The Perrin-McClintock Resolvent, Solvable Quintics and Plethysms

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1 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

2 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 ( ) 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

3 Example 1: f(x) = x a 2 x 3 + 5a 4 x + a 5 with a 4 = 4a 2 2 K(f, v) = (3a a 2 a 2 5)x 6 125a 4 2a 5 x 5 y + (4080a a 2 2a 2 5)x 4 y a 5 2a 5 x 3 y 3 + (960a a 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) = ( 498x6 5900x 5 y 22662x 4 y x 3 y x 2 y xy y 6 ) Example 3: f(x) = x 5 8x 4 + 5x 3 6x 2 + 8x 4 K(f, v) = x x 5 y x 4 y x 3 y x 2 y xy y 6 3

4 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

5 Properties of the Perrin-McClintock Resolvent The Hessian cubic covariant g SL(2, C), g = ( ) 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

6 ( 5 2 g = 17 7 ) f = 4x x 2 y + 3 ( 6)xy 2 + ( 1)y 3 g f = 42624x x 2 y 10779xy y 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

7 Properties of the Perrin-McClintock Resolvent Property 3: solvable quintics Theorem. Let f(x) = a 0 x 5 + 5a 1 x a 2 x a 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) = ( 498x6 5900x 5 y 22662x 4 y x 3 y x 2 y xy y 6 ) R f (x) = ( 498x6 5900x x x x x ) 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) = x x 5 y x 4 y x 3 y x 2 y xy y 6 R f (x) = x x x x x x 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

8 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 a 2 x 3 + 5a 4 x + a 5 with a 4 = 4a 2 2 R f (x) = (3a a 2 a 2 5)x 6 125a 4 2a 5 x 5 + (4080a a 2 2a 2 5)x a 5 2a 5 x 3 + (960a a 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 γx δx 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

9 Example 3: Brioschi quintics [Elia] f(x) = x 5 10zx z 2 x z 2 R f (x) = ( z z 6 )x z 6 x 5 + ( 15z z 7 )x z 7 x 3 + ( 95z z 8 )x 2 +(z z 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

10 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

11 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

12 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

13 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

14 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

15 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

16 Problems. 1. no SL 2 (C) action ( a b c d ) x i = (dx i b)/( cx i + a) ( ) 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 a 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 b σbp p σ i = ith elementary symmetric function. D(f) = max{b 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

17 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 e 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 e ( x 5 + βỹ 5 ) e5 ỹ d e5 5 = c (e) x e1 1 ỹd e x e5 5 ỹd e5 5 17

18 (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 e1 λỹ1) ỹ d e (λ x e5 λỹ5) ỹ d e5 5 = 1 c(e) x e1 1 λ ỹd e 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

19 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

20 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

21 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

22 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

23 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 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

24 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

25 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 a a a a 5, U 5 : 1 a 12 a 13 a 14 a a 23 a 24 a a 34 a a , 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

26 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

27 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

28 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

29 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

30 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] 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

31 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

32 References Abdesselam, Abdelmalek; Chipalkatti, Jaydeep On Hilbert covariants. Canad. J. Math. 66 (2014), no. 1, Dickson, L. E. Resolvent sextics of quintic equations. Bull. Amer. Math. Soc. 31 (1925), no. 9-10, Dummit, D. S. Solving solvable quintics. Math. Comp. 57 (1991), no. 195, Elia, Michele Solvable Brioschi quintics over Q. JP J. Algebra Number Theory Appl. 6 (2006), no. 1, Gay, David A. Characters of the Weyl group of SU(n) on zero weight spaces and centralizers of permutation representations. Rocky Mountain J. Math. 6 (1976), no. 3, Gutkin, Eugene. Schur-Weyl duality and representations of permutation groups.(english summary) The orbit method in geometry and physics (Marseille, 2000), , Progr. Math., 213, Birkhäuser Boston, Boston, MA, Howe, Roger (GL n,gl m ) -duality and symmetric plethysm. Proc. Indian Acad. Sci. Math. Sci. 97 (1987), no. 1-3, (1988). Howe, Roger Perspectives on invariant theory: Schur duality, multiplicity-free actions and beyond. The Schur lectures (1992) (Tel Aviv), 1 182, Israel Math. Conf. Proc., 8, Bar-Ilan Univ., Ramat Gan, Littlewood, Dudley E. The theory of group characters and matrix representations of groups. Reprint of the second (1950) edition. AMS Chelsea Publishing, Providence, RI, Macdonald, I. G. Symmetric functions and Hall polynomials. Second edition. With contributions by A. Zelevinsky. Oxford Mathematical Monographs. Oxford Science Publications. The Clarendon Press, Oxford University Press, New York, Malfatti, G. De AEquationibus Quadrato-cubicis Disquisitio Analytica, Siena transactions 1771 McClintock, E. Analysis of quintic equations, American J Math,Vol. 8, No. 1, Sep., 1885,

33 McClintock, E. Further researches in the theory of quintic equations, American Journal of Math, vol. 20, No. 2, Apr., 1898, pp Perrin, Sur les cas de resolubilite par radicaux de l equation du cinquieme degree Bulletin de la SMF Tome pp Pommerening, Klaus Ordered sets with the standardizing property and straightening laws for algebras of invariants. Adv. in Math. 63 (1987), no. 3,

Describing Resolvent Sextics of Quintic Equations. By Frank D. Grosshans

Describing Resolvent Sextics of Quintic Equations By Frank D. Grosshans L. E. Dickson, Bulletin A.M.S. 31, 1925, 515-523 RESOLVENT SEXTICS OF QUINTIC EQUATIONS BY L. E. DICKSON 1. Introduction. The object