Faithful actions of braid groups by twists along ADE-configurations of spherical objects
aa r X i v : . [ m a t h . K T ] S e p Faithful actions of braid groups by twists along ADE-configurations ofspherical objects
ANYA NORDSKOVA
L. Euler International Mathematical Institute, Saint Petersburg State University, 14thLine V.O., 29B, St. Petersburg 199178, Russia.
YURY VOLKOV
Department of Mathematics and Computer Science, Saint Petersburg State University,Universitetskaya nab. 7-9, St. Peterburg, Russia.
Abstract.
We prove that the actions of the generalized braid groups on enhanced tri-angulated categories, generated by spherical twist functors along ADE-configurations of ω -spherical objects, are faithful for any ω = 1. Introduction
Spherical twists along spherical objects are a prominent type of autoequivalences of tri-angulated categories. The notion of a spherical twist functor was first introduced in [18] byP. Seidel and R. Thomas in connection with the Kontsevich’s homological mirror symmetryprogram. Their original motivation was to look at the autoequivalences of the derived cate-gory of coherent sheaves on a variety that would arise as counterparts of generalized Dehntwists via mirror symmetry. In a general setting, a spherical twist is a functor constructedin a particular way from a spherical object, an object of a triangulated category whose Extalgebra is the same as the cohomology of a sphere.The theory of Seidel and Thomas received a lot of attention and developed rapidly innumerous works of other mathematicians. For instance, in [1] the notion of a spherical twistalong a spherical functor was introduced, generalizing spherical twists along spherical ob-jects. In [17], it was established that any triangulated autoequivalence is in fact a twist alongsome spherical functor. The theory of spherical twists constructed from spherical sequenceswas developed in [6]. Groups that can be generated by two spherical twists constructed fromspherical sequences were described in [19]. Since the appearance of [18] in 2001, sphericaltwists have proved to be a useful tool in algebraic geometry and beyond. Among their appli-cations are categorifications of Braid groups [14], Bridgeland stability conditions manifolds[10] and derived Picard groups [20].Let Γ be a simply-laced Dynkin diagram. Seidel and Thomas showed that spherical twistsalong a so-called Γ-configuration of ω -spherical objects satisfy braid relations of type Γmodulo natural isomorphisms, hence induce an action of an Artin group (generalized braidgroup) B Γ on the triangulated category in question. In the same paper they showed that for E-mail addresses : [email protected], wolf86 666@list . ORCID : 0000-0003-3592-4273 (first author), 0000-0003-0704-7912 (second author) .The work was supported by RFBR according to the research project 18-31-20004 and by the President’sProgram “Support of Young Russian Scientists” (project number MK-2262.2019.1). The first author was alsosupported by “Native Towns”, a social investment program of PJSC “Gazprom neft”. The second authorwas in part supported by Young Russian Mathematics award.
Key words:
Artin groups, spherical twists, triangulated categories, derived Picard groups primary 18E30, 20F36, secondary 16E35, 14F05 ≥ A n this action is faithful. By [6], their result can be also extended to the case ω = 0. The paper [18] also provides an example when the action is not faithful for ω = 1.Later several more results on the faithfulness have appeared. In [2], C. Brav and H. Thomasproved that the braid group action is faithful for ω = 2 and all Γ. Recently Y. Qiu and J.Woolf generalised their result to ω ≥ ω ≥ A n , D n , E , E , E and all ω = 1, including the case ω < ω − spherical objects withnegative ω , but they are also worth considering (for instance, see [9], [3], [4], [5]), despitebeing somewhat more exotic than those of non-negative CY dimension.The proofs we present for ω ≥ ω ≤ ω = 2 by Brav and Thomas [2], which then allows us to extendit to arbitrary ω ≥
2. After some preparation, this takes us only about two pages, and theproof we obtain for the case ω = 2 in particular is shorter and more elementary than inthe original exposition of [2], in particular, it does not rely on the existence of the Garsidestructure. On the other hand, proving the faithfulness in the case ω ≤ ω = 0,which we originally aimed for, having a particular application to representation theory inmind, happens to be the hardest to tackle.The structure of the paper is as follows. Section 2 introduces basic notions crucial in thepaper, i.e. we define spherical objects, spherical twist functors, configurations of sphericalobjects, etc. Section 3 contains the main result of the paper as well as the the key lemmaon which its proof is based on and some useful technical observations. At the end of thissection it is also explained how the key lemma implies the main result. The rest of thepaper is mainly devoted to proving the key lemma. Section 4 provides the proof in the case ω ≥
2. In Section 5 we outline the proof of the key lemma in the case ω ≤ ω ≤
0, the factorization. Section 8 contains the secondstep, to which we refer as ”braiding”. This finishes the proof of the key lemma and hencethe main theorem of the paper. The statement obtained in Section 8 deals with words inthe generalized braid monoids and might be of interest not only in connection with braidgroup actions on triangulated categories. In the last section one application of the mainresult of this paper is presented. Namely, we show how it may be used to make the firstyet crucial step towards the description of the derived Picard groups of representation-finiteselfinjective algebras.
Acknowledgements.
We would like to thank Alexandra Zvonareva for inspiring discus-sions and useful remarks. 2.
Preliminaries
Throughout this paper D is a triangulated category linear over a field k and with a fixedenhancement. For example, it can be an algebraic triangulated category in the sense of Keller(see [12]), i.e. the stable category of some Frobenius category. In this case we are equippedwith functorial cones of natural transformations of exact functors, and for every object X of D there is the derived Hom-complex functor RHom( X, − ) : D → D (k) and its right adjoint − ⊗ X : D (k) → D , where D (k) denotes the unbounded derived category of k-vector spaces. et Hom k ( A, B ) denote Hom D ( A, B [ k ]) and Hom ∗ ( A, B ) = L k ∈ Z Hom k ( A, B ). Forelements g ∈ Hom k ( A, B ) and f ∈ Hom l ( B, C ), we will write f g for the element f [ k ] g ∈ Hom k + l ( A, C ). Definition 2.1. (Seidel, Thomas [18]) Let ω ∈ Z . An object P ∈ D is called ω -spherical if(i) dim k Hom ∗ ( X, P ) < ∞ for any object X ∈ D .(ii) Hom ∗ ( P, P ) ∼ = k [ t ] / ( t ) as graded k − algebras, where deg ( t ) = ω .(iii) Hom ∗ ( P, X ) × Hom ∗ ( X, P ) ◦ −→ Hom ∗ ( P, P ) / h Id P i ∼ = kis a perfect pairing for any X ∈ D (defined by the composition).Fix some undirected graph Γ. We will assume that Γ is an ADE Dynkin diagram, butpart of our arguments can be transferred to more general cases. We represent the set Γ ofvertices of Γ as a disjoint union of sets V and V in such a way that each edge of Γ has oneendpoint in V and the other one in V . By N ( A ) for A ⊆ Γ we denote the set all neighborsof all vertices in A (in other words, N ( A ) is formed by j ∈ Γ such that s j s k = s k s j forsome k ∈ A ).Following Brav and Thomas ([2]), we now define a Γ-configuration of spherical objects. Definition 2.2.
A collection of ω -spherical objects { P i } i ∈ Γ enumerated by vertices of Γ isa Γ − configuration if for any i = j
1) Hom ∗ ( P i , P j ) is one-dimensional if i ∈ N ( j );2) Hom ∗ ( P i , P j ) = 0 if i N ( j ).Fix some integers ω and ω such that ω + ω = ω . To simplify our proofs we willassume that ω , ω ≤ ω ≤ ω , ω ≥ ω ≥
2. For example, we can simply take ω = ω = ω if ω is even and ω = ω +12 , ω = ω − if ω is odd. It follows from the definitionof ω -spherical object that after shifting the objects P i forming a Γ-configuration, one mayassume that Hom ∗ ( P i , P j ) is concentrated in degree ω u , where u is such that i belongs to V u . The index u in the notations ω u and V u will be always taken modulo 2. Definition 2.3. (Seidel, Thomas [18]) Let P be an ω -spherical object. The spherical twist functor t P along P is defined by t P ( X ) = cone ( P ⊗ RHom(
P, X ) counit −−−−→ X ) Remark. t P is indeed a functor since we have functorial cones of natural transformations ofexact functors. Definition 2.4.
The Artin group (generalized braid group) B Γ of type Γ is generated by s i , i ∈ Γ subject to the braid relations s i s j s i = s j s i s j for i, j adjacent in Γ and s i s j = s j s i for i, j not adjacent in Γ. The braid monoid B +Γ is a monoid given by the same generatorsand relations.We are now going to recall some crucial facts about spherical twists and spherical objects(see [18]) that we will actively use throughout the paper.1) If P is spherical, then t P is an autoequivalence of D with a quasi inverse t ′ P definedby t ′ P ( X ) = cone ( X unit −−−→ P ⊗ RHom(
P, X ) ∗ )[ − , where ∗ is the usual duality on the category D (k).2) If P is ω -spherical, then t P ( P ) = P [1 − ω ].3) For { P i } i ∈ Γ forming a Γ-configuration, the functors t P i satisfy braid relations oftype Γ up to a natural isomorphism. In particular, for spherical P, Q not adjacentin their Γ-configuration (equivalently Hom ∗ ( P, Q ) = 0), t P ( Q ) = Q . In other words,there is a group homomorphism F : B Γ −→ Aut( D )where Aut( D ) is a group of autoequivalences of D modulo natural isomorphisms.For α ∈ B Γ , we denote F ( α ) by t α . . Main result
We have just defined an action of the braid group B Γ of type Γ on the category D . Themain result of this paper says that this action is faithful for every ω = 1. The rest of thispaper is mainly devoted to proving this claim: Theorem 1. If Γ is a simply-laced Dynkin diagram, ω = 1 and { P i } i ∈ Γ is a Γ − configuration of ω − spherical objects in an enhanced triangulated category D , then theaction of B Γ on D generated by the spherical twists t P i is faithful. As we have already mentioned in the Introduction, the action for ω = 1 can be not faithful(see [18]).Since the case | Γ | = 1 is clear, we will assume hereafter that Γ has at least two vertices.We will use the following auxiliary notation in our proof. For each i ∈ V u and j ∈ N ( i ), wefix some generator of Hom ω u ( P i , P j ) and denote it by γ i,j . We also introduce the morphisms ρ i,j : P j → t i P j and ξ i,j : t i P j → P i [1 − ω u ] via the triangle P i [ − ω u ] γ i,j [ − ω u ] −−−−−−→ P j ρ i,j −−→ t i P j ξ i,j −−→ P i [1 − ω u ] . We follow the same general strategy to prove Theorem 1 in the two cases ω ≤ ω ≥
2. Nevertheless, the proof of the case ω ≥ ω = 2 byBrav and Thomas [2] and then adapt it for ω ≥
2. The proof we obtained for ω = 2 basingon the ideas of [2] not only allows us to then extend it to arbitrary ω ≥
2, but also hasthe advantage of being shorter than the original version of Brav and Thomas, as we havealready mentioned in the Introduction.Let Λ = n L i =1 P i . We will also write t i instead of t P i for brevity.We define the minimal and the maximal nonzero degree of an object in D : Definition 3.1.
Let T ∈ D . We say that min ( T ) (respectively max ( T )) is the minimal (respectively the maximal ) nonzero degree of T if Hom k + ω (Λ , T ) = 0 for k ≤ min ( T ) − k ≥ max ( T ) + 1) and Hom min ( T )+ ω (Λ , T ) (respectively Hom max ( T )+ ω (Λ , T ))is nonzero. Definition 3.2.
Let T be an object of D . We say that P j with j ∈ Γ is a direct summandof T r if there exists a nonzero f ∈ Hom r + ω ( P j , T ) such that f γ k,j = 0 for any k ∈ N ( j ). Amorphism f satisfying this condition will be referred to as long . In other words, a nonzeromorphism f : P j → T is called long if the induced morphism Hom ∗ ( P k , f ) : Hom ∗ ( P k , P j ) → Hom ∗ ( P k , T ) is zero for any k = j . We also say that P j is a direct summand of T [ a,b ] if P j is a direct summand of T c for some c ∈ [ a, b ]. Remark.
Let T ∈ D and i ∈ V u . If Hom r + ω ( P i , T ) = 0 for some r ≤ min ( T ) − ω u +1 −
1, then P i is a direct summand of T r . Indeed, any composition of the form P j → P i [ ω u +1 ] → T [ r + ω + ω u +1 ] is zero by the definition of the minimal nonzero degree.Moreover, if Hom min ( T )+ ω ( P j , T ) = 0 for every j ∈ N ( i ) and Hom min ( T )+ ω − ω u +1 ( P i , T ) = 0,then P i is a direct summand of T min ( T ) − ω u +1 , because in this case any composition of theform P j → P i [ ω u +1 ] → T [ min ( T ) + ω ] is zero too. Analogously, if Hom r + ω ( P i , T ) = 0 forsome i ∈ V u and r ≥ max ( T ) − ω u +1 + 1, then P i is a direct summand of T r .We denote t α (Λ) by T α . Let min ( α ) and max ( α ) denote the minimal and the maximalnonzero degrees of T α respectively. Following [2], we will deduce the faithfulness of the braidgroup action from the injectivity of the induced monoid homomorphism B +Γ −→ Aut( D ). Inturn, to prove the injectivity of the aforementioned monoid homomorphism, we require atool that would allow us to find a leftmost factor of the reduced expression of α ∈ B +Γ usingonly information about T α . This tool is presented in the following key lemma. Lemma 1.
Let α ∈ B +Γ , α = 1 . For u ∈ { , } let I u,α = [ min ( α ) , min ( α ) − ω u +1 ] inthe case ω ≤ and I u,α = [ max ( α ) − ω u +1 + 1 , max ( α )] in the case ω ≥ . Then for any ∈ { , } and any j ∈ V u such that the corresponding object P j is a direct summand of ( T α ) I u,α , the word α can be written as α = s j α ′ for some α ′ ∈ B +Γ with l ( α ) = l ( α ′ ) + 1 . Before proving Lemma 1 we are going to show how it easily implies our main result.However, first we are going to introduce and prove some rather technical facts regardingthe way spherical twists affect the minimal and the maximal degree of an object. Thesestatements will be required in our considerations throughout the paper. In this technicalpart we will consider the cases ω ≤ ω ≥ ω ≥ Lemma 2.
Let ω ≤ and T be an object of D . Let m be the minimal nonzero degree of T .1) If Hom r ( P i , t − i T ) = 0 , then m ≤ r − .2) For k ∈ V u , r ≤ m − ω u +1 and i = k , P k is a direct summand of ( t − i T ) r if and onlyif P k is a direct summand of T r .3) The minimal nonzero degree of t i T belongs to [ m − ω, m ] and P k can be a directsummand of T r with r < m only if k = i .Proof.
1) As spherical twists are autoequivalences,0 = Hom r ( P i , t − i T ) ∼ = Hom r ( t i P i , T ) = Hom r ( P i [1 − ω ] , T ) ∼ = Hom r − ω ( P i , T ) . Hence the minimal nonzero degree m of T is not greater than r − k / ∈ N ( i ). Since t i is an autoequivalenvce and t i P k = P k in this case,we have an isomorphism t i : Hom r + ω ( P k , t − i T ) ∼ = Hom r + ω ( P k , T ). It is sufficient toshow that f : P k → t − i T [ r + ω ] is long if and only if t i f is long. Pick some l ∈ N ( k ).If l N ( i ), then t i P l = P l and we have Im Hom ∗ ( P l , t i f ) = t i Im Hom ∗ ( P l , f ). Since t i is an autoequivalence, we have Hom ∗ ( P l , t i f ) = 0 if and only if Hom ∗ ( P l , f ) = 0.If l ∈ N ( i ), then i ∈ V u and we have a triangle P l ρ i,l −−→ t i P l ξ i,l −−→ P i [1 − ω u ] . Since Hom ∗ ( P i , P k ) = 0, we have γ l,k = gρ i,l for some g ∈ Hom ω u +1 ( t i P l , P k ). Hence,if ( t i f ) γ l,k = 0, then f ( t − i g ) = 0, and Hom ∗ ( P l , f ) = 0. On the other hand, if f γ l,k = 0, then the composition P l ρ i,l −−→ t i P l t i ( fγ l,k ) −−−−−→ T [ r + ω + ω u +1 ]is nonzero, because r + ω − ≤ m + ω u − < m , and henceHom r + ω + ω u +1 ( P i [1 − ω u ] , T ) ∼ = Hom r +2 ω − ( P i , T ) = 0 . Then we have Hom ∗ ( P l , t i f ) = 0. Thus, f is long if and only if t i f is long.Suppose now that k ∈ N ( i ). Consider the triangle(1) P k ρ i,k −−→ t i P k ξ i,k −−→ P i [1 − ω u +1 ]For any nonzero f : P k → t − i T [ r + ω ], one has t i f ρ i,k = 0, because r + ω u +1 − ≤ m −
1, the minimal nonzero degree of T is m , and henceHom r + ω ( P i [1 − ω u +1 ] , T ) = 0. Since ( t i f ) ρ i,k γ i,k [ − ω u +1 ] = 0, it remains to showthat ( t i f ) ρ i,k γ l,k [ − ω u +1 ] = 0 for l ∈ N ( k ) \ { i } if f is long. Indeed, suppose it isnonzero and apply t − i . Since l / ∈ N ( i ), we get a nonzero morphism P l [ − ω u +1 ] −→ P k f −→ t − i T [ r + ω ], which contradicts f being a long morphism.Now pick a long morphism f ′ : P k → T [ r + ω ]. Then f ′ γ i,k [ − ω u +1 ] = 0 by thedefinition of a long morphism, and hence f ′ = ( t i f ) ρ i,k for some f : P k → t − i T [ r + ω ].We have 0 = f γ i,k : P i → t − i T [ r + ω + ω u +1 ], because otherwise the minimal nonzerodegree of T would be not greater than r + ω + ω u +1 − ≤ m + ω − < m by the firstassertion of the current lemma. Pick some l ∈ N ( k ) \ { i } . Since Hom ∗ ( P l , P i ) = 0,the morphism t i γ l,k : P l → t i P k [ ω u +1 ] factors through ρ i,k [ ω u +1 ], and hence equals i,k γ l,k modulo a nonzero scalar factor. Since ( t i f ) ρ i,k is long, one has ( t i f ) ρ i,k γ l,k =0. Applying t − i , we see that f γ l,k = 0 as well. Thus, there exists a long morphismfrom P k to T [ r + ω ] if and only if there exists a long morphism from P k to t − i T [ r + ω ].3) First we show that the minimal nonzero degree of t i T is not greater than m . Thereexists some k ∈ Γ such that P k is a direct summand of T m . If k = i and theminimal nonzero degree of t i T is greater than m , then P k is a direct summand of( t i T ) m by the second assertion of this lemma and we get a contradiction. If k = i ,then Hom m + ω ( P i , T ) = 0 and the minimal nonzero degree of t i T is not greater than m + ω − < m by the first assertion of this lemma. Thus, the minimal nonzero degreeof t i T is not greater than m .Now suppose P k is a direct summand of ( t i T ) r for r < m , k = i . If k / ∈ N ( i ), then0 = Hom r + ω ( P k , t i T ) ∼ = Hom r + ω ( P k , T ), which contradicts the minimal nonzerodegree of T being m . If k ∈ N ( i ), then there exists a long morphism P k f −→ t i T [ r + ω ],in particular, f γ i,k = 0. Consider the triangle P i [ ω u − t − i γ i,k [ − ω u +1 ] −−−−−−−−−−→ t − i P k t − i ρ i,k −−−−→ P k . Since t − i ( f γ i,k [ − ω u +1 ]) = 0, the morphism t − i f factors through P k and the minimalnonzero degree of T is not greater than r < m which is impossible.Now it follows from what we have already established that if the minimal nonzerodegree of t i T equals d < m , then Hom d + ω ( P i , t i T ) = 0. Applying t − i , we getHom d +1 ( P i , T ) = 0, and hence d ≥ m − ω . (cid:3) Lemma 3.
Let ω ≥ , T be an object of D and h be its maximal nonzero degree.1) If Hom r ( P i , t − i T ) = 0 , then h ≥ r − .2) Suppose that k ∈ Γ \ (cid:0) { i } ∪ N ( i ) (cid:1) . Then Hom r + ω ( P k , t − i T ) = 0 if and only if Hom r + ω ( P k , T ) = 0 .3) Suppose that r ≥ h − ω u +1 + 2 and k ∈ V u , k = i . If Hom r + ω ( P k , t − i T ) = 0 , then Hom r + ω ( P k , T ) = 0 .4) Suppose that r ≥ h − ω u +1 + 1 and k ∈ V u , k = i . If Hom r + ω ( P k , T ) = 0 , then Hom r + ω ( P k , t − i T ) = 0 .5) The maximal nonzero degree of t i T is not less than h .Proof.
1) Just as in Lemma 2, we get 0 = Hom r − ω ( P i , T ), and hence h ≥ r − k
6∈ { i } ∪ N ( i ) one hasHom r + ω ( P k , t − i T ) = Hom r + ω ( t − i P k , t − i T ) ∼ = Hom r + ω ( P k , T ) .
3) The case k N ( i ) is already considered. Suppose now that k ∈ N ( i ).Consider the triangle (1). For any nonzero f : P k → t − i T [ r + ω ], one has0 = ( t i f ) ρ i,k : P k → T [ r + ω ], because r + ω u +1 − ≥ h + 1. The required assertionnow follows.4) Due to the assertions already proven, it remains to consider the case k ∈ N ( i ).Consider the triangle (1) and pick a nonzero morphism f ′ : P k → T [ r + ω ]. Since r + ω u +1 ≥ h + 1, one has f ′ γ i,k [ − ω u +1 ] = 0, and hence f ′ = ( t i f ) ρ i,k for somenonzero f : P k → t − i T [ r + ω ]. This finishes the proof.5) There exists some k ∈ Γ such that Hom h + ω ( P k , T ) = 0. If k = i and the maximalnonzero degree of t i T is less than h , then Hom h + ω ( P k , t i T ) = 0 by the previous asser-tions of this lemma, hence we get a contradiction. If k = i , then Hom h + ω ( P i , T ) = 0and the maximal nonzero degree of t i T is not less than h + ω − > h by the firstassertion of this lemma. Thus, the required assertion is valid in all cases. (cid:3) Now we are ready to deduce Theorem 1 from Lemma 1.
Proof of Theorem 1.
According to [2, Proposition 2.3], a group homomorphism B Γ −→ G is injective if and only if the induced monoid homomorphism B +Γ ֒ → B −→ G is injective. ence, in our case it is sufficient to show that B +Γ −→ Aut( D ) is injective. Assume that it isnot. Choose two words α , β with the smallest sum of lengths l ( α ) + l ( β ) among all pairs ofwords with coinciding images in Aut( D ) and α = β . In particular, t α (Λ) = T α ∼ = T β = t β (Λ)in D . Thus, for any u ∈ { , } , I u,α = I u,β and P i ( i ∈ V u ) is a direct summand of ( T α ) I u,α if and only if it is a direct summand of ( T β ) I u,β . First assume that one of α and β is 1, say α . Then min ( α ) = 0 if ω ≤ max ( α ) = 0 if ω ≥
2. But since β = 1, we have l ( β ) ≥ min ( β ) < ω ≤ max ( β ) > ω ≥
2. Thus wemay assume that α = 1, β = 1.By Lemma 1, there exists i ∈ Γ such that α = s i α ′ and β = s i β ′ . Obviously, the images of α ′ and β ′ also coincide in Aut( D ), and since l ( α ′ ) + l ( β ′ ) = l ( α ) + l ( β ) − < l ( α ) + l ( β ) we get α ′ = β ′ . But then α = s i α ′ = s i β ′ = β , which contradictsthe assumption that α = β . (cid:3) The proof of Lemma 1 for ω ≥ ω ≥ max ( α ) by h α . Observe that if ω ≥
2, then P j with j ∈ V u is a direct summand of ( T α ) [ h α − ω u +1 +1 ,h α ] if and only if Hom r + ω ( P j , T α ) = 0 for some r ∈ [ h α − ω u +1 + 1 , h α ], due to the remark rightafter the definition of a direct summand. In fact, we used the notion of a direct summand inthe form it was first provided only to unify the two cases ω ≤ ω ≥ ω ≥ Lemma 4. If Hom r + ω ( P j , T α ) = 0 for some r ∈ [ h α − ω u +1 + 1 , h α ] , then α is left-divisibleby s j .Proof. We argue by contradiction. Take α ∈ B +Γ not satisfying the required condition andof minimal length. It is clear that l ( α ) >
0, and hence α can be presented as α = s i β forsome i ∈ Γ and β ∈ B +Γ with l ( β ) < l ( α ). In particular, the statement of the lemma holdsfor β and all its right factors. Without loss of generality we may assume that i ∈ V .Since the assertion of the lemma fails for α , there exists some j ∈ V u ( u = 0 ,
1) such thatHom h + ω ( P j , T α ) = 0 for some h ∈ [ h α − ω u +1 + 1 , h α ], but α is not left-divisible by s j . It isclear that j = i . We may assume without loss of generality that if Hom r + ω ( P k , T α ) = 0 forsome k ∈ V u and r > h , then α is left-divisible by s k .1) First observe that β = s j γ for some γ ∈ B +Γ with l ( γ ) = l ( α ) −
2. Indeed, we haveHom h + ω ( P j , T β ) = 0 and h ∈ [ h β − ω u +1 + 1 , h β ] by Lemma 3. Then β is left-divisibleby s j because the assertion of the lemma holds for β .2) Now we show that j ∈ N ( i ) ⊂ V and, in particular, h ∈ [ h α − ω + 1 , h α ]. Indeed,this easily follows from the previous claim, because in the case j N ( i ) one has α = s i s j γ = s j s i γ which contradicts the choice of j .3) Note thatHom h − ω +1+ ω ( P i , T γ ) ∼ = Hom h + ω ( t i t j P i [ ω − , t i t j T γ ) = Hom h + ω ( P j , T α ) = 0 .
4) The next step is to establish that Hom r + ω ( P k , T γ )=0 for any k ∈ V and r > h .Suppose for a contradiction that Hom r + ω ( P k , T γ ) =0 for some k ∈ V and r > h .According to Lemma 3 we have h γ ≤ h β ≤ h α and r ≥ h + 1 ≥ h α − ω + 2. ThenHom r + ω ( P k , T α ) = 0, also by Lemma 3. Thus s k divides α on the left by the choiceof h , and hence we have α = s k α ′ for some α ′ ∈ B +Γ with l ( α ′ ) = l ( α ) −
1. Note thatLemma 3 again implies that Hom h + ω ( P j , T α ′ ) = 0 and h ∈ [ h α ′ − ω + 1 , h α ′ ]. Sincethe assertion of Lemma 1 is valid for α ′ , s j divides α ′ on the left. Since s j and s k commute, s j also divides α on the left which contradicts the choice of j . ) Let θ = γ and ∆ = ∅ . We repeat the following procedure. If at some moment h θ p is not greater than h , we define θ = θ p and ∆ = ∆ p . If h θ p is greater than h , we pick some k such that Hom h θp + ω ( P k , T θ p ) = 0 and define θ p +1 = s − k θ p and∆ p +1 = ∆ p ∪ { k } . We have θ p +1 ∈ B +Γ and l ( θ p +1 ) = l ( θ p ) −
1, because the assertionof the lemma holds for θ p . Now recall that Hom r + ω ( P l , T θ ) = 0 for any r > h and l ∈ V by 4). Since h ≥ h α − ω + 1 ≥ h θ p − ω + 1 for any p ≥ r + ω ( P l , T θ p ) = 0 for any r > h , l ∈ V and p ≥ p . Therefore ∆ p ⊂ V for any p ≥
0. Note also that k ∈ ∆ p , θ p = θ impliesHom h θp + ω ( P k , T θ p ) ∼ = Hom h θp + ω Y l ∈ ∆ p t l P k , Y l ∈ ∆ p t l T θ p = Hom h θp +2 ω − ( P k , T θ ) = 0 , because h θ p + ω − > h + ω − ≥ h α + ω > h α ≥ h θ . Hence whenever the maximalnonzero degree of T θ p is greater than h , we obtain ∆ p +1 by adding a new element to∆ p . Thus, this process is bound to terminate and as a result we get the factorization γ = (cid:18) Q k ∈ ∆ s k (cid:19) θ with ∆ ⊂ V and θ ∈ B +Γ such that l ( θ ) = l ( γ ) − | ∆ | and h θ ≤ h .6) If i ∈ ∆, then s i divides γ on the left. In this case s i s j s i = s j s i s j divides α on theleft, and hence so does s j which contradicts the choice of j .7) If i ∆, then Hom h − ω +1+ ω ( P i , T θ ) = 0 by Lemma 3. Since the assertion of thelemma is true for θ , s i divides θ on the left. Then γ is again left-divisible by s i andwe get a contradiction just as before. (cid:3) Remark.
Note that in the case ω = 2 the number h appearing in the proof is automaticallyequal to h α . This allows to omit the steps 4–6 in the proof and make it even shorter.5. Outline of the proof of Lemma 1 for ω ≤ ω ≤
0. Untilthe end of the paper we will assume that we are in this situation though some statementswill be true for any integer ω . From here on we denote min ( α ) by m α .Similarly to the case ω ≥
2, we argue by contradiction, and the beginning of this proof isthe same, i.e. we again pick α ∈ B +Γ not satisfying the assertion of Lemma 1 and of minimallength.As before, α can be presented as α = s i β for some i ∈ Γ and β ∈ B +Γ with l ( β ) < l ( α )and the statement of Lemma 1 holds for β and all its right factors. Without loss of generalitywe may assume that i ∈ V . Since Lemma 1 fails for α , there exists some j ∈ V u ( u = 0 , P j is a direct summand of ( T α ) [ m α ,m α − ω u +1 ] while α is not divisible by s j on theleft. It is clear that j = i . Let m ∈ [ m α , m α − ω u +1 ] be the minimal degree such that P j is adirect summand of ( T α ) m . We may assume without loss of generality that if r < m and P k with k ∈ V u is a direct summand of ( T α ) r , then α is divisible by s k on the left. Lemma 5. j ∈ N ( i ) ⊂ V .2) If P k with k ∈ V is a direct summand of ( T β ) r with r ≤ m , then r = m and k ∈ N ( i ) .3) If P k is a direct summand of ( T β ) m for some k ∈ V \ { j } , then s l does not divide α on the left for any l = i .4) Hom r ( P i , T β ) = 0 for r ≤ m + ω . Proof.
1) Suppose that j / ∈ N ( i ). It follows from Lemma 2 that P j is a direct summandof ( T β ) m and m ∈ [ m β , m β − ω u +1 ]. Since the statement of Lemma 1 holds for β , wehave α = s i β = s i s j β ′ = s j s i β ′ for some β ′ ∈ B +Γ with l ( β ′ ) = l ( α ) −
2. Thus, j doesnot produce a contradiction to the statement of Lemma 1, which in turn contradictsthe choice of j . Thus, j ∈ N ( i ) ⊂ V and, in particular, we have m ∈ [ m α , m α − ω ]. ) Pick some k ∈ V such that P k is a direct summand of ( T β ) r with r ≤ m . Then P k is a direct summand of ( T α ) r by Lemma 2. If r < m , then s k divides α on the leftby the definition of m . If r = m but k N ( i ), then s k divides α on the left by theargument from the proof of the first issue. In any case, we have α = s k α ′ for some α ′ ∈ B +Γ with l ( α ′ ) = l ( α ) −
1. Note that Lemma 2 implies again that P j is a directsummand of ( T α ′ ) m and m ∈ [ m α ′ , m α ′ − ω ]. Since the assertion of Lemma 1 is validfor α ′ , s j divides α ′ on the left. Since s j and s k commute, s j divides also α on theleft that contradicts the choice of j .3) Suppose that s l divides α on the left for some l = i . Then s l commutes with at leastone of the elements s j and s k . The argument from the proof of the second item showsthat s j or s k divides α on the left. Since s k and s j commute, the same argumentshows that s j divides α on the left in any case. This contradicts the choice of j .4) Suppose that Hom r ( P i , T β ) = 0 for some r ≤ m + ω . By Lemma 2 one has m α ≤ r − ≤ m + ω −
1. But in this case m [ m α , m α − ω ], which yields contradiction. (cid:3) Set s A = Q k ∈ A s k , t A = Q k ∈ A t k for any A ⊆ V u with some u ∈ { , } . Define σ u for u ∈ Z by σ = m and σ u +1 = σ u + 1 − ω u +1 . In other words, σ u = m + u (2 − ω ), σ u +1 = m + u (2 − ω ) + 1 − ω u +1 . Let∆ − = ∅ , ∆ = { i } , ∆ = { j ∈ V : P j is a direct summand of ( T β ) m } . If | ∆ | ≥
2, then α cannot be written as α = s k α ′ with l ( α ′ ) = l ( α ) − k ∈ Γ \ { i } byLemma 5. Hence, we need to show that either | ∆ | = 1 and α is left-divisible by s j where j is the unique element of ∆ or | ∆ | ≥ α is left-divisible by s k with some k = i . Withthis end in view, we will employ the following scheme:Step I: Factorization.
First we are going to construct a presentation of α of a particularform.We start with the presentation α = s ∆ s ∆ β = s i s ∆ β obtained earlier. Set χ ( i ) = 1. We continue the process inductively and obtain a presentation of the form α = s ∆ s ∆ . . . s ∆ q s y e β satisfying the following conditions: for 1 ≤ u ≤ q :(1) α = s ∆ . . . s ∆ u β u for some β u ∈ B +Γ with l ( β u ) = l ( α ) − u P v =0 | ∆ v | (2) ∆ u − ⊆ ∆ u ⊆ N (∆ u − ).(3) P l is not a direct summand of ( T β u ) [ σ u − +1 ,σ u − ] for any l ∈ V u .(4) The minimal nonzero degree of T β u is not smaller than σ u − + 1.(5) For any l ∈ ∆ u , P l is a direct summand of ( t l T β u ) [ σ u − +1 ,σ u − ] .(6) χ u ( k ) = P t ∈ N ( k ) χ u − ( t ) − χ u − ( k ) > k ∈ ∆ u , where we set for conve-nience χ v ( t ) = 0 if v < t ∆ v .Moreover, y ∈ ∆ q − , s y divides β q on the left and χ q +1 ( y ) = X t ∈ N ( y ) ∩ ∆ q χ q ( t ) − χ q − ( y ) = 0 . Note that the sets ∆ and ∆ defined earlier satisfy the required conditions. Indeed,one has ∆ ⊆ N ( i ) = N (∆ ) by Lemma 5. The minimal nonzero degree of T β is notsmaller than m α ≥ m + ω = σ − + 1 by Lemma 2. It follows from the same lemmaand the fact that P j is a direct summand of ( T β ) m that P j is a direct summand of (cid:16) t − \{ j } T β (cid:17) m = ( t j T β ) m for any j ∈ ∆ . Moreover, for any r ≤ m = σ , P k with k ∈ V \ ∆ is not a direct summand of ( T β ) r by Lemma 5, and hence is not a directsummand of ( T β ) r by Lemma 2, and P k with k ∈ ∆ cannot be a direct summandof ( T β ) r , becauseHom r + ω ( P k , T β ) ∼ = Hom r +2 ω − ( t ∆ P k [ ω − , T β ) = Hom r +2 ω − ( P k , T β ) = 0due to the inequality r + ω − ≤ m + ω − ≤ m β + ω − < m β . Finally, we clearlyhave χ ( j ) = 1 > j ∈ ∆ . hus, it is sufficient to show that if we have sets ∆ , . . . , ∆ p such that the propertiesabove are satisfied for any 1 ≤ u ≤ p , then we either can construct ∆ p +1 in such away that the properties above will be satisfied for u = p + 1 or find y ∈ ∆ p − suchthat s y divides β p on the left and P t ∈ N ( y ) ∩ ∆ p χ p ( t ) = χ p − ( y ). We introduce all thenecessary technical tools and discuss this step in detail in Section 7.Step II: Braiding.
Once a presentation for α of the form s ∆ s ∆ . . . s ∆ p s y e β is obtained, itremains to show that either ∆ = { j } for some j such that s j divides s ∆ s ∆ . . . s ∆ p s y on the left or | ∆ | ≥ s k with k ∈ Γ \ { i } divides ∆ s ∆ . . . s ∆ p s y on the left. Note that if ∆ = { j } , then χ ( i ) = 0, and hence our presentation isof the form α = s i s j s i e β = s j s i s j e β . Thus, at this point it is enough to consider thecase | ∆ | ≥ k ∈ Γ \ { i } can be pulled to the very left of thesubword s ∆ s ∆ . . . s ∆ p s y , applying a sequence of braid and commutator relations.This step is discussed in Section 8.6. Two-term objects
In this section we introduce the notion of the a two-term object in D and discuss someof their properties. The facts we prove in the current section will be required to fulfill thefactorization process announced above. Definition 6.1.
An object X of a triangulated category D with a fixed Γ-configuration of ω -spherical objects { P j } j ∈ Γ is called two-term if there exists a triangle X [ − β X −−→ M j ∈ V u P x j j [ − ω u ] ϕ X −−→ M k ∈ V u +1 P x k k α X −−→ X in D for some u ∈ { , } and some x j ≥ j ∈ Γ ). A two-term object X is called right-proper if f ϕ X = 0 for any split epimorphism f : L k ∈ V u +1 P x k k → P l with l ∈ V u +1 and iscalled left-proper if ϕ X g [ − ω u ] = 0 for any split monomorphism g : P l → L j ∈ V u P x j j with l ∈ V u . For a two-term object X we also define lsupp( X ) = { j ∈ V u | x j = 0 } andrsupp( X ) = { k ∈ V u +1 | x k = 0 } .For example, P l is a left-proper two-term object with α P l = id P l and β P l = ϕ P l = 0 forany l ∈ Γ . Moreover, lsupp( P l ) = ∅ and rsupp( P l ) = { l } . It is not difficult to show thatany two-term object X can be represented in the form X = X ′ ⊕ L k ∈ V u +1 P r k k , where X ′ is aright-proper two-term object. Remark.
The notion of a two-term object in some sense generalizes the notion of a two-termpartial tilting complex to the setting of triangulated categories with a Γ-configuration ofspherical objects.
Lemma 6.
Let X be a two-term object as in the definition above.1) The following three conditions are equivalent: • X is right-proper; • if l ∈ V u +1 and g : P l → L k ∈ V u +1 P x k k is not a split monomorphism, then α X g = 0 ; • dim k Hom ∗ ( X, P l ) = P k ∈ N ( l ) x k for any l ∈ V u +1 .2) The following three conditions are equivalent: • X is left-proper; • if l ∈ V u and f : L j ∈ V u P x j j → P l is not a split epimorphism, then f [ − ω u ] β X = 0 ; • dim k Hom ∗ ( X, P l ) = P k ∈ N ( l ) x k for any l ∈ V u . roof. We will prove only the first assertion, the second one can be deduced by dual argu-ments. Fix some l ∈ V u +1 . One hasdim k Hom ∗ ( X, P l ) = dim k Hom ∗ M j ∈ V u P x j j , P l + dim k Hom ∗ M k ∈ V u +1 P x k k , P l ! − k Im ϕ ∗ X = X k ∈ N ( l ) x k + 2 ( x l − dim k Im ϕ ∗ X ) , where ϕ ∗ X : Hom ∗ L k ∈ V u +1 P x k k , P l ! → Hom ∗ L j ∈ V u P x j j , P l ! is the map induced by ϕ X .Note that the set Hom ∗ L k ∈ V u +1 P x k k , P l ! = Hom ∗ ( P x l l , P l ) is a direct sum of two spacesof dimension x l of which the first is annihilated by ϕ X and the second is formed by splitepimorphisms. Thus, dim k Im ϕ ∗ X ≤ x l and the equality holds for all l ∈ V u +1 preciselywhen X is right-proper. Thus, X is right-proper if and only if dim k Hom ∗ ( X, P l ) = P k ∈ N ( l ) x k for any l ∈ V u +1 . Note that dim k Hom ∗ ( X, P l ) = dim k Hom ∗ ( P l , X ) because P l is spherical.Analogous arguments show that dim k Hom ∗ ( P l , X ) = P k ∈ N ( l ) x k for any l ∈ V u +1 if and onlyif α X g = 0 for any g : P l → L k ∈ V u +1 P x k k that is not a split monomorphism. (cid:3) The first crucial fact about the set of two-term objects is that it is stable under certainautoequivalences of D . For ∆ ⊆ V u let us set t +∆ = t ∆ [ ω u +1 −
1] and t − ∆ = t − [1 − ω u +1 ]. Lemma 7.
Let X be as above.1) If X is right-proper and rsupp( X ) ⊆ ∆ ⊆ V u +1 , then t +∆ X is a left-proper two-termobject with the defining triangle of the form t +∆ X [ − β t +∆ X −−−→ M k ∈ V u +1 P x ′ k k [ − ω u +1 ] ϕ t +∆ X −−−−→ M j ∈ V u P x j j α t +∆ X −−−−→ t +∆ X, where x ′ k = P j ∈ N ( k ) x j − x k for k ∈ ∆ and x ′ k = 0 for k ∈ V u +1 \ ∆ .2) If X is left-proper and lsupp( X ) ⊆ ∆ ⊆ V u , then t − ∆ X is a right-proper two-termobject with the defining triangle of the form t − ∆ X [ − β t − ∆ X −−−→ M k ∈ V u +1 P x k k [ − ω u +1 ] ϕ t − ∆ X −−−−→ M j ∈ V u P x ′ j j α t − ∆ X −−−−→ t − ∆ X, where x ′ j = P k ∈ N ( j ) x k − x j for j ∈ ∆ and x ′ j = 0 for j ∈ V u \ ∆ .Proof. We will prove only the first assertion, as before, the second one can be deduced bydual arguments.Let us fix some l ∈ V u +1 . By Lemma 6, Hom ∗ ( P l , X ) has dimension P k ∈ N ( l ) x k , andhence is generated by (1) x l compositions P l → L k ∈ V u +1 P x k k α X −−→ X , where the first arrowranges over x l linearly independent direct inclusions, and (2) x ′ l maps P l → X [ ω −
1] thatafter composition with β X [ ω ] give x ′ l linearly independent maps from P l to L j ∈ V u P x j j [ ω u +1 ]annihilated by ϕ X [ ω ]. Taking all these morphisms for all l together we get a map M k ∈ V u +1 P x k k [ ω u − ⊕ M k ∈ V u +1 P x ′ k k [ − ω u +1 ] (cid:16) α X [ ω u − γ (cid:17) −−−−−−−−−−−−−→ X [ ω u − hose cone is isomorphic to t +∆ X . Applying the octahedral axiom to the composition (cid:0) α X [ ω u − γ (cid:1) ◦ (cid:18) id Z (cid:19) = α X [ ω u − Z denotes L k ∈ V u +1 P x k k [ ω u − L k ∈ V u +1 P xkk [ ω u − L k ∈ V u +1 P xkk [ ω u − ⊕ L k ∈ V u +1 P x ′ kk [ − ω u +1 ] L k ∈ V u +1 P x ′ kk [ − ω u +1 ] L k ∈ V u +1 P xkk [ ω u − X [ ω u − L j ∈ V u P xjj t +∆ X t +∆ X idZ ! ( αX [ ωu − γ ) ϕt +∆ XαX [ ωu − βX [ − ωu ] αt +∆ XβX [ − ωu ] αt +∆ X Hence, t +∆ X indeed has the required form. A direct inclusion g : P l → L k ∈ V u +1 P x ′ k k suchthat the composition P l [ − ω u +1 ] g [ − ω u +1 ] −−−−−−→ M k ∈ V u +1 P x ′ k k [ − ω u +1 ] ϕ t +∆ X −−−−→ M j ∈ V u P x j j is zero would give a linear dependence between x ′ l components of the morphism P x ′ l l [ − ω u +1 ] → L j ∈ V u P x j j which are linearly independent by our construction. Thus, t +∆ X is left-proper. (cid:3) Next we need to study the behaviour of some relations between two-term objects withrespect to autoequivalences t ± ∆ . Definition 6.2.
Let X = cone L j ∈ V u P xjj [ − ω u ] ϕX −−→ L k ∈ V u +1 P xkk ! and Y = cone L j ∈ V u P yjj [ − ω u ] ϕY −−→ L k ∈ V u +1 P ykk ! be two-term objects. We will call X a two-term subobject of Y if there exist split monomor-phisms ι u : L j ∈ V u P x j j → L j ∈ V u P y j j and ι u +1 : L k ∈ V u +1 P x k k → L k ∈ V u +1 P y k k such that ι u +1 ϕ X = ϕ Y ι u [ − ω u ]. The two-term subobject X of Y is called trivial if either X = 0 or both of themaps ι u , ι u +1 are isomorphisms. Otherwise X is called a nontrivial two-term subobject of Y .We will say that a morphism f : X → Y [ ω ] is a right socle morphism if it can be presentedin the form f = α Y [ ω ] f ′ for some f ′ : X → L k ∈ V u +1 P y k k [ ω ] such that for any split epimor-phism g : L k ∈ V u +1 P y k k → P l with l ∈ V u +1 the morphism g [ ω ] f ′ α X : L k ∈ V u +1 P x k k → P l [ ω ] isnot a split epimorphism anymore. Remark.
The second condition in the definition of a right socle morphism is valid auto-matically if X is right-proper or ω = 0. Moreover, if ω = 0 and X is left-proper, then anymorphism of the form X f −→ Y [ ω ] is automatically right socle. It follows from the fact thatHom D ( P k , P j [1 + ω u +1 ]) = 0 for any k ∈ V u +1 , j ∈ V u and g [1 − ω u ] β X [1] = 0 for any g : L j ∈ V u P x j j → L j ∈ V u P y j j [ ω ]. In fact, the definition of a right socle morphism is introduced tocover the case ω = 0 which nevertheless is of special interest for us in view of an applicationto the derived Picard groups of algebras. Most of the assertions about right socle morphismswe provide below are trivial for ω = 0. Lemma 8.
Let X and Y be as above. Suppose also that both X and Y are right-proper. If X is a nontrivial two-term subobject of Y , then t +∆ X is a non-trivial two-term subobject of t +∆ Y for any rsupp( Y ) ⊆ ∆ ⊆ V u +1 . roof. It is clear that rsupp( X ) ⊆ rsupp( Y ), and hence both objects t +∆ X and t +∆ Y aretwo-term by Lemma 7. Recall that due to the proof of Lemma 7 one has t +∆ X = cone M k ∈ V u +1 P x ′ k k [ − ω u +1 ] ϕ t +∆ X −−−−→ M j ∈ V u P x j j where x ′ l = 0 for any l ∈ V u +1 \ ∆ and for any l ∈ ∆ the components of the map ϕ t +∆ X | P x ′ ll constitute the basis of Ker Hom D ( P l , ϕ X ). The morphism ϕ t +∆ Y satisfies analogous condi-tions. P l L k ∈ V u +1 P x ′ k k [ − ω u +1 ] L j ∈ V u P x j j L k ∈ V u +1 P x k k [ ω u ] L k ∈ V u +1 P y ′ k k [ − ω u +1 ] L j ∈ V u P y j j L k ∈ V u +1 P y k k [ ω u ] g ϕ t +∆ X ι ′ u +1 ϕ X [ ω u ] ι u ι u +1 [ ω u ] ϕ t +∆ Y ϕ Y [ ω u ] Since ϕ Y [ ω u ] ι u ϕ t +∆ X = 0, there exists a map ι ′ u +1 : L k ∈ V u +1 P x ′ k k → L k ∈ V u +1 P y ′ k k such that ι u ϕ t +∆ X = ϕ t +∆ Y ι ′ u +1 [ − ω u +1 ]. If ι ′ u +1 is not a split monomorphism, then for some l ∈ V u +1 there exists a direct inclusion g : P l → L k ∈ V u +1 P x ′ k k such that ι ′ u +1 g is not a direct inclusion.In this case we have ι u ϕ t +∆ X g = ϕ t +∆ Y ι ′ u +1 g = 0, and hence ϕ t +∆ X g = 0. This contradicts t +∆ X being left-proper, which has been proved in Lemma 7. This contradiction implies that ι ′ u +1 is a split monomorphism, and hence we are done. (cid:3) Lemma 9.
Let X and Y be as above. Then any right socle morphism f : X → Y [ ω ] factorsthrough some right socle morphism X ′ → Y [ ω ] , where X ′ is a two-term object such that rsupp( X ′ ) ⊂ rsupp( Y ) .Proof. Let f = α Y [ ω ] f ′ : X → Y [ ω ] be a decomposition of the right socle mor-phism f and set ∆ = rsupp( X ) \ rsupp( Y ). By ι we denote the direct inclusion L k ∈ ∆ P x k k ֒ → L k ∈ V u +1 P x k k and by π the split epimorphism L k ∈ V u +1 P x k k ։ L k ∈ V u +1 \ ∆ P x k k . SinceHom D L k ∈ ∆ P x k k , L k ∈ V u +1 P y k k [ ω ] ! = 0, one has f ′ α X ι = 0, and hence f ′ factors throughsome morphism f ′′ : X ′ → L k ∈ V u +1 P y k k [ ω ], where X ′ = cone ( α X ι ). Note that by the octahe-dral axiom one has X ′ ∼ = cone M j ∈ V u P x j j [ − ω u ] πϕ X −−−→ M k ∈ V u +1 \ ∆ P x k k , i.e. X ′ is a two-term object of the required form. j ∈ V u P x j j [ − ω u ] L j ∈ V u P x j j [ − ω u ] L k ∈ ∆ P x k k L k ∈ V u +1 P x k k L k ∈ V u +1 \ ∆ P x k k L k ∈ ∆ P x k k X X ′ L k ∈ V u +1 P y k k [ ω ] P l [ ω ] ϕ X πϕ X ι πα X α X ′ α X ι f ′ f ′′ g [ ω ] It remains to show that the morphism α Y [ ω ] f ′′ is right socle. Suppose that g : L k ∈ V u +1 P y k k → P l is a split epimorphism. We need to show that g [ ω ] f ′′ α X ′ is not asplit epimorphism. Observe that π is an epimorphism and gf ′ α X is not a split epimorphismsince f is right socle. On the other hand, we have f ′′ α X ′ π = f ′ α X by construction. Hence g [ ω ] f ′′ α X ′ is not a split epimorphism as well. (cid:3) Lemma 10. If X is right-proper and l ∈ V u +1 , then any right socle morphism from P l to X [ ω ] is zero.Proof. Let f = α X [ ω ] f ′ : P l → X be a right socle morphism. It follows from the definition ofa right socle morphism that f ′ is not a split monomorphism. Then one has f = α X [ ω ] f ′ = 0by Lemma 6. (cid:3) Lemma 11.
Let X and Y be as above. Suppose that both X and Y are right-proper and rsupp( X ) ⊂ rsupp( Y ) . Then for any right socle f : X → Y [ ω ] and any rsupp( Y ) ⊆ ∆ ⊆ V u +1 , the morphism t +∆ f : t +∆ X → t +∆ Y [ ω ] is right socle too.Proof. To simplify the notations we consider here only the case ω = 0. Other cases are obvi-ous due to the remark after the definition of a right socle morphism. Note that ω u = ω u +1 = 0by our assumption. Due to the proof of Lemma 7 there is a triangle t +∆ X [ − ψβ t +∆ X −−−−−−→ M k ∈ V u +1 P x k k [ − ⊕ M k ∈ V u +1 P x ′ k k (cid:16) α X γ (cid:17) −−−−−−−→ X [ − α t +∆ X β X −−−−−−→ t +∆ X emerging from the definition of a spherical twist. One has an analogous triangle for Y too. Observe that since f factros through α Y , we have β Y [1] f = 0. Then t +∆ f satisfies thecondition ( t +∆ f ) α t +∆ X β X = α t +∆ Y β Y f [ −
1] = 0 . Hence t +∆ f factros through (cid:18) ψ [1] β t +∆ X [1] (cid:19) , i.e. β t +∆ Y [1] t +∆ f = ( β t +∆ Y [1] f ′ ) (cid:18) ψ [1] β t +∆ X [1] (cid:19) for somemorphism f ′ : L k ∈ V u +1 P x k k ⊕ L k ∈ V u +1 P x ′ k k [1] → t +∆ Y. Since β t +∆ Y [1] f ′ is annihilated by ϕ t +∆ Y [1]and t +∆ Y is left-proper, all components of β t +∆ Y [1] f ′ are not isomorphisms, i.e. all morphisms L k ∈ V u +1 P x k k ⊕ L k ∈ V u +1 P x ′ k k [1] → P l [1] constituting β t +∆ Y [1] f ′ are not split epimorphisms. Notethat β t +∆ X [1] annihilates all morphisms of the form L k ∈ V u +1 P x ′ k k [1] → P l [1] with l ∈ V u +1 that are not split epimorphisms by the left-properness of t +∆ X . ince Hom D L k ∈ V u +1 P x k k , L k ∈ V u +1 P y k k [1] ! = 0, we have β t +∆ Y [1] t +∆ f = 0, and hence t +∆ f = α t +∆ Y f ′′ for some f ′′ : t +∆ X → L j ∈ V u P y j j . Suppose now that gf ′′ α t +∆ X is a split epimorphismfor some g : L j ∈ V u P y j j → P l . Then gf ′′ α t +∆ X β X = 0, and hence also f ′′ α t +∆ X β X = 0. Onthe other hand, we have proved earlier that α t +∆ Y f ′′ α t +∆ X β X = ( t +∆ f ) α t +∆ X β X = 0. Then f ′′ α t +∆ X β X factors as ϕ t +∆ Y θ for some morphism θ : X [ − → L k ∈ V u +1 P y ′ k k .Since Hom D L k ∈ V u +1 P x k k [ − , L k ∈ V u +1 P y ′ k k ! = 0, θ factors through β X , and hence wehave f ′′ α t +∆ X β X = ϕ t +∆ Y θ = ϕ t +∆ Y f ′′′ β X for some f ′′′ : L j ∈ V u P x j j → L k ∈ V u +1 P y ′ k k . Since allcomponents of ϕ t +∆ Y f ′′′ are not isomorphisms, the morphism g ( f ′′ α t +∆ X − ϕ t +∆ Y f ′′′ ) is a splitepimorphism, and hence ( f ′′ α t +∆ X − ϕ t +∆ Y f ′′′ ) β X cannot be zero. We have a contradiction,and this finishes the proof of the lemma. (cid:3) Factorization
We return to the context of Lemma 1. Until the end of the paper we assume that ω ≤ σ u = m + u (2 − ω ), σ u +1 = m + u (2 − ω ) + 1 − ω . Now suppose that wehave sets ∆ , . . . , ∆ p for some p ≥ = { i } for some i ∈ V . Let us recallthat the numbers χ u ( k ) (0 ≤ u ≤ p , k ∈ ∆ u ) are defined inductively by χ ( i ) := 1 and χ u ( k ) := P t ∈ N ( k ) χ u − ( t ) − χ u − ( k ) for u ≥
1, where we set for convenience χ v ( t ) = 0 if v < t ∆ v . Suppose also that the following conditions hold for 1 ≤ u ≤ p :(1) α = s ∆ . . . s ∆ u β u for some β u ∈ B +Γ with l ( β u ) = l ( α ) − u P v =0 | ∆ v | (2) ∆ u − ⊆ ∆ u ⊆ N (∆ u − ), where we set ∆ − = ∅ for convenience.(3) P l is not a direct summand of ( T β u ) [ σ u − +1 ,σ u − ] for any l ∈ V u .(4) The minimal nonzero degree of T β u is not smaller than σ u − + 1.(5) For any l ∈ ∆ u , P l is a direct summand of ( t l T β u ) [ σ u − +1 ,σ u − ] .(6) χ u ( k ) > k ∈ ∆ u .We want to continue the process and construct ∆ p +1 in such a way that the conditions(1) and (2) are fulfilled for u = p + 1 and, whenever the condition (6) is valid for u = p + 1,then so are the conditions (3)–(5). To this purpose, define C u := t − ∆ u . . . t − ∆ P i = ( t − u . . . t − P i )[ σ u − m + ω u − ω ] . The first crucial fact that we will need is that C u is a two-term object of a certain form. Lemma 12.
For any ≤ u ≤ p there exists a triangle of the form C u [ − β u −−→ M j ∈ ∆ u − P χ u − ( j ) j [ − ω u − ] ϕ u −−→ M k ∈ ∆ u P χ u ( k ) k α u −−→ C u . Proof.
Since P i is a left-proper two-term object, the required triangle can be obtained byiterated application of Lemma 7. (cid:3) The next proposition elaborates on the properties of the objects C u that will allow us tofind direct summands of ( T β u ) [ σ u − +1 ,σ u ] . Proposition 1.
For any ≤ u ≤ p , the two-term object C u satisfies the following twoconditions:1) Hom r ( C u , T β u ) = 0 for any r ≤ ω u + σ u .2) For any nontrivial two-term subobject C ′ of C u , there exists a nonzero morphism f : C ′ → T β u [ r ] with ω u + σ u − + 1 ≤ r ≤ ω u + σ u such that f [ ω ] g = 0 for any rightsocle morphism g : C ′′ → C ′ [ ω ] . roof. We proceed by induction on u with the base case u = 0. Since C = P i , P i does nothave nontrivial two-term subobjects and Hom r ( P i , T β ) = 0 for any r ≤ m + ω by Lemma5, there is nothing left to prove in the base case. Now we prove 1) and 2) for u , assumingthey are true for u −
1. The first property of C u is clear, becauseHom r ( C u , T β u ) = Hom r ( t − ∆ u C u − , t − u T β u − ) ∼ = Hom r + ω u +1 − ( C u − , T β u − ) . Now we turn to the second property. Suppose that C ′ is a nontrivial two-term subobjectof C u . If C ′ is right-proper, then t +∆ u C ′ is a nontrivial two-term subobject of C u − = t +∆ u C u by Lemma 8. By the induction hypothesis, there is a nonzero morphism f : t +∆ u C ′ → T β u − [ r ]with ω u − + σ u − + 1 ≤ r ≤ ω u − + σ u − such that f [ ω ] g = 0 for any right socle morphism g : C ′′ → t +∆ u C ′ [ ω ]. Let us prove that the morphism t − ∆ u f : C ′ → T β u [ r + 1 − ω u − ] satisfiesthe required properties.Suppose for a contradiction that there exists a right socle morphism g : C ′′ → C ′ [ ω ] suchthat ( t − ∆ u f )[ ω ] g = 0. Then applying Lemmas 9 and 10 we can find such a g with right-proper C ′′ satisfying the condition rsupp( C ′′ ) ⊆ rsupp( C ′ ). Then t +∆ u g : t +∆ u C ′′ → t +∆ u C ′ [ ω ] is rightsocle by Lemma 11 and satisfies the condition f [ ω ] (cid:0) t +∆ u g (cid:1) = t +∆ u (cid:0) ( t − ∆ u f )[ ω ] g (cid:1) = 0 which isimpossible. This shows that t − u f is indeed the required morphism.It remains to consider the case when C ′ is not right-proper. In this case we may assumethat C ′ = P l for some l ∈ ∆ u . Let a be the minimal integer such that P l is a direct summandof ( t l T β u ) a . Note that a ∈ [ σ u − + 1 , σ u − ] by the properties (3)–(5) of ∆ u . Picking somelong morphism f : P l → t l T β u [ a + ω ] and applying t − l [1 − ω ], we get a nonzero morphism t − l f [1 − ω ] : P l → T β u [ a + 1]. Note that a + 1 ∈ [ σ u − + 2 , σ u − + 1] = [ ω u + σ u − + 1 , ω u + σ u ] . Thus, it remains to show that ( t − l f )[1] g = 0 for any right socle morphism g : C ′′ → P l [ ω ].Due to Lemma 9, we may assume that C ′′ has the form C ′′ = cone M j ∈ V u +1 P x j j [ − ω u +1 ] ϕ C ′′ −−−→ P xl for some integers x j , x . Note now that if C ′′ = P l , then t l g : P l [1 − ω ] → P l [1] is a right soclemorphism, and hence factors though P j [1 − ω u +1 ] for j ∈ N ( l ). Then f [1]( t l g ) = 0 by thedefinition of a long morphism. Hence, we may assume that C ′′ is right-proper. Then t l g is amorphism from cone P yl [1 − ω ] ϕ t + l C ′′ [1 − ω u +1 ] −−−−−−−−−−→ L j ∈ V u +1 P x j j [1 − ω u +1 ] ! to P l [1] by Lemma7. Note that the composition M j ∈ V u +1 P x j j [ − ω u +1 ] α t + l C ′′ [ − ω u +1 ] −−−−−−−−−→ t l C ′′ [ − t l g [ − −−−−→ P l f −→ t l T β u [ a + ω ]is zero by the definition of a long morphism. Hence f ( t l g )[ −
1] factors through some morphism θ : P yl [1 − ω ] → t l T β u [ a + ω ]. Since a + ω − < a , P l is not a direct summand of ( t l T β u ) a + ω − .If f [1]( t l g ) = 0, then θ = 0 and there is some j ∈ N ( l ) such that P j θ [ ω + ω u − ι [ ω u ] γ j,l −−−−−−−−−−−−→ t l T β u [ a + 2 ω + ω u −
1] is nonzero, where ι : P l → P yl is some split monomorphism. Notethat by property (4) and the choice of a , the minimal nonzero degree of t l T β u is not smallerthan min ( a, σ u − + 1) by Lemma 2. On the other hand, if f [1]( t l g ) = 0, then the minimalnonzero degree of t l T β u does not exceed a + ω + ω u − < a . Then we have σ u − + 1 ≤ a + ω + ω u − ≤ σ u − + ω + ω u − σ u − + ω, i.e. ω >
0, a contradiction. This shows that f [1]( t l g ) = 0, and hence ( t − l f )[1] g = 0, asrequired. (cid:3) et us now construct ∆ p +1 ⊆ V p +1 . We will do this in the following way. Set ∆ p +1 = ∅ .Suppose that we have defined the set ∆ cp +1 . Choose a pair l, a with l ∈ V p +1 \ ∆ cp +1 , a ∈ Z such that P l is a direct summand of (cid:16) t − cp +1 T β p (cid:17) a , and with a the minimal possible amongall such pairs. We define ∆ c +1 p +1 = ∆ cp +1 ∪ { l } and continue the process if a ∈ [ σ p − + 1 , σ p ].If either such an integer a does not exist or a > σ p , then we terminate the process, defining∆ p +1 := ∆ cp +1 . Now we are ready to prove the factorization theorem. Theorem 2.
The ∆ p +1 constructed as described above satisfies the following conditions:1. α = s ∆ . . . s ∆ p +1 β p +1 for some β p +1 ∈ B +Γ with l ( β p +1 ) = l ( α ) − p +1 P v =0 | ∆ v | ∆ p − ⊆ ∆ p ⊆ N (∆ p − ) .Moreover, if χ p +1 ( l ) > for any l ∈ ∆ p +1 , then the following conditions are satisfied aswell:3. P l is not a direct summand of ( T β p +1 ) [ σ p − +1 ,σ p ] for any l ∈ V p +1 .4. The minimal nonzero degree of T β p +1 is not smaller than σ p − + 1 .5. For any l ∈ ∆ p +1 , P l is a direct summand of ( t l T β p +1 ) [ σ p − +1 ,σ p ] .Proof. The first condition can be obtained applying Lemma 1 to the words β p , s − p +1 β p , . . . , s − p +1 β p , all of which are of strictly smaller length than α . Indeed, if ∆ c +1 p +1 =∆ cp +1 ∪ { l } , then P l is a direct summand of (cid:16) t − cp +1 T β p (cid:17) a for a ≤ σ p and it is sufficient toprove that the minimal nonzero degree of t − cp +1 T β p is not smaller than a + ω p ≤ σ p − + 1.If this is not the case, there exists some b ≤ min ( a − , σ p − ) and k ∈ Γ such that P k is adirect summand of ( t − cp +1 T β p ) b . Observe that k cannot belong to V p by the conditions (3),(4) with u = p and Lemma 2 and cannot belong to V u +1 \ ∆ cp +1 by the choice of a . On theother hand, for k ∈ ∆ cp +1 , one hasHom b + ω ( P k , t − cp +1 T β p ) ∼ = Hom b + ω ( t ∆ cp +1 P k , T β p ) ∼ = Hom b +2 ω − ( P k , T β p ) = 0 , because the minimal nonzero degree of T β p is not smaller than σ p − + 1 and b + ω − ≤ σ p − + ω − ≤ σ p − .Let us now prove the second condition. Suppose first that there exists some l ∈ ∆ p +1 \ N (∆ p ). Then it is clear that l N (∆ u ) for all 0 ≤ u ≤ p , in particular, l = i .Hence we have α = s ∆ . . . s ∆ p s l s − l β p = s l γ for some γ ∈ B +Γ with l ( γ ) = l ( α ) −
1. Thenthe assertion of Lemma 1 is valid for α , a contradiction.Suppose now that l ∆ p +1 for some l ∈ ∆ p − . By construction, this means that P l is not a direct summand of (cid:16) t − p +1 T β p (cid:17) [ σ p − +1 ,σ p ] . Let π denote the split epimor-phism L j ∈ ∆ p − P χ p − ( j ) j ։ P χ p − ( l ) l . The octahedral axiom applied to the composition π [1 − ω p +1 ] ◦ β C p [1] gives us the following diagram: L k ∈ ∆ p P χ p ( k ) k / / C ′ / / (cid:15) (cid:15) L j ∈ ∆ p − \{ l } P χ p − ( j ) j [1 − ω p +1 ] (cid:15) (cid:15) L k ∈ ∆ p P χ p ( k ) k / / C p / / (cid:15) (cid:15) L j ∈ ∆ p − P χ p − ( j ) j [1 − ω p +1 ] π [1 − ω p +1 ] (cid:15) (cid:15) P χ p − ( l ) l [1 − ω p +1 ] P χ p − ( l ) l [1 − ω p +1 ] Let ψ denote the morphism of P χ p − ( l ) l [ − ω p +1 ] to C ′ arising from this diagram. Since χ p − ( l ) > C ′ is a nontrivial two-term subobject of C p and Proposi-tion 1 can be applied. For some ω p + σ p − + 1 ≤ r ≤ ω p + σ p , we have a nonzero morphism f : C ′ → T β p [ r ] such that f [ ω ] g = 0 for any right socle morphism g : C ′′ → C ′ [ ω ]. SinceHom r ( C p , T β p ) = 0 by Proposition 1, the morphism f ψ ′ is nonzero for some component ′ : P l [ − ω p +1 ] → C ′ of the map ψ . We are going to prove that t − p +1 ( f ψ ′ )[ ω p +1 ] : P l → t − p +1 T β p [ r + ω p +1 ] is long. Since r − ω p ∈ [ σ p − + 1 , σ p ], this contradicts the assumptionthat P l is not a direct summand of (cid:16) t − p +1 T β p (cid:17) [ σ p − +1 ,σ p ] .Pick some r ∈ N ( l ). Since P r [1 − ω p ] is a right-proper two-term object with α P r [1 − ω p ] = ϕ P r [1 − ω p ] = 0 and β P r [1 − ω p ] = id P r [ ω p ] , we have t +∆ p +1 P r [1 − ω p ] ∼ = cone M j ∈ N ( r ) ∩ ∆ p +1 P j [ − ω p +1 ] → P r by Lemma 7. Note that the morphism ψ ′ [ ω ]( t +∆ p +1 γ r,l [1 − ω p ]) is right socle. Indeed, as thediagram above shows, ψ ′ factors through L k ∈ ∆ p P χ p ( k ) k and the map P r α t +∆ p +1 Pr [1 − ωp ] −−−−−−−−−−→ t +∆ p +1 P r [1 − ω p ] t +∆ p +1 γ r,l [1 − ω p ] −−−−−−−−−−→ P l [ ω p ] −→ M k ∈ ∆ p P χ p ( k ) k [ ω ]is not a split monomorphism, because it factors through P l [ ω p ]. Then ( f ψ ′ )[ ω ]( t +∆ p +1 γ r,l [1 − ω p ]) = 0, and hence t − p +1 ( f ψ ′ )[ ω ] γ r,l = 0. Thus, t − p +1 ( f ψ ′ )[ ω p +1 ] is a long morphism.Suppose now that χ p +1 ( l ) > l ∈ ∆ p +1 . We define C p +1 := t − ∆ p +1 C p . Then we haveHom r ( C p +1 , T β p +1 ) = 0 for r ≤ ω p +1 + σ p +1 (see the proof of the first part of Proposition1). Due to Lemma 12, we have C p +1 = cone M j ∈ ∆ p P χ p ( j ) j [ − ω p ] ϕ p +1 −−−→ M k ∈ ∆ p +1 P χ p +1 ( k ) k for some morphism ϕ p +1 . We prove the third property by contradiction. If P l with l ∈ ∆ p +1 is a direct summand of ( T β p +1 ) a for some a ≤ σ p , then there exists a morphism of L k ∈ ∆ p +1 P χ p +1 ( k ) k to T β p +1 [ a + ω ] annihilated by ϕ p +1 . This gives a nonzero morphism of C p +1 to T β p +1 [ a + ω ], which is impossible since a + ω ≤ ω + σ p < ω p +1 + σ p +1 . Observe thatthe minimal nonzero degree of T β p +1 is not smaller than σ p − +1 and P l with l ∈ V p +1 \ ∆ p +1 cannot be a direct summand of ( T β p +1 ) [ σ p − +1 ,σ p ] by the construction of ∆ p +1 . This finishesthe proof of the third property. The forth property follows from the third one, the property(3) with u = p and Lemma 2.It remains to prove the fifth property. Let l ∈ ∆ p +1 . There is some c such that l ∆ cp +1 and l ∈ ∆ c +1 p +1 . Then P l is a direct summand of (cid:16) t − cp +1 T β p (cid:17) a for some a ∈ [ σ p − + 1 , σ p ].Choose the minimal such a number a . It follows from the construction of ∆ p +1 , the property(3) with u = p and Lemma 2 that the minimal nonzero degree of t − cp +1 T β p is not smallerthan min ( a, σ p − + 1) (see the beginning of this proof). Then P l is a direct summand of (cid:18) t − p +1 \ ∆ c +1 p +1 t − cp +1 T β p (cid:19) a = ( t l T β p +1 ) a by Lemma 2 and we are done. (cid:3) If χ p +1 ( l ) = 0 for some l ∈ ∆ p +1 , we can write α = s ∆ · · · s ∆ p s l e β and get the requiredpresentation of α , as we have announced in the previous section. In this case we terminatethe process of factorization (step I) and go to braiding (step II), which is described inthe next section. Otherwise condition (6) is also satisfied for u = p + 1. Then write α = s ∆ s ∆ · · · s ∆ p s ∆ p +1 β p +1 and continue the process, applying the arguments above to u = p +2instead of u = p + 1. 8. Mesh braiding in B +Γ The goal of this section is to show that any word of the form s ∆ · · · s ∆ p s l as obtained inthe previous section is left-divisible by some s j in B +Γ with j = i . As we explained in Section5, this finishes the proof of Lemma 1. et Z Γ be a directed graph whose vertices are pairs ( n, j ) for every n ∈ Z , j ∈ V n . Thearrows of Z Γ are of the form (( n, j ) → ( n + 1 , i )) for every edge ( i, j ) of Γ. The graph Z Γis essentially the stable translation quiver associated to Γ, but with vertices enumerateddifferently. It is convenient to draw Z Γ in such a way that vertices with the same firstcoordinate form one “vertical slice”. Set p ( a ) = n , p ( a ) = j for a = ( n, j ) ∈ ( Z Γ) . Let Θ n denote the set p − ( n ) = { ( n, j ) ∈ ( Z Γ) | j ∈ V n } ⊂ ( Z Γ) .Now any word γ representing an element of B +Γ can be depicted as some set of ver-tices Λ γ of Z Γ in the following way. Let us choose a presentation γ = s Σ . . . s Σ p , whereΣ , . . . , Σ p are such that Σ k ⊆ V k and the convention s ∅ = 1 is used. Now we can defineΛ γ = { ( k, j ) | k ≥ , j ∈ Σ k } ⊂ ( Z Γ) . On the other hand, to any finite set Λ ⊂ ( Z Γ) onecan assign a word γ Λ in B +Γ . More precisely, set γ Λ = + ∞ Q k = −∞ s Σ k , where Σ k = Λ ∩ Θ k . It iseasy to see that γ Λ γ = γ . Example.
Let Γ = D with vertices enumerated in such a way that the vertex of degreethree is labeled 2. Let γ = s ( s s s ) s ( s s s ) s s . Then we have the following set Λ γ : p Figure 1
On the other hand, observe that there are infinitely many sets Λ such that γ Λ = γ . Forinstance, the following set of vertices of ( Z Γ) also corresponds to the word γ : p Figure 2
Now fix some finite Λ ⊂ ( Z Γ) . Definition 8.1.
Let a, b ∈ Λ. We say that there is a generalized mesh starting at a andending at b if p ( a ) = p ( b ) = t , p ( a ) < p ( b ) and ( k, t ) / ∈ Λ for every k with p ( a ) < k
We say that (Λ , θ ) satisfies mesh relations if θ ( b ) + θτ Λ ( b ) = X c ∈ mesh Λ ( τ Λ ( b ) ,b ) θ ( c )for every b ∈ ΛThe following statement follows immediately from the definitions:
Lemma 13.
Let γ be a word of the form s ∆ . . . s ∆ p s l as described in the previous section.For every ( n, j ) ∈ Λ γ set θ ( n, j ) = χ n ( j ) , θ ( −∞ , i ) = − , θ ( −∞ , j ) = 0 for every j = i .Then (Λ γ , θ ) satisfies mesh relations. Example:
The word s ( s s s ) s ( s s s ) s s in B + D as in the previous example is de-picted below together with the function θ = χ .
111 2 111 1 0 p Figure 3
It is natural to ask which finite subsets of ( Z Γ) correspond to words that are equal inthe braid monoid B +Γ .First let a = ( n, j ) ∈ Λ be such that b = ( n + 2 , j ) / ∈ Λ (respectively b = ( n − , j ) / ∈ Λ)and ( n +1 , k ) / ∈ Λ (respectively ( n − , k ) / ∈ Λ) for every k ∈ N ( j ). Set Λ = (cid:0) Λ \{ a } (cid:1) ∪{ b } . Inaddition, let θ ( b ) = θ ( a ) and θ | Λ \{ b } = θ | Λ \{ a } . We refer to this procedure as commutation. θ ( a ) θ ( a ) p n n + 1 n + 2 n n + 1 n + 2 Figure 4
The following statement is clear:
Lemma 14.
Commutation does not change the corresponding word in B +Γ , i.e. γ Λ = γ Λ . Inaddition, if (Λ , θ ) satisfies mesh relations, then so does (Λ , θ ) . Now braid relations in B +Γ can be also depicted in terms of vertices of Z Γ. We say that a, b, c ∈ ( Z Γ) form a braid if a = ( n, j ), b = ( n + 1 , k ), c = ( n + 2 , j ) for some n ∈ Z and some j, k ∈ Γ such that k ∈ N ( j ). Now let a, b, c ∈ Λ as above form a braid andsuppose in addition that ( n + 1 , t ) / ∈ Λ for any t ∈ N ( j ) \ { k } , ( n + 2 , l ) / ∈ Λ for any ∈ N ( k ) \ { j } and d = ( n + 3 , k ) / ∈ Λ. Set Λ = (Λ \ { a } ) ∪ { d } , θ | Λ \{ b,c,d } = θ | Λ \{ a,b,c } , θ ( b ) = θ ( c ) , θ ( c ) = θ ( b ) , θ ( d ) = θ ( a ). We refer to this procedure as braiding. θ ( a ) θ ( c ) θ ( b ) θ ( b ) θ ( c ) p n n + 1 n + 2 n + 3 n n + 1 n + 2 n + 3 θ ( a ) Figure 5
Lemma 15.
Braiding does not change the corresponding word in B +Γ , i.e. γ Λ = γ Λ . More-over, if (Λ , θ ) satisfies mesh relations, then so does (Λ , θ ) .Proof. First we establish that γ Λ = γ Λ . It is sufficient to show that s Σ n s Σ n +1 s Σ n +2 s Σ n +3 = s Σ n s Σ n +1 s Σ n +2 s Σ n +3 where Σ p = p (Θ p ∩ Λ) and Σ p = p (cid:0) Θ p ∩ Λ (cid:1) . Observe that Σ n =Σ n \ { j } , Σ n +1 = Σ n +1 , Σ n +2 = Σ n +2 and Σ n +3 = Σ n +3 ∪ { k } . Hence s Σ n s Σ n +1 s Σ n +2 s Σ n +3 = s Σ n s j s Σ n +1 \{ k } s k s j s Σ n +2 \{ j } s Σ n +3 = s Σ n s Σ n +1 \{ k } s j s k s j s Σ n +2 \{ j } s Σ n +3 = s Σ n s Σ n +1 \{ k } s k s j s k s Σ n +2 \{ j } s Σ n +3 = s Σ n s Σ n +1 s j s Σ n +2 \{ j } s k s Σ n +3 = s Σ n s Σ n +1 s Σ n +2 s Σ n +3 Now we will show that braiding respects mesh relations. Since θ differs from θ only on a, b, c and d , it is sufficient to check the relations only for generalized meshes these fourvertices take part in. There are two types of such generalized meshes: those that have oneof a, b, c and d as starting and/or ending vertices and those that do not.1) Note that τ Λ ( b ) = τ Λ ( b ) and mesh Λ ( τ Λ ( b ) , b ) = mesh Λ ( τ Λ ( b ) , b ) \ { a } . We have θτ Λ ( b ) + θ ( b ) = X x ∈ mesh Λ ( τ Λ ( b ) ,b ) θ ( x ) = X x ∈ mesh Λ ( τ Λ ( b ) ,b ) θ ( x ) + θ ( a )and θ ( a ) + θ ( c ) = θ ( b ), because a, b, c form a generlized mesh in Λ starting at a andending at c . Then θτ Λ ( b ) + θ ( b ) = θτ Λ ( b ) + θ ( c ) = X x ∈ mesh Λ ( τ Λ ( b ) ,b ) θ ( x ) . The remaining cases are either analogous to the one just discussed or obvious.2) Now consider some generalized mesh mesh Λ ( x, y ) in Λ such that x, y
6∈ { a, b, c, d } .If b ∈ mesh Λ ( x, y ), then d ∈ mesh Λ ( x, y ) and vise versa. The corresponding meshrelation remains valid after braiding, because d / ∈ Λ and θ ( b ) + θ ( d ) = θ ( c ) + θ ( a ) = θ ( b ). If c ∈ mesh Λ ( x, y ), then a ∈ mesh Λ ( x, y ) and vise versa. The correspondingmesh relation remains valid after braiding, because θ ( c ) = θ ( b ) = θ ( c ) + θ ( a ). (cid:3) The main result of this section is the following theorem that in view of the facts we havealready proved implies Lemma 1.
Theorem 3.
Let (Λ , θ ) satisfying mesh relations be such that θ is non-negative on Λ , van-ishing on precisely one vertex of Λ , θ ( −∞ , i ) = − for some i ∈ Γ and θ ( −∞ , k ) = 0 forall k ∈ Γ \ { i } . Then γ = γ Λ is left-divisible by some s j in B +Γ with j = i . To prove this theorem we will need the following lemma. emma 16. Let (Λ , θ ) be as in Theorem 3 and a ∈ Λ such that θ ( a ) = 0 , τ Λ ( a ) = −∞ .Then there exists a sequence of commutations and braidings turning (Λ , θ ) into some (Λ ′ , θ ′ ) such that |{ x ∈ Λ ′ | p ( x ) ≤ p ( b ) }| < |{ x ∈ Λ | p ( x ) ≤ p ( a ) }| , where b is the unique vertex of Λ ′ on which θ ′ vanishes. Let us first show how Lemma 16 implies Theorem 3.
Proof of Theorem 3.
It is sufficient to show that applying a sequence of commutations andbraidings we can transform Λ into some Λ ′ such that Θ n ∩ Λ ′ = { i } , where n is the smallestinteger such that Θ n ∩ Λ ′ = ∅ . Note that commutation and braiding do not change themultiset { θ ( x ) } x ∈ Λ of values of θ .Applying Lemma 16 several times, one can obtain a pair (Λ ′′ , θ ′′ ) satisfying mesh relationssuch that γ Λ ′′ = γ Λ and θ ′′ vanishes on (and only on) a vertex with the smallest firstcoordinate among all vertices of Λ ′′ . Let θ ′′ ( x ) = 0. Then γ Λ ′′ = γ Λ is divisible by s p ( x ) on the left. It remains to ascertain that p ( x ) = i . Indeed, since x is a vertex with thesmallest p ( x ) among all vertices in Λ ′′ , there is an infinite generalized mesh ending at x and starting at ( −∞ , j ) for some j ∈ Γ . Since (Λ ′′ , θ ′′ ) satisfies infinite mesh relations, 0 = θ ′′ ( −∞ , j ) + θ ′′ ( x ) = θ ′′ ( −∞ , j ). We immediately see that j = i , since θ ′′ ( −∞ , i ) = − (cid:3) Now we prove Lemma 16.
Proof of Lemma 16.
Let a = a = ( n, j ). Since there is only one vertex of Λ on which θ vanishes and τ Λ ( a ) ∈ Λ, θτ Λ ( a ) + θ ( a ) = θτ Λ ( a ) >
0. Hence mesh Λ ( τ Λ ( a ) , a ) = ∅ . Let a = ( n , t ) ∈ mesh Λ ( τ Λ ( a ) , a ) be any vertex with the smallest first coordinate among ver-tices of mesh Λ ( τ Λ ( a ) , a ). Then, possibly applying several commutations, one can assumethat τ Λ ( a ) = ( n − , j ). If | mesh Λ ( τ Λ ( a ) , a ) | = 1, then, also possibly applying severalcommutations, one can assume that τ Λ ( a ) , a , a form a braid. Then we can apply braidingto the triple ( τ Λ ( a ) , a , a ) after some commutations as described below and we are done.Suppose now that q = | mesh Λ ( τ Λ ( a ) , a ) | > q = | mesh Λ ( τ Λ ( a ) , a ) | >
0, since τ Λ ( a ) ∈ mesh Λ ( τ Λ ( a ) , a ). Supposethat q >
1. Let a = ( n , p ) ∈ mesh Λ ( τ Λ ( a ) , a ) be any vertex with the smallest firstcoordinate among vertices of mesh Λ ( τ Λ ( a ) , a ) \ { τ Λ ( a ) } . Then, possibly applying severalcommutations, one can assume that τ Λ ( a ) = ( n − , t ). τ Λ ( a ) a a τ Λ ( a ) a Figure 6
Continue in the same fashion to obtain a sequence of elements a , . . . , a k of Λ such that τ Λ ( a m ) ∈ mesh Λ ( τ Λ ( a m +1 ) , a m +1 ), a m +1 ∈ mesh Λ ( τ Λ ( a m ) , a m ) \ { τ Λ ( a m − ) } for every m =0 , . . . , k − τ Λ ( a k ) = −∞ . Now set q m := | mesh Λ ( τ Λ ( a k ) , a k ) | for m = 0 , . . . , k . Weclaim that if the sequence a , . . . , a k is maximal with respect to inclusion among sequencessatisfying this property, then q k = 1. Indeed, if q k >
1, then there is a vertex a k +1 withthe smallest first coordinate among vertices of mesh Λ ( τ Λ ( a k ) , a k ) \ { τ Λ ( a k − ) } . Because the equence we consider is maximal, τ Λ ( a k +1 ) = ( −∞ , p ( a k +1 )). Since θ ( −∞ , p ( a k +1 )) iseither 0 or −
1, we have θ ( a k +1 ) ≥ θτ Λ ( a k ). Then θ ( a k ) ≥ θτ Λ ( a k − ), since θ ( a k ) + θτ Λ ( a k ) = θ ( a k +1 ) + θτ Λ ( a k − ) + X y ∈ mesh Λ ( τ Λ ( a k ) ,a k ) \{ a k +1 ,τ Λ ( a k − ) } θ ( y )Continuing in the same way we get θ ( a ) ≥ θτ Λ ( a ). On the other hand, θτ Λ ( a ) = θτ Λ ( a )+ θ ( a ) > θ ( a ), because q >
1, a contradiction.Take a sequence a , . . . , a k maximal with respect to inclusion and satisfying the propertiesdescribed above with the smallest corresponding sequence of integers ( q , . . . , q k ) in thelexicographic order. Denote such a ( q , . . . , q k ) by seq (Λ). If seq (Λ) = (1), there is nothingto prove, as we remarked earlier. Now suppose that the statement is known for all (Λ ′ , θ ′ )with seq (Λ ′ ) < seq (Λ). We will show that there exists a sequence of commutations andbraidings that turns (Λ , θ ) into ( e Λ , e θ ) such that seq ( e Λ) < seq (Λ) in the lexicographic order.Observe that commutations do not change seq (Λ).Since q k = 1, mesh Λ ( τ Λ ( a k ) , a k ) = { τ Λ ( a k − ) } and, possibly applying several com-mutations, one can assume that τ Λ ( a k ) , τ Λ ( a k − ) , a k form a braid. Let τ Λ ( a k ) = ( n, j ), τ Λ ( a k − ) = ( n + 1 , l ) , a k = ( n + 2 , j ). We already have ( n + 1 , t ) / ∈ Λ for every t ∈ N ( j ) \ { l } . τ Λ ( a k ) τ Λ ( a k − ) a k a k − τ Λ ( a k − ) z w Figure 7.
ΛTo perform a braiding on τ Λ ( a k ) , τ Λ ( a k − ) , a k , we also need to have ( n + 3 , j ) / ∈ Λ and( n + 2 , v ) / ∈ Λ for every v ∈ N ( l ) \ { j } . To this end we first apply a sequence of com-mutations shifting all vertices x of Λ with p ( x ) ≥ n + 2, x = a k , to the right. Moreprecisely, the resulting set Λ is Λ ⊔ Λ , where Λ = { x ∈ Λ | p ( x ) ≤ n + 1 } ∪ { a k } ,Λ = { ( q + 2 , r ) | ( q, r ) ∈ Λ \ { a k } , q ≥ n + 2 } and θ | Λ = θ | Λ , θ ( q + 2 , r ) = θ ( q, r ) for( q + 2 , r ) ∈ Λ . It is clear that such a transformation can be obtained consequently applyingcommutations to all vertices of Λ \ Λ starting with those with the largest first coordinate. τ Λ ( a k ) τ Λ ( a k − ) a k z τ Λ ( a k − ) a k − w Figure 8.
ΛNow let ( e Λ , e θ ) be a pair obtained by braiding ( τ Λ ( a k ) , τ Λ ( a k − ) , a k ) in Λ. Clearly, q , . . . , q k − remain unchanged. However, the mesh ending at a k − now starts at τ e Λ ( a k − ) =( n + 3 , l ), and hence obviously does not contain the vertex a k = ( n + 2 , j ). We see that | mesh e Λ ( τ e Λ ( a k − ) , a k − ) | = q k − −
1. Since any sequence that starts with q , . . . , q k − − s smaller in the lexicographic order than the sequence ( q , . . . , q k − , q k ), we have seq ( e Λ) Consider Λ γ as in the previous example. 111 2 111 1 01 Figure 10. Thefirst application ofLemma 16 requiresjust one braiding. 111 2 11 0 11 1 Figure 11. Nowthe sequence ofmeshes starting at0 is (2 , 111 2 1 11 0 1 1 Figure 12. The second application of Lemma 16 begins with a sequenceof commutations. 11 1 2 1 11 0 1 1 Figure 13. Now three vertices can be braided and the sequence of meshesstarting at 0 is (1) again. Figure 14. Braiding the remaining three vertices in the sequence of meshesfinishes the second application of Lemma 16. Remark. One can prove the assertion converse to Theorem 3. Namely, if (Λ , θ ) satisfy-ing mesh relations is such that θ is non-negative on Λ, θ ( −∞ , i ) = − i ∈ Γ , θ ( −∞ , k ) = 0 for all k ∈ Γ \ { i } and γ Λ is left-divisible by some s j in B +Γ with j = i ,then θ vanishes on some vertex of Λ. Indeed, in this case (Λ , θ ) can be transformed usingcommutations and braidings into ( e Λ , e θ ) such that there is a vertex x = ( n, j ) ∈ e Λ with τ e Λ ( x ) = ( −∞ , j ) and mesh e Λ (cid:0) τ e Λ ( x ) , x (cid:1) = ∅ . Then the mesh relation corresponding to x implies e θ ( x ) = 0 and the assertion follows from the fact that commutations and braidingsdo not change the multiset of values of θ .9. An application to derived Picard groups For this section we assume that the field k is algebraically closed. All modules in thissection are assumed to be left unless explicitly stated otherwise. Let D Γ = D b (mod − Λ Γ ),the bounded derived category of finitely generated Λ Γ -modules, where Γ is a simply lacedDynkin diagram and Λ Γ is the trivial extension algebra of the Γ diagram with alternatingorientation. In other words, Λ Γ = kQ Γ /I Γ , where Q Γ is one of the following quivers: α * * β j j n − α n − * * nβ n − k k Figure 15. Γ = A n n − β n − v v α * * β j j n − α n − + + n − β n − k k α n − α n − (cid:25) (cid:25) nβ n − Z Z Figure 16. Γ = D n α (cid:10) (cid:10) α * * β j j α * * β j j α * * β J J β j j n − αn − * * nβn − k k Figure 17. Γ = E n ( n = 6 , , he ideal I Γ of kQ Γ is generated by all paths of length greater or equal to 2, except forthe paths of length 2 starting and ending at the same vertex, and the differences of any twopaths of length two starting at the same vertex.The k-algebra Λ Γ is finite-dimensional and symmetric. Let e i be the idempotent associatedwith the vertex i of the quiver Q Γ . Denote by P i = Λ Γ e i the corresponding indecomposableprojective module. Note that P i is a 0-spherical object of D Γ . Indeed, the first conditionis satisfied automatically. Since P i is projective, Ext m Λ Γ ( P i , − ) vanishes for every m = 0,and hence the second condition simply means that End Λ Γ ( P i ) ∼ = k[ t ] / ( t ). Finally, since Λ Γ is symmetric, the last condition is satisfied automatically as well due to an isomorphismof functors Hom( P i , − ) ∼ = Hom( − , P i ) ∗ . To sum up, it is now clear that { P i } i ∈ ( Q Γ ) is aΓ-configuration of 0-spherical objects of D Γ . Remark. In this less general setting we could have defined spherical twists in the followingway: Definition . (Grant, [7]) The spherical twist functor along P i is t P i : D Γ → D Γ t P i ( − ) = cone (Λ Γ e i ⊗ k e i Λ Γ m −→ Λ Γ ) ⊗ L Λ Γ − where m is the multiplication map m ( ae i ⊗ e i b ) = ab of Λ Γ -Λ Γ bimodules.It is easy to check that on objects it coincides with our original definition, which is t P i ( X ) = cone ( P i ⊗ k Hom( P i , X ) ev −→ X )with ev the obvious evaluation map. Indeed, since cone (Λ Γ e i ⊗ k e i Λ Γ m −→ Λ Γ ) is a two-termcomplex of right-projective bimodules, cone (Λ Γ e i ⊗ k e i Λ Γ m −→ Λ Γ ) ⊗ L Λ Γ X ∼ = cone (cid:0) (Λ Γ e i ⊗ k e i Λ Γ ⊗ Λ Γ X ) −→ (Λ Γ ⊗ Λ Γ X ) (cid:1) ∼ = cone (cid:0) P i ⊗ k ( e i Λ Γ ⊗ Λ Γ X ) −→ X (cid:1) ∼ = cone (cid:0) P i ⊗ k Hom( P i , X ) ev −→ X (cid:1) . Definition 9.2. (R. Rouquier, A. Zimmermann, [15]) Let A be a finite-dimensional algebra.The derived Picard group T rP ic ( A ) of A is the group of isomorphism classes of objects ofthe derived category of A ⊗ A op -modules, invertible under ⊗ LA . 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