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Generations and relations for rings of invariants Askarogullari, Murat Can 2016

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Generations and Relations for Rings of InvariantsbyMurat Can AskarogullariB.Sc., Bilkent University, 2014a thesis submitted in partial fulfillmentof the requirements for the degree ofMaster of Scienceinthe faculty of graduate and postdoctoralstudies(Mathematics)The University of British Columbia(Vancouver)April 2016c© Murat Can Askarogullari, 2016AbstractIn this thesis we first prove that the algebra of invariants for reductivegroups over the base field complex numbers are finitely generated. Thenwe focus on invariant algebras of finite groups. After showing the naturalrelation between invariant algebras and reflections in the group we provethe Chevalley-Shephard-Todd theorem. We conclude with classification ofcomplex finite reflection groups and some examples. Throughout the thesiswe follow similar arguments as in Springer[8] and Kane[6] where we give fulldetails on the arguments.iiPrefaceThe topic of this thesis and the roadmap were suggested by my supervisorProf. Zinovy Reichstein. This thesis surveys known results that are notoriginal under particular assumptions. The author tried to present andmodify it to make things more fluent and clear than they were written inthe original form according to the assumptions made in a way unique to theauthor.iiiTable of ContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . ivAcknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . v1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Reductive Groups and Complete Reducibility . . . . . . . . 33 Finite Generation of the Algebra of Invariants . . . . . . . 114 Poincare` Series . . . . . . . . . . . . . . . . . . . . . . . . . . 175 Invariant Algebras of Finite Reflection Groups . . . . . . . 276 Classification and Examples . . . . . . . . . . . . . . . . . . . 36Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42ivAcknowledgementsI wish to express my deepest thanks for my supervisor Prof. Zinovy Reich-stein for his guidance and support. I would also want to thank my parentsfor their support and love throughout the years.vChapter 1IntroductionThe roots of invariant theory goes back to 18th century. In the study ofquadratic binary forms, the discriminant, an algebraic invariant is used todistinguish between equivalent forms. After the introduction of homoge-neous coordinates by Moebius and Plucker at the beginning of 19th centurypeople became more interested in invariant theory. Classically invariant the-ory looks for elements of polynomial algebra over a base field that doesn’tchange under the action of a given linear algebraic group- the algebra ofinvariants. During the first decades of invariant theory, people were lookingfor particular invariants. For instance, one of the major problems was find-ing the algebra of invariants where SL2(C) acting on (d + 1)-dimensionalvector space of binary forms of degree d. Is this invariant algebra generatedby finitely many elements over C? This question is a basic question in in-variant theory which can be generalized and is proven for this case beforeHilbert. Positive answer to this question was established by P. Gordan in1868 [5]. However solving the same question in a more general setting, forinstance for SLn(C), was at that time very hard. Hilbert introduced newmethods in algebra to settle this question and this inspired the Hilbert’s14th problem. Hilbert’s 14th problem asks whether or not finite generationof invariants continues to hold for every linear representation G → GL(V )of every linear algebraic group G. In this thesis we work over the base fieldC and this result is proven by Hermann Weyl in 1926 for reductive algebraic1groups over C [3]. We show this result in chapter 3 largely following [7] and[2]. However, in general, the answer to Hilbert’s 14th problem is negative.The first counterexample was discovered by M.Nagata around 60 years later,he gave an example of non finitely generated algebra of invariants [4].Given a finite set of generators for the invariant algebra, it is naturalto ask about the natural relations among these generators. The answer tothis question can be quite non-trivial even for finite groups. The algebra ofinvariants is a free polynomial algebra (i.e., has an algebraically independentset of generators) if and only if G is generated by reflections. This theorem iscalled Chevalley-Shephard-Todd theorem. It was first proven by Shephardand Todd[2] then later by Chevalley[1]. This is another main theorem of thethesis. Again our exposition in Chapter 6 largely follows the approach in [8]and [6]. In chapter 7 we will define irreducible complex reflection groups andshow that every complex reflection group is a product of irreducible ones.Finitely generated irreducible complex reflection groups were first classifiedby Shephard and Todd. There are 37 cases; 34 exceptional cases and 3infinite families. We don’t give the whole table in here but rather give someexamples.2Chapter 2Reductive Groups andComplete ReducibilityNotation and PreliminariesLet G ≤ GL(V ) for a finite dimensional vector space V over the field C. Weconsider the action of G on algebra of polynomial functions on V and weare interested in the invariant algebras induced by this action.We will denote the polynomial functions on V by S(V ) or by S inter-changeably. Let dimV = n, we have S(V ) ' C[T1, T2, . . . , Tn] in the cate-gory of C-algebras, so S is a graded C-algebra. We denote the homogeneouselements of S of degree d (d ∈ N) by Sd.There is a natural left action of G on the vector space V since g ∈ GL(V )an invertible linear transformation. This action induces a left action on Sas follows: Let f ∈ S, v ∈ V , g ∈ G, define g.f by g.f(v) = f(g−1.v). If wehave g.f = f for all g ∈ G we say that f is G-invariant. The G-invariantpolynomials form a C-algebra which we denote by SG (or S(V )G). This iscalled the algebra of invariants. Note that for an arbitrary g ∈ G we haveg.Sd ⊂ Sd, meaning that Sd is a G-invariant C-subspace of S for all d ≥ 0.Note that S1 = V∗; therefore action induced on V ∗ is defined. We definesome of the actions G induces for the future reference in here.3Definition 2.1. Let G ⊂ GL(V ) and let W be a G-invariant subspace ofV ; in other words let g.W ⊂W for all g ∈ G. We define the induced actionon the quotient vector space V/W as follows; let [v] ∈ V/W denote theequivalence class of v ∈ V , then g.[v] = [g.v]. This action is well-definedbecause W is G-invariant.Definition 2.2. Assume G ⊂ GL(V ) and G ⊂ GL(W ). Then both V andW are G-modules. We define the action induced on Hom(V,W ) as follows;let h ∈ Hom(V,W ), then define g.h by(g.h)(v) = g.(h(g−1.v)), v ∈ VNow, to be able to define a linear algebraic group we equip GL(V ) witha topology as follows: Let E(V ) be the vector space of all C-linear transfor-mations of the vector space V . This vector space is isomorphic to Mn×n(C)and therefore also isomorphic to Cn2. We can define the Zariski topology onE(V ) using that isomorphism; with respect to the Zariski topology on E(V ),GL(V ) is an open subset of E(V ) because GL(V ) is the complement of thezero set of the determinant function which is a polynomial in n2 variables.However by adding one variable, we make GL(V ) into an affine algebraicvariety. We identify the GL(V ) with the following closed setZ(Tn2+1.det(T1, T2, . . . , Tn2)− 1) ⊂ E(V )× C := {(g, x) | g ∈ E(V ), x ∈ C}Define the map ψ : GL(V )→ Z(Tn2+1.det(T1, T2, . . . , Tn2)−1) where ψ(g) =(g, (det g)−1). This map is clearly injective and surjective. From now on,when we talk about topology on GL(V ) or any subset of it we talk abouttopology induced by the Zariski topology on E(V )× C.Remark 2.3. (i) The inverse map φ : GL(V ) → GL(V ) where φ(g) =g−1 is continuous, bijective and involutory, i.e. its inverse is itself;therefore defines a homeomorphism from GL(V ) to GL(V ).(ii) Fixing a ∈ GL(V ), define ψa : G→ G where ψa(g) = a.g. This is alsoa continuous bijection which has the inverse ψa−1 .4Definition 2.4. A group G is called a linear algebraic group if and only ifit is a closed subgroup of GL(V ) for some vector space V . That is, a linearalgebraic group is also an affine variety.Remark 2.5. Any finite subgroup of GL(V ) is linearly algebraic. This isbecause it represents finite set of points on E(V ) × C which is closed withrespect to the Zariski topology.Reductive GroupsThe aim of this section is to investigate the reductivity of the linear algebraicgroups. Reductive groups play a special role in invariant theory. As we shallsee in the next chapter, invariant algebras for actions of these groups arealways finitely generated.Definition 2.6. A linear algebraic group G is called reductive if for anyrational representation ρ : G → GL(V ) and any 0 6= w ∈ V such thatρ(g).w = w for all g ∈ G, there exists a linear element f ∈ S(V )G withf(w) 6= 0.Proposition 2.7. If G ⊂ GL(V ) is a finite group, then it is reductive.Proof. Let w ∈ V such that g.w = w for all g ∈ G. Consider a w∗ ∈ V ∗such that w∗(w) = 1. Such a w∗ is not unique. If we apply the averagingoperator over G to w∗ we get the element1|G|∑g∈Gg.w∗ which is in V ∗. Thisis the linear element we are looking for; observe that1|G|∑g∈Gg.w∗(w) =1|G|∑g∈Gw∗(g−1.w) =1|G|∑g∈Gw∗(w) = 1 6= 0Lemma 2.8. If pi : V → W is a homomorphism of G-modules, then thesubset of Hom(W,V ) consisting of sections of pi is G-invariant. That is,the set R := {r ∈ Hom(W,V ) |pi ◦ r = idW } is G-invariant.5Proof. Let g ∈ G, r ∈ R and w ∈ W be arbitrary. (g.r)(w) = g.(r(g−1.w)),therefore (pi◦(g.r))(w) = pi(g.(r(g−1.w))) = g.pi(r(g−1.w)) = g.(g−1.w) = w.Here we used the fact that pi is a morphism of G-modules and the definitionof R.Lemma 2.9. If pi : V → W is a surjective homomorphism of G-moduleswhere G is reductive, then there exists a section r ∈ R := {r ∈ Hom(W,V ) |pi ◦ r = idW } such that g.r = r for all g ∈ G. Note that we use the inducedaction introduced in Definition 2.2.Proof. From Lemma 2.8, g.r ∈ R for any g ∈ G and r ∈ R, thereforeSpanC(g.r | g ∈ G, r ∈ R ) = SpanCR. Let f ∈ (SpanCR)∗ be definedas f(∑ni=1 λiri) =∑ni=1 λi where λi ∈ C and ri ∈ R. This is a naturaldefinition because pi◦(∑ni=1 λiri) = (f(∑ni=1 λiri)) idW . Observe that f isG-invariant; let g ∈ G be arbitrary, (g.f)(∑ni=1 λiri) = f(∑ni=1 λi(g−1.ri)) =∑ni=1 λi = f(∑ni=1 λiri) since g−1.ri ∈ R for ri ∈ R. Let R′ := ker f then R′is a G-invariant subspace of SpanCR. This gives the following short exactsequence0→ R′ → SpanCR f−→ C→ 0This is a short exact sequence because clearly f is not equal to zero and R′is a subspace of SpanCR of codimension 1 by definition. Now, by the reduc-tivity of G there exists a G-invariant element r′ ∈ SpanCR = ((SpanCR)∗)∗such that r′(f) = f(r′) 6= 0; so we can assume that f(r′) = 1; this meansr′ ∈ R since pi ◦ r′ = f(r′)idW = idW . So r′ is a G-invariant section. Hencewe are done.Definition 2.10. Let ρ : G→ GL(V ) be a representation;(i) ρ is reducible if there exists a nonzero G-stable proper subspace W .That is if there exists {0} 6= W ( V such that ρ(G).W ⊂ W . Ifthere is no such proper subspace we say that the representation ρ isirreducible.6(ii) ρ is semi-simple if for any G-stable subspace W ⊂ V there is a com-plementary G-stable subspace W ′ such that V = W ⊕W ′.Proposition 2.11. The following properties of a representation ρ : G →GL(V ) are equivalent;(i) For any irreducible G-stable subspace W of V there exists a G-stablecomplementary subspace,(ii) ρ is semi-simple,(iii) V is a direct sum of G-stable irreducible subspaces.Proof. We will use induction on dimV . For the dimV = 0 equivalenceof the statements is clear. Assume statements are equivalent for smallerdimensions.(i)⇒ (ii) Let’s take a proper G-stable subspace W of V . Let U be a nonzeroirreducible G-stable subspace of W ; U is also a subspace of V and by ourassumption (i) we have a G-stable complementary subspace U ′ such thatV = U⊕U ′. W is a G-stable subspace and for any w ∈W we have w = u+u′where u ∈ U, u′ ∈ U ′ are unique, which implies w−u = u′ ∈W since U ⊂W .Therefore we have the decomposition W = U ⊕ (W ∩U ′). Note that W andU ′ are G-stable subspaces therefore W∩U ′ is also a G-stable subspace. Thuswe have a G-stable complementary subspace for U in W . Here U was anarbitrary irreducible subspace of W therefore the statement (i) holds for W .By the induction assumption this implies ρ|W is semi-simple. To show thatρ|W is semi-simple we only used the fact that W is a G-stable subspace;therefore similar arguments show that ρ|U ′ is semi-simple. In particular wehave a decomposition U ′ = (W ∩U ′)⊕U ′′ where U ′′ is a G-stable subspace.Thus we have a decomposition V = U⊕(W ∩U ′)⊕U ′′ = W⊕U ′′ concludingthat ρ is semi-simple.(ii)⇒ (iii) If V is irreducible we are done. If not, take an irreducible G-stablesubspace U , there exists a complementary G-stable subspace W such thatV = U ⊕W by (ii). We claim that W is completely reducible. To showthat, it is enough to prove that ρ|W is semi-simple because W is a proper7subspace so we can use the induction assumption. Now, let’s take a nonzeroG-stable subspace W ′ ⊂ W . Observe that W ′ ⊂ V and ρ is semi-simple byour assumption. So we have a complementary G-stable subspace W˜ suchthat V = W ′ ⊕ W˜ . W is a G-stable subspace and W ′ ⊂ W therefore wecan say the following: W = W ′ ⊕ (W ∩ W˜ ). We know that W and W˜ areG-stable therefore W ∩ W˜ is also G-stable. Thus we have found a G-stablecomplementary subspace for W ′ in W ; meaning that ρ|W is semi-simple.Using the induction assumption, W is completely reducible, but we haveV = U ⊕W where U is irreducible. We are done.(iii)⇒ (i) V = U1 ⊕ U2 ⊕ · · · ⊕ Um where U1, U2, . . . , Um are G-stable irre-ducible subspaces of V . If we take an irreducible G-stable subspace of W wewould have the restriction of usual projection maps pii|W : W → Ui. BothW and Ui are irreducible therefore we have either W ' Ui or W ∩ Ui = 0.WLOG assume W ' Ui for 1 ≤ i ≤ k and that dimW = n. That is, wehave W ⊂ U1⊕U2⊕· · ·⊕Uk. Note that pii’s are G-module isomorphisms for1 ≤ i ≤ k, so we can define fi := pi−1i : Ui → W which is again a G-moduleisomorphism. Using these maps defineF :U1 ⊕ U2 ⊕ · · · ⊕ Uk Wu1 + u2 + . . .+ uk → f1(u1) + f2(u2) + . . .+ fk(uk)This is an G-module homomorphism;g.F (u1 + u2 + . . .+ uk) = g.(f1(u1) + f2(u2) + . . .+ fk(uk))= g.f1(u1) + g.f2(u2) + . . .+ g.fk(uk)= f1(g.u1) + f2(g.u2) + . . .+ fk(g.uk)= F (g.u1 + g.u2 + . . .+ g.uk)= F (g.(u1 + u2 + . . .+ uk))Let 0 6= w ∈W and let w = u1+u2+. . .+un in the direct sum decomposition,then fi(ui) = w by definition. This gives F (w) = kw 6= 0; that is, W ∩kerF = {0} and F is surjective. Thus if W ′ := kerF then W ⊕ W ′ =8U1 ⊕ U2 · · · ⊕ Uk. Note that W ′ is a G-stable subspace because it is kernelof a G-module homomorphism. Hence, W ′ ⊕ Uk+1 ⊕ . . .⊕ Um is a G-stablecomplementary subspace for W in the vector space V . We are done.Definition 2.12. Let G ⊂ be a linear algebraic group, a rational represen-tation of G is a homomorphism ρ : G → GL(W ) that is also morphism ofalgebraic varieties.The following proposition tells us that every rational representation of areductive group possesses properties (i), (ii) and (iii) of Proposition 2.11.Proposition 2.13. A linear algebraic group G is reductive if and only ifany rational representation of G is semi-simple.Proof. ”⇐”Let ρ : G→ GL(V ) be a rational representation. By assumptionit is semi-simple. Let w ∈ W such that ρ(G).w = w then the subspace Cwof V is a G-stable subspace. By the semi-simplicity there exists a G-stablecomplementary subspace W of V such that V = Cw ⊕W . We can define alinear function f on V such that f |W = 0 and f(w) = 1 then f isG-invariant.This is because (ρ(g).f)(v) = f(ρ(g)−1.v) and if v ∈ W then ρ(g)−1.v ∈ Wfor all g ∈ G implying that f(v) = (ρ(g).f)(v) = 0; and if v ∈ Cw theng−1.v = v again for all g ∈ G implying that f(v) = ρ(g.f(v)) = 0. Thus weare done. G is reductive.”⇒” Conversely assume that G is a reductive group and let ρ : G→ GL(V )be a rational representation also let {0} 6= W ( V be a G-stable subspace.We want to show that there exists a G-invariant complement of W . Considerthe quotient vector space V/W . There is a canonical surjection pi : V →V/W . In fact, we can regard elements of V/W as equivalence classes of Vformed by the canonical surjection as follows; for v, w ∈ V [v] = [w] if andonly if pi(v) = pi(w). Let R = {r ∈ Hom(V/W, V ) |pi ◦ r = idV/W }. Observethat V = W ⊕ r(V/W ), direct sum of vector spaces, for any r ∈ R. Alsonote that we have [r([v])] = [v] for any v ∈ V since pi(r([v])) = pi(v) = [v]by definition of r and V/W . However, of course r(V/W ) is not necessarilyG-stable. Therefore if we can find an r ∈ R such that r(V/W ) is G-stablethen we would prove the implication that ρ is semi-simple. Note that the9induced action on Definition 2.1 gives another the rational representationρ : G→ GL(V/W ) as follows: ρ(g).[v] = [ρ(g).v] or equivalently ρ(g).pi(v) =pi(ρ(g).v). From Lemma 2.9 we know that there exists a G-invariant sectionr′ ∈ R i.e. g.r′ = r′ for any g ∈ G. We claim that r′(V/W ) is a G-stablesubspace. To see that this is correct, let v ∈ V , g ∈ G be arbitrary andconsider the action of g on r′([v]) ∈ r′(V/W ) as follows;ρ(g).(r′([v])) = ρ(g).(r′(ρ(g−1)ρ(g).[v])) = ρ(g).(r′(ρ(g−1).[ρ(g).v]))= (g.r′)([ρ(g).v]) = r′([ρ(g).v]) ∈ r′(V/W )Hence, ρ is a semisimple representation.10Chapter 3Finite Generation of theAlgebra of InvariantsRecall from the introduction that one of the main question was the following:Is algebra of invariants generated by finitely many elements over C? Theanswer is yes for a particular case;Theorem 3.1. If G ⊂ GL(V ) is a reductive linear algebraic group, thenS(V )G is a C-algebra of finite type.In this chapter we will prove this theorem. In fact this will be deducedfrom a more general statement. Let’s start with recalling some definitionsand notions that will appear throughout this section.Definition 3.2. A C-algebra of finite type is a C-algebra generated byfinitely many elements.Definition 3.3. A C-algebra A is graded if A is a direct sum of its C-subspacesA =⊕d≥0Adwhere AdAe ⊂ Ad+e for every d, e ≥ 0. Nonzero elements of Ad are calledhomogeneous of degree d.11Definition 3.4. An ideal I of the graded C-algebra A, that is not A itself,is called homogeneous ifI =⊕d≥0I ∩AdThat is if a =∑d≥0 ad where ad ∈ Ad lies in I then all ad lies in I.Definition 3.5. Let A be a subring B. The A-algebra B is said to befinite type over A if there are elements b1, b2, . . . , bs ∈ B such that B =A[b1, b2, . . . , bs]Definition 3.6. Let A be a subring of B. We say that b ∈ B is integralover A if f(b) = 0 for some monic f ∈ A[x]. In other words b ∈ B is integralover A if it is root of a monic polynomial which has coefficients from A. Bis integral over A if every element of B is integral over A.Lemma 3.7. Let B be a C-algebra and let A be a subalgebra. Assume thatB is finite type over C, and is integral over A. Then A is finite type overC.Proof. Let B = C[b1, b2, . . . , bs]. Each bi satisfies a monic polynomial equa-tion with coefficients in A as followsbnii + ai1bni−1i + ai2bni−2i + . . .+ aini = 0Let a1, a2, . . . , at be the set of elements appearing as a coefficient in theseequations. Define A′ := C[a1, a2, . . . , at]. Observe that A′ is a Noetherianring because A′ is a finitely generated C-algebra. By the choice of gener-ators of A′, B is integral over A′ which means that B is generated, as anA′-module, by finitely elements of type bh11 bh22 . . . bhss where 0 ≤ hi < ni.By construction A is an A′-submodule of B. We know that B is finitelygenerated module over a Noetherian ring A′ therefore every submodule ofB is finitely generated over A′. The relationship among A, A′ and B isillustrated in the diagram below. Thus A is a finitely generated A′-modulei.e. A is finite type over A′. The lemma follows.12B|A|A′|CLemma 3.8. If G ⊂ GL(V ) is a finite group then S(V ) is integral overS(V )G.Proof. Let f ∈ S(V ); consider the polynomial ∏g∈G(x − g.f) ∈ S(V )[x].Clearly f is a root of this polynomial. Also observe that coefficient of x|G|−iis given by(−1)i∑g1,g2,...,gi∈Gall distinct(g1.f)(g2.f) . . . (gi.f) ∈ S(V )GProposition 3.9. Let G ⊂ GL(V ) be a finite group. S(V )G is a C-algebraof finite type over C.Proof. By Lemma 3.8 we can take A = S(V ) and B = S(V )G in the lemma3.7. The result follows.Now, let G ⊂ GL(V ) be a linear algebraic group. Let I and J be two G-stable ideals in S(V ). Assume that I ⊂ J . Define A := S/I and B := S/J .There exists a canonical C-algebra homomorphism as followsφ : A Bk + I → k + J13Since I and J are G-stable ideals the action of G on S induces an actionon the algebras A and B as follows; g.(k + I) = g.k + I and g.(k + J) =g.k + J which is a linear group action. That is, every element of G inducesan automorphism on the algebras A and B. In that respect we have thefollowing property for all g ∈ G;φ ◦ g = g ◦ φ : A→ BIf I and J are homogeneous ideals then A and B are graded rings. SinceG stabilizes the subspace Sd; it also stabilizes Ad and Bd. Moreover theaction of G gives representations for the vector spaces Ad and Bd which arerational representations due to the fact that G is a linear algebraic group.Let AG and BG denote the algebra of invariants in A and B.Lemma 3.10. Using the notation and setup above, assume that G is lin-early reductive. If b ∈ BG then there exists a ∈ AG such that φ(a) = b.Equivalently, φ(AG) = BG.Proof. Let b ∈ BG. The case b = 0 is trivial. Let’s assume b 6= 0. Choosea1 ∈ A such that φ(a1) = b. Since G is a linear algebraic group, the actionof G on A is rational. Therefore the following set is a finite dimensionalsubspace of A;W := SpanC(g.a1|g ∈ G) ⊂ ADefine the following subspaceW ′ := SpanC(g.a1 − a1|g ∈ G) ⊂WFirst observe that W ′ is G-invariant; therefore by Proposition 2.13 it has aG-invariant complement W ′′. The equality W = Ca1 +W ′ is clear thereforedimW ′′ = 1. This implies that W ′′ ⊂ AG. Also observe that φ(g.a1− a1) =g.φ(a1) − φ(a1) = g.b − b = 0 but φ(g.a1) = g.φ(a1) = g.b = b 6= 0 bythe commutativity of φ and g and by the fact that b ∈ BG. So this meansthat there exists a ∈ W ′′ ⊂ AG such that φ(a) = φ(a1) = b. Hence we aredone.14Lemma 3.11. Let A be a graded C-algebra and let I ( A be a finitelygenerated homogeneous ideal. If A/I is finite type over C then so is A.Proof. Let I = y1A + y2A + . . . + ykA where all yi are homogeneous ofpositive degree. Let A/I = C[a1, a2, . . . , an] where ai ∈ A and ai is itsimage under canonical surjective homomorphism pi : A → A/I. Then weclaim that A = C[y1, y2, . . . , yk, a1, a2, . . . , an]. Let x0 ∈ A then x0 = y1x1 +y2x2 + . . .+ ykxk + p(a1, a2, . . . , an) where xi ∈ A and p ∈ C[T1, T2, . . . , Tn]a polynomial with n variables. For xi ∈ A again there is an expansionxi = y1xi1 + y2xi2 + . . . + ykxik + pi(a1, a2, . . . , an) where xij ∈ A. Wecan continue in this fashion; we know that it will stop because yi’s are ofpositive degree and as we iterate degrees of xij ’s get smaller, eventually willbecome zero. Starting from the last step where every element from A is acombination of ai’s and plugging expansions back in to previous steps, weget x = P (y1, y2, . . . , yk, a1, . . . , an) where P ∈ C[T1, T2, . . . , Tn+k]. Thusour claim holds.Theorem 3.12. Let G be a linearly reductive algebraic group. Let I ( Sbe a homogeneous G-stable ideal of S, then (S/I)G is a C-algebra of finitetype.Proof. Assume that the statement does not hold. By the Noetherianity ofS, we can take a maximal G-stable homogeneous ideal I of S where (S/I)Gis not of finite type over C. Notice that defining A := S/I, by maximality ofI, for any nonzero G-invariant homogeneous ideal J of A the algebra (A/J)Gmust be finitely generated over C. We have the canonical homomorphismψ : AG → (A/J)G that is surjective by Lemma 3.10. Therefore ψ(AG) is offinite type over C. It is easy to see that ψ(AG) ∼= AG/(AG∩J). Thus for anynonzero G-invariant homogeneous ideal J of A we can say that AG/(AG∩J)is of finite type over C.Now assume that a ∈ AG is homogeneous with positive degree. Considertwo cases in here; a is not a zero divisor in A or a is a zero divisor in A.Assume the first case. Consider ax ∈ aA ∩ AG for some x ∈ A, then15g.(ax) = (g.a)(g.x) = a(g.x) and g.(ax) = ax holds for all g ∈ G by ourchoice of ax. That is a(g.x− x) = 0 for all g ∈ G and a is not a zero divisortherefore x ∈ AG holds, proving that aA ∩ AG ⊂ aAG. The other inclusionis trivial. We conclude that aA ∩ AG = aAG. Taking the homogeneousideal aA, by the previous paragraph AG/(AG ∩ aA) = AG/(aAG) is finitelygenerated over C. Let AG/(aAG) = C[a1, a2, . . . , an] where ai ∈ AG and aiis its image under canonical surjective homomorphism. Then Lemma 3.11gives AG = C[a, a1, a2, . . . , an] which contradicts the definition of A. Thuswe get a contradiction if a is not a zero divisor in A.We may now assume that any homogeneous element of positive degreebelonging to AG is a zero divisor in A. Fix such an a and define Ia := {x ∈A | ax = 0}. Clearly this is a G-invariant and homogeneous ideal and weproved that it is nonzero. Therefore (A/Ia)G ∼= AG/(AG∩Ia). We claim that(A/Ia)G is isomorphic to (aA)G as AG-module. Define φ : (A/Ia)G → (aA)Gas follows; let b ∈ A such that b ∈ (A/Ia)G; let φ(b) = ab. By choice of b,g.b − b ∈ Ia implying that g.(ab) − ab = (g.a)(g.b) − ab = a(g.b) − ab =a(g.b − b) = 0, this verifies well-definedness. Let ab1 = ab2, then we havea.(b1−b2) = 0, yielding b1 = b2, thus φ is injective. Let ab ∈ AG, then goingbackwards in the same equations above we get b ∈ (A/Ia)G proving thesurjectivity of φ. Now let x ∈ AG, x.b = xb showing that φ is an AG-modulehomomorphism. As a result of the claim, (aA)G ∼= (A/Ia)G ∼= AG/(AG∩Ia)which shows that (aA)G is a finitely generated AG-module. Recall thatAG/(aA)G ∼= AG/(aA∩AG) was finite type over C by the arguments in thefirst paragraph, and we know that aA∩AG is a finitely generated AG-module.Therefore using Lemma 3.11 we conclude that AG is finitely generated overC. Again a contradiction. Hence statement holds.Corollary 3.13. If G ⊂ GL(V ) is a reductive linear algebraic group, thenS(V )G is a C-algebra of finite type.16Chapter 4Poincare` SeriesWe introduce Poincare` series of graded algebras. This will help us to un-derstand the structure of invariant algebra. We will prove Molien’s theoremand use it to study reflection groups.Throughout this chapter for all graded C-algebras M = ⊕d∈NMd we willassume that dimCMd <∞.Definition 4.1. For a graded algebra M = ⊕d∈NMd, Poincare` series of Mis given byPT (M) =∞∑i=0(dimCMi)TiProposition 4.2. Let M , M ′, M ′′ be graded C-algebras;i. PT (M ⊗CM ′) = PT (M)PT (M ′)ii. If 0 → M ′ → M → M ′′ → 0 is an exact sequence of C-algebraswhere M ′ is an ideal of M and homomorphisms of exact sequence arecompatible with degrees, thenPT (M) = PT (M′) + PT (M ′′)In particular, if M = M ′ ⊕M ′′ then equation above holds.17Proof. i. (M ⊗CM ′)d =∑i+j=kMi ⊗CM ′j . This impliesdimC(M ⊗CM ′)d) =∑i+j=ddimC(Mi) dimC(M′j)Thus, the equality follows.ii. By the assumption on homomorphisms, 0 → M ′d → Md → M ′′d →0 is also an exact sequence. This gives dimC(Md) = dimC(M′d) +dimC(M′′d ). Using this identity, the equality follows from definition.Example 4.3. Let M = C[f ] where deg(f) = d. ThenPT (M) = 1 + Td + T 2d + T 3d + . . . =11− T dExample 4.4. Let M = C[f1, f2, . . . fn] where deg(fi) = di. Then M =C[f1]⊗C[f2]⊗ · · · ⊗C[fn]. By Proposition 4.2 (i) and previous example wehavePT (M) =(11− T d1)(11− T d2). . .(11− T dn)From now on we will focus on finite groups, their invariant algebras. andPoincare` series of those invariant algebras. Therefore it is worth making afew remarks and showing some properties for this particular case. If G ⊂GL(V ) is finite then it is a linear algebraic group, see Remark 2.5. Moreoverit is reductive from Proposition 2.7. We also know that algebra of polynomialfunctions is integral over invariant algebra in that case from Lemma 3.8.For the Poincare` series of invariant algebra of finite groups, we have avery useful theorem telling us description of it and using the theorem we canextract information about structure of invariant algebras. In particular, thistheorem will help us to establish the natural connection between invariantalgebras and finite reflection groups.Theorem 4.5. Let V be an n-dimensional C-vector space and let G ⊂18GL(V ) be a finite group. Then we havePT (SG) =1|G|∑g∈G1det(In×n − gIT ) , IT := TIn×nTo prove the Theorem, first we need to introduce new definitions andprove some lemmas using those.Definition 4.6. A linear operator g ∈ GL(V ) induces an action on S(V ), inparticular it induces a linear operator on Sd. We define gd : Sd(V )→ Sd(V )to be this induced linear operator.Lemma 4.7. Given V and g ∈ GL(V ) as above, then the following equalityholds ∞∑d=0tr(gd)Td =1det(In×n − g−1IT )Proof. Choose a basis β = {t1, t2, . . . tn} for V such that [g]β is upper-triangular. Let λ−11 , λ−12 , . . . , λ−1n ∈ C be the diagonal entries of [g]β. Sucha basis exists because C is algebraically closed. Let β′ = {T1, T2, . . . , Tn}be the basis of V ∗ where Ti = t∗i . Clearly S(V ) = C[T1, T2, . . . , Tn]. Ob-serve that [g1]β′ = [g]−1β . This shows that [g1]β′ is also an upper-triangularmatrix with diagonal entries λ1, λ2, . . . , λn ∈ C. Now for Sd(V ) we havethe basis βd := {T e11 T e22 . . . T enn | e1 + e2 + . . . + en = d}. Consider [gd]βdwhere βd has lexicographic order; it is an upper-triangular matrix becauseg.Ti ⊂ SpanC(T1, T2, . . . , Ti). Moreover diagonal entries of [gd]βd is given by{λe11 λe22 . . . λenn | e1 + e2 + . . .+ en = d}. Thus we gettr(gd) =∑e1+e2+...+en=dλe11 λe22 . . . λennIt follows from that19∞∑d=0tr(gd)Td =∞∑d=0 ∑e1+e2+...+en=dλe11 λe22 . . . λennT d=(∑e1λe11 Te1)(∑e2λe22 Te2)· · ·(∑enλen1 Ten)=11− λ1T11− λ2T · · ·11− λnT =1det(In×n − g−1IT )Lemma 4.8. Let G be a finite group and let V be a G-module; then we havedimC VG =1|G|∑g∈Gtr(g)Proof. Here Maschke’s theorem applies therefore V is completely reducible.Completely reducibility can be also shown using Proposition 2.11. V G isthe multiplicity of the trivial 1-dimensional representation in V . Eachirreducible one dimensional subspace gives us trivial representation. Letε : G→ C be the character of trivial representation; ε(g) = 1 for all g ∈ G.If we let χ : G → C to be character of the corresponding representationρ : G → GL(V ) i.e. χ(g) := tr(g) for all g ∈ G, then taking the standardinner product of χ with ε gives the number of trivial representations in ρ.Thus we write〈χ, ε〉 = 1|G|∑g∈Gχ(g)ε(g) =1|G|∑g∈Gχ(g) =1|G|∑g∈Gtr(g) = dimC VGNow we can prove Theorem 4.5.Proof. It is already established that Sd(V ) is a G-module under the action20of gd where g ∈ G and d ∈ N. Therefore by Lemma 4.8 we have the identitydimSGd (V ) =1|G|∑g∈Gtr(gd)Substituting this in the Poincare` series and applying Lemma 4.7 gives thedesired identity.PT (S(V )G) =∞∑d=0dimSGd (V )Td =∞∑d=01|G|∑g∈Gtr(gd)Td=1|G|∑g∈G( ∞∑d=0tr(gd)Td)=1|G|∑g∈G1det(In×n − g−1IT )=∑g∈G1det(In×n − gIT )Definition 4.9. An element g ∈ GL(V ), which is not identity, is called areflection if it is diagonalizable, of finite order and fixes a hyperplane Hg ⊂ Vpointwise.Lemma 4.10. With the notation above,PT (S(V )G) =1|G|(1(1− T )n +Cn−1(1− T )n−1 + · · ·)where 2Cn−1 = number of reflections in G.Proof. PT (S(V )G) is a finite sum of rational polynomial functions. We con-sider the Laurent series expansion of PT (S(V )G) at 1 ∈ C. Let λ 6= 1; thenwe expand 11−λT around 1 as follows;11− λT =1(1− λ)− λ(T − 1) =11− λ11− λ1−λ(T − 1)=1(1− λ)(1 +λ1− λ(T − 1) +(λ1− λ)2(T − 1)2 + · · ·)21This shows that only contribution to the principal part of Laurent seriescomes from elements in G which has 1 in the diagonal of its upper-triangularmatrix form. Therefore PT (S(V )G) has a pole of order n at 1 coming fromIn ∈ G ⊂ GL(V ). Note that only contribution to the term 1(1−T )n comesfrom identity element in G. To see this, assume that there exists an ele-ment g ∈ G other than identity such that all diagonal entries in the upper-triangular form of its matrix is equal to 1. Then minimal polynomial of gis (x− 1)k = 0 for some integer k ≥ 2. However g|G| = In and (x− 1)k = 0does not divide x|G| − 1 = 0 which is a contradiction since x − 1 is notminimal polynomial of g. Arguing similarly, observe that for any g ∈ Gminimal polynomial of g divides x|G| − 1 which has |G| many distinct com-plex roots,namely roots of unity; therefore minimal polynomial of g doesnot contain any linear polynomial with multiplicity more than one. This,together with the expansion above shows that only contribution to the term1(1−T )n−1 in the Laurent expansion comes from reflections. Thus, for a re-flection s with the eigenvalue ζs 6= 1, that contribution is 11−ζs .This tellsusCn−1 =∑s∈Gs reflection11− ζsFurthermore observe that if s ∈ G is a reflection then so is s−1 ∈ G and wehave the identity11− ζs +11− ζ−1s= 1Note that if ζs = −1 then s = s−1 and this identity still holds. Moreoverthis also givesCn−1 =∑s∈Gs reflection11− ζ−1sThe lemma follows from these three identities2Cn−1 =∑s∈Gs reflection11− ζs +∑s∈Gs reflection11− ζ−1s=∑s∈Gs reflection122Here G is linearly reductive and by Theorem 3.12 SG is finitely gener-ated over C. Moreover, S is integral over SG therefore Noether normaliza-tion Theorem tells us that there exists n algebraically independent elementsf1, f2, . . . , fn such that SG is integral over C[f1, f2, . . . , fn]. This gives us anidea of the Poincare` series.Proposition 4.11. Let M be a graded C-algebra that is integral overC[f1, f2, . . . , fn] with fi’s algebraically independent and deg fi = di. Thenthere exists a polynomial h(T ) ∈ C[T ] such thatPT (M) = h(T )n∏i=111− T diProof. We prove this statement by induction on n. If n = 0 then PT (M) = 1which is a polynomial. Assume that the statement holds for algebras thatare integral over polynomial algebras with fewer than n generators. Considerthe C-algebra homomorphismpi : M →Ms → sfnThis gives C-algebra isomorphism M/kerpi ∼= Im pi. By definition of pi wethen have (M/kerpi)d ∼= (Im pi)d+dn . ThereforePT (M/kerpi) = T−dnPT (Im pi)We have exact sequences of C-algebras that are compatible with degrees0→ kerpi →M →M/kerpi → 00→ Im pi →M →M/Im pi → 0Clearly kerpi and Im pi are ideals of M so we can apply Proposition 4.2 part23(i). This and a previous remark givesPT (M) = PT (kerpi) + PT (M/ kerpi) = PT (kerpi) + T−dnPT (Im pi) (4.1)PT (M) = PT (Im pi) + PT (M/Im pi) (4.2)Multiplying (4.1) by T dn and subtracting from (4.2) gives(1− T dn)PT (M) = PT (M/Im pi)− T dnPT (kerpi) (4.3)We will show that we can use induction argument for the RHS of (4.3).Firstly we claim that M/Im pi is integral over C[f1, f2, . . . , fn−1] where fi ∈M/Im pi is the image of fi under the canonical projection map. Let s ∈ SG,it is integral over C[f1, f2, . . . , fn] by assumption which gives us a monicpolynomial∑ksi=0 aiXi where ai ∈ C[f1, f2, . . . , fn] such that s is a root.Using this polynomial, construct the monic polynomial∑ksi=0 aiXi; clearlys is a root for this and ai ∈ C[f1, f2, . . . , fn−1] since fn = 0 so our claimholds. Therefore induction hypothesis is valid for PT (M/Im pi). That is,there exists a polynomial h1(T ) ∈ C[T ] such thatPT (M/Im pi) = h1(T )n−1∏i=111− T diSimilarly, any element s ∈ kerpi is integral over C[f1, f2, . . . , fn−1] becausesfn = 0. Therefore induction hypothesis also holds for kerpi which givesPT (kerpi) = h2(T )n−1∏i=111− T difor some polynomial h2(T ) ∈ C[T ]. Plugging in last two expansions anddividing both sides of (4.3) by (1− T dn) gives us the desired result.Corollary 4.12. Let G ⊂ GL(V ) be finite and SG is integral over C[f1, f2,. . . , fn] with fi’s algebraically independent and deg fi = di. Then there exists24a polynomial h(T ) ∈ C[T ] such thatPT (SG) = h(T )n∏i=111− T diIn the next chapter we will show that SG is not just integral over C[f1, f2,. . . , fn] for some fi ∈ SG algebraically independent but also equal to it if Gis finitely generated reflection group. We will also show that the converse ofthis statement holds, thus we have a characterization. The following lemmahelps us to prove converse of the statement in the next chapter. We put itin this chapter because we use the arguments in this chapter.Lemma 4.13. Suppose SG = C[f1, f2, . . . , fn] where deg fi = di. Theni. |G| = d1d2 . . . dnii. the number of reflections in G is∑ni=1(di − 1)Proof. It follows from Example 4.4 thatPT (SG) =(11− T d1)(11− T d2). . .(11− T dn)=1(1− T )nn∏i=1(1 + T + T 2 + . . .+ T di−1)On the other hand from the Lemma 4.10PT (SG) =1|G|(1(1− T )n +Cn−1(1− T )n−1 + · · ·)and 2Cn−1 is number of reflections in G. Multiplying both expressions by(1− T )n gives the following equality1∏ni=1(1 + T + . . .+ Tdi−1)=1|G| (1 + Cn−1(1− T ) + . . .) (4.4)25Rewriting the same equality as follows|G| = (1 + Cn−1(1− T ) + Cn−2(1− T )2 + . . .) n∏i=1(1 +T +T 2 + . . .+T di−1)Plug in T = 1 and we get |G| = d1d2 . . . dnFor the second part take derivative of (4.4) and we get1∏ni=1(1 + T + T2 + . . .+ T di−1)(−n∑i=11 + 2T + · · ·+ (di − 1)T di−21 + T + · · ·+ T di−1)=1|G| (−Cn−1 − 2Cn−2(1− T )− · · · )Again evaluating both sides at T = 1 gives1d1d2 . . . dn(−n∑i=1di(di − 1)di)= − 1|G|Cn−1After using the identity in part i. and simplification this gives Cn−1 =∑ni=1(di − 1).26Chapter 5Invariant Algebras of FiniteReflection GroupsIn this chapter I denotes the ideal of S generated by homogeneous elementsof SG of positive degree.Lemma 5.1. Let (eα)α∈A be a set of homogeneous elements from S suchthat (eα)α∈A is a basis for the vector space S/I, then S is generated by(eα)α∈A as an SG-module.Proof. Let M be the SG-module generated by elements (eα)α∈A. We wantto show that M = S, in particular Md = Sd. We apply induction on d. Thebase case d = 0 is clear as M0 = S0 = C. Assume the lemma holds for Se,where e < d and let f ∈ Sd. Considering f ∈ S/I and our assumption, wecan expand f in S as follows;f =∑αcαeα +∑βfβiβwhere the sums are finite with cα ∈ C, iβ ∈ SG and fβ ∈ S homogeneouswith degree smaller than d. By the induction assumption fβ ∈M thereforef ∈Md.Definition 5.2. A finite group G ⊂ GL(V ) is called a finite reflection groupif it is generated by reflections.27Proposition 5.3. Let G be a finite reflection group and let y1, y2, . . . , ynbe homogeneous elements of S such that y1, y2, . . . , yn ∈ S/I linearly inde-pendent in S/I with the previous notation. Then y1, y2, . . . , yn are linearlyindependent over SG.The proof of this proposition relies on Lemma 5.5 which uses the maps∆s : S(V ) → S(V ) associated to reflections s in G. This map has a nicemultiplicative property. We deal with the maps first then prove the lemma.∆ mapsLet s ∈ G be a reflection; s fixes a hyperplane of V , call this Hs. Thereexists a nonzero element ls ∈ V ∗ such that ls(Hs) = 0 and such an ls isunique up to a nonzero constant. Let ζs be the eigenvalue of s differentfrom 1 and let vs ∈ V be the eigenvector of ζs. Now take ls so thats.v = v + ls(v)vsThat is, choose it so that ls(vs) = ζs− 1. Recall that if s is a reflection thenso is s−1 with eigenvalue ζ−1s different from 1. It follows thats−1.v = v − ζ−1s ls(v)vsLet f ∈ S and v ∈ V be arbitrary, then we have(s.f)(v) = f(s−1.v) = f(v − ζ−1s ls(v)vs) = f(v)− ζ−1s ls(v)f(vs)This shows ls divides s.f − f for all f ∈ S. This allows us to writes.f = f + ls · (∆s(f))This is how ∆s is defined. It is apparent from the definition that ∆s mapsSd into Sd−1.Lemma 5.4. For the map ∆s : S(V ) → S(V ) we have the multiplicative28identity for any f, g ∈ S∆s(fg) = f · (∆s(g)) + g · (∆s(f)) + ls · (∆s(f)) · (∆s(g))Proof. This follows froms.(fg) = (s.f) · (s.g) = (f + ls · (∆s(f)) · (g + ls · (∆s(g))Lemma 5.5. Let xi ∈ SG and yi ∈ S for 1 ≤ i ≤ m be homogeneouselements such that x1y1 + x2y2 + . . . + xmym = 0. If x1 /∈ SGx2 + SGx3 +. . .+ SGxm then y1 ∈ I.Proof. We will apply induction on d = deg y1. We will denote the Reynoldsoperator by P = |G|−1∑g∈G g. Consider the base case d = 0. We havey1 /∈ I because I is generated by elements which has positive degree. Forthe base case we are trying to prove contrapositive of the statement. Noticethat x1 = x2z2 + x3z3 + . . . + xmzm where zi = yi/y1. Applying Reynold’soperator on both sides we get; x1 = x2P (z2) + x3P (z3) + . . .+ xmP (zm) ∈SGx2 + SGx3 + . . . + SGxm, since P is a projection onto SG. Now assumethe assertion for lower degrees and let d = deg y1. We are trying to provethe statement itself. Fixing a reflection s ∈ G, apply the ∆s operator tothe equality x1y1 + x2y2 + . . . + xmym = 0. We have ∆s(xi) = 0 becauses.xi = xi; combining this with the multiplicative identity for ∆s we getx1∆s(y1) + x2∆s(y2) + . . .+ xm∆s(ym) = 0. deg ∆s(y1) < deg y1, thereforeby induction assumption ∆s(y1) ∈ I; implying that s.y1 − y1 is in I. Thisholds for any reflection s ∈ G. Consider g.y1 − y1; g = s1s2 . . . sk forsome reflections s1, s2, . . . , sk ∈ G using this we can rewrite g.y1 − y1 as thefollowing telescoping sum:s1s2 . . . sk−1(sk.y1 − y1) + s1s2 . . . sk−2.(sk−1y1 − y1) + . . .+ (s1y1 − y1) ∈ Iknowing that siy1 − y1 ∈ I. That is g.y1 − y1 ∈ I for any g ∈ G. ThereforeP (y1) − y1 ∈ I, which in turn implies that y1 ∈ I because P (y1) ∈ SG and29degP (y1) > 0. This establishes the claim.Proof of Proposition 5.3. We use induction on n. Base case is n = 1 andit is trivial. Assume that statement holds numbers smaller than n. Letx1y1 + x2y2 + · · · + xnyn = 0 where xi ∈ SG. By the previous lemma andthe fact that y1 /∈ I we are able to write x1 = z2x2 + z3x3 + · · ·+ znxn withzi ∈ SG. Thus the former equality becomesx2(y2 + z2y1) + . . .+ xn(yn + zny1) = 0Observe that y2 + z2y1 = y2 since deg z2 > 0 i.e. z2 ∈ I (otherwise weget a contradiction with the assumption that y1, y2, . . . , yn are linearly in-dependent in S/I). This tells us that y2 + z2y1, . . . , yn + zny1 are linearlyindependent and by induction assumption y2 + z2y1, . . . , yn + zny1 are lin-early independent over SG. That is x2 = x3 = · · · = xn = 0. This makesx1 = 0. Thus we are done.Theorem 5.6. The following are equivalent for a finite group G ⊂ GL(V )where dimV = n:1. G is a finite reflection group;2. S is a free graded module over SG with a finite basis;3. SG is generated by n algebraically independent homogeneous elements.Proof. 1 ⇒ 2 : Define the quotient field of S, call it K; and the quotientfield of SG, call it KG. We can view G as a finite group of automorphismson K. In this case KG is the fixed field of G therefore K is a finite extensionof KG (see [7] pg.264). Now let the set (eα)α∈A be as defined in Lemma 5.1;Proposition 5.3 together with Lemma 5.1 gives us that S is a free moduleover SG with basis (eα)α∈A. So we only need to show that the basis is finite.To see that, observe that (eα)α∈A is also a basis for K over KG. Let ab ∈ K,30a, b ∈ S be arbitrary; thenab=a∏g∈Gg 6=1Gg.b∏g∈Gg.bDenominator is in SG and nominator can be rewritten as finite SG-linearcombination of (eα)α∈A’s. That is, any element in K is a KG-linear combi-nation of (eα)α∈A’s. Moreover the set (eα)α∈A is clearly linearly independentover KG by Proposition 5.3. Here we already know that K is a finite exten-sion of KG therefore this information is enough to conclude that (eα)α∈A isa basis for K over KG. In fact degree of extension is |G|, i.e. basis consistsof |G| many elements(see [7] pg.264).2⇒ 3: S is integral over SG. It follows from Lemma 3.7 that SG is finitetype over C because S is finite type over C. Let {f1, f2, . . . , fm} be the set ofgenerators with minimum cardinality where every element is homogeneous.That is SG = C[f1, f2, . . . , fm]. Now we will prove that these elements arealgebraically independent.Assume that f1, f2, . . . , fm are not algebraically independent. Then thereexists a polynomial P (x1, x2, . . . , xm) ∈ C[x1, x2, . . . , xm] such thatP (f1, f2, . . . , fm) = 0Assume that P has minimal degree and that each monomial in P hasthe same weight. Note that weight of a monomial xe11 xe22 . . . xemm is given byd1e1 + d2e2 + . . .+ dmem where di = deg fi. We write S = C[T1, T2 . . . , Tn]and letPi =∂P∂xi(f1, f2, . . . , fm) (1 ≤ i ≤ m) and fij = ∂fi∂Tj(1 ≤ i ≤ m, 1 ≤ j ≤ n)Observe that Pi ∈ SG and not all of them are zero since P 6= 0.Consider all monomials xe11 . . . xeii . . . xemm in P with ei 6= 0; then ∂P∂xi issum of eixe11 . . . xei−1i . . . xemm and all this monomials have the same weight;31d1e1 + . . .+ di(ei − 1) + . . .+ dmem . Therefore Pi’s are sums of monomialsof the same weight evaluated at (f1, f2, . . . , fm); which tells us that Pi’s arehomogeneous elements of SG. Consider the ideal (P1, P2, . . . , Pm) ⊂ S andassume that {P1, P2 . . . , Pk} is a minimal set of generators for this ideal.WritePj =k∑i=1sijPi , sij ∈ S , k + 1 ≤ j ≤ m (5.1)If we differentiate P (f1, f2, . . . , fm) = 0 with respect to Tj , the chainrule givesm∑i=1Pifij = 0Substituting (5.1) for Pi where k + 1 ≤ i ≤ m we getk∑i=1Pi · (fij +m∑l=k+1silflj) = 0Putuij = fij +m∑l=k+1silflj (5.2)This givesk∑i=1Piuij = 0 (5.3)Now since S is a free SG-module, we can choose a homogeneous basis ersuch thatS =⊕rSGerIn particular, it can be arranged so that er = 1 for some r. That is, we canarrange so that SG is a summand in this direct sum.Now we claim that uij ∈ (f1, f2, . . . , fm) ⊂ S. We know that uij ∈ S,therefore we have the decompositionuij =∑rρijrer , ρijr ∈ SG (5.4)32Plug this expansion into (5.3)0 =k∑i=1Piuij =k∑i=1Pi · (∑rρijrer) =∑r(k∑i=1Piρijr)erThis tells us that∑ki=1 Piρijr = 0 for any j, r. Here ρijr are viewed aspolynomials with variables T1, T2, . . . , Tn and they can be taken to be ho-mogeneous. This tells us that for these polynomials, if the constant termis nonzero then they are constants. Observe that none of ρijr is constantbecause otherwise we get a contradiction with the minimality of the gen-erating set {P1, P2 . . . , Pk}. We see that ρijr ∈ SG = C[f1, f2, . . . , fm] andhas constant term zero, therefore it is equal to combination of f1, f2, . . . , fm,plugging this expansion of ρijr into the (5.4) we see that uij belongs to theideal of S generated by f1, f2, . . . , fm. Claim is proved.We have the Euler identity;difi =n∑j=1fijTj , where di = deg fiUsing uij and Euler identity again along the way, we can rewrite RHS asfollows;difi =n∑j=1(uijTj −m∑l=k+1silflj)Tj =n∑j=1uijTj −n∑j=1m∑l=k+1silfljTj=n∑j=1uijTj −m∑l=k+1sil · (n∑j=1fljTj) =n∑j=1uijTj −m∑l=k+1sil · (dlfl)=n∑j=1uijTj −m∑l=k+1dlsilflWe know that uij can be written as S-linear combination of f1, f2, . . . , fmand clearly the second term on the RHS is also S-linear combination off1, f2, . . . , fm. Therefore, expanding RHS and taking the degree di compo-33nent of both sides, we getfi =∑i 6=jλjfj , λj ∈ SWe have fi ∈ SG and er = 1 for some r. Now if we expand the coefficients λjwith respect to decomposition S =⊕rSGer and take the component lyingin the summand SG; we get the following relationfi =∑i 6=jλ¯jfj , λ¯j ∈ SGNote that none of the λ¯j ’s contains fi as a component because we only tookdegree di component of each side earlier. Thus, this relation contradictsthe choice of {f1, f2, . . . , fm}. This set is algebraically independent. Thequotient field of S is integral over the quotient field of SG and transcendencedegree former field is n, therefore m = n.3 ⇒ 1: Assume SG = C[f1, f2, . . . , fn] with di = deg fi. Let H be thesubgroup of G generated by reflections of G. H is a finite reflection groupand we proved that SH = C[g1, g2, . . . , gn] where ei = deg gi and gi’s arehomogeneous. We can assume thatd1 ≤ d2 ≤ . . . ≤ dn and e1 ≤ e2 ≤ . . . ≤ enWe claim that di ≥ ei for all i. To see that first observe SG ⊂ SH .Now assume di < ei, this implies that fi can be written as a polyno-mial in {g1, g2, . . . gi−1}. Since di−1 ≤ di, fi−1 can be written as a poly-nomial in {g1, g2, . . . , gi−1}. In general we find that there is an inclu-sion C[f1, f2, . . . , fi] ↪→ C[g1, g2, . . . , gi−1] which is a contradiction with{f1, f2, . . . , fi} being algebraically independent.Both G and H contain the same number of reflections by the choice ofH; and by the second part of Lemma 4.13 the number of reflections in Gand H respectively are∑ni=1(di − 1) =∑ni=1(ei − 1).Combining these two, we see that di = ei for all i. By the first part of34Lemma 4.13 |G| = ∏ni=1 di and |H| = ∏ni=1 ei. Having H ≤ G, we concludethat H = G. Thus G is a finite reflection group.35Chapter 6Classification and ExamplesProposition 6.1. Suppose G ⊂ GL(V ) is a finite reflection group, thenthere exists a decomposition V = V1 ⊕ V2 ⊕ · · · ⊕ Vr and a decompositionG = G1 × G2 × . . . × Gr such that Gi ⊂ GL(Vi) where Vi is an irreducibleG-invariant subspace and Gi is a finite reflection group.Proof. Let H := {g1, . . . , gn} where gi’s are reflections and G = 〈H〉. G is afinite group therefore reductive therefore V is completely reducible. i.e wehave G-invariant irreducible subspaces Vi such that V = V1 ⊕ V2 ⊕ · · · ⊕ Vr.Take a reflection g ∈ G with eigenvalues 1 and ζ. Observe that g.Vi = Vitherefore Vi should be generated by eigenvectors of g. If Vi doesn’t containthe eigenvector of g corresponding to the eigenvalue ζ, then it fixes Vi. Ob-serve also that there is only one irreducible subspace Vi that g doesn’t fix.For that Vi, g|Vi has eigenvalues +1, ζ corresponding to (dimVi − 1) and1 many eigenvectors respectively. That is, g|Vi ∈ GL(Vi) is a reflection inthat case. This property gives a partition on H. Define Gi := 〈{g ∈ H|g doesn’t fix Vi}〉. Clearly for 1 ≤ i ≤ r, Gi is a finite reflection groupand Gi ⊂ GL(Vi) by previous remark. Note that Gi ∩ G1G2..Ĝi..Gr = 1Gby construction. Take two arbitrary elements gi, gj such that gi ∈ Gi,gj ∈ Gj . We claim that they commute. Take an arbitrary element v ∈ V ,we have decomposition v = v1 + v2 + . . . + vr where vi ∈ Vi. Observethat gj fixes every component but vj and similarly gi fixes every componentbut vi because of the choice of gi and gj . Therefore we have (gigj).v =36gi.(gj .(v1 + v2 + . . .+ vr)) = gi.(v + gj .vj − vj) = v + gj .vj − vj + gi.vi − vi.Similarly we have (gjgi).v = gj .(gi.(v1 +v2 + . . .+vr)) = gj .(v+gi.vi−vi) =v + gi.vi − vi + gj .vj − vj . Thus gigj = gjgi. Next we claim that there is anisomorphism ϕ : G1×G2×. . .×Gr → G. Define ϕ(g1, g2, . . . gr) = g1g2 . . . gr.It is well defined; let ϕ(g1, g2, . . . gr) = g1g2 . . . gr = g and ϕ(g′1, g′2, . . . g′r) =g′1g′2 . . . g′r = g′ then gg′ = g1g′1g2g′2 . . . grg′r and g′g = g′1g1g′2g2 . . . g′rgr.This map is surjective; take any element g in G. It can be written asproduct of reflections from H i.e g = h1h2 . . . , hk for some k ∈ N andh1, h2 . . . hk ∈ H. Using the commutativity of the elements belonging to dif-ferent Gi’s, we can regroup h1, h2, . . . , hk so that we get g = g1g2 . . . gr wheregi ∈ Gi. ϕ is also injective; assume g = g1g2 . . . gr and also g = g′1g′2 . . . g′rthen multiplying both products by g′−11 g−12 . . . g−1r and using commutativ-ity gives g1g′−11 = g′2g−12 . . . g′rg−1r . LHS belongs to G1 whereas RHS be-longs to G2G3...Gr therefore both are equal to 1G which proves the injectiv-ity. Thus ϕ gives an isomorphism. To establish the last statement, defineφi(gi) := φ(gi)|Vi for g ∈ Gi. The map φi : Gi → GL(Vi) gives a reflectiongroup by a previous remark. Now take an arbitrary element v ∈ V , we havedecomposition v = v1 + v2 . . . + vr where vi ∈ Vi. Take an arbitrary g ∈ Gwe have decomposition g = g1g2 . . . gr where gi ∈ Gi. Therefore we have fol-lowing equalities: g.v = φ(g).v = φ(g1g2 . . . gr).v = φ(g1)φ(g2) . . . φ(gr).v =φ(g1)φ(g2) . . . φ(gr).(v1 + v2 . . .+ vr) = φ(g1).v1 +φ(g2)v2 . . . φ(gr)vr. Hencethe result follows.Definition 6.2. Let G ⊂ GL(V ) be a finite reflection group. If V is anirreducible G-module then we call G an irreducible finite reflection group.In that case the rank of G is given by dimC V .It follows from this proposition that any complex finite reflection groupis a product of irreducible complex finite reflection groups. So it is enoughto study irreducible complex finite reflection groups and their invariant alge-bras. This classification is due to Shephard and Todd in 1954 [2]. It consistsof 3 infinite families and 34 exceptional groups. The 3 infinite families aresymmetric groups as will be shown in the following proposition, cyclic groups37Z/mZ and imprimitive reflection groups which is defined below.Proposition 6.3. Let V = Cn and Sn permutes the coordinates (x1, x2,. . . , xn). Define the hyperplane W := {(x1, x2, . . . , xn) ∈ Cn|n∑i=1xi = 0}.Sn stabilizes W and therefore Sn ⊂ GLn−1(C). Using this inclusion, Snis a finite reflection group. Moreover S(W )Sn = C[f2, f3, . . . , fn] wheref2, f3, . . . , fn ∈ S(W ) with deg fi = i.Proof. By definition we have S(W ) ∼= C[x1, x2 . . . , xn]/〈x1 + x2 + . . .+ xn〉.Let pi : S(V ) ∼= C[x1, x2, . . . , xn]→ S(W ) ∼= C[x1, x2, . . . , xn]/〈x1+x2+. . .+xn〉 be the natural projection. We know that S(V )Sn = C[f1, f2, . . . , fn]where fi are the elementary symmetric polynomials of degree i (see [7]pg. 191). That is we have S(W ) ∼= C[x1, x2, . . . , xn]/〈f1〉. Let fi :=pi(fi). Note that we have σ.fi = σ.fi for any σ ∈ Sn. From this weconclude that C[f2, f3, . . . , fn] ⊂ S(W )Sn . Now let’s take an arbitraryelement f ∈ S(W )Sn . We claim that there exists a g ∈ pi−1(f) suchthat g ∈ S(V )Sn . Clearly f ∈ pi−1(f). Averaging f over Sn, we obtaing := 1|Sn|∑σ∈Sn σ.f . We had that σ.f = σ.f for any σ ∈ Sn thereforewe have g = f . We also have g ∈ S(V )Sn . So our claim holds. Weknow that g = P (f1, f2, . . . , fn) for some polynomial P ∈ C[T1, T2, . . . , Tn]which means g = P (0, f2, . . . , fn) = f i.e. f ∈ C[f2, f3, . . . , fn]. ThusS(W )Sn ⊂ C[f2, f3, . . . , fn] holds as well.Clearly deg fi = i. Hence we aredone.Proposition 6.4. Let W ⊂ Cn be as in the previous proposition. Defineφ : Sn × Z/2Z → GL(W ) where the action on Cn is defined as follows;Sn permutes coordinates (x1, x2, . . . , xn) and Z/2Z takes (x1, x2, . . . , xn) to±(x1, x2, . . . , xn). The group Sn × Z/2Z is generated by reflections only forn = 2, 3, 4.Proof. We know that Sn is generated by transpositions and we know fromthe Proposition 6.3 that they are reflections. Therefore (σ,+1) is a reflectionwhere σ is a transposition . This tells us that any (pi, 1) where pi ∈ Sn isproduct of reflections. Similarly, if we can find an element τ ∈ Sn such that38(τ,−1) is a reflection then (piτ−1,+1)(τ,−1) = (pi,−1) is product of reflec-tions for an arbitrary pi ∈ Sn . We conclude that Sn × Z/2Z is generatedby reflections if and only if it contains a reflection of the form (τ,−1) forsome τ ∈ Sn. Let τ ∈ Sn, (τ,−1) is a reflection if and only if φ(τ) ∈ GL(W )has characteristic polynomial (x − 1)n−2(x + 1). If we consider the usualrepresentation where the codomain is GLn(C) , the range would be plusand minus n-dimensional permutation matrices. (τ,−1) takes (c, c, . . . , c)to −(c, c, . . . , c) so τ being a reflection in the previous representation φ isequivalent to image of τ having a characteristic polynomial (x−1)n−2(x+1)2in the usual representation. It is easy to examine the characteristic poly-nomials of permutation matrices because two permutations are conjugate ifand only if they have the same cycle structure. Let σ ∈ Sn be an n-cycle,then it is conjugate to the cycle (12 . . . n) i.e. (12 . . . n) = κσκ−1 for someκ ∈ Sn therefore the images of (σ,−1) and ((12 . . . n),−1) in the usual rep-resentation are similar. Thus the characteristic polynomials of images arethe same. Let’s compute the characteristic polynomial of ((12 . . . n),−1);the image of ((12 . . . n),−1) in the usual representation is given by0 −1 0 . . . 00 0 −1 . . . 0.......... . ....0 0 0 . . . −1−1 0 0 . . . 0n×n(6.1)Therefore characteristic polynomial is given as follows:∣∣∣∣∣∣∣∣∣∣∣∣∣x 1 0 . . . 0 00 x 1 . . . 0 0.......... . .......0 0 0 . . . x 11 0 0 . . . 0 x∣∣∣∣∣∣∣∣∣∣∣∣∣= xn − (−1)n =xn − 1 if n is evenxn + 1 if n is odd (6.2)This can be used to write a general formula for every permutation in Sn.39If there exists cl many disjoint l-cycles in an arbitrary permutation pi ∈ Snthen pi is conjugate to(12 . . . l1)((l1 + 1) . . . l2) . . . ((ln−1 + 1) . . . ln) (6.3)for some l1, . . . , ln depending on cl’s. That is, if we have a disjoint l-cyclethen it appears as a l × l block matrix of the form (6.1) on the diagonal ofn× n matrix. Therefore the characteristic polynomial is given asn∏l=1(xl − (−1)l)cl (6.4)Now it only remains to analyse for which values of n we can get (x−1)n−2(x+1)2 with a suitable choose of cl’s using the (6.4).τ should only contain 2-cycles to make (τ,−1) a reflection. Otherwise wewould get a root different than ±1 in the characteristic polynomial of (τ,−1).Also observe that it cannot have more than two 2-cycles because in thatcase (x + 1)3 divides the characteristic polynomial. Now let’s look at theremaining possibilities case by case. Our first case is that τ consists of 22-cycles. τ cannot fix an element because otherwise (x + 1)3 divides thecharacteristic polynomial of (τ,−1). Therefore the only possibility where(τ,−1) is a reflection in this case is where n is equal to 4. Our second caseis that τ is a transposition. In this case if n > 3 then τ fixes at least 2elements which means (x2−1)(x+ 1)2 divides the characteristic polynomialtherefore (x + 1)3 divides the characteristic polynomial of (τ,−1). If nis 3 then (τ,−1) is a reflection because it has characteristic polynomial(x2 − 1)(x + 1) = (x + 1)2(x − 1). If n is 2 then (τ,−1) would also be areflection on GL1(C) trivially. Our third case is that τ fixes all elements. Ifn > 2 then again (x+ 1)3 is a factor in characteristic polynomial of (τ,−1)so it is not a reflection. If n is 2 then it is a reflection. We have alreadyestablished that if n is 2 then the given group is generated by reflections.We conclude that Sn×Z/2Z is generated by reflections if and only if n = 2,3 or 4, as desired.40Proposition 6.5. A permutation group G ⊂ Sn is generated by reflectionsif and only if it is direct product of symmetric groups of smaller dimension.Proof. The only reflections in Sn are transpositions. It is clear from thecharacteristic polynomials coming from the natural representation. So anyG ≤ Sn reflection group is generated by transpositions. Fixing such G definean equivalence relation on the set {1, 2 . . . , n} as follows: i ∼ j if and onlyif (ij) ∈ G. Let’s check the well-definedness. It is reflexive; (ii) = 1Sn ∈ G.Clearly (ij) = (ji) so it is a symmetric relation. Assume i ∼ j and j ∼ ki.e (ij), (jk) ∈ G then (ij)(jk)(ij) = (ik) belongs to G therefore we havej ∼ k. That is our relation is also transitive, thus an equivalence relation.This gives a partition on the set {1, 2 . . . , n}. Let O1, O2, . . . , Or be theset of all equivalence classes. If (ab) ∈ G then a, b ∈ Ok for some k with1 ≤ k ≤ r. Define Gk := 〈(ij) ∈ G | i, j ∈ Ok〉. Gk is a subgroup ofG. Let σi ∈ Gk, σj ∈ Gl, then we have σiσj = σjσi because any elementin Gk fixes all elements from all equivalence classes except Ok. Thereforewe have following three properties by definition and previous remarks; G =G1G2 . . . Gr; GkGl = GlGk; Gk ∩ G1G2..Ĝk..Gr = {1Sn}. These threetogether gives G ∼= G1×G2× . . .×Gr. Next we claim that Gk ∼= S|Ok|. LetOk = a1, a2, . . . , a|Ok|. Define the isomorphism as follows: ϕ : Gk → S|Ok|where ϕ(aman) = (mn). Clearly this is a bijective homomorphism of groups.Thus such a G is isomorphic to S|O1| × S|O2| × . . .× S|Or|41Bibliography[1] C.Chevalley, Invariants of a group in an affine ring, 77 (1955), 778–782.[2] J.A.Todd G.C.Shephard, Finite unitary reflection groups, 6 (1954), 274–304.[3] H.Weyl, Theorie der darstellung kontinuierlichen halb einfacher gruppendurch lineare transformationen, 24 (1926), 328–395.[4] M.Nagata, Invariants of a group in an affine ring, 3 (1964), 379–382.[5] P.Gordan, Beweis, dass jede covariante und invariante einer bina¨renform eine ganze funktion mit numerischen coeffizienten einer endlichenanzahl solcher formen ist, 69 (1868), 323–354.[6] R.Kane, Reflection groups and invariant theory, Springer-Verlag, NewYork City, 2001.[7] S.Lang, Algebra, 3rd ed., Springer-Verlag, New York City, 2003.[8] T.A.Springer, Invariant theory, Springer-Verlag, Berlin, 1977.42

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