Tilting Quivers for Hereditary Algebras

Abstract

Let A be a finite dimensional hereditary algebra over an algebraically closed field k. In this paper, we study the tilting quiver of A from the viewpoint of $\tau$-tilting theory. First, we prove that there exists an isomorphism between the support $\tau$-tilting quiver Q(s$\tau$-tilt A) of A and the tilting quiver Q(tilt $\overline{A}$) of the duplicated algebra $\overline{A}$. Then, we give a new method to calculate the number of arrows in the tilting quiver Q(tilt A) when A is representation-finite. Finally, we study the conjecture given by Happel and Unger, which claims that each connected component of Q(tilt A) contains only finitely many non-saturated vertices. We provide an example to show that this conjecture does not hold for some algebras whose quivers are wild with at least four vertices.

1. Introduction

Throughout this paper, all algebras are finite dimensional algebras over an algebraically closed field k and all modules are finitely generated left modules. For an algebra A, we denote by A-mod the category of A-modules and by proj A the category of projective A-modules. For an A-module M, $\mathrm{pd}M$ stands for the projective dimension of M and $\left|M\right|$ stands for the number of pairwise non-isomorphic indecomposable direct summands of M. The subcategory $\mathrm{add}M$ consists of direct summands of finite direct sums of M. We denote by ${\tau }_A$ the Auslander–Reiten translation of A and by D the standard duality functor ${\mathrm{Hom}}_k(-,k)$.

Tilting theory is an essential tool in representation theory of finite dimensional algebras. It usually deals with the problem of comparing a fixed module category with the category of modules over the endomorphism algebra of a tilting module. These two categories are closely related by tilting functors constructed from the tilting module. Thus, tilting modules play an important role in tilting theory. Recall that an A-module T is called a tilting module if the following three conditions are satisfied:

(1)

$\mathrm{pd}T\le 1$,

(2)

${\mathrm{Ext}}_A^1(T,T)=0$,

(3)

there exists an exact sequence $0\to A\to T_0\to T_1\to 0$ with $T_0,T_1\in \mathrm{add}T$.

For example, the regular module A is a tilting module and we always have $\left|T\right|=\left|A\right|$ for any tilting A-module T. An A-module M satisfying the above conditions (1) and (2) is called a partial tilting module and if, moreover, $\left|M\right|=\left|A\right|-1$, then M is called an almost complete tilting module. An indecomposable A-module X such that $M\oplus X$ is a tilting A-module is called a complement to the almost complete tilting module M. It is well-known that any almost complete tilting module has either one or two complements, and it has two non-isomorphic complements if and only if it is faithful; see [1] for details.

Let tilt A be the set of isomorphism classes of basic tilting A-modules. The tilting quiver Q(tilt A) of A introduced by Riedtmann and Schofield in [2] shows the relations between tilting modules with an almost complete tilting module as common direct summands. The vertices of Q(tilt A) are elements of tilt A. For two basic tilting A-modules $T_1$ and $T_2$, there is an arrow $T_1\to T_2$ if $T_1=X_1\oplus M$, $T_2=X_2\oplus M$ with $X_1$, $X_2$ indecomposable and there exists a non-split short exact sequence $0\to X_1\to M^{\prime }\to X_2\to 0$ with $M^{\prime }\in \mathrm{add}M$. The tilting modules $T_1$ and $T_2$ are said to be mutations of each other. The basic idea of mutation is to replace an indecomposable direct summand of a tilting module to obtain a new tilting module. However, mutations of tilting modules are not always possible, since any non-faithful almost complete tilting module has only one complement.

Recently, Adachi, Iyama and Reiten introduced $\tau$-tilting theory in [3], which completes the classical tilting theory from the viewpoint of mutation. Instead of tilting modules, they considered a larger class of modules called support $\tau$-tilting modules whose mutations are always possible. Support $\tau$-tilting modules are a generalisation of tilting modules. They are closely related to functorially finite torsion classes, two-term silting complexes and cluster tilting objects; see [3] for details. In particular, the support $\tau$-tilting quiver Q(s$\tau$-tilt A) of A is defined by mutations of support $\tau$-tilting modules and each vertex in such a quiver has exactly $\left|A\right|$ neighbours. Since the tilting quiver Q(tilt A) can be embedded into Q(s$\tau$-tilt A), we study the tilting quiver from the viewpoint of $\tau$-tilting theory.

Let A be a hereditary algebra and ${\mathcal{C}}_A$ be the cluster category of A. An important property in ${\mathcal{C}}_A$ is that any almost complete cluster tilting object has exactly two complements. In [4], the duplicated algebra $\overline{A}=\left(\begin{array}{cc}A& 0\\ DA& A\end{array}\right)$ of A was considered as a perfect context to view cluster tilting objects in ${\mathcal{C}}_A$ as tilting $\overline{A}$-modules. There exists a one-to-one correspondence between tilting $\overline{A}$-modules and cluster tilting objects in ${\mathcal{C}}_A$; see [4,5] for details. Moreover, a connection between cluster tilting objects in ${\mathcal{C}}_A$ and support $\tau$-tilting modules over cluster tilted algebras was established in [3]. In this paper, we relate tilting $\overline{A}$-modules to support $\tau$-tilting A-modules and prove the following result.

Theorem 1.

Let A be a hereditary algebra. Then there exists an isomorphism between the support τ-tilting quiver $Q($sτ-tilt $A)$ of A and the tilting quiver $Q$(tilt $\overline{A}$) of $\overline{A}$.

In fact, the support $\tau$-tilting modules over a hereditary algebra are exactly the support tilting modules whose numbers were calculated in [6]. Thus, Theorem 1 gives the number of tilting $\overline{A}$-modules when A is a hereditary algebra of Dynkin type.

Let $A=kQ$ be a hereditary algebra where Q is a Dynkin quiver. It is well-known that the number of basic tilting A-modules is independent of the orientation of Q. This implies that the number of vertices in Q(tilt A) is a constant for all Dynkin quivers of the same type. Moreover, Kase proved in [7] that the number of arrows in Q(tilt A) is also a constant. Considering Q(tilt A) as a subquiver of Q(s$\tau$-tilt A), we give a new method to calculate the number of these arrows. The following result was included in [7] and we provide a new proof for it.

Theorem 2.

Let $A=kQ$ be a hereditary algebra where Q is a Dynkin quiver. Then the number of arrows in $Q($tilt $A)$ does not depend on the orientation of Q.

Let T be a basic tilting A-module. If the number of arrows starting or ending at T in Q(tilt A) is less than $\left|A\right|$, then T is said to be non-saturated. It follows from [8], Theorem 3.5, that each connected component of Q(tilt A) contains a non-saturated vertex. Happel and Unger conjectured in [8] that each connected component of Q(tilt A) contains only finitely many non-saturated vertices. This conjecture is true for hereditary algebras of Dynkin and Euclidean type; see [9]. Moreover, it also holds for hereditary algebras of wild type whose quivers have two or three vertices; see [10]. We provide an example to show that the conjecture of Happel and Unger is not true for some wild quivers with at least four vertices.

Proposition  1.

Let $A=kQ$ be a hereditary algebra with

$$Q:1\leftleftarrows 2\leftarrow 3\to \cdots \to n(n\ge 4).$$

Then there exists a connected component of $Q($tilt $A)$ that contains infinitely many non-saturated vertices.

2. Preliminaries

In this section, we recall some results about $\tau$-tilting theory that are needed in our paper.

Definition 1

([3], Definition 0.1). Let A be an algebra.

(1) 

An A-module M is called τ-rigid if ${\mathrm{Hom}}_A(M,\tau M)=0$.

(2) 

An A-module M is called τ-tilting (respectively, almost complete τ-tilting) if M is τ-rigid and $\left|M\right|=\left|A\right|$ (respectively, $\left|M\right|=\left|A\right|-1$).

(3) 

An A-module M is called support τ-tilting if there exists an idempotent e in A such that M is a τ-tilting $(A/\langle e\rangle )$-module.

Example 1.

Let $A=kQ/\langle \beta \alpha \rangle$ be an algebra with $Q:1\stackrel{\alpha }{\to }2\stackrel{\beta }{\to }3$. Then $\begin{array}{c}1\\ 2\end{array}\oplus 1\oplus 3$ is a τ-tilting module and $\begin{array}{c}2\\ 3\end{array}\oplus 3$ is a support τ-tilting module.

The following result shows the relation between tilting modules and $\tau$-tilting modules.

Proposition  2

([11], Proposition 3.8). Let A be an algebra. Then an A-module M is tilting if and only if M is τ-tilting and $\mathrm{pd}M\le 1$.

Thus, $\tau$-tilting modules are a generalisation of tilting modules. In particular, $\tau$-tilting modules over a hereditary algebra coincide with tilting modules.

It is often convenient to view support $\tau$-tilting modules as certain pairs of A-modules.

Definition 2

([3], Definition 0.3). Let $(M,P)$ be a pair with M ∈ A-mod and $P\in \mathrm{proj}A$.

(1) We call $(M,P)$ a τ-rigid pair if M is τ-rigid and ${\mathrm{Hom}}_A(P,M)=0$.

(2) We call $(M,P)$ a support τ-tilting pair (respectively, almost complete support τ-tilting pair) if $(M,P)$ is a τ-rigid pair and $\left|M\right|+\left|P\right|=\left|A\right|$ (respectively, $\left|M\right|+\left|P\right|=\left|A\right|-1)$.

These notions are compatible with those in Definition 1. In Example 1, $(\begin{array}{c}2\\ 3\end{array}\oplus 3,\begin{array}{c}1\\ 2\end{array})$ is a support $\tau$-tilting pair and $(\begin{array}{c}2\\ 3\end{array},\begin{array}{c}1\\ 2\end{array})$ is an almost complete support $\tau$-tilting pair.

We say $(M,P)$ is basic if both M and P are basic. If M and P are direct summands of $M^{\prime }$ and $P^{\prime }$, respectively, we say $(M,P)$ is a direct summand of $(M^{\prime },P^{\prime })$. One of the main results in [3] is the following.

Theorem 3

([3], Theorem 2.18). Let A be an algebra. Then any basic almost complete support τ-tilting pair $(U,Q)$ for A is a direct summand of exactly two basic support τ-tilting pairs $(T,P)$ and $(T^{\prime },P^{\prime })$ for A.

In the above theorem, the support $\tau$-tilting module $T^{\prime }$ is called the left mutation of the support $\tau$-tilting module T if $\mathrm{Fac}{T}^{\prime }\subset \mathrm{Fac}T$. In Example 1, the almost complete support $\tau$-tilting pair $(\begin{array}{c}2\\ 3\end{array},\begin{array}{c}1\\ 2\end{array})$ is the direct summand of support $\tau$-tilting pairs $(\begin{array}{c}2\\ 3\end{array}\oplus 3,\begin{array}{c}1\\ 2\end{array})$ and $(\begin{array}{c}2\\ 3\end{array}\oplus 2,\begin{array}{c}1\\ 2\end{array})$. Moreover, the support $\tau$-tilting module $\begin{array}{c}2\\ 3\end{array}\oplus 2$ is the left mutation of the support $\tau$-tilting module $\begin{array}{c}2\\ 3\end{array}\oplus 3$. The support $\tau$-tilting quiver is defined by mutations of support $\tau$-tilting modules.

Definition 3

([3], Definition 2.29). Let sτ-tilt A be the set of isomorphism classes of basic support τ-tilting A-modules. The support τ-tilting quiver Q(sτ-tilt A) of A is as follows.

(1) The set of vertices is sτ-tilt A.

(2) There is an arrow from T to $T^{\prime }$ if $T^{\prime }$ is a left mutation of T.

According to Theorem 3, the support $\tau$-tilting quiver Q(s$\tau$-tilt A) is an $\left|A\right|$-regular quiver, that is, each vertex in Q(s$\tau$-tilt A) has exactly $\left|A\right|$ neighbours. On the other hand, there is a partial order ≤ on the set s$\tau$-tilt A given by $T^{\prime }\le T$ if $\mathrm{Fac}{T}^{\prime }\subseteq \mathrm{Fac}T$. It was proved in [3] that the Hasse quiver of the partially ordered set (s$\tau$-tilt A, $\le )$ coincides with the support $\tau$-tilting quiver Q(s$\tau$-tilt A). It is a generalisation of the classical result in [12], which claims that the Hasse quiver of the partially ordered set (tilt A, $\le )$ is the tilting quiver Q(tilt A).

Definition 4.

An A-module T is called support tilting (respectively, almost complete support tilting) if T is a tilting ($A/\langle e\rangle$)-module (respectively, almost complete tilting ($A/\langle e\rangle$)-module) for some idempotent e of A.

For a hereditary algebra, it follows from Proposition 2 that the support $\tau$-tilting modules are exactly the support tilting modules. Let $A=kQ$ be a hereditary algebra where Q is a Dynkin quiver with n vertices. The number of support tilting A-modules was calculated in [6], which is independent of the orientation of Q. Denote by $a_s\left(Q\right)(0\le s\le n)$ the number of basic support tilting A-modules with s non-isomorphic indecomposable direct summands. Obviously, the number of basic tilting A-modules is $a_n\left(Q\right)$. Let $a\left(Q\right)$ be the number of basic support tilting A-modules. Then we have $a\left(Q\right)={\sum }_{s=0}^n{a}_s\left(Q\right)$.

Theorem 4

([6], Theorem 1). Let $A=kQ$ be a hereditary algebra where Q is a Dynkin quiver with n vertices. Then we have the following results where $\left(\genfrac{}{}{0pt}{}{s}{t}\right)$ is the binomial coefficient.

Q	$A_n$	$D_n$	$E_6$	$E_7$	$E_8$
$a_n\left(Q\right)$	$\frac{1}{n+1}\left(\genfrac{}{}{0pt}{}{2n}{n}\right)$	$\frac{3n-4}{2n-2}\left(\genfrac{}{}{0pt}{}{2n-2}{n-2}\right)$	418	2431	$\mathrm{17,342}$
$a_{n-1}\left(Q\right)$	$\frac{2}{n+1}\left(\genfrac{}{}{0pt}{}{2n-1}{n-1}\right)$	$\frac{3n-4}{2n-3}\left(\genfrac{}{}{0pt}{}{2n-3}{n-1}\right)$	228	1001	4784
$a\left(Q\right)$	$\frac{1}{n+2}\left(\genfrac{}{}{0pt}{}{2n+2}{n+1}\right)$	$\frac{3n-2}{2n-1}\left(\genfrac{}{}{0pt}{}{2n-1}{n-1}\right)$	833	4160	25,080

3. Main Results

Throughout this section, let $A=kQ$ be a hereditary algebra where Q is a finite and acyclic quiver. We investigate the tilting quiver Q(tilt A) and the support $\tau$-tilting quiver Q(s$\tau$-tilt A) of A.

3.1. Tilting Modules over the Duplicated Algebra $\overline{A}$

Let $\mathcal{C}$ be a Krull–Schmidt 2-Calabi–Yau triangulated category. An object T in $\mathcal{C}$ is called cluster tilting if $\mathrm{add}T=\{X\in \mathcal{C}|{\mathrm{Hom}}_{\mathcal{C}}(T,X\left[1\right])=0\}$, where [1] is the shift functor of $\mathcal{C}$. The endomorphism algebra $\Lambda ={\mathrm{End}}_{\mathcal{C}}T$ is called the 2-Calabi–Yau tilted algebra of $\mathcal{C}$ whose module category is closely related to the triangulated category $\mathcal{C}$. Let c-tilt $\mathcal{C}$ be the set of isomorphism classes of basic cluster tilting objects in $\mathcal{C}$.

Theorem 5

([3], Theorem 4.1). There exists a bijection between the setc-tilt $\mathcal{C}$ and the set s$\tau$-tilt $\Lambda$.

The bijection in Theorem 5 induces an isomorphism between the cluster tilting graph G(c-tilt $\mathcal{C}$) and the underlying graph of Q(s$\tau$-tilt $\Lambda$). Let ${\mathcal{C}}_A$ be the cluster category of A. It is known that ${\mathcal{C}}_A$ is a 2-Calabi–Yau triangulated category and A is a cluster tilting object in ${\mathcal{C}}_A$ with ${\mathrm{End}}_{{\mathcal{C}}_A}A\cong A$. Since the cluster tilting graph G(c-tilt ${\mathcal{C}}_A$) is connected, so is the support $\tau$-tilting quiver Q(s$\tau$-tilt A) of A. However, the tilting quiver Q(tilt A) is not always connected.

Example 2.

Let $A=kQ$ be the Kronecker algebra with $Q:1\leftleftarrows 2$. Then the support τ-tilting quiver $Q(\mathrm{s}\tau$-$\mathrm{tilt}A)$ is as follows.

$$\cdots \to \begin{array}{c}222\\ 11\end{array}\oplus \begin{array}{c}22\\ 1\end{array}\to \begin{array}{c}22\\ 1\end{array}\oplus 2\to 2\to 0\leftarrow 1\leftarrow 1\oplus \begin{array}{c}2\\ 11\end{array}\to \begin{array}{c}2\\ 11\end{array}\oplus \begin{array}{c}22\\ 111\end{array}\to \cdots$$

Let $\overline{A}=\left(\begin{array}{cc}A& 0\\ DA& A\end{array}\right)$ be the duplicated algebra of A. According to [4], Theorem 10, there exists a bijection between the tilting $\overline{A}$-modules and cluster tilting objects in ${\mathcal{C}}_A$. Combining Theorem 5, we give the following relation between the tilting quiver Q(tilt $\overline{A}$) of $\overline{A}$ and the support $\tau$-tilting quiver Q(s$\tau$-tilt A) of A.

Theorem 6.

Let A be a hereditary algebra. Then there exists an isomorphism between the support τ-tilting quiver $Q$(sτ-tilt $A$) of A and the tilting quiver $Q$(tilt $\overline{A}$) of $\overline{A}$.

Proof. 

We set Σ₁ = {${\Omega }_{\overline{A}}^{-1}P|P$ as an indecomposable projective A-module}, where ${\Omega }_{\overline{A}}^{-1}P$ is the cosyzygy of P over $\overline{A}$. Let $\overline{P}$ be the direct sum of all non-isomorphic indecomposable projective-injective $\overline{A}$-modules. For a tilting $\overline{A}$-module T, it follows from [4] that we have a decomposition $T=T_1\oplus T_2\oplus \overline{P}$ with $T_1\in A$-mod and $T_2\in \mathrm{add}{\Sigma }_1$. According to Theorem 5 and [4], Theorem 10, there exists a bijection between the set s$\tau$-tilt A and the set tilt $\overline{A}$, sending the support $\tau$-tilting pair $(M,P)$ to the tilting $\overline{A}$-module $M\oplus {\Omega }_{\overline{A}}^{-1}P\oplus \overline{P}$. Then it suffices to prove that this bijection preserves the partial orders on s$\tau$-tilt A and tilt $\overline{A}$.

By [13], Lemma 15(2), the injective envelope of an A-module in $\overline{A}$-mod is projective-injective. Then, for a support $\tau$-tilting pair $(M,P)$, the $\overline{A}$-module ${\Omega }_{\overline{A}}^{-1}P$ is generated by $\overline{P}$ and this implies $\mathrm{Fac}(M\oplus {\Omega }_{\overline{A}}^{-1}P\oplus \overline{P})=\mathrm{Fac}(M\oplus \overline{P})$. It follows from [4], Lemma 2(b) that we can consider Ind A as a convex subcategory of Ind $\overline{A}$ that is closed under predecessors. Note that $\overline{P}$ is not in A-mod; there are no nonzero morphisms from $\overline{P}$ to M. Thus, for two support $\tau$-tilting A-modules $M_1$ and $M_2$, we have $\mathrm{Fac}{M}_1\subset \mathrm{Fac}{M}_2$ if and only if $\mathrm{Fac}(M_1\oplus \overline{P})\subset \mathrm{Fac}(M_2\oplus \overline{P})$. As a result, the Hasse quiver of s$\tau$-tilt A is isomorphic to the Hasse quiver of tlit $\overline{A}$. This completes our proof. □

Example 3.

Let $A=kQ$ be the Kronecker algebra with $Q:1\leftleftarrows 2$. Then the tilting quiver $Q\left(\mathrm{tilt}\overline{A}\right)$ is isomorphic to the support τ-tilting quiver in Example 2.

Note that the support $\tau$-tilting modules over a hereditary algebra coincide with the support tilting modules. We give the number of tilting $\overline{A}$-modules when A is a hereditary algebra of Dynkin type.

Proposition  3.

Let $A=kQ$ be a hereditary algebra where Q is a Dynkin quiver. Then the number of basic tilting $\overline{A}$-modules, denoted by |$\mathrm{tilt}\overline{A}$|, is as follows.

$$|\mathrm{tilt}\overline{A}|=\left\{\begin{array}{cc}\frac{1}{n+2}\left(\genfrac{}{}{0pt}{}{2n+2}{n+1}\right)\hfill & ifQ=A_n\hfill \\ \frac{3n-2}{2n-1}\left(\genfrac{}{}{0pt}{}{2n-1}{n-1}\right)\hfill & ifQ=D_n\hfill \\ 833\hfill & ifQ=E_6\hfill \\ 4160\hfill & ifQ=E_7\hfill \\ \mathrm{25,080}\hfill & ifQ=E_8\hfill \end{array}\right.$$

Proof. 

This result follows immediately from Theorem 4 and Theorem 6. □

3.2. The Number of Arrows in Tilting Quivers

Let $A=kQ$ be a hereditary algebra where Q is a Dynkin quiver. The number of tilting A-modules was calculated in [14] via cluster algebras. It does not depend on the orientation of Q and the number of vertices in Q(tilt A) is a constant for Dynkin quivers of the same type. Moreover, Kase proved in [7] that the number of arrows in Q(tilt A) is also independent of the orientation of Q. Note that Q(tilt A) can be embedded into Q(s$\tau$-tilt A), which is an $\left|A\right|$-regular quiver; we give a new method to calculate the number of arrows in Q(tilt A). Then we provide a new and straightforward proof for the following result.

Theorem 7

([7], Theorem 0.1). Let $A=kQ$ be a hereditary algebra where Q is a Dynkin quiver with n vertices. Then the number of arrows in $Q$(tilt $A$), denoted by $\left|Q\right($tilt ${A)}_1|$, does not depend on the orientation of Q. In particular, we have

$${|Q\left(\mathrm{tilt}A\right)}_1|=\left\{\begin{array}{cc}\left(\genfrac{}{}{0pt}{}{2n-1}{n+1}\right)\hfill & ifQ=A_n\hfill \\ (3n-4)\left(\genfrac{}{}{0pt}{}{2n-4}{n-3}\right)\hfill & ifQ=D_n\hfill \\ 1140\hfill & ifQ=E_6\hfill \\ 8008\hfill & ifQ=E_7\hfill \\ \mathrm{66,976}\hfill & ifQ=E_8\hfill \end{array}\right.$$

Proof. 

We consider Q(tilt A) as a subquiver of Q(s$\tau$-tilt A). By Theorem 3, each vertex in Q(s$\tau$-tilt A) has exactly n neighbours. For a tilting A-module T, its neighbours in Q(s$\tau$-tilt A) are either tilting modules $T^{\prime }$ or support $\tau$-tilting modules M with $\left|M\right|=n-1$. Note that a support $\tau$-tilting module M with $\left|M\right|=n-1$ is not sincere. As a non-sincere almost complete tilting module, it is connected with exactly one tilting module by an arrow in Q(s$\tau$-tilt A). The number of such support $\tau$-tilting modules is $a_{n-1}\left(Q\right)$. Thus, we have

$${|Q\left(\mathrm{tilt}A\right)}_1|=\frac{1}{2}(a_n\left(Q\right)\times n-a_{n-1}\left(Q\right)).$$

The result follows from Theorem 4 by direct calculation. □

Example 4.

Let $A=kQ$ be a hereditary algebra with $Q:1\to 2\to 3$. Then the tilting quiver $Q$(tilt $A$) is as follows and it has five arrows.

Moreover, we calculate the number of arrows in $Q$(tilt $\overline{A}$).

Proposition 4.

Let $A=kQ$ be a hereditary algebra where Q is a Dynkin quiver with n vertices. Then the number of arrows in Q(tilt $\overline{A}$), denoted by |Q(tilt $\overline{A}$)₁| is as follows.

$$|Q{\left(\mathrm{tilt}\overline{A}\right)}_1|=\left\{\begin{array}{cc}\frac{n}{2(n+2)}\left(\genfrac{}{}{0pt}{}{2n+2}{n+1}\right)\hfill & ifQ=A_n\hfill \\ \frac{n(3n-2)}{2(2n-1)}\left(\genfrac{}{}{0pt}{}{2n-1}{n-1}\right)\hfill & ifQ=D_n\hfill \\ 2499\hfill & ifQ=E_6\hfill \\ 14,560\hfill & ifQ=E_7\hfill \\ 100,320\hfill & ifQ=E_8\hfill \end{array}\right.$$

Proof. 

Since Q(sτ-tilt A) is an n-regular quiver, the number of arrows in Q(sτ-tilt A) is $\frac{1}{2}(a\left(Q\right)\times n)$. According to Theorem 6, the number of arrows in Q(tilt $\overline{A}$) is $\frac{1}{2}(a\left(Q\right)\times n)$ and the result follows from Theorem 4 by direct calculation. □

By Proposition 3 and Proposition 4, we obtain the numbers of vertices and arrows in the tilting quiver Q(tilt $\overline{A}$) of the duplicated algebra $\overline{A}$ when A is a hereditary algebra of Dynkin type.

3.3. Non-Saturated Vertices in Tilting Quivers

Let $A=kQ$ be a hereditary algebra where Q is a finite and acyclic quiver with n vertices. For a tilting A-module T, denote by $s\left(T\right)$ (respectively, $e\left(T\right)$) the number of arrows starting (respectively, ending) at T in $Q$(tilt $A$). Note that each vertex in Q(sτ-tilt A) has n neighbours. Since $Q$(tilt $A$) can be embedded into Q(sτ-tilt A), we have $s\left(T\right)+e\left(T\right)\le n$. A tilting A-module T is called saturated if $s\left(T\right)+e\left(T\right)=n$.

For an A-module X, we denote by dim X its dimension vector. The following result gives a sufficient and necessary condition for a tilting A-module to be saturated in $Q$(tilt $A$).

Proposition 5

([8], Proposition 3.2). Let T be a basic tilting A-module. It is saturated if and only if ${\left(\dim T\right)}_i\ge 2$ for all $1\le i\le n$.

Let $i_0$ be a source vertex in Q and $A={\oplus }_{i=1}^n{P}_i$ with $P_i$ indecomposable. Then we have (dim ${\oplus }_{i\ne i_0}{P}_i{)}_{i_0}=0$ and so (dim ${A)}_{i_0}=1$. It follows from Proposition 5 that A is not saturated. Thus, the tilting quiver $Q$(tilt $A$) contains a non-saturated vertex. Happel and Unger in [8] gave the following result.

Theorem 8

([8], Theorem 3.5). Each connected component of $Q($tilt $A)$ contains a non-saturated vertex.

Moreover, they continued to conjecture in [8] that each connected component of $Q$(tilt $A$) contains only finitely many non-saturated vertices. This conjecture for Dynkin and Euclidean quivers was proved to be true by Wang and Zhang in [9].

Theorem 9

([9], Theorem 5.5). Let $A=kQ$ be a hereditary algebra where Q is a Dynkin or Euclidean quiver. Then each connected component of $Q($tilt $A)$ contains only finitely many non-saturated vertices.

For a wild quiver with two or three vertices, we collect some results in [10] to show the following.

Theorem 10

([10]). Let $A=kQ$ be a hereditary algebra where Q is a wild quiver with two or three vertices. Then each connected component of $Q($tilt $A)$ contains only finitely many non-saturated vertices.

Proof. 

If Q is a wild quiver with two vertices, it is of the form 2 1 with at least three arrows. By [15], XVIII, Corollary 2.16, there are no regular tilting A-modules. Then all tilting A-modules are either preprojective or preinjective. According to Section 2.2 in [16], the tilting quiver $Q\left(\mathrm{tilt}A\right)$ is of the form

$$\circ \to \circ \to \circ \to \cdots$$

$$\circ \leftarrow \circ \leftarrow \circ \leftarrow \cdots$$

Thus, $Q\left(\mathrm{tilt}A\right)$ has two connected components, both of which contain exactly one non-saturated vertex.

Now assume Q is a wild quiver with three vertices. Let $T=T_1\oplus T_2\oplus T_3$ be a tilting A-module with $T_i$ indecomposable for $1\le i\le 3$. If T is a non-saturated vertex in $Q$(tilt $A$), then there exists an arrow $T\to (T_1\oplus T_2,P)$ in the support τ-tilting quiver $Q$(sτ-$\mathrm{tilt}A$) where P is an indecomposable projective A-module. Let e be the primitive idempotent of A corresponding to P. Then $T_1\oplus T_2$ is a tilting $(A/\langle e\rangle )$-module. Thus, each non-saturated tilting A-module contains a tilting $(A/\langle e\rangle )$-module as a direct summand.

If $A/\langle e\rangle$ is representation-finite, there are only finitely many tilting $A/\langle e\rangle$-modules. Then the number of non-saturated tilting A-modules containing tilting $(A/\langle e\rangle )$-modules is finite.

If $A/\langle e\rangle$ is representation-infinite, the quiver of $A/\langle e\rangle$ is of the form with at least two arrows. According to [17], IX, Proposition 5.6, there are only finitely many non-sincere indecomposable preprojective and preinjective A-modules. Thus, all but finitely many tilting $(A/\langle e\rangle )$-modules are regular A-modules. Thus, all but finitely many non-saturated tilting A-modules contain tilting $(A/\langle e\rangle )$-modules that are regular A-modules as direct summands. Assume $T_1\oplus T_2$ is a regular A-module. Let $T^{\prime }\ncong T$ be a tilting A-tilting module in a connected component Γ of $Q$(tilt $A$) containing T. By [10], Theorem 4.3, $T^{\prime }$ has at least two sincere direct summands. It follows from Proposition 5 that $T^{\prime }$ is saturated. Thus, Γ contains only one non-saturated vertex T. □

We generalise [16], Example 3.2 to show that the conjecture of Happel and Unger does not hold for some wild quivers with at least four vertices.

Proposition 6.

Let $A=kQ$ be a hereditary algebra with

$$Q:1\leftleftarrows 2\leftarrow 3\to \cdots \to n(n\ge 4).$$

Then there exists a connected component of $Q($tilt $A)$ that contains infinitely many non-saturated vertices.

Proof. 

Denote by $I_i$ (respectively, $P_i$) the indecomposable injective (respectively, projective) A-modules corresponding to the vertex i in Q. Let N be a tilting module over the Kronecker algebra $B=k(1\leftleftarrows 2)$ without nonzero projective direct summands. We claim that ${\tau }_{A}N\oplus I$ with $I={\oplus }_{j=3}^n{I}_j$ is a tilting A-module. In fact, it follows from [3], Lemma 2.1(b) and [17], IV, Corollary 2.14(a) that

$${\mathrm{Hom}}_A(N,{\tau }_{A}N)\cong {\mathrm{Hom}}_B(N,{\tau }_{B}N)\cong D{\mathrm{Ext}}_B^1(N,N)=0.$$

According to [17], IV, Corollary 2.14(b), we have

$${\mathrm{Ext}}_A^1({\tau }_{A}N\oplus I,{\tau }_{A}N\oplus I)\cong D{\mathrm{Hom}}_A(N,{\tau }_{A}N\oplus I)\cong D{\mathrm{Hom}}_A(N,{\tau }_{A}N)=0.$$

Thus, ${\tau }_{A}N\oplus I$ is a tilting A-module.

Since ${\mathrm{Hom}}_A(P_n,{\tau }_{A}N)\cong D{\mathrm{Ext}}_A^1(N,P_n)=0$, we have ${\left(\dim {\tau }_{A}N\right)}_n=0$ and then ${\left(\dim ({\tau }_{A}N\oplus I)\right)}_n=1$. This implies that ${\tau }_{A}N\oplus I$ is non-saturated by Proposition 5. Denote by $Q($tilt${}_{I}A)$ the subquiver of $Q($tilt $A)$ with vertices T such that T contains I as a direct summand. Then $Q$(tilt${}_{I}A$) is infinite, since there are infinitely many tilting B-modules. Note that I is not sincere; by [16], Theorem 1, $Q$(tilt${}_{I}A$) is of the form

$$\circ \to \circ \to \circ \to \cdots$$

$$\circ \leftarrow \circ \leftarrow \circ \leftarrow \cdots$$

Thus, at least one of the above connected components of $Q($tilt${}_{I}A)$ contains infinitely many non-saturated vertices ${\tau }_{A}N\oplus I$ where N is a tilting B-module. This completes our proof. □

4. Conclusions

In tilting theory, the central objects of study are tilting modules and the tilting quiver reveals the relations between these modules. In this paper, we investigate the tilting quiver from the viewpoint of τ-tilting theory, which is a generalisation of tilting theory. For a hereditary algebra A, we prove that its support τ-tilting quiver $Q$(sτ-tilt $A$) is isomorphic to the tilting quiver $Q$(tilt $\overline{A}$) of its duplicated algebra $\overline{A}$. Then, by the $\left|A\right|$-regular quiver Q(sτ-tilt A), we give a new method to calculate the number of arrows in $Q$(tilt $A$) when A is representation-finite. Finally, we provide an example to show that the conjecture given by Happel and Unger does not always hold for wild hereditary algebras. However, we could not find a necessary and sufficient condition for a wild hereditary algebra to satisfy this conjecture. Our next goal is to try to characterise the wild hereditary algebras each connected component of whose tilting quiver contains only finitely many non-saturated vertices.