group theory with applications

Notes on Group Theory

Mario Trigiante

November 18, 2010

Contents

1 Abstract Groups 51.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Definition of an Abstract Group . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.2.1 Some examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.2.2 Finite and Infinite Groups. . . . . . . . . . . . . . . . . . . . . . . . . 91.2.3 Generators and Defining Relations . . . . . . . . . . . . . . . . . . . 10

1.3 Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.4 Classes and Normal Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.4.1 Conjugation classes in Sn and Young Diagrams. . . . . . . . . . . . . 171.4.2 Conjugation Classes in Dn . . . . . . . . . . . . . . . . . . . . . . . . 201.4.3 Normal Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

1.5 Homomorphisms, Isomorphisms and Automorphisms . . . . . . . . . . . . . 221.5.1 The Automorphism Group . . . . . . . . . . . . . . . . . . . . . . . . 25

1.6 Product of Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271.7 Matrix Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

1.7.1 The Group GL(n,F) . . . . . . . . . . . . . . . . . . . . . . . . . . . 271.7.2 The Group SL(n,F) . . . . . . . . . . . . . . . . . . . . . . . . . . . 281.7.3 The Group O(n,F) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281.7.4 The Group O(p, q;F) . . . . . . . . . . . . . . . . . . . . . . . . . . . 311.7.5 The Unitary Group U(n) . . . . . . . . . . . . . . . . . . . . . . . . . 331.7.6 The Pseudo-Unitary Group U(p, q) . . . . . . . . . . . . . . . . . . . 341.7.7 The Symplectic Group Sp(2n,F) . . . . . . . . . . . . . . . . . . . . 351.7.8 The Unitary-Symplectic Group USp(2p, 2q) . . . . . . . . . . . . . . 36

2 Transformations and Representations 392.1 Linear Vector Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

2.1.1 Homomorphisms Between Linear vector Spaces . . . . . . . . . . . . 422.1.2 Inner Product on a Linear Vector Space over R . . . . . . . . . . . . 442.1.3 Hermitian, Positive Definite Product on a Linear Vector Space over C 452.1.4 Symplectic product . . . . . . . . . . . . . . . . . . . . . . . . . . . . 462.1.5 Example: Space of Hermitian matrices . . . . . . . . . . . . . . . . . 47

2.2 Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 482.2.1 Real Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

3

4 CONTENTS

2.2.2 Complex Spaces With Hermitian Positive Definite Metric . . . . . . . 502.2.3 Symplectic Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

2.3 Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512.3.1 Transformations on Mn . . . . . . . . . . . . . . . . . . . . . . . . . . 532.3.2 Homogeneous-Linear Transformations on Mn and Linear Transforma-

tions on Vn. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 572.3.3 Affine Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . 602.3.4 The volume preserving group SL(n,F) . . . . . . . . . . . . . . . . . 612.3.5 (Pseudo-) Orthogonal Transformations . . . . . . . . . . . . . . . . . 612.3.6 Unitary Transformations . . . . . . . . . . . . . . . . . . . . . . . . . 652.3.7 Homomorphism between SU(2) and SO(3) . . . . . . . . . . . . . . . 652.3.8 Symplectic Transformations . . . . . . . . . . . . . . . . . . . . . . . 672.3.9 Active Transformations in Different Bases . . . . . . . . . . . . . . . 68

2.4 Realization of an Abstract Group . . . . . . . . . . . . . . . . . . . . . . . . 682.4.1 Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 702.4.2 The Regular Representation and the Group Algebra . . . . . . . . . . 78

2.5 Some Properties of Representations . . . . . . . . . . . . . . . . . . . . . . . 832.5.1 Unitary Representations and Quantum Mechanics . . . . . . . . . . . 86

2.6 Schur’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 892.7 Great Orthogonality Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 922.8 Characters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 952.9 Operations with Representations . . . . . . . . . . . . . . . . . . . . . . . . 101

2.9.1 Direct Sum of Representations . . . . . . . . . . . . . . . . . . . . . . 1012.9.2 Direct Product of Representations . . . . . . . . . . . . . . . . . . . . 1032.9.3 Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

2.10 Representations of Products of Groups . . . . . . . . . . . . . . . . . . . . . 111

3 Constructing Representations 1133.1 Constructing Representations of Finite Groups . . . . . . . . . . . . . . . . . 113

3.1.1 Irreducible Representations of Sn and Young Tableaux . . . . . . . . 1163.2 Irreducible Representations of GL(n,C) . . . . . . . . . . . . . . . . . . . . . 1213.3 Product of Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1343.4 Branching of GL(n,C) Representations With Respect to GL(n− 1,C) . . . . 1353.5 Representations of Subgroups of GL(n,C) . . . . . . . . . . . . . . . . . . . 137

3.5.1 Representations of U(n) . . . . . . . . . . . . . . . . . . . . . . . . . 1423.5.2 Representations of (S)O(p, q) . . . . . . . . . . . . . . . . . . . . . . 144

4 References 153

Chapter 1

Abstract Groups

1.1 Introduction

The mathematical notion of group plays an important role in natural sciences as it providesa quantitative description of Symmetry. By symmetry we mean the property of an object(e.g. a crystal, a molecule or any other physical system) to “look the same” after undergoinga certain transformation. A symmetry transformation, in other words, brings the object tooverlap with itself so that the final configuration of the system is virtually indistinguishablefrom the initial one. The system is then said to be invariant with respect to its symmetrytransformations. For instance the symmetry transformations of a square are rotations aboutits center by multiples of 90o and reflections in two mutually perpendicular lines, while thoseof an equilateral triangle are rotations about its incenter by multiple of 120o and reflectionsin the three altitudes. In three dimensional Euclidean space a spherical object is symmetricwith respect to any rotation about any direction through its center and to reflections inany plane containing the center. The symmetry of the five regular polyhedrons, i.e. thetetrahedron, the cube, the octahedron, the dodecahedron and the icosahedron can be totallycharacterized by 3 distinct sets (or groups as we shall see) of transformations, consisting ofrotations and reflections, which are enough to capture the symmetry properties of variousthree dimensional structures: For instance the molecule of methane CH4 has the shape of atetrahedron in which the Hydrogen atoms are located at the four vertices and the Carbonatom at the center; The Pyrite (FeS2) is a mineral whose crystal may appear in cubic oroctahedral shapes; The ion B12H

2−12 has an icosahedral shape.

Given two symmetry transformations, A and B, of a system, we can consider the transfor-mation resulting from first performing B and then applying A to the resulting configuration.This transfromation is defined as the product A ·B of the two transformations A and B andis clearly a symmetry transformation (take as A and B two clockwise rotations by 90o of acube about its center, the product A · B will then be a clockwise rotations by 180o). Forany transformation A which maps an initial configuration of the system into a final one, wecan define its inverse A−1 which brings the system back to its initial configuration. If A is asymmetry transformation, so is A−1 (in the example of a square, if A is a 90o clockwise ro-tation, A−1 is a 90o anti–clockwise rotation, or, equivalently, a 270o clockwise rotation). We

5

6 CHAPTER 1. ABSTRACT GROUPS

can define the identity transformation I as the trivial transformation mapping every point inthe original configuration into itself. It clearly is a symmetry transformation of any object.Moreover, of A is a transformation, then it can be easily verified that: A · I = I · A = A.Finally, given three transformations A,B,C, if we first apply B ·C and then A to an initialconfiguration or first C and then A · B the resulting configuration is the same, namely theproduct of transformations is associative: A ·(B ·C) = (A ·B) ·C. A set of elements endowedwith a product operation satisfying the above properties is called a group. Therefore the setof all the symmetry transformations of an object, together with the product defined above,is a group called the symmetry group of the system.

The symmetries of a system have consequences on its physical properties. Considertwo point-like electric charges, such as the electron e− and the proton p+ in a Hydrogenatom. The electric field

−→E generated by p+ has spherical symmetry, namely symmetry with

respect to any rotation about the proton itself. This implies that the physical properties ofthe system, such as the energy, do not change if we rotate the electron around the protonkeeping their distance fixed. Consider now an electric point-charge q in the field

−→E generated

by other two point-charges q1, q2. The symmetry of the latter now is no longer spherical,but rather cylindrical, since it is invariant with respect to an arbitrary rotation about onlyone direction, which is the one connecting the two source-charges. As a consequence of thisall the physical properties of the system, like the energy, should not change if we rotate qaround the symmetry axis keeping its distance from it fixed.

As an other instance consider a crystal with its own symmetry group. Since the physicalproperties of any system are left unchanged under the action of a symmetry transformation,the refraction index along directions connected to one another by the action of elements ofthe symmetry group is the same.

In quantum mechanics, all physical properties of a system, like a sub-atomic particle,an atom or a molecule, are encoded in the notion of quantum state Ψ, represented by avector |Ψ〉 in an infinite-dimensional Hilbert space. The effect of a transformation A, whichmay be a rotation, a reflection or a translation in our three-dimensional Euclidean space, ormay also include time evolution or Lorentz boosts if we consider the larger four-dimensionalMinkowski space-time, is to map a state into a different one:

|Ψ〉 → |Ψ′〉 ≡ A · |Ψ〉 . (1.1.1)

If A is a symmetry transformation of the system, its action will not alter the energy: EΨ =EΨ′ . The presence of a symmetry implies the existence of more than one independent state fora given value of energy, namely degeneracy of the energy levels. The amount of degeneracywill depend on the maximal number of independent states related to a given one by theaction of the elements of the symmetry group. Such numbers, for different initial states, area mathematical feature of the group itself, i.e. the dimensions of its representations.

Understanding all the physical properties of a system in relation to its symmetries is oneof the main successes of group theory applied to natural sciences.

1.2. DEFINITION OF AN ABSTRACT GROUP 7

1.2 Definition of an Abstract Group

We have introduced the notion of group in relation to the symmetry transformations ofan object. Different objects may have symmetry transformations which, through acting ina different way, close groups having the same structure. Such structure, which we shalldefine more rigorously in a moment, is encoded in the definition of a corresponding abstractgroup. We say that different symmetry groups have the same structure if they are differentrealizations of a same abstract group. As an example consider the (proper) symmetry groupof a square, consisting of rotations about its center by integer multiples of π/2 and thesymmetry group of a parallelepiped with square basis, consisting of π/2–rotations about itsaxis. The two groups are represented by different elements, since their transformations acton different objects, but, nevertheless share the same structure and are different realizationsof the same abstract group to be denoted by C4. Let us start then by giving the definitionof an abstract group.

A set of objects G, on which a product operation · is defined, is said to be a group if thefollowing axioms are satisfied:

i G is closed with respect to ·: For any couple of elements g1, g2 in G there exist athird element g3 such that g3 = g1 · g2. In other words · is a map from G×G into G;

ii Existence of unit element: There exist an element e in G so that, for any g ∈ Gwe have: e · g = g · e = g;

ii Existence of the inverse element: For any element g of G there exist its inverseg−1 in G so that: g · g−1 = g−1 · g = e;

iii Associativity: For any three elements g1, g2, g3 of G the following property holds:g1 · (g2 · g3) = (g1 · g2) · g3. The result of this triple product is denoted by g1 · g2 · g3;

Using the associative property of the product, it is easy to verify that: (g1 ·g2)−1 = g−12 ·g−1

1 .Moreover the inversion is involutive, namely, for any g in G, g = (g−1)−1.

The correspondence between couples of elements g1, g2 of G and elements g3 = g1 · g2 inG through the product operation defines the structure of G.

In general, given two elements g1, g2 of G, the product g1 ·g2 may be different from g2 ·g1,namely the two elements may not commute. If any two elements of G commute the groupis said to be abelian.

A group G is said to be finite if it consists of finitely many elements: G ≡ g1, . . . , gn.The number n of its elements is called the order of G and is also denoted by |G|. Thestructure of a finite group G is conveniently represented by a multiplication table in whichthe first row and column contain the elements of the group in the same order g1, . . . , gn whilethe entry (i, j) is the element gi · gj.

Exercise 1.1: Show that the four complex numbers 1, i, −1, −i form a group of orderfour with respect to ordinary multiplication and write the multiplication table.


1.2.1 Some examples

The cyclic group: An example of finite group is the cyclic group which is the groupwhose elements are powers gk, k integer, of an element g called the generator of the group:

Cyclic group ≡ . . . , g−2, g−1, g0, g, g2, . . . . (1.2.1)

Let us define what we mean by power of an element g: If k is a positive integer thengk ≡ g · g . . . · g k-times (g2 ≡ g · g, g3 ≡ g · g · g and so on, so that gk ≡ gk−1 · g); If k isa negative integer gk ≡ (g−1)−k; Finally we define g0 ≡ e. We say that the cyclic group isfinite of period n, for some positive integer n if gn = e, so that the group will consist of thefollowing elements

Cyclic group of period n ≡ e = g0, g, g2, . . . , gn−1 , (1.2.2)

where we omitted the negative powers since g−1 = gn−1,g−2 = g2n−2 = gn · gn−2 = e · gn−2 =gn−2 and so on. The multiplication table of this group reads:

e g . . . gn−1

e e g . . . gn−1

g g g2 . . . e...

gn−1 gn−1 e . . . gn−2

From the symmetry of the table we may conclude that a cyclic group is abelian.Exercise 1.2: Consider the group Cn of rotations about a point in the plane by integer

multiples of 2 π/n (C3 is the symmetry group of an equilateral triangle, C4 of a square etc...).Show that the group Cn is cyclic of order n.

The permutation group: Given a set of n objects x1, . . . , xn a permutation is a one-to-one correspondence between each of them and other objects in the same set. If a permutationS maps x1 into xi1 , x2 into xi2 up to xn into xin , it is usually represented by the short-handnotation:

S ≡(x1 x2 . . . xnxi1 xi2 . . . xin

)or simply

(1 2 . . . ni1 i2 . . . in

), (1.2.3)

the order of elements in the first row is not important, what matters is the mapping. Weshall also denote the number ik corresponding to k through the permutation S by S(k) andsay that S maps 1 into the number S(1), 2 into S(2) and so on. Consider now the set Snof all possible permutations of a set of n objects. We can define on Sn a product operation:If a permutation S maps a generic element xk into the element xik = xS(k) and an otherpermutation T maps xik = xS(k) into xjk = xT (S(k)), the product T · S is the permutationmapping xk into xjk = xT (S(k)):

T · S =

(xi1 xi2 . . . xinxj1 xj2 . . . xjn

)·(x1 x2 . . . xnxi1 xi2 . . . xin

)≡(x1 x2 . . . xnxj1 xj2 . . . xjn

)=

=

(x1 x2 . . . xn

xT (S(1)) xT (S(2)) . . . xT (S(n))

). (1.2.4)


For instance if

S ≡(

1 2 3 43 4 2 1

)and T ≡

(1 2 3 44 2 3 1

), (1.2.5)

we have:

T · S ≡(

1 2 3 43 1 2 4

). (1.2.6)

Clearly this product is not commutative, since S · T =

(1 2 3 41 4 2 3

)6= T · S. We define

the identity permutation I as the trivial permutation mapping each object in the set intoitself:

I ≡(

1 2 . . . n1 2 . . . n

). (1.2.7)

From the definition of product one can easily verify for any permutation S that: S · I =I · S = S.

Given a permutation S mapping a generic element xk into the element xik , we define itsinverse S−1 as the permutation mapping xik back into xk:

S ≡(

1 2 . . . ni1 i2 . . . in

)⇒ S−1 ≡

(i1 i2 . . . in1 2 . . . n

). (1.2.8)

Clearly the following property holds: S · S−1 = S−1 · S = I. Finally one can verify that theproduct of permutations is associative. The set of all permutations of n objects, endowedwith the product defined above, forms therefore a (non-abelian) group Sn of order n!.

Exercise 1.3: Show that the following set of permutations:

T1 =

(1 2 3 41 2 3 4

), T2 =

(1 2 3 42 1 4 3

), T2 =

(1 2 3 44 3 2 1

),

T4 =

(1 2 3 43 4 1 2

), (1.2.9)

for a group and write the multiplication table. Is the group abelian?

1.2.2 Finite and Infinite Groups.

Let G be a finite group and g any of its elements. It follows that gk, for any integer k is stillan element of G. However, since G is finite, not all gk, for different values of k are different.In particular we must have for some integers ` and `′: g`

′= g`. Multiplying a number of

times both sides by g−1 and using the associative property of the product, we deduce thatgk = e for a certain k. Therefore, for any element g there exist an integer k, called the orderor period of g, such that: gk = e, and therefore gk−1 is the inverse of g.


A group G is called infinite if it is not finite. An example of infinite group is the set ofrational positive numbers with respect to ordinary product. An other example of infinitegroup is an infinite cyclic group, such as the group of integer multiples of a real number a,. . . ,−2 a, −a, 0, a, 2 a, . . ., on which the product operation is defined as the ordinary sumof real numbers.

A special instance of infinite groups are the Lie groups, which will be extensively studiedin the present course. The elements of a Lie group are functions of a number of continuousparameters g = g(α1, . . . , αn) (for this reason they are called continuous groups) and theirstructure is described by specifying the parameters of each g3 = g1 · g2 as analytic functionsof the parameters of g1 and g2. The number n of parameters is called dimension of the Liegroup. An instance of Lie group is the group SO(3) of rotations in the three dimensionalEuclidean space, each element R being uniquely defined by the three continuous parameterssuch as the Euler angles: R = R[θ, ϕ, ψ]. This group, together with the spatial reflections,forms the symmetry group O(3) of the sphere. A simpler example of Lie group is the groupSO(2) of planar rotations R[θ] about a point by an angle θ. This group, as opposed to SO(3),is abelian and its structure is defined by the relation: R[θ1] ·R[θ2] = R[θ1 + θ2].

The Group GL(n, R). An other example of Lie group is the group GL(n, R) of generallinear transformations on a n–dimensional linear vector space. Its elements are n× n non-singular matrices M[aij]

1 depending on n2 continuous parameters aij which are it own entries:

M[aij] ≡ (aij) =

a11 a12 . . . a1n

a21 a22 . . . a2n...

. . ....

an1 an1 . . . ann

. (1.2.10)

This is a group with respect to ordinary rows-times-columns matrix multiplication. Theidentity element of the group is the n×n identity matrix 1n. With any non-singular matrixwe can associate its inverse matrix and the matrix multiplication is associative. This groupis clearly not abelian since the product of matrices is not.

The structure of this Lie group is defined by expressing the entries of the product of twomatrices in terms of their entries:

M[cij] = M[aij] ·M[bij] , (1.2.11)

then cij =∑n

k=1 aik bkj.

1.2.3 Generators and Defining Relations

According to the very definition, a group G is totally defined once its elements are given,together with its structure, namely its multiplication table. The groupGmay be very large, if

1In the present chapter the elements of a generic matrix will be labeled by two lower indices (e.g. aij).In next chapter we shall consider transformations implemented by matrices and will give a meaning tovectors with upper and lower indices, according to their transformation property. The elements of matricesrepresenting transformations will have one lower and one upper index (e.g. M i

j or Rij) since their action

on any vector should give a vector of the same kind, i.e. with index in the same (upper or lower) position.


not infinite and writing the multiplication table may be arduous, if not impossible. Considerthe infinite cyclic group with generator g. Although infinite, defining it was relatively simplesince its generic element can be written as an integer power of the generator g. In generalwe can generalize this characterization of a group by defining a minimal set of elements of agroup g1, g2, . . . , g`, called the generators of the group, such that any group element canbe written as a sequence of products of these generators. The generators are like letters of analphabet and the group elements words (in fact a generic group element is also referred to asa word when it is represented as a sequence of generators). Clearly a group is abelian if andonly if its generators commute with one another: gi · gj = gj · gi, for any i, j = 1, . . . , `. Oneimplication is straightforward. Let us show that, if this property is true a group is abelian.Take two elements g and g′ of G. Each of them is represented by a word. Compute then theproduct g · g′. Using the commutativity property of the generators, we can permute themfreely in this product and compose the word representing g′ first and that representing gsecond, obtaining g · g′ = g′ · g. The group G is then abelian.

Finite cyclic groups, such as the rotation symmetry group Cn of regular polygons, havejust one generator. For Cn groups, the generator is a rotation r by 2 π/n. The dihedralgroups Dn consists of rotations and reflections: D3 is the symmetry group of an equliateraltriangle. It has order 6 and consists of rotations by multiples of 2π/3 (which themselvesclose a group C3), and a reflection in three directions containing the altitudes σ, α, β (seeFigure 1.1); D4 is the symmetry group of a square. It has order 8 and consists of rotations

Figure 1.1: Group D3

by multiples of π/2 (which themselves close a group C4), and reflections σ, α, β, γ in thedirections along the diagonals or connecting the midpoints of opposite sides (see Figure 1.2).While Cn have just one generator r, Dn have two generators: The rotation r by 2 π/n anda reflection σ. This means that any element of Dn can be expressed as products of r andσ, as it can be seen for the D3 and D4 examples in Figures 1.1 and 1.2. The group Dn in


Figure 1.2: Group D4

general consists of rotations by multiples of 2π/n, which themselves close the group Cn, andn reflections in directions at angles multiple of π/n. All these n reflections can be obtainedby first acting by one of them σ, and then applying the rotation r a number of times. Theorder of Dn is then 2n:

Dn : e, r, r2, . . . , rn−1, σ, r · σ, r2 · σ, . . . , rn−1 · σ . (1.2.12)

If we are only told that Dn had just two generators r, σ, we would construct an infinite set ofwords our of these two letters. We used the geometric properties of rotations and reflectionsto conclude that a generic word can always be reduced to one of the 2n in (1.2.12). Whatgeometry is telling us is that not all words correspond to distinct elements of the group.For instance, we perform three consecutive rotations r by an angle 2 π/3 we obtain theidentity rotation, i.e. r3 = e. Moreover if we perform twice a reflection we end up with theidentity transformation, namely σ2 = e. As a consequence of this, in D3, the words r3+k

and rk represent the same element. Similarly σ2+k and σk. These two properties of the D3

generators have the effect of cutting down the number of independent words, though are notenough to reduce them to six. We need to use the property that r · σ is itself a reflectionand thus (r · σ)2 = e. From this relation, the previous two and the defining axioms of thegroups, we find r ·σ = σ · r−1 = σ · r2, which allows us to permute the order of the rs and theσs and reduce a generic word to the form rk · σp, with k = 0, . . . , 2 and p = 0, 1. Similarly,for Dn, we can reduce the number of independent words by using the relations rn = e andσ2 = e and the fact that r · σ is a reflection, namely that (r · σ)2 = e. From these relationsit follows that r · σ = σ · rn−1, using which we can reduce a generic word to the form rk · σp,with k = 0, . . . , n− 1 and p = 0, 1, yielding the 2n independent elements in (1.2.12). Using


the three relations rn = e, σ2 = e and (r · σ)2 = e we can construct the multiplication tableof the group, namely they are enough to totally characterize its structure.

We can, in general, characterize the structure of any group by a set of defining relations.By defining relations we mean a minimal set of conditions in the form of Wk = e, Wk beingsuitable words, namely sequences of products of generators, which are sufficient to deducethe whole group structure. Any other relation satisfied by the elements of G can be deducedfrom the defining ones using the defining axioms of a group. For instance a cyclic group ofperiod n and generator g is defined by the single relation gn = e. Consider now the trivialgroup G = e. Also in this case the relation gn = e, for any integer n, is trivially satisfiedby all its elements. However gn = e, for n > 1 is not the defining relation, since the relationsgn = e, for n < 1 cannot be derived from it. In this case the defining relation is simplyg = e. Notice that, as opposed to a finite cyclic group the infinite cyclic group, also referredto as C∞, has no defining relation: Different words correspond to different group elements.Let us summarize our discussion so far. The task of defining the elements of a group isconsiderably simplified by assigning its generators, in terms of which all the elements arewritten as words. Then the structure of the group can be deduced from a set of definingrelations. The defining relations for the Cn and the Dn groups are:

Cn : rn = e ,

Dn : rn = e , σ2 = e , (r · σ)2 = e . (1.2.13)

From the last three relations we find r · σ = σ · rn−1. If n = 1, the first relation impliesr = e, namely D1 consists of the only reflection σ, apart from e, and is trivially abelian. Ifn = 2 we find r · σ = σ · r and the group is again abelian. The reader can convince himselfthat these two, namely D1 and D2 are the only abelian dihedral groups. Using the relations(1.2.13) we are able to construct the multiplication table of a Dn group. Let us give belowthose of D3 and D4:

D3 e r r2 σ α = r · σ β = r2 · σe e r r2 σ r · σ r2 · σr r r2 e r · σ r2 · σ σr2 r2 e r r2 · σ σ r · σσ σ r2 · σ r · σ e r2 r

α = r · σ r · σ σ r2 · σ r e r2

β = r2 · σ r2 · σ r · σ σ r2 r e

D4 e r r2 r3 σ α = r · σ β = r2 · σ γ = r3 · σe e r r2 r3 σ r · σ r2 · σ r3 · σr r r2 r3 e r · σ r2 · σ r3 · σ σr2 r2 r3 e r r2 · σ r3 · σ σ r · σr3 r3 e r r2 r3 · σ σ r · σ r2 · σσ σ r3 · σ r2 · σ r · σ e r3 r2 r

α = r · σ r · σ σ r3 · σ r2 · σ r e r3 r2

β = r2 · σ r2 · σ r · σ σ r3 · σ r2 r e r3

γ = r3 · σ r3 · σ r2 · σ r · σ σ r3 r2 r e


Notice that the above tables are not symmetric and thus the groups are not abelian.

1.3 Subgroups

A subset H of G is a subgroup of G if it is itself a group. Clearly any group G has assubgroups the one consisting of the unit element only e and G itself. Actually e is theonly order one subgroup of G. Indeed if there were an other order one subgroup g, forsome g ∈ G, it should contain all the powers g` of g up to its period k: gk = e. Since thesubgroup consists of the element g only, k = 1 and g = e.

Here we prove a necessary and sufficient condition for a subset H of G to be a subgroup:

Property 1.1: H is a subgroup of G if and only if for any two elements h1, h2 of H, h1 ·h−12

is in H.

One implication is simple. Let us now suppose that for any two elements h1, h2 of H,h1 · h−1

2 is in H and show that H is a subgroup of G, namely a group with respect to theproduct operation induced from G. Notice that, from the hypothesis, it immediately followsthat e = h1 ·h−1

1 is in H, for any h1 ∈ H. Moreover given a h1 ∈ H, being e ∈ H, h−11 = e·h−1

1

is in H as well. As a consequence of this, given any two elements h1, h2 of H, h1 · h2 is stillin H, namely that H is closed with respect to the product on G. Indeed for any h1, h2 ∈ H,since h−1

2 is still in H, it follows that h1 · h2 = h1 · (h−12 )−1 is by hypothesis in H.

Remark: If G is infinite, in order for a subset H of G to be a subgroup, it is not sufficientfor it to be closed with respect to the product. As an example consider the group G of al realnumbers except 0, with respect to ordinary multiplication. Consider now a positive numberx > 0 and the subset H of all numbers of the form xn, with n positive integer. H is closedwith respect to ordinary product, however it is not a group since, for n > 0, xn ∈ H but(xn)−1 = x−n is not in H. If G is finite, on the other hand, any subset of G closed withrespect to the product is a subgroup. Indeed, for any h ∈ H, h` is contained in H, for any `.Being G finite there exist an integer k such that hk = e, which implies that e and h−1 = hk−1

are contained in H as well.

Clearly the only subgroup of order one consists of the identity e only.

Dihedral groups: The dihedral group Dn has Cn as a subgroup. Listing the elements ofDn as in (1.2.12), the first n × n diagonal block of its multiplication table is precisely themultiplication table of its Cn subgroup.

Cyclic permutations: A cyclic permutation of n objects x1, . . . , xn, denoted by the sym-bol (x1 x2 . . . xn), or simply (1 2 . . . n) is a permutation mapping each element into thesubsequent up to xn which is mapped into x1:

(1 2 . . . n) ≡(

1 2 . . . n− 1 n2 3 . . . n 1

). (1.3.1)

1.3. SUBGROUPS 15

In this notation (1 2 . . . n) = (2 3 . . . n − 1 1) = . . .. A cyclic permutation of n objects isalso called a cycle of order n or a n–cycle. We can perform a cyclic permutation of n objectsa number of times. Clearly if we perform it n times we are back to the initial sequence,namely we obtain the identity permutation:

(1 2 . . . n)2 ≡(

1 2 . . . n− 2 n− 1 n3 4 . . . n 1 2

),

...

(1 2 . . . n)n = I . (1.3.2)

One can show that any permutation in Sn of n objects can be decomposed in product ofcyclic permutations acting on different subsets of the original set:(

1 2 . . . n− 1 ni1 i2 . . . in−1 in

)= (j1 . . . jk1) . . . (s1 . . . sk`) . (1.3.3)

For instance the reader can prove that (1 2 3)2 = (3 2 1), (1 2 3 4)3 = (4 3 2 1) and in general(1 2 . . . n)n−1 = (nn− 1 . . . 2 1). Here are other examples of this decomposition:(

1 2 3 43 4 1 2

)= (1 2 3 4)2 = (1 3) (2 4) , (1.3.4)(

1 2 3 4 5 64 6 5 1 3 2

)= (1 4) (3 5) (2 6) , (1.3.5)(

1 2 3 4 53 1 2 4 5

)= (1 3 2) (4) (5) ≡ (1 3 2) , (1.3.6)

where, for the sake of simplicity, we have suppressed in the last equation the 1–cycles (4), (5),since it is understood that all indices which do not enter the (n > 1) -cycles in the decompo-sition are mapped into themselves. Since each factor in (1.3.3) is itself a permutation actingnon trivially on different sets of objects, the factors commute and moreover

∑`i=1 ki = n.

These cycles define the structure of the permutation. For instance the structure of (1.3.6)consists of one 3-cycle and 2 1-cycles. Suppose that in a generic permutation (1.3.3) the1-cycles occur ν1 times, the 2-cycles ν2 times and so on up to the n-cycles which occur νntimes. Since each cycle acts on different sets of objects, we must have:

ν1 + 2 ν2 + . . .+ n νn = n . (1.3.7)

We shall characterize the structure of a permutation by means of the nplet of numbers(ν1, ν2, . . . , νn). For instance the structures of the permutations (1.3.4), (1.3.5) and (1.3.6)are (0, 2, 0, 0), (0, 3, 0, 0) and (2, 0, 1, 0) respectively.

A 2–cycle is also called transposition. Clearly the identity permutation I is the productof 1–cycles:

I = (1) (2) . . . (n) . (1.3.8)


The powers of a n–cycle form a subgroup of Sn which is clearly a cyclic group called groupof cyclic permutations.

A n–cycle can be decomposed in products of transpositions:

(1 2 . . . n) = (1n) (1n− 1) . . . (1 3) (1 2) = (1 2) (2 3) . . . (n− 1n) , (1.3.9)

so that any permutation can be expressed as product of transpositions. A permutation iseven or odd if it is decomposed into an even or odd number of transpositions respectively.The reader can prove that the identity I is an even permutation and that the set of alleven permutations form a subgroup of Sn. There is an equal number (n!

2) of even and odd

permutations. As an example let us give the explicit composition of S3:

S3 =

I, (1 2 3), (3 2 1) even(1 2), (1 3), (2 3) odd

. (1.3.10)

We leave as an exercise to the reader to show that we can choose as generators of thepermutation group Sn the set of n− 1 adjacent transpositions: (1 2), (2 3), . . . , (n− 1n).

Property 1.2: The order of any subgroup H of G is a sub-multiple of the order n of G.To show this let m be the order of a subgroup H of G. If m = n or m = 1, H would

coincide with G or with e respectively and the property would trivially hold. Suppose1 < m < n. Then there is an element g1 ∈ G which is not in H. Consider the set H · g1,called left coset of G, consisting of all the elements of the form h · g1 with h in H. Thisset has no elements in common with H, since, if there were an element of the form h · g1

contained in H, also h−1 · (h · g1) = g1 would be in H, which contradicts our assumption. The number of elements of H · g1 equals the order of H, since different elements of Hcorrespond to different elements of H · g1. To show this suppose there existed two distinctelements h1, h2 ∈ H such that: h1 ·g1 = h2 ·g1. Multiplying both sides of the equation to theright by g−1

1 this equality would become h1 = h2 which is against our assumption. The setH ⊕H · g1 has then 2m elements. If it coincided with G the property would be proven sincethis would imply that n = 2m. Suppose H⊕H ·g1 be strictly contained in G. Then there isan element g2 ∈ G which is not contained in H ⊕H · g1. Consider now the left coset H · g2.This set is disjoint from H and from H · g1, as it follows from the fact that if there existedtwo elements h1, h2 ∈ H such that h1 · g1 = h2 · g2, then, multiplying by h−1

2 both sides tothe left we would obtain g2 = (h−1

2 · h1) · g1 which is an element of H · g1, contradicting ourhypothesis on g2. If the set H ⊕H · g1 ⊕H · g2 is still strictly contained in G we iterate ourconstruction until we decompose G in the following direct sum:

G = H ⊕H · g1 ⊕H · g2 ⊕ . . .⊕H · gk , (1.3.11)

where g1, . . . , gk are distinct elements of G which are not in H and (H · gi) ∩ (H · gj) = ∅.Since the direct sum on the right hand side of the above equation has dimension km, wehave proven that n = km. The number k is called index of H in G. If n is prime G canonly have a subgroup of order one, which is e. Moreover if n > 1 is prime, G can only be

1.4. CLASSES AND NORMAL SUBGROUPS 17

cyclic, since otherwise it would contain the cyclic group generated by one of its elements asa subgroup of order 1 < m < n which will be a sub-multiple of n.

Following the same procedure, we could have decomposed G into the following directsum:

G = H ⊕ g1 ·H ⊕ g2 ·H ⊕ . . .⊕ gk ·H , (1.3.12)

where giH is called right coset and is the set of elements of G of the form gi ·h, with h ∈ H.We denote by G/H and HG the collections of the k right and left cosets respectively:

G/H ≡ H, H · g1 , H · g2 , . . . , H · gk ,HG ≡ H, g1 ·H , g2 ·H , . . . , gk ·H . (1.3.13)

1.4 Classes and Normal Subgroups

We say that two elements g1, g2 of G are conjugate to one another in G, and write g1 ∼ g2

iff there exist a third element g ∈ G such that:

g1 = g · g2 · g−1 , (1.4.1)

g1 is also called conjugate of g2 by g and we say that it is obtained through the adjoint actionof g on g2. It is straightforward to prove that ∼ is an equivalence relation, namely that it isreflexive, symmetric and transitive. The elements of G arrange in equivalence classes withrespect to ∼ (conjugation classes). Clearly e forms a single class and, if G is abelian, eachelement forms a different class.

1.4.1 Conjugation classes in Sn and Young Diagrams.

Let us consider the effect of a conjugation of a permutation. Let S and T be the followingtwo permutations:

S =

(1 2 . . . n

S(1) S(2) . . . S(n)

), T =

(1 2 . . . n

T (1) T (2) . . . T (n)

), (1.4.2)

and let us compute T · S · T−1:(S(1) S(2) . . . S(n)

T (S(1)) T (S(2)) . . . T (S(n))

)·(

1 2 . . . nS(1) S(2) . . . S(n)

)·(T (1) T (2) . . . T (n)

1 2 . . . n

)=

=

(T (1) T (2) . . . T (n)

T (S(1)) T (S(2)) . . . T (S(n))

), (1.4.3)

namely the conjugate of S by means of T is the permutation obtained by acting on the firstand second rows of S by T . For example:

S =

(1 2 3 4 5 64 6 5 1 3 2

), T =

(1 2 3 4 5 62 1 3 5 4 6

)then:

S ′ = T · S · T−1 =

(2 1 3 5 4 65 6 4 2 3 1

). (1.4.4)


If we write S as the product of cycles acting on different subsets of objects, as in (1.3.3), itsconjugate by T is then obtained by replacing each element in the cycle by their transformedby T . For instance:

S = (1 2 6) (3 5) , T = (1 6 4) then:

S ′ = T · S · T−1 = (2 4 6) (3 5) , (1.4.5)

since T maps 1 → 6, 6 → 4, 4 → 1, 3 → 3, 5 → 5. In this example both S and S ′

have the same permutation structure, namely decompose in the same kind of cycles. Inthe notation introduced in the previous section their structure is (ν1, ν2, ν3, ν4, ν5, ν6) =(1, 1, 1, 0, 0 0). It is not difficult to convince ourselves that conjugation in general does notalter the structure of a permutation. Moreover two permutations with the same structure areclearly conjugate to one another. We conclude that the conjugation classes of Sn are in oneto one correspondence with the possible permutation structures, i.e. are labeled by the nplets

(ν1, ν2, . . . , νn). Consider for instance S3, whose content is displayed in (1.3.10). In thiscase the group consists in three classes: (3, 0, 0), (1, 1, 0), (0, 0, 1), the first consisting ofone element (i.e. the identity), the second by three elements and the third by two elements.Consider as a further example the group S4 whose class content is summarized below:

S4 :

class elements

(4, 0, 0, 0) I = (1) (2) (3) (4)

(2, 1, 0, 0) (1 2), (1 3), (1 4), (2 3), (2 4), (3 4)

(0, 2, 0, 0) (1 2) (3 4), (1 3) (2 4), (1 4) (2 3)

(1, 0, 1, 0) (1 2 3), (3 2 1), (1 2 4), (4 2 1), (1 3 4), (4 3 1), (2 3 4), (4 3 2)

(0, 0, 0, 1) (1 2 3 4), (1 2 4 3), (1 3 4 2), (1 4 3 2), (1 4 2 3), (1 3 2 4)

(1.4.6)

We have 12 even elements in the classes (4, 0, 0, 0), (0, 2, 0, 0), (1, 0, 1, 0) and 12 odd inthe remaining classes.

The conjugation classes of Sn are then defined by the solutions to equation (1.3.7).Let us determine the number of elements within a generic class (ν1, ν2, . . . , νn) of Sn. The

problem can be restated as that of determining the number of distinct ways n objects can bedistributed in a number of boxes, of which ν1 can contain just one element, ν2 two elements. . . νn n elements. Each class (ν1, ν2, . . . , νn) is therefore in one to one correspondence withthe distinct partitions of n objects and can be represented as follows:

ν1 1−cycles︷︸︸︷(•) . . . (•)

ν2 2−cycles︷︸︸︷(••) . . . (••) . . .

νn n−cycles︷︸︸︷(• . . . •) . . . (• . . . •) , (1.4.7)

the objects being represented by the • symbol. Given a class, the number of its elements canbe computed as the number of permutations which change the element within the class. Thetotal number of permutations we can perform on the n objects is n!. Out of these, permuta-tions whose effect is to exchange cycles of the same order k will leave the element invariant.There are ν1! ν2! . . . νn! such permutations. Moreover cyclic permutations within each cycle


will not change the cycle and thus the element. For instance (1 2 3) = (2 3 1) = (3 1 2). Foreach of the νk k–cycles there are k such permutations and thus there are 1ν1 2ν2 . . . nνn morepermutations which leave a generic element of the class invariant. The total number of per-mutations which do not effect a single element of a given is therefore 1ν1 ν1! 2ν2 ν2! . . . nνnνn!and thus the number of elements nc in the class is:

nc =n!

1ν1 ν1! 2ν2 ν2! . . . nνnνn!. (1.4.8)

Applying this formula to S4 we find:

n(4 0 0 0) =4!

14 4!= 1 ,

n(2 1 0 0) =4!

14 2! 21 1!= 6 ,

n(0 2 0 0) =4!

22 2!= 3 ,

n(1 0 1 0) =4!

11 1! 31 1!= 8 ,

n(0 0 0 1) =4!

41 1!= 6 ,

in accordance with table 1.4.6.We can alternatively describe each partition of n, i.e. each class in Sn, by a new set of

indices (λ1, . . . , λn), where the positive indices λi are related to νi as follows:

λ1 = ν1 + ν2 + . . .+ νn ,

λ2 = ν2 + . . .+ νn ,... ,

λn = νn . (1.4.9)

By definition we have: λ1 ≥ λ2 ≥ . . . ≥ λn and, in virtue of eq. (1.3.7),∑n

i=1 λi = n.Moreover eqs. (1.4.9) are readily inverted to yield: ν1 = λ1−λ2, . . . , νn−1 = λn−1−λn, νn =λn. Each class (λ1, . . . , λn) is represented graphically by a Young diagram which consistsof n boxes, distributed in a number of rows of which, moving downwards, the first containsλ1 boxes, the second λ2 boxes and so on: Let us apply this formalism to the S3 and S4 caseand write the Young diagrams corresponding to their classes:

S3 :

(ν1, ν2, ν3) (λ1, λ2, λ3) Young diagrams

(3, 0, 0) (3, 0, 0)

(1, 1, 0) (2, 1, 0)

(0, 0, 1) (1, 1, 1)

(1.4.10)


Figure 1.3: Young diagram

S4 :

(ν1, ν2, ν3, ν4) (λ1, λ2, λ3, λ4) Young diagrams

(4, 0, 0, 0) (4, 0, 0, 0)

(2, 1, 0, 0) (3, 1, 0, 0)

(0, 2, 0, 0) (2, 2, 0, 0)

(1, 0, 1, 0) (2, 1, 1, 0)

(0, 0, 0, 1) (1, 1, 1, 1)

(1.4.11)

1.4.2 Conjugation Classes in Dn

Let us use the defining relations (1.2.13) for Dn to construct its conjugation classes. Fromthe relation (r · σ)2 = e we find r · σ = σ · r−1, which implies, upon multiplication to the leftby suitable powers of r that:

r2 · σ = r · σ · r−1 , r3 · σ = r · (r · σ) · r−1 . . .

r2k · σ = r · (r2k−2 · σ) · r−1 , r2k+1 · σ = r · (r2k−1 · σ) · r−1 . (1.4.12)


From the above relations we conclude that all reflections of the form r2k ·σ, for some positiveinteger k, are in the same class as σ, while all reflections r2k+1 · σ, are in the same class asr ·σ. If n is odd, n = 2k+1 for some k, and thus σ = r2k+1 ·σ is in the same class as r ·σ andthus all reflections belong to the same class. If n is even, reflections group into two classes:

n odd :

[σ] = r` · σ , ` = 0, 1, 2, n− 1 ,

n even :

[σ] = σ, r2 · σ, . . . , rn−2 · σ ,[r · σ] = r · σ, r3 · σ, . . . , rn−1 · σ . (1.4.13)

As far as the rotations are concerned, using the same relation r · σ = σ · r−1 = σ · rn−1 wefind that r = σ · rn−1 · σ, r2 = σ · rn−2 · σ and so on, so that r is in the same class as rn−1

(recall that σ−1 = σ), r2 in the same class as rn−2 and so on. If n = 2k, we have k − 1classes [r`] = r`, rn−`, ` = 1, . . . , k − 1 and a single class [rk] = rk. If n = 2k + 1, wehave k classes [r`] = r`, rn−`, ` = 1, . . . , k. In all cases we have the class [e] consisting ofthe identity element only. Summarizing:

D2k+1 : k + 2 classes

[e] , [r`] = r`, rn−` (` = 1, . . . , k)

[σ] = σ, r · σ, . . . , rn−1 · σ ,D2k : k + 3 classes

[e] , [r`] = r`, rn−` (` = 1, . . . , k − 1) , [rk] ,

[σ] = σ, r2 · σ, . . . , rn−2 · σ ,[r · σ] = r · σ, r3 · σ, . . . , rn−1 · σ . (1.4.14)

1.4.3 Normal Subgroups

Let H be a subgroup of G and, for a given element g of G let us define the set g ·H · g−1 asthe set of all elements of the form g · h · g−1, with h ∈ H. This set is actually a group itself.To show this let us take any two of its elements g1 = g · h1 · g−1, g2 = g · h2 · g−1 and showthat g1 · g−1

2 is in g ·H · g−1:

g1 · g−12 = (g · h1 · g−1) · (g · h2 · g−1)−1 = (g · h1 · g−1) · (g · h−1

2 · g−1) =

= g · (h1 · h−12 ) · g−1 ∈ g ·H · g−1 , (1.4.15)

since h1 · h−12 is in H.

H is said to be a normal or invariant subgroup of G iff, for any g ∈ G, g ·H · g−1 = H,i.e. for any h ∈ H, g · h · g−1 ∈ H. This implies that for normal subgroups the left and rightcosets coincide: g · H = H · g. Clearly H = e is a normal subgroup of G, since, for anyg ∈ G, g · e · g−1 = g · g−1 = e ∈ H.


Property 1.3: If H is a normal subgroup of G, the set G/H of cosets H · gi close a groupcalled the factor group.

To show this let us prove that we can consistently define a product in G/H. Given twoelements H · gi, H · gj of G/H, define (H · gi) · (H · gj) as the set of all elements of the form(h1 · gi) · (h2 · gj), with h1, h2 ∈ H. We can write this generic element in the following form:h1 · (gi · h2 · g−1

i ) · gi · gj. Since H is normal, h3 = gi · h2 · g−1i is still in H, and thus a generic

element of (H · gi) · (H · gj) can be written as (h1 · h2) · (gi · gj) ∈ H · (gi · gj). Similarly ageneric element h · (gi · gj) of H · (gi · gj) can be written as (h · gi) · (e · gj) ∈ (H · gi) · (H · gj).We conclude that:

(H · gi) · (H · gj) = H · (gi · gj) . (1.4.16)

Clearly the identity element with respect to this coset-product is H = H ·e itself. Eq. (1.4.16)also implies that the inverse of a generic coset H · g is the coset H · g−1. Associativity of thisproduct is also implied by the same property of the product in G.

Remark: Notice that if H is not a normal subgroup of G, the element (h1 · gi) · (h2 · gj),for different h1, h2 ∈ H would not belong to a same coset and thus the product on G/Hwould not be properly defined.

One may define normal elements of G and elements g′ ∈ G such that, for any g ∈ G, wehave g · g′ · g−1 = g′. As a consequence a normal element of G commutes with all elements ofG. Indeed, since for any g ∈ G, g · g′ · g−1 = g′, multiplying both sides by g to the right wefind: g · g′ = g′ · g. An example of normal element is the identity element e. The set C ⊂ Gof all normal elements of G is called the center of G. It is a subgroup of G. Indeed considerg′, g′′ ∈ C and show that g′ · g′′ −1 ∈ C. Take any g ∈ G:

g · (g′ · g′′ −1) · g−1 = (g · g′ · g−1) · (g · g′′ −1 · g−1) = (g · g′ · g−1) · (g · g′′ · g−1)−1 = g′ · g′′ −1 ,

which implies that g′ · g′′ −1 ∈ C. Since any element of C commutes with all the elements ofG, it will also commute with all the elements of C itself, namely C is abelian. C is clearlya normal subgroup of G.

1.5 Homomorphisms, Isomorphisms and Automorphisms

Given two groups G, G′ a mapping ϕ from G to G′ is a homomorphism iff, given anytwo elements g1, g2 in G: ϕ(g1·, g2) = ϕ(g1) · ϕ(g2). In other words a homomorphism ϕ“preserves the product”. The group G is said to be homomorphic to G′. We shall denote ahomomorphism of G into G′ by G→ G′. here are come properties of homomorphisms:

1. A first property of homomorphisms is that ϕ(e) = e′, e′ being the identity elementof G′. Indeed take an element g′ ∈ G′ which is the image of some element g of G,g′ = ϕ(g), and consider the product g′ · ϕ(e) = ϕ(g) · ϕ(e) = ϕ(g · e) = ϕ(g) = g′ · e′.Multiplying both sides by g′ −1 to the left we find: ϕ(e) = e′.

2. As a second property, for any g ∈ G, ϕ(g)−1 = ϕ(g−1). Indeed ϕ(g) · ϕ(g−1) =ϕ(g · g−1) = ϕ(e) = e′.

1.5. HOMOMORPHISMS, ISOMORPHISMS AND AUTOMORPHISMS 23

3. If H is a subgroup of G, H ′ = ϕ(H) is a subgroup of G′. Take two elements h′1 =ϕ(h1), h′2 = ϕ(h2) of H ′, image of h1, h2 ∈ H. We have h′1 · h′ −1

2 = ϕ(h1) · ϕ(h−12 ) =

ϕ(h1 · h−12 ) ∈ H ′, since h1 · h−1

2 is an element of H. As a consequence of this, since Gis in particular subgroup of itself, it follows that ϕ(G) is a subgroup of G′.

4. If H is the largest subset of G whose image though ϕ is a subgroup H ′ = ϕ(H) of G′,then H is a subgroup of G. Take two elements h1, h2 of H and evaluate ϕ(h1 · h−1

2 ) =ϕ(h1) · ϕ(h2)−1. This element is in H ′, since ϕ(h1), ϕ(h2) are and H ′ is a subgroup ofG′. Now, by assumption, H consists of all the elements of G whose image through ϕis in H ′. We then conclude that h1 · h−1

2 ∈ H and thus that H is a subgroup of G.

5. If H is a normal subgroup of G then H ′ = ϕ(H) is a normal subgroup of ϕ(G). TakeH normal subgroup of G, and consider a generic element g′ = ϕ(g) of ϕ(G). We havethat g′ ·H ′ · g′ −1 = ϕ(g ·H · g−1) = ϕ(H) = H ′.

6. If H is the largest subset of G such that its image though ϕ coincides with a normalsubgroup H ′ = ϕ(H) of G′, then H is a normal subgroup of G. Let now H ′ be a normalsubgroup of G′ and consider the subgroup g ·H · g−1 of G, for a given g ∈ G. Its imageϕ(g ·H · g−1) = g′ ·H ′ · g′ −1 = H ′. By assumption we conclude that g ·H · g−1 = H,namely that H is normal.

7. The set E ⊂ G of all the elements of G which map through ϕ into e′ is a normalsubgroup of G. This follows form the previous property, applied to H ′ = e′ which isa normal subgroup of G′.

8. A homomorphism is one-to-one if and only if the only element of G which is mappedinto the identity element e′ of G′ is the identity e. Let us start assuming the ho-momorphism to be one-to-one, which means that ϕ(g1) = ϕ(g2), g1, g2 ∈ G impliesg1 = g2. Then suppose ϕ(g) = e′. Since also ϕ(e) = e′ = ϕ(g) it follows by hy-pothesis that g = e. Let us now assume that the identity e of G is the only el-ement which is mapped into the identity e′ of G′ and suppose that two elementsg1, g2 ∈ G are mapped into the same element ϕ(g1) = ϕ(g2) ∈ G′. This implies thatϕ(g1 · g−1

2 ) = ϕ(g1) · ϕ(g2)−1 = ϕ(g1) · ϕ(g1)−1 = e′. By assumption we must haveg1 · g−1

2 = e, that is g1 = g2.

We shall restrict from now on to onto homomorphisms: ϕ(G) = G′. If the onto homo-morphism is also one-to-one it is called an isomorphism. The concept of isomorphism isimportant since two isomorphic groups have the same order and the same structure. Fromthis point of view two isomorphic groups can be considered as a same one. For isomorphicgroups we shall use the short-hand notation G ∼ G′.

Property 1.4: If G is homomorphic to G′, the G/E is isomorphic to G′.Let us first define the correspondence between G/E and G′, by showing that all the

elements of a coset E · g ∈ G/E map into a single element g′ = ϕ(g) of G′. Indeed, for anye ∈ E, we have ϕ(e · g) = ϕ(e) ·ϕ(g) = e′ · g′ = g′. We can then define ϕ(E · g) as g′ = ϕ(g).


Let us now show that the correspondence is one-to-one, namely that if ϕ(E · g1) = ϕ(E · g2),we must have E ·g1 = E ·g2. Indeed let ϕ(E ·g1) = ϕ(g1) = g′1 = g′ = g′2 = ϕ(g2) = ϕ(E ·g2).Then g1 and g2 are two elements of G which map into g′. This implies that g1 · g−1

2 is inE since ϕ(g1 · g−1

2 ) = ϕ(g1) · ϕ(g2)−1 = g′ · g′ −1 = e′. As a consequence of this we haveE · g1 = E · (g1 · g−1

2 ) · g2 = E · g2. We have thus shown that the mapping ϕ : G/E → G′

is one-to-one. We need now to show that it preserves the product: Let g′1 = ϕ(E · g1) andg′2 = ϕ(E ·g2). The product (E ·g1) ·(E ·g2) is the coset E ·(g1 ·g2) and its image through ϕ is:ϕ(E · (g1 ·g2)) = ϕ(g1 ·g2) = g′1 ·g′2. This proves that ϕ : G/E → G′ is also a homomorphism.

Property 1.5: Let H be a normal subgroup of G, then G is homomorphic to the factorgroup G/H.

Let us define ϕ : G→ G/H as the mapping which associates with each element of G thecoset it belongs to:

ϕ : g ∈ G → ϕ(g) = H · g ∈ G/H . (1.5.1)

Clearly this mapping is not one to one. Let us show it is a homomorphism:

ϕ(g1 · g2) = H · (g1 · g2) = H ·H · (g1 · g2) = (H · g1) · (g−11 ·H · g1) · g2 =

= (H · g1) · (H · g2) = ϕ(g1) · ϕ(g2) , (1.5.2)

where we have used the property that H be normal. Finally ϕ is onto since any coset inG/H can be written as H · g and thus is the image of g through ϕ.

Therefore if we have a homomorphism G→ G′, since G/E ∼ G′ we can reduce it to thehomomorphism G→ G/E.

A homomorphism of G onto itself is called an endomorphism while an isomorphism of Gonto itself is called an automorphism. A meromorphism ϕ is a one-to-one endomorphism ofG which is not necessarily onto. In particular a proper meromorphism is a meromorphismwhich is not an automorphism, namely such that ϕ(G) ( G. If G admits a proper meromor-phism it is necessarily infinite. Indeed is G were finite, being a meromorphism a one-to-onecorrespondence, ϕ(G) would have the same order as G and thus would coincide with G.

Exercise 1.4: Consider the infinite set G of all the integer powers of a positive numbera: G = ak, k ∈ Z. Show that it is a group and show that the mapping ϕ : G → G suchthat ϕ(ak) = a2 k is a proper meromorphism.

Example : Let us show that the groups S3 and D3 are isomorphic. Consider the followingmapping:

ϕ : D3 → S3 ,

ϕ(e) = I , ϕ(r) = (1 2 3) , ϕ(r2) = (3 2 1) , ϕ(σ) = (1 2) ,

ϕ(r · σ) = (1 3) , ϕ(r2 · σ) = (2 3) .(1.5.3)

The reader can easily show that ϕ is a homomorphism. Being one-to-one and onto, it is anisomorphism.

1.5. HOMOMORPHISMS, ISOMORPHISMS AND AUTOMORPHISMS 25

1.5.1 The Automorphism Group

Consider two automorphisms α1, α2 of a group G. They are in particular functions of Ginto itself and thus we can consider their composition α2 α1 as the mapping of G into itselfsuch that associates with any g ∈ G the element α2 α1(g) = α2(α1(g)). Let us show thatα2 α1 is still an automorphism. It is a homomorphism since for any g1, g2 ∈ G:

α2 α1(g1 · g2) = α2(α1(g1 · g2)) = α2(α1(g1) · α1(g2)) = α2(α1(g1)) · α2(α1(g2)) =

= α2 α1(g1) · α2 α1(g2) . (1.5.4)

Now let us prove that α2 α1 is onto. Consider g′′ ∈ G. Being α1, α2 onto, there exist g′ ∈ Gsuch that α2(g′) = g′′ and g ∈ G such that α1(g) = g′. This implies that there exist g ∈ Gsuch that g′′ = α2(g′) = α2(α1(g)) = α2 α1(g) and thus α2 α1 is onto. Let us prove nowthat it is one-to one. Consider g1, g2 ∈ G such that α2 α1(g1) = α2 α1(g2). This meansthat α2(α1(g1)) = α2(α1(g2)). Being α2 one-to-one we have that α1(g1) = α1(g2), and, beingα1 one-to-one, it follows that g1 = g2.

We can consider now the set of all automorphisms of G endowed with the product .Let us show that it is a group, called the automorphism group AG of G. To this end wedefine the identity automorphism 1 as the trivial function which maps each element of Ginto itself: 1(g) = g, for any g ∈ G. From the definition we can convince ourselves that,given any automorphism α: α 1 = 1 α = α. The function 1 is trivially an automorphism.Since an automorphism α is one-to-one and onto, it can be inverted. Its inverse α−1 is thensuch that α−1 will map g ∈ G into g′ ∈ G iff α(g′) = g. The composition of α with itsinverse function is clearly the identity automorphism:

α α−1 = α−1 α = 1 . (1.5.5)

Let us show that α−1 is still an automorphism. Let g′1 = α(g1) and g′2 = α(g2) be twoelements of G. By definition, α−1(g′1) = g1 and α−1(g′2) = g2. Then α−1(g′1 · g′2) is defined asthe only element of G which is mapped through α into g′1 · g′2. This element is g1 · g2, and soα−1(g′1 · g′2) = g1 · g2 = α−1(g′1) ·α−1(g′2). Since α−1 is, by construction, one-to-one and onto,it is an isomorphism.

We still need to prove that the product is associative. Given a generic element g ∈ Gwe find:

α3 (α2 α1)(g) = α3(α2 α1(g)) = α3(α2(α1(g))) = α3 α2(α1(g)) =

= (α3 α2) α1(g) . (1.5.6)

Let us now consider, for a given h ∈ G the correspondence αh:

αh : g ∈ G → h · g · h−1 ∈ G . (1.5.7)

Clearly αe = 1. Let us show that αh is an automorphism. It is a homomorphism since,given two elements g1, g2 ∈ G, we have αh(g1 · g2) = h · (g1 · g2) · h−1 = (h · g1 · h−1) · (h ·g2 · h−1) = αh(g1) · αh(g2). αh is one to one: Indeed suppose αh(g1) = αh(g2), which means


h · g1 · h−1 = h · g2 · h−1. Multiplying both sides by h−1 to the left and by h to the right wefind g1 = g2. Finally αh is onto: For any g′ ∈ G there exist g = h−1 · g′ · h ∈ G such thatg′ = αh(g).

αh is called inner automorphism of G. The set of all the αh, for all h ∈ G, is a group.To show this consider αh1 and αh2 , with h1, h2 ∈ G and let us show that

αh2 αh1 = αh2·h1 , (1.5.8)

by computing this product on a generic g ∈ G:

αh2 αh1(g) = αh2(αh1(g)) = h2 · (h1 · g · h−11 ) · h−1

2 = (h2 · h1) · g · (h2 · h1)−1 = αh2·h1(g) .

From this it follows that α−1h = αh−1 since αh αh−1 = αh−1·h = αe = 1. Given any two inner

automorphisms αh1 and αh2 the product of the first times the inverse of the second is still aninner automorphism. Indeed αh1 α−1

h2= αh1 αh−1

2= αh−1

2 ·h1. We have then proved that the

inner automorphisms close a subgroup IG of AG called the group of inner automorphisms.

Property 1.6: IG is a normal subgroup of AG.Let β ∈ Ag be an automorphism of G and αh ∈ IG. Let us show that β αh β−1 ∈ IG.

To this end we compute this function on a generic element g ∈ G:

β αh β−1(g) = β αh(β−1(g)) = β(αh(β−1(g))) = β(h · β−1(g) · h−1) =

= β(h) · g · β(h)−1 = αβ(h)(g) . (1.5.9)

This shows that β αh β−1 = αβ(h) ∈ IG.We can define a correspondence of G into its group of inner automorphisms IG which

maps a generic element g ∈ G into αg. In virtue of equation (1.5.8) this is a homomorphism:G→ IG. It is not however one-to-one. Indeed for any h in the center C of G, αh = 1:

αh(g) = h · g · h−1 = g = 1(g) , (1.5.10)

where we have used the property that any element of C commutes with any other elementof G. In fact C contains all the elements of G which are mapped into the identity of IG. Invirtue of Property 1.4 we have that G/C ∼ IG.

Example 1.1: Consider an infinite cyclic group G = ak consisting of the integer powersof a generator a. If σ is an endomorphisms of G, it will map a into an other element of thegroup a`, for some integer ` (we shall suppose ` 6= 0). Being σ a homomorphism, it will mapak into σ(a)k = a` k. Therefore the most general endomorphism of a infinite cyclic grouphas the following action σ(ak) = a` k, for some integer `. This endomorphism is one-to-one,namely it is a meromorphism. Clearly σ(G) is a proper subgroup of G if ` 6= ±1, as thereader can show. The only automorphisms then correspond to the cases in which ` = ±1.AG is an order 2 group. Since G is abelian, it coincides with its center: G = C. As aconsequence IG ∼ G/C is an order one group consisting of the identity 1 alone.

1.6. PRODUCT OF GROUPS 27

1.6 Product of Groups

Given two groups G1 and G2 we define their product G1×G2 as a group consisting of couplesof elements (g1, g2), g1 ∈ G1 and g2 ∈ G2, on which the following product is defined:

(g1, g2), (g′1, g′2) ∈ G1 ×G2 , (g1, g2) · (g′1, g′2) ≡ (g1 · g′1, g2 · g′2) ∈ G1 ×G2 . (1.6.1)

From the above definition, if we denote by e1, e2 are the unit elements of G1 and G2 respec-tively, it follows that:

• The identity element of G1 × G2 is (e1, e2) since, for any (g1, g2) ∈ G1 × G2: (g1, g2) ·(e1, e2) = (g1 · e1, g2 · e2) = (g1, g2) = (e1, e2) · (g1, g2);

• For any element (g1, g2) ∈ G1 × G2, its inverse (g1, g2)−1 is (g−11 , g−1

2 ): (g1, g2) ·(g−1

1 , g−12 ) = (g1 · g−1

1 , g2 · g−12 ) = (e1, e2) = (g−1

1 , g−12 ) · (g1, g2);

• The associative property of the product in G1 ×G2 follows from the same property ofthe product in the two factors.

Consider the set of elements (g1, e2), g1 ∈ G1, of G1×G2. This subset is a subgroup, as thereader can easily verify, which corresponds to the product G1 × e2. The correspondence:

g1 ∈ G1 −→ (g1, e2) ∈ G1 × e2 , (1.6.2)

is an isomorphism: G1 ∼ G1 × e2 ⊂ G1 × G2. Similarly we can define the subgroupe1×G2 of G1×G1 and show that G2 ∼ e1×G2 ⊂ G1×G2, that is G2 is isomorphic toe1×G2. The two subgroups G1×e2 and e1×G2 commute, namely a generic elementof the former commutes with any element of the latter:

(g1, e2) ∈ G1 × e2 , (e1, g2) ∈ e1 ×G2 :

(g1, e2) · (e1, g2) = (g1 · e2, g2 · e2) = (e1 · g2, e2 · g2) = (e1, g2) · (g1, e2) .

1.7 Matrix Groups

We end this section by defining the main continuous matrix groups that we shall consider inthis course. They represent the main instances of Lie groups. Let F denote either the real orthe complex numbers. A matrix groups over F is a set of square matrices with entries in F,endowed with the ordinary rows-times-columns product, and satisfying certain properties.

1.7.1 The Group GL(n,F)

We have already defined GL(n,F) as the group of all n×n non-singular matrices with entriesin F, that is GL(n,R) and GL(n,C) are the groups of all n × n real and complex matricesrespectively. Since real numbers are in particular complex numbers, namely R ⊂ C, realmatrices are particular instances of complex matrices and therefore GL(n,R) is a subgroup


of GL(n,C). Its elements are defined by restricting to the complex matrices in GL(n,C)with real entries. GL(n,R) is said to be a real form of GL(n,C).

As we shall see in next chapter, this group described a generic change of basis in alinear vector space, or a generic linear homogeneous transformation on a space of points. Ageneric element GL(n,F) depends on its entries which are n2 continuous parameters in F:M = M[aij] = (aij). If F = R, the parameters will be n2 real, if F = C, there will be n2

complex parameters, that is 2n2 real.

1.7.2 The Group SL(n,F)

SL(n,F) is the set of n× n matrices with entries in F with unit determinant:

SL(n,F) = S = (Sij) : det(S) = 1 . (1.7.1)

Clearly SL(n,F) ⊂ GL(n,F). Let us show that SL(n,F) is actually a subgroup of GL(n,F).Consider S = (Sij) and T = (Tij) in SL(n,F), let us prove that S · T−1 is still in SL(n,F).

By assumption det(S) = det(T) = 1. Therefore det(S · T−1) = det(S)det(T)

= 1. This proves

that S · T−1 is an element of SL(n,F). The group SL(n,F) describes, as we shall illustratein next chapter, special linear transformations, which are volume preserving homogeneouslinear transformations on an n-dimensional space over F. As for the GL groups, SL(n,R)is a real form of SL(n,C). A generic element of SL(n,F) depends on its entries which aren2 continuous parameters in F, subject to the condition det(S) = 1. In the SL(n,C) case,det(S) = 1 is one condition on a complex number which implies two real conditions onthe 2n2 real parameters, leaving a generic element of the group to depend on 2n2 − 2 freeparameters. As for SL(n,R), det(S) = 1 is one real condition on n2 real parameters andthus the number of real independent parameters defining a generic element of the group isn2 − 1.

1.7.3 The Group O(n,F)

The group O(n,F) is defined as the group of n × n orthogonal matrices with entries in F,namely of matrices S = (Sij) satisfying the property:

ST · S = 1n , (1.7.2)

or, in components

n∑k=1

Ski Skj = δij , (1.7.3)

where ST denotes the transposed matrix of S, δij is 1 for i = j and 0 for i 6= j, 1n = (δij)is the n× n identity matrix. Multiplying both sides of equation (1.7.2) by S−1 to the right,we find that the inverse of any orthogonal matrix coincides with its transposed: S−1 = ST .Moreover one also finds, using the property that S · S−1 = S−1 · S = 1n, that S · ST = 1n.

1.7. MATRIX GROUPS 29

Computing the determinant of both sides of eq. (1.7.2) we deduce that det(S)2 = 1, thatis det(S) = ±1. To show that O(n,F) is a group, let us show it is a subgroup of GL(n,F).Consider two elements of S and T of O(n,F), let us show that S ·T−1 = S ·TT satisfies eq.(1.7.2):

(S ·T−1)T · (S ·T−1) = (S ·TT )T (S ·TT ) = (T · ST ) · (S ·TT ) = T · (ST · S) ·TT =

= T ·TT = 1n . (1.7.4)

We define the group of special orthogonal matrices SO(n,F) as the subset of orthogonalmatrices, with entries in F, which have unit determinant: det(S) = 1. Clearly SO(n,F) isa subgroup of SL(n,F). More precisely: SO(n,F) = O(n,F)

⋂SL(n,F). For the sake of

simplicity we shall denote simply by (S)O(n) the group (S)O(n,R).Exercise 1.5: Prove that SO(n,F) is a subgroup of O(n,F), while the subset of orthog-

onal matrices with determinant −1 is not.Exercise 1.6: Prove that SO(n,F) is a normal subgroup of O(n,F).As we shall see in next chapter, special orthogonal transformations describe rotations

in a n-dimensional space of points (e.g. the Euclidean space) over F, while orthogonaltransformations with determinant −1 describe reflections in the same space.

Exercise 1.7:

• Prove that the matrix

S[α] =

(cos(α) sin(α)− sin(α) cos(α)

), (1.7.5)

is in SO(2). In fact S[α] represents a generic element of this group and, as we shall seeit describes a rotation about a point by an angle α in the Euclidean plane. The readercan show that the structure of this group is described by the relation S[α1] S[α2] =S[α1 +α2], which represents the simple fact that the result of two consecutive rotationson the plane about a point is itself a rotation about the same point by an angle whichis the sum of the angles defining the two rotations.

• Show that

R =

(1 00 −1

), (1.7.6)

is in O(2) but not in SO(2). As we shall see, this matrix represents a reflection in theY axis in the Euclidean plane: x→ −x, y → y.

• Finally prove that the matrix

S =

0 0 10 1 01 0 0

, (1.7.7)

is in O(3) but not in SO(3).


Exercise 1.8: Prove that the matrix

S =

1√3

1√3

1√3

1√2− 1√

20

1√6

1√6− 2√

6

, (1.7.8)

is in SO(3).Consider the defining condition (1.7.3) for the real group O(n,R) and take the equations

corresponding to j = i:

n∑k=1

(Ski)2 = 1 . (1.7.9)

Since Sij are real numbers, the left hand side is a sum of positive numbers which has to beequal to one. The above equation implies then that each term in the sum be less or equalthan one, i.e. |Sij| ≤ 1. We shall call bounded a matrix whose entries, as functions of freeparameters, are all, in modulus, bounded from above. We conclude that any element ofO(n,R) is a bounded matrix. A group of matrices is called compact if all its elements arebounded. For this reason O(n,R) is a compact group.

Using equation (1.7.3), we can count the number of free continuous parameters a genericO(n,F) matrix depends on, i.e. the dimension of the group. This number would be given bythe n2 parameters in F of a generic matrix in GL(n,F), minus the number of independentconditions in eq. (1.7.3). We see that conditions (1.7.3) are symmetric in i, j: If we inter-

change the values of i and j the condition is the same. This leaves us with n(n+1)2

conditionson numbers in F. The number of free parameters will then be:

n2 − n(n+ 1)

2=

n(n− 1)

2parameters in F, (1.7.10)

O(n,R) will then be described by n(n−1)2

real parameters, O(n,C) by n(n−1) real parameters.Since the orthogonality condition (1.7.3) already implies that det(S) = ±1, the conditionof unit determinant, defining the SO(n,F) subgroup, is just a restriction on the sign of thedeterminant which does not lower the number of independent parameters.

Relation between O(n) and SO(n) Since SO(n) is a normal subgroup of O(n) we canconstruct the factor group O(n)/SO(n). This is done noticing that any two orthogonalmatrices O1, O2 with determinant −1 are connected by the left (or right) action of a SO(n)element S: O2 = S O1. Indeed it suffices to take S = O2 O−1

1 . Then take any orthogonalmatrix O with determinant −1. We will take for instance O = diag(−1,+1,+1, . . . ,+1), sothat O2 = 1n. We can decompose O(n) in the following cosets:

O(n) = SO(n) + SO(n) O . (1.7.11)

O(n)/SO(n) is an order 2 group, isomorphic to 1n, O, which is a cyclic group of period 2,also denoted by Z2.


1.7.4 The Group O(p, q;F)

A generalization of the concept of orthogonal group is the group O(p, q;F), p + q = n,consisting of pseudo-orthogonal n × n matrices with entries in F. A pseudo-orthogonalmatrix S = (Sij) is defined by the following property:

ST ηp,q S = ηp,q , (1.7.12)

or, in components

p∑k=1

SkiSkj −n∑

k=p+1

SkiSkj = (ηp,q)ij , (1.7.13)

where ηp,q ia s diagonal matrix with p (+1) and q (−1) diagonal entries:

ηp,q = diag(

p︷︸︸︷+1, . . . ,+1,

q︷︸︸︷−1, . . . ,−1) . (1.7.14)

We leave to the reader to show that this condition defines a group, subgroup of GL(n,F).In particular we can easily derive the following relations:

S−1 = ηp,q ST ηp,q , S ηp,q ST = ηp,q , det(S) = ±1 . (1.7.15)

If p = n and q = 0 or p = 0 and q = n, ηp,q becomes proportional to 1n and thereforeO(p, q;F) = O(n,F). Consider the real matrix I which we write in blocks as follows:

I =

(0q,p 1q1p 0p,q

), (1.7.16)

where 0p,q is the p×q matrix made of zeros. The matrix I is orthogonal: I IT = 1n. MoreoverI has the property that I ηp,q IT = −ηq,p. Therefore if S is in O(p, q;F), S′ = I S IT is inO(q, p;F), as the reader can easily verify. In other words we can write:

O(q, p;F) = I O(p, q;F) IT , (1.7.17)

namely the two groups are isomorphic, they have the same structure.Now consider the pseudo-orthogonal group over the complex numbers: O(p, q;C) and

define the following complex matrix:

Cp,q = diag(

p︷︸︸︷+1, . . . ,+1,

q︷︸︸︷i, . . . , i) . (1.7.18)

Clearly C−1p,q is the matrix C∗p,q whose entries are the complex conjugate of the corresponding

entries of Cp,q. The reader can also verify that Cp,q Cp,q = C−1p,q C−1

p,q = ηp,q. If S is an orthogonalcomplex matrix, S′ = Cp,q S C−1

p,q is in O(p, q;C). Indeed in the equation ST S = 1n we canexpress S in terms of S′:

1n = (Cp,q S′T C−1p,q ) (C−1

p,q S′ Cp,q) = Cp,q S′T ηp,q S′ Cp,q , (1.7.19)


where we have used the symmetry property of Cp,q: CTp,q = Cp,q. Multiplying both sides ofthe above equation, to the left and to the right by C−1

p,q , we find

ηp,q = S′T ηp,q S′ , (1.7.20)

namely that S′ ∈ O(p, q;C). We conclude that the orthogonal group over the complexnumbers and the pseudo-orthogonal one, for any partition p, q of n, are isomorphic throughthe adjoint action of the complex matrix Cp,q:

O(n;C) = C−1p,q O(q, p;C) Cp,q . (1.7.21)

Notice that we cannot say that O(n;R) and O(q, p;R) are isomorphic, since the adjointaction of the matrix Cp,q would transform a real matrix into a complex one. To summarizewe can write the following isomorphisms:

O(q, p;F) ∼ O(p, q;F) ,

O(n;C) ∼ O(p, q;C) , (1.7.22)

Just as we did for the orthogonal group, we can define the group of special pseudo-orthogonaltransformations SO(q, p;F) as the subgroup of O(q, p;F) consisting of all the matrices withunit determinant.

In what follows we shall use the short-hand notation for the groups over the real numbers:O(p, q) ≡ O(p, q;R), SO(p, q) ≡ SO(p, q;R).

An example of pseudo-orthogonal group over the real numbers is the Lorentz group O(1, 3)describing the most general transformations between two inertial reference frames.


T[β] =

(cosh(β) sinh(β)sinh(β) cosh(β)

), (1.7.23)

is in SO(1, 1). In fact T[β] represents a generic element of this group.Consider the defining condition (1.7.13) for the real group O(p, q;R) and take the equa-

tions corresponding to j = i:

p∑k=1

(Ski)2 −

n∑k=p+1

(Ski)2 =

+1 i = 1, . . . , p−1 i = p+ 1, . . . , n

. (1.7.24)

Being the left hand side a sum of terms with indefinite sign, the above condition no longerimplies that each term be bounded from above. In fact the entries of a O(p, q) matrix, forp or q different from 0, are in general, in absolute value, not bounded from above, see theexample in eq. (1.7.23). The group O(p, q), for pq 6= 0, is then non-compact, as it containsunbounded matrices.

The counting of free parameters for the pseudo-orthogonal group is the same as for theorthogonal one.


1.7.5 The Unitary Group U(n)

The unitary group U(n) is the subgroup of GL(n,C) consisting of n × n unitary matrices.A unitary matrix U is defined by the property:

U†U = 1n , (1.7.25)

or, in components,

n∑k=1

U∗ki Ukj = δij , (1.7.26)

where the hermitian -conjugate U† of U is defined as the complex conjugate of the transposedof U: U† ≡ (UT )∗. The reader can easily verify that U−1 = U† and that therefore U U† =1n. Moreover, computing the determinant of both sides of (1.7.25) and using the propertythat det(U†) = det(U)∗, we find

|det(U)|2 = det(U)∗ det(U) = 1 , (1.7.27)

that is the determinant of a unitary matrix is just a phase: det(U) = ei ϕ. Finally, just aswe did for the orthogonal and pseudo-orthogonal groups, it is straightforward to check thatU(n) is indeed a subgroup of GL(n,C).

The group U(1) consists of complex numbers c (i.e. 1× 1 matrices) with unit modulus,i.e. c∗ c = 1. In other words U(1) consists of phases c = ei ϕ.

Define now the group SU(n) of special unitary matrices, namely of unitary matrices withunit determinant. The reader can show that it is a normal subgroup of U(n). Let now U bea unitary matrix and let the phase ei ϕ be its determinant. We can write U as follows:

U = einϕ S , (1.7.28)

where S ∈ SU(n) and einϕ ∈ U(1).

Remark: Equation (1.7.28) shows that we can always write a unitary matrix as the

product of a U(1) element times a SU(n) element. Similarly, given a phase einϕ in U(1) and

a SU(n) matrix S we can define a unique U(n) matrix by equation (1.7.28). In other wordswe can define a correspondence:

(einϕ, S) ∈ U(1)× SU(n) −→ e

inϕ S ∈ U(n) . (1.7.29)

This correspondence is a homomorphism, however it is not an isomorphism since it is n-to-one, as the reader can verify by showing that all the elements of the form (e

in

(ϕ+2π k), e−2π i kn S),

k = 1, . . . , n, are mapped into the same element of U(n). One can say that U(1)× SU(n) iscontained n-times inside U(n).

Unitary groups are important in quantum mechanics, since they describe transformationson quantum states and operators representing physical observables.



S =1√2

(1n i1n1n −i1n

), (1.7.30)

is in U(2n) and determine its U(1) and SU(2n) components. The above matrix is also calledCayley matrix.

Exercise 1.11: Prove that the matrix:

U[σ, α, β, γ] =

(cos(σ) ei(γ+α) sin(σ) ei(γ+β)

− sin(σ) ei(γ−β) cos(σ) ei(γ−α)

), (1.7.31)

is in U(2) and determine its U(1) and SU(2) components. In fact one can show that themost general element of U(2) can be written in the form (1.7.31) and thus that it depends onfour real parameters. Equation (1.7.31) then defines a parametrization of the generic U(2)element, the parameters being σ, α, β, γ.

Consider now the defining condition (1.7.26) for i = j:

n∑k=1

|Uki|2 = 1 . (1.7.32)

The left hand side is the sum of positive terms which should be equal to one, implying thateach term is bounded from above by 1: |Uik| ≤ 1. We conclude that the modulus of eachentry of a unitary matrix is bounded from above and thus the corresponding group is saidto be compact.

Let us count the dimension of U(n). We start from the 2n2 real parameters of a genericn × n complex matrix and subtract the number of independent real conditions implied by(1.7.26). For each couple (i, j) with i 6= j we have a complex condition, i.e. 2 real conditions.Since the couples (i, j) and (j, i) yield the same real conditions (one expresses the vanishingof a complex number, the other the vanishing of its complex conjugate) we have in total2n(n − 1)/2 = n(n − 1) real conditions coming from the i 6= j entries of eq. (1.7.26). Thediagonal entries i = j are real and yield n real conditions. In total we have n(n−1)+n = n2

real conditions. The number of free real parameters a U(n) depends on is:

2n2 − n2 = n2 . (1.7.33)

The determinant of a unitary matrix U is a phase eiϕ, depending on a continuous realparameter ϕ. Setting it to 1 suppresses then one real parameter, leaving, for a genericelement of SU(n), n2 − 1 real independent parameters.

1.7.6 The Pseudo-Unitary Group U(p, q)

The notion of unitary group is extended to that of pseudo-unitary group U(p, q), p+ q = n,consisting of complex n× n matrices satisfying the pseudo-unitary requirement:

U† ηp,q U = ηp,q . (1.7.34)


If p = n or q = n the group reduces to U(n). The reader can easily verify that:

U−1 = ηp,q U† ηp,q , U ηp,q U† = ηp,q , det(U) = ei ϕ . (1.7.35)

Moreover, using the matrix I, we can prove the isomorphism U(p, q) ∼ U(q, p), just as wedid for the pseudo-orthogonal group.

Finally we define the subgroup

SU(p, q) = S ∈ U(p, q)|det(S) = 1 . (1.7.36)

Just as we did for the unitary case, one can write the isomorphism: U(p, q) ∼ U(1)×SU(p, q).The counting of free parameters for U(p, q) and SU(p, q) goes in the same way as for the

unitary group, yielding n2 real parameters for the former and n2 − 1.

Remark: Real (pseudo-) unitary matrices are (pseudo-) orthogonal. Indeed if M is areal matrix M† = MT and conditions (1.7.34) and (1.7.25) become (1.7.12) and (1.7.2)respectively. Therefore O(p, q) is a subgroup of U(p, q), that is real (pseudo-) orthogonalmatrices are in particular (pseudo-) unitary.

1.7.7 The Symplectic Group Sp(2n,F)

The symplectic group Sp(2n,F) consists of 2n× 2n matrices with entries in F satisfying thecondition:

ST Ω S = Ω , (1.7.37)

where

Ω =

(0n 1n−1n 0n

), (1.7.38)

0n being the n× n zero-matrix. In components equation (1.7.37) reads:

n∑k,`=1

Ski Ωk` S`j = Ωij . (1.7.39)

A matrix satisfying condition (1.7.37) is said symplectic. The matrix Ω has the followingproperties: ΩT = −Ω, Ω2 = −12n.

From (1.7.37) we immediately find: S−1 = −Ω ST Ω and S Ω ST = Ω, namely ST issymplectic as well. We leave to the reader to show that condition (1.7.37) defines a subgroupof GL(2n,F).

If we write a symplectic matrix S in terms of four n× n blocks:

S =

(A BC D

), (1.7.40)


equation (1.7.37) implies the following conditions:

AT C = CT A , AT D−CT B = 1n , BT D = DT B . (1.7.41)

Symplectic matrices are relevant in the Hamiltonian description of a system with n degrees offreedom described by the generalized coordinates (qi) = (q1, . . . , qn). In fact they representthe effect of canonical transformations on the vector (dqi, dpi), i = 1, . . . , n (such symplecticmatrix representing a canonical transformation in general is not constant, but its entries willdepend on the phase space coordinates).

Since (Ωij) is an antisymmetric matrix, the diagonal entries i = j of equation (1.7.39) areidentically zero and thus imply no condition. The only independent conditions come fromthe (i, j), i 6= j, entries. The equations from the (i, j) and the (j, i) entries are the same,leaving n(n− 2)/2 independent conditions on numbers in F. The number of free parametersis therefore:

(2n)2 − 2n(2n− 1)

2=

2n(2n+ 1)

2= n (2n+ 1) parameters in F , (1.7.42)

which means that Sp(2n,R) and Sp(2n,C) have n(2n+1) and 2n(2n+1) free real parametersrespectively.

Exercise 1.12: Show that the matrix

S = diag(

n︷︸︸︷ex, . . . , ex,

n︷︸︸︷e−x, . . . , e−x) , (1.7.43)

with x real, is in Sp(2n,R). Notice that its diagonal entries, as x varies in R, are not boundedfrom above. Since the group Sp(2n,R) has an unbounded element, it is non-compact.

1.7.8 The Unitary-Symplectic Group USp(2p, 2q)

The unitary-symplectic group USp(2p, 2q), with p+ q = n is defined as the set of symplectic2n× 2n complex matrices U = (Uij) which satisfy the pseudo-unitarity conditions:

UT Ω U = Ω ,

U† η2p,2q U = η2p,2q . (1.7.44)

where now the ± signs in the diagonal of η2p,2q are arranged as follows:

η2p,2q =

1p 0p,q 0p 0p,q0q,p −1q 0q,p 0q0p 0p,q 1p 0p,q0q,p 0q 0q,p −1q

. (1.7.45)

The group USp(2p, 2q) can then also be written as the following intersection:

USp(2p, 2q) = U(2p, 2q)⋂

Sp(2n; C) . (1.7.46)

If either p or q are zero, the group USp(2n), being contained inside U(n) is compact.


Summary Let us summarize below the dimensions and other features of the matrix groupsdiscussed above (p and q in the table are both considered non-vanishing):

Group Real dimension Compact (c.)/non-compact (n.c.)GL(n,C) 2n2 n.c.GL(n,R) n2 n.c.SL(n,C) 2n2 − 2 n.c.SL(n,R) n2 − 1 n.c.

(S)O(n,C) n(n− 1) n.c.(S)O(n) 1

2n(n− 1) c.

(S)O(p, q; C) n(n− 1) n.c.(S)O(p, q) 1

2n(n− 1) n.c.

U(n) n2 c.U(p, q) n2 n.c.SU(n) n2 − 1 c

SU(p, q) n2 − 1 n.c.Sp(2n,C) 2n(2n+ 1) n.c.Sp(2n,R) n(2n+ 1) n.c.USp(2n) n(2n+ 1) c.

USp(2p, 2q) n(2n+ 1) n.c.

Chapter 2

Transformations and Representations

2.1 Linear Vector Space

We define a linear vector space V over the real numbers R (or complex numbers C) acollection of objects, to be called vectors and denoted by boldface symbols, on which twooperations are defined: a summation + between vectors and a product · of a number timesa vector, such that:

1. (V, +) is an abelian group:

• For any V1 and V2, their sum is still a vector: V1 + V2 ∈ V ; (closure)

• There exist the null vector 0 such that, for any vector V: 0 + V = V + 0 = V;(null vector)

• For any vector V, there exist the opposite vector −V such that, V + (−V) =−V+V = 0; The sum W+(−V) will be simply denoted by W−V; (oppositevector)

• For any three vectors V1, V2, V3, we have: V1 + (V2 + V3) = (V1 + V2) + V3;(associativity)

• For any two vectors V1, V2, we have: V1 + V2 = V2 + V1; (commutativity)

2. The following additional axioms are satisfied:

i For any a in R (or C) and V ∈ V : a ·V ∈ V ;

ii For any a, b in R (or C) and V ∈ V : a · (b ·V) = (a b) ·V;

iii For any V ∈ V : 1 ·V = V · 1 = V;

iv For any a in R (or C) and V1, V2 ∈ V : a · (V1 + V2) = (a ·V1) + (a ·V2);

v For any a, b in R (or C) and V ∈ V : (a+ b) ·V = (a ·V) + (b ·V)

Real (or complex) numbers are called scalars. For the sake of simplicity hereafter the symbol· in writing the product between a scalar and a vector will be omitted: a ·V→ aV.

39

40 CHAPTER 2. TRANSFORMATIONS AND REPRESENTATIONS

A set of n vectors, e1, . . . , en, are said to be linearly independent iff the only vanishingcombination of them yielding the null vector is the one with vanishing coefficients:

a1 e1 + . . .+ an en = 0 ⇔ a1 = a2 = . . . = an = 0 . (2.1.1)

A vector space V is said to be n–dimensional, and is denoted by Vn, if the maximal numberof linearly independent vectors is n. This means that, if we denote by (ei) = (e1, . . . , en) aset of n linearly independent vectors, called a basis of Vn, the vectors consisting of these nplus any other vector en+1 are not linearly independent. As a consequence of this any vectorV can be expressed as a unique combination of the n vectors ei:

V =n∑i=1

V i ei = V 1 e1 + . . .+ V n en . (2.1.2)

The unique set of coefficients (V i) = (V 1, . . . , V n) associated with V are called the com-ponents of V in the basis (ei). They are conventionally labeled by an upper index (noticethat V 2 is the second component of V and should not be confused with the square of somenot well defined quantity V !) for reasons to be clarified in the sequel. The components of Vare real or complex numbers depending on whether Vn is defined over R or C. For the timebeing let us introduce Einstein’s summation convention, which will be extensively used inthe present course: When in a formula a same index appears in an upper and lower position,summation over that index is understood. For instance in the expression V i ei the indexi appears in an upper (as label of the vector components) and lower (as label of the basisvectors) position. Summation is then understood and we should read: V i ei ≡

∑ni=1 V

i ei.Notice that if we make a linear combination of two vectors V = V i ei and W = W i ei withreal (or complex) coefficients, the components of the resulting vector will be the same linearcombinations of the components V i and W i of two vectors:

aV + bW = a (V i ei) + b (W i ei) = (a V i ei) + (bW i ei) = (a V i + bW i) ei , (2.1.3)

where the reader should realize that we have first used properties iv and ii and then v of thescalar-vector product. It will be of considerable help to describe vectors as column vectors,namely to identify each of them with the column vector whose entries are the componentsrelative to a given basis:

V = V i ei ≡

V 1

V 2

...V n

. (2.1.4)

We shall often denote the column vector representing V, in a given basis, by the same symbolV: V ≡ (V i). The advantage of such a description is that all vector operations can nowbe translated into matrix operations. Consider for instance the problem of computing thecomponents of a linear combinations (2.1.3) of two vectors:

aV + bW ≡ a

V 1

V 2

...V n

+ b

W 1

W 2

...W n

=

a V 1 + bW 1

a V 2 + bW 2

...a V n + bW n

= (a V i + bW i) . (2.1.5)

2.1. LINEAR VECTOR SPACE 41

Identifying a vector V with the column consisting of its components can be done only if thebasis is fixed once for all. In this notation the basis elements are represented by the followingcolumn vectors:

e1 =

10...0

, e2 =

01...0

. . . en =

00...1

. (2.1.6)

If we are considering two different bases (ei) and (e′i), a same vector V = V i ei = V i ′ e′i,will be represented with respect to them by two different column vectors: V = (V i) andV′ = (V i ′). In this case we will keep in mind that the column vectors V and V′ representthe same abstract vector V.

Figure 2.1: Vector V in V3 and point P in E3

Example 2.1: A familiar example of linear vector space over the real numbers is thecollection of vectors in our three dimensional Euclidean space E3. These vectors are definedas arrows connecting two points A and B in E3 ad denoted by

−→AB. We know how to multiply

any such vector by a real number or how to sum two of them. The result is still a vectorin E3, namely there exist a couple of points of E3 connected by it. In particular the nullvector is the vector connecting a point to itself. They close the linear vector space V3. Givena basis of three vectors e1, e2, e3 we can decompose, in a unique way, any vector V alongthem, see Figure 2.1:

V = V 1 e1 + V 2 e2 + V 3 e3 ≡

V 1

V 2

V 3

. (2.1.7)


Vectors in V3 are then associated with couples of points of E3. This means that, if wearbitrarily fix a point O in E3, called the origin, we can uniquely associate with any otherpoint P the vector r ≡ −→OP ∈ V3, called the position vector of P with respect to O. Theorigin O in E3 and a basis (ei) of V3 define a reference frame (RF). The components of rrelative to (ei) are the coordinates of P in the chosen RF, see Figure 2.1:

r = x e1 + y e2 + z e3 ≡

xyz

. (2.1.8)

The coordinates (x, y, z) will also be denoted by (xi) = (x1, x2, x3). One of the definingproperty of an Euclidean space is also the notion of distance between two points, namelythat of metric.

2.1.1 Homomorphisms Between Linear vector Spaces

Consider two vector spaces Vn and Wm over F (which can be either R or C) of dimension nand m respectively. A mapping S:

S : V ∈ Vn −→W = S(V) ∈ Wn , (2.1.9)

is said to be a homomorphism between the two linear vector spaces if, for any V1, V2 ∈ Vnand α, β ∈ F:

S(αV1 + βV2) = αS(V1) + β S(V2) , (2.1.10)

namely if S preserves all the operations defined on Vn. A homomorphism is also called alinear mapping on Vn with values in Wm. S is onto if S(Vn) = Wm and is one-to-one if,S(V1) = S(V2) implies V1 = V2. The latter condition is equivalent to saying that thezero vector 0 of Vn is the only vector which is mapped into the zero-vector of Wm. S is anisomorphism if it is one-to-one and onto.

If the two spaces Vn and Wm have the same dimension, n = m, they are isomorphic.Consider indeed a basis (ei) for Vn and (fi) for Wn. Then define a mapping S between thetwo spaces such that for any i = 1, . . . , n, S(ei) ≡ fi. The reader can verify that S is anisomorphism. Vice-versa, if Vn and Wm are isomorphic, they have the same dimension. Letus prove this property. Take indeed a basis (ei) of Vn and consider the set of n vectors (fi)in Wm where fi ≡ S(ei). These vectors are linearly independent since, if αi fi = 0, using eq.(2.1.10) we have that S(αi ei) = 0 and thus, being S one-to-one, αi ei = 0, which impliesαi = 0 for any i = . . . , n. Suppose now (fi) is not a maximal system of linearly independentvectors in Wm. This means that there exist a vector f ∈ Wm such that α f + αi fi = 0implies α = αi = 0. Since S is onto, there exist e ∈ Vn, such that S(e) = f . We havethen that S(α e + αi ei) = 0, or , equivalently, α e + αi ei = 0 implies α = αi = 0, whichcontradicts the assumption that (ei) is a basis of Vn. A homomorphism of Vn onto Vn iscalled an operator or endomorphism on Vn. The space of all homomorphisms between Vnand Wm is denoted by Hom(Vn, Wm). The space of all endomorphisms on Vn is denoted by


End(Vn) ≡ Hom(Vn, Vn). If a homomorphism is onto and one-to-one it is called isomorphismand its space denoted by Isom(Vn, Wm), where m = n. Isomorphisms of a linear vector spaceon itself are called automorphisms and span a space denoted by Aut(Vn) ≡ Isom(Vn, Vn).Transformations on a linear vector space Vn will be described by automorphisms on Vn.

The action of a homomorphism S between Vn and Wm is represented by a m× n matrixMS as follows. Consider a basis (ei) for Vn, i = 1, . . . , n. Consider the set of vectorse′i ≡ S(ei) of Wm. Since S is not necessarily one-to-one, (e′i) are not necessarily linearlyindependent. We can nevertheless expand each of them in components with respect to thebasis (fa) of Wm, a = 1, . . . ,m. Let us denote by MS

aj the component of e′i along fa:

e′i ≡ S(ei) = MSaj fa . (2.1.11)

Let us compute S on a vector V = V i ei ∈ Vn:

S(V) = S(V i ei) = V i e′i = (MSai V

i) fa = W a fa . (2.1.12)

The components W a of the vector W ≡ S(V) ∈ Wm are then expressed as follows:

W a = MSai V

i . (2.1.13)

Representing these vectors in terms of column vectors:

W ≡

W 1

...Wm

, V ≡

V 1

...V n

, (2.1.14)

relation (2.1.13) can be written in the following matrix notation:

W ≡ S(V) = MS V , (2.1.15)

where MS denotes the m× n matrix MS ≡ (MSaj) which, acting on the n-vector V yields

the m-vector W. For the sake of simplicity, the matrix MS, representing the action ofthe homomorphism S, will also be denoted by the corresponding boldface letter S. If S ∈Aut(Vn), MS ≡ (MS

ij) will be a n × n invertible matrix, namely an element of GL(n,F)1.

In fact there is a one to one correspondence between S ∈ Aut(Vn) and GL(n,F).Just as for the groups we can define the product S · T of two endomorphisms S, T on a

linear vector space Vn as the endomorphism defined on a generic vector V as follows:

S · T (V) ≡ S(T (V)) . (2.1.16)

Let (ei) be a basis of Vn, so that S(ei) = MSji ej and T (ei) = MT

ji ej. Then

S · T (ei) = S(T (ei)) = S(MTji ej) = MT

jiMS

kj ek = (MS MT )ki ek , (2.1.17)

1Recall that we have denoted by GL(n,R) or GL(n,C) the group of all n × n real or complex matricesequipped with the ordinary row-times-column product.


so that the matrix MS·T , representing the action of S · T , is the product MS MT of thecorresponding matrices in the same order: On a column vector V = (V i):

S · T (V) = MS MT V . (2.1.18)

If S, T are two automorphisms, S ·T is still an automorphism. It is straightforward to provethat Aut(Vn), with this product, is a group. Moreover the correspondence:

S ∈ Aut(Vn) −→ MS = (MSij) ∈ GL(n,F) , (2.1.19)

is a homomorphism since, as shown above, MS·T = MS MT . It is onto: Any matrix Mdefines an endomorphism S on Vn, such that MS = M. Finally it is one-to-one, since an×n matrix M uniquely defines an endomorphism on Vn: Two endomorphisms representedby the same matrix M will define the same correspondence between vectors, according to(2.1.19), and thus coincide. The correspondence (2.1.19) is an isomorphism between Aut(Vn)and GL(n,F):

Aut(Vn) ∼ GL(n,F) , (2.1.20)

so that we can characterize GL(n,F) as the group of automorphisms, or linear transforma-tions, on a n-dimensional linear vector space over F.

2.1.2 Inner Product on a Linear Vector Space over RLet us return now to the general discussion of linear vector spaces and define on them somenew structures.

On a linear vector space Vn over R we define an inner or scalar product between vectorsas the mapping:

(·, ·) : Vn × Vn −→ R , (2.1.21)

which associates with any couple of vectors V and W a real number (V, W). This mappinghas to be symmetric and bilinear, i.e. linear in any of its arguments:

1. For any V and W in Vn: (V, W) = (W, V) (symmetry);

2. For any V1, V2 and W in Vn and b ∈ R: (aV1 + bV2, W) = a (V1, W) + b (V2, W)(linearity).

In virtue of its symmetry, linearity in the first argument implies linearity in the secondas well. If now we write two generic vectors in components with respect to a basis (ei),V = V i ei, W = W i ei, using the bilinear property of the inner product, we are able towrite:

(V, W) = (V i ei, Wj ej) = V i (ei, W

j ej) = V i (ei, ej)Wj = V i gijW

j , (2.1.22)


where the matrix g = (gij) ≡ ((ei, ej)) is a symmetric n × n real matrix called the metric.In matrix notation the scalar product (2.1.22) can be written in terms of row-times-columnmatrix products:

(V, W) = VT g W , (2.1.23)

where, in the right hand side, we have denoted by V and W the columns corresponding tothe two vectors, so that VT , namely the transposed of the column vector V, is actually arow vector: VT ≡ (V 1, V 2, . . . , V n). We define the norm squared of a vector V the innerproduct of the vector with itself:

||V||2 ≡ (V, V) = V i gijWj = VT g V . (2.1.24)

We have not required that g = (gij) be a positive defined matrix so far, and thus ||V||2 is notnecessarily a positive number. We will just require g to be non-singular, namely: det(g) 6= 0.As we shall show in the next sections, we can always find a basis (ei) with respect to whichthe matrix g has a diagonal form with diagonal entries ±1:

g = ηp,q ≡ diag(

p︷︸︸︷+1, . . . ,+1,

q︷︸︸︷−1, . . . ,−1) , (2.1.25)

where p + q = n. The signature q − p of g is a feature of the inner product. This basis iscalled orthonormal.

2.1.3 Hermitian, Positive Definite Product on a Linear VectorSpace over C

Consider now a vector space Vn over C. A hermitian positive definite product is defined asa mapping:

(·, ·) : Vn × Vn −→ C , (2.1.26)

which associates with any couple of vectors V and W a complex number (V, W) and whichsatisfied the following properties:

1. For any V and W in Vn: (V, W) = (W, V)∗ (hermitian);

2. For any V1, V2 and W in Vn and b ∈ R: (W, aV1 + bV2) = a (W, V1) + b (W, V2)(linearity);

3. For any V in Vn: ‖V‖2 ≡ (V, V) ≥ 0, the inequality is saturated only if V = 0(positive definiteness);

From the first two properties it follows that (aV1 +bV2, W) = a∗ (V1, W)+b∗ (V2, W). Ifnow we write two generic vectors in components with respect to a basis (ei), V = V i ei, W =W i ei, using the above properties of the hermitian product, we are able to write:

(V, W) = (V i ei, Wj ej) = V i ∗ (ei, W

j ej) = V i ∗ (ei, ej)Wj = V i ∗ gijW

j ,(2.1.27)


where the matrix g = (gij) ≡ ((ei, ej)) is now hermitian, namely it satisfies the property:gij = (gji)

∗. In matrix notation, if we denote by M† the hermitian conjugate of a matrix M,obtained by taking the complex conjugate of the transposed of M, we can write: g = g†.From the general theory of matrices we know that g, being hermitian, has real eigenvalues.We can also rewrite (2.1.22) in the following form:

(V, W) = V† g W , (2.1.28)

where V† ≡ (V 1 ∗, . . . , V n ∗). Hermitianity of the product in particular implies that, for anyV ∈ Vn, (V, V) = (V, V)∗ ∈ R. The fact that the product is positive definite furtherrequires that gij be a positive definite matrix, namely have real positive eigenvalues.

A complex linear vector space equipped with a hermitian, positive definite scalar productis also called a Hilbert space.

2.1.4 Symplectic product

Consider a 2n-dimensional linear vector space V2n on R. A symplectic product on V2n isdefined as a mapping:

(·, ·) : V2n × V2n −→ R , (2.1.29)

which associates with any couple of vectors V and W a real number (V, W) and whichsatisfied the following properties:

1. For any V and W in Vn: (V, W) = −(W, V) (skew-symmetry);

2. For any V1, V2 and W in Vn and b ∈ R: (W, aV1 + bV2) = a (W, V1) + b (W, V2)(linearity).

The above properties imply bi-linearity of the product. If now we write two generic vectors incomponents with respect to a basis (ei), V = V i ei, W = W i ei, using the bilinear propertyof the inner product, we are able to write:

(V, W) = (V i ei, Wj ej) = V i (ei, W

j ej) = V i (ei, ej)Wj = V i ΩijW

j , (2.1.30)

where Ω = (Ωij) ≡ ((ei, ej)) is a skew-symmetric 2n × 2n real matrix called symplecticmatrix: Ωij = −Ωji.

We can always choose a basis (ei) such that, if we split its elements into the first n, ea,labeled by a and the last n, en+a, we have:

(ea, eb) = (ea+n, eb+n) = 0 , (ea, eb+n) = −(eb+n, ea) = δab , (2.1.31)

where δab is equal to 1 if a = b, to 0 if a 6= b. In this basis the symplectic matrix reads:

Ω =

(0n 1n−1n 0n

), (2.1.32)

where 0n is an n × n matrix made of zeros, while 1n is the n × n identity matrix. In thisbasis, the symplectic product between two vectors (2.1.30) reads:

(V, W) = VT Ω W = V aW n+a − V n+aW a . (2.1.33)


2.1.5 Example: Space of Hermitian matrices

Let us show that the space of hermitian n × n matrices, is a vector space over the realnumbers. A hermitian matrix H ≡ (Hij) is a complex n× n matrix satisfying the property

H† = H ⇔ Hij = (Hji)∗ . (2.1.34)

We have already come across an example of hermitian matrix: The hermitian metric g =(gij). The reader can easily verify that if H1 and H2 are two hermitian matrices, also anylinear combination of them, aH1 + bH2, is. Therefore the space of hermitian matrices is avector space with respect to the ordinary sum of matrices and product times real numbers.To determine its dimension we should find the maximal number of free real parameters ageneric hermitian matrix depends on. These will be identified with the components of ageneric hermitian matrix relative to some basis. We start from the 2n2 real parameters ofa generic complex n× n matrix. The condition (2.1.34), for i = j implies that the diagonalentries of H are real numbers (Hii = (Hii)

∗). The same condition for i 6= j implies that theentries below the diagonal are the conjugate of the symmetric entries above the diagonal.The independent entries of a generic H consist therefore of the n real diagonal componentsand the real and imaginary parts of n(n − 1)/2 complex entries above the diagonal. Theirnumber is n + 2 × n(n − 1)/2 = n2. We conclude that the space of hermitian matrices hasreal dimension n2.

As an example consider the hermitian 2 × 2 matrices. The most general such matriceshas the following form:

H =

(a b− i c

b+ i c d

), (2.1.35)

a, b, c, d being real numbers. The space has dimension 4 and we can choose as a basis thefollowing set of hermitian matrices:

e1 = 12 , e2 = σ1 , e3 = σ2 , e4 = σ3 , (2.1.36)

where σi are the Pauli matrices:

σ1 =

(0 11 0

), σ2 =

(0 −ii 0

), σ3 =

(1 00 −1

). (2.1.37)

The generic matrix (2.1.35) is expressed in the basis (2.1.36) as follows:

H =a+ d

2e1 + b e2 + c e3 +

a− d2

e4 = H i ei . (2.1.38)

We can also define a positive definite inner product over the space of 2×2 hermitian matrices:

(H1, H2) ≡ 1

2Tr(H1 H2) . (2.1.39)


The reader can verify that with respect to the above inner product the basis (2.1.36) isortho-normal: (ei, ej) = δij. The components of H can be computed as H i = Hj δij =Hj (ej, ei) = (H, ei).

Exercise 2.1: Show that traceless hermitian matrices (H† = H, Tr(H) = 0) form a3-dimensional vector space, of which the Pauli matrices (σi) form a basis.

Exercise 2.2: Consider now anti-hermitian matrices. These are complex n× n matricesA = (Aij) satisfying the condition: A† = −A, or, in components, Aij = −(Aji)

∗. Show thatanti-hermitian matrices form a linear vector space over the real numbers, just as hermitianmatrices do. Show that the dimension of the algebra is still n2 and that, for n = 2 we canchoose as a basis the following matrices: fi = i ei, i = 1, 2, 3, 4, ei being defined in (2.1.36).

Let us end this subsection by mentioning some properties of the Pauli matrices, whichthe reader can easily verify:

σi σj = δij 12 + i3∑

k=1

εijk σk , (2.1.40)

σi σj σk = i εijk 12 + δij σk − δik σj + δjk σi , (2.1.41)

where εijk is different from zero only if i 6= j 6= k 6= i, ε123 = 1 and is totally antisymmetric,so that εijk = +1 or −1 depending on whether (i, j, k) is an even or odd permutation of(1, 2, 3) respectively. The following property of εijk, used to derive eq. (2.1.41) from (2.1.40)holds:

3∑k=1

εijk ε`nk = δi` δjn − δin δj` . (2.1.42)

Exercise 2.3: Compute σ1 σ2 σ3.

2.2 Spaces

Let us consider a space of points Mn (like the three dimensional Euclidean space) with whichthere is a linear vector space Vn associated in the same way as a three-dimensional vectorspace V3 is associated with the Euclidean space E3: To each couple of points A, B in Mn

there corresponds a unique vector V =−→AB of Vn connecting them. Mn is then said to be

a n–dimensional real or complex space of points depending on whether Vn is a linear vectorspace over R or C. The three-dimensional Euclidean space is a real space of points.

If we fix an origin O in Mn we can associate with each point P ∈ Mn a unique positionvector r =

−→OP and a unique set of n coordinates (xi) representing the components of r

relative to a basis (ei) of Vn:

r = xi ei ≡

x1

x2

. . .xn

. (2.2.1)

2.2. SPACES 49

The coordinates xi are real or complex numbers depending on whether Mn is a real orcomplex space respectively. Given two points P and P ′ of Mn, with position vectors r andr′ respectively, the relative position vector ∆r of P ′ with respect to P is defined as follows:

∆r ≡ r′ − r = (∆xi) ei =

∆x1

∆x2

. . .∆xn

, (2.2.2)

where ∆xi ≡ xi ′−xi. If the two points are infinitely close to one another, the relative positionvector will be denoted by dr and represents an infinitesimal shift and its components arethe differentials of the coordinates:

∆r ≡ (dxi) ei =

dx1

dx2

. . .dxn

. (2.2.3)

2.2.1 Real Metric Spaces

If Vn is endowed with an inner product, we can use it to define a metric on Mn: The squareddistance d(P, P ′)2 between two points P and P ′ is the squared norm of the relative positionvector:

d(P, P ′)2 ≡ ||∆r||2 = (∆xi) gij (∆xj) = ∆rT g ∆r . (2.2.4)

If the two points are infinitely close, their squared distance is also denoted by ds2 and reads:

ds2 ≡ d(P, P ′)2 ≡ ||dr||2 = dxi gij dxj = drT g dr . (2.2.5)

With respect to an orthonormal basis (2.1.25) this infinitesimal distance reads:

ds2 = (dx1)2 + . . .+ (dxp)2 − (dxp+1)2 − . . .− (dxn)2 . (2.2.6)

The space Mn, equipped with this notion of distance, is a metric space, characterized by thesignature p− q of g and will therefore be denoted by M (p,q). If n = 3, q = 0 and p = n themetric is positive definite (g is a positive definite matrix having all positive eigenvalues) andthe space M (3,0) is the familiar Euclidean space E3. In this case, if (x, y, z) and (x′, y′, z′)are the coordinates of P and P ′ respectively, equation (2.2.5) reads:

d(P, P ′)2 ≡ (x′ − x)2 + (y′ − y)2 + (z′ − z)2 ≥ 0 , (2.2.7)

which is the usual expression derived from Pythagoras’ theorem. Fro the above expressionit is clear that, being g positive definite, d(P, P ′) = 0 if and only if P = P ′. In general||V||2 ≥ 0 and is zero if and only of V = 0. Since the norm squared of a vector of V3,in this case, coincides with its length squared, i.e. the distance squared between its end


points, an orthonormal basis is a basis of mutually orthogonal vectors of unit length. Werecover in this case the familiar geometric definition of scalar product between two vectors:(V, W) = ||V|| ||W|| cos(θ), θ being the angle between the two vectors.

If n = 4, q = 3 and p = 1 the corresponding space M (1,3) is called Minkowski’s space.In the special theory of realativity a point in M (1,3) represents a physical event which tookplace at a point of coordinates x, y, z of our Euclidean space, at a time t. Its coordinatesare labeled by an index µ = i− 1 = 0, . . . , 3, so that x0 is related to time, x0 = c t, c beingthe speed of light, while the remaining three coordinates x1, x2, x3 are identified with thespatial coordinates x, y, z of the event. The distance between two nearby events reads:

ds2 ≡ dxµ gµν dxν = (dx0)2 − (dx1)2 − (dx2)2 − (dx3)2 = c2 dt2 − dx2 − dy2 − dz2 .

(2.2.8)

The whole special theory of relativity is based on the requirement that this distance be thesame whatever inertial RF we use for studying the events and derives from the assumptionthat the speed of light c be the same in any inertial RF. ds2 can now be positive, negative ornull, since the metric g = (gµν) has indefinite signature. From this it follows that non-nullvectors can have vanishing norm squared and, for the same reason, two distinct points canhave vanishing distance. A vector V = V µ eµ ≡ (V µ) of V4 with positive or negative normsquared are called time-like or space-like respectively, while vectors with vanishing normare called light-like. Similarly we talk about time-like, space-like or light-like distances if∆r = (∆xµ) is time-like, space-like or light-like.

2.2.2 Complex Spaces With Hermitian Positive Definite Metric

Consider a complex space Mn whose associated linear vector space Vn is endowed with ahermitian, positive definite, inner product. This product on Vn induces a metric on Mn, sothat the squared distance d(P, P ′)2 between two points P and P ′ is defined as the squarednorm of the relative position vector:

d(P, P ′)2 ≡ ‖∆r‖2 = (∆xi)∗ gij (∆xj) = ∆r† g ∆r . (2.2.9)

With respect to an ortho-normal basis (ei) of Vn, gij = (ei, ej) = δij, the above distancereads:

d(P, P ′)2 ≡n∑i=1

|∆xi|2 = |∆x1|2 + . . .+ |∆xn|2 . (2.2.10)

This distance, being the sum of squares, vanishes only if each term in the sum is zero, namelyif the two points coincide. If P, P ′ are infinitely close, so that their relative position vectoris described by the infinitesimal shift vector dr = (dxi) their distance d(P, P ′), also denotedby ds2, reads:

ds2 = d(P, P ′)2 = ‖dr‖2 = dr† dr =n∑i=1

|dxi|2 = |dx1|2 + . . .+ |dxn|2 , (2.2.11)

where we have used the ortho-normal basis g = 12.

2.3. TRANSFORMATIONS 51

2.2.3 Symplectic Spaces

Next we consider spaces of points M2n whose geometric properties are encoded in a linearvector space V2n over R on which a symplectic product is defined. Choosing a RF on M2n

with respect to which the symplectic matrix has the form (2.1.32) and, given two infinitesimaldisplacements dr = (dxi), dr′ = (dxi ′), we can compute their symplectic product:

(dr, dr′) = dxi Ωij dxj ′ = drT Ω dr′ =

n∑a=1

(dxa dx′ a+n − dxa+n dx′ a) . (2.2.12)

An example of symplectic space is the phase space of a Hamiltonian system with n degreesof freedom. Each point is this space represents a mechanical configuration of the systemand is described by n generalized coordinates qa and n conjugate momenta pa (for a pointparticle n = 3 and qa are the three spatial coordinates x, y, z, while pa the components ofthe momentum vector px, py, pz): x

a = qa, xn+a = pa. The infinitesimal displacement vectorr is then computed between two nearby configurations and the symplectic product (2.2.12)between two infinitesimal displacements reads:

(dr, dr′) = dxi Ωij dxj ′ = dqa dp′a − dpa dq′ a , (2.2.13)

where now Einstein’s summation convention was used for the index a. The relevance of thissymplectic structure for Hamiltonian systems relies in the fact that canonical transforma-tions, including time evolution, namely the correspondence between the configurations ata time t and at a time t′ > t, are represented on dr by symplectic transformations whichpreserve the product (2.2.13).

2.3 Transformations

In the present section we discuss transformations on a point-field. By point-field we mean anarbitrary collection of elements (which can be either finite or infinite) which we call points.An example of point-fields are the metric spaces M (p,q) introduced in the previous section.These are sets of infinite points, think about the three dimensional Euclidean space. Wewill also consider a vector space as a point-field, where the points are now vectors. A groupitself, finite or infinite, can be regarded as a point-field. Our notion of point-field so far israther generic.

Let M be a point-field. A transformation S on M is a correspondence of M into itself,which maps each point P of M into its image P ′ = S(P ):

S : P ∈M −→ P ′ = S(P ) ∈M . (2.3.1)

A transformation is then defined by assigning its value on any point of M . Given twotransformations S, T , we define the product transformation T · S as the transformationresulting from the consecutive action onM of S and T . Therefore if S maps P into P ′ = S(P )and T maps P ′ into P ′′ = T (P ′), T · S will map P into P ′′ = T (S(P )). We define the


identity transformation I as the transformation mapping each point into itself: I(P ) = P .If we restrict to one-to-one, onto correspondences, these can be inverted in M . Given atransformation S we define its inverse S−1 as the mapping between a point P ′ and the onlypoint P of which P ′ is the image through S:

S : P → P ′ ⇒ S−1 : P ′ → P . (2.3.2)

Clearly the inverse of the inverse of a transformation is the transformation itself: (S−1)−1 =S. By definition, the consecutive action of a transformation and its inverse on any pointwill yield the point itself, namely the product of any transformation by its inverse is theidentity transformation: S · S−1 = S−1 · S = I. If we have three transformations the readercan convince himself that their product is associative: R · (T · S) = (R · T ) · S. We concludethat the set of all transformations of a point-field M , endowed with the product definedabove, is a group called group of transformations of the point-field, or symmetry group ofthe point-field, and denoted by TM .

Example 2.2 For example consider a set of n objects: x1, . . . , xn. Any one to one transfor-mation on this point-field is described by a permutation. Indeed consider a transformationS. Its image on x1 will be one of the points of the same set, call it xi1 : S(x1) = xi1 . SimilarlyS(x2) = xi2 , which cannot coincide with xi1 being the correspondence one-to-one, and soon up to S(xn) = xin , where xin is different from xi1 , . . . , xin−1 . The action of S will then

totally be described by the permutation: S ≡(x1 . . . xnxi1 . . . xin

).

Example 2.3 We can consider as an example of point-field the Euclidean plane E2 orspace E3. These are real metric spaces, in which the metric is positive definite. Amongall the transformations in TE2 or TE3 , of particular interest are those transformations whichleave distances between points invariants. These are congruences and consist in translations,rotations about a point or an axis (in E3) and reflections. Rotations and reflections are alsocalled proper congruences. Congruences will preserve the shape and size of any collection ofpoints (e.g. a triangle, a rectangle etc...) in E2 or E3. We may also consider more generaltransformations which preserve the shape but not the size of a geometrical object. These arecalled similitudes and include, besides translations, rotations and reflections, also dilations.

Example 2.4 Let us consider M to be the familiar Euclidean plane E2. We can considera subset N of points in E2, like a geometrical object (e.g. an equilateral triangle, a square,or any polygon), and look for its symmetry group TN . As we have shown in last chapter, foran equilateral triangle, TN is the non abelian group D3, containing rotations and reflections,for a square D4 and so on for all the regular polygons...In any case TN is a subgroupof TE2 . A geometrical object of finite size can only have as symmetry transformationsrotations and reflections. Indeed a finite size object is always contained inside a circleof finite radius. If a translation t or a dilation d were a symmetry of the object, theirconsecutive actions tn and dn, for any n would be symmetries as well. But for n sufficiently


Figure 2.2: Symmetry of a decoration under a translation by its period a.

large the resulting transformation, being it a translation or a dilation, would bring all thepoints of the object outside the circle which originally contained it and thus tn and dn

cannot be symmetries, in contrast with our assumption. It can be shown that the groupsCn and Dn exhaust all possible proper congruences on the Euclidean plane. Only infinitelyextended objects with periodic shape, like a lattice, can have a translation as symmetry.An example are certain ornaments or regular tilings, see Figure 2.2. Similar argumentscan be given for geometrical objects in the three dimensional Euclidean space E3 as well.Proper congruences (i.e. rotations and reflections), act on En as linear homogeneous, metricpreserving, transformations and, as we shall see, are contained in the Lie group O(n). On E2

the group O(2) can be considered as the limit n→∞ of Dn, also denoted by D∞, and is thesymmetry group of a circle, containing rotations about a point by any angle and reflectionsin any straight line containing that point. The group SO(2), as we shall see, is abelian andcontains just rotations about a point by any angle. It can be identified with the group C∞.

2.3.1 Transformations on Mn

As point-field we can consider the spaces of points Mn discussed in Section 2.2. With respectto a RF, each point is described by its n coordinates (xi). The action of a transformation Smapping a point P of coordinates xi into a point P ′ of coordinates xi ′ is then totally definedby expressing the latter as functions of the coordinates of P : xi ′ = f iS(x1, x2, . . . , xn). Thefunctions f iS(x1, x2, . . . , xn), which we shall denote for the sake of simplicity by f iS(xj), candepend, in general, non-linearly on their arguments. If we consider the effect of S on therelative position vector dr = (dxi) between two nearby points of coordinates xi and xi+dxi,it can be deduced by expressing the relative position between the two image points (which


are still infinitely close to one another):

dxi ′ = f iS(xj + dxj)− f iS(xj) = df iS =∂f iS∂xj

dxj =∂xi ′

∂xjdxj = MS

ij dx

j , (2.3.3)

or, in matrix notation:

dr′ = MS dr , (2.3.4)

Where we have used the definition of the total differential of a function. The transformationS acts by means of a coordinate-dependent matrix MS ≡ (MS

ij), which is the Jacobian of

the transformation, on the column vector dr of coordinate differentials (dxi). The identitytransformation I maps xi into xi ′ = xi, namely f iI(x

k) = xi and the corresponding actionon the coordinate differentials is represented by the n× n identity matrix: M i

I j = δij, or, inmatrix notation, MI = 1n.

Consider now two transformations S and T whose action on the coordinates is definedby the functions f iS and f iT respectively:

S : xi −→ xi ′ = f iS(xj) ,

T : xi −→ xi ′ = f iT (xj) , (2.3.5)

which means that, if S maps xi into xi ′ = f iS(xj) and T xi ′ into xi ′′ = f iT (xj ′) = f iT (f jS(xk))effect of T · S is represented by the composites of the functions f iT and f jS:

T · S : xi −→ xi ′ = f iT (f jS(xk)) . (2.3.6)

The action of T · S on coordinate differentials is readily computed using the property ofderivation of composite functions:

dxi ′′ =∂f iT∂xj ′

dxj ′ =∂f iT∂xj ′

∂f jS∂xk

dxk = MTijMS

jk dx

k , (2.3.7)

or, in matrix notations:

dr′′ = MT MS dr = MT ·S dr . (2.3.8)

If the action of S is described by a set of functions f iS such that xi ′ = f iS(xj), its inverseS−1 will be described by the inverse functions f iS−1 = f−1 i

S , such that: f iS−1(xj ′) = xi =

f−1 iS (f jS(xk)). The Jacobian of this transformation is the inverse of the matrix MS: MS−1 =

M−1S . Indeed 1n = MI = MS−1·S = MS−1 MS.So far we have described the effect of a transformation on Mn as a coordinate transfor-

mation: xi → xi ′ = f i(xj). Coordinates are like the “names” we give to the points of Mn

in a given RF. We can view a same transformation either as a change of our description ofa same point, namely a change of the coordinate system we use (passive description of atransformation), or as a true mapping between different points in a fixed coordinate system(active description of a transformation). The two points of view are complementary andlead to the same relations between the final and initial coordinates. In the former xi ′ andxi are different descriptions of a same point in space, that is the transformation affects theRF we use, while in the latter xi ′ and xi are the coordinates, relative to a same RF, oftwo different points which are mapped one into the other by the transformation (this is theview-point we originally adopted when introducing the notion of transformation).


Figure 2.3: Active an passive description of a rotation in the plane.

Example 2.5 As a simple example let us consider a clockwise rotation by an angle θ ona plane, about a point O, see Figure 2.3. The plane is the two-dimensional Euclidean spaceE2, associated with the linear vector space V2, containing all the vectors on the plane. Usingthe active description of the transformation, a point P with coordinates x, y is mapped in apoint P ′ whose coordinates x′, y′ are related to x, y as follows:

(x, y) −→x′ = f 1(x, y) = cos(θ)x+ sin(θ) yy′ = f 2(x, y) = − sin(θ)x+ cos(θ) y

. (2.3.9)

Notice that the coordinates of the new point P ′ are linear and homogeneous functions ofthe coordinates of P through coefficients depending on the angle θ. We can indeed rewrite(2.3.9) in the following matrix form:

r′ = M r , M =

(cos(θ) sin(θ)− sin(θ) cos(θ)

). (2.3.10)

It can be easily verified that M is also the Jacobian of the transformation:

M =

(∂f1

∂x∂f1

∂y∂f2

∂x∂f2

∂y

), (2.3.11)


and therefore defines the effect of the rotation on infinitesimal displacements as well: dr′ =M dr. Let us now consider the same transformation from the passive point of view andobserve that the same eqs. (2.3.9) can be interpreted as relating the coordinates of a samepoint P relative to a basis ei (x, y) to those (x′, y′) relative to a new basis e′i, obtainedfrom the original one through an counter-clockwise rotation by the same angle θ about theorigin. To see this let us write the position vector of P in the basis (ei):

−→OP = x e1 + y e2.

Notice that, since we are changing the basis for the vectors, we should not identify theposition vector with the column containing the coordinates, but rather write it explicitly interms of the basis elements. To avoid confusion we denote by

−→OP the position vector, which

is associated with the point P and by r the column vector consisting of the componentsof−→OP in the basis (ei). The same position vector has the following form in the new basis

−→OP = x′ e′1 +y′ e′2. Consider next the relation between the two bases by writing each elementof the new basis in components with respect to the old one:

e′1 = cos(θ) e1 + sin(θ) e2 ,

e′2 = − sin(θ) e1 + cos(θ) e2 . (2.3.12)

Substituting the above formulas in−→OP we find:

−→OP = x′ (cos(θ) e1 + sin(θ) e2) + y′ (− sin(θ) e1 + cos(θ) e2) = x e1 + y e2 .(2.3.13)

Equating the components of e1 and e2 the relation between the old and the new coordinatesfollows:

x = cos(θ)x′ − sin(θ) y′

y = sin(θ)x′ + cos(θ) y′ . (2.3.14)

Finally, inverting the above relations we obtain the transformation law (2.3.9). In bothdescriptions the transformation is realized as the action of a same matrix M on the columnvector of coordinates r.

We can consider now a function F on Mn with values in R:

F : P ∈Mn −→ F (P ) ∈ R . (2.3.15)

This function can be described as a function of the n coordinates (xi) of a generic pointP ∈ Mn: F (P ) ≡ F (x1, x2, . . . , xn) = F (xi). The gradient ∇F of F is defined as therow vector consisting of the partial derivatives of F (x1, x2, . . . , xn) with respect to eachcoordinate:

∇F ≡(∂F

∂x1,∂F

∂x2, . . . ,

∂F

∂xn

)≡(∂F

∂xi

), (2.3.16)

such that the differential dF of F , which expresses the difference in value of F between twoinfinitely close points is expressed as the row-times-column product of ∇F and dr:

dF = F (xi + dxi)− F (xi) =∂F

∂xidxi =

(∂F

∂x1, . . . ,

∂F

∂xn

) dx1

...dxn

= ∇F dr .(2.3.17)


Consider now a transformation S mapping xi into xi ′ = f iS(xj). From the passive view-point,the two sets of coordinates are different descriptions of a same point. Since the value of Fdepends only on the point in which it is evaluated, the expression of F as a function of thetwo set of coordinates will in general be different: Its expression F ′(xi ′) in terms of the xi ′

is obtained from F (xi) by expressing xi as functions of the xi ′, through the inverse of thefunctions f iS:

F ′(xi ′) = F (P ) = F (xi(xj ′)) . (2.3.18)

The components of the gradient transform as follows:

∂F

∂xi→ ∂F ′

∂xi ′=∂xj

∂xi ′∂F

∂xj=∂F

∂xjM−1

Sji , (2.3.19)


∇′F ′ = ∇F M−1S . (2.3.20)

We have come across two kind of objects so far: Objects represented by column vectors,whose components are labeled by an upper index, like the vectors in Vn V = (V i) or aninfinitesimal displacement dr = (dxi) in Mn, and objects like the gradient of a function∇F = ( ∂F

∂xi), which are represented by row-vectors whose components are labeled by a lower

index. Under a transformation the former transform as in (2.3.3) and are called contravariantvectors, while the latter transform as in (2.3.19) and are called covariant vectors.

2.3.2 Homogeneous-Linear Transformations on Mn and Linear Trans-formations on Vn.

Let us generalize our example to generic homogeneous-linear transformations on Mn.

Passive picture: These transformations, in the passive picture, originate from a genericchange of basis of Vn: (ei)→ (e′i), keeping the origin O of the RF fixed. The position vector−→OP of a same point P in the two bases read:

−→OP = xi ei = xi ′ e′i . (2.3.21)

In the first basis the position vector is described by the column vector r = (xi) consistingof the coordinates relative to the initial RF, while in se second basis it is represented bythe column vector r′ = (xi ′), consisting of the new coordinates of P . Let us express eachelement of the new basis in terms of the old one:

e′i = Dji ej , (2.3.22)

where Dji represents the jth component of the ith vector e′i with respect to the old basis.

Equation (2.3.22) can be easily written in matrix notation by arranging the basis elementsin row vectors and writing:

(e′1, e′2, . . . , e′n) = (e1, e2, . . . , en) D , (2.3.23)


where D is the matrix (Dij), i and j being the row and column indices respectively. Since

both (ei) and (e′i) are systems of linearly independent vectors, the matrix D is non-singular:

det(D) 6= 0. In the example 2, the matrix D = (Dij) =

(cos(θ) − sin(θ)sin(θ) cos(θ)

). Let us

substitute eq. (2.3.22) into (2.3.21):

−→OP = xi ei = xi ′Dj

i ej . (2.3.24)

Equating each component with respect to the old basis we find a relation between the oldand new coordinates:

xi = Dij x

j ′ ⇔ r = D r′ . (2.3.25)

Since D is non-singular, the above system of n linear equations in n unknowns xi ′ can besolved by multiplying both sides to the left by D−1. We then find:

r′ = D−1 r ⇔ xi ′ = f i(xj) = D−1 ij x

j . (2.3.26)

We can readily compute the Jacobian matrix of the above transformation: M = (M ij) =

(∂xi ′

∂xj) = ( ∂f

i

∂xj) = (D−1 i

j) = D−1. The matrix D represents therefore the Jacobian matrix

of the inverse transformation: D = ( ∂xi

∂xj ′). In terms of M let us rewrite the transformation

properties of the basis elements and the coordinate differentials:

e′i = Dji ej = M−1 j

i ej =∂xj

∂xi ′ej , (2.3.27)

dxi ′ = M ij dx

j =∂xi ′

∂xjdxj , (2.3.28)


(e′1, e′2, . . . , e′n) = (e1, e2, . . . , en) M−1 , (2.3.29)

dr′ = M dr , (2.3.30)

Comparing (2.3.29) with (2.3.19) we see that, under a linear-homogeneous transformation,the basis elements transform as components of a covariant vector, and are indeed labeled bya lower index.

Active picture: If we adopt on the other hand the active point of view, points and vectorsare transformed into different points and vectors respectively. In this representation, we willderive the effect of a homogeneous linear transformation S on Mn from the action of a lineartransformation on the corresponding linear vector space Vn. A linear transformation over avector space Vn is an invertible linear function of Vn onto Vn, that is an automorphism on Vn,as defined in Section 2.1.1. As discussed in Section 2.1.1, with respect to a given basis (ei)of Vn, the action of an automorphism S on Vn is completely described by a n× n invertiblematrix MS = (MS

ij), whose generic entry MS

ij represents the ith component of the vector


S(ej), i, j = 1, . . . , n. The action of S on the column vector V ≡ (V i), given in eq. (2.1.15),is expressed as the following matrix product:

V→ V′ = S(V) = MS V , (2.3.31)

or, in components,

V i → V ′ i = MSij V

j . (2.3.32)

The transformation S on Vn induces on the space of points Mn a transformation, to bedenoted by the same letter S, which maps a point P described, with respect to a givenRF, by the vector

−→OP = xi ei, into a point P ′ described, in the same RF by the vector−−→

OP ′ ≡ S(−→OP ) = xi ′ ei:

−→OP

S−→−−→OP ′ ≡ S(

−→OP ) . (2.3.33)

using eq. (2.3.32) we can relate the components of the transformed vector, i.e. the coordi-nates of the new point P ′, to the coordinates of the original point P :

xi ′ = MSij x

j ⇔ r′ = MS r . (2.3.34)

We have described a homogeneous linear transformation on Mn, in the active picture, asoriginating from a linear transformation (an automorphism) on the corresponding linearvector space Vn. Notice that in the two descriptions (active and passive) the relation betweenthe final and initial coordinates is the same, their interpretation is different: In the formerr = (xi) and r′ = (xi ′) are the coordinates of a same point relative to different bases,in the latter the two vectors define the coordinates of different points relative to the samebasis. In particular, the matrix MS implementing the action of S on the coordinates (2.3.34)and on the components of a generic vector (2.3.32) is the constant Jacobian matrix of the

transformation MSij ≡ ∂xi ′

∂xj, so that just as dxi, also the components of a generic vector V i,

relative to (ei), transform as contravariant quantities: V i ′ = MSij V

j = ∂xi ′

∂xjV j. From now

on we shall not make any difference between homogeneous linear transformations on Mn andthe corresponding linear transformations on the corresponding Vn: Whatever we say for theformer holds true for the latter and vice-versa.

We have learned that the action of a linear homogeneous transformation on the coordinatevector is described by a constant n × n, non-singular matrix M ≡ (M i

j), which coincideswith its Jacobian matrix. The same transformation can be viewed as a linear transformation,an automorphism, on the corresponding linear vector space Vn and is represented on thecomponents of a generic vector by the action of the same constant matrix M. Let usshow that the set of linear homogeneous transformations is a subgroup of the group oftransformation TMn on Mn. Take two linear homogeneous transformations T, S, let us showthat T ·S−1 is still linear and homogeneous. Suppose S maps r into r′ = MS r, S−1 will mapr′ into r = M−1

S r′ and thus is realized by the matrix MS−1 = M−1S . Suppose now T maps r

into r′′ = MT r, then T · S−1 will map r′ into r′′ = MT r = (MT M−1S ) r′. We conclude that


T · S−1 is still a linear homogeneous transformation realized by the matrix MT M−1S , and

thus that linear homogeneous transformations close a subgroup of TMn . In Section 2.1.1 wehave shown that the group of automorphisms on Vn is isomorphic to GL(n,F), where F isR or C depending on whether Vn is a real or complex linear vector space. We conclude thatwe can identify GL(n,F) with the group of linear homogeneous transformations on Mn or,equivalently the group of linear transformations on Vn and that GL(n,F) is a subgroup ofTMn .

2.3.3 Affine Transformations

A more general form of transformations of Mn and on Vn are the affine ones which include,besides homogeneous linear transformations also translations. Affine transformations are nothomogeneous since the 0-vector in Vn is not mapped into itself or, equivalently, the originO of the RF in Mn is not invariant. Let us describe an affine transformation on Mn in thepassive view-point. Consider two generic coordinate systems in Mn: One with origin O andbasis (ei), the other with origin O′ and basis (e′i). Let the two bases be related as in eq.

(2.3.27). A point P is described by the vector−→OP = xi ei with respect to the former and

by−−→O′P = xi ′ e′i with respect to the latter RF. Let

−−→O′O = xi0 e′i be the position vector of O′

relative to O. From the relation:

−−→O′P =

−→OP +

−−→O′O , (2.3.35)

we derive the following relation between the new and old coordinates of P

xi ′ e′i = xi ei + xi0 e′i = xiM ji e′i + xi0 e′i . (2.3.36)

Equating the components of the vectors on the right and left hand side we find

xi ′ = xiM ji + xi0 . (2.3.37)

Let us introduce the column vectors of coordinates r′ ≡ (xi ′), r ≡ (xi), r0 ≡ (xi0) and thesquare non-singular matrix M ≡ (M i

j), so that we can rewrite the coordinate transformationas follows:

r′ = M r + r0 . (2.3.38)

The affine transformation, to be denoted by the couple (M, r0), is described by the n × nmatrix M and by the n-component vector r0 describing its translational component O → O′.Since r0 is a constant vector, independent of the point P , the coordinate differentials dr =(dxi) transform only through the homogeneous linear component of the affine transformation,according to (2.3.28), as the reader can verify by differentiating both sides of (2.3.38). Recallthat in the passive description adopted so far r′ and r are column vectors describing the samepoint P in two different RFs. In the active description they describe two different pointsP, P ′ with respect to the same RF.


On Vn we define an affine transformation as a correspondence between two column vectorsV and V′ of the form (2.3.38):

V′ = M V + a . (2.3.39)

This transformation will be denoted by the couple (M, a), a being a column vector a ≡ (ai).If r0, or a, describing the translation, are the zero-vector, the affine transformation reducesto a homogeneous linear transformation. Affine transformations close a group, as the readercan verify by applying two consecutive transformations on a same vector. Indeed the productof two affine transformations, the identity element and the inverse have the following form:

(M1, a1) · (M2, a2) = (M1 M2,M1 a2 + a1) ,

I = (1n, 0) ,

(M, a)−1 = (M−1,−M−1 a) . (2.3.40)

The homogeneous linear transformations on Mn close a subgroup of the affine transformationgroup.

Of particular interest in physics are the affine transformations on the Minkowski space-time M (1,3), which preserve the invariant distance ds2 in (2.2.8). These transformationsconsist in a combination of a Lorentz transformation which represents their homogeneouslinear component and of a translation, and close a group called the Poincare group.

2.3.4 The volume preserving group SL(n,F)

Consider now transformations on Mn, space of points over F, which preserve the volumeelement dV = dnx = dx1 . . . dxn. Under a generic transformation S mapping xi → xi ′(xj)the volume element transforms with the Jacobian, namely with the determinant of theJacobian matrix:

dV → dV ′ = J(xi) dV , J(xi) ≡ det

(∂xi ′

∂xj

)= det(MS) . (2.3.41)

Volume preserving transformations are those transformations under which dV ′ = dV andthus are described by a Jacobian matrix M with unit determinant: det(MS) = 1.

For general transformations the Jacobian matrix depends on the point through the co-ordinates (xi). For homogeneous linear transformations the Jacobian matrix is constant.Volume preserving homogeneous linear transformations on Mn are therefore represented byn×n matrices with entries in F and unit determinant. These transformations therefore closethe group SL(n,F).

2.3.5 (Pseudo-) Orthogonal Transformations

Transformation property of a metric Consider a real metric space M (p,q). On it thenotion of metric (i.e. of distance) is defined by means of a symmetric scalar product (·, ·)


on the linear vector space Vn over R. Consider two infinitely close points and let dr = (dxi)denote the components of the relative position vector. The squared distance ds2 betweenthe two points is given by eq. (2.2.5). Consider now a transformation S under which drtransforms with the corresponding Jacobian matrix MS as in eq. (2.3.4). In the passivedescription the points do not change and neither does their distance. We can thus write thesame distance squared with respect to dxi and dxi ′ as follows:

ds2 = dxi gij dxj = dxk ′

∂xi

∂xk ′gij

∂xj

∂x` ′dx` ′ = dxk ′ g′k` dx

` ′ , (2.3.42)


ds2 = drT g dr = dr′T(M−T

S g M−1S

)dr′ = dr′T g′ dr′ . (2.3.43)

With respect to the new coordinate differentials, the distance is expressed in terms of a newmetric g′ = (g′ij) related to the old one g = (gij) as follows:

g′k` ≡∂xi

∂xk ′gij

∂xj

∂x` ′⇔ g′ ≡M−T

S g M−1S . (2.3.44)

Equation (2.3.44) defines the transformation property of a metric with respect to a generictransformation.

A transformation S is said to be metric preserving if ds2 has the same expression interms of the new and old coordinates. This means that g′ij = gij, that is:

gk` =∂xi

∂xk ′gij

∂xj

∂x` ′⇔ g = M−T

S g M−1S ⇔ g = MT

S g MS . (2.3.45)

Covariant components of a vector Consider now a generic transformation S : xi →xi ′(xj) on M (p,q) not necessarily homogeneous linear or metric preserving. The metric trans-forms as in (2.3.44) while the components V = (V i) of a generic vector transform as con-

travariant quantities: V i ′ = M ij V

j = ∂xi ′

∂xjV j. Let us define the following quantities:

Vi ≡ gij Vj , (2.3.46)

called covariant components of the vector, obtained by lowering the upper index of thecontravariant components by means of the metric gij. Let us prove that these quantitiestransform indeed as covariant quantities, namely as the components of a gradient (2.3.19):

Vi ≡ gij Vj → V ′i = g′ij V

j ′ =

(∂xk

∂xi ′gk`

∂x`

∂xj ′

) (∂xj ′

∂xsV s

)=∂xk

∂xi ′gkj V

j =∂xk

∂xi ′Vk ,

where we have used the usual property of Jacobian matrices: ∂x`

∂xj ′∂xj ′

∂xs= δ`s.

If we write a vector as V = V i ei, using the linearity and the symmetry properties of theinner product we find:

(V, ei) = V j (ej, ei) = gij Vj = Vi . (2.3.47)


Consider an Euclidean space Mn = En in which the metric gij is positive definite, andwe can choose an ortho-normal basis in which gij = δij (p = n, q = 0 case). Recallingthe geometrical meaning of the scalar product on En, Vi are nothing but the orthogonalprojection of the vector V along the direction ei and coincide with V i.

The scalar product between two vectors can also be written in the following way:

(V, W) = V i gijWj = VjW

j = V iWi . (2.3.48)

It is natural to think of the covariant components of a vector as entries of a row vector, sothat the inner product (2.3.48) can be written in a nice matrix notation:

(V, W) = (V1, V2, . . . , Vn)

W 1

W 2

...W n

= (W1, W2, . . . ,Wn)

V 1

V 2

...V n

. (2.3.49)

Notice that, under any transformation:

ViWi = V ′i W

i ′ , (2.3.50)

that is the expression (2.3.48) of the scalar product in terms of the covariant-contravariantcomponents of the two vectors is the same before and after the transformation. Its expressionin terms of the covariant or contravariant components only is not, as previously explained.

(Pseudo)-Orthogonal transformations Consider now homogeneous linear transforma-tions on M (p,q). We have learned that we can always choose an orthonormal basis vectors ofVn in the definition of the RF so that g = ηp,q. The constant real matrix MS representingthe action of a metric preserving homogeneous linear transformation S will then satisfy thefollowing property:

M−TS ηp,q M−1

S = ηp,q ⇔ MTS ηp,q MS = ηp,q . (2.3.51)

The matrix MS is then a pseudo-orthogonal matrix.We conclude that metric preserving linear homogeneous transformations on a real space

M (p,q) close the a group O(p, q).In an Euclidean space, in which p = n and ηp,q = 1n, it is easy to show that, in an

ortho-normal basis, if M is an orthogonal transformation on Vn (and thus MT is), for anyvectors V and W:

(V, M W) = (MT V, MT M W) = (MT V, W) , (2.3.52)

where we have used MT M = 1n.As an example consider the space M (1,1) on which the metric reads:

ds2 = (dx0)2 − (dx1)2 = drT g dr , (2.3.53)


where the metric g =

(1 00 −1

). Consider the transformation represented by the following

constant matrix M:

M =

(cosh(a) sinh(a)sinh(a) cosh(a)

), (2.3.54)

so that

dx0 ′ = cosh(a) dx0 + sinh(a) dx1 ,

dx1 ′ = sinh(a) dx0 + cosh(a) dx1 . (2.3.55)

This transformation is a symmetry of the metric (2.3.53) as the reader can verify by substi-tuting the expression of dxµ in terms of dxµ ′ and substituting them in the metric ds2. Oneindeed finds:

ds2 = (dx0 ′)2 − (dx1 ′)2 , (2.3.56)

that is the metric in the new coordinate differential has the same expression as in the oldones. The matrix M is in fact in SO(1, 1).

Consider now the action of linear transformations on Vn. Given two vectors, whosecomponents with respect to an orthonormal basis are described by V = (V i) and W = (W i)respectively, their scalar product reads:

(V, W) = VT ηp,q W . (2.3.57)

The value of the scalar product depends only on the two vectors on which it is computedand not on their description. In the passive representation of a linear transformation, thevalue of the scalar product is the same if computed in terms of the components V and W ofthe two vectors in the old basis or in terms of the components V′ = M V and W′ = M Wof the same vectors in the new basis:

(V, W) = VT ηp,q W = V′T M−T ηp,q M−1 W′ . (2.3.58)

For a generic matrix M the product has a different expression in terms of the old andnew components: (V, W) = VT ηp,q W 6= V′T ηp,q W′ = (V′, W′). The transformationpreserves the inner product iff the expression of the product in terms of the old components(V, W) coincides with that in terms of the new ones:

(V′, W′) = (V, W) , (2.3.59)

forany V and W, or, in other words:

(M V, M W) = (V, W) . (2.3.60)

Using (2.3.58) the above condition amounts to requiring that:

MT ηp,q M = ηp,q , (2.3.61)

namely that M ∈ O(p, q). We conclude that the orthogonal (or pseudo-orthogonal) group isthe group of linear transformations on Vn preserving a symmetric inner product.


2.3.6 Unitary Transformations

Consider now a complex space of points Mn on whose complex linear vector space Vn ahermitian inner product (·, ·) is defined. Consider a linear transformation on Vn, representedby a constant complex matrix M = (M i

j), and two vectors with (complex) componentsV = (V i), W = (W i) before and V′ = M V, W′ = M W after the transformation. Thetransformation preserves the hermitian inner product if the latter has the same form whenexpressed in terms of the old and new components, for any couple of vectors:

(V,W) = V i ∗ gijWi = V i ′ ∗ gijW

i ′ = (V′,W′) = (M V, M W) . (2.3.62)

Substituting in the above equation the expression of the new components in terms of the oldones and requiring the equality to hold for any (V i) and (W i) we find:

(M ik)∗ gijM

j` = gk` ⇔ M† g M = g . (2.3.63)

If we choose an ortho-normal basis, gij = δij, and thus g = 1n, equation (2.3.63) becomes:

M†M = 1n , (2.3.64)

that is M is a unitary matrix. The group U(n) consists of all the transformations whichleave a hermitian inner product invariant.

From eq. (2.3.62), written in a ortho-normal basis, it follows that, if M is a unitarymatrix (and thus M† is), for any V and W in Vn:

(V, M W) = (M†V, M†M W) = (M†V, W) . (2.3.65)

2.3.7 Homomorphism between SU(2) and SO(3)

At this point we are ready to consider an important instance of homomorphism between Liegroups: That between SU(2) and SO(3). We define a mapping between elements of the twogroups as follows. Consider an element (2 × 2 complex matrix) S = (Sab), a, b = 1, 2, ofSU(2) and its adjoint action on the Pauli matrices defined in (2.1.37): S−1 σi S = S† σi S,i, j = 1, 2, 3. Since the Pauli matrices form a basis for hermitian traceless matrices, resultingmatrix is still hermitian traceless:

(S† σi S)† = S† σ†i S = S† σi S , Tr(S† σi S) = Tr(SS† σi) = Tr(σi) = 0 . (2.3.66)

Therefore S† σi S can be expanded in the basis (σi). Let us denote by R[S]ij the components

along σi of S† σi S:

S† σi S = R[S]ij σj . (2.3.67)

Since R[S] ≡ (R[S]ij) is a 3× 3 matrix, we have thus defined a correspondence which maps

a 2 × 2 matrix S of SU(2) into a 3 × 3 matrix R[S]. We want to show first that thiscorrespondence is a homomorphism, namely that R[S1 S2]i

j = R[S1]ik R[S2]k

j:

(S1 S2)† σi (S1 S2) = S†2 (S†1 σi S1) S2 = R[S1]ik (S†2 σk S2) = R[S1]i

k R[S2]kj σj =

= (R[S1] R[S2])ij σj . (2.3.68)


Let us prove now that the matrix R[S] is real by computing the hermitian-conjugate of bothsides of eq. (2.3.67) and using the property that the left hand side is hermitian:

S† σi S = (S† σi S)† = (R[S]ij)∗ σj . (2.3.69)

Since the components associated with any vector (in this space vectors are hermitian ma-trices!) are unique, comparing (2.3.69) to (2.3.67) we find: (R[S]i

j)∗ = R[S]ij. Using the

definition of inner product for hermitian 2× 2 matrices we can write

R[S]ij =

1

2Tr[(S† σi S)σj] , (2.3.70)

Finally let us show that the matrix R[S] is orthogonal. To this end we use the generalproperty of homomorphisms that: R[S−1] = R[S]−1 and write

R[S]−1ij = R[S†]i

j =1

2Tr[(Sσi S

†)σj] =1

2Tr[(S† σj S)σi] = R[S]j

i , (2.3.71)

where we have used the cyclic property of the trace. We conclude that R[S]−1 = R[S]T ,which means that R[S] ∈ O(3). Let us show that R[S] ∈ SO(3), namely that det(R[S]) = 1.To show this let us use the property that σ1 σ2 σ3 = i12. Then, from unitarity of S it followsthat:

12 = S† S = −iS† σ1 σ2 σ3 S = −i (S† σ1 S) (S† σ2 S) (S† σ3 S) =

= −i (R[S]1i σi) (R[S]2

j σj) (R[S]3k σk) = −i R[S]1

iR[S]2j R[S]3

k (σi σj σk) .(2.3.72)

Now use eq. (2.1.41) to rewrite σi σj σk. Notice that the terms with the δ matrix do not con-tribute because of the orthogonality property of R: R[S]k

iR[S]`j δij =

∑3i=1R[S]k

iR[S]`i =

δk`, which is zero if k 6= `. The only term in σi σj σk which contributes to the summation isi εijk 12, and therefore we can rewrite eq. (2.3.72) as follows:

R[S]1iR[S]2

j R[S]3k εijk 12 = 12 . (2.3.73)

We recognize in the sum R[S]1iR[S]2

j R[S]3k εijk the expression of the determinant of a

matrix in terms of its entries and therefore we conclude that:

det(R[S]) = 1 , (2.3.74)

namely that R[S] ∈ SO(3). We have thus defined a homomorphism between SU(2) andSO(3):

S ∈ SU(2)R−→ R[S] ∈ SO(3) . (2.3.75)

This homomorphism is two-to-one. Indeed, recalling the notation introduced in Section 1.5of the first chapter, let us compute the normal subgroup E of SU(2) consisting of all theelements which are mapped into the identity 13 of SO(3):

S ∈ E ⊂ SU(2) ⇔ S† σi S = σi (2.3.76)


The last condition implies that S commutes with all the Pauli matrices, i.e. Sσi = σi S, forall i = 1, 2, 3. The reader can convince himself that this condition can only be satisfied byS = µ12. From 1 = det(S) = µ2 it follows that µ = ±1. We conclude that E = 12, −12,namely E is the cyclic group of period 2, also denoted by Z2. Since σi are also SU(2)matrices, E = Z2 is also the center C of SU(2). From Property 1.4 of Section 1.5 it followsthat

SU(2)/E = SU(2)/Z2 ∼ SO(3) . (2.3.77)

Exercise 2.4: Consider the most general SU(2) matrix given in Exercise 1.11 of the firstchapter:

S[σ, α, β] =

(cos(σ) ei α sin(σ) ei β

− sin(σ) e−i β cos(σ) e−i α

), (2.3.78)

and let us construct the corresponding generic element R[σ, α, β] of SO(3). Observe that ifα→ α±π and β → β±π, the matrix in (2.3.78) changes sign: S→ −S. However, accordingto our previous discussion, the corresponding SO(3) matrix should remain unchanged, sinceboth S and −S correspond to the same orthogonal matrix. Using (2.3.70) prove that:

R[σ, α, β] =

cos(2α) cos2(σ)− cos(2β) sin2(σ) cos2(σ) sin(2α) + sin(2β) sin2(σ) − cos(α+ β) sin(2σ)−cos2(σ) sin(2α) + sin(2β) sin2(σ) cos(2α) cos2(σ) + cos(2β) sin2(σ) sin(α+ β) sin(2σ)

cos(α− β) sin(2σ) sin(α− β) sin(2σ) cos(2σ)

.(2.3.79)

We see that R[σ, α±π, β±π] = R[σ, α, β]. R[σ, α, β] describes the most general rotation ofa system of Cartesian axes in the three dimensional Euclidean space: e′i = R[σ, α, β]i

j ej. Itsparameters σ, α, β are related to the well known Euler angles (θ, ψ, φ) as follows: σ = 1

2θ,

α = 12

(ψ + φ) and β = 12

(ψ − φ), with θ ∈ [0, π], ψ, φ ∈ [0, 2π].

2.3.8 Symplectic Transformations

Let M2n be a space of points on whose linear vector space V2n over F, a symplectic productis defined. Consider a linear transformation over V2n, represented by the 2n × 2n matrixM = (M i

j) with entries in F, which leaves the product invariant. Just as for the symmetricand hermitian inner product, this means that, for any two vectors V = (V i) and W = (W i):

(M V, M W) = (V, W) . (2.3.80)

Using the definition (2.1.30), the above condition implies:

M ik ΩijM

j` = Ωk` ⇔ MT Ω M = Ω . (2.3.81)

Choosing a basis of V2n in which Ω = (Ωij) has the form (2.1.32), condition (2.3.81) impliesthat M is a symplectic matrix. The group Sp(n,F) consists of all the transformations whichleave a skew-symmetric (i.e. symplectic) product invariant.


2.3.9 Active Transformations in Different Bases

Consider a linear vector space Vn over F and two bases (ei), (fi), related by a transformationmatrix A = (Aij):

fi = Aj i ej . (2.3.82)

Now consider an active transformation S mapping a generic vector of components V = (V i)relative to (ei) into a new vector of components V = (V i) = MS V relative to the samebasis. We wish to describe the action of S relative to the basis (fi). Consider a vectorwith components V i ′ relative to (fi). Its components relative to (ei) are V i = Aij V

j ′.This vector is mapped by means of S into an other one with components V i = MS

ij V

j =MS

ij A

jk V

k ′ relative to (ei). Now let us express the new vector back in the basis (fi). Thecorresponding components will be V i ′ = A−1 i

j Vj = (A−1 MS A)ij V

j ′. The transformationS will then map a vector with components V′ = (V i ′) relative to (fi) into a different vectorwith components V′ = (V i ′) relative to the same basis, where:

V′ =(A−1 MS A

)V′ . (2.3.83)

To summarize, if the active action of a linear transformation S in the basis (ei) is representedby a matrix MS, the action of the same transformation in the basis (fi) related to the formeras in (2.3.82) is represented by the matrix A−1 MS A, obtained from MS through the adjointaction of the matrix A, also called similarity transformation. In other words, the effect ofa change of basis on the matrix representing an active transformation is described by asimilarity transformation.

2.4 Realization of an Abstract Group

Given an abstract group G a realization of G on a point-field M is a mapping T betweenthe elements of G and elements of TM , i.e. transformations on M :

T : g ∈ G −→ T [g] ∈ TM , (2.4.1)

such that, given two elements g1 g2 ∈ G, their product corresponds to the product of thetransformations associated with each of them:

T [g1 · g2] = T [g1] · T [g2] . (2.4.2)

In other words a realization is a homomorphism between an abstract group G and thegroup TM of transformations on M . As we have learned in the first chapter, when discussingabout homomorphisms between groups, the following properties hold: T [e] = I and T [g−1] =T [g]−1.

The realization is called faithful if it is one-to-one, namely if T [g1] = T [g2] implies g1 = g2.As shown in Section 1.5, this is equivalent to requiring that T [g] is the identity transformationI only if g = e.

2.4. REALIZATION OF AN ABSTRACT GROUP 69

Let now T be a faithful realization of G, then the image T [G] of G through T is agroup of transformations which is isomorphic to G (G ∼ T [G]), namely they have the samestructure. With an abuse of notation we shall call T [G], and not just the mapping T , afaithful realization of G. Given a transformation group G we can always define an abstractgroup G of which G is the image through a faithful realization T : G = T [G] ∼ G. If we havetwo transformation groups G and G′ which are isomorphic G ∼ G′, and thus share the samestructure, we can view them as the faithful realizations of a same abstract group. Indeed theabstract groups associated with G and G′ are themselves isomorphic, but, being abstractgroups made just of symbols, they can be identified. G and G′, on the other hand, in spiteof being isomorphic, cannot be identified since they are represented in general by differentobjects (transformations on a point-field). The notion of faithful realization allows us then tocapture the common structure of all isomorphic transformation groups into a single abstractgroup. We can then view an abstract group as an abstraction of a transformation group inwhich the single transformations are replaced by a symbol, and which has its same structure.For instance we can associate with each linear transformation group (GL(n,F), O(n,F), U(n)etc...) on a linear vector space its own abstract group to be denoted by the same symbol.As we shall see such abstract groups can have various other faithful realizations which aredifferent from the transformation groups we started from and which we used to define them(defining realizations).

Let us fist pose the question: Given an abstract group G, does it always admit a faithfulrealization over some point-field? The answer is positive. Indeed we can take as point-fieldthe group G itself, take an element a ∈ G and consider the mapping T [a] of G into itself:

g ∈ G −→ T [a](g) ≡ a · g ∈ G (2.4.3)

T is called left translation and is one-to-one. Indeed suppose for some a ∈ G T [a] is theidentity mapping T [a] = I. This means that, for any g ∈ G, T [a](g) ≡ a ·g = g. Multiplyingboth sides to the right by g−1 we find a = e. It is a homomorphism: Take a1, a2 ∈ G, forany g ∈ G, T [a1 · a2](g) ≡ (a1 · a2) · g = a1 · (a2 · g) = T [a1](T [a2](g)) = (T [a1] · T [a2])(g).We conclude that T is a faithful realization of G.

Let us give now an example of a non-faithful realization of G. For any a ∈ G let us defineT [a] as the mapping of G into itself which consists in the conjugation with respect to a:

g ∈ G −→ T [a](g) = αa(g) ≡ a · g · a−1 ∈ G . (2.4.4)

We have considered this realization in last chapter, αa being an inner automorphism of G,and we have shown that it is indeed a homomorphism. However it is not faithful in generalsince all the elements of the center C of G are mapped into the identity transformation:T [C] = I. The group of inner automorphisms IG of G is a faithful realization of G/C.

In general, given a realization T of an abstract group G, if we denote by E the normalsubgroup of G which is mapped into the identity transformation, T [E] = I, see Section1.5, as proven in the same section, the factor group G/E is isomorphic to T [G], G/E ∼ T [G],namely T is a faithful realization of G/E.

Given an abstract group G the mapping between any of its elements and the identitytransformation I on some point-field is clearly a non-faithful realization called the identity


representation and denoted by 1. In this case E = G and G/E consists of the identity cosetonly.

Property 2.1: Any finite group G of order n is realized by a subgroup of the permutationgroup Sn.

To show this let us define the permutation which realizes a generic group element g ∈ G.Let g1, g2, . . . , gn the generators of G and consider the action on them of the left translationT [g] by g. This transformation maps a generic element gi into T [g](gi) = g · gi which mustcoincide with some other element gki . This correspondence is one-to-one, as we have alreadyshown. The effect of T [g] on the group elements is therefore to permute them and so we candescribe it by means of a permutation:

g ∈ G −→ T [g] ≡(

g1 g2 . . . gng · g1 g · g2 . . . g · gn

)=

(g1 g2 . . . gngk1 gk2 . . . gkn

)∈ Sn .

(2.4.5)

If g, g′ ∈ G, the realization of their product is the product of the permutations realizingeach of them. In formulas:

T [g · g′] ≡(

g1 g2 . . . gn(g · g′) · g1 (g · g′) · g2 . . . (g · g′) · gn

)=

=

(g′ · g1 g′ · g2 . . . g′ · gn

g · (g′ · g1) g · (g′ · g2) . . . g · (g′ · gn)

)·(

g1 g2 . . . gng′ · g1 g′ · g2 . . . g′ · gn

)=

= T [g] · T [g′] . (2.4.6)

This shows that the correspondence (2.4.5) is a homomorphism and thus a realization. It isfaithful since the left translation is.

2.4.1 Representations

A realization of an abstract group G in terms of homogeneous linear transformations on aspace of points Mn, or, equivalently transformations on a linear vector space Vn, is calledn-dimensional representation. A representation D is therefore a mapping which associateswith each element g of the group G an invertible transformation (automorphism) D[g] on Vn,named the representation space or base space, whose action is described by a n × n matrixD[g] ≡ (D[g]ij), i, j = 1, . . . , n:

g ∈ GD−→ D[g] ∈ Aut(Vn) , (2.4.7)

where

D[g] : V ∈ Vn → D[g](V) = D[g] V ∈ Vn , (2.4.8)


or, in components,

V i D[g]−→ V ′ i = D[g]ij Vj . (2.4.9)

We shall characterize a representation D equivalently as the homomorphism D between anabstract group G and GL(n,F) which associates with each g ∈ G an invertible matrix D[g] ≡(D[g]ij) ∈ GL(n,F). Being D a homomorphism we must gave that, for any g1, g2 ∈ G:

D[g1 · g2] = D[g1] · D[g2] ⇔ D[g1 · g2] = D[g1] D[g2] . (2.4.10)

Depending on whether the representation space is defined over the real or complex numbers,i.e. if F equals R or C, the representation is said to be real or complex.

Example 2.6: Any cyclic group G of finite period n and generator g admits the followingfaithful representation on C (which is a one dimensional vector space over C):

g −→ D[g] = e2πin ⇔ gk −→ D[gk] = D[g]k = e

2πikn . (2.4.11)

The group D[G] = e 2πikn , k = 1, . . . , n, representing the generic cyclic group G is denoted

by Zn. An example of G is the group Cn of rotations about a point on the plane by anglesmultiple of 2π/n. Each of these rotations are realized by a phase in Zn.

Given a representation D of a group G over a representation space Vn and a non singularmatrix A, also the mapping D = A−1 D A, defined as:

g −→ D[g] = A−1 D[g] A , (2.4.12)

is a representation. As shown in Subsection 2.3.9, the adjoint action of a non singular matrixA represents the effect on the matrix representing a transformation over Vn, of a change ofbasis. Therefore D can be viewed as representing the abstract elements of G by means ofthe same transformations as D but expressed in a different basis of Vn. The representationsD and D, related by the adjoint action of a non-singular matrix, are called equivalent.

We can consider representing a group G in terms of transformations acting on covariantvectors (Vi). We have learned that, if a column vector of contravariant components V = (V i)transform through the left action of a matrix M = (M i

j), a covariant vector (Vi), seen as arow vector, transforms through the right action of M−1, as in the basis elements in (2.3.29),or, as a column vector, through the left action of M−T . Therefore if we have a representationD of G by means of transformations on contravariant vectors:

g ∈ G −→ D[g] : V i → D[g]ij Vj , (2.4.13)

we can associate with it a representation D′ as transformations on covariant vectors:

g ∈ G −→ D′[g] = D[g]−T : Vi → D′[g]ij Vj = (D[g]−T )i

j Vj . (2.4.14)

As previously mentioned, we can associate with each matrix group G defined in section 1.7,an abstract group G, by associating with each matrix M in G a unique symbol in g ∈ G


such that M = D[g], and requiring G to have the same structure as G. By construction thiscorrespondence:

g ∈ G −→ D[g] ∈ G , (2.4.15)

is a representation, which defines the abstract group G and thus is called defining represen-tation or fundamental representation. Representations of Lie groups are often denoted bythe boldface of their dimension, so that the defining representations D in terms of n × nmatrices are dented by n. The same group G will have infinite other representations. Wewill denote the abstract group associated with the matrix group by the same symbol, sothat, for instance SU(n) will also denote the abstract group associated with the group ofn× n special unitary matrices.

Example 2.7: The homomorphism (2.3.75) between SU(2) and SO(3) is an example ofrepresentation in which a generic element S of SU(2) is represented by a real 3×3 orthogonalmatrix R[S]. As previously discussed, the subgroup E which is mapped into the identitymatrix 13 is Z2 = 12, −12, so that R is a faithful representation of SU(2)/E = SU(2)/Z2.R is a representation of SU(2) over the Euclidean space E3 and is denoted by the boldface ofits dimension: R ≡ 3. This is clearly different from the defining representation 2 of SU(2),which is complex.

Example 2.8: Given a representation D of a group G over a space Vn over F (which canbe either R or C), an example of non-faithful representation is the determinant of D, overV1 = F:

g ∈ G −→ det(D[g]) ∈ F . (2.4.16)

It clearly is a representation since det(D[g1 · g2] = det(D[g1] D[g2]) = det(D[g1]) det(D[g2]).Consider the representation of a group G in terms of unitary matrices, unitary represen-

tation, on contravariant vectors V i:

g ∈ G −→ D[g] = (D[g]ij) ∈ U(n) , (2.4.17)

The corresponding representation on covariant vectors D′ associates with each g ∈ G amatrix D′[g] = (D′[g]i

j) = D[g]−T . Using the unitarity of D[g], i.e. D[g]−1 = D[g]† wefind D[g]−T = (D[g]†)T = D[g]∗ = ((D[g]ij)

∗), that is the matrices D′[g] are simply thecomplex conjugate of the matrices D[g] and the representation D′ is also denoted by Dand called the representation conjugate to D. We learn that, if a vector V = (V i) in acomplex linear vector space Vn transforms as a contravariant vector with respect to a groupof unitary transformations, its complex conjugate V∗ = (V i ∗) transform as a covariantvector: (V i)∗ = Vi. As far as orthogonal representations are concerned, since D[g]−T = D[g],covariant and contravariant vectors transform in the same representation: D′ = D.

Remark: The only unitary group whose defining representation n coincides with itsconjugate n is SU(2). Indeed the matrix D[g] = S[θ, α, β] corresponding to a generic SU(2)


element g in the defining representation D ≡ 2 is given in eq. (2.3.78) as a function of thethree parameters defining g. The reader can verify that:

S[σ, α, β]∗ = σ−12 S[σ, α, β]σ2 , (2.4.18)

namely the defining representation is equivalent to its conjugate: 2 ∼ 2. The matrix Adefining the equivalence is the unitary matrix σ2.

Consider a representation D of an abstract group G over a linear vector space Vn. Asubspace Vk of Vn, for some k < n is called invariant or stable if the action of the transfor-mations in D[G], representing G, maps vectors in Vk into vectors in Vk. A representationis said to be irreducible if the space Vn it acts on does not admit an invariant subspaceaside from the trivial one consisting of the zero-vector 0. A representation which is notirreducible is said to be reducible. Consider a reducible representation D over Vn and let Vkbe an invariant subspace. We can consider a basis (ei) of Vn so that the first k elements ea,a = 1, . . . , k, span Vk, while the last n− k elements e`, which are not in Vk, will be labeledby an index ` = k + 1, . . . , n. The subspace Vk consist of column vectors with componentsV a only along the ea vectors:

V ≡

V 1

...V k

V k+1

...V n

=

(V a

V `

)∈ Vn , V ∈ Vk ⇔ V ≡

V 1

...V k

0...0

=

(V a

0

)∈ Vk .

(2.4.19)

The transformation D[g] representing a generic G-element g has to map a column vector ofthe form V′ in (2.4.19) into a vector of the same form. It therefore, as a n× n matrix, willhave the following characteristic block structure:

g ∈ G −→ D[g] =

(Dk[g] B[g]k,n−k0n−k,k An−k[g]

), (2.4.20)

The action of D[g] on a generic Vn vector V = (V i) = (V `, V a) is:

V → V′ = D[g] V ⇔V ` ′ = A[g]``′ V

`′

V a ′ = B[g]a` V` +Dk[g]ab V

b . (2.4.21)

If V ∈ Vk, we have V ` = 0 and therefore also V′ will have components only in Vk: V` ′ = 0,

V a ′ = Dk[g]ab Vb. The k × k block Dk[g], for different g in G, defines a k-dimensional

representation Dk of G over the smaller subspace Vk.The representation D is called completely reducible if there exist a basis of Vn on which

the matrix representation D[g] of a generic g ∈ G has a block diagonal form, namely a form(2.4.20) with B[g]k,n−k = 0k,n−k. In this case Vn can be written as the direct sum of twoinvariant subspaces Vk and Vn−k: Vn = Vk ⊕ Vn−k. The block An−k[g], for a generic g ∈ G,


defines a (n − k)-dimensional representation Dn−k of G over Vn−k. We shall then say thatthe original representation D decomposes into the direct sum of Dk and Dn−k and write:

D → Dk ⊕Dn−k . (2.4.22)

The representations Dk, Dn−k may still be completely reducible, and thus can be furtherdecomposed into lower dimensional representations. We can iterate the above procedureuntil we end up with irreducible representations : Dk1 , . . . ,Dk` , where

∑`i=1 ki = n. This

corresponds to finding a basis in which the matrix representation under D of a genericelement g ∈ G has the following block structure:

D[g] =

Dk1 [g] 0 · · · 0

0 Dk2 [g] · · · 0...

. . ....

0 0 · · · Dk` [g]

. (2.4.23)

We say that the original representation D is completely reducible into the irreducible repre-sentations Dki and write:

D −→⊕i=1

Dki ≡ Dk1 ⊕Dk2 ⊕ . . .Dk` . (2.4.24)

Correspondingly the representation space Vn of D has been decomposed into the direct sumof spaces Vki on which Dk1 [g] act:

Vn = Vk1 ⊕ Vk2 ⊕ . . . Vk` , (2.4.25)

and the basis (ei) in which D[g], for any g ∈ G, has the form (2.4.23), is the union of basesin each of the subspaces: (ei) = (ea1 , ea2 , . . . , ea`), where ai = 1, . . . , ki and (eai) is a basisof Vki .

To summarize, a representation D is completely reducible if it is equivalent to a repre-sentation which, for any group element, has the matrix block diagonal form (2.4.23).

Example 2.9: Consider the representation of the group SO(2) as rotations in the XYplane of the three dimensional Euclidean space:

x → x′ = x cos(θ) + y sin(θ) ,

y → y′ = −x sin(θ) + y cos(θ) ,

z → z′ = z .

The above transformation is implemented by a 3× 3 real matrix Dz[g(θ)], representing thecorresponding abstract element g(θ) of SO(2)

g(θ) ∈ SO(2) −→ Dz[g(θ)] = Rz[θ] =

cos(θ) sin(θ) 0− sin(θ) cos(θ) 0

0 0 1

. (2.4.26)


The above representation is clearly completely reducible since there is a subspace of theEuclidean space, namely the Z axis, which is left invariant and the orthogonal complement,which is the XY plane, also defines an invariant subspace. The action along the Z axis is theidentity representation 1 while the action on the XY plane, defined by the first 2× 2 blockof the above matrix, is the defining the (irreducible) representation 2 of SO(2). Denotingthe representation D by its dimension 3 we will then write:

3 −→ 1⊕ 2 . (2.4.27)

Similarly we can define other two 3-dimensional representations of SO(2), Dx, Dy, describingrotations about the X and the Y axes:

Dx[g(θ)] = Rx[θ] =

1 0 00 cos(θ) sin(θ)0 − sin(θ) cos(θ)

, (2.4.28)

Dy[g(θ)] = Ry[θ] =

cos(θ) 0 − sin(θ)0 1 0

sin(θ) 0 cos(θ)

, (2.4.29)

The three representations Dx, Dy, Dz are equivalent as the reader can show by verifyingthat: Dz = A−1

1 Dx A1 = A−12 Dy A2, where:

A1 =

0 0 10 −1 01 0 0

, A2 =

−1 0 00 0 10 1 0

, (2.4.30)

and therefore can be identified with the same (reducible) representation 3 of SO(2).Question: Are A1 and A2 is O(3)? Are they in SO(3)?

Example 2.10: The group D4 Let us now consider a representation of the symmetrygroup D4 of a square, discussed in Section 1.2.3. Let the square lie on the XZ plane, becentered in the origin, and have its sides parallel to the axes. We wish to represent theaction of the D4 elements in terms of matrices acting on the three-dimensional Euclideanspace. Since all elements of D4 are expressed as products of its two generators r, σ, it isconvenient to start representing the action of these two transformations. Recall that r is a90o clockwise rotation about the Y axis. It is illustrated in Figure 2.4 and represented bythe following homogeneous linear transformation on the three coordinates:

x′ = −z , y′ = y , z′ = x . (2.4.31)

Writing r′ = D[r] r, we deduce the form of the matrix associated with this transformation:

D[r] =

0 0 −10 1 01 0 0

. (2.4.32)


Figure 2.4: Active and passive representation of a clockwise rotation r by 90o about the Yaxis

As expected, since this is a particular rotation about the Y axis, it coincides with Ry[θ] in(2.4.29), for θ = π

2. As far as the reflection σ is concerned, the corresponding homogeneous

linear transformation reads:

x′ = −z , y′ = y , z′ = −x , (2.4.33)

from which it follows that:

D[σ] =

0 0 −10 1 0−1 0 0

. (2.4.34)

Notice that D[e] = D[σ2] = D[σ]2 = 13. Since the elements of D4 are e, r, r2, r3, σ, α =(r · σ), β = (r2 · σ), γ = (r3 · σ), using the homomorphism property of a representation, wemay deduce the matrix representation of all the D4 elements, which we list below:

D[e] =

1 0 00 1 00 0 1

, D[r] =

0 0 −10 1 01 0 0

, D[r2] = D[r]2 =

−1 0 00 1 00 0 −1

,

D[r3] =

0 0 10 1 0−1 0 0

, D[σ] =

0 0 −10 1 0−1 0 0

, D[α] =

1 0 00 1 00 0 −1

,

D[β] =

0 0 10 1 01 0 0

, D[γ] =

−1 0 00 1 00 0 1

. (2.4.35)


Figure 2.5: Active representation of the reflection σ in the z + x = 0 plane

Let us notice that all rotations (e, r, r2, r3) are represented by SO matrices. In fact therotations close the subgroup C4 of SO(2), and are all obtained by computing the SO(2)matrix in (2.4.29) for θ = 0, π

2, π, 3π

2. The reflections are still orthogonal, but have negative

determinant, and thus are only in O(2). Moreover the matrices in (2.4.35) have a character-istic block structure (2.4.23), provided we order the basis elements as (e2, e1, e3), namelywe perform a change of basis described by:

A =

0 1 01 0 00 0 1

. (2.4.36)

The representation is therefore completely reducible in a representation D2 acting on thesubspace e1, e3, that is on the XZ plane, and a one-dimensional representation acting one2, that is on the Y axis, which is the trivial or identity representation. Let us write, for thesake of completeness, the D4 elements in the two dimensional representation:

D2[e] =

(1 00 1

), D2[r] =

(0 −11 0

), D2[r2] =

(−1 00 −1

),

D2[r3] =

(0 1−1 0

), D2[σ] =

(0 −1−1 0

), D2[α] =

(1 00 −1

),

D2[β] =

(0 11 0

), D2[γ] =

(−1 00 1

). (2.4.37)

The identity representation of D4 is usually denoted by A1, while the two dimensional oneby E. Another representation of D4 is the determinant of the above matrices in the D2


representation, usually denoted by the symbol A2:

A2 : det(D2[e]) = det(D2[r]) = det(D2[r2]) = det(D2[r3]) = 1 ,

det(D2[σ]) = det(D2[α]) = det(D2[β]) = det(D2[γ]) = −1 . (2.4.38)

The determinant representation, just like the trivial identity representation, is real one-dimensional. Rotations are represented by the number +1, reflections by −1. It clearly isnon-faithful, so we are losing information about the group. For D4 the reader can verify thatwe have two further one-dimensional representations, denoted by B1 and B2 respectively:

B1 : DB1 [e] = DB1 [r2] = DB1 [r · σ] = DB1 [r

3 · σ] = 1 ,

DB1 [r] = DB1 [r3] = DB1 [σ] = DB1 [r

2 · σ] = −1 , (2.4.39)

B2 : DB2 [e] = DB2 [r2] = DB2 [σ] = DB2 [r

2 · σ] = 1 ,

DB2 [r] = DB2 [r3] = DB2 [r · σ] = DB2 [r

3 · σ] = −1 . (2.4.40)

Therefore for D4 we have found four one-dimensional representations A1, A2, B1, B2 andone two-dimensional one E.

Exercise 2.5: Using eq. (2.4.28), write the matrix representation of the rotations in Cnabout the X axis.

Example 2.11: Consider the group U(1). We can take as the defining representation theone in which the elements of the group are described by a phase ei ϕ, ϕ ∈ (0, 2 π). Theelements of the abstract group U(1) are then functions g(ϕ) of the continuous parameterϕ. We can define, for any integer n, a one-dimensional representation D(n) of U(1) over thecomplex numbers as follows:

g(ϕ) ∈ U(1) −→ D(n)[g(ϕ)] ≡ ei nϕ . (2.4.41)

The reader can verify that this is indeed a representation.

2.4.2 The Regular Representation and the Group Algebra

Let us define, for a generic order-n finite group G its regular representation. We have seen inthe previous section that any group admits a faithful realization by means of left-translationson the group itself. We can associate with this realization a matrix representation of G. Letus denote by g0, g2, . . . , gn−1 the elements of G, g0 = e. Given a generic group element g,the action of T [g] on any element gi is the unique element gki = g · gi and is described bythe permutation (2.4.5). Let us associate each element gi of G with an element egi of abasis of a n-dimensional real vector space VR. Define the matrix DR[g] ≡ (DR[g]j i) whoseentries are all zero except DR[g]ki i which is 1. The matrix DR[g], for any g ∈ G, defines alinear transformation on the vector space VR, to be denoted by DR[g], which maps the basiselement egi into the element eg·gi = egki :

egig−→ DR[g](egi) ≡ eg·gi = egki = DR[g]j i egj . (2.4.42)


All the basis elements egi are constructed by acting on the same element eg0 ≡ ee by thecorresponding transformation DR[gi]: egi = egi·e = DR[gi](ee). Writing a generic vectorV = V i egi as the column vector V = (V i) of components V i, namely identifying the basiselements with the following column vectors:

eg0 = ee =

10...00

, · · · , egn =

00...01

, (2.4.43)

the action of the linear transformation DR[g] on a generic vector V is represented as follows:

DR[g](V) = V iDR[g]j i egj ≡ DR[g] V . (2.4.44)

Clearly DR[e]ij = δij and it is the only matrix with diagonal entries. To see this supposethere existed a g ∈ G such that DR[g]ij has a diagonal entry, say DR[g]kk (no summationover k). This means that g · gk = gk, which implies g = e.

Let us show that the correspondence DR between elements g ∈ G and transformationsDR[g] ∈ Aut(VR) is a homomorphism. Take two generic elements g, g′ ∈ G:

DR[g · g′](egi) = eg·g′·gi = DR[g](DR[g′](egi)) = (DR[g] · DR[g′])(egi) , (2.4.45)

which implies that DR[g ·g′] = DR[g] ·DR[g′]. This implies an analogous relation between thecorresponding matrices: DR[g·g′] = DR[g] DR[g′]. The correspondence DR between elementsg ∈ G and the n× n matrix DR[g] is then a n-dimensional representation called the regularrepresentation of G. This representation is clearly faithful and, as we shall see, completelyreducible. Notice now that DR is an orthogonal representation (i.e. real unitary). Indeed ifthe left translation by g is defined by the permutation gi → gki and thus the non-vanishingentries of the matrix DR[g] are DR[g]ki i = 1, the action g−1 is represented by the inversepermutation gki → gi and thus the non-vanishing entries of the corresponding matrix areDR[g−1]iki = 1. Therefore DR[g−1] = DR[g]T . Being DR a homomorphism, we then haveDR[g−1] = DR[g]−1 = DR[g]T , which proves the orthogonality of DR.

We can easily construct the matrices DR[g] directly from the multiplication table of thegroup. To this end we write the multiplication table corresponding to the chosen ordering(g0, g2, . . . , gn−1) of the group elements. We then permute the columns of the table so thatthe unit element e only appears along the diagonal (for instance in the multiplication tableof D3, given in Section 1.2.3, we should permute the second with the third column). DR[g]is then the matrix obtained from the resulting table by replacing the element g by 1 and allthe other elements by 0. In the D3 case we then find, for the elements e, r, σ the followingmatrices:

DR[e] = 16 , DR[r] =

0 0 1 0 0 01 0 0 0 0 00 1 0 0 0 00 0 0 0 0 10 0 0 1 0 00 0 0 0 1 0

, DR[σ] =

0 0 0 1 0 00 0 0 0 0 10 0 0 0 1 01 0 0 0 0 00 0 1 0 0 00 1 0 0 0 0

.


Exercise 2.6: Compute the regular representation of the generators r, σ of D4.Given two linear transformations DR[g], DR[g′] on VR, representing the corresponding

elements of G, we can consider the linear function on VR defined by an arbitrary linearcombination of the two:

(αDR[g] + βDR[g′])(V) ≡ αDR[g′](V) + βDR[g′](V) = (αDR[g] + βDR[g′]) V ,

where the matrix describing the action of the linear combination is given by the same combi-nation of the matrices associated with each transformation. In general such linear combina-tion is not invertible and thus should not be regarded as a transformation, i.e. an automor-phism on VR, but just an endomorphism2. The linear transformation αDR[g] + βDR[g′] ingeneral does not correspond to any group element either. As opposed to the abstract groupelements of G, on which no sum or multiplication by numbers is defined, we can define thesame operations on the corresponding linear transformations on VR. Consider the space ofall the linear transformations DR[gi] on VR, defining the regular representation and extendit with all the linear combinations of these transformations. We end up with a space AR[G]of endomorphisms on VR, which is closed with respect to linear combinations of its elementsand therefore is itself a linear vector space. By construction, a basis of this space consists ofthe linear transformations DR[gi] : VR → VR and therefore AR[G] is n-dimensional. A genericelement R of AR[G] is a endomorphism on VR which is expressed as a linear combination ofthe n basis elements DR[gi]: R = RiDR[gi] =

∑g∈GR(g)DR[g], where Ri ≡ R(gi):

AR[G] = R ∈ End(VR)|R = RiDR[gi] =∑g∈G

R(g)DR[g] . (2.4.46)

Similarly we can define the vector space AR[G] consisting of linear combinations of the n×nmatrices DR[gi]:

AR[G] = R n× n matrix|R = Ri DR[gi] =∑g∈G

R(g) DR[g] . (2.4.47)

In contrast with a generic linear vector space, on AR[G] a product operation is definedbetween the basis elements DR[gi]. Indeed, in virtue of eq. (2.4.45), if the left-translationby gi in G maps gj into gi · gj = gkj , we have:

DR[gi] · DR[gj] = DR[gi · gj] = DR[gkj ] = DR[gi]kj DR[gk] . (2.4.48)

An analogous relation holds between the corresponding matrices:

DR[gi] DR[gj] = DR[gi]kj DR[gk] . (2.4.49)

We can extend on the whole vector space AR[G] the product operation · by requiring it tobe linear in both its arguments (bi-linear), namely that, for any S1, S2, S, R1, R2, R ∈ AR[G]:

(αS1 + β S2) ·R = αS1 ·R + β S2 ·R ,S · (αR1 + β R2) = αS ·R1 + β S ·R2 , (2.4.50)

2To show this consider the combination DR[g]−DR[g], which maps all vectors V into the zero vector 0and is represented by the zero-matrix 0. It clearly is not invertible.


α, β being arbitrary numbers. Therefore, if R = RiDR[gi] and S = Sj DR[gj], we define theproduct of the two transformations as the following linear transformation R · S:

R · S = (RiDR[gi]) · (Sj DR[gj]) = RiDR[gi] · (Sj DR[gj]) = Ri Sj DR[gi] · DR[gj] =

= (Ri SjDR[gi]kj)DR[gk] .

We can also express the components of the product using the equivalent representation ofthe elements of AR[G], R =

∑g∈GR(g)DR[g], S =

∑g′∈G S(g′)DR[g′]:

R · S =∑g∈G

∑g′∈G

R(g)S(g′)DR[g] · DR[g′] =∑g∈G

∑g′∈G

R(g)S(g′)DR[g · g′] =

=∑g∈G

∑g′′=g·g′∈G

R(g)S(g−1 · g′′)DR[g′′] =∑g′′∈G

(∑g∈G

R(g)S(g−1 · g′′)

)DR[g′′] .

A linear vector space on which a bi-linear product is defined is called an algebra. Thespace AR[G], equipped with the product · is therefore an algebra, called the group algebraassociated with the group G. More specifically it is an associative algebra since the productis associative, as it can be easily verified. Similarly also the space AR[G] is an associativealgebra. Equations (2.4.48) and (2.4.49) express the product of two basis elements as alinear combination of the same basis elements and are called the structure equations of thealgebras AR[G] and AR[G]. They uniquely define the product of any two elements withineach algebra and are characterized by the constants DR[gi]

jk, called the structure constants of

the algebra. The spaces AR[G] and AR[G] have the same dimension n and the same structureconstants. They are isomorphic algebras since the correspondence RiDR[gi] → Ri DR[gi]is an isomorphism between the two vector spaces which preserves the product operation.Therefore we shall use either AR[G] and AR[G] to describe the same group algebra.

The product on AR[G] has an identity element which is the identity transformation DR[e]itself: DR[gi] · DR[e] = DR[e] · DR[gi] = DR[gi]. Moreover it is associative since the productof transformations is, being represented by product of matrices.

Notice that the whole representation space VR of the regular representation can be gen-erated by acting on the single vector ee by the algebra AR[G]. Indeed to generate a genericvector V = V i egi ∈ VR it suffices to act on ee by the element R = V iDR[gi] of AR[G]:R(ee) = V iDR[gi](ee) = V i egi = V.

The correspondence between R = V iDR[gi] ∈ AR[G] and V = V i egi ∈ VR is an iso-morphism between the linear vector spaces AR[G] and VR, the difference between the twobeing that on the former a product operation is defined while on the latter it is not. Wecan view AR[G] itself as the representation space of the regular representation in whichDR[gi] ∈ AR[G] plays the role of egi ∈ VR. Indeed, just as DR[gi] : egj → DR[gi](egj) = egi·gj ,we can define the action of AR[G] on itself: DR[gi] : DR[gj]→ DR[gi]DR[gj] = DR[gi · gj].

Finally let us consider a generic representation D of G on a m-dimensional linear vectorspace Vm. The elements gi ∈ G, i = 0, . . . , n = |G| − 1, are represented by linear transfor-mations D[gi] ∈ Aut(Vm). Just as we did for linear transformations on VR we can consider


linear combinations of automorphisms on Vm and define the linear vector space AD[G] withbasis elements D[gi], consisting of endomorphisms on Vm of the form: R = RiD[gi].

AD[G] = R ∈ End(Vm)|R = RiD[gi] , (2.4.51)

On this space a bilinear product is defined from the product of two generic basis elements:

D[gi] · D[gj] = D[gi · gj] = D[gkj ] = DR[gi]kj D[gk] , (2.4.52)

which promotes AD[G] to an algebra. Similarly we can define the matrix description of theabove algebra:

AD[G] = R = Ri D[gi] , (2.4.53)

characterized by the same structure:

D[gi] D[gj] = DR[gi]kj D[gk] , (2.4.54)

and thus isomorphic to AD[G]. Notice that AD[G] has the same dimension and structureas AR[G] and thus the two are isomorphic algebras. In fact AR[G] is a particular instanceof AD[G] when the representation D is the regular one: D = DR. Just as we did forgroups, we can capture the common structure of isomorphic algebras by defining an abstractalgebra A[G] consisting of symbols, uniquely characterized by its dimension and structure,and say that two isomorphic algebras of endomorphisms are different representations of asame abstract algebra. AR[G] and AD[G] are then different representations, over VR andVm respectively, of a same abstract algebra A[G].

Let us define A[G] as a n-dimensional vector space, n = |G|, whose basis elements arein correspondence with the elements of the group G and will be, with an abuse of notation,denoted by the same symbols: g0, g1, . . . , gn. A generic element of A ∈ A[G] is then expressedas a linear combination of the gi:

A[G] ≡ A =n∑i=1

Ai gi =∑g∈G

A(g) g . (2.4.55)

Let a bi-linear product · on A[G] be defined, which maps couples of elements of A[G] intotheir product which is still an element of A[G]: For any A,B ∈ A[G], A·B ∈ A[G]. Being bi-linear, this product is totally defined by the product of the basis elements, which is expressedin components relative to the same basis through the structure constants. The product ofthe basis elements is in turn defined by the product of the corresponding elements in G,namely by the structure of G itself: Suppose the left-translation by the element gi maps gjinto gkj , then

gi · gj = gkj = DR[gi]kj gk . (2.4.56)

Notice that the last equality in the above formula would be meaningless if gi, gj, gk wereelements of G, since on G there is no notion of sum, and thus the summation over k on

2.5. SOME PROPERTIES OF REPRESENTATIONS 83

the right hand side would be undefined. The product between two elements A = Ai gi andB = Bj gj of A[G] reads:

A ·B = AiBj gi · gj = (AiBj DR[gi]kj) gk . (2.4.57)

The structure constants DR[gi]kj in (2.4.56), defining the product on A[G], are the same

as those for AD[G] in (2.4.52). A[G] is an associative algebra called the abstract groupalgebra associated with G. Given a representation D of G on Vm, we define a correspondingrepresentation of its algebra A[G] as the following mapping between elements of A[G] andelements of AD[G], which are endomorphisms on Vm:

A = Ai giD−→ D[A] ≡ AiD[gi] ∈ AD[G] , (2.4.58)

or, equivalently, the mapping D between element of A[G] and m×m matrices in AD[G]:

A = Ai giD−→ D[A] ≡ Ai D[gi] ∈ AD[G] . (2.4.59)

By construction D is a homomorphism between the two linear vector spaces which preservesthe product:

D[A ·B] = D[A] · D[B] , (2.4.60)

and thus is a homomorphism between the two algebras. The same for D. Just as for groups,we shall not make any difference between D and D, they define the same representation.

We have then learned that a representation of G induces a representation of its abstractalgebra A[G]. The reverse is also true: A representation D of A[G] induces a representationof G. We just need to associate with a generic element gi of G the automorphism D[gi]corresponding to the basis element gi of A[G].

Just as it was the case for AR[G] and in general for AD[G], we can view A[G], whichas the representation space of the regular representation in which gi ∈ A[G] plays therole of egi ∈ VR. Indeed, we can associate with each gi ∈ G an automorphism DR[gi] onA[G] whose action on an element A = Aj gj is represented by the left-multiplication bythe corresponding element gi of A[G]: A → DR[gi](A) ≡ gi · A. In other words, just asDR[gi] : egj → DR[gi](egj) = egi·gj = DR[gi]

kj egk , we can define the action of DR[gi] on A[G]

in a similar fashion DR[gi] : gj → DR[gi](gj) ≡ gi gj = DR[gi]kj gk, so that it is still described

by the same n× n matrix DR[gi] ≡ (DR[gi]kj).

The abstract group algebra A[G] is then also the representation space of the regularrepresentation of the algebra itself. With any A = Ai gi ∈ A[G] we associate the endomor-phism DR[A] = AiDR[gi] on A[G], whose action on a generic element B ∈ A[G] is the leftmultiplication by A: B → DR[A](B) = AiDR[gi](B) = Ai (gi ·B) = A ·B.

2.5 Some Properties of Representations

Let us prove some important properties of representations of finite groups:


Property 2.2: Any representation of a finite group is equivalent to a unitary one.Consider a finite group G of order n, which has a representation D over a complex vector

space Vn on which a hermitian, positive definite, scalar product (·, ·)0 is defined. We wantto show that we can always define a hermitian scalar product (·, ·) with respect to which Dis a unitary representation, that is, for any g ∈ G and for any vectors V and W:

(D[g] V, D[g] W) = (V, W) . (2.5.1)

Define the scalar product (·, ·) on two generic vectors V and W as follows:

(V, W)def.=

1

n

∑g∈G

(D[g] V, D[g] W)0 . (2.5.2)

We can easily show that the product defined above is hermitian. In particular it clearly islinear with respect to the second argument. Moreover:

(W, V) =1

n

∑g∈G

(D[g] W, D[g] V)0 =1

n

∑g∈G

(D[g] V, D[g] W)∗0 = (V, W)∗ .(2.5.3)

The positive definiteness of (·, ·) follows from the same property of (·, ·)0 since (V, V), beingthe sum of positive numbers (D[g] V, D[g] V)0, is itself positive, and moreover it vanishesonly if each term in the sum vanish, which is the case only if V = 0.

Let us show that with respect to this product D is unitary:

(D[g] V, D[g] W) =1

n

∑g∈G

(D[g · g′] W, D[g · g′] V)0 =

=1

n

∑g·g′∈G

(D[g · g′] W, D[g · g′] V)0 =1

n

∑g′′∈G

(D[g′′] W, D[g′′] V)0 =

= (V, W) . (2.5.4)

To complete the proof, we should show that D is equivalent to a unitary representationrelative to the original scalar product (·, ·)0. Let us define the metrics associated with thetwo products: g0

ij ≡ (ei, ej)0 and gij ≡ (ei, ej), relative to a given basis (ei). These twomatrices g0 ≡ (g0

ij) and g ≡ (gij) are both hermitian positive definite. From the generaltheory of matrices it follows that there exist a (in general not unitary) complex matrixM such that g = M† g0 M or, equivalently (M−1)† g M−1 = g0. Let us show that therepresentation D ≡M D M−1 is unitary with respect to g0. Choose a generic g ∈ G:

D[g]† g0 D[g] = (M−1)†D[g]† (M† g0 M) D[g] M−1 = (M−1)†D[g]† g D[g] M−1 =

= (M−1)† g M−1 = g0 , (2.5.5)

where we have used unitarity of D with respect to g, namely that D[g]† g D[g] = g. Thiscompletes the proof.


Remark: If the representation D is defined over a real space Vn with a symmetric scalarproduct (·, ·)0, by the same token, we can define a new symmetric product (·, ·) with re-spect to which D is real orthogonal and, following the same procedure as above, define arepresentation equivalent to D which is orthogonal with respect to (·, ·)0. Therefore any realrepresentation of a finite group is equivalent to an orthogonal one.

Property 2.3: Any reducible representation of a finite group G is completely reducible.Let D be a reducible representation of G over a complex vector space Vn equipped

with a hermitian scalar product. In virtue of the previous property, we can define a newhermitian scalar product (·, ·) with respect to which D is unitary. By assumption thereexists a subspace Vk of Vn, for some k < n, which is stable with respect to the action ofthe transformation D[g], for any g ∈ G, i.e. such that D[G](Vk) ⊂ Vk. Let us choose anortho-normal basis (ei) with respect to (·, ·), and define with respect to (·, ·) the orthogonalcomplement Vn−k of Vk in Vn, as the space consisting of the vectors in Vn which are orthogonalto all vectors in Vk:

Vn−k ≡ V ∈ Vn | (V, W) = 0 for any W ∈ Vk . (2.5.6)

Vn−k is clearly a vector space, since the linear combination of any two vectors orthogonal toall vectors in Vk has itself this property. Let us show that Vn−k is stable with respect to theaction of G defined by D. To prove this, it suffices to show that, if V ∈ Vn−k, then, for anyg ∈ G, D[g] V ∈ Vn−k. This is the case if D[g] V is orthogonal to all vectors in Vk. Let usthen take a generic vector W in Vk:

(D[g] V, W) = (V, D[g]†W) = (V, D[g−1] W) = 0 , (2.5.7)

where we have used eq. (2.3.65) and the fact that D[g−1] W is still in Vk by assumption. Wehave thus shown that Vn can be decomposed into two disjoint subspaces Vk and Vn−k eachof which is stable with respect to D[G]. As a consequence of this, if we choose a basis for Vnwhose first k elements are in Vk and last n− k elements are in Vn−k, in this basis the matrixform of D[g], for any g ∈ G, is block diagonal. In other words D is completely reducible intotwo representations Dk and Dn−k acting on Vk and Vn−k respectively:

D −→ Dk ⊕Dn−k . (2.5.8)

This completes the proof.Let us mention, without proving, that the above two properties also hold for compact

Lie groups (such as O(n) or U(n)):

Property 2.4: Any representation of a compact Lie group is equivalent to a unitary one.This for instance does not apply to non-compact groups, like the Lorentz group, which

do not admit any unitary representations aside from the trivial one.

Property 2.5: Any reducible representation of a compact Lie group is completely re-ducible.


2.5.1 Unitary Representations and Quantum Mechanics

In quantum mechanics all information about the physical state of a system is encoded in themathematical notion of quantum state. A quantum state is a vector in a Hilbert space H ,which, as previously mentioned, is a complex (possibly infinite dimensional) linear vectorspace equipped with a hermitian, positive definite, scalar product. It is denoted by thesymbol |ψ〉 (instead of the boldface letter V that we have been using so far), while the scalarproduct (|ψ1〉 , |ψ2〉) between two states |ψ1〉 , |ψ2〉 is denoted by 〈ψ1|ψ2〉. One usually worksin a ontho-normal basis, with respect to which, a state |ψ〉 can be represented as a columnvector consisting of its components and 〈ψ| its hermitian conjugate, namely the row vectorconsisting of the complex conjugate components of |ψ〉: 〈ψ| ≡ |ψ〉†. This notation for thevectors representing quantum states was originally introduced by Dirac who named |ψ〉 aket vector and its hermitian conjugate 〈ψ| a bra vector. Given two states |ψ1〉 , |ψ2〉, thequantity:

P (ψ1, ψ2) ≡ |〈ψ2|ψ1〉|2

‖ψ2‖2‖ψ2‖2, (2.5.9)

where ‖ψi‖2 ≡ 〈ψi|ψi〉 > 0, represents the probability of finding, upon measurement, asystem which is initially prepared in a state |ψ1〉, in a state |ψ2〉, characterized by a definitevalue of the quantity we are measuring. P (ψ1, ψ2) is also called transition probability betweenthe states |ψ1〉 to |ψ2〉. Clearly the result of a measurement, described by the probabilityP (ψ1, ψ2), is not affected if the two states are multiplied by complex number: |ψi〉 →α |ψi〉. Physical states are then represented by vectors in H modulo multiplicative complexnumbers, which are unphysical. They are, in other words, described by rays in H , defined asone-dimensional subspaces α |ψ〉 of H consisting of vectors differing from a given one bya multiplicative number. If we normalize a quantum state to unit norm ‖ψ‖2 ≡ 〈ψ|ψ〉 = 1,the multiplicative complex number is fixed up to a phase.

Of particular relevance in quantum mechanics are operators S, i.e. linear functions, onH . The action of S on a vector |ψ〉 is denoted by |S ψ〉 and is represented by the action ofa (possibly infinite dimensional) matrix MS on the corresponding vector components:

|S ψ〉 ≡ S(|ψ〉) = MS |ψ〉 . (2.5.10)

Since S is not necessarily a transformation, MS need not be invertible. Clearly the actionof S on bra vectors is represented by the right action of the hermitian conjugate matrix:

〈S ψ| ≡ |S ψ〉† = 〈S ψ|M†S . (2.5.11)

In quantum mechanics observables, i.e. measurable physical quantities as energy, momen-tum, angular momentum, spin etc..., generically denoted by O, are represented by hermitianoperators O. An operator O is hermitian if its matrix, in an ortho-normal basis, is hermitian:M†O = MO. If O is hermitian, then, for any couple of states |ψ1〉 , |ψ2〉, we have:

〈ψ1|O ψ2〉 = 〈ψ1|MO |ψ2〉 = (〈ψ1|M†O) |ψ2〉 = 〈Oψ1|ψ2〉 , (2.5.12)


and therefore, on a same state |ψ〉:

〈ψ|O ψ〉 = 〈Oψ|ψ〉 = 〈ψ|O ψ〉∗ , (2.5.13)

that is 〈ψ|O ψ〉 is a real number. The number:

〈O〉 ≡ 〈ψ|O ψ〉‖ψ‖2

∈ R , (2.5.14)

represents the expectation value of the observable O on the state |ψ〉, the most likely valuethat a measure of O will give on that state. With an abuse of notation, to simplify life, weshall use the same symbol for the operator O and for the corresponding matrix: MO = O,

so that |O ψ〉 = O |ψ〉. A state |ψλ〉 with a definite value λ of the observable is an eigen-stateof O, to the eigen-value λ:

O |ψλ〉 = λ |ψλ〉 . (2.5.15)

Since eigen-vectors to different eigen-values are orthogonal, that is 〈ψλ|ψλ′〉 = 0 if λ′ 6= λ.This represents the experimental fact that, if a system is prepared in a state with a definitevalue λ of O, we can be sure that any further measure of O on the system will give the sameresult, namely the probability P (ψλ, ψλ′) of measuring the system in a state with a differentvalue λ′ of the same observable is zero.

Consider now a linear transformation on the space of states H , described by an invertibleoperator S. It may represent the effect on the states of a change in the space-time RF, dueto rotations, reflections, Lorentz boosts etc... If a system is described in a state |ψ〉 by anobserver in a RF, it will in general be described by a different state |ψ′〉 = |S ψ〉 with respectto the transformed RF3. What all observers should agree on is the transition probability Pbetween two states, defined in (2.5.9), namely we should have:

P (ψ1, ψ2) = P (ψ′1, ψ′2) = P (S ψ1, S ψ2) . (2.5.16)

This is the case only if the scalar product between any two states |ψ1〉 , |ψ2〉 is invariantin modulus: |〈S ψ2|S ψ1〉| = |〈ψ2|ψ1〉|. A theorem by Wigner, which we are not goingto prove, states that the phases of the vectors representing physical states can always bechosen so that a transformation preserving transition probabilities be either realized as aunitary transformation, 〈S ψ2|S ψ1〉 = 〈ψ2|ψ1〉, or as an anti-unitary one, 〈S ψ2|S ψ1〉 =〈ψ2|ψ1〉∗ (transformations involving time-reversal). We shall restrict ourselves to unitarytransformations only. Suppose we have an abstract group G of transformations (like therotation group or the Lorentz group), whose action on the states is described by a unitaryrepresentation D, which means that, for any element g of the abstract group G:

|ψ〉 −→ |g ψ〉 = D[g] |ψ〉 , (2.5.17)

3We are always going to describe transformations of quantum states from the active point of view: Statesare mapped into different states.


and, moreover, for any couple of states |ψ1〉 , |ψ2〉:

〈g ψ2|g ψ1〉 = 〈ψ2|D[g]†D[g] |ψ1〉 = 〈ψ2|ψ1〉 , (2.5.18)

which implies that D[g]†D[g] = D[g] D[g]† = 1, for any g ∈ G.Of course the expectation value of an observable will in general change since, for instance,

the position of a particle depends on the RF in which it is measured:

〈O〉 g∈G−→ 〈O〉′ = 〈ψ|D[g]† OD[g] |ψ〉‖ψ‖2

. (2.5.19)

From the above formula it is clear that we can either think of the transformations as actingon the states, as in (2.5.17), or on the operator O:

O −→ O′ = D[g]† OD[g] , (2.5.20)

the result on the expectation value of O is the same. For instance if O is the position vectorr = (xi) of a particle and the transformation group is SO(3), a generic rotation g ∈ SO(3) onr is represented by the action of the 3×3 real matrix R[g] ≡ (R[g]ij) in (2.3.79): r′ = R[g] r.This is the transformation property we expect for the expectation value of the position on anystate of the particle. Let D be the unitary representation of SO(3) on the space of quantumstates, and r ≡ (xi) the hermitian operator associated with r. The expected position 〈r〉 ofa particle in a state |ψ〉 is given by eq. (2.5.14) and should transform under g ∈ SO(3) as in(2.5.19):

〈xi〉 g∈G−→ 〈xi〉′ = 〈ψ|D[g]† xi D[g] |ψ〉‖ψ‖2

= R[g]ij 〈xj〉 , (2.5.21)

which implies that the operator r should transform as follows:

xi −→ x′ i = D[g]† xi D[g] = R[g]ij xj . (2.5.22)

Let us emphasize that D and R are different representations of the same rotation group:The former describes the effect of rotations on the vectors representing physical states, thelatter describes rotations on three dimensional space vectors.

We may have transformations on H which are not the effect of any change in the space-time RF. These are called internal transformations and are still represented by unitaryoperators on physical states. Internal transformations have an important role in the theoryof elementary particles and of fundamental interactions.

Consider a group G of transformations which do not involve time (this excludes forinstance Lorentz transformations other than rotations, i.e. Lorentz boosts), like rotations, orthe permutation of particles in a system of particles, or, in general, internal transformations.If these transformations represent symmetries of the system, the energy of the system shouldnot be affected. The energy is described by the Hamiltonian H, which corresponds, in

2.6. SCHUR’S LEMMA 89

quantum mechanics, to a hermitian operator H. If G is a symmetry group then, in virtueof eq. (2.5.20), we should have that, for any g ∈ G:

H −→ D[g]† H D[g] = H , (2.5.23)

namely, using the unitarity property of D[g], the Hamiltonian operator should commute withD[g]: [H, D[g]] = 0 for any g ∈ G. Suppose the system is in a state |ψE〉 of definite energyE. For what we have previously said, |ψE〉 is the eigen-state of H to the eigenvalue E:

H |ψE〉 = E |ψE〉 . (2.5.24)

There may be more than one eigen-state to the same eigenvalue E. In this case one speaksof degeneracy of the energy level E. Let HE denote the space spanned by all the states withdefinite energy value E, namely the eigen-space of H corresponding to the eigenvalue E. Itclearly is a subspace of H and its dimension measures the degeneracy of the energy level E.Let us show that it is a representation space for G, namely that it is invariant with respectto the action of G. Take a vector |ψE〉 and act on it by a generic transformation g ∈ G:|ψE〉 → |g ψE〉 = D[g] |ψE〉. Let us show that the transformed state is still in HE:

H |g ψE〉 = H D[g] |ψE〉 = D[g] H |ψE〉 = E |g ψE〉 . (2.5.25)

The dimension of HE is therefore the dimension of a representation of G, which, if we excludethe so-called accidental degeneracy, is also irreducible. In these general cases we can thendeduce an important physical property of a system, such as the degeneracy of its energylevels, merely from group theoretical considerations.

2.6 Schur’s Lemma

Consider two representations D1 and D2 of an abstract group G, acting on the linear vectorspaces Vn and Wm, respectively. Consider a homomorphism T ∈ Hom(Vn, Wm), whose actionon the (column) n-vectors in Vn is represented by the m× n matrix T ≡ (T ai), as explainedin Section 2.1.1: T (V) = T V. We define an intertwiner between the representations D1

and D2 a homomorphism T ∈ Hom(Vn, Wm) such that, for any g ∈ G, the correspondingmatrix T satisfies the following condition:

T D1[g] = D2[g] T . (2.6.1)

These maps are also called G-homomorphisms and their space is denoted by HomG(Vn, Wm).

Property 2.5 (Schur’s Lemma): Let D1 and D2 be two irreducible representations ofG on the linear vector spaces Vn and Wm respectively, and let T be an intertwiner betweenthem:

i) If D1 and D2 are inequivalent, T ≡ 0 ( i.e. T ≡ 0);


ii) If D1 and D2 are equivalent (so that Vn and Wm are isomorphic, i.e. m = n), then, forsome number c, T = cA, where the n × n matrix A defines the equivalence betweenthe representations: D2 = A−1 D1 A.

Let us prove property (i) first. Define E as the subspace of Vn spanned by the vectors whichare mapped by T into the zero-vector (also called the kernel of T or KerT ):

E ≡ V ∈ Vn|T (V) = T V = 0 ∈ Wm . (2.6.2)

It can be easily shown that E is a G-invariant subspace. Indeed consider V ∈ E and a genericg ∈ G. Then also D1[g] V is in E since T (D1[g] V) = T D1[g] V = D2[g] T V = D2[g] 0 = 0.Since, by assumption, D1 is irreducible, either

(a) E = 0 or (b) E = Vn . (2.6.3)

In case of (b), all vectors of Vn are mapped into the zero-vector of Wn, namely T = 0, i.e.T ≡ 0, and (i) is proven. In case of (a), T is one-to-one. Let us show that the spaceT (Vn) ⊂ Wm is G-invariant:

For any V ∈ Vn and g ∈ G : D2[g]T (V) = D2[g] T V = TD1[g] V = T (D1[g] V) ∈ T (Vn) .

Since D2 is irreducible and T (Vn) can not, by assumption, be 0and therefore we musthave: T (Vn) = Wm, namely T is onto. We conclude that T is an isomorphism and thus,from eq. (2.6.1) we deduce that D1 and D2 are equivalent: D1 = T−1 D2 T, contradictingour assumption. This completes the proof of (i).

Let us prove now proposition (ii). By assumption D1 and D2 are equivalent: D1 =A−1 D2 A, for some non singular matrix A. Using this equivalence, eq. (2.6.1) can bere-written as follows:

T D1[g] = A D1[g] A−1 T ⇔ (A−1 T) D1[g] = D1[g] (A−1 T) , (2.6.4)

which means that the matrix A−1 T intertwines D1 with D1. Let us focus now on this n×nmatrix A−1 T which maps n-vectors of Vn into n-vectors of Vn. Under a change of basisof Vn A−1 T will transform by a similarity transformation. From general matrix theory, weknow that any matrix can be reduced, through a similarity transformation, into its Jordannormal form. This normal form is a block diagonal matrix:

J =

J1 0. . .

0 J`

, (2.6.5)

where the blocks Jk, k = 1, . . . , `, have the following form:

Jk =

λk 1 0

λk. . .. . . 1

0 λk

. (2.6.6)

2.6. SCHUR’S LEMMA 91

From eq. (2.6.6) it follows that

Jk

10...0

= λk

10...0

, (2.6.7)

from which it that A−1 T must have at least one eigenvalue. Let us call it c and thecorresponding eigenvector Vc. Define the homomorphism T ′ ∈ End(Vn) described by thematrix:

T′ ≡ A−1 T− c1n . (2.6.8)

Since both matrices A−1 T and c1n intertwine D1 with D1, also T′ does. Define E = KerT ′.Since D1 is irreducible, also for E we have the options (2.6.3). Notice however that T ′ cannot be an isomorphism, i.e. E cannot be 0, since there exist the vector Vc on which:

T ′(Vc) = T′Vc = A−1 T Vc − cVc = 0 . (2.6.9)

We conclude that T ′ = 0, that is A−1 T = c1n. This concludes the proof of proposition (ii).As a corollary of Schur’s lemma, we have

Properties 2.6:

iii Let D be an irreducible n-dimensional representation of a group G. A matrix T whichcommutes with all matrices D[g], for any g ∈ G, is proportional to the identity matrix1n;

iv Let D be a n-dimensional representation of a group G. If there exist a matrix T whichcommutes with all matrices D[g], for any g ∈ G, and is not proportional to the identitymatrix 1n, then D is reducible;

Proposition (iii) is a particular case of Schur’s lemma, in which D1 = D2 and thus A = 1n.From proposition (ii) it follows that T = c1n, for some number c.

Schur’s lemma and its corollary, provide us with a criterion for telling if a representationis reducible and, in some cases, to determine its irreducible components: Suppose we findan operator T on Vn which commutes with all the matrices D[g] representing the action ofa group G on the same space. The matrix representation T of T will then have the form:

T =

c1 1k1 0. . .

0 cs 1ks

, (2.6.10)

where c1, . . . , cs are different numbers and the corresponding eigen-spaces Vk1 , . . . , Vks ofT correspond to different representations Dk1 , . . . ,Dks of G. If G is a symmetry groupof a quantum mechanical system, we know, from our discussion in Section 2.5.1, that the


hamiltonian operator H commutes with the action D[G] of G on the states, i.e. it intertwinesD with itself. Its matrix representation on the states will then have the form (2.6.10),where c1, . . . , cs (s may be infinite!) are the energy levels of the system, and k1, . . . ks theirdegeneracy. If there is no accidental degeneracy, Dk1 , . . . ,Dks are irreducible representationsof G. If Dki is not irreducible, this may indicate that there exist a larger symmetry groupG′ containing G, whose action on Vki is irreducible. On other words accidental degeneraciesmay signal the existence of a larger symmetry of the system.

Property 2.7: All irreducible representations of an abelian group are one-dimensional.Let G be an abelian group and D an irreducible n-dimensional representation of G.

Consider a generic element g ∈ G. The abelian property of G implies that, for any g′ ∈ G,D[g] D[g′] = D[g′] D[g], that is D[g] intertwines the representation D with itself. In virtueof Schur’s lemma, being D irreducible, we must have D[g] = d[g] 1n, for some numberd[g] associated with g. In an irreducible representation the matrix elements can all beproportional to the identity matrix only if the representation is one-dimensional, namely ifn = 1. In Section 2.8 we shall give, for finite groups, an alternative proof of this property.

2.7 Great Orthogonality Theorem

We prove in this section an important property of unitary representations of finite groups.

Property 2.8 (Great Orthogonality Theorem): Let G be a finite group and D, D′

two irreducible unitary representations of dimensions n and m respectively. With respect tosome bases of the corresponding representation spaces Vn and Wm, each element g ∈ G isthen represented by the matrices D[g] ≡ (D[g]ij) and D′[g] ≡ (D′[g]ab), i, j = 1, . . . , n anda, b = 1, . . . ,m. The following properties hold:

i) If D and D′ are inequivalent,∑g∈G

(D[g]ij

)∗D′[g]ab = 0 , (2.7.1)

for any i, j = 1, . . . , n and a, b = 1, . . . ,m;

ii) If D = D′, ∑g∈G

(D[g]ij

)∗D′[g]k` =

|G|nδki δ

j` , (2.7.2)

for any i, j, k, ` = 1, . . . , n, |G| denoting the order of G.

Let us start proving proposition (i). Under the hypotheses of the theorem, we take a generica rectangular m× n matrix X ≡ (Xa

i) and define the following matrix M ≡ (Mai):

Mdef.=

∑g∈G

D′[g] X D[g]−1 , (2.7.3)

2.7. GREAT ORTHOGONALITY THEOREM 93

or, in components,

Mai

def.=

∑g∈G

D′[g]abXbj D[g]−1 j

i . (2.7.4)

Let us show that M is an intertwiner between D′ and D. Take a generic h ∈ G:

D′[h] M =∑g∈G

(D′[h] D′[g]) X D[g]−1 =∑g∈G

D′[h · g] X D[g]−1 (D[h]−1 D[h]) =

=∑g∈G

D′[h · g] X D[h · g]−1 D[h] =∑g∈G

D′[g] X D[g]−1 D[h] = M D[h] .(2.7.5)

In virtue of Schur’s lemma, if D and D′ are inequivalent, M = 0. Let us construct the set ofnm matrices Mi

a ≡ [(Mia)bj], where a, i label the matrix, while b, j are the matrix indices,

out of the following matrices Xia ≡ [(Xi

a)bj] = [δij δ

ba]. Each of these matrices is zero:

Mia =

∑g∈G

D′[g] Xia D[g]−1 = 0m,n , (2.7.6)

or, writing the (b, j)-component of the above matrix:

0 = (Mia)bj =

∑g∈G

D′[g]bc δik δ

caD[g]−1 k

j =∑g∈G

D′[g]baD[g]−1 ij =

∑g∈G

(D[g]j i

)∗D′[g]ba ,

where in writing last equality we have used unitarity or the representation D[g]−1 = D[g]†.Let us now prove proposition (ii). By assumption D = D′. Schur’s theorem implies that

M = c1n:

M =∑g∈G

D[g] X D[g]−1 = c1n . (2.7.7)

Taking the trace of the left hand side of the above equation we find:

Tr (M) =∑g∈G

Tr(D[g] X D[g]−1

)=∑g∈G

Tr(D[g]−1 D[g] X

)=∑g∈G

Tr (X) = |G|Tr (X) .

Since the trace of the right hand side is cTr (1n) = n c, we find:

c =|G|n

Tr (X) . (2.7.8)

Now insert this result in (2.7.7), written in components:

M ij

def.=

∑g∈GD

′[g]ikXk`D[g]−1 `

j = |G|n

Tr (X) δij = |G|nXk

` δ`k δ

ij . (2.7.9)

The above equation can be rewritten in the following form:

Xk`

(∑g∈G

D′[g]ik D[g]−1 `j −|G|nδ`k δ

ij

)= 0 . (2.7.10)


Since (Xk`) is an arbitrary matrix, we conclude that the expression in brackets has to be

zero for any k, `, namely that:

∑g∈G

D′[g]ik(D[g]j`

)∗=|G|nδ`k δ

ij , (2.7.11)

where we have again used unitarity of D.Let us denote by D(s), s = 1, 2, . . ., all unitary irreducible representations of G, so that

different values of s correspond to inequivalent representations. Let ns denote the dimensionof D(s). We can express both properties (i) and (ii), namely the whole great orthogonalitytheorem, by the following single compact formula:

∑g∈G

D(s)[g]ik

(D(s′)[g]j`

)∗=|G|ns

δ`k δij δ

ss′ , (2.7.12)

If we represent the set of three indices s, i, j of D(s)[g]ik by a single index I, so that D[g](I) ≡D(s)[g]ik, equation (2.7.12) is rewritten in the following form:

∑g∈G

D[g](I) D[g](I′) =|G|nk

δII′, (2.7.13)

which expresses the orthogonality of a set of “vectors” D[g](I), labeled by the index I. Theseare vectors in the finite dimensional linear vector space HG of complex valued functions onthe group G. This space has dimension given by the order |G| of the finite group, since anyfunction f over G is completely defined by specifying its values on each of the |G| elementsof the group, namely by specifying its |G| components f(g). Given two functions f, f ′ on Gthe following scalar product is defined:

(f, f ′)def.=

∑g∈G

f(g)∗ f ′(g) . (2.7.14)

With respect to this product eq. (2.7.13) expresses the ortho-normality of the set of functionsD[g](I): The great orthogonality theorem states the mutual orthogonality of the matrixelements of unitary irreducible representations of G vectors in HG. (D[g](I)) provide a basisfor the space HG, so that any function f ∈ HG can be written as follows:

f(g) =∑I

cI D[g](I) =∑s

ns∑i,j=1

csij D(s)[g]ij . (2.7.15)

Remark: A great orthogonality theorem can be stated also for compact Lie groups, likeO(n) or U(n). In this case, however, the group is not finite set of elements, since its ele-ments are function of continuous parameters. Clearly we need to replace the sum over the

2.8. CHARACTERS 95

group elements by an integral over the group by suitably defining a (G-invariant) integrationmeasure dg:

∑g∈G −→

∫Gdg. The great orthogonality theorem then states that:∫

G

dgD(s)[g]ik

(D(s′)[g]j`

)∗=

Vol(G)

nsδ`k δ

ij δ

ss′ . (2.7.16)

where the volume of G is defined as Vol(G) =∫Gdg. Again we can define a space of

complex valued functions on G, with inner product (f, f ′) ≡∫Gdg f(g)∗ f ′(g), with respect

to which the matrix elements of unitary irreducible representations form a basis. We shall notelaborate further on this case. We just mention, as an example, the case of the group U(1),whose elements are functions g(ϕ) of a continuous angular parameter ϕ ∈ (0, 2π). For anyinteger s we have defined the representation D(s)[g(ϕ)] = ei s ϕ. We can take as integration

measure on the group dϕ, so that the volume of the group is Vol(U(1)) =∫Gdg =

∫ 2π

0dϕ =

2 π. Then eq. (2.7.16) is easily verified in this simple case, since the indices i, j, k, ` all runover a single value (ns ≡ 1 for any representation s):∫

G

dgD(s)[g]D(s′)[g]∗ =

∫ 2π

0

dϕei (s−s′)ϕ = 2π δss

′=

Vol(U(1))

nsδss′. (2.7.17)

Clearly complex valued functions over U(1) are functions f(ϕ) of ϕ ∈ (0, 2π), i.e. functionsover the circle. They form a linear vector space HU(1), which is infinite dimensional. Fromfunctional analysis we know that ei s ϕ, s = 0,±1,±2, . . ., form an ortho-normal basis offunctions for this space. Any function f(ϕ) can be expanded (the Fourier expansion) in thisbasis. Are there other irreducible U(1)-representations? If there existed an other irreducibleU(1)-representation D, it should still be one dimensional D[ϕ] = (D[ϕ]) and should be,according to our analysis, orthogonal to all the ei s ϕ, which can not be the case since we canFourier expand D[ϕ] in the basis ei s ϕ.

2.8 Characters

Given a finite dimensional representation D of a group G we define the character χ(g) of anelement g ∈ G in the representation D the quantity:

χ(g)def.= Tr(D[g]) = D[g]ii , (2.8.1)

where summation over i is understood. The set of all characters χ(g)|g ∈ G is called theset of characters of D. From the cyclic property of the trace we have that Tr(A−1 D[g] A) =Tr(D[g]), which implies that:

a) Equivalent representations have the same characters;

b) All elements of a same conjugation class have the same characters.


If D(s) are unitary irreducible representations of a finite group G (whatever we say for afinite group can be extended to compact Lie groups), from eq. (2.7.12), setting i = k andj = ` and summing over i and j, we find

(χ(s), χ(s′)) =∑g∈G

χ(s)(g)∗ χ(s′)(g) = |G| δss′ . (2.8.2)

That is the characters χ(s)(g), as complex valued functions over G, and thus elements ofHG, are mutually orthogonal. But characters have the same value over the elements of asame conjugation class. If we denote by C0, C1, C2, . . . , Cp−1 the conjugation classes of G,containing d0, d1, . . . , dp−1 elements respectively (we use the convention to identify C0 withthe class [e], so that d0 = 1), and by χ(s)(Ci) the value of χ(s) on the elements in Ci, theorthogonality relation (2.8.2) can be rewritten as follows:

p−1∑i=0

di χ(s)(Ci)

∗ χ(s′)(Ci) = |G| δss′ . (2.8.3)

Let the number of unitary irreducible representations of G be denoted by pr, so that theindex s runs from 0 to pr − 1, 0 denoting the identity representation. The characters shouldthen be thought of as pr complex valued functions over the set of p conjugation classes of G.These functions span a p-dimensional complex vector space, to be denoted by HC . On thisspace we can define a hermitian scalar product, such that, for any α, β ∈ HC :

(α, β) ≡∑g∈G

α(g)∗ β(g) =

p−1∑i=0

di α(Ci)∗ β(Ci) . (2.8.4)

Eq. (2.8.3) can then be viewed as an orthogonality relation over this space. Since in a pdimensional linear vector space there can be at most p mutually orthogonal (and thus linearlyindependent) vectors, there can be no more than p independent functions χ(s). Recall that slabels the unitary irreducible representations of the group. We conclude that, for any finitegroup G:

number of unitary irreducible representations ≤ number p of conjugation classes ,

that is pr ≤ p. We shall prove below a general property which states that the above twonumbers, for finite groups, are actually equal.

On the trivial (identity) representation, which we label with s = 0, χ(0)(g) = 1 for anyg ∈ G. Applying eq. (2.8.2) to the representations s 6= 0 and s′ = 0, we find:∑

g∈G

χ(s)(g) =

p−1∑i=0

di χ(s)(Ci) = 0 , (2.8.5)

that is the sum, over the group elements, of all characters of a given representation is zero.Writing (2.8.3) for s′ = s we find:

p−1∑i=0

di χ(s)(Ci)

∗ χ(s)(Ci) = |G| . (2.8.6)

2.8. CHARACTERS 97

The above property holds only for irreducible representations and thus gives a criterion forthe reducibility of a representation.

Suppose we have a reducible n-dimensional representation D of G which decomposesinto irreducible ones so that the representation D(s) occurs a number as of times in thedecomposition:

D −→pr−1⊕s=0

as times︷︸︸︷D(s) ⊕ . . .⊕D(s) . (2.8.7)

We have∑pr−1

s=0 ns as = n, where ns is the dimension of D(s). The trace of the n× n matrixD[g], for any g ∈ G, is the sum of the traces of each of its blocks, which consist in as blocksD(s)[g], for all irreducible representations D(s). We then find that, for any conjugation classCi:

χ(g) =

pr−1∑s=0

as χ(s)(g) . (2.8.8)

As anticipated, let us state and prove the following property:

Property 2.9: A finite group G has as many inequivalent irreducible unitary represen-tations as the number of conjugation classes.

To prove this, let us show that χ(s)(g) is a basis for HC , namely that any other elementα(g) of HC , which is linearly independent of the χ(s) or, equivalently, which is orthogonalto all the χ(s), must be zero4. Consider α(g) an element of HC . This means that α is acomplex valued function on G and moreover, for any g, h ∈ G, α(h−1 · g · h) = α(g), namelythe value of α is a function of the conjugation classes only. Suppose that α is orthogonal toall the χ(s), that is, for any irreducible representation s:∑

g∈G

α(g)∗ χ(s)(g) = 0 . (2.8.9)

This implies that α is orthogonal to the character χ of any representation of G, since thescalar product is bi-linear and χ can be expanded as a linear combination of the charactersχ(s). Let us show that, given a representation D of G over a vector space Vn, using α we canconstruct an intertwiner Sα,D between D and itself. Indeed let Sα,D be a linear operator onVn whose action on a generic vector V is represented as follows:

Sα,D(V) = Sα,D V ≡∑g∈G

α(g)∗D[g] V . (2.8.10)

4Let V and W be two elements of a complex linear vector space Vn on which a hermitian scalar product

(·, ·) is defined. If the two vectors are linearly independent the vector W ≡W − (V,W)‖V‖2 V is non zero. The

reader can show that (W,V) = 0. Conversely the reader can easily show that if V and W are orthogonal,they are linearly independent.


Let us show that, for any g ∈ G, Sα,D D[g] = D[g] Sα,D. For any V ∈ Vn and h ∈ G:

Sα,D D[h] V =∑g∈G

α(g)∗D[g] D[h] V =∑g∈G

α(g)∗D[g · h] V =

=∑

h·g′·h−1∈G

α(h · g′ · h−1)∗D[(h · g′ · h−1) · h] V =∑g′∈G

α(g′)∗D[h · g′] V =

= D[h] Sα,D V . (2.8.11)

From the arbitrariness of V we conclude that Sα,D is a matrix which intertwines the repre-sentation D with itself. If D is one of the irreducible representations D(s) of G, by Schur’slemma, Sα,D(s) = λs 1, where 1 is the identity matrix over the representation space of D(s),whose dimension we denote by ns. We can express λs as follows:

λs =1

nsTr(Sα,D(s)) =

1

ns

∑g∈G

α(g)∗ χ(s)(g) . (2.8.12)

Using the orthogonality assumption (2.8.9) we deduce that λs = 0, namely that, for anyirreducible representation Sα,D(s) = 0. Notice that a generic representation D is completely

reducible in combinations of D(s) and thus the matrix Sα,D will in general have, in a suitablebasis, a block diagonal form, each block being equal to Sα,D(s) . It then follows that, for anyrepresentation D, Sα,D = 0, that is:∑

g∈G

α(g)∗D[g] = 0 . (2.8.13)

Consider then the regular representation DR. The |G| × |G| matrices DR[g], for differentg in G, have non vanishing entries in different positions (i, j) and thus are |G| linearlyindependent matrices. Equation (2.8.13), applied to the regular representation, then impliesthat, for any g ∈ G, α(g) = 0, that is α is the zero-vector in HG, and thus that χ(s) form abasis of the same space. This shows that the number of irreducible representations (rangeof s) actually equals the number p of conjugation classes.

As an example of the above property let us consider the group D4. In Section 1.4.2, wehave found for D4 five conjugation classes: [e], [r`] = r, r3, [r2] = r2, [σ] = σ, β, [α] =α, γ. In Section 2.4.1, we have found for D4 precisely 5 irreducible representations: E,A1, A2, B1, B2. As we shall show below these are all the inequivalent irreducible unitaryrepresentations of D4.

Because of the above property, we will take the label s to run from 0 (corresponding tothe trivial representation) to p− 1.

Consider the decomposition (2.8.7) of a generic representation into the pr = p unitaryirreducible representations. Using the orthogonality relation (2.8.2) we can single out themultiplicities as in (2.8.7) by multiplying both sides times χ(s)(g)∗ and summing over g ∈ G.In formulas:

1

|G|∑g∈G

χ(g)χ(s)(g)∗ =1

|G|

p−1∑s′=0

∑g∈G

as′ χ(s′)(g)χ(s)(g)∗ =

1

|G||G|

p−1∑s′=0

as′ δss′ = as .(2.8.14)

2.8. CHARACTERS 99

Let us now compute the quantity χ(g)∗ χ(g), for a given g ∈ G, using eq. (2.8.8):

χ(g)∗ χ(g) =

p−1∑s,s′=0

as as′χ(s)(g)∗χ(s′)(g) . (2.8.15)

If we sum over g ∈ G and use the orthogonality property (2.8.2), we find:

∑g∈G

χ(g)∗ χ(g) =

p−1∑s,s′=0

as as′∑g∈G

χ(s)(g)∗χ(s′)(g) = |G|p−1∑s,s′=0

as as′δss′ = |G|

p−1∑s=0

(as)2 ,

that is:

p−1∑i=0

di χ(Ci)∗ χ(Ci) = |G|

p−1∑s=0

(as)2 . (2.8.16)

Comparing eqs. (2.8.6) to (2.8.16) we can state the following

Property 2.10: The sum of the squared moduli |χ(Ci)|2 of the characters over the conju-gation classes, weighted by the order di of each class, equals the order |G| of the group onlyif the representation is irreducible, otherwise it is an integer multiple of it.

Let us apply this property to the 3-dimensional representation D of D4 given in (2.4.35).Suppose we want to tell wether this representation is reducible or not. Let us compute theset of characters for each group element:

χ(e), χ(r), χ(r2), χ(r3), χ(σ), χ(α), χ(β), χ(γ) = 3, 1,−1, 1, 1, 1, 1, 1 .

or, equivalently, for each class:

χ([e]) = 3 , χ([r]) = 1 , χ([r2]) = −1 , χ([σ]) = 1 , χ([α]) = 1 . (2.8.17)

Compute the sum of the squares of χ(g), over the group elements:∑g∈G

χ(g)2 = 32 + 12 + (−1)2 + 12 + 12 + 12 + 12 + 12 = 16 = 2× 8 =

= 2× |D4| > |D4| . (2.8.18)

According to our criterion the three dimensional representation of D4 is therefore reducible.This was indeed shown by inspection of the explicit form of the matrices in this representa-tion. We have seen that:

D → E ⊕ A1 . (2.8.19)

Since the two representations E and A1 occur in the above decomposition just once, wehave

∑1s=0 a

2s = 1 + 1 = 2 and this explains the factor 2 on the right hand side of (2.8.18),

according to the general property (2.8.16).


Let us now consider the regular representation, whose dimension equals the order |G|of the group. Since the only matrix DR[g] which has diagonal entries, and therefore a nonvanishing trace, is DR[e] = 1|G|, the only non vanishing character of this representation isχR(e) = |G|.

Exercise 2.7: Apply the criterion expressed by Property 2.10 and prove that the regularrepresentation is reducible.

Applying (2.8.14) to the regular representation, the only contribution to the sum in theleft hand side, comes from g = e. Since χ(s)(e) = ns we find:

as =1

|G|χR(e)χ(s)(e)∗ = ns . (2.8.20)

This is telling us that each irreducible representation enters the decomposition of the regularrepresentation a number of times precisely equal to its dimension. From

∑p−1s=0 ns as = |G|

we finally find (Burnside Theorem):

p−1∑s=0

n2s = |G| , (2.8.21)

that is the sum of the squared dimensions of each irreducible representation of a finite groupequals the order of the group itself. In particular, for abelian groups we have as manyclasses as elements, p = |G|, which implies that, for any s, ns = 1, that is all irreduciblerepresentations are one-dimensional.

Let us come back to the D4 example. In this case the order of the group is |D4| = 8.If we sum the square of the dimensions of the irreducible representations E, A1, A2, B1, B2

that we found, recalling that E is two -dimensional and all the other are one-dimensional,we obtain:

4∑s=0

n2s = 1 + 1 + 1 + 1 + 22 = 8 = |D4| . (2.8.22)

The above sum already saturates the order of the group. According to our analysis, therecan not be other irreducible representations for D4, than those listed above.

Character Table. Characters are very useful since they allow to decompose representa-tions of a finite group into their unitary irreducible components according to (2.8.7). Tothis end we need to classify the set of characters χ(s)(Ci)i=1,...,p for each irreducible rep-resentation D(s). The set of characters χ(Ci)i=0,...,p−1 of a given representation D thendecompose in a unique way into the characters of its irreducible components, according toeq. (2.8.8), allowing us to determine the multiplicities as, for any s, and thus the whole de-composition (2.8.7). The characters for each irreducible representation are usually collectedin the character table, which has the following form:

2.9. OPERATIONS WITH REPRESENTATIONS 101

G [e] C1 C2 · · ·D(0) 1 1 1 · · ·D(1) χ(1)([e]) = n1 χ(1)(C1) χ(1)(C2) · · ·D(2) χ(2)([e]) = n2 χ2)(C1) χ(2)(C2) · · ·

......

......

For instance let us use the explicit form of the D4 irreducible representations (2.4.37),(2.4.38), (2.4.39), (2.4.40), to construct its character table:

D4 [e] r, r3 r2 σ, β α, γA1 1 1 1 1 1A2 1 1 1 −1 −1B1 1 −1 1 −1 1B2 1 −1 1 1 −1E 2 0 -2 0 0

The reader can verify, with reference to the above table, the orthogonality property (2.8.2).

Using this character table we can easily work out the decomposition of the regular rep-resentation of D4, whose character set is:

χR(e) = |D4| = 8 , χR(g 6= 2) = 0 . (2.8.23)

The only combination of the characters in Table 2.8 which reproduce (2.8.23), according tothe general formula (2.8.8) correspond to the following decomposition:

DR → 2× E ⊕ A1 ⊕ A2 ⊕B1 ⊕B2 , (2.8.24)

that is, as expected, all the irreducible representations occur in the decomposition of theregular one with a multiplicity given by their dimension (for instance E is two-dimensionaland occurs twice in (2.8.24)).

2.9 Operations with Representations

Let us define some operations which allow to construct new representations of a given groupfrom given ones.

2.9.1 Direct Sum of Representations

Consider two representations D1 and D2 of a groupG over the spaces Vn andWm respectively.If we denote by (ei) and (fa), i = 1, . . . , n and a = 1, . . . ,m, bases of Vn and Wm, respectivelyrecall that the direct sum Vn⊕Wm is defined as the (n+m)-dimensional vector space spannedby the basis (e′I) ≡ (e1, . . . , en, f1, . . . , fm), I = 1, . . . , n+m. A vector V′ in Vn⊕Wm is then


represented by the following column vector:

V′ = V ′ I e′I =

V ′ 1...

V ′n

V ′n+1

...V ′n+m

=

(V i

W a

), (2.9.1)

where V = V i ei and W = W a fa are the components of V′ along the subspaces Vn and Wm

respectively. Under the action of a group element g ∈ G, the component vectors transformas follows:

V −→ D1[g] V , W −→ D2[g] W , (2.9.2)

which implies that the whole vector V′ transforms as follows:

V′ −→(D1[g]ij V

j

D2[g]abWb

)=

(D1[g] 0

0 D2[g]

)V′ = D[g] V′ . (2.9.3)

It is straightforward to show that the correspondence D:

g ∈ GD−→ D[g] ≡

(D1[g] 0

0 D2[g]

), (2.9.4)

is a representation of G acting on the space Vn ⊕ Wm, that is D[g] = (D[g]IJ) is a (n +m)× (n+m) matrix, called direct sum of the representations D1 D2, and is denoted by thesymbol:

D = D1 ⊕D2 . (2.9.5)

We may generalize the above construction to define the direct sum D = D1 ⊕D2 . . . ⊕Dk

of the representations D1, D2, . . .Dk of G acting on the vector spaces Vn1 , Vn2 , . . . , Vnk as

the representation acting on the n-dimensional space Vn1 ⊕ Vn2 ⊕ . . .⊕ Vnk , n =∑k

`=i n` bymeans of the matrices

g ∈ GD−→ D[g] ≡

D1[g] 0 · · · 0

0 D2[g] · · · 0...

. . ....

0 0 · · · Dk[g]

. (2.9.6)

By construction D is reducible and its character χ(g) is the sum of the characters χi(g) ofeach representation Di: χ(g) =

∑ki=1 χi(g)


2.9.2 Direct Product of Representations

Consider two representations D1 and D2 of a groupG over the spaces Vn andWm respectively.As usual let us denote by (ei) and (fa), i = 1, . . . , n and a = 1, . . . ,m, bases of Vn and Wm,respectively. We define the direct product Vn⊗Wm a (nm)-dimensional vector space spannedby the basis (ei⊗ fa), where the ⊗ operation is defined on Vn×Wn and is bilinear in its twoarguments:

(aV1 + bV2)⊗W = a (V1 ⊗W) + b (V2 ⊗W) ,

V ⊗ (aW1 + bW2) = a (V ⊗W1) + b (V ⊗W2) , (2.9.7)

for any a, b numbers, Vi, V ∈ Vn and Wi, W ∈ Wm. Using this property we can writethe direct product of two vectors V = V i ei ∈ Vn and W = W a fa ∈ Wm as the vectorwhose components are the product of the components of the two vectors: V ≡ V ⊗W =V iW a ei⊗ fa. We can represent V = V⊗W only by means of its components V iW a, whichcan either be arranges in a n×m matrix, or in a nm vector whose components are labeledby the couple of indices I = (i, a): (V I) = (V 1W 1, V 2W 1, . . . V n−1Wm, V nWm). We willwrite V ⊗W = (V I) ≡ (V iW a). Under the action of an element g ∈ G, this product willtransform as follows:

V ⊗W −→ (D1 ⊗D2)[g] (V ⊗W) ≡ (D1[g] V)⊗ (D2[g] W) , (2.9.8)

or, in components:

V ⊗W = (V I) ≡ (V iW a) −→ (V ′ I) ≡ (V i ′W a ′) = (D1[g]ij Vj D2[g]abW

b) =

= (D1[g]ij D2[g]ab Vj, W b) =

= ((D1 ⊗D2)[g]iajb VjW b) =

= ((D1 ⊗D2)[g]IJ VJ) =

= (D1 ⊗D2)[g] (V ⊗W) , (2.9.9)

The nm× nm matrix (D1 ⊗D2)[g] = ((D1 ⊗D2)[g]iajb) ≡ (D1[g]ij D2[g]ab), for any g ∈ G,defines a representation of G called direct product (or Kronecker product ) of the representa-tions D1 and D2 and denoted by the symbol D1⊗D2. A generic element of Vn⊗Wm can notin general be written as the product of two vectors and has the form F = F ia ei⊗ fa ≡ (F I).It will transform, under the action of a g ∈ G, as the product of two vectors, according tothe product representation:

F −→ F′ = (D1 ⊗D2)[g] F , (2.9.10)

which means that each if its indices will transform under g according to the correspondingrepresentation:

F ia −→ F ia ′ = (D1 ⊗D2)[g]iajb Fjb = D1[g]ij D2[g]ab F

jb . (2.9.11)


The character χ(D1⊗D2)(g) of a product of two representations of a group g is obtained bytracing the matrix ((D1 ⊗D2)[g]iajb):

χ(D1⊗D2)(g) =n∑i=1

m∑a=1

(D1 ⊗D2)[g]iaia = (n∑i=1

D1[g]ii)(m∑a=1

D2[g]aa) = χ(D1)(g)χ(D2)(g) .

As for the direct sum of representations, we can extend the notion of direct product toan indefinite number of representations D1, D2, . . .Dk of G acting on the vector spacesVn1 , Vn2 , . . . , Vnk as the representation acting on the (n =

∏k`=1 n`)-dimensional space Vn1 ⊗

Vn2 ⊗ . . . ⊗ Vnk . A generic element of this product space F ≡ (F a1 a2... ak) has componentslabeled by kplets of indices (a1 a2 . . . ak), a1 = 1, . . . , n1, a2 = 1, . . . , n2 and so on, andtransforms as the product of components of vectors in each factor space V a1 V a2 . . . V ak :

F a1 a2... ak −→ F a1 a2... ak ′ = (D1 ⊗D2 ⊗ · · ·Dk)[g]a1 a2... ak b1 b2... bk Fb1 b2... bk ≡

≡ D1[g]a1b1 D2[g]a2b2 · · ·Dk[g]ak bk Fb1 b2... bk . (2.9.12)

The above transformation property defines the product representation D1 ⊗D2 ⊗ · · ·Dk:

F −→ F′ = (D1 ⊗D2 ⊗ . . .Dk)[g] F , (2.9.13)

In general the product of two or more representations of a same group is reducible, as weshall show in explicit examples.

So far we have considered objects with two or more indices, like F a1 a2... ak , which transformas product of contravariant components of vectors (i.e. all the indices are upper indices).We can generalize our analysis to objects of the form F ≡ (F a1 a2... ap

ap+1 ap+2... ak) whosecomponents have p-upper and (q = k − p)-lower indices and transform as products of p-contravariant and q-covariant components V a1 · · ·V apWap+1 · · ·Wak . Take for instance ouroriginal example of two representations D1 and D2 acting on two vector spaces Vn andWm. We have defined a quantity F = (F ia), whose components transform as productsV iW a of contravariant components V i and W a. We could have defined a quantity F =(F i

a) transforming as the product V iWa of contravariant components V i and covariantcomponents Wa. Recall that, if W a transform in the representation D2, the correspondingcovariant components Wa transform in the representation D′2 whose matrices are related tothose of D2 as follows: D′2[g] ≡ (D′2[g]a

b) = D2[g]−T

Wa −→ D′2[g]abWb = D2[g]−1 b

aWb . (2.9.14)

The quantity F = (F ia), under the action of a g ∈ G, will then transform as follows

F ia −→ F ′ ia = D1[g]ij D2[g]−1 b

a Fjb = (D1 ⊗D′2)[g]ia

bj F

jb , (2.9.15)

or, in compact form:

F −→ F′ = (D1 ⊗D′2)[g] F . (2.9.16)


The above transformation law defines the representation D1 ⊗ D′2: (D1 ⊗ D′2)[g]iabj ≡

D1[g]ij D2[g]−1 ba. The reader can verify that, if F = (F i

a) and G = (Gia) are two objects

transforming according to (2.9.15), any linear combination of them αF + βG ≡ (αF ia +

β Gia) will transform in the same way. Just like objects with two upper indices, also objects

with one upper and one lower indices span a linear vector space, which is the representationspace of D1 ⊗D′2.

Consider now the more general situation, mentioned above, of objects of the form F ≡(F a1 a2... ap

ap+1 ap+2... ak). Suppose the contravariant indices a1 a2 . . . ap transform with the rep-resentations D1, . . . ,Dp respectively and the covariant ones ap+1 ap+2 . . . ak with the repre-sentations D′ap+1

, . . . ,D′k, where a1, . . . , ak are indices of different vector spaces of dimensionsn1, . . . , nk. The quantities F will transform in the representation D1⊗· · ·Dp⊗D′ap+1

⊗D′ak :

F −→ F′ = (D1 ⊗ · · ·Dp ⊗D′p+1 ⊗D′k)[g] F , (2.9.17)

for any g ∈ G, where the above action is defined as follows:

F a1... apap+1... ak −→ F ′ a1... apap+1... ak =

= D1[g]a1b1 · · ·Dp[g]apbp Dp+1[g]−1 bp+1ap+1 · · ·Dk[g]−1 bk

ak Fb1... bp

bp+1... bk .

(2.9.18)

Just as for quantities with one covariant and one contravariant indices, quantities with q-covariant and p-contravariant indices form a linear vector space which is the representationspace of D1 ⊗ · · ·Dp ⊗D′p+1 ⊗ · · ·D′k.

The irreducible identity representation 1 is one-dimensional and all group elements arerepresented by the number 1. The direct product any representation times 1 is therefore therepresentation itself:

D⊗ 1 = 1⊗D = D . (2.9.19)

2.9.3 Tensors

Of particular interest is the case in which all the indices of the quantity F ≡ (F a1... apb1... bq)

refer to the same linear vector space Vn, so that a1, . . . , ap, b1, . . . , bq = 1, . . . , n. Thiscorresponds, in our previous analysis, to the case in which all the representations coincidewith a same one acting on Vn: D1 = D2 = Dk = D. The quantity F ≡ (F a1... ap

b1... bq),transforming in the representation:

D⊗p ⊗D′ ⊗p ≡p︷︸︸︷

D⊗ · · · ⊗D ⊗q︷︸︸︷

D′ ⊗ · · · ⊗D′ , (2.9.20)

is called a tensor of the representation D of G or simply a tensor (we have denoted by D⊗p

the p-fold product of the representation D). A tensor of this kind, with p-contravariant andq-covariant indices, is called type (p, q) tensor, while the total number of indices k = p + q


defines the rank of the tensor. Under the action of a group element g ∈ G the tensor Ftherefore transforms as follows:

F a1... apap+1... ak −→ F ′ a1... apap+1... ak =

= D[g]a1b1 · · ·D[g]apbp D[g]−1 bp+1ap+1 · · ·D[g]−1 bk

ak Fb1... bp

bp+1... bk .

(2.9.21)

All type (p, q)-tensors span a linear vector space, which we denote by V (p,q), which is therepresentation space of (2.9.20). Since tensors have been defined in relation to their trans-formation property (2.9.21) with respect to a certain group G, we talk about G-tensors.We have already come across tensors. Consider the metric g = (gij) defining a symmet-ric inner product on a real linear vector space. Under a generic linear transformation g inG = GL(n,R), the metric transforms as in (2.3.44):

gij −→ g′ij = D[g]−1 kiD[g]−1 `

j gk` . (2.9.22)

Comparing (2.9.22) with (2.9.21) we conclude that g = (gij) transforms as a type (0, 2)tensor. The Kronecker delta δji is an example of type-(1, 1) invariant GL(n,C)-tensor.Indeed if we transform it according to (2.9.21) we find:

δji −→ δ′ ji = D[g]−1 kiD[g]j` δ

`k = δji . (2.9.23)

We can define the inverse metric g−1 = (gij): gij gjk = δik. This defining condition of theinverse metric is invariant under a generic change of the RF, implemented by a genericelement of G = GL(n,R), provided gij transforms as a type-(2, 0) tensor

g′ ij g′jk = D[g]isD[g]jtD[g]−1 `j D[g]−1 r

k gst g`r = D[g]isD[g]−1 r

k δ`t g

st g`r =

= D[g]isD[g]−1 rk δ

sr = δij , (2.9.24)

where we have used the property D[g]jtD[g]−1 `j = δ`t . Other examples are the contravariant

components of vectors (V i) or the covariant ones (Vi), which are type (1, 0) and type (0, 1)tensor respectively. As opposed to the Kronecker delta, the metric is not an invariant tensorwith respect to a generic linear transformation, as we have seen, but it is invariant onlywith respect to transformations in G = O(r, s) (if the metric has r positive and s negativeeigenvalues). In this case we can go to an ortho-normal basis in which g = η(r,s) and say thatη(r,s) is a type (0, 2) invariant O(r, s)-tensor. We can define the following type-(0, n) tensor:

εi1i2...in =

0 two or more indices have the same value1 (i1i2 . . . in) even permutation of (1, 2, . . . , n)−1 (i1i2 . . . in) odd permutation of (1, 2, . . . , n)

. (2.9.25)

This definition makes sense only if we restrict to a transformation group which leaves the ten-sor invariant, otherwise it should refer to a specific RF. Such group is the group G = SL(n,R)of volume preserving transformations (i.e. of n × n real matrices with unit determinant).Indeed if we transform (2.9.25) under a generic g ∈ G we find:

εi1i2...in −→ D[g]−1j1i1 · · ·D[g]−1jn

in εj1j2...jn = det(D[g]−1) εi1i2...in = εi1i2...in , (2.9.26)


where we have used the property that det(D[g]−1) = 1. The tensor εi1i2...in is a type-(0, n)invariant SL(n,R)-tensor. Similarly we can define a type-(n, 0) invariant SL(n,R)-tensorεi1...in whose components are defined by the same rules (2.9.25).

Example: The defining representation of the (proper) Lorentz group SO(1, 3) has twoinvariant tensors: A type-(0, 2) η(1,3) = (ηµν) = diag(1,−1,−1,−1), which is the Minkowskimetric, and the type-(0, 4) tensor εµνρσ.

Exercise: Find a type (2, 0) invariant Sp(2n,R) tensor.Given a type-(p, q) tensor F a1...ap

b1...bq and a type-(r, s) tensor Ga1...arb1...bs we can con-

struct a type-(p+ r, q + s) tensor T as the product of the two:

T a1...ap+r b1...bq+s ≡ F a1...apb1...bq G

ap+1...ap+rbq+1...bq+s . (2.9.27)

The reader can show that the quantity defined above transforms indeed as a type-(p+r, q+s)tensor. We have seen examples of this general property when we originally constructed type-(2, 0) tensors (V i V j) as products of two contravariant components of vectors, namely of twotype-(1, 0) tensors (V i). Since product of tensors are still tensors, we say that the space ofall tensors, of whatever rank and type, close an algebra named tensor algebra.

Let us define a further operation on tensors. Given a type-(p, q) tensor F a1...apb1...bq , we

can consider only the components in which one upper and one lower index (say ap and bq)have the same value and sum over them, namely we compute

∑na=1 F

a1...ap−1 ab1...bq−1 a =

F a1...ap−1 ab1...bq−1 a (summation over a in the right hand side is understood). The result is a

type-(p− 1, q − 1) tensor G:

Ga1...ap−1b1...bq−1 ≡ F a1...ap−1 a

b1...bq−1 a, (2.9.28)

Let us show that the above object is actually a type-(p− 1, q − 1) tensor:

Ga1...ap−1b1...bq−1 −→ D[g]a1c1 · · ·D[g]acp D[g]−1 d1

b1 · · ·D[g]−1 dqa F

c1... cpd1... dq =

= D[g]a1c1 · · ·D[g]ap−1cp−1 D[g]−1 d1

b1 · · ·D[g]−1 dq−1bq−1 δ

dqcp F

c1... cpd1... dq

= D[g]a1c1 · · ·D[g]ap−1cp−1 D[g]−1 d1

b1 · · ·D[g]−1 dq−1bq−1 F

c1... cp−1 ad1... dq−1 a

= D[g]a1c1 · · ·D[g]ap−1cp−1 D[g]−1 d1

b1 · · ·D[g]−1 dq−1bq−1 G

c1...cp−1d1...dq−1 .

We say that Ga1...ap−1b1...bq−1 is obtained from F a1...ap

b1...bq by contracting the index ap withbq, or tracing over the indices ap and bq. We could have contracted any upper index aiwith any lower one bj of the same tensor F a1...ap

b1...bq thus obtaining p q independent tracesrepresented by different type-(p−1, q−1) tensors. A type-(1, 1) tensor (F i

j) can be thoughtof as a n × n matrix. In this case if we contract i with j, we are computing the trace ofthe matrix F i

i ≡∑n

i=1 Fii. The type (p, q) tensor F a1...ap

b1...bq could have been a productof two or more lower rank tensors and the indices over which we are contracting, ap andbq, could have belonged to different tensors. The above proof tells us that the result is stilla type-(p − 1, q − 1) tensor. As an example consider the metric (gij) and a contravariantvector V = (V k). The product of the two (gij V

k) is a type-(1, 2) tensor. If we contract


the index k with the index j we end up with a type-(0, 1) tensor (gij Vj) which is nothing

but the vector Vi of covariant components of V. We say that we have lowered the index jof the contravariant vectors by means of the metric tensor gij. In general, using the metricor its inverse, we can construct, out of a type-(p, q) tensor either a type-(p − 1, q + 1) or atype-(p+ 1, q − 1) tensor as follows:

F a1...apb1...bq −→

Ga1...ap−1

b1...bq bq+1 ≡ gbq+1a Fa1...ap−1a

b1...bq

Ga1...ap ap+1b1...bq−1 ≡ gap+1b F a1...ap

b1...bq−1b, (2.9.29)

in the first case we have lowered the index bq+1 using the metric, in the second we have raisedthe index ap+1 using the inverse metric.

From this discussion it is clear that the representation (2.9.20) of G = GL(n,C) withrespect to which a type-(p, q) transform is reducible. Indeed we can take subspaces V (p−`,q−`),` = 1, . . . ,min(q, p), of its representation space spanned by tensors of the form

V (p−`,q−`) ≡ F a1...ap−`b1...bq−` δ

ap−`+1

bq−`+1· · · δapbq ⊂ V (p,q) , (2.9.30)

which are invariant under the action of G, as the reader can prove. The components of ageneric type-(p, q) over the V (p−`,q−`) subspace are obtained by taking all possible contrac-tions of ` upper with ` lower indices. For instance the component of a type-(1, 1) (F i

j) inV (1,1) over the subspace of type-(0, 0) tensors (scalars) consists of the trace of the corre-sponding matrix F k

k δij. As a second example consider a type-(2, 2) tensor F = (F ij

k`). It

will be the sum of a trace-less component F0 = (F ij0 k`) in the space V (2,2), i.e. a tensor

whose traces are all vanishing, a component F1 in the space V (1,1) and a scalar componentF2 in V (0,0):

F ijk` = F ij

0 k` + F ij1 k` + F ij

2 k` , (2.9.31)

where

F ij1 k` = H i

k δj` +M i

` δjk +N j

k δi` + P j

` δik ,

F ij2 k` = F δik δ

j` +Gδjk δ

i` . (2.9.32)

The tensors H ik, M

ik, N

ik, P

ik are all traceless ( H i

i = M ii = N i

i = P ii = 0) and can be

expressed in terms of the four different traces of F ijk`.

Exercise: Show that:

H ij =

1

(n+ 3)(n− 1)[(n+ 2)F ik

1 jk − F ik1 kj − F ki

1 jk − F ki1 kj] ,

M ij =

1

(n+ 3)(n− 1)[−F ik

1 jk + (n+ 2)F ik1 kj − F ki

1 jk − F ki1 kj] ,

N ij =

1

(n+ 3)(n− 1)[−F ik

1 jk − F ik1 kj + (n+ 2)F ki

1 jk − F ki1 kj] ,

P ij =

1

(n+ 3)(n− 1)[−F ik

1 jk − F ik1 kj − F ki

1 jk + (n+ 2)F ki1 kj] ,


F =1

n(n2 − 1)[nF ij

2 ij − F ij2 ji] ,

G =1

n(n2 − 1)[−F ij

2 ij + nF ij2 ji] , (2.9.33)

Let us show now that the representation of GL(n,C) over type-(2, 0) tensors (and overtype-(0, 2) tensors) is reducible. We can decompose the space V (2,0) = (F ij) of type-(2, 0)

tensors, which is n2-dimensional, into two lower dimensional subspaces V(2,0)S and V

(2,0)A

consisting of symmetric and anti-symmetric tensors of the same kind:

V (2,0) = V(2,0)S ⊕ V (2,0)

A ,

V(2,0)S = (F ij)|F ij = F ji , V

(2,0)A = (F ij)|F ij = −F ji . (2.9.34)

The dimensions of these two subspaces is the number of independent parameters symmetricand anti-symmetric tensors depend on. Consider a n × n symmetric matrix (F ij). All itsentries below the diagonal are equal to the symmetric ones above the diagonal, so that theindependent entries consist in the n diagonal entries and the 1

2n(n − 1) entries above the

diagonal. The dimension of V(2,0)S is therefore n+ 1

2n(n−1) = 1

2n(n+ 1). Similarly the only

independent entries of an anti-symmetric matrix F ij are the entries above the diagonal, sothat the dimension of V

(2,0)A is 1

2n(n−1). The reader can show that dim(V

(2,0)A )+dim(V

(2,0)S ) =

dim(V (2,0)) = n2.Let us show that these subspaces are invariant under GL(n,C). Consider a type-(2, 0)

tensor with definite symmetry property, i.e. F ij = ±F ji and let us act on it by means of ageneric linear transformation g ∈ GL(n,C):

F ′ ij = D[g]i`D[g]jk F`k = ±D[g]i`D[g]jk F

k` = ±D[g]j`D[g]ik F`k = ±F ′ ij ,(2.9.35)

which shows that symmetric (anti-symmetric) tensors are transformed into symmetric (anti-

symmetric) tensors and thus V(2,0)S and V

(2,0)A are invariant subspaces. They correspond to

irreducible representations. A generic tensor (F ij) in V (2,0) is decomposed into its compo-

nents in V(2,0)S and V

(2,0)A as follows:

F ij = F ijA + F ij

S ,

F ijA =

1

2(F ij − F ji) ∈ V

(2,0)A ,

F ijS =

1

2(F ij + F ji) ∈ V

(2,0)S . (2.9.36)

The same decomposition can be given for type-(0, 2) tensors. Let us show that if we have atype-(0, 2) and a type-(2, 0) with opposite symmetry, their contraction (i.e. cross-contractionover all their indices) is zero. To show this consider a symmetric F ij

S and an antisymmetricFA ij tensors of the two types and compute their contraction:

F ijS FA ij = F ji

S FA ij = −F jiS FAji = −F ij

S FA ij = 0 . (2.9.37)


We would have obtained the same result had we considered and anti-symmetric tensor F ijA

and a symmetric one FS ij of the opposite type.

Consider now (G = O(n))-tensors of type-(2, 0) and let us show that V(2,0)S is further

reducible. We use now the property that, in an ortho-normal basis, the type-(2, 0) tensor

gij = δij is G-invariant. Using this tensor we can split V(2,0)S into two orthogonal subspaces:

V(2,0)T (trace tensors) of V

(2,0)TL (traceless tensors), defined as follows:

V(2,0)T = F ij = F δij , V

(2,0)TL = F ij|F ij = F ji , F ij δij = 0 , (2.9.38)

the former is clearly one dimensional while the latter has dimensions dim(V(2,0)S ) − 1 =

12n(n−1)−1. Let us show that these two subspaces are O(n)-invariant. The proof for V

(2,0)T

is trivial. As for V(2,0)TL , take a symmetric traceless tensor F ij. Under a O(n) transformation

it is mapped into a symmetric tensor F ′ ij, which we show below to be still traceless:

F ′ ij δij = D[g]ikD[g]j` Fk` δij = F ij δij = 0 , (2.9.39)

where we have used the orthogonality property of D[g]: D[g]ikD[g]j` δij = δk`. Notice thatanti-symmetric tensors can not be decomposed, as we did for symmetric ones, into a traceand a traceless part, since they are all traceless to start with. This is sue to the fact that δijis symmetric and the contraction of a symmetric and an anti-symmetric tensor is zero.

The space V (2,0) is then decomposed into three invariant subspaces:

V (2,0) = V(2,0)T ⊕ V (2,0)

TL ⊕ V (2,0)A . (2.9.40)

Correspondingly a tensor (F ij) in V (2,0) splits into its components along the three subspaces:

F ij = F ijT + F ij

TL + F ijA ,

F ijA =

1

2(F ij − F ji) ∈ V

(2,0)A ,

F ijT =

1

nF δij ∈ V

(2,0)T ,

F ijTL = F ij

S −1

nF δij ∈ V

(2,0)TL , (2.9.41)

where F ≡ F k` δk`.

Example: Consider the group O(3) and type-(2, 0) O(3)-tensors (F ij) which span a (32 =9)-dimensional space V (2,0). This representation is the two-fold product of the definingrepresentation 3 of the group: 3⊗ 3. This representation is reducible into one acting on theantisymmetric tensors, of dimension 1

23(3− 1) = 3, an other acting on symmetric traceless

tensors, of dimension 12

3(3 + 1) − 1 = 5 and a last one acting on the trace tensors, ofdimension 1. Denoting these irreducible representations by the boldface of their dimensionswe can write the following decomposition:

3⊗ 3 → 1⊕ 5⊕ 3 . (2.9.42)

Notice that, since for O(n) D = D′, that is covariant and contravariant vectors transformin the same representation, O(n) makes no difference between upper and lower indices, thus(F ij), (F i

j) and (Fij) transform in the same representation.

2.10. REPRESENTATIONS OF PRODUCTS OF GROUPS 111

2.10 Representations of Products of Groups

Consider a representation D1 of a group G on a linear vector space Vn and a representationD2 of a different group G′ on a linear vector space Wm. Generic vectors V = V i ei ∈ Vn andW = W a fa ∈ Wm in the two representation spaces transform under the action of elementsg ∈ G and g′ ∈ G′, respectively as follows:

V −→ D1[g] V , W −→ D2[g′] W . (2.10.1)

Let us define now the action of the couple of elements (g, g′) ∈ G × G′ on the vectorV⊗W ≡ V iW a ei⊗ fa ∈ Vn⊗Wm as the simultaneous, but independent, action of the twogroup elements on the respective vectors V and W:

V ⊗W −→ (V′ ⊗W′) = (D1 ×D2)[(g, g′)] (V ⊗W) ≡ (D1[g] V)⊗ (D2[g] W) ,

(2.10.2)

or, in components:

V iW a −→ V ′ iW ′ a = (D1 ×D2)[(g, g′)]iajb VjW b ≡ D1[g]ij D2[g′]ab V

jW b .(2.10.3)

The correspondence between elements (g, g′) in G × G′ and nm × nm matrices D1 ×D2)[(g, g′)] ≡ ((D1 × D2)[(g, g′)]iajb) = (D1[g]ij D2[g′]ab) is a representation of the prod-uct of the two groups. Consider indeed the action of the product of two elements (g1, g

′1)

and (g2, g′2) of G×G′ on a same vector V⊗W. Recalling the definition of product on G×G′

(i.e. that (g1, g′1) · (g2, g

′2) ≡ (g1 · g2, g

′1 · g′2)), we can write:

(D1 ×D2)[(g1, g′1) · (g2, g

′2)](V ⊗W) = (D1 ×D2)[(g1 · g2, g

′1 · g′2)](V ⊗W) =

= (D1[g1 · g2] V)⊗ (D2[g′1 · g′2] W) =

= (D1[g1] D1[g2] V)⊗ (D2[g′1] D2[g′2] W) =

= (D1 ×D2)[(g1, g′1)] [(D1[g2] V)⊗ (D2[g′2] W)] =

= (D1 ×D2)[(g1, g′1)] (D1 ×D2)[(g2, g

′2)] (V ⊗W) ,

for any V ∈ Vn and W ∈ Wn, which proves that

(D1 ×D2)[(g1, g′1) · (g2, g

′2)] = (D1 ×D2)[(g1, g

′1)] (D1 ×D2)[(g2, g

′2)] , (2.10.4)

namely that D1 ×D2 is a homomorphism on G × G′ and thus a representation. It can beshown that , if D1 and D2 are irreducible D1×D2 is an irreducible representation of G×G′.

In Section 2.9.2 we have denoted by the same symbol D1 × D2, the product of tworepresentations of a same group, defined on products of vectors by eq. (2.9.9). Let usemphasize here the conceptual difference between the product of two representations of asame group G and the product of representations of two different groups G, G′, which is arepresentation of G × G′. The former describes the simultaneous action of a single groupelement g ∈ G on the two vectors in the tensor product V⊗W, according to eq. (2.9.9). Thelatter describes the independent action of two elements g, g′ on the corresponding vectors inthe tensor product V ⊗W, according in eq. (2.10.2) . The former is generally reducible,the latter is irreducible if the two representations are.

Chapter 3

Constructing Representations

3.1 Constructing Representations of Finite Groups

Let us now address the problem of constructing the irreducible representations of a finitegroup G. In Section 2.4.2 we have introduced the notion of regular representation and ofabstract group algebra A[G]. We have later shown that the regular representation of afinite group is completely reducible into the sum of all the irreducible representations D(s),s = 0, . . . , p− 1, with multiplicities given by their dimensions ns:

DR →p−1⊕s=0

ns D(s) . (3.1.1)

This amounts to saying that the representation space of the regular representation can bewritten as the direct sum of the representation spaces of each irreducible representationin (3.1.1). As shown in Section 2.4.2, we can choose as representation space of DR ei-ther VR, with basis (egi)i=0,...,n−1, or, equivalently, the group algebra A[G] itself, with basis(gi)i=0,...,n−1. With each element gi ∈ G we associate the automorphism DR[gi] on A[G]defined as the left action of the algebra element gi on A[G]. Corresponding to the decom-position (3.1.1) A[G] will decompose into the direct sum of the representation spaces B(s) ofeach irreducible representation D(s):

A[G] →p−1⊕s=0

ns B(s) = B1 ⊕ B2 . . .⊕ B` , (3.1.2)

where we have denoted by different symbols (Bk)k=1,...,` the spaces of all the irreduciblerepresentations in (3.1.2). Clearly ` =

∑p−1s=0 ns. Since each Bk is invariant with respect to

the action of G, for each B ∈ Bk we must have that DR[gi](B) ≡ gi · B ∈ Bk, or, in otherwords, for any gi ∈ G:

DR[gi](Bk) = gi · Bk = Bk . (3.1.3)

From the bi-linearity of the product it follows that Bk is invariant with respect to the left-multiplication by any element of A = Ai gi ∈ A[G], represented by the action of DR[A] =

113

114 CHAPTER 3. CONSTRUCTING REPRESENTATIONS

AiDR[gi]: A · Bk = Ai (gi · Bk) = Bk. Since Bk is a subspace of A[G], it will be invariantwith respect to the left-multiplication by its own elements. This means that, if B1, B2 ∈ Bk,B1 ·B2 ∈ Bk. The space Bk is then close with respect to the product · induced by A[G], andthus it is an algebra itself. We say that Bk is a subalgebra of A[G]. Moreover it is invariantwith respect to the left-product by any element of A[G]:

∀A ∈ A[G] : A · Bk = Bk . (3.1.4)

This property makes the subalgebra Bk a left-ideal ofA[G]. We conclude that the space of anyrepresentation in which DR decomposes is defined by a left-ideal ofA[G]. The converse is alsotrue: Any left-ideal of A[G] defines the space of a representation of G in the decompositionof DR, being it invariant with respect to the action of all the automorphisms DR[gi]. Ofcourse, just as any representation of a finite group is completely reducible into smallerrepresentations, A[G] can always be written as the direct sum of the corresponding left-ideals: If B ⊂ A[G] is a left-ideal, its orthogonal complement B′, such that A[G] = B⊕B′, isstill a left-ideal. The two left-ideals may be still decomposable into direct sums of left-ideals.This occurs if and only if the corresponding representations or G are further reducible. Thisimplies that the left-ideals (Bk)k=1,...,` in (3.1.2), corresponding to irreducible representations,are themselves not further reducible into smaller left-ideals. We say that they are minimalleft-ideals. We have reduced the problem of decomposing the regular representation intoirreducible representations to that of finding finding the minimal left-ideals in A[G].

Any element A ∈ A[G] can be written, in a unique way, as the sum of its componentswithin each left-ideal in (3.1.2):

A = A1 + A2 + . . .+ A` , (3.1.5)

where Ak ∈ Bk. In particular we can decompose the identity element e = g0:

e = e1 + e2 + . . .+ e` . (3.1.6)

Let us take a generic element A ∈ A[G] and use the property A · e = A. Decomposing eaccording to (3.1.6) we find:

A = A · e = A · (e1 + e2 + . . .+ e`) = A · e1 + . . .+ A · e` . (3.1.7)

Since Bk are left-ideals, A ·ek ∈ Bk, but, being the decomposition (3.1.5) unique, we concludethat

A1 = A · e1 , . . . , A` = A · e` . (3.1.8)

In particular, if A ∈ Bk, A · ek = A and A · ek′ = Ak′ = 0, if k′ 6= k. We conclude thateach component ek is the identity element of the corresponding ideal Bk. Moreover, takingA = ek we find:

e2k ≡ ek · ek = ek , ek · ek′ = 0 , k′ 6= k . (3.1.9)

3.1. CONSTRUCTING REPRESENTATIONS OF FINITE GROUPS 115

An element A ofA[G] which equals its own square, A2 = A, is called idempotent. The identityelement ek of each minimal left-ideal Bk of A[G] is, according to eq. (3.1.9), idempotent.

Clearly the identity element e′ of any left-ideal B of A[G] is idempotent: (e′)2 = e′.Moreover we can write any element of a left-ideal B as the product of an element of A[G]times the identity element e′ of B. On the other hand, being B a left-ideal, the product ofany element of A[G] times e′ is in B. Therefore we can obtain any left-ideal through theleft-action of A[G] on its identity element: B = A[G] · e′. We say that the identity elemente′ ∈ B generates the left-ideal B.

Suppose we find an idempotent element a ∈ A[G], a2 = a. Let us show that the spaceA[G] · a ≡ A · a|A ∈ A[G], obtained through the left action of A[G] on a, is a left-idealof A[G] of which a is the identity element. Clearly this space is a left-ideal since, for anyA ∈ A[G], A · (A[G] · a) = A[G] · a. Given a generic element A · a ∈ A[G] · a we find(A · a) · a = A · a2 = A · a. This shows that a is the identity element of A[G] · a.

There is therefore a one-to-one correspondence between left-ideals of A[G] and idempotentelements of the same algebra.

Suppose now that the identity element e′ of a left-ideal B ⊂ A[G] can in turn be writtenas the sum of two idempotent elements: e′ = e1 + e2, with e2

1 = e1, e22 = e2, e1 · e2 = 0.

These elements will generate two disjoint subalgebras (B · e1, B · e2), of B so that we canwrite: B = B1 ⊕ B2 = B · e1 ⊕ B · e2. Indeed any element B = B · e′ = B · e1 + B · e2

of B can be written as the sum of a component B · e1 in B · e1 and a component B · e2 inB · e2. The two subalgebras are disjoint since, if there existed a common non-zero elementelement B ∈ (B ·e1)∩ (B ·e2) of the two algebras, being B in the fist subalgebra we can writeB = B · e1. On the other hand, being B an element of the second subalgebra we can alsowrite B = B · e1 = (B · e1) · e2 = B · (e1 · e2) = 0. This proves that the subspaces B · e1, B · e2

have just the zero-vector in common and thus are disjoint. Each of them is a left-ideal ofA[G], since B · e1 = A[G] · e′ · e1 = A[G] · e1 and similarly B · e2 = A[G] · e2. The left-ideal Bis therefore not-minimal and the corresponding representation of G is not irreducible sinceit decomposes into smaller representations on the two invariant subspaces.

On the other hand, if B is not minimal, it can be decomposed into the direct sum ofleft-ideals and its identity element decomposes into the sum of the identity elements of eachsubalgebra, which are themselves idempotent. We conclude that, a left-ideal B is minimal,and thus defines an irreducible representation if and only if its idempotent identity elementdoes not decompose into the sum of two or more idempotent elements. An idempotentelement with this property is called primitive idempotent element and our discussion showsthat there is a one-to-one correspondence between minimal left-ideals of A[G] and primitiveidempotent elements. Being each left ideal Bk in (3.1.2) minimal, their idempotent identityelements ek are primitive.

The problem of decomposing the regular representation into irreducible components canthus be reduced to the problem of finding primitive idempotent elements of A[G].

For what we have said each minimal left-ideal Bk is generated by its own primitiveidempotent element: Bk = A[G] · ek. From eq. (3.1.2) it follows that there are as manyprimitive idempotent elements of A[G] as irreducible representations in the decompositionof the regular one, namely ` =

∑p−1s=0 ns.


To summarize, in order to find the irreducible representations of a finite group G, we needto determine the primitive idempotent elements (ek)k=1,...,` of A[G]. For each such elementsek the space of the corresponding irreducible representation is just A[G]·ek. Actually we maygeneralize our search to elements a which are only idempotent up to a factor, namely suchthat a2 = µ a. To such an element there correspond a unique idempotent element a′ = a

µ.

3.1.1 Irreducible Representations of Sn and Young Tableaux

Let us apply our analysis to the permutation group1 Sm, of order n = m!. From thegeneral theory we know that there are as many irreducible representations of Sm as con-jugation classes. Each conjugation class, see Section 1.4.1, is defined by a partition (λ) ≡(λ1, λ2, . . . , λm) of m:

λ1 + λ2 + · · ·+ λm = m , λ1 ≥ λ2 ≥ · · · ≥ λm , (3.1.10)

and is represented graphically by a Young diagram. Young has identified, for the permutationgroup, a complete set of primitive idempotent (up to a factor) elements of A[Sm], calledYoung elements. Let us start giving two of them. One is the symmetrizer :

S =∑P∈Sm

P , (3.1.11)

the other is the antisymmetrizer :

A =∑P∈Sm

δ(P )P , (3.1.12)

where δ(P ) is +1 if the permutation P is even, −1 if it is odd.

Exercise 3.1: Show that δ(P · S) = δ(P ) δ(S), and also that δ(P ) = δ(S) δ(S · P ).

Let us show that these two elements of the algebra of Sm are idempotent up to a factor.Let us start showing that, for any S ∈ Sm:

S ·S = S , S ·A = δ(S) A . (3.1.13)

We indeed have:

S ·S =∑P∈Sm

S · P =∑

S′=S·P∈Sm

S ′ = S , (3.1.14)

S ·A =∑P∈Sm

δ(P )S · P =∑

S′=S·P∈Sm

δ(S) δ(S ′)S ′ = δ(S) A . (3.1.15)

1In this case of the permutation group we shall denote, as we have always done, the identity element byI: e = I.


Using the above properties we find:

S 2 =∑S∈Sm

S ·S =∑S∈Sm

S = m! S ,

A 2 =∑S∈Sm

δ(S)S ·A =∑S∈Sm

δ(S)2 A = m! A ,

S ·A =∑S∈Sm

S ·A = (∑S∈Sm

δ(S))A = (m!

2− m!

2) A = 0 . (3.1.16)

The corresponding idempotent elements are e1 = S√m!

and e2 = A√m!

. In virtue of eqs.

(3.1.14) and (3.1.15), the corresponding minimal left-ideals are one-dimensional since

B1 = A[Sm] ·S = S ·S |S ∈ Sm = S ,B2 = A[Sm] ·A = S ·A |S ∈ Sm = A , (3.1.17)

that is they correspond to one-dimensional representations of Sm, the former being theidentity representation, the latter being called the antisymmetric representation. Notice thatS and A are clearly primitives since the corresponding left-ideals, being one-dimensional,are minimal.

Example 3.1: Consider S2 = I, (1 2). Its algebra is two dimensional, generated by basiselements denoted by the same symbols as the corresponding group elements. A genericelement of A ∈ A[S2] reads:

A = A0 I + A1 (1 2) . (3.1.18)

The symmetrizer and antisymmetrizer read:

S = I + (1 2) ,

A = I − (1 2) . (3.1.19)

Example 3.2: Consider S3. Its elements are:

S3 = g0, g1, g2, g3, g4, g5 = I, (1 2), (1 3), (2 3), (1 2 3), (3 2 1) . (3.1.20)

Each of them corresponds to a basis element of the group algebra A[S3], denoted by thesame symbol. A generic element of A ∈ A[S3] has then the form:

A = A0 I + A1 (1 2) + A2 (1 3) + A3 (2 3) + A4 (1 2 3) + A5 (3 2 1) . (3.1.21)

The symmetrizer and antisymmetrizer read:

S = I + (1 2) + (1 3) + (2 3) + (1 2 3) + (3 2 1) ,

A = I − (1 2)− (1 3)− (2 3) + (1 2 3) + (3 2 1) . (3.1.22)


Each conjugation class (λ) of Sm is described by a Young diagram and is associated with anirreducible representation. A ns–dimensional irreducible representation occurs ns times inthe decomposition of the regular representation and thus corresponds to ns different primitiveidempotent elements, one for each copy of the same representation. Young gave a rule toconstruct these ns primitive idempotent elements corresponding to a given diagram. Westart defining a Young tableaux which is obtained by filling the boxes of a Young diagramwith different numbers from 1 to m in an arbitrary order. For example consider the Youngdiagram corresponding2 to (λ) = (4, 3, 2), and define the following tableaux:

1 4 6 72 3 89 5

(3.1.23)

The numbering of the boxes is arbitrary. Let us now denote by P the permutations of them numbers in the tableaux (in the above example m = 9), which leave the rows globallyinvariant, namely whose effect is to permute the numbers within each row. Examples ofpermutations P for the tableaux (3.1.23) are: (1 4 6 7), (1 4), (4 7), (2 3 8), (3 8), (9 5), . . ..The permutation (1 2), for example, is not of this kind, since it acts on numbers belonging totwo different rows. Define, in correspondence with a given Young tableau, the symmetrizer :

P =∑P

P ∈ A[Sm] . (3.1.24)

For instance, the symmetrizer corresponding to the tableau

1 23 is

P = I + (1 2) ∈ A[S3] . (3.1.25)

Let Q generically denote those permutations which leave the columns globally invariant,namely which act on numbers lying on a same column of a given tableaux, and define theantisymmetrizer:

Q =∑Q

δ(Q)Q ∈ A[Sm] . (3.1.26)

For instance, the antisymmetrizer corresponding to the tableau

1 23 is

Q = I − (1 3) ∈ A[S3] . (3.1.27)

Since two or more numbers can not lie on a row and a column at the same time, the Pand the Q permutations form disjoint sets. The P and the Q permutations are products of

2We use the convention of writing only the first non vanishing λi’s, the remaining being zero, so that, forinstance, (4, 3, 2, 0, 0, 0, 0, 0, 0) ≡ (4, 3, 2).


permutations on each row and column respectively and close two different subgroups of Sm.The Young element corresponding to the given Young tableau is defined as:

Y(λ) ≡ Q ·P . (3.1.28)

For instance, for the tableau

1 23 the Young element is:

Y 1 23

= [I − (1 3)] · [I + (1 2)] ∈ A[S3] . (3.1.29)

Clearly the elements S and A are the Young elements corresponding to the diagrams(m, 0, . . . , 0) and (1, 1, . . . , 1) respectively (all tableaux defined out of these two diagramscorrespond to the same Young elements: S and A respectively). We shall not prove ingeneral that Y(λ) is idempotent up to a factor, we shall restrict ourselves to some examples.In defining a Young tableau, we assigned numbers to boxes. Of course we can choose adifferent numbering. This amounts to performing a permutation on the numbers in theoriginal tableau. Let S be a permutation in Sm, the effect of S on the Young elementcorresponding to a tableau is:

YS−→ Y ′ = S · Y · S−1 , (3.1.30)

where YS is the Young element corresponding to the tableau obtained from the original oneby performing S on its labels. For instance:

Y 2 31

= (1 2 3) Y 1 23

(1 2 3)−1 . (3.1.31)

We need to identify those tableaux which define primitive idempotent elements, such that,if Y and Y ′ are two such elements Y · Y ′ = 0, namely one can not be obtained one fromthe other by the left action of the algebra: Y ′ 6= A · Y for some A ∈ A[Sm]. Each ofthem will generate a different irreducible representation of Sm. For a given Young diagram,such elements are defined by standard tableaux. A standard tableau is obtained by fillingthe corresponding diagram starting from the first row, from left to right, with the numbers1, 2, . . . ,m, respecting the following condition: Each number on the ith line should be greaterthan the number on the same column of the (i − 1)th line. A general property, which weshall not prove, states that, for a given Young diagram, any Young tableaux can be obtainedas (combinations of) left-products of group algebra elements on standard ones, and differentstandard tableaux can not be obtained one from an other in this way. The tableau (3.1.23)for instance is not standard since the 3 on the second row is smaller than the correspondingnumber (4) on the first. Examples of standard tableaux of the same type are:

1 2 3 45 6 78 9

,1 2 3 75 6 48 9

, · · · (3.1.32)

The standard tableaux for the diagram (2, 1) of S3 are Y = 1 23 and Y ′ = 1 3

2 . Standardtableaux corresponding to the same Young diagrams can, on the other hand, be obtained


one from the other through a conjugation by a certain permutation. Therefore they generatedifferent copies of the same irreducible representation of Sm. It turns out that the number ofstandard tableaux corresponding to a given diagram coincides with the dimension of the cor-responding representation, which is consistent with the property that a given representationenters the decomposition of the regular one a number of times equal to its dimension.

Let us discuss the S3 case in detail. The Young diagrams are three and are given in

Table 1.4.10. The first and the last diagrams have just one standard tableaux each

while the second diagram have two standard tableaux. The standard tableaux and thecorresponding Young elements, are:

S ↔ 1 2 3 , A ↔123

, Y ↔ 1 23

, Y ′ ↔ 1 32

, (3.1.33)

where

S = I + (1 2) + (1 3) + (2 3) + (1 2 3) + (3 2 1) ,

A = I − (1 2)− (1 3)− (2 3) + (1 2 3) + (3 2 1) ,

Y = [I − (1 3)] · [I + (1 2)] = I − (1 3) + (1 2)− (1 2 3) ,

Y ′ = [I − (1 2)] · [I + (1 3)] = I + (1 3)− (1 2)− (3 2 1) , (3.1.34)

Exercise 3.2: Verify that:

S 2 = 6 S , A 2 = 6 A , Y 2 = 3 Y , (Y ′)2 = 3 Y ′ ,

S ·A = A ·S = 0 , S · Y = Y ·S = 0 , S · Y ′ = Y ′ ·S = 0 ,

A · Y = Y ·A = 0 , A · Y ′ = Y ′ ·A = 0 , Y · Y ′ = Y ′ · Y = 0 . (3.1.35)

Exercise 3.3: Express the identity element I as a linear combination of the S , A , Y , Y ′.Let us verify that Y generates a two dimensional representation of S3, namely that

A[S3] · Y is a two-dimensional space. To this end we act on Y to the left by the S3

elements, namely we compute gi · Y :

I · Y = Y = I − (1 3) + (1 2)− (1 2 3) ,

(1 2 3) · Y = (1 3)− (2 3) + (1 2 3)− (3 2 1) ,

(1 3) · Y = −I + (1 3)− (1 2) + (1 2 3) = −Y ,

(1 2) · Y = I + (1 2)− (2 3)− (3 2 1) = Y + (1 2 3) · Y ,

(2 3) · Y = −(1 3) + (2 3)− (1 2 3) + (3 2 1) = −(1 2 3) · Y ,

(3 2 1) · Y = ((1 2) · (1 3)) · Y = −(1 2) · Y = −Y − (1 2 3) · Y . (3.1.36)

We see that all the elements gi · Y can be expressed as combinations of two basis elementsf1 ≡ Y and f2 ≡ (1 2 3) ·Y . This means that A[S3] ·Y is a two-dimensional space on whichthe following two-dimensional representation D of S3 act:

D[gi](fa) = D[gi]ba fb ≡ gi · fa . (3.1.37)

3.2. IRREDUCIBLE REPRESENTATIONS OF GL(N,C) 121

Where the matrix representation D[gi] = (D[gi]ba) of each group element reads:

D[I] = 12 , D[(1 2)] =

(1 01 −1

), D[(1 3)] =

(−1 10 1

),

D[(2 3)] =

(0 −1−1 0

), D[(1 2 3)] =

(0 −11 −1

), D[(3 2 1)] =

(−1 1−1 0

).

Exercise 3.4: Following the same procedure as above, show that the space A[S3] · Y ′ isgenerated by Y ′, (3 2 1) · Y ′ and thus is two-dimensional.

Exercise 3.5: Construct the two-dimensional matrix representation on the subalgebragenerated by Y ′, in the basis (Y ′, (3 2 1) · Y ′).

Exercise 3.6: Prove that Y 3 21

= Y − (3 2 1) · Y ′.

To summarize, we have shown for S3 that the regular representation decomposes asfollows:

DR → ⊕ ⊕ 2× , (3.1.38)

which implies that S3 has two one-dimensional representations and a two-dimensional ir-reducible representation. Correspondingly the algebra A[S3] decomposes into the followingdirect sum of minimal left-ideals:

A[S3] = A[S3] ·S ⊕A[S3] ·A ⊕A[S3] · Y ⊕A[S3] · Y ′ , (3.1.39)

where the last two spaces are two dimensional and are generated by the Young elementscorresponding to the standard tableaux 1 2

3 and 1 32 . They define the two copies of the

representation in eq. (3.1.38).Exercise 3.7: Construct for the group S4 the Young tableaux corresponding to the Young

diagrams in Table 1.4.11. What is the dimension of the irreducible representation correspond-ing to each diagram?

3.2 Irreducible Representations of GL(n,C)Consider the linear space V (p,0) of type-(p, 0) tensors T = (T i1i2···ip), the indices ii, . . . , iprunning from 1 to n, where V (p,0) represents the product of p copies of a n-dimensionalcomplex linear vector space Vn: V (p,0) = Vn ⊗ Vn ⊗ · · · ⊗ Vn. Recall that under a GL(n,C)transformation D = (Di

j)3 the tensor transforms as follows:

T i1i2···ip −→ T ′ i1i2···ip = Di1j1 D

i2j2 . . . D

ipjp T

j1j2···jp . (3.2.1)

Symbolically we shall represent the action of GL(n,C) on the tensor by the operator D⊗p ≡D⊗ . . .⊗D, so that the above transformation will read

T −→ T′ = D⊗p T . (3.2.2)

3Here, for the sake of simplicity, we denote by the symbol D the matrix D[g] corresponding to a genericelement g ∈ GL(n,C).


Let us define the action of a permutation S ∈ Sp on type-(p, 0) tensors represented by an

operator S on V (p,0) which maps a type-(p, 0) tensor T into a type-(p, 0) tensor ST defined,in components, as follows:

(ST)i1···ip ≡ T iS(1)···iS(p) , (3.2.3)

where S maps the label k into S(k). The correspondence between S and the operator S onV (p,0) is a homomorphism. To show this we need to prove that the operator correspondingto the product of two permutations coincides with the product of the operators associated

with each of the two: S · P = S · P , for any two S, P ∈ Sp. Consider the action of theproduct:

(S · PT)i1···ip = T iS(P (1))···iS(P (p)) . (3.2.4)

Consider now the consecutive action of the two operators:

(S (P T))i1···ip = (PT)iS(1)···iS(p) = T iS(P (1))···iS(P (p)) . (3.2.5)

This proves that the mapping S ∈ Sp → S ∈ Aut(V (p,0)) is a homomorphism, namely arepresentation of the permutation group over type-(p, 0) tensors. For instance:

((1 2) T)i1i2 = T i2i1 , ((1 2 3) T)i1i2i3 = T i2i3i1 . (3.2.6)

Suppose we want to compute ((1 3) (1 2) T)i1i2i3 , we have ((1 3) (1 2) T)i1i2i3 = ((1 2) T)i3i2i1 =

T i2i3i1 = ((1 2 3) T)i1i2i3 . Notice that the effect of (1 2) on a tensor is to permute the fist andthe second indices, whatever they are. The effect of a permutation on a tensor is just to re-shuffle its components, changing their labels. This representation induces a representation ofthe whole algebraA[G] associated with the group Sp on the type-(p, 0) tensors: An algebra el-

ement A =∑

S∈Sp A(S)S act on tensors by means of the endomorphism A =∑

S∈Sp A(S) S.

Let us show that the action of a GL(n,C)-transformation and of a permutation on a tensorcommute, namely that, if T′ = D⊗p T,

ST′ = (ST)′ , (3.2.7)

that is S and D⊗p are commuting operators. Let us write the tensor on the left hand sidein components:

(ST′)i1...ip = T ′ iS(1)...iS(p) = DiS(1)jS(1) . . . D

iS(p)jS(p) T

jS(1)...jS(p) =

= DiS(1)jS(1) . . . D

iS(p)jS(p) (ST)j1...jp . (3.2.8)

Now we can just re-shuffle the D matrices on the right hand side and write:

(ST′)i1...ip = Di1j1 . . . D

ipjp (ST)j1...jp = (ST)′ i1...ip , (3.2.9)

which proves eq. (3.2.7). Since the endomorphism A, representing a generic element ofthe algebra A[Sp], is a combination of S, S ∈ Sp, it will also commute with the action of


GL(n,C) on tensors. In particular the action of the primitive idempotent (up to a factor)Young elements Y(λ) commutes with D⊗p:

Y(λ) (D⊗p T) = D⊗p (Y(λ) T) . (3.2.10)

As a consequence of this the image of V (p,0) through Y(λ) is invariant under GL(n,C):

D⊗p (Y(λ) V(p,0)) = Y(λ) (D⊗p V (p,0)) = Y(λ) V

(p,0) . (3.2.11)

Given a tensor T = (T i1...ip), and a Young tableau (λ), what does Y(λ) T represent? Considerp = 2. In this case the Young elements are:

S = Y 1 2 = I + (1 2) , A = Y 12

= I − (1 2) . (3.2.12)

Since S 2 = 2 S and A 2 = 2 A , the idempotent elements are e1 = 12S and e2 = 1

2A , so

that the identity I decomposes in the corresponding left-ideals as follows:

I = e1 + e2 =1

2S +

1

2A . (3.2.13)

The same decomposition holds for the corresponding action on the tensors T = (T i1i2):

T = I T =1

2(S T) +

1

2(A T) = TS + TA . (3.2.14)

We have decomposed T into two components TS = (T i1i2S ) and TA = (T i1i2A ):

T i1i2S =1

2(S T)i1i2 =

1

2[(I + (1 2)) T]i1i2 =

1

2(T i1i2 + T i2i1) ,

T i1i2A =1

2(A T)i1i2 =

1

2[(I − (1 2)) T]i1i2 =

1

2(T i1i2 − T i2i1) , (3.2.15)

corresponding to the symmetric and antisymmetric part of T. The images S V (p,0) andA V (p,0) of V (p,0), through the symmetrizer and antisymmetrizer respectively, represent thesubspaces of symmetric and antisymmetric type-(2, 0) tensors respectively and, as discussedin the previous sections, they indeed define irreducible representations of GL(n,C). Theaction of the two elements S , A correspond to two distinct symmetry operations whichcan be applied to a type-(2, 0) tensor: The symmetrization and the antisymmetrization withrespect to its indices. In general the action of a Young element Y(λ) on a tensor representa maximal set of symmetry operations which can be applied to the tensor. By maximal wemean that any further symmetry operation on the resulting tensor would give either zeroor the tensor itself. For instance if we further symmetrize the tensor TS with respect to

its indices we obtain the tensor itself (this is due to the fact that e1 = 12S is idempotent

and thus e1 TS = e21 T = TS), while if we antisymmetrize it we get zero (this being related

to the property e2 · e1 = 0 which implies e2 TS = e2 e1 T = 0). We conclude that e1


and e2 are then projection operators into the symmetric and antisymmetric part of a type-(2, 0) tensor. Consider now a type (p, 0) tensor T = (T i1...ip). What is a maximal set ofsymmetry conditions that we can impose on the tensor? We can group the indices intosubsets containing λ1, λ2, . . . , λp indices (take λ1 ≥ λ2 ≥ . . . ≥ λp), so that λ1 + . . .+λp = p.We can then symmetrize the tensor to be symmetric with respect to the indices within thesame group. This amounts to construct a tensor which is totally symmetric with respectto the interchange of any couple of indices within each subset. Each component of thisnew tensor are obtained by summing together all the components of the original one whichare obtained from one another through the action of permutations P of indices within theeach group. In other words if we define the Young tableau (λ) = (λ1, λ2, . . . , λp) whoserows are filled with the labels of the indices belonging to each group, recalling that thesymmetrizer P is the sum of all the permutations of the objects within each group, the

symmetrization procedure described above amounts to applying the operator P to theoriginal tensor. For instance we can group the three indices of T = (T i1i2i3) into twosubsets (i1, i3) and (i2) (in this example λ1 = 2, λ2 = 1, λ3 = 0). Symmetrizing the

tensor with respect to i1 and i3 yields the tensor N i1i2i3 = (PT)i1i2i3 = T i1i2i3 + T i3i2i1 ,

where P = I + (13) is the symmetrizer corresponding tot he Young diagram 1 32

, which

is symmetric in i1 and i2. After having symmetrized the tensor with respect to the indiceswithin each subset, we can still impose antisymmetry conditions. Of course these conditionscan not involve indices within a same subset, since only the zero-tensor can be antisymmetricand symmetric with respect to a same couple of indices. We can impose antisymmetryconditions among indices belonging to different sets. Since the order of indices within eachset is irrelevant, we can place the labels of these indices in the first column of the Youngtableau (λ). After having performed this first antisymmetrization, we can choose otherindices within different subsets, place them in the second column of the Young tableau andproceed antisymmetrize the tensor with respect to them. Antisymmetrizing a tensor withrespect to a subset of indices amounts to defining a new tensor whose components are thesum of all components of the original tensors obtained one from another by a permutation ofthe indices within this subset, with a plus sign if the permutation is even and a minus sign ifit is odd. We iterate this procedure until we can apply no further symmetry condition. This

antisymmetrization procedure then amounts to applying to the tensor P T, obtained fromthe first symmetrization, the anitsymmetrizer operator Q corresponding to the same tableau.In other words the resulting tensor is obtained from the original one by applying to it the

operator Y(λ) = Q P. Each Young tableau defines therefore a maximal set of symmetryoperations which can be implemented to a tensor by the action of the corresponding Youngelement. In our example, on the tensor N i1i2i3 , which is symmetric with respect to (i1, i3), wecan still antisymmetrize with respect to the couple (i1, i2) and obtain the tensor M i1i3,i2 =

(Q PT)i1i2i3 = N i1i2i3 − N i2i1i3 = T i1i2i3 + T i3i2i1 − (T i2i1i3 + T i3i1i2), being Q = I − (12).Notice that the antisymmetrization spoils the symmetry property of the tensor N i1i2i3 withrespect to the indices in each subset, so that M i1i3,i2 6= M i3i1,i2 . There is no further symmetrycondition that we can impose on the tensor. Such tensor is then called irreducible and isdenoted by M i1i3,i2 , where we divide the subsets of indices by commas. Any tensor can be


written as the sum of irreducible components. Each component is obtained by acting onthe tensor by means of a primitive idempotent element of the algebra A[Sp]. Indeed takefor example a generic tensor T = (T i1i2i3) and lets act on it by means of the Young elementY 1 3

2

= Y ′:

(Y ′T)i1i2i3 = [(I − (1 2)) (I + (1 3)) T]i1i2i3 = T i1i2i3 + T i3i2i1 − T i2i1i3 − T i3i1i2 = M i1i3,i2 .

The reader can verify that M i1i3,i2 = −M i2i3,i1 . The effect of the symmetrizer P = (I+(1 3))in the above Young element is to symmetrize with respect to (i1, i3), while the effect of theantisymmetrizer Q = I − (1 2) is to antisymmetrize with respect to (i1, i2). In summary,each irreducible tensor, denoted by the symbol M , corresponds to a Young tableau whosefirst row is filled with the labels of the first set of λ1 indices, the second row, with the labelsof the second set of λ2 indices and so on, and is obtained by acting on a generic tensor bymeans of the corresponding Young element, which can then be thought of, up to a coefficient,as a projection operator. Consider for instance the tableau:

1 3 6 8 92 4 7 115 1012 (3.2.16)

The corresponding irreducible tensor is denoted by:

M i1i3i6i8i9,i2i4i7i11,i5i10,i12 . (3.2.17)

The Young element corresponding to the above tableau Y = Q P projects any type-(12, 0)

tensor into the irreducible component (3.2.17) as follows: The symmetrizer P symmetrizesthe tensor with respect to the indices within each row, while the effect of the antisymmetrizerQ is to antisymmetrize the resulting tensor with respect to the indices within each column.As a result, the tensor M is antisymmetric with respect to the indices within each column.For instance the tensor in eq. (3.2.17) is antisymmetric in the subsets of indices (i1, i2, i5, i12),(i3, i4, i10), (i6, i7) and (i8, i11).

For a given Young tableau (λ) the space Y(λ) V(p,0) consists of irreducible tensors and

defines a representation of GL(n,C). Since the only process of reduction of a tensor whichis consistent with the free action of GL(n,C) on its indices and commutes with it is the

symmetrization procedure described above, and since the tensors in each subspace Y(λ) V(p,0)

are subject to a maximal set of symmetrization conditions, defined by the corresponding

tableau (λ), Y(λ) V(p,0) defines an irreducible representation of GL(n,C).

We conclude that each Young tableau defines an irreducible representation of GL(n,C).The fundamental representation D acts on contravariant vectors (V i) which are type-

(1, 0) tensors, it is associated with the Young diagram 1 which coincides with its diagram. We shall identify D with the corresponding Young diagram and represent it by a single

box: D ≡ . The p-fold product of fundamental representations, acting on the type-(p, 0)


tensors in V (p,0) will be then denoted by:

p︷︸︸︷D⊗ · · · ⊗D =

p︷︸︸︷⊗ · · · ⊗ (3.2.18)

The decomposition of this product into irreducible representations correspond to the decom-

position of its representation space V (p,0) into irreducible tensors within Y(λ) V(p,0), for all

Young tableaux (λ). This decomposition is effected as follows: Suppose, for a given tableau(λ), Y 2

(λ) = µλ Y(λ). The corresponding idempotent element e(λ) is e(λ) = 1µλ

Y(λ). The repre-

sentation I on V (p,0) of the identity element I splits into the sum of the representations e(λ)

of the primitive idempotent elements e(λ) corresponding to all standard Young tableau:

I =∑

Young tableaux (λ)

e(λ) =∑

Young tableaux (λ)

1

µλY(λ) .

Applying both sides on V (p,0) we obtain the following decomposition:

V (p,0) = I V (p,0) =∑

Young tableaux (λ)

(e(λ) V

(p,0))

=∑

Young tableaux (λ)

(1

µλY(λ) V

(p,0)

).

(3.2.19)

The effect of e(λ) is to project the tensors in V (p,0) into the irreducible component correspond-ing to the diagram (λ). Equation (3.2.19) then defines the decomposition of the product(3.2.18) into irreducible representations. Let us apply this procedure to type- (3, 0) tensorsand decompose the 3-fold product ⊗ ⊗ of the fundamental representation of GL(,C).In Section 3.1.1 we have found four standard Young tableaux, the corresponding Youngelements being:

S = Y 1 2 3 , A = Y 123

, Y = Y 1 23

, Y ′ = Y 1 32

. (3.2.20)

These define four primary idempotent elements: e1 = 16S , e2 = 1

6A , e3 = 1

3Y , e4 = 1

3Y ′, .

The identity element decomposes as follows:

I =1

6S +

1

6A +

1

3Y +

1

3Y ′ . (3.2.21)

The space of type- (3, 0) tensors decompose correspondingly into the following subspaces

V (3,0) = I V (3,0) =1

6S V (3,0) +

1

6A V (3,0) +

1

3Y V (3,0) +

1

3Y ′ V (3,0) , (3.2.22)

each of which is acted on by a corresponding irreducible representation of GL(n,C). Let usanalyze their content. Consider a generic tensor T = (T i1i2i3) and project it into the abovesubspaces by acting on it with the corresponding idempotent element ei:

T = I T =4∑i=1

e1T =1

6S T +

1

6A T +

1

3Y T +

1

3Y ′T . (3.2.23)


where:

(e1T)i1i2i3 =1

6(S T)i1i2i3 =

1

6(T i1i2i3 + T i2i3i1 + T i3i1i2 + T i2i1i3 + T i3i2i1 + T i1i3i2) =

= M i1i2i3 , (3.2.24)

(e2T)i1i2i3 =1

6(A T)i1i2i3 =

1

6(T i1i2i3 + T i2i3i1 + T i3i1i2 − T i2i1i3 − T i3i2i1 − T i1i3i2) =

= M i1,i2,i3 , (3.2.25)

(e3T)i1i2i3 =1

3(Y T)i1i2i3 =

1

3(T i1i2i3 − T i3i2i1 + T i2i1i3 − T i2i3i1) =

= M i1i2,i3 , (3.2.26)

(e4T)i1i2i3 =1

3(Y ′T)i1i2i3 =

1

3(T i1i2i3 − T i2i1i3 + T i3i2i1 − T i3i1i2) =

= M i1i3,i2 , (3.2.27)

(3.2.28)

We will then write the corresponding decomposition of the 3-fold product of the fundamentalrepresentation of GL(n,C), as follows:

⊗ ⊗ → 1 2 3 ⊕123⊕ 1 2

3⊕ 1 3

2≡ ⊕ ⊕ 2× . (3.2.29)

The above decomposition is also written representing each diagram by the correspondingλ-indices:

(1)⊗ (1)⊗ (1) → (3)⊕ (1, 1, 1)⊕ 2× (2, 1) . (3.2.30)

Each Young diagram (λ) defines a kind of tensor with maximal symmetry and thus corre-

sponds to an irreducible representation D(λ) of GL(n,C)4: For instance corresponds to

the representation with tensors of the kind M ··,·. Different Young tableaux corresponding tothe same diagram, define different copies of the same representation in the decomposition ofa generic type-(p, 0) tensor, they just differ from the choices of the indices of the T- tensor onwhich the symmetry operations are applied and thus lead to the same GL(n,C)-irreduciblerepresentation since, with respect to the action of this group, all indices are on an equalfooting. Consider a k-row Young diagram (λ) ≡ (λ1, . . . , λk), with

∑ki=1 λk = p, p being

the total number of indices of the corresponding irreducible tensor. Of course we know thatk ≤ p, but we have a more fundamental bound on k directly related to the group: k ≤ n.Indeed recall that the tensor corresponding to a given Young tableau is antisymmetric withrespect to the indices in each column. Therefore the length of a column can not exceedn boxes, otherwise we will have a tensor which is antisymmetric in a number k of indiceswhich is greater than their range n. Such tensor would have vanishing components since twoor more of the k indices should have the same value and a tensor which is antisymmetric

4We shall also denote the representation D(λ) simply by (λ).


in two indices have vanishing entry when the indices have the same value. Therefore theaction of a Young element Y corresponding to a tableau with more than n rows annihilates

a GL(n,C)-tensor T: Y T = 0. We shall represent a k-row Young diagram (λ1, . . . , λk),defining a GL(n,C)-irreducible representation, also by the n-plet (λ1, . . . , λn), where all λi,with k < i < len are zero.

Let us now address the problem of finding the dimension of a GL(n,C)-irreducible repre-sentation described by a given Young diagram. We should write all independent componentsof the corresponding irreducible tensor M . Consider for instance M i1i2,i3 . We should keep inmind that these irreducible tensors are antisymmetric with respect to indices which are inthe same position within different subsets, namely in the same columns of the correspondingYoung tableaux. For instance M i1i2,i3 = −M i3i2,i1 . Therefore components of the kind M ij,i

are zero. We can convince ourselves that the independent components are:

M11,2 =1

3

(2T 112 − T 121 − T 211

),

M11,3 =1

3

(2T 113 − T 131 − T 311

),

M12,2 =1

3

(T 122 + T 212 − 2T 221

),

M13,3 =1

3

(T 133 + T 313 − 2T 331

),

M22,3 =1

3

(2T 223 − T 232 − T 322

),

M23,3 =1

3

(T 233 + T 32,3 − 2T 332

),

M12,3 =1

3

(T 123 − T 321 + T 213 − T 231

),

M13,2 =1

3

(T 132 − T 231 + T 312 − T 321

). (3.2.31)

Any other component is a combination of the ones listed above. Consider or instance M21,3.This component coincide with the difference M12,3 −M13,2. The independent componentsin (3.2.31) can be represented by the following symbols:

1 12

, 1 13

, 1 22

, 1 33

, 2 23

, 2 33

, 1 23

, 1 32

. (3.2.32)

The above diagrams, called standard components, should not be confused with the Youngtableaux, since the boxes are filled with values of the indices and not with their labels! Thestandard components corresponding to a given Young diagram represent the independent en-tries of the corresponding irreducible tensor and their number therefore gives the dimensionof the representation. For instance we have found 8 standard components for the represen-

tation which is therefore eight dimensional. There is a simple recipe to construct the

standard components for a given Young diagram and thus to deduce the dimension of thecorresponding representation: Fill the Young diagram with values of the indices, so that,


in reading each row from left to right these values are non-decreasing (they may be equal),and in reading each column from top to bottom, they are in a strictly increasing order. For

instance the components 3 12

, 2 13

are not standard. Using this rationale let us compute

the dimensions of certain irreducible representations.

Totally antisymmetric type-(p, 0) tensor. Consider the representation described by asingle column of p-boxes, namely by the diagram (1, 1, . . . , 1):

...

, (3.2.33)

When constructing the standard components, the values in each box have to be all differentand arranged in a strictly increasing order from top to bottom. Each standard component isthen defined by a set of distinct p values i1, . . . ip of the indices. The number of componentsis thus given by the number of sets of p distinct numbers out of n total values, namely by(np

). We then find:

dim(1, 1, . . . , 1) =

(np

)=

n!

p! (n− p)!. (3.2.34)

Totally symmetric type-(p, 0) tensor. The corresponding diagram (p) consists of pboxes in a single row:

· · · . (3.2.35)

To construct the standard components we need to fill them with p out of n, not necessarilydistinct, values i1, . . . , ip in a non-decreasing order:

i1 ≤ i2 ≤ . . . ≤ ip ≤ n . (3.2.36)

This implies the following relation among the values:

i1 < i2 + 1 < . . . < ip + p− 1 ≤ n+ p− 1 . (3.2.37)

namely each choice of values i1, . . . , ip corresponds to a single choice of strictly increasingvalues i1, i2 + 1, . . . , ip + p− 1 out of n+ p− 1. The total number of possibilities is still givenby a binomial:

dim(p) =

(n+ p− 1

p

)=

(n+ p− 1)!

p! (n− 1)!. (3.2.38)


The Gun-diagram (2, 1): . Let us consider now the diagram (2, 1). When construct-

ing the standard components, we have to distinguish three cases:

i) Three distinct values i < j < k;

ii) Two distinct values i < j.

Clearly we cannot use a single values since there is a column of two boxes. Let us considerthe two cases separately:

i) We have

(n3

)possible choices of 3 distinct numbers i < j < k out of n. For each

choice we can arrange them as follows:

i jk

, i kj

. (3.2.39)

We have therefore 2×(n3

)diagrams for this case.

ii) We have

(n2

)distinct couples of values i < j out of n possible ones. For each couple

we have the following diagrams:

i jj

, i ij

, (3.2.40)

for a total number of 2×(n2

)standard components.

The dimension of the representation is then:

dim(2, 1) = 2×(n3

)+ 2×

(n2

)=n (n2 − 1)

3. (3.2.41)

For n = 3 we find the eight diagrams in eq. (3.2.31).

The representation (2, 1, . . . 1) of type-(p, 0) tensors. The diagram of this representa-tion is:

...

, (3.2.42)

and consists of p-boxes arranges in a first column of p − 1-boxes and a second single boxcolumn. This generalizes the gun diagram corresponding to the p = 3 case. In constructingthe corresponding standard components we can either arrange p distinct values i1 < i2 <


. . . < ip in the boxes or p− 1 values i1 < i2 < . . . < ip−1. In the former case, for each of the(np

)choices of i1 < i2 < . . . < ip out of n, we have the following p− 1 possibilities:

i1 i2i3...ip

,

i1 i3i2...ip

. . .

i1 ipi2...

ip−1

, (3.2.43)

yielding (p−1)×(np

)standard components. As for the latter case, for each of the

(n

p− 1

)choices of i1 < i2 < . . . < ip−1 out of n, we have the following p− 1 arrangements:

i1 i1i2...

ip−1

,

i1 i2i2...

ip−1

. . .

i1 ip−1

i2...

ip−1

, (3.2.44)

yielding (p − 1) ×(

np− 1

)standard components. The dimension of the representation is

therefore:

dim(2, 1, . . . 1) = (p− 1)×(np

)+ (p− 1)×

(n

p− 1

)=

(p− 1)(n+ 1)!

p!(n− p+ 1)!.(3.2.45)

Notice that for p = 3 we find (3.2.41).

The box diagram (2, 2): .

When filling the diagram with numbers out of n, we have the following choices:

i) Four distinct values i < j < k < `;

ii) Three distinct values i < j < k;

iii) Two distinct values i < j.

Case i). For each of the

(n4

)choices of i < j < k < `, we have the following two

components:

i jk `

, i kj `

. (3.2.46)

Case ii). For each of the

(n3

)choices of i < j < k, we have the following three components:

i ij k

, i jj k

, i jk k

. (3.2.47)


Case ii). For each of the

(n2

)choices of i < j, we have the following component:

i ij j

. (3.2.48)

Summing up the numbers found above we get:

dim(2, 2) = 2×(n4

)+ 3×

(n3

)+

(n2

)=n2 (n2 − 1)

12. (3.2.49)

The rifle diagram (3, 1): . Let us list the possible cases with the corresponding

standard components:

i) Four distinct values i < j < k < `. The independent components are

i j k`

, i k `j

, i j `k

. (3.2.50)

ii) Three distinct values i < j < k. The independent components are

i i jk

, i i kj

, i k kj

, i j jk

, i j kj

, i j kk

. (3.2.51)

iii) Two distinct values i < j. The independent components are

i i ij

, i j jj

, i i jj

. (3.2.52)

We conclude that:

dim(3, 1) = 3×(n4

)+ 6×

(n3

)+ 3×

(n2

)=n (n+ 2)(n2 − 1)

8. (3.2.53)

A general formula. let us give for completeness a general formula for computing thedimension of a GL(n,C)-irreducible representation. Define the Cayley van der Monde de-terminant:

D(x1, x2, . . . , xn) ≡∏i<j

(xi − xj) . (3.2.54)

The dimension of the representation (λ1, λ2, . . . , λn) is:

dim(λ1, λ2, . . . , λn) ≡ D(`1, `2, . . . , `n−1, `n)

D(n− 1, n− 2, . . . , 1, 0), (3.2.55)


where ì ≡ λi + n− i. The reader can verify that:

D(n− 1, n− 2, . . . , 1, 0) = (n− 1)!(n− 2)! . . . 2! . (3.2.56)

Let us apply this formula to the rifle diagram (3, 1). In this case λ1 = 3, λ2 = 1 and all λi,for 2 < i ≤ n = 0. The numbers (ì) are:

`1 = 3 + n− 1 = n+ 2 , `2 = 1 + n− 2 = n− 1 , ì = n− i (i = 3, . . . , n) .

The reader can easily verify that:

D(`1, `2, . . . , `n−1, `n) =1

8(n+ 2)!(n− 1)!(n− 3)!(n− 4)! . . . 2!1! . (3.2.57)

From (3.2.55) we then find:

dim(3, 1) =(n+ 2)!(n− 1)!(n− 3)!(n− 4)! . . . 2!1!

8 (n− 1)!(n− 2)! . . . 2!=

(n+ 2)!

8 (n− 2)!=

1

8n (n+ 2) (n2 − 1) .

Exercise 3.8: By by directly applying formula (3.2.55), prove that:

• For n = 3, dim(7, 4, 2) = 42;

• For n = 4, dim(2, 2, 2) = 10;

• For n = 5, dim(2, 2, 2) = 50;

• For n = 3, dim(6, 6) = 28;

Now we are able, for instance, to compute the dimension of the irreducible representationsappearing on the right hand side of the decomposition (3.2.30). Representing the irreduciblerepresentations of GL(n,C) by the boldface of their dimensions, eq. (3.2.30) can also bewritten as follows:

3⊗ 3⊗ 3 → 1⊕ 10⊕ 8⊕ 8 . (3.2.58)

Exercise 3.9: Prove that type-(4, 0) GL(3,C)-tensors decompose as follows:

⊗ ⊗ ⊗ → ⊕ 3× ⊕ 2× ⊕ 3× , (3.2.59)

or,

3⊗ 3⊗ 3⊗ 3 → 151 ⊕ 3× 152 ⊕ 2× 6⊕ 3′ , (3.2.60)

where 3 = , 3′ = , 151 = (totally symmetric tensor) and 152 = .

Notice that the totally antisymmetric tensor is not present since the number of its indiceswould exceed their range.


3.3 Product of Representations

Suppose we have two GL(n,C)-irreducible tensors and multiply them together. Supposeone is of type-(p1, 0) and the other of type-(p1, 0). The product is again a tensor with asmany upper indices and the sum of the number of indices of the two factors, namely it isa type-(p1 + p2, 0) tensor. This tensor will in general be reducible. We have learned thattype-(p, 0) tensors decompose in irreducible representations which are in correspondencewith irreducible representations of Sp. Therefore the representations in the decompositionof the product will correspond to irreducible representations of Sp1+p2 which acts on eachcomponent of the product by shuffling all the p1 + p2 indices. Let us learn how, using Youngdiagrams, we can decompose the product of two irreducible representations into irreducibleones. Suppose the representations of the two tensors be described by the Young diagrams(λ) and (λ′) of Sp1 and Sp2 respectively. In which representations of Sp1+p2 does the product(λ) ⊗ (λ′) decompose? Consider a simple case in which (λ) = (3, 1) and (λ′) = (2, 1) of S4

and S3 respectively. There is a simple figurative rule to perform such decomposition. Fillthe first row of the left diagram with the letter a, the second with b the third with c and soon. It is convenient to choose the left diagram, to be filled with letters, as the simplest (i.e.smaller) of the two. In our example we will have:

⊗ a ab

. (3.3.1)

Now construct all possible (viable) Young diagrams by attaching the boxes filled with a tothe right diagram, with the only condition that two a-boxes should not lie on a same column.In our example we have the following diagrams:

a a a

a

a

a a a

a

a . (3.3.2)

Next we attach the b-boxes to each of the above diagrams to construct new viable Youngdiagrams. In doing this, however, we should not just follow the rule previously given forthe a-boxes, but also the following restriction: In reading each row from right ro left andeach column from top to bottom, the fist symbol we encounter should appear in the sequenceat least as many times as the second, the second symbol we encounter at least as manytimes as the third and so on. This last restriction defines sequences of symbols called latticepermutations. For instance if, reading a row from right to left, we find the sequence baa,the fist symbol we encounter is b and appears less times than the second a. This is not alattice permutation and the corresponding diagram should be discarded. Next we attach thec-boxes following the same restrictions given for the a and the b-boxes and so on. In ourexample we find:

a a

b

a a

b

a

a b

a

a

b

a

b

a

a

a

b

a a

b

3.4. BRANCHINGOF GL(N,C) REPRESENTATIONSWITH RESPECT TO GL(N−1,C)135

a

a b

a

a

b (3.3.3)

These are the representations of S7 which appear in the decomposition of the product of thetwo tensors. Suppose now the group we are considering is GL(4,C), so that n = 4. We cancompute, using the general formula (3.2.55), the dimensions of the representations involvedin the product and of those appearing in the decomposition:

= 45 , = 201 , = 224 , = 120 ,

= 1401 , = 1402 , = 202 ,

= 60 , = 36 , = 203 , (3.3.4)

The decomposition of the product (3.3.1) then reads:

⊗ → ⊕ ⊕ ⊕ 2× ⊕ ⊕ ⊕ ⊕ ,

or, equivalently:

45⊗ 201 → 224⊕ 120⊕ 1401 ⊕ 2× 1402 ⊕ 202 ⊕ 60⊕ 36⊕ 203 .

We can apply the above procedure to compute the 3–fold product ⊗ ⊗ of the funda-mental representations of GL(3,C):

⊗ ⊗ = ( ⊗ a )⊗ =

a ⊕ a

⊗ =(

⊕)⊗ a =

= a ⊕ a ⊕a

⊕ a = ⊕ 2× ⊕ . (3.3.5)

3.4 Branching of GL(n,C) Representations With Re-

spect to GL(n− 1,C)Consider a contravariant vector V = V i ei = (V i) of Vn, in the fundamental representationD ≡ n ≡ of GL(n,C). It is a type-(1, 0) tensor whose components are listed in a column


vector:

V =

V 1

...V n−1

V n

. (3.4.1)

The representation space Vn can be split into the direct sum of a subspace Vn−1 spannedby the vectors (ea), a = 1, . . . , n − 1, parametrized by the first n − 1 components V a in(3.4.1) and a one dimensional subspace V1 = en, parametrized by the nth component V n

in (3.4.1): Vn = Vn−1 ⊕ V1. We can consider the subgroup of GL(n,C) which has, throughthe representation D, a block diagonal action on the vector (3.4.1), in which the first blockis a (n−1)× (n−1) matrix acting on V a and the last block is 1. With respect to this actionVn−1 and V1 are invariant subspaces and thus n is, by construction, no longer irreduciblewith respect to this subgroup. The largest subgroup of GL(n,C) which has this action isGL(n − 1,C). The fundamental representation n of the former decomposes into the directsum of the fundamental representation n− 1 (acting on Vn−1) and the identity representation1 of the latter (the identity representation acting on V1). In other words, with respect toGL(n− 1,C):

GL(n,C) → GL(n− 1,C) ,

n → (n− 1)⊕ 1 ⇔ → ⊕ n , (3.4.2)

where n represents the nth component V n of the tensor V i on which the identity repre-sentation acts. The component V n is called a singlet with respect to GL(n − 1,C). Thissubgroup in other words does not “see” the nth component of the vector. A decompositionof a representation of a group with respect to a subgroup is called branching law.

How does a generic irreducible GL(n,C)-tensor branch with respect to GL(n − 1,C)?

Consider for instance a tensor (M i1i2,i3) in the representation . Its components split into

the following subsets: Mab,c, Man,b, Man,n, Mab,n (recall that (M i1i2,i3) is antisymmetric ini1, i3, so that i1 and i3 can not have the same value and that the only independent componentsare the standard ones). As the index i splits into (a, n), the standard components of thecorrespondent irreducible representation split as follows:

i jk

→ a bc⊕ a n

c⊕ a n

n⊕ a b

n. (3.4.3)

Notice that the index n can appear only at the bottom-end of each column. Since GL(n−1,C)

does not “see” indices with value n, the components a bc

, a nc

, a nn

, a bn

correspond to the

representations , , , of GL(n− 1,C), so that the following branching rule holds:

GL(n,C) → GL(n− 1,C) ,

→ ⊕ ⊕ ⊕ . (3.4.4)

In general, given an irreducible GL(n,C)–representation, its branching with respect toGL(n − 1,C) is obtained by grouping the corresponding standard components according

3.5. REPRESENTATIONS OF SUBGROUPS OF GL(N,C) 137

to the number and position of the boxes with the value n, keeping in mind that this valuecan appear at most once at the bottom-end of each column. As a second example considerthe representation (3, 1) of GL(n,C):

GL(n,C) → GL(n− 1,C) ,

→ ⊕n

⊕ n ⊕n n

⊕n

n ⊕n n

n ≡

≡ ⊕ ⊕ ⊕ ⊕ ⊕ . (3.4.5)

If (λ) ≡ (λ1, . . . , λn) is an irreducible GL(n,C)–representation (as usual, if p < n, we defineλi = 0 for p < i ≤ n), generalizing the above argument, one can show that it branches inthe following sum of GL(n− 1,C)–representations (λ′1, . . . , λ

′n−1):

GL(n,C) → GL(n− 1,C) ,

(λ1, . . . , λn) →⊕

(λ′1, . . . , λ′n−1) , (3.4.6)

where the direct sum is extended over values of the indices λ′a satisfying the following in-equalities:

λ1 ≥ λ′1 ≥ λ2 ≥ λ′2 ≥ . . . ≥ λ′n−1 ≥ λn . (3.4.7)

3.5 Representations of Subgroups of GL(n,C)So far we have been considering type (p, 0) GL(n,C)-tensors only. We may extend ouranalysis to more general mixed tensors having p contravariant and q covariant indices. Inthis case a first decomposition into lower-rank traceless tensors is described by eq. (2.9.30).Within each subspace V (p−`,q−`), traceless tensors are then further reduced by applying thedecomposition described in the previous sections to the upper and lower indices separately,namely by applying Young symmetrizers acting on two kinds of indices. In particular ir-reducible mixed tensor T = (T i1...ipj1...jq) will in general transform in a representation D(λ)

with respect to the upper indices and in a representation D′ (λ′) with respect to the lower

ones. The tensor as a whole, will then be acted on by the representation D(λ) × D′ (λ′),

simply denoted by (λ), (λ′), of the group GL(n,C). For instance an irreducible tensor inthe representation (2, 1), (3, 2, 1) will have the form:

M i1i2,i3j1j2j3,j4j5,j6 . (3.5.1)

Let us now address the issue of branching GL(n,C)- irreducible representations with respectto some of its subgroups. The action of an element g ∈ GL(n,C) on a type-(p, 0) tensor isdefined by the transformation:

(D(λ)[g])i1...ipj1...jp = (D[g]i1j1 · · ·D[g]ipjp)(λ) , (3.5.2)


where the subscript (λ) means that the indices i1 . . . ip are to be symmetrized/antisymmetrizedaccording to the tableau (λ). For instance the action on a type- (2, 0) antisymmetric tensor

(T ij), belonging to the representation (λ) = (1, 1) = , is given by:

(D( )

[g])ijk` =1

2(D[g]ikD[g]j` −D[g]i`D[g]jk) . (3.5.3)

From eq. (3.5.2) we see that the entries of D(λ)[g] are homogeneous polynomials of degree p inthe entries of D[g] = (D[g]ij). The representation D(λ) is reducible with respect to a subgroupG ∈ GL(n,C) if some of the non-vanishing entries of D(λ)[g], or linear combinations of them,vanish for all g ∈ G. Consider the subgroup G = GL(n,R). Recall that, from a standardtheorem of algebra, if a complex polynomial vanishes for all real values of its variables, it willvanish identically. Therefore if D(λ) were a reducible GL(n,R)-representation, one or moreentries of D(λ)[g] would vanish for any real values of the variables D[g]ij (i.e. g ∈ GL(n,R)),implying that the same entries would be zero for any GL(n,C) matrix. We conclude thatirreducible GL(n,C)–representations are irreducible also with respect to GL(n,R).

Consider now the subgroup SL(n,C). The only condition satisfied by the entries D[g]ijwhen g belongs to this subgroup is det(D[g]) ≡ 1. This does not imply the vanishing on anyentry of D(λ)[g] on the subgroup. To show this suppose that some entry, to be denoted by

P(p)k [g], of D(λ)[g] vanish for g ∈ SL(n,C). A generic n×n matrix M can always be written as

the product of the nth root of its determinant times a unimodular matrix: M = m1n S, where

m = det(M) and det(S) = 1. Therefore, for any g ∈ GL(n,C), there exists a g′ ∈ SL(n,C)

such that D[g] = m1n D[g′], where m = det(D[g]). The entries of P

(p)k [g], of D(λ)[g] that

vanish on SL(n,C), are homogeneous polynomials of degree p in the entries of D[g]. Thisimplies that they will vanish on a generic g ∈ GL(n,C), since, using their homogeneity, we

have P(p)k [g] = m

pn P

(p)k [g′] = 0, being g′ ∈ SL(n,C). We conclude that no entry of D(λ)[g] can

vanish on the subgroup SL(n,C) only and thus that if D(λ) is also irreducible with respect toSL(n,C). Using the same theorem of algebra mentioned earlier, we can prove that irreducibleGL(n,C)–representations are irreducible not just for SL(n,C) also with respect to SL(n,R).Similar arguments can be used to show that the same holds true for the U(n) and SU(n)subgroups of GL(n,C): Irreducible representations of the latter are irreducible also withrespect to the formers. As we shall see this is no longer true for the subgroups (S)O(p, q)and Sp(n,C) (n even) of GL(n,C) which will be dealt with separately.

Consider now a totally antisymmetric type (p, 0)-tensor T ≡ (T i1...ip). It spans theirreducible representation (λ) = (λ1, λ2, . . . , λp) = (1, 1, . . . , 1) referred to the contravariantindices, to be denoted by the symbol D∧p (recall that D is the representation acting oncontravariant Vn vectors). With it we can associate a totally antisymmetric quantity ∗T =(∗Ti1...in−p) transforming in the product of the representation det(D) times the representationof totally antisymmetric type-(0, n − p) tensors, to be denoted by D′ ∧(n−p) (recall thatD′ ≡ D−T is the representation acting on covariant vectors (Vi)) and defined as follows:

∗Ti1...in−pdef.≡ 1

p!εi1...in−p j1...jp T

j1...jp . (3.5.4)


The above quantity will also be denoted by the short-hand notation: ∗T ≡ εT. The corre-spondence ∗ between T and ∗T is called dualization. Similarly if F = (Fi1...ip) is a type-(0, p)tensor transforming in the representation D′∧p, we can associate with it a type-(n − p, 0)tensor ∗F using the contravariant symbol εi1...in :

∗F i1...in−pdef.≡ 1

p!εi1...in−p j1...jpFj1...jp . (3.5.5)

To show that ∗T transforms in the representation det(D)×D′ ∧(n−p) let us take a g ∈ GL(n,C)and use the following matrix property:

D[g]i1j1 · · ·D[g]injn εi1...in = det(D[g]) εj1...jn . (3.5.6)

The above property can also be represented symbolically as follows:

D∧n[g] ε = det(D[g]) ε . (3.5.7)

Contracting the first n − p indices on both sides of (3.5.6) by D[g]−1 j`i` and recalling that

D′[g] ≡ D[g]−T we find:

εi1...in−p j1...jp D[g]j1k1 · · ·D[g]jpkp = det(D[g])D′[g]i1`1 · · ·D′[g]in−p

`n−p ε`1...`n−p k1...kp .

(3.5.8)

Using the symbolic notation introduced above we will also write:

εD∧p[g] = det(D[g]) D′∧(n−p)[g] ε . (3.5.9)

As T transforms into T′ ≡ (T ′ i1...ip) ≡ D∧p[g] T, defined by:

T ′ i1...ip = D[g]i1j1 · · ·D[g]ipjp Tj1...jp , (3.5.10)

The quantity ∗T transforms into ∗T′ whose components are given by:

∗T ′i1...in−p =1

p!εi1...in−p j1...jp T

′ j1...jp =1

p!εi1...in−p j1...jp D[g]j1k1 · · ·D[g]jpkp T

k1...kp =

=1

p!det(D[g])D′[g]i1

`1 · · ·D′[g]in−p`n−p ε`1...`n−p k1...kp T

k1...kp

= det(D[g])D′[g]i1`1 · · ·D′[g]in−p

`n−p ∗T`1...`n−p , (3.5.11)

where we have used eq. (3.5.8). The above derivation in our symbolic notation is straight-forward:

∗T′ = εT′ = εD∧p[g] T = det(D[g]) D′∧(n−p)[g] εT = det(D[g]) D′∧(n−p)[g] ∗T .

Similarly the dual ∗F ≡ εF of a totally antisymmetric type-(0, p) tensor F, as defined in eq.(3.5.5), transforms in the det(D)−1 ×D∧(n−p). This is easily shown by using the propertyεD′ ∧p[g] = det D[g]−1 D∧(n−p)[g] ε, deduced from eq. (3.5.6) by contracting the last p indices


with D[g]−1 j`i` . Quantities like ∗T or ∗F, transforming in the product of a power of det(D)

times a tensor representation, are called tensor densities.

If we restrict to SL(n,C) transformations we have det(D[g]) ≡ 1, i.e. det(D) is the iden-tity representation, and ∗T is then still a tensor transforming in the D′∧(n−p) representation.The mapping (3.5.4) between type-(p, 0) and type-(0, n − p) totally antisymmetric tensors,T and ∗T ≡ εT, is a homomorphism (as the reader can verify) between two linear vector

spaces with the same dimension

(np

)=

(n

n− p

), i.e. it is an isomorphism. We say then

that the corresponding SL(n,C)-representations D∧p and D′ ∧(n−p) are dual to each otherand are in fact equivalent. In particular this implies that D∧n is equivalent to the identityrepresentation 1. Denoting the one-dimensional identity representation 1 also by • and theYoung diagrams defining irreducible representations D′ (λ) on covariant indices by crossedboxes, we can write:

p

...∼ n− p

××...

×

n

...∼ • . (3.5.12)

From this it follows that: In a Young diagram describing a SL(n,C)-representation we canremove all n-box columns, since they amount to multiplying the irreducible representationobtained deleting that column by the identity representation. In other words, attaching toa Young diagram any number of n-box columns, we obtain equivalent representations:

(λ1, . . . , λn) ∼ (λ1 + k, . . . , λn + k) ∼ • k ⊗ (λ1, . . . , λn) . (3.5.13)

In tensor language, if a SL(n,C)-tensor is antisymmetric in n indices, such indices can bethought of as belonging to an ε- tensor multiplying a lower rank tensor. For instance theSL(3,C)-tensor M i1i2,i3i4,i5 , which is antisymmetric in i1, i3, i5 can be written in the followingfactorized form:

M i1i2,i3i4,i5 ∼ εi1i3i5 M i2,i4 . (3.5.14)

The tensors M i1i2,i3i4,i5 and M i2,i4 define two equivalent SL(3,C)-representations. The tensorM i1i2,i3i4,i5 is antisymmetric in the subsets of indices (i1, i3, i5) and (i2, i4). We can think ofdualizing with respect to each subset to obtain a dual tensor transforming in the dualrepresentation:

∗Mk ≡1

2εki2i4

1

3!εi1i3i5 M

i1i2,i3i4,i5 . (3.5.15)


Since ∗Mk transforms as a covariant vector in × , we can write

∼ × , (3.5.16)

or simply (2, 2, 1) ∼ ∗(1), where the ∗ symbol indicates that the equivalence is defined throughthe duality operation.

We can generalize the above correspondence and associate with any SL(n,C) contravari-ant tensor (λ) ≡ (λ1, . . . λn), an equivalent covariant one, in the representation (λ′) ≡(λ′1, . . . λ

′n), obtained by dualizing the indices in each column of tableau (λ). We will call the

diagram (λ′) dual to (λ) and write:

(λ1, . . . λn) ∼ ∗(λ′1, . . . λ′n) . (3.5.17)

Let us define the relation between λi and λ′i. The reader can verify that a Young diagram(λ1, . . . λn) has λn column with n boxes, λn−1 − λn columns with n − 1 boxes, and so onup to λ1 − λ2 one-box columns. The dual diagram will then have λ1 − λ2 = λ′n−1 − λ′ncolumns with n − 1 boxes, λ2 − λ3 = λ′n−2 − λ′n−1 columns with n − 2 boxes and so on upto λn−1 − λn = λ′1 − λ′2 one-box columns. There will be no n-box columns, which impliesλ′n = 0. We have thus deduced the following relations

λ′n = 0 ,

λ1 − λ2 = λ′n−1 − λ′n ,λ2 − λ3 = λ′n−2 − λ′n−1 ,

...

λn−1 − λn = λ′1 − λ′2 , (3.5.18)

which are solved by λ′i = λ1 − λn−i+1, i = 1, . . . , n, so that:

(λ1, . . . λn) ∼ ∗(λ′1, . . . λ′n) = (λ1 − λn, . . . , λ1 − λ2, 0) . (3.5.19)

There is a geometrical way of constructing the dual of a given Young diagram. One shouldfirst inscribe the diagram (λ) in the upper half of a rectangle with the horizontal side con-sisting of λ1 boxes and the vertical one of n-boxes. The complement of the diagram in therectangle is the dual diagram (λ′) upsidedown. Consider, for example, the contravariantrepresentation (7, 6, 5, 2, 1) of GL(7,C) and inscribe it in a 7×7 square, as illustrated below:

×× ×

× × × × ×× × × × × ×

× × × × × × ×× × × × × × × (3.5.20)

The dual diagram is (λ′) = (7, 7, 6, 5, 2, 1, 0).What we have said for SL(n,C) also holds for its real form SL(n,R).


3.5.1 Representations of U(n)

In the previous chapter we have learned that the complex conjugate of a contravariant vectortransforms under a unitary representation as a covariant one: (V i)∗ ≡ Wi. This is becausethe unitarity condition implies D′ = D. This property generalizes easily to type-(p, q) tensorsT ≡ (T i1...ipj1...jq): The complex conjugate of a type-(p, q) tensor is a type-(q, p) tensor:

(T i1...ipj1...jq)∗ = F j1...jq

i1...ip . (3.5.21)

In particular complex conjugation defines a pairing between covariant and contravarianttensors which is however not an equivalence: The complex conjugate of a type-(p, 0) tensor

in the representation D(λ) is a type-(0, p) one in the representation D′ (λ) = D(λ)

. If weconsider the special unitary group SU(n), which is a subgroup of SL(n,C), the dualityoperation represents a further pairing between covariant and contravariant tensors and the

type-(0, p) tensor in D(λ)

will be the dual of a type-(n − p, 0) tensor in the representation

D(λ′) = ∗D(λ)

. For instance, in the SU(3) case D(1) ∼ D′ (1) is equivalent through duality to

D(1,1). In diagrams:

∼ × ∼ , (3.5.22)

or, in tensor language:

(V i)∗ = Wi =1

2εijkM

j,k . (3.5.23)

The SU(3) representation (2, 1) ≡ is self-dual, namely it coincides with its dual. In this

case we can write:

∼ × ×× ∼ , (3.5.24)

or, in tensor components:

(M i1i2,i3)∗ = Mi1i2,i3 =1

2 3!εi1i3j2 εi2j1j3 M

j1j2,j3 . (3.5.25)

For SU(2) the following relation holds:

∼ × ∼ , (3.5.26)

or, in tensor components:

(V i)∗ = Wi = εijMj . (3.5.27)

In fact SU(2) is the only special unitary group for which the fundamental representation2 = is equivalent to its conjugate, as we have already shown earlier. Let us now observe


that any Young diagram of SU(2) is equivalent to a single-row diagram (λ1), since any double-box column can be erased. The corresponding SU(2)-representation is usually labeled bythe half-integer j ≡ λ1

2and denoted by D(j), so that:

D( 12

) ≡ , D(1) ≡ , D( 32

) ≡ . (3.5.28)

The dimension of a representation D(j) is:

dim(D(j)

)=

(2 + λ1 − 1

λ1

)= λ1 + 1 = 2 j + 1 . (3.5.29)

A tensor in D(j) is totally symmetric in λ1 = 2 j indices, each taking the values 1, 2: M i1...i2j .A mixed tensor will transform with respect to SU(2) in a representation D(j) ⊗D′ (j

′) andwill have the form: M i1...i2j

j1...j2j′. We can apply the rule defied in Section 3.3 to decompose

the product of two SU(2)-irreducible representations D(j1) and D(j2). Consider for instancethe case j1 = 5

2and j2 = 1. We have:

⊗ → ⊕ ⊕ =

= ⊕ ⊕ .

D( 52

) ⊗D(1) → D( 72

) ⊕D( 52

) ⊕D( 32

) . (3.5.30)

We can generalize the above result and write the following decomposition:

D(j1) ⊗D(j2) =⊕k

D(k) , k = |j1 − j2|, |j1 − j2|+ 1, . . . , j1 + j2 . (3.5.31)

We have discussed in the previous chapter the relation between SU(2) and the rotation groupSO(3) in the three-dimensional Euclidean space: The latter is somewhat contained twice inthe former. This implies in particular that SU(2) has more representations than SO(3). Infact the two share only those representations with integer j (for instance the fundamentaltwo-dimensional representation of SU(2) is not an SO(3)-representation). In this sense SU(2)represents a generalization of the three-dimensional rotation group. In quantum mechanicsphysical systems with definite angular momentum are described by states transforming ina SO(3)-representation defined by an integer value of j, which measures the correspondingquantized value of angular momentum in units of ~. SU(2)-representations with half-integerj describe an other physical property of a particle called spin, no longer related to spatialrotations (it rather describes a kind of rotation in some internal space). By total angu-lar momentum we shall refer to both ordinary angular momentum and spin, and it thuscorresponds to a representation D(j) of SU(2).

If we have two interacting systems with total angular momenta j1 ~ and j2 ~ respectively,the total system will be described by a state transforming in the product of the represen-tations describing each system, namely in D(j1) ⊗ D(j2). The decomposition (3.5.31) thenprovides the irreducible representations in which the total system transforms and is calledsum rule of angular momenta.


Whatever we said for (S)U(n) also applies to representations of (S)U(p, n− p).Let us consider irreducible representations of SL(3,R), SU(3) and SU(2, 1). The columns

of Young diagrams for these groups can consist of at most two boxes. However couplesof antisymmetric covariant (contravariant) indices can be dualized to single contravariant(covariant) indices. Consider for instance the irreducible tensor M i1i2i3i4,i5i6i7 . It has mixedsymmetry. However if we dualize the antisymmetric couples of indices we end up with atensor defining an equivalent representation, which is totally symmetric in its upper andlower indices separately:

M ′ i4j1j2j3 ≡

1

8εj1i1i5 εj2i2i6 εj3i3i7 M

i1i2i3i4,i5i6i7 . (3.5.32)

We can represent graphically this transformation as follows:

→ × × × . (3.5.33)

In general we can state the following property:

Property 3.1: All irreducible tensors with respect to the groups SL(3,R), SU(3) andSU(2, 1) can be represented as mixed tensors which are totally symmetric in each kind ofindices. The representation can be therefore totally characterized by a couple of integers(p, q) representing the number of contravariant and covariant indices respectively.

3.5.2 Representations of (S)O(p, q)

While the only operations which commute with the action of the GL(n,C) group on a tensorwith only one kind of indices (covariant or contravariant) are permutations on the order ofits indices, so that irreducible tensors are tensors on which a maximal set of symmetriza-tion/antisymmetrization operations have been applied, when we restrict ourselves to theaction of the (S)O(p, q) group, there exists a further operation on tensors which commuteswith it: The contraction of two indices of the same kind. We have shown in the GL(n,C)case that contractions of upper with lower indices in a tensor yield lower rank tensors whichdefine invariant subspaces. As far as the group O(n) is concerned, we have the invarianttensors δij (type- (2, 0)) and δij (type- (0, 2)). Given a type-(p, q) tensor T = (T i1...ipj1...jq)the contraction of two upper indices amounts to multiplying T by δij and then contractingi and j with the two contravariant indices i` and ik:

T [`k]i1...i`−1i`+1...ik−1ik+1...ipj1...jq ≡ T i1...i`−1ii`+1...ik−1jik+1...ip

j1...jq δij =

=n∑i=1

T i1...i`−1ii`+1...ik−1iik+1...ipj1...jq .

The quantity T[`k] is a type- (p−2, q) tensor, as we can easily show by transforming T underg ∈ O(n) (let us denote D[g] simply by D):

T [`k]i1...ip−2j1...jq −→ T [`k]′ i1...ip−2

j1...jq =n∑i=1

T ′i1...i`−1ii`...ik−2iik...ip−2j1...jq =


=

n∑i=1

Di1k1 · · ·Di

m · · ·Dis · · ·D

ip−2

kp−2D−1 j1

m1 · · ·D−1 jqmq T

k1...m...s...kp−2m1...mq =

= Di1k1 · · ·D

ip−2

kp−2D−1 j1

m1 · · ·D−1 jqmqT

[`k]k1...kp−2m1...mq ,

where we have used the property of orthogonal matrices:∑n

i=1 DimD

is = δms. Similarly,

using the invariant tensor δij we can trace over two lower indices obtaining a type-(p, q− 2)tensor T[`k]:

T[`k]i1...ip

j1...jq−2 ≡n∑j=1

T i1...ipj1...j`−1jj`...jk−2jjk...jq−2 . (3.5.34)

There are p(p − 1)/2 independent traces T[`k] and q(q − 1)/2 traces T[`k] that can be con-structed out of a type-(p, q) tensor T. These tensors can be further traced, so that the spaceV (p,q) naturally splits into the direct sum of subspaces V (p−2`,q−2k) which are invariant underO(n) and whose elements have the form:

V (p−2`,q−2k) = F (p−2`)(q−2k)

`︷︸︸︷δ(2) · · · δ(2)

k︷︸︸︷δ(2) · · · δ(2) ⊂ V (p,q) , (3.5.35)

where F (p−2`)(q−2k) denotes a traceless type (p−2`, q−2k) tensor, δ(2) = (δij) and δ(2) = (δij).

A type- (p, q) tensor can then be written, in a unique way, as the sum of its components oneach of these subspaces. Let us consider for instance a type-(3, 0) tensor T = (T ijk). It willbe written as the sum of a traceless tensor of the same type T0 = (T ijk0 ) and a componentin V (1,0):

T ijk = T ijk0 + T ijk1 ,

T ijk1 = H i δjk + F j δik +Gk δij , (3.5.36)

where H i, F i, Gi are expressed in terms of the traces of T through the conditions:

T [12] i =n∑k=1

T kki = H i + F i + nGi ,

T [13] i =n∑k=1

T kik = H i + nF i +Gi ,

T [23] i =n∑k=1

T ikk = nH i + F i +Gi , (3.5.37)

which yield:

H i =1

(n− 1)(n+ 2)[(n+ 1)T [23] i − T [13] i − T [12] i] ,

F i =1

(n− 1)(n+ 2)[−T [23] i + (n+ 1)T [13] i − T [12] i] ,

Gi =1

(n− 1)(n+ 2)[−T [23] i − T [13] i + (n+ 1)T [12] i] , (3.5.38)


The components T0 and T1 are mutually orthogonal in the sense that:

n∑i=1

n∑j=1

n∑k=1

T ijk0 T ijk1 = 0 . (3.5.39)

We have defined so far two kinds of traces or contractions: One between upper and lowerindices which commutes with GL(n,C) transformations and an other involving couples ofindices of the same kind (both upper or lower), which commutes with the action of theorthogonal group only. When reducing a type-(p, q) tensor into irreducible components withrespect to O(n), we first need to restrict to components which are traceless with respectto the first kind of contraction, then we further decompose such components into tensorswhich are traceless with respect to the second kind of contraction as well. We shall calltraceless a tensor on which both kinds of contractions are zero. On these tensors we canapply symetrization and anti-symmetrization operations defined by Young tableaux (Youngsymmetrizers), as we did for the GL(n,C)-tensors, to derive the irreducible components. Ofcourse this procedure makes sense since any permutation of indices takes a traceless tensorinto an other traceless one.

Example 3.3: Let us apply this procedure to an example that we have already workedout in the previous chapter. Consider a type-(2, 0) tensor F = (F ij) and decompose it in atraceless and trace part:

F ij = F ij0 + F δij , (3.5.40)

where∑n

i=1 Fii0 = 0. We easily find F = 1

n

∑ni=1 F

ii. Next we apply to the traceless part,in the representation × = D⊗2, symmetrization and anti-symmetrization operators toconstruct its irreducible components according to the decomposition:

× → ⊕ ,

F ij0 → 1

2(F ij

0 + F ji0 ) +

1

2(F ij

0 − Fji0 ) = F ij

S + F ijA . (3.5.41)

The O(n) irreducible components of F are a singlet F in the representation 1, the trace, F ijS

in the representation n (n + 1)/2− 1 and F ijA in the representation n (n− 1)/2.

Example 3.4: Let us consider a type-(2, 1) tensor T = (T ijk). Let us first decompose itinto a traceless and trace part:

T ijk = T ij0 k + F i δjk +Hj δik +Gk δij . (3.5.42)

Exercise 3.10: Find F i, H i, Gi in terms of the traces of T.The traces F i, H i, Gi can not be further reduced with respect to O(n) and thus define

three representations n (recall that for orthogonal representations upper and lower indicesare on an equal footing since D = D and thus n = n). Consider now the traceless part T ij0 k.


In applying symmetrization and anti-symmetrization operations we should recall that withrespect to the orthogonal group upper and lower indices of the fundamental representation nare on an equal footing and thus the problem is the same as that of decomposing a tracelesstensor T ijk0 into its irreducible components. We find:

T0 → ⊕ ⊕ 2× . (3.5.43)

where now the dimensions of each of the above O(n) irreducible representations is obtainedfrom the dimension of the corresponding GL(n,C) by subtracting a the dimensions of traces.Take, for instance, the rank three symmetric GL(n,C) tensor M ijk transforming accordingto the representation . It contains a O(n)-invariant trace

∑ni=1M

iij transformingin the fundamental representation n. The dimension of the totally symmetric irreducibleO(n)-representation is therefore given by the dimension of the corresponding GL(n,C)-representation minus n. Similarly, for the gun diagram, a GL(n,C)-tensor M ij,k containsa O(n)-invariant trace

∑ni=1 M

ii,j in the representation n. We can therefore determine thedimensions of the O(n)-irreducible representations in (3.5.43):

dim [ ] =n(n+ 1)(n+ 2)

6− n ,

dim[ ]

=n(n2 − 1)

3− n ,

dim

[ ]=

n(n− 1)(n− 2)

6,

(3.5.44)

where we have used the fact that a totally anti-symmetric tensor has no trace part. We cannow write the decomposition of a type-(2, 1) tensor. It is nothing but the decomposition ofthe product of three fundamentals n⊗ n⊗ n = ⊗ ⊗ :

⊗ ⊗ → ⊕ ⊕ 2× ⊕ 3× ,

n⊗ n⊗ n →(

n(n + 1)(n + 2)

6− n

)⊕(

n(n− 1)(n− 2)

6

)⊕ 2×

(n(n2 − 1)

3− n

)⊕

⊕3× n ,

where the 3× n part act on the traces F i, H i, Gi.When we write the Young diagrams for the traceless part of a tensor, the following

important property, which we are not going to prove, should be taken into account:

Property 3.2: The Young element, defined by a diagram in which the sum of the lengthsof the first two columns exceeds n, when applied to a traceless tensor, gives zero.


In the decomposition of the traceless part of a tensor, the only irreducible componentswhich appear correspond therefore to admissible diagrams in which the total number of

boxes in the first two columns does not exceed n. For example is not admissible for

O(3) and is not admissible for O(2).

Admissible diagrams can be grouped in pairs of associate diagrams, Y and Y ′, in whichthe length a of the first column of Y does not exceed n

2boxes, a ≤ n/2, and the first column

of Y ′ consists of b = n−a boxes and all the other columns coincide with those of Y . Let c bethe length on boxes of the second column of both diagrams. We have that n/2 ≥ a ≥ c andtherefore a+ c ≤ n, namely Y is admissible. On the other hand a ≥ c so that n− a ≤ n− cand thus (n− a) + c ≤ n, namely Y ′ is admissible as well. Examples of associate diagramsare, for O(5):

Y = ↔ Y ′ = ,

Y = ↔ Y ′ = . (3.5.45)

If n is even, n = 2 ν, we may have diagrams Y which coincide with Y ′. These are calledself-associate diagrams and their first column consists of ν boxes.

To construct admissible diagrams we can first construct Y and then its associate counter-part Y ′. Since Y does not contain more than n/2 rows, it can be characterized by ν integers(µ1, . . . , µν), µ1 ≥ µ2 ≥ . . . ≥ µν and

∑νi=1 µi = p, p being the rank of the tensor (total

number of indices), where n = 2 ν if n is even and n = 2 ν + 1 if n is odd. The integers µiare nothing but the lengths in boxes of each row of Y . To any solution to the equation:

µ1 + µ2 + . . .+ µν = p , (3.5.46)

there corresponds a diagram Y . Knowing Y we determine Y ′.The relevance of this construction is apparent is we consider the group SO(n) which, as

opposed to O(n), is unimodular, namely is a subgroup of SL(n,R). With respect to this groupthe dualization operation maps tensors into tensors and defines a correspondence betweenequivalent representations. Since tensors in Y ′ are obtained from those in Y by dualizingover the indices in the first column, with respect to SO(n) associate diagrams are equivalentrepresentations. Self-associate diagrams yield two non-equivalent representations. Considerfor instance a totally antisymmetric rank ν tensor F = (F i1...iν ) with respect to SO(2ν). Sincethis group makes no difference between upper and lower indices, covariant and contravarianttensors belong to the same representation space. In particular the dualization operation mapsrank-ν tensors into rank-ν tensors. Moreover ∗∗F = F, namely the ∗ operation is involutive.This means that the space of totally anti-symmetric rank-ν tensors splits into two eigen-spaces of ∗ of equal dimension, to the eigenvalues ±1. These sub-spaces are spanned by thefollowing components of F:

F(±) ≡ 1

2(F± ∗F) ,


F (±) i1...iν ≡ 1

2

(F i1...iν ± 1

ν!εi1...iνj1...jν Fj1...jν

). (3.5.47)

The two eigen-spaces are invariant under the action of SO(n) and define two inequivalentirreducible representations. F(±) are called self-dual and anti-self-dual components of F.

All that we said for O(n) applies to the more general O(p, q) groups. The only differenceis that traces are defined using the invariant metric ηp,q:

T [`k]i1...i`−1i`+1...ik−1ik+1...ipj1...jq ≡ T i1...i`−1ii`+1...ik−1jik+1...ip

j1...jq (ηp,q)ij =

=

p∑i=1

T i1...i`−1ii`+1...ik−1iik+1...ipj1...jq −

−n∑

i=p+1

T i1...i`−1ii`+1...ik−1iik+1...ipj1...jq ,

the same for the traces T[`k] of a covariant tensor.Consider the group SO(3). In this case ν = 1 and thus all representations are defined

by a single integer µ1, which we shall also denote by j, and all diagrams Y are single-row.This means that the representation space of irreducible SO(3)-representations D(µ1) = D(j)

is described by totally symmetric tensors. Consider a generic irreducible representation,described by the single j–box row. The corresponding GL(3,R) representation has dimension(

3 + j − 1j

)= (j+2)(j+1)

2and acts on totally symmetric rank-µ1 tensors F = (F i1...ij). To

obtain the dimension of the SO(3) representation we have to subtract to this number thecontribution of the traces. A generic trace F[`k] is a rank j − 2 totally antisymmetric tensorand thus its vanishing implies j (j−1)

2conditions on the entries of F. We conclude that the

dimension of the generic SO(3)-irreducible representation (j) is:

dim(D(j)

)=

(j + 2)(j + 1)

2− j (j − 1)

2= 2 j + 1 . (3.5.48)

A vector V = (V i) in V3 is acted on by the representation j = 1 of SO(3), of dimension 3.The traceless product of the coordinates of a particle, xi xj − 1

3δij ‖r‖2, is a rank-2 tensor,

corresponding to the representation j = 2 of dimension 5.

Relation between SO(3) and SU(2) representations. Now we can make our previousstatements about the relation between SO(3) and SU(2) representations more precise. Letus consider the Pauli matrices σi = [(σi)

ab], i = 1, 2, 3, a, b = 1, 2, defined in (2.1.37). We can

lower the first index of these matrices using the εab tensor and define the three symmetricmatrices (γi)ab ≡ εac (σi)

cb :

(γ1)ab = −i ε σ1 = −i σ3 ,

(γ2)ab = −i ε σ2 = 12 ,

(γ3)ab = −i ε σ3 = i σ1 ,


Now consider a type-(0, 2) symmetric tensor F = (Fab) in the representation D′ (j=1) = × × .F is a complex 2 × 2 matrix on whose complex entries we can impose a linear conditionwhich commutes with the action of SU(2). This condition reduces the space spanned bytensors of this type to a lower dimensional SU(2)-invariant sub-space on which the irreduciblerepresentation acts. Recalling that (Fab)

∗ is a type-(0, 2) tensor with two upper indices, thiscondition reads:

Fab = εac εbd (Fcd)∗ ↔ F = −εF∗ ε . (3.5.49)

The most general tensor satisfying the above reality condition has the following form:

F =

(f + i h i ei e f − i h

), (3.5.50)

where e, f, h are real parameters. Therefore tensors F satisfying (3.5.49) depend on 3 realparameters and thus span a three-dimensional linear vector space over the real numbers ((γi)being a basis of such space), namely the corresponding representation is three dimensionaland real. Let us show that this representation is the fundamental 3 of SO(3). We noticethat Fab in (3.5.50) can be expanded in the symmetric matrices (γi)ab:

Fab = F i (γi)ab = −h (γ1)ab + f (γ2)ab + e (γ3)ab , (3.5.51)


F = −i F i ε σi . (3.5.52)

Let now S = S[σ, α, β] = (Sab) be a generic SU(2)-transformation. The tensor F transformsas follows:

Fab → F ′ab = S−1 ca S−1 d

b Fcd , (3.5.53)


F → F′ = S−T F S−1 . (3.5.54)

Using (3.5.52) we can write:

F′ = −i F i S−T ε σi S−1 . (3.5.55)

Now we use the property (2.4.18) which, taking into account that ε = i σ2, reads: S∗ = −εS ε.From this it follows that εS = S−T ε, where we have used the unitarity property of S, andwe can write:

F′ = −i Fi εSσi S−1 = −i Fi ε (R−1ijσj) = (Rj

i Fi) γj = F ′ j γj . (3.5.56)

Where R = R[S] = (Rij) is the orthogonal transformation corresponding to the unitary

matrix S, as defined in Section 2.3.7, and we have used the orthogonality property R−1 = RT ,


which, in components, reads R−1ij = RT

ij = Rj

i. We see that the components F i ofF transform in the fundamental 3 of SO(3). In other words the matrices (γi)ab define achange of basis between the two-times symmetric representation of SU(2) and the definingrepresentation of SO(3) so that the two representations coincide:

of SU(2) ≡ of SO(3) . (3.5.57)

From this we deduce that all the SU(2)-representations D(j), defined by integer j, coincidewith the corresponding representations of SO(3).

Chapter 4

References

• Lie Groups, Lie Algebras & Some of Their Applications , Robert Gilmore, John Wileyand Sons, 1974

• Group Theory and its Application to Physical Problems , Morton Hamermesh, Addison-Wesley Series in Physics.

• Prof. ’t Hooft’s Lecture Notes on Discrete Groups,http://www.phys.uu.nl/˜ sahlmann/teaching/lecture notes/discrete groups lecture notes.pdf

• Symmetry, Hermann Weyl, Princeton Science Library.

• The Theory of Groups and Quantum Mechanics, Hermann Weyl, Dover.

153

group theory with applications

Documents

group gln

unitary group

unitary representations

c representations

representations of subgroups

group sln

group algebra782

representations of sop