Tilting Modules and Exceptional Sequences for a Family of Dual Extension Algebras

We provide a classification of generalized tilting modules and full exceptional sequences for a certain family of quasi-hereditary algebras, namely dual extension algebras of path algebras of uniformly oriented linear quivers, modulo the ideal generated by paths of length two, with their opposite algebra. An important step in the classification is to prove that all indecomposable self-orthogonal modules (with respect to extensions of positive degree) admit a filtration with standard subquotients or a filtration with costandard subquotients. Furthermore, we prove that that every generalized tilting module, not equal to the characteristic tilting modules, admits either a filtration with standard subquotients or a filtration with costandard subquotients. Since the algebras studied in this article admit a simple-preserving duality, combining these two statements reduces the problem to classifying generalized tilting modules admitting a filtration with standard subquotients. To finalize the classification of generalized tilting modules we develop a combinatorial model for the poset of indecomposable self-orthogonal modules with standard filtration with respect to the relation arising from higher extensions. This model is also used for the classification of full exceptional sequences.


Introduction
Since its introduction in [BB80;HR82], tilting theory has become an important part of the representation theory of finite-dimensional algebras.A basic classification problem in tilting theory is to classify all generalized tilting modules over a given algebra.In general, this problem is very difficult.Some instances of where this problem and its generalizations have been studied can be found in [Ada16; BK04; MU93; PW20; Yam12].
Quasi-hereditary algebras were first defined in [Sco87].In [CPS88], highest weight categories were introduced as a category theoretical counterpart of certain structures in the representation theory of complex semisimple Lie algebras.In the same paper, [CPS88], the quasi-hereditary algebras were characterized as exactly those finite-dimensional algebras whose module categories are highest weight categories.Examples of quasi-hereditary algebras include hereditary algebras, Schur algebras, algebras of global dimension two and algebras describing blocks of BGG category O.
The chief protagonists of the representation theory of quasi-hereditary algebras are the standard and costandard modules, as well as the associated subcategories, F (∆) and F (∇), of the module category, consisting of those modules which admit a filtration by standard and costandard modules, respectively.Importantly, Ringel showed in [Rin91] that (F (∆), F (∇)) is a homologically orthogonal pair.Moreover, Ringel showed that for any quasi-hereditary algebra, there exists a generalized tilting module T , called the characteristic tilting module, whose additive closure equals exactly F (∆) ∩ F (∇).
Originally studied in [Bon89], and related to tilting theory, are exceptional modules and full exceptional sequences over a given algebra.For a quasi-hereditary algebra, the sequences of standard and costandard modules are examples of full exceptional sequences.The papers [HP19; PW20] provide a classification of the full exceptional sequences over the Auslander algebra of k[x]/(x n ) and its quadratic dual, respectively.
A large family of quasi-hereditary algebras is constituted by the dual extension algebras defined by Xi in [Xi94].These were introduced as an example of a class of BGG algebras.BGG algebras are quasi-hereditary algebras admitting a simple preserving duality on their module categories, see [Irv90].Xi's original construction was soon generalized and has been studied in greater detail in [DX94; DX96; LW15; LX17; Wu09; Xi95; Xi00].The algebras studied in this paper are examples of dual extension algebras.
The following is a brief description of the main results of this article.
(A) Let A n be the uniformly oriented linear quiver with n vertices, let k an algebraically closed field and set Let Λ n be the dual extension algebra A (A n , A op n ).We obtain a classification of generalized tilting modules over the algebra Λ n .This is achieved through the following steps.First, we show that any generalized tilting module is contained in F (∆) or in F (∇).Using the simple-preserving duality, this reduces the problem to the classification of generalized tilting modules in F (∆).Then, we consider the set of isomorphism classes of indecomposable self-orthogonal modules with a standard filtration.We provide a combinatorial description of the non-zero extensions of positive degree between the modules in this set in terms of a certain transitive relation.This interpretation allows us to classify generalized tilting modules contained in F (∆) as the maximal anti-chains with respect to this fixed relation.
Comparing the results with the first author's article, [PW20], in which the same problems are studied for another class of algebras, we see that the classification of generalized tilting modules is more complicated in the current case, while the classification of full exceptional sequences is identical in the two cases, in the sense that the form of the sequences is the same.
The present article is organized as follows.In Section 2, we briefly introduce the algebras Λ n , which are our objects of study, and recall some results on their quasi-hereditary structure.In Section 3, we classify the indecomposable Λ n -modules using the results from [BR87; WW85] on the classification of indecomposable modules over special biserial algebras.We also introduce some notation which is important for the readability of the subsequent sections.Section 4 contains a classification of the selforthogonal indecomposable Λ n -modules.In Section 5, we classify the generalized tilting modules over Λ n .Section 6 contains the classification of full exceptional sequences over Λ n .

Background
Throughout the rest of the article, let k be an algebraically closed field.Let A n be the uniformly oriented linear quiver 1 G G 2 G G . . .G G n − 1 G G n , for some n ∈ Z >1 and denote by kA n the corresponding path algebra.Let A n be the quotient of kA n by the ideal (rad kA n ) 2 .Finally, we define Λ n to be the dual extension algebra of A n with its opposite algebra, A op n , that is, Λ n = A(A n , A op n ).Then, Λ n is given by the quiver Let Λ n -mod denote the category of finite-dimensional left Λ n -modules.Throughout the rest of the article, we take "module" to mean left module.The algebra Λ n has a simple-preserving duality on its module category, denoted by ⋆, induced by the antiautomorphism given by swapping the arrows α i and α ′ i in the quiver of Λ n .Let L(i), where 1 ≤ i ≤ n, denote the simple Λ n -module corresponding to the vertex i.Let P (i) and I(i) denote its projective cover and injective envelope, respectively.Definition 1. [CPS88] Let Λ be a finite-dimensional algebra.Let {1, . . ., n} be an indexing set for the isomorphism classes of simple Λ-modules and let < be a partial order on {1, . . ., n}.The algebra Λ is said to be quasi-hereditary with respect to < if there exist modules ∆(i), where i ∈ {1, . . ., n}, called standard modules, satisfying the following.
(QH2) There is a surjection ∆(i) ։ L(i) whose kernel admits a filtration with subquotients L(j), where j < i.
This is equivalent to the existence of modules ∇(i), where i ∈ {1, . . ., n}, called costandard modules, satisfying the following.
It is easy to see that Λ n is quasi-hereditary with respect to the natural ordering on {1, . . ., n}.Indeed, (QH1) and (QH2) are easily verified with standard and costandard modules as below.Note that results from [Xi94] show that the dual extension algebra Λ n = A(A n , A op n ) is quasi-hereditary.Throughout, let F (∆) denote the full subcategory of Λ n -mod consisting of those modules which admit a filtration by standard modules.Similarly, let F (∇) denote the full subcategory consisting of those modules which admit a filtration by costandard modules.Since Λ n is quasi-hereditary, Theorem 5 from [Rin91] guarantees that there exists a basic module T , called the characteristic tilting module, such that the additive closure of T , denoted by add T , equals F (∆) ∩ F (∇).Moreover, T has the same number of non-isomorphic indecomposable summands as the number of isomorphism classes of simple modules, and we write The indecomposable direct summand T (k) of T is uniquely determined (up to isomorphism) by the property that it belongs to F (∆) ∩ F (∇) and that there exists a monomorphism ∆(k) ֒→ T (k), whose cokernel admits a filtration by standard modules.
We conclude this section by drawing the Loewy diagrams of the structural modules of Λ n and stating some of their elementary properties.

Indecomposable Λ n -modules
In order to classify indecomposable Λ n -modules, we use the fact that the algebra Λ n is a string algebra.For these algebras, the classification is known.Definition 4. [WW85] Let Λ = kQ I be the quotient of the path algebra of the quiver Q = (Q 0 , Q 1 ) by some admissible ideal I.For an arrow α ∈ Q 1 , denote by s(α) and t(α) the source and target vertex of α, respectively.Then Λ is called special biserial if the following hold.
(SB1) For each vertex i, there are at most two arrows with i as its source, and at most two arrows with i as its target.
If, in addition, the ideal I is generated by zero relations, Λ is called a string algebra.
We immediately note that Λ n is a string algebra for all n ∈ Z >1 .For special biserial algebras and string algebras the classification of indecomposable modules is known, see [BR87; WW85].There exist two classes of indecomposable modules, the so-called string modules and band modules.We will show that in the case of Λ n , there are no band modules and therefore a complete list of the indecomposable Λ n -modules is given by the string modules.
We follow closely the notation of [WW85].Let L = (L 0 , L 1 ) denote the quiver . . .ar−1 r ar r + 1 , r ≥ 0 where the a i are arrows with either orientation.Define a map ε : Similarly, we denote by Z = (Z 0 , Z 1 ) the quiver where the b i are arrows with either orientation and where i denotes the congruence class of i modulo r.Again, define ε : Let Q be some quiver.A quiver homomorphism w : for all i.The restriction of v (or u) to a connected linear subquiver L ′ of L (or Z) is called a subwalk or a subpath (or, a subtour or subcircuit ).Definition 5. [WW85] Fix a quiver Q and an admissible ideal I ⊂ kQ.A walk v : L → Q is called a V -sequence if the following hold.
(VS1) Each subpath of v does not belong to I.
Similarly, a tour u : Z → Q is called a primitive V -sequence if the following hold.
(VS3) The tour u is not a circuit and each subpath of u does not belong to I.
(VS5) There is no automorphism σ = id of Z, permuting the vertices cyclically such that u = u • σ.
In [WW85], the authors show that we can obtain all indecomposable modules over a special biserial algebra from V -sequences and primitive V -sequences.These correspond exactly to the string and band modules, respectively.In Proposition 7, we will see that there are no primitive V -sequences u : Z → Q, and consequently, no band modules.To obtain an indecomposable module from a V -sequence v : L → Q, consider the following representation of the bound quiver (Q, I).At each vertex x ∈ Q 0 , we put the vector space k if x is in the image of v, and the zero space otherwise.At each arrow x α G G y ∈ Q 1 , we put the identity map on the vector space k if α is in the image of v, and the zero map otherwise.This representation is then equivalent to a kQ/I-module.
However, an indecomposable module M does not arise from a unique V -sequence.In fact, the indecomposable module M corresponding to a V -sequence v : In this situation, we say that the V -sequences v and v ′ are isomorphic.There are only two possibilities for such an isomorphism σ.The first possibility is that L = L ′ and σ = id.The second possibility is that σ acts on the vertices of L ′ by σ(i) = r + 2 − i, that is, σ swaps the vertex 1 and the vertex r + 1, the vertex 2 and the vertex r, and so on.Here, we must have an Example 6.Let L be the quiver 1 Then L ′ as described above is the quiver . In a picture, the isomorphism σ : L ′ → L is as follows.
Proposition 7. Let Q be the quiver of Λ n and let I ⊂ kQ be the ideal generated by the relations in Section 1.Then, there are no primitive V -sequences u : Z → Q.Moreover, the only V -sequences v : L → Q, up to isomorphism, are of one of the following forms.
(a) We have ε(a i ) = ε(a i+1 ) and v(i + 1) = v(i) + 1 for all i.The V -sequence is given by the following picture.
, where v(i) = j, and consider the subquiver of L. There are four cases, depending on the values of ε(a i ) and ε(a i+1 ).
. A similar argument as in the previous case shows that this is mapped to the following subquiver.
), and we get the following picture.
and the following picture.
By similar arguments as in the previous case, we get one of the following two possible pictures.
Let u : Z → Q be a primitive V -sequence and let a be such that u(a) = j ≤ u(s) for all s ∈ Z 0 .It follows that u(a + 1) = j + 1 = u(a − 1).Then, to not contradict (VS4), the subquiver and so on.In this case ε(a s ) = ε(a s+1 ) for all s ≥ i.In particular, ε(a s ) = ε(a s+1 ) can occur at most once in a V -sequence.There are two cases.
However, any V -sequence of the latter type is isomorphic to one of the former type.This situation corresponds to part (a) of the statement of the proposition.
(b) We have ε(a s ) = ε(a s+1 ), for some s, ε( However, any V -sequence of the latter type is isomorphic to one of the former type.This situation corresponds to part (b) of the statement of the proposition.
Definition 8. Define Ω(i, j, k), where i, j ≤ k, to be the (up to isomorphism unique) indecomposable Λ n -module with the following Loewy diagram.
(a) If k ≡ i mod 2 and k ≡ j mod 2: Note that, in case (a) and (b), we may have i = k, and, in case (a) and (c), we may have j = k.For all i, j, k, the simple preserving duality maps the module Ω(i, j, k) to the module Ω(j, i, k).
Proposition 9.The set {Ω(i, j, k) | 1 ≤ i, j ≤ k ≤ n} is a complete and irredundant list of isomorphism classes of indecomposable Λ n -modules.There are, in total, n(n+1)(2n+1) Proof.By [WW85], all indecomposable Λ n -modules arise from V -sequences or primitive V -sequences.Using Proposition 7 we know that there are no primitive V -sequences, and what the possible Let v : L → Q be a V -sequence, of the form (b) in Proposition 7, such that v(1) = i.Then the indecomposable module corresponding to v is Ω(i, i + 2s − r, i + s).Thus, we see that any V -sequence gives rise to a (unique) module of the form Ω(i, j, k), and that any such module may be obtained from a V -sequence.This proves the first part of the proposition.
Every choice of i, j and k such that 1 ≤ i, j ≤ k ≤ n yields a unique module Ω(i, j, k).For a fixed k, there are k choices of i, and k choices of j, which implies that there are k 2 non-isomorphic modules Ω(i, j, k), with k fixed.The total number of non-isomorphic indecomposable Λ n -modules is therefore For any subset X ⊆ {1, 2, . . ., n}, we denote by P X the direct sum Should X be the empty set, we define P ∅ := 0. We use similar notation for such direct sums of other structural modules.For a, b ∈ {1, . . ., n}, with a ≤ b, we fix the following notation.

Self-orthogonal indecomposable modules
Having described the indecomposable Λ n -modules, the next step towards describing the generalized tilting Λ n -modules is determining which indecomposable Λ n -modules are self-orthogonal, as these will be candidates for inclusion in generalized tilting modules.For details on generalized tilting modules, we refer to Subsection 5.1.Throughout the following sections we will make frequent use of various dimension shifting arguments.We record the most common one in the following lemma.
Lemma 12. Let M and N be finite-dimensional Λ n -modules.Consider the following two short exact sequences, where K is the kernel of the projective cover P ։ M , and C is the cokernel of the injective envelope N ֒→ I: Proof.By applying Hom Λn ( , N ) to the first sequence, we get the long exact sequence for all k ≥ 2. If we instead apply Hom Λn (M, ) to the second sequence, a similar argument implies that ), for all k ≥ 2. By combining these results, we obtain Lemma 13.The module Ω(i, j, k) is contained in F (∆) if the following conditions are met.
(i) If i = 1 and j = 1, then Ω(i, j, k) ∈ F (∆) if and only if i ≡ k mod 2 and j ≡ k mod 2.
Let us instead draw the module Ω(2, 4, 6): 3 y y r r r r 7 7 ▲ ▲ ▲ ▲ 5 y y r r r r Here, we see that there is no way to remedy the composition factor L(4) contained in the top of Ω(2, 4, 6), preventing a standard filtration.The rest is similar.
Lemma 14.The module Ω(i, j, k) is contained in F (∇) if the following conditions are met.
(i) If i = 1 and j = 1, then Ω(i, j, k) ∈ F (∇) if and only if i ≡ k mod 2 and j ≡ k mod 2.
(iii) If i = 1 and j = 1, then Ω(i, j, k) ∈ F (∇) if and only if i ≡ k mod 2. (iv Proof.This follows from Lemma 13 by using the simple-preserving duality. Corollary 15.For all 1 ≤ k ≤ n, we have T (k) = Ω(1, 1, k), where T (k) denotes the kth indecomposable summand of the characteristic tilting module.
Lemma 16.Let M = Ω(i, j, k) and let K be the kernel of the projective cover P ։ M .
(i) If k = i then P ∼ = P (j,k] and if k > i then (ii) The form of the module K is given by the following table.
Lemma 17.The kernel of the projective cover of M has a simple direct summand if and only if Proof.This is easily checked by observing that the combinations of i, j and k which yield a simple direct summand of K, according to Lemma 16, are exactly those for which M does not have a ∆-filtration, according to Lemma 13.
Lemma 18.The cokernel of the injective envelope of M has a simple direct summand if and only if M / ∈ F (∇).
Proof.This follows from Lemma 17 by using the simple-preserving duality.
Proof.This follows from repeated application of Lemma 16.
Lemma 20.Let i and j be such that not both are equal to 1. Then Ext m Λn (L(i), L(j)) = 0 for Proof.Assume that i < j.By Lemma 19, the module P (j) appears at position j − i of the projective resolution of L(i), yielding a non-zero extension.The case j < i follows by using the simple-preserving duality.For i = j > 1, the same lemma tells us that P (i) appears in the second position of the projective resolution of L(i), which proves the claim.
Proposition 21.If an indecomposable Λ n -module M has neither a standard filtration nor a costandard filtration, then M is not self-orthogonal.
Proof.If M = Ω(i, j, k) is not simple (this case was covered in Lemma 20) and has neither a standard nor a costandard filtration, then i, j > 1 and either i ≡ k mod 2 and j ≡ k mod 2, or i ≡ k mod 2 and j ≡ k mod 2. This follows from Lemma 13 and Lemma 14.
Let K denote the kernel of the projective cover P ։ M and C the cokernel of the injective envelope M ֒→ I.By Lemma 17 and Lemma 18 both K and C have a simple direct summand.
• If i ≡ k mod 2 and j ≡ k mod 2, then L(j − 1) is a direct summand of K and L(i − 1) is a direct summand of C.
• If i ≡ k mod 2 and j ≡ k mod 2, then L(i − 1) is a direct summand of K and L(j − 1) is a direct summand of C.
Unless both i and j are equal to 2, using Lemma 12 together with Lemma 20, we get If i = j = 2, then M = Ω(2, 2, k) and the kernel K of the projective cover is equal to The beginning of the projective resolution of L(1) looks as follows: 2) is a submodule of M and there is no homomorphism from P (1) to M , we have a non-zero extension of degree 1 from L(1) to M .But this implies that there is a non-zero extension of degree two from M to itself, as L(1) is a direct summand of K and, by Lemma 12, we have Ext 2 Λn (M, M ) ∼ = Ext 1 Λn (K, M ).This proves the claim.
Proof.It is clear that Hom Λn (L(x), M ) = 0, for all 1 ≤ x < min(i, j), since M does not have L(x) as a composition factor for such x.It is also clear that Ext m Λn (L(1), M ) = 0, for m > 0, since L(1) ∈ F (∆) and M ∈ F (∇).This proves the claim for x = 1.Consider x such that 2 ≤ x < min(i, j).In this case, we have the short exact sequence and by Lemma 12 we have the following equality: Since neither L(x) nor L(x − 1) are composition factors of M this equality reduces to dim Ext 1 Λn (L(x), M ) = dim Hom Λn (∆(x + 1), M ).If 2 ≤ x < min(i, j) − 1, then clearly Hom Λn (∆(x + 1), M ) = 0, since L(x + 1) does not occur as composition factors in M .When x = min(i, j) − 1 we also have Hom Λn (∆(x + 1), M ) = 0. To see this, note that any non-zero homomorphism f : ∆(x + 1) → M annihilates rad(∆(x + 1)) = L(x) since L(x) does not occur as a composition factor in M .This implies that the image of f is isomorphic to L(x + 1).But M has no such submodule since the (unique) composition factor L(x + 1) is contained in the top of M , a contradiction.It follows that Ext 1 Λn (L(x), M ) = 0 for all x < min(i, j).
Assume that |i − j| = 1.Consider the surjection T (k) ։ M (which is unique up to a scalar) and denote the kernel of this projection by K.If i, j > 1, then the kernel K can be written as a direct sum of two indecomposable modules U ⊕ L, where U = Ω(1, i − 1, i − 1) and L = Ω(1, j − 1, j − 1).If i = 1 and j = 2, or i = 2 and j = 1, then the kernel K is equal to L(1).This gives us a short exact sequence By applying Hom Λn ( , M ), we get a long exact sequence Note that the only common composition factor of K and M is L(x), where x = min(i, j).However, L(x) is not a submodule of M , which means that there are no non-zero homomorphisms from K to M .Together with the fact that Ext 1 Λn (T (k), M ) = 0, this implies Ext 1 Λn (M, M ) = 0.
If i = 1 and j = 2, or i = 2 and j = 1, then Ext m Λn (M, M ) ∼ = Ext m−1 Λn (K, M ) = 0, for all m ≥ 2, since K = L(1) ∈ F (∆). Now assume that i, j > 1.By Lemma 16, the kernel J of the projective cover of Using Lemma 22 and the fact that M ∈ F (∇), this implies that Ext m−2 Λn (J, M ) = 0.It now follows that Λn (J, M ) = 0, for all m ≥ 2. Note that the second isomorphism is obtained by applying Lemma 12.This proves the claim that, if |i − j| = 1, then Ω(i, j, k) is self-orthogonal.
The beginning of the projective resolution of L(i − 1) looks as follows: Since L(i−2) is a submodule of M there is a homomorphism f : P (i−2)⊕P (i) → M such that the image of f is isomorphic to L(i − 2).Furthermore, the composition f • d 2 is equal to the zero homomorphism.However, we cannot have f = g • d 1 any homomorphism g : P (i − 1) → M .Indeed, the kernel of the unique (up to scalar) non-zero map from P (i − 1) to M is equal to L(i − 1).The image of g • d 1 therefore contains the submodule L(i).Since the image of f is isomorphic to L(i − 2) the two maps cannot be equal.This means that Ext 1 Λn (L(i − 1), M ) 0, which implies that Ext 2 Λn (M, M ) = 0.
By Lemma 16, the kernel K of the projective cover of M is equal to , where L(x) is interpreted as zero if x < 1.By dualizing Lemma 16, the cokernel C of the injective envelope of N is equal to ℓ−1 x=j+1 ∇(x) ⊕ L(j − 1) ⊕ L(j), where L(x) is interpreted as zero if x < 1.Unless i = j = 1, using Lemma 12 together with Lemma 20, we get Ext m Λn (M, N ) ∼ = Ext m−2 Λn (K, C) = 0, for some m > 0. Assume that i = j = 1.Since i = 1, as mentioned before, the kernel K of the projective cover of M is equal to k−1 x=2 ∆(x) ⊕ L(1).The beginning of the projective resolution of L(1) looks as follows: 2) and we a non-split short exact sequence ∆(2) ֒→ P (1) ։ L(1).
This, together with Lemma 12, implies that there is a non-zero extension of degree two from M to N .
If ℓ > 2, there is a homomorphism f : P (2) → N whose image is isomorphic to L(2) and annihilates rad P (2).Since the image of the differential d 2 is contained in rad P (2), we have f • d 2 = 0.However, the unique (up to scalar) homomorphism g : P (1) → N is such that L(2) does not occur as a composition factor in the image, so that we must have f = g • d 1 .Therefore, we have a non-zero extension of degree one from L(1) to N .But this, together with Lemma 12, implies that there is a non-zero extension of degree two from M to N .This proves the claim.

5.3.
A strict partial order on F (∆).We know that the only generalized tilting module contained in F (∆) ∩ F (∇) is the characteristic tilting module.Moreover, we have shown that if M ∈ F (∇), N ∈ F (∆) and M, N / ∈ F (∆) ∩ F (∇), then Ext m Λn (M, N ) = 0 for some m ≥ 1.This implies that any generalized tilting module must be contained in either F (∆) or F (∇). Together, these statements imply that it is enough to find the basic generalized tilting modules contained in F (∆), which do not equal the characteristic tilting module.All basic generalized tilting modules will then be these modules, their duals (with respect to the simple preserving duality), and the characteristic tilting module.
Let D n denote the set of indecomposable self-orthogonal modules in F (∆). Next, we define the relation ≺ E on D n given by M ≺ E N if and only if Ext m (M, N ) = 0 for some m ≥ 1.Note that it is not clear from the definition whether this relation is transitive.
Let Q n be the set of pairs (i, k), with 1 ≤ i ≤ k ≤ n.We want these pairs to encode the indecomposable self-orthogonal modules contained in F (∆).Let M (i, k) be the module This defines a bijection ϕ : We define the following relation on For k = 1, 2, . . .we have the additional relations Now define ≺ to be the transitive closure of ≺ 0 .This defines a strict partial order on Q n .By a strict partial order on a set, we mean a relation which is transitive, asymmetric and irreflexive.The set Q n is naturally graded by deg(i, k) = i, and from the definition it is clear that this grading, together with the strict partial order ≺, makes Q n into a graded poset.
The aim of the next subsection is to prove the following theorem: Theorem 28.The bijection ϕ : Q n → D n defined above is an order isomorphism between (Q n , ≺) and (D n , ≺ E ).
Before proving the theorem, let us briefly consider its implications.Let n = 4.We have the following Hasse diagram for the graded poset (Q n , ≺).
(1, 1) We view the above picture as a graph, with the vertices (i, k) corresponding to indecomposable selforthogonal modules in F (∆).We draw an edge (i, k) → (j, ℓ) if and only if (i, k) ≺ (j, ℓ), which, as we will show, holds if and only if there is a non-zero extension between the corresponding modules.
With the theorem established, we will be able to find all generalized tilting modules over Λ n which are contained in F (∆) by finding all anti-chains of length n in the above graph.To see this, note that an antichain of length n corresponds exactly to a self-orthogonal module of finite projective dimension, having n indecomposable summands.Such a module is a generalized tilting module by, Corollary 26.

5.4.
Proof of Theorem 28.Let M, N be two Λ n -modules.Consider the following two exact sequences, where K is the kernel of the projective cover P ։ M and C is the cokernel of the injective envelope N ֒→ I: If M is the module corresponding to (i, k), where k > i, then, by Lemma 16, we have . The module corresponding to (i, i) is ∆(i) and, in this case, P = P (i) and K = ∆(i + 1).If N is the module corresponding to (j, ℓ), then, by looking at N ⋆ and using Lemma 16, we find that Recall that, by Lemma 12, we have the following.
), for all k ≥ 3. Using the above, together with Lemma 16, we get the following equations for the dimensions of extension spaces from M to N . where where m ≥ 3, X = {i, i + 1, . . ., k − 1}, if k > i, and X = {i}, if k = i.Note that for the last equality we have used the fact that Ext m Λn (∆(x), ∇(y)) = 0, for all m > 0. These equations indicate that we need to find the dimensions of various homomorphism spaces, as well as determining when we have a non-zero extension of positive degree from a standard module to a simple module, or not.
Lemma 29.We have the following dimensions for the homomorphism spaces from a projective module to M (i, k) and from a standard module to M (i, k), respectively: Proof.Since dim Hom Λn (P (x), M (i, k)) is equal to the number of composition factors L(x) in M (i, k), the first result follows immediately from the definition of M (i, k).If x < i−1, or x > k, then ∆(x) and M (i, k) do not have any common composition factors, and there cannot exist any non-zero homomorphisms from ∆(x) to M (i, k).For x = k, we note that ∆(k) is a submodule of M (i, k), and, since there is only one composition factor L(k) in M (i, k), it is clear that the inclusion of ∆(k) into M (i, k) is the only homomorphism.
Now assume that i − 1 ≤ x ≤ k − 1.We claim that, for each such x, there is exactly one homomorphism from ∆(x) to M (i, k) (up to a scalar), namely the homomorphism that sends the top of ∆(x) to the (unique) submodule L(x) ⊂ M (i, k) and annihilates the radical of ∆(x).If x = 1, or equivalently, if ∆(x) is not simple, the only other possibility would be an inclusion of ∆(x) into M (i, k), but for such x, the module ∆(x) is not a submodule of M (i, k).
The following lemma can be proven using the main theorem in [CB89], but for pedagogical reasons we give a more elementary proof.
• dim Hom Λn (M (i, k), M (j, ℓ)) = min(k, ℓ) − max(i, j − 1), if i ≤ ℓ, j < k and, in addition, one of the following hold: • dim Hom Λn (M (i, k), M (j, ℓ)) = min(k, ℓ) − max(i, j − 1) + 1, if i ≤ ℓ, j < k and, in addition, one of the following hold: Proof.First we note that we have the following formulas for the top of M (i, k) and the socle of M (j, ℓ): Observe that, if i > ℓ or j > k, then no direct summand of top(M (i, k)) is a composition factor of M (j, ℓ), so dim Hom Λn (M (i, k), M (j, ℓ)) = 0.If j = k (and thus i ≤ ℓ), then dim Hom Λn (M (i, k), M (j, ℓ)) = 1, and a basis vector in Hom Λn (M (i, k), M (j, ℓ)) is given by the homomorphism defined by mapping Assume throughout the rest of the proof that i ≤ ℓ and j < k.Then, each composition factor L(x), belonging to the top of M (i, k), can be mapped to the composition factor L(x), contained in the socle of M (j, ℓ), for each x such that max(i, j − 1) We will now investigate whether or not there is an additional homomorphism f that does not belong to the subspace spanned by the homomorphisms previously mentioned.To get such a homomorphism f , some composition factor L(x), contained in the top of M (i, k), must be mapped to the (unique) composition factor L(x) not contained in the socle of M (j, ℓ).This means that either L(x) is contained in the top of M (j, ℓ), or that x = ℓ.
Assume that f is such a homomorphism.Since the image of f is a submodule, the composition factors L(x − 1) and L(x + 1), contained in the radical of M (j, ℓ), must also belong to the image of f , assuming that they are composition factors of M (j, ℓ).This means that the composition factors L(x − 1) and L(x + 1), contained in the radical of M (i, k), do not belong to the kernel of f .But the kernel is a submodule, so any composition factors having arrows to L(x − 1) or L(x + 1) in the Loewy diagram of M (i, k) also do not belong to the kernel of f .Repeating these two arguments we will either reach a contradiction, meaning that there are no more homomorphisms, or come to the conclusion that there is (up to scalar) exactly one more homomorphism.
We will investigate whether or not there exists x such that a homomorphism f could map L(x), contained in the top of M (i, k), to the (unique) composition L(x) not contained in the socle of M (j, ℓ).This will depend on the parameters i, j, k, ℓ, x.
• Assume that j ≥ i + 2. Repeating the arguments above we find that one of the composition factors L(j − 1) or L(j − 2), depending on the parity of x and j, both contained in the top of M (i, k), are not contained in the kernel of f .This is a contradiction.Indeed, L(j − 2) is not a composition factor of M (j, ℓ), so L(j − 2) must be contained in the kernel of f .The other case corresponds to x and j having different parity.In this case the (unique) composition factor L(j − 1) of M (j, ℓ) does not belong to the same arm as the composition factor L(x) in question.This implies that even if L(j − 1) would be contained in the image of f , the map would not commute with the action of Λ n , and is therefore not a homomorphism.Therefore, in this case, we have dim Hom Λn (M (i, k), M (j, ℓ)) = min(k, ℓ) − max(i, j − 1).
• Assume that j = i + 1.First we consider x such that x ≡ i mod 2. Then the composition factor L(x) that belongs to the top of M (i, k) is contained in the same arm as the (unique) composition factor L(i − 1) of M (i, k).Similarly, the composition factor L(x) that does not belong to the socle of M (j, ℓ) is contained in the same arm as the composition factor L(i + 1) that belongs to the socle of M (j, ℓ).If f would map L(x) contained in the top of M (i, k) to the (unique) composition factor L(x) not contained in the socle of M (j, ℓ), using previous arguments leads to a similar contradiction as in the case when j ≥ i + 2. This implies that f must map the composition factor L(x), for x ≡ i mod 2, that belongs to the top of M (i, k), to the socle of M (j, ℓ).
Next we consider x such that x ≡ i mod 2. Then the composition factor L(x) that belongs to the top of M (i, k) is contained in the same arm as the composition factor L(i) that belongs to the socle of M (i, k).Similarly, if x = ℓ, the composition factor L(x) that does not belong to the socle of M (j, ℓ) is contained in the same arm as the composition factor L(i) that belongs to the socle of M (j, ℓ).If x = ℓ, then it follows that L(i) ⊂ soc(M (j, ℓ)) belongs to the lower arm of M (j, ℓ).We have five cases.
− Suppose ℓ < k and i ≡ ℓ mod 2. This implies that L(x) ⊂ upp(M (j, ℓ)) and x = ℓ.If f would map L(x) contained in top of M (i, k) to the (unique) composition factor L(x) not contained in the socle of M (j, ℓ), then, in the same way as before, we see that L(ℓ + 1) ⊂ M (i, k) is not contained in the kernel of f , which is a contradiction since L(ℓ + 1) is not a composition factor of M (j, ℓ).Therefore, in this case, we have dim Hom Λn (M (i, k), M (j, ℓ)) = min(k, ℓ) − max(i, j − 1).
− Suppose ℓ < k and i ≡ ℓ mod 2. Then low(M (j, ℓ)) occurs as a quotient of M (i, k), giving us a homomorphism.Using the same argument as above we see that this is the only possibility for f .Therefore, in this case, we have dim Hom Λn (M (i, k), M (j, ℓ)) = min(k, ℓ) − max(i, j − 1) + 1.
− Suppose ℓ > k and i ≡ ℓ mod 2.Then, upp(M (i, k)) is isomorphic to a submodule of M (j, ℓ), giving us a homomorphism.Using the same argument as above we see that this is the only possibility for f .Therefore, in this case, we have dim Hom Λn (M (i, k), M (j, ℓ)) = min(k, ℓ) − max(i, j − 1) + 1.
, then also L(x) ⊂ upp(M (j, ℓ)).Using the same arguments as before leads to a similar contradiction as in the case j ≥ i + 2. If, instead, L(x) ⊂ low(M (i, k)) then also L(x) ⊂ low(M (j, ℓ)) and the situation is similar to the previous case.Therefore, in this case, we have dim Hom Λn (M (i, k), M (j, ℓ)) = min(k, ℓ) − max(i, j − 1).
• Assume that j ≤ i.We have three cases.
− Suppose ℓ < k.Depending on the parity of ℓ − i, either Ω(ℓ, i − 1, ℓ) or Ω(ℓ, i, ℓ) is a quotient of M (i, k) and this quotient is isomorphic to the lower arm of M (j, ℓ).This defines a homomorphism f from M (i, k) to M (j, ℓ).Using the same arguments as above, we see that this is the only possibility for f .Therefore, in this case, we have dim Hom Λn (M (i, k), M (j, ℓ)) = min(k, ℓ) − max(i, j − 1) + 1.
− Suppose ℓ > k.In this case, we see that the upper arm of M (i, k) is isomorphic to a submodule of M (j, ℓ), and f can be chosen as (a scalar multiple of) the obvious embedding.Using the same argument as above, we see that this is the only possibility for f .Therefore, in this case, we have dim Hom Λn (M (i, k), M (j, ℓ)) = min(k, ℓ) − max(i, j − 1) + 1.
− Suppose ℓ = k.In this case, we see that M (i, k) ⊂ M (j, ℓ) is a submodule, and consequently, f can be chosen as (a scalar multiple of) the obvious embedding.Using the same argument as above, we see that this is the only possibility for f .Therefore, in this case, we have dim Hom Λn (M (i, k), M (j, ℓ)) = min(k, ℓ) − max(i, j − 1) + 1. Proof.Splicing the short exact sequences ∆(x + 1) ֒→ P (x) ։ ∆(x), it follows that, at position m in the projective resolution of ∆(x), we find the projective module P (x+m).This means that, if y ≤ x, then Ext m Λn (∆(x), L(y)) = 0, for all m > 0. At position y − x, we find the projective module P (y), which surjects onto L(y), giving rise to a non-zero extension.It is also clear that Ext m Λn (∆(x), L(y)) = 0, for m = y − x, since there is no homomorphism from P (m + x) to L(y) for such m.
With these results in hand, we can now determine between which pairs of modules in D n all extensions of positive degree vanish.
Proposition 35.Let (i, k) and (i + 1, ℓ) be such that (i, k) Proof.By Lemma 32 and Proposition 33 all extensions of degree two or higher from M (i, k) to M (i+1, ℓ) vanish, as well as all extensions of positive degree from M (i + 1, ℓ) to M (i, k).That the remaining extensions vanish can be proved using the same strategy as for Lemma 32, by using Lemma 29 and 30 as well as Equation (1).
Proof of Theorem 28.By applying Proposition 33, Proposition 34, Proposition 35 and Proposition 36, it is now clear that (i, k) ≺ (j, ℓ) if and only if there is a non-zero extension from M (i, k) to M (j, ℓ).
• Suppose T = T ′ ⊕ M (i, n).If T ′ is the characteristic tilting module over Λ n−1 , then T would be the characteristic tilting module over Λ n , contradicting our assumption.If for some x ≡ i mod 2, then T contains the summand M (i + 1, n − 1), which is a contradiction as there is a non-zero extension from M (i, n) to M (i + 1, n − 1).Therefore, we must have for some 1 ≤ i ≤ n − 2 and i < x ≤ n − 2 such that x ≡ i mod 2.
• Suppose T = T ′ ⊕ M (i + 1, n).If T ′ is the characteristic tilting module over Λ n−1 , then i = 1 and we have a non-zero extension from M (1, n − 2) to M (2, n). If for some x ≡ i mod 2, then T contains the summand M (i, n − 1), which is a contradiction as there is a non-zero extension from M (i, n − 1) to M (i + 1, n).Therefore, we must have Assume instead that i ≡ n mod 2.
• Suppose T = T ′ ⊕ M (i, n).If T ′ is the characteristic tilting module over Λ n−1 , then T would be the characteristic tilting module over Λ n , contradicting our assumption.If for some x ≡ i mod 2, then T contains the summand M (i + 1, n − 2), which is a contradiction as there is a non-zero extension from M (i, n) to M (i + 1, n − 2).Therefore, we must have for some 1 ≤ i ≤ n − 2 and i < x ≤ n − 1 such that x ≡ i mod 2.
• Suppose T = T ′ ⊕ M (i + 1, n).If T ′ is the characteristic tilting module over Λ n−1 , then i = 1 and we have a non-zero extension from M (1, n − 1) to M (2, n). If for some x ≡ i mod 2, then T contains the summand M (i, n − 2), which is a contradiction as there is a non-zero extension from M (i, n − 2) to M (i + 1, n).Therefore, we must have for some 1 ≤ i ≤ n − 2 and i < x ≤ n − 1 such that x ≡ i mod 2.
This finishes the proof.
We remark that L(1) is a direct summand in exactly one basic generalized tilting module, namely the characteristic tilting module.

Exceptional sequences
Let Λ be a finite-dimensional k-algebra.Recall, see [Bon89], that an indecomposable Λ-module M is called exceptional provided that • End Λ (M ) ∼ = k; • Ext i Λ (M, M ) = 0, for all i > 0. A sequence M = (M 1 , . . ., M k ) of Λ-modules is called an exceptional sequence provided that • each M i is exceptional; • Ext i Λ (M x , M y ) = 0, for all 1 ≤ y < x ≤ k and all i ≥ 0. An exceptional sequence is called full (or complete) if it generates the derived category.In particular, this means that it must contain at least n modules, where n is the number of isomorphism classes of simple Λ-modules.Indeed, suppose an exceptional sequence N contains k < n modules.Let A ⊂ D b (Λ) be the triangulated subcategory generated by N. Passing to the Grothendieck group K 0 (A), we see that any mapping cone of a homomorphism between modules in N equals a linear combination of modules in N, implying that rank K 0 (A) ≤ k < n = rank K 0 (D b (Λ)).Thus, such an exceptional sequence cannot be full.
Proof.By Proposition 23, we know that Ω(i, j, k) is self-orthogonal exactly when i = j = 1 or |i − j| = 1.For such i, j, the module Ω(i, j, k) is neither standard nor costandard if and only if k > max(i, j).In this case, there is one composition factor L(k − 1) in the top of Ω(i, j, k) and another composition factor L(k − 1) in the socle of Ω(i, j, k).This means that there is an endomorphism of Ω(i, j, k) sending the composition factor L(k − 1) in the top to the composition factor L(k − 1) in the socle.Thus dim End Λn (Ω(i, j, k)) > 1 and Ω(i, j, k) is therefore not an exceptional module.
For any quasi-hereditary algebra both standard and costandard modules are self-orthogonal and have trivial endomorphism algebras, and are therefore exceptional modules.
Proof.By Proposition 45, the standard and costandard modules are the only exceptional modules, so no other modules may be included in an exceptional sequence.We observe that the only non-zero homomorphisms, not equal to a scalar multiple of the identity, between standard and costandard modules are the following.
• We may map the top of ∆(k) to the socle of ∇(k).
• We may map the top of ∇(k) to the socle of ∆(k).
• We may map the top of ∆(k) to the socle of ∆(k + 1).
• We may map the top of ∇(k) to the socle of ∇(k − 1).This implies that an exceptional sequence cannot contain both ∆(k) and ∇(k) for k = 1.Further, as a full exceptional sequence must contain n modules, it has to be of the form ( * ), up to ordering.By Proposition 27, there is a non-zero extension from each costandard module to any standard module.This implies that any costandard module must come before any standard module in an exceptional sequence.By Theorem 28, or more specifically, by Proposition 34 and Proposition 36, there is a non-zero extension from ∆(i) to ∆(j), for every i < j.Using the simple preserving duality, we find that there is a non-zero extension from ∇(i) to ∇(j), for every i > j.This implies that any full exceptional sequence must be of the form ( * ), as it must contain at least n modules.
Left to prove is that a sequence M = (M 1 , . . ., M n ) of the form ( * ) actually is a full exceptional sequence.That Hom Λn (M x , M y ) = 0, for x > y, is clear from the first part of the proof.Furthermore, for every quasi-hereditary algebra, the following equality holds.Ext m (∆(i), ∆(j)) ∼ = Ext m (∇(j), ∇(i)) ∼ = Ext m (∆(k), ∇(ℓ)) = 0, for i > j, any k, ℓ and m > 0. From this it follows that any sequence of the form ( * ) is exceptional.
Left to show is that a sequence of the form ( * ) generates the derived category (as a triangulated category).We recall that any triangulated category is closed under taking kernels of epimorphisms and cokernels of monomorphisms.As the set of simple modules generate the derived category, it suffices to show that we can obtain the simple modules L(i) by performing these operations on the modules in our sequence.We proceed by induction on i = 1, 2, . . ., n.The basis for the induction is clear, since L(1) is included in any sequence of the form ( * ).Next, assume that L(i − 1) can be obtained from our sequence.By assumption, the sequence contains either ∆(i) or ∇(i).In the first case, L(i) is the cokernel of the inclusion L(i − 1) ֒→ ∆(i).In the second case, L(i) is the kernel of the projection ∇(i) ։ L(i − 1).This shows that any sequence of the form ( * ) is a full exceptional sequence.