The Perrin-McClintock Resolvent, Solvable Quintics and Plethysms

The Perrin-McClintock Resolvent, Solvable Quintics and Plethysms Frank Grosshans In his seminal paper of 1771, Lagrange found that certain polynomials of degree 6 called resolvents could be used to determine whether a quintic polynomial was solvable in radicals. Among the various resolvents later discovered, the Perrin-McClintock resolvent has some particularly noteworthy properties. We shall discuss these properties and their application to solvable quintics. The properties suggest that the Perrin-McClintock resolvent may be unique. We discuss this question and relate it to the representation theory of the general linear group, especially zero weight spaces and plethysms of a special form. 1

Properties of the Perrin-McClintock Resolvent Property 1: a polynomial function d = 1, 2,... f(x, y) = d i=0 = a 0 x d + ( ) d a i x d i y i, i (a 0, a 1,..., a d ) ( ) d a 1 x d 1 y + 1 a i C ( ) d a 2 x d 2 y 2 +... + 2 ( ) d a d y d d V d is vector space over C spanned by all such f(x, y) {( λ A 2 = C 2 = µ )} The Perrin-McClintock resolvent a polynomial, K : V 5 A 2 C 6 K(a 0, a 1, a 2, a 3, a 4, a 5 ; x, y) = κ j (a 0, a 1, a 2, a 3, a 4, a 5 )x 6 j y j j=0 ( x R f (x) = K(f, 1 ) ) 2

Example 1: f(x) = x 5 + 10a 2 x 3 + 5a 4 x + a 5 with a 4 = 4a 2 2 K(f, v) = (3a 6 2 + a 2 a 2 5)x 6 125a 4 2a 5 x 5 y + (4080a 7 2 15a 2 2a 2 5)x 4 y 2 +1000a 5 2a 5 x 3 y 3 + (960a 8 2 + 70a 3 2a 2 5)x 2 y 4 + (128a 6 2a 5 + a 2 a 3 5)xy 5 Example 2: f(x) = x 5 + 5x 4 + 9x 3 + 5x 2 4x 5 K(f, v) = 1 80000 ( 498x6 5900x 5 y 22662x 4 y 2 41320x 3 y 3 36254x 2 y 4 6860xy 5 + 8150y 6 ) Example 3: f(x) = x 5 8x 4 + 5x 3 6x 2 + 8x 4 K(f, v) = 5681513x 6 + 22679884x 5 y 42714844x 4 y 2 + 6325088x 3 y 3 +16299792x 2 y 4 18575936xy 5 + 5294016y 6 3

Properties of the Perrin-McClintock Resolvent Property 2: a covariant {( a b SL(2, C), g = c d ) } : ad bc = 1 action on V d g x = dx by g y = cx + ay f = d i=0 ( ) d a i x d i y i g f = i d i=0 ( ) d a i (dx by) d i ( cx + ay) i i action on A 2 ( ) ( ) a b λ = c d µ ( aλ + bµ cλ + dµ ) The covariant property (1) K(g f, g v) = K(f, v) for all g SL(2, C), f V d, v A 2 (2) coeffi cients of x and y terms form irreducible representation of SL(2, C) ( 1 (3) source of covariant is K(f, 0 ) ) completely determines K 4

Properties of the Perrin-McClintock Resolvent The Hessian cubic covariant g SL(2, C), g = ( 5 2 17 7 ) action on V 3 g x = 7x 2y g y = 17x + 5y f = a 0 x 3 + 3a 1 x 2 y + 3a 2 xy 2 + a 3 y 3 g f = a 0 (7x 2y) 3 + 3a 1 (7x 2y) 2 ( 17x + 5y) + 3a 2 (7x 2y)( 17x + 5y) 2 + a 3 ( 17x + 5y) 3 action on A 2 ( ) ( 5 2 λ 17 7 µ ) = ( 5λ + 2µ 17λ + 7µ H(a 0, a 1, a 2, a 3 ; x, y) = 1 36 Det ( ) 2 f x 2 2 f x y 2 f x y 2 f y 2 ) = (a 0 a 2 a 2 1)x 2 + (a 0 a 3 a 1 a 2 )xy + (a 1 a 3 a 2 2)y 2 5

( 5 2 g = 17 7 ) f = 4x 3 + 3 5x 2 y + 3 ( 6)xy 2 + ( 1)y 3 g f = 42624x 3 + 37128x 2 y 10779xy 2 + 1043y 3 v = ( 8 3 ) ( 46 ; g v = 157 ) The covariant property (1) H(g f, g v) = H(f, v) for all g G, f V d, v A 2 H(f, v) = H(4, 5, 6, 1; 8, 3) = 2881 H(g f, g v) = H( 42624, 12376, 3593, 1043; 46, 157) = 2881 (2) coeffi cients of x and y terms form irreducible representation of SL(2, C) (a 0 a 2 a 2 1), (a 0 a 3 a 1 a 2 ), (a 1 a 3 a 2 2) ( 1 (3) source of covariant is H(f, 0 completely determines H ) ) = a 0 a 2 a 2 1 algebraic meaning: H(f, v) = 0 for all v A 2 if and only if there is a linear form, say g = ax + by, such that f = g 3. For example, f(x, y) = 64x 3 144x 2 y + 108xy 2 27y 3. H(f, v) 0, f(x, y) = (4x 3y) 3 [Abdesselam and Chipalkatti] 6

Properties of the Perrin-McClintock Resolvent Property 3: solvable quintics Theorem. Let f(x) = a 0 x 5 + 5a 1 x 4 + 10a 2 x 3 + 10a 3 x 2 + 5a 4 x + a 5 be an irreducible quintic polynomial in Q[x]. Then f(x) is solvable in radicals if and only if R f (x) has a rational root or is of degree 5. Example 2: f(x) = x 5 + 5x 4 + 9x 3 + 5x 2 4x 5 K(f, v) = 1 80000 ( 498x6 5900x 5 y 22662x 4 y 2 41320x 3 y 3 36254x 2 y 4 6860xy 5 + 8150y 6 ) R f (x) = 1 80000 ( 498x6 5900x 5 22662x 4 41320x 3 36254x 2 6860x + 8150) has root 1/3. Hence, f(x) is solvable in radicals. Example 3: f(x) = x 5 8x 4 + 5x 3 6x 2 + 8x 4 K(f, v) = 5681513x 6 + 22679884x 5 y 42714844x 4 y 2 + 6325088x 3 y 3 + 16299792x 2 y 4 18575936xy 5 + 5294016y 6 R f (x) = 5681513x 6 + 22679884x 5 42714844x 4 + 6325088x 3 + 16299792x 2 18575936x + 5294016 does not have a rational root. Hence, f(x) is not solvable in radicals. Get elegant way to find solutions in radicals Cayley to McClintock (McClintock, p.163): "McClintock completes in a very elegant manner the determination of the roots of the quintic equation...." 7

Properties of the Perrin-McClintock Resolvent Property 4: global information Problem: use resolvents to obtain global information about solvable quintics Example 1 (Perrin): f(x) = a 0 x 5 + 10a 2 x 3 + 5a 4 x + a 5 with a 4 = 4a 2 2 R f (x) = (3a 6 2 + a 2 a 2 5)x 6 125a 4 2a 5 x 5 + (4080a 7 2 15a 2 2a 2 5)x 4 +1000a 5 2a 5 x 3 + (960a 8 2 + 70a 3 2a 2 5)x 2 + (128a 6 2a 5 + a 2 a 3 5)x has root 0. Hence, f(x) is solvable in radicals. Example 2: the McClintock parametrization Have mapping ϕ, a rational function, ϕ : A 4 Q A 4 Q (p, r, w, t) (γ, δ, ε, ζ) (γ, δ, ε, ζ) identified with f(x) = x 5 + 10γx 3 + 10δx 2 + 5εx + ζ The polynomial f(x) is solvable (its resolvent R f (x) has t as a root). inverse map exists, rational function need R f (x) to have rational root t diffi culty: if quintic factors, t may be complex or irrational real so don t quite parametrize all solvable quintics 8

Example 3: Brioschi quintics [Elia] f(x) = x 5 10zx 3 + 45z 2 x z 2 R f (x) = ( z 5 + 128z 6 )x 6 + 400z 6 x 5 + ( 15z 6 46080z 7 )x 4 +40000z 7 x 3 + ( 95z 7 51840z 8 )x 2 +(z 7 + 1872z 8 )x 25z 8 If z is a non-zero integer, then f(x) is solvable in radicals. Example 4: subject to certain explicitly defined polynomials not vanishing, if f 0 is an irreducible quintic such that R f0 has a root t 0 R, then every Euclidean open neighborhood of f 0 contains a solvable quintic. 9

Dickson s Factorization Action of S 5 S 5 : symmetric group on 5 letters Action of S 5 on polynomials f(x 1, x 2, x 3, x 4, x 5 ) Z[x 1, x 2, x 3, x 4, x 5 ] σ S 5 σ f = f(x σ1, x σ2, x σ3, x σ4, x σ5 ) Example f = x 1 x 2 x 1 x 3 + x 2 x 3 x 1 x 4 x 2 x 4 + x 3 x 4 + x 1 x 5 x 2 x 5 x 3 x 5 + x 4 x 5 σ = (132) σ f = x 3 x 1 x 3 x 2 + x 1 x 2 x 3 x 4 x 1 x 4 + x 2 x 4 + x 3 x 5 x 1 x 5 x 2 x 5 + x 4 x 5 10

Dickson s Factorization The group F 20 S 5 : symmetric group on 5 letters F 20 : subgroup of S 5 generated by (12345) and (2354) S 5 = 6 τ i F 20 i=1 τ 1 = (1), τ 2 = (12), τ 3 = (13), τ 4 = (23), τ 5 = (123), τ 6 = (132) Theorem. Let f(x) Q[x] be an irreducible quintic. Then f(x) is solvable in radicals if and only if its Galois group is conjugate to a subgroup of F 20. Problem: extend resolvent program to polynomials of higher degree. replaces F 20?) (What 11

Dickson s Factorization Malfatti s resolvent Φ(x 1, x 2, x 3, x 4, x 5 ) = x 1 x 2 x 1 x 3 + x 2 x 3 x 1 x 4 x 2 x 4 + x 3 x 4 + x 1 x 5 x 2 x 5 x 3 x 5 + x 4 x 5 = (x 1 x 5 )(x 2 x 5 ) + (x 2 x 5 )(x 3 x 5 ) + (x 3 x 5 )(x 4 x 5 ) (x 2 x 5 )(x 4 x 5 ) (x 4 x 5 )(x 1 x 5 ) (x 1 x 5 )(x 3 x 5 ) Properties: (1) homogeneous of degree 2 in x 1, x 2, x 3, x 4, x 5 (2) for any β C, Φ(x 1 +β, x 2 +β, x 3 +β, x 4 +β, x 5 +β) = Φ(x 1, x 2, x 3, x 4, x 5 ) (3) highest power to which any x i appears is 1 (4) (12345)Φ = Φ (2354)Φ = Φ Note: Malfatti resolvent, put Φ 2 = Θ R(x) = (x Θ)(x τ 2 Θ)(x τ 3 Θ)(x τ 4 Θ)(x τ 5 Θ)(x τ 6 Θ) polynomial in a i s rational root if and only if f(x) solvable in radicals for resolvents of this form, lowest possible degree in Θ rediscovered by Jacobi (1835), Cayley (1861), Dummit (1991) 12

Dickson s Factorization Roots of the resolvent S 5 = 6 τ i F 20 i=1 τ 1 = (1), τ 2 = (12), τ 3 = (13), τ 4 = (23), τ 5 = (123), τ 6 = (132) The Malfatti resolvent Φ(x 1, x 2, x 3, x 4, x 5 ) = x 1 x 2 x 1 x 3 + x 2 x 3 x 1 x 4 x 2 x 4 + x 3 x 4 + x 1 x 5 x 2 x 5 x 3 x 5 + x 4 x 5 = (x 1 x 5 )(x 2 x 5 ) + (x 2 x 5 )(x 3 x 5 ) + (x 3 x 5 )(x 4 x 5 ) (x 2 x 5 )(x 4 x 5 ) (x 4 x 5 )(x 1 x 5 ) (x 1 x 5 )(x 3 x 5 ) Ψ(x 1, x 2, x 3, x 4, x 5 ) = (x 1 x 2 x 3 x 4 x 5 )Φ(1/x 1, 1/x 2, 1/x 3, 1/x 4, 1/x 5 ) where does Ψ come from? need highest power to which a root appears in Φ is 1 homogeneous of degree 3 in x 1, x 2, x 3, x 4, x 5 Perrin-McClintock resolvent: for i = 1, 2, 3, 4, 5, 6, put Φ i = τ i Φ, Ψ i = τ i Ψ 6 K(f; v) = a 6 0 ((τ i Φ)x (τ i Ψ)y) i=1 6 6 = a 6 0( (τ i Φ)) (x ((τ i Ψ)/(τ i Φ))y) i=1 i=1 13

Constructing resolvents Setting I: covariants Find covariants of the form with K(f; v) = a m 0 6 ((τ i Φ)x (τ i Ψ)y) i=1 (1) Φ(x 1, x 2, x 3, x 4, x 5 ) homogeneous of degree w 2(mod 5) in x 1, x 2, x 3, x 4, x 5 (2) highest power to which a root appears in Φ is d = 2w+1 5 (3) for any β C, Φ(x 1 +β, x 2 +β, x 3 +β, x 4 +β, x 5 +β) = Φ(x 1, x 2, x 3, x 4, x 5 ) (4) (12345)Φ = Φ (2354)Φ = Φ Ψ(x 1, x 2, x 3, x 4, x 5 ) = (x 1 x 2 x 3 x 4 x 5 ) d Φ( 1/x 1, 1/x 2, 1/x 3, 1/x 4, 1/x 5 ) Recall: source determines covariant. source is a m 0 6 ((τ i Φ) i=1 14

Constructing resolvents Setting II: polynomials in roots For w 2(mod 5), put d = 2w+1 5. Find Φ(x 1, x 2, x 3, x 4, x 5 ) (1) homogeneous of degree w in x 1, x 2, x 3, x 4, x 5 (2) the highest power to which any x i appears in Φ is d (3) for any β C, Φ(x 1 +β, x 2 +β, x 3 +β, x 4 +β, x 5 +β) = Φ(x 1, x 2, x 3, x 4, x 5 ) (4) (12345)Φ = Φ (2354)Φ = Φ For covariant, need w 2(mod 5) Malfatti is only such polynomial of degree 2 15

Problems. 1. no SL 2 (C) action ( a b c d ) x i = (dx i b)/( cx i + a) ( 0 1 1 0 ) x i = 1/x i 2. the highest power to which any x i appears in Φ is d f(x 1,..., x p ) Z[x 1,..., x p ], f = λ (a) x a1 1... xap p R(f) = max{a i : 1 i p, λ (a) 0} If f(x 1,..., x p ) is symmetric in x 1,..., x p, then f(x 1,..., x p ) = τ (b) σ b1 1... σbp p σ i = ith elementary symmetric function. D(f) = max{b 1 +... + b p : τ (b) 0} Theorem. If f(x 1,..., x p ) is symmetric in x 1,..., x p, then R(f) = D(f). 16

Constructing resolvents Setting III: matrix variables Translation: For d 1(mod 2), find matrix polynomials F (( )) x1 x 2 x 3 x 4 x 5, ỹ 1 ỹ 2 ỹ 3 ỹ 4 ỹ 5 (1) F is homogenous of degree d in each column, i.e., F = c (e) x e1 1 ỹd e1 1... x e5 5 ỹd e5 5 (2) F is left U-invariant, i.e., F (( 1 β 0 1 ) ( )) x1 x 2 x 3 x 4 x 5 = ỹ 1 ỹ 2 ỹ 3 ỹ 4 ỹ F 5 ( ) x1 x 2 x 3 x 4 x 5 ỹ 1 ỹ 2 ỹ 3 ỹ 4 ỹ 5 or c(e) ( x 1 + βỹ 1 ) e1 ỹ d e1 1... ( x 5 + βỹ 5 ) e5 ỹ d e5 5 = c (e) x e1 1 ỹd e1 1... x e5 5 ỹd e5 5 17

(3) F has left T -weight 1, i.e., 5d 2(e 1 + e 2 + e 4 + e 4 + e 5 ) = 1 or F (( λ 0 0 1/λ ) ( )) ( ) x1 x 2 x 3 x 4 x 5 = ỹ 1 ỹ 2 ỹ 3 ỹ 4 ỹ 1 F x1 x 2 x 3 x 4 x 5 λ 5 ỹ 1 ỹ 2 ỹ 3 ỹ 4 ỹ 5 or c(e) (λ x 1 + 1 e1 λỹ1) ỹ d e1 1... (λ x 5 + 1 e5 λỹ5) ỹ d e5 5 = 1 c(e) x e1 1 λ ỹd e1 1... x e5 5 ỹd e5 5 (4) S 5 acts on vector variables by permuting columns (12345) F = F (2354) F = F F (( x2 x 3 x 4 x 5 x 1 ỹ 2 ỹ 3 ỹ 4 ỹ 5 ỹ 1 )) (( = F x1 x 2 x 3 x 4 x 5 ỹ 1 ỹ 2 ỹ 3 ỹ 4 ỹ 5 )) (( )) (( )) x1 x F 3 x 5 x 2 x 4 = ỹ 1 ỹ 3 ỹ 5 ỹ 2 ỹ F x1 x 2 x 3 x 4 x 5 4 ỹ 1 ỹ 2 ỹ 3 ỹ 4 ỹ 5 18

Constructing resolvents equivalences Setting I: find covariants of the form K(f; v) = a m 0 6 ((τ i Φ)x (τ i Ψ)y) i=1 Setting II: find Φ(x 1, x 2, x 3, x 4, x 5 ) Setting III: find matrix polynomials F Definition. For i = 1, 2, let K i : V 5 A 2 C be covariants as in Setting I. Say K 1 K 2 if and only if there are µ, ρ C[V 5 ] SL2(C) with µk 1 (x, y) = ρk 2 (x, y). Let Φ and Φ be as in Setting II. Say Φ Φ if and only if Ψ Φ = Ψ Φ. Setting I and Setting II: 6 6 K = a m 0 ((τ i Φ)x (τ i Ψ)y), K = a m 0 ((τ i Φ )x (τ i Ψ )y) i=1 K K if and only if Φ Φ i=1 Setting II and Setting III: there is vector space isomorphism between Φ homogeneous of degree w F homogeneous of degree 2w + 1 also, have algebra homomorphism [ ] x1 x have mapping Ω : C 2 x 3 x 4 x 5 C[x ỹ 1 ỹ 2 ỹ 3 ỹ 4 ỹ 1, x 2, x 4, x 4, x 5 ] 5 x i x i, ỹ i 1 19

Constructing resolvents finitely generated modules R w C[x 1, x 2, x 3, x 4, x 5 ] For w 0(mod 5), w 0, R w is vector space spanned by all linear combinations Φ of products (x i1 x j1 )... (x iw x jw ) such that (1) each x i appears 2w 5 times in every product (2) (12345)Φ = Φ, (2354)Φ = Φ R = R w M w C[x 1, x 2, x 3, x 4, x 5 ] For w 2(mod 5), w 0, M w is vector space spanned by Φ(x 1, x 2, x 3, x 4, x 5 ) (1) homogeneous of degree w in x 1, x 2, x 3, x 4, x 5 (2) the highest power to which any x i appears in Φ is 2w+1 5 (3) for any β C, Φ(x 1 +β, x 2 +β, x 3 +β, x 4 +β, x 5 +β) = Φ(x 1, x 2, x 3, x 4, x 5 ) (4) (12345)Φ = Φ, (2354)Φ = Φ M = M w 20

Theorem. (a) R is finitely generated C-algebra. (b) = Φ 2 Ψ 7 Φ 7 Ψ 2 0 and is in R. (c) Φ M, Φ = r1 Φ 2 + r2 Φ 7 (d) M is finitely generated R-module (e) dim Q(R) M R Q(R) = 2 21

Poincaré series Hilbert - Serre theorem Recall R = R w, is finitely generated C-algebra M = M w, M w polynomials as in Setting II is finitely generated R-module Poincaré series: P (M, t) = w 2(mod 5) dim M w Theorem (Hilbert, Serre, applied here). Let γ be the number of generators of R. Then f(t) P (M, t) = γ (1 t di ) i=1 for suitable positive integers d i and f(t) Z[t]. Problem: determine P (M, t). Determine dim M w. 22

Poincaré series GL m GL n duality to understand: (2) F is left U-invariant (3) F has left T -weight 1 T r GL r ; subgroup consisting of diagonal matrices U r GL r ; subgroup consisting of upper triangular matrices, 1 s on diagonal A highest weight of an irreducible polynomial representation of GL r with respect to the Borel subgroup T r U r is a character of the form χ = e 1 χ 1 + + e r χ r where e 1... e r 0. If e l is the last non-zero e i, we say that the highest weight χ has depth l. Theorem (GL m GL n duality) [Howe, Section 2.1.2]. Let U and V be finite-dimensional vector spaces over C. The symmetric algebra S(U V ) is multiplicity-free as a GL(U) GL(V ) module. Precisely, we have a decomposition S(U V ) = ρ D U ρ D V D of GL(U) GL(V )-modules. Here D varies over all highest weights of depth at most min{dimu, dimv }. 23

Translation M 2,5 : the algebra consisting of all 2 5 matrices with entries in C. GL 2 acts on M 2,5 by left multiplication: g m = gm for all g GL 2 and m M 2,5. GL 5 acts on M 2,5 by right multiplication: g m = mg 1 for all g GL 5 and m M 2,5. These actions commute and give an action of G = GL 2 GL 5 on M 2,5 and C[M 2,5 ]. M 2,5 A 2 (A 5 ) C[M 2,5 ] S((A 2 ) A 5 ) Suppose that d 1(mod 2), 5d = 2w + 1 and that (2) F is left U-invariant (3) F has left T -weight 1 then: the terms F = ṽ V D appear when ṽ: highest weight vector of irreducible representation GL 2, highest weight (w + 1)χ 1 + wχ 2 ρ D V is irreducible representation of GL 5, highest weight (w + 1)χ 1 + wχ 2 Note. can explicitly construct the invariants F in terms of determinants using Young diagrams and straightening [Pommerening]. 24

Poincaré series Zero weight space to understand: (1) F is homogenous of degree d in each column recall: w 2(mod 5), 5d = 2w + 1 T 5 = a 1 0 0 0 0 0 a 2 0 0 0 0 0 a 3 0 0 0 0 0 a 4 0 0 0 0 0 a 5, U 5 : 1 a 12 a 13 a 14 a 15 0 1 a 23 a 24 a 25 0 0 1 a 34 a 35 0 0 0 1 a 45 0 0 0 0 1, ρ : GL 5 GL(V ) V 0 = {v V : ρ(t)v = (a 1 a 2 a 3 a 4 a 5 ) e v}= 0 weight space of V translation: F C[M2,5 ], t T 5, m = (v 1,..., v 5 ) M 2,5 (t F )(v 1,..., v 5 ) = F ((v 1,..., v 5 )t) = F (a 1 v 1,..., a 5 v 5 ) = (a 1 a 2 a 3 a 4 a 5 ) d F (v1,..., v 5 ) 25

Proposition. Let w 2(mod 5) and d = 2w+1 5. Let ρ : GL 5 GL(V ) be the irreducible representation having highest weight (w + 1)χ 1 + wχ 2. The vector space consisting of all F C[M 2,5 ] such that (1) F is homogenous of degree d in each column, (2) F is left U-invariant, (3) F has left T -weight 1 is isomorphic to the 0-weight space of V. 26

Poincaré series S 5 action on zero weight space to understand: (4) (12345) F = F, (2354) F = F S 5 acts on 0 weight space, V 0 V 0 = m χ V χ V χ runs over all irreducible representations of S 5 m χ is multiplicity with which V χ appears in V 0 S 5 has 7 irreducible representations [5], [41], [32], [31 2 ], [2 2 1], [21 3 ], [1 5 ] ρ : F 20 {±1} ρ(12345) = 1 ρ(2354) = 1. ρ appears with multiplicity 1 in both [3 2] and [1 5 ]. It does not appear in any of the other 5 irreducible representations. 27

Proposition. Let w 2(mod 5) and d = 2w+1 5. Let ρ : GL 5 GL(V ) be the irreducible representation having highest weight (w + 1)χ 1 + wχ 2. The vector space consisting of all F C[M 2,5 ] such that (1) F is homogenous of degree d in each column, (2) F is left U-invariant, (3) F has left T -weight 1, (4) (12345) F = F and (2354) F = F is isomorphic to the vector space consisting of vectors v in the 0-weight space of V which satisfy (12345)v = v and (2354)v = v. The dimension of this vector space is the sum of the multiplicities with which [1 5 ] and [3 2] appear in the representation of S 5 on the 0-weight space of V. 28

Poincaré series plethysms [Littlewood, p. 204: "induced matrix of an invariant matrix"] ρ : GL n GL m (irreducible representation) σ : GL m GL p (irreducible representation) (σ ρ) : GL n GL p (reducible representation) process to decompose into irreducibles, plethysm [Gay, Gutkin] µ : representation of S 5 corresponding to [1 5 ] or [3 2]. Consider H = S d S d S d S d S d. Then, N S5d (H)/H S 5. µ representation of S 5, is representation of N S5d (H) the multiplicity with which µ = [1 5 ] or [3 2] appears in the representation of S 5 on V 0 is the multiplicity with which [(w + 1) w] appears in the representation µ S 5d of S 5d induced from µ This is a plethysm [Macdonald, pp.135/6] denoted by [1 5 ] [d] (resp. [3 2] [d]). 29

There are special features of this plethysm which greatly simplify the usual calculations. For example, we obtain the following results: w multiplicity of [1 5 ] multiplicity of [3 2] 2 0 1 7 0 1 12 0 2 17 0 4 22 1 6 27 1 8 32 1 11 507 425 2176 10842 195843 980298 From the standpoint of solving equations, the representation [1 5 ] is not interesting; the corresponding resolvent is a(x by) 6. Theorem. Let w 2(mod 5) and d = 2w+1 5. Let ρ : GL 5 GL(V ) be the irreducible representation having highest weight (w + 1)χ 1 + wχ 2. The vector space consisting of all F C[M 2,5 ] such that (1) F is homogenous of degree d in each column, (2) F is left U-invariant, (3) F has left T -weight 1, (4) (12345) F = F and (2354) F = F is isomorphic to the vector space consisting of vectors v in the 0-weight space of V which satisfy (12345)v = v and (2354)v = v. The dimension of this vector space is the sum of the multiplicities with which [1 5 ] and [3 2] appear in the representation of S 5 on the 0-weight space of V. The dimension can be found by calculating the plethysms [1 5 ] [d] and [3 2] [d]. 30

Using the Theorems and plethysm considerations, can show there are infinitely many non-equivalent covariants of Perrin-McClintock type (Setting I). It also seems likely that there are infinitely many non-equivalent covariants of Perrin-McClintock type for which Ψ/Φ is fixed by F 20 and not by S 5 so we get resolvents for deciding solvability. 31

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