Invariant Generators of the Symmetry Groups ofRegular n-gons and Platonic Solids

Thomas Goller

May 2011

This thesis represents my own work in accordance with University regulations.

Abstract

Let G be a finite group of invertible linear transformations of a vector space V of finitedimension n over a field K of characteristic 0. The elements of G extend to ring automor-phisms of K[V ], the ring of polynomial functions on V , by acting as change of variables.The ring of invariants K[V ]G consists of those polynomials in K[V ] that are fixed by allthe elements of G. A famous theorem states that G is generated by pseudoreflectionsif and only if K[V ]G is a polynomial ring in n homogeneous generators. We will give acomplicated elementary proof of this nice result.

Knowing that K[V ]G has n algebraically independent generators is wonderful, butfinding these generators is even better. That is precisely what we will do for the sym-metry groups of the regular n-gons in R2 and the platonic solids in R3. We will studythese symmetry groups geometrically, proving – among other things – that they aregenerated by reflections. Then we will use intuitive geometric reasoning to find sets ofinvariant polynomials, and finally we will draw on algebraic rigor to prove that these setsof polynomials generate the rings of invariants K[V ]G.

Acknowledgements

I want to thank my adviser, Professor Janos Kollar. He suggested this topic, was patientwith my slow progress, and kept me on the right track without seizing control. Best ofall, he had a nice proof for every messy argument I proposed, and an insightful suggestioneach time I got stuck.

Contents

1 Introduction 31.1 Setup and Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Major Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Summary of Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.4 Simple Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.5 Proving Polynomials Generate K[V ]G . . . . . . . . . . . . . . . . . . . . 8

1.5.1 Step 1: Check the degrees . . . . . . . . . . . . . . . . . . . . . . 81.5.2 Step 2: Check n linear independences . . . . . . . . . . . . . . . . 81.5.3 Alternate method: field extensions . . . . . . . . . . . . . . . . . 9

1.6 Summary of What is to Come . . . . . . . . . . . . . . . . . . . . . . . . 10

2 Geometric Objects of Study 122.1 Regular n-gons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.2 The Platonic Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.3 Adjacency Graphs of the Platonic Solids . . . . . . . . . . . . . . . . . . 142.4 Dual Polyhedra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3 Symmetry Groups 163.1 Isometries that Fix the Origin . . . . . . . . . . . . . . . . . . . . . . . . 163.2 Symmetry Groups of Regular n-gons . . . . . . . . . . . . . . . . . . . . 17

3.2.1 Characterization, order, and degree signature of Dn . . . . . . . . 173.2.2 Dn is a reflection group . . . . . . . . . . . . . . . . . . . . . . . . 19

3.3 Symmetry Groups of the Platonic Solids . . . . . . . . . . . . . . . . . . 193.3.1 Group order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.3.2 Number of reflections . . . . . . . . . . . . . . . . . . . . . . . . . 213.3.3 Degree signature . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.3.4 Proving the reflections generate . . . . . . . . . . . . . . . . . . . 22

3.4 Finite Orbits of Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.4.1 Action of G on linear polynomials . . . . . . . . . . . . . . . . . . 243.4.2 An invariant polynomial . . . . . . . . . . . . . . . . . . . . . . . 24

4 Invariants of n-gon Symmetry Groups 264.1 Finding Polynomial Invariants . . . . . . . . . . . . . . . . . . . . . . . . 264.2 Proving the Invariants Generate . . . . . . . . . . . . . . . . . . . . . . . 27

1

4.3 Further Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274.3.1 Another way to find a generator . . . . . . . . . . . . . . . . . . . 274.3.2 Proving generation with field extensions . . . . . . . . . . . . . . 28

5 Invariants of Platonic Solid Symmetry Groups 305.1 Half Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315.2 Invariants of the Tetrahedral Group T . . . . . . . . . . . . . . . . . . . 315.3 Invariants of the Octahedral Group O . . . . . . . . . . . . . . . . . . . . 325.4 Invariants of the Icosahedral Group I . . . . . . . . . . . . . . . . . . . . 345.5 Proving the Invariants Generate . . . . . . . . . . . . . . . . . . . . . . . 36

6 Proof of the Main Theorem 386.1 Setup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386.2 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386.3 Main Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

6.3.1 Proving (i) =! (ii) . . . . . . . . . . . . . . . . . . . . . . . . . . 416.3.2 Proving (ii) =! (iii) . . . . . . . . . . . . . . . . . . . . . . . . . 416.3.3 Proving (iii) =! (i) . . . . . . . . . . . . . . . . . . . . . . . . . . 436.3.4 Proving (ii) "! (iv) . . . . . . . . . . . . . . . . . . . . . . . . 47

2

Chapter 1

Introduction

1.1 Setup and Definitions

Let V be a vector space of finite dimension n over a field K of characteristic 0. Let K[V ]denote the ring of polynomial functions on V with coe!cients in K. Picking a dual basisx1, . . . , xn for V !, we can write

K[V ] = K[x1, . . . , xn] = K # V ! # S2(V !)# S3(V !)# · · · ,

where Sd(V !) denotes the dth symmetric power of V !, which in the chosen basis is simplythe homogeneous polynomials of degree d in the xi with coe!cients in K. This directsum exhibits the natural grading of K[V ] by polynomial degrees.

Now let V ! !! V be an invertible linear transformation of V . The action of ! onV extends to a ring automorphism of K[V ] defined as a change of variables

!(f)(v) := f(!"1(v)), f $ K[V ], v $ V.

We will abuse notation by writing ! for both the linear transformation and the ring auto-morphism K[V ] ! !! K[V ] , since the meaning should be clear from the context. We use

the inverse linear transformation !"1 in the definition in order to ensure “associativity”:

namely if K[V ]!," !! K[V ] are automorphisms defined as above, then

"(!(f)) = "(f % !"1) = f % !"1 % ""1 = f % ("!)"1 = ("!)(f).

Definition 1. Let G be a finite group of invertible linear transformations of V , and con-sider its elements as ring automorphisms of K[V ], by the definition above. An invariantof G is a polynomial f $ K[V ] such that !(f) = f for all ! $ G. We write K[V ]G forthe subring of K[V ] consisting of all invariants of G and call it the ring of invariantsof G.

Note that the constant polynomials are always invariants since they are una"ectedby a change of variables.

3

Definition 2. An invertible linear transformation V ! !! V is a pseudoreflectionif all but one of its eigenvalues are 1, namely if ! fixes a hyperplane – an (n & 1)-dimensional subspace of V – pointwise. A pseudoreflection ! is called a reflection if itis of order 2, namely its non-1 eigenvalue is &1. We say a finite group G of invertiblelinear transformations of V is a pseudoreflection group (resp. reflection group) ifit is generated by pseudoreflections (resp. reflections).

We finish the section with several remarks.

• Every pseudoreflection in a finite group G has finite order, so the non-1 eigenvalueis a root of unity. In particular, if V has K = R as the base field then everypseudoreflection in G is a reflection.

• G being a finite group of invertible linear transformations of a finite-dimensionalvector space V is the same as having a faithful representation G #' GL(V ) ofthe isomorphism class G of G. We will not use the latter notation or discussrepresentation theory.

• We assume that the field K has characteristic 0 for simplicity, since the symmetrygroups we will study act on the vector spaces R2 and R3. The discussion in thissection still makes sense in finite characteristic p, and the Main Theorem we willintroduce in the next section holds as long as p and |G| are relatively prime, thoughallowing finite characteristic makes the proof slightly more complicated (see [2], [3],and [5] for details).

1.2 Major Results

A crucial result in the study of the rings of invariants of pseudoreflection groups datesback to the 1950s work of G. C. Shephard and T. A. Todd [6] and C. Chevalley [3].Chapter 6 will be dedicated to proving the following nice theorem.

Main Theorem. Let G be a finite group of invertible linear transformations of an n-dimensional vector space V over a field K of characteristic 0. Then the following areequivalent:

(i) G is generated by pseudoreflections.

(ii) K[V ]G is a polynomial ring in n = dimV homogeneous generators.

(iii) K[V ] is a free module over K[V ]G with a basis consisting of |G| homogeneous poly-nomials.

Prior to Chapter 6, we will focus on examples of the implication (i) =! (ii); (iii)is included mainly for completeness. Our goal is to determine sets of homogeneousgenerators for the rings of invariants of certain groups of symmetries that are generated bypseudoreflections. For this it is useful to know the degrees of the n generators mentionedin (ii) of the Main Theorem, which turn out to be uniquely determined by G, up toreordering.

4

Definition 3. The degree signature of a pseudoreflection group G is the orderedn-tuple (d1, . . . , dn) of degrees of any set of n algebraically independent homogeneousgenerators of K[V ]G, where d1 ( · · · ( dn. It is well-defined, namely every set ofhomogeneous generators has these degrees, by the equivalence of (i) and (ii) in Proposition13 (c) of Chapter 6.

We now present two identities that are useful in determining the degree signature andalso give a concrete connection to pseudoreflections. The proof is given in Proposition12 in Chapter 6 .

Proposition. Let G be a finite group of invertible linear transformations generated bypseudoreflections, with degree signature (d1, . . . , dn). Then

|G| =n!

i=1

di and |s(G)| =n"

i=1

(di & 1),

where s(G) is the set of pseudereflections of G.

Before returning to a discussion of the invariant generators, we first summarize theprevious definitions and then give examples to illustrate the results presented in thissection.

1.3 Summary of Notation

In the rest of this chapter and in the following chapters, we will assume the followingunless otherwise specified:

• V an n-dimensional vector space over a field K of characteristic 0.

• G a finite group of invertible linear transformations of V generated by pseudore-flections.

• s(G) the set of all pseudoreflections in G.

• K[V ] = K[x1, . . . , xn] the ring of polynomial functions on V , where {x1, . . . , xn} isa basis for V !.

• K[V ]G the ring of invariants, generated as a polynomial ring by n algebraicallyindependent invariants f1, . . . , fn, where deg fi = di and (d1, . . . , dn) is the degreesignature of G.

5

1.4 Simple Examples

Example 1 (Trivial group). For any finite-dimensional vector space V of dimension n,let G be the pseudoreflection group generated by the empty set of pseudoreflections,namely the trivial group consisting of just the identity linear transformation on V . ThenK[V ]G = K[V ] is generated by the n algebraically independent elements x1, . . . , xn, the

dual basis for V !. The degree signature is (

n# $% &1, . . . , 1 ), and we confirm that |G| = 1 ='n

i=1 1 and |s(G)| = 0 =(n

i=1(1& 1). The basis for K[V ] as a free module over K[V ]G

is of course just {1}.

Example 2. Let V = R and G be the group generated by the reflection R ! !! Rdefined by !(v) := &v for all v $ R, which fixes the origin hyperplane. Since ! has order2, G = {id, !}. Since ! is a ring automorphism of R[V ] = R[x1],

!(xd1) = !(x1)

d = (&x1)d = (&1)dxd

1,

so xd1 is an invariant if and only if d is even. Thus x2

1 generates the ring of invariants,namely R[V ]G = R[x2

1], so the degree signature is (2). Note that |G| = 2 and |s(G)| =2& 1. A homogeneous basis for R[V ] as a free R[V ]G-module is {1, x1}.Example 3. In Example 2, we can change the field R to C without a"ecting the rest ofthe discussion. Note that the linear transformation C ! !! C defined by !(z) := &z forall z $ C is still a reflection since the origin is still a hyperplane in C. But now we canvisualize C as a plane, on which ! acts as rotation by $ radians:

Re

Im

z = a+ bi

!(z) = &a& bi

The next example will show why this is interesting.

Example 4 (“Non-example”). We now give an example where G is a finite group oflinear transformations of V that is not a pseudoreflection group, and where K[V ]G is nota polynomial ring. Take V = R2 with the standard basis {e1, e2} and let G be generatedby the rotation r by $ radians about the origin, namely r(e1) = &e1 and r(e2) = &e2.Letting x denote the e1-coordinate and y denote the e2-coordinate, the action of r in thexy-plane looks just like the action of the pseudoreflection ! of Example 3 in the complex“plane” C:

x

y

v = (a, b)

r(v) = (&a,&b)

6

But r is not a pseudoreflection since both its eigenvalues are &1; indeed, its only fixedpoint is the origin, which is not a hyperplane since the vector space is two-dimensional,whereas in Example 3 the complex “plane” C was a one-dimensional vector space. SoG = {1, r} is not generated by pseudoreflections.

To determine the ring of invariants, we compute

r(xd11 xd2

2 ) = (&x1)d1(&x2)

d2 = (&1)d1+d2xd11 xd2

2 ,

so that the monomial xd11 xd2

2 is invariant if and only if d1 + d2 is even. Thus R[V ]G isgenerated by x2

1, x1x2, and x22, which are not algebraically independent since (x2

1)(x22) =

(x1x2)2, whence R[V ]G = R[x21, x1x2, x2

2] is not a polynomial ring. R[V ] is not a freeR[V ]G-module; for instance {1, x1, x2} is not a basis since (x1x2) · x1 = (x2

1) · x2.

Example 5 (Symmetric groups). Let V be Kn with basis {e1, . . . , en}. Let G be the

group generated by reflections Kn (ij)!! Kn defined for 1 ( i < j ( n by (ij)(ei) := ej,

(ij)(ej) := ei, and (ij)(ek) := ek for k )= i, j. If we heed the suggestive notation andthink of these reflections as transpositions, then it is clear that G is isomorphic to thesymmetric group Sn on n letters, which is generated by its transpositions. The invariantsof K[V ] are those polynomials “symmetric” in the xi, and it is a well-known fact (see547-551 of [1]) that all such polynomials are generated by the n algebraically independentelementary symmetric polynomials

Pr :="

1#i1<···<ir#n

xi1 · · · xir

for 1 ( r ( n. Each Pr is of degree r, so the degree signature is (1, 2, . . . , n). We confirmthat |G| = n! =

'nr=1 r and that |s(G)| =

)n2

*=

(nr=1(r & 1). We refrain from trying to

write down a basis for K[V ] as a free K[V ]G-module.

Example 6 (Some pseudoreflection groups that are not reflection groups). Let V = C, letr * 2 be an integer, and let G be the group generated by the pseudoreflection C ! !! Cdefined by !(z) := e2#i/r · z for each z $ C, which fixes the hyperplane z = 0. We alreadysaw the case r = 2 in Example 3, but for r * 3 the eigenvalue e2#i/r, a primitive rthroot of unity, is non-real, whence ! is a pseudoreflection that is not a reflection. We haveG = {id, !, !2, . . . , !r"1} and all the non-identity elements are pseudoreflections sinceeach one fixes only 0. If we visualize C as a plane, then ! corresponds to rotation by 2#

rradians. For instance, when r = 4:

Re

Im

z!(z) = e#i/2z

7

Since ! generates G, a polynomial f $ K[V ] is invariant if and only if !(f) = f . Recallingthat !(f) = f % !"1, we compute

!(xd1) = (e"2#i/rx1)

d = e"2d#i/rxd1,

so that xd1 is an invariant if and only if d + 0 (mod r). Thus xr

1 generates the ring ofinvariants, namely K[V ]G = K[xr

1], so the degree signature is (r). Note that |G| = r and|s(G)| = r&1. A homogeneous basis for K[V ] as a K[V ]G-module is {1, x1, x2

1, . . . , xr"11 }.

1.5 Proving Polynomials Generate K[V ]G

In the following chapters we will study the rings of invariants of more complicated reflec-tion groups. For each such group G, our two main goals will be to

(i) Determine a set of n homogeneous invariant polynomials g1, . . . , gn for K[V ]G withno obvious dependence relations;

(ii) Prove g1, . . . , gn generate K[V ]G.

Of course, we can only do (ii) if we chose polynomials in (i) that really are generators. Wewill see in later chapters that geometric features of the symmetry groups we are studyingwill yield sets of polynomials that will turn out to be generators. For now, we describeour approach to (ii): an easy method for proving that the polynomials generate K[V ]G.

1.5.1 Step 1: Check the degrees

Our first test, after making sure there are no obvious dependence relations among thehomogeneous invariants g1, . . . , gn, is to check whether the g1, . . . , gn have degrees match-ing the degree signature. But first we need to know the degree signature! If we know |G|and |s(G)|, which we will compute geometrically, as well as n& 2 of the dis in the degreesignature, then the system of two equations

|G| =n!

i=1

di, |s(G)| =n"

i=1

(di & 1)

allows us to solve for the remaining two unknowns. We will be restricting our attentionto groups acting on vector spaces of dimensions two and three, so the assumption thatwe know n& 2 of the dis will not be as demanding as it may initially sound.

1.5.2 Step 2: Check n linear independences

If our n homogeneous invariants g1, . . . , gn have degrees matching the degree signature,then showing they are algebraically independent is enough to prove that they generateK[V ]G (Prop. 13 (c)). Unfortunately, it seems di!cult to prove algebraic independence:in principle, we have to rule out dependence relations in every positive degree. Naively,

8

this requires checking an infinite number of cases. Fortunately, this brute force method ofruling out dependence relations is actually quite useful. Instead of focusing on algebraicindependence, we can directly show g1, . . . , gn are generators using the fact that theirdegrees match the degree signature, as long as the gi have no dependence relations inany of the degrees of the degree signature. Since d1 ( · · · ( dn, another way of sayingthis is that we just need to check gi /$ K[g1, . . . , gi"1] for each 1 ( i ( n, a finite numberof cases. We now formally state and prove this claim.

Proposition 1. Let (d1, . . . , dn) be the degree signature of G. If g1, . . . , gn are homoge-neous invariants such that deg gi = di and gi /$ K[gi, . . . , gi"1] for each 1 ( i ( n, thenthe gi are a set of homogeneous generators for K[V ]G.

Proof. By definition of degree signature we can choose homogeneous invariant genera-tors f1, . . . , fn for K[V ]G such that deg fi = di for each i. Since the gi are invariants,K[g1, . . . , gn] , K[V ]G = K[f1, . . . , fn]. Now we use the following lemma to deduce thatK[g1, . . . , gn] = K[f1, . . . , fn].

For the following lemma, ignore the notation of Section 1.3.

Lemma 1. Suppose g1, . . . , gn are homogeneous polynomials of degrees d1 ( · · · ( dn,such that gi /$ K[g1, . . . , gi"1] for each i. Suppose f1, . . . , fn are homogeneous polynomialswith deg fi = di for each i, such that K[g1, . . . , gn] , K[f1, . . . , fn]. Then K[g1, . . . , gn] =K[f1, . . . , fn].

Proof. Induction on n. If n = 0 the result is trivial. So suppose n > 0 and let r < nbe the maximum integer such that dr < dn, where we set r = 0 if d1 = · · · = dn.By the inductive assumption, K[g1, . . . , gr] = K[f1, . . . , fr]. By choice of r, we havedr+1 = · · · = dn > dr. Since each gi $ K[f1, . . . , fn], we can choose Pi $ K[f1, . . . , fr]and ai,j $ K for each 1 ( j ( n& r such that

gr+i = ai,1fr+1 + · · ·+ ai,n"rfn + Pi

for each 1 ( i ( n & r. Since gi /$ K[g1, . . . , gi"1] for each i and K[g1, . . . , gr] =K[f1, . . . , fr], the (n & r) (n & r)-tuples (ai,1, . . . , ai,n"r) for 1 ( i ( n & r are linearlyindependent as vectors in the K-vector space Kn"r. But this means that we can solvefor the fr+i in terms of the gr+i, namely there exist Qi $ K[f1, . . . , fr] and bi,j $ K suchthat

fr+i = bi,1gr+1 + · · ·+ bi,n"rgn +Qi

for each 1 ( i ( n & r. Thus fr+i $ K[g1, . . . , gn] for each 1 ( i ( n & r, whenceK[g1, . . . , gn] = K[f1, . . . , fn].

1.5.3 Alternate method: field extensions

There are also methods for proving algebraic independence or generation directly, forinstance using field extensions. We will only give a sketch of the ideas involved. Ex-tending the action of G to automorphisms of the field of fractions K(V ) of K[V ] as in

9

Section 6.2, we obtain a Galois (hence finite degree |G|) extension K(V )/K(V )G, whereK(V )G denotes the G-fixed subfield (Prop. 5). Moreover, K(V )G is the same as the fieldof fractions of the ring of invariants K[V ]G (Prop. 5), so that if f1, . . . , fn are homoge-neous generators of K[V ]G, then the fi also generate K(V )G as a field extension over K,whence K(x1, . . . , xn)/K(f1, . . . , fn) is a degree |G| extension. Now if g1, . . . , gn are n ho-mogeneous invariants, then by the additivity of both regular degrees and transcendencedegrees of the extensions

K(V ) = K(x1, . . . , xn)

degree |G|

tr. deg. n

K(V )G = K(f1, . . . , fn)

K(g1, . . . , gn)

K

we see that

(i) g1, . . . , gn are algebraically independent if and only if K(x1, . . . , xn)/K(g1, . . . , gn)is algebraic (i.e. has transcendence degree 0);

(ii) g1, . . . , gn generate K[V ]G if and only if K(x1, . . . , xn)/K(g1, . . . , gn) is finite ofdegree |G|.

When n is small and the gi are relatively simple, it is not too di!cult to computethe degree of the field extension K(x1, . . . , xn)/K(g1, . . . , gn) by splitting it up into nextensions in which the xi are added one at a time. We will give an example of thismethod in 4.3.2, but otherwise we will use the easier method of 1.5.2.

1.6 Summary of What is to Come

In this chapter, we introduced the basic notation and the main results pertaining tothe ring of invariants, which we will prove in Chapter 6. We also described the methodwe will use in Chapters 4 and 5 to prove that sets of polynomials generate the rings ofinvariants of certain symmetry groups. In the coming chapters, we will do the following:

• In Chapter 2 we introduce the geometric objects – regular n-gons and the platonicsolids – that we will be studying.

• In Chapter 3 we describe key properties of the symmetry groups of the regularn-gons and the platonic solids.

• In Chapter 4 we find generators for the rings of invariants of the symmetry groupsof regular n-gons.

10

• In Chapter 5 we find generators for the rings of invariants of the symmetry groupsof the platonic solids.

• In Chapter 6 we give a detailed proof of the Main Theorem.

11

Chapter 2

Geometric Objects of Study

This chapter is a brief introduction to the regular n-gons and the platonic solids. We willneed a basic geometric understanding in order to study the symmetry groups of thesegeometric objects in Chapter 3.

2.1 Regular n-gons

For n * 3, we consider the regular n-gon in the xy-plane, centered at the origin, withvertices on the unit circle at

)cos 2k#

n , sin 2k#n

*for each integer 0 ( k ( n & 1. Some

examples are given in the diagram below.

x

y

n = 3

x

y

n = 4

x

y

n = 5

x

y

n = 6

x

y

n = 7

x

y

n = 8

The First Six Regular n-gons

For now, we note only that regularity of the n-gon ensures that the n-gon is symmetricwith respect to each vertex and each edge. Intuitively, the regular n-gon “looks the same”from the perspective of each vertex, as well as from the perspective of each edge.

12

2.2 The Platonic Solids

A platonic solid is a convex polyhedron with all faces congruent regular polygons, suchthat the number of faces meeting at a vertex is the same for each vertex. One canprove by elementary geometry (see Coxeter 5, [4]) that there are only five platonic solids,namely:

Tetrahedron Cube Octahedron

Dodecahedron Icosahedron

The Platonic Solids

Like the regular n-gons, the platonic solids have the useful regularity property that theyare symmetric with respect to their vertices, edges, and faces.

Some basic characteristics of the platonic solids are:

Platonic Solid Faces Vertices Edges Face Shape

tetrahedron 4 4 6 triangle

cube 6 8 12 square

octahedron 8 6 12 triangle

dodecahedron 12 20 30 pentagon

icosahedron 20 12 30 triangle

The drawings above show the platonic solids centered at the origin in the framex

y

z

for R3, with the following coordinates for the vertices (Coxeter 52-53, [4]), where % =$5+12

is the golden ratio:

13

Platonic Solid Vertex Coordinates

tetrahedron (1, 1, 1), (&1,&1, 1), (1,&1,&1), (&1, 1,&1)

cube (±1,±1,±1)

octahedron (±1, 0, 0), (0,±1, 0), (0, 0,±1)

dodecahedron (±1,±1,±1), (0,± 1$ ,±%), (± 1

$ ,±%, 0), (±%, 0,± 1$)

icosehedron (0,±1,±%), (±1,±%, 0), (±%, 0,±1)

2.3 Adjacency Graphs of the Platonic Solids

If we are mainly interested in studying the arrangement of faces, vertices, and edges,then we can distort the faces of a solid to draw an adjacency graph. We can choose aface or a vertex as the central point of the graph; we will see later why both versions ofthe graph are useful.

Tetrahedron Cube Octahedron

Dodecahedron Icosahedron

Adjacency Graphs of the Platonic Solids

The straight lines in an adjacency graph correspond to the edges of the solid, withvertices at the intersections. The faces of the solid correspond to the regions boundedby lines in the adjacency graph. The outer circle denotes a single vertex if it intersectsstraight lines, while otherwise it indicates that there is a face containing all the verticesand edges on the border of the subgraph contained within the circle. For instance, theouter circle in the first tetrahedron graph represents a vertex with three incident edges,while the outer circle in the first cube graph indicates an additional face containing thefour vertices and four edges on the boundary of the inner subgraph, like the lid of a box.

One of the useful features of an adjacency graph drawn with radial symmetry, likeall the graphs above, is that a straight line through the center of the face-centered (resp.vertex-centered) graph corresponds to the intersection of the platonic solid with a planethrough the origin that intersects the center of a face (resp. a vertex).

14

2.4 Dual Polyhedra

Given any polyhedron, we can form its dual polyhedron by switching the roles of facesand vertices. The octahedron is the dual of the cube: if we place a vertex in the center ofeach face of the cube and add an edge between a pair of vertices when the correspondingfaces of the cube share an edge, we get the octahedron. The dual of a dual is the originalpolyhedron up to scaling – for instance, in the diagram below the inner cube is smallerthan the outer cube by a factor of 3 – so the cube is also the dual of the octahedron.

Duality of the Cube and Octahedron

Likewise, the dodecahedron and the icosahedron are dual, while the tetrahedron is dual toitself. Duality is important in our discussion because, as we will see in the next chapter,dual platonic solids share a group of symmetries.

15

Chapter 3

Symmetry Groups

We wish to study the groups of symmetries of the regular n-gons in R2 and the platonicsolids in R3. A symmetry of a geometric object drawn in Rn is an isometry of Rn thatpreserves the object as a whole, not necessarily pointwise. The collection of all symmetriesof an object forms a group called the group of symmetries or the symmetry groupof that object. For instance, as we will see, the group of symmetries of a regular n-gonin R2 consists of n rotations and n rotations composed with a reflection.

3.1 Isometries that Fix the Origin

An isometry or rigid motion Rn M !! Rn of the vector space Rn is a distance pre-serving map, namely

-Mv &Mw- = -v & w- for every v $ Rn.

In R2, every isometry is composed of a translation, a rotation about a point, and possiblya reflection about a line (Artin 157-158, [1]). Intuitively, if we imagine the plane as aninfinite sheet of paper lying on an infinitely large table, isometries correspond to slidingthe paper around, rotating it about any point, flipping it over, or any combination ofthese actions. Similarly, isometries of R3 are composed of translations, rotations aboutlines, and reflections across planes.

Isometries need not be linear transformations; consider for instance translation by anonzero vector. However, there is an equivalence (Artin 127, [1]) between

• isometries of Rn that fix the origin and

• orthogonal linear transformations of Rn,

where a linear transformation Rn T !! Rn is orthogonal if T"1 = T t. Since the regularn-gons and platonic solids are centered at the origin, the isometries that fix these geomet-ric objects must fix the origin, hence every symmetry we will be considering is an orthog-onal linear transformation. The fact that the symmetries are linear transformations is

16

crucial since the Main Theorem requires that G be a group of linear transformations. Wewill use the fact that every symmetry is orthogonal in 3.4.1, when we discuss a methodfor finding invariants of a group of symmetries.

3.2 Symmetry Groups of Regular n-gons

Let n * 3 and let Dn denote the symmetry group of the regular n-gon in the xy-planewith vertices on the unit circle at

)cos 2k#

n , sin 2k#n

*for each integer k $ {0, 1, . . . , n& 1},

as in Section 2.1. Namely, Dn consists of those isometries of R2 that fix the regular n-gonas a whole. Our choice of notation suggests a connection to the dihedral groups, andindeed, we will show that Dn, the symmetry group of the regular n-gon, is isomorphic tothe dihedral group of order 2n. The key facts we will prove are:

Group Order Number of Reflections Degree Signature

Dn 2n n (2, n)

3.2.1 Characterization, order, and degree signature of Dn

Let ! be rotation about the origin by 2#n radians, so that !k is rotation about the origin

by 2k#n radians, and let " be reflection across the x-axis (the hyperplane y = 0). Then "

and the powers of ! are isometries that preserve the regular n-gon, namely symmetriesof the n-gon. Let vk denote the vertex with coordinates

)cos 2k#

n , sin 2k#n

*. We have

already noted that the isometries of R2 that fix the regular n-gon must fix the origin.Moreover, the isometries fixing the n-gon are precisely those that map every vertex ofthe n-gon to another vertex. With this in mind, we focus attention on the coordinatesof one vertex, say (1, 0) for v0. Any isometry !% of R2 that fixes the n-gon must move v0to some vk. Thus the composition !"k % !% fixes (1, 0) as well as the origin, hence fixesthe x-axis pointwise, so it must be either the identity or reflection across the x-axis " .So !"k % !% = " l for l $ {0, 1}, which implies !% = !k % " l. Thus the symmetries of theregular n-gon are the elements

Dn =+!k" l : with k $ {0, 1, . . . , n& 1} and l $ {0, 1}

,.

We can confirm that these elements are all unique by writing ! and " as matrices. Inthe standard basis, they are

! =

-

.cos 2#n & sin 2#

n

sin 2#n cos 2#

n

/

0 , " =

-

.1 0

0 &1

/

0 .

Now by inspection, we see that the matrices

!k" l =

-

.cos 2k#n (&1)l+1 sin 2k#

n

sin 2k#n (&1)l cos 2k#

n

/

0

17

are distinct for integers k $ {0, 1, . . . , n& 1}, l $ {0, 1}.Thus we have shown that Dn has order 2n and is generated by !, rotation about the

origin by 2#n radians, and " , reflection across the x-axis. We can check (algebraically or

geometrically) that "! = !n"1" , so that we can also write

Dn = .!, " : !n = " 2 = 1, "! = !n"1"/.

Thus, as our notation suggests, the group of symmetries of the regular n-gon is isomorphicto the dihedral group of order 2n.

When n = 3, the elements of D3 are:

id ! !2

" !" !2"

D3, the Symmetry Group of the Equilateral Triangle

For each symmetry, we have circled the vertex that is the image of (1, 0); note that itis determined by the power of !, as in the discussion above. Rotations are denoted bycurving dotted lines arranged in a circle, and reflections by a dotted line representingthe fixed hyperplane, together with dotted arcs indicating which vertices are being in-terchanged (the remaining vertex is fixed). We have also labeled each element by itsexpression in the generators !, " . Note in particular that each element in the bottom rowis a reflection. Indeed, as we show in the next section, it is true for general n that eachof the elements !k" of Dn is a reflection. Since there are n elements of the form !k" , wethen know |Dn| = 2n, |s(Dn)| = n, and 2& 2 = 0 of the degrees in the degree signature,so that as discussed in 1.5.1, we can solve the system of equations

2n = |Dn| = d1d2, n = |s(Dn)| = d1 + d2 & 2

to deduce that d1 = 2 and d2 = n, whence the degree signature is (2, n).

18

3.2.2 Dn is a reflection group

The first thing to note is that there are n lines dividing a regular n-gon into mirrorimages. When n is odd, each of these lines intersects a vertex and the midpoint of anedge. When n is even, half the lines intersect two vertices, while the other half intersectthe midpoints of two edges. For n $ {3, 4, 5, 6}, these lines are

Reflection Hyperplanes of Some Regular n-gons

Since there are n possible reflection hyperplanes, Dn has exactly n reflections, namely|s(Dn)| = n. Since a reflection is orientation-reversing, meaning (algebraically) that ithas negative determinant and (geometrically) that it changes the clockwise order of thevertices of the regular n-gon, we see that the n reflections must coincide with the nelements !k" , which also have negative determinant and change the clockwise order ofthe vertices. Since half the elements of Dn are reflections and the identity id is not areflection, the reflections generate Dn by the following simple lemma.

Lemma 2. Let G be a finite group and S a subset of G such that |S| * |G|2 and id /$ S.

Then the elements of S generate G.

Proof. Let H be the subgroup of G generated by S. Then id $ H, so |H| > |S| * |G|2 .

But the order of a subgroup must divide the order of the group, whence |H| = |G|.

3.3 Symmetry Groups of the Platonic Solids

We now shift our focus to the platonic solids in R3, with vertices at the points given inSection 2.2. As with the two-dimensional symmetries of the regular n-gons, the isometriesthat fix a platonic solid are precisely those that fix its set of vertices, or, equivalently,its set of centers of faces. Since taking the dual of a platonic solid interchanges thevertex set with the set of centers of faces (more precisely, one of the sets will di"er by aconstant multiple), we see that the dual pairs of platonic solids have the same groups ofsymmetries. Thus the five platonic solids have only three distinct groups of symmetries:

• the tetrahedral group T of symmetries of the tetrahedron,

• the octahedral group O of symmetries of the cube and octahedron, and

• the icosahedral group I of symmetries of the dodecahedron and icosahedron.

In the following sections, we will confirm these basic properties:

19

Group Order Number of Reflections Degree Signature

T 24 6 (2,3,4)

O 48 9 (2,4,6)

I 120 15 (2,6,10)

Then we will show that the groups T , O, and I are generated by their reflections.

3.3.1 Group order

We proceed in the spirit of the two-dimensional case, where we characterized each sym-metry of the regular n-gon by the image of the “base position” (1, 0) and whether toreflect across the hyperplane y = 0. But instead of starting from scratch, we can buildon the two-dimensional result.

Let G denote the symmetry group of a platonic solid. Choose a face of the platonicsolid, calling its center the “base position”, and letting # denote the plane in R3 con-taining that face. Let !% $ G be a symmetry fixing the base position. Since !% also fixesthe origin, it in fact fixes the line through the origin and the base position pointwise.This line is perpendicular to the plane #, so !% is completely determined by how it actsin the plane #. But the intersection of # with the platonic solid is a regular n-gon, so !%

corresponds to a dihedral symmetry in the plane #! Explicitly, this correspondence is

dihedral rotationin #

"" !!rotation in R3 about the line containingthe origin and the base position

dihedral reflectionin # across a line &

"" !! reflection in R3 across the planecontaining the origin and the line &

Now, for each face f of the platonic solid, fix a symmetry !f $ G that moves thecenter of the face f to the base position. Such symmetries exist since there are rotationsthat move the center of a face to the center of any other face. If ! $ G is a symmetry ofthe platonic solid, then it must move the base position to the center of some face f , so!f % ! fixes the base position. Thus !% := !f % ! corresponds to a dihedral symmetry inthe plane #, so the symmetries of the platonic solid are

G =+!"1f !% : !% corresponds to a dihedral symmetry in #

,.

These elements are distinct by construction, so we see that the order of G is the numberof faces of the platonic solid times the order of Dn, where n is the number of verticeson each face of the platonic solid, namely each face is a regular n-gon. Recalling that|Dn| = 2n, we get the formula

|G| = (# faces) 0 2 0 (# vertices on each face),

which gives the group orders listed in the table.

20

3.3.2 Number of reflections

When G is a group of symmetries of a platonic solid, the reflections of G correspond toplanes through the origin that divide the solid into mirror images. We will refer to suchplanes as reflection planes. It is geometrically clear that a reflection plane must do atleast one of the following two things:

(a) contain an edge of the solid, or

(b) split a face of the solid into mirror images;

and that every plane through the origin that does (a) or (b) corresponds to a reflection.Choosing an edge to be contained or a line splitting a face into mirror images fullydetermines the reflection plane since the reflection plane must also pass through theorigin. So to obtain |s(G)| it su!ces to count the number of reflection planes thatcontain an edge, and add the number of reflection planes not yet counted that split aface into mirror images.

For the tetrahedron, every reflection plane that splits a face also contains an edge,and every reflection plane containing an edge contains exactly one edge. We can visualizethese two facts by looking at straight lines on the adjacency graphs of the tetrahedron:

Thus the number of reflections is equal to the number of edges of the tetrahedron, namely|s(T )| = 6.

For the cube, every reflection plane that contains an edge contains exactly one pair ofopposite edges, and every reflection plane that splits a face diagonally contains an edge.There are also reflection planes that split four faces into rectangles and contain no edges.

Since the cube has eight edges, there are four reflection planes containing edges. Eachof the reflection planes not containing an edge splits four faces into rectangles, and thereare two ways to split each face into a rectangle, for a total of 6·2

4 = 3 planes. Thus|s(O)| = 6 + 3 = 9.

21

For the dodecahedron, each reflection plane that splits a face contains two edges, andevery reflection plane that splits an edge contains exactly two edges:

Since the dodecahedron has 30 edges, |s(I)| = 302 = 15.

3.3.3 Degree signature

To compute the degree signature (d1, d2, d3) when G is the symmetry group of a platonicsolid, we again use

|G| = d1d2d3, |s(G)| = d1 + d2 + d3 & 3,

but in order to solve the system of two equations we need to know one of the di. Weclaim that d1 = 2 for each of the platonic solid symmetry groups. To see this, we notefirst that K[V ]G contains no linear polynomials since the only point of V fixed by G isthe origin (we explain why this is true in 3.4.1). There is, however, always a degree 2invariant, namely the squared distance from the origin x2 + y2 + z2. Since K[V ]G cancontain a degree 2 homogeneous polynomial only if one of its generators has degree atmost 2, we see that d1 = 2, which allows us to solve for d2 and d3. If G = T , then wehave the equations d2d3 = 24

2 = 12 and d2 + d3 = 6 + 3 & 2 = 7, whence d2 = 3 andd3 = 4. If G = O, then d2d3 =

482 = 24 and d2 + d3 = 9+ 3& 2 = 10, whence d2 = 4 and

d3 = 6. And if G = I then d2d3 =1202 = 60 and d2+ d3 = 15+3& 2 = 16, whence d2 = 6

and d3 = 10.

3.3.4 Proving the reflections generate

We have not yet proven the crucial fact that the reflections in T , O, and I generate thefull symmetry groups. This is less obvious than for the symmetries of the regular n-gon,since for the platonic solid symmetry groups the number of reflections is much less thanhalf the order of the group. However, the proofs that T , O, and I are reflection groupsare still easy if we use the method for enumerating the symmetries of a platonic solidfrom 3.3.1.

There we chose the center of a face of the platonic solid to be the “base position” andlet # denote the plane of R3 containing that face. For each face f , we fixed a symmetry!f $ G that moves the center of f to the base position. Then we showed that thesymmetries of the platonic solid are precisely the elements

G =+!"1f !% : !% corresponds to a dihedral symmetry in #

,.

22

Now, as mentioned in 3.3.1, the symmetries corresponding to the dihedral reflectionsin the plane # are reflections in G. Since the reflections of Dn generate Dn, the reflectionsin G generate all the symmetries !% corresponding to the dihedral symmetries in #. Thusto show that G is generated by reflections, it su!ces to show that we can choose !f foreach face f such that the !"1

f are generated by reflections. Equivalently, it su!ces toprove the existence of !f that are themselves generated by reflections, which is the contentof the following lemma.

Lemma 3. For any two faces f, f % of a platonic solid, there is a symmetry generated byreflections that moves the center of f to the center of f %.

The idea of the proof is simple: if f and f % are adjacent, then the reflection across theplane through their common edge moves the center of f to the center of f %. If f and f %

are not adjacent, then a finite sequence of such reflections will do the trick. Making thisidea rigorous takes slightly more work.

Proof. We define a “face path” on the platonic solid to be a sequence of faces {f0, f1, . . . , fr},such that fi is adjacent to fi+1 for each 0 ( i ( r & 1. We say that the face path startsat f0, ends at fr, and has length r. We then define the distance d(f, f %) between twofaces to be the minimum length among all face paths starting at f and ending at f %. Acrucial fact is that d(f, f %) is finite.

Now we can prove the result by induction on d(f, f %). If d(f, f %) = 0, then f andf % are the same face and the identity element is a symmetry generated by reflectionsthat moves the center of f to the center of f %. So assume r := d(f, f %) > 0. Let{f = f0, f1, . . . , fr"1, fr = f %} be a face path of minimum length starting at f and endingat f %. Then d(f, fr"1) = r & 1, so by the inductive assumption there is a symmetry !generated by reflections that moves the center of f to the center of fr"1. But fr"1 andfr = f % share an edge, and there is a reflection hyperplane containing that edge (see3.3.2), so there is a reflection " moving the center of fr"1 to the center of fr. Thus" % ! is a symmetry that moves the center of f to the center of f % and is generated byreflections.

3.4 Finite Orbits of Points

We conclude the chapter by introducing the method we will use to find the invariantgenerators for the groups of symmetries of the regular n-gons and the platonic solids.

Given a geometric object in Rn with a finite group of symmetries G, we can look atthe orbit $ of a point of Rn under the action of G. The order of $ is at most |G|, whichin particular means that |$| is finite, so we call $ a finite orbit of a point. Moreover,each element of G permutes the elements of $, and we will be able to use this informationto extract a homogeneous polynomial invariant of degree |$|.

We will assume that all symmetries of the geometric object fix the origin, as in thecases of the regular n-gons and the platonic solids. Thus each symmetry is orthogonal by

23

the discussion in Section 3.1. Moreover, the origin is a fixed point of G, but it is trivialand we will exclude it when talking about finite orbits of points.

We pause to give a few relevant examples of finite orbits of points. For regular n-gons,the set of n vertices and the set of n midpoints of edges are finite orbits of points. Forthe platonic solids, the sets of vertices, midpoints of edges, and centers of faces are eachfinite orbits of points.

3.4.1 Action of G on linear polynomials

Let’s explicitly write down how G acts on linear polynomials. We will consider the casewhen V = R2 for notational convenience, though the results extend automatically to Rn.Let x, y be the standard dual basis of V !, so that we write R[V ] = R[x, y] for the ringof polynomials on V . We first consider homogeneous linear polynomials 'x + (y, for

', ( $ R not both zero, which are linear functionals on V . If

1ab

2$ V , then

('x+ (y)

1ab

2= 'a+ (b,

so we will write linear polynomials as row vectors)' (

*and use normal matrix multi-

plication to represent evaluation at a vector in V . The point is that if ! $ G is given asa matrix in the basis for V , then we can write

)!)' (

**1ab

2:=

)' (

*1!"1

1ab

22=

1!

1'(

22t 1ab

2,

where the second equality follows from the associativity of matrix multiplication and the

orthogonality of !. By looking at the bijection)' (

*1

1'(

2between homogeneous

linear polynomials in R[V ] and nonzero vectors in V , we see that ! acts on linear polyno-

mials in the same way it acts on points, namely !)' (

*1 !

1'(

2. Thus finite orbits of

nonzero points correspond to finite orbits of linear polynomials. In particular, invariantlinear polynomials are in bijection with nonzero fixed points of V , a fact we used in 3.3.3.

3.4.2 An invariant polynomial

Let $ =

31a1b1

2, . . . ,

1arbr

24be a finite orbit of a point. Consider the polynomials

P!,i := aix+ biy for 1 ( i ( r. Since elements of G permute the elements of $, they alsopermute the polynomials P!,i. But the elements of G are ring automorphisms of R[V ],which in particular ensures they are multiplicative, so they permute the factors of thepolynomial

P! :=r!

i=1

P!,i =r!

i=1

(aix+ biy).

24

Thus P! is an invariant polynomial, and it is homogeneous of degree |$|.Note that although it is tempting to try to obtain a degree 1 homogeneous polynomial

by taking the sum of the P!,i, the sum tends to vanish, while the product cannot vanishsince the points in $ are nonzero.

25

Chapter 4

Invariants of n-gon SymmetryGroups

In this chapter we let V = R2 and G = Dn, the group of symmetries of the regularn-gon in the plane. We have shown that Dn is generated by its reflections, so by theMain Theorem the ring of invariants K[V ]G is a polynomial ring, with two homogeneousgenerators of degrees 2 and n. We will find two homogeneous invariant polynomials ofthese degrees, and then prove they generate the ring of invariants K[V ]G.

4.1 Finding Polynomial Invariants

Recall that the degree signature of Dn is (2, n). As in the case of the symmetry groupsof the platonic solids, there is an automatic degree 2 invariant, namely the distance fromthe origin squared g1 := x2 + y2.

Next we use the finite orbit of a point method to find a homogeneous invariant of de-gree n. Set ) := 2#

n . The n vertices of the regular n-gon, with coordinates (cos k), sin k))for k $ {0, 1, . . . , n&1}, are the finite orbit of any one vertex, so the homogeneous degreen polynomial

g2,n :=!

0#k#n"1

)cos(k)) · x+ sin(k)) · y

*

is an invariant. Note that g2,n is guaranteed to have degree n since each of the factors isa homogeneous polynomial of degree 1, which in turn is true since the points in the orbitof vertices corresponding to the factors of g2,n are nonzero. The invariant polynomialsobtained this way for n $ {3, 4, 5, 6}, multiplied by constants to be monic in the highestpower of x, are given in the table below.

26

n Vertex Coordinates Invariant Polynomial g2,n

3 (1, 0),5&1

2 ,±$32

6x3 & 3xy2

4 (±1, 0), (0,±1) x2y2

5 (1, 0),

1$5"14 ,±

22(5+

$5)

4

2,

1"$5"14 ,±

22(5"

$5)

4

2x5 & 10x3y2 + 5xy4

6 (±1, 0),5±1

2 ,±$32

6x6 & 6x4y2 + 9x2y4

4.2 Proving the Invariants Generate

We want to prove that K[V ]G = R[g1, g2,n]. By Proposition 1 in 1.5.2, since the degreesof g1 and g2,n match the degree signature it su!ces to show that g1 /$ R and g2,n /$ R[g1].The first statement is obvious. So is the second when n is odd, for then g2,n has odddegree, while R[g1] has only polynomials of even degree since g1 has even degree.

So suppose n is even. Then we need to show that g2,n is not a linear multiple of

gn/21 = (x2 + y2)n/2, since the latter are all the polynomials of degree n in R[g1]. Notethat since we chose each regular n-gon to contain the vertex (1, 0), one of the factors inthe product defining g2,n has no y term. Thus g2,n has no yn term. But (x2 + y2)n/2 does

have a yn term, so gn/21 and g2,n are linearly independent.

4.3 Further Remarks

4.3.1 Another way to find a generator

Another way to find a homogeneous degree n invariant polynomial is to think of the nth

power map on the complex plane. We associate the vector

1ab

2$ R2 with the complex

number a+ ib, and consider the composition1ab

23' a+ ib 3' (a+ ib)n.

Let ) := 2#n . This map is invariant under the rotations !k by the angles k) since the nth

power map on C maps eik%z to zn for all integers k; for instance

(ei%(a+ ib))n = ein%(a+ ib)n = e2#i(a+ ib)n = (a+ ib)n.

So the complex polynomial (x+ iy)n is fixed by the rotations !k, but it is not a candidateto be an invariant of G since it is not a real polynomial. However, its real and imaginaryparts Re(x + iy)n and Im(x + iy)n, which are real polynomials, are also fixed by the!k. When we take into account the reflection " across the x-axis, which corresponds to

27

complex conjugation in the complex plane, we see that " leaves the real part fixed, whilechanging the sign of the imaginary part. Alternatively, note that " fixes precisely thoseterms in which y has even degree, while changing the sign of any term in which y hasodd degree. Since Re(x + iy)n consists of precisely those terms of (x + iy)n in which yhas even degree, it is fixed by " as well. So the degree n homogeneous polynomial

g2 := Re(x+ iy)n

is fixed by ! and " and therefore an invariant. It is easy to see that g2 and (x2 + y2)n/2

are linearly independent, whence x2 + y2 and g2 are also generators for K[V ]G.We obtain the following non-obvious fact as a corollary:

Corollary 1. Let n * 3, ) := 2#n , and write gn :=

'0#k#n"1

)cos(k)) · x + sin(k)) · y

*.

Then gn is a real multiple of Re(x + iy)n if n is odd, while if n is even then gn is a reallinear combination of Re(x+ iy)n and (x2 + y2)n/2.

Also, since K[V ]G has no homogeneous linear invariants, the sum(

0#k#n"1

)cos(k)) ·

x+ sin(k)) · y*must vanish, whence we have proved the following identities:

Corollary 2. Let n * 3 and set ) := 2#n . Then

"

0#k#n"1

cos(k)) ="

0#k#n"1

sin(k)) = 0.

4.3.2 Proving generation with field extensions

Let g1 = x2+y2 and g2 = Re((x+ iy)n), as before. By the discussion in 1.5.3, a su!cient(and also necessary) condition for the homogeneous invariants g1 and g2 to generate thering of invariants is that the field extension

R(V ) = R(x, y)d

R(g1, g2)

has degree d = |Dn| = 2n. It su!ces to prove that d ( 2n since R(g1, g2) is a subfieldof the G-fixed subfield and R(x, y) has degree 2n as an extension of the G-fixed subfield,whence d * 2n.

We split the extension into two extensions by first adding x, then y:

R(x, y)

R(g1, g2)(x) = R(x, y2)

R(g1, g2).

28

The top extension has degree 2 since y has minimal polynomial t2 & y2 over R(x, y2). Toshow that the bottom extension has degree ( n, which is all we need, it su!ces to showthere is a degree n polynomial with coe!cients in R(g1, g2) that vanishes at x.

Set D :=7n2

8and write

g2 = Re(x+ iy)n = Re

9n"

k=0

1n

k

2xn"k(iy)k

:=

D"

k=0

(&1)k1n

2k

2xn"2ky2k.

Note that every term of g2 has y to an even power. Thus we can add an appropriatemultiple of xn"2D(x2 + y2)D to g2 to eliminate the term of highest degree in y. Indeed,by inductively eliminating the remaining term of highest degree in y with an appropriatemultiple of xn"2k(x2+y2)k, we see that linear multiples of xn, xn"2(x2+y2), . . . , xn"2D(x2+y2)D generate g2 over R. The degree n linear dependence obtained this way shows thatx satisfies a polynomial of degree at most n with coe!cients in R[g1, g2], which is whatwe wanted.

29

Chapter 5

Invariants of Platonic SolidSymmetry Groups

In this chapter we let V = R3 and G be the tetrahedral group T , the octahedral groupO, or the icosahedral group I. The method of finding invariants by using finite orbits ofpoints will yield invariants with degrees that are slightly too large, for recall from Section3.3 that the symmetry groups have

Group Order Degree Signature

T 24 (2, 3, 4)

O 48 (2, 4, 6)

I 120 (2, 6, 10)

and from Section 2.2 that the platonic solids have

Platonic Solid Faces Vertices Edges

tetrahedron 4 4 6

cube 6 8 12

octahedron 8 6 12

dodecahedron 12 20 30

icosehedron 20 12 30

We know one invariant polynomial is distance squared from the origin x2 + y2 + z2,which accounts for the d1 = 2 in each degree signature. For T we need to get two moreinvariants with degrees d2 = 3 and d3 = 4, yet the only degrees we can get using thefinite orbit of a vertex, midpoint of an edge, or center of a face are in the set {4, 6}.Similarly, for O we need two invariants with degrees d2 = 4 and d3 = 6, yet the onlyavailable degrees are {6, 8, 12}. And for I we need two invariants with degrees d2 = 6and d3 = 10, but we only have {12, 20, 30} as the orders of our orbits. However, notethat in each case di has the property that either

• di is in the set of orders of the orbits, or

30

• 2di is in the set of orders of the orbits.

We will show that when the former fails, there is an orbit of order 2di with a half orbitof cardinality di that also gives rise to an invariant polynomial, whence we will obtaininvariant polynomials of degrees matching the degree signature.

5.1 Half Orbits

We investigate a su!cient condition for half of a finite orbit of a point, which we call ahalf orbit, to yield an invariant polynomial. If $ is a finite orbit of a point, then thefirst thing to look for is that the points of $ are symmetric with respect to the origin,namely that if (a, b, c) $ $, then the opposite point (&a,&b,&c) $ $. For then if $% 4 $is a half orbit with no two points opposite, we have

P! = ±(P!!)2

in the notation of 3.4.2, and P!! is our candidate polynomial of half the degree of P!,which may turn out to be an invariant. Now we look at the orbit G($%) of $% under theaction of G on subsets of $ of cardinality |!|

2 , namely half orbits of $. $% has no twopoints opposite, and the action of G on $% preserves this property. Moreover, $% containsexactly one point from each pair of opposite points of $, hence so does each element ofG($%). Thus the polynomials corresponding to the half orbits in G($%) are the same upto sign. The polynomials corresponding to two such half orbits $%

1,$%2 have the same sign

if and only if the number of points $%1 does not have in common with $%

2 is even, sinceeach non-common point leads to a factor of &1 in the polynomial. Thus we have shownthat:

Proposition 2. Let $ be a finite orbit of a point consisting of pairs of opposite points.Let $% denote a half orbit of non-opposite points. Then P!! is an invariant polynomial ifand only if |$% & !($%)| is even for each ! $ G.

5.2 Invariants of the Tetrahedral Group T

Recall that T has degree signature (2, 3, 4), that x2+y2+z2 is an invariant, and that thetetrahedron has four vertices, four faces, and six edges, which yield invariants of degreesfour, four, and six, respectively. The centers of faces lie on lines through a vertex andthe centroid of the tetrahedron, so the invariant polynomial corresponding to centers offaces will be a constant multiple of the polynomial corresponding to the orbit of vertices.Thus we really have just three invariants of degrees two, four, and six, where the sixis too large by a factor of 2. So we would like to show that taking half of the orbit ofmidpoints of edges will still yield an invariant polynomial.

The orbit of six midpoints of edges $ := {(±1, 0, 0), (0,±1, 0), (0, 0,±1)} is symmetricwith respect to the origin. Set $% := {(1, 0, 0), (0,&1, 0), (0, 0, 1)}, which is a half orbit

31

of non-opposite points, all of which lie on a single face of the tetrahedron:

$% on the Tetrahedron

Then the orbit G($%) of $% consists of four elements corresponding to the four faces ofthe tetrahedron:

$% The other three half orbits in G($%) on the tetrahedron

Each of these triples of points has an odd number of points in common with $%, whence|$% & !($%)| is even for each ! $ G, so the polynomial

P!! = x(&y)(z) = &xyz

is an invariant of degree 3.Now we compute the invariant of degree 4. The orbit of vertices is {(1, 1, 1), (&1,&1, 1),

(1,&1,&1), (&1, 1,&1)} (which by the way is not symmetric with respect to the origin),with corresponding invariant

(x+ y + z)(&x& y + z)(x& y & z)(&x+ y & z)

= 2(x4 + y4 + z4)& (x2 + y2 + z2)2.

Since x2+ y2+ z2 is itself invariant, we will take x4+ y4+ z4 as our degree 4 invariant tosimplify notation. Likewise, we multiply P!! by &1 and the result is still invariant. Wewill prove later that

x2 + y2 + z2, xyz, x4 + y4 + z4

generate the ring of invariants of T .

5.3 Invariants of the Octahedral Group O

O has degree signature (2, 4, 6) with x2+y2+ z2 as a degree 2 invariant. The octahedronhas eight faces, six vertices, and 12 edges, while its dual polyhedron the cube has sixfaces, eight vertices, and 12 edges.

32

We begin by looking at the six centers of faces of the cube (vertices of the octahedron){(±1, 0, 0), (0,±1, 0), (0, 0,±1)}. Though this orbit is symmetric with respect to theorigin, a half orbit of non-opposite points corresponds to the centers of faces that have acommon vertex, and two such sets of centers can di"er at only one point:

5 55

55

5

Two half orbits of centers of faces of the cube

So we must take the full orbit of six centers of faces, obtaining the invariant polynomialx2y2z2.

Now we look at the orbit of eight vertices of the cube $ := {(±1,±1,±1)} (whichare also a constant multiple of the centers of faces of the octahedron), with the intentof finding a half orbit yielding an invariant polynomial. Again, we have symmetry withrespect to the origin. Let $% be a half orbit of four vertices that lie on single face of thecube (the centers of four faces of the octahedron that meet at a vertex) and are hencenon-opposite.

55

55

The points corresponding to $%

The orbitG($%) of $% contains six elements corresponding to the faces of the cube (verticesof the octahedron):

5555 55 55

555

5 5555

555

5 55

55

The orbit G($%) on the octahedron

Letting the first image denote $%, we note that every half orbit in G($%) di"ers from $%

at an even number of points, namely |$% & !($%)| is even for each ! $ G. So letting $%

correspond to the face of the cube on which the x-coordinate is uniformly 1, we get theinvariant

P!! = (x+ y + z)(x+ y & z)(x& y + z)(x& y & z)

= 2(x4 + y4 + z4)& (x2 + y2 + z2)2,

33

where as before we will take x4 + y4 + z4 for simplicity. We will prove shortly that theinvariants

x2 + y2 + z2, x4 + y4 + z4, x2y2z2

generate the ring of invariants of O.

5.4 Invariants of the Icosahedral Group I

We know that I has degree signature (2, 6, 10) and that x2+y2+z2 is a degree 2 invariant.The dodecahedron has 12 faces, 20 vertices, and 30 edges, while the icosahedron has 20faces, 12 vertices, and 30 edges. We will show that taking half orbits of the vertices ofthe dodecahedron (faces of the icosahedron) and of the vertices of the icosahedron (facesof the dodecahedron) yields invariants of the right degrees.

The 12 vertices of the icosahedron are $ := {(0,±1,±%), (±1,±%, 0), (±%, 0,±1)},where % = 1+

$5

2 is the golden ratio. Clearly $ is symmetric with respect to the origin.We choose a non-opposite half orbit by selecting one vertex of the icosahedron and itsfive neighboring vertices, for instance $% := {(1,±%, 0), (%, 0,±1), (0,±1,%)}, which wecan visualize as

$% on the Icosahedron

The orbit G($%) of $% contains 12 elements, one for each face of the dodecahedron, whichcan have the following possible arrangements up to radial symmetry:

555

55 5

Type 1

555

55

5

Type 2

5555

5

5

Type 3

55

55

55

Type 4

G($%) on the dodecahedron, up to radial symmetry

Letting $% be the set of type 1, we note that the set of type 1 di"ers from any set of type2 at two vertices, from any set of type 3 at four vertices, and from the set of type 4 atall six vertices. Thus |$% & !($%)| is even for each ! $ G, so the polynomial

P := P!! = (x+ %y)(x& %y)(%x+ z)(%x& z)(y + %z)(&y + %z)

= (x2 & %2y2)(%2x2 & z2)(&y2 + %2z2).

34

is invariant.The 20 vertices of the dodecahedron are (±1,±1,±1), (0,± 1

$ ,±%), (± 1$ ,±%, 0), (±%, 0,± 1

$),

where % = 1+$5

2 is the golden ratio, and they correspond to the centers of faces of theicosahedron. We choose a non-opposite half orbit of ten vertices of the dodecahedron byselecting the five vertices that lie on a face, as well as the five vertices adjacent to theoriginal five vertices. For instance

$% :=

3(±1,±1, 1),

1%, 0,±1

%

2,

11

%,±%, 0

2,

10,±1

%,%

24,

which we can visualize as

$% on the Dodecahedron

The orbit G($%) of $% contains one element for each of the 12 faces of the dodecahedron.Viewed as centers of faces on the adjacency graph of the icosahedron, these elements are,up to radial symmetry:

5 555

5555 5

5

Type 1

5555

5

5

5

5 5 5

Type 2

5 55

5

5

5 55

55

Type 3

5555

5 5

55 55

Type 4

G($%) on the icosahedron, up to radial symmetry

The set $% of type 1 di"ers from type 2 on four faces, from type 3 on six faces, and fromtype 4 on all 10 faces. Thus |$% & !($%)| is even for each ! $ G, so the polynomial

P!! := (x+ y + z)(x& y + z)(&x+ y + z)(&x& y + z)

1%x+

1

%z

2

1%x& 1

%z

211

%x+ %y

211

%x& %y

211

%y + %z

21&1

%y + %z

2

is invariant. To simplify notation we eliminate % in the denominators, obtaining theinvariant

Q := %6 · P!! =)2(x4 + y4 + z4)& (x2 + y2 + z2)2

*(%4x2 & z2)(x2 & %4y2)(&y2 + %4z2).

We will show below that the polynomials

x2 + y2 + z2, P, Q

generate the ring of invariants of I.

35

5.5 Proving the Invariants Generate

Proposition 3. The following table of invariant generators for the rings of invariantsof the tetrahedral, octahedral, and icosahedral groups is correct:

Group Generators of the Ring of Invariants

T x2 + y2 + z2 xyz x4 + y4 + z4

O x2 + y2 + z2 x4 + y4 + z4 x2y2z2

I x2 + y2 + z2 P Q

where

P := (x2 & %2y2)(%2x2 & z2)(&y2 + %2z2);

Q :=)2(x4 + y4 + z4)& (x2 + y2 + z2)2

*(%4x2 & z2)(x2 & %4y2)(&y2 + %4z2).

Proof. We use Proposition 1 from 1.5.2, noting that the degrees of the proposed genera-tors match the degree signatures of T , O, and I.

For the tetrahedral group T , it su!ces to check that

(a) xyz /$ R[x2 + y2 + z2], and

(b) x4 + y4 + z4 /$ R[x2 + y2 + z2, xyz].

For (a), we note simply that R[x2 + y2 + z2] has no polynomials of odd degree. For (b),it su!ces to check that x4 + y4 + z4 is not a linear multiple of (x2 + y2 + z2)2, which isclear since the latter has mixed terms such as 2x2y2.

For the octahedral group O, we need to check that

(a) x4 + y4 + z4 /$ R[x2 + y2 + z2], and

(b) x2y2z2 /$ R[x2 + y2 + z2, x4 + y4 + z4].

We just argued (a) for the tetrahedral group above. For (b), there is no linear dependencebetween (x2 + y2 + z2)3, (x2 + y2 + z2)(x4 + y4 + z4), and x2y2z2 since the 30 3 matrix

PolynomialCoe!cient of

x6 x4y2 x2y2z2

(x2 + y2 + z2)3 1 3 6

(x2 + y2 + z2)(x4 + y4 + z4) 1 1 0

x2y2z2 0 0 1

is invertible.Finally, for the icosahedral group I, we want to show that

(a) P /$ R[x2 + y2 + z2], and

(b) Q /$ R[x2 + y2 + z2, P ].

36

For (a), note that the 20 2 matrix

PolynomialCoe!cient of

x6 x4y2

(x2 + y2 + z2)3 1 3

P 0 &%2

is invertible. Similarly, for (b), the matrix

PolynomialCoe!cient of

x10 x8y2 x8z2

(x2 + y2 + z2)5 1 5 6

(x2 + y2 + z2)2 · P 0 &%2 %4

Q 0 &%4 %8

is invertible since dividing the second row by %2 and the third row by %4 makes the secondentries of these rows equal, but makes the third entries %2 )= %4.

This concludes our discussion of specific groups of symmetries. In the next chapter,we present a proof of the Main Theorem.

37

Chapter 6

Proof of the Main Theorem

In this final chapter we will give a detailed proof of the Main Theorem stated in Section1.2. The tools used in the proof are elementary, but several of the arguments are fairlycomplicated nonetheless. Some of the key ideas involved in the proof are induction onhomogeneous degrees, the relation between the ring of invariants and the G-fixed subfield,and studying the poles of Poincare series.

6.1 Setup

The following notation is assumed unless otherwise specified:

• V a vector space of dimension n over a field K of characteristic 0.

• G a finite group of invertible linear transformations of V .

• K[V ] = K[x1, . . . , xn] the ring of polynomial functions on V .

• K[V ]G the ring of invariants.

• K(V ) the field of fractions of K[V ], with G-fixed subfield K(V )G.

• s(G) the set of all pseudoreflections in G.

• I the ideal of K[V ] generated by all homogeneous elements of K[V ]G of strictlypositive degrees.

6.2 Background

We begin by establishing some well-known results that we will need.

Proposition 4.

(a) K[V ] is integral over K[V ]G.

38

(b) K[V ]G is finitely generated as an algebra over K.

(c) There exist homogeneous f1, . . . , fr $ K[V ]G that give an ideal basis for I (i.e. thefi generate I and none of the fi are in the ideal generated by all the others).

Proof. (a): For any f $ K[V ], the polynomial

Pf (t) :=!

!&G

(t& !(f))

is monic with coe!cients that are symmetric polynomials in the elements of G appliedto f , hence invariants, and Pf (f) = 0 (look at the factor corresponding to the identityof G).

(b): By (a), each xi satisfies an equation of the form

xdii + ai,1x

di"1i + · · ·+ ai,di = 0, ai,j $ K[V ]G.

Set A := K[{ai,j}]. Then K[V ] is integral over A, hence K[V ] is a finitely generatedA-module. Since A is a Noetherian ring, K[V ] is a Noetherian A-module, thus the A-submodule K[V ]G is finitely generated over A. Combining the generators of K[V ]G overA with the {ai,j} gives a finite generating set for K[V ]G as an algebra over K.

(c): By (b), let K[V ]G = K[a1, . . . , am]. Since a polynomial f $ K[V ] is invariantif and only if each homogeneous component of f is invariant (G is a group of lineartransformations), we may assume the ai to be homogeneous of strictly positive degrees.Thus I = (a1, . . . , am) * K[V ], and removing redundant elements, we may assume the{ai} are an ideal basis.

Recall that the elements of G are invertible linear transformations of V , which alsogive ring automorphisms of K[V ] by the definition

!(f)(v) := f(!"1(v)), ! $ G, f $ K[V ], v $ V.

This same definition allows us to view ! $ G as field automorphisms of the field offractions K(V ) of K[V ], simply by considering f $ K(V ) instead of f $ K[V ]. Thisputs the tools developed for studying field extensions at our disposal, which will be veryuseful due to the following proposition.

Proposition 5.

(a) The G-fixed subfield K(V )G of K(V ) coincides with the field of fractions of K[V ]G.

(b) If g1, . . . , gr generate K[V ]G as a K-algebra, then g1, . . . , gr also generate K(V )G

as a field extension of K.

(c) K(V ) is a Galois extension of K(V )G of degree |G|.

39

Proof. (a): For any f/g $ K(V ), we may assume g $ K[V ]G since we can write

f

g=

f ·'

id '=!&G !(g)'

!&G !(g).

But if f/g $ K(V )G and g $ K[V ]G, then clearly f $ K[V ]G as well.(b): Follows immediately from (a).(c): K(V )G is the fixed field of the group of automorphisms G.

6.3 Main Theorem

We begin with a definition that will serve as a stepping stone in our proof of the MainTheorem.

Definition 4. Let us say that G has the property P if the following holds:

(P): For any g1, . . . , gm $ K[V ]G such that g1 /$ (g2, . . . , gm) * K[V ]G, if h1, . . . , hm $K[V ] are homogeneous polynomials such that h1g1 + · · ·+ hmgm = 0, then h1 $ I.

Main Theorem. Let G be a finite group of invertible linear transformations of an n-dimensional vector space V over a field K of characteristic 0. Then the following areequivalent:

(i) G is generated by pseudoreflections.

(ii) G has property P .

(iii) K[V ]G is a polynomial ring in n = dimV homogeneous generators.

(iv) K[V ] is a free module over K[V ]G with a basis consisting of |G| homogeneous poly-nomials.

We will prove the theorem by showing that (i) =! (ii) =! (iii) =! (i) and (ii) "! (iv),in that order. The proofs (i) =! (ii) =! (iii) and (ii) =! (iv) are based on Chevalley[3], while the proof (iii) =! (i) is from Neusel [5].

Definition 5. We will make use of the averaging projection K[V ] #G!! K[V ]G , defined

by

$G(f) :=1

|G|"

!&G

!(f), f $ K[V ].

40

6.3.1 Proving (i) =! (ii)

Proposition 6 ((i) =! (ii)). If G is generated by pseudoreflections, then G has propertyP .

Proof. Assume G is generated by pseudoreflections. Suppose g1, . . . , gm $ K[V ]G suchthat g1 /$ (g2, . . . , gm) *K[V ]G, and that h1, . . . , hm $ K[V ] are homogeneous and satisfy

h1g1 + · · ·+ hmgm = 0. (5)

We need to prove that h1 $ I.We use induction on deg h1. If deg h1 = 0, then applying $G to (5) yields

$G(h1)g1 + · · ·+ $G(hm)gm = 0,

which implies $G(h1) = 0, since otherwise g1 $ (g2, . . . , gm) * K[V ]G, a contradiction.But this means h1 = 0 $ I.

So assume d := deg h1 > 0 and that the claim holds for all homogeneous polynomialsof degree < d. If ! $ G is a pseudoreflection fixing the hyperplane (L = 0), then !"1

fixes the same hyperplane. Thus !(hi)& hi vanishes on (L = 0) for each i, so there existhomogeneous polynomials hi such that !(hi)& hi = hiL. Applying ! to (5), subtracting(5), and canceling the common factor L, we see that

h1g1 + · · ·+ hmgm = 0,

whence h1 $ I by the inductive hypothesis. Thus !(h1) + h1 (mod I), and since thisholds for each pseodoreflection in G, it holds for any ! $ G since G is generated bypseodoreflections. Thus $G(h1) + h1 (mod I), so since h1 homogeneous of strictly pos-itive degree and $G(h1) $ K[V ]G together imply $G(h1) $ I, we see that h1 $ I asdesired.

6.3.2 Proving (ii) =! (iii)

We break up the proof into three main steps, given by the following proposition.

Proposition 7 ((ii) =! (iii)). Suppose G has property P . Let f1, . . . , fr $ K[V ]G behomogeneous invariants giving an ideal basis of I. Then

(a) the fi are algebraically independent over K.

(b) K[f1, . . . , fr] = K[V ]G.

(c) r = n.

Proof of (a). Write di for the degree of fi. Assume the claim is false and let H(y1, . . . , yr)be a polynomial with coe!cients in K such that H(f1, . . . , fr) = 0 is a nontrivial relation

41

of minimum homogeneous degree D in the xi. Each monomial yk11 · · · ykrr of H(y1, . . . , ym)satisfies

k1 · d1 + · · ·+ kr · dr = D.

Since H is not constant the partial derivatives +yiH are not all zero. Set

Hi := +yiH(f1, . . . , fr) (1 ( i ( r),

which are in K[V ]G and either vanish or are homogeneous of degree D & di. The Hi arenot all zero since the +yiH are not all zero and since H(f1, . . . , fr) is a relation among thefi of minimum degree. After reordering, we may assume H1 )= 0 and that all the Hi arein the ideal generated in K[V ]G by the first s * 1 of them, but that none of H1, . . . , Hs

belongs to the ideal generated by the other Hj (1 ( j ( s) in K[V ]G. Write

Hs+j =s"

i=1

gj,iHi (1 ( j ( r & s),

where each gj,i $ K[V ]G is 0 or homogeneous of degree (D&ds+j)&(D&di) = di&ds+j * 0.Di"erentiating the relation H(f1, . . . , fr) = 0 in xk, the chain rule yields

r"

i=1

Hi · +xkfi = 0 (1 ( k ( n),

and substituting for all the Hs+j yields a relation in H1, . . . , Hs in which the coe!cientof Hi is

Ci,k := +xkfi +

r"s"

j=1

gj,i+xkfs+j (1 ( i ( s; 1 ( k ( n).

The coe!cient Ci,k is either 0 or homogeneous of degree di&1 since deg gj,i+deg +xkfs+j =

di & ds+j + ds+j & 1 = di & 1. Since the Hi are in K[V ]G and none of H1, . . . , Hs belongsto the ideal in K[V ]G generated by the other Hj, the fact that G has property P allowsus to conclude that each Ci,k is in I. Since f1, . . . , fr is an ideal basis for I, this meansthat

Ci,k =r"

l=1

ai,k,lfl (1 ( i ( s; 1 ( k ( n),

for homogeneous ai,k,l $ K[V ] that are each 0 or of degree di & dl & 1 * 0. Now we usethe identity +x1f · x1 + · · ·+ +xnf · xn = (deg f) · f to write

difi +r"s"

j=1

gj,ids+jfs+j =n"

k=1

Ci,kxk =r"

l=1

bi,lfl,

where each bi,l $ (x1, . . . , xn) *K[V ] is 0 or of degree di & dl * 1, so that in particularbi,i = 0. But since di is invertible, this means that fi belongs to the ideal generated bythe other fj, which is a contradiction. Thus f1, . . . , fr are algebraically independent.

42

Proof of (b). Since a polynomial is invariant if and only if its homogeneous componentsare invariant (G is a group of linear transformations), it su!ces to prove that any homo-geneous g $ K[V ]G is in K[f1, . . . , fr]. We proceed by induction on deg g = d. If d = 0,then g $ K , K[f1, . . . , fr].

Suppose d * 1 and any homogeneous invariant of degree < d is in K[f1, . . . , fr]. If gis invariant and deg g = d, then g $ I, so

g = a1f1 + · · ·+ arfr, ai $ K[V ]. (5)

We may assume the ai are homogeneous with deg ai + deg fi = d when ai )= 0. Hitting(5) with the projection $G, we may also assume each ai $ K[V ]G. Since the fi arenot constant, each ai must have degree < d, so the inductive assumption completes theargument.

Proof of (c). Let f1, . . . , fr be homogeneous invariants that form an ideal basis of I,which are algebraically independent by (a) and which generate K[V ]G by (b). By Propo-sition 5 (b), f1, . . . , fr generate K(V )G as a field extension of K, hence K(V )G/K hastranscendence degree r. Now, K(V )/K(V )G is Galois of finite degree |G| (in particularalgebraic), so the two fields have the same transcendence degree n as extensions of K,whence r = n.

6.3.3 Proving (iii) =! (i)

We begin by introducing Poincare series, which will play a key role in this proof.

Definition 6. A graded vector space of finite type over a field K is a K-vectorspace W =

;i(0 Wi, where each Wi is a K-vector space of finite dimension. If W is a

graded vector space of finite type, then the Poincare series of W is the formal powerseries

P (W, t) ="

i(0

dimWi · ti.

Note that P (W, 1) = dimW if the dimension is finite. Also, Poincare series are additiveunder direct sums and multiplicative under tensor products: if W and W % are gradedvector spaces of finite type, then

P (W #W %, t) = P (W, t) + P (W %, t)

andP (W 6W %, t) = P (W, t) · P (W %, t).

The graded vector spaces we will consider are polynomial rings over the field K,graded by polynomial degree. If f is a homogeneous polynomial of degree d, then

P (K[f ], t) = 1 + td + t2d + · · · = 1

1& td,

from which we deduce, using multiplicativity under tensor powers, that:

43

Proposition 8. Let K[f1, . . . , fr] be a polynomial ring in r homogeneous generators,with di the degree of fi. Then

P (K[f1, . . . , fr], t) =r!

i=1

1

1& tdi.

Now if ! $ G, then ! acts on both V and on K[V ]. Unless otherwise specified, theaction is assumed to be on V , as in the determinant in the proposition below. We writetr(!, K[V ]j) to denote the trace of ! acting on K[V ] restricted to the graded piece K[V ]j.

Proposition 9. If ! $ G, then)"

j=0

tr(!, K[V ]j) · tj =1

det(1& !"1t). (5)

Proof. Neither side of (5) is changed by extending the field, so we may assume K isalgebraically closed. Then we can choose a basis for V such that ! is diagonal witheigenvalues ,1, . . . ,,n. Letting x1, . . . , xn be the dual basis for V !, the eigenvalues of !acting on V ! are ,"1

1 , . . . ,,"1n , so the eigenvalues of ! acting on K[V ]j, which has a basis

consisting of the degree j monomials in the xi, are all the products of j not necessarilydistinct ,"1

i s. Thus the left side of (5) is equal to9 )"

j=0

,"j1 tj

:· · ·

9 )"

j=0

,"jn tj

:=

n!

i=1

1

1& ,"1i t

=1

det(1& !"1t),

as claimed.

Proposition 10 (Molien’s theorem).

P (K[V ]G, t) =1

|G|"

!&G

1

det(1& !"1t)=

1

|G|"

!&G

1

det(1& !t).

Proof. Recall the projection K[V ] #G!! K[V ]G defined by 1

|G|(

!&G !. Since $G induces

projections K[V ]j !! K[V ]Gj of the graded pieces, the trace tr($G, K[V ]j) of $G re-

stricted to K[V ]j is equal to dimK[V ]Gj . Thus

P (K[V ]G, t) =)"

j=0

(dimK[V ]Gj ) · tj

=)"

j=0

tr($G, K[V ]j) · tj

=)"

j=0

91

|G|"

!&G

tr(!, K[V ]j)

:· tj.

The result now follows by interchanging the order of summation and applying Proposition9.

44

Proposition 11. The Laurent expansion of the Poincare series P (K[V ]G, t) at t = 1begins as follows:

P (K[V ]G, t) =1|G|

(1& t)n+

|s(G)|2|G|

(1& t)n"1+ · · · ,

where the remaining terms have a pole of order at most n& 2 at t = 1.

Proof. The term 1det(1"!t) has a pole at t = 1 of order equal to the multiplicity of the

eigenvalue 1 for !. For the identity id $ G, this multiplicity is n, while for any otherelement of G the multiplicity is ( n & 1. By Proposition 10, P (K[V ]G, t) is a rationalfunction in t with only possible poles at the eigenvalues of the elements of G. By whatwe just discussed, (1 & t)nP (K[V ]G, t) no longer has a pole at t = 1 and is thereforeholomorphic in a neighborhood of t = 1. So P (K[V ]G, t) is holomorphic in that neigh-borhood except at the pole t = 1, hence it makes sense to consider the Laurent series ofP (K[V ]G, t) at t = 1.

The coe!cient of the term 1/(1 & t)n is determined solely by id, so it is 1/|G|.The coe!cient of the term involving 1/(1 & t)n"1 can be computed by multiplyingP (K[V ]G, t) & 1

|G|1

(1"t)n by (1 & t)n"1 and evaluating at t = 1. The only terms notkilled are those corresponding to ! $ G that have eigenvalue 1 with multiplicity preciselyn&1, namely the terms corresponding to the pseudoreflections. The contribution of eachpseudoreflection ! to the coe!cient of the term involving 1/(1& t)n"1 is

1

|G|(1& t)n"1 1

det(1& !t)

<<<<t=1

=1

|G|1

1& ,!,

where ,! )= 1 is the only non-1 eigenvalue of !. If ! has order 2, then ,! = &1, whencethe contribution of ! is 1/2|G|. Otherwise !"1 is also a pseudoreflection with ,!"1 = ,"1

! ,and we have the identity

1

1& ,!+

1

1& ,"1!

=(1& ,!) + (1& ,"1

! )

2& ,! & ,"1!

= 1,

so once again we can think of each of !, !"1 as contributing 1/2|G|. Thus the totalcontribution of all the pseudoreflections is |s(G)|/2|G|, hence this is the coe!cient of1/(1& t)n"1 in the Laurent expansion.

Proposition 12. Suppose K[V ]G is a polynomial ring generated by n homogeneous in-variants f1, . . . , fn of degrees d1, . . . , dn. Then

|G| =n!

i=1

di, |s(G)| =n"

i=1

(di & 1).

Proof. Combining Propositions 8 and 11, we get

n!

i=1

1

1& tdi=

1|G|

(1& t)n+

|s(G)|2|G|

(1& t)n"1+ · · ·

45

as a Laurent expansion at t = 1. Multiplying both sides by (1& t)n gives

n!

i=1

1

1 + t+ t2 + · · ·+ tdi"1=

1

|G| + (1& t)|s(G)|2|G| + · · · , (5)

where the remaining terms are divisible by (1& t)2. Substituting t = 1 in (5) shows that'ni=1 di = |G|. To prove the second identity, we will simply di"erentiate (5) by t and

then evaluate at t = 1. On the right side, we get &|s(G)|/2|G|. Now we will compute thederivative of the left side. Set Ti(t) := 1 + t+ t2 + · · ·+ tdi"1 for notational convenience.Di"erentiating the left side using product rule and chain rule yields

&n"

i=1

)1 + 2t+ 3t2 + · · ·+ (di & 1)tdi"2

*

Ti(t)2 ·

'j '=i Tj(t)

.

Substituting t = 1, note that Ti(1) = di and that 1+ 2+ · · ·+ (di & 1) = di(di"1)2 for each

i, so the above yields

&n"

i=1

di(di"1)2

di ·'n

j=1 di= & 1

2|G|

n"

i=1

(di & 1).

Thus |s(G)| =(n

i=1(di & 1), as claimed.

We are almost ready to prove that (iii) =! (i), but we will need a general fact:

Proposition 13. Let K[y1, . . . , ym] be a polynomial ring. Let

{g1, . . . , gr} of degrees d1 ( · · · ( dr

{h1, . . . , hs} of degrees e1 ( · · · ( es

be algebraically independent sets of homogeneous polynomials in K[y1, . . . , ym], such thatK[g1, . . . , gr] , K[h1, . . . , hs]. Then

(a) r ( s ( m;

(b) di * ei for each 1 ( i ( r;

(c) if r = s then the following are equivalent:

(i) K[g1, . . . , gr] = K[h1, . . . , hr];

(ii) di = ei for all 1 ( i ( r;

(iii)'r

i=1 di ='r

i=1 ei.

46

Proof. (a): This follows immediately from algebraic independence and the additivity ofthe transcendence degrees of the field extensions

K(y1, . . . , ym) 7 K(h1, . . . , hs) 7 K(g1, . . . , gr) 7 K.

(b): Since each gi $ K[h1, . . . , hs], gi is generated by the hj with degree no largerthan di. The stated claim will follow if we can show that for each k > 0,

#{di : di ( k} ( #{ei : ei ( k}.

Suppose this is not true for some k0 > 0. Then there is some i0 > 0 such thatd1 ( · · · ( di0 ( k0 and e1 ( · · · ( ei0"1 ( k0, with ei0 > k0. But then K(g1, . . . , gi0) ,K(h1, . . . , hi0"1), contradicting the additivity of the transcendence degrees of these ex-tensions over K.

(c): Since K[g1, . . . , gr] , K[h1, . . . , hr], we have K[g1, . . . , gr] = K[h1, . . . , hr] "!Pt(K[g1, . . . , gr]) = Pt(K[h1, . . . , hr]). Noting that Pt(K[g1, . . . , gr]) =

'ri=1(1 & tdi)"1

and Pt(K[h1, . . . , hr]) ='r

i=1(1 & tei)"1 now gives (i) "! (ii). (ii) =! (iii) is trivial,and the converse follows immediately from (b).

We are finally ready to prove that (iii) =! (i) in the Main Theorem.

Proposition 14 ((iii) =! (i)). Suppose K[V ]G is a polynomial ring in the n homogeneousgenerators f1, . . . , fn. Then G is generated by pseudoreflections.

Proof. Write di for the degree of fi, with d1 ( · · · ( dn. Let H be the subgroup of Ggenerated by all pseudoreflections of G. SinceH is generated by pseudoreflections, K[V ]H

is a polynomial ring with homogeneous generators h1, . . . , hn of degrees e1 ( · · · ( en.Since H ( G, K[V ]G , K[V ]H , so each fi $ K[h1, . . . , hn]. By Proposition 13 (b),ei ( di for each i. But |s(G)| = |s(H)| implies

((di& 1) =

((ei& 1) by Proposition 12,

whence ei = di for each i. Thus'

di ='

ei, which implies |G| = |H| by Proposition 12,whence G = H as claimed.

6.3.4 Proving (ii) "! (iv)

We break up the proof of (ii) =! (iv) into five steps.

Proposition 15 ((ii) =! (iv)). Suppose G has property P .

(a) K[V ]G is a polynomial ring in n homogeneous generators f1, . . . , fn, where the fiform an ideal basis of I.

(b) Let h1, . . . , hs $ K[V ] be homogeneous with residue classes modulo I linearly inde-pendent over K in K[V ]/I. Then h1, . . . , hs are linearly independent over K(V )G.

(c) K[V ]/I is a finite-dimensional vector space over K and has a basis given by theresidues of homogeneous polynomials A1, . . . , At $ K[V ].

(d) K[V ] is a finitely-generated free K[V ]G-module generated by A1, . . . , At.

47

(e) t = |G|.

Proof of (a). This is Proposition 7.

Proof of (b). Suppose ai $ K(V )G for each 1 ( i ( s such that

a1h1 + · · ·+ ashs = 0. (5)

We want to prove that each ai = 0. By clearing denominators and restricting to homo-geneous parts, we may assume each ai $ K[V ]G and homogeneous, such that for each iand some constant D, either ai = 0 or deg ai + deg hi = D.

Let di denote the degree of fi. Consider the sequence {Sj}j(1 of monomials fk11 · · · fkn

n

arranged in increasing order of degree k1 ·d1+ · · ·+kn ·dn, with S1 = 1. Since ai $ K[V ]G,we can write

ai ="

j(1

,ijSj, ,ij $ K and ,ij = 0 when deg ai )= deg Sj.

Substituting into (5), we get

"

j(1

bjSj = 0,

9bj =

s"

i=1

,ijhi

:, (55)

where each bj is either 0 or homogeneous of degree D & deg Sj * 0.Assume for contradiction that the ,ij are not all 0. Then let N * 1 be the maximum

integer such that ,ij = 0 for all 1 ( i ( s and all j < N . By the algebraic independenceof the fi, SN does not belong to the ideal generated in K[V ]G by the Sj with j > N , sosince G has property P we deduce from the first relation in (55) that bN $ I. But thenthe equation for bN in (55), together with the linear independence of the residue classesof hi modulo I, imply that ,iN = 0 for all 1 ( i ( s, contradicting the maximality of N .So the ,ij are all 0, whence the ai are all 0, as desired.

Proof of (c). K(V )/K(V )G is a Galois extension of finite degree |G|. By (b), this impliesthat K[V ]/I has finite dimension ( |G|, say t. Since K[V ]/I inherits the grading ofK[V ], we may choose a homogeneous basis for K[V ]/I, and let A1, . . . , At $ K[V ] behomogeneous polynomials whose residues modulo I give the basis.

Proof of (d). We will show that for each P $ K[V ], there are unique invariants gi $K[V ]G such that

P = g1A1 + · · ·+ gtAt,

namely that A1, . . . , At generate K[V ] freely as a K[V ]G-module. For uniqueness, wenote simply that the Ai are linearly independent over K[V ]G by (b). For existence, wemay assume P is homogeneous of degree d * 0, and we will use induction on d. Assumethat all polynomials of K[V ] of degree < d can be expressed as linear combinations of

48

the Ai with coe!cients in K[V ]G. Since the residues of the Ai are a basis of K[V ]/I, wecan choose ,1, . . . ,,t $ K such that

P % := P & (,1A1 + · · ·+ ,tAt) $ I,

where P % is either 0 or homogeneous of degree d. Since the fi form an ideal basis of I by(a), choose homogeneous Q1, . . . , Qn $ K[V ] such that

P % = Q1f1 + · · ·+Qnfn.

Since each Qi is either 0 or has degree < d, applying the inductive assumption to the Qi

and substituting into the defining equation for P % gives the desired result.

Proof of (e). Each element of K(V ) may be written as the quotient of a polynomial byan invariant. Thus by (d), we see that the Ai form a basis of K(V ) as a field extensionover K(V )G, whence t = |G|.

At last, we are ready to prove the final link in the chain of implications, which willcomplete the proof of the Main Theorem.

Proposition 16 ((iv) =! (ii)). Suppose K[V ] is a free module over K[V ]G with basisconsisting of |G| homogeneous polynomials. Then G has property P .

Proof. Let A1, . . . , A|G| $ K[V ] be the basis of homogeneous polynomials. Supposeg1, . . . , gm $ K[V ]G such that g1 /$ (g2, . . . , gm) * K[V ]G and that h1, . . . , hm $ K[V ]homogeneous such that

h1g1 + · · ·+ hmgm = 0. (5)

We want to show that h1 $ I. Let

hi = 'i1A1 + · · ·+ 'i|G|A|G|, 'ij $ K[V ]G homogeneous

be the unique expression of each hi in the module basis. Substituting into (5) gives therelation

(1A1 + · · ·+ (|G|A|G| = 0,

9(i =

m"

j=1

'jigj $ K[V ]G:

whence we must have (i = 0 for each i since A1, . . . , A|G| are a basis. The condition thatg1 /$ (g2, . . . , gm) implies that each '1i must be either 0 or have strictly positive degree,and thus '1i $ I for each i. So each term of the above expression for h1 is in I, whenceh1 $ I.

49

Bibliography

[1] M. Artin. Algebra. Prentice Hall, New Jersey, 1991.

[2] D. J. Benson. Polynomial invariants of finite groups. Cambridge University Press,1993.

[3] C. Chevalley. Invariants of finite groups generated by reflections. Amer. J. of Math.77 (1955), 778-782.

[4] H. S. M. Coxeter. Regular polytopes (2nd ed.) The Macmillan Company, New York,1963.

[5] M. D. Neusel and L. Smith. Invariant theory of finite groups. American MathematicalSociety, 2002.

[6] G. C. Shephard and J. A. Todd. Finite unitary reflection groups. Canad. J. of Math.6 (1954), 274-304.

[7] T. A. Springer. Invariant theory. Springer Lecture Notes in Mathematics 585.Springer-Verlag, Berlin/New York, 1977.

50