NAKAYAMA ORIENTED PULLBACKS AND STABLY HEREDITARY ALGEBRAS

JESSICA LÉVESQUE

Abstract. We introduce a new class of algebras, the Nakayama oriented pullbacks, obtained from pullbacks of surjective morphisms of algebras A ↠ C and B ↠ C. We prove that such a pullback is tilted when A and B are hereditary. We also show that stably hereditary algebras respecting the clock condition are Nakayama oriented pullbacks, and we use results about these pullbacks to show when is a stably hereditary algebra tilted or iterated tilted.

Introduction.

Pullbacks of rings and algebras have been studied from many points of view (see, for instance, [1, 8, 16, 17, 22, 25]), but not from the tilting point of view. In this paper, we consider a particular class of pullbacks of algebras over an algebraically closed field K, the Nakayama oriented pullbacks. These are pullbacks of surjective morphisms of algebras A ↠ C and B ↠ C with C a hereditary Nakayama algebra. We construct a tilting module T over this kind of pullback, and we compute the endomorphism algebra End T of this module when B is hereditary (see 2.4.6). A first consequence of this result is the principal theorem of this paper:

Theorem . Let R be a Nakayama oriented pullback of K-algebra surjective morphisms A ↠ C and B ↠ C. Suppose that A and B are hereditary. Then R is tilted.

On the other hand, we show that a stably hereditary algebra respecting the clock condition, that is, the number of clockwise oriented relations on each cycle of its bound quiver equals the number of counterclockwise oriented relations, can be expressed as a Nakayama oriented pullback. As a second consequence of 2.4.6 we give a new proof of theorem 2.6 of [19].

This paper consists of three sections. The first is devoted to preliminaries, the second to Nakayama oriented pullbacks, and the third to stably hereditary algebras.

1. Preliminaries

1.1. Notation. All algebras in this paper are basic, associative, finite dimensional algebras with identities over a fixed algebraically closed field K, and all modules are finitely generated right modules. For an algebra A, we denote by mod A its module category, by ind A a full subcategory of mod A consisting of a complete set of representatives of the isomorphism classes of indecomposable objects in mod A, and by proj A the full subcategory of ind A consisting of the projective objects.

1991 Mathematics Subject Classification. 16G20, 16G70, 16S50.

Key words and phrases. pullback, stably hereditary algebras, tilted and iterated tilted algebras, tilting module, double-zero, clock condition.

Given an A-module M, we denote by pd M its projective dimension and by id M its injective dimension.

We recall that a quiver Q is defined by a set of points Q_{0} and a set of arrows Q_{1}. A relation from x ∈ Q_{0} to y ∈ Q_{0} is a linear combination of paths from x to y of length at least two. Let KQ be the path algebra of Q, and let I denote the ideal of KQ generated by a set of relations; then the pair (Q, I) is called a bound quiver. A relation ρ = ∑_{i=1}^{m} λ_{i}w_{i} in I (where the λ_{i} are non-zero scalars and the w_{i} are paths) with m ≥ 2 is called minimal if there is no proper non-empty subset J ⊂ {1, ..., m} such that ∑_{J} λ_{i}w_{i} is also a relation in I, and is called monomial or zero-relation if it equals a path (m = 1). We say that a monomial relation is of minimal length if it does not contain a proper subpath which is also in I. It is well-known that if A is a basic and connected finite dimensional K-algebra, then there exists a connected bound quiver (Q_{A}, I) such that A ≅ KQ_{A}/I (see [5]). For a point a in the quiver of A, we denote by P (a) the corresponding indecomposable projective A-module, and by I(a) the corresponding indecomposable injective Amodule. Given an A-module M, we denote by Supp M the full bound subquiver of Q_{A} generated by the points a such that Hom_{A}(P (a), M) ≠ 0. We say that A is triangular whenever its quiver Q_{A} has no oriented cycles.

For an arrow α of Q, we denote by s(α) its source, by t(α) its target and by α^{−1} its formal inverse of source s(α^{−1}) = t(α) and of target t(α^{−1}) = s(α). A walk in Q is a sequence w = c_{1}...c_{n} with c_{i} an arrow or the formal inverse of an arrow such that t(c_{i}) = s(c_{i+1}) for all i such that 1 ≤ i < n. A walk w in Q is called reduced if w = c_{1}...c_{n} with c_{i} ≠ c_{i+1}^{−1} for all i such that 1 ≤ i < n. It is called a non-zero walk if it does not contain any zero-relation. Finally, a reduced walk is called a double-zero if it contains exactly two zero-relations, and moreover, they point in the same direction in w. The double-zero has been used for the classification of tilted and quasi-tilted special biserial algebras, string algebras and gentle algebras [2, 7, 13, 14, 15, 18].

For general properties of the category mod A of finitely generated right Amodules, we refer the reader to [5, 23]. For tilting theory, tilted and iterated tilted algebras, we refer the reader to [3, 10, 12, 23]. Throughout this paper, we assume all the tilting modules to be multiplicity-free.

1.2. The Bound Quiver of a Pullback. Let A = KQ_{A}/I_{A} and B = KQ_{B}/I_{B} be bound quiver algebras, and let Q_{C} be a subquiver of Q_{A} and Q_{B} such that i) every connected component of Q_{C} is full and convex in Q_{A} and Q_{B}, ii) the restrictions of I_{A} and I_{B} to KQ_{C} are the same (that is, I_{A} ∩ KQ_{C} = I_{B} ∩ KQ_{C}). We denote I_{A} ∩ KQ_{C} = I_{B} ∩ KQ_{C} by I_{C}. Let e_{C} = ∑_{i∈(Q_{C})_{0}} e_{i} be the identity of C. Then C = KQ_{C}/I_{C} ≅ e_{C}Ae_{C} ≅ e_{C}Be_{C} is a common quotient of A and B.

We have the canonical projections

\begin{aligned}f_{A} : & A & ↠ & C & , & f_{B} : & B & ↠ & C \\ & a & ↦ & e_{C}ae_{C} & & & b & ↦ & e_{C}be_{C} \end{aligned}

which are well defined since every connected component of Q_{C} is full and convex in Q_{A} and Q_{B}. The algebra R = {(a, b) ∈ A × B | e_{C}ae_{C} = e_{C}be_{C}} is the pullback of f_{A} and f_{B}. The following lemma describes the bound quiver of R in terms of the bound quivers of A, B and C. Similar descriptions are done in [16] and [17].

Lemma 1.2.1. Let R be the pullback of f_{A} and f_{B}. Let Q_{R} = Q_{A} ⨿_{Q_{C}} Q_{B} be the pushout of the inclusion morphisms Q_{C} ↪ Q_{A} and Q_{C} ↪ Q_{B}, and I_{R} = I_{A} + I_{B}+ < ρ̄ > with ρ̄ the set of paths v of Q_{R} such that v = β_{1}v′β_{2} with v′ a path, and

β_{1} ∈ (Q_{A})_{1}\(Q_{C})_{1} and β_{2} ∈ (Q_{B})_{1}\(Q_{C})_{1}

or

β_{1} ∈ (Q_{B})_{1}\(Q_{C})_{1} and β_{2} ∈ (Q_{A})_{1}\(Q_{C})_{1}.

Then R ≅ KQ_{R}/I_{R}. Moreover, Q_{A}, Q_{B} and Q_{C} are convex in Q_{R}.

Proof. Let R′ = KQ_{R}/I_{R}. Let e_{A} = ∑_{a∈(Q_{A})_{0}} e_{a} and e_{B} = ∑_{b∈(Q_{B})_{0}} e_{b} be the identities of A and B respectively.

Consider the following map:

\begin{aligned}Φ : & R′ & → & R \\ & x & ↦ & (e_{A}xe_{A}, e_{B}xe_{B}) . \end{aligned}

Then Φ is well defined. Indeed, e_{C}(e_{A}xe_{A})e_{C} = e_{C}xe_{C} = e_{C}(e_{B}xe_{B})e_{C}, and Φ(I_{R}) = 0 since by hypothesis no path of ρ̄ is a path of Q_{A} or of Q_{B}.

It is straightforward to verify that Φ is a K-algebra isomorphism. The last statement follows easily from the structure of (Q_{R}, I_{R}) and the convexity of the connected components of Q_{C} in Q_{A} and Q_{B}. □

Example 1.2.2. Suppose that we have

[figure: 8; (Q_{A}, I_{A}) =; 7; 5; 4; ,; 6; (Q_{C}, I_{C}) =; 7; 5; 4]

[figure: 2; 1; and (Q_{B}, I_{B}) =; 7; 5; 4; .; 3]

Then the bound quiver of the pullback R of the canonical projections of KQ_{A}/I_{A} and KQ_{B}/I_{B} over KQ_{C}/I_{C} is

[figure: 8; 2; 1; (Q_{R}, I_{R}) =; 7; 5; 4; .; 6; 3]

Since Q_{A}, Q_{B} and Q_{C} are full and convex in Q_{R}, it is easily seen that every indecomposable A-module, B-module or C-module has a natural R-module structure. Therefore, from now on, when we have an R-module that is isomorphic to a A, B or C-module, we will refer to it as a A, B or C-module. We can assume without loss of generality that ind C is contained in ind A and in ind B and similarly that ind A and ind B are contained in ind R.

2. Nakayama Oriented Pullbacks

In this section we study a particular type of pullback R of K-algebras morphisms A ↠ C and B ↠ C. We show that the Auslander-Reiten quiver of this kind of pullback has very nice properties, and we construct a tilting R-module from tilting modules over A, B and C. Finally, we conclude that R is tilted when A and B are hereditary.

2.1. Bound Quiver.

Definition 2.1.1. Let R ≅ KQ_{R}/I_{R} be the pullback of the K-algebra surjective morphisms f_{A} : A ↠ C and f_{B} : B ↠ C as seen in 1.2. We say that R is a Nakayama oriented pullback if its bound quiver (Q_{R}, I_{R}) satisfies the following conditions:

(i) There is no path from (Q_{C})_{0} to (Q_{A})_{0}\(Q_{C})_{0} and from (Q_{B})_{0}\(Q_{C})_{0} to (Q_{C})_{0} in Q_{R}.

(ii) C is a hereditary Nakayama algebra and the connected components Q_{C_{1}}, Q_{C_{2}}, ..., Q_{C_{r}} of Q_{C} are of the form Q_{C_{i}} = a_{i, t_{i}} → a_{i, t_{i} − 1} → ⋯ → a_{i, 3} → a_{i, 2} → a_{i, 1} with 1 ≤ i ≤ r and t_{i} ≥ 1.

(iii) In Q_{C}, only sources are target of arrows of (Q_{A})_{1}\(Q_{C})_{1}, and only sinks are

sources of arrows of (Q_{B})_{1}\(Q_{C})_{1}. (iv) No minimal (or of minimal length) relation of R has its origin in (Q_{C})_{0}.

The last condition ensures that the projective dimension of C as a B-module (and so as an R-module) is at most one. This is a necessary condition for the existence of a tilting C-module that is also a partial tilting B-module. We need such a module to construct a tilting R-module in 2.3. (The first three conditions imply that C is always a projective A-module.)

For the remainder of this section, we suppose that R is a Nakayama oriented pullback of the surjective morphisms f_{A} : A ↠ C and f_{B} : B ↠ C, and we use the notations of the definition 2.1.1.

Remarks 2.1.2. These conditions and 1.2.1 imply that:

(i) For all i ∈ (Q_{A})_{0}, I(i)e_{A} = I(i), (ii) For all j ∈ (Q_{B})_{0}, P (j)e_{B} = P (j), (iii) For all i ∈ (Q_{A})_{0}\(Q_{C})_{0}, P (i)e_{A} = P (i), (iv) For all j ∈ (Q_{B})_{0}\(Q_{C})_{0}, I(j)e_{B} = I(j).

Examples 2.1.3. (1) In 1.2.2, condition (i) of 2.1.1 is not satisfied since there is an arrow from 7 to 6, whereas 7 ∈ (Q_{C})_{0} and 6 ∈ (Q_{A})_{0}\(Q_{C})_{0}.

(2) All conditions of 2.1.1 are satisfied in the following case:

[figure: 7; Q_{A} =; 5; 4; 3; ,; 6; 1; Q_{B} =; 5; 4; 3; ,; 2]

\xymatrix{Q_{C} = & 5 \ar[r] & 4 \ar[r] & 3}

[figure: 7; 1; and (Q_{R}, I_{R}) =; 5; 4; 3; .; 6; 2]

2.2. Auslander-Reiten Quiver. The category of modules mod R of a Nakayama oriented pullback R has a particular structure: every indecomposable R-module is isomorphic to an A- or to a B-module, and Γ_{R} is the pushout of the inclusion morphisms Γ_{C} ↪ Γ_{A} and Γ_{C} ↪ Γ_{B} (see 2.2.4 below).

This first lemma is essentially due to Igusa, Platzeck, Todorov and Zacharia (see [16], Lemma 3.5). This result is given in a context where A, B and C are connected, but this condition is not necessary. So the result and the proof are the same in our context.

Lemma 2.2.1. [16] indR = indA ∪ indB and indA ∩ indB = indC. □

The next result follows easily from the proof of 2.2.1 (see [16]).

Corollary 2.2.2. Let M ∈ indA\indC. Then top M ∈ mod A\ mod C. Dually, let N ∈ indB\indC. Then soc M ∈ mod B\ mod C. □

The following results allow us to describe the structure of the Auslander-Reiten quiver of R. We recall that ν denotes the Nakayama functor and τ denotes the Auslander-Reiten translation.

Lemma 2.2.3.

(i) Let M ∈ indA be an A-module that is not A-projective. Then τ_{R}M ≅ τ_{A}M. (ii) Let M ∈ ind B be a B-module that is not B-injective. Then τ_{R}^{−1}M ≅ τ_{B}^{−1}M. (iii) Let M ∈ ind C be a C-module that is not C-projective. Then τ_{B}M ≅ τ_{C}M. (iv) Let M ∈ ind C be a C-module that is not C-injective. Then τ_{A}^{−1}M ≅ τ_{C}^{−1}M.

(v) Let M ∈ ind A be a projective A-module that is not R-projective. Then τ_{R}M ≅ τ_{B}M.

(vi) Let M ∈ ind B be an injective B-module that is not R-injective. Then τ_{R}^{−1}M ≅ τ_{A}^{−1}M.

Proof. (i) Since M is not A-projective, it cannot be R-projective.

Let P_{1} → P_{0} → M → 0 be a minimal projective presentation of M in mod R. By 2.2.2 and condition (i) of 2.1.1, we have P_{0} = ⨁_{i∈I}(e_{i}R)^{m_{i}} and P_{1} = ⨁_{j∈J} (e_{j}R)^{n_{j}} with I ⊆ (Q_{A})_{0} and J ⊆ (Q_{A})_{0} (if M ∈ ind C, this follows from the fact that since M is not A-projective, it is not C-projective, and so the top of the kernel of P_{0} → M → 0 is in mod C).

Then

P′_{1} = ⨁_{j∈J}(e_{j}Re_{A})^{n_{j}} → P′_{0} = ⨁_{i∈I}(e_{i}Re_{A})^{m_{i}} → M_{A} → 0

is a minimal projective presentation of M in mod A. So we have the following exact sequences:

0 → τ_{R}M → ν_{R}P_{1} → ν_{R}P_{0} in mod R
and 0 → τ_{A}M → ν_{A}P′_{1} → ν_{A}P′_{0} in mod A.

By 2.1.2 (i) we have

\begin{aligned}& ν_{R}P_{0} & = & ν_{R}(⨁_{i∈I} e_{i}R^{m_{i}}) & ≅ & ⨁_{i∈I} I(i)_{R}^{m_{i}} & ≅ & ⨁_{i∈I} I(i)_{A}^{m_{i}} & ≅ & ν_{A}P′_{0} \\ and & ν_{R}P_{1} & = & ν_{R}(⨁_{j∈J} e_{j}R^{n_{j}}) & ≅ & ⨁_{j∈J} I(j)_{R}^{n_{j}} & ≅ & ⨁_{j∈J} I(j)_{A}^{n_{j}} & ≅ & ν_{A}P′_{1} . \end{aligned}

Since these isomorphisms are functorial, we have τ_{R}M ≅ τ_{A}M. (ii) Dual of the proof of (i), using 2.1.2 (ii). (iii) Follows from (i), by setting Q_{A} = Q_{C} (and so B = R). (iv) Follows from (ii), by setting Q_{B} = Q_{C} (and so A = R).

(v) Let M be an indecomposable A-module which is A-projective but not Rprojective. Then M ≅ P (i)e_{A} with i ∈ (Q_{C})_{0}. Therefore M has a non-projective B-module structure. Therefore there exists an indecomposable B-module M′ such that M ≅ τ_{B}^{−1}M′. But it follows from (ii) that τ_{B}^{−1}M′ ≅ τ_{R}^{−1}M′. So τ_{R}M ≅ τ_{R}(τ_{R}^{−1}M′) ≅ M′ ≅ τ_{B}(τ_{B}^{−1}M′) ≅ τ_{B}M. (vi) Similar to the proof of (v). □

It follows from this lemma that τ_{A} and τ_{B} are defined in the same way over the common objects of ind A and ind B, which are the modules in ind C. Also, we see that τ_{R} is defined for all modules in ind A and ind B. Since ind R = ind A ∪ ind B and ind C = ind A ∩ ind B, we naturally have the following corollary.

Corollary 2.2.4. The Auslander-Reiten quiver Γ_{R} is the pushout of the quiver inclusions Γ_{C} ↪ Γ_{A} and Γ_{C} ↪ Γ_{B} with τ_{R} completely determined by τ_{A} and τ_{B}.

Proof. By 2.2.1, we know that ind R = ind A ∪ ind B and ind A ∩ ind B = ind C. It follows from 2.2.3 (iii) and (iv) that Γ_{C} is a full subquiver of Γ_{A} and Γ_{B}, and by 2.2.3 (i) and (ii) that Γ_{A} and Γ_{B} are full subquivers of Γ_{R}.

The definition of τ_{R} follows directly from 2.2.3.

It is now easy to see that Γ_{R} = Γ_{A} ⨿_{Γ_{C}} Γ_{B}, which completes the proof. □

2.3. Tilting Module. In this subsection, we construct a tilting module over R using tilting modules over C, A and B. More precisely, we first take a tilting Cmodule T′′, which we complete to a tilting A-module T = N ⊕ T′′ and to a tilting B-module T′ = N′ ⊕ T′′. Here, we see the importance of the condition stating that there is no minimal relation of R that has its origin in Q_{C} (2.1.1 (iv)); without this condition the existence of a module like T′ is not ensured.

So let T = N ⊕T′′ and T′ = N′⊕T′′ be tilting modules over A and B respectively, with T′′ tilting over C.

Lemma 2.3.1. Every indecomposable direct summand of N is in ind A\ ind C, and every indecomposable direct summand of N′ is in ind B\ ind C.

Proof. Since T′′ is tilting over C, it has t isomorphism classes of indecomposable summands, and their dimension vectors form a basis of K_{0}(C) (see [12], Proposition 3.2). If N has an indecomposable summand in ind C, its dimension vector is a linear combination of the t dimension vectors of the indecomposable summands of T′′; it gives t + 1 linearly dependent dimension vectors in K_{0}(C), and so in K_{0}(R), a contradiction. Therefore every indecomposable summand of N is in ind A\ ind C. Similarly, every indecomposable summand of N′ is in ind B\ ind C. □

Recall that Q_{C} has r connected components of type 𝔸_{n}, and KQ_{C} is a hereditary Nakayama algebra. So for each of these components there is an indecomposable C-module which is both injective and projective. We denote by Q_{1}, Q_{2}, ..., Q_{r} these projective-injective C-modules (Q_{i} = P (a_{i,t_{i}})e_{C}). For each i between 1 and r, the module Q_{i} is necessarily a summand of T′′ (see [3]).

Lemma 2.3.2. Let M be an indecomposable direct summand of N. Then pd M_{R} ≤ 1. Moreover, if f : P ↠ M is a projective cover of M in mod R, then, for all j ∈ (Q_{C})_{0}, e_{j}Re_{A} is not a direct summand of the kernel L of f.

Proof. Since M is a summand of N, we have M ∈ ind A\ ind C (2.3.1). Hence, by 2.2.2, top M is in mod A\ mod C, and so by 2.1.2 (iii) we have P e_{A} = P. Therefore the kernel L of f : P ↠ M is an A-module, and so Le_{A} = L.

Suppose that pd M_{R} > 1. Since T is tilting over A, we have pd M_{A} ≤ 1. So L has an indecomposable summand which is A-projective but not R-projective. Therefore there exists j ∈ (Q_{C})_{0} such that e_{j}Re_{A} is not an R-projective summand of L. Let L = L′ ⊕ e_{j}Re_{A}.

We have the following exact sequence in mod A:

\xymatrix{0 \ar[r] & τ_{A}M \ar[r]^{f} & ν_{A}L \ar[r]^{g} & ν_{A}P}

with ν_{A}L ≅ (ν_{A}L′ ⊕ I(j)) and ν_{A}P ≅ ⨁_{l∈J} I(l) (here, J ⊆ (Q_{A})_{0}\(Q_{C})_{0} since top M is in mod A\ mod C ).

Since j ∈ (Q_{C})_{0} and J ⊆ (Q_{A})_{0}\(Q_{C})_{0}, the C-submodule I(j)e_{C} of I(j)e_{A} is a submodule of Ker g = Imf. So I(j)e_{C} is a submodule of τ_{A}M.

However, there exists i between 1 and r such that we have an epimorphism Q_{i} ↠ I(j)e_{C} (where Q_{i} is the module P (a_{i,t_{i}})e_{C} which is both C-projectice and C-injective). Hence Hom_{A}(Q_{i}, τ_{A}M) ≠ 0; this contradicts the fact that T is tilting over A (since Q_{i} and M are both summands of T). So pd M_{R} ≤ 1.

Therefore, L is R-projective. Since for all j ∈ (Q_{C})_{0} we have e_{j}Re_{A} ≠ e_{i}R (with i the index of Q_{i}), e_{j}Re_{A} cannot be a summand of L. □

Lemma 2.3.3. We have Hom_{R}(N′, τ_{R}N) = 0.

Proof. Let M be an indecomposable summand of N. If M is R-projective, we have Hom_{R}(N′, τ_{R}M) = 0. Otherwise, it follows from 2.3.2 that pd M_{R} = 1. Let P be the projective cover of M, and L be the kernel of a projective cover P ↠ M. Then L is R-projective, and so L = ⨁_{j∈J} e_{j}R with J ⊆ (Q_{A})_{0}\(Q_{C})_{0} (see 2.3.2).

Therefore we have the following exact sequence in mod R:

\xymatrix{0 \ar[r] & τ_{R}M \ar[r] & ν_{R}L ≅ ⊕_{j∈J} I(j) \ar[r] & ν_{R}P .}

Since J ⊆ (Q_{A})_{0}\(Q_{C})_{0}, the support of ν_{R}L is completely contained in Q_{A}\Q_{C}, and so is the support of τ_{R}M. As N′ ∈ ind B\ ind C, we have Hom_{R}(N′, τ_{R}M) = 0. Therefore Hom_{R}(N′, τ_{R}N) = 0. □

It is now possible to prove the principal result of this subsection.

Theorem 2.3.4. Let R ≅ KQ_{R}/I_{R} be a Nakayama oriented pullback of K-algebras surjections f_{A} : A ↠ C and f_{B} : B ↠ C. Let T = N ⊕ T′′ and T′ = N′ ⊕ T′′ be tilting modules over A and B respectively, with T′′ tilting over C. Then T′′′ = N ⊕ T′′ ⊕ N′ is tilting over R.

Proof. Since T, T′ and T′′ are tilting over A, B and C respectively, it follows from 2.3.1 that T′′′ has |(Q_{R})_{0}| isomorphism classes of indecomposable summands, which is exactly the number of isomorphism classes of simple R-modules.

By 2.3.2, we have pd N_{R} ≤ 1. It follows from 2.1.2 (ii) and from the fact that T′ = T′′ ⊕ N′ is tilting over B that pd N′_{R} ≤ 1 and pd T′′_{R} ≤ 1.

We have

Ext_{R}^{1}(T′′′, T′′′) = Ext_{R}^{1}(N ⊕ T′′ ⊕ N′, N ⊕ T′′ ⊕ N′)
≅ D Hom_{R}(N ⊕ T′′ ⊕ N′, τ_{R}(N ⊕ T′′ ⊕ N′)).

By 2.3.3 we have Hom_{R}(N′, τ_{R}N) = 0. By 2.2.4 we have Hom_{R}(N, τ_{R}N′) = 0 and Hom_{R}(N, τ_{R}T′′) = 0.

Since N ∈ ind A\ ind C, the module N is A-projective if and only if it is R-projective (2.1.2 (iii)). So by 2.2.3 and 2.2.4 we have D Hom_{R}(N, τ_{R}N) ≅ D Hom_{A}(N, τ_{A}N) = 0 since T_{A} is tilting over A.

Similarly, we find D Hom_{R}(T′′, τ_{R}N) ≅ D Hom_{A}(T′′, τ_{A}N) = 0.

The other steps to show that Ext_{R}^{1}(T′′′, T′′′) = 0 are done similarly using the fact that T′_{B} is tilting over B. □

Examples 2.3.5. (1) Let T′′ be a tilting module over C which is partial tilting over A and over B. By Bongartz’ lemma, there exist an A-module N and a Bmodule N′ such that T′′ ⊕ N and T′′ ⊕ N′ are tilting over A and B respectively. Then it follows from 2.3.4 that N ⊕ T′′ ⊕ N′ is tilting over R.

(2) Let A, B and C be hereditary algebras whose ordinary quivers are the following:

\xymatrix{ &  & 3 \ar[r] & 2 &  &  & 3 \ar[r] & 2 \ar[drr] \\ Q_{A} = & 7 \ar[ur] \ar[dr] &  &  & , & Q_{B} = &  &  &  & 1 \\  &  & 6 \ar[r] & 5 \ar[r] & 4 &  & 6 \ar[r] & 5 \ar[r] & 4 \ar[ur]}

[figure: and Q_{C} =; 3; 2; 6; 5; 4; .]

We easily see that T′′ = I(2)e_{C} ⊕ S(3) ⊕ I(4)e_{C} ⊕ I(5)e_{C} ⊕ S(6) is tilting over C, T′′ ⊕ P (7) is tilting over A and T′′ ⊕ I(1) is tilting over B. Therefore by 2.3.4 the module P (7) ⊕ T′′ ⊕ I(1) is tilting over R, and the bound quiver of R is the following:

[figure: 3; 2; 7; 1; .; 6; 5; 4]

(3) Let A be an algebra, x be a source of Q_{A} such that S(x) is partial tilting over A, and T be a tilting A-module that has S(x) as a summand (Bongartz’ lemma ensures the existence of such a module T). Consider the one-point extension R = A[S(x)] of A by S(x), and let y be the new vertex added to Q_{A} to obtain Q_{R}. It is clear that R is a Nakayama oriented pullback, and it follows from 2.3.4 that P (y) ⊕ T is tilting over R (since P (y) ⊕ S(x) is tilting over the hereditary algebra whose ordinary quiver is y → x).

2.4. The case where B is hereditary. We now consider the particular case where B is hereditary. We construct an R-module using tilting modules over C, A and B, and we show that it is tilting over R using 2.3.4. The particularity of this module is that the minimal relations of its endomorphism algebra are the minimal relations of A (we "remove" the other relations). We then conclude that R is tilted when A is also hereditary.

So let

T′′ = ⨁_{i∈(Q_{C})_{0}} P (i)e_{C}, N = ⨁_{i∈(Q_{A})_{0}\(Q_{C})_{0}} P (i) and N′ = ⨁_{i∈(Q_{B})_{0}\(Q_{C})_{0}} I(i).

We clearly see that T′′ is a C-module, N an A-module and N′ a B-module.

Lemma 2.4.1. T′′ is tilting over C and partial tilting over A and B.

Proof. We have T′′_{C} ≅ C as a C-module, so it is tilting over C. Moreover, T′′ is A-projective, and so is partial tilting over A.

Since B is hereditary, the projective dimension of T′′ over B is at most one. As T′′ is tilting over C, we have Ext_{C}^{1}(T′′, T′′) = 0. Since C is full and convex in B, we also have Ext_{B}^{1}(T′′, T′′) = 0. □

Lemma 2.4.2. T′′ ⊕ N is tilting over A.

Proof. This follows from the fact that T′′ ⊕ N ≅ A as an A-module. □

Lemma 2.4.3. T′′ ⊕ N′ is tilting over B.

Proof. Let T′ = T′′ ⊕ N′. Since B is hereditary, we have pd T′_{B} ≤ 1. It is clear that the number of isomorphism classes of indecomposable summands of T′ is equal to the rank of K_{0}(B).

Using the B-injectivity of N′ and 2.4.1, we have

Ext_{B}^{1}(T′′ ⊕ N′, T′′ ⊕ N′) ≅ Ext_{B}^{1}(T′′ ⊕ N′, T′′)
≅ Ext_{B}^{1}(T′′, T′′) ⊕ Ext_{B}^{1}(N′, T′′)
= Ext_{B}^{1}(N′, T′′)
≅ D Hom_{B}(T′′, τ_{B}N′).

Let M be an indecomposable summand of N′. Suppose that τ_{B}M is a C-module. Then τ_{B}M cannot be C-injective because C-injective modules are also B-injective. Therefore, it follows from 2.2.3 (iv) that τ_{A}^{−1}(τ_{B}M) ≅ τ_{C}^{−1}(τ_{B}M). Moreover, by 2.2.4, we have τ_{B}M ≅ τ_{R}M and τ_{A}^{−1}(τ_{B}M) ≅ τ_{R}^{−1}(τ_{B}M). So

τ_{C}^{−1}(τ_{B}M) ≅ τ_{A}^{−1}(τ_{B}M)
≅ τ_{R}^{−1}(τ_{R}M)
≅ M.

However, τ_{C}^{−1}(τ_{B}M) is a C-module, which is a contradiction since M is in ind B\ ind C. Hence τ_{B}M cannot be a C-module, and neither can every indecomposable summand of τ_{B}N′. Therefore, it follows from 2.2.2 that soc(τ_{B}N′) is in mod B\ mod C. So since T′′, as a C-module, does not have any composition factor in mod B\ mod C, we have Hom_{B}(T′′, τ_{B}N′) = 0.

Hence Ext_{B}^{1}(T′, T′) = 0, and T′ is tilting over B. □

Proposition 2.4.4. Let T = N ⊕ T′′ ⊕ N′. Then T is tilting over R.

Proof. This follows from 2.4.1, 2.4.2, 2.4.3, and 2.3.4. □

Now, let us consider a particular quiver construction, which we need in order to describe Q_{End T}.

Definition 2.4.5. Let Q be a quiver and Q′′ a full subquiver of Q. The difference of quivers Q\Q′′ is the subquiver of Q such that (Q\Q′′)_{1} = Q_{1}\Q′′_{1} and

(Q\Q′′)_{0} = (Q_{0}\Q′′_{0}) ∪ \{ x ∈ Q′′_{0} | \begin{matrix}there exists α ∈ (Q\Q′′)_{1} such that \\ x = s(α) or x = t(α) \end{matrix} \} .

Now, consider the difference Q_{B}\Q_{C}. Then, in Q_{B}\Q_{C}, we identify the vertices that are sinks in Q_{C} to the sources t_{1}, t_{2}, ..., t_{r} of Q_{C}, and we denote this new quiver by Q_{BC}.

For instance, if

[figure: Q_{B} =; 7; 6; 5; 2; 1; 4; 3; and; Q_{C} =; 7; 6; 5; ,; 4; 3]

then

\xymatrix{Q_{B}\Q_{C} = & 5 \ar[dr] &  &  & and & Q_{BC} = & 7 \ar[dr] &  & . \\  &  & 2 \ar[r] & 1 &  &  &  & 2 \ar[r] & 1 \\  & 3 \ar[ur] &  &  &  &  & 4 \ar[ur]}

Finally, we consider the quiver Q_{A} ⨿_{Q_{C_{S}}} Q_{BC} where Q_{C_{S}} = •_{t_{1}} •_{t_{2}} ⋯ •_{t_{r}} (the sources of the connected components of Q_{C}).

If we continue the previous example with

\xymatrix{ &  & 7 \ar[r] & 6 \ar[r] & 5 \\ Q_{A} = & 9 \ar[dr] \ar[ur] &  &  & , \\  &  & 8 \ar[r] & 4 \ar[r] & 3}

we obtain

[figure: 7; 6; 5; Q_{A} ⨿_{Q_{C_{S}}} Q_{BC} =; 9; 2; 1; .; 8; 4; 3]

We easily see that |(Q_{R})_{0}| = |(Q_{A} ⨿_{Q_{C_{S}}} Q_{BC})_{0}|. Actually, from now on, we identify the vertices of Q_{A} to those which naturally correspond to them in the quiver Q_{A} ⨿_{Q_{C_{S}}} Q_{BC}, and we do the same for the vertices of Q_{B} that are not in Q_{C}.

Theorem 2.4.6. Let T = N ⊕ T′′ ⊕ N′. The endomorphism algebra End T admits a presentation (Q_{End T} , I_{End T}) with Q_{End T} = Q_{A} ⨿_{Q_{C_{S}}} Q_{BC} and I_{End T} = I_{A}.

Proof. Similar to the proof of [19] (2.3), and done in [20] (3.4.5). □

As a direct consequence, we obtain the main result of this paper.

Theorem 2.4.7. Let R ≅ KQ_{R}/I_{R} be a Nakayama oriented pullback of K-algebra surjective morphisms A ↠ C and B ↠ C. Suppose that A and B are hereditary. Then R is tilted. □

Example 2.4.8. Consider the bound quiver of 2.3.5 (2). By 2.4.4, the module T = I(1) ⊕ S(2) ⊕ 3 2 ⊕ S(4) ⊕ 5 4 ⊕ 6 5 4 ⊕ P (7) is tilting over R, and by 2.4.6 and 2.4.7, End T is hereditary with

\xymatrix{ &  & 3 \ar[r] \ar[dr] & 2 \\ Q_{End T} = & 7 \ar[ur] \ar[dr] &  & 1 & . \\  &  & 6 \ar[r] \ar[ur] & 5 \ar[r] & 4}

3. Iterated Tilted and Tilted Stably hereditary algebras

3.1. The bound quiver of a stably hereditary algebra. In our main results, we use some properties of the bound quiver of a stably hereditary algebra that are easy to identify. This subsection is therefore devoted to the bound quiver of a stably hereditary algebra.

Let R = KQ/I be a stably hereditary algebra. Then, by [6], we have I = I_{Σ_{R}} with

Σ_{R} = { x ∈ Q_{0} | S(x) is a non-projective submodule of R }

and

I_{Σ_{R}} = ⟨ αβ | t(α) ∈ Σ_{R} ⟩.

That is, I_{Σ_{R}} is the ideal generated by all paths αβ with t(α) = s(β) ∈ Σ_{R}. In particular, R is a monomial algebra (that is, I_{Σ_{R}} is generated by monomial relations).

Definitions 3.1.1. A cycle C in a bound quiver (Q, I) satisfies the clock condition if the number of clockwise oriented relations on C equals the number of counterclockwise oriented relations. We say that (Q, I) satisfies the clock condition if all cycles in (Q, I) satisfy the clock condition.

The following theorem, due to Skowroński, allows us to characterize the bound quiver of a stably hereditary algebra using the clock condition. For the notion of special cycle in G_{eRe}, we refer the reader to [24].

Theorem 3.1.2. [24] Let S be an algebra tilting equivalent to a hereditary or to a canonical algebra. Then for any idempotent e of S, any special cycle C in G_{eSe} satisfies the clock condition. In particular, Q_{S} has no oriented cycles. □

Actually, in our context of stably hereditary algebras, the above statement can be reformulated to say that if R is tilting equivalent to a hereditary algebra, then any cycle of (Q, I_{Σ_{R}}) satisfies the clock condition. Therefore, if (Q, I_{Σ_{R}}) does not satisfy the clock condition, R is not iterated tilted. Hence, from now on, we suppose that R is a stably hereditary algebra such that its bound quiver (Q, I_{Σ_{R}}) satisfies the clock condition. In particular, R is a triangular algebra.

We want to decompose Q into maximal subquivers which do not contain any relation, that is, the ordinary quivers of the algebras R_{1}, ..., R_{n} such that R is stably equivalent to R_{1} × R_{2} × ... × R_{n} (see [6]).

Let α ∈ Q_{1}, and let Q_{α} be the subquiver of Q such that

(Q_{α})_{1} = \{ β ∈ Q_{1} | \begin{matrix}there exists a non-zero walk w \\ such that w = α^{∗}w′β^{∗} where \\ α^{∗} ∈ \{α, α^{−1}\}, β^{∗} ∈ \{β, β^{−1}\} \end{matrix} \}

and

(Q_{α})_{0} = { s(β), t(β) ∈ Q_{0} | β ∈ (Q_{α})_{1} } .

Remarks 3.1.3.

(i) Since R is a stably hereditary algebra, we easily see that Q_{α} is a full subquiver of Q. Moreover, since (Q, I_{Σ_{R}}) satisfies the clock condition, Q_{α} is convex in Q.

(ii) Since (Q, I_{Σ_{R}}) satisfies the clock condition, every walk w containing zerorelations that all point in the same direction in w (and these are the only relations on w)

[figure: w = •; p_{1}; •; •; p_{2}; •; •; p_{s}; •]

is such that p_{1} ≠ p_{2} ≠ ... ≠ p_{s}. In particular, every double-zero in (Q, I_{Σ_{R}}) contains two distinct vertices of Σ_{R}.

(iii) Let α ∈ Q_{1}, and β ∈ (Q_{α})_{1}. Then Q_{α} = Q_{β}. (iv) If Q_{α} ≠ Q_{β}, then (Q_{α})_{1} ∩ (Q_{β})_{1} = ∅ and (Q_{α})_{0} ∩ (Q_{β})_{0} ⊆ Σ_{R}. On the

other hand, for all x ∈ Σ_{R}, there exist α, β ∈ Q_{1} such that Q_{α} ≠ Q_{β} and x ∈ (Q_{α})_{0} ∩ (Q_{β})_{0}.

(v) For all α ∈ Q_{1} and x ∈ (Q_{α})_{0}\Σ_{R}, we have I(x) ∈ ind KQ_{α} and P (x) ∈ ind KQ_{α}.

Let α_{1}, α_{2}, ..., α_{t} ∈ Q_{1} be such that Q_{α_{i}} ≠ Q_{α_{j}} when i ≠ j, and such that for all β ∈ Q_{1}, there exists i ∈ {1, ..., t} such that β ∈ (Q_{α_{i}})_{1}. We set I(R) = {α_{1}, α_{2}, ..., α_{t}}.

Then Q_{1} = ⋃_{i=1}^{t}(Q_{α_{i}})_{1} and Q_{0} = ⋃_{i=1}^{t}(Q_{α_{i}})_{0}. It follows from the remarks above and from [6] that R is stably equivalent to KQ_{α_{1}} × KQ_{α_{2}} × ... × KQ_{α_{t}}.

Example 3.1.4. Let R be a stably hereditary algebra whose bound quiver (Q, I_{Σ_{R}}) is the following:

[figure: 5; β_{5}; 3; 10; β_{10}; 9; β_{9}; β_{8}; 4; β_{3}; 6; β_{6}; β_{2}; 2; β_{1}; 1; .; 8; β_{7}; 7; β_{4}]

A possible choice for I(R) is I(R) = {β_{1}, β_{2}, β_{3}, β_{10}}, and then we have

[figure: Q_{β_{10}} =; 10; β_{10}; 9; ,; Q_{β_{3}} =; 5; β_{5}; 3; ,; 4; β_{3}]

[figure: 5; Q_{β2} =; 9; β_{9}; β_{8}; 4; 6; β_{6}; β_{2}; 2; 8; β_{7}; 7; β_{4}; and; Q_{β_{1}} =; 2; β_{1}; 1; .]

In the following lemma, we show that when a stably hereditary algebra respects the clock condition, it is always possible to describe it as a pullback of surjective algebra morphisms.

Lemma 3.1.5. Let R ≅ KQ/I_{Σ_{R}}. There exist K-algebras A and C and arrows α_{i1}, α_{i2}, ..., α_{is} ∈ I(R) such that R is the pullback of surjective morphisms of K-algebras A ↠ C and KQ_{α_{i1}} × KQ_{α_{i2}} × ... × KQ_{α_{is}} ↠ C.

Proof. Let α_{i1}, α_{i2}, ..., α_{is} ∈ I(R) be such that (Q_{α_{ij}})_{0} ∩ (Q_{α_{ij′}})_{0} = ∅ when j ≠ j′. Let B = KQ_{α_{i1}} × KQ_{α_{i2}} × ... × KQ_{α_{is}}, and let Q_{C} be the quiver such that (Q_{C})_{0} = Σ_{R} ∩ (Q_{B})_{0} (and (Q_{C})_{1} = ∅). Let I(R)^{∗} = I(R)\{α_{i1}, α_{i2}, ⋯, α_{is}} (it follows from 3.1.3 (iv) that I(R)^{∗} ≠ ∅). Let C = KQ_{C}, and A = KQ_{A}/I_{A} with Q_{A} the full subquiver of Q generated by ⋃_{α_{j}∈I(R)^{∗}}(Q_{α_{j}})_{0} and I_{A} =< αβ | t(α) ∈ Σ_{R}\(Q_{B})_{0} >.

We easily see that the connected components of Q_{C} are full and convex in Q_{A} and Q_{B}, and that C is a common quotient of A and B (see 3.1.3 (iv)). Let R′ be the pullback of the canonical projections A ↠ C and B ↠ C. Then Q_{R′} = Q_{A} ⨿_{Q_{C}} Q_{B} (see 1.2.1), and R′ = KQ_{R′}/I_{R′} with I_{R′} generated by I_{A} and by the set ρ̄ (as described in 1.2.1).

But it follows from 3.1.3 that the quiver of R is exactly Q_{A} ⨿_{Q_{C}} Q_{B}, and

I_{Σ_{R}} = < αβ | t(α) ∈ Σ_{R} >
= < {αβ | t(α) ∈ Σ_{R}\(Q_{B})_{0}} ∪ {αβ | t(α) ∈ Σ_{R} ∩ (Q_{B})_{0}} >
= < {αβ | t(α) ∈ Σ_{R}\(Q_{B})_{0}} ∪ ρ̄ >
= I_{R′}.

So R ≅ R′ and R is the pullback of the canonical projections A ↠ C and B ↠ C. □

In this case, we say that R = AΠ_{C}B is the pullback associated to KQ_{α_{i1}} × KQ_{α_{i2}} × ... × KQ_{α_{is}}.

We now describe (Q, I_{Σ_{R}}) in a way showing that there always exist α_{i1}, α_{i2}, ..., α_{is} ∈ I(R) such that the pullback associated to KQ_{α_{i1}} × KQ_{α_{i2}} × ... × KQ_{α_{is}} is a Nakayama oriented pullback which respects all the conditions of 2.4 (with B = KQ_{α_{i1}} × KQ_{α_{i2}} × ... × KQ_{α_{is}}). This description also allows us to state a sufficient and necessary condition for (Q, I_{Σ_{R}}) to contain a double-zero.

So, first, let

V_{1} = { α_{i} ∈ I(R) | there exists β ∈ (Q_{α_{i}})_{1} such that t(β) ∈ Σ_{R} }

and

V_{2} = { α_{i} ∈ I(R) | there exists β ∈ (Q_{α_{i}})_{1} such that s(β) ∈ Σ_{R} } .

In 3.1.4, we have V_{1} = {β_{2}, β_{10}} and V_{2} = {β_{1}, β_{2}, β_{3}}.

Remark 3.1.6. Let α_{i}, α_{j} ∈ V_{2}\V_{1} with α_{i} ≠ α_{j}. Then (Q_{α_{i}})_{0} ∩ (Q_{α_{j}})_{0} = ∅. Indeed, if the intersection is non-empty, it follows from 3.1.3 (iv) that there exists x ∈ Σ_{R} which is contained by both Q_{α_{i}} and Q_{α_{j}}. But this contradicts the fact that α_{i} and α_{j} are both in V_{2}\V_{1}.

Lemma 3.1.7. Let R ≅ KQ/I_{Σ_{R}}. Then there exists α_{i} ∈ V_{2}\V_{1}.

Proof. Let α_{i1} ∈ I(R). If α_{i1} ∈ V_{2}\V_{1}, we are done.

If this is not the case, then Q_{α_{i1}} contains an arrow β_{i1} such that t(β_{i1}) ∈ Σ_{R}.

[figure: •; β_{i1}; p_{i1}; β′_{i1}; •]

Let α_{i2} ∈ I(R)\{α_{i1}} such that β′_{i1} ∈ (Q_{α_{i2}})_{1}. If α_{i2} ∈ V_{2}\V_{1}, we are done. Otherwise, Q_{α_{i2}} contains an arrow β_{i2} such that t(β_{i2}) ∈ Σ_{R}, and so there is a double-zero of the form

[figure: •; β_{i1}; p_{i1}; β′_{i1}; •; •; β_{i2}; p_{i2}; β′i2; •; (p_{i1} ≠ p_{i2} by 3.1.3).]

We then consider α_{i3} ∈ I(R)\{α_{i1}, α_{i2}} such that β′_{i2} ∈ (Q_{α_{i3}})_{1}. We repeat the argument, and since |I(R)| and |Σ_{R}| are finite, the statement follows by induction. □

Lemma 3.1.8. The bound quiver (Q, I_{Σ_{R}}) contains no double-zero if and only if V_{1} ∩ V_{2} = ∅.

Proof. If V_{1} ∩V_{2} ≠ ∅, there exists α_{i} ∈ I(R) such that Q_{α_{i}} contains an arrow whose target is in Σ_{R}, and an arrow whose source is in Σ_{R}. This implies the existence of a double-zero in (Q, I_{Σ_{R}}).

On the other hand, if (Q, I_{Σ_{R}}) contains a double-zero,

[figure: •; p_{1}; β_{1}; •; •; β_{2}; p_{2}; •]

we see that there exists a non-zero walk containing the arrows β_{1} and β_{2}. Hence there exists α_{i} ∈ I(R) such that β_{1}, β_{2} ∈ (Q_{α_{i}})_{1}, and so α_{i} ∈ V_{1} ∩ V_{2}. □

Lemma 3.1.9. Let α_{i1}, α_{i2}, ..., α_{is} ∈ V_{2}\V_{1}.Then there exist K-algebras A and C such that R is the pullback of morphisms A ↠ C, KQ_{α_{i1}} × KQ_{α_{i2}} × ⋯ × KQ_{α_{is}} ↠ C, and this pullback is Nakayama oriented.

Proof. The first statement was proven in 3.1.5.

We now have to verify that R, as the pullback associated to KQ_{α_{i1}} × KQ_{α_{i2}} × ⋯ × KQ_{α_{is}}, satisfies the conditions of 2.1.1.

The condition saying that there is no path from (Q_{C})_{0} to (Q_{A})_{0}\(Q_{C})_{0} nor from (Q_{B})_{0}\(Q_{C})_{0} to (Q_{C})_{0} is satisfied since α_{i1}, α_{i2}, ..., α_{is} ∈ V_{2}\V_{1}.

We easily see that the conditions (ii) and (iii) of 2.1.1 are satisfied.

The last condition (which says that there is no minimal relation of R having its origin in (Q_{C})_{0}) is satisfied since α_{i1}, α_{i2}, ..., α_{is} ∈ V_{2}\V_{1} and B is hereditary. □

Hence, in the context of this lemma, R satisfies the conditions of 2.4 since B = KQ_{α_{i1}} × KQ_{α_{i2}} × ⋯ × KQ_{α_{is}} is hereditary. So we can use the results of 2.4 to construct a tilting R-module in the next subsection.

3.2. Tilting module. To show that a stably hereditary algebra which respects the clock condition is iterated tilted, we need a particular tilting module T, and we give its construction in 3.2.1. We see in 3.2.2 that End T is also a stably hereditary algebra which respects the clock condition. Moreover, the bound quiver of End T contains fewer relations. This is the key to the proof that R is iterated tilted.

Lemma 3.2.1. Let J ⊆ {1, ..., t} be such that for all j ∈ J, α_{j} ∈ V_{2}\V_{1}. Let Q_{J_{0}} = ∪_{j∈J}(Q_{α_{j}})_{0}. Then

T = ( ⨁_{i∈Q_{0}\Q_{J_{0}}} P (i)) ⊕ ( ⨁_{p∈Σ_{R}∩Q_{J_{0}}} S(p)) ⊕ ( ⨁_{k∈Q_{J_{0}} \Σ_{R}} I(k))

is a tilting R-module.

Proof. Follows from 3.1.9 and 2.4.4. □

Lemma 3.2.2. Let J ⊆ {1, ..., t} be such that for all j ∈ J, α_{j} ∈ V_{2}\V_{1}. Let Q_{J_{0}} = ∪_{j∈J}(Q_{α_{j}})_{0}, and

T = ( ⨁_{i∈Q_{0}\Q_{J_{0}}} P (i)) ⊕ ( ⨁_{p∈Σ_{R}∩Q_{J_{0}}} S(p)) ⊕ ( ⨁_{k∈Q_{J_{0}} \Σ_{R}} I(k)).

Then End T ≅ KQ/I_{Σ_{End T}} with Σ_{End T} = Σ_{R}\Q_{J_{0}}, and (Q, I_{Σ_{End T}}) respects the clock condition.

Proof. The first statement follows from 3.1.9 and 2.4.6. Actually, we have End T ≅ KQ_{End T} /I_{Σ_{End T}} with Q_{End T} = Q_{A} ⨿_{Q_{C_{S}}} (Q_{BC}) and I_{Σ_{End T}} = I_{A} with A, B and C as described in the proof of 3.1.9. But when R is stably hereditary, we easily see that Q_{A} ⨿_{Q_{C_{S}}} (Q_{BC}) = Q, and so Q = Q_{End T}.

Finally, the construction of (Q, I_{Σ_{R}}) and the fact that for all j ∈ J, α_{j} ∈ V_{2}\V_{1}, imply that the bound quiver (Q, I_{Σ_{End T}}) respects the clock condition. □

Corollary 3.2.3. If (Q, I_{Σ_{R}}) contains no double-zero, then R is tilted of type Q.

Proof. If (Q, I_{Σ_{R}}) contains no double-zero, then it follows from 3.1.8 that V_{1} ∩V_{2} = ∅.

Let x ∈ Σ_{R}. It follows from 3.1.3 (iv) that there exist α_{i}, α_{j} ∈ I(R) such that x is contained in both Q_{α_{i}} and Q_{α_{j}}. So, by 3.1.6, for all x ∈ Σ_{R}, there exists α_{j} ∈ V_{2} such that x ∈ (Q_{α_{j}})_{0} ∩ Σ_{R}. Therefore, for the construction of the module T as seen in 3.2.2, we take all the α_{j} contained in V_{2} (since V_{1} ∩ V_{2} = ∅), and so Q_{J_{0}} = Σ_{R}. Hence End T is hereditary since Σ_{End T} = Σ_{R}\Q_{J_{0}}. So R is tilted. □

The proof of the following lemma is similar to those of [18] (2.3) and [14] (2.6), which are done in the contexts of gentle and special biserial algebras respectively.

Lemma 3.2.4. If (Q, I_{Σ_{R}}) contains a double-zero, then R is not tilted.

Proof. Suppose that (Q, I_{Σ_{R}}) contains a double-zero of the form

[figure: 1; 2; 3; •; •; t − 2; t − 1; t]

with t ≥ 4.

Since R is monomial, if t = 4, then, by [9](1.2), gl. dim R > 2, hence R is not tilted.

Thus, suppose that t ≥ 5, and let M be the indecomposable R-module of support

3 • • t − 2

such that M(x) = K for all x such that 3 ≤ x ≤ t − 2 (this indecomposable module exists since R is monomial).

Let s be the source of Supp M such that there exists a path from s to t − 2 in Supp M. Since R is monomial, we see that the kernel of the canonical morphism P (s) → M has a non-projective direct summand and hence pd M > 1.

Similarly, one proves that id N > 1. Thus, by [11](III 2.3), R is not quasi-tilted, and therefore is not tilted. □

It is now possible to prove the main result of this section:

Theorem 3.2.5. Let R = KQ/I_{Σ_{R}} be a stably hereditary algebra. Then (i) R is iterated tilted if and only if (Q, I_{Σ_{R}}) satisfies the clock condition. In this

case the type of R is Q. (ii) R is tilted if and only if (Q, I_{Σ_{R}}) satisfies the clock condition and does not

contain any double-zero.

Proof. The first statement follows from 3.1.7, 3.2.1, 3.2.2, from [24](cor.1) and from the fact that |Σ_{R}| is finite. The second statement follows from 3.2.3 and 3.2.4. □

We easily obtain the following corollary, which, in particular, proves a conjecture of Dieter Happel saying that an algebra R = KQ/I with Q a tree and such that rad^{2} R = 0 is iterated tilted.

Corollary 3.2.6. Let R = KQ/I_{Σ_{R}} be a stably hereditary algebra with Q a tree. Then R is iterated tilted of type Q, and is tilted if and only if (Q, I_{Σ_{R}}) does not contain any double-zero. □

Let R = KQ/I be an algebra with Q a tree and such that I is generated by paths of length two, and such that (Q, I) does not contain any double-zero. If R is not stably hereditary, then it is not tilted in general, as is shown in the following example.

Example 3.2.7. Let (Q, I) be the following quiver:

\xymatrix{4 \ar[ddr]^{β_{1}} \\ 5 \ar[dr]_{β_{2}} &  & 1 \\  & 3 \ar[ur]^{α_{1}} \ar[dr]_{α_{2}} \\ 6 \ar[ur]^{β_{3}} &  & 2 \\ 7 \ar[uur]_{β_{4}}}

bound by β_{1}α_{1} = 0, β_{1}α_{2} = 0, β_{4}α_{1} = 0 and β_{4}α_{2} = 0. Then R = KQ/I is isomorphic to R_{1} = H[S(3)][S(3)] with H the hereditary algebra with ordinary quiver

\xymatrix{5 \ar[dr]_{β_{2}} &  & 1 \\  & 3 \ar[ur]^{α_{1}} \ar[dr]_{α_{2}} \\ 6 \ar[ur]^{β_{3}} &  & 2 & .}

By the proof of [4](3.2), the component of Γ_{A_{1}} containing S(3) is not directed. Hence, it follows from [21](3.7) that for A_{1} to be tilted, this component should be quasi-serial or obtained from a quasi-serial translation quiver by ray or coray insertions, which is not the case. Therefore A_{1} = H[S(3)][S(3)] is not tilted.

Example 3.2.8. Let R = KQ/I_{Σ_{R}} be the stably hereditary algebra seen in 3.1.4:

[figure: 5; β_{5}; 3; 10; β_{10}; 9; β_{9}; β_{8}; 4; β_{3}; 6; β_{6}; β_{2}; 2; β_{1}; 1; .; 8; β_{7}; 7; β_{4}]

We have Σ_{R} = {2, 4, 5, 9}, and {β_{1}, β_{2}, β_{3}, β_{10}} is a possible choice for I(R). Hence V_{1} = {β_{2}, β_{10}}, V_{2} = {β_{1}, β_{2}, β_{3}} and V_{2}\V_{1} = {β_{1}, β_{3}}, Q_{J_{0}} = (Q_{β_{1}})_{0} ∪ (Q_{β_{3}})_{0} = {1, 2, 3, 4, 5}, Σ_{R} ∩ Q_{J_{0}} = {2, 4, 5} and thus by 3.2.1

T_{1} = ( ⨁_{i=6}^{10} P (i)) ⊕ S(5) ⊕ S(4) ⊕ S(2) ⊕ I(3) ⊕ I(1)

is a tilting R-module. Moreover, by 3.2.2 we have End T_{1} ≅ KQ/I_{Σ_{End T_{1}}} with Σ_{End T_{1}} = Σ_{R}\(Σ_{R} ∩ (Q_{β_{1}})_{0}) = {9}. Therefore (Q, I_{Σ_{End T_{1}}} ) is the following bound quiver:

[figure: 5; β_{5}; 3; 10; β_{10}; 9; β_{9}; β_{8}; 4; β_{3}; 6; β_{6}; β_{2}; 2; β_{1}; 1; .; 8; β_{7}; 7; β_{4}]

We see that End T_{1} is stably hereditary. A possible choice for I(R) is {β_{1}, β_{10}}, and then V_{1} = {β_{10}}, V_{2} = {β_{1}} and thus V_{1} ∩ V_{2} = ∅. Hence by 3.1.8 (Q, I_{Σ_{End T_{1}}} ) does not contain any double-zero, and therefore it follows from 3.2.3 that End T_{1} is a tilted algebra of type Q, and so R is iterated tilted of type Q.

Acknowledgments. This paper is part of my Ph.D. thesis. I would like to thank Professor I. Assem for his precious help and direction, and Professors D. Zacharia and S. Liu for useful discussions. I would also like to thank the N.S.E.R.C. (Gov. of Canada), F.C.A.R (Gov. of Québec) and I.S.M. (Institut des Sciences Mathématiques) for financial support.

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Jessica Lévesque Département de SIMQG Université de Sherbrooke Sherbrooke, Québec J1K 2R1, Canada E-mail: jlevesque@adm.usherbrooke.ca