Document [original]

Thick subcategories for quiver

representations

Nikolay Dimitrov Dichev

A thesis presented for the degree of

Doktor der Naturwissenschaften (Dr. rer. nat.)

Institute of Mathematics

Faculty of Computer Science, Electrical Engineering and

Mathematics

University of Paderborn

Germany

June 2009

Thick subcategories for quiver representations

Nikolay Dimitrov Dichev

Submitted for the degree of

Doktor der Naturwissenschaften (Dr. rer. nat.)

June 2009

Abstract

The central objects of investigation in this thesis are the thick subcategories as

well as the exact abelian extension closed subcategories of the category of quiver rep-

resentations. A full additive subcategory Cof an abelian category Ais called thick,

provided that Cis closed under taking direct summands, kernels of epimorphisms,

cokernels of monomorphisms and extensions. The category Cis called exact abelian

if it is abelian, the embedding functor preserves exact sequences, hence closed under

arbitrary kernels and cokernels.

First we consider the category of locally nilpotent representations over the path

algebra of the cyclic quiver. We show that any thick subcategory is exact abelian.

Then we give a combinatorial description of thick subcategories via non-crossing

arcs on the circle and using generating functions, we calculate their number. Fur-

thermore, we establish a bijection between thick subcategories with a projective

generator, thick subcategories without a projective generator, support-tilting and

cotilting modules. Then we study exact abelian extension closed subcategories for

Nakayama algebras, and we find a recursive formula for their number.

For a finite and acyclic quiver, we consider the category of its quiver representa-

tions. We show that any thick subcategory generated by preprojective or preinjective

representations is exact abelian. Then we specialise to Euclidian quiver case and we

verify that any thick subcategory is exact abelian. Furthermore, we extend a result

of Ingalls and Thomas and we give a complete combinatorial classification of thick

subcategories in that case.

For a hereditary algebra A, we consider the tilted algebra B= EndA(TA), where

TAis a tilting module. We establish a bijection between the exact abelian extension

and torsion closed subcategories of mod Aand the exact abelian extension closed

subcategories of mod B.

Dicke Unterkategorien f¨ur K¨ocherdarstellungen

Nikolay Dimitrov Dichev

Dissertation zur Erlangung des akademischen Grades

Doktor der Naturwissenschaften (Dr. rer. nat.)

Juni 2009

Zusammenfassung

Die vorliegende Arbeit besch¨aftigt sich mit den dicken, sowie mit den exak-

ten abelschen und erweiterungsabgeschlossenen Unterkategorien der Kategorie der

Darstellungen eines K¨ochers. Eine volle additive Unterkategorie Ceiner abelschen

Kategorie Aheißt dick, falls Cabgeschlossen ist unter Bildung von direkten Sum-

manden, Kernen von Epimorphismen, Kokernen von Monomorphismen und Er-

weiterungen. Die Kategorie Cheißt exakt abelsch, falls sie abelsch ist und der

Einbettungsfunktor exakte Folgen erh¨alt, insbesondere ist Cdann abgeschlossen

bez¨uglich Bildung von beliebigen Kernen und Kokernen.

Zun¨achst untersuchen wir die Kategorie der lokal nilpotenten Darstellungen ¨uber

der Wegealgebra eines zyklischen K¨ochers. Wir zeigen, dass eine dicke Unterkate-

gorie exakt abelsch ist. Hiernach beschreiben wir kombinatorisch die dicken Un-

terkategorien durch nicht kreuzende B¨ogen auf einem Kreis und mit Hilfe der erzeu-

genden Funktionen berechnen wir ihre Anzahl. Weiterhin zeigen wir Bijektionen

zwischen den dicken Unterkategorien mit projektivem Generator, den dicken Un-

terkategorien ohne projektiven Generator, Tr¨agerkipp- und Kokippmoduln. Dann

untersuchen wir die exakten abelschen und erweiterungsabgeschlossenen Unterkate-

gorien f¨ur Nakayama Algebren und finden eine rekursive Formel f¨ur ihre Anzahl.

Danach wenden wir uns der Kategorie der Darstellungen endlicher azyklischer

K¨ocher zu. Wir zeigen, dass dicke Unterkategorien, die von pr¨aprojektiven oder

preinjektiven Darstellungen erzeugt werden, exakt abelsch sind. Wir untersuchen

euklidische K¨ocher im Speziellen und zeigen, dass dicke Unterkategorie exakt abelsch

sind. Dann erg¨anzen wir ein Ergebnis von Ingalls und Thomas zu einer vollst¨andige

kombinatorische Klassifikation der dicken Unterkategorien f¨ur diesen Fall.

F¨ur eine erbliche Algebra Abetrachten wir die gekippte Algebra B= EndA(TA),

wobei TAKippmodul ist. Wir zeigen eine Bijektion zwischen den exakten abelschen

erweiterungs- und torsionsabgeschlossenen Unterkategorien von mod Aund den ex-

akten abelschen erweiterungsabgeschlossenen Unterkategorien von modB.

Acknowledgements

I would like to thank my advisor Henning Krause for his support of the thesis. Also

for the patience and encouraging words in the moments of uncertainty. I express

my gratitude to Yu Ye, for his responsiveness and for his constant encouragements.

I am grateful to him for his ideas, which led to the results in Chapter 3. Special

thanks to Dirk Kussin and Hugh Thomas for being my referees; in particular I

thank Hugh Thomas for the patience to read the draft of my thesis, for corrections,

comments and recommendations he gave, which significantly improved the thesis. I

would like to thank Xiao-Wu Chen for the helpful discussions concerning Chapter 4.

Thanks to Marcel Wiedemann for proofreading parts of this thesis, as well as for his

and Claudia K¨ohler’s help on writing die Zusammenfassung. I thank the members

of Paderborn Representation Theory Group for the friendly atmosphere during my

research.

Thanks to PACE(Paderborn Institute for Advanced Studies in Computer Science

and Engineering) and in particular IRTG(International Research Training Group

“Geometry and Analysis of Symmetries”) for the financial support during my stud-

ies, as well as for the organized cultural events and soft skills training. Also for the

generous travel allowance, which favours my participation to numerous conferences

and workshops.

I thank my father Dimitˆar, my brother Georgi, and especially my mother Tonka,

for the constant moral support, and although far away from me, their encouraging

words helped me to overcome the difficult moments during my stay in Paderborn. I

thank my colleagues and friends Carsten Balleier, Indrava Roy and his wife Saumya,

Stefan Wolf for the nice moments we spent together. Finally, I thank my fianc´ee

Radoslava for her true love, for the energy and for the inspiration that she is giving

me.

Contents

Abstract ii

Zusammenfassung iii

Acknowledgements iv

1 Introduction 1

2 Thick subcategories for cyclic quivers 5

2.1 Cyclic quivers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.2 Orthogonal sequences and thick subcategories . . . . . . . . . . . . . 9

2.3 Cotilting, support-tilting modules and thick subcategories . . . . . . 15

2.4 Number of thick subcategories . . . . . . . . . . . . . . . . . . . . . . 26

2.5 Lattice of thick subcategories . . . . . . . . . . . . . . . . . . . . . . 29

2.6 Nakayama algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

2.6.1 Self-injective Nakayama algebras . . . . . . . . . . . . . . . . 33

3 Thick subcategories for hereditary algebras 37

3.1 Thick subcategories generated by preprojective modules . . . . . . . . 37

3.2 Thick subcategories for Euclidean quivers . . . . . . . . . . . . . . . . 42

3.2.1 Classification of thick subcategories . . . . . . . . . . . . . . . 51

3.3 Thick subcategories are exact abelian . . . . . . . . . . . . . . . . . . 53

4 Exact abelian extension closed subcategories for tilted algebras 55

4.1 Torsion pairs, tilting modules and tilted algebras . . . . . . . . . . . . 55

4.2 Exact abelian extension closed subcategories for tilted algebras . . . . 60

Appendix 68

A Basic and auxiliary results 68

A.1 Quivers and their representations . . . . . . . . . . . . . . . . . . . . 68

A.2 Hereditary algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

Contents vi

A.3 The Auslander-Reiten quiver . . . . . . . . . . . . . . . . . . . . . . . 72

A.4 Two homological facts . . . . . . . . . . . . . . . . . . . . . . . . . . 74

Bibliography 76

Chapter 1

Introduction

In investigations of the structure and properties of algebras (resp. their modules), it

is often essential to have a concrete realisation of a given algebra (resp. their mod-

ules). In general, the aim of the representation theory of algebras is to develop tools

for such realisations. Due to work of Gabriel [Gab1], each finite dimensional algebra

over an algebraically closed field kcorresponds to a graphical structure, called a

quiver, and conversely, each quiver, more precisely its associated path algebra, corre-

sponds to an associative k-algebra, which has an identity and it is finite dimensional

under some conditions. In fact, using the quiver associated to an algebra A, it

is possible to visualise a finitely generated A-module as a quiver representation, a

family of finite dimensional k-vector spaces, connected by linear maps.

In the thesis, we deal mostly with finite dimensional hereditary algebras. An

algebra is hereditary, if any submodule of a projective module (=a module with

basis vectors) is projective. In fact, any such algebra is realised by the path algebra

of a finite and acyclic quiver. The working environment for us is the module category

of a finite dimensional (hereditary) algebra, that is the category of finite dimensional

vector spaces with scalars from the algebra.

Quiver-theoretical techniques provide a convenient way to visualise finite dimen-

sional algebras. However, actually to compute the indecomposable modules and

the homomorphisms between them, we need other tools. For a finite dimensional

algebra A, there is a special quiver, called the Auslander-Reiten quiver of mod A,

that combinatorially encodes the building blocks of mod A, namely the indecom-

posable modules and the irreducible morphisms. It can be considered as a first

approximation of the module category of a finite dimensional algebra.

The central objects of our study are thick and exact abelian extension closed

subcategories of a module category of an algebra (or equivalently the category of its

quiver representations).

A full additive subcategory Cof an abelian category Ais called thick, provided

Chapter 1. Introduction 2

that Cis closed under taking direct summands, kernels of epimorphisms, cokernels

of monomorphisms and extensions. Cis called exact abelian if it is abelian, the

embedding functor preserves exact sequences, hence closed under arbitrary kernels

and cokernels. From the definition it follows that an exact abelian subcategory is

thick if and only if it is closed under taking extensions, and a thick subcategory is

exact abelian if and only if it is closed under taking arbitrary kernels. The latter is

true since if Cis thick, and X, Y are objects in C,

X

f//Y

Ker f

-

yIm f

.

zCoker f

then Ker f∈ C ⇔ Im f∈ C ⇔ Coker f∈ C.

The study of exact abelian extension closed subcategories was highlighted by

recent work of Colin Ingalls and Hugh Thomas. They establish a large class of bi-

jections involving them, which give a relation to important objects of representation

theory of finite dimensional algebras, as well as a relation to recently developing

cluster algebras and cluster categories.

Theorem 1.0.1 [IT] Let Qbe a finite acyclic quiver. There are bijections between

the following objects:

•clusters in the acyclic cluster algebra with initial seed Q;

•isomorphism classes of basic cluster-tilting objects in the cluster category;

•isomorphism classes of basic support-tilting objects in mod kQ;

•torsion classes in mod kQ with a projective generator;

•exact abelian extension closed subcategories in mod kQ with a projective gen-

erator.

Further, a connection with derived categories was found by Kristian Br¨uning in

his thesis.

Theorem 1.0.2 [Br1] There is a bijection between thick subcategories in Db(mod kQ)

and exact abelian extension closed subcategories in mod kQ.

In this thesis, the study of exact abelian extension closed subcategories of a

hereditary abelian category is continued. I shall give a brief account of my work by

outlining the obtained results.

Chapter 1. Introduction 3

The core work of the thesis is contained in chapters 2, 3 and 4. Each chapter

begins with a short introduction. In order to be self-contained, all the facts needed

(with appropriate references) are also exposed within the chapter.

In chapter 2, we consider the path algebra k˜

∆nof the cyclic quiver,

∆n: 1 //2//3//... //n.

and two (full, additive) subcategories of category of their representations, namely

Tnthe category of locally nilpotent representations and Tnthe category of nilpotent

representations. We comment that the category Tnplays an important rˆole in the

representation theory of algebras of infinite representation type, since it describes

(connected) components of the Auslander-Reiten quiver of their module categories.

In proposition 2.2.10, we observe that every thick subcategory in Tnis exact

abelian. After that, in proposition 2.2.13 we give a combinatorial classification of

thick subcategories via establishing a bijection with the non-crossing arcs on the

circle. Further, in proposition 2.4.2 using generating functions we calculate their

number.

The main result in the chapter is a bijection involving thick subcategories.

Theorem 1.0.3 There is a bijective correspondence between:

•isomorphism classes of support-tilting objects in Tn;

•thick subcategories in Tnwith a projective generator;

•thick subcategories in Tnwithout a projective generator;

•isomorphism classes of cotilting objects in ˜

Tn.

At the end, we classify exact abelian extension closed subcategories for a class of

algebras, called Nakayama algebras, which are quotients of the path algebra of the

cyclic quiver. In proposition 2.6.13, we give a recursive formula for their number.

We comment that the found formula is a generalisation of the recursive formula for

the Catalan numbers.

The results in Chapter 3 are joint work with Yu Ye. For a finite and acyclic

quiver Q, we consider its path algebra kQ. We step on a result of Crawley-Boevey

[CB1, Lemma 5], which says that any thick subcategory of mod kQ generated by an

exceptional sequence (a special sequence of indecomposable kQ-modules) is exact

abelian. In proposition 3.1.10, we construct for a thick subcategory C ⊆ mod kQ

generated by preprojective modules, an exceptional sequence that generates C. After

that we specialise to the module category of kQ, where Qis an Euclidian quiver.

We introduce reduction techniques, some of which work in a more general context

Chapter 1. Introduction 4

(see proposition 3.2.12), which enable us to prove that any thick subcategory in

mod kQ is exact abelian (theorem 3.2.14). Further, by a result of Colin Ingalls and

Hugh Thomas [IT, Theorem 1.1], there is a bijection between non-crossing partitions

associated to Qand exact abelian extension closed subcategories with a projective

generator in mod kQ. As one observes, there are exact abelian extension closed

subcategories without a projective generator (for instance the tubes in the regular

component of the Auslander-Reiten quiver of mod kQ). So we use results from the

second chapter and combining with the above cited theorem, we give a complete

classification.

Theorem 1.0.4 Let kbe an algebraically closed field, Qan Euclidian quiver and C

a connected exact abelian extension closed subcategory of mod kQ.

(i) [IT] If Chas a projective generator, then Ccorresponds to a non-crossing

partition of type Q.

(ii) If Chas no projective generator, then Ccorresponds to a configuration of

non-crossing arcs covering the circle.

At the end of the chapter, we present a very elegant proof, due to Dieter Vossieck,

that every thick subcategory of a hereditary abelian category is exact abelian.

In chapter 4 we deal with tilted algebras, an important class of algebras which

have been extensively studied in [Bo] and [HaR]. For a finite dimensional hereditary

algebra A, there is the concept of a tilting module TA, which can be thought of as

being close to the Morita progenerator. If we consider the k-algebra B= EndA(TA),

then the categories mod Aand mod Bare reasonably close to each other. The

algebra B= EndA(TA) is called tilted algebra. The benefit of tilted algebras is that

when the representation theory of an algebra Ais difficult to study directly, it may

be convenient to replace Awith the simpler algebra B= EndA(TA), and then to

reduce the problem on mod Ato a problem on mod B.

The main result in the chapter is a classification of exact abelian extension closed

categories for tilted algebras.

Theorem 1.0.5 Let Abe a finite dimensional hereditary k-algebra, TAa basic tilt-

ing module and B= EndA(TA). Then there is a bijection between the exact abelian

extension and torsion closed subcategories of mod Aand the exact abelian extension

closed subcategories of mod B.

The thesis end with an Appendix, where some basic facts, relevant to all chapters,

are collected.

Chapter 2

Thick subcategories for cyclic

quivers

This chapter is dedicated to study thick subcategories for the category of locally

nilpotent cyclic quiver representations. We establish a bijection involving thick

subcategories, cotilting and support-tilting objects of that category. Further, we

present a combinatorial classification of thick subcategories as well as we calculate

their number. At the end, we investigate the exact abelian extension closed cate-

gories for algebras which are quotients of the path algebra of the cyclic quiver.

2.1 Cyclic quivers

In the whole chapter kis an algebraically closed field. We begin with very general

framework and consider categories which are k-linear, small abelian, Hom-finite,

hereditary and satisfy Serre duality. Following [Ln], we recall shortly all these con-

cepts and then specialise to particular examples of such categories, which are target

of our investigations.

Let Tbe an abelian k-linear category. Recall that k-linearity of Tmeans that

the morphism groups are k-vector spaces, and that composition

Hom(Y, Z)×Hom(X, Y )→Hom(X, Z),(g, f)7→ gf,

is k-bilinear for all objects X, Y and Zfrom T.

We recall the notion of an abelian category. By definition, a sequence 0 →

→Bv

→C→0 is called short exact if for each object Xof Tthe induced se-

quence 0 →Hom(X, A)Hom(X,u)

−→ Hom(X, B)Hom(X,v)

−→ Hom(X, C) is exact and dually

for each object Yof Tthe sequence 0 →Hom(C, Y )Hom(v,Y )

−→ Hom(B, Y )Hom(u,Y )

−→

Hom(A, Y ) is exact. For Tto be abelian, one requires two things:

2.1. Cyclic quivers 6

(1) For every morphism Af

→Bthere exist two short exact sequences 0 →Kα

→

Aβ

→C→0 and 0 →Cγ

→Bδ

→D→0 such that fis obtained from the

commutative diagram below:



β??

~f//B

α>>



0.

(2) Thas finite direct sums, which implies the uniqueness of the additive struc-

ture.

We impose on Tsome finiteness assumptions: Tis a small category, that is,

the objects of Tform a set, and Tis Hom-finite, that is, all morphism spaces

HomT(X, Y ) are finite dimensional over k.

The properties of Tso far imply that Tis a Krull-Schmidt category.

Proposition 2.1.1 Each abelian Hom-finite k-category is a Krull-Schmidt cate-

gory, that is,

(i) each indecomposable object from Thas a local endomorphism ring, and

(ii) each object from Tis a finite direct sum of indecomposable objects.

We assume that the category Tis hereditary, that is, the extensions Extn

T(X, Y )

vanish in degrees n≥2 for all objects X, Y from T, see also A.2. Later we shall use

that exact abelian subcategory of a hereditary category is again hereditary.

We continue with strengthening the heredity condition, namely, we assume the

existence of an equivalence τ:T → T and of natural isomorphism

Ext1

T(X, Y )∼

→DHomT(Y, τX)

for all objects X, Y from T. The consequences of a Serre duality are of major

importance:

Proposition 2.1.2 Assume that Tis an abelian k-category which is Hom-finite

and satisfies Serre duality. Then the following holds:

2.1. Cyclic quivers 7

(i) Tis an Ext-finite hereditary category without non-zero projectives or injec-

tives.

(ii) Thas almost split sequences with τacting as the Auslander-Reiten

translation. That is, for each indecomposable object Xthere is an almost-

split sequence 0→τX →E→X→0.

We assume that Tis a length category, that is, each object of Thas finite length.

An object Uof an abelian category is called uniserial if the subobjects of Uare

linearly ordered by inclusion and form a finite chain

0 = U0⊆U1⊆ · · · ⊆ Uℓ−1⊆Uℓ=U.

If all indecomposables in an abelian length category Uare uniserial, we call U

uniserial category.

The following theorem, due to Gabriel, unifies all notions used up-to-now.

Theorem 2.1.3 [Gab2, Proposition 8.3] Let Tbe a Hom-finite hereditary length

category with Serre duality. Then Tis uniserial. Moreover, for the indecomposable

objects ind Tof T, we have ind-T=Fλ∈ITλ, where the Auslander-Reiten quiver of

Tλis of the form ZA∞/(τn), where n∈N0.

Therefore the Auslander-Reiten quiver of Tdecomposes into stable tubes, where for

convenience ZA∞is also viewed as a tube of an infinite period.

Now, we introduce the main example of our investigation in this chapter, namely

the categories that satisfy all the conditions of the Gabriel’s theorem. Before that,

we refer the reader to A.1 for recalling basic facts about quivers and their represen-

tations. We consider the path algebra k˜

∆nof the cyclic quiver:

∆n: 1 //2//3//... //n.

Let R=R˜

∆nbe the two-sided ideal generated by all arrows of ˜

∆n. A k˜

∆n-module

Mis R-nilpotent (nilpotent for short) if for each m∈Mthere exist ℓ≥0 such

that Rℓ.m = 0. If ℓ=ℓ(m) depends on m, we say that Mis locally R-nilpotent

(locally nilpotent for short). We denote by nrep(k ˜

∆n) the category of nilpotent and

by NRep(k ˜

∆n) the category of locally nilpotent modules over k˜

∆n. If we consider

the category of finite dimensional locally nilpotent modules over k˜

∆n, we notice

that it is the same as nrep(k ˜

∆n). The argument is the following: Trivially, every

nilpotent module is locally nilpotent. Now, if ℓ(M)<∞, then for any ℓ > ℓ(M),

Rℓannihilates M.

Remark 2.1.4 If we consider the subquiver

∆n: 1 //2//... //n,

2.1. Cyclic quivers 8

of ˜

∆n, then we have a fully-faithful embedding mod k∆n֒→mod k˜

∆n. The quiver

∆nis the directed AnDynkin quiver. The construction of the Auslander-Reiten

quiver of mod k∆nis well-known, see A.3 for more details.

Example 2.1.5 The Auslander-Reiten quiver of mod k∆3.

r r r

T1T2T3

T1[2] T2[2]

T1[3]



@@R

@@R@@R

We comment that the category of nilpotent modules plays an important rˆole in

the representation theory of algebras of infinite representation type, since it describes

(connected) components of the Auslander-Reiten quiver of their module categories.

For convenience, from now onwards, we denote with Tnthe category of nilpotent

modules and with ˜

Tnthe category of locally-nilpotent modules over k˜

∆n. The

following proposition collects all the properties of Tnso far.

Proposition 2.1.6 Tnis Hom-finite hereditary length uniserial category with Serre

duality.

The number of isoclasses of simple objects of an abelian category Ais called the

rank of Aand we denote it by rk(A). In Tnwe have nsimple modules, and we de-

note them with T1, T2,...,Tn. Since Tnis an uniserial category, any indecomposable

object is uniquely determined by its socle and length. We set Ti[ℓ] to be the inde-

composable module with socle Tiand length ℓ. Recall that the simple composition

factors of a module Xis called the support of Xand it is denoted by supp(X).

The construction of the Auslander-Reiten quiver of Tnis well-known, see [R2,

Chapter 4.6]. As mentioned in the Gabriel’s theorem, it is of the form ZA∞/(τn).

@@@

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T1[4]

T1[5]

T2[2]

T2[3]

T2[4]

T2[5]

T3[2]

T3[3]

T3[4]

T3[5]

T1[2]

T1[3]

T1[4]

T1[5]

T2[2]

T2[3]

T2[4]

T2[5]

T3[2]

T3[3]

T3[4]

T3[5]

T1[2]

T1[3]

Figure 2.1: AR-quiver of T3

2.2. Orthogonal sequences and thick subcategories 9

We consider another category related to k˜

∆n, namely the category of locally

finite modules over the completion algebra k[[ ˜

∆n]] of k[˜

∆n] = k˜

∆n. First, recall

that k[[ ˜

∆n]] = lim

←k[˜

∆n]/Ri, where Ris the same as before. A module is locally

finite if it is a filtered colimit of finite length modules. For more details, we refer to

the paper of [BKr, Section 2]. Now, we point out the following result.

Theorem 2.1.7 [CY, Main Theorem] The category of locally finite modules over

k[[ ˜

∆n]] is equivalent to NRep(k ˜

∆n).

In [BKr], the classification of indecomposable objects in the category of locally

finite modules over k[[ ˜

∆n]] and hence in NRep(k˜

∆n) is made and we shall use it later.

We refer the reader to the paper [RV], where complete classification of categories

sharing the same properties as Tnis made.

2.2 Orthogonal sequences and thick subcategories

From now onwards, Tnwill be a tube of rank n. We begin with recalling the following

lemma.

Lemma 2.2.1 [Happel-Ringel] Let Hbe a hereditary abelian category. Assume that

X, Y ∈ H are indecomposable objects and Ext1

H(Y, X) = 0. Then any non-zero

morphism f:X→Yis either monomorphism or epimorphism.

The proof can be found in [AS, Chapter VIII.2, Lemma 2.5]. Now, we make the

following observation.

Lemma 2.2.2 Let ζ: 0 →X→Y→Z→0be a non-split short exact sequence

with X, Z indecomposables in Tn. Then Yhas at most two indecomposable sum-

mands.

Proof: Let Y=Y1L···LYn,n≥3 be the decomposition of Yinto indecom-

posable modules. Since Z is uniserial and g:Y→Zis an epimorphism, then at

least one of gi’s (gi:Yi→Z,i= 1,...,n) is an epimorphism, say g1. Consider the

following diagram:

Y1g1

f199

f%%

KZ,

Y−˜g

where ˜

Y=Y2⊕ · · · ⊕ Yn. The sequence ζis short exact hence the square above

is both push-out and pull-back. By the property of the pull-back, we have that ˜

is an epimorphism and Ker g1∼

=Ker ˜

f. But Ker g1is indecomposable, then so is

X/ Ker ˜

f∼

=˜

Yand hence n≦2. 2

2.2. Orthogonal sequences and thick subcategories 10

Remark 2.2.3 Let ζbe as above. Then we have two cases for Y:

(1) Yis indecomposable. Then Soc(Y) = Soc(X), Top(Y) = Top(Z) and surely

ℓ(Y) = ℓ(X) + ℓ(Z).

(2) Y=Y1⊕Y2. Then ˜

Y=Y2,˜

f=f2, ˜g=g2. Since 0 = Ker f1∩Ker f2

and f2is not monomorphism (since then f2will be an isomorphism and the

sequence will split), we have that f1is monomorphism and using the push-out

property, so is g2. So in this case we have f1, g2are monomorphisms and f2, g1

are epimorphisms and hence Top(X) = Top(Y2) and Soc(Y2) = Soc(Z). We

make another conclusion: Given 0 6=f:Xi→Xj,fis neither monomorphism

nor epimorphism, then the following short exact sequence is non-split:

Y1g1

0//Xi

f177

f2&&

f//________ Xj//0,

Im fg2

Now, we prove the following lemma.

Lemma 2.2.4 Let Xbe indecomposable in Tn. Then EndTn(X)∼

=k[x]/(xt+1),

where t=⌈ℓ(X)−1

n⌉.

Proof: We notice that for any 0 6=f:X→X, which is neither monomorphism

nor epimorphism, we have Xπ

։Im fi

֒→Xwith Im f= Soc(X) = Top(X), which

yields a non-split short exact sequence of the form 0 →X→Im f⊕Y2→X→0.

Now, if ℓ(X)≤n, then it is straightforward to see that EndTn(X)∼

=k. Now, let

ℓ(X)> n. Since Xis uniserial, there are exactly ℓ(X)−1 indecomposable modules

with length smaller than ℓ(X), which have a socle Soc(X). Since Tnis n-periodic,

then we have t=⌈ℓ(X)−1

n⌉modules with the same top and socle as X. At last, we

notice that for s= 1,...,t we have πs

1=πs, where Xπ1

։Y1

π2

։Y2

π3

։· · · πt

։Yt,

which yields immediately EndTn(X)∼

=k[x]/(xt+1). 2

An object Xin an abelian category Ais called a point if EndA(X) is a division

ring. Two objects X, Y in Aare orthogonal if HomA(X, Y ) = HomA(Y, X) =

0. For example any two simple objects in Aare orthogonal. A sequence E=

(X1,...,Xk) is called an orthogonal sequence if any pair (Xi, Xj) for i6=jis

orthogonal.

We comment that if Xis a point in Tnwith ℓ(X)< n, then Xdoes not have self-

extensions, and hence by [Br1, Lemma 6.3.4], add(X) is an exact abelian extension

closed subcategory of Tn.

Corollary 2.2.5 In Tnthe points are all indecomposable modules with length less

or equal n.

2.2. Orthogonal sequences and thick subcategories 11

Let Ebe a set of pairwise orthogonal points in A. If Ais an object of A, then

an E-filtration of Ais given by a sequence of subobjects

0 = A0⊆A1⊆A2⊆ · · · ⊆ An=A,

with Ai/Ai−1∈Efor 1 ≤i≤n. We denote by U(E) the full subcategory of A

consisting of all objects of Awith an E-filtration. The next theorem, due to Ringel,

explains why orthogonal sequences are important.

Theorem 2.2.6 [R1, Theorem 2] Let Ebe a set of pairwise orthogonal points in

A. Then U(E)is an exact abelian subcategory which is closed under extensions, and

the set Eis the set of all simple objects in U(E).

For any subset S∈ Tn, we define Thick(S) to be the smallest thick subcategory

of Tncontaining S, and call it the thick subcategory generated by S.

We associate to every thick subcategory Cin Tna sequence of indecomposable

modules, called reduced sequence as follows: For every simple module Ti, i =

1,...,n of Tnwe take an indecomposable module Xi∈ C such that Soc(Xi) = Ti

and Xihas minimal length or Xi= 0 if such module does not exist.

Proposition 2.2.7 Let E= (X1,...,Xk)be a reduced sequence in Tn. Then Eis

orthogonal and each Xiis a point.

Proof: Let Xi, Xj∈E(i6=j) and f:Xi→Xjbe a non-zero morphism. By

construction of E,fcan not be a monomorphism. Also, it can not be epimor-

phism, since Soc(Ker f) = Soc(Xi) and Ker fhas smaller length. Now, fis neither

monomorphism nor epimorphism and by lemma 2.2.1 we have a non-split exact se-

quence: 0 →Xi→Y→Xj→0 with Y∈ C, since Cis closed under extensions.

By the remark 2.2.3, we have Y=Y1LY2and Y2֒→Xj, which contradicts the

minimal choice of Zand hence f= 0. Now, assume that EndTn(Xi) is not isomor-

phic to k. Then by proposition 2.2.4 we have that ℓ(Xi)> n and hence we have

a non-split short exact sequence 0 →Xi→Y→Xi→0,with Y=Y1⊕Im f.

Clearly, Im f∈ C and ℓ(Im f)< ℓ(Xi), which shows that for all Xi∈Ewe have

ℓ(Xi)≤nand hence EndTn(Xi)∼

=k.2

Proposition 2.2.8 In Tnthere is a bijection between orthogonal sequences and thick

subcategories.

Proof: Let Cbe an arbitrary thick subcategory and E= (E1,...,Ek) be its

associate reduced sequence. By proposition 2.2.7, we have that Eis orthogonal.

Obviously, Thick(E)⊆ C has reduced sequence E. We claim that Thick(E) = C.

To verify this, take X∈ C with minimal length such that X /∈Thick(E) and consider

0//Soc(X)//X//X/ Soc(X)//0.

2.2. Orthogonal sequences and thick subcategories 12

Then Soc(X)∈Thick(E) and ℓ(X/ Soc(X)) < ℓ(X) implies X/ Soc(X)∈Thick(E).

Since Thick(E) is closed under extensions, then Xmust be in Thick(E). Thus, we

justified that any thick subcategory is uniquely determined by its reduced sequence.

Now, we take an arbitrary orthogonal sequence E′= (E′

1,...,E′

k) and consider

Thick(E′). We claim that the reduced sequence of Thick(E′) is E′. By definition,

we have HomTn(E′

i, E′

j) = 0 for i6=j. If we have non-zero extensions among E′

i’s, say

0→E′

i→E′′

ij →E′

j→0, and 0 →E′

j→E′′

jk →E′

k→0, then Thick(E′′

ij, E′′

jk) =

Thick(E′

i, E′

j, E′

k), since Top(E′′

ij) = Top(E′

j), Soc(E′

j) = Soc(E′′

jk), and hence E′

would appear in the middle term of the extension E′′

ij by E′′

jk. We conclude that it is

not possible to obtain an indecomposable module with length smaller than ℓ(E′

i) for

i= 1,...,k in Thick(E′

1, . . . , E′

k). Hence E′is the reduced sequence of Thick(E′).2

Corollary 2.2.9 The number of thick subcategories in Tnis finite.

Proof: As noticed in proposition 2.2.7, there is no module with length greater than

nwhich belongs to a reduced sequence, since this module has a self-extension and

the middle term has a direct summand with smaller length. Since there are finite

number of points in Tn, there are finitely many reduced sequences as well as thick

subcategories. 2

Recall that an abelian category Cis connected, if any decomposition C=

C1`C2into abelian categories implies C1= 0 or C2= 0.

Theorem 2.2.10 Any thick subcategory of Tnis exact abelian. More precisely, for

any connected thick subcategory Cof Tn,Cis either equivalent to mod k∆sor to a

tube of rank s, where s≤n.

Proof: Take a thick subcategory C ⊆ Tnand its reduced sequence E. We show

that Thick(E) = U(E). Then using the result of Ringel, U(E) is exact abelian

and hense so is C= Thick(E). Obviously, Thick(E)⊆ U(E). Now, since U(E) is

uniserial, for the indecomposable object M∈ U(E) we take its composition series in

U(E): M⊇M1⊇M2⊇. . . ⊇Mt−1⊇Mt= 0. Consider the short exact sequence

0→Mt−1→Mt−2→Mt−2/Mt−1→0. We have that Mt−1and Mt−2/Mt−1

are simples, hence are in Thick(E) and since the latter is closed under extensions,

we have that Mt−2∈Thick(E). Using the same argument along the composition

series of M, we conclude that M∈Thick(E). Hence U(E) = Thick(E). For

the last part of the theorem: Note that since Cis hereditary, there exists a finite

and connected quiver Q, such that C∼

=mod kQ. Moreover Cis uniserial, hence

by [AS, Chapter V.3, Theorem 3.2], C∼

=mod k∆sor C∼

=Ts, for some s≤n.2

Example 2.2.11 The indecomposable modules of Thick(Ti[n]) are Ti[kn] (k∈N).

In fact, Thick(Ti[n]) ∼

=T1.

2.2. Orthogonal sequences and thick subcategories 13

As we have shown, any thick subcategory Cof Tnis exact abelian and therefore, it

is uniquely determined by its simple objects. The simple modules of Care among

the points of T, which have length at most n. Hence they lie in the n×n“square”:

a part of the Auslander-Reiten quiver containing the points of Tn.

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T1[2] T2[2] T3[2]

T1[3] T2[3] T3[3]

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123

(1,2) (2,3) (3,1)

(1,3) (2,1) (3,2)

Figure 2.2: The points in T3

Now, we visualise thick subcategories in Tnin the following way. We place

1,2,...,n on the circle, which represents the simples of Tn. Since each point is

uniquely determined by its socle and its top, we associate to a point in Tnan arc

from the circle with start-point its socle and end-point its top; the direction is

clockwise. For a point X, set the length of the arc(X) to be ℓ(X) and simply denote

the arc(X) by the ordered couple (s, ℓ), where Soc(X) = sand ℓ(X) = ℓ. The simple

objects Tkare represented by the singleton (k). We call two arcs non-crossing, if

they do not intersect.

(1)(25)(34) (35)(62)

Figure 2.3: Non-crossing arcs on the circle

We say that the arcs arc(Xi)(i∈I)cover the circle, if each simple module Ti

belongs to the union of supp(Xi).

Now, we interpret the morphisms between modules in Tnin terms of arcs. Let

X1, X2be points, f:X1→X2be a morphism between them and arc(X1),arc(X2)

be their associated arcs.

(1) If fis a monomorphism, then X1and X2have the same socle and hence

arc(X1) and arc(X2) have the same start-point. If fis an epimorphism, then X1

and X2have the same top and hence arc(X1) and arc(X2) have the same end-point.

(2) If fis neither monomorphism nor epimorphism, then Top(X1) = Top(Im f),

Soc(Im f) = Soc(X2) and hence the arcs intersect. Note that by Happel-Ringel’s

lemma we have Ext1

Tn(X2, X1)6= 0.

2.2. Orthogonal sequences and thick subcategories 14

(3) From (1) and (2) we conclude, that if f:X1→X2is a non-zero morphism,

then the corresponding arcs intersect. We notice that if arc(X1), arc(X2) intersect,

then HomTn(X1, X2)6= 0 or(and) HomTn(X2, X1)6= 0. The latter is true, since the

two arcs intersect in a point Z(algebraic meaning) with Top(X1) = Top(Z) and

Soc(X2) = Soc(Z) (or vice versa), and hence we have 0 6=f:X1։Z ֒→X2(or

vice versa). We conclude that HomTn(X1, X2) = HomTn(X2, X1) = 0 if and only if

the arcs representing these modules are non-crossing.

Example 2.2.12 In T4we consider the modules X1=T1[3] and X2=T2[3]. Then

arc(X1),arc(X2) intersect, and hence there is a non-zero morphism between T1[3]

and T2[3], namely T1[3] ։T2[2] ֒→T2[3].

Arc(X )2

Arc(X )1

T1T2T3T4

T [3]1T [3]2

T [2]2

(4) Later we shall use that if Ext1

Tn(X1, X2) = Ext1

Tn(X2, X1) = 0, then either one

of the arc contains the other or there is at least one point between arc(X1),arc(X2)

and at least one point between arc(X2),arc(X1).

Now, having in mind proposition 2.2.8, we get immediately the following propo-

sition.

Proposition 2.2.13 There is a bijection between non-crossing arcs on the circle

with npoints and thick subcategories in Tn.

Example 2.2.14 In T3we consider the thick subcategory C1= Thick(T1[3], T2).

Note that T1[3] and T2are simples in C. Then arc(T1[3]) = (1,3) and arc(T2) = (2).

(13)(2)

(31)(2)

T [3]1

T1T3

T [2]3

Figure 2.4: Non-crossing arcs and thick subcategories

2.3. Cotilting, support-tilting modules and thick subcategories 15

Now, consider C2= Thick(T3[2], T2). It is easy to see that the simple objects in C

are T2and T3[2]. Moreover, arc(T3[2]) = (3,1). Note that different arc orientations,

represent different points.

2.3 Cotilting, support-tilting modules and thick

subcategories

Let Abe an abelian category. We say that Ahas a finite generator, if there is

an object P∈ A with ℓ(P)<∞such that for each indecomposable object X∈ A

there exist an integer d≥0 and an epimorphism Pd→X. If the category Ahas a

finite generator P, we shall write A= Gen(P).

We say that Ais bounded, if each indecomposable object X∈ A has a bounded

length, that is, there is k∈Nsuch that ℓ(X)< k. For instance, in Tnany thick

subcategory equivalent to mod k∆s, for s≤nis bounded. If in Athere are inde-

composable objects with arbitrary lengths, then we say that Ais unbounded. For

example, Tsis unbounded thick for any natural number s.

Recall that the simple composition factors of a module is called the support of

the module. For instance, supp(Ti[k]) = {Ti, Ti+1,...,Ti+k−1}, where the indices are

taken modulo nand we identify T0with Tn. The next lemma elucidates the above

notions.

Lemma 2.3.1 Let Cbe a thick subcategory of Tn.

(i) If the simple objects Xiof Chave pairwise disjoint supports, then Cis bounded

if and only if Pk

i=1 ℓ(Xi)< n.

(ii) Cis bounded if and only if supp(C)⊂ {T1,...,Tn}.

Proof: (i) Since Xi’s have pairwise disjoint supports, then it is equivalent to say

that arc(Xi), i = 1,...,k do not intersect on the circle, and therefore the sum

of the lengths of all these arcs is at most n. Now, suppose that the sum of the

length is n, or equivalently that all arcs cover the circle. Then we have a non-

split short exact sequence 0 →X1→Y1→X2→0 with Y1indecomposable,

HomTn(Y1, Xi) = HomTn(Xi, Y1) = 0 for i= 3,...,k,ℓ(Y1) = ℓ(X1) + ℓ(X2) and

Y1belongs to C, since the latter is closed under extensions. Then the sequence

(Y1, X3,...,Xk) is orthogonal. We apply the same argument for Y1and X3and

following that procedure, at the end we obtain an indecomposable object Ykwith

Soc(Yk) = Soc(X1) and ℓ(Yk) = Pk

i=1 ℓ(Xi) = n, which belongs to C. We conclude

that Cis not bounded. Now if Cis unbounded, then there is an indecomposable

2.3. Cotilting, support-tilting modules and thick subcategories 16

module Xwith length ≥n, and hence the sum of the lengths of the simples, that

appear in the composition series of X(which are among Xi’s) is ≥n.

(ii) Consider the thick subcategory C′of Cgenerated by simples X′

iof Cwith

supp(X′

i)∩supp(X′

j) = ∅for i6=j. By construction, C′is obtained from Cby

removing a finite number of its bounded thick subcategories. Hence C′is bounded

if and only if Cis bounded. Moreover, supp(C) = supp(C′). Now, C′satisfies

the conditions of (i), hence C′is bounded if and only if Pk

i=1 ℓ(X′

i)< n, which is

equivalent to say that supp(C′)⊂ {T1,...,Tn}. The claim follows. 2

From the proposition follows that Cis unbounded if and only if supp(C) =

{T1,...,Tn}. Then, having in mind proposition 2.2.13, we immediately get the

following corollary.

Corollary 2.3.2 There is a bijection between unbounded thick subcategories in Tn

and non-crossing arcs on the circle with npoints that covers the circle.

For a thick subcategory Cof Tnwe define a new category, namely C⊥={X∈

Tn|HomTn(C, X) = Ext1

Tn(C, X) = 0}and call it right perpendicular of C.

Similarly one defines ⊥C={X∈ T | HomTn(X, C) = Ext1

Tn(X, C) = 0}and call

it left perpendicular of C. We refer the reader to [GLn] and [Sc] for detailed

exposition of perpendicular categories.

The following proposition is from [GLn, Proposition 1.1].

Proposition 2.3.3 Let Ibe a system of objects in an abelian category A. Then the

category I⊥right perpendicular to Iis closed under the formation of kernels and

extensions. If additionally, proj.dim I ≤ 1, then I⊥is an exact subcategory of A;

i.e., I⊥is abelian and the inclusion I⊥→ A is exact.

Definition 2.3.4 Let Abe a finite dimensional k-algebra. A finitely presented

module T∈mod Ais a partial-tilting module if

(T1) the projective dimension of Tis at most 1;

(T2) Ext1

A(T, T) = 0.

If additionally,

(T1) there is an exact sequence 0 →A→T0→T1→0 with each Ti∈add(T),

then Tis called a tilting module.

A tilting module is called basic if each indecomposable direct summand occurs

exactly once in a direct sum decomposition.

In [AS, Chapter VI.4, Corollary 4.4], an alternative characterisation of a tilting

module is given.

2.3. Cotilting, support-tilting modules and thick subcategories 17

Proposition 2.3.5 Let Abe a finite dimensional hereditary algebra. A finitely

presented module T∈mod Ais a tilting module if

(T1) Ext1

A(T, T) = 0, and

(T2) the number of pairwise non-isomorphic indecomposable summands of Tequals

the number of pairwise non-isomorphic simple modules.

A partial tilting Amodule Cis called support-tilting, if it is tilting as an A/ ann(C)

module. For instance, any simple Amodule is support-tilting. The following propo-

sition clarifies the notion of a support-tilting module:

Proposition 2.3.6 [IT, Proposition 2.5] Suppose that Cis a support-tilting A-

module. Then the number of distinct indecomposable direct summands of Cis the

number of distinct simples in its support.

For the support-tilting modules in Tn, we have the following:

Lemma 2.3.7 Let Cbe a support-tilting module. Then supp(C)⊂(T1, . . ., Tn).

Proof: Suppose that supp(C) = (T1,...,Tn). Then Chas nindecomposable

direct summands with no extensions between them. But then each indecompos-

able summand of Chas length less then nand hence there is at least two in-

decomposable summands of C, say Ci, Cjwith different socle and top, such that

supp(Ci)∩supp(Cj)6=∅. The latter implies that there is an extension between

them, see lemma 2.2.2, which is not possible. Hence supp(C)⊂(T1,...,Tn). 2

Example 2.3.8 In mod k∆3consider the module C=C1⊕C2. The minimal

s q q

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subquiver on which Cis supported is k∆2and Cis tilting as a k∆2-module. Hence

Cis a support-tilting module.

We continue with pointing out a relation between support-tilting modules and exact

abelian extension closed categories. Let Qbe a finite acyclic quiver and kQ be its

associated path algebra and mod kQ is the category of finite dimensional modules

over kQ. The following two theorems are from [IT, Section 2.2, 2.3]. We indicate

that there the term wide subcategories refers to exact abelian extension closed

subcategories in our notations.

Recall that a torsion class is a full subcategory of an abelian category A, which

is closed under direct summands, quotients and extensions. We say that an object

Pin Ais Ext-projective if Ext1

A(P, X) = 0 for any X∈ A.

2.3. Cotilting, support-tilting modules and thick subcategories 18

Theorem 2.3.9 In mod kQ there is a bijection between torsion classes with a finite

generator and basic support-tilting modules.

The bijection is realised as follows:

•Let Cbe a support-tilting object. Then Gen(C) is a torsion class having a

finite generator.

•Let Cbe a torsion class with a finite generator and let Cbe the direct sum of

its indecomposable Ext-projectives. Then Cis support-tilting.

Example 2.3.10 We consider again mod k∆3. The module C=C1⊕C2is support-

tilting and Gen(C) = U(C1, C2/C1) is exact abelian extension closed. Conversely,

the Ext-projectives of U(C1, C2/C1) are C1and C2.

s q q

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Proposition 2.3.11 In mod kQ there is a bijection between torsion classes with a

finite generator and exact abelian extension closed categories with a finite generator.

The bijection is given as follows:

•Let Cbe a torsion class. Then α(C) = {X∈ C | for all (g:Y→X)∈

C,Ker g∈ C} is exact abelian extension closed.

•If Cis exact abelian extension closed, then Gen(C) is a torsion class.

Example 2.3.12 We consider again mod k∆3. Then C= add(C1⊕C2) is a torsion

class and α(C) = add(C1) is exact abelian extension closed. Conversely, Gen(C1) =

q s q

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We make a connection between support-tilting modules and bounded thick sub-

categories in Tn. We comment that the results discussed so far are not directly appli-

cable to Tn, since the settings are different (the quiver is assumed to be acyclic). Let

Tbe an arbitrary support-tilting module in Tn. Then by proposition 2.3.7, supp(T)

is a proper subset of {T1, . . ., Tn}, say supp(T) = {T1,...,Tk}, for some k < n.

Then Thick(T1,...,Tk)∼

=mod k∆k, and since mod k∆khas a projective generator,

then any support-tilting Tn-module is inside some bounded thick subcategory. In

that sense, the following theorem is true.

2.3. Cotilting, support-tilting modules and thick subcategories 19

Theorem 2.3.13 In Tnthere is a bijection between basic support-tilting modules

and bounded thick subcategories.

We enlarge the settings and consider the category of locally nilpotent modules

over k˜

∆n. The classification of indecomposable objects of category of locally finite

modules over k[[ ˜

∆n]] and hence in ˜

Tnis known, since these categories are equivalent,

see [CY, Main Theorem]. Following [BKr, Section 2] we recall it shortly. For each

simple object Tiand each ℓ∈Nwe have a chain of monomorphisms:

Ti=Ti[1] ֒→Ti[2] ֒→ · · ·

and denote by Ti[∞] the Pr¨ufer module defined to be lim

→Ti[ℓ]. Note that each

Pr¨ufer module is indecomposable injective and End˜

Tn(Ti[∞]) ∼

=k[[t]].

Lemma 2.3.14 [BKr, Lemma 2.1] Every non-zero object in ˜

Tnhas an indecom-

posable direct factor, and every indecomposable object is of the form Ti[ℓ]for some

simple Tiand some ℓ∈N∪ {∞}.

The lemma tells us that ind ˜

Tn= ind Tn∪ {Pr¨ufer modules}. This allows us

to visualise the Pr¨ufer modules via “extending” the part of the Auslander-Reiten

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T1[2] T2[2] T3[2]

T1[3] T2[3] T3[3]

T1[∞]T2[∞]T3[∞]

r r r

Figure 2.5: The points in ˜

quiver, containing the points in Tn(see figure 2.2) with nextra vertices. In that way,

we represent the points in ˜

Tn.

Remark 2.3.15 After knowing the indecomposables in ˜

Tn, it is not difficult to

show that in ˜

Tn, Thick(Ti[∞]) = Thick(Ti[n]). To see this, we notice that for k∈N,

we have 0 →Ti[kn]→Ti[∞]πk

→Ti[∞]→0 and since there are no extensions

between Pr¨ufer modules, any indecomposable object in Thick(Ti[∞]) is of the form

Ti[kn] for some k∈N∪ {∞}. Now, having in mind that lim

→Ti[k] = lim

→Ti[kn], the

equality above holds. We conclude that simple objects of any thick subcategory in

Tnare among the points of Tnand hence for any thick subcategory ˜

Cof ˜

Tn, we have:

ind ˜

C= ind C ∪ {all Ti[∞]|Ti[n]∈ C}, where C=˜

C ∩ Tn.

Next, we recall the definition of a cotilting object for any abelian category A. To

this end, we fix an object Tin A. We let Prod(T) denote the category of all direct

2.3. Cotilting, support-tilting modules and thick subcategories 20

summands in any product of copies of T. The object Tis called cotilting object if

the following holds:

(1) the injective dimension of Tis at most 1;

(2) Ext1

A(T, T) = 0;

(3) there is an exact sequence 0 →T1→T0→Q→0 with each Tiin Prod(T)

for some injective cogenerator Q.

In this paper, we shall use an alternative characterisation of a cotilting module,

see [BKr, Lemma 1.2], which resembles the one we have for a tilting module.

Proposition 2.3.16 Let Tbe an object in ˜

Tnwithout self-extensions.

(1) Tdecomposes uniquely into a coproduct of indecomposable objects having local

endomorphism rings.

(2) Tis a cotilting object if and only if the number of pairwise non-isomorphic

indecomposable direct summands of Tequals n.

Now, we recall the following lemma.

Lemma 2.3.17 [GLn, Lemma 1.2] Let Iand Tbe systems of objects of an abelian

category A. Then:

(i) I ⊂ T ⇒ T ⊥⊂ I⊥.

(ii) I ⊂⊥(I⊥).

(iii) I⊥= (⊥(I⊥))⊥.

For an indecomposable module X∈ Tn, set

compl(X) := {Ti[∞]|Ext1

Tn(Ti[∞], X) = 0}.

We shall use that every morphism between Pr¨ufer objects is an epimorphism, and

that there are no morphisms from Pr¨ufer modules to modules in Tn.

Lemma 2.3.18 Let X, Y be indecomposables with no self-extensions in Tn. Then:

(i) #{Ti[n]⊆X⊥}=n−ℓ(X).

(ii) supp(X)⊆supp(Y)⇔ {Ti[n]|Ti[n]∈Y⊥} ⊆ {Ti[n]|Ti[n]∈X⊥}.

(iii) #{Ti[∞]|Ext1

Tn(Ti[∞], X)6= 0}= #{compl(X)}=ℓ(X).

2.3. Cotilting, support-tilting modules and thick subcategories 21

(iv) supp(X)⊆supp(Y)⇔compl(Y)⊆compl(X).

(v) supp(X)∩supp(Y) = ∅ ⇔ compl(X)∩compl(Y) = ∅.

Proof: After relabeling the simples, we may assume that X=T1[ℓ]. Note that

both X, Y have lengths < n.

(i) Suppose 0 6=f:Ti[n]→Xfor some i∈ {1,...,n}. Since ℓ(X)< ℓ(Ti[n]) =

n, then fis not a monomorphism. It is immediate to check that if fis an epi-

morphism, then Ext1

Tn(X, Ti[r]) = 0 and HomTn(X, Ti[n]) = 0, hence Ti[n]∈X⊥,

where i=ℓ(X) + 1. If fis neither monomorphism nor epimorphism, then we

have Ext1

Tn(X, Ti[n]) 6= 0 hence Ti[n] is not in X⊥. Therefore if Ti[n]∈X⊥for

some i∈ {1,...,n}, then either HomTn(Ti[n], X) = 0 or Top(Ti[n]) = Top(X),

which is the same to say that arc(Ti[n]) and arc(X) are either non-crossing or

have the same end-point. Hence the number of Ti[n], which are in X⊥equals

n−ℓ(X). Later we shall use that the indices of all Ti[n]⊆X⊥are from the

set i∈I={ℓ(X) + 1, ℓ(X) + 2,...,n−1, n}, and we shall visualise this set as an

arc on the circle with consequent integral points.

(ii) Now, since ℓ(X)≤ℓ(Y), then due to (i), the indices ifor which {Ti[n]⊆X⊥}

are from the set I={ℓ(X)+1, ℓ(X)+2,...,n−1, n}, which obviously contains the

set I′={ℓ(Y) + 1, ℓ(Y) + 2,...,n−1, n}. Since I′is the index set of all i’s such

that {Ti[n]⊆Y⊥}, the proof follows.

(iii) Let 0 6=f:T1[ℓ]→Ti[∞]. If fis a mono, then Ext1

Tn(Tℓ+1[∞], X)6= 0 since

0→T1[ℓ]→T1[∞]→Tℓ+1[∞]→0 is a non-split exact sequence. Now, exactly as

in remark 2.2.2 (ii), any proper epimorphism f:T1[ℓ]։Tk[ℓ+ 1 −k], k = 2,...,ℓ

yields a non-split short exact sequence

0→T1[ℓ]→T1[∞]⊕Tk[ℓ+ 1 −k]→Tk[∞]→0,

and hence a Pr¨ufer module Tk[∞] with Ext1

Tn(Tk[∞], X)6= 0. It is straightforward

to check that the other Pr¨ufer modules do not have extensions with X. Therefore

#{Ti[∞]|Ext1

Tn(Ti[∞], X)6= 0}= #{Y∈ Tn|Top(Y) = Top(X)}+ 1 =

(ℓ(X)−1) + 1 = ℓ(X).

(iv) From (iii) we have that compl(X) = {T2[∞], T3[∞],...,Tℓ[∞], Tℓ+1[∞]}

and hence compl(X) = {Tℓ+2[∞], Tℓ+3[∞],...,Tn[∞], T1[∞]}. Since supp(X) =

{T1, T2,...,Tℓ}, we notice that the indices of Pr¨ufer modules in compl(X) are shifted

by one (modulo n) the indices of simples in supp(X). Then supp(X)⊆supp(Y)⇔

compl(X)⊆compl(Y)⇔compl(Y)⊆compl(X).

(v) Follows immediately from (iv). 2

Now, we are able to prove the following theorem.

Theorem 2.3.19 In ˜

Tnthere is a bijection between cotilting modules and support-

tilting modules.

2.3. Cotilting, support-tilting modules and thick subcategories 22

Proof: Recall that a module X∗is cotilting if and only if it has nindecomposable

summands and has no self-extensions. Note that every cotilting module has at

least one direct summand which is Pr¨ufer module, since otherwise, we would have

that T∗is support-tilting with supp(X∗) = (T1,...,Tn), which is impossible, see

lemma 2.3.7.

Let X=⊕t

i=1Xibe a support-tilting module having t < n indecomposable

summands. First we show that Xcan be completed by Pr¨ufer modules in a unique

way to a cotilting module. The statement will follow at once if we show that there

are exactly n−tPr¨ufer modules in compl(X). The quiver, on which Xis supported,

is a disjoint union of kquivers (1 ≤k < n) of type ∆si(i= 1,...,k) and since

Xis support-tilting, we have Pk

i=1 si=t. Then Thick(X) is a disjoint union of

categories of type Ci= mod k∆si. Take X∗=⊕k

i=1X∗

ito be a submodule of Xsuch

that each X∗

iis indecomposable and supp(X∗

i) = supp(Ci). Then by construction of

X∗we have supp(X) = supp(X∗) and supp(X∗

i)∩supp(X∗

j) = ∅(i6=j). Now, for

appropriate iand j, we have supp(Xj)⊆supp(X∗

i) and having in mind property

(iv), we get compl(X) = ∩t

j=1 compl(Xj) = ∩k

i=1 compl(X∗

i) = compl(X∗). Now,

taking into account that compl(X∗

j)∩compl(X∗

i) = ∅, we have #{compl(X)}=

#{compl(X∗)}=n−#{compl(X∗)}=n−Pk

i=1 #{compl(X∗

i)}=n−Pk

i=1 ℓ(X∗

=n−Pk

i=1 si=n−t.

Let Y∗=Y1⊕· · ·⊕Yk⊕Yk+1⊕· · ·⊕Ynbe a cotilting module and Y=Y1⊕· · ·⊕Yk

be a submodule of Ysuch that Y1,...,Ykare in Tnand Yk+1,...,Ynare Pr¨ufer

modules. We show that Yis support-tilting. First we have that Yhas no self-

extensions. Then it is sufficient to show that the number of simple modules of

supp(Y) is k. Now, since Yhas ksummands, we have #{supp(Y)} ≥ k. But

if #{supp(Y)}> k, then n−k= #{compl(Y)}=n−#{compl(Y)}=n−

#{supp(Y)}< n −k, which is impossible. Hence #{supp(Y)}=kand Yis a

support-tilting module. 2

Example 2.3.20 Consider the support-tilting module X=T1⊕T1[2] in T3from

example 2.3.10. Then supp(X) = {T1, T2}, Thick(T1, T2)∼

=mod k∆2=C1. Now,

take X∗=T1[2]. Then supp(X∗) = supp(C1), #{compl(X∗)}= 3−ℓ(X∗) = 3−2 =

1 and compl(T1[2]) = {T1[∞]}. Hence the module T1⊕T1[2] ⊕T1[∞] is cotilting.



r r ru

T1[2]

T1[∞]

We return to thick subcategories in Tn. The following proposition relates bounded

and unbounded thick categories.

2.3. Cotilting, support-tilting modules and thick subcategories 23

Proposition 2.3.21 Let Cbe a thick subcategory in Tn. If Cis bounded (resp.

unbounded), then C⊥is unbounded (resp. bounded).

Proof: Let Cbe bounded. First we assume that for the set of simples {X1, X2,...,Xk}

of Cwe have supp(Xi)∩supp(Xj) = 0 for i6=j. We show that there is a module

X∈ C⊥with ℓ(X)≥n, which implies that C⊥is unbounded. By lemma 2.3.18(i),

we have that for each Xiwith ℓ(Xi) = ki< n there are n−kimodules of length

nin X⊥

i. From the same lemma we have that, if, say arc(X1) = (1, ℓ), then the

consequent integral points on the circle (ℓ+ 1,...,n), which could be interpreted as

an arc, represent the indecomposable modules with length n, which are in X⊥

1. We

call such an arc an integral-arc and for a module Xi, we denote it by AXi.

Now, disjoint supports of the simples implies that Pk

i=1 ℓ(Xi)< n, hence any

two integral-arcs intersect and cover the circle, since ℓ(AXi) + ℓ(AXj) = n−ℓ(Xi) +

n−ℓ(Xi)> n. Moreover, it is not possible that one integral-arc to be contained in

other, say AXj⊂AXi, since this would imply that supp(Xi)⊂supp(Xj), which is

impossible. Then all such integral-arcs have a non-zero intersection and therefore,

there is an indecomposable module of length nwhich is in C⊥.

Now, if the supports of the simples of Care not disjoint, then we form a thick

subcategory C∗⊆ C as in lemma 2.3.1(ii) with supp(C∗) = supp(C) and with the

property that the simples of C∗have disjoint supports. Then from the discussions

above follows that there is a module of length nwhich is in (C∗)⊥. Now, if Xiis

a simple module of C, which is not in C∗, then by the construction of C∗there is a

simple module X∗

iof C∗such that supp(Xi)⊆supp(X∗

i). Using lemma 2.3.18(ii),

any indecomposable module with length n, which is in (X∗

i)⊥, is in (Xi)⊥. The last

argument implies that the intersection C∗∩ C contains an indecomposable module

with length n.

Let Cbe an unbounded thick. Take a module X∈ C with ℓ(X)≥n. Since there

is no module Yin Tnwith ℓ(Y)≥nsuch that Ext1

Tn(X, Y ) = 0, then C⊥has no

indecomposable modules of length greater then nand thus, it is bounded. 2

Remark 2.3.22 In the same way, one can show that forming the left perpendicular

category transforms bounded to unbounded thick subcategories and vice versa.

For a thick subcategory C ∈ Tndefine τkC(k∈Z) to be the full subcategory of Tn

whose indecomposable objects are the τk-shifts of the indecomposables of C. Also,

set C⊥0:= Cand define inductively C⊥k= (C⊥k−1)⊥if k > 0 and C⊥k=⊥(C⊥k+1 ) if

k < 0.

Before we prove the next proposition, we restate [CB1, Lemma 5].

Lemma 2.3.23 Let Qbe a finite and an acyclic quiver with nvertices and let Cbe

an exact abelian extension closed subcategory of mod kQ. Then rk(C) + rk(C⊥) = n.

2.3. Cotilting, support-tilting modules and thick subcategories 24

Proposition 2.3.24 Let C ⊆ Tnbe a thick subcategory. Then:

(i) rk(C⊥) + rk(C) = n.

(ii) ⊥(C⊥) = (⊥C)⊥=C.

(iii) C⊥=⊥τC.

(iv) C⊥k=Cfor some k∈N.

Proof: (i) By proposition 2.3.21, we have that either Cor C⊥is bounded, so we

may assume that Cis bounded. Without loss of generality we also may assume that

C ⊆ k∆n−1. Denote by C⊥

∆n−1=C⊥∩mod k∆n−1. By previous lemma, we have that

rk(C⊥

∆n−1) + rk(C) = rk(mod ∆n−1) = n−1. Let Xbe an indecomposable module

with Soc(X) = Tnand ℓ(X) = n. Then since mod k∆n−1⊂supp(X), we have

X∈ C⊥. We claim that C⊥= Thick(C⊥

∆n−1, X). The inclusion ”⊇” is obvious. Let

S= Thick(S1,...,Sk) be the set of simples of C⊥. We show that Sis contained in

Thick(C⊥

∆n−1, X). Suppose Soc(Ei) = Soc(X) for some i. Then X/Eibelongs to both

C⊥and mod k∆n−1and hence to C⊥

∆n−1. But then Simust be in Thick(C⊥

∆n−1, X).

If Soc(Si)∈ {T1,...,Tn−1}but Si/∈mod k∆n−1, then supp(X)∩supp(Si)6= 0 and

hence Ext1

Tn(Si, X)6= 0. Then one of the middle term is a submodule of Siand in

the same time must be in C⊥

∆n−1, which contradicts the assumption that Siis simple.

Hence C⊥= Thick(C⊥

∆n−1, X), which means that C⊥= Thick(S′, X), where S′is

the set of simples of C⊥

∆n−1. Now, having in mind that add(X)∼

=T1, then we have

rk(C) + rk(C⊥) = rk(C) + rk(C⊥

∆n−1) + rk(add(X)) = rk(C) + (n−1) −rk(C) + 1 = n.

(ii) By lemma 2.3.17, we have that C ⊂ ⊥(C⊥). But since rk(⊥(C⊥)) = n−(n−

rk(C)) = rk(C), we have C=⊥(C⊥).

(iii) Using the Auslander-Reiten formula, for Xindecomposable in C⊥we have

0 = Ext1

Tn(C, X) = DExt1

Tn(C, X) = HomTn(X, τC) and 0 = HomTn(C, X) =

HomTn(τC, τX) = DExt1

Tn(X, τC) = Ext1

Tn(X, τC).

(iv) Since for every simple module Siof Cwe have τnSi= ((S⊥

i)⊥)n=S⊥2n

i=Si,

then applying perpendicular category 2ntimes to C, we return to C.2

Now, we establish a link between unbounded and bounded thick subcategories.

Theorem 2.3.25 In Tnforming the right perpendicular (resp. left perpendicular)

category induces a bijection between bounded and unbounded thick subcategories.

Proof: For an arbitrary bounded (unbounded) thick subcategory Cof Tn, we

have that C⊥is unbounded (bounded). Then using lemma 2.3.24(ii), we have that

⊥(C⊥) = (⊥C)⊥=C, which yields the bijection. 2

Remark 2.3.26 The last theorem gives us an argument, that in Tn, as well as in

mod kQ for QDynkin quiver, there is k∈Zsuch that C⊥k=C. In fact, for any

2.3. Cotilting, support-tilting modules and thick subcategories 25

hereditary category C, we have ⊥(C⊥) = (⊥C)⊥=C. Therefore, in both cases right

perpendicular is a bijection. Now, having in mind that the number of exact abelian

extension closed categories in Tnand in mod kQ is finite, the claim follows.

Example 2.3.27 The example illustrates lemma 2.3.24(iv) for n= 3. The thicken

points represent the simples of the respective thick subcategory. Note that rk(C) +

rk(C⊥) = 3. We comment that in general, the period does not equals n.

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For completeness, we collect all the bijections established so far.

Theorem 2.3.28 Let Tnbe the category of nilpotent modules and ˜

Tnbe the category

of locally nilpotent modules over the path algebra k˜

∆n. Then there are bijections

between:

(1) support-tilting objects in Tn;

(2) bounded thick subcategories in Tn;

(3) unbounded thick subcategories in Tn;

(4) cotilting objects in ˜

Tn.

Proof: Let X=⊕k

i=1Xibe a support-tilting object. Schematically the bijections

are described as follows:

α(Gen(X))

))

∩k

i=1 compl(Xi)



|⊥



Ext -proj. in Gen(C)

X∗

⊕X∗

i,ℓ(X∗

i)<n

C⊥

⊥|

Theorem 2.3.13 justifies the bijection between (1) and (2).

2.4. Number of thick subcategories 26

(1) ⇒(2). The category Gen(X) is a torsion class. Then α(Gen(X)) = {Y∈

Gen(X)|for all (g:Z→Y)∈Gen(X),Ker g∈Gen(X)}is exact abelian exten-

sion closed with a finite generator, that is, a bounded thick subcategory in Tn.

(2) ⇒(1). Given a bounded thick subcategory C ⊂ Tn, then Gen(C) is a torsion

class and the direct sum of Ext-projectives in that torsion class is a support-tilting

module.

Theorem 2.3.25 establishes bijection between (2) and (3).

(2) ⇔(3). Let Cbe a bounded thick subcategory. Then C⊥( respectively ⊥C) is

unbounded thick and using the perpendicular on the other side, we return to C.

Theorem 2.3.19 yields the bijection between (1) and (4).

(1) ⇒(4). We complete the support-tilting module Xto a cotilting module X∗

in a unique way just by taking the intersection of all compl(Xi).

(4) ⇒(1). Let X∗=⊕n

i=1X∗

ibe a cotilting module. Then the direct sum of all

X∗

isuch that ℓ(X∗

i)< n is a support-tilting module. 2

Example 2.3.29 In T3consider the support-tilting module X=T2[2] ⊕T3. Then

X∗=T2[2] ⊕T3⊕T2[∞] is a cotilting module in ˜

T3,C=α(Gen(X)) = Thick(T2[2])

is bounded thick and C⊥= Thick(T1[3], T2) is unbounded thick.

(13)(2)

(23)

Figure 2.6: Bijections in ˜

At the end of the chapter, we list the rest of bijections in ˜

T3.

2.4 Number of thick subcategories

By a result of Colin Ingalls and Hugh Thomas [IT, Section 3.3], there is a bijection

between exact abelian extension closed subcategories with a projective generator in

mod kQ and non-crossing partitions of type Q, where Qis a Dynkin or an Euclidean

quiver. The number of non-crossing partitions of type Q, where Qis Dynkin quiver

is well known. We shall use that when Q= ∆ntheir number is Cn+1, where

2.4. Number of thick subcategories 27

Cn=1

n+12n

nis the nth Catalan number, although in the section 2.6, we give

another proof.

Let Cbe a thick subcategory in Tnand E= (S1,...,Sk) be the set of its simple

modules. For a simple module T1of Tn, define

roof(C) = 





Si∈E , T1∈supp(Si) and ℓ(Si) maximal

0, T1/∈supp(Si) for any Si∈E

ht(C) = ℓ(roof(C)).

Lemma 2.4.1 Let Xbe an indecomposable module in Tnwith ℓ(X) = ℓ, 1< ℓ ≤n

and T1∈supp(X).

(i) #{C | roof(C) = X}=Cℓ−1.Cn−ℓ+1;

(ii)#{C | ht(C) = ℓ}=ℓ.Cℓ−1.Cn−ℓ+1.

Proof: (i) Let Cbe a thick subcategory with roof(C) = X. Without loss of

generality we may assume that Soc(X) = T1. Then supp(X) = {T1,...,Tℓ}and

Thick(T1,...,Tℓ)∼

=mod k∆ℓ, if ℓ < n or Thick(T1,...,Tℓ)∼

=Tn, if ℓ=n. Now,

since Xis a simple module in C, then HomTn(X, Y ) = HomTn(Y, X) = 0, where Yis

another simple in C. Hence supp(Y)j{T2,...,Tℓ−1}or supp(Y)j{Tℓ+1, . . . , Tn}.

Denote by C1=k∆ℓ−2and C2=k∆n−ℓthe thick subcategories generated by

{T2,...,Tℓ−1}and {Tℓ+1, . . . , Tn}. It is immediate to see that there are neither

homomorphism nor extensions between these two categories. Then any thick sub-

category Cwith roof(C) = Xmust be of the form C= Thick(X, C∗), where C∗is

thick in C1⊕ C2, see the figure below. But since C1and C2are disjoint, then the

number of thick subcategories in C1⊕ C2is exactly Cℓ−1.Cn−ℓ+1.

r r r r r r r r r

n1 2 ℓ−1ℓ ℓ + 1 r

ℓ+ 2













... ...

C1C2

n−1n1

Figure 2.7: Thick subcategories with roof(C) = X

(ii) The length of Xstays invariant under the τtranslate hence all thick subcat-

egories Ci∈ Tnwith ht(Ci) = ℓ(X) are shifts of C, that is, Ck=τk(C) for appropriate

k. Since roof(Ck) = τk(X), then T1∈supp(τk(X)) if and only if k= 1, . . . , ℓ(X).

Hence by (i) #{C | ht(C) = ℓ}=ℓ.#{C | roof(C) = X}=ℓ.Cℓ−1.Cn−ℓ+1.2

Proposition 2.4.2 The number of thick subcategories in Tnis 2n

n.

Proof: Let Cbe a thick subcategory. Since ht(C) varies from 0 to n, by previous

lemma we have: #{C ∈ Tn}=Pn

i=0 #{C | ht(C) = i}=Cn+Pn

i=1 i.Ci−1.Cn−i+1 =

2.4. Number of thick subcategories 28

Cn−Tn.C0+Pn+1

i=1 Ti−1.Cn−i+1 =Cn−Tn+An, where Tn= (n+ 1)Cn=2n

nand

An=Pn

k=0 Tk.Cn−k. Here we used that {C ∈ Tn|ht(C) = 0}={C ∈ Tn|supp(C) =

(T2,...,Tn)}={C ∈ Tn| C ∈ Thick(T2,...,Tn)}and since Thick(T2, . . . , Tn)∼

mod k∆n−1, we have #{C ∈ Tn|ht(C) = 0}=Cn.

Now, consider the following power series:

c(x) = C0+C1x+C2x2+···+Cnxn+···=P∞

i=0 Cixi,Ci=1

i+12i

i,

t(x) = T0+T1x+T2x2+···+Tnxn+···=P∞

i=0 Tixi,Ti=2i

i.

From the Cauchy product of c(x) and t(x) follows that a(x) = c(x)t(x) has

An(n= 0,1,2,...) as coefficients, that is,

a(x) = A0+A1x+A2x2+···+Anxn+···=P∞

i=0 Aixi.

It is a classical result that the power series expansions of 1−√1−4x

2xand 1

√1−4xare

exactly c(x) and t(x).Then

a(x) = c(x)t(x) = 1

2x(1

√1−4x−1) = 1

2x(T1x+T2x2+···) = T1

2+T2

2x+···+

Tn+1

2xn+···=P∞

i=0 Aixiand comparing the coefficients, we have An=Tn+1

Hence #{C ∈ Tn}=Cn−Tn+Tn+1

2=Tn=2n

n.2

Corollary 2.4.3 The number of cotilting modules in ˜

Tn, support-tilting modules,

bounded and unbounded thick subcategories in Tnis 2n−1

n.

Proof: By theorem 2.3.28, we have a bijection between bounded, unbounded thick

subcategories and support-tilting modules in Tnand cotilting modules in ˜

Tn, hence

their number is equal. Now previous proposition tells us that the number of all thick

subcategories is 2n

nand since the number of bounded thick equals the number of

unbounded thick, their number is half of the number of all thick subcategories, that

is, 1

22n

n=1

(2n)!

n!n!=1

(2n−1)!2n

n!(n−1)!n=2n−1

n.2

Remark 2.4.4 In fact, the number of cotilting objects in ˜

Tnis already known,

see [BKr, Theorem D].

The following result, first shown by Gabriel, is folklore in the tilting theory. We

present another proof by pointing out a connection between certain exact abelian

extension closed subcategories and basic tilting modules in mod k∆n.

Proposition 2.4.5 The number of tilting modules in mod k∆nis Cn.

Proof: First we comment that in mod k∆nany thick subcategory is exact abelian.

Let S1,...,Snbe the simple and P1,...,Pnbe the indecomposable projective k∆n-

modules. Now, in mod k∆nthere is an indecomposable projective-injective module

and we denote it by Pn. We shall use again that the number of thick subcategories in

mod k∆nis Cn+1. First we show that the number of thick subcategories that contain

Pnis Cn. Let Cbe a thick subcategory with Pn∈ C. Then Pn⊆ C ⇔ C⊥⊆P⊥

nand

since C⊥is thick, then the number of thick subcategories of P⊥

nis the same as the

2.5. Lattice of thick subcategories 29

number of thick subcategories that contain Pn. But since P⊥

n=U(S1,...,Sn−1) =

mod k∆n−1, the claim follows.

Now, we show that there is a bijection between thick subcategories that con-

tain Pnand tilting modules in mod k∆nand the proof shall follow. From theo-

rem [IT, Theorem 1.1], in mod k∆nwe have a bijection between thick subcategories

and support-tilting modules. Now, let Cbe a thick subcategory containing Pn. Then

since supp(Pn) = {S1,...,Sn}, the corresponding support-tilting module is tilting.

Conversely, let Tbe a tilting module. Then we have the following exact sequence:

0→AA→T′

A→T′′

A→0 with T′, T′′ ∈add(T) and A=k∆n. Since Pnis also

injective, it follows that it is a direct summand of T′and hence of T. Now, Gen(T)

is a torsion class and the corresponding thick subcategory α(Gen(T)) (see proposi-

tion 2.3.11) contains Pn, since any morphism f:X→Pnwith Xindecomposable

in Gen(T) is a monomorphism, hence Ker f= 0 ∈Gen(T). The proof follows. 2

2.5 Lattice of thick subcategories

As we observed, the set of thick subcategories in Tnis finite. We consider the poset

(L, ≤) formed by subsets of the set of thick subcategories in Tn. In fact, it is not

difficult to see that (L, ≤) is a lattice: We notice that intersection of any two thick

subcategories C1,C2is again thick, so we have naturally defined meet in L, namely

C1∧ C2:= C1∩ C2. The join of any two thick subcategories is defined to be the meet

of all thick subcategories that contains both of them.

(1)(2)(3)

(12)(3) (13)(2) (2)(3) (2)(31) (21)(3) (1)(2) (1)(23) (1)(32) (1)(3)

(13)

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!!!!!!!!!





Figure 2.8: The lattice of thick subcategories in T3

Proposition 2.5.1 The set of thick subcategories in Tnforms a lattice. Moreover τ

induces a lattice isomorphism and forming the right perpendicular category induces

a lattice anti-isomorphism.

2.6. Nakayama algebras 30

Proof: The first statement follows from the discussions above. For the rest: As we

noticed, τrelabels the simples within Tn, hence it yields a lattice isomorphism. If we

apply right perpendicular on the set of thick subcategories, then it yields a bijection,

see theorem 2.3.25. To show that it is a lattice anti-isomorphism, we check that

meets and joints in (L, ≤) are sent to joints and meets in (L⊥,≤). By lemma 2.3.17,

we have that right perpendicular is order reversing, that is, C1≤ C2⇒ C⊥

2≤ C⊥

1. If

C1=X1∨X2· · · ∨ Xk, where Xi’s are the simples of C1, then we claim that C⊥

X⊥

1∧X⊥

2· · ·∧X⊥

k. First Xi≤ C1implies C⊥

1≤X⊥

iand hence C⊥

1≤X⊥

1∧X⊥

2· · ·∧X⊥

If Y≤X⊥

1∧X⊥

2· · · ∧ X⊥

kis an arbitrary module, then Y≤X⊥

ifor every iand

applying left perpendicular we get Xi≤⊥Y. Then C1=X1∨X2· · · ∨ Xk≤⊥Y

and therefore, Y≤ C⊥

1. Since Ywas arbitrary, then we get X⊥

1∧X⊥

2· · · ∧ X⊥

k≤ C⊥

and the claim follows. We derive that C=C1∨ C2⇒ C⊥=C⊥

1∧ C⊥

2. In the same

way, one shows that for the meet we have C=C1∧ C2⇒ C⊥=C⊥

1∨ C⊥

2. The proof

follows. 2

2.6 Nakayama algebras

In this section, we consider certain algebras, which are quotients of k∆nand k˜

∆n.

We naturally generalise the methods used in the previous sections in order to classify

the exact abelian extension closed subcategories for these algebras.

Definition 2.6.1 An algebra Ais said to be left serial (resp. right serial) if

every indecomposable projective left (resp. right) A-module is uniserial. It is called

Nakayama algebra if it is both right and left serial.

We point out that Nakayama algebras are well studied. We recall certain facts for

Nakayama algebras, but we refer to [AS, Chapter V] for complete reference to the

subject.

Definition 2.6.2 An algebra Ais called basic, if eiA6=ejAfor all i6=j, where

{e1,...,en}is its complete set of primitive orthogonal idempotents. We say that an

algebra Ais connected, if Ais not a direct product of two algebras.

Theorem 2.6.3 A basic and connected algebra Ais a Nakayama algebra if and only

if its ordinary quiver QAis one of the following quivers:

(a) ∆n: 1 //2//3//... //n;

(b) ˜

∆n: 1 //2//3//... //n

kk.

The quotients of Nakayama algebras are again Nakayama.

2.6. Nakayama algebras 31

Proposition 2.6.4 Let Abe an algebra, and Jbe a proper ideal of A. If Ais

Nakayama algebra, then A/J is also Nakayama algebra.

Example 2.6.5 The algebra k∆h

n=k∆n/Ih(h≥1), where Iis the two-sided ideal

generated by all arrows of ∆nis a Nakayama algebra.

It is not difficult to construct the Auslander-Reiten quiver for the module category

over Nakayama algebras. The technique is explained in [AS, Chapter V.4]. We give

an example.

Example 2.6.6 The AR-quiver of mod k∆3

S1[3]



S2[3]



S3[3]



S4[3]



S1[2]

;;

xoo



S2[2]

;;



ooS3[2]

;;



ooS4[2]

;;



ooS5[2]



ooS2

;;

ooS3

;;

ooS4

;;

ooS5

;;

ooS6

We consider the exact abelian extension closed subcategories in mod k∆h

n. Since

mod k∆h

n⊆mod k∆nand k∆nis representation finite, then k∆h

nis also represen-

tation finite and hence there are finite number of exact abelian extension closed

categories in mod k∆h

n. We denote by ∆h

ntheir number.

Let Abe a Nakayama algebra with simple objects S1,...,Sn, where n= rk(mod A).

Any exact abelian extension closed category Cin mod Ais uniquely determined by

its simple objects S∗

i, that is, C=U(S∗

1,...,S∗

k) for k≤n. Since C ⊆ mod A⊆

mod k∆n⊂nrep(k ˜

∆n) and the embedding functor is exact, we deduce that in

mod Athere is a bijection between orthogonal sequences and exact abelian exten-

sion closed categories, as we established for nrep(k ˜

∆n), see theorem 2.2.8. As we

did in the previous section, in order to count the number of exact abelian extension

closed subcategories, we count the number of orthogonal sequences in mod A. For

a simple module S1of mod A, define

roof(C) = 





S∗

i, S1∈supp(S∗

i) and ℓ(S∗

i) maximal

0, S1/∈supp(S∗

i) for any simple S∗

i∈E

ht(C) = ℓ(roof(C)).

Proposition 2.6.7 Consider the algebra k∆h

n. Then

∆h

n=∆h

n−1+

h−1

i=0

Ci.∆h

n−i−1,(2.1)

where Ci=1

i+12i

iis the ith Catalan number.

2.6. Nakayama algebras 32

Proof: Let Xbe an indecomposable module with ℓ(X) = ℓ≥2 and S1∈

supp(X). We show that

#{C | roof(C) = X}= #{C | ht(C) = ℓ}=∆ℓ−2.∆h

n−ℓ.(2.2)

For the first equality: By definition Xis simple in Cand therefore there is no other,

simple module Y∈ C that contains T1in its support since this yields Soc(X) =

Soc(Y) and hence a monomorphism between Xand Y, which is impossible. For the

second equality, since the simples are orthogonal, from the AR-quiver of mod k∆h

we notice that all indecomposable objects Yin mod k∆h

nsuch that Homk∆h

n(X, Y ) =

Homk∆h

n(Y, X) = 0 are contained in two regions (see the figure below):

•the triangle-shaped region - the part of the AR-quiver that contains all inde-

composable objects with support from the set {T2,...,Tℓ−1};

•the trapezium-shaped region - the part of the AR-quiver that contains all

indecomposable objects with support from the set {Tℓ+1,...,Tn}.

r r r r r r r r

1 2 ℓ−1ℓ ℓ + 1ℓ+ 2













r r

. . . . . .

∆ℓ−2∆h

n−ℓ

n−1n

Figure 2.9: The orthogonal points of X

Then U(T2,...,Tℓ−1)∼

=mod k∆ℓ−2,U(Tℓ+1,...,Tn)∼

=mod k∆h

n−ℓand hence

the number of orthogonal sequences in these subcategories is ∆ℓ−2and ∆h

n−ℓand

(2.2) follows at once. If ℓ(X) = 0 or ℓ(X) = 1, then evidently all orthogonal to X

must be in U(T2,...,Tn)∼

=mod ∆h

n−1. Now, since #{C ∈ mod k∆h

n}=Ph

i=0 #{C |

ht(C) = i}, we obtain the following formula:

∆h

n=∆h

n−1+∆h

n−1+

i=2

∆i−2.∆h

n−i=∆h

n−1+

h−1

i=0

∆i−1.∆h

n−i−1,(2.3)

where we set ∆i= 1 for i < 0. If h=n, then mod k∆n

n= mod k∆nand hence

∆n

n=∆nand recurrent formula reads:

∆n=∆n−1+∆n−1+

n−1

i=1

∆i−1.∆n−i−1,(2.4)

with ∆0= 1 and ∆1= 2. On the other hand, it is a classical result that for n≥1

the Catalan numbers are defined via the following recurrent formula:

Cn+1 =

i=0

CiCn−i=C0.Cn+C0.Cn+

n−1

i=1

CiCn−i,(2.5)

2.6. Nakayama algebras 33

where C0= 1. Comparing with (2.4), we conclude that ∆i=Ci+1 and having in

mind (2.3), we obtain (2.1). 2

Remark 2.6.8 When h=n, we obtain that the number of exact abelian extension

closed subcategories in mod k∆nis Cn+1. Hence the formula could be interpreted

as a generalisation of the recursive formula for the Catalan numbers. For detailed

reference to Catalan numbers, we point out [RSt].

We present a table of the number of exact abelian extension closed subcategories

in mod k∆h

n. The numbers in bold are the Catalan numbers.

n\h0 1 2 3 4 5 6 7 8

011 1 1 1 1 1 1 1

1 1 22 2 2 2 2 2 2

2 1 4 55 5 5 5 5 5

3 1 8 12 14 14 14 14 14 14

4 1 16 29 37 42 42 42 42 42

5 1 32 70 98 118 132 132 132 132

6 1 64 169 261 331 387 429 429 429

7 1 128 408 694 934 1130 1298 1430 1430

8 1 256 985 1845 2645 3317 3905 4430 4862

2.6.1 Self-injective Nakayama algebras

Definition 2.6.9 An algebra Ais called self-injective, if the left module AAis an

injective A-module.

Theorem 2.6.10 Let Abe a basic and connected algebra, which is not isomorphic

to k. Then Ais a self-injective Nakayama algebra if and only if A∼

=k˜

∆n/Rh, for

some h≥2, where

∆n: 1 //2//3//... //n

with n≥1and Ris the two-sided ideal generated by all arrows of ˜

∆n.

Example 2.6.11 The construction of the AR-quiver of mod k˜

∆h

nis well-known,

see [AS, Chapter V.4]. Here is an example for mod k˜

∆3

S5[3]



S6[3]



S1[3]



S2[3]



S3[3]



S4[3]



... S6[2]

;;



ooS1[2]

;;



ooS2[2]

;;



ooS3[2]

;;



ooS4[2]

;;



ooS5[2]



...

ooS1

;;

ooS2

;;

ooS3

;;

ooS4

;;

ooS5

;;

ooS6

2.6. Nakayama algebras 34

We classify the exact abelian extension closed categories in mod k˜

∆h

n. First we

recall that in nrep(k ˜

∆n) the points are all indecomposable modules with length

less or equal n. Now, since mod k˜

∆h

n⊆nrep(k ˜

∆n) is a full embedding, the points

in mod k˜

∆h

nare all indecomposables Xwith ℓ(X)≤k= min{n, h}. Since each

exact abelian extension closed subcategory of mod k˜

∆h

nis uniquely determined by

its simple objects, which are points, we conclude that all these simples must lie in

mod ˜

∆k

n⊆mod ˜

∆h

n. Having in mind these observations and theorem 2.2.8, together

with proposition 2.4.2, we have immediately:

Corollary 2.6.12 There is a bijection between exact abelian extension closed sub-

categories of mod k˜

∆h

nand non-crossing arcs with length at most k= min{n, h}on a

circle with npoints. Moreover, the number of exact abelian extension closed subcat-

egories of mod k˜

∆h

n, where h≥nis equal to the number of exact abelian extension

closed subcategories in nrep(k ˜

∆n), which equals 2n

n.

Denote by ˜

∆h

nthe number of exact abelian extension closed categories in mod k˜

∆h

Proposition 2.6.13 In mod k˜

∆h

nwe have the following recursive formula:

∆h

n=∆h

n−1+

h−1

i=1

Ti−1.∆h

n−i,(2.6)

where Tn=2n

nis the central binomial coefficient.

Proof: Let Cbe an exact abelian extension closed subcategory in mod ˜

∆h

nand

let Xbe an indecomposable with ℓ(X) = ℓ. The proof mimics the proof of propo-

sition 2.6.7. The only difference is that #{C | ht(C) = ℓ}=ℓ.#{C | roof(C) = X}.

To verify that, we notice that all exact abelian extension closed subcategories Ci

with ht(Ci) = ℓare of the form Ci=τi(C), i= 1,...,ℓ. The latter is true

…

...

L-1 L+1 2

…L-1

...

N-1

Figure 2.10: ℓ-times shifts of thick subcategories

since all indecomposable modules with the same length must lie on the same τ-

orbit and hence roof(Ci) = τi(roof(C)) and ht(C) = ht(Ck). It is exactly ℓ-times

2.6. Nakayama algebras 35

since X1∈supp(τiX) for i= 1,...,ℓ. We conclude that #{C | ht(C) = ℓ}=

ℓ.Cℓ−1.∆h

n−i=Tℓ.∆h

n−i.2

Here is a table of the number of exact abelian extension closed subcategories in

mod k˜

∆h

n. The numbers in bold are the central binomial coefficients.

n\h0 1 2 3 4 5 6 7 8

011 1 1 1 1 1 1 1

1 1 2222222 2

2 1 4 66 6 6 6 6 6

3 1 8 14 20 20 20 20 20 20

4 1 16 34 50 70 70 70 70 70

5 1 32 82 132 182 252 252 252 252

6 1 64 198 354 504 672 924 924 924

7 1 128 478 940 1430 1920 2508 3842 3432

8 1 256 1154 2498 4078 5646 7326 9438 12870

Remark 2.6.14 In fact, using similar arguments as in the last proposition, one

gets the following recursive formula:

Proposition 2.6.15 Consider the algebra k˜

∆h

n. Then

∆h

n=˜

∆h

n−1+

h−1

i=1

Ci−1.˜

∆h

n−i,(2.7)

where Cnis the nth Catalan number.

If we compare (2.6) and (2.7), we notice that the last formula is more coherent

in a sense that it involves terms from the same type.

2.6. Nakayama algebras 36

We illustrate the bijections established in theorem 2.3.28. The thicken points (•)

represent the indecomposable direct summands of the cotilting and support-tilting

modules and the simples of the thick subcategories. For simplicity, we do not set

labels of the indecomposable modules, but we refer to figure 2.5.

Cotilting Support-tilting Bounded Unbounded

modules modules thick thick

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Chapter 3

Thick subcategories for hereditary

algebras

For a finite and acyclic quiver Q, we consider its path algebra kQ. We step on a

result of Crawley-Boevey [CB1, Lemma 5], which says that any thick subcategory of

mod kQ generated by an exceptional sequence is exact abelian. We construct for a

thick subcategory C ⊆ mod kQ generated by preprojective modules, an exceptional

sequence that generates C.

Next, we specialise to the module category of kQ, where Qis an Euclidian quiver.

Its path algebra is an example of representation-infinite hereditary algebra, for which

the classification of indecomposable modules is well-known. We introduce reduction

techniques, some of which work in a more general settings, which enable us to prove

that any thick subcategory in mod kQ is exact abelian.

By a result of Colin Ingalls and Hugh Thomas [IT, Theorem 1.1], there is a

bijective correspondence between non-crossing partitions associated to Q(Qis an

Euclidian quiver) and exact abelian extension closed subcategories with a projective

generator in mod kQ. As one observes, there are exact abelian extension closed

subcategories without a projective generator (for instance the tubes in the regular

component of the Auslander-Reiten quiver of mod kQ). So we use results from the

second chapter, and combining with the above cited theorem, we give a complete

combinatorial classification of thick subcategories in mod kQ.

The results in this chapter are joint work with Yu Ye.

3.1 Thick subcategories generated by preprojec-

tive modules

From now on, we assume that Qis a finite and acyclic quiver and kan is algebraically

closed field. We begin with recalling some facts for the structure of the Auslander-

3.1. Thick subcategories generated by preprojective modules 38

Reiten quiver of mod kQ. As a reference, we point out [AS, Chapter VIII.2].

Definition 3.1.1 Let Abe an arbitrary (not necessarily hereditary) k-algebra, and

Γ(mod A) the Auslander-Reiten quiver of A.

(a) A connected component Pof Γ(mod A) is called preprojective if Pis acyclic

and, for any indecomposable module Min P, there exist t≥0 and a∈(QA)0

such that M∼

=τ−tP(a). An indecomposable A-module is called preprojec-

tive if it belongs to a preprojective component of Γ(mod A), and an arbitrary

A-module is called preprojective if it is a direct sum of indecomposable pre-

projective A-modules.

(b) A connected component Qof Γ(mod A) is called preinjective if Qis acyclic

and, for any indecomposable module Min Q, there exist s≥0 and b∈(QA)0

such that M∼

=τsI(b). An indecomposable A-module is called preinjective if

it belongs to a preinjective component of Γ(mod A), and an arbitrary A-module

is called preinjective if it is a direct sum of indecomposable preinjective A-

modules.

Proposition 3.1.2 Let Qbe a finite, connected, and acyclic quiver, and let A=

kQ. The quiver Γ(mod A)contains a preprojective P(A)and preinjective Q(A)

component.

Now we look at the structure of the preprojective (preinjective) component of

Γ(mod A). Let M, N be two indecomposable A-modules. A path in mod Afrom M

to Nof length tis a sequence:

M=M0

→M1

→M2··· ft

→Mt=N

where all the Miare indecomposable, and all fiare non-zero nonisomorphisms. In

this case, Mis called a predecessor of Nin mod A. Dually, one has a definition

of a successor. We have the following proposition.

Proposition 3.1.3 [AS, Chapter VIII.2, Proposition 2.1] Let Abe arbitrary (not

necessarily hereditary) algebra.

(a) Let Pbe a preprojective component of the quiver Γ(mod A)and Mbe an inde-

composable module in P. Then the number of predecessors of Min Pis finite

and any indecomposable A-module Lsuch that HomA(L, M)6= 0 is a prede-

cessor of Min P. In particular, HomA(L, M) = 0 for all but finitely many

indecomposable A-modules L.

3.1. Thick subcategories generated by preprojective modules 39

(b) Let Qbe a preinjective component of the quiver Γ(mod A)and Nbe an inde-

composable module in Q. Then the number of successors of Nin Qis finite and

any indecomposable A-module Lsuch that HomA(N, L)6= 0 is a successor of N

in Q. In particular, HomA(N, L) = 0 for all but finitely many indecomposable

A-modules L.

A path from an indecomposable A-module to itself, is a sequence on non-zero noni-

somorphisms between indecomposables of the form

M=M0

→M1

→M2··· ft

→Mt=M,

is called a cycle in mod A. Then the previous proposition says that, in case of

modules lying in preprojective or preinjective components, these module-theoretical

notions can be expressed graphically.

Proposition 3.1.4 [AS, Chapter VIII.2, Corollary 2.6] Let Abe an arbitrary (not

necessarily hereditary) k-algebra.

(a) Let Pbe preprojective component of Γ(mod A)and Mbe an indecomposable

module in P. Then

(i) any predecessor Lof Min mod Ais preprojective and there is a path in P

from Lto M, and

(ii) Mlies on no cycle in mod A.

(b) Let Qbe preinjective component of Γ(mod A)and Nbe an indecomposable mod-

ule in Q. Then

(i) any successor Nof Lin mod Ais preinjective and there is a path in Q

from Nto L, and

(ii) Nlies on no cycle in mod A.

AkQ-module Xis called exceptional, provided that Xis indecomposable and

Ext1

kQ(X, X) = 0. Examples of exceptional modules are the simple modules. The

following lemma gives more examples.

Lemma 3.1.5 [AS, Chapter VIII.2, Lemma 2.7] Let Abe an arbitrary (not nec-

essarily hereditary) k-algebra and Mbe an indecomposable preprojective, or prein-

jective, A-module. Then EndAM∼

=kand Ext1

A(M, M) = 0.

Let E= (X1,...,Xr) be a sequence of kQ-modules. Then Eis exceptional se-

quence of length r, if all Xiare exceptional and HomkQ(Xj, Xi) = 0 for 1 ≤i <

j≤rand Ext1

kQ(Xj, Xi) = 0 for 1 ≤i≤j≤r. If requals the number of

3.1. Thick subcategories generated by preprojective modules 40

vertices of Q, then Eis called complete. Recall that Eis called orthogonal, if

HomkQ(Xi, Xj) = 0 for any i6=j. We refer the reader to papers of [R3] and [CB1],

where exceptional sequences are studied in great details.

We notice that since Qhas no oriented cycles, we can relabel the vertices of Q

such that (Sn, Sn−1,...,S1) forms an exceptional sequence. In that case, it is imme-

diate to check that the sequence (P1, P2,...,Pn) of projectives is also exceptional.

So, from now on, we assume that we label the vertices of Qin such a way, that the

above sequences are exceptional.

As before, we denote by Thick(S) the smallest thick subcategory containing S,

where Sis an arbitrary set of kQ-modules. We shall frequently use the following

lemma.

Lemma 3.1.6 [CB1, Lemma 5]) Let Ebe an exceptional sequence of length rin

mod kQ. Then Thick(E)is equivalent to the category of representations of a quiver

Q(E)with rvertices and no oriented cycles. The functor mod kQ(E)֒→mod kQ is

exact and induces isomorphism on both Hom and Ext. Moreover, any exceptional

sequences in mod kQ can be enlarged to a complete sequence.

In other words Thick(E), for Eexceptional, is an exact abelian extension closed

subcategory of mod kQ.

After recalling these facts, we start with our investigation. Let {S1, . . ., Sn}

be the complete set of simple kQ-modules, and {P1,...,Pn}and {I1, . . . , In}the

corresponding indecomposable projective and injective modules.

We consider the preprojective component P={τmPi, m ≤0,1≤i≤n}of the

Auslander-Reiten quiver of mod kQ. The structure of P(see theorem 3.1.3) allows

us to introduce a total order on Pas follows: τmPi≺τnPjif m > n or m=nand

i < j. Obviously, HomkQ(X1, X2) = 0 for any X1≻X2in P. For any X1, X2∈ P,

the distance d(X1, X2) between X1and X2is defined to be the supremum of the

lengths of paths starting in X1and terminating at X2in the Auslander-Reiten quiver

of kQ, and 0 when no such a path exists.

Example 3.1.7 We consider a part of the preprojective component of Γ(mod kQ).

P2//τ−kP2g1

))

R//

K//Pk//τ−kP1

f166

fn((

Q//τ−kPk//τ−k−1P1//

Pn//τ−kPn

l//

Now, d(τ−kP1, τ−kPi) = 1, d(τ−kP1, τ−k−1P1) = 2 and d(τ−k−tPi, τ−kP1) = 0 for

k, t ∈Nand i= 2,...,n.

The following facts are easily derived from the definition.

3.1. Thick subcategories generated by preprojective modules 41

Lemma 3.1.8 Let Qbe a quiver and X, Y ∈ P.

(i) If X≻Y, then d(X, Y )=0.

(ii) If HomkQ(X, Y )6= 0 and X6=Y, then d(X, Y )≥1; if Ext1

kQ(Y, X)6= 0,

then d(X, Y )≥2.

(iii) For any given X∈ P and d > 0, there exist only finitely many Y∈ P, such

that 0< d(X, Y )≤dor 0< d(Y, X)≤d.

By induction on distance, we get the following useful lemma.

Lemma 3.1.9 Let S⊆ P be a set of kQ-modules and Z∈ P. Then there exists

a set S∗⊆ P, such that Thick(S∗, Z) = Thick(S, Z),Ext1

kQ(Z, X) = 0 for any

X∈S∗, and HomkQ(X, Z)6= 0 for any X∈S∗\S.

Proof: Set d(S;Z) = sup({d(X, Z)|X∈S, Ext1

kQ(Z, X)6= 0}) if Ext1

kQ(Z, X)6= 0

for some X∈S, and 0 otherwise. We use induction on d(S;Z). By lemma 3.1.8,

d(S;Z) = 0 or d(S;Z)≥2. If d(S;Z) = 0, then Ext1

kQ(Z, X) = 0 for all X∈Sand

hence we may take S∗=S. So, we may assume that d(S;Z)>0.

For any X∈Ssuch that Ext1

kQ(Z, X)6= 0, we fix a non-split short exact

sequence 0 →X→ ⊕l

i=1X⊕ni

i→Z→0, where Xi’s are indecomposable and

pairwise non-isomorphic. Set SX;Z={X1,...,Xl}. By construction, we have

Thick(X, Z) = Thick(SX;Z, Z) and d(SX;Z;Z)≤d(X;Z)−1. The last equality

holds since d(X, Z)≥d(X, Y ) + d(Y, Z) for any X, Y and Z∈ P, provided that

d(X, Z)6= 0, d(X, Y )6= 0 and d(Y, Z)6= 0.

Now, we take

S′={X∈S|Ext1

kQ(Z, X) = 0} ∪ [

X∈S,Ext1

kQ(Z,X)6=0

SX;Z.

We showed that d(S′;Z)< d(S;Z) and Thick(S′, Z) = Thick(S, Z). Clearly,

HomkQ(X, Z)6= 0 for any X∈S′\S. Now, repeat the argument for S′, and

after finite steps we get S∗⊂ P with the desired properties. 2

Proposition 3.1.10 Let S⊂ P be a set of kQ-modules. Then there exists an

exceptional sequence E, such that Thick(S) = Thick(E). As a consequence, any

thick subcategory generated by preprojective modules is exact abelian.

Proof: We use induction on the total order on P. First we assume that Sis a finite

set. We construct the required exceptional sequence Ein the following way.

We take a maximal element Z1in S. This can be done since Sis a finite

set. Set S′=S\ {Z1}. Now, we know that HomkQ(Z1, X) = 0 for any X∈

S′. By lemma 3.1.9, there exists S1⊆ P, such that Thick(S1, Z1) = Thick(S),

Ext1

kQ(Z1, X) = 0 for any X∈S1, and moreover, HomkQ(Z, X) = 0 for any X∈S1.

3.2. Thick subcategories for Euclidean quivers 42

In other words, Thick(S1, Z1) = Thick(S) and Thick(S1)⊆Z⊥

1, where as usual

Z⊥

1={X∈mod kQ |HomkQ(Z1, X) = Ext1

kQ(Z1, X) = 0}.

Set Z1to be the last term of Eand repeat the argument on S1to get an ascending

sequence E={...,Z2, Z1}in Pwith respect to the order we defined before, such

that Thick(E) = Thick(S) and Thick(...,Zn+2, Zn+1)⊆Zn⊥for any n≥1. Since

for any Z∈ P, there exist only finitely many X∈ P with X≺Z, we will stop

after finite steps, which means that Eis a finite sequence. By construction, Eis an

exceptional sequence and Thick(E) = Thick(S).

Now, let Sbe an arbitrary subset of P. Since there exist only countably many

preprojective modules, we assume that S={X1, X2,...}. Set Si={X1, X2, . . . , Xi}

and Ci= Thick(Si) for any i≥1. We complete the proof by showing that

Thick(S) = Ckfor some k. Otherwise, assume that there exists an ascending se-

quence 1 = r1< r2< r3<···, such that

C1=Cr1$Cr2$Cr3$···.

We showed that each Cri= Thick(Eri), for some exceptional sequence Eri, and hence

Criis isomorphic to the finite dimensional module category of some quiver. Now, fix

a complete sequence F1in Cr1. The latter can be enlarged to a complete sequence

F2in Cr2, and do this repetitively to get an exceptional sequence Fifor any i. We

know that each Fiis an exceptional sequence in mod kQ. Since the length of an

exceptional sequence is at most n, we know that there exists k, such that Fi=Fk

for any i≥k, which contradicts the assumption that Cri6=Cri+1 for any i.2

Remark 3.1.11 Dually, we can prove that if S⊆ Q is a set of kQ-modules, then

there exists an exceptional sequence E⊆ Q such that Thick(E) = Thick(S). Hence

any thick subcategory generated by preinjective modules is exact abelian.

Corollary 3.1.12 Let Qbe a Dynkin quiver. Then any thick subcategory in mod kQ

is exact abelian.

Proof: Since any indecomposable module in mod kQ is preprojective, the claim

follows. 2

3.2 Thick subcategories for Euclidean quivers

As we have seen, thick hereditary categories generated by preprojective or preinjec-

tive modules are exact abelian. But in general not all modules of finite dimensional

algebras are preinjective or preprojective, as we see from the following proposition.

3.2. Thick subcategories for Euclidean quivers 43

Proposition 3.2.1 [AS, Chapter VIII.2, Corollary 2.10] Let Abe a representation

infinite algebra. Then there exists an infinite family of pairwise non-isomorphic

indecomposable A-modules that are neither preprojective nor preinjective.

Therefore, it may happen that the exact abelian subcategories in mod Aare not

generated only by preprojective or preinjective modules.

Definition 3.2.2 Let Abe an arbitrary (not necessarily hereditary) k-algebra. A

connected component Cof Γ(mod A) is called regular component, if Ccontains

neither projective nor injective modules. An indecomposable A-module is called

regular indecomposable, if it belongs to a regular component of Γ(mod A) and

an arbitrary A-module is called regular, if it is a direct sum of indecomposable

A-modules. A non-zero regular module having no proper regular submodules is said

to be regular simple.

For any regular module X, there exists a chain

X=X0)X1)···)Xℓ−1=)Xℓ= 0

of regular submodules of Xsuch that Xi−1/Xiis simple regular for any iwith

1≥i≥ℓ, and ℓis called the regular length of X, which we denote by rℓ(X).

Let Abe a representation-infinite hereditary algebra. We denote by R(A) the

family of all the regular components of Γ(mod A) and by add(R(A)) the full sub-

category of mod Awhose objects are all the regular A-modules.

The following proposition tells us more about Hom-spaces between different com-

ponents in Γ(mod A).

Proposition 3.2.3 [AS, Chapter VIII.2, Corollary 2.13] Let Abe as before and

L, M and Nbe three indecomposable A-modules.

(a) If Lis preprojective and Mis regular, then HomA(M, L) = 0.

(b) If Lis preprojective and Nis preinjective, then HomA(N, L) = 0.

The picture visualises the shape of the Auslander-Reiten quiver of mod A.

The proposition above is more briefly expressed by writing:

3.2. Thick subcategories for Euclidean quivers 44

HomA(R(A),P(A)) = 0, HomA(Q(A),P(A)) = 0, HomA(Q(A),R(A)) = 0. Us-

ing the Auslander-Reiten formula and the previous proposition, we get immediately:

Ext1

A(P(A),R(A)) = 0,Ext1

A(Q(A),R(A)) = 0,Ext1

A(P(A),Q(A)) = 0.

The behaviour of the Auslander-Reiten translate τon the regular component is

recorded in the following proposition, see [AS, Chapter VIII.2, Corollary 2.14].

Proposition 3.2.4 Let Abe representation-infinite hereditary algebra. The the

Auslander-Reiten translations τand τ−1, induce mutually inverse equivalences of

categories

add(R(A)) τ//add(R(A))

τ−1

oo.

There are few cases of infinite-dimensional algebras in which the regular component

is well-known. Examples of such algebras are tame hereditary algebras, which

are the path algebras of the quivers, whose underling graph are Euclidian diagrams

(one point extensions of Dynkin diagrams, see A.2). We list the Euclidian quivers,

the dotted lines shows how these diagrams are obtained from the Dynkin diagrams.

•

pppppppp

An:• • ... • • (n≥1)

•

<•



Dn:• • ... • • (n≥4)

•



•

<<<<<<<<

•



•

E6:•••••

•

E7:•___ ••••••

•

E8:•••••••___ •

The index refers to the number of points minus one (thus ˜

Anhas n+ 1 points).

3.2. Thick subcategories for Euclidean quivers 45

In the next theorem, we collect the basic properties of the module category over

the path algebra of Euclidian type. Before that we need the following two definitions.

Definition 3.2.5 Two components Cand C′of the Auslander-Reiten quiver of an

algebra Ais said to be orthogonal if HomA(C,C′) = 0 and HomA(C′,C) = 0, that

is, HomA(C, C′) = 0 and HomA(C′, C) = 0, for any module C∈ C and any module

C′∈ C′.

Definition 3.2.6 Let T={Ti}i∈Λbe a family of stable tubes and (m1, . . ., ms) a

sequence of integers with 1 ≤m1≤ · · · ≤ ms. We say that Tis of tubular type

(m1,...,ms) if Tadmits stubes Ti1,...,Tisof ranks m1,...,ms, respectively, and

the remaining tubes Tiof T, with i /∈ {i1,...,is}, are homogeneous, that is, of

rank 1.

Theorem 3.2.7 [SS, Chapter XII.3] Let Qbe an acyclic quiver whose underlying

graph Qis Euclidean, and A=kQ be the path algebra of Q.

(a) The Auslander-Reiten quiver Γ(mod A)of Aconsists of the following three types

of components:

•a preprojective component P(A)containing all indecomposable projective

modules,

•a preinjective component Q(A)containing all indecomposable injective mod-

ules, and

•a unique P1(k)-family

TQ={T Q

λ}λ∈P1(k)

of pairwise orthogonal tubes, in the regular part R(A)of Γ(mod A).

(b) The tubes are exact abelian extension closed subcategories of mod A. Any inde-

composable regular module is uniserial.

•mQ= (p, q)if Q=˜

Am,m≥1,p= min{p′, p′′}, and q= max{p′, p′′},

where p′and p′′ are the numbers of counterclockwise-oriented arrows in Q

and clockwise-oriented arrows in Q, respectively,

•mQ= (2,2, m −2), if Q=˜

Dmand m≥4,

•mQ= (2,3,3), if Q=˜

E6,

•mQ= (2,3,4), if Q=˜

E7, and

3.2. Thick subcategories for Euclidean quivers 46

•mQ= (2,3,5), if Q=˜

E8.

In other words, in Euclidean quiver case, kernels and cokernels of morphisms be-

tween regular modules are again regular and there are neither homomorphisms nor

extensions between different tubes. The number of non-homogeneous tubes is finite.

From now on, we assume that Qis an Euclidean quiver. We adopt some notations

from the previous chapter. We let Trbe the tube of rank rin the regular component

of mod kQ. We denote by {T1, T2,...,Tr}the set of simples of Tr, and assume

that τ(Ti) = Ti−1for any 1 ≤i≤r, where as before indices are taken modulo

r. Since the tubes are uniserial categories, any indecomposable object in Tris

uniquely determined by its socle and length. As in the first chapter, Ti[ℓ] denotes

the indecomposable object with socle Tiand length ℓ. Recall that the regular simple

composition factors of a regular module is called the regular support. For example,

Ti[ℓ] has support {Ti, Ti+1, . . . , Ti+ℓ−1}.

In this section, we aim to prove that any thick subcategory of mod kQ is exact

abelian. First, we restate theorem 2.2.10 and lemma 2.3.1 from the previous chapter.

Proposition 3.2.8 Let Trbe a tube of rank rin mod kQ. Then any thick subcat-

egory of Tris exact abelian in Trand hence in mod kQ. More precisely, for any

connected thick subcategory Cof Tr,

(1) there exists a sequence {Ti1[ℓ1],...,Tis[ℓs]} ⊆ Trof indecomposable objects

with ik+ℓk=ik+1 for any kand ℓ1+ℓ2+···+ℓs≤r;

(2) Cis either equivalent to mod kAsfor the Dynkin quiver of directed Astype, or

to a tube of rank s; moreover, Cis equivalent to a tube if and only if ℓ1+···+ℓs=r.

Before proving the next proposition, we recall the following fact. Let Rbe an

indecomposable module in mod kQ and let Tr⊆ R be the unique tube of rank r

that contains R. If rℓ(R)< r, then Ris exceptional.

Proposition 3.2.9 Let S⊆ P be an arbitrary set and E= (X1,...,Xk)⊆ Tran

exceptional sequence with pairwise disjoint regular supports. Then there exists an

exceptional sequence E′⊆ P ∪ Trsuch that Thick(S, E) = Thick(E′).

Proof: Since Xi’s have pairwise disjoint regular supports, we see that Eis orthog-

onal, that is, HomkQ(Xi, Xj) = 0 for any 1 ≤i6=j≤k. To prove the proposition,

we use the induction on the sum of the lengths of Xi’s.

If Ext1

kQ(Xi, P ) = 0 for any P∈Sand Xi∈E, then by applying proposi-

tion 3.1.10, we have an exceptional sequence F⊆ P such that Thick(F) = Thick(S).

Since S⊆E⊥, then Thick(S)⊆E⊥, and hence E′= (F, E) forms an exceptional

sequence and Thick(E′) = Thick(S, E).

3.2. Thick subcategories for Euclidean quivers 47

Now, assume that there exists some P∈Sand Xi∈Esuch that Ext1

kQ(Xi, P )6=

0. Taking a non-split short exact sequence 0 →P→P1⊕R→Xi→0 with P1

preprojective and Rregular (the middle term can be written in this way since it

has no preinjective direct summands). Notice that Ker(R→Xi)⊆Pand since

Ker(R→Xi) is again regular, it follows that Ker(R→Xi) = 0. Moreover since the

sequence is non-split, Ris a proper submodule of Xi.

We set S1to be the union of Sand the indecomposable direct summands of P1

and E1= (X1,...,Xi−1, R, Xi+1,...,Xk) if R6= 0, or to be the union of Sand the

indecomposable direct summands of P1and E1= (X1,...,Xi−1, Xi+1, . . . , Xk) if

R= 0. It follows that Thick(S, E) = Thick(S1, E1), and again E1is an orthogonal

sequence with pairwise disjoint regular supports. In both cases the total sum of

lengths of elements of E1is strictly less than the one of E, since in case R6= 0,

by construction Ris a proper submodule of Xi. Repeat the argument, we get to

the case such that S⊆E⊥after finite step, and the conclusion follows. Now, by

the inductional hypothesis, there is an exceptional sequence E′⊆ P ∪ Tr, such that

Thick(E′) = Thick(S, E) = Thick(S1, E1). The proof follows. 2

As a consequence, we get the following corollary.

Corollary 3.2.10 Let P∈ P and R∈ R such that Ext1

kQ(R, P)6= 0. Then there

exists an exceptional sequence Esuch that Thick(P, R) = Thick(E).

Proof: Let Trbe the unique tube with rank r, which contains R. Assume that

the simple objects T1, T2, . . . , Trof Trare ordered in such a way that τTi=Ti−1.

Without loss of generality, we assume that R=T1[ℓ] for some ℓ.

There are three cases.

Case 1. If 1 < ℓ < r, then Ris an exceptional module, and hence there exists

an exceptional sequence Esuch that Thick(P, R) = Thick(E) by proposition 3.2.9.

Case 2. ℓ=mr, for m≥1. Then Thick(T1[mr]) = Thick(T1[r]). Without

loss of generality, we may assume that ℓ=r. By assumption, Ext1

kQ(R, P)6= 0

and we may take a non-split short exact sequence 0 →P→P1⊕R1→R→0.

With the same argument as in the proof of proposition 3.2.9, we can show that

Thick(P, R) = Thick(P, P1, R1) with R1a proper regular submodule of R. Now, R1

has no self extensions and again using proposition 3.2.9 there exists an exceptional

sequence Esuch that Thick(E) = Thick(P, P1, R1) = Thick(P, R).

Case 3. ℓ=rm +s, for some 1 ≤s < r. Then Thick(T1[rm +s]) =

Thick(T1[s], Ts+1[r−s]). Since T1[ℓ] has a filtration with factors T1[s] and Ts+1[r−s]

and Ext1

kQ(T1[ℓ], P)6= 0, we have Ext1

kQ(T1[s], P)6= 0 or Ext1

kQ(Ts+1[r−s], P)6= 0.

First, assume that Ext1

kQ(T1[s], P)6= 0. We take a non-split short exact sequence

0→P→P1⊕R1→T1[s]→0. By the same argument as before, we get that R1is a

proper regular submodule of T1[s], and hence {R1, Ts+1[r−s]}forms an exceptional

3.2. Thick subcategories for Euclidean quivers 48

sequence. Now, by applying proposition 3.2.9, we get that there exists an exceptional

sequence Esuch that Thick(E) = Thick(P, P1, R1) = Thick(P, R).

The same argument works for the case that Ext1

kQ(Ts+1[r−s], P)6= 0, which

completes the proof. 2

As in the previous chapter, we shall frequently use the Happel-Ringel’s lemma,

so we recall it here.

Lemma 3.2.11 (Happel-Ringel) Let Hbe a hereditary abelian category. Assume

that X, Y are indecomposable objects in Hand Ext1

H(Y, X) = 0. Then any non-zero

morphism f:X→Yis either monomorphism or epimorphism.

In Chapter 1 we discussed, that any thick subcategory Cof an abelian category

Ais exact abelian if and only if Cis closed under kernels (or equivalently closed

under images, or closed under cokernels). In the next proposition we prove that any

thick subcategory in mod kQ, where Qis an Euclidian quiver is closed under kernels,

and hence it is exact abelian. As we shall see in theorem 3.3.1, this statement holds

true for any abelian hereditary category. But the proof, we shall present, is explicit

and we shall use it in the next chapter to prove the main theorem there. We also

refer the reader to A.4, where basic homological facts, which are frequently used in

the proof, are collected.

Proposition 3.2.12 Let Qbe an Euclidian quiver, Xand Y kQ-modules and

f:X→Ya non-zero morphism between them. Then Ker f∈Thick(X, Y ).

Proof: We use induction on d= dim(X) + dim(Y), where the dimension is over k.

Clearly, the assertion holds in case either Xor Yis simple. Now, assume that the

assertion is true for any morphism f:X′→Y′with dim(X′) + dim(Y′)< d.

First, we assume that Xis decomposable and write X=X1⊕X2with X1and

X2non-zero. Then we have the following commutative diagram:

0//X1

(1,0)



X1⊕X2

1)//

f=(f1

f2)



X2//



0//YId //Y//0//0.

Applying the snake lemma, we get the exact sequence

0→Ker f1→Ker f→X2→Coker f1→Coker f→0.

From dim(X1)<dim(X) follows by induction that Coker f1∈Thick(X1, Y )⊆

Thick(X, Y ). Now, since Coker f= Coker(X2→Coker f1) and dim(X2)<dim(X),

dim(Coker f1)≤dim Y, we get Coker f∈Thick(X2,Coker f1)⊆Thick(X, Y ), and

hence Ker f∈Thick(X, Y ).

3.2. Thick subcategories for Euclidean quivers 49

The dual version of the above argument shows that if Yis decomposable, then

Ker f∈Thick(X, Y ).

Now, we may assume that both Xand Yare indecomposable. If Xand Y

are preprojective (preinjective), then by proposition 3.1.10 (remark 3.1.11) we have

Ker f∈Thick(X, Y ). If Xand Yare regular, then by proposition 3.2.8 we have

Ker f∈Thick(X, Y ). Having in mind proposition 3.2.3, the only cases left are:

Case 1. X=P∈ P and Y=R∈ R.

Case 2. X=R∈ R and Y=Q∈ Q.

Case 3. X=P∈ P and Y=Q∈ Q.

We proceed with a case-by-case analysis.

Case 1. P∈ P,R∈ R and 0 6=f:P→R.

If Ext1

kQ(R, P) = 0, then by Happel-Ringel’s lemma, fis either injective or

surjective, so in both cases Ker f∈Thick(R, P ).

Now, we assume that Ext1

kQ(R, P)6= 0. Applying corollary 3.2.10, we show that

Thick(P, R) is exact abelian, and hence Ker f∈Thick(R, P).

Case 2. Dual to Case 1.

Case 3. P∈ P,Q∈ Q and 0 6=f:P→Q.

Again by Happel-Ringel’s lemma, we may assume that Ext1

kQ(Q, P )6= 0, so let

ηbe a non-split short exact sequence η: 0 →P→M→Q→0. There are two

possibilities: (i) Mis indecomposable or (ii) Mis decomposable, and we deal with

these cases separately.

(i) Mis indecomposable.

By proposition 3.1.10 and remark 3.1.11, if Mis either preprojective or prein-

jective, then Thick(P, Q) is exact abelian and therefore Ker f∈Thick(P, Q).

Now, assume that Mis regular. If Mhas no self-extensions, we can apply

proposition 3.2.9 to get that Thick(P, Q) = Thick(P, M) is exact abelian, and

hence Ker f∈Thick(P, Q). If Mhas self-extensions, by applying the functor

HomkQ(M, −) on 0 →P→M→Q→0, we get an exact sequence Ext1

kQ(M, P)→

Ext1

kQ(M, M)→0, which forces that Ext1

kQ(M, P )6= 0. By corollary 3.2.10, we

have that Thick(P, M) is exact abelian and hence Ker f∈Thick(P, Q).

(ii) Mis decomposable.

Suppose M=M1⊕M2for some M1, M26= 0. The proof that Ker f∈Thick(P, Q)

is divided into 3 steps.

Step 1. Let 0 →P(g1

g2)

→M1⊕M2

(h1,h2)

→Q→0 be a non-split short exact

sequence. If one of Ker g1,Ker g2,Coker h1,Coker h2is non-zero and contained in

Thick(P, Q), then so is Ker f.

3.2. Thick subcategories for Euclidean quivers 50

First, assume that 0 6= Ker g1∈Thick(P, Q). We have the following commuta-

tive diagram:

0//Ker g1i//

f◦i



Pπ//



U//



0//QId //Q//0//0.

By the snake lemma, we get a long exact sequence

0→Ker(f◦i)→Ker f→U→Coker(f◦i)→Coker f→0.

Since Qis indecomposable, we claim that Ker g1is a proper submodule of P.

Otherwise, if Ker g1=P, then Im h1∼

=M1and Im h2∩Im h1= 0, and hence Q∼

Im h1⊕Im h2. Since Qis indecomposable, we have that Im h2= 0, M1∼

=Q,P∼

=M2

and the short exact sequence splits. This leads to a contradiction. Hence we have

dim(Ker g1)<dim(P) and by induction hypothesis on the dimensions, Ker(f◦i)∈

Thick(Ker g1, Q)⊆Thick(P, Q) and Coker(f◦i)∈Thick(P, Q). Moreover Ker g16=

0 implies that dim(U)<dim(P), together with the facts that dim(Coker(f◦i)) ≤

dim(Q) and Coker f= Coker(U→Coker(f◦i)), it follows from that Coker f∈

Thick(U, Coker(f◦i)) ⊆Thick(P, Q).

A dual version works in case that 0 6= Coker h1∈Thick(P, Q) by using the

commutative diagram

0//0//



PId //



P//

π◦f



0//Vi//Qπ//Coker h1//0

and the snake lemma. The other cases are treated the same.

Step 2. Let 0 →P(g1,g2)

−→ M1⊕M2

(h1

h2)

→Q→0 be a non-split short exact se-

quence. If min{dim M1,dim M2}<max{dim P, dim Q}, then Ker f∈Thick(P, Q).

First assume that dim M1<dim P. By the induction hypothesis on the dimen-

sion, follows that Ker h1∈Thick(M1, Q)⊆Thick(P, Q). Therefore, if Coker h16= 0,

then Ker f∈Thick(P, Q) by Step 1. Now, assume that Coker h1= 0.

By the property of push-out and pull-back, we know that Ker g1= 0 if and only

if Ker h2= 0 and Coker g1= 0 if and only if Coker h1= 0. Now, Coker h1= 0

implies that Coker g2= 0 and hence Ker g2∈Thick(P, M2)⊆Thick(P, Q). By Step

1, to show that Ker f⊆Thick(P, Q), it suffices to show that Ker g26= 0. In fact,

if Ker g2= 0, then g2is an isomorphism and hence the short exact sequence splits,

which gives a contradiction and the assertion follows.

A dual version of the above argument works for the case dim M1<dim Q.

Step 3. Let 0 →P(g1,g2)

−→ M1⊕M2

(h1

h2)

→Q→0 be a non-split short exact sequence

with dim(P) = dim(M1) = dim(M2) = dim(Q). We claim that Ker f∈Thick(P, Q).

3.2. Thick subcategories for Euclidean quivers 51

Applying Step 2 we may assume that M1and M2are both indecomposable.

If both M1and M2are preinjective, then Thick(P, Q) = Thick(M1, M2, Q) and

hence is exact abelian by remark 3.1.11, which implies that Ker f∈Thick(P, Q).

Otherwise if one of them, say M1, is preprojective or regular, then by Case 1, we know

that Ker g1∈Thick(P, Q). We claim that Ker g16= 0. Otherwise the assumption

dim(P) = dim(M1) implies that g1is an isomorphism and hence the short exact

sequence splits. It follows that Ker f∈Thick(P, Q) by Step 1.

So far, we have shown that if Mis not indecomposable, we are either in the

situation of Step 2 or Step 3, and in both cases Ker f∈Thick(P, Q).

Now, we have shown that Ker f∈Thick(X, Y ) holds for any f:X→Ywith

dim(X) + dim(Y) = d, which finishes the proof. 2

As we already discussed, any thick category closed under arbitrary kernels is

exact abelian. The previous proposition gives us immediately the following result.

Corollary 3.2.13 Let Cbe a thick subcategory in mod kQ. Then Cis exact abelian.

Let us summarize the results obtained so far.

Theorem 3.2.14 Let kbe an algebraically closed filed, Qa finite quiver and kQ its

path algebra.

(i) Let Sbe a set of kQ-modules with S⊆ P or S⊆ Q, where Pand Qdenote

the preprojective and preinjective component of mod kQ respectively. Then Thick(S)

is exact abelian.

(ii) If Qis either Dynkin or Euclidean quiver, then any thick subcategory of

mod kQ is exact abelian.

3.2.1 Classification of thick subcategories

A result by Ingalls and Thomas says that for an Euclidean quiver Q, there exists a

one-to-one correspondence between the non-crossing partitions associated to Qand

the “finitely generated wide subcategories” [IT, Theorem 1.1] of mod kQ. Note that

the “wide subcategories” refer to the exact abelian extension closed subcategories

in our sense, and “finitely generated” means that the subcategory has a projective

generator. As we shall prove, any thick subcategory of kQ has either a projective

generator, or consists of regular modules.

Theorem 3.2.15 Let kbe an algebraically closed field and Qan Euclidean quiver.

Let Cbe a thick subcategory of mod kQ. Then at least one of the following holds:

(i) There exists an exceptional sequence E, such that C= Thick(E).

(ii) Any object in Cis regular.

3.2. Thick subcategories for Euclidean quivers 52

Proof: In the previous section, we showed that any thick subcategory of mod kQ

is exact abelian. In particular, Cis exact abelian, and we can consider its simple

objects. Let SPbe a complete set of simples in C, which are preprojective in mod kQ.

By using the order we defined on P, we know that SPforms an exceptional sequence.

We denote by SRand SQthe set of simples which are regular and preinjective

respectively.

We claim that if SP∪SQ6=∅, then we can make SQ∪SR∪SPinto an exceptional

sequence.

Without loss of generality, we may assume that SP6=∅and P=τ−lPj∈SP,

where Pjis a projective module and l≥0. First we show that in this case, SRis

finite. Note that in Euclidean case, dimk(HomkQ(Pj, R)) −dimk(Ext1

kQ(Pj, R)) >0

for any module Rwhich appears in some homogeneous tube, see [CB2, Lemma 7.2],

since the dimension vector of any such regular module is a multiple of δ, the min-

imal imaginary root associated to Q. But Ext1

kQ(Pj, R) = 0, so we conclude that

HomkQ(Pj, R)6= 0. Since SPand SRare both sets of simples in the category C, then

there are no non-zero morphisms between different elements in these sets. Hence

the elements of SRare from the non-homogeneous tubes. From theorem 3.2.7, we

know that there are finitely many non-homogeneous tubes in Euclidean quiver case,

and since the number of elements in SRfrom one tube is not greater than the rank

of the tube, we conclude that SRis finite.

Next, assume that E={X1, X2,...,Xt}=SR∩ Trfor some non-homogeneous

tube Trof rank r. We show that Eforms an exceptional sequence after some

reordering. Let {T1, T2, . . . , Tr}be the complete set of regular simple modules in Tr,

and again assume that τTi=Ti−1. The indices are taken modulo rand we identify

T0=Tr.

Since the dimension vector of T=⊕r

i=1Tiequals δ, see [CB2, Lemma 9.3],

again by [CB2, Lemma 7.2], we have that HomkQ(Pj, T)6= 0. Therefore there

exists some Tisuch that HomkQ(Pj, Ti)6= 0. Moreover, for any object Xin Tr

which has Tias a composition factor, HomkQ(Pj, X)6= 0. This is equivalent to say

that HomkQ(τ−lPj, X)6= 0 for any object in Trwhich has Ti+las a composition

factor. Since Pand all Xi’s are simples in C, we have HomkQ(P, Xi) = 0 for any

1≤i≤t, which forces that the regular support of {X1, X2,...,Xt}to be contained

in {T1,...,Tr} \ {Ti+l−1}. Now, it is not difficult to show that Eis an exceptional

sequence, since the subcategory Thick({T1,...,Tr}\{Ti+l−1}) is equivalent to the

module category of the quiver of directed Ar−1type.

Since for each non-homogeneous tube Tr, we showed that SR∩ Trforms an

exceptional sequence, it follows that SRforms an exceptional sequence since there

exists no extensions between different tubes. Combined with the fact Ext1

kQ(P,Q) =

Ext1

kQ(P,R) = Ext1

kQ(R,Q) = 0, we have that SQ∪SR∪SPforms an exceptional

3.3. Thick subcategories are exact abelian 53

sequence.

Now, we assume that both SPand SQare empty, and in this case, any objects

Xin Chas a filtration with factors in SRand hence Xis regular. We are done. 2

Remark 3.2.16 (1) Notice that there are thick subcategories of mod kQ, which

consist of regular modules and are generated by exceptional sequences. All these

subcategories are given by direct sums of bounded thick subcategories of non-

homogeneous tubes, which we classified in the previous chapter.

(2) The thick subcategories generated by exceptional sequences coincide with

the so called “finitely generated wide subcategories”, as defined in [IT]. In fact, a

thick subcategory generated by an exceptional sequence is isomorphic to the module

category of some quiver, and hence has a projective generator. Conversely, if a thick

subcategory Cis not generated by any exceptional sequence, then by the last theorem

and proposition 3.2.8, Chas a tube as a direct summand, and clearly a tube has no

finite projective generator. We comment that all these categories refer to unbounded

thick subcategories, which we classified in the previous chapter.

Now, having in mind these remarks, the result of Colin Ingalls and Hugh Thomas

[IT, Theorem 1.1] and theorem 2.2.13, we also obtain the combinatorial classification

of thick subcategories in mod kQ.

Corollary 3.2.17 Let kbe an algebraically closed field, Qan Euclidian quiver and

Ca connected thick subcategory in mod kQ.

(i) If Chas a projective generator, then Ccorresponds to a non-crossing partition

of type Q.

(ii) If Chas no projective generator, then Ccorresponds to a configuration of non-

crossing arcs covering the circle.

3.3 Thick subcategories are exact abelian

We finish this chapter with pointing out a very elegant proof due to Dieter Vossieck,

that any thick subcategory Cof an abelian hereditary category His exact abelian.

Theorem 3.3.1 Let Hbe a hereditary abelian category and C ⊆ H be a thick sub-

category. Then Cis exact abelian.

Proof: Let X, Y be arbitrary objects in Cand fbe a non-zero morphism:

X

f//Y

Ker f

-

yIm f

.

zCoker f.

3.3. Thick subcategories are exact abelian 54

Consider the following short exact sequences:

ψ: 0 →Ker f→Xπ

։Im f→0

ξ: 0 →Im fi

֒→Y→Coker f→0.

Now, apply the functor HomH(Coker f, −) to ψ. Since His hereditary, then the

long exact sequence terminates at Ext2-terms:

· · · → Ext1

H(Coker f, X)Ext1

H(Coker f,π)

−→ Ext1

H(Coker f, Im f)→0.

We get that Ext1

H(Coker f, π) is surjective, and hence there is η∈Ext1

H(Coker f, X)

such that Ext1

H(Coker f, π)(η) = ξ:

η:

Ext1

H(Coker f,π)



0//X//



E//



Coker f//



ξ:0//Im fi//Y//Coker f//0

But then Yis the push-out of Im fπ

ևX ֒→E(see A.4) and therefore the sequence

0→X→Im f⊕E→Y→0 is short exact, see [AS, A.5., Proposition 5.2].

Now, since Cis closed under extensions and direct summands, we have that Im f

is in C. As we already discussed, if Cis closed under arbitrary images, then it is

automatically closed under arbitrary kernels and cokernels. The proof follows. 2

Chapter 4

Exact abelian extension closed

subcategories for tilted algebras

Tilting theory is one of the main tools in the representation theory of finite dimen-

sional algebras. The main idea of the tilting theory is that when the representation

theory of an algebra Ais difficult to study directly, it may be convenient to replace

Awith another simpler algebra Band to reduce the problem on Ato a problem on

B. It is possible to construct a module TA, called a tilting module, which can be

thought of as being close to the Morita progenerator such that, if B= EndA(TA),

then the categories mod Aand mod Bare reasonably close to each other and there

is a natural way to pass from one category to the other.

In this chapter we study exact abelian extension closed categories for tilted al-

gebras. We show that there is a bijection between the exact abelian extension and

torsion closed subcategories of mod A, where Ais a hereditary algebra and the exact

abelian extension closed subcategories of the module category of its tilted algebra

B= EndA(TA).

4.1 Torsion pairs, tilting modules and tilted alge-

bras

In this section, we collect same facts from tilting theory, which we shall use later.

The reference for all facts is [AS, Chapter VI].

Definition 4.1.1 A pair (T,F) of full subcategories of mod Ais called a torsion

pair if the following conditions are satisfied:

(a) HomA(M, N) = 0 for all M∈ T ,N∈ F.

(b) HomA(M, −)|F= 0 implies M∈ T .

4.1. Torsion pairs, tilting modules and tilted algebras 56

Definition 4.1.2 A subfunctor tof the identity functor in mod Ais called an idem-

potent radical if, for every module MA,t(tM) = tM and t(M/tM) = 0.

We recall that a subfunctor of the identity functor on mod Ais a functor t: mod A→

mod Athat assigns to each module Ma submodule tM ⊆Msuch that each ho-

momorphism M→Nrestricts to a homomorphism tM →tN. The following

proposition gives us characterisation of torsion and torsion-free classes.

Proposition 4.1.3 (a) Let Tbe a full subcategory of mod A. The following condi-

tions are equivalent:

(i) Tis a torsion class of some torsion pair (T,F)in mod A.

(ii) Tis closed under images, direct sums, and extensions.

(iii) There exists an idempotent radical tsuch that T={M|tM =M}.

(b) Let Fbe a full subcategory of mod A. The following conditions are equivalent:

(i) Fis a torsion-free class of some torsion pair (T,F)in mod A.

(ii) Fis closed under submodules, direct products, and extensions.

(iii) There exists an idempotent radical tsuch that F={N|tN = 0}.

The idempotent radical tattached to a given torsion pair is called the torsion

radical. It follows from the definition that for any module MA, we have tM ∈ T

and M/tM ∈ F. The uniqueness follows from the next proposition, which also says

that any module can be written in a unique way as the extension of a torsion-free

module by a torsion module.

Proposition 4.1.4 Let (T,F)be a torsion pair in mod Aand Mbe an A-module.

There exists a short exact sequence

0→tM →M→M/tM →0

with tM ∈ T and M/tM ∈ F. This sequence is unique in a sense that, if 0→

M′→M→M′′ →0is exact with M′∈ T and M′′ ∈ F, then the two sequences

are isomorphic.

The short exact sequence 0 →tM →M→M/tM →0 is called the canonical

sequence for M.

Corollary 4.1.5 Every simple module is either torsion or torsion-free.

4.1. Torsion pairs, tilting modules and tilted algebras 57

A torsion pair (T,F) such that each indecomposable A-module lies either in T

or in Fis called splitting. Splitting torsion pairs are characterised as follows.

Proposition 4.1.6 Let (T,F)be a torsion pair in mod A. The following conditions

are equivalent:

(a) (T,F)is splitting.

(b) For each A-module M, the canonical sequence for Msplits.

Next, we recall the definition of a tilting module.

Definition 4.1.7 Let Abe an algebra. An A-module Tis called a partial tilting

module if the following two conditions are satisfied:

(T1) the projective dimension of Tis at most 1.

(T2) Ext1

A(T, T) = 0.

A partial tilting module Tis called a tilting module, if it also satisfies the

following additional condition:

(T3) There exists a short exact sequence 0 →AA→T′

A→T′′

A→0 with T′, T′′ in

add(T).

A tilting module is called basic, if each indecomposable direct summand occurs

exactly once in its direct sum decomposition.

Let Tbe an arbitrary A-module. We define Gen(T) to be the class of all modules

Min mod Agenerated by T, that is, the modules Msuch that there exists an integer

d≥0 and an epimorphism Td→Mof A-modules. Dually, we define Cogen(T) to

be the class of all modules Nin mod Acogenerated by T, that is, the modules N

such that there exist an integer d≥0 and a monomorphism N→Tdof A-modules.

Proposition 4.1.8 Let TAbe a partial tilting module. The following are equivalent:

(a) TAis a tilting module.

(b) Gen(T) = T(T) = {MA|Ext1

A(T, M) = 0}is a torsion class in mod Awith

corresponding torsion-free class Cogen(τT) = F(T) = {MA|HomA(T, M) =

0}.

For a given tilting module, we introduce a new class of algebras.

Definition 4.1.9 Let Abe a finite dimensional, hereditary k-algebra and TAbe a

tilting module. The k-algebra EndA(TA) is called a tilted algebra.

4.1. Torsion pairs, tilting modules and tilted algebras 58

The following proposition tells us what is the effect of any tilting module TAon

mod B.

Proposition 4.1.10 Let Abe an algebra. Any tilting A-module TAinduces a tor-

sion pair X(TA),Y(TA)in the category mod B, where B= EndA(TA)and

X(TA) = {XB|HomB(X, DT) = 0}={XB|X⊗BT= 0},

Y(TA) = {YB|Ext1

B(Y, DT) = 0}={YB|TorB

1(Y, T) = 0}.

The next theorem, known as Brenner-Butler theorem or a tilting theorem, is a

milestone in the tilting theory.

Theorem 4.1.11 (Brenner-Butler) Let Abe an algebra, TAbe a tilting module, B=

EndA(TA), and (T(TA),F(TA)),X(TA),Y(TA)be induced torsion pairs in mod A

and mod B, respectively. Then TAhas the following properties:

(a) BTis a tilting module, and the canonical k-algebra homomorphism A→End(BT)op

defined by a a7→ (t7→ ta)is an isomorphism.

(b) The functors HomA(T, −)and −⊗BTinduce quasi-inverse equivalences between

T(TA)and Y(TA).

A(T, −)and TorB

1(−, T)induce quasi-inverse equivalences be-

tween F(TA)and X(TA).

The following proposition asserts that the composition of any two of the four

functors HomA(T, −), Ext1

A(T, −), − ⊗BTand TorB

1(−, T), which are not quasi-

inverse to each other, vanishes.

Proposition 4.1.12 (a) Let Mbe an arbitrary A-module. Then

(i) TorB

1(HomA(T, M), T) = 0.

(ii) Ext1

A(T, M)⊗BT= 0.

(iii) The canonical sequence of Min (T(TA),F(TA)) is

0→HomA(T, M)⊗BT→M→TorB

1(Ext1

A(T, M), T)→0.

(b) Let Xbe an arbitrary B-module. Then

(i) HomA(T, TorB

1(X, T)) = 0.

(ii) Ext1

A(T, X ⊗BT) = 0.

4.1. Torsion pairs, tilting modules and tilted algebras 59

(iii) The canonical sequence of Xin (X(TA),Y(TA)) is

0→Ext1

A(T, TorB

1(X, T)) →X→HomA(T, X ⊗BT)→0.

We introduce two types of tilting modules.

Definition 4.1.13 Let Abe an algebra, TAbe a tilting module, and B= EndA(TA).

Then

(a) TAis said to be separating if the induced torsion pair (T(TA),F(TA)) in mod A

is splitting, and

(b) TAis said to be splitting if the induced torsion pair (X(TA),Y(TA)) in mod B

is splitting.

The next proposition tells us when a tilting module is separating or splitting.

Proposition 4.1.14 Let Abe an algebra, TAbe a tilting A-module, and B=

EndA(TA)

(a) TAis separating if and only if pd X= 1 for every XB∈ X (TA).

(b) TAis splitting if and only if id N= 1 for every NA∈ F(TA).

We have immediately the following corollary.

Corollary 4.1.15 If Ais hereditary, then every tilting module TAis splitting. If

additionally Bis hereditary, then TAis separating.

We finish this section with a very useful proposition that gives us a relation between

Ext-spaces of mod Aand mod B.

Proposition 4.1.16 Let Abe an algebra, TAbe a tilting module, and B= EndA(TA).

If M∈ T (TA)and N∈ F(TA), then, for any j≥1, there is an isomorphism

Extj

A(M, N)∼

=Extj−1

B(HomA(T, M),Ext1

A(T, N)).

In particular if Ais hereditary, we have

Ext1

A(M, N)∼

=HomB(HomA(T, M),Ext1

A(T, N)).

4.2. Exact abelian extension closed subcategories for tilted algebras 60

4.2 Exact abelian extension closed subcategories

for tilted algebras

As we already discussed in the previous chapters, an exact abelian subcategory is

thick if and only if it is closed under extensions and also a thick subcategory is exact

abelian if and only if it is closed under arbitrary kernels.

In this section, we shall use another characterisation of these two types of cate-

gories. The first proposition is from [Hov], but for completeness we write the proof

here. We always assume that the subcategories we are considering are full additive

and closed under direct summands.

Proposition 4.2.1 A full additive subcategory Cof an abelian category Ais exact

abelian extension closed if and only if for every exact sequence

M1→M2→M3→M4→M5

the object M3is in Cif the objects M1, M2, M4, M5are in C.

Proof: Let C ⊂ A be exact abelian subcategory and

M1→M2→M3→M4→M5

be exact in A. If M1=M5= 0 and M2and M4are in C, then M3is in Csince

Cis exact abelian. Therefore Cis closed under extensions. If M1=M2= 0 and

M4=M5= 0, then Cis closed under kernels and cokernels.

Conversely, let C ⊂ A be closed under extensions, kernels and cokernels and let

M1→M2→M3→M4→M5

be exact with M1, M2, M4, M5∈ C. Then C= Coker(M1→M2) and K=

Ker(M4→M5) are in C. We obtain the following diagram:

M1//M2//

BM3//

BM4//M5

{

Therefore M3is an extension of Cand Kand hence it is in C.2

Immediately from the definition of a thick category, we get the following propo-

sition.

4.2. Exact abelian extension closed subcategories for tilted algebras 61

Proposition 4.2.2 A full additive subcategory Cof an abelian category Ais thick

if and only if for every short exact sequence

0→M1→M2→M3→0

in A, if any two of its (non-zero) terms are in C, then the third one is also in C.

From now on, we assume that Ais a finite dimensional hereditary k-algebra

and we denote by mod Athe category of finite dimensional A-modules. Let TAbe

a basic tilting module, B= EndA(TA) be its tilted algebra and (T(TA),F(TA)),

X(TA),Y(TA) be the induced torsion pairs in mod Aand mod B, respectively. Since

gl.dim A≤1, by corollary 4.1.15, the torsion pair in mod Bis splitting. We set

F= HomA(T, −), F′= Ext1

A(T, −), G=− ⊗BTand G′= TorB

1(−, T).

Definition 4.2.3 A full additive subcategory Cin mod Ais called torsion closed

if M∈ C implies tM ∈ C.

Note that if Cis a thick (or an exact abelian) category, which is torsion closed, then

for any object M∈ C we have M/tM ∈ C.

In this section, we show that there is a bijection between exact abelian exten-

sion and torsion closed subcategories in mod Aand exact abelian extension closed

subcategories in mod B. In two separate lemmas, we prove each of the directions in

the bijection. We denote as before Thick(S) to be the smallest thick subcategory

that contains S, where Sis a set of modules. Also if Aand Bare abelian categories,

C ⊆ A a full subcategory and Fa functor from Ato B, then set F(C) to be the full

subcategory of Bconsisting of objects isomorphic to F(C), for C∈ C.

Lemma 4.2.4 Let Cbe an exact abelian extension and torsion closed subcategory

in mod A. Then the full subcategory

M={M∈mod B|M=M′⊕M′′ , M′∈HomA(T, C), M′′ ∈Ext1

A(T, C)}

in mod Bis exact abelian and extension closed.

Proof: We divide the proof into two steps. First, we show that Mis thick, and then

we show that it is closed under arbitrary kernels. We comment that by definition,

Mis closed under direct summands.

Step 1. Mis thick subcategory in mod B. The torsion pair in mod Bis splitting,

hence any indecomposable object is either torsion or torsion-free. Take an arbitrary

short exact sequence in mod B:

0→Z1→Z2→Z3→0,

4.2. Exact abelian extension closed subcategories for tilted algebras 62

where Zi=Xi⊕Yi,Xi∈ M ∩ X (TA), Yi∈ M ∩ Y(TA). We show that if any two of

its terms are in M, then the third one is also in M, and then by proposition 4.2.2,

the claim shall follow. We apply the functor G=− ⊗BTto the above sequence,

and get the following exact sequence in mod A:

0→G(X1)→G(X2)→G(X3)→G′(Y1)f

→G′(Y2)g

→G′(Y3)→0.

If, say Z1, Z2are in M, then G(X1), G(X2), G′(Y1), G′(Y2) are in Cand since the

latter is exact abelian extension closed, from proposition 4.2.1 we get the exact

sequence

G(X1)→G(X2)→G(X3)→G′(Y1)→G′(Y2),

and we conclude that G(X3)∈ C. Having in mind that Cis closed under kernels

and images, then Ker fand G′(Y2)/Ker f∼

=G′(Y3) are in C. We conclude that Z3

is in M. The other cases are treated in the same way. This gives an argument for

Mto be thick.

Step 2. We prove that if Z1, Z2are arbitrary objects in Mand f:Z1→Z2a

non-zero morphism between them, then Ker fis in M. We show that we can reduce

the proof to one the following cases:

Case 1. Ker f∈ M, where f:X1→X2and X1, X2∈ M are torsion objects.

Case 2. Ker f∈ M, where f:Y1→Y2and Y1, Y2∈ M are torsion-free objects.

Case 3. Ker f∈ M, where f:Y1→X1and Y1, X1∈ M are torsion-free and

torsion objects.

To see that, we use a similar argument as in proposition 3.2.12 of the previous

chapter. We use induction on d= dim(Z1) + dim(Z2), where the dimension is over

k. Clearly, the assertion holds when d= 1. Now, assume that the assertion is true

for any morphism f′:Z′→Z′′ with dim(Z′) + dim(Z′′)< d,Z′, Z′′ ∈ M. Since

mod Bis splitting, we write Zi=Xi⊕Yi(i= 1,2) with Xi∈ X (TA) and Yi∈ Y(TA)

non-zero and X1⊕Y1

→X2⊕Y2, where f=f11

f21

f22 (f12 = HomB(X1, Y2) = 0).

Then we have the following commutative diagram:

0//X1

(1,0) //

f11



X1⊕Y1

1)//



Y1//



0//Z2

Id //Z2//0//0.

Applying the snake lemma(see A.4), we get the exact sequence

0→Ker f11 →Ker f→Y1→Coker f11 →Coker f→0.

From dim(X1)<dim(Z1), follows that Coker f11 ∈Thick(X1, Z2)⊆Thick(Z1, Z2).

Since Coker f= Coker(Y1→Coker f11), combined with dim(Y1)<dim(Z1) and

4.2. Exact abelian extension closed subcategories for tilted algebras 63

dim(Coker f11)≤dim(Z2), we get Coker f∈Thick(Y1,Coker f11)⊆Thick(Z1, Z2),

and hence Ker f∈Thick(Z1, Z2). Since Coker f11 ∈Thick(Z1, Z2)⇔Ker f11 ∈

Thick(Z1, Z2), by the inductional hypothesis we have that Ker f11 ∈ M implies

Ker f∈ M. Having in mind that f11 :X1→X2is a morphism between torsion-free

modules, we land to Case 1.

Now, if happens that Z1=Y1is a torsion-free module, then we consider the

following diagram:

0//0//



Id //

(f21,f22)



Y1//

f22



0//X2

(1,0) //X2⊕Y2

1)//Y2//0

Then, as we did above, we apply the snake lemma, and by induction we conclude

that Coker f22 ∈ M(⇔Ker f22 ∈ M) implies Coker f∈ M(⇔Ker f∈ M). Then

we land to Case 2, since f22 :Y1→Y2is a morphism between torsion-free modules.

The last possible case is when Z1=Y1is a torsion-free module and Z2=X1is

a torsion module.

Case 1. Let Z1, Z2are arbitrary torsion objects in M. For convenience, we

write Z1=X1and Z2=X2. From the definition of Mfollows, that M ∩ X (TA) =

Ext1

A(T, C) = Ext1

A(T, C ∩ F(TA)). Then an arbitrary morphism 0 6=f:X1→X2

is of the form

X1= Ext1

A(T, C1)f=Ext1(T,f′)

−→ Ext1

A(T, C2) = X2,

where f′:C1→C2, and C1, C2are in C ∩ F(TA). We show that Ker f∈ M.

Consider the following diagram in mod A:

C1π′

f′=i′◦π′

//C2

Ker f′

-

;;

xIm f′

-

i′<<

yCoker f′.

Since Cis an abelian subcategory, we have that Ker f′,Im f′and Coker f′are in C,

and hence Ext1

A(T, Ker f′), Ext1

A(T, Im f′),Ext1

A(T, Coker f′) and HomA(T, Coker f′)

are in M. Moreover, since C1, C2are torsion-free objects in mod A, then Ker f′≤C1

and Im f′≤C2are also torsion-free. Now, we apply the functor F= HomA(T, −)

to the two short exact sequences above

0→Im f′i′

→C2→Coker f′→0 (mod A)

0→F(Coker f′)→F′(Im f′)F′(i′)

→F′(C2)→F′(Coker f′)→0 (mod B)

0→Ker f′→C1

π′

→Im f′→0 (mod A)

0→F′(Ker f′)→F′(C1)F′(π′)

−→ F′(Im f′)→0 (mod B)

(F′= Ext1

A(T, −)) and transfer to mod B. We obtain the following diagram:

4.2. Exact abelian extension closed subcategories for tilted algebras 64

X1=F′(C1)

F′(π′)

f=F′(f′)//F′(C2) = X2

F′(Im f′).

F′(i′)77

We comment that in general F′(i′) is not a monomorphism. Now f=F′(f′) =

F′(i′◦π′) = F′(i′)◦F′(π′). We have that Ker F′(π) is in M, since F′(π′) is an

epimorphism and Mis closed under kernels of epimorphisms. From the second

short exact sequence, we get Ker F′(i′)∼

=F(Coker f′)∈ M, hence Ker F′(i′) is in

M. Applying lemma A.4.2 to X1

F′(π′)

։F′(Im f′)F′(i′)

→X2, we get:

0→Ker F′(π′)→Ker f→Ker F′(i′)→Coker F′(π′)→Coker f→Coker F′(i′)→0.

Now, since Coker F′(π′) = 0, and Mis closed under extensions, we have that Ker f

is in M.

Case 2. Let Z1, Z2∈ Y(TA). The case is analogous to the first case.

Case 3. Let Z1∈ X(TA) and Z2∈ Y(TA). Write Z1=X1and Z2=Y1and

denote by F1=G′(X1) and T1=G(Y1). Since Ais hereditary, by proposition 4.1.16

we have Ext1

A(T1, F1)∼

=HomB(Y1, X1). Let η∈Ext1

A(T1, F1) be the extension that

corresponds to the morphism f:Y1→X1. We apply the functor Fto η

η: 0 →F1→M→T1→0 (mod A)

0→F(M)→Y1

→X1→F′(M)→0 (mod B)

and transfer to mod B. Since Cis closed under extensions and F1, T1∈ C, then

M∈ C and hence F(M)∼

=Ker f∈ M. This finishes the proof in that case as well

as of the lemma. 2

The next lemma deals with the reverse direction.

Lemma 4.2.5 Let Mbe an exact abelian extension closed subcategory in mod B.

Then the full subcategory

C={M∈mod A|HomA(T, M)∈ M and Ext1

A(T, M)∈ M}

in mod Ais exact abelian extension and torsion closed.

Proof: We recall that for an arbitrary module Min mod A, we have HomA(T, M) =

HomA(T, tM) and Ext1

A(T, M) = Ext1

A(T, M/tM), hence we can write

C={M∈mod A|HomA(T, tM)∈ M and Ext1

A(T, M/tM)∈ M}.

First, we show that Cis torsion closed. We have that M∈ C if and only if

F(tM)∈ M and F′(M/tM)∈ M. Now, t(tM) = tM and tM/t(tM) = 0, hence

F(t(tM)) = F(tM)∈ M and F′(tM/t(tM)) = 0, which implies that tM ∈ C.

4.2. Exact abelian extension closed subcategories for tilted algebras 65

Next we show that Cis thick. We notice that by definition Cis closed under

direct summands. Now, an arbitrary short exact sequence in mod Ais of the form:

0→Z1→Z2→Z3→0,

where Zi=Ti⊕Ri⊕Fi(i= 1,2,3), Ti∈ T (TA), Fi∈ F(TA) and Riis neither

torsion nor torsion-free. We use the functor Fand get the following exact sequence

in mod B:

0→F(T1)⊕F(tR1)→F(T2)⊕F(tR2)→F(T3)⊕F(tR3)→F′(F1)⊕

F′(R1/tR1)f∗

→F′(F2)⊕F′(R2/tR2)→F′(F3)⊕F′(R3/tR3)→0.

By proposition 4.2.2, we have three cases to consider, but since they are treated

the same, we give the details for only one of the cases, namely assume that Z1, Z2∈

C. We show that Z3∈ C. We have that F(Ti)⊕F(tRi) and F′(Fi)⊕F′(Ri/tRi)

(i= 1,2) are in M. If we consider the first five non-zero terms of the exact sequence

above, then by proposition 4.2.1, we have that F(T3)⊕F(tR3)∈ M. Since Mis

closed under images and cokernels, we have that Im f∗∈ M, and hence F′(F2)⊕

F′(R2/tR2)/Im f∗∼

=F′(F3)⊕F′(R3/tR3) is also in M. Now, all of the modules

F(T3), F(R3) = F(tR3), F′(R3) = F′(R3/tR3) and F′(F3) are in M, hence T3,R3,

F3and Z3are in C.

To finish the proof, we use theorem 3.3.1. Since C ⊆ mod A, and Ais a hereditary

algebra, then Cis exact abelian. 2

We are now able to prove the main theorem in this chapter.

Theorem 4.2.6 Let Abe a finite dimensional hereditary k-algebra, TAa basic tilt-

ing module and B= EndA(TA). Then the assignments:

7→ M ={M∈mod B|M=M′⊕M′′ , M′∈HomA(T, C), M′′ ∈Ext1

A(T, C)}

7→ C ={M∈mod A|HomA(T, M)∈ M and Ext1

A(T, M)∈ M}

induce mutually inverse bijections between:

•exact abelian extension and torsion closed subcategories in mod A, and

•exact abelian extension closed subcategories in mod B.

Proof: The previous two lemmas showed that the assignments are defined properly.

We verify that (j◦i)(C) = C, for arbitrary exact abelian extension and torsion closed

subcategory C ⊆ mod Aand (i◦j)(M) = M, for arbitrary exact abelian extension

closed subcategory M ⊆ mod B, and the claim shall follow.

(1) (j◦i)(C) = C.

”⊇” Take C∈ C arbitrary. Then HomA(T, C)∈HomA(T, C) = HomA(T, C ∩

T(TA)) = M ∩ Y(TA)⊆ M =i(C). In the same way Ext1

A(T, C)∈ M =i(C).

Then C∈(j◦i)(C).

4.2. Exact abelian extension closed subcategories for tilted algebras 66

”⊆” Take C∈(j◦i)(C). Then both HomA(T, C) and Ext1

A(T, C) are in i(C) =

M. Note that by construction (j◦i)(C) is torsion closed, hence both tC and C/tC

are in (j◦i)(C). We have that HomA(T, C) = HomA(T, tC)∈ M ∩ Y(TA) =

HomA(T, C) = HomA(T, C ∩ T (TA)) and therefore by theorem 4.1.11(b) tC ∈ C ∩

T(TA). In the same way we have that C/tC ∈ C ∩ F(TA). Taking the canonical

sequence for C, and having in mind that Cis extension closed, we have that C∈ C.

(2) (i◦j)(M) = M. Since the torsion pair (X(TA),Y(TA)) in mod Bis splitting,

any M∈mod Bis of the form M=M1⊕M2, where M1is a torsion-free module

and M2is a torsion module.

”⊇” Take an arbitrary object M∈ M. Then there are objects X, Y ∈mod A

such that HomA(T, tX) = M1and Ext1

A(T, Y/tY ) = M2. But then HomA(T, tX)∈

HomA(T, j(M)) = {HomA(T, Z)|Z∈mod A, such that HomA(T, Z)∈ M and

Ext1

A(T, Z)∈ M}, since Ext1

A(T, tX) = 0, that is, M1∈(i◦j)(M). Similarly,

Ext1

A(T, Y/tY )∈Ext1

A(T, j(M)), that is, M2∈(i◦j)(M). Hence M∈(i◦j)(M).

”⊆” Take an arbitrary object Min (i◦j)(M)⊆mod B. Then M=M1⊕M2,

where M1= HomA(T, X) and M2= Ext1

A(T, Y ), for X, Y ∈j(M)⊆mod A. By

definition of j(M), both HomA(T, X) and Ext1

A(T, Y ) are in M, and hence M1and

M2are in M.2

Before we give a corollary of the theorem, we need to recall some facts from the

theory of quiver representations.

Let Q= (Q0, Q1, s, t) be a finite, connected, and acyclic quiver and let n=|Q0|.

For every point a∈Q0, we define a new quiver σaQ= (Q′

0, Q′

1, s′, t′) as follows: All

the arrows of Qhaving aas a source or as target are reversed, all others arrows

remain unchanged. An admissible sequence of sinks in a quiver Qis defined to

be a total ordering (a1, . . . , an) of all points in Qsuch that:

(i) a1is a sink in Q, and

(ii) aiis a sink in σai−1. . . σa1Qfor every 2 ≤i≤n.

We have the following proposition.

Proposition 4.2.7 [AS, Chapter VII.5, Proposition 5.2] Let Qand Q′be two trees

having the same underlying graph. There exists a sequence i1,...,itof points of Q

such that σit. . .σi1Q=Q′.

Let Abe a finite dimensional hereditary k-algebra, which we assume that is

not simple. There exists an algebra isomorphism A∼

=kQA, where QAis a finite,

connected, and acyclic quiver. Then there exists a sink a∈(QA)0that is not a

source, so that the simple A-module S(a)Ais projective and non-injective. Consider

4.2. Exact abelian extension closed subcategories for tilted algebras 67

the following module in mod A

T[a]A=τ−1S(a)⊕(M

b6=a

P(b)).

It is not difficult to check that it is a tilting module, see [AS, Chapter VI.2.8], which

is called APR-tilting.

We may ask whether there a connection between path algebras kQAand k(σaQA),

where σais a reflection at the sink afor the quiver QA. The following proposition

gives an answer to that question.

Proposition 4.2.8 [AS, Chapter VII.5, Proposition 5.3] Let Abe a basic heredi-

tary and non-simple algebra, abe a sink in its quiver QA, and T[a]be the APR-tilting

A-module at a. Then the algebra B= End T[a]Ais isomorphic to k(σaQA).

Now, we prove the following proposition.

Proposition 4.2.9 Let Qbe a finite acyclic quiver, abe a sink and σaQbe the

reflected at aquiver Q. Then there is a bijection between exact abelian extension

closed subcategories in mod kQ and mod k(σaQ).

Proof: By theorem 4.2.6, we have a bijection between exact abelian extension and

torsion closed subcategories in mod kQ and exact abelian extension closed subcate-

gories in mod k(σaQ). But since k(σaQ) is hereditary, then the APR-tilting module

is separating. Then the torsion pair (T(TA), F(TA)) in mod kQ splits and hence

every exact abelian extension closed subcategory in mod kQ is also torsion closed.2

Corollary 4.2.10 Let Qand Q′be a finite acyclic quivers, having the same under-

lying graph but with different orientations. Then there is a bijection between exact

abelian extension closed subcategories in mod kQ and mod kQ′.

Proof: By proposition 4.2.7, we have that there exists a sequence i1, . . . , itof

points of Qsuch that σit. . . σi1Q=Q′. Set σi0Q=Q. By proposition 4.2.9,

we have that for k= 0, . . . , t −1, there is a bijection between exact abelian ex-

tension closed subcategories of mod k(σik...σi0Q) and mod k(σik+1 . . . σi0Q). Since

mod k(σit. . . σi1Q) = mod kQ′, the proof follows. 2

Remark 4.2.11 We point out that Kristian Br¨uning showed the same result, see

[Br2, Corollary 5.6]. There he established a bijection between thick subcategories of

bounded derived category Db(A) and exact abelian extension closed subcategories in

A, where Ais a hereditary abelian category. Then using the fact that Db(mod kQ)∼

Db(mod kQ′), see [Ha, Proposition 4.5], the claim follows.

Appendix A

Basic and auxiliary results

A.1 Quivers and their representations

The reference for this section is [AS, Chapter II.1, Chapter III.1].

Aquiver Q= (Q0, Q1, s, t) is a quadruple consisting of two sets: Q0(whose

elements are called vertices) and Q1(whose elements are called arrows), and two

maps s, t :Q1→Q0which associate to each arrow α∈Q1its source s(α)∈Q0and

its target t(α)∈Q0respectively.

We denote a quiver Q= (Q0, Q1, s, t) simply by Q. A subquiver of a quiver

Q= (Q0, Q1, s, t) is a quiver Q′= (Q′

0, Q′

1, s′, t′) such that Q′

0⊆Q0,Q′

1⊆Q1and

the restrictions sQ′

1,tQ′

1of s, t to Q′

1are respectively equal to s′, t′. Such a subquiver

is called full if Q′

1equals the set of all those arrows in Q1whose source and target

both belong to Q′

Example A.1.1 The quiver

∆n: 1 //2//3//... //n

is a subquiver of the quiver

∆n: 1 //2//3//... //n

kk,

and a full subquiver of the quiver

∆n+1 : 1 //2//... //n//n+ 1 .

A quiver Qis said to be finite if Q0and Q1are finite sets. The quiver Qis said

to be connected if its underlying graph is connected.

Let Q= (Q0, Q1, s, t) be a quiver and a, b ∈Q0. A path of length ℓ≥1 with

source aand target bis a sequence

(a|α1, α2,...,αℓ|b),

A.1. Quivers and their representations 69

where αk∈Q1for all 1 ≤k≤ℓ, and we have s(α1) = a,t(αk) = s(αk+1) for each

1≤k≤ℓ, and finally t(αℓ) = b. To each point a∈Qa path of length ℓ= 0 is

called the trivial path at aand it is denoted by ǫa. Thus, the paths of lengths 0

and 1 are in bijective correspondence with the elements of Q0and Q1, respectively.

A path of length ℓ≥1 is called cycle whenever its source and target coincide. A

quiver is called acyclic if it contains no cycles.

Let Qbe a quiver. The path algebra kQ of Qis the k-algebra whose underlying

k-vector space has as its basis the set of all paths (a|α1, α2,...,αℓ|b) of length

ℓ≥0 in Qand such that the product of two basis vectors (a|α1, α2,...,αℓ|b) and

(c|β1, β2,...,βk|d) of kQ is defined by

(a|α1, α2,...,αℓ|b)(c|β1, β2,...,βk|d) = δbc(a|α1, α2,...,αℓ, β1, β2,...,βk|d),

where δbc is the Kronecker delta:

δbc =





0 if t(αℓ)6=s(β1)

1 if t(αℓ) = s(β1).

In other words, the product of two paths α1, α2,...,αℓand β1, β2,...,βkis equal

to zero if t(αℓ)6=s(β1) and is equal to the composed path α1, α2,...,αℓβ1, β2,...,βk

if t(αℓ) = s(β1). The product of basis elements is then extended to arbitrary elements

of kQ by distributivity.

Lemma A.1.2 Let Qbe a quiver and kQ be its path algebra. Then

(a) kQ is an associative algebra;

(b) kQ has an identity element if and only if Q0is finite;

We point out that for a given path algebra kQ, there is a direct sum decompo-

sition

kQ =kQ0⊕kQ1⊕kQ2⊕ · · · ⊕ kQℓ...

of the k-vector space kQ, where, for each ℓ≥0, kQℓis the subspace of kQ generated

by the set Qℓof all paths of length ℓ. It is easy to see that (kQn)(kQm)⊆kQn+m

for all n, m ≥0, which shows that kQ is a graded algebra.

Definition A.1.3 Let Qbe a finite and connected quiver. The two-sided ideal of

the path algebra kQ generated (as an ideal) by the arrows of Qis called the arrow

ideal of kQ and is denoted by RQ.

A.1. Quivers and their representations 70

Note that there is a direct sum decomposition

RQ=kQ1⊕kQ2⊕ · · · ⊕ kQℓ⊕ · · ·

of the k-vector space RQ, where kQℓis the subspace of kQ generated by the set

Qℓof all paths of length ℓ. In particular, the underlying k-vector space of RQis

generated by all paths in Qof length ℓ≥1. This implies that, for each ℓ≥1,

Rℓ

Q=M

m≥ℓ

kQm

and therefore Rℓ

Qis the ideal of kQ generated, as a k-vector space, by the set of all

paths of length ≥ℓ.

Definition A.1.4 Let Qbe a finite quiver. A representation Mof Qis defined

by the following data:

(1) To each point ain Q0is associated a k-vector space Ma;

(2) To each arrow α:a→bin Q1is associated a k-linear map φα:Ma→Mb.

Such a representation is denoted as M= (Ma, φa)a∈Q0,α∈Q1, or simply M= (Ma, φα).

It is called finite dimensional if each vector space Mais finite dimensional.

Let M= (Ma, φα) and M′= (M′

a, φ′

α) be two representations of Q. A rep-

resentation morphism f:M→M′is a family f= (fa)a∈Q0of k-linear maps

(fa:Ma→M′

a)a∈Q0that are compatible with the structure maps φαthat is, for

each arrow α:a→b, we have φ′

afa=fbφαor equivalently, the following square is

commutative:

φα//



M′

φ′

α//M′

Let f:M→M′and g:M′→M′′ be two morphisms of representations of Q,

where f= (fa)a∈Q0and g= (ga)a∈Q0. Their composition is defined to be the family

gf = (gafa)a∈Q0. Then gf is easily seen to be a morphism from Mto M′′. We have

defined a category Repk(Q) of k-linear representations of Q. We denote by repk(Q)

the full subcategory of rep(Q) consisting of the finite dimensional representations.

Lemma A.1.5 Let Qbe a finite quiver. Then Repk(Q)and repk(Q)are abelian

k-categories.

A.2. Hereditary algebras 71

A.2 Hereditary algebras

In the whole thesis, we considered hereditary algebras and their module categories.

Here we recall what a hereditary algebra is, and we point out the connection with

the quiver representations. The reference is [AS, Chapter VIII]. In this section kis

an algebraically closed field.

Definition A.2.1 Let Abe a finite dimensional k-algebra. The following assertions

are equivalent:

•Ais hereditary;

•Submodules of projective modules are projective;

•The global dimension of Ais at most one;

•Exti

A(M, N) = 0 for all A-modules Mand Nand for all i≥2.

Example A.2.2 If Qis a finite, connected, and acyclic quiver, then the algebra

A=kQ is hereditary.

The next theorem relates modules and representations.

Theorem A.2.3 Let Abe a basic and connected finite dimensional hereditary k-

algebra. There exists a finite and acyclic quiver QAsuch that

Mod A∼

−→ Repk(Q)

is k-linear equivalence of categories, that restricts to an equivalence of categories

mod A∼

−→ repk(Q).

Definition A.2.4 A finite dimensional k-algebra Ais said to be representation-

finite (or an algebra of finite representation type) if the number of the iso-

morphism classes of indecomposable finite dimensional right A-modules is finite. A

k-algebra Ais called representation-infinite (or an algebra of infinite repre-

sentation type) if Ais not representation-finite.

By a result of Gabriel, representation-finite hereditary algebras are classified.

Theorem A.2.5 (Gabriel) Let Qbe a finite, connected, and acyclic quiver; kbe

an algebraically closed field; and A=kQ be the path k-algebra of Q. The algebra

Ais representation finite if and only if the underlying graph Qof Qis one of the

Dynkin diagrams An,Dn, with n≥4,E6,E7, and E8.

A.3. The Auslander-Reiten quiver 72

An:•••... • • (n≥1)

•

Dn:•••... •(n≥4)

•



•

E6:•••••

•

E7:••••••

•

E8:•••••••

A.3 The Auslander-Reiten quiver

In this section, we turn to the structure theory of the module category. Fix a finite

dimensional hereditary k-algebra A. There is a special quiver, called Auslander-

Reiten quiver, that combinatorially encodes the building blocks of mod A, namely

the indecomposable modules and the irreducible morphisms.

First, we recall he following fundamental theorem, that reduces the study of

modules to indecomposable modules.

Theorem A.3.1 (Krull-Schmidt) Let Abe a finite dimensional k-algebra. For

a finitely generated A-module Mthere are indecomposable A-modules M1,...,Mn

such that M∼

=⊕n

i=1Mi. Furthermore, the modules M1,...,Mnare unique up to

permutation.

Definition A.3.2 Let Abe a finite dimensional k-algebra. A morphism of A-

modules f:M→Nis an irreducible morphism, if

(i) fis neither a section nor a retraction, and

(ii) if f=f1◦f2, then either f1is a retraction or f2is a section.

Denote by Irr(M, N) the k-vector space of irreducible morphisms from Mto N.

As for objects the study of morphisms is reduced to the study of irreducible ones.

A.3. The Auslander-Reiten quiver 73

Theorem A.3.3 Let Abe a finite dimensional k-algebra of a finite representation

type. Every morphism between finitely presented indecomposable A-modules that is

not invertible is a finite sum of finite compositions of irreducible maps.

Definition A.3.4 The Auslander-Reiten quiver (AR-quiver) Γ(A) of the algebra

Ahas as vertices the isomorphism classes of indecomposable modules. The arrows

from [M] to [N] correspond bijectively to a k-basis of the vector space of irreducible

maps Irr(M, N). The quiver Γ(A) is locally finite in the sense that every vertex

has only finitely many neighbors. The Auslander-Reiten quiver is equipped with an

extra structure: the translate. It is a bijective map

τ: Γ(A)\Proj(A)→Γ(A)\Inj(A),

where Proj(A) and Inj(A) denote the sets of isomorphism classes of indecomposable

projective and injective modules, respectively.

The following notion is central for the structure of the AR-quiver.

Definition A.3.5 A short exact sequence

0→L→M→N→0

is called almost split or an Auslander-Reiten sequence (AR-sequence), if L

and Nare indecomposable and the maps L→Mand M→Nare irreducible. The

following theorem describes the relation between an indecomposable module Nand

its translate τN.

Theorem A.3.6 Let Abe a finite dimensional algebra over an algebraically closed

field k. For every indecomposable non-projective A-module Nthere is an AR-

sequence

0→τN →

i=1

Mni

i→N→0

in which ni≥0and the modules Miare pairwise non-isomorphic indecomposable.

Furthermore, ni= dimkIrr(Mi, N) = dimkIrr(τN, Mi).

We finish this section with the following important formula that expresses the

translate homologically.

Theorem A.3.7 (Auslander-Reiten formulas) Let Abe a hereditary algebra,

and M, N be A-modules. There exist functorial isomorphisms

τM ∼

=DExt1

A(M, A)and τ−1M∼

=Ext1

A(DM, A).

Moreover,

Ext1

A(M, N)∼

=DHomA(N, τM)and Ext1

A(M, N)∼

=DHomA(τ−1N, M).

A.4. Two homological facts 74

We comment that these formulas are valid in a larger context. We refer to [Kr] for

a very elegant proof of these formulas.

A.4 Two homological facts

In the last section, we recall two standard facts from the homological algebra. In

this section Ris an associative ring.

Lemma A.4.1 (Snake Lemma) Consider a commutative diagram of R-modules

of the form

A//



B//



C//



0//A′//B′//C′.

If the rows are exact, there is an exact sequence

Ker f→Ker g→Ker h→Coker f→Coker g→Coker h.

Moreover, if A→Bis a monomorphism, then so is Ker f→Ker g, and if B′→C′

is an epimorphism, then so is Coker g→Coker h.

The proof can be found in [Wb, Chapter 1].

Corollary A.4.2 If we have maps Aψ

→Bφ

→Cof R-modules, then there is an

exact sequence

0→Ker ψ→Ker φψ →Ker φ→Coker ψ→Coker φψ →Coker φ→0.

Proof: Applying the snake lemma to the following commutative diagram

0//



AId //



A//

φψ



0//Ker φ//Bφ//Im φ,

we get 0 →Ker ψ→Ker φψ →Ker φ→Coker ψ→Im φ/ Im φψ →0, since B→

Im φis an epimorphism. Then using the third isomorphism theorem for the modules

Im φψ ≤Im φ≤C, we obtain 0 →Im φ/ Im φψ →C/ Im φψ →C/ Im φ→0, and

gluing with the above sequence, the claim follows. 2

Proposition A.4.3 [ARS, Chapter 1, Proposition 2.6] Let

Af//

f′



B′g′//C

be a commutative diagram of morphisms between R-modules.

A.4. Two homological facts 75

(a) The following are equivalent

(i) The diagram is a push-out diagram;

(ii) The induced sequence A(f

−f′)

→B`B′(g,g′)

→C→0is exact;

(iii) In the induced exact commutative diagram

Af//

f′



B//



Coker f//



B′g′

//C//Coker g′//0

his an isomorphism.

(b) The following are equivalent

(i) The diagram is a pull-back diagram;

(ii) The induced sequence 0→A(f

−f′)

→B`B′(g,g′)

→Cis exact;

(iii) In the induced exact commutative diagram

0//Ker f//



Af//

f′



0//Ker g′//B′g′

//C

his an isomorphism.

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