Group Theory Crash Course

Group Theory Crash Course M. Bauer WS 10/11 1 What is a group? In physics, symmetries play a central role. If it were not for those you would not find an advantage in polar coordinates for the Kepler problem and there would be no seperation ansatz to help you solving the Schrödinger equation of the Hydrogen atom. The mathematical structures behind symmetries are groups. For more complex theories, such as QCD, where there is no analytic solution known, identifying the underlying symmetry group helps a lot for understanding the theory. There have even been Nobel prizes for guessing the right symmetry group of electroweak interactions 1, at least that should motivate you to understand group theory. Let us start with the definition: A group (M, ) is a set M with a group multiplication, with the properties E Existence and uniqueness (or closure): g 1, g 2, g 3 M, g 1 g 2 = g 3 lies in M. A Associativity: g 1, g 2, g 3 M, (g 1 g 2 ) g 3 = g 1 (g 2 g 3 ). N Neutral element: e M, so that g M, g e = e g = g. I Inverse element: g M there is a g 1 M, so that g 1 g = g g 1 = e. (C Commutativity g 1, g 2 M, g 1 g 2 = g 2 g 1 ). If the last requirement is fulfilled the group is called abelian. A subgroup H = (N, ) is a subset N M, which is a group with the same group multiplication as G = (M, ). One writes H < G. Also, Abelian groups can only have abelian subgroups. Above, Group multiplication does not have to be literally multiplication. One can take any binary operation (addition, matrix multiplication, composition of functions, etc.). Further, we distinguish finite and infinite groups: A finite group is a group with a finite number of elements. The number of elements is called the order of the group G. If the number of elements is infinite, the group is called infinite group. In this case group elements are labelled by one or more parameters instead of an index like in the definition above 2, g i g(a) 1 Weinberg, Glashow, Salam 1979 2 Just like the transition from many particles- to field theory, where you can describe a stretched string as the limes of infinitely many harmonic oscillators and finally exchange their label by the parameter of the field. 1

The number of (real) parameters is called the dimension of the group. Z 2 = ({1, 1}, ). The group structure (that is the result of all possible combinations of elements) can be expressed through a multiplication or Cayley table. Now for a few examples. Lets consider a unilateral triangle and the symmetry transformations which project it onto itself. Besides the identity, there are rotations about the central point of 120 and 240 which leave the triangle invariant. If we name these operations d and d 2, we already found ourselves the alternating group 3 A 3 = ({e, d, d 2 }, ), where denotes the composition of operations. We can go on and add reflections on the bisecting lines of the angles, lets call them s 1, s 2, s 3 respectively. We have found an even larger group, namely the symmetric group 4 S 3 = ({e, d, d 2, s 1, s 2, s 3 }, ). Both groups are obviously finite and therefore their respective group structure (that is the instruction how to combine all possible group elements) can be represented in a so called multiplicative or Cayley table, see Figure 1. You will have noticed that A 3 is a subgroup of S 3. We won t find any larger symmetry groups by looking at a triangle. We can of course go on and draw squares and hexagons and more and more complicated polygons and step by step classify the finite groups. However, since everyone of us knows how to compute π, it should not be surpising that if we instead draw the outer circle of the triangle above and look at its symmetry group we will find a larger group which will include many of these finite polygon groups as subgroups. This group is called unitary group of dimension one or circle group U(1) = ({d(φ)}, ). Here, d(φ) denotes the generalization of d, d 2 above, denoting rotations about any angle φ. The index has been traded for a parameter. Since it includes rotations about any angle, U(1) is an infinite group. Most probably you are used to identify U(1) = ({e i φ }, ) and the above notation looks clumsy to you. But this is already a representation of U(1), which should be well distinguished from the abstract group, as we will learn in Section 3. Exercise [1pt] Which of the groups A 3, S 3 is a subgroup of U(1) and why? Instead of drawing the multiplication table, one can define a function g k = g i g j k = f(i, j), with i, j, k {1..., G }. Here, f is called composition function and gives the index of the group element which is given by the product of the ith with the jth group element. This is a non-continuous function mapping {1,..., G } R to itself. While multiplication tables cannot be generalized to infinite groups, the composition function can, g(c) = g(a) g(b) c = f(a, b), with a, b, c R. 3 The alternating groups A n contain the n!/2 even permutations of n elements. 4 The symmetric groups S n contain the n! permutations of n elements. 2

e d d 2 s 1 s 2 s 3 e e d d 2 s 1 s 2 s 3 d d d 2 e s 3 s 1 s 2 d 2 d 2 e d s 2 s 3 s 1 s 1 s 1 s 2 s 3 e d d 2 s 2 s 2 s 3 s 1 d 2 e d s 3 s 3 s 1 s 2 d d 2 e Figure 1: Unilateral triangle and its symmetry groups A 3 and S 3. The red part denotes the Cayley table of A 3 embedded into the Cayley table of S 3. The properties of the composition function now allow for further classification of infinite groups. It can be not continuous 5, continuous or even analytical 6. In the last two cases one speaks of continuous or Lie groups, respectively. Let us look at an example: (Z, +) < (Q, +) < (R, +) These groups are all infinite, and Q is in addition continuous, but no Lie group, while R is a Lie group. Just think of the smoothness of the function which describes the combination of elements of these groups. In physics, Lie groups are the most important groups. Because only Lie groups give rise to conserved Noether currents. In addition, finite groups are typically subgroups of some Lie group. In the next section we will therefore put our focus on Lie groups. But before that, I want to fit in one more definition: Groups with a bounded parameter space are called compact. Compact groups are in a way generalizations of finite groups (all finite groups are compact) and their representation theory is very well understood. On the other hand, if a group is not compact a lot of nice theorems do not hold and since non-compact groups are of physical relevance it is important to be able to handle them. 5 To be mathematically strict, with a discrete topology even finite groups are continuous(those are then called discrete groups). What I mean by continuous is continuous with respect to the standard topology of R. 6 That is, f can be represented by a convergent Taylor series. 3

Let me further introduce the notion of a matrix group. Matrix groups are sets of N N dimensional matrices over some field (R, C) with matrix multiplication as group multiplication. All matrix groups are subgroups of GL(N, F), which is the general linear group, the group of all invertible matrices over the field F. Matrix groups are probably what you are familiar with and know as a group. The reason is that a lot of groups can be represented by matrix groups. Namely, Every finite group is isomorphic to a matrix group. Every compact Lie group is isomorphic to a matrix group (Peter-Weyl Theorem). We will therefore generalize the definition and speak of a matrix group whenever the group at hand is isomorphic to a matrix group. This holds also for most non-compact Lie-groups. There exist counterexamples, but they are rare. I will give one later on, but it needs more math to understand it. Also, the group multiplication is always matrix multiplication for matrix groups and therefore I will just define the set of matrices from now on. 2 Lie Groups & Algebras First of all, Lie groups are differentiable manifolds. Instead of defining them as infinite groups with an analytic composition function one can define them as manifolds with a group structure. This endows them with a vivid geometric meaning. Therefore it is worth to work through a bit more mathematical formalism, in order to understand these statements. Let us start with manifolds. I will not give the rigorous mathematical definition. For our purpose a manifold is a space M that looks like R n if one zooms in, but on a large scale it can be curved. One has than a number of charts, maps φ : M R n that map those locally flat regions to R n. In this notation n is called the dimension of the manifold. The dimension of a Lie group, which we have defined earlier as the number of parameters the group elements depend on, is also the dimension of the corresponding manifold. As an example let us look at U(1) again. Each element of the U(1) can be identified with a point on a circle. Since the circle is a one-dimensional manifold, the dimension of U(1) is one. We might as well count the parameters and find that it depends only on one parameter, the angle. Dont get confused here, the N in GL(N, F) is not the same as the dimension of the group/ the corresponding manifold. A general invertible N N - matrix rather depends on N 2 parameters, so that n = N 2. We can get a better understanding of the connection of these parameters and the dimension of the group manifold if we introduce the notion of a tangent space. Imagine a point x on a manifold M. If one collects all possible curves γ : R M on the manifold going through that point x, and builds their respective tangential vectors v at γ(0) = x, v = d dt γ(t) t=0 T x M they span a vector space attached at x with the same dimension as the manifold. That is the tangential space T x M. An illustration is given in Figure 2. 4

TxM υ x γ(t) M Figure 2: Illustration of the definition of a tangent space. For all manifolds we care about (that would be semi- Riemann or Riemann manifolds) one can locally define an exponential map. An exponential map is a map which goes from the tangent space at some point x to the underlying manifold, exp : T x M M v exp x (v) = γ v (0). That means, if you have a tangential vector in T x M the exponential map gives you the corresponding curve on M in a neighborhood of the point x. For manifolds formed by matrix groups M = G, the exponential map can be defined as the matrix exponential exp V = k=0 V k k! = 1 + V + 1 2 V 2 +... Now you should recall how we represented the elements of U(1) in section 1, which is isomorphic to the group of 1 1 complex matrices with norm 1. For the unit circle in the complex plane, the tangent space at each point is the imaginary line, which is R (if we take out an i). Writing a vector in R as its component times the only basis vector 1, we see that all elements of U(1) can be written as g(φ) = exp(i φ 1), with φ R. More general, elements of an arbitrary matrix group g G, which are close to the identity (since exp is only locally defined) can always be written as g = exp(i n α a T a ), (T a are matrices here), a=1 where n is the number of basis vectors T a of T e G and equals the dimension of the Lie group. Because of the closure requirement, every group element can then be generated (by repeated multiplication) out of those in the vicinity of the identity. Therefore these basis vectors are called infinitesimal generators of the Lie group.and calls the basis vectors T a generators. 5

z z y y x x Figure 3: Illustration of the non-commutativity of SO(3). Starting point is the red vector. The green vector is the result of first performing a rotation of 45 about the the z axis and than a rotation of 45 about the y axis, while the blue vector is the result when we first rotate about the y axis and than about the z axis (all rotations counterclockwise). Again, for the U(1) that means, that if we choose an element close to the identity, that is φ = 0 + ɛ with ɛ 1, we can write g = exp(i φ) = exp(i ɛ) exp(i ɛ)... exp(iɛ) = exp(i [ɛ + ɛ +... ɛ]), which justifies the statement above, but is pretty boring, since there is just one generator. Let me quote another example. The group of all rotations in R 3 is isomorphic to the matrix group SO(3) = { R GL(3, R) R T R = 1 det(r) = +1 }, the special orthogonal group in 3 dimensions. As a matrix group it is embedded in GL(3, R). One calls it special, because it is a subgroup of the orthogonal group in 3 dimensions, O(3), where the requirement det(r) = 1 is dropped. We know that any arbitrary rotation in R 3 can be described by successively rotating about all 3 axes (as soon as a coordinate system is chosen): R(φ, θ, ψ) = R x (φ)r y (θ)r z (ψ) = exp(i φt x ) exp(i θt y ) exp(i ψt z ) = exp(i 3 α a T a ). We also know, that rotating first about the x-axes and than about the y-axes is not the same as rotating first about the y- and than about the x-axes as shown in Figure 2, which simply translates into SO(3) being non-abelian. In general: R(α)R(β)R 1 (α)r 1 (β) = R(θ) 1. Upon expanding the group elements this yields (from now on I will use Einstein notation), 1 α a β b (T a T b T b T a ) + O(α 2, β 2 ) = 1 + iθ c T c + O(θ 2 ), [T a, T b ] = i f abc T c. a=1 6

Where [, ] is the commutator and the symbols f abc are called structure constants and are antisymmetric in the first two indices. It should be clear that the commutator only vanishes if the group is abelian. Exercise [3pt] What can you say about the generators of SO(3) just from the group definition? (Hint: Write R T R = 1 and det(r)=1 in exponential form.) Find real 3 3 matrices which have the properties you just derived (choose a basis for T e SO(3)), and compute the structure constants f abc of SO(3). With the bilinear operation [, ] : T e G T e G T e G we have found additional structure on the tangent vectors of the Lie group. A vector space with such an inner product is called an algebra. The commutator of matrices is in addition antisymmetric and since associativity of the matrix product on group level implies associativity of multiplication on the level of the generators, the Jacobi identity holds. We can therefore define: A Lie algebra is a vector space V which is closed under a bilinear operation [, ] : V V V, such that 1. [v, w] = [w, v] v, w V 2. The Jacobi identity [v, [w, z]] + [z, [v, w]] + [w, [z, v]] = 0 holds. Viewed in the geometrical context this tells us that not all differentiable manifolds are Lie groups. Only if the tangent space can be made a Lie algebra, the manifold is a Lie group. An example for a manifold which is no Lie group is the 2-dimensional sphere pictured in Figure 2. Even though the elements of the algebra and the elements of the group are both represented by n n matrices you must keep in mind that you are dealing with different algebraic and geometric objects here. For example the closure requirement makes sure that the matrix product of two elements of the group does not lead you out of the group, the same does not hold for the elements of the algebra. On the other hand, since the Lie algebra is a flat vector space and thus is isomorphic to R n it is closed under addition. That does not hold for the Lie group. Just think of two matrices A, B SO(3) having determinant 1. The sum of them has determinant 2 and is no element of SO(3). I want to introduce a few more geometrical terms. One says a manifold M is pathconnected if for every two points x, y M on a manifold there is a continuous curve γ : [0, 1] M with γ(0) = x and γ(1) = y. The number of connected components is the number of segments of the manifold which cannot be joined by such a curve. Let us look at the Lie groups we have discussed so far. The circle and therefore U(1) is connected. What about SO(3)? In this case and for many others where we cannot simply draw the manifold and look at the picture we will have to find another way to tell about their properties. A conveniant way is to look at the determinant as a continuous function from the manifold/group into the real numbers, det : SO(3) R. One can show that continuous functions preserve the number of connected components (as well as they preserve compactness) and since det(r) = 1 R SO(3) and 1 has one connected component and is compact we can deduce that the same holds for SO(3). 7

What about orthogonal matrices which are not special? For those we have O(3) = {M GL(3, R) M T M = 1} det(m T M) = 1 det(m) = ±1. We thus find that O(3) has two connected components (but is still a compact group). I introduced these concepts because what I said earlier in this chapter is not (entirely) true. You can only generate those elements of the Lie group from its algebra, which are in the connected component of the identity. If you don t believe me try to generate the parity P = diag( 1, 1, 1) with infinitesimal rotations using the traceless generators you found in the last exercise. On the other hand, as long as the manifold is not totally disconnected (and then its no longer a continuous group), one can generate every element of the connected component of the one and add the additional discrete transformations to go from there to any other connected component of the group and thus generate any element of the group. Therefore one can write O(3) = SO(3) P SO(3) or O(3) = SO(3) {1, P }. Bonus Introduce simply connected,fundamental group, universal cover and how one lie algebra can generate different Lie groups. Introduce 1 Z 2 SU(2) SO(3) 1. 3 Representations In physics, when we work with groups we almost always work with representations of groups. In general, a representation of an abstract group G on some vector space V of dimension d is defined as a linear map D : G GL(d, V ), D(g) GL(d, V ) : V V which satisfies the Homomorphism property (it should preserve the structure of the group) D(g 1 )D(g 2 ) = D(g 1 g 2 ). Note, that GL(d, V ) is not the general linear group in which the matrix group G is imbedded. This would be GL(N, V ), where n = N 2 denotes the number of generators (or the dimension of the group manifold), while d is the number of basis vectors of the vector space the group is represented on. In a representation, the exponential map reads D(g) = exp ( i d α a D(T a ) ) where D(T a ) is the representation of the Lie algebra elements. In addition, a representation is called faithful, if it is injective. That is, D(g i ) D(g j ) g i g j (the indices are just for conveniance, all this holds also for continuous groups). 8 a=1

Let us have a look at a few examples. Consider the 2-dimensional representation of U(1) on C 2, given by ( ) D(g) GL(2, C 2 e iφ 0 ), D(g) =. 0 e iφ In this example, the representation is faithful, but n = 1 and d = 2. Another example is the trivial representation, which sends all group elements to the identity. This can never be a faithful representation, if the group is anything else besides the trivial group G = {e}. Let me give another example for a representation which is not faithful, the representation of O(3) on R, D(R) GL(1, R), D(R) = det(r), where n = 3 and d = 1. Every element in one connected component is send to the same D(g). That would however be a faithful representation of Z 2 = {1, 1}. A very important representation is the fundamental representation, which is the smallest dimensional representation which is faithful. For the fundamental representation, d = N. The example above is the fundamental representation of Z 2. The fundamental representation for the SO(3) is given by the real 3 3 matrices for which R T R = 1 and det R = 1. Active rotations about the 3 axes are then given by 1 0 0 cos θ 0 sin θ cos ψ sin ψ 0 R x (φ) = 0 cos φ sin φ, R y (θ) = 0 1 0, R z (ψ) = sin ψ cos ψ 0. 0 sin φ cos φ sin θ 0 cos θ 0 0 1 Exercise [3pt] Derive the fundamental representation of the generators with these matrices via the exponential map, n R = exp(i α a D(T a )) D(T a ) = i R α a a=1 and check what you found in the last exercise (your result should agree up to a similarity transformation). You will probably protest now and say that we defined the abstract matrix groups SO(3) in section 2 as the same thing which I now try to introduce as the fundamental representation. And you are right. The general abstract definition would have been: R SO(N) are those matrices which preserve the euclidean scalar product on the vector space R N and lie in the connected component of the identity, for y i R i j x j y i δ ij y j = R i l x l δ ij R j k xk = x l (R T ) i l δ ij R j k xk! = x i δ ij x j (R T ) i l δ ij R j k = δ lk. If one now chooses a representation 7 of R N, these matrices can explicitly be given as a representation of the group on this vector space. The fundamental representation is one 7 That means, choose a basis for R N. 9

possibility. But since all properties can be derived from the fundamental representation, it is also called defining representation in the literature. If we define a group via its fundamental representation, we can write for the according vector space with dimension dim V = d! = N for a vector x V, D(g) x = g x, and D(T a ) = T a. Further, one defines the conjugate representation (or dual representation) D(g), for which also d = N, by D(g) x = g x, where the denotes complex conjugation. It follows immediately for the conjugate representation of the generators, exp(iα a T a ) = exp( iα a T a ) D(T a ) = (T a ). For real groups like SO(3) the fundamental and the conjugate representations are isomorph. For U(1) the only generator is also real, so nothing new here. How about the unitary group in N dimensions? In a representation independent way it is defined as the transformations which leave the scalar product on C N invariant. Defining it via its fundamental representation we can write U(N) = {U GL(N, C) U U = 1}. An arbitrary element of the Lie algebra is given by a complex N N matrix and therefore the conjugate and fundamental representations should (in general) not be the same. Exercise [1pt] Ask yourself if the structure constants depend on the representation. What would you expect? (Please look up what Homomorphism property means, if you are unsure). Compute the commutator of the fundamental [D(T a ), D(T b )], and of the conjugate representation [D(T a ), D(T b )] of the generators. Can you derive if the structure constants are in general real of complex? There is yet one more important representation you must know, the adjoint representation. Let me introduce it via an example. Consider the two-dimensional complex vector space C 2. We denote its elements as ψ C 2 and the corresponding adjoint vector as ψ ψ. The scalar product L = ψψ is invariant under the fundamental representation U U(2). If we put a diagonal, universal (that means m 11 = m 22 ) matrix M in there L stays invariant ψ U MU ψ m U 1U = M. How does this change if we replace M by a general matrix A. We can directly read off that L stays invariant if A transforms as A UAU. 10

Since the space of complex 2 2 matrices is isomorph to C 4, we can then define the adjoint representation as a 4 4 matrix D ad (g) acting on the C 4 vector A through the action of the fundamental representation on the corresponding 2 2 matrix D ad (g)a = gag 1, The following steps can be a bit confusing if you have never seen it before, so I recommend careful reading from now on. Again, A denotes a complex 2 2 matrix. Since we know that the fundamental representation of the generators D(T b ) of the Lie algebra of U(2) form a basis for M 2 2 (C) (the tangent space), we can write A as A = 4 A b D(T b ). i=1 It has nothing to do with the group itself. We just decompose A into components times basis vectors. On the other hand, A as a C 4 column-vector can be decomposed into the basis vectors of C 4, A = A i e i, and we can treat it as the same object. Now we put the latter into the left hand side of the definition of the adjoint representation, the first into the right hand side and expand both sides (suppressing higher orders and applying Einstein notation), So that D ad (g) A b e b = U A b D(T b ) U exp ( i α c D ad (T c ) ) A b e b = exp ( i α a D(T a ) ) A b D(T b ) exp ( i α a D(T a ) ) ( 1 + αc D ad (T c ) ) A b e b = ( 1 + i α a D(T a ) ) A b D(T b ) ( 1 i α a D(T a ) ) = A b D(T b ) + i α a A b [D(T a ), D(T b )] = A α a A b f abc T c ( Dad (T c ) ) ab = i f abc = i f bac, since the structure constants are real (check your last exercise). We can generalize the result from U(2) to U(N) and see that the adjoint representation has dimension d = n, where n is the dimension of the Lie group. Let me repeat, the dimension of the fundamental/conjugate representation is given by d = N, whereas the dimension of the adjoint representation is n, the dimension of the Lie group (the number of parameters the group depends on). There is a convention to call the representations by the dimension of the vector spaces they are defined on. The fundamental representation of the SO(3) is called 3 as is the fundamental representation of SU(3). The conjugate representation of SU(3) would be 3, the adjoint representation 8, and the trivial 1. You get the idea. A few more definitions: A representation is called unitary, if the representation matrices are unitary. Note, that that does not mean that the group has to be unitary. Further, two 11

representations D 1 and D 2 of the same group are called equivalent representations, if there exists an invertible linear map S, so that This implies for the generators, D 2 (g) = S D 1 (g) S 1. D 2 (T a ) = S D 1 (T a ) S 1, a = 1,..., n. We can then quote the important theorem, that the representations of every finite group and every compact Lie group are equivalent to a unitary representation. Again, compact Lie groups play a special role. Keep in mind that that also implies that we cannot always find a unitary representation if the group is not compact. Let me give another example. Consider the special unitary group in two dimensions, SU(2) = {U U(2) det U = 1}. Analogue to the exercise you did in the last chapter for SO(3), we can use the fundamental representation to determine number and form of the generators, U = exp (i n α a T a ). We know that n is less then N 2 = 8, since SU(2) < GL(2, C), which just means that the generators will be complex 2 2 matrices. From a=1 U U = 1 exp(i(t a T a )) T a = T a det(u) = 1 Tr(T a ) = 0, we can deduce that they are hermitian, which takes away 4 parameters (the trace is real and the non-diagonal entry is self-conjugated), and traceless, which is one more condition, leaving us with 3 parameters. And everyone of you knows a basis for the space of hermitian, traceless complex 2 2 matrices, the Pauli matrices: ( ) ( ) ( ) σ 1 0 1 =, σ 2 0 i =, σ 3 1 0 =. 1 0 i 0 0 1 Therefore, every element of the SU(2) in the fundamental representation can be written as ( ) 3 σ a D(g) = exp i θ a, 2 a=1 where the factor 1/2 is a convention. D(T a ) = (D(T a )) and thus For the conjugate representation we know that D(T a ) = (σ a ) = { σ 1, σ 2, σ 3}. 12

If you know remind that σ 2 { σ 1, σ 2, σ 3} σ 2 = { σ 1, σ 2, σ 3}, you can check that the fundamental and the conjugate representations of SU(2) are equivalent, using S = σ 2 and the equivalence relation ( ) 3 S D(g) S 1 = σ 2 σ a exp i θ a σ 2 = D(g) or 2 2. 2 This is a speciality of SU(2), which does not hold for higher SU(N) s. a=1 As promised earlier, every other representation can be constructed from the fundamental representation. In order to see this we need to generalize the notion of direct sums and Kronecker (tensor) products from matrices to representations. A direct sum of a vector space is a way to get a new big vector space from two (or more) smaller vector spaces. One denotes the direct sum by V = V 1 V 2. Let me demonstrate the direct sum of two vector spaces of column vectors. Consider a vector x V 1, which has dimension n and y V 2 which has dimension m. Than, the direct sum x y V = V 1 V 2 can be written as x 1 x 1. 0.. x y = x n + 0 y = x n 1 y n + m 1 0... 0 y m and the resulting vector space V has dimension n + m. Consider now a quadratic matrix A acting on x and a quadratic matrix B acting on y, via y m x A x and y B y. The obvious way to define the direct sum is ( ) A 0 n m A B =, 0 m n B what becomes clear by acting on the direct sum of the vectors, ( ) ( ) A 0 n m x (A B) ( x y) = = (A x) (B y). 0 m n B y Keep in mind that there are n 2 independent matrices on V 1 and m 2 independent matrices on V 2, but (n + m) 2 independent matrices on V = V 1 V 2. The remaining 2nm matrices cannot be written as a direct sum. Now, replace matrices by representations and you know what the direct sum of two representations is. 13

Excersice and B. Express det(a B) and Tr(A B) through the determinants and traces of A The tensor product is another way to construct a bigger vector space out of two or more smaller ones. I will also explain this with an example. Let us take the same vector spaces as above and build W = V 1 V 2. In order to simplify notation, let me set n = 2 and m = 3. Then, the direct product of our two vectors reads x 1 y 1 ( ) ( ) x 1 y 2 x ( ) 1 x 1 y 1 x 1 y 2 x 1 y 3 x y = y 1 y 2 y 3 = x 1 y 3 x 2 x 2 y 1 x 2 y 2 x 2 y 3 x 2 y n m,, 1 x 2 y 2 x 2 y 3 where the last two notations are equivalent. Proceeding with the latter, the tensor product (also called Kronecker product) of the two matrices acting on these vectors is defined by ( ) b 11 b 12 b 13 a 11 a 12 A 2 2 B 3 3 = b 21 b 22 b 23 a 21 a 22 b 31 b 32 b 33 a 11 b 11 a 11 b 12 a 11 b 13 a 12 b 11 a 12 b 12 a 12 b 13 a 11 b 21 a 11 b 22 a 11 b 23 a 12 b 21 a 12 b 22 a 12 b 23 = a 11 b 31 a 11 b 32 a 11 b 33 a 12 b 31 a 12 b 32 a 12 b 33 a 21 b 11 a 21 b 12 a 21 b 13 a 22 b 11 a 22 b 12 a 22 b, 13 a 21 b 21 a 21 b 22 a 21 b 23 a 22 b 21 a 22 b 22 a 22 b 23 a 21 b 31 a 21 b 32 a 21 b 33 a 22 b 31 a 22 b 32 a 22 b 33 so that (A B) ( x y) = (A x) (B y). You can convince yourself that the last line holds by explicitly writing it out. The point is that the direct sum as well as the tensor product of representations are again representations. Consider now a representation D(g) on some vector space W (this might well be the tensor product of two vector spaces). If it is possible to find a sub vector space V W, which is left invariant by the operation of D(g), D(g) x V x V and g G, the representation is said to be reducible. That means it can be reduced. On the level on the matrix you can find an equivalence transformation, so that ( ) D V (g) α(g) D V (g) : V V D(g) = D W \V (g) : W \ V W \ V 0 D W \V g α(g) : W \ V V 14

If one cannot find such a transformation, that means the vector space on which the representation acts does not have a (non-trivial) invariant subspace, the representation is called irreducible. And if you can decompose a vector space into a direct sum of sub vector spaces W = V 1 V 2 which all transform under irreducible representations, W is called completely reducible. For the representation matrices above this implies that α(g) = 0, W \ V = V 2 and all representations D V1 (g) and D V2 (g) are irreducible. In general a completely reducible representation can be brought in a blockdiagonal form, D 1 (g) D 2 (g) = D a Db D c... where all representations on the blockdiagonal are irreducible and depend on the same group element g. Not all representations are completely reducible. However, there is a powerful theorem which guarantees that every (reducible) unitary representation is completely reducible. One more reason why compact groups are so well understood. Quantum numbers are labels of representations, be it charge or spin or mass. They tell us under which representation of the corresponding group (through Noethers Theorem) the field which describes the respective particle transforms. There is an abuse of language since people say that a particle is in the fundamental representation (for example), which means that the field is a vector of the vector space on which the fundamental representation acts. When we want to know the quantum numbers of bound states (protons, atoms, etc.) we built the tensor product of the representations of the involved fundamental particles (which transform under irreducible representations) and decompose the product into a direct sum of irreducible representations, D 1 D 2 = D a D b D c..., where the right hand side decomposes in a unique way. Thus, in some sense, you can see irreducible representations as the prime numbers of representation theory. It makes therefore sense that fundamental particles transform according to them, since we expect everything we see to be build up from elementary particles, just as every representation is built up from irreducible representations (and every number from prime numbers). When I say, build the tensor product of irreducible representations, I mean irreducible representations of the group. However, as it is often the case, the whole procedure can be simplified by going to the level of algebras. It is way easier to identify the generators then the group element in some representation. We can expand the exponential map again and 15

use that D 1 (g) D 2 (g) = (1 + i α a D 1 (T a )) (1 + i α a D 2 (T a )) + O(α 2 ) = (1 D 1 (T a ) + D 2 (T a ) 1) iα a + O(α 2 ) which we decompose into a direct sum and write it as an exponential again. In that sense you can understand the term, addition of angular momenta in the case of SU(2). Let me illustrate this with a few examples. If we describe a coupled system of two charged particles, for example an electron and a proton, we know already (since we consider charge a conserved quantum number) that the outcome carries charge zero. This symmetry is related to the invariance of the lagrangian under U(1) through Noethers theorem. In representation theory what we have done by reading off the possible charges of the final state corresponds to building the direct product of the representations under which our input particles transform and compute the possible irreducible representations of this direct product, what gives us the possible final state charges. All irreducible representations of U(1) are of dimension 1, because it is an abelian group 8 and can therefore be written as D l (g) = exp(i l φ) l Z. That l must be in Z follows from the fact that φ is 2 π periodic. Therefore two representations are the the same if D l (φ 1 ) = exp((i l φ)) and D l (φ 2 ) = exp((i l φ 2 )) = exp(i l(φ 1 + k 2 π) with k Z exp(2 π l k) = 0 l Z. Because every representation is one-dimensional, one does not use the dimension to label the representations but l itself. Our electron is thus in the representation l = 1 and our proton transforms under the representation l = 1. the direct product of the generators is easily computed and confirms our result, l 1 + 1 (l) = l + l = 0 which is the trivial representation and by itself already irreducible, so we have not to convert it into a direct sum. The right hand side is a one-dimensional block-diagonal matrix, a number. Let us now consider the spin of said particles. Electrons and protons carry spin 1/2 and you know that the resulting system can either have spin 1 or spin 0 from the addition of angular momenta. This however is another example for representation theory at work. Both fields transform under the fundamental representation of SU(2) (which is what spin 1/2 means), and we know its generators are given by the pauli matrices. We can now build the direct 8 This is a consequence of Schurs Lemma. Suppose that we have a representation of an Abelian group of dimensionality d is greater than one. Suppose furthermore that these matrices are not all unit matrices (for, if they were, the representation would already be reducible to the d-fold direct sum of the identical representation.) Then, since the group is Abelian, and the representation must reflect this fact, any nonunit matrix in the representation commutes with all the other matrices in the representation. According to Schur s First Lemma, this renders the representation reducible. Hence, all the irreducible representations of an Abelian group are one-dimensional. 16

product of these generators which gives us the representation under which the final state transforms, 0 1 1 0 1 0 0 1 J x =σ x 1 + 1 σ x = 1 0 0 1 0 1 1 0 0 i i 0 i 0 0 i J y =σ y 1 + 1 σ y = i 0 0 i 0 i i 0 2 0 0 0 0 0 0 0 J z =σ z 1 + 1 σ z = 0 0 0 0 0 0 0 2 With the help of the a unitary matrix 9 U, these can be transformed into a direct sum, that is we choose the appropriate basis vectors in the product space and get the generators in blockdiagonal form, 0 0 0 0 U J x U 0 0 1 0 = 0 1 0 1 0 0 1 0 0 0 0 0 U J y U 0 0 i 0 = 0 i 0 i 0 0 i 0 0 0 0 0 U J z U 0 1 0 0 = 0 0 0 0 0 0 0 1 which is the generator of the trivial representation in the upper left corner and the generators of SO(3) in the second block. They agree with what we have found earlier, if you choose a 9 Derived in the appendix 4, 17

equivalence transformation and make the J z generator diagonal. We have found, that the hydrogen atom can transform as a Singlet or as a Triplet state. In short one writes, 2 2 = 1 3. For larger systems and groups, this calculation becomes more and more complicated. There is a systematic way to decompose a tensor product into a direct sum, making use of socalled Young tableaus. I will not introduce them here, but you should definitely read up on it. We will rather go a more constructive way and see what we can conclude from the symmetry properties of the product spaces the representations act on. We will restrict ourself to the orthogonal and unitary groups, since every group we care about here is at least locally isomorph to those. An element of the product space of two (or more) copies of the same vector space is the most common way to define a (second rank) tensor. From the discussion above, you can see that this is a special case of a tensor product. Take for example a vector space V with dimv = 3, v 1 ) v 1 w 1 v 1 w 2 v1w 3 T = v w = (w 1 w 2 w 3 = v 2 w 1 v 2 w 2 v 2 w 3 for v, w V v 2 v 3 v 3 w 1 v 3 w 2 v 3 w 3 which is also called a dyadic product in older textbooks. From this expression, we can understand why people often define tensors as objects which transform as tensors. Or, as an object which transforms as if it is equal to the product T ij v i w j. Therefore, let us assume v i = D(g) ij v j transforms under the fundamental representation of SO(N). Then T ij T ij = D(g) il D(g) jm T lm. If we decompose this tensor into a symmetric and an antisymmetric part, S ij = 1/2(T ij + T ji ) and A ij = 1/2(T ij T ji ), we see that a transformation of those stays symmetric or asymmetric 10. Thus, the product representation acting on T ij is reducible, both the antisymmetric and the symmetric rank 2 tensors transform among themselves and are thus invariant subspaces. In addition, it follows from the orthogonality of the representation matrices that the trace of a symmetric tensor, Tr(T ) = δ ij T ij, also transforms into itself, δ ij T ij = δ ij D(g) il D(g) jm T lm = (D(g) T ) li δ ij D(g) jm T lm = δ lm T lm. The symmetric part can thus be further reduced and we end up with three irreducible subspaces, the antisymmetric tensors, the symmetric traceless tensors and the trace. As an example, for SO(3) this tells us 3 3 = 1 3 5, which reflects the well-known result of two spin-one particles coupling to a spin 0, 1 or 2 object. 10 You can easily read off that D(g) il D(g) kj S kl and D(g) il D(g) kj A kl still have the same symmetry properties. 18

Exercise [3pt] Find a formula to decompose general products N N for SO(N). In the case of SU(N) we have to distinguish between vectors transforming under the fundamental and under the conjugate transformation and tensors in the product space of combinations of those. We denote a vector in V by v i and v i v i = D(g) i jv j and a complex conjugate vector in V as v i, transforming as v i v i = D(g) i jv j = (D(g)) j i v j. Remember that in terms of the representation matrices D(g) = D(g). We can thus introduce the notion v i for a vector transforming as v i. We distinguish them by the position of the indices. We can now have rank 2 tensors transforming as Tj i v i v j v i v j as well as those which transfrom udner the direct product of two fundamental T ij or two conjugate representations T ij. The trace is only an invariant subspace of the Tj i type tensors. One can see this by explicite computation, if TrT δj k T j k, one has δ k j T j k = δk j D(g) j l (D(g) ) h kt l h = D(g) k l (D(g) ) h k T l h = δ h l T l h, while δ ij T ij is not invariant. Therefore, we have just the symmetric and antisymmetric tensors as irreducible components of the product representation acting on T ij and consequentially for SU(3), 3 3 = 3 6. On the other hand. for the representation acting on T j i the trace is the only invariant subspace. There is really no point in looking for a symmetry by exchanging indices which belong to different vector spaces. But since we are talking about a compact group, we can assume a unitary representation and therefore the rest is also an invariant subspace. In fact it turns out to be the adjoint representation. Let us again take SU(3) as an example, 3 3 = 1 8. We can go on and derive the irreducible components from the symmetry properties of the tensors on which the product representations act. For example, for SU(3), if we want to know how 3 3 3 decomopses, we have to consider the object T ijk with 3 3 independent components. We can reduce it first in the subspace of tensors which are symmetric or antisymmetric under exchange of 2 indices, T ijk = T i[j k] i{j k} + T which have 3 times the components you found in the last exercise (for every possible value of i). And we can further take out the space of total antisymmetric tensors as an irreducible subspace of T i{j k}, which is only one tensor, and the subspace of total symmetric tensors of T i[j k] with 10 components 11 i[j k]. These have to be substracted from the dimensions of T and T i{j k} and if you made no mistake in the last exercise, we end up with 3 3 3 = 1 8 8 10. This can become arbitrarily complicated if you want, but you should really look into Young tabloux for more complicated products. 11 A general symmetric tensor of rank r over a vector space of dimension N has ( ) N+r 1 r elements. 19

4 Physics Explain gauge groups and SU(2) SO(3) and SL(2, C) L +. First and only appendix Compute the matrix in section 3 and bring counterexamples for section 2. Questions about the problems: mbauer@thep.physik.uni-mainz.de 20