τ-Tilting modules over one-point extensions by a simple module at a source point
aa r X i v : . [ m a t h . R T ] F e b τ -Tilting modules over one-point extensionsby a simple module at a source point ∗ Hanpeng Gao
Department of Mathematics, Nanjing University, Nanjing 210093, Jiangsu Province, P.R. China
Abstract
Let B be an one-point extension of a finite dimensional k -algebra A by a simple A -module at a source point i . In this paper, we classify the τ -tilting modules over B . Moreover, it is shown that there are equations | τ -tilt B | = | τ -tilt A | + | τ -tilt A/ h e i i| and | s τ -tilt B | = 2 | s τ -tilt A | + | s τ -tilt A/ h e i i| . As a consequence, we can calculate the numbers of τ -tilting modules and support τ -tilting modules over linearly Dynkin type algebras whose square radical are zero. : 16G20, 16G60. Key words : support τ -tilting modules, one-point extensions, Dynkin type algebras. As a generalization of tilting module, the concept of support τ -tilting modules is intro-duced by Adachi, Iyama and Reiten[2]. They are very important in representation theoryof algebras because they are in bijection with some important objects including functo-rially finite torsion classes, 2-term silting complexes, cluster-tilting objects. It is veryinteresting to calculate the number of support τ -tilting modules over a given algebra.For Dynkin type algebras ∆ n , the numbers of tilting modules and support tiltingmodules were first calculated in [6] via cluster algebras, and later in [8] via representationtheory.Recall that a finite-dimensional k -algebra is said to be a N akayama algebra if ev-ery indecomposable projective module and every indecomposable injective module has aunique composition series. May authors calculate the numbers of τ -tilting modules andsupport τ -tilting modules over Nakayama algebras. In particular, for square radical zeroNakayama algebra Λ n with n simple modules, there are the following recurrence relations(see, [1, 3, 5]), | τ -tilt Λ n | = | τ -tilt Λ n − | + | τ -tilt Λ n − | and | s τ -tilt Λ n | = 2 | s τ -tilt Λ n − | + | s τ -tilt Λ n − | . E-mail: [email protected] ∗ This work was partially supported by NSFC (Grant No. 11971225).
1n this paper, we consider τ -tilting modules and support τ -tilting modules over theone-point extension B of A by a simple A -module at a source point i . We will show thatthere is a bijection τ -tilt B τ -tilt A a τ -tilt A/ h e i i . We also get the following equations, | τ -tilt B | = | τ -tilt A | + | τ -tilt A/ h e i i| and | s τ -tilt B | = 2 | s τ -tilt A | + | s τ -tilt A/ h e i i| . As an application , we can calculate the numbers of τ -tilting modules and support τ -tiltingmodules over linearly Dynkin type algebras whose square radical are zero.Throughout this paper, all algebras will be basic, connected, finite dimensional k -algebras over an algebraically closed field k . For an algebra A , we denote by mod A thecategory of finitely generated left A -modules and by τ the Auslander-Reiten translationof A . Let P i be the indecomposable projective module and S i the simple module of A corresponding to the point i for i = 1 , , · · · , n . For M ∈ mod A , we also denote by | M | the number of pairwise nonisomorphic indecomposable summands of M and by add M the full subcategory of mod A consisting of direct summands of finite direct sums of copiesof M . For a set X , we denote by | X | the cardinality of X . For two sets X, Y , X ` Y means the disjoint union. Let A be an algebra. we recall the definition about support τ -tilting modules. Definition 2.1. ([2, Definition 0.1])
Let M ∈ mod A .(1) M is τ -rigid if Hom A ( M, τ M ) = 0 .(2) M is τ -tilting if it is τ -rigid and | M | = | A | .(3) M is support τ -tilting if it is a τ -tilting A/AeA -module for some idempotent e of A . We will denote by τ -tilt A (respectively, s τ -tilt A ) the set of isomorphism classes ofbasic τ -tilting (respectively, support τ -tilting) A -modules.Let X ∈ mod A . The one-point extension of A by X is defined as the following matrixalgebra B = (cid:18) A X k (cid:19) with the ordinary matrix addition and the multiplication, and we write B := A [ X ] with a the extension point.Let A be an algebra with a source point i , in this paper, we always assume B := A [ S i ].In this case, P a is an indecomposable projective-injective B -module and S a is a simpleinjective B -module by [4, Proposition 2.5(c)]. Lemma 2.2.
Let M ∈ mod B . If M is τ -rigid, then M ⊕ P a is also. Moreover, if M is τ -tilting, then it have P a as a direct summand. roof. Since S i is simple, there are only two indecomposable B -modules P a , S a whichhave S a as a composition factor and they are injective, we have τ M have no S as acomposition factor.Thus, Hom A ( P a , τ M ) = 0 and, we get M ⊕ P a is τ -rigid. If M is τ -tilting, then it is maximal τ -rigid by [2, Theorem 2.12]. Hence, it have P a as a directsummand. Theorem 2.3.
There is a bijection τ - tilt B ←→ τ - tilt A a τ - tilt A/ h e i i . Proof.
Let M ∈ τ -tilt A ` τ -tilt A/ h e i i . If M ∈ τ -tilt A , then τ M has no S a as a compo-sition factor since the vertex a is a source in B , and hence Hom B ( P a , τ M ) = 0. Therefore, M ⊕ P a is a τ -tilting B -module since it is τ -rigid and | M ⊕ P a | = | M | + 1 = | A | + 1 = | B | .If M ∈ τ -tilt A/ h e i i , then M has no S i as a composition factor and τ M has no S a as acomposition factor. Note that there is an almost split sequence 0 → S i → P a → S a → τ S a = S i . Thus,Hom B ( M ⊕ P a ⊕ S a , τ ( M ⊕ P ⊕ S a )) = Hom B ( M ⊕ P a ⊕ S a , τ M ⊕ S i ) = 0 . So, M ⊕ P a ⊕ S a is a τ -tilting B -module.Conversely, Let M ∈ τ -tilt B . Then we decompose M as M = P a ⊕ N by Lemma2.2. If N has no S a as direct summand, the N is a τ -tilting B/ h e a i ( ∼ = A )-module, thatis, N ∈ τ -tilt A . If N has S a as direct summand, then we decompose N as N = S a ⊕ L where L has no S a as a composition factor. We claim that L has no S i as a compositionfactor. Otherwise, there is a summand K of L such that the top of K is S i since i is asource point. In particular, Hom B ( L, S i ) = 0. This impliesHom B ( M, τ M ) = Hom B ( L ⊕ P a ⊕ S a , τ L ⊕ S i ) = 0 . This is a contradiction. Hence, L is a τ -tilting A/ h e i i -module , that is, L ∈ τ -tilt A/ h e i i . Corollary 2.4.
All τ -tilting B -modules are exactly those forms P a ⊕ M and P a ⊕ S a ⊕ M where M and M are τ -tilting modules over A and A/ h e i i respectively. The above Corollary give a relation about | τ -tilt B | and | τ -tilt A | . Corollary 2.5.
We have | τ - tilt B | = | τ - tilt A | + | τ - tilt A/ h e i i| . Let A be an algebra and M ∈ mod A . M is called a (classical) tilting module if(1) The projective dimension of M is at most one.(2) Ext A ( M, M ) = 0.(3) | M | = | A | .Hence, an A -module M is tilting if and only if it is a τ -tilting and its projectivedimension is at most one by the Auslander-Reiten formula. The set of all tilting A -modules will be denoted by tilt A . 3 orollary 2.6. Let A be an algebra with a source i . Assume that i is not a sink and B := A [ S i ] . All tilting B -modules are exactly those forms P a ⊕ M where M is a tiltingmodule over A . In particular, | tilt B | = | tilt A | .Proof. By Corollary 2.4, All τ -tilting B -modules are exactly those forms P a ⊕ M and P a ⊕ S a ⊕ M where M and M are τ -tilting modules over A and A/ h e i i respectively. Notethat the projective dimension of M as A -module is equal to the projective dimension of M as B -module since a is a source of B . Hence P a ⊕ M is a tilting B -module if andonly if M is a tilting A -module.Since there is an exact sequence 0 → S i → P a → S a → B , we have theprojective dimension of S a is at least two since S i is not projective when i is not a sink.Hence P a ⊕ S a ⊕ M is not tilting. Thus, | tilt B | = | tilt A | . Example 2.7.
Let B be a algebra given by the quiver31 / / sss % % ❑❑❑ = 0. Assume that A is the path algebra given by the quiver 3 2 o o / / B = A [2]. There are five τ -tilting A -modules as follows (there are exactly alltilting- A -modules since A is hereditary) ,
24 23 4 4 , ,
24 23 4 2 , . We only have one τ -tilting A/ h e i -module . Hence, we get all τ -tilting B -modules byCorollary 2.4.
12 3 23 4 4 ,
12 24 23 4 4 ,
12 3 23 4 23 ,
12 24 23 4 2 ,
12 2 23 4 23 ,
12 1 3 4 . By Corollary 2.6,
12 3 23 4 4 ,
12 24 23 4 4 ,
12 3 23 4 23 ,
12 24 23 4 2 ,
12 2 23 4 23 are all tilting B -modulesNext, we will consider the relationship between s τ -tilt B and s τ -tilt A . We need thefollowing notions.Let A be an algebra. The support τ -tilting quiver (or Hasse quiver) H( A ) of A is definedas follows (more detail can be found [2, Definition 2.29]) • vertices : the isomorphisms classes of basic support τ -tilting A -modules. • arrows: from a module to its left mutation.It is well known that H( A ) is a poset. Let N be a subposet of H( A ) and N ′ := H( A ) \N .We define a new quiver H( A ) N from H( A ) as follows. • vertices : vertices in H( A ) and N + where N + is a copy of N . • arrows: { a → a | a → a ∈ N ′ } ` { n → a | n → a , n ∈ N , a ∈ N ′ } ` { n → n , n +1 → n +2 | n → n ∈ N } ` { a → n +1 | a → n , n ∈ N , a ∈N ′ } ` { n +1 → n | n ∈ N } . (cid:15) (cid:15) * * ❱❱❱❱❱❱❱❱❱ a (cid:15) (cid:15) + + ❲❲❲❲❲❲❲❲❲❲❲ n (cid:15) (cid:15) n +1 " " ❊❊ x x rrrr n ' ' ❖❖❖❖❖ n +2 z z ✈✈ n s s ❣❣❣❣❣❣❣❣❣ n r r ❢❢❢❢❢❢❢❢❢❢❢ a a H( A ) H( A ) N . Suppose that A is an algebra with an indecomposable projective-injective module Q .Let A := A/ soc( Q ) and N := { N ∈ s τ -tilt A | Q/ soc( Q ) ∈ add N and Hom A ( N, Q ) = 0 } . The following Lemma can be found in [1, Theorem 3.3].
Lemma 2.8.
Let A be an algebra with an indecomposable projective-injective module Q .Then there is an isomorphism of posets H( A ) ←→ H ( A ) N . Applying this result to the algebra B , we have the following Proposition 2.9.
Let N := { S a ⊕ L | L ∈ s τ - tilt A/ h e i i} . Then there is an isomorphismof posets H( B ) ←→ H ( A × k ) N . Proof.
Take Q = P a which is an indecomposable projective-injective B -module. Sincesoc P a ∼ = S i , we have B = B/S i ∼ = A × k and P a /S i ∼ = S a . We only need to show N = { S a ⊕ L | L ∈ s τ -tilt A/ h e i i} in Lemma 2.8. Note that N = { N ∈ s τ -tilt B | Q/ soc( Q ) ∈ add N and Hom B ( N, Q ) = 0 } = { N ∈ s τ -tilt( A × k ) | S a ∈ add N and Hom B ( N, P a ) = 0 } = { S a ⊕ L | L ∈ silt A and Hom B ( L, P a ) = 0 } = { S a ⊕ L | L ∈ silt A and Hom A ( L, S i ) = 0 } . Since i is a source point of A , this implies L has no S i as a composition factor and henceit is exactly a support τ -tilting A/ h e i i -module. Thus, N = { S a ⊕ L | L ∈ s τ -tilt A/ h e i i} . Corollary 2.10.
We have | s τ - tilt B | = 2 | s τ - tilt A | + | s τ - tilt A/ h e i i| . Proof.
According to the definition of H ( A × k ) N , we have | H ( A × k ) N | = | H ( A × k ) | + |N | = 2 | H( A ) | + |N | = 2 | s τ -tilt A | + | s τ -tilt A/ h e i i| . Therefore, | s τ -tilt B | = | H( B ) | = 2 | s τ -tilt A | + | s τ -tilt A/ h e i i| by Proposition 2.9.5ow, it is easy to draw the quiver of H( B ) from the quiver of H( A ) as follows.H( A ) → H( A × k ) → H ( A × k ) N ∼ = H( B ) . Example 2.11.
Let A be a finite dimensional k -algebra given by the quiver2 −→ . Considering the one-point extension of A by the simple module corresponding to the point2, the algebra B = A [2] is given by the quiver3 α −→ β −→ αβ = 0. We can get the Hasse quiver H( B ) of B as follows where N isremarked by red and N + by blue.
21 1 (cid:15) (cid:15) { { ✈✈✈✈
21 1 (cid:15) (cid:15) } } ③③③ (cid:15) (cid:15) w w ♣♣♣♣♣ o o
21 1 (cid:15) (cid:15) } } ③③③
32 21 1 (cid:27) (cid:27) ✼✼✼✼✼✼✼✼✼ y y ttt o o
21 2 (cid:15) (cid:15)
21 2 (cid:15) (cid:15) (cid:15) (cid:15) o o
21 2 (cid:15) (cid:15)
32 21 2 (cid:15) (cid:15) o o (cid:15) (cid:15) (cid:15) (cid:15) (cid:15) (cid:15) o o (cid:15) (cid:15) (cid:15) (cid:15) o o (cid:15) (cid:15) o o ❍❍❍❍❍❍ ! ! ❉❉❉❉❉ ' ' ◆◆◆◆◆◆◆◆ o o ! ! ❉❉❉❉❉
32 2 * * ❚❚❚❚❚❚❚❚❚❚❚❚ o o o o o o o o H( A ) / / H( A × k ) / / H( B )The linearly Dynkin type algebras be the following quivers. A n : n / / n − / / · · · / / / / D n : n / / n − / / · · · / / ; ; ✇✇✇✇ ●●●● A n := kA n / rad and D n := kD n / rad . Applying our results, we can giverecurrence relations about the numbers of τ -tilting modules and support τ -tilting modulesover A n and D n . Theorem 2.12.
Let Λ n be an algebra ( A n or D n ). Then we have(1) | τ - tilt Λ n | = | τ - tilt Λ n − | + | τ - tilt Λ n − | . (2) | s τ - tilt Λ n | = 2 | s τ - tilt Λ n − | + | s τ - tilt Λ n − | . Proof.
Since Λ n is the one-point extension of Λ n − by simple module S n − and Λ n − / h e n − i ∼ =Λ n − . Now, the result follows from Corollary 2.5 and Corollary 2.10. Corollary 2.13. | tilt A n | = 2 ( n > .(2) | τ - tilt A n | = (1+ √ n +1 − (1 −√ n +1 √ · n +1 . (3) | s τ - tilt A n | = (1+ √ n − (1 −√ n √ . (4) | tilt D n | = 5 . (5) | τ - tilt D n | = (2 √ − √ n − +(2 √ −√ n − √ · n − . (6) | s τ - tilt D n | = (3 √ − √ n − +(3 √ −√ n − √ . Example 2.14.
We give some examples of the numbers of τ -tilting modules and support τ -tilting modules over A n and D n in the following tables. n | τ -tilt A n | | s τ -tilt A n | n | τ -tilt D n | | s τ -tilt D n |
32 78 118 454 1026 2506 6038
Acknowledgements
The author would like to thank Professor Zhaoyong Huang for helpful discussions. Thiswork was partially supported by the National natural Science Foundation of China (No.11971225).
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