Projective resolutions of associative algebras and ambiguities
aa r X i v : . [ m a t h . K T ] S e p PROJECTIVE RESOLUTIONS OF ASSOCIATIVE ALGEBRAS ANDAMBIGUITIES
SERGIO CHOUHY AND ANDREA SOLOTAR
Abstract.
The aim of this article is to give a method to construct bimodule resolutionsof associative algebras, generalizing Bardzell’s well-known resolution of monomial algebras.We stress that this method leads to concrete computations, providing thus a useful tool forcomputing invariants associated to the considered algebras. We illustrate how to use it bygiving several examples in the last section of the article. In particular we give necessary andsufficient conditions for noetherian down-up algebras to be 3-Calabi-Yau.
Hochschild cohomology, resolution, homology theory.1.
Introduction
The invariants attached to associative algebras and in particular to finite dimensional algebras,have been widely studied during the last decades. Among others, Hochschild homology andcohomology of diverse families of algebras have been computed.The first problem one faces when computing Hochschild (co)homology is to find a convenientprojective resolution of the algebra as a bimodule over itself. Of course, the bar resolution isalways available but it is almost impossible to perform computations using it.M. Bardzell provided in [Ba] a bimodule resolution for monomial algebras, that is, algebras A = kQ/I with k a field, Q a finite quiver and I a two-sided ideal which can be generated bymonomial relations; in this situation, the set of classes in A of paths in Q which are not zero isa basis of A . Moreover, this resolution is minimal. A simple proof of the exactness of Bardzell’scomplex has been given by E. Sk¨oldberg in [Sk], where he provided a contracting homotopy. Ofcourse, having such a resolution does not solve the whole problem, it is just a starting point.The non monomial case is more difficult, since it involves rewriting the paths in terms of abasis of A . Different kinds of resolutions for diverse families of algebras have been provided inthe literature. For augmented k -algebras, Anick constructed in [An] a projective resolution of theground field k . The projective modules in this resolution are constructed in terms of ambiguities(or n -chains), and the differentials are not given explicitly. In practice, it is hard to make thisconstruction explicit enough in order to compute cohomology. For quotients of path algebrasover a quiver Q with a finite number of vertices, Anick and Green exhibited in [AG] a resolutionfor the simple module associated to each vertex, generalizing the result of [An], which deals withthe case where the quiver Q has only one vertex. Also, Y. Kobayashi in [Kob] proposes a methodto construct a resolution which seems not to be extremely useful.One may think that the case of binomial algebras is easier than others, but in fact it is notquite true since it is necesssary to keep track of all reductions performed when writing an elementin terms of a chosen basis of the algebra as a vector space. This work has been supported by the projects UBACYT X475, PIP-CONICET 2012-2014 11220110100870,PICT 2011-1510 and MathAmSud-GR2HOPF. The first author is a CONICET fellow. The second author is aresearch member of CONICET.
In this article we construct in an inductive way, given an algebra A , a projective bimoduleresolution of A , which is a kind of deformation of Bardzell’s resolution of a monomial algebraassociated to A . For this, we use ideas coming from Bergman’s Diamond Lemma and from thetheory of Gr¨obner bases. The resolution we give is not always minimal, but we prove minimalityfor various families of algebras.In the context of quotients of path algebras corresponding to a quiver with a finite number ofvertices, our method consists in constructing a resolution whose projective bimodules come fromambiguities present in the rewriting system. Of course there are many different ways of choosinga basis, so we must state conditions that assure that the rewriting process ends and that it isefficient.One of the advantages of doing this is that, once a bimodule resolution is obtained, it is easyto construct starting from it a resolution of any module on one side and, in particular, to recoverthose constructed in [An] and [AG] for the case of the simple modules associated to the verticesof the quiver.To deal with the problem of effective computation of these resolutions, Theorem 4.1 belowgives sufficient conditions for a complex defined over these projective bimodules to be exact. Wewill be, in consequence, able to prove that some complexes are resolutions without following theprocedure prescribed in the proof of the existence theorem.Briefly, we do the following: given an algebra A = kQ/I we compute a bimodule resolutionof A from a reduction system R for I which satifies a condition we denote ( ♦ ). We prove thatsuch a reduction system always exists, but we also show in an example that it may not be themost convenient one. In particular the resolution obtained may not be minimal.Applying our method we recover a well-known resolution of quantum complete intersections,see for example [BE] and [BGMS]. We also construct a short resolution for down-up algebraswhich allows us to prove that a noetherian down-up algebra A ( α, β, γ ) is 3-Calabi-Yau if andonly if β = − A is graded by the length of paths,and it has a reduction system satisfying the conditions requiered for minimality of the resolution,then A is N -Koszul if and only if the associated monomial algebra A S is N -Koszul.We have just seen a recent preprint by Guiraud, Hoffbeck and Malbos [GHM] where theyconstruct a resolution that may be related to ours.We are deeply indebted to Mariano Su´arez-Alvarez and Eduardo Marcos for their help inimproving this article. We also thank Roland Berger, Quimey Vivas and Pablo Zadunaisky fordiscussions and comments. 2. Preliminaries
In this section we give some definitions, present some basic constructions and we also proveresults that are necessary in the sequel.Let k be a field and Q a quiver with a finite set of vertices. Given n ∈ N , Q n denotes the setof paths of length n in Q and Q ≥ n the set of paths of length at least n , that is, Q ≥ n = S i ≥ n Q i .Whenever c ∈ Q n , we will write | c | = n . If a, b, p, q ∈ Q ≥ are such that q = apb , we say that ROJECTIVE RESOLUTIONS OF ASSOCIATIVE ALGEBRAS AND AMBIGUITIES 3 p is a divisor of q ; if, moreover, a = 1, we say that p is a left divisor of q and analogously for b = 1 and right divisor . We denote t , s : Q → Q the usual source and target functions. Given s ∈ Q ≥ and a finite sum f = P i λ i c i ∈ kQ such that c i ∈ Q ≥ and t ( s ) = t ( c i ), s ( s ) = s ( c i )for all i , we say that f is parallel to s . Let E := kQ be the subalgebra of the path algebragenerated by the vertices of Q .Given a set X and a ring R , we denote h X i R the left R -module freely spanned by X .Let I be a two sided ideal of kQ , A = kQ/I and π : kQ → A the canonical projection. Weassume that π ( Q ∪ Q ) is linearly independent.We recall some terminology from [B] that we will use. A set of pairs R = { ( s i , f i ) } i ∈ Γ where s i ∈ Q ≥ , f i ∈ kQ is called a reduction system . We will always assume that a reduction system R = { ( s i , f i ) } i ∈ Γ satisfies the following conditions • for all i , f i is parallel to s i and f i = s i . • s i does not divide s j for i = j .Given ( s, f ) ∈ R and a, c ∈ Q ≥ such that asc = 0 in kQ , we will call the triple ( a, s, c ) a basic reduction and write it r a,s,c . Note that r a,s,c determines an E -bimodule endomorphism r a,s,c : kQ → kQ such that r a,s,c ( asc ) = af c and r a,s,c ( q ) = q for all q = asc .A reduction is an n -tuple ( r n , . . . , r ) where n ∈ N and r i is a basic reduction for 1 ≤ i ≤ n .As before, a reduction r = ( r n , . . . , r ) determines an E -bimodule endomorphism of kQ , thecomposition of the endomorphisms corresponding to the basic reductions r n , . . . , r .An element x ∈ kQ is said to be irreducible for R if r ( x ) = x for all basic reductions r . Wewill omit mentioning the reduction system whenever it is clear from the context. A path p ∈ Q ≥ will be called reduction-finite if for any infinite sequence of basic reductions ( r i ) i ∈ N , there exists n ∈ N such that for all n ≥ n , r n ◦ · · · ◦ r ( p ) = r n ◦ · · · ◦ r ( p ). Moreover, the path p willbe called reduction-unique if it is reduction-finite and for any two reductions r and r ′ such that r ( p ) and r ′ ( p ) are both irreducible, the equality r ( p ) = r ′ ( p ) holds. Definition 2.1.
We say that a reduction system R satisfies condition ( ♦ ) for I if • the ideal I is equal to the two sided ideal generated by the set { s − f } ( s,f ) ∈R , • every path is reduction-unique and • for each ( s, f ) ∈ R , f is irreducible.The reason why we are interested in these reduction systems is the following lemma, which isa restatement of Bergman’s Diamond Lemma. Lemma 2.2.
If the reduction system R satisfies ( ♦ ) for I , then the set B of irreducible pathssatisfies the following properties,(i) B is closed under divisors,(ii) π ( b ) = π ( b ′ ) for all b, b ′ ∈ B with b = b ′ ,(iii) { π ( b ) : b ∈ B} is a basis of A .Remark . In view of Lemma 2.2, we can define a k -linear map i : A → kQ such that be i ( π ( b ))) = b for all b ∈ B . We denote by β : kQ → kQ the composition i ◦ π . Notice that if p isa path and r is a reduction such that r ( p ) is irreducible, then r ( p ) = β ( p ). In the bibliography, β ( p ) is sometimes called the normal form of p . Definition 2.3. If R is a reduction system satisfying ( ♦ ) for I , we define S := { s ∈ Q ≥ :( s, f ) ∈ R for some f ∈ kQ } . Remark . Notice that:(1) S is equal to the set { p ∈ Q ≥ : p / ∈ B and p ′ ∈ B for all proper divisors p ′ of p } .(2) If s and s ′ are elements of S such that s divides s ′ , then s = s ′ . PROJECTIVE RESOLUTIONS OF ASSOCIATIVE ALGEBRAS AND AMBIGUITIES (3) Given q ∈ Q ≥ , q is irreducible if and only if there exists no p ∈ S such that p divides q . Definition 2.4.
Given a path p and q = P ni =1 λ i c i ∈ kQ with λ , . . . , λ n ∈ k × and c , . . . , c n ∈ Q ≥ , we write p ∈ q if p = c i for some i , or, in other words, when p is in the support of q .Given p, q ∈ Q ≥ we write q p if there exist n ∈ N , basic reductions r , . . . , r n and paths p , . . . p n such that p = q , p n = p , and for all i = 1 , . . . , n − p i +1 ∈ r i ( p i ). Lemma 2.5.
Suppose that every path is reduction-finite with respect to R .(i) If p is a path and r a basic reduction such that p ∈ r ( p ) , then r ( p ) = p .(ii) The binary relation is an order on the set Q ≥ which is compatible with concatenation,that is, satisfies that q p implies aqc apc for all a, c ∈ Q ≥ such that apc = 0 in kQ .(iii) The binary relation satisfies the descending chain condition.Proof. (i) The hypothesis means that r ( p ) = λp + x with λ ∈ k × and p / ∈ x . If x = 0 or λ = 1,then r acts nontrivially on p and so it acts trivially on x . Since the sequence of reductions( r, r, · · · ) stabilizes when acting on p , there exists k ∈ N such that λ k p + kx = r k ( p ) = r k +1 ( p ) = λ k +1 p + ( k + 1) x . As a consequence, λ = 1 and x = 0. (ii) It is clear that is a transitive and reflexive relation and that it is compatible withconcatenation. Let us suppose that it is not antisymmetric, so that there exist n ∈ N , paths p , . . . , p n +1 and basic reductions r , . . . , r n such that p i +1 ∈ r i ( p i ) for 1 ≤ i ≤ n and p n +1 = p .Suppose that n is minimal. There exist x , . . . , x n ∈ kQ and λ , . . . , λ n ∈ k × such that r i ( p i ) = λ i p i +1 + x i with p i +1 / ∈ x i . Notice that since n is minimal, r i ( p i ) = p i and then r i acts triviallyon every path different from p i , for all i .Let us see that p i / ∈ x j for all i = j. Since the sequence p , . . . , p n +1 = p is cyclic, it is enough to prove that p / ∈ x j for all j .Suppose that p ∈ x j for some j ∈ { , . . . , n } . Since p i +1 / ∈ x i for all i and p n +1 = p , it followsthat j = n , and by part (i), j = 1. Let u k = p k and t k = r k for 1 ≤ k ≤ j and u j +1 = p . Noticethat u k +1 ∈ t k ( u k ) for 1 ≤ k ≤ j and u j +1 = u . Since j < n this contradicts the choice of n . Itfollows that p i / ∈ x j for all i, j. One can easily check that this implies r n ◦ · · · ◦ r ( p ) = λp + x with p i / ∈ x for all i . Now,define inductively for i > n , r i := r i − n . The sequence ( r i ) i ∈ N acting on p never stabilizes, whichcontradicts the reduction-finiteness of the reduction system R . (iii) Suppose not, so that there is a sequence ( p i ) i ∈ N of paths and a sequence of basic reductions( t i ) i ∈ N such that p i +1 ∈ t i ( p i ). Since is an antisymmetric relation, p i = p j if i = j .Let i = 1. Suppose that that we have constructed i , . . . , i k such that i < · · · < i k , p i k ∈ t i k − ◦· · ·◦ t ( p ) and p j / ∈ t i k − ◦· · ·◦ t ( p ) for all j > i k . Set X k = { i > i k : p i ∈ t i k ◦· · ·◦ t i ( p ) } .By the inductive hypothesis, there is x ∈ kQ and λ ∈ k × such that t i k − ◦ · · · ◦ t i ( p ) = λp i k + x with p i k / ∈ x . Since we also know that p i k +1 ∈ t i k ( p i k ), and p i k +1 / ∈ t i k − ◦ · · · ◦ t i ( p ) itfollows that p i k +1 ∈ t i k ( p i k ) + x . Also, t i k ◦ · · · ◦ t i ( p ) = λt i k ( p i k ) + t i k ( x ) = λt i k ( p i k ) + x , so p i k +1 ∈ t i k ◦ · · · ◦ t i ( p ). Therefore X k is not empty. We may define i k +1 = max X k , because X k is a finite set.This procedure constructs inductively a strictly increasing sequence of indices ( i k ) k ∈ N with p i k ∈ ˜ p i k := t i k − ◦ · · · ◦ t i ( p ) for all k ∈ N . The set { t i k − ◦ · · · ◦ t i ( p ) : k ∈ N } is thereforeinfinite. This contradicts the reduction-finiteness of R . (cid:3) The converse to Lemma 2.5 also holds, that is, if R is a reduction system for which is apartial order satisfying the descending chain condition, then every path is reduction-finite. In ROJECTIVE RESOLUTIONS OF ASSOCIATIVE ALGEBRAS AND AMBIGUITIES 5 other words, the order captures most of the properties we require R to verify, and it will beimportant in the next sections.The following characterization of the relation is very useful in practice. Lemma 2.6. If p, q are paths, then q p if and only if p = q or there exists a reduction t suchthat p ∈ t ( q ) .Proof. First we prove the necessity of the condition. Let n ∈ N , r , . . . , r n and p , . . . , p n be asin the definition of , and suppose that n is minimal. Let ˜ p = p and for each i = 1 , . . . , n − p i +1 = r i (˜ p i ). Notice that the minimality implies that r i ( p i ) = p i . Let us first show that(1) if i > j then p i / ∈ ˜ p j . Suppose otherwise and let ( i, j ) be a counterexample with j minimal. We will prove that in thissituation, p l ∈ ˜ p l for all l < j . We proceed by induction on l . By definition, p ∈ ˜ p . Suppose1 ≤ l < j − p l ∈ ˜ p l . Then we have p l +1 ∈ r l ( p l ) and, since l < j , p l +1 / ∈ ˜ p l . Write˜ p l = λp l + x with x ∈ kQ and p l / ∈ x . Since r l acts nontrivially on p l , it acts trivially on x ; itfollows that r l (˜ p l ) = λr l ( p l ) + x and so p l +1 ∈ r l (˜ p l ) = ˜ p l +1 . In particular p j − ∈ ˜ p j − . Since p i / ∈ ˜ p j − and p i ∈ ˜ p j , we must have p i ∈ r j − ( p j − ).Now, let m = n + j − i , t k = r k and u k = p k if k ≤ j −
1, and t k = r i + k − j and u k = p i + k − j if j ≤ k ≤ m . One can check that u = q , u n + j − i = p and that u k +1 ∈ t k ( u k ) for all k = 1 , . . . , m −
1. Since m < n this contradicts the choice of n . We thus conclude that (1)holds.We can use the same inductive argument as before to prove that p i ∈ ˜ p i for all 1 ≤ i ≤ n .Denoting t = ( r n , . . . , r ), observe that p ∈ t ( q ).Let us now prove the converse. Let t = ( t m , . . . , t ) be a reduction such that p ∈ t ( q ) and m isminimal, and let us proceed by induction on m . Notice that if m = 1 there is nothing to prove.If t i is the basic reduction r a i ,s i ,c i , let p i = a i s i c i . Using the same ideas as above one can showthat if u = q and u / ∈ t i ( p i ) for each 1 ≤ i ≤ m , then u / ∈ t l ◦ · · · ◦ t ( q ) for each 0 ≤ l ≤ m .Since p ∈ t ( q ) either p = q or there exists i ∈ { , . . . , m } such that p ∈ t i ( p i ). In the first case q p . In the second case, we know that p i p and we need to prove that q p i . Since m isminimal, t i ( t i − ◦ · · · ◦ t ( q )) = t i − ◦ · · · ◦ t ( q ) and then p i ∈ t i − ◦ · · · ◦ t ( q ). The result nowfollows by induction because i − < m . (cid:3) Proposition 2.7. If I ⊆ kQ is an ideal, then there exists a reduction system R which satisfiescondition ( ♦ ) for I . We will prove this result by putting together a series of lemmas.Let ≤ be a well-order on the set Q ∪ Q such that e < α for all e ∈ Q and α ∈ Q .Let ω : Q → N be a function and extend it to Q ≥ defining ω ( e ) = 0 for all e ∈ Q and ω ( c n · · · c ) = P ni =1 ω ( c i ) if c i ∈ Q and c n · · · c is a path. Given c, d ∈ Q ≥ we write that c ≤ ω d if • ω ( c ) < ω ( d ), or • c, d ∈ Q and c ≤ d , or • ω ( c ) = ω ( d ), c = c n · · · c , d = d m · · · d ∈ Q ≥ and there exists j ≤ min ( | c | , | d | ) such that c i = d i for all ∈ { , . . . , j − } and c j < d j .Notice that the order ≤ ω is in fact the deglex order with weight ω , and it has the followingtwo properties: PROJECTIVE RESOLUTIONS OF ASSOCIATIVE ALGEBRAS AND AMBIGUITIES (i) If p, q ∈ Q ≥ and p ≤ ω q , then cpd ≤ ω cqd for all c, d ∈ Q ≥ such that cpd = 0 and cqd = 0in kQ .(ii) For all q ∈ Q ≥ the set { p ∈ Q ≥ : p ≤ ω q } is finite.It is straightforward to prove the first claim. For the second one, let { c i } i ∈ N be a sequence in Q ≥ such that c i +1 ≤ ω c i for all i . If c i ∈ Q for some i , then it is evident that the sequencestabilizes, so let us suppose that { c i } i ∈ N is contained in Q ≥ and c i +1 < ω c i for all i ∈ N . Since( ω ( c i )) i ∈ N is a decreasing sequence of natural numbers, it must stabilize, so we may also supposethat ω ( c i ) = ω ( c j ) for all i, j and that the lengths of the paths are bounded above by some M ∈ N . By definition of ≤ ω , we know that the sequence of first arrows of elements of { c i } i ∈ N forms a decreasing sequence in ( Q , ≤ ), which must stabilize because ( Q , ≤ ) is well-ordered.Let N ∈ N be such that the first arrow of c i equals the first arrow of c j for all i, j ≥ N . If c i = c in i · · · c i , and we denote c ′ i = c in i · · · c i , then { c ′ i } i ≥ N is a decreasing sequence in ( Q ≥ , ≤ ω )with | c ′ i | = M − i . Iterating this process we arrive to a contradiction. Definition 2.8.
Consider as before a well-order ≤ on Q ∪ Q and ω : Q → N , and ≤ ω beconstructed from them. If p ∈ kQ and p = P ni =1 λ i c i with λ i ∈ k × , c i ∈ Q ≥ and c i < ω c forall i = 1, we write tip ( p ) for c . If X ⊆ kQ , we let tip ( X ) := { tip ( x ) : x ∈ X \ { }} .Consider the set S := Mintip ( I ) = { p ∈ tip ( I ) : p ′ / ∈ tip ( I ) for all proper divisors p ′ of p } . Notice that if s and s ′ both belong to S and s = s ′ , then s does not divide s ′ . For each s ∈ S ,choose f s ∈ kQ such that s − f s ∈ I , f s < ω s and f s is parallel to s .Describing the set tip ( I ) is not easy in general. We comment on this problem at the beginningof the last section, where we compute examples. Lemma 2.9.
Let ≤ ω and S be as before. The ideal I equals the two sided ideal generated by theset { s − f s } s ∈ S , which we will denote by h s − f s i s ∈ S .Proof. It is clear that h s − f s i s ∈ S is contained in I . Choose x = P ni =1 λ i c i ∈ I with λ i ∈ k × and c i ∈ Q ≥ . We may suppose that c = tip ( x ), so that c ∈ tip ( I ). There is a divisor s of c suchthat s ∈ tip ( I ) and s ′ / ∈ tip ( I ) for all proper divisor s ′ of s and s ∈ S by definition of S . Let a, c ∈ Q ≥ with asc = c .Define x ′ := af s c + P ni =2 λ i c i . We have x = λ c + P ni =2 λ i c i = λ a ( s − f s ) c + x ′ , so that x ′ ∈ I and, by property (i) of the order ≤ ω , we see that c > tip ( x ′ ). We can apply this procedure againto x ′ and iterate: the process will stop by property (ii) and we conclude that x ∈ h s − f s i s ∈ S . (cid:3) Lemma 2.10.
Let ≤ ω and S be as before. The set R := { ( s, f s ) } s ∈ S is a reduction system suchthat every path is reduction-unique.Proof. Since s > ω tip ( f s ) for all s ∈ S , properties (i) and (ii) guarantee that every path isreduction-finite. We need to prove that every path is reduction-unique. Recall that π is thecanonical projection kQ → kQ/I . Let p be a path. Since I = h s − f s i s ∈ S , we see that π ( r ( p )) = π ( p ) for any reduction r . Let r and t be reductions such that r ( p ) and t ( p ) are both irreducible.Clearly, π ( r ( p ) − t ( p )) = π ( p ) − π ( p ) = 0, so that r ( p ) − t ( p ) ∈ I . If this difference is not zero,then the path d = tip ( r ( p ) − t ( p )) can be written as d = asc with a, c paths and s ∈ S . It followsthat the reduction r a,s,c acts nontrivially either on r ( p ) or on t ( p ), and this is a contradiction. (cid:3) This lemma implies that for each s ∈ S , there exists a reduction r and an irreducible element f ′ s such that r ( f s ) = f ′ s . Consider the reduction system R ′ := { ( s, f ′ s ) : s ∈ S } . The setof irreducible paths for R clearly coincides with the set of irreducible paths for R ′ and, since π ( s − f ′ s ) = π ( s − f s ) = 0, we have that h s − f ′ s i s ∈ S ⊆ I . From Bergman’s Diamond Lemma it ROJECTIVE RESOLUTIONS OF ASSOCIATIVE ALGEBRAS AND AMBIGUITIES 7 follows that I = h s − f ′ s i s ∈ S . We can conclude that the reduction system R ′ satisfies condition( ♦ ), thereby proving Proposition 2.7.It is important to emphasize that different choices of orders on Q ∪ Q and of weights ω will give very different reduction systems, some of which will better suit our purposes thanothers. Moreover, there are reduction systems which cannot be obtained by this procedure, asthe following example shows. Example . Consider the algebra A = k h x, y, z i / ( x + y + z − xyz )and let R = { ( xyz, x + y + z ) } . Clearly this reduction system does not come from a monomialorder and neither from a monomial order with weights. It is not entirely evident but this reductionsystem satisfies ( ♦ ).Finally, we define a relation (cid:22) on the set k × Q ≥ := { λp : λ ∈ k × , p ∈ Q ≥ } ∪ { } as the leastreflexive and transitive relation such that λp (cid:22) µq whenever there exists a reduction r such that r ( µq ) = λp + x with p / ∈ x . We state 0 (cid:22) λp for all λp ∈ k × Q ≥ . Lemma 2.11.
The binary relation (cid:22) is an order satisfying the descending chain condition andit is compatible with concatenation.Proof.
The second claim is clear. In order to prove the first claim, let us first prove that if p ∈ Q ≥ is such that there exists a reduction r with r ( p ) = λp + x and p / ∈ x , then λ = 1and x = 0. Suppose not. For r a basic reduction, this has already been done in Lemma 2.5.If r is not basic, then r = ( r n , . . . , r ) with r i basic and n ≥
2. Let r ′ = ( r n , . . . , r ). Since p ∈ r ( p ) = r ′ ( r ( p )), there exists p ∈ r ( p ) such that p ∈ r ′ ( p ). By the previous case, we obtainthat p / ∈ r ( p ), so p = p . As a consequence of Lemma 2.6, we know that p p since p ∈ r ( p )and that p p since p ∈ r ′ ( p ). This contradicts the antisymmetry of .It is an immediate consequence of the previous fact that given a path p and a reduction t ,(2) if t ( λ p ) = λ p + x with p / ∈ x , then λ = λ .Let λ , . . . , λ n +1 ∈ k × , p , . . . , p n +1 ∈ Q ≥ , x , . . . , x n ∈ kQ and reductions t , . . . , t n be suchthat t i ( λ i p i ) = λ i +1 p i +1 + x i , p i +1 / ∈ x i and λ n +1 p n +1 = λ p . This implies that p i p i +1 foreach 1 ≤ i ≤ n and p n +1 = p . Since is antisymmetric, it follows that p i = p for all i and (2)implies that λ i = λ for all i . We thus see that (cid:22) is antisymmetric.Let now ( λ i p i ) i ∈ N be a sequence in k × Q ≥ and ( t i ) i ∈ N a sequence of reductions such that t i ( λ i p i ) = λ i +1 p i +1 + x i with p i +1 / ∈ x i . Then p i p i +1 for all i and since satisfies thedescending chain condition there exists i such that p i = p i for all i ≥ i . Observation (2)implies then that λ i = λ i for all i ≥ i , so that the sequence ( λ i p i ) i ∈ N stabilizes. (cid:3) If x = P ni =1 λ i p i ∈ kQ with λ i ∈ k × and λp belongs to k × Q ≥ , we write x (cid:22) λp if λ i p i (cid:22) λp for all i . If in addition x = λp we also write x ≺ p . The following simple fact is key to provingeverything that follows. Corollary 2.12.
Given a path p , its normal form β ( p ) is such that β ( p ) (cid:22) p . Moreover, β ( p ) ≺ p if and only if p / ∈ B .Proof. There is a reduction r such that β ( p ) = r ( p ) = P ni =1 λ i p i . It is clear that λ i p i (cid:22) p for all i , so that β ( p ) (cid:22) p . The last claim follows from the fact that β ( p ) = p if and only if p ∈ B . (cid:3) PROJECTIVE RESOLUTIONS OF ASSOCIATIVE ALGEBRAS AND AMBIGUITIES Ambiguities
Given an algebra A = kQ/I and a reduction system R satisfying ( ♦ ) for I , there is a monomialalgebra associated to A defined as A S := kQ/ h S i and equipped with the canonical projection π ′ : kQ → A S . The set π ′ ( B ) is a k -basis of A S . The algebra A S is a generalization of thealgebra A mon defined in [GM]: in that article, the order is necessarily monomial.From now on we fix the reduction system R satisfying condition ( ♦ ). Notice that in thissituation we can suppose without loss of generality, that S ⊆ Q ≥ .The family of modules {P i } i ≥ appearing in the resolution of A as A -bimodule will be in bijec-tion with those appearing in Bardzell’s resolution of the monomial algebra A S . More precisely,we will define E -bimodules k A i for i ≥ −
1, such that the former will be { A ⊗ E k A i ⊗ E A } i ≥− while the latter will be { A S ⊗ E k A i ⊗ E A S } i ≥− . The resolution will start as usual: A − = Q , A = Q and A = S .For n ≥ A n will be the set of n -ambiguities of R . We will next recall the definition of n -ambiguity – or n -chain according to the terminology used in [Sk], [An], [AG] and to Bardzell’s[Ba] associated sequences of paths , and we will take into account that the sets of left n -ambiguitiesand right n -ambiguities coincide. This fact is proved in [Ba] and also in [Sk]. See [GZ] too. Definition 3.1.
Given n ≥ p ∈ Q ≥ ,(1) the path p is a left n - ambiguity if there exist u ∈ Q , u , . . . , u n irreducible paths suchthat(i) p = u u · · · u n ,(ii) for all i , u i u i +1 is reducible but u i d is irreducible for any proper left divisor d of u i +1 .(2) the path p is a right n - ambiguity if there exist v ∈ Q and v , . . . , v n irreducible pathssuch that(i) p = v n · · · v ,(ii) for all i , v i +1 v i is reducible but dv i is irreducible for any proper right divisor of v i +1 . Proposition 3.2.
Let n, m ∈ N , p ∈ Q ≥ . If u , ˆ u ∈ Q and u , . . . u n , ˆ u , . . . , ˆ u n are pathsin Q such that both u , . . . , u n and ˆ u , . . . , ˆ u n satisfy conditions (1i) and (1ii) of the previousdefinition for p , then n = m and u i = ˆ u i for all i , ≤ i ≤ n .Proof. Suppose n ≤ m . It is obvious that u = ˆ u , since both of them are arrows. Notice that kQ = T kQ kQ , that is the free algebra generated by kQ over kQ , which implies that either u u divides ˆ u ˆ u or ˆ u ˆ u divides u u , and moreover u u , ˆ u ˆ u ∈ A = S . Remark 2.3.1 saysthat u u = ˆ u ˆ u . Since u = ˆ u , we must have u = ˆ u . By induction on i , let us suppose that u j = ˆ u j for j ≤ i . As a consequence, u i +1 · · · u n = ˆ u i +1 · · · ˆ u m .If i + 1 = n , this reads u n = ˆ u n · · · ˆ u m , and the fact that u n is irreducible and ˆ u j ˆ u j +1 isreducible for all j < m implies that m = n and u n = ˆ u n . Instead, suppose that i + 1 < n .From the equality u i +1 · · · u n = ˆ u i +1 · · · ˆ u m we deduce that there exists a path d such that u i +1 = ˆ u i +1 d or ˆ u i +1 = u i +1 d . If u i +1 = ˆ u i +1 d and d ∈ Q ≥ , we can write d = d d with d ∈ Q . The path ˆ u i +1 d is a proper left divisor of u i +1 and by condition (1ii) we obtain that u i ˆ u i +1 d is irreducible. This is absurd since u i ˆ u i +1 d = ˆ u i ˆ u i +1 d by inductive hypothesis, andthe right hand term is reducible by condition (1ii). It follows that d ∈ Q and then u i +1 = ˆ u i +1 .The case where ˆ u i +1 = u i +1 d is analogous. (cid:3) Corollary 3.3.
Given n, m ≥ − , A n ∩ A m = ∅ if n and m are different. Just to get a flavor of what A n is, one may think about an element of A n as a minimal propersuperposition of n elements of S . ROJECTIVE RESOLUTIONS OF ASSOCIATIVE ALGEBRAS AND AMBIGUITIES 9
We end this section with a proposition that indicates how to compute ambiguities for aparticular family of algebras.
Proposition 3.4.
Suppose S ⊂ Q . For all n ≥ , A n = { α . . . α n ∈ Q n +1 : α i ∈ Q for all i and α i − α i ∈ S } Moreover, given p = α . . . α n ∈ A n , we can write p as a left ambiguity choosing u i = α i , for all i , and as a right ambiguity choosing v i = α n − i Proof.
We proceed by induction on n . If n = 1 we know that A = S in which case there isnothing to prove. Let u · · · u n u n +1 ∈ A n +1 and suppose that the result holds for all p ∈ A n .Since u · · · u n belongs to A n we only have to prove that u n +1 ∈ Q and that u n u n +1 ∈ S . Weknow that u n ∈ Q , that u n +1 is irreducible and that u n u n +1 is reducible. As a consequence,there exist s ∈ S and v ∈ Q ≥ such that u n u n +1 = sv . Moreover, u n d is irreducible for anyproper left divisor d of u n +1 , so the only possibility is v ∈ Q . We conclude that u n u n +1 belongs to S . Since S ⊆ Q and u n ∈ Q , we deduce that u n +1 ∈ Q . This proves that A n +1 ⊆ { α · · · α n ∈ Q n +1 : α i ∈ Q for all i and α i − α i ∈ S } .The other inclusion is clear. (cid:3) The resolution
In this section our purpose is to construct bimodule resolutions of the algebra A . We achievethis in Theorems 4.1 and 4.2: in the first one we construct homotopy maps to prove that a givencomplex is exact, while in the second one we define differentials inductively.We will make use of differentials of Bardzell’s resolution for monomial algebras, so we beginthis section by recalling them. Keeping the notations of the previous section, note that the kQ -bimodule kQ ⊗ E k A n ⊗ E kQ is a k -vector space with basis { a ⊗ p ⊗ c : a, c ∈ Q ≥ , p ∈ A n , apc =0 in kQ } .As we have already done for A , we define a k -linear map i ′ : A S → kQ such that be i ′ ( π ′ ( b ))) = b for all b ∈ B , and we denote by β ′ : kQ → kQ the composition i ′ ◦ π ′ .Given n ≥ −
1, let us fix notation for the following k -linear maps: π n := π ⊗ id k A n ⊗ π, π ′ n := π ′ ⊗ id k A n ⊗ π ′ ,i n := i ⊗ id k A n ⊗ i, i ′ n := i ′ ⊗ id k A n ⊗ i ′ ,β n := i n ◦ π n , β ′ n := i ′ n ◦ π ′ n . Consider the following sequence of kQ -bimodules, · · · f / / kQ ⊗ E k A ⊗ E kQ f / / kQ ⊗ E k A ⊗ E A f / / kQ ⊗ E kQ f − / / ∼ = (cid:15) (cid:15) kQ / / kQ ⊗ E k A − ⊗ E kQ where(i) f − ( a ⊗ b ) = ab ,(ii) if n is even, q ∈ A n and q = u · · · u n = v n · · · v are respectively the factorizations of q asleft and right n -ambiguity, f n (1 ⊗ q ⊗
1) = v n ⊗ v n − · · · v ⊗ − ⊗ u · · · u n − ⊗ u n , (iii) if n es odd and q ∈ A n , f n (1 ⊗ q ⊗
1) = X apc = qp ∈A n − a ⊗ p ⊗ c. The maps f n induce, respectively, A -bimodule maps δ n : A ⊗ E k A n ⊗ E A → A ⊗ E k A n − ⊗ E A where δ n := π n − ◦ f n ◦ i n , and A S -bimodule maps δ ′ n : A S ⊗ E k A n ⊗ E A S → A S ⊗ E k A n − ⊗ E A S defined by δ ′ n := π ′ n − ◦ f n ◦ i ′ n . Observe that δ − and δ ′− are respectively multiplication in A and in A S .The algebra A S is monomial. The following complex provides a projective resolution of A S as A S -bimodule [Ba]: · · · δ ′ / / A S ⊗ E k A ⊗ E A S δ ′ / / A S ⊗ E k A ⊗ E A S δ ′ / / A S ⊗ E A S δ ′− / / A S / / . We will make use of the homotopy that Sk¨oldberg defined in [Sk] when proving that this complexis exact. We recall it, but we must stress that our signs differ from the ones in [Sk] due to thefact that he considers right modules, while we always work with left modules.Given n ≥ −
1, the morphism of kQ − E -bimodules S n is defined as follows.For n = − S − : kQ → kQ ⊗ E k A − ⊗ E kQ is the kQ − E -bimodule map given by S − ( a ) = a ⊗
1, for a ∈ kQ .For n ∈ N , S n : kQ ⊗ E k A n − ⊗ E kQ → kQ ⊗ E k A n ⊗ E kQ is given by S n (1 ⊗ q ⊗ b ) = ( − n +1 X apc = qbp ∈A n a ⊗ p ⊗ c. Let s ′ n = π ′ n ◦ S n ◦ i ′ n − . The family of maps { s ′ n } n ≥− verifies the equalities s ′ n ◦ δ ′ n + δ ′ n − ◦ s ′ n − = id A S ⊗ E k A n ⊗ E A S for n ≥ s ′− ◦ δ ′− = id A S ⊗ E k A − ⊗ E A S . Next we define some sets that will be useful in the sequel. For any n ≥ − µq ∈ k × Q ≥ ,consider the following subsets of kQ ⊗ E k A n ⊗ E kQ : • L (cid:22) n ( µq ) := { λa ⊗ p ⊗ c : a, c ∈ Q ≥ , p ∈ A n , λapc (cid:22) µq } , • L ≺ n ( µq ) := { λa ⊗ p ⊗ c : a, c ∈ Q ≥ , p ∈ A n , λapc ≺ µq } , and the following subsets of A ⊗ E k A n ⊗ E A : • L (cid:22) n ( µq ) := { λπ ( b ) ⊗ p ⊗ π ( b ′ ) : b, b ′ ∈ B , p ∈ A n , λbpb ′ (cid:22) µq } , • L ≺ n ( µq ) := { λπ ( b ) ⊗ p ⊗ π ( b ′ ) : b, b ′ ∈ B , p ∈ A n , λbpb ′ ≺ µq } . Remark . We observe that f n +1 ( x ) ∈ hL (cid:22) n ( µq ) i Z , for all x ∈ L (cid:22) n +1 ( µq ), and S n ( x ) ∈ hL (cid:22) n ( µq ) i Z , for all x ∈ L (cid:22) n − ( µq ) . Moreover, the only possible coefficients appearing in the linear combinations are +1 and − ROJECTIVE RESOLUTIONS OF ASSOCIATIVE ALGEBRAS AND AMBIGUITIES 11
Theorem 4.1.
Set d − := δ − and d := δ . Given N ∈ N and morphisms of A -bimodules d i : A ⊗ E k A i ⊗ E A → A ⊗ E k A i − ⊗ E A for ≤ i ≤ N . If(1) d i − ◦ d i = 0 for all i , ≤ i ≤ N ,(2) ( d i − δ i )(1 ⊗ q ⊗ ∈ hL ≺ i − ( q ) i k for all i ∈ { , . . . , N } and for all q ∈ A i ,then the complex A ⊗ E k A N ⊗ E A d N / / · · · d / / A ⊗ E k A ⊗ E A d / / A ⊗ E A d − / / A / / is exact. Theorem 4.2.
There exist A -bimodule morphisms d i : A ⊗ E k A i ⊗ E A → A ⊗ E k A i − ⊗ E A for i ∈ N and d − : A ⊗ E A → A such that(1) d i − ◦ d i = 0 , for all i ∈ N ,(2) ( d i − δ i )(1 ⊗ q ⊗ ∈ hL ≺ i − ( q ) i Z for all i ≥ − and q ∈ A i . We will carry out the proofs of these theorems in the following section.5.
Proofs of the theorems
We keep the same notations and conditions of the previous section. We start by proving sometechnical lemmas.
Lemma 5.1.
Given n ≥ , the following equalities hold(1) δ n ◦ π n = π n − ◦ f n ,(2) δ ′ n ◦ π ′ n = π ′ n − ◦ f n . The proof is straightforward after the definitions.Next we prove three lemmas where we study how various maps defined in Section 4 behavewith respect to the order.
Lemma 5.2.
For all n ∈ N and µq ∈ k × Q ≥ , the images by π n of L (cid:22) n ( µq ) and of L ≺ n ( µq ) arerespectively contained in hL (cid:22) n ( µq ) i Z and in hL ≺ n ( µq ) i Z .Proof. Given n ∈ N , µq ∈ k × Q ≥ and x = λa ⊗ p ⊗ c ∈ L (cid:22) n ( µq ), where a, c ∈ Q ≥ and p ∈ A n ,suppose β ( a ) = P i λ i b i and β ( c ) = P j λ ′ j b ′ j . Since β ( a ) (cid:22) a and β ( c ) (cid:22) c , then λ i b i (cid:22) a and λ ′ j b ′ j (cid:22) c for all i, j . This implies λλ i λ j b i pb ′ j (cid:22) λapc (cid:22) µq and so λλ i λ ′ j π ( b i ) ⊗ p ⊗ π ( b ′ j ) belong to L (cid:22) n ( µq ) for all i, j . The result follows from the equalities π n ( x ) = λπ ( a ) ⊗ p ⊗ π ( c ) = λπ ( β ( a )) ⊗ p ⊗ π ( β ( c )) = X i,j λλ i λ ′ j π ( b i ) ⊗ p ⊗ π ( b ′ j ) . The proof of the second part is analogous. (cid:3)
Corollary 5.3.
Let n ≥ − and µq ∈ k × Q ≥ . Keeping the same notations of the proof of theprevious lemma, we conclude thati) if x ∈ L (cid:22) n ( µq ) , then λπ ( a ) xπ ( c ) ∈ hL (cid:22) n ( λµaqc ) i Z ,ii) if x ∈ L ≺ n ( µq ) , then λπ ( a ) xπ ( c ) ∈ hL ≺ n ( λµaqc ) i Z . Lemma 5.4.
Given n ∈ N and µq ∈ k × Q ≥ , there are inclusions i) δ n ( L (cid:22) n ( µq )) ⊆ hL (cid:22) n − ( µq ) i Z ,ii) δ n ( L ≺ n ( µq )) ⊆ hL ≺ n − ( µq ) i Z , iii) s n ( L (cid:22) n − ( µq )) ⊆ hL (cid:22) n ( µq ) i Z ,iv) s n ( L ≺ n − ( µq )) ⊆ hL ≺ n ( µq ) i Z .Proof. From x = λπ ( b ) ⊗ p ⊗ π ( b ′ ) ∈ L (cid:22) n ( µq ), with b, b ′ ∈ B and p ∈ A n , we get i n ( x ) = λb ⊗ p ⊗ b ′ .This element belongs to L (cid:22) n ( µq ) and this implies that f n ( λb ⊗ p ⊗ b ′ ) belongs to hL (cid:22) n − ( µq ) i Z ,by Remark 4.0.1. As a consequence of Lemma 5.2 we obtain that δ n ( x ) = π n − ( f n ( λb ⊗ p ⊗ b ′ ))belongs to hL (cid:22) n − ( µq ) i Z . The proofs of the other statements are similar. (cid:3) Lemma 5.5.
Given n ≥ − and µq ∈ k × Q ≥ , if x = λa ⊗ p ⊗ c ∈ L (cid:22) n ( µq ) is such that π ′ n ( x ) = 0 ,then π n ( x ) ∈ hL ≺ n ( µq ) i Z . Proof.
By hypothesis we get that 0 = π ′ n ( x ) = π ′ ( a ) ⊗ p ⊗ π ′ ( c ). The only possibilities are π ′ ( a ) = 0 or π ′ ( c ) = 0, this is, a / ∈ B or c / ∈ B , namely β ( a ) ≺ a or β ( c ) ≺ c .Writing β ( a ) = P i λ i b i and β ( c ) = P j λ ′ j b ′ j , we deduce that λλ i λ ′ j b i pb j ≺ µq for all i, j . Asa consequence, P i,j λλ i λ ′ j π ( b i ) ⊗ p ⊗ π ( b ′ j ) ∈ hL ≺ n ( µq ) i Z .The proof ends by computing π n ( x ) = π n ( β ( x )) = π n ( X i,j λλ i λ ′ j b i ⊗ p ⊗ b ′ j ) = X i,j λλ i λ ′ j π ( b i ) ⊗ p ⊗ π ( b ′ j ) . (cid:3) The importance of the preceding lemmas is that they guarantee how differentials and mor-phisms used for the homotopy behave with respect to the order. This is stated explicitly in thefollowing corollary.
Corollary 5.6.
Given n ≥ , µq ∈ k × Q ≥ and x ∈ L (cid:22) n ( µq ) , the following facts hold:(1) δ n − ◦ δ n ( x ) ∈ hL ≺ n − ( µq ) i Z ,(2) x − δ n +1 ◦ s n +1 ( x ) − s n ◦ δ n ( x ) ∈ hL ≺ n ( µq ) i Z .Proof. Let us first write x = λπ ( b ) ⊗ p ⊗ π ( b ′ ) with b, b ′ ∈ B and x ′ := i n ( x ) = λb ⊗ p ⊗ b ′ . Lemma5.1 implies that δ n − ◦ δ n ( x ) = δ n − ◦ δ n ◦ π n ( x ′ ) = δ n − ◦ π n − ◦ f n ( x ′ ) = π n − ◦ f n − ◦ f n ( x ′ ) . By Remark 4.0.1, f n − ◦ f n ( x ′ ) ∈ L (cid:22) n − ( µq ). Next, by Lemma 5.5, in order to prove that δ n − ◦ δ n ( x ) ∈ hL ≺ n − ( µq ) i Z , it suffices to verify that π ′ n − ◦ f n − ◦ f n ( x ′ ) = 0, which is in facttrue using Lemma 5.1, and the fact that ( A S ⊗ E k A • ⊗ E A S , δ ′• ) is exact.In order to prove (2), we first remark that if k ∈ N and y ∈ hL (cid:22) k ( µq ) i Z , then i ′ k ◦ π ′ k ( y ) − i k ◦ π k ( y ) ∈ hL ≺ k ( µq ) i Z . Indeed, let us write y = λa ⊗ p ⊗ c ∈ L (cid:22) k ( µq ). In case a ∈ B and c ∈ B , there are equalities i ′ k ◦ π ′ k ( y ) = y = i k ◦ π k ( y ), and so the difference is zero. If either a / ∈ B or c / ∈ B , then π ′ k ( y ) = 0 and in this case Lemma 5.5 implies that π k ( y ) ∈ hL ≺ k ( µq ) i Z . So, i k ◦ π k ( y ) ∈ hL ≺ k ( µq ) i Z and the difference we are considering belongs to hL ≺ k ( µq ) i Z .Fix now x = λπ ( b ) ⊗ p ⊗ π ( b ′ ) and x ′ = i n ( x ) = λb ⊗ p ⊗ b ′ , with b, b ′ ∈ B .Since x ′ = i ′ n ◦ π ′ n ( x ′ ), x − δ n +1 ◦ s n +1 ( x ) − s n ◦ δ n ( x ) = π n ( x ′ ) − π n ( f n +1 ◦ i n +1 ◦ π n +1 ◦ S n +1 ( x ′ )) − π n ( S n ◦ i n − ◦ π n − ◦ f n ( x ′ )) . ROJECTIVE RESOLUTIONS OF ASSOCIATIVE ALGEBRAS AND AMBIGUITIES 13
The previous comments and Remark 4.0.1 allow us to write that π n ◦ f n +1 ◦ ( i ′ n +1 ◦ π ′ n +1 − i n +1 ◦ π n +1 ) ◦ S n +1 ( x ′ ) ∈ hL ≺ n ( µq ) i Z ,π n ◦ S n ◦ ( i ′ n − ◦ π ′ n − − i n − ◦ π n − ) ◦ f n ( x ′ ) ∈ hL ≺ n ( µq ) i Z . It is then enough to prove that π n ( x ′ − f n +1 ◦ i ′ n +1 ◦ π ′ n +1 ◦ S n +1 ( x ′ ) − S n ◦ i ′ n − ◦ π ′ n − ◦ f n ( x ′ )) ∈ hL ≺ n ( µq ) i Z , but π ′ n ( x ′ − f n +1 ◦ i ′ n +1 ◦ π ′ n +1 ◦ S n +1 ( x ′ ) − S n ◦ i ′ n − ◦ π ′ n − ◦ f n ( x ′ ))= π ′ n ( x ′ ) − δ ′ n +1 ◦ s ′ n +1 ( π ′ n ( x ′ )) − s ′ n ◦ δ ′ n ( π ′ n ( x ′ ))= 0 . Finally, we deduce from Lemma 5.5 that π n ( x ′ − f n +1 ◦ i ′ n +1 ◦ π ′ n +1 ◦ S n +1 ( x ′ ) − S n ◦ i ′ n − ◦ π ′ n − ◦ f n ( x ′ )) ∈ hL ≺ n ( µq ) i Z . (cid:3) Next we prove another technical lemma that shows how to control the differentials.
Lemma 5.7.
Fix n ∈ N , let R be either k or Z .(1) If d : A ⊗ E k A n ⊗ E A → A ⊗ E k A n − ⊗ E A is a morphism of A -bimodules such that ( d − δ n )(1 ⊗ p ⊗ ∈ hL ≺ n − ( p ) i R for all p ∈ A n , then given x ∈ hL (cid:22) n ( µq ) i R , ( d − δ n )( x ) ∈hL ≺ n − ( µq ) i R for all µq ∈ k × Q ≥ .(2) If ρ : A ⊗ E k A n ⊗ E A → A ⊗ E k A n +1 ⊗ E A is a morphism of A − E -bimodules suchthat ( ρ − s n )(1 ⊗ p ⊗ π ( b )) ∈ hL ≺ n +1 ( pb ) i R , for all p ∈ A n and b ∈ B , then for all x ∈ hL (cid:22) n ( µq ) i R , ( ρ − s n )( x ) belongs to hL ≺ n +1 ( µq ) i R for all µq ∈ k × Q ≥ .Proof. Given µq ∈ k × Q ≥ and x ∈ hL (cid:22) n ( µq ) i R , let us see that ( d − δ n )( x ) ∈ hL ≺ n − ( µq ) i R . Itsuffices to prove the statement for x = λπ ( b ) ⊗ p ⊗ π ( b ′ ) ∈ L (cid:22) n ( µq ).By hypothesis, ( d − δ n )(1 ⊗ p ⊗
1) belongs to hL ≺ n − ( p ) i R , so ( d − δ n )( x ) equals λπ ( b )( d − δ n )(1 ⊗ p ⊗ π ( b ′ ) and it belongs to hL ≺ n − ( λbpb ′ ) i R ⊆ hL ≺ n − ( µq ) i R , using Corollary 5.3.The second part is analogous. (cid:3) Next proposition will provide the remaining necessary tools for the proofs of Theorem 4.1 andTheorem 4.2.
Proposition 5.8.
Fix n ∈ N . Suppose that for each i ∈ { , . . . , n } there are morphisms of A -bimodules d i : A ⊗ E k A i ⊗ E A → A ⊗ E k A i − ⊗ E A , and morphisms of A − E -bimodules ρ i : A ⊗ E k A i − ⊗ E A → A ⊗ E k A i ⊗ E A . Denote d − = µ and define ρ − : A → A ⊗ E A as ρ ( a ) = a ⊗ .If the following conditions hold,(i) d i − ◦ d i = 0 for all i ∈ { , . . . , n } ,(ii) ( d i − δ i )(1 ⊗ q ⊗ ∈ hL ≺ i − ( q ) i R for all i ∈ { , . . . , n } and for all q ∈ A i ,(iii) for all i ∈ {− , . . . , n − } and for all x ∈ A ⊗ E k A i ⊗ E A , x = d i +1 ◦ ρ i +1 ( x ) + ρ i ◦ d i ( x ) ,(iv) ( ρ i − s i )(1 ⊗ q ⊗ π ( b )) ∈ hL ≺ i ( qb ) i R for all i ∈ { , . . . , n } , for all q ∈ A i and for all b ∈ B ,then:(1) If d n +1 : A ⊗ E k A n +1 ⊗ E A → A ⊗ E k A n ⊗ E A is a map satisfying the following conditions:(i) d n ◦ d n +1 = 0 ,(ii) ( d n +1 − δ n +1 )(1 ⊗ q ⊗ ∈ hL ≺ n ( q ) i R , then there exists a morphism ρ n +1 : A ⊗ E k A n ⊗ E A → A ⊗ E k A n +1 ⊗ E A of A − E bimodules such that(a) for all x ∈ A ⊗ E k A n ⊗ E A , x = d n +1 ◦ s n +1 ( x ) + s n ◦ d n ( x ) (b) for all q ∈ A n and for all b ∈ B , ( ρ n +1 − s n +1 )(1 ⊗ q ⊗ π ( b )) ∈ hL ≺ n +1 ( qb ) i R .(2) there exists a morphism of A -bimodules d n +1 : A ⊗ E k A n +1 ⊗ E A → A ⊗ E k A n ⊗ E A such that(i) d n ◦ d n +1 = 0 ,(ii) ( d n +1 − δ n +1 )(1 ⊗ q ⊗ ∈ hL ≺ n ( q ) i R .Proof. In order to prove (2), fix q ∈ A n +1 . By Lemma 5.4, δ n +1 (1 ⊗ q ⊗
1) belongs to hL (cid:22) n ( q ) i Z and using Lemma 5.7, ( d n − δ n )( δ n +1 (1 ⊗ q ⊗ hL ≺ n − ( q ) i R . Corollary 5.6 tells usthat δ n ◦ δ n +1 (1 ⊗ q ⊗
1) is in hL ≺ n − ( q ) i Z . We deduce from the equality d n ( δ n +1 (1 ⊗ q ⊗ δ n ◦ δ n +1 (1 ⊗ q ⊗
1) + ( d n − δ n )( δ n +1 (1 ⊗ q ⊗ d n ( δ n +1 (1 ⊗ q ⊗ hL ≺ n − ( q ) i R .Let us define ˜ d n +1 : A × k A n +1 × A → A ⊗ E k A n ⊗ E A by˜ d n +1 ( a, q, c ) = aδ n +1 (1 ⊗ q ⊗ c − aρ n ( d n ( δ n +1 (1 ⊗ q ⊗ c, for a, c ∈ A , q ∈ A n +1 . The map ˜ d n +1 is E -multilinear and balanced, and it induces a uniquemap d n +1 : A ⊗ E k A n +1 ⊗ E A → A ⊗ E k A n ⊗ E A. It is easy to verify that d n +1 is in fact a morphism of A -bimodules.Putting together the equality ρ n = s n + ( ρ n − s n ) and Lemmas 5.4 and 5.7, we obtain that( d n +1 − δ n +1 )(1 ⊗ q ⊗
1) = − ρ n ◦ d n ◦ δ n +1 (1 ⊗ q ⊗
1) belongs to hL ≺ n ( q ) i R . Moreover, given x ∈ A ⊗ E k A n − ⊗ E A , x = d n ◦ ρ n ( x ) + ρ n − ◦ d n − ( x ), choosing x = d n ( δ n +1 (1 ⊗ q ⊗ d n ◦ δ n +1 (1 ⊗ q ⊗
1) = d n ◦ ρ n ◦ d n ◦ δ n +1 (1 ⊗ q ⊗ d n ◦ d n +1 = 0.For the proof of (1), fix q ∈ A n and b ∈ B . Using Lemmas 5.4 and 5.7, we deduce that theelement 1 ⊗ q ⊗ π ( b ) − ρ n ◦ d n (1 ⊗ q ⊗ π ( b ))= 1 ⊗ q ⊗ π ( b ) − ρ n ◦ δ n (1 ⊗ q ⊗ π ( b )) − ρ n ◦ ( d n − δ n )(1 ⊗ q ⊗ π ( b ))differs from 1 ⊗ q ⊗ π ( b ) − ρ n ◦ δ n (1 ⊗ q ⊗ π ( b )) by elements in hL ≺ n ( qb ) i R . We will write that(id − ρ n ◦ δ n + ρ n ◦ ( d n − δ n ))(1 ⊗ q ⊗ π ( b )) ≡ id − ρ n ◦ δ n (1 ⊗ q ⊗ π ( b )) mod hL ≺ n ( qb ) i R . Also, (id − ρ n ◦ δ n )(1 ⊗ q ⊗ π ( b )) ≡ (id − s n ◦ δ n )(1 ⊗ q ⊗ π ( b )) mod hL ≺ n ( qb ) i R ≡ δ n +1 ◦ s n +1 (1 ⊗ q ⊗ π ( b )) mod hL ≺ n ( qb ) i R ≡ d n +1 ◦ s n +1 (1 ⊗ q ⊗ π ( b )) mod hL ≺ n ( qb ) i R . We deduce from this that there exists a unique ξ ∈ hL ≺ n ( qb ) i R such that(id − ρ n ◦ d n )(1 ⊗ q ⊗ π ( b )) = d n +1 ◦ s n +1 (1 ⊗ q ⊗ π ( b )) + ξ. It is evident that ξ belongs to the kernel of d n .The order (cid:22) satisfies the descending chain condition, so we can use induction on ( k × Q ≥ , (cid:22) ).If there is no λp ∈ k × Q ≥ is such that λp ≺ qb , then ξ = 0 and we define ρ n +1 (1 ⊗ q ⊗ π ( b )) = ROJECTIVE RESOLUTIONS OF ASSOCIATIVE ALGEBRAS AND AMBIGUITIES 15 s n +1 (1 ⊗ q ⊗ π ( b )). Inductively, suppose that ρ n +1 ( ξ ) is defined. The equality d n ( ξ ) = 0 impliesthat ξ = d n +1 ◦ ρ n +1 ( ξ ) and(id − ρ n ◦ d n )(1 ⊗ q ⊗ π ( b )) = d n +1 ( s n +1 (1 ⊗ q ⊗ π ( b )) + ρ n +1 ( ξ )) . We define ρ n +1 (1 ⊗ q ⊗ π ( b )) := s n +1 (1 ⊗ q ⊗ π ( b )) + ρ n +1 ( ξ ).Lemmas 5.4 and 5.7 assure that ρ n +1 ( ξ ) belongs to hL ≺ n +1 ( qb ) i R , and as a consequence ρ n +1 (1 ⊗ q ⊗ π ( b )) − s n +1 (1 ⊗ q ⊗ π ( b )) ∈ hL ≺ n +1 ( qb ) i R . (cid:3) We are now ready to prove the theorems.
Proof of Theorem 4.1.
We will prove the existence of an A − E -bimodule map ρ : A ⊗ E k A − ⊗ E A → A ⊗ E k A ⊗ E A satisfying d ◦ ρ + ρ − ◦ d − = id, where d − = µ and ρ − ( a ) = s − ( a ) = a ⊗ a ∈ A . Once this achieved, we apply Proposition 5.8 inductively with R = k , for all n such that 0 ≤ n ≤ N −
1, obtaining this way an homotopy retraction of the complex A ⊗ E k A N ⊗ E A d N / / · · · d / / A ⊗ E A d − / / A / / b = b k · · · b ∈ B , with b i ∈ Q , 1 ≤ i ≤ k , s (1 ⊗ π ( b )) = − X i π ( b k · · · b k − i +1 ) ⊗ b k − i ⊗ π ( b k − i − · · · b ) . On one hand 1 ⊗ π ( b ) − π ( b ) ⊗ ⊗ π ( b ) − s − ( d − (1 ⊗ π ( b ))) and on the other hand theleft hand term equals δ ( s (1 ⊗ π ( b ))), yielding 1 ⊗ π ( b ) − s − (1 ⊗ π ( b )) = δ ( s (1 ⊗ π ( b )). Byhypothesis, ( d − δ )(1 ⊗ π ( b )) belongs to hL ≺− ( b ) i k , and so there exists ξ ∈ hL ≺− ( b ) i k such that1 ⊗ π ( b ) − s − ( d − (1 ⊗ π ( b ))) = d ( s (1 ⊗ π ( b ))) + ξ. It follows that d − ( ξ ) = 0. Suppose first that there exists no λp ∈ k × Q ≥ such that λp ≺ b .In this case ξ = 0 and we define ρ (1 ⊗ π ( b )) = s (1 ⊗ π ( b )). Inductively, suppose that ρ ( ξ ) is defined for any ξ such that d − ( ξ ) = 0. Since in this case ξ = d ( ρ ( ξ )), we set ρ (1 ⊗ π ( b )) := s (1 ⊗ π ( b )) + ρ ( ξ ). (cid:3) Proof of Theorem 4.2.
It follows from the proof of Theorem 4.1 that1 ⊗ π ( b ) = ( s − ◦ d − + δ ◦ s )(1 ⊗ π ( b ))and so s − ◦ d − + δ ◦ s = id A ⊗ E A . Setting d := δ , the theorem follows applying Proposition5.8 for R = Z . (cid:3) We finish this section showing that this construction is a generalization of Bardzell’s resolutionfor monomial algebras.
Proposition 5.9.
Given an algebra A , let ( A ⊗ E k A • ⊗ E A, d • ) be a resolution of A as A -bimodule such that d • satisfies the hypotheses of Theorem 4.1. If p ∈ A n is such that r ( p ) = 0 or r ( p ) = p for every reduction r , then for all a, c ∈ kQ , d n ( π ( a ) ⊗ p ⊗ π ( c )) = δ n ( π ( a ) ⊗ p ⊗ π ( c )) . Proof.
By hypothesis, there exists no λ ′ p ′ ∈ k × Q ≥ such that λ ′ p ′ ≺ p , so L ≺ n − ( p ) = { } and d n (1 ⊗ p ⊗
1) = δ n (1 ⊗ p ⊗ a, c ∈ kQ we deduce from the previous equality that d n ( π ( a ) ⊗ p ⊗ π ( c )) − δ n ( π ( a ) ⊗ p ⊗ π ( c )) = π ( a )( d n (1 ⊗ p ⊗ − δ n (1 ⊗ p ⊗ π ( c ) = 0 . (cid:3) Corollary 5.10.
Suppose the algebra A = kQ/I has a monomial presentation. Choose a reduc-tion system R whose pairs have the monomial relations generating the ideal I as first coordinateand as second coordinate. In this case, the only maps d verifying the hypotheses of Theorem . are those of Bardzell’s resolution. Morphisms in low degrees
In this section we describe the morphisms appearing in lower degrees of the resolution.Let us consider the following data: an algebra A = kQ/I and a reduction system R satisfyingcondition ♦ .We start by recalling the definition of δ and δ − . For a, c ∈ kQ , α ∈ Q , δ − : A ⊗ E A → A, δ − ( π ( a ) ⊗ π ( c )) = π ( ac ) and δ : A ⊗ E k A ⊗ E A → A ⊗ E A, δ ( π ( a ) ⊗ α ⊗ ⊗ ( c )) = π ( aα ) ⊗ π ( c ) − π ( a ) ⊗ π ( αc ) . Definition 6.1.
We state some definitions. • Let φ : kQ → A ⊗ E k A ⊗ E A be the unique k -linear map such that φ ( c ) = n X i =1 π ( c n · · · c i +1 ) ⊗ c i ⊗ π ( c i − · · · c )for c ∈ Q ≥ , c = c n · · · c with c i ∈ Q for all i , 1 ≤ i ≤ n . • Given a basic reduction r = r a,s,c , let φ ( r, − ) : kQ → A ⊗ E k A ⊗ E A be the unique k -linear map such that, given p ∈ Q ≥ φ ( r, p ) = ( π ( a ) ⊗ s ⊗ π ( c ) , if p = asc ,0 if not.In case r = ( r n , . . . , r ) is a reduction, where r i is a basic reduction for all i , 1 ≤ i ≤ n ,we denote r ′ = ( r n , . . . , r ) and we define in a recursive way the map φ ( r, − ) as theunique k -linear map from kQ to A ⊗ E k A ⊗ E A such that φ ( r, p ) = φ ( r , p ) + φ ( r ′ , r ( p )) . • Finally, we define an A -bimodule morphism d : A ⊗ E k A ⊗ E A → A ⊗ E k A ⊗ E A bythe equality d (1 ⊗ s ⊗
1) = φ ( s ) − φ ( β ( s )) , for all s ∈ A . Next we prove four lemmas necessary to the description of the complex in low degrees.
Lemma 6.2.
Let us consider p ∈ Q ≥ and x ∈ kQ such that x ≺ p . For any reduction r theelement φ ( r, x ) belongs to hL ≺ ( p ) i Z .Proof. We will first prove the result for x = µq ∈ k × Q ≥ . The general case will then follow bylinearity. Fix x = µq ∈ k × Q ≥ . We will use an inductive argument on ( k × Q ≥ , (cid:22) ).To start the induction, suppose first that there exists no µ ′ q ′ ∈ k × Q ≥ and that µ ′ q ′ ≺ µq = x .In this case, every basic reduction r a,s,c satisfies either r a,s,c ( x ) = x or r a,s,c = 0. In the first case, asc = q and so φ ( r a,s,c , x ) = 0. In the second case, asc = q , so φ ( r a,s,c , x ) = µπ ( a ) ⊗ s ⊗ π ( c ).Given an arbitrary reduction r = ( r n , . . . , r ) with r i basic for all i , there are three possiblecases.(1) r ( x ) = x and n > r ( x ) = x and n = 1,(3) r ( x ) = 0. ROJECTIVE RESOLUTIONS OF ASSOCIATIVE ALGEBRAS AND AMBIGUITIES 17
Denote r ′ = ( r n , . . . , r ) as before and r = r a,s,c . In case 1), φ ( r, x ) = φ ( r ′ , x ). In case 3), φ ( r, x ) = φ ( r , x ) = 0 . Finally, in case 2), φ ( r, x ) = φ ( r , x ) = µπ ( a ) ⊗ s ⊗ π ( c ). UsingLemma 5.2, we obtain that in all three cases φ ( r, x ) ∈ hL ≺ ( p ) i Z .Next, suppose that x = µq and that the result holds for µ ′ q ′ ∈ k × Q ≥ such that µ ′ q ′ ≺ µq = x .Let us consider r, r and r ′ as before. Again, there are three possible cases:(1) asc = q ,(2) asc = q and n > asc = q and n = 1.Case 3) is immediate, since in this situation φ ( r, x ) = 0. The second case reduces to the otherones, since φ ( r, x ) = φ ( r ′ , x ) In the first case, φ ( r, x ) = µπ ( a ) ⊗ s ⊗ π ( c ) + φ ( r ′ , r ( x )) . We know that r ( x ) ≺ x , and we may write it as a finite sum r ( x ) = P i µ i q i . Using theinductive hypothesis, we deduce that φ ( r, x ) ∈ hL ≺ ( p ) i Z . (cid:3) Lemma 6.3.
For all x ∈ A ⊗ E k A ⊗ E A , x belongs to the kernel of δ ◦ d ( x ) .Proof. Since these maps are morphisms of A -bimodules, we may suppose x = 1 ⊗ s ⊗
1, with s ∈ A . A direct computation gives δ ( d (1 ⊗ s ⊗
1) = δ ( φ ( s ) − φ ( β ( s ))) = π ( s ) ⊗ − ⊗ π ( s ) − π ( β ( s )) ⊗ +1 ⊗ π ( β ( s )) = 0 . (cid:3) Lemma 6.4.
Given a, c ∈ Q ≥ and p = P ni =1 λ i p i ∈ kQ , with p i ∈ Q ≥ for all i , we obtain theequality φ ( apc ) = φ ( a ) π ( pc ) + π ( a ) φ ( p ) π ( c ) + π ( ap ) φ ( c ) . The proof is immediate using the definition of φ and k -linearity of φ and π .Next we prove the last of the preparatory lemmas. Lemma 6.5.
Given p ∈ Q ≥ and a reduction r = ( r n , . . . , r ) , with r i a basic reduction for all i such that ≤ i ≤ n , there is an equality d ( φ ( r , p )) = φ ( p ) − φ ( r ( p )) . Proof.
We will prove the result by induction on n . We will denote r i = r a i ,s i ,c i .For n = 1, there are two cases. The first one is when p = a s c . In this situation, r ( p ) = r ( p ) = p , φ ( r , p ) = 0 and so the equality is trivially true. In the second case, p = a s c , φ ( r , p ) = π ( a ) ⊗ s ⊗ π ( c ) and r ( p ) = r ( p ) = a β ( s ) c . Moreover, d ( φ ( r , p )) + φ ( r ( p )) = d ( π ( a ) ⊗ s ⊗ π ( c )) + φ ( a β ( s ) c )= π ( a ) φ ( s ) π ( c ) − π ( a ) φ ( β ( s )) π ( c ) + φ ( a β ( s ) c ) . Using Lemma 6.4, the last term equals φ ( a ) π ( β ( s ) c ) + π ( a ) φ ( β ( s )) π ( c ) + π ( a β ( s )) φ ( c ) , so the whole expression is π ( a ) φ ( s ) π ( c ) + φ ( a ) π ( β ( s ) c ) + π ( a β ( s )) φ ( c )= π ( a ) φ ( s ) π ( c ) + φ ( a ) π ( s c ) + π ( a s ) φ ( c ) , and using again Lemma 6.4, this equals φ ( p ).Suppose the result holds for n −
1. As usual, we denote r ′ = ( r n , . . . , r ). Since r ( p ) = r ′ ( r ( p )), d ( φ ( r, p )) + φ ( r ( p )) = d ( φ ( r , p )) + d ( φ ( r ′ , r ( p ))) + φ ( r ′ ( r ( p )))= d ( φ ( r , p )) + φ ( r ( p ))= φ ( p ) . (cid:3) Consider now an element p ∈ A . By definition we write p = u u u = v v v where u u and v v are paths in A dividing p . Suppose r = r a,s,c is a basic reduction such that r ( p ) = p .We deduce that either s = u u or s = v v . For an arbitrary reduction r = ( r n , . . . , r ), we willsay that r starts on the left of p if r = r a,s,c , s = u u and asc = p , and we will say that r startson the right of p if r = r a,s,c , s = v v and asc = p . Proposition 6.6.
Let { r p } p ∈A and { t p } p ∈A be two sets of reductions such that r p ( p ) and t p ( p ) belong to k B , r p starts on the left of p and t p starts on the right of p . Consider d : A ⊗ E k A ⊗ E A → A ⊗ E k A ⊗ E A the map of A -bimodules defined by d (1 ⊗ p ⊗
1) = φ ( t p , p ) − φ ( r p , p ) .The sequence A ⊗ E k A ⊗ E A d / / A ⊗ E k A ⊗ E A d / / A ⊗ E k A ⊗ E A δ / / A ⊗ E A δ − / / A / / is exact.Proof. To check that d is well defined, consider the map ˜ d : A × k A × A → A ⊗ E k A ⊗ E A defined by ˜ d ( x, p, y ) = xφ ( t p , p ) y − xφ ( r p , p ) y , for all x, y ∈ A , which is clearly multilinear;taking into account the definition of φ , it is such that ˜ d ( xe, p, y ) = ˜ d ( x, ep, y ) and ˜ d ( x, pe, y ) =˜ d ( x, p, ey ) for all e ∈ E , so it induces d on A ⊗ E k A ⊗ E A .The sequence is a complex: • δ − ◦ δ = 0 and δ ◦ d = 0 follow from Lemma 6.3. • Given p ∈ A , d ( d (1 ⊗ p ⊗ d ( φ ( t p , p ) − φ ( r p , p )). Using Lemma 6.5, this lastexpression equals φ ( p ) − φ ( t p ( p )) − φ ( p ) + φ ( r p ( p )), which is, by Remark 2.2.1, equalto − φ ( β ( p )) + φ ( β ( p )), so d ◦ d = 0.It is exact: • We already know that this is true at A and at A ⊗ E A . • Given s ∈ A , d (1 ⊗ s ⊗ − δ (1 ⊗ s ⊗
1) belongs to hL ≺ ( s ) i k : indeed, notice that δ (1 ⊗ s ⊗
1) = φ ( s ), and φ ( β ( s )) belongs to hL ≺ ( s ) i k since β ( s ) ≺ s . It follows that d (1 ⊗ s ⊗ − δ (1 ⊗ s ⊗
1) = − φ ( β ( s )) ∈ hL ≺ ( s ) i k . • Given p ∈ A , we will now prove that ( d − δ )(1 ⊗ p ⊗
1) belongs to hL ≺ ( p ) i k . We maywrite p = u u u = v v v , as we did just before this proposition and thus δ (1 ⊗ p ⊗
1) = π ( v ) ⊗ v v ⊗ − ⊗ u u ⊗ π ( u ). Besides, if r p = ( r n , . . . , r ) and t p = ( t m , . . . , t )with t i and r j basic reductions, the fact that r p starts on the left and t p starts on theright of p gives( d − δ )(1 ⊗ p ⊗
1) = φ ( t ′ p , t ( p )) − φ ( r ′ p , r ( p )) , where t ′ p = ( t m , . . . , t ) and r ′ p = ( r n , . . . , r ). Since t ( p ) ≺ p and r ( p ) ≺ p , Lemma6.2 allows us to deduce the result.Finally, Theorem 4.1 implies that the sequence considered is exact. (cid:3) Remark . Given a ∈ A = Q , we have that L ≺− ( a ) = ∅ , so for any morphism of A -bimodules d : A ⊗ E k A ⊗ E A → A ⊗ E k A − ⊗ E A such that ( d − δ )(1 ⊗ a ⊗
1) belongs to hL ≺− ( a ) i k , itmust be d = δ . ROJECTIVE RESOLUTIONS OF ASSOCIATIVE ALGEBRAS AND AMBIGUITIES 19
On the other hand, given s ∈ A , write β ( s ) = P mi =1 λ i b i . Let r = r a,s ′ ,c be a basic reductionsuch that r ( s ) = s . We must have s ′ = s and a, c ∈ Q must coincide with the source and targetof s , respectively. In other words, the only basic reduction such that r ( s ) = s is r a,s,c with a and c as we just said, and in this case r ( s ) = β ( s ) ∈ k B .In this situation { λq ∈ k × Q ≥ : λq ≺ s } = { λ b , . . . , λ m b m } , and writing b i = b n i i · · · b i with b ji ∈ Q , L ≺ ( s ) = N [ i =1 { λ i π ( b n i i · · · b i ) ⊗ b i ⊗ , . . . , λ i ⊗ b n i i ⊗ π ( b n i − i · · · b i ) } . If d : A ⊗ E k A ⊗ E A → A ⊗ E k A ⊗ E A verifies ( d − δ )(1 ⊗ s ⊗ ∈ L ≺ ( s ) and δ ◦ d ( s ) = 0for all s ∈ A , then there exists γ ji ∈ k such that d (1 ⊗ s ⊗
1) = φ ( s ) − m X i =1 n i X j =1 γ ji λ i π ( b n i i · · · b j +1 i ) ⊗ b ji ⊗ π ( b j − i · · · b i ) . From this, applying δ and reordering terms we can deduce that γ ji = 1 for all i, j . We concludethat the unique morphism with the desired properties is d .7. Examples
In this section we construct explicitly projective bimodule resolutions of some algebras usingthe methods we developed in previous sections.Given an algebra A = kQ/I , we proved in Lemmas 2.9 and 2.10 that it is always possibleto construct a reduction system R such that every path is reduction-unique. However, it is notalways easy to follow the prescriptions given by these lemmas for a concrete algebra. Moreover,the reduction system obtained from a deglex order ≤ ω may be sometimes less convenient thanother ones. In fact, describing the set tip ( I ) is not in general an easy task.Bergman’s Diamond Lemma is the tool we use to effectively compute a reduction system inmost cases. Next we sketch this procedure, which is also described in [B], Section 5.The two sided ideal I is usually presented giving a set { x i } i ∈ Γ ⊆ kQ of generating relations. Ifwe fix a well-order on Q ∪ Q , a function ω : Q → N and consider the total order ≤ ω on Q ≥ , wecan easily write x i = s i − f i , and we can eventually rescale x i so that s i is monic, with s i > ω f i for all i and define the reduction system R = { ( s i , f i ) } i ∈ Γ . Every path p will be reduction-finitewith respect to R . Bergman’s Diamond Lemma says that every path is reduction-unique if andonly if for every path p ∈ A there are reductions r, t with r starting on the left and t starting onthe right of p such that r ( p ) = t ( p ). This last situation is described by saying that p is resolvable .The set A is usually finite and so there is a finite number of conditions to check.In case there exists a non resolvable ambiguity p ∈ A , choose any two reductions r, t startingon the left and on the right respectively with r ( p ) and t ( p ) both irreducible. The element r ( p ) − t ( p ) belongs to I \ { } . We can write r ( p ) − t ( p ) = s − f with f < ω s and add the element( s, f ) to our reduction system, and so p is now resolvable. New ambiguities may now appear, soit is necessary to iterate this process, which may have infinitely many steps, but we will arriveto a reduction system R satisfying condition ( ♦ ).Next we give an example to illustrate this procedure, which will be also useful to exhibit acase where another reduction system found in an alternative way is better that the prescribedone. Example . Consider the algebra of Example 2.10.1. Let x < y < z and ω ( x ) = ω ( y ) = ω ( z ) = 1. The ideal I is presented as the two sided ideal generated by the element x + y + z − xyz . We see that z = tip ( z − ( xyz − x − y )), so we start considering the reductionsystem R = { ( z , xyz − x − y ) } . Notice that A = { z } . If we apply the reduction r z,z , to z we obtain zxyz − zx − zy which is irreducible. On the other hand, if we apply thereduction r ,z ,z to z we obtain xyz − x z − y z which is also irreducible and different fromthe first one. The difference between them is xyz − x z − y z − zxyz + zx + zy , so we add( xyz , x z + y z + zxyz − zx − zy ) to the reduction system R . Notice that now the set A is { z , xyz } . Applying reductions on the left and on the right to the element xyz we obtainagain two different irreducible elements and, proceeding as before, we see that we have to addthe element ( y z , − x z − z xyz + z x + z y + xyxyz − xyx − xy ) to our reduction system R . We obtain the new ambiguity y z which is not difficult to see that it is resolvable. Thus,the reduction system R = { ( z , xyz − x − y ) , ( xyz , x z + y z + zxyz − zx − zy ) , ( y z , − x z − z xyz + z x + z y + xyxyz − xyx − xy ) } , satisfies condition ( ♦ ).There is another reduction system for this algebra, namely R = { ( xyz, x + y + z ) } . Let usdenote A n and A n the respective set of n -ambiguities. Notice that z ( n +1) ∈ A n for n odd and z n +1 ∈ A n for n even, so A n is not empty for all n ∈ N . On the other hand, A n is empty for all n ≥
2. We conclude that using R we will obtain a resolution of length 2, with differentials givenexplicitely by Proposition 6.6, and using R the resolution obtained will have infinite length.This shows how different can the resolutions from different reduction systems be.Notice that R cannot be obtained by the procedure described above by any choice of orderon Q ∪ Q and weight ω . The algebra A = k < x, y, z > / ( xyz − x − y − z ) is in fact a3-Koszul algebra. Indeed, denoting by V the k -vector space spanned by x, y, z and by R the onedimensional k -vector space spanned by the relation xyz − x − y − z , it is straightforward that R ⊗ V ⊗ V ∩ V ⊗ V ⊗ R = { } , and so the intersection is a subset of V ⊗ R ⊗ V . Theorem 2.5 of [Be1] guarantees that A is3-Koszul.The resolution we obtain from the reduction system R is the Koszul resolution, since it isminimal, see Theorem 8.1. As we shall see, this is a particular case of a general situation.7.1. The algebra counterexample to Happel’s question.
Let ξ be an element of the field k and let A be the k -algebra with generators x and y , subject to the relations x = 0 = y , yx = ξxy . Choose the order x < y with weights ω ( x ) = ω ( y ) = 1 and fix the reduction system R = { ( x , , ( y , , ( yx, ξxy ) } . The set B of irreducible paths is thus { , x, y, xy } . It is easyto verify that A = { x , yx , y x, y } and that all paths in A are reduction-unique. Bergman’sDiamond Lemma guarantees that R satisfies ( ♦ ).The only path of length 2 not in S is xy ; Proposition 3.4 implies that for each n , A n is theset of paths of lenght n + 1 not divisible by xy , A n = { y s x t : s + t = n + 1 } . Lemma 7.1.
The following complex provides the beginning of an A -bimodule projective resolutionof the algebra A A ⊗ E k A ⊗ E A d / / A ⊗ E k A ⊗ E A d / / A ⊗ E k A ⊗ E A δ / / A ⊗ E A δ − / / A / / ROJECTIVE RESOLUTIONS OF ASSOCIATIVE ALGEBRAS AND AMBIGUITIES 21 where d is the A -bimodule map such that d (1 ⊗ x ⊗
1) = x ⊗ x ⊗ ⊗ x ⊗ x,d (1 ⊗ y ⊗
1) = y ⊗ y ⊗ ⊗ y ⊗ y,d (1 ⊗ yx ⊗
1) = y ⊗ x ⊗ ⊗ y ⊗ x − ξx ⊗ y ⊗ − ξ ⊗ x ⊗ y and d is the A -bimodule morphism such that d (1 ⊗ y ⊗
1) = y ⊗ y ⊗ − ⊗ y ⊗ y,d (1 ⊗ y x ⊗
1) = y ⊗ yx ⊗ ξ ⊗ yx ⊗ y + ξ x ⊗ y ⊗ − ⊗ y ⊗ x,d (1 ⊗ yx ⊗
1) = y ⊗ x ⊗ − ⊗ yx ⊗ x − ξx ⊗ yx ⊗ − ξ ⊗ x ⊗ yd (1 ⊗ x ⊗
1) = x ⊗ x ⊗ − ⊗ x ⊗ x. Proof.
We apply Proposition 6.6 to the following sets { r p } p ∈A of left reductions and { t p } p ∈A of right reductions, where r y = r ,y ,y , r y x = r ,y ,x , t y = r y,y , , t y x = ( r x,y , , r ,yx,y , r y,yx, ) ,r yx = ( r ,x ,y , r x,yx, , r ,yx,x ) , r x = r ,x ,x , t yx = r y,x , , t x = r x,x , . (cid:3) One can find an A -bimodule resolution of A in [BGMS] and in [BE]; the authors also computethe Hochschild cohomology of A therein. We recover this resolution with our method.Given q ∈ A n , there are s, t ∈ N such that s + t = n + 1 and q = y s x t . Suppose q = apc with p = y s ′ x t ′ ∈ A n − and a, c ∈ Q ≥ . Since s + t = n + 1 and s ′ + t ′ = n , either a belongs to Q and c = x or a = y and c ∈ Q . As a consequence of this fact, the maps δ n : kQ ⊗ E k A n ⊗ E kQ → kQ ⊗ E k A n − ⊗ E A are δ n (1 ⊗ y s x t ⊗
1) = y ⊗ y s − x t ⊗ − n +1 ⊗ y s x t − ⊗ x, if s = 0 and t = 0, y ⊗ y n ⊗ − n +1 ⊗ y n ⊗ y, if t = 0, x ⊗ x n ⊗ − n +1 ⊗ x n ⊗ x, if s = 0,Moreover, given a basic reduction r = r a,s,c , the fact that s belongs to S = { x , y , yx } impliesthat r ( y s x t ) is either 0 or ξy s − xyx t − . Considering the reduction system R , if s = 0 and t = 0,then L ≺ n − ( y s x t ) = { ξ s x ⊗ y s x t − ⊗ , ξ t ⊗ y s − x t ⊗ y } . In case s = 0 or t = 0, the set L ≺ n − ( y s x t ) is empty.The computation of d − δ suggests the definition of the maps d n : A ⊗ E k A n ⊗ E A → A ⊗ E k A n − ⊗ E A as follows d n (1 ⊗ y s x t ⊗
1) = δ n (1 ⊗ y s x t ⊗
1) + ǫ ( ξ s x ⊗ y s x t − ⊗ ξ t ⊗ y s − x t ⊗ y )where ǫ denotes a sign depending on s, t, n . The equality d n − ◦ d n = 0 shows that making thechoice ǫ = ( − s does the job.Finally, Theorem 4.1 shows that the complex · · · / / A ⊗ E k A n ⊗ E A d n / / · · · d / / A ⊗ E k A ⊗ E A d / / A ⊗ E A d − / / A / / with d n (1 ⊗ y s x t ⊗
1) = y ⊗ y s − x t ⊗ − n +1 ⊗ y s x t − ⊗ x + ( − s ξ s x ⊗ y s x t − ⊗ − s ξ t ⊗ y s − x t ⊗ y, for s > t >
0, and d n (1 ⊗ y n +1 ⊗
1) = y ⊗ y n ⊗ − n +1 ⊗ y n ⊗ y,d n (1 ⊗ x n +1 ⊗
1) = x ⊗ x n ⊗ − n +1 ⊗ x n ⊗ x, is a projective bimodule resolution of A .Again, the algebra A is Koszul, see for example [Be2] and the resolution obtained using ourprocedure is the Koszul resolution, which is the minimal one, see Theorem 8.1.7.2. Quantum complete intersections.
These algebras generalize the previous case. Insteadof the relations x = 0 = y , yx = ξxy , we have x n = 0 = y m , yx = ξxy , where n and m arefixed positive integers, n, m > A . Consider the order x < y with weights ω ( x ) = ω ( y ) = 1.The set of 2-ambiguities associated to the reduction system R = { ( x n , , ( y m , , ( yx, ξxy ) } is A = { y m +1 , y m x, yx n , x n +1 } , and the set of irreducible paths is B = { x i y j ∈ k h x, y i : 0 ≤ i ≤ n − , ≤ j ≤ m − } . We easily check that every path in A is reduction-unique and usingBergman’s Diamond Lemma, we conclude that R satisfies ( ♦ ), Also, A = S = { y m , yx, x n } and A = { y m , y m +1 x, y m x n , yx n +1 , x n } .Denote by ϕ : N → N the map ϕ ( s, n ) = s n if s is even, s − n + 1 if s is odd.Given N ∈ N , the set of N -ambiguities is A N = { y ϕ ( s,m ) x ϕ ( t,n ) : s + t = N + 1 } . We willsometimes write ( s, t ) instead of y ϕ ( s,m ) x ϕ ( t,n ) ∈ A N .We first compute the beginning of the resolution. Lemma 7.2.
The following complex provides the beginning of a projective resolution of A as A -bimodule: A ⊗ E k A ⊗ E A d / / A ⊗ E k A ⊗ E A d / / A ⊗ E k A ⊗ E A δ / / A ⊗ E A δ − / / A / / where d and d are morphisms of A -bimodules given by the formulas d (1 ⊗ x n ⊗
1) = n − X i =0 x i ⊗ x ⊗ x n − − i ,d (1 ⊗ y m ⊗
1) = m − X i =0 y i ⊗ y ⊗ y m − − i ,d (1 ⊗ yx ⊗
1) = 1 ⊗ y ⊗ x + y ⊗ x ⊗ − ξ ⊗ x ⊗ y − ξx ⊗ y ⊗ d (1 ⊗ y m +1 ⊗
1) = y ⊗ y m ⊗ − ⊗ y m ⊗ y,d (1 ⊗ y m x ⊗
1) = m − X i =0 ξ i y m − − i ⊗ yx ⊗ y i + ξ m x ⊗ y m ⊗ − ⊗ y m ⊗ xd (1 ⊗ yx n ⊗
1) = y ⊗ x n ⊗ − n − X i =0 ξ i x i ⊗ yx ⊗ x n − − i − ξ n ⊗ x n ⊗ y,d (1 ⊗ x n +1 ⊗
1) = x ⊗ x n ⊗ − ⊗ x n ⊗ x. Proof.
It is straightforward, using Proposition 6.6 applied to the set { r p } p ∈A of left reductions,where r y m +1 = r ,y m ,y , r y m x = r ,y m ,x ,r yx n = ( r ,x n ,y , . . . , r x,yx,x n − , r ,yx,x n − ) r x n +1 = r ,x n ,x , ROJECTIVE RESOLUTIONS OF ASSOCIATIVE ALGEBRAS AND AMBIGUITIES 23 and the set { t p } p ∈A of right reductions, where t y m +1 = r y,y m , , t y m x = ( r x,y m , , . . . , r y m − ,yx,y , r y m − ,yx, ) ,y yx n = r y,x n , , t x n +1 = r x,x n , . (cid:3) Of course we want to construct the rest of the resolution. Denote ( s, t ) = y ϕ ( s,m ) x ϕ ( t,n ) ∈ A N .We will first describe the set L ≺ N − ( s, t ). There are four cases, depending on the parity of s, t and N . With this in view, it is useful to make some previous computations that we list below.(1) For s even, for all j , 0 ≤ j ≤ m − y ϕ ( s,m ) = y m − − j y ϕ ( s − ,m ) y j .(2) For s odd, y ϕ ( s,m ) = yy ϕ ( s − ,m ) = y ϕ ( s − ,m ) y. (3) For t even, for all i , 0 ≤ i ≤ n − x ϕ ( t,n ) = x i x ϕ ( t − ,n ) x n − i − ,(4) For t odd, x ϕ ( t,n ) = xx ϕ ( t − ,n ) = x ϕ ( t − ,n ) x . First case: N even, s even, t odd, L ≺ N − ( s, t ) = { ξ ϕ ( t,n ) j y m − − j ⊗ ( s − , t ) ⊗ y j } m − j =1 ∪ { ξ ϕ ( s,m ) x ⊗ ( s, t − ⊗ } . Second case: N even, s odd, t even, L ≺ N − ( s, t ) = { ξ ϕ ( t,n ) ⊗ ( s − , t ) ⊗ y } ∪ { ξ ϕ ( s,m ) i x i ⊗ ( x, t − ⊗ x n − − i } n − i =1 . Third case: N odd, s even, t even, L ≺ N − ( s, t ) = { ξ ϕ ( t,n ) j y m − − j ⊗ ( s − , t ) ⊗ y j } m − j =1 ∪ { ξ ϕ ( s,m ) i x i ⊗ ( s, t − ⊗ x n − − i } n − i =1 . Fourth case: N , s and t odd, L ≺ N − ( s, t ) = { ξ ϕ ( t,n ) ⊗ ( s − , t ) ⊗ y, ξ ϕ ( s,m ) x ⊗ ( s, t − ⊗ } . Remark . We observe that, analogously to the case n = m = 2,( d − δ )(1 ⊗ ( s, t ) ⊗
1) = ( − s X u ∈L ≺ ( s,t ) u, ( d − δ )(1 ⊗ ( s, t ) ⊗
1) = ( − s X u ∈L ≺ ( s,t ) u. Proposition 5.8 for R = Z guarantees that there exist A -bimodule maps d N : A ⊗ E k A N ⊗ E A → A ⊗ E k A N − ⊗ E A such that ( d N − δ N )(1 ⊗ ( s, t ) ⊗ ∈ hL ≺ N − ( s, t ) i Z and, most important, thecomplex ( A ⊗ E k A • ⊗ E A, d • ) is a projective resolution of A as A -bimodule.We are not yet able at this point to give the explicit formulas of the differentials.In order to illustrate the situation, let us describe what happens for N = 3. We know afterthe mentioned proposition that there exist t , t ∈ Z such that d (1 ⊗ y m +1 x ⊗
1) = d (1 ⊗ (3 , ⊗ δ (1 ⊗ (3 , ⊗
1) + t ξ ⊗ (2 , ⊗ y + t ξ x ⊗ (3 , ⊗ y ⊗ y m x ⊗ ⊗ y m +1 ⊗ x + t ξ ⊗ y m x ⊗ y + t ξ x ⊗ y m +1 ⊗ . Of course, d ◦ d = 0. It follows from this equality that t = t = −
1. This example motivatesthe following lemma, stated in terms of the preceding notations.
Lemma 7.3.
The A -bimodule morphisms d N : A ⊗ E k A N ⊗ E A → A ⊗ E k A N − ⊗ E A definedby the formula d N (1 ⊗ ( s, t ) ⊗
1) = δ N (1 ⊗ ( s, t ) ⊗
1) + ( − s X u ∈L ≺ N − ( s,t ) u satisfy the hypotheses of Thm. 4.1.Proof. It is straightforward. (cid:3)
We gather all the information we have obtained about the projective bimodule resolution of A in the following proposition. Proposition 7.4.
The complex of A -bimodules ( A ⊗ E k A • ⊗ E A, d • ) , with A N = { y ϕ ( s,m ) x ϕ ( t,n ) : s + t = N + 1 } and differentials defined as follows is exact.(1) For N even, s even and t odd, d N (1 ⊗ ( s, t ) ⊗
1) = y m − ⊗ ( s − , t ) ⊗ m − X j =1 ( − s ξ ϕ ( t,n ) j y m − − j ⊗ ( s − , t ) ⊗ y j + ( − N +1 ⊗ ( s, t − ⊗ x + ( − s ξ ϕ ( s,m ) x ⊗ ( s, t − ⊗ . (2) For N even, s odd and t even, d N (1 ⊗ ( s, t ) ⊗
1) = y ⊗ ( s − , t ) ⊗ − s ξ ϕ ( t,n ) ⊗ ( s − , t ) ⊗ y + ( − N +1 ⊗ ( s, t − ⊗ x n − + n − X i =1 ( − s ξ ϕ ( s,m ) i x i ⊗ ( s, t − ⊗ x n − − i (3) For N odd, s and t even, d N (1 ⊗ ( s, t ) ⊗
1) = y m − ⊗ ( s − , t ) ⊗ m − X j =1 ( − s ξ ϕ ( t,n ) j y m − − j ⊗ ( s − , t ) ⊗ y j + ( − N +1 ⊗ ( s, t − ⊗ x n − + n − X i =1 ( − s ξ ϕ ( s,m ) i x i ⊗ ( s, t − ⊗ x n − − i (4) For N , s and t odd, d N (1 ⊗ ( s, t ) ⊗
1) = y ⊗ ( s − , t ) ⊗ − s ξ ϕ ( t,n ) ⊗ ( s − , t ) ⊗ y + ( − N +1 ⊗ ( s, t − ⊗ x + ( − s ξ ϕ ( s,m ) x ⊗ ( s, t − ⊗ . Again, we obtain the minimal resolution of A , even for n = 2 or m = 2, when the algebra isnot homogeneous.7.3. Down-up algebras.
Given α, β, γ ∈ k , we will denote A ( α, β, γ ) the quotient of k h d, u i bythe two sided ideal I generated by relations d u − αdud − βud − γd = 0 ,du − αudu − βu d − γu = 0 . Down-up algebras have been deeply studied since they were defined in [BR]. We can mention thearticles [CM], [BW],[BG], [CS], [CL], [KK], [KMP], [Ku1], [Ku2], [P1], [P2], [P3], in which theauthors prove diverse properties of down-up algebras. It is well known that they are noetherianif and only if β = 0 [KMP]. They are graded with dg ( d ) = 1, dg ( u ) = −
1, and they are filteredif we consider d and u of weight 1. If γ = 0 they are also graded by this weight.Down-up algebras are 3-Koszul if γ = 0, and if γ = 0, they are PBW deformations of 3-Koszulalgebras [BG].Little is known about their Hochschild homology and cohomology, except for the center, de-scribed in [Z] and [Ku1]. We apply our methods to construct a projective resolution of A as A -bimodule, and then use this resolution to compute H • ( A, A e ) and prove that in the noe-therian case, A ( α, β, γ ) is 3-Calabi-Yau if and only if β = −
1. Moreover, in this situation we
ROJECTIVE RESOLUTIONS OF ASSOCIATIVE ALGEBRAS AND AMBIGUITIES 25 exhibit a potential Φ( d, u ) such that the relations are in fact the cyclic derivatives ∂ u Φ and ∂ d Φ,respectively.We briefly recall that a d -Calabi-Yau algebra is an associative algebra such that there is anisomorphism f of A -bimodules Ext iA e ( A, A e ) ∼ = (cid:26) i = d , A if i = d .where the A -bimodule outer structure of A e is used for the computation of Ext iA e ( A, A e ), whilethe isomorphism f takes account of the inner bimodule structure of A e . Bocklandt proved in [Bo]that graded Calabi-Yau algebras come from a potential and Van den Bergh [VdB] generalizedthis result to complete algebras with respect to the I -adic topology.We fix a lexicographical order such that d < u , with weights ω ( d ) = 1 = ω ( u ). The reductionsystem R = { ( d u, αdud + βud + γd ) , ( du , αudu + βu d + γu ) } has B = { u i ( du ) k d j : i, k, j ∈ N } as set of irreducible paths and A = { d u } ; using Bergman’s Diamond Lemma we see that R satisfies condition ( ♦ ). Also, A = { d, u } and A n = ∅ for all n ≥
3. The set B is the k -basisalready considered in [BR].The reductions r d u = ( r u,d u, , r ,d u,u ) and t d u = ( t ,du ,d , t d,du , ) are respectively leftand right reductions of d u .In view of Proposition 6.6 and observing that δ − is in fact an epimorphism and that A = ∅ ,the following complex gives a free resolution of A as A -bimodule: → A ⊗ E kd u ⊗ E A → d A ⊗ E ( kd u ⊕ kdu ) ⊗ E A → d A ⊗ E ( kd ⊕ ku ) ⊗ E A → δ A ⊗ E A → δ − A → where d (1 ⊗ d u ⊗
1) = 1 ⊗ d ⊗ du + d ⊗ d ⊗ u + d ⊗ u ⊗ − α (1 ⊗ d ⊗ ud + d ⊗ u ⊗ d + du ⊗ d ⊗ − β (1 ⊗ u ⊗ d + u ⊗ d ⊗ d + ud ⊗ d ⊗ − γ ⊗ d ⊗ ,d (1 ⊗ du ⊗
1) = 1 ⊗ d ⊗ u + d ⊗ u ⊗ u + du ⊗ u ⊗ − α (1 ⊗ u ⊗ du + u ⊗ d ⊗ u + ud ⊗ u ⊗ − β (1 ⊗ u ⊗ ud + u ⊗ u ⊗ d + u ⊗ d ⊗ − γ ⊗ u ⊗ , and d (1 ⊗ d u ⊗
1) = d ⊗ du ⊗ β ⊗ du ⊗ d − ⊗ d u ⊗ u − βu ⊗ d u ⊗ . As we have proved in general, the map d takes into account the reductions applied to theambiguity. Proposition 7.5.
Suppose that β = 0 . The algebra A ( α, β, γ ) is -Calabi-Yau if and only if β = − .Proof. We need to compute Ext • A e ( A, A e ). We apply the functor Hom A e ( − , A e ) to the previousresolution, and we use that for any finite dimensional vector space V which is also an E -bimodule,the space Hom A e ( A ⊗ E V ⊗ E A, A e ) is isomorphic to Hom E e ( V, A e ), and this last one is, in turn,isomorphic to A ⊗ E V ∗ ⊗ E A . All the isomorphisms are natural. The explicit expression of thelast isomorphism is, fixing a k -basis { v , . . . , v n } of V and its dual basis { ϕ , . . . , ϕ n } of V ∗ , A ⊗ E V ∗ ⊗ E A → Hom E e ( V, A e ) a ⊗ ϕ ⊗ b [ v ϕ ( v ) b ⊗ a ]with inverse f P i,j b ij ⊗ ϕ i ⊗ a ij , where f ( v i ) = P j a ij ⊗ b ij .After these identifications, we obtain the following complex of k -vector spaces whose homologyis Ext • A e ( A, A e )0 → A ⊗ E A δ ∗ → A ⊗ E ( kD ⊕ kU ) ⊗ E A d ∗ → A ⊗ E ( kD U ⊕ kDU ) ⊗ E A d ∗ → A ⊗ E kD U ⊗ E A → , where { D, U } denotes the dual basis of { d, u } and, accordingly, we denote with capital lettersthe dual bases of the other spaces .The maps in the complex are, explicitely: δ ∗ (1 ⊗
1) = 1 ⊗ D ⊗ d − d ⊗ D ⊗ ⊗ U ⊗ u − u ⊗ U ⊗ d ∗ (1 ⊗ U ⊗
1) = 1 ⊗ D U ⊗ d − αd ⊗ D U ⊗ d − βd ⊗ D U ⊗ u ⊗ DU ⊗ d + 1 ⊗ DU ⊗ du − αdu ⊗ DU ⊗ − α ⊗ DU ⊗ ud − βud ⊗ DU ⊗ − βd ⊗ DU ⊗ u − γ ⊗ DU ⊗ .d ∗ (1 ⊗ D ⊗
1) = du ⊗ D U ⊗ u ⊗ D U ⊗ d − αud ⊗ D U ⊗ − α ⊗ D U ⊗ du − βd ⊗ D U ⊗ u − β ⊗ D U ⊗ ud − γ ⊗ D U ⊗ u ⊗ DU ⊗ − αu ⊗ DU ⊗ u − β ⊗ DU ⊗ u .d ∗ (1 ⊗ DU ⊗
1) = 1 ⊗ D U ⊗ d + βd ⊗ D U ⊗ ,d ∗ (1 ⊗ D U ⊗
1) = − u ⊗ D U ⊗ − β ⊗ D U ⊗ u. Consider the following isomorphisms of A -bimodules ψ : A ⊗ E A → A ⊗ E kd u ⊗ E A,ψ (1 ⊗
1) = 1 ⊗ d u ⊗ ,ψ : A ⊗ E ( kD ⊕ kU ) ⊗ E A → A ⊗ E ( kd u ⊕ kdu ) ⊗ E Aψ (1 ⊗ D ⊗
1) = 1 ⊗ du ⊗ , and ψ (1 ⊗ U ⊗
1) = 1 ⊗ d u ⊗ ψ : A ⊗ E ( kD U ⊕ kDU ) ⊗ E A → A ⊗ E ( kd ⊕ ku ) ⊗ E A,ψ (1 ⊗ D U ⊗
1) = 1 ⊗ u ⊗ , and ψ (1 ⊗ DU ⊗
1) = 1 ⊗ d ⊗ ψ : A ⊗ E kD U ⊗ E → A ⊗ E Aψ (1 ⊗ D U ⊗
1) = 1 ⊗ . It is straightforward to verify that the following diagram commutes, thus inducing isomor-phisms between the homology spaces of both horizontal sequences: / / A ⊗ E A δ ∗ / / ψ (cid:15) (cid:15) A ⊗ E ( k A ) ∗ ⊗ E A d ∗ / / ψ (cid:15) (cid:15) A ⊗ E ( k A ) ∗ ⊗ E A d ∗ / / ψ (cid:15) (cid:15) A ⊗ E ( kA ) ∗ ⊗ E A / / ψ (cid:15) (cid:15) / / A ⊗ E k A ⊗ E A d / / A ⊗ E k A ⊗ E A d / / A ⊗ E k A ⊗ E A d / / A ⊗ E A / / where d is given by d (1 ⊗ d u ⊗
1) = 1 ⊗ du ⊗ d − d ⊗ du ⊗ − u ⊗ d u ⊗ ⊗ d u ⊗ u.d is d (1 ⊗ d u ⊗
1) = 1 ⊗ d ⊗ du − βd ⊗ d ⊗ u − βd ⊗ u ⊗ − α (1 ⊗ d ⊗ ud + d ⊗ u ⊗ d + du ⊗ d ⊗ − β ( − β − ⊗ u ⊗ d − β − u ⊗ d ⊗ d + ud ⊗ d ⊗ − γ ⊗ d ⊗ d (1 ⊗ du ⊗
1) = − β ⊗ d ⊗ u − βd ⊗ u ⊗ u + du ⊗ u ⊗ − α (1 ⊗ u ⊗ du + u ⊗ d ⊗ u + ud ⊗ u ⊗ − β (1 ⊗ u ⊗ ud − β − u ⊗ u ⊗ d − β − u ⊗ d ⊗ − γ ⊗ u ⊗ ROJECTIVE RESOLUTIONS OF ASSOCIATIVE ALGEBRAS AND AMBIGUITIES 27 and d is d (1 ⊗ u ⊗
1) = − β ⊗ u − u ⊗ , d (1 ⊗ d ⊗
1) = 1 ⊗ d + βd ⊗ , From this we deduce that HH ( A, A e ) ∼ = A ⊗ E A/ (Im d ). Let σ be the algebra automorphismof A defined by σ ( d ) = − βd , σ ( u ) = − β − u . Recall that A σ is the A -bimodule with A asunderlying vector space and action of A ⊗ k A op given by: ( a ⊗ b ) · x = axσ ( b ), that is, it istwisted on the right by the automorphism σ .It is easy to see that if β = 0 then A σ ∼ = A ⊗ E A/ (Im d ) ∼ = HH ( A, A e ) as A -bimodules.If β = 0 then the action on the left by u on HH ( A, A e ) is zero and then A ≇ HH ( A, A e )since the action on the left by u on A is injective. We conclude after a short computation that HH ( A, A e ) ∼ = A if and only if β = −
1. Notice that for β = − A . As a consequence, A is 3-Calabi-Yau if and only if β = −
1. In this case the potential Φ equals d u + α dudu + γdu . For β = 0 , −
1, we shall seein a forthcoming article that A is twisted 3-Calabi-Yau algebra [BSW], coming from a twistedpotential. (cid:3) Final remarks
We have studied some examples of algebras, in particular of N -Koszul algebras for which wemanaged to obtain the minimal resolution using our methods. This fact can be stated in generalas follows. Theorem 8.1.
Given an algebra A = kQ/I such that(1) there is a reduction system R = { ( s i , f i ) } i for I satisfying ( ♦ ) with s i and f i homogeneousof length N ≥ for all i ,(2) for all n ∈ N , the length of the elements of A n is strictly smaller that the length of theelements of A n +1 .The resolutions of A as A -bimodule obtained using Theorem 4.1 and Theorem 4.2 are minimal.Proof. Let ( A ⊗ E k A • ⊗ E A, d • ) be a resolution of A as A -bimodule obtained using Theorem4.1 or Theorem 4.2. Denote by | c | the length of a path c ∈ Q ≥ . Condition (1) guarantees thatfor all paths p, q such that λp (cid:22) q for some λ ∈ k × , we have | p | = | q | . Let n ≥ q ∈ A n and λπ ( b ) ⊗ p ⊗ π ( b ′ ) ∈ L ≺ n − ( q ). Since p ∈ A n − , condition (2) says that | p | < | q | . On the otherhand, λbpb ′ ≺ q and then | bpb ′ | = | q | . We deduce that b ∈ Q ≥ or b ′ ∈ Q ≥ . As a consequence,Im( d n ) is contained in the radical of A ⊗ E k A n − ⊗ E A and therefore the resolution of A isminimal. (cid:3) Remark . The conclusion holds in a more general situation, which includes Example 7.2.It is sufficient to have a reduction system satisfying (1) and such that the ambiguities p thatappear when reducing a given n + 1-ambiguity q are of length strictly smaller than the length of q . Remark . In Example 7.0.1, the reduction system R satisfies the conditions of Theorem8.1, while R does not satisfy (2).Notice that if R is a reduction system for an algebra for which there is a non-resolvableambiguity, then, even if we complete it like we did in Example 7.0.1, the resolutions obtainedusing Theorem 4.1 and Theorem 4.2 will not be minimal.We end this article proving a generalization of Prop. 8 of [GM] and a corollary. Proposition 8.2.
Let A = kQ/I , where Q is a finite quiver, kQ is the path algebra graded bythe length of paths and I a homogeneous ideal with respect to this grading, contained in Q ≥ .Let R be a reduction system satisfying conditions (1) and (2) of Theorem 8.1 and let A S bethe associated monomial algebra. The algebra A S is N -Koszul if and only if A is an N -Koszulalgebra.Proof. The projective bimodules appearing in the minimal resolution of A S are in one–to–onecorrespondence with those appearing in the resolution of A , so either both of them are generatedin the correct degrees or none is. (cid:3) This proposition, together with Proposition 3.4 and Thm. 3 of [GH] give the following result.