from solving polynomial equations to groups

8132019 From Solving Polynomial Equations to Groups

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Math55a (Fall 2011) Yum-Tong Siu 1

FROM SOLVING CUBIC AND QUARTIC EQUATIONS

TO INTRODUCTION OF GROUPS

These notes start with the discussion of classical formulas for solvingquadratic cubic and quartic equations and explain the historic role whichgroups play in the study of formulas for solving polynomial equations Wewill consider polynomial equations with coefficients in C but the discussionworks also for Q and Q with some obvious modifications Before talkingabout the role of groups we derive here the classical formulas for solvingquadratic cubic and quartic equations

Quadratic Equation We first look at the well-known formula for solving thequadratic equation obtained by completion of squares Given a quadraticequation ax2 + bx + c = 0 we can complete the square on the right-hand sideby writing

ax2 + bx + c = a

1048616x +

b

2a

10486172

minus b2 minus 4ac

4a

so that the two roots x1 x2 of ax2 + bx + c = 0 can be obtained from

a

1048616x +

b

2a

10486172

= b2 minus 4ac

4a

giving

x1 x2 = minusb plusmnradic

b2 minus 4ac

2a

The use of the completion of squares seems to be a very ingenious trickActually a more straightforward way of obtaining the formula is to use anew variable t which is related to the original variable x by translation withx = t + γ for some γ to be determined so that the quadratic equation ax2 +bx + c = 0 in terms of the new variable t would not have the degree-one termThis is done by expanding

ax2

+ bx + c = a (t + γ )2

+ b (t + γ ) + c = at2

+ 2aγt + aγ 2

+ bt + bγ + c

so that we can set 2aγ + b = 0 which is the same as γ = minus b2a

With thischoice of γ

ax2 + bx + c = at2 + aγ 2 + bγ + c = at2 + b2

4a minus b2

2a + c = at2 minus b2 minus 4ac

4a




giving again upon expressing t in terms of x the same formula

ax2 + bx + c = a

1048616x +

b

2a

10486172

minus b2 minus 4ac

4a

which we obtained earlier by the seemingly very ingenious trick of completionof squares

Cubic Equation Next we look at the classical solution for solving the cubicequation ax3 + bx2 + cx + d = 0 We can use the change of variables bytranslation x = t + γ to get rid of the second-highest-degree term with a judicious choice of the number γ just as in the case of the quadratic equation

ax3 + bx2 + cx + d = a (t + γ )3 + b (t + γ )2 + c (t + γ ) + d

= a852008

t3 + 3γt2 + 3γ 2t + γ 3852009

+ b852008

t2 + 2γt + γ 2852009

+ c (t + γ ) + d

= at3 + (3aγ + b) t2 +852008

3aγ 2 + 2bγ + c852009

t +852008

aγ 3 + bγ 2 + cγ + d852009

Note that this formula is the same as the Taylor expansion of the functionax3 + bx2 + cx + d in terms of t = x minus γ so that

(i) the 0-degree term aγ 3 + bγ 2 + cγ + d is the value of the function ax3 +bx2 + cx + d at x = γ

(ii) the coefficient of t is the value of the first derivative 3aγ 2 + 2bγ + c of the function ax3 + bx2 + cx + d at x = γ

(iii) the coefficient 3aγ + b of t2 is 12

times the value of the second derivativeof the function ax3 + bx2 + cx + d at x = γ and

(iv) the coefficient a of t3 is 13

times the value of the third derivative of thefunction ax3 + bx2 + cx + d at x = γ

We should set 3aγ + b = 0 so that γ = minus b3a

Dividing by a we end up with

t3 + pt + q = 0

where

p = 3aγ 2 + 2bγ + c

a =

3a852008minus b

3a

8520092+ 2b

852008minus b3a

852009+ c

a




q =

aγ 3 + bγ 2 + cγ + d

a =

a 852008minus b3a852009

3+ b 852008minus

b3a852009

2+ c 852008minus

b3a852009+ d

a

Unlike the case of the quadratic equation for the cubic equation this specialform t3 + pt + q = 0 is not of much help So far for both the quadraticequation and the cubic equation the tool which we have introduced is thechange of variable by translation This tool up to this point has been usedto eliminate the second-highest-order term For the cubic equation this maynot be the best way to utilize this tool of change of variable by translation

It turns out that the best way to use this tool is to trade the vanishing of

the second-highest-order term for some form of symmetry We start out with

the special form t3 + pt + q = 0 and make a change of variable by translationwith the amount of translation to be determined So we set t = u + v whereu is the new variable to replace t and v is the amount of translation to bedetermined Plugging t = u + v into t3 + pt + q = 0 we get

t3 + pt + q = (u + v)3 + p (u + v) + q = u3 + 3u2v + 3uv2 + v3 + p (u + v) + q

which we rewrite as

t3 + pt + q =

852008u3 + v3

852009+ (u + v) (3uv + p) + q

to highlight the symmetric functions u + v uv u3 + v3 in the two variablesof degrees 1 2 and 3 respectively

We are going to choose the amount of translation u so that the resultingequation is as simple as possible (ie with as many vanishing terms aspossible) and at the same time retain the symmetry in u and v For thatpurpose we set 3uv + p = 0 and get the following two equations

(lowast)

u3 + v3 + q = 0

3uv + p = 0

in the two unknowns u and v These two equations are symmetric in u

and v This is what we mean by using this tool of change of variable bytranslation to trade the vanishing of the second-highest-order term for some

form of symmetry This trade holds the key to the derivation of the classicalformula for the solution of the cubic equation




We now take the cube of the second equation of () to get

(lowastlowast)

u3 + v3 + q = 0

u3v3 = minus p3

27 = 0

so that from the data on the product and sum of the two unknowns u3 andv3 we can solve them from the quadratic equation

X 2 + qX minus p3

27 = 0

to get

u3 v3 = minusq

2 plusmn991770

q 2

4 +

p3

27

By taking the cubic roots of u3 v3 we get nine possible solutions but usingthe second equation 3uv + p = 0 of (lowast) we narrow the number down to only3 because a choice of a cubic root of u3 forces us to choose a cubic root of v3 by 3uv + p = 0 The final formula for the three roots of x are u + v with

u = 3

1057306 minusq

2 +

991770 q 2

4 +

p3

27

v = 3

1057306 minusq

2 minus991770

q 2

4 +

p3

27

subject to 3uv + p = 0 in the choice of the cubic roots of u3 and v3 In otherwords

x1 x2 x3 = 3

1057306 minusq

2 +

991770 q 2

4 +

p3

27 +

p

3 3

991770 minus q

2 +radic

q2

4 + p3

27

Our later discussion for the solution of a polynomial equation of degreen will be formulated in terms of solving the equation

xn minus σ1xnminus1 + σ2xnminus2 minus + middot middot middot + (minus1)nminus1σnminus1x + (minus1)nσn = 0

to get the roots x1 middot middot middot xn when σ1 middot middot middot σn are the n elementary symmetricfunctions of x1 middot middot middot xn In order to fit the classical formula for the solution




of the cubic equation into the general discussion of such a formulation we

rewrite as cubic equation ax3 + bx2 + cx + d = 0 as x3minusσ1x2 + σ2xminus σ3 = 0In other words a = 1 b = minusσ1 c = σ2 and d = minusσ3 Thus we have theformula

p = minusσ2

1

3 + σ2

q = minus2σ31

27 + σ1σ2

3 minus σ3

In terms of σ1 σ2 σ3 the three roots x1 x2 x3 of x3 minus σ1x2 + σ2x minus σ3 = 0are given by

minus1

21048616minus2σ3

1

27 +

σ1σ2

3 minus σ31048617+1057306 1

41048616minus2σ3

1

27 +

σ1σ2

3 minus σ310486172

+

1

271048616minusσ

21

3 + σ210486173

1

3

+ p

3

minus1

2

1048616minus2σ3

1

27+

σ1σ2

3minus σ3

1048617+

1057306 1

4

1048616minus2σ3

1

27+

σ1σ2

3minus σ3

10486172

+ 1

27

1048616minusσ

2

1

3+ σ2

10486173

minus

1

3

Quartic Equation Now we look at the classical solution for solving the quarticequation x4minusσ1x3+σ2x2minusσ3x+σ4 = 0 where σ1 σ2 σ3 σ4 are the elementarysymmetric functions of its roots x1 x2 x3 x4 In order to keep track of theaction of permutations on the four roots

x1 x2 x3 x4

we introduce the

following linear change of variables

y0 = 12 (x1 + x2 + x3 + x4)

y1 = 12 (x1 minus x2 + x3 minus x4)

y2 = 12 (x1 + x2 minus x3 minus x4)

y3 = 12 (x1 minus x2 minus x3 + x4)

The reason for this change of variables is that when x1 x2 x3 x4 is permuted

a new variable y j (for j = 1 2 3) is changed to plusmnyk for some k = 1 2 3because each of the last three new variables y1 y2 y3 is defined by partitioningthe four old variables x1 x2 x3 x4 into two pairs and use +1 as the coefficientfor one of the two pairs and use minus1 as the coefficient for the other pair Thisproperty of the new variables y0 y1 y2 y3 makes it very easy to keep track of which monomials (y0) j0 (y1) j1 (y2) j2 (y3) j3 are invariant under a given subset




of the set of all permutations of

x1 x2 x3 x4

The reason why we use the

coefficient 12 in the definition of the new variables is that when the linearchange of variables is put into a matrix form

y0

y1

y2

y3

= A

x1

x2

x3

x4

where A is the 4 times 4 matrix

1

2

1 1 1 11 minus

1 1 minus

11 1 minus1 minus11 minus1 minus1 1

the matrix A is symmetric with A2 equal to the identity matrix In otherwords the matrix A is an orthogonal matrix

The monomials (y1)2 (y2)2 (y3)2 are permutated among themselves byany permutation of x1 x2 x3 x4 Let τ 1 τ 2 τ 3 be the three elementary sym-metric functions of (y1)2 (y2)2 (y3)2 Then τ 1 τ 2 τ 3 are invariant under anypermutation of x1 x2 x3 x4 and therefore can be explicitly expressed poly-nomials of the elementary symmetric functions σ1 σ2 σ3 σ4 of x1 x2 x3 x4

Actually these polynomials are independent of σ1 by degree considerationsbut this point is not important to our discussion By the formula for solvingcubic equations we can express (y1)2 (y2)2 (y3)2 in terms of τ 1 τ 2 τ 3 whichin turn are polynomials of σ1 σ2 σ3 σ4 By taking square roots we haveformulas for y1 y2 y3 in terms of σ1 σ2 σ3 σ4 Since y0 = 1

2σ1 we can now

obtain x1 x2 x3 x4 in terms of y1 y2 y3 y4 This concludes our derivation of the formula to solve a quartic equation

Schematics for Solving Polynomial Equations We now discuss the solution of the polynomial equation


where the coefficients σ1 middot middot middot σn are the n elementary symmetric functionsof its n roots x1 middot middot middot xn The rule for the formula of the solution is that onlyrational functions and radicals are allowed in the formula The formula ex-presses x1 middot middot middot xn in terms of its elementary symmetric functions σ1 middot middot middot σn




by using a finite number of steps each of which is either the formation of

rational functions or the formation of radicals What we start out with isthe functions σ1 middot middot middot σn of the variables x1 middot middot middot xn each of which is fullysymmetric under the set of all permutations of x1 middot middot middot xn The final re-sult is x1 middot middot middot xn each of which possesses no symmetry at all with respectto the set of all permutations of x1 middot middot middot xn In between each step yieldssome functions of x1 middot middot middot xn each of which may be partially symmetric withrespect to the permutations of x1 middot middot middot xn that is invariant under a certainsubset of the set of all permutations of x1 middot middot middot xn We will discuss about theuse of such partial symmetries

The schematics for the formula of solving a polynomial equation are as

follows Let τ (0)1 middot middot middot τ (0)m0 be σ1 middot middot middot σn We allow two kinds of processesOne is the use of rational functions and the other is the use of radicalsFrom τ

(0)1 middot middot middot τ

(0)m0

we use one of the two processes to get another set of

rational functions τ (1)1 middot middot middot τ

(1)m1 of x1 middot middot middot xn Then in the next step from

τ (1)1 middot middot middot τ

(1)m1

we use one of the two processes to get another set of rational

functions τ (2)1 middot middot middot τ

(2)m2

of x1 middot middot middot xn We keep on doing this until we reach

the final step where the functions τ (ℓ)1 middot middot middot τ

(ℓ)mℓ

are simply x1 middot middot middot xn Werepresent this in the following diagram

σ1

middot middot middot σn

darrτ (1)1 middot middot middot τ (1)m1


darrmiddotmiddotmiddot

darrτ (ℓminus1)

1 middot middot middot τ (ℓminus1)mℓminus1

darrx1 middot middot middot xn




When we go from τ ( j)1

middot middot middot τ

( j)mj to τ

( j+1)1


( j+1)mj+1

in

τ ( j)1 middot middot middot τ ( j)mj

darrτ ( j+1)1 middot middot middot τ ( j+1)

mj+1

we use only one of the two kinds of processes

Use of Partial Symmetries We are using two processes in the formula of solving polynomial equations One is the formation of rational functions andthe other is the formation of radicals We are going to change our schematics

by suppress the processes involving only the formation of rational functionsLet F (x1 middot middot middot xn) be the set of all rational functions of x1 middot middot middot xn with coef-ficients in F where F is C For τ 1 middot middot middot τ m isin F (x1 middot middot middot xn) let F (τ 1 middot middot middot τ m)be the subset of F (x1 middot middot middot xn) consisting of all elements which can be writtenas rational functions of τ 1 middot middot middot τ m We now redo our schematics as follows

F (σ1 middot middot middot σn) = F 983080

τ (0)1 middot middot middot τ (0)m0

983081darr

F

983080τ (1)1 middot middot middot τ (1)m1983081darr

F 983080


983081darrmiddotmiddotmiddotdarr

F 983080

τ (ℓminus1)1 middot middot middot τ (ℓminus1)

mℓminus1

983081

darrF (x1 middot middot middot xn) = F

983080τ (ℓ)1 middot middot middot τ (ℓ)mℓ

983081

where τ (ν ) j is a rational function of x1 middot middot middot xn (for 1 le mν ) and

983080τ (ν ) j

983081κνjbelongs to F

983080τ (ν minus1)1 middot middot middot τ

(ν minus1)mνminus1

983081 for some positive integer κνj Note that




κνj = 1 is allowed The process of the formation of rational functions is in-

corporated into the use of the set F 983080

τ (ν )1 middot middot middot τ (ν )mν

983081of all rational functional

functions of τ (ν )1 middot middot middot τ

(ν )mν The process of formation of radicals is expressed

by 983080τ (ν ) j

983081κνj isin F 983080

τ (ν minus1)1 middot middot middot τ (ν minus1)

mνminus1

983081for some positive integer κνj

It turns out that in writing down the schematics for the classical formulas

for the roots of the cubic and quartic equations the set F 983080

τ (ν minus1)1 middot middot middot τ


983081is equal to the subset of F (x1

middot middot middot xn) consisting of all elements which are

invariant under a certain subset E of the set S n of all permutations of the nelements x1 middot middot middot xn) We introduce the notation

F 983080


983081= F (x1 middot middot middot xn)E

to express this fact The set S n of all permutations of the n elementsx1 middot middot middot xn) is the same as the set of all self-bijections of x1 middot middot middot xn) Forsuch bijections we have the law of composition of maps and we can take theinverse of any such map and also we have the identity map The associativelaw holds when we compose maps In general when we have a set endowedwith a law of composition of its elements such that the associative law holdsand the inverse of every element and an identity element exit we call such aset a group Any subset of a group which is closed under the law of compo-sition is called a subgroup The group S n is called the symmetric group on n

elements and is also called the permutation group on n elements When wehave

F 983080



by replacing E by the smallest subgroup of S n containing E we can assumewithout loss of generality that E is a subgroup of S n In the case of theclassical formulas for the roots of the cubic and quartic equations (and also

for the quadratic equation) we have

F 983080


983081= F (x1 middot middot middot xn)Gν

for subgroups Gν of S n with the inclusion relations

1 = Gℓ sub Gℓminus1 sub middot middot middot sub G1 sub G0 = S n




where

1

here denotes the group consisting of only one element which is

its identity element Sometimes the group 1 of only one element is alsodenoted simply by 1 (or 0 or 0 in conjunction with other groups whichare commutative that is satisfy the commutative law) We now describeexplicitly the subgroups


used in the formulas for the roots of the quadratic the cubic and the quarticequations

Subgroups for Quadratic Equation For the formula for the roots of thequadratic equation we have ℓ = 1 and

1 = G1 sub G0 = S 2

andF (σ1 σ2)

darrF (x1 x2) = F

983080τ (1)1 τ

(1)2

983081where

τ (1)1 = σ1 983080τ

(1)2 983081

2

= σ21 minus 4σ2

so that κ11 = 1 and κ12 = 2 The roots x1 x2 are given by

x1 x2 = τ

(1)1 plusmn τ

(1)2

2

Subgroups for Cubic Equation Let A3 be the subgroup of S 3 consisting of allpermutations of x1 x2 x3 with sign +1 that is all cyclic permutations Ingeneral An is the subgroup of S n consisting of all permutations of x1 middot middot middot xnwith sign +1 The group An is called the alternating group on n elementsFor the formula for the roots of the cubic equation we have ℓ = 2 and

1

= G2

subG1

subG0 = S 3

with G1 = A3 Before we describe τ (ν ) j and κνj we first recall the following

notations

p = minusσ21

3 + σ2

q = minus2σ31

27 + σ1σ2

3 minus σ3




from from the classical formula for solving the cubic equation We would like

to introduce some new notations too Let ϵ = minus12 +

radic

32 i so that ϵ is the cubicroot of unity satisfying ϵ3 = 1 Let

y1 = x1 + x2 + x3

y2 = x1 + ϵx2 + ϵ2x3

y3 = x1 + ϵ2x2 + ϵx3

so that the effect of elements of S 3 sends y2 y3 to

983163ϵ jy2 ϵky3

983165 for some

0

le j k

le2 making the tracking of the invariance of polynomials in y1 y2 y3

rather simple The monomial y2y3 and the polynomials (y2)3 + (y3)3 areinvariant under S 3 and can therefore be expressed in terms of σ1 σ2 σ3Explicitly the expressions are

y2y3 = (σ1)2 minus 3σ2 = minus3 p

(y2)3 + (y3)3 = 2 (σ1)3 minus 9σ1σ2 + 27σ3 = minus27q

so that

983080y2

3 9830813

983080y3

3 9830813

= minus p3

27

983080y2

3 9830813

+

983080y3

3 9830813

= minusq

which yields 983080y2

3

9830813983080y3

3

9830813= minusq

2 plusmn991770

q 2

4 +

p3

27

as roots of the quadratic equation

X 2 + qX minus p3

27 = 0

We now can describe τ (ν ) j and κνj in

F (σ1 σ2 σ3)

darrF 983080

τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4

983081darr

F (x1 x2 x3) = F 983080

τ (2)1 τ

(2)2 τ

(2)3

983081




We define

τ (1)1 = σ1 τ

(1)2 = q τ

(1)3 =

991770 q 2

4 +

p3

27 τ

(1)4 = p

so that

τ (1)2 =

983080y2

3

9830813+983080y3

3

9830813and τ

(1)3 =

1

2

1048616983080y2

3

9830813minus983080y3

3

98308131048617

Note that the addition of τ (1)4 = p = minus1

3y2y3 to F

983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4

983081is to make sure that F 983080τ

(1)1 τ

(1)2 τ

(1)3 τ

(1)4 983081 contains enough elements of

F (x1 x2 x3) to make it equal to F (x1 x2 x3)A3 Since

F (x1 x2 x3) = F (σ1 p q )

it follows that we can use κ11 = κ12 = κ14 = 1 and κ13 = 2 with

983080τ (1)3

9830812=

q 2

4 +

p3

27 isin F (σ1 p q ) = F (x1 x2 x3)

SinceF (x1 x2 x3) = F (y1 y2 y3)

to finish the justification of the diagram as a description of the formula forsolving the cubic equation we need only observe that we can set τ (2) j = y j for

j = 1 2 3 with

τ (2)1 = τ

(1)1

983080τ (2)2

9830813=

1

2τ (1)2 + τ

(1)3

983080τ (2)3

9830813

= 1

2τ (1)2 minus τ

(1)3

so that κ21 = 1 and κ22 = κ23 = 3

Subgroups for Quartic Equation Let K 4 be the subgroup of S 4 consisting of the four elements 1 (12) (34) (13) (24) (14) (23) It is called the Klein 4-

group which is characterized abstractly as the group of 4 elements

1 a b c

with the following law of composition

a2 = b2 = c2 = 1 ab = ba = c bc = cb = a ac = ca = b

For the formula for the roots of the quartic equation we have ℓ = 3 and

1 = G3 sub G2 sub G1 sub G0 = S 4




with G1 = A4 and G2 = K 4 Before we describe τ (ν ) j and κνj we first recall

the following notations

y0 = 12 (x1 + x2 + x3 + x4)




Let τ 1 τ 2 τ 3 be the three elementary symmetric functions of (y1)2 (y2)2 (y3)2That is

τ 1 = (y1)2 + (y2)2 + (y3)2

τ 2 = (y1)2 (y2)2 + (y1)2 (y3)2 + (y2)2 (y3)2

τ 3 = (y1)2 (y2)2 (y3)2

Embedded inside this description for solution of the quartic equation is theformula for solving for (y1)2 (y2)2 (y3)2 in terms of their three elementarysymmetric functions τ 1 τ 2 τ 3 For this embedded formula for solution of the

cubic equation we need to introduce the corresponding definition

P = minusτ 21

3 + τ 2

Q = minus2τ 31

27 + τ 1τ 2

3 minus τ 3

which is obtained by replacing p qσ1 σ2 σ3 respectively by P Q τ 1 τ 2 τ 3 inthe formula for p and q in terms of σ1 σ2 σ3 Let

z 1 = (y1)2 + ϵ (y2)2 + ϵ2 (y3)2

z 2 = (y1)2 + ϵ2 (y2)2 + ϵ (y3)2

where as before ϵ = minus12

+radic 32

i is the cubic root of unity We are now ready




to describe τ (ν ) j and κνj in

F (σ1 σ2 σ3 σ4)

darrF 983080

τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5

983081darrdarr

F 983080

τ (2)1 τ

(2)2 τ

(2)3 τ

(2)4

983081darr

F (x1 x2 x3 x4) = F 983080

τ (3)1 τ (3)2 τ (3)3 τ (3)4983081

First we remark that τ 1 τ 2 τ 3 isin F (σ1 σ2 σ3 σ4) due to their invariance underS 4 As a consequence P Q isin F (σ1 σ2 σ3 σ4) from the explicit expressionsof P and Q in terms of τ 1 τ 2 τ 3 In particular

Q2

4 +

P 3

27 isin F (σ1 σ2 σ3 σ4)

Again as above from

z 1z 2 = (τ 1)2

minus3τ 2 =

minus3P

(z 1)3 + (z 2)3 = 2 (τ 1)3 minus 9τ 1τ 2 + 27τ 3 = minus27Q

so that 983080z 1

3

9830813 983080z 2

3

9830813

= minusP 3

27983080z 1

3

9830813+983080z 2

3

9830813= minusQ

which yields 983080z 1

3

9830813983080z 2

3

9830813

= minusQ

2 plusmn991770

Q2

4 +

P 3

27


X 2 + QX minus P 3

27 = 0

We define

τ (1)1 = σ1 τ

(1)2 = τ 1 τ

(1)3 = Q τ

(1)4 =

991770 Q2

4 +

P 3

27 τ

(1)5 = P




so that

τ (1)3 =

983080z 1

3

9830813+983080z 2

3

9830813and τ

(1)4 =

1

2

1048616983080z 1

3

9830813minus983080z 2

3

98308131048617

Note that the addition of τ (1)5 = P = minus1

3z 1z 2 to F

983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5

983081is to make sure that F

983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5

983081contains enough elements of

F (x1 x2 x3 x4) to make it equal to F (x1 x2 x3 x4)A4 We can use κ11 =κ12 = κ13 = κ15 = 1 and κ14 = 2 with

983080τ (1)4 983081

2

= Q2

4

+ P 3

27 isinF (x1 x2 x3 x4)

We defineτ (2)1 = σ1 τ

(2)2 = τ 1 τ

(2)3 = z 1 τ

(2)4 = z 2

Since 983080τ (2)3

9830813=

1

2τ (1)3 + τ

(1)4

983080τ (2)4

9830813=

1

2τ (1)3 minus τ

(1)4

it follows that we can use κ21 = κ22 = 1 and κ23 = κ24 = 3 Finally we let

τ (3)1 = y0 τ

(3)2 = y1 τ

(3)3 = y2 τ

(3)4 = y3

and κ31 = 1 and κ32 = κ33 = κ34 = 2 Then F (x1 x2 x3 x4) = F (y0 y1 y2 y3)

and983080

τ (3) j

983081κ3j is a C-linear combination of τ 1 z 1 z 2 and therefore belongs to

F 983080

τ (2)1 τ

(2)2 τ

(2)3 τ

(2)4

983081for j = 2 3 4

Comparison of Subgroups for Cubic Equation to those for Quartic Equation

For the cubic equation we have the chain of subgroups

1 sub A3 sub S 4

whereas for the quartic equation we have the chain of subgroups

1 sub

K 4 sub

A4 sub

S 4

The relation between these two chains of subgroups can be geometrically il-lustrated by the following picture of a quadrilateral with vertices P 1 P 2 P 3 P 4so that the line joining a pair of vertices intersects the line joining the re-maining pair of vertices intersects at a point Q j with j = 1 2 3 from differentchoices of the two disjoint pairs of vertices




Any permutation among the four vertices P 1 P 2 P 3 P 4 results in a permu-tation among the three points Q1 Q2 Q3 This defines a mapping S 4

rarr S 3

which respects the laws of composition (and is known as a group homomor-phism) The set of points of S 4 which is mapped to the identity of S 3 (thatis the kernel of the group homomorphism) is the Klein 4-group K 4 Whenthe group homomorphism S 4 rarr S 3 is restricted to the subgroup A4 of S 4 weget a homomorphism A4 rarr A3 whose kernel is the Klein 4-group K 4 Thesequence

1 rarr A3 rarr S 4

comes from the sequenceK 4 rarr A4 rarr S 4

by taking the quotient by K 4




giving again upon expressing t in terms of x the same formula

ax2 + bx + c = a

1048616x +

b

2a

10486172

minus b2 minus 4ac

4a

which we obtained earlier by the seemingly very ingenious trick of completionof squares

Cubic Equation Next we look at the classical solution for solving the cubicequation ax3 + bx2 + cx + d = 0 We can use the change of variables bytranslation x = t + γ to get rid of the second-highest-degree term with a judicious choice of the number γ just as in the case of the quadratic equation

ax3 + bx2 + cx + d = a (t + γ )3 + b (t + γ )2 + c (t + γ ) + d

= a852008

t3 + 3γt2 + 3γ 2t + γ 3852009

+ b852008

t2 + 2γt + γ 2852009

+ c (t + γ ) + d

= at3 + (3aγ + b) t2 +852008

3aγ 2 + 2bγ + c852009

t +852008

aγ 3 + bγ 2 + cγ + d852009

Note that this formula is the same as the Taylor expansion of the functionax3 + bx2 + cx + d in terms of t = x minus γ so that

(i) the 0-degree term aγ 3 + bγ 2 + cγ + d is the value of the function ax3 +bx2 + cx + d at x = γ

(ii) the coefficient of t is the value of the first derivative 3aγ 2 + 2bγ + c of the function ax3 + bx2 + cx + d at x = γ

(iii) the coefficient 3aγ + b of t2 is 12

times the value of the second derivativeof the function ax3 + bx2 + cx + d at x = γ and

(iv) the coefficient a of t3 is 13

times the value of the third derivative of thefunction ax3 + bx2 + cx + d at x = γ

We should set 3aγ + b = 0 so that γ = minus b3a

Dividing by a we end up with

t3 + pt + q = 0

where

p = 3aγ 2 + 2bγ + c

a =

3a852008minus b

3a

8520092+ 2b

852008minus b3a

852009+ c

a




q =

aγ 3 + bγ 2 + cγ + d

a =

a 852008minus b3a852009

3+ b 852008minus

b3a852009

2+ c 852008minus

b3a852009+ d

a






which we rewrite as

t3 + pt + q =

852008u3 + v3

852009+ (u + v) (3uv + p) + q



(lowast)

u3 + v3 + q = 0

3uv + p = 0








(lowastlowast)

u3 + v3 + q = 0

u3v3 = minus p3

27 = 0


X 2 + qX minus p3

27 = 0

to get

u3 v3 = minusq

2 plusmn991770

q 2

4 +

p3

27


u = 3

1057306 minusq

2 +

991770 q 2

4 +

p3

27

v = 3

1057306 minusq

2 minus991770

q 2

4 +

p3

27


x1 x2 x3 = 3

1057306 minusq

2 +

991770 q 2

4 +

p3

27 +

p

3 3

991770 minus q

2 +radic

q2

4 + p3

27









p = minusσ2

1

3 + σ2

q = minus2σ31

27 + σ1σ2

3 minus σ3


minus1

21048616minus2σ3

1

27 +

σ1σ2

3 minus σ31048617+1057306 1

41048616minus2σ3

1

27 +

σ1σ2

3 minus σ310486172

+

1

271048616minusσ

21

3 + σ210486173

1

3

+ p

3

minus1

2

1048616minus2σ3

1

27+

σ1σ2

3minus σ3

1048617+

1057306 1

4

1048616minus2σ3

1

27+

σ1σ2

3minus σ3

10486172

+ 1

27

1048616minusσ

2

1

3+ σ2

10486173

minus

1

3


x1 x2 x3 x4

we introduce the


y0 = 12 (x1 + x2 + x3 + x4)










x1 x2 x3 x4



y0

y1

y2

y3

= A

x1

x2

x3

x4


1

2

1 1 1 11 minus

1 1 minus





2σ1 we can now













(0)m0





(1)m1



(2)m2



(ℓ)mℓ


σ1





darrτ (ℓminus1)








( j)mj to τ

( j+1)1


( j+1)mj+1

in



mj+1






983081darr

F


F 983080



F 983080


mℓminus1

983081



983081


983080τ (ν ) j














by 983080τ (ν ) j

983081κνj isin F 983080


mνminus1









F 983080





F 983080





F 983080








where

1






1 = G1 sub G0 = S 2

andF (σ1 σ2)

darrF (x1 x2) = F

983080τ (1)1 τ

(1)2

983081where

τ (1)1 = σ1 983080τ

(1)2 983081

2

= σ21 minus 4σ2


x1 x2 = τ

(1)1 plusmn τ

(1)2

2


1

= G2

subG1

subG0 = S 3


notations

p = minusσ21

3 + σ2

q = minus2σ31

27 + σ1σ2

3 minus σ3






radic


y1 = x1 + x2 + x3

y2 = x1 + ϵx2 + ϵ2x3

y3 = x1 + ϵ2x2 + ϵx3


983163ϵ jy2 ϵky3

983165 for some

0

le j k





so that

983080y2

3 9830813

983080y3

3 9830813

= minus p3

27

983080y2

3 9830813

+

983080y3

3 9830813

= minusq


3

9830813983080y3

3

9830813= minusq

2 plusmn991770

q 2

4 +

p3

27


X 2 + qX minus p3

27 = 0


F (σ1 σ2 σ3)

darrF 983080

τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4

983081darr

F (x1 x2 x3) = F 983080

τ (2)1 τ

(2)2 τ

(2)3

983081




We define

τ (1)1 = σ1 τ

(1)2 = q τ

(1)3 =

991770 q 2

4 +

p3

27 τ

(1)4 = p

so that

τ (1)2 =

983080y2

3

9830813+983080y3

3

9830813and τ

(1)3 =

1

2

1048616983080y2

3

9830813minus983080y3

3

98308131048617


3y2y3 to F

983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4


(1)1 τ

(1)2 τ

(1)3 τ



F (x1 x2 x3) = F (σ1 p q )


983080τ (1)3

9830812=

q 2

4 +

p3

27 isin F (σ1 p q ) = F (x1 x2 x3)



j = 1 2 3 with

τ (2)1 = τ

(1)1

983080τ (2)2

9830813=

1

2τ (1)2 + τ

(1)3

983080τ (2)3

9830813

= 1

2τ (1)2 minus τ

(1)3

so that κ21 = 1 and κ22 = κ23 = 3



1 a b c










y0 = 12 (x1 + x2 + x3 + x4)





τ 1 = (y1)2 + (y2)2 + (y3)2

τ 2 = (y1)2 (y2)2 + (y1)2 (y3)2 + (y2)2 (y3)2

τ 3 = (y1)2 (y2)2 (y3)2



P = minusτ 21

3 + τ 2

Q = minus2τ 31

27 + τ 1τ 2

3 minus τ 3


z 1 = (y1)2 + ϵ (y2)2 + ϵ2 (y3)2

z 2 = (y1)2 + ϵ2 (y2)2 + ϵ (y3)2


+radic 32






F (σ1 σ2 σ3 σ4)

darrF 983080

τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5

983081darrdarr

F 983080

τ (2)1 τ

(2)2 τ

(2)3 τ

(2)4

983081darr

F (x1 x2 x3 x4) = F 983080

τ (3)1 τ (3)2 τ (3)3 τ (3)4983081


Q2

4 +

P 3

27 isin F (σ1 σ2 σ3 σ4)

Again as above from

z 1z 2 = (τ 1)2

minus3τ 2 =

minus3P


so that 983080z 1

3

9830813 983080z 2

3

9830813

= minusP 3

27983080z 1

3

9830813+983080z 2

3

9830813= minusQ


3

9830813983080z 2

3

9830813

= minusQ

2 plusmn991770

Q2

4 +

P 3

27


X 2 + QX minus P 3

27 = 0

We define

τ (1)1 = σ1 τ

(1)2 = τ 1 τ

(1)3 = Q τ

(1)4 =

991770 Q2

4 +

P 3

27 τ

(1)5 = P




so that

τ (1)3 =

983080z 1

3

9830813+983080z 2

3

9830813and τ

(1)4 =

1

2

1048616983080z 1

3

9830813minus983080z 2

3

98308131048617


3z 1z 2 to F

983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5


983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5



983080τ (1)4 983081

2

= Q2

4

+ P 3



(2)2 = τ 1 τ

(2)3 = z 1 τ

(2)4 = z 2

Since 983080τ (2)3

9830813=

1

2τ (1)3 + τ

(1)4

983080τ (2)4

9830813=

1

2τ (1)3 minus τ

(1)4


τ (3)1 = y0 τ

(3)2 = y1 τ

(3)3 = y2 τ

(3)4 = y3


and983080

τ (3) j


F 983080

τ (2)1 τ

(2)2 τ

(2)3 τ

(2)4

983081for j = 2 3 4



1 sub A3 sub S 4


1 sub

K 4 sub

A4 sub

S 4






rarr S 3


1 rarr A3 rarr S 4






q =

aγ 3 + bγ 2 + cγ + d

a =

a 852008minus b3a852009

3+ b 852008minus

b3a852009

2+ c 852008minus

b3a852009+ d

a






which we rewrite as

t3 + pt + q =

852008u3 + v3

852009+ (u + v) (3uv + p) + q



(lowast)

u3 + v3 + q = 0

3uv + p = 0








(lowastlowast)

u3 + v3 + q = 0

u3v3 = minus p3

27 = 0


X 2 + qX minus p3

27 = 0

to get

u3 v3 = minusq

2 plusmn991770

q 2

4 +

p3

27


u = 3

1057306 minusq

2 +

991770 q 2

4 +

p3

27

v = 3

1057306 minusq

2 minus991770

q 2

4 +

p3

27


x1 x2 x3 = 3

1057306 minusq

2 +

991770 q 2

4 +

p3

27 +

p

3 3

991770 minus q

2 +radic

q2

4 + p3

27









p = minusσ2

1

3 + σ2

q = minus2σ31

27 + σ1σ2

3 minus σ3


minus1

21048616minus2σ3

1

27 +

σ1σ2

3 minus σ31048617+1057306 1

41048616minus2σ3

1

27 +

σ1σ2

3 minus σ310486172

+

1

271048616minusσ

21

3 + σ210486173

1

3

+ p

3

minus1

2

1048616minus2σ3

1

27+

σ1σ2

3minus σ3

1048617+

1057306 1

4

1048616minus2σ3

1

27+

σ1σ2

3minus σ3

10486172

+ 1

27

1048616minusσ

2

1

3+ σ2

10486173

minus

1

3


x1 x2 x3 x4

we introduce the


y0 = 12 (x1 + x2 + x3 + x4)










x1 x2 x3 x4



y0

y1

y2

y3

= A

x1

x2

x3

x4


1

2

1 1 1 11 minus

1 1 minus





2σ1 we can now













(0)m0





(1)m1



(2)m2



(ℓ)mℓ


σ1





darrτ (ℓminus1)








( j)mj to τ

( j+1)1


( j+1)mj+1

in



mj+1






983081darr

F


F 983080



F 983080


mℓminus1

983081



983081


983080τ (ν ) j














by 983080τ (ν ) j

983081κνj isin F 983080


mνminus1









F 983080





F 983080





F 983080








where

1






1 = G1 sub G0 = S 2

andF (σ1 σ2)

darrF (x1 x2) = F

983080τ (1)1 τ

(1)2

983081where

τ (1)1 = σ1 983080τ

(1)2 983081

2

= σ21 minus 4σ2


x1 x2 = τ

(1)1 plusmn τ

(1)2

2


1

= G2

subG1

subG0 = S 3


notations

p = minusσ21

3 + σ2

q = minus2σ31

27 + σ1σ2

3 minus σ3






radic


y1 = x1 + x2 + x3

y2 = x1 + ϵx2 + ϵ2x3

y3 = x1 + ϵ2x2 + ϵx3


983163ϵ jy2 ϵky3

983165 for some

0

le j k





so that

983080y2

3 9830813

983080y3

3 9830813

= minus p3

27

983080y2

3 9830813

+

983080y3

3 9830813

= minusq


3

9830813983080y3

3

9830813= minusq

2 plusmn991770

q 2

4 +

p3

27


X 2 + qX minus p3

27 = 0


F (σ1 σ2 σ3)

darrF 983080

τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4

983081darr

F (x1 x2 x3) = F 983080

τ (2)1 τ

(2)2 τ

(2)3

983081




We define

τ (1)1 = σ1 τ

(1)2 = q τ

(1)3 =

991770 q 2

4 +

p3

27 τ

(1)4 = p

so that

τ (1)2 =

983080y2

3

9830813+983080y3

3

9830813and τ

(1)3 =

1

2

1048616983080y2

3

9830813minus983080y3

3

98308131048617


3y2y3 to F

983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4


(1)1 τ

(1)2 τ

(1)3 τ



F (x1 x2 x3) = F (σ1 p q )


983080τ (1)3

9830812=

q 2

4 +

p3

27 isin F (σ1 p q ) = F (x1 x2 x3)



j = 1 2 3 with

τ (2)1 = τ

(1)1

983080τ (2)2

9830813=

1

2τ (1)2 + τ

(1)3

983080τ (2)3

9830813

= 1

2τ (1)2 minus τ

(1)3

so that κ21 = 1 and κ22 = κ23 = 3



1 a b c










y0 = 12 (x1 + x2 + x3 + x4)





τ 1 = (y1)2 + (y2)2 + (y3)2

τ 2 = (y1)2 (y2)2 + (y1)2 (y3)2 + (y2)2 (y3)2

τ 3 = (y1)2 (y2)2 (y3)2



P = minusτ 21

3 + τ 2

Q = minus2τ 31

27 + τ 1τ 2

3 minus τ 3


z 1 = (y1)2 + ϵ (y2)2 + ϵ2 (y3)2

z 2 = (y1)2 + ϵ2 (y2)2 + ϵ (y3)2


+radic 32






F (σ1 σ2 σ3 σ4)

darrF 983080

τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5

983081darrdarr

F 983080

τ (2)1 τ

(2)2 τ

(2)3 τ

(2)4

983081darr

F (x1 x2 x3 x4) = F 983080

τ (3)1 τ (3)2 τ (3)3 τ (3)4983081


Q2

4 +

P 3

27 isin F (σ1 σ2 σ3 σ4)

Again as above from

z 1z 2 = (τ 1)2

minus3τ 2 =

minus3P


so that 983080z 1

3

9830813 983080z 2

3

9830813

= minusP 3

27983080z 1

3

9830813+983080z 2

3

9830813= minusQ


3

9830813983080z 2

3

9830813

= minusQ

2 plusmn991770

Q2

4 +

P 3

27


X 2 + QX minus P 3

27 = 0

We define

τ (1)1 = σ1 τ

(1)2 = τ 1 τ

(1)3 = Q τ

(1)4 =

991770 Q2

4 +

P 3

27 τ

(1)5 = P




so that

τ (1)3 =

983080z 1

3

9830813+983080z 2

3

9830813and τ

(1)4 =

1

2

1048616983080z 1

3

9830813minus983080z 2

3

98308131048617


3z 1z 2 to F

983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5


983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5



983080τ (1)4 983081

2

= Q2

4

+ P 3



(2)2 = τ 1 τ

(2)3 = z 1 τ

(2)4 = z 2

Since 983080τ (2)3

9830813=

1

2τ (1)3 + τ

(1)4

983080τ (2)4

9830813=

1

2τ (1)3 minus τ

(1)4


τ (3)1 = y0 τ

(3)2 = y1 τ

(3)3 = y2 τ

(3)4 = y3


and983080

τ (3) j


F 983080

τ (2)1 τ

(2)2 τ

(2)3 τ

(2)4

983081for j = 2 3 4



1 sub A3 sub S 4


1 sub

K 4 sub

A4 sub

S 4






rarr S 3


1 rarr A3 rarr S 4







(lowastlowast)

u3 + v3 + q = 0

u3v3 = minus p3

27 = 0


X 2 + qX minus p3

27 = 0

to get

u3 v3 = minusq

2 plusmn991770

q 2

4 +

p3

27


u = 3

1057306 minusq

2 +

991770 q 2

4 +

p3

27

v = 3

1057306 minusq

2 minus991770

q 2

4 +

p3

27


x1 x2 x3 = 3

1057306 minusq

2 +

991770 q 2

4 +

p3

27 +

p

3 3

991770 minus q

2 +radic

q2

4 + p3

27









p = minusσ2

1

3 + σ2

q = minus2σ31

27 + σ1σ2

3 minus σ3


minus1

21048616minus2σ3

1

27 +

σ1σ2

3 minus σ31048617+1057306 1

41048616minus2σ3

1

27 +

σ1σ2

3 minus σ310486172

+

1

271048616minusσ

21

3 + σ210486173

1

3

+ p

3

minus1

2

1048616minus2σ3

1

27+

σ1σ2

3minus σ3

1048617+

1057306 1

4

1048616minus2σ3

1

27+

σ1σ2

3minus σ3

10486172

+ 1

27

1048616minusσ

2

1

3+ σ2

10486173

minus

1

3


x1 x2 x3 x4

we introduce the


y0 = 12 (x1 + x2 + x3 + x4)










x1 x2 x3 x4



y0

y1

y2

y3

= A

x1

x2

x3

x4


1

2

1 1 1 11 minus

1 1 minus





2σ1 we can now













(0)m0





(1)m1



(2)m2



(ℓ)mℓ


σ1





darrτ (ℓminus1)








( j)mj to τ

( j+1)1


( j+1)mj+1

in



mj+1






983081darr

F


F 983080



F 983080


mℓminus1

983081



983081


983080τ (ν ) j














by 983080τ (ν ) j

983081κνj isin F 983080


mνminus1









F 983080





F 983080





F 983080








where

1






1 = G1 sub G0 = S 2

andF (σ1 σ2)

darrF (x1 x2) = F

983080τ (1)1 τ

(1)2

983081where

τ (1)1 = σ1 983080τ

(1)2 983081

2

= σ21 minus 4σ2


x1 x2 = τ

(1)1 plusmn τ

(1)2

2


1

= G2

subG1

subG0 = S 3


notations

p = minusσ21

3 + σ2

q = minus2σ31

27 + σ1σ2

3 minus σ3






radic


y1 = x1 + x2 + x3

y2 = x1 + ϵx2 + ϵ2x3

y3 = x1 + ϵ2x2 + ϵx3


983163ϵ jy2 ϵky3

983165 for some

0

le j k





so that

983080y2

3 9830813

983080y3

3 9830813

= minus p3

27

983080y2

3 9830813

+

983080y3

3 9830813

= minusq


3

9830813983080y3

3

9830813= minusq

2 plusmn991770

q 2

4 +

p3

27


X 2 + qX minus p3

27 = 0


F (σ1 σ2 σ3)

darrF 983080

τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4

983081darr

F (x1 x2 x3) = F 983080

τ (2)1 τ

(2)2 τ

(2)3

983081




We define

τ (1)1 = σ1 τ

(1)2 = q τ

(1)3 =

991770 q 2

4 +

p3

27 τ

(1)4 = p

so that

τ (1)2 =

983080y2

3

9830813+983080y3

3

9830813and τ

(1)3 =

1

2

1048616983080y2

3

9830813minus983080y3

3

98308131048617


3y2y3 to F

983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4


(1)1 τ

(1)2 τ

(1)3 τ



F (x1 x2 x3) = F (σ1 p q )


983080τ (1)3

9830812=

q 2

4 +

p3

27 isin F (σ1 p q ) = F (x1 x2 x3)



j = 1 2 3 with

τ (2)1 = τ

(1)1

983080τ (2)2

9830813=

1

2τ (1)2 + τ

(1)3

983080τ (2)3

9830813

= 1

2τ (1)2 minus τ

(1)3

so that κ21 = 1 and κ22 = κ23 = 3



1 a b c










y0 = 12 (x1 + x2 + x3 + x4)





τ 1 = (y1)2 + (y2)2 + (y3)2

τ 2 = (y1)2 (y2)2 + (y1)2 (y3)2 + (y2)2 (y3)2

τ 3 = (y1)2 (y2)2 (y3)2



P = minusτ 21

3 + τ 2

Q = minus2τ 31

27 + τ 1τ 2

3 minus τ 3


z 1 = (y1)2 + ϵ (y2)2 + ϵ2 (y3)2

z 2 = (y1)2 + ϵ2 (y2)2 + ϵ (y3)2


+radic 32






F (σ1 σ2 σ3 σ4)

darrF 983080

τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5

983081darrdarr

F 983080

τ (2)1 τ

(2)2 τ

(2)3 τ

(2)4

983081darr

F (x1 x2 x3 x4) = F 983080

τ (3)1 τ (3)2 τ (3)3 τ (3)4983081


Q2

4 +

P 3

27 isin F (σ1 σ2 σ3 σ4)

Again as above from

z 1z 2 = (τ 1)2

minus3τ 2 =

minus3P


so that 983080z 1

3

9830813 983080z 2

3

9830813

= minusP 3

27983080z 1

3

9830813+983080z 2

3

9830813= minusQ


3

9830813983080z 2

3

9830813

= minusQ

2 plusmn991770

Q2

4 +

P 3

27


X 2 + QX minus P 3

27 = 0

We define

τ (1)1 = σ1 τ

(1)2 = τ 1 τ

(1)3 = Q τ

(1)4 =

991770 Q2

4 +

P 3

27 τ

(1)5 = P




so that

τ (1)3 =

983080z 1

3

9830813+983080z 2

3

9830813and τ

(1)4 =

1

2

1048616983080z 1

3

9830813minus983080z 2

3

98308131048617


3z 1z 2 to F

983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5


983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5



983080τ (1)4 983081

2

= Q2

4

+ P 3



(2)2 = τ 1 τ

(2)3 = z 1 τ

(2)4 = z 2

Since 983080τ (2)3

9830813=

1

2τ (1)3 + τ

(1)4

983080τ (2)4

9830813=

1

2τ (1)3 minus τ

(1)4


τ (3)1 = y0 τ

(3)2 = y1 τ

(3)3 = y2 τ

(3)4 = y3


and983080

τ (3) j


F 983080

τ (2)1 τ

(2)2 τ

(2)3 τ

(2)4

983081for j = 2 3 4



1 sub A3 sub S 4


1 sub

K 4 sub

A4 sub

S 4






rarr S 3


1 rarr A3 rarr S 4








p = minusσ2

1

3 + σ2

q = minus2σ31

27 + σ1σ2

3 minus σ3


minus1

21048616minus2σ3

1

27 +

σ1σ2

3 minus σ31048617+1057306 1

41048616minus2σ3

1

27 +

σ1σ2

3 minus σ310486172

+

1

271048616minusσ

21

3 + σ210486173

1

3

+ p

3

minus1

2

1048616minus2σ3

1

27+

σ1σ2

3minus σ3

1048617+

1057306 1

4

1048616minus2σ3

1

27+

σ1σ2

3minus σ3

10486172

+ 1

27

1048616minusσ

2

1

3+ σ2

10486173

minus

1

3


x1 x2 x3 x4

we introduce the


y0 = 12 (x1 + x2 + x3 + x4)










x1 x2 x3 x4



y0

y1

y2

y3

= A

x1

x2

x3

x4


1

2

1 1 1 11 minus

1 1 minus





2σ1 we can now













(0)m0





(1)m1



(2)m2



(ℓ)mℓ


σ1





darrτ (ℓminus1)








( j)mj to τ

( j+1)1


( j+1)mj+1

in



mj+1






983081darr

F


F 983080



F 983080


mℓminus1

983081



983081


983080τ (ν ) j














by 983080τ (ν ) j

983081κνj isin F 983080


mνminus1









F 983080





F 983080





F 983080








where

1






1 = G1 sub G0 = S 2

andF (σ1 σ2)

darrF (x1 x2) = F

983080τ (1)1 τ

(1)2

983081where

τ (1)1 = σ1 983080τ

(1)2 983081

2

= σ21 minus 4σ2


x1 x2 = τ

(1)1 plusmn τ

(1)2

2


1

= G2

subG1

subG0 = S 3


notations

p = minusσ21

3 + σ2

q = minus2σ31

27 + σ1σ2

3 minus σ3






radic


y1 = x1 + x2 + x3

y2 = x1 + ϵx2 + ϵ2x3

y3 = x1 + ϵ2x2 + ϵx3


983163ϵ jy2 ϵky3

983165 for some

0

le j k





so that

983080y2

3 9830813

983080y3

3 9830813

= minus p3

27

983080y2

3 9830813

+

983080y3

3 9830813

= minusq


3

9830813983080y3

3

9830813= minusq

2 plusmn991770

q 2

4 +

p3

27


X 2 + qX minus p3

27 = 0


F (σ1 σ2 σ3)

darrF 983080

τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4

983081darr

F (x1 x2 x3) = F 983080

τ (2)1 τ

(2)2 τ

(2)3

983081




We define

τ (1)1 = σ1 τ

(1)2 = q τ

(1)3 =

991770 q 2

4 +

p3

27 τ

(1)4 = p

so that

τ (1)2 =

983080y2

3

9830813+983080y3

3

9830813and τ

(1)3 =

1

2

1048616983080y2

3

9830813minus983080y3

3

98308131048617


3y2y3 to F

983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4


(1)1 τ

(1)2 τ

(1)3 τ



F (x1 x2 x3) = F (σ1 p q )


983080τ (1)3

9830812=

q 2

4 +

p3

27 isin F (σ1 p q ) = F (x1 x2 x3)



j = 1 2 3 with

τ (2)1 = τ

(1)1

983080τ (2)2

9830813=

1

2τ (1)2 + τ

(1)3

983080τ (2)3

9830813

= 1

2τ (1)2 minus τ

(1)3

so that κ21 = 1 and κ22 = κ23 = 3



1 a b c










y0 = 12 (x1 + x2 + x3 + x4)





τ 1 = (y1)2 + (y2)2 + (y3)2

τ 2 = (y1)2 (y2)2 + (y1)2 (y3)2 + (y2)2 (y3)2

τ 3 = (y1)2 (y2)2 (y3)2



P = minusτ 21

3 + τ 2

Q = minus2τ 31

27 + τ 1τ 2

3 minus τ 3


z 1 = (y1)2 + ϵ (y2)2 + ϵ2 (y3)2

z 2 = (y1)2 + ϵ2 (y2)2 + ϵ (y3)2


+radic 32






F (σ1 σ2 σ3 σ4)

darrF 983080

τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5

983081darrdarr

F 983080

τ (2)1 τ

(2)2 τ

(2)3 τ

(2)4

983081darr

F (x1 x2 x3 x4) = F 983080

τ (3)1 τ (3)2 τ (3)3 τ (3)4983081


Q2

4 +

P 3

27 isin F (σ1 σ2 σ3 σ4)

Again as above from

z 1z 2 = (τ 1)2

minus3τ 2 =

minus3P


so that 983080z 1

3

9830813 983080z 2

3

9830813

= minusP 3

27983080z 1

3

9830813+983080z 2

3

9830813= minusQ


3

9830813983080z 2

3

9830813

= minusQ

2 plusmn991770

Q2

4 +

P 3

27


X 2 + QX minus P 3

27 = 0

We define

τ (1)1 = σ1 τ

(1)2 = τ 1 τ

(1)3 = Q τ

(1)4 =

991770 Q2

4 +

P 3

27 τ

(1)5 = P




so that

τ (1)3 =

983080z 1

3

9830813+983080z 2

3

9830813and τ

(1)4 =

1

2

1048616983080z 1

3

9830813minus983080z 2

3

98308131048617


3z 1z 2 to F

983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5


983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5



983080τ (1)4 983081

2

= Q2

4

+ P 3



(2)2 = τ 1 τ

(2)3 = z 1 τ

(2)4 = z 2

Since 983080τ (2)3

9830813=

1

2τ (1)3 + τ

(1)4

983080τ (2)4

9830813=

1

2τ (1)3 minus τ

(1)4


τ (3)1 = y0 τ

(3)2 = y1 τ

(3)3 = y2 τ

(3)4 = y3


and983080

τ (3) j


F 983080

τ (2)1 τ

(2)2 τ

(2)3 τ

(2)4

983081for j = 2 3 4



1 sub A3 sub S 4


1 sub

K 4 sub

A4 sub

S 4






rarr S 3


1 rarr A3 rarr S 4







x1 x2 x3 x4



y0

y1

y2

y3

= A

x1

x2

x3

x4


1

2

1 1 1 11 minus

1 1 minus





2σ1 we can now













(0)m0





(1)m1



(2)m2



(ℓ)mℓ


σ1





darrτ (ℓminus1)








( j)mj to τ

( j+1)1


( j+1)mj+1

in



mj+1






983081darr

F


F 983080



F 983080


mℓminus1

983081



983081


983080τ (ν ) j














by 983080τ (ν ) j

983081κνj isin F 983080


mνminus1









F 983080





F 983080





F 983080








where

1






1 = G1 sub G0 = S 2

andF (σ1 σ2)

darrF (x1 x2) = F

983080τ (1)1 τ

(1)2

983081where

τ (1)1 = σ1 983080τ

(1)2 983081

2

= σ21 minus 4σ2


x1 x2 = τ

(1)1 plusmn τ

(1)2

2


1

= G2

subG1

subG0 = S 3


notations

p = minusσ21

3 + σ2

q = minus2σ31

27 + σ1σ2

3 minus σ3






radic


y1 = x1 + x2 + x3

y2 = x1 + ϵx2 + ϵ2x3

y3 = x1 + ϵ2x2 + ϵx3


983163ϵ jy2 ϵky3

983165 for some

0

le j k





so that

983080y2

3 9830813

983080y3

3 9830813

= minus p3

27

983080y2

3 9830813

+

983080y3

3 9830813

= minusq


3

9830813983080y3

3

9830813= minusq

2 plusmn991770

q 2

4 +

p3

27


X 2 + qX minus p3

27 = 0


F (σ1 σ2 σ3)

darrF 983080

τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4

983081darr

F (x1 x2 x3) = F 983080

τ (2)1 τ

(2)2 τ

(2)3

983081




We define

τ (1)1 = σ1 τ

(1)2 = q τ

(1)3 =

991770 q 2

4 +

p3

27 τ

(1)4 = p

so that

τ (1)2 =

983080y2

3

9830813+983080y3

3

9830813and τ

(1)3 =

1

2

1048616983080y2

3

9830813minus983080y3

3

98308131048617


3y2y3 to F

983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4


(1)1 τ

(1)2 τ

(1)3 τ



F (x1 x2 x3) = F (σ1 p q )


983080τ (1)3

9830812=

q 2

4 +

p3

27 isin F (σ1 p q ) = F (x1 x2 x3)



j = 1 2 3 with

τ (2)1 = τ

(1)1

983080τ (2)2

9830813=

1

2τ (1)2 + τ

(1)3

983080τ (2)3

9830813

= 1

2τ (1)2 minus τ

(1)3

so that κ21 = 1 and κ22 = κ23 = 3



1 a b c










y0 = 12 (x1 + x2 + x3 + x4)





τ 1 = (y1)2 + (y2)2 + (y3)2

τ 2 = (y1)2 (y2)2 + (y1)2 (y3)2 + (y2)2 (y3)2

τ 3 = (y1)2 (y2)2 (y3)2



P = minusτ 21

3 + τ 2

Q = minus2τ 31

27 + τ 1τ 2

3 minus τ 3


z 1 = (y1)2 + ϵ (y2)2 + ϵ2 (y3)2

z 2 = (y1)2 + ϵ2 (y2)2 + ϵ (y3)2


+radic 32






F (σ1 σ2 σ3 σ4)

darrF 983080

τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5

983081darrdarr

F 983080

τ (2)1 τ

(2)2 τ

(2)3 τ

(2)4

983081darr

F (x1 x2 x3 x4) = F 983080

τ (3)1 τ (3)2 τ (3)3 τ (3)4983081


Q2

4 +

P 3

27 isin F (σ1 σ2 σ3 σ4)

Again as above from

z 1z 2 = (τ 1)2

minus3τ 2 =

minus3P


so that 983080z 1

3

9830813 983080z 2

3

9830813

= minusP 3

27983080z 1

3

9830813+983080z 2

3

9830813= minusQ


3

9830813983080z 2

3

9830813

= minusQ

2 plusmn991770

Q2

4 +

P 3

27


X 2 + QX minus P 3

27 = 0

We define

τ (1)1 = σ1 τ

(1)2 = τ 1 τ

(1)3 = Q τ

(1)4 =

991770 Q2

4 +

P 3

27 τ

(1)5 = P




so that

τ (1)3 =

983080z 1

3

9830813+983080z 2

3

9830813and τ

(1)4 =

1

2

1048616983080z 1

3

9830813minus983080z 2

3

98308131048617


3z 1z 2 to F

983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5


983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5



983080τ (1)4 983081

2

= Q2

4

+ P 3



(2)2 = τ 1 τ

(2)3 = z 1 τ

(2)4 = z 2

Since 983080τ (2)3

9830813=

1

2τ (1)3 + τ

(1)4

983080τ (2)4

9830813=

1

2τ (1)3 minus τ

(1)4


τ (3)1 = y0 τ

(3)2 = y1 τ

(3)3 = y2 τ

(3)4 = y3


and983080

τ (3) j


F 983080

τ (2)1 τ

(2)2 τ

(2)3 τ

(2)4

983081for j = 2 3 4



1 sub A3 sub S 4


1 sub

K 4 sub

A4 sub

S 4






rarr S 3


1 rarr A3 rarr S 4











(0)m0





(1)m1



(2)m2



(ℓ)mℓ


σ1





darrτ (ℓminus1)








( j)mj to τ

( j+1)1


( j+1)mj+1

in



mj+1






983081darr

F


F 983080



F 983080


mℓminus1

983081



983081


983080τ (ν ) j














by 983080τ (ν ) j

983081κνj isin F 983080


mνminus1









F 983080





F 983080





F 983080








where

1






1 = G1 sub G0 = S 2

andF (σ1 σ2)

darrF (x1 x2) = F

983080τ (1)1 τ

(1)2

983081where

τ (1)1 = σ1 983080τ

(1)2 983081

2

= σ21 minus 4σ2


x1 x2 = τ

(1)1 plusmn τ

(1)2

2


1

= G2

subG1

subG0 = S 3


notations

p = minusσ21

3 + σ2

q = minus2σ31

27 + σ1σ2

3 minus σ3






radic


y1 = x1 + x2 + x3

y2 = x1 + ϵx2 + ϵ2x3

y3 = x1 + ϵ2x2 + ϵx3


983163ϵ jy2 ϵky3

983165 for some

0

le j k





so that

983080y2

3 9830813

983080y3

3 9830813

= minus p3

27

983080y2

3 9830813

+

983080y3

3 9830813

= minusq


3

9830813983080y3

3

9830813= minusq

2 plusmn991770

q 2

4 +

p3

27


X 2 + qX minus p3

27 = 0


F (σ1 σ2 σ3)

darrF 983080

τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4

983081darr

F (x1 x2 x3) = F 983080

τ (2)1 τ

(2)2 τ

(2)3

983081




We define

τ (1)1 = σ1 τ

(1)2 = q τ

(1)3 =

991770 q 2

4 +

p3

27 τ

(1)4 = p

so that

τ (1)2 =

983080y2

3

9830813+983080y3

3

9830813and τ

(1)3 =

1

2

1048616983080y2

3

9830813minus983080y3

3

98308131048617


3y2y3 to F

983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4


(1)1 τ

(1)2 τ

(1)3 τ



F (x1 x2 x3) = F (σ1 p q )


983080τ (1)3

9830812=

q 2

4 +

p3

27 isin F (σ1 p q ) = F (x1 x2 x3)



j = 1 2 3 with

τ (2)1 = τ

(1)1

983080τ (2)2

9830813=

1

2τ (1)2 + τ

(1)3

983080τ (2)3

9830813

= 1

2τ (1)2 minus τ

(1)3

so that κ21 = 1 and κ22 = κ23 = 3



1 a b c










y0 = 12 (x1 + x2 + x3 + x4)





τ 1 = (y1)2 + (y2)2 + (y3)2

τ 2 = (y1)2 (y2)2 + (y1)2 (y3)2 + (y2)2 (y3)2

τ 3 = (y1)2 (y2)2 (y3)2



P = minusτ 21

3 + τ 2

Q = minus2τ 31

27 + τ 1τ 2

3 minus τ 3


z 1 = (y1)2 + ϵ (y2)2 + ϵ2 (y3)2

z 2 = (y1)2 + ϵ2 (y2)2 + ϵ (y3)2


+radic 32






F (σ1 σ2 σ3 σ4)

darrF 983080

τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5

983081darrdarr

F 983080

τ (2)1 τ

(2)2 τ

(2)3 τ

(2)4

983081darr

F (x1 x2 x3 x4) = F 983080

τ (3)1 τ (3)2 τ (3)3 τ (3)4983081


Q2

4 +

P 3

27 isin F (σ1 σ2 σ3 σ4)

Again as above from

z 1z 2 = (τ 1)2

minus3τ 2 =

minus3P


so that 983080z 1

3

9830813 983080z 2

3

9830813

= minusP 3

27983080z 1

3

9830813+983080z 2

3

9830813= minusQ


3

9830813983080z 2

3

9830813

= minusQ

2 plusmn991770

Q2

4 +

P 3

27


X 2 + QX minus P 3

27 = 0

We define

τ (1)1 = σ1 τ

(1)2 = τ 1 τ

(1)3 = Q τ

(1)4 =

991770 Q2

4 +

P 3

27 τ

(1)5 = P




so that

τ (1)3 =

983080z 1

3

9830813+983080z 2

3

9830813and τ

(1)4 =

1

2

1048616983080z 1

3

9830813minus983080z 2

3

98308131048617


3z 1z 2 to F

983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5


983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5



983080τ (1)4 983081

2

= Q2

4

+ P 3



(2)2 = τ 1 τ

(2)3 = z 1 τ

(2)4 = z 2

Since 983080τ (2)3

9830813=

1

2τ (1)3 + τ

(1)4

983080τ (2)4

9830813=

1

2τ (1)3 minus τ

(1)4


τ (3)1 = y0 τ

(3)2 = y1 τ

(3)3 = y2 τ

(3)4 = y3


and983080

τ (3) j


F 983080

τ (2)1 τ

(2)2 τ

(2)3 τ

(2)4

983081for j = 2 3 4



1 sub A3 sub S 4


1 sub

K 4 sub

A4 sub

S 4






rarr S 3


1 rarr A3 rarr S 4








( j)mj to τ

( j+1)1


( j+1)mj+1

in



mj+1






983081darr

F


F 983080



F 983080


mℓminus1

983081



983081


983080τ (ν ) j














by 983080τ (ν ) j

983081κνj isin F 983080


mνminus1









F 983080





F 983080





F 983080








where

1






1 = G1 sub G0 = S 2

andF (σ1 σ2)

darrF (x1 x2) = F

983080τ (1)1 τ

(1)2

983081where

τ (1)1 = σ1 983080τ

(1)2 983081

2

= σ21 minus 4σ2


x1 x2 = τ

(1)1 plusmn τ

(1)2

2


1

= G2

subG1

subG0 = S 3


notations

p = minusσ21

3 + σ2

q = minus2σ31

27 + σ1σ2

3 minus σ3






radic


y1 = x1 + x2 + x3

y2 = x1 + ϵx2 + ϵ2x3

y3 = x1 + ϵ2x2 + ϵx3


983163ϵ jy2 ϵky3

983165 for some

0

le j k





so that

983080y2

3 9830813

983080y3

3 9830813

= minus p3

27

983080y2

3 9830813

+

983080y3

3 9830813

= minusq


3

9830813983080y3

3

9830813= minusq

2 plusmn991770

q 2

4 +

p3

27


X 2 + qX minus p3

27 = 0


F (σ1 σ2 σ3)

darrF 983080

τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4

983081darr

F (x1 x2 x3) = F 983080

τ (2)1 τ

(2)2 τ

(2)3

983081




We define

τ (1)1 = σ1 τ

(1)2 = q τ

(1)3 =

991770 q 2

4 +

p3

27 τ

(1)4 = p

so that

τ (1)2 =

983080y2

3

9830813+983080y3

3

9830813and τ

(1)3 =

1

2

1048616983080y2

3

9830813minus983080y3

3

98308131048617


3y2y3 to F

983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4


(1)1 τ

(1)2 τ

(1)3 τ



F (x1 x2 x3) = F (σ1 p q )


983080τ (1)3

9830812=

q 2

4 +

p3

27 isin F (σ1 p q ) = F (x1 x2 x3)



j = 1 2 3 with

τ (2)1 = τ

(1)1

983080τ (2)2

9830813=

1

2τ (1)2 + τ

(1)3

983080τ (2)3

9830813

= 1

2τ (1)2 minus τ

(1)3

so that κ21 = 1 and κ22 = κ23 = 3



1 a b c










y0 = 12 (x1 + x2 + x3 + x4)





τ 1 = (y1)2 + (y2)2 + (y3)2

τ 2 = (y1)2 (y2)2 + (y1)2 (y3)2 + (y2)2 (y3)2

τ 3 = (y1)2 (y2)2 (y3)2



P = minusτ 21

3 + τ 2

Q = minus2τ 31

27 + τ 1τ 2

3 minus τ 3


z 1 = (y1)2 + ϵ (y2)2 + ϵ2 (y3)2

z 2 = (y1)2 + ϵ2 (y2)2 + ϵ (y3)2


+radic 32






F (σ1 σ2 σ3 σ4)

darrF 983080

τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5

983081darrdarr

F 983080

τ (2)1 τ

(2)2 τ

(2)3 τ

(2)4

983081darr

F (x1 x2 x3 x4) = F 983080

τ (3)1 τ (3)2 τ (3)3 τ (3)4983081


Q2

4 +

P 3

27 isin F (σ1 σ2 σ3 σ4)

Again as above from

z 1z 2 = (τ 1)2

minus3τ 2 =

minus3P


so that 983080z 1

3

9830813 983080z 2

3

9830813

= minusP 3

27983080z 1

3

9830813+983080z 2

3

9830813= minusQ


3

9830813983080z 2

3

9830813

= minusQ

2 plusmn991770

Q2

4 +

P 3

27


X 2 + QX minus P 3

27 = 0

We define

τ (1)1 = σ1 τ

(1)2 = τ 1 τ

(1)3 = Q τ

(1)4 =

991770 Q2

4 +

P 3

27 τ

(1)5 = P




so that

τ (1)3 =

983080z 1

3

9830813+983080z 2

3

9830813and τ

(1)4 =

1

2

1048616983080z 1

3

9830813minus983080z 2

3

98308131048617


3z 1z 2 to F

983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5


983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5



983080τ (1)4 983081

2

= Q2

4

+ P 3



(2)2 = τ 1 τ

(2)3 = z 1 τ

(2)4 = z 2

Since 983080τ (2)3

9830813=

1

2τ (1)3 + τ

(1)4

983080τ (2)4

9830813=

1

2τ (1)3 minus τ

(1)4


τ (3)1 = y0 τ

(3)2 = y1 τ

(3)3 = y2 τ

(3)4 = y3


and983080

τ (3) j


F 983080

τ (2)1 τ

(2)2 τ

(2)3 τ

(2)4

983081for j = 2 3 4



1 sub A3 sub S 4


1 sub

K 4 sub

A4 sub

S 4






rarr S 3


1 rarr A3 rarr S 4












by 983080τ (ν ) j

983081κνj isin F 983080


mνminus1









F 983080





F 983080





F 983080








where

1






1 = G1 sub G0 = S 2

andF (σ1 σ2)

darrF (x1 x2) = F

983080τ (1)1 τ

(1)2

983081where

τ (1)1 = σ1 983080τ

(1)2 983081

2

= σ21 minus 4σ2


x1 x2 = τ

(1)1 plusmn τ

(1)2

2


1

= G2

subG1

subG0 = S 3


notations

p = minusσ21

3 + σ2

q = minus2σ31

27 + σ1σ2

3 minus σ3






radic


y1 = x1 + x2 + x3

y2 = x1 + ϵx2 + ϵ2x3

y3 = x1 + ϵ2x2 + ϵx3


983163ϵ jy2 ϵky3

983165 for some

0

le j k





so that

983080y2

3 9830813

983080y3

3 9830813

= minus p3

27

983080y2

3 9830813

+

983080y3

3 9830813

= minusq


3

9830813983080y3

3

9830813= minusq

2 plusmn991770

q 2

4 +

p3

27


X 2 + qX minus p3

27 = 0


F (σ1 σ2 σ3)

darrF 983080

τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4

983081darr

F (x1 x2 x3) = F 983080

τ (2)1 τ

(2)2 τ

(2)3

983081




We define

τ (1)1 = σ1 τ

(1)2 = q τ

(1)3 =

991770 q 2

4 +

p3

27 τ

(1)4 = p

so that

τ (1)2 =

983080y2

3

9830813+983080y3

3

9830813and τ

(1)3 =

1

2

1048616983080y2

3

9830813minus983080y3

3

98308131048617


3y2y3 to F

983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4


(1)1 τ

(1)2 τ

(1)3 τ



F (x1 x2 x3) = F (σ1 p q )


983080τ (1)3

9830812=

q 2

4 +

p3

27 isin F (σ1 p q ) = F (x1 x2 x3)



j = 1 2 3 with

τ (2)1 = τ

(1)1

983080τ (2)2

9830813=

1

2τ (1)2 + τ

(1)3

983080τ (2)3

9830813

= 1

2τ (1)2 minus τ

(1)3

so that κ21 = 1 and κ22 = κ23 = 3



1 a b c










y0 = 12 (x1 + x2 + x3 + x4)





τ 1 = (y1)2 + (y2)2 + (y3)2

τ 2 = (y1)2 (y2)2 + (y1)2 (y3)2 + (y2)2 (y3)2

τ 3 = (y1)2 (y2)2 (y3)2



P = minusτ 21

3 + τ 2

Q = minus2τ 31

27 + τ 1τ 2

3 minus τ 3


z 1 = (y1)2 + ϵ (y2)2 + ϵ2 (y3)2

z 2 = (y1)2 + ϵ2 (y2)2 + ϵ (y3)2


+radic 32






F (σ1 σ2 σ3 σ4)

darrF 983080

τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5

983081darrdarr

F 983080

τ (2)1 τ

(2)2 τ

(2)3 τ

(2)4

983081darr

F (x1 x2 x3 x4) = F 983080

τ (3)1 τ (3)2 τ (3)3 τ (3)4983081


Q2

4 +

P 3

27 isin F (σ1 σ2 σ3 σ4)

Again as above from

z 1z 2 = (τ 1)2

minus3τ 2 =

minus3P


so that 983080z 1

3

9830813 983080z 2

3

9830813

= minusP 3

27983080z 1

3

9830813+983080z 2

3

9830813= minusQ


3

9830813983080z 2

3

9830813

= minusQ

2 plusmn991770

Q2

4 +

P 3

27


X 2 + QX minus P 3

27 = 0

We define

τ (1)1 = σ1 τ

(1)2 = τ 1 τ

(1)3 = Q τ

(1)4 =

991770 Q2

4 +

P 3

27 τ

(1)5 = P




so that

τ (1)3 =

983080z 1

3

9830813+983080z 2

3

9830813and τ

(1)4 =

1

2

1048616983080z 1

3

9830813minus983080z 2

3

98308131048617


3z 1z 2 to F

983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5


983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5



983080τ (1)4 983081

2

= Q2

4

+ P 3



(2)2 = τ 1 τ

(2)3 = z 1 τ

(2)4 = z 2

Since 983080τ (2)3

9830813=

1

2τ (1)3 + τ

(1)4

983080τ (2)4

9830813=

1

2τ (1)3 minus τ

(1)4


τ (3)1 = y0 τ

(3)2 = y1 τ

(3)3 = y2 τ

(3)4 = y3


and983080

τ (3) j


F 983080

τ (2)1 τ

(2)2 τ

(2)3 τ

(2)4

983081for j = 2 3 4



1 sub A3 sub S 4


1 sub

K 4 sub

A4 sub

S 4






rarr S 3


1 rarr A3 rarr S 4






where

1






1 = G1 sub G0 = S 2

andF (σ1 σ2)

darrF (x1 x2) = F

983080τ (1)1 τ

(1)2

983081where

τ (1)1 = σ1 983080τ

(1)2 983081

2

= σ21 minus 4σ2


x1 x2 = τ

(1)1 plusmn τ

(1)2

2


1

= G2

subG1

subG0 = S 3


notations

p = minusσ21

3 + σ2

q = minus2σ31

27 + σ1σ2

3 minus σ3






radic


y1 = x1 + x2 + x3

y2 = x1 + ϵx2 + ϵ2x3

y3 = x1 + ϵ2x2 + ϵx3


983163ϵ jy2 ϵky3

983165 for some

0

le j k





so that

983080y2

3 9830813

983080y3

3 9830813

= minus p3

27

983080y2

3 9830813

+

983080y3

3 9830813

= minusq


3

9830813983080y3

3

9830813= minusq

2 plusmn991770

q 2

4 +

p3

27


X 2 + qX minus p3

27 = 0


F (σ1 σ2 σ3)

darrF 983080

τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4

983081darr

F (x1 x2 x3) = F 983080

τ (2)1 τ

(2)2 τ

(2)3

983081




We define

τ (1)1 = σ1 τ

(1)2 = q τ

(1)3 =

991770 q 2

4 +

p3

27 τ

(1)4 = p

so that

τ (1)2 =

983080y2

3

9830813+983080y3

3

9830813and τ

(1)3 =

1

2

1048616983080y2

3

9830813minus983080y3

3

98308131048617


3y2y3 to F

983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4


(1)1 τ

(1)2 τ

(1)3 τ



F (x1 x2 x3) = F (σ1 p q )


983080τ (1)3

9830812=

q 2

4 +

p3

27 isin F (σ1 p q ) = F (x1 x2 x3)



j = 1 2 3 with

τ (2)1 = τ

(1)1

983080τ (2)2

9830813=

1

2τ (1)2 + τ

(1)3

983080τ (2)3

9830813

= 1

2τ (1)2 minus τ

(1)3

so that κ21 = 1 and κ22 = κ23 = 3



1 a b c










y0 = 12 (x1 + x2 + x3 + x4)





τ 1 = (y1)2 + (y2)2 + (y3)2

τ 2 = (y1)2 (y2)2 + (y1)2 (y3)2 + (y2)2 (y3)2

τ 3 = (y1)2 (y2)2 (y3)2



P = minusτ 21

3 + τ 2

Q = minus2τ 31

27 + τ 1τ 2

3 minus τ 3


z 1 = (y1)2 + ϵ (y2)2 + ϵ2 (y3)2

z 2 = (y1)2 + ϵ2 (y2)2 + ϵ (y3)2


+radic 32






F (σ1 σ2 σ3 σ4)

darrF 983080

τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5

983081darrdarr

F 983080

τ (2)1 τ

(2)2 τ

(2)3 τ

(2)4

983081darr

F (x1 x2 x3 x4) = F 983080

τ (3)1 τ (3)2 τ (3)3 τ (3)4983081


Q2

4 +

P 3

27 isin F (σ1 σ2 σ3 σ4)

Again as above from

z 1z 2 = (τ 1)2

minus3τ 2 =

minus3P


so that 983080z 1

3

9830813 983080z 2

3

9830813

= minusP 3

27983080z 1

3

9830813+983080z 2

3

9830813= minusQ


3

9830813983080z 2

3

9830813

= minusQ

2 plusmn991770

Q2

4 +

P 3

27


X 2 + QX minus P 3

27 = 0

We define

τ (1)1 = σ1 τ

(1)2 = τ 1 τ

(1)3 = Q τ

(1)4 =

991770 Q2

4 +

P 3

27 τ

(1)5 = P




so that

τ (1)3 =

983080z 1

3

9830813+983080z 2

3

9830813and τ

(1)4 =

1

2

1048616983080z 1

3

9830813minus983080z 2

3

98308131048617


3z 1z 2 to F

983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5


983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5



983080τ (1)4 983081

2

= Q2

4

+ P 3



(2)2 = τ 1 τ

(2)3 = z 1 τ

(2)4 = z 2

Since 983080τ (2)3

9830813=

1

2τ (1)3 + τ

(1)4

983080τ (2)4

9830813=

1

2τ (1)3 minus τ

(1)4


τ (3)1 = y0 τ

(3)2 = y1 τ

(3)3 = y2 τ

(3)4 = y3


and983080

τ (3) j


F 983080

τ (2)1 τ

(2)2 τ

(2)3 τ

(2)4

983081for j = 2 3 4



1 sub A3 sub S 4


1 sub

K 4 sub

A4 sub

S 4






rarr S 3


1 rarr A3 rarr S 4








radic


y1 = x1 + x2 + x3

y2 = x1 + ϵx2 + ϵ2x3

y3 = x1 + ϵ2x2 + ϵx3


983163ϵ jy2 ϵky3

983165 for some

0

le j k





so that

983080y2

3 9830813

983080y3

3 9830813

= minus p3

27

983080y2

3 9830813

+

983080y3

3 9830813

= minusq


3

9830813983080y3

3

9830813= minusq

2 plusmn991770

q 2

4 +

p3

27


X 2 + qX minus p3

27 = 0


F (σ1 σ2 σ3)

darrF 983080

τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4

983081darr

F (x1 x2 x3) = F 983080

τ (2)1 τ

(2)2 τ

(2)3

983081




We define

τ (1)1 = σ1 τ

(1)2 = q τ

(1)3 =

991770 q 2

4 +

p3

27 τ

(1)4 = p

so that

τ (1)2 =

983080y2

3

9830813+983080y3

3

9830813and τ

(1)3 =

1

2

1048616983080y2

3

9830813minus983080y3

3

98308131048617


3y2y3 to F

983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4


(1)1 τ

(1)2 τ

(1)3 τ



F (x1 x2 x3) = F (σ1 p q )


983080τ (1)3

9830812=

q 2

4 +

p3

27 isin F (σ1 p q ) = F (x1 x2 x3)



j = 1 2 3 with

τ (2)1 = τ

(1)1

983080τ (2)2

9830813=

1

2τ (1)2 + τ

(1)3

983080τ (2)3

9830813

= 1

2τ (1)2 minus τ

(1)3

so that κ21 = 1 and κ22 = κ23 = 3



1 a b c










y0 = 12 (x1 + x2 + x3 + x4)





τ 1 = (y1)2 + (y2)2 + (y3)2

τ 2 = (y1)2 (y2)2 + (y1)2 (y3)2 + (y2)2 (y3)2

τ 3 = (y1)2 (y2)2 (y3)2



P = minusτ 21

3 + τ 2

Q = minus2τ 31

27 + τ 1τ 2

3 minus τ 3


z 1 = (y1)2 + ϵ (y2)2 + ϵ2 (y3)2

z 2 = (y1)2 + ϵ2 (y2)2 + ϵ (y3)2


+radic 32






F (σ1 σ2 σ3 σ4)

darrF 983080

τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5

983081darrdarr

F 983080

τ (2)1 τ

(2)2 τ

(2)3 τ

(2)4

983081darr

F (x1 x2 x3 x4) = F 983080

τ (3)1 τ (3)2 τ (3)3 τ (3)4983081


Q2

4 +

P 3

27 isin F (σ1 σ2 σ3 σ4)

Again as above from

z 1z 2 = (τ 1)2

minus3τ 2 =

minus3P


so that 983080z 1

3

9830813 983080z 2

3

9830813

= minusP 3

27983080z 1

3

9830813+983080z 2

3

9830813= minusQ


3

9830813983080z 2

3

9830813

= minusQ

2 plusmn991770

Q2

4 +

P 3

27


X 2 + QX minus P 3

27 = 0

We define

τ (1)1 = σ1 τ

(1)2 = τ 1 τ

(1)3 = Q τ

(1)4 =

991770 Q2

4 +

P 3

27 τ

(1)5 = P




so that

τ (1)3 =

983080z 1

3

9830813+983080z 2

3

9830813and τ

(1)4 =

1

2

1048616983080z 1

3

9830813minus983080z 2

3

98308131048617


3z 1z 2 to F

983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5


983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5



983080τ (1)4 983081

2

= Q2

4

+ P 3



(2)2 = τ 1 τ

(2)3 = z 1 τ

(2)4 = z 2

Since 983080τ (2)3

9830813=

1

2τ (1)3 + τ

(1)4

983080τ (2)4

9830813=

1

2τ (1)3 minus τ

(1)4


τ (3)1 = y0 τ

(3)2 = y1 τ

(3)3 = y2 τ

(3)4 = y3


and983080

τ (3) j


F 983080

τ (2)1 τ

(2)2 τ

(2)3 τ

(2)4

983081for j = 2 3 4



1 sub A3 sub S 4


1 sub

K 4 sub

A4 sub

S 4






rarr S 3


1 rarr A3 rarr S 4






We define

τ (1)1 = σ1 τ

(1)2 = q τ

(1)3 =

991770 q 2

4 +

p3

27 τ

(1)4 = p

so that

τ (1)2 =

983080y2

3

9830813+983080y3

3

9830813and τ

(1)3 =

1

2

1048616983080y2

3

9830813minus983080y3

3

98308131048617


3y2y3 to F

983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4


(1)1 τ

(1)2 τ

(1)3 τ



F (x1 x2 x3) = F (σ1 p q )


983080τ (1)3

9830812=

q 2

4 +

p3

27 isin F (σ1 p q ) = F (x1 x2 x3)



j = 1 2 3 with

τ (2)1 = τ

(1)1

983080τ (2)2

9830813=

1

2τ (1)2 + τ

(1)3

983080τ (2)3

9830813

= 1

2τ (1)2 minus τ

(1)3

so that κ21 = 1 and κ22 = κ23 = 3



1 a b c










y0 = 12 (x1 + x2 + x3 + x4)





τ 1 = (y1)2 + (y2)2 + (y3)2

τ 2 = (y1)2 (y2)2 + (y1)2 (y3)2 + (y2)2 (y3)2

τ 3 = (y1)2 (y2)2 (y3)2



P = minusτ 21

3 + τ 2

Q = minus2τ 31

27 + τ 1τ 2

3 minus τ 3


z 1 = (y1)2 + ϵ (y2)2 + ϵ2 (y3)2

z 2 = (y1)2 + ϵ2 (y2)2 + ϵ (y3)2


+radic 32






F (σ1 σ2 σ3 σ4)

darrF 983080

τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5

983081darrdarr

F 983080

τ (2)1 τ

(2)2 τ

(2)3 τ

(2)4

983081darr

F (x1 x2 x3 x4) = F 983080

τ (3)1 τ (3)2 τ (3)3 τ (3)4983081


Q2

4 +

P 3

27 isin F (σ1 σ2 σ3 σ4)

Again as above from

z 1z 2 = (τ 1)2

minus3τ 2 =

minus3P


so that 983080z 1

3

9830813 983080z 2

3

9830813

= minusP 3

27983080z 1

3

9830813+983080z 2

3

9830813= minusQ


3

9830813983080z 2

3

9830813

= minusQ

2 plusmn991770

Q2

4 +

P 3

27


X 2 + QX minus P 3

27 = 0

We define

τ (1)1 = σ1 τ

(1)2 = τ 1 τ

(1)3 = Q τ

(1)4 =

991770 Q2

4 +

P 3

27 τ

(1)5 = P




so that

τ (1)3 =

983080z 1

3

9830813+983080z 2

3

9830813and τ

(1)4 =

1

2

1048616983080z 1

3

9830813minus983080z 2

3

98308131048617


3z 1z 2 to F

983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5


983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5



983080τ (1)4 983081

2

= Q2

4

+ P 3



(2)2 = τ 1 τ

(2)3 = z 1 τ

(2)4 = z 2

Since 983080τ (2)3

9830813=

1

2τ (1)3 + τ

(1)4

983080τ (2)4

9830813=

1

2τ (1)3 minus τ

(1)4


τ (3)1 = y0 τ

(3)2 = y1 τ

(3)3 = y2 τ

(3)4 = y3


and983080

τ (3) j


F 983080

τ (2)1 τ

(2)2 τ

(2)3 τ

(2)4

983081for j = 2 3 4



1 sub A3 sub S 4


1 sub

K 4 sub

A4 sub

S 4






rarr S 3


1 rarr A3 rarr S 4








y0 = 12 (x1 + x2 + x3 + x4)





τ 1 = (y1)2 + (y2)2 + (y3)2

τ 2 = (y1)2 (y2)2 + (y1)2 (y3)2 + (y2)2 (y3)2

τ 3 = (y1)2 (y2)2 (y3)2



P = minusτ 21

3 + τ 2

Q = minus2τ 31

27 + τ 1τ 2

3 minus τ 3


z 1 = (y1)2 + ϵ (y2)2 + ϵ2 (y3)2

z 2 = (y1)2 + ϵ2 (y2)2 + ϵ (y3)2


+radic 32






F (σ1 σ2 σ3 σ4)

darrF 983080

τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5

983081darrdarr

F 983080

τ (2)1 τ

(2)2 τ

(2)3 τ

(2)4

983081darr

F (x1 x2 x3 x4) = F 983080

τ (3)1 τ (3)2 τ (3)3 τ (3)4983081


Q2

4 +

P 3

27 isin F (σ1 σ2 σ3 σ4)

Again as above from

z 1z 2 = (τ 1)2

minus3τ 2 =

minus3P


so that 983080z 1

3

9830813 983080z 2

3

9830813

= minusP 3

27983080z 1

3

9830813+983080z 2

3

9830813= minusQ


3

9830813983080z 2

3

9830813

= minusQ

2 plusmn991770

Q2

4 +

P 3

27


X 2 + QX minus P 3

27 = 0

We define

τ (1)1 = σ1 τ

(1)2 = τ 1 τ

(1)3 = Q τ

(1)4 =

991770 Q2

4 +

P 3

27 τ

(1)5 = P




so that

τ (1)3 =

983080z 1

3

9830813+983080z 2

3

9830813and τ

(1)4 =

1

2

1048616983080z 1

3

9830813minus983080z 2

3

98308131048617


3z 1z 2 to F

983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5


983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5



983080τ (1)4 983081

2

= Q2

4

+ P 3



(2)2 = τ 1 τ

(2)3 = z 1 τ

(2)4 = z 2

Since 983080τ (2)3

9830813=

1

2τ (1)3 + τ

(1)4

983080τ (2)4

9830813=

1

2τ (1)3 minus τ

(1)4


τ (3)1 = y0 τ

(3)2 = y1 τ

(3)3 = y2 τ

(3)4 = y3


and983080

τ (3) j


F 983080

τ (2)1 τ

(2)2 τ

(2)3 τ

(2)4

983081for j = 2 3 4



1 sub A3 sub S 4


1 sub

K 4 sub

A4 sub

S 4






rarr S 3


1 rarr A3 rarr S 4







F (σ1 σ2 σ3 σ4)

darrF 983080

τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5

983081darrdarr

F 983080

τ (2)1 τ

(2)2 τ

(2)3 τ

(2)4

983081darr

F (x1 x2 x3 x4) = F 983080

τ (3)1 τ (3)2 τ (3)3 τ (3)4983081


Q2

4 +

P 3

27 isin F (σ1 σ2 σ3 σ4)

Again as above from

z 1z 2 = (τ 1)2

minus3τ 2 =

minus3P


so that 983080z 1

3

9830813 983080z 2

3

9830813

= minusP 3

27983080z 1

3

9830813+983080z 2

3

9830813= minusQ


3

9830813983080z 2

3

9830813

= minusQ

2 plusmn991770

Q2

4 +

P 3

27


X 2 + QX minus P 3

27 = 0

We define

τ (1)1 = σ1 τ

(1)2 = τ 1 τ

(1)3 = Q τ

(1)4 =

991770 Q2

4 +

P 3

27 τ

(1)5 = P




so that

τ (1)3 =

983080z 1

3

9830813+983080z 2

3

9830813and τ

(1)4 =

1

2

1048616983080z 1

3

9830813minus983080z 2

3

98308131048617


3z 1z 2 to F

983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5


983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5



983080τ (1)4 983081

2

= Q2

4

+ P 3



(2)2 = τ 1 τ

(2)3 = z 1 τ

(2)4 = z 2

Since 983080τ (2)3

9830813=

1

2τ (1)3 + τ

(1)4

983080τ (2)4

9830813=

1

2τ (1)3 minus τ

(1)4


τ (3)1 = y0 τ

(3)2 = y1 τ

(3)3 = y2 τ

(3)4 = y3


and983080

τ (3) j


F 983080

τ (2)1 τ

(2)2 τ

(2)3 τ

(2)4

983081for j = 2 3 4



1 sub A3 sub S 4


1 sub

K 4 sub

A4 sub

S 4






rarr S 3


1 rarr A3 rarr S 4






so that

τ (1)3 =

983080z 1

3

9830813+983080z 2

3

9830813and τ

(1)4 =

1

2

1048616983080z 1

3

9830813minus983080z 2

3

98308131048617


3z 1z 2 to F

983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5


983080τ (1)1 τ

(1)2 τ

(1)3 τ

(1)4 τ

(1)5



983080τ (1)4 983081

2

= Q2

4

+ P 3



(2)2 = τ 1 τ

(2)3 = z 1 τ

(2)4 = z 2

Since 983080τ (2)3

9830813=

1

2τ (1)3 + τ

(1)4

983080τ (2)4

9830813=

1

2τ (1)3 minus τ

(1)4


τ (3)1 = y0 τ

(3)2 = y1 τ

(3)3 = y2 τ

(3)4 = y3


and983080

τ (3) j


F 983080

τ (2)1 τ

(2)2 τ

(2)3 τ

(2)4

983081for j = 2 3 4



1 sub A3 sub S 4


1 sub

K 4 sub

A4 sub

S 4






rarr S 3


1 rarr A3 rarr S 4







rarr S 3


1 rarr A3 rarr S 4



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