Department of Science and Technology
An Introduction to Invariant
Theory
¨
Orebro University
Department of Science and Technology Matematik C, 76 – 90 ECTS
An Introduction to Invariant Theory
Alberto Daniel January 5, 2017
Handledare: Holger Schellwat Examinator: Niklas Eriksen
Sj¨alvst¨andigt arbete, 15 hp Matematik, C–niva, 76 – 90 hp
Acknowledgements
I would like to thank the many people who have helped me with criticism and suggestions for this project, and in particular Holger Schellwat for reading, correcting, and commenting on the many preliminary drafts, as well as for many fruitful discussions.
I would like to express my thanks to the Mathematics Department at the ¨Orebro University for its hospitality, generosity and teachings during the autumn semester 2016/2017. Specific thanks are due to the Linnaeus Palme programme for the wonderful opportunity they gave me to study in Sweden.
I would like to express my thanks to the mathematics department at the Eduardo Mondlane university for many things they did for my studies.
I also grateful to Sloane whose paper [Slo 77] provided the initial moti-vation for this project.
Abstract
This work is an attempt to explain in some detail section III D of the pa-per: N.J.A. Sloane. ”Error-correcting codes and Invariant Theory [Slo 77]: New application of a Nineteenth-Century Technique”. For that, we will be concerned with polynomial invariants of finite groups which come from a group action. We will introduce the basic notions of Invariant Theory to give an almost self-contained proof of Molien’s theorem, and also present applications on linear codes.
Contents
1 Invariant Theory 13
1.1 Preliminaries . . . 13
1.2 Getting started . . . 15
1.3 Ordering on the Monomials . . . 19
1.4 The division algorithm . . . 21
1.5 Monomial Ideals and Dickson’s Lemma . . . 22
1.6 Symmetric polynomials . . . 24
1.7 Finite generation of invariants . . . 27
1.8 Molien’s Theorem . . . 32
Introduction
The purpose of this project is to give an almost self contained introduction and clarify the proof of an amazing theorem of Molien, as presented in [Slo 77]. In this paper, we discuss invariant theory of finite groups. We begin by giving some preliminaries for invariants and we prove the fundamental theorem on symmetric functions.
We shall prove the fundamental results of Hilbert and Noether for in-variant rings of finite linear groups. We will also derive the Molien’s formula of the ring of invariants. We will show, through examples, that the Molien’s formula helps us to see how many Invariants are linearly independent under finite group.
Invariant Theory is concerned with the action of groups on rings and the invariants of the action. Here we will restrict ourselves to the left actions of finite linear groups on homogeneous polynomials rings with entries in the complex number C. In the 19th century it was found that the set of all homogeneous invariants under group G, that we will denote by RG, could be described fully by finite set of generators for several suggestive special cases of G. It soon became clear that the fundamental problem of Invariant Theory was to find necessary and sufficient conditions for RG to be finitely generated. In this paper we will give the answer of this problem.
Chapter 1
Invariant Theory
1.1
Preliminaries
1.1.1 Definition. The group G acts on the set X if for all g ∈ G, there is a map
G × X → X, (g, x) 7→ g.x such that
1. ∀x ∈ X, ∀g, h ∈ G : h.(g.x) = (hg).x. 2. ∀x ∈ X : 1.x = x.
1.1.2 Remark. An action of G on X may be viewed as a homomorphism G → S(X) = {θ : θ : X → X bijective}.
Hence, we may write g(x) instead of g.x. We will use both notations. When X is a vector space we get the following.
1.1.3 Definition. A linear representation of a finite group G is a homo-morphism G −→ L(V, V ).
Here, L(V, V ) is the space of linear mappings from a vector space V to itself. The dimension of the vector space V is known as the dimension of the representation.
Once we have chosen a basis for V the elements of L(V, V ) can be inter-preted as n×n matrices, where n is the dimension of the representation. The condition that the representation must be a homomorphism now becomes the condition
∀g, h ∈ G : [g][h] = [gh].
The multiplication on the left is matrix multiplication while the product on the right of this equation is multiplication in the group. This condition has the consequence that
1.1.4 Definition. Let V be a finite dimensional complex inner product space and T : V → V linear. Then T is called unitary if
∀u, v ∈ V : hT (u), T (v)i = hu, vi.
1.1.5 Proposition. In this context, with [T ] ∈ Mat(n, n, C) standard ma-trix w.r.t orthonormal basis. Then T is unitary ⇐⇒ [T ]∗[T ] = I, where [T ]∗= [T ]>⇐⇒ [T ] has orthonormal rows ⇐⇒ [T ] has orthonormal columns. Proof. For vectors u, v of the same dimension, we have that u · v := u>v where the right-hand term is just matrix multiplication.
Let [T ] be defined as follows, with each vi∈ Cn, vi = [T (xi)]: [T ] = v1 | · · · | vn
Now, note that [T ]> is equal to almost the same thing, except the columns are now the rows, and we have turned them up. it’s equal to:
[T ]>= v>1 .. . v>n
Now we just multiply by [T ], we obtain
[T ]>[g] = v>1 .. . v>n v1 | · · · | vn = v>1v1 · · · v>1vn .. . ... v>nv1 · · · v>1vn But, since u · v = u>v, we can simplify this nightmarish mess.
= v1v1 · · · v1vn .. . ... vnv1 · · · v1vn
Since [T ] has orthonormal rows or columns, then vi · vj = 1 if i = j and vi· vj = 0 if i 6= j. Replacing these values in the matrix, we get:
[T ]>[T ] = 1 0 · · · 0 0 1 · · · 0 .. . ... ... 0 0 · · · 1 Thus, [T ]>[T ] = I ⇐⇒ [T ]>[T ] = I.
Now we start looking at the ring R = C[x1, · · · , xn] of complex polyno-mials in n variables.
1.1.6 Definition. The polynomial f ∈ R is homogeneous of degree d if f (tv) = tdf (v) for all v ∈ V and t ∈ C.
Cleary, the product of polynomials f1 and f2 homogeneous of degrees d1 and d2 respectively, is again homogeneous polynomial of total degree d1+ d2 since (f1f2)(tv) = f1(tv)f2(tv) = td1f1(v)td2f2(v) = td1+d2(f1f2)(v). Now, it follows that every homogeneous polynomials f can be written as a sum
f = X
|α|=d aαxα
where α = (α1, · · · , αn) ∈ Nn, aα ∈ C and xα:= xα11· · · xαnn. Every polyno-mial f over C can be expressed as a finite sum of homogeneous polynopolyno-mials. The homogeneous polynomials that make up the polynomial f are called the homogeneous components of f .
1.1.7 Example. 1. f (x, y) = x2+ xy + y2 is a homogeneous polynomial of degree 2.
2. f (x) = x3+ 1 is not a homogeneous polynomial.
3. f (x, y, z) = x3+ xyz + yz2+ x2+ xy + yz + y + 2x + 6 is a polynomial that is the sum of four homogeneous polynomials: x3+ xyz + yz2 (of degree 3), x2+ xy + yz (of degree 2), y + 2x (of degree 1) and 6 (of degree 0).
Let R be the algebra of polynomials in the variables x1, · · · , xn with coefficients in C, i.e, R = C[x1, · · · , xn]. Note that f ∈ R is homogeneous if and only if all their components are the same total degree.
In terms of the basis x = (x1, · · · , xn), we have Rd = C[x1, · · · , xn]d consisting of all homogeneous polynomials of degree d.
1.1.8 Definition. Rd:= {f ∈ R : f is homogeneous of degree d}. 1.1.9 Proposition. R =L
d∈NRd.
Since the monomials x1, · · · , xn of degree one form a basis for R1, it follows that their products x2 := all monomials of total degree 2 form a basis for R2, and, in general, the monomials xd11· · · xdnn for d1+ · · · + dn= d form a basis xd of Rd.
1.2
Getting started
The ring R = C[x1, · · · , xn] may be viewed as a complex vector space where multiplication by scalars is multiplication with constant polynomials.
We want to make the connection between vector spaces and rings much clear by viewing the variables as linear forms.
From now on V denotes a complex inner product space with orthonormal standard basis e = (e1, · · · , en) (only here ei denote standard basis vectors) and V∗ = L(V, C) its algebraic dual space with orthonormal basis x = (x1, · · · , xn) satisfying ∀1 ≤ i, j ≤ n : xi(ej) = δij.
Thus, V∗ = hx1, · · · , xni and at the same time R = C[x], that is, the variables are linear forms. In particular, R1 = V∗ (and R0 ≈ C) and R is an algebra.
From now on we also fix a finite group G acting unitarily on V∗, i.e. for every g ∈ G, the mapping V∗ → V∗, f 7→ g.f (or f 7→ g(f )) is unitary. Using coordinates we get [g.f ]x,x= [g]x,x[f ]xor shorter [g.f ] = [g][f ], where
[g] := [g]x,x=
[g(x1)]x | · · · | [g(xn)]x is its standard matrix, which is unitary, i.e. [g]−1 = [g]∗ = [g]>.
1.2.1 Example. Let G = D2 = {1, ρ, σ, τ } = hρ, σ; ρ2 = σ2 = (σρ)2 = 1i. V∗ = hx, yi
σ.x = −x, σ.y = y, ρ.x = x, ρ.y = −y =⇒ [1] = I = 1 0 0 1 , [σ] = −1 0 0 1 , [ρ] = 1 0 0 −1 , [τ ] = −1 0 0 −1 f (x, y) = 2x + 3y ⇐⇒ [f ] = 2 3 gives [σ(f )] = [σ][f ] = −1 0 0 1 2 3 = −2 3 , i.e (σ(f ))(x, y) = −2x + 3y = f (σ.x, σ.y), From τ = σρ we conclude [τ ] = [σρ] = [σ ◦ ρ] = [σ][ρ] = −1 0 0 1 1 0 0 −1 = −1 0 0 −1
In terms of representation theory we get a unitary representation T1 : G → U (V∗), g 7→ Tg1 : V∗ → V∗, f 7→ Tg1(f ) = g.f. Since R1 = V∗ we can write this as
T1 : G → U (R1), g 7→ Tg1: R1→ R1, f 7→ Tg1(f ) = g.f. In the previous example, Tσ1(2x + 3y) = −2x + 3y.
1.2.2 Definition. Let g ∈ G and d ≥ 0. Define Tgd: Rd→ Rd by 1. Tg0 : R0→ R0, C → C : d = 0
3. For d ≥ 1 and xi1· · · xid put Tg(xi1· · · xid) = d Y j=1 Tg(xij) and extend linearly so that Tg d X j=1 ajxαj = d X j=1 ajTg(xαj), where αj ∈ Nn, x = (x1, · · · , xn). Finally, for f = f1+ · · · + fd∈ R with
∀1 ≤ j ≤ d : fj ∈ Rd put Tg: R → R, f 7→ Tg(f ) := d X j=1
Tgj(fj).
1.2.3 Lemma. For every g ∈ G, Tg : R → R is an algebra automorphism preserving the grading, i.e. ∀d ∈ N : Tg(Rd) ⊆ Rd.
Proof. For g ∈ G, c ∈ C and f, f0 ∈ R. Let define f and f0 as f := f1+ · · · + fd and f0 := f10 + · · · + fd0,
where ∀1 ≤ i ≤ d : fi ∈ Ri. Then, using the Definition 1.2.2 we get 1. Tg(f + f0) = d X i=1 Tgi(fi+ fi0) = d X i=1 Tgi(fi) + d X i=1 Tgi(fi0) = Tg(f ) + Tg(f0). 2. Tg(f · f0) = Tg X 1≤i,j≤d fifj0 = X 1≤i,j≤d Tg(fifj0) = X 1≤i≤d Tgi(fi)Tgj(f 0 j) = d X i=1 Tgi(fi) d X j=1 Tgj(fj0) = Tg(f ) · Tg(f0). 3. Tg(cf ) = d X i=1 Tgi(cfi) = d X i=1 cTgi(fi) = cTg(f ). 4. By part 2. it is clear that the grading is preserved.
5. To show that Tgd(f ) 7→ g.f is bijective it is enough to show that this mapping is injective on Rd. Let f ∈ Rd, then f ∈ ker(Tgd) implies that Tgd(f ) = g.f = 0 ∈ Rd, that is ∀x ∈ V∗ : f ([g]x) = 0. Since G acts on V∗ this implies that ∀x ∈ V∗ : f (x) = 0, so f is the zero mapping. Since our ground field has characteristic 0, this implies that f is the zero polynomial, which we view as an element of every Rd. Thus ker(Tgd) = {0} and hence the result.
1.2.4 Example. Let g(x) = √1 2 1 1 −1 1 x y and f (x, y) = x2− y2, then g(f ) = f ([g]x) = 1 √ 2(x + y) 2 − 1 √ 2(−x + y) 2 = x2+ 2xy. Setting Ag := [Tg1]x,x, then Ag[d]:= [Tgd]xd,xd is the d–th induced matrix
in [Slo 77], because Tg(f · f0) = Tg(f ) · Tg(f0). Also, if f, f0 are eigenvectors of T1
g corresponding to the eigenvalues w, w0, then f · f0 is an eigenvector of Tg2with eigenvalue w ·w0, because Tg(f ·f0) = Tg(f )·Tg(f0) = (wf )·(w0f0) = (ww0)(f · f0). All this generalizes to d > 2.
For f ∈ R and g ∈ G we say that f is an invariant of g if g.f = f and that f is a (simple) invariant of G if ∀g ∈ G : g.f = f . Finally, we call
RG:= {f ∈ R : ∀g ∈ G : g.f = f } the algebra of invariants of G.
1.2.5 Proposition. RG is a subalgebra of R.
Proof. For x∈ V∗, g ∈ G, c ∈ C, and f, f0 ∈ RG we check
1. g(f + f0)(x) = (g(f ) + g(f0))(x) = (f + f0)(x), thus (f + f0) ∈ RG. 2. g(f · f0)(x) = (g(f ) · g(f0))(x) = (f · f0)(x), thus f · f0 ∈ RG. 3. g(cf )(x) = (cg(f ))(x) = (cf )(x) hence cf ∈ RG
1.2.6 Proposition. Let f be a polynomial over C. Then, f is invariant under group G if and only if its homogeneous components are invariants. Proof. Note that we can write f as the sum of homogeneous polynomials, i.e. f = fd1 + · · · + fds, where fdi is homogeneous polynomial of degree di.
Now, let fd1, · · · , fds be invariants, then
∀g ∈ G : g(f ) = g(fd1+ · · · + fds) = g(fd1) + · · · + g(fds) = fd1+ · · · + fds = f
Thus, f is invariant under group G.
Conversely, let f be invariant, then for all g ∈ G g(f ) = g(fd1 + · · · + fds)
f = g(fd1) + · · · + g(fds)
fd1 + · · · + fds = g(f d1) + · · · + g(fds)
Since g.Rd = Rd, we have fd1 = g(fd1), · · · , fds = g(fds). Thus,
In other words have the following result:
1.2.7 Proposition. The algebra of invariants of G is naturally graded as RG =M
d∈N RGd,
where RGd = {f ∈ Rd : ∀g ∈ G : g.f = f }, called the d–th homogeneous component of RG.
1.2.8 Definition (Molien series). Viewing RGd as a vector space, we define ad:= dimCRGd,
the number of linearly independent homogeneous invariants of degree d ∈ N, and
ΦG(λ) := X d∈N
adλd, the Molien series of G.
Thus, the Molien series of G is an ordinary power series generating func-tion whose coefficients are the numbers of linearly independent homogeneous invariants of degree d. The following beautiful formula gives these numbers, its proof is the aim of this paper.
1.2.9 Theorem (Molien, 1897). ΦG(λ) := 1 |G| X g∈G 1 det(I − λ[g])
1.3
Ordering on the Monomials
In order to begin studying nice bases for ideals (for more details see [Cox 91]), we need a way of ordering monomials. We also want the ordering structure to be consistent with polynomial multiplication. This is formalized in the following definition.
1.3.1 Definition. A monomial order is a recipe for comparing two mono-mials in a polynomial ring R = C[x1, · · · , xn] with the following properties:
1. It is a total order: ∀α, β ∈ Nn, α 6= β: xα> xβ or xα < xβ; 2. It is a multiplicative order:
xα> xβ =⇒ xα+γ= xαxγ > xβ+γ, α, β, γ ∈ Nn
3. It is a well-order: every nonempty set (of monomials) has a minimal element.
1.3.2 Definition (Lexicographic Order ). Let xα:= xα1
1 · · · xαnn and xβ := xβ1
1 · · · x βn
n . We say α >lex β if there is k such that αk > βk and αi = βi for all i < k. We will write xα >lex xβ if α >lexβ.
1.3.3 Example. 1. (1, 2, 0) >lex(0, 3, 4) since α1= 1 > 0 = β1. 2. (3, 4, 5) >lex(3, 1, 9) since α2= 4 > 1 = β2 and α1= β1 = 3. 3. x1 >lex x2 >lexx3>lex· · · >lexxn since
(1, 0, · · · , 0) >lex (0, 1, 0, · · · , 0) >lex· · · >lex(0, · · · , 0, 1). In other words, the variable x1 is considered as most important, if it can not distinguish two monomials, then its role is passed to x2 etc. 4. x21 >lex x1x42 >lexx1x32>lexx1x2x53 >lex· · ·
1.3.4 Definition. Fixing a monomial ordering on C[x1, · · · , xn]. We write every polynomial f in C[x1, · · · , xn] in decreasing order of monomials in f , i.e.,
f = a1xα(1)+ · · · + anxα(k),
where ai6= 0 in C and xα(1)>lexxα(2) >lex· · · >lex xα(k) for α(i) ∈ Nn. 1. The multidegree of f is multideg(f ) = |α(1)| = α1(1) + · · · + αn(1). 2. The leading coefficient of f is LC(f ) = a1.
3. The leading monomial of f is LM(f ) = xα(1).
4. The leading term of f is LT(f ) = a1xα(1) = LC(f )LM(f ).
1.3.5 Example. Let f = 4xy2z + 4z2 − 5x3+ 7x2z2 ∈ C[x, y, z]. Then, with respect to lex ordering, one has
f = −5x3+ 7x2z2+ 4xy2z + 4z2.
In this case LC(f ) = −5, LM(f ) = x3, LT(f ) = −5x3 and multideg(f ) = 3. 1.3.6 Example. Let f := x1x2+ x2x3+ x23 ∈ R2 then LM(f ) = x1x2. 1.3.7 Lemma. (xα+weaker terms)(xβ+weaker terms) = xα+β+weaker terms. Proof. This follows from the following obvious observation: If xα >lex xβ, then for any monomial xγ, xα+γ >lex xβ+γ.
Recycling the lemma 1.3.7, we conclude that the leading monomial of product is product of leading monomial, that is: LM(f1f2) = LM(f1)LM(f2).
1.4
The division algorithm
For polynomials in one variable C[x1], the unique monomial order is given by
1 < x1 < x21< · · · ,
so that, for f ∈ C[x1], its leading term LT(f ) is simply the monomial of highest degree. Thus the usual division algorithm for polynomials in C[x1] can be write f uniquely in the form
f = qg + r, where no term of r is divisible by LT(f ).
In the division algorithm for polynomials in one variable, for the input of a divisor and a dividend we are guaranteed a unique and well defined output of a quotient and remainder. However, in the case of multivariate polynomials, the quotients and remainder depend on the monomial ordering and on the order of the divisors in the division. The division algorithm in the multivariate case allows us to divide f ∈ R by g1, · · · , gs ∈ R, so that we can express f in the form
f = q1g1+ · · · + qsgs+ r
where the remainder r should satisfy no term of r is divisible by any of LT(g1), · · · , LT(gs). The strategy is to repeatedly cancel the leading term off by subtracting off an appropriate multiple of one of the gi. However, the result of the division algorithm fails to be unique for multivariate polyno-mials because there may be a choice of divisor at each step.
The division Algorithm 1. Start with q1 = q2 = · · · = qi = r = 0.
2. If f = 0, stop. Otherwise, for each i = 1, · · · , s check if LT(gi) divides LT(f ). If so, replace f by f −LT(gLT(f )
i), add
LT(f )
LT(gi) to qiand then return to
the beginning of 2). If LT(gi) does not divide LT(f ) for any i continue to 3.
3. Add LT(f ) to r, replace f by f − LT(f ), and then return to the beginning of 2.
Since lex order is a well ordering and the multidegree of f is reduced in each iteration, after finitely many steps, the process must stop.
Recall that an ideal I = hf1, · · · , fti ⊂ R generated by f1, · · · , ft∈ R is the set of all elements of the form h1f1+ · · · + htft, where h1, · · · , ht∈ R. Now, if the remainder when f is divided by g1, · · · , gs is zero, then clearly f is in the ideal generated by g1, · · · , gs.
1.4.1 Example. Let us divide f = x2y + y3by g1 = xy + y2 and g2= x − y, using lex order with x > y
LT(f ) LT(g1) = x 2y xy = x = q1, f0= f − q1g1= x2y + y3− x(xy + y2) = y3− xy2 q2 = LT(f0) LT(g2) = −xy 2 x = −y 2, f00= f0− q2g2 = y3− xy2+ xy2− y3= 0 Since f00= 0 we stop, so r = 0 and hence
f = q1g1+ q2g2+ r
= x(xy + y2) + (−y2)(x − y) + 0 = x(xy + y2) + (−y2)(x − y)
1.4.2 Example. We will divide f = x3y2+ xy + x + 1 by g1 = x3+ 1 and g2= y2+ 1 using lex order with x > y.
q1 = LT(f ) LT(g1) = x 3y2 x3 = y 2 f0 = f − q1g1 = x3y2+ xy + x + 1 − y2(x3+ 1) = xy + x − y2+ 1. The first low terms, xy and x are not divisible by the leading term of g2, and so these go to the remainder components.
q2 =
xy + x − y2+ 1 y2+ 1 = −1
f0− q2g2− (xy + x) = xy + x − y2+ 1 − (−y2− 1) − (xy + x) = 2 And so this term is sent to the remainder component and we have total remainder r = xy + x + 2. Thus, we have
x3y2+ xy + x + 1 = y2(x3+ 1) + (−1)(y2+ 1) + (xy + x + 2).
1.5
Monomial Ideals and Dickson’s Lemma
1.5.1 Definition. A monomial ideal is an ideal generated by a set of mono-mials.
This is, I is a monomial ideal, if there is a subset A ⊂ Nn such that I consists of all polynomials which are finite sums of the form P
α∈Ahαxα, where hα ∈ R. We write I = hxα: α ∈ Ai. For example
I = hx5y2z, x2yz2, xy3z2i ⊂ R is a monomial ideal.
1.5.2 Proposition. Let I = hxα : α ∈ A ⊆ Nni be a monomial ideal. Then a monomial xβ ∈ I if and only if xβ is divisible by xα for some α ∈ A. Proof. If xβ is a multiple of xα for some α ∈ A, then xβ ∈ I by definition of ideal. Conversely, if xβ ∈ I, then
xβ = s X
i=1
fixα(i),
where fi∈ R and α(i) ∈ A. If we expand each fi as a linear combination of monomials, we see that every term on the right side of equation is divisible by some xα. Hence the left hand side xβ must have the same property.
The following lemma follows from [Cox 91] and [CJ 10].
1.5.3 Lemma (Dickson’s lemma). Every monomial ideal I = hxα : α ∈ A ⊆ Nni (i.e., ideal generated by monomials) is finitely generated.
Proof. We prove the lemma by induction on the number of variables. If there is a single variable x = x1, then I = hxα1, α ∈ A ⊆ Ni. Let β be the smallest element in this list. Then for any α ∈ A, xα1 = xβ1xγ1 for some γ ∈ N. This says that I is generated by the single element xβ1.
Now we assume that any monomial ideal in n − 1 variables generated by a subset A ⊆ Nn−1 is generated by a finite set of monomials, and we may take these monomials from the set A.
We write xn = y so C[x1, · · · , xn−1, y] = C[x1, · · · , xn]. We can write any polynomial in the form xαym for some α ∈ Nn−1 and m ∈ N. Let J ⊆ C[x1, · · · , xn−1] be the ideal of C[x1, · · · , xn−1] generated by all the xα so that xαym ∈ I for some m ∈ N. By the induction hypothesis J is generated by a finite set of α ∈ Nn−1. Call this set B∗ with |B∗| = s. This is just one part of the induction hypothesis. The second part tells us that we may take these xα so that there is a mj ∈ N such that
xαymj ∈ I
Let m = max{m1, m2, · · · , ms} so that xαymj ∈ B∗. Now let B = {xαym : α ∈ B∗}.
For each k with 0 ≤ k ≤ m − 1, let Jk denote the ideal of C[x1, · · · , xn−1] generated by xα for which xαyk∈ I. By induction this is finitely generated by elements xαk(i), that is J
k= hxαk(1), · · · , xαk(sk)i. Let Bk= {xαk(i)yk}. We claim that I is generated by
B ∪ B0∪ B1∪ · · · ∪ Bm−1.
We prove the claim: Let xβyt ∈ I. If t ≥ m, there is an α ∈ B∗ so that xα divides xβ, furthermore, since t ≥ m we have xαymj divides xβyt. One the
other hand, suppose that t < m. In this can there is an element of Bt that divides xβyt. We have shown that the ideal I has a finite set of generators. Call this set of generators H.
We show that we can find a finite set of generators in the set xα, α ∈ Nn. We have implicitly changed notation here. The notation xα now refers to a monomial in the complete set of variables x1, · · · , xn and not just the first n − 1 variables. For each element xα in H there is some element xβ in Nn for which xβ divides xα. The set of such xβ generates I and is a subset of xβ, β ∈ Nn.
1.6
Symmetric polynomials
Let the group Sn act on the polynomial ring R by permuting the variables. The polynomials invariant under this action of Snare called symmetric poly-nomials. Using matrix representation of Snwe have the following definition. 1.6.1 Definition. A polynomial in R is called symmetric if for any permu-tation matrix g, g.f = f .
1.6.2 Definition. In R, the elementary symmetric functions e1, · · · , enare defined by: e1 := x1+ · · · + xn e2 := x1x2+ · · · + x1xn+ x2x3+ · · · + x2xn+ · · · + xn−1xn .. . ... ek := X 1≤i1<···<ik≤n xi1· · · xik en := x1· · · xn.
The invariant ring of Sn in this representation is generated by the al-gebraically independent invariants e1, · · · , en and we denote by RSn := C[e1, · · · , en]. This result is proved in the following theorem.
The proof of the following is following [Cox 91].
1.6.3 Theorem. Every symmetric polynomial in R can be written uniquely as a polynomial in e1, · · · , en and are linearly independents.
Proof. We will use lexicographic order with x1 > x2 > · · · > xn. Given a non-zero f ∈ RSn, LT(f ) = axα, where α = (α
1, · · · , αn) ∈ Nn and a ∈ C∗, we first claim that α1 > α2· · · > αn. To prove this, suppose that αi < αi+1 for some i. Let β = (· · · , αi+1, αi, · · · ). Since axαf is a term of f , it follows that axβ is a term of f (· · · , xi+1, xi, · · · ). But f is symmetric, so that f (· · · , xi+1, xi, · · · ) = f , and hence, axx
β
is a term of f . This is impossible since β > α under lexicographic order, and our claim is proved.
Now let
f0:= eα1−α2
1 e
α2−α3
2 · · · eαnn.
To compute the leading monomial of f0, first note that LM(ek) = x1· · · xk for 1 ≤ k ≤ n. Hence LM(f0) = LM(eα1−α2 1 e α2−α3 2 · · · e αn n )
= LM(e1)α1−α2LM(e2)α2−α3· · · LM(en)αn = xα1−α2
1 (x1x2)α2−α3· · · (x1· · · xn)αn = xα1
1 · · · xαnn = xα
It follows that f and αf0 have the same leading term. Subtracting αf0 from f therefore cancels the monomial axα, thus LM(f − af0) < LM(f ) whenever f − af0 6= 0.
Now set f1 := f − af0 and note that f1 is symmetric since f and af0 are. Hence, if f1 6= 0, we can repeat the above process to form f2 := f1− a1f10, where a1 is constant and f10 is a product of powers of e1, · · · , en, defined as above.
Further we know that LM(f2) < LM(f1) when f2 6= 0. Continuing in this way, we get the sequence f, f1, f2, · · · with LM(f ) > LM(f1) > LM(f2) > · · ·
Since lex order is a well ordering, the sequence must be finite, hence fs+1= 0 for some s and it follows that
f = af0+ f1
= af0+ a1f10 + f2 ..
.
= af0+ a1f10 + · · · + asfs0
which shows that f is a polynomial in the symmetric polynomials. For uniqueness, suppose that we have f ∈ RSn which can be written
f := φ1(e1, · · · , en) = φ2(e1, · · · , en)
Hence, φ1 and φ2 are polynomials in n variables, say y1, · · · , yn. We need to prove that φ1 = φ2∈ C[x1, · · · , xn]. Let φ = φ1− φ2, then φ(e1, · · · , en) = 0 in C[x1, · · · , xn] and we need to prove that φ = 0 in C[y1, · · · , yn]. Suppose that φ 6= 0. If we write φ = X
β
aβyβ, then φ(e1, · · · , en) is a sum of the polynomials φβ = aβeβ11· · · e
βn
n , where β = (β1, · · · , βn). Now LM(φβ) = aβLM(e1)β1LM(e2)β2· · · LM(en)βn
= aβxβ11(x1x2) β2· · · xβn n = aβxβ11+···+βnx β2+···+βn 2 · · · xβn
As the map θ : Nn7→ Nn where
θ(α1, α2, · · · , αn) = (α1+ · · · + αn, α2+ · · · + αn, · · · , αn)
is injective, the φβ’s have distinct leading monomials. It follows that LT(φβ) can’t be cancelled and thus φ(e1, · · · , en) 6= 0 in C[x1, · · · , xn]. This is a contradiction. Hence RSn = C[e
1, · · · , en]. See for instance [Cox 91], chapter 7.
This theorem clarifies how to generalise the invariants ring of permuta-tion group Sn. If we have the permutation group S3, the invariants ring is generalized by elementary invariants e1, e2 and e3, i.e. RS3 := C[e1, e2, e3] where e1:= x + y + z, e2 := xy + xz + yz and e3 := xyz.
Note that the proof of Theorem 1.6.3 gives an algorithm for writing the invariants of permutation group Sn in terms of the elementary invariants e1, · · · , en.
1.6.4 Example. We write x3+ y3+ z3 as a polynomial in the elementary invariants ei. Since the leading monomial is x3y0z0 we subtract e31e02e03 and are left with −3(x2y + x2z + xy2+ xz2+ y2z + yz2) − 6(xyz).
The leading monomial is now x2y, so we add 3e2−11 e12e03. This leaves 3xyz = 3e1−11 e1−12 e13, which is reduced to zero in the next step.
This way we obtain x3+ y3+ z3 = e31− 3e1e2+ 3e3.
1.6.5 Definition. The polynomial pk = xk1+ · · · + xkn for k = 1, 2, · · · are called the power sum symmetric polynomials.
1.6.6 Theorem. Let n ∈ N∗ and k ∈ N. Then
pk= e1pk−1− e2pk−2+ · · · + (−1)kek−1p1+ (−1)k+1kek if k ≤ n. Proof. Let y, z be indeterminate. Then
(y − x1)(y − x2) · · · (y − xn) = yn− e1yn−1+ e2yn−2+ · · · + (−1)nen Put y = 1z to get
(1 − x1z)(1 − x2z) · · · (1 − xnz) = 1 − e1z + e2z2+ · · · + (−1)nenzn=: e(z) Consider the generating function of x1, · · · , xn
p(z) = p1z + p2z2+ p3z3+ · · · = ∞ X k=1 pkzk = ∞ X k=1 n X i=1 xkizk= n X i=1 ∞ X k=1 (xiz)k = n X i=1 xiz 1 − xiz
Since e(z) = (1 − x1z)(1 − x2z) · · · (1 − xnz), e0(z) = n X i=1 −xie(z) 1 − xiz and hence p(z) = n X i=1 xiz 1 − xiz = −ze 0(z) e(z) This implies that
p(z)e(z) = −z(−e1+ e2· 2z − e3· 3z2+ · · · + (−1)nnenzn−1) = e1z − 2e2z2+ 3e3z3+ · · · + (−1)n+1nenzn. If k ≤ n, equating the coefficient of zk we get
(−1)k+1kek= pk− e1pk−1+ pk−2e2+ · · · + (−1)kp1ek−1 Hence
pk = e1pk−1− e2pk−2+ · · · + (−1)k+1kek.
1.6.7 Theorem. Every symmetric polynomial in C[x1, · · · , xn] can be writ-ten as a polynomial in the power sums p1, · · · , pn, where pk= xk1+ · · · + xkn for k ∈ {1, · · · , n}.
Proof. We prove by induction on k that ek is a polynomial in p1, · · · , pn. This is true for k = 1 since e1 = p1. If we assume that that for some r, r > 1 any et, t ∈ {1, · · · , r − 1}, can be written as a polynomial in the pk where pk= xk1+ · · · + xkn, for k ∈ {1, · · · , n}, then from Newton’s identities:
er= (−1)r+1 1
r(pr− e1pr−1+ e2pr−2+ · · · + (−1) r−1e
rp1).
which can be written since char(C) = 0, and er is a polynomial in the pk, where pk = xk1 + · · · + xkn, for k ∈ {1, · · · , n}. Hence, by induction, any of ek is a polynomial in the pk, and finally, with the fundamental theorem of symmetric polynomials, it is also true for any symmetric polynomial.
1.7
Finite generation of invariants
We have seen that RSn is generated by n elementary symmetric polynomials.
We now show that this property is shared by RG for all finite subgroups G of GL(n, C).
1.7.1 Definition. Let G ≤ GL(V ), the Reynolds Operator of G is the map ρ : R → R, ρ(f ) = 1
|G| X g∈G
One can think of ρ(f ) as averaging the effect of G on f . This operator is the method of creating the invariants under G and we prove that ρ(f ) is invariant in the following proposition.
1.7.2 Proposition. Let ρ be the Reynolds operator of G. 1. If f ∈ R then ρ(f ) ∈ RG.
2. ρ ◦ ρ = ρ
3. If f ∈ RG and f0 ∈ R then ρ(f f0) = f ρ(f0). Proof. 1. Let h ∈ G. Then,
h(ρ(f )) = ρ(f ([h]v)) = 1 |G| X g∈G g(f ([h]v)) = 1 |G| X g∈G g(h(f (v)) = 1 |G| X g∈G (g ◦ h)(f (v) = 1 |G| X u∈G u(f (v)) = ρ(f (v)) Hence, ∀f ∈ R : ρ(f ) ∈ RG. 2. Let f ∈ R, we check ρ(ρ(f )) = 1 |G| X g∈G g(ρ(f )) = 1 |G| X g∈G g 1 |G| X h∈G h(f ) ! = 1 |G| X g∈G 1 |G| X h∈G g(h(f )) = 1 |G| X g∈G 1 |G| X h∈G (g ◦ h)(f ) = X g∈G 1 |G| 1 |G| X u∈G u(f ) = |G| · 1 |G| 1 |G| X u∈G u(f ) = ρ(f ). Hence ∀f ∈ R, (ρ ◦ ρ)(f ) = ρ(f ).
3. Let f ∈ RG and f0 ∈ R, then ∀g ∈ G : g(f ) = f . Thus, ρ(f f0) = 1 |G| X g∈G g(f f0) = 1 |G| X g∈G (g(f ) · g(f0)) = 1 |G| X g∈G f · g(f0) = f · 1 |G| X g∈G g(f0) = f ρ(f0).
One nice aspect of this proposition is that it gives us a way of creating invariants.
1.7.3 Definition. Let R be a ring. If M 6= R is an ideal of R such that whenever N is an ideal of R containing M , then N = M or N = R is called an maximal ideal of R.
1.7.4 Definition. Let LM(f ) be the leading monomial of a non-zero func-tion f , then the initial ideal of I ⊆ R is
in(I) = hLM(f ) : f ∈ Ii.
1.7.5 Theorem (Hilbert’s basis theorem). Every ideal in the polynomial ring R = C[x1, · · · , xn] is finitely generated.
Proof. Let I ⊆ R be a non-zero ideal of R, then by Dickson’s lemma 1.5.3, its initial ideal is finitely generated:
in(I) = hm1, · · · , msi.
Now we can show that each monomial miis the leading term of some gi∈ I. Certainly
mi= h1LT(g1) + · · · + htLT(gt)
for some finite set of gj ∈ I. So mi ∈ hLT(g1), · · · , LT(gs)i and hence mi will be divisible by one of the monomials, say LT(gj). This means that mi = xαLT(gj) for some monomial xα. By the properties of a monomial order xαLT(gj) = LT(xαgj) and xαgj ∈ I. So we see that mi = LT(gi) for some gi∈ I.
So far we have shown that in(I) = hLT(g1), · · · , LT(gs)i for some finite set of polynomials g1, · · · , gs ∈ I. Finally, we must show that the set of polynomials in the initial ideal does in fact generate the ideal. Clearly, the hg1, · · · , gsi ⊆ I since each gi ∈ I. Now suppose there is an f ∈ I but not in hg1, · · · , gsi. Dividing this f by the set of gis gives
f = h1g1+ · · · + hsgs+ r,
where the remainder r, has no monomials divisible by LT(gi). Now, r = f − h1g1− · · · − hsgs,
so that the leading monomial of the remainder is in the ideal in(I) and hence LT(r) ∈ hLT(g1), · · · , LT(gs)i this gives a contradiction unless r = 0. Of course when r = 0 we have that f ∈ hg1, · · · , gsi and so we may conclude that I = hg1, · · · , gsi. Then, the theorem is proved.
1.7.6 Theorem (Hilbert’s finiteness theorem). Let G ≤ GL(n, C) be a matrix group, then RG is a finitely generated C-algebra.
Proof. Let M be the maximal ideal of RG generated by homogeneous ele-ments of positive degree. Since every ideal of C[x1, · · · , xn] has a finite set of generators, the M has finitely many generators. Let these be homogeneous elements f1, · · · , fs∈ M .
We claim that RG= C[f1, · · · , fs]. Let f ∈ R be homogeneous of degree d. Apply induction on d. If d = 0, f ∈ C. Suppose that d > 0. Then f ∈ M . Hence ∃f10, · · · , fs0 ∈ R such that f = f0
1f1+ · · · + fs0fs. Apply Reynolds operator in G to get f = ρ(f10)f1+ · · · + ρ(fs0)fs. We may assume that fi0 are homogeneous. Then deg(ρ(fi0)) = deg(fi0) = deg(f ) − deg(fi) < deg(f ). Since ρ(fi0) are of smaller degree than deg(f ), by induction ρ(f10), · · · , ρ(fs0) ∈ C[f1, · · · , fs]. Hence f ∈ C[f1, · · · , fs]. For more details of this proof see [DK 95].
Let xα := xα1
1 · · · xαnn, where x ∈ Cn and α := (α1, · · · , αn) ∈ Nn. The following result is along [Cox 91].
1.7.7 Theorem (Noether). Let G 6 GL(n, C) be a finite matrix group, then C[x1, · · · , xn]G= C[ρ(xβ) : |β| ≤ |G|]. Proof. Let f =X α cαxα∈ RG. Then, f = ρ(f ) = ρ(X α cαxα) = X α cαρ(xα).
Hence, every invariant polynomial is a C-linear combination of the ρ(xα). We now shall prove that for all α, ρ(xα) ∈ C[ρ(xβ) : |β| ≤ |G|].
Let (x1+ · · · + xn)k= X |α|=k aαxα where aα = k α = α k! 1!···αn!.
If A ∈ G is a matrix, let Ai denote the i-th row of A. We can define (Ax)α:= (A1x)α1· · · (Anx)αn.
In this notation, we have ρ(xα) = 1 |G| X A∈G (A1x)α1· · · (Anx)αn = 1 |G| X A∈G (Ax)α. We introduce n new variables u1, · · · , un, and we now have
(u1A1x + · · · + unAnx)k= X |α|=k
aα(Ax)αuα.
If we sum both members over all A ∈ G, we get X A∈G (u1A1x + · · · + unAnx)k= X |α|=k aα X A∈G (Ax)α ! uα = X |α|=k bαρ(xα)uα where bα = |G|aα.
Let UA= u1A1x + · · · + unAnx, and Sk= X A∈G UAk, then we have Sk= X |α|=k bαρ(xα)uα
By theorem 1.6.7, any polynomial in the UA and symmetric in the UA can be written as a polynomial in the Sk, 0 ≤ k ≤ |G|. Hence,
X |α|=k
bαρ(xα)uα= Sk = F (S1, · · · , S|G|) for some polynomial F ∈ C[Y1, · · · , Y|G|]. Thus,
X |α|=k bαρ(xα)uα= F ( X |β|=1 bβρ(xβ)uβ, · · · , X |β|=|G| bβρ(xβ)).
Expanding the last expression and equaling the coefficients of uα, it follows that bαρ(xα) is a polynomial in the ρ(xβ) with |β| ≤ |G|.
Since char(C) = 0, |G| 6= 0 and bα = |G|aα 6= 0, we have ρ(xα) in the desired form.
Therefore, ∀α ∈ Nn : ρ(xα) ∈ C[ρ(xβ : |β| ≤ |G|] and since ρ(R) = RG, the result is proven.
1.7.8 Example. Consider the cyclic matrix group C4⊂ GL(2, C) of order 4 generated by A = 0 −1 1 0
By Noether’s theorem we can take many monomials and average the Reynolds operator to find some invariants.
ρ(f ) = 1 |C4| X g∈C4 g(f ) = 1
4[f (x, y) + f (−y, x) + f (−x, −y) + f (y, −x)] . Now, ρ(x2) = 1 4x 2+ (−y)2+ (−x)2+ y2 = 1 2(x 2+ y2), ρ(xy) = 1
4[xy + (−y)x + (−x)(−y) + y(−x)] = 0, ρ(x3y) = 1 4x 3y + (−y)3x + (−x)3(−y) + y3(−x) = 1 2(x 3y − xy3), ρ(x2y2) = 1 4x 2y2+ (−y)2x2+ (−x)2(−y)2+ y2(−x)2 = x2y2. Thus, x2+ y2, x3y − xy3, x2y2∈ RC4.
So far we know how to find the invariants of finite group and by Hilbert’s finiteness theorem the set of invariants is finitely generated, now our problem is to find the linearly independent invariants, or at least how many they are.
1.8
Molien’s Theorem
While the Noether’s theorem is nice and constructive, it is rather unwieldy. For instance, to compute the n generating invariants of the permutation representation of Sn, we would need to compute the Reynolds operator for every single monomial of degree less than or equal to n!, and no one wants to do that. There is however, a way to see in advanced how many linearly independent invariants of a given degree are, and it is due to Molien. 1.8.1 Theorem. The number of linearly independent invariants of G of degree 1 is given by the sum of the traces of the matrices of G,
a1 = |G|1 Pg∈GTr([g]). Proof. Let S : V∗ → V∗, S(f ) = 1 |G| X g∈G g(f ). Since V∗ ⊂ R and by Proposition 1.7.2 we have S ◦ S = S, that is S2= S ⇐⇒ S2− S = 0. Thus, S satisfies t(t − 1) = 0, and it follows that [S] is diagonaizable, i.e.
[S0] = [T ][S][T ]−1.
Now we can change coordinates, and it follows that [S0] corresponds to matrix with 0 and 1 as eigenvalues. Hence, the diagonal entries of [S0] are 0 or 1. [S0] = 1 0 . .. 1 0 . .. 0 0
Each 1 on the diagonal corresponds to a linearly independent invariant so the trace of this matrix count the number of linearly independent invariants of degree 1, and Tr([S0]) = Tr([T ][S][T ]−1) = Tr([S][T ]−1[T ]) = Tr([S]). Finally, the trace is a linear operator so the trace of a sum is the sum of the traces and hence the result.
For finite matrix groups G, it is possible to compute the Molien series directly, without prior knowledge about generators. This is captured in the following beautiful theorem of Molien
1.8.2 Theorem (Molien 1897). Suppose that ad is the number of linearly independent homogeneous invariants of G with degree d, and
be the generating function for this sequence, then,
ΦG(λ) = |G|1 Pg∈G det(I−λ[g])1 .
Proof. The number of degree d invariants ad is given by the sum of the traces of the matrices [g][d] divided by the order of G, i.e,
ad= 1 |G| X g∈G Tr([g][d])
To prove the theorem all we need to do is to compare this sum with the one given in the theorem.
Since G is finite, gk = id for some k positive integer, then g satisfies tk− 1 = 0, which has n distinct roots over C, hence the minimal polynomial of g over C has no repeated roots and hence g is diagonalizable. Now given g ∈ G, let w1, w2, ..., wn be its eigenvalues. Since [g] is diagonalizable it follows that there exist n linearly independent eigenvectors v1, v2, ..., vn be-longing to w1, w2, ..., wnrespectively. A basis of C[x1, · · · , xn]dconsists of all monomials of degree d namely xd1
1 · · · xdnn where d1+ · · · + dn. If x1, · · · , xn are eigenvectors of Tg1 corresponding to the eigenvalues w1, · · · , wn, then xd1
1 · · · xdnn are eigenvectors of Tgd with eigenvalues wd11· · · wdnn, because Tg(xd11· · · xdnn) = Tg(xd11) · · · Tg(xdnn)
= (Tg(x1))d1· · · (Tg(xn))dn = (w1x1)d1· · · (wnxn)dn = wd1
1 · · · wdnn· xd11· · · xdnn.
Hence, even the induced operators are diagonalizable. Now, for this action of g we have
Tr([g][d]) = X d1+...+dn=d
wd1
1 · · · wdnn
Since [g] is diagonalizable, i.e. exists P ∈ GL(n, C) invertible such that
[g] = P DP−1, where D := diag(w1, ..., wd) then
det(I − λ[g]) = det(P P−1− λP DP−1) = det(P (I − λD)P−1)
= det(P ) det(I − λD) det(P−1)
So 1 det(I − λ[g]) = 1 (1 − λw1) · · · (1 − λwn) = 1 (1 − λw1) · · · 1 (1 − λwn) = (1 + λw1+ λ2w12+ · · · ) · · · (1 + λwn+ λ2wn2+ · · · ) The result follows by computing the coeficient of λdto be X
d1+···+dn=d
wd1
1 · · · wndn.
1.8.3 Example. Consider de matrix group V4 = {
±1 0
0 ±1
} ⊂ GL(2, C). The Molien’s series is calculated as follows
A1 := 1 0 0 1 , det(I − λA1) = (1 − λ)2 A2 := −1 0 0 −1 , det(I − λA2) = (1 + λ)2 A3 := −1 0 0 1 , det(I − λA3) = 1 − λ2 A4 := 1 0 0 −1 , det(I − λA4) = 1 − λ2 Hence ϕ(λ) = 1 4 1 (1 − λ)2 + 1 (1 + λ)2 + 2 1 − λ2 = 1 (1 − λ)2(1 + λ)2 = 1 (1 − λ2)2 = 1 + 2λ2+ 3λ4+ · · ·
Hence there are 2 independent invariants of degree 2.
Now, ρ(f ) = 14{f (x, y) + f (−x, −y) + f (−x, y) + f (x, −y)}. Hence ρ(x2) = 1 4{x 2+ (−x)2+ (−x)2+ x2} = 1 4 · 4x 2 = x2, ρ(y2) = 1 4{y 2+ (−y)2+ y2+ (−y)2} = 1 4 · 4y 2= y2.
Thus α := x2, β := y2 ∈ C[x, y]V4 and are linearly independent, hence
C[x, y]V4 = C[α, β].
1.8.4 Example. Our second example will be the permutation representa-tion of S3 as usual. We can simplify the calculation a little using the fact that det(I − λAα) is constant on a conjugancy class. So we have
det(I − λ[e]) = det 1 − λ 0 0 0 1 − λ 0 0 0 1 − λ = (1 − λ)3,
det(I − λ[σ]) = det 1 −λ 0 −λ 1 0 0 0 1 − λ = (1 − λ)(1 − λ2), and finally, det(I − λ[τ ]) = det 1 −λ 0 0 1 −λ −λ 0 1 = (1 − λ3), where e is the identity, σ is (12) and τ is (123). Assembling these results gives,
1 |G| |G| X α=1 1 det(I − λAα) = 1 6 1 (1 − λ)3 + 3 (1 − λ)(1 − λ2 + 2 (1 − λ3) = (1 − λ 2)(1 − λ3) + 3(1 − λ)2(1 − λ3) + 2(1 − λ)3(1 − λ2) 6(1 − λ)3(1 − λ2)(1 − λ3) = 1 (1 − λ)(1 − λ2)(1 − λ3) so, ΦS3(λ) = 1 (1 − λ)(1 − λ2)(1 − λ3),
this means that the ring of invariants is generated by three elements, a linear invariant, a degree 2 invariant and a degree 3 invariant. To see this in detail we can expand the first few terms in the generating function,
ΦS3(λ) = (1 + λ + λ
2+ · · ·)(1 + λ2+ λ4+ · · ·)(1 + λ3+ λ6+ · · ·) = 1 + λ + 2λ2+ 3λ3+ 4λ4+ · · ·
So there is one linear invariant, as we have seen this is I1 = x + y + z. There are two degree 2 invariants, one must be I12 = (x + y + z)2 the square of the linear invariant. The other is a new invariant which could be I2 = x2+ y2+ z2 or xy + xz + yz but not both since these are not linearly independent I12 − I2 = 2(xy + xz + yz). The three cubic invariants will be I13, I1I2 and a new invariant say I3 = x3+ y3+ z3. All the polynomial are now accounted for as sums and products of these three invariants, for example the four degree 4 invariants are I14, I12I2, I22 and I1I3.
1.9
Linear codes
We assume some familiarity with Coding Theory, for example [Bie 04]. A linear code is a linear subspace C ⊆ Fnq, where Fq is the field of q elements.
1.9.1 Definition. The weight of a word x ∈ Fnq is the number of nonzero positions in x, that is w(x) := |{i : xi = 1}|.
The Hamming distance d(u, v) between two words is defined as the num-ber of positions in which u and v differ: d(u, v) := |{i : ui 6= vi}|. And the minimum distance d of a non-trivial code C is given by
d := min{d(u, v) : u, v ∈ C, u 6= v}.
A code C ⊆ Fnq is called an [n, k, d]-code if the dimension of C is equal to k and the smallest Hamming distance between two distinct codewords is equal to d.
Let C ⊂ Fnq be an [n, k, d] linear code. Much information about a code, including the parameters d and k, can be read of from its weight enumerator WC. This is a polynomial in x, y and homogeneous of degree n.
Put: Ai:= |{x ∈ C : w(x) = i}| for i ∈ N and
WC(x, y) := n X
i=0
Aixn−iyi∈ C[x, y]
the weight enumerator polynomial, satisfying WC⊥(x, y) = 1
qkWC(x + (q − 1)y, x − y) (Mac Williams). For certain classes of self-dual binary [n, n/2, d] codes this implies
WC(x, y) = WC x + y √ 2 , x − y √ 2 , WC(x, y) = WC(x, iy). But this means that WC is invariant under the group
G = 1 √ 2 1 1 1 −1 , 1 0 0 i ≤ GL(2, C), |G| = 192.
Let ζ = e2πi8 be a primitive 8th root of unity. The group G defined above is
equal to the set of matrices ζk 1 0 0 α , ζk 0 1 α 0 , ζk√1 2 1 β α αβ
where α, β ∈ {1, i, −1, −i} and k = 0, · · · , 7, see [Slo 77]. Using Molien’s theorem, we can find the Molien’s series.
A computation [GAP] gives φG(λ) =
1
This suggests that the invariant ring is generated by two algebraically in-dependent polynomials f1, f2 homogeneous of degrees 8 and 24 respectively. Using Reynolds operator we can compute two invariants f1 and f2. By theorem 1.7.7 we can average f (x, y) = x8 to find invariant of degree 8.
ρ(f ) = 1 192 192 X γ=1 f (Aγx), Aγ∈ G
8 · 4 = 32 matrices in each line P32
γ=1f (Aγx) ζk 1 0 0 1 , ζk 1 0 0 −1 , ζk 1 0 0 i , ζk 1 0 0 −i 32x8 ζk 0 1 1 0 , ζk 0 1 −1 0 , ζk 0 1 i 0 , ζk 0 1 −i 0 32y8 ζk 1√ 2 1 1 1 1 , ζk 1√ 2 1 1 −1 −1 , ζk 1√ 2 1 1 i i , ζk 1√ 2 1 1 −i −i 2(x + y)8 ζk 1√ 2 1 −1 1 −1 , ζk 1√ 2 1 −1 −1 1 , ζk 1√ 2 1 −1 i −i , ζk 1√ 2 1 −1 −i i 2(x − y)8 ζk 1√ 2 1 i 1 i , ζk 1√ 2 1 i −1 −i , ζk 1√ 2 1 i i −1 , ζk 1√ 2 1 i −i 1 2(x + iy)8 ζk 1√ 2 1 −i 1 −i , ζk 1√ 2 1 −i −1 −i , ζk 1√ 2 1 −i i 1 , ζk 1√ 2 1 −i −i 1 2(x − iy)8 ρ(f ) = 1 192[32x
8+ 32y8+ 2(x + y)8+ 2(x − y)8+ 2(x + iy)8+ 2(x − iy)8]
= 1 192(40x 8+ 560x4y4+ 40y4) = 40 192(x 8+ 14x4y4+ y8)
Thus θ := x8+ 14x4y4+ y8 is an invariant of degree 8 in group G defined as above. Using the same way we find the invariant of degree 24,
ϕ0 := x24+ 759x16y8+ 257x12y12+ 759x8y16+ y24. Actually it is easier to work with
ϕ = θ 3− ϕ0
42 = x
4y4(x4− y4)4
So the invariant ring is generated by θ and ϕ, which implies the following powerful theorem on the weight enumerators of even self-dual codes. 1.9.2 Theorem (Gleason). The weight enumerator of an even self-dual code is a polynomial in θ := x8+ 14x4y4+ y8 and ϕ := x4y4(x4− y4)4.
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Univer-sit¨at Basel, October 15, 1995.
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