Representation Theory

Home , Adjoint representation, Cartan subalgebra

Jock McOrist

Lecture notes, perpetually in progress MATM035

Department of Mathematics University of Surrey Guildford GU2 7XH, United Kingdom

Symmetries are a powerful method for easily understanding properties of otherwise complicated mathematical and physical objects. Group theory is a branch of mathematics developed to understand symmetries, however it often leads to complicated abstract quantities. Representation theory turns such abstract algebraic concepts into linear transformations of vector spaces; the linearity making the system much easier to solve. In doing, representation theory can unveil deep symmetry properties of physical systems as well as leading to powerful and compact solutions to otherwise difficult and intractable problems. There will be on average 3 lectures per week. There are two unassessed courseworks and one class test. These will be marked and returned to you. The coursework and tests are important feedback for you, telling you how you are doing in the course, what your strengths and weaknesses are, and what type of questions to expect in the exam. All important information will be disseminated in the usual channels via SurreyLearn. There is a syllabus on SurreyLearn, and I encourage to look at it. It describes the course content, what you are expected to know, and all of the assessments. The lecture notes are structured to closely follow what we do in class, but I cannot emphasise enough the value of attending lectures. It is where all important information is disseminated. I will give explanations complementary to that written in the notes, and where appropriate fill in omitted steps in deriving formulae. There are also many worked examples throughout the notes. There are exercises at the end of the chapter which you should work through as a minimum. Some of these exercises may be used as unassessed coursework, marked and returned to you for feedback. Key concepts are boldfaced when they are introduced. Many ideas and examples in representation theory are developed in the exercises. They are also an important platform to practice proving results. There will be solutions to some (but not necessarily all) questions, and I will release them throughout the semester once you have had a serious attempt at the questions. There are also many wonderful textbooks out there which I encourage you to read through. I first learnt representation theory from Georgi’s book ‘Lie Algebras in Particle Physics’ and some of the material here are based on his explanations. Although the course is largely self-contained, we will draw heavily on your knowledge of group theory and linear algebra. In particular you need to be familiar eigensystems, diagonalisation of matrices, properties of inner products, traces and vector spaces. Finally, my email [email protected] is always on and door (05AA04) is always open, so if you have any questions or comments then I will try to respond as quickly as possible.

i Contents

1. Introduction ...... 1 1.1. Why representation theory? ...... 1 2. Symmetries and ﬁnite groups ...... 4 2.1. Some preliminary deﬁnitions ...... 4 2.2. Some examples of groups ...... 5 2.2.1. Abelian groups ...... 5 2.2.2. Non-abelian groups ...... 6 2.3. Subgroups and cosets ...... 8 2.4. Conjugacy classes ...... 10 2.5. Automorphisms ...... 10 2.6. Exercises ...... 12 3. Representation theory basics ...... 14 3.1. What is representation theory? ...... 15 3.2. Some examples ...... 16 3.3. Reducible and irreducible representations ...... 19 3.4. Useful theorems ...... 20 3.5. Characters ...... 23 3.6. Tensor Products ...... 24 3.7. Exercises ...... 26 4. Lie groups and Lie algebras ...... 29 4.1. Lie groups ...... 29 4.2. Examples of Lie groups ...... 31 4.3. Multiplication and Lie algebras ...... 34 4.4. Adjoint representation ...... 38 4.5. An inner product and compact algebras ...... 38 4.6. Subalgebras, simple and semi-simple algebras ...... 39 4.7. Exercises ...... 41 5. The key example: su(2)...... 44 5.1. From the group SU(2) to the algebra su(2) ...... 44 5.2. Highest weight construction of representations of su(2) ...... 45 5.3. Examples ...... 47 5.4. Exercises ...... 49 6. Roots and weights ...... 51 6.1. Weights ...... 51 6.2. The adjoint representation ...... 52 6.3. Roots ...... 54 6.4. Raising and lowering operators ...... 55 6.5. Heaps and heaps of su(2)s ...... 56 6.6. Example: su(3)...... 57 6.7. Exercises ...... 61 7. Simple roots ...... 63

ii 7.1. Positive weights ...... 63 7.2. Simple roots ...... 64 7.3. Constructing the algebra ...... 69 7.4. Exercises ...... 72 8. Dynkin Diagrams ...... 74 8.1. Dynkin Diagrams ...... 74

8.2. Example: g2 ...... 75 8.3. Cartan Matrix ...... 76

8.4. Constructing the g2 algebra ...... 83 8.5. Example: c3 ...... 85 8.6. Fundamental weights ...... 85 8.7. Example: more representations su(3) ...... 87 8.8. Complex conjugation ...... 89 8.9. Exercises ...... 92 9. A tour of compact simple Lie algebras and their physical applications ...... 93 ∼ 9.1. su(N) = aN−1 ...... 93 ∼ 9.2. The so(2N) = dN algebra ...... 95 ∼ 9.3. so(2N + 1) = bN ...... 97 ∼ 9.4. sp(2N) = cN ...... 98 9.5. Exotic Lie algebras: e6, e7, e8, f4, and g2 ...... 99 9.6. Exercises ...... 101

iii 1. Introduction 1

1. Introduction

1.1. Why representation theory?

To understand the importance and relevance of representation theory in mathematics and physics, we first must ask the question: why group theory? The answer to this comes from trying to understanding real world phenomena through mathematics, also known as physics. It often amounts to solving difficult, if not intractable equations. For example, most equations of motion that arise in classical dynamics, as you may have seen in Lagrangian and Hamiltonian Dynamics (MAT3008), are complicated non-linear partial dif- ferential equations. The properties of these equations are often elusive, even numerically. There is no known formalism for systematically solving the equations, and analytic solutions are rare. Similar conclusions arise in modern mathematical physics: studies of quantum dynamics (small scale phenomena) and large scale phenomena (general relativity) involve equations that are ridiculously hard to solve. So if life is so tough, how can we make progress? A remarkably successful approach comes from the study of symmetries. Symmetries can be an incredible labour saving device – they are often shortcuts to understanding physical systems even before understand exactly what the systems are! For example, even before knowing how to define a model of the hydrogen atom, using spherical symmetry (and some other symmetries), we are able to determine the orbits of electrons and their spectra. Group theory is the study of such symmetries. In studying the real world, the information we often extract from symmetries is not group theory itself, but rather a representation of the group. For example, a hydrogen atom is spherically symmetric, meaning that it looks the same as you change the angles θ, φ that appear in spherical polar coordinates (r, θ, φ). Experimentally, we do not observe this symmetry directly, despite cartoons you may have seen representing the hydrogen atom as spherical ball. Instead this symmetry is seen experimentally in studying the atom in various excited states, as shown in 1.1. What scientists have noticed is that the excited states are observed in different representations of the group SO(3). Representations are the physically important thing to study. Mathematically speaking, a representation of a group is a way of expressing a group in terms of linear transformation of a vector space. That is, we turn abstract group elements, whose properties are often rather opaque, into matrices, something far more intuitive. We will see what this means in later chapters, but suffice it to say for now, the problem is now vastly simpler. So we study representations of groups because they are simpler than groups themselves and because this is what turns up in the real world. The representation theory of groups divides up according to the type of group being studied:

• Finite groups: these are groups with a ﬁnite number of elements. The representation 1. Introduction 2

Figure 1.1: A numerical picture of the hydrogen atom in various excited states. The plots are really three-dimensional, and what is illustrated are two-dimensional slices. The excited states are labelled by the three integers at the top, and each uniquely corresponds to a representation of a Lie group, including that of SO(3). (Technically speaking, these are numerical solutions of Schrodinger’s wave equation. The plots are probability densities arising from the resulting wavefunction.)

theory is important to crystallography, some areas of quantum mechanics and geometry.

• Compact groups: groups that are continuous but of bounded domain. Representation theory of compact groups is important as many of the results of ﬁnite group representation theory generalise in a nice way.

• Lie groups: these are the most prominent category of continuous groups. Many results of compact groups apply to Lie groups. The representation theory of Lie groups is important in many areas of physics and chemisty, as well as mathematics.

• Linear algebraic groups: these are Lie groups but not over the ﬁelds R or C but over 1. Introduction 3

a finite field. The representation theory of these groups is far less well-understood, and quite difficult involving sophisticated tools in algebraic geometry.

• Non-compact groups: non-compact groups are too broad and opaque in structure to construct any systematic form of representation theory. Several special cases have been studied on an ad hoc basis.

Clearly, group representation theory is a huge subject, and an incredibly active area of research within mathematics. We cannot possibly hope to cover the entire of the subject. Instead our aim is to give you a flavour for how representation theory can be used, particularly in understanding finite groups and Lie groups, underlined above, as well as Lie algebras, which are essentially a linearisation of Lie groups, and so more tractable to study. These groups are the most well-understood in mathematics, and also the most relevant to the physical world. At the end of the course, I hope to have given you at least a small taste as to how much of a labour saving device representation theory and symmetries are in solving difficult problems. 2. Symmetries and finite groups 4

2. Symmetries and ﬁnite groups

We ﬁrst review some basic deﬁnitions and theorems of groups, many of which you will have seen before, and some examples to illustrate the concepts.

2.1. Some preliminary deﬁnitions

Deﬁnition 2.1. A group G is a set with a composition rule (or multiplication law) for assigning to every ordered pair of elements f, g ∈ G a third element fg satisfying four axioms:

(i) If f, g ∈ G then h = fg ∈ G (closure).

(ii) For f, g, h ∈ G then f(gh) = (fg)h (associativity).

(iii) There is an identity (or unit) element e such that for all f ∈ G, ef = fe = f (identity).

(iv) Every element f ∈ G has an inverse f −1 such that ff −1 = f −1f = e (inverse).

For a group G, we can draw up a multiplication table:

e g1 g2 ···

e e g1 g2 ···

g1 g1 g1g1 g1g2 ···

g2 g2 g2g1 g2g2 ··· ......

In each row and column each element appears exactly once. To see this, we prove by −1 contradiction. Assume that ggi = ggk with gi 6= gk. Then, we can multiply by the inverse g : −1 gi = g ggk = gk, a contradiction.

Definition 2.2. If G consists of a finite number of elements, then G is a finite group. Oth- erwise it is infinite. If it is finite, then the number of elements is the order of the group.

Deﬁnition 2.3. An abelian group is one in which the multiplication law is commutative:

g1g2 = g2g1 ∀ g1, g2 ∈ G. (2.1)

Otherwise the group is non-abelian.

Deﬁnition 2.4. A left group action of G on a set S is a set of maps on S indexed by group elements: φ : G × S → S, (g, s) → φ(g, s). (2.2)

We denote φ(g, s) = g · s. It must satisfy two axioms: 2. Symmetries and ﬁnite groups 5

(i) (gh) · s = g · (h · s) for all g, h ∈ G and s ∈ S (compatibility).

(ii) e · s = s for the identity element e ∈ G and for all s ∈ S (identity).

We could have equally well defined a right group action; analogous comments would have applied. A group action is what allows us to connect the symmetries of an object and a corresponding group. The group action is typically an operation on an object such as a transformation or reflection. The transformation is a symmetry if it leaves the object invariant,. For example, S might be a square, and the actions of the group are rotations by π/2, π, 3π/2, 2π around the centre of the square. These transformations form a finite group labelled Z4.

2.2. Some examples of groups

2.2.1. Abelian groups

1. The set of all integers Z is an inﬁnite abelian group where the composition law is addition. The identity element is 0. The inverse of an integer n is simply −n.

2. An example of a ﬁnite abelian group is Z2. This has order 2 with elements Z2 = {e, a}. Every element in a group has an inverse. This leaves only one possibility: that a−1 = a and a2 = e. Hence, its multiplication table is

e a e e a a a e

The group Z2 has a group action on geometric objects via the symmetry operation known as parity. This is essentially reﬂection. The statement that a2 = e means that if you

reﬂect twice, you get back to where you started. For example, Z2 has a group action on a square, which is to reﬂect it about a symmetry axes.

3. Z3 is another ﬁnite group this time with order 3. We write Z3 = {e, a, b} and its multiplication table is

e a b e e a b a a b e b b e a 2. Symmetries and ﬁnite groups 6

Z3 is abelian as ab = ba = e.

4. The n-dimensional vector space Vn forms an inﬁnite abelian group where the composition law is the usual addition of vectors.

To see this, we need to check the axioms. The addition of two vectors v, w ∈ Vn is also a

n-dimensional vector: v + w ∈ Vn, so the composition law is closed. It is also associative:

(v + w) + u = v + (w + u) where v, w, u ∈ Vn. The identity element is the zero vector −1 v = 0. Every vector has an inverse given by v = −v ∈ Vn so that v − v = −v + v = 0.

5. The set of rational numbers p/q where p, q ∈ Z and p, q 6= 0 forms a group where the composition law is the usual multiplication of numbers. Proving this is an exercise.

2.2.2. Non-abelian groups

1. The permutation group on ordered set of n objects Sn acts by permuting the order of

the n objects in some way. The group Sn is non-abelian for n > 2. For simplicity consider

S3. It has order 6 with elements identity e and

a1 = (1, 2, 3), a2 = (3, 2, 1), (2.3) a3 = (1, 2), a4 = (2, 3), a5 = (3, 1).

a1 is a cyclic permutation, reordering the positions as 1 → 2, 2 → 3 and 3 → 1 and is

illustrated in Figure 2.2; a2 is the inverse of a1; a3 interchanges 1 and 2; a4 interchanges 2

and 3; a5 interchanges 3 and 1. Multiplication is deﬁned as applying the transformations

consecutively: a3a4 means interchange 2 and 3 and then 1 and 2. We can draw up the multiplication table

e a1 a2 a3 a4 a5

e e a1 a2 a3 a4 a5

a1 a1 a2 e a5 a3 a4

a2 a2 e a1 a4 a5 a3

a3 a3 a4 a5 e a1 a2

a4 a4 a5 a3 a2 e a1

a5 a5 a3 a4 a1 a2 e

To see that S3 is non-abelian note that a1a3 = a5 while a3a1 = a4.

2. The set of all non-singular n × n matrices over C is a group under multiplication of matrices. This group is called the general linear group and denoted GL(n, C). It is inﬁnite 2. Symmetries and ﬁnite groups 7

a 1

Figure 2.1: The action of a1 ∈ S3 on a set of three distinct objects S = {moon, sun, star}.

and non-abelian. The identity element is the identity matrix   1 0 0 ...   0 1 0 ...      I = 0 0 1    . . . ..  . . . .    1

3. The symmetry group of an equilateral triangle D3 is a non-abelian group. The group elements are

D3 = {e, a, b, c, d, f},

where e is the identity element and the group action of each of the remaining elements are: a, b, c reﬂect about the symmetry axes A, B, C as shown in Figure 2.2; d rotation by 2π/3 about the centre; f rotation by 4π/3.

The multiplication table is constructed by inspection

e a b c d f e e a b c d f a a e d f b c b b f e d c a c c d f e a b d d c a b f e f f b c a e d 2. Symmetries and ﬁnite groups 8

B C

Figure 2.2: An equilateral triangle with its symmetry axes labelled A, B, C about which the triangle may be reﬂected.

2.3. Subgroups and cosets

Deﬁnition 2.5. A subgroup H of a group G is a subset H ⊆ G that is a group under the same composition law as G. Some trivial examples of subgroups of any group G are H = {e} and H = G. If H 6= {e} and H 6= G then it is said to be a proper subgroup. An example of a proper subgroup is

Z3 = {e, a1, a2} ⊂ S3. To see this, we note from the multiplication table that a2a1 = a1a2 = e and so if g, h ∈ Z3 then gh ∈ Z3 meaning the composition law is closed. Associativity is satisfied −1 −1 as it is inherited from G while we also have e ∈ Z3. Finally, a2 = a1, a1 = a2, and so the inverse axiom is satisfied. Z3 is a subgroup. In fact, to check that H is a subgroup, all we need to check is that the composition law is closed: g, h ∈ H then gh ∈ H and that e ∈ H. The remaining axioms are automatically satisfied, and proving this is left as an exercise. Subgroups can be used to divide up a group into subsets called cosets.

Deﬁnition 2.6. Given g ∈ G,A left-coset of the subgroup H ⊆ G is deﬁned as

gH = {gh | h ∈ H}. (2.4)

So we take all the elements in H and multiply them on the left by g.A right-coset Hg is deﬁned analogously, where the multiplication occurs on the right.

Remarks:

1. If g ∈ H then it is a simple exercise to see that gH = Hg = H. 2. Symmetries and ﬁnite groups 9

2. If g∈ / H, then gH is not a subgroup as it does not contain the identity element.

3. The element g of the coset gH is called the representative of the coset. It is not unique, and any other element f = gh, for some h ∈ H would suﬃce. To see this note that fH = ghH = gH.

4. We can expand a ﬁnite group G into a union of its cosets

l [ G = gkH, gk ∈ G. k=1

Deﬁnition 2.7. H ⊂ G is a normal subgroup if the left- and right-cosets agree. That is, for all g ∈ G we have gH = Hg. (2.5)

This statement means that for every g ∈ G, and h1 ∈ H there exists a h2 ∈ H such that −1 gh1 = h2g, or equivalently h2 = gh1g . Equivalently, H is normal if and only if

g−1Hg = H, ∀ g ∈ G. (2.6)

In the exercises you will show that Z3 is in fact a normal subgroup of S3.

Deﬁnition 2.8. A group that contains no proper normal subgroups is said to be simple. The group is semi-simple if none of its normal subgroups are abelian.

Deﬁnition 2.9. If H is a normal subgroup then set of all cosets of G, denoted G/H is in fact a group and called a coset space. The multiplication law is

(g1H)(g2H) = g1h1g2h2 | h1, h2 ∈ H . (2.7) and this is equivalent to

−1 (g1H)(g2H) = (g1g2)(g2 Hg2H) = (g1g2)H, (2.8)

−1 where the second equality follows from H being normal: g2 Hg2 = H. With this multiplication law, we can check the group axioms for G/H:

(i) closure follows from (2.8): (g1H)(g2H) = (g1g2)H.

(ii) associativity follows from that of G:(g1H)(g2Hg3H) = g1(g2g3)H = (g1g2)g3H = (g1Hg2H)(g3H).

(iii) The identity element is H = eH. That is, H(gH) = (eH)(gH) = (eg)H = gH. 2. Symmetries and ﬁnite groups 10

(iv) Inverse is g−1H = (gH)−1 which follows from (g−1H)(gH) = (g−1g)H = H.

In the exercises you are invited to show that the coset space S3/Z3 = Z2. Note that ord(G/H) = ord(G)/ord(H).

Deﬁnition 2.10. The centre ZG of a group G is the set of all elements that commute with G:

0 0 0 ZG = {g ∈ G | gg = g g ∀ g ∈ G}. (2.9)

The identity element e ∈ ZG and so ZG is always non-empty. Note that G is abelian if and only if ZG = G. Furthermore, ZG is always abelian and is an normal subgroup of G. You will show some of these results in the exercises.

2.4. Conjugacy classes

Deﬁnition 2.11. Let f, g ∈ G. They are said to conjugate elements if there is an x ∈ G such that f = x−1gx.

Conjugacy is an equivalence relation meaning we can deﬁne a set, called a conjugacy class: n o −1 [f] = x fx x ∈ G .

The choice of label for the conjugacy class, called the representative element, is ambiguous: if f = x−1gx then [f] = [g]. It is a simple exercise to see that [e] = {e} while if the group is abelian, then [f] = {f} for all f ∈ G. Conjugacy classes are mutually exclusive meaning if [f] and [g] are two diﬀerent conjugacy classes then [f] ∩ [g] = ∅. The group can then be decomposed into a mutually exclusive union of conjugacy classes [˙ k G = [Ca], a=1 where [C1],..., [Ck] are the conjugacy classes. As an example, the conjugacy classes of S3 are

[e] = {e}, [a1] = {a1, a2}, [a3] = {a3, a4, a5}.

2.5. Automorphisms

Deﬁnition 2.12. An isomorphism is a one-to-one and onto map

ψ : G1 → G2, where G1,G2 are two groups, and ψ preserves the multiplication laws of the groups:

ψ(g1g2) = ψ(g1)ψ(g2), ∀ g1, g2 ∈ G. 2. Symmetries and ﬁnite groups 11

Deﬁnition 2.13. An automorphism is an isomorphism where G1 = G2. That is, an automorphism is a one-to-one mapping of a group G onto itself preserving the multiplication law.

Deﬁnition 2.14. An inner automorphism φg is a map

−1 φg : G → g Gg, (2.10) for some ﬁxed g ∈ G. To see this map is an automorphism we ﬁrst check it preserves the multiplication law. Given g1, g2 ∈ G: −1 −1 −1 φ(g1g2) = g g1g2g = g g1gg g2g = φ(g1)φ(g2).

Secondly, Im φ ⊂ G so it is a mapping of G into itself. If g0 ∈ Ker φ then φ(g0) = g−1g0g = e. This implies g0 = e, and so Ker φ = {e} with φ a one-to-one mapping, and we are done. In some sense, the inner automorphism measures the degree of ‘non-abelian’ of the group.

That is, if G is abelian then φg is the identity map for all g ∈ G. Similarly, it reduces to the identity map on the centre of the group: φg(ZG) = ZG. For elements that are non-commuting,

φg conjugates them by g.

Deﬁnition 2.15. An outer automorphism is an automorphism that cannot be written as g−1Gg for any g ∈ G. 2. Symmetries and ﬁnite groups 12

2.6. Exercises

1. (a) Show that a group G contains only one identity element

(b) Show that the inverse of an element a ∈ G is unique.

2. Show that the set of rational numbers p/q, where p, q 6= 0, form a group where the composition law is the usual multiplication of numbers.

3. Show that if H ⊂ G is a subgroup if and only if (i) the inherited composition law is closed i.e. g, h ∈ H then gh ∈ H; and (ii) e ∈ H.

4. Show that if H ⊂ G is a normal subgroup then the ord(H) | ord(G), where ord(G) denotes the order of the group G.

5. Consider G = S3, the permutation group of 3 objects.

(a) Show that Z3 = {e, a1, a2} ⊂ S3 is a normal subgroup.

(b) Show that the set {e, a4} is a subgroup of S3, and that it is not a normal subgroup.

6. (a) Show that G is abelian if and only if ZG = G.

(b) Show that ZG is abelian and a normal subgroup. (c) Show that if H is a subgroup of G and a union of conjugacy classes then H is normal.

7. The symmetry group of a square, D4, is the set of group elements whose group action leave

the square invariant (looking the same). The group elements are D4 = {e, a, b, c, d, f, g, h} and their actions are:

• e is the identity and does nothing;

• a, b, c, d are reﬂections about the axes A, B, C, D shown in Figure 2.3

• f, g, h are clockwise rotations by π/2, π, 3π/2.

(a) Compute the multiplication table of D4. (b) Determine if the group is abelian.

(d) Find which of these subgroups correspond to symmetry groups of rectangles. 2. Symmetries and ﬁnite groups 13

(e) Show that the conjugacy classes of D4 are

[C1] = {e}, [C2] = {g}, [C3] = {f, h}, [C4] = {a, b}, [C5] = {c, d}.

(f) Construct a multiplication table for these conjugacy classes

C A D

Figure 2.3: A square with its symmetry axes labelled A, B, C, D about which the square may be reﬂected. 3. Representation theory basics 14

3. Representation theory basics

As we alluded to earlier, the fundamental connection between groups and symmetries is the idea of a group action. We can elucidate this connection further using representation theory. It provides a framework for speciﬁc techniques aimed at understanding and exploiting symmetries of a mathematical or physics object, whether it be the triangle or the Hydrogen atom. In the remainder of this course we will make extensive use of various results from linear algebra, which we will cite as we need them. Some comments on notation. We will adopt Dirac’s braket notation for vectors acted on by representations.1 An N-component column vector is denoted by a ket, |vi, while a row vector is denoted hw|. The two are related by hermitian conjugation: |vi† = hv|. The inner product of a pair of vectors is then hv|wi:

N X ∗ hw|vi = wi vi i=1 In this course, unless otherwise stated, the index i can be written as a superscript or subscript, there is no distinction. An N × N matrix M multiplying a column vector is denoted M|vi. Given an orthonormal basis of column vectors |e1i,..., |eN i, the (i, j)-component of the matrix M with respect to this basis is given by

Mij = hei|M|eji. (3.1)

The ﬁrst index always denotes the row, while the second index always denotes the column. Again, the indices can be written as superscripts or subscripts, there is no distinction.

Orthonormality of the basis vectors can be stated using the Kronecker delta symbol δij:

hei|eji = δij, where δij = 1 if i = j and 0 otherwise. Notice that this is the same as the (i, j)th entry of the identity matrix

δij = hei|I|eji.

We are using the summation convention in which a repeated index means they are im- plicitly summed over unless stated otherwise. For example

N X ∗ ∼ ∗ wi vi = wi vi i=1 The repeated index is often referred to as a dummy index. This is to emphasise that just as PN when we write a summation symbol j=1 we can equally well use any letter as an index, when

1You may have seen this notation already if you took quantum mechanics. 3. Representation theory basics 15 we write repeated indices we can use any letter we like e.g. i, j or p — the answer is the same:

∼ ∗ ∼ ∗ ∼ ∗ hw|vi = wi vi = wj vj = wpvp

Exercises 1 and 2 of the notes has some examples to help familiarise yourself with this notation.

The matrix components Mij depend on the basis vectors we chose. To make this clearer, let us distinguish the linear operator M and its matrix presentation with respect to |eii by denoting

e Mij = hei|M|eji

Now suppose we choose a diﬀerent orthonormal basis |f1i,..., |fN i. Then the same linear operator M has a diﬀerent presentation with respect to this new basis:

f Mij = hfi|M|fji.

What is the relationship between M f and M e? Firstly, the two bases are related to each other by a change of basis:

† |fji = |ekiSkj, hfi| = Silhel|

† † † hfi|fji = Sikhek|eliSlj = SikδklSlj = SikSkj = δij.

We recognise the last equality as S†S = I, where I is the identity matrix. Hence, S† = S−1. f Returning to our expression for Mij:

f † † e Mij = Silhel|M|ekiSkj = SilMlkSkj

As S† = S−1 we see the two presentations of the linear operator M are related via a similarity transformation: M f = S−1M eS.

3.1. What is representation theory?

Deﬁnition 3.1. A representation of a group G is a mapping Γ: G → End V of the elements of G to a set of linear operators End V acting on a vector space V with the following properties

1. Γ(e) = I, where I is the linear operator corresponding to the identity operation;

2. Γ(g1)Γ(g2) = Γ(g1g2), meaning the group multiplication law is preserved. 3. Representation theory basics 16

We will always take V to be a vector space, so the linear operators in question are actually matrices. The identity operator I is then the identity matrix   1 0 0 ...   0 1 0 ...      I = 0 0 1    . . . ..  . . . .    1 and composition of operators Γ(g1)Γ(g2) is simply matrix multiplication. There are more general representations than V being a vector space, but we will not consider them here.

Remarks:

• The dimensionality of the matrices {Γ(g)} is the dimension of the representation.

• If Γ(g) 6= Γ(h) whenever g 6= h then the representation is said to be faithful. This means

the number of elements in the representation {Γ(g1),..., Γ(gn)} is the same as the order of the group n = ord(G).

• Given a representation Γ of a group we may generate an inﬁnite number of other representations by a similarity transformation:

−1 −1 −1 {S Γ(g1)S, S Γ(g2)S, . . . , S Γ(gn)S},

where S is an invertible matrix. This corresponds to a change of basis and we regard the two representations as equivalent.

3.2. Some examples

2 1. The group Z2 = {e, a} with a = 1 has a representation ! ! 1 0 0 1 Γ2(e) = , Γ2(a) = . (3.2) 0 1 1 0

As Γ(a)2 = Γ(e), this is indeed a representation. The representation is two-dimensional and it is also faithful. A three-dimensional representation is given by     1 0 0 0 1 0     Γ3(e) = 0 1 0 , Γ3(a) = 1 0 0 . (3.3)     0 0 1 0 0 1

Checking this is an exercise. 3. Representation theory basics 17

2. (a) A representation of Z3 is

Γ(e) = 1, Γ(a) = e2πi/3, Γ(b) = e4πi/3. (3.4)

The representation is faithful and 1-dimensional.

(b) A 3-dimensional representation of Z3 is given by       1 0 0 0 0 1 0 1 0       Γ3(e) = 0 1 0 , Γ3(a) = 1 0 0 , Γ3(b) = 0 0 1 . (3.5)       0 0 1 0 1 0 1 0 0 We construct this using a trick. Take the three group elements {e, a, b}, and let them correspond to the three basis vectors:       1 0 0       |ei = 0 , |ai = 1 , |bi = 0 . (3.6)       0 0 1 while the hermitian conjugates are denoted he| = 1 0 0 , ha| = 0 1 0 , hb| = 0 0 1 . (3.7)

Now for any g1, g2 ∈ G, let

Γ(g1)|g2i = |g1g2i. (3.8) This deﬁnes a representation, called the regular representation. It is easier to

relabel the column vectors in (3.6) as |e1i, |e2i, |e3i and row vectors as he1|, he2|, and

he3|. From (3.1), the (i, j) component of Γ(g) is then given by

[Γ(g)]ij = hei|Γ(g)|eji. (3.9)

It is an exercise to check that (3.9) gives (3.5).

The equations (3.8) and (3.9) hold generally and not just for Z3. For example, Z2 has a regular representation given by (3.2). The relation (3.9) is a simple but general way of going back and forth between linear operators and matrices, and we will use it constantly. Furthermore, we can use (3.8), (3.9) for any ﬁnite group G: deﬁne a

vector space Vn whose dimension n = ord(G) is the order of the group. For each

group element gi we associate a basis vectors |eii, and Γ(gi) deﬁnes a representation via (3.9).

3. The general linear group GL(n, C), the set of invertible matrices with matrix multiplication, is itself a representation. This is because they form a vector space, and obey the axioms of a representation. We refer to this representation as the deﬁning representation. 3. Representation theory basics 18

4. A 2-dimensional representation of D3, the symmetry group of the equilateral triangle is given by

! 1 0 Γ(E) = , 0 1 ! √ ! √ ! 1 0 1 −1 3 1 −1 − 3 Γ(A) = , Γ(B) = √ , Γ(C) = √ , (3.10) 0 −1 2 3 1 2 − 3 1 √ ! √ ! 1 −1 3 1 −1 − 3 Γ(D) = √ , Γ(F ) = √ . 2 − 3 −1 2 3 −1

5. The one-dimensional representation of a group G deﬁned by

Γ(g) = 1 ∀ g ∈ G,

is the trivial representation. Every group has a trivial representation. It is not faithful.

6. The permutation group of n ordered objects Sn = {x1, . . . , xn}, can be written in terms of cycles, where a cycle is a cyclic permutation of a subset. Each cycle is written as a set of numbers in parenthesis, indicating which items are to be permuted. For example, (1)

means x1 → x1 while (1372) means x1 → x3, x3 → x7, x7 → x2, x2 → x1.A k-cycle has k-elements in the given parenthesis. So (1372) is a 4-cycle, (12) is a 2-cycle and (1) is a 1-cycle. A group element can consist of multiple cycles, but no number can appear more than once. For example (12)(234) is not a valid permutation, but (12)(34) is valid. The identity element has just 1-cycles, e = (1)(2) ... (n). Where convenient, we can omit 1-cycles as they have trivial action. So the group element (12)(3)(4) could be written (12).

An arbitrary element has kj j-cycles where

n X jkj = n. j=1

The deﬁning representation of Sn is where the objects being permuted are the basis vectors of an n-dimensional vector space

|e1i,..., |eni.

For example, if g = (12), the representation acts as

Γ(g)|e1i = |e2i, Γ(g)|e2i = |e1i, Γ(g)|eαi = |eαi, α = 2, . . . , n. 3. Representation theory basics 19

and in matrix notation   0 1 0 ···   1 0 0 ···   Γ(g) =   . 0 0 1 ··· . . . .  . . . ..

3.3. Reducible and irreducible representations

We can take two separate representations Γ1, Γ2 of a group G a form a new representation

! Γ1(g) 0 Γ(g) = ∀ g ∈ G. (3.11) 0 Γ2(g)

This block diagonal structure (3.11) can be written as a direct sum

Γ(g) = Γ1(g) ⊕ Γ2(g).

The representation Γ is not minimal as we can extract lower dimensionality representations Γ1, Γ2. What we want is a minimal representation of a group, so that there is no way of decomposing into such a direct sum, and without reference to its matrix structure. A vector subspace {0} ⊂ W ⊂ V is called Γ-invariant if Γ(g)|wi ∈ W for all g ∈ G and for all vectors |wi ∈ W . This is denoted as Γ(g)W = W . If we deﬁne a projector P : V → W such that P 2 = P then we can express W being Γ-invariant as P Γ(g)P = Γ(g)P . For example, the representation (3.11) leaves a vector subspace with projector given by:

( !) ! |w1i 1 0 W = ,P = |0i 0 0

This leads us to some important deﬁnitions.

Deﬁnition 3.2. If a representation Γ of a group G has a Γ-invariant vector subspace then it is reducible.

Deﬁnition 3.3. If a representation Γ is not reducible then it is called irreducible.

For example, an irreducible representation of the permutation group S3 is given by ! √ ! √ ! 1 0 1 −1 − 3 1 −1 3 Γ(e) = , Γ(a1) = √ , Γ(a2) = √ , 0 1 2 3 −1 2 − 3 −1 ! √ ! √ ! (3.12) −1 0 1 1 3 1 1 − 3 Γ(a3) = , Γ(a4) = √ . Γ(a5) = √ . 0 1 2 3 −1 2 − 3 −1 3. Representation theory basics 20

Deﬁnition 3.4. A representation Γ is semi-simple if it can be expressed as a direct sum

r M i Γ(g) = aiΓ (g) ∀ g ∈ G, (3.13) i=1

i where each of the representations Γ are irreducible and ai is the number of times of the matrix Γi(g) appears in Γ(g). Recall that a unitary matrix U is one that satisﬁes U † = U −1, where U † is the hermitian conjugate of the matrix (recall that hermitian conjugation is complex conjugation and transposition U † = (U ∗)t). We also have the notion of a unitary representation.

Deﬁnition 3.5. If Γ(g) are unitary matrices for all g ∈ G, then the representation Γ is unitary.

3.4. Useful theorems

In the following knowing the theorems is examinable, but you will not be required to know the proofs.

Theorem 3.1. Every representation Γ of a ﬁnite group G is equivalent, via a similarity transformation, to a unitary representation.

Proof. Suppose Γ is a representation of a ﬁnite group G. Construct the matrix

X S = Γ(g)†Γ(g). g∈G

Then S is hermitian and positive semi-deﬁnite; that is, it can be diagonalised and all its eigenvalues are non-negative

  λ1 0 ··· −1   S = U  0 λ2 ··· U, λi ≥ 0 ∀ i.  . . .  . . ..

In fact, all of the eigenvalues are positive: λi > 0. To see this, suppose we have a zero eigenvalue. Then, there is an eigenvector |vi such that S|vi = 0. But, from the deﬁnition of S:

X X 0 = hv|S|vi = hv|Γ(g)†Γ(g)|vi = |Γ(g)|vi|2. g∈G g∈G

The only way to satisfy this equality is if Γ(g)|vi = 0 for all g ∈ G. This is a contradiction though as we could choose g = e with Γ(e)|vi = |vi= 6 0. 3. Representation theory basics 21

Given this we can construct a hermitian matrix, which is roughly speaking the square root of the matrix S (i.e. we take the square root of the eigenvalues):

 1/2  λ1 0 ··· −1  1/2  X = U  0 λ2 ··· U,  . . .  . . .. and as none of the diagonal elements vanish, X is invertible . It has the property that X2 = S. Remarkably if we deﬁne a new representation Γ0 through a similarity transformation:

Γ0(g) = X Γ(g)X−1, ∀ g ∈ G.

We show Γ0 is a unitary representation. First,

Γ0(g)†Γ0(g) = X−1Γ(g)†SΓ(g)X−1

Second, ! X Γ(g)†SΓ(g) = Γ(g)† Γ(h)†Γ(h) Γ(g) h∈G X X = Γ(hg)†Γ(hg) = Γ(h)†Γ(h) h∈G h∈G = S = X2. where the middle line follows as the summation runs over all elements in the group. Therefore,

Γ0(g)†Γ0(g) = I, ∀ g ∈ G, and Γ0 is a unitary representation.

Consequently, we will only consider unitary representations hereon in.

Theorem 3.2. Every representation Γ of a ﬁnite group G is semi-simple.

Proof. By the previous theorem we only need to consider unitary representations. If the representation Γ is irreducible then we are done. So we assume Γ is reducible acting on a vector space V . Then, there exists a Γ-invariant subspace W with projector P such that

P Γ(g)P = Γ(g)P (3.14) for all g ∈ G. Taking the hermitian conjugate we ﬁnd P Γ(g)†P = P Γ(g)†. Unitarity implies Γ(g)† = Γ(g)−1 = Γ(g−1) and so we have P Γ(g)P = P Γ(g) (3.15) 3. Representation theory basics 22 for all g ∈ G. But this implies (1 − P )Γ(g)(1 − P ) = Γ(g)(1 − P ) for all g ∈ G. Thus 1 − P projects on an invariant subspace Wf, and so V = W ⊕ Wf. We can repeat this process (induct- ively) on W and Wf until we completely reduce the representation. As G is ﬁnite this process must stop, leaving with a block diagonal representation: V = ⊕iWi. Hence, Γ is semi-simple.

How can we determine if a representation is irreducible? Shur’s Lemmas2 allow us to determine this.

Theorem 3.3 (Schur’s Lemma 1). If Γ1(g)A = AΓ2(g) for all g ∈ G where Γ1 and Γ2 are inequivalent, irreducible representations, and A is a linear map, then A = 0.

Proof. Let V be the vector space upon which Γ1, Γ2 act. Suppose there is a vector subspace W ⊆ V such that A|wi = 0 for |wi ∈ W . Let P be the projector P : V → W onto this subspace, so that AP = 0. But then

AΓ2(g)P = Γ1(g)AP = 0 ∀ g ∈ G.

This means W is Γ2-invariant as Γ2W ⊆ W . However, Γ2 is irreducible, and so either W = V or

W = {|0i}. If W = V then A = 0. An analogous argument utilising Γ1 being irreducible shows that if there is a |vi ∈ V such that hv|A = 0, then either |vi = 0 or hv|A = 0 for all hv| ∈ V . So we have two cases: (i) A annihilates a vector on either the left or right, in which case A = 0 and we are done; (ii) there is no vector which annihilates A. In that case A must be an invertible square matrix. It is square because if the number of rows were greater than the number of columns then there would be a vector that annihilates A on the right. Its determinant is non-zero as there are no zero eigenvalues and so it is invertible. However, this means

−1 Γ2 = A Γ1(g)A, and Γ2, Γ1 are related by a similarity transformation, a contradiction. Hence, case (ii) is impossible, and A = 0.

The other half of Schur’s lemma covers the case where Γ1, Γ2 are equivalent representations.

In this case, by a change of basis, we may as well take Γ1 = Γ2 = Γ.

Theorem 3.4 (Schur’s Lemma 2). If Γ(g)A = AΓ(g) for all g ∈ G where Γ is a ﬁnite dimen- ∗ sional irreducible representation then A = I for some λ ∈ C . 2For historical reasons Shur’s Lemma is referred to as ‘lemma’, though it is so important it should really be referred to as theorem. 3. Representation theory basics 23

Proof. As Γ is finite dimensional acting a finite dimensional vector space V , the polynomial det(A − λI) = 0 has at least one solution and we can find a corresponding eigenvector |vi such that A|vi = λ|vi. This means

Γ(g)(A − λI) = (A − λI)Γ(g), ∀ g ∈ G and (A − λI)|vi = 0.

The argument of the previous result now applies as combining these two equations we ﬁnd (A − λI)Γ(g)|vi = 0, and hence the subspace annihiliated by A − λI is left Γ-invariant. Irreducibility then implies this subspace must actually be V , and hence A − λI = 0.

Remarks:

• Schur’s Lemma mean if we can ﬁnd a matrix that commutes with all the elements of the representation then that matrix is proportional to identity.

• The form of the basis states of an irreducible representation are unique up to a rescaling by a non-zero complex number. If we were to change basis using a similarity transformation A−1Γ(g)A = Γ(g) for all g ∈ G then Schur’s lemma implies A ∝ I.

• Furthermore, if we demand the change of basis is unitary then |λ| = 1, and so the complex number is a phase λ = eiφ.

3.5. Characters

In this section we omit the proofs, and simply state the results. Some of the proofs are left as exercises, and others take us too far aﬁeld form our main goal.

The character χΓ of a representation Γ are the traces of the matrices:

χΓ(g) = Tr Γ(g). (3.16)

As the Tr obeys a cyclic property: Tr AB = Tr BA characters χΓ are unchanged by similarity transformations Tr S−1Γ(g)S = Tr Γ(g). Characters are also constant over conjugacy classes:

−1 χΓ(g hg) = χΓ(h). (3.17)

α ∼ As χ is invariant over conjugacy classes, we denote χa = χΓα ([Ca]) for the conjugacy class [Ca] in the representation Γα. The character of an irreducible representation is unique, and so give a neat way of distin- guishing inequivalent representations. That is, if Γ1 and Γ2 are inequivalent representations then χΓ1 6= χΓ2 . 3. Representation theory basics 24

There are two orthogonality theorems. The ﬁrst says if we consider two inequivalent unitary irreducible representations, denoted Γα and Γβ then

X α ∗ β h αβ Γ (g)ijΓ (g)kl = δikδjlδ . (3.18) nα g∈G

α where h = ord(G) and nα is the dimension of the representation Γ . The second is the analogue for characters, the character orthogonality theorem:

k X α β ∗ X α β ∗ αβ χ (g)χ (g) = haχa χb = hδ . (3.19) g∈G a=1

Here k is the number of conjugacy classes with the ath class having ha elements. If Γ is a semi-simple representation then it immediately follows that

n X α χ(g) = aαχ (g). α=1

Turning this around using the character orthogonality theorem, the number of times the irreducible representation Γα appears in Γ is

k 1 X 1 X a = χ(g)χα(g)∗ = h χ χα∗. α h h a a a g∈G a=1

A necessary and suﬃcient condition then for Γ to be irreducible is that

k X 2 X 2 |χ(g)| = ha|χa| = h. (3.20) g∈G a=1

This gives us a criterion for deciding if an representation is irreducible, and it depends only on the characters and order of the group, both relatively straightforward quantities to compute.

3.6. Tensor Products

We can put representations together to form new representations. We have already seen this through the direct sum (3.11), though it is somewhat trivial. A more interesting way to proceed is to construct a tensor product representation.

Suppose Γ1 is an m-dimensional representation with basis vectors |e1i,..., |emi acting on a vector space V1 and Γ2 is an n-dimensional representation with basis vectors hf1|,..., hfn| acting on the vectors space V2. Then, we can make an mn-dimensional vector space V1 ⊗ V2 called the tensor product space whose basis vectors are deﬁned by the tensor product hei| ⊗ hfp| for i = 1, . . . , m and p = 1, . . . , n. The order here matters, and so there are mn-basis vectors. On 3. Representation theory basics 25

this new vector space V1 ⊗ V2 we can construct a representation by taking the tensor product of the corresponding matrices:

ΓΓ1⊗Γ2 (g) = Γ1(g) ⊗ Γ2(g). (3.21)

It is an exercise to prove that this is a representation. It is not in general an irreducible representation though. 3. Representation theory basics 26

3.7. Exercises

1. (a) Suppose we have a column vector ! a |vi = , a, b ∈ C, b

then write an expression for hv|vi in terms of a, b. (b) Start with the basis vectors ! ! 1 0 |e1i = |e2i = 0 1

Suppose we have linear operator M with a matrix presentation with respect to this basis as ! e a b M = , a, b, c, d ∈ C. (3.22) c d Compute e Mij = hei|M|eji, for i, j = 1, 2.

Check that this basis is orthonormal. (d) By computing f Mij = hfi|M|fji, for i, j = 1, 2. construct a matrix M f . (e) Find a similarity transformation S that relates M f to M e given in (3.22).

2. The aim of this exercise is to get familiar with the summation convention in which a repeated index denotes it is summed. For example, two vectors hv| = (v1, . . . , vN ) and ∗ ∗ hw| = (w1, . . . , wN ) have inner product

N X ∗ ∼ ∗ hw|vi = vawa = vawa. a=1

(a) A matrix M is denoted in terms of indices as Mab, where a, b = 1,...,N and a is the row index, while b denotes the column index. Show that if M,N are square matrices then matrix multiplication P = MN is given by

Pab = MacNcb. 3. Representation theory basics 27

(b) Show that the transpose of a matrix has components

t [M ]ab = Mba.

(c) The Kronecker delta symbol δ delta symbol satisﬁes δab = 1 if a = b and vanishes otherwise. Show that ∗ hv|vi = vavbδab.

(d) Show that the trace of a pair of square matrices M,N satisﬁes

Tr MN = MabNba = MabNcdδbcδda.

(e) Show that Tr M t = Tr M

using the summation convention.

3. (a) Show that (3.3) is a faithful representation of Z2.

(b) Show that Γe2 is also a representation of Z2, where ! ! 1 0 0 −i Γe2(e) = , Γe2(a) = . (3.23) 0 1 i 0

4. (a) Show that (3.8) deﬁnes a representation. (b) Using (3.9), derive the expressions in (3.3) and (3.5).

(d) Find a representation of D4, analogous to (3.10).

5. Prove equation (3.17).

6. (a) Let Γi, Γj be two representations. Prove the character orthogonality relation, which states that X i j ∗ χ (g)χ (g) = hδij. (3.24) g∈G where χi(g), χj(g) are the characters for the representations Γi, Γj. (b) Using the fact every semi-simple representation can be bought into the form (3.13) it follows that n X i χ(g) = aiχ (g), i=1 3. Representation theory basics 28

i where ai is the degeneracy of the representation Γ . Show ai is given by 1 X a = χ(g)χi(g)∗, (3.25) i h g∈G

where χ is the character of Γ and h = ord(G).

7. Prove the tensor product (3.21) deﬁnes a representation. 4. Lie groups and Lie algebras 29

4. Lie groups and Lie algebras

Thus far we have primarily looked at ﬁnite groups, whose elements are labelled by discrete numbers e.g. G = {g1, . . . , gn}. We want to extend our notion of group to include group elements that depend smoothly on a set of parameters e.g.:3

N M g(α) ∈ G, α ∈ C , R .

The parameters can be complex or real. The corresponding dimension is simply the number of independent parameters. The simplest example of a Lie group we have already met: GL(m, C). This group has m2 complex parameters, and so its complex dimension is m2. Groups of this type are known as Lie groups and were developed by Sophus Lie, a 19th century Norwegian mathematician. In this course we are going to study representations of Lie groups and their closely related cousins, Lie Algebras—the linearisation of a Lie group. Both are of central importance to huge range of areas within modern mathematics and physics. Our brief discussion of Lie groups and Lie algebras here certainly does not do this large and fascinating area of mathematics justice, as there are lots of interesting facts beyond the scope of our course. Instead we focus on the bare necessities in order to study representations of Lie algebras in later chapters.

4.1. Lie groups

A Lie group is a group that is also a smooth manifold whose composition law is compatible with this smoothness. The manifold is described by the set of parameters α mentioned above, and we need to connect these to the group elements. The way to do this is to start with the identity element, an element common to all groups. We want to parameterise our group elements so that α = 0 corresponds to identity g(0) = e. To that end, we assume the group elements in the neighbourhood of e are described by a function of N real parameters, αa, for a = 1,...,N such that

g(α)|α=0 = e.

We will always assume the Lie group G has a representation, Γ, and that

Γ(α)|α=0 = I.

We can Taylor expand in the neighbourhood of e for δα 1:

a Γ(δα) = I + iδαaX + ..., (4.1)

3Smooth here means that g(α) and g(β) are close to each other iﬀ α and β are close to each other. 4. Lie groups and Lie algebras 30 for some set of matrices Xa, and don’t forget we are using the summation convention. The matrices Xa can be written

a ∂ X = −i Γ(α)|α=0. (4.2) ∂αa The inclusion of the i means that if the representation is unitary then the matrices Xa will be hermitian.

Figure 4.1: A Lie group can be visualised a smooth manifold where points on the manifold are group elements. There is a distinguished point e. We are interested in representations of these group elements so that Γ(e) = I. A small movement in the neighbourhood of e is represented a a by iδαaX , where X are a set of matrices, known as the generators.

Deﬁnition 4.1. The set of matrices Xa for a = 1,...N are a representation of the generators of the Lie group. The generators allow us to describe the group elements through the exponential map. As we ~ move away from e in some ﬁxed direction in parameter space described by δα = (δα1, . . . , δαN ), a we parameterise the group elements by exponentiating iδαaX giving a matrix exponential:

a Γ(α) = exp(iαaX ), (4.3)

1 2 where for a square matrix M we have exp(M) = I +M + 2! M +··· . This gives an expression for elements g(α) of the Lie group G in terms of the representation Γ. We will always be interested in representations of the group, not the abstract group itself. 4. Lie groups and Lie algebras 31

4.2. Examples of Lie groups

The most prominent example of Lie groups are the matrix groups. We will now list a few examples. Showing many of the properties of these Lie groups are left as exercises at the end of the chapter, and are an excellent way to get familiar with Lie groups.

1. GL(m, C): the general linear group over C. These are m × m invertible matrices (non- zero determinant). There is a complex parameter for every entry in the matrix, and so 2 2 GL(m, C) is m -complex dimensional. We can write this as dimC GL(m, C) = m . For example, ( ! ) a b GL(2, C) = M = a, b, c, d ∈ C and det M 6= 0 . c d

All matrix other matrix groups are subgroups of GL(m, C).

We can also study the general linear group over the reals given by GL(m, R), where the entries of the matrices are now real numbers. This has real dimension m2 and can be 2 written dimR GL(m, R) = m . 2 Note, just as C can be viewed as a real manifold R with real dimension dimR C = 2, we 2 can view GL(m, C) as a real manifold with real dimension dimR GL(m, C) = 2m .

2. The special linear group SL(m, C) over complex numbers are elements of GL(m, C) such 2 that det M = 1. This group is m − 1 complex dimensional. SL(m, R) is deﬁned analogously except it is over the reals.

3. The unitary group U(m) is the group of unitary matrices over the complex numbers:

n † o U(m) = M ∈ GL(m, C) M M = I . (4.4)

These are matrices that preserve the inner product of m-dimensional complex column m 0 vectors |vi, |wi ∈ C so that if |vi = M|vi then

hv0|w0i = hv|wi. (4.5)

Matrices M satisfy (4.5) if and only if they are in the set (4.4). To see this ﬁrst suppose M ∈ U(m). Then expand the left hand side of (4.5) in terms of M:

hv|M †M|wi = hv|wi.

Hence, (4.5) is satisﬁed as M †M = I. For the converse, suppose M satisﬁes (4.5). This m equation applies for any |vi, |wi ∈ C . Denote |e1i,..., |eni as the canonical orthonormal 4. Lie groups and Lie algebras 32

m basis for C so that hei|eji = δij, where δij is the Kronecker delta symbol δ which satisﬁes

δij = 1 if i = j and vanishes otherwise. In this basis the (i, j) entry of any matrix N is

hei|N|eji = Nij. Apply to (4.5):

† † hei|M M|eji = (M M)ij = δij.

The (i, j)-entry of M †M is the same as the identity matrix and

M †M = I. (4.6)

To calculate the dimension of U(m), or any Lie group, we need to subtract the number of independent equations that arise from (4.6) from the number of unknowns. Let us start with a warm-up example of m = 2. A generic matrix is ! a b M = c d

where a, b, c, d ∈ C. Each complex number is composed of two real numbers and so we start with 2×4 = 8 real parameters. These are reduced by imposing the unitary condition (4.6): ! ! |a|2 + |b|2 ac¯ + bd¯ 1 0 = . ac¯ + ¯bd |c|2 + |d|2 0 1 Two important features: the diagonal elements are real and the (1, 2)-component is the complex conjugate of the (2, 1)-component. So we get two real equations and one complex equation: |a|2 + |b|2 = 1, |c|2 + |d|2 = 1, ac¯ + bd¯= 0.

The complex equation reduces the real parameters by two, the real equations by one each leaving 8 − 1 − 1 − 2 = 4 real parameters. The Lie group U(2) is 4 real dimensional. The same logic applies for U(m). We start with 2m2 real parameters. These are reduced by (4.6). As M †M is hermitian the (j, i) entry is the complex conjugate of the (i, j) entry. The diagonal elements are real giving m real independent equations, and while the upper 1 triangular elements give 2 m(m − 1) independent complex equations. The real dimension of U(m) is 2m2 − m − m(m − 1) = m2. Finally, det U †U = | det U|2 = det I = 1 and so det U = eiφ for some φ.

4. The special unitary group SU(m) are unitary matrices with det M = 1:

n † o SU(m) = M ∈ GL(m, C) M M = I and det M = 1 . (4.7)

det M = 1 reduces the real dimension by 1, ﬁxing eiφ = 1. 4. Lie groups and Lie algebras 33

5. The orthogonal group O(m) is in some sense the real analogue of the unitary group. It is the set t O(m) = M ∈ GL(m, R) M M = I . (4.8)

It is an exercise to show that these are all matrices that preserve the inner product on m 0 m-dimensional real vectors. If M ∈ O(m) and vectors |vi, |wi ∈ R so that |vi = M|vi, |wi0 = M|wi then hw0|v0i = hw|vi. (4.9)

m t Of course, as |wi ∈ R , hermitian conjugation is simply transposition so that |wi = hw|. Taking the determinant of the equation M tM = I, we ﬁnd4 det(M tM) = (det M)2 = 1. Hence, orthogonal matrices have det M = ±1.

1 The real dimension of O(m) is 2 m(m − 1).

6. Special orthogonal group SO(m). These are orthogonal matrices with the additional con- straint that det M = 1:

t SO(m) = M ∈ GL(m, R) M M = I with det M = 1. . (4.10)

1 The real dimension of SO(m) is also 2 m(m − 1).

Remarks:

• Notice that every matrix group written above has been written in its deﬁning representation or fundamental representation. The groups have other representations, which we will return to in later chapters.

• We can assign a nice geometric interpretation to some of these groups. An n-dimensional real sphere can be described as the locus of

2 2 2 2 a1 + a2 + ··· + an = R , ai ∈ R.

We can phrase this equation in vector notation by deﬁning   a1   a2   n |ai =  .  ∈ R ,  .    an

4Recall that det AB = det A det B and det At = det B for square matrices A, B. 4. Lie groups and Lie algebras 34

so that the deﬁning equation of the sphere can be written as

ha|ai = R2.

Recall that a linear transformation (e.g. rotation/scaling) can be represented as an invertible matrix action on the vector |ai:

|a0i = M|ai,

where M is a n × n invertible matrix. The symmetry group of the sphere are the group of linear transformations of |ai that preserve the deﬁning equation. This means

ha0|a0i = ha|ai = R2.

For (4.9) we see that M ∈ O(n). Hence, the symmetry group of the sphere is O(m). Geometrically, the matrix M is an invertible map that takes any given point on the sphere to some other point on the sphere. The geometric interpretation of SO(m) is that it is the group of rotations that preserves orientation. For example, this rules out reﬂections.

• Similarly, the group U(n) is the symmetry group of the n-complex dimensional sphere deﬁned by 2 2 2 2 |a1| + |a2| + ··· + |an| = R , a1, . . . , an ∈ C.

• As a manifold U(1) = {eiφ|φ ∈ [0, 2π)} =∼ S1. We will see below that SU(2) =∼ S3. The special orthogonal groups are SO(2) =∼ S1, SO(3) =∼ S3 while SO(4) =∼ S3 × S3. The global topological of the Lie groups is clearly interesting. We will focus instead on the local behaviour of the Lie groups, studying the behaviour of group elements in the neighbourhood of identity.

4.3. Multiplication and Lie algebras

The local behaviour of a Lie group is dictated by its Lie algebra. The Lie algebra is concerned with answering the question: if g(α), g(β) ∈ G then can we find a g(γ) such that g(α)g(β) = g(γ)? Or, in terms of representations Γ(α)Γ(β) = Γ(γ). The answer is in general difficult. But if g(α), g(β) are in the neighbourhood of the identity element, then we can use the Lie algebra to find an answer. If the matrices all commute, i.e. XaXb = XbXa for all a, b, then multiplication of group elements is straightforward a b a eiαaX eiβbX = ei(αa+βa)X . 4. Lie groups and Lie algebras 35

That is, Γ(α)Γ(β) = Γ(α + β). But this equation does not hold in general. To see this, Taylor expand around identity:

a b 1 1 eiαaX eiβbX = 1 + iα Xa + iβ Xa − α Xaβ Xb − (α Xa)2 − (β Xa)2 + ··· , (4.11) a a a b 2 a 2 a and compare with

a b 1 eiαaX +iβbX = 1 + iα Xa + iβ Xa − (α Xa + β Xa)2 + ··· . (4.12) a a 2 a a The two expressions diﬀer by terms proportional to the commutator [Xa,Xb] =∼ XaXb−XbXa. To ﬁnd an expression for the multiplication of group elements we start with the fact the group is smooth. This means there must be a γa such that

a b a eiαaX eiβbX = eiγaX

a 1 2 We solve for γaX by using the Taylor expansion log(1 + K) = K − 2 K + ··· :

a b a iαaX iβbX iγaX = log 1 + (e e − 1)

a b 1 a b = (eiαaX eiβbX − 1) + (eiαaX eiβbX − 1)2 + ··· 2 1 1 1 = iα Xa + iβ Xa − α Xaβ Xb − (α Xa)2 − (β Xa)2 + (α Xa + β Xa)2 + ··· a a a b 2 a 2 a 2 a a 1 = iα Xa + iβ Xa − [α Xa, β Xb] + ··· , (4.13) a a 2 a b Now, [Xa,Xb] is itself a generator and so

[Xa,Xb] = if abcXc, (4.14) for some constants f abc which satisfy f abc = −f bac. These are called the structure constants. The structure constants are very important as it turns out they entirely deﬁne the group multiplication law to all orders.

Deﬁnition 4.2. The matrix generators satisfy an important identity

[Xa, [Xb,Xc]] + [Xb, [Xc,Xa]] + [Xc, [Xa,Xb]] = 0, (4.15) called the Jacobi identity. Using (4.14) it is an exercise to show that this gives a condition on the structure constants

f bcdf ade + f cadf bde + f abdf cde = 0. (4.16)

To linear order in X, γa = αa + βa. In the exercises you will check this works to quadratic order in X, showing 1 γ = α + β − f abcα β . (4.17) c c c 2 a b 4. Lie groups and Lie algebras 36

[X,Y]

X Y

Lie group Lie algebra

Figure 4.2: A cartoon of a Lie algebra being a local description of a Lie group. The Lie group being in general a complicated manifold, with curvature; the Lie algebra is in some sense the local behaviour of the Lie group. The Lie algebra pertains to the neighbourhood of identity and is illustrated by the circle colour yellow. Two generators, X,Y , elements of a vector space, are pictured. Their commutator [X,Y ] is also an element of the vector space.

The interesting thing is that γ can be determined in terms of just α, β and f abc to all orders in the Taylor series expansion provided the generators form an algebra under commutation. This leads us to an important deﬁnition.

Deﬁnition 4.3. Let G be a Lie group. Then its Lie algebra g is an N-dimensional vector space spanned by its generators {X1,...,XN } together with a commutation operation [Xa,Xb] such that [Xa,Xb] + [Xb,Xa] = 0 and it satisﬁes the Jacobi identity. A representation Γ of the Lie group G furnishes a natural representation of the Lie algebra Xa via the parametrisation (4.3). In this case the commutation relation is given by (4.14).

Remarks:

• We will always denote a Lie group using upper case G and its corresponding Lie algebra in lower case gothic font g. 4. Lie groups and Lie algebras 37

• Intuitively a Lie algebra describes the local multiplication structure of a Lie group. That is, for small variations of parameters around the identity element.

• We have deﬁned a Lie algebra via a matrix representation of its generators Xa. There are deﬁnitions of Lie algebras which do not reference representations, but this is beyond the scope of this course.

• The structure constants f abc are the same for all representations because they are ﬁxed purely by group multiplication and smoothness.

• We give an example of a Lie algebra so(3) corresponding to the Lie group SO(3) in Figure 4.3.

To sum up, studying Lie groups is diﬃcult thanks to its opaque multiplication law. We will approach an easier problem, and study the representations of the corresponding Lie algebra. This will allow us to shed light on local properties of the group in the neighbourhood of the identity element.

Figure 4.3: An example: the Lie algebra so(3) corresponding to the rotation group SO(3). As discussed above, SO(3) are the group of symmetry transformations that preserve the sphere and its orientation. The Lie algebra is given by considering inﬁntesimal rotations. In this case there are two possibilities: a rotation along longitude X and a rotation along latitude Y . These are vectors as shown, and together with the commutator [X,Y ] = XY − YX form a Lie algebra. 4. Lie groups and Lie algebras 38

4.4. Adjoint representation

A representation common to all Lie algebras is known as the adjoint representation. We deﬁne a set of N matrices

[T a]bc = −if abc, a = 1,...,N. (4.18) where on the left hand side the index b refers to the row and c refers to the column of the matrix T a. We need to check the commutation relation is satisfied in order for this to be a bona fide representation. Using the Jacobi identity (4.16) we find

[T a,T b] = if abcT c. and so (4.18) is indeed a representation. The dimension of the representation is simply N, the number of real parameters required to describe the corresponding Lie group. For example, the Lie algebra su(2) has an adjoint representation given by       0 0 0 0 0 i 0 −i 0 1   2   3   [T ] = 0 0 −i , [T ] =  0 0 0 , [T ] = i 0 0 . (4.19)       0 i 0 −i 0 0 0 0 0

4.5. An inner product and compact algebras

We would like to construct an inner product on the generators of a Lie algebra in the adjoint 0a a b representation. Under a linear transformation of the generators: X = Lb X we have

0a 0b a b dec c a b dec −1 c 0f 0 abc 0c [X ,X ] = iLdLef X = iLdLef Lf X = f X .

Hence, under the linear transformation the structure constants transform as

0 abc a b −1 f def f = LdLeLc f .

The matrices [T a] also transform

a b a b d e −1 f [T ]c = LdLe[T ]f Lc .

Recall the b and c refer to row and column indices respectively, and we can write this equation in terms of matrix multiplication:

a a d −1 T = Ld (LT L ). 4. Lie groups and Lie algebras 39

The matrix is undergoing a generalised similarity transformation. Notice that the trace is invariant under part of this transformation

a b a d −1 b e −1 a e d e Tr T T = Tr (LdLT L )(LeLT L ) = LdLbTr (T T ).

We can construct a new matrix K whose (a, b) entry is the number Tr T aT b. The matrix K is a type of inner product. What this result shows is that under a similarity transformation, the matrix K undergoes a similarity transformation. By choosing an appropriate L, we can therefore diagonalise K. When all the eigenvalues are positive the algebra is said to be compact, and one can show Tr T aT b = λδab, λ > 0. (4.20)

When this is the case, it is an exercise to show that

f abc = −iλ−1Tr [T a,T b]T c. (4.21) and are completely antisymmetric

f abc = f bca = f cab = −f bac = −f acb = −f cba. (4.22)

Unless we state otherwise, we will only consider compact algebras.

4.6. Subalgebras, simple and semi-simple algebras

Deﬁnition 4.4. Let g be a Lie algebra. Then h ⊂ g is a subalgebra if it is a vector subspace5 and closed under the commutation operation (i.e. X,Y ∈ h then [X,Y ] ∈ h.)

Deﬁnition 4.5. Let g be a Lie algebra. An invariant subalgebra h ⊂ g is a subset of a generators h ⊂ g that go into themselves under commutation with any other generator in g:

h = {X ∈ g | [X,Y ] ∈ h ∀ Y ∈ g}.

Two trivial examples of invariant subalgebras as the algebra itself, g and {|0i}. When expo- tentiated, an invariant subalgebra generates a normal subgroup.

Deﬁnition 4.6. An algebra g which has no non-trivial invariant subalgebras is called simple. There are special invariant subalgebras that we would like to isolate. These are abelian invariant subalgebra with a single generator X. When they are compact, they generate the group U(1) = {eiθ | θ ∈ [0, 2π)} corresponding to phase transformations. The fact the algebra is abelian means the generator X commutes with all elements in the algebra.

5Recall W ⊂ V is a vector subspace if 0 ∈ W , and is closed under addition of vectors and multiplication by a scalar. 4. Lie groups and Lie algebras 40

Deﬁnition 4.7. An algebra without any abelian invariant subalgebras is called semi-simple. A semi-simple algebra is built by putting together simple algebras. In these algebras every generator has a non-zero commutator with some other generator. From hereon in, unless we state otherwise, we will study semi-simple algebras and their representations by unitary matrices. 4. Lie groups and Lie algebras 41

4.7. Exercises

a 0 −1 1. (a) Consider a Lie group G, with representation Γ(a) = exp(iαaX ). Let Γ = S ΓS be a representation obtained by a similarity transformation on Γ. Show that the generators transform as: Xa0 = S−1XaS.

(b) By studying (4.14) under the similarity transformation, show that the structure constants f abc do not change under a similarity transformation.

2. Show that the structure constants deﬁned in (4.14) satisfy f abc = −f bac.

3. (a) Show that each lines of (4.13) holds using the definition of a exp(M) for a square matrix M. (b) Show (4.17) by starting with (4.13) and making use of both the definition of the inner product for a compact Lie algebra (4.20) as well as the structure constants defined in (4.14).

4. (a) Show that (4.15) can be written as

[Xa, [Xb,Xc]] = [[Xa,Xb],Xc] + [Xb, [Xa,Xc]] . (4.23)

(b) Show the “product rule”

[Xa,XbXc] = [Xa,Xb]Xc + Xb[Xa,Xc] .

5. In the adjoint representation, show (4.21) and (4.22).

6. In this exercise we will show that an invariant subalgebra h ⊂ g generates an normal subgroup H ⊂ G, where H = {eiX | X ∈ h}. Recall, a normal subgroup means that if h ∈ H, and g ∈ G then g−1hg ∈ H.

(a) Suppose X ∈ g, Y ∈ h and let h = eiX and g = eiY . By Taylor expanding X0() = e−iY XeiY and then setting = 1 show that

0 g−1hg = eiX ,

where 1 X0 = e−iY XeiY = X − i[X,Y ] − [Y, [Y,X]] + ··· . 2 4. Lie groups and Lie algebras 42

(b) Each derivative in the Taylor expansion brings another commutator. Using the deﬁn- ition of an invariant subalgebra show then that eiX0 ∈ H. (c) Conclude that g−1hg ∈ H for all g ∈ G and h ∈ H.

7. (a) Show that the set of orthogonal matrices O(m) as deﬁned in (4.10) are equivalent to the set of matrices that preserve the inner product (4.9). (b) By studying the condition M tM = I for m = 2, show that the components a, b, c, d of the matrix ! a b M = c d satisfy the equations

a2 + b2 = 1, c2 + d2 = 1, ac + bd = 0.

By counting the number of matrix elements and comparing with the number of equations show that O(2) depends on one parameter. (c) Using cos θ2 + sin θ2 = 1, show that a general solution to M tM = I can be written as ! ! sin θ cos θ sin θ cos θ M = or M = . − cos θ sin θ cos θ − sin θ

Which solution is a element of SO(2) and why? (d) By writing out an arbitrary M ∈ O(m) as   a11 a12 . . . a1m   a21 a22 . . . a2m  M =   ,  .. .  a31 a32 . .    . . . . amm

t how many equations for the aij appear in the matrix identity M M = I? Using this determine the dimension of O(m). Hint: the number of entries in the (upper) oﬀ-diagonal part of a matrix is:   0 a12 . . . a1m   0 0 . . . a2m    .. .  0 0 . .  . .  . . 0

1 is 2 m(m − 1). 4. Lie groups and Lie algebras 43

1 (e) Show that the dimension of SO(m) is also 2 m(m − 1).

8. By using analogous reasoning given in section 4.2., which showed that O(n) is the symmetry group of the real sphere, show that U(n) is the symmetry group of the n-complex dimensional sphere deﬁned by

2 2 2 2 |a1| + |a2| + ··· + |an| = R , a1, . . . , an ∈ C.

Show that U(1) is the symmetry group of the real circle x2 + y2 = R2. In terms of polar coordinates, what is the generator of u(1)?

9. Show that the adjoint representation of su(2) is given by (4.19) using (4.18).

10. Carry out the expansion of γ in (4.17) to third order in α, β, i.e. one order beyond what is discussed. 5. The key example: su(2) 44

5. The key example: su(2)

Central to the representations of compact Lie algebras is su(2). For this reason, we need to study representations of su(2). In fact, we will be able to do quite a lot: we will be able to classify all the representations of su(2).

5.1. From the group SU(2) to the algebra su(2)

T he group SU(2) is deﬁned as: ( ! ) a b −1 † SU(2) = U = , a, b, c, d ∈ C U = U , det U = 1 . c d

Any SU(2) matrix can be written in its deﬁning representation as

i ψ i φ ! e 2 cos θ/2 −e 2 sin θ/2 U(θ, φ, ψ) = , φ, ψ ∈ [0, 4π), θ ∈ [0, 2π). (5.1) − i φ − i ψ e 2 sin θ/2 e 2 cos θ/2 where φ, ψ, θ are the three parameters of SU(2).6 The identity element corresponds to θ = ψ = 0. To determine the generators of the corresponding Lie algebra, su(2), we need to Taylor expand around the identity element, meaning δθ, δψ 1. Recall to linear order

sin δθ = δθ + ..., cos δθ = 1 + . . . , eiδψ = 1 + iδψ + ....

Substituting into (5.1) and keeping only linear order terms we ﬁnd, ! ! 1 0 i δψ δφ − iδθ 1 2 3 U(θ, φ, ψ) = + + ··· = I2 + i(δφT + δθT + δψT ) + ··· , 0 1 2 δφ + iδθ −δψ where ! ! ! 1 0 1 1 0 −i 1 1 0 T 1 = ,T 2 = ,T 3 = , (5.2) 2 1 0 2 i 0 2 0 −1 are the generators of su(2) in the deﬁning representation. and are referred to as the Pauli matrices. The structure constants f abc are given by computing the commutators of generators. In the exercises you will show that they satisfy

[T a,T b] = iabcT c, for a, b, c = 1, 2, 3. (5.3) where the epsilon symbol is completely antisymmetric and takes non-zero values 123 = 231 = 312 = 1 and 132 = 213 = 321 = −1. We see that the structure constants for su(2) are simply the epsilon symbol: f abc = abc.

6The normalisation of 1/2 is a common convention which we follow. 5. The key example: su(2) 45

5.2. Highest weight construction of representations of su(2)

In this section we use the ‘highest weight’ construction to study su(2). We will ﬁnd a representation Γ is classiﬁed by a half–integer j called its spin. We denote the matrices of the representation as

Γa =∼ Γ(T a) , so that [Γa, Γb] = iabcΓc . (5.4)

We would like to diagonalise as many of the matrices Γa as possible. By a suitable choice of basis, we can always diagonalise at least one matrix, say Γ3. A key fact of linear algebra, is that to diagonalise two or more matrices simultaneously they need to mutually commute [Γa, Γb] = 0. There are no such matrices in su(2). So Γ3 is the only diagonal matrix. Now consider Γ3. Its eigenvectors are called states7 of the representation and denoted by |mi where m is its eigenvalue Γ3|mi = m|mi .

The vector space is finite dimensional and so there is only a finite number of eigenvectors. Hence, there is an eigenvector with largest eigenvalue j, and we often label this the spin–j representation. The state |ji is called the highest weight state. Without loss of generality, take |ji to have unit norm: kjk2 = hj|ji = 1. We define a pair of matrices called raising and lowering operators as: 1 Γ± = √ (Γ1 ± iΓ2). 2 The commutation relations take the form

[Γ3, Γ±] = ±Γ±, [Γ+, Γ−] = Γ3. (5.5)

Using this equation, and given a state |mi, we notice that

Γ3Γ±|mi = (m ± 1)Γ±|mi. (5.6)

Hence Γ±|mi is also an eigenvector whose eigenvalue is m ± 1. If we deﬁne |m − 1i = Γ−|mi then by repeatedly applying Γ− to |ji we can generate a ‘tower of states’:

|ji , |j − 1i , |j − 2i , · · · |qi ,

7This terminology comes from quantum mechanics and quantum ﬁeld theory, in which representation theory plays a key role. 5. The key example: su(2) 46 where as the vector space is ﬁnite dimensional, the process of lowering must stop at some state |qi where q is the smallest eigenvalue. To determine q, we need to consider the action Γ+. We already know that Γ+|mi has eigenvalue m + 1 and so is proportional to |m + 1i:

+ Γ |mi = Nm|m + 1i .

We can compute the normalisation factor Nm using the commutation relations

+ Nm|m + 1i = Γ |mi = Γ+Γ−|m + 1i (5.7) = (Γ−Γ+ + Γ3)|m + 1i

= (Nm+1 + m + 1)|m + 1i.

This gives a recurrence relation Nm = Nm+1 + m. By assumption, we cannot raise the highest + weight state so Γ |ji = 0 and hence Nj = 0. We can then solve the relation to give 1 1 N = j(j + 1) − m(m + 1) m 2 2 The state |qi satisﬁes the property that Γ−|qi = 0. Hence

+ − − + 3 Γ Γ |qi = (Γ Γ + Γ )|qi = (Nq + q)|qi = 0 .

This has two solutions: q = j + 1 and q = −j. The former contradicts the starting assumption of j being the largest eigenvalue. Hence the lowest weight state is |−ji:

|ji , |j − 1i , |j − 2i , · · · |−ji , (5.8)

The representation Γ uniquely speciﬁed by the positive half-integer j. This being so, we will refer to this as the spin-j representation. There are 2j + 1 eigenvectors of Γ3 in total. So we learn that the spin j representation is 2j + 1–dimensional. It turns out this construction gives rise to irreducible representations of su(2), a fact that we will not prove, but will use extensively. 0 It is straightforward to check that the states above are orthogonal hm|m i = δm,m0 . However, they are not normalised to unity in general kmk2 = hm|mi= 6 1. To see this, note that (Γ+)† = Γ− − meaning hm|Γ = Nmhm + 1| and so

kj − 1k2 = hj|Γ+Γ−|ji = hj|(Γ−Γ+ + Γ3)|ji = j kjk2 . √ We can rescale |ji so that kjk = 1. The above equation implies kj − 1k = j, and so |j − 1i is not a normal vector except in the special case of j = 1. For states with eigenvalue m 6= ±j the same calculation gives another recurrence relation

2 + − − + 3 2 2 km − 1k = hm|Γ Γ |mi = hm|(Γ Γ + Γ )|mi = Nm km + 1k + m kmk . (5.9) 5. The key example: su(2) 47

We can deﬁne an orthonormal basis by 1 |gmi = |mi , kmk + − 3 With respect to this basis, the Γj , Γj , Γj have entries given by:

0 3 h^j, m |Γj |^j, mi = mδm,m0 , r 0 + 1 h^j, m |Γ |^j, mi = δ 0 (j + m + 1)(j − m), j m ,m+1 2 (5.10) r 0 − 1 h^j, m |Γ |^j, mi = δ 0 (j + m)(j − m + 1). j m ,m−1 2 Using Γ1 = √1 (Γ+ + Γ−) and Γ2 = √−i (Γ+ − Γ−) we ﬁnd j 2 j j j 2 j j

0 1 1 p 1 p h^j, m |Γ |^j, mi = δ 0 (j + m + 1)(j − m) + δ 0 (j + m)(j − m + 1), j 2 m ,m+1 2 m ,m−1 0 2 −i p i p h^j, m |Γ |^j, mi = δ 0 (j + m + 1)(j − m) + δ 0 (j + m)(j − m + 1), (5.11) j 2 m ,m+1 2 m ,m−1 0 3 h^j, m |Γj |^j, mi = mδm,m0 . In the above we have labelled each of the basis vectors by

|^j, mi, (5.12) for m = j, . . . , −j. We will do this when we are dealing with more than one su(2) representation. From now on we will assume we work with the orthonormal basis and drop the tilde.

5.3. Examples

The key examples are j = 1/2 and j = 1. The spin-1/2 representation is the smallest non-trivial representation and is two-dimensional. The basis vectors are

|e1i = |1/2, 1/2i, |e2i = |1/2, −1/2i

a corresponding to m = 1/2, −1/2. The (i, j)th component of the matrices [Γ1/2]ij are given by the inner product (3.1) and the formulae (5.11): ! ! ! 1 1 0 1 2 1 0 −i 3 1 1 0 Γ1/2 = , Γ1/2 = , Γ1/2 = . (5.13) 2 1 0 2 i 0 2 0 −1 This is exactly the deﬁning representation (5.2), which makes sense as its also two-dimensional and so the two should be the same up to a similarity transformation. The next non-trivial representation is the spin-1 representation which is 3-dimensional. The generators are       0 1 0 0 −i 0 1 0 0 1 1   2 1   3 1   Γ = √ 1 0 1 , Γ = √ i 0 −i , Γ = √ 0 0 0  . (5.14) 1 2   1 2   1 2   0 1 0 0 i 0 0 0 −1 5. The key example: su(2) 48

It is straightforward to check that these matrices satisfy the su(2) algebra (5.4), and doing so is left as an exercise.

Remarks:

• The construction we have just described nicely generalises to any compact Lie algebra, as we will shortly see.

• The eigenvalues of Γ3 are called the weights, and the construction is the highest weight construction because it starts with the highest weight vector. We will use the highest weight construction extensively in the remaining chapters.

• The highest weight construction for a general Lie algebra in the next chapters utilises the spin-1/2, spin-1 and spin-3/2 representations. 5. The key example: su(2) 49

5.4. Exercises

1. In this exercise, we show that any U ∈ SU(2) can be written as (5.1).

(a) Suppose the matrix U is written as ! a b U = . c d

By studying U −1 = U † and det U = 1 show that

d =a, ¯ c = −¯b.

2 2 i ψ (b) We have det U = 1. Show that this implies |a| + |b| = 1, and that a = e 2 cos θ/2 i φ and b = e 2 sin θ/2 parameterises all possible solutions, when φ, ψ ∈ [0, 4π) and θ ∈ [0, 2π). Hence, conclude that U may be written in the form (5.1). (c) Show that θ = ψ = 0 corresponds to the identity element. (d) Show that the generators of su(2) in (5.2) satisfy (5.3).

2. The generators of the group SU(2) in its deﬁning representation are given by (5.2).

1 2 (a) Using exp(M) = I + M + 2! M + ··· for a square matrix M, show that the group 1 3 element g3(θ) = exp(iθ 2 T ) can be written as ! eiθ/2 0 g3(θ) = 0 e−iθ/2

This deﬁnes a 1-dimensional subgroup, U(1) ⊂ SU(2). (b) Construct the corresponding Lie subalgebra u(1), by constructing its generator(s) and commutation relations. Is it an invariant subalgebra?

1 1 1 2 (c) Construct analogous expressions for g1(φ) = exp(iφ 2 T ) and g2(ψ) = exp(iψ 2 T ).

(d) Does the subset {g2(φ) | φ ∈ [0, 4π)} ⊂ SU(2) form a group? Why?

3. (a) Show that (Γ+)† = Γ−. (b) Show the raising and lowering operators Γ± and Γ3 satisfy the commutation relations (5.5). (c) By using (5.10) show that the recursion relation (5.9) is satisﬁed. Hint: (5.10) implies that r (j − m + 1)(j + m) Γ−|gmi = |m^− 1i . 2 5. The key example: su(2) 50

Compare with Γ−|mi = |m − 1i to find a relation between kmk , km − 1k. An analogous calculation determines a relation between kmk , km + 1k. Use both of these to show that the recursion relation is satisfied. (d) Can you find a direct method to solve the recursion relation (5.9)?

4. (a) Show that the formulae (5.11) for j = 1 give the matrices in (5.14). (b) Show that the matrices (5.14) satisfy the su(2) algebra (5.4). (c) Construct the generators T 1,T 2,T 3 in the 4-dimensional spin-3/2 representation.

5. There are two abelian subgroups of U(2). One is the U(1) subgroup described in the previous question. Find the other U(1)-subgroup.

6. Show that the adjoint representation of su(2) is given by using (4.18). Consider the adjoint representation of su(2), given in (4.19), and the spin-1 representation constructed in (5.14). Compare the two: why are the two diﬀerent, and how are they related to each other?

7. Construct the spin-3/2 representation of su(2), and check it satisﬁes the commutation relations. 6. Roots and weights 51

6. Roots and weights

We want generalise the highest weight construction to a simple compact Lie algebra g. Our strategy is

1. Find the largest possible set of commuting generators Hi, i = 1, . . . , m. These are analogous to the generators Γ3. We will use their eigenvalues to label the basis vectors for the representation.

2. The remaining generators will be raising and lowering operators, just like Γ± for su(2). These operators are used to deﬁne a set of basis vectors for the representation, each of

which is an eigenvector of all the Hi.

3. Each Hi with a pair of raising/lowering operators corresponds to a subalgebra su(2) ⊂ g. The technology developed in the previous chapter will allow us to build up any representation of g.

6.1. Weights

Suppose we have a representation Γ of the Lie algebra g, with the dimension of Γ given by M. The largest possible set of commuting hermitian generators is the Cartan subalgebra.

Deﬁnition 6.1. For a representation Γ let Hi for i = 1, . . . , m correspond to elements of the

Cartan subalgebra. The Hi are known as the Cartan generators and satisfy

† Hi = Hi , [Hi,Hj] = 0. (6.1)

The integer m is the rank of the algebra.

We will choose the Hi so that they satisfy this relation:

Tr HiHj = kΓδij, i, j = 1, . . . , m. (6.2)

The algebra su(2) is rank 1, as there is only one element of the Cartan subalgebra, which in our choice of basis of the last chapter is given by X3. The algebra su(3) is rank 2. We will explore this in more detail shortly, but for now just know that in the deﬁning representation     1 0 0 1 0 0 1   1   H1 = 0 −1 0 ,H2 = √ 0 1 0  . (6.3) 2   2 3   0 0 0 0 0 −2

The Cartan generators Hi will allow us to construct a label for the basis vectors of the representation Γ. As all the Hi commute with each other, we can ﬁnd a basis |e1i,..., |eM i so 6. Roots and weights 52

that all the Hi are simultaneously diagonal. In this basis, each of the |epi are eigenvectors of the matrices H1,H2,...,Hm:

p Hi|epi = µi |epi, p = 1, . . . M, i = 1, . . . m, no sum on p , (6.4)

p p p p where µi is the eigenvalue. For each p, we deﬁne an m-dimensional vector µ = (µ1, . . . , µm) called the weight vector. The weight vectors form a labelling system for the basis vectors:

p |epi ⇔ |µ i

The weight vectors live in an m-dimensional vector space which is called the weight space. It is easy to get confused at this point as we have two separate vector space. There is the representation Γ, which is an M-dimensional vector space. There is also an m-dimensional weight space.8 Each of the generators of the representation corresponds uniquely to a particular weight vector. Let us come back su(3). In this case, the deﬁning representation has three basis vectors       1 0 0       |e1i = 0 , |e2i = 1 , |e3i = 0 .       0 0 1

Each one is an eigenvector of H1,H2 in (6.3), and the corresponding weight vectors are 1 1 1 2 1 1 3 1 |e1i ↔ µ = , √ , |e2i ↔ µ = − , √ , |e1i ↔ µ = 0, − √ , (6.5) 2 2 3 2 2 3 3

Hence, we can label the three basis vectors through their weight vectors:

|µ1i, |µ2i, |µ3i.

6.2. The adjoint representation

The adjoint representation, as deﬁned by (4.18), is rather important. As [T a]bc = −if abc, the rows and columns are the same index as the labels for the generators T a. In this case, we can also label the basis vectors by the generators themselves! That is,

|T ai, a = 1,...,N.

Linear combinations of vectors corresponds to linear combinations of generators:

α|T ai + β|T bi = |αT a + βT bi. (6.6)

8The dimension of the weight space is representation independent. 6. Roots and weights 53

We deﬁne an inner product on basis vectors |T ai: 1 hT a|T bi = Tr (T a †T b), (6.7) λ where λ = kΓ for the adjoint representation. The vectors |T ai are linearly independent and so a standard result of linear algebra states the identity matrix can be written as

N X I = |T cihT c|. c=1 Using this, (6.6) and (4.18), we can calculate the action of a generator on any basis vector:

T a|T bi = |T cihT c|T a|T bi = |T ci[T a]cb. (6.8)

From this we conclude that T a acts via the adjoint representation T a. Furthermore, as [T a]cb = if abc, we get

T a|T bi = if abc|T ci = |if abcT ci (6.9) = |[T a,T b]i.

This is a nice result: the action of a generator in the adjoint representation is given by commutation. Returning to our example of su(2), there are three basis vectors, which we label as |T 1i, |T 2i, |T 3i. The action of say T 1 is

T 1|T 1i = |[T 1,T 1]i = 0,T 1|T 2i = |[T 1,T 2]i = i|T 3i,T 1|T 3i = |[T 1,T 3]i = −i|T 2i.

Result of linear algebra: if M any matrix and |eii is the ith basis vector, then with respect to this basis M|eii describes the ith column of M:

M|eii = Mji|eji,Mji the (i, j) component of M.

Hence, the three equations above determine the three columns of T 1 with respect to the basis labelled by |T ai. It is an exercise to repeat this for T 2,T 3, and check with the answers below.       0 0 0 0 0 i 0 −i 0 1   2   3   [T ] = 0 0 −i , [T ] =  0 0 0 , [T ] = i 0 0 .       0 i 0 −i 0 0 0 0 0

It is reassuring that these answers are consistent with (4.19). 6. Roots and weights 54

6.3. Roots

The adjoint representation has basis vectors which can be labelled by either the generators themselves |T ai or by their weights. The weights of the adjoint representation are called roots, and carry with them intrinsic information about g itself and not just its representation.

Applying the deﬁnition of a weight, we see that the roots are the eigenvalues of the Hi acting on the basis vectors |T ai:

a a a a Hi|T i = µi |T i = |[Hi,T ]i, no sum. (6.10)

In the second equality, the “no sum” indicates there is no summation over the repeated index a. As [Hi,Hj] = 0 the basis vectors |Hii have zero weight vectors:

Hi|Hji = |[Hi,Hj]i = 0.

Because of (6.2) and (6.7) the vectors |Hii are orthonormal:

hHi|Hji = δij. (6.11)

The remaining basis vectors, i.e. those not corresponding to Cartan generators, have non-zero weight vectors, α = (α1, . . . , αm), with

Hi|Eαi = αi|Eαi, αi ∈ R.

9 The eigenvalues are real as the matrices Hi are hermitian (exercise). This means the matrices

Eα satisfy

[Hi,Eα] = αiEα. (6.12) Just as Γ±, the raising and lowering operators of su(2), were not hermitian satisfying Γ+ † = − Γ , the Eα are also not hermitian. To see this, take the hermitian conjugate of (6.12):

† † [Hi,Eα] = −αiEα. (6.13)

Thus, we identify † Eα = E−α. (6.14) If we take two distinct root vectors α 6= β then the corresponding vectors are orthonormal: m ∼ Y hEβ|Eαi = δαβ = δαiβi . (6.15) i=1

The fact they are orthogonal comes from them having distinct eigenvalues under Hi while we get orthonormality by an appropriate normalisation of the eigenvectors. Proving this is left as an exercise.

9In the adjoint representation all the generators can be speciﬁed uniquely by a weight vector α without need for any additional labels. 6. Roots and weights 55

6.4. Raising and lowering operators

As you might have guessed, the generators E±α are going to be used as raising and lowering operators, just as Γ± were for su(2). Recall the situation for su(2). The algebra is rank 1 with a single Cartan generator Γ3. All representations are speciﬁed by a single number, the highest weight j, with corresponding basis vector |ji. All other basis vectors have weights j, . . . , −j, with Γ± changing m → m ± 1 as shown by (5.6). Now for the Lie algebra g with representation Γ, consider the basis vector |µi, where µ =

(µ1, . . . , µm) is its weight vector. Then,

HiE±α|µi = [Hi,E±α]|µi + E±αHi|µi (6.16) = (µi ± αi)E±α|µi.

If E±α|µi is non-zero, then it has weight vector µ ± α, given by adding or subtracting the root vector α to the weight vector. 10 The equation (6.16) is true for all representations and in particular for the adjoint representation. Consider the vector Eα|E−αi, which has weight α − α = 0 and so is an element of the

Cartan subalgebra. Hence, it is a linear combination of the Hi:

Eα|E−αi = βi|Hii = |βiHii = |[Eα,E−α]i.

11 Using this equation, (6.11) and the cyclicity property of the trace we can compute βi:

βi = hHi|Eα|E−αi 1 = Tr (H [E ,E ]) λ i α −α 1 = Tr (E [H ,E ]) (6.17) λ −α i α 1 = α Tr (E E ) λ i −α α

= αi.

We conclude that

[Eα,E−α] = αiHi. (6.18)

This should remind you of the su(2) commutation relation [Γ+, Γ−] = Γ3. We will now exploit this to construct any representation of a compact simple Lie algebra.

10 + E±α|µi = 0 is analogous to the situation in su(2) with, say, the highest weight vector |ji, where Γ |ji = 0. 11 Recall this says for a series of square matrices Tr A1 ...An = Tr A2A3 ...AnA1 = ··· . 6. Roots and weights 56

6.5. Heaps and heaps of su(2)s

The previous section tells us that for each pair of root vectors ±α we can deﬁne an su(2) sub-algebra of g with generators

1 1 E =∼ E ,E =∼ α · H. (6.19) ± |α| ±α 3 |α|2

These satisfy an algebra

[E3,E±] = ±E±, [E+,E−] = E3, (6.20) which is the algebra of su(2) given in (5.5). Checking this is an exercise. This means each state |µi in the representation Γ is also a state of a representation of the su(2) sub-algebra (6.19) deﬁned by the root α. We have already classiﬁed representations of su(2) so we can make use of the technology developed in the previous chapter.

Two important facts: if α is a root then there is a unique generator Eα; if α is a root, then kα is a root if and only if k = ±1. Proving these are left as exercises.

For any weight µ of a representation Γ the E3 eigenvalue is α · µ E |µi = |µi , α2 := |α|2 . (6.21) 3 α2

If we label the eigenvalue m, then this implies µ can be viewed as a state |mi in the representation of the su(2) algebra in (6.19). Hence, m is half integral and so

2α · µ ∈ . α2 Z

Let us suppose this highest weight state of this representation is j. Then there is a non-negative integer p such that (E+)p|µi is the highest weight vector, i.e.

(E+)p+1|µi = 0. and this spin of (E+)p|µi is j. We also know that E+ raises the eigenvalue by α we have

α · µ + p = j. (6.22) α2

There is also a non-negative q such that

(E−)q|µi= 6 0, (E−)q+1|µi = 0, so that the weight µ − qα is the lowest weight vector of the spin-j representation. The E3 eigenvalue of this vector is α · µ − q = −j. (6.23) α2 6. Roots and weights 57

Combining (6.22)-(6.23) gives a master equation for weights: 2α · µ = −(p − q). (6.24) α2 This equation, together with (6.22), (6.23) may not look like much, but it turns out they have a lot of information hidden within them waiting to be exploited. The ﬁrst bit of exploitation is to apply (6.24) to a pair of root vectors α, β twice. That is,

ﬁrst start with the su(2) algebra (6.19) deﬁned by E±α on the vector |βi. Then (6.24) gives α · β 1 = − (p − q). (6.25) α2 2

Now consider the su(2) algebra (6.19) deﬁned by E±β and the vector |αi. Then (6.24) gives

β · α 1 = − (p0 − q0). (6.26) β2 2 Multiplying these two equations together gives a formula for the angle between the root vectors α, β: (α · β)2 (p − q)(p0 − q0) cos2 θ = = . (6.27) αβ α2β2 4 0 0 As (p − q)(p − q ) ∈ Z there are four interesting possibilities:

0 0 1.( p − q)(p − q ) = 0 and θαβ = π/2;

0 0 2.( p − q)(p − q ) = 1 and θαβ = π/3, 2π/3;

0 0 3.( p − q)(p − q ) = 2 and θαβ = π/4, 3π/4;

0 0 4.( p − q)(p − q ) = 3 and θαβ = π/6, 5π/6.

0 0 The possibility (p − q)(p − q ) = 4 corresponds θαβ = 0, π and neither are interesting: θαβ = 0 corresponds to α = β and θαβ = π corresponds to β = −α, both roots are in the same su(2) subgroup. So unlike one’s initial expectation, the root vectors cannot just orient themselves in any old way: the angles between them can only take four possible choices!

6.6. Example: su(3)

Let us illustrate some of these ideas using su(3). This the algebra of the group SU(3), and is generated by eight 3 × 3 traceless hermitian matrices. In fact, this is true in general: su(n) is generated by traceless hermitian matrices. To see this, consider U(α) ∈ SU(n) given by

U(α) = exp(iαaXa). 6. Roots and weights 58

Then, U(α)† = exp(−iαa(Xa)†) = U(α)−1 iff (Xa)† = Xa. So Xa are hermitian. Traceless follows from det U(α). We can compute det U(α) in any basis. We choose one in which αaXa is diagonal   µ1 0 0 ···    0 µ2 0 ··· MαaXaM −1 = D =   ,  ..   0 0 .    . . . . µn Then, n Y a a det(U(α)) = det eiD = eiµj = eiTr D = eiTr α X . j=1 Hence, det U(α) = 1 for all α if and only Tr Xa = 0. For su(3), the traceless hermitian matrices we make use of are the Gell-Mann matrices which are defined as:     0 1 0 0 −i 0     λ1 = 1 0 0 , λ2 = i 0 0 ,     0 0 0 0 0 0     1 0 0 0 0 1     λ3 = 0 −1 0 , λ4 = 0 0 0 ,     0 0 0 1 0 0     (6.28) 0 0 −i 0 0 0     λ5 = 0 0 0  , λ6 = 0 0 1 ,     i 0 0 0 1 0     0 0 0 1 0 0   1   λ7 = 0 0 −i , λ8 = √ 0 1 0  .   3   0 i 0 0 0 −2 The generators of su(3) are conventionally defined as 1 T a = λa, (6.29) 2 and are defining representation of su(3). The generators satisfy 1 Tr T aT b = δab. (6.30) 2 Notice, the first three generators generate an su(2) sub-algebra. Unlike su(2), there are two matrices that are diagonal T 3,T 8 and these form a basis for the Cartan sub-algebra of su(3):

 1    2 0 0 √ 1 0 0 1 3   2 8 3   H = T = 0 − 1 0 ,H = T = 0 1 0  . (6.31)  2  6   0 0 0 0 0 −2 6. Roots and weights 59

Hence, su(3) is rank 2. We want to calculate the weights of the deﬁning representation. Recall the weights are labels 1 m for the basis vectors given by the eigenvalues Hi|µi = µi|µi with |µi = (µ , . . . , µ ) where m is the rank of the algebra. As su(3) is rank 2, the weights are two-dimensional vectors. As the Hi are already diagonal, the eigenvectors of the Hi are simply the canonical vectors shown below together with the weight vectors:

  1 √ (1)   (1) |µ i = 0, µ = (1/2, 3/6),   0   0 √ (2)   (2) |µ i = 1 µ = (−1/2, 3/6), (6.32)   0   0 √ (3)   (3) |µ i = 0 µ = (0, −1/ 3).   1

We can plot the weight vectors on the plane as showin in Figure 6.1.

H 2

H 1

Figure 6.1: The weight vectors of the deﬁning representation of su(3). 6. Roots and weights 60

To recap:

• An n-dimensional representation Γ acts on a a n-dimensional vector space, with generators being n × n matrices. e.g. the deﬁning representation for su(3) is three-dimensional.

• the generators Hi help to deﬁne a separate vector space, the weight space. The dimension of this vector space is the rank of the algebra, the number of Cartan generators, and is the same for all representations. e.g. su(3) has a weight space that is two-dimensional.

• Each representation is deﬁned by a set of weights. The weights are labels for the basis vectors of the representation. So the number of weights is given by the dimension of the representation. The deﬁning representation of su(3) is three-dimensional and has three

weights µ1, µ2, µ3 plotted in Figure (6.1).

• The roots are the weights of the adjoint representation. Their role is to take us between

the weight vectors as E±α|µi = |µ ± αi with µ → µ ± α for a root α.

For su(3), the generators corresponding to the roots are

1 1 2 E±1,0 = √ (T ± iT ), 2 1 E √ = √ (T 4 ± iT 5), (6.33) ±1/2,0± 3/2 2 1 E √ = √ (T 6 ± iT 7). ∓1/2,0± 3/2 2 The roots form a regular hexagon as plotted in Figure 6.2 with the two elements of the Cartan subalgebra at the origin (as they have zero weight). 6. Roots and weights 61

H 2

H 1

Figure 6.2: The root vectors of su(3).

6.7. Exercises

1. (a) Show that a hermitian matrix H has real eigenvalues. Hint: one way to show this is to consider an eigenvector |vi and the reality properties of the combination hv|H|vi. (b) Using (a) show that if α and β are distinct root vectors then

hEβ|Eαi = 0.

Hint: use αi 6= βi for some i and consider hEβ|Hi|Eαi.

± 2. Given the deﬁnition of E ,E3 show (6.20) holds.

3. We will show that if α is a root then there is only one corresponding generator Eα. We will do this by contradiction.

0 Suppose that there two generators Eα and Eα. Without loss of generality we may choose them to be orthogonal (Gram-Schmidt):

0 1 † 1 hE |E i = Tr E E = Tr E E 0 = 0. α α λ α α λ −α α

(a) Show that E−|Eαi is a linear combination of the Hi. 0 (b) Show that hHi|E−|Eαi = 0 for all i, concluding that the coeﬃcients in (a) all vanish and so 0 E−|Eαi = 0.

(d) By considering the eigenvalue of E3, and the fact the lowest weight vector of an su(2)

representation cannot have a positive E3 eigenvalue, show that equations (b) and (c) lead to a contradiction.

4. Show that [Eα,Eβ] is proportional to Eα+β. What happens if α + β is not a root?

5. From the previous question, for two roots α, β such that α + β is a root, the generators satisfy

[Eα,Eβ] = NEα+β.

for some non-zero N. Calculate

[Eα,E−α−β].

6. Calculate f147 and f458 in su(3).

7. (a) Show that T 1,T 2,T 3 in (6.28)-(6.29) generate an su(2) subalgebra of su(3). (b) Every irreducible representation of su(3) must also form a representation of su(2) ⊂ su(3). That is, picking an su(2) ⊂ su(3) subalgebra, the corresponding generators of the su(3) representation form a representation of this su(2) subalgebra. However, it may not be irreducible. Do the T 1,T 2,T 3 form an irreducible representation of su(2)? If not, can you decompose the representation into a direct sum of irreducible representations. 7. Simple roots 63

7. Simple roots

The roots take us between weights through the identification E±α|µi ∼ |µ ± αi, and this is analogous to the relation in su(2) which amounts to Γ±|mi ∼ |m ± 1i. However, because α is a vector, we need to establish a notion of a positive root so that we can meaningfully talk about a raising operator. That will allow us to establish a notion of ‘highest weight’ and ‘lowest’ weight. In doing so, we will find a minimal set of roots out of which we can build the entire algebra; this minimal set of roots are known as the simple roots. In this chapter will often refer to su(3) as a guiding example, and the weights of the defining representation are drawn in Figure 6.1

7.1. Positive weights

For the remainder, we will ﬁx a basis for the Cartan sub-algebra. All statements we make will turn out to be independent of the choice of basis making the choice redundant.

Definition 7.1. A weight µ = (µ1, . . . , µm) is positive if its first non-zero component is positive. It is negative if its first non-zero component is negative. This definition is not unique, but has the property we desire: each root is either positive or negative (and not both!). If µ is positive then −µ is negative. √ For su(3) only one is positive. As shown in Figure 7.1 the weight |µ1i = (1/2, 3/6) is √ √ positive, while |µ2i = (−1/2, 3/6) and |µ3i = (0, −1/ 3) are both negative.

Deﬁnition 7.2. Weights are ordered according to the rule:

µ > ν if µ − ν is positive. (7.1)

The zero weight vectors belong to the Cartan sub-algebra and are neither positive, negative consistent with their action not changing the weight of a vector |µi. Recall that the roots are the weights of the adjoint representation. Therefore, roots themselves can be positive or negative or zero. Furthermore, each root corresponds to a generator of the algebra:

|αi ←→ Eα.

Pick a positive root α. Then, in the notation of (6.19), +α corresponds to a raising operator

E+, while −α is a lowering operator E−.

Deﬁnition 7.3. The highest weight µ of any representation has the property that we cannot raise it, and so all generators corresponding to positive roots must annihilate the corresponding vector |µi. The roots of su(3) are illustrated in Figure 7.2. Those to the right of the y-axis are positive, those to the left are negative. 7. Simple roots 64

H 2

is positive is negative H 1

is negative

√ Figure 7.1: The weights of the deﬁning representation of su(3). |µ1i = (1/2, 3/6) is positive, √ √ while |µ2i = (−1/2, 3/6) and |µ3i = (0, −1/ 3) are both negative.

7.2. Simple roots

There are often more positive roots than the rank of the algebra, as is the case in Figure 7.2. Therefore, some of the positive roots are redundant: they can be obtained through a linear combination of a minimal set of roots known as the simple roots.

Definition 7.4. The simple roots are positive roots which cannot be written as a linear combination of the other positive roots with all coefficients being positive. Once we know the simple roots, we will be able to construct all of the remaining roots of a Lie algebra. We first derive a series of facts.

1. If α 6= β are simple roots then α − β is not a root.

One of the pair of simple roots must be larger, let us take it to be α. Then α − β is a positive root. But this is a contradiction on α being simple as it is a sum of two positive roots: α − β and β.

2. If α−β is not a root then neither is β −α. Hence, if we consider the adjoint representation 7. Simple roots 65

H 2

negative positive

negative positive H 1

negative positive

Figure 7.2: The roots of su(3) showing if they are positive or negative.

then

E−α|Eβi = 0,E−β|Eαi = 0.

If we consider the two su(2) algebras deﬁned in (6.19) through positive roots α and β, then

this equation implies |Eβi and |Eαi are lowest weight vectors. Hence, applying the master equation for each of these su(2) algebras, c.f. (6.25)-(6.26), the integers q, q0 vanish: α · β 1 β · α 1 = − p, = − p0. (7.2) α2 2 β2 2 As α and β are vectors in a m-dimensional vector space, the angle between the vectors is 0 given by the inner product α · β = |α||β| cos θαβ. Using (7.2), we can relate p, p to the angle between the roots α, β: √ pp0 β2 p cos θ = − , = . (7.3) αβ 2 α2 p0

3. The angle between any pair of simple roots satisﬁes π ≤ θ < π. 2 The ﬁrst inequality follows from cos θ being negative in (7.3). The second inequality follows as all the simple roots are positive. 7. Simple roots 66

4. The simple roots {αi} are linearly independent. To see this, suppose a linear combination of simple roots vanishes: m X γ = λiαi = 0, λi ∈ R. (7.4) i=1

As all the simple roots are positive, if all the coeﬃcients, λi have the same sign then γ = 0

only if λi = 0 for all α. If there are some coeﬃcients with positive sign and some with negative sign then we can write X X γ = µ − ν, µ = λiαi, ν = − λjαj,

λi>0 λj <0 so that both µ, ν are both positive. Then, the norm of the vector γ cannot vanish:

kγk2 = γ · γ = (µ − ν)2 = µ2 + ν2 − 2(µ · ν) ≥ µ2 + ν2 > 0.

The penultimate inequality follows from (α · β) ≤ 0 for any pair of simple roots, see equation (7.3) above.

Thus, γ = 0 if and only if λα = 0 and the simple roots are linearly independent.

5. At this point, we could have the number of simple roots being less than m. But in fact, the simple roots span the m-dimensional vector space. To see this, note that if it were not true then there would be some vector ξ orthogonal to all the simple roots. As all the roots are linear combinations of the simple roots, it implies ξ is orthogonal to all the roots. This implies

[ξiHi,Eγ] = (ξ · γ)Eγ = 0, for all roots γ.

Hence, ξiHi commutes with all generators of the algebra. But this is a contradiction of the assumption of the algebra being simple.

Thus the number of simple roots equals the rank of the algebra m.

With these facts we are now in a position to construct all the roots of the algebra from the simple roots.

Theorem 7.1. We can determine all the roots of a Lie algebra g starting from just simple roots.

Proof. Here we will denote the simple roots by α, β. The simple roots span the m-dimensional root space. However, in principle we do not know which linear combinations of the simple roots are actually roots. This is what we aim to determine. All the positive roots have the form

m m X X γk = kiαi, ki > 0 , k = kj, (7.5) i=1 j=1 7. Simple roots 67

where we sum over all the simple roots α and ki are non-negative integers. We call the integer k the level of the root. The burden of our proof is to determine which of the combinations in (7.5) are not roots. We do this by induction on the level k. For k = 1, equation (7.5) simply lists the simple roots and the statement is trivially true. In the adjoint representation, this means we can list states |Eβi for each of the simple roots β.

All roots at level k = 2 are of the form β + αi for some pair of simple roots β and αi. In terms of the adjoint representation, this means we act with the matrix Eαi on one of the states

|Eβi. If Eαi |Eβi = 0 then β + αi is not a root. Otherwise it is a root. The induction hypothesis extends this to a level k = L with L > 1 a positive integer. That is, we assume we know all the roots at level k ≤ L, and in particular how the state |Eγk i, with level of γ being k, is obtained out of acting with the matrices Eβ on some starting vector |Eαi i. P In other words, we know the coeﬃcients ki in the expression γk = i kiαi.

Let γL be a positive root at level L and consider the set of vectors n o

Eα|γLi α is a simple root ,

These all have level k = L + 1. Using (6.25), the state Eα|γLi has 2α · γ p = q − L . (7.6) α2 Recall that p, q tell us the maximum number of raising/lowering operators that can be applied p+1 q+1 until (Eα) |γLi = 0 and (E−α) |γLi = 0. As a lowering operator E−α always reduces the level k, by the inductive hypothesis we will always know q. Thus, we can always determine p using (7.6). If p > 0 then Eα|γLi 6= 0 and so γL + α is a root with Eα|γLi ∝ |γL + αi. If p ≤ 0 12 the Eα|γLi = 0 and so γL + α is not a root.

Thus we can always ﬁnd all the roots γL+1 by acting on all the γL with simple roots, and know level-by-level, how they are built up out of raising and lowering operators.

Let us demonstrate the inductive procedure by returning to the case of level k = 2. If αi 6= β, then β − αi is not a root by our facts above, and so in the the master formula we have q = 0 and can solve for p: 2αi · β p = − 2 . (αi)

12 The only way we could fail in this procedure is if there is a positive root γL+1 which is not the sum of a root γL and some simple root. This is impossible. Such a vector always has E−α|γL+1i = 0 (if it were non-zero we could apply Eα and get γL+1 back as a sum of γL and a simple root). So |γL+1i is a lowest weight vector of every SU(2) representation associated with all the simple roots α. This requires all their E3 eigenvalues to satisfy

α · γL+1 ≤ 0 for all α. In turn, this implies

2 X γL+1 = kiα · γL+1 ≤ 0, i a contradiction. 7. Simple roots 68

Thus if αi · β = 0 then p = 0 and αi + β is not a root. Otherwise p > 0 and αi + β is a root.

If αi = β then p = 0 as 2αi is not a root, while the formula gives q = 2, reﬂecting the vectors

|E−αi i, |0i, |Eαi i form a spin-1 representation under the su(2) generated by Eαi . Its appropriate at this point to revisit the example of su(3) to illustrate how this all works. Recall the positive roots for su(3) are the positive weights for the adjoint representation. These are to the right of the y-axis, as shown in 7.2. su(3) has two simple roots: √ √ α1 = (1/2, 3/2), α2 = (1/2, − 3/2).

We already know that these can be used to compute the remaining positive root (1, 0) = α1 +α2 as shown in Figure 7.3. Note, the coeﬃcients −α1 −α2 −α1 − α2 in α1 + α2 are both positive, as required by (7.4), and this root is level k = 2. We want to check this using the inductive reasoning given above and (7.6).

H 2

H 1

Figure 7.3: The simple roots of su(3) are α1 and α2. These can be used to build all the positive roots, such as (1, 0) = α1 + α2.

Firstly, 1 α2 = α2 = 1, α · α = − , 1 2 1 2 2 Secondly, we apply it to (7.6) 2α1 · α2 2α1 · α2 2 = 2 = −1, α1 α2 7. Simple roots 69

and so p = 1 for both Eα1 |α2i and Eα2 |α1i telling us that α1 +α2 is indeed a root and Eα1+α2 ∝

Eα1 |α2i ∝ Eα2 |α1i is the corresponding generator.

Is 2α1 + α2 a root? Let’s ﬁnd out. The generator would be Eα1 |α1 + α2i, and so applying (7.6): 2α1 · (α1 + α2) 2(1 − 1/2) p = q − 2 = 1 − = 0, α1 1 and so 2α1 + α2 is not a root, consistent with what we already knew from Figure 7.3! Similarly, we can also check 2α1: 2α1 · α1 p = 1 − 2 = −1, α1 and so 2α1 is not a root.

7.3. Constructing the algebra

These tools are more powerful than just checking which positive combinations of simple roots are roots. We can actually construct the entire algebra! We need to go back to our derivation of the master equation for a weight vector µ of any representation. This derives from a formula

α · µ α · µ + p = j, − q = −j, α2 α2

−1 −2 which follows because under the su(2) subalgebra generated by E± = E±α|α| , E3 = α·H|α| the weight vector µ transforms in the spin j representation. p is the max number of raising operators E+ we can apply to |µi, while q is the max number of lowering operators E− we can apply. Adding the equations together p + q = 2j.

Let’s demonstrate the procedure for su(3); the general case will then follow. The root diagram is illustrated in Figure 7.4, and recall

2 2 1 α1 · α2 α1 · α2 1 α1 = α2 = 1, α1 · α2 = − , 2 = 2 = − . 2 α1 α2 2

We have already determined that p = 1, q = 0 for Eα1 acting on |α2i and vice versa. Let’s begin by computing Eα1+α2 using the su(2) subalgebra corresponding to Eα1 . This has p + q = 1, and so it is a spin-1/2 representation. The state |Eα2 i is properly normalised by assumption, and 1 corresponds to a lowest weight vector |1/2, −1/2i in the spin- 2 representation. To ﬁnd |Eα1+α2 i we look at the Eα1 in two ways: ﬁrst through its action on the root:

+ 1 E |Eα2 i = Eα1 |Eα2 i = |[Eα1 ,Eα2 ]i = N|Eα1+α2 i, (7.7) α1 7. Simple roots 70

H 2

H 1

Figure 7.4: The roots of su(3) in terms of the simple roots α1, α2.

where N is a normalisation to be ﬁxed. Secondly, we think of E+ as a raising operators on the state |1/2, −1/2i in (5.12) so that we can use the normalisation of (5.10) to deduce:

iη + e E |Eα i = E+|1/2, −1/2i = √ |E1/2,1/2i. 2 2 The phase factor comes from the fact (5.10) tells us only about the norm. Identifying the two equations, the state eiη √1 |E i must be |E i. Hence, N = √1 eiη. Choose η = 0, we 2 1/2,1/2 α1+α2 2 can express the positive root α1 + α2 as a commutator of the simple roots: √ √ |Eα1+α2 i = 2|[Eα1 ,Eα2 ]i =⇒ Eα1+α2 = 2[Eα1 ,Eα2 ]. (7.8)

We can now compute any su(3) commutator using the Jacobi identity. For example, √ [E−α1 ,Eα1+α2 ] = 2[E−α1 , [Eα1 ,Eα2 ]] √ √ = 2[[E−α1 ,Eα1 ],Eα2 ] = 2[−α1 · H,Eα2 ] (7.9) √ 1 = 2α1 · α2Eα = √ Eα . 2 2 2 It is an exercise to show that 1 [E−α ,Eα +α ] = −√ Eα . (7.10) 2 1 2 2 1 7. Simple roots 71

Interesting thing about this commutator is the phase: it can be (−1) relative to (7.9). This technique is clearly generalisable to a general algebra. For each root γ it is a part of an spin-j representation associated to the su(2) algebra generated by each simple root α. We denote the corresponding vector in the spin-j representation by |m, ji. Then, we can view Eα|Eγi in two ways:

Eα|Eγi = |[Eα,Eγ]i = N|Eα+γi,E+|m, ji = Nm,j|m + 1, ji.

We see N = Nm,j and can then identify the commutation relation [Eα,Eγ] = Nm,jEα+γ. The Jacobi identity can then be used to determine all the remaining commutation relations. This completely determines the algebra. 7. Simple roots 72

7.4. Exercises

1. Show (7.10).

2. (a) Consider two Lie groups G1,G2 in their deﬁning representations. By writing g1 ∈

G1, g2 ∈ G2 as a a i i g1 = exp(iθ X1 ), g2 = exp(iφ X2),

deﬁne the direct product group G = G1 × G2. (b) The group G has a Lie algebra associated to it g. By considering the generators of

G and their commutation relations, show that g = g1 ⊕ g2.

3. Consider now the group G = SU(2) × SU(2).

(a) Using the result of the previous exercise, list the generators of the Lie algebra for G. (b) What is the rank of g?

Γi is the deﬁning representation for gi. What is the dimension of the deﬁning representation of su(2) ⊕ su(2)? (d) Write down expressions for the generators of g in block diagonal form. Find ex-

pressions for the elements of the Cartan subalgebra H1,H2,...,Hm in the deﬁning representation and m is the rank of the algebra. (e) Find the weights of the fundamental representation of g. Plot the weights in weight space. Hint: write down the eigenvectors for the Cartan generators in the previous question, and use this to ﬁnd the eigenvalues.. (f) Deduce expressions for the non-zero roots. Add these to your diagram.

4. The algebra so(4) is rank 2, and the generators in their deﬁning representation are imaginary antisymmetric 4 × 4 matrices.

(a) There are 2 Cartan generators H1,H2 with components i [H ] = − δ δ − δ δ , p = 1, 2. (7.11) p jk 2 j,2p−1 k,2p k,2p−1 j,2p

Write this out in matrix form. Check that H1,H2 are mutually commuting and therefore part of the Cartan subalgebra.

k k (b) The 4 eigenvectors of Hp are come in pairs: |e i and |−e i, where k = 1, 2 and

k k k [|±e i]j = δj,2k−1 ± iδj,2k,Hp|±e i = ±δkp|±e i. (7.12)

Show these are eigenvectors of H1,H2. Find the corresponding weight vectors. 7. Simple roots 73

(d) The roots connect the weights. Deduce an expression for them ﬁrst in terms of νj and then plot them in weight space. Identify the positive roots. Compare with your answer for su(2) ⊕ su(2). 8. Dynkin Diagrams 74

8. Dynkin Diagrams

8.1. Dynkin Diagrams

A Dynkin diagram is a shorthand for writing down the simple roots. They exists as the angles between simple roots is highly constrained as we saw in (6.27). Each simple root is indicated by a node, drawn as a circle. Pairs of nodes are connected by lines, and the number of lines depends on the angle θ between the corresponding simple roots. There are four possibilities for a pair of roots and these are shown in Figure 8.1.

if the angle is

Figure 8.1: The possible Dynkin diagrams for a pair of simple roots with possible angles π/2, 2π/3, 3π/4, 5π/6.

Dynkin showed that a simple Lie algebra the roots have at most two possible lengths. Hence, we denote this on the diagram by showing the shorter root as hatched.13 The Dynkin diagrams for su(2) and su(3) are shown in Figure 8.2.

Figure 8.2: The Dynkin diagrams for su(2) and su(3).

13This is not necessarily standard notation in the literature, but is used frequently. 8. Dynkin Diagrams 75

8.2. Example: g2

We now come to our ﬁrst Lie algebra that does not derive from the classical matrix groups. The algebra g2 has simple roots √ α1 = (0, 1), α2 = ( 3/2, −3/2). (8.1)

We compute 2α1 · α2 2α1 · α2 2 = −3, 2 = −1. (8.2) α1 α2

The angle between α1, α2 is computed using (6.27) and is √ 3 cos θ = − , θ = 5π/6. α1α2 2 α1α2 The corresponding Dynkin diagram can now be drawn, as shown in Figure 8.3, with one of the 2 2 nodes hatched to show α1 < α2.

Figure 8.3: The Dynkin diagram for g2. The node is hatched to illustrate this root is shorter. It turns out this is the unique Dynkin diagram with three lines.

We now construct all the roots of g2. We do this by following the algorithm outlined in the previous chapters. The key points are that for each simple root αi, there is an su(2) subalgebra of g, of which Eαi is proportional to a raising operator; given a state |µi acting with a generator

Eαi results in either zero (if p = 0 in the master formula) or in a new state |µ + αii, and we conclude that µ+αi is a root. The master formula, therefore, is key to determining which linear combinations of α1, α2 are actually roots.

We start with level k = 1, of which we have the two states |α1i, |α2i for the simple roots.

Note, we can view these are given by acting with Eα1 ,Eα2 on |0i respectively.

For level k = 2, we have two options. First, Eα1 could act on |α2i. In that case,

2α1 · α2 2 = −3, α1 and so because α1 − α2 is not a root q = 0 in the master formula giving p = 3. Hence,

α1 + α2 , 2α1 + α2, 3α1 + α2 . 8. Dynkin Diagrams 76

The second option is Eα2 acting on |α1i. Then,

2α1 · α2 2 = −1, α2 which as before we have q = 0 in the master formula and conclude p = 1. Hence, we rederive

α1 + α2 as a root, but also learn that α1 + 2α2 is not a root. Hence, we have found all possible roots at levels k = 2 and k = 3.

At level k = 4 we have already determined 3α1 + α2 is a root. The only other possibilities are 2α1 + 2α2 and α1 + 3α2. The former is not a root as 2α1 + 2α2 = 2(α1 + α2) and we have already proved there are no non-trivial multiples of roots. The latter is not a root as it would be obtained by raising α1 + 2α2, but this is not a root. Hence, there is only one root at k = 4.

At level k = 5, all roots must come from raising 3α1 + α2. The possibilities are therefore

4α1 + α2 and 3α1 + 2α2. We already know that 4α1 + α2 is not a root. So only need to check

3α1 + 2α2. This would be obtained by acting with Eα2 on |3α1 + α2i. Applying the master equation for this 2α2 · (3α1 + α2) 2 = −1 = −(p − q). α2

As 3α1 is not a root, we have q = 0. Hence, p = 1 and so we have one level 5 root 3α1 + 2α2.

At level k = 6 roots, the possibilities are 4α1+2α2 and 3α1+3α2. Clearly, neither 3α1+3α2 =

3(α1 + α2) nor 4α1 + 2α2 = 2(2α1 + α2) are roots. Hence, there are no roots at level k = 6 and the algorithm terminates. The positive roots of g2 are therefore

α1 , α2 , α1 + α2 , 2α1 + α2 , 3α1 + α2 , 3α1 + 2α2 .

The negative roots are the minus of all these. We can plot these in Figure 8.4. It is in fact helpful to remember how we constructed these roots by inserting lines to show the raising operations, as illustrated in Figure 8.5. Starting with |α2i, we generated the remaining positive roots.

8.3. Cartan Matrix

Given a Lie algebra g of rank m, with simple roots α1, ··· , αm. Pick a simple root αj. For each of the simple roots αi, there is an su(2) algebra, deﬁned in (6.19), and we can compute an integer qi − pi using the master formula (6.21):

2H · αi 2αj · αi 2E3|αji = 2 |αji = 2 |αji = (qi − pi)|αji. (8.3) αi αi

As E3 eigenvalues are always half-integers, qi − pi is always an integer. This formula holds for i = 1, ··· , m, and so we can associate to αj a vector of integers ~q − ~p = [q1 − p1, . . . , qm − pm]. 8. Dynkin Diagrams 77

H 2

H 1

Figure 8.4: The roots for g2.

We can deﬁne a matrix A whose jth row is this vector of integers. This is called the Cartan matrix. Said diﬀerently, the (j, i)th entry is

2αj · αi Aji = 2 . (8.4) αi

The diagonal entries Aii = 2 for all i. There are two ways to see this. The uninformative way 2 is to simply note αi · αi = αi in (8.4). The informative way is to note that |αji is a highest weight vector for a spin-1 representation of the su(2) whose raising and lowering operators are 14 proportional to E±αj . We know that |αii is a highest weight vector because 2αi is not a root and so Eαi |αii = 0. We can apply the lowering operator twice to |αii before we get to the lowest weight vector |−αii:

|αii −→ |0i −→ |−αii.

In the master formula, the state |αii has p = 0 and q = 2 giving q − p = 2. The oﬀ-diagonal entries of A record the angles between the simple roots, and tell us their relative lengths. They can take possible values 0, −1, −2, −3. This is more powerful than the

14Recall, that the spin-1 representation is 3-dimensional with states labelled by their spins as

|1i → |0i → |−1i.

The states |1i and |−1i are the highest and lowest weight states respectively. 8. Dynkin Diagrams 78

H 2

H 1

Figure 8.5: The generation of the positive roots for g2 consists of starting with |α2i and then acting with raising operators. The resulting roots generated are |α1 + α2i, |2α1 + α2i, |3α1 + α2i,

|4α1 + α2i, |3α1 + 2α2i.

Dynkin diagram, as it contains information about lengths of roots. Finally, the Cartan matrix is invertible as the simple roots αi are complete, and so the rows are linearly independent. For su(3), the Cartan matrix is ! 2 −1 A = (8.5) −1 2 while for the g2 algebra ! 2 −1 A = (8.6) −3 2 Now pick any positive root γ. It is expanded in terms of the simple roots X γ = kjαj , j with the ki positive integers. The master equation (6.24) for the su(2) algebra associated to αi gives m m 2γ · αi X 2αj · αi X qi − pi = = kj = kjAji, (8.7) α2 α2 i j=1 i j=1 8. Dynkin Diagrams 79

Hence, acting with Eαj increases the level of γ and changes q − p:

|γi → Eαj |γi ∝ |γ + αji, kj → kj + 1, qi − pi → qi − pi + Aji, for i = 1, . . . , m.

At ﬁrst we labelled roots by their root vectors. It is in fact, easier, to label each root by its vector of integers q−p. This will then allow us to use the Cartan matrix to dramatically simplify the algorithm for generating the positive roots. We illustrate this for su(3). We worked out, using the master formula, that the roots occur at levels k = 0, 1, 2 and are Hi, α1, α2 and α1 +α2 respectively. We now rederive this, but at each level assigning the roots their new labelling system by a vector of integers. We go level–by–level:

• k = 0, the only generators are the Cartan elements. They have vanishing E3 eigenvalues for all the su(2) algebras, and so q − p = 0:

k = 0 : [0, 0] Hi. (8.8)

• k = 1, we have the simple roots α1, α2.

k = 1 : [2, −1] [−1, 2] α1, α2 (8.9) k = 0 : [0, 0] Hi

The vector α1 is to the left, and corresponds to acting with Eα1 on |0i, and so we add the

ﬁrst row of A. Similarly, α2 is to the right and we add the second row of A.

• k = 2: For the roots at level k = 1, we know the values of q, and write them above the vectors as follows15 q = 2 0 0 2

k = 1 : [2, −1] [−1, 2] α1, α2 (8.10)

k = 0 : [0, 0] Hi We can then compute the values of p:

p = 0 1 1 0 q = 2 0 0 2 (8.11) k = 1 : [2, −1] [−1, 2] α1, α2

k = 0 : [0, 0] Hi

For each non-zero p, we can construct a root at level k = 2 by acting with the corresponding simple root, and adding the corresponding row of A:

k = 2 : [1, 1] α1 + α2

k = 1 : [2, −1] [−1, 2] α1, α2 (8.12)

k = 0 : [0, 0] Hi

If we are going from α1 = [2, −1] to α1 + α2 = [1, 1] we are acting with α2 and we add the second row of A to [2, −1] to give [1, 1]. Computing the values for p, q is easy, as we know them already know every from k = 1 in (8.11):

p = 0 0 q = 1 1

k = 2 : [1, 1] α1 + α2 (8.13)

k = 1 : [2, −1] [−1, 2] α1, α2

k = 0 : [0, 0] Hi The procedure now terminates as we see that p = 0 for both entries, and so this is the highest weight state.

The procedure is algorithmic, as everything you need to know to go from level k = l to k = l+1 is in the diagram. We can also insert the corresponding negative levels (corresponding to negative roots). The result is shown in Figure 8.6. There are su(2) representations lurking here. We add in lines to illustrate the lowering and raising. The lines join components corresponding to the simple root with which we are raising or lowering. For roots of su(3), there are spin-1/2 and spin-1 representations. For example, the su(2) associated with α1 acts on the following roots as a spin-1 representation

q − p : [2, −1] ←→E± [0, 0] ←→E± [−2, 1] (8.14) spin–1 : |1i ←→Γ± |0i ←→Γ± |−1i

We see the ﬁrst entry of q − p is exactly 2m in |1, mi. As a second example, α2 generates a 1 spin- 2 representation on the following roots:

q − p : [2, −1] ←→ [1, 1] (8.15) 1 1 spin–2 : | i ←→ |− i 2 2

As its α2, the second component of q − p is identiﬁed with 2m.

We can repeat the algorithm for g2. The answer is shown in Figure 8.7. Let’s reiterate the procedure to derive this result. 8. Dynkin Diagrams 81

k = 1 2 -1 -1 2

k = 0 0 0

k = -1 1 -2 -2 1

Figure 8.6: The roots for su(3), labelled by their eigenvalues [q1 − p1, q2 − p2]. Both positive and negative roots are shown. The lines start and end on the component corresponding to the simple root with which we are raising or lowering.

1. There are two simple roots α1, α2, and each one generates an su(2) algebra. Each root is then part of a spin-j representation associated to each of those su(2)s.

2. Start with level k = 0. These always have ~q − ~p = [0, 0], and correspond to the elements in the Cartan subalgebra.

3. The k = 1 roots are always the simple roots.

~q − ~p for αl is always the lth row of A. We draw lines indicating how we got from k = 0

to k = 1: for αl the line begins and ends on the lth component of each ~q − ~p.

4. The k = 2 roots are constructed by raising the k = 1 roots.

Start with α1 = [2, −1]. As it’s a simple root, it is a highest weight vector associated to

the su(2) generated by α1. We identify it with |1, 1i in a spin-1 representation. So we

cannot raise it with Eα1 .(p1 = 0, q1 = 2.)

With respect to the su(2) associated to α2, q2 − p2 = −1. We know q2 = 0, and so p2 = 1. 1 1 Hence, we can raise it once, and we identify it with | 2 , − 2 i. A lowest weight vector with 1 1 m = − 2 is a spin- 2 representation. Raising it once, the level k = 2 root [−1, 1] is given 8. Dynkin Diagrams 82

0 1

3 -1

1 0

-1 1

2 -1 -3 2

0 0

Figure 8.7: The positive roots for g2, labelled by their eigenvalues [q1 − p1, q2 − p2].

by adding the 2nd row of the Cartan matrix A to [2, −1]. We draw an line beginning and

ending on the second component to indicate we used the su(2) associated to α2. It is an exercise to apply a similar logic to the root [−3, 2]. The end result is [−3, 2] 3 3 is a lowest weight vector of a spin-3/2 with respect to α1, identiﬁed with | 2 , − 2 i. It is the highest weight vector of the spin-1 representation with respect to α2. Hence, we can

raise it with Eα1 three times, generating roots at k = 2, 3, 4. Each time the line is drawn beginning and ending on the ﬁrst component.

5. Level k = 3.

There is only one root [−1, 1]. From the knowledge in the previous step, we know that 1 3 with respect to α1, we have q1 = 1, p1 = 2, and so identify with |− 2 , 2 i. We can act with

Eα1 to generate [1, 0]. With respect to α2, we see q2 = 1 and p2 = 0, and it is a highest 1 weight vector of a spin- 2 representation. Hence, we cannot raise it with respect to α2.

6. Level k = 4. 8. Dynkin Diagrams 83

1 3 There is only one root at k = 3: [1, 0]. With respect to α1, it is the state | 2 , 2 i, and we can raise it once more to give [3, −1]. With respect to α2, we have q2 = p2 = 0, and so we

cannot raise with respect to α2.

7. Level k = 5:

At level k = 4 there is only one root [3, −1]. With respect to α1, it is highest weight, and

so we cannot raise it with α1. With respect to α2, we have q2 = 0 and so identify p1 = −1. 1 1 It is identiﬁed with |− 2 , 2 i, and so we can raise it once to give [0, 1].

8. Level k = 6:

At level k = 5, there is only one root [0, 1]. With respect to α1, we see that p1 = q1 = 0,

and so we cannot raise it with α1. With respect to α2, p2 = 0 it is the highest weight

vector and so cannot be raised anymore. Whenever p1 = p2 = 0, we have ﬁnished, and so stop.

I hope you get the pattern. As we go through the roots, level-by-level, the key is to consider the su(2) representation associated to a simple root αi; identify which vector |j, mi it belongs to, and raise it if appropriate, adding the ith row of A to ~q − ~p. If its a highest weight vector, so pi = 0, it cannot be raised any more. Once we end up with a root which has p1 = ... = pm = 0, we are ﬁnished and so stop.

8.4. Constructing the g2 algebra Not assessable

We will do another example shortly, but for the moment let’s stick with g2. We have all the roots, but lets actually construct the generators of the algebra. This will amount to constructing the adjoint representation of g2.

The two raising operators for α1, α2 are:

+ + 1 E = Eα ,E = √ Eα . (8.16) 1 1 2 3 2

Starting with |Eα2 i we know that p = 3, q = 0 with respect to the su(2) generated by α1, and 3 3 so is identiﬁed with | 2 , − 2 i. Applying α1 raising operator

Eα1 |Eα2 i = |[Eα1 ,Eα2 ]i r3 = |3/2, −1/2i 2 (8.17) r3 =∼ |E i. 2 α1+α2 8. Dynkin Diagrams 84

The last line is a definition, and the normalisation is fixed by the definition of the raising operator. Applying the raising operator again, gives us a definition for |E2α1+α2 i:

√ r3 √ |[E , [E ,E ]]i = 2 |3/2, 1/2i =∼ 3|E i, (8.18) α1 α1 α2 2 2α1+α2 and a third time gives

r √ 3 ∼ 3 3 |[Eα , [Eα , [Eα ,Eα ]]]i = 3|3/2, 3/2i = √ |E3α +α i = √ |1/2, −1/2i, (8.19) 1 1 1 2 2 2 1 2 2

1 where the last equality follows from the vector being the lowest weight vector of the spin- 2 representation generated by α . This means we can raise the state with E+ = √1 E : 2 2 3 α2 √ √ √ + 3 3 3 ∼ 3 3 |[Eα , [Eα , [Eα , [Eα ,Eα ]]]]i = 3E √ |1/2, −1/2i = |1/2, 1/2i = |E3α +2α i 2 1 1 1 2 2 2 2 2 1 2 (8.20) Putting it together we get expressions for the positive roots in terms of multiple commutators of generators for the simple roots:

r2 E = [E ,E ], α1+α2 3 α1 α2 1 E2α +α = √ [Eα , [Eα ,Eα ]], 1 2 3 1 1 2 √ (8.21) 2 E = [E , [E , [E ,E ]]], 3α1+α2 3 α1 α1 α1 α2 2 E3α +2α = √ [Eα , [Eα , [Eα , [Eα ,Eα ]]]], 1 2 3 3 2 1 1 1 2 Together with the Jacobi identity, this is enough to determine the commutation relations of the entire algebra. As an example, we can check that E−α1 acts like a lowering operator as we would expect on Eα1+α2 :

r2 [E ,E ] = [E , [E ,E ]] −α1 α1+α2 3 −α1 α1 α2 r2 = [[E ,E ],E ] 3 −α1 α1 α2 r2 = − [α · H,E ] (8.22) 3 α2 r2 = − α · α E 3 1 2 α2 r2 = E 3 α2

This is what we would expect for E− acting on |3/2, −1/2i. 8. Dynkin Diagrams 85

This works in general. The commutator [Eγ,Eφ] for two roots γ, φ involves multiple commutator of negative simple roots with a multiple commutator of positive simple root generators. Use the Jacobi identity to rearrange them so the positive and negative simple roots are paired up

[Eαi ,E−αi ], and eat each other. We then end up with multiple commutators of purely positive roots, or multiple commutators purely negative roots or a Cartan generator.

8.5. Example: c3

2 2 Lets study the algebra associated to the Dynkin diagram in Figure 8.8. where α1 = α2 = 1,

Figure 8.8: The Dynkin diagrams for c3.

α2 = 2. The Cartan matrix is 3   2 −1 0   −1 2 −1 (8.23)   0 −2 2 The construction of the positive roots is shown in Figure 8.9. The algebra is rank 3, and so the diagram is really three-dimensional. We are lucky in this case — the digram can be projected to being two-dimensional. If one were to work out the commutation relations for the generators, as we did for g2, you would ﬁnd they are identical to the algebra so(7). So we have discovered the roots for the algebra so(7) without knowing anything about the fact they come from 7-dimensional orthogonal matrices.

8.6. Fundamental weights

Let’s now turn to a arbitrary representation Γ. It has a highest weight µ which satisﬁes the property that µ + γ is not a weight for any positive root γ.

In fact, µ is a highest weight vector of an irreducible representation if and only if µ + αi is not a weight for all simple roots αi. To see this, ﬁrst suppose µ+αi is not a weight for all simple roots. Then,

Eαj |µi = 0 for j = 1, . . . , m. P But, any positive root is the sum of the simple roots with positive coeﬃcients γ = kjαj where kj ≥ 0. Hence, Eγ|µi = 0. Conversly, if Eγ|µi = 0 for all positive roots γ, we can simply pick 8. Dynkin Diagrams 86

2 0 0

0 1 0

1 -1 1 -2 2 0

1 1 -1 -1 0 1

2 -1 0 -1 2 -1 0 -2 2

0 0 0

Figure 8.9: The positive roots for c3, labelled by their eigenvalues [q1 − p1, q2 − p2, q3 − p3].

γ = αi, and the result follows. If we refer back to the master equation, then αj acting on |µi has p = 0 meaning 2αj · µ 2 = lj . αj where lj are non-negative integers. The integers lj completely determine µ as the αj are linearly independent. Hence, we are free to label the highest weight state |µi by the vector of integers

[l1, ··· , lm]. These integers are called the Dynkin coefficients. As for representations of su(2), starting from the highest weight state |µi we can construct the remaining states of the irreducible representation by acting with negative simple roots (lowering operators). Therefore, each highest weight state |µi gives rise to an irreducible representation, and the process is unique. The previous equation shows there is a one-to-one correspondence between Dynkin coefficients and highest weights. Hence, we arrive at a powerful statement: every representation of a rank m algebra is labelled by a set of m non-negative integers, [l1, . . . , lm]. It is not just the highest weight state that can be labelled by Dynkin coefficients. All other states in the representation can also be labelled by vectors of integers, though they may be 8. Dynkin Diagrams 87 negative. In fact, when the representation is the adjoint representation these are precisely the integers we have been using to construct the roots of algebras in this chapter, e.g. Figure 8.9. It is useful to note that the rows of the Cartan matrix are the Dynkin coefficients of the simple roots.

We can choose a basis vectors µk satisfying

2αj · µk 2 = δjk. (8.24) αj for k = 1, . . . , m. These are called the fundamental weights. Every highest weight vector µ can be written m X µ = ljµj. (8.25) j=1 1 We can build a representation with highest weight µ by tensoring l representations of µ1 with 2 l of µ2 and so on. The representation is in general reducible, but we can always pick out the irreducible representation of interest by applying lowering operators to µ. j For each fundamental weight µj there is an irreducible representation Γ , which is called the fundamental representation. The deﬁning representation is an example of a fundamental representation.

8.7. Example: more representations su(3)

We now return to su(3) and use the Dynkin coeﬃcients to construct other representations of su(3). Recall the simple roots are √ √ α1 = (1/2, 3/2), α2 = (1/2, − 3/2).

i i The fundamental weights µi satisfy µi · αj = 0 if i 6= j. If we set µi = (a , b ), then this implies √ ai = ± 3bi. The normalisation is ﬁxed by (8.24) giving √ √ µ1 = (1/2, 3/6), µ2 = (1/2, − 3/6). (8.26)

Now, µ1 is the highest weight of the defining representation of su(3) generated by the Gell-Mann √ matrices (6.28) as discussed in chapter 6.6. To see this note that starting from µ1 = (1/2, 3/6), the fact µ1·α2 = 0 means we can only use E−α1 to lower the vector. Doing so, using the procedure in the last chapter the lowering operators E−α1 , E−α2 fill out the remaining weights as shown in Figure 8.10. This is to be compared with the diagram we constructed in Figure 6.1 – they are exactly the same! We can write the fundamental representation in terms of its Dynkin coefficients as

µ = [1, 0] 8. Dynkin Diagrams 88

H 2

H 1

Figure 8.10: The weight vectors of the deﬁning representation of su(3).

1 0

-1 1

0 -1

Figure 8.11: Constructing the fundamental representation with Dynkin coeﬃcients starting from the fundamental weight µ1 or [1, 0] and applying lowering operators.

A diﬀerent way to construct the representation is to use the Dynkin coeﬃcients and subtract appropriate rows of the Cartan matrix (8.5) as we did for roots of c3 in Figure 8.9. To start, we see that [1, 0] is the highest weight state of a spin-1/2 representation of the algebra generated by α1; hence we can lower once to get µ−α1 = [−1, 1]. This state is now highest weight state of a spin-

1/2 representation with the algebra generated by α2. Hence, lower once: µ − α1 − α2 = [0, −1]. 8. Dynkin Diagrams 89

This is now a lowest weight state and we are ﬁnished. This is illustrated in Figure 8.11. The construction is completely general. The fundamental representation of su(n) in terms of Dynkin coeﬃcients is µ = [1, 0,..., 0].

What about the representation corresponding to the other fundamental weight µ2 or [0, 1]? We construct it using Dynkin coeﬃcients analogous to [1, 0]. The answer is shown in Figure

8.12. The weights are µ2, µ2 − α2, µ2 − α2 − α1. In weight space the weights are, respectively, √ √ √ (1/2, − 3/6), (0, 3/3), (−1/2, − 3/6) , and are shown in Figure 8.13 Let us construct the representation [2, 0]. The highest weights state is √ 2µ1 = (1, 1/ 3). (8.27)

The remaining weights are constructed in Figure 8.14 in the language of Dynkin coeﬃcients. There are six weights, and so the representation is 6-dimensional. The weights are illustrated in weight space in Figure 8.15.

8.8. Complex conjugation

Notice the fundamental representations corresponding to µ1 and µ2 are closely related: the weight diagrams are mirror images of each other. We show this explicitly by putting the two diagrams together in Figure 8.16. It turns out there is a really good reason for this: complex conjugation!

0 1

1 -1

-1 0

Figure 8.12: Constructing the representation for µ2 or [0, 1] in the language of Dynkin coeﬃ- cients. Starting from [0, 1] the remaining vectors are given by applying lowering operators. 8. Dynkin Diagrams 90

H 2

H 1

Figure 8.13: The weights for the representation [0, 1] plotted in the root vector space.

∗ If matrices Ta are generators of some representation Γ of a Lie algebra, then the object −Ta have the same commutation relations:

∗ ∗ ∗ ∗ ∗ ∗ [−Ta , −Tb ] = [Ta ,Tb ] = ([Ta,Tb]) = ifabc(−Tc ).

∗ So −Ta form another representation of the Lie algebra, called the complex conjugate of the ∗ representation Γ, and is denoted Γ. The Cartan generators of Γ are −Hi . As Hi is hermitian, ∗ Hi has the same eigenvalues as Hi. So if µ is a weight of Γ then −µ is a weight of Γ. The highest weight of Γ is minus the lowest weight of Γ. This is precisely what Figure 8.16 illustrates: −µ2 is the lowest weight vector of the representation [1, 0]. In general, a representation [n, m] has complex conjugate representation [m, n]. If m = n, then the representation is real. 8. Dynkin Diagrams 91

2 0

0 1

1 -1 -2 2

-1 0

0 -2

Figure 8.14: The weight diagrams for the representation [2, 0] of su(3).

H 2

H 1

Figure 8.15: The weights for the representation [2, 0] of su(3) in weight space. 8. Dynkin Diagrams 92

H 2

H 1

Figure 8.16: The weight diagrams for representations with highest weights µ1 (coloured circles) and µ2 (open circles).

8.9. Exercises

1. (a) Starting from the highest weight [2, 0] for su(3) construct the weights of the repres-

entation by acting with the appropriate lowering states E−αi . Show they correspond to those in Figure 8.14. (b) Plot the weights for (0, 2) representation, and show it corresponds to Figure 8.15. Describe the relation of the weights to the (2, 0) representation.

2. Construct the representation [3, 0] for su(3) in an analogous fashion. How many dimensions does it have?

3. Construct a diagram showing the Dynkin coeﬃcients of the [1, 1] representation, and a diagram in weight space showing the weights.

4. Construct the weights for the fundamental representation [1, 0] for g2. What is the dimension of this representation?

5. Using the g2 algebra derived in the text, and the Jacobi identity, construct an expression

for the commutator [E−α2 ,E3α1+2α2 ]. In addition, compute [E−α1 ,E3α1+2α2 ].

6. Given the expression for the Cartan matrix A for c3, show that its Dynkin diagram is indeed Figure 8.8. 9. A tour of compact simple Lie algebras and their physical applications 93

9. A tour of compact simple Lie algebras and their physical applications

In the final part of the notes we will discuss some of the technology we have developed for other Lie algebras, and point out some of the place they appear in modern mathematical physics. This is by no means doing either the classification of Lie algebras nor their physical applications justice, but I hope you get the flavour for what the algebras are like and in some of the scientific fields they are useful.

∼ 9.1. su(N) = aN−1

Recall this is the Lie algebra for the group SU(N), unitary N×N matrices with unit determinant. The title of this subsection has given the label to this algebra originally described by Dynkin of aN−1. For most of the algebras in this section we will do this, as it is still used in the literature today. We have already dealt with su(3) and su(2) extensively; we want to generalise this discussion to su(N). The generators are normalised as 1 Tr T T = δ . (9.1) a b 2 ab The group has rank m = N − 1 as there are N − 1 independent traceless diagonal real matrices. The Cartan generators are taken to be

p ! 1 X [Hp]ij = p δikδjk − pδi,p+1δj,p+1 . (9.2) 2p(p + 1) k=1 So for example, p = 1, 2, 3 are     1 0 0 ··· 1 0 ···   1   1 0 1 0 ··· H1 = 0 −1 ··· ,H2 = √   , 2   12 0 0 −2 ··· . . ..   . . . . . . .  . . . ..   1 0 0 0 ··· (9.3)   0 1 0 0 ···   1   H3 = √ 0 0 1 0 ··· . 24     0 0 0 −3 ··· . . . . .  ......

The remaining generators are raising and lowering operators. We take the raising operators to √ have a single non-zero oﬀ-diagonal entry, 1/ 2 in the upper half triangle; the lowering operators are the transpose of this. This gives N 2 − N matrices, which together with the N − 1 Cartan generators gives a total of N 2 −1 matrices. Reassuringly, this is exactly the dimension of su(N). 9. A tour of compact simple Lie algebras and their physical applications 94

The weights are N − 1-dimensional vectors:

p ! j 1 X [ν ]p = [Hp]jj = p δjk − pδj,p+1 (no sum) . (9.4) 2N(N + 1) k=1 This satisfy N − 1 1 (νj)2 = , νi · νj = − , for i < j. (9.5) 2N 2N Explicitly ! 1 1 1 ν1 = , √ , ··· , 2 2 3 p2(N − 1)N ! 1 1 1 ν2 = − , √ , ··· , 2 2 3 p2(N − 1)N ! 1 1 (9.6) ν3 = 0, −√ , ··· , 3 p2(N − 1)N . . ! N − 1 νN = 0, 0, ··· , − p2(N − 1)N

For su(N), its convenient to choose a diﬀerent deﬁnition of positivity: a positive weight is one in which the last non-zero component is positive. This has the virtue of

ν1 > ν2 > ··· > νN . (9.7)

The raising and lowering operators E±α take us from one weight to another, so the roots are

i i+1 αi = ν − ν , i = 1,...,N − 1. (9.8)

All roots have length 1 and satisfy 1 α · α = δ − δ . (9.9) i j ij 2 i,j+1 Hence, the Dynkin diagram for su(N) is shown in Figure 9.1.

The fundamental weights µj are j X µj = νk, (9.10) k=1 and can be checked to satisfy 2αi · µj 2 = δij. (9.11) αi The fundamental weight µ1 is the highest weight for the deﬁning representation. 9. A tour of compact simple Lie algebras and their physical applications 95

...

Figure 9.1: The Dynkin diagram for su(N). There are N − 1 circles (nodes).

Physically su(N) features extensively in Grand Uniﬁed Theories (GUTs) and the standard model of particle physics. These are both types of quantum ﬁeld theories with a symmetry group that is a product of unitary groups. The standard model has a symmetry group featuring

SU(3) × SU(2) × U(1).

All particles in the standard model are in representations of this group, and are the vectors within the vector space of the representation. A miraculous theoretical breakthrough is to realise that almost all the particles of the standard model (and there are lots of them) can be really be thought of as representations of a bigger group su(5). This works because su(5) ⊃ su(3) ⊕ su(2) ⊕ u(1), and under this subgroup the representation theory works out just right. (To see this explicitly we need to take a detour and discuss tensor methods, which is beyond our time remit). The unitary group appears in many other places of theoretical physics. It features heavily in string theory, which describes elementary particles as vibrating strings; in quantum mechanics through the simple harmonic oscillator and therefore within quantum chemistry.

∼ 9.2. The so(2N) = dN algebra

The algebra is generated by imaginary antisymmetric 2N × 2N matrices of which N(2N − 1) are linearly independent. There are N Cartan generators H1,...,HN with components

[Hp]jk = −i δj,2p−1δk,2p − δk,2p−1δj,2p , p = 1,...,N. (9.12)

Notice this is an example where the Cartan generators are not diagonal. The rank of so(2N) is k k therefore N. The 2N eigenvectors of Hp are come in pairs: |e i and |−e i, where k = 1,...,N and k k k [|±e i]j = δj,2k−1 ± iδj,2k,Hp|±e i = ±δkp|±e i , no sum. (9.13)

By collecting the eigenvalues, the positive weight vectors µk have components

k [µ ]r = δkr, r = 1,...,N. (9.14) 9. A tour of compact simple Lie algebras and their physical applications 96

That is, the are unit vectors in the N-dimensional Cartan space. Do not get confused between the vector space upon which so(2N) acts (which is 2n-dimensional); and the vector space in which the roots and weights live (which is m-dimensional). The vector ek above lives in the latter. The roots connect pairs of weights ±µi, ±µj with i 6= j in all possible ways, and so are

± µj ± µk, j 6= k. (9.15)

The positive roots are of the form µj ± µk for j < k while the simple roots are

µj − µj+1, j = 1,...,N − 1, µN−1 + µN .

Note if N = 1 there is a single node. If N = 2 there are two nodes, and the two simple roots µ1 +µ2 and µ1 −µ2 have an angle π/2 and so the Dynkin diagram are two non-connected nodes. The Dynkin diagram is shown in Figure 9.2. The group so(2N) appears is any physical

...

Figure 9.2: The Dynkin diagram for so(2N). There are N circles (nodes), one for each simple root. problem involving some type of rotational symmetry of a system. The symmetry may be external – i.e. associated to a geometric symmetry, such as the object being spherical in nature – or it may be internal, a hidden symmetry, such as a gauge symmetry. Such internal symmetries turn out to be responsible for inter-particle forces, such as the electromagnetic force, the strong and weak nuclear forces. You may come across these forces in a course on quantum field theory. We mentioned in the previous subsection that it is possible to use the group SU(5), and its corresponding algebra su(5), to repackage the standard model of particle physics into a single grand unified theory. All the particles, which at a macroscopic level look different, turn out to be simply different representations of the algebra su(5). There is more than one way to skin a cat however: the group SO(10) and its algebra so(10) also do the same job, and more. It repackages all the field content, as well as some extra fields such as a right handed neutrino, into representations of so(10). If they both do the job, which one is correct? At the point of 9. A tour of compact simple Lie algebras and their physical applications 97 writing, the answer is nobody knows!. Its an open research question, both theoretically, and being pursued experimentally at the Large Hadron Collider (LHC) in Switzerland.

∼ 9.3. so(2N + 1) = bN

Interestingly the even and odd orthogonal groups have a diﬀerent Lie algebraic structure, and so we have discussed them separately. We choose the Cartan generators as for so(2N), as the rank of the algebra is still N. The weights are still of the form ±ek together with a zero weight. The roots connect these weights, and are of the same form as before with additional roots connecting the zero weight. Thus the roots are of the form

±ej ± ek, j 6= k, and ± ej.

The positive roots are ej ± ek for j < k and ej. The simple roots are

ej − ej+1, for j = 1,...,N − 1 and eN . The Dynkin diagram is shown in Figure 9.3.

...

Figure 9.3: The Dynkin diagram for so(2N + 1). The node associated to eN is hatched to show the root is shorter than the others.

The SO(2N) and SO(2N +1) groups, as well as their algebras so(2N) and so(2N +1) have a special representation called the spinor representation. This representation is hugely important in the physical world: spinor representations described fermions. Fermions are particles such as electrons, protons and neutrons. They have all sorts of interesting physical properties, such as half-integer spin. This means instead of rotating a particle by 2π radians, you need to rotate it by 4π radians before you get back to where you started from. This rotational property, intuitively bizarre, is reponsible for the Pauli exclusion principle. This says you cannot have two 9. A tour of compact simple Lie algebras and their physical applications 98 fermions (i.e. electrons) near each other travelling with the same momentum. This is responsible for special properties of semi-conductors and the conductivity of metals. Neeedless to say, the impact of these properties has been huge in a range of ﬁelds such as material sciences, quantum chemistry and particle physics. All of this is due to representation theory. Amazing stuﬀ.

∼ 9.4. sp(2N) = cN

The group of sympletic matrices is a matrix group that we have not yet met. Denoted Sp(2N), it consists of 2N × 2N unitary matrices that satisfy

U tJU = J, (9.16) where ! 0 I J = N , −IN 0 a and IN is the N × N identity matrix. The algebra consists of 2N × 2N hermitian matrices T satisfying JT a + (T a)tJ = 0,

a To see this, recall that passing to the algebra consists of writing U = exp(iαaT ) in (9.16) where T a are the generators, and Taylor expanding the expotential. The previous equation is then the ﬁrst non-trivial equation to arise. In the fundamental representation, the generators take the form

I2 ⊗ A, σ1 ⊗ S1, σ2 ⊗ S2, σ3 ⊗ S3, (9.17) where σa are the Pauli matrices, A an antisymmetric matrix, while S1,S2,S3 are symmetric matrices. It is an exercise to check that these close under commutation. The subset

I2 ⊗ A, σ3 ⊗ S3, when S3 are traceless generate an su(N) subalgebra of sp(2N) with generators ! Ta 0 ∗ 0 −Ta

This is a reducible representation [1, 0] ⊕ [0, 1], i.e. the direct sum of the fundamental representation and its complex conjugate. We take the ﬁrst N − 1 elements of the Cartan subalgebra of su(N) to be of this form, and the ﬁnal element to be √ HN = σ3 ⊗ I/ 2N. 9. A tour of compact simple Lie algebras and their physical applications 99

All the generators of su(N) commute with the HN and so the roots of the subalgebra have j k HN = 0. Already know the other components are given by the su(N) root vectors ν − ν ; the other root corresponds to the matrices (σ1 ± iσ2) ⊗ Skl where

[Skl]ij = δikδjl + δilδjk.

If we take nn+1 to be a unit vector orthogonal to all the νj for j = 1,...,N then the roots can be written as ! r 2 νj − νk j 6= k, ± νj + νk + νn+1 , n while the positive roots are ! r 2 νj − νk j < k, νj + νk + νn+1 , n and the simple roots ! r 2 νj − νj+1, 2νn + νn+1 , n for j = 1, . . . , n − 1. The Dynkin diagram is shown in Figure 9.4.

...

Figure 9.4: The Dynkin diagram for sp(2N). The su(N) roots are shorter, and are hashed to show this.

Recall, we already met this algebra in the guise of c3 in Chapter 8, where we constructed the roots. The sp(2N) algebra is closely connected with the quarternions, a mathematical object 3 4 that generalises the basis vectors i, j, k of R to R .

9.5. Exotic Lie algebras: e6, e7, e8, f4, and g2

What Dynkin originally did was classify all the possible compact Lie algebras. The classification is beyond the scope of our course, but intuitively you can get a flavour for it. He essentially 9. A tour of compact simple Lie algebras and their physical applications 100 wrote down all the types of Dynkin diagrams you can think of; our of those Dynkin diagrams only a subset of them are consistent, and give root systems that make sense. Amazingly, there are a finite number of types of root systems available. We have already met some of them: su(N), so(2N), so(2N + 1) and sp(2N). These are known as the classical Lie algebras, probably due to their close connection with matrix groups. The exotic Lie algebras do not have any simple matrix group connections. They are e6, e7, e8, f4, and g2. We have already met g2 and its roots. We will not describe the other exotic Lie algebras in any great detail but do list their Dynkin diagrams in Figure 9.5 along with the other compact Lie algebras. It is an impressive result: these are the only compact Lie algebras we can write down. There is nothing else. For reference, the angles between simple roots are redrawn in Figure 9.6 below. Physically, these exotic groups appear in particle physics, this time in the context of string theory. The group E8 is central to constructing string theories, and can function as a ‘supped- up’ grand unified theory. It is supped-up, because in addition to the strong and weak nuclear forces, it includes gravity. This is a highly non-trivial feat: for over a 100 years mathematicians and physicists have tried coming up with quantum theories of gravity, and everyone of them has failed except in the pursuit of string theory. The precise experimental meaning of this is not at all clear though is an active area of research both at the LHC and also in more distant contexts through the study of the early universe cosmology, black holes and dark matter. 9. A tour of compact simple Lie algebras and their physical applications 101

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Figure 9.5: The Dynkin diagrams for all the compact Lie algebras, including the exotic Lie algebras.

9.6. Exercises

1. Some results on su(N):

(a) Show the weights deﬁned in (9.4) satisfy (9.5).

(b) Show (9.6). 9. A tour of compact simple Lie algebras and their physical applications 102

if the angle is

Figure 9.6: The possible Dynkin diagrams for a pair of simple roots with possible angles π/2, 2π/3, 3π/4, 5π/6.

2 (c) Show the roots of su(N) have length 1: αi = 1 and satisfy (9.9). (d) From this, show the Dynkin diagram is Figure 9.1.

(e) Using the fact the roots αi = νi − νi+1, show the fundamental weights µj satisfy (9.11).

2. Some results on so(2N)

(a) Show the eigenvectors of the Cartan generator Hp are indeed for (9.13). Construct the generators and eigenvectors explicitly for so(4). (b) Now construct the generators for so(7). Show that the Dynkin coeﬃcients for the

roots of so(7) are the same as c3 constructed in the previous chapter.