GROUP THEORY PRIMER. New terms: tensor, rank k tensor, Young tableau, Young diagram, hook, hook length, factors over hooks rule

GROUP THEORY PRIMER New terms: tensor, rank k tensor, Young tableau, Young diagram, hook, hook length, factors over hooks rule 1. Tensor methods for su(n) To study some aspects of representations of a Lie algebra, it is sometimes useful to consider the space of vectors on which the Lie algebra elements operate. We called this space V to distinguish it from the Lie algebra itself, which is also a vector space L. These vectors should have the same dimension as the dimension of the representation. We often dealt with the components of the vectors v a. The way that a Lie algebra element acts on the components of a vector follows from the way that the vector transforms under the Lie group, where we can consider an infinitesimal transformation (1) v i v i + δv i = v i + λ a [T a ] i jv j α That is, the generators act by the usual rules of multiplying a vector by a matrix. () δ a v i = [T a ] i jv j where the repeated index j is assumed to be summed over. We have also distinguished the first and second index of T a by putting the first index up and the second index down. This distinction is an aid for index book-keeping, it does not change the meaning of the generators as matrices that we have discussed in previous chapters. As well as vectors, we could consider objects with more indices, which are called tensors, T i 1i...i k. A tensor of rank k transforms as δ a T i 1,i...i k = [T a ] i 1 T j 1,i...i k + [T a ] bi j T i 1,j...i k +... + [T a ] i k jk T i 1,i...j (3) k or, alternatively, we could think of the operation as being implemented by a large matrix, [δ a ] i 1i...i k j 1 j...j k = [T a ] i 1 δ i j... δ i k jk + δ i 1 [T a ] i j... δ i k jk +... + δ i 1 δ i j... [T a ] i k (4) jk Here, we have assumed that all of the indices of the tensor transform under the same representation and we have called the matrices corresponding to the generators in that representation T a. It is straightforward to generalize to the situation where different indices transform in different representations. The indices i l would then 1

GROUP THEORY PRIMER run over the range, 1,,..., D l where D l is the dimension of its representation and of the dimension of the matrix T a in each term in equation (4) would be the one that is appropriate for the representation of the index that it acts on. It is easy to show, by explicitly taking the commutator of the objects in equation (4), that this transformation law is a representation of the Lie algebra, that is, that (5) δ a δ b δ b δ a = if abc δ c where f abc are the structure constants in the Lie algebra (6) [T a, T b ] = if abc T c We could ask whether this is a reducible or an irreducible representation. The answer is, it can be either, depending on the nature of the tensor. To see this, consider a two-index tensor, T i 1,i where both indices transform in the same representation. Let us assume there that the Lie algebra is su(n) and the representation is the first fundamental representation. 1 When the tensor has two indices, it transforms in a reducible N -dimenisonal representation of the Lie algebra. We can decompose the tensor into its symmetric and anti-symmetric parts, T i 1,i = 1 [ T i 1,i + T i,i 1 ] 1 [ + T i 1,i T i,i 1 ] If it is already symmetric or anti-symmetric then one of the above parts would vanish. Let us assume that both parts are non-zero. Then, consider the transformation of the symmetric part, 1 [ δ a T i 1,i + T i,i 1 ] 1 [ = [Ta ] i 1 T j 1,i + [T a ] i j T i 1,j + [T a ] i T j 1,i 1 + [T a ] i 1 j T i,j ] = [T a ] i 1 [ 1 T j 1,i + T i,j 1 ] + [Ta ] i 1 [ j T i 1,j + T j,i 1 ] = ( ) [T a ] i 1 δ i j + δ i 1 [T a ] i 1 [ j T j 1,j + T j,j 1 ] Similarly for the anti-symmetric part, 1 [ δ a T i 1,i T i,i 1 ] ( ) = [Ta ] i 1 δ i j + δ i 1 [T a ] i 1 [ j T j 1,j T j,j 1 ] We see that the symmetric part of the tensor transforms like a tensor and that the anti-symmetric part also transforms like a tensor. Moreover, the symmetric and anti-symmetric parts of the tensor do not mix with each other under the Lie algebra transformations. This means that the original tensor transforms under a reducible representation that can be block diagonalized by separating it into its symmetric and anti-symmetric components. Thus, from the original N - dimensional representation of the algebra acting on the tensor, we have block diagonalized to find two smaller 1 This is the representation with highest weight µ 1 obeying α i µ 1 / α i = δ 1i.

GROUP THEORY PRIMER 3 representations, the N(N+1) -dimensional symmetric one and the N(N 1) -dimensional anti-symmetric one. There is the question as to whether these smaller representations can be block diagonalized further, that is, do the symmetric and anti-symmetric parts themselves transform under reducible representations of the Lie algebra? The answer turns out to be no, at least in the case of su(n), they are irreducible. We can think of the symmetrization and anti-symmetrization of the two-index tensor that we have done in a group theoretical context. The essential observation is that the symmetric and anti-symmetric tensors transform under irreducible representations of the symmetric group S acting on the two indices of the tensor. S is the two-element group S = {e, σ} with σ = e. e is the identity and σ interchanges the tensor indices. Its character table is S {e} {σ} A 1 1 1 A 1-1 This two element group has two irreducible representations, the trivial symmetric representation A 1 and the anti-symmetric representation A. In general, the operation of the Lie algebra on a rank k tensor, as defined in equation (3) commutes with the operation of the symmetric group S k on the k indices of the tensor. We can always generalize the symmetrization and anti-symmetrization of the two-index tensor to the the analogous procedure for a rank k tensor which would write it as a sum of tensors which transform under irreducible representations of the symmetric group, S k, acting on the indices. It turns out that the tensors which lie in irreducible representations of the symmetric group will also carry irreducible representations of the su(n) Lie algebras. Let us begin to understand this with another example. Consider the three-index tensor T ink. The set of all permutations of its three indices ai, j, k constitutes the symmetric group, S 3. Remember that S 3 is a six-element group which has three distinct conjugacy classes and three inequivalent irreducible representations. It had generators c and b with the relations c 3 = e, b = e and (bc) = e. The character table is S 3 {e} {c} 3{b} A 1 1 1 1 A 1 1-1 E -1 0 There are two one-dimensional representations, A 1 and A, and one two-dimensional representation, E. We must find the linear combination of tensor with permuted indices which transform in these representations under permutations of the indices.

4 GROUP THEORY PRIMER The completely symmetric and completely anti-symmetric are easy to find. The completely symmetric one is mapped onto itself by any permutation and it corresponds to the representation A 1. It is made by taking a symmetric linear combination of the tensor with all of the possible permutations of its indices: T ijk A 1 = 1 [ (7) T ijk + T jki + T kij + T kji + T jik + T ikj] 6 Under any permutation of the indices, ijk P (i)p (j)p (k), T P (i)p (j)p (k) A 1 = 1 T ijk A 1 TA ink 1 transforms in the representation A 1 of S 3. In the one-dimensional anti-symmetric representation A, the generators c and b of S 3 are represented by c = 1 and b = 1. The elements of the conjugacy class of b are odd permutations in that they require an odd number of exchanges of neighbouring indices. The elements of the conjugacy class of c are even permutations. Thus, we must find a tensor which is multiplied by 1 any time we interchange two indices. It is easy to find this. We simply add together the tensors with all permutations of the indices and we put a minus one in front of those which are obtained by an odd number of permutations of the indices in the starting order: T ijk A = 1 [ T ijk + T jki + T kij T kji T jik T ikj] 6 This is anti-symmetrization of the rank 3 tensor. Under any permutation of the indices, abc P (i)p (j)p (k), T = D (A) (P )T ijk A P (i)p (j)p (k) where D (A) (P ) = ( 1) deg(p ), A P (i)p (j)p (k) TA = ( 1) deg(p ) T ijk T ijk A 1 transforms in the representation A 1 of S 3. What about the two-dimensional representation, E? For this, we appeal to the fact that the representations of the symmetric group are in one-to-one correspondence with Young tableaux. This will give us a handy way of constructing the tensor which has this and other transformation properties under the symmetric group.. Young Tableaux A Young tableau is an array of horizontal rows of neighbouring boxes, examples of which are displayed in figures 1 and, which are constructed according to the following rules For the symmetric group S k, there are k boxes. The rows of boxes of the array are aligned at the left-hand margin. Reading the array from the top to the bottom, the horizontal rows of boxes are of non-increasing lengths. A

GROUP THEORY PRIMER 5 Figure 1. A Young tableau is an array of horizontal lines of intersecting boxes which is constructed according to a set of rules. The lines are aligned on the left-hand-side of the tableau. The lengths of the rows is non-decreasing from the top to the bottom. Strictly speaking, the examples depicted in figures 1 and are called Young diagrams and they become Young tableaux only when certain numbers are inserted into the boxes, but we will not need this distinction here, and we will call all such objects Young tableaux, which is the plural of the term Young tableau which we shall use for s single object. The set of all legal Young tableaux with k boxes are in one-to-one correspondence with the inequivalent irreducible representations of the symmetric group S k. We will not prove this here, but will simply give a plausibility argument. For this, we recall that the elements of the group S k can be decomposed into cycles. Moreover, those elements which have the same cycle decomposition reside in the same conjugacy class. The conjugacy classes are thus in one-to-one correspondence with distinct cycle decompositions. Conjugacy classes are also in one-to-one correspondence with irreducible representations. Thus, we have a one-to-one mapping between distinct cycle decompositions and irreducible representations. Distinct cycle decompositions are characterized completely by the lengths of the cycles, that is, by a partition of the integer k. Such a partition is a non-decreasing set of positive integers λ 1, λ,... such that λ 1 + λ +... = k. These sets of integers are in one-to-one correspondence with the possible distinct Young tableaux, drawn according to the rules written above. Examples of tableaux are depicted in figures 1 and. The irreducible representations of the symmetric group S k are therefore also in one-to-one correspondence with the Young tableaux with k boxes.

6 GROUP THEORY PRIMER Figure. A second example of a Young tableau. A legitimate tableau can have rows of equal lengths or columns of equal lengths. Not only do the Young Tableaux give us a quick pictorial view of the irreducible representations of S k, but, together with Young symmetrization, they give us a way of reducing a rank k tensor so that it transforms under the corresponding representation. To implement Young symmetrization, we put the indices of the tensor into the boxes of the Young tableau as in figure 5. Then, we follow Young symmetrization procedure. This first symmetries the indices which appear in all of the rows. Afterward, it anti-symmetrizes on the indices that appear in the columns. What it produces is a tensor which transforms in the corresponding representation of S k when its indices are permuted. These representations are sometimes one dimensional, like the completely symmetric or completely anti-symmetric, but they can also be multidimensional. In the latter case, there are inequivalent permutations of the indices so that tensors of a given type transform into each other. To illustrate, consider again the three-index tensor. We have already discussed the completely symmetric combination TA abc 1 and the completely anti-symmetric combination TA abc. These correspond to the one-dimensional representations which are associated with the two Young tableaux depicted in figures 3 and 4. The symmetric tableau has three boxes in a horizontal line. The anti-symmetric tableau has three boxes in a vertical line. The other representation of S 3 is two-dimensional. We get it by beginning with the tableau in figure 5. We have placed the indices in the boxes of the tableau. Then, we take the tensor and symmetrize on the indices which occur in rows: T ijk 1 [ T ijk + T kji]

GROUP THEORY PRIMER 7 Figure 3. The row tableau corresponding to the trivial representation of S 3. The indices of the tensor are completely symmetrized. Figure 4. The column tableau corresponding to the one-dimensional representation, A, of S 3 where e = 1, c = 1, b = 1. The indices of the tensor are completely anti-symmetrized. Then, we subsequently anti-symmetrize in the indices which occurred in the column of the tableau: T ijk E = 1 [ T ijk + T kji T jik T kij] 4 E is a two dimensional representation of S 3. This is manifest in the fact that different permutations of the indices give inequivalent tensors. There are only two linearly independent combinations due to the fact that the tensor is anti-symmetric, and that its symmetric permutations obey T ijk E + T jki E + T kij E = 0 as can be confirmed by direct substitution. This leaves two independent combinations which form the doublet, E. 3. Irreducible representations of su(n) Remember that a the Lie algebra is characterized by a defining representation, whose weights were ξ 1, ξ,..., ξ N. We chose an ordering convention for weights where

8 GROUP THEORY PRIMER Figure 5. The tableau corresponding to the two-dimensional representation, E, of S 3. The indices which occur in the rows are symmetrized and, subsequently, the indices which occur in the columns are anti-symmetrized. they are ordered as ξ 1 > ξ >... > ξ N. Simple roots of the algebra were taken as the r = N 1 positive vectors α 1 = ξ 1 ξ, α = ξ ξ 3,..., α N 1 = ξ N 1 ξ N. The fundamental weights, µ i, are the highest weights of the fundamental representations. They are the unique vectors which satisfy the equations α i µ j / α j = δ ij. A solution of these equations is µ 1 = ξ 1 µ = ξ 1 + ξ... N µ N 1 = ξ i These are the highest weights of the N 1 fundamental representations. Moreover, the highest weight of any representation can be written as ν = N 1 i=1 i=1 n k µ k where n k are non-negative integers. We can denote the representations by the sequence of integers, [n 1, n,..., n N 1 ]. In this array of integers, n i corresponds to how many times the highest weight can be lowered by the i th simple root α i. (Since it

GROUP THEORY PRIMER 9 is a highest weight, it cannot be raised.) In this notation, the fundamental representations are [1, 0, 0,..., 0], [0, 1, 0,..., 0],..., [0, 0, 0,..., 1]. The adjoint representation is [1, 0,..., 0, 1]. The weights ξ 1,..., ξ N are simultaneous eigenvalues of the Cartan sub-algebra. It is very useful to think about the eigenvectors which correspond to these sets of eigenvalues, v 1 to ξ 1, v to ξ, etc. These obey Hv 1 = ξ 1 v 1 Hv = ξ v...... Hv N = ξ N v N These eigenvectors constitute N ortho-normal basis vectors for the vector space on which the generators, T a of this N-dimensional, defining representation operate. 3 Just as we can find eigen-vectors of the Cartan sub algebra, which are simultaneous eigenvalues of all of the Cartan generators, we can also find eigentensors. To do this, we consider a particular type of tensor which is made from a direct product of the eigen-vectors, i ˆT m1 m...m k = v m1 v m... v mk, ˆT 1 i...i k m 1 m...m k = v i 1 m1 v i m... v i k (8) mk The set of all of these tensors, for all values of the labels i 1, i,..., i k, forms a complete basis for the space of tensors of rank k. Remember that the defining representation of the Lie algebra would act on this tensor as δ a ˆT i 1 i...i k m 1 m...m k = [T a ] i 1 ˆT j 1 i...i k m 1 m...m k + [T a ] i j ˆT i 1 j...i k m 1 m...m k +... + [T a ] in j k ˆT i 1 i...j k m 1 m...m k Then, the Cartan sub-algebra generators acting on this tensor as δ r ˆTm1 m...m k = H r v m1 v m... v mk +v m1 H r v m... v mk +...+v m1 v m... H r v mk ( = ξm1 + ξ m +... + ξ ( mk) v m1 v m... v mk = ξm1 + ξ m +... + ξ mk) ˆT m1 m...m k r r That is, the weight of the tensor in the direct product representation is just the sum of the weights of the eigenvectors from which the direct product is made. This The highest weight of the adjoint representation is µ 1 + µ N 1 = ξ 1 + ( ξ 1 +... ξ N 1 ) = ξ 1 ξ N = α 1 + α +... + α N=1 where we have remembered that N i=1 ξ i = 0. The reader might check this with su(3) where we saw that the highest weight of the adjoint representation was α 1 + α. 3 The vectors v l are orthogonal because they are non-degenerate simultaneous eigenvectors of the hermitian matrices H. The equation for them is linear so they can be normalized to form an orthonormal set, v m v n = δ mn.

10 GROUP THEORY PRIMER gives us a basis of N k tensors which are eigenvalues of the Cartan generator. The eigenvalues ξ m1 + ξ m +... + ξ mk are the weights of this reducible direct product representation. The highest weight in the direct product representation is kξ 1 and the eigenvector with that weight is i ˆT 11...1 = v 1 v 1... v 1, ˆT 1 i...i k 11...1 = v i 1 1 v i 1... v i k (9) 1 This eigenvector is non-degenerate and as well as being the highest weight of the tensor representation, it is also the highest weight of the irreducible representation [k, 0, 0,..., 0]. We therefore expect that the direct product representation must contain one copy of the irreducible representation [k, 0,..., 0] as well as other irreducible representations. Note that ˆT i 1i...i k 11...1 in equation (9) this is a completely symmetric tensor in its indices i 1, i,..., i k. Since it is composed of a direct product of highest weight vectors, the state in equation (9) cannot be raised at all. It can be lowered, but only by the simple root α 1, by which it can be lowered k times. Remembering that α 1 = ξ 1 ξ, so that the result of lowering ξ 1 is ξ 1 α 1 = ξ 1 ( ξ 1 ξ ) = ξ, we know that operating E α 1 on v 1 v 1... v 1 produces (10) E α 1 v 1 v 1... v 1 v v 1... v 1 + v 1 v... v 1 +... + v 1 v 1... v which has weight (k 1) ξ 1 + ξ. It is a tensor with coefficients given by T i 1i...i k 1...1 + T i 1i...i k 1...1 +... + T i 1i...i k 11... is also completely symmetric in the indices i 1, i,..., i k. The other weights of this representation can be found by further operating E α s on the vectors kξ 1, kξ 1 α 1 = (k 1) ξ 1 + ξ, ξ 1 α 1 = (k ) ξ 1 +ξ,..., kξ 1 k α 1 = kξ. Doing this will have used up only one of the k independent linear combinations of the basis tensors which have k 1 1 s and one, the completely symmetrized one. There are k 1 other states with the same weight. A basis for these k 1 states could be (11) v v 1 v 1... v 1 v 1 v v 1... v 1 v v 1 v 1... v 1 v 1 v 1 v... v 1... v v 1 v 1... v 1 v 1 v 1 v 1... v These tensors are all orthogonal to the one in equation (10). We expect that they are the highest weights of representations whose highest weight is [k, 1, 0,..., 0]. Indeed, v 1, since it is the eigenvector corresponding to the highest weight of the

GROUP THEORY PRIMER 11 Figure 6. The tableau corresponding to the [k, 1, 0,..., 0] representation of su(n) and associated with the indices on the basis tensor given in equation (11) is displayed. defining representation is annihilated by all E α with positive roots α, so it cannot be raised at all. The other vector, v, since it corresponds to the second highest weight, it can only be raised back to v 1 by the application of E α1. However, E α1 applied to any of the vectors in equation (11) must produce zero because of the relative minus sign. For example, and therefore (1) E α1 [v v 1 v 1... v 1 v 1 v v 1... v 1 ] v 1 v 1 v 1... v 1 v 1 v 1 v 1... v 1 = 0 E αi [v v 1 v 1... v 1 v 1 v v 1... v 1 ] = 0 and [v v 1 v 1... v 1 v 1 v v 1... v 1 ]. That is, the states in (11) cannot be raised at all. Linear combinations of them must therefore be the highest weights of k 1 copies of the irreducible representation of su(n) that is labeled [k, 1, 0, 0,..., 0]. Generally, the representation [n 1, n,..., n k ] corresponds to a Young tableau with n 1 columns with 1 box, n columns with two boxes, and so on. We can see this by finding the highest weight that is allowed for such a Young tableau. We do this by filling the top row with the states corresponding to the highest weight, v 1, the second row with the states corresponding to the second highest weight, v, (no v 1 s are allowed in the second row because of the anti-symmetrization of the columns and the fact that first row already has a v 1 in all cases), the third row with v 3, and so on to get the highest possible sum of the weight vectors. Then a columns of height q will have the vectors v 1, v,..., v q and will contribute ξ 1 + ξ +... + ξ q = µ q to the highest weight. This is the q th fundamental weight.

1 GROUP THEORY PRIMER Infobox 3.1 3 3 3 in su(3) As a concrete example, we can reconsider the direct product of three threedimensional representation defining representations of su(3). These are the first fundamental representations with highest weight µ 1 = ξ 1. The state vector associated with it is v 1 and Hv 1 = ξ 1 v 1. Similarly, Hv = ξ v and Hv 3 = ξ 3 v 3. The vectors are normalized so that v i v j = δ ij. The simple roots are α 1 = ξ 1 ξ and α = ξ ξ 3, or E α1 = N 1 v 1 v, E α = N v v 3 Here, we have formed a matrix v v 1 from the product of a vector and its hermitian conjugate. This matrix is non-zero only when it operators on a state containing a non-zero projection onto v 1 and it produces a state proportional to v. In this notation, the diagonal Cartan sub-algebra matrices are H = ξ 1 v 1 v 1 + ξ v v + ξ 3 v 3 v 3 We can get the normalization of E αi from α 1 H = [ ] E α1, E α 1 = N1 (v 1 (v v )v 1 v (v 1v 1 )v ( = α 1 ξ1 v 1 v 1 + ξ v v + ξ ) 3 v 3 v 3 = 1 v 1v 1 1 v v (where, we know that α 1 ξ 1 / α 1 = 1 and α 1 ( ξ 1 + ξ )/ α 1 = 0 since ξ 1 and ξ 1 + ξ are fundamental weights and we take α 1 = 1). Then, with a phase convention, N 1 = 1. By a similar argument N = 1. Now, let us consider the direct product of three of these three-dimensional representations. A basis tensor is (13) v i1 v i v i3 and it has weight ξ i1 + ξ i + ξ i3. There are twenty-seven linearly independent basis tensors, and the representation is 7-dimensional. The highest weight is the sum of the highest weights of each of the three representations in the direct product, that is, 3 ξ 1, and the state associated with it is v 1 v 1 v 1. We can expect the direct product representation whose states are (13) to be reducible. It should contain a number of irreducible representations. The unique state of the reducible representation with the highest weight should also be the highest weight of an irreducible representation which is contained inside the irreducible one. We should be able to find the states of this irreducible representation by lowering the highest weight. It is the highest weight of the [3, 0] representation )

GROUP THEORY PRIMER 13 and it can be lowered one, two or three times by the simple root α 1 = ξ 1 ξ. This produces the weights ξ + ξ, ξ 1 + ξ, and 3 ξ. These weights can then be lowered by the simple root α = ξ ξ 3. This produces the family of weights 3ξ 1 ξ 1 + ξ ξ 1 + ξ 3 ξ 1 + ξ ξ1 + ξ + ξ 3 ξ1 + ξ 3 3 ξ ξ + ξ 3 ξ + ξ 3 3 ξ 3 where going down in the table is lowering by α 1 and going to the right is lowering with α. These are the weights of the ten dimensional representation of su(3). The highest weight state, v 1 v 1 v 1, is a totally symmetric tensor. Lowering it produces symmetric tensors, for example E α 1 (v 1 v 1 v 1 ) = (E α 1 v 1 ) v 1 v 1 + v 1 (E α 1 v 1 ) v 1 + v 1 v 1 (E α 1 v 1 ) (14) v v 1 v 1 + v 1 v v 1 + v 1 v 1 v which has weight ξ 1 + ξ. There are three states with this weight and the above lowering has produced the totally symmetric linear combination of them. If we continue to lower, we sweep out the ten dimensional representation made by the totally symmetric tensors with three indices. This representation corresponds to the Young tableau in figure 3 (which also corresponds to the completely symmetric (trivial) representation of the symmetric group S 3 ). In the ten dimensional representation, we have used up one of the tensors with the second highest weight, the totally symmetric combination given in (14). The two other states should be highest weight states of further irreducible representations which are contained in the direct product. The weight is ξ 1 + ξ = µ 1 + µ, and there are two linearly independent vectors, so there must be two representations, both the [1, 1] adjoint representation. Two states with this second highest weight which are orthogonal to the totally symmetric state, and each other, are (15) and (16) v 1 v 1 v v v 1 v 1 v 1 v v 1 v v 1 v 1 v 1 v v 1 Our choice of these states is motivated by the symmetry that we expect a representation with Young tableau depicted in figure 5 which corresponds to the representation [1, 1]. Neither of the states (15) or (16) can be raised by a positive root, so they are highest weight states. Their weight is ξ 1 + ξ = µ 1 + µ, so they can be lowered

14 GROUP THEORY PRIMER once by the simple root α 1 and once by the simple root α. Taking the second of these, (16), and lowering it alternatively by the simple roots α, α 1, α,... yields the sequence of states v v 1 v 1 v 1 v v 1 v 3 v 1 v 1 v 1 v 3 v 1 v 3 v v 1 v v 3 v 1 + v 3 v 1 v v 1 v 3 v v 3 v 1 v 3 v 1 v 3 v 3 v 3 v v 3 v v 3 v 3 (17) 0 Now but with the sequence of lowerings by the simple roots α 1, α, α 1,... instead yields v v 1 v 1 v 1 v v 1 v v 1 v v 1 v v v 3 v 1 v v 1 v 3 v + v v 1 v 3 v 1 v v 3 v 3 v v v v 3 v v 3 v v 3 v v 3 v 3 (18) 0 Given that the first and last states in the sequences in (17) and (18) are identical, together they contain eight linearly independent states. Similarly, if we begin with the other highest weight, (15), and lower with α, α 1, α,... we find v 1 v 1 v v v 1 v 1 v 1 v v 1 v 1 v 1 v 3 v 3 v 1 v 1 v 1 v 3 v 1 v v 1 v 3 v 3 v v 1 v v 3 v 1 + v 1 v v 3 v 3 v 1 v v 1 v 3 v v 3 v 1 v 3 v 3 v 3 v 1 v 1 v 3 v 3 v 3 v v 3 v 3 v 3 v v v 3 v 3 0 (19)

and with α 1, α, α 1,... we find v 1 v 1 v v v 1 v 1 v 1 v v 1 v v 1 v v v v 1 + v 1 v v v v 1 v v v v 1 v 1 v v GROUP THEORY PRIMER 15 v 3 v 1 v v 3 v v 1 + v 1 v 3 v + v v 1 v 3 v v 3 v 1 + v 1 v v 3 v 3 v v + v v v 3 v v 3 v v 3 v 3 v + v 3 v v 3 + v v 3 v 3 0 (0) There are also eight linearly independent states in the sequences (19) and (0), as there were in (17) and (18). All of the states in the second set, (19) and (0), are orthogonal to all of the states in the first set (17) and (18). Both sets of eight states have the same weights, α 1 ξ 1 + ξ ξ1 + ξ 0 α ξ 1 + ξ 3 two of ξ 1 + ξ + ξ 3 ξ + ξ 3 0 ξ1 + ξ 3 ξ + ξ 3 where, we move to the right in the table by lowering with α 1 and down by lowering with α. Each set of states form a basis for the vector space on which the generators of an eight dimensional adjoint representation [1, 1] of su(3) operate. In the above we have found the ten-dimensional and the two eight-dimensional irreducible representations of su(3) which are contained in the direct product of three three-dimensional representations. In them, we have not quite used up all of the states in the original direct product space. Amongst the six possible zero weight states, we have only used five of them. Orthogonal to all of the ones that we have used is the completely anti-symmetric state v 1 v v 3 + v v 3 v 1 + v 3 v 1 v v 3 v v 1 v v 1 v 3 v 1 v 3 v This state cannot be raised or lowered by any of the simple roots. It has weight ξ 1 + ξ + ξ = 0 and it is a singlet of the Lie algebra. In a language that is sometimes used, the direct product of three threedimensional representations is equal to the direct sum of one ten, two eight and one one-dimensional representation, 3 3 3 = 10 8 8 1

16 GROUP THEORY PRIMER Figure 7. Each box in a Young tableau is associated with a unique hook, the set of all boxes to the right of it plus all boxes below it plus the box itself. The number of boxes is defined as the hook length. The tableau that is depicted here has the hook lengths drawn in the boxes corresponding to the hooks. 3.1. Dimension of a representation of S k. We have argued that the set of Young tableaux with k boxes are in one-to-one correspondence with the irreducible representations of the symmetric group S k. There is an easy formula for computing the dimension of the representation of S k. The dimension is given by the following ratio: (1) dim D = all hooks k! (hook lengths) A hook length is defined for each box in a Young tableau. It is the number of boxes that occur directly to the right of it plus the number of boxes that occur directly below it plus one, the box itself. An example of a Young tableau with hook lengths filled into the boxes is depicted in figure 7. 3.. Dimension of representation of su(n). We have also argued that the irreducible representations of su(n) are in one-to-one correspondence with Young tableaux which have columns of no more than N 1 boxes. Each Young tableau corresponds to a highest-weight representation of the su(n) Lie algebra. There is a simple algorithm for finding the dimension of this representation, beginning with the Young tableau. It is called the factors over hooks rule. To do this, we take the tableau and fill the boxes with integers following the rules: (1) N appears in the top left box. () the box that is immediately below a given box has an integer that is one lower than that in the given box

GROUP THEORY PRIMER 17 (3) the box that is immediately to the right of a given box has an integer that is one unit larger than that in the given box. These rules can be used to fill all of the boxes of a tableau with integers. Then, the dimension of the irreducible representation of su(n) is () dim D = boxes (integer in each box as found above) (hook lengths) all hooks 3.3. The algorithm for decomposing a direct product into irreducible representations of su(n). Young tableaux give us a useful algorithm for decomposing reducible direct product representations of su(n) into direct sums of irreducible representations. Consider two irreducible representations of su(n) with correspond to Young tableaux τ 1 and τ, respectively. Then the algorithm is (1) Write the tableaux τ 1 and τ, with the simpler one to the right. () Place the letter a in every box in the first row of τ 1, b in every box in the second row, and so on until all of the boxes in the tableau have been filled. (3) Take the boxes of τ, with their labels one at a time and add them to τ 1 in all possible places. After each box is added, discard all of the objects which are produced and which do not obey the following rules: (a) Each object must itself be a legal Young tableau. (b) In each column of the object, the letters which labeled the boxes of τ must appear only once. (c) For every box in the object, consider region of the object that is defined by the box itself, plus the portion of its row to the right of it, plus the portion of its column above it, plus the region of the object above and to the right of the row and column. If k a is the number of boxes in this region labeled by a, q b labeled by b and so on, it is required that n a n b.... (4) Repeated identically shaped tableaux are considered distinct only if they have different patterns of the labels. Then they can be replaced by one tableau times their multiplicity. If they have the same pattern of labels, they can be replaced by one tableau. (5) Columns with N boxes can be dropped from the tableaux. A simple example is illustrated in figure 8.

18 GROUP THEORY PRIMER Figure 8. The direct product of two fundamental representations [1, 0, 0,..., 0] results in the direct sum of the representations [, 0, 0,..., 0] and [0, 1, 0,..., 0]. We can check the dimensions of the representations, N N = N(N+1 + N(N 1) where the dimensions of [, 0, 0,..., 0] and [1, 1, 0,..., 0] are easily computed using the factors over hooks rule of the previous section. Remember that [q 1, q,...] corresponds to a tableau with q 1 columns of length 1, q columns with length, and so on. Figure 9. The direct product of three fundamental representations [1, 0, 0,..., 0] results in the direct sum of the representations [3, 0, 0,..., 0], two copies of [1, 1, 0,..., 0, 0] and [0, 0, 1, 0..., 0]. We can check the dimensions of the representations, N N N = N(N+1)(N+) + N(N+1)(N 1)) + N(N 1)(N ) where the dimensions of 6 3 6 [3, 0, 0,..., 0], [1, 1, 0,..., 0, 0] and [0, 0, 1, 0..., 0] are easily computed using the factors over hooks rule of the previous section.