[j1,j2](−I) = (−1)2(j1+j2)I. (9.81)
Clearly, D[j1,j2] can be a rep of L↑+ only if j1 + j2 is an integer.
Next we consider the connection between D[j,0] and D[0,j]:
D[j,0](A) = exp(−~χ · ~S [j]) exp(−i~θ · ~S [j]),
D[0,j](A) = exp(~χ · ~S [j]) exp(−i~θ · ~S [j]). (9.82)
Now consider A†−1. A† is obtained from A by the replacement of ~θ by −~θ, A−1 by ~θ → −~θand ~χ → −~χ. So, taken together, A†−1 is obtained by the replacement ~χ → −~χ while ~θis left unchanged. This is exactly the connection between D[j,0] and D[0,j]:
D[j,0](A) = D[0,j](A†−1). (9.83)
There is another way of expressing this connection. The reps D[j] of SU(2) have theproperty:
D∗[j] = V [j]D[j]V [j]−1, (9.84)
where V [j] is a unitary matrix. So, D∗ ∼= D. As D[j] is given by
D[j](~θ) = exp(−i~θ · ~S [j]), (9.85)
it follows that (see also the discussion on conjugate reps after Example 40)
~S∗[j] = −V [j]~S [j]V [j]−1. (9.86)
For an arbitrary element of SL(2,C) we thus find:
V [j]D[0,j](A)V [j]−1 = V [j] exp(~χ · ~S [j]) exp(−i~θ · ~S [j])V [j]−1
= exp(−~χ · ~S∗[j]) exp(i~θ · ~S∗[j])
= D[j,0]∗(A). (9.87)
106
Hence D[0,j](A) ∼= D[j,0]∗(A).If we restrict ourselves to rotations, U †−1 = U , we find the normal Clebsch-Gordan
series:
D[j1,j2](U) = D[j1](U)⊗D[j2](U)
= D[j1−j2](U)⊕ · · · ⊕D[j1+j2](U). (9.88)
The case j = 12is special: D[1/2,0](A) = A. From Eq. (9.83) it follows thatD[0,1/2](A) =
A†−1. In the case of a pure Lorentz transformation such thatp = (m,~0) is transformed
to pµ we find (see also Eq. (9.74)):
D[1/2,0] =p0 +m− ~σ · ~p√2m(p0 +m)
D[0,1/2] =p0 +m− ~σ · ~p√2m(p0 +m)
=p0 +m+ ~σ · ~p√2m(p0 +m)
(9.89)
9.4 Casimir operators of the Lorentz group
In the previous section we encountered the operators ~C2 and ~D2, which commute withany of the generators C i and Di. These are therefore Casimir-operators of the Lorentzgroup. Here we provide some more details.
Consider the generators Mµν of the Lorentz group. Instead of using three-vectors oneshould also be able to express the Casimir operators in terms of these generators Mµν .Indeed, one can show that the operators C1 and C2 defined as
C1 =1
2MµνMµν ,
C2 =1
4ǫµναβM
µνMαβ , (9.90)
are the Casimir operators of the Lorentz group. The fact that there are two quadraticCasimir operators is a consequence of sl(2,C) ∼= su(2) ⊕ su(2), i.e. the complexificationof the algebra of the Lorentz group is the direct sum of two rank-1 algebras.
Expressing C1 and C2 in terms of the operators ~J and ~K,
J i =1
2ǫijkM jk,
Ki = M0i, (9.91)
and in terms of ~C and ~D,
~C =1
2( ~J + i ~K),
~D =1
2( ~J − i ~K), (9.92)
yields
C1 = ~J2 − ~K2 =1
2
(~C2 + ~D2
),
C2 = 2 ~J · ~K =1
2i
(~C2 − ~D2
). (9.93)
It is now clear why the irreps of the Lorentz group are labelled by two numbers j1 and j2.
107
9.5 Exercises
Exercise 9.1
Verify that L00(Λ1Λ2) ≥ 1 if L0
0(Λ1) ≥ 1 and L00(Λ2) ≥ 1.
Exercise 9.2
Show that the Lorentz transformation in Eq. (9.27) can be obtained by exponentiation ofK3 in Eq. (9.34).
Exercise 9.3
Verify that Eq. (9.35) for µ = 0 and ν = 3 yields K3 in Eq. (9.34).
Exercise 9.4
Prove that the mapping φ in Eq. (7.35) is a two-to-one homomorphism of SU(2) to SO(3)by using Eq. (9.53).
Exercise 9.5
Show that the operators C1 and C2 are Casimir operators of the Lorentz group.
108
Chapter 10
The Poincare group
In this chapter we discuss the Poincare group (often also called the inhomogeneous Lorentzgroup) and its representations.
10.1 The Poincare Group
The definition of the Poincare group is almost trivial if it is considered as the extensionof the Lorentz group with the translations in space and time. The latter are generated bythe four-momentum operators P µ. It is useful to define a new notation: a transformationbelonging to the Poincare group is denoted by (Λ|a), where a is a four-vector specifyinga translation and Λ is a Lorentz transformation. In accordance with the characterisationwritten down in Sect. 9.1.2 we write for the Lorentz transformation Λ:
(Λ|0) = exp(−i ~χ · ~K
)exp
(−i ~θ · ~J
); (10.1)
a translation along the four-vector aµ is given by:
(1|a) = exp(−iaµP µ). (10.2)
As we have seen in Part I the Poincare group is a semi-direct product group, P = T ⊙L.Every element can be written as
(Λ|a) = (1|a)(Λ|0). (10.3)
Hence, a general Poincare transformation is given by:
(Λ|a) = exp (−iaµP µ) exp(−i ~χ · ~K
)exp
(−i ~θ · ~J
). (10.4)
As we will see below P µ andMµν do not commute, so one cannot write a general Poincaretransformation as one exponential, which is expected for a non-compact group.
It is immediately clear that, because translations in space and time commute, the gen-erators P µ should commute too. Because the argument of the exponential in Eq. (10.2)must be an invariant, the P µ must themselves be four-vectors. This property can be ex-pressed in a commutation relation involving the momentum operators and the generators
109
of the Lorentz group. For the sake of completenes we give here the relevant commutationrelations:
[P µ, P ν] = 0,
[Mµν , P σ] = i(P µgνσ − P νgµσ). (10.5)
In terms of the boosts and angular momentum operators, the second of these relationsreads:
[Kj , P 0] = −iP j ,
[Kj , P k] = iP 0gjk = −iP 0δjk,
[J l, P 0] = 0,
[J l, P k] = iǫlkmPm. (10.6)
The first of these equations is the reason why physical states are not labeled with eigen-values of for instance ~K2 and K3, as opposed to ~J2 and J3: ~K does not correspond to aconserved quantity.
We have seen that the Poincare transformations can be considered as combinationsof Lorentz transformations and translations in space-time. Therefore we can split thePoincare group into four pieces in the same vein as the Lorentz group: P ↑
+, P↑−, P
↓+ and
P ↓−.
10.2 Invariants: Mass and Spin
In order to label irreps of the Poincare group one wants to identify the Casimir operators.From the commutation relations given above it is clear that the Casimir operators of theLorentz group (cf. Eq. (9.90)) are not Casimir operators of the Poincare group.
In the case of the Poincare group the first Casimir operator is the mass operator M ,which is the four-momentum squared:
M2 = PµPµ = E2 − ~P 2. (10.7)
Applying Eq. (10.5) it is an easy matter to verify that M2 is indeed invariant.The second Casimir operator is the square of the Pauli-Lubanski operator W µ. The
latter is the generator of all Lorentz transformations that leave the four-momentum P ν
invariant. W µ can be expressed in terms of the generators Mµν and P µ as follows:
W µ = −12ǫµνρσPνMρσ. (10.8)
The symbol ǫµνρτ denotes the fully antisymmetric tensor in four dimensions normalizedaccording to: ǫ0123 = ǫ1230 = 1. If we rewrite Eq. (10.8) in terms of ~P , ~J and ~K we find:
W 0 =3∑
j=1
PjJj = ~P · ~J, (10.9)
and~W = P 0 ~J − ~P × ~K = E ~J − ~P × ~K. (10.10)
110
Because the tensor ǫ is fully antisymmetric, the vector W µ is orthogonal to the mo-mentum:
WµPµ = 0. (10.11)
This fact can be used to express, e.g. W 0 in terms of E, ~P and ~W . One finds then
W 0 =~W · ~PE
= ~J · ~P . (10.12)
Apparently, W 0 is proportional to the helicity λ, which is defined as
λ = ~J · P ≡ ~J · ~P/|~P |. (10.13)
The commutation relations of the components of W and P are easily derived if use ismade of the definition of W and the commutators of Mµν and P σ. One finds that W andP commute:
[W µ, P ν] = 0. (10.14)
Next we establish the fact that W is a four-vector, which means that it has the samecommutation relations with Mµν as P :
[Mµν ,W σ] = i(W µgνσ −W νgµσ). (10.15)
One can derive these relations in a straightforward manner employing the explicitforms for M0k, M jk, W 0 and W l. Then one finds easily:
[ ~J,W 0] = 0, (10.16)
and[J j ,W k] = iǫjklW
l. (10.17)
These two relations mean that W 0 is a rotational scalar and that ~W is a vector.The commutation relations of boost operators with the components of W are:
[Kj ,W 0] = −iW j,
[Kj ,W k] = −iW 0δjk. (10.18)
These relations tell us indeed thatW behaves in the same way as P with respect to boosts.An immediate consequence of this property is that W 2 ≡ WµW
µ is invariant under anyPoincare transformation:
[Mµν ,WσWσ] = 0. (10.19)
So we have found the two Casimir operators for the Poincare group M2 and W 2. Theproof that these are the only two Casimir operators can be found in the literature.
Finally, we write the commutation relations that hold for the components of W :
[W µ,W ν] = −iǫµναβWαPβ, (10.20)
which can be translated into:
[W 0,W l] = iǫlmkWmP k,
[W k,W j] = iǫkjl(WlP 0 −W 0P l). (10.21)
In the rest frame (~P = ~0 and P 0 = m) the above are commutation relations between thecomponents of an angular momentum vector. In a general frame the relation to spin ismore complicated as will be shown in Sect. 10.3.2.
Next we will see that W µ are the generators of Lorentz transformations that leave thefour-momentum P ν invariant, which form the so-called little group.
111
10.3 Representations of the Poincare group
The discussion of the reps of the Poincare group can be made rather technical by referringto the theory of induced reps. We shall not do so, and refer for details to the literature[Cornwell 1984]. Here we will introduce the irreps of the translation group T4, which isgenerated by the four-momentum operator P µ. Next we shall discuss the concept of a“little group”, whose generators are formed by the Pauli-Lubanski operators W µ.
10.3.1 Translations and the little group
Because T4 is Abelian, which is a consequence of the fact that the components of P µ
mutually commute, all its irreps are one-dimensional. They can be written as (see alsoEq. (10.2)):
(1|a) = exp (−iaµP µ) . (10.22)
The carrier space of this rep is called momentum space. We denote the states by |p; σ〉.They are the eigenstates of the four-momentum P :
P µ|p; σ〉 = pµ|p; σ〉, (10.23)
the pµ’s being the eigenvalues. The index σ is an as yet unspecified degeneracy label.Now suppose that p is a fixed four-vector. Then the little group L(p) is defined as the setof all elements of L which leave p invariant, i.e. if Λ ∈ L(p), then the corresponding 4× 4matrix Lµ
ν(Λ) has the property:
pµ = Lµν(Λ)p
ν . (10.24)
Evidently, L(p) is a subgroup of L. To find its generators consider an infinitesimal Lorentztransformation acting on pα:
Lβαp
α ≈ pβ − i
2ωµν(M
µν)βαpα = pβ + ωβ
αpα, (10.25)
where we used Eq. (9.35). To satisfy the condition that pα remains unchanged (ωβαp
α = 0),we write ωµν = −ǫµνρσsρpσ, for an arbitrary spacelike four-vector sρ. In this way oneobtains
L = exp
(− i2ωµνM
µν
)= exp (isµW
µ) . (10.26)
This demonstrates that the components of W µ generate the little group L(p).
Now let Λ vary over the whole of L. The set of all four-vectors p(Λ) defined as
pµ(Λ) = Lµν(Λ)p
ν (10.27)
is called the orbit O(p) of p.Since L(p) includes the identity of the Lorentz group, one deduces that L(p) is a
subgroup of the proper orthochronous Lorentz group L↑+. From this one can show that
there are six different types of orbits:(i) p2 > 0, p0 > 0, p = (m,~0);(ii) p2 > 0, p0 < 0, p = (−m,~0);
112
(iii) p2 = 0, p0 > 0, p = (p, 0, 0, p);(iv) p2 = 0, p0 < 0, p = (−p, 0, 0, p);(v) p2 < 0, p = (0, 0, 0, p), p 6= 0;(vi) p2 = 0, p0 = 0, p = (0, 0, 0, 0).The idea of an orbit is important for the following reason: if p and p′ are four-vectors
belonging to the same orbit, O(p), then the little groups L(p) and L(p′) are isomorphic.Consequently, in the case of the Poincare group we only need to consider six possibledistinct little groups.
Because the physical interpretation of the momentum four-vector p does not allow forthe case p0 < 0, there are only three physically interesting orbits.
vacuum Orbit (vi) is a single state. It has the property that all generators of the Poincaregroup are represented by one-dimensional null matrices. Consequently, all group elementsare represented by identity matrices. So, this state has momentum and angular momen-tum equal to zero and is invariant under any Poincare transformation. We interpret it asthe vacuum.
massive case Orbit (i) has characteristic representative p = (m,~0). This is a time-like four-vector. The little group consists of the space rotations, so its covering groupis SU(2). The reps are the well-known D[j] with j = 0, 1/2, 1, 3/2 · · ·. The label σ inEq. (10.23) is interpreted as the magnetic quantum number j3. But in general the statescannot be labelled by j3, since [J l, P k] = iǫlkmPm. The states will be labelled using thePauli-Lubanski vector W µ instead (see Sect. 10.3.2).
A Poincare transformation is given in this case by:
U [j]([Λ|a])|p; σ〉 = exp(−iΛp · a)j∑
τ=−j
|Λp; τ〉 D[j]([(p← Λp) Λ (p←
p)|0])τσ. (10.28)
The argument of the D-function in this expression is the so-called Wigner rotation.The sequence of three pure Lorentz transformations from rest to p, from p to p′ and
finally back to rest, defines a rotation, the Wigner-rotation:
UW (n, θ) ≡ (p← Λp) Λ(p←
p). (10.29)
One can use the generalization of Eq. (9.74)
(p′ ← p) =(p′0 + p0)− (~p ′ − ~p) · ~σ√(p′0 + p0)2 − (~p ′ − ~p)2
, (10.30)
to write
UW (n, θ) =[m+ p′0 + ~p ′ · ~σ] [p′0 + p0 − (~p ′ − ~p) · ~σ] [m+ p0 − ~p · ~σ]
2m√
(p′0 +m)(p0 +m)[(p′0 + p0)2 − (~p ′ − ~p)2]. (10.31)
This can be rewritten as a rotation:
UW (n, θ) =[(p′0 +m)(p0 +m) + ~p ′ · ~p] + i(~p ′ × ~p)~σ√(p′0 +m)(p0 +m)[(p′0 + p0)2 − (~p ′ − ~p)2]
, (10.32)
with
n = ~p× ~p ′/|~p× ~p ′|, θ = 2 arctan|~p× ~p ′|
(p′0 +m)(p0 +m) + ~p ′ · ~p. (10.33)
113
One thus obtains:
(p′ ← p) |p; σ〉 = (p′ ← p) (p← p) |p; σ〉
= (p′ ← p) (
p← p′) (p′ ← p) (p←
p) |p; σ〉= (p′ ←
p) UW (n, θ) |p; σ〉
= (p′ ← p)
j∑
τ=−j
D[j](UW )τσ |p; τ〉
=
j∑
τ=−j
D[j](UW )τσ |p′; τ〉.
massless case Orbit (iii), p = (p, 0, 0, p), p > 0. This four-vector is light-like. In orderto identify the little group in this case, it is useful to go back to SL(2,C). The hermitianmatrix that represents p has the form:
P = pµσµ =
[0 00 2p
]. (10.34)
The elements of SL(2,C) that leave P invariant have the form:
A =
[a11 0a21 1/a11
], |a11|2 = 1. (10.35)
It is an easy matter to show that such matrices form a group.We choose the following parametrization:
a11 = exp(iφ/2), a21 = (ξ + iη) exp(−iφ/2), (10.36)
such that fromA(ξ1, η1;φ1)A(ξ2, η2;φ2) = A(ξ3, η3;φ3), (10.37)
one obtains the following relations:
φ3 = φ1 + φ2,
ξ3 = ξ1 + cos φ1 ξ2 − sinφ1 η2,
η3 = η1 + sinφ1 ξ2 + cosφ1 η2. (10.38)
These are the relations for the two-dimensional Euclidean transformations, which formthe group called E(2), sometimes also called ISO(2). We interpret φ as a rotation angleand (ξ, η) as a translation vector in R
2. Differently put, taking derivatives of A w.r.t.φ, ξ, η, evaluated at the origin, yields matrices iλ,N1,−N2 respectively, that satisfy
[λ,N1] = iN2,
[λ,N2] = −iN1,
[N1, N2] = 0, (10.39)
which are the commutation relations of the generators of E(2).
114
The covering group of those transformations is a semi-direct product of translationsin R
2, T2, and rotations1 exp(iφ/2), 0 ≤ φ < 4π. Because T2 is Abelian, we can write itsirreps as
χ[k1,k2] = exp(i(k1x1 + k2x2)), k1, k2 real. (10.40)
(Apparently, this is the two-dimensional analog of the plane wave states exp(ip · x/~).)There are two different types of orbit:
(a) k21 + k22 > 0, which corresponds to an infinite-dimensional irrep;(b) k1 = k2 = 0, that corresponds to the identity rep of T2.In case (a) one has for each value of k21 + k22 (the eigenvalue of W 2) infinitely many
possible values for (k1, k2). In case (b) the little group consists of the two-dimensionalrotations only. Its unitary reps are:
D[j](φ) = exp(i jφ/2), j = 0,±1/2,±1, · · · . (10.41)
So, for every value of j there exists a one-dimensional rep, D[j].
10.3.2 Spin states
In the previous section we have seen that there are only three orbits of physical relevance.However, since the vacuum has P µ = 0,W µ = 0, there is no need to discuss that case anyfurther. Here we will study the spin labels of the other two states more systematically.For this one has to consider the Casimir operator W 2 in more detail.
massive case Orbit (i)
For any state |p; σ〉 we can replace P µ by its eigenvalue pµ. In the rest frame p =p =
(m,~0), and Eqs. (10.21) reduce to
[W 0,W l] = 0,
[W k,W j] = iǫjklWlm. (10.42)
The latter are the commutation relations between the components of an angular momen-tum three-vector. However, we do not want to restrict our discussion to the rest frame.Therefore we are guided by the transversality property Eq. (10.11) to introduce threevectors n[i], i = 1, 2, 3 that have the following properties:
a) transversality:n[i] · p = nµ[i]pµ = 0; (10.43)
b) orthogonality:n[i] · n[j] = nµ[i]nµ[j] = gij = −δij ; (10.44)
c) closure:3∑
i=1
nµ[i]nν [i] = −(gµν − pµpν
p2
); (10.45)
1Note that we are actually dealing with a double cover of SO(2) ∼= U(1), i.e. a projective rep of U(1).The simply connected or universal covering group of U(1) is R+, the noncompact multiplicative group ofpositive real numbers.
115
d) handedness:ǫµναβn
µ[i]nν [j]nα[k]pβ = mǫijk. (10.46)
The last property may require some discussion. First, we notice that both sides areLorentz scalars, so if this relation holds in some reference frame, it holds in any frame.Now write Eq. (10.46) in the rest frame, where p = (m,~0). Then we see that in order tofulfil this equation we must have
nµ[i] =nµ[i] = δµi, (10.47)
which means that in the rest frame the n’s are just the unit vectors along the spacecoordinate axes. It is easy to verify that Eqs. (10.43) - (10.45) are fulfilled by the choice(10.47). So we see that Eq. (10.46) expresses the fact that the vectors
n[i] are the unit
vectors of a right handed coordinate frame. The vectors n[1], . . . , n[3] and p form a set ofmutually orthogonal vectors that is usually called a vierbein.
The next step is to project W onto these vectors:
W [i] = W · n[i] =Wµnµ[i]. (10.48)
Upon substituting Eq. (10.45) we find the converse relation:
W µ = −∑
i
nµ[i]W [i]. (10.49)
The commutation relations of theW [i]’s follow easily from those of theW µ’s (see Eq. (10.20)):
[W [j] , W [k] ] = [−Wµnµ[j] , −Wνn
ν [k] ]
= nµ[j]nν [k] [Wµ,Wν ]
= −iǫµναβ nµ[j]nν [k] W αP β
= iǫµναβ nµ[j]nν [k]
∑
l
nα[l]P βW [l]
= im ǫjklW [l] (10.50)
The final step is to introduce the vector ~S with components:
Si =~
mW [i], (10.51)
that satisfies the canonical commutation relations for a spin operator. It is called theintrinsic spin of the particle considered. In order to obviate any misunderstanding westress that ~S is not a vector in the ordinary sense: it is a set of Lorentz scalars thatdepend on the choice of the vectors n[i]. Only for the case that n[i] =
n[i], Eq. (10.47),
can one deduce from Eq. (10.17) that [J j, Sk] = iǫjklSl. So we find that the operator ~S is
a true three-dimensional vector if defined in the particle rest frame.The Si form the generators of an SU(2) subgroup that belongs to the maximal set of
commuting operators. We also find that
W 2 = −m2
~2
∑
i
(Si)2, (10.52)
116
so the eigenvalues of W 2 are −m2s(s+ 1), with s = 0, 12, 1, . . .
For particles with mass m > 0 the states |p; σ〉 are thus written as |m, s; ~p, λ〉, so
P µ|m, s; ~p, λ〉 = pµ|m, s; ~p, λ〉,
P 2|m, s; ~p, λ〉 = m2|m, s; ~p, λ〉,
S3|m, s; ~p, λ〉 = ~λ|m, s; ~p, λ〉,~S2|m, s; ~p, λ〉 = ~
2s(s+ 1)|m, s; ~p, λ〉.
Here λ is the helicity and the third equation holds for a particular choice of the vierbein.
massless case Orbit (iii): p2 = 0We treat only the physically important case where the nontrivial elements of the littlegroup are two-dimensional rotations. (This is case (b) at the end of Sect. 10.3.1, since case(a) would require a continuous spin quantum number.) The generators of the translationsin two dimensions, N1 and N2, have eigenvalue zero. The rotations are related to thehelicity λ = ~J · ~P/|~P | = W 0/|~P |.
Again a vierbein is defined. In the Lorentz frame where p = p this is:
p = p = (1, 0, 0, 1)
n[1] = (0, 1, 0, 0)
n[2] = (0, 0, 1, 0)
s = (s0, 0, 0, s3), s0 > |s3|. (10.53)
In a general Lorentz frame, where p = (p, 0, 0, p), the corresponding four-vectors are foundby applying a suitable Lorentz transformation.
Next we expand W µ on this vierbein:
W µ = N1nµ[1] +N2n
µ[2] + λP µ +Wssµ. (10.54)
Because W µ is transverse: W ·P = 0, and P 2 = 0, P · s 6= 0, one finds easily that Ws = 0.The commutation relations for the components of W , Eq. (10.21), reduce in the presentcase to Eq. (10.39). As mentioned before these relations are the commutation relationsof the generators of the group E(2) of two-dimensional Euclidean transformations. Thisresult was to be expected as we already knew that the W µ form the generators of thelittle groups of the Poincare group.
The decomposition Eq. (10.54) also implies W 2 = N21 + N2
2 . In order to have finitedimensional reps we require W 2 = 0. This means that physical states in the massless caseare not labelled by the spin quantum number s as was the case for massive states, but bythe helicity λ instead, which can have any half-integer or integer value. As we have seenin Eq. (10.41) irreps D[j] are only one dimensional for massless particles. Therefore, wecan equate j and λ.
In the case where the space inversion operators are included in the symmetry group,D[j] and D[−j] together form one irrep. The states |p; j〉 and |p;−j〉 (j > 0), are inter-preted as states of a massless particle with spin j and helicity j and −j resp.
117
Basis states, denoted by |p;λ〉, are defined by:
W 0|p;λ〉 = λp0|p;λ〉,
Ni|p;λ〉 = 0, (i = 1, 2).
The helicity for massless particles is invariant under rotations (clearly, since λ = ~J · ~P/|~P |)and boosts. The latter is true, even though
[Kj ,W 0] = −iW j . (10.55)
This can be seen from for instance
[Kj ,W 0] |p;λ〉 = −iW j |p;λ〉 = −iλP j |p;λ〉 = λ[Kj, P 0] |p;λ〉, (10.56)
from which it follows that
W 0(Kj |p;λ〉
)= λP 0
(Kj |p;λ〉
). (10.57)
For Λ = (p′ ← p) generated by Kj , one thus finds
W 0 |p′;λ′〉 = λP 0 |p′;λ′〉 = λp′0 |p′;λ′〉, (10.58)
so one concludes that λ′ = λ.If p and p′ are two light-like four-momenta differing by a rotation about the direction
of ~p only, then to p and p′ there correspond two different vierbeins. Consequently, |p′;λ〉and |p;λ〉 are different. If the rotation mentioned before is over an angle φ, one finds:
|p′;λ〉 = |p;λ〉eiλφ = |p;λ〉D[|λ|]λλ ((
p← p′)Λ(p←
p)). (10.59)
Here there is no summation over the helicities, since the helicity for massless particles isinvariant under boosts and rotations. This is different from rotations of massive states,where states p = (E, ~p) obtained from p = (E, 0, 0, p) by a rotation are given by
|p;λ〉 = U(R(θ, φ))|p;λ〉
=∑
λ′
D[s](R(θ, φ))λ′λ|p;λ′〉.
10.4 Exercises
Exercise 10.1
Verify that M2 and W 2 and not the operators C1 and C2 of Eq. (9.90) are the Casimiroperators of the Pioncare group.
Exercise 10.2
Let L be the Lie algebra of E(2).(i) Show that L contains an invariant subalgebra.(ii) Show that the three generators of Lorentz transformations J3, J1+K2 and J2−K1
form a basis of L.(iii) Write down an explicit four-dimensional representation of these generators.(iv) Use the quadratic Casimir operator C = N2
1 + N22 to show whether this rep is
irreducible.
118
Appendix A
Bibliography
This second part of the lectures are based on the books listed below.
H.F. Jones Groups, Representations and Physics,Institute of Physics Publishing, Second Ed., 1998
J.S. Cornwell Group Theory in Physics, Vols I and II,Academic Press, 1984
The set of two volumes by Cornwell contains most of the proofs of the theorems on Liegroups and Lie algebras addressed in the lectures notes.
M. Hamermesh Group Theory and its Application to Physical Problems,Addison Wesley, 1962 and also: Dover Pubns; Reprint Ed. (1989)
W. Miller Symmetry groups and their applications,Academic Press, 1972
W.-K. Tung Group theory in physics: an introduction to symmetry principles,
group representations, and special functions in classical and
quantum physics,World Scientific, 2003
119
Group theory topics relevant for particle physics are covered extensively in:
D.B. Lichtenberg Unitary Symmetry and Elementary Particles,Academic Press, 1970
H. Georgi Lie Algebras in Particle Physics,Benjamin/Cummings, 1982
H.J. Lipkin Lie Groups for Pedestrians,Dover Publications (Reprint of the original from 1966)
R.N. Cahn Semi-simple Lie algebras and their representations,Benjamin/Cummings, 1984
T.-P. Cheng & L.-F. Li Gauge theory of elementary particle physics,Oxford University Press, 1984
Mathematical treatments of reps of Lie groups and Lie algebras are:
T. Brocker & T. tom Dieck Representations of compact Lie groups,Springer Verlag, 1985
A.L. Onishchik & E.B. Vinberg Lie groups and Algebraic groups,Springer Verlag, 1990
W. Fulton & J. Harris Representation theory: a first course,Springer Verlag, 1991
B.C. Hall Lie groups, Lie algebras, and representations,Springer Verlag, 2003
S. Hassani Mathematical Physics: a Modern Introduction to
Its Foundations, Springer Verlag, 1999
120
Group Theory in Physics
Part III, Representations of Semi-Simple Lie algebras
Ben Bakker and Daniel Boer
Fourth Edition, January 2009
Group Theory in Physics
Part III, Representations of Semi-Simple Lie algebras
Authors: Ben Bakker and Daniel BoerDepartment of Physics and Astronomy
Vrije Universiteit, Amsterdam
Fourth Edition, January 2009
Cover illustration: M.C. Escher, Cirkellimiet III, 1959Weblocatie: www.worldofescher.com
i
Contents
1 Root systems 11.1 Simple and Semi-Simple Lie Algebras . . . . . . . . . . . . . . . . . . . . . 11.2 Complexification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Cartan Subalgebras and Roots . . . . . . . . . . . . . . . . . . . . . . . . 51.4 Weyl Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.5 Classification of simple complex Lie algebras . . . . . . . . . . . . . . . . . 111.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2 Irreps of Semi-Simple Lie Algebras 152.1 Weights . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.2 Weights of A2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.3 Highest weights and fundamental weights . . . . . . . . . . . . . . . . . . . 202.4 Quarks and the hadron spectrum . . . . . . . . . . . . . . . . . . . . . . . 212.5 Casimir operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
3 Global properties of Lie groups 273.1 Homotopy groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.2 Projective spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293.3 Universal covering groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 293.4 Norm preserving Lie groups . . . . . . . . . . . . . . . . . . . . . . . . . . 30
3.4.1 Symplectic groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
4 Bibliography 34
ii
Chapter 1
Root systems
In Part II of the lecture notes on “Lie Groups and Lie algebras” we have studied aspectsof the irreps of several Lie groups, in particular of su(n). Unitary symmetries play animportant role in particle physics, but these are not the only Lie algebras of importancefor physics. Therefore, in this chapter we will study so-called simple and semi-simpleLie algebras in a systematic way, namely via their characteristic root systems whichcompletely determine the structure and the irreps of the algebras. In this way all simplecomplex Lie algebras have been classified. This is mainly the work of W. Killing, E.Cartan, H. Weyl and E.B. Dynkin. We will go into sufficient detail such that their resultscan be understood and appreciated. But as before, we shall often state the main resultswithout proofs, and focus instead on important examples.
Chapter 3 is not part of the lectures or the exam, but is included solely for studentsinterested in learning some aspects about the global properties of Lie groups.
1.1 Simple and Semi-Simple Lie Algebras
Definition simple Lie group
A Lie group G of dimension greater than 1 is said to be simple if it does not possess aproper invariant Lie subgroup.
Definition simple Lie algebra
A Lie algebra L is said to be simple if it is not Abelian and does not possess a properinvariant Lie subalgebra.
Definition semi-simple Lie group
A Lie group G is said to be semi-simple if it does not possess an Abelian invariant Liesubgroup.
Definition semi-simple Lie algebra
A Lie algebra L is said to be semi-simple if it does not possess an Abelian invariant Liesubalgebra.
In these definitions Abelian Lie groups and algebras have been explicitly or implicitlyexcluded from being simple or semi-simple. In particular, it is important to realize thatthe group of phase transformations U(1) is not considered to be simple according to theabove definition. The reason for excluding such Abelian groups (and algebras) is thatif ai is a generator of such a group, then all structure constants with an index i will
1
vanish. Hence, the structure constants do not tell us anything about Abelian subgroupsor subalgebras. The classification to be discussed in this chapter is based for a largepart on the use of structure constants and the adjoint representation. The classificationis therefore restricted to simple and semi-simple Lie algebras (which necessarily havedimension larger than one, since all one-dimensional Lie algebras are Abelian).
At first sight it seems a difficult task to determine whether a given Lie group (algebra)is simple or semi-simple. However, Cartan found a simple criterion using the followingconcept named after W. Killing.
Definition Killing form
The Killing form B(a, b) corresponding to any two elements a and b of a Lie algebra L isdefined by:
B(a, b) = Tr [ad(a)ad(b)]. (1.1)
In the adjoint rep a basis element ap is represented by the matrix ad(ap)kj = fkpj, where
fkpj are the structure constants of the Lie algebra. Now consider two elements a and bfrom L:
a =∑
p
αpap, b =∑
q
βqaq. (1.2)
Then the matrix product of a and b in the adjoint rep is
[ad(a)ad(b)]ij =∑
p
∑
q
∑
k
αp[ad(ap)]ikβq[ad(aq)]kj
=∑
p
∑
q
αpβq
∑
k
f ipkf
kqj . (1.3)
Now take the trace of this matrix:
Tr [ad(a)ad(b)] =∑
p,q
(∑
k,j
f jpkf
kqj
)αpβq
=∑
p,q
gpqαpβq. (1.4)
Here we definedgmn = gnm =
∑
k,l
fkmlf
lnk. (1.5)
Theorem Cartan’s criterionThe Lie algebra L is semi-simple iff its Killing form is nondegenerate, i.e., B(a, c) =B(b, c), ∀c, implies a = b, or equivalently, iff det g 6= 0.
Remark
For a semi-simple Lie algebra one can view gmn as the metric in a linear space where theinner product of the vectors a and b, which is not necessarily positive definite, is given bythe Killing form B(a, b). The tensor gmn is called the Cartan metric tensor.
For a semi-simple Lie algebra one can use the Killing form to define orthogonal sub-spaces of the Lie algebra.
2
TheoremEvery semi-simple Lie algebra is a direct sum of simple Lie algebras, i.e. L = L1⊕ . . .⊕Lk
for some k.
For a semi-simple Lie algebra det g 6= 0, so g has an inverse. We shall denote theelements of g−1 by gkj:
gikgkj = δij . (1.6)
We can use g to raise or lower indices. For example,
fijk = f lijglk. (1.7)
The constants fijk are fully antisymmetric. From the above equation it is clear that fijkis antisymmetric under the interchange of i and j. To show that it is for instance alsoantisymmetric under the interchange of j and k, one can use the Jacobi identity to obtain:
fijk = f lijf
nlmf
mkn
= −f lmjf
nlif
mnk + f l
mifnljf
mnk. (1.8)
By writing this same equation for fikj and renaming the indices that are summed over(different renaming for both terms), one arrives at fikj = −fijk.
TheoremA connected semi-simple Lie group is compact iff the Killing form of its Lie algebra isnegative definite, i.e. B(a, a) < 0 ∀a.Remark
By considering Ji = −iai as the generators, the corresponding Killing form (and Cartanmetric tensor) will be positive definite. This is often done in physics applications in orderto deal only with Hermitian generators, and can be done without problems. Note howeverthat assigning a factor −i to a subset of the generators ai, will turn the Lie algebra of acompact group into one of a noncompact group and should thus be done with care. Thefollowing example illustrates this point.
Example su(2) and sl(2,R)The basis we chose for su(2) was ap = (i/2)σp (i.e. all elements of su(2) are real linearcombinations of the ap). The structure constants are fk
ij = −ǫijk. The Killing form istherefore
gmn =∑
k,l
ǫmlkǫnkl = −2δmn. (1.9)
The group SL(2,R) consists of the real 2×2 matrices with determinant equal to unity.A basis for the algebra sl(2,R) is
b1 =1
2
[0 −1
−1 0
], b2 =
1
2
[0 1
−1 0
], b3 =
1
2
[1 00 −1
]. (1.10)
It is easy to compute the structure constants and the Killing form:
g11 = g33 = 2, g22 = −2, (1.11)
all other elements being zero.
3
The Killing form of SU(2) (and of SO(3)) is negative definite, the one of SL(2,R) isnot. This confirms the earlier results that SU(2) and SO(3) are compact, and SL(2,R)is not.
The groups SU(2) and SL(2,R) are semi-simple as the above shows, but they are infact also simple. Their Lie algebras are three-dimensional and if there would be a properinvariant Lie subalgebra, that algebra or its complement would be one-dimensional andtherefore, Abelian, which would imply the groups are not semi-simple to begin with. Theonly proper Abelian invariant subgroup of both groups is the center (the maximal Abelianinvariant subgroup), which being a finite group is not a Lie group.
The next theorem simplifies the classification of semi-simple Lie algebras, as one canrestrict oneself to the classification of simple Lie algebras.
TheoremEvery compact Lie algebra L can be decomposed according to
L = Z(L)⊕ S = Z(L)⊕ S1 ⊕ . . .⊕ Sk, (1.12)
for some integer k and where Z(L) denotes the center of L, S is a semi-simple Lie algebraand Si are simple Lie algebras.
The following three theorems are about representations of semi-simple Lie algebras.
TheoremThe adjoint rep of a simple Lie algebra is irreducible.
To see that this is true, we will assume the contrary. Recall that the carrier space Vof the adjoint rep of L is L itself and that ad(a) acts on V via ad(a)b = [a, b], ∀b ∈ V . Ifthe adjoint rep would be reducible, then there exists an invariant subspace V ′ of V , suchthat V ′ 6= V and ad(a)b′ ∈ V ′, ∀a ∈ L, ∀b′ ∈ V ′. This means that [a, b′] ∈ V ′, ∀a ∈ L,∀b′ ∈ V ′ ⊂ L. Therefore, V ′ is a proper invariant subalgebra of L, which means L is notsimple. Hence, the adjoint rep is irreducible.
TheoremThe adjoint rep of a semi-simple Lie algebra is faithful, i.e. ad(a) = ad(b) ⇔ a = b.
Theorem Weyl’s theoremEvery reducible rep of a semi-simple Lie algebra is fully reducible (decomposable).
This theorem, taken together with the theorem that every semi-simple Lie algebra isa direct sum of simple Lie algebras, means that the analysis of reps of semi-simple groupscan be restricted to the irreps of simple groups.
1.2 Complexification
The method Cartan employed to find the complete classification of the simple Lie algebrasmade use of the idea of an eigenvalue equation of the form
[a, x] = λx. (1.13)
As, even for real operators, the eigenvalues may be complex, one needs to consider acomplex Lie algebra or the so-called complexification of a real Lie algebra. The completeclassification is of all simple complex Lie algebras.
4
Thus far we dealt only with real Lie algebras. Although the basis elements of a Liealgebra L may be complex, we were considering only real linear combinations of the basiselements. If in the expansion
a =∑
p
αpap (1.14)
only real numbers αp are allowed, the algebra L is called real. This is an importantconcept, as it determines the dimension of the algebra. The dimension n is determinedas the largest integer n such that
n∑
p=1
αpap = 0 (1.15)
implies αp = 0, p = 1, . . . , n. In what follows we shall need to consider complex algebras.If we simply allow in sums like Eq. (1.14) complex coefficients αp we obtain the complexi-
fication of L, denoted by L (not to be confused with the algebra of the universal coveringgroup).
Definition real form
A real form of a complex Lie algebra L′ is a real Lie algebra L whose complexification Lis isomorphic to L′.
To a complex Lie algebra there may be associated different real forms. Consider againsl(2,R) and su(2). The basis for sl(2,R) is given in Eq. (1.10), for su(2) ak = (i/2)σk.Then we have b1 = ia1, b2 = a2 and b3 = ia3, so the complexification of sl(2,R) is thesame as the one for su(2). In other words, sl(2,R) and su(2) are different real forms of
the same complex Lie algebra: sl(2,R) ∼= su(2), but clearly sl(2,R) 6∼= su(2).
TheoremIf L is the complexification of the real Lie algebra L, then L is semi-simple iff L is semi-simple. If L is simple, then L is also simple.
Example sl(2,C)The Lie group SL(2,C) is the set of all complex 2 × 2 matrices with determinant equalto unity. The Lie algebra sl(2,C) consists of all traceless complex 2× 2 matrices. Thereare six basis elements if sl(2,C) is viewed as a real Lie algebra. We choose the form
a1 =
[1 00 −1
], a2 =
[0 10 0
], a3 =
[0 01 0
],
a4 =
[i 00 −i
], a5 =
[0 i0 0
], a6 =
[0 0i 0
]. (1.16)
Clearly, ak+3 = iak, k = 1, 2, 3, so sl(2,C) is six-dimensional as a real Lie algebra, butonly three-dimensional as a complex one (it is then su(2)), as the six ak are not linearlyindependent over the complex numbers.
The real Lie algebra sl(2,C), which is itself a simple Lie algebra, has a complexification
that is not simple, but rather semi-simple: sl(2,C) ∼= su(2)⊕ su(2).
1.3 Cartan Subalgebras and Roots
Now we will address the other tools employed to achieve the classification of all simplecomplex Lie algebras. It leans heavily on the following theorem due to Cartan.
5
TheoremLet L be semi-simple. Choose an element a of L such that it has the maximal number ofdifferent eigenvalues λ (recall [a, x] = λx). Then only the eigenvalue 0 is degenerate. Thedegree of degeneracy of this eigenvalue is characteristic for the Lie algebra. It is denotedby l and called the rank of the Lie algebra L.
Remark
The element a may be non-unique. The number of eigenvalues an element of a Lie algebraof finite dimension n may have is necessarily less than or equal to the dimension. So thereis a maximum number of distinct eigenvalues. An eigenvalue zero of a corresponds to anelement that commutes with a.
Now we keep the element a fixed. Consider the l independent eigenvectors, say hi (i =1, . . . , l) with eigenvalues zero;
[a, hi] = 0. (1.17)
As a commutes with itself, it must belong to the subspace H spanned by the hi. So wecan redefine the basis of H such that a = h1. Using the Jacobi identity one can provethat H is a subalgebra: it is called the Cartan subalgebra.
Next we consider the n − l non-degenerate eigenvalues and the corresponding eigen-vectors of h1:
[h1, eα] = α1eα. (1.18)
Using the Jacobi identity we find
[h1, [hi, eα]] = [hi, [h1, eα]] + [[h1, hi], eα] = α1[hi, eα]. (1.19)
So for any i, [hi, eα] is an eigenvector of h1 with the same eigenvalue α1. As the latter isnon-degenerate, [hi, eα] must be proportional to eα:
[hi, eα] = αieα. (1.20)
Apparently, every eigenvector of h1 is also an eigenvector of the other hi, so all hi can bediagonalized simultaneously. This means that they must commute: the Cartan subalgebra
H is Abelian (but note that it is not an invariant subalgebra). A full characterization ofan eigenvector eα is given by the l eigenvalues α1, . . . , αl. This l-tuple can be consideredas a vector α in an l-dimensional space usually denoted by ∆. The vector α is called aroot.
Upon using the Jacobi identity one finds then
[hi, [eα, eβ]] = (αi + βi)[eα, eβ ]. (1.21)
So we have the following alternatives:(i) α + β is a root. If β = −α then α + β = 0 and [eα, eβ ] ∈ H . If α + β 6= 0, then
α+β is again a root, different from α and β. The product [eα, eβ ] is proportional to eα+β:[eα, eβ] = Nαβeα+β.
(ii) α + β is not a root. Then [eα, eβ] = 0.It is possible to choose the set hi, eα as a new basis for the full Lie algebra, called
the standard or Cartan-Weyl basis. The structure constants then become
fαij = fk
ij = f jiα = 0,
6
fβiα = δαβαi,
(α + β)f iαβ = 0,
(γ − α− β)f γαβ = 0. (1.22)
Next we calculate the Cartan metric tensor in this basis. Using the structure constantsgiven in Eq. (1.22) we find
gij =∑
α
αiαj,
gα−α = 2αifiα−α + fα+β
αβ fβ−αα+β, (1.23)
all other components vanish. If one normalizes the gα−α by a suitable normalization ofthe roots α such that
gα−α = 1, (1.24)
then the Cartan metric tensor becomes
g =
gij ∅0 11 0
∅ . . .
(1.25)
and det g 6= 0 implies that det[gij] 6= 0. In addition, we state without proof (cf. Cornwell,Chap. 13 for details) that for a semi-simple Lie algebra, the hi can be normalized suchthat gij = δij and the root space becomes a real, Euclidian vector space. In physicsapplication the basis for the Cartan subalgebra is usually chosen in that way.
Recapitulating, the Cartan-Weyl basis of the Lie algebra is:
[hi, hj ] = 0,
[hi, eα] = αieα,
[eα, e−α] = αihi ≡ hα,
[eα, eβ] =
Nαβeα+β if α + β is a root 6= 00 if α + β is not a root
(1.26)
It is important to note that nonzero roots are nondegenerate (if eα and e′α correspond tothe same root, then eα = e′α) and roots are not proportional apart from α and −α.
Example su(2)In Cartan’s terminology the complexification of su(2) is called A1. It consists of all com-plex linear combinations of the matrices ak. The Cartan subalgebra is one-dimensional,as the three elements a1, a2 and a3 do not commute. Take for H the element h1 = −ia3and define the ladder operators
e± =i(a1 ± ia2)√
2, (1.27)
then we obtain[h1, e±] = ±e±, [e+, e−] = h1. (1.28)
7
e h +e- 1
Figure 1.1: The roots of A1 = su(2).
The roots are one-dimensional vectors, so we can picture them on the real axis as inFig. 1.1, which is called a root diagram.
Example su(3)The complexification of su(3) is called A2. The usual basis, defined by Gell-Mann, is
λ1 =
0 1 01 0 00 0 0
, λ2 =
0 −i 0i 0 00 0 0
, λ3 =
1 0 00 −1 00 0 0
,
λ4 =
0 0 10 0 01 0 0
, λ5 =
0 0 −i0 0 0i 0 0
,
λ6 =
0 0 00 0 10 1 0
, λ7 =
0 0 00 0 −i0 i 0
, λ8 =
1√3
1 0 00 1 00 0 −2
. (1.29)
The structure constants can be computed easily. Traditionally one writes
[λj, λk] =
8∑
l=1
2ifjklλl. (1.30)
The f ’s are real and fully antisymmetric. The λ’s are hermitian. Alternatively one maytake ak = iλk as a basis for su(3) and A2. The Cartan metric tensor is
gjk = B(aj , ak) = −12δjk, (1.31)
so det g 6= 0, which implies that su(3) is semi-simple. It is in fact simple, but it does havesu(2) (non-invariant) subalgebras. Clearly, λ1, λ2, λ3 generate such an su(2) subalgebra,but also for instance, λ2, λ5, λ7. Every rep of SU(3) must therefore also be a rep ofSU(2). However, the irreps of SU(3) are not necessarily irreducible under the subalgebra.To understand this better, consider the easier example of the subgroup of SU(2) formedby the diagonal matrices diag(eiφ, e−iφ). This is an Abelian (non-invariant) subgroupequivalent to SO(2). Clearly, this 2× 2 matrix rep does not form an irrep of this Abeliansubgroup.
The SU(3) generators are conventionally defined by Ta = λa/2, such that Tr(TaTb) =δab/2. It is also customary to use T3 and T8 as the basis elements of the Cartan subalgebraof su(3), h1 = T3, h2 = T8, because they are already diagonal. One can find the roots bya straightforward computation. The eigenvectors are
e±α1= (T1 ± iT2) /
√2, e±α2
= (T6 ± iT7) /√2, e±α3
= (T4 ± iT5) /√2, (1.32)
8
and the root vectors are
α1 = (1, 0), α2 = (−1/2,√3/2), α3 = (1/2,
√3/2). (1.33)
Fig. 1.2 displays the root diagram of A2, where the open circle indicates that there aretwo vanishing roots.
∗
∗
∗
∗
∗∗
α 1
α 2 α 3
Sα2
Sα2Sα
β
1Sα1Sα2Sα1
Sα2Sα1
Sα1 ββ
β
β β
Figure 1.2: The roots of A2 = su(3).
Remark
The rank of su(n) is n− 1.
1.4 Weyl Group
In this section we shall discuss the possible forms of root diagrams.
Take two distinct roots α and γ such that α + γ is not a root. One then obtains
[e−α, eγ] = e′γ−α,[e−α, e
′γ−α
]= e′γ−2α, . . . ,
[e−α, e
′γ−qα
]= e′γ−(q+1)α, . . . ,
[e−α, e
′γ−sα
]= 0
(1.34)for some integer s, because the algebra is finite. (We write e′γ to denote a vector that isproportional to eγ .) Now consider the product
[eα, e′γ−(q+1)α] = λq+1e
′γ−qα (1.35)
and use the Jacobi identity to compute λq+1:
[eα, [e−α, e′γ−qα] ] = [ [eα, e−α], e
′γ−qα] + [e−α, [eα, e
′γ−qα] ]. (1.36)
In the standard basis defined above we have [eα, e−α] = αihi, so it follows that
λq+1e′γ−qα = αi[hi, e
′γ−qα] + λq[e−α, e
′γ−(q−1)α]
= αi(γi − qαi)e′γ−qα + λqe
′γ−qα (1.37)
henceλq+1 = α · γ − qα · α + λq. (1.38)
9
We use the notationα · β =
∑
i
αiβi =∑
i
αiβi. (1.39)
Since α+γ is not a root, λ0 vanishes. Now we can solve the recurrence relation Eq. (1.38)and find
λq = q α · γ − 1
2q(q − 1) α · α. (1.40)
Now Eq. (1.34) implies λs+1 = 0, which gives the relation
(s+ 1) α · γ = 1
2s(s+ 1) α · α, (1.41)
or2α · γα · α = s, s non-negative integer. (1.42)
The vector defined as αi ≡ 2αi/(α · α) is often called a co-root.The situation we just discussed can be formulated in a theorem. It is actually valid
for any combination of roots, not only for roots such that their sum is not a root.
TheoremIf α and β are two arbitrary roots, then the ratio 2(α·β)/(α·α) is an integer (0,±1,±2,±3)and β − 2(α · β)/(α · α)α is also a root.
The operation on roots just described is called a Weyl reflection. It can be visualizedin the space of l dimensions as a reflection in the hyperplane perpendicular to the vectorα. So we define
Sα β = β − 2α · βα · αα. (1.43)
It is easy to verify that S2α β = β. Furthermore,
Sα α = −α, Sα β · Sα γ = β · γ. (1.44)
The reflections map roots onto one another, so they must form a finite group: theWeyl group.
The Weyl reflections can be used to establish limitations on the angles between rootvectors. Consider two roots α and β and suppose that the angle between them has theproperty 0 ≤ θαβ ≤ π/2. Our theorem says
2α · β = mα · α = nβ · β (1.45)
for some integers m and n. Suppose m ≥ n, then
cos2 θαβ =(α · β)2
α · α β · β =mn
4. (1.46)
As the cosine is limited in magnitude we find four possibilities
n = m = 0 θαβ = π/2 β · β/α · α arbitrary
n = 1, m = 1 θαβ = π/3 β · β = α · αn = 1, m = 2 θαβ = π/4 β · β = 2α · αn = 1, m = 3 θαβ = π/6 β · β = 3α · α. (1.47)
10
The cases n = 0, m 6= 0 and n = m = 2 do not occur, the former leading to α · α = 0,the latter to α = β. As α is a nondegenerate root, these conditions are not fulfilled. Thecase n = 1, m = 4 is also impossible, as it leads to β = 2α which is again not possible.So we can summarize the situation in the following table:
θαβ π/6 π/4 π/3 π/2 2π/3 3π/4 5π/6 π‖β‖/‖α‖ √
3√2 1 arbitrary 1
√2
√3 1
In this table we denoted the length of a root vector α with ‖α‖.Example The Weyl group of A2.The three roots are given by Eq. (1.33). The Weyl reflections are
Sα1α1 = −α1, Sα2
α1 = α1 + α2, Sα1α2 = α1 + α2, Sα2
α2 = −α2. (1.48)
Recall that α1 + α2 = α3. By considering the products of Sα1and Sα2
one finds that thefull Weyl group W is W = E, Sα1
, Sα2, Sα1
Sα2, Sα2
Sα1, Sα1
Sα2Sα1
. This group of ordersix is not Abelian. Fig. 1.3 depicts the roots α1, α2, α3 of A2 and the Weyl reflections theygenerate on a (fictitious) root β.
α 1
α 2 α 3
Sα1Sα2β
Sα1
Sα2β
Sα1Sα2
Sα1 β Sα2Sα1 β
∗
∗ ∗
∗
∗
∗β β
Figure 1.3: Illustration of Weyl reflections.
1.5 Classification of simple complex Lie algebras
Due to the symmetry properties of the root diagram (the Weyl group), one does notneed to consider all roots in order to fix the structure of the algebra. This leads to thenotion of simple roots. One can order the roots lexicographically relative to a chosen basis(therefore this is not a unique procedure, but any choice will do). First note that not allcomponents of a root can be zero. So there must be a first element that does not vanish.
Definition positive root
A root α = (α1, . . . , αl) is said to be positive if its first non-vanishing component ispositive.
We can now define the ordering of roots and the notion of simple roots.
11
Definition order of roots
A root β is said to be greater than a root α if the vector β − α is positive.
Definition simple root
A root α is said to be a simple root if it is positive and cannot be written as the sum oftwo positive roots.
TheoremIf the rank of a Lie algebra is l, then there are l linearly independent simple roots.
TheoremIf α and β are two simple roots and α 6= β, then α− β is not a root and α · β ≤ 0.
Definition Cartan matrix
The Cartan matrix A of L is an l × l matrix whose elements are defined in terms of thesimple roots αi as
Aij =2αi · αj
αi · αi
. (1.49)
Clearly, Aii = 2, and if i 6= j, then Aij can be either 0,−1,−2 or −3. For example,the Cartan matrix for the complexification of su(2) is simply the 1 × 1 matrix A = (2)and of su(3) (with simple roots α3 and −α2) the 2× 2 matrix
A =
(2 −1−1 2
). (1.50)
Hence, it is easy to construct the Cartan matrix if one knows the root system of L. Butthe converse is also possible. So, the Cartan matrix also allows one to obtain the rootsystem and hence the irreps of L (see next chapter).
We will not discuss the classification of all simple complex Lie algebras, but onlymention that it is often presented in terms of so-called Dynkin diagrams, where simpleroots are represented by open circles and the maximum of |Aij | and |Aji| determines thenumber (0, 1, 2, 3) of connecting lines between the circles of roots α and β. There areseveral conventional ways of denoting the shorter root if |Aij| 6= |Aji|: in some cases anarrow is added to a connecting line pointing towards the shorter root, sometimes closedcircles are employed and sometimes the values of |Aij | and |Aji| are given explicitly.
This graphical representation completely determines the Cartan matrix and hence theroot system and the irreps. There are only a relatively small number of different typesof Dynkin diagrams. They are divided into classical and exceptional simple Lie algebras.There are countably infinite classical ones:
An = su(n+ 1) for n ≥ 1,Bn = so(2n+ 1) for n ≥ 1,Cn = sp(n) for n ≥ 1,Dn = so(2n) for n ≥ 3.
When written in this way there is some redundancy, since so(3) ∼= su(2) ∼= sp(1), sp(2) ∼=so(5) and so(6) ∼= su(4). Note that so(4) is not included, as it is not simple, as we haveseen in Part II of these lectures.
There are only 5 exceptional ones, which are called G2, F4, E6, E7 and E8.
12
Algebra dimension rank
su(n) n2 − 1 n− 1so(2n+ 1) (2n+ 1)n n
sp(n) (2n+ 1)n nso(2n) (2n− 1)n nG2 14 2F4 52 4E6 78 6E7 133 7E8 248 8
For further details we refer to the literature. As a final remark on this topic, wemention a theorem that is essential for undoing the complexification.
TheoremEvery complexification of a semi-simple Lie algebra is the complexification of a singlecompact semi-simple Lie algebra. More explicitly: for each semi-simple complex Liealgebra, the set of elements
ihi, eα + e−α, i(eα − e−α), (1.51)
where i ∈ 1, . . . , l and α is taken from the set of positive roots, form the basis of a compactsemi-simple real Lie algebra, called the compact real form of L.
The compact real form of A1 is su(2) and of A2 is su(3). To systematically obtainnoncompact real forms one uses so-called inner and outer involutive automorphisms, whichare beyond the scope of these lectures.
1.6 Exercises
Exercise 1.1
Show that fijk defined in Eq. (1.7) is fully antisymmetric.
Exercise 1.2
Show that the matrices in Eq. (1.10) form a basis for sl(2,R).
Exercise 1.3
Verify Eq. (1.11).
Exercise 1.4
Show that the Cartan subalgebra is indeed a subalgebra.
Exercise 1.5
Verify Eqs. (1.22) and Eqs. (1.23).
13
Exercise 1.6
Verify Eq. (1.28).
Exercise 1.7
Verify Eq. (1.33).
Exercise 1.8
Prove that if α and β are two simple roots and α 6= β, then α − β is not a root andα · β ≤ 0.
14
Chapter 2
Irreps of Semi-Simple Lie Algebras
2.1 Weights
Due to the fact that we restrict to compact Lie groups, the search for irreps of semi-simpleLie algebras is very much simplified. The simplification is that the irreps of compactgroups can be taken unitary and the basis elements aj of the Lie algebra (the generators)to be anti-hermitian. In physics it is customary to use the hermitian operators Xj = −iaj .In terms of these operators we find
[Xj , Xk] = if ljkXl. (2.1)
Let D be a rep of the Lie algebra, then the Lie-product becomes a commutator. We shallfor the sake of brevity, drop the D and use Hj for the elements of the Cartan subalgebrain the rep D and Eα for the other elements of the Lie algebra.
The Hj commute pairwise, so they can be diagonalized simultaneously. Let |m〉 be aneigenvector of all Hj :
Hj |m〉 = mj|m〉, (2.2)
then it is natural to consider the eigenvalues as components of an l-dimensional vector(m1, . . . , ml). If we work in the Cartan-Weyl basis (see Eq. (1.26)), the vectors belong toan l-dimensional Euclidian space.
Definition weight vector
Let H1, . . . , Hl be the hermitian matrices representing the basis of a Cartan algebra insome rep. Then the l-tuples (m1, . . . , ml) of eigenvalues corresponding to simultaneouseigenvectors of H1, . . . , Hl are said to be weight vectors (weights) of the representation.
As the Hj are hermitian, all the eigenvalues mj are real. There is a connection betweenthe weights and the roots. In the adjoint representation the generators Xi themselves areused as basis elements, here denoted as |Xj〉, for the rep. First we note that in the adjointrepresentation one has that
Xa|Xb〉 =∑
c
|Xc〉〈Xc|Xa|Xb〉 =∑
c
|Xc〉[ad(Xa)]cb =
∑
c
ifabc|Xc〉 = |∑
c
ifabcXc〉 = |[Xa, Xb]〉. (2.3)
15
This leads to
Hi|Hj〉 = 0, (2.4)
Hi|Eα〉 = αi|Eα〉. (2.5)
Hence, the roots are the weights of the adjoint representation.
TheoremIf |m〉 is an eigenvector of the Hj corresponding to the weight m, then Eα|m〉 is also aneigenvector. It corresponds to the weight m+ α.
Proof
The proof of this theorem is elementary. Recall that the Lie-product is a commutator.We have
[Hj, Eα] = αjEα, (2.6)
so[Hj , Eα]|m〉 = αjEα|m〉. (2.7)
Working out the commutator gives
HjEα|m〉 −EαHj |m〉 = αjEα|m〉. (2.8)
HenceHjEα|m〉 = (mj + αj)Eα|m〉. (2.9)
2
This proof reminds us of the proof that J±|j,m〉 ∝ |j,m ± 1〉 for eigenstates of J2
and J3. Indeed, the Eα play the role of ladder operators, the Hj act as operators thatdetermine the different eigenstates. To make this even more transparent one can use theCartan-Weyl basis (see Eq. (1.26)) to show that the operators
E± ≡ E±α/|α|, E3 ≡ α ·H/|α|2, (2.10)
for any pair of roots ±α, satisfy the su(2) subalgebra
[E3, E±] = ±E±, [E+, E−] = E3. (2.11)
Strictly speaking, this should be called an A1 subalgebra (see Eq. (1.27)), since one isdealing with the complexification of su(2) (even if one would take these commutationrelations as those of a real Lie algebra, it would not be isomorphic to su(2), but toanother real form of the complexification of su(2), namely the noncompact real Lie algebrasl(2,C)).
For such “su(2)” subalgebras we know all the irreps (Chapter 5 of Part c) and thereforewe know from
E3|m〉 = α ·mα2
|m〉, (2.12)
that (α ·m)/α2 must be a half-integer or integer. In general, for a rank l Lie algebra onecan characterize all states by the irreps of the su(2) subalgebras of the l simple roots.
As the eigenvectors of theHj are the basis functions of the irreps, that are characterizedby weight vectors, one would like to construct weight diagrams. The symmetries of the
16
weight diagrams are the same as for the root diagrams. The following theorem establishesthat the Weyl reflections Sα are the symmetries of the weight diagrams.
TheoremIf m is a weight of an irrep and α a root, then the ratio 2(m · α)/(α ·α) is an integer andm− 2(m · α)/(α · α)α is also a weight of that irrep.
2.2 Weights of A2
Fig. 1.2 is a root diagram, but therefore also an example of a weight diagram, since theroots are the weights of the adjoint rep. Fig. 1.2 is in this case the weight diagram of theirrep of A2 of dimension 8. In general, since A2 is of rank 2, its irreps are characterizedby pairs of integers (p, q), the j-values of the two su(2) subalgebras of the simple rootsα3 and −α2.
The conventional naming scheme that physicists use is the following:
T± =√2E±α1
,
U± =√2E±α2
,
V± =√2E±α3
,
T3 = H1,
Y =2√3H2. (2.13)
Recall that E±αistands for D(e±αi
) and Hi = D(hi). In the previous chapter we alreadydiscussed these quantities in the defining rep. Now we consider the case of a generalirrep (hence T3 should not be identified with the 3× 3 matrix T3 = λ3/2 of the previouschapter, except for the defining rep). The factor 2/
√3 is introduced in Y , for reasons of
convention, but one has to keep in mind that this changes the metric in the root space(gij 6= δij in this way). In particle physics T3 is the third component of the isospin operator
and Y is called the hypercharge, with the electric charge given by Q = T3 + Y/2. In thisway hadrons (see next section) all have electric charges in units of 1.
The above operators satisfy the following commutation relations:
[T3, T±] = ±T±, [Y, T±] = 0,[T3, U±] = ∓U±/2, [Y, U±] = ±U±,[T3, V±] = ±V±/2, [Y, V±] = ±V±,[T+, T−] = 2T3, [T+, V+] = 0,[U+, U−] = 3Y/2− T3, [T+, U−] = 0,[V+, V−] = 3Y/2 + T3, [U+, V+] = 0,[T+, V−] = −U−, [T+, U+] = V+,[U+, V−] = T−, [T3, Y ] = 0.
(2.14)
One can draw all states (each weight vector corresponds to a state) in a two-dimensionaldiagram, with on the y-axis the eigenvalues y of Y and on the x-axis the eigenvalues t3of T3. This leads to the following scheme (see also figure 2.1):
• T+ raises t3 by 1 unit and leaves y unchanged
• U+ lowers t3 by 1/2 unit and raises y by 1 unit
17
• V+ raises t3 by 1/2 unit and raises y by 1 unit
T3
Y
V+
T
U-
+
Figure 2.1: Action of the operators V+, T+, U− in the (t3, y) space of states.
We define a maximum state |max〉 such that
T+|max〉 = V+|max〉 = U−|max〉 = 0 (2.15)
Also, there are two integers p and q, such that
(V−)p+1|max〉 = 0 = (U+)
q+1|max〉 (2.16)
One can show (see for example S. Gasiorowicz, “Elementary Particle Physics”, Wiley1966) that
T3|max〉 = 1
2(p+ q)|max〉, Y |max〉 = 1
3(p− q)|max〉 (2.17)
The irreps are graphically displayed as a figure with a hexagonal boundary in the t3, yplane, with three sides having p t3-units and three with q y-units, as given in Fig. 2.2.Fig. 1.2 corresponds to the (1, 1) irrep of A2. And like Fig. 1.2, every figure correspondingto an irrep obeys the symmetries of the Weyl group. The dimensions of the irreps aregiven by Weyl’s formula
d = (p+ 1)(q + 1)
(p + q
2+ 1
). (2.18)
This can be calculated explicitly, by considering the degeneracy of each of the states insidethe hexagon. The degeneracy is determined by the raising and lowering operators andfor the irreps of A2 the degeneracy follows the following rules: sites on the boundary ofthe hexagon are singly occupied, on the next layer they are doubly occupied, etc., untila triangular layer is reached after which the multiplicity does not increase anymore, butremains q+1 for p > q and p+1 for q > p. Thus for reps with p = 0 or q = 0, which havetriangular weight diagrams from the start, each site is singly occupied. Note that dueto the factor 2/
√3 introduced in Y with respect to T8, the triangle is not an equilateral
triangle, unless one draws it in (t3, t8) space. For this reason and because we are going
18
q
p
q
p
q
p
Y
T3
Figure 2.2: Boundary of the A2 irrep (p, q).
to discuss inner products between weights and roots, we will return to the mathematicsconvention and consider T8 again instead of Y .
The Gell-Mann matrices λ1, . . . , λ8, Eq. (1.29), are written in the three-dimensionaldefining rep of A2. The weights of this rep are easily found by considering again the twocommuting operators T3 = λ3/2 and T8 = λ8/2, that represent the Cartan subalgebra.The weights of T3 and T8 in this representation are easily found, since these are simplythe diagonal entries ([T3]ii, [T8]ii) (no summation over i): (1/2,
√3/6), (−1/2,
√3/6) and
(0,−1/√3). These form an equilateral triangle, which is the “hexagon” (1, 0), one of the
two simplest (nontrivial) irreps. It is depicted in Fig. 2.3 with weight m[1] ≡ (1/2,√3/6)
and its Weyl reflections. This irrep is called the fundamental triplet (see next section)and denoted by its dimension as 3. Note that Q is not an integer for these three states.
m[1]
m[1]
S α 1S α 2
m[1]
S α 1
t 3
t 8
Figure 2.3: The fundamental triplet of su(3).
Of course, there is also another equilateral triangle, namely the (0, 1) irrep. This irrepis called the fundamental anti-triplet and denoted by 3∗. It is depicted in Fig. 2.4.
The irreps (1, 0) and (0, 1) are inequivalent conjugate irreps. If an irrep and itsconjugate irrep are equivalent, i.e. if there exists a nonsingular matrix S such thatSTjS
−1 = −T ∗j for all j, then the Tj form what is called a real rep. An example is
provided by su(2). In the fundamental rep (also called defining rep) Tj = σj/2, S = σ2
(since σ2 is an antisymmetric matrix, the reps of su(2) are usually referred to as be-
19
m[2]
S α 2
m[2]
m[2]
S α 1
t
t
8
3
Figure 2.4: The fundamental anti-triplet of su(3).
ing pseudo-real, but here we will not motivate that distinction). This is another way ofviewing the fact that for su(2) the 3 and 3∗ are equivalent.
2.3 Highest weights and fundamental weights
As we have seen explicitly, it may happen that some of the weights are degenerate: onespeaks of multiple weights. A non-degenerate weight is called simple.
TheoremEvery weight m can be written in terms of the simple roots as
m =
l∑
i=1
µiαi, (2.19)
where the coefficients µi are real and rational.
Definition positive weight, highest weight
A weight m is said to be positive if its first non-vanishing coefficient µi is positive. If theweight M is such that M > m (i.e. M −m is positive) for every other weight m of therep, then M is called the highest weight.
ExampleIn the previous section we discussed the (1, 0) and (0, 1) irreps of A2, the two simplest(nontrivial) irreps. The weights m[1] = (1/2,
√3/6) and m[2] = (1/2,−
√3/6), are the
highest weights of the 3 and 3∗ irreps. These are also called the fundamental weights andbelow we explain this in more detail.
All other weights can be obtained from the highest weight by acting with the raisingand lowering operators of the su(2) subalgebras of the simple roots or simply by actingwith the Weyl group.
TheoremThe highest weight of an irrep is simple.
TheoremTwo irreps are equivalent iff they have the same highest weight.
Therefore, an irrep is completely specified by its highest weight.
20
Definition fundamental weights
The l weights Mi that satisfy, for all simple roots αj ,
2Mi · αj
αj · αj
= δij (2.20)
are called the fundamental weights.
The highest weight rep can be written as M =∑l
i=1 niMi, with ni nonnegative inte-gers, called Dynkin coefficients, determined as
2M · αj
αj · αj
= nj . (2.21)
Moreover, every set of such nonnegative integers determines an irrep of L with highestweight M =
∑l
i=1 niMi.
The fundamental representations are those reps that have one Dynkin coefficient equalto 1 and all others equal to zero. In the previous section we called the inequivalentconjugate irreps (1, 0) and (0, 1) the fundamental triplet and anti-triplet. For (1, 0) weobtain the highest weight m[1] = (1/2, 1/3). In terms of the simple roots α3 = (1/2,
√3/2)
and −α2 = (1/2,−√3/2), one obtains: m[1] = 2/3α3 + 1/3(−α2), which is indeed a
fundamental weight as one can check by using Eq. (2.20). Similarly for m[2] = 1/3α3 +2/3(−α2).
The method of using weight diagrams to represent irreps gets very complicated in caseof su(n) for n ≥ 4. A more convenient way is by exploiting Young tableaux as we haveseen.
2.4 Quarks and the hadron spectrum
There exist three light quarks (up, down and strange) and three heavy ones (charm,bottom and top). Here we will restrict to the three light ones u, d, s. The quantumnumbers these quarks carry are as follows. The quarks are spin-1/2, therefore, in aj = 1/2 irrep of SU(2). They carry an isospin or flavor quantum number and are in thefundamental triplet irrep of flavor SU(3) (see figure 2.5, where the quarks and anti-quarksare depicted by their eigenvalues of the operators (T3, T8)). In addition, the quarks carrya color quantum number and are also in the fundamental triplet irrep of color SU(3).
Hadrons are colorless states (singlets of color SU(3)) consisting of combinations ofquarks and anti-quarks, which are in the 3 and 3∗ of color SU(3), respectively. Thesimplest hadrons are mesons, which consist of one quark and one anti-quark. Recall thatfor SU(3): 3∗ ⊗ 3 = 8⊕ 1 or in Young tableau language
⊗ = ⊕ (2.22)
This shows that tensor products of a quark and an anti-quark states indeed allow for aone-dimensional invariant subspace. The mesons transform according to this 1 irrep, theyare color singlets. One can also apply the above Clebsch-Gordan decomposition to flavor
21
Q=−233Q=−−1
d u
s
3Q=−13Q=−−2
u d
s
−
−
−
Figure 2.5: The fundamental triplet and anti-triplet of su(3) identified with the three lightquarks and antiquarks. Q indicates the electric charge, recall that Q = T3 + T8/
√3 =
T3 + Y/2.
su− sd− 0K
_K * 0_
π0 0ρ
Q=
Q=−1
Q= 0 1
K
+
K
−
−
ρρ
+** 0
φ ω
K*
us
Q=
Q=−1
Q= 0
uddu
ds −
−
−
1
−uudd
ss
−
−
−
Q=
Q=−1
Q= 0
π
1
K+
π+
K 0
’ηη−
−K
Figure 2.6: The pseudoscalar and vector meson octets. The singlet states are also includedin these diagrams.
SU(3). It follows that mesons can either form an octet or a singlet. The pseudoscalar(spin-0) meson octet and vector (spin-1) meson octet are depicted in figure 2.6.
Another type of hadrons are baryons, which consist of three quarks. For SU(3) thetensor product of three quark states can be decomposed as follows:
3⊗ 3⊗ 3 = 10⊕ 8⊕ 8⊕ 1 (2.23)
or in Young tableau language
⊗ ⊗ = ⊕ ⊕ ⊕ (2.24)
Indeed, this shows that there is a color singlet irrep present in the decompositions. Whenapplied to flavor SU(3) the above decomposition shows that baryons can form decupletstates, octet states and singlet states. It turns out that protons and neutrons (both spin-1/2 and isospin-1/2) belong to an octet and the ∆’s (four spin-3/2 and isospin-3/2 states)to the decuplet of flavor SU(3), see figure 2.7.
22
Q=−1 Q= 0 Q= 1
dds
dss uss
uds uus
uududd
sss
ddd uuu
Q= 2 Q=−1 Q= 0 Q= 1 Q= 0
Σ
Ξ
Σ Σ
∆∆∆ ∆Q= 2
Ξ
Ω
+++0−
*−
* 0 * +
*−
−
* 0
Q=
Q=−1
1
n p
ΣΣ
Σ
Ξ Ξ
−
0 +
0−
Λ
Figure 2.7: The baryon decuplet and octet. Note that the left hand figure includes boththe decuplet and the octet.
In 1961 Murray Gell-Mann (and independently Yuval Ne’eman) came up with thisclassification of the hadrons and in 1963 Gell-Mann (and independently George Zweig)postulated the existence of the underlying quarks. One of the hadrons in this classification,the Ω−, was not known at the time, but by assuming this (approximate) flavor SU(3)symmetry its mass could be predicted. In 1964 this particle was experimentally observed,with more or less the predicted mass. This was a great success for the flavor SU(3)picture and a clear demonstration of the power of group theory in physics. Gell-Mannreceived the Nobel prize in 1969 for the “classification of elementary particles and theirinteractions”.
In 2003 the discovery of a new type of hadron was reported: a so-called pentaquark.Pentaquarks consist of 4 quarks and an anti-quark. Irrespective of whether such pen-taquarks indeed exist, one can perform the exercise of classifying them according to irrepsof flavor SU(3). It turns out that the smallest irrep in which the lightest pentaquark state(called the Θ+), with flavor content uudds, fits is the anti-decuplet 10∗. Note that onlythe three states at the corner can be distinguished unambiguously from a three quarkbaryon, since the anti-quark cannot annihilate with any of the four other quarks. Thethree pentaquarks at the corner of the triangle are also called exotic baryons. The Θ+
is reported to have a mass of 1.6 times the proton mass, but its existence is still underdebate (roughly as many experiments see it as ones that don’t see it in the data). Peoplehave considered several other types of multi-quark states, such as hadrons consisting oftwo quarks and two-antiquarks, but these have not been observed (yet).
Considering also the charm quark, one may think of a similar classification using flavorSU(4). However, this is less meaningful, due to the large mass difference of the charmquark with respect to the three light quarks. One can classify the states according to irrepsof SU(4) (leading to three dimensional pictures), but SU(4) transformations do not yieldthe right masses and mass corrections cannot be taken into account in a perturbative way(the correction may be as large in magnitude as some of the hadron masses themselves).
23
Q=−1
−ddssu− −[Ξ ] [Ξ ]+
[Θ ]+
udsss
−
−
−
−
−
−−
−−−
−
−
−
−
−
−
Q= Q= 10Q=−2
uddss
ddsssuudsu
ddssd udssd
−uussd
uudsd
uuddduudss
uudds
uuddu
uddsuuddsd
udssu uussu
uusss
Figure 2.8: The pentaquark anti-decuplet. Octet states are also indicated.
2.5 Casimir operators
A Casimir operator of second order is a quadratic operator (i.e. a linear combination ofquadratic products of generators), with the property that it commutes with all elementsof the Lie algebra. Casimir (1931) showed a general procedure of how to construct suchan operator. We will discuss this in more detail in this section. For Lie algebras of rankl > 1, higher order Casimir operators can be constructed.
TheoremLet ai (i = 1, . . . , n) be a basis of a semi-simple Lie algebra and V be the carrier spaceof some rep of L whose linear operators are Φ(a) (a ∈ L). Then the second-order Casimiroperator C2 is given by
C2 =n∑
p,q=1
gpqΦ(ap)Φ(aq), (2.25)
where gpq is the inverse of the Cartan metric tensor. C2 is independent of the choice ofbasis and it commutes with Φ(a) for all a ∈ L. For an irrep it holds that C2 = λp1, whereλp is a constant that depends on the irrep p and 1 is the dp × dp identity matrix, wheredp is the dimension of the irrep p. For the adjoint rep, C2 equals the (n2 − 1)× (n2 − 1)identity matrix, which means λ = 1.
Example su(2)The basis for the defining j = 1/2 rep of su(2) is ap = (i/2)σp. The Cartan metric tensorwas found to be gmn = −2δmn. Therefore,
C2 =1
8
3∑
p,q=1
δpqσpσq =3
81. (2.26)
For general irreps one obtains:
C2 =1
2
3∑
p=1
D[j](J2p ) ≡
1
2D[j](J2) =
1
2j(j + 1)1. (2.27)
24
Indeed, we find that for the adjoint rep j = 1: C2 = 1.
Example su(3)The Cartan metric tensor for su(3) is
gjk = B(aj , ak) = −12δjk, (2.28)
with ai = iλi. This leads to the second-order Casimir operator
C2 =1
3
(1
2T+, T−+
1
2U+, U−+
1
2V+, V−+ T 2
3 +3
4Y 2
). (2.29)
Now calculate its eigenvalue for the maximum state |max〉 of the irrep (p, q):
〈max |C2|max〉 = 1
3
(1
3(p2 + pq + q2) + p+ q
)(2.30)
For (1, 0) this leads to C2 =491 and for (1, 1): C2 = 1.
Physicists often define the Casimir operator for su(n) as nλp (which is n times theeigenvalue of the Casimir operator C2) and denote it for the fundamental rep by CF orC2(F ), and for the adjoint rep by CA or C2(A). One obtains: CF = (n2 − 1)/2n andCA = n, which can be derived as described below.
Denote the n2 − 1 Hermitian generators of su(n) by T a and Dp(T a) by T ap . We
define T ap T
ap = C2(p)1, such that Tr(T a
p Tap ) = C2(p)dp, where dp is the dimension of
the irrep. In the fundamental irrep physicists define Tr(T aFT
bF ) = δab/2. This leads to
C2(F ) = (n2 − 1)/(2dF ) = (n2 − 1)/(2n).Next we consider the product rep: Dp ⊗ Dq. The generators for this rep are T a
p⊗q ≡T ap ⊗ 1+ 1⊗ T a
q , such that
T ap⊗qT
ap⊗q = T a
p Tap ⊗ 1+ 2T a
p ⊗ T aq + 1⊗ T a
q Taq , (2.31)
and
Tr(T ap⊗qT
ap⊗q
)= Tr
(T ap T
ap
)dq + dpTr
(T aq T
aq
)= (C2(p) + C2(q)) dpdq. (2.32)
Now we apply this to n⊗ n∗ = (n2
− 1)⊕ 1, leading to
2C2(F )dFdF = C2(A)dA + 0. (2.33)
This yields C2(A) = n.
TheoremFor every semi-simple Lie algebra L of rank l with a basis Xi, there exists a set of lCasimir operators in the form of polynomials in D(Xi), whose eigenvalues characterizethe irreps of L.
If one defines br = (g−1)rsas, where g−1 is the inverse of the Cartan metric tensorgrs = fn
rmfmsn, then
C2 =
n∑
p,q,m,n=1
fnpmf
mqnΦ(bp)Φ(bq). (2.34)
25
The higher-order Casimir operators are constructed in a similar way:
Cj =
n∑
p1,q1,p2,q2,...
f q2p1q1
f q3p2q2
. . . f q1pjqj
Φ(bp1) . . .Φ(bpj ). (2.35)
But in practice, for instance for su(n), one does not construct all these Casimir operators;instead one labels the irreps by the j-values of the su(2) subalgebras of the l simpleroots. Note that in Chap. 8.3 of Part II the irreps of su(n), which has rank n − 1 areindeed labelled by n− 1 numbers, i.e. by ν1, . . . , νn−1. For su(3) p and q discussed abovecorrespond to ν1 and ν2, with Eq. (2.18) corresponding to Eq. (8.26) in the lecture notesof Part II.
2.6 Exercises
Exercise 2.1
Show that the weights m[1] = (1/2,√3/6) and m[2] = (1/2,−
√3/6), are the highest
weights of the 3 and 3∗ irreps, when α3 and −α2 are the simple roots. And show thatthey are also the fundamental weights.
Exercise 2.2
Show that for an irrep C2 as defined in Eq. (2.25) is a constant times the identity matrix.
Exercise 2.3
Verify Eq. (2.29).
26
Chapter 3
Global properties of Lie groups
N.B. This chapter is not part of the lectures or the exam.
3.1 Homotopy groups
In this section we consider once more the property of ‘simply connectedness’ of a Liegroup or a manifold in general. It is a topological property defined in terms of continuousmaps T : X → Y , between two manifolds X and Y . The case we considered before wasthat X is a closed line interval X = [0, 1], with the end points identified (T (0) = T (1)),in other words, X = S1. Here we will also consider the more general case, where X = Sn.
Two continuous maps T1 and T2 are called ‘homotopic’ if they can be continuousdeformed into each other.
Definition continuously deformable
Let T s(x) be a continuous map for all s ∈ [0, 1] and x ∈ X . If T s(x) is a continuousfunction of s for all x, we say that for all s1 and s2 in [0, 1] the paths T s1(x) and T s2(x)are homotopic, i.e. continuously deformable into each other.
If two maps are homotopic one writes T1 ∼ T2, because homotopy is an equivalencerelation. If T1 ∼ T2 and T2 ∼ T3, then one can establish T1 ∼ T3 as follows. Take onefamily of maps Ss(x), such that S0(x) = T1(x) and S1(x) = T2(x), and another family ofmaps Us(x), such that U0(x) = T2(x) and U1(x) = T3(x), then construct a new family ofmaps V s(x):
V s(x) =
S2s(x) for 0 ≤ s ≤ 1/2U2s−1(x) for 1/2 ≤ s ≤ 1
(3.1)
Homotopically equivalent maps form an equivalence class which is indicated by T.The set of such classes can be turned into a group. Here we give the example of X = [0, 1],with the end points identified. In this way one restricts to continuous maps which satisfyT (0) = T (1) = y0, a fixed point in Y , then the group can be formed as follows. The groupcomposition law is defined as T1 ∗ T2 = T1 · T2, where
(T1 · T2)(x) =
T1(2x) for 0 ≤ x ≤ 1/2T2(2x− 1) for 1/2 ≤ x ≤ 1
(3.2)
The identity E is the class of maps that are null-homotopic, i.e. continuously deformableto a constant map. The inverse of T is T−1, where T−1(x) = T (1−x). It is straight-
27
forward to check that the composition law is independent of the choice of representativemaps one chooses from the classes.
The group formed in this way for X = S1 is called the first homotopy group or thefundamental group, π1(Y ). Consider again the notion of simply connectedness.
Definition simply connected
A connected Lie group is said to be simply connected, if every closed path, i.e. a path T (t)such that T (0) = T (1), is contractible to a point (i.e., all closed paths are null homotopic).
This means that if a manifold Y is simply connected, then π1(Y ) contains only theidentity E. This is often written as π1(Y ) = 0. If π1(Y ) 6= 0, then it cannot be simplyconnected. A compact Lie group one can fully characterize by its Lie algebra (determiningthe local properties of the group) and by its fundamental group (determining the globalproperties).
Some nontrivial examples:
• Consider the space Y = R2\(0, 0). It is easy to verify that π1(Y ) = Z. The
composition law in this case is addition. For Y ′ = R3\(0, 0) on the other hand, it
holds that π1(Y′) = 0.
• π1(S1) = π1(U(1)) = π1(SO(2)) = Z
• We already have established that SU(2) is simply connected, such that π1(SU(2)) =0, and that SO(3) was not: π1(SO(3)) = π1(SU(2)/Z2) = Z2.
• π1(SU(n)) = 0, ∀n ≥ 2 and π1(SO(n)) = Z2, ∀n ≥ 3
If one generalizes X = S1 to X = Sn, then one can construct in an analogous way thenth homotopy group πn(Y ). To explicitly construct the composition law, it is useful torealize that the n-dimensional sphere Sn is topologically equivalent to the n-dimensionalcube In = (x1, . . . , xn) ∈ R
n|0 < xi < 1, with the faces of the cube identified andequivalent to a fixed point of Sn.
It turns out to be quite nontrivial to calculate πn(Y ) for an arbitrary manifold Yand here we will list only those results that one encounters most frequently in physicsapplications. For instance, they are very important for determining possible topologi-cally nontrivial vacuum solutions in quantum field theories (such as solitons, vortices,monopoles, instantons).
Without proof we state the following results:
• πn(Sn) = Z
• Since SU(2) was shown to have a parameter space that is S3, it follows thatπ3(SU(2)) = Z.
• From the embedding of SU(2) into SU(n) for n ≥ 3 is follows that π3(SU(n)) = Z.Clearly π3(U(1)) = 0.
• π3(SO(3)) = Z also and from its embedding into higher SO(n) one may expectπ3(SO(n)) = Z, which turns out to be true for n > 4. However, for SO(4) which isisomorphic to SO(3)⊗ SO(3), one finds π3(SO(4)) = Z⊗ Z.
28
• π2(Y ) = 0 for any Lie group Y
• If π1(G) = 0 for a connected compact Lie group G and H is a subgroup of G, thenπ2(G/H) = π1(H). For instance, π2(SU(2)/U(1)) = π1(U(1)) = Z.
• π1(SU(n)/Zn) = Zn
3.2 Projective spaces
In this section we collect a few results concerning frequently encountered quotients of Liegroups, such as the n-dimensional sphere Sn and complex n-dimensional projective spaceCPn.
The group SO(n) is the group of rotations in n dimensions. Consider the unit vectoren = (0, . . . , 0, 1). Under rotations it either transforms into another point of the sphereSn−1 (each point of the sphere can thus be viewed as representing a particular rotation)or it stays invariant. Dividing out from SO(n) the subgroup SO(n − 1) that leaves eninvariant then allows one to establish the diffeomorphism (also indicated by the symbol∼= here)
SO(n)/SO(n− 1) ∼= Sn−1. (3.3)
From this one can show that the n-dimensional sphere Sn is not a Lie group in general,because SO(n− 1) is generally not an invariant subgroup of SO(n). Only for n = 1 andn = 3 they are Lie groups, namely S1 ∼= U(1) and S3 ∼= SU(2).
The sphere Sn−1 in Rn is also diffeomorphic to the quotient of orthogonal groups:
Sn−1 ∼= O(n)/O(n− 1) (3.4)
If one further identifies the antipodes of Sn−1, one arrives at RPn−1, the real (n− 1)-dimensional projective space, which is the space of one-dimensional subspaces (calledrays) in R
n.Similarly, one has the complex (n − 1)-dimensional projective space CPn−1, which is
the space of two-dimensional subspaces in Cn. It has the following diffeomorphism:
CPn−1 ∼= SU(n)/U(n− 1) (3.5)
From the previous section it can be concluded that π2(CPn−1) = Z and hence that CPn−1
is not a Lie group for any n. For n = 2: CP1 ∼= S2.
3.3 Universal covering groups
We have already discussed the concept of a universal covering group. For the exampleswe have in mind this notion is straightforward. However, there are complications notmentioned earlier. The problem is that the universal covering group need not be a linearLie group, although for a compact or Abelian Lie group it is.
We will give one example as a warning. It can be shown (see the course “GroupTheory”) that SL(2,C) is the universal covering group of SO(3, 1), which means thattheir (real) Lie algebras are isomorphic. As mentioned before, the complexification of
29
sl(2,C) ∼= so(3, 1) is semi-simple: sl(2,C) ∼= su(2) ⊕ su(2). The complexification ofsl(2,R) is sl(2,C) and π1(SL(2,R)) = Z, whereas π1(SL(2,C)) = 0.
Suppose G is the universal covering group of SL(2,R), then expect G/Z ∼= SL(2,R).Now consider its complexification: G/Z ∼= SL(2,C). However, this leads to a contradic-tion, since π1(SL(2,C)) = 0 and π1(G) = 0 by definition of a universal covering group,but π1(G/Z) = Z 6= π1(SL(2,C)) = 0.
The solution to this problem lies in the fact that we have assumed that the universalcovering group of a noncompact group like SL(2,R) is a linear Lie group. But in factG cannot possess any faithful, finite dimensional rep D : G → GL(n,C). Therefore,G/Z 6∼= SL(2,R).
As said, for compact Lie groups there is no problem. The universal covering groupcan always be embedded in U(n) and its complexification is then a subgroup of GL(n,C),which is unique up to isomorphism. For the noncompact group SO(3, 1) the universalcovering group happens to be a linear Lie group (SL(2,C)), but in general one cannotexpect that as we have seen from the counter-example.
We end this section by listing the universal covering groups of some frequently encoun-tered compact groups. Since the groups SU(n) and Sp(n) (see next section) are simplyconnected, they are their own universal covering groups. The universal covering groupsof SO(n) are denoted by Spin(n), for n ≥ 3. For n = 2: SO(2) ∼= U(1) the universal cov-ering group is R+. In contrast, the groups Spin(n), for n ≥ 3, are all compact. Withoutproof we give the following results:
• Spin(3) = SU(2)
• Spin(4) = SU(2)⊗ SU(2)
• Spin(5) = Sp(2)
• Spin(6) = SU(4)
The universal covering groups of U(n) ∼= U(1)⊗ SU(n) are R+ ⊗ SU(n).
3.4 Norm preserving Lie groups
A metric of a vector space defines a norm via an inner product. Once this is defined onecan consider all those linear transformations that preserve such an inner product. This isin fact how O(n) can be defined, namely as those elements of GL(n,R) that preserve theinner product of two real, n-dimensional vectors:
〈x|y〉 =∑
i
xiyi =∑
i,j
δijxiyj. (3.6)
This can be extended to complex n-dimensional vectors:
〈x|y〉 =∑
i
x∗i yi =
∑
i,j
δijx∗i yj. (3.7)
The elements of GL(n,C) that preserve the latter inner product form the group U(n).
30
In physics one also encounters less trivial metrics, such as the Minkowski space-timein 3+1 dimensions. The basic elements of this space are the four-vectors x. These vectorshave contravariant (denoted by upper indices) components xµ and covariant components(lower indices) xµ. The relation between them is given by the metric tensor g in thefollowing way:
xµ = gµνxν , xµ = gµνxν , µ, ν ∈ 0, 1, 2, 3. (3.8)
The geometry of Minkowski space is fully determined by specifying the metric tensor. Ithas the form:
gµν = gµν = diag (1,−1,−1,−1). (3.9)
The inner product is now given by
〈x|y〉 =∑
µ,ν
gµνxµyν (3.10)
and the group of transformations that preserve this inner product is called the Lorentzgroup O(3, 1). In an analogous way, the groups O(m,n) can be defined and needless tosay these are all non-compact. Clearly, O(n) ∼= O(0, n).
3.4.1 Symplectic groups
Now we discuss another example of Lie groups defined via invariance of an inner productarising from a less trivial metric. Consider real 2n-dimensional vectors and the anti-symmetric inner product defined by
gij =
(0 I−I 0
)
ij
, (3.11)
where I is the n× n unit matrix. The set of all 2n× 2n matrices A that leave the anti-symmetric inner product invariant forms the real symplectic group Sp(2n,R), a subgroupof GL(2n,R). It is easy to show that a 2n × 2n matrix A is in Sp(2n,R) iff ATJA = J ,where J is the conventional name for the metric g given above in Eq. (3.11). The Liealgebra consists of 2n× 2n matrices a that satisfy Ja+ aTJ = 0.
Analogously, Sp(2n,C) can be defined. Perhaps somewhat unexpectedly it is definedas
Sp(2n,C) = A ∈ GL(2n,C)|ATJA = J, (3.12)
instead of with A†JA = J . Furthermore, one defines the symplectic group Sp(n) =Sp(2n,C) ∩ U(2n). The reason for these definitions has to do with the following scheme:
O(n) ⊂ GL(n,R) invarianceofEuclideaninnerproduct
U(n) ⊂ GL(n,C) invarianceofHermitianinnerproduct
Sp(n) ⊂ GL(n,H) invarianceofsymplecticinnerproduct
The latter line needs further explanation. It involves the quaternion algebra H, which canbe written in many different ways. Most familiar is the definition as a real vector spaceH, with the four basis elements 1, i, j, k, which satisfy:
i2 = j2 = k2 = −1,
ij = −ji = k, jk = −kj = i, ki = −ik = j. (3.13)
31
Quaternions are then of the form a + bi + cj + dk, where a, b, c, d ∈ R. As a real vectorspace it splits into R and the space of pure quaternions bi+ cj + dk, which is isomorphicto R
3, via q = (b, c, d). In the notation (r, q) conjugation means (r,−q). The norm can bedefined as N(r, q) = r2 − q2 = a2 + b2 + c2 + d2, which means q2 is non-positive. A 2× 2matrix rep of the 4-dimensional algebra is formed by the 2× 2 unit matrix and the Paulimatrices: i = iσ3, j = iσ2, k = iσ1. In this way any quaternion can be written as a matrix
(a b
−b∗ a∗
), a, b ∈ C, with |a|2 + |b|2 = 1. (3.14)
One sees that this is the same as SU(2).Now we define the quaternion group Sp(1) as
Sp(1) = h ∈ H|N(h) = 1, (3.15)
which consists of the matrices given in Eq. (3.14). The standard isomorphism of H ∼= R4,
implies that Sp(1) ∼= S3 ∼= SU(2).Elements U ∈ SU(2) in the defining irrep satisfy UTJU = J . This follows from the
equivalence of the defining rep and its complex conjugate in SU(2): σ2σjσ−12 = −σ∗
j forall j. Since σ2 = −iJ , it follows that JUJ = U∗ = (UT )−1.
One can also view the quaternions as a complex vector space. Now the basis consistsof two elements 1, j, which satisfy the multiplication rules:
j2 = −1, aj = ja∗, for a ∈ C (3.16)
The standard isomorphism between C2 and H is:
C2 → H, (a, b) 7→ a + bj. (3.17)
Conjugation in H now means: a+ bj 7→ a∗− bj. The norm is then written as N(a+ bj) =|a|2 + |b|2.
A convenient way of satisfying the noncommutative multiplication rules is to write thetwo basis elements 1, j as 2× 2 matrices:
1 =
(1 00 1
), j =
(0 1−1 0
), (3.18)
and such that
a+ bj =
(a b
−b∗ a∗
). (3.19)
Conjugation in H now means Hermitian conjugation. This is the reason we will arrive ata subgroup of U(2n) below.
The standard isomorphism H ∼= C+ Cj ∼= C2 induces also:
Hn ∼= C
n + Cnj ∼= C
n ⊕ Cn ∼= C
2n (3.20)
Now one wants to find the subgroup of GL(n,H) that preserves the norm, which on Hn
is simply defined as the sum of norms on each of the n H’s: N(h) =∑n
i=1N(hi). Thesymplectic group Sp(n) is thus defined as:
Sp(n) = A ∈ GL(n,H)|N(Ah) = N(h). (3.21)
32
This implies that Sp(n) is the quaternionic unitary group U(n,H). Its elements A viewedas a map from C
2n → C2n belong to U(2n). Next we want to translate this into a direct
requirement on A.First we note that left multiplication by j can be written as:
j : C2n → C2n, a + bj 7→ −b∗ + a∗j, (3.22)
and right multiplication by j as:
j : C2n → C2n, a+ bj 7→ −b+ aj. (3.23)
If we define complex conjugation by the operation c, then one finds (as may have beenexpected) cjh = hj. Any operator A that satisfies N(Ah) = N(h), combined withN(hj) = N(cjh) = N(h), should satisfy Acj = cjA. Since Ac = cA∗, it follows A∗j = jA.For A is unitary, this implies AT jA = j. This is the motivation for the definition ofSp(2n,C) using AT rather than A†.
The relevance of the symplectic groups for physics is in the application to Hamiltoniandynamics. For a discussion on this we refer to Hassani (see its sections 25.3.3 and 26.6).
The Lie algebras of the groups Sp(2n,R), Sp(2n,C) and Sp(n) all have the samecomplexification Cn = sp(n). The real Lie algebra sp(n) is the compact real form of Cn,whereas sp(2n,R) and sp(2n,C) are non-compact. Also, the groups Sp(2n,C) and Sp(n)are simply connected, but π1 (Sp(2n,R)) = Z.
33
Chapter 4
Bibliography
The lectures are based on the books listed below.
H.F. Jones Groups, Representations and Physics,Institute of Physics Publishing, Second Ed., 1998
H. Bacry Lectures on Group Theory and Particle Physics,Gordon and Breach, 1977
A.D. Barut & R. Raczka Theory of Group Representations and Applications,Wrold Scientific, 1986
A rather comprehensive source on group theory is:
J.S. Cornwell Group Theory in Physics, vols I and II,Academic Press, 1984
J.S. Cornwell Group Theory in Physics, An Introduction,Academic Press, 1997
The set of two volumes contains much more than the subject matter of the lecture notes.It contains most of the proofs of the theorems on Lie groups and Lie algebras addressedin the lectures notes. The second reference is a condensed and abridged version of thesevolumes.
M. Hamermesh Group Theory and its Application to Physical Problems,Addison Wesley, 1962 and also: Dover Pubns; Reprint Ed. (1989)
This work is more accessible than Cornwell, and is particularly recommended for thetreatment of the symmetric groups. It is, however, much more limited in scope thanCornwell. One should keep in mind that at the time Hamermesh published his book thequark model did not exist yet.
34
Classic textbooks on group theory and its applications in quantum mechanics are:
E.P. Wigner Gruppentheorie und ihre Anwendung auf die Quantenmechanik der
Atomspektren, Friedr. Vieweg & Sohn, 1931
H. Weyl The Theory of Groups and Quantum Mechanics,Dover Publications (Reprint of the original from 1928)
Group theory topics relevant for particle physics are covered extensively in:
D.B. Lichtenberg Unitary Symmetry and Elementary Particles,Academic Press, 1970
H. Georgi Lie Algebras in Particle Physics,Benjamin/Cummings, 1982
H.J. Lipkin Lie Groups for Pedestrians,Dover Publications (Reprint of the original from 1966)
R.N. Cahn Semi-simple Lie algebras and their representations,Benjamin/Cummings, 1984
T.-P. Cheng & L.-F. Li Gauge theory of elementary particle physics,Oxford University Press, 1984
Mathematical treatments of reps of Lie groups and Lie algebras are:
T. Brocker & T. tom Dieck Representations of compact Lie groups,Springer Verlag, 1985
A.L. Onishchik & E.B. Vinberg Lie groups and Algebraic groups,Springer Verlag, 1990
W. Fulton & J. Harris Representation theory: a first course,Springer Verlag, 1991
B.C. Hall Lie groups, Lie algebras, and representations,Springer Verlag, 2003
S. Hassani Mathematical Physics: a Modern Introduction to
Its Foundations, Springer Verlag, 1999
35
