{'fraction_non_alphanumeric': 0.05916918883520062, 'fraction_numerical': 0.012071010106373565, 'mean_word_length': 3.4699367088607596, 'pattern_counts': {'":': 0, '<': 19, '<?xml version=': 0, '>': 7, 'https://': 0, 'lorem ipsum': 0, 'www.': 0, 'xml': 0}, 'pii_count': 0, 'substrings_counts': 0, 'word_list_counts': {'cursed_substrings.json': 25, 'profanity_word_list.json': 0, 'sexual_word_list.json': 0, 'zh_pornsignals.json': 0}}
Classification of abelian hereditary directed categories satisfying Serre duality

In an ongoing project to classify all hereditary abelian categories, we provide a classification of Ext-finite directed hereditary abelian categories satisfying Serre duality up to derived equivalence. In order to prove the classification, we will study the shapes of the Auslander-Reiten components extensively and use appropriate generalisations of tilting objects and coordinates, namely partial tilting sets and probing of objects by quasi-simples.


Introduction
Let k be an arbitrary algebraically closed field. In this paper we classify, up to derived equivalence, k-linear Ext-finite directed hereditary categories satisfying Serre duality (see below for definitions). In this way we accomplish a step in the ongoing classification project of hereditary categories. As an Ext-finite hereditary category may be viewed as a homological generalization of a smooth projective curve, one of the motivations for this classification project is non-commutative algebraic geometry (see e.g. [12]).
Our classification is a natural complement to the following results.
(2) In [6] Reiten and Van den Bergh classify Ext-finite Noetherian hereditary categories satisfying Serre duality. The hereditary categories considered in (2) fall naturally into several classes. One class contains mild generalizations of smooth projective curves and the other classes are in some way associated to quivers.
In an early version of [6] it was conjectured that the hereditary categories appearing in [6] constitute in fact a complete classification of Ext-finite hereditary categories with Serre duality up to derived equivalence. This conjecture was quickly shown to be false by Ringel [10] who constructed counterexamples. Since these counterexamples are directed it was very natural for us to try to classify directed hereditary categories.
To precisely state our classification result we now give some definitions. Let A be a k-linear abelian category.
(1) We say that A is hereditary if Ext 2 A (X, Y ) = 0 for X, Y arbitrary objects in A. (2) We say that A is Ext-finite if dim Ext i A (X, Y ) < ∞ for all i. (3) Assume that A is Ext-finite. We say that A satisfies Serre duality [3] if there is an autoequivalence F of D b (A) such that there are isomorphisms of k-vector spaces If A is hereditary then if A has Serre duality it has almost split sequences, and the converse is true if A has neither projectives nor injectives [6]. (4) Assume that A is a Krull-Schmidt category (e.g. if A is Hom-finite). Then A is directed if there is no cycle of maps between indecomposable objects X 0 → X 1 → · · · → X n → X 0 which are neither zero, nor isomorphisms. It follows in particular that Ext 1 A (X, X) = 0 for any indecomposable object X. Thus any indecomposable object is rigid. In fact: experience suggests that the entire structure of directed categories is very rigid and that their structure is entirely controlled by combinatorics. Our classification will be stated in terms of the representation theory of certain partially ordered sets. If P is a partially ordered set then we may view it as a category where an arrow between objects x, y ∈ P exists if and only if x < y. The category Rep(P) of P-representations is the category of covariant functors from P to k-vector spaces.
To every element of P there are naturally associated an indecomposable projective object as well as an indecomposable injective object. We call such objects standard projective and injective objects. We say that a representation is finitely presented if it has a finite presentation by finite direct sums of standard projectives. Cofinitely presented is defined dually. We define rep cfp (P) as the full subcategory of Rep(P) consisting of representations which are both finitely and cofinitely presented. The following is a special case of Proposition 4.2.
Proposition 1.1. Assume that P is a forest (i.e. P does not contain x, y, z, t such that x < y < t and x < z < t but y, z are incomparable). Then rep(P) is a hereditary abelian category.
Theorem 1.2. A connected directed hereditary category A satisfying Serre duality is derived equivalent to rep cfp (P) where P is either a Dynkin quiver, A L or D L .
The categories rep cfp (A L ) with L = T → × Z for |T | > 2 were precisely the above mentioned counterexamples constructed by Ringel in [10].
The categories occurring in Theorem 1.2 have rather attractive Auslander-Reiten quivers (see §4). If P = A L then the Auslander-Reiten quivers of rep cfp (A L ) and its derived category have the form: PSfrag replacements L A A [1] In this picture the triangles and squares are symbolic representations for ZA ∞ and ZA ∞ ∞ -components, respectively (see below).
If P = D L then the Auslander-Reiten quivers of rep cfp (D L ) and its derived category have the form: PSfrag replacements L A A [1] In this picture the triangles with a double base are symbolic representation for ZD ∞ -components (see below).
The proof of Theorem 1.2 is quite involved and consists of a number of steps which we now briefly sketch.
Step 1. The following result (see Theorem 5.1) is used a various places. Since it does not depend on Serre duality it may be of independent interest. Theorem 1.3. Let A be a directed abelian Ext-finite hereditary category. Then dim Ext i A (X, Y ) ≤ 6 for all indecomposable X, Y ∈ A and i = 0, 1. Furthermore if either of the vector spaces Hom A (X, Y ) or Ext A (Y, X) is zero then the other is at most 1-dimensional.
Step 2. For the rest of the proof we assume that A is a connected directed hereditary category satisfying Serre duality and we put C = D b (A). Our first aim is to identify the shapes of the connected components of the Auslander-Reiten quiver of C. Since such a component is a stable translation quiver it must be of the form ZB/G where B is an oriented tree and G is acting on ZB ( [8]). Using an appropriate generalization of the theory of sectional paths ( [1]) and directedness we deduce: (1) All components are standard, i.e. all maps are linear combinations of compositions of irreducible ones. In particular all relations can be obtained from the mesh relations. (2) |G| = 1 and furthermore one of the following is true: B is Dynkin, B = A ∞ , B = A ∞ ∞ or B = D ∞ (these may be characterized as the trees not containing non-Dynkin diagrams).
If C has a component ZB with B Dynkin then from connectedness it follows easily that C ∼ = D b (rep(B)). So below we exclude this case.
Step 3. The next step is to understand the maps between different components. Let K be an Auslander-Reiten component of C. Since we know all morphisms in K we may select a partial tilting set ( §5) in K which generates K. In this way we construct a partially ordered set P together with an exact embedding D b (rep(P)) → C whose essential image contains K (and its shifts). The fact that this essential image usually also contains other components allows us to obtain information on the interaction between different components.
Step 4. Now we develop the probing technique ( §6.1). Let us say that an indecomposable object S in a ZA ∞ -component is quasi-simple if the middle term of the right Auslander-Reiten triangle built on S is indecomposable. Using the technique developed in Step 3 we prove that most indecomposable objects in C have precisely two distinct quasi-simples mapping to it and these quasi-simples identify the object uniquely.
Step 5. Our next observation is that we if we have a morphism X → Y in C we can often determine the quasi-simples mapping to its cone by knowing the quasi-simples mapping to X and Y . This gives us a hold on the triangulated structure of C.
Step 6. Finally we use the information gathered in the previous steps to construct a tilting set in C. For example if C has no ZD ∞ -components then this tilting set is where S is an arbitrarily chosen quasi-simple in C. The structure of this tilting set allows us finally to complete the proof of Theorem 1.2.
Acknowledgment I wish to thank Michel Van den Bergh for many useful discussions as well as for contributing some ideas.

Notations and conventions
Throughout this paper, fix an algebraically closed field k of arbitrary characteristic. All algebras and categories are assumed to be k-linear.
If a is a small pre-additive category, we denote by Mod(a) the abelian category of all left modules, thus all covariant functors from a to Mod(k), the category of vectorspaces over k. The category mod(a) is the full subcategory of Mod(a) consisting of finitely presented objects. We will interpret an algebra as a pre-additive category consisting of one object.
For a category C, we will write ind C for a chosen set of representatives of non-isomorphic indecomposable objects of C. If E is an object of C, then ind E is a chosen set of representatives of non-isomorphic indecomposable direct summands of E.
The ring Z is the ring of integers. We will write Z 0 for Z \ {0}. If Q is a quiver, a poset, an algebra, or more generally, a pre-additive category, we will write Q • for the dual quiver, poset, algebra, or pre-additive category.
Given a category C, the Auslander-Reiten quiver of C is defined as follows: the vertices of C are the isomorphism classes [X] of indecomposable objects X. There is an arrow [X] → [Y ] if there is an irreducible morphism between X → Y . If K is a component of the Auslander-Reiten quiver of C, also called an Auslander-Reiten component of C, we will often, by abuse of notation, write X ∈ K and X = Y , when we mean [X] ∈ K and [X] = [Y ], respectively.
For an abelian category or pre-additive category A, we will write K b A for the category of bounded complexes modulo homotopy relations.
If A is abelian, we write D b A for the bounded derived category of A. The category D b A has the structure of a triangulated category. Whenever we use the word "triangle" we mean "distinguished triangle".
We will say a Krull-Schmidt abelian or triangulated category A is connected if for all X, Y ∈ ind A there is a finite, non-oriented path of non-zero morphisms between X and Y or between X and Y [z] for a z ∈ Z, respectively.

Preliminaries
Let k be an algebraically closed field. A k-linear abelian or triangulated category A is Ext-finite if for all objects X, Y ∈ Ob(A) one has that dim k Ext i (X, Y ) < ∞ for all i ∈ N. We say that A is hereditary if Ext i (X, Y ) = 0 for all i ≥ 2.
If A is an abelian category, we will say that A satisfies Serre duality if there exists an auto- is natural in X and Y and where (−) * denotes the vector space dual. Such a functor is necessarily exact ( [3]).
It has been proven in [6] that an abelian category A has a Serre functor if and only if the category D b A has Auslander-Reiten triangles. In that case the action of the Serre functor on objects coincides with τ [1], where τ is the Auslander-Reiten translation.
We will continue with some general remarks concerning directed categories. Recall that a category C is directed if all indecomposable objects are directing, thus if for all indecomposable As a result, we may conclude that rad(X, X) = 0 and thus Hom(X, X) = k.
Mostly, we will work on the bounded derived category D b A of a directed abelian hereditary category A. It is standard that there is a full embedding A → D b A by identifying objects of A with complexes of D b A concentrated in degree zero. We will identify A as a subcategory of D b A by this embedding. In the hereditary case, every element of D b A can be written as a finite direct it is easy to check that the category D b A is directed as well.
We shall formulate certain restrictions on the Hom-sets of directed categories as a lemma. Proof. Since A is hereditary and X is indecomposable, we know that X is contained in A[z] considered as full subcategory of D b A, for a certain z ∈ Z. Without loss of generality, assume X to be contained in A[0]. Since Hom(X, Y ) = 0 and A is hereditary, we may assume either , we have Hom(X, Y [z]) = 0 for all z ∈ Z 0 . Indeed, the case z < 0 is clear and the case z > 1 follows from hereditariness. Thus assume Hom(X, built on a nonzero morphism X → Y [1] yields a path from Y to X, contradicting directedness. We will continue by proving Hom(Y, X[z + 1]) = 0 for all z ∈ Z 0 . This is clear for z < −1, follows from directedness for z = −1 and from hereditariness when z > −1.
Note that in the case X, Y ∈ A[0] we have only used that there exists a path from X to Y to prove that Hom(X, is not necessarily indecomposable for i = 1, . . . , n. Write g = (g 1 , . . . , g n ) and f = (f 1 , . . . , f n ) with maps f i : X → Y i and g i : Y i → Z. Then the following statements are true.
(1) If g i were invertible, then g would be a split epimorphism and h would be zero.
(2) If f i were zero, then consider the following morphisms of triangles Since the two compositions on the left are isomorphisms, so too must the composition Y i → Z → Y i . We may conclude that Y i is a direct summand of Z. But since the split monomorphism g i : Y i → Z is not an isomorphism (by 1.), this would contradict the indecomposability of Z. We will also need the following fact.
Lemma 4.1. Assume that f is fully faithful and assume that P is a summand of an object of the form n i=1 b(B i , −) with B i in the essential image of f . Then the canonical map b ⊗ a P a → P is an isomorphism.
Proof. P is given by an idempotent e in ⊕ i,j b(B i , B j ). Hence we may write P as the cokernel of The result now follows easily from the right exactness of b ⊗ a −.
Let M be in Mod(a). We will say that M is finitely generated if M is a quotient of finitely generated projectives. Similarly we say that M is finitely presented if M has a presentation where P, Q are finitely generated projectives. It is easy to see that these notions coincides with the ordinary categorical ones.
Dually we will say that M is cofinitely generated if it is contained in a cofinitely generated injective. Cofinitely presented is defined in a similar way.
The categorical interpretation of the latter notions is somewhat less clear. However if a is Homfinite then both finitely and cofinitely presented representations correspond to each other under duality (exchanging a and a • ).
We say that a small pre-additive category a is semi-hereditary if the finitely presented objects mod(a) in Mod(a) form an abelian and hereditary category. Following proposition shows that semi-hereditariness is a local property. Proposition 4.2. Let a be a small pre-additive category such that any full subcategory of a with a finite number of objects is semi-hereditary. Then a is itself semi-hereditary.
Proof. As usual it is sufficient to prove that the kernel of a map between finitely generated projectives p : P → Q in Mod(a) is finitely generated projective and splits off.
Since a finitely generated projective a-module is a summand of an a-module of the form ⊕ n i=1 a(A i , −) we may without of loss of generality assume that p is a map of the form Such a map is given by a collection of maps p ji : B j → A i . Let b be the full subcategory of a containing the objects (A i ) i , (B j ) j and let F be the filtered collection of full subcategories of a containing b and having a finite number of objects.
For c ∈ F let K c be the kernel of the map given by the same (p ji ) ij . Put K = a ⊗ b K b . By hypotheses K b is finitely generated and a summand of and it follows that the analogous facts are true for K. So to prove the proposition it is sufficient to prove that K is the kernel of p.
That K is the kernel of p can be checked pointwise. Hence it is sufficient to show it for an arbitrary c ∈ F . Since it is easy to see that ( Remark 4.4. It is not true that a filtered direct limit of semi-hereditary rings is semi-hereditary. A counterexample is given in [2].
Let P be a partially ordered set. We associate to P a pre-additive category kP as follows: the object of kP are the elements of P and For i ≤ j denote the element of (kP)(i, j) corresponding to 1 ∈ k by (i, j). Composition of maps in kP is defined as (j, k)(i, j) = (i, k). We write Rep(P) for Mod(kP) and rep(P) for mod(kP). The objects of Rep(P) are often called P-representations.
In this article, we will mainly be interested in the category of finitely presented and cofinitely presented representations of a poset P, which we will denote by rep cfp (P).
Note that if in Rep(P) the finitely generated projectives are cofinitely presented, we have rep cfp (P) ∼ = rep(P).
We will say that a poset is a forest if for all i, j ∈ P such that i < j the interval is totally ordered. It is clear that a subposet of a forest is a forest. Now, assume that Q is a poset which is a forest. Then any finite subposet Q 0 of Q is still a forest and hence Rep(Q 0 ) is hereditary. It follows from Proposition 4.2 that rep(Q) and also rep cfp (Q) are hereditary abelian categories.
We will now proceed to define two posets of special interest. We will say that a poset is locally discrete if no element is an accumulation point. Thus a linearly ordered poset is locally discrete if and only if for each non-maximal element i there exists an immediate successor i + 1 and for each non-minimal element i there exists an immediate predecessor i − 1. If L is a linearly ordered poset, we will denote by D L the set {Q 1 , Q 2 } · ∪ L endowed with a poset structure induced by the relation X < Y ⇔ X, Y ∈ L and X < L Y , or X ∈ {Q 1 , Q 2 } and Y ∈ L In analogy with the notation used for Dynkin quivers, we will also write A L for L. For the rest of this section, we will assume L to be a locally discrete linearly ordered set with no extremal elements, thus not having a maximal nor a minimal element.
4.1. The category rep cfp (A L ). These categories have already been considered in [10]. In this section we will recall some results.
For all i, j ∈ A L with i ≤ j we will write It is easily seen that A i,j is cofinitely presented, thus it is an indecomposable object of rep cfp (A L ). Following lemma will classify all objects of rep cfp (A L ). Proof. We first prove that A i,j is indecomposable. It is easy to see that the number of indecompos- Conversely, let X be an indecomposable object of rep(A L ). Since X is finitely presented in Rep(A L ), we may choose finitely many projectives generating a full subcategory A of Rep(A L ) containing X, such that the embedding i : A → Rep(A L ) is right exact. This subcategory A is equivalent to rep(A n ) for a certain n ∈ N, hence for certain i, j ∈ A n . We may assume f = 0, since otherwise X would be projective in A and in rep cfp (A L ), and hence will not have a cofinite presentation in rep cfp (A L ). It is easily seen that rep cfp (A L ) is connected. We need only check that rep cfp (A L ) is directed and satisfies Serre duality. First Since each X i is has a finite presentation in Rep(A L ), we may choose finitely many projectives generating a full subcategory A of Rep(A L ) containing every X i . Since A is equivalent to the directed category mod(A n ) for a certain n ∈ N, this gives the required contradiction.
Since rep cfp (A L ) has neither projectives nor injectives (for every indecomposable A i,j there is a non-split epimorphism A i,j+1 → A i,j , and a non-slit monomorphism A i,j → A i−1,j ), we know by [6] that the existence of a Serre functor on D b (rep cfp (A L )) is equivalent to the existence of Auslander-Reiten sequences in rep cfp (A L ). Let A i,j be an indecomposable object of rep cfp (A L ). We claim that the exact sequence is an Auslander-Reiten sequence. To illustrate this, let Y be an indecomposable object of rep cfp (A L ) and choose finitely many projectives generating a full subcategory A of Rep(A L ) containing Y and the exact sequence (1). It is clear that A is equivalent to the category rep(A n ) for a certain n ∈ N and that the short exact sequence (1) is an almost split exact sequence in A. Hence all morphisms Y → A i,j and A i−1,j−1 → Y factor through the middle term. This shows that the exact sequence (1) is an Auslander-Reiten sequence in rep cfp (A L ).
Finally, we will give the Auslander-Reiten quiver of rep cfp (A L ). Therefore, let T be any linearly ordered set and consider the poset L = T → × Z defined by endowing T × Z with the lexicographical ordering. Thus, for all t, t ′ ∈ T and z, z ′ ∈ Z, we have It is readily seen that L is a locally discrete linearly ordered set with no extremal elements and, conversely, that every such ordered set is constructed in this way. For every t ∈ T the Auslander-Reiten quiver of rep cfp (A L ) has a ZA ∞ -component as given in Figure 1. With two distinct elements t < t ′ correspond a ZA ∞ ∞ -component as given in Figure 2.

4.2.
The category rep cfp (D L ). This section closely parallels the previous one, although some arguments are slightly more elaborate. For all i, j ∈ D L with i ≤ j we will write where in the definition of B i,j we assume i = j.
It is easy to see that A i,j , A 1 j , A 2 j and B i,j are also cofinitely presented, hence they are objects of rep cfp (D L ). In following lemma we prove that these are all indecomposable objects.  With two distinct elements t < t ′ correspond two ZA ∞ ∞ -component as given in Figures 5 and 6.

Partial Tilting Sets
In this section we shall assume A is a k-linear abelian Ext-finite category, not necessarily satisfying Serre duality. We will say that the set In the rest of this article, we will often use next theorem. Recall that a category is called Karoubian if the category has finite direct sums and idempotents split. A small pre-additive category a is coherent if the finitely presented objects mod(a) in Mod(a) form an abelian category.
Theorem 5.1. Let A be a k-linear abelian category, {P i } i∈I a partial tilting set of D b A and a the pre-additive category given by {P i } i∈I as a full subcategory of D b A. Assume that a • is Karoubian and coherent, and that mod(a • ) has finite global dimension, then there is a full exact embedding Proof. Due to the conditions on the pre-additive category a, we know that D b (mod a • ) is equivalent to K b a. It thus suffices to construct a full and exact embedding K b a → D b A. It is well known that the category Ind A of left exact contravariant functors from A to Mod k is a k-linear Grothendieck category and that the Yoneda embedding of A into Ind A is a full and exact embedding. By [5,Prop. 2.14], this embedding extends to a full and exact embedding D b A → D b Ind A. As a Grothendieck category, Ind A has enough injectives and we may, by [7, Prop. 10.1], consider the full and exact embedding K b a → D b Ind A that extends the embedding of a in D b Ind A. Induction over triangles shows that the essential image of K b a lies in D b A.
Next, let A be a directed k-linear abelian hereditary Ext-finite category, not necessarily satisfying Serre duality. Given a pair of indecomposable objects, X and Y , in D b A we will use Theorem 5.1 to find a full and exact subcategory B of D b A, such that X, Y ∈ B and B ∼ = D b (mod A) for a certain finite dimensional k-algebra A.
Proof. In order to ease notation, write Hom(X, Y ) = V . We will prove that (ind E) ∪ {X} is a partial tilting set by applying Hom-functors to triangle (2). Out of the long exact sequence given by Hom(X, −) and directedness we deduce that Hom(X, E[z]) = 0 for all z = 0, 1. For z = 1, consider the following exact sequence Since the map Hom(X, X ⊗ V ) → Hom(X, Y ) is an isomorphism and since due to directedness Hom(X, (X ⊗ V )[1]) = 0, we have Hom(X, E[1]) = 0. Thus, Hom(X, E[z]) = 0, for all z ∈ Z 0 . Next, the functor Hom(−, X) yields a long exact sequence from which we easily deduce that Finally, we are left to prove that Hom(E, E[z]) = 0, for all z ∈ Z 0 . To this end, we first apply the functor Hom(Y, −) to triangle (2). Using Lemma 3.1 to see that Hom(Y, Y [z]) = 0 and Hom(Y, X[z + 1]) = 0 for all z ∈ Z 0 , we may deduce that Hom(Y, E[z + 1]) = 0 for all z ∈ Z 0 . One may now readily see that the long exact sequence one acquires by applying Hom(−, E) to triangle (2) yields Hom(E, E[z]) = 0 for all z ∈ Z 0 . This proves the assertion. Theorem 5.3. Let A be a directed abelian Ext-finite k-linear hereditary category and let X and Y be two indecomposable objects of A, then dim Hom(X, Y ) ≤ 6 and dim Ext(X, Y ) ≤ 6. If Ext(Y, X) = 0, respectively Hom(Y, X) = 0, then dim Hom(X, Y ) ≤ 1, respectively dim Ext(X, Y ) ≤ 1.
Proof. We will work on the derived category D b A. Possibly by renaming Y [1] to Y , it suffices to prove that dim We may assume Hom(X, Y ) = 0. This shows that A is a representation-directed algebra. Let P and Q be the projective objects of mod A corresponding to E and X, respectively, under i. Since i is exact, we know R = cone(P → Q ⊗ Hom(X, Y )) corresponds to Y under i.
By applying the functor Hom In this section, let A be a connected directed hereditary abelian k-linear Ext-finite category satisfying Serre duality, and write C = D b A. Since A satisfies Serre duality, the bounded derived category C = D b A is stable. Due to [8], we may state that the only possible components of the Auslander-Reiten quivers are of the form ZB/G, where G is an admissible subgroup of Aut(ZB) and B is an oriented tree. We wish to show that, in our context, the only possible components are ZA ∞ , ZA ∞ ∞ , ZD ∞ , or ZQ where Q is Dynkin, and that all such components are standard, i.e. rad ∞ (X, Y ) = 0 when X and Y are contained in the same Auslander-Reiten component. We will need some preliminary results, starting with a definition from [1]. . . , n} modulo n. Let · · · → A i → A i+1 → · · · be a sequence of irreducible morphisms between indecomposables with each index in I. The sequence is said to be sectional if τ A i+2 ∼ = A i whenever both i and i + 2 are in I. The corresponding path in the Auslander-Reiten quiver is said to be a sectional path.  Next, assume the assertion has been proven for n ∈ {1, 2, . . . , k − 1}, we wish to prove the case n = k. First of all note that since Hom [1], and using the fact that dim Hom(A, A k−1 ) = 1, one has dim Hom(A, M ) = 1. Since M has both A k−2 and τ A k as directs summands and it has already been proven that dim Hom(A, A k−2 ) = 1, it follows easily that Hom(A, τ A k ) = 0, and thus Hom(A k , A[1]) = 0. It has already been noted that Hom(A, A k ) = 0 since they lie in the same sectional sequence and thus, by    We will now discuss the form of the components that can occur in the Auslander-Reiten quiver of the category C. Recall from [8] that a stable component K from the Auslander-Reiten quiver of C is covered by π : ZB → K, where B is defined as follows : fix a vertex X from K, then the vertices of B are defined to be all (finite, non-trivial) sectional paths of K starting at X, and there is an arrow in B from the sectional path X → · · · → Y to the sectional path X → · · · Y → Z. With these definitions, it is clear that B is a tree with a unique source. There also is a morphism f : B → K by mapping a sectional path X → · · · → Y to Y . This morphism f extends to the covering π : ZB → K of translation quivers given by In the following lemma, we will prove that the map π is injective, such that ZB ∼ = K. Lemma 6.4. Every component of the Auslander-Reiten quiver of C is isomorphic to ZB, as stable translation quivers, for a certain oriented tree B with a unique source.
We will consider the sectional paths where we may without loss of generality assume that A i = τ k B j for 1 ≤ i ≤ n−1, for 1 ≤ j ≤ m−1 and for all k ∈ Z. We will consider two separate cases. First, assume that z ′ − z ≥ n. In that case, we have a path from τ z ′ −z Y to τ n Y and a path If z ′ − z < n, then we find two different sectional paths We are now ready to prove the main theorem of this section.
Theorem 6.5. Let A be a directed hereditary abelian k-linear Ext-finite category satisfying Serre duality. Each component of the Auslander-Reiten quiver of C = D b A is both (1) standard, and (2) of the form ZA ∞ , ZA ∞ ∞ , ZD ∞ or ZQ, where Q is a quiver of Dynkin type. We will split the proof of this theorem over the next two lemmas. In the next lemma, we will denote by d(a, b) the usual graph-theoretical distance between vertices a and b. Lemma 6.6. Each component of the Auslander-Reiten quiver of C is standard.
Proof. Let X and Y be two indecomposable objects in an Auslander-Reiten component K of C. In order to prove rad ∞ (X, Y ) = 0 we will write K as ZB such that X corresponds to (0, b) where b is the source of the tree B. We will consider two cases.
The first case is where Y has coordinates (n, v Y ) with n ≥ 0 and v Y ∈ B. If n = 0 then rad ∞ (X, Y ) = 0 as a consequence of Proposition 6.3. Now assume n > 0. If rad ∞ (X, Y ) = 0, then consider the Auslander- is finite, iteration gives rad ∞ (X, Z) = 0 for a certain Z with coordinates (0, v Z ). This contradicts Proposition 6.3.
The last case is where Y has coordinates (−n, v Y ) with n > 0. We will proceed by induction on n to prove that Hom(X, Y ) = 0. First, we consider n = 1, thus let Y have coordinates (−1, v Y ). We will use a second induction argument on d = d(b, v Y ). If d = 0, then b = v Y and there is a path from Y to X. Directedness then implies Hom(X, Y ) = 0. Now, assume d ≥ 1. Choose a direct predecessor v Z of v Y . Let Z ∈ K be the indecomposable object corresponding to the coordinates (−1, v Z ), then Y is a direct summand of M where M is defined by the Auslander-Reiten triangle Z → M → τ −1 Z → Z [1].
By the induction on d above we know that Hom(X, Z) = 0 and by Proposition 6.3 that dim Hom(X, τ −1 Z) = 1 so we infer that dim Hom(X, M ) = 1 if X ∼ = τ −1 Z, and dim Hom(X, M ) = 0 if X ∼ = τ −1 Z. Thus there is at most one indecomposable direct summand X ′ of M such that Hom(X, X ′ ) = 0. This needs to be an indecomposable object lying on a sectional path from X to We will now proceed with the induction on n, thus let Y have coordinates (−n, v Y ) with n > 1. Also in this case, we will use a second induction argument on d = d(b, v Y ). As above, if d = 0, then b = v Y and there is a path from Y to X. Directedness then implies Hom(X, Y ) = 0. Thus we may assume d ≥ 1. Let v Z be a direct predecessor of v Y in the tree B and let Z ∈ K be the indecomposable object corresponding to the coordinates (−n, v Z ). Again, Y is a direct summand of M where M is defined by the Auslander-Reiten triangle Z → M → τ −1 Z → Z [1].
For the statement of the next proposition, we will say that a subquiver Q of a stable translation quiver T is a section if the embedding of quivers Q → T lifts to an isomorphism ZQ ∼ = T as stable translation quivers, hence Q contains exactly one object from every τ -orbit of T and if X ∈ Q and there is an arrow X → Y in T , then either Y ∈ Q or τ Y ∈ Q. Proposition 6.8. Let K be a component of the Auslander-Reiten quiver of C, and let Q be a section of K. Then the vertices of Q form a partial tilting set.
Proof. Let X and Y be two indecomposable objects from the section Q. We must show that Hom(X, Y [z]) = 0 for z ∈ Z 0 . Without loss of generality, assume X ∈ Ob A[0]. It is clear that there are i, j ∈ N such that there is a sectional path from τ i Y to X and from X to τ −j Y and thus, by Proposition 6.3 and the fact that A is hereditary, we may conclude [1]. Hence Hom(X, Y [z]) = 0 for z < −1 and z > 2. We will proceed to show that Hom(X, Y [z]) = 0 for z ∈ {−1, 1, 2}.
Therefore, we will show there exists an indecomposable Z ∈ K such that there are sectional paths from Z to both X and Y . Indeed, let n ∈ N be the smallest natural number such that there is a path τ n Y = A 0 → A 1 → · · · → A m = X.
Note that such a path is necessarily sectional and n ≤ m. By turning the first n arrows one gets a path from A n to Y and a path from A n to X which are sectional by minimality of n. Hence let Z = A n . First, we will prove Hom(X, Having proved Theorem 6.5, we now turn our attention to the possible shapes of the Auslander-Reiten components. First we will discuss a tool we will be developing and using in the next sections.
6.1. A word on probing. In this section, we describe the probing technique. Following results summarize the results of §6.2, §6.3, §6.4, and §6.5.
As usual, A is a connected directed abelian hereditary k-linear Ext-finite category satisfying Serre duality, and we write C = D b A. We have proven in Theorem 6.5 that the only occurring Auslander-Reiten quivers are of the form ZQ where Q is either A ∞ , A ∞ ∞ , D ∞ or a Dynkin quiver. In Proposition 6.13 will be proven that if C has an Auslander-Reiten component ZQ where Q is a Dynkin quiver, then this is the only component of C. Since we are interested in the connection between different components, we will exclude such Auslander-Reiten components from this section.
We will start our discussion with a definition. Let U and U ′ be Auslander-Reiten components. We will say U maps to U ′ if there is an object X ∈ U and Y ∈ U ′ such that Hom(X, Y ) = 0.
It will turn out that the ZA ∞ -components, also called wings, are the building blocks of the category C. We consider the following map. Proposition 6.9. The map φ comp is injective. Also consists of a single wing.
Proof. If U is a ZA ∞ -component, a ZA ∞ ∞ -component, or a ZD ∞ -component, then the shape of φ comp is a direct consequence of Propositions 6.15 and 6.17, Propositions 6.18 and 6.22, or Propositions 6.24 and 6.28, respectively. Injectivity of φ comp follows from Propositions 6.17, 6.23 and 6.29.
We now turn our attention from the components to the objects. Again, we start with a definition.
We will say that an indecomposable object S ∈ ind C is a peripheral object if the middle term of the right Auslander-Reiten triangle is indecomposable. A peripheral object lying in a wing is a quasi-simple object.
Quasi-simple objects will be used to, in a certain sense, give coordinates to objects of C much like wings can be used as coordinates for components. We define the function φ obj : {indecomposables of C} → {sets of quasi-simples of C} X → {S | S is a quasi-simple that maps non-zero to X} Proposition 6.10. Let X be an indecomposable object lying in an Auslander-Reiten component U. We have the following properties.
(1) For all W ∈ φ comp (U), there is an S ∈ φ obj (X) such that S ∈ W.
(2) The set φ obj (X) consists of two elements, except if X is a peripheral object from a ZD ∞component, then φ obj (X) has only one element. (3) The fiber of φ obj (X) consists of one element, except when X is a peripheral object in a ZD ∞ -component, then the fiber of φ obj (X) consists of two elements. (4) If S ∈ φ obj (X) then dim Hom(S, X) = 1. (5) If S is a quasi-simple and f : S → X a non-zero non-invertible morphism, then the map g Proof.
(1) This follows from Propositions 6.15, 6.18 and 6.24. (2) First assume that X is not a peripheral object from a ZD ∞ -component. Propositions 6.15, 6.18 and 6.24 yield that there are at least two different quasi-simple objects mapping non-zero to X. Propositions 6.17, 6.22 and 6.28 imply that these are unique. If X is a peripheral object from a ZD ∞ -component, then Proposition 6.24 yields that there is at least one quasi-simple object mapping non-zero to X. Finally, Proposition 6.28 then shows this quasi-simple is unique.
(3) Again, first assume X is not a peripheral object of a ZD ∞ -component.
If U is a ZA ∞ ∞ -component, then Proposition 6.9 yields φ comp (U) = {V, W}, with V = W[z] for all z ∈ Z. Now, (1) and (2) yield that φ obj (X) = {S, T } with S ∈ V and T ∈ W. By Proposition 6.18 we see that the restricted function φ obj | ZA ∞ ∞ is injective. If U is a ZD ∞ -component, then Proposition 6.9 yields φ comp (U) = {V} and (1) and (2) imply that φ obj (X) = {S, T } with S, T ∈ V. We may now use Proposition 6.24 to see that the restricted function φ obj | ZD∞ is injective.
We may now conclude that the fiber of φ obj (X) consists of only one object, X. Now, assume X is a peripheral object of a ZD ∞ -component U. We have already shown that there is a unique quasi-simple object, S, such that Hom(S, X) = 0 and S ∈ W where φ comp (U) = {W}. Since φ comp is injective, U is the only ZD ∞ -component where W maps to. Proposition 6.24 now yields that the fiber of φ obj (X) = S consists of exactly two objects, both peripheral objects of U. (4 & 5) These are immediate consequences of Propositions 6.15, 6.18 and 6.24. Proposition 6.10(5) will be used in combination with following lemma from [1], adapted to the triangulated case.
and a morphism h : Z ′ → Z. If f is irreducible, then g factors through h or vice versa, thus there exists a t : Z ′ → Y such that h = gt or an s : Y → Z ′ such that g = hs.
Proof. Consider the morphism of triangles Because f is irreducible, we know that u is split mono (and thus v split epi) or w split epi. In the former case there exists a t : Z ′ → Y such that h = gt while in the latter there is a morphism s : Y → Z ′ such that g = hs.  We know that C does not have any ZD ∞ -components, and thus, by Proposition 6.10(2), that all objects have exactly two quasi-simples mapping non-zero to them. Since there are exactly four quasi-simples mapping to M , we may conclude that M has exactly two direct summands, M 1 and M 2 .
Using Proposition 6.10(1), we may infer that there are two possibilities, either φ obj (M 1 ) = In the former case, Proposition 6.10(3) yields M 1 ∼ = A −1,1 and M 2 ∼ = A 0,3 . Lemma 3.2 then implies that there exists a non-zero morphism from A −1,1 to M 1 . This morphism is necessarily an isomorphism; we conclude that A −1,1 → M is a split monomorphism and hence that f = 0. A contradiction.

6.2.
A ZQ-component with Q a Dynkin quiver. We first consider a category C = D b A whose Auslander-Reiten quiver has a ZQ-component where Q is a Dynkin quiver. Note that the categories rep(Q) and rep cfp (Q) are equivalent. Following proposition shows we may exclude these components from our further discussion of the other components. Proof. Proposition 6.8 yields that the section Q in the ZQ-component K is a partial tilting set. Using the exactness of the Serre functor F , it is easily seen that the full and exact embedding ∆ : D b (rep(Q • )) → C given by Theorem 5.1 commutes with Serre duality. Thus those indecomposable objects in the essential image of ∆ are exactly those whose isomorphism class lie in K. We claim that for all X in the essential image of ∆ and for all Y ∈ Ob C, we have rad ∞ (X, Y ) = 0. Indeed, if rad ∞ (X, Y ) = 0 then, for all n ∈ N, there would be an X n ∈ K such that rad n (X, X n ) = 0. Yet, this is not true in D b (rep(Q • )).
Hence, since A is connected, C can consist of only one component. We conclude C ∼ = D b (rep(Q • )).
We give a further result in this context. Proposition 6.14. Let A be a hereditary category. If D b A ∼ = D b (rep cfp (Q)) where Q is a Dynkin quiver, then A ∼ = rep(Q ′ ) where the quiver Q ′ is a tilt of Q.
Proof. First note that D b (rep cfp (Q)) and thus also D b A are directed and Ext-finite. Since A and rep cfp (Q) are hereditary, we have Hence ind A is finite. We will now show this implies that A has enough projectives. Indeed, let X ∈ Ob(A). If X is not projective, there exists an object M = ⊕ i M i where M i is indecomposable for all i, such that there is a non-split epimorphism M → X. Since A is directed and ind A is finite, every sequence of non-split epimorphisms · · · → M → X needs to be finite and we deduce the existence of a projective object P that admits a non-split epimorphism P → X.
Consider the object P = ⊕ j P j where P j ranges through all projectives of ind A. We see that P is a generator, and hence A ∼ = mod(A) with A = End(P ).
Since A is hereditary and of finite representation type, A needs to be Morita equivalent to the path algebra of a Dynkin quiver Q ′ .
Finally, note that the Auslander-Reiten quiver of D b A and D b rep cfp (Q), are equal to ZQ ′ and ZQ, respectively, hence Q ′ is a tilt of Q.

6.3.
A ZA ∞ -component. A ZA ∞ -component will often be called a wing and when we encounter such a component in a derived category, we will represent it by a triangle as shown in Figure 7. . . .
and let K be a ZA ∞ -component of C, then the smallest full and exact subcategory of C containing K is equivalent to D b rep Q, and the embedding ∆ : D b rep Q → C commutes with Serre duality. Hence ∆ maps Auslander-Reiten components to Auslander-Reiten components.
Proof. Consider within K the quiver Q • as in Figure 8. We will denote the indecomposable corresponding to the vertex i of Q • by P i . Invoking Proposition 6.8 and Theorem 5.1, we may consider a full and exact embedding ∆ : D b rep Q → C which we claim to commute with the Serre functor.
Considering the exactness of ∆, and the connection between the Auslander-Reiten translation τ and the Serre functor F , it is easy to see ∆F P i ∼ = F ∆P i for all i ∈ N. Since the Serre functor is exact and commutes with ∆ on generators of D b rep Q, it will commute with ∆.
We still need to check whether ∆ maps Auslander-Reiten components to Auslander-Reiten components essentially surjective, i.e. if an indecomposable of a component is in the essential image of ∆, then so is every indecomposable of that component. To this end, consider an indecomposable object C in C in the essential image of ∆ such that there is an irreducible D → C where D is not in the essential image of ∆ (the dual case where there is an irreducible C → D is completely analogue). If C is in the essential image of ∆, then so is τ C since ∆ commutes with F . We may consider the Auslander-Reiten triangles  Since ∆ is full, faithful and exact, we know that End(N ) ∼ = End(N ′ ), hence N and N ′ consists of the same number of indecomposable summands and there must be an indecomposable direct summand D ′ of N ′ such that ∆(D ′ ) ∼ = D.
Remark 6.16. The category rep Q occurring in the proof has been described in [6]. We may sketch the bounded derived category D b rep Q as shown in Figure 9 where we have marked the abelian subcategory rep Q with gray. Note that D b rep Q ∼ = D b rep cfp Z.
Proposition 6.17. Let V and W be wings. If V maps to W, then W = V or W = V [1].

Proof. It is clear that
To prove that V does not map to W otherwise, we start by fixing a notation. Let Q be the quiver Since V ∼ = ZQ ∼ = W as stable translation quivers, we may label the vertices of V by V m,n and the vertices of W by W m,n for m ∈ Z and n ∈ N as is illustrated in Figure 10.
One sees easily that Assume Hom(V m,n , W i,j ) = 0, then there are p and q in N with m ≤ p ≤ m + n and i ≤ q ≤ i + j such that Hom(V p,0 , W q,0 ) = 0, and hence Hom(τ k V p,0 , τ k W q,0 ) ∼ = Hom(V p−k,0 , W q−k,0 ) = 0 for

6.4.
A ZA ∞ ∞ -component. In this part, we will discuss the ZA ∞ ∞ -component. A ZA ∞ ∞ -component in a derived category will be drawn as in Figure 11. Proposition 6.18. Let Q be the quiver and let K be a ZA ∞ ∞ -component of C, then the smallest full and exact subcategory of C containing K is equivalent to D b rep Q, and the embedding ∆ : D b rep Q → C commutes with Serre duality. Hence ∆ maps Auslander-Reiten components to Auslander-Reiten components.
Proof. The construction of the functor ∆ is similar to the construction in the proof of Proposition 6.15; one now finds the quiver Q • within the ZA ∞ ∞ -component K.
Remark 6.19. The category rep Q occurring in the proof has been discussed in [6]. We may sketch the derived category D b rep Q as in Figure 12 where we have filled the abelian subcategory rep Q with gray. Note that the category rep cfp ({0, 1} → × Z) is derived equivalent to rep Q.
Since the embedding ∆ : D b rep Q → C maps Auslander-Reiten components to Auslander-Reiten components, we may define the map φ ZA ∞ ∞ : {ZA ∞ ∞ -components of C} → {pairs of ZA ∞ -components of C} by mapping a ZA ∞ ∞ -components K to the pair of wings within the essential image of ∆ that map non-zero to K. Remark 6.20. Note that the map φ ZA ∞ ∞ does not depend on the choice of the partial tilting set in the proof of Proposition 6.18. Remark 6.21. It will follow from the following Propositions that φ ZA ∞ ∞ = φ comp | ZA ∞ ∞ . Proposition 6.22. If a wing W maps to a ZA ∞ ∞ -component K, then W ∈ φ ZA ∞ ∞ (K). Proof. We start by fixing an X ∈ K. An argument analogous to the proof of Proposition 6.17 shows that there are only finitely many quasi-simple objects of W that map non-zero to X. We may choose a quasi-simple S from W such that Hom(τ −1 S, X) = 0 and Hom(S, X) = 0. Now, consider the triangle S → X → C → S [1]. Applying the functor Hom(X, −), we may conclude easily that dim Hom(X, C) = 1.
Next, consider the Auslander-Reiten triangle X → Y ⊕ Y ′ → τ −1 X → X [1]. Since the morphism X → C factors through Y ⊕ Y ′ , we may assume there exists a morphism X → Y such that the composition X → Y → C is non-zero. This gives rise to the following morphism of triangles y y Since X → Y is an irreducible morphism between indecomposable objects, Proposition 6.18 yields that T is a quasi-simple object from a wing V ∈ φ ZA ∞ ∞ (K). The induced morphism T → S is easily seen to be non-zero. Proposition 6.17 yields W = V or W = V [1]. By Proposition 6.18 we may exclude the latter. We conclude W ∈ φ ZA ∞ ∞ (K). Proposition 6.23. The map . Fix a quasi-simple object S from the component V and a quasi-simple object T from the component W. Proposition 6.18 yields a unique indecomposable object X from K and a unique indecomposable object X ′ from K ′ such that Hom(S, X), Hom(T, X), Hom(S, X ′ ) and Hom(T, X ′ ) are all non-zero. Furthermore, it follows from Proposition 6.18 that all these Hom-spaces are 1-dimensional and from Proposition 6.22 that these are the only quasi-simples mapping to X or X ′ . We wish to prove that X ∼ = X ′ .
Since by Proposition 6.18 the map X → Y 1 occurring in the triangle S → X → Y 1 → S[1] is irreducible, we may use Lemma 6.11 to see there is a morphism X → X ′ or X ′ → X. Thus without loss of generality, we may conclude there is a commutative diagram Analogously, considering the triangle T → X → Y 2 → T [1] and directedness gives a morphism X → X ′ and we obtain the commuting diagram It easily follows there is a morphism f : X → X ′ such that both compositions S → X → X ′ and T → X → X ′ are non-zero. In order to prove that f is an isomorphism, we will use quasi-simples to probe the object M = cone(f : X → X ′ ), i.e. we will look which quasi-simple objects map to M . The triangle built on f , extended with all the quasi-simple objects mapping to each of its objects looks like y y We will show that no quasi-simple object U may map to M . Seeking a contradiction, assume that f ′ : U → M is such a non-zero morphism. We will first consider the case where the composition U → M h → X [1] is zero. In this case the map U → M factors though g : X ′ → M , and U would be isomorphic to either S or T . But since then dim Hom(U, X ′ ) = 1, we may further → X ′ [1] is non-zero, a contradiction. Using Propositions 6.15 and 6.18, and already using Proposition 6.24 from the next section, we see that every non-zero object of C has at least one quasi-simple object mapping to it. The cone M thus has to be the zero object, establishing the fact that X and X ′ are isomorphic, and thus that K = K ′ . 6.5. A ZD ∞ -component. In this part, we will discuss the ZD ∞ -component within the directed category. Mostly, the proofs are analogous to the case of a ZA ∞ ∞ -component. We will draw a ZD ∞ -component of a derived category as a triangle with a doubled base as in Figure 13 Proposition 6.24. Let Q be the quiver and let K be a ZD ∞ -component of C, then the smallest full and exact subcategory of C containing K is equivalent to D b rep Q, and the embedding ∆ : D b rep Q → C commutes with Serre duality. Hence ∆ maps Auslander-Reiten components to Auslander-Reiten components.
Proof. The proof is analogue to the proof of Proposition 6.15. One finds the quiver Q • within the ZD ∞ -component K.
Remark 6.25. The category rep Q occurring in Proposition 6.24 has been discussed in [6]. We may sketch the derived category D b rep Q as in Figure 14 where we have filled the abelian subcategory rep Q with gray. Note that the category rep cfp D Z is derived equivalent to the category rep Q.
As in discussion of the ZA ∞ ∞ -component, the embedding ∆ : D b rep Q → C maps Auslander-Reiten components to Auslander-Reiten components. Hence, we may define the map by mapping a ZD ∞ -components K to the set of wings within the essential image of ∆ that map non-zero to K.
Remark 6.26. Note that the map φ ZD∞ does not depend on the choice of the partial tilting set in the proof of Proposition 6.24.
The proofs of the following propositions are analogous to the proofs of the corresponding properties in our discussion of the ZA ∞ ∞ -component.

Classification
Let A be a connected directed hereditary abelian k-linear Ext-finite category satisfying Serre duality, and write C = D b A. In this section we will prove our main theorem. We will proceed as follows. First, we will define a partial tilting set S → . Then Theorem 5.1 will give a full and exact embedding D b (mod S • → ) → D b A. We will proceed to prove that this embedding is an equivalence of triangulated categories. Finally, the classification will follow from the shape of the poset S → . We start by defining S → .
Choose a quasi-simple object S in a wing W. We will consider two cases. First, assume that S does not map to two peripheral objects Q 1 S and Q 2 S of a ZD ∞ -component or, equivalently, there is no ZD ∞ -component K such that W ∈ φ ZD∞ (K). In this case just let S → be the set of indecomposable objects X such that there exists a map from S to X, thus Secondly, assume S does map to two peripheral objects Q 1 S and Q 2 S of a ZD ∞ -component or, equivalently, there is a ZD ∞ -component K such that W ∈ φ ZD∞ (K). Then, let S → be the set of indecomposable objects X such that there exists a map from S to X and a map from X to Q 1 S or Q 2 S , thus S → = {X ∈ ind C| Hom(S, X) = 0 and Hom(X, Q 1 S ⊕ Q 2 S ) = 0}. This defines the full subcategory S → of C. We will define a poset structure by X ≤ Y ⇔ Hom(X, Y ) = 0 and by identifying isomorphic objects. Before proving in Lemma 7.3 that this does indeed define a poset structure, we will give two examples.
Therefore, we will fix following notation. Let P 1 and P 2 be posets. The poset P 1 · P 2 has P 1 · ∪ P 2 as underlying set and We may draw S → within D b (rep cfp (A L )) as in Figure 15 where, as usual, the abelian category rep cfp (A L ) has been filled with gray. Note that in Examples 7.1 and 7.2 the set S → is of the form A • L or D • L where L is a bounded locally discrete linearly ordered set, i.e. a locally discrete linearly ordered set with both a minimal and a maximal element. It is easily seen that for such linearly ordered posets we have L ∼ = (N) · (T → × Z) · (−N) for a certain linearly ordered set T . Following lemma will classify all possible posets that may occur as S → . For X, Z ∈ S → we will write Proof. We start with the case where S does not map to peripheral objects of a ZD ∞ -component and wish to prove that S → ∼ = A • L where L is a bounded locally discrete linearly ordered set. The relation ≤ in S → is obviously reflexive. The fact that it is antisymmetric follows from directedness. In order to prove transitivity and linearly ordered, it suffices to prove that Hom(X, Y ) = 0 or Hom(Y, X) = 0. For all X, Y ∈ S → , we may consider the commutative diagram where the bottom line is a triangle and g : X → C is irreducible as is shown in Proposition 6.10(5). Lemma 6.11 now yields that Hom(X, Y ) = 0 or Hom(Y, X) = 0, thus S → is a linearly ordered poset. Even more so, if X ≤ Y we may assume that the composition S → X → Y is non-zero. Using Serre duality, it is easily seen that S → = [S, τ S [1]], thus S → is bounded.
To prove that S → is locally discrete, assume Z ∈ S → is a non-minimal element. We need to prove there exists a finite set A ⊆ S → such that, for all X ∈ S → with X < Z, there is a Y ∈ A with X ≤ Y < Z.
Therefore, consider the Auslander-Reiten triangle τ Z → M → Z → τ Z [1]. Write M = ⊕ i M i where M i is indecomposable. We have already proven that there exists a non-zero morphism from X to Z such that the composition S → X → Z is non-zero. Since X → Z factors through M , it is clear that there exists an M i ∈ ind M such that M i ∈ S → and X ≤ M i < Z, thus we have shown A = S → ∩ ind M .
The case where Z is a non-maximal element is analogous. Hence S → is locally discrete. We now turn our attention to the case where S does map to two peripheral objects of a ZD ∞component and wish to prove that S → ∼ = D • L where L is a bounded locally discrete linearly ordered set.
Note that the left Auslander-Reiten triangles built on Q 1 S and Q 2 S both have the same indecomposable middle term N . It is straightforward to check that S → ∼ = [S, N ] · {Q 1 S , Q 2 S }. Analogous to the first part of the proof, one shows that [S, N ] is a bounded locally discrete linearly ordered set, thus S → ∼ = D • L where L = [S, N ] • . We will now prove most technical results needed to prove our main result, Theorem 7.7. In particular, we will construct an embedding D b (rep(S • → )) → C and prove that it is an equivalence of categories.
Lemma 7.4. The set S → is a partial tilting set.
Proof. We need to prove that Hom(X, Y [z]) = 0 for all z ∈ Z 0 , and all X, Y ∈ S → . If X and Y are the peripheral objects Q 1 S and Q 2 S then the assertion follows easily. Indeed, both are then contained within the same ZD ∞ -component, and all the maps within such component are known from Proposition 6.24.
Since we may now assume that either X or Y is not isomorphic to Q 1 S or to Q 2 S , Lemma 7.3 yields that we may assume Hom(X, Y ) = 0 and hence, by Lemma 3.1, we need only to prove that Hom(Y, X[1]) = 0.
Let f ∈ Hom(Y, X [1]) and consider the triangle We will now probe M to prove M ∼ = X ⊕ Y and hence f = 0. First, consider the case where X = Q 1 S and Y = Q 2 S . Since Y ∈ S → , we have Hom(S, Y ) = 0, and consequently Hom(S, Y [1]) = 0. We conclude that the composition S → X y y Lemma 7.6. Let X, Y ∈ Ob C be indecomposables objects, and let X → Y → Z → X[1] be the triangle built on a non-zero morphism X → Y . If Hom(Y, X[1]) = 0 then Z is indecomposable.
Proof. Using Lemma 3.2 it should be clear that Z has at most dim Hom(Y, Z) direct summands. Applying the functor Hom(Y, −) to the triangle X → Y → Z → X [1], and using that dim Hom(Y, Y ) = 1 and Hom(Y, X[1]) = 0, it follows easily that dim Hom(Y, Z) = 1 and thus that Z is indecomposable.
We now prove our main result.
Theorem 7.7. A connected directed hereditary abelian k-linear Ext-finite category A satisfying Serre duality is derived equivalent to rep cfp (P) where P is either a Dynkin quiver, A L , or D L where L is a locally discrete linearly ordered set without maximum or minimum.
Proof. From Theorem 6.5 we know that the only components of the Auslander-Reiten quiver of D b A are of the form ZA ∞ , ZA ∞ ∞ , ZD ∞ or ZQ, where Q is a quiver of Dynkin type. First, assume C has an Auslander-Reiten component of the form ZQ where Q is a Dynkin quiver. It has been proven in Proposition 6.13 that D b A ∼ = D b (rep cfp (Q • )) Thus we may now assume that the Auslander-Reiten quiver of D b A does not have a component of the form ZQ where Q is a Dynkin quiver. The only possible components then are of the form ZA ∞ , ZA ∞ ∞ or ZD ∞ . It follows from Proposition 6.9 that there is at least one wing W. Fix a quasi-simple S ∈ W and consider the set S → . Lemma 7.5 yields that D b A ∼ = D b (rep(S • → )). From Lemma 7.3 we know that S → ∼ = A • L ′ or S → ∼ = D • L ′ where L ′ = (N) · (T ′ → × Z) · (−N) for a certain linearly ordered set T ′ . In the first case, we will give a poset A L such that D b A is equivalent to D b (rep cfp A L ); in the second case we will give a poset D L such that D b A is equivalent to D b (rep cfp D L ).
Thus, first assume S → ∼ = A • L ′ . We will now consider the category Remark 7.8. It follows from the proof of Theorem 7.7 that L can also be chosen to be a bounded linearly ordered set (i.e. L has a maximal and a minimal element). It then follows in particular that A is derived equivalent to a hereditary category which has both enough projectives and injectives.
Remark 7.9. It is proven in Proposition 6.14 that if, in the statement of Theorem 7.7, P is a Dynkin quiver, then A is equivalent to mod kQ for a certain Dynkin quiver Q.