# τ-Complexity and Tilting Modules

Authors: Lijing Zheng; Chonghui Huang; Qianhong Wan
Journal: Advances in Mathematical Physics
 (2016)

---

## Abstract

LetA be a finite dimensional algebra over an algebraic closed field k. In this note, we will show that if T is a separating and splitting tilting A-module, then τ-complexities of A and B are equal, where B
=
End
A
(
T
).


---

## Body

## 1. Introduction

Background. Tilting theory plays an important role in the modern representation theory of algebras. Let A be a finite dimensional algebra over a field k and T a tilting A-module. It is well known that A and End
A
(
T
) are derived equivalent. The endomorphism algebra of a tilting module preserves many significant invariants, for example, the center of an algebra, the number of nonisomorphic simple modules, the Hochschild cohomology groups, and Cartan determinants. In particular, if T is a separating and splitting tilting A-module (see the definition in Section 2), then End
A
(
T
) preserves representation dimension [1].On the other hand,τ-complexity (see the definition in Section 2) is an important invariant in the representation theory of algebras. With τ-complexity, Bergh and Oppermann described the classification of hereditary algebras and studied the classification of cluster tilted algebra [2].However, the precise value ofτ-complexity of a given algebra is not known in general, and it is hard to compute even for small examples. One possible way is to compare τ-complexities of “nicely” related algebras.Question. Suppose B is the endomorphism algebra of a tilting module T over an algebra A. What is the relationship between τ-complexities of A and B?Note that in generalA and B do not have the same τ-complexities, since there are examples where A is representation finite while B is representation infinite. Our main result in this paper is the following theorem.Theorem 1.
LetT be a tilting module over a finite dimensional k-algebra A, with B
=
End
A
⁡
(
T
). If T is separating and splitting, then c
τ
A
=
c
τ
B, where c
τ
A, c
τ
B denote τ-complexity of A and B, respectively.Organization. This paper is organized as follows. In Section 2, we shall give the proof of our main result Theorem 1. In Section 3, we shall give two examples to illustrate our results.


## 2. Proof of the Main Theorem

Throughout this paper,k is an algebraically closed field, A is a finite dimensional k-algebra. Denote by m
o
d
A the category of finitely generated left A-modules, P
(
m
o
d
A
) the full subcategory of m
o
d
A consisting of all projective objects in m
o
d
A, and g
l
·
d
i
m
A the global dimension of A. D
≔
Hom
k
(
-
,
k
) denotes the standard duality functor between m
o
d
A and m
o
d
A
op. Given a left A-module M, add
M denotes the full subcategory of m
o
d
A consisting of all direct summands of finite direct sums of copies of M.Torsion Pair. A pair (
T
,
F
) of full subcategories of m
o
d
A is called a torsion pair, if the following conditions are satisfied: (
1
)  Hom
A
(
M
,
N
)
=
0 for all M
∈
T, N
∈
F; (
2
)  Hom
A
M
,
-
F
=
0 implies M
∈
T; (
3
)  H
o
m
A
(
-
,
N
)
T
=
0 implies N
∈
F. A torsion pair (
T
,
F
) is called splitting if each indecomposable A-module lies either in T or in F.A moduleT is called a tilting module if the following three conditions are satisfied: (
1
)  p
d
T
≤
1; (
2
)  E
x
t
A
1
(
T
,
T
)
=
0; (
3
) there exists a short exact sequence: 0
→
A
→
T
1
→
T
2
→
0, with T
1
,
T
2
∈
add
⁡
T
.It is well known thatT
A induces a torsion pair (
T
T
,
F
T
) in m
o
d
A, and a torsion pair (
X
T
,
Y
T
) in m
o
d
B. T is said to be separating if the induced torsion pair (
T
T
,
F
T
) in m
o
d
A is splitting and said to be splitting if the induced torsion pair (
X
T
,
Y
T
) in m
o
d
B is splitting.The following lemma is crucial in this paper.Lemma 2.
LetA be a finite dimensional algebra and T a tilting A-module. Let F
=
H
o
m
A
(
T
,
-
), G
=
E
x
t
A
1
(
T
,
-
). Then the following assertions hold.(1)
For anyM
∈
T
T,
(a)
there exists a constantc
>
0, such that d
i
m
k
M
≤
c
d
i
m
k
F
M.
(b)
there exists a constantλ
>
0, such that d
i
m
k
F
M
≤
λ
d
i
m
k
M.
(2)
For anyM
∈
F
T,
(a)
there exists a constantc
′
>
0, such that d
i
m
k
M
≤
c
′
d
i
m
k
G
M.
(b)
there exists a constantλ
′
>
0, such that d
i
m
k
G
M
≤
λ
′
d
i
m
k
M.Proof.
The proof is similar to [3, Lemma  3.1, Lemma  3.3]. There exists a short exact sequence of the form 0
→
A
→
T
1
→
T
2
→
0 where T
1
,
T
2
∈
add
T since T is a tilting A-module. Denote by t the number of indecomposable summands of T
1
⊕
T
2. Given a module M, apply Hom
A
(
-
,
M
) to the short exact sequence above to obtain the long exact sequence: ∗
0
⟶
Hom
A
T
2
,
M
⟶
Hom
A
T
1
,
M
⟶
Hom
A
A
,
M
⟶
E
x
t
A
1
T
2
,
M
⟶
E
x
t
A
1
T
1
,
M
⟶
0
.
(1)
(a)
Assume thatM is a torsion module. Then E
x
t
A
1
(
T
2
,
M
)
=
0 in the long exact sequence because T
2 is in add
T and M is torsion. We obtain the short exact sequence 0
→
Hom
A
(
T
2
,
M
)
→
Hom
A
(
T
1
,
M
)
→
Hom
A
(
A
,
M
)
→
0. Noting that Hom
A
(
A
,
M
)
≅
M, we have d
i
m
k
M
≤
t
d
i
m
k
Hom
(
T
,
M
)
=
t
d
i
m
k
F
M.
(b)
For any finitely generatedA-module M, we have d
i
m
k
F
M
≤
d
i
m
k
T
·
d
i
m
k
M [3, Remark  3.2], we set λ
=
d
i
m
k
T, and then the assertion follows immediately.
(2)
(a)
Assume thatM is torsion-free. In the long exact sequence ∗ above, we now have Hom
A
(
T
1
,
M
)
=
0 since M is torsion-free. We thus have the short exact sequence 0
→
Hom
A
(
A
,
M
)
→
E
x
t
A
1
(
T
2
,
M
)
→
E
x
t
1
1
(
T
1
,
M
)
→
0
. Noting that Hom
A
(
A
,
M
)
≅
M, we have d
i
m
k
M
≤
d
i
m
k
E
x
t
A
1
(
T
2
,
M
)
≤
d
i
m
k
E
x
t
A
1
(
T
t
,
M
)
≤
t
d
i
m
k
E
x
t
A
1
(
T
,
M
)
=
t
d
i
m
k
G
M
.
(b)
For any finitely generatedA-module M, we have d
i
m
k
G
M
≤
d
i
m
k
T
·
d
i
m
k
M
·
(
d
i
m
k
A
)
2 [3, Remark  3.2], we set λ
′
=
d
i
m
k
T
·
(
d
i
m
k
A
)
2, and then the assertion follows immediately.LetT be a tilting A-module, B
=
End
A
(
T
). Denote by τ
A, τ
B the Auslander-Reiten translation in m
o
d
A and m
o
d
B, respectively. Let M
∈
m
o
d
A; the τ
A-complexity of M is defined as follows:(1)
c
τ
A
M
=
inf
⁡
t
∈
N
∣
∃
λ
∈
R
such  that
d
i
m
k
τ
A
i
M
≤
λ
i
t
-
1
for
i
≫
0
.When no such t
∈
N exists, we say that τ
A-complexity of M is infinite and write c
τ
A
(
M
)
=
∞
. And τ-complexity of the algebra A is defined to be the supreme of τ
A-complexities of all the finitely generated A-modules, which will be denoted by c
τ
A.Proof of Theorem1.
Step  1 (c
τ
A
≤
c
τ
B). We will show that, for each indecomposable A-module M, there exists N
∈
m
o
d
B such that c
τ
A
(
M
)
=
c
τ
B
(
N
).
Case  1 (0
≠
M
∈
F
T). In this case, by [4, Chapter VI, Proposition  1.7], for any i
>
0, τ
A
i
M
∈
F
T.
Case  1.1. There exists an integer i
0
>
0, such that τ
A
i
0
(
M
)
=
0. Then c
τ
A
(
M
)
=
c
τ
A
(
τ
A
i
0
M
)
=
0
. Let N
=
0
∈
m
o
d
B; then c
τ
A
(
M
)
=
c
τ
B
(
N
)
.
Case  1.2. For any integer i
>
0, τ
A
i
M
≠
0. In this case, we will show that c
τ
A
(
M
)
=
c
τ
B
(
G
M
) and the assertion holds. By [4, Chapter VI, Lemma  5.3(b)], τ
B
i
(
G
M
)
≅
G
(
τ
A
i
M
). Set c
0
=
c
τ
A
(
M
) and d
0
=
c
τ
B
(
G
(
M
)
), respectively. By definition, there exist constants λ
>
0 and λ
′
>
0 such that d
i
m
k
τ
A
i
M
≤
λ
i
c
0
-
1 and d
i
m
k
τ
B
i
G
M
≤
λ
′
i
d
0
-
1. By Lemma 2(2)(b), there exists a constant c such that d
i
m
k
τ
B
i
G
M
=
d
i
m
k
G
(
τ
A
i
M
)
≤
c
d
i
m
k
τ
A
i
M
≤
c
λ
i
c
0
-
1. Therefore d
0
≤
c
0. On the other hand, by Lemma 2(2)(a), there exists a constant c
′ such that d
i
m
k
τ
A
i
M
≤
c
′
d
i
m
k
τ
B
i
(
G
M
)
≤
c
′
λ
′
i
d
0
-
1. It implies that c
0
≤
d
0, and hence c
τ
A
(
M
)
=
c
τ
B
(
G
M
).
Case  2 (0
≠
M
∈
T
T).
Case  2.1. For any i
≥
0, τ
A
i
M
∉
F
T. In this case, 0
≠
τ
A
i
M
∈
T
T. By [4, Chapter VI, Lemma  5.3(a)], we have that τ
B
i
(
F
M
)
≅
F
(
τ
A
i
M
) for any i
≥
0. By Lemma 2(1) and the similar argument of Case  1, we have c
τ
A
(
M
)
=
c
τ
B
(
F
(
M
)
).
Case  2.2. There exists an integer i
0
>
0, such that τ
A
i
0
(
M
)
∈
F
T
. Let i
0 be the minimal integer such that τ
A
i
0
(
M
)
∈
F
T. And then 0
≠
τ
A
i
0
-
1
(
M
)
∈
T
T. Since T is separating and splitting, we obtain that τ
A
i
0
-
1
(
M
)
∈
add
T by [4, Chapter VI, Proposition  1.11]. And we have the following two subcases:(i)
τ
A
i
0
(
M
)
=
0. In this case, c
τ
A
(
M
)
=
0. Set N
=
0
∈
m
o
d
B; then c
τ
A
(
M
)
=
c
τ
B
(
N
)
.
(ii)
0
≠
τ
A
i
0
(
M
)
∈
F
T. By Case  1, there exists N
∈
m
o
d
B, such that c
τ
A
(
τ
A
i
0
M
)
=
c
τ
B
(
N
), and then c
τ
A
(
M
)
=
c
τ
A
(
N
).
Step  2 (c
τ
A
≥
c
τ
B). We will show that for any indecomposable B-module X, there exists N
∈
m
o
d
A such that c
τ
B
(
X
)
=
c
τ
A
(
N
), and the assertion holds.
Case  1 (0
≠
X
∈
X
T). In this case, there exists 0
≠
N
∈
F
T such that X
≅
G
(
N
)
. If N belongs to Case  1.1 in Step  1, then denote i
0 by the minimal positive integer such that τ
A
i
0
N
=
0. Then τ
A
i
0
-
1
N
≅
P
≠
0, where P is an indecomposable projective A-module and does not belong to add
T (if P belongs to add
T, then P
∈
T
T
∩
F
T, contradiction!). Now, by [4, Chapter VI, Lemma  5.3(b)] we have that τ
B
i
0
-
1
(
X
)
≅
E
x
t
A
1
(
T
,
τ
A
i
0
-
1
N
)
≅
E
x
t
A
1
(
T
,
P
). Therefore, by [4, Chapter VI, Lemma  4.9], τ
B
i
0
(
X
)
≅
τ
B
E
x
t
A
1
(
T
,
P
)
≅
Hom
A
(
T
,
E
(
top
P
)
), where E
(
top
P
) denotes the injective hull of top
P. Clearly, 0
≠
E
(
top
P
)
∈
T
T
. If E
(
top
P
) belongs to Case  2.1 in Step  1, then c
τ
A
(
E
(
top
P
)
)
=
c
τ
B
(
F
(
E
(
top
P
)
)
)
=
c
τ
B
(
τ
B
i
0
(
X
)
)
=
c
τ
B
(
X
). Otherwise, E
(
top
P
) belongs to Case  2.2 in Step  1; then there exists j
0
>
0 such that τ
A
j
0
-
1
(
E
(
top
P
)
)
∈
add
T. By [4, Chapter VI, Lemma  5.3(a)], we know τ
B
j
0
-
1
(
τ
B
i
0
(
X
)
)
≅
τ
B
j
0
-
1
(
Hom
A
(
T
,
E
(
top
P
)
)
)
≅
Hom
A
(
T
,
τ
A
j
0
-
1
E
(
top
P
)
). Therefore, τ
B
j
0
+
i
0
-
1
(
X
) is a projective B-module. It implies that c
τ
B
(
X
)
=
0
.
Case  2 (X
∈
Y
T). In this case, there exists 0
≠
N
∈
T
T such that Y
≅
Hom
A
(
T
,
N
).
IfN belongs to Case  2.1 in Step  1, then c
τ
B
(
X
)
=
c
τ
B
(
F
(
N
)
)
=
c
τ
A
(
N
).
OtherwiseN belongs to Case  2.2 in Step  1; that is, there exists a positive integer i
0 such that τ
A
i
0
(
N
)
∈
F
T and τ
A
i
0
-
1
(
N
)
∉
F
T
. In this case, τ
A
i
0
-
1
N
∈
add
T. By [4, Chapter VI, Lemma  5.3(a)], τ
B
i
0
-
1
(
X
)
≅
Hom
A
(
T
,
τ
A
i
0
-
1
N
)
. Therefore, τ
B
i
0
-
1
(
X
) is a projective B-module, and c
τ
B
(
X
)
=
0
.Proposition 3 (See [2, Proposition  3.1]).
LetA be a finite dimensional hereditary algebra. Then the following assertions hold.(1)
A is of finite representation type if and only if c
τ
A
=
0.
(2)
A is of tame representation type if and only if c
τ
A
=
2.
(3)
A is of wild representation type if and only if c
τ
A
=
∞.Combining Theorem1 and Proposition 3, we have the following corollary.Corollary 4.
LetA be a finite dimensional hereditary algebra and T a separating and splitting A-module, with B
=
End
A
(
T
). If B is a hereditary algebra, then A and B are of the same representation type.Remark 5.
LetA be a finite dimensional hereditary algebra and T a APR-tilting module. It is well known that B
=
End
(
T
) is a hereditary algebra. Therefore, they are of the same representation type.


## 3. Two Examples

In this section, we shall give two examples to illustrate our result.Example 1.
LetA be the path algebra of the Euclidean quiver Q:(2)
Consider the indecomposableA-modules: T
1, T
2, T
3, and T
4, respectively:(3)
ThenT
=
T
1
⊕
T
2
⊕
T
3
⊕
T
4 is a tilting module, and B
=
End
A
(
T
) is the algebra given by the following quiver(4)
bound byα
β
=
0, γ
δ
=
0. This example was given as Example  4.8(a) in [4, Chapter VIII]. Clearly, T is splitting, but it is not separating. A is a hereditary algebra and of tame representation type; then c
τ
A
=
2
. However, B is of finite representation type and of finite global dimension, c
τ
B
=
0
.Example 2.
LetA be the algebra (over a field k) given by the quiver(5)
with relationsα
β
=
σ
γ
=
η
δ
. This example was also given in [1].LetA
T
=
τ
-
1
S
4
⊕
P be the APR-tilting module coresponding to the vertex 4. Then A
T is separating, but it isn’t splitting. By [5], B
=
End
A
(
T
) is given by the quiver(6)with relationsα
β
=
0, α
′
β
′
=
0. Since A is of finite global dimension and finite representation type, c
τ
A
=
0, however, B is of infinite representation type, c
τ
B
>
0
.



---

*Source: 5181730-2016-06-07.xml*