Mutation of tilting bundles of tubular type

TL;DR

This paper proves the connectedness of the tilting graph for coherent sheaves on tubular weighted projective lines, using mutations and slope changes of tilting sheaves, providing new insights into their structure.

Contribution

It demonstrates the connectedness of the tilting graph for coherent sheaves on tubular weighted projective lines and explores slope changes under mutations.

Findings

Connectedness of the tilting graph is established.

Mutation affects slopes of tilting sheaves.

Provides an alternative proof for tilting graph connectedness.

Abstract

Let $\mathbb{X}$ be a weighted projective line of tubular type and $\operatorname{coh}\mathbb{X}$ the category of coherent sheaves on $\mathbb{X}$. The main purpose of this note is to show that the subgraph of the tilting graph consisting of all basic tilting bundles in $\operatorname{coh}\mathbb{X}$ is connected. This yields an alternative proof for the connectedness of the tilting graph of $\operatorname{coh}\mathbb{X}$. Our approach leads to the investigation of the change of slopes of a tilting sheaf in $\operatorname{coh}\mathbb{X}$ under (co-)APR mutations, which may be of independent interest.

Equations

$$L(\mathbf{p}):=⟨{\vec{x}}_1,\cdots ,{\vec{x}}_t|p_1{\vec{x}}_1=\cdots =p_t{\vec{x}}_t=:\vec{c}⟩$$

$$S(\mathbf{p},\lambda )=k[u,v,x_1,\cdots ,x_t]/(x_i^{p_i}-{\lambda }_i^{\prime }u-{\lambda }_i^{{}^{\prime \prime }}v|i=1,\cdots ,t)$$

$$p_{\mathbf{\lambda }}:{\mathbb{P}}^1\to \mathbb{N},\mu \mapsto \{\begin{array}{ll}{p}_i& if\mu ={\lambda }_{i}forsomei;\\ 1& else.\end{array}$$

$$g_{\mathbb{X}}=1+\frac{1}{2}((t-2)p-{\Sigma }_{i=1}^{t}p/p_i).$$

$$\mathrm{rank}\mathcal{O}(\vec{x})=1,\mathrm{deg}\mathcal{O}(\vec{x})=\delta (\vec{x})$$

$$\mu E:=\mathrm{deg}E/\mathrm{rank}E\in \mathbb{Q}\cup \{\infty \}.$$

$$\mu (E(\vec{x}))=\mu E+\delta (\vec{x}).$$

$$\overline{\chi }(X,Y)=rk(X)rk(Y)(p(1-g_{\mathbb{X}}))+(\mu Y-\mu X)$$

$$\overline{\chi }(X,Y)=\mu Y-\mu X.$$

$$\overline{\chi }(X,Y)=\mu Y-\mu X=q^{\prime }-q\ne 0.$$

$$0\ne \overline{\chi }(X,Y)={\Sigma }_{j=0}^{p-1}\chi ({\tau }^{j}X,Y)$$

$$\mu Y-\mu X=\overline{\chi }(X,Y)=p\chi (X,Y)=p({\dim }_k{\mathrm{Hom}}_{\mathbb{X}}(X,Y)-{\mathrm{Ext}}_{\mathbb{X}}^1(X,Y)).$$

$${\mathrm{Aut}}_{\mu }({\mathcal{D}}^b(\mathbb{X}))=⟨[1]⟩\times {\mathrm{Aut}}_{\mu }(\mathrm{coh}\mathbb{X}),{\mathrm{Aut}}_{\mu }(\mathrm{coh}\mathbb{X})={\mathrm{Pic}}_0\mathbb{X}⋉\mathrm{Aut}(\mathbb{X}),$$

$$\mu ({\sigma }_1(E))=\mu E+1and\mu ({\sigma }_2(E))=\frac{\mu E}{1+\mu E}$$

$$G({\mathcal{C}}_{\mu (F(X))}[l])=G({\mathcal{C}}_{\mu (F(X)[-l])})[l]\subset {\mathcal{C}}_{\mu (G\circ F(X))}[-l][l]={\mathcal{C}}_{\mu X}.$$

$${\mathrm{Hom}}_{{\mathcal{D}}^b(\mathbb{X})}({\mathcal{C}}_{\mu (F(X))}[l],{\mathcal{C}}_{\mu (F(Y))}[m])\cong {\mathrm{Hom}}_{{\mathcal{D}}^b(\mathbb{X})}(F({\mathcal{C}}_{\mu X}),F({\mathcal{C}}_{\mu Y}))\cong {\mathrm{Hom}}_{\mathbb{X}}({\mathcal{C}}_{\mu X},{\mathcal{C}}_{\mu Y})\ne 0,$$

$$0\ne {\mathrm{Hom}}_{{\mathcal{D}}^b(\mathbb{X})}({\mathcal{C}}_{\mu (F(X))}[l],{\mathcal{C}}_{\mu (F(Y))}[m])$$

$${\mathrm{Hom}}_{{\mathcal{D}}^b(\mathbb{X})}(\tau X[-1],\overline{T}[-1])\cong {\mathrm{Hom}}_{\mathbb{X}}(\tau X,\overline{T})=0.$$

$$0\to T_1^{\ast }\to B\stackrel{g}{\to }{T}_1\to 0$$

$$\mu (F(\tau T_1^{\ast }[-1]))<\mu (F(T_1))\le \mu (F(T_i))\le \mu (F(T_j))\le \mu (F(T_n)).$$

$$\phi (\mu (\mathcal{O}(-{\vec{x}}_t)))=\phi (-1)=m+\frac{a-c}{b-d}<m+\frac{c}{d}=\phi (\infty )$$

$$\mu Y=\mu (\overline{\phi }(S))=\phi (\mu S)=\phi (\infty )=m+\frac{c}{d}>m+\frac{a}{b}=\mu X.$$

Full text

1. Introduction

Tilting quiver of a hereditary abelian category was introduced by Happel and Unger [HU], which encodes information of mutations of tilting objects. Its underlying graph is called the tilting graph. The connecetedness of tilting graphs for hereditary catetgories is a basic problem in tilting theory, which has important application in the categorification theory of cluster algebras (cf. [BMRRT, CK, BG] for instance). Among others, building on the connectedness of tilting graphs established by Happel and Unger [HU], Buan et al. [BMRRT] established the connectedness of cluster-tilting graphs of cluster categories, which implies that there is a one-to-one correspondence between indecomposable rigid objects of a cluster category and cluster variables of the associated cluster algebra. The connectedness of (cluster)-tilting graphs also play a fundamental role in the representation-theoretic approach to the denominator conjecture in cluster algebras (cf. [GP, FG1]).

Let $\mathcal{H}$ be a connected hereditary abelian category over an algebraically closed field $k$ and $\mathcal{G}(\mathcal{T}_{\mathcal{H}})$ the tilting graph of $\mathcal{H}$. According to Happel’s classification [H], each connected hereditary abelian category with tilting objects over $k$ is derived equivalent to $\operatorname{\mathsf{mod}}\nolimits kQ$ for a finite acyclic quiver $Q$ or the category $\operatorname{\mathsf{coh}}\mathbb{X}$ of coherent sheaves on a weighted projective line $\mathbb{X}$. If $\mathcal{H}$ is derived to $\operatorname{\mathsf{mod}}\nolimits kQ$ for an acyclic quiver $Q$ which is not of wild type, Happel and Unger [HU] obtained an explicitly characterization of the connectedness of $\mathcal{G}(\mathcal{T}_{\mathcal{H}})$. Moreover, they also proved that $\mathcal{G}(\mathcal{T}_{\mathcal{H}})$ is connected provided that $Q$ is of wild type and $\mathcal{H}$ does not contain non zero projective objects. Weighted projective lines are classified as domestic type, tubular type and wild type according to their genus. It is known that $\operatorname{\mathsf{coh}}\mathbb{X}$ of domestic type is derived to $\operatorname{\mathsf{mod}}\nolimits kQ$ for an acyclic quiver $Q$ of tame type, while $\operatorname{\mathsf{coh}}\mathbb{X}$ is never derived to a hereditary algebra when $\mathbb{X}$ is of tubular or wild type. It was conjectured by Happel and Unger [HU] that the tilting graph $\mathcal{G}(\mathcal{T}_{\mathbb{X}})$ of $\operatorname{\mathsf{coh}}\mathbb{X}$ is connected for a tubular and wild weighted projective line $\mathbb{X}$. This has been confirmed by Barot, Kussin and Lenzing [BKL] for tubular type and by Fu and Geng [FG2] for wild type.

Let $\mathbb{X}$ be a weighted projective line over $k$. Denote by $\operatorname{\mathsf{vect}}\mathbb{X}$ the full subcategory of $\operatorname{\mathsf{coh}}\mathbb{X}$ consisting of vector bundles. Recently, the subcategory $\operatorname{\mathsf{vect}}\mathbb{X}$ has obtained a lot of attention (cf. [KLM1, KLM2, L, CLR]). A tilting object $T$ of $\operatorname{\mathsf{coh}}\mathbb{X}$ lying in $\operatorname{\mathsf{vect}}\mathbb{X}$ is called a tilting bundle. Denote by ${\mathcal{G}}(\mathcal{T}^{v}_{\mathbb{X}})$ the subgraph of the tilting graph ${\mathcal{G}}(\mathcal{T}_{\mathbb{X}})$ consisting of all basic tilting bundles in $\operatorname{\mathsf{coh}}\mathbb{X}$. We are interested in whether the subgraph ${\mathcal{G}}(\mathcal{T}^{v}_{\mathbb{X}})$ is also connected. When $\mathbb{X}$ is of domestic type, the Auslander-Reiten quiver of $\operatorname{\mathsf{vect}}\mathbb{X}$ is equivalent to $\mathbb{Z}Q$ with $Q$ is an extended Dynkin quiver. By suitable $\operatorname{\mathsf{APR}}$ mutations and co-$\operatorname{\mathsf{APR}}$ mutations, one can get the subgraph ${\mathcal{G}}(\mathcal{T}^{v}_{\mathbb{X}})$ is connected (cf. [BKL]). When $\mathbb{X}$ is of tubular type, the Auslander-Reiten quiver of $\operatorname{\mathsf{vect}}\mathbb{X}$ is consisted of standard tubes. In this note, by investigating some properties of mutations on tilting objects, we find that ${\mathcal{G}}(\mathcal{T}^{v}_{\mathbb{X}})$ is also connected, i.e. we get the following main results:

Theorem 1.1**.**

(see Theorem 5.13) Let $\mathbb{X}$ be a weighted projective line of tubular type. The subgraph ${\mathcal{G}}(\mathcal{T}^{v}_{\mathbb{X}})$ of the tilting graph ${\mathcal{G}}(\mathcal{T}_{\mathbb{X}})$ consisting of all the basic tilting bundles in $\operatorname{\mathsf{coh}}\mathbb{X}$ is connected.

As an easy consequence, Theorem 1.1 yields an alternative proof for the connectedness of the tilting graph ${\mathcal{G}}(\mathcal{T}_{\mathbb{X}})$. In order to establish the main result, we introduce APR muation and co-APR mutation on tilting objects in $\operatorname{\mathsf{coh}}\mathbb{X}$. For an object $E\in\operatorname{\mathsf{coh}}\mathbb{X}$, denote by $\mu E$ the slope of $E$. The change of slopes of indecomposable direct summands of a tilting object under mutation plays a center role, which is of independent interest.

Theorem 1.2**.**

(see Theorem 4.8) Let $\mathbb{X}$ be a weighted projective line of tubular type, $T=\bigoplus_{i=1}^{n}T_{i}$ be a tilting sheaf in $\operatorname{\mathsf{coh}}\mathbb{X}$ with $\mu T_{i}\leq\mu T_{j}$ for $1\leq i<j\leq n$. For $1\leq k\leq n$, let $\mu_{T_{k}}(T)=T_{k}^{*}\oplus(\bigoplus_{i\neq k}T_{i})$ be the mutation of $T$ at $T_{k}$. Then

• (1)

$\mu T_{1}\leq\mu T_{1}^{*}\leq\mu T_{n}$* and $\mu T_{1}\leq\mu T_{n}^{*}\leq\mu T_{n}$.*

• (2)

If $T_{k}$ is a first object of $T$, then $\mu T_{k}\leq\mu T_{k}^{*}\leq\mu T_{n}$.

• (3)

If $T_{k}$ is a last object of $T$, then $\mu T_{k}\geq\mu T_{k}^{*}\geq\mu T_{1}$.

The paper is organized as follows. In Section 2, we recall some definitions and properties of coherent sheaves on a weighted projective line. We then focus on weighted projective lines of tubular type. For a weighted projective line $\mathbb{X}$ of tubular type, we discuss the tilting objects in $\operatorname{\mathsf{coh}}\mathbb{X}$, the automorphisms of bounded derived category $\mathcal{D}^{b}(\mathbb{X})$ of $\operatorname{\mathsf{coh}}\mathbb{X}$ and the perpendicular category of some quasi-simple rigid object in $\operatorname{\mathsf{coh}}\mathbb{X}$ in Section 3. In Section 4, for some special direct summands of a tilting object in $\operatorname{\mathsf{coh}}\mathbb{X}$, we discuss how the slopes changed under mutations. In particular, Theorem 1.2 is proved. We introduce APR mutation and co-APR mutation for tilting objects and proved that the APR mutation and co-APR mutation on a tilting bundle are tilting bundles. The proof of Theorem 1.1 is presented in Section  5.

Notation. For a weighted projective line $\mathbb{X}$ with weight sequence $(p_{1},p_{2},\cdots,p_{t})$, we denote by $\operatorname{\mathsf{ind}}\mathbb{X}$ the set of all indecomposable objects in $\operatorname{\mathsf{coh}}\mathbb{X}$, $\operatorname{\mathsf{ind}}\mathcal{D}^{b}(\mathbb{X})$ the set of all indecomposable objects in $\mathcal{D}^{b}(\mathbb{X})$. For an indecomposable object $X$, we denote by $\mu X$ the slope of $X$. Denote by $p=\operatorname{\mathsf{lcm}}(p_{1},\cdots,p_{t})$ the least common multiple of $p_{1},p_{2},\ldots,p_{t}$. For an indecomposable object $Z$ in a standard tube, we denote by $\mathcal{T}_{Z}$ the tube where $Z$ lies in, $\mathcal{W}_{Z}$ the wing determined by Z, $q.l.Z$ the quasi-length of $Z$ in $\mathcal{T}_{Z}$.

2. Preliminary

2.1. Coherent sheaves associated to a weighted projective line

Let $\mathbb{X}=\mathbb{X}(\mathbf{p},\bm{\lambda})$ be a weighted projective line attached to a weight sequence $\mathbf{p}=(p_{1},\dots,p_{t})$ of positive integers $p_{i}$ and a parameter sequence $\bm{\lambda}=(\lambda_{1}\dots,\lambda_{t})$ of pairwise distinct elements of $\mathbb{P}_{1}(k)$. Denote by $L(\mathbf{p})$ the rank one additive group

[TABLE]

and $S(\mathbf{p},\lambda)$ the $L(\mathbf{p})$-graded commutative algebra

[TABLE]

where $\deg x_{i}=\vec{x}_{i}$ and $\lambda_{i}=[\lambda_{i}^{\prime}:\lambda_{i}^{{}^{\prime\prime}}]\in\mathbb{P}^{1}$. In [GL1], Geigle-Lenzing have associated to each weighted projective line $\mathbb{X}(\mathbf{p},\bm{\lambda})$ a category $\operatorname{\mathsf{coh}}\mathbb{X}$ of coherent sheaves on $\mathbb{X}$, which is the quotient category of finitely generated $L(\mathbf{p})$-graded $S(\mathbf{p},\bm{\lambda})$-modules by the Serre subcategory of finite length modules. Geigle-Lenzing showed that $\operatorname{\mathsf{coh}}\mathbb{X}$ is a connected, $k$-linear hereditary abelian category with finite dimensional $\operatorname{\mathsf{Hom}}$ and $\operatorname{\mathsf{Ext}}$ spaces. The free module $S(\mathbf{p},\bm{\lambda})$ yields a structure sheaf ${\mathcal{O}}$, and shifting the grading gives twists $E(\vec{x})$ for any $E\in\operatorname{\mathsf{coh}}\mathbb{X}$ and $\vec{x}\in L(\mathbf{p}).$ Moreover, they showed that, putting $\vec{w}:=\Sigma_{i=1}^{t}(\vec{c}-\vec{x}_{i})-2\vec{c}$, $\operatorname{\mathsf{coh}}\mathbb{X}$ has Serre dual in the form $\mathbb{D}\operatorname{\mathsf{Ext}}^{1}_{\mathbb{X}}(E,F)=\operatorname{\mathsf{Hom}}_{\mathbb{X}}(F,\tau E)$ for all $E,F\in\operatorname{\mathsf{coh}}\mathbb{X}$ where $\tau E=E(\vec{w})$.

Denote by $\operatorname{\mathsf{vect}}\mathbb{X}$ the full subcategory of $\operatorname{\mathsf{coh}}\mathbb{X}$ consisting of vector bundles, i.e. torsion-free sheaves, and by $\operatorname{\mathsf{coh}}_{0}\mathbb{X}$ the full subcategory consisting of sheaves of finite length, i.e. torsion sheaves. Each coherent sheaf is a direct sum of a vector bundle and a finite length sheaf. Each vector bundle has a finite filtration by line bundles and there is no nonzero morphism from $\operatorname{\mathsf{coh}}_{0}\mathbb{X}$ to $\operatorname{\mathsf{vect}}\mathbb{X}$.

There are ordinary simple sheaves $S_{\mu}$( $\mu\in\mathbb{P}^{1}\backslash\bm{\lambda}$ ) and exceptional simple sheaves $S_{i,j}$($1\leq i\leq t,1\leq j\leq p_{i}$). It is known that $S(\vec{x})=S,\operatorname{\mathsf{Hom}}_{\mathbb{X}}({\mathcal{O}},S)\neq 0$ for each ordinary simple sheaf $S$. For the exceptional simple sheaves, for each $1\leq i\leq t$, there is exactly one integer $j(1\leq j\leq p_{i})$ such that $\operatorname{\mathsf{Hom}}_{\mathbb{X}}({\mathcal{O}},S_{i,j})\neq 0$.

Let

[TABLE]

be the weight function.

Lemma 2.1**.**

[GL1]** The category $\operatorname{\mathsf{coh}}_{0}\mathbb{X}$ of sheaves of finite length is an exact abelian, uniserial subcategory of $\operatorname{\mathsf{coh}}\mathbb{X}$ which is stable under Auslander-Reiten transation. The components of the Auslander-Reiten quiver of $\operatorname{\mathsf{coh}}_{0}\mathbb{X}$ form a family of pairwise orthogonal standard tubes $(\mathcal{T}_{\mu})_{\mu\in\mathbb{P}^{1}}$ with rank $rk\mathcal{T}_{\mu}=p_{\bm{\lambda}}(\mu)$.

2.2. Classifications of weighted projective line

The genus $g_{\mathbb{X}}$ of a weighted projective line $\mathbb{X}$ is defined by

[TABLE]

A weighted projective line of genus $g_{\mathbb{X}}<1$($g_{\mathbb{X}}=1$, resp. $g_{\mathbb{X}}>1$) will be called of domestic(tubular, resp. wild) type.

The domestic weight types are, up to permutation, $(q)$ with $q\geq 1$, $(q_{1},q_{2})$ with $q_{1},q_{2}\geq 2$, $(2,2,n)$ with $n\geq 2$, $(2,3,3),(2,3,4),(2,3,5)$, whereas the tubular weight types are, up to permutation, $(2,2,2,2),(3,3,3),(2,4,4)$ and $(2,3,6)$.

2.3. Slopes of coherent sheaves

Denote by $K_{0}(\mathbb{X})$ the the Grothendieck group of $\operatorname{\mathsf{coh}}\mathbb{X}$, it is a free abelian group of rank $n=2+\Sigma_{i=1}^{t}(p_{i}-1)$. Recall from  [GL1], there are two $\mathbb{Z}$-linear forms, rk and $\deg$ on $K_{0}(\mathbb{X})$, called the rank and degree. In particular, for any $\vec{x}\in L(\bf{p})$,

[TABLE]

with $\delta:L(\mathbf{p})\rightarrow\mathbb{Z}$ defined by $\delta(\vec{x}_{i})=p/p_{i}$ where $p=\operatorname{\mathsf{lcm}}(p_{1},\cdots,p_{t})$ is the least common multiple of $\bf{p}$. For each object $E\in\operatorname{\mathsf{coh}}\mathbb{X}$, the slope of $E$ is defined as

[TABLE]

It is obvious that $\mu({\mathcal{O}}(\vec{x}))=\delta(\vec{x})$. By [L, Lemma 2.5], for each vector bundle $E\in\operatorname{\mathsf{coh}}\mathbb{X}$ and $\vec{x}\in L(\bf{p})$, we have

[TABLE]

An indecomposable object $E\in\operatorname{\mathsf{coh}}\mathbb{X}$ is called semistable if for each non-trivial subbundle $E^{\prime}$ of $E$, we have $\mu E^{\prime}\leq\mu E$. For each $q\in\mathbb{Q}\cup\{\infty\}$, we denote by $\mathcal{C}_{q}$ the full subcategory of $\operatorname{\mathsf{coh}}\mathbb{X}$ consisting of all semistable coherent sheaves of slope $q$. In particular, $\mathcal{C}_{\infty}$ is the full subcategory consisting of the objects of finite length. For $q,q^{\prime}\in\mathbb{Q}\cup\{\infty\}$ such that $q<q^{\prime}$, according to [GL1, Proposition 5.2], we know that $\operatorname{\mathsf{Hom}}_{\mathbb{X}}(\mathcal{C}_{q}^{\prime},\mathcal{C}_{q})=0$. Hence, for any objects $X\in\mathcal{C}_{q},Y\in\mathcal{C}_{q^{\prime}}$, we have $\operatorname{\mathsf{Ext}}^{1}_{\mathbb{X}}(X,Y)\cong\mathbb{D}\operatorname{\mathsf{Hom}}_{\mathbb{X}}(Y,\tau X)=0$ since $\tau X$ is still in $\mathcal{C}_{q}$.

2.4. Tilting sheaf and mutation

A sheaf $M\in\operatorname{\mathsf{coh}}\mathbb{X}$ is called rigid if $\operatorname{\mathsf{Ext}}^{1}_{\mathbb{X}}(M,M)=0$. A sheaf $T\in\operatorname{\mathsf{coh}}\mathbb{X}$ is a tilting sheaf if $T$ is rigid and for any object $X\in\operatorname{\mathsf{coh}}\mathbb{X}$ the condition $\operatorname{\mathsf{Hom}}_{\mathbb{X}}(T,X)=0=\operatorname{\mathsf{Ext}}_{\mathbb{X}}^{1}(T,X)$ implies that $X=0$. Moreover, if $T$ is a vector bundle, i.e. $T$ has no direct summand of finite length, then we call $T$ a tilting bundle. It is known the canonical tilting sheaf $T_{can}=\bigoplus\limits_{0\leq\vec{x}\leq\vec{c}}{\mathcal{O}}(\vec{x})$ is a tilting bundle in $\operatorname{\mathsf{coh}}\mathbb{X}$.

Note that the number of (pairwise non-isomorphic) indecomposable direct summands of any tilting sheaf equals the rank $n=2+\Sigma_{i=1}^{t}(p_{i}-1)$ of the Grothendieck group $K_{0}(\mathbb{X})$. A rigid sheaf with $n-1$ (pairwise non-isomorphic) indecomposable direct summands is called an almost complete tilting sheaf. An indecomposable sheaf $E^{\prime}$ is called a complement of the almost complete tilting sheaf $E$ if $E\oplus E^{\prime}$ is a tilting sheaf.

Definition 2.2**.**

Let $T=\overline{T}\oplus X$ and $T^{\prime}=\overline{T}\oplus X^{*}$ be two tilting objects in $\operatorname{\mathsf{coh}}\mathbb{X}$ where $X$ and $X^{*}$ are indecomposable. We call $T^{\prime}$ is the mutation of $T$ at $X$, denoted by $T^{\prime}=\mu_{X}(T)$.

By [Hü], each almost tilting sheaf in $\operatorname{\mathsf{coh}}\mathbb{X}$ have precisely two complements. In particular, for a given tilting sheaf $T$, we may mutate $T$ at any indecomposable direct summand of $T$ to obtain a new tilting sheaf. Moreover, we have

Lemma 2.3**.**

[Hü, Proposition 2.6, 2.8]** Let $T=T_{k}\oplus\overline{T}$ be a tilting sheaf in $\operatorname{\mathsf{coh}}\mathbb{X}$ with $T_{k}$ an indecomposable direct summand. Let $T_{k}\xrightarrow{f_{k}}B$ a minimal left $\operatorname{\mathsf{add}}\nolimits\overline{T}$-approximation and $B^{\prime}\xrightarrow{g_{k}}T_{k}$ a minimal right $\operatorname{\mathsf{add}}\nolimits\overline{T}$-approximation. Then

• (1)

$f_{k}$* (resp. $g_{k}$) is either mono or epi.*

• (2)

$f_{k}$* is mono (resp. epi) iff $g_{k}$ is mono (resp. epi).*

• (3)

let $T_{k}^{*}=\operatorname{\mathsf{ker}}\nolimits g_{k}\oplus\operatorname{\mathsf{cok}}\nolimits f_{k}$, then $T_{k}^{*}\oplus\overline{T}$ is again a tilting sheaf in $\operatorname{\mathsf{coh}}\mathbb{X}$.

As a direct consequence, we obtain

Lemma 2.4**.**

Let $\overline{T}$ be an almost tilting sheaf in $\operatorname{\mathsf{coh}}\mathbb{X}$, $T_{k}$ and $T_{k}^{*}$ are two complements of $\overline{T}$. Then either there is an exact sequence $0\rightarrow T_{k}\xrightarrow{f}B\xrightarrow{}T_{k}^{*}\rightarrow 0$ or an exact sequence $0\rightarrow T_{k}^{*}\rightarrow B^{\prime}\xrightarrow{g}T_{k}\rightarrow 0$, where $f$ is the minimal left $\operatorname{\mathsf{add}}\nolimits\overline{T}$-approximation and $g$ is the minimal right $\operatorname{\mathsf{add}}\nolimits\overline{T}$-approximation. Moreover, if $\mu(T_{k})<\mu(T_{k}^{*})$, the exact sequence $0\rightarrow T_{k}\rightarrow B\xrightarrow{g}T_{k}^{*}\rightarrow 0$ holds.

3. Tubular type

Let $\mathbb{X}$ be a tubular weighted projective line with weight sequence $(p_{1},\cdots,p_{t})$, set $p:=\operatorname{\mathsf{lcm}}(p_{1},\cdots,p_{t})$.

3.1. Shape of tubular hereditary category

Let $\mathbb{X}$ be a tubular weighted projective line. By [GL1], each indecomposable coherent sheaf in $\operatorname{\mathsf{coh}}\mathbb{X}$ is semistable. Moreover, we have

Theorem 3.1**.**

[GL1, Theorem 5.6]** For each $q\in\mathbb{Q}\cup\{\infty\}$, $\mathcal{C}_{q}$ is equivalent to $\operatorname{\mathsf{coh}}_{0}\mathbb{X}$, i.e. the components of the Auslander-Reiten quiver of $\mathcal{C}_{q}$ form a family of pairwise orthogonal standard tubes $(\mathcal{T}_{\mu})_{\mu\in\mathbb{P}^{1}}$ with rank $rk\mathcal{T}_{\mu}=p_{\lambda}(\mu)$.

In the following of this note, for an indecomposable object $Z\in\operatorname{\mathsf{coh}}\mathbb{X}$, we denote by $\mathcal{T}_{Z}$ the tube where $Z$ lies in, $\mathcal{W}_{Z}$ the wing determined by $Z$, $q.l.Z$ the quasi-length of $Z$. If $q.l.Z=1$, we say that $Z$ is quasi-simple. We call the smallest positive integer $d$ such that $\tau^{d}Z=Z$ the $\tau$-period of $Z$. A tube with rank one is called a homogeneous tube.

It is known that every line bundle is quasi-simpe with integer slope and the $\tau$-period of a line bundle is $p$. Let $X$ be an indecomposable object of $\operatorname{\mathsf{coh}}\mathbb{X}$ lying in a tube with rank $r$, it is clear that $X$ is rigid if and only if $q.l.X\leq r-1$.

3.2. Riemann-Roch theorem

For a weighted projective line $\mathbb{X}$ of arbitrary type, the Euler form on $K_{0}(\mathbb{X})$ is given on sheaves by $\chi(X,Y)=\dim_{k}\operatorname{\mathsf{Hom}}_{\mathbb{X}}(X,Y)-\dim_{k}\operatorname{\mathsf{Ext}}_{\mathbb{X}}^{1}(X,Y)$. The weighted form of Riemann-Roch theorem ([GL1, 2.9], [M2, 3.1.14]) states that

[TABLE]

where $\overline{\chi}(X,Y)=\Sigma_{j=0}^{p-1}\chi(\tau^{j}X,Y)$ is the averaged Euler form. In particular, if $\mathbb{X}$ is of tubular type, then

[TABLE]

Hence, one can get:

Lemma 3.2**.**

[L, Proposition 3.2]** Let $\mathbb{X}$ be a tubular weighted projective line. For $q,q^{\prime}\in\mathbb{Q}\cup\infty$ and $q\neq q^{\prime}$, $\operatorname{\mathsf{Hom}}_{\mathbb{X}}(\mathcal{C}_{q},\mathcal{C}_{q^{\prime}})\neq 0$ if and only if $q<q^{\prime}$.

Proof.

If $q>q^{\prime}$, it is known that $\operatorname{\mathsf{Hom}}_{\mathbb{X}}(\mathcal{C}_{q},\mathcal{C}_{q^{\prime}})=0$. Hence, if $\operatorname{\mathsf{Hom}}_{\mathbb{X}}(\mathcal{C}_{q},\mathcal{C}_{q^{\prime}})\neq 0$ and $q\neq q^{\prime}$, we must have $q<q^{\prime}$.

Now suppose that $q<q^{\prime}$. Let $X\in\mathcal{C}_{q},Y\in\mathcal{C}_{q^{\prime}}$ be two indecomposable sheaves. Then we have

[TABLE]

Since for any integer $j$, $\tau^{j}X\in\mathcal{C}_{q}$, $\operatorname{\mathsf{Ext}}^{1}_{\mathbb{X}}(\tau^{j}X,Y)\cong\mathbb{D}\operatorname{\mathsf{Hom}}_{\mathbb{X}}(Y,\tau^{j+1}X)=0$, then

[TABLE]

Therefore, $\operatorname{\mathsf{Hom}}_{\mathbb{X}}(\mathcal{C}_{q},\mathcal{C}_{q^{\prime}})\neq 0$ since $\tau^{j}X\in\mathcal{C}_{q}$ for any $j$. ∎

It is easy to get that $\overline{\chi}(X,Y)$ is also equal to $\Sigma_{j=0}^{p-1}\chi(X,\tau^{j}Y)$. Hence, if $X$ or $Y$ lies in a homogenous tube, we have

[TABLE]

Then it is easy to get

Lemma 3.3**.**

Let $\mathbb{X}$ be a tubular weighted projective line, $X,Y\in\operatorname{\mathsf{ind}}\mathbb{X}$, where $Y$ lies in a homogenous tube. If $\mu X<\mu Y$, then $\operatorname{\mathsf{Hom}}_{\mathbb{X}}(X,Y)\neq 0$, while if $\mu X>\mu Y$, then $\operatorname{\mathsf{Hom}}_{\mathbb{X}}(Y,X)\neq 0$.

3.3. Some basic properities of tilting objects in tubular cases

Let $\mathbb{X}$ be a tubular weighted projective line. Similar as the properties of wings in the situation of modules [SS1, Chapter X, Lemma 2.8, Lemma 2.9](cf. [M2, Lemma 8.3.4] for wild type), one can get

Lemma 3.4**.**

Let $X,Z$ be two indecomposable object in $\operatorname{\mathsf{coh}}\mathbb{X}$ and $X\notin\mathcal{W}_{Z}$. If $\operatorname{\mathsf{Hom}}_{\mathbb{X}}(X,Z)=0$, then for any object $Y\in\mathcal{W}_{Z}$, $\operatorname{\mathsf{Hom}}_{\mathbb{X}}(X,Y)=0$. Dually, if $\operatorname{\mathsf{Hom}}_{\mathbb{X}}(Z,X)=0$, then for any object $Y\in\mathcal{W}_{Z}$, $\operatorname{\mathsf{Hom}}_{\mathbb{X}}(Y,X)=0$.

According Lemma 3.4, one can get

Lemma 3.5**.**

Let $T$ be a tilting sheaf in $\operatorname{\mathsf{coh}}\mathbb{X}$, $Z$ be an indecomposable direct summand of $T$. Denote by $N$ the direct sum of all the summands of $T$ lying in the wing $\mathcal{W}_{Z}$ and $T=T^{\prime}\oplus N$. Then

• (1)

$N$* is a tilting object in $\mathcal{W}_{Z}$.*

• (2)

for any tilting object $N^{\prime}$ in $\mathcal{W}_{Z}$, $T^{\prime}\oplus N^{\prime}$ is a tilting sheaf in $\operatorname{\mathsf{coh}}\mathbb{X}$.

• (3)

$T$* has a quasi-simple direct summand lying in $\mathcal{W}_{Z}$.*

Proof.

For $(1)$, it is easy to get that $N$ is partial tilting in $\mathcal{W}_{Z}$. Choose an object $M\in\mathcal{W}_{Z}$ such that $N\oplus M$ is tilting in $\mathcal{W}_{Z}$. Since $\operatorname{\mathsf{Hom}}_{\mathcal{H}}(T^{\prime},\tau Z)=\mathbb{D}\operatorname{\mathsf{Ext}}_{\mathbb{X}}^{1}(Z,T^{\prime})=0$, by Lemma 3.4, we have $\operatorname{\mathsf{Ext}}_{\mathbb{X}}^{1}(M,T^{\prime})=\mathbb{D}\operatorname{\mathsf{Hom}}_{\mathcal{H}}(T^{\prime},\tau M)=0$ since $\tau M\in\mathcal{W}_{\tau Z}$. Similarly, we have $\operatorname{\mathsf{Ext}}_{\mathbb{X}}^{1}(T^{\prime},M)=0$, then we get a partial tilting object $T^{\prime}\oplus N\oplus M=T\oplus M$. Since $T$ is tilting, then $M\in\operatorname{\mathsf{add}}\nolimits T$. By $M\in\mathcal{W}_{Z}$, $M\in\operatorname{\mathsf{add}}\nolimits N$. Therefore, $N$ is a tilting object in $\mathcal{W}_{Z}$.

For $(2)$, similarly as $(1)$, we have $T^{\prime}\oplus N^{\prime}$ is partial tilting. Since $|N|=|N^{\prime}|$, $|T^{\prime}\oplus N^{\prime}|=|T^{\prime}|+|N^{\prime}|=|T^{\prime}|+|N|=|T^{\prime}\oplus N|=|T|$. Hence, $T^{\prime}\oplus N^{\prime}$ is a tilting sheaf in $\operatorname{\mathsf{coh}}\mathbb{X}$.

For $(3)$, suppose $q.l.Z=l$. It is known that the wing $\mathcal{W}_{Z}$ is equivalent to $\operatorname{\mathsf{mod}}\nolimits A_{l}$, where $A_{l}$ is the hereditary algebra given as the path algebra of a quiver of Dynkin type $A_{l}$ with linear orientation. Hence $N$ can be seen as a tilting module in $\operatorname{\mathsf{mod}}\nolimits A_{l}$. It is known that each tilting $A_{l}$-module has at least one simple $A_{l}$-module, so $N$ has a quasi-simple direct summand lying in $\mathcal{W}_{Z}$. Then we can get the result. ∎

Lemma 3.6**.**

[M1, Proposition 4.2]** Each tilting sheaf $T$ in $\operatorname{\mathsf{coh}}\mathbb{X}$ has an indecomposable direct summand whose $\tau$-period equals $p.$

By Lemma 3.6 and Lemma 3.5, one can get

Lemma 3.7**.**

Each tilting sheaf $T$ in $\operatorname{\mathsf{coh}}\mathbb{X}$ has a quasi-simple direct summand whose $\tau$-period equals $p.$

For a line bundle, we have

Lemma 3.8**.**

[M1, Lemma 4.1]** Let $T$ be a tilting bundle in $\operatorname{\mathsf{coh}}\mathbb{X}$. Then for any line bundle $L$, either $\operatorname{\mathsf{Ext}}^{1}_{\mathbb{X}}(T,L)=0$ or $\operatorname{\mathsf{Hom}}_{\mathbb{X}}(T,L)=0$.

3.4. Some automorphisms of $\mathcal{D}^{b}(\mathbb{X})$

Let $\mathbb{X}$ be a weighted projective line. Denote the finite bounded derived category of $\operatorname{\mathsf{coh}}\mathbb{X}$ by $\mathcal{D}^{b}(\mathbb{X})$, the translation functor of $\mathcal{D}^{b}(\mathbb{X})$ by $[1]$. Since $\operatorname{\mathsf{coh}}\mathbb{X}$ is hereditary, every $E\in\operatorname{\mathsf{ind}}(\mathcal{D}^{b}(\mathbb{X}))$ belongs to $\operatorname{\mathsf{coh}}\mathbb{X}[l]$ for some integer $l$. Hence each self-equivalence of $\operatorname{\mathsf{coh}}\mathbb{X}$ can induce an automorphism of $\mathcal{D}^{b}(\mathbb{X})$ naturally. Denote by $\operatorname{\mathsf{Pic}}(\mathbb{X})$ the Picard group which can be identified with the grading group $L(\mathbf{p})$, $\operatorname{\mathsf{Pic}}_{0}(\mathbb{X})$ be the subgroup of $\operatorname{\mathsf{Pic}}{\mathbb{X}}$ consisting of all degree-preserving shifts, $\operatorname{\mathsf{Aut}}(\operatorname{\mathsf{coh}}\mathbb{X})$ the automorphism group of $\operatorname{\mathsf{coh}}\mathbb{X}$. Define the automorphism group $\operatorname{\mathsf{Aut}}\mathbb{X}$ of $\mathbb{X}$ as the subgroup of $\operatorname{\mathsf{Aut}}(\operatorname{\mathsf{coh}}\mathbb{X})$ of automorphisms $F$ fixing the structure sheaf, i.e. satisfying $F({\mathcal{O}})={\mathcal{O}}$.

For an object $E\in\operatorname{\mathsf{coh}}\mathbb{X}$ and any $i\in\mathbb{Z}$, we say $\mu(E[i])=\mu E$. An automorphism $F$ of $\mathcal{D}^{b}(\mathbb{X})$ is called slope-preserving if $\mu(F(E))=\mu E$ for any $E\in\operatorname{\mathsf{ind}}\mathcal{D}^{b}(\mathbb{X})$. Denote by $\operatorname{\mathsf{Aut}}_{\mu}(\operatorname{\mathsf{coh}}\mathbb{X})$ and $\operatorname{\mathsf{Aut}}_{\mu}(\mathcal{D}^{b}(\mathbb{X}))$ of slope-preserving automorphism of $\operatorname{\mathsf{coh}}\mathbb{X}$ and $\mathcal{D}^{b}(\mathbb{X})$.

Lemma 3.9**.**

[LM2, Proposition 4.1, Proposition 4.4]** Let $\mathbb{X}$ be a weighted projective line, then

[TABLE]

where $\langle[1]\rangle\cong\mathbb{Z}$ is the subgroup generated by the translation functor of $\mathcal{D}^{b}(\mathbb{X})$.

3.4.1. Tubular case

Assume that $\mathbb{X}$ is a weighted projective line of tubular type. In this subsection, we are going to discuss some automorphisms of $\mathcal{D}^{b}(\mathbb{X})$. For more details, we refer to [LM1, LM2, M2].

Theorem 3.10**.**

[M2, Theorem 5.2.6]** For each $q,q^{\prime}\in\mathbb{Q}\cup\{\infty\}$, there is an automorphism $\Phi_{q^{\prime},q}$ of $\mathcal{D}^{b}(\mathbb{X})$ such that $\mathcal{C}_{q}$ is mapped to $\mathcal{C}_{q^{\prime}}$. Moreover, this functors satisfy $\Phi_{q^{{}^{\prime\prime}},q}=\Phi_{q^{{}^{\prime\prime}},q^{\prime}}\Phi_{q^{\prime},q}$ and $\Phi_{q,q}=Id$.

Let $B_{3}$ be the braid group on three strands, i.e. a group with generators $\sigma_{1},\sigma_{2}$ and with relations $\sigma_{1}\sigma_{2}\sigma_{1}=\sigma_{2}\sigma_{1}\sigma_{2}$. Then we have

Theorem 3.11**.**

[LM2, Theorem 6.3]** $\operatorname{\mathsf{Aut}}(\mathcal{D}^{b}(\mathbb{X}))\cong(\operatorname{\mathsf{Pic}}_{0}(\mathbb{X})\ltimes\operatorname{\mathsf{Aut}}(\mathbb{X}))\ltimes B_{3}.$

By [LM2, M2], there are automorphisms $\sigma_{1},\sigma_{2}$ of $\mathcal{D}^{b}(\mathbb{X})$ such that

[TABLE]

for any $E\in\operatorname{\mathsf{ind}}\mathcal{D}^{b}(\mathbb{X})$ and the subgroup generated by $\sigma_{1},\sigma_{2}$ is isomorphic to $B_{3}$. There are also an group epimorphism $f:B_{3}\rightarrow\operatorname{\mathsf{SL}}(2,\mathbb{Z})$ defined by f(\sigma_{1})=\left(\begin{array}[]{cc}1&1\\ 0&1\end{array}\right) and f(\sigma_{2})=\left(\begin{array}[]{cc}1&0\\ 1&1\end{array}\right). According to Theorem 3.11, one can get the following lemma easily.

Lemma 3.12**.**

For any linear fractional transformation $\varphi_{w}(x)=\frac{cx+a}{dx+b}$ with w=\left(\begin{array}[]{cc}c&a\\ d&b\end{array}\right)\in\operatorname{\mathsf{SL}}(2,\mathbb{Z}), there is an automorphism $\overline{\varphi}_{w}$ of $\mathcal{D}^{b}({\mathbb{X}})$, such that $\overline{\varphi}_{w}$ coincides with the action of $\varphi_{w}$ on slopes, i.e. for any $E\in\operatorname{\mathsf{ind}}\mathcal{D}^{b}(\mathbb{X})$, $\mu(\overline{\varphi}_{w}(E))=\varphi_{w}(\mu E)$.

Dually, by Lemma 3.9 and Theorem 3.11, one can get:

Lemma 3.13**.**

[LM2]** Let $\overline{\varphi}$ be an automorphism of $\mathcal{D}^{b}({\mathbb{X}})$, then there exists a linear fractional transformation $\varphi(x)=\frac{cx+a}{dx+b}$ with \left(\begin{array}[]{cc}c&a\\ d&b\end{array}\right)\in\operatorname{\mathsf{SL}}(2,\mathbb{Z}), such that the action of $\varphi$ on the slopes coincides with $\overline{\varphi}$.

Using Lemma 3.9 and Theorem 3.10, it is easy to get

Lemma 3.14**.**

For any two quasi-simple rigid objects $X,Y\in\operatorname{\mathsf{coh}}\mathbb{X}$ with same $\tau$-period and slope, there is a slope-preserving automorphism $F$ of $\mathcal{D}^{b}(\mathbb{X})$ such that $F(X)=Y$.

According to Lemma 3.2, one can get

Lemma 3.15**.**

Let $F\in\operatorname{\mathsf{Aut}}(\mathcal{D}^{b}(\mathbb{X}))$, $X,Y\in\operatorname{\mathsf{ind}}\mathbb{X}$ such that $\mu X<\mu Y$. Suppose that $F(X)\in\operatorname{\mathsf{coh}}\mathbb{X}[l],F[Y]\in\operatorname{\mathsf{coh}}\mathbb{X}[m]$ for some integers $l,m$. Then either $m=l$ or $m=l+1$. Moreover, if $m=l$, then $\mu(F(X))<\mu(F(Y))$. Else if $m=l+1$, then $\mu(F(X))>\mu(F(Y))$.

Proof.

Firstly, we claim that $F(\mathcal{C}_{\mu X})=\mathcal{C}_{\mu(F(X))}[l]$.

Let $X^{\prime}$ be an indecomposable object lying in $\mathcal{C}_{\mu X}$. Suppose that $F(X^{\prime})\in\operatorname{\mathsf{coh}}\mathbb{X}[l^{\prime}]$ for some integer $l^{\prime}$. Since $\mu X=\mu X^{\prime}$, by Lemma 3.13, we have $\mu(F(X))=\mu(F(X^{\prime}))$.

Let $Z$ be an indecomposable object lying in a homogenous tube in $\operatorname{\mathsf{coh}}\mathbb{X}$ such that $\mu Z<\mu X$. By Lemma 3.13, we have $\mu(F(X))\neq\mu(F(Z))$. By Lemma 3.3, we have $\operatorname{\mathsf{Hom}}_{\mathbb{X}}(Z,X)\neq 0$ and $\operatorname{\mathsf{Hom}}_{\mathbb{X}}(Z,X^{\prime})\neq 0$. Then $\operatorname{\mathsf{Hom}}_{\mathcal{D}^{b}(\mathbb{X})}(F(Z),F(X))\neq 0$ and $\operatorname{\mathsf{Hom}}_{\mathcal{D}^{b}(\mathbb{X})}(F(Z),F(X^{\prime}))\neq 0$. Since $\operatorname{\mathsf{coh}}\mathbb{X}$ is hereditary and $\mu(F(X^{\prime}))=\mu(F(X))\neq\mu(F(Z))$, we must have $l^{\prime}=l$. Then $F(X^{\prime})\in\mathcal{C}_{\mu(F(X))}[l]$. Hence, $F(\mathcal{C}_{\mu X})\subset\mathcal{C}_{\mu(F(X))}[l]$. Let $G$ be the automorphism of $\mathcal{D}^{b}(\mathbb{X})$ such that $G\circ F=Id$. Since any automorphism of $\mathcal{D}^{b}(\mathbb{X})$ is commuted with the translation functor $[1]$, by $F(X)[-l]\in\operatorname{\mathsf{coh}}\mathbb{X}$, $G(F(X)[-l])=X[-l]\in\operatorname{\mathsf{coh}}\mathbb{X}[-l]$, one can get

[TABLE]

Therefore, we must have $F(\mathcal{C}_{\mu X})=\mathcal{C}_{\mu(F(X))}[l]$ since $G\circ F=Id$. Similarly, we also have $F(\mathcal{C}_{\mu Y})=\mathcal{C}_{\mu(F(Y))}[m]$.

In the next, we are going to prove the Lemma.

Since $\mu X<\mu Y$, by Lemma 3.13, we have $\mu(F(X))\neq\mu(F(Y))$ while by Lemma 3.2, we have $\operatorname{\mathsf{Hom}}_{\mathbb{X}}(\mathcal{C}_{\mu X},\mathcal{C}_{\mu Y})\neq 0$. Then

[TABLE]

then $m=l$ or $m=l+1$.

If $m=l$, then

[TABLE]

Since $\mu(F(X))\neq\mu(F(Y))$, by Lemma 3.2, $\mu(F(X))<\mu(F(Y))$.

If $m=l+1$, then

[TABLE]

Then by $\mathcal{C}_{\mu(F(X)}$ is stable under Auslander-Reiten translation $\tau$, we have

[TABLE]

Since $\mu(F(X))\neq\mu(F(Y))$, by Lemma 3.2, $\mu(F(Y))<\mu(F(X))$. ∎

The following lemma is useful to construct matrices in $\operatorname{\mathsf{SL}}(2,\mathbb{Z})$.

Lemma 3.16**.**

For integers $a,b$ such that $0<a<b$ and $(a,b)=1$, there are integers $c,d$ such that $bc-ad=1$ and $0<c\leq d<b$ and $c\leq a$.

Proof.

Since $(a,b)=1$, so there are integers $c_{0},d_{0}$ such that $bc_{0}-ad_{0}=1$. Then $b(c_{0}+ma)-a(d_{0}+mb)=1$ for any integer $m$. Choose an integer $m$ such that $0<d_{0}+mb\leq b$, let $c=c_{0}+ma,d=d_{0}+mb$. If $d=b$, then $1=bc-ad=b(c-a)\neq 1$ since $b\geq 2$, a contradiction. So $0<d<b$. By $d>0$ and $bc-ad=1$, we have $c>0$. If $c>a$ or $c>d$, then $bc-ad>1$. Hence, $c\leq d$ and $c\leq a$. ∎

3.5. Perpendicular category

Let $\mathcal{A}$ be an abelian category and $\mathcal{S}$ be a system of objects in $\mathcal{A}$. The right perpendicular category $\mathcal{S}^{\perp}$ of $\mathcal{S}$ is defined as the full subcategory of all objects $M\in\mathcal{A}$ which satisfy both $\operatorname{\mathsf{Hom}}_{\mathcal{A}}(S,M)=0$ and $\operatorname{\mathsf{Ext}}_{\mathcal{A}}^{1}(S,M)=0$ for all $S\in\mathcal{S}$. Similarly, one can define the left perpendicular category ${}^{\perp}\mathcal{S}$ of $\mathcal{S}$.

Lemma 3.17**.**

[Hü]** Let $\mathbb{X}$ be a weighted projective line and $X\in\operatorname{\mathsf{coh}}\mathbb{X}$ be an indecomposable rigid vector bundle, then ${}^{\perp}X$ and $X^{\perp}$ are module category of some finite dimensional hereditary algebra with $n-1$ simple objects where $n$ is the rank of $K_{0}(\mathbb{X})$.

Lemma 3.18**.**

[GL2]** Let $\mathbb{X}$ be a weighted projective line of tubular type with weight sequence $(p_{1},p_{2},\cdots,p_{t})$ and $S_{i}\in\operatorname{\mathsf{coh}}\mathbb{X}$ be an exceptional simple sheaf lying in a tube with rank $p_{i}$, then ${}^{\perp}S_{i}$ and $S_{i}^{\perp}$ are the hereditary category of coherent sheaves of domestic type of $(p_{1},\cdots,p_{i}-1,\cdots,p_{t})$.

Let $\mathbb{X}$ be a weighted projective line. Let $Z$ be an indecomposable object in $\mathcal{D}^{b}(\mathbb{X})$, denote by $Z^{\perp_{\mathcal{D}}}$ the full subcategory of all objects $M\in\mathcal{D}^{b}(\mathbb{X})$ such that $\operatorname{\mathsf{Hom}}_{\mathcal{D}^{b}(\mathbb{X})}(Z[i],M)=0$ for any $i\in\mathbb{Z}$. Since $\operatorname{\mathsf{coh}}\mathbb{X}$ is hereditary, it is easy to get that $Z^{\perp_{\mathcal{D}}}$ in $\mathcal{D}^{b}(\mathbb{X})$ is the derived category of $Z^{\perp}$ in $\operatorname{\mathsf{coh}}\mathbb{X}$. Similarly, we can define ${}^{\perp_{\mathcal{D}}}Z$.

Proposition 3.19**.**

Let $\mathbb{X}$ be a tubular weighted projective line and $X\in\operatorname{\mathsf{coh}}\mathbb{X}$ a quasi-simple rigid vector bundle, then $X^{\perp}$ is the module category of some connected hereditary algebra of tame type. Similar result holds true for ${}^{\perp}X$.

Proof.

Let $S\in\operatorname{\mathsf{coh}}\mathbb{X}$ be an exceptional simple sheaf such that $S$ has same $\tau$-period with $X$. By Theorem 3.10 and Lemma 3.14, we may choose an automorphism $F$ of $\mathcal{D}^{b}(\mathbb{X})$ such that $F(S)=X$. Consequently, $S^{\perp_{\mathcal{D}}}$ is equivalent to $X^{\perp_{\mathcal{D}}}$. Note that $S^{\perp_{\mathcal{D}}}$ is the derived category of $S^{\perp}$ and $X^{\perp_{\mathcal{D}}}$ is the derived category of $X^{\perp}$. By Lemma 3.17 and Lemma 3.18, $X^{\perp}$ is the module category of some hereditary algebra $H$, $S^{\perp}$ is the category of coherent sheaves of domestic type. Since the category of coherent sheaves of domestic type is derived equivalent to the module category of some connected hereditary algebra $A$ of tame type, the two hereditary algebras $H$ and $A$ are derived equivalent. Hence, $H$ is also a connected hereditary algebra of tame type. ∎

For a tame hereditary algebra $A$, it is well-known that there is no nonzero map from the preinjective component to the regular and the preprojective components and there is no nonzero map from the regular component to the preprojective component. And it is known that every preprojective $A$-module has a nonzero map to each homogenous tube and every homogenous tube has a nonzero map to each preinjective $A$-module, then using Lemma 3.3 and Proposition 3.19, one can get:

Lemma 3.20**.**

Let $\mathbb{X}$ be a tubular weighted projective line, $X$ be a quasi-simple rigid vector bundle in $\operatorname{\mathsf{coh}}\mathbb{X}$. Let $Z$ be an indecomposable object in $\operatorname{\mathsf{coh}}\mathbb{X}$ such that $Z\in X^{\perp}$ (resp. ${}^{\perp}X$).

• (1)

If $\mu Z<\mu X$, then $Z$ lies in the preprojective component of $X^{\perp}$ (resp. ${}^{\perp}X$).

• (2)

If $\mu Z=\mu X$, then $Z$ lies in the regular component of $X^{\perp}$ (resp. ${}^{\perp}X$).

• (3)

If $\mu Z>\mu X$, then $Z$ lies in the preinjective component of $X^{\perp}$ (resp. ${}^{\perp}X$).

Proof.

It is clear that the homogenous tubes in the subcategory $\mathcal{C}_{\mu X}$ belong to $X^{\perp}$ and they are still homogenous in $X^{\perp}$. Choose one in these homogenous tubes, denoted by $\mathcal{T}_{thin}$.

If $\mu Z<\mu X$, by Lemma 3.3, $\operatorname{\mathsf{Hom}}_{\mathbb{X}}(Z,\mathcal{T}_{thin})\neq 0$. Then $\operatorname{\mathsf{Hom}}_{X^{\perp}}(Z,\mathcal{T}_{thin})\neq 0$, hence $Z$ must lie in the preprojective component of $X^{\perp}$.

If $\mu Z>\mu X$, by Lemma 3.3, $\operatorname{\mathsf{Hom}}_{\mathbb{X}}(\mathcal{T}_{thin},Z)\neq 0$. Then $\operatorname{\mathsf{Hom}}_{X^{\perp}}(\mathcal{T}_{thin},Z)\neq 0$, hence $Z$ must lie in the preinjective component of $X^{\perp}$.

Now suppose $\mu Z=\mu X$. If $Z\in\mathcal{T}_{thin}$, then obviously $Z$ lies in the regular component of $X^{\perp}$. Else $\operatorname{\mathsf{Hom}}_{\mathbb{X}}(\mathcal{T}_{thin},Z)=0=\operatorname{\mathsf{Hom}}_{\mathbb{X}}(\mathcal{T}_{thin},Z)$, then $\operatorname{\mathsf{Hom}}_{X^{\perp}}(\mathcal{T}_{thin},Z)=0=\operatorname{\mathsf{Hom}}_{X^{\perp}}(Z,\mathcal{T}_{thin})$. Since the tame hereditary algebra whose module category is equivalent to $X^{\perp}$ is connected, $Z$ lies in the regular component of $X^{\perp}$. ∎

For a tame hereditary algebra $A$, a preprojective tilting $A$-module is a tilting $A$-module lying in the preprojective component of $\operatorname{\mathsf{mod}}\nolimits A$. Similarly, we can define a preinjective tilting $A$-module.

Lemma 3.21**.**

Let $\mathbb{X}$ be a tubular weighted projective line and $X\in\operatorname{\mathsf{coh}}\mathbb{X}$ a quasi-simple rigid vector bundle.

• (1)

Let $M$ be a preprojective tilting module in $X^{\perp}$, then $X\oplus M$ is a tilting object in $\operatorname{\mathsf{coh}}\mathbb{X}$.

• (2)

Let $N$ be a preinjective tilting module in ${}^{\perp}X$, then $X\oplus N$ is a tilting object in $\operatorname{\mathsf{coh}}\mathbb{X}$.

Proof.

We prove $(1)$ and the proof of $(2)$ is similar.

Since $M\in X^{\perp}$, $\operatorname{\mathsf{Ext}}^{1}_{\mathbb{X}}(X,M)=0$. Note that, as $M$ lies in the preprojective component of $X^{\perp}$, each direct summand of $M$ has slope smaller than $\mu X$ by Lemma 3.20. As a consequence, $\operatorname{\mathsf{Ext}}_{\mathbb{X}}^{1}(M,X)=0$. Therefore, we have $\operatorname{\mathsf{Ext}}^{1}_{\mathbb{X}}(X\oplus M,X\oplus M)=0$. Since $|M|=n-1$, hence $|X\oplus M|=n$, then $X\oplus M$ is a tilting object in $\operatorname{\mathsf{coh}}\mathbb{X}$. ∎

4. Bundle-mutation

4.1. $\operatorname{\mathsf{APR}}$ mutation and co-$\operatorname{\mathsf{APR}}$ mutation

Let $\mathcal{H}$ be a hereditary abelian category with tilting objects.

Definition 4.1**.**

Let $T=\bigoplus_{i=1}^{n}T_{i}$ be a tilting object in $\mathcal{H}$ where $T_{i}\in\operatorname{\mathsf{ind}}\mathcal{H}$ for each $i$, $T_{k}$ and $T_{l}$ be two indecomposable direct summands of $T$.

• (1)

If $\operatorname{\mathsf{Hom}}_{\mathbb{X}}(T_{i},T_{k})=0$ for any $i\neq k$, we call $T_{k}$ a first object of $T$.

• (2)

If $\operatorname{\mathsf{Hom}}_{\mathbb{X}}(T_{l},T_{i})=0$ for any $i\neq l$, we call $T_{l}$ a last object of $T$.

Definition 4.2**.**

For a tilting object $T$ in $\mathcal{H}$, we call the mutation at a first object of $T$ an $\operatorname{\mathsf{APR}}$ mutation and the mutation at a last object of $T$ a co-$\operatorname{\mathsf{APR}}$ mutation.

The following Lemma is easy to get (cf. [ASS, Chapter VIII, Theorem 4.5] and [P, BKL]).

Lemma 4.3**.**

Let $\mathcal{H}$ be the module category of a tame hereditary algebra, $T$ and $T^{\prime}$ be two tilting modules in $\mathcal{H}$.

• (a)

Suppose that $T^{\prime}$ is obtained from $T$ by an $\operatorname{\mathsf{APR}}$ mutation or a co-$\operatorname{\mathsf{APR}}$* mutation. If $T$ is preprojective, then $T^{\prime}$ is also preprojective, while if $T$ is preinjective, then $T^{\prime}$ is also preinjective.*

• (b)

If $T$ is preprojective, then by $\operatorname{\mathsf{APR}}$ mutations, $T$ can be transformed into a slice lying in the preprojective component. While if $T$ is preinjective, then by co-$\operatorname{\mathsf{APR}}$* mutations, $T$ can be transformed into a slice lying in the preinjective component.*

• (c)

If both $T$ and $T^{\prime}$ are preprojective tilting modules in $\mathcal{H}$ or preinjective tilting modules in $\mathcal{H}$, then $T$ can be get from $T^{\prime}$ by $\operatorname{\mathsf{APR}}$ mutations and co-$\operatorname{\mathsf{APR}}$* mutations.*

Then by Lemma 3.21, we have:

Lemma 4.4**.**

Let $\mathbb{X}$ be a tubular weighted projective line, $X$ be a quasi-simple rigid vector bundle. Then

• (1)

the $\operatorname{\mathsf{APR}}$ mutation on a preprojective tilting module in $X^{\perp}$ induces an $\operatorname{\mathsf{APR}}$ mutation in $\operatorname{\mathsf{coh}}\mathbb{X}$ naturally;

• (2)

the co-$\operatorname{\mathsf{APR}}$* mutation on a preinjective tilting module in ${}^{\perp}X$ induces a co-$\operatorname{\mathsf{APR}}$ mutation in $\operatorname{\mathsf{coh}}\mathbb{X}$ naturally.*

4.2. Minimal and maximal direct summands of a tilting sheaf

Let $\mathbb{X}$ be a tubular weighted projective line, for some special direct summands of a tilting sheaf, we give the following definitions for convenience.

Definition 4.5**.**

Let $T=\bigoplus_{i=1}^{n}T_{i}$ be a tilting sheaf in $\operatorname{\mathsf{coh}}\mathbb{X}$ where $T_{i}\in\operatorname{\mathsf{ind}}\mathbb{X}$ for each $i$. Let $T_{k}$, $T_{l}$ are two indecomposable direct summands of $T$.

• (1)

If $\mu T_{k}\leq\mu T_{i}$ for any $1\leq i\leq n$, we say that $T_{k}$ is a minimal direct summand of $T$. Moreover, if $\mu T_{k}<\mu T_{i}$ for any $i\neq k$, we say that $T_{k}$ is the only minimal direct summand of $T$.

• (2)

If $\mu T_{l}\geq\mu T_{i}$ for any $1\leq i\leq n$, we say that $T_{l}$ is a maximal direct summand of $T$. Moreover, if $\mu T_{l}>\mu T_{i}$ for any $i\neq l$, we say that $T_{l}$ is the only maximal direct summand of $T$.

• (3)

Let $T_{k}$ be a minimal direct summand of $T$, $T_{l}$ be a maximal direct summand of $T$, we call the interval $[\mu T_{k},\mu T_{l}]$ the slope range of $T$.

Note that the first object of a tilting sheaf may not be a minimal direct summand, and a minimal direct summand of a tilting sheaf also may not be a first object. But the only minimal direct summand of $T$ must be a first object.

Lemma 4.6**.**

Let $T=X\oplus\overline{T}$ be a tilting sheaf in $\operatorname{\mathsf{coh}}\mathbb{X}$ where $X\in\operatorname{\mathsf{ind}}\mathbb{X}$.

• (1)

If $X$ is the only maximal direct summand of $T$, then $T^{\prime}=\tau X[-1]\oplus\overline{T}$ is a tilting complex in $\mathcal{D}^{b}(\mathbb{X})$.

• (2)

If $X$ is the only minimal direct summand of $T$, then $\tau^{-1}X[1]\oplus\overline{T}$ is a tilting complex in $\mathcal{D}^{b}(\mathbb{X})$.

Proof.

We only prove $(1)$ here, and $(2)$ could be deduced similarly.

For $(1)$, since $X$ is the only maximal direct summand of $T$, $\mu(\tau X)=\mu X>\mu T_{i}$ for any direct summand $T_{i}$ of $\overline{T}$, then

[TABLE]

Then using $T=X\oplus\overline{T}$ be a tilting sheaf in $\operatorname{\mathsf{coh}}\mathbb{X}$, one can get that $\operatorname{\mathsf{Hom}}_{\mathcal{D}^{b}(\mathbb{X})}(T^{\prime},T^{\prime}[l])=0$ for any $l\neq 0$ easily. Since the number of non-isomorphic indecomposable direct summands of $T^{\prime}$ is $n$, we conclude that $T^{\prime}$ is a tilting complex in $\mathcal{D}^{b}(\mathbb{X})$ by [M2, Lemma 9.1.3]. ∎

4.3. The behavior of slopes under mutations

Let $\mathbb{X}$ be a tubular weighted projective line. In this subsection, we investigate how the slope changed when we make mutations at some special direct summands of a tilting sheaf.

Lemma 4.7**.**

Let $T=T_{k}\oplus\overline{T}$ be a tilting object in $\operatorname{\mathsf{coh}}\mathbb{X}$ with an indecomposable direct summand $T_{k}$. Let $\mu_{T_{k}}(T)=T_{k}^{*}\oplus\overline{T}$ be the mutation of $T$ at $T_{k}$. If there is another direct summand of $T$, denoted by $Z$, such that $T_{k}\in\mathcal{W}_{Z}$, then $T_{k}^{*}\in\mathcal{W}_{Z}$. Otherwise, $T_{k}^{*}$ and $T_{k}$ have different slopes.

Proof.

Assume that there is a direct summand $Z$ of $T$ such that $T_{k}\in\mathcal{W}_{Z}$. Assume moreover that $q.l.Z=l$. By Lemma 3.5, the summands of $T$ lying in $\mathcal{W}_{Z}$ can be viewed as a tilting module in $\operatorname{\mathsf{mod}}\nolimits A_{l}$, where $A_{l}$ is the path algebra of a quiver of Dynkin type $\mathrm{A}_{l}$ with linear orientation. Note that, as $Z$ is both a projective and injective object in $\mathcal{W}_{Z}$, the tilting modules in $\mathcal{W}_{Z}$ always have $Z$ as a direct summand. Since $Z$ is a sincere $A_{l}$-module, every almost tilting module in $\mathcal{W}_{Z}$ containing $Z$ as a direct summand has two complements in $\mathcal{W}_{Z}$. Hence if $T_{k}\in\mathcal{W}_{Z}$ and $Z\ncong T_{k}$, we can deduce that $T_{k}^{*}\in\mathcal{W}_{Z}$ by Lemma 3.5.

Now assume that there is no such direct summand. By Lemma 2.4, there is an exact sequence $0\rightarrow T_{k}\xrightarrow{f}B\rightarrow T_{k}^{\prime}\rightarrow 0$ or $0\rightarrow T_{k}^{*}\rightarrow B^{\prime}\xrightarrow{g}T_{k}\rightarrow 0$ in $\operatorname{\mathsf{coh}}\mathbb{X}$, where $f$ is a minimal left $\operatorname{\mathsf{add}}\nolimits\overline{T}$-approximation and $g$ is a minimal right $\operatorname{\mathsf{add}}\nolimits\overline{T}$-approximation. Suppose $\mu T_{k}^{*}=\mu T_{k}$. Then each direct summand of $B$ and $B^{\prime}$ must have the same slope as $T_{k}$. If $T_{k}^{*}$ and $T_{k}$ belong to different tubes, by $\operatorname{\mathsf{Hom}}_{\mathbb{X}}(\mathcal{T}_{T_{k}},\mathcal{T}_{T_{k}^{*}})=0=\operatorname{\mathsf{Hom}}_{\mathbb{X}}(\mathcal{T}_{T_{k}^{*}},\mathcal{T}_{T_{k}})$, neither of the exact sequences $0\rightarrow T_{k}\xrightarrow{f}B\rightarrow T_{k}^{\prime}\rightarrow 0$ or $0\rightarrow T_{k}^{*}\rightarrow B^{\prime}\xrightarrow{g}T_{k}\rightarrow 0$ will hold. So $T_{k}^{*}$ lies in the same tube as $T_{k}$, then there must be one direct summand of $B$ or $B^{\prime}$, denoted by $Z$, such that $T_{k}$ has a monomorphism to $Z$ or $Z$ has an epimorphism to $T_{k}$. In any case, $T_{k}\in\mathcal{W}_{Z}$, which contradicts with the assumption. Therefore, $T_{k}$ and $T_{k}^{*}$ have different slopes. ∎

Theorem 4.8**.**

Let $T=\bigoplus_{i=1}^{n}T_{i}$ be a tilting sheaf in $\operatorname{\mathsf{coh}}\mathbb{X}$ with $\mu T_{i}\leq\mu T_{j}$ for $1\leq i<j\leq n$. For $1\leq k\leq n$, let $\mu_{T_{k}}(T)=T_{k}^{*}\oplus(\bigoplus_{i\neq k}T_{i})$ be the mutation of $T$ at $T_{k}$. Then

• (1)

$\mu T_{1}\leq\mu T_{1}^{*}\leq\mu T_{n}$* and $\mu T_{1}\leq\mu T_{n}^{*}\leq\mu T_{n}$.*

• (2)

If $T_{k}$ is a first object of $T$, then $\mu T_{k}\leq\mu T_{k}^{*}\leq\mu T_{n}$.

• (3)

If $T_{k}$ is a last object of $T$, then $\mu T_{k}\geq\mu T_{k}^{*}\geq\mu T_{1}$.

Proof.

For $(1)$, if $\mu T_{1}^{*}<\mu T_{1}$, by Lemma 2.4, there is an exact sequence

[TABLE]

in $\operatorname{\mathsf{coh}}\mathbb{X}$, where $g$ is a minimal right $\operatorname{\mathsf{add}}\nolimits(\bigoplus_{i=2}^{n}T_{i})$-approximation. Since $\mu T_{1}\leq\mu T_{j}$ for $1\leq j\leq n$, all direct summands of $B$ must belong to the same tube with $T_{1}$. Then there is a direct summand $Z$ of $B$ such that $Z$ has an epimorphism to $T_{1}$, then $T_{1}\in\mathcal{W}_{Z}$. By Lemma 4.7, $T_{1}^{*}\in\mathcal{W}_{Z}$. This contradicts the assumption that $\mu T_{1}^{*}<\mu T_{1}$. Hence $\mu T_{1}^{*}\geq\mu T_{1}$.

If $\mu T_{1}^{*}>\mu T_{n}$, then $T_{1}^{*}$ is the only maximal direct summand of the tilting sheaf $T_{1}^{*}\oplus(\bigoplus_{i=2}^{n}T_{i})$. By Lemma 4.6, we know $\tau T_{1}^{*}[-1]\oplus(\bigoplus_{i=2}^{n}T_{i})$ is a tilting complex in $\mathcal{D}^{b}(\mathbb{X})$. By Theorem 3.10, we can choose an automorphism $F$ of $\mathcal{D}^{b}(\mathbb{X})$ such that $F(T_{n})\in\mathcal{C}_{\infty}$. Hence, $F(T)$ and $F(\tau T_{1}^{*}[-1]\oplus(\bigoplus_{i=2}^{n}T_{i}))$ are two tilting objects in $\mathcal{D}^{b}(\mathbb{X})$. Since for any $1\leq i<j\leq n$, $\mu T_{i}\leq\mu T_{j}\leq\mu T_{n}<\mu T_{1}^{*}=\mu(\tau T_{1}^{*})$, by Lemma 3.15, we have $F(\tau T_{1}^{*})\in\operatorname{\mathsf{coh}}\mathbb{X}[1]$, i.e. $F(\tau T_{1}^{*}[-1])\in\operatorname{\mathsf{coh}}\mathbb{X}$ and $F(T_{i})\in\operatorname{\mathsf{coh}}\mathbb{X}$ and

[TABLE]

Now, we have $F(\tau T_{1}^{*}[-1]\oplus(\bigoplus_{i=2}^{n}T_{i}))$ and $F(T)$ are two tilting objects in $\operatorname{\mathsf{coh}}\mathbb{X}$. Moreover, by $F(T_{1})$ and $F(\tau T_{1}^{*}[-1])$ are two complements of the almost tilting object $F(\bigoplus_{i=2}^{n}T_{i})$. By $\mu(F(T_{i}))\leq\mu(F(T_{j}))$ for any $1\leq i<j\leq n$, we have $\mu(F(\tau T_{1}^{*}[-1]))\geq\mu(F(T_{1}))$ in $\operatorname{\mathsf{coh}}\mathbb{X}$, which is in a contradiction with $\mu(F(\tau T_{1}^{*}[-1]))<\mu(F(T_{1}))$. Therefore, we must have $\mu T_{1}^{*}\leq\mu T_{n}$. Using similar proof, we can get the results for $\mu T_{n}^{*}$.

For $(2)$, by Lemma 2.4, there is an exact sequence $0\rightarrow T_{k}\xrightarrow{f}B\rightarrow T_{k}^{*}\rightarrow 0$ or $0\rightarrow T_{k}^{*}\rightarrow B^{\prime}\xrightarrow{g}T_{k}\rightarrow 0$ in $\operatorname{\mathsf{coh}}\mathbb{X}$, where $f$ is a minimal left $\operatorname{\mathsf{add}}\nolimits(\bigoplus_{i\neq k}T_{i})$-approximation and $g$ is a minimal right $\operatorname{\mathsf{add}}\nolimits(\bigoplus_{i\neq k}T_{i})$-approximation. Since $T_{k}$ is the first object of $T$, then only the exact sequnece $0\rightarrow T_{k}\rightarrow B\xrightarrow{g}T_{k}^{*}\rightarrow 0$ can happen. So $\mu T_{k}\leq\mu T_{k}^{*}$.

If $\mu T_{k}^{*}>\mu T_{n}$, similar to the case $(1)$, one can get a contradiction. Hence, $\mu T_{k}^{*}\leq\mu T_{n}$,

The proof of $(3)$ is similar to $(2)$ and we omit it here. ∎

4.4. Bundle-mutations

Let $\mathbb{X}$ be a tubular weighted projective line. Let $T$ and $T^{\prime}$ be two tilting objects in $\operatorname{\mathsf{coh}}\mathbb{X}$ such that $T^{\prime}$ is a mutation of $T$, if $T,T^{\prime}$ are both tilting bundles, we call such mutation a bundle-mutation.

Let $T$ and $T^{\prime}$ be tilting bundles in $\operatorname{\mathsf{coh}}\mathbb{X}$, we say that $T$ can be transformed into $T^{\prime}$ by bundle-mutations if there is a sequence of tilting bundles $M_{0}=T,M_{1},\cdots,M_{l-1},M_{l}=T^{\prime}$ such that $M_{i}$ is a mutation of $M_{i-1}$ for any $1\leq i\leq l$. Note that if $T$ can be transformed into $T^{\prime}$ by bundle-mutations, then $T^{\prime}$ also can be transformed into $T$ by bundle-mutations.

As a direct consequence of Theorem 4.8, we have

Lemma 4.9**.**

Both the $\operatorname{\mathsf{APR}}$ mutation and co-$\operatorname{\mathsf{APR}}$ mutation on a tilting bundle in $\operatorname{\mathsf{coh}}\mathbb{X}$ are bundle-mutations.

Then combining Lemma 3.20, Lemma 3.21 and Lemma 4.4, one can get

Lemma 4.10**.**

Let $X$ be quasi-simple rigid vector bundle in $\operatorname{\mathsf{coh}}\mathbb{X}$. Then the $\operatorname{\mathsf{APR}}$ mutation and co-$\operatorname{\mathsf{APR}}$ mutation in $X^{\perp}$ on a preprojective tilting module induce bundle-mutations in $\operatorname{\mathsf{coh}}\mathbb{X}$, while the co-$\operatorname{\mathsf{APR}}$ mutation in ${}^{\perp}X$ on a preinjective tilting module which is a vector bundle in $\operatorname{\mathsf{coh}}\mathbb{X}$ is also a bundle-mutation in $\operatorname{\mathsf{coh}}\mathbb{X}$.

5. Connectedness of tilting bundles

Let $\mathbb{X}$ be a tubular weighted projective line with weight sequence $(p_{1},\cdots,p_{t})$, up to permutation, we assume that $p_{i}\leq p_{j}$ for $i<j$. Set $p:=\operatorname{\mathsf{lcm}}(p_{1},\cdots,p_{t})$. Recall that the tubular weight types are, $(2,2,2,2),(3,3,3),(2,4,4)$ and $(2,3,6)$. It follows that $p=p_{t}$, $\delta(\vec{x}_{t})=p/p_{t}=1$ and $\mu({\mathcal{O}}(m\vec{x}_{t}))=\delta(m\vec{x}_{t})=m\delta(\vec{x}_{t})=m$ for any integer $m$. Denote by $T_{can}=\bigoplus_{0\leq\vec{x}\leq\vec{c}}{\mathcal{O}}(\vec{x})$ the canonical tilting bundle in $\operatorname{\mathsf{coh}}\mathbb{X}$.

5.1. Some tilting bundles

Lemma 5.1**.**

Let $X$ be a rigid vector bundle in $\operatorname{\mathsf{coh}}\mathbb{X}$, then there is a tilting bundle with $X$ as a direct summand.

Proof.

Since $X$ is a rigid object in $\operatorname{\mathsf{coh}}\mathbb{X}$, we can choose a tilting object $T$ in $\operatorname{\mathsf{coh}}\mathbb{X}$ with $X$ as a direct summand. If $T$ is a tilting bundle, we are done. Else suppose that $X_{1},\cdots,X_{s}(s\geq 1)$ are all the indecomposable direct summands of $T$ which lie in $\mathcal{C}_{\infty}$ and $q.l.X_{i}\leq q.l.X_{i+1}$ for $1\leq i\leq s-1$. Let $T^{\prime}=\mu_{X_{1}}\cdots\mu_{X_{s}}(T)$. By Lemma 4.7, $T^{\prime}$ has no direct summand lying in $\mathcal{C}_{\infty}$, so $T^{\prime}$ is a tilting bundle. Since $X$ is a vector bundle, $X\neq X_{i}$ for any $1\leq i\leq s$, $X$ is still a direct summand of $T^{\prime}$. Hence, $T^{\prime}$ is the tilting bundle we needed. ∎

Lemma 5.2**.**

Let $X$ be a quasi-simple rigid vector bundle in $\operatorname{\mathsf{coh}}\mathbb{X}$, then

• (a)

there is a tilting bundle such that $X$ is the only minimal direct summand;

• (b)

there is a tilting bundle such that $X$ is the only maximal direct summand;

• (c)

if the $\tau$-period of $X$ is $p$, then there is a tilting bundle $T$ such that $X$ is a direct summand of $T$ and $\mu X$ is neither minimal nor maximal among the slopes of all the direct summands of $T$.

Proof.

For $(a)$, note that ${}^{\perp}X$ is the module category of a tame hereditary algebra. By Lemma 3.20, it is easy to get that there are only finite indecomposable objects lying in ${}^{\perp}X\cap\mathcal{C}_{\infty}$. Hence, we can choose a preinjective tilting module $M_{1}$ in ${}^{\perp}X$ such that $M_{1}$ is a vector bundle in $\operatorname{\mathsf{coh}}\mathbb{X}$. Then by Lemma 3.20 and Lemma 3.21, $X\oplus M_{1}$ is a tilting bundle in $\operatorname{\mathsf{coh}}\mathbb{X}$ such that $X$ is the only minimal direct summand of this tilting bundle.

For $(b)$, consider $X^{\perp}$, then the proof is similar to $(a)$.

For $(c)$, we will discuss it in the following two cases.

Case 1: $X$ is a line bundle ${\mathcal{O}}(\vec{x})$ for some $\vec{x}\in L(\bf p)$. Let $T=T_{can}(\vec{x}-\vec{x}_{t})$, then $T$ is a tilting bundle, ${\mathcal{O}}(\vec{x}-\vec{x}_{t}),{\mathcal{O}}(\vec{x}+\vec{x}_{t})$ are direct summands of $T$ and $\mu({\mathcal{O}}(\vec{x}-\vec{x}_{t}))<\mu({\mathcal{O}}(\vec{x}))<\mu({\mathcal{O}}(\vec{c}+\vec{x}+\vec{x}_{t}))$, hence $T_{can}(\vec{x}-\vec{x}_{t})$ is the tilting bundle we needed.

Case 2: $X$ is not a line bundle. Since each line bundle in $\operatorname{\mathsf{coh}}\mathbb{X}$ has integer slope, we can suppose that $\mu X=m+\frac{a}{b}$ for some integers $m,a,b$ such that $(a,b)=1$ and $0<a<b$. Then there are positive integers $c,d$ such that $bc-ad=1,0<c\leq d<b,c\leq a$. Let $\varphi(x)=\frac{(c+dm)x+(a+bm)}{dx+b}$. By Lemma 3.12, there is an automorphism of $\mathcal{D}^{b}({\mathbb{X}})$, denoted by $\overline{\varphi}$, such that $\overline{\varphi}$ coincides with the action of $\varphi$ on slopes. Since $X$ is a quasi-simple object in $\operatorname{\mathsf{coh}}\mathbb{X}$ with $\tau$-period $p$, we may assume that $\overline{\varphi}({\mathcal{O}})=X$ by Lemma 3.14. Since $T_{can}(-\vec{x}_{t})$ is a tilting object in $\operatorname{\mathsf{coh}}\mathbb{X}$, $\overline{\varphi}(T_{can}(-\vec{x}_{t}))$ is a tilting object in $\mathcal{D}^{b}(\mathbb{X})$. By

[TABLE]

and ${\mathcal{O}}(-\vec{x}_{t})$ is a minimal direct summand of $T_{can}(-\vec{x}_{t})$, by Lemma 3.15, all direct summands of $\overline{\varphi}(T_{can}(-\vec{x}_{t}))$ have slopes smaller than $\varphi(\infty)=m+\frac{c}{d}$, hence $\overline{\varphi}(T_{can}(-\vec{x}_{t}))$ is a tilting object in $\operatorname{\mathsf{coh}}\mathbb{X}$, moreover, a tilting bundle in $\operatorname{\mathsf{coh}}\mathbb{X}$. As ${\mathcal{O}}$ is a direct summand of $T_{can}(-\vec{x}_{t})$, $X=\overline{\varphi}({\mathcal{O}})$ is a direct summand of $\overline{\varphi}(T_{can}(-\vec{x}_{t}))$. Hence, the tilting bundle $\overline{\varphi}(T_{can}(-\vec{x}_{t}))$ is what we wanted. ∎

5.2. Connectedness in some special cases

Lemma 5.3**.**

Let $T,T^{\prime}$ be two tilting bundles in $\operatorname{\mathsf{coh}}\mathbb{X}$ such that $T$ and $T^{\prime}$ have a common quasi-simple direct summand $X$. Then $T$ can be transformed into $T^{\prime}$ by bundle-mutations if $X$ satisfies one of the following two conditions:

• (1)

$X$* is the only minimal direct summand of $T$ and $T^{\prime}$.*

• (2)

$X$* is the only maximal direct summand of $T$ and $T^{\prime}$.*

Proof.

We give a proof for the condition $(1)$ and the proof for the condition $(2)$ is similar.

Denote by $T=X\oplus\overline{T}$ and $T^{\prime}=X\oplus\overline{T^{\prime}}$. According to Lemma 3.20, both $\overline{T}$ and $\overline{T^{\prime}}$ lie on the preinjective component of ${}^{\perp}X$. According to Lemma 4.3, by co-$\operatorname{\mathsf{APR}}$ mutations, $T$ can be transformed into a slice $\Sigma$, $T^{\prime}$ also can be transformed into a slice $\Sigma^{\prime}$. Choose a slice $\Sigma^{\prime\prime}$ on the preinjective component such that $\operatorname{\mathsf{Fac}}\Sigma^{\prime}\subset\operatorname{\mathsf{Fac}}\Sigma^{\prime\prime}$ and $\operatorname{\mathsf{Fac}}\Sigma\subset\operatorname{\mathsf{Fac}}\Sigma^{\prime\prime}$. Then by co-$\operatorname{\mathsf{APR}}$ mutations, $\Sigma$ and $\Sigma^{\prime}$ can be transformed into $\Sigma^{\prime\prime}$. By Lemma 4.10, all the mutation appeared above are bundle-mutations, hence, $T$ and $T^{\prime}$ can be transformed into each other by bundle-mutations. ∎

Lemma 5.4**.**

Let $T$ and $T^{\prime}$ be two tilting bundles in $\operatorname{\mathsf{coh}}\mathbb{X}$. Let $X$ and $X^{\prime}$ be two quasi-simple objects with same slope but belong to different tubes in $\operatorname{\mathsf{coh}}\mathbb{X}$. If $X$ and $X^{\prime}$ satisfy one of the following two conditions:

• (1)

$X$* is the only minimal direct summand of $T$ and $X^{\prime}$ is the only minimal direct summand of $T^{\prime}$,*

• (2)

$X$* is the only maximal direct summand of $T$ and $X^{\prime}$ is the only maximal direct summand of $T^{\prime}$;*

then $T$ can be transformed into $T^{\prime}$ by bundle-mutations.

Proof.

We give a proof of for the condition $(1)$ and omit the proof for the condition $(2)$.

Denote by $\mathcal{H}^{\prime}=^{\perp}X$. Note that, as $\mu X=\mu X^{\prime}$ and $X,X^{\prime}$ belong to different tubes, we have $\operatorname{\mathsf{Ext}}^{1}_{\mathbb{X}}(X^{\prime},X)=\operatorname{\mathsf{Hom}}_{\mathbb{X}}(X^{\prime},X)=0$. Consequently, $X^{\prime}\in\mathcal{H}^{\prime}$ is a quasi-simple regular module in $\mathcal{H}^{\prime}$ by Lemma 3.20. It is known that the left perpendicular category ${}^{\perp_{\mathcal{H}^{\prime}}}X^{\prime}$ of $X^{\prime}$ in $\mathcal{H}^{\prime}$ is also a module category of a tame hereditary algebra (cf. [SS2]). Since in $\operatorname{\mathsf{coh}}\mathbb{X}$, there are only finite indecomposable objects lying in $\mathcal{C}_{\infty}\cap^{\perp}X$, we can choose a tilting module $T_{v}$ in the preinjective component of ${}^{\perp_{\mathcal{H}^{\prime}}}X^{\prime}$ such that $T_{v}$ is a vector bundle in $\operatorname{\mathsf{coh}}\mathbb{X}$. It is easy to get that $T^{\prime\prime}=X\oplus X^{\prime}\oplus T_{v}$ is a tilting bundle in $\operatorname{\mathsf{coh}}\mathbb{X}$ such that $X$ and $X^{\prime}$ are minimal direct summands of $T^{\prime\prime}$. Moreover, $T^{\prime\prime}$ admits no mininal direct summands other than $X$ and $X^{\prime}$. Let $M_{1}=\mu_{X^{\prime}}(T^{\prime\prime})$, $M_{2}=\mu_{X}(T^{\prime\prime})$. By Theorem 4.8 and Lemma 4.7, $M_{1}$ and $M_{2}$ are also tilting bundles, and $X$ is the only minimal direct summand of $M_{1}$, $X^{\prime}$ is the only minimal direct summand of $M_{2}$. By Lemma 5.3, $T$ can be obtained from $M_{1}$ by bundle-mutations, $T^{\prime}$ can be obtained from $M_{2}$ by bundle-mutations. Hence $T$ can be obtained from $T^{\prime}$ by bundle-mutations. ∎

Lemma 5.5**.**

Let $T$ be a tilting bundle in $\operatorname{\mathsf{coh}}\mathbb{X}$ with a quasi-simple direct summand $X$, then

• (1)

$T$* can be transformed into a tilting bundle such that $X$ is the only minimal direct summand of the new tilting bundle by bundle-mutations.*

• (2)

$T$* can be transformed into a tilting bundle such that $X$ is the only maximal direct summand of the new tilting bundle by bundle-mutations.*

Proof.

Let $T_{min}$ be a minimal direct summand of $T$, $T_{max}$ be a maximal direct summand of $T$. Note that $\mu(T_{min})\neq\mu(T_{max})$.

Firstly, we claim that if $\mu X<\mu(T_{max})$, then $(1)$ holds and if $\mu X>\mu(T_{min})$, then $(2)$ holds.

Now suppose that $\mu X<\mu(T_{max})$. Consider the right perpendicular category $T_{max}^{\perp}$ of $T_{max}$ in $\operatorname{\mathsf{coh}}\mathbb{X}$. Denote by $\mathcal{H}^{\prime}=T_{max}^{\perp}$. $\mathcal{H}^{\prime}$ is the module category of a tame hereditary algebra. Since $\mu X<\mu(T_{max})$, by Lemma 3.20, $X$ lies in the preprojective component of $\mathcal{H}^{\prime}$. Then it is easy to get that the right perpendicular category $X^{\perp_{\mathcal{H}^{\prime}}}$ of $X$ in $\mathcal{H}^{\prime}$ is a hereditary category of finite type. So the set $\{N\in\operatorname{\mathsf{ind}}\mathbb{X}|\mu N<\mu X,\operatorname{\mathsf{Ext}}_{\mathbb{X}}^{1}(X,N)=0=\operatorname{\mathsf{Ext}}^{1}_{\mathbb{X}}(T_{max},N)\}\subset X^{\perp_{\mathcal{H}^{\prime}}}$ is finite. Hence by $\operatorname{\mathsf{APR}}$ mutations, $T$ can be transformed into a new tilting bundle $L$ such that $X$ is a minimal direct summand of $L$. If $X$ is the only minimal direct summand, then we are done. Else suppose that $\{L_{i}|1\leq i\leq s\}$ are all minimal direct summands of $L$. Assume moreover that $q.l.L_{i}\leq q.l.L_{i+1}$ for $1\leq i\leq s-1$. Since $X$ is quasi-simple, we may assume that $L_{1}=X$. Then let $T^{\prime}=\mu_{L_{2}}\cdots\mu_{L_{s}}(L)$, by Theorem 4.8 and Lemma 4.7, $X$ is the only minimal direct summand of $T^{\prime}$. By Lemma 4.10 and Theorem 4.8, all the mutations appeared above are bundle-mutations. Hence, $T$ can be transformed into a tilting bundle $T^{\prime}$ such that $X$ is the only minimal direct summand of $T^{\prime}$ by bundle-mutations.

For the case $\mu X>\mu(T_{min})$, we may apply a similar proof and we omit it here.

Now we separate the remaining proof into three cases.

Case 1: $\mu(T_{min})<\mu X<\mu(T_{max})$. It follows from the claim directly.

Case 2: $\mu X=\mu(T_{min})$. Note that, as $\mu(T_{min})\neq\mu(T_{max})$, we have $\mu X<\mu(T_{max})$. It follows from the above claim that $T$ can be transformed into a tilting bundle $M$ such that $X$ is the only minimal direct summand of $M$ by bundle-mutations. In order to prove $(2)$, it suffices to prove that $M$ can be transformed into a tilting bundle such that $X$ is the only maximal direct summand of the new tilting bundle by bundle-mutations. We will discuss it in the following two cases.

Subcase 2.1: the $\tau$-period of $X$ is $p$. By Lemma 5.2, there is a tilting bundle $T^{\prime\prime}$ such that $X$ is a direct summand of $T^{\prime\prime}$ with $\mu X$ is neither minimal nor maximal among the slopes of all the direct summands of $T^{\prime\prime}$. By Case 1, $T^{\prime\prime}$ can be transformed into a tilting bundle $M^{\prime}$ such that $X$ is the only minimal direct summand of $M^{\prime}$ by bundle-mutations and also can be transformed into a tilting bundle $M^{\prime\prime}$ such that $X$ is the only maximal direct summand of $M^{\prime\prime}$ by bundle-mutations. Now $X$ is the only minimal direct summand of $M$ and $M^{\prime}$, by Lemma 5.3, $M$ can be transformed into $M^{\prime}$ by bundle-mutations. In sum, by bundle-mutations, $M$ can be transformed into $M^{\prime\prime}$ such that $X$ is the only maximal direct summand of $M^{\prime\prime}$.

Subcase 2.2: the $\tau$-period of $X$ is not $p$. Choose a quasi-simple rigid object $X^{\prime}\in\mathcal{C}_{\mu X}$ such that the $\tau$-period of $X^{\prime}$ is $p$. Then we have $X$ and $X^{\prime}$ belong to different tubes. Let $L$ be a tilting bundle such that $X^{\prime}$ is the only minimal direct summand of $L$. By Lemma 5.4, we know that $L$ can be obtained from $M$ by bundle-mutations. By Subcase 2.1, $L$ can be transformed into a tilting bundle $L^{\prime}$ such that $X^{\prime}$ is the only maximal direct summand of $L^{\prime}$ by bundle-mutations. Choose a tilting bundle $T^{\prime}$ such that $X$ is the only maximal direct summand of $T^{\prime}$, then again by Lemma 5.4, we know that $T^{\prime}$ can be obtained from $L^{\prime}$ by bundle-mutations. In sum, by bundle-mutations, $M$ can be transformed into $T^{\prime}$ such that $X$ is the only maximal direct summand of $T^{\prime}$.

Case 3: $\mu X=\mu(T_{max})$. The proof is similar to the Case 2 and we omit it here. ∎

The following result is a direct consequence of Lemma 5.5 and Lemma 5.3.

Lemma 5.6**.**

Let $T$ and $T^{\prime}$ be two tilting bundles in $\operatorname{\mathsf{coh}}\mathbb{X}$ such that $T$ and $T^{\prime}$ contain a common quasi-simple direct summands $X$, then $T$ can be obtained from $T^{\prime}$ by bundle-mutations.

Lemma 5.7**.**

Let $T$ and $T^{\prime}$ be two tilting bundles in $\operatorname{\mathsf{coh}}\mathbb{X}$, $X$ be a quasi-simple direct summand of $T$ and $X^{\prime}$ be a quasi-simple direct summand of $T^{\prime}$. If $X\oplus X^{\prime}$ is rigid, then $T$ can be obtained from $T^{\prime}$ by bundle-mutations.

Proof.

Since $X\oplus X^{\prime}$ is rigid, by Lemma 5.1, we can choose a tilting bundle $T^{\prime\prime}$ containing $X\oplus X^{\prime}$ as direct summands. Then $X$ is a common direct summand of $T$ and $T^{\prime\prime}$, while $X^{\prime}$ is a common direct summand of $T^{\prime}$ and $T^{\prime\prime}$. By Lemma 5.6, $T$ can be transformed into $T^{\prime}$ by bundle-mutations. ∎

5.3. Connectedness in general case

In this subsection, we will prove that any tilting bundle can be obtained from the canonical tilting bundle $T_{can}$ by bundle-mutations.

Lemma 5.8**.**

Let $T=T_{can}(\vec{x})$ be a tilting bundle in $\operatorname{\mathsf{coh}}\mathbb{X}$, then $T$ can be obtained from $T_{can}$ by bundle-mutations.

Proof.

If $\vec{x}=\vec{x}_{i}$ for some $1\leq i\leq t$, then ${\mathcal{O}}(\vec{x}_{i})$ is a common quasi-simple direct summand of $T_{can}$ and $T_{can}(\vec{x}_{i})$. By Lemma 5.6, $T$ can be obtained from $T_{can}$ by bundle-mutations. For the general case, one obtains the result by induction on the degree of $\vec{x}$.

∎

Lemma 5.9**.**

Let $T$ be a tilting bundle in $\operatorname{\mathsf{coh}}\mathbb{X}$ such that one direct summand of $T$ is a line bundle, then $T$ can be obtained from $T_{can}$ by bundle-mutations.

Proof.

Without loss of generality, we may assume that ${\mathcal{O}}(\vec{x})$ is a direct summand of $T$. Then ${\mathcal{O}}(\vec{x})$ is a common quasi-simple direct summand of $T$ and $T_{can}(\vec{x})$. By Lemma 5.6 and Lemma 5.8, $T$ can be obtained from $T_{can}$ by bundle-mutations. ∎

Lemma 5.10**.**

Let $T$ be a tilting bundle in $\operatorname{\mathsf{coh}}\mathbb{X}$ such that the slope range of $T$ contained an integer, then $T$ can be obtained from $T_{can}$ by bundle-mutations.

Proof.

Denote this integer by $m$. Let $L$ be the line bundle ${\mathcal{O}}(m\vec{x}_{t})$, then $\mu L=\mu({\mathcal{O}}(m\vec{x}_{t}))=m$. By Lemma 3.8, we have either $\operatorname{\mathsf{Ext}}^{1}_{\mathbb{X}}(T,L)=0$ or $\operatorname{\mathsf{Hom}}_{\mathbb{X}}(T,L)=0$.

Case 1: $\operatorname{\mathsf{Ext}}^{1}_{\mathbb{X}}(T,L)=0$. By Lemma 3.5, we can choose a quasi-simple direct summand of $T$, denoted by $X$, such that $\mu X\geq m$. If $X$ is a line bundle, by Lemma 5.9, $T$ can be obtained from $T_{can}$ by bundle-mutations. If $X$ is not a line bundle, then $X$ and $L$ belong to different tubes since all quasi-simple objects in the tube which $L$ lies in are line bundles. Then by $\mu L\leq\mu X$, we must have $\operatorname{\mathsf{Ext}}^{1}_{\mathbb{X}}(L,X)=0$. Hence, $X\oplus L$ is rigid. Let $T^{\prime}$ be a tilting bundle containing $L$ as a direct summand. By Lemma 5.7, $T$ can be obtained from $T^{\prime}$ by bundle-mutations. We conclude that $T$ can be obtained from $T_{can}$ by bundle-mutations by Lemma 5.9.

Case 2: $\operatorname{\mathsf{Hom}}_{\mathbb{X}}(T,L)=0$. Then $\operatorname{\mathsf{Ext}}_{\mathbb{X}}^{1}(\tau^{-1}L,T)\cong\mathbb{D}\operatorname{\mathsf{Hom}}_{\mathbb{X}}(T,L)=0$. Notice that $\tau^{-1}L$ is a line bundle and $\mu(\tau^{-1}L)=\mu L=m$. Similar to the Case 1, we can obtain the desired result. ∎

Lemma 5.11**.**

Let $X$ be an indecomposable quasi-simple rigid object in $\operatorname{\mathsf{coh}}\mathbb{X}$. Suppose the $\tau$-period of $X$ is $p$ and $\mu X=m+\frac{a}{b}$ for some integers $m,a,b$, where $(a,b)=1,0<a<b$, then there are positive integers $c,d$ and a quasi-simple rigid object $Y$ in $\operatorname{\mathsf{coh}}\mathbb{X}$ with $\tau$-period $p$ such that $bc-ad=1,0<c\leq d<b,c\leq a$, $\mu Y=m+\frac{c}{d}$ and $X\oplus Y$ is rigid in $\operatorname{\mathsf{coh}}\mathbb{X}$.

Proof.

Since $(a,b)=1,0<a<b$, by Lemma 3.16, there are positive integers $c,d$ such that $bc-ad=1,0<c\leq d<b,c\leq a$. Let $\varphi(x)=\frac{(c+dm)x+(a+bm)}{dx+b}$, since \left(\begin{array}[]{cc}c+dm&a+bm\\ d&b\end{array}\right)\in\operatorname{\mathsf{SL}}(2,\mathbb{Z}), by Lemma 3.12, there is an automorphism of $\mathcal{D}^{b}({\mathbb{X}})$, denoted by $\overline{\varphi}$, such that $\overline{\varphi}$ coincides with the action of ${\varphi}$ on slopes. Note that $\varphi(0)=m+\frac{a}{b}$. Since $X$ is a quasi-simple rigid object in $\operatorname{\mathsf{coh}}\mathbb{X}$ with $\tau$-period $p$ and $\mu X=m+\frac{a}{b}$, by Lemma 3.14, we may assume that $\overline{\varphi}({\mathcal{O}})=X$. It is clear that there is a simple object $S$ in $\operatorname{\mathsf{coh}}\mathbb{X}$ such that the $\tau$-period of $S$ is $p$ and ${\mathcal{O}}\oplus S$ is rigid. Let $Y=\overline{\varphi}(S)$, then we have

[TABLE]

By Lemma 3.15, we deduce that $Y\in\operatorname{\mathsf{coh}}\mathbb{X}$. Since ${\mathcal{O}}\oplus S$ is rigid, we conclude that $X\oplus Y$ is a rigid object in $\operatorname{\mathsf{coh}}\mathbb{X}$. ∎

Lemma 5.12**.**

Let $T$ be a tilting bundle such that the slope range of $T$ does not contain an integer, then by bundle-mutations, $T$ can be transformed into a tilting bundle $T^{\prime}$ such that the slope range of $T^{\prime}$ contains an integer.

Proof.

By Lemma 3.6, let $X_{0}$ be an indecomposable direct summand of $T$ such that the $\tau$-period of $X_{0}$ is $p$. By Lemma 3.5, we may assume that $X_{0}$ is quasi-simple. Suppose that $\mu(X_{0})=m+\frac{a_{0}}{b_{0}}$ for some integers $m,a_{0},b_{0}$, where $(a_{0},b_{0})=1,0<a_{0}<b_{0}$. By Lemma 5.11, there are positive integers $a_{1},b_{1}$ and a quasi-simple rigid object $X_{1}$ in $\operatorname{\mathsf{coh}}\mathbb{X}$ with $\tau$-period $p$ such that $b_{0}a_{1}-a_{0}b_{1}=1,0<a_{1}\leq b_{1}<b_{0},a_{1}\leq a_{0}$, $\mu X_{1}=m+\frac{a_{1}}{b_{1}}$ and $X_{0}\oplus X_{1}$ is rigid in $\operatorname{\mathsf{coh}}\mathbb{X}$. Since $0<a_{1}\leq b_{1}<b_{0}$, we can continue this process until $a_{k}=b_{k}$ for some positive integer $k$. Denote the quasi-simple rigid objects appeared in this process by $X_{i}(1\leq i\leq k)$, then we have $\mu(X_{i})=m+\frac{a_{i}}{b_{i}}$ and $X_{i-1}\oplus X_{i}$ is rigid in $\operatorname{\mathsf{coh}}\mathbb{X}$ for each $1\leq i\leq k$. For each $1\leq i\leq k$, we choose a tilting bundle $T_{i}$ such that $X_{i}$ is a direct summand of $T_{i}$. Denote by $T_{0}=T$. By Lemma 5.7, $T_{i-1}$ can be transformed into $T_{i}$ by bundle-mutations for each $1\leq i\leq k.$ Therefore, $T$ can be transformed into the tilting bundle $T_{k}$ by bundle-mutations where the direct summand $X_{k}$ of $T_{k}$ satisfies that $\mu(X_{k})=m+\frac{a_{k}}{b_{k}}=m+1$ is an integer. ∎

5.4. Connectedness of the subgraph

Denote by $\mathcal{T}_{\mathbb{X}}$ be the set of all basic tilting objects in $\operatorname{\mathsf{coh}}\mathbb{X}$. By [HU], $\mathcal{T}_{\mathbb{X}}$ admits a partial order $\leq$. The tilting graph ${\mathcal{G}}(\mathcal{T}_{\mathbb{X}})$ of $\operatorname{\mathsf{coh}}\mathbb{X}$ is the Hasse diagram of the poset $(\mathcal{T}_{\mathbb{X}},\leq)$. Equivalently, the tilting graph $\mathcal{G}(\mathcal{T}_{\mathbb{X}})$ of $\operatorname{\mathsf{coh}}\mathbb{X}$ has as vertices the isomorphism classes of basic tilting objects in $\operatorname{\mathsf{coh}}\mathbb{X}$, while two vertices $T$ and $T^{\prime}$ are connected by an edge if and only if they differ by precisely one indecomposable direct summand. Denote by $\mathcal{T}^{v}_{\mathbb{X}}$ the set of all basic tilting bundles in $\operatorname{\mathsf{coh}}\mathbb{X}$, ${\mathcal{G}}(\mathcal{T}^{v}_{\mathbb{X}})$ the subgraph of the tilting graph ${\mathcal{G}}(\mathcal{T}_{\mathbb{X}})$ consisting of all basic tilting bundles in $\operatorname{\mathsf{coh}}\mathbb{X}$.

Theorem 5.13**.**

Let $\mathbb{X}$ be a tubular weighted projective line. Then the subgraph ${\mathcal{G}}(\mathcal{T}^{v}_{\mathbb{X}})$ of the tilting graph is connected.

Proof.

Let $T$ be a basic tilting bundle in $\operatorname{\mathsf{coh}}\mathbb{X}$, it is enough to prove that $T$ can be obtained from $T_{can}$ by bundle-mutations. If the slope range of $T$ contained an integer, by Lemma 5.10, $T$ can be obtained from $T_{can}$ by bundle-mutations.

If the slope range of $T$ does not contain an integer, by Lemma 5.12, $T$ can be transformed into a tilting bundle $T^{\prime}$ by bundle-mutations such that the slope range of $T^{\prime}$ contains an integer. Then by Lemma 5.10, $T$ can be obtained from $T_{can}$ by bundle-mutations. Therefore, ${\mathcal{G}}(\mathcal{T}^{v}_{\mathbb{X}})$ is connected. ∎

Theorem 5.13 yields an alternative proof for the connectedness of the tilting graph of ${\mathcal{G}}({\mathcal{T}}_{\mathbb{X}})$.

Corollary 5.14**.**

[BKL]** Let $\mathbb{X}$ be a tubular weighted projective line. Then the tilting graph ${\mathcal{G}}(\mathcal{T}_{\mathbb{X}})$ of $\operatorname{\mathsf{coh}}\mathbb{X}$ is connected.

Proof.

By Theorem 5.13, it suffices to prove that a tilting object with direct summands of finite length can be transformed to a tilting bundle by mutations. Let $T$ be a tilting object in $\operatorname{\mathsf{coh}}\mathbb{X}$ with $X_{1},\cdots,X_{s}(s\geq 1)$ are all the indecomposable direct summands of $T$ which lie in $\mathcal{C}_{\infty}$. Suppose that $q.l.X_{i}\leq q.l.X_{i+1}$ for $1\leq i\leq s-1$. Let $T^{\prime}=\mu_{X_{1}}\cdots\mu_{X_{s}}(T)$. By Lemma 4.7, $T^{\prime}$ has no direct summand lying in $\mathcal{C}_{\infty}$, i.e. $T^{\prime}$ a tilting bundle. ∎

Acknowledgments

The author is grateful to Changjian Fu for his many helpful discussions and suggestions.