The Perrin-McClintock Resolvent, Solvable Quintics and Plethysms - PDF document

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 , . . . � d � � d a i x d − i y i , f ( x, y ) = a i ∈ C i i =0 � d � � d � � d � a 0 x d + a 2 x d − 2 y 2 + . . . + a 1 x d − 1 y + a d y d = 1 2 d ↔ ( a 0 , a 1 , . . . , a 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 κ j ( a 0 , a 1 , a 2 , a 3 , a 4 , a 5 ) x 6 − j y j K ( a 0 , a 1 , a 2 , a 3 , a 4 , a 5 ; x, y ) = j =0 � � x R f ( x ) = K ( f, ) 1 2

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

Properties of the Perrin-McClintock Resolvent Property 2: a covariant �� a � � b SL (2 , C ) , g = : ad − bc = 1 c d action on V d g · x = dx − by g · y = − cx + ay � d � � d � � d � d a i x d − i y i → g · f = a i ( dx − by ) d − i ( − cx + ay ) i f = i i i =0 i =0 action on A 2 � � � � � � a b λ aλ + bµ · = c d µ 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) coefficients 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 � � 5 2 g ∈ SL (2 , C ) , g = 17 7 action on V 3 g · x = 7 x − 2 y g · y = − 17 x + 5 y a 0 x 3 + 3 a 1 x 2 y + 3 a 2 xy 2 + a 3 y 3 → f = a 0 (7 x − 2 y ) 3 + 3 a 1 (7 x − 2 y ) 2 ( − 17 x + 5 y ) + 3 a 2 (7 x − 2 y )( − 17 x + 5 y ) 2 + a 3 ( − 17 x + 5 y ) 3 g · f = action on A 2 � � � � � � 5 2 λ 5 λ + 2 µ · = 17 7 µ 17 λ + 7 µ � � ∂ 2 f ∂ 2 f ∂x 2 1 ∂x∂y H ( a 0 , a 1 , a 2 , a 3 ; x, y ) = 36 Det ∂ 2 f ∂ 2 f ∂x∂y ∂y 2 1 ) x 2 + ( a 0 a 3 − a 1 a 2 ) xy + ( a 1 a 3 − a 2 = ( a 0 a 2 − a 2 2 ) y 2 5

� 5 � 2 g = 17 7 f = 4 x 3 + 3 × 5 x 2 y + 3 × ( − 6) xy 2 + ( − 1) y 3 g · f = − 42624 x 3 + 37128 x 2 y − 10779 xy 2 + 1043 y 3 � � � � 8 46 v = ; g · v = 3 157 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) coefficients 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 ) = a 0 a 2 − a 2 (3) source of covariant is H ( f, 1 0 completely determines H 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 ) = 64 x 3 − 144 x 2 y + 108 xy 2 − 27 y 3 . H ( f, v ) ≡ 0 , f ( x, y ) = (4 x − 3 y ) 3 [Abdesselam and Chipalkatti] 6

Properties of the Perrin-McClintock Resolvent Property 3: solvable quintics Theorem . Let f ( x ) = a 0 x 5 + 5 a 1 x 4 + 10 a 2 x 3 + 10 a 3 x 2 + 5 a 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 + 5 x 4 + 9 x 3 + 5 x 2 − 4 x − 5 80000 ( − 498 x 6 − 5900 x 5 y − 22662 x 4 y 2 − 41320 x 3 y 3 − 36254 x 2 y 4 1 K ( f, v ) = − 6860 xy 5 + 8150 y 6 ) 80000 ( − 498 x 6 − 5900 x 5 − 22662 x 4 − 41320 x 3 − 36254 x 2 1 R f ( x ) = − 6860 x + 8150) has root 1/3. Hence, f ( x ) is solvable in radicals. Example 3 : f ( x ) = x 5 − 8 x 4 + 5 x 3 − 6 x 2 + 8 x − 4 K ( f, v ) = − 5681513 x 6 + 22679884 x 5 y − 42714844 x 4 y 2 + 6325088 x 3 y 3 + 16299792 x 2 y 4 − 18575936 xy 5 + 5294016 y 6 R f ( x ) = − 5681513 x 6 + 22679884 x 5 − 42714844 x 4 + 6325088 x 3 + 16299792 x 2 − 18575936 x + 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): a 0 x 5 + 10 a 2 x 3 + 5 a 4 x + a 5 f ( x ) = 4 a 2 with a 4 = 2 5 ) x 6 − 125 a 4 2 a 5 x 5 + (4080 a 7 R f ( x ) = (3 a 6 2 + a 2 a 2 2 − 15 a 2 2 a 2 5 ) x 4 2 a 5 x 3 + (960 a 8 5 ) x 2 + (128 a 6 +1000 a 5 2 + 70 a 3 2 a 2 2 a 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, A 4 Q → A 4 ϕ : Q ( p, r, w, t ) → ( γ, δ, ε, ζ ) x 5 + 10 γx 3 + 10 δx 2 + 5 εx + ζ ( γ, δ, ε, ζ ) identified with f ( x ) = 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 difficulty: 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 − 10 zx 3 + 45 z 2 x − z 2 ( − z 5 + 128 z 6 ) x 6 + 400 z 6 x 5 + ( − 15 z 6 − 46080 z 7 ) x 4 R f ( x ) = +40000 z 7 x 3 + ( − 95 z 7 − 51840 z 8 ) x 2 +( z 7 + 1872 z 8 ) x − 25 z 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 f 0 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 σ 1 , x σ 2 , x σ 3 , x σ 4 , x σ 5 ) 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) � 6 S 5 = 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. (What replaces F 20 ?) 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