1 g 1,gg 2 xg 1. 2 g 1,...,gg n xg 1. n g 1 } = {gxg 1,gg 1 x(gg 1 ) 1,gg 2 x(gg 2 ) 1,...,gg n x(gg n ) 1 } = {x,g 1 xg 1. 2,...

Transcription

1 Physics 5 Solution Set Spring 7. A finite group G can be decomposed into conjugacy classes C k. (a) Construct the set C k gc kg, which is obtained by replacing each element x C k by gxg. Prove that C k = C k. The elements of a finite group G (consisting of n + elements) will be denoted by G = {e,g,g,...,g n } where e is the identity element. If x C k, then the elements of C k are given by: C k = {x,g xg,g xg,...,g n xgn }, () where we keep only distinct elements in the set C k and discard any duplicate elements. The set of elements in C k gc kg, for a fixed element g G, is then given by: C k = {gxg,gg xg g,gg xg g,...,gg n xg n g } = {gxg,gg x(gg ),gg x(gg ),...,gg n x(gg n ) } = {x,g xg,g xg,...,g nxg n } = C k, after using the rearrangement lemma in the penultimate step to reorder the elements of C k. (b) Suppose that D (i) (g) is the ith irreducible (finite-dimensional) matrix representation of the finite group G. For a fixed class C k, prove that D (i) jl (g) = N k χ (i) (C k )δ jl, () n g C i k where n i is the dimension of the ith irreducible representation of G, N k is the number of elements in the kth conjugacy class and χ (i) (C k ) is the irreducible character corresponding to the kth conjugacy class. Define the matrix A (i) k g C k D (i) ( g). () Then, D (i) (g)a (i) k = g C k D (i) (g)d (i) ( g) = g C k D (i) (g g) = g C k D (i) (g gg )D (i) (g), where we have used the fact that the matrix representation D (i) (g) must obey the group multiplication law, D (i) (g g ) = D (i) (g )D (i) (g ). Using the result of part (a), it follows that D (i) (g gg )D (i) (g) = D (i) (g )D (i) (g) = A (i) k D(i) (g), g C k g C k after noting that C k = C k and g is a dummy summation variable.

2 Thus, we have established that for any i and k, D (i) (g)a (i) k = A(i) k D(i) (g), for all g G. Hence, Schur s second lemma applies, and it follows that for some complex constant λ (i) k Using eq. (), we can rewrite eq. (4) as A (i) k = λ(i) k I, (4) (which can depend on i and k), where I is the identity matrix. D (i) jl g C k (g) = λ(i) k δ jl. (5) To determine the constant λ (i) k, we set j = l in eq. (5) and sum over j, which yields: N k χ (i) (C k ) = n i λ (i) k, where N k, n i and χ (i) (C k ) are defined in the statement of the problem. Solving for λ (i) k then completes the derivation of eq. (). (c) Starting from the completeness result that is satisfied by the matrix elements of the irreducible matrix representations of G and using the result of part (b), derive the completeness relation for the irreducible characters, N k O(G) χ (i) (C k )[χ (i) (C l )] = δ kl, (6) i where O(G) is the order of the group G (i.e. the number of elements of G), and the sum is taken over all inequivalent (finite-dimensional) irreducible representations. The completeness relation is given by: O(G) i m,n n i D (i) mn (g)d(i) mn (g ) = δ gg, (7) where O(G) is the order of the group G (i.e., the number of elements of G). The sum over i runs over all inequivalent irreducible representations of G. We now sum eq. (7) over g C k and g C l and make use of eq. (), which yields: O(G) i m,n n i N k n i χ (i) (C k ) N l n i χ (i) (C l ) δ mn δ mn = N k δ kl, (8) since there are N k elements in the class C k. Noting that δ mn δ mn = δ mm = n i, it follows that the factors of n i cancel in eq. (8). Thus, after summing over m and n, we obtain the completeness relation for the irreducible characters given in eq. (6).

3 (d) Using the orthogonality and the completeness relations satisfied by the irreducible characters, prove that the number of inequivalent irreducible representations of G is equal to the number of conjugacy classes. The orthogonality relation satisfied by the irreducible characters was derived in class, O(G) n c k= The completeness relation derived in part (c) is given by N k O(G) N k [χ (i) (C k )] χ (j) (C k ) = δ ij. (9) n irr χ (i) (C k )[χ (i) (C l )] = δ kl. () i= If we set i = j in eq. (9) and sum over i, we obtain n irr n irr = = i= n c k= N k O(G) n irr n c [χ (i) (C k )] χ (i) (C k ) = = n c, i= after interchanging the order of the summation and using eq. (). Hence, the number of inequivalent irreducible representations of G is equal to the number of conjugacy classes. The same conclusion can be obtained by regarding the characters as vectors in class space, whose dimension is given by the number of distinct classes, n c. Different vectors are labeled by i. The orthogonality relation then implies that there are n irr mutually orthogonal vectors living in the n c -dimensional class vector space, where n irr are the number of inequivalent irreducible representations. But, the completeness relation implies that any vector in the class space can be expressed as a linear combination of the n irr class vectors. That is, the n irr class vectors form a complete set of mutually orthogonal vectors, which span the class space. It then immediately follows that n irr = n c, since the number of vectors in a complete set of mutually orthogonal vectors is equal to the dimension of the vector space. k=. Consider the transformations of the triangle that make up the dihedral group D. The elements of this group are D = {e,r,r,d,rd,r d}, with the group multiplication law determined by the relations r = e, d = e and dr = r d, where e is the identity element. In class, the following two-dimensional representation matrices for r,d D n were given, ( ) ( ) cos(π/n) sin(π/n) r =, d =. () sin(π/n) cos(π/n) Setting n =, one can construct a two-dimensional matrix representation of D. (a) Consider the six-dimensional function space W consisting of polynomials of degree in two real variables (x,y): f(x,y) = ax +bxy +cy +dx+ey +h, ()

4 where a,b,...,h are complex constants. We can view (a,b,...,h) as a six-dimensional vector that lives in a vector space which is isomorphic to W. If we perform a transformation of (x,y) under D according to the two-dimensional representation obtained from eq () with n =, then the polynomial f(x,y) given by eq. () transforms into another polynomial. That is, the vector (a,b,...,h) transforms under D according to a six-dimensional representation. Compute the 6 6 matrices that represent the elements of D. Determine which irreducible representations of D are contained in this six-dimensional representation and their corresponding multiplicities. One can rewrite eq. () as f(x,y) = z T Az +z T B +h, where ( ) ( x a z =, A = b ) ( ) d y b c, B =. e Under a D transformation, z D(g)z, where D(g) is the corresponding matrix obtained by using the two dimensional matrix representation given in eq. () with n =. It follows that under a D transformation, f(x,y) z T A z +z T B +h, where In particular, if A = D(g) T AD(g), B = D(g) T B. ( ) d d D(g) =, d d then we find after some algebra, a d d d d a b d d d d +d d d d b c d = d d d d c d d d e d d e f f, under the action of D. Using the specific matrices given in eq. (), we obtain the following six-dimensional representation of D, e = 4 4 4, r =, 4

5 r = , d =, rd = 4 4, r d = We know that D possesses one two-dimensional and two inequivalent one-dimensional irreducible representations. To compute the number of times a given irreducible representation appears in the six-dimensional representation obtained above, we make use of the formula derived in class, n c n i = k= χ (i) (C k )χ(c k ) N k O(G), () where the sum is taken over the n c possible classes of D, N k is the number of elements in the kth class, and O(G) is the number of elements of G (the order of the group). We can read off the irreducible characters, χ (i) (C k ) from the character table of D = S obtained in class: i dimension C = {e} C = {d,rd,r d} C = {r,r } where i labels one of the three possible irreducible representations of D. For the sixdimensional matrix representation obtained above, Applying eq. (), it follows that: χ(c ) = 6, χ(c ) =, χ(c ) =. n =, n =, n =. That is, the six-dimensional representation of D obtained above contains the trivial representation twice and the two-dimensional irreducible representation twice. 5

6 (b) Identify the irreducible invariant subspaces of W under D. Check that your result is consistent with the results of part (b). The irreducible invariant subspaces are easily identified. The space spanned by ( ) clearly corresponds to the trivial representation. By inspection of the six-dimensional representation matrices given above, it follows that the space spanned by ( ) and ( ) corresponds to the irreducible two-dimensional representation given in eq. (). The final task is to decompose the remaining three-dimensional subspace into its irreducible components. Recall that ( a A = D(g) T AD(g), where A = b ) b c. Then, if we choose a = c and b =, it follows that A = A since in this case A = ai, where I is the identity matrix. Thus, it follows that ( ) spans an invariant space, which is thus one-dimensional. We can now construct two linearly independent vectors that are orthogonal to the four vectors already identified. By inspection, these vectors can be chosen to be ( ) and ( ). In light of the result of part (b), these two vectors must also span an invariant subspace under the action of D. This is indeed the case, as one can easily check by applying the six 6 6 matrices that represent the elements of D to an arbitrary linear combination of ( ) and ( ). Thus, we have explicitly identified two one-dimensional and two two-dimensional irreducible subspaces of the vector space C 6 under the action of the six-dimensional representation of D obtained above.. Consider the dihedral group D 4 treated in problem 4 of Problem Set. The elements of this group are D 4 = {e,r,r,r,d,rd,r d,r d} with the group multiplication law determined by the relations r 4 = e, d = e and dr = r d, where e is the identity element. (a) Write out the conjugacy class multiplication table. Using the results of problem 4(b) of Solution Set, the conjugacy classes of D 4 are C = {}, C = {r,r }, C = {r }, C 4 = {d,r d} and C 5 = {rd,r d}. (4) Using the multiplication table of D 4, previously obtained in the solution to part (a) of problem 4 of Solution Set, one immediately obtains the following class multiplication table. C C C C 4 C 5 C C C C C 4 C 5 C C C +C C C 5 C 4 C C C C C 4 C 5 C 4 C 4 C 5 C 4 C +C C C 5 C 5 C 4 C 5 C C +C To savespace, all vectorswill henceforth be specified by rowvectorsratherthan the usual column vectors Since D T = D for the representation given in eq. () and A = ai, we have A = D aid = ai = A. 6

7 (b) Determine explicitly the matrices of the regular representation. We rewrite the group multiplication table, previously obtained in the solution to part (a) of problem 4 of Solution Set, so that the group elements are listed in the first column and the corresponding inverses are listed in the first row. r r r d rd r d r d r r r d rd r d r d r r r r rd r d r d d r r r r r d r d d rd r r r r r d d rd r d d d rd r d r d r r r rd rd r d r d d r r r r d r d r d d rd r r r r d r d d rd r d r r r The matrix of the regular representation corresponding to the element g D 4 is then obtained from the multiplication table above by replacing every appearance of g with, and filling up the rest of the corresponding matrix with zeros. That is, =, r =, r =, r =, d =, rd =, 7

8 r d =, r d =. (c) Using the two dimensional matrix representation given in eq. () with n = 4, verify that the group multiplication table of D 4 is preserved. Prove that this representation is irreducible. For n = 4, eq. () yields: ( ) ( ) ( ) ( ) =, r =, r =, r =, ( ) ( ) ( ) ( ) d =, rd =, r d =, r d =. (5) One easily checks that the representation matrices exhibited in eq. (5) satisfy the D 4 group multiplication table. To show that eq. (5) is an irreducible representation of D 4, we must prove that there is no basis in which the above matrices are reduced to block diagonal form. If such a basis existed, then we could simultaneously diagonalize the matrices that represent r and rd. But these elements do not commute and thus are not simultaneously diagonalizable. For completeness, I now provide two other proofs that the above two-dimensional representation exhibited in eq. (5) is irreducible. (i) In class, we proved that a necessary and sufficient condition for a representation D(g) with characters χ(c k ) TrD(g) [for g C k ] to be irreducible is n irr N k χ(c k ) = O(G), (6) k where N k is the number of elements in conjugacy class C k, n irr is the number of inequivalent irreducible representations of G, and O(G) is the order of the group G. Employing eqs. (4) and(5),wecanimmediatelyenumeraten k andtheχ(c k )forthetwo-dimensionalirreducible representation of D 4, C C C C 4 C 5 N k χ(c k ) and check that eq. (6) is satisfied with O(D 4 ) = 8. 8

9 (ii) One can check explicitly that if AD(g) = D(g)A for all g D 4, then A is a multiple of the identity. For this problem, it is enough to check that for an arbitrary matrix A, if Ar = ra and Ad = da, with r and d given by the matrices listed in eq. (5), then A = c for some complex number c (where is the identity matrix). Hence, by Schur s second lemma, D(g) is an irreducible representation of D 4. (d) Construct the character table for the irreducible representations of D 4. First,weneedtospecifyalltheirreduciblerepresentationsofD 4. Wealreadyhaveidentifieda two-dimensional irreducible representation in part (c). Moreover, the trivial one-dimensional representation in which all elements of the group are represented by the matrix () is always present. We now make use of Part of Burnside s theorem, n irr n i = O(G), (7) i= where n i is the dimension of the ith inequivalent irreducible representation, and Part which states that n irr = n c, where n irr the number of inequivalent irreducible representations and n c is equal to the number of conjugacy classes. Applying Burnside s theorem to D 4, we have O(D 4 ) = 8, and it follows that D 4 must have four inequivalent one-dimensional representations. The possible one-dimensional representations can be determined by inspection. In particular, we know that d = and dr = r d. Since d and r are represented by matrices, it follows that as matrices, d and r commute. Then dr = r d implies that r =. Thus, the four one dimensional representations correspond to: r = (), d = (), r = (), d = ( ), r = ( ), d = (), r = ( ), d = ( ). (8) The character of a representation is equal to the trace of the corresponding representation matrix. Moreover, the character of all elements of a given class are equal. Thus, we can immediately write down the character table: dimension C C C C 4 C 5 The zeros in the last row of the character table are easy to understand. One can obtain equivalent two-dimensional irreducible representations by multiplying the irreducible representation exhibited in eq. (5) by either ± according to the sign of r and d given by the four possible choices given in eq. (8). Since the characters of equivalent representations are equal, the characters of the two-dimensional irreducible representation corresponding to classes C, C 4 and C 5 must be equal to their negatives, and hence must be zero. One can also check this character table by verifying that the class orthogonality relations are satisfied. By definition, all inequivalent one-dimensional representations of a group are irreducible. 9

10 4. Suppose that D is an irreducible n-dimensional representation of a finite group G, and D () is a (nontrivial) one-dimensional representation of G. Prove that the direct product D D () is an irreducible representation of G. I will provide two different proofs. Proof : Suppose that D (i) is an irreducible representation and D (j) is a one-dimensional irreducible representation. Then, no invertible matrix S exists such that ( ) A(g) B(g) SD (i) (g)s =. C(g) Considerthematrixrepresentation, D (i j) D (i) (g)d (j) (g).sinced (j) (g)isone-dimensional, we see that D (j) (g) is simply a (complex) number. Recalling that all matrix representations of a finite group are equivalent to unitary representations, it follows that D (j) (g) must be a complex number of unit modulus (since all complex numbers commute). That is, D (j) (g) =. (9) Thus, we conclude that no invertible matrix S exists such that: ( ) D SD (i j) (g)s = D (j) (g)sd (i) (g)s = (j) (g)a(g) D (j) (g)b(g) D (j) (g)c(g). It immediately follows that D (i j) (g) is irreducible. Proof : Using eq. (9), it follows that any one-dimensional representation D (j) of a finite group must satisfy: χ (j) (g) = D (j) (g) =, where χ (i) (g) TrD (i) (g) is the character of g G for the ith irreducible representation. We also showed in class that for a direct product representation, χ (i j) (g) = χ (i) (g)χ (j) (g). Hence, if D (j) is a one-dimensional representation, χ (i j) (g) = χ (i) (g). () The necessary and sufficient condition for a representation of a finite group to be irreducible is that n irr N k χ(c k ) = O(G), () k= where N k is the number of elements in class C k and O(G) is the order of the group. If the χ (i) satisfy eq. (), then the χ (i j) must also satisfy eq. () in light of eq. (). Hence D (i j) must be an irreducible representation of G.

11 5. Given a normal subgroup N of a group G, a representation D G/N of the quotient group G/N can be lifted to give a representation D G of the full group G by the following definition: D G (g) D G/N (gn). () That is, each element of the group g G is assigned the matrix D G/N of the coset gn to which it belongs. Verify that if D G/N is a representation of G/N, then D G (g) is indeed a representation of the group G. To prove that D G (g) is a representation of G, we must show that the group multiplication law is preserved. That is, D G (g )D G (g ) = D G (g g ). () This is easily verified using the definition given in eq. (). First, we recall that the group multiplication law for G/N is: (g N)(g N) = g g N. Thus, if D G/N is a representation of G/N, it follows that: Hence, eqs. () and (4) implies that: D G/N (g N)D G/N (g N) = D G/N (g g N). (4) D G (g )D G (g ) = D G/N (g N)D G/N (g N) = D G/N (g g N) = D G (g g ), which confirms eq. (). One can also check that D(e) = I and D G (g ) = [D G (G)]. For example, since D G/N is a representation of G/N and en = N is the identity of the subgroup G/N, it follows from eq. () that D G (e) = I. Likewise, using the fact that the inverse of gn is g N, it follows that D G (g ) = D G/N (g N) = [D G/N (gn)] = [D G (g)]. 6. (a)displayallthestandardyoungtableauxofthepermutationgroups 4. Fromthisresult, enumerate the inequivalent irreducible representations of S 4 and specify their dimensions. At the top of the next page, all the standard Young tableaux for S 4 are listed, where each irreducible representation corresponds to a different Young diagram. There are five possible Young diagrams involving four boxes, corresponding to five possible partitions of 4: 4, +, ++, + and +++. Hence there are five inequivalent irreducible representations of S 4. The number of standard Young tableaux corresponding to a given Young diagram is equal to the dimension of the corresponding irreducible representation. As a check, we can use eq. (7) to verify that = 4 = 4!, which is equal to the order of the group S 4 as required.

12 standard Young tableaux dimension (b) Show that the normal subgroup {e,()(4),()(4),(4)()} of S 4 is isomorphic to D. Using this result, prove that D = S4 /D. The multiplication table for {e,()(4),()(4),(4)()} is given by: e ()(4) ()(4) (4)() e e ()(4) ()(4) (4)() ()(4) ()(4) e (4)() ()(4) ()(4) ()(4) (4)() e ()(4) (4)() (4)(4) ()(4) ()(4) e We recognize this as the multiplication table of the group D = Z Z. 4 One can check that D = {e,()(4),()(4),(4)()} is a normal subgroup of S4. That is, for any element h D and g S 4, one can verify that ghg D. Next, we examine the cosets of the form gd, for g S 4. First, we have ed = D = {e, ()(4), ()(4), (4)()}. (5) 4 There are only two possible finite groups of four elements: Z 4 and D = Z Z. Note that Z 4 is a cyclic group with one generator. However, the multiplication table for {e,()(4),()(4),(4)()} is clearly not a cyclic group with one generator. By the process of elimination, the only possible conclusion is that the multiplication table corresponds to the group D = Z Z.

13 Using the multiplication table of D, it then follows that: ()D = {(), (4), (4), (4)}, (6) ()D = {(), (4), (4), (4)}, (7) ()D = {(), (4), (4), (4)}, (8) ()D = {(), (4), (4), (4)}, (9) ()D = {(), (4), (4), (4)}. () All other cosets coincide with one of the six specified above. Since D is a normal subgroup of S 4, it follows that S 4 /D is a group of six elements. There are only two possible groups of six elements: Z 6 which is an abelian cyclic group and D = S which is non-abelian. It is straightforward to check that the group multiplication law of the cosets given above is non-abelian. Since the multiplication rule of the cosets is given by (g D )(g D ) = g g D, it follows that [()D ][()D ] = ()()D = ()D, () [()D ][()D ] = ()()D = ()D, () so that [()D ][()D ] [()D ][()D ]. Hence, S 4 /D is non-abelian, and we conclude that D = S4 /D One can also verify directly that D = S4 /D by checking that there exists a one-to-one correspondence of the multiplication table of the cosets, S 4 /D = {ed, ()D, ()D, ()D, ()D, ()D }, and the multiplication table of D = S = {e,(),(),(),(),()}. Indeed, one can identify the elements of D with those of S 4 /D as follows. Recall that D can be formally defined as: D = {d k r l d = e, r = e, dr = r d}. Thus, we can identify d = ()D and r = ()D. In particular, ()() = e, ()()() = e, ()() = ()()(), where we have chosen the simplest representative elements from the cosets ()D and ()D. It then follows that r = ()D, rd = ()D, r d = ()D, and of course e = ed. This completes the explicit identification of the elements of D with the cosets gd.

14 (c) Using the two-dimensional irreducible representation of D given in class and the result of problem 5, construct a two-dimensional representation of S 4 and determine its characters. Is the latter an irreducible representation of S 4? We can now make use of the two-dimensional representation of D given by eq. () with n =, ( ) ( ) e =, r =, r = ( ) ( d =, rd = ), r d = ( ) ( ) Note that r corresponds to an active counterclockwise rotation by π/ radians about an axis perpendicular to the rotation plane. Using the results of problem 5, we can lift this representation to S 4. The resulting representation is unfaithful, since four elements of S 4 are mapped into the same matrix. In particular, e, ()(4), ()(4), (4)() (), (4), (4), (4) (), (4), (4), (4)) (), (4), (4), (4) (), (4), (4), (4) (), (4), (4), (4) ( ), () ( ), (4) ( ) ( ) ( ) ( ), (5), (6), (7). (8) The characters are obtained by taking the trace of the corresponding representation matrices. Since elements of the same conjugacy class possess the same characters, we first determine the conjugacy classes of S 4. Recall that all elements of S n of a given conjugacy class possess the same cycle structure (which are in one-to-one correspondence with the possible Young 4,.

15 diagrams). Hence, C = {e}, C = {(), (), (), (4), (4), (4)}, C = {(), (), (4), (4), (4), (4), (4), (4)}, C 4 = {()(4), ()(4), (4)()}, C 5 = {(4), (4), (4), (4), (4), (4)} Comparing the elements of the conjugacy classes with the cosets given in eqs. (5) (), it follows that: elements of ed belong to classes C and C 4, elements of ()D belong to class C, elements of ()D belong to classes C, elements of ()D belong to classes C and C 5, elements of ()D belong to classes C and C 5, elements of ()D belong to classes C and C 5. Thus, using the two-dimensional representation of S 4 obtained in eqs. () (8), it follows that the corresponding characters are given by: χ(c ) = χ(c 4 ) =, χ(c ) =, χ(c ) = χ(c 5 ) =. (9) We can check whether the two-dimensional representation of S 4 obtained in eqs. () (8) is irreducible by employing the theorem that states that a representation is irreducible if and only if n irr N k χ(c k ) = O(G), k= where N k is the number of elements in class C k and O(G) is the order of the group. For S 4, we have O(G) = 4! = 4 and N =, N = 6, N = 8, N 4 =, N 5 = 6. The representation given by the matrices in eqs. () (8) is irreducible if: ( ) + +6? = 4. Indeed, 4+8+ = 4, so that we conclude that the irreducible (simple) characters corresponding to the two-dimensional irreducible representation of S 4 are given by eq. (9). (d) Using the known one-dimensional representations of S 4 and the results of parts (a) and (c), construct the character table for the group S 4. Determine any unknown entries in the character table by using the orthonormality and completeness relations for the irreducible characters. Using this technique, all entries of the character table can be uniquely determined up to a sign ambiguity in some of the entries. 5

16 Using all known information (up to this point), the character table for S 4 is given by: irreps C C C C 4 C 5 a b c d e f g h Note that the dimension d of the irreducible representation (irrep) is equal to the character of the identity element, which corresponds to conjugacy class C, since the matrix representation of the identity is the d d identity matrix. Moreover, the totally symmetric one-dimensional representation corresponds to representing each permutation by, and the totally antisymmetric one-dimensional representation corresponds to representing each permutation by ( ) p, which is + for even permutations (i.e., classes C, C and C 4 ), and for odd permutations (i.e., classes C and C 5 ). This immediately yields the first two lines of the character table above. The third line of the character table was obtained in eq. (9). The numbers a, b,...,h represent the presently unknown entries of the character table, which we shall determine by employing the orthogonality and completeness relations for the simple characters, n c k= N k χ (i) (C k )χ (j) (C k ) = O(G)δ ij, n irr χ (i) (C k )χ (i) (C l ) = O(G) δ kl, N k i= where k =,,...,n c labels the conjugacy classes, N k is equal to the number of group elements in conjugacy class C k, i =,,...n irr labels the irreps, and O(G) is equal to the number of elements in the group G. According to a theorem proved in class, the simple characters of S n are all real. Thus, we can ignore the complex conjugation in the orthogonality and completeness relations. First, we make use of the orthogonality of row 4 with rows, and, respectively. Using N =, N = 6, N = 8, N 4 = and N 5 = 6, we obtain three relations, +6a+8b+c+6d =, 6a+8b+c 6d =, 6 8b+6c =. Solving these three equations yields d = a, c = and b =. Next, we make use of the orthogonality of row 5 with rows, and, respectively. By an identical analysis, we obtain h = e, g = and f =. Next, we make use of the orthogonality of rows 4 and 5 to obtain 9+6ae+8bf +cg +6dh =. 6

17 Using the relations previously obtained, it follows that ae =. Finally, we make use of the completeness relation for columns and, which yields: 5 ++a+e =, which implies that a = e. Combining this with ae = yields a = e =. Hence, a = e = ±, where the sign ambiguity is not yet resolved. 6 Thus, the character table of S 4 is given by: irreps C C C C 4 C 5 ± ± The sign ambiguity above will be resolved in part (e) of this problem. (e) Resolve the sign ambiguity of part (d). One possible approach is to construct the matrix representative of the transposition ( ) corresponding to the three-dimensional irreducible representation of S 4. By taking the trace of this matrix, complete the character table of S 4. Using the method discussed in class, I shall determine an explicit irreducible three-dimensional matrix representation for the element ( ) C. The Young element corresponding to is given by 4 (4) Y = [ e+( ) ][ e+( 4)+( 4 ) ( ) ( 4) ( 4) ] = e+( 4)+( 4 ) ( ) ( 4) ( 4) +( )+( 4 )+( 4 ) ( ) ( 4 ) ( )( 4). (4) Since eq. (4) corresponds to a three-dimensional irreducible representation of S 4, we need to find two additional elements of the group algebra of S 4 that span the irreducible threedimensional subspace of the regular representation. Since ( ) and ( 4) do not appear in 5 We could also make use of the orthogonality relation for row 4 (or 5) by itself. This yields a = e =, which when combined with ae = yields a = e = ±. 6 Note that the completeness relation for column (or 5) alone yields, +++a +e = 4 6. Combining this with a = e, it again follows that a = e =. 7

18 eq. (4), two other possible elements of the group algebra that span the irreducible threedimensional subspace of the regular representation are: and ( )Y = ( )+( 4)+( 4 ) ( ) ( 4)( ) ( 4) +( )+( )( 4)+( 4 ) ( ) ( 4 ) ( 4 ), (4) ( 4)Y = ( 4)+( 4)+( 4 ) ( 4) ( )( 4) ( 4 ) +(4 )+( )( 4)+( ) ( ) ( 4 ) ( 4 ). (4) Thus, we choose the basis, { Y, ( )Y, ( 4)Y }, for the irreducible three-dimensional subspace of the regular representation corresponding to eq. (4). We represent these basis vectors by Y =, ( )Y =, ( 4)Y =. With respect to this basis choice, any element g S 4 can be represented by the matrix D[g] = gy g( )Y g( 4)Y, (44) where the three respective columns of the matrix representation of g are indicated above. We now compute the three-dimensional matrix representation of g = ( ). Using all we need to compute is: ( )( ) = ( ), ( )( 4) = ( 4), ( )Y = ( )+( 4 )+( 4 ) ( ) ( 4 ) ( )( 4) = Y, +e+( 4)+( 4 ) ( ) ( 4) ( 4) ( )( )Y = ( )Y = ( )+( 4)+( 4)( ) ( ) ( 4 ) ( 4) +( )+( 4)+( 4) e ( 4 ) ( 4) = Y ( )Y, ( )( 4)Y = ( 4)Y = ( 4)+( 4 )+( )( 4) ( 4) ( 4) ( 4 ) +( 4)+( 4)+( ) e ( 4) ( 4 ) = Y ( 4)Y. That is, ( )Y =, ( )( )Y =, ( )( 4)Y =. 8

19 Thus, eq. (44) yields: D[( )] =. The trace of D[( )] yields the character, χ(c ) = Tr D[( )] =, for the threedimensional irreducible representation corresponding to the Young diagram, That is, in the notation of part (e), we have e = and therefore a =. The character table for S 4 is now complete: irreps C C C C 4 C 5 7. (a) Derive the following properties of the Pauli matrices σ (σ, σ, σ ): (i) σ i σ j = Iδ ij +iǫ ijk σ k, (ii) σ σσ = σ, (iii) exp( iθˆn σ/) = I cos(θ/) iˆn σ sin(θ/), where I is the identity matrix. By direct calculation, σ = σ = σ = I, where I ( ). Moreover, σ σ = iσ, σ σ = iσ, σ σ = iσ. These results are summarized by one equation, σ i σ j = Iδ ij +iǫ ijk σ k, (45) where there is an implicit sum over the repeated index k. Next, it is a simple exercise of matrix multiplication to show that σ σσ = σ by verifying that σ σ i σ = σ i for i =,,. 9

20 Finally, we compute k= e iθˆn σ/ = k= k! ( ) k iθˆn σ. Using eq. (45), it follows that (ˆn σ) = ˆn iˆn j σ i σ j = I, since ˆn is a unit vector. Thus, for any positive integer k, (ˆn σ) k = I and (ˆn σ) k+ = ˆn σ. Hence, ( ) k e iθˆn σ/ iθ ( ) k+ iθ = I ˆn σ = (k)! (k +)! = Icos(θ/) iˆn σsin(θ/). k= (b) In the angle-and-axis parameterization of SO(), a rotation by an angle θ about an axis that points along the unit vector ˆn is represented by an SO() matrix given by R ij (ˆn,θ) = exp( iθˆn J ) ij, with (ˆn J ) ij iǫ ijk n k. By convention, we assume that θ π, and the axis ˆn can point in any direction. Evaluate R ij explicitly and show that R ij (ˆn,θ) = n i n j +(δ ij n i n j )cosθ ǫ ijk n k sinθ. (46) To evaluate the matrix R(ˆn,θ), we compute R(ˆn,θ) = e iθˆn J = where (ˆn J ) ij iǫ ijk n k. Note that k= k! ( iθˆn J) k, (47) (ˆn J ) ij = ǫ ilkn k ǫ ljm n m = (δ ij δ km δ im δ jk )n k n m = δ ij n i n j, after employing the well-known ǫ-tensor identity and noting that δ km n k n m = for the unit vector ˆn. Next, we compute: Thus, for any positive integers k, (ˆn J ) ij = i(δ il n i n l )ǫ ljk n k = iǫ ijk n k = (ˆn J) ij. (ˆn J ) k = ˆn J, (ˆn J ) k = (ˆn J). Inserting these results in eq. (47), we obtain: R ij (ˆn,θ) = δ ij (ˆn J) ij k= (k +)! (iθ)k+ +(ˆn J) ij = δ ij i(ˆn J) ij sinθ +(ˆn J) ij (cosθ ) = δ ij ǫ ijk n k sinθ +(δ ij n i n j )(cosθ ) = n i n j +(δ ij n i n j )cosθ ǫ ijk n k sinθ. k= (k)! (iθ)k

21 (c) Verify the formula: e iθˆn σ/ σ j e iθˆn σ/ = R ij (ˆn,θ)σ i. To evaluate e iθˆn σ/ σ j e iθˆn σ/, we first note that this quantity is a traceless hermitian matrix. The traceless condition follows from: Tr(e iθˆn σ/ σ j e iθˆn σ/ ) = Trσ j =, after cyclically permuting the terms inside the parenthesis. Hermiticity is also easily demonstrated given that σ j = σ j. Since an arbitrary traceless hermitian matrix can always be written as a real linear combination of σ, σ and σ, it follows that e iθˆn σ/ σ j e iθˆn σ/ = R ij (ˆn,θ)σ i, (48) for some real coefficients R ij. To determine these coefficients, we multiply eq. (48) by σ k and take the trace of the resulting equation. Using Tr(σ i σ j ) = δ ij [which immediately follows after taking the trace of eq. (45)], it follows that: R ij = Tr( e iθˆn σ/ σ j e iθˆn σ σ i ). (49) This is easily evaluated using the results of part (a). R ij = Tr { [cos(θ/) iˆn σsin(θ/)]σ j [cos(θ/) iˆn σsin(θ/)]σ i } = cos (θ/)trσ i σ j + sin (θ/)tr(ˆn σσ iˆn σσ j ) + icos(θ/)sin(θ/)tr(σ jˆn σσ i ˆn σσ j σ i ). (5) We proceed to work out the three traces. The first trace has already been obtained, Tr(σ i σ j ) = δ ij. The second trace is given by: Tr(ˆn σσ iˆn σσ j ) = n k n l Tr(σ k σ i σ l σ j ) = n k n l Tr( [ (Iδ ki +iǫ kim σ m )(Iδ lj +iǫ ljn σ n ) ] = n k n l δ ki δ lj n k n l ǫ kim ǫ ljn δ mn = n i n j n k n l (δ kl δ ij δ kj δ il ) = 4n i n j δ ij, after using TrI =, Trσ i =, and Tr(σ i σ j ) = δ ij. Finally, the third trace is given by: Tr(σ jˆn σσ i ˆn σσ j σ i ) = Tr [ (ˆn σ(σ i σ j σ j σ i ) ] = iǫ ijk Tr [ˆn σσ k ] = 4iǫijk n k, after making use ofthe commutation relations, [σ i, σ j ] = iǫ ijk σ k ; thelatter isaconsequence of eq. (45). Inserting these traces back into eq. (5) yields, R ij = cos (θ/)+sin (θ/)(n i n j δ ij ) sin(θ/)cos(θ/)ǫ ijk n k.

22 Finally, using the well-known trigonometric identities, sinθ = sin(θ/)cos(θ/), cos (θ/) = (+cosθ), sin (θ/) = ( cosθ), we arrive at: R ij = δ ij +(δ ij n i n j )cosθ ǫ ijk n k sinθ, which are the matrix elements of the rotation matrix obtained in part (b). (d) The set of matrices exp( iθˆn σ/) constitutes the defining representation of SU(). Prove that this representation is pseudoreal. To prove that D(θ) exp( iθˆn σ/) = cos(θ/) iθˆn σ sin(θ/) is a pseudoreal representation of SU(), we must first prove that D(θ) and D (θ) are equivalent representations. Noting that D (θ) = exp(iθˆn σ /) = cos(θ/)+iθˆn σ sin(θ/), it is straightforward to check that σ D(θ)σ = D (θ). (5) This is a consequence of properties (i) and (ii) of part (a). Since σ = σ, we can rewrite eq. (5) as D (θ) = S D(θ)S, where S = σ. Consequently, D(θ) and D (θ) are equivalent representations. To show that D(θ) is pseudoreal, we can employ the theorem (proved in class) that states that if D = ADA where A A = I then D is pseudoreal. In the present case, A = σ and σ σ = I, which confirms that D(θ) is pseudoreal. ADDED NOTE: A direct proof that D(θ) is pseudoreal. If D(θ) is pseudoreal, then no basis exists in which D(θ) = cos(θ/) iθˆn σsin(θ/) is real. That is, no invertible matrix S exists such that S D(θ)S is real. This requirement is equivalent to the condition that no invertible matrix S exists such that is σs is real. That is, no basis exists in which the iσ k are simultaneously real. Without loss of generality, one can assume that dets = by rescaling the elements of S (since the overall determinant factor cancels in is σs). One way to prove that no such matrix S exists is by writing ( ) ( ) a b d b S =, S =, c d c a where ad bc = and then computing is σ k S for k =,,. Assuming that all the matrix elements of is σ k S are real, one quickly reaches a contradiction. Admittedly, this is not a very elegant proof. If S is unitary, then I can assume that its determinant is equal to one without loss of generality (by rescaling the elements of S as above). Then S is an SU() matrix which can be expressed as S = exp(iθˆn σ/). In this case, we can use the results of part (c) to obtain is σ k S = ie iθˆn σ/ σ k e iθˆn σ/ = ir jk σ j, (5)

23 which is simultaneously real for k =,, if R k = R k =. But this is impossible since the R kj arematrix elements of an orthogonal matrix which cannot possess a row of zeros. Taking S to be unitary means that S D(θ)S is unitary, so this last argument establishes that no real representation exists that is unitarily equivalent to D(θ). However, this last argument does not yield the stronger result that no real representation exists (whether unitary or not) that is equivalent to D(θ). A direct proof of the latter result, which is responsible for the theorem cited in the paragraph following eq. (5), is given below. Suppose a nonsingular matrix S exists such that A S iσ k S, where A is a real matrix. (5) Then taking the complex conjugate of this equation (using A = A ) yields which can be rewritten as S iσ k S = S iσ ks, iσ k = (SS ) iσ k SS. Using property (ii) of part (a) of this problem, σ k = σ σ k σ, then yields: SS σ iσ k = iσ k SS σ. (54) Schur s second lemma states that if ZD(θ) = D(θ)Z for all θ, then Z is a multiple of the identity. Using D(θ) = cos(θ/) iθˆn σ sin(θ/), Schur s second lemma then implies that if Ziσ k = iσ k Z for k =,,, then Z is a multiple of the identity. Using eq. (54) with Z = SS σ and noting that (σ ) = I, it follows that SS = kσ, where k is a non-zero complex number. (55) Note that k = is not allowed since S and σ are invertible matrices. Taking the complex conjugate of eq. (55) and multiplying the result by by eq. (55) yields which simplifies to SS S S = k σ σ, I = k I. No complex k exists that satisfies this equation, so we have a contradiction. Therefore, our initial assumption that the matrix A defined in eq. (5) is real must be false. Thus, there exists no nonsingular matrix S such that S iσ k S is real for k =,,. This completes the proof that no basis exists in which the iσ k are simultaneously real.