Invariants of (-1)-Skew Polynomial Rings under Permutation Representations
aa r X i v : . [ m a t h . R A ] M a y INVARIANTS OF ( − -SKEW POLYNOMIAL RINGSUNDER PERMUTATION REPRESENTATIONS E. KIRKMAN, J. KUZMANOVICH AND J.J. ZHANG Introduction
Let k be a base field of characteristic zero (unless otherwise stated) and let k q [ x ]denote the q -skew polynomial ring k q [ x , . . . , x n ] that is generated by { x i } ni =1 andsubject to the relations x j x i = qx i x j for all i < j , where q is a nonzero elementin k . In previous work [KKZ1]-[KKZ4] we have studied the invariant theory ofnoncommutative Artin-Schelter regular (or AS regular, for short) algebras such as k q [ x ] under linear actions by finite groups G . We have shown that often the clas-sical invariant theory of the commutative AS regular algebra k [ x ] := k [ x , . . . , x n ]extends to noncommutative AS regular algebras in some analogous way. In thispaper we consider the case where G is a group of permutations of { x i } ni =1 acting onthe ( − k − [ x ], which is generated by { x i } ni =1 and subjectto the relations(E0.0.1) x i x j = − x j x i for all i = j . We have chosen to consider k − [ x ] because any permutation of { x i } ni =1 preserves the relations (E0.0.1), and hence extends to an algebra automorphism of k − [ x ]; the only q -skew polynomial algebras k q [ x ] with this property are the caseswhen q = ±
1. Hence any subgroup of the symmetric group S n acts on both k [ x ]and k − [ x ] as permutations, and our main focus is on the ring of invariants k − [ x ] G when G is a subgroup of S n .The study of the fixed subring k [ x ] G under permutation groups G of the commu-tative indeterminates { x i } ni =1 has a long and distinguished history. Gauss showedthat when G is the full symmetric group S n , invariant polynomials could be ex-pressed uniquely in terms of the n symmetric polynomials [Ne, Theorem 4.13];the symmetric polynomials are algebraically independent so that k [ x ] G is itself apolynomial ring. This result was generalized to other groups (so-called “reflectiongroups”) by Shephard-Todd [ST] and Chevalley [Ch] in the 1950s. It follows from[KKZ2, Theorem 1.1] that k − [ x ] G will not be an AS regular algebra, even for aclassical reflection group like the symmetric group. However, we will show that A G is always an AS Gorenstein domain [Theorem 1.5], while k [ x ] G is not always Goren-stein [Example 1.6]. In [CA] algebra generating sets for k − [ x ] S n , the invariantsunder the full symmetric group [Theorem 3.10], and for k − [ x ] A n , the invariantsunder the alternating group A n [Theorem 4.10] have been produced. We will showthat for both the full symmetric group [Theorem 3.12] and the alternating group Mathematics Subject Classification.
Key words and phrases.
Skew polynomial ring, permutation group, symmetric function,Hilbert series, fixed subring, complete intersection. [Theorem 4.15] the fixed subring is isomorphic to an AS regular algebra R moduloa central regular sequence of R (what we call a “classical complete intersection”in [KKZ4]). Moreover, we generalize some results for upper bounds on the de-grees of algebra generators for k − [ x ] G [Theorems 2.5 and 2.6] from results in thecommutative case.One motivation for this study was to consider the theorem of Kac-Watanabe[KW], and independently of Gordeev [G1], that provides a necessary condition forany finite group, not necessarily a permutation group, to have the property that k [ x ] G is a complete intersection (the condition is that G be a group generatedby so-called “bireflections”). This theorem of Kac-Watanabe-Gordeev was a firststep toward the (independent) classification of finite groups G , acting linearly asautomorphisms of k [ x ], such that k [ x ] G is a complete intersections that was provenby Gordeev [G2] and Nakajima [N1, N2, NW]. We verify that an analogous resultholds for k − [ x ] and subgroups of the symmetric group S n for n ≤ k − [ x ]: if G is a group of permutationsof the { x i } ni =1 that is generated by quasi-bireflections then k − [ x ] G is a classicalcomplete intersection [Theorem 5.4] (this result is not true for the commutativepolynomial ring k [ x ] [Example 5.5]).These fixed rings of k − [ x ] under permutation subgroups produce a tractableclass of AS Gorenstein domains that possess a variety of properties; in many casestheir generators have combinatorial descriptions and their Hilbert series can bedescribed explicitly. The following table summarizes results presented in this paperand gives a comparison between the results of k − [ x ] G with that of k [ x ] G for anysubgroup { } 6 = G ⊂ S n :Statements about A G when A = k [ x ] when A = k − [ x ]Being AS Gorenstein Not always AlwaysBeing AS regular Sometimes Never cci + ( A S n ) 0 ⌊ n ⌋ deg H A G ( t ) ≤ − n − n Bound for degrees of generators max { n, (cid:0) n (cid:1) } (cid:0) n (cid:1) + ⌊ n ⌋ ( ⌊ n ⌋ + 1)KWG theorem holds Yes ConjectureConverse of KWG holds No Yeswhere KWG stands for Kac-Watanabe-Gordeev. NVARIANTS OF ( − Definitions and basic properties
An algebra A is called connected graded if A = k ⊕ A ⊕ A ⊕ · · · and A i A j ⊂ A i + j for all i, j ∈ N . The Hilbert series of A is defined to be H A ( t ) = X i ∈ N (dim A i ) t i . Definition 1.1.
Let A be a connected graded algebra.(1) We call A Artin-Schelter Gorenstein (or
AS Gorenstein , for short) if thefollowing conditions hold:(a) A has injective dimension d < ∞ on the left and on the right,(b) Ext iA ( A k, A A ) = Ext iA ( k A , A A ) = 0 for all i = d , and(c) Ext dA ( A k, A A ) ∼ = Ext dA ( k A , A A ) ∼ = k ( l ) for some integer l . Here l iscalled the AS index of A .If in addition,(d) A has finite global dimension, and(e) A has finite Gelfand-Kirillov dimension,then A is called Artin-Schelter regular (or
AS regular , for short) of dimen-sion d .(2) If A is a noetherian, AS regular graded domain of global dimension n and H A ( t ) = (1 − t ) − n , then we call A a quantum polynomial ring of dimension n .Skew polynomial rings k q [ x ], where q ∈ k × := k \ { } , with deg x i = 1 arequantum polynomial rings and also Koszul algebras. Next we recall from [KKZ1]the definition of a noncommutative version of a reflection. If A is a connectedgraded algebra, let Aut( A ) denote the group of all graded algebra automorphismsof A . If g ∈ Aut( A ), then the trace function of g is defined to beTr A ( g, t ) = ∞ X i =0 tr( g | A i ) t i ∈ k [[ t ]] , where tr( g | A i ) is the trace of the linear map g | A i . Note that Tr A ( g,
0) = 1 andthat the trace of the identity map is the Hilbert series of the algebra A . The traceof a graded algebra automorphism of a Koszul algebra can be computed from theKoszul dual using the following result. Lemma 1.2. [JiZ, Corollary 4.4]
Let A be a Koszul algebra with Koszul dual algebra A ! . Let g ∈ Aut( A ) and g τ be the induced dual automorphism of A ! . Then Tr A ( g, t ) = (Tr A ! ( g τ , − t )) − . Definition 1.3.
Let A be an AS regular algebra such that H A ( t ) = 1(1 − t ) n f ( t )where f (1) = 0. Let g ∈ Aut( A ).(1) [KKZ1, Definition 2.2] Then g is called a quasi-reflection of A ifTr A ( g, t ) = 1(1 − t ) n − q ( t ) E. KIRKMAN, J. KUZMANOVICH AND J.J. ZHANG for q (1) = 0. If A is a quantum polynomial ring, then H A ( t ) = (1 − t ) − n .In this case g is a quasi-reflection if and only if(E1.3.1) Tr A ( g, t ) = 1(1 − t ) n − (1 − λt )for some scalar λ = 1. Note that we have chosen not to call the identitymap a quasi-reflection.(2) [KKZ4, Definition 3.6(b)] Then g is called a quasi-bireflection of A ifTr A ( g, t ) = 1(1 − t ) n − q ( t )for q (1) = 0.When A is noetherian and AS Gorenstein and g is in Aut( A ), the homologicaldeterminant of g , denoted by hdet g , is defined in [JoZ, Definition 2.3]. When A = k [ x ], the homological determinant of g is the inverse of determinant of thelinear map, induced by g on the degree one piece A = L ni =1 kx i of A , and, moregenerally, it is defined using a scalar map induced on the local cohomology of A ;see [JoZ] for details. The homological determinant is a group homomorphismhdet : Aut( A ) → k × . When A is AS regular, the conditions of the following theorem are satisfied by [JiZ,Proposition 3.3] and [JoZ, Proposition 5.5], and hdet g can be computed from thetrace of g , as given in the following result. Lemma 1.4. [JoZ, Lemma 2.6]
Let A be noetherian and AS Gorenstein and let g ∈ Aut( A ) . If g is k -rational in the sense of [JoZ, Definition 1.3] , then the rationalfunction Tr A ( g, t ) has the form Tr A ( g, t ) = ( − n (hdet g ) − t − ℓ + lower terms when it is written as a Laurent series in t − . The following result is not hard to prove.
Theorem 1.5.
Let G be any subgroup of the symmetric group S n acting on k − [ x ] as permutations. (1) The fixed subring k − [ x ] G is an AS Gorenstein domain. (2) If G = { } , then k − [ x ] G is not AS regular.Proof. (1) The trace of any transposition g = ( i, j ) in S n can be computed usingthe Koszul dual ( k − [ x ]) ! by Lemma 1.2, which is isomorphic to k [ x , . . . , x n ] / ( x , . . . , x n ) , and found to beTr A ( g, t ) = 1(1 + t )(1 − t ) n − = ( − n − t − n + lower terms . It follows from Lemma 1.4 that the homological determinant of g is 1. Since S n isgenerated by transpositions, hdet g = 1 for all g ∈ S n .By the last paragraph, hdet g = 1 for all g ∈ G . The assertion follows from [JoZ,Theorem 3.3].(2) Since k − [ x ] is a quantum polynomial ring, any quasi-reflection g has thehomological determinant λ = 1 where λ is given in (E1.3.1). Since hdet g = 1 NVARIANTS OF ( − for all g ∈ G , G contains no quasi-reflection (also see Lemma 1.7(4) below). Theassertion follows from [KKZ2, Theorem 1.1]. (cid:3) The analogous theorem is not true in the commutative case. As we mentionedin the introduction, k [ x ] S n is isomorphic to the commutative polynomial ring k [ x ],which is AS regular. Hence Theorem 1.5(2) fails for k [ x ]. The next example showsthat Theorem 1.5(1) fails for k [ x ]. Example 1.6.
Set n = 4. Let G = h (1 , , , i be the cyclic subgroup of S gener-ated by the 4-cycle (1 , , , G contains no reflections, and has elements ofdeterminant −
1, so B + := k [ x , x , x , x ] G cannot be Gorenstein by [Wa]. Or, onecan also use Molien’s Theorem to check that the Hilbert series of the fixed subring B + is t + t − t + 1(1 − t ) (1 + t ) (1 + t ) , which does not have the symmetry property required in Stanley’s criteria [S2, The-orem 4.4] for B + to be Gorenstein.Note that for the same subgroup G , but acting on the noncommutative ring k − [ x , . . . , x ], the Hilbert series of the fixed ring B − := k − [ x , . . . , x ] G is( t − t + 1)( t − t + 3 t − t + 3 t − t + 1)(1 − t ) (1 + t ) (1 + t ) , which has the symmetry property, and hence is AS Gorenstein by a noncommutativeversion of Stanley’s criteria [JoZ, Theorem 6.2], as well as by Theorem 1.5(1).The trace of any permutation is computed as follows. Lemma 1.7.
Let S n act on A = k − [ x ] as permutations and g ∈ S n . (1) If g is an m -cycle, then Tr A ( g, t ) = 1(1 + ( − t ) m )(1 − t ) n − m . (2) If g = ν i · · · ν i k µ j · · · µ j ℓ a product of disjoint cycles of length i p and j p ,with ν i p odd permutations and µ j p even permutations, then Tr A ( g, t ) = 1(1 + t i ) · · · (1 + t i k )(1 − t j ) · · · (1 − t j ℓ )(1 − t ) n − ( i + ··· + i k + j + ··· + j ℓ ) . (3) The only quasi-bireflections of k − [ x ] in S n are the two-cycles and three-cycles. (4) Permutation groups (namely, subgroups of S n ) contain no quasi-reflections.Proof. (1) This follows from Lemma 1.2 and direct computations.(2) This follows from part (1) and a graded vector space decomposition of k − [ x ].(3,4) These are consequences of part (2). (cid:3) In [KKZ4] we introduced several possible generalizations of a commutative com-plete intersection. We review these notions here.
Definition 1.8.
Let A be a connected graded noetherian algebra. E. KIRKMAN, J. KUZMANOVICH AND J.J. ZHANG (1) We say A is a classical complete intersection (or a cci ) if there is a connectedgraded noetherian AS regular algebra R and a sequence of regular normalhomogeneous elements { Ω , . . . , Ω n } of positive degree such that A is iso-morphic to R/ (Ω , . . . , Ω n ). The minimum such n is called the cci-number of A and denoted by cci ( A ).(2) We say A is a hypersurface if cci ( A ) ≤ A is a complete intersection of noetherian type (or an nci ) if theExt-algebra Ext ∗ A ( k, k ) := L i ≥ Ext iA ( A k, A k ) is noetherian.(4) We say A is a complete intersection of growth type (or a gci ) if the Ext-algebra Ext ∗ A ( k, k ) has finite Gelfand-Kirillov dimension.(5) We say A is a weak complete intersection (or a wci ) if the Ext-algebraExt ∗ A ( k, k ) has subexponential growth.In [KKZ4] we showed that a property of all of these kinds of complete intersec-tions is the cyclotomic Gorenstein property defined below. Definition 1.9.
Let A be a connected graded noetherian algebra.(1) We say A is cyclotomic if its Hilbert series H A ( t ) is a rational function p ( t ) /q ( t ) for some coprime polynomials p ( t ) , q ( t ) ∈ Z ( t ) and the roots of p ( t ) and q ( t ) are roots of unity.(2) We say A is cyclotomic Gorenstein if the following conditions hold(i) A is AS Gorenstein;(ii) A is cyclotomic. Theorem 1.10. [KKZ4, Theorem 3.4]
Let A be R G for some noetherian Auslanderregular algebra R and a finite subgroup G ⊂ Aut( R ) . If A is any of the kinds ofcomplete intersection in Definition 1.8, then it is cyclotomic Gorenstein. We note that in Example 1.6 although the fixed subring A G is AS Gorenstein,it is not any of the kinds of generalized “complete intersection” of Definition 1.8since its Hilbert series has zeros that are not roots of unity.The following theorem of Kac-Watanabe-Gordeev is one of the motivations forthis paper. Theorem 1.11. [KW, G1]
Let G be a finite group acting linearly on k [ x ] . If k [ x ] G is a complete intersection, then G is generated by bireflections. A noncommutative version of Kac-Watanabe-Gordeev Theorem holds for skewpolynomial rings k q [ x ] when q = ± k − [ x ] theonly unknown case. In this paper we will prove some partial results for this specialskew polynomial ring. We note that in Example 1.6 the trace of a four-cycle actingon k − [ x , . . . , x ] is 1 / (1 + t ), which is not a quasi-bireflection, supporting ageneralization of the Kac-Watanabe-Gordeev Theorem.To conclude this section we compute the automorphism group Aut( k − [ x ]). Lemma 1.12. (1) g ∈ Aut( k − [ x ]) if and only if g ( x i ) = a i x σ ( i ) for some σ ∈ S n and { a i } ni =1 ⊂ k × , namely, Aut( k − [ x ]) = ( k × ) n ⋊ S n . (2) If g is of the form in part (a), then hdet g = Q ni =1 a i .Proof. (a) Every diagonal map g : x i → a i x i , for ( a , · · · , a n ) ∈ ( k × ) n , extendseasily to a unique graded algebra automorphism of k − [ x ]. And we have alreadyseen that S n is a subgroup of Aut( k − [ x ]) such that S n ∩ ( k × ) n = { } . Thus NVARIANTS OF ( − ( k × ) n ⋊ S n ⊂ Aut( k − [ x ]). By [KKZ2, Lemma 3.5(e)], Aut( k − [ x ]) ⊂ ( k × ) n ⋊ S n .The assertion follows.(b) If g ∈ ( k × ) n , or g ( x i ) → a i x i for ( a , · · · , a n ) ∈ ( k × ) n , then it is easy to seethat hdet g = Q ni =1 a i . If g ∈ S n , then hdet g = 1 by the proof of Theorem 1.5(a).The assertion follows by the fact hdet is a group homomorphism. (cid:3) Upper bound for the algebra generators
In this section we show that Broer’s and G¨obel’s upper bounds on the degrees ofminimal generating sets of k [ x ] G , for arbitrary subgroup G ⊂ S n , have analoguesin this context. In this section we do not assume that char k = 0.The Noether upper bound on the degrees of generators does not hold for k − [ x ],as k − [ x , x ] S requires a generator of degree 3 [Example 3.1]. More generally onecan ask if the degrees of generators of k − [ x ] G are bounded above by | G | times thedimension of the representation of G . Broer’s degree bound [DK, Proposition 3.8.5]states that when f i are primary invariants, i.e. f i , for 1 ≤ i ≤ n , are algebraicallyindependent and k [ x ] G is a finite module over k [ f , . . . , f n ], then k [ x ] G is generatedas an algebra by homogeneous invariants of degrees at mostdeg( f ) + · · · + deg( f n ) − n. (The above statement is not true when n = 2 and g : x → x , x → − x . In thiscase f = x , f = x . Therefore we need to assume n ≥ G (not necessarily a permutation group) when thegiven hypotheses are satisfied [Lemma 2.2].Let A be any connected graded algebra. Define d A to be the maximal degreeof A ≥ / ( A ≥ ) . Then A is generated as an algebra by homogeneous elements ofdegree at most d A . Lemma 2.1.
Let A be a noetherian connected graded AS Gorenstein algebra and B and C be graded subalgebras of A such that C ⊂ B ⊂ A . Assume that (i) A = B ⊕ D as a right graded B -modules, (ii) A is a finitely generated right C -module, and (iii) There is a noetherian AS regular algebra R and a surjective graded algebramap φ : R → C and gldim R = injdim A .Then (1) φ is an isomorphism and A C is free. (2) d B ≤ max { d C , l C − l A } where l A and l C are AS index of A and C respec-tively.Proof. (1) Let n = injdim A . Induced by the composite map f : R → C → A wehave a convergent spectral sequence [WZ, Lemma 4.1],Ext pA (Tor Rq ( A, k ) , A ) = ⇒ Ext p + qR ( k, A ) . Since A R is finitely generated and R is right noetherian, Tor Rq ( A, k ) is finite dimen-sional for all q . Thus Ext pA (Tor Rq ( A, k ) , A ) = 0 for all p = injdim A = n . The abovespectral sequence collapses to the following isomorphismsExt nA (Tor Rq ( A, k ) , A ) ∼ = Ext n + qR ( k, A ) . E. KIRKMAN, J. KUZMANOVICH AND J.J. ZHANG
For any q >
0, Ext nA (Tor Rq ( A, k ) , A ) ∼ = Ext n + qR ( k, A ) = 0. Since Tor Rq ( A, k ) is finitedimensional, A is AS Gorenstein, we obtain thatdim Tor Rq ( A, k ) = dim Ext nA (Tor Rq ( A, k ) , A ) = 0for all q >
0. Hence A R is projective, whence free, as R is connected graded. As aconsequence, f : R → A is injective. This implies that φ is an isomorphism. Since φ is an isomorphism and A R is free, A C is free.(2) Now we identify R with C . By part (1), A is a finitely generated free C -module. Since A = B ⊕ D , both B and D are projective, whence free, graded right C -modules. Pick a C -basis for B and D , say V B ⊂ B and V D ⊂ D . Then we have B = V B ⊗ C and D = V D ⊗ C . Therefore A = V A ⊗ C where V A = V B ⊕ V D . Hence H A ( t ) = H V A ( t ) H C ( t ) = ( H V B ( t ) + H V D ( t )) H C ( t ) , and H B ( t ) = H V B ( t ) H C ( t ) . Since B = V B ⊗ C , B is generated by V B and C as a graded algebra. Thus we have d B ≤ max { deg H V B ( t ) , d C } ≤ max { deg H V A ( t ) , d C } . It remains to show that deg H V A ( t ) = − l A + l C . First, as H A ( t ) = H V A ( t ) H C ( t ),we have deg H V A ( t ) = deg H A ( t ) − deg H C ( t ). Recall that C is noetherian and ASregular. Since A is a finite module over C , H A ( t ) is rational and the hypotheses(1 ◦ , 2 ◦ , 3 ◦ ) of [JoZ, Theorem 6.1] hold. By the proof of [JoZ, Theorem 6.1] (we arenot using the hypothesis that A is a domain), H A ( t ) = ± t l A H A ( t − )where l is the AS index of A . Since H A ( t ) is a rational function such that H A (0) = 1,the above equation forces that(E2.1.1) deg H A ( t ) = − l A . Similarly, deg H C ( t ) = − l C . The assertion follows. (cid:3) The degree of algebra generators of B is bounded by l C − l A when d C ≤ l C − l A ,which is easy to achieve in many cases. The following lemma is a generalization ofBroer’s upper bound [DK, Proposition 3.8.5]. Lemma 2.2 (Broer’s Bound) . Let A be a quantum polynomial algebra of dimension n and C an iterated Ore extension k [ f ][ f ; τ , δ ] · · · [ f n ; τ n , δ n ] . Assume that (1) B = A H where H is a semisimple Hopf algebra acting on A , (2) C ⊂ B ⊂ A and A C is finitely generated, and (3) deg f i > for at least two distinct i ’s.Then d A H ≤ l C − l A = n X i =1 deg f i − n. Proof.
Since H is semisimple, A = B ⊕ D by [KKZ3, Lemma 2.4(a)] where B = A H .Let R = C . Then the hypotheses Lemma 2.1(i,ii,iii) hold. By Lemma 2.1, d B ≤ max { d C , l C − l A } . It is clear that l A = n . By induction on n , one sees that H C ( t ) = Q ni =1 (1 − t deg fi ) .By (E2.1.1), l C = − deg H C ( t ) = P ni =1 deg f i . Now it suffices to show that d C ≤ NVARIANTS OF ( − P ni =1 deg f i − n . For the argument sake let us assume that deg f i is increasing as i goes up. So d C = deg f n . Now n X i =1 deg f i − n = n X i =1 (deg f i − ≥ deg f n − − f n − ≥ deg f n . The assertion follows. (cid:3)
This result applies to subgroups G ⊂ S n acting on k − [ x ].Let C be any commutative algebra over k and let n be a positive integer. Define D be the algebra generated by C and { y , · · · , y n } subject to the relations(E2.2.1) [ y i , c ] = 0for all c ∈ C , and(E2.2.2) y i y j + y j y i = c ij for 1 ≤ i < j ≤ n , where { c ij | ≤ i < j ≤ n } is a subset of the subalgebra C [ y , · · · , y n ] (which is in the center of D ). Lemma 2.3.
Retain the above notation. Then (1) σ : ( y i
7→ − y i ∀ ic c ∀ c ∈ C extends uniquely to an algebra automorphism of D , and (2) Let { w , · · · , w n } be a subset of C [ y , · · · , y n ] . Then φ : ( y i w i ∀ ic ∀ c ∈ C extends uniquely to a σ -derivation of D .Proof. (a) Since D is generated by C and { y i } ni =1 , the extension of σ is unique. Itis clear that the extension of σ preserves relations (E2.2.1) and (E2.2.2).(b) Since D is generated by C and { y i } ni =1 , the extension of φ , using the σ -derivation rule, is unique. For any c ∈ C , using the fact φ ( c ) = 0, we have φ ([ y i , c ]) = φ ( y i ) c − σ ( c ) φ ( y i ) = w i c − cw i = 0 . For any i , δ ( y i ) = σ ( y i ) δ ( y i ) + δ ( y i ) y i = − y i δ ( y i ) + δ ( y i ) y i = 0 . As a consequence, δ ( c ij ) = 0. Now φ ( y i y j + y j y i − c ij ) = φ ( y i ) y j + σ ( y i ) φ ( y j ) + φ ( y j ) y i + σ ( y j ) φ ( y i )= w i y j − y i w j + w j y i − y j w i = 0 . So the extension of φ is a σ -derivation. (cid:3) We need a lemma on symmetric functions of k − [ x ]. For every positive integer u , let P u denote the u th power sum P ni =1 x ui ∈ k − [ x ]. Let C be the subalgebraof k − [ x ] generated by P , P , · · · , P n − , P n , C be the subalgebra of k − [ x ] gen-erated by P , P , P , · · · , P n − , P n . Define P ′ i = P i is i is odd and P ′ i = P i if i iseven. Let C be the subalgebra of k − [ x ] generated by P ′ , P ′ , · · · , P ′ n − , P ′ n . Notethat C contains P i for all i . Lemma 2.4.
Retain the above notation. (1) k − [ x ] is a finitely generated free module over the central subalgebra C . (2) If u is even, then P u P v = P v P u for any v . (3) If u and v are odd, then P u P v + P v P u = 2 P u + v . (4) If u is odd, then P u = P u . (5) C ⊂ C ⊂ C ⊂ k − [ x ] S n ⊂ k − [ x ] G . (6) C is isomorphic to an iterated Ore extension R := k [ P , P , · · · , P ⌊ n ⌋ ][ P ][ P ; τ , δ ] · · · [ P n ′ ; τ n ′ , δ n ′ ] where n ′ = 2 ⌊ n − ⌋ + 1 .Proof. (1) The algebra k − [ x ] is a finitely generated module over k [ x , · · · , x n ] and k [ x , · · · , x n ] is finitely generated over C = k [ P , P , · · · , P n ] where each P i is the i th power sum of the variables { x , · · · , x n } . Therefore k − [ x ] is finitely generatedover C . By the proof of Lemma 2.1(1), k − [ x ] is free over C .(2,3,4) By direct computations.(5) If i is odd, ( P ′ i ) = ( P i ) = P i , and if i is even, P ′ i = P i . So C ⊂ C . Therest is clear.(6) For odd integers i < j , part (3) says that P j P i + P i P j = 2 P i + j . We can easily determine the automorphisms τ j and derivations δ j by using Lemma2.3. As a consequence, there is a surjective map φ : R → C . Also gldim R = n =gldim k − [ x ]. By the proof of Lemma 2.1(1), C ∼ = R . (cid:3) Theorem 2.5 (Broer’s Bound for k − [ x ] G ) . Let G be a subgroup of S n acting on k − [ x ] naturally. Suppose | G | does not divides char k . Then d ( k − [ x ] G ) ≤ n ( n −
1) + ⌊ n ⌋ ( ⌊ n ⌋ + 1) ∼ n . Proof.
The assertion can be checked directly for n = 1 ,
2. Assume now that n ≥ A := k − [ x ] and C be C as in Lemma 2.4(6). Then C is a subalgebra of A G for any G ⊂ S n . Since | G | does not divides char k , H := kG is semisimple. Notethat deg P i = i . Hence all hypotheses in Lemma 2.2 are satisfied. By Lemma 2.2, d A G ≤ X i =1 deg f i − n = 12 n ( n + 1) + ⌊ n ⌋ ( ⌊ n ⌋ + 1) − n = 12 n ( n −
1) + ⌊ n ⌋ ( ⌊ n ⌋ + 1) . (cid:3) This bound is sharp when n = 2 [Example 3.1]. For larger n , we have no examplesto show this bound is sharp – and it probably is not sharp.Next we consider a generalization of the G¨obel bound [Go]. If G is a group ofpermutation of { x i } ni =1 acting as automorphisms on k [ x ] then G¨obel’s Theoremstates that k [ x ] G is generated by the n symmetric polynomials (or the power sums)and “special polynomials”. Let O G ( X I ) represent the orbit sum of X I under G .“Special polynomials” are all G -invariants of the form O G ( X I ), where λ ( I ) =( λ i ), the partition associated to I (i.e. arranging the elements of I in weaklydecreasing order), has the properties that the last part of the partition λ n = 0, and λ i − λ i +1 ≤ i . It follows that an upper bound on the degree of a minimalset of generators of k [ x ] G for any n -dimensional permutation representation of G is max { n, (cid:18) n (cid:19) } . In this context the G¨obel bound can be a sharp bound, as it iswhen the alternating group A n acts on k [ x ]. A similar idea works for k − [ x ], see[CA, Corollary 3.2.4]. But we consider a modification of S n . NVARIANTS OF ( − Let c S n be the group S n ⋊ {± } n , where {± } n is the subgroup of diagonalactions x i → a i x i for all i , where a i = ± Theorem 2.6 (G¨obel’s Bound for k − [ x ] G ) . Let G be a subgroup of c S n . Then d k − [ x ] G ≤ n , and d k [ x ] G ≤ n . Proof.
Let A be k − [ x ] or k [ x ]. Let C = k [ P , P , · · · , P n ]. Then A is a finitelygenerated free module over C such that C ⊂ A G . By Lemma 2.2, d A G ≤ X i deg f i − n = X i i − n = n ( n + 1) − n = n . (cid:3) [CA, Corollary 3.2.4] is a consequence of the above theorems.3. Invariants under the full symmetric group S n Some results in this and the next section have been proved in [CA]. We repeatsome of the arguments for completeness.We consider the ring of invariants k − [ x ] S n under the full symmetric group S n .Gauss proved that k [ x ] S n is generated by the n elementary symmetric functions σ k for 1 ≤ k ≤ n , each of which is an orbit sum (sum of all the elements in the S n -orbit) of the given monomials. Recall that, for each 1 ≤ k ≤ n , σ k ( x , . . . , x n ) = X i
Let A = k − [ x , x ] and let G = h g i = S for g = (1 , σ = x x is not invariant, and, moreover, P is not a generator because P = P ; it is easy to check that there are no other invariants of degree 2. We willshow that the invariants are generated by P = x + x and P = x + x , or by S = P = x + x and S = x x + x x . In this example the maximal degreeof a minimal set of generators is 3 [Theorem 2.5], which is larger than the orderof the group | G | (the “Noether bound” [No] guarantees the maximal degree of aminimal set of generators of k [ x ] G is ≤ | G | ). In the case of either set of generators,the generators are not algebraically independent, and the ring of invariants is notAS regular, but AS Gorenstein [Theorem 1.5]; and we will show that it is a cci ina couple ways. First, it is a hypersurface in the AS regular algebra B generated by x, y with relations xy = y x and x y = yx : A S ∼ = B (2 x − x y − yx + 4 y ) (where P x and P y ). Second, it is a factor of the iterated Ore extension C = k [ a, b ][ x ][ y ; τ, δ ], where τ is the automorphism of k [ a, b, x ] defined by τ ( a ) = a, τ ( b ) = b, τ ( x ) = − x , and δ is a τ -derivation of k [ a, b, x ] defined by δ ( a ) = δ ( b ) = 0and δ ( x ) = 2 b : A S ∼ = C ( x − a, y − c ) . Here { x − a, y − c } for c = (3 ab − a ) / C . In this isomorphism P a, P b, P x, P y, since we have therelations [Lemma 2.4] P P + P P = 2 P P = P P = P = P P − P ( P − P ) / . The aim of this section is to prove the analogous result for arbitrary n . Wefirst repeat the analysis from [CA] and show that there are two sets of algebragenerators of k − [ x ] S n : the n odd power sums P , P , · · · , P n − and the n elementsfor 1 ≤ k ≤ n : S k = X x i x i · · · x i k − x i k =: O S n ( x x · · · x k − x k )where the sum is taken over all distinct i , . . . , i k with i < i < · · · < i k − , and O S n represents the sum of the orbit under the full symmetric group; we call theseelements S k the “super-symmetric polynomials” since they play the role that thesymmetric functions play in the commutative case. Hence the maximal degree of aset of minimal generators for the full ring of invariants k − [ x ] S n is 2 n − k − [ x ] can be written as the form ± x i x i · · · x i n n , where thesign is due to the fact that these x i s are ( − I denote the index( i k ) := ( i , · · · , i n ) and let X I denote the monomial x i x i · · · x i n n . Throughout let G be a subgroup of S n unless otherwise stated. Define stab G ( X I ) = { g ∈ G | g ( X I ) = X I in k − [ x ] } . For any permutation σ ∈ G , stab G ( X I ) and stab G ( x i σ (1) x i σ (2) · · · x i n σ ( n ) ) are conju-gate to each other. As a consequence, | stab G ( X I ) | = | stab G ( x i σ (1) x i σ (2) · · · x i n σ ( n ) ) | . Definition 3.2.
Let λ ( m ) = ( λ , λ , · · · , λ n ) be a partition m , where λ i are weaklydecreasing and λ i ≥
0. Let X λ be the monomial x λ x λ · · · x λ n n . The G -orbit sumof the monomial X λ of (total) degree m is defined by O G ( X λ ) = O G ( x λ x λ · · · x λ n n ) = 1 | stab G ( X λ ) | X g ∈ G x λ g (1) x λ g (2) · · · x λ n g ( n ) . In this section we take G = S n and in the next G = A n . Remark 3.3.
We divide by the order of the stabilizer of X λ so that each elementof the orbit is counted only once. Throughout we will compare monomials using thelength-lexicographical order: for I = ( i k ) and J = ( j k ) we say X I < X J if P i k < P j k , or if P i k = P j k , and if k is the smallest index for which i k = j k then i k < j k ;when considering elements of the same degree this order is the lexicographical orderon the exponents with x > x > . . . > x n . Hence we will denote the S n -orbit sumby O S n ( X I ), where X I is the leading term of the orbit sum under the (length)-lexicographic order and so I is a partition, and we call O S n ( X I ) the S n -orbit sum NVARIANTS OF ( − corresponding to the partition I . We refer to the entries in I as the “parts” of thepartition (so a part may be 0).The following lemma is easily verified. Lemma 3.4. [CA, Theorem 2.1.3]
Let G be a finite subgroup of S n . Then any G -invariant is a sum of homogeneous G -invariants and homogeneous invariants arelinear combinations of G -orbit sums. Lemma 3.5. [CA, Lemma 2.2.2] A S n -orbit sum corresponding to a partition λ ( m ) = ( λ , λ , · · · , λ n ) is zero if and only if it has repeated odd parts. Hence anon-zero S n -orbit sum corresponds to a partition with no repeated odd parts.Proof. An orbit sum O S n ( X I ) is zero if and only if the S n -orbit of X I consists ofmonomials and their negatives, i.e. σX I = − X I for some σ ∈ S n . In order for σX I = − X I there must be a repeated exponent. Consider a monomial of the form x e · · · x e j j · · · x e k k · · · x e n n where e j = e k and both are odd. We claim that when thetransposition ( j, k ) is applied to this monomial we get the same monomial but witha negative sign. We induct on k − j . If k − j = 1 then the result is clear. Henceassume that result is true for k − j < ℓ and we prove it for k − j = ℓ . We writethe monomial as x e · · · x e j j · · · x e k − k − x e k k · · · x e n n and consider the case when e k − isodd and the case when e k − is even. When e k − is odd then ( j, k ) applied to themonomial yields x e · · · x e k k · · · x e k − k − x e j j · · · x e n n = − x e · · · x e k k · · · x e j j x e k − k − · · · x e n n which by induction is = x e · · · x e j j · · · x e k k x e k − k − · · · x e n n = − x e · · · x e j j · · · x e k − k − x e k k · · · x e n n . When e k − is even then ( j, k ) applied to the monomial yields x e · · · x e k k · · · x e k − k − x e j j · · · x e n n = x e · · · x e k k · · · x e j j x e k − k − · · · x e n n which by induction is = − x e · · · x e j j · · · x e k k x e k − k − · · · x e n n = − x e · · · x e j j · · · x e k − k − x e k k · · · x e n n . Hence σX I = − X I , and so for any τ X I in the S n -orbit of X I we have − τ X I = τ σX I is in the orbit of X I , and hence the S n -orbit sum of X I is zero.Clearly when indices with even exponents of the same value are permuted nosign change occurs, and so the orbit sum will not be zero unless there is at leastone repeated odd exponent. (cid:3) By Lemma 3.5 the set of elements in k − [ x ] S n of degree k has a vector spacebasis corresponding to the partitions of k into at most n parts with no repeatedodd entries. We next will show that both the sets S k and P k − for k = 1 , . . . , n (corresponding to the partitions (2 , . . . , , , , . . . ,
0) and (2 k − , , . . . ,
0) of 2 k − k − [ x ] S n . Lemma 3.6.
Let I = ( λ k ) be a partition where no λ i are both equal and odd. Theleading term of O S n ( x λ x λ · · · x λ n n ) S k is x λ +21 · · · x λ k − +2 k − x λ k +1 k x λ k +1 k +1 · · · x λ n n . Proof.
By our assumption on I the orbit of X I does not contain another elementwith the same entries as X I . Clearly x λ +21 · · · x λ k − +2 k − x λ k +1 k x λ k +1 k +1 · · · x λ n n is a sum-mand of the product of O S n ( x λ x λ · · · x λ n n ) S k . This product of orbits can bewritten as a linear combination of S n -orbit sums; let O S n ( X E ) be one of theseorbit sums. The entries of the any such partition E are obtained from the partition I = ( λ , · · · , λ n ) by adding 2 to k − I , adding 1 to one entry of I , andplacing these entries into numerical order. It is clear that the largest such partition E that can be obtained in this manner is ( λ +2 , . . . , λ k − +2 , λ k +1 , λ k +1 , · · · , λ n ),and the leading term of this S n -orbit sum occurs in the product of orbits onlyonce. (cid:3) The following lemma follows essentially as in Gauss’s proof for k [ x ] S n ; the super-symmetric polynomials S k ∈ k − [ x ] S n play the role of the symmetric polynomials σ k in k [ x ] S n . Lemma 3.7.
Suppose that f = 0 is a S n -invariant with leading term x λ x λ · · · x λ n n of degree m where at least one λ k odd. Then there is a positive integer k , a partition λ ∗ ( m − k + 1) = ( λ ∗ , . . . , λ ∗ n ) of m − k + 1 , and a c ∈ k × such that f − c O S n ( x λ ∗ · · · x λ ∗ n n ) S k has leading term of smaller degree than f . As a consequence, the fixed subring k − [ x ] S n is generated as an algebra by the n elements S k , for k = 1 , . . . , n , andinvariants with all even powers, k [ x , · · · , x n ] S n .Proof. I = ( λ i ) is a partition and hence is weakly decreasing. Let k be the largestindex with λ k odd, and let I ∗ = ( λ − , λ − , . . . , λ k − − , λ k − , λ k +1 , . . . , λ n ) . We claim that I ∗ is a weakly decreasing sequence. First note that since λ k is odd, λ k ≥
1, and for ℓ ≥ k + 1 the λ ℓ are even and weakly decreasing, so for ℓ ≥ k + 1we have λ k ≥ λ ℓ + 1 ≥ λ ℓ +1 + 1, and the final n − k + 1 entries of I ∗ are weaklydecreasing. Next, since λ k is odd and there are no repeated odd exponents in anonzero S n -orbit sum, we have λ k − ≥ λ k + 1 and λ j − − ≥ λ j − − ≥ λ k − ≤ j ≤ k , so the first k entries of I ∗ are weakly decreasing. Hence by Lemma3.6 we have x λ x λ · · · x λ n n = ( x λ − · · · x λ k − − k − x λ k − k · · · x λ n n )( x · · · x k − x k )is the leading term in O S n ( X I ∗ ) S k , and if c is the coefficient of the leading termof f then c O ( X I ∗ ) S k − f has smaller order leading term. Furthermore O ( X I ∗ )also has smaller order. Since there are only a finite number of smaller orders, thealgorithm must terminate when all exponents are even. (cid:3) Since the central subring k [ x , . . . , x n ] of k − [ x ] is a commutative polynomialring and S n acts on it as permutations, the invariants k [ x , . . . , x n ] S n are generatedby either the even power sums P , · · · , P n or the n symmetric polynomials in thesquares; in particular, if ρ i := σ i ( x , · · · , x n ) for the elementary symmetric function σ i , then k [ x , · · · , x n ] S n = k [ ρ , ρ , . . . , ρ n ]. Since P k ∈ k [ ρ , ρ , . . . , ρ n ] , each P k can be expressed as a polynomial in the elementary symmetric functions, say(E3.7.1) P k = f k ( ρ , ρ , . . . , ρ n ) . NVARIANTS OF ( − Next we show that k [ x , . . . , x n ] S n is contained in the algebra generated by the n odd power sums P , . . . , P n − , and k [ x , . . . , x n ] S n is contained in the algebragenerated by the n super-symmetric polynomials S k . Lemma 3.8.
The fixed subring k [ x , . . . , x n ] S n is contained in the algebra gener-ated by either the odd power sums P , · · · , P n − or by the super-symmetric poly-nomials S , · · · , S n in k − [ x ] .Proof. We obtain the even power sums from the odd ones as follows: P = P , andmore generally(E3.8.1) P i = ( P P i − + P i − P ) / ≤ i ≤ n . Also(E3.8.2) ρ j = O S n ( x · · · x j ) = ( S S j + S j S ) / (2 j )for all 1 ≤ j ≤ n . (cid:3) The next argument follows as in the case of k [ x ] [S1, p. 4]. Given a monomial X I , we define λ ( I ), the partition associated with X I , to be the elements of I listedin weakly decreasing order (i.e. the partition associated to O S n ( X I )). We define atotal order on the set of monomials as X I < X J if the associated partitions havethe property that λ ( I ) is lexicographically larger than λ ( J ), or, if the partitions areequal, when I is lexicographically smaller than J . As an example for n = 3 anddegree = 4 x < x < x < x x < x x < x x < x x < x x < x x < x x < x x < x x < x x x < x x x < x x x . In the case of k [ x ], where all partitions represent basis elements in the subringof invariants, in a given degree k ≤ n the “largest” partition is (1 , . . . , , . . . , k, . . . , k − [ x ], for monomialsthat correspond to nonzero invariants there are no repeated odd parts, so for odddegrees 2 k − ≤ n −
1, the partition (2 , . . . , , , , . . . ,
0) is “largest” under thisorder, and while the partition (2 k − , . . . ,
0) is smallest, and x k − n is the smallestmonomial of degree 2 k −
1. Furthermore in a product of power sums P i P i · · · P i k the leading monomial will be cx i x i · · · x i k k for some nonzero integer c when the i j are weakly decreasing. Lemma 3.9.
The fixed subring k − [ x ] S n is generated by the n odd power sums P , . . . , P n − .Proof. By Lemma 3.8 the even power sums are generated by the odd power sums P , . . . , P n − , so it suffices to show invariants are generated by power sums P k for k ≤ n −
1. By Lemmas 3.7 and 3.8 the S k are algebra generators of k − [ x ] S n ,so it suffices to show they can be expressed in terms of power sums. Hence itsuffices to describe an algorithm that writes an invariant f ∈ k − [ x ] S n of degree ≤ n − f as ax i x i · · · x i n n for some a ∈ k × . The exponents of the leading term are weakly decreasing, andeach is ≤ n −
1. The element f − ac P i P i · · · P i n has the same total degree as f ,but its leading term is less than that of f . Since there are only a finite number of monomials of smaller order for a fixed degree, the algorithm terminates with f written in terms of power sums of degree ≤ n − (cid:3) The following theorem of Cameron Atkins follows from the lemmas above, andgives us two choices of algebra generators for k − [ x ] S n . It is often convenient tochoose the power sums, since they have fewer summands. Theorem 3.10. [CA, Theorems 2.2.6 and 2.2.8]
The fixed subring k − [ x ] S n isgenerated by either the set of the n odd power sums P , · · · , P n − or the set of the n super-symmetric polynomials S , · · · , S n . We next show that the AS Gorenstein domain k − [ x ] S n is a cci. First we haveto construct a suitable AS regular algebra.Let R = k [ p , p , . . . , p n ] be a commutative polynomial ring, and let a i = f i ( p , p , . . . , p n ) where the f i are the polynomials of (E3.7.1). Consider thefollowing iterated Ore extension B = k [ p , . . . , p n ][ y : τ , δ ] · · · [ y n : τ n , δ n ]where coefficients are written on the left, R = k [ p , . . . , p n ] is a commutative poly-nomial ring, τ j is the automorphism of k [ p , . . . , p n ][ y : τ , δ ] · · · [ y j − : τ j − , δ j − ]defined by τ j ( y i ) = − y i for i < j and τ j ( r ) = r for r ∈ k [ p , . . . , p n ], and δ j is the τ j -derivation δ j ( y i ) = 2 a i +2 j − with δ j ( r ) = 0 for all r ∈ k [ p , . . . , p n ].By Lemma 2.3, δ k are τ k -derivation for all k where ( τ k δ k ) appeared in the defi-nition of B .We grade B by setting degree( p i ) = 2 i and degree( y i ) = 2 i −
1. With this gradingthe Hilbert series of B is given by H B ( t ) = 1(1 − t )(1 − t ) · · · (1 − t n − )(1 − t n ) . The algebra B is an AS regular algebra of dimension 2 n . Let r i = y i − a i − foreach i = 1 , , . . . , n ; it is easy to see that r i is a central element of B . Lemma 3.11.
The sequence { r , r , . . . , r n } is a central regular sequence in B .Proof. First we note that the r i are central since a i and y i are central y i y j = y i ( − y j y i + p i + j ) = − y i y j y i + y i p i + j = − ( y i y j + p i + j ) y i = y j y i . Since B is a domain, r = 0 is regular in B .Let B i = k [ p , . . . , p n ][ y : τ , δ ] · · · [ y i : τ i , δ i ] and let B i = B i / ( r , r , . . . , r i ) B i .Now consider the algebra C i = B i [ y i +1 : τ i +1 , δ i +1 ] · · · [ y n : τ n , δ n ], where the τ j and δ j are the induced maps. These maps are well-defined since for j > i and k ≤ i, τ j ( r k ) = r k and δ j ( r k ) = 0. Note that B = B i [ y i +1 : τ i +1 , δ i +1 ] · · · [ y n : τ n , δ n ],and hence every element of B can be written in the form P I b I y I where b I ∈ B i , I =( e i +1 , e i +2 , . . . , e n ) is a nonnegative integral vector, and y I = y e i +1 i +1 y e i +2 i +2 · · · y e n n . Thealgebra B/ ( r , r , . . . , r i ) B is isomorphic to the algebra C i under the map X I b I y I + h r , r , . . . , r i i B X I ¯ b I y I where ¯ b I denotes reduction mod ( r , r , . . . , r i ) B i . Now the standard polynomialdegree argument in C i shows that the image of r i +1 is regular in C i . (cid:3) We now can prove that k − [ x ] S n ∼ = B/ ( r , r , . . . , r n ) where, by Lemma 3.11,each r i is central in B and regular in B/ ( r , r , . . . , r i − ). NVARIANTS OF ( − Theorem 3.12.
The algebra k − [ x ] S n is a cci.Proof. By Definition 3.2 and Lemma 3.5 k − [ x ] S n as a graded vector space hasa basis of orbit sums of monomials having no repeated odd exponents. Hence itsHilbert series is the same as the generating function for the restricted partitionshaving no repeated odd parts. By Proposition 5.1 of the Appendix this Hilbertseries is given by D n ( t ) = (1 − t )(1 − t )(1 − t ) · · · (1 − t n − )(1 − t )(1 − t )(1 − t ) · · · (1 − t n − )(1 − t n ) . Let ρ i = σ i ( x , x , . . . , x n ) where σ i is the i th elementary symmetric polynomial.Then the algebra k [ x , x , . . . , x n ] S n = k [ ρ , ρ , . . . , ρ n ] is a commutative polyno-mial ring. By Theorem 3.10, k − [ x ] S n is generated as an algebra by the odd powersums, and hence k − [ x ] S n = k [ ρ , ρ , . . . , ρ n ][ P , P , . . . , P n − ] . Consider the iterated Ore extension B constructed above and define a map φ : B −→ k − [ x ] S n by φ ( p i ) = ρ i and φ ( y j ) = P j − . Note that φ preserves degree.Clearly φ takes R = k [ p , p , . . . , p n ] isomorphically onto k [ ρ , ρ , . . . , ρ n ] , and bothsubrings are central. In the iterated Ore extension B , we have for i < j that y j y i + y i y j = 2 a i +2 j − = f i +2 j − ( p , p , . . . , p n ) . Calculation in k − [ x ] S n shows that P j − P i − + P i − P j − = 2 P i +2 j − = 2 f i +2 j − ( ρ , ρ , . . . , ρ n );hence φ ( y j ) φ ( y i ) + φ ( y i ) φ ( y j ) = 2 φ ( a i +2 j − ) . Hence the skew extension relations are preserved, and we conclude that φ is agraded ring homomorphism. Since the odd power sums P , P , . . . , P n − generate k − [ x ] S n as an algebra by Theorem 3.10, the homomorphism φ is an epimorphism.Calculation yields0 = P i − − P i − = P i − − f i − ( ρ , ρ , . . . , ρ n )= φ ( y i − a i − ) = φ ( r i ) . Hence the ideal ( r , r , . . . , r n ) ⊆ ker( φ ), and φ induces a graded ring homomor-phism ¯ φ : B/ ( r , r , . . . , r n ) −→ k − [ x ] S n . Since for each i the degree of r i is 4 i − { r , r , . . . , r n } is a regular sequence,the Hilbert series of ¯ B = B/ ( r , r , . . . , r n ) is given by H ¯ B ( t ) = (1 − t )(1 − t )(1 − t ) · · · (1 − t n − )(1 − t )(1 − t )(1 − t ) · · · (1 − t n − )(1 − t n ) . This shows that ¯ φ is an isomorphism. (cid:3) Definition 3.13.
Let A be a connected graded noetherian algebra.(1) We say A is a classical complete intersection + (or a cci + ) if there is aconnected graded noetherian AS regular algebra R with H R ( t ) = Q ni =1 (1 − t di ) and a sequence of regular normal homogeneous elements { Ω , . . . , Ω n } ofpositive degree such that A is isomorphic to R/ (Ω , . . . , Ω n ). The minimumsuch n is called the cci + -number of A and denoted by cci + ( A ). (2) Let A be cyclotomic (e.g., A is cci). The cyc -number of A , denoted by cyc ( A ), is defined to be v if the Hilbert series of A is of the form H A ( t ) = Q vs =1 (1 − t m s ) Q ws =1 (1 − t n s )where m s = n s ′ for all s and s ′ .Clearly we have cci + ( A ) ≥ cci ( A ). It is a conjecture that every noetherian ASregular algebra has Hilbert series of the form Q ni =1 (1 − t di ) . If this conjecture holds,then being cci + is equivalent to being cci and cci + ( A ) = cci ( A ). One can easilyshow that the expression of H A ( t ) in Definition 3.13(2) is unique (as we assumethat m s = n s ′ for all s, s ′ ). It follows from the definition that cci + ( A ) ≥ cyc ( A ).Finally we would like to calculate cci + ( k − [ x ] S n ). Theorem 3.14. cci + ( k − [ x ] S n ) = cyc ( k − [ x ] S n ) = ⌊ n ⌋ .Proof. First we prove the claim that cci + ( k − [ x ] S n ) ≤ ⌊ n ⌋ .Let C be the subalgebra of k − [ x ] S n defined before Lemma 2.4. By Lemma2.4(6), it is isomorphic to the iterated Ore extension k [ P , P , · · · , P ⌊ n ⌋ ][ P ][ P ; τ , δ ] · · · [ P n ′ ; τ n ′ , δ n ′ ]where n ′ = 2 ⌊ n − ⌋ + 1. By Lemma 2.4(5), C contains P i for all i ≥
1. Let F n ′ := C , and for any odd integer n ′ < j ≤ n −
1, we inductively constructa sequence of iterated Ore extensions F j = F j − [ P j , τ j , δ j ] where τ j is defined by τ j ( P s ) = ( − s P s for all even s and for all odd s ≤ j −
2, and where the τ j -derivation δ j is defined by δ j ( P s ) = ( s is even2 P s + j if s is odd . . It follows from theinduction and Lemma 2.3 that τ j is an automorphism of F j − and δ j is a τ j -derivation of F j − . Therefore F j (and whence F n − ) is an iterated Ore extension(which is a noetherian AS regular algebra with Hilbert sires of the form ( Q ni =1 (1 − t d i )) − ). Let u s = P s − − P s − for all integers from s = ⌊ n − ⌋ + 2 to s = n . Theproof of Lemma 3.11 shows that { u ⌊ n − ⌋ +2 , · · · , u n } is a central regular sequenceof F n − . It is easy to see that F n − / ( u ⌊ n − ⌋ +2 , · · · , u n ) ∼ = k − [ x ] S n . Therefore cci + ( k − [ x ] S n ) ≤ n − ( ⌊ n − ⌋ + 1) = ⌊ n ⌋ and we proved the claim.By Theorem 3.12 H k − [ x ] S n ( t ) = H ¯ B ( t ) = (1 − t )(1 − t )(1 − t ) · · · (1 − t n − )(1 − t )(1 − t )(1 − t ) · · · (1 − t n − )(1 − t n )= Q ns = ⌊ n − ⌋ +2 (1 − t s − ) Q ⌊ n ⌋ j =1 (1 − t j ) Q ni =1 (1 − t i − )which is an expression satisfying the condition in Definition 3.13(2). Hence cyc ( k − [ x ] S n ) = ⌊ n ⌋ . The assertion follows from the claim and the fact cci + ( A ) ≥ cyc ( A ). (cid:3) NVARIANTS OF ( − Invariants under A n First let us review the classical case. Let A n be the alternating group. Anyelement of k [ x ] A n can be written uniquely as h + Dh , where h and h aresymmetric polynomials and D is the “Vandermonde determinant” D = D ( x , · · · , x n ) = Y i 3. Asa general setup, we are working with the noncommutative algebra k − [ x ] unlessotherwise stated. Again there is an overlap between [CA] and this section.The trace of a 3-cycle g acting on k − [ x ] is alsoTr k − [ x ] ( g, t ) = 1(1 − t )(1 − t ) n − , hence A n is generated by quasi-bireflections of k − [ x ]. The aim of this section is toshow that k − [ x ] A n is a cci, which is consistent with the conjectured generalizationof the Kac-Watanabe-Gordeev Theorem. Here the smallest degree antisymmetricpolynomial is O A n ( x x · · · x n − ), and the subring of invariants k − [ x ] A n is gen-erated by O A n ( x x · · · x n − ), and either the n − S , . . . , S n − or the power sums P , . . . , P n − , and so an upper bound on the de-grees of generators of k − [ x ] A n is 2 n − 3. We will show that the Hilbert series of k − [ x ] A n is given by(E4.0.1) H k − [ x ] A n ( t ) = (1 + t )(1 + t ) · · · (1 + t n − )(1 + t n )(1 + t n − )(1 − t )(1 − t ) · · · (1 − t n ) . We construct invariants under A n as O A n ( X I ), the sum of the orbit of a mono-mial X I under A n [Definition 3.2]; we note that the number of terms in this sumis the index of the A n -stabilizer of X I in A n . Lemma 4.1. [CA, Lemma 4.1.1] If there is an odd permutation that stabilizes X I then f = O A n ( X I ) is also invariant under the full symmetric group S n .Proof. Since the index of the subgroup stab A n ( X I ) in stab S n ( X I ) is less thanor equal to [ S n : A n ] = 2, if there is an odd permutation that stabilizes X I then the index [ stab S n ( X I ) : stab A n ( X I )] = 2, and the order of the orbit X I under S n = [ S n : stab S n ( X I )] is the same as the order of the orbit of X I under A n = [ A n : stab A n ( X I )], so the orbit sum of X I under S n is the same as thatunder A n ; hence O A n ( X I ), the orbit sum of X I under A n , is S n -invariant. (cid:3) Here is an immediate consequence. Corollary 4.2. If I = ( i j ) with at least 2 indices i j = i k , an even number, then O A n ( X I ) is an S n -invariant. In particular if there are at least 2 indices i j = i k = 0 then O A n ( X I ) is an S n -invariant. Lemma 4.3. [CA, Lemma 4.1.2] An A n -orbit sum O A n ( X I ) = 0 if and only if I has at least two indices i j = i k an even number, and two indices i r = i s an oddnumber.Proof. If I has repeated even indices then O A n ( X I ) = O S n ( X I ) by Corollary 4.2.Since I has repeated odd indices then O S n ( X I ) = 0 by Lemma 3.5.Conversely, suppose that O A n ( X I ) = 0, then X I and − X I are in the A n -orbitof X I , hence in the S n -orbit of X I . Hence for every τ X I in the S n -orbit of X I we also have − τ X I in the S n -orbit, and so the S n orbit sum is 0, which forces atleast two indices to have the same odd value by Lemma 3.5. We have τ X I = − X I for an even permutation τ . Write τ as a product of disjoint cycles τ = ν · · · ν m µ · · · µ k where the ν i are odd permutations and the µ j are even permutations. Note thatsince τ X I = − X I exponents in I must be constant over the support of each cycle.Suppose there are no repeated even indices in I , so that all repeats are of oddindices. Hence for each µ j = ( a , · · · , a s j +1 ), an even cycle, µ j can be writtenas an even number of transpositions, interchanging variables with the same oddexponent. By the proof of Lemma 3.5 each of these transpositions maps X I to − X I , and hence µ j X I = X I , For similar reasons each ν i X I = − X I . It followsthat τ X I = ν · · · ν m µ · · · µ k X I = X I , a contradiction. Hence I must also containat two indices with the same even number. (cid:3) Note that A n -orbit sums do not necessarily correspond to partitions, e.g. when n = 4 the orbit sums O A n ( x x x x ) and O A n ( x x x x ) are different (and O A n ( x x x x ) + O A n ( x x x x ) = O S n ( x x x x )).Adapting the classical definition, an element g ∈ k − [ x ] is called symmetric (respectively, antisymmetric ) if τ ( g ) = g (respectively, τ ( g ) = − g ) for every oddpermutation τ ∈ S n . Note that g is symmetric if and only if g is S n -invariant. If g is antisymmetric, then g is A n -invariant. The following lemma follows easily. Lemma 4.4. Let f, g, h be elements in k − [ x ] . NVARIANTS OF ( − (1) Linear combinations of antisymmetric invariants are antisymmetric. Henceif f + g and g are antisymmetric invariants, then f is an antisymmetricinvariant. (2) If f = gh with g an antisymmetric invariant and h a symmetric invariant,then f is an antisymmetric invariant. (3) If f = gh with f and g antisymmetric invariants then h a symmetric in-variant. The following lemma follows as in the case of k [ x ] and the proof is omitted. Lemma 4.5. [CA, Theorem 4.1.4] If f is an A n -invariant and σ is the transposition (1 , then σf = τ f for any odd permutation τ . Furthermore f + σf is symmetricand f − σf is antisymmetric. As a consequence, each invariant f ∈ k − [ x ] A n canbe be written uniquely as the sum of a symmetric invariant and an antisymmetricinvariant. Example 4.6. For n ≥ A n -orbits are antisymmetric (e.g. O A n ( x x x ) and O A n ( x x x )) and an-tisymmetric elements can be constructed from the lemma above (e.g. f − σf for f = O A n ( x x x )).For the rest of this section, we assume that n ≥ A is trivial. In the case k − [ x ] we have the two antisymmetric orbit sums given in the lemma below; theorbit sums of these monomials are symmetric polynomials when A n acts on k [ x ]. Lemma 4.7. The A n orbit sums O A n ( x x · · · x n ) = x x · · · x n and O A n ( x x · · · x n − ) are both antisymmetric A n -invariants. And O A n ( x x · · · x n − ) is the smallest de-gree antisymmetric invariant.Proof. It is easy to show that x x · · · x n is an antisymmetric A n -invariant, whence O A n ( x x · · · x n ) = x x · · · x n . So we focus on O A n ( x x · · · x n − ).We note that O A ( x x ) = x x − x x + x x . For n ≥ , n − , n ) to x x · · · x n − we obtain(1 , n − , n )( x x · · · x n − ) = x x · · · x n − x n = − x x · · · x n − x n and similarly(1 , n − , n )( x x · · · x n − ) = x x · · · x n − x n x n − = x x · · · x n − x n − x n and (1 , j, n )( x x · · · x n − ) = x x · · · x j − x n x j +1 · · · x n = ( − n − j x x · · · x j − x j +1 · · · x n so that the n monomials with j th missing variable occur in the A n -orbit with thesign ( − n − j . Since x · · · x n − has repeated odd exponents we have seen that themonomials in the S n -orbit of x · · · x n − occur with both plus and minus signs,and the S n -orbit sum of x x · · · x n − is 0. Hence the S n -orbit of x · · · x n − has2 n elements, and so the S n -stabilizer of x · · · x n − has ( n − / n − / { , . . . , n − } stabilize x · · · x n − somust constitute its stabilizer. Hence the stabilizer in A n must also have ( n − / elements, and hence the A n -orbit of x · · · x n − must be the n elements we havecomputed, and hence O A n ( x · · · x n − ) = ( x · · · x n − ) − ( x · · · x n − x n ) + ( x · · · x n − x n − x n )+ · · · + (( − n − x x · · · x n ) . Then to see the effect of any transposition ( i, j ) on this orbit sum, consider asummand of the orbit sum that contains both i and j and note, as in the argumentabove, that the transposition ( i, j ) changes the sign of this term; since any elementin an orbit represents the orbit, any transposition reverses the sign on the A n -orbitsum of x · · · x n − , and hence O A n ( x x · · · x n − ) is an antisymmetric A n -invariant.There can be no smaller degree antisymmetric A n -invariant since any smallerdegree monomial X I must have at least two zero entries in I , hence O ( X I ) mustbe S n -symmetric, and so no linear combination of such orbits can be antisymmetric. (cid:3) The antisymmetric orbit sum O A n ( x · · · x n ) can be generated from the super-symmetric polynomials and O A n ( x · · · x n − ). Lemma 4.8. The antisymmetric orbit sum O A n ( x · · · x n ) = x · · · x n is gener-ated by the super-symmetric polynomial P = S = O S n ( x ) = O A n ( x ) and theantisymmetric orbit sum O A n ( x · · · x n − ) as follows O A n ( x · · · x n ) = 12 n ( O A n ( x · · · x n − ) S + ( − n − S O A n ( x · · · x n − )) . Proof. Computing O A n ( x · · · x n − ) O A n ( x )= ( x x · · · x n − − x x · · · x n − x n + · · · + ( − n − x · · · x n )( x + · · · + x n )we see that the monomial x · · · x n occurs n times (each with positive sign) as asummand in this product when expanded, and x · · · x n − x = ( − n − x · · · x n − so the respective orbits sums occur in the expanded product. Since there are n monomials in the product O A n ( x · · · x n − ) O A n ( x ), and n ( n − 1) summands in O A n ( x x · · · x n − ) these orbit sums account for all the terms, and so O A n ( x · · · x n − ) O A n ( x ) = n O A n ( x · · · x n ) + ( − n − O A n ( x x · · · x n − ) . Similarly O A n ( x ) O A n ( x · · · x n − ) = O A n ( x x · · · x n − ) + ( − n − n O A n ( x · · · x n ) , and the result follows. (cid:3) Next we note that the super-symmetric polynomial S n = O S n ( x · · · x n − x n ) canbe generated by antisymmetric invariants O A n ( x · · · x n ) and O A n ( x · · · x n − ). Lemma 4.9. The super-symmetric polynomial S n = O S n ( x · · · x n − x n ) can begenerated by antisymmetric invariants O A n ( x · · · x n ) and O A n ( x · · · x n − ) as fol-lows O S n ( x · · · x n − x n ) = ( − ( n − n − / ( O A n ( x · · · x n − ))( O A n ( x · · · x n )) . NVARIANTS OF ( − Proof. The monomial x · · · x n − x n is stabilized by (1 , 2) so S n = O S n ( x · · · x n − x n ) = O A n ( x · · · x n − x n ) , and S n = n X i =1 x · · · x i − x i x i +1 · · · x n . This expression is a sum of n terms, each with x · · · x n as a factor. Consider theproduct ( O A n ( x · · · x n − ))( x · · · x n ), and observe when this product is expandedone term is( x · · · x n − )( x · · · x n ) = ( − n − ( x x · · · x n − )( x · · · x n )= ( − n − ( − n − ( x x · · · x n − )( x · · · x n )= ( − ( n − n − / ( x · · · x n − x n ) , the last equality holding by induction. Since ( O A n ( x · · · x n − ))( x · · · x n ) is an in-variant, the entire orbit sum of this monomial must occur as terms in this expandedproduct, accounting for the n terms in S n yielding the result. (cid:3) Here we are ready to prove a result of Cameron Atkins [CA]. Theorem 4.10. [CA, Theorem 4.2.7] The fixed subring k − [ x ] A n is generated bythe super-symmetric polynomials S , · · · , S n − and the antisymmetric A n -invariant O A n ( x · · · x n − ) (or the odd power sums P , P , . . . , P n − and O A n ( x · · · x n − ) ).Proof. By Lemma 4.8 S = P and O ( x · · · x n − ) generate x x · · · x n , which inturn by Lemma 4.9 generate S n . By Theorem 3.10 S , S , . . . , S n generate all thesymmetric invariants. Hence it suffices to show that any antisymmetric A n -invariant f can be obtained.We will induct on the degree of f , noting that the result is true in degrees ≤ n − O ( x · · · x n − ) is the only antisymmetric A n -invariant of degree ≤ n − X I be the leading term of f under the length-lexicographic order. If σ is antransposition σf = − f also has leading term X I . Hence by applying transpositions,we may assume that f has leading term X I where I is weakly decreasing (and hencecorresponds to a partition). We can write f as a linear combination of distinct orbitsums f = P c I O A n ( X I ) where X I is the highest degree monomial in the orbit, andwhere c I ∈ k [Lemma 3.4]. Since f is antisymmetric σf = P c I σ O A n ( X I ) = − f so that 2 f = f − ( − f ) = P c I ( O A n ( X I ) − σ O A n ( X I )). Hence without loss ofgenerality we may assume that f = O A n ( X I ) − σ ( O A n ( X I )), with X I the leadingterm of f , and with I = ( λ i ) weakly decreasing; (since X I is the leading term of f , and f is antisymmetric, σ ( O A n ( X I )) = O A n ( X I )). Since x · · · x n is centraland symmetric, we can factor it out of f , obtaining an antisymmetric invariant ofsmaller degree. Hence we may assume without loss of generality that λ n = 0 or 1.If λ n = 1, then each x i occurs in all terms of f , so we can factor out ( x · · · x n )from f and write f = h ( x · · · x n ) for some A n -invariant h . It follows that h issymmetric, and we are done. Hence, assume that λ n = 0 and I = ( λ , . . . , λ n − , I . The lowest order possible for I is when I = ( λ , , · · · , n ≥ 3, we have λ n − = 0. If λ n − = 0, then thetransposition τ = ( n − , n ) stabilizes X I and hence O ( X I ) is S n -invariant [Lemma4.1]. Consequently, f = 0 and we are done. Therefore we can assume that λ i = 0for all i = 1 , · · · , n − 1. Let I ∗ = ( λ − , λ − , . . . , λ n − − , decreasing sequence, and let h = O A n ( X I ∗ )+ σ O A n ( X I ∗ ), which is S n -invariant (itis possible that O ( X I ) itself is S n -invariant – e.g. if λ n − = 1 or I ∗ has two evenentries that are equal – in this case h = 2 O A n ( X I ∗ )). Let g = h O A n ( x · · · x n − ),which is an antisymmetric A n -invariant that is a product of a S n -invariant and O A n ( x · · · x n − ). We claim that ± f is a summand of g and that all other termshave lower order; by induction these claims will complete the proof. Notice thatthe terms g and g occur in g where g = ( x λ − x λ − · · · x λ n − − n − )( x · · · x n − ) g = ( x λ − x λ − · · · x λ n − − n − )( x · · · x n − ) , and hence their A n -orbit sums occur in g . Note that g = ± X I and g = ± σX I and σg = − g , and hence ± f is a summand of g . Finally notice that X I is clearlythe leading term of g and so all the other terms of g are of lower order. Hence f ± g is antisymmetric of lower order, hence of the desired form by induction.The argument of Lemma 3.9 shows that S , S , . . . , S n − can be obtained from P , P , . . . , P n − . (cid:3) In the above proof we have shown that antisymmetric invariants correspond topartitions I 7→ O A n ( X I ) − σ O A n ( X I )for any odd permutation σ . This antisymmetric invariant will be non-zero if andonly if 0 = O A n ( X I ) is not S n -invariant, i.e. O A n ( X I ) has no odd permutationsstabilizing it. By the lemma below this is equivalent to I having no repeated evenindices (by Lemma 4.3 this condition also assures O A n ( X I ) = 0.) Lemma 4.11. Let X I be the highest degree lexicographic ordered term in the A n -orbit of X I . Then σ O A n ( X I ) = O A n ( X I ) for an odd permutation σ if and only if I has at least two entries λ j = λ k that are an even number (including ).Proof. If λ j = λ k is even then ( j, k ) X I = X I so S n = A n ∪ A n ( j, k ) and the A n -orbit of X I is the same as the S n -orbit of X I so ( j, k ) O A n ( X I ) = O A n ( X I ), and,in fact, any permutation stabilizes the orbit sum.Conversely, suppose that there is an odd permutation σ with σ O A n ( X I ) = O A n ( X I ). Since σX I is in the A n -orbit of X I we must have σX I = τ X I for τ aneven permutation. Hence τ − σX I = X I so X I is stabilized by an odd permutation.Suppose that I has no repeated even entries, and write σ = ν · · · ν m +1 µ · · · µ k asa product of disjoint cycles, where ν i are odd permutations and µ i are even. Notingthat entries of I in the support of each cycle must be constant and all repeatedentries are assumed to be odd, we see that each µ i X I = X I because µ i is theproduct of an even number of transpositions of variables with the same odd expo-nents and so each transposition changes the sign; since there are an even numberof sign changes µ i X I = X I . However ν i X I = − X I since ν i is the product of anodd number of interchanges of variables to the same odd power, and hence resultsin an odd number of sign changes. Hence σX I = ν · · · ν m +1 µ · · · µ k X I = ν · · · ν m +1 X I = ( − m +1 X I = − X I , contradicting σX I = X I . Hence I must have at least one repeated even entry. (cid:3) NVARIANTS OF ( − We note that in the commutative case the antisymmetric nonzero invariants O A n ( X I ) − σ O A n ( X I ) that corresponding to a partition I are those with all entriesof I distinct.We next compute the Hilbert series for k − [ x ] A n and use it to show that k − [ x ] A n is a cci. For specific values of n the coefficients of these series do not seem to be inthe Online Encyclopedia of Integer Sequences . Lemma 4.12. The Hilbert series of k − [ x ] A n is given by H k − [ x ] A n ( t ) = (1 − t )(1 − t )(1 − t ) · · · (1 − t n − )(1 + t n )(1 + t n − )(1 − t )(1 − t )(1 − t ) · · · (1 − t n − )(1 − t n )(1 + t n − ) . Proof. By remarks above in each dimension the invariants are vector space directsums of the symmetric invariants and the antisymmetric invariants, so the Hilbertseries H k − [ x ] A n ( t ) for the invariants under A n is the sum of H k − [ x ] S n ( t ) and thegenerating function S n ( t ) for s n ( k ), the number of partitions of k with at most n parts having no repeated even parts (not even 0). By Proposition 6.3 of theAppendix we have S n ( t ) = D n ( t ) t n − (1 + t )(1 + t n − ) . Hence H k − [ x ] A n ( t ) = D n ( t ) + S n ( t ) = D n ( t ) + D n ( t ) t n − (1 + t )(1 + t n − )= D n ( t ) (cid:18) t n − (1 + t )(1 + t n − ) (cid:19) = D n ( t ) (1 + t n )(1 + t n − )(1 + t n − )= (1 − t )(1 − t )(1 − t ) · · · (1 − t n − )(1 + t n )(1 + t n − )(1 − t )(1 − t )(1 − t ) · · · (1 − t n − )(1 − t n )(1 + t n − ) . Canceling yields the expression in equation (E4.0.1). (cid:3) Consider the algebras given by B n − = k [ p , · · · , p n ][ y : τ , δ ] · · · [ y n − : τ n − , δ n − ] B n +1 = B n − [ y n +1 ; τ n +1 ] B n +2 = B n +1 [ y n +2 ; τ n +2 , δ n +2 ] . For i ≤ n − τ i and δ i as for the algebra B considered in the previoussection (note that B is not a subalgebra of C since y n is not adjoined). Define the τ n +1 by letting it be the identity on R = k [ p , · · · , p n ] and τ n +1 ( y i ) = ( − n − y i for i ≤ n − 1. Then τ n +1 extends uniquely to an algebra automorphism of B n − .Define the algebra automorphism τ n +2 of D n by letting it be the identity on R andletting τ n +2 ( y i ) = ( ( − n y i if i ≤ n − , ( − n +1 y n +1 if i = n + 1 . The derivation δ n +2 is given by letting δ n +2 ( a ) = 0 for all a ∈ R , δ n +2 ( y i ) =( − n − na i − y n +1 for i ≤ n − 1, and δ n +2 ( y n +1 ) = 0 . Recall that a i − = f i − ( p , p , . . . p n ) where f i − is given by (E3.7.1). Lemma 4.13. Retain the above notation. (1) τ n +2 is an algebra automorphism of B n +1 . (2) δ n +2 is a τ n +2 -derivation of B n +1 .Proof. (1) It is straightforward to check that τ n +2 is an algebra automorphism of B n +1 .(2) The relations of B n +1 are of the form y i a − ay i = 0 , ∀ i = 1 , · · · , n − , n + 1 , a ∈ Ry i y j + y j y i = 2 a i +2 j − , ∀ ≤ i, j ≤ n − ,y n +1 y i + ( − n y i y n +1 = 0 , ∀ i = 1 , · · · , n − . The proof of δ n +2 preserving the relations y i a − ay i = 0 is similar to the proof ofLemma 2.3(2). Now we show that δ n +2 preserves other relations. For i, j ≤ n − δ n +2 ( y i y j + y j y i − a i +2 j − )= δ n +2 ( y i ) y j + τ n +2 ( y i ) δ n +2 ( y j ) + δ n +2 ( y j ) y i + τ n +2 ( y j ) δ n +2 ( y i )= ( − n − na i − y n +1 y j + ( − n y i ( − n − na j − y n +1 + ( − n − na j − y n +1 y i + ( − n y j ( − n − na i − y n +1 = 0For i ≤ n − 1, we have δ n +2 ( y n +1 y i + ( − n y i y n +1 )= τ n +2 ( y n +1 ) δ n +2 ( y i ) + ( − n δ n +2 ( y i ) y n +1 = ( − n +1 y n +1 ( − n − na i − y n +1 + ( − n ( − n − na i − y n +1 y n +1 = 0 . (cid:3) The above lemma verifies that δ n +2 is a τ n +2 -derivation. Let C = B n +2 . Thealgebra C is AS regular of dimension 2 n + 1. Grade C by letting degree( y i ) = 2 i − i ≤ n − 1, degree( y n +1 ) = n , and degree( y n +2 ) = n − . Then the Hilbert seriesof C is given by H C ( t ) = 1(1 − t )(1 − t ) · · · (1 − t n − )(1 − t n − )(1 − t n )(1 − t n )(1 − t n − ) . Since A n ≤ S n , the algebra k [ x , x , . . . , x n ] S n is a subalgebra of k − [ x ] A n . Then k [ x , x , . . . , x n ] S n = k [ ρ , ρ , . . . , ρ n ] , a commutative polynomial ring where ρ i = σ i ( x , x , . . . , x n ) and σ i is the i th elementary polynomial. Observe that(E3.12.1) O A n ( x · · · x n − ) = ±O A n (( x · · · x n − ) ) = ±O A n ( x · · · x n − )because( x · · · x k · · · x n − )( x · · · x k − x k +1 · · · x n )= ( − n − ( x · · · x k · · · x n − x n )( x · · · x k − x k +1 · · · x n − )= ( − n ( − ( n − k )+( k − ( x · · · x k − x k +1 · · · x n − x n )( x · · · x k − x k x k +1 · · · x n − )= − ( x · · · x k − x k +1 · · · x n − x n )( x · · · x k − x k x k +1 · · · x n − ) , so the orbits of the cross-terms cancel out, leaving only an orbit in the x i thatis symmetric. Hence we can write O A n ( x x · · · x n − ) = g ( ρ , ρ , . . . , ρ n ) for apolynomial g . Similarly, ( x x · · · x n ) = ± x x · · · x n = h ( ρ , ρ , . . . , ρ n ) for apolynomial h . NVARIANTS OF ( − As in the previous section let for i ≤ n − r i = y i − a i − . Let(E4.13.1) b = g ( p , p , . . . , p n )and(E4.13.2) b = h ( p , p , . . . , p n )and consider two additional relations r n +1 = y n +1 − b and r n +2 = y n +2 − b .The proof of the following lemma is the same as that of Lemma 3.11. Lemma 4.14. The sequence { r , r , . . . , r n − , r n +1 , r n +2 } is a central regular se-quence in C . We are now ready to show that k − [ x ] A n is a cci. Theorem 4.15. The algebra k − [ x ] A n is a cci.Proof. Note that O A n ( x x · · · x n − ) and x x · · · x n are elements of k − [ x ] A n .Consider the algebra C constructed above and define a map φ : C −→ k − [ x ] A n as follows: for i ≤ n let φ ( p i ) = ρ i ; for i ≤ n − φ ( y i ) = P i − ; let φ ( y n +1 ) = x x · · · x n ; and let φ ( y n +2 ) = O A n ( x x · · · x n − ) . Note that φ takes k [ p , p , . . . , p n ] isomorphically onto k [ ρ , ρ , . . . , ρ n ]. In the proof of Theorem 3.12it was shown that φ preserves the skew polynomial relations associated to y i for i ≤ n − 1. Calculating shows that ( x x · · · x n ) P i − = ( − n − ( x x · · · x n ) P i − ,and hence φ preserves the relation associated to y n +1 . Further calculation showsthat O A n ( x x · · · x n − ) P i − = ( − n P i − O A n ( x x · · · x n − )+ ( − n − nP i − · ( x x · · · x n ) . Since O A n ( x x · · · x n − )( x x · · · x n ) = ( − n − ( x x · · · x n ) O A n ( x x · · · x n − )and P i − = f i − ( ρ , ρ , . . . , ρ n ), the relation associated to y n +2 is preserved by φ . Hence φ is a graded ring homomorphism. The homomorphism φ is onto byTheorem 4.10. By (E3.12.1)0 = O A n ( x x · · · x n − ) − O A n ( x x · · · x n − )= O A n ( x x · · · x n − ) − g ( ρ , ρ . . . . , ρ n ) = φ ( y n +2 − b ) = φ ( r n +2 ) . Similarly, φ ( r n +1 ) = φ ( y n +1 − b ) = 0 . As in the proof of Theorem 3.12 φ ( r i ) = 0 for i ≤ n − 1. Hence ( r , r , . . . , r n − , r n +1 , r n +2 ) ⊆ ker( φ ), and φ induces a graded ringhomomorphism ¯ φ : C −→ k − [ x ] A n where C = C/ ( r , r , . . . , r n − , r n +1 , r n +2 ) . We have degree( r i ) = 4 i − i ≤ n − 1, degree( r n +1 ) = 2 n , and degree( r n +2 ) =2 n − 2. Since { r , r , . . . , r n − , r n +1 , r n +2 } is a regular sequence, the Hilbert seriesof C is given by H C ( t ) = (1 − t )(1 − t )(1 − t ) · · · (1 − t n − )(1 − t n )(1 − t n − )(1 − t )(1 − t ) · · · (1 − t n − )(1 − t n )(1 − t n )(1 − t n − )= (1 − t )(1 − t )(1 − t ) · · · (1 − t n − )(1 − t n )(1 − t n − )(1 − t )(1 − t ) · · · (1 − t n − )(1 − t n )(1 − t n )(1 − t n − ) (1 − t n − )(1 − t n − )= (1 − t )(1 − t )(1 − t ) · · · (1 − t n − )(1 + t n )(1 + t n − )(1 − t )(1 − t )(1 − t ) · · · (1 − t n − )(1 − t n )(1 + t n − ) . This is the Hilbert series of k − [ x ] A n , and hence the ring homomorphism ¯ φ is anisomorphism as desired. The assertion follows. (cid:3) Theorem 4.16. ⌊ n ⌋ = cyc ( k − [ x ] S n ) ≤ cci + ( k − [ x ] S n ) ≤ ⌊ n ⌋ + 1 . Proof. First we prove the claim that cci + ( k − [ x ] S n ) ≤ ⌊ n ⌋ + 1.Following the proof of Theorem 3.14, let C be the subalgebra of k − [ x ] S n definedbefore Lemma 2.4, which is (isomorphic to) the iterated Ore extension k [ P , P , · · · , P ⌊ n ⌋ ][ P ][ P ; τ , δ ] · · · [ P n ′ ; τ n ′ , δ n ′ ]where n ′ = 2 ⌊ n − ⌋ +1. Let F n − be the iterated Ore extension defined in the proofof Theorem 3.14. (We are not going to use F n − , instead we will define two newalgebras H n − and H n +1 .) Recall from the proof of Theorem 4.15 that p i is theimage of P i for all i = 1 , · · · , n . By Lemma 2.4(5), P i are in C for all i . Define H n − = F n − [ Q n − ; φ n − ] where φ n − : P i ( − i ( n − P i for all even i andall odd i ≤ n − P i appeared in F n − . It is easy to check that φ n − is an algebraautomorphism of F n − and therefore H n − is an iterated Ore extension. Define H n +1 = H n − [ Q n +1 ; φ n +1 , λ n +1 ] where φ n +1 is an algebra automorphism de-termined by φ n +1 : ( P i ( − in P i for even i or odd i ≤ n − Q n − ( − n +1 Q n − (seethe proof of Lemma 4.13(1)), and φ n +1 -derivation λ n +1 is determined by λ n +1 : P i i is even and i ≤ nP i ( − n +1 nQ n − f i − ( P , · · · , P n ) if i is odd and i ≤ n − Q n − , where f i − is given by (E3.7.1). Similar to the proof of Lemma 4.13(2), one canshow that λ n +1 is a φ n +1 -derivation, therefore H n +1 is an iterated Ore extension.Let u s = P s − − P s − for all integers from s = ⌊ n − ⌋ + 2 to s = n − 1. Let u n +1 be Q n − − b where b ∈ C ⊂ C is defined in (E4.13.2). Let u n +2 be Q n +1 − b where b ∈ C ⊂ C is defined in (E4.13.1).The proof of Lemma 3.11 (see also Lemma 4.14) shows that { u ⌊ n − ⌋ +2 , · · · , u n − , u n +1 , u n +2 } is a central regular sequence of H n +1 . It is straightforward to see that H n +1 / ( u ⌊ n − ⌋ +2 , · · · , u n − , u n +1 , u n +2 ) ∼ = k − [ x ] S n . Therefore cci + ( k − [ x ] S n ) ≤ n + 1 − ( ⌊ n − ⌋ + 1) = ⌊ n ⌋ + 1 and we proved the claim.By Theorem 4.15 H k − [ x ] A n ( t ) = H k − [ x ] S n ( t ) (1 + t n )(1 + t n − )1 + t n − = Q ns = ⌊ n − ⌋ +2 (1 − t s − ) Q ⌊ n ⌋ j =1 (1 − t j ) Q ni =1 (1 − t i − ) (1 − t n − )(1 − t n )(1 − t n − )(1 − t n − )(1 − t n )(1 − t n − )= Q n − s = ⌊ n − ⌋ +2 (1 − t s − ) Q ⌊ n ⌋ j =1 (1 − t j ) Q n − i =1 (1 − t i − ) (1 − t n )(1 − t n − )(1 − t n )(1 − t n − )= Q n − s = ⌊ n − ⌋ +2 (1 − t s − ) Q ⌊ n − ⌋ j =1 (1 − t j ) Q n − i =1 (1 − t i − ) (1 − t ⌊ n − ⌋ +1) )(1 − t n )(1 − t n − )which is an expression satisfying the condition in Definition 3.13(2). Hence ⌊ n ⌋ = cyc ( k − [ x ] S n ) ≤ cci + ( k − [ x ] S n ) ≤ ⌊ n ⌋ + 1 . NVARIANTS OF ( − (cid:3) Question 4.17. Let A be either k − [ x ] S n and k − [ x ] A n . Let E ( A ) be the Ext-algebra Ext ∗ A ( k, k ).(1) Is E ( A ) noetherian?(2) What is the GK-dimension of E ( A )?5. Converse of Kac-Watanabe-Gordeev Theorem Kac-Watanabe-Gordeev showed that when k [ x ] G is a complete intersection then G must be generated by classical bireflections. We next prove the converse of thisresult for k − [ x ] G when G ⊂ S n and note that the converse is not true for k [ x ] G . ByLemma 1.7(3) a quasi-bireflection must be a 2-cycle or a 3-cycle. We conclude byshowing that for subgroups G of S acting on k − [ x , x , x , x ], the fixed subring k − [ x , x , x , x ] G is a cci if and only if G is generated by quasi-bireflections, andwhen G is not generated by quasi-bireflections, k − [ x , x , x , x ] G is not cyclotomicGorenstein, hence k − [ x , x , x , x ] G is not any of the kinds of complete intersec-tions described in Definition 1.8. The following result on permutation groups maybe well-known, but is included for completeness. Proposition 5.1. Let G be a subgroup of S n . (1) If G is generated by 3-cycles, then G is an internal direct product of alter-nating groups. (2) If G is generated by 3-cycles and 2-cycles, then G is an internal directproduct of alternating and symmetric groups. We first prove some lemmas. Let X be any subset of { i } ni =1 := { , · · · , n } . Weuse S X for the full symmetric group of X . Proof. Suppose that G is generated by 3-cycles and 2-cycles. We may assumethat G = h τ , τ , . . . , τ ℓ i where τ , τ , . . . , τ ℓ are all of the 3-cycles and 2-cycles in G . Let X = { , , . . . , n } . We will show that there are disjoint nonempty subsets X , X , . . . , X k of X such that G = G × G ×· · ·× G k where G i is the alternating orsymmetric group on X i . Given a permutation σ define M ( σ ) = { x ∈ X : σ ( x ) = x } ,the set of elements that are moved by σ . Let Y = [ σ ∈ G M ( σ ) and define a relation ∼ on Y by x ∼ y if there exists 3-cycles and/or 2-cycles σ , σ , . . . , σ m such that x ∈ M ( σ ) , y ∈ M ( σ m ) and M ( σ i ) ∩ M ( σ i +1 ) = ∅ for i = 1 , , . . . , m − . In this casewe say that there is a path from x to y . It is easy to see that ∼ is an equivalencerelation on Y . Let X , X , . . . , X k be the equivalence classes. We view the X i asthe path connected components of Y . Clearly either M ( τ j ) ⊆ X i or M ( τ j ) ∩ X i = ∅ for all i, j . Let G i = h τ j : M ( τ j ) ⊆ X i i .Case 1: Suppose that G is generated by 3-cycles. It will be sufficient to showthat each G i is an alternating group. Furthermore, there is no loss of generalityin assuming that there is one component Y . We will induct on ℓ . If | Y | = 3,(the smallest possible) then G = h τ i ∼ = A . If | Y | = 4, we may assume that Y = { , , , } , τ = (1 , , 3) and τ = (2 , , |h τ ih τ i| = 9 and G must be all of A . Inductively assume that whenever G = h τ , τ , . . . , τ ℓ i has onecomponent Y with 4 ≤ | Y | = s ≤ n , then G ∼ = A s . We may let Y = { , , . . . , s } . Now suppose that G ′ = h τ , τ , . . . , τ ℓ +1 i where τ ℓ +1 is a 3-cycle, and Y ′ = ℓ +1 [ M ( τ j ) is connected. Let τ i , τ i , . . . , τ i m be a maximal path in Y ′ . Then ∪ j = i m M ( τ j ) mustbe connected, for otherwise, we could extend the path. Hence there is no loss ofgenerality in assuming that τ ℓ +1 is such that Y = [ i = ℓ +1 M ( τ i ) is connected with | Y | = s . Let G = h τ , τ , . . . , τ ℓ i . There are two subcases.Case 1.1: | Y ′ | = s + 1 . We may assume, renumbering if necessary, that τ ℓ +1 =( s − , s, s + 1) . We will show that G ′ contains all elements that are products oftwo disjoint 2-cycles. The set of all such generates a normal subgroup of A s +1 ,and hence we would have G ′ = A s +1 . By induction we have all disjoint products( i, j )( k, ℓ ) where i, j, k, ℓ ≤ s. If i, j ≤ s − 1, then ( i, j )( s − , s )( s − , s, s + 1) =( i, j )( s, s +1). Then ( s − , k )( i, j ) · ( i, j )( s, s +1) = ( s − , k )( s, s +1) . The conjugation( k, s, ℓ )( i, j )( s, s + 1)( k, ℓ, s ) = ( i, j )( ℓ, s + 1) gives the remaining products. Thus G ′ = A s +1 and the result follows by induction.Case 1.2: | Y ′ | = s + 2 . We may assume that τ ℓ +1 = ( s, s + 1 , s + 2) . By induction G is A s and we have the following chain from 1 to s − , , , (2 , , , . . . , ( s − , s − , s − . Computing(1 , s − , s )( s, s + 1 , s + 2)(1 , s − , s ) = ( s − , s + 1 , s + 2) , and ( s − , s + 1 , s + 2) ∈ G ′ . We have that Y ′′ = { , , . . . , s − } ∪ { s + 1 , s + 2 } is a connected component, and by induction G ′′ = h A s − , ( s − , s + 1 , s + 2) i is acopy of A s +1 . Then G = h G ′′ , τ ℓ +1 i is the alternating group A s +2 by Case 1.1.Case 2: Once again there is no loss of generality in assuming that there is oneconnected component. We may also suppose that G contains at least one 2-cycle byCase 1. Again the proof is by induction on ℓ . If | Y | = 2 , the result is clear. Since(1 , , 3) = (1 , , | Y | = 3, then G = S . Inductively assume thatwhenever G = h τ , τ , . . . , τ ℓ i with 3 ≤ | Y | ≤ n then G is a symmetric group. Nowsuppose that G ′ = h τ , τ , . . . , τ ℓ +1 i with Y ′ = ℓ +1 [ M ( τ j ) connected. Again we mayassume that Y = [ j = ℓ +1 M ( τ j ) is connected with | Y | = s ≤ n . Then by inductionor by case 1 we have that G = h τ , τ , . . . , τ ℓ i is either a symmetric group or analternating group (if all τ i for i ≤ ℓ are 3-cycles). We have two subcases.Case 2.1: τ ℓ +1 is a 3-cycle. By the argument in Case 1, G ′ contains the fullalternating group. Since G ′ must also contain a 2-cycle, it is the full symmetricgroup.Case 2.2: τ ℓ +1 is a 2-cycle. Without loss of generality we may assume that τ ℓ +1 = ( s, s + 1) . As noted, G is either the symmetric group or the alternatinggroup. In this case G ′ must contain(1 , s − , s )( s, s + 1) = (1 , s − , s, s + 1) . Squaring yields that ( s − , s + 1 , s ) ∈ G ′ . By Case 1.1, G ′ contains the fullalternating group. Since it also contains a 2-cycle, it must be the full symmetricgroup.The result follows by induction. (cid:3) NVARIANTS OF ( − Let A and B be two graded algebra. Define A ⊗ − B be the Z -graded twist ofthe tensor product A ⊗ B by the twisting system σ := { σ i,j = Id i ξ j − | ( i, j ) ∈ Z } where ξ − maps a ⊗ b ( − | a | + | b | a ⊗ b for all a ⊗ b ∈ A ⊗ B . The following lemmasare easy to check. Lemma 5.2. Retain the above notation. (1) A ⊗ − B = A ⊗ B as Z -graded vector spaces. (2) Identifying A with A ⊗ ⊂ A ⊗ − B and identifying B with ⊗ B ⊂ A ⊗ − B .Then A and B are subalgebras of A ⊗ − B , and the algebra A ⊗ − B is equalto the vector space generated by the products AB ( and BA respectively ) . (3) Under the identification in part (2), ab = ( − | a | | b | ba for all a ∈ A and b ∈ B . Lemma 5.3. Let m < n . (1) k − [ x , · · · , x m ] ⊗ − k − [ x m +1 , · · · , x n ] ∼ = k − [ x , · · · , x n ] . (2) If G ⊂ Aut( A ) and G ⊂ Aut( B ) , then ( A ⊗ − B ) G × G = A G ⊗ − B G . (3) [KKZ3, Lemma 2.7] If A and B are AS regular, then so is A ⊗ − B . (4) Suppose A = R/ (Ω , · · · , Ω m ) and B = C/ ( f , · · · , f d ) where R and C are AS regular and { Ω i } mi =1 and { f j } dj =1 are regular normal sequences ofpositive even degrees. If R ⊗ C is noetherian, then A ⊗ − B is a factor ringof a noetherian AS regular algebra modulo a regular normal sequences ofpositive even degrees. As a consequence, A ⊗ − B is a cci. For any subset X of [1 , · · · , n ], let S X denote the symmetric group of X (allpermutations of X ). Theorem 5.4. If G is a subgroup of S n generated by quasi-bireflections, then k − [ x ] G is a cci.Proof. We use induction on n . Suppose the assertion holds for G ⊂ S m for all m ≤ n − 1. Now let G be a subgroup of S n generated by quasi-bireflections. If G is { } , the assertion is trivial. If G = S n or A n , the assertion follows fromTheorems 3.12 and 4.15. Otherwise, by Proposition 5.1, there is a disjoint union X ∪ Y = [1 , · · · , n ] such that G is a product of G and G , where G and G aresubgroups S X and S Y respectively, and further G is either S X or A X and G isgenerated by quasi-bireflections of k − [ x i | i ∈ Y ] (or equivalently, 2- or 3-cycles of S Y ). By induction, both A G and B G are cci, where A = k − [ x i | i ∈ X ] and B = k − [ x i | i ∈ Y ]. It follows from Lemma 5.2 and 5.3 that k − [ x ] G ∼ = A G ⊗ − B G isa cci. (cid:3) The following example shows that for k [ x ] permutation groups generated byclassical bireflections need not have a fixed ring that is a complete intersection. Example 5.5. Let S act on A := k [ x , x , x , x , x ] by permuting the vari-ables. Let G = h (1 , , , (2 , , i . These two generators are classical bire-flections. Note that (1 , , · (2 , , 5) = (1 , , , , h (1 , , , (1 , , , , i is a copy of the dihedral group D of order 10 and is in fact all of G . Using Molien’s Theorem we have H A G ( t ) = 110 (cid:18) − t ) + 5(1 − t ) (1 + t ) + 41 − t (cid:19) = t − t + 2 t − t + 1(1 − t ) (1 − t ) (1 − t ) . The numerator is an irreducible polynomial that is not cyclotomic; in fact, none ofits zeros are roots of unity. Hence A G cannot be a complete intersection.We conclude by computing the invariants of A = k − [ x , x .x , x ] under eachof the subgroups of S . In this case the conjectured generalization of the Kac-Watanabe-Gordeev Theorem becomes both necessary and sufficient. We show that A H is a cci if and only if H is generated by quasi-bireflections (i.e. 2-cycles or3-cycles); when H is not generated by quasi-bireflections A H is not cyclotomicGorenstein – hence not any kind of complete intersection by Theorem 1.10. Example 5.6. For the following subgroups H of S we consider the fixed subring A H . We show that A H is either a cci or not cyclotomic Gorenstein (and hence noneof the kinds of complete intersection we considered in Definition 1.8). • If H is the full symmetric group or the alternating group, both generatedby quasi-bireflections, we have shown that A H is a cci. Similarly, cyclicsubgroups generated by a 2-cycle (so isomorphic to S ) or by a 3-cycle (soisomorphic to A ) are also easily seen to give ccis when they act on A = k − [ x , x , x , x ] (we showed they did when they acted on A = k − [ x , x ]and A = k − [ x , x , x ] and the results extend by fixing the remainingvariable(s)). • Let H be the subgroup of order 2 generated by an element that is a productof two disjoint 2-cycles, e.g. (12)(34); this subgroup is not generated byquasi-bireflections of A (it is generated by a bireflection of k [ x , x , x , x ]).Molien’s Theorem shows that the Hilbert series of A H is1 − t + 4 t − t + t (1 − t ) (1 + t ) which has zeros that are not roots of unity. Hence A H is not cyclotomicGorenstein. • We have already noted (Example 1.6) that the subgroup H generated by a4-cycle is not generated by quasi-bireflections, and that the invariants A H are not cyclotomic Gorenstein. • Let H be the Klein-Four subgroup generated by two disjoint 2-cycles (e.g. H = h (12) , (34) i ). Then H is generated by quasi-bireflections of A , thegenerators of A H are x + x , x + x , x + x , x + x , Hilbert series of A H is 1 − t + t (1 − t )(1 + t ) , and A H ∼ = k [ p , p , q , q ][ y ][ y ; τ , δ ][ z ; τ , δ ][ z ; τ , δ ] h y − a , y − a , z − b , z − b i , where p , p (resp., q , q ) correspond to the first two symmetric polynomi-als in x , x (resp., x , x ), y , y (resp., z , z ) correspond to x + x , x + x (resp., x + x , x + x ). NVARIANTS OF ( − • The Klein-Four subgroup of even permutations H = { , (12)(34) , (13)(24) , (14)(23) } , which is not generated by quasi-bireflections of A . The Hilbert series of A H is 1 − t + 5 t − t + t (1 − t ) (1 + t ) , so A H is not cyclotomic Gorenstein. • Let H be a subgroup S of order 6. Then H is isomorphic to the symmetricgroup S , without loss of generality of the form H = h (123) , (12) i . Thisgroup is generated by quasi-bireflections, and A H is a complete intersection(we showed this for k − [ x , x , x ] and the extension to A is not difficult). • Let H be a dihedral group of order 8 (a Sylow-2 subgroup of S ). Then H is of the form D = { , (1234) , (13)(24) , (1432) , (13) , (24) , (12)(34) , (14)(23) } , so not generated by quasi-bireflections. The Hilbert series of the fixedsubring is 1 − t + 5 t − t + 5 t − t + 5 t − t + t (1 − t ) (1 + t )(1 + t ) = (1 − t + t )(1 − t + 2 t − t + 2 t − t + t )(1 − t ) (1 + t )(1 + t ) so A H is not cyclotomic Gorenstein.Note: It might be nice to know degrees of generators and how they compare to n = 16. Question 5.7. For H a subgroup of S n , is k − [ x ] H a cci if and only if H isgenerated by quasi-bireflections? 6. Appendix In this section we find generating functions for the class of restricted partitionshaving no repeated odd parts and the class having no repeated even parts. It isincluded since we were unable to find them in the literature.Let d n ( k ) be the number of partitions of k with at most n parts having norepeated odd parts. Make the convention that d n (1) = 1 and d n ( ℓ ) = 0 for ℓ < . Let D n ( t ) be the corresponding generating function D n ( t ) = ∞ X k =0 d n ( k ) t k . There is only one way to partition k into 1 part, so D ( t ) = 1 + t + t + t + · · · + t k + · · · = 11 − t = 1 − t (1 − t )(1 − t )We will now try to find a recurrence relation for d n ( k ) . We will write a partition P of k having at most n parts as P = p , p , . . . , p n where p ≥ p ≥ . . . ≥ p n and k = p + p + · · · + p n . Let D n,k = {P = p , p , . . . , p n : with no repeated odd parts } . Then we have D n,k = {P : p n = 0 } ∪ d {P : p n = 1 } ∪ d {P : p n ≥ } . • Clearly |{P : p n = 0 }| = d n − ( k ) . • If p n = 1, consider the association P 7→ P ′ = p − , p − , . . . , p n − − , . Since p n − > p n = 1 , this will be a partition of k − − n − 1) = k − n + 1 . Since parity is preserved there will be no repeated odd parts, and every suchpartition of k − n + 1 can occur in this manner. Hence |{P : p n = 1 }| = d n − ( k − n + 1) . • If p n ≥ , consider the association P 7→ P ′ = p − , p − , . . . , p n − . This will be a partition of k − n with no repeated odd parts. Once againevery such partition can occur in this manner. Hence |{P : p n ≥ }| = d n ( k − n ) . This yields the following recurrence relation d n ( k ) = d n − ( k ) + d n − ( k − n + 1) + d n ( k − n ) . In terms of generating functions we have D n ( t ) = D n − ( t ) + D n − ( t ) t n − + D n ( t ) t n . This gives the recurrence D n ( t ) = D n − ( t ) (1 + t n − )(1 − t n )= D n − ( t ) (1 − t n − )(1 − t n − )(1 − t n ) . Using this last recurrence relation a simple induction argument proves the fol-lowing Proposition. Proposition 6.1. The generating function D n ( t ) for the number of partitions withat most n parts having no repeated odd parts is given by D n ( t ) = (1 − t )(1 − t )(1 − t ) · · · (1 − t n − )(1 − t )(1 − t )(1 − t ) · · · (1 − t n − )(1 − t n ) . Remark 6.2. We note using the Online Encyclopedia of Integer Sequences (http://oeis.org/)for specific values of n we found that D n ( t ), the Hilbert series of k − [ x ] S n , is alsothe Hilbert series of the invariants of A = k [ y . . . , y n ] ⊗ E ( e , . . . , e n ) under theaction of S n , where k is any field of characteristic not equal to two, the degree ofeach y i = 2, E ( e , . . . , e n ) is the exterior algebra on elements e i of degree 1, and S n acts on both k [ y . . . , y n ] and E ( e , . . . , e n ) by permutations. (See [AM, pp.110-11]). We note that one can filter k − [ x ] by letting I be the ideal generated by { x , . . . , x n } . Then the associated graded algebragr( k − [ x ]) = k − [ x ] /I ⊕ I/I ⊕ I /I ⊕ · · · ⊕ I m /I m +1 ⊕ · · · NVARIANTS OF ( − is isomorphic as a graded algebra to A under the map that associates y i x i + I and e i x i + I . Further the action of S n on k − [ x ] extends to an action ongr( k − [ x ]), and A S n ∼ = gr( k − [ x ]) S n ∼ = gr( k − [ x ] S n ) . Since gr( k − [ x ] S n ) has the same Hilbert series as k − [ x ] S n , it follows that D n ( t )is the Hilbert series of k − [ x ] S n .Let s n ( k ) be the number of partitions of k with at most n parts having norepeated even parts (not even repeated 0 parts), and let S n ( t ) be the correspondinggenerating function. The purpose of this section is to find S n ( t ).First we briefly consider a slight variation. Let w n ( k ) be the number of partitionsof k with exactly n nonzero parts having no repeated even parts, and let W n ( t ) bethe corresponding generating function. Let P be such a partition. Correspond to P the partition P 7→ P ′ = p − , p − , . . . , p n − 1. This will be a partition of k − n with at most n parts having no repeated odd parts, and any such partitioncan occur in this manner. Hence w n ( k ) = d n ( k − n ), and W n ( t ) = t n D n ( t ) . Let S n,k be the collection of all partitions of k with at most n parts having norepeated even parts. Then we have S n,k = {P : p n = 0 } ∪ d {P : p n = 1 } ∪ d {P : p n ≥ } . Since there are no repeated empty parts, the partitions in the first set will bepartitions having exactly n − |{P : p n = 0 }| = w n − ( k ) . For eachpartition P in the second set we correspond P 7→ P ′ = p , p , . . . , p n − , 0, whichwill be a partition of k − n − |{P : p n = 1 }| = w n − ( k − . Similar to the no repeated odd case we see that |{P : p n ≥ }| = s n ( k − n ) . Thisgives the recurrence relation s n ( k ) = w n − ( k ) + w n − ( k − 1) + s n ( k − n )In terms of generating functions we have S n ( t ) = W n − ( t ) + W n − ( t ) t + S n ( t ) t n , and(E6.2.1) S n ( t ) = W n − ( t ) (1 + t )(1 − t n ) = D n − ( t ) (1 + t ) t n − )(1 − t n ) . Summarizing we have the following Proposition. Proposition 6.3. If S n ( t ) is the generating function for the number of partitionshaving at most n parts with no repeated even parts, then S n ( t ) = (1 − t )(1 − t )(1 − t ) · · · (1 − t n − ) t n − (1 + t )(1 − t )(1 − t )(1 − t ) · · · (1 − t n − )(1 − t n )(1 + t n − ) . and S n ( t ) = D n ( t ) t n − (1 + t )(1 + t n − ) . Proof. From (E6.2.1) we have S n ( t ) = D n − ( t ) t n − (1 + t )(1 − t n )= (cid:18) (1 − t )(1 − t ) · · · (1 − t n − )(1 − t )(1 − t ) · · · (1 − t n − ) (cid:19) (cid:18) t n − (1 + t )(1 − t n ) (cid:19) = (cid:18) (1 − t ) · · · (1 − t n − )(1 − t ) · · · (1 − t n − ) (cid:19) (cid:18) t n − (1 + t )(1 − t n ) (cid:19) (1 − t n − )(1 + t n − )(1 − t n − )(1 + t n − )= (1 − t )(1 − t )(1 − t ) · · · (1 − t n − ) t n − (1 + t )(1 − t )(1 − t )(1 − t ) · · · (1 − t n − )(1 − t n )(1 + t n − )= D n ( t ) t n − (1 + t )(1 + t n − ) . (cid:3) Acknowledgments. Some of the results (e.g., Theorems 3.10 and 4.10) in thispaper appeared in the Master Thesis [CA] of James Cameron Atkins, under thedirection of E. Kirkman. The authors thank James Cameron Atkins for the analysisgiven in his Thesis on which Sections 3 and 4 of this paper are based. E. Kirkmanwas partially supported by the Simons Foundation (grant no. 208314) and J.J.Zhang was partially supported by the National Science Foundation (NSF DMS0855743). References [AM] A. Adem and R. J. Milgram, “Cohomology of Finite Groups”, Grundlehren der mathema-tischen Wissenschaften Vol 309, Springer-Verlag, Berlin, 1994.[Be] D.J. Benson, Polynomial invariants of finite groups, London Mathematical Society LectureNote Series, 190. Cambridge University Press, Cambridge, 1993.[CA] J. Cameron Atkins, The invariant theory of k − [ x , · · · , x n ] under premutation reprensen-tations, Mater Thesis, Wake Forest Unibersity, 2012.[Ch] C. Chevalley, Invariants of finite groups generated by reflections, Amer. 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Box 7388, Wake Forest University,Winston-Salem, NC 27109 E-mail address : [email protected] Kuzmanovich: Department of Mathematics, P. O. Box 7388, Wake Forest University,Winston-Salem, NC 27109 E-mail address : [email protected] Zhang: Department of Mathematics, Box 354350, University of Washington, Seattle,Washington 98195, USA E-mail address ::