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London Mathematical Society Lecture Note Series. 119
Triangulated Categories in the Representation Theory of Finite Dimensional Algebras Dieter Happel University of Bielefeld
i1 11
r11 1 :
V
J
The right of the University of C-1-bridge to print and ,ell all manner f hook, m-granted by Henry V3// in 134. The University has printed and published tominuou,ly sime 1584.
CAMBRIDGE UNIVERSITY PRESS Cambridge
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© Cambridge University Press 1988 First published 1988 Library of Congress cataloging in publication data
Happel, Dieter, 1953 Triangulated categories in the representation theory of finite dimensional algebras. (London Mathematical Society lecture note series; 119) Bibliography: p. Includes index
1. Categories (Mathematics) 2. Representations of algebras 3. Modules (Algebra) I. Title II. Series QA169.H36 1988
512'.55
87- 31100
British Library cataloguing in publication data
Happel, Dieter Triangulated categories in the representation theory of finite dimensional algebras. - (London Mathematical Society lecture note series, ISSN 0076-0552; 119). 1. Categories (Mathematics) 1. Title II. Series 512'.55 QA 169
ISBN 0 521 33922 7 Transferred to digital printing 2002
TABLE OF CONTENTS
Preface
CHAPTER
I:
Triangulated categories 1. 2. 3. 4. 5.
CHAPTER
CHAPTER
II:
III:
CHAPTER
IV:
V:
1
10
24 31
43
Repetitive algebras
57
1. 2. 3. 4. 5.
57 59 70 74 89
t-categories Repetitive algebras Generating subcategories The main theorem Examples
Tilting theory 1. 2. 3. 4. 5. 6. 7.
CHAPTER
Foundations Frobenius categories Examples Auslander-Reiten triangles Description of some derived categories
1
Grothendieck groups of triangulated categories The invariance property The Brenner-Butler Theorem Torsion theories Tilted algebras Partial tilting modules Concealed algebras
93
95 103 113 118 133 141
149
Piecewise hereditary algebras
152
1. 2. 3. 4. 5. 6. 7.
152 160 164
Piecewise hereditary algebras Cycles in mod k13 The representation-finite case Iterated tilted algebras The general case The Dynkin case The affine case
171 175
180 188
Trivial extension algebras
196
1. Preliminaries 2. The representation-finite case 3. The representation-infinite case
196 199 201
References
203
Index
207
PREFACE
The aim of these notes is to show that the concept of triangulated categories might be useful for studying modules over finitedimensional k-algebras, where k is a field which for technical reasons usually will be assumed to be algebraically closed. On the other hand we try to show that certain triangulated categories become quite accessible if one applies methods of representation theory emerged in recent years. Let A be an abelian category, then the most famous example of bounded of a triangulated category is the derived category Db(A) complexes over A. Usually we will be interested in the case that A is the category mod A of finitely generated left modules over a finitedimensional k-algebra. In this case the category Db(A) = Db(mod A) may be identified with the homotopy category K-,b (AP) of complexes bounded above over the finitely generated projective left A-modules having bounded cohomology. Since their introduction in the sixties they have turned out quite useful in algebraic geometry and homological algebra. Examples for this can be found in duality theory (Hartshorne (1966), Iversen (1986)) or in the fundamental work on perverse sheaves by Bernstein, Beilinson, Deligne (1986).
The concept of derived categories goes back to suggestions of Grothendieck. We refer to Grothendieck (1986) for some information about the motivations and developments. The formulation in terms of triangulated categories was achieved by Verdier in the sixties and an account of this is published in Verdier (1977). We will assume that the reader is familiar with the concept of localization which is needed in the construction of the derived category. Information about this may be found for example in Verdier (1977), Hartshorne (1966) or Iversen (1986). In recent years there has been quite some interest in the structure of the category mod A of finitely generated left modules over a finite-dimensional k-algebra A.(Unless stated otherwise the term module always refers to a finitely generated left A-module). We will assume some elementary facts from the general theory which are easily accessible in textbooks such as Anderson, Fuller (1974) or Curtis, Reiner (1962), (1981). For more andvanced results we refer to Ringel (1984). Also we adopt the categorical language of Mac Lane (1971) and use rather standard results from homological algebra for which Cartan, Eilenberg (1956) serves as a reference. There are two classes of finite-dimensional k-algebras which have been studied quite intensively. These are the finite-dimensional
hereditary k-algebras (i.e. submodules of projective modules are projective) and the selfinjective finite-dimensional k-algebras (i.e. the algebra considered as a module is an injective module). It seems that there is a strong relationship between these two classes of algebras. Some evidence for this is contained in chapter V. We now turn to a description of the content of the various chapters.
Chapter I serves as an introduction to the theory of triangulated categories. We introduce Frobenius categories and show that the associated stable categories admit a triangulated structure. It is customary in representation theory to associate a quiver (directed graph) 1I (a) to an additive category a. In a lot of cases this quiver has the additional structure of a translation quiver. A celebrated example is the case of a = mod A. Motivated by the work of Auslander and Reiten we introduce Auslander-Reiten triangles in a triangulated category and show that Db(A) has Auslander-Reiten triangles in case A has finite global dimension. This result endows 1r(Db(A)) with the structure of a translation quiver. If A is a hereditary finite dimensional k-algebra this quiver is calculated explicitly. In chapter II we show that the derived category Db(A) for a finite-dimensional k-algebra of finite global dimension can be interpreted in quite some different way. There is a classical construction to associate with a given finite-dimensional k-algebra A a finite-dimensional selfinjective k-algebra. The injective cogenerator DA = Homk(A,k) admits a natural A-bimodule structure. So we may form the trivial extenis a Z-graded algesion algebra T(A) of A by DA. The algebra T(A) bra and the category modZT(A) of finitely generated Z-graded T(A)modules with morphisms of degree zero is a Frobenius category. (For different descriptions of modZT(A) we refer to chapter 11.2). In particular we obtain that the stable category modT(A) is a triangulated category. The main theorem now asserts that there exists a full and faithful functor F Db(A) - modET(A) preserving the triangulated structure which is dense if A has finite global dimension. :
The third chapter gives a self-contained approach to tilting theory. Some of the history is recalled in remarks preceding this chapter. It is independent of chapter II and most of chapter I. We have included most of the theoretical results available. In section 5 some of the results are stated without proof in order to avoid some overlap with Ringel (1984). For related results, not treated explicitly, we have included references and some comments. We regret any incompleteness in those. In chapter IV we concentrate on a rather special class of finite-dimensional k-algebras. We assume that their derived categories have a predescribed form, namely they are equivalent to the derived category of a finite-dimensional hereditary k-algebra. This has a lot of consequences. For example a representation-finite (there are up to isomorphism only finitely many indecomposable modules) algebra in this class has the property that the indecomposable modules are uniquely (up to isomorphism) determined by their composition factors. The main result of this chapter asserts that such an algebra is already tiltable (see IV.4 for a definition) to a hereditary finite-dimensional k-algebra.
ix
The final chapter relates the results from 11.4 and IV.5 to give a description of the representation-finite trivial extension algebras.
With a few the needed results are exceptions are usually hope that this and the for further studies in
exceptions (for example 11.5, 111.5, IV.6, V.2) contained in the references given above. These not essential in the subsequent developments. We additional references will stimulate the reader this area.
It is my pleasure to thank I. Assem, S. Brenner, C.M. Ringel and L. Unger for animating discussions on various aspects of the theory presented here. My special thanks go to Mrs Kdllner for her splendid job in typing the manuscript. Also I thank Cambridge University Press for publishing it in this series and for checking the English.
CHAPTER
TRIANGULATED CATEGORIES
I
Foundations
1.
The basic reference for triangulated categories and derived categories is the original article of Verdier (1977). Also Hartshorne (1966), Beilinson, Bernstein and Deligne (1982)
and Iversen (1986) give
introductions to these concepts.
Let us remind the reader of our convention that the composition of morphisms
denoted by
f
: X -+ Y
and
g
: Y - Z
K
in a given category
is
fg. Unless otherwise stated, we adopt the categorical language
of Mac Lane (1971). In particular, our additive categories have finite direct sums.
Let
1.1
C
C. The automorphism
of
inverse of
T
be an additive category and
an automorphism
is usually called the translation functor. The
T
is denoted by
given by objects
T
T . A sextuple
X,Y,Z E C
and morphisms
(X,Y,Z,u,v,w)
in
u : X - Y, v : Y -, Z
is
C
and
w : Z -. TX. A more suggestive notation of sextuples is: X
u
(The following notation for a sextuple is also used sometimes in the literature: Z
W
X
u Y
2
w
But keep in mind that
is a morphism from
to
Z
TX. We will not use
this notation, except in the situation when we explain the terminology of the axiom
(TR4)
of a triangulated category). (X,Y,Z,u,v,w)
A morphism of sextuples from (X',Y',Z',u',v',w')
is a triple
to
of morphisms such that the
(f,g,h)
following diagram commutes:
Xu-Y f1
g
,
0 Z w TX
v v,
l
hl
U,
X'
Tfl
-Z' w-TJXL'
YI*
If in this situation
,
f,g
h
and
are isomorphisms in
C,
the morphism is then called an isomorphism. T
A set
of sextuples in
is called a triangulation of
C
if the following conditions are satisfied. The elements of
T
C
are then
called triangles.
Every sextuple isomorphic to a triangle is a triangle.
(TRI)
Every morphism u
:
X -+ Y
in
(TR2)
(Y,Z,TX,v,w,-Tu)
If
can be embedded into a triangle
(X,X,O,1X,o,o)
(X,Y,Z,u,v,w). The sextuple
the identity morphism from
C
X
to
is a triangle. (1X
X).
is a triangle, then
(X,Y,Z,u,v,w)
is a triangle.
(TR3)
(X',Y',Z',u',v',W')
Given two triangles and morphisms
fu' = ug, there exists a morphism
f
(X,Y,Z,u,v,w) : X -+ X', g
(f,g,h)
and
: Y - Y'
Consider triangles
such that
from the first triangle to the
second. (TR4)
denotes
(The octahedral axiom)
(X,Y,Z',u,i,i'), (Y,Z,X',v,j,j')
and
3
(X,Z,Y',u.v,k,k'). Then there exist morphisms
f
Z' - Y', g
:
:
Y' -+ X'
such that the following diagram commutes and the third row is a triangle.
T Y'
TI u
T g
TX'
-* X
X
T3
uv
,z
v
Y
i
k
z, f i'l
The additive category tor
T
and a triangulation
T
Ti
g
> X'
TZ'
k'
LX
C
TY
1X,
Y'
___
TX
X'
_
TX
together with a translation func-
is called a triangulated category.
A different way of displaying the axiom
is the fol-
(TR4)
lowing: Given triangles as above. Then there exist morphisms g
:
Y' - X'
triangle and
f
:
Z' -+ Y',
such that the hatched face of the following octahedron is a if = vk, fk' = i', kg = j
and
k'Tu = gj'. (Observe that
the last conditions can be expressed by the requirement that a morphism from the first triangle to the third and that morphism from the third triangle to the second).
(1X,v,f)
(u,1Z,g)
is a
is
4
C = (C,T,T), C' _ (C,T',T')
Let
An additive functor
F
:
be triangulated categories.
C -+ C' is called exact if there exists an inver-
tible natural transformation
a
(FX,FY,FZ,Fu,Fv,FwaX)
T' whenever
is in
If an exact functor
such that
FT -, T'F
:
F
:
(X,Y,Z,u,v,w)
C - C'
is in
T.
is an equivalence of cate-
gories, we call it a triangle-equivalence. C
and
are then called
C'
triangle-equivalent. The following lemma is straightforward.
LEMMA.
Let
be a quasi-inverse of
F
:
be a triangle-equivalence and let
C - C'
F. Then
G
is a triangle-equivalence.
An additive functor H : C--4A C
to an abelian category
tor, if whenever
... - H(T1X)
A
from a triangulated category
is called a (covariant) cohomological func-
(X,Y,Z,u,v,w)
H(Tu)
G
is a triangle, the long sequence
H(Tw)
H(Tv)
--; H(T'Y)
H(Ti+1X) -
i H(T1Z)
is exact. The additive functor H
:
cohomological functor, if whenever
...
C -A is called a (contravariant) (X,Y,Z,u,v,w)
is a triangle, the long
sequence
H(Tu)
H(Tv)
... - H(T1+1X) -; H(T'Z) -- H(T1Y)
H(Tw) ---
H(T1(X) -
...
is exact.
1.2
In the following proposition we collect some elementary
properties.
PROPOSITION. (X,Y,Z,u,v,w) (a)
Let
C
be a triangulated category,
be a triangle and M be an object in uv = vw = 0.
C. Then
5
(b)
Hom C(M,-)
(c)
Let
and
(f,g,h)
are cohomological functors.
Hom C(-,M)
be a morphism of triangles from
(X,Y,Z,u,V,W)
(X',Y',Z',u',v',w'). If
to
are isomorphisms, then so is
Proof: uv = o. (TR3)
By
In virtue of
(a)
we know that
(TR2)
f
and
g
h.
it is enough to show that
(TR2)
is a triangle. By
(Y,Z,TX,v,w,-Tu)
the following diagram can be completed to a commutative diagram:
Y -°'Z WTX vI
1Z
-*TY IT
I
4
0 Z Z>Z -0 0 --4TZ
(-Tu)Tv = 0, hence
In particular
We show that
(b)
uv = 0.
is exact. The other assertion
HomC(M,-)
follows dually. It is enough to show that
Hom(M,Tlu)
Hom(M,T1v)
Hom(M,T'X)
is
exact. By
(a)
g E Hom(N,T'Y)
Hom(M,TLY)
--> Hom(M,T'Z)
we know that
0. So let
gTly = 0. By
such that
and
(TRI)
(TR2)
we obtain that
the rows of the following diagram are triangles T
,
0
M IT 1g
1
hTu -
v
T1+lg.
) T i+1M
-1 T_1+14
T i+1M
T-1 gv = 0. By
Hence
T 1+1g
to
Y -- Zi By assumption
0
0
w
(TR3)
(T1-lh)T1u = g.
TX
-Tu
there exists
-a TY
h
such that
6
Consider the following commutative diagram:
(c)
Xu3Y v > Z w--
are isomorphisms.
g
f
We apply the cohomological functor
Hom(Z',-)
commutative diagram which has exact rows by
and obtain the following (b).
s Hom(Z',Y) - Hom(Z',Z) -+ Hom(Z',TX)
Hom(Z',X) Hom(Z',f)
Hom(Z',g)
Hom(Z',h)
' Hom(Z',TY)
Hom(Z',Tf)
Hom(Z',Tg)
Hom(Z',X') - Hom(Z',Y') ---k Hom(Z',Z') --} Hom(Z',TX') - Hom(Z',TY')
Since
Hom(Z',f), Hom(Z',g), Hom(Z',Tf)
we infer that
Hom(Z',h)
m E Hom(Z',Z)
ly, apply the cohomological functor
such that
Hom(Z',Tg)
such that
Kph = 1Z,. Converse-
Hom(-,Z). As above we conclude that
is an isomorphism. In particular there exists hi = 1Z. Hence
1.3
are isomorphisms,
is an isomorphism.
In particular there exists
Hom(h,Z)
and
h
E Hom(Z',Z)
is an isomorphism.
The following lemma shows that the converse of
(TR2)
is satisfied. This shows that we are dealing with the classical definition of a triangulated category.
LEMMA.
(X,Y,Z,u,v,w)
(TR2)
(Y,Z,TX,v,w,-Tu)
is a triangle, then
is a triangle.
Proof:
applying
If
If
that
(Y,Z,TX,v,w,-Tu)
is a triangle we obtain by
(TX,TY,TZ,-Tu,-Tv,-Tw)
is a triangle. By
(TRI)
7
(TX,TY,TZ',-Tu,-Tv',-Tw')
(X,Y,Z',u,v',w'), hence
there exists a triangle
(TR2), Now consider the following diagram:
is a triangle by
-Tv -Tu
TX
TY
TZ
-Tw
2
) T X
h II
II
_
The morphism by
h
exists by
_
r
TY Tv TZ
TX Tu
' _
r
II
2 -> TX
which moreover is an isomorphism
(TR3)
1.2 c). So we obtain the following commutative diagram, where the
second row is a triangle and the vertical morphisms are isomorphisms.
X u Y ->Z w' TX IT h
u' Y -IL- Z' ---' TX
X
By
(TRI)
we infer that the first row is a triangle.
(X,Y,Z,u,v,w)
Let
LEMMA.
1.4
C
be a triangulated category and
be a triangle. The following are equivalent:
(i)
w = o.
(ii)
u
is a section.
(iii)
v
is a retraction.
Proof:
The existence of
If u'
consider the following morphism of triangles.
w = o
is guaranteed by
V-+
U
X
(TR3).
Z W }TX o
IUt
X
Thus
uu' = 1X, hence
Conversely, if
u
u
1XX
o10-°
TX
is a section.
is a section, there exists
u'
such that
(1X,u',o)
8
is a morphism from the triangle
wlTX = o, hence w - o.
(X,X,0,1X,o,o). In particular
In the same way one can show that
LEMMA.
1.5
(X,Y,Z,u,v,w)
If
f' (ii) If
Proof:
Hom(W,v),
g : Y - Y'
f
C
(iii)
are equivalent.
be a triangulated category and
f
By
and
is not a retraction, then there exists
: W - Z
: W -+ Y
such that f'v = f. is not a retraction, then
: W - Z
1.2 (b)
Let
C
fw - o.
we know that
Hom(W,w),
Hom(W,Z)
LEMMA.
1.6
(X,Y,Z,u,v,w)
Let
and
(i)
be a triangle. The following are equivalent:
(i)
Hom(W,Y)
to the triangle
(X,Y,Z,u,v,w)
Hom(W,TX)
is exact.
be a triangulated category and be triangles in
(X',Y',Z',u',v',w')
C. Let
be a morphism. Then the following are equivalent: ugv' = o.
(i)
(ii) There exists a morphism
(f,g,h)
from the-first triangle
to the second.
Hom(X,T Z') = 0
If these conditions are satisfied and are uniquely determined by
Proof:
an exact sequence
to the second triangle and obtain
HomC(X,-)
f
: X -e X'
such that
fu' = ug. By
(TR3)
we
(i) * (ii).
1.2 a.
shows that also
h
and
Hom(X,T Z') - Hom(X,X') -. Hom(X,Y') - Hom(X,Z'). If
Conversely, if
by
f
g.
We apply
ugv' - o, there exists obtain
then
If f
(f,g,h)
is a morphism then
Hom(X,T Z') = 0, then Hom(X,u') is uniquely determined by
Hom(TX,Z') = 0. Thus
Hom(v,Z')
g. If
ugv' = fu'v' = 0
is a monomorphism, which
Hom(X,T Z') = 0, then
is a monomorphism, which shows
9
that
h
is uniquely determined by
LEMMA.
1.7
(X,Y,O)u,o,o)
If
C
be a triangulated category. Then
is a triangle if and only if
Proof:
isomorphic to
Let
g.
If
u
is an isomorphism.
is an isomorphism, then
u
(X,Y,O,u,o,o)
(Y,Y,O,1Y,o,o). Thus the assertion follows from
(X,Y,O,u,o,o)
is
(TRI).
is a triangle, we consider the following morphism of
triangles
x
u 'Y-° -'0 Y
Y
Thus the assertion follows from
Y --) 0 - TY 1.2.
10
Frobenius categories
2.
Let
2.1
be an additive category embedded as a full and
B
extension-closed subcategory in some abelian category
A with terms in
set of exact sequences in the pair
be the
S. Following Quillen (1973)
is called an exact category. A morphism
(B,S)
S
u
:
X - Y
in
is called a proper monomorphism if there exists an exact sequence
8
0 .. X + Y - Z -+ 0
in S. A morphism v : Y - Z
epimorphism if there exists an exact sequence An object epimorphisms g
A. Let
:
P - Y
P
v : Y - Z
such that
g : Y -' I
is called a prope,
0 -. X -, Y + Z -' 0
in
S.
is called S-projective if for all proper
and morphisms f
I
u : X -, Y
such that
B
B
:
P - Z
B
in
there exists
f = gv.
An object
monomorphisms
in
in
B
in
S-injective if for all proper
is called
and morphisms
f
:
X -. I
in
B
there exists
f = ug.
The following two assertions are straightforward.
LEMMA.
Let
(B,S)
be an exact category. Then the following
are equivalent: (i)
P E B
(ii)
All exact sequences
is
S-projective. 0 -' X -' Y -
P -, 0
in
S
split.
It
LEMMA.
Let
be an exact category. Then the following
(B,S)
are equivalent: I E B
(i)
is
S-injective.
(ii) All exact sequences 0 -, I - Y -+ Z -+ 0 Let
(B,S)
with
v : P -+ Z
an
P
an
I
S-injective in
the
has
(8,S)
there exists a proper epimorphism B. We say that
(B,S)
has enough u : X - I
B.
is called a Frobenius category if
(B,S)
S-projectives and enough
has enough
split.
S
there exists a proper monomorphism
An exact category (B,S)
Z E B
S-projective in
X E B
S-injectives if for with
be an exact category. We say that
S-projectives if for all
enough
in
S-projectives coincide with the
S-injectives and if moreover
S-injectives.
In the sequel we follow closely the approach of Heller
2.2
(1960) which also contains a proof of the properties of the suspension functor mentioned below. Let X,Y Y
in in
we denote by
B B
(B,S)
B
morphic in
is the category whose objects are the objects of
B
and
X,Y, the set of morphisms from
X
to
Y
is denoted
Hom(X,Y) = Hom8(X,Y)/I(X,Y). The residue class of a
u : X -+ Y
such that B.
to
asso-
LEMMA.
S
X B
Hom(X,Y), and
be in
of morphisms from
S-injective. The stable category
given two objects
morphism
the subgroup
I(X,Y)
which factor over an
ciated with
by
be a Frobenius category. For a pair of objects
is denoted by
Let
I
M. 7r it
7r I
0 - X u. I' -+ X' -. 0 and 0 - X + I" -+ X" i 0
I',I"
are
S-injective. Then
X'
and
X"
are iso-
12
Proof: and
I"
are
Since
and
p'
are proper monomorphisms and
if'
I'
S-injective we obtain the following commutative diagram such
that the rows belong to
S.
, X',0
0-+X E >I' o _} X
u,
X"
I"
fI
A
II
o
,
g,
I
,R,
0--+X u-+ I' -'VX' -0 Thus
u'(ff'-l1,) - 0. So there exists
h
X'
:
I'
such that
Tr'h = ff' - li,. Therefore Tr'hTr' = (ff'-1I,)Tr' Tr'gg'-Tr' = TrI(gg'-1X,). gg'-1X, = hTr'. In other words
Thus we have that
we can show that Let
g'p,, = 1X,,.
X E B. We denote by
[X]
the isomorphism class of
in B. Moreover let 0 -. X -. I -. X' -. 0 be in S-injective. We assume that for all yX
:
gg' = 1X,. Similarly,
[X] - [X']. The preceding
depend on the choice of
X E B
S
X
such that I is
there is a bijection
lemma shows that this assumption does not
0 -+ X -. I r X' - 0. (It is easily seen that this
assumption is satisfied in all our applications.)
0 -+ X 11M
For all
X E B
IM 70)
TX -*0
TX = yX(X). Let
monomorphism and
f
: X - Y I(Y)
is
we now choose elements
in B with I(X) an S-injective such that be a morphism in
B. Since
y(X)
S-injective we obtain a morphism
is a proper I(f)
such
that
fli(Y) = u(X)I(f). In particular we obtain a morphism
that
7r(X)T(f) = I(f)Tr(Y). It is easily seen that the residue class of
T(f)
in
T
B
does not depend on the choice of
as a functor from
B
to
T(f)
such
I(f). Thus we may consider
B. The following proposition is straightfor-
13
ward. PROPOSITION.
Remark:
is an automorphism of
T
S.
Without the existence of a bijection yX
as above
we still may choose for X E 8 elements 0 - X - I(X) -+ S(X) - 0 in with
I(X)
that
S
an
S
S-injective. The same construction as above then yields
may be considered as a functor on
S. This time providing us
with a selfequivalence. We claim that any two such choices yield
isomor-
phic functors. In fact suppose that two such assignments have been made.
I'(X) ML X
0 -+ X
are
S'(X) -+ 0
S-injective. Since
S-injective we obtain
This induces a morphism We know that a(X)
a(X)
:
in
S
I(X) -I'(X)
a(X)
such that
and
I(X)
is a proper monomorphism and
p(X)
fX
I(X) n(X S(X) -+ 0 and
0 -+ X u(X
So we have for all X E S elements
:
such that
S(X) + S'(X)
I'(X)
I'(X)
p(X)fX = u'(X).
such that
n(X)a(X) - fX7r'(X).
is an isomorphism. Thus it remains to be seen that
is a natural transformation. In fact, let
f
: X - Y. This induces
the following two commutative diagrams:
0 1
uM
00
0
f }
1
X
lu(Y)
IM
I(() f )I 71(Y)
71 (X)
S(X)
is
S(f)+S(Y)
Y
lu'(Y)
11,M
I' (X) I f )-}III' (Y) n'(Y)
711(X)
S'(X) f)-+S'(Y)
I
I
I
I
0
0
0
0
14
We have morphisms u(X)fX = u'(X) a(X)
:
fX
and
S(X) -+ S'(X)
and
a(Y)
such that
such that
factors over an
S(f)a(Y)-a(X)S'(f)
u(X)(I(f)fY fXI'(f)) = 0
such that
S(X) -, I'(Y)
I(Y) - I'(Y)
T(Y)a(Y) = fYwr'(Y). We claim that
S-injective. Observe that :
:
S(Y) -+ S'(Y)
:
S(f)a(Y) = a(X)S'(f). We show that
g
fY
and
I(X) -. I'(X)
u(Y)fy = u'(Y). Moreover we have morphisms
and
IT(X)a(x) = fXar'(X)
:
Thus there exists
¶r(X)g = I(f)fy-fXI'(f). Now
wr(X)gTr' (Y) = (I(f)fy-fXI' (f))lr'(Y) = I(f)fy ' (Y)-f XI' (f)1r' (Y) _ I(f)7r(Y)a(Y)-fxlr'(X)S'(f) = ,r(X)S(f)a(Y)-ir(X)a(X)S'(f) =
7r(X)(S(f)a(Y)-a(X)S'(f)). In particular
2.3 f
:
u
Let
B. Then
of a proper monomorphism
Proof:
be a Frobenius category and
(B,S)
be a morphism in
X -, Y
class
LEMMA.
gir'(Y) = S(f)a(Y)-a(X)S'(f).
f
is isomorphic to the residue
u.
Indeed, given a morphism
per monomorphism X x I(X)
of
X
into an
clearly isomorphic to the residue class of
:
X - Y
S-injective
of
B
and a pro-
I(X), f
is
(f,x). On the other hand
(f,x)
contains the short exact sequence C
f
0 -> X --:Y 0 I(X)
S
(-A C -+ 0
where
is the pushout occurring in the following commutative diagram with
rows in
S.
O -X x >I(X) - TX >0 f
y
0 -)Y 2.4
Let
A
C , TX -)0
be an abelian Frobenius category such that every
object has finite length. This assumption ensures us that the classical Krull-Schmidt theorem is valid in
A.
15
LEMMA. and f
f
f * 0
satisfies
in
f = gh. Since g'
:
P -+ X
and morphisms
A
in
P
gg'
g
in
A. Then there exists a proand
: X -+ P
End X
is nilpotent, since
X,Y
Let
:
P -+ Y
such that
be in
f = (gg')f. Since
X
is
is not an isomorphism. There-
X
of
gg'
h
is an epimorphism, there exists
f
h = g'f. In particular
such that
COROLLARY. tive. If
Y E A. Then the residue class of
f = 0
is projective and
P
not projective the endomorphism fore
indecomposable and not projective
A.
Suppose that
Proof:
jective
A
be in
an epimorphism for some
X - Y
:
X
Let
is local; a contradiction.
A
Hom(X,Y) * 0, then there exists
indecomposable and not projecZ E A
such that
Hom(X,Z) * 0 * Hom(Z,Y).
Let
Proof:
f = gh
such that
be the domain of and
h
g
0 * f E Hom(X,Y). Consider a factorization of
is a monomorphism and
h. By the previous lemma and its dual we infer that
yield non-zero morphisms in
2.5
is an epimorphism. Let
h
Let
be a Frobenius category and
(B,S)
category. We define a set
A.
T
of sextuples in
B. Let
u E HomB(X,Y). Consider the following diagram in
X - u) Y
B:
B
the stable
X,Y E B
and
Z
g
16
where
Cu
is the pushout of Since
A
is in S
0 - X i I(X) R TX -, 0
and
I(X)
S-injective.
is
x.
is closed under extensions in some abelian category
8
Cu
the pushout
u
and
coincides with the pushout in
B
in
of the form X 4 Y 1 Cu 4 TX
and its image in
A. A sextuple
will be called stan-
B
dard. U
w
v-
A sextuple X - Y - Z - TX of objects and morphisms in T
lies in
if it is isomorphic in
B
to a standard sextuple.
2.6
T
is a triangulation of
THEOREM.
We check the axioms from
Proof: (TRI)
The set
By definition T
B
B.
I.I.
is closed under isomorphisms and every morphism
can be embedded into a triangle. Clearly the sextuple
X
0 >TX
X +X
(TR3)
lies in
T.
Let us first consider the case of standard triangles. Consider the
following two standard sextuples:
Xu xI I(X)
X' u Y'
-
(X,
u
f
Then there exists
a
that
:
and :
g
such that
I(X) - Y'
I(X) -. I(X')
:
I(X) -. Cu,
such that
such that
fu' = ug
such that
gv'
in
B.
ug = fu' + xa. We have mor-
fx' = xIf
Ifx' = xTf. So we obtain morphisms
Ifu' + av'
lwt
TX' - TX'
TX
and two morphisms
) u-+ IVu'
x'
lv
TX
If
I
and
Cu
x
phisms
x'1
jv
and
Tf : TX -+ TX'
Y - Cu,
and
ugv' = fu'v' + xav' - fx'u'+ xav'
such
17
x(Ifu' + av'). This yields a morphism h : Cu -. Cu, and
uh = Ifu' + av', for
Cu
is a pushout. We claim that
For this it is enough to show that
vwTf = 0
the first observe that
we have
uwTf = xTf = Ifx' Thus
such that
vhw' = vwTf and
vh = gv'
hw' = wTf.
uhw' = uwTf. For
and
vhw' = gv'w' = 0
For the second
Ifuw = Ifu'w' + av'w' = (Ifu' + av')w' = uhw'. is a morphism of triangles.
(f,g,h)
The general case is easily deduced from the previous one. In fact, let
in T
(X,Y,Z,u,vv,w)
and
f,g
(X',Y',Z',u',v',w')
and
two morphisms such that
ug
fu'
be arbitrary elements in
B. Then we have
isomorphisms to the corresponding standard triangles. Using the first part we obtain the following commutative diagram, where the rows are in and
h1,h2
T
are isomorphisms:
Z+TX
Y
X
11
1
h' -w
-v
II
u
'l
Cu
-> Y
X
TX
Tf
X'-- Y'-----* Cul
TV
h2 u_, X'
Then
(f,g,h1hh21)
gv'h21 - gv'
(TR2)
and
1 Y'
,
II
w TX'
Z'
is a morphism of triangles. In fact, vvh1hh21 = vhh21 h1hh2'w' - hlhw' = h1!Tf a wTf.
It is easily seen that it suffices to consider the case of a stan-
u dard triangle. Suppose that y y
0 - Y - I(Y) -. TY - 0 be in exists
I
v
Iu
such that
_v
w_
X -Y - Cu - TX S
xIu = uy
with I(Y)
is a standard triangle. Let
being S-injective. There
and there exists
Tu
such that I y - xTu. u
18
Define
thus
f
:
Cu -+ I(Y)
y = vf, Iu = uf. Since
uwTu, we infer that
by using the pushout property:
vfy = yy = o = vwTu, and
fy = wTu, for
Cu
ufy = Iuy = xTu =
is a pushout.
In this way, we obtain the following commutative diagram: 0
0
0 --j Y -°
C
w
U
)
TX
0
>
TX
0
(fw)
y
1(, ,0,
1
yI
C0 1
0 -'DI(Y) - I(Y)(DTX '`-Tu)
TY
TY
1
1
0
0
The upper two rows are exact, therefore the middle one is the induced exact sequence of the upper one, in particular the left upper square is a pushout. Therefore, Y
v
\-Tu/
0 TX ----QTY
Cu
v_
which is obviously isomorphic in (TR4)
B
to
w_
Y -, Cu -, TX
is a standard sextuple, -Tu -,
TY.
Again it is enough to consider the case of standard triangles.
Suppose we are given three standard sextuples:
19
X
u
°Y
i
x
I(X)
u
v
Y
-* Z
y
Z'
j
Y
xlI
-
Z
k
xIi
I(Y) v X'
,
w
X
I(X)
and
x
J,I
k'
i TX
TX
TX
TX
TY
TY
with w = uv. Let us replace
0 -+ Y
I(Y),TY,y,y
as follows: We have the sequence
Z' -+ TX y 0 in S, and consider a sequence
0 -' Z' - I(Z') * TZ' -> 0
in
S
with
I(Z')
being
S-injective. Consider
the following commutative diagram of exact sequences in 0
0-aY
'
A:
0
TX-;0
Z' z
i I 0 --> Y 1=-r I (Z') --+ M --* 0
Since
B
second row belongs to
TZ'
TZ'
1
1
0
0
is closed under extensions in S. Thus instead of
A
we infer that the we may
0 -. Y $ I(Y) y TY - 0
take 0 -> Y-+ I(Z') -+ M -+ 0. Changing the notation, we may assume that y = it. Since I
uy = uit = xi, we denote
y = uty. The identity map
u
I(Y) = I(Z')
ui
by
Iu, and define by
can be denoted by
it = y idZ, = yIi, and there is the induced map i = Iii = yTi.
I(Y) = I(Z')
Ti : TY -> TZ'
and
xTu =
Ii, for such that
20
xw = wk - uvk, there exists
Since
of = w and
f
:
such that
Z' - Y'
if = vk, using the pushout property of V. Similarly, wj = uvj - uyv - uity - xuty, shows that there
exists
g
pushout property of
and
and
kg = j, using this time the
Y'.
We claim that ifg = itv
wg - utv
such that
Y' - X'
:
fg = tv. For this it is enough to show that
ufg = utv. In fact, ifg = vkg = vj = yv = itv, and
ufg - wg - utv. Altogether, we obtain the following commutative diagram:
v -Z kl
l
I
I(X) u * Z' f Y'
with
of = w
it =y xI
tl
TX
I(Z1)
_
gl
kg = j
X'
Let us first check the various relations to be satisfied. By construction
if = vk
and
kg = j.
Let us show next that show that
ifk' = ii'
and
fk' = V. For this it is enough to
ufk' - ui'
In fact, ifk' = vkk' = 0 = ii', and
ufk'
Finally let us show that to show that
kk'Tu = kgj', and
out property of
Y'. In fact
using the pushout property of V. wk' - x = ui'.
k'Tu = gj'. For this it is enough
wk'Tu = wgj'
using this time the push-
kk'Tu - 0 = jj' = kgj', and wgj' = ufgj' _
utvj' = uty = xTu = wk'Tu.
It remains to be seen that
Z' * Y'-4 X' 1'1-1-TV
dard sextuple. Since V is a pushout of
it
and
v, and
is a stanY'
is a push-
21
out of
i
and
v, it follows that the next diagram is a pushout
Z'
f
Y'
R
Ig
I (Z' ) Recall that
R = yTi, thus
v-+X'
R = yTi - vj'Ti.
Therefore we obtain the following commutative diagram with columns in
S:
R)
-
jg
1(Z') v 'X'
1
!JT1
TZ'
TZ'
Hence
is a standard sextuple.
(Z',Y',X',f,g,j'Ti)
This finishes the proof of the theorem.
2.7
(B,S)
Let
shows how elements in
S
be a Frobenius category. The following lemma
give rise to triangles in
lowing two exact sequences 0 - X -u, ,Y 0 --, Y Z' I(Y) Z' TY -+ 0
Z -+ 0
B. Consider the fol-
and
in S, with I(Y) being S-injective. They induce
the commutative diagram with exact rows
o -X
a
0 -; X
uy
'Y
v
'
)I(Y)
p
, TX -0
22
LEMMA.
X-Y-Z
-, TX
With the notation above it follows that
belongs to T. By construction, the following square is a pullback:
Proof:
Yv>Z
iIp -r TX
I(
,
This induces the following diagram with exact rows and columns: 0
0
0-X u ->Y° 1
1
J1Ii
Z
0
0 --*1 (Y) (Ol),}Z®I(Y) \O-I+z
-r 0
(vi)
uyl
t
i(-w P/
pl
TX
TX
1
1
0
0
The assertion now follows as in the proof of
2.8
ditive functor
Let
(B,S)
S'-injectives then
F
lemma follows directly from tion functor on
-o .O
is contained in
0 -s X + Yvi Z -s 0
B
be Frobenius categories. An ad-
is called exact if
F : B -s 8'
0 - F(X) -F(u)F(Y) -F(v)F(Z)
into
and (8',S')
(]'R2).
is contained in S' S. If
whenever
transforms
F
S-injectives
induces a functor
F
2.7. Denote by
(resp. T') the transla-
(resp. B').
T
:
B - B'. The following
23
LEMMA. ries
B
and
B'
Let
F
such that
be an exact functor between Frobenius categoF
transforms
S-injectives into
S'injec-
tives. If there exists an invertible natural transformation a
:
FT - T'F
then
F
is an exact functor of triangulated categories.
24
Examples
3.
There are two main examples of Frobenius categories we want to present here and which will be important in the following developments. Also we include some remarks on derived categories.
Let
3.1
be a field and
k
Note that we do not require that that R2
R
R an associative
is finite-dimensional over R2 = R
R has a unit element. However, we do require that denotes the subspace of
R
k-algebra.
generated by all products
k
and (here,
r1r2, with
r1,r2E R). Moreover, we usually will assume that there exists a complete set
{ex
I
x E I}
of pairwise orthogonal idempotents in
pleteness means that
R =
M
R, a (left) R-module M
RM = M
(as above, RM
generated by all elements of the form
to the usual one on M If
potents, and M
{ex
I
rm, with
elements
RM = M
is equivalent
is a complete set of pairwise orthogonal idem-
R-module, then, as a vector-space M decomposes in
the form M = ® ex M. An if there exist
r E R, m E M). Note
to be unitary..
x E I}
is an
is always supposed
denotes the subspace of
R has a unit element, the condition
that in case
(the com-
® ex Rey). x,y
Given an algebra to satisfy the condition
R
R-module M
is said to be finitely generated n m1,...,mn E M such that M = F Rm i=1
25
We denote by Mod R
the category of all
R-modules, by mod R
the full
subcategory of all finitely generated ones. In the sequel a module will always be finitely generated.
R
The algebra
is said to be locally bounded if there exists
a complete set of pairwise orthogonal primitive idempotents such that
Rex
are finite-dimensional over
exR
and
R
Assume that
k
is locally bounded, and let
{ex
for all {e x
I
will be denoted by
composable projective
R-modules. Similarly, let
For a locally bounded algebra R-module has finite length. In particular
rad P(x)
denotes the radical of
mod R
R-modules.
is an abelian category. for
P(x). Also in this case
x E I, where
mod R R
has
a Frobe-
is locally bounded, and the indecomposable projective
R
R-modules coincide with the indecomposable injective
mod R
R-module
R, any finitely generated
enough projectives and enough injectives. We call an algebra
case
be
I(x) = Homk(exR,k);
R-modules are of the form P(x)/rad P(x)
nius algebra if
x E I.
P(x). One obtains in this way all possible inde-
one obtains in this way all possible indecomposable injective
The simple
x E I}
x E I)
a complete set of pairwise orthogonal primitive idempotents. The Rex
I
R-modules. In this
is a Frobenius category.
Let us conclude this example by some more specific ones. Let
R be a finite-dimensional
is a Frobenius algebra if and only if
RR
k-algebra. Then obviously is an injective
The last condition is for example satisfied if group algebra of a finite group over the field
R
R-module.
R
is the
k.
There is a classical construction to associate to an arbitrary finite-dimensional
k-algebra A
trivial extension of
a Frobenius algebra
A. For this denote by
T(A), called the
Q = Homk(A,k). Q
A-A-bimodule structure in the obvious way: given
admits an
a',a" E A and
q E Q,
26
then
k-linear map which sends
is the
a'qa"
Using this we can define
a E A
q(a"aa').
to
T(A). The underlying vectorspace of
T(A) = A®Q,
and the multiplication is given by
(a,q)(a',q') = (aa',aq' + qa')
a,a' E A, q,q' E Q.
for
Then it is straightforward to check that
is a Frobenius
T(A)
algebra.
Let
3.2
a
be an additive category with splitting idem oe = e2 E Homa(X,X)
tents. Recall that an idempotent are morphisms
p
Y -* X, p
:
pp = lY
such that
: X -. Y
is split if there and
pp = e.
This is equivalent to the condition that the endomorphism ring
X
of an indecomposable object
a
in
End (X)
is a local ring.
If these conditions are satisfied
a
will be called a Krull-
Schmidt category. The theorem of Krull-Schmidt holds in a Krull-Schmidt category asserting the uniqueness (up to order) for direct decompositions.
over i
d
a
is by definition a collection of objects i
i
= dX : X i X
bounded below if X1 = 0
above if
i+1
such that
X1 = 0
i i+1
d d
i object
X o
i
0
for all but finitely many
X
such that
o
* 0
and morphisms i
i
X' = (X ,d )
is
and bounded
i < o
i > o. It is bounded if it is a stalk complex if
X' _ (X',dl)
and
7L
X1 = 0
for all
i * i . The
0
is then called the stalk. Suppose that
Then the width w(X') s
= 0. A complex
for all but finitely many
is bounded above and below. A complex there exists
X1
i
i
X' = (X 'dX)iE
A differential complex or simply a complex
of
X1 = 0 X'
is called the deviation of
for
i < r
and
s < i
is by definition equal to X'
and is denoted by
and
xr * 0 * Xs.
s-r+1. If
d(X').
s < o,
27
and
X' =(X1,dX)
If
Y' = (Y1,dY)
is a sequence of morphisms
phism f': X' -. Y'
are two complexes a morf1
a
of
X1 -, Y1
:
such
difl+1
i E X. These morphisms are composed in an
for all
= fldi
that
obvious way.
Denote by
(resp. C (a), resp.
a, by
the category of complexes over
C(a)
Cb(a))
C+(a)
the full subcategories of complexes bounded
below (resp. above, resp. above and below). There is a full embedding of X
object
a
of
into the stalk complex
We will identify this complex with a
If
:
X' - Y'
of
H1(X')
X' = (X',dl)
(a)
(resp.
X° = X.
with
X.
are defined for
A
then
X' E C(a). And a morphism
are isomorphisms for all
-$b C
which sends each
C(a)
is called a quasi-isomorphism if the induced mor-
C(a)
H1(u'):H1(X') - H1(Y')
phisms
into
is a full subcategory of an abelian category
the cohomology objects u'
a
C+'b(a))
the full subcategories of
i. We denote by
c _(a) (resp. C+(a))
of complexes with only finitely many non-zero cohomology objects. We claim that all introduced categories of complexes are Frobenius categories and that the correspodning stable categories coincide with the classical homotopy categories. First of all let us recall that two morphisms are called homotopic for all
i
(f' - g')
if there exists morphisms
f',g' :
ki
:
Xi
X' - Y' Yi-1
_
such that fi - gi = diki+l + kidi-I
X In the following let over
C
Y
be one of the introduced categories of complexes
a. We are going to construct the set
S
of exact sequences. Let X®Z01
(1,0)
X,Z
be in
a. An exact sequence of the form
0 + X
and isomorphic ones, are said to be split exact.
*Z - 0
28
S be the set of all exact sequences
Let
Y' , Z' - 0
0 -, X'
are split exact.
Yi g- Zi - 0
0 i Xi f
such that the exact sequences
C
in
i
i
X
Let
a
be in
and denote by
P1(X)1 = P1(X)i+1 = X, P1(X)J = 0
with
Then it is easily seen that Ii(X)'
the complex in
j * i,i-1
and
di i
C
P1(X)'
with
for
P1(X)'
j * i,i+1
C
= IV
d1 P'(X)
and
X
S-projective. Dually, denote by
is
I1(X)1 = I1(X)1-I = X,I1(X)j = 0
- 1X. Then it follows that
I
the complex in
I1(X)'
for
S-injective
is
I(X) and that the
S-injectives coincide with the S-projectives. Let us show
that there exist enough and denote by
I(X')'
S-injectives. Indeed, let the complex
be in
X' = (X1,d1)
C
0 I1(X1)'. Next we define a morphism i
X' -. I(X')'
by
x1 : X1 y X1 0 X1+I
easily seen that
x'
is a proper monomorphism and we have an exact
x'
.
with
xl
(I id1). It is
x'
sequence in (TX,)i =
S
Xi+1
of the following form: 0 - X' y I(X')' - T(X')' - 0 where di+1
i
(d
X
TX)
The following assertion is left to the reader. Let
morphism from
X'
if and only if
to
f'
Y'
in
C. Then
f'
factors over an
ciated with
be a
S-injective
is homotopic to zero.
Altogether we see that the stable category nius category
f'
C of the Frobe-
is nothing else but the homotopy category K
(C,S)
asso-
C. Moreover, the above calculation shows that the classical
shift functor coincides with the suspension functor. From the description of standard sextuples in Frobenius categories we can easily deduce the classical definition of the mapping cone Cf
for a morphism
f'
:
X. -. Y', This is the complex
Cf = ((TX')1 0 Y',d' ) with differential f
29 i+1
dX
f
i+1
dY
0
X1+1 0 Y1
X' E C
For instance if
satisfies
the associated truncated complex
is
X1 = 0
i < o
for
(X" = 0
induces a morphism from
dX
i > 1)
Xi+2 0
for
T X°
to
Yi+1
and, if X"
i < o, dX, = dX X'*
is
for
whose mapping cone
X, .
For later reference we include some remarks on derived
3.3
categories, but refer for a deeper account and proofs of the statements mentioned here to (Verdier (1977), Hartshorne (1966) or Iversen (1986)). Let
A be an abelian category and denote by
of bounded complexes over
category obtained from
A. The category
Kb(A)
Db(A) the derived category is the triangulated
Db(A)
by localizing with respect to the set of
quasi-isomorphisms. Thus there is an exact functor such that
Db(A)
is an isomorphism in
4A(f')
QA
:
whenever
Kb(A) i Db(A) f'
is a quasi-
Cb(A). Moreover, any other functor having this property
isomorphism in factors through
QA.
In most of the situations
A will always be the category
mod A of finite-dimensional (left)-modules over a finite-dimensional A. In this case we simply denote
k-algebra
tional assumptions
AP
(resp.
tive
A1)
Db(A)
.
Kb(mod A)
Db(A). Under addi-
mod A with objects the projec-
injective) A-modules. Now, Kb(AP)
(resp.
by
becomes a quite accessible category. Denote by
the full subcategory of
(resp.
Kb(AI))
is a
Kb(mod A). Thus we obtain functors
full subcategory of Kb(AP)
Db(A)
1
Db(A)
and
Kb(AI)
i Kb(mod A) i Db(A).
If
A has
finite global dimension the composition of these functors are triangleequivalences. If
A has infinite global dimension the situation is a little
30
bit more complicated. In this case equivalent, and also
There is a full embedding of each object
X
of
mod A
K
and
K+'b(A1)
and
Db(A)
Db(A)
-,b (AP)
are triangle-equivalent.
mod A
X' =
X. Let
Hom
which sends
(Xi, di)
with
X,Y E mod A. Using
projective resolutions and the triangle-equivalence of it is easily seen that for
Db(A)
into
into the stalk complex
X = X. We will identify this complex with
Db(A)
are triangle-
K- b
(AP)
and
i > o
(T1X,Y) = 0
Db(A)
and
Hom
(X,TiY) = Extl(X,Y).
A
Db(A)
Later we will need also the following easy fact. Let be an exact sequence in
mod A
the corresponding element. Then
and
Y y Z - 0
(Z,TX) = Ext'(Z,X)
w E Hom
wDb
v
0 - X
(A)
X 4 Y -+ Z H. TX
A
is a triangle in
be
b
D (A).
31
4.
Auslander-Reiten triangles
4.1
Let
be a triangulated category such that
C
k-vectorspace for all
is a finite-dimensional
X,Y E C
HomC(X,Y)
and assume that
the endomorphism ring of an indecomposable object is local. Thus
C
is
a Krull-Schmidt category. This will be assumed throughout this section.
A triangle
X u+ Y -°+ Z -w+ TX
in
C
is called an Auslander-
Reiten triangle if the following conditions are satisfied: (ARI)
X,Z
(AR2)
w # 0.
(AR3)
If
are indecomposable.
f
: W -+ Z
: W - Y
f'
We will say that indecomposable objects
Z E C
is not a retraction, then there exists
such that
C
f'v = f.
has Auslander-Reiten triangles if for all
there is a triangle satisfying the condi-
tions above.
We refer to (AR2)
and
and
1.5
for conditions equivalent to
(AR3).
4.2
Let
1.4
LEMMA.
X y Y 2+7 Z 4 TX
(Selfduality for Auslander-Reiten triangles)
be an Auslander-Reiten triangle. If
not a section, then there exists
f'
: Y -+ W with
f
uf' = f.
: X -. W
is
32
Proof:
into a triangle T -W'
By
the morphism
(TRI)
X f W 4 W' h TX. Using
T hX f W
: X -+ W can be embedded
f
we see that
(TR2)
is again a triangle. We apply the octrahedral
axiom to the composition
and obtain the following diagram of
(-T h)u
triangles:
T W' ' T -W' (-T h)u
I-T h
I
x
u
if
w
Z
Y
TX
jr t
ITf
t
wTf
11
Y' ? ' Z
W
TX
Ig
1,
W' ----- W' is a retraction, then
If
t2
t;
with
tltj = 1W. Now define So assume that
t2
:
Y' - Y
with
f' = rte. Then
1.4. So there exists
uf' - urt; = ft1t; = f.
is not a retraction. Then there exists
t2
t2v = t2
is a section by
t1
by
(AR3).
Consider the following morphism of triangles by
(f
exists
(TR3)):
X u tY -v +Z w -'-TX ITf
II
-Z -TW wTf
t2
r
1t2 X
Since
f
u
is not a section and
is nilpotent. Hence there exists Therefore:
0Y
X
v
'Z
ITf
w yTX
is indecomposable, we infer that n E NC
such that
(ff')n = 0.
ff
33
Y v Z - TX
u
X
10
ji
1(t.tpnl
10
Z w '-TX
X u +Y is a morphism of triangles. But then
w = o
gives the required contra-
diction.
A morphism h
4.3
between objects
trary additive category is called irreducible if nor a retraction but for any factorization h2
section or
and
Z1
Z2
of an arbi-
is neither a section,
h
h = hI h2
either
is a
h,
is a retraction.
Let
PROPOSITION.
X 4 Y Y Z 39 TX
be an Auslander-Reiten
triangle. (i)
Given
(ii)
u
(iii)
If
g
:
is irreducible, there is a section
Z1 - Z
: Z1-'Y with f=gv.
If
(iv)
are irreducible morphisms.
v
and f
it is unique up to isomorphism of triangles.
Z
f
:
is irreducible, there is a retraction
X - XI
g YX with Proof:(i) Let triangle. Since By
(TR3)
v'
X'
1
f = ug.
Y' Y' Z Yk' TX' be an Auslander-Reiten
is not a retraction there exists
g
with
v' = gv.
we obtain a morphism of triangles:
X' u - Y' X
u
'Y
v
.
vI Z w TX
is not an isomorphism we obtain a morphism
If
f
by
4.2. But
w = w'Tf
= w'Tu'Tf' = 0
TX'
Z
f'
with
u'f' _
gives a contradiction. Thus
f f
is
34
an isomorphism and so is
1.2c.
We will show that
(ii)
u = hih2. If
a factorization
uh; = hl. By
with
by
g
is not a section, there exists
h1
Y -° -* Z w TX
v
11
Il
Ih
1F12
X
h
Z ,I
is not an isomorphism, then w = hw = 0
tion we obtain
f
g
:
:
' TX
by
1.5, a contradiction.
Tl
be irreducible. Since
Z1 -' Z
Z1 - Y
w
is a retraction.
h2
Thus (iii)
hl
we obtain a morphism of triangles:
(TR3)
X
If
is irreducible. In fact, consider
u
with
v
f = gv. As
f
is not a retrac-
is not a retraction, g
is a section.
This is dual to
(iv)
4.4
assume that g : Y - Y
is
LEMMA.
End(Z)
X 4 Y i Z 4 TX
be a triangle in
is local and that w * o. Let
be morphisms such that
fu = ug. If
f
f
: X -+ X
C
and
and
is an isomorphism so
g.
Proof: gles
Let
(iii).
(h
exists by
Consider the following commutative diagram of trian(TR3)):
XuY
v
I
Z
w
TX
19
jf
X
jh
u
). Yv
Clearly it is enough to show that If not there exists
Therefore
n E K
o = hnw = w(Tf)n
h
such that
Z
ITf
w =TX
is an isomorphism ((TR2)
hn = 0, for
End Z
and
is local
gives the required contradiction.
1.2c).
35
4.5
We will relate Auslander-Reiten triangles to source and
sink morphisms.
category
u : X -, Y
By definition a morphism
in a triangulated
is called a source morphism if the following three conditions
C
are satisfied: (i)
u
(ii)
If
is not a section. f : X - M
such that (iii)
If
g
is not a section there exists
f'
: Y -' M
f = uf'.
: Y -, Y
ug = u, then
satisfies
g
is an auto-
morphism.
Dually, a morphism v : Y - Z
in a triangulated category
C
is called a sink morphism if the following three conditions are satisfied: (i)
v
(ii)
If f'
(iii)
If
is not a retraction. f
: M -. Z
: M i Y g
is not a retraction there exists
such that
: Y -' Y
f = f'v.
satisfies
gv = v, then
g
is an auto-
morphism.
Note that if there exists a source morphism
X
has to be indecomposable. Indeed, 0 * X
and let
X = X1 ® X2
pi
pi = ugi, i = 1,2, for
u(g1,g2)
u
is not a section,
be a non-trivial decomposition. We denote by
the canonical projection that
since
u : X 1 Y, then
: X -+ Xi, i = 1,2. There exists p i
is not a section. Thus
gives a contradiction, for
u
gi
pi
such
1X = (p1,p2) _
is not a section. Clearly we
have an analogous result for sink morphisms. We will say that
C
has source morphism (or has sink mor-
phisms) if for any indecomposable object in
C, there exists a source
morphism (or a sink morphism respectively). Let
by
4.2
and
4.4
X
Y _+ Z -+ TX
we infer that
be an Auslander-Reiten triangle. Then u
is a source morphism and that
v
is
36
a sink morphism. We will show that the converse holds. For this let
be a source morphism in
Let
C.
X Y* Y X Z w+ TX
be a triangle in
We claim that this is an Auslander-Reiten triangle. Since section we see that
(AR2)
holds. The dual version of
is satisfied. So it remains to show that
(AR3)
There exists an indecomposable summand, say
w
morphism
w1
of
from
to
Z. Let
to
Z1
Z2. Let
X P M °4
u
4.2
C.
is not a shows that
is indecomposable.
Z1, such that the component
is non-zero. Denote by Z = Z1 ® Z2
Z
X it Y
q
and let
p
the canonical inclusion
be the projection from
Z
w Z1 -1 TX be a triangle. Using the defining proper-
ty of a source morphism and that
u'
is not a section we obtain the fol-
lowing commutative diagram of triangles:
XuY
-* Z w + TX
f'1 X
u = u'f = uf'f
Since
is an automorphism of In particular, g
u
T. M
g'1 V'
Z1
Xu0Y
v --4.
we infer that
f'f
Z, for
Z w I TX
is an automorphism. Thus
(1X,f'f,g'g)
is a retraction. Thus
. TX
g
g'g
is a morphism of triangles.
is an isomorphism and
is
Z
indecomposable.
A similar proof shows the analogous result if we start with a sink morphism.
4.6
a field left
k. By
Let A be a finite-dimensional algebra with
I
over
mod A we have denoted the category of finite-dimensional
A-modules,by
AP
(reap.
A1)
the full subcategory of projective
(resp. injective) A-modules. It is well-known and easy to see that
AP
37
and
are equivalent under the Nakayama functor
AI
where
v = D HomA(-,AA),
denotes the duality on mod A with respect to the base field k.
D
A quasi-inverse of
v
is given by
HomA(D(AA),-). There is an in-
v
ap : D Hom(P,-) -, Hom(-,VP). Equivalently,
vertible natural transformation
X E mod A, there is a vector-space duality
for each
Hom(P,X) x Hom(X,vP) - k, (E,n) - (9In) for all morphisms
(iEIn) = (lJnv(n))
Let
THEOREM.
such that in
y
mod A
and all
A be a finite-dimensional
global dimension. Then the derived category
and
(EzIn) = n
in
AP.
k-algebra of finite
has Auslander-Reiten
Db(A)
triangles.
The Nakayama functor
v
induces an equivalence of
triangulated categories again denoted by
v
between
and an invertible natural transformation
aP.
Proof:
In fact, if
E
is defined by
(-1)10,1 In1).
i EZ Since A has finite global dimension
equivalent to
Kb(AP)
written in the form
and to
9 E D Hom(P',P')
P'
is contained in
P'
End(P')
whenever the morphism
can be
Db(A). Let
which vanishes on the
f
of
P'
to
Db(A)
vP'
such that
is not a retraction.
This implies that
a
Db(A)
p(lp.) = 1. We consider the image
and satisfies
ap.(9); it is a non-zero linear map from fap.(q) = 0
is triangle-
Kb(AP).
is indecomposable in
be the linear form on
rad End(P')
Db(A)
Kb(AI). Thus an object in
P', where
Now assume that
radical
Kb(AI)
: D Hom(P',-) - Hom(-,vP').
Hom(P',X') x Hom(X',vP') - k, (F',n') - Q *171')
Win*) =
and
Kb(mod A), the associated duality
is an object of
X'
Kb(AP)
P(9)- + P'
r
VP'
38
satisfies the axiom
(AR3). Therefore this triangle is an Auslander-Reiten
triangle.
Our motivation for studying Auslander-Reiten triangles
4.7
comes from the useful concept of Auslander-Reiten sequences. These are by definition non-split exact sequences over a finite-dimensional and
(AR3)
0 -, X - Y -' Z ' 0
of modules
k-algebra satisfying the conditions
(ARI)
4.1. The basic existence theorem (Auslander, Reiten (1975))
of
asserts that for an indecomposable non-projective A-module
Z
there
exists such an'Auslander-Reiten sequence. Let
X i Y v Z 1 0
0
(Z,TX) = Extl(Z,X) Db(A) A following are equivalent:
w E Hom
M
X
be an Auslander-Reiten sequence and
be the corresponding element. Then the
is an Auslander-Reiten triangle in
Y y Z -+ TX
Db (A).
(ii)
inj.dim X < 1, proj.dim Z < 1. 0
(iii)
HomA(Z,P) = 0
for any injective
I, and
A-module
for any projective
A-module
P.
The proof is an easy consequence of
4.5
and some well-known facts of
Auslander-Reiten sequences (compare
2.4
of Ringel (1984)).
4.8
For further applications of
4.5
some more terminology for a Krull-Schmidt category
we have to introduce
a which also will
be needed later.
A path in (0
a
and non-zero, non-invertible morphisms
(0
A path will be denoted by
are said to belong to the path, and exists a path
Xi
is a sequence of indecomposable objects
(Xo9X19 ...,Xr)
in
r
fi
: Xi -' Xi+1
(X oXI,...,Xr). The objects
Xi
is called its length. If there
a, we write
Xo 4 Xr; if
r >
I
39
we write
X0
Xr. (Note that the relation l usually is not antisymme-
tric, even modulo isomorphism. If
Xo = Xr, the path is called
and
a directed if it does not contain any cycle. An inde-
a cycle. We call composable object to a cycle in
r > 0
X
a
of
X
is called directing, if
a.
Let
{Xi}i
be a complete set of representatives from the
E 1
isomorphism classes of indecomposable objects in the full subcategory of
a
whose objects are
a. We denote by
Xi, for
Next we will define the quiver r = r(a) category
ind a
i E I.
of a Krull-Schmidt
a.
X,Y
Let
be objects in
set of morphisms of the form some object
M
in a (rad(N,M)
Irr(X,Y) = rad(X,Y)/rad2(X,Y)
the
If
rad2(X,Y)
is given by the
f E rad(X,M), g E rad(M,Y)
denotes the subspace of N
Hom(X,Y), and let
a. Then
with
fg
non-invertible morphisms from
Homa(N,M)
for of
M). We denote by
to
End(X)-End(Y)-subbimodule of
dXY = dimkIrr(X,Y).
X
irreducible map from Irr(X,Y)
are indecomposable, then
Y
and
ducible if and only if
bimodule
does not belong
f E rad(X,Y) X
to
Y
f
: X - Y
is irre-
rad2(X,Y). Thus there exists an
if and only if
Irr(X,Y) * 0; thus the
is a measure for the multiplicity of irreducible maps
and it is called the bimodule of irreducible maps.
By definition, the vertices of the quiver r = r(a) isomorphism classes quiver has
d Y
[X]
of the indecomposable objects
arrows from
[X]
to
X
of
are the
a. The
[Y].
The following lemma shows that the knowledge of AuslanderReiten triangles in a triangulated category of
C
r(C). The proof is taken from Ringel (1984).
carries the information
40
Let
LEMMA.
and M be indecomposable objects in
X
Reiten triangles. Let
be a triangulated category with Auslander-
C
r
X
an Auslander-Reiten triangle. Let
11 Y -* Z -+ TX
Y =
and
C
d
Yi'
0
with Yi
i=1
Yi + Yj
indecomposable and if
We have given morphisms
Proof:
I < j < di. By
and
for some
M ed Yi
we know that
4.3
: Y - M with
f
if and only
di - dXM.
uij : X - Yi
for
I
4.3
: X -' M. By
f = ug. Thus there exists
Irr(X,M) * 0
with M - Yi
i
k-basis of
for some
of
for some
I < j < di
for
uij
i. i.
form a
Irr(X,Yi).
f E rad(X,Yi) .rad2(X,Yi). Then there exists
For this let g
uij
we
we obtain a retraction
For the remaining part we may assume that M ^t Yi We claim that the residue classes
< i < r
is irreducible. We infer that
uij
Irr(X,M) * 0. If
implies that
i
have an irreducible morphism g
Irr(X,M) * 0
i. Moreover, in this case
for some
M CU Yi
i * j. Then
for
f = ug =
such that
: Y -, Yi
F utgtj . Then
is an endomorphism
gij
Rrj
of
Yi, thus
+ gij
gij = cjly
for some
cj E k
and
gij E rad(Yi,Yi).
1
Since all
gRj, with
rad(YR,Yi), and all
t * j, are in
utj E rad(X,YR
it follows that:
uij cj (mod rad2(X,Yi)),
f
I ukj gRj
t 1j
thus
generate
ui1,.... Uid
Irr(X,Yi)
as a
k-vectorspace.
1
ei - dimkIrr(X,Yi). Then
Let
be a
k-basis of
Irr(X,Y1), with
hij
ei < di. Let
b.
.
E k. Let
h = (h.
map given by the matrix with Then
u-hb E rad2(X,Y), say
g E rad(N,Y). By
4.2
is
I
X
h.ISb.IS
e.
s=1 d.
X i ® Y.1, b. : Y 1
Y 1
the
e.
:
r
i=1
si-th entry equal to b, and isj
u-hb - fg
there exist
< j < ei, ei -
u..
: X 1 Yi, thus r
for some
hij,
h'
for some and
f'
b =
f E rad(X,N),
such that
*
i=1
h = uh',
b.. 1
41
f = uf', and therefore
u = hb + fg = u(h'b + f'g).
By
we have that
4.4
an automorphism,for
h'b + f'g
is an automorphism of
g E rad(N,Y). As a consequence, b
ei < di, we see that
thus, using that
Y, thus
ei - di
actually is an isomorphism. We infer that
h'b
is
is a retraction,
for all
u;,,...,u;a
i, and that
b
k-basis of
is a i
Irr(X,Yi). This finishes the proof.
4.9
A translation quiver
(locally finite) quiver
(ro,r1)
r - (ro,r1,T)
denotes the vertex set, r1
(ro
the set of arrows) together with an injective map on a subset
such that for any
ro c ro
number of arrows from y TZ
y. The map
to
r
to
z
is given by a
r
:
ro -+ ro
z E r,,, and any
defined
y E ro
the
is equal to the number of arrows from
is called the translation.
The vertices in
ro
which do not belong to
projective vertices, those not belonging to the image of injective vertices. If
denotes
ra - ro
ro
are called
r
are called
the translation quiver is called stable.
A morphism of translation quivers Y
_,
r -+ r'
:
is a morphism of quivers
which commutes with the translations.
I Given a translation quiver r
is given by an injective map
all arrows for
a
:
a
: a -, b
with
b
a
:
a polarization of
r = (ro,rl,r)
ri -, r1
where
ri
not projective, such that
is the set of a(a)
:
rb - a
a -+ b.
i If
r
has no multiple arrows, there is a unique polarization.
With these definitions we obtain the following result.
COROLLARY.
Let
finite global dimension. Then
A be a finite-dimensional b r(D(A))
k-algebra of
has the structure of a stable
42
translation quiver.
Proof:
gory and that
P'
Observe that
be as in
(r(Db(A)),t)
4.10
Db(A) - Kb(AP)
tP' = T vP'. It follows from 4.3
4.5. We define
is a stable translation quiver.
For later computations the following formula is useful.
PROPOSITION.
Let
finite global dimension and
D Hom b
A
be a finite-dimensional
Clearly
T
(T2i-1X',Y') cw Hom b
(X',tT2iX')
for all
T. So
and
P2 ne, Y'
TT2iX'
with
TT2iP1 ad
= vT
P;,P2 E Kb(AP). 2i-1
Pi. And the
isomorphism is induced by the invertible natural transformation
a
T
(compare
4.6).
i.
D (A)
Let P c X'
commutes with
k-algebra of
X',Y' E Db(A). Then
D (A) Proof:
is a Krull-Schmidt cate-
2i-1
.
P1
43
5.
Description of some derived categories
In this section we want to apply our previous considerations to consider one example in more detail. We will gather a lot of information on the derived category of a hereditary, basic finite-dimensional
k-algebra A which is of great importance for the further developments. For simplicity we will assume throughout that the field
k
is algebrai-
cally closed. Although this assumption is not really needed it simplifies the exposition.
5.1
In this section we collect some information on the combi-
natoric approach to the representation theory of finite-dimensional k-algebras. A basic reference for this is Gabriel (1980), which contains also a proof of the main result of this section.
i Let
A be a quiver (i.e. a directed graph). By
its set of vertices and by by
s(a)
(resp.
e(a))
Finally we denote by A
&1
its set of arrows. For
A0
a E Al
we denote we denote
the starting point (resp. end point) of the underlying graph of
A.
Certain graphs will play a special role.
a.
44
Dynkin graphs
(i)
d1
.
n
o-o-o ... o-o--o
(n
vertices)
0-0-0
(n
vertices)
o-o-o
(n+1
vertices)
o-<
(n+1
vertices)
0 It
I o-0--o-0-0 0 IE7:
0--0-0-0 I0
IE8:
0-0-6-0-0-0-0
affine graphs
(ii)
A
:
n
ID
>--O
I
IE 6:
-o-b-o-o--o I0
m
7
:
o
0
IE 8 :
(iii)
an infinite graph
9L :
o-o-o-v ......
45
(If we drop the assumption on the field being algebraically closed the other graphs occuring for example in Lie theory (Bourbaki, 1968) would complete the list above). A finite graph which is neither a Dynkin graph nor an affine graph is called a wild graph.
A path of length a quiver fying
t > I
A is of the form
from a vertex
(xja1,...)atly)
(e(ai) =s(ai+1) for all
I
(from
x
one from
x), denoted by
to
x
to
x
with arrows
< i < t, s(a1) = x
addition, we also define for any vertex
x
to vertex
x
and
ai
y
in
satis-
e(at) = Y. In
in A a path of length
0
(xix). A path of length greater or equal to
is called a cycle.
If there exists a path from x is a predecessor of x -, y, we say that
successor of
y, and x
y
y
to
a successor of
in
A, we say that
x
x. If there is an arrow
is a direct predecessor of
y, and
y
a direct
x.
We define the path algebra of
A
as being given by the
y k -vector-space with basis the set of all paths in
A. The product of two
composable paths is defined to be the corresponding composition, the product of two non-composable paths is, by definition zero. The path algebra
of A
is denoted by
kA. In
kA, we denote by kA
the ideal generated
by all arrows.
For example if
kA
is the algebra of
nxn
A = o+-o+-o ... o+-o+---o
of
vertices), then
upper triangular matrices over
In this way, we obtain an associative unit element if and only if
(n
A0
k.
k-algebra which has a
is finite. In this case the unit element
kA is given by
and only if
A
(xlx). The algebra kA is finite-dimensional if Z xEA0 -. is finite and A does not contain a cycle.
Let A be a finite quiver without oriented cycle. Then it is easily seen that the path algebra kA
is hereditary. Conversely, let
A
46
i
k-algebra, then there exists
be a hereditary, basic finite-dimensional
y
a finite quiver A without oriented cycle such that A me kA. In fact, be the additive category of finitely generated projective
a
let
A-modules then A = r(a). More generally, let
a
and let
k-algebra
A be a basic, locally bounded
be the additive category of finitely generated projective
A-modules. Let A = r(a). Then there exists a surjective algebra homomorphism
p
: kA -* A
and a two-sided ideal
-*+ n - + 2 (kA ) c I c (kA )
n > 2
for some
I
kA
of
with
such that the induced map
-1-
kA/I -+ A
is an algebra isomorphism (Gabriel (1980)). The pair with
(A,I)
will be called the bound quiver associated
A.
Given vertices
x,y E Ao, a finite linear combination F cww
w with x
to
cw E k, where
w
are paths of length greater or equal to two from
-A. Any ideal
y, is called a relation on
-.+ 2
I c :(kA )
can be gene-
rated, as an ideal, by relations; if it is generated as an ideal by the set
{pili}
of relations, we write Given a quiver
the form
w-w', where
I =
.
A, a commutativity relation is a relation of are paths having the same starting point, and
w,w'
the same end point. A relation of the form w
(given by a single path
is called a zero relation. If
is a path in A we also write
w = (xla1,...,aQly)
We consider the following two examples. By the algebra of
nxn matrices over
k.
a 0 0 e 0 (1)
A={ O d
cc
0
0 0 0 c
E M4(k)}
Mn(k)
we denote
w)
47
Then the bound quiver
associated with A
(A,I)
oa+o-o, I =
A =
a 0 0 c e
Then the bound quiver
is given by
0 a 0 0 0
0 0 b 0 0
0 0 0 a 0
0 0 0 d b
E M5(k)I
associated with A
(A,I)
A = (Ao,AI)
is given by
a (contravariant) representation
i V = (V(x),V(a))
spaces
V(x), for all
for any arrow
f(x)
a
k, a map
over
A
of
for any arrow
I L(L)
:
a path in
V
k-linear maps V(a)f(x) = f(y)V'(a)
A.
V(w)
V(w)
A. If
is a
k-linear map from p = E cww, if
w w occuring in
(Note that, by definition all paths
and a fixed end point, say
E cwV(w) E Homk(V(y),V(x))). w Let A be a quiver and
I = . We denote by
L((A,I))
satisfying the relations
pi
w = (xjai,...,atly)
is
the composition
satisfies the relation
x
A
i
be a representation of
So
ting point, say
: V(y) - V(x),
V(a)
in A. In this way, we obtain the category
x -, y
A, we may denote by
V
k-vector-
are two representations of
x E Ao, such that
for all
V(w) =
We say that
V,V'
: V -+ V is given by
of representations of Let
AI' If
in
f = (f(x))
a
k-linear maps
x E Do, and
x -> y
: V(x) - V'(x),
is given by finite-dimensional
k
over
I
V(y)
to
V(x).
cwV(w) = 0.
w p
have a fixed star-
y, thus
be a two sided-ideal in
kt, with
the category of representations of A
for all
i.
48
A be a basic
The following result is easily established. Let
y k-algebra and
finite-dimensional
be the bound quiver associated
(A,I)
I with
A. Then mod A and
are equivalent.
L((A,I))
We need some additional notation for certain modules occuring
A be a basic locally bounded
quite frequently. Let {ex,x E J}
Then
k-algebra. Let
be a complete set of primitive orthogonal idempotents in
P(x) = Aex, for
x E J, is an indecomposable projective
A.
A-module.
We obtain in this way a complete set of representattives from the isomorphism classes of indecomposable projective the radical of
P(x). Then
Clearly, S(x)
x E J
for
A-modules. Let
S(x) = P(x)/rad P(x)
rad P(x)
is a simple
be
A-module.
is a complete set of representatives from the
A-modules. We denote by
isomorphism classes of simple the indecomposable injective
A-module whose socle is
I(x) = Homk(exA,k)
S(x). Again, we
obtain in this way a complete set of representatives from the isomorphism classes of indecomposable injective
be the bound quiver associated with
Let
ax
has vertices
A-modules.
x E J. We assume that
for
A. Then p
is mapped to
(axIsx)
ex
1 under the surjection projective call
p
: kA -, A. We call
P(x) = P(ax)
A-module associated with the vertex
S(x) = S(ax) (resp.
I(x) = I(ax))
of A
(i.e. there is no arrow
I(ax) = S(ax)
arrow
a E A
5.2
if and only if such that
Let
a E AI ax
A. Similarly we
of
the simple (resp. indecomposable
injective) A-module associated with the vertex the following observation. P(ax) = S(ax)
ax
the indecomposable
ax
of
A. Clearly we have
if and only if such that
is a sink of
ax
is a source
e(a) = ax). Dually,
A
(i.e. there is no
s(a) - ax).
A be a hereditary, basic finite-dimensional k-alge-
1 bra
(i.e. the path algebra
kA
of a finite quiver without cycle).
49
LEMMA.
Then
Let
X'
Db(ks).
be an indecomposable object in
is isomorphic to a stalk complex with indecomposable stalk.
X'
Proof :
Since. D (kA)
is equivalent to
Kb(-*I), it is enough
Kb(PI)
is isomorphic to some
to show that each indecomposable object of
... 0 - Ij dj
I
Ij+J -, 0 ... where
Let
is surjective.
d3
be indecomposable in Kb(kf). Applying
I'
sary, we may assume that
if neces-
has the form:
I'
°
dl
with
g
Consider a factorization
T
I° --
h X - I
1
0
* 0.
-+
in mod kA with
d°
of
g
sur-
I h
jective and
injective. Since
kA
is hereditary it follows that
X
is
I an injective
X 0 C
kA-module and
i i. I
h
n mod M. Since
is a section. Thus we have an isomorphism
= 0 we obtain an isomorphism of
hd
complexes:
u II
II
... 0 -- I°®0 Since
I'
I
X®C
o
o
udf
0
o 02
II
0®I2
0d
II
0013 ^+
is indecomposable we conclude that
... O-'I°4X-+0 ... Kb(I)
go 00
...
or
is zero in
(i.e. acyclic). In the second case
... 0 - I° i X - 0 ...
I'
is isomorphic to
in Kb(-I). In the first case, we are reduced to
a complex of smaller width.
Note that if
Db(A)
for a finite-dimensional
satisfies the assertion of the lemma then A
k-algebra A
is hereditary.
50
5.3
Let
COROLLARY.
Xo
a cycle in
Db(kA). Then each
Xi
X. E mod kA
and some fixed
n E Z.
fo-rX
f rr1
... -* Xr-1
is isomorphic to
Xo
be
for some
TnX.
1
5.4
The computation of the Auslander-Reiten triangles is di-
vided into two steps. First let
for some
Z' = T1Z
and some indecomposable
i E 7L
H non-projective u
kA-module
Z. Then we have the Auslander-Reiten sequence
v
1-4 0 -+ X -. Y - Z -. 0 ending in Z. Let w E Ext kA A(Z,X) = Hom
(Z,TX)
_+
Db(ks)
be the corresponding element. Then we obtain a triangle Tv
Tu
'T1Z
It is straightforward that the properties
X.
(AR1), (AR2)
and
(AR3)
of
are satisfied.
4.1
Let us now turn to the case is the indecomposable projective
lowing
P(a)
Z' = T1P(a), i E 7L, where
kA-module associated with the vertex
A; for simplicity, we will assume that
of
Ti
T1X =.T1Y
i = o. Denote by
a
the fol-
E
A):
kA-module (considered as a (contravariant) representation of
E(x), x E Ao, is the vector-space freely generated by the paths of the form
p: x -4 ... - a or q: a comparable with
a
in the order
a
x - y
is an arrow of
the composed path ding as
q
-A
ap; if
has the form
(so we have
-' x
and
<
x
if
is not
1 defined by the arrows of
y < a, E(a)
y > a, E(a) q'a
E(x) = 0
or not.
:
maps
E(y) -. E(x) q
onto
q'
A); if
maps or
p
0
onto
accor-
SI
The paths (resp. the non-trivial paths) stopping at a generate
module of
which is identified with
E
P(a) rad R(a)
P(a)
P(a)). The quotient
of
with the quotient
I(a)
I(a)
of
(resp. with the radical
E/P(a)
tified with the indecomposable injective
associated with
by its socle
k l
(I(a),P(a)) = Hom
(I(a),TP(a))
Db
(resp.
p
I(a), by
P(a) -+ E --
and
(k0)
p'E ExtI (I(a),P(a)) = Hom
k
a
S(a)). i
p E Ext
is iden-
E/P(a))
(resp.
I(a)
By w we denote the composition
sub-
a
(I(a),T(P(a))
Db (M)
the extensions associated
i
with the exact sequences 0 -1- P(a) -> E -- I(a) -. 0 and
(i denotes an inclusion, p a projection).
0 -, P(a) ii E --pi I(a) -+ 0 LEMMA.
The sextuple associated with the sequence [T_p,-T_n')}T
(*)
TI(a)
T.
I) P(a)
I(a)®P(a)
I(a)
is an Auslander-Reiten triangle.
Proof:
Clearly, I(a) = vP(a). By the proof of theorem
The Auslander-Reiten triangle starting at
tP(a) = TI(a)
has
w
4.6, as
last morphism. So it suffices to verify that the sextuple of our lemma is a triangle. This directly follows from the diagram below, where an arbitrary module, P c P
two submodules of
E, I
and
E
denotes
I, the quotients
d
E/P
and
E/P
respectively; by
in degrees different from
0
and
(X d Y] I
we denote a complex vanishing
which has
X
as
o-component, Y
I-component. For the other notation, see the particular case above.
as
52
CO w] [0 - I] CO 1] [P w I]
[0 - P]
[1 0]
0]
[P
to P]
[P 4E] - (ESP
El 101 1
By construction, the first line is a triangle, and the vertical morphisms are ouasi-isomorphisms of Kb(mod kA). Since
[P®P - E]
TP 0 I, the first line is isomorphic in
phic to
is quasi-isomor-
b -0 D (kA)
to the following
sequence: n
P - I [p'n ]-1I ® TP The assertion now follows from The triangle
kp. Denote by
will be called a connecting triangle.
(*)
structure of P(Db(kA)). Let a copy
lei
TP.
(TR2).
Using the result of
5.5
Tl
r r
4.8
is is now easy to derive the
be the Auslander-Reiten quiver of for
i E Z, by
r
the quiver obtained
of
from the disjoint union
i module
in
I(a)
arrow from
b
to
ri
by adding an arrow from the injective
a
to the projective module in
PROPOSITION.
5.6
11 ti iE Z
in Ii+1
P(b)
from each
A.
The quiver
r(Db(kA))
Since the structure of r
is
r.
is known (Gabriel (1980),
Ringel (1978),Ringel (1980)) we can obtain a more precise description. But first we have to recall a construction of Riedtmann (1980). Let
A be
a quiver (not necessarily finite). We will construct a translation quiver
ZA as follows: (ZA)o - Z x Ao. The number of arrows from to
(m,y) E (ZA)o
equals the number of arrows from x
to
(n,x) E (Za)o y
if
n - m
53
and equals the number of arrows from y are no arrows otherwise. The translation (ZA)o If
A
by
T((n,x)) = (n-1,x). Then
to T
(ZA,T)
x
if
m = n+1
and there
is defined on the whole of is a stable translation quiver.
is a tree, then ZZ does not depend (up to isomorphism of trans-
lation quivers) on the particular orientation of
A. In this situation we
sometimes will simply denote it by ZA. Let us give some examples: a)
Z 46
b)
Z M4
c)
ZA1
Let
r E IN, we denote by Z -/r
the stable translation qui-
ver which is obtained from Z A_ by identifying any vertex with
Trx, and any arrow x - y
in Z& with the arrow
translation quiver 2-/r is called a stable tube of rank
x E ZAm
Trx -+ rry. The r.
54
For example
71 A_/3
is given by the following translation
quiver, where the identification is along the dotted lines.
With these definitions we have the following result.
A = kA
Let
COROLLARY.
be a finite-dimensional hereditary
k-algebra. -+
(i)
(ii)
If
A
If
A
is an affine graph then the components of
-'
,
-, b
b
r(D (kA)) = ZA.
is a Dynkin graph then
r(D (kA))
are of the form ZA and 7°/r for some
r E IN. (iii)
If
A
is a wild graph then the components of
r(Db(kg))
are of the form ZA and 7A.-
_,
5.6
Let
r _7 = (r,T)
be a translation quiver. Following Riedt-
mann (1980), (see also Bongartz and Gabriel (1981)) we recall the definition of the mesh category associated with define the path category of vertices of to
x
y to
r. And given
is given by the
F. For this we have first to
r. This is the category whose objects are the
x,y E r0, the
k-space of morphisms from
x
k-vector-space with basis the set of all paths from
y. The composition of morphisms is induced from the usual composi-
tion of paths.
Next we define the mesh ideal in the path category of the ideal generated by the elements
r
as
55
mz =
E -
yEz with
a non-projective vertex (mz
z
The category
E a(a) a
a:y-z
is called a mesh relation).
is defined as the quotient category of the
k(r)
-.,
path category of
by the mesh ideal.
r
PROPOSITION.
Let A be a Dynkin graph. Then
ind Db(ks)
is
X
equivalent to
k(7lA).
Proof:
By
5.5
both categories have the same quiver. Using -)
5.4
it is easy to see that we can represent the arrows of ZA by irre-
ducible morphisms of globally stable under
-. F
:
b
T. This provides us with a full and dense functor
1
-4
k(7A) -* ind D (kA). Let
may assume that jective
F(x) = TiP
kA-module
F(y) = T1Y
which satisfy the mesh relations and are
ind Db(kA)
x,y E k(ZA). Since for some
tbmk(ZA)(x,y) - Hom
implies that
commutes with
T, we
and some indecomposable pro-
i E 7L
Homk($)(x,y) * 0
P. Under these assumptions
for some indecomposable
F
implies
Y. But then Riedtmann (1980)
kA-module
(P,Y) = Hom b . (F(x),F(y)). Thus D (kA)
F
is faithful.
5.7
The following considerations allow us to introduce'the
notion of type for the finite-dimensional
k-algebras investigated in
chapter IV. For this we have to define reflections (compare Bernstein, Gel'fand, Ponomarev (1973), or Gabriel (1980)).
Let A be a finite graph and A
a quiver without cycle. Let
i x E A0
be a sink. Then we define a new quiver
underlying graph as follows: is an arrow in in case quiver
a(A)
x * b. Let
if
-i
in case
x E A0
a :
x = b
a - b and
ax(A)
having the same
is an arrow in A then a t x a - b
is an arrow in
t -+
ax(A)
be a source then we dually define a new
ax(A). These two operations are called reflections. Let
A1, A2
56
be quivers without cycles and underlying graphs equal to
can be obtained from A2
AI
vertices aX
x1,...,x
by a sequence of reflections if there exists such that
is a sink or source of
(A2 ). If
r
I
from A2
xi
AI = aX
and
(A2)
i-I
of A
A. We say that
AI
can be obtained
I
by a sequence of reflections and a quiver isomorphism we write
1 " 2. Let us consider two examples. A = Ak
Let
then
and
AI =
If A
equal to
A
where
AI + A2
5
is a tree and
2 =
AI, A2
be quivers with underlying graph
then AI - A2. The following lemma is a straightforward exercise.
1 LEMMA.
YAI
and ZA2
are isomorphic as translation quivers
if and only if AI - A2. COROLLARY. then
AI
If
Db(kAI)
is triangle-equivalent to
Db(kA2),
A2.
y Proof:
By
5.5
the components of .y
to Z, or
7
"fir
are isomorphic to
morphic as translation quivers.
71A. Thus
r(Db(kA))
72AI
not isomorphic -,
and 7A2 are iso-
CHAPTER
1.
1.1
REPETITIVE ALGEBRAS
II
t-categories
There is a rather useful notion in the theory of triangu-
lated categories which was introduced by Beilinson, Bernstein, Deligne (1982).
A
V
t-category is a triangulated category
full subcategories
VSO
V>n
and such that for
and
endowed with two
which are closed under isomorphisms
V>_O
D
and
= T n(VlO)
the following
= T
three conditions are satisfied: (1)
For
X E VSO
(2) V° - V<1 (3)
For that
and
and
X E V
V>-1
Y E
V-1
we have that
c V-0.
there is a triangle
B' E V5O
and
H
H
V. Denote by
is called the heart of the
B" E V'-1.
the full subcategory
of a
of
is a V.
t-structure. The following fact (which we (1982): the heart
t-structure is an abelian category.
1.2
and
(V`-°,V'-°)
V`O n V'-O
will not use)is shown in Beilinson, Bernstein, Deligne H
such
B' - X -+ B" -. TB'
Under these conditions, we say that the pair t-structure on
Hom(X,Y) = 0.
Db(A)
We include an example. Let
be the derived category of
full subcategory of
Db(A)
A
A. Let
be an abelian category (resp. VSO)
V<-O
formed by the objects
X.
such that
be the
58
H1(X') = 0
i > o
for
respectively). The properties
(i < o
are easily verified. For the third property we denote by
(2)
i < o,
T>1X' E D>-1. The short exact sequence of complexes
and
T<0 X' - X' -T
0 -
for
the
T
T>1X' = X'/T< V. Clearly
and zero otherwise. Set
(T<0 X')° = ker dX T<0 X' E D-O
(T0X')1 = X1
with
X' _ (X1,dX)
subcomplex of
and
(1)
>1
yields the required triangle.
X* - 0
A second example will be given in the next section.
a be an additive category and
Let
1.3
subcategories of
a. The pair
torsion pair) on
a
D
X E T
and
Y E F.
Hom(X,Y) = 0
for all
Y E F
then
x E T.
Hom(X,Y) = 0
for all
X E T
then
Y E F.
(ii)
If
(iii)
If
Let
D
be a triangulated category with is a torsion theory on
(D')
Proof:
such that
is called a torsion theory (or
for all
Hom(X,Y) = 0
(Dfo,D-O). Then
By definition
Hom(X,Y) = 0
this property. So let
x E D<-O
for
with
traction and for
and to
X
DSO
B" E
D>1,
belongs to
We infer that
DSO. Dually, let
B" E D>1. We infer that
D.
and
Y E
D>1
Hom(X,Y) = 0
X E DSO. Again we have a triangle
D>1.
and
D?1.
v
This proves
are maximal with
for all
Y E
The
B' t X 1 B" - TB'
v = o. Thus
Y E D with
B' y Y # B" i TB'
u = o. Thus
t-structure
are full subcategories
t-structure gives us a triangle
and
B' E D$O
X E D with
e-1
and
D`-°
(i). Thus it remains to be seen that
third property of
be two full
if the following three conditions are satisfied:
(i)
LEMMA.
of
(T,F)
T,F
u
is a re-
Hom(X,Y) = 0
with
is a section and
B' E DSO Y
belongs
59
2.
2.1
Repetitive algebras
Let
city we assume that
A be a finite-dimensional A
is basic and
k-algebra. For simpli-
is algebraically closed. By
k
mod A we have denoted the category of finitely generated left Denote by Q
the standard duality on
D = Homk(-,k)
is the minimal injective cogenerator. It is an
a',a" E A
and
a E A
cp(a"aa').
to
cp E Q
then
a'(pa"
A-modules.
mod A and let
Q = DA.
A-bimodule: given
k-linear map which sends
is the
Let us construct the repetitive algebra
A, as proposed by
Hughes and Waschbisch (1983). It will be a Frobenius algebra and always infinite-dimensional (except in the trivial case
A = 0
which we ex-
clude).
The underlying vectorspace of
A
is given by
A = ( 0 A) 0 ( 0 Q), iE7L we denote the elements of course with almost all
A by
ai,ipi
iE7
(ai''Pi)i' where
ai E A, (pi .E Q, of
being zero. The multiplication is defined
by
(ai,(pi)i
(bi,$i)i = (aibi,ai+1'pi + Oibi)i'
60
Clearly
A
k-algebra.
is a locally bounded
(A more suggestive interpretation is to consider A as the doubly infinite matrix algebra, without identity
I.. Ai-1
A =
Ai
Qi-1
Ai+1
Qi
0
in which matrices have only finitely many non-zero entries, Ai = A placed on the main diagonal, Qi = Q
is
i E Z, all the remaining en-
for
tries are zero, and the multiplication is induced from the canonical maps
A ®A Q - Q, Q ®A A - Q and the zero map Q ®A Q -+ 0-) A-modules can be written in the
It is easily seen that the following way:
M = (Mi,fi)i, where the
finitely many being zero, the (1 0 fi)fi+l = 0
such that
are
fi
for all
Mi
A-modules, all but
are
A-linear maps
fi
i E Z. Instead of
:
Q ®A Mi -+ Mi+l
(Mi,fi)i
we
also write
... M-2 ... M_2
or simply
f'2 -
...
M_ l
M-1
f'1 -
f'
0
f 2'
M2 .. .
...
M0
,..
Mo -, Ml ,.. M
Mi
i if we do not want to specify the maps f.
Using this description of h : M = (Mi'fi) - N = (N.,gi) of
A-linear maps
for all
i E Z:
hi : M. - Ni
A-modules a morphism
between
A-modules is a sequence
h - (hi)
such that the following diagram commutes
61
f
1
Q OA Mi --} Mi+1 hi+1
1 ®h gi
Q 0A Ni
Ni+1
Sometimes it is convenient to use the adjoint description of A-modules. Thus an
A-module M can also be written as
where the
A-modules, all but finitely many being zero, the
are
Mi
are
A-linear maps
for all
fi
fi
: Mi - HomA(Q,Mi+1) such that fi HomA(Q,fi+1) = 0
i E Z. Instead of
... M_2
(Mi,fi)
Z
f 2
M_1
... M-2 -
or simply
M = (Mi)fi),
M_1
we also write
f.1
-
if we do not want to specify the maps
M 1°
M1
fl
Mo - M1
M
- M
fi.
It will always be clear which description we use.
Using this description of h :M = (Mi,fi) - N = (Ni,gi) of
A-linear maps
for all
A-modules a morphism
between A-modules is a sequence
hi : Mi - Ni
such that the following diagram commutes
i E Z: f.
Mi
1
hi
HomA(Q,Mi+I) HomA(Q,hi+1)
g Ni
h = (hi)
1
1 HomA(Q,Ni+1)
We will give an alternative description in
2.4.
62
2.2
Let
LEMMA.
A be a finite-dimensional
A be the repetitive algebra associated with
k-algebra and
A. Then mod A
is a Fro-
benius category.
Proof:
Since
mod A has enough projective
A-modules and has
A-modules it is enough to show that the projective
enough injective
A-modules coincide with the injective
A-modules. This is rather easy.
Indeed, the indecomposable projective
A-modules are given by f.
M = ... 0
where (so
M.
~ Mi+I
is an indecomposable injective
Mi+I M.
-
is an indecomposable projective
A-module, N. = Hom(Q,Mi+I)
A-module) and
fi = IM
.
Of
1
course, M
A-module. All this fol-
is also an indecomposable injective
lows immediately from the considerations in
I. 3.1.
It is easily seen that the suspension functor may be chosen so as to be an automorphism on the stable category
mod A
we see that
M E mod A
which sends
I. 2.6
is a triangulated category.
We have a canonical embedding of
2.3
mod A. Using
onto
(Mi,fi)
where
mod A
Mo = M
and
into
mod A
Mi = 0
for
i * 0. LEMMA.
functor
mod A -, mod A
Proof: If
f = 0
in
is a full embedding.
Indeed, let
mod A
In this case, f in
The composition of this embedding with the canonical
then
f
M,N
be A-modules and f:M -+ N be a map.
factors through an injective
A-module.
actually factors through an injective envelope of
mod A. In particular we obtain in
mod A
M
63
0
ti
M
... HoMA(Q,I)~ 1
with
f2 = 0, hence
implies
I -
0
f = flf2. Thus
and
Hom(Q,f2) =0
f = 0.
Let
LEMMA.
f
M
A-injective envelope of
an
I
1 N
-
...
I0
I I
0
0
M,N E mod A
i E Z. Then
and
Hom(M,T1N) c Ext1A(M,N).
First we show that
Proof: 0
Hom(M,T1N) = ExtA(M,N). Let 2
2
0 -N u ->I°(N) a II(N) d I2(N) a ... N
of
considered as
I3(N) = (Mi)
all
A-module. It is easily seen that we may choose
j. Then
MI = 0
such that
= (Mi,fi)iE 7L
A-module for all
is an injective
be an injective resolution
j. Let
for
p = (p1)
i > o. Note that Mo and
°
J°(N) e *JI(N) a -J2 (N) a considered as
0 ---*N
2
1
V = u
for
Extll(M,N) = Extj(M,T1-'N). But clearly
Hom(M,T'N) = ExtA(M,T1-IN), as an easy argument shows. Let
and
dJ = (di)
be an injective resolution of
....
A-module. It is easily seen that we may choose e3 = do
,
for all
N
Jj(N) = Mo
j. In particular we obtain that the
0
following two complexes are isomorphic.
Hom"(M,dI)
Hom"(M,d°)
HomA(M,I°(N) I
o
HomA(M,II(N))
`
HomA(M,I2(N)) - ...
M J 2 HomA(M,e°) HomA(M,eI) HomA(M,J (N)) - >... ; HomA(M,J (N)) I
HomA(M,J (N)
In particular
)
Ext'(M,N) = ker HomA(M,d1)/im
ker HomA(M,e1)/im HomA(M,e1-')
ExtA(M,N).
Homm(M,di-1)
64
we have quoted a corresponding property
1.3.3
Note that in
M,N E mod A
Db(A). For
of the derived category
and
i E Z we have that
(M,N) = Extl(M,N).
Hom
A
Db(A)
These simple observations will turn out to be very useful in the later developments of this chapter.
Let us pause for a moment and turn to an alternative
2.4
description of
mod A
for the repetitive algebra
introduced the trivial extension algebra
jective cogenerator of
Q. (Chapter V
of
we have
1.3.1
A by the minimal in-
is devoted to a more detailed study
T(A)). We recall the definition.
is the following finite-dimensional
T(A)
additive structure is
A $ Q
(a,p)
for
T(A)
A. In
a,b E A
k-algebra, whose
and the multiplication is defined by
(b,ip) _ (ab,a$ + pb)
p,i E Q).
and
(A more suggestive interpretation is to consider the elements of (0 a), where
given by matrices
a E A, 'p E Q
T(A)
as
and the multiplication is
the ordinary matrix multiplication).
The algebra of
A 0 Q
T(A)
is a Z-graded algebra, where the elements
are the elements of degree
ments of degree
1. We denote by
0
and those of
mod7T(A)
0 0 Q
the ele-
the category of finitely gene-
rated Z-graded T(A)-modules with morphisms of degree zero.
LEMMA.
mod A and
moclT(A)
Proof:
Let
A be a finite-dimensional
are equivalent.
This is straightforward.
k-algebra. Then
65
2.5
we will give some examples of repetitive alge-
2.6
In
bras. For the computation needed we have to introduce the concept of a
A be a finite-dimensional
one-point extension. Let a (left)
A-module. The one-point extension
definition the finite-dimensional
A[M]
of
k-algebra
A by M
and M is by
k-algebra:
aEA, mEM, AEk}
A[M] = {(o A)
with multiplication a a)(o' m:)
An
For example,let matrices over
am,AX'ma1
- (ao,
(o
k. Then An
be the algebra of
operates on an
n-dimensional vector-space by
left multiplication. We denote this An -module by fication shows that
(nxn)-upper triangular
M. Then an easy veri-
An[M] = An+,.
From the definition of the one-point extension we see that a necessary condition for a finite-dimensional form
M
A[M]
for some finite-dimensional
then let
P(S)
in
B
P(S) = Be. Let
generated by
e)
and
it is easily checked that
to be of the
A-module
B-module. This clearly is
admits a simple injective module
B
be a projective cover of
potent such that
B
k-algebra A and an
is that there exists a simple injective
sufficient. Indeed, if an algebra
k-algebra
S
A = B/<e>
and let (<e>
M = rad P(S). Then M
e E B
S
be an idem-
the two-sided ideal is an
A-module and
B = A[M].
Dually, we also consider one-point coextensions. Again, given a finite-dimensional coextension [MIA
of
k-algebra A and an
A by M
A-module M
the one-point
is by definition the finite-dimensional
k-algebra:
[M]A = ((0 a) , 11
to E DM, A E k, a E Al 1
66
with the obvious multiplication.
Then A
2.6
We will give now some examples of repetitive algebras.
(a)
Let
A be the algebra of upper triangular
is the path algebra of the quiver
vertices. Let
nxn-matrices.
with
.... o-o
A+ = o+o
n
be the following quiver
.. 0E0*-0*-0*10 ...... 0+0+0+0+0 ...
with set of vertices the integers Z, for any vertex arrow kA
as
:
a -, a-l. Let
generated by
I
a E Z, there is one
be the two-sided ideal in the path algebra
as aa_I " ' aa_n_lfor
a E Z. Then it is an easy compu-
tation using successive one-point extensions and one-point coextensionsto show
that
A
kg by the ideal
is isomorphic to the factor algebra of
I.
2
(b)
A be the quiver
Let
and
I
the two-sided
Y
ideal in the path algebra kX generated by
tor algebra of kZ by Let
F3a. We denote by
A
the fac-
I.
be the following quiver:
-3
Y-2
-2
Yo
0
Y2
2
Y4
with set of vertices the integers Z, for any vertex two arrows
as : a -+ a-1
ideal in the path algebra
and
ya
:
a i a-2. Let
I
a E Z
there are
be the two-sided
kk generated by
aaaa-1' aa'a-1 - Yaaa-1' Ya1a-2
for
a E Z.
Using successive one-point extensions and one-point coextensions one can show that
A
is isomorphic to the factor algebra of
kA by the ideal
I.
67
2.7
In
we have defined the notion of a
1.1
t-structure on
a triangulated category.
A be a finite-dimensional
Let
PROPOSITION.
mod A has a natural
the triangulated category
Consider the full subcategory
M>O
(resp.
mod A formed by the objects which admit a decomposition such that
is projective-injective and
Z
n > o). We claim that formed by the objects
M>1
= T(M?O)
Yri 0
f'1
f°
is adjoint to
n < o
mod A
in
(resp. for M>°
: Y1 - Hom(Q,Y0).
mod A under isomorphisms.
belongs to it if
Therefore, in order to prove that
T -V
enough to consider the case where
Vn = 0
P = (Pnign)
yIom(Q,Y1)
f_1
Indeed, this subcategory is obviously closed in
has a projective cover
Y 0 Z
such that the induced sequence
Y = (Yn,fn)
mod A, where
for
of
M-O)
is the full subcategory of
Q 0A Y-1 - yYo is exact in
o (M- ,M- )
t-structure
mod A.
with heart equivalent to
Proof:
k-algebra. Then
in
for
mod A
V E MAO, it is
n < o. In this case such that
Pn = 0
V for
g
n < o
and
PO
°
}HomA(Q,P1)
shared by all subobjects of in
is injective. These two conditions are
P, in particular by the kernel of
mod A, which is isomorphic to
T
Conversely, suppose that is exact. In order to prove that morphic object of
1V
mod A.
and that the sequence above
Y E M-O
Y E M-1, we may replace
Y
mod A, hence restrict to the case where
n < o. We then choose an injective hull Pn = HomA(Q,In)
in
for
In
P -, V
of
Yn
in
by an iso-
Yn = 0
mod A
for
and set
n > 1. The injection
f
omA(Q,Y1) -3-HomA(Q,II) = P1
Yo
phism
e
below, so that
Y
obviously extends to the monomor-
is isomorphic to
T (coker e), where
68
coker e E M>-°.
f
fN
-
0 e_1
Y0
Y2
Y1
e0
ell
ell
1
-
0
....
go
...
1
®1
gl
®P
3
®1
P2
0
1
2
Now consider a morphism h : X - Y, where h = 0
Y E M-1. In order to prove that Xn = 0
for
n > o
and
ho = 0, hence
plies
Yn - 0
in
X E M'50
mod A, we may suppose that
n < o. The diagram below then im-
for
h = 0.
h
0
0 -Y0 This proves condition
HomA(Q,Y1)
t-structure. The inclusion
of a
(1)
is clear from the above considerations. The inclusion from dual arguments. It implies
M
mod A, we construct a triangle
in
by setting
and
B" E M'1
and
B" - im f0, Bn = 0
f_I
: X-1 -. HomA(Q,XO)
map by
f_1. Let
u
the projection from into
and
Bn = Xn and
Bn = 0 for
onto
follows is
B' E MAO
n < o, Bo = ker f0
for
n > o. Note that
HomA(Q,ker f0). We denote this induced
be the inclusion of X0
C M'-°
X = (Xn,fn) such that
B' -$ X - B" -+ TB'
B" = Xn
maps to
M<-1 C MG°
c M-<1. Finally, if
and
M>1
ker f0
in f0, and let
u
into
X0, and let x
the inclusion of
in f0
HomA(Q,XI). Altogether we obtain the following exact sequence of
A-modules:
be
69
..
...
0
0 1
f-2
X_2
f-I
1
X-I
I
ker fo
,..
l
f0
0
0
I
I
0
f-2 -2 X-2
I X-I
f -1
x0
l
I
-
... B' E M`-°
B" E
M'-1.
Now
I
-
-
and using the characterization of 1.2.7
This finishes the proof.
1
...
l X2 ...
1X2
IXI fI
u
im f o
0
that
f
-
XI
n
Clearly
0
..
1IX-
IIX-2 ...
-
0
X 2 ...
XI
M>I
yields the required triangle in
I ... above we see mod A.
70
3. Generating subcategories
3.1
category of
Let
C
be a triangulated category and
C. We say that
is a generating subcategory (or simply
a
which is closed under isomorphisms and contains Let
for all
Tla
C
a
A
generates the derived category
coincides with
C C.
which is closed under isomorphims and con-
A
is an abelian category then obviously
Db(A). Another example will be important
to us and is just a reformulation of the results in
AI
of
i E Z.
For example if
finite-dimensional
C'
C. We denote by a the
a be a full subcategory of
smallest full subcategory of
well as
be a full sub-
C) if the smallest full triangulated subcategory
generates
tains
a
1.3.3. Let
k-algebra of finite global dimension. Then
generate the derived category
A be a AP
as
Db(A ).
The concept of generating subcategories is useful for proving that exact functors between triangulated categories are equivalences (compare
Let
3.4). For this it is necessary to define the notion of distance.
a c C be a generating subcategory. Inductively we define full sub-
categories
a(i)
of
C
for
i E IN U{0}. Set
i > o, be the full subcategory of
a(o) = Z. Let
formed by all objects
C
occur in a triangle whose remaining terms are in
U
a'(i), for X
which
a(i) (to be precise:
j
(L,M,N,u,v,w) is a triangle in
C
and two of the objects belong to
71
U
a'(i)). Then set
the third will belong to
a(i)
a(i) = a'(i).
j
then we define the distance
X E C
C = U a(i). Let
is a generating subcategory we see that
a c C
d(X,a)
X
of
a
to
i
d(X,a) = min{i
to be
If
X E a(i)}.
I
then
X E a
i E Z. Moreover, for
for all
d(T1X,a) = 0 then
d(T1X,a) = r. Let
X E C
d(X,a) = r. Then there exists a triangle
X1u, X2 -V, Xw, TX1
such that
d(X,a) = r
we clearly have
X E C
d(Xi,a) < r
i = 1,2. In fact, d(X,a) = r
for
Thus there exists
i E Z
such that
X E a(r).
implies that
X' E a'(r). Thus
and
X = T'X'
with
X'
occurs in a triangle whose remaining terms have smaller distance. By we may assume that
(TR2)
exists a triangle
Xi
i = 1,2. Applying
(TR2)
tain atriangle X1
-u,,
X' X2'
_+ ,
occurs in the third position. Thus there X' -+ TXi
d(Xia) < r
such that
for
sufficiently often if necessary we finally ob-
-u+ X2 -V, X w, TX1
with X. = TSX!
s E Z and
for some
i = 1,2.
3.2. We now determine generating subcategories of
PROPOSITION.
Let
A
be a finite-dimensional
finite global dimension. Then mod A
mod A.
k-algebra of
is a generating subcategory of
mod A.
Proof:
which contains X E C
that
if
Let
C
be a full triangulated subcategory of
mod A and is closed under isomorphisms. First we show X E M>-O
denotes the
(where (M-O,M-°)
mod A). We proceed by induction on If
e(X) = 0, X
to
C. So we may suppose that
and
mod A
e(X) = min{e E IN
is isomorphic to an object of e(X) = e >
1
mod A
and that
t-structure on for
: Xn = 0
n > e}.
and thus belongs
Y E C
if
Y E
M>°
e(Y) < e. We then proceed by induction on the injective dimension
id(Xe)
of
Xe
in
mod A . A minimal injection
i
:
Xe -I l
into an in-
72
jective
A-module I extends to a morphism
denotes the projective-injective such that in = 0
n * e,e-I. Clearly
if
(e(C) < e
id(Xe) = 0
if
follows that
Proof:
r = id(X)
Let
A
Let
for
C = coker j
MAO c C. Our propo-
be a finite-dimensional
k-algebra of
is a generating subcategory of
be a full triangulated subcategory of
C
mod A
and
id(Xe) > 0). It
AI and is closed under isomorphisms. By
enough to show that on
if
J
(3).
finite global dimension then AI
which contains
C, and so does
X E C. By duality, we obtain that
COROLLARY.
mod A, where
e(ker j) < e, we finally have that
sition now follows from axiom
3.3
of
Je = I, Je-I = HomA(Q,I)
id(Ce) < id(Xe)
im j E C. Since
ker j E C, hence
: X -, J
belongs to
J
and
j
is containd in
X E mod A. For
o
r
3.2
mod A.
mod A
it is
C. For this we use induction there is nothing to show. So
let X E mod A with id(X) = r > 1. Let 0 -+ X - I -+ Y -+ 0 be exact in mod A with
id(Y) < r. The above exact sequence yields a
I E AI. Then
triangle X -+ I -, Y -, TX in mod A. As I that
and
X E C.
We also note the dual assertion. Let AP
is a generating subcategory of
3.4
(resp. T')
Let
C
and
X,X' E a (b)
is also dense.
and all
If
F
:
C
C -+ C'
(resp. C'). Let
a c C
then F(a)
F
T
be a
be an exact functor.
HomC(X,T'X') F Hom C,(F(X),T'1F(X'))
i E 71
If moreover
be as above. Then
mod A.
the translation functor on
LEMMA. (a)
A
be triangulated categories with
C'
generating subcategory and let
all
belong to C, we infer
Y
for
is full and faithful.
is a generating subcategory then
F
73
Proof:
transformation
Since
a : FT -+ T'F. To prove
double induction on d(Y,a)
d(Y,a)
and
d(X,a)
let
(a)
X,Y E C. We proceed by d(X,a) = 0 =
(compare 3.1). If
the assertion follows from the assumption. Assume that the asser-
tion is true for
with
is exact there exists an invertible natural
F
with
X,Y E C
d(X,a) = 0
and
d(Y,a) = r. Then there exists a triangle
Y E C
d(Y,a) < r. Let
in
Y1 + Y2 _+ Y w+ TY1
C
d(Y1,a) < r for i = 1,2. Since F is exact this yields a F(u) u) F(Y2) F( F(w)ayl * T'F(Y1) F(Y1) F(Y) in C'. Applying
such that
the cohomological functors
HomC(X,-)
and
yields the
HomC,(F(X),-)
following diagram with exact rows:
If2
Ifi
HomC(X,TY2)
HomC(X,TY,)
RomC(X,Y2) -- HomC(X,Y)
HomC(X,YI)
f5
f4
lf3
Homc,(F(XX),F(Yj)) - HomC,(F(X),F(Y2)) - Homc,(F(X),F(Y)) - Homc,(F(X),TrF(YI)) + HomC.(F(X).T.F(Y2))
By induction it follows that
are isomorphisms, hence
f1,f2,f4
and
f5
f3. The remaining part of the proof of
(a)
is dual.
To prove d(X,F(a)). If
let
(b)
d(X,F(a)) = 0
X E C'. We proceed by induction on there is nothing to show, for
F
commutes
with the translation functors up to isomorphism. Assume that the assertion is true for
X E C'
with
d(X,F(a)) < r
d(Yi,F(a)) < r. By induction there exist
such that
u' E HomC(X1,X2) X1
-11
such that
r
r
X2 °.
u
X1,X2 E C
i = 1,2
for
and
with in
C'
and
F(u') = u.
r
X -+
TX1
F(v')
F(X1) + F(X2) -+ F(X) finally apply
F(Xi) = Yi
Y E C'
Y1 u Y2 °i Y _* T'Y1
d(Y,F(a)) = r. Then there exists a triangle
Let
and let
(TR3)
be a triangle in F(w')aX1 4 T'F(X1)
to conclude that
This finishes the proof of the lemma.
C. So we obtain a triangle in
C', for
F(X) - Y, hence
is exact. We
F
F
is dense.
74
The main theorem
4.
A be a finite-dimensional
Let
A
k-algebra and
its asso-
ciated repetitive algebra. In this section we will construct a full and faithful functor
F
A has finite global dimension
Db(A) -, mod A. If
:
F
will be even an equivalence of triangulated categories. The scheme of the construction will be outlined in
4.2, while the single steps in the con-
struction will occupy the remaining sections.
There exists an exact functor
LEMMA.
4.1
and a monomorphism
u
:
id -, I
such that
I(X)
I
: mod A 1 mod A
is injective for each
X E mod A.
For each
Proof:
X = (Xn,fn) E mod A, we define
I(X) _ (Indn)
by
where the left
A-module structure on
In = Homk(Q,Xn+1) 0 Homk(A,Xn)
dule structure of
is induced by the right
In
do : Homk(Q,Xn+1) - HomA(Q,Homk(A,Xn+i))
mapping
p(X)
y(X)n = (fn'En)' where
En : Xn -. Homk(A,Xn)
maps
is easily seen that
and
onto
q -
I
p
n(X)
:
I(X) -, S(X)
fines an exact functor
S
by
X i I(X)
:
x
onto
a - ax. It
satisfy the assertions above.
With the notation above, we set note by
A-mo-
is the canonical isomorphism
(a -' tp(aq)). We define
p
(0 dnl, `0 0),
do =
A, and where
and
Q
and
S(X) = coker p(X)
and de-
the canonical projection. This clearly de-
: mod A -, mod A mapping injectives onto injec-
75
tives. Thus
by
mod A - mod A which we will denote
induces a functor
S
S.
We have defined an automorphism
on mod A which serves
T
us as a translation functor for the triangulated category exists an invertible natural transformation
y
:
mod A. There
T. We will use this
S
fact in the final part of the construction.
4.2 Cb(mod A)
Let
be the full subcategory of the category
A)
of bounded complexes which is formed by the complexes vanishing
in positive degrees. The translation functor and the mapping cone of a morphism
f'
is defined on
T
in
C50(mod
A)
is contained in
A)
A).
i > o, denote by
For A)
with objects
We identify
C[0,0]
with
X' = (Xn,dn)
yields a functor properties of
F'
F'. In
i
such that :
of
4.5
C-O(mod A)
the diagram below. In F
:
Kb(mod A) -, mod A
such that
Xn = 0
for
n < -i.
we will construct in
4.3
functors
= Fi_1
for
i > 1. This
FiI
C[-i±1,0] C5O(mod A) -4 mod A. In
4.4
we will derive certain
we will show that the composition of
the projection mod A -' mod A K-O(mod A)
the full subcategory of
mod A.
By induction on Fi : C[-i,0] - mod A
C[-i,o]
F'
with
factors over the residue category
module homotopy. This defines the functor 4.6
we extend this functor to a functor
and show in
4.8
to isomorphisms, hence yields a functor
that it maps quasi-isomorphisms Db(A) 4 mod A.
The scheme of the construction is depicted in the following diagram:
76
id
mod A = C[0,0]_
mod A
1
C(-i'01 - mod A
F
CEO (mod A)
I K5O(mod A) I
F
Kb(mId A)
mod A
I
Db(A)
Let
4.3
mod A, we define Suppose that
F
0
Fi_1
:
C[-i+1,0] - mod A
n > o, X'n = Xn
that
X'n = 0
Then
T X'
is contained in
from
T -X'*
to
defined on
for
X°
T X'',X°
X'' = (X'n,dn,)
n < o and
whose mapping cone is and
into
dXl
and
the complex such
dX, = dX
for
X. The functor
Fi-1
F
i-1
(e')
X
Fi-1(X°)
u(Fi-1(T X")) I(Fi-1 (T X")) ir(Fi-1 (T X"))
S(Fi-1 (T X"))
i-1
eX
is
eX. Consider the following pushout diagram in
Fi-1 (T X'')
Fi(X-) = CF
n < -l.
induces a morphism
mod A:
Then we set
mod A.
is already constructed. Let
for
C[-i+1,0]
with
C[0,0]
mod A
to be the canonical embedding of
C[-i,0]. Denote by
be in
X' = (Xn,dn)
i = o. Using the identification of
(e )
X
uX CF
1 I
- (eX 1
vX
S(Fi-1 (T X"))
77
Let
f'
:
induces a morphism f" n > o
f'n = fn
and
be a morphism in
X' -+ Y' (f'
for
X,*
n)
- Y''
:
C[-i,O]. Then
f' = (f n)
f'n = 0
such that
for
n < o. In particular we obtain the following
commutative diagram with objects and morphisms contained in
C[-i+1,0]:
eX
TX"
X°
f°
T -f'* eY
) Y°
T -Y'*
By induction and the definition of
on objects we obtain the following
Fi
diagram: F1_I(eX)
Fi-1(X°)
F(T X')
Fi-I(f°)
F. I(T f") (Fi_I(T X")) +
Pi_I(T Y")
uX
l-1
Y) 1
(Y°)
e
I(Fi-I(T X")) X Fi (X')
(Fi-1(T Y"))
"Y
I(Fi_I(T f")) n(Fi_I (T X"))
I(Fi_(T Y")) Y-- Fi(Y')
vX
vY
S(F i_1(T X"))
.(Fi-1(T Y"))
S(Fi_1(T X")) S(F
S(Fi_I(T f
We have morphisms Fi_1(f°)uY
:
Fi_1(X°) --+ Fi(Y') Fi_1(eX)Fi_1(fo)uy
unique
I(Fi-1(T f " )) eY
:
S(Fi_1(T Y'*))
(i i ")
I(Fi-1(T X'')) -+ Fi(Y1)
and
such that
= 11(Fi_1(T X "))I(Fi_1(T F " ))eY. Thus there exists a
h : Fi(X') -# Fi(Y')
such that
eXh = I(Fi_1(T f " ))ey. Then we set
uXh = Fi_1(f°)uy
and
Fi(f') = h. Observe that
Fi(f')vy = vX S(Fi_1(T Y " )). For this it is enough to show that uX Fi(f')vy = uivX S(fi-1(T Y'*))
and
eX Fi(f')vy = eX vX S(Ii-1(T Y ")).
The first follows from uX Fi(f')vY = Fi_1(f°)uy vy = 0. For the second
78
we have that
eX Fi(f')vY = I(Fi-1(T f "))eY vy =
I(Fi-1 (T f"))r(Fi-1 (T Y")) = r(Fi-1 (TX' ))S(Fi-1 (T f")) _ S(Fi-1(TY'')).
eX vY
It is a straightforward verification that from
C[-i,0]
to
i =
I
and
k-linear functor
mod A.
Next we show by induction on If
defines a
F.
X' E C[0,0], then T-X"
der the following pushout diagram in
i
that
= F. i-1'
F.
1 C[-i+1,0] vanishes, and we have to consi-
mod A:
0
F0(X°)
1
II
0 - F1(X°) 0
Thus
F1(X') = F (X'). If
f*
0
is a morphism in
vanishes, and it is easily seen that
C[0,0], then
F1(f') = F(f').
Suppose that the assertion is true for X' E C[-i+1,0]. Then
T f "
j < i Fi(X')
T -X'* E C[-i+2,0]. We compute
by means of the following two pushout diagrams in
and let and
mod A:
Fi-2(e') Fi-2(T X'')
Fi-2(X°)
u(Fi-2(T X'*))
I(Fi-2(T
S(Fi-2(T X''))
Fi1(X')
S(Fi-2(T X"))
Fi-1(X')
79
Fi-1 (eX) Fi-2(T X'') u(Fi-1 (T X'*))
I(Fi-1 (T X")) -
S(Fi-1 (T X'*))
Fi_2(T X'') = Fi_1(T X''), Fi_2(X°) = Fi_1(X°)
By induction
Fi_2(eX) =
Therefore
Fi_1(eX).
Fi(f') = Fi_1(f')
see that
S(Fi-1 (T X' ))
LEMMA.
and
Fi_1(X*) = Fi(X'). It is also easy to
for a morphism
The functor
F' = lim F.
f' E C[-i+1,0].
satis-
A) -* mod A
:
i
fies
F'IC[-i,0] = Fi
morphism
r1
:
that
r1
i > o
and is associated with a canonical iso-
F'T =, SF'.
Proof:
struct
for
The first assertion is clear. So it remains to con-
: F'T - SF'. Let
X' E C[-i,0], but
X* E
A); then there exists X' = T ((TX')')
X' ff C[-i+1,0]. Clearly
e.X = 0. We consider the following pushout diagram in
i
such and
mod A:
0 -* 0
I(Fi(X')) - Fi+1(TX') II v
i S(Fi(X'))
Note that
Fi+1(TX*) = F'(TX')
and that
S(Fi(X*))
S(Fi(X')) = S(F'(X')). We set
80
n(X') = v. Clearly
Let
4.4
is an invertible natural transformation.
n
X. E C5O(mod A). We consider the associated standard
e
sextuple
T X" X >Xo X' -°sX". LEMMA. F'(ek)
F
(T-X
X'E C50(mod A), the associated sequence
If
F (X
o)
F'(u'). F (X')
F'(v' )*F (X ')
is a standard
sextuple in mod A. In particular 0 -
F'(u)-
`F'(X°)
sequence in
F' (X.)
F' v')
F'(X ') -0 is a short exact
mod A.
There exists
Proof: and
F'(u') = Fi(u')
i
such that
F'(X') = Fi(X')
F'(v') = Fi(v'). By definition of
consider the following commutative diagrams in
Fi
we have to
mod A. We consider the
diagram which has to be constructed in order to compute
0 -
and
Fi(u'):
F.1(XO) IFi-1(Xo)
Fi-1(T X ")
IIFi(Xo).
Fi(u )
i
- Fi-1(X°)
i
X
uX
I F.(X')
I(Fi-1T X''))
0
1
vX
0 II
S(Fi-1(T X''))
S(Fi-1(T X''))
In particular
uX = F1(u').
The second case is similar.
LEMMA.
Let
0 -+X' u 4Y'
quence of complexes contained in
Z'° 0
be a short exact se-
A). Then also
81
0 -.. F'(X') F'(u ), F'(Y') -rF'(Z') -oO is a short exact sequence in
mod A.
Proof:
If
Y' E C[-i,0]
for some
X',Z' E C[-i,0]. We proceed by induction on exact sequence is an exact sequence of
cal embedding of mod A that
F'
C[-j,0]
into
mod A
i > o, then
i. For
A-modules. As
i = o F
0
the above
is the canoni-
the assertion follows. Now assume
transforms short exact sequences of complexes contained in for
j < i
to short exact sequences in
mod A. Let
0 --+ X, u > Y. - >Z' -p0 be a short exact sequence with Y' E CC-i,01. Consider the following commutative diagram of exact sequences
0
0
1
0 -k 0 u -0 X
1
>
Y
0 0
0
>
1
X'.n
0
k 0
1
1
0->X' u Y' 0 -->
Z
Z' --k0
,. -y.Y,.v 1 ,.- I
-> Z0
I
I
I
0
0
0
By the preceding lemma we know that
F'
diagram above to short exact sequences of
transforms the columns of the
A-modules. Clearly the same
holds for the first row. Consider the exact sequence
0 -FT X''
T U"
-Y'' T T
" T-Z" v+
0
which is contained in
C[-i+1,0]. By induction we have that
0
F' (T X") F'
(T
F' (T F' (T Z") ->0
is exact. Therefore also
0 -SF'(T
X'.) SF'(T u'*)- SF'(T Y'.) SF'(T v'*), SF'(T Z'')
-0
82
is exact, which is isomorphic to
0 ->F'(X'') F,(u )I F'(Y' ) F'(
)- F'(Z'') -- 0. Thus the third row
of the diagram above is transformed under
to an exact sequence. This
F'
yields the assertion.
4.5
We denote by
the residue category of
K-O(mod A)
modulo homotopy.
C-O(mod A)
LEMMA.
F'
- mod A
F5O : e-O(mod A)
induces a functor
associated with a canonical isomorphism O : FOOT =,
Proof:
ject in
It is enough to show that a projective-injective ob-
A)
module in
is transformed under
mod A. Let
into a projective-injective
F'
be projective-injective. We may
X' E C-O(mod A)
assume that
X'
is indecomposable. Applying
assume that
X'
is of the form
if necessary, we may
T
1- X°
... 0 -, X
-4 0 ...
.
But then
is computed by means of the following pushout diagram in
F'(X')
X1
X0
1
J_1
I(X
- F'(X')
)
1-
S(X-1) 1
S(X
1)
Thus
F'(X') =, I(X 1), which is a projective-injective module in
Thus
F'
formation
factors over
As noted before
mod A denoted by
Denote by
K-o(mod A). Clearly
mod A.
induces a natural trans-
p
C5°.
4.6
gory
mod A
S'
induces a functor on the stable cate-
S
S. It turns out that
a quasi-inverse of
S
on
S
is a selfequivalence.
mod A and by
a
:
S'S - id
an
83
invertible natural transformation. We also choose an invertible natural transformation
S
:
id -
SS'.
We inductively construct an invertible natural transformation ar
:
S'rsr - id
al = a
i < r. Then we define
structed for ar
r > 1. Let
for
and suppose that
r',r > o (1)
is con-
ar(X) =
Clearly
is an invertible natural transformation. Let
that for
ai
ao = id. It follows
we have:
ar+r'(X) =
Later we will need two consequences of this formula: (2)
S,r(as-r(Sr(y)))
(3)
S's(ar1s(Ss(Y)))
LEMMA. F E
:
Proof:
such that
s-r > o
=
for
r-s > o
FIK
Let
r > o
Sit(X)(ar(F
X*).
we have isomorphims
S't(X)+rF:5o(Tt(X)+1x.) ->F(x').
:
on morphisms as follows. Let
Kb(mod A). If
t(Y) > t(X)
t(Y) < t(X)
f'
:
X. - Y'
be
we define with
F(f') = If
t(X) > 0
be minimal with this property.
t(X)
t(X) F
Note that for
a morphism in
EIK
X' E Kb(mod A). Then there exists
F(X') = S'
We define
F
A) =
Tt(X)X. E K-
Then we define
k-linear functor
and an invertible natural transformation
such that
SF
for
There exists a
Kb(mod A) -, mod A
: FT -
=
r = t(Y)-t(X).
then we define s = t(X)-t(Y).
F(f') =
Observe that for Clearly
F
is
t(X) = t(Y)
k-linear and
both definitions coincide. F(1X') = IF(X')'
84
Let us show that
F
preserves the composition of morphisms.
X',Y',Z' E Xb(mod A)
For this let
and
f'
: X' -' Y', g'
:
Y' -' Z'
two morphisms. We consider the following cases: t(Z) < t(Y) < t(X).
a)
Let
s - t(X) - t(Y), s' - t(Y)-t(Z). Thus
s+s' = t(X)-t(Z).
By definition
by
(1)
S,t(Z)
=
by naturality of
by definition
as
F(f')
and
F(g')
=
b)
Let
t(Y) < t(Z) < t(X).
s = t(X)-t(Y), r - t(Z)-t(Y). Thus
By definition
by naturality of =
by
(2)
a
s-r
s-r = t(X)-t(Z).
be
85
= r
by definition of
F(f')
and
F(g')
=
c)
Let
t(y) < t(X) < t(z).
s = t(X)-t(Y), r = t(Z)-t(Y). Thus
r-s - t(z)-t(x) > 0.
By definition
r
=
by naturality of
r- s
=
by
1
r s
(3)
r
by definition of
F(f')
and
F(g')
F(g').
= F(f')
The remaining three cases are similar.
Next we will define . For this let If
(X') _
t(X) = 0, let
S,t(X)-1F<0(Tt(X)X')
F(TX') =
SS1S1t(X)-1F<0(Tt(X)X').
where
S
Clearly
:
Let
show that
and
t(TX') = t(X)-1 > 0. Then SS't(X)F
SF(X') =
Then we define
!;(X') =
_
0(S,t(X)-1F<0(Tt(X)X.)),
is the chosen invertible natural transformation.
id y SS'
!;(X')
Otherwise
X' E Kb(mod A).
is an isomorphism. f'
:
X' - Y'
be a morphism in
Kb(mod A). We have to
We present the case
t(X) > t(Y).
86
The other case is similar. Let
s = t(X)-t(Y). Then S't(Y)-1(as(F-O(Tt(Y)Y.)))
(Tt(Y)Y.)).
SS't(Y)(as(F$O(Tt(Y)Y*))) =
SS't(Y)(as(F-O(Tt(Y)Y.))) =
is an invertible natural transformation such that
E
'
'IK
and clearly also
FIK
FRO. This finishes
the proof of the lemma.
4.7
mation
y
:
In
4.1
we have chosen an invertible natural transfor-
S :, T. Recall that
T
is the chosen automorphism which serves
us as a translation functor for the triangulated category ticular we obtain an invertible natural transformation
mod A. In parFT
with
TF
_ C (yF) . PROPOSITION.
Proof:
F
is an exact functor of triangulated categories.
We have noticed before that
F
commutes with
T
up
to isomorphism. Clearly we may restrict to the triangles constructed 1.2.7
from
and contained in 4.4.
4.8
F(f')
K:50(mod A). But then the assertion follows
LEMMA.
Let V : X' -a-Y'
is an isomorphism in
mod A.
be a quasi-isomorphism. Then
87
Proof:
Applying
T
C
contained in
triangle associated with
Kb(mod A). Since
F(X') F(f.) F(Y') -+ F(Cf.) - TF(X')
a triangle
mod A. By
in
morphism, we have
is acyclic.Thus it remains to show that
A). Let
Z'
f'
it
F'(Z')
A-module whenever V is an acyclic complex
is a projective-injective in
mod A. Since
1.1.7
is a quasi-iso-
F(Cf.) = 0
Cf.
is
is exact, we obtain
F
is enough to show that
in
f'
be the standard
X' f rY' -, Cf. - TX' in
f'
if necessary, we may assume that
be an acyclic complex in
(mod A). We may
assume that
d-1
do 0 -, Zn _Z Zn-1
Z'
-
... -s
Z
Z-1
Z° - 0
...
Consider the following exact sequence of complexes:
0
= ... 00-00 .....
1
1
U
= ...
1
, --,Z
n -j- Zn
1
1
Idn
... 0 -
Z
11n
n
dZ yZn-I
V' _ ... 0 -.. 0 --im1 1
1
0
where
1
-dZ_I
1
u
1
-s Zn-2
0 -,0
0 ...
.....
is the canonical factorization of
projective-injective
1
0 ...
n- I
= ... 0 -, 0 -. 0 --,0
Tru
1
0
dZ-.Zn 2 .....
Inductively we may assume that
By
0- 0- 0 ...
dZ-1.
F'(U')
and
F'(V')
are
A-modules. The start of the induction being
4.5.
4.4 we infer that 0 - F' (U') -, F' (Z') - F' (V') -. 0 is a short exact
sequence of
A-modules. As the end terms of this short exact sequence are
projective-injective
A-modules, so is the middle term.
88
4.9
THEOREM.
The exact functor
F : Kb(mod A) - mod A
triangulated categories induces an exact functor
F
Db(A) - mod A of
:
triangulated categories which is full and faithful such that If
of
id. F(mod A =
A has finite global dimension, this induced functor is a triangle-
equivalence.
Proof:
The existence of an exact functor
F : Db(A) -, mod A
follows from the universal property of the derived category using and
4.8. We use
3.4
to show the remaining properties. Clearly
is a generating subcategory of for
X,Y E mod A and
This follows from ful. If
and
Db(A)
(X,T1Y)
1.3.3. This shows that
A has finite global dimension we know by
F(mod A) = mod A
is a generating subcategory of
dense in this situation.
mod A
Db(A). Thus it remains to be shown that
i E Z we have Hom
2.3
4.7
F
3.2
F Hom(F(X),T1F(Y)). is full and faiththat
mod A. Thus
F
is also
89
Examples
5.
We now turn to an example of a repetitive algebra.
5.1
Let
V
be an
algebra on
(n+1)-dimensional
k-vectorspace and
A
the exterior
V which we consider with its usual grading. Let
modZA
be
the category of finitely generated Z-graded A-modules with morphisms of degree zero. We will construct a locally bounded Frobenius algebra A such
that
mod A ce modZA. Let t be the following quiver:
a
a
n
a
n
a
n
with set of vertices the integers Z, for any vertex n+1 arrows
path algebra
a E Z, there are
ai = aia) : a -+ a+1, 0 < i < n. Let ? be the ideal in the
generated by
k
a(a)a(a+1),
i
n
i
i
a(a)a(a+l)
J
j
for all
i
i,j
with
i <
and all
Denote by A the factor algebra of a locally bounded
kZ by the ideal
j a E Z.
T. Clearly A is
k-algebra which moreover is a Frobenius algebra. Note
that the indecomposable A-projective ponding to the vertex
P
with
top P
the simple corres-
a E t coincides with the indecomposable x injective
90
I
with
the simple corresponding to the vertex
soc I
a-n-1. Moreover
mod A - modZA.
it is easily checked that
Next we want to show that A - A for some finite-dimensional k-algebra
A. Given
a < b
A
(1,I)
where I
a < x < b. We
A
is given by
is the following quiver:
a(l) 0
(n-2) 0
(n-1) 0
n-2
n a(n-2)
a(n-1)
n I
the restriction of
the finite-dimensional algebra Ao n. Thus
the bound quiver
and
7L, denote by Aa,b
I with vertices satisfying
to the full subquiver of denote by
in
n
is generated by
a(a)a(a+1) a(a)aja+1) +
for all
with
i,j
We claim that simple projective module
posable injective we see that
i < j
a(a)a(a+1)
and
A cd A. For this observe that
PA(o). Let
IA(o)
A-module. Then clearly
Ao,b, for
0 < a < n-1.
A has a unique
be the corresponding indecomA[IA(o)J = Ao n+1' Inductively,
b > n, is obtained from Ao,b-1
extension, using the indecomposable injective
as a one-point
Ao,b_1-module
I(b-n-1).
Using the dual process of successive one-point coextensions we finally see that
A = A.
As an immediate application of triangle-equivalent to
4.9
we see that
modZA. This special case of
4.9
Db(A)
is
follows also
from the work of Bernstein, Gel'fand, Gel'fand (1978) and Beilinson (1978).
A more detailed account can also be found in Gel'fand (1984). It is beyond the scope of this book to go into the details of this work, but we will
91
briefly indicate their approach. In the work of Bernstein, Gel'fand, Gel'fand the category of graded modules over the exterior algebras was reVn
lated to the category IPn(C), where
C
of vector bundles over the projective space
denotes the field of complex numbers. More precisely
they obtained the following result. Let coherent sheaves on
coh(IPn(C)) be the category of
IPn (C). Then they show that
Db(coh(IPn (C)))
molLA. The result of Beilinson can be formulated
triangle-equivalent to
as follows. Db(coh(IPn(C)))
Db(A). Thus com-
is triangle-equivalent to
bining both results we obtain the above mentioned special case of In other words, using
is
4.9.
we see that both approaches are quite similar
4.9
in the sense that one can derive one result from the other.
5.2
In Happel, Ringel (1986) a description of the derived
category of a tubular algebra was given. Here we want to give a rather sketchy account of the main result, for is is beyond the scope of this work to give the technical details. We will restrict to the case of canonical tubular algebras which were introduced and studied intensively by Ringel (1984). They are easily defined by their associated bound quivers. a)
The canonical algebras of type
(2,2,2,2)
are defined
by the quiver
and the ideal generated by is some fixed element in
aa' + ae '
k-,[0,1}
+ yy', aa' + X
(for different
'
X,A'
tain non-isomorphic algebras, the only exceptions being
+ 65',
where
A
we usually obA' = 1-A,
92
a-1
and
The canonical algebras of type
b)
are defined by
(p,q,r)
the quiver
and the ideal generated by the only tubular ones are those of type Let
C
be a canonical tubular algebra. If
(2,2,2,2), (3,3,3), (4,4,2), or 2,3,4
or
6
(6,3,2), let us denote by
C
be a component of the quiver of
shown in Happel, Ringel (19$6) that d.
C
and
(6,3,3).
is of type d
the number
respectively. Let
of
(3,3,3), (4,4,2)
C
Db(mod C). Then it is
is a stable tube of rank a divisor
CHAPTER
III
TILTING THEORY
In this chapter we try to give a rather full account of the tilting theory which emerged in recent years. The history of tilting is
quite interesting and we take this opportunity to collect some of the important steps. The first approach to tilting appeared in the proof of Gabriel's theorem given by Bernstein, Gel'fand and Ponomarev (1973). In this article the so-called Coxeter functors were introduced. The usage
was still quite restricted as it only dealt with finite-dimensional hereditary k-algebras. A few years later this approach was generalized by Auslander, Pla'zek and Reiten (1979). In this paper a rather special 'tilting
module' was introduced which nowadays is called an APR-tilting module and still is a very useful tilting module as we will see in the later develop-
ments. The decisive step in the building up of this theory was the work of Brenner and Butler (1980). A completely different point of view was presented. This work contained the first axiomatic description of a tilting mo-
dule and several fundamental results were presented. Their basic approach gives still the main direction in tilting theory. Later the conditions of
Brenner and Butler were relaxed by Rappel and Ringel (1982) and the present state of most of the theory was formulated. There are several papers dealing with simplifications of the proofs. (Bongartz (1981), Hoshino (1983)) or with special aspects of the theory (Hoshino (1982), Sma16 (1984) and Assem (1984)). A more general approach to tilting was presented by Miyashita (1985) by relaxing the defining conditions even more.
94
In Happel (1986) we realized that the derived category of bounded complexes over the category of finitely generated left modules over a finite-dimensional
k-algebra
ded the global dimension of
A
A is invariant under tilting provi-
is finite. This result was immediately
generalized by Cline, Parshall and Scott (1986). Since most of the applications of tilting theory in the representation theory of finite dimensional
k-algebras deal with algebras of finite global dimension we have de-
cided to present here the more restricted result which avoids the use of derived functors. We consider this result as the beginning of tilting theory and we will start from there deriving most of the results in tilting theory which have been previously established. A further generalization of the invariance principle is contained in Rickard (1987).
95
Grothendieck groups of triangulated categories
1.
The material discussed here follows closely Grothendieck (1977) and Ringel (1984). k denotes an algebraically closed field.
Let
1.1
C
be a triangulated category. Let
abelian group generated by representatives objects in
C. We denote by
[X]
factor group
in
of the isomorphism classes of F
be the
0
for all triangles
C. The Grothendieck group
K0(C)
is by definition the
F/F0.
A Z-valued function called additive if in
X i Y -* Z -, TX
1.2
Db(A)
be the free
such a representative. Let
subgroup generated by [X]-[Y]+[Z] X . Y - Z -, TX
F
defined on the objects of
a
a(X) - a(Y) + a(Z) = 0
We now turn to the case where
tion of the Grothendieck group of
C
k-algebra
A. Let
is
for all triangles
C. The condition implies that
of a basic finite-dimensional
C
F
a(X) = -a(TX).
is the derived category
A. We recall the defini-
be the free abelian group ge-
neratedby representatives of the isomorphism classes of objects in We denote by ted by
[X]
[X]-[Y]+[Z]
such a representative. Let for all exact sequences
The Grothendieck group
K0(A)
F0
mod A.
be the subgroup genera-
0 - X - Y -. Z - 0
is by definition the factor group
in
mod A.
F/Fo.
There are only finitely many isomorphism classes of simple A-modules, and we have denoted them by
S(i),
1
< i < n, a complete set of
96
(representatives of the isomorphism classes of) simple corresponding set of indecomposable projective
A-modules. The
A-modules was denoted by
and the corresponding set of indecomposable injective
P(i)
was denoted by
(compare
I(i)
A-modules
1.5.1).
X has a chain of submodules
Any module
0=X0CXIC...aXr=X Xj/Xj_1
with
being simple for all
tion series, the modules the length of
X
tor of
or also by
Xj/Xj_1
j; such a chain is called a composithe composition factors and
X. The number of times
occurs as a composition fac-
S(i)
in a given composition series is given by
dimkHomA(P(i),X),
dimkHomA(X,I(i)). This is a simple restatement of the classi-
cal Jordan-Holder theorem. We will denote this number by this way, we define the dimension vector n'
]NO
dim X
(dim X)i *0
for all
The Jordan-Holder theorem implies that s(i)
basis element of
of the modules
S(i)
K0(A)
[X], the image of
F) under the canonical map
Proof:
K0(A)
with
A-module
X
is called
K0(A).
is a free abelian group
K0(A)
LEMMA. The canonical embedding of an isomorphism of
as an element of
dim is obtained by using
F - Ko(A). Using this basis we may identify given any isomorphism class
(dim X)i. In
i.
A different way of introducing
with basis the images
X
of
which we will consider as a row vector. The
sincere, if
r = 1XI
under the canonical map
with Zn. In this way,
[X]
(considered as a
F - K0(A)
mod A
is just
into
Db(A)
dim X.
induces
K0(Db(A)).
Using the remark in
1.3.3
we clearly obtain an injecKO(Db(A)). Using the pre-
tive map of abelian groups from
K0(A)
into
vious remarks and the fact that
mod A
is a generating subcategory of
97
of
Db(A)
this map also is surjective.
1.3
An elementary but very useful observation is that the
Grothendieck group
is endowed with a bilinear form in case
K0(A)
A has
finite global dimension. This is not the most general condition, but sufficient for us and will thus be assumed for the rest of this section. We have denoted by
P(1),...,P(n)
a complete set of represen-
tatives of the isomorphism classes of indecomposable projective Let
C = CA be the Cartan matrix of
A-modules.
A. This is an integer-valued matrix
with entries
Cij = dimk HomA(P(i),P(j))
Thus the
j-th column of
C
is
(I < i, j < n).
dim P(j) - p(j)t, where
t
denotes the
transpose.
Using the equivalence of functor
v
and the fact that
(1.4.6)
(I(1),...,I(n)
AP
and
AI
given by the Nakayama
vP(i) = I(i)
for all
i
a complete set of representatives of the isomorphism clas-
ses of indecomposable injective
A-modules such that
soc I(i)
P(i)/rad P(i)) we have
(dim I(i))
Thus the
= dimk HomA(I(i),I(j)) = dimk HomA(P(i),P(j)).
i-th row of
LEMMA.
Let
global dimension then
Proof:
dule
X
If
is given by
C
dim I(i) = q(i).
A be a finite-dimensional CA
A
k-algebra of finite
is invertible over Z.
has finite global dimension, then every
has a finite projective resolution
0 -P r _,P r+1 -+... -.P1 -P0 -X-O,
A-mo-
98
r Y (-1) dim P _j ; in particular dim X is an integral =o < i < n. Hence, the linear combination of the dimension vectors p(i),
dim X =
thus
I
s(i) = dim S(i)
basis vectors tions of the
p(i),
of
K0(A)
are integral linear combina-
I < i < n. In this way, we obtain that
CA
is inver-
tible.
Observe that
CA may be invertible over $
even if
A does
not have finite global dimension. An easy example is given by the bound quiver
(A,I)
D= a Q
I= . Y
Then the Cartan matrix is given by
(2 1) 1
so is invertible over
7L, but
1
gl dim ks/I = .
<-,->A on
The matrix Ct defines a bilinear from K0(A) =
71n
<x,y>A = x C t yt. The corresponding quadratic form
by
XA(x) = <x,x>A
is called the Euler characteristic of
The bilinear form pretation. Suppose that
Let
LEMMA.
<-,->A has the following homological inter-
A has finite global dimension.
X,Y
by
A-modules. Then
A =
Since
Proof:
nd X = d the
for some
A-module
X
A.
A
E (-1) 1 dimk Ext11(X,Y). i>o
has finite global dimension we have that
d E IN. We proceed by induction on
is a projective
decomposable. So assume that
d. For
A-module. We may assume that
X = P(j)
for some
I
< j < n. Let
d = 0
X
is in-
99
(yl,...,yn) = y = dim Y.
A = A = P(j)C tyt = s(j)yt = yj = dimk HomA(P(j),Y).
Exti(P(j),Y) = 0
Since
for
the assertion follows.
i > o
So assume the assertion is true for all Let
X be an
A-module such that
X with
pd X < d.
pd X = d. Consider an exact sequence
0-1. X' with
projective. Then
P
pd X' < d. Applying
Hom(-,Y)
to this short
exact sequence yields a long exact sequence
... - Ext1(X,Y) -+ Ext1(P,Y) -+ Ext1(X',Y) -+ ... Calculating dimensions we obtain F
(-1)1 dimk ExtA(X,Y)
i>o
Y (-1)1 dimk ExtA(P,Y) - F (-1)1 dimk ExtA(X',Y). i>o i>o
By induction, the right hand side is equal to
A - A = A = A.
This finishes the proof.
1.4
to
Db(A). If
Using
1.2
we can transfer the concepts introduced above
X' _ (X1,d1) E Db(A), we obtain
dim X. =
f (-1)1 dim X1.
iEZ Since dimj
X'
of
is bounded, the sum is finite. This shows that each component
dim
is an additive function on the objects of
Again we assume that
A
has finite global dimension.
Db(A).
100
Let
LEMMA.
X*,Y' E Db(A). Then
A =
F (-1)1 dimk Hom b
Proof: of
w(Y')
X'
w(X') = I
with
We proceed by induction on the width Y'. Suppose that
and
tion follows from
w(X')
and
w(Y') < r. Let
Y° * 0. We consider
X',Y' E Db(A)
Y' = (Y3,d3)
X',Y' E Db(A)
with
satisfies
as a mapping cone of
Y'
and
w(X') = w(Y') = 1. Then the asser-
1.3. Suppose the assertion holds for
w(Y') = r. We may assume that and
(X',T1Y').
D (A)
iEZ
1.3.3). In particular, dim Y' = dim Y'' + dim Y°
w(X') =
Y3 = 0
I
for
T -Yo - Y"
and
j < o
(compare
and so
A = A + A.
Hom
Applying
Db(A)
...
to the triangle
(X',-)
HomDb(A)(X',T Y°) - HomDb
T -Yo -Y'' - Y' - Y°
(A)(X'.Y ") -. Hom (X'.Y') -' Hom Db (A) Db(A)
yields
(X',Y°) - ...
Therefore I (-1)1 dim Hom b
(X',TtY') -
D (A)
iEZ
I (-1)' dim Hom
b D (A)
iEZ
(X',T'Y°) +
E (-1)1 dim Horn b (X',T1Y " ). D (A)
WE
By induction the right hand side is equal to
A + A.
If the assertion holds for
w(X') < s
is achieved by considering
X'
Hom
and
w(Y') = r
the induction step
as a mapping cone and applying
(-,Y,).
Db(A) 1.5
We say that f
:
Let
K°(A)
A and and
K0(B)
B
be basic finite-dimensional
k-algebras.
are isometric if there exists an isometry
K0(A) - K0(B), i.e. a linear bijection such that
101
<x,y>A = <xf,yf>B
x,y E K0(A).
for all
A
Let
LEMMA.
and
is a triangle-equivalence, then
Proof:
has finite global dimension.
B
We denote by
a quasi-inverse to
G
S(i), S(j)
to be a triangle-equivalence. Let enough to show that there exists an
Hom
r
0
which we know
F
B-modules. It is
be simple
E IN, independent
of
(G(S(i)),TrG(S(j)), there exists Db (A) (G(S(i)),TrG(S(j))) = 0 for r > r i., since
r .. E IN
(S(i),TrS(j)) .=, Hom
Db(B) with Hom
S(i), S(j),
ExtB(S(i),S(j)) _
r > ro. As
for all
ExtB(S(i),S(j)) = 0
such that
F :Db(A) -, Db(B)
A has finite global dimension. If
bras and assume that
k-alge-
be basic finite-dimensional
B
1] A has finite
Db (A)
r0 = max r...
global dimension. Then the assertion follows for
i,3
Let
PROPOSITION.
-
Db(B)
:
Db(A)
f
:
K0(A) -, K0(B)
particular, A
and
and
B
be basic finite-dimensional
A has finite global dimension. If
k-algebras and assume that F
A
is a triangle-equivalence, there exists an isometry such that B
dim F(X') = (dim X')f
for
X' E Db(A). In
have the same number of simple modules up to iso-
morphism.
Proof:
By the preceding lemma it follows that
global dimension. So
<-,->B
is defined. Again let
B
has finite
P0),-..,P(n)
be a
complete set of representatives of the isomorphism classes of indecomposable projective obtain for
A-modules. Let
I < i,j < n
Xi = F(P(i))
for
I
< i < n. Then we
that
B = B =
F (-1) t dimk Hom b
D (B)
t>o =
F (-1)t dimk
t>o
k
(F(P(i)),TtF(P(j))
(P(i),TtP(j)) _
Db (A)
102
= dimk HomA(P(i),P(j))A.
=
We obtain a linear map x
by defining for
K0(A) -+ K0(B)
f
n
n Iii dim P(i)
the value for
xf =
F
p. dim X:.
i=1
i=1
<x,y>A = <xf,yf>B
By the previous calculation we infer that for
x,y E K0(A). Since
<-,->A
is non-degenerate it follows that
injective. Using a quasi-inverse to
To show the equality it is enough to consider the case tion on the width of
w(X')
F
we obtain that
we may assume that
with
X' _ (X1,dl)
X' E Db(A)
X' E Kb(AP). We again proceed by induc-
of X. For
X' E Kb(AP)
is
is surjective. for
dim F(X') _ (dim X')f
w(X') =
I
this is the definition
f. So we may assume that this equality holds for
w(X') < r. Let
f
f
X' E Kb(AP)
w(X') = r. Applying satisfies
X1 = 0
T
for
X° * 0. Then we have a triangle T X° -, X'* -i X' -. X° in
is exact and f is linear we obtain by using induction dim F(X') = dim F(X ") + dim F(X°) = (dim X'')f + (dim X°)f _ (dim X')f.
This finishes the proof of the proposition.
with
if necessary, i < o
and
Kb (AP) . As
F
103
2. The invariance property
A be a finite-dimensional
Let
k-algebra which we suppose to
be of finite global dimension throughout this section. Let M be an Aadd M
module. We denote by
M. Then we obtain a natural func-
objects the direct sums of summands of tor tor
:
p
- Db(A)
Kb(add M)
into
Kb(add M)
2.1
which is the composition of the embedding funcand the localization functor
Kb(mod A)
- Db (A)
Qb : Kb (mod A)
mod A having as
the full subcategory of
.
LEMMA.
Ext'(M,M) = 0
If
i > o, then
for all
pp
is
full and faithful.
M*,M2 E Kb(add M). Applying
Let
Proof:
M2 = 0
may assume that
i < o
for
double induction on the widths of then there exists i = o, then
Hom Kb(add M)
such that
i E 71
if necessary, we
MZ = M2 * 0. We proceed by
and
and M. If
M*
T
M' = T1M1
w(M*) = w(M2) = 1,
for some
M1 E add M. If
(M*,M2) = Horn b (M',M2). Otherwise Hom b K (add M) D (A) (M',M') = 0 for i > o (M* M*) = 0 and Horn 1
i
= ExtA (M1,M2) If
T M2 - M2' _, MZ
Db(A)
2
for
1
2
and
i < o, and the assertion follows by assumption.
w(MP = M2
cohomological functors
I
where
and M2'
w(M2) - r, then we consider the triangle is the truncated complex. We apply the
Hom
(M',-)
Kb(add M)
1
and
Horn
Db(A)
(M',-) 1
to this
104
triangle. Using induction and the Hom b
(M*,M2)
Horn b
K (add M)
5-lemma we infer that under
p.
D (A) The remaining part of the proof is dual.
2.2
We say that an
(M-codim(X) < -)
A-module
X
M-codimension
has finite
if there exists an exact sequence
M° -Ml -... -+Ms -0 with MlE addM for o
i > o. Let
pd M < r -*
Ms-1
X
Let M be an be an
A-module such that
A-module of finite
M-codimension. Then
implies that there is an exact sequence
-. Ms - 0
0
d
We choose such an exact sequence with
ds-1
o < i < s-1. It follows that
for
s-1
Ms-,0
minimal and set
s
i > o. Let
P
HomA(-,M)
Let
j > 1.
s. So
s < r.
ExtI(M,M) = 0
A-module. If
M-codim(P) <
B = End M. Then
pd MB <
In fact, apply
to the exact sequence 0 - AA - M° -4 M1 - ... -' Ms i 0. This
gives a finite projective resolution of projective
A-module such that
be an indecomposable projective
M-codim(AA) < -, then
Proof:
Let M be an
for
Ext1(M,Ka-1) = 0. Therefore
is a retraction. This contradicts the minimality of
LEMMA.
K1 = ker d1
Exti(M,K1+1) = Ext-+1(M,K1)
s > r. Then it follows that
2.3
, ...
By assumption there exists an exact sequence
d M1 d
Now assume
0 - X -. M° - M1
such that s < r.
Proof:
for
ExtI(M,M) = 0
A-module. Then
P - Ae
MB. Let
P
be an indecomposable
for some primitive idempotent
Let 0- Qt-' .. - Q0- eM -, 0 be a projective resolution of
eM
e E A.
considered
105
B-module. This implies that
as right
for
i > o
M-codim(P) <
Let M be an
2.4
LEMMA.
and
B = End M. If
A-module such that
M-codim(AA) < m and
ExtA(M,M) = 0
pd AM < r, then
pd MB < r. Proof:
2.5
for
i > o. If
This follows immediately from
Let M be an
LEMMA.
2.2
2.3.
and
A-module such that
M-codim(AA) < -, then the functor
p
:
ExtI(M,M) = 0
Kb(add M) -+ DB(A)
is dense.
Proof:
A has finite global dimension, Db(A)
Since
triangle-equivalent to also
M-codim(P) < -
fore
AP belongs to the image of
for a projective
generating subcategory of
2.6 for
Proof: 1
1M.
M be an
satisfies
M-codim(AA) < by
above. There-
2.3
AP
is exact and
is a
is dense.
9
A-module such that
be acyclic. Then
ExtI(M,M) = 0 in
M' =4 0'
K (add M).
if necessary, we may assume that the com-
T 1
M
0
for
i > o. We have to show that
is homotopic to the zero morphism. Or equivalently we have to con-
struct
A-linear maps Let
0-
1
M' _ (M ,d )
plex
Applying
p
P
we infer that
M' E K (add M)
i > o. Let
A-module
T. Since
Db(A)
Let
LEMMA.
1.3.3). As
(compare
Kb(AP)
is
kl
: M1 _' M1-1
K1 = ker d1
K1-1 + Ml-1 - K1
for
I
= d1k1+I + kldl-1.
i < o. So we obtain exact sequences
-' 0 for i < o. Applying
Ext3(M,K1) = Extj+1(M,Kl-1)
Ext1(M,K1) = 0
for
such that
for all
HomA(M,-)
yields that
j > 1. It follows that
i < o.
We now turn to the construction of the
A-linear maps
k1.
k1+1
For
i > o
we set
kl = 0. Suppose we have constructed
such that
106
I
k1+1d1 + dl+1 kl+2.
=
Consider the following diagram:
di-1
Mi-1
Mi+1
dl
Mi
NK-'ZI Now fl
o
0
0
(1-dlk1+l)dl = d'(1-kl+ldl) = dl(d1+1k1+2) = 0. Thus there exists : M1 - K1
M1-1
kl : M1 -
exists
fl11
such that
= 1-dlkl+l. Since k'w'-l.
f' =
such that
Extl(M,K1-l) = 0, there
Thus we have
diki+1 + kidi-1 = diki+l + kiri-I i = diki+l + flue = diki+l + 1-dlkl+1 = 1.
This finishes the proof.
2.7
We recall from
full subcategory of
that
C_'
b(add M)
denotes the
of complexes with only finitely many non-
C _(add M)
zero cohomology groups. By
1.3.2
K 'b (add M)
we have denoted the associated
homotopy category.
LEMMA. for
i > o. Let
and
M2
M be an
A-module such that
M' E K b(add M). Then
M' = (M',d1)
Applying satisfies
H1(M') = 0
j < o. So we obtain exact sequences
for
j < o.
Let
with Mi E Kb(add M)
if necessary, we may assume that the com-
T
for
j < o
M' 1M1'' ® M2
Ext1(M,M) = 0
is acyclic.
Proof:
plex
Let
and t < o
Applying
HomA(M,-)
i > 1. It follows that and apply
HomA(-,Kt)
yields
for
Ki = ker d3
i < o. Let
0 - K3
yI
Mi
Exti(M,Ka+I)
Exti(M,Ki) = 0
for
i
K3+1 - 0
Exti+1(M,Ki) i > I
and
to the above exact sequence. This
for
j < o.
107
yields the following long exact sequence
...
i
t
j
i+1
t
i
+ ExtA(M ,K ) - ExtA(K ,K ) - ExtA
jo+1
jo
1
K
ExtA(K
Extl+1(Mj,Kt)
A
.+
Exti(Kj,Kt) = Exti+l(K3+I,Kt)
In particular we see that j,t < o. Suppose that
j+1 t ,K ) (K
jo = -m+1. Then it follows that
gl dim A = m. Let m-I
0. Note that
Ext
i > o,
for
t o m (Mo /Ko ,K) (K ,K t) -, Ext
for
M°/K° -, im d°, we infer that
t > o. Since
Extm+1(M1/im d°,Kt)
Extm(im d°,Kt) =,
t > o. Therefore
for
+1
0 -1, K ° u + M 0 n MJ°
exists
-+
K ° -r0 is a split exact sequence. Thus there such that
K,°
j and K ° is an isomorphism. This shows that the fol-
+1
(v170)
M,°
:
K,°
pj°p' =
I
lowing determines an isomorphism of complexes o I
j°-2 ...
M
j+1 M °
dM
M
°
M
-:M°-K°®K°
+1, Let
(a °
M* _
0)
j +2
--. M °
j°+l
11
0
j +1
j
+1
i
-1
j -I
dM
j0+1
u
be the complex with
dM
II
j0+2
Mi = 0
for
j
M1° = K °
Mi = M1
for
i > jo, dM
...
Ir °)
(1
jo 2
j°+1
j°
dM
_+ Mjo-I
= 0
for
+i 1<
jo'
j
i < j °, dM° = u °
and
1
= dM
dM
for
i > jo. Then
M. E Kb(add M). Let
M2 = (M2, dM)
j°
complex with
M2 = 0
for j -1
d
M
= 0
for
i > jo, dM2
i > j0, M2
be the
2
1
j
= K °, M2 = M1
for
i < jo,
i -1
= 7 °
and
dM2 = dM
for
i < jo-l. Then
2
M2
is acyclic. Moreover we have seen that
assertion.
M' =,M $ M. This proves the
108
M be an
Let
2.8
B = End M. Then we obtain
A-module and
from mod A
F = HomA(M,-)
G = M 0B from mod B
mod A whose restrictions to
BP
to
mod B
to
a pair of adjoint functors
add M
and and
induce inverse equivalences. This functors trivially extend to inverse
triangle-equivalences, again denoted by K (add M)
and
F
G, between
and
Kb(add M)
K_( P), whose restrictions to B
Kb(BP)
and
are again inverse triangle-equivalences. This information is depicted in the following diagram:
' BP
add M 4 n
n ' Kb(BP)
Kb(add M)I
n
n
K (add M)4--' K (BP) LEMMA.
i > o
for of
F
Let
M be an
and assume that the
and
K 'b(add M)
to
G
A-module such that
Ext1A(M,M) = 0
M-codim(AA) < -. Then the restrictions and
K-1
b (B P)
are inverse triangle-equi-
valences.
Proof:
M' E K b(add M)
K 'b(BP) for for
It is enough to show that and
F(M')
is contained in
G(P')
b
P' E K_' (B P). The first assertion follows from
P. _ (P1,dl) E K_'
H1(P") = 0
for
b
(B P). Applying
i < o
K 1(G(P')) = TorB(MB,BX) G(P') E K_'
b
(add M).
T
K b(add M)
2.7. Let
if necessary we may assume that
and How) * 0. Let for
is contained in
i > o. By
2.4
BX = coker d°. Clearly we infer that
109
2.9 for
i > o
LEMMA.
Let
M be an
and assume that the
A-module such that
M-codim(AA) < W. Then
ExtI(M,M) = 0
B = End M has
finite global dimension.
Let
Proof:
jective resolution of have that G(P-)
M'
be a simple
BS
considered as object of
BS
G(P') E K b(add M). By
2.6
pd BS < .. Thus
which shows that
2.10
M be an
and
Db(B)
By
2.1
and
Proof:
2.9
we have that
we infer that F(M*) E Kb(BP),
A-module such that M-codim(AA) <
Let
are triangle equivalent.
2.5
we have that and
are triangle-equivalent. Moreover, Kb(add M) equivalent. By
2.8 we
K 'b(BP). By
has finite global dimension.
B
Db(A)
be a pro-
P'
F(M*). But
P'
and assume that the
i > o
for
B = End M. Then
Let
THEOREM.
2.7
and
Mi E Kb(add M). Therefore
with
ExtA(M,M) - 0
B-module and let
Kb(BP)
and
Kb(add M)
Kb(BP)
Db(B)
Db(A)
and
are triangle-
are triangle-equi-
valent, hence the assertion.
2.11
Ext1(M,M) = 0
for
COROLLARY.
M be an
A-module such that
and assume that the
i > o
B = End M. Then A and
Let
B
M-codim(AA) < -. Let
have the same number of isomorphism classes
of simple modules.
Proof:
2.12
This follows from 2.10
and
1.5.
Combining the results of chapter II
with
2.10
imme-
diately gives the following corollary. We point out that a similar result has been obtained by Wakamatsu (1986) using a result of Tachikawa, Wakamatsu (1986).
110
Let M be an
COROLLARY. i > o
for
mod A and
and assume that the
mod B
2.13
Db(B)
M-codim(AA) < -. Let
Let
constructed in
B = End M. Then
be the triangle-equivalence from
F
is triangle-equivalent to
on
Kb(AI). We consider the functor
it is easily seen that
Db(A)
to
2.10. We want to give an explicit description of
Kb(AI)
Hom(M,-)
ExtI(M,M) = 0
are triangle-equivalent.
As
of
A-module such that
F
it is enough to describe
Db(A)
Hom(M,-)
Qb
F
Kb(AI) - Kb(mod B). Then
:
can be described on
and the localization functor
F.
Kb(AI) :
as the composition
Kb(mod B) - Db(B). This
description will turn out to be quite useful in the next section.
We conclude this section by some examples of modules
2.14
A
satisfying the conditions discussed in this section. To be precise let be now an arbitrary finite-dimensional
k-algebra and
M
an
A-module
satisfying the conditions:
(i)
pdM
(ii)
Ext
(iii)
M-codim(AA) < -.
We remark that
(iii)
i (M,M)
M
is a faithful
If
A
A-module, for
AA
M.
Trivial examples of modules satisfying conditions (iii)
(b)
i > o
for
implies that
is clearly cogenerated by (a)
= 0
(i), (ii)
and
are provided by the Morita nrogenerators. has finite global dimension, then any injective cogenerator
satisfies these conditions. (c)
If
A
is selfinjective, then any module
M
satisfying these con-
ditions is a Morita progenerator. (d)
The following example due to Assem shows that the converse of
(c)
III
is not true. Consider the following algebra A given by the bound -1
quiver
(A,I).
and
I
be the ideal generated by all
paths of length two.
Then it is easily seen that an dimension if and only if
X
A-module
X
has finite projective
is a projective module. But
A
is not
selfinjective. (e)
Let
be a basic finite-dimensional
A
directed. Let
P(1),...,P(n)
is directed we may assume that
AP
A-modules.
is simple projective.
P(1)
We consider the Auslander-Reiten sequence starting at
0 -+ P(1) -
is
be a complete set of representatives
from the isomorphism classes of indecomposable projective Since
AP
k-algebra such that
P(1).
0 P(i) i T P(1) - 0. i E I
M1 = T P(1), Mi = P(i) for 2 < i < n. Then it is easily seen n that M = 0 Mi satisfies the conditions (i), (ii) (iii) and Set
i=1
for
r = 1. This construction is due to Auslander, Platzeck and
Reiten (1979) and will turn out to be important in chapter
IV. We
will also use the notation M = M(P(1)). (f)
Let us give a more specific example of the preceding construction. Suppose
A
is given by the bound quiver
where
(A,I)
2
,
I =
.
3
Then
P(1)
is simple projective. It is a straightforward calculation
112
End M(P(1)) 4 kA where
that
(g)
The following example shows that there exist modules (i), (ii)
and
(iii)
such that
End M = A, but M
M
satisfying
is not a Morita
progenerator. Consider the algebra given by the bound quiver
(A,I)
a
Or- _1b
I = .
,
Then it is easily checked that
M = S(2) 0 P(2)
provides such an
example. (h)
We now turn to the question whether there always exists modules satisfying the conditions above which are not projective. A necessary condition is that there exist non-projective modules
M satisfying
pd M < oo. We will show that this is also sufficient. Let M be a non-projective module satisfying indecomposable projective u
:
pd M < W. Then there exists an
P, a projective
P -. Q which is not a section. Let
k-basis of
HomA(P,Q)
is injective and let
and let
Q
and a monomorphism
u = f1'121...Ifr
be a
f = (f1,...,fr) : P -+ Qr. Then
X = cok f. Clearly, pd X = 1. We obtain an
f exact sequence 0 - P - Qr -. X -+ 0. By construction Hom(f,Q) surjective. Thus
Using
6.1
f
ExtA(X,Q) = 0, which implies
now proves the assertion.
ExtA(X,X) - 0.
is
113
3.
The Brenner-Butler Theorem
Let
A be a finite-dimensional
k-algebra which we suppose
to be of finite global dimension throughout this section. Let M be an A-module satisfying the conditions
(i)
pdM
(ii)
Exti(M,M) = 0
(iii)
M-codim(AA) <
Let
for
i > o
B = End M. Then we may consider M
and denote it by MB, so we consider M we obtain a (left)
B-module
as
D(MB). Now
as a right
B-module,
A-B-bimodule. Dualizing
D(MB)
is in fact a
MB,
B-A-bimo-
dule. Thus there is a canonical ring homomorphism A . End(D(MB)). If this
is an isomorphism, we will identify A and 2.10
that the pair of adjoint functors
between mod A and Db(A)
and
mod B
End(D(MB)). We have seen in
F = HomA(M,-)
and
G = M 0B
induce inverse triangle-equivalences between
Db(B), again denoted by
F
and
G. In this section we study
these triangle-equivalences on certain full subcategories of
mod A and
mod B. Using a different approach these results are also contained in Miyashita.
3.1
and
(iii). Then
LEMMA. D(MB)
Let M be an satisfies
A-module satisfying
(i)*
id D(MB) < r,
(i), (ii)
114
ExtB(D(MB),D(MB)) = 0
(ii)*
D(MB) - dim(D(BB)) < -
(iii)*
0 -, Ns -,
Ns-1
and
(i.e. there exists an exact sequence
.+ ... -, N° -, D(BB) -, 0 with N1 E add D(MB)). Moreover, canonically.
A -, End(D(MB))
We have seen in
Proof:
2.4, that
D(AA) E All thus
HomA(AMB,Homk(AA,k)) = HomA(AMB,D(AA)). But D(MB) = F(D(AA)). Now
pd MB < r. This clearly
D(MB) = Homk(MB,k) =, Homk(A ®A AMB,k) =,
(i)*. Observe that
shows
i > o
for
ExtB(D(MB),D(MB)) ., Hom b
(D(MB),T1D(MB))
D (B) (D(AA),T'D(AA)) =, ExtA(D(AA),D(AA)) = 0 for i > o. This shows D (A) (ii)*. The same argument also shows that A =, End(D(MB)) canonically.
Hom b
To show
0 , Pr ,
(iii)* we consider a projective resolution of Pr-1
-+ ... -, P° -, M -1-0. Applying
MB codim(BB) < -. Dualizing then yields
Let
3.2.
full subcategory of for
with objects
(iii)*.
be a non-negative integer. We denote by
i
mod A with objects
X
functors
satisfying
F1 = ExtA(M,-) : mod A -, mod B
E.
the
Exti(AM,AX) = 0
the full subcategory of
Ti
TorB(MB,BY) = 0
satisfying
Y
yields that the
HomA(-,M)
j * i. Correspondingly, we denote by
mod B
M, say
and
G1
and
Ti
:
for
j * i. We have
TorB(MB,-)
: mod B -,
mod A.
THEOREM.
The categories
the restrictions of the functors
F1
Ei
(these restrictions are
G1
and
are equivalent under
mutually inverse to each other).
Proof:
ween
Db(A)
and
Let
Db(B)
F,G
constructed in
as full subcategories of seen that
F I E
be the inverse triangle-equivalences bet-
Db(A)
and
1
Db(B)
Ei
and
Ti
respectively. It is easily
= T'G'IT . From this it follows that
and
= T _
2.10. We consider
GIT
1
1
115
has values in
F1IE.
and
T.
has values in
G1IT
Ei. Now
2.10
fini-
1
1
shes the proof.
We want to give some examples illustrating the distribu-
3.3
tion of these subcategories inside (a)
mod A
and mod B.
A be the path algebra of the quiver
Let
A
Then the Auslander-Reiten quiver of
is easily computed as
If we choose a module M with indecomposable summands corresponding to the vertices marked as the conditions
(i), (ii)
it is easily seen that and
The Auslander-Reiten quiver of
As
pd M < I
mod A
and
(iii)
for
M satisfies
r = 1.
B = End M has the following shape:
we are only interested in the subcategories To TI
of
mod B
E0,EI
of
respectively. The indecomposable modules
lying in these subcategories are depicted in the following figures:
(here
E0
is marked by (= and
EI
by Q )
116
(b)
is marked by ® and
To
(here
Let us again consider the algebra by
by
T1
of the previous example. Then
B
the minimal injective cogenerator
2.14 b)
conditions
and
(i), (ii)
modules lying in the subcategories
E0,EI
EI
(c)
and those in
x
by
by
E2
E2
and
depicted in the Auslander-Reiten quiver of
of
mod B
are
B.
are marked by , those in
E0
(here the indecomposable modules in
satisfies the
r = 2. The indecomposable
for
(iii)
D(BB)
*)
A similar example is as follows: Let
A be given by the bound quiver
y where
(A, 1)
a2
al
A _ of 1
Let
of
2
3
ono,n ... n-1 a
a3
0(
4
I =
L 2.-1 ,
M be the minimal injective cogenerator
D(AA). Then M satis-
fies the conditions
(i), (ii)
E0 = add D(AA)
En-1 = add S(n), while all other
and
conditions
PROPOSITION.
(i), (ii)
and
(iii)
for
r = n-I. Then Ei
do not
A-module.
contain any indecomposable
3.4
and
2 - -
Let M be an
B = End M. Then
and let
(iii)
A-module satisfying the
gl dim A-r < gl dim B < gl dim A+r.
Proof:
similar. Let
We show
BX be a
jective resolution of
gl dim B < gl dim A+r. The other assertion is
B-module and choose
P' E Kb(BP)
BX. Then M' = AMB ® P'
lies in
a minimal pro-
Kb(add M). Then
B
M. _ (M1,4)
satisfies
Mi = 0
for
i > o
and
i < -pd BX. Clearly
117
Hi(M') = 0
for
i < -r, for
pd MB < r. Since
P*
tive resolution, the arguments used in the proof of
is a minimal projec2.7
show that
pd BX < gl dim A+r.
3.5
We give an example to show that these bounds are best
possible. Consider the algebra A given by the bound quiver al
-+
a2
and 2
Then
4
3
I =
5
gl dim A = 4. Let M = S(3) 0 P(1) ® P(2) ® P(3) ® P(4). Then M
satisfies the conditions
and
(i), (ii)
B = End M is given by the bound quiver al
a2
1
gl dim B = 2
(iii) (A',I')
for
r - 2. Then
where
a4
a3
+- and I' _
A' = o
Then
where
a4
a3
A = 1
(A,I)
2
3
4
5
as it is easily verified.
.
118
4.
Torsion
theories
We now turn to the investigation of tilting modules. Let be a finite-dimensional
A
gl dim A = m. An A-module
k-algebra such that
AM satisfying (i)
pd AM < 1
(ii)
ExtA(M,M) = 0
(iii)
M-codim(AA) < 1.
is called a tilting module. This is of course a special case of the conditions
(i), (ii)
and
(iii)
considered in the previous sections. The
main advantage of tilting modules is that they induce torsion theories on
mod A and
mod B, where
B = End M. The investigation of these torsion
theories is the objective of this section.
4.1
A pair
(T,F)
a torsion theory on mod A
of full subcategories of
mod A
is called
if the following three conditions are satis-
fied: (a)
HomA(X,Y) = 0
(b)
If
HomA(X,Y) = 0
for all
X E T, then
Y E F
(c)
If
HomA(X,Y) = 0
for all
Y E F, then
X E T
for
X E T
and
In this case, the modules belonging to
modules, those in
F
Y E F
T
are called torsion
are called torsionfree modules (with respect to the
119
given torsion theory Let
(T,F).
be a torsion theory on mod A
(T,F)
A-module. Then there exists a unique submodule such that
and
be an
AZ
T
belonging to
Z' c Z
Z/Z' E F. This unique submodule is usually denoted by
t(Z).
Thus we are given a canonical exact sequence
0 - t(Z) -. Z - Z/t(Z) -. 0 . If these sequences are split exact sequences for all
Z E mod A we say that
(T,F)
splits. This is precisely the case when
Extl(Y,X) = 0
x E T
and
for all
Y E F. Or equivalently, all indecompo-
A-modules are either torsion or torsionfree.
sable
Note that in a given torsion theory are closed under extensions in
T
mod A, F
both subcategories
(T,F)
is closed under submodules and
is closed under factor modules. In the situations we will encounter, the modules in
always be generated by a fixed module inside
4.2
and
LEMMA.
ExtA(M,M) = 0. Let
F = {Y E mod A
Let
T.
M be an A-module satisfying
T = {X E mod A
HomA(M,Y) = 0}. Then
I
X
is generated by
(T,F)
T will
pd AM <
1
and
M}
defines a torsion theory on
mod A.
Proof: let
X E T
map
c
and
We verify the conditions
Y E F. Since
: Mr -. X. Applying
from the definition of with
HomA(X,Y) = 0
a module. Since Let
E11 ...5Er
X
X
(a), (b)
and
(c). For this
is generated by M we have a surjective
HomA(-,Y)
yields
(a). Condition
follows
(b)
F. Thus it remains to be seen that a module
for all
Y E F
is generated by
does not belong to
be a basis of
HomA(M,X)
F
M. So let
we have that and let
X
X
be such
HomA(M,X) * 0.
120
C = (el,...,er)t : Mr - X. This gives rise to an exact sequence it
0 --- im e-} X --* cok a -+ 0 We claim that e
cok e E F. Let
is surjective and
: Mr - I
: M - cok c. Since
p
also
pd M < I
ExtA(M,e)
is surjective.
Thus we can construct the following commutative diagram with exact rows (the second row is the sequence induced by seond under
9, the third a preimage of the
ExtA(M,e)).
x
0 - in a -} X -'- cok a --> 0 0 - im C -> Z -1
1
0-fMr Since
ExtA(M,M) = 0
exists that
: M -+ X
T'
p'
maps into
such that
in
--> 0
cp'Tr. By construction of
P
in e, thus
e, we see
rp = 0. Therefore, cok e E F. But by as-
O. Therefore, cok e = 0. Or equivalently, X
In
3.2
we have introduced subcategories
associated with a module M of section
trivial for
M
is
M.
4.3
(iii)
--+ Z'
we have that the last sequence splits. Thus there
sumption, HomA(X,cok e)
generated by
-% M - 0
3. If
i > 1. Let
M
satisfying the conditions
Ei, 0 < i < r,
(i), (ii)
is a tilting module then clearly
(T(M),F(M))
E i
and is
be the torsion theory constructed
4.2.
and
LEMMA.
T(M) = E0
Proof:
By definition
F(M) = E1.
E0 = {X E mod A
I
ExtA(M,X) = 0}
and
121
El - (Y E mod A I HomA(M,Y) = 0). This shows that
X
X E T(M). Since e
(i)
and
is generated by M we have a surjective map
for some
: Mr -o .X
r > o. Applying HomA(M,-)
of a tilting module show that
(ii)
M-codim(AA) < -
0 * X E Eo. Since
F(M) = E1. Let
and using properties
we have an exact
pd AM < I
and
Eo. Let
X belongs to
0 - AA - M° - Ml - 0 with M°,Ml E add M by 2.2.
sequence
(*)
Applying
HomA(-,X)
X E Eo
and using
yields the following exact sequence
0 - HomA(MI,X) - HomA(M°,X) -. HomA(AA,X) -, 0
This shows that and let
HomA(M,X) * 0. Let
el,...,er
be a basis of
HomA(M,X)
Mr -+ X. This yields two exact sequences
e= (el,...,er)t
0-4 ker a -+Mr - ime -+0
-+coke-+0 . Clearly
ExtI(M,im e) = 0 = ExtA(M,cok e). By construction,
HomA(M,im e) -+ HomA(M,X)
applying e
is surjective, thus
HomA(-,cok e)
to
HomA(AA,cok c) = 0, therefore
yields
(*)
HomA(M,cok e) - 0. Now
is surjective. Or equivalently, X belongs to
4.4
Let
LEMMA.
M be a tilting module and
module in the torsion theory
sequence
0 - Mm -+ Proof:
M-m+l
e l,...,er be a basis of is surjective, but also
X
a torsion
(T(M,F(M)). Then there exists an exact
- ... -+
m-1
-+
X E T(M). Then
Let
T(M).
M° -+ X -+ 0 X
HomA(M,X). Then
HomA(M,e)
with Ml E add M. M. Let
is generated by
c = (e l,...,er)t
r :
0 M-+ X i-1
is surjective. Thus, ExtA(M,ker e) = 0, r
showing that
ker e E T(M). Set
M° -
® i=1
applying
2.7
yields the assertion.
M. Iterating this procedure and
122
Let
4.5
AM be a tilting module and
troduced in
3.2
clearly
is trivial for
Ti
To = {BX nally
I
To
is denoted by
mod B. Since
of
Ti
T1 = {BY
and that and
y(M)
then
(X(M),y(M))
mod B. The following is a restatement of
The categories
der the restrictions of the functors
MB OB BY = 01. Traditio-
is denoted by
T1
It is easily seen that
THEOREM.
pd MB < 1
i > 1. Recall that
TorB(MB,BX) = O}
this notation. on
subcategories
B = End M. We have in-
is a torsion theory
3.2.
and
T(M)
X(M). We will keep
y(M)
are equivalent un-
F = HomA(AMB,-) and
G = AMB QB
these restrictions are mutually inverse to each other, and the categories F(M)
and
are equivalent under the restrictions of the functors
X(M)
G' = TorB(AMB,-)
and
F' = ExtA(AMB,-)
(again, these restrictions are
mutually inverse to each other).
4.6
definition of
Let
AM be a tilting module and
T(M), it is obvious that the injective cogenerator
T(M). We have seen before
belongs to
Denoting as usual by
PA(a)
tive envelope of a simple
is an injective
in this case
(3.1)
that
A-module
S(a). Then
D(AA)
F(D(AA)) = D(MB).
the projective cover, by
CONNECTING LEMMA. F(IA(a))
B = End M. Using the
IA(a)
the injec-
IA(a) E T(M).
TBF(IA(a)) = F'(PA(a)). In particular,
B-module if and only if
PA(a) E add M, and
vF(PA(a)) = F(IA(a)).
Proof:
(*)
By
2.3
we have an exact sequence
0 i P(a) -+ 140
with M ,M1 E add M. Applying
111
M1
HomA(-,AM)
10 we obtain an exact sequence
123
Hom(r,M)
1
0 - HomA(M M)
o
- HomA(M M) i HomA(P(a),M) - 0,
and this a projective resolution of the right D = Homk(-,k)
Applying the duality functor
B-module
HomA(P(a),M).
yields an injective resolu-
tion
D Hom(r,M)
o
D HomA(P(a),M)
Observe that
1
D HomA(M
0 -, D HomA(P(a),M) -+ D HomA(M ,M)
M) i 0.
HomA(M,I(a)) = F(I(a)). Now
TBF(I(a)) = cok v D Hom(r,M) = cok Hom(M,r) = ExtI(M,P(a)) = F'(P(a)). Of course, F(I(a))
F'(P(a)) = 0. But clearly
thus if and only if if
P(a) E add M. In this case
4.7 COROLLARY. P(a)
and
Let
belong to
I(a)
P(a) E T(M)
TBF(I(a)) = 0,
if and only
vBF(P(a)) = D HomB(F(P(a)),BB)
D HomA(P(a),AM)
D HomB(F(P(a)),F(AM))
If
is injective if and only if
AM
HomA(AM,I(a)) = F(I(a)).
be a tilting module and
add M, then
F(I(a))
injective, and conversely, every indecomposable
B = End M.
is projective and
B-module which is projec-
tive and injective is of this form.
The first assertion follows immediately from
Proof. Let
BX
we know that
for some indecomposable direct summand Mi
BX = HomA(M,I(a))
Mi Z I(a), for
T(M)
and
4.8 LEMMA. any
A-module
AX
Proof. 0 -' P
1
B-module. Then
be an indecomposable projective and injective
BX = HomA(M,Mi)
-, P° -' M -
Let
Y(M)
M
for some
I(a)
with
4.6.
of
M. By
4.6
P(a) E add M. Thus
are equivalent. This finishes the proof.
be an
A-module with
we have an isomorphism
pd M < 1. Then for
D ExtI(M,X) i HomA(X,TM).
Choose a minimal projective resolution 0. By definition of
T
In particular, we obtain an exact sequence
we have that
TM = D ExtA(M,AA).
124
0 - TM - D HomA(P 1,AA) - D HomA(P°,AA). We also have an exact sequence 0 -+ D Ext1(M,X) -oD HomA(P 1,X) - D HomA(P°,X). Recall that for any projective
A-module
there is an invertible natural transformation
P
aP : HomA(X,D HomA(P,AA)) -+ D HomA(P,X). This gives the following commuta-
tive diagram with exact columns:
0
0 1
a D Ext1(M,X)
HomA(X,t M) i
P_l
1
HomA(X,D HomA(P
I
-1
a
Mi
<--P
D HomA(P°,X)
is an isomorphism.
Let
4.9 LEMMA. Let
AA)
I
a o
HomA(X,D HomA(P°,AA))
Thus
i
D HomA(P
,AA))
AM be a tilting module and
B = End M.
M. Then
be an indecomposable non-projective direct summand of
TMi E F(M)
is an injective
F'(TMi)
and
composable injective
Proof.
This shows that
B-modules contained in
By
4.8
we know that
belongs to
TMi
D HomB(F(Mi),F(M)) = vF(Mi). But B-module, thus
B-module. Moreover, all indeX(M)
are of this form.
HomA(M,TMi)
D ExtI(Mi,M).
F(M). Now F'(TM1 2; D HomA(Mi,M) _, F(Mi)
is an indecomposable projective
vF(Mi)) is an indecomposable injective
B-module. The
remaining part is similar.
4.10 COROLLARY.
Then
F(M) _
AT
I
Proof.
AY
Let
ATM
be a tilting module and
is cogenerated by
Clearly, if
HomA(M,Y) = 0. Conversely, let
AY
B - End M.
TM}.
is cogenerated by
AY E F(M). Then
F'(AY)
TM, then
belongs to
X(M).
125
Let
jective envelope. Then using
be exact in mod B
0 - F'(AY) -, I -.Q -, 0
4.9
Q
and
I
belong to
with
minimal in-
a
I
X(M). Applying
and
G'
now yields the assertion.
In a similar way one can show that
4.11
Y(M) _ {BX
BX
is cogenerated by
X(M) = {BY
BY
is generated by
D(MB)} T D(MB)}
D(MB) = F(D(AA)). For the second use
For the first use that the connecting lemma.
Let
4.12 THEOREM. Then
splits if and only if all
(X(M),Y(M))
Clearly, (X(M),Y(M))
Proof.
ExtI(Y,X) = 0
Hom
AM be a tilting module and
for all
X E X(M)
for
Db(Y,TX) = 0
and
X E X(M)
X E F(M)
B = End M. id X < 1.
satisfy
splits if and only if
Y E Y(M). Or equivalently, Y E Y(M). Let
and
be the triangle-
G
(B)
equivalence from and
G(X) E TF
Db(B) for
2.10. Then
Y E Y(M). Therefore
and
X E X(M)
splits if and only if
contructed in
Db(A)
to
Ext2(Y',X') = 0
for
X' E F(M)
(X(M),Y(M))
and
This shows the "if-part" of the assertion. Conversely, let 2
ExtA(Y',X') = 0
and suppose be exact with Then
I0,I1
ExtA(L1,L°)
injective
X' E F(M)
0 -+ X'
d0
,
I°
I1
L° = cok u, L1 = cok d°.
L1 E T(M), for
T(M)
is closed
II E T(M). Therefore, ExtA(L1,L°) = 0, thus
is injective, showing that
4.13 COROLLARY.
ditray finite-dimensional
Y' E T(M).
11
Y' E T(M). Let
A-modules. Let
ExtA(L1,X'). But
under factor modules and
splits.
for all
G(Y) E T
L°
id X' < 1.
Let
AM be a tilting module and
k-algebra and
B = End M. Then
A
a here-
(X(M),Y(M))
126
4.14 We now turn to the special case of tilting modules considered in
A be a finite-dimensional
2.14(e). Let
a simple projective
A-module
S. (Observe
case). Then we have constructed in we call an an
B = End M(S)
F(M(S)) = add S
and that
splits. For a representation-finite algebra A
Let
id AS >
(ii)
End(M(S))
(T(M(S)),F(M(S))) n(A)
A-modules.
A be a representation-finite algebra admit-
ting a simple projective module (i)
which
will be called
let us denote by
the number of isomorphism classes of indecomposable
COROLLARY.
M(S)
be the torsion theory on mod A.
(T(M(S)),F(M(S)))
Then it is easily seen that
that this need not to be the
2.14(e) a tilting module
APR-tilting module. The algebra
APR-tilt. Let
k-algebra admitting
S. Then the following are equivalent:
I
is representation-infinite or
n(End M(S))l> n(A).
We point out that the first alternative in occurs. Let
A be given by the'bound quiver
A
Then
A
and
=
tilting module
where
APR-tilt is not.
Next we present a theorem due to Assem
be a finite-dimensional
AM
THEOREM. (X(M),Y(M))
k-algebra such that
AP
(T(M),F(M))
If for all splitting tilting modules
A
( 1987). Let
A
is directed. We call a
splitting if the torsion theory
splits, then
actually
I = <Sa-dy,sa-nE>
is representation-finite, while the unique
4.15
theory
(A,I)
(ii)
is hereditary.
splits.
AM the torsion
127
Proof.
AP
Since
is directed, we may assume that
A
S(i) = top P(i). Assume that
i > j. Let
HomA(P(i),P(j)) = 0
are ordered in such a way that
P(1),...,P(n)
A-module S(i)
exists a simple
is not hereditary. Then there id S(i) > 1. We choose
such that
for
i
minimal with respect to the order above and this property. Let
i P1 =
n P2 =
and
0 P(j)
® P(j). Note, pd S(j) < I
for
I
< j < i. We
j=i+1
j=1
claim that
AM = TAP1 0 P2
property that
is a splitting tilting module having the does not split.
(X(M),(Y(M))
AM
First we show that
is a tilting module.
Note that by assumption summand and any submodule of add P1. Let
f
:
or equivalently
is again projective and belongs to
be an
D(AA) i P1
tive, hence a summand of
P1
A-linear map. Then
D(AA). Thus
pd M <
im f = 0
we know by
1
im f
is projec-
HomA(D(AA),P1) = 0,
and
pdAM < 1.
pd T P1 < 1, hence
Since
has no injective indecomposable
P1
4.8
that
ExtA(M,M) -,
D HomA(AM,TAM) = D HomA(TAIP1 ® P2,P1) = D HomA(TAIPl,P1) = 0, for let 0 * f E HomA(TA'P1,P1), then mand of
im f
is projective and thus a direct sum-
TAIP1, an absurdity.
We refer to
6.2
for completing the proof that
AM
is a
tilting module.
Next we show that an indecomposable
(T(M),F(M))
A-module not belonging to
or equivalently, HomA(Y,P1) * 0. Thus then
splits. For this let
HomA(P2,Y) = 0
and
Y
Y
be
T(M). Thus, ExtA(M,Y) * 0,
is a direct summand of
HomA(T AP1,Y) = 0, thus
Y
belongs to
P1. But F.
Moreover, F = add P1. To complete the proof, it is enough to find a module such that
id X > 1
(compare
4.12). Let
X E F
X = P(i). Then it is clear that
id X > 1. In fact, consider 0 -* P -, P(i) -o.S(i) -o-0 exact, with
128
P E add P1, (by construction
pd S(i) < 1). By assumption we have that
Ext2(-,P(i)) i Ext2(-,S(i)).
id P < 1, therefore
In the remaining part of this section we reproduce a
4.16
result of Hoshino
( 1982)
given torsion theory
(T,F)
giving a rather useful criterion whether a
mod A coincides with a torsion theory
on
given by a tilting module. For related results we refer to Assem and Small
( 1984 ).
Let
C
sions. An object ExtI(X,C) = 0,
be a full subcategory of
X E C
Let
mod A closed under exten-
is called Ext-projective (Ext-injective) if
(ExtI(C,X) = 0)
LEMMA.
X E F
( 1984)
for
C E C.
be a torsion theory on mod A. Then
(T,F)
is Ext-projective if and only if
X=, P/t(P)
for some projective
A-module P.
Proof.
Let
P
be a projective
A-module and let
0 - t(P) - P - P/t(P) -+ 0 be the canonical exact sequence. Let Y E F. Applying HomA(-,Y)
yields HomA(t(P),Y) - ExtA(P/t(P),Y) -. ExtA(P,Y).
Thus, ExtA(P/t(P),Y) = 0, hence Conversely, let
P/t(P)
X E F
is
Ext-projective.
be Ext-projective and let
: P -+ X
e
be a projective cover. We consider again the canonical exact sequence
0 - t(P) - P - P/t(P) - 0. Let Then
a
is surjective. Since
e X
: P/t(P) -s X the map induced by e. is Ext-projective we infer that
c
is
a retraction.
4.17 LEMMA.
X E T
Let
(T,F)
be a torsion theory on mod A. Then
is Ext-projective if and only if
TX E F. Dually, X E F
is Ext-
129
injective if and only if
T X E T.
We only prove the first assertion. The second follows
Proof.
by duality. Let X E T be Ext-projective and 0 - TX -' E -' X -' 0 be the Auslander-Reiten sequence. Let 0 -' t(TX) -. TX ' TX/t(TX) -b O be the canonical exact sequence. Since
is injective. Thus
Ext1 (X, n)
n
X
is Ext-projective we infer that
is a section, for 0 -' TX - E - X - 0
is an Auslander-Reiten sequence. But then clearly
y E T. Then
Conversely, let image of
D HomA(Y,TX). But
ExtA(X,Y)
HomA(Y,TX) = 0
for
belongs to
TX
F.
is an epimorphic is a torsion
(T,F)
theory.
4.18 LEMMA. Let that
D(AA) E T. If
tion of
T
I1
be a torsion theory on
is Ext-projective, then
We may assume that
Proof.
0 -' TX '' Io
x E T
(T,F)
X
mod A such
pd X < 1.
is not projective. Let
be a minimal injective presentation of TX. By defini-
we obtain the following exact sequence
1
0-'v (TX) -'vIo-'vI1 -,X-'0. TX E F
Since
and
D(AA) E T we infer that
COROLLARY. that
Let
(T,F)
V -(TX) = 0. Thus
be a torsion theory on
pd X < 1.
mod A
such
D(AA) E T. Let M be the direct sum of all Ext-projectives belon-
ging to
T. Then M
4.19
satisfies
We refer to
pd M < I
6.2
and
Ext1(M,M) = 0.
for a proof of the following criterion
that a module satisfying the conditions of the last corollary is a tilting module. Let
be the number of isomorphism classes of simple
n r
and let M =
0 Mil i=1
A-modules
n.
be a direct sum decomposition of
M
such that
Mi
130
is indecomposable with and
r = n, then M
Let
pd M < 1, Ext1(M,M) = 0
or
T
mod A
be a torsion theory on
(T,F)
and either
D(AA) E T
i * j. If
for
is a tilting module.
THEOREM. that
N. ; Mj
only contain finitely many isomor-
F
A-modules. Let M be the direct sum of
phism classes of indecomposable
T. Then M
all Ext-projectives belonging to
is a tilting module such
(T(M),F(M)) _ (T,F).
that
Proof.
4.18
By
and the previous remark we have to show T. Ob-
that in these situations there exist enough Ext-projectives in
T
serve that
and
are Krull-Schmidt categories. It is easily seen
F
of Ringel (1984)) that a finite Krull-Schmidt category has
2.2
(compare
sink and source maps. Suppose now that and
such
T
has sink maps there exist a least
T
T
morphism) in and let
s
D(AA) c T
is finite. Since
Ext-projectives (up to iso-
n
by the following lemma. Suppose now that
F
is finite
be the number of isomorphism classesof indecomposable projecT. Thus
Since
has source maps we infer that
F
contains at least
n-s
Ext-
injectives by the following lemma. Thus
T
contains at least
n-s
Ext-
F
contains
Ext-projectives (up to isomorphism).
tives in
F
n-s
projectives of the form provided by tains at least
s+n-s = n
4.17. Altogether we see that
T
con-
Ext-projectives.
(T(M),F(M)) _ (T,T). This finishes the proof of the
Clearly theorem.
LEMMA. (i)
Let Let
(T,F)
X E T
sink map in (ii)
Let
Y E F
map in
be a torsion theory on be not Ext-projective and T, then
E -4 X
be a
ker g E T.
be not Ext-injective and
F, then
mod A.
cok f E F.
Y + E
be a source
131
We prove
Proof.
(i). The second assertion follows by duality.
g
f
We have the exact sequence 0 - K = ker g - E 4X -+ 0 in mod A. Suppose K ( T and let 0 -> t(K) Since
K/t(K) E F
phism
d
K/t(K) - 0 be the canonical exact sequence.
V1K 7-T
E,X E T we obtain that the connecting homomor-
and
is injective. In particular,
: HomA(K,K/t(K)) -, Ext1A(X,K/t(K))
we have that the second row of the following diagram does not split:
0
f
K Tr
I
E
hl E.'
0 --- K/t(K) -p Now h
is surjective and
E' E T. Since
infer that there exists hh'g = hg' = g
:
implies that
X
Since
l X -} 0 II
hh' E Aut(E), for n
T we
is a sink map in
g
such that
E' -, E
is injective, hence
particular, h
g,
is closed under factor modules, hence
T
h'
X -+ 0
9
is not a retraction and
g'
g
g' = big. Now g
is a sink map. In
is an isomorphism, or
is not Ext-projective, there exists u v
K E F.
Y E T
such
that ExtA(X,Y) * 0. Let 0 -. Y - E" - X -+ 0 be an non-split exact sequence. Since there exists
g"
is not a retraction, E" E T
v
E" - E
:
such that
and
g
is a sink map,
g"g = v. Thus we obtain the follow-
ing diagram with exact rows: u v 0 -*Y--+ E" -+X -r 0
h'I
''I
1
fg 1
O --r K -r E Clearly
I
g
1
I
1
X -> 0
h' = 0(Y E T,K E F). Thus the lower sequence splits, a contradic-
tion.
4.20
ditions. Let
We conclude this section
(T,F)
be a torsion theory on
by some comments on the conmod A. If
(T,F)
is the
132
(T(M),F(M))
torsion theory D(AA)
belongs to
for some tilting module
M, then clearly
T.
Let us give an example of a torsion theory (T(M),F(M))
which is not given by
(T,F)
for some tilting module
be a hereditary, representation-infinite finite-dimensional Let
T be the full subcategory of
A-modules,and let
F
on mod A M. Let
A
k-algebra.
mod A formed by the preinjective
be the full subcategory of
mod A formed by those
A-modules having no non-zero preinjective summand. Then it follows from well-known properties of
mod A
clearly cannot be of the form
that
is a torsion theory, which
(T,F)
for some tilting module
(T(M),F(M))
M,
for T does not contain any Ext-projective indecomposable module.
Let us finally give an example of a tilting module M such that
T(M)
A = kA where
are not finite. Let
F(M)
and
A tilting module M having the desired property is given by the module whose indecomposable summands have the following dimension vectors
1
I
0
1
0 0
I
I
I
I
0
0
0
0
0
0
1
0 0
0
0
0
1
10, 1 0,
1
1, 1 0,
1
and
00
0 0
133
Tilted algebras
5.
5.1
and let
k-algebra
A be a hereditary, finite-dimensional
Let
AM be a tilting module. Then
B = End M
is called a tilted
algebra. This section is devoted to an investigation of tilted algebras. We follow closely Ringel (1984),Ringel (1986).We omit most of the proofs. The following is a simple consequence of
PROPOSITION.
Let
B. Then
tilted algebra
Y(M)
(X(M),Y(M))
4.13.
be the torsion theory on the
is closed under predecessors, and
is closed under successors. In particular, Y(M) X(M)
is closed under
is closed under
X(M)
tB, and
TB
As a consequence, given an Auslander-Reiten sequence in either all terms are in left hand term is in
Y(M), or all terms are in
Y(M), the right hand term in
mod B
X(M), or finally, the
X(M). There are only
a few Auslander-Reiten sequences of the last form, and they can be described explicitely and will be called connecting sequences.
LEMMA.
in mod B
with
A-projective X - F(I(i)) form
If
0 -),X -, Y -* Z -, 0
is an Auslander-Reiten sequence
X E Y(M), Z E X(M). Then there exists an indecomposable
P = P(i)
for some
i
with
P ( add M
such that
and the Auslander-Reiten sequence starting at
X
0 -. F(I(i)) -, F(I(i)/S(i)) 0 F'(rad P(i)) - F'(P(i)) - 0.
is of the
134
This follows easily from
Proof.
5.2
1.5.4, I. 4.7
and
2.10.
For the characterization of tilted algebras let us introS. Let
duce the concept of a slice S
A full subcategory
of
mod B
B
be a finite-dimensional
k-algebra.
closed under direct sums and direct
summands is called a slice if the following conditions are satisfied: (a)
is sincere (i.e. there exists
S
HomA(P(i),S))* 0 B-modules (B)
then also
X
If
(y)
P(i)).
AM
and in
X,TX
belong to
S.
are indecomposable, f : X - S
and
S E S, then either
and
T XES.
THEOREM.
So,S1 E S,
is indecomposable and not projective, then at
X,S
If
So < X < S1, and
X E S).
most one of (d)
such that
for all indecomposable projective
is path closed (i.e. if
S
S E S
Let
A
X E S
is not injective
X
or
irreducible,
be a hereditary finite-dimensional k-algebra,
a tilting module with
End M = B. Then
add F(D(AA))
is a slice
mod B. Conversely, any slice in a module category occurs in this way. For a proof we refer to Let
be a slice in
S
composable summands of lander-Reiten quiver and the module
S
S
4.2
of Ringel (1984).
mod B, say
S = add S. Then the inde-
belong to a single component
B. In this case we say that
C
is called a slice module. Note that
C
of the Aus-
contains a slice End S
S
is heredi-
tary.
5.3
bras
As Ringel (1984) shows certain finite-dimensional k-alge-
A admit slices in quite a natural way. Let
A-module. Then we define
S(-+ X)
X
be an indecomposable
to be the additive category generated
135
by all indecomposable
A-modules
Y
Y < X, and such that there is
with
A-module Z
no indecomposable non-projective
satisfying
Y < T Z
and
Z < X. Dually, let posable
be the additive category generated by all indecom-
S(X -)
A-modules
with
Y
sable non-projective
X < Y, and such that there is no indecompo-
A-module
X < TZ
satisfying
Z
We refer to the appendix of
Ringel (1984)
and
Z < Y.
for a proof of
the following lemma.
LEMMA. S(X -p)
and
Let
S(- X)
mod A. More generally, if
are slices in
decomposable and contained in
Y
A-module. Then both
be a sincere directing
X
S(X -'), then
S(-+ Y)
is indecomposable and contained in S(- X), then
5.4
COROLLARY. Let
with a sincere directing
Proof.
5.5
Apply
Let
A
A
5.3
S(Y -)
is a slice.
k-algebra
is a tilted algebra.
5.2.
to
be a finite-dimensional
P, S, Q of
have given full subcategories
is in-
is a slice; and if
A be a finite-dimensional
A-module. Then
Y
k-algebra. Suppose we
mod A
each closed under
direct sums and direct summands, such that the following conditions are satisfied: (a)
The indecomposable
A-modules are contained in
PusuQ. (b)
(c)
HomA(X,Y) = 0
for
for
X E S, Y E P.
Let
X E P
S E S and
Y E Q
and
f
: X -, S, h
:
S -1,Y
and
g
X E Q, Y E P; for
X E Q, Y E S;
: X -+ Y. Then there exists
such that f = gh.
136
S
Following Ringel we call mod A, separating
from
P
gory of Q = {X
I
be a finite-dimensional k-algebra such
B
contains a slice S = add S. Then S
mod B
that
Q.
Let
PROPOSITION.
mod B, separating
P - {X
X is generated by
TBIS}.
Proof.
By
BS = F(D(AA)) = D(MB). By
cogenerated by
and
BS}
Y E Y(M)
is cogenerated by
TBS}
from
AM with
B - End M. Moreover, we have
4.11 we know that
X(M) - {X
I
Y(M) _ {X
X is generated by
I
TBS}
X
is
is a split-
mod B. It is easily seen that any indecomposable
ting torsion theory on B-module
X
I
is a separating subcat-
there exists a hereditary finite-dimensional
5.2
k-algebra A and a tilting module that
a separating subcategory of
which does not belong to
From this the properties
(a)
and
P
is a direct summand of
S.
of a separating subcategory are
(b)
easily deduced.
So it remains to verify (c). Let there exists an
A-module
0 1 X' i 10 i II - 0
Y E Q. Then clearly
HomB(-,Y)
to
(*)
such that
F(X') - X. Let
be an injective resolution of V. Applying
yields an exact sequence Let
X' E T(M)
X E P. Then X E Y(M). So
(*)
0 - X -
S° -, S1 -0 0
ExtI(S,Y) = 0, for
with
(X(M),Y(M))
F
S°,S1 E add S.
splits. Apply
yields the following exact sequence.
0 - HomA(S1,Y) - HomA(S°,Y) - HomA(X,Y) - 0. This shows that also property (c)
is satisfied.
5.6
k-algebra and and only if
PROPOSITION. Let
AM add M
Proof.
A be a hereditary finite-dimensional
a tilting module. Then
B = End M
is hereditary if
is a slice.
If
add M
is a slice, then clearly
B = End M
is
137
hereditary. Conversely, if in
mod B. Let
F
:
Let
AM
add AM
is a slice.
-p
for some finite quiver A without oriented cycles.
A _, kA
B
is a slice
be a triangle-equivalence with
be a tilting module with
then clearly
add BB
A be a basic, hereditary finite-dimensional
Let
Z k-algebra. So
is hereditary, then
Db(B) -+ Db(A)
F(BB) = AM. We infer that
5.7
B
B = End M. If
Z A
is a Dynkin quiver
is representation-finite.
PROPOSITION.
Let
A
sentation-finite if and only if
be an affine quiver. Then AM
is repre-
B
contains a non-zero preprojective
and a non-zero preinjective direct summand.
Proof.
Let us denote by
P, (R,Q respectively) the additive
categories generated by all indecomposable preprojective (regular, preinjective respectivly) A-modules. If
AM
does not contain a non-zero pre-
projective summand then all modules in to the torsion theory
P
(T(M),F(M)), hence
are torsionfree with respect B
is representation-infinite.
does not contain a non-zero preinjective summand then all modules
If
AM
in
Q are torsion modules, hence If
AM
contains a non-zero preprojective summand
for some indecomposable projective we claim that
F(M)
is representation-finite.
B
A-modules
HomA(T-rP(a),X) = 0
X E F(M) we have that X
is of the form
A-module
o < s < r
where Aa
and some
r > o. Then
contains only finitely many isomorphism classes of in-
decomposable A-modules. Since
P(b)
P(a)
T-rP(a),
and some
or
TsP(b)
TrX
for an indecomposable for some indecomposable
is a module over
is obtained from A by deleting the vertex
is representation-finite. Therefore, F(M)
isomorphism classes of indecomposable
Aa = kAa
a. Note that
Aa
contains only finitely many
A-modules. Dually, if
AM
contains
138
non-zero preinjective summand, then
T(M)
isomorphism classes of indecomposable
contains only finitely many
A-modules. This finishes the proof.
While the situation for affine quivers is solved in a satisfactory way by the proposition above, the general question is still open.
5.8
Let
AM
Let A be an affine quiver and
PROPOSITION.
be a tilting module. Then AM
A = M.
contains a non-zero preinjective
or preprojective direct summand.
n
Let M =
Proof.
r
0 Mil
be a direct summand sum decomposi-
i=1
tion of M with
indecomposable and
Mi
have seen before that
dim Mi,
< i < n
1
M. !Z M.
generate
for
i * j. Then we
K0(A). It is well-
known (compare Dlab, Ringel (1976) or Ringel (1984))that the dimension vectors of regular
A-modules generate a proper subgroup of
K0(A). This
proves the assertion.
Let us write down explicitely the alternatives for a tilting
d
-,
module AM where
A
A = kA and
an affine quiver. Let
Mp (Mr,Mq
respectively) the direct sum of the preprojective (regular, preinjective) direct summands of
M.
(i)
M = M p 0 Mr
Mr * 0.
(i)*
M = Mr 0 Mq
Mr * 0.
(ii)
M=Mp0Mq
Mp*0*Mq.
(ii)'
M = Mp 0 Mr 0 Mq,
Mr * 0.
(iii)
M = Mp
(iii)* M = Mq
,
The corresponding tilted algebras in case (i) studied in great detail in examples for the cases
(ii)
4.9
and
and
(i)*
are
of Ringel (1984). Below we will give two (ii)', while the cases
(iii)
and
139
(iii)*
are studied in section 7.
The computation that the following two finite-dimensional k-algebras given by bound quivers are examples for the cases (ii)'
and
(ii)
is left to the reader. (a)
example for the case
(ii)
I = <ar>
This gives a tilted algebra obtained from
(b)
example for the case
kb, with
(ii)'
This gives a tilted algebra obtained from
k&, with
A' _
5.9
Let
A be a wild quiver. We have already pointed out
that there is no analogue to
may exist tilting modules
5.7
available. In contrast to
5.8
there
AM with only regular direct summands. In fact
one has the following theorem due to Ringel (1986a), which we state without proof.
THEOREM.
Let
A be a connected wild quiver with at least
three vertices. Then there exists a tilting module lar direct summands.
_6
with only regu-
140
AM with only
Let us give an example of a tilting module regular summands.
i
y
We consider A = k0, where A _ There exist uniquely determined (up to isomorohism)indecomposable
A-modules
dim M2 = (3,4,4)
M1, M2, M3 and
with dimension vectors
dim M3 = (12,15,14). This follows from a theorem
of Kac (1980) and the fact that
gk-A(dim Mi) -
may also show that M = M1 ® M2 0 M3
5.10
dim Ml = (4,5,5),
is an
I
for
1
A-tilting module.
We point out that for certain quivers
classification of the tilted algebras has the relevant references. If
< i < 3. One
0
a complete
been obtained. We just include
A = An we refer to Assem (1982); if
we refer to Conti (1986); and if
A = IDn
A = An we refer to Assem-Skowronski
(1986).
For classification results of certain subclasses of tilted algebras we refer to some remarks in section 7.
141
6. Partial tilting modules Let
dimension. An
A be a finite-dimensional A-module satisfying
k-algebra of finite global
pd AM < I
and
ExtI(M,M) - 0
is
called a partial tilting module. In this section we present several equivalent conditions on a partial tilting module to be a tilting module. The first result is due to Bongartz (1981) and the second follows ideas of Brenner and Butler (1980). Following ideas of Geigle and Lenzing (1986) we introduce the perpendicular category. As a main result we present that
A
is a tilted algebra if
End AM
AM
is a hereditary algebra and
a
partial tilting module.
6.1
LEMMA.
Let
AM
exists a tilting module AM 0 AM' M"
summand
of
0 -+ AA
+
C
AM
such that any indecomposable direct
is projective or satisfies
M'
Let
Proof. 11
be a partial tilting module. Then there
ExtA(AM,AA), and let
E1,...,Em be a basis of
M 0
AM -. 0
be the pushout along the diagonal map
i=1
(AA)m -' A of the exact sequence
0 Ei. We claim that
ting module. For this it is enough to show that Applying
HomA(M",M) * 0.
HomA(AM,-)
M' 0 M
is a til-
ExtA(M' 0 M,M' 0 M) = 0.
to the sequence above yields
m HomA(AM,
0 i=1
AM) - 6 -* ExtA(AM,AA)
ExtA(AM,AM') -> 0,
142
and
is surjective by construction, thus
6
ExtI(AM,AM') = 0. Applying
to the sequence above yields
HomA(-,AM)
ExtI(® AM,AM) --k Extj(AM',AM) -± ExtA(AA,AM).
Thus
ExtA(AM',AM) = 0.
Finally, applying
HomA(-,AM')
to the sequence above yields
ExtA(® AM,AM') ---> ExtA(AM',AM') -} ExtA(AA,AM).
Thus
ExtA(AM',AM') = 0. Let
be a direct summand of
M"
HomA(M",M) = 0. Thus
lies in the kernel of
M"
summand of the image of
satisfying e, thus
M"
is a direct
u, and therefore is projective. r
6.2
M'
COROLLARY.
Let
AM =
n.
0 Mil, with
Mi
be indecomposable
i=1
and
M.
M.
i * j, be a partial tilting module. Then the following
for
two conditions are equivalent: (a)
AM
(b)
r
is a tilting module.
equals the number of isomorphism classes of simple
A-modules.
Proof.
rk K0(B) = r. By which shows
AM
If
2.10
and
1.5
we infer that
B = End M, then
rk Ko(A) = rk K0(B),
(b).
Conversely let (b). Then by
is a tilting module and
6.1
AM be a partial tilting module satisfying
there exists
M'
such that M 0 M'
module. By the previous part of the proof we infer that
M
is a tilting M' E add M. Thus
is a tilting module.
6.3
Let
A be a hereditary finite-dimensional algebra. This
will be assumed for the rest of this section.
143
Let
LEMMA.
AM be a partial tilting module. The the follo-
wing two conditions are equivalent: (a)
AM
(b)
any non-zero
is a tilting module.
HomA(X,M) * 0
If
Proof.
AM
with
X
1
Ext A(X,X) = 0
satisfies either
ExtI(X,M) * 0.
or
is a tilting module, we clearly have an exact
sequence 0 -+ M-1 -+ M° -+ D(AA) -+ 0 with M1 ,M° E add M. Applying now gives
HomA(X,-)
(b).
f1,...,fr
Conversely, let let
: AA .+ Mr
f
be given by
be a basis of
HomA(AA,AM), and
f = (fI ...,fr). We claim that
is in-
f
jective.
Consider the two exact sequences:
0-+K= ker f -+AA -+im f -+0 0 -+ im f -+ Mr -+cok f = Q - 0.
and
Applying
HomA(-,M) = (-,M)
yields
0 -+ (im f,M) -+ (AA,M) -, (K,M) -+ ExtA(im f,M) - ExtI (AA,M) -+ Ext1A(K,M) -+ 0
0 -+ (Q,M) -+ (Mr,M) -+ (im f,M) -+ ExtA(Q,M) Thus
ExtA(im f,M) = 0. Since
K
-+
(K,M) = 0. Thus
-+
Ext1 (im f,M) -+ 0
is projective we also have
Ext1A(K,K) = 0 = ExtA(K,M). By construction tive, hence
ExtA(Mr,M)
(im f,M) -+ (AA,M)
K = 0. This shows that
f
is surjec-
is injective.
So we have an exact sequence O_O' AA
Now
(*)
Mr
g
ExtI(M,Q) = 0, since
Ext1(Q,M) = 0, since to
-fi
now yields
(Mr,M) -+ (AA,M)
Ext1(Q,Q) = 0.
Q-+0
Q
(*). is generated by
M, and
is surjective. Applying
HomA(Q,-)
144
add M.
Q * 0, then HomA(Q,M) * 0
If
be a basis of
belongs to
Q
We claim that
and consider
HomA(Q,M)
by assumption. Let
hl,...,hs
h = (hl,...,hs) : Q -. Ms. This
yields the following two exact sequences:
(+)
0-pimh- Ms -Q' -0
(++)
0-+ker h -'Q ExtI(ker h,M) = ExtA(im h,M) = 0. By con-
We conclude that
struction HomA(im H,M) -, HomA(Q,M)
HomA(ker h,M) = 0. Applying
is surjective, thus
HomA(ker h,-)
0 -, ExtI(ker h,ker h) -+ ExtI(ker h,Q) to
ExtA(Q,Q) = 0, we infer that
h
shows that
(++)
is exact, and applying
ExtA(Q,Q) - ExtI(ker h,Q) - 0
shows
(++)
to
HomA(-,Q)
is exact. As
ExtI(ker h,ker h) = 0, or equivalently that
is injective.
So consider the exact sequence:
0 -1. Q
Since
ExtA(Q',M) = 0.
Q
6.4 an
Let
HomA(Q',-)
Applying
U
belongs to
(-)
to
Q' -i 0
(-).
is surjective, we see that (*)
yields that
is a split exact sequence, which im-
add M. This finishes the proof.
We now introduce the perpendicular category. Let
A-module satisfying n
Msh
HomA(Ms,M) -+ HomA(Q,M)
ExtI(Q',Q) = 0. In particular plies that
-0
ExtI(M,M) - 0, End M = k
and
be the number of isomorphism classes of simple
be the full subcategory of
HomA(M,X) - 0 = ExtA(M,X). termined by
M.
U
mod A with objects
X
M be
HomA(M,AA) = 0. A-modules. Let satisfying
is called the perpendicular category de-
145
There exists a hereditary finite-dimensional
PROPOSITION.
A0 with
algebra
that
U =, mod
isomorphism classes of simple
n-1
Ao-modules such
Ao. It is a straightforward calculation to show that
Proof.
is an abelian, exact and extension-closed full subcategory of Let
E
A-module constructed in
be the
such that
6.1
U
mod A.
E 0 M
is a til-
ting module. Thus we have an exact sequence
(*)
0
where m - dim ExtI(M,AA).
Applying RomA(M,-)
to
0 - HomA(M,E) - HomA(M,Mm) By construction
HomA(M,E) = 0, and Clearly
HomA(-,X)
ExtA(M,AA) - ExtA(M,E) - 0.
E
X
E ® M on
satisfies
Extj(E,X) - 0
has
mod A. Thus
X
X
X E U, for apply
E. Indeed, E 0 M
n-1
is generated by
is a
E 0 M. But
is in fact generated by U = mod End E. Let
E.
A. = End E.
isomorphism classes of simple modules.
Next we show that for each X E U
0 - E 1 - E° -. X -r 0 with E 1,E° E add E. Let RomA(E,X). Then
for
is clearly a torsion module for the torsion theory
In other words we see that Ao
is also injective.
is relative projective.
is generated by
X
HomA(M,X) = 0 now shows that
Then clearly
6
E E U.
X E U, then
tilting module and
ExtI(M,E) = 0. As
we see that
(*). In particular, E
to If
induced by
d
is surjective, whence
6
dimk HomA(M,Mm) = dimk ExtI(M,AA) Thus
yields
(*)
f = (fl,...,fr)t : Er - X
there is an exact sequence
f1,...,fr be a basis of
is surjective. Applying
146
ker f E U. By construction we have that
shows that
HomA(M,-)
HomA(E,Er) - HomA(E,X) And also
is surjective, which shows that
ExtI(ker f,E) = 0. Thus
ExtA(E,ker f) = 0.
ker f E add E. This shows that
Ao
is
again hereditary, thus finishing the proof.
6.5
B = End P. Then
AM
decomposable
AM
If
A' = A/AeA. Then AM
mod A' -+ mod A. Note that
n If
gory
U
ExtA(X,M) _
X = Ae
for some primitive idem-
is again a hereditary finite-
A'
lies in the image of the canonical embedding
A'
has
isomorphism classes of simple
n-1
A-modules.
is not projective, we consider the perpendicular cate-
X
X. Then
k-algebra
Ao-modules. Clearly partial
and
is the number of isomorphism classes of simple
determined by
dimensional
there exists an in-
End X = k.
dimensional algebra and
modules, if
6.3
Ext1(X,X) = 0
satisfying
is projective, then
e E A. Let
potent
is a tilting module there is nothing to show.
X
0 = HomA(X,M). Clearly
X
be a partial tilting module and
is not a tilting module. By
A-module
If
AM
is a tilted algebra.
B
Proof.
So assume that
Let
COROLLARY.
A0
U =, mod A0
having
for some hereditary finite-
n-1
isomorphism classes of simple
U
and can thus be considered as a
AM belongs to
Ao-tilting module.
Thus in both cases we have reduced the number of isomorphism classes of simple modules. Iterating this procedure we finally reach a hereditary finite-dimensional tified with a
k-algebra
C-tilting module
CM
C
such that
such that
AM
can be iden-
B = End AM = End CM. This
proves the assertion.
6.6 If
A
Let
A be a hereditary finite-dimensional
is basic, we know that
A =, kA
k-algebra.
for some finite quiver A without
147
oriented cycles. We want to give two examples of the perpendicular category. For the representation theoretic terminology and results occuring here we refer to Ringel (1984). -r
Let
(a)
M be a non-regular indecomposable
Then M automatically satisfies End M = k. Up to duality we may assume that
Ext
M
(M,M) = 0
r
such that
and
is a preprojective
module. Then there exists an indecomposable projective and a non-negative integer
k&-module.
kA-module
M = T-rP(i). Let
quiver obtained from A by deleting the vertex
A'
kAP(i)
be the
i. Then one may show
that the perpendicular category determined by M
is equivalent to
-, mod kA'.
Let M be a regular indecomposable
(b)
ing
Ext1(M,M) = 0
and
A
(b1)
kA-module satisfy-
End M = k.
is an affine quiver.
Our assumptions imply that the position of
M
in the Auslan-
.4
der-Reiten quiver of some
kA has to be in a tube of the form Z &_/r
r > 1. In particular we see that A * c(a
for
Then the structure of ,
the perpendicular category determined by M can be read off the results in
4.7
of Ringel (1984). We refrain here from dealing with those tech-
nicalities in general. Instead we will give one particular example.
Let A =
1p.
-o
M be the indecomposable module with dimension vector
and
Then the perpendicular category determined by M
is equivalent to
0
mod kA'
where
(b2)
As
A
is a wild quiver.
Here the situation is much more complicated as in the previous
148
cases and a complete solution is unknown. So we have to be content with
A be the quiver
Let
an example.
Let M be an indecomposable
kA-module with dimension vector X5+-5. Then M satisfies Ext1(M,M) = 0
and
End M = k
and is a regular
Then the perpendicular category determined by M mod kA', where
Clearly the
A'
kZ-module (compute
TMS).
is equivalent to
kA-module X = T rM
for
r > o
1
still satisfies
Ext (X,X) = 0
and
End X = k
and of course is a regular
kZ-module. Then the perpendicular category determined by
y lent to
mod kAr, where Ar = c
x is equiva-
al :
and
s = 4+2r.
In this way we see that the dimension of the algebra given by the perpendicular category is not bounded.
149
7.
7.1
Concealed algebras
A be a basic, connected, hereditary finite-dimen-
Let
k-algebra. So
sional
assume in addition that
rent from fin, ID n,
IE6
A !Z kA
A
for some finite connected quiver
is representation-infinite; so A
, ]E7 , M8 . A finite-dimensional
A. We is diffe-
k-algebra
B
is called a concealed algebra if there exists a preprojective tilting
AM
module then
B
such that
B - End AM. If in addition, A is an affine quiver,
is called a tame concealed algebra. Thus, concealed algebras
are quite special tilted algebras. In this section we will prove a result
of Ringel (1986), which gives a characterization of those algebras. For a more detailed discussion of concealed algebras we refer to
4.3
of
Ringel (1984). We point out that a classification of the tame concealed algebras was obtained by Happel and Vossieck (1983). This classification has turned out to be quite useful. We refer to Bongartz (1984) and Bautista, Gabriel, Rojter and Salmeron (1985). Also we remark that Bongartz (1984a) obtained by using a different characterization the same classification.
Finally, let us remark that Unger (1986) has obtained a complete classification of the concealed algebras arising from minimal wild _, -4 quiver s A. These are precisely those quivers A which are wild but every
I proper subquiver of A
is an affine or Dynkin quiver. It turns out that
150
these algebras e E B
are minimal wild, in the sense that for all idempotents
B
dimensional algebra into
B/BeB
the factor algebra
mod C
is not wild; where we call a finite
wild if there exists an embedding of
C
mod k<x,y>
is the free (non-commutative) algebra with two
(k<x,y>
generators).
7.2
A finite-dimensional
THEOREM.
k-algebra
is a con-
B
cealed algebra if and only if there exist two different components of the Auslander-Reiten quiver of
If
Proof.
containing a slice.
is a concealed algebra, then a proof of the
B
assertion can be found in a finite-dimensional
B
of Ringel (1984). Conversely, let
4.3
k-algebra with slices
S
and
S', which belong to
different components of the Auslander-Reiten quiver of
B. Then
necessarily connected and representation-infinite. Moreover, by know that let
is a tilted algebra. Let
B
A = End S. Then A
over, Db(B)
Db(A)
(compare
and by
Db(A). We denote by
R[i], for
form
S
and
C[i], for
i E Z, the component of
C[i]. We may assume that
r(Db(A))
S'. Clearly, the indecompo-
By
5.5
C(j].
j * o.
we know that
mod B, say separating
of the
i = o. Similarly, the indecomposable sum-
mands of S' belong to some fixed component of 1(Db(A)) of the form It is easily seen that
i E Z
ri
belong to some fixed component of r(Db(A))
S
1.5.5
ri.
be a slice module for
S'
sable summands of
we
be a slice module for
containing projective modules in
containing regular modules in Let
is
5.2
and we may use some fixed identification. In
r(Db(A))
1.5.5)
B
is hereditary and representation-infinite, More-
we have determined the quiver of the component of
S
be
B
P
from
S
Q. Let
is a separating subcategory of S
be an indecomposable summand
151
of
S'. If
belongs to
Si
Q, then
an indecomposable projective HomB(P,S) * 0, for ty (c)
S
B-module with
P E P U S
is sincere. Therefore
lar way that
belongs to
j = 1. If
S;
belongs to
S
C[-1]
or
indecomposable projective
or
P, then we show in a simi-
C[0], and since
B-module belongs to
if necessary, we may assume that
mod A
into
Db(A)
A-module. This shows that
S'
C[0]
B-modules lie in
Thus all indecomposable projective
embedding of
HomB(S,S;) * 0. There-
is sincere, any indecomposable projective R[-1]
BB
be
and use proper-
j = -1. Without loss of generality, we may assume Since
P
HomB(P,Sj) * 0. Then
of a separating subcategory to establish
fore we see that
as an
HomB(P,S;) * 0. Indeed, let
j = 1.
B-module
is sincere, any or
R[0]
or
C[1].
C[0]. Applying
T
lies in the image of the canonical
and is preprojective when considered B
is a concealed algebra.
Finally, let us remark that the Auslander-Reiten quiver of a tilted algebra contains at most two components which contain a slice.
CHAPTER
1.
Piecewise hereditary algebras
We call a finite-dimensional
1.1
hereditary if
PIECEWISE HEREDITARY ALGEBRAS
IV
k-algebra A piecewise
is triangle-equivalent to
Db(A)
Db(kA)
for some finite
quiver A without oriented cycle. We recall that A is uniquely determined up to the relation - introduced in
I. 5.7. In this section we
present some general facts about piecewise hereditary algebras. But first we need to recall some elementary facts for hereditary finite-dimensional k-algebras.
LEMMA. Let
1.2
bra and let
X1,X2,X3
tive and that and linear maps
g
be
B-modules. Suppose that
X2 - X3
:
h1
:
be a hereditary finite-dimensional
B
X1 -1-Y
f: X1 -+ X2
and
h2 : Y -+ X3
such that
(-h8
2
0--->X1 ----------- ). X2
-+0
is exact.
Proof. Consider the following exact sequence
Since
B
0-x
is surjec-
is injective. Then there exists a module
(f h1)
(*)
k-alge-
92,, X
is hereditary, ExtB(X3/X2,f)
X/X -0. is surjective. Let
Y
153
h
be a preimage of
Ext$(X3/X2,X1). Then we obtain the following
in
(*)
commutative diagram of exact sequences
h
Y -- X3/X2 - 0
0 ---} X 1
fi
Ih2
0 -+ X2-' with
h1
injective and
X3---
II
X3/X2 --- 0 h2
surjective. By construction we have that -h8
(f h1)
0-+X1
2 :X211 Y --)X
is exact.
LEMMA.
1.3
bra and let
X,Y
B
be a hereditary finite-dimensional k-alge-
be indecomposable
ExtI(Y,X) - 0. Then
h with
B-modules. Suppose that
0 * h E HomB(X,Y)
Proof. Let zation of
Let
f
is either injective or surjective.
0 * h E HomB(X,Y). Let X+ Z g Y surjective and
g
injective. By
be a factori1.2
we obtain
an exact sequence
(f h) By assumption this sequence splits. By Krull-Schmidt we infer that isomorphic to
X or isomorphic to
Y. In the first case we have that
is injective while in the second case we have that
1.4
Z
h
is
h
is surjective.
We also note the following immediate consequence for an
indecomposable module
X over a hereditary finite-dimensional
k-algebra B.
154
ExtI(X,X) = 0, then End X
If
LEMMA. Let
1.5
ExtB(Xi,Xi) = 0
then
be a hereditary finite-dimensional k-alge-
be indecomposable
X1, X2
bra and let
B
k
for
< i < 2
1
and
ExtI(X2,X1) = 0. If
HomA(XI,X2) * 0
ExtI(X1,X2) = 0.
ExtB(X2,X1) = 0
Since
Proof.
is either injective or surjective by duces a surjection Ext B(X1,X2) = 0. If
Ext1(X X) i Ext1(X ,X ). So 20 2
1.3. If
f
B
2
1
Ext1(X ,X) = 0 B
is surjective, f
in-
Ext1(X1,X1) = 0
implies
induces a surjection
is injective, f
f
f E HomB(XI,X2)
a non-zero map
ExtI(X1,X1) - ExtI(X1,X2). So
1
B
B-modules such that
2
implies
2
Ext1(X ,X) = 0. B
2
1
The proof of the next lemma is due to Ringel. We point
1.6
out that a different proof might be obtained by using elementary arguments from algebraic geometry.
Let
LEMMA.
0- X 4 Y
be a finite-dimensional k-algebra. Let
C
be a non-split exact sequence in
Z 1 0
mod C. Then
dim End (Y) < dim End(X 0 Z).
Proof.
with W
and
in = Bu}
is surjective and
Thus
T(y) - ryp Let
*(y) = Uyw. Clearly paw = 0
shows that
a i
defined by
k-linear map
there exists
: Y i X
S
c
y
:
p
p(y) _ 7rp. Since is injective. Let
such that
Z -' X
r
a = Su. Now
such that
xy = d.
is surjective.
be the
: End(Y) -, HomC(X,Z)
R
a E End(X)
pa = an = 0}. Let
I
shows that there exists
pa = 0
there exists
is injective we infer that
p
ax = 0
p6p
Then
R = {a E End(Y)
S = {a E End(Y) be the
: HomC(Z,X) -' S
a E S. Since
Consider
is contained in
shows that there exists
k-linear map defined by
ker P. Conversely, let $
such that
y E ker p.
ya = Sp. Thus
155
R = ker gyp. a E R we choose
For
and
S
k-linear map
an = irl'. This defines a
ip
S'
such that
and
ua = F3p
: R -End(X)x End(Z)
by
Clearly, ker ,y = S.
We claim that a E R a'
such that
with
is not surjective. Otherwise there exists
ry(a) = (1X,0). Thus
a = a'u. We infer that
ua7r - 0
implies that there exists
in contrast to the assumption
Pa' = 1X
that 0 - X - Y - Z -+ 0 does not split. This shows that dim R - dim S < dim End(X) + dim Z. We have seen before that dim End(Y) - dim R < dimk HomC(X,Z).
dim S = dimkHomC(Z,X)
and
Combining these results show the
assertion.
k-algebra and let End(Xi)
and
k
X1,X2
for
I
be
B
be a hereditary finite-dimensional
B-modules such
that
ExtB(Xi,Xi) = 0,
< i < 2, HomB(XI,X2) = HomB(X2,X1) = 0, ExtB(X2,X1) = 0
= 1. If
dimkExtI(X1,X2)
extension then
Let
LEMMA.
1.7
End(E) :, k
(*) and
0 -+ X2 -+ E -+ X1 -+ 0
is a non-split
ExtI(E,E) = 0. In particular, E
is inde-
composable.
Proof.
sequence
It follows from
that
End(E) =, k. The exact
ExtI(X1,E) = 0 = ExtI(X2,E) = 0. Applying
yields
(*)
1.6
Ext (-,E)
gives now the assertion.
1.8
follows from
A be a piecewise hereditary algebra of type
Let
III. 1.5
that
A has finite global dimension. In particu-
lar, the bilinear form <x,y>A = x CAt
yt
is defined on
is isometric to the corresponding bilinear form on III. 1.5). In property.
1.11
o. It
we will see that
K0(k-')
K0(A) =,71n
which
(compare
<-,->A has a rather surprising
156
let
X,Y
be indecomposable
Exti(X,X) - 0
(ii)
If
Proof.
Db(kZ). By
and some
for
i > I
occur in a cycle of
X,Y
mod A, Exti(X,Y) = 0
i > 1.
(i)
Denote by
I. 5.2, F(X) : TJX'
a triangle-equivalence from
F
for an indecomposable
Db(A)
kX-module
X'
j E Z. Then
ExtA(X,X) - Horn
(X,T'X)
b
(ii)
that
Horn b
If
F(X)
occur in a cycle of
X,Y and
some indecomposable
y (F(X),T1F(X)) eeExtkS(X',X') - 0
D (kg)
D (A)
for i > 1.
1. 5.3
A-modules. Then
(i)
for
to
A be a piecewise hereditary algebra and
Let
LEMMA.
1.9
F(Y)
kAX-modules
mod A, it follows from
are isomorphic to X'
and
T3X'
and
and some
Y'
TjY'
for
j E Z. Thus
(X,T'Y) =, Hom ExtI(X,Y) = Horn (X',T"L') - Ext'-(X',Y') = 0 A Db(A) Db(kA) kA
for
i > 1.
1.10
LEMMA.
A piecewise hereditary algebra A
algebra of a finite-dimensional hereditary
Proof. Assume indecomposable projective
from
Db(A)
to
k-algebra.
P0 - P1 - P2 -, ... - Pr A-modules. Denote by
Db(kZ). Using
I. 5.3
and
is a factor
F
F
P0
is a cycle of
a triangle-equivalence
we obtain a cycle
X0 - X1 -, X2 -, ... - Xr = X0 of indecomposable kZ-modules satisfying and thus
by
Moreover, we know by
Ext
(Xi,X.) - 0
1.3
that a map occuring in the cycle is either injective or surjective.
End(X1) =, k
1.4.
Clearly, either all maps are injective, or all maps are surjective (since otherwise we obtain a proper surjection followed by a proper injection
whose composition is non-zero but neither injective nor surjective, con-
157
1.3). Thus all maps have to be isomorphisms, in contrast to the
trary to
definition of a cycle.
S,S'
let
such that
Let A be a piecewise hereditary algebra and
LEMMA.
1.11
A-modules. Then there exists at most one
be simple
ExtA(S,S') * 0.
Proof.
1.10
By
we may assume that
Exti(S,S') * 0 we may assume that by
such that
TAX, F(S') = Tj'X'.
F(S)
Exti(S,S') * 0
and
Clearly
implies X,X'
for all
and integers
Let
j,j' E Z
Now ExtA(S,S') _+ Extk-0
j(X,X'),
0 < j'+i-j < I.
i. Then
for some
Extkt(X',X) - 0
consideratijns above. Now the assertion follows from
EXAMPLE.
j. Denote
Db(kA). As before there
to X'
and if
satisfy Exta(X,X) - Exta(X',X') = 0. Now
ExtA(S,S') * 0
suppose that
Db(A)
X and
exist indecomposable k0-modules
S * S'
Extk(S',S) = 0
a triangle-equivalence from
F
i > 0
C
by the
1.5.
be given by the bound quiver
(X,I)
I =
Let
S
be the simple associated with the vertex
simple associated with the vertex
I
and
S'
be the
4. An easy computation shows that
Ext2(S,S') * 0 * Ext3(S,S').
We infer that
C
is not piecewise hereditary.
REMARK. S,S'
be simple
Let
A be a piecewise herediary algebra and let
A-modules with
A - 0
then
158
i. This follows immediately from
for all
Ext1A(S,S') = 0
and
1.11
III. 1.3.
Let A be a finite quiver without oriented cycle. Let and let
H[i]
plexes
X'
be the full subcategory of isomorphic to
posable object Let
note by
F
sume that
F(Y) E H[o]. If For
mod A. Let
Ui
is contained in
X E ind A
H[i]
i > o
for
be the full subcategory of
X
then
Proof.
Let
satisfying
A-module Y with normalized.
F
of
consisting of the in-
F(X) E H[i]. If
is normalized
F
U.. 1
X,Y E Ui
j > 1.
we know that
X E Ui
kZ-module
(X,TdY)
F(X) = T1X'
Y E ind A. Then and some
Y'
Hom
for some
F(Y) = TrY'
r > o. Let
(T'X',Tr+iY')
for
j > o, then
Extr+j-1(X',Y').
Db(kX)
Db(A)
then
Ui
pd X < i+1. Moreover, if
X E Ui. Then
kA-module V. Let
ExtA(X,Y) * 0
mod A
is contained in some
Since
Extj(X,Y) = Hom
A
X
for
some indecomposable
an indecompo-
X
and
we introduce now certain subcategories
i > o
ExtA(X,Y) = 0
indecomposable
H[i].
if necessary, we may as-
T
satisfies these conditions we call
F
LEMMA.
If
is contained in precisely one
A be a piecewise hereditary algebra of type A and de-
A-modules
decomposable
then
X E mod kA. We know that an indecom-
for
a triangle-equivalence. Applying F(X)
consisting of the com-
A-module and that there exists an indecomposable
sable
and
Db(11)
of
X'
T1X
Db(kA)
i E 7L
0 < r+j-i < 1. In particular, j < i+1. Or equi-
valently, pd X < i+1.
Using similar arguments as before one can show that HomA(X,Y) * 0
for
X E Ui
and
Y E Uj
implies
0 < j-i < 1, and
Extr(X,Y) * 0
for
X E Ui
and
Y E Uj
implies
0 < r+j-i < 1.
159
LEMMA.
Each
Proof.
Let
contains simple
U.
X E Ui
is not simple. There exists a 0 i S i X -, X/S -, 0
with
previous remark we infer that summand of
X/S
such that
be of minimal length and suppose that
X
non-split exact sequence a simple
S
A-modules.
A-module. By assumption and the
S E Ui-1. Let
X'
be an indecomposable
ExtI(X',S) * 0. Again we infer that
By the previous remark we must have
X'E Ui+1'
0 < I+i-1-(i+1) < 1, a contradiction.
160
2.
Cycles in mod kkk.
Throughout this section let
A a kk for some finite quiver I
without oriented cycles.
Let
THEOREM.
C
closed under extensions and direct summands. If also contains an indecomposable
Proof. (1)
If
X,Y E C
(2)
If
C
mod A which is
be a full subcategory of
A-module
Z
contains a cycle, C
C
such that
End Z * k.
The proof is divided into the following steps: and
f E HomA(X,Y), then
im f E C
by
1.2.
contains a cycle, it contains an even cycle, i.e. a cycle of
the form
X2n-1
where all
f2i
are injective and all
f2i+1
surjective.
This follows from M. In the sequel, we suppose that among all even cycles the given one has minimal length implies
n > 2.
2n. We also assume that
End(X1) - k
for all
i. This
161
(3)
We may assume that
In fact, if RomA(Xi,-)
ExtI(X.,Xi) - 0
End(Xi) - k
and End(E) cu k[X] /(X2)
(4)
HomA(X,,Xi) = 0
Suppose there exists summands of Y
in f
if
shows that
E i
is indecomposable,
(1)
all indecomposable
finishing the proof.
2 < i < 2n.
0 * f E HomA(XO,Xi). By
belong to
in C and linear maps
i.
is a non-split extension. Applying
0 1 Xi -. Ei i Xi -o.0
and using
for all
C. So there exists an indecomposable A-module
X0 f . Y f Xi
with
injective. This yields a cycle of length less than
f'
2n
surjective and in
f"
C, contradic-
ting the minimality of the given cycle. In the sequel, we suppose that
dim X0 + dimkHomA(XI,X3)
is
smaller or equal to the corresponding sum of any other even cycle of length
(5)
2n.
Each non-zero
In fact, suppose that there exists HomA(X0,X2) - 0
jective. As
exists an indecomposable X0
f Y' f XI
exact sequence striction to
with
by
A-module f'
0 * f E HomA(X0,XI) (4)
Y' E C
is non-zero. Therefore
the minimality of
and
there is a diagram
there
f"
f0
injective. Consider the is injective its re-
HomA(u,X1) * 0. In particu-
dim Y' < dim Xo, contradicting
dim X0 + dim HomA(XI,X3).
1.2
(1)
and linear maps
0 + ker V_ -Xo Y' + 0. As ker f'
which is not in-
is not surjective. By
f
surjective and
lar, dim HomA(Y',XI) < dimA(X0,XI)
By
is injective.
f E HomA(X0,XI)
162
(g f1)
such that (+) 0 - X
f J
Y 0 X2 -r X3 - 0 is exact. Let
r
0
Y =
be a decomposition of
Y.
Y
into indecomposables. Let
and
gi
i=1
be the corresponding components of
hi
hi * 0
r > 2. In particular
(6)
and
g
for all
h.
i.
The first follows from the minimality of the given cycle. Suppose for one
Y. c ker1 = im(g f1). Thus
i. Then
Y.
hi = 0
is a direct summand
2
im(g f1). But
of
im(g f1)
X1
is indecomposable. Hence the sequence
splits, contradicting again the minimality of the given cycle.
Each non-zero
(7)
In fact, apply
f E HomA(X0,Yi)
HomA(Xo,-)
to r
HomA(Xo,XI) =s HomA(Xo,Y) =
is injective.
(+). By
(4)
it follows that
HomA(X0,Yi). Let
8
u E HomA(Xo,X,)
be the
i=1
preimage of
f. Thus
and injective by
(5). Therefore we have that
is injective.
f = ug
Choose some that
u * o
hi
i
such that
f0gi * 0. Since
2n
is neither surjective nor injective. By
ExtI(X3,X3) = 0. Thus
(+)
is minimal we infer (3)
we have that
induces the exact sequence
0 -i HomA(X3,X3) - HomA(X2 ® Y,X3) - HomA(XI,X3) - 0. Denoting
by
dimkHomA(M,N)
<M,N>, we infer that
r
<XI,X3> =
F + <X2,X3> - I > j=1
+
F j*i
163
>
since
* 0
for all
j
by
(6).
It follows that the cycle
f
where
is an indecomposable direct summand of
Z'
tion to the minimality of
im hi
is a contradic-
be a full subcategory of
mod A which is
dim X0 + <X1'X3>.
This finishes the proof of the theorem.
COROLLARY.
Let
C
closed under extensions and direct summands. If also contains an indecomposable
Proof.
using
A-module
Z
C
contains a cycle, C
such that
ExtA(Z,Z) * 0.
This follows immediately from the theorem above by
1.4.
REMARK.
proof of
1.10.
Using the theorem above one obtains an alternative
164
The representation-finite case
3.
In this section we study representation-finite piecewise hereditary algebras. First we prove a generalization of a result due to Ringel (1983). The proof is a simple modification of his proof adapted to our situation.
A be a finite-dimensional
Let
is called a brick if
M be an indecomposable a brick
Z
such that
A-module
Z
End Z = k.
Let
THEOREM.
3.1
k-algebra. An
A be a piecewise hereditary algebra and
A-module which is not a brick. Then M contains ExtI(Z,Z) * 0.
Proof. It is enough to produce an indecomposable proper sub-
X of
module that
M such that
ExtI(X,X) * 0. Let
has minimal length. Then
im f = S
S
0 * f E End M be such
is indecomposable. If
1
Ext (S,S) * 0, we set
X c N = ker f Such an
X = S. Otherwise, we choose an indecomposable
of minimal length such that
X exists by
(1)
I
Ho
below. We will show in
(SIX). (2)
that
ExtI(X,X) * 0. s
(1)
Let
N =
0
Ni
with
Ni
indecomposable. Denote by
pi
the cano-
i=1
nical projection from exact sequences:
N
to
Ni. Consider the following diagram of
165
f
0-
i
0
The lower sequence does not split. Otherwise, Ni mand of that
ExtI(S,Ni) * 0
HomA(S,Ni) * 0
Let
(2)
g
M. Therefore
for at least one
0 * g E HomA(S,X). Since
is injective. ExtA(S,S) = 0
S
for all
would be a direct sumand
i
S c ker f
implies
i.
has minimal length we infer that
implies that
g
is not bijective. Con-
sider the exact sequence
(*) The exact sequence
induces an exact sequence
(*)
ExtI(S,S)
Therefore
-
ExtI(S,X)
-
ExtI(S,Q).
ExtI(S,Q) * 0. So there is a non-split extension
0 -.Q -, E - S - 0. This induces a sequence Q uI. E' -° S $ X - Q for each indecomposable summand
E'
of
E. Since
Q
is indecomposable by
(3)
below, u
are non-zero and non-invertible, we obtain a cycle of that
by
Ext2(Q,X) = 0
1.9(ii). Applying
ExtI(-,X)
and
mod A. We infer to
(*)
yields
the exact sequence
ExtA(X,X)
Therefore
-+
ExtI(S,X) -' Ext2(Q,X).
ExtI(X,X) * 0. r
(3)
Suppose
Q =
0 i=1
Qi
with
Qi
indecomposable and
v
r > 1. We may
166
assume that
ExtA(S,Q1) * 0. Denote by
the inclusion from
p1
to
Q1
Q.
Consider the induced sequence:
0 --} S -} Y --p Q 1 --- 0
0 -+ S --> X ---} Q -+ 0
(*)
X
The upper sequence does not split, for Ext2(S,S) = 0
by
is indecomposable. Since
the exact sequence
1.9(i)
ExtA(S,Y) -ExtA(S,Q1) -Ext2(S,S) t
ExtA(S,Y) * 0. Let
yields
Y =
0
with
Y.
Yi
indecomposable. As
i=1
does not split
(**)
ExtA(S,Y.) * 0
Hence
and
HomA(S,Yi) * 0
i. So there exists
j
with
HomA(S,Y.) * 0. But this contradicts the choice of
X.
is indecomposable.
Q
3.2
By duality we obtain the following result.
THEOREM. Let be an indecomposable module
for all
Z
A be a piecewise hereditary algebra and
A-module which is not a brick. Then M has a factor
which is a brick and satisfies
3.3
M
ExtA(Z,Z) * 0.
The following lemma is due to Ringel. For a more sophis-
ticated result in this direction we refer to
3.1
of Ringel (1984). For
our purposes this simplified version suffices.
LEMMA.
a brick satisfying
Let
B
be a finite-dimensional
ExtB(X,X) * 0
and
k-algebra and
Ext2(X,X) = 0.
Then
B
X be
is repre-
sentation-infinite.
Proof. Xd
for
d > o
We will construct inductively indecomposable B-modules
and non-split extensions
-
167
IT
11
Ed : 0 - Xd_1
dim.HomB(Xo,Xd) = Set
X0 - 0 for d >
Xd and
I
X0 = X
Ext2(Xo,Xd) = 0 and let
such that
I
for
d > 0.
be a non-split extension in
EI
ExtB(X0,X0), say u1
Apply
HomB(Xo,-)
Ext2(X0,X0) = 0
't 1
0 - Xo -- XI --> Xo-- 0.
EI
E1. So we obtain by using
to
End Xo = k
and
the following exact sequence d
0 - k - HomB(Xo,X1) - k -, ExtI(Xo,Xo) - ExtI(Xo,X1) - ExtI(Xo,Xo) - 0. Since
6
dimkHomB(XO,XI) = 1. Since
is injective we infer that
not split we infer that
does
E1
is indecomposable. Observe that
XI
Ext2(Xo,XI) = 0. n
ud1'
Now suppose that is constructed with
Ext2(X ,X B
o
d-1
)
Xd_l
= 0. Since
Ed_I
0 -} Xd_2
Ext2(X B
X
o' d-2
) = 0
d-1' Xo -- 0
dimkHomB(X0,Xd_,) =
indecomposable and
Ed E ExtI(Xo,Xd_1)
is surjective. Let
Xd_I
we have that
be a preimage of
Extl(X B
o
:
0 - Xo
'
Apply
HomB(Xo,-)
to
XI -} X0 -- 0 r
lid-I
Ed :
al
Ill
T
Pd
ITd
0 - Xd-1' Xd ' Xo -> 0 I
,Ir
d-1
E1. Thus we
obtain the following diagram of non-split extensions:
E1
and
I
11'
Ed. This gives the following exact sequence
)
168
0 -+ HomB(XO'Xd-I)
HomB(XX,Xd) 6
By induction, dimkHomB(Xo,Xd_I) = I Since
6
Xd
ExtB(X'Xd)
and by assumption
Ed
which also
does not split. By induction,
Ext2(X0,X0)
0
hence
0.
This shows that ryly large length, hence
3.4
A
End X0 = k.
dimkHomB(XO,Xd) = I
is indecomposable, for
Ext2(Xo,Xd_,) = 0
If
and by assumption
is injective we infer that
shows that
- HomB(Xo,X0)
B
COROLLARY.
admits indecomposable modules of arbitra-
B
is representation-infinite.
Let
A be a piecewise hereditary algebra.
is representation-finite then every indecomposable
A-modules is
a brick.
Proof.
M be an indecomposable
Let
a brick then M contains a brick By
1.9(i)
the brick
diction using
3.5
Z
B
a full subcategory of
be a finite-dimensional
mod B. The subcategory
U
Let
be a finite subcategory of
Let
3.1.
U
k-algebra and let
U
is called finite if
be U
finitely many indecomposable B-modules.
A be a piecewise hereditary algebra and let
mod A closed under extensions and sub-
modules (or factor modules). Then
Proof.
by
is a brick.
contains (up to isomorphism) only
COROLLARY.
ExtI(Z,Z) * 0
such that
is not
Ext2(Z,Z) = 0. So we obtain a contra-
satisfies
3.3; hence M
Let
Z
A-module. If M
U
is directed.
X0 -+ XI - ... - Xr - Xo be a cycle of indecom-
169
posable
A-modules with
equivalence from
Db(A)
Xi E U
Db(kA). It follows from
to
F(X0) - F(X1) - ... - F(Xr) = F(X0) ks-modules. Let
composable
I < i < r. Denote by
for
a triangle-
F
I. 5.3
that
may be considered as a cycle of inde-
be the smallest full subcategory of
C
mod ka closed under extensions and direct summands containing
0 < i < r. Then for Y E C
(*)
Y' E U. In fact, let
such that
F(Y') me Y
0 - Y1
YI = F(ZI)
u
there exists an A-module
for
with
Y'
YI,Y2 E C
F(Xi)
and
v
Y - Y2 - 0 be exact in mod k;. We may assume that
and
for some
Y2 = F(Z2)
A-modules
and
ZI
Z1,Z2 E U. We have
Z2
such that
Ho Db(A)(Z2,TZ1)
ExtA(Z2,Z1). Let
w E Hom b
1 (Y2,TY1)
be the element corresponding to
D (kg) to
(*). Then w
F(w')
f
0 -ZI -, Y'
assumption
8
Db(k0)
2
29
TZ ). Let 1
be the corresponding element in
Z2 - 0
Y' E U. So we obtain a triangle F(ZI) F(f
Z1
-f+
F(Z2) F(w
F(Y')
ExtA(Z2,Z1). By 8
Y'
which by construction is isomorphic to
In particular section
(Z
Db(A)
Db(A). Thus also in
w' E Hom
for some
TZI
F(TZI)
in
is a triangle
YI -Y - Y2
w TYI.
F(Y') -, Y. The assertion now follows from the theorem in
using
3.4. Observe that
3.4
may be applied, for
U
is
closed under extensions and submodules or factor modules. Moreover, the modules constructed in
3.6
3.3 are niltiple extensions of a fixed module.
COROLLARY.
Let
hereditary algebra. Then mod A
Proof.
3.7
racteristic
XA
With
REMARK.
A be a representation-finite piecewise is directed.
U - mod A
In
the result follows from
III. 1.3
we have introduced the Euler cha-
for a finite-dimensional
we have defined the dimension vector
3.5.
k-algebra
dim X
for an
A
and in
A-module
is a representation-finite piecewise hereditary algebra and
III. 1.2
X. If
A
X an inde-
170
composable
A-module then
3.4
shows that
XA(dim X) = 1.
171
Iterated tilted algebras
4.
4.1
In this section we are going to define two classes of
finite-dimensional
k-algebras which are related by a sequence of tilting
modules to a finite-dimensional hereditary k-algebra. For this let p be a
kZ be the associated path
finite quiver without oriented cycle and let
A
algebra. Let
be a finite-dimensional
k-algebra. We say that
tiltable to k if there exists triples
(Ai,A M1,Ai+1= End M1)0 1
such that
A = Ao, kZ = Am and
lows that
A
A M1
A
is
-
< i< m
is an Ai tilting module. It fol-
is a piecewise hereditary if
A
is tiltable to
I. Using
this observation we see that Z is uniquely determined up to the relation ^-
introduced in
is of type
I. 5.7. If
A
is tiltable to
kZ we will say that
A
A.
The following trivial lemma is stated for the sake of completeness.
LEMMA.
Let
M be a k0 tilting module and
corresponding tilted algebra. Then
Proof.
we know that
By
III. 3.1
B
is tiltable to
we have that
D(MB) = F(D(kA)) where
F
B = End M
I.
kA =, End D(MB). Moreover,
is the triangle equivalence
from
Db(kZ)
to
that
D(MB)
is a tilting module. In fact, pd D(MB) < 1 follows from
Db(A)
constructed in
the
III. 2.10. This immediately shows 1.12,
172
ExtB(D(MB),D(MB)) - 0
4.2
A. We will not carry out the computations.
A be given by the bound quiver
Let
0
o
D(MB)-codim(BB) < I
and
Let us include some examples of algebras tiltable to
for some quiver (a)
But
III. 3.1
III. 5.5.
follows from
11
follows from
o
o
with
Ra = 6y = 0.
is not a tilted algebra, although
gl dim A = 2.
kZ, where A - A5.
is tiltable to
(b)
Y
A be given by the bound quiver
Let a
B 6
Y1,
e
with
fa = S'a' = 6a'-ay = c3'-06 = 0
It
(c)
Let
A be given the bound quiver
with
ar
for
aiai-i = Sisi-1 - 0 1
< i < r.
Br
Then A
4.3
of type Z if
is tiltable to
kA, where
We have noticed before that
A
is tiltable to
A = 12r-1'
A
is piecewise hereditary
kA. In particular we know the quiver
t(Db(A)). In the following two examples we have marked the vertices of I?(Db(A))
which correspond to the indecomposable
A-modules by * .
173
(a)
Let A be given by the bound quiver a1
a2
a3
a4
as
0
with
Then A
a5a4 = a4a3 = a3a2 = a2a1 = 0. is tiltable to
kS, where A = A6. Therefore
i(Db(A)) - ZA6.
(b)
Let A be given by the bound quiver
with
Then A
is tiltable to
a tilted algebra. Therefore
4.4
$a-6y = 0.
k0, where A = ID 4. A
is in fact
?(Db(A)) - Z D4 .
A more restricted concept is that of iterated tilted
algebras. Let A be a finite-dimensional
k-algebra. We say that
A
is
an iterated tilted algebra of type Z if there exists a finite quiver without oriented cycles and triples
(Ai,A M1,Ai+1 = End M1 )0< 1
such that A = Ao, k0 = Am and
A M1
i< m
is an Ai-tilting module having the
1
property that the torsion theory on
(T(M1),F(M1))
mod Ai. We call the passage from
step.
A.
to
(compare
Ai+1
III. 4.2) splits
a splitting tilting
174
If
tiltable to
A
is an iterated tilted algebra of type Z then A
is
kX and clearly every tilted algebra is an iterated tilted
algebra. Examples of iterated tilted algebras are given by the examples in
4.2.
REMARKS.
4.5
The concept of iterated tilted algebras was
introduced by Assem, Rappel (1981). In this article the iterated tilted algebras of type 9 n have been classified. Afterwards a lot of classification results were obtained. The case wronski (1986a). The discrete cases
ID n
IE6'
was settled by Assem, Sko79 IE8
of course could be
handled by using a computer. The case in was solved by Assem, Skowronski (1986). If
A
is an affine diagram different from A then a description
of the representation-infinite iterated tilted algebras of type contained in Assem, Skowronski (1986b) in
4.9
A
is
in a spirit similar to the results
of Ringel (1984). A surprisingly simple procedure to decide for
a given finite-dimensional
k-algebra whether it is iterated tilted of
Dynkin type was obtained by Brenner (1986). Almost nothing is known if is a wild quiver.
175
The general case
5.
For the proof of the main result in this section we need
5.1
a criterion to identify those complexes sional
X'
in
Db(A)
k-algebra A which are contained in the image of the canonical mod A
embedding
into
be an object in
mod A
ding of
Db(A).
A be a finite-dimensional
Let
LEMMA.
Db(A). Then into
X'
X'
is in the image of the canonical embed-
if and only if
Db(A)
k-algebra and let
Hom
(T1 A, X') = 0
A
D(A) b all
for a finite-dimen-
for
i * 0.
Proof.
mod A
into
If
then
Db(A)
(T1
Horn
is in the image of the. canonical embedding of
X'
AA,X
is isomorphic to some
X'
o: Horn
Db (A)
Conversely, let
Db (A)
(T1
AA,X)
A-module
ocExtA 1( A,X) = 0 A
be such that
X' E Db(A)
Hom b
X. Thus
for
i * 0.
(T1AA,X') =0
D (A)
for all
i * 0. Then
P' = (PJ,d3)
where
X' P-
it is enough to show that
is quasi-isomorphic to a bounded above complex is a projective
H1(P') = 0
A-module for all
i * 0. The complex
for all
given by
... 3 P- 2
d2 -+ P
d
1
1
d°
j. Clearly
d1
->. P°-> P 1-i P2 3 .. .
P'
is
176
Let
i * -1
be a projective
P
and let
This induces a morphism
dl+l
A-module mapping onto
ker
which
of complexes V : T1+1P -' P'
f' = (fl)
by assumption is homotopic to zero. Thus there exists
such
g : P -, P1
gdl = fi+1. In particular, ker di+1 aim d1.
that
The following is easily obtained by induction from
5.2
III. 6.4. Let
A be a finite quiver without oriented cycle having n
vertices. Let
X be an
X X=
= (Y E mod A
r
ExtA(X,Y) = 0)
I HomA(X,Y)
with
®
A = ks-module. Then the perpendicular category is defined. Suppose
for
indecomposable, Xi ¢, Xj
X.
i*j
ri > I
and
i=1 ExtA(Xi,Xi) = 0
satisfies
for
1 <,i < r
and
X. E X.
i < j. Then
)CLa
mod H where n-r
CM
is a finite-dimensional hereditary
k-algebra with
simple modules up to isomorphism.
5.3
Let
H
C
We will also need the following fact from
be a finite-dimensional
a tilting module with
valence
k-algebra of finite global dimension and
D = End M. Then there exists a triangle-equi-
G : Db(D) - Db(C)
5.4
III. 2.13.
such that
G(DD) = CM.
The following theorem is a result due to Happel, Rickard,
Schofield (1986). Let
be a finite-dimensional
B
k-algebra which is ba-
sic and connected and A a finite quiver without oriented cycle.
THEOREM.
B
If
is piecewise hereditary of type
is an iterated tilted algebra of type
Proof.
Let
Db(B)
to
B
0.
A - kZ and denote by
triangle-equivalence from
p, then
F
Db(A). By
a normalized
(1.12)
G we denote a quasi-
r
inverse to X.
F. Let
X' = F(BB), then
with X0 * 0 * Xr. Note that
X.
X. _
0
j-0
T3X.
for some
A-modules
is not necessarily indecomposable.
177
As
G(X.) E Uj
(compare
HomA(Xi,XJ) - 0
for
i * j
and
r = 0, then
If
and projective we infer that
1.12)
ExtI(Xi,X.) - 0
for
i+1 $ j.
is a tilting module, thus A -End X0
X' = X0
is a tilted algebra and the assertion follows from
4.4.
So we may assume inductively that the theorem holds in all cases where either
r
takes a smaller value or
takes the same value
r
X0 has fewer indecomposable direct summands.
and
Note that
X0
is a partial tilting module and a tilting
module in the perpendicular category where
is a finite-dimensional hereditary
H
(X1 0 ... 0 X
r
for in
(X1 0 ... 0 Xr)1 =, mod H - mod A,
)
1
with mod H. Let
mod H. Since
HQ
induced by
(T(XX,F(X0))
k-algebra. We will identify
HQ be the minimal injective cogenera-
is a torsion module in the torsion theory on
X0
mod H we have an exact sequence
0 - X"0 -, X'0 -, Q - 0 in mod H. This yields a triangle X" -, X'0 -, Q -, TX"0 0 in
Hom (T1Xto this triangle shows that D b(A) such that 5.1. Let G(Q) = BY and let BB - P 0 P'
Db(A). Applying by
G(Q) E mod B F(P) - X0.
We claim that
is a tilting module such that
BM - BY 0 P'
splits on mod B.
(T(M),F(M))
It follows from
1.12
it suffices to show
ExtI(M,M) - 0
sable direct summand
P.
of
pd BM < 1. In order to show that
that
ExtI(Y,Pi) - 0
P'. But
for each indecompo-
ExtI(Y,P.)
(Y'TP.) ..,
Horn b
D(B) (Q,TFP .). The latter term is zero, for Hom 1 Db (A) j > I
1.12). Using
(compare
III. 6.2
FP . E H[j]
1
finally shows that
for some
BM
is a
tilting module.
Let Since
P'
infer that
is
Z
be an indecomposable
B-projective we have that
B-module with
ExtI(M,Z) $ 0.
ExtI(Y,Z) * 0. By
F(Z) E H[0). It also follows that
1.12
we
F(Z) E mod H, in particular
178
Hom(P',Z) = 0. Since This shows that
ExtI(Y,Z) * 0
F(Z)
H-injective. But then
is not
HomB(Y,Z) = 0. Thus
showing that
HomH(Q,F(Z)) = 0,
is a torsion free module in the
Z
BM
induced by
(T(M),F(M))
torsion theory
ExtI(Q,F(Z)) * 0.
we infer that
on
mod B.
C = End M. Then we obtain a triangle-equivalence
Let
r F'
:
F'(CC) = Q 0
such that
Db(C) -, Db(A)
dl
T1Xi. Let
be a simple
CS
i=1
projective such that
is a direct summand of
F'(S)
C * k we see that
nected and we may also assume that
APR-tilting module CM(S)
injective. Thus the
We have seen before that the torsion theory splits. Let
mod C
by
CM(S)
of
S. We shall prove below that
on
:
Db(D)
-+
is defined (compare 111.2.14). induced
be the Auslander-Reiten translate
T S
F'(T S) E H[1]. This of course finishes
Db(A)
F"(DD)
such that
of indecomposable direct summands in number for
is con-
is not simple
S
(T(M(S)),F(M(S)))
=
F"
C
D = End M(S). Then we obtain a triangle-equiva-
the proof. In fact, let lence
Q. Since
r
0 T1Y. i=o
where the number
is less than the corresponding
Yo
Xo
Let 0 -. S -+ P" -. T S -. 0 be the Auslander-Reiten sequence starting in in
S
and let
Db(C). Thus also
Db(A). Applying
S -+ P" -. T S -+ TS
F'(S) -+ F'(P") -, F'(T S) -, TF'(S)
Hom
(
Db (A) Hom b
A,-)
(AA,T F'(T S)) -. Hom b
(
A,F'(S)) -4 Hom
A
Hom
(
such that
F'(P"')
(AA,F'(P")).
D (A)
A,T F'(T S)) * 0. Otherwise
A D(A) b
(
A,F'(P"))
A
D(A) b Db (A) injection V(S) u F'(P" where P"
(AA,F'(S)) - Hom b
D (A)
It is enough to show that
is a triangle in
to this triangle yields the exact sequence
A
D (A)
Hom
be the corresponding triangle
is injective. This induces an
P"' is the largest direct summand of
is a direct summand of
Q. By construction
u
F'(S) - F'(P"')
is an injective map of injective
H-modules. Thus
u
is
a section, contradicting the fact that 0 -+ S - P" - T S - 0 is non-split.
179
5.5
COROLLARY.
Let
A be a finite-dimensional
k-algebra
and S a finite quiver without oriented cycle. Then the following are equivalent. (i)
A
is piecewise hereditary of type
(ii)
A
is an iterated tilted algebra of type
(iii)
A
is tiltable to
REMARK.
implication
(iii)
5.6 and
A*
A.
kA.
So far there does not exist a direct proof of the (ii).
COROLLARY.
Let
A
the opposite algebra. Then
type 9 if and only if
A*
be a finite-dimensional
A
k-algebra
is an iterated tilted algebra of
is an iterated tilted algebra of type .
180
The Dynkin case
6.
6.1
algebra of type to
Let I be a Dynkin quiver and A a piecewise hereditary A. In this section we will show A can be transformed
kG by using only
b D (A)
APR-tilting modules. We know that the quiver of
i
mod A
coincides with ZA. In particular we know that
Euler characteristic of
X
A. If
is an indecomposable
XA(dim X) - 1. We have denoted before by
n(A)
classes of indecomposable
K0(A)
A-modules. As
we know that r0(A) - #{x E Zn be denoted by
rz. Clearly
I
XA(x) - 1)
XA be the
3.6. Let
tation-finite and directed without reference to
is represen-
A-module then
the number of isomorphism is isometric to
K0(kX)
does not depend on A
n(A) < r0(A). The values for
rZ
and will
are well-
known and are given as follows
r . = n(n+1), rID - n(2n-2), r .- 72, rE - 126 6 7 n n 6.2
8
Let A be a Dynkin quiver and
THEOREM.
hereditary algebra of type
A
a piecewise
A. Then there exist triples
(A. ,AM1,Ai+1 = End M1)0 < i< m an
and, rE - 240.
such that A = A0, kZ = Am
and
A Ml
is
APR-tilting module.
Proof.
Therefore A
We have noticed before that
admits an
APR-tilting module
M(S)
mod A
is directed.
for some simple projec-
181
tive
A-module S. Then
B - End M(S)
is again piecewise hereditary of
type Z and hence directed such that from
that
III. 4.13
A
If
exists a simple
n(B) > n(A)
if and only if
A-module S
A = A0
and we know
idAS > 1.
is hereditary there is nothing to show. Otherwise there such that
easily seen that there exist triples such that
rZ > n(B) > n(A)
and A M1
idAS > 1. In this situation it is
(Ai,AM1,Ai+1 = End M)0< i
m
is an APR-tilting module of the form
1
M1 = M(S') S
id S1 -1
with
and
Am
ml
ml
with
> 1. Therefore
id S
admits a simple projective module ml
1
n(End(M(S
))) > n(Am ). This process
has to stop and this is precisely the case when we reach In the situation above we will say that by a finite sequence of
APR-tilts to
A can be transformed
ki.
We include two examples which show that
6.3
hold for arbitrary quivers
kp.
6.2
does not
S.
Let A be the algebra defined by the bound quiver
(a)
a
Then
End M(S) ., A, where
the other hand
A
M(S)
is the unique
is a tilted algebra of type
Note that the opposite algebra
A*
APR-tilting module. On ]-
where
may be transformed by an
APR-tilt
to a hereditary algebra. This phenomena will be proved in the next section.
182
Let A be the algebra defined by the bound quiver
(b)
Blal = 82a2 = 83a3 = B04
It may be shown that neither A nor sequence of
A*
may be transformed by a finite
APR-tilts to a hereditary algebra. On the other hand A
is
a tilted algebra of type A where
Let A be a finite-dimensional
6.4
bound quiver
(r,I). For a vertex
subquiver of r with vertices path from A-module
to
j
P(i)
i
in
j
i E r0
k-algebra given by the
we denote by
ri
the full
for which there exists a non-trivial
I. We say that the indecomposable projective
associated with the vertex
i
has a separated radical if
the supports of each pair of non-isomorphic indecomposable summands of rad P(i)
are contained in two different connected components of
Finally we say that
A satisfies the
(S)-condition if all indecomposable
A-modules have separated radical. This condition was intro-
projective
duced by Bautista, Larrion, Salmeron (1983). For the topological implications of this condition we also refer to Bretscher, Gabriel (1983). Using this and
3.5
of Assem
LEMMA.
dimensional
Let
(1983)
one obtains
A be a representation-finite connected finite-
k-algebra. Let
S
be a simple projective
A-module and
183
B = End M(S). If dition, so does
B
A
is
It is easily seen that the following result holds.
be a connected finite-dimensional
projective
(S)-con-
A.
REMARK.
Let
is representation-finite and satisfies the
A-module and let
k-algebra. Let
B = End M(S). If
mod B
S
be a simple
is directed, so
mod A.
The converse is not true as the following example shows. Let
A be given the bound quiver
with
Let
M(S)
be the unique
APR-tilting module. Then
presentation-finite, but
COROLLARY.
mod B
B = End M(S)
is not directed, while
be a Dynkin quiver and
Let
Then A satisfies the
hereditary algebra of type
6.6
c6 =y6=Ra=0.
mod A
is reis.
A be a piecewise (S)-condition.
In the remaining part of this section we will present
the classification of the iterated tilted algebras of type
An. This
mainly serves as an illustration of the material developed so far. For the formulation of the additional information we need, we have to introduce some further notation. The translation quiver Z A has as vertices the pairs
(i,j)
this choice in the case where
with
n = 6.
i E Z
and
j E (An)o. We sketch
184
Let
k(ZA n)
ZA1. For
be the mesh category associated with the translation quiver
x E ZA n
we define a function
fx
on the vertices of
7I n by
(1979)).
x = (i,j) E Z . The support S
LEMMA. Let
of
fx
is given
by S = {y = (i',j') E Z i < i' < i+j-1, i+j < i'+j' < n+1}.
Denote by Lx
for
x - (i,j)-E 71A
the subset of ZA
n
given
Lx = {y= (i',j') EZn I V = i, 1 <j'
Lx
is just the union of the two diagonals through
x.
The preceding lemma has an easy consequence.
Let
COROLLARY.
type n and let where A - An. Let x,y E ZAn
F
A be a piecewise hereditary algebra of
be a triangle-equivalence from P,P'
correspond to
be indecomposable projective P,P'
under
F
and if
Db(A)
to
Db(kS)
A-modules. If
HomA(P,P') * 0
or
185
HomA(P,P') * 0
6.7
then
y E L. Cn
We are going to define a class
of finite-dimensio-
nal algebras as follows. A finite-dimensional,basic and connected algebra
A given by a bound quiver
(?,I)
belongs to
Cn
if the following con-
ditions are satisfied (i)
r
is a tree with
(ii)
I
can be generated by paths of length two.
(iii)
Every point has at most four neighbors.
(iv)
If a point has four neighbors, then
n
with
subquiver of (v)
vertices.
a6 - Yd = 0
is a full bound
(I1,I).
If a point has three neighbors, then
0
0
o
-o
with a8=o or
0'-
is a full bound subquiver of
-
O
as - o
(I?,I).
It is easily seen that a finite-dimensional, basic and connected algebra
A given by a bound quiver
(I1,I)
belongs to
Cn
if
(I1,I)
is a full
bound subquiver with n vertices of the following infinite quiver bound by all possible relations of the form
a8 = o
and
Ba = o.
186
Let
THEOREM.
k-algebra. Then
A
A be a finite-dimensional, basic and connected
is an iterated tilted algebra of type An
if and only
if AECn Proof.
A E Cn S
The proof rests on a few observations. Note that if
is hereditary then A =, kI, where
a simple projective
A-module and
it is straightforward that proof of
6.2
M(S)
A = A. Next let the
A E Cn
and
APR-tilting module. Then
End M(S) E Cn. Using the arguments as in the
this shows that if
A belongs to
Cn, then
A
is an itera-
ted tilted algebra of type n Conversely, let and let
F
A = n. Let
A be an iterated tilted algebra of type An
be a triangle equivalence from Db(A)
to
Db(M), where
P(1),...,P(n) be a complete set of representatives from the
187
isomorphism classes of the indecomposable projective and
Xi - F(P(i))
6.6
lary in
xi
be the corresponding vertex in 7! n. Using the corol-
the properties (ii), (iii), (iv) and (v)
Let A be given by the bound quiver show that
r
A-modules. Let
easily follow.
(r,I). It remains to
is a tree. Otherwise ? contains a full subquiver Q which
is a (non-oriented) cycle. Since
_
has no oriented cycle,
always
contains at least one source and one sink. In particular, Q has the fol-
lowing form where w denotes a path and
w'
a sequence of arrows not
necessarily forming a path. a1
w
b1
0 a o
b
>6Y a2
Since < 1
I
w'
b2
can be generated by paths of length
we infer that
rad P(b)
2
and
dimkHomA(P(a),P(b))
is decomposable. In particular
to have separated radical in contrast to
6.5.
P(b)
fails
188
The affine case
7.
In this section we study the case where A is an affine
7.1
quiver. We follow Rappel (1987). In nents
C[i]
form C[i] forming
R[i]
and
of
I1(Db(kX))
III. 7.2 for
we have introduced compo-
i E Z. The components of the
will be called transjective components while the components
R[i]
will be called regular components. Observe that
H[i] c C[i] U R[i] U C[i+1]
(compare
1.12). In the proof of the fol-
lowing proposition we freely use the structure of in Dlab, Ringel (1976)
mod kA as contained
or Ringel (1984).
PROPOSITION.
Let A be a representation-infinite piecewise
hereditary algebra of affine type A and
F : Db(A)
-
Db(1A)
a triangle
equivalence. (a)
taining
F(P) (b)
posable
There exists a unique transjective component for some indecomposable projective
con-
A-module P.
There are (up to isomorphism) only finitely many indecom-
A-modules (c)
C[i]
X
such that
F(X) f C(i] U R[i] U C[i+1].
C[i], R[i], C[i+1]
each contains (up to isomorphism)
infinitely many indecomposable objects of the form F(X)
Proof.
for
The existence is a simple reformulation of
The uniqueness follows from
5.1
X E mod A.
III. 5.8.
and the fact that a preprojective kX
189
module X
has the property that
finite subcategory of also yields
{Y E mod kZ I Hom.(X,Y) = 0}
(a), and the last argument
mod a. This proves
(b).
Let
be a complete set of representatives of
P(1),...,P(n)
the isomorphism classes of indecomposable projective
X* = F(Pj)
is a
for
I
A-modules. Let
< j < n.
Suppose X t C[i]. Then clearly
dim X is not homogeneous,
for
0 - ExtI(P(j),P(j)) = Hom A
for some indecomposable regular 5.1
R[i]
Db
Ext I M,X.)
(a)
ks-module Xi. Therefore it follows from
that the complexes in the tubes of rank are contained in the image of
F
{Y E mod kZ gory of
Y
I
of the tubular family
restricted to
The remaining assertions in fact that a non-sincere
I
(c)
mod A.
follow from 5.1
and the
kZ-module R has the property that
preinjective, HomkX(R,Y) = 0}
is not a finite subcate-
mod kX.
7.2
Let A be a representation-infinite piecewise hereditary
algebra of affine type A and let
P(i1),...,P(ir)
be a complete set of
representatives from the isomorphism classes of projective F(P(ii))
having the property that
mined transjective component
A-modules
is contained in the uniquely deter-
C[i], where
F
is the triangle-equivalence
r
from
7.1. Let
A' = End
0 P(ij). j=1
COROLLARY.
7.3
LEMMA.
A'
Let
is a concealed algebra.
s
be a natural number. Then there exist
(up to isomorphism) only finitely many basic finite-dimensional
bras having
s
simple modules and directed module category.
k-alge-
190
Proof.
We proceed by induction on
s. For
s =
I
there is
exactly one algebra having these properties. Let A be a basic finitek-algebra with
dimensional
simple modules and directed module cate-
s
gory. Then there exists a basic finite-dimensional
B
with
simple modules and directed module category. Moreover, there exists
s-1
a
k-algebra
B-module M such that
A
is the one-point extension of
B
by M
(note that an epimorphic image of an algebra A with directed module r
category has again a directed module category). Let M =
0 Mi
with
M.
i=1
A
indecomposable. As
is representation-finite we infer that
r < 3.
Thus the result follows from the induction hypothesis.
LEMMA.
7.4
type
Z. If S
A be a piecewise hereditary algebra of
Let
is not a Dynkin quiver, then A may be transformed by a
APR-tilts to a representation-infinite piecewise hereditary
sequence of
algebra of type
A.
Using the same argument as in
Proof.
follows from
III. 4.13. Observe that the number of simple
and
7.3
this immediately
6.2
modules is stable under the tilting process.
7.5
A. A component
the Auslander-Reiten quiver of projective component if each indecomposable
C
of
C
r
is called a
re-
does not contain an oriented cycle and for
A-module
composable projective
7.6
k-algebra and let t be
A be a finite-dimensional
Let
there exists
X E C
A-module
P
such that
t E N
and an inde-
X = t tP.
For the rest of this section we assume that A
is an
affine quiver. Let A be a representation-infinite piecewise hereditary algebra of type By
7.1
and let
F
:
Db(A)
-
Db(kA)
be a triangle-equivalence.
there exists a unique transjective component
C[i]
of
r(Db(kZ))
191
A-modules under
containing images of indecomposable projective may assume that
F. We
i = o. In the sequel we will choose our triangle-equiva-
lences in such a way that they satisfy this property.
A
If
jo = jo(A)
jective
is not a concealed algebra, there exists an integer
such that
A-module
R[jo]
P, but
R[j]
A-module P'
posable projective
for some indecomposable pro-
F(P)
contains
F(P')
does not contain j < jo. Let
and
that F(P)
be the number
r(A)
of isomorphism classes of indecomposable projective
for an indecom-
A-modules
such
P
U R (j). j >o LEMMA.
of
APR-tilts to a concealed algebra
jo(B) = 0. Moreover
A may be transformed by a sequence
jo < o, then
If
or an algebra
B
B
satisfying
r(B) = r(A).
Let
Proof.
X
decomposable objects
be the full subcategory of
V
satisfying
F(X) E
U
mod A with in-
(C[j] U R[j-1]). By
7.1
j
V
and submodules. It is easily seen that with injective also
is contained in
T X
is closed under extensions
X
V. Moreover
contained in Reiten quiver
3.5
PA
Let
of S
and not
contains (up to such that
it follows that the indecomposable
A-modules
form a preprojective component
V
V
V
A-modules X
isomorphism) only finitely many indecomposable F(X) 4 C[o]. Using
contained in
C
of the Auslander-
A.
be a simple projective A-module such that
F(S) E R[jo] and let 0 - S - P -. T S -4 0 be the Auslander-Reiten sequence starting at Note that
S.
F(T S)
If
r(End(M(S))) = r(A). So we may assume that
T S E V, so that
F(T S) E R[jo]. Observe that
ponent of the form Zk./s
R[jo] there is nothing to show.
F(S)
for some
and s >
F(T S) 1
are contained in a com-
and that such a component con-
192
tains only finitely many
FI
:
Db(A1)
-+
the image of
Db(kZ). It follows from 5.1 restricted to
F1
III. 3.2. Since
Al
F"
Db(A2)
to
F1(S1)
and let TF(S)
Db(kXX). Then
be the
F2
and
lies
TF1(S1)
mod A2. In particular we see that
APR-tilts we obtain a finite-dimensional
admitting a simple projective
At
k-algebra,
such that
S1
A2 = End(M(S1))
restricted to
after a finite number of
Ft(St)
A1-module
F(S). Let
triangle-equivalence from
is in
TF(S) E R[jo+1]
is a factor of a finite-dimensional
in the same component as
k-algebra
that
mod A1. This may also be seen from
there exists a simple projective
are in the image of
End Y =. k.
and consider the triangle equivalence
Al = End(M(S))
Let
such that
Y
kXA-modules
lies in the same component as
Ar-module St
Ft(t St) f R[jo]. This
but
F(S)
such that
finishes the proof of the lemma.
7.7
Let A be a concealed algebra of type
LEMMA.
APR-tilts to a.
A may be transformed by a sequence of
Let
Proof.
We may assume that
F : Db(A) -, Db(a)
F(P) E C[0]
is an indecomposable projective
Z. Then
be a triangle equivalence. ks-module if
and preprojective as
P(l),...,P(n)
A-module. Let
P
be a com-
plete set of representatives of the isomorphism classes of indecomposable projective
A-modules and
X. = F(P(i))
for
I
< i < n
the correponding
k- modules which we identify with the corresponding vertices in Z. Let
<X1...,x >
be the convex hull of
X1,...,X
occurring in the paths of Zp which start in
n 1 < i, j < n. If
X =
9 Xi
is a slice in
(i.e. the vertices of X.
i
and stop in
mod kA
X.
J
for
there is nothing to
i=1
show. So we may assume that such that choose
io
X
is not a slice. Thus there exists
t Xi E <X1...,X>, where so that
t
is minimal in
Xi 0
is the translation on {X1,.... Xn}
(i.e. no
Xi
Z2. We X.
is a
193
proper predecessor of
in Z,). The last assumption implies that
Xi 0
E <11,...,Xn> -{Xi }. Clearly, the indecomposable projective 0 0 A-module P(i0) is simple. Let Xi - F(T P(i0)). Then Xi E ZA (compare T Xi
the proof of
B = End(M(P(I0))) and
7.6). Let
induced triangle-equivalence. Let
F'
Q(1),...,Q(n)
:
Db(B)
for
I
}
0
<Xl,...,Xn>.
j *i
such that and
{Xi }
Hom
(X.' ,X.) * 0, then Db(kk) to c <Xl,...,Xn>'{Xi }. On the
°
E <X1,...,Xn> 0 contrary, if Hom
B-modules. Let
. {Xi }. We claim that
0
X!
the
< i < n. Then we may assume that
{Y1,.... Yn} _ {X1,.... Xn,X!
If there is an index
Db(kg)
be representatives of
the isomorphism classes of indecomposable projective Yi = F'(Q(i))
-
0
0
(X.'
b(k HomA(T P(i0),P(j)) = 0
io
X.)- 0
for all
j* i0, then
for all
j* i0. We infer that
idA(Pi ) < 1
and
o
clearly pdAT P(i0)
1. Let 0 - P(i0) -, E - T P(i0) - 0 be the Auslan-
der-Reiten sequence starting in
P(i0). By
I. 4.7
we have that
P(i0) -, E -+ T P(i0) -, TP(i0) is an Auslander-Reiten triangle in
Db(A). Hence
-,F (E) -,X! -, TX.
X.
0
0
0
Db(kp). In particular, Xi
is an Auslander-Reiten triangle in
- T Xi 0
Therefore,
E <X1,...,Xn>
X1
0 {Xi }
}, or
. {x.
0
C <X1,...,Xn> -
0
which proves our claim.
0
Iterating this process proves the assertion.
7.8
For the formulation of the next result we need some ad-
ditional terminology. Let D = HomA(-,k)
A be a finite-dimensional
k-algebra and let
be the standard duality. Thus for a left module
AX we
194
have that
is a right module. An
D(AX)
co-tilting module if
A. Then A may be transformed to
Proof.
A
finite. If
bra A*
7.6
7.6. By
we may assume that
satisfies the assumptions of
F*(P) 4 R[j]
sequence of
words
APR-tilts
is representation-in-
7.7. Otherwise let
be a triangle-equivalence which satisfies the assump-
triangle-equivalence. Let Then
A
we may assume that
7.4
is a concealed algebra we apply
D b(A) -, Db(kZ)
tions of
By
kZ by a sequence of
APR-co-tilts.
followed by a sequence of
:
is
A be a piecewise hereditary algebra of affine
Let
THEOREM.
F
EndAM
APR-
APR-co-tilt.
called an
type
is called an
APR-tilting module and
is an
D(AM)
AM
A-module
for
jo(A) = 0. The opposite algeF*
7.6. Let
be the corresponding A*-module.
be an indecomposable projective
P
j > o. Thus
A*
may be transformed by a finite
APR-tilts to a concealed algebra
B
of type
S. In other
A may be transformed to a concealed algebra of type p by a
finite sequence of bra of type
APR-co-tilts. With
B
also
B*
X. Thus the assertion follows by using
is a concealed alge7.7.
We conclude this section by an example which shows the
7.9
distribution of the images of the indecomposable projective under a triangle-equivalence
F
-
Db(A)
:
A-modules
Db(kA).
A be given by the following bound quiver
Let
Y a
a1
1
E
0'
(q-2)'
b
aiLl
a
p-1
...
ono 2
I
YI
with
ap-Iy = y'aq-1 = 0
aiai-I = 0
for
and
I < i < q-1
and
aiai+1 = 0
for
I
< i
195
It may be shown quite easily that
A
is piecewise hereditary of type S
(just constract suitable tilting modules!) where
aq-I
(q-1)'
(q-2)'
0 .....
Of
f
.....
p-1 B
P-11
Let
P(1),...,P(p-1),P(a),P(b),P(O'),...,P(q-2)')
projective
be the indecomposable
A-modules.
We consider the following a-modules. Let indecomposable projective and
b. Let
dim Eo =
vertex
Eo
i'
a--modules associated with the vertices
for
be the a
be the indecomposable regular k- module with E'1 = S(i'), the simple
1
Q(a),Q(b)
I < i < q-2
ciated with the vertex
i
for
and I
li-module associated with the
Ei = S(i), the simple < i < p-1. Note that
ki-module assoare regular
Ei,Ei
kA modules.
Then a triangle-equivalence such that
0 < i < q-2
F : Db(A) s D (kA')
F(P(a)) = Q(a), F(P(b)) = Q(b), F(P(i')) and
F(P(i)) = T 1Ei
for
I < i < p-I.
T1E!
may be chosen for
CHAPTER V
TRIVIAL EXTENSION ALGEBRAS
Preliminaries
1.
A be a basic, connected finite-dimensional algebra over
Let
an algebraically closed field extension algebra
k. In
I. 3.1
we have defined the trivial
T(A). In this chapter we will give a summary of some
results dealing with
T(A). We refer to Possum, Griffith, Reiten (1975)
and Tachikawa (1986) for further details and further references.
The following observation is due to Iwanaga, Wakamatsu
1.1
(1980).
T(A)
Proof.
It is enough to show that
isomorphic as cp
:
T(A)
and
D(T(A))
are
A-bimodules (see Auslander, Platzeck, Reiten (1977)). Let
T(A) -, D(T(A))
a,a' E A and phism of
is a symmetric algebra.
LEMMA.
be given by
[p(a,q)](a',q') = q(a') + q'(a)
for
is an isomor-
q,q' E Q. Then it is easily checked that
A-bimodules.
1.2
In
II. 2.4 we have defined a grading on
over we have noticed that
modZT(A)
and
is the repetitive algebra associated with be the forgetful functor. Then
0
mod A A. Let
T(A). More-
are equivalent, where ¢
:
A
modZT(A) - mod T(A)
is a Galois covering with Galois group
Z. For the covering theoretic results needed here we refer to Bongartz,
197
Gabriel (1981), Gabriel (1981) and Dowbor, Skowronski (1986). Let
B
(A,I). We say that
quiver
M. = {j E A0
1
k-algebra with associated bound
be a locally bounded
is locally support-finite if the set
B
with
3 M E ind B
HomB(P(i),M) * 0
is finite for all
i E Ao. We say that
finite if for all
i E A0
HomB(P(j),M)* 0}
is locally representation-
there exist only finitely many (up to isomor-
M such that
B-modules
phism) indecomposable
B
and
HomB(P(i),M) * 0.
We will need the following facts of the forgetful functor
A
If
is locally support-finite then
T(A)-module is gradable. Moreover A and only if
4
is dense. In other words, every
is locally representation-finite if
is representation-finite.
T(A)
It is also known that
preserves Auslander-Reiten sequences.
4
Consequently, the quotient of the Auslander-Reiten quiver P(A)
Galois group Z
Ii'(A)
under the
is a set of complete connected components of the Auslanr(T(A)). Let
der-Reiten quiver quiver of
¢.
(resp. r(T(A)))
rs(A)
(resp. Ii's(T(A)))
be the full sub-
formed by the non-projective vertices.
This is called the stable Auslander-Reiten quiver. It is easily seen that the triangulated categories triangles and that
mod A
Ts(A) = P(mod A)
It follows from Riedtmann (1980) that
Iis(A) = ZA
in
have Auslander-Reiten
Ps(T(A)) = r(mod T(A)).
(see also Rappel, Preiser, Ringel (1980)) if
A
is locally representation-
is then called the Cartan class of
T(A).
The following is an immediate consequence of the result
II. 4.9, III. 2.10
COROLLARY. iA
mod T(A)
and that
for a Dynkin graph A
finite. The graph A
1.3
and
and
IV. 5.5.
Let
A be a finite-dimensional
k-algebra and
a finite quiver without oriented cycles. Then the following are equi-
valent:
198
(i)
A
(ii)
mo(h(A)
REMARK.
is an iterated tilted algebra of
and mod$T(kX)
.
We point out that the preceding result is false for
given by the bound quiver
a -::p
(j=D
as = $a = 0. Then clearly
a
A
1
are triangle-equivalent.
the ungraded module categories. For example let A =
but
type
is not an iterated tilted algebra.
and T(A)
A be
199
The representation-finite case
2.
In this section we present a characterization of the representation-finite trivial extension algebras. We refer to Assem, Rappel, Roldan (1984), Bretscher, Laser, Riedtmann (1981), Hoshino (1982a), Hughes, Waschbusch (1983), Tachikawa (1980) for different approaches and alternative proofs.
2.1
nal T(A)
k-algebra, and let
A be a basic, connected finite-dimensio-
Let
THEOREM.
T(A)
be the trivial extension algebra. Then
is representation-finite of Cartan class
an iterated tilted algebra of Dynkin type
Proof.
we know that
Db(A)
tion Z of mod A
A
Let
is isomorphic to ZA. Since
Ps(T(A))
T(A)
II. 4.9). Thus
is a proper quotient of T(A)
rs(A)
rs(A)
is representation-finite.
is representation-finite of Cartan
A, we know by the remarks in section
1
to Z. It is easily seen that this implies that some orientation p of
A. Then
for some orienta-
Db(k0)
(compare
is finite. Thus
Conversely, if class
Db(A)
is
gl dim A < . we have that
II. 2.10). Since
(compare
A
A.
is triangle-equivalent to
Pa(T(A))
if and only if
A be an iterated tilted algebra of type
is triangle-equivalent to
we infer that
A
that
rs(A)
mod A
is isomorphic
Db(kX)
for
A. It follows from Auslander, Reiten (1977) that
200
mod A
this equivalence is a triangle-equivalence. As
mod A
fer that
mod A
we see that
(II. 2.3)
gl dim A <
Thus
Altogether we see that
is directed. In particular we in-
mod A and
Db(A)
and
Db(A)
Db(kAA)
Let
are triangle-equivalent.
are equivalent as triangulated
categories. Thus the assertion follows from
2.2
embeds into
IV. 5.5.
A be a finite-dimensional
or
IV. 6.2..
k-algebra such that
is representation-finite. Then it was shown in Hughes, Waschbiisch
T(A)
(1983) that there exists a tilted algebra phic to
Let
B
T(A)
is isomor-
to be hereditary. We leave the proof to the reader.
A be given by the following bound quiver
0+ _ P with T(A)
such that
T(B). We will include an example which shows that we cannot
always choose
Then A
B
is a tilted algebra of type
1E
Y$a = 0. 6'
The Auslander-Reiten quiver of
is given as follows. The numbers indicate the length of the inde-
composable modules. The projective-injective modules correspond to the encircled ones. The identification is along the dotted lines.
I
1\ '3\ T1\ 4 1\ /'\ 6
2 I
4
\ , ;,4.\,
\3
III
-
1/\2/';-
\5
/ / \5/ I
201
3.
The representation-infinite case
While the situation for representation-finite trivial extension algebras is well understood there is little known in the representation-infinite case. Here we will generalize a result of Tachikawa (1980). We also mention an article of Assem, Nehring, Skowronski (1986) which characterizes certain representation-infinite trivial extension algebras.
LEMMA.
3.1
Let
A be an iterated tilted algebra of type
for some finite quiver Z without oriented cycle. Then A
p,
is locally
support-finite.
By
Proof.
triangle-equivalence
and
from mod A
F
quiver associated with A-module with
III. 2.10
A. Let
we know that there is a
II. 4.9 to
i E 90
Db(kg). Let
posable
Applying
Homm(P(i),M) * 0. Since the indecomposable projective
HomA(P(i)/soc P(i),M) * 0. By
A-module T
be the bound
M be an indecomposable
and let
A-modules have bounded length we may assume that Thus also
(&,I)
such that
Z
M
is not projective.
I. 2.4 there exists an indecom-
Hom(P(i)/soc P(i),Z) * 0 * Hom(Z,M).
if necessary we may assume that
F(P(i)/soc P(i)) E H[0]
2
(compare
IV. 1.12). Thus
HomA(P(j),M) * 0.
F(M) E
U H[s]. Let j E Go s=o Arguing as above we infer that
and suppose that
2
F(P(j)/soc P(j)) E
U
s= -2
H[s]. Since
H[s], for
s E 7L, contains up to iso-
202
morphism only finitely many indecomposable objects of the form F(P(r)/soc P(r))
r E Do we infer that
for
is locally support-finite.
A be an iterated tilted algebra of
Let
THEOREM.
3.2
A
type A for some finite quiver A without oriented cycle. Then the stable Auslander-Reiten quiver ? s(T(A)) T(A)
is isomorphic to
?(Db(kA))/.
By
Proof.
of the trivial extension algebra
we may apply the remarks of section
3.1
It is easily checked that
1.
generates the Galois group for the covering
T2t
Db(A) - mod T(A).
3.3 s(T(A))
There exist trivial extension algebras
is a quotient of
oriented cycle such that For example let T(A)
T(A) % T(B)
for an iterated tilted algebra
is the four-dimensional local algebra
Clearly
T(A) W T(B)
?s(T(A))
situation of
7L2x
Z2.
1(Db(kA))
with A = Qom. In the
and A not a Dynkin quiver there exist two components
of the form ZA in one such
k[x,y]/(x2,y2), which in
for an iterated tilted algebra B. It is easy to see
is a quotient of 3.2
B.
k[x]/(x2). Then
is the group algebra of Klein's four-group
char k = 2
such that
for some finite quiver A without
A be the algebra of the dual numbers
case of
that
1(Db(kZ))
T(A)
component.
rs(T(A))
while in this example there exists only
203
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207
INDEX additive function affine graphs APR-co-tilt APR-co-tilting module APR-tilt APR-tilting module Auslander-Reiten triangle
bilinear form bimodule of irreducible maps bounded bounded above bounded below bound quiver brick canonical exact sequence canonical tubular algebra Cartan matrix cohomological functor cohomology objects commutativity relation complex concealed algebra contravariant cohomological functor contravariant representation covariant cohomological functor cycle
95 44 194 194 126 126
70 95
heart homotopic homotopy category
57 27 27
31
98 39 26 26 26 46 164 119
irreducible morphism isometric isometry iterated tilted algebra Krull-Schmidt category locally bounded k-algebra locally representation-finite locally support-finite
33 100 100 173 26
25 197 197
91
97 4
27
46 4 149 4
47
mapping cone 28 54 mesh category mesh ideal 54 149 minimal wild quiver 2 morphism of sextuples morphism of translation quivers 41 Nakayama functor 37 normalized triangle-equivalence 158
4
39
derived category deviation differential complex dimension vector directed directing direct predecessor direct successor Dynkin graphs
29 26 26 96 39 39 45 45 44
enough S-injectives enough S-projectives Euler characteristic exact category exact functor Ext-injective Ext-projective
11
finite M-codimension finite subcategory Frobenius algebra Frobenius category
generating subcategory Grothendieck group
one-point coextension one-point extension partial tilting module path path algebra path category perpendicular category piecewise hereditary algebra polarization predecessor preprojective component proper epimorphism proper monomorphism
65 65 141
38 45 54 144 152 41
45 190 10 10
11
98 10 4
128 128 104 168
25 11
quasi-isomorphism quiver of a Krull-Schmidt category
27 39
relation repetitive algebra
46 59
(S)-condition separated radical separating subcategory
182 182 136
208
sextuple sincere S-injective sink sink morphism slice source source morphism split exact sequence splitting idempotent splitting tilting step S-projective stable category stable translation quiver stalk stalk complex standard sextuple successor suspension functor
I
96 10
48 35 134
48 35 27 26 173 10 11
41
26 26 16
45 11
tame concealed algebra t-category tiltable tilted algebra tilting module torsion pair torsion theory translation translation functor translation quiver triangle triangle-equivalence triangle-equivalent triangulated category triangulation trivial extension algebra truncated complex tubular algebra t-structure
149
57 171
133 118 58 58 41 1
41 2
4 4 3 2
25 29 91
57
width
26
zero relation
46
