Actions of Small Groups on Two-Dimensional Artin-Schelter Regular Algebras
aa r X i v : . [ m a t h . R A ] O c t ACTIONS OF SMALL GROUPS ON TWO-DIMENSIONALARTIN-SCHELTER REGULAR ALGEBRAS
SIMON CRAWFORD
Abstract.
In commutative invariant theory, a classical result due to Auslander says thatif R = k [ x , . . . , x n ] and G is a finite subgroup of Aut gr ( R ) ∼ = GL( n, k ) which contains noreflections, then there is a natural graded isomorphism R G ∼ = End R G ( R ). In this paper, weshow that a version of Auslander’s Theorem holds if we replace R by an Artin-Schelter regularalgebra A of global dimension 2, and G by a finite subgroup of Aut gr ( A ) which contains noquasi-reflections. This extends work of Chan–Kirkman–Walton–Zhang. As part of the proof,we classify all such pairs ( A, G ), up to conjugation of G by an element of Aut gr ( A ). In all butone case, we also write down explicit presentations for the invariant rings A G , and show thatthey are isomorphic to factors of AS regular algebras. Introduction
Throughout let k be an algebraically closed field of characteristic 0. A classical theorem ofAuslander is the following: Theorem 1.1.
Let R = k [ x , . . . , x n ] and let G be a finite subgroup of Aut gr ( R ) = GL( n, k ) .Consider the graded ring homomorphism φ : R G → End R G ( R ) , φ ( rg )( s ) = r ( g · s ) . Then φ is an isomorphism if and only if G is small. By small , we mean that G contains no reflections, in the sense described in Section 2.2. Forexample, if G is a finite subgroup of SL(2 , k ) acting on k [ u, v ] (meaning that k [ u, v ] G is a Kleiniansingularity), then G is small and therefore the map φ is an isomorphism.Recently, a number of authors have studied when noncommutative generalisations of thisresult hold. More specifically, one can replace the polynomial ring R by an Artin-Schelter (AS)regular algebra A and consider the action of a finite group G of graded automorphisms on A ,and ask whether the natural map φ : A G → End A G ( A ) , φ ( ag )( b ) = a ( g · b ) , is an isomorphism. If this is the case, then we say that the Auslander map is an isomorphismfor the pair ( A, G ). More generally, one can replace G by a finite dimensional semisimple Hopfalgebra H , and ask the same question. In [2], the authors provided a useful criterion for deter-mining when the Auslander map is an isomorphism, which allows one to check specific examplesby hand. We state only the group-theoretic version of their result below; technical terminologyis defined in Section 2. Theorem 1.2 ([2, Theorem 0.3]) . Let G be a finite group acting on an AS regular, GK-Cohen-Macaulay algebra A with GKdim A > . Let g = P g ∈ G g , viewed as an element of A G . Then Date : October 31, 2019.2010
Mathematics Subject Classification. the Auslander map is an isomorphism for the pair ( A, G ) if and only if GKdim (cid:0) ( A G ) / h g i (cid:1) GKdim A − . Using this, the Auslander map has been shown to be an isomorphism in the following cases: • Actions of small subgroups of Aut
Lie ( g ) on universal enveloping algebras of finite dimen-sional Lie algebras U ( g ), [1, Theorem 0.4]; • Actions by finite groups on noetherian graded down-up algebras, [1, Theorem 0.6] and [9,Theorem 4.3]; • Permutation actions on k − [ x , . . . , x n ], [9, Theorem 2.4]; • Actions of semisimple Hopf algebras on AS regular algebras of dimension 2, such that theaction has trivial homological determinant, [6, Theorem 4.1].The last of these examples can be viewed as a noncommutative generalisation of the fact thatthe Auslander map is an isomorphism for Kleinian singularities. Here, the “trivial homologicaldeterminant” condition serves as a noncommutative analogue of requiring that the finite subgroupof Aut gr ( k [ u, v ]) = GL(2 , k ) lie inside SL(2 , k ). In fact, it is conjectured that the Auslander mapis an isomorphism whenever the action has trivial homological determinant; the result of Chan–Kirkman–Walton–Zhang establishes this in the dimension 2 case.The notion of smallness has a noncommutative generalisation, and so it is natural to askwhether a version of Theorem 1.1 holds in the noncommutative setting. One of the main resultsof this paper is to show that one direction of this result generalises to the noncommutative settingin dimension 2: Theorem 1.3 (Theorem 4.3, Theorem 4.8) . Suppose that A is a two-dimensional AS regularalgebra and that G is a small subgroup of Aut gr ( A ) . Then the Auslander map is an isomorphismfor the pair ( A, G ) . If a finite group G acts on a two dimensional AS regular algebra with trivial homologicaldeterminant, then one can check that G is small (in fact, this follows from Theorem 1.4 below),and so the above theorem can be viewed as an extension of [6, Theorem 4.1].Our method for proving Theorem 1.3 is to first classify all possible actions of small groups ontwo-dimensional AS regular algebras and to then verify that the criterion of Theorem 1.2 is metin each case. The following is the main result of Section 3, where k q [ u, v ] and k J [ u, v ] are thequantum plane and Jordan plane, respectively, whose definitions can be found in Section 2.3.We also write ω n for a primitive n th root of unity. Theorem 1.4 (Theorem 3.13) . Suppose that A is a two-dimensional AS regular algebra whichis not commutative and that G is a small subgroup of Aut gr ( A ) . Then, up to conjugation of G by an element of Aut gr ( A ) , the possible pairs ( A, G ) are as follows: Case
A G
Generators Conditions(i) k q [ u, v ] n (1 , a ) (cid:18) ω n ω an (cid:19) q = 1 , a < n , gcd( a, n ) = 1 . (ii) k − [ u, v ] G n,k (cid:18) ω n ω − n (cid:19) , (cid:18) ω k ω k (cid:19) k , gcd( n, k ) = 1 . (iii) k J [ u, v ] n (1 , (cid:18) ω n ω n (cid:19) n > . We remark that if A is a commutative AS regular algebra of dimension n then it is isomorphicto k [ x , . . . , x n ]. In this case, the classification of small subgroups when n = 2 (i.e. of small CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 3 subgroups of GL(2 , k )) can be found in [4].Establishing the reverse implication in Theorem 1.3 appears to be more difficult, since if onewishes to utilise Theorem 1.2, it is necessary to show that GKdim (cid:0) ( A G ) / h g i (cid:1) > G Aut gr ( A ) is not small. In general, this is more difficult to show than showing that an algebrahas GK dimension 0 (i.e. that it is finite dimensional). However, the reverse implication appearsto hold based off the examples we have computed.The final part of this paper is devoted to writing down presentations for the invariant rings A G , where A and G appear in the classification of Theorem 1.4. There is a long-standing beliefthat invariant rings arising from group actions (or, more generally, Hopf actions) on AS regularalgebras can be written as factors of AS regular algebras. Assuming this is true, this gives astrong motivation for writing down a presentation for A G , as from this one might be able toconstruct new examples of AS regular algebras. For cases (i) and (ii), we are able to write theinvariant rings as factors of AS regular algebras: Theorem 1.5 (Corollary 5.13, Corollary 6.6) . Let ( A, G ) be a pair from case (i) or case (iii) ofTheorem 1.4. Then A G is isomorphic to a factor of an AS regular algebra. For case (i), this is not a particularly surprising result, and the AS regular algebra of which A G is a factor is a quantum polynomial ring. Case (iii) is more interesting: the AS algebra ofwhich A G is a factor is a quantisation of a Poisson structure on a commutative polynomial ring.These rings were first studied in [16].For the remaining case, case (ii) of Theorem 1.4, even writing down a set of generators for theinvariant ring A G is a nontrivial task. When n or k is even, A G is commutative, and turns outto be a ring which is already well understood. If instead n and k are both odd, then A G is notcommutative, and we are only able to write down a set of generators. Theorem 1.6.
Suppose that A = k − [ u, v ] and G = G n,k , where n and k are coprime and k . (1) (Proposition 3.12) The invariant ring A G is commutative if and only if n or k is even. (2) (Propositions 7.2, 7.3, 7.8, and 7.9) If A G is commutative, then it is a surface quotientsingularity of type A or type D . (3) (Theorems 8.13 and 8.19) If A G is not commutative, then there is a formula to write downits generators which depends on the Hirzebruch-Jung continued fraction expansion of n ( n + k ) . In the setting of the above theorem, it is difficult to write down an explicit presentation forthe invariant rings A G when they are not commutative. We anticipate that they can always bewritten as a factor of an AS regular algebra, and show that this is the case for a specific example.Writing down presentations for arbitrary n and k , and studying further properties of these ringsin general, is the topic of work in progress.We now explain how our results fit into a broader picture. Let R = k [ u, v ] and let G be a smallsubgroup of GL(2 , k ). In this case, Spec R G is a (generically non-Gorenstein) surface quotientsingularity. There are strong connections between properties of the rings R G , representationtheory of the group G , and geometric properties of Spec R G . The results in this paper can beviewed as a first step towards better understanding a noncommutative version of this setup: byCorollary 3.14, the invariant rings A G under consideration are generically non-Gorenstein (theyare Gorenstein precisely when G acts with trivial homological determinant). Moreover, theyare graded isolated singularities in the sense of [17]. It is natural to ask how far the analogycan be pushed: for example, how much of the Auslander–Reiten theory for commutative surfacequotient singularities holds in the noncommutative setting? Can one identify analogues of “ex-ceptional curves” in a noncommutative resolution of A G ? Answering versions of these questions CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 4 is the subject of work in progress.
Organisation of the paper.
This paper is organised as follows. In Section 2, we fix thenotation used throughout this paper, and recall some basic definitions and results. In Section 3we prove Theorem 1.4, and then in Section 4 we use this to prove Theorem 1.3. The remainderof the paper is dedicated to determining the invariant rings A G : we treat case (iii) from Theorem1.4 in Section 5, case (i) in Section 6, and case (ii) in Sections 7 and 8.2. Preliminaries
Conventions.
Throughout k will denote an algebraically closed field of characteristic 0. Let R be an N -graded ring. We write gr- R (respectively, R -gr) for the category of finitely generated Z -graded right (respectively, left) R -modules with degree-preserving morphisms. Given M ∈ gr- R ,we define M [ i ] to be the graded module which is isomorphic to M as an ungraded module, butwhich satisfies M [ i ] n = M i + n . If M, N ∈ gr- R then End R ( M, N ) has a natural grading given by L i ∈ Z End gr- R ( M, N [ i ]) as graded vector spaces. In this paper, we shall use right modules unlessotherwise stated. We write i . dim M for the injective dimension of M ∈ mod- R , and gl . dim R forthe global dimension of R . We write ω n for a primitive n th root of unity.2.2. Actions of Small Groups on Polynomial Rings.
We begin by recalling some of thedefinitions and basic results from commutative algebraic geometry which we seek to generalise toa noncommutative setting. Suppose that R := k [ x , . . . , x n ] and that g ∈ Aut gr ( R ) = GL( n, k ).Then g is said to be a reflection if the fixed subspace { v ∈ k n | gv = v } has dimension n − G is a finite subgroup of Aut gr ( R ), then it is said to be small if it contains no reflections. Asstated in the introduction, it is a classical result due to Auslander that the natural map φ : R G → End R G ( R ) , φ ( rg )( s ) = r ( g · s )is an isomorphism if and only if G is small.The classification of small groups is easy to write down in the two dimensional case. Thesewere classified by Brieskorn in [4], but we follow the notation of [18], which lends itself morenaturally to the study of the invariant rings. An abridged version of the classification is asfollows:Type Group Generators Conditions A n (1 , a ) (cid:18) ω n ω an (cid:19) a < n , gcd( a, n ) = 1. D D m,q ω q ω − q ! , (cid:18) ω m − q ) ω m − q ) (cid:19) < q < m , gcd( m, q ) = 1.We remark that the generators we have chosen differ from those that appear elsewhere in theliterature, but it is straightforward to show that these yield the same groups. In both cases, the“type” refers to the dual graph of the exceptional divisor in the minimal resolution of Spec R G .There are three other infinite families that we have omitted since they will play no role in thispaper, called types T , O , and I .Given a non-negative rational number x , its Hirzebruch-Jung continued fraction expansion is CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 5 the unique representation of x in the form x = a − a −
1. . . − a n =: [ a , a , . . . , a n ]where the a i are all integers and a > a , . . . , a n >
2. Somewhat surprisingly, ring-theoretic, representation-theoretic and geometric properties of the invariant rings R G are con-trolled by the Hirzebruch-Jung continued fraction expansion of certain rational numbers whichdepend on n and a for type A , or m and q for type D . We briefly summarise some of the resultsthat will be needed later in this paper regarding the invariant rings R G in these two cases.2.2.1. Type A . Fix integers a and n satisfying 1 a < n and gcd( a, n ) = 1, and set G = n (1 , a ).Write [ α , . . . , α N ] for the Hirzebruch-Jung continued fraction expansion of na . By [18, Satz 8],the dual graph of the minimal resolution of of Spec R G is as follows: • • · · · • •− α − α − α N − − α N where each vertex corresponds to a curve isomorphic to P , and the label − α i give the self-intersection numbers of the corresponding curve. In the case when a = n − G SL(2 , k )), Spec R G is a Kleinian singularity, and the Hirzebruch-Jung continued fraction expan-sion of nn − is [2 , , . . . , n − A Kleinian singularityconsists of a chain of P , each having self-intersection − R G using Hirzebruch-Jung continuedfractions. Write nn − a = [ β , . . . , β d − ] . Now define two series of integers i , . . . , i d and j , . . . , j d as follows: i = n, i = n − a and i k = β k − i k − − i k − for 3 k d,j = 0 , j = 1 and j k = β k − j k − − j k − for 3 k d. Then, by [18, Satz 1], the invariant ring R G is minimally generated by the d elements x k := u i k v j k , k d. One can also use the β k to write down a minimal set of relations between the x i [18, Satz 8]: x k − x k +1 = x β k − k for 2 k d − ,x k x ℓ = x β k − k +1 x β k +1 − k +2 . . . x β ℓ − − ℓ − x β ℓ − − ℓ − for 2 k + 1 < ℓ − d − . We draw attention to the fact that, on the right hand side of the relation on the second line,the first and last exponents have the form β m −
1, while the remaining exponents have the form β m − CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 6
Type D . Now fix integers m and q satisfying 1 < q < m and gcd( m, q ) = 1, and set G = D m,q . Write [ α , . . . , α N ] for the Hirzebruch-Jung continued fraction expansion of mq . By[18, Satz 8], the dual graph of the minimal resolution of of Spec R G is as follows: •• • • · · · •− α − α − α N − − A case. If q = m − R G is a type D Kleiniansingularity, and α i = 2 for all i .As in the type A case, one is able to write down a minimal set of generators for R G usingHirzebruch-Jung continued fractions. Write mm − q = [ β , . . . , β d − ] , and define three series of integers r , . . . , r d − , s , . . . , s d − , and t , . . . , t d − as follows: s = 1 , s = 1 s = β , s k = β k − s k − − s k − for 4 k d − ,t = β , t = β − t = β ( β − − , t k = β k − t k − − t k − for 4 k d − ,r k = ( m − q ) t k − qs k for 1 k d − , where the entries to the right of the vertical line only exist when d >
3, which happens if andonly if q < m −
1, if and only if D m,q is not a subgroup of SL(2 , k ). By [18, Satz 2], the invariantring R G is minimally generated by the d elements x k := ( u qs k + ( − t k v qs k )( uv ) r k , k d − , and x d := ( uv ) n − q ) . As in the type A case, a minimal set of relations can be written down using the β k ; see [18, Satz8].2.3. Definitions and Basic Results.
Suppose that A is a k -algebra. We say that A is connectedgraded if there is a direct sum decomposition A = L i > A i such that A = k and A i · A j ⊆ A i + j .If, moreover, A is finitely generated as a k -algebra, then it is called finitely graded . If this is thecase, then dim k A i < ∞ for all i , and if M ∈ gr- A then dim k M i < ∞ for all i ∈ Z . We thendefine the Hilbert series of M (which of course allows M = A A ) to be the formal Laurent serieshilb M := X i ∈ Z (dim k M i ) t i . Given an N -graded algebra A , its n th Veronese is the subring A ( n ) := L i > A ni .The algebras of interest in this paper are particular examples of finitely graded k -algebraswhich have additional properties. These algebras are defined as follows. Definition 2.1.
Let A be a finitely graded k -algebra, and also write k = A/A > for the trivialmodule. We say that A is Artin-Schelter Gorenstein (or AS Gorenstein) of dimension d if:(1) i . dim A A = i . dim A A = d < ∞ , and(2) Ext i gr- A ( k A , A A ) ∼ = (cid:26) i = d A k [ ℓ ] if i = d as left A -modules, for some integer ℓ .We call ℓ the AS index of A . If moreover(3) gl . dim A = d , and CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 7 (4) A has finite GK dimension,then we say that A is Artin-Schelter regular (or AS regular) of dimension d .If A is a commutative AS regular algebra, then it is a polynomial ring. All known AS regularalgebras are noetherian domains, and this is conjectured to always be the case. AS regularalgebras are viewed as noncommutative analogues of commutative polynomial rings since theyshare many ring-theoretic and homological properties.In dimension 2, which is the case of interest to us, it is easy to classify all AS regular algebrasthat are generated in degree one; up to isomorphism, they are the quantum plane and Jordanplane , which have respective presentations: k q [ u, v ] = k h u, v ih vu − quv i , q ∈ k × , and k J [ u, v ] = k h u, v ih vu − uv − u i . Given an AS regular algebra A , we are interested in actions of finite subgroups G of Aut gr ( A )on A . In dimension 2, all such groups can be viewed as subgroups of GL(2 , k ), and in this casethe action of g = (cid:0) a bc d (cid:1) on u and v in the algebras above is given by g · u = au + cv, g · v = bu + dv. In particular, we are interested in actions by groups which contain no quasi-reflections , so wenow recall the relevant definitions:
Definition 2.2.
Suppose that A is finitely graded and let M ∈ gr- A (so that M is left-bounded,in the sense that M i = 0 for i ≪ g be a graded endomorphism of M . Then the trace of g on M is Tr M ( g ) := X i ∈ Z tr( g | M i ) t i ∈ k J t, t − K , where tr( g | M i ) is the usual trace of the linear map g | M i : M i → M i .Now assume that A is AS regular, and that its Hilbert series has the formhilb A = 1(1 − t ) n f ( t ) , where f (1) = 0 (and hence GKdim A = n ). We say that g ∈ Aut gr ( A ) is a quasi-reflection ifTr A ( g ) = 1(1 − t ) n − p ( t ) , where p (1) = 0. We say that a finite subgroup G Aut gr ( A ) is small if it contains no quasi-reflections.We remark that, if A is a commutative polynomial ring, then g is a quasi-reflection if and onlyif g is a reflection in the classical sense. In the dimension 2 case at hand, since both k q [ u, v ] and k J [ u, v ] have Hilbert series (1 − t ) − , a graded automorphism g is a quasi-reflection if its tracehas the form − t )(1 − λt ) for some λ = 1.We give a brief example which demonstrates the dependency of the trace on the algebra A . Example 2.3.
Let h = (cid:0) (cid:1) . Then h can be viewed as a graded automorphism of both R = k [ u, v ] and A = k − [ u, v ]. Noting that both R and A have k -bases [ n > { u n − i v i | i n } , it is straightforward to calculate thatTr R ( h ) = 1 + t + t + t + · · · = 11 − t = 1(1 − t )(1 + t ) , CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 8 Tr A ( h ) = 1 − t + t − t + · · · = 11 + t . Hence h is a quasi-reflection when it acts on R , but it is not a quasi-reflection when it acts on A .We also note that the trace can be used to determine the Hilbert series of an invariant ring,giving a result which can be viewed as a noncommutative version of Molien’s Theorem: Theorem 2.4 ([11, Lemma 5.2]) . Suppose that A is finitely graded and G is a finite subgroupof Aut gr ( A ) . Then hilb A G = 1 | G | X g ∈ G Tr A ( g ) . Initially, one might think that focusing attention on only small groups is quite restrictive.However, from the perspective of invariant theory, one can always assume that a finite subgroup G of Aut gr ( A ) is small. The proof of the following lemma is contained in the proof of [14,Proposition 1.5 (a)], where we note that the authors assume that A is noetherian and AS regular,but these properties are not needed to establish what follows. Lemma 2.5.
Suppose that A is finitely graded. If G is a finite subgroup of Aut gr ( A ) , then thereexists a small group G ′ Aut gr ( A ) such that A G ∼ = A G ′ . If we restrict to the case where G is small, one can easily detect whether the invariant ring A G is AS Gorenstein. To be able to state this result, we first recall a definition: Definition 2.6.
Suppose that A is AS regular and g ∈ Aut gr ( A ). Then Tr A ( g ) has a seriesexpansion in k (( t − )) of the formTr A ( g ) = ( − d c − t − ℓ + lower order terms , for some c ∈ k , where ℓ is as in Definition 2.1. We call this constant c the homological determinantof g , which we denote by hdet( g ). If hdet( g ) = 1 for all g ∈ G , then we say that the action of G on A has trivial homological determinant .It is shown in [12, Proposition 2.5] that this assignment gives rise to a group homomorphismhdet : G → k × . It is also shown in [12, Section 2] that if A is the commutative polynomial ring k [ x , . . . , x n ],then hdet( g ) = det( g ) for all g ∈ GL( n, k ), justifying the terminology.We remark that the homological determinant is usually defined using local cohomology (see[12, Definition 2.3]), but the above definition is equivalent to the usual one by [12, Lemma 2.6],and is sufficient for our needs.In dimension 2, by [5, Theorem 2.1] it is straightforward to calculate the homological deter-minant of a graded automorphism of an AS regular algebra. Lemma 2.7.
Suppose that A is a two-dimensional AS regular algebra, so either A = k q [ u, v ] or A = k J [ u, v ] , and let g ∈ Aut gr ( A ) . Lift the action of G on A to an action on the free algebra k h u, v i . Let r be the defining relation for A so that k r is a one-dimensional submodule of k h u, v i .Then g acts as scalar multiplication by hdet( g ) on k r . This allows us to state the result to which we previously alluded:
Theorem 2.8 ([13, Theorem 4.10]) . Suppose that A is AS regular and that G is a finite subgroupof Aut gr ( A ) which is small. Then A G is AS Gorenstein if and only if hdet( g ) = 1 for all g ∈ G . In the next section, we will determine the small subgroups of Aut gr ( A ) when A is AS regularof dimension 2. By the above theorem, it is then easy to check when the resulting invariant ringsare AS Gorenstein. CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 9 Small Subgroups of
Aut gr ( A ) when A is AS Regular of Dimension 2 In this section, we will determine the small subgroups of Aut gr ( A ) when A is AS regular ofdimension 2, i.e. A is either a quantum plane or the Jordan plane. In general, the gradedautomorphism group of A is strictly contained in GL(2 , k ), and so it will turn out that there arerelatively few possibilities for G . We remark that, if G and H are finite subgroups of Aut gr ( A )which are conjugate, then A G and A H are isomorphic, so we only need to classify the groups upto conjugation.Throughout this section we will assume that A is not commutative, i.e. if A = k q [ u, v ], then q = 1. We will also assume that G is nontrivial. The cases when A = k q [ u, v ] ( q = −
1) or A = k J [ u, v ] are easiest to analyse, and so we begin with these. The final case, when A = k − [ u, v ],is the most involved, and it is to this case that the majority of this section is devoted.3.1. Actions of Small Groups on the Jordan Plane.
It is straightforward to check that thegraded automorphism group of the Jordan plane A = k J [ u, v ] consists of automorphisms of theform u au, v bu + av, where a ∈ k × and b ∈ k . Viewed as a subgroup of GL(2 , k ), we therefore haveAut gr ( A ) = ((cid:18) a b a (cid:19) (cid:12)(cid:12)(cid:12)(cid:12) a ∈ k × , b ∈ k ) . (3.1)We now wish to identify all small subgroups of Aut gr ( A ). By [15, p. 7], it is known thatAut gr ( A ) contains no quasi-reflections, and so every finite subgroup of Aut gr ( A ) is a small group,so we only need to classify the finite subgroups of Aut gr ( A ).From (3.1), it is clear that an element of Aut gr ( A ) has finite order if and only if b = 0 and a is a root of unity. We summarise the above discussion in the following lemma: Lemma 3.2.
Let A = k J [ u, v ] . Then the small subgroups of Aut gr ( A ) are n (1 ,
1) := (cid:28)(cid:18) ω n ω n (cid:19)(cid:29) , where n > . Actions of Small Groups on the Quantum Plane, q = ± . Now let A = k q [ u, v ],where q = ± q = 1 is excluded to ensure that A is not commutative). In this case, aneasy calculation shows that every graded automorphism of A is of the form u au, v dv, where a, b ∈ k × ; that is, Aut gr ( A ) = ((cid:18) a d (cid:19) (cid:12)(cid:12)(cid:12)(cid:12) a, d ∈ k × ) . (3.3)We now wish to determine all small subgroups of Aut gr ( A ). Clearly an element of Aut gr ( A )has finite order if and only if a and d are both (possibly distinct) roots of unity. Moreover, notingthat the trace of an element g of the form given in (3.3) is independent of q , we find thatTr A ( g ) = Tr k [ u,v ] ( g ) = 1(1 − at )(1 − dt ) , CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 10 and so g is a quasi-reflection if an only if precisely one of a or d is equal to 1.Now let G be a small subgroup of Aut gr ( A ), so that the only element of G with at least one1 on the diagonal is the identity. Consider the map φ : G → k × , (cid:18) a d (cid:19) a. This map has trivial kernel (since if a = 1 then necessarily d = 1), and the image is a finitesubgroup of k × , which is necessarily cyclic of order n , say. Therefore G ∼ = im φ is cyclic of order n . It is straightforward to show that this forces G to be generated by an element of the form g = (cid:18) ω n ω an (cid:19) , for some integers a and n satisfying 1 a < n . To ensure that G contains no quasi-reflections,we also require gcd( a, n ) = 1. Again, we summarise our findings: Lemma 3.4.
Let A = k q [ u, v ] , where q = ± . Then the small subgroups of Aut gr ( A ) are n (1 , a ) := (cid:28)(cid:18) ω n ω an (cid:19)(cid:29) , where a < n and gcd( a, n ) = 1 . Actions of Small Groups on the ( − -Quantum Plane. Throughout this subsection,let A = k − [ u, v ]. We now seek to classify all small subgroups of Aut gr ( A ). We first note thatAut gr ( A ) = ((cid:18) a d (cid:19) (cid:12)(cid:12)(cid:12)(cid:12) a, d ∈ k × ) ∪ ((cid:18) bc (cid:19) (cid:12)(cid:12)(cid:12)(cid:12) b, c ∈ k × ) ∼ = ( k ∗ ) ⋊ C , where we identify automorphisms with elements of GL(2 , k ) in the usual way. We call elementsin the second set antidiagonal .For k J [ u, v ] and k q [ u, v ] ( q = ± gr ( A ), not just up to conjugation. For k − [ u, v ] it is more convenient to classifythe small subgroups of Aut gr ( A ) up to conjugation which, by Lemma 2.5, has no ill-effects fromthe perspective of invariant theory.Note that conjugation by an element of Aut gr ( A ) has no effect on the diagonal elements in G , so the presence (or absence) of antidiagonal elements in G is not affected by conjugation bydiagonal elements.Our first step is to identify precisely when an element of Aut gr ( A ) is a quasi-reflection. Lemma 3.5.
Suppose that g = (cid:0) a bc d (cid:1) ∈ Aut gr ( A ) . If g is diagonal, then Tr A ( g ) = 1(1 − at )(1 − dt ) , while if g is antidiagonal, then Tr A ( g ) = 11 + bct . In particular, if g is diagonal (respectively, antidiagonal), then it is a quasi-reflection if and onlyif a = 1 or d = 1 , but not both (respectively, bc = − ).Proof. The trace of a diagonal automorphism was noted in Section 3.2, so suppose that g = (cid:0) bc (cid:1) is antidiagonal. By [11, Corollary 4.4], since A is Koszul, the trace of g on A can be computedusing the Koszul dual A ! of A . It is easy to check that A ! ∼ = k [ x, y ] / h x , y i , and that the inducedmap g ! acts via g ! ( x ) = cy , g ! ( y ) = bx . From this one calculatesTr A ! ( g ! ) = 1 + bct , CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 11 and then [11, Corollary 4.4] says thatTr A ( g ) = 11 + bc ( − t ) = 11 + bct . Finally, since g is a quasi-reflection if and only if its trace is given by1(1 − t )(1 − λt )for some λ = 1, the final claim follows. (cid:3) To begin the classification, suppose that G is a small subgroup of Aut gr ( A ). If G containsno antidiagonal elements then every non-identity element is a diagonal matrix. By the sameargument as in Section 3.2, G = n (1 , a ), where 1 a < n and gcd( a, n ) = 1.So now suppose that G contains at least one antidiagonal element. There exists a short exactsequence of groups 1 → G ∩ SL(2 , k ) → G det −−→ h ω m i → , for some m >
2, where G ∩ SL(2 , k ) is a finite subgroup of SL(2 , k ). However, since G containsno quasi-reflections, by Lemma 3.5 all of its antidiagonal elements have determinant not equal to1, so G ∩ SL(2 , k ) consists of diagonal matrices of determinant 1. Therefore the only possibilityis that G ∩ SL(2 , k ) = (cid:28)(cid:18) ω n ω − n (cid:19)(cid:29) for some integer n > n = 1, this group is trivial); call the generator of this group g . Since G/ ( G ∩ SL(2 , k )) is cyclic, it follows that there exists an antidiagonal element h ∈ G such that G = h g, h i . If h = (cid:0) bc (cid:1) has (necessarily even) order 2 k then bc is a primitive k th root of unity,and conjugating by (cid:16) √ b √ c (cid:17) yields (cid:16) √ bc √ bc (cid:17) = (cid:16) ω k ω k (cid:17) . So we can assume h = (cid:16) ω k ω k (cid:17) for some k > n and k give rise to small subgroups of Aut gr ( A ): for example, if k = 2 then h = (cid:16) ω ω (cid:17) , which is a quasi-reflection. We now seek to eliminate those n and k which result in groups G containing quasi-reflections. Henceforth, it will be convenient to let ω be a primitive (2 nk )th root of unity, so that w k is a primitive n th root of unity and w n is aprimitive (2 k )th root of unity, and to write G n,k = h g, h i , where g = ω k ω − k ! and h = (cid:18) ω n ω n (cid:19) . (3.6) Lemma 3.7.
Suppose that G = G n,k is as in (3.6) and k ≡ . Then G contains aquasi-reflection.Proof. If k ≡ k is odd and so h k = ω nk ω nk ! . The product of the antidiagonal elements of h k is ω nk = −
1, and hence h k is a quasi-reflectionby Lemma 3.5. (cid:3) Lemma 3.8.
Suppose that G = G n,k is as in (3.6) . Then G contains no quasi-reflections if andonly if k and gcd( n, k ) . CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 12
Proof. ( ⇒ ) Lemma 3.7 already shows that if k ≡ G contains a quasi-reflection, sowrite d = gcd( n, k ) and suppose d >
2. Setting i = nd and j = kd , direct calculation gives g i h j = ω nj + ki ) ω nj − ki ) ! . Now, 2( nj + ki ) = 2( nkd + knd ) = nkd < nk , where the inequality follows since d >
2. Therefore ω nj + ki ) = 1. On the other hand, 2( nj − ki ) = 2( nkd − knd ) = 0, so ω nj − ki ) = 1. In particular,by Lemma 3.5, g i h j is a quasi-reflection.( ⇐ ) Now assume that k n, k )
2, and consider a diagonal element of G ,which therefore has the form g i h j for some i and j . As above, g i h j = ω nj + ki ) ω nj − ki ) ! . Now suppose ω nj + ki ) = 1, so nj + ki = nkr for some integer r . Rearranging and dividingthrough by d = gcd( n, k ), we find kd i = nd ( kr − j ), so that nd divides kd i . Since nd and kd arecoprime this implies that nd | i . Therefore n | di , where d is either 1 or 2, so n | i . Since nj + ki = nkr , we have nj − ki = nkr − ki = k ( nr − i ). By the above argument, we seethat n divides the term nr − i , and hence 2( nj − ki ) = 2 k ( nr − i ) is an integer multiple of2 nk . Therefore ω nj − ki ) = 1, and so g i h j is the identity. A similar analysis shows that if ω nj − ki ) = 1 then also ω nj + ki ) = 1, and so G contains no diagonal elements which are quasi-reflections.Now consider an antidiagonal element of G , which must have the form g i h j +1 for some i and j . Direct calculation gives g i h j +1 = ω (2 j +1) n +2 ki ω (2 j +1) n − ki ! and the product of the diagonal elements is ω n (2 j +1) . By Lemma 3.5, it follows that g i h j +1 is aquasi-reflection if and only if ω n (2 j +1) = −
1, which happens if and only if 2(2 j + 1) = k (2 r + 1)for some integer r . Since k g i h j +1 is not a quasi-reflection. (cid:3) We have therefore determined precisely when G n,k contains no quasi-reflections. However,some values of n and k give rise to the same groups: Lemma 3.9.
In the notation of (3.6) , if n ≡ and k ≡ , then G n,k = G n ,k .Proof. Letting ω be a primitive (2 nk )th root of unity, we need to show that * ω k ω − k ! , (cid:18) ω n ω n (cid:19)+ = * ω k ω − k ! , (cid:18) ω n ω n (cid:19)+ . The fact that the right hand side is contained in the left hand side is obvious. For the reverseinclusion, it suffices to show that (cid:16) ω k ω − k (cid:17) lies in the right hand side. Indeed, we have ω k ω − k ! n +24 ω n ω n ! k = ω k ( n +2) ω − k ( n +2) ! ω nk ω nk ! = ω k ω − k ! , as required. (cid:3) CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 13
By Lemma 3.9, if k is even, we can assume that n is odd. However, since Lemma 3.8 also tellsus that if G contains no quasi-reflections then necessarily gcd( n, k ) n and k can be assumedto be coprime. To summarise, we have shown the following: Theorem 3.10.
Let A = k − [ u, v ] . Then, up to conjugation by elements of Aut gr ( A ) , the smallsubgroups of Aut gr ( A ) are n (1 , a ) = (cid:28)(cid:18) ω n ω an (cid:19)(cid:29) , where a < n and gcd( a, n ) = 1 , and G n,k = (cid:28)(cid:18) ω n ω − n (cid:19) , (cid:18) ω k ω k (cid:19)(cid:29) , where gcd( n, k ) = 1 and k . We record the following lemma, which is easy to prove:
Lemma 3.11.
Suppose that G n,k is as in Theorem 3.10. Then this group has order nk andhas the presentation G n,k ∼ = h g, h | g n , h k , hg = g n − h i . It turns out that the invariant rings A G n,k are frequently not commutative. While we postponewriting down formulas for the generators and relations until sections 7 and 8, the following showswhich choices of n and k lead to invariant rings that are not commutative. Proposition 3.12.
Let G = G n,k be as above. Then A G is commutative if and only if n or k iseven.Proof. ( ⇐ ) First suppose that at least one (and hence exactly one) of n and k is even. If n is even, then g n/ = (cid:16) − − (cid:17) , while if k is even then h k = (cid:16) − − (cid:17) . In either case, call thiselement α . Then A G ⊆ A h α i = k [ u , v , uv ] ∼ = k [ x, y, z ] h xy + z i , where the last ring is commutative. Therefore A G is commutative.( ⇒ ) As before, it will be convenient to let ω be a primitive (2 nk )th root of unity, and toassume ω n = ω k and ω k = ω n . Now suppose that both n and k are odd, and choose an oddinteger m such that mk > n . Set i = mk + n and j = mk − n , both of which are positive integers.Moreover, since i − j = n is odd, one of i and j is odd and the other is even. Also observe that i + j = mk . Now set a = u i v j − u j v i and b = u i v j − u j v i . Then g · a = ω k ( i − j ) u i v j − ω k ( j − i ) u j v i = ω nk u i v j − ω − nk u j v i = u i v j − u j v i = a, so a is invariant under the action of g . Moreover, noting that u i and v j commute since one of i and j is even, h · a = ω n ( i + j ) v i u j − ω n ( i + j ) v j u i = ω mnk ( u j v i − u i v j )= − ω mnk ( u i v j − u j v i ) = ω ( m +1) nk a = a, so that a is also invariant under the action of h and hence lies in k − [ u, v ] G . Similarly, one canshow that b ∈ A G .Finally, we show that a and b do not commute, showing that A [ u, v ] G is not commutative.Assuming that i is even and j is odd, we have ab = ( u i v j − u j v i )( u i v j − u j v i ) CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 14 = u i v j u i v j − u i v j u j v i − u j v i u i v j + u j v i u j v i = u i v j + u i +3 j v i + j − u i + j v i +3 j + u j v i while ba = ( u i v j − u j v i )( u i v j − u j v i )= u i v j u i v j − u i v j u j v i − u j v i u i v j + u j v i u j v i = u i v j + u i + j v i +3 j − u i +3 j v i + j + u j v i . Therefore ab − ba = 2 u i +3 j v i + j − u i + j v i +3 j = 0 . If instead i is odd and j is even, then a similar calculation shows that ab − ba = 2 u i + j v i +3 j − u i +3 j v i + j = 0 , and so in either case a and b do not commute, as claimed. (cid:3) Summary of the classification.
We now summarise our findings from this section.
Theorem 3.13.
Suppose that A is a two-dimensional AS regular algebra which is not commu-tative and that G is a small subgroup of Aut gr ( A ) . Then, up to conjugation of G by an elementof Aut gr ( A ) , the possible pairs ( A, G ) are as follows: Case
A G
Generators Conditions(i) k q [ u, v ] n (1 , a ) (cid:18) ω n ω an (cid:19) q = 1 , a < n , gcd( a, n ) = 1 . (ii) k − [ u, v ] G n,k (cid:18) ω n ω − n (cid:19) , (cid:18) ω k ω k (cid:19) k , gcd( n, k ) = 1 . (iii) k J [ u, v ] n (1 , (cid:18) ω n ω n (cid:19) n > . Using Lemma 2.7 and Theorem 2.8, the following is easy to check:
Corollary 3.14.
Suppose that ( A, G ) is a pair from Theorem 3.13. Then the action has trivialhomological determinant, and hence A G is AS Gorenstein, if and only if: • Case (i): a = n − ; • Case (ii): k = 1 ; • Case (iii): n = 2 . The cases in which the action has trivial homological determinant are precisely those consid-ered in [6], with the exception of ( n, k ) = (2 ,
1) in case (ii). We will see later (Remark 7.4) that,from the perspective of invariant theory, its omission from [6] is not too important. We alsoremark that, in case (ii) when k = 1, the 2-dimensional representation of our groups is differentfrom those appearing in [6], but they are conjugate under Aut gr ( A ). This causes a change insome signs when writing down generators for the invariant rings, but has no other effects. CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 15 Auslander’s Theorem
The goal of this section is to show that the Auslander map is an isomorphism for any smallsubgroup of Aut gr ( A ), where A is AS regular of dimension 2. That is, we wish to show that if( A, G ) is a pair from Theorem 3.13, then the natural homomorphism φ : A G → End A G ( A ) , φ ( ag )( b ) = a ( g · b )is an isomorphism. We use the same strategy as was used in, for example, [6, 1, 9] to establishsimilar isomorphisms; namely, we show that the conditions of the following result holds, whichwe restate from the introduction: Theorem 4.1 ([2, Theorem 0.3]) . Let G be a finite group acting on an AS regular, GK-Cohen-Macaulay algebra A with GKdim A > . Let g = P g ∈ G g , viewed as an element of A G . Thenthe Auslander map is an isomorphism for the pair ( A, G ) if and only if GKdim (cid:0) ( A G ) / h g i (cid:1) GKdim A − . It is well-known that the algebras k q [ u, v ] and k J [ u, v ] satisfy the hypotheses on A given inthis theorem. Since in our setting GKdim A = 2, we need to show that GKdim( A G ) / h g i = 0,i.e. that ( A G ) / h g i is finite dimensional.We split the proof into two parts. We first assume that G is a diagonal subgroup of Aut gr ( A ),so that we are considering either case (i) or case (iii) from Theorem 3.13. We will then considercase (ii), which requires a more involved argument.4.1. Diagonal groups.
Suppose that A is either k q [ u, v ] ( q = 1) or k J [ u, v ] and that G = 1 n (1 , a ) = (cid:28)(cid:18) ω ω a (cid:19)(cid:29) , where ω n = 1, 1 a < n , and gcd( a, n ) = 1. (Necessarily a = 1 if A = k J [ u, v ].) Write g for thegiven generator of n (1 , a ), so that g = 1 + g + · · · + g n − . We note that ω satisfies the following identity: n X j =0 ω ij = (cid:26) n if i ≡ n, . (4.2) Theorem 4.3.
Suppose that A and G are as above. Then ( A G ) / h g i is finite dimensional. Inparticular, the Auslander map is an isomorphism for the pair ( A, G ) .Proof. Let 0 j n −
1. The element u n − − j gu j lies in h g i , and u n − − j gu j = n − X i =0 ω ij u n − g i . Summing over all j , we have h g i ∋ n − X j =0 u n − − j gu j = n − X i =0 (cid:18) n − X j =0 ω ij (cid:19) u n − g i . Using (4.2), P n − j =0 ω ij is equal to 0 unless i = 0. Therefore u n − ∈ h g i . Similarly, h g i ∋ n − X j =0 v n − − j gv j = n − X i =0 (cid:18) n − X j =0 ω aij (cid:19) v n − g i . CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 16
Since a is coprime to n , ai ≡ n if and only if i ≡ n , and therefore as abovewe find that v n − ∈ h g i . It follows that A > n − is contained in the ideal h g i , and hence( A G ) > n − ⊆ h g i . Therefore ( A G ) / h g i is finite dimensional, as claimed. (cid:3) Nondiagonal groups.
We now consider case (ii), so A = k − [ u, v ] and G = G n,k , where n and k are positive coprime integers with k ω be aprimitive (2 nk )th root of unity, and to think of G as being generated by the elements g and h ,where g = ω k ω − k ! , h = (cid:18) ω n ω n (cid:19) . By Lemma 3.11, this is a group of order 2 nk and has a presentation given by G ∼ = h g, h | g n , h k , hg = g n − h i . A complete list of elements of G is n g i h j (cid:12)(cid:12)(cid:12) i n − , j k − o ⊔ n g i h j +1 (cid:12)(cid:12)(cid:12) i n − , j k − o , where the first set in this disjoint union consists of diagonal elements, and the second set consistsof antidiagonal elements. We will frequently make use of this fact.In this subsection, we show that the Auslander map is an isomorphism for the pair ( A, G )using a sequence of lemmas. We first define some elements of A G that will play an importantrole in the proof. For any integer ℓ , define G ℓ = n − X i =0 k − X j =0 ω ℓ ( nj + ki ) g i h j , H ℓ = n − X i =0 k − X j =0 ω ℓ ( n (2 j +1) − ki ) g i h j +1 . In the following lemma, we will show (see part (5)) that these elements satisfy G u ℓ = u ℓ G ℓ and H u ℓ = v ℓ H ℓ for any ℓ >
0, which motivates their definition. We also record a number of other useful proper-ties:
Lemma 4.4.
Let ℓ be an integer (in parts (5) and (6), also assume that ℓ > ). (1) G ℓ = G ℓ + nkr for any integer r . (2) H ℓ = ( − nr H ℓ + nkr for any integer r . (3) P nk − ℓ =0 G ℓ = nk . (4) g = G + H . (5) G u ℓ = u ℓ G ℓ and H u ℓ = v ℓ H ℓ . (6) Since n and k are coprime, there exist integers a, b with an + bk = 1 . Set m = an − bk , whichis uniquely determined modulo nk . Then G v ℓ = v ℓ G mℓ and H v ℓ = u ℓ H mℓ .Proof. (1) Since ω nk = 1, G ℓ + nkr = n − X i =0 k − X j =0 ω ℓ + nkr )( nj + ki ) g i h j = n − X i =0 k − X j =0 ω ℓ ( nj + ki ) ω nkr ( nj + ki ) g i h j = n − X i =0 k − X j =0 ω ℓ ( nj + ki ) g i h j = G ℓ . CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 17 (2) In a similar vein to the above, we have H ℓ + nkr = n − X i =0 k − X j =0 ω ( ℓ + nkr )( n (2 j +1) − ki ) g i h j +1 = n − X i =0 k − X j =0 ω ℓ ( n (2 j +1) − ki ) ω n kr (2 j +1) − nk ri g i h j +1 = n − X i =0 k − X j =0 ω ℓ ( n (2 j +1) − ki ) ω ( nr ) kr g i h j +1 = ( − nr H ℓ . (3) We have nk − X ℓ =0 G ℓ = n − X i =0 k − X j =0 nk − X ℓ =0 ω ℓ ( nj + ki ) g i h j . Writing ε = ω , which is a primitive nk th root of unity, the term in parentheses is nk − X ℓ =0 ε ℓ ( nj + ki ) = (cid:26) nk if nj + ki ≡ nk, n and k are coprime, the only way for nj + ki to be divisible by nk is if j is divisible by k and i is divisible k . Since 0 i < n and 0 j < k , this happens if and only if i = 0 = j , so nk − X ℓ =0 ε ℓ ( nj + ki ) = (cid:26) nk if i = 0 = j, P nk − ℓ =0 G ℓ = nk , as claimed.(4) This is clear.(5) These are both true by construction; indeed, since 2 j is even, G u ℓ = n − X i =0 k − X j =0 g i h j u ℓ = n − X i =0 k − X j =0 ω ℓnj g i u ℓ h j = n − X i =0 k − X j =0 ω ℓnj +2 ℓki u ℓ g i h j = u ℓ G ℓ . The calculation showing that H u ℓ = v ℓ H ℓ is similar, after noting that, since 2 j + 1 is alwaysodd, h j +1 u ℓ = ω ℓn (2 j +1) v ℓ h j +1 .(6) This time we provide the calculation for H v ℓ , with the calculation for G v ℓ being similar.We have H v ℓ = n − X i =0 k − X j =0 g i h j +1 v ℓ = n − X i =0 k − X j =0 ω ℓn (2 j +1) g i u ℓ h j +1 = n − X i =0 k − X j =0 ω ℓn (2 j +1)+2 ℓki u ℓ g i h j +1 . We now consider the exponent of ω . Working modulo 2 nk , we have ℓn (2 j + 1) + 2 ℓki ≡ ℓ ( n (2 j + 1) + 2 ki )( an + bk ) ≡ ℓ ((2 j + 1) an + 2 bk i + bnk ) ≡ ℓ ((2 j + 1)( m + bk ) n + 2( an − m ) ki + bnk ) ≡ ℓ ( mn (2 j + 1) − mki + 2 bnk ) ≡ mℓ ( n (2 j + 1) − ki ) . CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 18
Therefore H v ℓ = n − X i =0 k − X j =0 ω mℓ ( n (2 j +1) − ki ) u ℓ g i h j +1 = u ℓ H mℓ , as claimed. (cid:3) Our goal now is to show that both u s and v s lie in h g i for some s ≫
0, which will allow us toshow that ( A G ) / h g i is finite dimensional. As a first step towards this, we have the followinglemma, where we will often invoke the properties established above without mention. Lemma 4.5. (1) ( u nk + v nk ) G ∈ h g i or ( u nk − v nk ) G ∈ h g i . (2) Let m be as in Lemma 4.4 (6) . Fix ℓ ∈ { , . . . , nk − } , and let r ∈ { , . . . , nk − } be suchthat mℓ ≡ r mod nk . Then ( u ℓ + r + v ℓ + r ) G ℓ ∈ h g i or ( u ℓ + r − v ℓ + r ) G ℓ ∈ h g i .Proof. (1) Both of the following elements lie in h g i : gu nk = G u nk + H u nk = u nk G n,k + v nk H nk = u nk G + δv nk H ,v nk g = v nk G + v nk H , where δ ∈ {± } . Adding or subtracting one from the other, depending on the sign of δ , we seethat the claim holds.(2) Again, the following elements both lie in h g i : u r gu ℓ = u r G u ℓ + u r H u ℓ = u ℓ + r G ℓ + u r v ℓ H ℓ ,v ℓ gv r = v ℓ G u r + v ℓ H u r = v ℓ + r G mr + δu r v ℓ H mr , for some δ ∈ {± } (the sign of δ depends on r as in Lemma 4.4 (2), as well as the fact that vu = − uv ). Since m = an − bk as in the statement of Lemma 4.4 (6), working modulo 2 nk wehave m ≡ ( an − bk ) ≡ a n − nkab + b k ≡ a n + 2 nkab + b k ≡ ( an + bk ) ≡ . Therefore mr ≡ m ℓ ≡ ℓ , so that v ℓ gv r = v ℓ + r G ℓ + δu r v ℓ H ℓ . The claim now follows by addingor subtracting the above two elements, depending on the sign of δ . (cid:3) Remark . The above proof shows that 2 nk divides m −
1, and hence m must be odd. Thereforewe find that ℓ and r , as in part (2) of the statement, have the same parity, and hence the exponent ℓ + r is even. This will be used later. Lemma 4.7. (1) u k v k G ∈ h g i . (2) u ( n +1) k v ( n +1) k ∈ h g i .Proof. (1) First suppose that k is odd. Then gu k v k = n − X i =0 2 k − X j =0 g i h j u k v k + n − X i =0 2 k − X j =0 g i h j +1 u k v k = n − X i =0 2 k − X j =0 ω nkj g i u k v k h j + n − X i =0 2 k − X j =0 ω nk (2 j +1) g i v k u k h j +1CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 19 = n − X i =0 2 k − X j =0 ω nkj u k v k g i h j + n − X i =0 2 k − X j =0 ω nk (2 j +1) v k u k g i h j +1 = u k v k G + v k u k H = u k v k G − u k v k H , where the final equality follows from the fact that u k and v k ( − k is odd. Ifinstead k is even (and hence k ≡ u k/ v k/ gu k/ v k/ = u k/ v k/ n − X i =0 2 k − X j =0 ω nkj u k/ v k/ g i h j + n − X i =0 2 k − X j =0 ω nkj + nk v k/ u k/ g i h j +1 = u k/ v k/ (cid:16) u k/ v k/ G + ω nk v k/ u k/ H (cid:17) = u k v k G − u k v k H , where the final equality holds since ω nk = −
1, and u k/ and v k/ commute since k is even.Therefore in either case we have u k v k ( G − H ) ∈ h g i . Since obviously u k v k ( G + H ) = u k v k g ∈ h g i , the result follows.(2) Fixing ℓ ∈ { , . . . , nk − } , by part (1) and by Lemma 4.4 (5), we have h g i ∋ u nk − ℓ v nk u k v k G u ℓ = ± u ( n +1) k v ( n +1) k G ℓ . By Lemma 4.4 (3), the sum of these elements (with appropriate sign choices) is equal to u ( n +1) k v ( n +1) k , and hence this element lies in h g i , as claimed. (cid:3) We are now in a position to prove the main result of this subsection:
Theorem 4.8.
Suppose that A = k − [ u, v ] and G = G n,k are as above. Then ( A G ) / h g i isfinite dimensional. In particular, the Auslander map is an isomorphism for the pair ( A, G ) .Proof. We first show that u s − v s ∈ h g i for some positive integer s . By Lemma 4.7 (2), thereexists an even integer t such that u t v t ∈ h g i (and hence u t v t G ℓ ∈ h g i for all ℓ ∈ { , . . . , nk − } ).As in Lemma 4.5 (2), if we fix ℓ ∈ { , . . . , nk − } , then there exists a positive integer r suchthat ( u ℓ + r + v ℓ + r ) G ℓ ∈ h g i or ( u ℓ + r − v ℓ + r ) G ℓ ∈ h g i . Multiplying this element on the left by u ℓ + r − v ℓ + r or u ℓ + r + v ℓ + r , respectively, and noting that u ℓ + r and v ℓ + r are both central in A since ℓ + r is even (Remark 4.6), we have ( u ℓ + r ) − v ℓ + r ) ) G ℓ ∈ h g i . Now multiplying this elementon the left by ( u ℓ + r ) + v ℓ + r ) ) shows that ( u ℓ + r ) − v ℓ + r ) ) ∈ h g i . Repeating this process inthe obvious way, for each ℓ we can find an even integer s ℓ > t such that ( u s ℓ − v s ℓ ) G ℓ ∈ h g i . Set s = t + max { s ℓ | ℓ nk − } . Now, for any ℓ ∈ { , . . . , nk − } , since s ℓ > t and s − s ℓ > t , we have u s − s ℓ v s ℓ G ℓ , v s − s ℓ u s ℓ G ℓ ∈ h g i .Therefore h g i ∋ ( u s − s ℓ + v s − s ℓ )( u s ℓ − v s ℓ ) G ℓ + u s − s ℓ v s ℓ G ℓ − v s − s ℓ u s ℓ G ℓ = u s G ℓ − u s − s ℓ v s ℓ G ℓ + v s − s ℓ u s ℓ G ℓ − v s G ℓ + u s − s ℓ v s ℓ G ℓ − v s − s ℓ u s ℓ G ℓ = ( u s − v s ) G ℓ . Summing over ℓ and applying Lemma 4.4 (3) shows that u s − v s ∈ h g i .Since s > t we have u s v s ∈ h g i , and so it follows that u s = u s ( u s − v s ) + u s v s ∈ h g i , and similarly v s ∈ h g i . Hence A > s is contained in h g i , and so the same is true of ( A G ) > s .In particular, ( A G ) / h g i is finite dimensional, as claimed. (cid:3) CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 20
Corollaries.
We conclude this section by noting some corollaries that follow from resultsin the literature. Following [19], if R is a noetherian graded algebra, then R is a graded isolatedsingularity if gl . dim(qgr R ) < ∞ , where qgr R = gr- R/ tors- R . If R is commutative, then thisdefinition generalises the usual commutative definition of R being an isolated singularity. By [17,Theorem 2.13, Lemma 2.12], it follows that the algebras A G are graded isolated singularities.Since the Auslander map is an isomorphism for each of the pairs ( A, G ) appearing in Theorem3.13, we are also able to deduce the existence of bijections between objects in various modulecategories related to A and G . We note that M ∈ gr- A is called initial if it is generated in degree0 and M < = 0. Theorem 4.9 ([7, Lemma 1.6, Proposition 2.3, Proposition 2.5, Corollary 4.5]) . Suppose that ( A, G ) is a pair from Theorem 3.13. Then there are natural bijections between isomorphismclasses of the following objects: • Simple k G -modules; • Indecomposable direct summands of A as a left A G -module; • Indecomposable, finitely generated, projective, initial left A G -modules; • Indecomposable, finitely generated, projective, initial left
End A G ( A ) -modules; and • Indecomposable maximal Cohen-Macaulay left A G -modules, up to a degree shift. A Presentation for Invariants of the Jordan Plane
The remainder of this paper is devoted to providing presentations for the invariant rings A G ,where the pair ( A, G ) is as in Theorem 3.13. The most interesting case is case (iii) when A G isnoncommutative, which is analysed in Section 8.In this section, we write A = k J [ u, v ], and let G be a small subgroup of Aut gr ( A ). By Lemma3.2, G = n (1 ,
1) for some n >
2. It is easy to see that invariant ring k J [ u, v ] G consists of thoseelements lying in degrees which are multiples of n , so k J [ u, v ] G is simply the n th Veronese of k J [ u, v ]. The main goal of this section is to provide a presentation for this invariant ring. Wewill in fact show that these rings can be written as factors of AS regular algebras.To achieve this, we take advantage of the fact that G also acts as a small group on thecommutative ring k [ u, v ], and that the resulting invariant rings k [ u, v ] G are well understood. Inparticular, by the results in Section 2.2.1, if we write x i := u n − i v i , for 0 i n, then these elements generate k [ u, v ] G . Moreover, a minimal set of relations between the x i isgiven by x i x j = x i − x j +1 for all i, j satisfying 1 i j < n. (5.1)We note that we will occasionally swap between viewing u and v as elements of k J [ u, v ] andviewing them as elements of the commutative ring k [ u, v ], but we will always make it clear whichring we are working inside.For the remainder of this section, for α ∈ k and k ∈ N we use the notation (cid:18) αk (cid:19) := α ( α − . . . ( α − k + 1) k ! . If α is an integer, then this definition agrees with the usual definition of the binomial coefficientwhen α > k , while (cid:0) αk (cid:1) = 0 when 0 α < k . One useful identity that this notation satisfies is( − k (cid:18) αk (cid:19) = (cid:18) k − α − k (cid:19) , (5.2) CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 21 which we will use later on.We first need to write down a formula for the commutator of monomials in k J [ u, v ] in termsof the standard basis { u i v j | i, j > } . Lemma 5.3.
In the Jordan plane k J [ u, v ] , the following relations hold: (1) For all positive integers j , vu j = u j v + ju j +1 . (2) For all positive integers i , v i u = i X k =0 k ! (cid:18) ik (cid:19) u k +1 v i − k . (3) For all positive integers i and j , v i u j = i X k =0 k ! (cid:18) j + k − k (cid:19)(cid:18) ik (cid:19) u j + k v i − k . Proof.
Parts (1) and (2) can be found in [10]. Presumably part (3) is also known, but we havebeen unable to find a reference; for completeness, we provide the proof, which is routine butlengthy.We proceed by induction on i . The case i = 1 is the statement of part (2), so now fix i > i . Then v i u j = v · v i − u j = v i − X k =0 k ! (cid:18) j + k − k (cid:19)(cid:18) i − k (cid:19) u j + k v i − − k = i − X k =0 k ! (cid:18) j + k − k (cid:19)(cid:18) i − k (cid:19)(cid:0) u j + k v + ( j + k ) u j + k +1 (cid:1) v i − − k = i − X k =0 k ! (cid:18) j + k − k (cid:19)(cid:18) i − k (cid:19) u j + k v i − k + i − X k =0 k !( j + k ) (cid:18) j + k − k (cid:19)(cid:18) i − k (cid:19) u j + k +1 v i − ( k +1) = u j v i + i − X k =1 k ! (cid:18) j + k − k (cid:19)(cid:18) i − k (cid:19) u j + k v i − k + i − X k =1 ( k − j + k − (cid:18) j + k − k − (cid:19)(cid:18) i − k − (cid:19) u j + k v i − k + ( i − i + j − (cid:18) i + j − i − (cid:19) u i + j = u j v i + i − X k =1 k ! (cid:18) j + k − k (cid:19)(cid:18) i − k (cid:19) + ( k − j + k − (cid:18) j + k − k − (cid:19)(cid:18) i − k − (cid:19)! u j + k v i − k + i ! (cid:18) i + j − i (cid:19) u i + j = u j v i + i − X k =1 k ! (cid:18) j + k − k (cid:19)(cid:18) i − k (cid:19) + k ! (cid:18) j + k − k (cid:19)(cid:18) i − k − (cid:19)! u j + k v i − k CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 22 + i ! (cid:18) i + j − i (cid:19) u i + j = u j v i + i − X k =1 k ! (cid:18) j + k − k (cid:19) (cid:18) i − k (cid:19) + (cid:18) i − k − (cid:19)! u j + k v i − k + i ! (cid:18) i + j − i (cid:19) u i + j = u j v i + i − X k =1 k ! (cid:18) j + k − k (cid:19)(cid:18) ik (cid:19) u j + k v i − k + i ! (cid:18) i + j − i (cid:19) u i + j = i X k =0 k ! (cid:18) j + k − k (cid:19)(cid:18) ik (cid:19) u j + k v i − k , where the second equality follows from the inductive hypothesis, the third equality follows frompart (1), and where we have suppressed a number of routine binomial coefficient calculations.Hence the result follows by induction. (cid:3) Now fix n and let G = n (1 , k [ u, v ], we write x i := u n − i v i , for 0 i n, now viewed as elements of k J [ u, v ]. It will actually be more convenient to work with certainscalar multiples of the x i ; to this end, we also define y i := ( − i i ! x i = ( − i i ! u n − i v i , for 0 i n. (5.4)We first verify that these elements generate the invariant ring k J [ u, v ] G , which is straightforward: Lemma 5.5.
The elements y i for i n (or equivalently, the x i ) generate k J [ u, v ] G .Proof. We prove the statement for the x i . As observed previously, k J [ u, v ] G is simply the n thVeronese of k J [ u, v ], so it suffices to show that we can write any element of the form u i v j , with i + j ≡ n , as a product of the x i . By Euclid’s algorithm, i = an + r and j = bn + s forsome integers a, b, r, s , where 0 r, s < n . Therefore u i v j = ( u n ) a u r v s ( v n ) b . Working modulo n , 0 ≡ i + j ≡ ( an + r ) + ( bn + s ) ≡ r + s , and since 0 r, s < n , it follows that either r = 0 = s or r + s = n . In the former case, u i v j = x a x bn , while in the latter case, u i v j = x a x s x bn . It followsthat the x i generate k J [ u, v ] G . (cid:3) We now wish to give a formula for the commutator y j y i − y i y j . We will make use of theChu-Vandermonde identity, which states that for any α, β ∈ k and any non-negative integer n , n X k =0 (cid:18) αk (cid:19)(cid:18) βn − k (cid:19) = (cid:18) α + βn (cid:19) . (5.6) Proposition 5.7. In k J [ u, v ] , j X k =0 (cid:18) n − ik (cid:19) y j − k y i = i X ℓ =0 (cid:18) n − jℓ (cid:19) y i − ℓ y j for j > i. (5.8) Remark . Observe that the relation (5.8), can be rewritten as y j y i − y i y j = i X ℓ =1 (cid:18) n − jℓ (cid:19) y i − ℓ y j − j X k =1 (cid:18) n − ik (cid:19) y j − k y i , and so we do have a formula for the commutator y j y i − y i y j . CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 23
Proof of Proposition 5.7.
We claim that both sides of (5.8) are equal to( − i + j i ! j ! u n − i − j v i + j . First consider the left hand side of (5.8). Using Lemma 5.3, we have j X k =0 (cid:18) n − ik (cid:19) y j − k y i = j X k =0 ( − i + j − k i !( j − k )! (cid:18) n − ik (cid:19) u n − j + k v j − k u n − i v i = j X k =0 j − k X m =0 ( − i + j − k m ! i !( j − k )! (cid:18) n − ik (cid:19)(cid:18) n − i + m − m (cid:19)(cid:18) j − km (cid:19) u n − i − j + k + m v i + j − k − m = j X k =0 j − k X m =0 ( − i + j − k i !( j − k − m )! (cid:18) n − ik (cid:19)(cid:18) n − i + m − m (cid:19) u n − i − j + k + m v i + j − k − m = j X r =0 ( − i + j i !( j − r )! X k + m = r k j ( − − k (cid:18) n − ik (cid:19)(cid:18) n − i + m − m (cid:19) u n − i − j + r v i + j − r = j X r =0 ( − i + j + r i !( j − r )! r X k =0 ( − r − k (cid:18) n − ik (cid:19)(cid:18) n − i + r − k − r − k (cid:19) u n − i − j + r v i + j − r ( . ) = j X r =0 ( − i + j + r i !( j − r )! r X k =0 (cid:18) n − ik (cid:19)(cid:18) i − nr − k (cid:19) u n − i − j + r v i + j − r ( . ) = j X r =0 ( − i + j + r i !( j − r )! (cid:18) r (cid:19) u n − i − j + r v i + j − r = ( − i + j i ! j ! u n − i − j v i + j . An entirely similar calculation shows that i X ℓ =0 (cid:18) n − jℓ (cid:19) y i − ℓ y j = ( − i + j i ! j ! u n − i − j v i + j , and so we have the claimed equality. (cid:3) In addition to the above relations, one should expect to have relations which are similar tothose given in (5.1). This is indeed the case, as we now show.
Proposition 5.10. In k J [ u, v ] , iy i y j = ( j + 1) y i − y j +1 − (cid:0) n − − ( j − i ) (cid:1) y i − y j for all i, j satisfying i j < n. Proof.
It is easy to see that it is equivalent to show x i x j = x i − x j +1 + (cid:0) n − − ( j − i ) (cid:1) x i − x j for all i, j satisfying 1 i j < n. We verify this equality using Lemma 5.3 (3). Again, we suppress some of the routine calculationsinvolving binomial coefficients. We have x i − x j +1 = u n − i +1 v i − u n − j − v j +1 = i − X k =0 k ! (cid:18) n − j + k − k (cid:19)(cid:18) i − k (cid:19) u n − i − j + k v i + j − k , CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 24 and similarly, x i − x j = i − X k =0 k ! (cid:18) n − j + k − k (cid:19)(cid:18) i − k (cid:19) u n − i − j + k +1 v i + j − k − , = i X k =1 ( k − (cid:18) n − j + k − k − (cid:19)(cid:18) i − k − (cid:19) u n − i − j + k v i + j − k . Therefore x i − x j +1 + (cid:0) n − − ( j − (cid:1) x i − x j = u n − i − j v i + j + i − X k =1 c k u n − i − j + k v i + j − k + (cid:0) n − − ( j − i ) (cid:1) ( i − (cid:18) n − j + i − i − (cid:19) u n − j v j , where c k = k ! (cid:18) n − j + k − k (cid:19)(cid:18) i − k (cid:19) + (cid:0) n − − ( j − i ) (cid:1) ( k − (cid:18) n − j + k − k − (cid:19)(cid:18) i − k − (cid:19) . It is easily shown that c k = k ! (cid:18) n − j + k − k (cid:19)(cid:18) ik (cid:19) for all 1 k i −
1, and that (cid:0) n − − ( j − i ) (cid:1) ( i − (cid:18) n − j + i − i − (cid:19) = i ! (cid:18) n − j + i − i (cid:19) . Hence x i − x j +1 + (cid:0) n − − ( j − (cid:1) x i − x j = u n − i − j v i + j + i − X k =1 k ! (cid:18) n − j + k − k (cid:19)(cid:18) ik (cid:19) u n − i − j + k v i + j − k + i ! (cid:18) n − j + i − i (cid:19) u n − j v j = i X k =0 k ! (cid:18) n − j + k − k (cid:19)(cid:18) ik (cid:19) u n − i − j + k v i + j − k . This last expression is easily seen to be equal to x i x j , whence the result. (cid:3) This allows us to give a presentation for the invariant ring k J [ u, v ] G . The proof uses the theoryof (noncommutative) Groebner bases, so we briefly recall some terminology and results. We orderthe words in the free algebra k h X i i using degree lexicographic order with X < X < · · · < X n .Given an ideal K of k h X i i , we write Lead( K ) for the ideal generated by all the leading wordsof elements of K . A Groebner basis of K is a subset of K such that the leading terms of thissubset generate Lead( K ). In this case, we have hilb k h X i i /K = hilb k h X i i / Lead( K ). Theorem 5.11.
Let n > and G = n (1 , . Then k J [ u, v ] G has the presentation k J [ u, v ] G ∼ = k h X , X , . . . , X n i I where the ideal of relations I is generated by j X k =0 (cid:18) n − ik (cid:19) X j − k X i = i X ℓ =0 (cid:18) n − jℓ (cid:19) X i − ℓ X j for j > iiX i X j = ( j + 1) X i − X j +1 − (cid:0) n − − ( j − i ) (cid:1) X i − X j for all i, j satisfying i j < n. CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 25
Proof.
Writing y i = ( − i i ! u n − i v i for the generators of k J [ u, v ] G , Propositions 5.7 and 5.10 showthat there is a surjection k h X , X , . . . , X n i I ։ k J [ u, v ] G , X i y i . (5.12)To show that this is in fact an isomorphism, it suffices to show that k h X i i /I and k J [ u, v ] G havethe same Hilbert series. Since the above map is a surjection, certainlyhilb k h X i i /I > hilb k J [ u, v ] G . By equation (5.1), we know that k [ u, v ] G can be presented as k [ u, v ] G ∼ = k h X , X , . . . , X n i J , where J is the ideal of relations with generators X j X i = X i X j for j > i,X i X j = X i − X j +1 for all i, j satisfying 1 i j < n. Here we present the commutative ring k [ u, v ] G as the factor of a free algebra with, in particular,the obvious commutativity relations X j X i = X i X j , so as to more easily draw comparisonswith our claimed presentation for k J [ u, v ] G . With respect to degree lexicographic ordering, theelements on the left hand side of the relations above are the leading words. The generators of J form a Groebner basis, so hilb k h X i i /J = hilb k h X i i / Lead( J ) . Now observe that the relations which define I can be written as X j X i = − j X k =1 (cid:18) n − ik (cid:19) X j − k X i + i X ℓ =0 (cid:18) n − jℓ (cid:19) X i − ℓ X j for j > iX i X j = j + 1 i X i − X j +1 − n − − ( j − i ) i X i − X j for all i, j satisfying 1 i j < n. The leading words are those which appear on the left hand side of each equality, and these areprecisely the leading words for the relations in J . Therefore if we find a Groebner basis for J , itfollows that Lead( I ) ⊇ Lead( J ). Hence we havehilb k h X i i /I = hilb k h X i i / Lead( I ) hilb k h X i i / Lead( J ) = hilb k h X i i /J. However, since k [ u, v ] G and k J [ u, v ] G have the same Hilbert series by Molien’s Theorem, and wehave an inequality k h X i i /J ∼ = k [ u, v ] G , we findhilb k h X i i /I hilb k h X i i /J = hilb k [ u, v ] G = hilb k J [ u, v ] G . Therefore hilb k h X i i /I = hilb k J [ u, v ] G and so the map in (5.12) is an isomorphism. (cid:3) Corollary 5.13.
Let n > and G = n (1 , . Then k J [ u, v ] G is a factor of an AS regular algebraof dimension n + 1 .Proof. Define B := k [ X , . . . , X n ] /K , where K is the ideal of relations generated by j X k =0 (cid:18) n − ik (cid:19) X j − k X i = i X ℓ =0 (cid:18) n − jℓ (cid:19) X i − ℓ X j for j > i. By [16, Proposition 3.15], B is isomorphic to an algebra denoted R ( n, a ), where a = − n , whichis an AS regular algebra of dimension n + 1 by [16, Proposition 3.8]. The result then followsfrom Theorem 5.11. (cid:3) CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 26
We end this section with an example:
Example 5.14.
In this example, we set a = X , b = X , c = X , . . . when writing downpresentations for the invariant rings k J [ u, v ] G . When n = 2, the presentation for k J [ u, v ] G givenin Theorem 5.11 is k J [ u, v ] G ∼ = k h a, b, c i (cid:28) ba + 2 a − ab, ca + 2 ba + a − accb + b − bc, b − ac + ab (cid:29) . The algebra B = k h a, b, c i (cid:28) ba + 2 a − ab, ca + 2 ba + a − accb + b − bc (cid:29) , is AS regular, and Ω = b − ac + ab is a regular normal element in B . Observe that an equivalentpresentation for k J [ u, v ] G is the following: k J [ u, v ] G ∼ = k h a, b, c i (cid:28) ba − ab + 2 a , ca − ac + 2 ab − a cb − bc + b , b − ac + ab (cid:29) , and that if we make the change of variables a a ′ , b
7→ − b ′ , c c ′ , then we obtain the presentation given in [6, Theorem 5.2].Now suppose that n = 3. Then, by Theorem 5.11, k J [ u, v ] G ∼ = k h a, b, c, d i * ba + 3 a − ab, ca + 3 ba + 3 a − ac, da + 3 ca + 3 ba + a − adcb + 2 b + ab − bc − ac, db + 2 cb + b − bd, dc + c − cdb − ac + 2 ab, bc − ad + ac, c − bd + 2 bc + . If we omit the three relations in the last row, then the resulting algebra B is AS regular.6. Invariant Rings for Diagonal Subgroups of
Aut gr ( k q [ u, v ])In this section, we fix A = k q [ u, v ] and G = 1 n (1 , a ) = (cid:28)(cid:18) ω ω a (cid:19)(cid:29) , where ω n = 1, 1 a < n , and gcd( a, n ) = 1. We wish to write down a presentation for theinvariant ring A G . Since G acts on A in the same way as on k [ u, v ], one should expect therelations in A G to be q -deformed versions of those in k [ u, v ] G ; we will show that this is in factthe case. The proof is quite straightforward if one uses the results in [18], with the only realdifficulty stemming from keeping track of the powers of q that appear in the relations.Recall from Section 2 that if we write nn − a = [ β , . . . , β d − ] . and define two series of integers i , . . . , i d and j , . . . , j d as follows, i = n, i = n − a and i k = β k − i k − − i k − for 3 k d,j = 0 , j = 1 and j k = β k − j k − − j k − for 3 k d, CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 27 then the invariant ring k [ u, v ] G is minimally generated by the d elements x k := u i k v j k , k d. Moreover, a minimal set of relations between the x k is as follows: x k − x k +1 = x β k − k for 2 k d − ,x k x ℓ = x β k − k +1 x β k +1 − k +2 . . . x β ℓ − − ℓ − x β ℓ − − ℓ − for 2 k + 1 < ℓ − d − . (6.1)Now view the x k as elements of A G . (As in the previous subsection, we will often swap betweenviewing the x k as elements of k [ u, v ] G and of A G .) Using the same arguments as for k [ u, v ] G ,one can show that the x k generate A G . It remains to determine the relations between them.In contrast with the commutative case, when viewed as elements of A G , in general the x k donot commute. Instead, we have the following: Lemma 6.2.
Suppose that k < ℓ d . Then x ℓ x k = q i k j ℓ − i ℓ j k x k x ℓ . Proof.
Since vu = quv , direct calculation using the definition of the x k gives x ℓ x k = u i ℓ v j ℓ u i k v j k = q i k j ℓ u i k + i ℓ v j k + j ℓ ,x k x ℓ = u i k v j k u i ℓ v j ℓ = q i ℓ j k u i k + i ℓ v j k + j ℓ . Therefore, x ℓ x k = q i k j ℓ − i ℓ j k q i ℓ j k u i k + i ℓ v j k + j ℓ = q i k j ℓ − i ℓ j k x k x ℓ , as claimed. (cid:3) The remaining relations are q -deformed versions of those from (6.1). As mentioned previously,the only difficulty in establishing the relations is keeping track of the powers of q . To this end,we record an easy lemma, the proof of which is omitted: Lemma 6.3.
Let a, b, c, a , . . . , a m , b , . . . , b m be non-negative integers. Then, in A , we have ( u a v b ) c = q abc ( c − u ac v bc , ( u a v b ) . . . ( u a m v b m ) = q r u P a i v P b i , where r = m X i =2 i − X j =1 a i b j . Using this, we can prove the following:
Lemma 6.4. In A , the following relations hold: x β k − k = q i k j k β k − ( β k − − − i k +1 j k − x k − x k +1 for k d − ,q r kℓ x k x ℓ = x β k − k +1 x β k +1 − k +2 . . . x β ℓ − − ℓ − x β ℓ − − ℓ − for k + 1 < ℓ − d − , where r kℓ = ℓ − X m = k +1 i m j m ( β m − − δ m,k +1 + δ m,ℓ − )( β m − − δ m,k +1 + δ m,ℓ − )+ ℓ − X m = k +2 m − X r = k +1 i m j r ( β m − − δ m,ℓ − )( β r − − δ r,k +1 ) − i ℓ j k , and where δ s,t is the Kronecker delta. CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 28
Proof.
To prove the first relation, observe that, by Lemma 6.3, x β k − k = ( u i k v j k ) β k − = q i k j k β k − ( β k − − u i k β k − u j k β k − . On the other hand, x k − x k +1 = u i k − v j k − u i k +1 v j k +1 = q i k +1 j k − u i k − + i k +1 v j k − + j k +1 Now, viewing the x i as elements of k [ u, v ] G , we know that x β k − k = x k − x k +1 , and hence i k β k − = i k − + i k +1 and j k β k − = j k − + j k +1 . (In fact, these identities follow directly from the definition of the i k and j k given at the beginningof this section.) Therefore x β k − k = q i k j k β k − ( β k − − u i k β k − u j k β k − = q i k j k β k − ( β k − − − i k +1 j k − q i k +1 j k − u i k − + i k +1 v j k − + j k +1 = q i k j k β k − ( β k − − − i k +1 j k − x k − x k +1 , as claimed.For the second relation, fix k and ℓ , and for ease of notation write γ m = β m − − δ m,k +1 + δ m,ℓ − for k + 1 m ℓ −
1. Notice that γ m is precisely the exponent of x m appearing on the righthand side of the claimed relation. Then, using Lemma 6.3, x γ k +1 k +1 x γ k +2 k +2 . . . x γ ℓ − ℓ − x γ ℓ − ℓ − = ( u i k +1 v j k +1 ) γ k +1 ( u i k +2 v j k +2 ) γ k +2 . . . ( u i ℓ − v j ℓ − ) γ ℓ − ( u i ℓ − v j ℓ − ) γ ℓ − = q a ( u i k +1 γ k +1 v j k +1 γ k +1 )( u i k +2 γ k +2 v j k +2 γ k +2 ) . . . ( u i ℓ − γ ℓ − v j ℓ − γ ℓ − )( u i ℓ − γ ℓ − v j ℓ − γ ℓ − )= q a q b u c v d , where a = ℓ − X m = k +1 i m j m γ m ( γ m − , b = ℓ − X m = k +2 m − X r = k +1 i m j r γ m γ r ,c = ℓ − X s = k +1 i s γ s , d = ℓ − X s = k +1 j s γ s . Observe that a + b − i ℓ j k = r kℓ . On the other hand, x k x ℓ = q i ℓ j k u i k + i ℓ u j k + j ℓ . Again, if we viewthe x i as elements of k [ u, v ] G , then we know that x k x ℓ = x β k − k +1 x β k +1 − k +2 . . . x β ℓ − − ℓ − x β ℓ − − ℓ − , andhence we have the identities c = i k + i ℓ and d = j k + j ℓ . Therefore, in A G , x γ k +1 k +1 x γ k +2 k +2 . . . x γ ℓ − ℓ − x γ ℓ − ℓ − = q a + b u c v d = q a + b − i ℓ j k q i ℓ j k u i k + i ℓ u j k + j ℓ = q a + b − i ℓ j k x k x ℓ , as claimed. (cid:3) This allows us to give a presentation for the invariant ring A G . The proof of this result issimilar to that of Theorem 5.11, and is therefore omitted. CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 29
Theorem 6.5.
Let G = n (1 , a ) , where a < n and gcd ( a, n ) = 1 . Let q ∈ k × . Then k q [ u, v ] G has the presentation k q [ u, v ] G ∼ = k h x , . . . , x d i I where the ideal of relations I is generated by x ℓ x k = q i k j ℓ − i ℓ j k x k x ℓ for k < ℓ d,x β k − k = q i k j k β k − ( β k − − − i k +1 j k − x k − x k +1 for k d − ,q r kℓ x k x ℓ = x β k − k +1 x β k +1 − k +2 . . . x β ℓ − − ℓ − x β ℓ − − ℓ − for k + 1 < ℓ − d − , where r kℓ is as in Lemma 6.4. Since the factor of k h x , . . . , x d i by the ideal generated by the first line of relations in Theorem6.5 is a quantum polynomial ring, which are well known to be AS regular, the following isimmediate: Corollary 6.6.
Let G = n (1 , a ) , where a < n and gcd ( a, n ) = 1 . Let q ∈ k × . Then A G isa factor of an AS regular algebra of dimension d . Commutative Invariant Rings of the Form k − [ u, v ] G n,k In this section, let n and k be positive coprime integers where k G = G n,k = h g, h i , where g = ω k ω − k ! , h = (cid:18) ω n ω n (cid:19) , (7.1)and where ω is a primitive (2 nk )th root of unity. We have already seen that the invariant ring k − [ u, v ] G is commutative if and only if n or k is even, so we will assume that this is the case inthis subsection. When this is the case, we will show that the rings k − [ u, v ] G are commutativequotient singularities, and hence are well-understood. More precisely, we shall see that everytype D singularity can be obtained in this way, as well as some type A singularities.7.1. Cyclic quotient singularities.
We first consider the type A cases, which occur when n
2. If n = 1 then k has to be even, and since k k ≡ n = 2, then since n and k are coprime, it follows that k is odd. Proposition 7.2.
Suppose that n = 1 and k is even, so k ≡ . Then k − [ u, v ] G n,k ∼ = k [ u, v ] k (1 ,k +1) , so that k − [ u, v ] G n,k is a cyclic quotient singularity.Proof. In this case ω k = 1, so g is the identity and G n,k = h h i . Observe that h k = (cid:16) − − (cid:17) ,which is the generator of (1 , k − [ u, v ], set x = ω k/ ( u − v ) , y = 2 uv, z = u + v , which generate the commutative invariant ring k − [ u, v ] h k and satisfy the single relation x + y + z = 0. We also have h k +1 · x = ω k +2 ω k/ ( v − u ) = ω ω k/ ( u − v ) = ω x,h k +1 · y = ω k +2 (2 vu ) = ω k +2 ω k (2 uv ) = ω y,h k +1 · z = ω k +2 ( v + u ) = ω k +2 z, CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 30 where h k/ is also a generator for G .Now consider the group k (1 , k + 1) with generator γ = (cid:16) ω ω k +1 (cid:17) , acting on the commutativering k [ u, v ] where, as above, ω k = 1. Notice that γ k = (cid:16) − − (cid:17) and that the invariant ring k [ u, v ] γ k is generated by x = ω k/ ( u + v ) , y = u − v , z = 2 uv, where we emphasise that these elements live inside k [ u, v ]. These satisfy x + y + z = 0 and γ · x = ω x, γ · y = ω y, γ · z = ω k +2 z. Therefore we have a chain of isomorphisms k − [ u, v ] G ,k = (cid:16) k − [ u, v ] h h k i (cid:17) h h k/ i ∼ = (cid:18) k [ x, y, z ] h x + y + z i (cid:19) h h k/ i ∼ = (cid:18) k [ x, y, z ] h x + y + z i (cid:19) h γ i ∼ = (cid:16) k [ u, v ] h γ k i (cid:17) k (1 ,k +1) = k [ u, v ] k (1 ,k +1) , as claimed. (cid:3) Proposition 7.3.
Suppose that n = 2 and k is odd. Then k − [ u, v ] G n,k ∼ = k [ u, v ] k (1 , k +1) , so that k − [ u, v ] G n,k is a cyclic quotient singularity.Proof. Here we have that ω k = 1 and g = (cid:18) − − (cid:19) , h = (cid:18) w w (cid:19) . Observe that the invariant ring k − [ u, v ] h g i is generated by x = ω k ( u − v ) , y = 2 uv, z = u + v , and these satisfy the relation x + y + z = 0. Now define ℓ = (cid:26) k + 1 if k ≡ , k + 1 if k ≡ , so that ℓ/ ℓ ≡ k + 2 mod 4 k . Moreover, ℓ/ k and so h h i = h h ℓ/ i . Then h ℓ/ · x = ω ℓ ω k ( v − u ) = ω k +2 ω k ω k ( u − v ) = ω x,h ℓ/ · y = ω ℓ (2 vu ) = ω k +2 ω k (2 uv ) = ω y,h ℓ/ · z = ω ℓ ( u + v ) = ω k +2 z. Now instead consider the group k (1 , k + 1) with generator γ = (cid:16) ω ω k +1 (cid:17) , where ω k = 1,acting on the commutative ring k [ u, v ]. Since γ k = (cid:16) − − (cid:17) , the invariant ring k [ u, v ] γ k isgenerated by x = w k ( u + v ) , y = u − v , z = 2 uv, and these satisfy x + y + z = 0. Also, we have γ · x = ω x, γ · y = ω y, γ · z = ω k +2 z, CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 31 and so there is a chain of isomorphisms k − [ u, v ] G ,k = (cid:16) k − [ u, v ] h g i (cid:17) h h ℓ/ i ∼ = (cid:18) k [ x, y, z ] h x + y + z i (cid:19) h h ℓ/ i ∼ = (cid:18) k [ x, y, z ] h x + y + z i (cid:19) h γ i ∼ = (cid:16) k [ u, v ] h γ k i (cid:17) k (1 , k +1) = k [ u, v ] k (1 , k +1) , as claimed. (cid:3) In both cases, one can obtain an explicit presentation of k − [ u, v ] G n,k using the results fromSection 2.2.1. In particular, every cyclic quotient singularity of the form k [ u, v ] k (1 , k +1) with k > k − [ u, v ] G n,k . Remark . Suppose that R = k [ x , · · · , x n ] and G and H are small subgroups of Aut gr ( A ) ∼ =GL( n, k ). By [3, Theorem 1.9], if R G ∼ = R H , then G and H are conjugate. However, we can usethe results from above to show a version of this result need not hold if we replace R by an ASregular algebra.Indeed, let A = k − [ u, v ] and consider G , = (cid:28)(cid:18) − − (cid:19) , (cid:18) − − (cid:19)(cid:29) . Proposition 7.3 tells us that k − [ u, v ] G , ∼ = k [ u, v ] (1 , , which is the coordinate ring of a type A Kleinian singularity. Explicitly, the elements x = ( u − v )( u + v )( u − v )( u + v ) , x = ( u + v )( u − v )( u + v )( u − v ) , z = ( u + v ) generate the invariant ring and give rise to an isomorphism k − [ u, v ] G , ∼ = k [ x, y, z ] h xy − z i . However, using Theorem 6.5 one can show that k − [ u, v ] (1 , is also the coordinate ring of atype A Kleinian singularity, and so we have an isomorphism k − [ u, v ] G , ∼ = k − [ u, v ] (1 , . The groups G , and (1 ,
3) are both small but are certainly not isomorphic (the former isisomorphic to the Klein four group, while the latter is cyclic), let alone conjugate.7.2.
Type D singularities. It remains to consider the cases where n >
3. In Section 2.2, werecalled the definition of the group D m,q , where m and q are positive coprime integers with1 < q < m . In this subection, it will be convenient to let ω be a primitive 4 q ( m − q )th root ofunity, and to write D m,q = * ω m − q ) ω − m − q ) ! , (cid:18) ω q ω q (cid:19)+ , (7.5)We now seek to show that, if n > k − [ u, v ] G n,k is commutative, then there is a one-to-onecorrespondence between the invariant rings k − [ u, v ] G n,k and k [ u, v ] D m,q .To be more precise, first define S := { ( n, k ) | gcd( n, k ) = 1 , n > , n − k ≡ , k } ,T := { ( m, q ) | < q < m, gcd( m, q ) = 1 } . CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 32
By Theorem 3.10 and Proposition 3.12, the set S consists of all pairs of integers ( n, k ) with n > k − [ u, v ] G n,k is commutative, while the set T consists of all pairs of integers( m, q ) which can be used to define the group D m,q . The aim of this subsection is to construct abijection θ : S → T such that k − [ u, v ] G n,k ∼ = k [ u, v ] D θ ( n,k ) . (7.6)We begin by constructing the map θ . Lemma 7.7.
The maps θ : S → T, θ ( n, k ) = ( ( n + k , n ) if n is odd ( ⇔ k is even),( n + k, n ) if n is even ( ⇔ k is odd), η : T → S, η ( m, q ) = (cid:26) ( q, m − q )) if m − q ≡ , (2 q, m − q ) if m − q ≡ , are well-defined, mutually inverse bijections.Proof. We first show that θ is well-defined. Let ( n, k ) ∈ S , and suppose that n is odd and k iseven, so that θ ( n, k ) = ( n + k , n ) =: ( m, q ). Since q = n > n + k > n , we have 1 < q < m .Moreover, since gcd( n, k ) = 1, we have gcd( m, q ) = gcd( n + k , n ) = gcd( k , n ) = 1. Therefore θ ( n, k ) ∈ T . If instead n is even, so that k is odd, we have θ ( n, k ) = ( n + k, n ) =: ( m, q ). Here n > n >
2, meaning that 1 < q < m . As before, gcd( m, q ) = 1 follows from the fact thatgcd( n, k ) = 1. Again we deduce that θ ( n, k ) ∈ T .To show that η is well-defined, let ( m, q ) ∈ T , and first suppose that m − q ≡ m and q are both odd. We then have η ( m, q ) = ( q, m − q )) =: ( n, k ). Now, since1 < q < m and q is odd, we have that n = q >
3. Moreover, since n and k have differentparities, n − k ≡ m − q ≡ k = 2( m − q ) ≡ n, k ) = gcd( q, m − q )) = gcd( q, m ) = gcd(1 , m ) = 1, where the penultimate equalityfollows since q and m are both odd. It follows that η ( m, q ) ∈ S .If instead m − q ≡ η ( m, q ) = (2 q, m − q ) =: ( n, k ). Since q >
2, we have n = 2 q >
3, and clearly 2 q and m − q have different parities, so n − k ≡ m − q ≡ k = m − q n − q is odd,gcd( n, k ) = gcd(2 q, n − q ) = gcd( q, n − q ) = gcd( n, q ) = 1. Therefore η ( m, q ) ∈ S .Finally, it is straightforward to check that θ and η are mutual inverses after noting that, ifthe pair ( n, k ) satisfies the conditions in the first (respectively, second) line in the definition of θ , then the pair ( m, q ) = θ ( n, k ) satisfies the conditions in the first (respectively, second) line inthe definition of η . (cid:3) We now seek to establish the isomorphism (7.6). We split the proof of this into two casesdepending on the parity of n (and hence of k ). The proof when n is odd is straightforward dueto the fact that G n,k = D θ ( n,k ) in this case. Proposition 7.8.
Suppose that ( n, k ) ∈ S and that n is odd and k is even (hence k ≡ ).Then there is an isomorphism k − [ u, v ] G n,k ∼ = k [ u, v ] D θ ( n,k ) . Proof.
Let ( m, q ) := θ ( n, k ), so m = n + k and q = n . Observe that 4 q ( m − q ) = 2 nk , and hencethe roots of unity appearing in (7.1) and (7.5) are the same; we write ω for this common root ofunity. Since m − q = k and q = n , we have G n,k = * ω k ω − k ! , (cid:18) ω n ω n (cid:19)+ , D θ ( n,k ) = * ω k ω − k ! , (cid:18) ω n ω n (cid:19)+ . CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 33
Clearly G n,k ⊆ D θ ( n,k ) . Moreover, we have | D m,q | = 4 q ( m − q ) = 2 nk = | G n,k | , and hence G n,k = D θ ( n,k ) . Notice that both groups contain (cid:0) w n w n (cid:1) k = (cid:16) − − (cid:17) =: α , and that k − [ u, v ] h α i = k [ u , − v , uv ] ∼ = k [ x, y, z ] h xy − z i , k [ u, v ] h α i = k [ u , v , uv ] ∼ = k [ x, y, z ] h xy − z i , where the induced actions of G n,k and D θ ( n,k ) on k [ x, y, z ] / h xy − z i are the same. Therefore, k − [ u, v ] G n,k = (cid:0) k − [ u, v ] h α i (cid:1) G n,k ∼ = (cid:0) k [ u, v ] h α i (cid:1) D θ ( n,k ) = k [ u, v ] D θ ( n,k ) , as claimed. (cid:3) We now turn our attention to the case when n is even, where the proof is more involved. Proposition 7.9.
Suppose that ( n, k ) ∈ S and that n is even and k is odd. Then there is anisomorphism k − [ u, v ] G n,k ∼ = k [ u, v ] D θ ( n,k ) . Proof.
Let ( m, q ) := θ ( n, k ), so that m = n + k and q = n . Then 4 q ( m − q ) = 2 nk and so, asin the proof of Proposition 7.8, the roots of unity appearing in (7.1) and (7.5) are the same, andwe write ω for this common root of unity. Since m − q = k and q = n , we have G n,k = * ω k ω − k ! , (cid:18) ω n ω n (cid:19)+ , D θ ( n,k ) = * ω k ω − k ! , ω n/ ω n/ !+ . Writing a and b for the generators of D θ ( n,k ) in the order given above, we have k [ u, v ] D θ ( n,k ) = (cid:0) k [ u, v ] h a i (cid:1) h b i . Notice that, since b has order 4 k and gcd(4 k, k + 2) = gcd( − , k + 2) = 1 (usingthat k is odd), we have h b i = h b k +2 i and hence k [ u, v ] D θ ( n,k ) = (cid:0) k [ u, v ] h a i (cid:1) h b k +2 i . Now, if we set x = u n , y = v n , and z = uv , then we obtain an isomorphism k [ u, v ] h a i ∼ = k [ x, y, z ] h xy − z n i . The action of b k +2 on this ring is as follows: b k +2 · x = ω n ( k +2) y, b k +2 · y = ω n ( k +2) x, b k +2 · z = ω n ( k +2) z. (7.10)We now turn our attention to the invariant ring k − [ u, v ] G n,k . Clearly if we write g and h for the generators of G n,k in the order given above, then k − [ u, v ] G n,k = (cid:0) k − [ u, v ] h g i (cid:1) h h i . Wesplit our analysis into two cases. First suppose that n ≡ x = u n , y = v n , and z = uv , we obtain an isomorphism k − [ u, v ] h g i ∼ = k [ x, y, z ] h xy − z n i . Here, the fact that n ≡ xy − z n rather than xy + z n . Theaction of h on this ring is given by h · x = ω n y, h · y = ω n x, h · z = ω n vu = ω n + nk uv = ω n ( k +2) z. (7.11)Since n ≡ n is an even integer and so12 n ( k + 2) ≡ n + n nk ≡ n mod 2 nk. Therefore, comparing the actions given in (7.10) and (7.11), we obtain isomorphisms k − [ u, v ] G n,k = (cid:0) k − [ u, v ] h g i (cid:1) h h i ∼ = (cid:18) k [ x, y, z ] h xy − z n i (cid:19) h h i CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 34 = (cid:18) k [ x, y, z ] h xy − z n i (cid:19) h b k +2 i ∼ = (cid:0) k [ u, v ] h a i (cid:1) h b k +2 i = k [ u, v ] D θ ( n,k ) . If instead n ≡ x = u n , y = − v n , and z = uv , then we obtain an isomorphism k − [ u, v ] h g i ∼ = k [ x, y, z ] h xy − z n i . (This time the fact that n ≡ y = − v n ensures that the correct relation is xy − z n .)The action of h in this case is given by h · x = ω n v n = ω n + nk y, h · y = − ω n u n = ω n + nk x, h · z = ω n ( k +2) z. (7.12)Since n ≡ n is an odd integer and so12 n ( k + 2) ≡ n + n nk ≡ n + nk mod 2 nk. Hence, comparing the actions given in (7.10) and (7.12), and arguing as in the previous case, weobtain an isomorphism k − [ u, v ] G n,k ∼ = k [ u, v ] D θ ( n,k ) . (cid:3) Remark . The proof of Proposition 7.8 shows that G n,k ∼ = D θ ( n,k ) when n is odd. On theother hand, when n is even, G n,k and D θ ( n,k ) have the following presentations: G n,k = h g, h | g n = 1 = h k , hg = g n − h i , D θ ( n,k ) = h a, b | a n = 1 , a n/ = b k , ba = a n − b i , and one can show that these groups are not isomorphic.8. Noncommutative Invariant Rings of the Form k − [ u, v ] G n,k Throughout, A = k − [ u, v ] and G = G n,k as in Theorem 3.10, and we write g amd h for thegenerators of G as in (7.1). We now seek to obtain generators for the invariant rings A G whenthis ring is not commutative; by Proposition 3.12, this happens precisely when n and k are bothodd, and we will assume that this is the case throughout this section.We begin by identifying a k -basis for the invariants. Proposition 8.1. A k -basis for A G is given by n u i v j + ( − ( i +1)( j +1)+1 u j v i (cid:12)(cid:12)(cid:12) i, j, > , i − j ≡ n and i + j ≡ k o . Proof.
By [13], the Reynolds operator ρ : A → A G , ρ ( a ) = X x ∈ G x · a, is surjective. Therefore, to find a k -basis for A G it suffices to evaluate ρ on the k -basis for A given by { u i v j | i, j > } . Fixing i, j >
0, we have n − X ℓ =0 g ℓ · ( u i v j ) = n − X ℓ =0 ω kℓ ( i − j ) u i v j = (cid:26) nu i v j if i − j ≡ n, , using (4.2) and noting that ω k is a primitive n th root of unity. Similarly, using the fact that ω n is a primitive k th root of unity k − X m =0 h m · ( u i v j ) = k − X m =0 h m · ( u i v j ) + k − X m =0 h m +1 · ( u i v j ) CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 35 = k − X m =0 ω mn ( i + j ) u i v j + k − X m =0 ω (2 m +1) n ( i + j ) v i u j = k − X m =0 ω mn ( i + j ) u i v j + ( − ij ω n ( i + j ) k − X m =0 ω mn ( i + j ) u j v i = ( k (cid:16) u i v j + ( − ij ω n ( i + j ) u j v i (cid:17) if i + j ≡ k, , = ( k (cid:16) u i v j + ( − ( i +1)( j +1)+1 u j v i (cid:17) if i + j ≡ k, , since if i + j ≡ k , then ( − ij ω n ( i + j ) = ( − ij ( − i + j = ( − ( i +1)( j +1)+1 . Therefore,since G = { h m g ℓ | m k − , ℓ n − } , we have ρ ( u i v j ) = k − X m =0 n − X ℓ =0 h m g ℓ · ( u i v j )= ( nk (cid:16) u i v j + ( − ( i +1)( j +1)+1 u j v i (cid:17) if i − j ≡ n and i + j ≡ k, . It follows that A G has the claimed k -basis. (cid:3) We now write the k -basis from the above proposition in a modified form, which will make iteasier to write down a set of generators. Without loss of generality, we may assume that i > j .By our assumptions on i and j , we have i − j = ns and i + j = kt for some non-negative integers s, t . Clearly i = ( kt + ns ) and j = ( kt − ns ), and we also have u i v j + ( − ( i +1)( j +1)+1 u j v i = ( u i − j + ( − ( i − j ) j +( i +1)( j +1)+1 v i − j )( uv ) j = ( u i − j + ( − i v i − j )( uv ) j . (8.2)If i = j , then (8.2) is nonzero if and only if i ≡ ≡ j mod 2 k , and so the elements we obtain arepowers of ( uv ) k . Otherwise i > j , in which case (8.2) is equal to( u ns + ( − ( kt + ns ) v ns )( uv ) ( kt − ns ) = ( u ns + ( − r + ns v ns )( uv ) r , where r = ( kt − ns ). We summarise our findings below: Proposition 8.3. A k -basis for A G is given by n ( uv ) ki (cid:12)(cid:12)(cid:12) i > o ∪ n ( u ns + ( − r + ns v ns )( uv ) r (cid:12)(cid:12)(cid:12) r > , s, t > , r + ns = kt o . We now seek to use Proposition 8.3 to find a set of generators for A G . The form of thegenerators depends on whether n > k or n < k , and so we need to divide the argument into thesetwo cases. The only case which this does not consider is when n = 1 = k , but this is covered by[15, Remark 2.6].8.1. The case n > k . First suppose that n > k . Write n ( n + k ) = [ γ , . . . , γ d − ]for the Hirzebruch-Jung continued fraction expansion of n ( n + k ) (obviously one can write thisfraction as nn + k , but the given presentation is in lowest terms). Since n > k , we note that γ = 2. CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 36
Also define β i = (cid:26) γ + 1 if i = 3 ,γ i if i = 3 . Define three series of integers r , . . . , r d − , s , . . . , s d − , and t , . . . , t d − as follows: s = 1 s = 1 s i = β i s i − − s i − for 3 i d − ,t = 2 β + 1 t = 2 β − t i = β i t i − − t i − for 3 i d − ,r i = ( kt i − ns i ) for 1 i d − , where the entries to the right of the vertical line only exist when d >
2, which happens if andonly if k >
3, if and only if the action of G on A has nontrivial homological determinant. Alsoobserve that the r i obey the same recurrence relation as the s i and t i . We now collect someproperties of these series: Lemma 8.4. If k > then [ β , β , . . . , β d − ] = kβ − ( n − k ) kβ − ( n + k ) . Proof.
Let α = [ γ , γ , . . . , γ d − ] = [ β − , β , . . . , β d − ], so that the quantity we wish to deter-mine is α + 1. Then, since β = 2,2 nn + k = 2 − β − α = α (2 β − − αβ − , which rearranges to give α = kkβ − ( n + k ) . Therefore, [ β , β , . . . , β d − ] = α + 1 = k + kβ − ( n + k ) kβ − ( n + k ) = kβ − ( n − k ) kβ − ( n + k ) , as claimed. (cid:3) Lemma 8.5.
The r i , s i , and t i have the following properties: (1) r = kβ − ( n − k ) , r = kβ − ( n + k ) , r d − = 1 and r d − = 0 . Moreover, r > r > · · · > r d − > r d − . (2) r i s i +1 − r i +1 s i = k and r i t i +1 − r i +1 t i = n for i d − . (3) s d − = k and t d − = n .Proof. (1) Using the definition of the r i , r = 12 (cid:0) k (2 β + 1) − n (cid:1) = kβ −
12 ( n − k ) r = 12 (cid:0) k (2 β − − n (cid:1) = kβ −
12 ( n + k ) . Then, since r r = [ β , β , . . . , β d − ] and the r i satisfy r i = β i − r i − − r i − for 3 i d −
1, itfollows from general theory that r > r > · · · > r d − > r d − , with r d − = 1 and r d − = 0.(2) These are both easy to prove by induction. CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 37 (3) To determine the value of s d − , we use parts (1) and (2): s d − = 1 · s d − − · s d − = r d − s d − − r d − s d − = k. Similarly, we can use the identity r d − t d − − r d − t d − = n to show that t d − = n . (cid:3) We also require the following two technical lemmas:
Lemma 8.6.
Suppose that the triple ( r, s, t ) satisfies r + ns = kt and that r < k and s, t > . If rs > r i s i for some i with i d − , then i > and r > r i − .Proof. Let ( r, s, t ) be as in the statement. We first show that i >
1. Observing that β = ⌈ n + k k ⌉ ,we need to show that rs r s = k (cid:24) n + k k (cid:25) −
12 ( n − k ) . If s >
2, then rs r < k k = k (cid:18) n + k k (cid:19) −
12 ( n − k ) k (cid:24) n + k k (cid:25) −
12 ( n − k ) , as required. So now suppose that s = 1. Write n uniquely in the form n = ka − b where a is oddand b ∈ { , , . . . , k − } . Then k (cid:24) n + k k (cid:25) −
12 ( n − k ) = k (cid:24) a + 12 − b k (cid:25) − k ( a − − b k ( a + 1)2 − k ( a − − b k + b . Rearranging the equality 2 r + n = kt gives r = k (cid:0) t − a (cid:1) + b . Since n and k are both odd, t isalso odd, so t − a is an integer. But since r < k , the previous equality forces t − a to be at most1. Therefore r = k (cid:18) t − a (cid:19) + b k + b k (cid:24) n + k k (cid:25) + 12 ( k − n ) . (8.7)This shows that rs r s .Now assume that rs > r i s i for some i (where necessarily i > r i and s i satisfy r i − s i − r i s i − = k for 2 i d −
1, it follows that r i s i < r i − s i − for2 i d −
1, so we may assume that i has been chosen minimal subject to rs > r i s i . In particular,we have r i s i < rs r i − s i − . (8.8)Seeking a contradiction, assume that r < r i − , so that r i − − r >
1. Consider the equations2 r + ns = kt and 2 r i + ns i = kt i ;multiplying the former by r i and the latter by r and then subtracting gives n ( r i s − rs i ) = k ( r i t − rt i ) . (8.9)Since n and k are coprime, this forces rs i − r i s to be a multiple of k . Since (8.8) implies that rs i − r i s >
0, we must therefore have rs i − r i s > k . Combining this with Lemma 8.5 (2), we findthat rs i − r i s > r i − s i − r i s i − . This rearranges to gives r i ( s i − − s ) > s i ( r i − − r ) > , CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 38 where the second inequality follows since r i − − r > s i − > s .Then, using Lemma 8.5 (2), rs = rs i − r i sss i + r i s i > kss i + r i s i > ks i s i − + r i s i = k + r i s i − s i s i − = r i − s i s i s i − = r i − s i − . Hence rs > r i − s i − , contradicting the minimality of i , so we must have r > r i − . (cid:3) Remark . Suppose that s = 1 and ( r, s, t ) satisfy the hypotheses of Lemma 8.6, and let a and b be as in the above proof. It is straightforward to check that a = 2 β −
1. Before equation(8.7), we saw that t − a was an integer which was at most 1. If t − a −
1, then r = k (cid:18) t − a (cid:19) + b − k + ( k −
1) = − , contradicting the fact that r >
0. It follows that t − a is either 0 or 1, i.e. t = a = 2 β − t or t = a + 2 = 2 β + 1 = t . This in turn forces r = r or r = r , respectively. We will requirethis fact in the proof of Theorem 8.13. Lemma 8.11.
Suppose that the triple ( r, s, t ) satisfies r + ns = kt with r < k , s, t > .Then there exist non-negative integers c , . . . , c d − such that r = d − X i =1 c i r i , s = d − X i =1 c i s i , t = d − X i =1 c i t i . (8.12) Proof.
We prove this by induction on r . If r = 0, then the triple ( r, s, t ) satisfies ns = kt and so,since n and k are coprime, s = mk and t = mn for some non-negative integer m . By Lemma 8.5,we know that r d − = 0, s d − = k , and t d − = n , and so the claim follows by setting c d − = m and c i = 0 for i = d − r >
1. Since r > r > · · · > r d − , choose i minimal subject to r − r i > r i satisfies r i r < r i − . By Lemma 8.6, it follows that rs r i s i , and hence rs i r i s . Since r i r , this also forces s i s . Then, using (8.9),0 n ( r i s − rs i ) = k ( r i t − rt i ) kr ( t − t i ) , and so t > t i .Now set ( r ′ , s ′ , t ′ ) = ( r − r i , s − s i , t − t i ), which is a triple satisfying r ′ , s ′ , t ′ >
0, and2 r ′ + ns ′ = kt ′ . If s ′ = t ′ = 0 then necessarily r ′ = 0, and so setting c i = 1 and c j = 0 for j = i gives the result. If s ′ , t ′ >
0, then applying the induction hypothesis to the triple ( r ′ , s ′ , t ′ )gives a decomposition of the form (8.12), and hence also one for the triple ( r, s, t ). It remains toexclude the possibility that exactly one of s ′ and t ′ is 0. Indeed, if t ′ = 0 then 2 r ′ + ns ′ = 0, andso r ′ = 0 = s ′ . If instead s ′ = 0, then 2 r ′ = kt ′ , and since k is odd we have r ′ = mk and t ′ = 2 m for some non-negative integer m . We claim that necessarily r ′ < k , which will force m = 0, andhence r ′ = 0 = t ′ . To see this, first notice that if r < k then this is immediate, so suppose that r > k . As noted in the proof of Lemma 8.6, β = ⌈ n + k k ⌉ , and so β < n + k k + 1. It follows that r = kβ −
12 ( n − k ) < k (cid:18) n + k k + 1 (cid:19) −
12 ( n − k ) = 2 k, and since r = r − k , we also have r < k . Therefore, if r > k , then exactly one of r and r satisfies 0 r − r i < k , and this is our choice for r i . Therefore r ′ = r − r i < k , which finishesthe proof. (cid:3) We are finally able to write down a list of generators for A G . CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 39
Theorem 8.13.
Suppose n > k . Then the elements x i = (cid:0) u ns i + ( − r i + ns i v ns i (cid:1) ( uv ) r i , i d − , and x d = ( uv ) k generate A G .Proof. Since 2 r i + ns i = kt i , the x i have the form given in Proposition 8.3 and hence areelements of A G . So now suppose that x = (cid:0) u ns + ( − r + ns v ns (cid:1) ( uv ) r is an element of A G , where2 r + ns = kt ; we wish to show that x can be generated by x , . . . , x d . Since x d = ( uv ) k , we mayassume that r < k . We prove by induction on s that this is possible.When s = 1, Remark 8.10 shows that the only possibilities are ( r, s, t ) = ( r , s , t ) or ( r, s, t ) =( r , s , t ), and so x = x or x = x . So now suppose that s > s . By Lemma 8.11, there exist non-negative integers c i such that r = d − X i =1 c i r i , s = d − X i =1 c i s i , t = d − X i =1 c i t i . If P d − i =1 c i = 1, then r = r i , s = s i , and t = t i for some i , and hence x = x i . Else P d − i =1 c i > a i , b i > a i + b i = c i for all i , and P d − i =1 a i , P d − i =1 b i > r a = d − X i =1 a i r i , s a = d − X i =1 a i s i , t a = d − X i =1 a i t i , and define r b , s b , t b analogously. Without loss of generality, we may assume that s a > s b . Observethat r = r a + r b , s = s a + s b , and t = t a + t b , and that 2 r a + ns a = kt a and 2 r b + ns b = kt b .Therefore the elements x a := (cid:0) u ns a + ( − r a + ns a v ns a (cid:1) ( uv ) r a and (cid:0) u ns b + ( − r b + ns b v ns b (cid:1) ( uv ) r b lie in A G , and since s a , s b < s , they can be generated by the x i by the induction hypothesis. Wethen have x a x b = (cid:0) u ns a + ( − r a + ns a v ns a (cid:1) ( uv ) r a (cid:0) u ns b + ( − r b + ns b v ns b (cid:1) ( uv ) r b = ± (cid:0) u ns a + ( − r a + ns a v ns a (cid:1)(cid:0) u ns b + ( − r b + ns b v ns b (cid:1) ( uv ) r = ± ( u ns + ( − r + ns v ns )( uv ) r ± (( − r b + ns b u ns a v ns b + ( − r a + ns a v ns a u ns b )( uv ) r = ± x ± (cid:0) u n ( s a − s b ) u ns b v ns b + ( − r + ns + n s b v n ( s a − s b ) u ns b v ns b (cid:1) ( uv ) r = ± x ± (cid:0) u n ( s a − s b ) + ( − r + ns + ns b v n ( s a − s b ) (cid:1) ( uv ) r + ns b = ± x ± (cid:0) u n ( s a − s b ) + ( − r + ns b + n ( s a − s b ) v n ( s a − s b ) (cid:1) ( uv ) r + ns b , (8.14)where we use ± to indicate that one can keep track of the powers of − r ′ , s ′ , t ′ ) = ( r + ns b , s a − s b , t ), and it is easy to check that2 r ′ + ns ′ = kt ′ . In particular s ′ < s , and so the inductive hypothesis tells us that this element isgenerated by the x i . Since the same was true of x a and x b , it follows that x can be generatedby the x i , completing the proof. (cid:3) Example 8.15.
First suppose that n > k = 1, so that G n,k is the dihedral groupof order 2 n . Then the Hirzebruch-Jung continued fraction expansion of n ( n + k ) is n ( n + 1) = 2 − ( n + 1) = [2 , ( n + 1)] . We therefore obtain the following β -, r -, s -, and t -series: CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 40 i β i ( n + 1) r i s i t i n + 2 n By Theorem 8.13, the invariant ring A G is generated by the three elements x = ( u n + v n ) uv, x = ( u n − v n ) , x = ( uv ) . This case has previously been considered in [6]. Indeed, the above generators agree with thosegiven in [6, Table 3 (d)], up to a change of sign for one generator, which is due to the fact thatthe representation of G we consider is conjugate to the representation used in [6].For a more involved example, suppose instead that n = 17 and k = 11. The Hirzebruch-Jungcontinued fraction expansion of n ( n + k ) is then175 = [2 , , , , , . We therefore obtain the following β -, r -, s -, and t -series: i β i r i
19 8 5 2 1 0 s i t i A G is given by the following seven elements: x = ( u + v )( uv ) , x = ( u − v )( uv ) , x = ( u − v )( uv ) ,x = ( u − v )( uv ) , x = ( u + v )( uv ) , x = ( u − v ) uv, x = ( uv ) . It is natural to ask whether one can use the preceding results to give presentations for theinvariant rings k − [ u, v ] G n,k . While it is possible to achieve this for specific examples by hand orby using computer assistance, we have been unable to find a way to write down a presentationfor arbitrary n and k . We give an example to highlight some of the difficulties, even for small n and k . Example 8.16.
Suppose that n = 7 and k = 3. The Hirzebruch-Jung continued fractionexpansion of n ( n + k ) is 75 = [2 , , . By Theorem 8.13, A G is generated by a := ( u − v )( uv ) , b := ( u + v )( uv ) , c := u − v , d := ( uv ) . One can verify that a, b, c, d satisfy the relations ba + ab + 4 d , ca + ac − b − d , cb + bc − b d,da − ad, db − bd, dc − cd,a + b d, ab + cd + bd , ac + abd − b . Therefore, if we define B := k h a, b, c, d i (cid:28) ba + ab + 4 d , ca + ac − b − d , cb + bc − b dda − ad, db − bd, dc − cd (cid:29) , CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 41 and let I be the following two-sided ideal of B , I := h a + b d, ab + cd + bd , ac + abd − b i , then there is a surjection of graded algebras φ : B/I ։ A G . We first claim that B is AS regular. To show this, we first determine the Hilbert series of B . Ordering the words in k h a, b, c, d i using degree lexicographic order with a < b < c < d , itis straightforward to check that the overlap relations between the leading terms of the relationsdefining B resolve. Therefore B has a k -basis given by { a i b j c ℓ d m | i, j, ℓ, m > } , and so hilb B = 1(1 − t )(1 − t )(1 − t )(1 − t ) . In particular, d is a right nonzerodivisor, and since it is central, it is a regular central element.Finally the ring B/ h d i = k h a, b, c ih ba + ab, ca + ac − b , cb + bc i is easily seen to be AS regular of dimension 3, so B is AS regular of dimension 4 by [8, Proposition2.1].We now claim that the map φ is an isomorphism; since we already know that this map is asurjection, it suffices to check that B/I and A G have the same Hilbert series. Using Molien’sTheorem, a computer calculation can be used to verify thathilb A G = 1 − t − t − t + t + t (1 − t )(1 − t )(1 − t )(1 − t ) . On the other hand, to determine the Hilbert series of
B/I , one can check that the followingsequence of graded right B -modules is exact,0 → B [ − ⊕ B [ − ψ −→ B [ − ⊕ B [ − ⊕ B [ − η −→ B → B/I → , where the maps η and ψ are given by left multiplication by the following matrices: η = (cid:0) a + b d ab + cd + bd ac + abd − b (cid:1) , ψ = − b bd + ca − b d a . This implies thathilb
B/I = (1 − t − t − t + t + t ) hilb B = 1 − t − t − t + t + t (1 − t )(1 − t )(1 − t )(1 − t ) , and hence that we have an isomorphism B/I ∼ = A G . In the above example, in particular we saw that A G was a factor of an AS-regular algebra.We expect that this is always the case. CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 42
The case n < k . Now suppose that n < k . As with the case when n > k , write n ( n + k ) = [ γ , . . . , γ d − ]Since n < k , we note that γ = 1. Also define β i = (cid:26) γ + 2 if i = 2 ,γ i if i = 2 . Define three series of integers r , . . . , r d , s , . . . , s d , and t , . . . , t d as follows: s = 1 s = 1 s i = β i − s i − − s i − for 3 i d,t = 3 t = 1 t i = β i − t i − − t i − for 3 i d,r i = ( kt i − ns i ) for 1 i d. In particular, note that the r i obey the same recurrence relation as the s i and t i .At this point, the proof proceeds in a similar fashion to the case when n > k . We now stateanalogues of the results for that case, only providing proofs when the details are sufficientlydifferent. Lemma 8.17.
We have (3 k − n ) ( k − n ) = [ β , β , . . . , β d − ] . Lemma 8.18.
The r i , s i , and t i have the following properties: (1) r = (3 k − n ) , r = ( k − n ) , r d − = 1 and r d = 0 . Moreover, r > r > · · · > r d − > r d . (2) r i s i +1 − r i +1 s i = k and r i t i +1 − r i +1 t i = k for i d − . (3) s d = k and t d = n . The statement of Lemma 8.11 also holds when n < k (and this can be proved using an analogueof Lemma 8.6) which allows us to write down generators for A G in this case: Theorem 8.19.
Suppose n < k . Then the elements x i = (cid:0) u ns i + ( − r i + ns i v ns i (cid:1) ( uv ) r i , i d generate A G .Proof. The only difference between the proof of this theorem and that of Theorem 8.13 is thatwe must first show that the claimed generators generate ( uv ) k , before proceeding as we didbefore. Since s = 1 = s , and r = (3 k − n ) and r = ( k − n ), we have x = ( u n + ( − (3 k + n ) v n )( uv ) (3 k − n ) and x = ( u n + ( − ( k + n ) v n )( uv ) ( k − n ) . Direct calculation gives x x = ( − (3 k − n ) (cid:0) u n + ( − ( k + n ) u n v n − v n (cid:1) ( uv ) k − n x x = ( − ( k − n ) (cid:0) u n + ( − (3 k + n ) u n v n − v n (cid:1) ( uv ) k − n . Since (3 k − n ) and ( k − n ) have different parities, we have x x + x x = 4 u n v n ( uv ) k − n = ( − n − uv ) k , and so the x i generate ( uv ) k . The remainder of the proof is similar to that of Theorem 8.13. (cid:3) Acknowledgements.
The author is a postdoctoral fellow at the University of Waterloo and issupported by the NSERC. The author is grateful to Jason Bell and Michael Wemyss for helpfuldiscussions, and to Susan J. Sierra for suggesting that the algebras in Section 5 might be factorsof the algebras studied in [16].
CTIONS OF SMALL GROUPS ON TWO-DIMENSIONAL AS REGULAR ALGEBRAS 43
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