On periodic stable Auslander-Reiten components containing Heller lattices over the symmetric Kronecker algebra
aa r X i v : . [ m a t h . R T ] S e p ON PERIODIC STABLE AUSLANDER–REITEN COMPONENTSCONTAINING HELLER LATTICES OVER THE SYMMETRICKRONECKER ALGEBRA
KENGO MIYAMOTO
Abstract.
Let O be a complete discrete valuation ring, K its quotient field, and let A be the symmetric Kronecker algebra over O . We consider the full subcategory of thecategory of A -lattices whose objects are A -lattices M such that M ⊗ O K is projective A ⊗ O K -modules. In this paper, we study Heller lattices of indecomposable periodicmodules over the symmetric Kronecker algebra. As a main result, we determine theshapes of stable Auslander–Reiten components containing Heller lattices of indecompos-able periodic modules over the symmetric Kronecker algebra. Contents
Introduction 2Acknowledgement 31. Preliminaries 41.1. Orders and lattices 41.2. Valued stable translation quivers 41.3. The stable AR quiver for the category of lattices over an O -order 71.4. Indecomposable modules over special biserial algebras 102. The Heller lattices over the symmetric Kronecker algebra 102.1. Indecomposable modules over A λ = ∞ . 193.1. The almost split sequence ending at Z λn Z λn Z λn λ = ∞ . Z ∞ n Z ∞ n Z ∞ n Mathematics Subject Classification.
Key words and phrases. almost split sequences, the stable Auslander–Reiten quiver, Heller lattices,tree classes.The author is a Research Fellow of Japan Society for the Promotion of Science (JSPS). This work wassupported by Japan Society for the Promotion of Science KAKENHI 18J10561.
Introduction
In representation theory of algebras, we often use Auslander–Reiten theory to analysevarious additive categories and prove many important combinatorial and homologicalproperties with the help of the theory, for example [ARS, ASS, H, I3, Li2, Y]. In the casefor the category of lattices over an order, see [A2, Bu, I2, K2, RoS].Let O be a complete discrete valuation ring with a uniformizer ε and K the quotientfield of O . An O -algebra A is an O -order if A is free of finite rank as an O -module. Wewrite A for the induced algebra A ⊗ O ( O /ε O ). For an O -order A , a right A -module M is called an A -lattice if M is free of finite rank as an O -module. We denote by latt - A the full subcategory of the module category mod - A consisting of A -lattices. Accordingto [A2], the category latt - A admits almost split sequences if and only if A is an isolatedsingularity, that is, A ⊗ O K is a semisimple K -algebra. In this case, one can find someresults on the shapes of Auslander–Reiten quivers, for example [K2, Lu, Wi].When A is not an isolated singularity, we have to consider a suitable full subcategoryof latt - A which admits almost split sequences. It follows from [AR, Theorem 2.1] that M ∈ latt - A appears at the end term of an almost split sequence if and only if M satisfiesthe condition ( ♮ ): M ⊗ O K is projective as an A ⊗ O K -module. ( ♮ )Here, the full subcategory of latt - A consisting of A -lattices which satisfy the condition ( ♮ )is denoted by latt ( ♮ ) - A . When A is symmetric, that is, A is isomorphic to Hom O ( A, O ) as( A, A )-bimodules, the category latt ( ♮ ) - A admits almost split sequences. Thus, Ariki, Kaseand the author defined the concept of the stable Auslander–Reiten quiver for latt ( ♮ ) - A ,and found several conditions to be satisfied on the shapes of stable periodic componentswith infinitely many vertices by using Riedtmann’s structure theorem. As an anotherrestriction, the author proved that the tree class of any stable component is one of infiniteDynkin diagrams or Euclidean diagrams when A is of finite representation type [M].However, the shapes of stable components of an O -order seems to be largely unknown,and there are only few concrete examples since it is difficult to compute almost splitsequences in general.Let A be a symmetric O -order. To get a new example of stable components for latt ( ♮ ) - A , we consider a special kind of A -lattices called Heller lattices , which is defined to bedirect summands of the first syzygy of an indecomposable A -module as an A -module.There are two reasons why we consider Heller lattices. The first reason is that theyalways belong to latt ( ♮ ) - A . Thus, the category latt ( ♮ ) - A admits some stable componentscontaining indecomposable Heller lattices. We call such components Heller components of A . Another reason is that Heller lattices of a group algebra play important roles inmodular representation theory. For a p -modular system ( K , O , κ ) for a finite group G ,Heller lattices over O G were studied by Kawata [K1, K2]. It follows from [K1, Theorem4.4] that Heller lattices over O G provide us with certain relationship between almostsplit sequences for latt - O G and mod - κG , namely he showed that if 0 → A → B → Z M → Z M ofan indecomposable κG -module M , then the induced exact sequence0 → A ⊗ O κ → B ⊗ O κ → Z M ⊗ O κ → N THE HELLER LATTICES 3 is the direct sum of the almost split sequence ending at M and a split sequence (see also[P, Corollary 5.8]). They motivate us to study Heller lattices when A is an arbitrary sym-metric O -order. In [AKM], we study Heller lattices over truncated polynomial rings, anddetermined the shapes of stable components containing indecomposable Heller lattices.This is the first example of stable Auslander–Reiten component containing Heller latticeswhen A ⊗ O K is not semisimple.In this paper, we consider the symmetric Kronecker algebra A = O [ X, Y ] / ( X , Y ).Then, the Auslander–Reiten quiver of A consists of a unique non-periodic component,which contains the simple A -module, and infinitely many homogeneous tubes [ARS, ASS,SY1]. In [M], I studied Heller lattices of indecomposable non-periodic A -modules, andshowed that latt ( ♮ ) - A admits a unique non-periodic Heller component containing them, andit is of the form Z A ∞ . In this article, we focus on the remaining Heller lattices, and we willshow that they are indecomposable. It is well-known that such homogeneous tubes areclassified by the projective line P ( κ ) [ARS, SS]. Hence, Heller lattices of indecomposableperiodic A -modules are parametrized by Z > × P ( κ ). We denote by Z λm the Heller latticeassociated with ( m, λ ) ∈ Z > × P ( κ ). The main result is the following. Main Theorem (Theorems 3.8 and 4.10) . Let O be a complete discrete valuation ring, κ the residue field and A = O [ X, Y ] / ( X , Y ). Suppose that κ is algebraically closed. Let CH ( Z λm ) be the stable Auslander–Reiten component for latt ( ♮ ) - A containing Z λm . Then,the following statements hold.(1) Assume that Char ( κ ) = 2, then CH ( Z λm ) ≃ Z A ∞ / h τ i for all λ ∈ P ( κ ).(2) Assume that Char ( κ ) = 2, then CH ( Z λm ) ≃ (cid:26) Z A ∞ / h τ i if λ = 0 or ∞ , Z A ∞ / h τ i otherwise.Moreover, the Heller lattice Z λm appears on the boundary of CH ( Z λm ).This paper consists of four sections. In Section 1, we recall some notions includingalmost split sequences, stable Auslander–Reiten quivers and some results from [A1, AKM,Ri]. In Section 2, we give a complete list of Heller lattices of A = O [ X, Y ] / ( X , Y ), andexplain their properties including the indecomposability and the periodicity/aperiodicity.In Section 3, we consider the case λ = ∞ and determine the shape of the stable Auslander–Reiten component containing Z λm . Moreover, we show that every Heller lattice Z λm appearson the boundary of the Heller component of A . Note that, by using Riedtmann’s structuretheorem, it is not necessary to construct all almost split sequences to determine the shapeof Heller components of A . In fact, we only construct the almost split sequences endingat Z λm . Finally, we consider the case λ = ∞ in Section 4. Acknowledgement
My heartfelt appreciation goes to Professor Susumu Ariki (Osaka University) who pro-vided helpful comments and suggestions. I would also like to thank Professor ShigetoKawata (Nagoya City University), Professor Michihisa Wakui (Kansai University), Pro-fessor Ryoichi Kase (Okayama University of Science) and Dr. Qi Wang (Osaka university)whose meticulous comments were an enormous help to me.
KENGO MIYAMOTO Preliminaries
Throughout this paper, we use the following conventions.(1) k is an algebraically closed field and Λ is a finite-dimensional algebra over k .(2) All modules are right modules unless otherwise noted. For a finite-dimensional algebraΛ, we denote by mod -Λ the category of finite dimensional Λ-modules.(3) Tensor products are taken over O .(4) For an additive category C , let C be the projectively stable category of C .(5) The symbol δ i,j means the Kronecker delta.(6) The identity matrix of size n and the zero matrix of size n are denoted by n and n ,respectively.1.1. Orders and lattices.
First of all, we recall some terminologies on orders and lat-tices, for example see [I]. Throughout this paper, O denotes a complete discrete valuationring with a uniformizer ε , and κ is the residue field and K is the quotient field. We set D = Hom O ( − , O ). An O -algebra is called an O -order if it is free of finite rank as an O -module. For an O -order A , an A -module M is called a lattice if M is free of finiterank as an O -module. Let A be an O -order. Then, A is called Gorenstain if D ( A ) is aprojective A -module, and A is said to be symmetric if D ( A ) ≃ A as ( A, A )-bimodules.We note that the definitions of O -orders and lattices are different from Auslander’s sense[A1, Chapter I, Section 7]. However, it is obvious that if A is a symmetric O -order, then A is an O -order in Auslander’s sense [A1, Chapter III, Section 1]. We write latt - A forthe full subcategory of mod - A consisting of A -lattices. Then, we define latt ( ♮ ) - A to bethe full subcategory of latt - A consisting of A -lattices M such that M ⊗ K is projectiveas an A ⊗ K -module. Then, the category latt ( ♮ ) - A is enough projective and closed underdirect summands. In addition, if A is symmetric, the category latt ( ♮ ) - A is closed underextension. Given a pair of A -lattices M and N , we denote by Hom A ( M, N ) the O -moduleof all A -module homomorphisms from M to N . We write A for the finite dimensionalalgebra A ⊗ κ . For an O -order A , the syzygy functors on latt - A and mod - A are denotedby Ω and e Ω, respectively. Clearly, the following lemma holds.
Lemma 1.1.
Let R be an O -order and M an indecomposable R -lattice. If f ∈ End R ( M )is surjective, then f is an isomorphism. Moreover, the set of non-surjective endomorphismsof M coincides with the radical of the endomorphism ring of M .1.2. Valued stable translation quivers.
In this subsection, we recall notations onstable translation quivers. A quiver Q = ( Q , Q , s, t ) is a quadruple consisting of twosets Q and Q , and two maps s, t : Q → Q . Each element of Q and Q is called a vertexand an arrow, respectively. For an arrow α ∈ Q , we call s ( α ) and t ( α ) the source and thetarget of α , respectively. We understand that quivers are directed graphs. We write Q forthe underlying graph of Q . Given two quivers Q and ∆, a quiver homomorphism f : Q → ∆ is a pair of maps f : Q → ∆ and f : Q → ∆ such that ( s × t ) ◦ f = ( f × f ) ◦ ( s × t ).From now on, we assume that quivers have no multiple arrows, that is, the map ( s × t )is injective. Let ( Q, v ) be a pair of a quiver Q and a map v : Q → Z ≥ × Z ≥ . For anarrow x → y of Q , we write v ( x → y ) = ( d xy , d ′ xy ), and we understand that there is noarrow from x to y if and only if d xy = d ′ xy = 0. Then, ( Q, v ) is called a valued quiver if d x,y = 0 if and only if d ′ x,y = 0, and the values of the map v are called valuations . If v ( x → y ) = (1 ,
1) for all arrows x → y of Q , then v is said to be trivial . We usually omit N THE HELLER LATTICES 5 to write trivial valuations. For each vertex x ∈ Q , we set x + = { y ∈ Q | x → y ∈ Q } , x − = { y ∈ Q | y → x ∈ Q } . Note that a quiver is determined by the sets x + . A quiver Q is locally finite if x + ∪ x − is afinite set for any x ∈ Q . A translation quiver is a triple ( Q, Q ′ , τ ) of a locally finite quiver Q , a subset Q ′ ⊂ Q and an injective map τ : Q ′ → Q satisfying x − = ( τ x ) + . If Q ′ = Q and τ is bijective, the translation quiver is said to be stable . Then, we write ( Q, τ ) for thestable translation quiver, simply. Let C be a full subquiver of a stable translation quiver( Q, τ ). Then, C is a (connected) component if the following three conditions are satisfied.(i) C is stable under the quiver automorphism τ .(ii) C is a disjoint union of connected components of the underlying undirected graph.(iii) There is no proper subquiver of C that satisfies (i) and (ii).A quiver homomorphism f from a translation quiver ( Q, Q ′ , τ ) to a translation quiver(∆ , ∆ ′ , τ ′ ) is a translation quiver homomorphism if f ◦ τ = τ ′ ◦ f is satisfied on Q ′ .It is easily seen that τ induces a translation quiver automorphism when ( Q, Q ′ , τ ) isstable, and we use the same letter τ . In this paper, we denote by Aut τ ( Q ) the set ofall translation quiver automorphisms of ( Q, τ ). Let Q and ∆ be two stable translationquivers. A surjective translation quiver homomorphism f : Q → ∆ is a covering if f | x + gives a bijection between x + and ( f ( x )) + .For a stable translation quiver ( Q, τ ) and a subgroup G ⊂ Aut τ ( Q ), we define thetranslation quiver homomorphism π G : Q → Q/G by π G ( x ) = Gx for x ∈ Q . Asubgroup G ⊂ Aut τ ( Q ) is admissible if each G -orbit intersects x + ∪ { x } in at most onevertex and x − ∪ { x } in at most one vertex, for any x ∈ Q . Then, the map π G is covering. Definition 1.2. A valued stable translation quiver is a triple ( Q, v, τ ) such that(i) (
Q, v ) is a valued quiver,(ii) (
Q, τ ) is a stable translation quiver,(iii) v ( τ y → x ) = ( d ′ xy , d xy ) for each arrow x → y .Given a valued quiver ( Q, v ), one can construct the valued stable translation quiver( Z Q, ˜ v, τ ) as follows [Ri]. • ( Z Q ) := Z × Q . • ( n, x ) + := { ( n, y ) | y ∈ x + } ∪ { ( n − , z ) | z ∈ x − } . • ˜ v (( n, x ) → ( n, y )) = ( d xy , d ′ xy ), ˜ v (( n − , y ) → ( n, x )) = ( d ′ xy , d xy ) . • τ (( n, x )) = ( n − , x ).We write it simply Z Q . Note that Z Q has no loops whenever Q has no loops. Thefollowing theorem is well-known and it is effective to describe the structure of stabletranslation quivers [Ri]. Theorem 1.3 (Riedtmann’s structure theorm) . Let (
Q, τ ) be a stable translation quiverwithout loops and C a connected component of ( Q, τ ). Then, there exist a directed tree T and an admissible group G ⊆ Aut ( Z T ) such that C ≃ Z T /G as stable translationquivers. Moreover, T is uniquely determined by C , and the admissible group is unique upto conjugation.In Theorem 1.3, the underlying undirected tree T is called the tree class of C .Let ( Q, τ ) be a connected stable translation quiver. A vertex x ∈ Q is called periodic if x = τ k x for some k >
0, where τ k is the composition of k copies of τ . It is well-known KENGO MIYAMOTO that if there is a periodic vertex in Q , then all vertices of Q are periodic [HPR]. Indeed,if x is a periodic vertex in Q , then there is a positive integer and n x such that τ n x x = x .Since ( Q, τ ) is a stable translation quiver, τ n x induces a permutation on the finite set x + ,and so some power of τ n x stabilizes x + elementwise. Hence, all vertices in x + are periodic.It follows that all vertices are periodic. In this case, ( Q, v, τ ) is called periodic . Definition 1.4.
Let I be a set. A Cartan matrix on I is a function C : I × I → Z satisfying the following properties.(i) For all i ∈ I , C ( i, i ) = 2.(ii) C ( i, j ) ≤ j = i , and for each i , we have that C ( i, j ) < j ∈ I .(iii) C ( i, j ) = 0 if and only if C ( j, i ) = 0.Let ( Q, v ) be a connected valued quiver without loops. Then, (
Q, v ) gives rise to aCartan matrix on Q : C ( x, y ) = x = y, − d ′ x,y if there is an arrow x → y, − d y,x if there is an arrow y → x, Definition 1.5.
Let C be a Cartan matrix on I . A subadditive function for C is a function ℓ : I → Q > such that it satisfies X y ∈ I C ( x, y ) ℓ ( y ) ≥ x ∈ I . A subadditive function ℓ is called additive if the equality holds for all x ∈ I .We say that a connected valued quiver Q admits a subadditive function when there existsa subadditive function for a Cartan matrix on Q . Remark 1.6.
Let (
Q, v, τ ) be a connected valued translation quiver without loops, andlet T be the tree class of Q . If a function ℓ : Q → Q > satisfies ℓ ( τ x ) = ℓ ( x ) and2 ℓ ( x ) ≥ X y → x in T d yx ℓ ( y ) + X x → y in T d ′ xy ℓ ( y ) , then the restriction ℓ | T is a subadditive function for a Cartan matrix on T .The following result is well-known. Theorem 1.7 ([HPR]) . Let (∆ , v ) be a connected valued quiver without loops. If ∆admits a subadditive function ℓ , then the following statements hold.(1) The underlying undirected graph ∆ is either a finite or infinite Dynkin diagram or aEuclidean diagram.(2) If ℓ is not additive, then ∆ is either a finite Dynkin diagram or A ∞ .(3) If ℓ is additive, then ∆ is either an infinite Dynkin diagram or a Euclidean diagram.(4) If ℓ is unbounded, then ∆ is A ∞ . N THE HELLER LATTICES 7
The stable AR quiver for the category of lattices over an O -order. In thissubsection, we recall the definitions of almost split sequences and the stable Auslander–Reiten quiver for latt ( ♮ ) - A , see [AKM] for details. Let A be a Gorenstain O -order. Amorphism f : M → N in latt - A which is neither a section nor retraction is called irre-ducible if f = f ◦ f in latt - A , then either f is a section or f is a retraction. Definition 1.8.
A short exact sequence 0 → L f −−→ M g −−→ N → latt - A is called an almost split sequence ending at N if the following two conditions are satisfied:(i) The morphisms f and g are irreducible in latt - A .(ii) The A -lattices L and N are indecomposable.Note that it follows from [A1, Proposition 4.4] that each almost split sequence isuniquely determined by the third term, and it is also uniquely determined by the firstterm if it exists. Here, we set τ ( M ) = L and τ − ( L ) = N , and we call both τ and τ − AR translations . Theorem 1.9 ([AR, Theorems 2.1, 2.2]) . Assume that A ⊗ K is self-injective. Then, latt ( ♮ ) - A has almost split sequences. Moreover, almost split sequences in latt ( ♮ ) - A are alsothose in latt - A .It is natural to ask how we compute almost split sequences. Proposition 1.10 ([AKM, Proposition 1.15]) . Let A be a Gorenstein O -order, M anindecomposable A -lattice in latt ( ♮ ) - A , and let p : P → M be the projective cover of M .Given an endomorphism ϕ : M → M , we obtain the pullback diagram along p and ϕ :0 Ker( ν ( p )) E M
00 Ker( ν ( p )) ν ( P ) ν ( M ) 0 / / / / / / / / / / / / ν ( p ) / / / / (cid:15) (cid:15) ϕ (cid:15) (cid:15) Here, ν = D (Hom A ( − , A )) is the Nalayama functor. Then, the following statements areequivalent.(1) The upper short exact sequence is the almost split sequence ending at M .(2) The following three conditions hold.(i) The morphism ϕ does not factor through ν ( p ).(ii) Ker( ν ( p )) is an indecomposable A -lattice.(iii) For all f ∈ rad End A ( M ), the morphism ϕ ◦ f factors through ν ( p ).Moreover, any almost split sequence ending at M is given in this way.Recall that O is a complete discrete valuation ring. Corollary 1.11. If A is a Gorenstein O -order, then we have a functorial isomorphism τ ≃ Ω ν . In particular, if A is symmetric, then there is a functorial isomorphism τ ≃ Ω. Proof . Let M be an A -lattice in latt ( ♮ ) - A and let Q q −→ P p −→ M → M . Then, it follows from [AKM, Lemma 1.13, Remark 1.14]that we have the following exact sequences in latt - A :0 −→ D Coker( p ∗ ) −→ ν ( P ) ν ( p ) −−→ ν ( M ) −→ −→ Coker( p ∗ ) −→ Q ∗ −→ Tr( M ) −→ KENGO MIYAMOTO
Since the lower sequence is the projective cover of Tr( M ), we have D ΩTr( M ) = τ ( M ).The upper exact sequence implies that D Coker( p ∗ ) = Ω( ν ( M )). Therefore, we have τ ≃ D ΩTr ≃ Ω ν . (cid:3) Remark 1.12. As A is a Gorenstein O -order, the Nakayama functor ν : latt - A → latt - A isan autofunctor, and latt - A is a Frobenius category. Hence, latt - A is a triangulated categorywith the shift functor Ω − . Then, we have a triangulated equivalence ν : latt - A ∼ −→ latt - A ,and the Auslander–Reiten translation τ is represented by Ω ν by [H]. Definition 1.13.
Let A be an O -order and M be an indecomposable A -module. We calleach direct summand of Ω( M ) a Heller lattice of M . Note that Ω( M ) may not be anindecomposable A -lattice. Lemma 1.14.
Any Heller lattices belong to latt ( ♮ ) - A . Proof . This is the assertion of [AKM, Remark 1.12]. (cid:3)
The following proposition is used in this paper everywhere.
Proposition 1.15 ([K1, Proposition 4,5]) . Let A be an O -order and L an indecomposable A -lattice and let 0 → τ L → E → L → L . Assume that L is not a direct summand ofany Heller lattices. Then, the induced exact sequence0 → τ L ⊗ κ → E ⊗ κ → L ⊗ κ → Lemma 1.16.
Suppose that A is a symmetric O -order. Then, for any non-projective A -lattice M , there is an isomorphism τ ( M ) ⊗ κ ≃ e Ω( M ⊗ κ ). Proof . Let M be an A -lattice and π : P → M the projective cover. Let Q ⊗ κ → M ⊗ κ be the projective cover. Then rank Q ≤ rank P . On the other hand, it lifts to Q → M and it is an epimorphism by Nakayama’s lemma. Thus, we have rank Q = rank P and P ⊗ κ is the projective cover of X ⊗ κ . Therefore, we have τ ( M ) ⊗ κ ≃ e Ω( M ⊗ κ ) asobjects in the stable module category mod - A . Since the functor − ⊗ κ is exact on latt - A ,the assertion follows. (cid:3) Definition 1.17.
Let A be a symmetric O -order.(1) The stable Auslander–Reiten quiver for latt ( ♮ ) - A is the valued quiver defined as follows: • The set of vertices is a complete set of isoclasses of non-projective indecomposable A -lattices in latt ( ♮ ) - A . • We draw a valued arrow M ( a,b ) −−→ N whenever there exist irreducible morphisms M → N , where the valuation ( a, b ) means:(i) a is the multiplicity of M in the middle term of the almost split sequenceending at N .(ii) b is the multiplicity of N in the middle term of the almost split sequencestarting at M .The stable Auslander–Reiten quiver for latt ( ♮ ) - A is denoted by Γ s ( A ) in this paper.(2) A component of Γ s ( A ) containing an indecomposable Heller lattice Z is said to be a Heller component of A , and denoted by CH ( Z ). N THE HELLER LATTICES 9
By the definition, we note that a component C of Γ s ( A ) does not have multiple arrows,and τ M and τ − M exist for each vertex M of C by Theorem 1.9. Thus, ( C , τ ) is a valuedstable translation quiver. Note that there are possibilities that C has loops [Wi].Let A be a symmetric O -order and C a periodic component of Γ s ( A ) without loops.Assume that C has infinitely many vertices. Let T be the tree class. For a vertex X ∈ C ,we define R ( X ) by R ( X ) := n X − X i =0 rank( τ i X ) n X , where n X is the smallest positive integer k such that X ≃ τ k X . Then, for each X ∈ T ,the inequality(1.17.1) X Y → X d Y,X rank( Y ) ≤ rank( X ) + rank( τ X )implies that R satisfies(1.17.2) 2 R ( X ) ≥ X Y ∈ X − ∩ T d Y,X R ( Y ) + X Y ∈ X + ∩ T d ′ X,Y R ( Y )for all X ∈ T . This is shown as follows. By the definition of R , it is a τ -invariant function.Let n = Q Y → X n Y . Then, we have n X n − X k =0 X Y → X d τ k Y,τ k X rank( τ k Y ) ! = X Y → X n X n − X k =0 ( d Y,X rank( τ k Y ))= X Y → X d Y,X n X nn Y n Y − X k =0 rank( τ k Y )= X Y → X d Y,X n X n R ( Y ) . On the other hand, we have n X n − X k =0 (rank( τ k X ) + rank( τ k +1 X )) = 2 n X nn X n X − X k =0 rank( τ k X ) = 2 n X n R ( X ) . Thus, the inequality (1.17.1) yields the inequality (1.17.2) since C is a valued stabletranslation quiver. Therefore, R| T is a subadditive function on T . By Theorem 1.7, thefollowing proposition holds. See [AKM] for details. Proposition 1.18 ([AKM, Theorem 1.27]) . Let A be a symmetric O -order and C aperiodic component of Γ s ( A ). Assume that Γ s ( A ) has infinitely many vertices. Then, oneof the following statements holds.(1) If C has loops, then C \ { loops } = Z A ∞ / h τ i . Moreover, the loops only appear on theboundary of C .(2) If C has no loops, then the tree class of C is one of infinite Dynkin diagrams.For X ∈ latt ( ♮ ) - A , we define D ( X ) to be the number of non-projective indecomposabledirect summands of X ⊗ κ . Lemma 1.19.
Let C be a component of Γ s ( A ). For an indecomposable Heller lattice Z ∈ C , we denote by E Z the middle term of the almost split sequence ending at Z . If D satisfies 2 D ( Z ) ≥ D ( E Z )for any indecomposable Heller lattice Z ∈ C , then the restriction of D to the tree class of C is subadditive. In particular, D| T is additive if and only if the equalities hold for any Z . Proof . The assertion follows from [M, Lemmas 3.2, 3.3]. (cid:3)
Indecomposable modules over special biserial algebras.
For a finite-dimensional algebra Λ, let Q be the Gabriel quiver, I the admissible ideal such thatΛ ≃ k Q/ I . Then, Λ is called special biserial if the following two conditions are satisfied.(i) For each vertex x of Q , ♯x + ≤ ♯x − ≤ α of Q , there exist at most one arrow β such that αβ / ∈ I and atmost one arrow γ such that γα / ∈ I .A typical examples are Brauer graph algebras. In fact, the class of Brauer graph alge-bras coincides with the class of symmetric special biserial algebras [ES]. Thus, symmetricspecial biserial algebras are one of important classes of algebras in representation theory.Special biserial algebras are of tame representation type and all finite-dimensional inde-composable modules are classified into “string modules” and “band modules” [BR, WW].For the definitions of string modules and band modules, for example see [HL]. Theorem 1.20 ([WW, (2.3) Proposition]) . Let Λ be a special biserial algebra. Then,the disjoint union of the set of string modules, the set of band modules and the set of allprojective-injective modules corresponding to the binomial relations forms a complete setof isoclasses of finite-dimensional indecomposable Λ-modules.2.
The Heller lattices over the symmetric Kronecker algebra
In this section, we consider the symmetric Kronecker algebra A := O [ X, Y ] / ( X , Y ),that is, it is the bound quiver algebra over O defined by the following quiver and relations:1 Y f f X ; X = Y = 0 , XY − Y X = 0 . From this section to end of this paper, we assume that κ is algebraically closed. Then, a d -dimensional A -module M is of the form M = κ d M i i M , where M and M are square matrices of size d which commute and have square zero[ASS, SY1]. To simplify, we denote by ( d, M , M ) the A -module M . Throughout thissection, for a positive integer n , we denote by e , . . . e n the standard basis of O n andwe adopt e , Xe , Y e , XY e , . . . , e n , Xe n , Y e n , XY e n as an O -basis of A n . We call this O -basis of A n the standard basis of A n . N THE HELLER LATTICES 11
Indecomposable modules over A . In this subsection, we give a complete list ofHeller lattices. By Theorem 1.20, all finite-dimensional indecomposable A -modules areclassified into string modules, band modules and projective-injective modules.First, the unique indecomposable projective-injective module A is given by , , . Now, we present a complete list of the other finite-dimensional indecomposable A -modules,which are denoted by M ( m ), M ( − m ), M ( λ ) n , where m ∈ Z ≥ , n > λ lies on theprojective line P ( κ ) = κ ⊔ {∞} .(i) The string module M ( m ) := M (( β ∗ β ) m ) ( m ∈ Z ≥ ) is given by the formula: M ( m ) = m + 1 , m m +1 m m +1 · · · · · · , m m +1 · · · · · · m m +1 (ii) The string module M ( − m ) := M (( β β ∗ ) m ) ( m ∈ Z ≥ ) is given by the formula: M ( − m ) = m + 1 , m +1 m m m , m +1 m m m (iii) The string module M (0) n := M (( β β ∗ ) n − β ) ( n ∈ Z > ) is given by the formula: M (0) n = (cid:18) n, (cid:18) n n n n (cid:19) , (cid:18) n n J (0 , n ) n (cid:19)(cid:19) (iv) The string module M ( ∞ ) n := M ( β ( β ∗ β ) n − ) ( n ∈ Z > ) is given by the formula: M ( ∞ ) n = (cid:18) n, (cid:18) n n J (0 , n ) n (cid:19) , (cid:18) n n n n (cid:19)(cid:19) (v) Let V be a finite-dimensional indecomposable left κ [ x, x − ]-module. Assume that V is represented by x J ( λ, n ) with respect to a basis of V for some λ ∈ κ × and n ∈ Z > . The band module M ( λ ) n := N ( β ∗ β , V ) is given by the formula: M ( λ ) n = (cid:18) n, (cid:18) n n n n (cid:19) , (cid:18) n n J ( λ, n ) n (cid:19)(cid:19) Lemma 2.1.
The set of the A -modules { M ( m ) | m ∈ Z } ⊔ { M ( λ ) n | λ ∈ P ( κ ) , n ∈ Z ≥ } ⊔ { A } forms a complete set of isoclasses of finite-dimensional indecomposable modules over A . Proof . The assertion follows from Proposition 1.20. See also [M] for a construction. (cid:3)
For simplicity, we visualize an A -module as follows: • Vertices represent basis vectors of the underlying κ -vector spaces. • Arrows of the form −→ represent the action of X , and represent the action of Y . • If there is no arrow (resp. dotted arrow) starting at a vertex, then X (resp. Y )annihilates the corresponding basis element.By using this notation, the indecomposable modules listed above are represented as fol-lows:1. A = e Xe Y e XY e ( ( PPP ♥♥♥♥♥♥ ( ( PPP ♥♥♥♥♥ . M ( m ) = u v v m − u m − ...... u m v v m ...... / / ❴❴❴❴ ❣❣❣❣❣❣❣ / / ❴❴❴❴ / / ❴❴❴ ❣❣❣❣❣❣❣❣❣ M ( − m ) = u v m − u m ... u v u m +1 v m v ... ❣❣❣❣ ❣❣❣❣ ❣❣❣❣❣ / / / / / / M (0) n = u v n − u n − ... u v u n v n v ... / / / / ❣❣❣❣❣ / / / / ❣❣❣❣ M ( ∞ ) n = u u v v n − u n − ...... u n v v n ...... / / ❴❴❴❴ / / ❴❴❴❴ ❣❣❣❣❣❣❣❣ / / ❴❴❴❴ / / ❴❴❴ ❣❣❣❣❣❣❣❣❣ M ( λ ) n = u v n − u n − ... u v u n v n v ... / / / / ❥❥❥❥❥ λ ⑤ ✉ ❴ ■ ❇ / / λ > > ❇ ■ ❴ ✉ ⑤ / / ❥❥❥❥ λ > > ❇ ■ ❴ ✉ ⑤ Here, u i v i v i − ♦♦♦♦♦ λ > > ❇ ■ ❴ ✉ ⑤ in the picture 6 means Y u i = λv i + v i − .From now on, as a κ -basis of a non-projective indecomposable module over A , we adoptthe above κ -basis. Remark 2.3 ([ARS, ASS, Erd, SY1]) . Almost split sequences for mod - A are known tobe as follows:0 −→ M ( − −→ A ⊕ M (0) ⊕ M (0) −→ M (1) −→ −→ M ( n − −→ M ( n ) ⊕ M ( n ) −→ M ( n + 1) −→ n = 00 −→ M ( λ ) −→ M ( λ ) −→ M ( λ ) −→ λ ∈ P ( κ )0 −→ M ( λ ) n −→ M ( λ ) n − ⊕ M ( λ ) n +1 −→ M ( λ ) n −→ n > , λ ∈ P ( κ ) Lemma 2.4.
For all λ ∈ P ( κ ) and n ∈ Z > , there is an isomorphism e Ω( M ( λ ) n ) ≃ M ( − λ ) n , e Ω( M ( ∞ ) n ) ≃ M ( ∞ ) n . Proof . For λ ∈ P ( κ ) and n >
0, we define a map π λn : ( A ) n → M ( λ ) n by π λn : e i u i .Then, π λn is the projective cover of M ( λ ) n as an A -module. First, we assume that λ = ∞ . N THE HELLER LATTICES 13
In this case, the kernel of π λn is given by κ ( Y e − λXe ) ⊕ κXY e ⊕ κ ( Y e − λXe − Xe ) ⊕ κ ( − XY e ) ⊕ · · ·⊕ κ ( Y e n − λXe n − Xe n − ) ⊕ κ ( − n XY e n , and it is isomorphic to M ( − λ ) n in mod - A . Next, we consider λ = ∞ case. A κ -basis ofthe kernel of π ∞ n is given by κXe ⊕ κXY e ⊕ κ ( Xe − Y e ) ⊕ κ ( − XY e ) ⊕ · · ·⊕ κ ( Xe n − Y e n − ) ⊕ κ ( − n XY e n , and it is isomorphic to M ( ∞ ) n in mod - A . In the both cases, the isomorphisms are liftedin mod - A since the kernels have no A as a direct summand. (cid:3) Properties of Heller lattices.
Let M be a non-projective indecomposable A -module listed in Lemma 2.1. For each m and λ , the projective cover of M as an A -module π M is given by π M : A m −→ M, e i u i if M ≃ M ( m ), m > A m +1 −→ M, e i u i if M ≃ M ( − m ), m > A −→ M, e u if M ≃ M (0), A n −→ M, e i u i if M ≃ M ( λ ) n , n > λ ∈ P ( κ ).The author studied the Heller lattices of M ( m ) for m ∈ Z and determined the shapeof the unique non-periodic Heller component containing them. Theorem 2.5 ([M, Proposition 2.7, Theorem 3.1]) . For each m ∈ Z , let Z m be the kernelof π M ( m ) . Then the following statements hold.(1) There is an isomorphism Z m ⊗ κ ≃ M ( m − ⊕ M ( m ).(2) The Heller lattice Z m is indecomposable.(3) There is an isomorphism τ Z m ≃ Z m − .(4) The Heller component containing Z m is isomorphic to Z A ∞ .(5) The Heller lattice Z m appears on the boundary of the component.In this paper, we focus on the remaining Heller lattices. For n ∈ Z > and λ ∈ P ( κ ),we define the Heller A -lattice Z λn to be the kernel of π M ( λ ) n . We use the following notations: • For the Heller lattice Z λn ( λ = ∞ ), we define a λ , a λ , a λ , a λ , a λ , a λ , a λ , a λ , ... ... ... ... a λ n − , a λ n − , a λ n − , a λ n − , a λ n , a λ n , a λ n , a λ n , = εe εXe ( Y e − λXe ) XY e εe εXe ( Y e − λXe − Xe ) XY e ... ... ... ... εe n − εXe n − ( Y e n − − λXe n − − Xe n − ) XY e n − εe n εXe n ( Y e n − λXe n − Xe n − ) XY e n when n >
1, and if n = 1, we define( a λ , , a λ , , a λ , , a λ , ) = ( εe , εXe , ( Y e − λXe ) , XY e ) . We understand that a λ , j = 0 for j = , , , . Then, X and Y act on Z λn as follows. If n >
1, then X a λ i , j = (cid:26) a λ i , j + if j = , , Y a λ i , j = ε a λ i , + λ a λ i , + a λ i − , if j = , ε a λ i , if j = , − λ a λ i , − a λ i − , if j = , n = 1, then X a λ , j = (cid:26) a λ , j + if j = , ,0 otherwise, Y a λ , j = ε a λ , + λ a λ , if j = , ε a λ , if j = , − λ a λ , if j = , • For the Heller lattice Z ∞ n , we define b , b , b , b , b , b , b , b , ... ... ... ... b n − , b n − , b n − , b n − , b n , b n , b n , b n , = εe Xe ( Y e − Xe ) XY e εe εXe ( Y e − Xe ) XY e ... ... ... ... εe n − εXe n − ( Y e n − − Xe n ) XY e n − εe n εXe n εY e n XY e n when n >
1, and if n = 1, we define( b , , b , , b , , b , ) = ( εe , Xe , εY e , XY e ) . Then, X and Y act on Z ∞ n as follows. If n >
1, then X b i , j = ε b , if i = j = , b i , if i = , j = , b i , if i = n , j = ,ε b n , if i = n , j = , Y b i , j = ε b i , + b i + , if i = n , j = , b n , if i = n , j = , b , if i = , j = ,ε b i , if i = , j = , − b i + , if i = n , j = , N THE HELLER LATTICES 15 If n = 1, then X b , j = (cid:26) ε b , j + if j = , ,0 otherwise, Y b , j = (cid:26) b , j + if j = , ,0 otherwise.It is straightforward to prove the following lemma. Lemma 2.7.
We use the lexicographical order on { ( i , j ) | i = , . . . , n , j = , , , } . Then,the sets { a λ i , j | i = , . . . , n , j = , , , } and { b i , j | i = , . . . , n , j = , , , } form an(ordered) O -basis of Z λn and Z ∞ n , respectively.In this paper, when we consider the matrix representation of an A -module, we alwaysuse these O -bases, and we denote these O -bases by B λn and B ∞ n , respectively. Proposition 2.8. (1) For each λ ∈ κ and n >
0, the Heller lattice Z λn is indecomposable.(2) For each n >
0, the Heller lattice Z ∞ n is indecomposable.(3) For each λ ∈ κ and n >
0, there is an isomorphism Z λn ⊗ κ ≃ M ( λ ) n ⊕ M ( − λ ) n as A -modules.(4) For each n >
0, there is an isomorphism Z ∞ n ⊗ κ ≃ M ( ∞ ) ⊕ n as A -modules. Let e X , e Y and ee Y be square matrices of size 4defined by e X := e Y := λ ε ε − λ ee Y := − Then, the representing matrices of the actions of X and Y on Z λn with respect to the O -basis B λn are of the form: X = e X e X . . . e X e X Y = e Y ee Y e Y ee Y . . . e Y ee Y e Y ∈ Mat(4 n, n, O )Obviously, the Heller lattice Z λ is indecomposable since Z λ ⊗ K ≃ A ⊗ K . We provethat idempotents of End A ( Z λn ) are only n and n . Let M = ( m i,j ) be an idempotent ofEnd A ( Z λn ). We partition M into n blocks of size 4 ×
4, and denote by M i,j ∈ Mat(4 , , O )the ( i, j )-block of M and by α i,j the (4 i − , j − M . The equalities M X = XM and M Y = Y M yield that the block M ij is of the form M i,j = d i,j m i − , j − d i,j c i,j m i − , j − d i,j m i, j − m i − , j − m i, j − d i,j , where(2.9.1) d i,j = m , if i = j = 1, m , + ε P j − k =1 α k,k +1 if i = j > ,ε P jk =1 α i − j − k,k if n ≥ i > j ≥ ,m , j − if n ≥ j > i = 1 ,m , j − i )+1 + ε P i − k =1 α k,j − i +1+ k if n ≥ j > i > ,c i,j = i = n, j = 1, α i,j if i = n, − P j − k =1 α i − j + k,k if n = i ≥ j > . Here, we have to choose each element m k,l in M i,j in such a way that the equation M Y = Y M holds. By comparing the (1 , M and M , we have the equation m , = m , + ε n − X k =1 m , k +1 m k − , . We write x for the coset in the residue field κ = O /ε O represented by x ∈ O . The aboveequation implies that m , is either 0 or 1.Assume that m , = 0. Then, the element d i,i belongs to ε O for all i by (2.9.1). Bycomparing the (1 , k + 1)-entries of M and M , we have(2.9.2) m , k +1 = m , m , k +1 + k X l =1 m , l +1 d l +1 ,l +1 + ε n − X l = k +1 m , l +1 P ( l )for some P ( l ) ∈ O , and hence m , k +1 ∈ ε O for all k . From (2.9.2), m , k +1 belongs to ε t O for all t >
0. It implies that m , k +1 = 0 for all k . Therefore, the first row of M iszero. By comparing the (5 , M and M , the following equation holds: εm , = (cid:26) ε m , if n = 2, ε m , + ε P n − k =1 m , k +7 d k +2 , if n > n = 2, m , = 0 because 1 − εm , is invertible. Therefore, we have: M = m , m , m , m , m , m , m , m , m , m , m , εm , m , εm , m , − m , m , εm , m , m , m , m , εm , m , m , m , By M = M , all elements of M must be .In the other case, first we prove that the (4 k − M is zero for all k = 1 , , . . . , n by induction on k . By comparing the (2 , s − M and M , the followingequations hold:(2.9.3) m , s − = n − X l =1 m , l +3 d l +1 ,s , s = 1 , , . . . , n. N THE HELLER LATTICES 17
Since the first row of M is zero, each d l +1 ,s of the right hand side of (2.9.3) belongs to ε O and so is m , s − for all s = 1 , , . . . , n . Thus, for s = 1 , , . . . , n , the element m , s − lies on ε t O for all t >
0. It implies that m , s − = 0 for all s = 1 , , . . . , n . Then, the(2 , s − M and M yield m , s − = n − X l =1 m , l +5 d l +2 ,l , s = 1 , , . . . , n. As each d l +2 ,l belongs to ε O , so is m , s − for all s = 1 , , . . . , n . It implies that theelement m , s − lies on ε t O for t >
0, and hence the second row of M is zero.Assume that the statement holds for 2 ≤ t ≤ k −
1, we will show the statement for k .Then, by the induction hypothesis, we have m k − , s − = n − X l =1 m k − , l +3 d l +1 ,s , s = 1 , , . . . , n. Thus, we obtain m k − , s − = 0 and m k − , s − = n − X l =1 m k − , l +5 d l +2 ,l , s = 1 , , . . . , n by similar arguments to the proof of the case of k = 1. It implies that the (4 k − M is zero for all k = 2 , . . . , n .Since the first and the (4 k − M are zero for all k , the ( i, j )-block of M isof the form M i,j = m i − , j − m i, j − m i − , j − m i, j − . Therefore, we obtain M = n by comparing each entry of M and M .Next we assume that m , = 1. Then, n − M is an idempotent whose (1 , ε O and M = n follows. We have finished the proof of (1). Let X ( a,b ) , Y ( a,b ) , Y be square matrices of size 4 defined by X ( a,b ) := a b Y ( a,b ) := a b Y := − , where a , b ∈ { , ε } . Then, the representing matrices of the actions of X and Y on Z ∞ n with respect to the O -basis B ∞ n are of the form: X = X ( ε, X (1 , . . . X (1 , X (1 ,ε ) Y = Y ( ε, Y Y ( ε,ε ) . . . Y ( ε,ε ) Y Y (1 ,ε ) Lemma 2.11.
The endomorphism ring of Z ∞ n is subset of ( m i,j ) i,j ∈ Mat(4 n, n, O ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) m i,i = m i +1 ,i +1 for all 1 ≤ i ≤ n − ,m i,j = 0 for i < j whenever ( i, j ) = (2 , , (2 , , (4 , ,
7) or (8 , . Proof . The proof is straightforward. (cid:3)
Let M be an idempotent of the endomorphism ring of Z ∞ n . It follows from Lemma 2.11that M must be either the zero matrix or the identity matrix by comparing all entries of M with those of M . Therefore, the A -lattice Z ∞ n is indecomposable. For any n >
0, we define A -submodules of Z λn ⊗ κ by Z ( λ, n,
1) :=
Span κ { a λ i , , a λ i , | i = , . . . , n } ,Z ( λ, n,
2) :=
Span κ { a λ i , , a λ i , | i = , . . . , n } . Then, Z λn ⊗ κ is decomposed into Z ( λ, n, ⊕ Z ( λ, n,
2) as A -modules. Define A -homomorphisms f λ,n : M ( λ ) n → Z ( λ, n,
1) and f λ,n : M ( − λ ) n → Z ( λ, n,
2) by f λ,n ( u i ) = a λ i , , f λ,n ( v i ) = a λ i , , f λ,n ( u i ) = ( − i + a λ i , , and f λ,n ( v i ) = ( − i + a λ i , . As these morphisms are isomorphisms, we have the assertion.
For any n >
0, we put Z ( ∞ , n,
1) :=
Span κ { b i , , b j , , b n , | i = , . . . , n , j = , . . . , n } ,Z ( ∞ , n,
2) :=
Span κ (cid:26) b , , b i , , b n , , b j , (cid:12)(cid:12)(cid:12)(cid:12) i = , . . . , n − , j = , . . . , n − (cid:27) . Then, one can show that Z ( ∞ , n, ≃ Z ( ∞ , n, ≃ M ( ∞ ) n . Proposition 2.15.
For λ ∈ P ( κ ) and n >
0, the following statements hold.(1) If λ = ∞ , there exists an isomorphism τ Z λn ≃ Z − λn .(2) If λ = ∞ , there exists an isomorphism τ Z ∞ n ≃ Z ∞ n . Proof . (1) The map π n,λ defined by π n,λ : A n −→ Z λn e i (cid:26) a λ k , if i = 2 k − k = , , . . . , n , a λ k , if i = 2 k , k = , , . . . , n N THE HELLER LATTICES 19 is the projective cover of Z λn as an A -module. Its kernel τ Z λn is given by O ( εe − Y e + λXe ) ⊕ O ( εXe − XY e ) ⊕ O ( Y e + λXe ) ⊕ O XY e n M k =2 (cid:18) O ( − k − ( εe k − Y e k − + λXe k − + Xe k − ) ⊕ O ( − k − ( εXe k − XY e k − ) ⊕ O ( − k − ( Y e k + λXe k + Xe k − ) ⊕ O ( − k − XY e k (cid:19) . Then, the actions X and Y on τ Z λn coincide with those on Z − λn .(2) We define an A -module homomorphism by π n, ∞ : A n −→ Z ∞ n e i b , if i = 1 , b , if i = 2, b k , if i = 2 k + 1, k = , , . . . , n − , b k , if i = 2 k , k = , , . . . , n . Then, the π n, ∞ is the projective cover of Z ∞ n , and an O -basis of the kernel of π n, ∞ is givenas follows. If n = 1, then the kernel of π , ∞ is O ( − Xe + εe ) ⊕ O Xe ⊕ O ( − XY e + εY e ) ⊕ O XY e , and it is isomorphic to Z ∞ . If n = 2, then the kernel of π , ∞ is O ( − XY e + εe ) ⊕ O Xe ⊕ O ( − Xe + Y e ) ⊕ O XY e ⊕O ( − Y e + Xe + εe ) ⊕ O ( − XY e + εXe ) ⊕ O ( XY e + εY e ) ⊕ O XY e , and it is isomorphic to Z ∞ . Suppose that n ≥
3. Then an O -basis of the kernel of π n, ∞ is given by O ( εe − Xe ) ⊕ O Xe ⊕ O ( Y e − Xe ) ⊕ O XY e ⊕ O ( εe + Xe − Y e ) ⊕ O ( εXe − XY e ) ⊕ O ( Y e + Xe ) ⊕ O XY e n − M k =2 (cid:18) O ( − k +1 ( εe k +1 + Xe k +1) − Y e k ) ⊕ O ( − k +1 ( εXe k +1 − XY e k ) ⊕ O ( − k +1 ( Y e k +1 + Xe k +3 ) ⊕ O ( − k +1 XY e k +1 (cid:19) ⊕ O ( − n ( εe n − + Xe n − Y e n − ) ⊕ O ( − n ( εXe n − − XY e n − ) ⊕ O ( − n ( εY e n − + XY e n ) ⊕ O ( − n XY e n − . Then, it is easy to check that the actions X and Y on the kernel of π n, ∞ coincide withthose on Z ∞ n . (cid:3) The case λ = ∞ . The almost split sequence ending at Z λn . Throughout this subsection, we assumethat λ = ∞ . Lemma 3.1.
An endomorphism ρ ∈ End A ( Z λn ) is determined by ρ ( a λ , ) , . . . , ρ ( a λ n , ). Proof . Let ρ ∈ End A ( Z λn ). For any k = , , . . . , n , since ρ is an A -module homomor-phism, we have Xρ ( a λ k , ) = ρ ( X a λ k , ) = ρ ( a λ k , ) and ρ ( a λ k , ) = ε − XY ρ ( a λ k , ). Assumethat n = 1. In this case, Y ρ ( a λ k , ) = ερ ( a λ k , ) − λρ ( a λ k , ) holds. Thus, ρ ∈ End A ( Z λn ) isdetermined by ρ ( a λ , ). Now, we assume that n >
1. Then, we have ρ ( a λ k , ) = (cid:26) ε − ( Y ρ ( a λ , ) + λρ ( a λ , )) k = ,ε − ( Y ρ ( a λ k , ) + λρ ( a λ k , ) + ρ ( a λ k − , )) k = . This completes the proof of the lemma. (cid:3)
Lemma 3.2.
Let ρ ∈ radEnd A ( Z λn ). If we write ρ ( a λ k , ) = n X l = c ( k ) l , a λ l , + A ( k ) , A ( k ) ∈ Span O { a λ i , j | j = } , where c ( k ) l , ∈ O , then the following statements hold.(1) det( c ( k ) l , ) l , k ∈ ε O .(2) c ( k ) n , ∈ ε O for all k = , , . . . , n . Proof . (1) Let ρ ∈ radEnd A ( Z λn ). Assume that(3.2.1) ρ ( a λ k , ) = n X l = c ( k ) l , a λ l , + A ( k )as above. We show that if the matrix C := ( c ( k ) l , ) l , k is invertible, then ρ is surjective. As XY a λ l , = ε a λ l , holds for all l = , . . . , n , we have( ρ ( a λ , ) , . . . , ρ ( a λ n , )) = ( a λ , , . . . , a λ n , ) C . Thus, a λ , , . . . , a λ n , are contained in the image of ρ . By (3.2.1), we have ρ ( a λ k , ) = n X l = c ( k ) l , a λ l , + XA ( k ) . For each k , since XA ( k ) belongs to Span O { a λ l , | l = , . . . , n } , there exists x ( k ) ∈ Span O { a λ l , | l = , . . . , n } such that ρ ( x ( k )) = XA ( k ). Hence, we have( ρ ( a λ , − x ( )) , . . . , ρ ( a λ n , − x ( n ))) = ( a λ , , . . . , a λ n , ) C . Therefore, a λ , , . . . , a λ n , belong to the image of ρ . Finally, we show that a λ , , . . . , a λ n , belongto the image of ρ . By the equation (3.2.1), we have Y ρ ( a λ k , ) = c ( k ) , ( ε a λ , + λ a λ , ) + n X l = c ( k ) l , ( ε a λ l , + λ a λ l , + a λ l − , ) + Y A ( k )= n X l = ε c ( k ) l , a λ l , + n X l = λ c ( k ) l , a λ l , + n − X l = c ( k ) l + , a λ l , + Y A ( k ) . N THE HELLER LATTICES 21
On the other hand,
Y ρ ( a λ k , ) is ερ ( a λ k , ) + ρ ( λ a λ k , ) + ρ ( λ a λ k − , ). Let y ( k ) ∈ Z λn such that ρ ( y ( k )) = P nl = λ c ( k ) l , a λ l , + P n − = c ( k ) l + , a λ l , + Y A ( k ). Then, we have(3.2.2) ερ ( a λ k , ) = n X l = ε c ( k ) l , a λ l , + ρ ( − λ a λ k , − a λ k − , + y ( k )) . Put z ( k ) = − λ a λ k , − a λ k − , + y ( k ). By the construction of z ( k ), we note that ρ ( z ( k )) belongsto Span O { a λ i , , a λ i , | i = , . . . , n } . Since the restriction of ρ to Span O { a λ i , , a λ i , | i = , . . . , n } is a bijection from Span O { a λ i , , a λ i , | i = , . . . , n } to itself, the equation (3.2.2) implies thatthere exists z ′ ( k ) ∈ Z λn such that z ( k ) = εz ′ ( k ). Then, we have( ρ ( a λ , − z ′ ( )) , . . . , ρ ( a λ n , − z ′ ( n ))) = ( a λ , , . . . , a λ n , ) C . This completes the proof of the statement (1).(2) The statement for n = 1 is clear by (1). In order to prove this statement for n > f ( Y a λ k , − λX a λ k , − X a λ k − , ) in two ways. Since Y a λ k , = ε a λ k , + λ a k , + a λ k − , and a λ k , = X a λ k , , we have(3.2.3) ρ ( Y a λ k , − λX a λ k , − X a λ k − , ) = εf ( a λ k , ) . Now, we assume that A ( k ) = P nl = ( c ( k ) l , a λ l , + c ( k ) l , a λ l , + c ( k ) l , a λ l , ). For k > , the left-handside of (3.2.3) is n − X l = ( c ( k ) l + , − c ( k − ) l , ) a λ l , − c ( k − ) n , a λ n , + ε n X l = c ( k ) l , a λ l , + n − X l = ( ε c ( k ) l , − λ c ( k ) l , − c ( k − ) l , − c ( k ) l + , ) a λ l , + ( ε c ( k ) n , − λ c ( k ) n , − c ( k − ) n , ) a λ n , . Thus, the coefficients c ( k ) l + , − c ( k − ) l , ( l = , . . . , n − ) and c ( k − ) n , belong to ε O . It impliesthat strictly lower triangular entries of the matrix C belong to ε O . On the other hand,by the statement (1), we havedet C ≡ c ( ) , · · · c ( n ) n , + X e = σ ∈ S n c (( σ ( )) , · · · c ( σ ( n )) n , ≡ ε O , where S n is the symmetric group of degree n and e is its identity element. Hence, c ( k ) k , ≡ ε O for some k . Since c ( k + ) k + , − c ( k ) k , ∈ ε O ( k = , . . . , n − ), the assertion follows. (cid:3) For each n >
1, we define an endomorphism Φ λn : Z λn → Z λn by a λ k , (cid:26) a λ n , if k = n , Z λn is given by π n,λ : A n −→ Z λn e i (cid:26) a λ k , if i = 2 k − , k = , , . . . , n , a λ k , if i = 2 k , k = , , . . . , n . Lemma 3.3.
Let Φ λn be the endomorphism of Z λn as above. Then, the following statementshold. (1) Φ λn does not factor through π n,λ .(2) For any ρ ∈ radEnd A ( Z λn ), Φ λn ρ factors through π n,λ . Proof . (1) Suppose that Φ λn factors through the map π n,λ . Let ψ = ( ψ , . . . , ψ n ) : Z λn → A n such that Φ λn = π n,λ ψ . Put ψ k ( a λ i , ) = a ( i ) k, + a ( i ) k, X + a ( i ) k, Y + a ( i ) k, XY.
By comparing coefficients in π n,λ ψ ( a λ k , ) with those in Φ λn ( a λ k , ), we have the followingequations:(3.3.1) εa ( i )2 s − , + a ( i )2 s, − λa ( i )2 s, − a ( i )2 s +2 , = 0 if s = n, (3.3.2) εa ( i )2 n − , + a ( i )2 n, − λa ( i )2 n, = (cid:26) i = n , . On the other hand, as ψ k ( a λ i , ) = Xψ k ( a λ i , ), it follows from εψ k ( a λ i , ) = Y ψ k ( a λ i , ) − λψ k ( a λ i , ) − ψ k ( a λ i − , ) that(3.3.3) εψ k ( a λ i , ) = − ( λa ( i ) k, − a ( i − ) k, ) X + a ( i ) k, Y + ( a ( i ) k, − λa ( i ) k, − a ( i − ) k, ) XY, where a ( ) k, = 0, 1 ≤ k ≤ n and ≤ i ≤ n . In order to obtain a contradiction, we showthat a ( n )2 n, − λa ( n )2 n, ∈ ε O . By the equation (3.3.3), this is equivalent to a ( n − )2 n, ∈ ε O .The equation (3.3.1) implies that a ( n − )2 n, ∈ ε O if and only if a ( n − )2 n − , − λa ( n − )2 n − , ∈ ε O . Byrepeating this procedure, we deduce that the claim is equivalent to a ( )2 , − λa ( )2 , ∈ ε O .However, a ( )2 , − λa ( )2 , ∈ ε O follows from the equation (3.3.3). Now, we obtain1 = εa ( n )2 n − , + a ( n )2 n, − λa ( n )2 n, ∈ ε O , a contradiction.(2) Let ρ ∈ radEnd A ( Z λn ). We put ρ ( a λ k , ) = n X i = ( c ( k ) i , a λ i , + c ( k ) i , a λ i , + c ( k ) i , a λ i , + c ( k ) i , a λ i , ) . Lemma 3.2 yields that there exists f ( k ) n . ∈ O such that ε f ( k ) n . = c ( k ) n , for each k . We definean A -module homomorphism ψ : Z λn → A n by ψ ( a λ k , ) = (0 , . . . , , f ( k ) n , XY, ψ is well-defined and Φ λn ρ ( a λ k , ) = c ( k ) n , a λ n , = π n,λ ψ ( a λ k , ). (cid:3) Summing up, we have obtained the following proposition.
Proposition 3.4.
Consider the following pull-back diagram:0 Z − λn E λn Z λn Z − λn A n Z λn / / / / / / / / / / / / π n,λ / / / / (cid:15) (cid:15) Φ λn (cid:15) (cid:15) Then, the upper exact sequence is the almost split sequence ending at Z ∞ n . Proof . The statement follows from Proposition 1.10 and Lemma 3.3. (cid:3)
N THE HELLER LATTICES 23
The middle term of the almost split sequence ending at Z λn . We denote by E λn the middle term of the almost split sequence ending at Z λn . By Proposition 3.4, the A -lattice E λn is of the form E λn = { ( x, y ) ∈ A n ⊕ Z λn | π n,λ ( x ) = Φ λn ( y ) } . Then, an O -basis of the A -lattice E λn is given as follows: E λn = O ( εe − λXe − Y e ) ⊕ O ( εXe − XY e ) ⊕ O ( Y e + λXe ) ⊕ O ( XY e ) n M k =2 (cid:18) O ( εe k + λXe k − − Y e k − + Xe k − ) ⊕ O ( εXe k − XY e k − ) ⊕ O ( Y e k + λXe k + Xe k − ) ⊕ O ( XY e k ) (cid:19) n − M k = (cid:18) O a λ k , ⊕ O a λ k , ⊕ b λ k , ⊕ O b λ k , (cid:19) ⊕ O ( a λ n , − Xe n ) ⊕ O a λ n , ⊕ O a λ n , ⊕ O a λ n , Lemma 3.5.
The following statements hold.(1) There is an isomorphism E λn ⊗ κ ≃ M ( λ ) n − ⊕ M ( λ ) n +1 ⊕ M ( − λ ) ⊕ n .(2) We have an isomorphism ( τ E λn ) ⊗ κ ≃ M ( − λ ) n +1 ⊕ M ( − λ ) n +1 ⊕ M ( λ ) ⊕ n .(3) E λn is a non-projective indecomposable A -lattice. Proof . (1) We define A -submodules of E λn ⊗ κ as follows. E ( λ, n ) := Span κ ( εe − λXe − Y e ) , ( εXe − XY e )( εe k + λXe k − − Y e k − + Xe k − ) , ( εXe k − XY e k − ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) k = 2 , . . . , n E ( λ, n ) := Span κ (cid:8) a λ k , , a λ k , (cid:12)(cid:12) k = , . . . , n (cid:9) E ( λ, n ) := Span κ ( Y e + λXe ) , ( XY e ) , ( Y e k + λXe k + Xe k − − a λ k − , ) , ( XY e k − a λ k − , ) , ( a λ n , − Xe n ) , a λ n , , (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) k = k = 2 , . . . , n E ( λ, n ) := Span κ (cid:8) a λ k , , a λ k , (cid:12)(cid:12) k = , . . . , n − (cid:9) Then, there are isomorphisms E ∞ n ⊗ κ = E ( λ, n ) ⊕ E ( λ, n ) ⊕ E ( λ, n ) ⊕ E ( λ, n ) and E ( λ, n ) ≃ E ( λ, n ) ≃ M ( − λ ) n , E ( λ, n ) ≃ M ( λ ) n +1 , E ( λ, n ) ≃ M ( λ ) n − . (2) This follows from Lemmas 1.16, 2.4 and the statement (1).(3) Suppose that E λn is decomposable. We write E λn = E ⊕ E with E = 0 = E as A -lattices. Then, the ranks of the A -lattices E and E are divisible by four. The statement(1) implies that E ⊗ κ ≃ M ( − λ ) ⊕ n and E ⊗ κ ≃ M ( λ ) n +1 ⊕ M ( λ ) n − . Assume that n isodd. If n = 1, then, E is not isomorphic to any Heller lattice, and it is indecomposable.Let 0 → τ E → Z − λn ⊕ W → E → E . By Lemma 2.4, we have τ E ⊗ κ ≃ e Ω( M ( λ ) n +1 ) ≃ M ( − λ ) n +1 . On the other hand, theinduced sequence0 → ( τ E ⊗ κ ) → ( Z − λ ⊗ κ ) ⊕ ( W ⊗ κ ) → ( E ⊗ κ ) → n >
1. Then, E ⊗ κ ≃ M ( λ ) n − ⊕ M ( λ ) n +1 and E is indecomposable. Indeed, if E = E , ⊕ E , with E , = 0 = E , as A -lattices and E , ⊗ κ ≃ M ( λ ) n +1 , then we have a splitable exactsequence 0 −→ M ( − λ ) n +1 −→ W ⊕ M ( λ ) n ⊕ M ( − λ ) n −→ M ( λ ) n +1 −→ W ∈ mod- A ⊗ κ , a contradiction. Thus, E is indecomposable. Then, theindecomposability of E λn follows by the same method as in the proof of n = 1.Assume that n is even. Then, E is indecomposable since the rank of any directsummand of E λn is divisible by four. In this case, we can prove the indecomposability of E λn by using similar arguments. (cid:3) Corollary 3.6.
For any n > λ ∈ κ , the Heller component CH ( Z λn ) has no loops.3.3. The Heller component containing Z λn . In this subsection, we determine theshape of CH ( Z λn ). Lemma 3.7.
Let C be a component of stable Auslander–Reiten quiver of A . Then, C hasinfinitely many vertices. Proof . The assertion follows from [AKM, Proposition 1.26] and Theorem 2.5 (cid:3)
Theorem 3.8.
Let O be a complete discrete valuation ring, κ its residue field and A = O [ X, Y ] / ( X , Y ). Assume that κ is algebraically closed, and λ = ∞ .(1) If the characteristic of κ is 2, then CH ( Z λn ) ≃ Z A ∞ / h τ i .(2) If the characteristic of κ is not 2, then CH ( Z n ) ≃ Z A ∞ / h τ i if λ = 0, otherwise CH ( Z λn ) ≃ Z A ∞ / h τ i .Moreover, any Heller lattice Z λn appears on the boundary of CH ( Z λn ). Proof . Lemma 3.5 implies that every Heller lattice Z λn appears on the boundary of CH ( Z λn )(= CH ( Z − λn )). It follows from Proposition 1.18 and Lemma 3.7 that the treeclass T of CH ( Z λn ) is one of A ∞ , B ∞ , C ∞ , D ∞ or A ∞∞ .Let F be the middle term of the almost split sequence ending at E λn . Then, F is thedirect sum of Z − λn and an A -lattice F λn . By Proposition 1.15, we have F λn ⊗ κ ≃ M ( λ ) n +1 ⊕ M ( λ ) n − ⊕ M ( − λ ) n +1 ⊕ M ( − λ ) n − ⊕ M ( λ ) n ⊕ M ( − λ ) n . Suppose that F λn is not indecomposable. Then, there is an indecomposable direct sum-mand W of F λn such that the almost split sequence ending at W is of the form 0 → τ W → E − λn → W →
0. As rank( E λn ) = 8 n , we have rank( W ) = 4 n . If W is a Heller lattice, then W ⊗ κ must be isomorphic to M ( λ ) n ⊕ M ( − λ ) n . Then, F λn /W is indecomposable, and itis not a Heller lattice by Proposition 2.8. Let 0 → τ ( F λn /W ) → E − λn ⊕ G → F λn /W → F λn /W . Then, the induced exact sequence0 → τ F λn /W ⊗ κ → E − λn ⊗ κ ⊕ G ⊗ κ → F λn /W ⊗ κ → W is not a Heller lattice. This implies that the inducedexact sequence 0 → τ W ⊗ κ → E − λn ⊗ κ → W ⊗ κ → N THE HELLER LATTICES 25 splits. However, this situation does not occur for any W . Therefore, F λn is an indecom-posable A -lattice, and T = A ∞ . (cid:3) The case λ = ∞ . The almost split sequence ending at Z ∞ n . In this subsection, we study the almostsplit sequence ending at Z ∞ n . We see that the following lemmas hold as the case of Z λn . Lemma 4.1.
An endomorphism ρ ∈ End A ( Z ∞ n ) is determined by ρ ( b , ) , . . . , ρ ( b , n ). Proof . Since ρ is an A -module homomorphism, we have Xρ ( b , ) = ρ ( X b , ) = ερ ( b , ),and hence ρ ( b , ) = ε − Xρ ( b , ) follows. For k = , we have Xρ ( b k , ) = ρ ( b k , ). Next for ≤ k ≤ n − , the equation Y ρ ( b k , ) = ερ ( b k , ) + ρ ( b k + , )implies that ρ ( b k , ) = ε − ( Y ρ ( b k , ) − Xρ ( b k + , )), and for k = n , we have ρ ( b n , ) = Y ρ ( b n , ). Finally, ρ ( b k , ) is ε − Y ρ ( b k − , ). (cid:3) Lemma 4.2.
Let ρ ∈ radEnd A ( Z ∞ n ). If we write ρ ( b k , ) = n X l = d ( k ) l , b l , + B ( k ) , B ( k ) ∈ Span O { b i , j | j = } , where d ( k ) l , ∈ O , then the following statements hold.(1) det( d ( k ) l , ) l , k ∈ ε O .(2) d ( k ) n , ∈ ε O for all k = , , . . . , n . Proof . (1) We show that any ρ such that the matrix D := ( d ( k ) l , ) l , k is invertible is surjective.As XY b l , = ε b l , holds for l = , . . . , n , we have( ρ ( b , ) , . . . , ρ ( b n , )) = ( b , , . . . , b n , ) D . Hence, b , , . . . , b n , are contained in the image of ρ .Assume that n = 1. By acting X to the both sides of ρ ( b , ) = d ( ) , b , + B ( ), we have ερ ( b , ) = ε d ( ) , b , + XB ( ) . Thus, we get ε d ( ) , b , = ερ ( b , ) − εt b , for some t ∈ O since XB ( ) ∈ ε O b , . It impliesthat b , = ρ (( d ( ) , ) − b , − ( d ( ) , ) − t b , )) . By letting Y act on the both sides of ρ ( b , ) = d ( ) , b , + B ( ), we have ρ ( b , ) = d ( ) , b , + Y B ( ) = d ( ) , b , + s b , = d ( ) , b , + ρ ( s d ( ) , b , )for some s ∈ O since Y A ( ) ∈ O b , , and hence b , = ρ (( d ( ) , ) − b , − ( a d ( ) , ) − s b , )).Therefore, the morphism ρ is surjective.Next, we assume that n >
1. We note that Xρ ( b k , ) = (cid:26) ερ ( b , ) if k = ,ρ ( b k , ) if k = , Y ρ ( b k , ) = (cid:26) ρ ( ε b k , + b k + , ) if k = n ,ρ ( b n , ) if k = n , X n X l = d ( k ) l , b l , + B ( k ) ! = ε d ( k ) , b , + n X l = d ( k ) l , b l , + XB ( k ) ,Y n X l = d ( k ) l , b l , + B ( k ) ! = n − X l = d ( k ) l , ( ε b l , + b l + , ) + d ( k ) n , b n , + Y B ( k ) , and we also note that XB ( k ) and Y B ( k ) belong to Span O { b i , | i = , . . . , n } .Assume that k = . Then, the equality ερ ( b , ) = ε d ( ) , b l , + n X l = d ( ) l , b , + XB ( ) . implies that d ( ) l , ( l = , , . . . , n ) are in ε O and XB ( ) ≡ ε O . Thus, there exists x ( ) ∈ Z ∞ n such that ερ ( x ( )) = XB ( ). If k > , then, for each k , there exists x ( k ) ∈ Z ∞ n such that ρ ( x ( k )) = XB ( k ). Therefore, it is easy to see that( ρ ( b , − x ( )) , . . . , ρ ( b n , − x ( n ))) = ( b , , . . . , b n , ) d ( ) , ε d ( ) , ε d ( n ) , ε − d ( ) , d ( ) , d ( n ) , ... ... · · · ... ε − d ( ) , d ( ) n , d ( n ) n , . Since the determinant of the rightmost matrix in the above equation equals to det D , eachelement b i , belongs to the image of ρ .For each k = , , . . . , n , let y ( k ) and z ( k ) be elements of Z ∞ n such that ρ ( y ( k )) = Y B ( k )and ρ ( z ( k )) = P n − = d ( k ) l , b l + , . Then, we have the equations ερ ( b k , ) = n − X l = ε d ( k ) l , b l , + d ( k ) n , b n , + ρ ( y ( k ) + z ( k ) − b k + , ) for k = , , . . . , n − and ρ ( b n , − y ( n ) − z ( n )) = n − X l = ε d ( n ) l , b l , + d ( n ) n , b n , . As ρ ( y ( k )+ z ( k ) − b k + , ) belongs to Span O { b i , , b i , | i = , . . . , n } , ρ ( y ( k )+ z ( k ) − b k + , ) ≡ ε O . Since the restriction of ρ to Span O { b i , , b i , | i = , . . . , n } is a bijection from Span O { b i , , b i , | i = , . . . , n } to itself, one can define w ( k ) ∈ Z ∞ n by ρ ( w ( k )) := (cid:26) ε − ρ ( y ( k ) + z ( k ) − b k + , ) if k = n ,ρ ( y ( k ) + z ( k )) if k = n . This gives the following equation:( ρ ( b , − w ( )) , . . . , ρ ( b n , − w ( n ))) = ( b , , . . . , b n , ) d ( ) , d ( n − ) , ε d ( n ) , ... · · · ... ... d ( ) n − , d ( n − ) n − , ε d ( n ) n − , ε − d ( ) n , ε − d ( n − ) n , d ( n ) n , Since the determinant of the rightmost matrix in the above equation equals to det D , eachelement b i , belongs to the image of ρ . Therefore, the A -morphism ρ is surjective. N THE HELLER LATTICES 27 (2) The statement for n = 1 is clear by (1). In order to prove this statement for n >
1, we compute ρ ( Y b k , − X b k + , ), for k = , , . . . , n − , in two ways. Set W ( k ) = Y B ( k ) − XB ( k + ). Since Y b k , = ε b k , + b k + , and b k , = X b k , , we have ρ ( Y b k , − X b k + , ) = ερ ( b k , ) . On the other hand, we have ρ ( Y b k , − X b k + , ) = − ε d ( k + ) , b , + n X l = ( d ( k ) l − , − d ( k + ) l , ) b l , + n − X l = ε d ( k ) l , b l , + d ( k ) n , b n , + W ( k ) . Thus, we have d ( k ) l − , − d ( k + ) l , ≡ ε O , d ( k ) n , ≡ ε O l = , . . . , n , k = , . . . , n − . This means the strictly lower entries of the matrix D belong to ε O . By the statement (1),det D ≡ d ( ) , d ( ) , · · · d ( n ) n , + X e = σ ∈ S n sgn( σ ) d ( σ ( )) , · · · d ( σ ( n )) n , ≡ d ( ) , d ( ) , · · · d ( n ) n , mod ε O , where S n is the symmetric group of degree n and e is its identity element. Now, the claimis clear. (cid:3) Recall that the projective cover of Z ∞ n is given by π n, ∞ : A n −→ Z ∞ n e i b , if i = 1 , b , if i = 2, b k , if i = 2 k + 1, k = , , . . . , n − , b k , if i = 2 k , k = , , . . . , n . Now, for each n ≥
1, we define an endomorphism Φ ∞ n : Z ∞ n → Z ∞ n by b k , (cid:26) b n , if k = n ,0 otherwise.Clearly, Φ ∞ n gives an endomorphism of Z ∞ n . First, we construct the almost split sequenceending at Z ∞ by using Φ ∞ . Lemma 4.3.
Let Φ ∞ : Z ∞ → Z ∞ as above. Then, the following statements hold.(1) Φ ∞ does not factor through π , ∞ .(2) For any ρ ∈ radEnd A ( Z ∞ ), Φ ∞ ρ factors through π , ∞ . Proof . (1) Suppose that there exists ψ = ( ψ , ψ ) : Z ∞ → A ⊕ A such that Φ ∞ = π , ∞ ψ .Then, we have(4.3.1) b , = Φ ∞ ( b , ) = π , ∞ ψ ( b , ) = ψ , ∞ ( b , ) b , + ψ ( b , ) b , . If we put ψ ( b , ) = a + a X + a Y + a XY, ψ ( b , ) = b + b X + b Y + b XY, where a , . . . , a , b , . . . , b ∈ O , the rightmost side of (4.3.1) equals to a b , + ( εa + b ) b , + a b , + ( εa + b ) b , . Thus, we have ψ ( b , ) = − εa + b X + (1 − εa ) Y + b XY . Multiplying X to ψ ( b , ),we have εψ ( b , ) = Xψ ( b , ) = − εa X + (1 − εa ) XY, a contradiction.(2) Let ρ ∈ rad End A ( Z ∞ ). We write ρ ( b , ) = α b , + B ( ), where α ∈ O and B ( ) ∈ Span O { b , , b , , b , } . By Lemma 4.2, α = εα ′ for some α ′ ∈ O . Define an A -modulehomomorphism ψ : Z ∞ → A ⊕ A by ψ ( b , ) = α ′ XY e . Then, since π , ∞ ( α ′ XY e ) = α ′ XY b , = εα ′ b , , we have Φ ∞ f ( b , ) = α b , = π , ∞ ψ ( b , ). (cid:3) From now on, we construct the almost split sequence ending at Z ∞ n for n ≥ Lemma 4.4.
Let Φ ∞ n : Z ∞ n → Z ∞ n as above. Then, the following statements hold.(1) Φ ∞ n does not factor through π n, ∞ .(2) For any ρ ∈ radEnd A ( Z ∞ n ), Φ ∞ n ρ factors through π n, ∞ . Proof . (1) Suppose that there exists ψ = ( ψ k ) k =1 ,..., n : Z ∞ n → A n such that Φ ∞ n = π n, ∞ ψ . We put ψ l ( b k , ) = a ( k ) l, + a ( k ) l, X + a ( k ) l, Y + a ( k ) l, XY.
Then, we notice that, for all k = , . . . , n and l = 1 , . . . , n , a ( k ) l, belongs to ε O since XY b k , = ε b k , for all k = , . . . , n . By comparing the coefficient of b n , in Φ ∞ n ( b n , ) withthat in π n, ∞ ψ ( b n , ), we have εa ( n )2 n, − a ( n )2 n − , = 1 . In order to obtain a contradiction weshow that a ( n )2 n − , ∈ ε O .For s = , . . . , n and t = , . . . , n − , by comparing the coefficient of b t , in Φ ∞ n ( b s , )with that in π n, ∞ ψ ( b s , ), we obtain the following equations: εa ( s )1 , + a ( s )2 , + a ( s )3 , = 0 t = , (4.4.1) − a ( s )2 t − , + εa ( s )2 t, + a ( s )2 t +1 , = 0 t > . (4.4.2)On the other hand, for t = 1 , . . . , n , the following equations hold: ψ t ( b s , ) = Xψ t ( b s , ) = a ( s ) t, X + a ( s ) t, XY s = ( ∗ ) εψ t ( b s , ) + ψ t ( b s + , ) = Y ψ t ( b s , ) = a ( s ) t, Y + a ( s ) t, XY s = n ( ∗∗ )In particular, it follows from ( ∗ ) that ψ n − ( b n , ) = a ( n )2 n − , X + a ( n )2 n − , XY holds. As a ( n )2 n − , ∈ ε O , a ( n )2 n − , belongs to ε O if and only if ψ n − ( b n , ) belongs to εA . It is equivalentto a ( n − )2 n − , ∈ ε O by the equation ( ∗∗ ). Then, it follows from the equation (4.4.2) that a ( n − )2 n − , ∈ ε O if and only if a ( n − )2 n − , ∈ ε O . By repeating this procedure, we deduce that a ( n )2 n − , ∈ ε O if and only if a ( )3 , ∈ ε O . Since εψ ( b , ) = Xψ ( b , ) = a ( )2 , X + a ( )2 , XY , a ( )2 , belongs to ε O . It implies that a ( )3 , ∈ ε O by (4.4.1).(2) Let ρ ∈ rad End A ( Z ∞ n ). We put ρ ( b k , ) = n X l = d ( k ) l , b l , + B ( k ) , where B ( k ) ∈ Span O { b i , j | j = } . By Lemma 4.2, there are e ( k ) n , such that d ( k ) n , = ε e ( k ) n , .Define an A -module homomorphism ψ = ( ψ k ) k =1 ,..., n : Z ∞ n → A n by ψ ( b k , ) =(0 , . . . , , e ( k ) n , XY ). Then, it is easy to check that Φ ∞ n ρ ( b k , ) = d ( k ) n , b n , = π n, ∞ ψ ( b k , ). (cid:3) Summing up, we obtain the following proposition.
N THE HELLER LATTICES 29
Proposition 4.5.
Consider the following pull-back diagram:0 Z ∞ n E ∞ n Z ∞ n Z ∞ n A n Z ∞ n / / / / / / / / / / / / π n, ∞ / / / / (cid:15) (cid:15) Φ ∞ n (cid:15) (cid:15) Then, the upper exact sequence is the almost split sequence ending at Z ∞ n . Proof . The statement follows from Proposition 1.10 and Lemmas 4.3 and 4.4. (cid:3)
The middle term of the almost split sequence ending at Z ∞ n . In this subsec-tion, we study the middle term of the almost split sequence ending at Z ∞ n , say E ∞ n , andexplain some properties of E ∞ n . Lemma 4.6. (1) An O -basis of E ∞ is given by O ( εe − Xe ) ⊕O Xe ⊕O ( εY e − XY e ) ⊕O XY e ⊕O ( b , + Y e ) ⊕O b , ⊕O b , ⊕O b , . (2) There is an isomorphism E ∞ ⊗ κ ≃ M ( ∞ ) ⊕ ⊕ M ( ∞ ) .(3) We have an isomorphism ( τ E ∞ ) ⊗ κ ≃ M ( ∞ ) ⊕ ⊕ M ( ∞ ) .(4) E ∞ is a non-projective indecomposable A -lattice. Proof . (1) Straightforward.(2) We put E ( ∞ , := Span κ { ( εe − Xe ) , ( εY e − XY e ) } ,E ( ∞ , := Span κ { b , , b , } ,E ( ∞ , := Span κ { ( Xe ) , ( XY e ) , ( b , + Y e ) , b , } . Then, it is easy to check that E ( ∞ , ≃ E ( ∞ , ≃ M ( ∞ ) and E ( ∞ , ≃ M ( ∞ ) .(3) This follows from Lemmas 1.16, 2.4 and the statement (2).(4) Suppose that E ∞ is decomposable. We write E ∞ = E ⊕ E as A -lattices with E = 0 = E . Then, the ranks of E and E are divisible by four. Thus, one can assumethat E ⊗ κ ≃ M ( ∞ ) ⊕ , E ⊗ κ ≃ M ( ∞ ) , and E and E are indecomposable. Then,the A -lattice E is not isomorphic to any Heller lattices by Theorem 2.5 and Proposition2.8. Let 0 → τ E → Z ∞ ⊕ W → E → E . Byapplying − ⊗ κ , the induced sequence0 → τ E ⊗ κ → Z ∞ ⊗ κ ⊕ W ⊗ κ → E ⊗ κ → (cid:3) By the definition of E λ , we have E ∞ = O ( εe − Xe ) ⊕ O ( Xe ) ⊕ O ( Xe − Y e ) ⊕ O ( XY e ) ⊕ O ( εe + Xe − Y e ) ⊕ O ( εXe − XY e ) ⊕ O ( εY e + XY e ) ⊕ O ( XY e ) ⊕ O b , ⊕ O b , ⊕ O b , ⊕ O b , ⊕ O ( b , − Y e ) ⊕ O b , ⊕ O b , ⊕ O b , . Lemma 4.7.
The following statements hold.(1) There is an isomorphism E ∞ ⊗ κ ≃ ⊕ M ( ∞ ) ⊕ ⊕ M ( ∞ ) ⊕ M ( ∞ ) .(2) We have an isomorphism ( τ n E ∞ ) ⊗ κ ≃ M ( ∞ ) ⊕ ⊕ M ( ∞ ) ⊕ M ( ∞ ) . (3) E ∞ is a non-projective indecomposable A -lattice. Proof . (1) We put E ( ∞ , := Span κ { ( εe − Xe ) , ( εXe − XY e ) , ( εe + Xe − Y e ) , ( εY e + XY e ) } ,E ( ∞ , := Span κ { b , , b , , b , , b , } ,E ( ∞ , := Span κ { ( Xe ) , ( Xe − Y e − b , ) , ( XY e ) , ( XY e − b , ) , ( b , − Y e ) , b , } E ( ∞ , := Span κ { b , , b , } Then, it is easy to check that E ( ∞ , ≃ E ( ∞ , ≃ M ( ∞ ) , E ( ∞ , ≃ M ( ∞ ) and E ( ∞ , ≃ M ( ∞ ) .(2) This follows from Lemmas 1.16, 2.4 and the statement (1).(3) Suppose that E ∞ is decomposable. We write E ∞ ≃ E ⊕ E as A -lattices with E =0 = E . Then, we may assume that E ⊗ κ ≃ M ( ∞ ) ⊕ and E ⊗ κ ≃ M ( ∞ ) ⊕ M ( ∞ ) .Note that the A -lattice E is not isomorphic to any Heller lattices and it is indecomposable.Let 0 → τ E → Z ∞ ⊕ W → E → E . It followsfrom Lemma 2.4 that ( τ E ) ⊗ κ ≃ e Ω( M ( ∞ ) ⊕ M ( ∞ ) ) ≃ M ( ∞ ) ⊕ M ( ∞ ) . Then,the induced sequence 0 → τ E ⊗ κ → ( Z ∞ ⊗ κ ) ⊕ ( W ⊗ κ ) → E ⊗ κ → (cid:3) From now on, we assume that n >
2. Then, an O -basis of the A -lattice E ∞ n is given asfollows: E ∞ n = O ( εe − Xe ) ⊕ O ( Xe ) ⊕ O ( Y e − Xe ) ⊕ O ( XY e ) ⊕ O ( εe + Xf − Y e ) ⊕ O ( εXe − XY e ) ⊕ O ( Y e + Xe ) ⊕ O ( XY e ) n − M k =1 (cid:18) O ( εe k +3 + Xe k +4 − Y e k +2 ) ⊕ O ( εXe k +3 − XY e k +2 ) ⊕ O ( Y e k +3 + Xe k +5 ) ⊕ O ( XY e k +3 ) (cid:19) ⊕ O ( εe n − + Xe n − Y e n − ) ⊕ O ( εXe n − − XY e n − ) ⊕ O ( εY e n − + XY e n ) ⊕ O ( XY e n − ) n − M k = (cid:18) O b k , ⊕ O b k , ⊕ b k , ⊕ O b k , (cid:19) ⊕ O ( b n , − Y n − ) ⊕ O b n , ⊕ O b n , ⊕ O b n , Lemma 4.8.
The following statements hold.(1) There is an isomorphism E ∞ n ⊗ κ ≃ M ( ∞ ) ⊕ n ⊕ M ( ∞ ) n +1 ⊕ M ( ∞ ) n − .(2) We have an isomorphism ( τ E ∞ n ) ⊗ κ ≃ M ( ∞ ) ⊕ n ⊕ M ( ∞ ) n − ⊕ M ( ∞ ) n +1 .(3) E ∞ n is a non-projective indecomposable A -lattice. N THE HELLER LATTICES 31
Proof . (1) The statement is true for n = 1 , n >
2. We define A -submodules of E ∞ n ⊗ κ as follows. E ( ∞ , n ) := Span κ ( εe − Xe ) , ( εXe − XY e )( εe k +1 + Xe k +2 − Y e k − ) , ( εXe l +3 − XY e l +2 ) , ( εY e n − + XY e n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) k = 1 , . . . , n − ,l = 1 , . . . , n − E ( ∞ , n ) := Span κ (cid:26) b , , b k , , b l , (cid:12)(cid:12)(cid:12)(cid:12) k = , . . . , n − , l = , . . . , n (cid:27) E ( ∞ , n ) := Span κ Xe , XY e , ( Y e − Xe + b , ) , ( Y e k +1 + Xe k +3 + b k + , ) , ( XY e l +1 − b l + , ) , ( b n , − Y e n − ) , b n , , (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) k = k = 1 , . . . , n − ,l = l = 1 , . . . , n − E ( ∞ , n ) := Span κ (cid:26) b s , , b t , (cid:12)(cid:12)(cid:12)(cid:12) s = , . . . , n − , t = , . . . , n (cid:27) Then, it is easy to check that E ∞ n ⊗ κ = E ( ∞ , n ) ⊕ E ( ∞ , n ) ⊕ E ( ∞ , n ) ⊕ E ( ∞ , n ) ,E ( ∞ , n ) ≃ E ( ∞ , n ) ≃ M ( ∞ ) n ,E ( ∞ , n ) ≃ M ( ∞ ) n +1 ,E ( ∞ , n ) ≃ M ( ∞ ) n − . (2) This follows from Lemmas 1.16, 2.4 and the statement (1).(3) We can prove the indecomposability of E λn by using similar arguments of the proofof the case λ = ∞ . (cid:3) Corollary 4.9. CH ( Z ∞ n ) = CH ( Z ∞ m ) whenever n = m . Moreover, CH ( Z ∞ n ) has no loops.4.3. The Heller component containing Z ∞ n .Theorem 4.10. Let O be a complete discrete valuation ring, κ its residue field and A = O [ X, Y ] / ( X , Y ). Assume that κ is algebraically closed. Then, CH ( Z ∞ n ) ≃ Z A ∞ / h τ i .Moreover, the Heller lattice Z ∞ n appears on the boundary of CH ( Z ∞ n ). Proof . Lemmas 4.6, 4.7 and 4.8 imply that every Heller lattice Z ∞ n appears on theboundary of CH ( Z ∞ n ). It follows from Proposition 1.18 and Lemma 3.7 that the tree class T of CH ( Z ∞ n ) is one of A ∞ , B ∞ , C ∞ , D ∞ or A ∞∞ .Let F be the middle term of the almost split sequence ending at E ∞ n . Then, F is thedirect sum of Z ∞ n and an A -lattice F ∞ n . By Proposition 1.15, we have F ∞ n ⊗ κ ≃ M ( ∞ ) ⊕ n +1 ⊕ M ( ∞ ) ⊕ n − ⊕ M ( ∞ ) ⊕ n . Suppose that F ∞ n is not indecomposable. Then, there is an indecomposable direct sum-mand W of F ∞ n such that the almost split sequence ending at W is of the form 0 → τ W → E ∞ n → W →
0. As rank( E ∞ n ) = 8 n , we have rank( W ) = 4 n . If W is a Heller lattice, then W ⊗ κ must be isomorphic to M ( ∞ ) n ⊕ M ( ∞ ) n . Then, F ∞ n /W is indecomposable, and it is not a Heller lattice by Proposition 2.8. Let 0 → τ ( F ∞ n /W ) → E ∞ n ⊕ G → F ∞ n /W → F ∞ n /W . Then, the induced exact sequence0 → τ F ∞ n /W ⊗ κ → E ∞ n ⊗ κ ⊕ G ⊗ κ → F ∞ n /W ⊗ κ → W is not a Heller lattice. This implies that the inducedexact sequence 0 → τ W ⊗ κ → E ∞ n ⊗ κ → W ⊗ κ → W . Therefore, F ∞ n is an indecom-posable A -lattice, and T = A ∞ . (cid:3) Appendix A. Dynkin and Euclidean diagrams
We list Dynkin and Euclidean diagrams. The following labelled undirected graphs arecalled finite Dynkin diagrams . A n • • • · · · • • B n • • • · · · • • (1 , C n • • • · · · • • (2 , D n • • •• · · · • • ♦♦♦♦ ❖❖❖❖ ( n vertices and n ≥ E • • • •• • E • • • •• • • E • • • •• • • • F • • • • (2 , G • • (3 , The following are infinite Dynkin diagrams . A ∞ • • • · · · B ∞ • • • · · · (1 , C ∞ • • • · · · (2 , D ∞ • • •• · · · ♦♦♦♦ ❖❖❖❖ A ∞∞ · · · • • • · · · N THE HELLER LATTICES 33
The following are
Euclidean diagrams . e A n • • • · · ·• • • ❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥ ❖❖❖❖❖❖❖❖❖❖❖ e B n • • • · · · • • (1 ,
2) (2 , e C n • • • · · · • • (2 ,
1) (1 , e D n • •• · · · • •• ♦♦♦♦ ❖❖❖❖ ♦♦♦♦ ❖❖❖❖ g BC n • • • · · · • • (1 ,
2) (1 , g BD n • • · · · • •• (1 , ♦♦♦♦ ❖❖❖❖ g CD n • • · · · • •• (2 , ♦♦♦♦ ❖❖❖❖ e E • • • ••• • e E • • • •• • • • e E • • • • •• • • • e F , • • • • • (2 , e G , • • • (1 , e G , • • • (3 , e A , • • (1 , e A , • • (2 , Here, we note that e A is a single loop with one vertex and e A is the underlying graph ofthe Kronecker quiver. References [A1] M. Auslander, Functors and morphisms determined by objects, Proc. Conf. on RepresentationTheory, Philadelphia, Lecture Notes in Pure and Applied Math. , Marcel Dekker, 1978.[A2] M. Auslander, Isolated singularities and existence of almost split sequences, Representation TheoryII (Ottawa, Ont., 1984), Lecture Note in Math., vol. 1178, Springer-Berlin (1986), 194–242.[A3] M. Auslander, Rational singularities and almost split sequences, Trans. Amer. Math. Soc. (1986), no. 2, 511–531.[AKM] S. Ariki, R. Kase and K. 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