The Invariant Theory of K−1[X1,X2,...,Xn ] Under Permutation Representations

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THE INVARIANT THEORY OF K−1[X1,X2,...,XN ] UNDER PERMUTATION REPRESENTATIONS

J. CAMERON ATKINS

A Thesis Submitted to the Graduate Faculty of

WAKE FOREST UNIVERSITY GRADUATE SCHOOL OF ARTS AND SCIENCES

in Partial Fulﬁllment of the Requirements

for the Degree of

MASTER OF ARTS

Mathematics

May 2012

Winston-Salem, North Carolina

Approved By:

Ellen Kirkman, Ph.D., Advisor

Sarah K. Mason, Ph.D., Chair W. Frank Moore, Ph.D. Acknowledgments

I would like to thank my parents for being so patient with me through my adoles- cents. When I decided to pursue a path in mathematics, they supported me all the way, and they continue to support me as I head to the University of South Carolina. My siblings Jared and Whitney treated me as the younger of the three, and over the past years we have grown to truly love each other. I feel lucky to have the relationship we share and encouragement they give me. I would also like to thank Melissa Bechard. She and I began studying mathematics our freshmen year at James Madison Univer- sity. We struggle and we ﬁght through problems together, and if nothing else, I have fun exploring math with her. Lastly, I would like to thank my advisor Dr. Kirkman. Out of everyone she is the most deserving of my praise. As a professor, I have never learned more from one individual. Her rigor from the classroom reappeared in her guidance as an advisor, and I am a better mathematician for it.

ii Table of Contents

Acknowledgments ...... ii

Abstract ...... iv

Chapter 1 Introduction ...... 1 1.1 Groups and Rings ...... 1 1.2 Group Representation ...... 2

1.3 Invariant Theory of k[x1, x2 . . . , xn] Under Permutation Matrices . . . 2

1.4 The Non-Commutative Ring k−1[x1, x2, ··· , xn] ...... 5

Chapter 2 Orbit Sums and the Symmetric Group...... 7 2.1 Orbit Sums ...... 7 2.2 Symmetric Group ...... 9

Chapter 3 G¨obel’s Bound ...... 19 3.1 Special Monomials ...... 19 3.2 G¨obel’s Bound ...... 21

Chapter 4 The Alternating Group...... 24 4.1 Antisymmetric Invariants ...... 24 4.2 RAn Generators ...... 28

Chapter 5 Examples With G¨obel’s Bound with 4 indeterminates...... 33 5.1 Alternating Z2 Subgroup ...... 33 5.2 Alternating Klein IV Subgroup ...... 36 5.3 Non-Alternating Klein IV Subgroup ...... 38

Bibliography ...... 42

Appendix A 5.1 ...... 43

Appendix B 5.2 ...... 46

Vita...... 48

iii Abstract

Let k be a field of characteristic zero, and let R = k−1[x1, ··· , xn] denote the skew- polynomial ring with xjxi = −xixj for i 6= j. The symmetric group Sn acts on R as permutations of the {xi}. Given a subgroup G of permutations in Sn we consider the problem of finding algebra generators for RG, the subring of invariants under G G. We find generators for R when G is the full symmetric group Sn and when G is the alternating group An. We describe an algorithm that produces generators of G R for a general subgroup G of Sn. The algorithm produces a an upper bound on the degrees of algebra generators of RG, generalizing the Göbel bound for invariants under permutation actions on k[x1, ··· , xn].

iv Chapter 1: Introduction

1.1 Groups and Rings

Let G be a group and S be a set. We deﬁne a left group action on S by G to be G × S → S, where (g, s) = gs, and if e is the identity of G the action satisﬁes es = s and g(hs) = (gh)s for all g, h ∈ G and for all s ∈ S. We call the set S a G-set. The orbit of element s ∈ S under a group G is the set of elements t in S such that there exists some g ∈ G with gs = t. We denote the orbit of an element s ∈ S under a group G to be OrbG(s). A stabilizer of element s ∈ S is an element of g ∈ G such that gs = s. The set of all stabilizers of an element s ∈ S from G forms a subgroup of G denoted StabG(s). We let Sn denote the symmetric group on n elements and

An be the alternating group.

A ring is a set R with two binary operations + and · which are called addition and muliplication respectively. The ring under addition forms an abelian group. The ring under multiplication satisfies the left and right distribution laws, i.e. a(b + c) = ab + ac and (a + b)c = ac + bc, as well as being associative, i.e. a(bc) = (ab)c.A field is a commutative ring with unity where every non-zero element is a unit. The characteristic of a field, k, is the smallest positive integer n such that ∀r ∈ k, nr = 0. If no positive integer exists we say the ring has characteristic zero. Throughout this thesis k is always a field of characteristic zero. We can construct polynomials with n-indeterminates with coefficients from a field k denoted k[x1, x2, . . . , xn] or R. Let

I I = (i1, i2, . . . , in) be a vector of length n of non-negative integers, and let X denote

i1 i2 in the monomial x1 x2 ··· xn .

1 1.2 Group Representation

A linear representation of a group G is a homomorphism ϕ : G → Gl(n, k), where k is any field, and Gl(n, k) is the set of n × n invertible matrices with entries from k. A representation, ϕ, is called faithful if the kernel of ϕ is only the identity of G. The −1 0 example ϕ : → Gl(2, ), with ϕ(1) = defines a representation fo , Z4 R 0 1 Z4 where ϕ(j) = ϕ(1 + 1 + ··· + 1) (one is added to itself j times). This representation is not fiathful to since ker(ϕ) = {0, 2}. On the other hand, ρ : Z3 → Gl(2, C) 2πi e 3 0 where ρ(1) = defines a faithful representation since ker(ρ) = {0}.A 0 1 permutation matrix is denoted as Mσ, and is the matrix where the entries mi,j = 1 if σ(j) = i and mi,j =0 otherwise. For example consider S4 and the permutation σ = (1, 2, 3). Then the permutation matrix is

 0 0 1 0   1 0 0 0  Mσ =  .  0 1 0 0  0 0 0 1

The representation of the symmetric group Sn with φ : Sn → Gl(n, R) where φ(σ) =

Mσ is called a permutation representation of Sn. This representation is faithful. In this thesis we explore the permuation representation of the symmetric group as well as the subgroups of the symmetric group under the same representation.

1.3 Invariant Theory of k[x1, x2 . . . , xn] Under Permutation Ma- trices

For any subgroup G of Sn, represented as permutation matrices, we deﬁne a G-action on the set of monomials. If σ is a permutation of {1, 2, . . . , n} then we deﬁne

i −1 i −1 i −1 i1 i2 in i1 i2 in σ (1) σ (2) σ (n) σ(x1 x2 ··· xn ) = xσ(1)xσ(2) ··· xσ(n) = x1 x2 ··· xn .

2 This action is extended to R = k[x1, ··· , xn] by

X I X I σ aI X = aI σ(X ). I I

Given any subgroup G of Sn and an element f ∈ R, we say f is invariant under G if ∀g ∈ G, g(f) = f. The set of elements of R that are invariant under G forms a subring of R that we denote by RG. This thesis concerns ﬁnding algebra generators

G G for the invariant subring R , i.e. ﬁnding elements f1, ··· , fm of R such that any invariant element f ∈ RG can be written as sum of the form

X i1 i2 im f = aI (f1 f2 ··· fm ) for aI ∈ k.

I=(i1,i2,...,im)

One way to produce invariants is to form orbit sums. For any monomial XI , let

I I OG(X ) denote the sum of the elements in the G-orbit of X , namely

I X I OG(X ) = g(X ). I OrbG(X )

Any orbit sum will be an invariant in RG. Moreover, it can be shown that any element of RG is a linear combination of orbit sums [2]. When G is the full symmetric group

Sn an orbit sum can be represented by its leading term under the lexicographic order with x1 > x2 > ··· > xn; the exponent sequence of that leading term will be weakly decreasing and hence form a partition into n parts of the sum of the exponents. Hence, orbit sums under Sn correspond with partitions into n parts. When G = Sn Gauss proved that the n elementary symmetric polynomials are a generating set for RSn , where the elementary symmetric polynomials and the corresponding partitions are:

e1(x1, . . . , xn) = OSn (x1) = x1 + x2 + ··· + xn ↔ (1, 0, 0, ··· 0),

X e2(x1, . . . , xn) = OSn (x1x2) = xixj ↔ (1, 1, 0, ··· 0), i6=j

3 X e3(x1, . . . , xn) = OSn (x1x2x3) = xixjxk ↔ (1, 1, 1, 0, ··· 0), i,j,k distinct

. .

en(x1, . . . , xn) = OSn (x1x2 ··· xn) = x1x2 ··· xn ↔ (1, 1, 1, ··· 1).

Power sums form another algebra generating set for the symmetric functions RSn , where for 1 ≤ i ≤ n the n power sums are deﬁned by

i i i i pi = OSn (x1) = x1 + x2 + ··· + xn ↔ (i, 0, 0, ··· 0).

Newton proved the Newton Symmetric Formulas, which are relations between the power sums and the elementary symmetric polynomials, and Waring gave a formula expressing the power sums in terms of the elementary symmetric polynomials [2]. It has also been shown that either set of algebra invariants is algebraically independent (i.e. the invariant subring is a polynomial ring). Hence, invariants under the full symmetric group can be written uniquely in terms of the elementary symmetric polynomials or the power sums [3]. Under other subgroups of the symmetric group, the invariant subring is generally not a polynomial ring. In other words, for a particular set of generators of an invariant subring, RG, polynomials in RG are not written uniquely in terms of the generators.

The Vandermonde Determinant ∇n = Πi>j(xi − xj) is invariant under An but not

Sn. It has been shown that the elementary symmetric polynomials along with the

Vandermonde Determinant are algebra generators of RAn [3]. More recently the problem of describing algebra generators for RG, where G is a subgroup of Sn under the permutation representation, was considered. In his 1996 thesis Manfred Göbel determined a set of algebra generators for RG and an upper bound for their degrees, now called the Göbel bound. Göbel’s generating set is usually not

4 a minimal generating set, but the Göbel bound can be sharp. The Göbel bound is very useful because it provides the largest degree of orbit sums one needs to consider in finding an algebra generating set for RG. Göbel’s generating set consists of the elementary symmetric polynomials along with orbit sums of called “special monomials.” An exponent sequence, I, is called special if rewriting I in a weakly decreasing order denoted λ(I) where λ(I) = (λ1, λ2, . . . , λn) satisfies the following:

1) λi − λi+1 ≤ 1

2) λn = 0.

n The Gobel¨ bound is the maximum of n and . Since the Vandermonde Determi- 2 n nant has degree , the G¨obel bound is sharp for G = A for n ≥ 3 [1]. 2 n

1.4 The Non-Commutative Ring k−1[x1, x2, ··· , xn]

We deﬁne the ring R = k−1[x1, x2, ··· , xn] to be the ring over a ﬁeld k of characteristic zero with n indeterminates with the property xixj = −xjxi when i 6= j. For example

2 2 3 consider the ring k−1[x1, x2, x3] and the element x2x1 − 3x1x3 + 4x3x2 = −x1x2 −

2 3 2 n m nm m n 3x1x3 + 4x2x3. In general xi xj = (−1) xj xi . The elements of this ring are called ”skew-polynomials” instead of polynomials since these elements have the property xixj = −xjxi when i 6= j. We note that permutations of the indeterminates xi extend to automorphisms of R = k−1[x1, x2, ··· , xn] because any transposition of

{1, 2, ··· , n} preserves the relations xixj = −xjxi. For a ﬁxed scalar q ∈ k there are other noncommutative algebras kq[x1, x2, ··· , xn], with deﬁning relations xixj = qxjxi for i > j. However, the only q for which all transpositions preserve these relations are when q = 1 (the classical commutative case) and q = −1 (the case considered

5 here). This thesis provides results for k−1[x1, x2, . . . , x2] that are analogous to those described for k[x1, x2, . . . , xn] in Section 1.3.

6 Chapter 2: Orbit Sums and the Symmetric Group

Let k be a ﬁeld of characteristic zero, and adjoin a ﬁnite number of indeterminates with the property xixj = −xjxi when i 6= j. We denote this ring as

R = k−1[x1, x2, . . . , xn]. Let G be a subgroup of a permutation group acting faith- fully on the ring with g(c) = c, for any c in k and g(xi) = xj. We extend g to an automorphism of k−1[x1, . . . , xn] by deﬁning g on monomials by

i1 in i1 in g(x1 ··· xn ) = g(x1) ··· g(xn) , and deﬁning g on the skew-polynomials ring by

X I X I g aI X = aI g(X ). I I

Unlike the commutative case, we note that in the skew-polynomial case to rewrite g(XI ) with the variables in ascending order may require interchanging variables, and

5 7 7 5 hence may introduce sign changes. For example, x3x2 = −x2x3. If f in R satisﬁes g(f) = f for all g in G, then we say f is invariant under G. These invariants form a subring denoted RG.

2.1 Orbit Sums

In this section we produce orbit sums as invariants, and we show that any invariant is a linear combination of orbit sums. The lemma below reduces the search for invariants to the search for homogeneous invariants, i.e. skew-polynomials where all the terms have the same total degree.

Lemma 2.1.1. Homogeneous components of invariants under a linear action are invariants themselves.

7 Proof. Since g does not change the total degree of a monomial, homogeneus compo- Pm nents of f are invariant. That is, if f = i=1 fni where the terms of fni all have the same degree, then g(f) = f implies g(fni ) = fni .

Deﬁnition 1. An orbit sum of a monomial XI under the action of a group G is the sum of the distinct monomials in the orbit of XI under G. An orbit sum is denoted

I OG(X ). Note that

I X I X I OG(X ) = g(X ) 6= g(X ). I OrbG(X ) g∈G

If we sum over every element in the group some terms in the orbit sum may have a higher multiplicity than one.

S3 7 3 Example 2.1.2. Consider the ring k−1[x1, x2, x3] and the monomials x1x2 and

2 2 7 3 x1x2x3. The orbit sum of x1x2 is

7 3 7 3 7 3 3 7 3 7 7 3 3 7 OS3 (x1x2) = x1x2 + x1x3 − x1x2 − x1x3 + x2x3 − x2x3.

2 2 The orbit sum of x1x2x3 is

2 2 2 2 2 2 2 2 OS3 (x1x2x3) = x1x2x3 + x1x2x3 + x1x2x3.

Note the orbit sum of a monomial under a group action G is invariant under G.

Theorem 2.1.3. Every invariant is a linear combination of orbit sums.

Proof. Let f be in RG, and by Lemma 2.1.1 we can assume without loss of generality that f is homogeneous. Let

X I f = aI X with aI 6= 0. I

8 We want to show that f is a linear combination of orbit sums; i.e.

X J f = aJ OG(X ). J

Let’s consider some element g ∈ G acting on f. Since f is invariant, we have g(f) = f. If we look at g acting on some particular summand we see

I I I g(aI X ) = g(aI )g(X ) = aI g(X ), and

i −1 i −1 i −1 I i1 i2 in i1 i2 in g (1) g (2) g (n) g(X ) = g(x1 x2 ...xn ) = xg(1)xg(2)...xg(n) = ±x1 x2 ...xn .

The plus or minus sign comes from the possible sign change to put variables into

I P I ascending order. Since f is invariant g(X ) is a term in I aiX , and since distinct I P I powers are linearly independent the coeﬃcient of g(X ) in I aI X must be ag(I). I Hence aI = ag(I). Thus all monomials in the orbit of X will have coeﬃcient aI , and

X J f = aJ OG(X ) J is the sum of orbit sums.

2.2 Symmetric Group

In this section we describe two families of polynomials that generate RSn as an algebra. These families are analogous to the power sums and the elementary symmetric polynomials in the commutative case that generate subring of invariants of k[x1, x2, . . . , xn] under the permutation action of the full symmetric group.

In the non-commutative ring k−1[x1, x2, . . . , xn], unlike the commutative case, some orbit sums are equal to zero.

9 Example 2.2.1. Consider k−1[x1, x2] and G = S2.

OS2 (x1x2) = x1x2 + x2x1

= x1x2 − x1x2

= 0.

This raises the question of when an orbit sums is zero. Later in this thesis we deﬁne algorithms that reduce the exponent sequence of particular monomials. Thus we have to be careful that we do not reduce an orbit sum to the zero polynomial. Since the set of monomials of degree d are linearly independent over the vector space of all polynomials of degree d, if O(XI ) = 0 then for any term XJ occuring in orbit of XI , the monomial −XJ must also occur in the orbit.

Note that G ⊆ Sn acts on a monomial by

i −1 i −1 i −1 I i1 i2 in i1 i2 in g (1) g (2) g (n) g(X ) = g(x1 x2 ··· xn ) = xg(1)xg(2) ··· xg(n) = ±x1 x2 ··· xn , so when g acts on the monomial and the variables are arranged in ascending order,

−1 I the exponent sequence goes from I to g (I). Also OG(X ) is a sum of X to powers

−1 that are all g (I) for g ∈ G. If G = Sn then the exponent sequences are all the

I permutations of I. Hence it is convenient to represent the orbit sums by OSn (X ) where I is a partition. Thus each orbit sum over Sn corresponds to a partition of

Pn I j=1 ij into n parts for I = {ij}. Futhermore, X (where I is a partition) is the leading term under the lexicographic order, where x1 > x2 > ··· > xn, of the orbit sum.

I Lemma 2.2.2. An orbit sum, OSn (X ), is zero if and only if there are at least two repeated odd entries in I.

Proof. ⇐) Since I is a partition, the pair of repeated odd entries can be assumed to

I be adjacent. Let ij = ij+1 be odd. Consider the transposition (j, j + 1) acting on X .

10 I i1 i2 ij ij in (j, j + 1)X = (j, j + 1)(x1 x2 ...xj xj+1...xn )

i1 i2 ij ij in = x1 x2 ...xj+1xj ...xn

i1 i2 ij ij in = −x1 x2 ...xj xj+1...xn

= −XI .

Decompose the group Sn into a union of distinct left cosets of the subgroup H = h(j, j + 1)i, so

[ Sn = σH.

σ∈Sn

The orbit of XI under H is {XI , −XI }. The image of XI under the two elements in the coset, σH, are σ(XI ) and −σ(XI ). For any element, XJ , in the orbit of XI ,its negative, −XJ , is also in the orbit of XI ; therefore, the orbit sum is zero.

I ⇒) Let the orbit sum OSn (X ) = 0. Since the orbit sum is zero each term needs to pair with its negative. If (−XI ) is to occur in the orbit, then some nonidentity

I I element σ ∈ Sn must be an element of Stab(I) and σ(X ) = −(X ). Therefore, there must be at least one pair of repeated entries in I. Assume I = (i1, i2, ..., in) has no repeated odd entries, so every σ from Stab(I) acting on I must ﬁx every odd entry. Thus, every σ in Stab(I) acting on (XI ) is a permutation of the even powered indeterminates. Thus, for any σ from Stab(I), σ(XI ) = XI . So −XI cannot occur in the orbit, hence the orbit sum cannot be zero. Therefore, there must be a repeated pair of odd entries in the exponent sequence for the orbit sum to be zero. Hence,

I OSn (X ) = 0 if and only if there are two repeated odd entries.

Lemma 2.2.3. The largest monomial under the lexicographic order in the product of

I J two non-zero orbits sums, OSn (X ) and OSn (X ), has degree I + J, where I and J

I+J are partitions. Also the orbit sum OSn (X ) 6= 0.

11 I J I J I+J Proof. Assume OSn (X ) 6= 0 and OSn (X ) 6= 0. (X )(X ) = ±X is a term in

I J I J OSn (X )OSn (X ). If you consider any other term, σ(X ) × τ(X ), either I or J will be permuted or both will be permuted. A permutation of a partition (P) gives an exponent sequence that is less than or equal to (P) under the lexicographic order. Since we are dealing with orbit sums, each term may only appear once, implying that both exponent sequences may not be ﬁxed by the permutations. So one of the exponents, σ(I) or τ(J) must be lower lexicographically than I or J respectively. Thus

I J I+J σ(I)+τ(J)

I+J I J be lower lexicographically. The monomial ±X occurs once in OSn (X )OSn (X ) and all other summands are lower lexicographically, implies that ∓XI+J cannot occur.

I+J Hence, OSn (X ) 6= 0.

Now we introduce two sets of invariants: the n odd power sums and the n elementary skew-symmetric polynomials.

i Deﬁnition 2. The n odd power sums are P := {OSn (x1)|1 ≤ i ≤ 2n − 1 and i is odd}. We denote a particular polynomial from P as pi where i determines the degree.

I Deﬁnition 3. The n elementary skew-symmetric polynomials are OSn (X ) where I can be any of the following sequences: (1,0,...,0),(2,1,0,...,0), (2,2,1,0,...,0),...,(2,2,...,2,1).

I We denote a particular skew-polynomial from the set as Sa = OSn (X ), where a de- notes the position where the 1 is in the sequence.

Example 2.2.4. Consider R = k−1[x1, x2, x3]. The 3 odd power sums for R are p1 =

3 3 3 3 5 5 5 5 OS3 (x1) = x1 +x2 +x3, p3 = OS3(x1) = x1 +x2 +x3, and p5 = OS3 (x1) = x1 +x2 +x3.

2 The 3 elementary skew-symmetric polynomials are OS3 (x1) = x1+x2+x3, OS3 (x1x2) =

2 2 2 2 2 2 2 2 2 2 2 2 2 2 x1x2 +x1x3 +x1x2 +x2x3 +x1x3 +x2x3, and OS3 (x1x2x3) = x1x2x3 +x1x2x3 +x1x2x3.

12 i The n odd power sums are the analog to the n power sums, {OSn (x )|1 ≤ i ≤ n}, in the commutative case. The n elementary skew-symmetric polynomials are the analog to the elementary symmetric polynomials in the commutative case.

2 2 2 Sn Lemma 2.2.5. k[x1, x2, ..., xn] is contained in the algebra generated by P and also in the algebra generated by the n elementary skew-symmetric polynomials.

2 2 2 2 Sn Proof. Since xi is in the center of R, k[x1, x2, ..., xn] is commutative and

2 2 2 Sn Sn k[x1, x2, ..., xn] ' k[y1, y2, ..., yn] where the set {yi, y2, . . . , yn} is algebraically

Sn independent. This ring k[y1, y2, ..., yn] is generated as an algebra either by the power sums or the n elementary symmetric polynomials [3]. Therefore, the ring

2 2 2 Sn i k[x1, x2, ..., xn] is generated by {OSn (x1) where 2 ≤ i ≤ 2n and i is even} or

2 2 2 {OSn (x1x2...xa)|1 ≤ a ≤ n}.

2 2 2 Sn Sn Since k[x1, x2, ..., xn] is contained in k−1[x1, x2, ··· , xn] , we show that it is

2m 2 2 2 Sn generated by P. Let OSn (x1 ) where 1 ≤ m ≤ n be a generator for k[x1, x2, ..., xn] .

1 2m We claim that 2 (p1p2m−1 + p2m−1p1) = OSn (x1 ). First we compute: n 2m−1 2m−1 X 2m−1 p1p2m−1 = (x1 + ... + xn)(x1 + ... + xn ) = xixj . 1≤i,j≤n

n th X 2m−1 Consider the j term in xixj when we arrange the variables in ascending 1≤i,j≤n order.

2m If i = j then the term is xi .

2m−1 If i < j then the term is xixj .

2m−1 2m−1 If i > j then the term is xixj = −xj xi. On the other hand,

n 2m−1 2m−1 X 2m−1 P2m−1P1 = (x1 + ... + xn )(x1 + ... + xn) = xj xi. 1≤j,i≤n

13 n th X 2m−1 Consider the ji term in xj xi. 1≤j,i≤n

2m If i = j then the term is xi .

2m−1 If i > j then the term is xj xi.

2m−1 2m−1 If i < j then the term is xj xi = −xixj .

n n ! 1 X 2m−1 X 2m−1 2m 2m 2m So 2 xixj + xj xi = x1 + ... + xn = OSn (x1 ). 1≤i,j≤n 1≤j,i≤n

2 2 2 Sn Hence, k[x1, x2, . . . , xn] is contained in the algebra generated by P. Now we

2 2 2 Sn show k[x1, x2, ··· , xn] is contained in the algebra generated by the n elementary skew-symmetric polynomials.

I 2 2 2 2 2 2 Sn Let OSn (X ) = OSn (x1x2 . . . xa) be a generator of k[x1, x2, ..., xn]. We claim

I 1 that OSn (X ) = 2a (S1Sa + SaS1). 3 2 2 2 2 2 First we want to show S1Sa = OSn (x1x2x3 . . . xa−1xa) + aOSn (x1x2 . . . xa). We are considering Sn-invariants, so we know that every Sn orbit sum is deﬁned by monomials with exponent sequences that are partitions. Thus we consider every partition that can be constructed from permutations of (1, 0, ··· , 0) summed with permutations of (2, 2, ··· , 2, 1, 0, ··· , 0). There are only 3 such partitions, which are (3, 2, ··· , 2, 1, 0, ··· , 0), (2, ··· , 2, 0, ··· , 0), and (2, ··· , 2, 1, 1, 0, ··· , 0). The partition (2, ··· , 2, 1, 1, 0, ··· , 0) has a repeated odd entry, thus by Lemma 2.2.2 this orbit sum is trivial. Now we must determine the multiplicity as well as the sign of the other orbit sums. The partition (3, 2, ··· , 2, 1, 0, ··· , 0) is the largest lexicographically in terms of S1Sa, so by the proof of Lemma 2.2.3 it occurs once, and its sign is positive,

2 2 3 2 2 i.e. (x1)(x1 ··· xa−1xa) = x1x2 . . . xa−1xa. The ﬁnal partition (2, ··· , 2, 0, ··· , 0) oc-

2 2 2 2 2 curs when (x1)(x1x2 ··· xa), (x2)(x1x2x3 ··· xa),...,

2 2 (xa)(x1 ··· xa−1xa). Since the ﬁrst term xi only commutes past variables with even exponents, the sign of the monomial is always positive. Thus the partition (2, ··· , 2, 0, ··· , 0)

14 3 2 2 2 2 2 occurs a times. Hence, S1Sa = OSn (x1x2x3 . . . xa−1xa) + aOSn (x1x2 . . . xa). Sim- ilarly, SaS1 produces the same partitions with the same multiplicity, but the partition (3, 2, ··· , 2, 1, 0, ··· , 0) will be negative since x1 must commute past exactly

3 2 2 2 one variable with an odd power. Therefore, SaS1 = −OSn (x1x2x3 . . . xa−1xa) +

2 2 2 1 2a 2 2 2 2 2 2 aO.Sn (x1x2 . . . xa). Thus, 2a (S1Sa + SaS1) = 2a OSn (x1x2 . . . xa) = OSn (x1x2 . . . xa).

2 2 2 Sn Sn Hence, k−1[x1, x2, . . . , xn] ⊆ k−1[x1, x2, . . . , xn] is contained in the algebra generated by either P or the elementary skew-symmetric polynomials.

Sn Theorem 2.2.6. The ring k−1[x1, x2, ..., xn] is generated as an algebra by the n elementary skew-symmetric polynomials, {Sa}.

Proof. We proceed by induction on the lexicographic order of the leading monomial of an invariant. Let f be an invariant with leading monomial of degree zero. Hence f is an element of the ﬁeld, so it can be written as a trivial polynomial in the elementary skew-symmetric polynomials. Let cXI be the leading monomial of f where I is a partition and is not of degree zero. We will subtract products of elementary skew- symmetric polynomials and invariants of lower leading length lexicographic order from f to remove the highest lexicographic term. Assume there are no odd entries in I.

I According to Lemma 2.2.5, OSn (X ) can be written as a linear combination of powers

I of elementary skew-symmetric polynomials. So the leading monomial in f −OSn (X ) is of lower degree than f. Assume I = (i1, ..., in) has at least one odd entry. Let

I ia be the last odd entry in the sequence. Since I is a partition and OSn (X ) is not

I∗ trivial, we have ij ≥ ia + 1 ≥ 2. for all ij, where j < a. Consider OSn (X )Sa where

∗ I = (i1 − 2, i2 − 2, ..., ia−1 − 2, ia − 1, ia+1, ..., in), which is also a partition of non-

I∗ I negative entries. By Lemma 2.2.3, the largest monomial in OSn (X )Sa is X and all

I I∗ other terms are lower lexicographically than X . So all terms of f − cOSn (X )Sa, for some c ∈ k, are lower lexicographically than f. Since the degree is ﬁnite and there are only a ﬁnite number of exponent sequences of lower lexicographic order than I,

15 this process terminates. Hence, any skew-polynomial f ∈ RSn can be written as elementary skew-symmetric polynomials.

S3 4 Example 2.2.7. Consider k−1[x1, x2, x3] and the skew-polynomial OS3 (x1x2) =

4 4 4 4 4 4 x1x2 + x1x3 + x1x2 + x2x3 + x1x3 + x2x3 . In the exponent sequence I = (4, 1, 0), the last odd entry is the second entry. So we subtract two from all the higher entries and

∗ 2 2 one from second entry, and we get I = (2, 0, 0). So we multiply OS3 (x1)OS3 (x1x2) =

4 4 4 4 4 4 3 2 3 2 2 3 2 3 3 2 2 3 x1x2 + x1x3 + x1x2 + x2x3 + x1x3 + x2x3 + x1x2 + x1x3 + x1x2 + x1x3 + x2x3 + x2x3 +

2 2 2 2 2 2 4 3 2 2 2 2x1x2x3 + 2x1x2x2 + 2x1x2x3 = OS3 (x1x2) + OS3 (x1x2) + 2OS3 (x1x2x3). In this

3 2 step we get an orbit OS3 (x1x2) that we also need to reduce. So we continue the

2 2 3 2 3 2 2 3 2 3 2 3 process and we consider OS3 (x1x2)OS3 (x1) = x1x2 + x1x3 + x1x2 + x1x3 + x2x3 +

3 2 2 2 2 2 2 2 3 2 2 2 4 x2x3 + x1x2x3 + x1x2x2 + x1x2x3 = OS3 (x1x2) + OS3 (x1x2x3). Hence, OS3 (x1x2) =

2 2 2 2 2 2 OS3 (x1)OS3 (x1x2) − OS3 (x1x2)OS3 (x1) − OS3 (x1x2x3). Lastly we need to write

2 2 OS3 (x1x2) in terms of the elementary skew-symmetric polynomials. From Lemma

2 2 1 2 2 4 2.2.5, OS3 (x1x2) = 2 (OS3 (x1)OS3 (x1x2)+OS3 (x1x2)OS3 (x1)). Therefore, OS3 (x1x2) = 2 2 1 2 2 1 2 2 2 OS3 (x1) OS3 (x1x2)− 2 OS3 (x1x2)OS3 (x1) − 2 OS3 (x1)OS3 (x1x2)OSn (x1)−OS3 (x1x2x3).

Sn We will show that P also generates k−1[x1, x2, . . . , xn] , but we need to introduce a new order on the set of degree d monomials. In the commutative case under the usual lexicographical order on monomials we have an order on orbit sums determined by the leading term. In degree d ≤ n the orbit sum with the largest leading term is the

d power polynomial Pd = OSn (x1), and the orbit sum with the smallest leading term is the elementary symmetric polynomial Sd = OSn (x1x2 ··· xd). Sturmfels deﬁned the order below and used it to prove the power sums generate the invariant subring when

Sn acts on the commutative polynomial ring k[x1, ··· , xn] [3]. Next we deﬁne this order and adapt his proof to the noncommutative ring k−1[x1, ··· , xn].

i1 i2 in j1 j2 jn Deﬁnition 4. A monomial (x1 x2 . . . xn ) ≺ (x1 x2 . . . xn ) if the partition

λ(i1, i2, . . . in) >lex λ(j1, j2, . . . , jn), or if λ(i1, i2, . . . in) =lex λ(j1, j2, . . . , jn).

16 This order forms a partial order on the set of monomials in R. Under ≺ the n elementary skew-symmetric polynomials are considered the largest. This is because for a ﬁxed degree d ≤ 2n − 1, the smallest degree d monomial in terms of the lexicographic order, is created by assigning exactly one 1 and assigning other exponents twos or zeros. If we assign more than one 1 then we create a trivial orbit sum. Similarly, the n odd power sums are the smallest under ≺ since they are the largest lexicographically.

Sn Theorem 2.2.8. The ring k−1[x1, x2, . . . , xn] is generated by the n odd power sums, P .

Proof. By Theorem 2.2.6 the n elementary skew-symmetric polynomials generate

RSn , so it suﬃces to show that they can be generated by the n odd power sums P. Hence, it is enough to show that we can express invariant skew-polynomials of total degree less than or equal to 2n − 1 in terms of P. Let f ∈ RSn with leading

I i1 i2 in Pn i term cX = x1 x2 ··· xn where t=1 it ≤ 2n − 1. Let Pi = OSn (x1), a power sum, where i can be even or odd. Consider the invariant P = Pi1 Pi2 ...Pin whose largest

i1 i2 in c monomial under is dx1 x2 . . . xn . Let f := f − d P . The largest monomial in f must be less than XI , since XI was the largest in f. This process terminates since f

Sn and f have the same degree. Thus the ring k−1[x1, x2, . . . , xn] is generated by the power sums. From Lemma 2.2.5, the power sum Pi where i is even can be expressed by the n odd power sums. Hence, f can be expressed by the n odd power sums.

Example 2.2.9. Consider the ring k−1[x1, x2, x3, x4] and rewriting the elementary

2 2 skew-symmetric polynomial OS4 (x1x2x3) in terms of odd power sums. The ﬁrst step

2 2 1 2 2 1 5 from the algorithm above is OS4 (x1x2x3) − 2 (OS4 (x1)) OS4 (x1) = − 2 OS4(x1) − 1 4 3 2 4 3 2 2 OS4(x1x2) − OS4(x1x2). Again, we need to rewrite OS4 (x1x2) and OS4 (x1x2) as 2 power sums, and we need to use Lemma 2.2.5 to write OS4 (x1) as odd power sums.

4 4 We continue the algorithm and see that the orbit sum OS4 (x1x2) = OS4 (x1)OS4 (x1)−

17 5 3 2 3 2 5 OS4 (x1) and OS4 (x1x2) = OS4 (x1)OS4 (x1) − OS4 (x1). We continue the algorithm

2 2 1 2 2 1 4 further and see that OS4 (x1x2x3) = 2 (OS4 (x1)) OS4 (x1) − 2 OS4 (x1)OS4 (x1) 3 2 5 2 2 − OS4 (x1)OS4 (x1) + OS4 (x1). Then by Lemma 2.2.5 we see OS4 (x1x2x1) =

1 2 2 1 3 3 2 (OS4 (x1) ) OS4 (x1) − 4 (OS4 (x1)OS4 (x1) + OS4 (x1)OS4 (x1)) OS4 (x1) − 3 2 5 OS4 (x1)OS4 (x1) + OS4 (x1).

2 2 Thus, OS4 (x1x2x3) can be written as odd power sums.

18 Chapter 3: G¨obel’s Bound

Now we want to determine a set of generators for RG where G is a subgroup of

Sn. If a skew-polynomial f from R is invariant under all elements of Sn, then f is invariant under G; thus RSn ⊆ RG. We will use the elementary skew-symmetric polynomials as part of our generating set

3.1 Special Monomials

Given a subgroup G of Sn we deﬁne a class of special monomials that will be used to produce a generating set RG.

Deﬁnition 5. A monomial, XI , is called special if the exponent sequence I has an associated partition λ(I) with:

i) λi(I) − λi+1(I) ≤ 2 for all 1 ≤ i ≤ n, and

ii) λn(I) is either 1 or 0.

Example 3.1.1. Consider the ring k−1[x1, x2, x3, x4, x5]. The sequences I = (1, 0, 3, 4, 2) and J = (2, 2, 1, 2, 0) are both special, where as the sequences K = (1, 3, 1, 0, 0) and L = (5, 4, 3, 2, 1) are not special.

For every monomial XI we associate a special monomial Red(XI ) through an iterative process generalized from G¨obel’s algorithm for k[x1, x2 . . . , xn] [1]. When

λi(I) − λi+1(I) > 2, we call this a gap in the exponent sequence. Then for all j ≤ i subtract 2 from λj(I). Repeat this procedure until there are no more gaps. This produces the special exponent sequence Red(I) and special monomial Red(XI ). XI and Red(XI ) diﬀer by a factor with even exponents and thus the factor is in the

19 center of R. We can write XI = Red(XI )XJ where XJ = XI /Red(XI ). Notice by construction that the ith largest entry in Red(I) is in the same position the ith largest entry in J. This implies the permutation λ that takes (I) to its associated partition λ(I) has the property λ(I) = λ ((Red(I)) + (J)) = λ(Red(I))+λ(J), where λ(J) and λ(Red(I)) are both partitions of J and Red(I).

I Example 3.1.2. Let R = k−1[x1, x2, ..., x7], and consider X where I = (0, 7, 4, 1, 0, 1, 7). The ﬁrst gap appears between 7 and 4. We reduce every entry that is greater than or equal to 7 by 2. We get (0,5,4,1,0,1,5). There is still a gap between 4 and 1 so we need to reduce again. We get (0,3,2,1,0,1,3). This new exponent sequence is now

I 7 4 7 I 3 2 3 J 4 2 4 special. Thus X = x2x3x4x6x7, Red(X ) = x2x3x4x6x7, and X = x2x3x7.

There is a concern that when reducing an exponent sequence I to Red(I) we can

I get a trivial orbit sum, i.e. OG(Red(X )) = 0.

I I Lemma 3.1.3. If OG(Red(X )) = 0 then OG(X ) = 0.

Proof. Let Red(XI ) be the associated special monomial of XI and XI = Red(XI )XJ .

I I I Assume OG(Red(X )) = 0, so ∃σ ∈ G such that σ(Red(X )) = −Red(X ). Thus, σ(Red(I)) = Red(I). Since σ(Red(I)) = Red(I), we have σ(J) = J. By construction, J is an exponent sequence where every entry is even. Therefore σ acting on (XJ ) cannot create a sign change. Hence, σ(XJ ) = XJ . Recall XI = Red(XI ))XJ , and consider:

σ(XI ) = σ(Red(XI )XJ )

= σ(Red(XI ))σ(XJ )

= −Red(XI )XJ

= −XI

I I I Since σ ∈ G, we have that −X ∈ OrbG(X ). Therefore, OG(X ) = 0.

20 I Lemma 3.1.4. For any monomial X and for every σ ∈ Sn,

(a) σ(Red(XI )) = Red(σ(XI ))

(b) σ(Red(XI )) = Red(XI ) if and only if σ(XI ) = XI .

Proof. (a) follows since permuting a sequence then reducing is the same as reducing then permuting. (b) If σ(XI ) = XI then σ(Red(XI )) = Red(σ(XI )) = Red(XI ). Conversely, if σ(Red(XI )) = Red(XI ) then Red(σ(XI )) = Red(XI ). Thus σ(XI ) = XI .

3.2 G¨obel’s Bound

We use the same order on the set of monomials that Derksen used to prove G¨obel’s theorem in the commutative case [1].

Definition 6. We say XI XJ if the first difference, i, between the associated

I J partitions λ(I) and λ(J) is λi(I) > λi(J). We say X X if either the ﬁrst diﬀerence, i, between λ(I) and λ(J) is λi(I) < λi(J), or if λj(I) = λj(J), for all j.

Example 3.2.1. Let XI , XJ , and XK be monomials with exponent sequences I = (0, 5, 1, 2), J = (5, 0, 1, 1), and K = (2, 5, 0, 1). The partitions of the exponent sequences are λ(I) = (5, 2, 1, 0), λ(J) = (5, 1, 1, 0), and λ(K) = (5, 2, 1, 0). The

ﬁrst diﬀerence in the partition of I and J is the second entry. λ2(I) = 2, λ2(J) = 1

I J I K so λ2(I) > λ2(J). Thus X X . Now X X since λj(I) = λj(K) for all j.

Theorem 3.2.2. Let XI be some monomial, and XJ = XI /Red(XI ). Every mono-

J I I I mial occurring in OSn (X )OG(Red(X )) − OG(X ) is lower than X under order. To prove this theorem we will prove:

I K K J I I (a) X X , ∀X occuring in OSn (X )Red(X ) − X .

21 J I We consider this case where OSn (X ) is multiplied by the monomial Red(X ), since

I I it is less cumbersome than considering OSn (X )OG(Red(X )). Then we utilize part (a) along with Lemma 3.1.3 to prove:

I K K J I I (b) X X , ∀X occuring in OSn (X )OG(Red(X )) − OG(X ).

I J I Proof. (a) Clearly X occurs in the sum, OSn (X )Red(X ). Let’s assume to the

K J I I K contrary that there exists X = σ(X )Red(X )such that X X , where σ ∈ Sn. We want to show that σ(XJ ) = XJ ; then (a) is proved.

I n ei J n di I K Let Red(X ) = Πi=1xi and X = Πi=1xi . Thus X and X have exponents ei + di

I and ei + dσ−1(i) respectively. We reorder X such that the exponent sequence is a

I λ1 λ2 λn partition, i.e. X = ±xτ(1)xτ(2) ··· xτ(n) for some permutation τ. By construction, the permutation τ that takes I to its partition is the same permutation that takes

J and Red(I) to their respective partitions. So we may assume that ei ≥ ei+1 and

I K di ≥ di+1. Let m be maximal with d1 = d2 = ... = dm. Since X X we know that σ must permute the set {1, 2, ..., m}. Similarly, we consider the second biggest

I K exponent, and maximal l such that dm+1 = ... = dm+l. Since X X we know that σ permutes {m + 1, m + 2, ..., m + l}. We continue this process and see that the

J J exponent sequence (d1, ..., dn) is invariant under σ. Thus σ(X ) = X . (b) For any σ ∈ G we have the following as a consequence of 3.2.2(a),

I K K σ(XI ) I I σ(X ) X for all X occurring in OSn ( Red(σ(XI )) )Red(σ(X )) − σ(X ).

Thus XI XK . By Lemma 3.1.3(a),

σ(XI ) I J I J I OSn ( Red(σ(XI )) )Red(σ(X )) = OSn (σ(X ))σ(Red(X )) = OSn (X )σ(Red(X )).

I K K J I I Therefore, X X for all X occuring in OSn (X )σ(Red(X )) − σ(X ).

22 By Lemma 3.1.3(b), the permutations σ ∈ G, such that σ(XI ) = XI are exactly those that ﬁx Red(XI ). So summing over the coset representation, σ, of the stabilizer

I I K J I I in G of X , we obtain X X for all terms in OG(X )OG(Red(X ))−OG(X ).

Z4 Example 3.2.3. Consider k−1[x1, x2, x3, x4] where < g >= Z4, and

 0 0 0 1   1 0 0 0  g =   .  0 1 0 0  0 0 1 0

Let I = (2, 0, 5, 1), and therefore, Red(I) = (2, 0, 3, 1) and J = (0, 0, 2, 0). So

I 2 4 3 2 2 3 2 5 2 3 3 OZ4 (Red(X ))OS4 (x1) = OZ4 (x1x3x4)+OZ4 (x1x2x3x4)+OZ4 (x1x3x4)+OZ4 (x1x3x4). I 2 I 4 3 2 2 3 Therefore, OZ4 (Red(X ))OS4 (x1) − OZ4 (X ) = OZ4 (x1x3x4) + OZ4 (x1x2x3x4) + 2 3 3 OZ4 (x1x3x4). The remaining exponent sequences are (4, 3, 1, 0), (3, 2, 2, 1), (3, 3, 2, 0). All of them are lower in comparison with I = (2, 0, 5, 1) under order.

G Corollary 3.2.4. The degree of the largest possible generator k−1[x1, x2, . . . , xn] is n2.

Proof. From G¨obel’s algorithm a set of generators of RG is the orbit sums of special monomials along with the n-elementary skew-symmetric polynomials. The largest elementary skew-symmetric polynomial in terms of total degree is 2n-1. The largest

I orbit sum of a special monomial, OG(X ), in terms of total degree has the associated partition λ(I) = (2n − 1, 2(n − 1) − 1, 2(n − 2) − 1,..., 5, 3, 1). Therefore, the degree

Pn 2 2 of the orbit sum is i=1(2i − 1) = n . The value n ≥ 2n − 1, ∀n ≥ 1. Thus, the largest possible generator of RG has degree n2.

We do not know if the bound is sharp.

23 Chapter 4: The Alternating Group.

An Now we want to determine a set of algebra generators for R . Since An ≤ Sn, we have RSn ⊆ RAn .

4.1 Antisymmetric Invariants

In this section we begin our consideration of invariants under the alternating group

An by discussing antisymmetric invariants. In the commutative case the set of an-

An tisymmetric invariants and the symmetric polynomials generate k[x1, x2, . . . , xn] [3]. Analogously, the set of antisymmetric skew-polynomials and symmetric skew-polynomials generate RAn .

I Lemma 4.1.1. An orbit sum OAn (X ) is invariant under Sn if and only if there exists some odd permutation τ such that τ(XI ) = XI .

I I I Proof. (⇒) Let OAn (X ) be invariant under Sn. Therefore OrbSn (X ) = OrbAn (X ),

I I I and |StabSn (X )| = 2|StabAn (X )|. Thus, StabSn (X ) must contain some permuta-

I tion that StabAn (X ) does not. Hence there is some odd permutation that stabilizes XI .

I I (⇐). Since [StabSn (X ): StabAn (X )] ≤ [Sn : An] = 2, so if some odd permu-

I I I tation σ stabilizes X , then |StabSn (X )| = 2|StabAn (X )|. Notice |Sn| = 2|An| =

I I I I I I 2|OrbAn (X )||StabAn (X )| = |OrbSn (X )||StabSn (X )| = 2|OrbSn (X )||StabAn (X )|.

I I I I Therefore, OrbAn (X ) = OrbSn (X ), and OSn (X ) = OAn (X ).

Let f be an An-invariant and let σ and τ be any two odd permutations. Then τ −1σ is even, so τ −1σ(f) = f, and hence σ(f) = τ(f).

24 I Lemma 4.1.2. The orbit sum OAn (X ) = 0 if and only if there are exponents ij = ik, with ik odd, and is = it, with it even.

I Proof. (⇒) Let OAn (X ) = 0. Similarly to the Sn case, we need some permutation,

I I I σ, from An such that σ(X ) = −X . If (−X ) is to occur in the orbit, then some

I I nonidentity element σ ∈ An must be an element of Stab(I) and σ(X ) = −(X ). If σ ﬁxes all the odd exponents then −XI cannot occur. Thus σ must permute at least one pair of exponents ij with ik where ij = ik, and ik is odd. Also since σ is an even permutation, σ must permute at least one other pair of exponents and not create a sign change. Thus, there must be some pair of exponents it = is, where it is even which σ permutes. Therefore, σ permutes j with k and t with s.

(⇐) Let the exponent sequence I have some indices ij = ik, an odd number, and

I is = it, an even number. Consider the even permutation (j, k)(s, t) acting on X . So (s, t) ﬁxes XI since it permutes two even numbers. Then (j, k)(XI ) = −Xi. Thus

I the orbit sum OAn (X ) = 0.

n−1 Corollary 4.1.3. The invariant OAn (Πi=1 xi) is the smallest in terms of total degree that is both an An-invariant and not a Sn-invariant.

n−1 Proof. The exponent sequence of OAn (Πi=1 xi) has no repeated evens, thus by Lemma

n−1 n−1 Sn 4.1.1. and 4.1.2. OAn (Πi=1 xi) 6= 0 and OAn (Πi=1 xi) ∈/ R . Any other orbit sum, f, that is lower in terms of degree will have at least two zeros in the exponent sequence.

n−1 Thus by Lemma 4.1.1, f is a Sn-invariant. Hence OAn (Πi=1 xi) is the smallest degree

An-invariant that is not a Sn-invariant.

Deﬁnition 7. A skew-polynomial f is called antisymmetric if τf = −f for every odd permutation in Sn.

Theorem 4.1.4. Every An-invariant decomposes uniquely into a sum of a symmetric and an antisymmetric skew-polynomial.

25 It suﬃces to show all An-invariants that are not Sn-invariant can be constructed in this manner. Let f ∈ RAn and not in RSn . Therefore, some odd permutation σ has the property σf 6= f. Consider g = f + σf and h = f − σf. It is clear that

1 2 (g + h) = f, so it remains to show that g is a Sn-invariant and h is antisymmetric.

Lemma 4.1.5. The An-invariant g is a Sn-invariant.

Proof. Let τ ∈ Sn − An, and consider τ(g) = τ(f) + τ(σ(f))

= σ(f) + f

= g.

Let τ ∈ An, and consider τ(g) = τ(f) + τ(σ(f))

= σ(f) + f

= g.

Thus g is a Sn-invariant.

Lemma 4.1.6. The An-invariant h is antisymmetric.

Proof. Let τ be some odd permutation, and consider τ(h) = τ(f) − τ(σ(f))

= τ(f) − f

= −(f − σ(f))

= −h.

Hence, h is antisymmetric.

Therefore every An-invariant is a sum of a Sn-invariant and an antisymmetric invariant.

26 Now we want to show the decomposition is unique.

Proof. Let A1 and A2 be antisymmetric and S1 and S2 be symmetric. Assume for contradiction that f = A1 + S1 and f = A2 + S2. So A1 + S1 = A2 + S2, thus

A1 − A2 = S2 − S1. Therefore A1 − A2 is a symmetric polynomial. Let σ be any odd permutation and consider σ acting on (A1 − A2).

σ(A1 − A2) = σ(A1) − σ(A2)

= −A1 + A2

= −(A1 − A2)

= A1 − A2.

Thus, A1 = A2, and hence S1 = S2. Therefore, f decomposes uniquely into the sum of an antisymmetric invariant and a symmetric invariant.

Lemma 4.1.7. Linear combinations of antisymmetric invariants are antisymmetric

Proof. Let f and g be antisymmetric, c, d ∈ k and let τ be any odd permutation

τ(cf + dg) = τ(cf) + τ(dg)

= cτ(f) + dτ(g)

= −cf − dg

= −(cf + dg).

If f and f + g are antisymmetric then g is antisymmetric too. Also if f = gh and f and g are antisymmetric then h is symmetric.

n−1 n Lemma 4.1.8. The invariants OAn (Πi=1 xi) and OAn (Πi=1xi) are antisymmetric.

27 n Proof. First we show OAn (Πi=1xi) is antisymmetric. To show this let’s show that any

n n n transposition, (i, j), acting on OAn (Πi=1xi) satisﬁes (i, j)OAn (Πi=1xi) = −OAn (Πi=1xi).

n n The invariant OAn (Πi=1xi) = Πi=1xi. Without loss of generality let i > j and let (i, j) be any transposition and compute:

n (i, j)OAn (Πi=1(xi) = (i, j)(x1x2 ··· xi ··· xj ··· xn)

= x1x2 ··· xj ··· xi ··· xn

2(j−i−1)+1 = (−1) (x1x2 ··· xi ··· xj ··· xn)

= −(x1x2 ··· xi ··· xj ··· xn)

n = −OAn (Πi=1(xi)

The fact that any odd permutation is the product of an odd number of transposi-

n n tions and that any transposition takes OAn (Πi=1(xi) to −OAn (Πi=1(xi), we have that

n OAn (Πi=1(xi) is antisymmetric. Recall that if f is symmetric, g is antisymmetric, and f = gh, then h is antisymmet-

n n−1 2 2 2 n ric. We note that ±OAn (Πi=1xi)OAn (Πi=1 xi) = OSn (x1x2 ··· xn−1xn), OAn (Πi=1(xi)

2 2 2 n−1 is antisymmetric, and OSn (x1x2 ··· xn−1xn) is symmetric. Therefore, OAn (Πi=1 (xi) is antisymmetric.

4.2 RAn Generators

We now proceed to the discussion of producing algebra generators of RAn .

Deﬁnition 8. An exponent sequence I = (ij) is a pseudo-partition if ij ≥ ij+1 for

1 ≤ j ≤ n − 2 and in−2 > in > in−1. We denote the pseudo-partition as λ(I).

A3 Example 4.2.1. Consider the ring k−1[x1, x2, x3] , and the orbit sums OA3 (x1) and

3 2 OA3 (x1x2x3).

The orbit sum OA3 (x1) = x1 + x2 + x3. The monomial x1 occurs in the orbit sum;

28 therefore, we can deﬁne the orbit sum with the partition (1, 0, 0).

3 2 3 2 2 3 2 3 3 2 Now the orbit sum OA3 (x1x2x3) = x1x2x3 +x1 +x2x3 −x1x2x3. The monomial x1x2x3 does not occur in the orbit sum so we cannot deﬁne this orbit sum with the partition

3 2 (3, 2, 1), yet the monomial x1x2x3 occurs in the obit sum. Hence we can deﬁne this orbit with the pseudo-parition (3, 1, 2).

Lemma 4.2.2. Every orbit sum under the alternating group is represented by a monomial XI where I is either a partition or a pseudo-partition.

Proof. For any exponent sequence I there exists some permutation σ such that

λ(I) I σ(I) = λ(I), a partition. If σ ∈ An then X is a summand in OAn (X ) and

I λ(I) OAn (X ) = OAn (X ). Otherwise σ ∈ Sn − An and therefore (n − 1, n)σ ∈

An. The permutation (n − 1, n)σ induces a pseudo-partition where acting on I,

(n − 1, n)σ(I) = (λ1(I), λ2(I), . . . , λn−2(I), λn(I), λn−1(I)) and λn(I) < λn−1(I)).

I λ(I) Similarly, OAn (X ) = OAn (X ). So for any orbit sum under the alternating group,

I OAn (X ), we can assume that the exponent sequence I of the representative monomial XI is a partition or a pseudo-partition.

The symmetric group Sn = An ∪ σAn for any odd permutation σ. Therefore

I I I I I OrbSn (X ) = OrbAn (X ) ∪ σOrbAn (X ). This implies that OSn (X ) = OAn (X ) or

I I I OSn (X ) = OAn (X ) + σOAn (X ).

There are two trivial cases when determining the algebra generators for RAn . The

ﬁrst trivial case is the subring RA1 = RS1 = R, and the second is RA2 = R. Both of these subrings are equal to R, and so we already know the generators for the subring.

n n Lemma 4.2.3. The invariant OAn (Πi=1xi) = Πi=1xi can be constructed by OAn (x1) n−1 and OAn (Πi=1 xi) for n ≥ 3.

2 3 Proof. First consider when n = 3. Compute OA3 (Πi=1xi)OA3 (x1) = 3OA3 (Πi=1xi) +

2 2 OA3 (x1x2) − OAn (x1x3), and

29 2 3 2 2 OA3 (x1)OA3 (Πi=1xi) = 3OA3 (Πi=1xi) − OA3 (x1x2) + OAn (x1x3). So,

1 2 2 3 6 (OA3 (Πi=1xi)OA3 (x1) + OA3 (x1)OA3 (Πi=1xi)) = Πi=1xi. Now let n ≥ 4, and notice that for any orbit sum of a monomial XI where I is sequence of n − 1 ones and 1 zero, the orbit sum is deﬁned by the partition λ(I) = (1, 1,..., 1, 0) and not the pseudo-partition λ(I) = (1, 1,..., 1, 0, 1). This is because n ≥ 4 so n −

1 > 2, thus, the even permutation (1,2)(n-1,n) acting on Xλ(I) = ±Xλ(I). There- fore, the only partitions and pseudo-partitions we need to count from the product

n−1 OAn (Πi=1 xi)OAn (x1) are (2, 1,..., 1, 0) and (1, 1,..., 1). The only way to get the

n−1 n−2 partition (2, 1,..., 1, 0) is by (Πi=1 xi)(x1). The sign of this monomial is (−1) . The

n−j j−1 n other sequence (1, 1,..., 1) occurs n many times, namely ((−1) Πi=1 xiΠi=j+1xi)(xj) for all 1 ≤ j ≤ n. The sign of any of these monomials is (−1)n−j(−1)n−j = 1.

n−1 n−2 2 n Hence, OAn (Πi=1 xi)OAn (x1) = (−1) OAn (x1x2 ··· xn−1)+nOAn (Πi=1xi). Similarly,

n−1 2 n−1 n OAn (xi)OAn (Πi=1 xi) = OAn (x1x2 ··· xn−1)+(−1) nOAn (Πi=1xi). When n ≥ 4 and 1 n−1 n−1 n n is even we get 2n OAn (Πi=1 xi)OAn (x1) − OAn (xi)OAn (Πi=1 xi) = Πi=1xi. When 1 n−1 n−1 n n is odd we get 2n OAn (Πi=1 xi)OAn (x1) + OAn (xi)OAn (Πi=1 xi) = Πi=1xi. Thus, n n n−1 OAn (Πi=1xi) = Πi=1xi can be constructed by OAn (x1) and OAn (Πi=1 xi).

2 2 2 Corollary 4.2.4. The Sn-invariant OSn (x1x2 ··· xn−1xn) can be generated by n−1 OAn (Πi=1 xi) and OAn (x1).

2 2 2 n n−1 Proof. The invariant OSn (x1x2 ··· xn−1xn) = ±OAn (Πi=1xi)OAn (Πi=1 xi). From Lemma

n n n−1 4.2.3, OAn (Πi=1xi) = Πi=1xi can be constructed by OAn (x1) and OAn (Πi=1 xi). There-

2 2 2 n−1 fore, the invariant OSn (x1x2 ··· xn−1xn) can be constructed by OAn (Πi=1 xi) and

OAn (x1).

I Lemma 4.2.5. All antisymmetric orbit sums OAn (X ) are represented by a monomial XI where I is a partition.

I Proof. Let OAn (X ) be an orbit sum that is an antisymmetric invariant. Then from

30 Lemma 4.2.1 I is either a partition or a pseudo-partition. If I is a partition then we are done. Otherwise I is a pseudo-partition. Then consider the odd permutation

I I λ(I) I (n, n − 1) acting on OAn (X ), (n, n − 1)OAn (X ) = −OAn (X ). Thus −OAn (X ) is represented by a monomial whose exponent sequence is a partition.

I Lemma 4.2.6. Every antisymmetric orbit sum OAn (X ) has a pair of repeated odd entries in I.

I Proof. Let OAn (X ) be an orbit sum that is an antisymmetric invariant. Therefore

I I any odd permutation, σ, has the property σOAn (X ) = −OAn (X ). Thus no odd

I I I permutations stabilize OAn (X ). By Lemma 4.1.1. OAn (X ) 6= OSn (X ). Thus

I I I OSn (X ) = OAn (X ) + σOAn (X ). Therefore:

I I I OSn (X ) = OAn (X ) + σOAn (X )

I I = OAn (X ) − OAn (X )

= 0.

Thus the orbit sum under Sn is trivial. Lemma 2.2.2 implies that the exponent sequence of XI has a pair of repeated odd entries.

Theorem 4.2.7. The ring RAn is generated as an algebra by the ﬁrst n − 1 skew-

n−1 symmetric polynomials and OAn (Πi=1 xi).

Proof. Since every invariant under An can be written as a linear combination of antisymmetric invariants and Sn-invariants, it suﬃces to show that any antisymmetric invariants f can be constructed in this manner. Let f be antisymmetric. Therefore, by Lemma 4.2.3, f has a leading monomial

I e X where I = (i1, i2, . . . , in) and I is a partition. If in 6= 0 then let I = (i1 −

I Ie 1, i2 − 1, . . . , in − 1). The fact that OAn (X ) 6= 0 implies OAn (X ) 6= 0. Therefore,

I n Ie OAn (X ) = ± (Πi=1xi) OAn (X ). So we may assume that in = 0.

31 ∗ I Consider I = (i1 − 1, i2 − 1, . . . , in−1 − 1, 0). Since OAn (X ) is antisymmetric then Lemma 4.2.6, implies that I has a pair of repeated odd entries. Thus I∗ has a

I∗ pair of repeated even entries. By Lemma 4.1.1. OAn (X ) is a Sn-invariant. Since

I OAn (X ) is not a Sn-invariant, by Lemma 4.1.1, the exponent sequence I has no repeated even entries. Thus I∗ has no repeated odd entries. Therefore by Lemma 4.1.2.

I∗ n−1 I∗ n−1 OAn (X ) 6= 0. Consider OAn (Πi=1 (xi))OAn (X ). Since OAn (Πi=1 (xi)) is antisym-

I∗ n−1 I∗ metric and OAn (X ) is a Sn-invariant, we can conclude that OAn (Πi=1 (xi))OAn (X )

n−1 I∗ is antisymmetric. Therefore, OAn (Πi=1 (xi))OAn (X ) has a leading monomial under the lexicographic order, which is XI , and all other monomials are smaller lexico-

I n−1 I∗ graphically. Thus, all summands occurring in OAn (X ) ± OAn (Πi=1 (xi))OAn (X ) are smaller lexicographically and are antisymmetric. Thus there are only a ﬁnite number of degree sequences less than I, so this process terminates. Hence, every antisymmetric orbit sum can be written in terms of the ﬁrst n − 1 elementary skew-

n−1 symmetric invariants and OAn (Πi=1 xi).

32 Chapter 5: Examples With G¨obel’s Bound with 4 indeterminates

The intent of this chapter is to compute explicit examples of some invariant subrings. We explore subrings of R = k−1[x1, x2, x3, x4], the skew-polynomial ring with four indeterminates, since all the other invariant subrings with three or less indeterminates have been computed by previous theorems in this thesis. Recall that G¨obel’s bound in the commutative case showed that for four indeterminates, generators of

G 4 R have degree less than or equal to 2 =6. Also, G¨obel’s bound was sharp in the

A4 commutative case since k[x1, x2, x3, x4] needs a generator of degree 6, namely the Vandermonde Determinant. In this chapter we explore G¨obel’s bound to determine how sharp the bound is in the non-commutative case.

5.1 Alternating Z2 Subgroup

Let Z2 be a subgroup of S4 deﬁned by:

 1 0 0 0   0 1 0 0     0 1 0 0   1 0 0 0  Z2 =   ,   .  0 0 1 0   0 0 0 1   0 0 0 1 0 0 1 0 

This Z2 subgroup is contained in A4 and is generated by g = (1, 2)(3, 4).

I Lemma 5.1.1. An orbit sum OZ2 (X ) = 0 if and only if I = (i1, i2, i3, i4) where i1 = i2, i3 = i4 and either i1 is even and i3 is odd or i1 is odd and i3 is even.

33 I I I Proof. ⇒) Let OZ2 (X ) = 0, so g(X ) = −X . Therefore, g(I) = I

g(I) = g(i1, i2, i3, i4)

= (i2, i1, i4, i3)

= (i1, i2, i3, i4)

i1 i1 i3 i3 i1 i3 i1 i1 i3 i3 Therefore, i1 = i2 and i3 = i4. So g(x1 x2 x3 x4 ) = (−1) (−1) (x1 x2 x3 x4 ). Hence, either i1 or i3 must be odd, but not both.

I ⇐). Consider OZ2 (X ) and let I = (i1, i1, i3, i3) and i1 is even and i3 is odd or vice I I I I I versa. Then OZ2 (X ) = X +g(X ) = X −X = 0. Thus the orbit sum is trivial.

To determine a set of generators for RZ2 we use G¨obel’s bound along with the generators from RA4 to bound the degree of potential generators. To further bound

2 2 2 the degree of potential generators we take OZ2 (x1x2), OZ2 (x3x4), and OZ2 (x1x2x3) to

I Z2 bound generators. Let OZ2 (X ) be a potential generator of R .

I 1. G¨obel’s bound bounds the degree of a generator of OZ2 (X ) to 16 where λ(I) =

(λ1(I), λ2(I), λ3(I), λ4(I)), λi − λi+1 ≤ 2 and λ4(I) = 1 or 0.

∗ 2. When λ4(I) = 1 we consider I = (i1 − 1, i2 − 1, i3 − 1, i4 − 1), and notice

I∗ I ±OZ2 (X )OA4 (x1x2x3x4) = OZ2 (X ). Therefore, we may assume that

λ4(I) = 0.

3. When λ3(I) = 2, the exponents of x1 and x2 are both at least 2, or the exponent

i1−2 i2−2 i3 i4 2 2 of x3 and x4 are at least two. Therefore, OZ2 (x1 x2 x3 x4 )OZ2 (x1x2) =

I i1 i2 i3−2 i4−2 2 2 I OZ2 (X ), or OZ2 (x1 x2 x3 x4 )OZ2 (x3x4) = OZ2 (X ). Therefore, we may

assume that λ3(I) ≤ 1.

I At this point we have reduced possible generators, OZ2 (X ), to have the associated partition of λ(I) = (λ1(I), λ2(I), 0, 0) or (λ1(I), λ2(I), 1, 0) and be special. We further

34 bound the case when λ3(I) = 1 and λ2(I) ≥ 2. We break this case into two subcases:

1) the exponents of x1 and x2 are both at least 2 or the exponent of x3 and x4 are at least 2, and 2) otherwise.

1. When the exponents of x1 and x2 are both at least 2 or the exponent of x3 and

i1−2 i2−2 i3 i4 2 2 I x4 are at least 2, then either OZ2 (x1 x2 x3 x4 )OZ2 (x1x2) = OZ2 (X ),

i1 i2 i3−2 i4−2 2 2 I or OZ2 (x1 x2 x3 x4 )OZ2 (x3x4) = OZ2 (X ).

2. Let at least one of the exponents of x1 and x2 be less than 2 and at least one of

I λ1(I) λ2(I) 0 the exponents of x3 and x4 be less than 2. Let X = ±xσ(1) xσ(2) xσ(3)xσ(4). Con-

I∗ λ1(I)−1 λ2(I)−1 sider ±X = xσ(1) xσ(2) . So both non-zero exponents cannot be the expo-

nents of x1 and x2 nor can they both be the exponents of x3 and x4. Therefore,

I∗ I∗ by Lemma 5.1.1, OZ2 (X ) 6= 0. Consider ±OZ2 (X )OZ2 (xσ(1)xσ(2)xσ(3)) =

I e1 e2 e3 e4 e1 e2 e3 e4 OZ2 (X )±OZ2 (x1 x2 x3 x4 ), where every ei > 0. So the orbit sums OZ2 (x1 x2 x3 x4 )

e1 e2 e3 e4 is reducible and thus, we can construct OZ2 (x1 x2 x3 x4 ). Therefore, we can

assume λ3(I) ≤ 1.

So possible partitions for exponent sequences of potential generators are from the set S = {(1, 0, 0, 0), (2, 0, 0, 0), (1, 1, 0, 0), (2, 1, 0, 0), (2, 2, 0, 0), (3, 1, 0, 0), (3, 2, 0, 0), (4, 2, 0, 0), (1, 1, 1, 0), (2, 1, 1, 0), and (3, 1, 1, 0)}.

2 2 2 Theorem 5.1.2. The set of invariants G = {OZ2 (x1x2), OZ2 (x3x4), OZ2 (x1x2x3), OA4 (x1),

2 2 2 2 Z2 OA4 (x1x2), OA4 (x1x2x3), OZ2 (x1), OZ2 (x1x3), OZ2 (x1x2) and OA4 (x1x2x3)} generate R .

I Proof. The remaining invariants, OZ2 (X ), where the associated partition λ(I) is an element of S are shown explicitly in the appendix, and thus, the theorem is proved.

35 5.2 Alternating Klein IV Subgroup

Let V be a Klein-4 subgroup of S4 generated by matrices with representation:

 1 0 0 0   0 1 0 0   0 0 1 0   0 0 0 1     0 1 0 0   1 0 0 0   0 0 0 1   0 0 1 0  V =   ,   ,   ,    0 0 1 0   0 0 0 1   1 0 0 0   0 1 0 0   0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0 

We will denote V as the set {(1), (12)(34), (13)(24), (14)(23)}. This particular V

A4 is a subgroup contained in A4, so we can use the generators for R to help bound the degree of generators needed for RV.

I I Lemma 5.2.1. The orbit sum OV(X ) = OA4 (X ) if and only if there exists some three cycle c such that c(XI ) = XI .

Before we can prove this claim we must prove a lemma ﬁrst.

Lemma 5.2.2. A4 can be generated by the set {(12)(34), (13)(24), (14)(23), c}for any three cycle c in A4.

Proof. Clearly the subgroup generated by the set {(12)(34), (13)(24), (14)(23), c} con- tains the four elements shown, the inverse of c, and the identity. This accounts for six elements. The group A4 has no subgroups of order 6 so {(12)(34).(13)(24), (14)(23), c} must generate A4.

I I Lemma 5.2.1. ⇒) Let OV(X ) = OA4 (X ), and assume for contradiction that every I I I I three cycle c has the property c(X ) 6= X . Since OV(X ) = OA4 (X ), we have that I I I OrbV(X ) = OrbA4 (X ). Thus, the StabA4 (X ) ⊆ {(12)(34), (13)(24), (14)(23)} = V. I I I This implies StabA4 (X ) = StabV(X ). We know |V| = |StabV||OrbV(X )|, and I I I I I |A4| = |StabA4 (X )||OrbA4 (X )|. Since OrbV(X ) = OrbA4 (X ) and StabA4 (X ) = I I I StabV(X ), we can substitute and get |V| = |StabA4 (X )||OrbA4 (X )| = |A4|. Hence,

36 |V| = |A4| implies 4=12, but 4 6= 12. Therefore some three cycle, c, must have the property c(XI ) = XI .

I I ⇐) Let c be a three cycle such that c(X ) = X . We also know A4 is gnerated

I 2 by the set {(12)(34), (13)(24), (14)(23), c}. So OrbA4 (X ) ⊆ V ∪ Vc ∪ Vc . Thus I I I I OrbA4 (X ) ⊆ OrbV(X ). V ≤ A4 therefore we have that OrbV(X ) ⊆ OrbA4 (X ). I I I I Hence, OrbV(X ) = OrbA4 (X ), and therefore OV(X ) = OA4 (X ).

I I Theorem 5.2.3. The orbit sum OV(X ) = 0 if and only if OA4 (X ) = 0.

I I I Proof. ⇒) Let OV(X ) = 0. Then there exists g ∈ V such that g(X ) = −X . I I Since V ≤ A4 implies that g ∈ A4, −X is a term occurring in OA4 (X ). Thus I OA4 (X ) = 0.

I ⇐) Let OA4 (X ) = 0. This implies that the exponent sequence I has a repeated

I I even and repeated odd entry. So any permutation σ in A4 that takes X to −X must interchange the even exponents and interchange the odd exponents. Therefore,

σ must be the product of two disjoint transpostitions within A4. So σ must also be

I I I in V. Therefore −X must occur in OV(X ). Thus OV(X ) = 0.

We want to determine the generators by bounding the degree of monomials. We can look explicitly at the associated partition of the exponent sequence on a monomial. G¨obel’s Bound bounds the degree of a monomial, XI , to 16 with an associated partition of the form λ(I) = (λ1, λ2, λ3λ4), where λi − λi+1 ≤ 2 and λ4 is either 1 or 0.

I When λ4 = 1, then we can factor out OV(x1x2x3x4) = x1x2x3x4 from OV(X ). I I∗ ∗ This results in OV(X ) = ±OV(x1x2x3x4)OV(X ), where λ(I ) = (λ1(I) − 1, λ2(I) −

I V 1, λ3(I)−1, 0). So we can assume that for any generator OV(X ) of R has λ4(I) = 0, which bounds the degree to 12.

37 ∗ When λ3 = 2 we consider λ(I ) = (λ1(I) − 1, λ2(I) − 1, 1, 0). Then, consider

I∗ I∗ I∗ I∗ OV(X )OV(x1x2x3) = OV(X )(x1x2x3) − OV(X )(x1x2x4) + OV(X )(x1x3x4) − I∗ ∗ OV(X )(x2x3x4). The fact that I has exactly three non-zero entries implies that I∗ exactly four monomials from OV(X )OV(x1x2x3) will have exactly three non-zero I entries in their exponent sequence. These four monomials will form OV(X ). Every I∗ other monomial in OV(X )OV(x1x2x3) will have an exponent sequence K where K has 4 non-zero entries. Thus, we can reduce them by the previous step. Now we can

I V assume that for any generator OV(X ) of R has λ3(I) ≤ 1, which bounds the degree to 9.

V 2 2 2 Theorem 5.2.4. The ring R is generated by G = {OS4 (x1), OS4 (x1x2), OS4 (x1x2x3),

2 2 3 3 OA4 (x1x2x3), OV(x1x2), OV(x1x3), OV(x1x2), OV(x1x3)}.

I Proof. The remaining invariants, OV(X ), where the associated partition λ(I) is an element of S are shown explicitly in the appendix, and thus, the theorem is proved.

5.3 Non-Alternating Klein IV Subgroup

Let V be a Klein-4 subgroup of S4 generated by matrices with representation:

 1 0 0 0   0 1 0 0   1 0 0 0   0 1 0 0     0 1 0 0   1 0 0 0   0 1 0 0   1 0 0 0  V =   ,   ,   ,    0 0 1 0   0 0 1 0   0 0 0 1   0 0 0 1   0 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0 

38 We will denote V as V = {(1), (1, 2), (3, 4), (1, 2)(3, 4)} and note that V is generated by (1, 2) and (3, 4).

Lemma 5.3.1. R(1,2) ∩ R(3,4) = RV.

Proof. Let f ∈ R(1,2) ∩ R(3,4). Thus (1, 2)f = f and (3, 4)f = f. Hence f ∈ RV.

Conversely, let f ∈ RV, so (1, 2)f = (3, 4)f = f. So f ∈ R(1,2) ∩ R(3,4). Therefore,

R(1,2) ∩ R(3,4) = RV.

Therefore, a natural choice for generators of RV would be the generators for R(1,2) that are also invariant under (3, 4), and the generators for R(3,4) that are also invariant

3 3 under (1, 2). These generators would be (x1 +x2) = g1,(x3 +x4) = g2,(x1 +x2) = g3,

3 3 and (x3 + x4) = g4. Let G = {g1, g2, g3, g4}.

I I Lemma 5.3.2. Given an orbit sum OV(X ), if i3 = i4 and i3 is odd, then OV(X ) = 0.

2 I i1 i2 i3 i3 i3 i1 i2 i3 i3 I Proof. Consider (3, 4)X = (3, 4)x1 x2 x3 x4 = (−1) )x1 x2 x3 x4 = −X .

I I I I I I I I So, OV(X ) = X + (1, 2)X + (3, 4)X + (1, 2)(3, 4)X = X + (1, 2)X − X − (1, 2)XI = 0.

Hence, we can assume that for any arbitrary polynomial f ∈ RV has no monomial

I X where i3 = i4 and i3 is odd.

Theorem 5.3.3. The ring RV is generated as an algebra by G.

Proof. ⊇) Clearly, every polynomial generated from G is invariant under V since the generators are invariant under both (1, 2) and (3, 4).

⊆) Let f ∈ RV. We can consider rewriting R as a polynomial ring with indeterminates x3 and x4, with coeﬃcients as polynomials in terms of x1 and x2, denoted k−1[x1, x2](x3, x4). Since f ∈ k−1[x1, x2](x3, x4), f can be written as:

39 X i j f = pi,j(x1, x2)x3x4, i,j

V where pi,j(x1, x2) is some polynomial in terms of x1 and x2. Since f ∈ R , (3, 4)f = f.

X i j (3, 4)f = (3, 4) pi,j(x1, x2)x3x4 i,j

X i j = (3, 4)pi,j(x1, x2)(3, 4)x3x4 i,j

X ij j i = (−1) pi,j(x1, x2)x3x4 i,j

= f.

ij This shows pi,j(x1, x2) = (−1) pj,i(x1, x2). Now consider (1, 2) acting on f:

X i j (1, 2)f = (1, 2) pi,j(x1, x2)x3x4 i,j

X i j = (1, 2)pi,j(x1, x2)(1, 2)x3x4 i,j

X i j = (1, 2)pi,j(x1, x2)x3x4 i,j

= f.

(1,2) Thus pi,j(x1, x2) = (1, 2)pj,i(x1, x2), hence pi,j(x1, x2) ∈ R , and pi,j(x1, x2) is gen-

40 erated by g1 and g3. Now we rewrite f as:

X i j j i X i i f = pi,j(x1, x2)x3x4 + pj,i(x1, x2)x3x4 + pi,i(x1, x2)x3x4. i>j i,i

X i j ij j i X i i = pi,j(x1, x2)x3x4 + (−1) pi,j(x1, x2)x3x4 + pi,i(x1, x2)x3x4. i>j i,i

X i j ij j i X i i = pi,j(x1, x2) x3x4 + (−1) x3x4 + pi,i(x1, x2)x3x4. i>j i,i

= f.

i j ij j i Let (3, 4) act on a term (x3x4 + (−1) x3x4).

i j ij j i ij j i 2ij i j (3, 4)(x3x4 + (−1) x3x4) = ((−1) x3x4 + (−1) x3x4)

i j ij j i = (x3x4 + (−1) x3x4).

i j ij j i (3,4) i j ij j i Thus,(x3x4 + (−1) x3x4) is an element of R , and (x3x4 + (−1) x3x4) is generated

i i i2 i i i i i i by g2 and g4. Now consider (3, 4)(x3x4) = (−1) x3x4 = x3x4. Therefore, (x3x4) is in

(3,4) R , and is generated by g2 and g4. Hence f is generated by G.

41 Bibliography

[1] H. Derksen, G. Kemper. Computational Invariant Theory, volume 130 Ency- clopaedia of Mathematical Sciences. Springer, Berlin, Gemany, 2002. Invariant Theory and Algebraic Transformation Groups.

[2] Mara D. Neusel. Invariant Theory, volume 36 of Student Mathematical Library. American Mathematical Society, Providence, Rhode Island, 2006.

[3] Bernd Sturmfels. Algorithms in Invariant Theory, 2nd edition. Springer- Verlag/Wien, Germany, 2008.

42 Appendix A: 5.1

The remaining invariants that need to be explicitly generated by G for Theorem 5.1.2 are listed here.

OZ2 (x1) ∈ G

OZ2 (x3) = OA4 (x1) − OZ2 (x1). 2 2 OZ2 (x1) = OZ2 (x1) 2 2 OZ2 (x3) = OZ2 (x3)

OZ2 (x1x3) ∈ G

OZ2 (x1x4) = OZ2 (x1)OZ2 (x3) − OZ2 (x1x3) −1 OZ2 (x1x2x3) = 2 (OZ2 (x1x3)OZ2 (x1) + OZ2 (x1)OZ2 (x1x3)) 2 −1 OZ2 (x1x3) = 2 (OZ2 (x1x3)OZ2 (x1) − OZ2 (x1)OZ2 (x1x3)) 1 OZ2 (x1x3x4) = 2 (OZ2 (x1x3)OZ2 (x3) + OZ2 (x3)OZ2 (x1x3)) −1 OZ2 (x1x2x3) = 2 (OZ2 (x1x3)OZ2 (x1) + OZ2 (x1)OZ2 (x1x3)) 2 1 OZ2 (x1x3) = 2 (OZ2 (x1x3)OZ2 (x3) − OZ2 (x3)OZ2 (x1x3)) 2 OZ2 (x1x2) ∈ G 2 3 3 3 2 OZ2 (x3x4) = OZ2 (x3) − OA4 (x1) + OZ2 (x1) − O(x1x2) 2 2 1 OZ2 (x1x4) = (OZ2 (x1) OZ2 (x3)) + 2 (OZ2 (x1x3)OZ2 (x1) − OZ2 (x1)OZ2 (x1x3)) 2 2 1 OZ2 (x1x4) = (OZ2 (x1)OZ2 (x3) ) − 2 (OZ2 (x1x3)OZ2 (x3) − OZ2 (x3)OZ2 (x1x3)) 2 1 OZ2 (x1x2x3) = 2 (OZ2 (x1x2x3)OZ2 (x1) + OZ2 (x1)OZ2 (x1x2x3)) 2 −1 OZ2 (x1x2x3) = 2 (OZ2 (x1x2x3)OZ2 (x1) − OZ2 (x1)OZ2 (x1x2x3)) 2 1 2 OZ2 (x1x3x4) = 2 (OZ2 (x1x3x4)OZ2 (x3) − OZ2 ((x3x4)OZ2 (x1)) 2 1 2 OZ2 (x1x2x4) = 2 (OZ2 (x1x2x4)OZ2 (x1) + OZ2 (x1x2)OZ2 (x3)) 2 OZ2 (x1x3x4) = OZ2 (x1x3x4)OZ2 (x1) − OZ2 (x1x2x3x4) 2 OZ2 (x1x2x3) = OZ2 (x1x2x3)OZ2 (x3) − OZ2 (x1x2x3x4) 2 2 OZ2 (x1x3x4) = −OZ2 (x1x3x4)OZ2 (x3) + OZ2 (x1x2x4) 2 2 2 OZ2 (x1x3) = −OZ2 (x1x3)

43 2 2 1 4 4 OZ2 (x1x4) = 2 (OZ2 (x1) − OA4 (x1)) 2 2 1 4 4 OZ2 (x3x4) = 2 (OZ2 (x3) − OA4 (x1) 2 2 2 OZ2 (x1x4) = −OZ2 (x1x4) 3 2 2 2 OZ2 (x1x2) = OZ2 (x1)OZ2 ((x1x2) − OZ2 (x1x2) 3 2 2 2 OZ2 (x3x4) = OZ2 (x3)OZ2 (x3x4) − OZ2 (x3x4) 3 2 2 OZ2 (x1x3) = OZ2 (x1)OZ2 (x1x3) − OZ2 (x1x2x3) 3 2 2 OZ2 (x1x4) = OZ2 (x1)OZ2 (x1x4) − OZ2 (x1x2x4) 3 2 2 OZ2 (x1x3) = OZ2 (x3)OZ2 (x1x3) − OZ2 ((x1x3x4) 3 2 2 OZ2 (x1x4 = OZ2 (x4)OZ2 (x1x4) − OZ2 (x1x3x4) 3 2 2 2 OZ2 (x1x2x3) = −OZ2 (x1x2)OZ2 (x1x3) + OZ2 (x1x2)OZ2 (x3) 3 2 2 2 OZ2 (x1x2x4) = −OZ2 (x1x2)OZ2 (x1x4) + OZ2 (x1x2)OZ2 (x4) 3 2 2 2 OZ2 (x1x3x4) = OZ2 (x3x4)OZ2 (x1x3) + OZ2 (x3x4)OZ2 (x2) 3 2 2 2 OZ2 (x1x3x4 = −OZ2 (x33x4)OZ2 (x1x4) − OZ2 (x3x4)OZ2 (x1) 3 2 2 2 OZ2 (x1x3x4) = −OZ2 (x1x3)OZ2 (x1x4) − OZ2 (x1x2x3) 2 2 2 2 OZ2 (x1x2x3) = −OZ2 (x2x4)OZ2 (x1x4) − OZ2 (x2x3x4) 3 2 2 2 OZ2 (x1x2) = OZ2 (x1x2)OZ2 (x1) 3 2 2 2 OZ2 (x3x4) = OZ2 (x3x4)OZ2 (x3) 3 2 2 2 2 2 OZ2 (x1x3) = OZ2 (x1)OZ2 (x1x3) − OZ2 (x1x2x3) 3 2 2 2 2 2 OZ2 (x1x4) = OZ2 (x1)OZ2 (x1x4) − OZ2 (x1x2x4) 2 3 2 2 2 2 OZ2 (x1x3) = OZ2 (x1x3)OZ2 (x3) − OZ2 (x1x3x4) 2 3 2 2 2 2 OZ2 (x1x4) = OZ2 (x1x4)OZ2 (x4) − OZ2 (x1x2x4) 4 2 2 2 2 OZ2 (x1x2) = OZ2 (x1x2)OZ2 (x1) 4 2 2 2 2 OZ2 (x3x4) = OZ2 (x3x4)OZ2 (x3) 4 2 2 2 2 2 OZ2 (x1x3) = OZ2 (x1x3) − OZ2 (x1x2x3x4) 4 2 2 2 2 2 OZ2 (x1x4) = OZ2 (x1x4) + OZ2 (x1x2x3x4) 2 2 2 2 2 2 OZ2 (x1x4) = OZ2 (x1x3) − OZ2 (x1x2x3x4)

44 2 4 2 2 2 2 OZ2 (x1x4) = OZ2 (x1x4) − OZ2 (x1x2x3x4)

45 Appendix B: 5.2

The remaining invariants that need to be explicitly generated by G for Theorem 5.2.4 are listed here.

OV(x1) ∈ G 2 2 OV(x1) = OV(x1)

3 A4 OV(x1) ∈ R 2 OV(x1x2) ∈ G 2 OV(x1x3) ∈ G 2 2 2 2 OV(x1x4) = OA4 (x1x2) − OV(x1x2) − OV(x1x3)

OV(x1x2x3) ∈ G 4 1 3 3 OV(x1) = 2 (OV(x1)OV(x1) + OV(x1)OV(x1))

4 2 2 2 = OV(x1) − OV(x1x2)OV(x1) − OV(x1x3)OV(x1) − OV(x1x4)OV(x1) 3 OV(x1x2) ∈ G 3 OV(x1x3) ∈ G 3 3 3 3 OV(x1x4) = OA4 (x1x2) − OV(x1x2) − OV(x1x3) 2 2 1 2 3 OV(x1x2) = 4 (OV(x1x2)OV(x1) + OV(x1)OV(x1x2)) 2 2 1 2 3 OV(x1x3) = 4 (OV(x1x3)OV(x1) + OV(x1)OV(x1x3)) 2 2 1 2 3 OV(x1x4) = 4 (OV(x1x4)OV(x1) + OV(x1)OV(x1x4)) 2 OV(x1x2x3) ∈ G 2 OV(x1x2x4) ∈ G 2 2 2 2 OV(x1x3x4) = OA4 (x1x2x3) − OV(x1x2x3) − OV(x1x2x4) 3 2 OV(x1x2) ∈ G 3 2 OV(x1x3) ∈ G 3 2 3 2 3 2 3 2 OV(x1x4) = OA4 (x1x2) − OV(x1x2) − OV(x1x3)

46 2 2 2 2 3 OV(x1x2x3) = OV(x1x2)OV(x1) − OV(x1x2) 2 2 2 2 3 2 OV(x1x2x3) = OV(x1x3)OV(x1) − OV(x1x2) 2 2 2 2 3 2 OV(x1x2x4) = OV(x1x4)OV(x1) − OV(x1x4) 3 2 2 2 2 2 2 OV(x1x2x3) = OV(x1x2x3)OV(x1) + OV(x1x2x3) − OV(x1x2x3) − OV(x1x2x3x4) 3 2 2 2 2 2 2 OV(x1x2x4) = OV(x1x2x4)OV(x1) + OV(x1x2x4) + OV(x1x2x3x4) − OV(x1x2x4) 3 2 2 2 2 2 2 OV(x1x3x4) = OV(x1x3x4)OV(x1) − OV(x1x2x3x4) + OV(x1x3x4) − OV(x1x3x4) 4 2 1 3 2 3 2 OV(x1x2) = 2 (OV(x1x2)OV(x1) + OV(x1)OV(x1x2)) 4 2 1 3 2 3 2 OV(x1x3) = 2 (OV(x1x3)OV(x1) + OV(x1)OV(x1x3)) 4 2 1 3 2 3 2 OV(x1x4) = 2 (OV(x1x4)OV(x1) + OV(x1)OV(x1x4)) 3 2 1 3 2 4 2 2 3 OV(x1x2x3) = 2 (OV(x1x2)OV(x1) − OV(x1x2) − OV(x1x2x3)OV(x1x2) − OV(x1x2x3x4)) 3 2 1 3 2 4 2 2 3 OV(x1x2x4) = 2 (OV(x1x2)OV(x1) − OV(x1x2) + OV(x1x2x3)OV(x1x2) − OV(x1x2x3x4)) 3 2 1 3 2 4 2 2 3 OV(x1x2x3) = 2 (OV(x1x3)OV(x1) − OV(x1x3) − OV(x1x2x3)OV(x1x3) − OV(x1x2x3x4)) 3 2 1 3 2 4 2 2 3 OV(x1x3x4) = 2 (OV(x1x3)OV(x1) − OV(x1x3) + OV(x1x2x3)OV(x1x3) − OV(x1x2x3x4)) 3 2 1 3 2 4 2 2 3 OV(x1x2x4) = 2 (OV(x1x4)OV(x1) − OV(x1x4) − OV(x1x2x3)OV(x1x4) + OV(x1x2x3x4)) 3 2 1 3 2 4 2 2 3 OV(x1x3x4) = 2 (OV(x1x4)OV(x1) − OV(x1x4) + OV(x1x2x3)OV(x1x4) + OV(x1x2x3x4))

47 Vita

J. Cameron Atkins

Date of Birth: March 23, 1988

Place of Birth: Richmond, Virginia

Education:

• Master of Arts in Mathematics, Wake Forest University, expected May 2012.

Thesis title: The Invariant Theory of k−1[x1, x2, . . . , xn] under Permuta- tion Representations

Advisor: Ellen Kirkman

• Bachelor of Science in Mathematics, James Madison Univeristy, May 2010.

Honors and Awards:

• Pi Mu Epsilon, Wake Forest University

Experience

• Graduate Teaching Assistantship, Wake Forest University

• Research Day, Winston-Salem, NC 2012, Poster Presentation

• MAA Conference, Petersburg, VA, 2010, Poster Presentation

• SUMS Conference, Harrisonburg, VA 2009, Poster Presentation