Symmetrized squares and cubes of the fundamental unirreps of Sp(2n, R)

Transcription

1 J. Phys. A: Math. Gen. 31 (1998) Printed in the UK PII: S (98) Symmetrized squares cubes of the fundamental unirreps of Sp(n, R) Jean-Yves Thibon, Frédéric Toumazet Brian G Wybourne Institut Gaspard Monge, Université de Marne-la-Vallée, rue de la Butte-Verte, Noisyle-Gr cedex, France Instytut Fizyki, Uniwersitet Miko laja Kopernika, ul. Grudzi adzka 5/7, Toruń, Pol Received 16 July 1997, in final form 10 November 1997 Abstract. We establish the complete decompositions of the symmetrized Kronecker squares of the basic harmonic representations of Sp(n, R) conjectured by Grudzinski Wybourne (Grudzinski K Wybourne B G 1996 J. Phys. A: Math. Gen ) give analogous complete results for the symmetrized cubes. Equivalence relationships are shown to exist between certain pairs of plethysms. 1. Introduction The non-compact symplectic group Sp(n, R) is well known as the dynamical group of the isotropic n-dimensional harmonic oscillator [1 finds significant applications in symplectic models of nuclei [1, 9 in the mesoscopic properties of quantum dots [3, 4, 13. A central problem in making applications is the resolution of Kronecker powers of the two fundamental irreducible unitary representations (unirreps), which, following King Wybourne [6, we shall designate as 1 (0) 1 (1), into their various symmetry types. This problem has recently been studied by Grudzinski Wybourne [ who, on the basis of numerical calculations, conjectured the explicit decompositions 1 (0) {} = 1 (4i) (1) 1 (0) {1 } = 1 ( + 4i) () 1 (1) {} = 1 ( + 4i) (3) 1 (1) {1 } = 1 (1 ) + i 1 1 (4i). (4) It was shown that the validity of such conjectures implies some unusual S-function identities which they duly established, but without establishing a rigorous proof of the conjectured equivalences. Herein we supply a formal proof of equations (1) to (4). Moreover, the methods introduced lead to the establishment of a complete resolution of the Kronecker cubes of the fundamental irreps of Sp(n, R). These results are then used to demonstrate the existence of a symmetry relationship linking plethysms of the 1 (0) unirrep to those of 1 (1) /98/ $19.50 c 1998 IOP Publishing Ltd 1073

2 1074 J-Y Thibon et al. Background notation Our notation is as in [10, 6 for characters, as in [11 for symmetric functions. The Sp(n, R) U(n) branching rule for the harmonic series representations 1 k(λ) is given [10, equation (3.9), [6, equation (5.6) by the equation: 1 k(λ) ǫ 1 k {{λ s } k N D N} N (5) where N = min(n, k), {λ s } k N is the signed sequence of terms ±{ρ} such that ±[ρ is equivalent to [λ under the modification rules of O(k) ( see [10). For example, the terms 1(m) arising in the right-h sides of (1) (4) have U(n) content for n : while 1(m) ǫ {({m} {m, }) D } (6) 1(1) ǫ {{1}D } (7) 1(1 ) ǫ { {1 }D }. (8) We will also need equation (8.18) of [6, giving the tensor product of two harmonic series representations of Sp(n, R) as 1 k(µ) 1 l(λ) = 1 (k + l)( ({µ s } k {λ s } l D) ) (9) k+l,n where ((λ)) k+l,n is defined to be zero unless λ 1 n λ 1 + λ k + l in which case (λ) is retained. For example 1 (0) 1 (0) = 1(i) (10) 1 (0) 1 (1) = 1(i + 1) (11) 1 (1) 1 (1) = 1(1 ) + i 1 1(i). (1) by Finally, the branching rule Sp(nk, R) Sp(n, R) O(k) [6, equation (4.3) is given λ 1 k(λ) [λ (13) where λ runs over partitions satisfying λ 1 +λ k λ 1 n. The harmonic representation nk of Sp(nk, R) has the following restrictions under the chain Sp(nk, R) Sp(n, R) O(k) Sp(n, R) 1: nk λ 1 k(λ) [λ k n. As pointed out by the second referee, the Kronecker products can also easily be obtained by means of theorem 3.1 of [7.

3 Fundamental unirreps of Sp(n, R) Decomposition of symmetrized squares In this section we give a proof of equations (1) (4). Let us first look at the U(n) content of the left-h-side of (1) (3). We know that under Sp(n, R) U(n) [ (ǫ 1 (0) {} ) [ 1 {m} {} = ǫh h m (ǫ 1 (1) {} ) [ 1 {m + 1} {} = ǫh h m+1 where for later convenience the right-h side has been rewritten in terms of symmetric functions following the notation of [8. Hence, the sum of the left-h sides correspond to the terms of even degree in the series [ h [σ 1 = h h m + h m+1 = h [ ( h m + ) ( h m ) [ h m+1 + h h m+1 which (up to the factor ǫ), describes the U(n) content of {}. Therefore, to compute the sum it is sufficient to evaluate 1 (0) {} + 1 (1) {} {} = 1 (0) {} + 1 (1) {} + 1 (0) 1 (1). (14) We can now resolve n into its symmetric antisymmetric part by regarding it as the restriction of 4n under Sp(4n, R) Sp(n, R) introducing the chain of groups Sp(4n, R) Sp(n, R) O() Sp(n, R) S. Indeed, under Sp(4n, R) Sp(n, R) S, we have ( ( ) 4n n {}) () + n {1 } (1 ). Under Sp(4n, R) Sp(n, R) O(), the branching rule (13) gives n 1(1 ) [1 + 1(i) [i. The restriction O(k) S k is given [5, 11 by [λ 1 {λ/g} (15) where G = {0} + {1} {1} involves only self-conjugate partitions, so that [1 1 ( {1 } + {1} ) = 1, for k =, [1 ( 1 ) S = (1 ) reduces to the alternating representation of S. For i 1 [i 1 ({i} + {i 1})

4 1076 J-Y Thibon et al where, for S, 1 {j} reduces to () for j even to (1 ) for j odd. Therefore, the restriction n Sp(n, R) S is equal to so that 1(1 ) (1 ) + 1(0) () + 1(i) [ () + (1 ) i 1 {} = 1(i) (16) {1 } = 1(1 ) + i 1 1(i). (17) If we consider the restriction of (14) to U(n), 1 (0) {} yields symmetric functions of degree d 0 mod[4, 1 (1) {} gives terms of degree d mod[4 1 (0) 1 (1) the terms of odd degree. From equations (16) (11) we have 1 (0) {} + 1 (1) {} = 1(i). (18) Comparing equations (18) (10), we obtain 1 (0) {1 } = 1 (1) {} (19) (since 1 (0) = 1 (0) {} + 1 (0) {1 }). In conclusion, consider the U(n) content of the difference 1 (0) {} 1 (0) {1 } = 1 (0) p (0) where p is the second power sum. Under restriction to U(n), the right-h side is given in terms of symmetric functions p [ǫ 1 h n = ǫ p [h n. n 0 n 0 This expression can be exped in the basis of Schur functions n 0 p [h n = n 0 n k=0 ( 1) k s 4n k,k. (1) Under restriction to U(n), the left-h-side of (0) corresponds to the even terms in the series obtained by taking the difference of the two expressions ( ( ( ) 1(4i) ǫ h 4i )σ 1 [h ǫ s 4i, )σ 1 [h i 1 d 0[4 ( ( ( ) 1(4i) ǫ h 4i )σ 1 [h ǫ s 4i+, )σ 1 [h We can extract these terms by taking the real part of [ σ 1 [h n 0(it) n h n (it) n s n, n d [4.

5 Fundamental unirreps of Sp(n, R) 1077 which, for t = 1, is equal to ( σ 1 [h [σ i σ i s 1 ) i s i s 1 + s 1. () We want to show that in two variables {x, y}, the symmetric functions coincide with the right-h side of (1). Indeed [ p h n = p [ σ1 + σ 1 n 0 = p [ 1 (1 x)(1 y) + 1 (1 + x)(1 + y) = 1 [ 1 (1 x )(1 y ) + 1 (1 + x )(1 + y ), using the specialization of σ 1 [ h (x, y) = σ 1 [ x + xy + y = σ i (x, y) = 1 (1 ix)(1 iy) so on, we see that the specialization S of () is equal to S = 1 [ 1 (1 x )(1 y ) + 1 (1 + x )(1 + y ) y + x + x 3 y + x y 3 + i (1 + x )(1 + y )(1 x )(1 xy)(1 y ) we now have the result that 1 (1 x )(1 xy)(1 y ) 1 (0) {} 1 (0) {1 } = 1(4i) + ). (3) 1(4i From equations (3) (10) we obtain (1) (), taking into account equations (19) (1), we obtain (3) (4). 4. Symmetrized cubes We shall now establish similar decompositions for the symmetrized cubes of the fundamental representations 1 (0) 1 (1). As above, we obtain 3 n by restriction of 6n through the chain Sp(6n, R) Sp(n, R) O(3) Sp(n, R) S 3 Sp(n, R). To obtain the plethysm n {µ}, we need to determine the restriction from O(3) to S 3. We first remark that in the restriction O(3) S 3 only the partitions [1 3, [m [m, 1 are retained. Since [m, 1 = [m [m S 3 = sgn [m S 3 (see [10), it is enough to determine [m S 3 (clearly, [1 3 sgn). Applying (15), we find [m 1 {m/g} = 1 {m} + 1 {m 1}.

6 1078 J-Y Thibon et al The value of 1 {m} for S k is equal to the coefficient of z m in the series ( ) X n 1 (1 z)h n = 1 z 1 z K j λ,1 n(z)s λ (X) j= (cf [11) where K λ,1 k(z) are the Kostka Foulkes polynomials. In particular, for n = 3, [m S 3 is the sum of the coefficients of z m z m 1 in 1 [ F(z) = z 3 (1 3 ) + (z + z)(1) + (3). (4) (1 z )(1 z 3 ) Writing we obtain 1 (1 z )(1 z 3 ) = n 0 ( F(z) = n 0 ( a n z )(3) n + after simplification, one finds that λ n a n z n (5) n 1 ( a n z n + a n 1 z )(1) n + [m S 3 = (a m + a m 1 )(3) + (a m 1 + a m + a m 3 )(1) + (a m 3 + a m 4 )(1 3 ) [m, 1 S 3 = (a m 3 + a m 4 )(3) + (a m 1 + a m + a m 3 )(1) + (a m + a m 1 )(1 3 ) Hence [1 3 S 3 = sgn = (1 3 ). 3n Sp(n) S 3 = 3 (m) E 1(m) + m 1 3 (m, 1) E (m) + 3 (13 ) (1 3 ) where E 1 (m) = (a m + a m 1 )(3) + (a m 1 + a m + a m 3 )(1) + (a m 3 + a m 4 )(1 3 ) E (m) = (a m 3 + a m 4 )(3) + (a m 1 + a m + a m 3 )(1) + (a m + a m 1 )(1 3 ). On the other h, since = 1 (0) + 1 (1), {3} = 1 (0) {3} + 1 (1) {3} + 1 (0) 1 (1) {} n 3 a n 3 z n )(1 3 ) + 1 (1) 1 (0) {} (6) {1} = 1 (1) {1} + 1 (0) {1} + 1 (0) 1 (1) {} + 1 (1) 1 (0) {} + 1 (0) 1 (1) {1 } + 1 (1) 1 (0) {1 } (7) {1 3 } = 1 (1) {13 } + 1 (0) {13 } + 1 (0) 1 (1) {1 } + 1 (1) 1 (0) {1 }. (8)

7 Fundamental unirreps of Sp(n, R) 1079 To extract the plethysms 1 (i) {λ} from these expressions, we rewrite the left-h-sides as (6), (7) (8), we manage to subtract the contribution of the crossed products 1 (i) 1 (i 1) {µ}, ( µ = ). From equations (1) (4), we obtain 1 (1) 1 (0) {} = 1 (1) 1(4i) (9) 1 (1) 1 (0) {1 } = 1 (1) 1(4i + ) (30) 1 (0) 1 (1) {} = 1 (0) 1(4i + ) (31) ( 1 (0) 1 (1) {1 } = 1 (0) 1(1 ) + ). (3) i 1 1(4i) From equation (9) we find for m 0 1 (1) 1(0) = 3 (i + 1) (33) 1 (1) 1(m) = 3 (m + i, 1) + 3 (m + i + 1). (34) Let p (m) = m+1, the number of partitions of m into parts not greater that, given by the generating function p (m)z m 1 = (1 z)(1 z ). Exping the right-h side of (9) by means of (33) (34), we get 1 (1) 1 (0) {} = 3 (i + 1) + 3 (4i + j, 1) + 3 (4i + j + 1) i 1 j 0 = 3 (i + 1) + p (m ) ( 3 (m, 1) + 3 (m + 1) ) m 1 where 4i + j = m p (m ) comes from the rearrangement 1 = 1 = p (m ). i 1, j 0 i+j=m k 0, j 0 k+j=m For equation (31) we have a similar expression. For m 0 1 (0) 1(m) = j 0 3 (m + j + 1, 1) + 3 (m + j) then 1 (0) 1 (1) {} = 3 (4i + j + 1, 1) + 3 (4i + j) j 1 = m 1 p (m 1) ( 3 (m + 1, 1) + 3 (m) ).

8 1080 J-Y Thibon et al Now it is possible to give a closed expression for 1 (0) {3} + 1 (1) {3} = {3} 1 (0) 1 (1) {} 1 (1) 1 (0) {} i.e. 1 (0) {3} + 1 (1) {3} = (a m + a m 1 ) 3 (m) + m 3 + a m 4 ) m 3(a 3 (m, 1) 3 (i + 1) p (k ) ( 3 (k, 1) + 3 (k + 1) ) k 1 k 1 p (k 1) ( 3 (k + 1, 1) + 3 (k) ). (35) We can now give the coefficients of 3 (m) 3 (m, 1) in (35). The coefficient of 3 (m) is equal to a m + a m 1 p (k 1) = a m + a m 1 a k 1 a k a k 3 for m = 4k a m + a m 1 p (k) = a m + a m 1 1 a k 1 a k a k 3 for m = 4k + a m + a m 1 1 p (k )) = a m + a m 1 1 a k a k 3 a k 4 for m = k + 1 the coefficient of 3 (m, 1) is equal to a m 3 + a m 4 p (k ) = a m 3 + a m 4 a k a k 3 a k 4 for m = k a m 3 + a m 4 p (k 1)) = a m 3 + a m 4 a k 1 a k a k 3 for m = k + 1. To separate the components of 1 (0) {3} 1 (1) {3}, we consider their U(n) contents. It is enough to see that the terms of 1 (0) {3} are those 3 (λ) with λ even the terms of 1 (1) {3} are those with λ odd: 1 (0) {3} = k 0(a 4k + a 4k 1 1 a k 1 a k a k 3 ) 3 (4k) + 4k+ + a 4k+1 1 a k 1 a k a k 3 ) k 0(a 3 (4k + ) + k + a k 3 a k 1 a k a k 3 ) k 0(a 3 (k + 1, 1) (36) 1 (1) {3} = k+1 + a k 1 a k a k 3 a k 4 ) k 0(a 3 (k + 1) + k 3 + a k 4 a k a k 3 a k 4 ) k 0(a 3 (k, 1). (37) The derivation of the other plethysms is similar. The right-h-side of (3) becomes 1 (0) 1 (1) {1 } = 3 (i + 1, 1) + i 1 3 (4i) + 3 (4i + j 1, 1) + 3 (4i + j) i 1 j 1

9 Fundamental unirreps of Sp(n, R) 1081 = 3 (i + 1, 1) + i 1 3 (4i) + p (m 3) ( 3 (m 1, 1) + 3 (m) ) m 3 for equation (30) we obtain 1 (1) 1 (0) {1 } = 3 (4i + j + 1) + 3 (4i + j, 1) j 1 = m 1 p (m 1) ( 3 (m, 1) + 3 (m + 1) ). Subtracting the contributions of the crossed products in (8) we obtain 1 (1) {13 } + 1 (0) {13 } = {1 3 } 1 (0) 1 (1) {1 } this gives 1 (1) 1 (0) {1 } 1 (0) {13 } + 1 (1) {13 } = (a m 3 + a m 4 ) 3 (m) + m + a m 1 ) m 4(a 3 (m, 1) + 3 (13 ) 3 (i + 1, 1) i 1 3 (4i) p (m 3) ( 3 (m 1, 1) + 3 (m) ) m 3 m 1 p (m 1) ( 3 (m, 1) + 3 (m + 1) ). We can now write down the coefficients of 3 (m) 3 (m, 1). coefficient of 3 (m) : a m 3 + a m 4 1 p (k 3) We first give the = a m 3 + a m 4 1 a k 3 a k 4 a k 5 for m = 4k a m 3 + a m 4 p (k ) = a m 3 + a m 4 a k a k 3 a k 4 for m = 4k + a m 3 + a m 4 p (k 1) = a m 3 + a m 4 a k 1 a k a k 3 for m = k + 1 then the coefficient of 3 (m, 1) : a m + a m 1 p (k 1) = a m + a m 1 a k 1 a k a k 3 for m = k a m + a m 1 1 p (k ) = a m + a m 1 1 a k a k 3 a k 4 for m = k + 1. Separating the terms as above, we obtain 1 (0) {13 } = k 0(a 4k 3 + a 4k 4 1 a k 3 a k 4 a k 5 ) 3 (4k) + 4k 1 + a 4k a k a k 3 a k 4 ) k 0(a 3 (4k + ) + k+1 + a k 1 a k a k 3 a k 4 ) k 0(a 3 (k + 1, 1) (38)

10 108 J-Y Thibon et al 1 (1) {13 } = k + a k 3 a k 1 a k a k 3 ) k 0(a 3 (k + 1) Finally + k 1(a k + a k 1 a k 1 a k a k 3 ) 3 (k, 1) + 3 (13 ). (39) 1 (1) {1} + 1 (0) {1} which yields = {1} 1 (0) 1 (1) {} 1 (1) 1 (0) {} 1 (0) 1 (1) {1 } 1 (1) 1 (0) {1 } 1 (0) {1} + 1 (1) {1} = 3 (m) (a m 1 + a m + a m 3 ) + 3 (m, 1) (a m 1 + a m + a m 3 ) m 1 3 (i + 1) p (k ) ( 3 (k, 1) + 3 (k + 1) ) p (k 1) ( 3 (4i) ) k 1 k 1 p (m 3) ( 3 (m 1, 1) + 3 (m) ) m 3 p (m 1) ( 3 (m, 1) + 3 (m + 1) ) m 1 hence the coefficient of 3 (m) is a m 1 + a m + a m 3 1 p (k 1) p (k 3) = a m 1 + a m + a m 3 1 a k 1 a k a k 3 a k 4 a k 5 for m = 4k (40) a m 1 + a m + a m 3 p (k) p (k ) = a m 1 + a m + a m 3 a k 1 a k a k 3 a k 4 for m = 4k + (41) a m 1 + a m + a m 3 p (k 1) p (k ) = a m 1 + a m + a m 3 a k 1 a k a k 3 a k 4 for m = k + 1 (4) that of 3 (m, 1) is a m 1 + a m + a m 3 p (k 1) p (k ) = a m 1 + a m + a m 3 a k 1 a k a k 3 a k 4 for m = k (43)

11 Fundamental unirreps of Sp(n, R) 1083 a m 1 + a m + a m 3 1 p (k 1) p (k ) = a m 1 + a m + a m 3 1 a k 1 a k a k 3 a k 4 for m = k + 1. (44) To separate the terms of 1 (0) {1} 1 (1) {1} we again make use of the U(n) content of 1 (0) {1} + 1 (1) {1}, which yields 1 (0) {1} = 4k 1 + a 4k + a 4k 3 1 a k 1 a k a k 3 k 0(a a k 4 a k 5 ) 3 (4k) + k 0(a 4k+1 + a 4k + a 4k 1 a k 1 a k a k 3 a k 4 ) 3 (4k + ) + k 0(a k + a k 1 + a k 1 a k 1 a k a k 3 a k 4 ) 3 (k + 1, 1) (45) 1 (1) {1} = k + a k 1 + a k a k 1 a k a k 3 a k 4 ) k 0(a 3 (k + 1) + k 0(a k 1 + a k + a k 3 a k 1 a k a k 3 a k 4 ) 3 (k, 1). (46) 5. Star equivalent Sp(n, R) unirreps plethysm Recall the modification rule for O(k) [λ = ( 1) c 1 [λ h h = λ 1 k 0 (47) where a continuous strip of boxes of length h, starting at the foot of the first column working up along the right edge ending in column c is removed from the Ferrers graph of (λ). The constraints on partitions (λ) corresponding to unirreps 1 k(λ) of Sp(n, R) allows us to restrict our attention to those cases corresponding to removing boxes from just the first column. Thus for the unirreps 1 k(λ) of Sp(n, R), we have (λ) (λ h) h = λ 1 k 0 (48) conversely we may add boxes to the first column of a stard partition (λ) for O(k) to produce a (λ + h) that is non-stard for O(k) but stard for Sp(n, R): (λ + h) (λ) h = k λ 1 0. (49) Unirreps of Sp(n, R) will be said to be star equivalent if 1 k(λ) = 1 k(λ) = 1 k(λ ± h) (50) which for brevity we shall write as 1 k(λ) 1 k(λ ± h) (51) where it is assumed that k n. Thus 1 (0) 1 (1) is a pair of star equivalent unirreps of Sp(n, R).

12 1084 J-Y Thibon et al Bearing in mind the preceding remarks definitions, consider the leading terms in the plethysm 1 (0) {3}: 3 (0) + 3 (4) + 3 (6) + 3 (8) + 3 (91) + 3 (10 ) + 3 (1 ) + 3 (13 1) + 3 (14 ) + 3 (15 1) + 3 (16 ) + 3 (17 1) + 3 (18 ) + 3 (19 1) + 3 (0 ) + 3 (1 1) + 3 ( ) + 3 (3 1). Replacing each unirrep by its star equivalent gives the leading terms of ( 1 (0) {3}) as 3 (13 ) + 3 (41) + 3 (61) + 3 (81) + 3 (9) + 3 (10 1) + 3 (1 1) + 3 (13 ) + 3 (14 1) + 3 (15 ) + 3 (16 1) + 3 (17 ) + 3 (18 1) + 3 (19 ) + 3 (0 1) + 3 (1 ) + 3 ( 1) + 3 (3 ) which are precisely the leading terms in 1 (0) {13 }: 3 (13 ) + 3 (41) + 3 (61) + 3 (81) + 3 (9) + 3 (10 1) + 3 (1 1) + 3 (13 ) + 3 (14 1) + 3 (15 ) + 3 (16 1) + 3 (17 ) + 3 (18 1) + 3 (19 ) + 3 (0 1) + 3 (1 ) + 3 ( 1) + 3 (3 ). Similarly, the leading terms in ( 1 (0) {1}) are 3 () + 3 (4) + 3 (51) + 3 (6) + 3 (71) + 3 (8) + 3 (91) + 3 (10 ) + 3 (11 1) + 3 (1 ) + 3 (13 1) (14 ) + 3 (15 1) (16 ) (17 1) (18 ) (19 1) (0 ) +3 3 (1 1) ( ) (3 1) which are identical to the leading terms in 1 (0) {1}: 3 () + 3 (4) + 3 (51) + 3 (6) + 3 (71) + 3 (8) + 3 (91) + 3 (10 ) + 3 (11 1) + 3 (1 ) + 3 (13 1) (14 ) + 3 (15 1) (16 ) (17 1) (18 ) (19 1) (0 ) (1 1) ( ) (3 1). The above results lead us to conjecture that 1 (0) {λ} 1 (1) {λ }. (5) That the conjecture holds for λ is evident from (1) to (4).That the conjecture holds for λ 3 follows by noting that 3 (m) 3 (m, 1) 3 (0) 3 (13 ).

13 Fundamental unirreps of Sp(n, R) 1085 Application of these star equivalences to the results obtained for the symmetrization of the cubes of the star equivalent pair 1 (0) 1 (1), together with inspection of the given coefficients, noting in particular (40) to (44), completes the verification of (5) for all (λ), 3. Proof that the conjecture (5) holds for all (λ) will be given in a later paper (with R C King) together with a proof of a more general star equivalence 1 k(µ) 1 k(µ ± h) {λ } k odd (53) 1 k(µ ± h) {λ} k even. (54) 6. Concluding remarks It is now possible to establish results for higher powers by applying the same methods to obtain the plethysms. But because longer partitions are involved the calculations rapidly become cumbersome. For example, the branching rule O(4) S 4 is obtained by examination of partitions of the form [1 4, [m, 1, 1, [m,, [m, 1 [m. One remarks that [m, 1, 1 = [m, 1 [1 4 = [0 one needs to examine only the partitions of the form [m, [m, 1, [m,. One applies the formula to give [m S 4, [m, 1 S 4, [m, S 4 which is equivalent to [m 1 {m/g} [m, 1 1 {m1/g} [m, 1 {m/g}. Now, to obtain the result, one writes that {m, j} = {m}{j} {m + 1}{j 1} finally 1 {m, j} = ( 1 {m}) ( 1 {j}) ( 1 {m + 1}) ( 1 {j 1}). These remarks are sufficient to give a method to compute the fourth power. The existence of the star equivalence operation for unirreps of Sp(n, R) implies certain one-to-one mappings between the N-particle states in even parity orbitals those in odd parity orbitals. Acknowledgments The work of BGW has been partially supported by a Polish KBN Grant. BGW is grateful to JYT for facilitating a visit to Institut Gaspard Monge where most of this work was done to Professor R C King (University of Southampton) for stimulating discussions hospitality at University of Southampton. References [1 Carvalho M J 1990 J. Phys. A: Math. Gen [ Grudzinski K Wybourne B G 1996 J. Phys. A: Math. Gen [3 Haase R W Johnson N F 1993 J. Phys. A: Math. Gen [4 Haase R W Johnson N F 1993 Phys. Rev. B [5 King R C, Luan Dehuai Wybourne B G 1981 J. Phys. A: Math. Gen [6 King R C Wybourne B G 1985 J. Phys. A: Math. Gen [7 Leung E Y Ton-That T 1995 Proc. Am. Math. Soc [8 Macdonald I G 1995 Symmetric Functions Hall Polynomials (Oxford Science Publications) nd edn [9 Rowe D J 1985 Rep. Prog. Phys

14 1086 J-Y Thibon et al [10 Rowe D J, Wybourne B G Butler P H 1985 J. Phys. A: Math. Gen [11 Scharf T, Thibon J-Y Wybourne B G 1993 J. Phys. A: Math. Gen [1 Wybourne B G 1974 Classical Groups for Physicists (New York: Wiley) [13 Wybourne B G 1994 Rept. Math. Phys. 34 9