J.
Indones.
Math.
Soc.
Vol.
No.
, pp.
93Ae130.
CHARACTERIZATION OF NAKAYAMA m-CLUSTER TILTED ALGEBRAS OF TYPE An Faisal and Intan Muchtadi-Alamsyah Algebra Research Group Institut Teknologi Bandung, fl107101@students.
id, ntan@math.
Abstract.
For any natural natural number m, the m-cluster tilted algebras are generalization of cluster tilted algebras.
These class algebras are defined as the endomorphism of certain object in m-cluster category called m-cluster tilting object.
Finding such object in the m-cluster category has become a combinatorial problem.
In this article we characterize Nakayama m-cluster tilted algebras of type An by geometric description given by Baur and Marsh.
Key words and Phrases: Cluster tilted algebras, cluster category, tilting object.
Nakayama algebra Abstrak.
Untuk setiap bilangan asli m, aljabar teralih m-kluster adalah generalisasi dari aljabar teralih kluster.
Kelas aljabar ini didefinisikan sebagai endomorfisma objek tertentu di kategori m-kluster yang disebut objek pengalih m-kluster.
Mencari objek tersebut dalam kategori m-kluster dapat menjadi masalah kombinatorial.
Dalam artikel ini dikarakterisasi aljabar Nakayama yang merupakan aljabar teralih m-kluster jenis An berdasarkan deskripsi geometris yang diberikan oleh Baur dan Marsh.
Kata kunci:
Nakayama.
aljabar teralih kluster, kategori kluster, objek pengalih, aljabar Introduction Let K be an algebraically closed field, and Q a finite acyclic quiver with n vertices.
Let Db (H) be a bounded derived category of mod H where H is a basic, finite dimensional hereditary algebra over K.
We can assume H as a path algebra KQ of some quiver Q.
The m-cluster category is the orbit category CH = Db (H)/E Oe1 .
where E is the Auslander-Reiten translation of D (H) and .
denotes m-th power of shift .
in the derived category Db (H).
The m-cluster category is triangulated 2000 Mathematics Subject Classification: 16G70.
Received: 27-04-2015, revised: 07-08-2016, accepted: 07-08-2016.
Faisal and I.
Muchtadi-Alamsyah .
and it is a Krull-Schmidt category .
These categories are generalization of cluster categories defined in .
and independently .
for the Dynkin type An case.
In m-cluster category we consider a class of objects called m-cluster tilting These objects have nice combinatorial properties.
By definition, an object T is an m-cluster tilting object if for any object X, we have X OO add T if only if ExtiC m (T.
X) = 0 for all i OO .
, 2, .
, .
The objects T always have exactly n indecomposable direct summands .
The endomorphism algebra Endop m (T ) is called m-cluster tilted algebra.
In this paper we investigate m-Cluster Tilted Algebras.
-CTA) of type An which are Nakayama algebras.
Nakayama algebra itself by its quiver is divided into two types, namely type An and cyclic.
In this paper we focus on m-CTAs which are Nakayama algebras of type An and all possible relations as from .
we have known all m-CTAs which are Nakayama algebras of type cyclic, see also .
order to do this we use the geometric description of m-cluster category type An in .
We will divide into three cases in the search of m-CTAs of type An .
We divide these two cases based on the relationship between m and n.
The first case is when m Ou n Oe 2, the second case is m < n Oe 2.
This article is organized as follows.
In Section 2 we describe the geometric description and the relations of Nakayama m-CTAs.
in Section 3 we give a characterization of Nakayama m-CTA of cyclic type.
in Section 4 we give a characterization of Nakayama m-CTA of acyclic type which will be divided into two Geometric Description and Relations in Nakayama m-CTAs The geometric description of m-cluster category type An in .
briefly representing indecomposable objects and arrows of the AR-quiver of m-cluster category in a regular gon.
The indecomposable object is described as a diagonal of a regular gon while an arrow between two indecomposable objects described as two diagonals that have a common endpoint.
From this geometric description we can also see the relations of quivers of the m-CTAs of type An .
Let Pm.
2 be .
-regular gon, m, n OO N, where its corner points are numbered clockwise from 1 to m.
A diagonal D of Pm.
2 can be denoted as a pair .
, .
Consequently, the diagonal .
, .
is the diagonal .
, .
We said a diagonal D of Pm.
2 is an m-diagonal if D divide Pm.
2 into two parts that is .
-gon and .
Oe .
-gon where j = 1, 2, .
, d n2 e.
For i 6= j, an arc Dij of Pm.
2 is a part of boundary that connect i to j Note that if j is a clockwise direct neighbor of i then arc Dij is an edge ij of Pm.
We always have two arcs Dij .
Dji .
Let em An be a quiver with the vertices are all m-diagonals of polygon Pm.
2 while arrows obtained in the following way: suppose D = .
, .
and D0 = .
, j 0 ) are m-diagonals which have a common vertex i in Pm.
2 then there is an arrow from D to D0 if D.
D0 Characterization of Nakayama m-CTA of type An together with arc from j to j 0 form .
-gon in Pm.
2 and D can be rotated clockwise to D0 about the common endpoint i.
Using this regular gon we can easily make a quiver of an m-CTA.
The set of indecomposable objects of a tilting object of m-cluster category of type An can be identified as the set of maximal m-diagonals in Pm.
2 and the number of direct summands of this object is always n.
Such a set is called an .
Oeangulation of Pm.
By definition, we can conclude that if X and Y are m-diagonals of a tilting object T that has a common endpoint then there is a path from TX and TY in the Auslander-Reiten(AR) quiver of m-cluster category where TX and TY are indecomposable objects associated to X and Y .
It is clear that the composition of the arrows in this path is not zero.
If there is no m-diagonal between X and Y in Pm.
2 then the composition of irreducible maps from TX to TY does not pass through another indecomposable object which is a direct summand of a tilting object T .
It means that there is an arrow from the point corresponding to X and Y in the quiver of m-CTA Endop (T ).
By the above argument we can define a quiver of an m-CTA independently from .
-angulation of Pm.
Let T = {T1 .
T2 , .
Tn } be an .
Define a quiver QT as follows: The vertices of QT are the numbers 1, 2, .
, n which are in bijective correspondence with the m-diagonals T1 .
T2 , .
Tn .
Given two vertices a, b of QT , there is an arrow from a to b if .
Ta and Tb have a common point in Pm.
2 , .
there is no m-diagonal of T between Ta and Tb and .
Ta can be rotated clockwise to Tb at the common endpoint.
Our first lemma characterize the possible forms of two m-diagonals in polygon Pm.
2 , correspond to a path of length two in the quiver of an m-CTA.
We have the following easy lemma.
Lemma 2.
Let H =Endop (T ) be an m-CTA with T is an m-cluster tilting object of CA If x Ie y Ie z is a path of length two in QH and Tx .
Ty .
Tz respectively are m-diagonals correspond to points x, y, z then .
Tx = .
1 , x2 ).
Ty = .
2 , x3 ).
Tz = .
3 , x4 ) with x4 in arc Dx3 x1 .
Tx = .
1 , x2 ).
Ty = .
2 , x3 ).
Tz = .
2 , x4 ) with x4 in arc Dx3 x2 , where xi 6= xj if i 6= j.
Proof.
Let Tx = .
1 , x2 ).
Since there is an arrow from x to y then Tx and Ty have a common endpoint.
Without loss of generality, suppose Ty = .
2 , x3 ).
Since there is an arrow from y to z then Ty and Tz have a common endpoint.
If x3 is a common endpoint of Ty and Tz then Tz = .
3 , x4 ) where x4 in arc Dx1 x3 , otherwise Tz will cross Tx .
If x2 is a common endpoint of Ty and Tz then Tz = .
2 , x4 ) where x4 in arc Dx3 x2 .
Let Q be a finite quiver without cycle and H = KQ/I where I is an admissible ideal of KQ.
If Q is not connected then the algebra H is not connected.
Indeed Faisal and I.
Muchtadi-Alamsyah let Q be Ythe collection of maximal connected subquivers of Q.
It can be shown that KQ0 /I 0 where I 0 is an ideal of Q0 , but then H is a finite direct product
Q0 OOQ
of some algebras.
Hence.
H is not connected.
In order to know the condition of an .
-angulation such that the quiver of m-cluster tilted algebra is connected, we have the following easy lemma.
Lemma 2.
Let T be an .
-angulation of Pm.
The graph generated by the diagonals in T is connected if only if the quiver QT is connected.
Let X = .
1 , x2 ) be a diagonal of Pm.
We may assume x2 > x1 .
Define the length of diagonal X to be the min.
2 Oe x1 , m.
2 x1 Oe x2 }.
Thus, the length of X is equal to the minimum of the number of sides between arc Dx1 x2 and Dx2 x1 .
An m-diagonal X of Pm.
2 is said to be short if its length Figure 1.
short m-diagonal is minimal, that is of length m 1.
An m-diagonal X is short if only if there is no m-diagonal whose endpoints are in smaller polygon divided by X.
Lemma 2.
Let T be an .
-angulation of Pm.
2 with n Ou 3.
If QT is cyclic then all m-diagonals in T are short.
Proof.
Let X be an m-diagonal of T which is not short .
Without loss of generality, let X = .
, x1 ) and X has length which is minimal among the diagonals in T which m.
2 are not short .
First, assume that x1 O The diagonal X will divide Pm.
2 into two smaller polygons P1 and P2 with P1 is the smallest polygon .
ee Figure .
Since X is not short and T is maximal, there exists an m-diagonal of T whose endpoints in arc Dx1 x2 .
By the same argument we also have another m-diagonal of T which divides the polygon P2 .
We then have that all m-diagonals in P1 are short by the minimality of X.
Since QT is connected there exists a short m-diagonal X1 of T in P1 that adjacent to X.
We may assume that X1 = .
, .
Now there exists a short m-diagonal that adjacent to X1 , namely X2 .
By the same argument we have a collection of short m-diagonals X1 = .
, a1 ).
X2 = .
1 , a2 ) .
Xk = .
kOe1 , ak ) where all of these are in P1 and maximal with respect Characterization of Nakayama m-CTA of type An Figure 2.
m-diagonal X to this property.
It follows that xk = x1 , otherwise there is no arrow which target is Xk in QT .
We describe this situation in the following figure Figure 3.
m-diagonals in P1 But now we have a path X1 Ie X Ie Xk in QT .
So there can be no further m-diagonals adjacent to X, which is a contradiction.
2 If x1 > we get similar proof for P2 since in this case P2 becomes the smallest polygon divided by X.
Lemma 2.
3 gives us a characterization of m-cluster tilting object such that the corresponding m-CTA is a Nakayama algebra of cyclic type.
We will find all m-cluster tilting objects in this form in the next section.
Now we look at the configuration of an .
-angulation T which QT is of An type.
Lemma 2.
Let T be an .
-angulation of Pm.
2 with n Ou 3.
If QT is of An type then T = TC O T1 O T2 O A A A O TrOe1 Faisal and I.
Muchtadi-Alamsyah for some r Ou 2 where .
p to rotatio.
TC = {.
, x1 ), .
1 , x2 ), .
, .
rOe1 , xr )} and all m-diagonals in TC are short.
T1 = {.
1 , y11 ), .
1 , y12 ), .
, .
1 , y1j1 )}, j1 Ou 0 T2 = {.
2 , y21 ), .
1 , y22 ), .
, .
1 , y2j2 )}, j2 Ou 0 TrOe1 = {.
rOe1 , yrOe1,1 ), .
1 , yrOe1,2 ), .
, .
rOe1 , yrOe1,jrOe1 )}, jrOe1 Ou 0 with y11 < y12 < A A A < y1j1 < y21 < A A A < y2j2 < A A A < ynOe1,jnOe1 .
Figure 4.
-angulation of T with QT = An Proof.
Let .
, x1 ) be an m-diagonal of Pm.
2 correspond to a source in QT .
We claim that .
, x1 ) is short.
If .
, x1 ) is not short then either there is an mdiagonal .
1 , .
with t > x1 or there is an m diagonal .
, .
with u > x1 .
ee Figure Consider the first case , if there is an m-diagonal .
1 , .
, we chose t maximal Figure 5.
m-diagonals .
1 , .
, .
such that t > x1 .
Then we have an arrow .
1 , .
Ie .
, x1 ), but it contradicts that .
, x1 ) is a source.
Second case, if there is an m-diagonal .
, .
we chose u minimal such that u > x1 .
Since .
, x1 ) is not short, there is either an m-diagonal .
1 , .
Characterization of Nakayama m-CTA of type An with 1 < a < x1 or an m-diagonal .
, .
with 1 < b < x1 .
We may assume that a is minimal and b maximal.
If there is a diagonal .
1 , .
then there is an arrow .
, .
Ie .
1 , .
It contradicts the fact that there is also an arrow .
, x1 ) Ie .
, .
So we can assume that there is a diagonal .
, .
It follows that there is an arrow .
, .
Ie .
, x1 ).
This is a contradiction since .
, x1 ) is a source.
Therefore .
, x1 ) is short, this proves our claim.
Let .
, x1 ) Ie .
1 , .
be the arrow starting in .
, x1 ) then z > 1.
Now there are two cases, either .
1 , .
is short or .
1 , .
is not short.
Figure 6.
m-diagonal .
1 , .
1 , .
is short.
If T = .
1 , .
is short then arc Dzx1 together with T is a smaller polygon divided by .
1 , .
Hence, there is no m-diagonal with endpoints in arc Dzx1 .
We also have that there is no m-diagonal .
1 , .
with 1 < y < z since otherwise the arrow .
, x1 ) Ie .
1 , .
will not exist.
1 , .
is not short.
If .
1 , .
is not short then there is no m-diagonal .
, .
with 1 < v < z.
Indeed, assume to the contrary that there is an m-diagonal .
, .
with 1 < v < z.
It follows that there is no m-diagonal .
1 , .
for z < u < x1 since otherwise there is also an arrow .
1 , .
Ie .
1 , .
If there is an mdiagonal .
, .
for z < l < x1 , and choose z maximal, then there is an arrow .
, .
Ie .
1 , .
, a contradiction.
Therefore there is no m-diagonal with endpoints in arc Dzx1 .
This is a contradiction since .
1 , .
is not short.
Hence there is no diagonal .
, .
Therefore arc D1z together with .
, x1 ) and .
1 , .
forms an .
-gon.
We describe condition 1 and 2 respectively as follows where the shaded polygons are m 2-gons and hence there is no m-diagonal in these Now we perform same analysis by consider the arrow starting at .
1 , .
Faisal and I.
Muchtadi-Alamsyah Indeed, in case .
1 , .
is short then the arrow starting at .
1 , .
1 , .
Ie .
, .
with 1 < w < z.
In case .
1 , .
is not short then the arrow starting at .
1 , .
1 , .
Ie .
1 , .
with 1 < w < x.
We have similar case for the third m-diagonal from the source which adjacent to .
1 , .
There are again two cases to consider, that is either this m-diagonal is short or not short.
These two cases will be similar to the condition 1 and 2 above.
We complete the proof by induction using the fact that the the next m-diagonal adjacent to the previous have two possibilities like condition 1 and 2.
Two cases in Lemma 2.
1 hold for any path of length two in the quiver of m-CTAs of type An .
For both cases the picture is as follows Figure 7.
m-diagonals correspond a path of length two Using the above lemma we can conclude that each path of length two in the quiver of m-CTAs of type An is one of these two cases.
Now we will see the composition of paths of length two in End (T ) O
= KQ/I
for both cases.
We have the following facts.
Lemma 2.
Let T = T1 Oi T2 Oi A A A Oi Tn be an m-cluster tilting object of CA Q be a quiver of m-CTA Endop (T ).
Suppose i Oe IejOe Ie k is a path of length two in Q corresponding to the m-diagonals Ti .
Tj .
Tk in Pm.
If Ti = .
1 , x2 ).
Tj = .
2 , x3 ).
Tk = .
3 , x4 ) with x4 in arc Dx3 x1 then the composition i Oe IejOe Ie k in Endop (T ) is zero.
If Tx = .
1 , x2 ).
Ty = .
2 , x3 ).
Tz = .
2 , x4 ) with x4 in arc Dx3 x2 then the composition i Oe IejOe Ie k in Endop (T ) is not zero.
Proof.
See .
Now we can identify the relation of connected Nakayama m-cluster tilted algebras using Lemma 2.
3, 2.
4 and 2.
Theorem 2.
Let H = KQ/I be a connected Nakayama m-cluster tilted algebra of CA An ideal I of H is generated by a relation of paths of length two.
Proof.
If Q is cyclic then by Lemma 2.
Q = QT where T is an .
-angulation such that all m-diagonals in T are short.
Therefore, every path of length two in Characterization of Nakayama m-CTA of type An QT is in case 1 of Lemma 2.
By Lemma 2.
5 all paths of length two is zero.
Q is of type An then by Lemma 2.
4 every path of length two is either case one or case two of Lemma 2.
It remains to prove that every path P = 1 2 .
with ` Ou 3 is not zero in H if every subpath of P is not zero in H.
It follows that every subpath of length two in P is case two of Lemma 2.
We may assume that T1 = .
, mr .
with 1 O r < n whose common endpoint with T2 and T3 is 1.
Hence.
Tj = .
, mrj .
for every j Ou 2 with r < ri < ri 1 for all i.
We have that T1 .
T2 , .
T` will be in the subquiver of em An as in Figure 8.
Since the Figure 8.
subquiver of em composition of irreducible morphism T1 Ie T2 Ie A A A Ie` is not zero in m-cluster category, we conclude that 1 2 .
` not zero in H.
This finishes the proof.
m-CTAs which are Nakayama Algebra of Cyclic Type In this section we will show that m-CTAs which are Nakayama algebras of cyclic type only occur if m = n Oe 2.
It means that there is no m-CTA whose quiver is cyclic when m 6= n Oe 2.
In addition, in m-CTA there is only one possibility relation that is relations of paths of length two.
More generally, m-CTAs which have cyclic quivers have been stated by Murphy in .
However, in this section we explain how to characterize m-CTAs which quivers are cyclic by using geometric description in .
The results in this section have been proved in .
We state again here with more structured proofs.
We show that if T = T1 Oi T2 Oi A A A Oi Tn then T is a m-cluster tilting object for m Ou n 2 where Ti Aos are m-diagonals described in Proposition 3.
The quivers of m-CTAs Endop (T ) have different forms for each case m = n Oe 2 and m > n Oe 2.
Indeed, for 1 O i O n Oe 1 diagonals Ti and Ti 1 have a common endpoint in Pm.
2 for m Ou n Oe 2.
It means that for every i, we have an arrow i Ie i 1 in Faisal and I.
Muchtadi-Alamsyah the quiver of Endop (T ).
Now consider m-diagonals Tn = .
mOe.
Oe.
, 2mOe.
Oe.
) and T1 = .
, m .
If m = n Oe 2 then Tn = .
m 3, m .
Hence.
Tn and T1 have a common endpoint .
in Pm.
Therefore there exists an arrow n Ie 1 in quiver of Endop (T ).
Thus, for m = n Oe 2 the quiver of m-cluster tilted algebra Endop (T ) is Figure 9.
Figure 9.
Quiver of Endop (T ) for m = n Oe 2 Proposition 3.
Let CA = Db (KAn )/Fm , where Fm = E Oe1 .
and m = n Oe 2.
Suppose that T1 = .
, m .
T2 = .
, nm .
and for 3 O i O n.
Ti = (.
Oe .
Oe .
)m Oe .
Oe .
, .
Oe .
Oe .
)m Oe .
Oe .
) .
T1 .
T2 , .
Tn are m-diagonals of Pm.
T = T1 Oi T2 Oi A A A Oi Tn is an m-cluster tilting object.
m-cluster tilted algebra Endop (T ) is isomorphic to KQ/I where Q is cyclic with n vertices and I is an ideal generated by all paths of length two.
Proof.
It is clear that if T1 = .
, m .
T2 = .
, nm .
and for 3 O i O n.
Ti = (.
Oe .
Oe .
)m Oe .
Oe .
, .
Oe .
Oe .
)m Oe .
Oe .
) then T1 .
T2 , .
Tn are m-diagonals of Pm.
For i = n we have that Tn = (.
Oe .
Oe .
)m Oe .
Oe .
, .
Oe .
Oe .
)m Oe .
Oe .
) = .
m Oe .
Oe .
, 2m Oe .
Oe .
Consider m-diagonals T1 .
T2 , .
Tn in Pm.
2 , see Figure 10.
Because T1 .
T2 .
Tn are not crossing each other then T is an m-cluster tilting object.
Let Q be a quiver of m-cluster tilted algebra Endop (T ), then there is only one arrow i Ie i 1 for every 1 O i O nOe1.
Since m = nOe2, we obtain that Tn = .
m 3, m .
and T1 = .
, m .
have a common endpoint.
Consequently, there is exactly one arrow n Ie 1 in Q.
It means that Q is a cyclic quiver with n vertices.
By Lemma 5 the composition of all paths of length two is zero.
Next we show that the m-CTA of type An whose quiver is cyclic is the algebra stated in Proposition 3.
Characterization of Nakayama m-CTA of type An Figure 10.
m-diagonals T1 .
T2 , .
Tn Proposition 3.
If T is an m-cluster tilting object of m-cluster category CA that the quiver of m-cluster tilted algebra End (T ) is connected and cyclic, then m = n Oe 2.
Moreover.
Endop (T ) = KQ/I with I an ideal generated by all paths of length two.
Proof.
Let Q be a quiver of m-cluster tilted algebra Endop (T ).
Suppose T = T1 Oi T2 Oi A A A Oi Tn ,we may assume {T1 .
T2 , .
Tn } is a set of maximal non-crossing m-diagonals in .
-gon Pm.
Assume that Q0 = {T1 .
T2 , .
Tn } the set of vertices of Q, and the set of arrows Q1 = .
, 2 , .
, nOe1 , n } with i : Ti Ie Ti 1 for every i OO .
, 2, .
, n Oe .
and n : Tn Ie T1 .
Consider any path of length two Tp Ie Tq Ie Tr in Q.
By Lemma 2.
3 Tq .
Tr .
Ts are short.
follows that Tq = .
1 , x2 ).
Tr = .
2 , x3 ).
Ts = .
3 , x4 ) can be described as in Figure By applying the above argument, the picture of m-diagonals T1 .
T2 , .
Tn in Figure 11.
m-diagonals correspond to Tq .
Tr and Ts Pm.
2 is Figure 12.
Faisal and I.
Muchtadi-Alamsyah Figure 12.
m-diagonals T1 .
T2 , .
Tn for m = n Oe 2 Since all Ti are short then the length of Ti is m 1.
Consequently, we have the .
A A A .
= m.
Therefore, .
n = m.
2 Ni n = m 2 For the last statement we apply Lemma 3.
Example 3.
Let m = 4 and n = 6 then m.
2 = 4.
2 = 30.
Consider 30-gon P30 , let T1 = .
, .
T2 = .
, .
T3 = .
, .
T4 = .
, .
T5 = .
, .
and T6 = .
, .
then T = T1 Oi T2 Oi T3 Oi T4 Oi T5 Oi T6 is a 4-cluster tilting object.
The picture of P30 together with the six m-diagonals is Figure 13.
-angulation T for m = 4 and n = 6 Characterization of Nakayama m-CTA of type An m-CTAs which are Nakayama Algebras with Acyclic Quivers In this section we will characterize m-CTA which are Nakayama algebras whose quivers are connected acyclic.
In other words, we find m-cluster tilting objects T = T1 Oi T2 Oi A A A Oi Tn such that Endop (T ) O = KQ/I where Q is nOe1 T3 Ie A A A Ie TnOe1 OeOeOeIe Tn .
T2 OeIe T1 OeIe Throughout.
Q is assumed to be the above quiver, unless otherwise specified.
We will also observe the relation in this type of m-CTA.
To do this we divide into two cases correspond to m and n.
These three cases are m Ou n Oe 2 and m < n Oe 2.
The following is the list of m-diagonals in Pm.
Table 1.
m-diagonals .
2m 2 3m 2 4m 2 .
2m 1 3m 1 nm 2 .
Oe .
m 1 (.
Oe .
Oe .
Oe .
m Oe 1.
nm 2 .
Oe .
m 1 .
Oe .
m 2 nm 1 .
Oe .
m 1 mOe1 .
Oe .
Oe .
m Oe 1 .
Oe .
m Oe 2 (.
Oe .
m Oe 2.
Oe .
m Oe 3.
Oe .
m Oe 1 .
Oe .
m Oe 2 .
Oe .
m Oe 1 .
Oe .
m Oe 2 .
Oe .
m Oe 1 .
Oe .
m Oe 2 nm Oe 1 .
m Oe 1 mOe3 2m Oe 3 .
Oe .
m Oe 2 nm Oe 2 .
m Oe 2 mOe4 (.
Oe .
m Oe .
Oe .
Oe .
)m Oe .
Oe .
Oe .
Oe .
)m Oe .
Oe .
Oe .
m Oe .
Oe .
Oe .
Oe .
)m Oe .
Oe .
Oe .
Oe .
)m Oe .
Oe .
Oe .
Oe .
)m Oe .
Oe .
Oe .
Oe .
)m Oe .
Oe .
nm Oe .
Oe .
Oe .
m Oe .
Oe .
m Oe .
Oe .
nm Oe .
Oe .
m Oe .
Oe .
m Oe .
Oe .
Oe .
m Oe 3 .
Oe .
m Oe 5 .
Oe .
)m Oe .
Oe .
Oe .
)m Oe i Faisal and I.
Muchtadi-Alamsyah From Table 1 we take m-diagonals which will be used as a direct summand of an m-cluster tilting object such that the quiver of m-CTA is An .
The following table lists some m-diagonals which will be used for our m-cluster tilting object.
Table 2.
m-diagonals of m-cluster tilting objects X1,1 = .
, 2m .
X2,1 = .
, 3m .
X3,1 = .
, 4m .
X1,2 = .
m 2, 2m .
X2,2 = .
m 2, 3m .
X3,2 = .
m 2, 4m .
XnOe2,1 = .
, .
Oe .
XnOe2,2 = .
m 2, .
Oe .
X1,3 = (.
Oe .
m 1, 2.
X2,3 = (.
Oe .
m 1, 3.
X3,3 = (.
Oe .
m 1, 4.
XnOe3,3 = (.
Oe .
m 1, .
Oe .
X1,i = (.
Oe .
Oe .
)m Oe .
Oe .
, 2m Oe .
Oe .
) X2,i = (.
Oe .
Oe .
)m Oe .
Oe .
, 3m Oe .
Oe .
) X3,i = (.
Oe .
Oe .
)m Oe .
Oe .
, 4m Oe .
Oe .
) XnOei,i = (.
Oe .
Oe .
)m Oe .
Oe .
, .
Oe i .
m Oe .
Oe .
) Throughout, for every 1 O i O n.
Ti is assumed to be the m-diagonal described in Proposition 3.
Case m Ou n Oe 2.
Recall that T1 = .
, m .
T2 = .
, nm .
and for 3 O i O n Oe t we have Ti = (.
Oe .
Oe .
)m Oe .
Oe .
, .
Oe .
Oe .
)m Oe .
Oe .
We have that all m-diagonals in the set T = {T1 .
T2 , .
TnOe1 .
Tn } are short.
In the case m = n Oe 2 the quiver of QT is a cyclic quiver and every path of length of two is a relation in the corresponding m-CTA.
We will prove that there is no m-CTA whose quiver is An and every path of length two is zero in the case m = n Oe 2.
But in the case m > n Oe 2 the quiver QT is a path and every path of length of two is a relation in the corresponding m-CTA.
Lemma 4.
Suppose that CA = Db (KAn )/Fm , where Fm = E Oe1 .
with m > n Oe 2.
T1 .
T2 , .
Tn are m-diagonals of Pm.
T = T1 Oi T2 Oi A A A Oi Tn is an m-cluster tilting object.
The m-cluster tilted algebra Endop (T ) is isomorphic to KQ/I where Q is nOe1 1 OeIe 2 OeIe 3 Ie A A A Ie .
Oe .
OeOeOeIe n.
and I is an ideal generated by all paths of length two.
Characterization of Nakayama m-CTA of type An Proof.
It is clear that T1 .
T2 , .
Tn are m-diagonals of Pm.
2 , where if i = n then Tn = (.
Oe .
Oe .
)m Oe .
Oe .
, .
Oe .
Oe .
)m Oe .
Oe .
) = .
m Oe .
Oe .
, 2m Oe .
Oe .
Observe that the picture of m-diagonals T1 .
T2 , .
Tn in Pm.
2 is Figure 14.
Since T1 .
T2 .
Tn are not crossing each other then T is an m-cluster Figure 14.
m-diagonals of T tilting object.
Let Q be the quiver of m-cluster tilted algebra Endop (T ), then there exists exactly one arrow Ti Ie Ti 1 for every 1 O i O n Oe 1.
If m > n Oe 2 then m Oe .
Oe .
> 0 and consequently m 2 m Oe .
Oe .
> m 2.
Hence.
Tn and T1 donAot have common endpoint.
In other words there is no arrow from Tn to T1 .
conclude Q is the quiver in the proposition.
Finally, by Lemma 2.
5 the composition of all paths of length two is zero.
Lemma 4.
Let m Ou n Oe 2 and T = T1 Oi T2 Oi A A A Oi TnOe1 Oi X1,i with 1 O i O n Oe 2 .
T is an m-cluster tilting object in CA .
If Q is a quiver of End (T ) then Q is nOe1 1 OeIe 2 OeIe 3 Ie A A A Ie n Oe 1 OeOeOeIe n.
If Aj = j j 1 for every 1 O j O n Oe 2 then Endop (T ) = KQ/I where I = hA1 .
A2 , .
AiOe1 .
Ai 1 , .
AnOe2 i.
Proof.
Suppose that T 0 = {T1 .
T2 , .
TnOe1 } then it is clear that T 0 is the set of m-diagonals that are not crossing each other in Pm.
We have that X1,1 = .
, 2m .
and X1,i = .
Oe .
Oe .
) Oe .
Oe .
, 2m Oe .
Oe .
) for 1 O i O n Oe 2 = m.
Hence, m 2 < 2m Oe .
Oe .
< 2m 3 It follows that the set T 0 O {X1,i } of m-diagonals in Pm.
2 is as in Figure We conclude that T is an m-cluster tilting object of CA From Figure 15 we Faisal and I.
Muchtadi-Alamsyah Figure 15.
m-diagonal T 0 O X1,i obtain easily that quiver of Endop (T ) is Q.
Note that m-diagonals Ti .
X1,i .
Ti 1 satisfy case 2, hence the composition Ai = i i 1 is not zero.
But all Aj with j 6= i is zero since the corresponding m-diagonals with Aj satisfy case 1.
We conclude Endop (T ) O = KQ/I, as required.
Lemma above gives us how to construct other m-cluster tilting objects which have different relations.
We know that the number of paths of length two in An is .
Oe .
, where the relations are A1 .
A2 , .
AnOe2 .
In Lemma 4.
2 ideal I is generated by a combination of .
Oe .
relations of paths of length two from .
Oe .
We can get the m-CTA Endop (T ) O = KQ/I where I generated by .
Oe .
relations of paths of length two from .
Oe .
relations by the following lemma.
Lemma 4.
Suppose that m Ou n Oe 2 and T = T1 Oi T2 Oi A A A Oi TnOe2 Oi X1,i Oi X2,j where 1 O i O j O n Oe 3 then T is an m-cluster tilting object of CA Furthermore.
O the algebra End (T ) = KQ/I where I generated by .
Oe .
relations of paths of nOe2 length two.
If T be the collection of such T then |T| = nOe4 Proof.
It is clear that m-diagonal T1 .
T2 , .
TnOe2 are not crossing each other in Pm.
Now we just need to consider m-diagonals X1,i and X2,j in Pm.
We have that X1,1 = .
, 2m .
X2,1 = .
, 3m .
X1,i = .
Oe .
Oe .
) Oe .
Oe .
, 2m Oe .
Oe .
) and X2,j = .
Oe .
Oe .
) Oe .
Oe .
, 3m Oe .
Oe .
) where i > 1 and j > 1.
It is easy to see that for i = 1 and j = 1, m-diagonals T1 .
T2 , .
TnOe2 .
X1,1 .
X2,1 are not crossing each other.
Next, we consider endpoints of X1,i and X2,j for every i Ou 1, j > 1 .
If i = j then 3m Oe .
Oe .
Oe .
m Oe .
Oe .
) = m = n Oe 2.
Since j O n Oe 3 then m 2 < m 4 O 2m Oe .
Oe .
< 3m Oe .
Oe .
O 3m 2 < 3m 4.
Characterization of Nakayama m-CTA of type An It follows that one end point of X1,i and X2,j is in arc Dm 2,3m 4 .
While other point both of X1,i and X2,j coincides with one of endpoint of T1 .
T2 , .
TnOe2 .
It turns out that X1,i is not crossing with T1 .
T2 , .
TnOe2 as well as also for X2,j .
It remains to prove that X1,i and X2,k are not crossing each other.
If i = 1 and j = 1 then it is clear that X1,1 and X2,1 are not crossing each other.
If i = 1 and 1 < j O nOe3 then X1,1 = .
, 2m .
and X2,j = .
Oe.
Oe.
)Oe.
Oe.
, 3mOe.
Oe.
) are not crossing each other.
If j Ou i > 1, we have X1,i = .
Oe .
Oe .
) Oe .
Oe .
, 2m Oe .
Oe .
) and X2,j = .
Oe .
Oe .
) Oe .
Oe .
, 3m Oe .
Oe .
Since m.
Oe .
Oe .
) Oe .
Oe .
O m.
Oe .
Oe .
) Oe .
Oe .
and 2m Oe .
Oe .
< 3m Oe .
Oe .
then X1,i and X2,j are not crossing each other.
We deduce that T1 .
T2 .
TnOe2 .
X1,i .
X2,j is the set of m-diagonals which are not crossing each other.
Thus.
T = T1 Oi T2 Oi A A A Oi TnOe2 Oi X1,i Oi X2,j is an m-cluster tilting object.
Observe that paths of length two X 0 Ie X1,i Ie X 00 and Y 0 Ie X2,j Ie Y 00 with X 0 .
Y 0 .
X 00 .
Y 00 are m-diagonals of T which satisfy case 2 in Lemma 2.
Beside these two paths, all other path of length two in quiver End(T ) satisfy case 1 in Lemma 2.
Furthermore, for such T there are exactly two paths of length two in Q which composition in Endop (T ) is not zero .
We can compute the number of such T by compute the number of all combinations .
, .
where 1 O i O n Oe 3 and i O j O n Oe 3.
Table 3.
Pair of .
, .
nOe3 nOe3 nOe3 nOe2 nOe3 nOe3 nOe2 nOe3 nOe3 The number of such T is 1 2 A A A .
Oe .
Oe .
= .
Oe .
! .
Oe .
Oe .
= .
Oe .
!2! We combine two lemmas above into a more general result, that is m-CTA Endop (T ) O = KQ/I where I is an ideal generated by .
Oe 2 Oe .
relations of paths of length two from .
Oe .
relations and 1 O t O n Oe 2.
Lemma 4.
Suppose that m Ou nOe2 and T = T1 OiT2 OiA A AOiTnOet OiX1,j1 OiX2,j2 Oi A A A Oi Xt,jt with 1 O j1 O j2 O A A A O jt O n Oe t Oe 1 and 1 O t O n Oe 2, then T is an m-cluster tilting object of CA The m-cluster tilted algebra Endop (T ) O = kQ/I where I is generated by .
Oe 2 Oe .
relations be the nOe2 collection of such T then |T| = nOe2Oet Faisal and I.
Muchtadi-Alamsyah Proof.
For t = 1 and t = 2, it has been proved in Lemma 4.
2 and Lemma In general, we have that m-diagonals T1 .
T2 , .
TnOet are not crossing each other in regular gon Pm.
Now consider m-diagonals X1,j1 .
X2,j2 , .
Xt,jt in Pm.
If m = n Oe 2 then TnOet = (.
m Oe n t 4, .
m Oe n t .
= (.
m t 2, .
m t .
We will see all cases of j1 , j2 , .
, jt in Pm.
To show this we first consider the case j1 = j2 = A A A = jt = 1 with the picture of this case in Pm.
2 is Figure 16.
m-diagonals of T in Lemma 4.
We get that X1,j1 = .
, 2m .
X2,j2 = .
, 3m .
XtOe1,jtOe1 = .
, tm .
Xt,jt = .
, .
The configuration of these m-diagonals in Pm.
2 can be illustrated as in Figure We will use that picture to see the other cases of j1 , j2 , .
, jt .
The upper line has .
Oe t Oe .
black dots while the bottom line has t black dots.
Let us observe the m-diagonal Xi,ji = .
i , yi ) where xi is one of the black dots on the upper line and yi one of the points .
ot necessarily black do.
on the bottom line.
We have that Xk,1 = .
, .
with 1 O k O t.
We can conclude that Xi,ji = .
i , yi ) where xi is the ji -th black dot on the upper line counted from the right-hand side, and yi = .
m 2 Oe .
i Oe .
= .
m 3 Oe ji .
Suppose that 1 O i O t Oe 1 and Xi,ji = .
i , yi ).
Xi 1,ji 1 = .
i 1 , yi 1 ) then yi = .
m 3 Oe ji < yi 1 = .
m 3 m Oe ji 1 .
Characterization of Nakayama m-CTA of type An Figure 17.
m-diagonals X1,1 .
X2,1 , .
Xt,1 Since ji O ji 1 O n Oe t Oe 1 O m then either xi = xi 1 or xi 1 Aos position is on the left of xi .
Moreover im 2 < xi O .
We describe this situation as in Figure 18.
Figure 18.
m-diagonals Xi,ji and Xi 1,ji 1 Since Xi,ji = .
i , yi ).
Xi 1,ji 1 = .
i 1 , yi 1 ) satisfy this condition.
ee Figure .
for every i then X1,j1 .
X2,j2 , .
Xt,jt are not crossing each other in Pm.
Finally we conclude that m-diagonals T1 .
T2 , .
TnOet .
X1,j1 .
X2,j2 , .
Xt,jt are not crossing each other in regular gon Pm.
2 , it proves that T is an m-cluster tilting Next we show the last statement.
Every m-diagonal Xi,ji represent one path of length two which is not zero in Endop (T ).
Hence, there exists .
Oe 2 Oe .
relations of paths of length two in Endop (T ).
Now we compute the number of Faisal and I.
Muchtadi-Alamsyah T in this theorem.
This number equal to the number of possibilities of t-tuple .
1 , j2 , .
, jt ) where 1 O j1 O j2 O A A A O jt O n Oe t Oe 1.
This problem is equivalent to counting the number of distinct shortest routes from point A to point B in the the following diagram :
Figure 19.
Map of routes from A to B Here ji interpreted as a step up to the i-th and for every ji there is .
Oe t Oe .
positions can be chosen.
It is easy to see that the number of distinct shortest route is combination .
Oe 2 Oe .
Oe .
, that is nOe2 nOe2Oet .
Oe .
! t!.
Oe 2 Oe .
! Proposition 4.
Let m = n Oe 2 and H = KQ/I where Q is quiver nOe1 1 OeIe 2 OeIe 3 Ie A A A Ie n Oe 1 OeOeOeIe n.
Let W = {Aj = j j 1 .
O j O n Oe .
and B OI W , |B| 6= n Oe 2.
If I = hBi then H is an m-CTA.
Proof.
If B = OI then I = 0, we choose T in Lemma 4.
4 with t = n Oe 2 hence we get Endop (T ) = KQ.
If |B| = k > 1, by Lemma 4.
4 we can choose T with t = n Oe 2 Oe k such that Endop (T ) O = H.
So far we have obtain some m-CTAs in case m = n Oe 2.
By Theorem 3.
1 it remains to find m-CTAs whose number of relations is n Oe 2.
But we will show that there is no such m-CTA.
Lemma 4.
If m = n Oe 2 then there is no m-cluster tilting object T of CA that End (T ) = KQ/I with I = hA1 .
A2 , .
AnOe1 .
AnOe2 i.
Characterization of Nakayama m-CTA of type An Figure 20.
m-diagonals T1 .
T2 , .
Tn Proof.
Let T1 .
T2 , .
Tn be m-diagonals corresponding to T , then by Lemma 2.
these m- diagonal in Pm.
2 should be as in Figure 20.
It means that xn 1 6= x1 or equivalently arc Dx1 xn 1 has at least one side.
Note that arc Dxi 1 xi has at least m 1 side.
If all Ti are short then without loss of generality, suppose that x1 = m 2 and x2 = 1.
Consequently.
T1 = .
, m .
T2 = .
, nm .
T3 = (.
Oe .
m 1, nm .
T4 = (.
Oe .
m 1, .
Oe .
and for 5 O i O n.
Ti = (.
Oe .
Oe .
)m Oe .
Oe .
, .
Oe .
Oe .
)m Oe .
Oe .
The number of sides in arc Dxn 1 x1 is .
n = mn n.
Hence, the number of sides in arc Dx1 xn 1 is m.
2 Oe .
= m Oe .
Oe .
However if m = n Oe 2 then there is no side in arc Dx1 xn 1 , a contradiction.
Now suppose that there exists Tj which is not short.
It follows that the number of sides in arc Dxn 1 x1 is more than .
If x is the number of sides in arc Dxn 1 x1 then x > mn n.
We have that .
2 Oe .
is the number of side in arc Dx1 xn 1 .
Consequently m.
2 Oe x < m.
2 Oe .
= m Oe .
Oe .
= 0 since m = n Oe 2, a contradiction.
We conclude that there is no such T .
We end this section by giving all m-CTAs which are Nakayama algebras with acyclic quiver in the case m Ou n Oe 2.
Proposition 4.
Let m = n Oe 2 and H O = KQ/I be an algebra with Q is nOe1 1 OeIe 2 OeIe 3 Ie A A A Ie n Oe 1 OeOeOeIe n.
The algebra H is an m-CTA of CA if only if I is generated by at most .
Oe .
paths of length two.
Proof.
Use Theorem 2.
Corollary 4.
5 and Lemma 4.
Faisal and I.
Muchtadi-Alamsyah Proposition 4.
Let m > n Oe 2 and H = KQ/I with Q is the quiver nOe1 3 Ie A A A Ie n Oe 1 OeOeOeIe n.
2 OeIe 1 OeIe Suppose that W = {Aj = j j 1 .
O j O n Oe .
and B OI W .
If I = hBi then H is an m-CTA.
Proof.
If B 6= W , we choose m-cluster tilting object T in Lemma 4.
If B = W then we choose the m-cluster tilting object T in Lemma 4.
Theorem 4.
Let m > n Oe 2 and H O = KQ/I be an algebra with Q is the quiver nOe1 3 Ie A A A Ie .
Oe .
OeOeOeIe n.
2 OeIe 1 OeIe The algebra H is an m-CTA of CA if only if I is generated by any collection of paths of length two.
Proof.
Apply Theorem 2.
6 and Proposition 4.
Case m < n Oe 2.
Just like in the two previous cases to characterize Nakayama m-CTA, in this case it is sufficient to simply consider the relations of path of length two that appear on this algebra.
If the number of relations is at most m, then there is m-cluster tilting object such that the corresponding m-CTA is Nakayama algebra.
If the ideal generated by more than m relations of paths of length two we have not been able to guarantee which algebras are Nakayama m-CTA.
This happens because we get different cases depending on the difference between m and n Oe 2 .
e denote by In the first part we put forward some Nakayama algebra which are not m-CTA in the case m < n Oe 2.
This class of algebra are given in Lemma 4.
Lemma 11.
Lemma 4.
12 and Lemma 4.
Next, we provide all the Nakayama m-CTA algebras which have at most m relation of path of length two in Lemma 4.
14 parts .
, .
and Lemma 4.
16 parts .
In Theorem 4.
18 we give a characterization of Nakayama m-CTA which have at most m relations.
In the last part we try to find the possibility of more than m relations of path of length two.
In Proposition 4.
there are Nakayama algebras with more than m relation which are not m-CTA for some certain condition of a.
We also give Nakayama algebras with more than m relation which are m-CTA for some certain condition in Proposition 4.
We begin by giving Nakayama algebras acyclic type which are not m-CTAs.
Lemma 4.
If m < n Oe 2 then there is no m-cluster tilting object T in CA that End (T ) O = KQ/I with Q is nOe1 1 OeIe 2 OeIe 3 Ie A A A Ie .
Oe .
OeOeOeIe n and I = hA1 .
A2 , .
AnOe3 .
AnOe2 i, where Ai = i i 1 for every i.
Proof.
We utilize the same methods as in the proof of Lemma 4.
If T1 .
T2 , .
Tn are m-diagonals correspond to T then by Lemma 2.
5, these n m-diagonals in Pm.
2 should be as Figure 21, and it turns out that arc Dx1 xn 1 at least Characterization of Nakayama m-CTA of type An Figure 21.
m-diagonals T1 .
T2 , .
Tn has one side.
Note that the number of sides in arc Dxi 1 xi is at least m 1.
Therefore, arc Dxn 1 x1 has at least .
Let x be the number of sides in arc Dxn 1 x1 , then x Ou mn n.
We also have that .
2 Oe .
is the number of sides in arc Dx1 xn 1 .
Therefore m.
2 Oe x O m.
2 Oe mn Oe n = m Oe .
Oe .
< 0, because m < n Oe 2, a contradiction.
The proof is complete.
Next lemma shows that the Nakayama algebra whose relations are m 1 consecutive relation paths of length two starting from Aa 1 is not m-CTA.
Lemma 4.
Suppose that m < nOe2 and a = nOe2Oem then there is no m-cluster tilting object T of CA such that Endop (T ) O = KQ/I with Q is nOe1 1 OeIe 2 OeIe 3 Ie A A A Ie .
Oe .
OeOeOeIe n and I = hAa 1 .
Aa 2 , .
AnOe3 .
AnOe2 i, where Ai = i i 1 for every i.
Proof.
Suppose that there exists such T .
By Lemma 2.
1, the configuration of mdiagonals correspond to T in Pm.
2 is as in Figure 22.
Hence we may write T = Y1 Oi Y2 Oi A A A Oi Ym 2 Oi X1 Oi X2 Oi A A A Oi Xa .
It follows that arc Dxa 1 ym 2 has at least one side.
By the definition of m-diagonal, arc Dyi 1 yi and arc Dx1 y1 have at least m 1 sides, while arc Dxj xj 1 has at least m side.
Hence, arc Dym 2 xa 1 has at least .
am = .
Oe 2 Oe .
m = m.
2 A contradiction since Pm.
2 has m.
2 sides and arc Dxa 1 ym 2 has at least one side.
Faisal and I.
Muchtadi-Alamsyah Figure 22.
m-diagonals Y1 .
Y2 , .
Ym 2 .
X1 .
X2 , .
Xa We have that Nakayama algebra with m consecutive relations of path of length two is not m-CTA of type An .
Lemma 4.
Suppose that m < nOe2 and a = nOe2Oem then there is no m-cluster tilting object T of CA such that Endop (T ) O = KQ/I with Q is nOe1 1 OeIe 2 OeIe 3 Ie A A A Ie .
Oe .
OeOeOeIe n and I = hAt .
At 1 , .
At mOe1 i where 1 O t O a, where Ai = i i 1 for every i.
Proof.
Assume that there exists such T , then we have m paths of length two in Endop (T ) whose composition is zero.
Therefore we need exactly m triplets of mdiagonals satisfy case 1 in Lema 2.
Since the quiver of Endop (T ) is a path then there exist .
m-diagonals in Pm.
2 , where the configuration is as in Figure Thus it remains a m-diagonals.
Because I = hAt .
At 1 , .
At mOe1 i then we should have .
Oe .
m-diagonals whose endpoint is y1 and the other endpoint in arc Dx1 xm 2 while the remaining .
Oe .
Oe .
) m-diagonals have one endpoint at ym 1 and the other point in arc Dx1 ym 2 .
More precisely, the picture of all m-diagonals should be like Figure 24.
From Figure 24, m-diagonals which correspond to T are T1 .
T2 , .
Tm 2 .
X1 .
X2 , .
XtOe1 .
Y1 .
Y2 , .
YaOet 1 with Xi = .
1 , xi 1 ) and Yj = .
m 1 , zj ).
Note that for every 1 O i O t Oe 1, arc Dxi xi 1 has at least m sides.
We also have that either arc Dxj xjOe1 or arc Dz1 ym 1 has at least m sides.
Hence, the number of sides in arc DzaOet 1 xt is at least .
Oe .
Oe t .
m = .
am = m.
2, this contradicts the fact that arc Dxt zaOet 1 has at least one side.
The following lemma states that Nakayama algebra with consecutive relations of path of length two ending in AnOe2 is not m-CTA of type An .
Characterization of Nakayama m-CTA of type An Figure 23.
m-diagonals T1 .
T2 , .
Tm 2 Figure 24.
m-diagonals T1 , .
Tm 2 .
X1 .
X2 , .
XtOe1 .
Y1 , .
YaOet 1 Lemma 4.
Suppose that m < nOe2 and a = nOe2Oem then there is no m-cluster such that Endop (T ) O tilting object T of CA = KQ/I with Q is nOe1 1 OeIe 2 OeIe 3 Ie A A A Ie .
Oe .
OeOeOeIe n and I = hAj 1 .
Aj 2 , .
AnOe3 .
AnOe2 i for every 0 O j O a, where Ai = i i 1 for every i.
Proof.
The cases j = 0 and j = a have been proved in Lemma 4.
10 and Lemma 4.
Now assume that 1 < j < a, then the picture of m-diagonals which corresponds to Faisal and I.
Muchtadi-Alamsyah T in Pm.
2 is Observe that arc DynOej x1 has at least .
Oe .
sides, while Figure 25.
m-diagonals T1 .
T2 , .
TnOej .
X1 , .
Xj arc Dx1 xj 1 has at least jm sides.
Thus, the number of sides in arc DynOej xj 1 is at .
Oe .
jm = mn Oe jm n Oe j jm = n.
Oe j.
Since j < a we have m.
Oe j > m.
Oe a = n.
Oe .
Oe 2 Oe .
= m.
This contradicts the fact that Pm.
2 has .
Lemma 4.
Suppose that m < n Oe 2, a = n Oe 2 Oe m and T = T1 Oi T2 Oi A A A Oi TnOet Oi X1,j1 Oi X2,j2 Oi A A A Oi Xt,jt with 1 O j1 O j2 O A A A O jt O min.
, n Oe t Oe .
and a O t O n Oe 2 then T is an m-cluster tilting object of CA If t = a and jt = 1 then the algebra End (T ) = KQ/I with Q is and I generated by all paths of length two in the cycle.
If t > a then Endop (T ) O = KQ/I with Q is nOe1 1 OeIe 2 OeIe 3 Ie A A A Ie .
Oe .
OeOeOeIe n and I generated by .
Oe 2 Oe .
relations of paths of length two from .
Oe .
relations of paths of length two.
If t = a and jt 6= 1 then Endop (T ) O = KQ/I with Q is the quiver in part .
and I generated by m relations of paths of length two where AnOe2 OO I.
Characterization of Nakayama m-CTA of type An Proof.
First, consider case t = a and jt = 1, we get j1 = j2 = A A A = jt = 1.
Consequently T = T1 Oi T2 Oi A A A Oi Tm 2 Oi X1,1 Oi X2,1 Oi A A A Oi Xa,1 .
We have that Tm 2 = .
2, m.
and Xa,ja = Xa,1 = .
, m.
, it follows that Tm 2 and Xa,1 have a common endpoint m.
Hence, the picture of m-diagonals that corresponds to T is as in Figure 26.
It is clear Figure 26.
m-diagonals T1 .
T2 , .
Tm 1 .
X1,1 , .
Xa,1 that the algebra Endop (T ) satisfies the first part of the lemma.
Furthermore, if t > a then .
m 2 t m Oe n Oe 2 > .
Therefore TnOet = (.
m 2 t m Oe n Oe 2, .
m 3 t m Oe n Oe .
and Xt,1 = .
, .
either are not crossing each other or have a common endpoint in Pm.
Since t Ou a 1 then min.
, n Oe t Oe .
= m 1 or min.
, n Oe t Oe .
= n Oe t Oe 1.
follows that 1 O j1 O j2 O A A A O jt O min.
, n Oe t Oe .
O m and then we may use the same way as in the proof of Lemma 4.
If t = a then T = T1 Oi T2 Oi A A A Oi Tm 2 Oi X1,j1 Oi X2,j2 Oi A A A Oi Xa,ja .
The fact that m-diagonals which correspond to T are not crossing each other can be obtained by the same argument as in the proof of Lemma 4.
Because 2 O ja O m we have that Xa,ja does not have a common endpoint neither with Tm 2 nor at the point am 2.
Thus, we have the quiver of Endop (T ) is An .
Next, we will prove that AnOe2 OO I.
Consider m-diagonals Tm .
Tm 1 and Tm 2 in Pm.
2 in Figure 27.
Since ji O m then there is no m-diagonal Xi,ji that have a common endpoint at ym 1 .
So there exists an irreducible map Tm Ie Tm 1 Ie Tm 2 .
Because at the point xm 2 there is only one m-diagonal Tm 2 then this irreducible map corresponds to the path nOe2 nOe1 in Q.
But this path satisfies case 1 in Lemma 1, hence by Lemma 2.
AnOe2 = 0 in Endop (T ).
Faisal and I.
Muchtadi-Alamsyah Figure 27.
m-diagonals T1 .
T2 , .
Tm 2 Figure 28.
m-diagonals of T for m = 2 and n = 8 Example 4.
Let m = 2 and n = 8 then a = 8Oe2Oe2 = 4 and m.
2 = 20.
All m-diagonals which correspond to T in Lemma 4.
14 are as in Figure 28 Characterization of Nakayama m-CTA of type An Lemma 4.
Suppose that m < n Oe 2, a = n Oe 2 Oe m and T = T1 OiA A AOiTm 2 OiX1,j1 OiX2,j2 OiA A AOiXk,jk OiXk 1,m 1 OiA A AOiXaOe1,m 1 OiXa,m 1 with 1 O j1 O j2 A A A O jk O m and 1 O k < a then T is an m-cluster tilting object of CA .
If jk = 1 then the algebra Endop (T ) O = KQ/I with Q is and I generated by all paths of length two in the cycle.
If jk 6= 1 and jk O m then the algebra Endop (T ) O = KQ/I with Q is nOe1 3 Ie A A A Ie .
Oe .
OeOeOeIe n 2 OeIe 1 OeIe and I generated by m relations of paths of length two with Ak m 1 .
Ak m 2 , .
AnOe3 .
AnOe2 6OO I and Ak m OO I.
Proof.
Note that for k = a.
T is the m-cluster tilting object in Lemma 4.
14 part Assume that k < a, if jk = 1 then T = T1 Oi A A A Oi Tm 2 Oi X1,1 Oi X2,1 Oi A A A Oi Xk,1 Oi Xk 1,m 1 Oi A A A Oi XaOe1,m 1 Oi Xa,m 1 .
We have that m-diagonal Xk,1 = .
, .
and Xk 1,m 1 = Xk 1,2 if m = 1 or Xk 1,m 1 = (.
Oe .
1 Oe .
)m Oe .
1 Oe .
, .
m Oe .
1 Oe .
) = (.
Oe .
m 3, .
if m 6= 1.
If m 6= 1 then Xk,1 and Xk 1,m 1 have a common endpoint at .
If m = 1 then k = 1 and hence Xk 1,m 1 = X2,2 = .
2, .
Xk,1 = X1,1 = .
, .
It turns out that Xk 1,m 1 and Xk,1 have a common endpoint if m = So the picture of m-diagonals which correspond to T = T1 Oi A A A Oi Tm 2 Oi X1,1 Oi X2,1 Oi A A A Oi Xk,1 Oi Xk 1,m 1 Oi A A A Oi XaOe1,m 1 Oi Xa,m 1 in Pm.
2 is as in Figure 29.
For jk 6= 1 the configuration of m-diagonals T1 .
T2 , .
Tm 2 and Xk 1,m 1 , .
XaOe1,m 1 .
XaOe1,m 1 in Pm.
2 is the same as in the Figure It remains to consider the position of X1,j1 .
X2,j2 , .
Xk,jk in Pm.
2 if jk 6= 1 and jk O m.
By the same arguments as in the proof of Lemma 4.
4 then for Xi,ji and Xi 1,ji 1 in Pm.
2 will be one of the following pictures in Figure If jk O m then the number of black dots on the top line that can be the end point of Xi,ji except point 1 is m .
ee Figure .
Consequently the leftmost black dot on the top line is .
We claim that the ideal I generated by m relations of paths of length two.
From Figure 29 we have that T2 .
T3 , .
Tm 1 are m-diagonals that correspond to a midpoint of a path of length two that satisfies case 1 in Lemma 2.
1 while others m-diagonal satisfy case 2 in Lemma 2.
So the number of relations that generate I is only m.
Faisal and I.
Muchtadi-Alamsyah Figure 29.
m-diagonals T1 , .
Tm 2 .
X1,1 .
X2,1 , .
Xk,1 .
Xk 1,m 1 , .
XaOe1,m 1 .
Xa,m 1 Figure 30.
m-diagonals Xi,ji .
Xi 1,ji 1 Note that m-diagonals Tm 1 .
Xk 1,m 1 .
Xk 2,m 1 , .
Xa,m 1 .
Tm 2 have a common endpoint at .
Therefore there exists a composition of irreducible Tm 1 Ie Xk 1,m 1 Ie Xk 2,m 1 Ie A A A Ie Xa,m 1 Ie Tm 2 .
Characterization of Nakayama m-CTA of type An Since there is no other m-diagonal whose one endpoint is .
m 3 and in the arc D.
m 2,.
m 3 then this composition of irreducible maps correspond to k m 1 k m 2 nOe2 nOe1 .
OeOeOeOeOeIe .
OeOeOeOeOeIe A A A Ie .
Oe .
OeOeOeIe .
Oe .
OeOeOeIe n.
We conclude that Ak m 1 .
Ak m 2 , .
AnOe3 .
AnOe2 6OO I.
The path k m k m 1 .
OeOeOeOeIe .
OeOeOeOeOeIe .
in Q correspond to the composition of irreducible maps X Ie Tm 1 Ie Xk 1,m 1 where X = Tm or X = Xk,m .
Because either m-diagonals Tm .
Tm 1 .
Xk 1,m 1 or Xk,m .
Tm 1 .
Xk 1,m 1 always satisfy case 1 in Lemma 2.
1, then Ak m OO I.
Example 4.
Let m = 3 and n = 7 then a = nOemOe2 = 2 and m.
2 = 26.
The figure of m-diagonals that correspond to T in Lemma 4.
16 for this case is Figure 31.
m-diagonals of T for m = 3 and n = 7 Lemma 4.
16 gives us the information of m-CTA from type An which is a Nakayama algebra of acyclic type and have m relations.
Therefore we can compute the number of m-CTA from type An which has less than or equal to m relations.
By the second part of Lemma 4.
14, the number of m-CTA which have less than m relations of paths of length two is nOe2 nOe2 nOe2 nOe2 a mOe2 mOe1 Next, the possibility of thenumber of m-CTAs that have exactly m relations nOe2 of path of length two is But, by Lemma 4.
12 there are a m-CTAs who have m relations which are not Nakayama algebras of acyclic type and from Lemma 13 we get one more this kind.
So the numberof m-CTAs which have m relations nOe2 and whose quiver is An for this case is at most Oe .
We compute the number of m-cluster tilting objects in Lemma 4.
14 part .
together with Lemma 16 part .
Since 1 O j1 O j2 A A A O jk < m and jk 6= 1 then for every k the number of m-cluster tilting objects is Oe 1.
Because 1 O k O a then the Faisal and I.
Muchtadi-Alamsyah total number of m-cluster tilting objects in Lemma 4.
14 part .
and Lemma 4.
is Oe a.
Using PascalAos identity it can be proved that n a 1 We know that a = n Oe 2 Oe m, hence Oea= 1 Oe .
Oe .
Oe .
m a 1 Oe .
nOe2 Oe .
nOe2Oem nOe2 Oe .
We conclude that all m-CTAs which are Nakayama algebras of acyclic type and have m relations of paths of length two are the algebras in Lemma 4.
14 part .
and Lemma 4.
16 part .
We write the results so far for the case m < n Oe 2 in the following theorem.
Theorem 4.
Let H O with m < n Oe 2, and let I = KQ/I be an m-CTA of CA be an ideal generated by less than or equal to m relations of paths of length two and Q is nOe1 1 OeIe 2 OeIe 3 Ie A A A Ie .
Oe .
OeOeOeIe n.
Suppose that W = {Aj = j j 1 .
O j O n Oe .
then the generator of I is one of the following .
B OI W for any B with 0 O |B| < m.
B OI W for any B with |B| = m and B 6= {At .
At 1 , .
At mOe1 } for every 1 O t O a 1.
Proof.
Apply Lemma 4.
11, 4.
12, 4.
13, 4.
14, 4.
Until here we have known all m-CTAs H = KQ/I with Q = An and I generated by at most m relations of path of length two for the case m < n Oe 2.
Characterization of Nakayama m-CTA of type An Next we will give some m-CTAs whose ideal is generated by more than m relations of paths of length two.
Proposition 4.
Suppose that m < n Oe 2 and km O a with 1 O k O .
Oe .
then there is no m-cluster tilting object T of CA such that Endop (T ) O = KQ/I with Q nOe1 3 Ie A A A Ie .
Oe .
OeOeOeIe n 2 OeIe 1 OeIe and ideal I generated by at least .
Oe 2 Oe .
paths of length two.
Proof.
Assume that such m-cluster tilting object T exists.
First assume that a Ou k.
Since 1 O k O .
Oe .
and I generated by at least .
Oe 2 Oe .
relations of paths of length two then there exist .
m-diagonals which configuration is as in Figure 32.
Observe that Dym 2 x1 has at least .
Hence Figure 32.
m-diagonals Y1 .
Y2 , .
Ym 2 arc Dx1 ym 2 has at least m.
2 Oe .
= am Now there are a m-diagonals which are not shown in the Figure 32.
Since I is generated by at least .
Oe 2 Oe .
paths of length two, there exist m-diagonals X1 .
X2 , .
XaOek together with .
m-diagonal in the Figure 32 such that the configuration as in the Figure 33.
Note that arc Dxym 2 at least has .
Oe .
Since a Ou k.
we get .
Oe .
= am .
Oe k.
) Ou am, a contradiction.
Now assume that km O a < k.
Consider Figure 32, we obtain that arc Dx1 x has at least .
Oe .
Hence k.
Oe a O k.
Oe km O k.
But there exist k m-diagonals of T besides Y1 .
Y2 , .
Ym 1 .
Ym 2 .
X1 .
X2 , .
XaOek .
Each of them has one endpoint outside arc Dx1 x and the other endpoint should be in arc Dx1 x and different from x1 , x.
Since arc Dx1 x has at most k sides then there exist two m-diagonals from these k m-diagonals whose common endpoint is Faisal and I.
Muchtadi-Alamsyah Figure 33.
m-diagonals Y1 .
Y2 , .
Ym 2 .
X1 .
X2 , .
XaOek in arc Dx1 x.
Consequently the quiver of Endop (T ) has a cycle, a contradiction.
This completes the proof.
Consider Proposition 4.
19 for the case k = a Oe 1.
If k = a Oe 1 then a Ou .
Oe .
Ni a O 1 m We get that a must be equal to 1.
If a = 1 or equivalently n Oe 2 = m 1 then by Lemma 4.
10 the ideal I is generated by at most m relations of paths of length two.
Proposition 4.
Suppose that 2 O m < n Oe 2, 1 < a = .
Oe 2 Oe .
< m and T = T1 OiT2 OiA A AOiTm 2 OiTm 3 OiTm 4 OiA A AOiTm 2 t OiX1,j1 OiX2,j2 OiA A AOiXaOet,jaOet with 1 O j1 O j2 O A A A O jaOet O m 1, 1 O t O a Oe 1 and jaOet > t then T is an m-cluster tilting object of CA .
if jsOe1 = 1 and js = m 1 for 1 O s O a Oe t then the algebra Endop (T ) O KQ/I where Q is and I generated by all paths of length two in the cycle and t paths of length two from the right.
If jaOet = t 1 then Endop (T ) O = KQ/I where Q is Characterization of Nakayama m-CTA of type An and I generated by all paths of length two in the cycle and t paths of length two in the path 1 Ie 2 Ie 3 Ie A A A Ie a Ie a 1.
Otherwise Endop (T ) O = KQ/I where Q is nOe1 3 Ie A A A Ie .
Oe .
OeOeOeIe n 2 OeIe 1 OeIe and I generated by .
relations of paths of length two with AnOetOe1 , .
AnOe3 .
AnOe2 OO Proof.
It is clear that m-diagonals which correspond to T1 .
T2 , .
Tm 2 t are not crossing each other in Pm.
Now consider case .
that is jsOe1 = 1 and js = m 1 for 1 O s O a Oe t.
We get that T = T1 Oi T2 Oi A A A Oi Tm 2 t Oi X1,1 Oi X2,1 Oi A A A Oi XsOe1,1 Oi Xs,m 1 Oi Xs 1,m 1 Oi A A A Oi XaOet,m 1 .
We have that X1,1 = .
, 2m .
X2,1 = .
, 3m .
XsOe1,1 = .
, sm .
Xs,m 1 = .
m 3, sm .
Xs 1,m 1 = .
m 3, .
XaOet,m 1 = .
m 3, .
Oe .
Tm 2 t = (.
Oe .
m m 2 Oe t, .
Oe t .
m m 3 Oe .
It follows that XsOe1,1 and Xs,m 1 have a common endpoint.
Since t O a Oe 1 < m then m diagonals Tm 2 t .
XaOet,m 1 are not crossing each other and do not have a common endpoint.
We get the figure of m-diagonals which correspond to T for this case is as in Figure 34.
Now we come to the case 2, let jaOet = t Figure 34.
m-diagonals of T Note that XaOet,t 1 = (.
Oe t .
m 3 Oe t, .
Oe m Oe t Oe .
m 2 Oe .
and Faisal and I.
Muchtadi-Alamsyah Tm 2 t = (.
Oe m Oe t Oe .
m 2 Oe t, .
Oe m Oe .
m 3 Oe .
It turns out that XaOet,t 1 and Tm 2 t have a common endpoint and T1 .
T2 , .
Tm 2 t .
XaOet,t 1 are not crossing each other in Pm.
We obtain the figure of m-diagonals T1 .
T2 , .
Tm 2 t .
XaOet,t 1 in Pm.
2 as in Figure 35.
It is easy to check that Figure 35.
m-diagonals T1 .
T2 , .
Tm 2 .
XaOet,t 1 X1,jt .
X2,j2 , .
XaOetOe1,jaOetOe1 are not crossing each other since t O a Oe 1 < m and 1 O j1 O j2 O A A A O jaOet = t 1.
We end the case m < n Oe 2 by the above proposition.
We have not been able to find all m-CTAs which is Nakayama algebra type An .
This is because many cases on the value of a have to be considered and have different characteristics in some cases of the value of a.
However.
Proposition 4.
19 gives some m-CTAs which are not Nakayama algebras in the case km O a with 1 O k O a Oe 1.
While Proposition 4.
20 part .
give some m-CTAs which are Nakayama algebras in the case 1 < a < m and have more than m relations.
A way to find all m-CTAs which are Nakayama algebras in this case is by investigating all m-CTAs in each case km O a where 1 O k O a Oe 1.
Example 4.
The following figure shows m-diagonals correspond to m-cluster tilting objects in Proposition 4.
20 in the case m = 4 and n = 9.
Characterization of Nakayama m-CTA of type An Figure 36.
m-diagonals of T for m = 4 and n = 9 Acknowledgement.
The first author would like to thank Prof.
Aslak Bakke Buan for many helpful discussions while preparing this paper.
This research is supported by IMHERE Sandwich Program 2012.
References