GROUP THEORY PRIMER. D(g 1 g 2 ) = D(g 1 )D(g 2 ), g 1, g 2 G. and, as a consequence, (2) (3)

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1 GROUP THEORY PRIMER New terms: representation, irreducible representation, completely reducible representation, unitary representation, Mashke s theorem, character, Schur s lemma, orthogonality theorem, character table, regular representation, crystal field splitting. 1. Representations A representation D of a group G is a homomorphism from G to GL(N, C), the group of invertible N N matrices with complex entries. For a given group element g, D(g) is an N N matrix. Being a homomorphism, the mapping has the property (1) and, as a consequence, () () D(g 1 g ) = D(g 1 )D(g ), g 1, g G D(e) = I D(g 1 ) = D 1 (g), g G where we use the notation I for the unit matrix. Clearly, these properties are preserved if we perform a simultaneous similarity transform D(g) SD(g)S 1 for all g G. Here, S is itself an invertible, complex N N matrix, that is, an element of GL(N, C). Clearly, SD(g)S 1 satisfies (1)-() if D(g) does. This means that, once we find one representation of G, any similarity transform of it is also a representation. Two representations are said to be equivalent representations if they are related by such a similarity transformation. Also, we must distinguish between reducible and irreducible representations. A representation is reducible if, using a similarity transformation, the matrices in the representation can put in the form, [ ] D D(g) = (1) (g) W (g) (4) 0 D (), g G (g) Note that, if D(g) in the above equation is a representation, then multiplying the matrices for two group elements results in [ ] [ ] D D(g 1 g ) = D(g 1 )D(g ) = (1) (g 1 ) W (g 1 ) D (1) (g ) W (g ) 0 D () (g 1 ) 0 D () (g ) 1

2 GROUP THEORY PRIMER [ ] D = (1) (g 1 )D (1) (g ) D (1) (g )W (g ) + W (g 1 )D () (g ) 0 D () (g 1 )D () (g ) [ ] D = (1) (g 1 g ) W (g 1 g ) 0 D () (g 1 g ) which implies D (1) (g 1 g ) = D (1) (g 1 )D (1) (g ) and D () (g 1 g ) = D () (g 1 )D () (g ), that is, D (1) (g) and D () (g) are also representations. A representation which is not reducible in this way is said to be irreducible. A representation is said to be completely reducible if it is equivalent to a representation of the form (4) but with W (g) = 0 g. Mashke s theorem, which we review in info box 1.1, states that, for a finite discrete group, all finite dimensional reducible representations are completely reducible. It is clear that the irreducible representations are the fundamental building blocks of all representations. If we know the irreducible representations of a group, we can always build up reducible representations by making block diagonal representations. Infobox 1.1 Mashke s Theorem Generally, a representation D(g) of a group G is reducible if, using a similarity transformation, it can be put in the upper triangular form [ ] D D(g) = 1 (g) W (g) (5) 0 D (), g G (g) The reasoning behind this definition comes from consider the vectors on which these matrices act. When D(g) has the upper triangular form (5) and D (1) (g) is a k k matrix, any vector of the form that only its first k entries are nonzero, v 1.. v k has the special property that D(g)v is also a vector of the same form, where only the first k components can be non-zero. This implies that the vector space on which D(g) acts has a subspace such that the action of all of matrices D(g) for all of the group elements g G produces vectors only in that subspace. It is said to be an invariant subspace, or a g-module. However, the same statement is

3 GROUP THEORY PRIMER not true for the complement of this subspace. Due to the action of W (g) in (5), vectors of the form v k+1.. v n do not retain their special nature. They would do so (and the vector space consists of two invariant subspaces) only if W (g) = 0 g G. Let us say that the whole vector space is V, the invariant subspace is V 1 and its complement is V. When W (g) 0 for some g G, the operation of D(g) on elements of V produces a linear combination of elements in V and elements in V 1. The idea of Mashke s theorem is to find a different basis where the space decomposes into orthogonal subspaces U 1 and U such that D(g) operating on vectors in U 1 produces vectors in U 1 exclusively and D(g) operating on vectors in U produced vectors in U exclusively. Such a change of basis is equivalent to a similarity transform and, if it can be done, it is also identical to finding a representation equivalent to (5) but where W (g) = 0 g G. To begin, we form a projection operator π onto the subspace V 1. It has the property that πv = v if v V 1 and πy = 0 if y V. In those spaces, where D(g) has the form in (5), [ ] I 0 π = 0 0 Then, from this projection operator, we form the object π G = 1 D(g)πD(g 1 ) where is the order of G, π G = 1 [ ] [ ] [ ] D (1) (g) W (g) I 0 D (1) (g 1 ) W (g 1 ) 0 D () (g) D () (g 1 ) g G [ ] I 1 = g G D(1) (g)w (g 1 ) 0 0 g G

4 4 GROUP THEORY PRIMER Squaring the matrix above, we see that this π G is a projection operator, π G = π G In particular, given its construction, we easily see that it commutes with D(g), D(g)π G = D(g) 1 D( g)πd( g 1 ) = 1 D( g)πd( g 1 ) g = 1 D( g)πd((g 1 g) 1 ) = 1 D( g)πd( g 1 g) = 1 D( g)πd( g 1 )D(g) = π G D(g) Whenever we have a projection operator, since π G (1 π G ) = 0 = (1 π G )π G, we can divide the vector space into two orthogonal subspaces, U 1 which consists of vectors in the kernel of 1 π G and U which consists of vectors in the kernel of π G. Any vector is always decomposable into a sum of vectors in each subspace, v = (1 π G )v + π G v. Now, we note that the action of the group, D(g) commutes with the projection. That is, if u U 1, then (1 Π G )u = 0 and (1 π G )D(g)u = D(g)(1 π G )u = 0, that is, the action of D(g) on a vector in U 1 results in a vector in U 1. By a similar argument, the operation of D(g) on a vector in U results in a vector in U. Consequently, in an orthonormal basis, D(g) has the form in (5), except with W = 0. This establishes Mashke s theorem. The fact about the group which was needed to prove this theorem is the existence of the projection operator π G = 1 D( g)πd( g 1 ) must exist. For a finite group, it certainly exists. It also exist for compact Lie groups where there is a well-defined integral over group elements and a finite volume of the group which replaces. Infobox 1. Unitary representations Theorem: Every representation of a finite discrete group is equivalent to a unitary representation. Proof: Consider a finite-dimensional representation D(g) of a finite discrete group G of order. Let us form the following matrix: A = 1 (6) D(g)D (g) g

5 GROUP THEORY PRIMER 5 Here, D is the hermitian conjugate of the matrix D, D ab = D ba. The sum over the elements of G. The matrix A has the property that A = A. It is a Hermitian matrix. Being a Hermitian matrix, it has real eigenvalues. Moreover, it can be diagonalized by a unitary transformation. (7) A = U a U 0 a a where a 1, a,... are real numbers. What is more, we can show that all of the eigenvalues are positive. If φ i is an eigenvector, it obeys b A abφ ib = a i φ ia. This implies a i = ab φ iaa ab φ ib a φ ia φ ia = 1 g abc φ iad ac (g)d bc (g)φ ib a φ ia φ ia = 1 c b D bc (g)φ ib g a φ ia φ ia 0 which is positive and can be zero only if D(g) = 0 g. It is therefore non-zero and positive. Given positivity of the eigenvalues, we can form the matrices a a A 1 = U 0 a a 1... U, A 1 = U 0 a a 1... U where U is the same matrix as occurs in (7) above. Clearly A 1 A 1 = I and A 1 A 1 = A. Also, (A 1 ) = A 1 and (A 1 ) = A 1. Then, we can perform the similarity transformation to find an equivalent representation D(g) = A 1 1 D(g)A, g G The statement is now that, for any g G, D(g) is a unitary matrix, that is, D(g) D (g) = I. Let us confirm this. First, note that D (g) = A 1 D (g)a 1. Then D(g) D (g) = A D(g)A A D (g)a 1 = A 1 D(g)AD (g)a 1 [ ] = A 1 1 D(g) D( g)d ( g) D (g)a 1 g

6 6 GROUP THEORY PRIMER qed. = A 1 [ ] = A 1 1 D(g)D( g)d ( g)d (g) A 1 g [ ] [ ] 1 D(g g)d (g g) A 1 = A 1 1 D( g)d ( g) g = A 1 A A 1 = I g A 1 In info box 1., we prove that any representation of a finite discrete group is equivalent to a unitary representation. The upshot of this theorem is that, when we are discussing finite discrete groups, we can restrict our attention to unitary representations. The proof of the theorem relies on the existence of a sensible sum over group elements in the expression in equation (6), A = 1 g D(g)D (g). It turns out that the theorem extends to infinite groups whenever a generalization of the sum can be sensibly defined. For some continuous groups called compact Lie groups, SO(N) and SU(N) are examples, the sum can be replaced by a particular integral over group elements and the order can be replaced by the volume of the group. In the case of those groups too, every finite dimensional representation of the group is equivalent to a unitary representations and the entire discussion of representations can be focused on finding unitary ones. This is fortunate as, in our example of a quantum mechanical state whose degeneracy was governed by a representation of a symmetry group, the induced representation was unitary. There are also simple examples of infinite groups where representations exist which are not unitary. An example of such an infinite group is the additive group of the integers. There, if the integers are n, D(n) = λ n is an example of a onedimensional representation. λ is a real number and this representation is not unitary and it is not equivalent to a unitary representation. If, instead of the infinite group, we consider the additive group of the integers modulo p, the representation would have to obey λ p = 1 which forces λ to be complex λ = exp (πik/p), and the representation of the finite discrete group must be unitary. Another example is the Poincare group, the symmetries of space-time under translations, rotations and Lorentz transformations. That group has infinite volume and it has no non-trivial finite dimensional representations.

7 GROUP THEORY PRIMER 7. Character Consider an N N matrix D ab. The trace of this matrix is defined as (8) TrD = N a=1 D aa The trace, defined in (8) is left unchanged by a similarity transformation. That is, TrD = Tr(SDS 1 ) for any invertible N N matrix S. To see this, we display the indices, Tr(SDS 1 ) = N abc=1 S ab D bc S 1 ca = N N ( Sca 1 S ab ) D bc = bc=1 a=1 N δ cb D bc = bc=1 N D bb = TrD where δ bc is the Kronecker delta function δ bc = 1 if b = c and δ bc = 0 of b c. It can be thought of as the unit matrix with indices, I bc. An interesting quantity which is invariant under similarity transformations of a representation is the character. For a group G = {g 1, g,..., g k }, with a representation D = {D(g 1 ), D(g ),..., D(g k )}, the character is defined as b=1 (9) χ = {TrD(g 1 ), TrD(g ),..., D(g k )} {χ(g 1 ), χ(g ),..., χ ( g k )} The trace of D(g) for all elements of a single conjugacy class are equal. Recall that a conjugacy class of g is the set of all distinct elements of the set {g 1 gg 1 1, g gg 1,..., g k gg 1 k } Since the representation is a homomorphism, the representation of the members of the conjugacy class are the distinct elements in {D(g 1 gg 1 1 ), D(g gg 1 ),..., D(g k gg 1 k )} = {D(g 1 )D(g)D(g 1 ) 1, D(g )D(g)D(g ) 1,..., D(g k )D(g)D(g k ) 1 } Then, since Tr(D(g i )D(g)D(g i ) 1 ) = TrD(g) = χ(g), every element of a conjugacy class has the same character. Thus, the argument of the character should be conjugacy classes rather than group elements, χ(c) = TrD(g) where g C Let us label inequivalent irreducible representations of a group G by the superscript µ of D (µ) (g). We will denote the character of irreducible representations by χ (µ) (C).

8 8 GROUP THEORY PRIMER. Schur s Lemmas and the Orthogonality Theorem Now, we are ready to discuss the most important formula what we are going to study in discrete group theory, the orthogonality theorem. This theorem is a straightforward consequence of Schur s lemmas which we will begin with. Schur s first lemma states that, if D (µ) (g) is an irreducible representation of a finite, discrete group, and if a matrix A exists so that D (µ) (g)a = AD (µ) (g) g G, then A is proportional to the identity, A = ai or A ab = aδ ab. We give a proof of this theorem in infobox.1. Schur s second lemma states that, if D (1) (g) and D () (g) are inequivalent irreducible representations of a finite discrete group G, with dimensions d 1 and d, respectively, the only solution of the equation BD (1) (g) = D () (g)b for the d d 1 - dimensional matrix B, is B = 0. We give a proof of this theorem in info box.. A direct consequence of the above two lemmas is the orthogonality theorem. Consider the set two irreducible representations, D (µ) (g) and D (ν) (g) of dimensions d µ and d ν, respectively, of a finite discrete group, G. The elements of the matrices D (µ) (g) and D (ν) (g) obey the following sum rule (10) 1 g G D (µ) ab (g)d(ν) cd (g 1 ) = 1 d µ δ µν δ ad δ bc where δ µν = 1 when the representations are identical, and it is zero otherwise. We give a proof of this theorem in info box 1. The orthogonality theorem in (1) has an immediate important consequence. We can think of the left-hand-side as the summation over the components of complex vectors. The components are labeled by the g in D (µ) ab (g). The different vectors are labeled by the other indices a, b, µ. This is an orthogonality relation for those vectors. However, the dimension of the space is, the order of the group. There cannot be more orthogonal vectors than there are dimensions in the space. Thus the total number of orthogonal vectors, which is µ d µ must be less than or equal to G, (11) d µ µ The left-hand-side of this inequality is a sum over all inequivalent irreducible representations of the finite discrete group of the squares of the dimensions of the representations. The right hand side is the order of the group. This means that, especially for smaller order groups, the number and dimension of the inequivalent representations is greatly restricted. We eventually prove that, for all finite discrete groups, the equality in equation (11) is actually saturated, that is, the equality holds. From now on, we will treat this equation as if it is an equality.

9 Infobox.1 Schur s First Lemma GROUP THEORY PRIMER 9 Lemma: If D (µ) (g) is an irreducible representation of a finite, discrete group, and D (µ) (g)a = AD (µ) (g) g G, then A is proportional to the identity, A = ai or A ab = aδ ab. Proof: Without loss of generality, we can assume that D (µ) (g) are unitary matrices. Then D (µ) (g 1 ) = D (µ) (g). Then D (µ) (g)a = AD (µ) (g) implies D (µ) (g)a = A D (µ) (g) and D (µ) (g 1 )A = A D (µ) (g 1 ) and, since it should apply to all group elements, we can replace g 1 by g and D (µ) (g)a = A D (µ) (g). Then, adding this expression to D (µ) (g)a = AD (µ) (g) we obtain D (µ) (g)(a + A ) = (A + A )D (µ) (g). Thus, we can take A to be a hermitian matrix. Then, we can always diagonalize a hermitian matrix. That is, we can find an equivalent representation where a A ab = 0 a a When A is diagonal, AD (µ) (g) = D (µ) (g)a implies (1) (a a a b )D (µ) ab (g) = 0 If any of the differences (a a a b ) are non-zero, equation (1) places restrictions on the matrix elements of D (µ) (g). If all of the eigenvalues are equal, all of the differences (a a a b ) vanish, A is proportional to the unit matrix and equation (1) is solved by any matrix D (µ) ab (g). If even one of the eigenvalues is different from the rest, say a 1 a i>1, then D (µ) 1i (g) = 0 and D(µ) i1 (g) = 0 g G. D(µ) 11 (g) is not restricted. The result is that the possible non-zero elements of D (µ) (g) must leave it block-diagonal, of the form D (µ) 11 (g) D (µ) 0 D (µ) (g) = (g) D (µ) (g)... 0 D (µ) (g) D (µ) (g)

10 10 GROUP THEORY PRIMER and it must therefore be reducible. If two eigenvalues, say a 1 and a are equal, and different from all the rest, D (µ) (g) must have the form D (µ) 11 (g) D (µ) 1 (g) D (µ) D (µ) 1 D (µ) (g) (g) = 0 0 D (µ) (g) D (µ) 4 (g) D (µ) 4 (g) D (µ) 44 (g) The multiplicity of sets of equal eigenvalues determine the size of allowed nonzero blocks in D (µ) (g). D (µ) (g) has only one block (the entire matrix) when all of the eigenvalues are equal. Otherwise, it has multiple blocks and the representation is reducible. Infobox. Schur s Second Lemma Lemma: If D (1) (g) and D () (g) are two inequivalent irreducible representations, with dimensions d 1 and d, respectively, the only solution of the following equation for the d d 1 matrix B, BD (1) (g) = D () (g)b is B = 0. Proof: Without loss of generality, we can restrict our attention to representations where both D (1) (g) and D () (g) are unitary matrices. If they are not unitary, they are equivalent to unitary representations in that similarity transformations exist, D(1) (g) = S 1 D (1) (g)s1 1 and D () (g) = S D () (g)s 1 so that, for the unitary representations, B D (1) = D () (g) B where B = S BS1 1. Since S1, S are non-singular matrices, if B is non-zero, so is B, and vice-versa. In the following, we shall drop the tildes. Then, if D (1) (g) and D () (g) are unitary matrices, BD (1) (g) = D () (g)b implies B D () (g) = D (1) (g)b. Multiplying BD (1) (g) = D () (g)b from the left by B and multiplying B D () (g) = D (1) (g)b from the right by B leads to B BD (1) (g) = B D () (g)b, B D () (g)b = D (1) (g)b B and B BD (1) (g) = D (1) (g)b B, g G We have identified a matrix, B B, which commutes with every matrix in an irreducible representation. By Schur s first lemma, it must be a multiple of the identity, B B = bi

11 GROUP THEORY PRIMER 11 If B is a square matrix (d 1 = d ), either b = 0 or det B 0, B is invertible and D (1) (g) and D () (g) are equivalent representations which contradicts the assumption that they are not equivalent. Thus, if B is a square matrix, b = 0, which is what we are trying to prove, and have now done so for the special case d 1 = d. Let us assume that d 1 > d and B is therefore not a square matrix. Then we can expand B to make it a square matrix by adding zeros. The result in matrix is B, given by B 11 B 1 B 1... B 1d1 B 1 B B... B d1 B 1 B D... B d1 B = B d 1 B d B d... B d d Certainly, if B B = bi then B B = bi But, det B = b d 1 and det B = 0. This is consistent only when b = 0. This tells us that B B = 0. Take the trace of this equation, B ab = 0 ab which can only be true of every element of B is zero, that is, if B = 0. This proves the theorem. Infobox. The Orthogonality Theorem Theorem: Consider two irreducible representations, D (µ) (g) and D (ν) (g) of dimensions d µ and d ν, respectively, of a finite discrete group, G. The matrix elements in the representations obey the following sum rule (1) 1 g G D (µ) ab (g)d(ν) cd (g 1 ) = 1 d µ δ µν δ ad δ bc where δ µν = 1 when the representations are identical, and it is zero otherwise. Proof:Let the dimension of D (µ) (g) be d µ and the dimension of D (ν) (g) be d ν. Consider the d µ d ν matrix X and form the following d µ d ν matrix, K µν = 1 D (µ) ( g)xd (ν) ( g 1 )

12 1 GROUP THEORY PRIMER Consider ( ) D (µ) (g)k µν = 1 D(µ) (g) D (µ) ( g)xd (ν) ( g 1 ) ( ) = 1 D (µ) (g)d (µ) ( g)xd (ν) ( g 1 ) ( ) = 1 D (µ) (g g)xd (ν) ( g 1 ) D (ν) (g 1 )D (ν) (g) ( ) = 1 D (µ) (g g)xd (ν) ( g 1 g 1 ) D (ν) (g) ( ) = 1 D (µ) (g g)xd (ν) ((g g) 1 ) D (ν) (g) = K µν D (ν) (g) Thus, K µν is a matrix with the property K µν D (µ) (g) = D (ν) (g)k µν. By Shur s second Lemma, either D (µ) (g) and D (ν) (g) are equivalent representations or K µν must be equal to zero. Thus K µν is non-zero only when µ = ν. Now, let us assume that µ and ν are the same representation. Then, by Schur s first lemma, K µν must be proportional to the unit matrix. Thus, we see that K µν ab = kδµν δ ab and, we can solve this equation for k to get where k = 1 TrK µµ = 1 Tr 1 D (µ) ( g)xd (µ) ( g 1 ) d µ d µ = 1 d µ 1 Tr ( D (µ) ( g)xd (µ) ( g 1 ) ) = 1 TrX d µ Thus, we see that k = 1 d µ TrX. This equation then implies 1 D (µ) ab ( g)x bcd (ν) cd ( g 1 ) = 1 δ µν δ ad d µ Taking a derivative of each side by X bc yields the orthogonality theorem 1 D (µ) ab ( g)d(ν) cd ( g 1 ) = 1 δ µν δ ad δ bc d µ.1. The example of D. As an example, let us consider the dihedral group D. Recall that it has two generators, c and b with the relations c = e, b = e and c X cc

13 GROUP THEORY PRIMER 1 (bc) = e. There are six group elements, {e, c, c, b, bc, bc }. The order D = 6 is six. What about the representations of this group. By equation (11), the squares of their dimensions must add up to six. And of course the dimensions are all positive integers which are greater than or equal to one. There are only two possibilities. The first is that there could be six inequivalent one-dimensional representations. The second is that there could be two one-dimensional representations and one two-dimensional representation. To see which of these it is, we can find the representations explicitly by finding the elements of SO() which corresponding to the group transformations. We remember that the group elements are the symmetries of an equilateral triange. If we place that triangle in the x-y plane with its centre a the origin and with one side parallel to the x-axis. The generator c is a counterclockwise rotation in the x y plane by angle π/ radians. The SO() matrix which would implement a rotation in the x-y plane (14) cos θ sin θ 0 R z (θ) = sin θ cos θ and a rotation in the y-z plane is this rotation of a vector is (15) R x (φ) = cos φ sin φ 0 sin φ cos φ Using θ = π/ radians and cos π/ = 1 and sin π/ = and φ = π radians, (16) (17) 0 D(c) = D(b) =

14 14 GROUP THEORY PRIMER Having found the generators, we can easily identify the entire representation (18) D(e) = , D(c) = , D(c ) = (19) D(b) = , D(bc) = , D(bc ) = Let us begin with the subgroup C = {e, c, c } for which we have the representation in equation (18). This is a reducible representation of C. The three matrices commute and they can be simultaneously diagonalized and put in the form (0) D(e) = , D(c) = 1 + i i 0, D(c ) = 1 i i Of course, this is entirely compatible with what we know about representations. C is an order three group. The inequality (11) indicates that it can have at most three one-dimensional representations. Indeed, in the above matrices we have found all three. C e c c A A i 1 i A 1 1 i 1 + i Now, if we consider the full representation of D given in equations (18) and (19), it is clear that it is reducible in that it is already block diagonal. However, the two dimensional representation which occurs in the upper block there cannot be further reduced. It is irreducible. The --components of the matrices also form a nontrivial one-dimensional representation. Thus, we have two one-dimensional representations and one two-dimensional representation, which is all that is allowed by the sum rule in equation (11). Thus, we have identified all of the irreducible representations of D, The trivial representation, A 1, e = 1, c = 1, c = 1, b = 1, bc = 1, bc = 1

15 A one-dimensional representation, A, GROUP THEORY PRIMER 15 e = 1, c = 1, c = 1, b = 1, bc = 1, bc = 1 And a two-dimensional representation, E, [ ] [ 1 0 D(e) =, D(c) = 1 ] [ 0 1, D(c ) = 1 ] 1 1 [ ] [ 1 0 D(b) =, D(bc) = 1 ] [, D(bc 0 1 ) = 1 ].. Abelian groups. Theorem: All irreducible representations of a finite discrete abelian group are onedimensional. Proof: Consider a representation D(g) of an finite discrete abelian group. Without loss of generality, we can assume that D(g) are unitary matrices. By the homomorphism property of the representation, these matrices must represent the group product. Therefore, given any two g 1, g G, since g 1 g = g g 1 the matrices obey D(g 1 )D(g ) = D(g )D(g 1 ), that is, the group elements are represented by a set of commuting unitary matrices. Commuting unitary matrices can be simultaneously diagonalized. We can therefore find a basis where all of the matrices D(g) are diagonal. By this reasoning, any representation is completely reducible to one-dimensional representations. Then, we also know from equation (11), an abelian group has exactly distinct irreducible representations. We have already seen an example of this with the representations of C which we discussed in the previous section. Since all irreducible representations are one-dimensional, and all elements of a finite group have finite order ( recall that, for each g G, there must exist a smallest positive integer p, called the order of g, such that g p = e), D(g) for a finite Abelian group must be a unimodular complex number which is a p th root of unity Characters The orthogonality theorem, (1), which we copy here for the reader s convenience: 1 D (µ) ab (g)d(ν) cd (g 1 ) = 1 δ µν δ ad δ bc d µ g G carries significant information about the character of a representation. To see this, we only need to take the traces of D (µ) (g) and D (ν) (g 1 ) and, accordingly, we put the indices b = a and d = c and sum over remaining a and c. The result is 1 χ (µ) (g)χ (ν) (g 1 ) = δ µν g G 1

16 16 GROUP THEORY PRIMER which, since the character of a given group element, χ (µ) (g), depends only on its conjugacy class, can be recast as a sum over classes, C, rather than group elements, 1 (1) n C χ (µ) (C)χ (ν) (C) = δ µν C where, we have assumed that the representation is unitary, so that χ (ν) (g 1 ) = χ (ν) (g) = χ (ν) (C). Also, n C, the number of elements in the conjugacy class C. What we have obtained (1) is an orthogonality sum of characters of irreducible representations as vectors whose indices are the conjugacy classes of the group. One immediate result is, since there cannot be more orthogonal vectors than the dimension of the space that the vectors live in, there is an inequality for the number of representations of a finite discrete group, () number of inequivalent irreducible representations number of classes This important inequality restricts the number of inequivalent irreducible representations. Indeed, this inequality turns out to be an equality. We will give a proof of this fact later. In the meantime, we will assume that it is so, that the inequality in () is actually an equality which fixes the number of inequivalent irreducible representations as equal to the number of conjugacy classes. 5. Decomposition of reducible representations A reducible representation is equivalent to a block diagonal one. Explicitly, a generic representation of a finite discrete group is equivalent to a representation of the form D (ν1) (g) D (ν) (g) D(g) = 0 0 D (ν) (g) D (ν4) (g)..., g G Then, since it is unchanged by similarity transformations, the character of the representation can be found from the block diagonal representation. The trace of a block diagonal matrix is given by the sum of the traces of the blocks. Thus, we find, χ D (g) = i χ ν i (g) for each individual group element. The sum on the right-hand-side is over the irreducible representations which are contained in D(g). Specific irreducible representations of G could be repeated in this sum. Clearly, each time the same irreducible

17 GROUP THEORY PRIMER 17 representation occurs, it would make exactly the same contribution to the character. Taking this into account, can write the above formula as χ D (g) = µ n µ χ µ (g) where the right-hand-side is now a sum over the irreducible representations of G, labeled by µ, and the non-negative integers n µ are the number of times each irreducible representation appears in D(g). Of course, the characters really depend on the conjugacy class in which a group element occurs, rather than individual group elements, so we should properly write the above formula as one where the arguments of the characters are classes () χ D (C) = µ n µ χ ν (C) In various practical problems, given a representation D(g), it might be of interest to find the integers n µ in equation (). The orthogonality relation for characters (1) can be used for this purpose, if we know the characters of D(g). For this purpose, we multiply equation () by χ (ν) (C), χ D (C)χ (ν) (C) = µ n µ χ µ (C)χ (ν) (C) We multiply by n C, the number of group elements in the class C and we sum both sides of the equation over the classes: n C χ D (C)χ (ν) (C) = n µ n C χ µ (C)χ (ν) (C) µ C We use the orthogonality equation (1) to simplify the right-hand-side of the above equation, n C χ D (C)χ (ν) (C) = n ν C where, we remind the reader that d ν is the dimension of the irreducible representation D (ν) (g) and is the order of G. Finally, we obtain the formula for n ν, n ν = 1 (4) n C χ D (C)χ (ν) (C) C If we know (1) the conjugacy classes C of the group G () the irreducible representations, labelled by ν, of the group G () the character χ (µ) (C) of each class in each irreducible representation (4) the character χ D (C) of the reducible representation C

18 18 GROUP THEORY PRIMER then, we are ready to compute the right-hand-side of the formula (4) and to find the integers n ν The example of D. Let us once again revisit our example of the dihedral group D. The generators are c and b and the relations are c = e, b = e and (bc) = e. The classes are [e] = {e}, [c] = {c, c }, [b] = {b, bc, bc } We have already found the three irreducible representations of this group. We remind the reader that they were A 1 : e = 1, c = 1, c = 1, b = 1, bc = 1, bc = 1 χ (A 1) ([e]) = 1, χ (A 1) ([c]) = 1, χ (A 1) ([b]) = 1 A : e = 1, c = 1, c = 1, b = 1, bc = 1, bc = 1 χ (A ) ([e]) = 1, χ (A ) ([c]) = 1, χ (A ) ([b]) = 1 E: [ ] [ 1 0 D(e) =, D(c) = 0 1 [ ] [ 1 0 D(b) =, D(bc) = 0 1 ] ], D(c ) = [ 1 1 ] [, D(bc ) = 1 ] χ (E) ([e]) =, χ (E) ([c]) = 1, χ (E) ([b]) = 0 Where, we have also listed the characters of the classes for each representation. We also need the number of elements of each class, n [e] = 1, n [c] =, n [b] = We are now ready to decompose any irreducible representation of D by computing the integers n µ. All we need is the character of that representation. 1

19 GROUP THEORY PRIMER The example of SO(). Remember that we can embed the group elements of D and some other discrete groups of interest in the rotation group, SO() by thinking of the group elements as the symmetry transformations of a regular object. In the case of D, the regular object is an equilateral triangle. In that case, the generator c corresponds to a rotation by π/ radians and the generator b a rotation by π radians. Also, we showed that, within SO(). the conjugacy class of a rotation depends only on the angle of the rotation. We can compute the character of an element of SO() by computing the character of any element where the rotation is by the same angle. This means we can take the rotation to be about the z-axis. The matrices corresponding to rotations about the z axis in the various representations of SO() are easy to find. They occur in the usual quantization of angular momentum in quantum mechanics where, it is generally the components of the angular momentum vector that are studied. There, one normally diagonalizes the z-component of angular momentum. The, using algebra one finds that the eigenvalues of L z in an irreducible representation. They are l, l + 1,..., l 1, l where l is a non-negative integer. Then, l l l L z = (l 1) l The SO() group element corresponding to a rotation about the z-axis is gotten by exponentiating the z-component of the angular momentum. That is, we can construct D(θ) for a rotation about the z-axis explicitly, as D(θ) = exp(iθl z ). The representation of this group element is thus e ilθ e i(l 1)θ D(θ) = 0 0 e i(l )θ e i(l 1)θ e ilθ The trace of this matrix gives the character of this group elements, and the character of any SO() element in the conjugacy class, that is, any SO() element where the rotation angle is θ, χ (l) (θ) = e ilθ + e i(l 1)θ e i(l 1)θ + e ilθ

20 0 GROUP THEORY PRIMER The sum on the right-hand-side of the above formula can be done. The result is (5) χ (l) (θ) = sin ( (l + 1 )θ) sin ( 1 θ) This is the formula for the character of an element of SO(). 5.. Crystal field splitting. Now, let us consider the following question. Consider an electron in an orbital of a spherically symmetric molecule. Because of the spherical symmetry, the quantum states of the electron can be classified by the orbital angular momentum quantum number l. Then, let us imagine that we take the spherically symmetric atom with the electron in the orbital l and immerse it in an environment, for example a crystal, where the SO() symmetry is reduced to D. The six elements of D are six particular elements of SO(). Even though the representation of SO() was irreducible, the representation of D could be (and almost surely will be) reducible. Even though no similarity transformation existed which would block diagonalize all SO() matrices, the elements of D are smaller in number and they may be block diagonalizable. In fact, we have already seen this in equations (18) and (19) where we found the elements of D embedded in the -dimensional representation of SO() (the representation with l = 1). They already were block diagonal, reducing to a and a 1 1 block. Since no irreducible representation of D has dimension greater than, we know that larger l representations must be reducible. A question that we could ask is how many of each irreducible representation of D are contained in an irreducible representation of SO() which is characterized by l. We can answer this by finding the characters of the D elements in the SO() representation. For this, we simply note that c is a rotation by π/. The SO() character is given by equation (5). Then, 1 l = 0 mod (6) χ (l) ([e]) = l + 1, χ (l) ([c]) = 0 l = 1 mod, χ (l) (b) = ( 1) l 1 l = mod Now, we are equipped to use the formula (4) to find the content of each irreducible representation of D in the representations of SO(). We copy (4) here for the reader s convenience, n ν = 1 n C χ D (C)χ (ν) (C) C Using = 6 and equation (6), this equation becomes n ν = 1 1 l = 0 mod 1 (l + 1) χ (µ) ([e]) + 0 l = 1 mod 6 1 l = mod χ(µ) ([c]) + ( 1) l χ (µ) ([b])

21 GROUP THEORY PRIMER 1 To use this formula, we need to plug in the characters of a particular irreducible representation of D. It is useful to summarize the characters of D in a table: D [e] [c] [b] (7) A A E 1 0 The columns are labeled by the conjugacy classes ( the number of elements in each class is also given). The rows are the inequivalent irreducible representations of the group. The entries in the rows and columns are the characters of the classes in those irreducible representations. Using this data, we see that an angular momentum state, that is a representation of SO() with quantum number l decomposes into representations of the group D with multiplicities n A1 = 1 1 l = 0 mod (l + 1) + 0 l = 1 mod 6 1 l = mod + ( 1)l = (1, 0, 1, 1,, 1,,,,, 4,...) n A = 1 1 l = 0 mod (l + 1) + 0 l = 1 mod 6 ( 1)l = (0, 1, 0,, 1,,,,, 4,,...) 1 l = mod n A = 1 1 l = 0 mod (l + 1) 0 l = 1 mod = (0, 1,,,, 4, 4, 5, 6, 6, 7,...) 6 1 l = mod where, in the last equalities of each of the three equations, we have quoted the results for l = (0, 1,,, 4, 5, 6, 7, 8, 9, 10,...). We see that, for example, an electron in the l = 5 state, which begins with an 11-fold degenerate wave-function, when subjected to a potential which has only D symmetry, generically has this degeneracy split so that there are 1 A 1, A s, and 4 E s of D. The representations of D will generically have different energies. This splitting of the energy level is depicted in figure The regular representation Every finite discrete group has a particular finite dimensional reducible representations which we can find explicitly. It is called the regular representation. The dimension of this representation is the order,, of the group and the elements of the matrices in the representation are labeled by group elements. Explicitly, for

22 GROUP THEORY PRIMER Figure 1. The figure depicts the generic splitting of an 11-fold degenerate l = 5 orbital into states which represent a D subgroup of SO(). The vertical direction is energy and the ordering of the representations is determined by the details of the quantum mechanical problem (and could be different from what is shown). elements g, g i, g j of a group G, (8) D ij (g) = { 1 if gi g 1 j = g 0 otherwise D(g) are -matrices. Also, it is clear that since g i g 1 j D ij (e) = δ ij = e only when g i = g j. In addition D ii (g e) = 0

23 GROUP THEORY PRIMER since if we require g i g 1 i = g the only possibility is g = e. Thus D(e) = I, as usual, and all of the diagonal elements of D(g) for any g e are equal to zero. This means that we always know the characters of the regular representation explicitly, (9) χ R ([e]) =, χ R (C [e]) = 0 The regular representation is reducible. It must be, as it has dimension G and we already know that the dimensions of irreducible representations are restricted by the formula µ d µ G which, since there is always a trivial one-dimensional irreducible representation D (0) (g) = 1, tells us that all other irreducible representations must have dimensions strictly less than. In fact, since we know the characters explicitly, and the only non vanishing character is for the class of the identity, we can use the formula (4), n ν = 1 n C χ D (C)χ (ν) (C) C to find the number of times that each irreducible representation appears in the regular representation. Explicitly n ν = 1 n [e]χ D ([e])χ (ν) ([e]) = 1 d ν = d ν since n [e] = 1 and since all other terms in the sum vanish. We see that the regular representation contains every irreducible representation, and the number of times that a given irreducible representation appears is equal to the dimension of that representation. Since all of these must fit in a -matrix, and each time the ν th representation occurs it takes up a d ν d ν block, we obtain the equality (0) d ν = ν This proves that the inequality what we deduced from the orthogonality theorem is actually an equality. The other equality that we found is the one that says the number of irreducible representations is less than or equal to the number of classes. This was deduced from the orthogonality formula for irreducible representations, that with a certain definition of their inner product, the charters of irreducible representations are orthonormal vectors and only so many such vectors fit in a space of a given dimension. This means that the rows of the character table displayed in equation (7) are each normalized and they are orthogonal to each other. We might also notice that the columns are orthogonal and they presumably also obey an orthogonality relation, like µ χ(µ) (C i )χ (µ) (C j ) δ ij. Such an equality would then tell us that the number of classes is less than or equal to the number of representations and the upshot is

24 4 GROUP THEORY PRIMER that the number of classes and the number of irreducible representations are equal. We will prove the orthogonality formula for representations in info box 6.1. Infobox 6.1 Character vector orthogonality Consider a conjugacy class C of a finite discrete group G and a unitary representation D(g). The matrices corresponding to the group elements in the conjugacy class transform into each other under conjugation: Recall that g and g are in the same class if there exists h G such that hgh 1 = g. If we consider the group elements in a given conjugacy class, C i = {g 1, g,..., g nci }, and an element h G, hc i h 1 must simply be a re-arrangement of the class. First of all hgh 1 C i h G, g C i. Moreover, conjugation of distinct elements of the class produces distinct elements of the class, hg i h 1 = hg j h 1 implies g i = g j and vice-versa, that is, conjugation is a one-to-one mapping of the class onto itself. Now, given an irreducible representation D (µ) (g) of the group, we can consider the following object, made, by summing over the elements in a class (1) D (µ) (C) = g C D (µ) (g) Since conjugation is one-to-one mapping of the class onto itself, this object obeys D (µ) (g)d (µ) (C)D (µ) (g 1 ) = D (µ) (C) which implies that D (µ) (g)d (µ) (C) = D (µ) (C)D (µ) (g), g G By Schur s first lemma, since D (µ) (C) is a matrix which commutes with every matrix in an irreducible representation, it must be a multiple of the identity D (µ) (C i ) = λ µ i I If we take a trace of both sides of this equation we find that () n Ci χ (µ) (C i ) = λ µ i d µ It is also clear that the product D (µ) (C i )D (µ) (C j ) is preserved by conjugation: D (µ) (g) [ D (µ) (C i )D (µ) (C j ) ] D (µ) (g 1 ) = [ D (µ) (g)d (µ) (C i )D (µ) (g 1 ) ] [ D (µ) (g)d (µ) (C j )D (µ) (g 1 ) ] = D (µ) (C i )D (µ) (C j ), g G This product must therefore consists of summations over full conjugacy classes, that is D (µ) (C i )D (µ) (C j ) = c ijk D (µ) (C k ) k

25 GROUP THEORY PRIMER 5 where the constants c ijk are non-negative integers which tell us how many times the class C k appears in the set {g i g j g i C i, g j C j }. These coefficients clearly depend on the group only, and not on the specific representations that were used to obtain them. This fact will be important shortly. Remembering that D (µ) (C i ) = λ µ i I in turn implies that λ i λ j = k c ijk λ k and, from (), n Ci χ (ν) (C i )n Cj χ (ν) (C j ) = c ijk d ν n Ck χ (ν) (C k ) Now, we sum both sides over the representation, ν to get () n Ci n Cj χ (ν) (C i )χ (ν) (C j ) = c ijk n Ck d ν χ (ν) (C k ) ν Now, from the regular representation, since it contains each irreducible representation d µ times, χ R (C i ) = d µ χ (µ) (C i ) µ Moreover, we know from the explicit construction of the representation that so we can conclude χ R ([e]) =, χ R (C i [e]) = 0 d µ χ (µ) (C i ) = µ ν { Ci = [e] 0 C i [e] Using this equation on the left-hand-side of () yields (4) n Ci n Cj χ (ν) (C i )χ (ν) (C j ) = c ij[e] ν Now we can ask when c ij[e] can be non-zero. For it to be non-zero, the identity e must appear in the set {g i g j g i C i, g j C j }. For this to be so, for some g i C i, g 1 i must be in C j. Assume that g 1, g C i. There exists h G such that g 1 = hg h 1. Then, with the same h, g1 1 = hg 1 h 1. If g 1 and g are in the same class then g1 1 and g 1 are the same class. There is a one-to-one mapping between the class which contains g and the class which contains g 1, obtained by simply associating all of the elements with their inverses. They could be either identical classes or distinct classes. For example, in the dihedral group, D, the class {c, c } is identical to the class of inverses. This is also true for the class {b, bc, bc }.

26 6 GROUP THEORY PRIMER Finally, in a unitary representation D (µ) (g 1 ) = D (µ) (g) and, if C j contains the inverses of the elements of C i, then χ (µ) (C j ) = χ (µ) (C i ). For example, in the abelian group C p which has one generator, c with the relation c p = e, each element is in a class by itself, so the classes are {e}, {c}, {c },..., {c p 1 } The inverse class to {c} is {c p 1 }, the inverse class of {c } is c p, etc. The irreducible representations of this group are all unitary and one-dimensional, D (k) (c) = e πik/p. Then, indeed, the inverse representation is obtained by conjugation, D (k) (c p 1 ) = D (k) (c). Using this information in equation (4) yields (5) χ (ν) (C i )χ (ν) (C j ) = δ ij n Ci ν This result tells us that the columns in the character table are also orthogonal. The characters of different classes are vectors in a space whose dimension is the number of irreducible representations. There cannot be more orthogonal vectors than the dimension of the space. This gives us the inequality number of classes number of irreducible representations Since we have previously derived this inequality in the other sense in equation (), we conclude that, for finite discrete groups, number of classes = number of irreducible representations

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