Advances in Mathematical Physics

Volume 2016, Article ID 5181730, 4 pages

http://dx.doi.org/10.1155/2016/5181730

## -Complexity and Tilting Modules

^{1}School of Mathematics and Physics, University of South China, Hengyang, Hunan 421001, China^{2}School of Mathematics and Statistics, Hunan University of Commerce, Changsha, Hunan 410205, China

Received 6 January 2016; Accepted 15 May 2016

Academic Editor: André Nicolet

Copyright © 2016 Lijing Zheng et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

Let be a finite dimensional algebra over an algebraic closed field . In this note, we will show that if is a separating and splitting tilting -module, then -complexities of and are equal, where .

#### 1. Introduction

*Background*. Tilting theory plays an important role in the modern representation theory of algebras. Let be a finite dimensional algebra over a field and a tilting -module. It is well known that and 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 is a separating and splitting tilting -module (see the definition in Section 2), then preserves representation dimension [1].

On the other hand, -complexity (see the definition in Section 2) is an important invariant in the representation theory of algebras. With -complexity, Bergh and Oppermann described the classification of hereditary algebras and studied the classification of cluster tilted algebra [2].

However, the precise value of -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 -complexities of “nicely” related algebras.

*Question*. Suppose is the endomorphism algebra of a tilting module over an algebra . What is the relationship between -complexities of and ?

Note that in general and do not have the same -complexities, since there are examples where is representation finite while is representation infinite. Our main result in this paper is the following theorem.

Theorem 1. *Let be a tilting module over a finite dimensional -algebra , with . If is separating and splitting, then , where , denote -complexity of and , 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, is an algebraically closed field, is a finite dimensional -algebra. Denote by the category of finitely generated left -modules, the full subcategory of consisting of all projective objects in , and the global dimension of . denotes the standard duality functor between and . Given a left -module , denotes the full subcategory of consisting of all direct summands of finite direct sums of copies of .

*Torsion Pair*. A pair of full subcategories of is called a torsion pair, if the following conditions are satisfied: for all , ; implies ; implies . A torsion pair is called splitting if each indecomposable -module lies either in or in .

A module is called a tilting module if the following three conditions are satisfied: ; ; there exists a short exact sequence: , with

It is well known that induces a torsion pair in , and a torsion pair in . is said to be separating if the induced torsion pair in is splitting and said to be splitting if the induced torsion pair in is splitting.

The following lemma is crucial in this paper.

Lemma 2. *Let be a finite dimensional algebra and a tilting -module. Let , . Then the following assertions hold.*(1)*For any ,(a) there exists a constant , such that .(b)there exists a constant , such that .*(2)

*For any ,(a)*

*there exists a constant , such that .*(b)*there exists a constant , such that .**Proof. *The proof is similar to [3, Lemma 3.1, Lemma 3.3]. There exists a short exact sequence of the form where since is a tilting -module. Denote by the number of indecomposable summands of . Given a module , apply to the short exact sequence above to obtain the long exact sequence: (1) (a)Assume that is a torsion module. Then in the long exact sequence because is in and is torsion. We obtain the short exact sequence . Noting that , we have .(b) For any finitely generated -module , we have [3, Remark 3.2], we set , and then the assertion follows immediately.(2) (a)Assume that is torsion-free. In the long exact sequence above, we now have since is torsion-free. We thus have the short exact sequence Noting that , we have (b)For any finitely generated -module , we have [3, Remark 3.2], we set , and then the assertion follows immediately.

*Let be a tilting -module, . Denote by , the Auslander-Reiten translation in and , respectively. Let ; the -complexity of is defined as follows:When no such exists, we say that -complexity of is infinite and write And -complexity of the algebra is defined to be the supreme of -complexities of all the finitely generated -modules, which will be denoted by .*

*Proof of Theorem 1. * *Step 1* (). We will show that, for each indecomposable -module , there exists such that .*Case 1* (). In this case, by [4, Chapter VI, Proposition 1.7], for any , .*Case 1.1*. There exists an integer , such that . Then Let ; then *Case 1.2*. For any integer , . In this case, we will show that and the assertion holds. By [4, Chapter VI, Lemma 5.3(b)], . Set and , respectively. By definition, there exist constants and such that and . By Lemma 2(2)(b), there exists a constant such that . Therefore . On the other hand, by Lemma 2(2)(a), there exists a constant such that . It implies that , and hence .*Case 2* ().*Case 2.1*. For any , . In this case, . By [4, Chapter VI, Lemma 5.3(a)], we have that for any . By Lemma 2(1) and the similar argument of Case 1, we have .*Case 2.2*. There exists an integer , such that Let be the minimal integer such that . And then . Since is separating and splitting, we obtain that by [4, Chapter VI, Proposition 1.11]. And we have the following two subcases:(i). In this case, . Set ; then (ii). By Case 1, there exists , such that , and then .*Step 2* (). We will show that for any indecomposable -module , there exists such that , and the assertion holds.*Case 1* (). In this case, there exists such that If belongs to Case 1.1 in Step 1, then denote by the minimal positive integer such that . Then , where is an indecomposable projective -module and does not belong to (if belongs to , then , contradiction!). Now, by [4, Chapter VI, Lemma 5.3(b)] we have that . Therefore, by [4, Chapter VI, Lemma 4.9], , where denotes the injective hull of . Clearly, If belongs to Case 2.1 in Step 1, then . Otherwise, belongs to Case 2.2 in Step 1; then there exists such that . By [4, Chapter VI, Lemma 5.3(a)], we know . Therefore, is a projective -module. It implies that *Case 2* (). In this case, there exists such that .

If belongs to Case 2.1 in Step 1, then .

Otherwise belongs to Case 2.2 in Step 1; that is, there exists a positive integer such that and In this case, . By [4, Chapter VI, Lemma 5.3(a)], Therefore, is a projective -module, and

*Proposition 3 (See [2, Proposition 3.1]). Let be a finite dimensional hereditary algebra. Then the following assertions hold.(1) is of finite representation type if and only if .(2) is of tame representation type if and only if .(3) is of wild representation type if and only if .*

*Combining Theorem 1 and Proposition 3, we have the following corollary.*

*Corollary 4. Let be a finite dimensional hereditary algebra and a separating and splitting -module, with . If is a hereditary algebra, then and are of the same representation type.*

*Remark 5. *Let be a finite dimensional hereditary algebra and a APR-tilting module. It is well known that is a hereditary algebra. Therefore, they are of the same representation type.

*3. Two Examples*

*3. Two Examples*

*In this section, we shall give two examples to illustrate our result.*

*Example 1. *Let be the path algebra of the Euclidean quiver : Consider the indecomposable -modules: , , , and , respectively:Then is a tilting module, and is the algebra given by the following quiver bound by , . This example was given as Example 4.8(a) in [4, Chapter VIII]. Clearly, is splitting, but it is not separating. is a hereditary algebra and of tame representation type; then However, is of finite representation type and of finite global dimension,

*Example 2. *Let be the algebra (over a field ) given by the quiver with relations This example was also given in [1].

*Let be the APR-tilting module coresponding to the vertex 4. Then is separating, but it isn’t splitting. By [5], is given by the quiver*

* with relations , . Since is of finite global dimension and finite representation type, , however, is of infinite representation type, *

*Competing Interests*

*Competing Interests*

*The authors declare that they have no competing interests.*

*Acknowledgments*

*Acknowledgments*

*Lijing Zheng’s work is partly supported by Natural Science Foundation of China no. 11271119 and supported by Hunan Provincial Natural Science Foundation of China no. 2016JJ6124. Chonghui Huang’s work is partly supported by Natural Science Foundation of China no. 11201220. Qianhong Wan’s work is partly supported by Hunan Provincial Natural Science Foundation of China no. 2016JJ6049.*

*References*

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