Cluster-tilted algebras and slices
aa r X i v : . [ m a t h . R T ] J u l Cluster-tilted algebras and slices
Ibrahim Assem , D´epartement de Math´ematiques, Universit´e de Sherbrooke, Sherbrooke (Qu´ebec),J1K 2R1, Canada
Thomas Br¨ustle , D´epartement de Math´ematiques, Universit´e de Sherbrooke, Sherbrooke (Qu´ebec),J1K 2R1, Canada and
Department of Mathematics, Bishop’s University,Lennoxville, (Qu´ebec), J1M 1Z7, Canada
Ralf Schiffler , Department of Mathematics and Statistics, University of Massachusetts atAmherst, Amherst, MA 01003-9305, USA
Abstract
We give a criterion allowing to verify whether or not two tilted algebras have thesame relation-extension (thus correspond to the same cluster-tilted algebra). Thiscriterion is in terms of a combinatorial configuration in the Auslander-Reiten quiverof the cluster-tilted algebra, which we call local slice.
Key words:
Local slice, cluster-tilted algebra, relation-extensions of tilted algebras
Email addresses: [email protected] (Ibrahim Assem), [email protected] (Thomas Br¨ustle), [email protected] (Ralf Schiffler). Partially supported by the NSERC of Canada and the University of Sherbrooke Partially supported by the NSERC of Canada and the universities of Sherbrookeand Bishop’s Partially supported by the University of Massachusetts
Preprint submitted to Elsevier 26 October 2018
Introduction
Cluster categories were introduced in [7] and, for type A , also in [11], as acategorical model allowing to understand better the cluster algebras of Fominand Zelevinsky [13]. Cluster-tilted algebras were defined in [11] for type A ,and in [8] for arbitrary hereditary algebras as follows: Let A be a hereditaryalgebra and ˜ T be a tilting object in the associated cluster category C A , that is,an object such that Ext C A ( ˜ T , ˜ T ) = 0 and the number of isomorphism classesof indecomposable summands of ˜ T equals the rank of the Grothendieck groupof A , then the algebra B = End C A ˜ T is called cluster-tilted. These algebrashave been studied by several authors, see, for instance, [1,8,9,10,12,17]. Inparticular, they were shown in [1] to be closely related to the tilted algebrasintroduced by Happel and Ringel in the early eighties [15]. Indeed, let C be atilted algebra, then the trivial extension ˜ C = C ⋉ Ext C ( DC, C ) of C by the C - C -bimodule Ext C ( DC, C ) is cluster-tilted, and every cluster-tilted algebrais of this form. Thus, we have a surjective map C ˜ C from tilted to cluster-tilted algebras. However, easy examples show that this map is not injective.Our objective in this paper is to give a criterion allowing to verify whetherfor two tilted algebras C and C , the corresponding cluster-tilted algebras ˜ C and ˜ C are isomorphic or not.Since tilted algebras are characterised by the existence of complete slices intheir Auslander-Reiten quiver (see, for instance, [15,20,19,21] or [3]), it is nat-ural to study the corresponding concept for cluster-tilted algebras. For thispurpose, we introduce what we call a local slice , by weakening the axioms ofcomplete slice (thus, in a tilted algebra, complete slices are local slices). Weshow that a complete slice in a tilted algebra C embeds as a local slice in ˜ C (and, in fact, any local slice in ˜ C is of this form).Our main theorem is the following: Theorem 1
Let B be a cluster-tilted algebra. Then a tilted algebra C is suchthat B = C ⋉ Ext C ( DC, C ) if and only if there exists a local slice Σ in mod B such that C = B/ Ann B Σ . We also show that cluster-tilted algebras have many local slices. In fact, all butat most finitely many indecomposable modules lying in the transjective compo-nent of the Auslander-Reiten quiver belong to a local slice. If the cluster-tiltedalgebra is of tree type (which is the case, for instance, if it is of Dynkin type orof Euclidean type distinct from ˜ A ), then this is the case for all indecomposablemodules in this component.We now describe the contents of our paper. In the first section, we introduce2he notion of local section in a translation quiver and in the second section westudy sections and local sections in the derived category. In the third section,we introduce the concept of a local slice and prove our main result, and insection four, we prove that cluster-tilted algebras of tree type have sufficientlymany local slices.We would like to thank David Smith for useful comments on the paper. Throughout this paper, all algebras are connected finite dimensional algebrasover an algebraically closed field k . For an algebra C , we denote by mod C thecategory of finitely generated right C -modules and by ind C a full subcategoryof mod C consisting of exactly one representative from each isomorphism classof indecomposable modules. When we speak about a C -module (or an inde-composable C -module), we always mean implicitly that it belongs to mod C (or to ind C , respectively). Also, all subcategories of mod C are full and so areidentified with their object classes. Given a subcategory C of mod C , we some-times write M ∈ C to express that M is an object in C . We denote by add C thefull subcategory of mod C with objects the finite direct sums of modules in C and, if M is a module, we abbreviate add { M } as add M . We denote the pro-jective (or injective) dimension of a module M as pd M (or id M , respectively).The global dimension of C is denoted by gl.dim. C and its Grothendieck groupby K ( C ). Finally, we denote by Γ(mod C ) the Auslander-Reiten quiver of analgebra C , and by τ C = D T r , τ − C = T r D its Auslander-Reiten translations.For further definitions and facts needed on mod C or Γ(mod C ), we refer thereader to [3]. We also need facts on the bounded derived category D b (mod C )of mod C , for which we refer to [14]. For translation quivers, we refer to [3,20]. Let (Γ , τ ) be a connected translationquiver. We recall that a path x = x → x → . . . → x t = y in Γ is called sectional if, for each i with 0 < i < t , we have τ x i +1 = x i − . A full connectedsubquiver Σ of Γ is said to be convex in Γ if, for any path x = x → x → . . . → x t = y in Γ with x, y ∈ Σ , we have x i ∈ Σ for all i . It is called acyclic if there is no cycle x = x → x → . . . → x t = x (with t >
0) which is entirelycontained in Σ. The following definition is due to Liu and Skowro´nski (see318,21] or else [3]).
Definition 2
Let (Γ , τ ) be a connected translation quiver. A connected fullsubquiver Σ of Γ is a section in Γ if:(S1) Σ is acyclic.(S2) For each x ∈ Γ , there exists a unique n ∈ Z such that τ n x ∈ Σ .(S3) Σ is convex in Γ . This definition is motivated by the study of tilted algebras. The well-knowncriterion of Liu and Skowro´nski asserts that, if C is an algebra, and Σ isa faithful section in a component of its Auslander-Reiten quiver such thatHom C ( X, τ C Y ) = 0 for all X, Y ∈ Σ , then C is tilted having Σ as completeslice (see [18,21,3]).We note that, if a translation quiver Γ contains a section, then Γ is acyclic. We need some weaker notions. The first one is the following.
Definition 3
Let (Γ , τ ) be a connected translation quiver. A connected fullsubquiver Σ of Γ is called a presection in Γ if it satisfies the following twoconditions:(P1) If x ∈ Σ and x → y is an arrow, then either y ∈ Σ or τ y ∈ Σ .(P2) If y ∈ Σ and x → y is an arrow, then either x ∈ Σ or τ − x ∈ Σ . The following Lemma collects the elementary properties of presections. Recallthat a translation quiver is called stable if there are neither projective, norinjective points in Γ.
Lemma 4
Let (Γ , τ ) be a connected translation quiver.(a) If Σ is a section in Γ , then Σ is a presection.(b) Any path entirely contained in a presection Σ is a sectional path.(c) If the translation quiver Γ is stable, then conditions (P1) and (P2) areequivalent.(d) If Σ is a presection in a stable translation quiver Γ , then Σ intersects every τ -orbit of Γ at least once. PROOF. (a) This is well-known (see, for instance, [3, VIII.1.4 p.304]).4b) Assume that Σ is a presection in Γ, and that x = x → x → . . . → x t = y is a path lying entirely in Σ. If this path is not sectional, then there exists aleast i with 0 < i < t and τ x i +1 = x i − . But, in this case, we have arrows x i − → x i and x i → τ − x i − = x i +1 with x i , x i − , x i +1 ∈ Σ , a contradiction.(c) Assume (P1) holds and that x → y is an arrow with y ∈ Σ . Since x is notinjective, there exists an arrow y → τ − x . Applying (P1) to the latter yieldsthat τ − x ∈ Σ or x = τ ( τ − x ) ∈ Σ . Thus (P2) holds. The converse is shownin the same way.(d) It suffices to prove that, if x ∈ Σ and y ∈ Γ lie in two neighbouring τ -orbits, then Σ intersects the τ -orbit of y . Since Γ is stable, there exists m ∈ Z such that there is an arrow x → τ m y . But then τ m y ∈ Σ or τ m +1 y ∈ Σ . ✷ For instance, let Γ be a stable tube. A ray in Γ is a presection but clearly nota section. On the other hand, if Γ is a component of type Z A ∞ , then a ray isa section. We now define an intermediate notion between those of presection and section.
Definition 5
Let (Γ , τ ) be a connected translation quiver. A presection Σ in Γ is called a local section if it satisfies moreover the following additionalcondition: Σ is sectionally convex, that is, if x = x → x → . . . → x t = y is asectional path in Γ such that x, y ∈ Σ , then x i ∈ Σ for all i . The following Lemma is immediate.
Lemma 6
Any section is a local section.
PROOF.
Indeed, any section is convex, and hence sectionally convex. Wethen apply Lemma 4 (b). ✷ The main result of this section is the following.
Proposition 7
Let Q be a finite acyclic quiver and Γ = Z Q . The followingconditions are equivalent for a connected full subquiver Σ of Γ such that | Σ | = | Q | : a) Σ is a section.(b) Σ is a local section.(c) Σ is a presection. PROOF.
Because of Lemma 6 (and the definition of local section), it sufficesto prove that (c) implies (a). Since Γ is a stable translation quiver, then,because of Lemma 4 (d), Σ intersects every τ -orbit of Γ at least once. Then, itfollows from the hypothesis that | Σ | = | Q | that Σ intersects each τ -orbit ofΓ exactly once. Since Σ is clearly acyclic (because Γ is), there only remains toprove convexity. Suppose that there exists a path x = x → x → . . . → x t = y in Γ with x, y ∈ Σ but x , . . . , x t − / ∈ Σ ( t ≥ τ x ∈ Σ . From the arrow τ x → τ x , we deduce that either τ x ∈ Σ or τ x ∈ Σ . Repeating this argument t times, we get that one of τ y, τ y, . . . , τ t y lies in Σ. This contradicts the fact that Σ cuts each τ -orbit exactly once,because y ∈ Σ . ✷ Let A denote a hereditary algebra, and D = D b (mod A ) denote the boundedderived category of mod A . We denote by τ D , τ − D the Auslander-Reiten trans-lations in D . We recall that the Auslander-Reiten quiver of D consists of twotypes of components: the regular (which correspond to the regular componentsof Γ(mod A ) and their shifts) and the transjective (which are the form Z Q ,where Q denotes the ordinary quiver of A ), see [14].We need the following result, known as Skowro´nski’s Lemma [22,3]. Lemma 8
Let C be an artin algebra. Assume that a C -module M is the directsum of m pairwise non-isomorphic indecomposable modules and is such that Hom C ( M, τ C M ) = 0 . Then m ≤ rk K ( C ) . The following Lemma is motivated by [21,19].
Lemma 9
Let Σ be a section in a connected component Γ of Γ( D ) then thefollowing conditions are equivalent:(a) Σ is convex in ind D , that is, if X = X → X → . . . → Y is a sequence ofnon-zero morphisms between indecomposable objects in D such that X, Y ∈ Σ , then X i ∈ Σ for all i .(b) Hom (
X, τ D Y ) = 0 for all X, Y ∈ Σ .(c) Hom ( τ − D X, Y ) = 0 for all
X, Y ∈ Σ .(d) Σ is finite.(e) | Σ | = rk K ( A ) . f ) Γ is a transjective component. PROOF.
The equivalence of ( b ) and ( c ) follows trivially from the fact that τ is an automorphism of D .( a ) implies ( b ). A non-zero morphism X → τ D Y induces a path X → τ D Y →∗ → Y in ind D . The hypothesis implies that both Y and τ D Y lie in Σ, acontradiction.( b ) implies ( f ). If Γ is not a transjective component, then we may assume with-out loss of generality that Γ is concentrated in degree zero. By Skowro´nski’sLemma 8, Σ is finite. Now this contradicts the fact that Σ intersects each τ D -orbit exactly once.( f ) implies ( e ). The number of τ D -orbits in a transjective component is exactlyrk K ( A ).( e ) implies ( d ). This is trivial.( d ) implies ( a ). Suppose that X = X f → X f → . . . f t → X t = Y is a path, wherethe f i are non-zero morphisms, the X i lie in Γ and X, Y ∈ Σ . Now, since Σ is finite, then Γ has finitely many τ D -orbits. Therefore, Γ is transjective. Thisimplies that the f i lie in a finite power of the radical of D . Therefore the pathabove can be refined to a path of irreducible morphisms. Convexity of Σ in Γthen implies that X i ∈ Σ for all i . ✷ Corollary 10
Let Σ be a connected full subquiver in a connected component Γ of Γ( D ) such that | Σ | = rk K ( A ) . The following are equivalent:(a) Σ is a section.(b) Σ is a local section.(c) Σ is a presection. PROOF.
Again, it suffices to prove that (c) implies (a). Since Γ is a stabletranslation quiver, then the presection Σ intersects each τ D -orbit of Γ at leastonce. Since, by hypothesis, | Σ | = rk K ( A ) < ∞ , then Γ is a transjectivecomponent. The statement then follows from Proposition 7. ✷ Local slices
We now define the main concept of this paper.
Definition 11
Let C be a finite dimensional algebra. A local slice Σ in mod C is a local section in a component Γ of Γ(mod C ) such that | Σ | = rk K ( C ) . Remark 12
Let C be an algebra, and Σ be a local slice in mod C .(a) Since local sections are presections, every path entirely contained in Σ issectional (because of Lemma 4 (b)). This implies that Σ is acyclic.(b) If Γ is a stable component of Γ(mod C ) and Σ ⊂ Γ then, by Lemma 4 (d), Σ intersects any τ C -orbit of Γ at least once. Since | Σ | < ∞ , this impliesthat Γ has only finitely many τ C -orbits. Example 13
Let C be a tilted algebra, and Γ be a connecting component of Γ(mod C ) . Then any complete slice in Γ is also a local slice. We shall provebelow that any cluster-tilted algebra has (many) local slices in its Auslander-Reiten quiver. Example 14
The following is an example of an algebra which is neither tilted,nor cluster-tilted, but whose Auslander-Reiten quiver contains a local slice. Let B be given by the quiver α v v nnnnnnnnnnnnnn β h h PPPPPPPPPPPPPP δ v v nnnnnnnnnnnnnn ǫ / / γ h h PPPPPPPPPPPPPP bound by α β = 0 , δ ǫ = 0 , ǫ γ = 0 and γ δ = 0 . The Auslander-Reiten quiver Γ(mod B ) of B is shown in Figure 1, where modules are represented by theirLoewy series and we identify the two copies of the underlined module . Here Σ =
32 1 , , , , and Σ ′ = , , , , are local slices.Note that neither Σ nor Σ ′ is a section, because both intersect twice the τ B -orbit of . It is an interesting question to identify the algebras which have localslices. (cid:31) (cid:31) ???????? (cid:31) (cid:31) ???????? (cid:31) (cid:31) ??????? ? ? (cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127) (cid:31) (cid:31) ???????? (cid:31) (cid:31) ?????? ? ? (cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127)
232 1 ? ? (cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127) (cid:31) (cid:31) ?????? ? ? (cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127) (cid:31) (cid:31) ???????? ? ? (cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127) (cid:31) (cid:31) ??????? ? ? (cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127) ? ? (cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127) (cid:31) (cid:31) ???????? ? ? (cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127) ? ? (cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127) Fig. 1. Auslander-Reiten quiver of Example 13
Let A be a hereditary algebra. The cluster category C A of A is defined asfollows. Let F be the automorphism of D = D b (mod A ) defined as the com-position τ − D [1], where τ − D is the Auslander-Reiten translation in D and [1] isthe shift functor. Then C A is the orbit category D /F , that is, the objects of C A are the F -orbits ˜ X = ( F i X ) i ∈ Z , where X ∈ D , and the set of morphismsfrom ˜ X = ( F i X ) i ∈ Z to ˜ Y = ( F i Y ) i ∈ Z isHom C A ( ˜ X, ˜ Y ) = M i ∈ Z Hom D ( X, F i Y ) . It is shown in [7,16] that C A is a triangulated category with almost split tri-angles. Furthermore, the projection functor π : D → C A is a functor oftriangulated categories and commutes with the Auslander-Reiten translationsin both categories. We refer to [7] for facts about the cluster category.An object ˜ T in C A is called a tilting object provided Ext C A ( ˜ T , ˜ T ) = 0 andthe number of isomorphism classes of indecomposable summands of ˜ T equalsrk K ( A ). The endomorphism algebra B = End C A ( ˜ T ) is then called a cluster-tilted algebra [8]. Of particular interest to us is the fact that the functorHom C A ( ˜ T , − ) : C A → mod B induces an equivalence C A / add ( τ ˜ T ) ∼ = mod B, where τ denotes the Auslander-Reiten translation in C A , see [8]. This resultentails several interesting consequences. For instance, it is shown it [17] thatany cluster-tilted algebra is 1-Gorenstein and hence of global dimension 1 or ∞ . For the convenience of the reader, we give here a short proof of this fact.9 roposition 15 Let B be a cluster-tilted algebra.(a) For every injective B -module I , we have pd I ≤ (and for every projective B -module P , we have id P ≤ ).(b) gl.dim. B ∈ { , ∞} . PROOF. (a) By [3, (IV.2.7) p.115], we need to prove that Hom B ( DB, τ B I ) =0. Now mod B ∼ = C A / add ( τ ˜ T ) (where A and T are as above) and, underthis equivalence, every injective ˜ C -module is the image of an object of theform τ ˜ T ∈ C A , where ˜ T ∈ add ˜ T . It thus suffices to show that, for every˜ T ∈ add ˜ T , we have Hom C A ( τ ˜ T , τ ˜ T ) = 0. But τ is an equivalence in C A ,hence the result follows from Hom C A ( ˜ T , τ ˜ T ) ∼ = Ext C A ( ˜ T , ˜ T ) = 0.(b) This is the proof of [17], but we include it for completeness. It suffices toprove that, for every B -module M , id M = d < ∞ implies pd M ≤
1. Thus,let 0 / / M / / I f / / I / / · · · f d / / I d / / K i = Im f i for every i . Then the exactsequence 0 → K d − → I d − → I d → K d − ≤ . An easyinduction yields pd M ≤ ✷ It is also shown in [8] that the equivalence C A / add ( τ ˜ T ) ∼ = mod B commuteswith the Auslander-Reiten translations in both categories. Let π denote thecomposition of the functors D π / / / / C A Hom C A ( ˜ T , − ) / / / / mod B, where π is, as above, the canonical projection. We notice that π commuteswith the Auslander-Reiten translations in both categories and also that, if X ∈ D , then π ( X ) = 0 if and only if X ∈ add ( τ D ˜ T ). With the above notations, we deduce the shape of the Auslander-Reiten quiverof C A and mod B . Let Q be the ordinary quiver of A . If A is representation-finite, then the Auslander-Reiten quiver Γ( C A ) is of the form Z Q/ < ϕ > ,where ϕ is the automorphism of Z Q induced by the functor F . Since Γ( C A ) isstable and has sections isomorphic to Q , we say that it is transjective. If, onthe other hand, A is representation-infinite, then Γ( C A ) consists of a uniquecomponent of the form Z Q , which we call transjective because it is the im-age under π of the transjective component of Γ( D ), and also of components10hich we call regular because they are the image under π of the regular com-ponents of Γ( D ). In both cases, we deduce Γ(mod B ) from Γ( C A ) by deletingthe | Q | points corresponding to the summands of τ ˜ T . In particular, Γ(mod ˜ C )always has a unique transjective component, deduced from that of Γ( C A ) uponapplying the functor Hom C A ( ˜ T , − ). Lemma 16
Let Γ be a component of the Auslander-Reiten quiver of a cluster-tilted algebra B . If Γ contains a local slice, then Γ is the transjective compo-nent. PROOF.
Assume that Γ is not transjective. Then Γ is either a stable tube ora component of type ZA ∞ , or is obtained from one of these by deleting finitelymany points. Also, since the functor Hom C A ( ˜ T , − ) : C A → mod B commuteswith the Auslander-Reiten translations, deleting these points will not changethe τ -orbits. Consequently, a local slice Σ in Γ lifts to a unique finite localsection ˜Σ in a regular component ˜Γ of Γ( C A ). In particular, ˜Γ is stable. Letthus ˜ X ∈ ˜Σ and ˜ X = ˜ X → ˜ X → . . . → ˜ X i → . . . be a sectional path ofirreducible morphisms (a ray) starting at X . Then ˜ X or τ ˜ X belongs to ˜Σ.By induction, for each i ≥
0, one of the objects τ i ˜ X i , τ i − ˜ X i , . . . , ˜ X i belongsto ˜Σ. Therefore ˜Σ is infinite, a contradiction. ✷ Lemma 17
Let Σ be a connected full subquiver of the transjective componentof the Auslander-Reiten quiver of a cluster-tilted algebra B arising from ahereditary algebra A and Σ be a connected full subquiver of D = D b (mod A ) such that π | Σ : Σ → Σ is bijective. Then Σ is a local slice in mod B if andonly if Σ is a section in D such that | Σ | = rk K ( A ) . PROOF.
Since both subquivers are full, then the bijection π | Σ induces anisomorphism of quivers. Assume that Σ is a local slice in mod B . We claimthat Σ is a presection in D . Assume that X → Y is an irreducible morphismin D with X ∈ Σ . Then we have two cases to consider:(1) If π ( Y ) = 0, then either π ( Y ) ∈ Σ or τ B π ( Y ) = π ( τ D Y ) ∈ Σ . Therefore Y ∈ Σ or τ D Y ∈ Σ .(2) If π ( Y ) = 0, then π ( τ D Y ) = 0 because Hom D ( Y , τ D Y [1]) = 0. But wehave an arrow X → Y which gives an arrow τ D Y → X , hence an arrow π ( τ D Y ) → π ( X ). Since π ( X ) ∈ Σ , then π ( τ D Y ) ∈ Σ and so τ D Y ∈ Σ .Because of Lemma 4 (c), this shows that Σ is a presection. Since π | Σ is abijection, we have | Σ | = | Σ | = rk K ( A ). By Corollary 10, Σ is a section.11onversely, assume that Σ is a section in D and that | Σ | = rk K ( A ). ByLemma 9, Σ lies in a transjective component of Γ( D ). Note that | Σ | = | Σ | and rk K ( A ) = rk K ( B ). Hence | Σ | = rk K ( B ).We show that Σ is sectionally convex. Let X = X → X → . . . → X t = Y bea sectional path with X, Y ∈ Σ . It lifts to a unique path X = X → X → . . . → X t = Y in D with X, Y ∈ Σ . Since Σ is a section, then this path issectional and all X i ∈ Σ . Applying π , we get that all X i lie in Σ.Finally, we show that Σ is a presection. Assume that X → Y is an irreduciblemorphism in mod B with X ∈ Σ . Let X = π | − ( X ) and choose Y in π − ( Y )such that we have an irreducible morphism X → Y in D . Then Y ∈ Σ or τ D Y ∈ Σ . Hence Y ∈ Σ or τ B Y ∈ Σ . Note that if τ D Y ∈ Σ then π ( τ D Y ) = 0 because π | Σ is a bijection. We treat similarly the case of anirreducible morphism X → Y with Y ∈ Σ . ✷ There is a close relation between tilted and cluster-tilted algebras. First, if B is a cluster-tilted algebra, then there exist a hereditary algebra A and a tilting A -module T such that B ∼ = End C A ( ˜ T ), see [7, 3.3]. Also, if A is a hereditaryalgebra and T is a tilting A -module so that the algebra C = End A ( T ) is tilted,the trivial extension ˜ C = C ⋉ Ext C ( DC, C ) (called the relation-extension of C )is cluster-tilted, and conversely, every cluster-tilted algebra is of this form, see[1]. Now, since ˜ C = C ⋉ Ext C ( DC, C ), then any C -module can be considered asa ˜ C -module under the standard embedding i : mod C → mod ˜ C . Note that, ingeneral, i does not preserve irreducible morphisms. We consider the completeslice Σ = add Hom A ( T, DA ) in mod C , where C = End T A (see, for instance,[3] or [20]) and denote its image as i (Σ) = Σ ′ in mod ˜ C . The following Lemmacollects the important properties of Σ ′ . Lemma 18
Let
Σ = Hom A ( T, DA ) and Σ ′ = i (Σ) , then(a) The image Σ ′ is a local slice in mod ˜ C .(b) i induces an isomorphism of quivers between Σ and Σ ′ .(c) Ann ˜ C Σ ′ ∼ = Ext C ( DC, C ) as C - C -bimodules. PROOF. (a) We have Hom C A ( ˜ T , g DA ) = Hom A ( T, DA ) because T is an A -module. Thus the image i (Σ) = Σ ′ of Σ is equal to the image of Σ under thecomposition mod C j / / D b (mod C ) φ ∼ = / / D b (mod A ) π / / mod ˜ C, j is the inclusion in degree zero and φ the equivalence − ⊗ LC T . Let Σ bea connected full subquiver of D b (mod A ) such that the restriction π | Σ : Σ → Σ ′ is bijective, that is, π (Σ) = Σ ′ = i (Σ). Using the fact that φ is a quasi-inverseof the derived functor R Hom A ( T, − ) : D b (mod A ) → D b (mod C ), we see thatΣ = φ ◦ j (Σ) is equal to DA which is a connected section in a transjectivecomponent of D b (mod A ). Since add ( τ T ) ∩ add ( DA ) = ∅ , then π ( I ) = 0 foreach I ∈ add DA , so that π | add DA : DA → Σ ′ is bijective. By Lemma 17, Σ ′ is a local slice in mod ˜ C .( b ) This follows from the fact that Σ ′ = πφj (Σ) and each one of j | Σ , φ and π | φj (Σ) preserves irreducible morphisms (in the case of π , this is becauseadd DA ∩ add τ T = ∅ ).( c ) By [1], we have an isomorphism Ext C ( DC, C ) ∼ = Hom D ( T, F T ) of C - C -bimodules. NowHom A ( T, DA ) · Hom D ( T, F T ) = Hom D ( T, DA ) · Hom D ( T, F T ) ⊂ Hom D ( T, F ( DA ))= Hom D ( T, A [2]) = 0 , because T is an A -module. Therefore Hom D ( T, F T ) ⊂ Ann ˜ C Σ ′ .Now let f : T → T be a non-zero morphism. Since its image is an A -module,then Hom A (Im f, DA ) = 0. Let g : Im f → DA be a non-zero morphismand denote by p : T → Im f the canonical epimorphism. Since DA is aninjective module, there exists g ′ : T → DA such that g ′ f = gp = 0. ThereforeAnn ˜ C Σ ′ ∩ Hom A ( T, T ) = 0. Since, as a k -vector space, ˜ C = Hom A ( T, T ) ⊕ Hom D ( T, F T ). We have dim ˜ C = dim Hom D ( T, T ) + dim Hom D ( T, F T ) ≤ dim Hom D ( T, T ) + dim Ann ˜ C Σ ′ , because dim Hom D ( T, F T ) ⊂ Ann ˜ C Σ ′ . SinceHom D ( T, F T ) and Ann ˜ C Σ ′ are in direct sum, then we have dim Hom D ( T, T )+dim Ann ˜ C Σ ′ = dim(Hom D ( T, T ) + Ann ˜ C Σ ′ ) ≤ dim ˜ C . Hence dim Ann ˜ C Σ ′ =dim Hom D ( T, F T ). This shows that Hom D ( T, F T ) and Ann ˜ C Σ ′ are equal assubspaces of ˜ C , and the statement follows. ✷ We are now ready for the proof of our main theorem.
Theorem 19
Let B be a cluster-tilted algebra. Then a tilted algebra C is suchthat B = C ⋉ Ext C ( DC, C ) if and only if there exists a local slice Σ in mod B such that C = B/ Ann B Σ . ROOF.
Necessity. It is well-known (see, for instance, [3]) that any com-plete slice Σ in mod C is of the form Σ = add Hom A ( T, DA ) for some hered-itary algebra A and some tilting A -module T . By Lemma 18 (a), Σ em-beds as a local slice in mod B (because B = ˜ C ). Moreover, by Lemma 18(c), we have Ann B Σ ∼ = Ext C ( DC, C ) as C - C -bimodules. Therefore C = B/ Ext C ( DC, C ) = B/ Ann B Σ.Sufficiency. Let B be cluster-tilted, and Σ be a local slice in mod B . Set C = B/ Ann B Σ. Because of the definition of a cluster-tilted algebra, there exist ahereditary algebra A and a tilting object ˜ T ∈ C A such that B = End C A ( ˜ T ). LetΣ be a connected component of the preimage π − (Σ) of Σ in D b (mod A ). Sincethe local slice Σ can only occur in the transjective component of Γ(mod B ), be-cause of Lemma 16, then Σ lies in a transjective component of Γ( D b (mod A )).Since Σ and Σ have the same number of points, then π | Σ : Σ → Σ is bijective,whence Σ is a section in Γ( D b (mod A )) such that | Σ | = rk K ( A ), by Lemma17. We may suppose without loss of generality that Σ = add DA .The fact that π | add DA : add DA → Σ is a bijection implies that the F -orbit τ D π − ( ˜ T ) in D b (mod A ) does not intersect add DA . Therefore, we haveadd π − ( ˜ T ) ∩ add A [1] = ∅ , because τ − D DA = A [1] in D b (mod A ). Thus, wecan choose a representative T in the F -orbit π − ( ˜ T ) such that T is an A -module and π ( T ) = ˜ T . Then T is a tilting A -module. Let C = End A ( T ) bethe corresponding tilted algebra. By [1], we have B = C ⋉ Ext C ( DC , C ). ByLemma 18 (c), we have Ext C ( DC , C ) = Ann B Σ. Thus C = B/ Ann B Σ = C . In particular, C is tilted and verifies B = C ⋉ Ext C ( DC , C ). ✷ Corollary 20
Let C be a tilted algebra and ˜ C be the corresponding cluster-tilted algebra. Then any complete slice in mod C embeds as a local slice in mod ˜ C and any local slice in mod ˜ C arises this way. PROOF.
This follows directly from Lemma 18 and Theorem 19. ✷ Our Theorem 19 actually gives a concrete way to compute the tilted algebra C starting from ˜ C . Given a local slice Σ in mod ˜ C , one computes its annihilatorusing the following result. Corollary 21
Let B be a cluster-tilted algebra and Σ be a local slice in mod B .Then Ann B Σ is generated (as an ideal) by arrows in the quiver of B . (cid:29) (cid:29) <<<<<<< (cid:29) (cid:29) <<<<<<< (cid:28) (cid:28) τ ˜ T (cid:28) (cid:28) B B (cid:6)(cid:6)(cid:6)(cid:6)(cid:6)(cid:6)(cid:6) τ ˜ T (cid:29) (cid:29) <<<<<<< A A (cid:2)(cid:2)(cid:2)(cid:2)(cid:2)(cid:2)(cid:2) / /
42 31 / /
42 3 (cid:29) (cid:29) <<<<<<< A A (cid:2)(cid:2)(cid:2)(cid:2)(cid:2)(cid:2)(cid:2) τ ˜ T / / / / B B (cid:6)(cid:6)(cid:6)(cid:6)(cid:6)(cid:6)(cid:6) (cid:28) (cid:28) τ ˜ T A A (cid:2)(cid:2)(cid:2)(cid:2)(cid:2)(cid:2)(cid:2) A A (cid:2)(cid:2)(cid:2)(cid:2)(cid:2)(cid:2)(cid:2) B B (cid:6)(cid:6)(cid:6)(cid:6)(cid:6)(cid:6)(cid:6) τ ˜ T Fig. 2. Auslander-Reiten quiver of Example 3.8
PROOF.
This follows from [4, 1.3] using that ˜ C is a trivial extension (hencea split extension) of C by the C - C -bimodule Ext C ( DC, C ) ∼ = Ann ˜ C Σ . ✷ We can be a bit more precise. Let ˜ C = k ˜ Q/ ˜ I . Since ˜ C = End C A ˜ T , there is abijection between the points x ∈ ˜ Q and the indecomposable summands ˜ T x of˜ T so that each arrow α : x → y in ˜ Q corresponds to a non-zero morphism f α ∈ Hom C A ( ˜ T y , ˜ T x ) = Hom D ( T y , T x ) ⊕ Hom D ( T y , F T x ). With this notation,Ann ˜ C Σ is generated by all arrows α : x → y such that f α ∈ Hom D ( T y , F T x ) . This indeed follows immediately from the isomorphismsHom D ( T, F T ) ∼ = Ext C ( DC, C ) ∼ = Ann ˜ C Σ . Note that, as shown in [2], the arrows α which generate Ann ˜ C Σ have to satisfycertain conditions. Moreover, if C = kQ/I , then I = ˜ I ∩ kQ . Let ˜ C be the cluster-tilted algebra (of type D ) given by the quiver2 β u u kkkkkkkkkkkkk ǫ / / α i i SSSSSSSSSSSSS γ u u kkkkkkkkkkkkk δ i i SSSSSSSSSSSSS bound by αβ = γδ , βǫ = 0, δǫ = 0, ǫα = 0, ǫγ = 0. The Auslander-Reitenquiver of ˜ C is of the form shown in Figure 2 where indecomposable modulesare represented by their Loewy series, and one identifies the two copies of theunderlined simple module 1 (and also the two copies of the modules 21 and 31 ).The entries τ ˜ T i indicate the position of τ ˜ T i in the cluster category. We find15 β | | yyyyyyyyyyy C α b b EEEEEEEEEEE γ | | yyyyyyyyyyy Σ =
42 31 ,
42 3 , , δ b b EEEEEEEEEEE β { { vvvvvvvvvv C ǫ / / = , , , δ d d HHHHHHHHHH C ǫ / / α d d HHHHHHHHHH γ { { vvvvvvvvvv Σ = , , , Fig. 3. Tilted algebras and corresponding local slices easily all the local slices hence all the tilted subalgebras of ˜ C which realise itas a relation-extension. There are only the three algebras C i shown in Figure3 each corresponding to a local slice Σ i . with the inherited relations in eachcase. In view of our main result, Theorem 19, it is reasonable to ask whether thereexist sufficiently many local slices in the Auslander-Reiten quiver of a cluster-tilted algebra. Since the latter is deduced from the Auslander-Reiten quiverof the cluster category by dropping finitely many points, and since local slicescan only occur in the transjective component, then all but at most finitelymany indecomposables lying in the transjective component of the cluster-tilted algebra belong to a local slice. That not necessarily all indecomposablesin the transjective component belong to a local slice is seen in the followingexample. 16 xample 22
Let A be the hereditary algebra given by the quiver u u kkkkkkkkkkkkk i i SSSSSSSSSSSSS o o and consider the tilting A -module T = 1 ⊕
31 21 ⊕ . Note that while T = 1 and T = 31 21 are projective A -modules, the module T = 31 is simple regularnon-homogeneous. The transjective component of the Auslander-Reiten quiverof ˜ C = End C A ˜ T is thus of the form · (cid:27) (cid:27) M ' ' OOOOOOOOO · ' ' NNNNNNNNNNN (cid:27) (cid:27) · (cid:27) (cid:27) . . . · B B (cid:6)(cid:6)(cid:6)(cid:6)(cid:6) % % JJJJJJJJJ • P @ @ (cid:0)(cid:0)(cid:0)(cid:0)(cid:0) & & NNNNNNNNN · B B (cid:6)(cid:6)(cid:6)(cid:6)(cid:6) % % JJJJJJJJJ · . . . · C C (cid:6)(cid:6)(cid:6)(cid:6)(cid:6) · • P @ @ (cid:1)(cid:1)(cid:1)(cid:1)(cid:1) · C C (cid:6)(cid:6)(cid:6)(cid:6)(cid:6) where P i = Hom C A ( ˜ T , T i ) , for i ∈ { , } , and the two • represent τ T and τ T . It is easily seen that the indecomposable ˜ C -module M = rad P lies on nolocal slice. We say that a cluster-tilted algebra End C A ( ˜ T ) is of tree type if the ordinaryquiver of A is a tree. Theorem 23
Let ˜ C be a cluster-tilted algebra of tree type. Then any inde-composable ˜ C -module lying in the transjective component lies on a local slice. PROOF.
Let ˜ M be an indecomposable ˜ C -module in the transjective compo-nent and M be an indecomposable object in π − ( ˜ M ); here, as in section 3, π denotes the composition of the natural functors D b (mod A ) π / / / / C A Hom C A ( ˜ T , − ) / / / / mod ˜ C .
We claim that M lies on a section Σ in D b (mod A ) such that | Σ | = rk K ( A )and Σ ∩ add τ ˜ T = ∅ .Before proving this claim, we show that the theorem follows from it. Indeed,under these conditions, π | Σ : Σ → π (Σ) is bijective and π (Σ) is a connected17ull subquiver of the transjective component of the Auslander-Reiten quiverof ˜ C . By Lemma 17, π (Σ) is a local slice. Obviously ˜ M = π ( M ) ∈ π (Σ).We now prove the claim. First, we fix some terminology. For any connectedfull subquiver T of Γ( D b (mod A )) such that T is a tree and M ∈ T , there isa unique reduced walk in T from M to any other point N ∈ T . We definethe distance d T ( M, N ) between M and N in T to be the number of arrows ofthis walk. Moreover, we define d T = rk K ( A ) if T ∩ add τ ˜ T = ∅ min { d T ( M, N ) | N ∈ T ∩ add τ ˜ T } if T ∩ add τ ˜ T = ∅ .Thus d T measures the distance between M and the set add τ ˜ T in the tree T . Note that d T ≥ M / ∈ add τ ˜ T and d T = rk K ( A ) if and only if T ∩ add τ ˜ T = ∅ .Now, let Σ be any section in D b (mod A ) containing M . By hypothesis, Σ isa tree. If d Σ = rk K ( A ) we are done. Suppose that d Σ < rk K ( A ). Considerthe set N = { N ∈ Σ ∩ add τ ˜ T | d Σ = d Σ ( M, N ) } and let N ∈ N . Considerthe unique reduced walk M = X X . . . X ( d Σ1 − = L X d Σ1 = N in Σ from M to N . Deleting the edge L N cuts Σ into two subtrees. LetΣ M be the subtree containing M (and L ) and let Σ N the subtree containing N . There are two cases to consider, according to the orientation of the arrowbetween L and N .(1) If L → N , we define Σ to be the full subquiver of Γ( D b (mod A )) havingas points those of Σ M ∪ τ Σ N .(2) If L ← N , we define Σ to be the full subquiver of Γ( D b (mod A )) havingas points those of Σ M ∪ τ − Σ N .By construction, Σ is a connected tree, it lies in the transjective componentand it intersects every τ -orbit exactly once.We now show that Σ is also convex. Assume that L → N (the proof incase L ← N is entirely similar). Then Σ has two subtrees Σ M and τ Σ N ,connected by the arrow τ N → L . We show first that these two subtreesare convex. Suppose that X = X → X → . . . → X s = Y is a path inΓ( D b (mod A ) with X, Y ∈ Σ M . By convexity of Σ , we have X i ∈ Σ for all i . Since there is exactly one walk from X to Y in Σ we actually have X i inΣ M for all i . Thus Σ M is convex. Similarly, τ Σ N is also convex. Now supposethat X = X → X → . . . → X s = Y is a path with X, Y ∈ Σ . If X ∈ Σ M ,then Y ∈ Σ M because of the structure of the transjective component, and18hen the convexity of Σ M implies the result. If X, Y ∈ τ Σ N we are done byconvexity of τ Σ N . Hence the only remaining case is when X ∈ τ Σ N and Y ∈ Σ M . Suppose that there is an i ∈ { , . . . , s − } such that X i / ∈ Σ . Wemay suppose, without loss of generality, that i = 1. Now τ Σ is a section suchthat | ( τ Σ ) | = rk K ( A ), hence, in particular, it is a presection. Therefore X ∈ τ Σ , X → X and X / ∈ τ Σ imply τ X ∈ τ Σ , hence X ∈ Σ . Thenall X i with 1 ≤ i ≤ s lie in Σ because Σ is convex. Since X / ∈ Σ , we have X ∈ Σ N . Then, since Y ∈ Σ M and the subtrees Σ N , Σ M are only joined bythe arrow L → N , there exists j such that X j = N and X j +1 = L . But thenthere is an arrow L ← N , a contradiction. This shows that Σ is convex andhence is a section in D b (mod A ) satisfying | (Σ ) | = rk K ( A ) . We repeat this construction for every element N of N . Note that, since N ∈ add τ ˜ T , neither τ N nor τ − N are in add τ ˜ T . In this way, we obtain, after k = |N | steps, a section Σ k +1 in Γ( D b (mod A )) that contains M , such that | (Σ k +1 ) | = rk K ( A ) and d Σ k +1 > d Σ . If d Σ k +1 = rk K ( A ) then Σ k +1 ∩ add ( τ ˜ T ) = ∅ and we are done. Otherwise,we repeat the construction with the set N k +1 = { N ∈ Σ k +1 | d Σ k +1 ( M, N ) = d Σ k +1 } . This algorithm will find the required section Σ in a finite number of steps. ✷ Corollary 24
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