Representations of SO(3) in C [X, Y, Z]

DEGREE PROJECT IN TECHNOLOGY, FIRST CYCLE, 15 CREDITS STOCKHOLM, SWEDEN 2019

Representations of SO(3) in C [x, y, z]

OTTO SELLERSTAM

KTH ROYAL INSTITUTE OF TECHNOLOGY SCHOOL OF ENGINEERING SCIENCES EXAMENSARBETE INOM TEKNIK, GRUNDNIVÅ, 15 HP STOCKHOLM, SVERIGE 2019

Representationer av SO(3) i C [x,y,z]

OTTO SELLERSTAM

KTH SKOLAN FÖR TEKNIKVETENSKAP Abstract This paper will present a concrete study of representations of the special orthogonal group, SO(3), a group of great importance in physics. Specifically, we will study a natural representation of SO(3) in the space of polynomials in three variables with complex coefficients, C[x, y, z]. We will find that this special case provides all irreducible representations of SO(3), and also present some corollaries about spherical harmonics. Some preparatory theory regarding abstract algebra, linear algebra, topology and measure theory will also be presented.

Sammanfattning Denna rapport kommer att redogöra en konkret undersökning av representationer av den speciella ortogonala gruppen, SO(3), en grupp med viktiga tillämpningar inom fysik. Mer specifikt kommer vi att studera en naturlig representation av SO(3) i rummet av polynom i tre variabler med komplexa koefficienter, C[x, y, z]. Vi kommer att se att alla irreducibla representationer av SO(3) dyker upp ur detta specialfall, samt presentera n˚agraföljdsatser ang˚aendeklotytefunktioner. Förberedande teori ang˚aende abstrakt algebra, linjär algebra, topologi och m˚atteorikommer även att redovisas.

1 Contents

1 Introduction 4

2 Some Algebraic Structures 5

3 Linear Algebra 8 3.1 Basic concepts ...... 8 3.2 Multilinear algebra ...... 11

4 Topology 15 4.1 Point-set topology ...... 15 4.1.1 Topological spaces and continuity ...... 15 4.1.2 Compactness ...... 16 4.2 Topological groups ...... 17

5 Measure Theory 19 5.1 Construction of the Lebesgue integral ...... 19 5.1.1 Lp-spaces ...... 20 5.2 Borel algebras and the Haar measure ...... 21

6 Representation Theory 23 6.1 Basic deﬁnitions ...... 23 6.2 Some important theorems ...... 24 6.3 Character theory ...... 26 6.4 Canonical decomposition of a representation ...... 30 6.5 From ﬁnite to compact groups ...... 30

7 The 3D Rotation Group SO(3) 32 7.1 Parameterization ...... 32 7.2 Conjugacy classes of SO(3) ...... 33 7.3 The Haar measure on SO(3) ...... 34

8 Representations of SO(3) 35 8.1 Representations in C[x, y, z]...... 35 8.1.1 Introduction ...... 35 8.1.2 The natural representation ...... 36 8.1.3 Characters of SO(3)...... 38 8.1.4 An inner product on C[x, y, z]...... 39 8.2 Tensor products of irreducible representations ...... 44

2 3 1 Introduction

The goal of representation theory is to reduce the study of complicated algebraic structures to the study of vector spaces; elements are represented as bijective linear operators on a vector space, which often helps to concretize the underlying algebraic structure. Some of the most common structures to study using representation theory include groups, associative algebras and Lie algebras. In this paper we will discuss an analysis of the 3D rotation group from a representation theoretic per- spective. We will go through some fundamental concepts of abstract and linear algebra, which will be used to define representations and to present some basic ideas from representation theory of finite groups. SO(3) is, however, not a finite group, so we will also need to discuss a framework of how to translate the ideas from the finite case to the infinite case. To discuss this framework, we also need to develop some concepts from topology and measure theory, since notions such as compactness and integration with respect to a measure are required.

4 2 Some Algebraic Structures

We start of by defining some basic algebraic structures, together with some examples. Definition 1 (Group). A group is a set G equipped with a binary operation ? : G × G → G which satisfies the following axioms for all a, b, c ∈ G. 1. Associativity: a ? (b ? c) = (a ? b) ? c 2. Identity: there exists an element e ∈ G such that a ? e = e ? a = a 3. Inverse: there exists an element a−1 ∈ G such that a−1 ? a = a ? a−1 = e When we want to emphasize both the underlying set G and the group operation ?, we will write (G, ?). A group where the elements commute, i.e. a ? b = b ? a for all a, b ∈ G, is called an abelian group, named after Niels Henrik Abel. Example 1. We provide some examples of groups. Notice that the first two examples are abelian, while the last one is not.

1. The integers Z under addition, (Z, +) form a group. Note however that the integers under multiplication, (Z, ·), does not, since the multiplicative inverse of for example 2 is 1/2, which is not an integer.

2. The rational numbers without the number 0 form a group under multiplication, denoted (Q − {0}, ·). 3. The general linear group GL(n) consisting of all invertible n × n-matrices with real entries is a group 2 under matrix multiplication. We will later on consider this group to be a subset of Rn . 4. The orthogonal group O(n) = {A ∈ GL(n): AT A = AAT = I} ⊂ GL(n) and the special orthogonal group SO(n) = {A ∈ O(n) : det(A) = 1} ⊂ GL(n) form subgroups of the general linear group, i.e. they are subsets of GL(n) that also form groups. Next we define a field, a structure which is a bit more complex than a group, since its definition involves two binary operations instead of one, together with more axioms.

Definition 2 (Field). A field is a set F together with two binary operations, addition + : F × F → F and multiplication · : F × F → F, that satisfies the following axioms for all a, b, c ∈ F. 1. Commutativity of addition: a + b = b + a 2. Associativity of addition:(a + b) + c = a + (b + c) 3. Additive identity: there exists an element 0 ∈ F such that a + 0 = a 4. Additive inverse: there exists an element −a ∈ F such that a + (−a) = 0 5. Commutativity of multiplication: a · b = b · a 6. Associativity of multiplication:(a · b) · c = a · (b · c) 7. Multiplicative identity: there exists an element 1 ∈ F such that 1 · a = a 8. Multiplicative inverse: for a 6= 0, there exists an element denoted a−1 such that a−1 · a = 1 9. Distributivity: a · (b + c) = (a · b) + (a · c) Notice that this definition can be summarized by saying that a field is an abelian group under addition with additive identity denoted 0, and that the nonzero elements form an abelian group under multiplication, with multiplicative identity denoted 1. The multiplication also needs to be distributive over the addition. Example 2. Next, we provide some examples of fields.

5 1. The rational numbers Q, the real numbers R and the complex numbers C together with regular addition and multiplication all form a field. The reader should be comfortable with the notion of a vector space, but we include the definition for the sake of completeness. Definition 3 (Vector space). A vector space over a field F is a set V together with two binary operations, addition + : V × V → V and scalar multiplication · : F × V → V that satisfies the following axioms for all x, y, z ∈ V and a, b ∈ F.1 1. Commutativity of addition: x + y = y + x 2. Associativity of addition:(x + y) + z = x + (y + z) 3. Additive identity: there exists a vector 0 such that 0 + x = x 4. Additive inverse: there exists a vector −x such that −x + x = 0 5. First distributive law:(a + b) · x = a · x + b · x 6. Second distributive law: a · (x + y) = a · x + a · y

7. Multiplicative identity: 1 · x = x, where 1 denotes the multiplicative identity in F 8. Relation to ordinary multiplication:(ab) · x = a · (b · x) Example 3. Again, we include some examples, also for the sake of completeness. 1. One of the most simple examples of a vector space is R3, the space in which we live, consisting of 3-tuples with vector addition and scalar multiplication deﬁned coordinate wise.

2. More generally, by letting F be an arbitrary field we can construct the set Fn, where n is some positive whole number, of n-tuples of elements of F. Next we define a special kind of vector space with additional structure. Definition 4. Let A be a vector space over a field F equipped with an additional binary operation × : A × A → A. We call A an algebra if it satisfies the following for all x, y, z ∈ A and a, b ∈ F. 1. Right distributivity:(x + y) × z = x × z + y × z 2. Left distributivity: x × (y + z) = x × y + x × z 3. Compatibility with scalars:(ax) × (by) = (ab)(x × y) A substructure is a subset of some algebraic structure, that itself is the same type of structure. Examples of substructures are 1. Subgroups, a subset of a group that is a group 2. Subfield, a subset of a field that is a field 3. Subspace, a subset of a vector space that is a vector space 4. Subalgebra, a subset of an algebra that is an algebra For algebras, we also have a more strict class of substructures: a left ideal of an algebra A is a subalgebra I ⊂ A such that aI = {ax : x ∈ I} ⊂ I for all a ∈ A. Analogously, a right ideal of A is a subalgebra I ⊂ A such that Ia = {xa : x ∈ I} ⊂ I for all a ∈ A.A two-sided ideal is an ideal that is both a left and a right ideal. For an algebra A we need to consider an ideal, and not just a subalgebra, I to define the quotient algebra A/I, or else we might run into problems. A central concept in the study of algebraic structures is the concept of homomorphisms. A homomorphism is a structure-preserving map between two algebraic structures of the same type, in the sense that is preserves the operations of the structures. We will define group homomorphisms and vector space homomorphisms, but it should be fairly clear how the concept carries over to the case of fields and algebras.

1 We will only consider the case where the ﬁeld F is equal to the complex numbers C.

6 Deﬁnition 5 (Group homomorphism). Let (G, ?) and (H, ) be groups. A function f : G → H is called a group homomorphism if f(a ? b) = f(a) f(b) for all a, b ∈ G.

Deﬁnition 6 (Vector space homomorphism). Let V and W be two vector spaces over the same ﬁeld F.A function f : V → W is called a vector space homomorphism (or linear map) if

f(u + v) = f(u) + f(v) f(cu) = cf(u) for all u, v ∈ V and c ∈ F. The set of all homomorphisms between two algebraic structures A and B is denoted Hom(A, B). A bijective homomorphism is called an isomorphism, and the set of all isomorphisms from A to B is denoted Iso(A, B). A homomorphism from an object to itself is called an endomorphism, and the set of all endomor- phisms on A is denoted End(A). A bijective endomorphism is called an automorphism, and the set of all automorphisms on A is denoted Aut(A). In conclusion, we have that

1. Hom(A, B) = {f : A → B : f is a homomorphism} 2. Iso(A, B) = {f ∈ Hom(A, B): f is bijective} 3. End(A) = Hom(A, A) 4. Aut(A) = Iso(A, A).

If there exists an isomorphism between two algebraic structures A and B, we say that they are isomorphic, and denote this as A =∼ B. In some sense, two isomorphic structures are the same. Even if they look diﬀerent, they behave in the same way. As an analogy, a regular chess board and a themed one are practically the same. They represent the same game and behave in the same way, even though they might look diﬀerent. A monopoly board, however, does not behave in the same way as a chess board.

7 3 Linear Algebra

3.1 Basic concepts In this section we will simply repeat some basic deﬁnitions and results from linear algebra that should be familiar to the reader. We start by describing a way to build new vector spaces with the help of old ones.

Deﬁnition 7. Let V amd W be two vector spaces over the same ﬁeld F. The external direct sum of V and W , denoted V ⊕ W , is the set of pairs (v, w), with v ∈ V and w ∈ W , with the following operations.

(v, w) + (v0, w0) = (v + v0, w + w0) c(v, w) = (cv, cw) for all v, v0 ∈ V , w, w0 ∈ W and c ∈ F .

Note the the direct sum of two vector spaces is yet another vector space.2

Deﬁnition 8. Let V be a vector space, and W1,W2 ⊂ V be subspaces of V such that any vector v ∈ V can be written uniquely as v = w1 + w2 where w1 ∈ W1 and w2 ∈ W2. Then V is said to be the internal direct sum of W1 and W2.

Theorem 1. Let V be a vector space that is equal to the internal direct sum of two subspaces W1,W2 ⊂ V . Then V is isomorphic to the external direct sum of W1 and W2.

Proof. We construct a linear map L : W1 ⊕W2 → V by L(w1, w2) = w1 +w2. It is clear that this map is both surjective and injective from the deﬁnition of the internal direct sum. It is therefore also an isomorphism. Since the concept of internal and external direct sum are isomorphic and therefore practically the same, we will write W1 ⊕ W2 for both the internal and external direct sum of W1 and W2. Suppose that V is a vector space and that W ⊂ V is a subspace of V . A subspace W 0 ⊂ V is called a complement of W in V if V = W ⊕ W 0. The mapping p which sends each v ∈ V to its component w ∈ W is called the projection of V onto W associated with the decomposition V = W ⊕ W 0. It follows that

range(p) = W p(w) = w, for all w ∈ W.

If p ∈ End(V ) is any linear operator on V satisfying the two properties above, we see that V is the direct sum of W and the kernel of p. That is, V = W ⊕ ker(p). We therefore have a bijective correspondence between the projections of V onto W and the complements of W in V .

Proposition 1. Let V be a vector space, and let V1,V2 ⊂ V be subspaces such that V = V1 ⊕ V2. Let L : V → W be an injective linear transformation to some other vector space W . Then

L(V ) = L(V1) ⊕ L(V2).

Proof. It is clear that any element w in L(V ) can be written as w = w1 + w2, where w1 ∈ L(V1) and w2 ∈ L(V2). We show that L(V1) ∩ L(V2) = {0}. Let a ∈ L(V1)∩L(V2). Then there exists a v1 ∈ V1 such that a = L(v1) and a v2 ∈ V2 such that a = L(v2). This means that L(v1) = L(v2). Since L is injective it must be the case that v1 = v2, but V1 ∩ V2 = {0}, so v1 = v2 = 0, and therefore also a = 0. Next we examine how we can remove a subspace from a space.

2In fact, the direct sum is the coproduct in the category of vector spaces.

8 Definition 9 (Quotient space). Let V be a vector space over a field F, and let W be subspace of V . We define the coset of x in W as x + W = {x + w : w ∈ W }. The quotient space V/W is defined to be the space all cosets

V/W := {x + W : x ∈ V }

with addition and scalar multiplication deﬁned by

1.( x + W ) + (x0 + W ) = (x + x0) + W 2. c(x + W ) = cx + W

for all x, x0 ∈ V and c ∈ F. It is easy to check that these operations are well-deﬁned, and that dim(V/W ) = dim(V )/ dim(W ).

Definition 10 (Norm). Let the field F either be the real numbers R or the complex numbers C, and let V be a vector space over F.A norm of V is a non-negative scalar-valued function || · || : V → [0, ∞) that satisfies the following properties for all x, y ∈ V and a ∈ F. 1. Triangle equality: ||x + y|| ≤ ||x|| + ||y||

2. Absolute scalability: ||ax|| = |a| · ||x|| 3. Positive-deﬁniteness: ||x|| = 0 ⇐⇒ x = 0

A vector space equipped with a norm is called n normed product space.

Definition 11 (Inner product). Let the field F either be the real numbers R or the complex numbers C, and let V be a vector space over F. An inner product on V is a map h·, ·i : V × V → F that satisfies the following axioms for all x, y, z ∈ V and a ∈ F 1. Conjugate symmetry: hx, yi = hy, xi, where the bar denotes the complex conjugate. 2. Linearity in the second argument:

hx, ayi = ahx, yi hx, y + zi = hx, yi + hx, zi

3. Positive-deﬁniteness: hx, xi ≥ 0 hx, xi = 0 ⇐⇒ x = 0

A vector space equipped with an inner product is called an inner product space. Note that every inner product space V is also a normed vector space by deﬁning the induced norm ||x|| := hx, xi for all x ∈ V . However, the converse is not always true; not every norm can be produced by an inner product. Let V be an inner product space equipped with the inner product h·, ·i, and let W ⊂ V be a subspace of V . The subspace of all element which are orthogonal to all elements in W is called the orthogonal complement of W in V , and is denoted by W ⊥. Written out, we have that

W ⊥ = {x ∈ V : hx, yi = 0 for all y ∈ W }.

It is clear that W ⊥ is a complement of W in V , that is, V = W ⊕ W ⊥. Suppose that we have two inner product spaces V and W , both over the same ﬁeld F, endowed with inner products h·, ·iV and h·, ·iW respectively. There is a natural way to use the (external) direct sum to

9 create a new inner product space V ⊕ W , by assigning it the inner product h·, ·iV ⊕W : V ⊕ W × V ⊕ W → F deﬁned by 0 0 0 0 h(v, w), (v , w )iV ⊕W = hv, v iV + hw, w iW .

This is then naturally extended to the sum of multiple product spaces. Notice that if V = V1 ⊕ V2 is the internal direct sum of two subspaces V1 and V2, each equipped with a special inner product, then the inner product deﬁnition is the same as saying

0 0 0 hv, v iV = hv1, v1iV1 + hv2, v2iV2

0 0 0 0 for all v, v ∈ V , where v1, v1 ∈ V1 and v2, v2 ∈ V2 is the projection of v and v onto V1 and V2 respectively.

Deﬁnition 12. Let V and W be two ﬁnite dimensional inner product spaces with inner products h·, ·iV and h·, ·iW respectively, and let A : V → W be a linear transformation between these spaces. The adjoint of A is a linear transformation A† : W → V such that

† hx, AyiW = hA x, yiV for all x ∈ W and y ∈ V . It is possible to show that the adjoint of a linear transformation always exists, and that it is unique.3 However, we will not prove it.

Theorem 2. Let V and W be two ﬁnite dimensional inner product spaces, and let A : V → W be a linear transformation. The relationship between the image of A and the kernel of its adjoint is given by

ker(A†) = A(V )⊥.

Proof. We have the following equivalences.

x ⊥ A(V ) ⇐⇒ hx, AyiW = 0, ∀y ∈ V † ⇐⇒ hA x, yiV = 0, ∀y ∈ V ⇐⇒ A†x = 0.

† The last equivalence is true because if hA x, yiV = 0 for all y ∈ V , then it it is also zero for the specific case † of y = A x. Definition 13 (Banach space). Let the field F either be the real numbers R or the complex numbers C, and ∞ let V be a vector space over F equipped with a norm || · ||. A sequence {xi}i=1 in V is said to be a Cauchy sequence, if for every positive real number > 0 there exists a positive integer N such that for all positive integers m, n ≥ N it holds that ||xm − ym||. ∞ V is said to be a Banach space if for every Cauchy sequence {xi}i=1 in V there exists an element x in V such that lim ||xn − x|| = 0. n→∞ Recall that every inner product space is also a normed space under the induced norm. Definition 14 (Hilbert space). An inner product space V equipped with the inner product h·, ·i that is also a Banach space with respect to the induced norm ||x|| = hx, xi for all x ∈ V is called a Hilbert space. We state the following proposition without proving it. Proposition 2. Every finite-dimensional normed vector space is a Banach space. Note that this also means that every finite dimensional inner product space is a Hilbert space.

3This is not necessarily true if we allow the vector spaces to be inﬁnite-dimensional.

10 Example 4. Let V and W be two ﬁnite-dimensional normed vector spaces over R or C with norms || · ||V and || · ||W respectively, and let Hom(V,W ) denote the vector space of all linear functions from V to W . We can equip this space with a norm called the operator norm. Let T ∈ Hom(V,W ) be a linear function from V to W , and deﬁne the operator norm of T as

||Av||W ||T ||op = sup . v∈V :v6=0 ||v||V

Since both V and W are ﬁnite-dimensional, so is Hom(V,W ).4 By Proposition 2, Hom(V,W ) equipped with the operator norm is therefore a Banach space.

3.2 Multilinear algebra In this section our main goal will be to define a new way to build new vector spaces out of old ones, namely the tensor product. The definition of the tensor product follows the one given in [Roman, 2008]. We begin by defining the notion of a bilinear map.

Definition 15. Let U, V and W be vector spaces over a field F, and let U ×V denote the Cartesian product of U and V as sets.5 A function f : U × V → W is said to be bilinear if it is linear in both variables separately, that is if f(αu + βu0, v) = αf(u, v) + βf(u0, v) and f(u, αv + βv0) = αf(u, v) + βf(u, v0) for all u, u0 ∈ U, v, v0 ∈ V and α, β ∈ F. We will now define a type of universality that will help us define the tensor product.

Deﬁnition 16. Let U × V be the Cartesian product of two vector spaces over a ﬁeld F, as sets. A pair (T, t : U × V → T ), where T is a vector space, is said to be universal for bilinearity if for any vector space W , and every bilinear map f : U × V → W , there is a unique linear transformation τ : T → W such that

f = τ ◦ t.

This can be summarized by the following commutative diagram.

U × V t T

∃!τ f W

Definition 17. Let V and W be vector spaces over a field F. Any universal pair for bilinearity (T, t : U × V → T ) is called the tensor product of V and W . The vector space T is denoted by V ⊗ W , and the image t(v, w) of any two vectors v ∈ V , w ∈ W is denoted v ⊗ w. This definition of the tensor product is rather abstract, but we will provide a more concrete, but coordinate dependant, construction of the tensor product. Again, let V and W be vector spaces, and let B = {ei : i ∈ I} and D = {fi : i ∈ J} be bases for V and W respectively, for some index sets I and J. A bilinear map t on V × W is uniquely determined by its value P P on the ”basis pairs” (ei, fj): if V 3 v = i∈I αiei and W 3 w = j∈J βJ fj then X X X X t(u, v) = t( αiei, βJ fj) = αiβjt(ei, fj) i∈I j∈J i∈I j∈J

4 This can be seen by identifying Hom(V,W ) with Mdim(V )×dim(W ), the space of all dim(V ) × dim(W )-matrices. 5It is important to note that U × V is just the Cartesian product of sets, and that we therefore do not have any algebraic structure on U × V . For example, expressions like (x, y) + (z, w) and α(x, y) are meaningless.

11 For each of the images t(ei, fj) we now invent a new symbol and write ei ⊗ fj and define T to be the vector space with the basis E = {ei ⊗ fj : i ∈ I, j ∈ J}. One can indeed verify that this construction satisfies the definition of the tensor product. If g : U × V → W is bilinear, then the function τ : T → W defined by

τ(ei ⊗ fj) = g(ei, fj) is a unique linear map satisfying the deﬁnition. An obvious consequence from the previous construction is the following. Proposition 3. For two ﬁnite-dimensional vector spaces V and W , we have that

dim(V ⊗ W ) = dim(V ) · dim(W ).

We can now define some important algebras, starting with the tensor algebra T (V ): an algebra on a vector space V over a field F with multiplication of elements being the tensor product ⊗. It is defined as ∞ M T (V ) = V ⊗n, n=0 where V ⊗n = V ⊗ V ⊗ · · · ⊗ V | {z } n times and V ⊗0 := F. Now let I ⊂ T (V ) be the ideal generated by elements of the form v ⊗ v for v ∈ V , that is the smallest ideal of T (V ) containing v ⊗ v for v ∈ V . We then define the exterior algebra of V , denoted Λ(V ), to be the quotient algebra Λ(V ) := T (V )/I. The exterior product of two elements of Λ(V ) is the product induced by the tensor product ⊗ on T (V ). More explicitly, first define the canonical surjection

π : T (V ) → Λ(V ) v 7→ v + I.

Then for any a, b ∈ Λ(V ) there are v, w ∈ T (V ) such that a = π(v) = v + I and b = π(w) = w + I. The exterior product of a and b is then deﬁned by

a ∧ b := π(v ⊗ w) = v ⊗ w + I.

It follows from the definition of the exterior product that it is anticommutative, which in turn implies that a ∧ a = 0 for all a ∈ Λ(V ). k The subspace Λ (V ) ⊂ Λ(V ) spanned by elements of the form v1 ∧ v2 ∧ · · · ∧ vk, where each vi ∈ Λ(V ), n is called the k-th exterior power of V . Suppose the V is finite dimensional of dimension n, and that {ei}i=1 k is a basis for V , then the set {ei1 ∧ ei2 ∧ · · · ∧ eik : 1 ≤ i1 < i2 < ··· < ik ≤ n} is a basis for Λ (V ). This shows that the dimension of Λk(V ) is n dim(Λk(V )) = . k Lastly we define the symmetric algebra, Sym(V ), by letting J ⊂ T (V ) be the ideal generated by elements of the form v ⊗ w − w ⊗ v, for all v ∈ V and w ∈ W .

Sym(V ) := T (V )/J.

Just as with the exterior power, we can construct the k-th symmetric power of V , by

Symk(V ) = T k(V )/J,

12 where T k(V ) = V ⊗k. We do not use any special symbol for multiplication in this algebra; instead we just type s1s2 for the product of two elements s1, s2 ∈ Sym(V ). n k If we again suppose that {ei}i=1 is a basis for V , then a basis for Sym (V ) is given by {ei1 ei2 ··· eik : 1 ≤ i1 ≤ i2 ≤ · · · ≤ ik ≤ n}. This shows that the dimension of the k-th symmetric power of a ﬁnite-dimensional vector space V is n + k − 1 dim(Symk(V )) = , k where n = dim(V ). Now deﬁne φ : V ⊗ V → Sym2(V ) ⊕ Λ2(V ) via

φ(ei ⊗ ej) = (ei ⊗ ej + J, ei ⊗ ej + I).

One can show that this deﬁnes an isomorphism, which gives that V ⊗ V =∼ Sym2(V ) ⊕ Λ2(V ). An interesting property of the tensor product that we will use later on is that it distributes over the direct sum. The proof of the following proposition is taken out of [Dummit and Foote, 2004]. Proposition 4. Let U, V and W be vector spaces. Then there are unique isomorphisms so that

(U ⊕ V ) ⊗ W =∼ (U ⊗ W ) ⊕ (V ⊗ W ) U ⊗ (V ⊕ W ) =∼ (U ⊗ V ) ⊕ (U ⊗ W ).

Proof. We start with the ﬁrst isomorphism. Consider the map

(U ⊕ V ) × W → (U ⊕ W ) ⊗ (V ⊗ W ) ((u, v), w) 7→ ((u ⊗ w), (v ⊗ w)).

This map is clearly bilinear, so by the universal property of the tensor product it induces a unique linear transformation τ :(U ⊕ V ) ⊗ W → (U ⊕ W ) ⊗ (V ⊗ W ) ((u, v) ⊗ w) 7→ ((u ⊗ w), (v ⊗ w)). We now want to deﬁne a map in the other direction. Consider the maps

U × W → (U ⊕ V ) ⊗ W (u, w) 7→ (u, 0) ⊗ w and V × W → (U ⊕ V ) ⊗ W (v, w) 7→ (0, v) ⊗ w.

Again, both these maps are bilinear, so they both induce unique linear maps φ1 and φ2 described by the commutative diagram U ⊗ W V ⊗ W φ1 φ2

(U ⊕ V ) ⊗ W where φ1(u ⊗ w) = (u, 0) ⊗ w and φ2(v ⊗ w) = (0, v) ⊗ w. Therefore, the map

τ ∗ :(U ⊕ W ) ⊗ (V ⊕ W ) → (U ⊕ V ) ⊗ W

(u ⊗ w1, v ⊗ w2) 7→ φ1(u, w1) + φ2(v, w2)

= (u, 0) ⊗ w1 + (0, v) ⊗ w2 is a well defined linear transformation. In fact, τ and τ ∗ are inverses of each other, which proves the first isomorphism in the theorem. The other one is proved analogously. Remark. Notice the power of the coordinate-free definition of the tensor product by the universal property.

13 Now suppose that we have two linear maps between vector spaces, L1 : V1 → W1 and L2 : V2 → W2. It is natural to ask if we can build a new linear map from L1 and L2, with inputs from V1 ⊗ V2 and outputs in W1 ⊗ W2. This can be done by deﬁning the tensor product of linear maps:

L1 ⊗ L2 : V1 ⊗ V2 → W1 ⊗ W2

v1 ⊗ v2 7→ L1(v1) ⊗ L2(v2).

Let L : V → W be a linear map, then in analogy with the tensor product of vector spaces we deﬁne the linear map L⊗n as L⊗n : V ⊗n → W ⊗n

v1 ⊗ v2 ⊗ · · · ⊗ vn 7→ L(v1) ⊗ L(v2) ⊗ · · · ⊗ L(vn).

n The map L⊗n can then be used to construct induced linear maps L∧n :Λn(V ) → Λn(W ) and LS : Symn(V ) → Symn(W ) by

∧n ⊗n L (v1 ∧ v2 ∧ · · · ∧ vn) = L(v1) ∧ L(v2) ∧ · · · ∧ L(vn) = L (v1 ⊗ v2 ⊗ · · · ⊗ vn) + I

and Sn ⊗n L (v1v2 ··· vn) = L(v1)L(v2) ··· L(vn) = L (v1 ⊗ v2 ⊗ · · · ⊗ vn) + J, where I and J is the relevant ideals from the deﬁnition of the two algebras.6,7

6It is clear that if dim(V ) = n, then dim(Λn(V )) = 1, i.e. Λn(V ) is one-dimensional. Let L ∈ End(V ) be an operator on V . Since Λn(V ) is one-dimensional, the operator L∧n corresponds to a scalar. After some closer analysis, one can conclude that this scalar corresponds to the usual determinant of L, that is, L∧n = det(L) · id. This can be taken as the deﬁnition of the determinant of a linear operator, thereby giving a coordinate-free deﬁnition. 7This can all be made more general and natural in the language of category.

14 4 Topology

4.1 Point-set topology 4.1.1 Topological spaces and continuity Continuity is the foundation of topology, and as a topological property it only relies on the concept of open sets. We will begin the preliminaries with the definition of a topological space. The following section is mostly based on the material presented in [Basener, 2006]. Definition 18. Let X be a set, and let τ be a collection of subsets of X. τ is a topology on X if it satisfies the following. 1. Any (finite or infinite) union of members of τ belong to τ 2. Any finite intersection of members of τ is in τ 3. Both X and the empty set ∅ is in τ The members of τ are called open sets, and the set X together with the topology τ forms a topological space. This is the topological definition of open sets. When we want to emphasize the specifics of a topology, we will write a topological space as the tuple (X, τ), otherwise we will just refer to the topological space by the symbol of the set (in this case X). One often defines the open sets of Rn with the help of open balls: a subset O ⊂ Rn is defined to be open n if there for every x ∈ O exists an r > 0 such that Br(x) = {y ∈ R : |y − x| < r} ⊂ O. One can show that this definition does indeed satisfy the three axioms in definition 18. We will now define the notion of continuity in a topological setting.

Deﬁnition 19. Let (X, τX ) and (Y, τY ) be two topological spaces. A function f : X → Y is said to be continuous if the inverse of every open set is open. That is if

−1 f (O) = {x ∈ X : f(x) ∈ O} ∈ τX , for all O ∈ τY .

This definition is a generalization of the regular -δ definition often used in calculus. Example 5. We will provide some examples of topological spaces. 1. Let X be a set, and define τ = P(X), where P(X) denotes the power set of X. This is called the discrete topology on X, and a space with the discrete topology is called a discrete space. Any function from a discrete space to another topological space is continuous. 2. A set X together with the topology τ = {X, ∅} is called an indiscrete space

3. Let (X, τX ) be a topological space, and let Y be a subset of X. We can deﬁne a natural topology on Y called the subspace topology, in which every subset of Y is open if it is the intersection of an open subset of X with Y . That is, τY = {Y ∩ O : O ∈ τX }.

4. Let (M, d) be a metric space. The metric topology is the topology induced by the metric by deﬁning open sets to be the union of open balls

Br(x) = {y ∈ X : d(x, y) < r},

where x ∈ X and r > 0. With this general deﬁnition of continuity, it is very easy to prove that the composition of two continuous functions is continuous. Proposition 5. Let X, Y and Z be topological spaces, and let f : X → Y and g : Y → Z be two continuous functions. Then the composition g ◦ f is also continuous.

15 Proof. Let O ⊂ Z be an open subset of Z. Then g−1(O) is open, hence also f −1(g−1(O)) is open. Therefore −1 −1 −1 we have that (g ◦ f) (O) = f (g (O)) ⊂ X is open. Next we define closed sets in a topological setting. Definition 20. Let X be a topological space. A subset A of X is closed if its complement Ac = X − A is open. Proposition 6. Let X and Y be topological spaces. A function f : X → Y is continuous if and only if f −1(A) ⊂ X is closed for all closed subsets A of Y . Proof. First let f : X → Y be continuous, and let A be a closed subset of Y , i.e. Ac is open. Then f −1(Ac) is open, hence f −1(Ac)c = f −1(A) is closed. Now let f : X → Y be a function such that f −1(A) ⊂ X is closed for all closed subsets A of Y , and let −1 c −1 c −1 O be an open subset of Y . We know that f (O ) = f (O) is closed, hence f (O) is open. Definition 21. Let X be a topological space, and x be a point in X. We say that a subset N ⊂ X is a neighborhood of x if there is an open set O ⊂ X such that

x ∈ O ⊂ N.

We call N an open neighborhood if N itself is open. Intuitively we see a neighborhood of a point x as all the points ”near” x. Notice, however, that the concept of ”near” does not really make sense in an arbitrary topological space, where no metric is defined. Before we move on to our next definition, we need to define a special function. Let A1,A2,...,An be Qn sets. The ith canonical projection is the map pi : j=1 Aj → Ai given by pi(x1, x2, . . . , xn) = xi. Next we define a topology which we typically endow Cartesian products of topological spaces, with the help of this canonical projection.

Definition 22. Let X1,X2,...,Xn be topological spaces. The product topology on the Cartesian product Qn X = i=1 Xi is the topology in which a subset O ∈ X is open if and only if pi(O) is open for each i, where pi : X → Xi is the canonical projection. Definition 23. A topological space X is said to be a Hausdorff space if for any two points x and y in X, there exist open neighborhoods Nx and Ny of x and y respectively, such that Nx ∩ Ny = ∅. Observe that every metric space is a Hausdorff space.

4.1.2 Compactness Here we will generalize the notion of compactness, which should be a concept the reader should recognize from calculus. The deﬁnition is a rather long one, but we will dive into it. Deﬁnition 24. Let (X, τ) be a topological space and let E ⊂ X be a subset of X. A collection C of open subsets of X is said to be an open cover of E if the union of all elements in C contain E, that is [ E ⊂ A. A∈C

A subcollection of an open cover that is itself a cover is called a subcover. That is, if C0 ⊂ C is a cover of E, then C0 is a subcover of C. A subcover is called a finite subcover if it contains only finitely many open sets. A subset F of X is said to be compact if every open cover of F has a finite subcover. The topological space (X, τ) is itself compact if every open cover of X has a finite subcover. We include a famous theorem, for the sake of completeness. However, we do not prove it.

Theorem 3 (Heine-Borel Theorem). A subset of Euclidean space Rn is compact if and only if it is closed and bounded.

16 Theorem 4. A closed subset of a compact space is compact.

Proof. Let X be a compact space and let A be a closed subset of X. Let C = {Oα} be a collection of open sets in X that covers A. Observe that X − A is open. The collection of open sets in X consisting of the sets in C together with the set X − A is an open cover of X. Since X is compact, this cover has a finite subcover 0 {O1,O2,...,On,X − A}. Then C = {O1,O2,...,On} is a finite subfamily whose union contains A. Although R is not compact, every point in R has a compact neighbourhood. We can generalize this property to arbitrary topological spaces. Definition 25. A topological space X is locally compact if every point in X has a compact neighbourhood.

4.2 Topological groups A topological group is a set with both topological and algebraic properties that fit nicely together. Definition 26. A topological group is a group (G, ?) that is also a Hausdorff space such that the group operations of multiplication and taking inverses are continuous. That is, the functions

µ : G × G → G (a, b) 7→ ab

and ι : G → G a 7→ a−1, where G × G is viewed as a topological space with the product topology, are continuous. The concept of homomorphisms carries over to the case of topological groups, however, we now require the function to be continuous. Example 6. Some examples of simple topological groups are the following. 1. Any group can be made into a topological group trivially by equipping it with the discrete topology. A group equipped with the discrete topology is called a discrete group.

2. Rn with the usual Euclidean topology (i.e. the topology induced by the ordinary Euclidean metric) is a topological group under addition. Proposition 7. The matrix group GL(n) is a topological group.

Proof. First recall that a function f : Rn → Rn given by

f(x) = (f1(x), f2(x), . . . , fn(x))

is continuous if and only if each fi is continuous. Now let A, B ∈ GL(n) be matrices. The ijth entry of the Pn product AB is given by the polynomial k=1 aikbkj. Thus, matrix multiplication is continuous. By Cramer’s rule (see [Lay, 2006]), the ijth entry of A−1 is given by

(ijth cofactor of A) , det(A)

which shows that taking inverses is continuous. We now take O(n),SO(n) ⊂ GL(n) to be subsets of GL(n) equipped with the subspace topology. This makes multiplication and taking inverses continuous in O(n) and SO(n) automatically. Thus we have the following corollary.

Corollary 1. The groups O(n) and SO(n) are topological groups.

17 Next we will consider compact groups, i.e. topological groups that are compact as topological spaces. It is easy to see that GL(n) is not compact. For example, the matrix cI is in GL(n) for all c ∈ R − {0} and 2 hence GL(n) is not bounded as a subsets of Rn . However, we have the following proposition. Proposition 8. The topological groups O(n) and SO(n) are compact.

Proof. To prove that O(n) is compact, we need to show that it is closed and bounded as a subset of Rn×n. For a matrix A ∈ O(n), the ijth entry of the equation AAT = I is given by

n X aikajk = δij. (1) k=1

For each i, j, let Sij denote the set of solutions to the polynomial equation above. Each Sij is closed, since the continuous inverse image of a point is closed. 8 Thereby O(n) is closed, since it is the intersection of all Tn Sij, O(n) = i,j=1 Sij. To show that O(n) is bounded, we observe that, putting i = j in equation (1) gives

n X aikaik = 1 k=1

which implies that |aij| ≤ 1. The group SO(n) is a closed subset of O(n), since it is the preimage of 1 under the continuous function det : O(n) → R, and hence also compact. A topological group that is locally compact as a topological space is called a locally compact group. We can also give other algebraic structures topological qualities.

Deﬁnition 27. Let the ﬁeld F either be the real numbers R or the complex numbers C.A topological vector space is a vector space V over F that is endowed with a topology such that vector addition + : V × V → V and scalar multiplication · : F × V → V are continuous functions.

8 n×n This comes from the fact the R is a Hausdorﬀ space, where every singleton set is closed, which can be proven as follows: let A be a Hausdorﬀ space, and let x ∈ A be a point in A. Then for every point y distinct from x there is an open set Oy containing y but not x. Then c [ {x} = Oy y∈A−{x} is open, hence {x} is closed.

18 5 Measure Theory

In this section we will cover some very basic measure theory and the construction of the Lebesgue integral. We will also deﬁne the Haar measure and state some important theorems; however, we will not provide proofs for the sake of brevity.9

Deﬁnition 28 (σ-algebra). Let S be some set, and let P(S) denote its power set. A subset S ⊂ P(S) is called a σ-algebra if it satisﬁes the following three properties.

1. S is in S

2. If A is in S , then so is its complement Ac ∞ S∞ 3. If each Ai in (Ai)i=1 is in S , then the countable union i=1 Ai is also in S . Notice that it follows from property 1. and 2. that ∅ is also in S . The elements of a σ-algebra are called measurable sets. If S is a set and S is a σ-algebra over S, then the tuple (S, S ) is called a measurable space.

Definition 29 (Measurable function). Let (S, S ) and (T, T ) be two measurable spaces. A function f : X → Y is called a measurable function if the inverse image of every measurable set is measurable. That is, if f −1(E) = {x ∈ S : f(x) ∈ E} ∈ S , for all E ∈ T . If f : S → T is a measurable function we will often write f :(S, S ) → (T, T ) to emphasize the dependency on the measurable spaces. Note the resemblance between the definition of a measurable function and a topologically continuous function. Next we define a way to measure the size of different sets, by the way of a function called a measure.

Deﬁnition 30 (Measure space). Let (S, S ) be a measurable space. A function µ : S → [−∞, ∞] is called a measure if it satisﬁes the following properties.

1. For all E in S , it holds that µ(E) ≥ 0 2. The value of the empty set is zero: µ(∅) = 0

3. For all countable collections of {Ei}i∈I , where I is some countable index set, of pairwise disjoint sets in it holds that S ! [ X µ Ei = µ(Ei) i∈I i∈I

If µ is a measure, the triple (S, S , µ) is called a measure space. A measure µ such that µ(S) = 1 is called a probability measure.

If (S, S , µ) is a measure space, a property P is said to hold µ-almost everywhere if there exists a set N ∈ S such that µ(N) = 0 and all x ∈ S − N have the property P .

5.1 Construction of the Lebesgue integral After all these deﬁnitions, one may wonder how all of this actually connects to the theory of integration. This is all answered by the Lebesgue integral, a construction that extends the ordinary Riemann integral to a larger class of functions. This version of the integral can be constructed in several ways, but we will construct it using so called simple functions, which are linear combinations of indicator functions:

9The main point of this section is to state that there is a unique way to integrate on a compact group.

19 Deﬁnition 31 (Indicator function). Let S be a set, and A ⊂ S be a subset of S. The indicator function of A is the function 1A : S → {0, 1} deﬁned by ( 1, if x ∈ A 1A(x) := 0, if x∈ / A.

The idea is then to start with a measure space (S, S , µ) and define the integral of a measurable function f : S → C with respect to the measure µ. Let S ∈ S be some measurable set. We start by assigning a value to the integral of an indicator function 1A, where A ∈ S is some measurable set, as Z 1A(x)dµ(x) := µ(A ∩ S). S Pn As previously mentioned, a linear combination of indicator functions i=1 ai1Ai , where ai are real numbers and the sets Ai ∈ S are measurable sets, is called a simple function. We extend the definition of the integral above to non-negative simple functions. When all the coefficients ai are non-negative, we set

n ! n n Z X X Z X ai1Ai (x) dµ(x) = ai 1Ai (x)dµ(x) = aiµ(Ai ∩ S), S i=1 i=1 S i=1 where the convention 0 · ∞ = 0 is used. We can now deﬁne the Lebesgue integral for arbitrary non-negative measurable functions f : S → R+ as Z Z f(x)dµ(x) := sup φ(x)dµ(x) : 0 ≤ φ ≤ f, φ is simple . S S

To handle the case of arbitrary real-valued functions f : S → R, we decompose f as f = f + − f − where ( f(x), if f(x) > 0 f +(x) = 0, otherwise ( −f(x), if f(x) < 0 f −(x) = 0, otherwise

R + R − We say that the Lebesgue integral of f exists if at least one of S f (x)dµ(x) and S f (x)dµ(x) is finite. In this case we define the integral of f to be Z Z Z f(x)dµ(x) := f +(x)dµ(x) − f −(x)dµ(x). S S S For complex valued functions f = g + ih we simply define Z Z Z f(x)dµ(x) := g(x)dµ(x) + i h(x)dµ(x). S S S As it turns out, this integral is equivalent to the Riemann integral for Riemann integrable functions.

5.1.1 Lp-spaces In this subsection we will construct Lp-spaces rigorously using the Lebesgue integral. We begin by letting (S, S , µ) be a measure space, and p be a real number such that p ∈ [0, ∞). Let Lp(S, S , µ) denote the vector space of all measurable functions f : S → C such that

1 Z p p ||f||p := |f(x)| dµ(x) < ∞ S

20 together with pointwise function multiplication. This space, when equipped with the function || · ||p, forms a seminormed vector space.10 To make it a normed space, we take the quotient space with respect to the kernel of || · ||p. We therefore deﬁne

N = {f : f = 0 µ-almost everywhere} = ker(|| · ||p) in order to construct the Lp-space as a quotient

Lp(S, S , µ) = Lp(S, S , µ)/N.

In Lp(S, S , µ), two functions f and g are identiﬁed if f = g µ-almost everywhere. Lp(S, S , µ) is a Banach space for all p ∈ [1, ∞), and for the special case of p = 2, we can deﬁne an inner product compatible with the norm || · ||2 as Z hf, giL2 = f(x)g(x)dµ(x). S This shows that L2(S, S , µ) is also a Hilbert space. Notice that the elements of Lp(S, S , µ) are not functions, but cosets of the form f(x)+N. It is, however, usual to abuse notation and speak of the elements as functions. We will do this for the rest of this text.

5.2 Borel algebras and the Haar measure Our goal is now to define a special kind of measure that will be of particular interest: the Haar measure. But before we define it, we need to go over a few preliminary definitions and results. The definitions are taken from [Gleason, 2010]. Suppose that we have a set X, and a collection of subsets of X (a topology on X, for example). It is possible to generate a σ-algebra from the collection of subsets.

Deﬁnition 32 (Generated σ-algebra). Let X be a set, and let C and D be two collections of subsets of X. Then for any collection of subsets E of X, we deﬁne the σ-algebra generated by E to be the smallest σ-algebra that contains E . This σ-algebra is denoted σ(E ), and is constructed as the intersection of all σ-algebras that contain E \ σ(E ) := {A ⊂ P(X): A is a σ-algebra and E ⊂ A }

Deﬁnition 33 (Borel σ-algebra). Let (X, τ) be a topological space. The Borel σ-algebra on X is the smallest σ-algebra containing all open subsets of X. That is, it is the σ-algebra σ(τ) generated by τ. A subset A ⊂ X of X is called a Borel subset of X if A ∈ σ(τ). If (X, τ) is a topological space, and µ is a measure on the measurable space (X, σ(τ)), we will call the measure space (X, σ(τ), µ) a topological measure space. If (X, τ) is Hausdorﬀ, we will call a measure µ on (X, σ(τ)) a Borel measure and (X, σ(τ), µ) a Borel measure space. A measurable function on a Borel space is called Borel measurable.

Deﬁnition 34 (Regular measure). Let (S, S , µ) be a Borel measure space. Then µ is said to be a regular Borel measure if 1. If K ⊂ S is a compact subset of S, then µ(K) < ∞

2. If A ∈ S , then µ(A) = inf{µ(O): A ⊂ O,O is open}.11

3. If O ⊂ S is open, then µ(O) = sup{µ(K): K ⊂ O,K is compact}.12 10A seminormed vector space is a vector space V equipped with a norm || · ||, that does not need to satisfy the property ||v|| = 0 if and only if v = 0, for v ∈ V 11This is referred to as outer regularity. 12This is referred to as inner regularity.

21 We finally arrive at the definition of the Haar measure. Definition 35 (Haar measure). Let G be a topological group. A left (right) Haar measure on G is a nonzero regular Boreal measure µ on G such that µ(gA) = µ(A)(µ(Ag) = µ(A)) for all g ∈ G and measurable subsets A ⊂ G of G.

Now we state some existence and uniqueness theorems for the Haar measure on locally compact groups. Theorem 5 (Existence). Let G be a locally compact group. Then there exists a left Haar measure on G. Theorem 6 (Uniqueness). Let G be a locally compact group, and let µ and µ0 be two left Haar measures on G. Then µ = aµ0 for some positive real number a ∈ R+. This theorem tells us that the left Haar measure on a locally compact group G is essentially unique, in the sense that two left Haar measures only diﬀer by a positive multiplicative constant. Proofs can be found in [Gleason, 2010]. Using the theory of Lebesgue integration, one can deﬁne an integral for all Borel measurable functions f on a topological group G with respect to a Haar measure µ. Z f(g)dµ(g). G If µ is a left Haar measure on G, then we have that Z Z f(hg)dµ(g) = f(g)dµ(g) (2) G G for all h ∈ G. This is clear for indicator functions, since Z Z Z −1 1A(hg)dµ(g) = 1h−1A(g)dµ(g) = µ(h A) = µ(A) = 1A(g)dµ(g) G G G because of the left invariance. We will make use of this integral frequently.

22 6 Representation Theory

The theory presented in this section mainly follows [Serre, 1977].

6.1 Basic deﬁnitions

Deﬁnition 36 (Representation). Let V be a vector space over the ﬁeld over complex numbers C, let Aut(V ) denote the group of automorphisms on V , and suppose that G is a group.13 A representation of G in V is a homomorphism ρ : G → Aut(V ). In other words, we associate with each element g ∈ G an operator ρ(g) ∈ Aut(V ) such that that ρ(gh) = ρ(g) ◦ ρ(h), for all g, h ∈ G.

(We will often write ρg instead of ρ(g) to avoid cluttering with parentheses.) Observe that this implies that

ρ(e) = id, ρ(g−1) = ρ(g)−1,

where e ∈ G denotes the identity element in G, and id ∈ Aut(V ) is the identity function. When ρ is given, V is called a representation space of G, however, we will often abuse notation and language and call V itself a representation of G when there is no chance of confusion. In the rest of this section, we will only consider representations in finite-dimensional vector spaces. This is not too limiting, as one can often decompose an infinite-dimensional vector space into finite-dimensional subspaces. We will now define how we can view two different representations as being practically equal. Definition 37. Let ρ : G → Aut(V ) and φ : G → Aut(W ) be two representations of the same group G. These representations are said to be isomorphic if there exsists a vector space isomorphism T ∈ Iso(V,W ) that ”transforms” one representation into the other, that is, it satisfies the following identity.

T ◦ ρg = φg ◦ T, for all g ∈ G.

We denote two isomorphic representations by ρ =∼ φ. We will see that most notions for representations only depend on the equivalence classes of representations. Suppose that G is a group and that V is a vector space, and let ρ : G → Aut(V ) be a representation of a G in V . A vector subspace W ⊂ V is said to be G-invariant if it is stable under the action of G, that is, if ρg(w) ∈ W for all w ∈ W . Suppose that W ⊂ V is a G-invariant subspace, then the restriction ρg|W of ρg to W is an element of Aut(W ), hence the function

ρW : G → Aut(W )

g 7→ ρg|W is a representation of G in W . This function is called a subrepresentation of ρ. Deﬁnition 38 (Direct sum of representations). Suppose that two representations ρ : G → Aut(V ) and φ : G → Aut(W ) are given. Then their (external) direct sum

ρ ⊕ φ : G → Aut(V ⊕ W )

is deﬁned by (ρ ⊕ φ)g(v, w) = (ρg(v), φg(w)), for all g ∈ G.

Now let ρ : G → Aut(V ) be a representation. If V1,V2 ⊂ V are two G-invariant subspaces of V such that V = V1 ⊕ V2 is equal to the (internal) direct sum of V1 and V2. Then it is easy to show that ρ is isomorphic ∼ V1 V2 to the (external) direct sum of its restriction to V1 and V2, that is ρ = ρ ⊕ ρ . Simply let T : V → V1 ⊕ V2 V1 V2 be deﬁned by T (v) = (v1, v2). It is clear that T is an isomorphism, and that T ◦ ρ = ρ ⊕ ρ ◦ T . We can also deﬁne the tensor product of two representations.

13Note that in the case where V is ﬁnite dimensional, Aut(V ) can be identiﬁed with the general linear group GL(n) (where n = dim(V )) once a basis for V has been chosen.

23 Deﬁnition 39. Let ρ : G → Aut(V ) and φ : G → Aut(W ) be two representations. The tensor product representation is given by the homomorphism

ρ ⊗ φ : G → Aut(V ⊗ W ) g 7→ ρ(g) ⊗ φ(g),

where ρ(g) ⊗ φ(g) is the tensor product of linear maps, as deﬁned in the section about multilinear algebra. A representation ρ ∈ Hom(G, Aut(V )) of a group G in a vector space V also induces representations of k G in Λk(V ) and Symk(V ). Simply deﬁne ρ∧k ∈ Hom(G, Aut(Λk(V ))) and ρS ∈ Hom(G, Aut(Symk(V ))) by

(ρ∧k)(g) = (ρ(g))∧k

k k (ρS )(g) = (ρ(g))S

k where (ρ(g))∧k and (ρ(g))S are the linear maps discussed in the section about multilinear algebra. It is now time to define a special class of representations, which, in some way, constitute the fundamental building blocks in representation theory. Definition 40 (Irreducible representations). A representation ρ : G → Aut(V ) of a group G is said to be irreducible if the only G-invariant subspaces of V are {0} and V . Suppose that ρ : G → Aut(V ) is a representation, and that W ⊂ V is a G-invariant subspace of V . If the restriction of ρ to W , ρW , is irreducible, we will often abuse notation and call the vector space W irreducible. This is to avoid cluttering, when there is no confusion about the underlying representation. It is clear that all one-dimensional representations are irreducible, since there are no proper non-zero subspaces. Definition 41 (Completely reducible). Let G be a group, and let V be a vector space. A representation Ln ρ : G → Aut(V ) is said to be completely reducible if V = i=1 Vi, where each Vi is a non-trivial G-invariant subspace of V , and each ρVi is irreducible. Definition 42. Let G be a group, and let V be a vector space. A representation ρ : G → Aut(V ) is said to be decomposable if V = V1 ⊕ V2 where V1 and V2 are two non-trivial G-invariant subspaces. Otherwise, ρ is called indecomposable.

6.2 Some important theorems It is time to state and prove some important theorems about group representations that will be used through- out this paper. Proposition 9. Let ρ : G → Aut(V ) be isomorphic to a decomposable representation. Then ρ is decomposable. Proof. Let φ : G → Aut(W ) be a decomposable representation that is isomorphic to ρ, and let T : V → W be an isomorphism such that T ◦ ρg = φg ◦ T for all g ∈ G. Suppose that W1,W2 are two non-trivial −1 −1 G-invariant subspaces of W such that W = W1 ⊕ W2. It is then clear that V = T (W1) ⊕ T (W2), and −1 −1 that both T (W1) and T (W2) are G-invariant. We then have the following two propositions for the other types of representations. The proofs of these propositions are analogous to that of Proposition 9, and are therefore omitted for the sake of brevity. Proposition 10. Let ρ : G → Aut(V ) be isomorphic to an irreducible representation. Then ρ is irreducible. Proposition 11. Let ρ : G → Aut(V ) be isomorphic to a completely reducible representation. Then ρ is completely reducible. Before we state our next theorem, recall the bijective correspondence between the projections of a vector space V onto a subspace W ⊂ V and the complements of W in V . We will from now on also assume that all groups are ﬁnite, unless stated otherwise.

24 Theorem 7. Let ρ : G → Aut(V ) be a representation of a ﬁnite group G in a vector space V , and let W ⊂ V be a G-invariant subspace of V . Then there exists a G-invariant complement W 0 of W in V . Proof. Let W 0 be an arbitrary complement of W in V , and let p be the corresponding projection of V onto W . We deﬁne the average operator p0 ∈ End(V ) by

1 X p0 = ρ ◦ p ◦ ρ−1(v). |G| g g g∈G

−1 −1 −1 Since W is G-invariant we have that p ◦ ρg (w) = ρg (w), why ρg ◦ p ◦ ρg (w) = w for all w ∈ W . This also gives that p0(w) = w for all w ∈ W . Thus p0 is a projection of V onto W , corresponding to some 0 0 −1 complement W of W . Computing ρh ◦ p ◦ ρh we ﬁnd that 1 X ρ ◦ p0 ◦ ρ−1 = ρ ◦ ρ ◦ p ◦ ρ−1 ◦ ρ−1 h h |G| h g g h g∈G 1 X = ρ ◦ p ◦ ρ−1 |G| hg hg g∈G 1 X = ρ ◦ p ◦ ρ−1 |G| g g g∈G = p0.

0 0 0 If now w ∈ W is an element of the complement of W in V , and g ∈ G, then p (w) = 0, hence p ◦ ρg(w) = 0 0 0 ρg ◦ p (w) = 0, that is, ρg(w) ∈ W , which shows that W is G-invariant. Remark. Let ρ : G → Aut(V ) be a representation, where G is a ﬁnite group and V is an inner product space, endowed with the inner product h·, ·i. We say that this inner product is G-invariant if hρg(x), ρg(y)i = hx, yi for all x, y ∈ V and g ∈ G.14 Suppose that h·, ·i is G-invariant, and that W ⊂ V is a G-invariant subspace of V . Then it is easy to show that the orthogonal complement W ⊥ of W in V is also G-invariant. Note that we can consider any representation ρ of a group G to be in an inner product space V endowed with a G-invariant inner product: just let φ : G → Aut(Cdim(V )) be a representation that is isomorphic to ρ, and deﬁne a new inner product (·, ·) on Cdim(V ) by X (x, y) = hφg(x), φg(y)i, g∈G

where h·, ·i is the standard inner product on Cdim(V ).15 The inner product (·, ·) is G-invariant. Thus another proof of the theorem is obtained. We can now prove Maschke’s theorem. Theorem 8 (Maschke’s theorem). Every representation of a ﬁnite group G is completely reducible.

Proof. Let ρ : G → Aut(V ) be a representation of G. We proceed by induction of dim(V ). If dim(V ) = 1, then ρ is irreducible, since no non-trivial proper subspaces exist. Suppose then that dim(V ) ≥ 2. If ρ is irreducible, we are done. Otherwise, because of Theorem 7, V can be decomposed into a direct sum of G-invariant subspaces as V = V1 ⊕ V2, where dim(V1), dim(V2) < dim(V ). By the induction hypothesis, V1 and V2 are both completely reducible, which proves the theorem. 14 This is equivalent to saying that ρg ∈ U(V ) for all g ∈ G, where U(V ) denotes the unitary group on V . 15 dim(V ) dim(V ) To construct a representation φ : G → Aut(C ) that is isomorphic to ρ, simply let T ∈ Iso(V, C ) be an arbitrary −1 isomorphism, and deﬁne φg = T ◦ ρg ◦ T .

25 6.3 Character theory Let ρ : G → Aut(V ) be a representation of a group G in a vector space V . The character of ρ is the function χρ : G → C deﬁned by 16 χρ(g) = tr(ρg). As we will see, this function encodes a lot of information about the representation ρ; one could say that it characterizes it.

Proposition 12. If χ is the character of a representation ρ ∈ Hom(G, Aut(V )), then 1. χ(e) = dim(V ) 2. χ(g−1) = χ(g), for all g ∈ G 3. χ(hgh−1) = χ(g), for all g, h ∈ G

Proof. We prove the three cases one by one. 1. Since ρ(e) = id we have that χ(e) = tr(id) = dim(V ) 2. By the previous remark, any representation can be considered unitary. This means that all eigenvalues λ1, . . . , λn have absolute value 1, and we have that

n n −1 −1 X −1 X χ(g ) = tr(ρg ) = λi = λi = tr(ρg) = χ(g) i=1 i=1

3. This follows directly from the cyclic property of the trace function.

−1 χ(hgh ) = tr(ρhρgρh−1 ) = tr(ρg) = χ(g).

Proposition 13. Let ρ ∈ Hom(G, Aut(V )) and φ ∈ Hom(G, Aut(W )) be two representations of G in vector spaces V and W respectively. We have that

1. χρ⊕φ = χρ + χφ ∼ 2. χρ = χφ if ρ = φ

3. χρ⊗φ = χρ · χφ Proof. Again, we prove the cases one by one.

1. Let ρ(g) and φ(g) be given in matrix form by Rρ(g) and Rφ(g). The matrix form of ρ ⊕ φ(g) is then

R (g) 0 R(g) = ρ 0 Rφ(g)

why χρ⊕φ(g) = tr(R(g)) = tr(Rρ(g)) + tr(Rφ(g)) = χρ(g) + χφ(g) for all g ∈ G. ∼ 2. Since ρ = φ, there exists an isomorphism T ∈ Iso(V,W ) such that T ◦ ρg = φg ◦ T for all g ∈ G. Again, by the cyclic property of the trace function we have that

−1 χρ(g) = tr(ρg) = tr(T ◦ φg ◦ T ) = tr(φg) = χφ(g). 16Note that this requires the representation to be ﬁnite-dimensional, or else we might run into problems computing the trace.

26 3. Suppose that λ1, λ2, . . . , λn and , µ2, . . . , µn are the eigenvalues of ρg and φg respectively. Then the n eigenvalues of ρg ⊗ φg are given by {λiµj}i,j=1, so

n n !  n  X X X χρ⊗φ(g) = tr(ρg ⊗ φg) = λiµj = λi  µj = tr(ρg) tr(φg) = χρ · χφ i,j=1 i=1 j=1

∼ Statement 2. in Proposition 13 actually goes both ways, that is, χρ = χφ if and only if ρ = φ. We will work our way towards proving this. However, we ﬁrst prove the following proposition regarding induced representations in Λ2(V ) and Sym2(V ). Proposition 14. Let ρ ∈ Hom(G, Aut(V )) be a representation of a group G in a vector space V , and let χ be its character. The characters of the induced representations of ρ in Λ2(V ) and Sym2(V ) are given by

1 2 2 χ ∧2 (g) = (χ(g) − χ(g )) ρ 2 1 2 2 χ 2 (g) = (χ(g) + χ(g )) ρS 2 for all g ∈ G, respectively.

n Proof. Let g ∈ G be an element of G, and choose a basis {ei}i=1 of V consisting of eigenvectors of ρg. This can be done, since ρg can be seen as unitary. We have that

ρg(ei) = λiei,

for some λi ∈ C, so n n X 2 X 2 χ(g) = λi, χ(g ) = λi . i=1 i=1 We also have that ∧2 ρg (ei ∧ ej) = λiλjei ∧ ej S2 ρg (eiej) = λiλjeiej, hence !2 X 1 X 1 X 2 χ ∧2 (g) = λ λ = λ − λ ρ i j 2 i 2 i i

1. If ρ φ, then f = 0 2. If V = W and ρ = φ, then f = λ · id is a scalar multiple of the identity function Proof. Suppose f 6= 0. For x ∈ ker(f), we have

f ◦ ρ(x) = φ ◦ f(x) = 0,

27 so ρ(x) ∈ ker(f), why ker(f) is G-invariant. Since ρ is irreducible, we have that ker(f) = V or ker(f) = {0}. The ﬁrst case implies that f = 0, so suppose that ker(f) = {0}. Now let y ∈ range(f), which means that there exists an x ∈ V such that f(x) = y. Thereby

f ◦ ρg(x) = φg ◦ f(x) = φg(y) so φg(y) ∈ range(f), hence range(f) = W or range(f) = {0}, since φ is irreducible. Since range(f) = {0} implies f = 0, we have that ker(f) = {0} and range(f) = W , which shows that f is an isomorphism. Now let V = W and ρ = φ, and let λ be an eigenvalue of f (there is at least one, since the representation is complex). Define f 0 = f − λ · id. Since λ is an eigenvalue of f, we have that ker(f 0) 6= {0}. We also have 0 0 0 that f ◦ ρg = φg ◦ f . The first part of the proof shows that these two properties are only possible if f = 0. This gives that f = λ · id. In the proceeding text we will keep the hypothesis that V and W are irreducible. Corollary 2. Let h ∈ Hom(V,W ) be any linear mapping of V into W . Define h ∈ Hom(V,W ) by

1 X h0 = φ−1 ◦ h ◦ ρ , |G| g g g∈G where we use the same notation as in the proof of Theorem 7. Then

1. ρ φ, then h0 = 0. 0 1 2. If V = W and ρ = φ, then h = dim(V ) tr(h) · id.

0 0 0 Proof. It is clear that φg ◦ h = h ◦ ρg. Applying Schur’s lemma, we see that case 1. gives that h = 0, and that case 2. gives that h0 = λ · id. In the latter case we also have that

1 X dim(V )λ = tr(h0) = tr(φ−1 ◦ h ◦ ρ ) = tr(h). |G| g g g∈G

1 We get that λ = dim(V ) tr(h). We now wish to rewrite Corollary 2 by translating ρ and φ into their matrix forms:

ρg = (aij(g)), φg = (bij(g)).

0 0 0 Let the linear maps h and h be deﬁned by the matrices h = (xij) and h = (xij). We have by the deﬁnition of h0 that 1 X x0 = b (g−1)x a (g). ij |G| il lk kj g∈G, k,l We have the two following corollaries for the two cases in Schur’s lemma:

Corollary 3. In case 1. of Corollary 2 we have that

1 X b (g−1)a (g) = 0 |G| il kj g∈G

for arbitrary i, j, k, l. Corollary 4. In case 2. of Corollary 2 we have that ( 1 X 1 , if i = j and k = l b (g−1)a (g) = dim(V ) |G| il kj g∈G 0, otherwise

28 We can now begin to investigate some orthogonality relations for characters. We begin by deﬁning the scalar product between two complex-valued functions on G. For φ, ψ ∈ CG, deﬁne 1 X hφ, ψi = φ(g)ψ(g). |G| g∈G

It is easy to see that this is a scalar product. Theorem 9. Let χ be the character of an irreducible representation ρ. We then have that ||χ||2 = hχ, χi = 1. Furthermore, if χ and χ0 are the characters of two non-isomorphic irreducible representations, then hχ, χ0i = 0.

Proof. We begin by showing that hχ, χi = 1. Let ρ be given in matrix form by ρg = (aij(g)). We have that P χ(g) = i aii(g), hence X X X hχ, χi = h aii(g), aj(g)i = haii(g), ajj(g)i. i j i,j

δij Corollary 4 gives that haii(g), ajj(g)i = dim(V ) , so

1 X dim(V ) hχ, χi = δ = = 1. dim(V ) ij dim(V ) i,j

The second part of the theorem is proved in the same way, but by using Corollary 3 instead of Corollary 4. Theorem 10. Let ρ ∈ Hom(G, Aut(V )) be a representation of G in a vector space V , with character φ. Suppose V decomposes into a direct sum of irreducible representations as

k M V = Wi. i=1

Then, if W is an irreducible representation with character χ, the number of Wi isomorphic to W is given by hφ, χi.

Proof. Let χi be the character of Wi. By Proposition 13 we have that

k X φ = χi. i=1

According to the previous theorem, we have that hχi, χi is either 1 or 0, depending on whether Wi is isomorphic to W or not. We therefore have that X hφ, χi = hχi, χi,

Wi is isomorphic to W

which proves the theorem

Corollary 5. The number of Wi isomorphic to W does not depend on the chosen decomposition of V .

Proof. Clearly, the number hφ, χi is not dependent on the decomposition. Corollary 6. Two representations are isomorphic if and only if they have the same character. That is, if ∼ ρ ∈ Hom(G, Aut(V )) and φ ∈ Hom(G, Aut(W )) are two representations, then χρ = χφ if and only if ρ = φ. ∼ Proof. We have already proven that if ρ = φ, then χρ = χφ. So on the contrary, suppose that χρ = χφ. Corollary 5 shows that both ρ and φ contain each given irreducible representation the same number of times.

29 The above results show that the study of representations often reduce to that of their characters. Let W1,...,Wk denote all the irreducible representations of a group G, and let χ1, . . . , χk denote their corresponding characters. Then each representation V of G decomposes into a direct sum

k ∼ M ⊕mn V = Wn . n=1

⊕mn where, analogously as for the tensor product, Wn is deﬁned as Wn ⊕ · · · ⊕ Wn and each mn is a non- | {z } mn times negative integer. We will now prove a very handy theorem that will frequently be used to check if a given representation is irreducible or not.

Theorem 11. If φ is the character of a representation ρ, then hφ, φi is a positive integer. Moreover, we have that hφ, φi = 1 if and only if ρ is irreducible. Proof. The above results show that the orthogonality relations imply that

k X 2 hφ, φi = mn, n=1

which is equal to 1 if and only if one of the mn’s is equal to 1 and the rest are 0.

6.4 Canonical decomposition of a representation The decomposition of a representation into irreducible representations is only unique up to isomorphism. We will now deﬁne a “coarser” decomposition of a representation that has the advantage of being unique. It is deﬁned as follows: Let ρ : G → Aut(V ) be a representation of a group G, and let χ1, . . . , χk be the characters of the irreducible representations W1,...,Wk of G. Suppose that V decomposes into irreducible representations as V = U1 ⊕ · · · ⊕ Um. For i = 1, . . . , k, let Vi be the direct sum of those U1,...,Um which are isomorphic to Wi. It is then clear that V = V1 ⊕ · · · ⊕ Vk. This is called the canonical decomposition of V . A proof of the following theorem can be found in [Serre, 1977].

Theorem 12. Keep the notation used above. The canonical decomposition V = V1 ⊕· · ·⊕Vk does not depend on the initially chosen decomposition of V into irreducible representations. Furthermore, the projection pi of V onto Vi associated with this decomposition is given by the formula

dim(Wi) X p = χ (g)ρ . i |G| i g g∈G

Thus the decomposition of a representation V can be done in two stages: ﬁrst determine the canonical decomposition V = V1 ⊕ · · · ⊕ Vk using the projection formula above, and then decompose each Vi into a direct sum of irreducible representations.

6.5 From ﬁnite to compact groups Up until now we have only considered representations of ﬁnite groups.17 We will now indicate how our preceding results carry over to arbitrary compact groups. As the representation theory of compact groups can become a bit messy, we will not provide rigorous proofs for all of our claims. We will, however, provide some ideas for some of the proofs. For proofs, see [Weyl, 1931]. One more requirement that eases our job is that we want the representations to be continuous.

17However, some of our results hold for arbitrary groups.

30 Deﬁnition 43. A representation ρ of a topological group G in a topological vector space V is said to be a continuous representation if the function G × V → V

(g, v) 7→ ρg(v) is continuous with respect to the product topology on G × V . We will now state the “complement theorem” and Maschke’s theorem in the setting of compact groups, together with a proof of the former. Theorem 13. Let ρ : G → Aut(V ) be a continuous representation of a compact group G in a vector space V , and let W ⊂ V be a G-invariant subspace of V . Then there exists a G-invariant complement W 0 of W in V . Proof. Let W 0 be an arbitrary complement of W in V , and let p be the corresponding projection of V onto W . We deﬁne the average operator p0 ∈ End(V ) by

p0 : V → V Z −1 v 7→ ρg ◦ p ◦ ρg (v)dµ(g) G

R −1 The integrand is integrable, since it is continuous, and we will use the notation p0 = G ρg ◦ p ◦ ρg dµ(g) for 18 −1 −1 −1 the sake of brevity. Since W is G-invariant we have that p ◦ ρg (w) = ρg (w), why ρg ◦ p ◦ ρg (w) = w for all w ∈ W . This also gives that p0(w) = w for all w ∈ W . Thus p0 is a projection of V into W , corresponding 0 0 −1 to some complement W of W . Computing ρh ◦ p ◦ ρh we ﬁnd that Z 0 −1 −1 −1 ρh ◦ p ◦ ρh = ρh ◦ ρg ◦ p ◦ ρg ◦ ρh dµ(g) G Z −1 = ρhg ◦ p ◦ ρhg dµ(g) G Z −1 = ρg ◦ p ◦ ρg dµ(g) G = p0, where we have used the property of the Haar measure given by equation (2). If now w ∈ W 0 is an element 0 0 0 of the complement of W in V , and g ∈ G, then p (w) = 0, hence p ◦ ρg(w) = ρg ◦ p (w) = 0, that is, 0 0 ρg(w) ∈ W , which shows that W is G-invariant. Theorem 14 (Maschke’s theorem). Every continuous representation of a compact group G is completely reducible. Theorem 10 and 11 together with Corollary 5 and 6 also carry over, as well as Theorem 12, but now the projection formula is given by Z pi(x) = dim(Wi) χ(g)ρg(x)dµ(g), G for x ∈ V .

18Notice, however, that this is for notational convenience only. We have not deﬁned how to integrate an operator. Though this might be resolved with the use of the Bochner integral, which extends the ordinary Lebesgue integral to functions that take values in Banach spaces.

31 7 The 3D Rotation Group SO(3)

We define SO(3) as the group of all the possible rotations in 3-dimensional space R3 about the origin. The product gh of two elements g and h in SO(3) is defined to be the rotation obtained by first applying the rotations h and then the rotation g. We know this forms a group since the product of two rotations is another rotation, every rotations has an inverse and the ”identity rotation” is to simply do nothing. The group is nonabelian, since rotations are generally non-commutative. Rotations are linear transformations on R3 and can therefore be represented by matrices once a basis for R3 has been chosen. If we choose an orthonormal basis, every rotation correspond to an orthogonal 3×3-matrix with determinant 1. These matrices are called special orthogonal matrices, why we denote the group by SO(3). We will often describe elements of SO(3) as matrices.

7.1 Parameterization There are a couple of different ways to parameterize the group. The simplest is probably the axis-angle representation, in which every non-trivial rotation is described uniquely by a rotation angle, t, and an axis around which we rotate.19 The axis of rotation can in turn be described by an azimuthal angle φ and a polar angle θ. To make the parameterization unique we restrict the ranges of the parameters: t ∈ [0, π) φ ∈ [0, 2π) θ ∈ [0, π]. An explicit expression for this parameterization is given by g : [0, π) × [0, 2π) × [0, π] → SO(3) defined by g(t, φ, θ) = QBQT , where  sin φ cos θ cos φ sin θ cos φ Q = − cos φ cos θ sin φ sin θ sin φ 0 − sin θ cos θ cos t − sin t 0 B = sin t cos t 0 . 0 0 1 This expression might seem very daunting, but we will seldom actually refer to it explicitly. Another parameterization of SO(3) is given by noticing that any rotation can be achieved by composing three so called elemental rotations, which are rotations about the axes of a coordinate system. The angles of these three elemental rotations are called Euler angles, and are typically denoted α, β, γ or φ, θ, ψ. With these a general rotation g in SO(3) can be written as cos α − sin α 0 1 0 0  cos γ − sin γ 0 g(α, β, γ) = sin α cos α 0 0 cos β − sin β sin γ cos γ 0 , 0 0 1 0 sin β cos β 0 0 1 that is, first a rotation about the z-axis, followed by a rotation about the x-axis and then another rotation about the z-axis, of angle α, β and γ respectively. The ranges of the Euler angles are α ∈ [0, 2π] β ∈ [0, π] γ ∈ [0, 2π]. While the parameterization based on Euler angles is interesting, we will find that the axis-angle representation is a much more natural choice for our purposes, since the variable t in this parameterization will, in some way, encapsulate all the information we need. We will therefore use the axis-angle representation of SO(3) in the rest of this report, and we will keep the notation t, θ and φ for the three parameters. We will often think about elements in SO(3) as functions of these three variables, and denote an element g ∈ SO(3) by g(t, φ, θ) to emphasize the variables.

19Note that the use of the word ”representation” is not the same as in representation theory.

32 7.2 Conjugacy classes of SO(3) We intend to study the characters of representations of SO(3). Recall that these only depend on the conjugacy classes.20 It is therefore important to characterize the conjugacy classes in order to ease future calculations. We start oﬀ with the following theorem, which, together with proof, is taken from [Sadun, 2001].

Theorem 15. Let A be an element of SO(3), then +1 is an eigenvalue of A. The other two eigenvalues are ±it e for some angle t. Let ξ1 be an eigenvector corresponding to the eigenvalue +1. Then A is a rotation by an angle ±t about the axis spanned by ξ1. Proof. The determinant is the product of the eigenvalues. If A has three real eigenvalues, they must all be +1 = e±i0, or two of them −1 = e±iπ and one +1. If A has a complex pair of eigenvalues, their product is eite−it = 1, and the last eigenvalue must therefore be +1. In conclusion, +1 must be an eigenvalue so we can ﬁnd a corresponding eigenvector, let us call it ξ1. If all the eigenvalues are +1, then A is the identity which we can see as a rotation by 0 radians about the vector ξ1. If the eigenvalues are +1, −1 and −1 then A preserves ξ1 but ﬂips the sign of any vector that is orthogonal to ξ1. This means that A must be a rotation by π radians about ξ1. Lastly, consider the case where the eigenvalues are not all real. Let ξ2 be the normalized eigenvector −it corresponding to the eigenvalue λ2 = e . We decompose this vector into a real and an imaginary part: 1 ξ2 = √ (vR + ivI ). 2 Since real matrices have eigenvectors that come in conjugate pairs, we know that the third eigenvector must be ξ = ξ = √1 (v − iv ) with corresponding eigenvalue λ = eit. We now show that ξ , v and v form 3 2 2 R I 3 1 R I an orthonormal basis for R3. Orthogonality comes from 1 0 = ξ1 · ξ2 = √ (ξ1 · vR + iξ1 · vI ) 2

0 = =(ξ2 · ξ3) = −vR · vI and normalization comes from 1 1 = =(ξ · ξ ) = (v · v + v · v ) 2 3 2 R R I I 1 0 = <(ξ · ξ ) = (v · v − v · v ). 2 3 2 R R I I 3 Therefore, B = {ξ1, vR, vI } forms an orthonormal basis for R . Assume without loss of generality that this basis is right-handed (otherwise we can just change the sign of ξ1). We ﬁnd that A is equal to

1 0 0  0 cos(t) − sin(t) 0 sin(t) cos(t) in the basis B, which shows that A is a rotation by an angle t about the axis spanned by ξ1. Since the trace of a matrix is equal to the sum of its eigenvalues, Theorem 15 gives that the trace of a matrix A in SO(3) is given by tr(A) = 1 + eit + e−it = 1 + 2 cos t. We will use this fact in the following theorem.

Theorem 16. Two elements A and B in SO(3) are conjugate if and only if their traces are equal, that is tr(A) = tr(B). Speciﬁcally, A and B are conjugate if and only if they have the same rotation angle t. 20A function on a group that only depends on the conjugacy classes is called a class function.

33 Proof. We ﬁrst prove that two conjugate elements have the same trace. Let A be an element of SO(3), and let B be an element conjugate to A in SO(3), that is, there exists a matrix X ∈ SO(3) such that B = XAX−1. We then have that

tr(B) = tr(XAX−1) = tr(AX−1X) = tr(A), by the cyclic property of the trace function. Now let A and B be elements of SO(3) with the same trace, i.e. tr(A) = tr(B). Since the trace of a matrix in SO(3) is given by 1+2 cos t, where t is the rotation angle of the matrix, we know that A and B are rotations through the same angle. Let u be the normalized eigenvector of A corresponding to the eigenvalue +1, that is Au = u. We can extend u to an orthonormal basis B = {u, v, w} and create an orthogonal matrix P by using these basis vectors as columns. Then the matrix P −1AP is either a rotation about the x-axis by t degrees, or by −t degrees. If the latter we simply switch the sign of any vector in B. We therefore have that −1 Rx(t) = P AP, where Rx(t) denotes a rotation about the x-axis by t-degrees. In the same fashion we can ﬁnd a matrix Q such that −1 Rx(t) = Q BQ, −1 −1 −1 which gives that B = (PQ ) AP Q . We have now characterized all the conjugacy classes of SO(3): they only depend on the rotation angle t.

7.3 The Haar measure on SO(3) Since SO(3) is a compact topological group, we know that there is a unique (up to multiple) Haar measure deﬁned on the group. What this measure looks like depends on our choice of parameterization. For example, if we parameterize the group by means of Euler angles we ﬁnd that the normalized Haar measure is given by

Z Z 2π Z 2π Z π 1 21 f(g)dµ(g) = 2 f(g(α, β, γ)) sin(β)dβ dγ dα. SO(3) 4π 0 0 0

But we already concluded that Euler angles are of no special interest, and that we therefore need to determine the Haar measure when using the axis-angle representation. This can be done using the expression above and simply changing variables. By doing this we get that the normalized Haar measure on SO(3) using the axis-angle representation is given by

Z 1 Z 2π Z π Z π f(g)dµ(g) = 2 f(g(t, φ, θ))(1 − cos(t)) sin(θ)dt dθ dφ, SO(3) 4π 0 0 0 that is dµ = (1 − cos t) sin θdθdφdt.

21This result is well known, and can be found in, for example, Naimark, A. Linear Representations of the Lorentz Group.

34 8 Representations of SO(3)

Before we dive into explicit representations of SO(3) we need some machinery. First of all we will need to consider an inner product on a space of functions from SO(3) to C. A natural space to consider is L2(SO(3), B, µ), where B is the Borel algebra and µ is the previously acquired Haar measure on SO(3). We have already seen how this can be done for compact groups, so using the acquired expression for the Haar measure we have Z hf, hiSO(3) = f(g)h(g)dµ(g) SO(3) (3) 1 Z 2π Z π Z π = 2 f(g(t, φ, θ))h(g(t, φ, θ))(1 − cos(t)) sin(θ)dt dθ dφ, 4π 0 0 0 for two functions f, h ∈ L2(SO(3), B, µ). We use the subscript SO(3) to specify that this is the inner product on the space L2(SO(3), B, µ), since we will introduce another frequently used inner product on another space later on. The expression in equation (3) can be shortened if we assume that f and h are class functions, by realizing that the conjugacy classes of SO(3) only depend on t, and that the class functions therefore also only depend on t. We introduce the shorthand notation f(t) = f(g(t, φ, θ)). With this notation we are free to integrate with respect to φ and θ, whereby we get 1 Z π hf, hiSO(3) = f(t)h(t)(1 − cos(t))dt. π 0

8.1 Representations in C[x, y, z] 8.1.1 Introduction

We will define C[x, y, z] as the polynomial ring in the variables x, y and z over the complex numbers. This ring is also a vector space under addition of polynomials, in which we will study representations of the group SO(3). The reader might notice that this is indeed an infinite dimensional vector space, and that this might be a problem. However, this space has the special property that it decomposes naturally into a sum of finite dimensional subspaces which are easy to study: ∞ M C[x, y, z] = Vd d=0

where Vd denotes the space of all homonogenous polynomials of degree d, i.e. all polynomials where the degree of each monomial in the polynomial sum is d. An example of such a space is V1 = hx, y, zi = i j k {ax + by + cz : a, b, c ∈ C}. The basis {x y z }i+j+k=d, where i, j, k ≤ 0, is probably the most simple basis for this space. We have the following proposition, regarding the dimension of Vd. d+2 Proposition 16. The dimension of the vector space Vd is nd = dim(Vd) = 2 . Proof. We prove this by a combinatorial argument: we wish to find the number of ways we can write i + j + k = d, where i, j, k ≥ 0. We find that we can chose i in d + 1 ways, and that when i is fixed we can chose j in d + 1 − i ways, and k in one way. This gives that

d X (d + 2)(d + 1) d + 2 n = (d + 1 − i) = = . d 2 2 i=0 i j k The basis {x y z }i+j+k=d might be the simplest one, however, there is a basis that is much easier to work with, at least for our purposes.

d nd Proposition 17. The set {(aix + biy + ciz) }i=1 forms a basis for Vd for suitable choices of ai, bi, ci where i = 1, . . . , nd. We will return to the applications of this basis, as well as provide a proof for the proposition, later.

35 8.1.2 The natural representation

There is a natural representation of SO(3) in C[x, y, z]; to construct this, we begin by changing our view as to see the variables x, y, z as coordinates of a point in three-dimensional space, R3. We will denote this point by (x, y, z). SO(3) then acts naturally on this point by rotating it. For an element A in SO(3), we denote the action of A on (x, y, z) by A(x, y, z) (note that this action is regular matrix multiplication). We now deﬁne the natural representation Π ∈ Hom(SO(3), Aut(C[x, y, z])) of SO(3) in C[x, y, z] by

ΠA(f(x, y, z)) = f(A(x, y, z)) where f(x, y, z) ∈ C[x, y, z]. We will provide an example for how this representation is deﬁned explicitly.

a11 a12 a13 Example 7. Let A = a21 a22 a23 be an element of SO(3). A then acts on the monomials x, y and z as a31 a32 a33

ΠA(x) = a11x + a12y + a13z

ΠA(y) = a21x + a22y + a23z

ΠA(z) = a31x + a32y + a33z.

More generally, it acts on a monomial xiyjzk as

i j k i j k ΠA(x y z ) = (a11x + a12y + a13z) (a21x + a22y + a23z) (a31x + a32y + a33z) .

We know that Π constitutes a representation, since

ΠAB(f(x, y, z)) = f(AB(x, y, z)) = ΠA(ΠB(f(x, y, z))) by the associative property of matrix multiplication, which shows that ΠAB = ΠA ◦ ΠB, i.e. that Π is a homomorphism from SO(3) to Aut(C[x, y, z]). Since Π does not change the degree of a homogeneous polynomial, it is clear that all the subspaces Vd are SO(3)-invariant. We will often focus on the restriction of Π to some arbitrary SO(3)-invariant subspace W ⊂ C[x, y, z], which we will denote ΠW . Note that this is an element of Hom(SO(3), Aut(W )). In the (d) V special case of W = Vd we will shorten the notation and write Π = Π d . We will also use the letter ψ to denote the characters of Π and write ψW for the character of Π restricted to some G-invariant subspace W ⊂ C[x, y, z]. Again, we will often abuse notation and call W a representation when we really are referring to ΠW . Since our main goal is to ﬁnd all irreducible representations of SO(3) in C[x, y, z], we start by examining how SO(3) acts on the subspaces Vd for small d, since these should be somewhat easy to study. The case of d = 0 is trivially irreducible, since it is one-dimensional, so let d = 1. All elements in SO(3) can be written cos t − sin t 0 as A = sin t cos t 0 in some basis (i.e. they are conjugate to a matrix of this form). This means that the 0 0 1 (1) character of Π is ψV1 (A) = 1 + 2 cos(t). We ﬁnd that Z π 2 1 2 ||ψV1 || = hψV1 , ψV1 iSO(3) = (1 + 2 cos(t)) (1 − cos(t))dt = 1, π 0 which means that Π(1) is irreducible by Theorem 11. To calculate the character of Π(2) we need to determine the diagonal of the representation’s matrix representation. We have the following mappings under Π.

x2 7→ cos2(t)x2 + ··· y2 7→ cos2(t)y2 + ··· z2 7→ z2 + ··· xy 7→ (cos2(t) − sin2(t))xy + ··· xz 7→ cos(t)xz + ··· yz 7→ cos(t)yz + ···

36 which gives that 2 2 ψV2 (t) = 3 cos (t) − sin (t) + 2 cos(t) + 1 = 4 cos2(t) + 2 cos(t) = 2 cos(2t) + 2 cos(t) + 2. This time, however, we ﬁnd that Z π 2 1 2 ||ψV2 || = hψV2 , ψV2 iSO(3) = (2 cos(2t) + 2 cos(t) + 2) (1 − cos(t))dt = 2 6= 1 π 0 which means that Π(2) is not irreducible. So we now wish to know what irreducible representations make (2) (2) 2 2 2 2 2 2 up Π . It is easily veriﬁable that Πg (x + y + z ) = x + y + z for all g ∈ SO(3), why the space hx2 + y2 + z2i is SO(3)-invariant. This result is important, as it provides a basis for our next lemma.

Lemma 1. Let W ⊂ C[x, y, z] be a ﬁnite-dimensional SO(3)-invariant subspace. Then (x2 + y2 + z2)W is 2 2 2 also SO(3)-invariant. Furthermore, ΠW is isomorphic to Π(x +y +z )W . Before we dive into the proof of this lemma, we will employ some new notation: since it is quite cumbersome to both write and read (x2 + y2 + z2)W , we deﬁne the operator L ∈ End(C[x, y, z]) by L(f(x, y, z)) = (x2 + y2 + z2)f(x, y, z) for some polynomial f(x, y, z) in C[x, y, z]. With this notation 2 2 2 we have that (x + y + z )W = L(W ). It is clear that L and Πg commute for all g in SO(3).

Proof. To prove that L(W ) is SO(3)-invariant, we wish to show that if f(x, y, z) ∈ W , then Πg(L(f(x, y, z))) ∈ L(W ) for all g in SO(3). From the deﬁnition of Π and L we have that

2 2 2 Πg(L(f(x, y, z))) = Πg((x + y + z )f(x, y, z)) = (x2 + y2 + z2)f(g(x, y, z)) = L(f(g(x, y, z))) ∈ L(W )

since W is SO(3)-invariant. W L(W ) n To show that Π is isomorphic to Π we show that their characters are equal. Let B = {pi(x, y, z)}i=1, n W L(W ) be a basis for W . Then C = {L(pi(x, y, z))}i=1 is a basis for L(W ). We examine how Π and Π act on the bases B and C respectively, starting with B:

n W X Πg (pi(x, y, z)) = pi(g(x, y, z)) = aji(g)pj(x, y, z) j=1

W where aij(g) is some complex number that depends on g. This means that the matrix of Πg in the basis n {pi(x, y, z)} is given by i=1   a11(g) a12(g) ··· a1n(g) a21(g) a22(g) ··· a2n(g) A(g) =    . . .. .   . . . .  an1(g) an2(g) ··· ann(g) and that the character of ΠW is n X ψW (g) = tr(A(g)) = aii(g). i=1 Likewise, for C we have that

n L(W ) X Πg (L(pi(x, y, z))) = L(pi(g(x, y, z))) = L(aji(g)pj(x, y, z)) j=1 n X = aji(g)L(pj(x, y, z)). j=1

37 L(W ) n From this we see that the matrix of Πg in the basis {L(pi(x, y, z))}i=1 is also given by A(g), and therefore that ψL(W )(g) = tr(A(g)) = ψW (g), L(W ) ∼ W which shows that Π = Π . Remark. Note that the second part of this proof requires that the subspace W is ﬁnite dimensional, since it rests on the trace of a matrix.

Corollary 7. Let W ⊂ C[x, y, z] be a ﬁnite-dimensional SO(3)-invariant subspace such that ΠW is irreducible. Then ΠL(W ) is also irreducible.

Proof. This follows directly from the fact that the two representations are isomorphic by Lemma 1.

We now return to ﬁnding the irreducible representations of V2. Since V0 is irreducible, so is L(V0) by 22 Corollary 7. L(V0) consists of homogoneous polynomials of degree 2, so it is a subspace of V2. We therefore have V2 = W2 ⊕ L(V0),

for some complement W2 ⊂ V2 of L(V0) in V2. Lemma 1 also gives that ψL(V0) = ψV0 , so

ψV2 (g) = ψW2 (g) + ψV0 (g) which means that

ψW2 (g) = ψV2 (g) − ψV0 (g) = 2 cos(2t) + 2 cos(t) + 2 − 1 = 2 cos(2t) + 2 cos(t) + 1.

1 R π 2 A quick calculation shows that ||ψW2 || = π 0 (2 cos(2t) + 2 cos(t) + 1) (1 − cos(t))dt = 1, and thereby that ΠW2 is irreducible. In general, since L(Vd−2) is an SO(3)-invariant subspace of Vd, Theorem 7 tells us that we can ﬁnd an SO(3)-invariant complement Wd of L(Vd−2) in Vd such that

Vd = Wd ⊕ L(Vd−2). (4)

We begin by asking the question if it is always true that the subspace Wd is irreducible? As it turns out, the answer is yes, but to prove it we need to go a bit more in depth about the characters.

8.1.3 Characters of SO(3) By diagonalizing an element A in SO(3) we will always nr able to obtain a matrix of the form

eit 0 0 D =  0 e−it 0 . 0 0 1 where t is the rotation angle of A. Since D is conjugate to A, we only need to study diagonal matrices of this form when we want to study the characters, since the characters are class functions and therefore only depend on the conjugacy classes. The action of D on V1 is given by its action on the basis vectors

it ΠD(x) = e x −it ΠD(y) = e y

ΠD(z) = z.

An explicit expression for this action’s extension to Vd is

n m d−n−m i(n−m)t n m d−n−m ΠD(x y z ) = e x y z ,

22Note that it is also irreducible by the simple fact that it is one-dimensional.

38 from which we see that the character of the representation Π restricted to Vd is d d−m ! X i(n−m)t X X i(n−m)t ψVd (A) = e = e . n+m≤d m=0 n=0 With the help of some closer analysis, this expression can be simpliﬁed to n X ψV2n (t) = n + 1 + 2(n − k + 1) [cos((2k − 1)t) + cos(2kt)] k=1 n (5) X ψV2n+1 (t) = n + 1 + (2n + 2) cos(t) + 2(n − k + 1) [cos(2kt) + cos((2k + 1)t)] k=1 where we have one expression for subspaces of even and odd dimension respectively. These expressions allow us to prove that Wd is irreducible.

Theorem 17. The subspaces Wd in equation (4) are irreducible. Proof. Equation (5) gives that

d X ψWd (g) = ψVd (g) − ψVd−2 (g) = 1 + 2 cos(kt). k=1 Some integral-crunching gives us that

d 1 Z π X ||ψ ||2 = hψ , ψ i = (1 + 2 cos(kt))2(1 − cos(t))dt = 1, Wd Wd Wd SO(3) π 0 k=1 W which shows that Π d is irreducible.

8.1.4 An inner product on C[x, y, z]

Before we continue, we return to our previous sub-problem of ﬁnding the irreducible representations of V2. We now know that a decomposition of V2 into irreducible representations is given by V2 = W2 ⊕ L(V0), and we know what the elements in L(V0) look like, but what are the elements of W2? With the use of the projection formula given by the continuous case of Theorem 12 we ﬁnd that

2 2 2 2 23 W2 = hx − y , x − z , xy, xz, yzi. 2 2 2 The expression above looks suspiciously like an orthogonal complement of L(V0) = hx + y + z i, where all basis vectors xiyjz2−i−j are mutually orthogonal. This invites us to try to define some kind of inner product on C[x, y, z] that satisfies this. We will do thise by first defining inner products on each of the subspaces Vd, and then combine these into one on the whole space C[x, y, z]. However, before we continue, we need to consider a special kind of differential operator. Let f(x, y, z) be a polynomial in C[x, y, z], which means that f(x, y, z) can be written as the finite sum (since it is a polynomial) X i j k f(x, y, z) = aijkx y z . (6) i,j,k where each i, j and k are non-negative integers and each aijk is some complex number. Define the differential ∂ ∂ ∂ operator f ∂x , ∂y , ∂z ∈ End(C[x, y, z]) by

∂ ∂ ∂ X ∂i ∂j ∂k f , , = a . (7) ∂x ∂y ∂z ijk ∂xi ∂yj ∂zk i,j,k where aijk is the same as in equation (6). This operator enables us to deﬁne a very useful inner product, as shown below. 23This computation is very involved, as one needs to use the general form of an element in SO(3) and integrate with respect to θ, φ and t. This was done using Wolfram Mathematica.

39 Proposition 18. Let f(x, y, z), g(x, y, z) be polynomials in Vd. The function h·, ·iVd : Vd × Vd → C deﬁned by 1 ∂ ∂ ∂ hf(x, y, z), g(x, y, z)i = f , , g(x, y, z), (8) Vd d! ∂x ∂y ∂z

∂ ∂ ∂ where f ∂x , ∂y , ∂z is the differential operator defined by equation (7), is an inner product on Vd. P i j k Proof. Assume that g(x, y, z) has the form g(x, y, z) = i+j+k=d bijkx y z . Since all the monomial terms in both f(x, y, z) and g(x, y, z) have degree d, the only terms that survive the action of the differential operator are those with the same ijk. In other words, we have that

∂ ∂ ∂ 1 X f , , g(x, y, z) = i!j!k!a b . (9) ∂x ∂y ∂z d! ijk ijk i+j+k=d With this expression, all three axioms for an inner product, i.e. conjugate symmetry, linearity in the second argument and positive-deﬁniteness, follow easily. i j k With this inner product, all monomials in Vd become mutually orthogonal, i.e. {x y z }i+j+k=d forms an orthogonal basis for Vd, since 1 ∂i ∂j ∂k i!j!k! hxiyjzk, xlymzni = xlymzn = δ δ δ .24 d! ∂xi ∂yj ∂zk d! il jm kn But this is not the only useful property of this inner product, as we will soon discover. However, before we proceed to this property, we need the following lemma.

Lemma 2. Let f(x, y, z) be a polynomial in Vd, and let α, β and γ be some complex numbers. Then d h(αx + βy + γz) , f(x, y, z)iVd = f(α, β, γ).

Proof. Since f(x, y, z) ∈ Vd, we know that

X i j k f(x, y, z) = aijkx y z . i+j+k=d Thereby we have that

d X d i j k h(αx + βy + γz) , f(x, y, z)iVd = aijkh(αx + βy + γz) , x y z iVd , i,j,k

d i j k and we see that we only need to study h(αx+βy +γz) , x y z iVd . With the use of the multinomial theorem, we can expand (αx + βy + γz)d as X d (αx + βy + γz)d = αk1 βk2 γk3 xk1 yk2 zk3 . k1, k2, k3 k1+k2+k3=d

d k1 k2 k3 The inner product h(αx + βy + γz) , x y z iVd then becomes k k k d k ∂ 1 ∂ 2 ∂ 3 d i j k X k1 2 k3 i j k h(αx + βy + γz) , x y z iVd = α β γ k k k (x y z ) k1, k2, k3 ∂x 1 ∂y 2 ∂z 3 k1+k2+k3=d 1 d j = αiβ γki!j!k! d! i, j, k j = αiβ γk.

25 And the rest follows easily. 24Note that one of the Kroncker deltas is superﬂuous, since i + j + k = l + m + n = d. 25 1 The proof of this lemma also shows why we included the factor d! in the deﬁnition of the inner product.

40 Before we go on, we will provide a proof for Proposition 17.

Proof of Proposition 17. Suppose that L1(x, y, z),L2(x, y, z),...,Lk(x, y, z) ∈ V1 are linear forms such that d k Li(x, y, z) = aix + biy + ciz for i = 1, . . . , k, and that the set {Li(x, y, z) }i=1 is linearly independent and that it is impossible to add another linearly independent element of this form. Then for any linear form d L(x, y, z) = ax + by + cz ∈ V1, the power L(x, y, z) can be written as a linear combination

k d X d L(x, y, z) = αiLi(x, y, z) . i=1

Now take any f(x, y, z) ∈ Vd. According to Lemma 2 we have that

k k d X d X f(a, b, c) = hL(x, y, z) , f(x, y, z)iVd = αihLi(x, y, z) , f(x, y, z)iVd = αif(ai, bi, ci), i=1 i=1 which shows that k ≥ dim(Vd). But we also have that k ≤ dim(Vd), so we can conclude that k = dim(Vd), d k and that the set {Li(x, y, z) }i=1 constitutes a basis for Vd. We are now ready to prove a general theorem with the help of the previous lemma.

Theorem 18. The inner product deﬁned by equation (8) is invariant under an orthogonal change of basis on V1. That is,

hf(A(x, y, z)), g(A(x, y, z))iVd = hf(x, y, z), g(x, y, z)iVd (10) for all f(x, y, z), g(x, y, z) ∈ Vd, where A is an orthogonal matrix.

d d Proof. We begin by showing that the inner product of two basis vectors (aix+biy+ciz) and (ajx+bjy+cjz) d nd in {(aix + biy + ciz) }i=1 is invariant under an orthogonal change of basis on V1. Let the orthogonal change of basis be given by the matrix   α1 α2 α3 A = β1 β2 β3 γ1 γ2 γ3 i.e. that x 7→ α1x + α2y + α3z

y 7→ β1x + β2y + β3z

z 7→ γ1x + γ2y + γ3z. The mapping of the two basis vectors then becomes

d d (aix + biy + ciz) 7→ (ai(α1x + α2y + α3z) + bi(β1x + β2y + β3z) + ci(γ1x + γ2y + γ3z)) d = ((aiα1 + biβ1 + ciγ1)x + (aiα2 + biβ2 + ciγ2)y + (aiα3 + biβ3 + ciγ3)z) d d (ajx + bjy + cjz) 7→ (aj(α1x + α2y + α3z) + bj(β1x + β2y + β3z) + cj(γ1x + γ2y + γ3z)) d = ((ajα1 + bjβ1 + cjγ1)x + (ajα2 + bjβ2 + cjγ2)y + (ajα3 + bjβ3 + cjγ3)z) which combined with Lemma 2 gives that

d d h(aix + biy + ciz) , (ajx + bjy + cjz) iVd = (aiα1 + biβ1 + ciγ1)(ajα1 + bjβ1 + cjγ1)

+ (aiα2 + biβ2 + ciγ2)(ajα2 + bjβ2 + cjγ2)

+ (aiα3 + biβ3 + ciγ3)(ajα3 + bjβ3 + cjγ3)

d d By simplifying the expression above, we see that aiaj +bibj +cicj = h(aix+biy +ciz) , (ajx+bjy +cjz) iVd , which shows that the inner product of two basis vectors is invariant under an orthogonal change of basis.

41 Now let f(x, y, z), g(x, y, z) ∈ Vd be given by

X i j k f(x, y, z) = aijkx y z i+j+k=d and X i j k g(x, y, z) = bijkx y z i+j+k=d respectively. We then have that ! X X i j k l m n hf(x, y, z), g(x, y, z)iVd = aijkblmnhx y z , x y z iVd i+j+k=d l+m+n=d

which combined with the previous result shows that hf(A(x, y, z)), g(A(x, y, z))iVd = hf(x, y, z), g(x, y, z)iVd . Corollary 8. The inner product deﬁned by equation (8) is invariant under the natural representation of SO(3) in C[x, y, z].

Proof. The natural representation of SO(3) in C[x, y, z] constitutes an orthogonal change of basis on V1. Since the inner product is invariant under the representation which we study, we know that the orthogonal ⊥ complement L(Vd−2) is a SO(3)-invariant subspace of Vd, that is,

⊥ Vd = L(Vd−2) ⊕ L(Vd−2). (11)

⊥ Since we concluded that the complement of L(Vd−2) in Vd has to be irreducible, we know that L(Vd−2) is irreducible (since it is the complement). We now wish to see if we can find a more explicit expression for the ⊥ space L(Vd−2) . To do this we start by defining a new differential operator.

Deﬁnition 44. Let f(x, y, z) be a polynomial in C[x, y, z]. Deﬁne the Laplace operator ∆ ∈ End(C[x, y, z]) by ∂2 ∂2 ∂2 ∆(f(x, y, z)) = + + f(x, y, z). ∂x2 ∂y2 ∂z2

We also deﬁne the restriction of this operator to the subspace Vd as ∆d : Vd → Vd−2 (for d ≥ 2). This restricted operator will be important, since we intend to study its kernel.

A polynomial f : U → C, where U is some open subset of R3, that satisﬁes Laplaces equation ∆f = 0 (i.e. is in the kernel ker(∆)) is called a harmonic polynomial. We will refer to the elements of ker(∆d) ⊂ ker(∆) as homogeneous harmonic polynomials. It is clear that any harmonic polynomial can be represented as a sum of homogeneous harmonic polynomials:

∞ M ker(∆) = ker(∆d). d=0

To shorten the notation, we will write Hd = ker(∆d). We can now provide the following theorem, which will ⊥ give some insight into what the elements of L(Vd−2) actually are.

Theorem 19. The orthognal complement of L(Vd−2) is equal to the space of all homogeneous harmonic polynomials of degree d, i.e. the kernel of the Laplace operator restricted to Vd:

⊥ L(Vd−2) = Hd.

42 ⊥ Proof. We begin by showing that Hd ⊂ L(Vd−2) . Let f(x, y, z) be an element of Hd, and let g(x, y, z) be 2 2 2 a polynomial in L(Vd−2), that is g(x, y, z) = (x + y + z )h(x, y, z) for some polynomial h(x, y, z) in Vd−2. We have that ∂ ∂ ∂ ∂2 ∂2 ∂2 hg(x, y, z), f(x, y, z)i = h , , + + f(x, y, z) = 0, Vd ∂x ∂y ∂z ∂x2 ∂y2 ∂z2

⊥ which shows that Hd ⊂ L(Vd−2) . ⊥ ⊥ We now show that L(Vd−2) ⊂ Hd. Now let f(x, y, z) be an element of L(Vd−2) and let g(x, y, z) = (x2 + y2 + z2)h(x, y, z) be deﬁned as earlier. This means that

∂ ∂ ∂ ∂2 ∂2 ∂2 0 = hg(x, y, z), f(x, y, z)i = h , , + + f(x, y, z). Vd ∂x ∂y ∂z ∂x2 ∂y2 ∂z2

But since this holds for all g(x, y, z) ∈ L(Vd−2) and therefore all h(x, y, z) ∈ Vd−2, it must be that case that ∂2 ∂2 ∂2 ⊥ ∂x2 + ∂y2 + ∂z2 f(x, y, z) = 0, i.e. f(x, y, z) ∈ Hd, which shows that L(Vd−2) = Hd.

† Remark. This theorem can also be proven by realizing that ∆d = (L|Vd−2 ) , where L|Vd denotes the restriction of L to the subspace Vd. That is,

hL|Vd−2 (f(x, y, z)), g(x, y, z)iVd = hf(x, y, z), ∆d(g(x, y, z))iVd−2

for all f(x, y, z) ∈ Vd−2 and g(x, y, z) ∈ Vd. Given this result, the theorem follows directly from Theorem 2, ⊥ † which says that L(Vd−2) = ker((L|Vd−2 ) ) = ker(∆d) = Hd. We have thereby shown that Vd = Hd ⊕ L(Vd−2), (12)

and that Hd is irreducible. Equation (12) tells us that

dim(Vd) = dim(Hd) + dim(L(Vd−2))

and the fact that dim(L(Vd−2) = dim(Vd−2) combined with Proposition 16 gives

d + 2 d = dim(H ) + , 2 d 2

that is d + 2 d dim(H ) = − d 2 2 = 2d + 1. We have thereby proved the following proposition.

Proposition 19. The dimension of Hd is dim(Hd) = 2d + 1.

We have now found a decomposition of every subspace Vd into irreducible subspaces.

Theorem 20. The subspace Vd decomposes into irreducible subspaces as

( d d 2 L 2 (H0) = L 2 (V0), d is even Vd = Hd ⊕ L(Hd−2) ⊕ L (Hd−4) ⊕ · · · ⊕ d−1 d−1 L 2 (H1) = L 2 (V1), d is odd.

Proof. We have that Vd decomposes into two subspaces

Vd = Hd ⊕ L(Vd−2).

Applying this repeatedly combined with Proposition 1 proves the theorem.

43 Suppose that W ⊂ C[x, y, z] is some arbitrary space of third degree polynomials with complex coefficients. We denote the restriction of all polynomials in W to the unit 2-sphere S2 = {(x, y, z) ∈ R3 : x2 +y2 +z2 = 1} 2 2 2 as W (S ). We see that the restriction of L to W (S ), which we denote L|W (S2) ∈ End(W (S )), is equal to 2 2 2 2 2 2 the identity function, since L|W (S2)(p(x, y, z)) = (x + y + z )p(x, y, z) = p(x, y, z) if x + y + z = 1, for all p(x, y, z) ∈ W (S2). This gives the following corollary. 2 Corollary 9. The restriction of Vd to the unit 2-sphere S decomposes as 2 2 2 2 Vd(S ) = Hd(S ) ⊕ Hd−2(S ) ⊕ Hd−4(S ) ⊕ · · · The harmonic homogeneous polynomials of degree d restricted to S2 are called spherical harmonics of m degree d. There are 2d + 1 linearly independent spherical harmonics of degree d, each denoted Yd (x, y, z), where −d ≤ m ≤ d. The reader might recognize these from a course in partial differential equations. Corollary 9 tells us that any polynomial in three variables restricted to S2 can be written uniquely as a linear combination of spherical harmonics. There is an interesting consequence of this, which depends on the Stone-Weierstrass theorem. Before we state the theorem, we introduce some new concepts: Let X be a compact Hausdorff space and let C(X, R) denote the algebra of continuous real-valued function on X. This algebra is endowed with the topology of uniform convergence, that is, the topology derived from the uniform metric d(f, g) = sup |f(x) − g(x)|. x∈X We say that a subalgebra A ⊂ X separates points if, for every two different points x, y ∈ X, there exist a function f ∈ A such that f(x) 6= f(y). Theorem 21 (Stone-Weierstrass). Let X be a compact Hausdorff space, and let A be a subalgebra of C(X, R) that contains a non-zero constant function. Then A is dense in C(X, R) if and only if it separates points. Proof. Eee [Young, 2006] A direct consequence of this theorem is that any continuous function on S2 can be approximated uniformly on S2, and therefore also in the space¡ L2(S2), by spherical harmonics. An interesting thing to note is that we have actually found all irreducible representations of SO(3). More precisely, there is no irreducible representation of SO(3) that is not isomorphic to Hd for some d ∈ N. We will state this as a theorem, but we will not prove it. Theorem 22. Let ρ ∈ Hom(SO(3), Aut(V )) be an irreducible representation of SO(3) in some vector space V . Then ρ is isomorphic to ΠHd for some non-negative integer d.

8.2 Tensor products of irreducible representations Now that we have acquired all the irreducible representations of SO(3), it is natural to ask if there is a slick way of decomposing tensor product of these. That is, we want to decompose

Hc ⊗ Hd into irreducible representations. We will do this by means of character theory. By Proposition 13 we know that the character χ of Hc ⊗ Hd is given by c ! d ! X X χ(t) = 1 + 2 cos(nt) 1 + 2 cos(nt) n=1 n=1 c ! d ! X X = eint eint n=−c n=−d c d X X = ei(n1+n2)t

n1=−c n2=−d c+d X = ψHn (t). n=|c−d|

44 This proves the following theorem. Theorem 23. The tensor product of two irreducible representations of SO(3) is decomposed as

c+d ∼ M Hc ⊗ Hd = Hn. n=|c−d|

2 2 It is also interesting to see how the spaces Λ (Hd) and Sym (Hd) decompose into irreducible representations. Again, we use character theory to tackle this problem. Proposition 14 gives that the character of 2 Sym (Hd) is given by

1 2 χ 2 (t) = (ψ (t) + ψ (2t)) Sym (Hd) 2 Hd Hd  d !2 d ! 1 X X = 1 + 2 cos(kt) + 1 + 2 cos(2kt) 2   k=1 k=1

 d !2 d ! 1 X X = eikt + e2ikt 2   k=−d k=−d d X = ψH2k (t). k=0 This shows that d 2 ∼ M Sym (Hd) = H2k = H0 ⊕ H2 ⊕ · · · ⊕ H2d. k=0 ∼ 2 2 Since Hd ⊗ Hd = Sym (Hd) ⊕ Λ (Hd), we also get that

d 2 ∼ M Λ (Hd) = H2k−1 = H1 ⊕ H3 ⊕ · · · ⊕ H2d−1. k=1

45 References

[Basener, 2006] Basener, W. F. (2006). Topology and its Applications. John Wiley & Sons. [Dummit and Foote, 2004] Dummit, D. S. and Foote, R. M. (2004). Abstract Algebra. John Wiley & Sons. [Gleason, 2010] Gleason, J. (2010). Existence and Uniqueness of Haar Measure.

[Lay, 2006] Lay, D. C. (2006). Linear Algebra and Its Applications. Pearson. [Roman, 2008] Roman, S. (2008). Advanced Linear Algebra. Springer. [Sadun, 2001] Sadun, L. (2001). Applied Linear Algebra. Prentice Hall. [Serre, 1977] Serre, J.-P. (1977). Linear Representations of Finite Groups. Springer.

[Weyl, 1931] Weyl, H. (1931). The Theory of Groups and Quantum Mechanics. Dover. [Young, 2006] Young, M. (2006). The Stone-Weierstrass Theorem.