(Ref: Schensted Part II) If we have an arbitrary tensor with k indices W i 1,,i k. we can act on it 1 2 k with a permutation P = = w ia,i b,,i l

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1 Chapter S k and Tensor Representations Ref: Schensted art II) If we have an arbitrary tensor with k indices W i,,i k ) we can act on it k with a permutation = so a b l w) i,i,,i k = w ia,i b,,i l. Consider the algebra A formed by taking arbitrary linear combinations of the different permutations, considered as operators acting on the space of k th rank tensors. This algebra can be constructed for any group, particularly finite groups, and is called the group algebra. this is not the Lie algebra!). Note that this sum of permutations makes sense only as operators on a vector space. It is not the composition of permutations. Also note that as A is an algebra, one can both add and multiply by composition) elements in A. Definition: An algebra consists of a vector space V over a field F, together with a binary operation of multiplication on the set V of vectors, such that for all a F and α, β, γ V, the following are satisfied:. aα)β = aαβ) = αaβ). α + β)γ = αγ + βγ. αβ + γ) = αβ + αγ V is an associative algebra over F if, in addition, 4 αβ)γ = αβγ) for all α, β, γ V.

2 . Last Latexed: April 5, 07 at 9:45 Joel A. Shapiro The group algebra is useful because it can extract the tensors of specified symmetry. First consider tensors of rank. Writing I = I + )) + I )) we can extract s ij a ij = I + ))wij = I ))wij and w ij = s ij + a ij is a decomposition into a symmetric tensor and an antisymmetric tensor. The action of the permutations commutes with the SUn) rotations on the tensors, so a constraint on a tensor of the form Aw = 0 for some A A, if it holds for one state of an irreducible representation of SUn), will hold on all states in that representation. Thus s and a are separate representations. Now consider a rank tensor w ijk, and define s ijk = 6 S w ijk a ijk = sign )w ijk 6 S These are the totally symmetric and totally antisymmetric parts of w, but it is not all of w. For example, suppose w = w =, w =, all other components zero. Then s ijk and a ijk are both zero. The rest is related to the two-dimensional representation of S see homework #, problem ). In general, there will be operators in A associated with the different irreducible representations of S k, which extract the corresponding irreducible representations of SUn). So we now turn to the problem of finding the irreducible representations of S k.. Irreducible Representations of S k We know in general that the number of irreducible representations is the number of conjugacy classes. So let us begin with that. Any element of S k can ) be written as a product of disjoint cycles. For 4 5 example, = )4 5). This factorization is unique re- 5 4

3 68: Last Latexed: April 5, 07 at 9:45 member ) = )) up to the order of the factors, which commute because they are disjoint cycles. ) k Under conjugation by = a cycle simply has its k elements permuted. Thus i j k) = i j k ). This is true for products of cycles as well. Thus two permutations whose descriptions in terms of disjoint cycles contain the same number of cycles of each length are conjugate, and only those are. We describe the conjugacy class of elements describable in terms of disjoint cycles, I k of length l k, as l i l i ). Including one-cycles for any element left unmoved, we have m i ml m = k. Example: S permutations class I ) ) = )), ) ), ), ) as well ), ) ) There is one conjugacy class for each partition of k. A partition of an integer k is an unordered set of positive integers, possibly with repeats, which add to k. Example: How many classes are there in S 5? 5); 4, );,, ) =, );, );, );, ); 5 ) answer: 7. Thus we also know that there are that many irreducible representations, although there is not a straightforward correspondance between the representations and the conjugacy classes. Define a Young graph for S k as a set of k boxes arranged, left-justified, in rows each of which is no longer than the preceeding. The lengths of the rows provide a partition of k. So The number of partitions of n is given by the partition function of number theory, pn). There are other things called partition functions, especially Z of statistical mechanics, which is different. The number-theory one, also called the integer partition function, arises also in counting states in string theory. It has the fascinating property that pk) has the generating function k= xk ) = k=0 pk)xk, where we say p0) =.

4 4. Last Latexed: April 5, 07 at 9:45 Joel A. Shapiro 5) 4, ), ), ), ), ) 5 ) There is one irreducible representation of S k corresponding to each Young graph. A Young tableau is a Young graph with the numbers,,, k inserted in the boxes in some order, for example τ = 4 5 For each tableau we define an element of the group algebra, τ =, where the sum is over those permutations which permute the numbers within each row but do not move them from one row to another. Here τ = [I + ) + 4) + 4) + 4) + 4 )][I + 5 )], which includes of the 0 permutations in S 5. We also associate Q τ = sign ) where the sum includes only permutations which permute numbers in the same column but don t move numbers from one column to another. Thus Q τ = [I 5)][I )]. Finally we define the Young operator Y τ = Q τ τ. We see that the way to get a totally symmetric rank 5 tensor is to apply Y to an arbitrary one while you get a totally antisymmetric tensor by applying Y, with the numbers in any order in the boxes.. The Y τ corresponding to any Young tableau τ is almost, but not quite, the element of the group algebra we want to extract irreducible representations. We find a related set of basis vectors in the group algebra by using the representations of S k. Define e η ij = l η k! S k Γ η ji ), where η is the Young graph corresponding to an irreducible representation of S k, and l η is the dimension of that representation. The sum is over all the permutations.

5 68: Last Latexed: April 5, 07 at 9:45 5 For, l η =, Γ =, and e = Y 4 5 which is also equal to any other Young operator for a tableau in. For, l η =, Γ = sign, e = Y. Other Young operators in differ. only in sign, e.g. Y = sign ) Y = Y The e s have some marvelous properties. They form vector spaces transforming as irreducible representations under S k separately from the right and from the left: For Q S k, Qe η ij = l η k! = l η k! Γ η ji Γ η mi Q) m R ) Q = l η k! Γ η jm Γ η mi Q) m R ) R = m Γ η jm Q ) Q Γ η mi Q)eη mj where we again used the rearrangement theorem. Thus Q acts just the way you ld expect for a basis vector e i of representation η to transform, for each fixed j. From the other side, e η ij Q = m Γη jm Q)eη im. We say that the set e η ij is a two sided ideal or invariant subalgebra) of the group algebra over S k. This gives the e s an interesting algebra: e η ij eη mn = l ηl η k!) = l ηl η k!), S k Γ η ji,r S k Γ η ji ) Γ η nm ) p ) Γ ) η np R Γ η pm )R = l η Γ η ij k! ) Γ η pm )e η pn by unitarity S k p = δ ηη δ jm e η in by the great orthogonality theorem We may also show that the diagonal elements e η ii form a decomposition of the identity. From the great orthogonality theorem transposed, δ GG = ijη l η k! Γη ij G )Γ η ij G) Note and Q are any elements of S k, and are not related to τ and Q τ defined earlier.

6 6. Last Latexed: April 5, 07 at 9:45 Joel A. Shapiro we can write the identity element of S k as I = G δ G,IG = ijηg l η k! Γη ij I)Γη ij G )G so = iηg l η k! Γη ii G )G = η i e η ii I = η i e η ii Thus the whole algebra is spanned by these two sided ideals. In particular, the Y τ are contained in the corresponding e η ij an l η dimensional algebra). In fact, the space spanned by e η ij is also spanned by Q is ij j, where Q i and i are the antisymmetrizers and symmetrizers of a set of standard tableaux for η, which means tableaux in which the numbers increase left to right in each row, and also top to bottom in each column. Thus and are standard tableaux, but, and are not. Here s ij is the permutation such that τ i = s ij τ j. Each of these spaces has dimension l η, with l η equal to the number of standard tableaux, so The dimension of Γ η is the number of standard tableaux of η. Counting all possibilities is tedious, so we have a magic formula in terms of hooks. For each box b in a Young graph with k boxes, define the hook of b, g b = plus the number of boxes directly to the right plus the number of boxes directly beneath. Then l η = k!. g b Example: In the Young graph I have placed the corresponding hooks this is not a Young tableau) l = b! = 0. It would be hard to count this explicitly. For our more reasonable case, gives l =! =.

7 68: Last Latexed: April 5, 07 at 9:45 7. Representations of SUn) We now turn to the extraction of arbitary representations of SUn). Georgi discusses the fundamental weights of SUn), and shows that an arbitrary representation can be found from a tensor product of an adequate number of defining representations. The problem is to extract from the tensor product of k defining representations N k the irreducible pieces. We have seen that this can be done by demanding that elements of the permutation group algebra vanish. If we impose e η iiw = 0 for all η and i save one, that is equivalent to projecting out our representation T a a a k = e η ii w)a a a k no sum on i for one particular representation η and one basis vector i. The different i generate equivalent representations. The different η s, however, each correspond to a different inequivalent) representation of SUn). Before doing more formal arguments, we will do an example. Consider three spin objects, or the tensor product of three defining representations of SU). We will extract from this 8 dimensional state space the piece e. From the problem you did for homework #, problem ), e = Γ e ) 6 = I + ) ) ) ) ) ), Let this act on the basis vectors which we expand as = e, = e.

8 8. Last Latexed: April 5, 07 at 9:45 Joel A. Shapiro v e v = = = = = = = = 0 Notice this only results in one state of J z = and one of J z =. So e projects out a -dimensional s = state. e would project out an orthogonal spin. Thus the tensor product of three spin / s is a spin / the totally symmetric part, e ) and two spin / representations, = Having completed this trivial but tedious example of the simple case of SU) and, we are ready for some abstract reasoning. Now we consider the general case of N k. The basis vectors which are mixed by the permutations are only those with the same number of indices equal to, and the same number equal to, etc.. Consider the subspace with r i of the indices equal to i, with r i = k, each r i =,...,N. This subspace S r is spanned by the basis vector e = e e e }{{} e e e }{{} N e N, }{{} r times r times r N times together with all permutations e, for S k. If all the indices are different, all r i = 0 or, all of the permutations are inequivalent, and we get a k! dimensional space. But if the r i s are not all, there is a subgroup S k with B e = e for B. In fact, = S r S r S rn. Let = B B which is a element of the group algebra A. Then while the subspace S r is spanned by { e S k } it is also spanned by { e S k }.

9 68: Last Latexed: April 5, 07 at 9:45 9 We now want to extract from S r the piece projected out by e η ii. The products {e η ii } for all are just sums of multiples of eη ij, for all j by p. 5) so we want to know the dimension of the space {e η ij j =, l η }. As e η is a two-sided ideal, this space is k eη ik b k, so the dimensionality depends on the constraints on b k. If they were all independent, they would form an l η dimensional space. But there are constraints. For B, B =. Let s be more explicit: e η ij = n e η in b nj = e η ij B = n b nj e η in B = nm Γ η nm B)eη im b nj. The e η im are linearly independent, so b nj = m b mjγ η mnb), for B. To find out how many degrees of freedom survive this constraint for each j, observe that Γ η mn B) forms a reducible representation of the subgroup. So we can write Γ η B) = U Γ ǫ B) U ǫ where Γ ǫ are irreducible representations of. Now if c = bu, b = bγ η B) = bu ) Γ ǫ B)U = c = c Γ ǫ B). The vector c breaks up into pieces for each representation ǫ, with c ǫ Γ ǫ B) = c ǫ for all B. This is possible for nonzero c only if ǫ is the identity representation, as the representations are irreducible. Therefore the dimensionality of the space e η iie is the number of times, γ η, that the identity representation of is contained in Γ η. But the number of times the representation i is contained in Γ η is γ η = a i=i = g B χ i B)χ η B) = g B χ η B) for i=identity, where g is the number of elements in, which is r i! Example: η =. From homework, and recalling χ = Tr Γ, χ = for B = I χ = 0 for B = ), ), or ) χ = for B = ), or )

10 0. Last Latexed: April 5, 07 at 9:45 Joel A. Shapiro Consider the space starting from e e e. = {I, )}, g =, γ η = + 0) = so e generates only one state from the three-dimensional space S r. From e e e, = S, g = 6, γ η = ) = 0 so we get no state 6 here. If all vectors are unequal, say e e e for SUn > ), = I, g =, γ η = =. For SUN), there are N states of the form e i e i e i, not contributing anything to η =. There are NN ) states e i e i e j with i j, each contributing one state, so from these we get NN ) states. There are also NN )N )/6 states of the form e i e j e k, with i < j < k, each contributing states, so the dimension of is NN ) + NN )N ) = NN ) = { for N = 8 for N = Let s work another example, for SUN). As we need the characters for this representation, let s take them from Schensted: χ = for I 4 ), [ element]; χ = 0 for, ), [8 elements]; χ = for, ), [6 elements]; χ = for, ), [ elements] and for 4), [6 elements]. Enumerating the basis states in the various partitions, and multiplying χb) by their number for those within, we find indices subspace γ 4, ), ), ) 4) η ) all i e i e i e i e i S )+6 ) =0 4 ) i j e i e i e i e j S )+0 ) = 6 ) i < j e i e i e j e j S S )+0 ) = 4 ) i j<k i e i e i e j e k S )+0 ) = ) all e i e j e k e l I) )+0 ) = Rewriting this with only the results for γ η to allow room for counting index choices and states, we have

11 68: Last Latexed: April 5, 07 at 9:45 indices subspace γ η # index choices # states all i e i e i e i e i S 4 0 N 0 i j e i e i e i e j S NN ) NN ) i < j e i e i e j e j S S NN )/ NN )/ i j<k i e i e i e j e k S NN )N )/ NN )N ) N all e i e j e k e l I) 4) N) 4 is So the total dimensionality of for SUN) NN ) 0 + NN ) + + NN )N ) NN )N )N ) + 4! = NN ) + N ) + ) N )N ) = 8 N + )! 8 N )!. For SU), N =, the total dimension is 5!/8!) = 5. We now know how to extract the irreducible representations or just to count their dimensionality. Now it is time for magic. The number γ η of states extracted by e η ii from S r, the space spanned by e r i i ) by all S k, is given by the number of ways one can place r s, r s, in the Young graph so that in each row the numbers do not decrease, and in each column they increase. This is called a permissible placement. To see how this works, let s check it out on for SUN). For r = 4 there is no way to avoid two s in the same column, so γ = 0. For r = and r =, the has to be in the second row, so there is only one way, γ =. For r = r =, the only possibility is, γ =. For r =, r = r =, we have and, so γ =. For r = r = r = r 4 =, we have 4, 4, 4, and γ =. Note in our previous method, it was clear that these numbers only depended on the set {r i } and not on the order. This is now not obvious.

12 . Last Latexed: April 5, 07 at 9:45 Joel A. Shapiro Consider r =, r =, r 4 =. Then we have 4 and 4, so again γ =, as for r =, r = r =. Now to count the dimensionality of an irreducible representation of SUN) belonging to the Young graph η, we must sum, over all choices r i, the corresponding γ η. But for each choice of r i the γ η is the number of ways of placing the indices in the graph in a permissible fashion. So the dimension of the full irreducible representation of SUN) is the number of ways of placing k integers, chosen from,,,...n repeats allowed) in a permissible fashion in η. Example for SU):,,,,,,,, for a total of 8 dimensions. There is an easier method of finding the dimensionality. For each box, associate the value N + column number row number). Then divide by the hook of that box. The dimension of the representation is the product of these quotients over all the boxes. Examples: Dim Dim = N = N N+ N = N N ) N N+ N+ 4 N+ )! = 8 N )! Note: If the first column of a graph has N boxes, the hook of each box in column is equal to the N +column number row number) of the last box in that row. Thus eliminating the first row does not change the dimension. In fact, it does not change the representation either. This is because a totally antisymmetric tensor with N indices is invariant. Thus in SU), =, as we saw in detail. It also means that for SUN), we needn t consider representations with N or more rows except perhaps to indicate the identity representation by one column of N boxes).