The cluster character for cyclic quivers

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by  Ming Ding

THE CLUSTER CHARACTER FOR CYCLIC QUIVERS
MINGDINGANDFANXU
0
1 Abstract. We define an analogue of the Caldero-Chapoton map ([CC]) for
0 theclustercategoryoffinitedimensionalnilpotentrepresentationsoveracyclic
2 quiver. We prove that it is a cluster character (in the sense of [Pa]) and
satisfiessomeinductiveformulasforthemultiplicationbetweenthegeneralized
n
clustervariables(theimagesofobjectsoftheclustercategoryunderthemap).
a Moreover,weconstructaZ-basisforthealgebrasgeneratedbyallgeneralized
J
clustervariables.

5
2
Contents

]
T 1. Introduction 1
R 2. The cluster character for cyclic quivers 2
. 3. Inductive multiplication formulas 6
h
4. A Z-basis for cyclic quivers 8
t
a Acknowledgements 10
m
References 10
[

1
v
0 1. Introduction
6
3 ClusteralgebraswereintroducedbyS.FominandA.Zelevinsky[FZ]inorderto
4 developacombinatorialapproachto study problemsoftotalpositivityin algebraic
.
1 groups and canonical bases in quantum groups. The link between cluster algebras
0 andrepresentationtheoryofquiverswerefirstrevealedin[MRZ]. In[BMRRT],the
0
authors introduced the cluster category of an acyclic quiver Q (a quiver without
1
oriented cycles) as the categorification of the corresponding cluster algebras. In
:
v order to show that a cluster category categorifiesthe involving cluster algebra,the
i
X Caldero-Chapoton map was defined by P. Caldero and F. Chapoton in [CC]. Let
C(Q) be the cluster category associated to an acyclic quiver Q (a quiver without
r
a oriented cycles). The Caldero-Chapotonmap of an acyclic quiver Q is a map
XQ :obj(C(Q))→Q(x ,··· ,x ).
? 1 n
The map was extensively defined by Y. Palu for a Hom-finite 2-Calabi-Yau trian-
gulated categories with a cluster tilting object ([Pa]).
As in [Ke], the cluster category can be defined for any small hereditary abelian
categorywithfinitedimensionalHom-andExt-spaces. Itisinterestingtostudythe
cluster categorieswithout cluster tilting objects andthe involving cluster algebras.
For example, the cluster category of a 1-Calabi-Yau abelian category contains no
cluster tilting objects (even no rigid objects).
Keywordsandphrases: cyclicquiver,clusteralgebra,Z-basis.
FanXuwassupportedbyAlexandervonHumboldtStiftungandwasalsopartiallysupported
bythePh.D.ProgramsFoundation ofMinistryofEducationofChina(No. 200800030058).
1
2 MINGDINGANDFANXU
In this paper, we will focus on the simplest example of the cluster category
without cluster tilting objects: the cluster category of a cyclic quiver. We first
define an analogue of the Caldero-Chapoton map for a cyclic quiver. We prove a
multiplication formula analogous to the cluster multiplication theorem for acyclic
cluster algebras ([CK], [XX]). As a corollary, the map is a cluster character in
the sense of [Pa]. Let A be the cyclic quiver with r vertices and C(A ) be its
r r
cluster category. Let AHebe the subalgebra of Q(x ,··· ,x ) generatedeby {X |
1 r M
M ∈ modCA } and EH be the subalgebra of AH generated by {X | M ∈
r M
modCA ,Exte1 (M,M)=0}.WecallthealgebraEHtheclusteralgebraofA .
r e r
e C(An) e
We show that AH coincides with EH and construct a Z-basis of EH.
2. The cluster character for cyclic quivers
Let k = C be the field of complex number. Throughout the rest part of this
paper, we fix a cyclic quiver Q = A , i.e, a quiver with oriented cycles, where
r
e
Q ={1,2,··· ,r}:
0
qqqqqqq88 r PPPPPPP((
1oo ··· oo r−1
We denote by modkQ the category of finite-dimensional nilpotent representations
ofkQ. Letτ betheAuslander-Reitentranslationfunctor. LetE ,··· ,E besimple
1 r
modules of the vertices 1,··· ,r, respectively. Set dimE = s for i = 1,··· ,r. We
i i
have τE = E ,··· ,τE = E The Auslander-Reiten quiver of kQ is a tube of
2 1 1 r
rank r with E ,··· ,E lying the mouth of the tube. For 1 ≤ i ≤ r, we denote
1 r
by E = E , and by E [n] the unique nilpotent indecomposable representation
i i+r i
with socle E and length n. Set E [0] = 0 for i = 1,··· ,r. We note that that any
i i
indecomposable kQ-module is of the form E [j] for i = 1,...,r and j ∈ N⊔{0}.
i
Let x = {x |i ∈ Q } is a family indeterminates over Z and set x = x for
i 0 i i+mr
1≤i≤r,m∈Z . Here we denote by Z =Z⊔{0}.
≥0 ≥0
Bydefinition,theclustercategoryC =C(Q)istheorbitcategoryDb(modkQ)/τ◦
[−1]. It is a triangulatedcategoryby [Ke, Theorem9.9]. Different fromthe cluster
categoryofanacyclicquiver,thesetofobjectsinC coincideswiththesetofobjects
in modkQ. Also, for any two indecomposable objects M,N ∈C, we have
Ext1(M,N)=Ext1 (M,N)⊕Ext1 (N,M).
C kQ kQ
It is possible that both of two terms in the right side don’t vanish. We denote by
h−,−i the Euler form on modkQ, i.e., for any M,N in modkQ,
hdimM,dimNi:=dim Hom (M,N)−dim Ext1 (M,N).
k kQ k kQ
It is well defined. Thus according to the Caldero-Chapoton map, we can similarly
define a map on modkQ
X :obj(modkQ)−→Z[x±1]
?
by mapping M to
XM =Xχ(Gre(M)) Y x−i he,sii−hsi,dimM−ei
e i∈Q0
where Gr (M) is the e-Grassmannian of M, i.e., the variety of finite-dimensional
e
nilpotent submodules of M with dimension vector e, and set X = 1. Here, we
0
need not assume that M is indecomposable by the following Proposition 2.2. The
fraction X is called a generalized cluster variable.
M
THE CLUSTER CHARACTERS FOR CYCLIC QUIVERS 3
Proposition 2.1. With the above notation, we have
n−1
x x x x
l+n l+n l+r−1 l+r−1
XEl[n] = x +X x x + x .
l l+k−1 l+k l+n−1
k=1
for n∈N and l=1,··· ,r.
Proof. It is known that all submodules of E[n] are E [0],E [1],··· ,E[n]. Set
l l l l
d =dimE [j]. By definition,
i,j i
n
XEl[n] =X Y x−i hdl,k,di,1i−hdi,1,dl+k,n−ki.
k=0i∈Q0
By the definition of the Euler form, we have
− d ,d − d ,d =−dim Hom(E [k],E )+dim Ext1(E [k],E )
(cid:10) l,k i,1(cid:11) (cid:10) i,1 l+k,n−k(cid:11) k l i k l i
−dim Hom(E ,E [n−k])+dim Ext1(E ,E [n−k]).
k i l+k k i l+k
If k =0, then x−hdl,k,di,1i−hdi,1,dl+k,n−ki = xl+n.
Qi∈Q0 i xl
If k =n, then x−hdl,k,di,1i−hdi,1,dl+k,n−ki = xl+r−1.
Qi∈Q0 i xl+n−1
If 0<k <n, then x−hdl,k,di,1i−hdi,1,dl+k,n−ki = xl+nxl+r−1. (cid:3)
Qi∈Q0 i xl+k−1xl+k
Proposition 2.2. (1) For M,N in modkQ, we have
X X =X .
M N M⊕N
(2) Let 0−→τM −→B −→M −→0 be an almost split sequence in modkQ, then
X X =X +1.
M τM B
Proof. The proof is similar to [CC, Proposition 3.6]. For (1), by definition, it is
enough to prove that for any dimension vector e, we have
χ(Gre(M ⊕N))= X χ(Grf(M))χ(Grg(N)).
f+g=e
Consider the natural morphism of varieties
f : Y Grf(M)×Grg(N)→Gre(M ⊕N)
f+g=e
defined by sending (M ,N ) to M ⊕N . Since f is monomorphism, we have
1 1 1 1
X χ(Grf(M))χ(Grg(N))=χ(Imf).
f+g=e
On the other hand, we define an action of C∗ on Gr (M ⊕N) by
e
t.(m,n)=(tm,t2n)
for t ∈ C∗ and m ∈ M,n ∈ N. The set of stable points is just Imf. Hence,
χ(Imf)=χ(Gr (M ⊕N)). This proves (1).
e
(2) Assume that M =E [n]. It is enough to prove:
i
X X =X X +1
E1[n] E2[n] E1[n+1] E2[n−1]
where 0 −→ E [n] −→ E [n−1]⊕E [n+1] −→ E [n] −→ 0 is an almost split
1 2 1 2
sequence in modkQ. The equation in (2) follows from the direct confirmation by
using Proposition 2.1. (cid:3)
4 MINGDINGANDFANXU
Let M,N be indecomposable kQ−modules satisfying that dim Ext1 (M,N)=
k kQ
dim Hom (N,τM) = 1. Assume that M = E [j], N = E [l]. Then in C(Q),
k kQ i k
there are just two involving triangles
E [l]→E →E [j]→τE [l]
k i k
and
E [j]→E′ →E [l]−→g τE [j]
i k i
where E ∼=E [i+j−k]⊕E [k+l−i] and E′ ∼=kerg⊕τ−1cokerg.
k i
Theorem 2.3. With the above notation, we have
XMXN =XE +XE′.
Proof. Assume that
XM =Xχ(Gre(M)) Y x−i he,sii−hsi,dimM−ei
e i∈Q0
and
XN =Xχ(Gre′(N)) Y xi−he′,sii−hsi,dimN−e′i
e′ i∈Q0
Then
XMXN =X Y x−i he+e′,sii−hsi,dimM+dimN−(e+e′)i.
e,e′ i∈Q0
Note that χ(Gr (L)) = 1 or 0 for any kQ-module L. Since Ext1 (M,N) 6= 0, we
e kQ
have a short exact sequence
0→E [l]−f→1 E −f→2 E [j]→0.
k i
Define a morphism of varieties
φ:Grd(E)→ G Gre(M)×Gre′(N)
e+e′=d
by sending (E1) to (f1−1(E1),f2(E1)). For (M1,N1)∈Gre(M)×Gre′(N), we con-
sider the natural map:
β′ :Ext1 (M,N)⊕Ext1 (M ,N )→Ext1 (M ,N)
kQ kQ 1 1 kQ 1
sending (ε,ε′) to ε −ε′ where ε and ε′ are induced by including M ⊆ M
M1 N M1 N 1
and N ⊆N, respectively and the projection
1
p :Ext1 (M,N)⊕Ext1 (M ,N )→Ext1 (M,N).
0 kQ kQ 1 1 kQ
Itiseasytocheckthat(M ,N )∈Imφifandonlyifp (kerβ′)6=0.Hence,wehave
1 1 0
XE = X Y xi−he+e′,sii−hsi,dimM+dimN−(e+e′)i.
e,e′;p0(kerβ′)6=0i∈Q0
Assume that
XE′ = X χ(Grd′1(K))χ(Grd′2(τ−1C)) Y x−i hd′1+d′2,sii−hsi,dimK+dimτ−1C−d′1−d′2i
d′1,d′2 i∈Q0
Set d∗ =dimM −dimτ−1C. We have
s ,dimK+dimτ−1C
(cid:10) i (cid:11)
= hs ,dimN −τ(d∗)+dimM −d∗i
i
= hs ,dimM +dimN −d∗i+hd∗,s i.
i i
THE CLUSTER CHARACTERS FOR CYCLIC QUIVERS 5
Hence, XE′ can be reformulated as
X Y x−hd′1+d′2+d∗,sii−hsi,dimM+dimN−(d′1+d′2+d∗)i
i
d′1,d′2i∈Q0
Since dim Hom (N,τM) = 1, there is only one element in PHom (N,τM)
k kQ kQ
with the representative g. We have a long exact sequence
g
0 // K // N //τM //C //0
Given submodules K ,C of K,C, respectively, we have the commutative diagram
1 1
g
0 //K //N // τM //C //0
OO OO
(cid:15)(cid:15) (cid:15)(cid:15) g′
0 //K/K1 //N/K1 //τM1 // C1 //0
where τM is the corresponding pullback. Define a morphism of varieties
1
φ′ : G Grd′(K)×Grd′(τ−1C)→ G Gre(M)×Gre′(N)
1 2
d′+d′+d∗=d′ e+e′=d′
1 2
by sending (K ,τ−1(C )) to (K ,M ). Checkingthe abovediagram,we know that
1 1 1 1
(M ,N )∈Imφ′ if and only if Hom (N/N ,τM )6=0. Therefore, we obtain
1 1 kQ 1 1
XE′ = X Y x−i he+e′,sii−hsi,dimM+dimN−(e+e′)i.
e,e′;HomkQ(N/N1,τM1)6=0i∈Q0
Consider the dual of β′:
β :Hom (N,τM )→Hom (N,τM)⊕Hom (N ,τM ).
kQ 1 kQ kQ 1 1
Then
(p0(kerβ′))⊥ =Imβ\Hom(N,τM)≃Hom(N/N1,τM1).
We obtain that
dim (p (kerβ′))+dim Hom(N/N ,τM )=dim Ext1 (M,N)=1
k 0 k 1 1 k kQ
Hence, any (M ,N ) belongs to either Imφ or Imφ′ for some d or d′. We complete
1 1
the proof. (cid:3)
Following the definition of a cluster character in [Pa], we can easily check the
following corollary.
Corollary 2.4. The Caldero-Chapoton map for a cyclic quiver is a cluster char-
acter.
We will construct some inductive formulasin the next section. For convenience,
we write down the following corollary.
Corollary 2.5. With the above notation, we have
(1) X X =X +X
Ei+n Ei[n] Ei[n+1] Ei[n−1]
(2) X X =X +X .
Ei Ei+1[n] Ei[n+1] Ei+2[n−1]
6 MINGDINGANDFANXU
3. Inductive multiplication formulas
Inthissection,wewillgiveinductivemultiplicationformulasforanytwogeneral-
izedclustervariablesonmodkQ. Notethattheseinductivemultiplicationformulas
are an analogue of those for tubes in [DXX] for acyclic cluster algebras.
Theorem 3.1. Let i,j,k,l,m and r be in Z such that 1 ≤ k ≤ mr+l,0 ≤ l ≤
r−1,1≤i,j ≤r,m≥0.
(1)When j ≤i, then
1)for k+i≥r+j, we have X X =X X +
Ei[k] Ej[mr+l] Ei[(m+1)r+l+j−i] Ej[k+i−r−j]
X X ,
Ei[r+j−i−1] Ek+i+1[(m+1)r+l+j−k−i−1]
2)for k + i < r + j and i ≤ l + j ≤ k + i − 1, we have X X =
Ei[k] Ej[mr+l]
X X +X X ,
Ej[mr+k+i−j] Ei[l+j−i] Ej[mr+i−j−1] El+j+1[k+i−l−j−1]
3)for other conditions, i.e, there are no extension between E [k] and E [mr+l],
i j
we have X X =X .
Ei[k] Ej[mr+l] Ei[k]⊕Ej[mr+l]
(2)When j >i, then
1)for k ≥j−i, we have X X =X X +
Ei[k] Ej[mr+l] Ei[j−i−1] Ek+i+1[mr+l+j−k−i−1]
X X ,
Ei[mr+l+j−i] Ej[k+i−j]
2)for k < j − i and i ≤ l + j − r ≤ k + i − 1, we have X X =
Ei[k] Ej[mr+l]
X X +X X ,
Ej[(m+1)r+k+i−j] Ei[l+j−r−i] Ej[(m+1)r+i−j−1] El+j+1[k+r+i−l−j−1]
3)for other conditions, i.e, there are no extension between E [k] and E [mr+l],
i j
we have X X =X .
Ei[k] Ej[mr+l] Ei[k]⊕Ej[mr+l]
Proof. We only prove (1) and (2) is totally similar to (1).
1) When k =1, by k+i≥r+j and 1≤j ≤i≤r =⇒i=r and j =1.
Then by Proposition 2.2 and Corollary 2.5, we have
X X =X +X .
Er E1[mr+l] Er[mr+l+1] E2[mr+l−1]
When k =2, by k+i≥r+j and 1≤j ≤i≤r =⇒i=r or i=r−1.
For i=r =⇒j =1 or j =2:
The case for i=r and j =1, we have
X X
Er[2] E1[mr+l]
= (X X −1)X
Er E1 E1[mr+l]
= X (X +X )−X
E1 Er[mr+l+1] E2[mr+l−1] E1[mr+l]
= X X +(X +X )−X
E1 Er[mr+l+1] E1[mr+l] E3[mr+l−2] E1[mr+l]
= X X +X .
E1 Er[mr+l+1] E3[mr+l−2]
The case for i=r and j =2, we have
X X
Er[2] E2[mr+l]
= (X X −1)X
Er E1 E2[mr+l]
= X (X +X )−X
Er E1[mr+l+1] E3[mr+l−1] E2[mr+l]
= X +(X +X X )−X
Er[mr+l+2] E2[mr+l] Er E3[mr+l−1] E2[mr+l]
= X +X X .
Er[mr+l+2] Er E3[mr+l−1]
For i=r−1=⇒j =1:
X X
Er−1[2] E1[mr+l]
= (X X −1)X
Er−1 Er E1[mr+l]
= X (X +X )−X
Er−1 Er[mr+l+1] E2[mr+l−1] E1[mr+l]
= (X +X )+X X −X
Er−1[mr+l+2] E1[mr+l] Er−1 E2[mr+l−1] E1[mr+l]
= X +X X .
Er−1[mr+l+2] Er−1 E2[mr+l−1]
THE CLUSTER CHARACTERS FOR CYCLIC QUIVERS 7
Now, suppose it holds for k ≤n, then by induction we have
X X
Ei[n+1] Ej[mr+l]
= (X X −X )X
Ei[n] Ei+n Ei[n−1] Ej[mr+l]
= X (X X )−X X
Ei+n Ei[n] Ej[mr+l] Ei[n−1] Ej[mr+l]
= X (X X +X X )
Ei+n Ei[(m+1)r+l+j−i] Ej[n+i−r−j] Ei[r+j−i−1] En+i+1[(m+1)r+l+j−n−i−1]
−(X X +X X )
Ei[(m+1)r+l+j−i] Ej[n+i−r−j−1] Ei[r+j−i−1] En+i[(m+1)r+l+j−n−i]
= X (X +X )
Ei[(m+1)r+l+j−i] Ej[n+i+1−r−j] Ej[n+i−r−j−1]
+X (X +X )
Ei[r+j−i−1] En+i[(m+1)r+l+j−n−i] En+i+2[(m+1)r+l+j−n−i−2]
−(X X +X X )
Ei[(m+1)r+l+j−i] Ej[n+i−r−j−1] Ei[r+j−i−1] En+i[(m+1)r+l+j−n−i]
= X X +X X .
Ei[(m+1)r+l+j−i] Ej[n+i+1−r−j] Ei[r+j−i−1] En+i+2[(m+1)r+l+j−n−i−2]
2) When k =1, by i≤l+j ≤k+i−1=⇒i≤l+j ≤i=⇒i=l+j.
Then by by Proposition 2.2 and Corollary 2.5, we have
X X =X X =X +X
Ei Ej[mr+l] El+j Ej[mr+l] Ej[mr+l+1] Ej[mr+l−1]
Whenk =2,byi≤l+j ≤k+i−1=⇒i≤l+j ≤i+1=⇒i=l+j ori+1=l+j:
For i=l+j, we have
X X
Ei[2] Ej[mr+l]
= X X
El+j[2] Ej[mr+l]
= (X X −1)X
El+j El+j+1 Ej[mr+l]
= (X +X )X −X
Ej[mr+l+1] Ej[mr+l−1] El+j+1 Ej[mr+l]
= X +X +X X −X
Ej[mr+l+2] Ej[mr+l] El+j+1 Ej[mr+l−1] Ej[mr+l]
= X +X X .
Ej[mr+j+2] El+j+1 Ej[mr+l−1]
For i+1=l+j, we have
X X
Ei[2] Ej[mr+l]
= X X
El+j−1[2] Ej[mr+l]
= (X X −1)X
El+j−1 El+j Ej[mr+l]
= (X +X )X −X
Ej[mr+l+1] Ej[mr+l−1] El+j−1 Ej[mr+l]
= X X +(X +X )−X
Ej[mr+l+1] El+j−1 Ej[mr+l] Ej[mr+l−2] Ej[mr+l]
= X X +X .
Ej[mr+l+1] El+j−1 Ej[mr+l−2]
Suppose it holds for k≤n, then by induction we have
X X
Ei[n+1] Ej[mr+l]
= (X X −X )X
Ei[n] Ei+n Ei[n−1] Ej[mr+l]
= (X X )X −X X
Ei[n] Ej[mr+l] Ei+n Ei[n−1] Ej[mr+l]
= (X X +X X )X
Ej[mr+n+i−j] Ei[l+j−i] Ej[mr+i−j−1] El+j+1[n+i−l−j−1] Ei+n
−(X X +X X )
Ej[mr+n+i−j−1] Ei[l+j−i] Ej[mr+i−j−1] El+j+1[n+i−l−j−2]
= (X +X )X
Ej[mr+n+i+1−j] Ej[mr+n+i−j−1] Ei[l+j−i]
+(X +X )X
El+j+1[n+i−l−j] El+j+1[n+i−l−j−2] Ej[mr+i−j−1]
−(X X +X X )
Ej[mr+n+i−j−1] Ei[l+j−i] Ej[mr+i−j−1] El+j+1[n+i−l−j−2]
= X X +X X .
Ej[mr+n+i+1−j] Ei[l+j−i] Ej[mr+i−j−1] El+j+1[n+i−l−j]
3) It is trivial by the definition of the Caldero-Chapoton map. (cid:3)
8 MINGDINGANDFANXU
4. A Z-basis for cyclic quivers
In this section, we will focus on studying the following set
B(Q)={X |Ext1 (R,R)=0}.
R kQ
We prove that B(Q) is a Z-basis of the algebra AH(Q) generated by all these
generalized cluster variables. We first give the following definition.
Definition 4.1. For M,N ∈ modkQ with dimM = (m ,··· ,m ) and dimN =
1 r
(s ,··· ,s ), we write dimM (cid:22)dimN if m ≤s for 1≤i≤r. Moreover, if there
1 r i i
exists some i such that m <s , then we write dimM ≺dimN.
i i
Remark 4.2. It is easy to see that dimE [n−1]≺dimE [n+1] and dimE [n−
i+2 i i
1]≺dimE [n+1] in Corollary 2.5.
i
Lemma 4.3. Let T ,T be kQ-modules such that dimT =dimT . Then we have
1 2 1 2
XT1 =XT2 + X aRXR
dimR≺dimT2
where R∈modkQ and a ∈Z.
R
Proof. Suppose T = T ⊕T ⊕···⊕T and dimT = (d ,d ,··· ,d ) where
1 11 12 1m 1 1 2 r
T (1 ≤ i ≤ m) are indecomposable regular modules with quasi-socle E and
1i i1
dimT =(d ,d ,··· ,d ) for 1≤i≤m. Thus, we can see that (d ,d ,··· ,d )=
1i 1i 2i ri 1 2 r
m (d ,d ,··· ,d ). By Corollary 2.5 and Theorem 3.1, we have
Pi=1 1i 2i ri
Xd1Xd2 ···Xdr
E1 E2 Er
m
= Y(XEi1XEi1+1XEi1+2···XEi1+d1i+···+dri−1)
i=1
m
= Y(XT1i + X aL′XL′)
i=1 dimL′≺dimT1i
= XT1 + X aLXL.
dimL≺dimT1
where aL′,aL are integers. Similarly we have
XEd11XEd22···XEdnn =XT2 + X bMXM
dimM≺dimT2
where b are integers.
M
Thus
XT1 + X aLXL =XT2 + X bMXM.
dimL′≺dimT1 dimM≺dimT2
Therefore, we have
XT1 =XT2 + X aRXR
dimR≺dimT2
where a are integers. (cid:3)
R
We explain the method used in Lemma 4.3 by the following example.
Example4.4. Consider r =4,X andX . Wecanseethat dim(E [4]⊕
E2[5] E1[4]⊕E2 1
E ) =dimE [5]=dim(E ⊕2E ⊕E ⊕E ) and satisfy the conditions in Lemma
2 2 1 2 3 4
THE CLUSTER CHARACTERS FOR CYCLIC QUIVERS 9
4.3. Thus, for X , we have
E1[4]⊕E2
X X2 X X = X X X X X
E1 E2 E3 E4 E1 E2 E3 E4 E2
= (X +1)X X X
E1[2] E3 E4 E2
= (X +X )X X +X X X
E1[3] E1 E4 E2 E3 E4 E2
= (X +X )X +X X X +X X X
E1[4] E1[2] E2 E1 E4 E2 E3 E4 E2
= X +X +(X +1)X +(X +1)X
E1[4]⊕E2 E1[2]⊕E2 E1[2] E4 E2[2] E4
= X +X +X +X +X +2X .
E1[4]⊕E2 E1[2]⊕E2 E1[2]⊕E4 E2[3] E2 E4
Similarly for X , we have
E2[5]
X X2 X X = X X X X X
E1 E2 E3 E4 E2 E3 E4 E1 E2
= (X +1)X X X
E2[2] E4 E1 E2
= (X +X )X X +X X X
E2[3] E2 E1 E2 E4 E1 E2
= (X +X )X +X X X +X X X
E2[4] E2[2] E2 E2 E1 E2 E4 E1 E2
= X +X +X +(X +1)X +(X +1)X
E2[5] E2[3] E2[2]⊕E2 E1[2] E2 E1[2] E4
= X +X +X +X +X +X +X .
E2[5] E2[3] E2[2]⊕E2 E1[2]⊕E2 E2 E1[2]⊕E4 E4
Hence,X =X +X −X ,wheredim(E [2]⊕E )≺dimE [5],dimE ≺
E1[4]⊕E2 E2[5] E2[2]⊕E2 E4 2 2 2 4
dimE [5].
2
Lemma 4.5.
X =X +2.
Ei[r] Ei+1[r−2]
Proof. According to Proposition 2.1. (cid:3)
Lemma 4.6. For any M,N ∈ modkQ, X X is a Z-linear combination of the
M N
elements in B(Q).
Proof. By Theorem 3.1, we know that X X must be a Z-linear combination of
M N
elements in the set
{X |Ext1 (T,R)=Ext1 (R,T)=0}
T⊕R kQ kQ
where R is 0 or any regular exceptional module and T is 0 or any indecomposable
regular module with self-extension.
By Lemma 4.3 and Lemma 4.5, we can easily find that X X is actually a
M N
Z-linear combination of elements in the set B(Q). (cid:3)
Proposition 4.7. Let Ω = {A = (a ) ∈ M (Z ) | a ···a 6= 0} where
ij r×r ≥0 i,r i,r+i−2
a = a for i ≥ 2 and s ∈ N. Let E(A,i) = Eai,1 ⊕···⊕Eai,r−1 for A ∈ Ω
i,r+s i,s i i+r−2
and i=1,··· ,r. Then the set B′(Q)={X |i=1,··· ,r,A∈Ω} is a linearly
E(A,i)
independent set over Z.
Proof. SupposethatthereexiststheidentityS := n(A,i)X =0
PA∈Ω0,i=1,···r E(A,i)
where Ω is a finite subset of Ω and n(A,i)6=0∈Z for i=1,··· ,r. Note that
0
i+r−2
x +x
XE(A,i) = Y ( j+1x j−1)ai,j−i+1
j
j=i
for i = 1,··· ,r. Define a lexical order by set x < x < x < ··· < x and
r 1 2 r−1
xa < xb if a < b. Set l (A) = max{a ,··· ,a } and l = max{l (A)} .
i i r 2,r−1 r,1 r r A∈Ω0
Then l 6= 0. Note that a is just the exponent of X in the expression
r i,r−i+1 Er
of X for i = 2,··· ,r. Then the expression of n(A,i)X
E(A,i) PA∈Ω0,i=1,···r E(A,i)
contains the unique part of the form L(x1,···,xr−1) which has the minimal exponent
xlrr
at x and L(x ,··· ,x ) is a Laurent polynomial associated to x ,··· ,x . In
r 1 r−1 1 r−1
10 MINGDINGANDFANXU
fact, L(x1,···,xr−1) isapartofthesum n(A,i)X . Note
xlrr Pi=2,···r,A∈Ω0;ai,r−i+1=lr E(A,i)
that the terms in this sum have a common factor Xlr , thus we have the following
Er
identity
X n(A,i)XE(A,i) =(∗)XElrr
i=2,···r,A∈Ω0;ai,r−i+1=lr
here we denote 1 n(A,i)X by (∗).
Xlr Pi=2,···r,A∈Ω0;ai,r−i+1=lr E(A,i)
Er
Nowwesetl (A)=max{a ,··· ,a }andl =max{l (A)} . Then
r+1 3,r−1 r,2 r+1 r A∈Ω0
l 6= 0. In the same way as above, we know that the expression of the term (∗)
r+1
containstheuniquepartoftheform L(x2,···,xr−1) whichhastheminimalexponentat
xlr+1
1
x =x andL(x ,··· ,x )isaLaurentpolynomialassociatedtox ,··· ,x .
1 r+1 2 r−1 2 r−1
Note that L(x2,···,xr−1)Xlr is actually a part of the following term
xlr+1 Er
1
X n(A,i)XE(A,i) =(∗∗)XElr1+1XElrr
i=3,···r,A∈Ω0;ai,r−i+1=lr,ai,r−i+2=lr+1
herewedenote 1 n(A,i)X by(∗∗).
Xlr+1Xlr Pi=3,···r,A∈Ω0;ai,r−i+1=lr,ai,r−i+2=lr+1 E(A,i)
E1 Er
Continue this discussion, we deduce that there exists some n(A,i)=0. It is a con-
tradiction. (cid:3)
Theorem 4.8. The set B(Q) is a Z-basis of the algebra AH(Q).
Proof. It is easy to prove that the elements in B′(Q) and elements in B(Q) have
a unipotent matrix transformation. Then by Proposition 4.7 and Lemma 4.6, we
know that B(Q) is a Z-basis of the algebra AH(Q). (cid:3)
Example 4.9. (1) Consider r =1, then we can calculate
X =2,X =3,X =4,··· ,X =n+1,···
E1 E1[2] E1[3] E1[n]
It is obvious that B(Q)={1}.
(2) Consider r =2, then we can calculate
2x 2nx
2 2
X = ,X =3,··· ,X = ,X =2n+1,···
E1 x E1[2] E1[2n−1] x E1[2n]
1 1
2x 2nx
1 1
X = ,X =3,··· ,X = ,X =2n+1,···
E2 x E2[2] E2[2n−1] x E2[2n]
2 2
It is obvious that B(Q)={Xm,Xn |m,n∈Z }.
E1 E2 ≥0
Acknowledgements
The authors are grateful to Professor Jie Xiao for helpful discussions.
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