Solving the simultaneous equation of the symmetric tetravariate polynomials and The roots of a quartic equation Oomori, Yasuhiro in Himeji City, Japan Dec.1, 2011 Abstract 1. S 4 2.
1. {α, β, γ, δ} (1) s 1 a 1 = s2 = a 2 s = a a n C: (1) s 4 = a 4 s 1 = α + β + γ + δ s 2 = αβ + αγ + αδ + βγ + βδ + γδ s (1.2) = αβγ + αβδ + αγδ + βγδ s4 = αβγδ 1. (1.2) (1) 2. Ex.1 (2) φ α + β + γ + δ αβ + βγ + γδ C αβγ + βγδ 4 [Φ]Φ = {α, β, γ, δ} αβγδ C 4 [α, β, ]: C[α, β, ]Descartes C[α, β, ]: C C: Ex.1 α β φ: = γ (2) δ
Fig.1 S 4 S 4 α sss 1 (α, β, γ, δ) β φ = γ T sss sss 2 (α, β, γ, δ) = φ sss (α, β, γ, δ) sss δ sss 4 (α, β, γ, δ) pppp 1 (s 1, s 2, s, s 4 ) sss 1 (s 1, s 2, s, s 4 ) α Change of variables by eq.(1) pppp 2 (s 1, s 2, s, s 4 ) T 1 sss the fffffffffff theeeee = φ oo ef ssssetesc pefsfessff pppp (s 1, s 2, s, s 4 ) sss sss 2 (s 1, s 2, s, s 4 ) β = sss (s 1, s 2, s, s 4 ) γ pppp 4 (s 1, s 2, s, s 4 ) sss 4 (s 1, s 2, s, s 4 ) δ T sym sym n poly n sol n Change of variables by eq(1) (1) the fundamental theorem of symmetric polynomial Fig.1 Ex.2 Ex. L A P n α A 11 α + A 12 β + A 1 γ + A 14 δ A 11 A 12 A 1 A 14 L β A A (φ) = A γ = 21 α + A 22 β + A 2 γ + A 24 δ A A = 21 A 22 A 2 A 24 A 1 α + A 2 β + A γ + A 4 δ A 1 A 2 A A 4 δ A 41 α + A 42 α + A 4 γ + A 44 δ A 41 A 42 A 4 A 44 L A A Ex.2
α α α n β β β (φ) = γ γ = n γ n : = φn n N: δ δ δ n P n n ʘ:Hadamard P n : Ex. L P n. A Ex.2 A L A 4. Ex. P n ψ P n ψ Ex.4 C ψa, C ψb {ψ ψ C 4 [Φ]} {ψ ψ C 4 [Φ]} α α α 2 αβ αβ P ψ a = αβγ, ψ a2 = αβγ, ψ a 2 ψ 2 α a = 2 β 2 un invetible α 2 β 2 γ 2 γδ γδ γ 2 δ 2 C ψa = ψ = ψ a, ψ a2 ψ 2 = ψ a 2 C ψb = ψ = ψ b, ψ b2 ψ 2 = ψ b 2 extension of definition P2 ψ a 2 ψ b 2 C 4 [Φ] = C ψa C ψb, C ψi C ψj = if i j Ex.4
5. Fig. C 4 [Φ] ψ a, ψ a P n ψ a ψ a ~ n C 4 [Φ] ψ a ψ a [ψ a ] n C 4 [Φ] n C 4 [Φ] n C 4 [Φ] n P n (C 4 [Φ]) C 4 [Φ] n n P n P P n ψ a n ψ a : ψ n a = ψ n a ψ a, ψ a C 4 [Φ] Φ = {α, β, γ, δ} n P P n(c 4 [Φ]) C 4 [Φ] n ; ψn a [ψ a ] n C 4 [Φ] n : = {[ψ] n ψ C 4 [Φ]} n P = full inverse of P n Fig. [ψ a ] n : = {ψ C 4 [Φ] ψ n ψ a } P n C 4 [Φ] C 4 [Φ] ; ψ a ψn a 5. φ S 4 Tab.1 S 4 Tab.1.2 S 4 Schönflies S 4 T d = {E, 8C, C 2, 6S 4, 6σ d } φ
Class S 4 Normal subgroup 1 α β γ δ = e α β γ δ (ab)(cd) α β γ δ β α δ γ α β γ δ γ δ α β α β γ δ δ γ β α V K α β γ δ β γ δ α α γ δ β α δ β γ α β γ δ β γ δ α γ β δ α δ β α γ, A 4 (abc) α β γ δ β γ δ α β δ γ α δ α γ β α β γ δ β γ δ α β γ α δ γ α β δ α β γ δ α γ β δ α β γ δ δ β γ α α β γ δ α β δ γ α β γ δ β α γ δ, (ab) α β γ δ α δ γ β α β γ δ γ β α δ α β γ δ β δ α γ α β γ δ γ α δ β α β γ δ δ γ α β α β γ δ γ δ β α, (abcd) α β γ δ β γ δ α α β γ δ δ α β γ a b c d a,b,c,d {α,β,γ,δ} Tab.1
T d S 4 class Permutation representation of S 4 with skeletal fomulae [E] α β γ δ α β γ δ [C 2 ] α β γ δ β α δ γ φ α β γ δ γ δ α β α β γ δ δ γ β α,φ,φ [C ] α β γ δ α γ δ β φ α β γ δ α δ β γ φ α β γ δ γ β δ α,φ α β γ δ δ β α γ φ α β γ δ β δ γ α φ α β γ δ δ α γ β φ α β γ δ β γ α δ,φ α β γ δ γ α β δ φ [σ d ] α β γ δ α γ β δ φ α β γ δ δ β γ α α β γ δ α β δ γ α β γ δ β α γ δ,φ,φ,φ, α β γ δ α δ γ β φ α β γ δ γ β α δ,φ [S 4 ] α β γ δ β δ α γ φ α β γ δ γ α δ β α β γ δ δ γ α β α β γ δ γ δ β α,φ,φ,φ, α β γ δ β γ δ α φ α β γ δ δ α β γ,φ φ = : Schönflies notation Tab.1.2
φ S 4 Ex.5 α β γ δ β γ S 4 α δ γ, δ α, S β 4 ooooo δ γ β α 0 0 0 1 1 0 0 0 0 0 0 0, 0 0 0 1 1 0 1 0 0 1 Ex.5 0 1, 0 0 0 0 1 0 α 1 0 β 0 0 γ = R φ (S 4 )φ 0 0 δ S 4 S 4 6. Ex.6 S 4 ψs 4 ψ ψ s S 4 R ψ(s) S 4 R ψ (S 4 ) α 2 + β 2 α 2 β 2 α β γ δ γ 2 + δ 2 0 0 1 1 α 2 + β 2 γ δ α β S 4 γ 2 γ 2 δ 2 0 0 1 1 1 1 α 2 0 0 α 2 β 2 2 2 γ 2 δ 2 β 2 1 1 0 0 2 2 δ 2 Ex.6 φ S 4 R φ(s) s S 4 Tab.1.
T d class [E] = I 0 0 1 0 4 = R φ (e) R φ (s)s S 4 T d 0 0 0 1 0 0 1 0 [C 2 ] 0 1 0 0 [C ] 0 0 0 1 0 0 0 1 0 1 0 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 1 Normal subgroup V K abelian A 4 [σ d ] [S 4 ] 0 1 0 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1, 0 1 0 0 0 0 0 1 0 0 1 0 0 0 1 0 Tab.1.
Fig.4 S 4 φ 0 0 0 1 R φ (s) 0 0 1 0 =,, T sss, s S 0 0 1 0 4 0 1 0 0 α β φ = γ δ sss 1 (α, β, γ, δ) sss 2 (α, β, γ, δ) = φ sss (α, β, γ, δ) sss sss 4 (α, β, γ, δ) Fig.4 T sss step1 φ(2) S 4 R φ (s) ; s S 4 α β φ = γ (2) δ 7. Ex.7 A A Λ(A) A A 11 A 12 A 1 A 14 λ Λ(A) = Λ λ Λ λ AΛ 1 A λ = Λ λ 21 A 22 A 2 A 24 Λ A 1 A 2 A A 1 0 λ λ = 2 0 0 Ad 4 0 0 λ 0 Λλ (A) A 41 A 42 A 4 A 44 0 0 0 λ 4 Ex.7
S 4 Fig.5 [E] [C 2 ] «Klein V K» R φ (V K ) () H H φ (4) H «4 Hadamard» 0 0 0 1 R φ (V K ) = R φ ([E], [C 2 ]) 0 0 1 0 =,,, 0 0 0 1 0 0 1 0 Fig.5 1 1 1 1 1 1 1 1 Λ(R φ (V K )) H = = H 1 1 1 1 4 H 4 4Hadamard () 1 1 1 1 α α + β + γ + δ L β α β + γ δ H (φ) = H γ = = φ α + β γ δ 1 (4) δ α β γ + δ φ 1 (4) S 4 R φ1 (s); s S 4 Tab.2 Klein V K step1 R φ1 (s) ; s S 4 R φ (s); s S 4 R φ1 (s); s S 4 (4.2)
T d class [E] = I 0 0 1 0 4 = R φ1 (e) R φ1 (s)s S 4 T d 0 1 0 0 [C 2 ] 0 0 1 0 0 0 0 1 [C ] 0 0 0 1, 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 1 0 Normal subgroup V K abelian A 4 [σ d ] [S 4 ] 0 0 0 1, 0 0 1 0 0 0 0 1 0 1 0 0 0 1 0 0 0 0 0 1 0 0 1 0 0 0 0 1, 0 0 1 0 0 0 0 1 0 1 0 0 0 1 0 0 0 0 0 1 0 0 1 0 Tab.2 R φ (s) L H R φ1 (s) = H 4 R φ (s)h 4 1 = Ad H4 R φ (s) s S 4 (4.2) H 4 1 = 1 4 H 4 H 4 :Hadamard matrix of order 4
8. Ex.8 Ex.9 1 0200 000 : rrrrrrr, ssssssss 0005 4000 Ex.8 0100 0010 bbbbbb, rrrrrrr, sssssess 0001 1000 Ex.9 step2 step1 φ 1 (4) R φ 1 (V K ) R φ1 (s); s S 4 R φ1 (s); s S 4 ±1 step1 φ 1 (4) n n=1 P 2 φ 1 () φ 2 (5) P 2 (φ 1 ) = (α + β + γ + δ) 2 (α β + γ δ) 2 (α + β γ δ) 2 = φ2 (5) (α β γ + δ) 2 φ 2 (5) S 4 R φ1 (s); s S 4 Tab. φ 2 R φ1 (s); s S 4 R φ2 (s); s S 4 (5.2)
T d class [E] = I 0 0 1 0 4 = R φ2 (e) R φ2 (s)s S 4 T d 0 1 0 0 [C 2 ] 0 0 1 0 0 0 0 1 [C ] 0 0 0 1, 0 0 0 1 0 0 1 0 0 0 0 1 0 0 0 1 0 0 1 0 Normal subgroup V K abelian A 4 [σ d ] [S 4 ] 0 0 0 1, 0 0 1 0 0 0 0 1 0 1 0 0 0 1 0 0 0 0 0 1 0 0 1 0 0 0 0 1, 0 0 1 0 0 0 0 1 0 1 0 0 0 1 0 0 0 0 0 1 0 0 1 0 Tab. R φ1 (s) P2 R φ2 (s) = R φ1 (s) R φ1 (s)s S 4 (5.2) ʘ: Hadamard product sign
φ 2 (5) S 4 R φ2 (S 4 ) Tab..2 R φ2 (S 4 ) S S 4 /V 4 step step2 φ 2 (5) S 4 R φ2 (S 4 ) S Tab..2 Tab. [E], [C 2 ] [C ] 4 A 4 R φ2 (A 4 ) R φ2 (S 4 ) R φ 2 (A 4 ) A (6) H φ 2 H φ (7) H (6) «Hadamard» I 1 : 1 0 1 1 1 Λ(R φ2 (A 4 )) H = 0 1 ω ω 2 = I 1 H H : Hadamard (6) 0 1 ω 2 ω ω: ω 2 + ω + 1 = 0 (α + β + γ + δ) 2 L (α β + γ δ) H (φ 2 ) 2 = H (α + β γ δ) 2 (α β γ + δ) 2 (α + β + γ + δ) 2 (α β + γ δ) = 2 + (α + β γ δ) 2 + (α β γ + δ) 2 (α β + γ δ) 2 + ω(α + β γ δ) 2 + ω 2 (α β γ + δ) 2 = φ (7) (α β + γ δ) 2 + ω 2 (α + β γ δ) 2 + ω(α β γ + δ) 2 φ (7) S 4 R φ (s) s S 4 Tab.4 R φ2 (s) s S 4 R φ (s) s S 4 (7.2) A R φ (s) s A 4 R φ (S 4 ) Tab.4.2
C₃ v class [E] = I 0 0 1 0 4 = R φ2 (e) R φ2 (S 4 ) S C Normal subgroup [C ] [σ v ] 0 0 1 0 0 0 1 0 0 0 1 0,, 0 0 1 0 0 0 1 0 A abelian Tab..2
T d class [E] = I 0 0 1 0 4 = R φ (e) R φ (s)s S 4 T d 0 1 0 0 [C 2 ] 0 0 1 0 0 0 0 1 [C ] 0 1 0 0 0 0 ω 2 0 0 0 ω 0 0 0 ω 0 0 0 ω 2, 0 0 0 0 ω 0 0 0 ω 2 0 0 0 ω 2 0 0 0 ω 0 1 0 0 0 0 ω 2 0 0 0 ω 0 0 0 ω 0 0 0 ω 2 0 0 0 0 ω 0 0 0 ω 2 0 0 0 ω 2 0 0 0 ω Normal subgroup V K abelian A 4 [σ d ] [S 4 ] 0 1 0 0 0 0 0 ω 2 0 0 0 ω 2, 0 0 0 ω 0 0 0 ω 0 0 ω 0 0 0 ω 0 0 0 ω 2 0 0 0 ω 2 0 0 1 0 0 0 0 0 1 0 0 1 0 0 1 0 0 0 0 0 ω 2 0 0 0 ω 2, 0 0 0 ω 0 0 0 ω 0 0 ω 0 0 0 ω 0 0 0 ω 2 0 0 0 ω 2 0 0 1 0 0 0 0 0 1 0 0 1 0 Tab.4 ω: cubic rootω 2 +ω+1=0 R φ2 (s) L H R φ (s) = H R φ2 (s)h 1 = Ad H R φ2 (s) s S 4 H : I 1 H H 1 = I 1 1 H (7.2) H 1 = I 1 1 H I 1 :Identity matrix of order 1 H :Complex Hadamard matrix of order
C₃ v class [E] = I 0 0 1 0 4 = R φ (e) R φ (S 4 ) S C Normal subgroup [C ] [σ v ] 0 0 ω 2 0 0 0 0 ω 0 0 ω 0 0 0 0 ω 2 0 0 0 ω 2, 0 0 ω 0, 0 0 0 ω 0 0 ω 2 0 0 0 1 0 A abelian Tab.4.2 ω: cubic rootω 2 +ω+1=0
step4 step φ (7) S 4 R φ (S 4 ) 1 1 step φ (7) n=1 P φ (7) 6 φ 4 (8) (α + β + γ + δ) 6 P {(α β + γ δ) (φ ) 2 + (α + β γ δ) 2 + (α β γ + δ) 2 } = {(α β + γ δ) 2 + ω(α + β γ δ) 2 + ω 2 (α β γ + δ) 2 } = φ4 (8) {(α β + γ δ) 2 + ω 2 (α + β γ δ) 2 + ω(α β γ + δ) 2 } φ 4 (8) S 4 R φ4 (s) s S 4 Tab.5 φ 4 R φ (s) s S 4 R φ4 (s) ; s S 4 (8.2) φ 4 (8) S 4 R φ 4 (S 4 )Tab.5.2 R φ4 (S 4 ) S 2 S /A step5 step4 φ 4 (8) S 4 R φ4 (S 4 ) S 2 Tab.5.2 R φ 4 (S 4 ) (9) H φ 4 H φ 5 (10) H (9) «2 Hadamard» Λ(R φ4 (S 4 )) H = = I 0 0 1 1 2 H 2 I 2: 2 (9) H 2 : 2 Hadamard 0 0 1 1
T d class [E] = I 0 0 1 0 4 = R φ4 (e) R φ4 (s)s S 4 T d 0 1 0 0 [C 2 ] 0 0 1 0 0 0 0 1 [C ] 0 1 0 0, 0 0 1 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0 1 Normal subgroup V K abelian A 4 [σ d ] [S 4 ] 0 1 0 0, 0 0 0 1 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 0 0 1 0 0, 0 0 0 1 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 0 Tab.5 R φ (s) P R φ4 (s)r φ (s) R φ (s) R φ (s)s S 4 (8.2) ʘ:Hadamard product sign
C s class [E] = I 0 0 1 0 4 = R φ4 (e) R φ4 (S 4 ) S 2 C s Normal subgroup [σ] 0 0 1 0 Tab.5.2
L H (φ 4 ) = H φ 4 (α + β + γ + δ) 6 {(α β + γ δ) 2 + (α + β γ δ) 2 + (α β γ + δ) 2 } [{(α β + γ δ) 2 + ω(α + β γ δ) 2 + ω 2 (α β γ + δ) 2 } = +{(α β + γ δ) + ω 2 (α + β γ δ) 2 + ω(α β γ + δ) 2 } ] = φ 5 (10) [{(α β + γ δ) 2 + ω(α + β γ δ) 2 + ω 2 (α β γ + δ) 2 } {(α β + γ δ) 2 + ω 2 (α + β γ δ) 2 + ω(α β γ + δ) 2 } ] φ 5 (10) S R φ5 (s); s S 4 Tab.6 R φ4 (s); s S 4 R φ5 (s); s S 4 (10.2) R φ5 (s); s S 4 R φ5 (S 4 ) Tab.6.2 step6 step5 φ 5 (10) S 4 R φ5 (s); s S 4 I 4 R φ5 ; (S 4 ) ±1 step5 φ 5 (10) n=1 P 2 φ 5 (10) 12 φ sss (11) P 2 (φ 5 ) = (α + β + γ + δ) 12 {(α β + γ δ) 2 + (α + β γ δ) 2 + (α β γ + δ) 2 } 6 [{(α β + γ δ) 2 + ω(α + β γ δ) 2 + ω 2 (α β γ + δ) 2 } + {(α β + γ δ) 2 + ω 2 (α + β γ δ) 2 + ω(α β γ + δ) 2 } ] 2 = φ sss (11) [{(α β + γ δ) 2 + ω(α + β γ δ) 2 + ω 2 (α β γ + δ) 2 } {(α β + γ δ) 2 + ω 2 (α + β γ δ) 2 + ω(α β γ + δ) 2 } ] 2 step1 step6 12 φ sss (11) S 4 R φsss (s); s S 4 Tab.7 R φsss (S 4 )Tab. 7.2 R φsss (s); s S 4 R φ5 (s); s S 4 (11.2) φ(2) S 4 φ sss (11)
T d class [E] = I 0 0 1 0 4 = R φ5 (e) R φ5 (s)s S 4 T d 0 1 0 0 [C 2 ] 0 0 1 0 0 0 0 1 [C ] 0 1 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0 1 Normal subgroup V K abelian A 4 [σ d ] [S 4 ] 0 1 0 0, 0 0 1 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0 1 0 1 0 0, 0 0 1 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0 1 Tab.6 R φ4 (s) L H" R φ5 (s) = H R φ4 (s)h 1 = Ad H R φ4 (s) s S 4 H : I 2 H 2 (10.2) H 1 = I 2 1 2 H 2 I 2 :Identity matrix of order 2 H 2 :Hadamard matrix of order 2
C s class [E] = I 0 0 1 0 4 = R φ5 (e) R φ5 (S 4 ) S 2 C s Normal subgroup [σ] 0 0 1 0 0 0 0 1 Tab.6.2
T d class [E] = I 0 0 1 0 4 = R φsss (e) R φsss (s)s S 4 T d 0 1 0 0 [C 2 ] 0 0 1 0 0 0 0 1 [C ] 0 1 0 0, 0 0 1 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0 1 Normal subgroup V K abelian A 4 [σ d ] [S 4 ] 0 1 0 0, 0 0 1 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0 1 0 1 0 0, 0 0 1 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0 1 Tab.7 R φ5 (s) P2 R (s) = R φsss φ 5 (s) R φ5 (s)s S 4 (11.2) ʘ: Hadamard product sign
R φsss (S 4 ) {e} = I 0 0 1 0 4 Tab.7.2
step7 12 φ sss (11) s n φ sss (11) φ o (12) o 1 o 2 φ o = o = o 4 o 1 (α, β, γ, δ) o 12 1 (α, β, γ, δ) o 2 (α, β, γ, δ) o = φ o (α, β, γ, δ) o φ sss : = 6 2 (α, β, γ, δ) o (α, β, γ, δ) (12) o 4 (α, β, γ, δ) o 4 (α, β, γ, δ) α + β + γ + δ (α β + γ δ) 2 + (α + β γ δ) 2 + (α β γ + δ) 2 [{(α β + γ δ) 2 + ω(α + β γ δ) 2 + ω 2 (α β γ + δ) 2 } + {(α β + γ δ) 2 + ω 2 (α + β γ δ) 2 + ω(α β γ + δ) 2 } ] [{(α β + γ δ) 2 + ω(α + β γ δ) 2 + ω 2 (α β γ + δ) 2 } {(α β + γ δ) 2 + ω 2 (α + β γ δ) 2 + ω(α β γ + δ) 2 } ] 2 (12.2) ω: ω 2 + ω + 1 = 0 φ o (12.2) {o 1, o 2, o, o 4 }(1) {s 1, s 2, s, s 4 } Fig.6 (1 ) Change of variables by eq.(1) o n the fffffffffff theeeee oo ø n o n (s 1, s 2, s, s 4 ) ef ssssetesc pefsfessff Fig.6
ø 1(s 1, s 2, s, s 4 ) = o 1 = α + β + γ + δ ø 2 (s 1, s 2, s, s 4 ) = o 2 = (α β + γ δ) 2 + (α + β γ δ) 2 + (α β γ + δ) 2 ø (s 1, s 2, s, s 4 ) = o = {(α β + γ δ) 2 + ω(α + β γ δ) 2 + ω 2 (α β γ + δ) 2 } +{(α β + γ δ) 2 + ω 2 (α + β γ δ) 2 + ω(α β γ + δ) 2 } (1) ø 4 (s 1, s 2, s, s 4 ) = o 4 = [{(α β + γ δ) 2 + ω(α + β γ δ) 2 + ω 2 (α β γ + δ) 2 } {(α β + γ δ) 2 + ω 2 (α + β γ δ) 2 + ω(α β γ + δ) 2 } ] 2 ø 1 = s 1 ø 2 = ø 2 1 8s 2 (1.2) ø = ø 2 c d ø 4 = ø 2 4c c = ø 2 2 1 ø 4 2 2 + 1 (s 4 1 4 16s 2 1 s 2 + 64s 1 s 256s 4 ) (1.) d = ø 2 27(s 1 4s 1 s 2 + 8s ) 2 (1) (1) φ sss (11) (1 2) (1) (1) (1) (1)(1.2), (1.2) (1 4) (1 ) s 1 a 1 = s2 = a 2 s = a a n Q (1) s 4 = a 4 O n ø n (s n = a n ) b 1 c(s n = a n ) (14) b 2 d (s n = a n ) (1 ) (14)(1 5) (14) (1 5) {α, β, γ, δ} (1 5) (1) (1)
a 1 = a α, β, γ, δ 2 = a = a 4 = α + β + γ + δ αβ + αγ + αδ + βγ + βδ + γδ αβγ + αβδ + αγδ + βγδ αβγδ 1 O 4 1 + 1 O 2 + 1 O 2 + O 4 1 O 4 1 1 O 2 + 1 O 2 + O 4 = 1 O 4 1 + 1 O 2 + 1 O 2 + O 4 1 O 4 1 1 O 2 + 1 O 2 + O 4 + 1 O 2 O 4 + 1 O 2 + ω 1 2 O + O 4 + 1 O 2 + ω 2 1 O 2 + O 4 + 1 O 2 O 4 + 1 O 2 + ω 1 2 O + O 4 1 O 2 + ω 2 1 O 2 + O 4 + 1 O 2 O 4 1 O 2 + ω 1 2 O + O 4 1 O 2 + ω 2 1 O 2 + O 4 + 1 O 2 O 4 1 O 2 + ω 1 2 O + O 4 + 1 O 2 + ω 2 1 O 2 + O 4 + ω 2 1 O 2 O 4 + ω 1 O 2 O 4 + ω 2 1 O 2 O 4 + ω 1 O 2 O 4 + ω 2 1 O 2 O 4 + ω 1 O 2 O 4 + ω 2 1 O 2 O 4,, + ω 1 O 2 O 4 ω: ccccc rrrr ω 2 + ω + 1 = 0 (15) O 1 = a 1 O 2 = O 2 1 8a 2 (15 2) O = O 2 b 1 b 2 O 4 = O 2 4b 1 b 1 = O 2 2 1 O 4 2 2 + 1 (a 4 1 4 16a 2 1 a 2 + 64a 1 a 256a 4 ) (15 ) b 2 = O 2 27(a 1 4a 1 a 2 + 8a ) 2,
step1 step6 9. L Λ P n O n Λ: = P n L Λ step1 step2step step4step5 step6 (1 6) O 2 H: = P 2 LH : = P LH (16) 2 : = P 2 LH" O H O H" 10. φ φ sss step1 step6 (17) T o : = O2 H" OH 2 OH (17) T o T sss step1 step6 Fig.6 Fig.6-2 S 4 Schönflies S Mulliken S 4 5 χ 4 S 1, χ 4 S 2, χ 4 S, χ 4 S 4, χ 4 5 Tab.8
φ ee.(2) T o O 2 H O H φ O2 H" 4 φ sss φ ee.(1) o ee.(5) ee.(8) ee.(11) ee.(12.2) φ 2 s 1 SSSSSSSS of ee.(1) ø 2 1 8s 2 φ o = ø 2 c d ee. (16) ø 2 4c ee. (1.2) VVVVVV tttttttttttttt O 2 H O H O H" φ 2 φ 4 S R φ4 (S 4 ) Tab.5.2 φ R φ (S 4 ) S 4 R φ2 (S 4 ) Tab.1. Tab..2 R φ2 (A 4 ) A R φ (S 4 ) D φ (V K ) R φ2 (S 4 ) R φ2 (A 4 ) MMMMMM rrrrrrrrrrrrrr eeeeeeeee kerr φ2 V K oos 4 kerr φ4 A 4 oos 4 R φ (S 4 ) R φ(s 4 ) R φ (V 4 ) R φ 2 (S 4 ) R φ 2 (S 4 ) R φ2 (A 4 ) R φ 4 (S 4 ) 2 φ sss S 2 R φsss (S 4 ) Tab.7.2 = {E 4 } 1 1 1 1 1 1 1 1 Λ(R φ (V K )) H = = H 1 1 1 1 4 H 4 4Hadamard () 1 1 1 1 H 1 = 1 4 H 4 I 1 : 1 0 1 1 1 Λ(R φ2 (A 4 )) H = 0 1 ω ω 2 = I 1 H H : Hadamard (6) 0 1 ω 2 ω ω: ω 2 + ω + 1 = 0 H 1 = I 1 1 H Λ R φ4 (S 4 ) H = = I 0 0 1 1 2 H 2 I 2: 2 (9) H 2 : 2 Hadamard 0 0 1 1 H 1 = I 2 1 H 2 2 Fig.6
φ R φ (S 4 ) T d Tab.1. O 2 H φ 2 R φ2 (S 4 ) C v Tab..2 O H φ 4 R φ4 (S 4 ) C s Tab.5.2 O2 H" φ sss R φsss (S 4 ) {e} Tab.7.2 0 0 0 0 0 0 0 0 Td 0 0 0 0 0 0 0 0 R φ (S 4 ) L H P 2 0A 1 0 0 0 0 0 0 0 0 0 Cv 0 0 0 0 0 0 R φ2 (S 4 ) L H P 0A 1 0 0 0 0 0A 0 0 1 0 0 0 0 Cs 0 0 0 0 R φ4 (S 4 ) L H" P 2 0A 1 0 0 0 0 0A 1 0 0 0 0 A0 1 0 0 0 0A 1 0 R φsss (S 4 ) = 4(A 1 ) 0A 1 0 0 0 0 0 0 0 0 0 T0 1 0 0 0 0 0 R φ1 (S 4 ) 0A 1 0 0 0 0 A0 1 0 0 0 0 0 0 E 0 0 0 0 R φ (S 4 ) 0A 1 0 0 0 0 0A 1 0 0 0 0 A0 1 0 0 0 0A 2 0 R φ5 (S 4 ) = (A 1 ) (T 1 ) = 2(A 1 ) (E) = (A 1 ) (A 2 ) (A 1 ) = {(1)} (A 2 ) = {(1), ( 1)} (E) = 1 0 1 0 ω2, 0, ω 0 1 1 0 0 ω2, 0 ω 0, 0 0 ω ω2, 0 ω ω 2 0 (T 1 ) = 1 1 0 1 0, 0 0 1, 1 0 0, 0 0 1, 1 0 0, 0 1 0, 0 0 1 1 0 0 1 1 0 0 1 1 0 1 0, 0 0 1, 1 0 0, 0 0 1, 1 0 0, 0 1 0, 0 0 1 1 0 0 1 1 0 0 1 1 0 1 0, 0 0 1, 1 0 0, 0 0 1, 1 0 0, 0 1 0, 0 0 1 1 0 0 1 1 0 0 1 1 0 1 0, 0 0 1, 1 0 0, 0 0 1, 1 0 0, 0 1 0 0 0 1 1 0 0 1 1 0 0 Fig.6.2
character Irreducible representation of S 4 martix groups χ 1 S 4 = (1 1 1 1 1) R χ1 S4 (S 4) = {(A 1 )} χ 2 S 4 = (1 1 1 1 1) R χ2 S4 (S 4) = {(A 2 )} χ S 4 = (2 2 1 0 0) R χ S4 (S 4) (E) χ 4 S 4 = ( 1 0 1 1) R χ4 S4 (S 4) (T 1 ) χ 5 S 4 = ( 1 0 1 1) R χ5 S4 (S 4) χ S 4(ψ) = (trr ψ (E) trr ψ (C ) trr ψ (C 2 ) trr ψ (S 4 ) trr ψ (σ d )) S 4 T d = {E, 8C, C 2, 6S 4, 6σ d } ψ C n [Φ] Φ = {α, β, γ, δ} χ = {ψ C n [Φ]χ S 4(ψ) = χχ C 5 } Tab.8
2. 1. (2 1) f(x) = x 4 + ox + px 2 + qq + r = 0o, p, q, r C(2 1) f(x) (2 1) f(x) = (x α)(x β)(x γ)(x δ) (2 2) Fundamental theorem of algebra α, β, γ, δ {ρ C f(ρ) = 0} (2 2) (2 1) (2 ) o = p = q = r = (α + β + γ + δ) αβ + αγ + αδ + βγ + βδ + γδ (αβγ + αβδ + αγδ + βγδ ) αβγδ Relation between coefficients and roots of a polynomial (2 ) (2 «) 4Viéta» «4 Viéta»(2 ) (2 1) (2 ) 1. (1.2)(1)(1.1)(1.2) (2 4) o = s 1 = α + β + γ + δ p = s 2 = αβ + αγ + αδ + βγ + βδ + γδ q = s = αβγ + αβδ + αγδ + βγδ r = s 4 = αβγδ (2.2) (2 )(2.2)(2 4.4) (2 5) (2 4) (2 6) (2 7) (2 1) (2 7)
o 1(α, β, γ, δ) = ø 1 (s 1, s 2, s, s 4 ) o 2 (α, β, γ, δ) = ø 2 (s 1, s 2, s, s 4 ) (2 4) o (α, β, γ, δ) = ø (s 1, s 2, s, s 4 ) o 4 (α, β, γ, δ) = ø 4 (s 1, s 2, s, s 4 ) s 1 s 2 s s4 = α + β + γ + δ = αβ + αγ + αδ + βγ + βδ + γδ = αβγ + αβδ + αγδ + βγδ = αβγδ (2 4.2) o 1 = α + β + γ + δ o 2 = (α β + γ δ) 2 + (α + β γ δ) 2 + (α β γ + δ) 2 o = {(α β + γ δ) 2 + ω(α + β γ δ) 2 + ω 2 (α β γ + δ) 2 } +{(α β + γ δ) 2 + ω 2 (α + β γ δ) 2 + ω(α β γ + δ) 2 } (2 4.) o 4 = [{(α β + γ δ) 2 + ω(α + β γ δ) 2 + ω 2 (α β γ + δ) 2 } {(α β + γ δ) 2 + ω 2 (α + β γ δ) 2 + ω(α β γ + δ) 2 } ] 2 ø 1 = s 1 ø 2 = s 2 1 8s 2 (2 4.4) ø = ø 2 c d ø 4 = ø 2 4c c = ø 2 2 1 ø 4 2 2 + 1 (s 4 1 4 16s 2 1 s 2 + 64s 1 s 256s 4 ) (2 4.4.2) d = ø 2 27(s 1 4s 1 s 2 + 8s ) 2 Ø n ø n (s 1 = o, s 2 = p, s = q, s 4 = r) C c(s 1 = o, s 2 = p, s = q, s 4 = r) (2 5) D d (s 1 = o, s 2 = p, s = q, s 4 = r) Ø 1 = α + β + γ + δ Ø 2 = (α β + γ δ) 2 + (α + β γ δ) 2 + (α β γ + δ) 2 Ø = {(α β + γ δ) 2 + ω(α + β γ δ) 2 + ω 2 (α β γ + δ) 2 } +{(α β + γ δ) 2 + ω 2 (α + β γ δ) 2 + ω(α β γ + δ) 2 } (2 6) Ø 4 = [{(α β + γ δ) 2 + ω(α + β γ δ) 2 + ω 2 (α β γ + δ) 2 } {(α β + γ δ) 2 + ω 2 (α + β γ δ) 2 + ω(α β γ + δ) 2 } ] 2
{α, β, γ, δ} = {ρ C f(ρ) = ρ 4 + oρ + pρ 2 + qρ + r = 0} 1 Ø 4 1 + 1 Ø 2 + 1 Ø 2 + Ø 4 1 Ø 4 1 1 Ø 2 + 1 Ø 2 + Ø 4 = 1 Ø 4 1 + 1 Ø 2 + 1 Ø 2 + Ø 4 1 Ø 4 1 1 Ø 2 + 1 Ø 2 + Ø 4 + 1 Ø 2 Ø 4 + 1 Ø 2 + ω 1 2 Ø + Ø 4 + 1 Ø 2 + ω 2 1 Ø 2 + Ø 4 + 1 Ø 2 Ø 4 + 1 Ø 2 + ω 1 2 Ø + Ø 4 1 Ø 2 + ω 2 1 Ø 2 + Ø 4 + 1 Ø 2 Ø 4 1 Ø 2 + ω 1 2 Ø + Ø 4 1 Ø 2 + ω 2 1 Ø 2 + Ø 4 + 1 Ø 2 Ø 4 1 Ø 2 + ω 1 2 Ø + Ø 4 + 1 Ø 2 + ω 2 1 Ø 2 + Ø 4 + ω 2 1 Ø 2 Ø 4 + ω 1 Ø 2 Ø 4 + ω 2 1 Ø 2 Ø 4 + ω 1 Ø 2 Ø 4 + ω 2 1 Ø 2 Ø 4 + ω 1 Ø 2 Ø 4 + ω 2 1 Ø 2 Ø 4,, + ω 1 Ø 2 Ø 4 ω: ccccc rrrr ω 2 + ω + 1 = 0 (2 7) Ø 1 = o Ø 2 = Ø 2 1 8p (2 7.2) Ø = Ø 2 C D Ø 4 = Ø 2 4C, C = Ø 2 2 C 2 D = Ø 2 27D 2 (2 7.) 2 C 2 = 1 Ø 4 2 2 + 1 4 (o4 16o 2 p + 64oo 256r) (2 7.4) D 2 = o 4oo + 8q
f(x) (2 1) (2 7) Tab.2-1 Tab.2-1 ΔΔ Δ (2 1) f(x) f(x) = x 4 + ox + px 2 + qq + r = 0(2 1) Δ = {(α β)(α γ)(α δ)(β γ)(β δ)(γ δ)} 2 : f(x) Ø 4 = Ø 2 4C = 4 6 Δ f (x) = 4x + ox 2 + 2pp + q(2 8) Δ = 1 4 ( o q + 2 2 o 2 p 2 + 2 2 ooo 2 5 p 2 2 q 2 ) f (x) D = Ø 2 27D 2 2 = 4 5 Δ f (x) = 12x 2 + 6oo + 2p(2 9) Δ = 1 12 (o2 8p) f (x) Ø 2 = Ø 2 1 8p = 12Δ f(x) Δ Δ f(x)
Multiplicity Discriminant Roots of f(x) double Ø 4 = 0 triple Ø 4 = 0 Ø = 0 two double Ø 4 = 0 Ø = 2Ø 2 quadruple Δ = 0 C = 0 Δ = 0 or C = 0 Δ = 0 C = 0 D = 0 C 2 = 0 D 2 = 0 1 Ø 4 1 + 1 Ø 2 + 2 1 Ø 2 + 2 1 Ø 2 1 Ø 2, 1 Ø 4 1 1 Ø 2 + 2 1 Ø 2, 1 Ø 4 1 + 1 Ø 2 + 2 1 Ø 2 2 1 Ø 2 1 Ø 2, 1 Ø 4 1 1 Ø 2 + 2 1 Ø 2 1 Ø 4 1 + 1 Ø 2, 1 Ø 4 1 1 Ø 2 1 4 Ø 1 1 Ø 2 1 Ø 4 1 1 Ø 2 1 Ø 4 1 + Ø 2, 1 Ø 4 1 Ø 2 1 Ø 4 1 Ø 2 1 Ø 4 1 + Ø 2 Ø 4 = 0 Ø = 0 Ø 2 = 0 Δ = 0 Δ = 0 Δ = 0 Δ: Discriminant of f(x) Δ : Discriminant of f (x) Δ : Discriminant of f (x) 1 Ø 4 1 1 Ø 4 1 1 Ø 4 1 1 Ø 4 1 C = (o 2 8p) 2 C 2 C 2 = 1(o2 8p) 2 + 1 (o4 16o 2 p + 64oo 256r) 4 4 D 2 = o 4oo + 8q Tab.2-1
Reference