An Introduction to Invariant Theory Alberto Daniel 2 Orebro¨ University Department of Science and Technology Matematik C, 76

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Department of Science and Technology

An Introduction to Invariant Theory Alberto Daniel 2 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¨andigtarbete, 15 hp Matematik, C–niva, 76 – 90 hp 4 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.

5 6 Abstract

This work is an attempt to explain in some detail section III D of the paper: 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.

7 8 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 1.9 Linear codes ...... 35

9 10 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 invariant 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.

11 12 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 homomorphism 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 [g−1] = [g]−1, [id] = I.

13 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 matrix 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. n Let [T ] be deﬁned as follows, with each vi ∈ C , 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:

 >  v1 >  .  [T ] =  .  > vn

Now we just multiply by [T ], we obtain

 >   > >  v1 v1 v1 ··· v1 vn >  .   . .  [T ] [g] =  .  v1 | · · · | vn =  . .  > > > vn vn v1 ··· v1 vn

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:

 1 0 ··· 0   0 1 ··· 0  [T ]>[T ] =    . . .   . . .  0 0 ··· 1

> Thus, [T ]>[T ] = I ⇐⇒ [T ] [T ] = I.

Now we start looking at the ring R = C[x1, ··· , xn] of complex polynomials in n variables.

14 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 d d d +d since (f1f2)(tv) = f1(tv)f2(tv) = t 1 f1(v)t 2 f2(v) = t 1 2 (f1f2)(v). Now, it follows that every homogeneous polynomials f can be written as a sum

X α f = aαx |α|=d

n α α1 αn where α = (α1, ··· , αn) ∈ N , aα ∈ C and x := x1 ··· xn . Every polynomial f over C can be expressed as a ﬁnite sum of homogeneous polynomials. 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}. L 1.1.9 Proposition. R = d∈N Rd.

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 d1 dn basis for R2, and, in general, the monomials x1 ··· xn 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.

15 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]x or 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] .

2 2 2 1.2.1 Example. Let G = D2 = {1, ρ, σ, τ} = hρ, σ; ρ = σ = (σρ) = 1i. V ∗ = hx, yi σ.x = −x, σ.y = y, ρ.x = x, ρ.y = −y =⇒ 1 0 −1 0 1 0 −1 0 [1] = I = , [σ] = , [ρ] = , [τ] = 0 1 0 1 0 −1 0 −1 2 f(x, y) = 2x + 3y ⇐⇒ [f] = gives 3 −1 0 2 −2 [σ(f)] = [σ][f] = = , 0 1 3 3 i.e (σ(f))(x, y) = −2x + 3y = f(σ.x, σ.y), From τ = σρ we conclude

−1 0 1 0 −1 0 [τ] = [σρ] = [σ ◦ ρ] = [σ][ρ] = = 0 1 0 −1 0 −1

In terms of representation theory we get a unitary representation

1 ∗ 1 ∗ ∗ 1 T : G → U(V ), g 7→ Tg : V → V , f 7→ Tg (f) = g.f.

∗ Since R1 = V we can write this as

1 1 1 T : G → U(R1), g 7→ Tg : R1 → R1, f 7→ Tg (f) = g.f.

1 In the previous example, Tσ (2x + 3y) = −2x + 3y. d 1.2.2 Deﬁnition. Let g ∈ G and d ≥ 0. Deﬁne Tg : Rd → Rd by 0 1. Tg : R0 → R0, C → C : d = 0 1 1 2. Tg : R1 → R1, f 7→ Tg (f) = g.f : d = 1

16 d Y 3. For d ≥ 1 and xi1 ··· xid put Tg(xi1 ··· xid ) = Tg(xij ) and extend j=1  d  d X αj X αj n linearly so that Tg  ajx  = ajTg(x ), where αj ∈ N , j=1 j=1 x = (x1, ··· , xn). Finally, for f = f1 + ··· + fd ∈ R with d X j ∀1 ≤ j ≤ d : fj ∈ Rd put Tg : R → R, f 7→ Tg(f) := Tg (fj). j=1

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, f 0 ∈ R. Let deﬁne f and f 0 as 0 0 0 f := f1 + ··· + fd and f := f1 + ··· + fd,

where ∀1 ≤ i ≤ d : fi ∈ Ri. Then, using the Deﬁnition 1.2.2 we get 1. d d d 0 X i 0 X i X i 0 Tg(f + f ) = Tg(fi + fi ) = Tg(fi) + Tg(fi ) i=1 i=1 i=1 0 = Tg(f) + Tg(f ).

2.   0 X 0 X 0 Tg(f · f ) = Tg  fifj = Tg(fifj) 1≤i,j≤d 1≤i,j≤d d d X i j 0 X i X j 0 = Tg(fi)Tg (fj) = Tg(fi) Tg (fj) 1≤i≤d i=1 j=1 0 = Tg(f) · Tg(f ).

d d X i X i 3. Tg(cf) = Tg(cfi) = cTg(fi) = cTg(f). i=1 i=1 4. By part 2. it is clear that the grading is preserved.

d 5. To show that Tg (f) 7→ g.f is bijective it is enough to show that this d mapping is injective on Rd. Let f ∈ Rd, then f ∈ ker(Tg ) implies d ∗ that Tg (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. d Thus ker(Tg ) = {0} and hence the result.

17 1.2.4 Example. Let 1 1 x g(x) = √1 and f(x, y) = x2 − y2, 2 −1 1 y then 1 2 1 2 g(f) = f([g]x) = √ (x + y) − √ (−x + y) = x2 + 2xy. 2 2

1 [d] d Setting Ag := [Tg ]x,x, then Ag := [Tg ]xd,xd is the d–th induced matrix 0 0 0 in [Slo 77], because Tg(f · f ) = Tg(f) · Tg(f ). Also, if f, f are eigenvectors 1 0 0 of Tg corresponding to the eigenvalues w, w , then f · f is an eigenvector of 2 0 0 0 0 0 Tg with eigenvalue w·w , because Tg(f ·f ) = Tg(f)·Tg(f ) = (wf)·(w f ) = (ww0)(f · f 0). 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, f 0 ∈ RG we check 1. g(f + f 0)(x) = (g(f) + g(f 0))(x) = (f + f 0)(x), thus (f + f 0) ∈ RG. 2. g(f · f 0)(x) = (g(f) · g(f 0))(x) = (f · f 0)(x), thus f · f 0 ∈ 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(fd1) + ··· + g(fds )

Since g.Rd = Rd, we have fd1 = g(fd1 ), ··· , fds = g(fds ). Thus, homogeneous components of polynomial f are invariant under group G.

18 In other words have the following result: 1.2.7 Proposition. The algebra of invariants of G is naturally graded as

G M G R = Rd , d∈N

G where Rd = {f ∈ Rd : ∀g ∈ G : g.f = f}, called the d–th homogeneous component of RG.

G 1.2.8 Definition (Molien series). Viewing Rd as a vector space, we define G ad := dimC Rd , the number of linearly independent homogeneous invariants of degree d ∈ N, and X d ΦG(λ) := adλ , d∈N the Molien series of G. Thus, the Molien series of G is an ordinary power series generating function 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).

1 X 1 Φ (λ) := G |G| det(I − λ[g]) g∈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 monomials 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.

19 α α1 αn 1.3.2 Deﬁnition (Lexicographic Order). Let x := x1 ··· xn and β β1 βn x := x1 ··· xn . We say α >lex β if there is k such that αk > βk and α β αi = βi for all i < k. We will write x >lex x if α >lex β.

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 >lex x3 >lex ··· >lex xn 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.

2 4 3 5 4. x1 >lex x1x2 >lex x1x2 >lex x1x2x3 >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., α(1) α(k) f = a1x + ··· + anx , α(1) α(2) α(k) n where ai 6= 0 in C and x >lex x >lex ··· >lex x for α(i) ∈ N .

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).

α(1) 4. The leading term of f is LT(f) = a1x = 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.

2 1.3.6 Example. Let f := x1x2 + x2x3 + x3 ∈ R2 then LM(f) = x1x2. 1.3.7 Lemma. (xα+weaker terms)(xβ+weaker terms) = xα+β+weaker terms.

α β Proof. This follows from the following obvious observation: If x >lex x , γ α+γ β+γ then for any monomial x , x >lex x . 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).

20 1.4 The division algorithm

For polynomials in one variable C[x1], the unique monomial order is given by 2 1 < x1 < x1 < ··· , 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 polynomials 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) LT(f) LT(f). If so, replace f by f − , add to qi and then return to LT(gi) LT(gi) 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.

21 2 3 2 1.4.1 Example. Let us divide f = x y +y by g1 = xy +y and g2 = x−y, using lex order with x > y

LT(f) x2y = = x = q1, LT(g1) xy 0 2 3 2 3 2 f = f − q1g1 = x y + y − x(xy + y ) = y − xy 0 2 LT(f ) −xy 2 q2 = = = −y , LT(g2) x 00 0 3 2 2 3 f = f − q2g2 = y − xy + xy − y = 0 Since f 00 = 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)

3 2 3 1.4.2 Example. We will divide f = x y + xy + x + 1 by g1 = x + 1 and 2 g2 = y + 1 using lex order with x > y. 3 2 LT(f) x y 2 q1 = = 3 = y LT(g1) x 0 3 2 2 3 2 f = f − q1g1 = x y + xy + x + 1 − y (x + 1) = xy + x − y + 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. xy + x − y2 + 1 q = = −1 2 y2 + 1 0 2 2 f − q2g2 − (xy + x) = xy + x − y + 1 − (−y − 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 monomials. This is, I is a monomial ideal, if there is a subset A ⊂ Nn such that P α I consists of all polynomials which are ﬁnite sums of the form α∈A hαx , α where hα ∈ R. We write I = hx : α ∈ Ai. For example I = hx5y2z, x2yz2, xy3z2i ⊂ R is a monomial ideal.

22 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

s β X α(i) x = fix , i=1 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 α there is a single variable x = x1, then I = hx1 , α ∈ A ⊆ Ni. Let β be α β γ the smallest element in this list. Then for any α ∈ A, x1 = x1 x1 for some β γ ∈ N. This says that I is generated by the single element x1 . 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 α we may take these x so that there is a mj ∈ N such that

xαymj ∈ I

α m ∗ Let m = max{m1, m2, ··· , ms} so that x y j ∈ 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 α (i) α (1) α (s ) α (i) k by elements x k , that is Jk = hx k , ··· , x k k i. Let Bk = {x k y }. 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

23 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 Sn are called symmetric polynomials. Using matrix representation of Sn we have the following deﬁnition. 1.6.1 Deﬁnition. A polynomial in R is called symmetric if for any permutation matrix g, g.f = f.

1.6.2 Deﬁnition. In R, the elementary symmetric functions e1, ··· , en are deﬁned by:

e1 := x1 + ··· + xn

e2 := x1x2 + ··· + x1xn + x2x3 + ··· + x2xn + ··· + xn−1xn . . . . X ek := xi1 ··· xik 1≤i1<···

The invariant ring of Sn in this representation is generated by the al- Sn gebraically independent invariants e1, ··· , en and we denote by R := 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 Sn α n a non-zero f ∈ R , LT(f) = ax , where α = (α1, ··· , αn) ∈ N and ∗ a ∈ C , we ﬁrst claim that α1 > α2 ··· > αn. To prove this, suppose that α αi < αi+1 for some i. Let β = (··· , αi+1, αi, ··· ). Since ax f is a term of β f, it follows that ax is a term of f(··· , xi+1, xi, ··· ). But f is symmetric, xβ so that f(··· , xi+1, xi, ··· ) = f, and hence, ax is a term of f. This is impossible since β > α under lexicographic order, and our claim is proved.

24 Now let 0 α1−α2 α2−α3 αn f := e1 e2 ··· en . 0 To compute the leading monomial of f , ﬁrst note that LM(ek) = x1 ··· xk for 1 ≤ k ≤ n. Hence

0 α1−α2 α2−α3 αn LM(f ) = LM(e1 e2 ··· en ) α1−α2 α2−α3 αn = LM(e1) LM(e2) ··· LM(en) α1−α2 α2−α3 αn = x1 (x1x2) ··· (x1 ··· xn) α1 αn = x1 ··· xn = xα

It follows that f and αf 0 have the same leading term. Subtracting αf 0 from f therefore cancels the monomial axα, thus LM(f − af 0) < LM(f) whenever f − af 0 6= 0. 0 0 Now set f1 := f − af and note that f1 is symmetric since f and af are. 0 Hence, if f1 6= 0, we can repeat the above process to form f2 := f1 − a1f1, 0 where a1 is constant and f1 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

0 f = af + f1 0 0 = af + a1f1 + f2 . . 0 0 0 = af + a1f1 + ··· + asfs 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 X β that φ 6= 0. If we write φ = aβy , then φ(e1, ··· , en) is a sum of the β β1 βn polynomials φβ = aβe1 ··· en , where β = (β1, ··· , βn). Now

β1 β2 βn LM(φβ) = aβLM(e1) LM(e2) ··· LM(en)

β1 β2 βn = aβx1 (x1x2) ··· xn β1+···+βn β2+···+βn β = aβx1 x2 ··· xn

25 As the map θ : Nn 7→ 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 Sn a contradiction. Hence R = C[e1, ··· , en]. See for instance [Cox 91], chapter 7.

This theorem clariﬁes how to generalise the invariants ring of permutation group Sn. If we have the permutation group S3, the invariants ring is S generalized by elementary invariants e1, e2 and e3, i.e. R 3 := 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 3 0 0 3 0 0 invariants ei. Since the leading monomial is x y z we subtract e1e2e3 and are left with −3(x2y + x2z + xy2 + xz2 + y2z + yz2) − 6(xyz). 2 2−1 1 0 The leading monomial is now x y, so we add 3e1 e2e3. This leaves 1−1 1−1 1 3xyz = 3e1 e2 e3, which is reduced to zero in the next step. 3 3 3 3 This way we obtain x + y + z = e1 − 3e1e2 + 3e3. k k 1.6.5 Deﬁnition. The polynomial pk = x1 + ··· + xn for k = 1, 2, ··· are called the power sum symmetric polynomials. 1.6.6 Theorem. Let n ∈ N∗ and k ∈ N. Then

k k+1 pk = e1pk−1 − e2pk−2 + ··· + (−1) ek−1p1 + (−1) kek if k ≤ n. Proof. Let y, z be indeterminate. Then

n n−1 n−2 n (y − x1)(y − x2) ··· (y − xn) = y − e1y + e2y + ··· + (−1) en 1 Put y = z to get 2 n n (1 − x1z)(1 − x2z) ··· (1 − xnz) = 1 − e1z + e2z + ··· + (−1) enz =: e(z)

Consider the generating function of x1, ··· , xn ∞ 2 3 X k p(z) = p1z + p2z + p3z + ··· = pkz k=1 ∞ n n ∞ X X k k X X k = xi z = (xiz) k=1 i=1 i=1 k=1 n X xiz = 1 − x z i=1 i

26 Since e(z) = (1 − x1z)(1 − x2z) ··· (1 − xnz), n X −xie(z) e0(z) = 1 − x z i=1 i and hence n 0 X xiz −ze (z) p(z) = = 1 − x z e(z) i=1 i This implies that

2 n n−1 p(z)e(z) = −z(−e1 + e2 · 2z − e3 · 3z + ··· + (−1) nenz ) 2 3 n+1 n = e1z − 2e2z + 3e3z + ··· + (−1) nenz . If k ≤ n, equating the coeﬃcient of zk we get

k+1 k (−1) kek = pk − e1pk−1 + pk−2e2 + ··· + (−1) p1ek−1 Hence k+1 pk = e1pk−1 − e2pk−2 + ··· + (−1) kek.

1.6.7 Theorem. Every symmetric polynomial in C[x1, ··· , xn] can be writ- k k ten as a polynomial in the power sums p1, ··· , pn, where pk = x1 + ··· + xn 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 k k pk = x1 + ··· + xn, for k ∈ {1, ··· , n}, then from Newton’s identities: 1 e = (−1)r+1 (p − e p + e p + ··· + (−1)r−1e p ). r r r 1 r−1 2 r−2 r 1 which can be written since char(C) = 0, and er is a polynomial in the pk, k k where pk = x1 + ··· + xn, 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 1 X ρ : R → R, ρ(f) = g(f). |G| g∈G

27 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 f 0 ∈ R then ρ(ff 0) = fρ(f 0). Proof. 1. Let h ∈ G. Then, 1 X 1 X h(ρ(f)) = ρ(f([h]v)) = g(f([h]v)) = g(h(f(v)) |G| |G| g∈G g∈G 1 X 1 X = (g ◦ h)(f(v) = u(f(v)) = ρ(f(v)) |G| |G| g∈G u∈G

Hence, ∀f ∈ R : ρ(f) ∈ RG.

2. Let f ∈ R, we check 1 X ρ(ρ(f)) = g(ρ(f)) |G| g∈G ! 1 X 1 X = g h(f) |G| |G| g∈G h∈G 1 X 1 X = g(h(f)) |G| |G| g∈G h∈G 1 X 1 X = (g ◦ h)(f) |G| |G| g∈G h∈G X 1 1 X 1 1 X = u(f) = |G| · u(f) = ρ(f). |G| |G| |G| |G| g∈G u∈G u∈G

Hence ∀f ∈ R, (ρ ◦ ρ)(f) = ρ(f).

3. Let f ∈ RG and f 0 ∈ R, then ∀g ∈ G : g(f) = f. Thus,

1 X 1 X ρ(ff 0) = g(ff 0) = (g(f) · g(f 0)) |G| |G| g∈G g∈G 1 X 1 X = f · g(f 0) = f · g(f 0) = fρ(f 0). |G| |G| g∈G g∈G

28 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 function 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 mi is 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 α α mi = x LT(gj) for some monomial x . By the properties of a monomial α α α order x LT(gj) = LT(x gj) and x gj ∈ 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.

29 Proof. Let M be the maximal ideal of RG generated by homogeneous elements 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. G We claim that R = 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. 0 0 0 0 Hence ∃f1, ··· , fs ∈ R such that f = f1f1 + ··· + fsfs. Apply Reynolds 0 0 0 operator in G to get f = ρ(f1)f1 + ··· + ρ(fs)fs. We may assume that fi 0 0 are homogeneous. Then deg(ρ(fi )) = deg(fi ) = deg(f) − deg(fi) < deg(f). 0 0 0 Since ρ(fi ) are of smaller degree than deg(f), by induction ρ(f1), ··· , ρ(fs) ∈ C[f1, ··· , fs]. Hence f ∈ C[f1, ··· , fs]. For more details of this proof see [DK 95].

α α1 αn n n Let x := x1 ··· xn , where x ∈ C and α := (α1, ··· , αn) ∈ N . The following result is along [Cox 91].

1.7.7 Theorem (Noether). Let G 6 GL(n, C) be a ﬁnite matrix group, G β then C[x1, ··· , xn] = C[ρ(x ): |β| ≤ |G|].

X α G Proof. Let f = cαx ∈ R . Then, α

X α X α f = ρ(f) = ρ( cαx ) = cαρ(x ). α α 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|]. k X α Let (x1 + ··· + xn) = aαx |α|=k k k! where aα = = . α α1!···αn! If A ∈ G is a matrix, let Ai denote the i-th row of A. We can deﬁne α α αn (Ax) := (A1x) 1 ··· (Anx) . 1 X 1 X In this notation, we have ρ(xα) = (A x)α1 ··· (A x)αn = (Ax)α. |G| 1 n |G| A∈G A∈G We introduce n new variables u1, ··· , un, and we now have

k X α α (u1A1x + ··· + unAnx) = aα(Ax) u . |α|=k

If we sum both members over all A ∈ G, we get ! X k X X α α X α α (u1A1x + ··· + unAnx) = aα (Ax) u = bαρ(x )u A∈G |α|=k A∈G |α|=k where bα = |G|aα.

30 X k Let UA = u1A1x + ··· + unAnx, and Sk = UA, then we have A∈G X α α Sk = bαρ(x )u |α|=k

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 α bαρ(x )u = Sk = F (S1, ··· ,S|G|) |α|=k for some polynomial F ∈ C[Y1, ··· ,Y|G|]. Thus, X α α X β β X β bαρ(x )u = F ( bβρ(x )u , ··· , bβρ(x )). |α|=k |β|=1 |β|=|G| Expanding the last expression and equaling the coeﬃcients of uα, it α β follows that bαρ(x ) is a polynomial in the ρ(x ) with |β| ≤ |G|. α Since char(C) = 0, |G|= 6 0 and bα = |G|aα 6= 0, we have ρ(x ) 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 0 −1 A = 1 0 By Noether’s theorem we can take many monomials and average the Reynolds operator to find some invariants. 1 X ρ(f) = g(f) |C4| g∈C4 1 = [f(x, y) + f(−y, x) + f(−x, −y) + f(y, −x)] . 4 Now, 1 1 ρ(x2) = x2 + (−y)2 + (−x)2 + y2 = (x2 + y2), 4 2 1 ρ(xy) = [xy + (−y)x + (−x)(−y) + y(−x)] = 0, 4 1 1 ρ(x3y) = x3y + (−y)3x + (−x)3(−y) + y3(−x) = (x3y − xy3), 4 2 1 ρ(x2y2) = x2y2 + (−y)2x2 + (−x)2(−y)2 + y2(−x)2 = x2y2. 4 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.

31 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,

1 P a1 = |G| g∈G Tr([g]). 1 X Proof. Let S : V ∗ → V ∗,S(f) = g(f). Since V ∗ ⊂ R and by |G| g∈G 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.  1 0   ..   .     1  [S0] =    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

P∞ d ΦG(λ) = d=0 adλ

32 be the generating function for this sequence, then,

1 P 1 ΦG(λ) = |G| g∈G det(I−λ[g]) .

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, 1 X a = Tr([g][d]) d |G| g∈G 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, ..., wn respectively. A basis of C[x1, ··· , xn]d consists of all d1 dn monomials of degree d namely x1 ··· xn where d1 + ··· + dn. If x1, ··· , xn 1 are eigenvectors of Tg corresponding to the eigenvalues w1, ··· , wn, then d1 dn d d1 dn x1 ··· xn are eigenvectors of Tg with eigenvalues w1 ··· wn , because

d1 dn d1 dn Tg(x1 ··· xn ) = Tg(x1 ) ··· Tg(xn ) d1 dn = (Tg(x1)) ··· (Tg(xn)) d1 dn = (w1x1) ··· (wnxn) d1 dn d1 dn = w1 ··· wn · x1 ··· xn .

Hence, even the induced operators are diagonalizable. Now, for this action of g we have

[d] X d1 dn Tr([g] ) = w1 ··· wn d1+...+dn=d Since [g] is diagonalizable, i.e. exists P ∈ GL(n, C) invertible such that

−1 [g] = PDP , where D := diag(w1, ..., wd) then det(I − λ[g]) = det(PP −1 − λP DP −1) = det(P (I − λD)P −1) = det(P ) det(I − λD) det(P −1) −1 = det(I − λD)det(PP ) = det(I − λD) = (1 − λw1) ··· (1 − λwn).

33 So 1 1 = det(I − λ[g]) (1 − λw1) ··· (1 − λwn) 1 1 = ··· (1 − λw1) (1 − λwn) 2 2 2 2 = (1 + λw1 + λ w1 + ··· ) ··· (1 + λwn + λ wn + ··· )

d X d1 dn The result follows by computing the coeﬁcient of λ to be w1 ··· wn . d1+···+dn=d

±1 0 1.8.3 Example. Consider de matrix group V = { } ⊂ GL(2, C). 4 0 ±1 The Molien’s series is calculated as follows

1 0 A := , det(I − λA ) = (1 − λ)2 1 0 1 1 −1 0 A := , det(I − λA ) = (1 + λ)2 2 0 −1 2 −1 0 A := , det(I − λA ) = 1 − λ2 3 0 1 3 1 0 A := , det(I − λA ) = 1 − λ2 4 0 −1 4 1 1 1 2 1 1 Hence ϕ(λ) = + + = = = 4 (1 − λ)2 (1 + λ)2 1 − λ2 (1 − λ)2(1 + λ)2 (1 − λ2)2 1 + 2λ2 + 3λ4 + ···

Hence there are 2 independent invariants of degree 2. 1 Now, ρ(f) = 4 {f(x, y) + f(−x, −y) + f(−x, y) + f(x, −y)}. Hence 1 1 ρ(x2) = {x2 + (−x)2 + (−x)2 + x2} = · 4x2 = x2, 4 4 1 1 ρ(y2) = {y2 + (−y)2 + y2 + (−y)2} = · 4y2 = y2. 4 4 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 representation 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  1 − λ 0 0  det(I − λ[e]) = det  0 1 − λ 0  = (1 − λ)3, 0 0 1 − λ

34  1 −λ 0  det(I − λ[σ]) = det  −λ 1 0  = (1 − λ)(1 − λ2), 0 0 1 − λ and ﬁnally,

 1 −λ 0  det(I − λ[τ]) = det  0 1 −λ  = (1 − λ3), −λ 0 1 where e is the identity, σ is (12) and τ is (123).

Assembling these results gives,

|G| 1 X 1 1 1 3 2 = + + |G| det(I − λA ) 6 (1 − λ)3 (1 − λ)(1 − λ2 (1 − λ3) α=1 α (1 − λ2)(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, 1 Φ (λ) = , S3 (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,

2 2 4 3 6 ΦS3 (λ) = (1 + λ + λ + ···)(1 + λ + λ + ···)(1 + λ + λ + ···) = 1 + λ + 2λ2 + 3λ3 + 4λ4 + ···

So there is one linear invariant, as we have seen this is I1 = x + y + z. 2 2 There are two degree 2 invariants, one must be I1 = (x + y + z) the square of the linear invariant. The other is a new invariant which could be 2 2 2 I2 = x + y + z or xy + xz + yz but not both since these are not linearly 2 independent I1 − I2 = 2(xy + xz + yz). The three cubic invariants will 3 3 3 3 be I1 ,I1I2 and a new invariant say I3 = x + y + z . All the polynomial are now accounted for as sums and products of these three invariants, for 4 2 2 example the four degree 4 invariants are I1 ,I1 I2,I2 and I1I3.

1.9 Linear codes

We assume some familiarity with Coding Theory, for example [Bie 04]. A n linear code is a linear subspace C ⊆ Fq , where Fq is the ﬁeld of q elements.

35 n 1.9.1 Definition. The weight of a word x ∈ Fq 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 number 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}.

n A code C ⊆ Fq 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. n Let C ⊂ Fq 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

n X n−i i WC (x, y) := Aix y ∈ C[x, y] i=0 the weight enumerator polynomial, satisfying

1 WC⊥ (x, y) = qk WC (x + (q − 1)y, x − y) (Mac Williams). For certain classes of self-dual binary [n, n/2, d] codes this implies

x + y x − y WC (x, y) = WC √ , √ ,WC (x, y) = WC (x, iy). 2 2

But this means that WC is invariant under the group

1 1 1 1 0 G = √ , ≤ GL(2, C), |G| = 192. 2 1 −1 0 i

2πi th Let ζ = e 8 be a primitive 8 root of unity. The group G deﬁned above is equal to the set of matrices

1 0 0 1 1 1 β ζk , ζk , ζk √ 0 α α 0 2 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 1 φ (λ) = . G (1 − λ8)(1 − λ24)

36 This suggests that the invariant ring is generated by two algebraically independent 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.

192 1 X ρ(f) = f(A x),A ∈ G 192 γ γ γ=1

P32 8 · 4 = 32 matrices in each line γ=1 f(Aγx) 1 0 1 0 1 0 1 0 ζk , ζk , ζk , ζk 32x8 0 1 0 −1 0 i 0 −i 0 1 0 1 0 1 0 1 ζk , ζk , ζk , ζk 32y8 1 0 −1 0 i 0 −i 0 1 1 1 1 1 1 1 1 ζk √1 , ζk √1 , ζk √1 , ζk √1 2(x + y)8 2 1 1 2 −1 −1 2 i i 2 −i −i 1 −1 1 −1 1 −1 1 −1 ζk √1 , ζk √1 , ζk √1 , ζk √1 2(x − y)8 2 1 −1 2 −1 1 2 i −i 2 −i i 1 i 1 i 1 i 1 i ζk √1 , ζk √1 , ζk √1 , ζk √1 2(x + iy)8 2 1 i 2 −1 −i 2 i −1 2 −i 1 1 −i 1 −i 1 −i 1 −i ζk √1 , ζk √1 , ζk √1 , ζk √1 2(x − iy)8 2 1 −i 2 −1 −i 2 i 1 2 −i 1

1 ρ(f) = [32x8 + 32y8 + 2(x + y)8 + 2(x − y)8 + 2(x + iy)8 + 2(x − iy)8] 192 1 = (40x8 + 560x4y4 + 40y4) 192 40 = (x8 + 14x4y4 + y8) 192 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

θ3 − ϕ0 ϕ = = x4y4(x4 − y4)4 42 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.

37 Bibliography

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[CJ 10] S. Cooper & B. Johnson, Power of Monomial Ideals, Lecture notes, 2010, (https://www.math.unl.edu/~s-bjohns67/Files/MonomialIdeals.pdf).

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