A note on abelian quotient categories

TL;DR

This paper explores conditions under which a subcategory of a triangulated category yields an abelian quotient, linking Serre functors and cluster tilting, and generalizes previous results by Beligiannis.

Contribution

It establishes a characterization of cluster tilting subcategories via abelian quotient categories and Serre functor conditions, extending prior work in the field.

Findings

X is cluster tilting iff C/X is abelian and S(X)=X[2]

Generalizes Beligiannis's results on abelian quotients

Provides new criteria for cluster tilting in triangulated categories

Abstract

Let C be a triangulated category with a Serre functor S and X a non-zero contravariantly finite rigid subcategory of C. Then X is cluster tilting if and only if the quotient category C/X is abelian and S(X)=X[2]. As an application, this result generalizes work by Beligiannis.

Equations

$$\mathcal{X}=\{M\in \mathcal{C}|{\mathrm{Hom}}_{\mathcal{C}}(\mathcal{X},M[1])=0\}=\{M\in \mathcal{C}|{\mathrm{Hom}}_{\mathcal{C}}(M,\mathcal{X}[1])=0\}.$$

$$\mathcal{X}=\{M\in \mathcal{C}|{\mathrm{Hom}}_{\mathcal{C}}(\mathcal{X},M[1])=0\}.$$

$$A\stackrel{u}{\to }B\stackrel{v}{\to }C\stackrel{w}{\to }A[1]$$

$${\mathrm{Hom}}_{\mathcal{C}}(A,B)\simeq D{\mathrm{Hom}}_{\mathcal{C}}(B,\mathbb{S}A)$$

$$\Omega C\stackrel{\alpha }{\to }X\stackrel{\beta }{\to }C\stackrel{\gamma }{\to }\Omega C[1]$$

$$\Omega C\stackrel{\overline{u}}{\to }A\stackrel{\overline{v}}{\to }B\stackrel{\overline{w}}{\to }C$$

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$$A\stackrel{f}{\to }B\stackrel{g}{\to }C\stackrel{h}{\to }A[1]$$

$${\text{mboxHom}}_{\overline{\mathcal{C}}}(C,M)\stackrel{\overline{g}}{\to }{\text{mboxHom}}_{\overline{\mathcal{C}}}(B,M)\stackrel{\overline{f}}{\to }{\text{mboxHom}}_{\overline{\mathcal{C}}}(A,M),$$

$$(u-tx)f=uf-ts=0.$$

$$\begin{array}[]{l}N\xrightarrow{~{}f~{}}X_{0}\xrightarrow{~{}g~{}}M\xrightarrow{~{}h~{}}N[1]\end{array}$$

$$\text{mbox}{Hom}_{\mathcal{C}}(\mathcal{X},X_0)\stackrel{\text{mbox}{Hom}_{\mathcal{C}}(\mathcal{X},g)}{\to }\text{mbox}{Hom}_{\mathcal{C}}(\mathcal{X},M)\to \text{mbox}{Hom}_{\mathcal{C}}(\mathcal{X},N[1])\to \text{mbox}{Hom}_{\mathcal{C}}(\mathcal{X},X_0[1])=0.$$

$$\text{mbox}{Hom}_{\mathcal{C}}(\mathcal{X},M[1])\simeq D\text{mbox}{Hom}_{\mathcal{C}}(M[1],\mathbb{S}\mathcal{X})\simeq D\text{mbox}{Hom}_{\mathcal{C}}(M,\mathbb{S}\mathcal{X}[-1])=D\text{mbox}{Hom}_{\mathcal{C}}(M,\mathcal{X}[1]),$$

$$\tau M\stackrel{f}{\to }C\stackrel{g}{\to }M\stackrel{h}{\to }\tau M[1]$$

$$\Omega M\stackrel{x}{\to }{X}_0\stackrel{y}{\to }M\stackrel{z}{\to }\Omega M[1]$$

$$\Omega M\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$\Omega M\stackrel{\overline{\varphi }}{\to }\tau M\stackrel{\overline{f}}{\to }C\stackrel{\overline{g}}{\to }M$$

$$\begin{array}[]{l}M\xrightarrow{~{}u=\tau^{-1}f~{}}\tau^{-1}C\xrightarrow{~{}v=\tau^{-1}g~{}}\tau^{-1}M\xrightarrow{~{}w=\tau^{-1}h~{}}M[1]\end{array}$$

$$\overline{\beta }=\overline{\delta }\circ \overline{v}=\overline{\delta }\circ \overline{\omega }\circ \overline{\beta }\text{and }\overline{v}=\overline{\omega }\circ \overline{\beta }=\overline{\omega }\circ \overline{\delta }\circ \overline{v}.$$

Full text

Key words: Triangulated categories; Cluster tilting subcategories; Abelian categories.

** 2010 Mathematics Subject Classification:** 18E30; 18E10.

1 Introduction

Let $\mathscr{C}$ be an additive category and $\mathscr{X}$ be a subcategory of $\mathscr{C}$ and let $M\in\mathscr{C}$. A morphism $f_{M}\colon X_{M}\to M$ is called a right $\mathscr{X}$-approximation of $M$ if $X_{M}\in\mathscr{X}$ and every morphism from an object in $\mathscr{X}$ to $M$ to factors through $f_{M}$. The subcategory $\mathscr{X}$ is said to be * contravariantly finite* in $\mathscr{C}$, if every object in $\mathscr{C}$ has a right $\mathscr{X}$-approximation. A left $\mathscr{X}$-approximation and a covariantly finite subcategory of $\mathscr{C}$ are dually defined. A contravariantly and covariantly finite subcategory is called functorially finite. For more details, see [AR].

Recall the notion of cluster tilting subcategories from [KZ, Definition 3.1].

Definition 1.1**.**

[KZ, Definition 3.1]* Let $\mathscr{C}$ be a triangulated category with a shift functor $[1]$.*

1. (1)

A subcategory $\mathscr{X}$ of $\mathscr{C}$ is called rigid if ${\rm Hom}_{\mathscr{C}}(\mathscr{X},\mathscr{X}[1])=0$. 2. (2)

A functorially finite subcategory $\mathscr{X}$ of $\mathscr{C}$ is called cluster tilting if

[TABLE]

Remark 1.2**.**

In fact, Koenig and Zhu [KZ, Lemma 3.2] indicate that a subcategory $\mathscr{X}$ of a triangulated categoru $\mathscr{C}$ is cluster tilting if and only if it is contravariantly finite in $\mathscr{C}$ and

[TABLE]

Koenig and Zhu gave a general framework from passing from triangulated categories to abelian categories by factoring out cluster tilting subcategories. More precisely, they proved the following.

Theorem 1.3**.**

[KZ, Theorem 3.3]* Let $\mathscr{C}$ be a triangulated category and $\mathscr{X}$ a cluster tilting subcategory of $\mathscr{C}$. Then the quotient category $\mathscr{C}/\mathscr{X}$ is an abelian category.*

A triangulated category is called connected if it cannot be decomposed into a direct sum of two non-zero triangulated subcategories. Beligiannis showed the following characterization of cluster tilting subcategories which complements, and was inspired by Koenig and Zhu.

Theorem 1.4**.**

[Be, Theorem 7.3]* Let $\mathscr{C}$ be a connected triangulated category with a Serre functor $\mathbb{S}$ and $\mathscr{X}$ a non-zero functorially finite rigid subcategory of $\mathscr{C}$. Then $\mathscr{X}$ is cluster tilting if and only if the quotient category $\mathscr{C}/\mathscr{X}$ is abelian and $\mathbb{S}(\mathscr{X})=\mathscr{X}[2]$.*

Our aim in this article is to give a simple short proof of the following more general result.

Theorem 1.5**.**

Let $\mathscr{C}$ be a triangulated category with a Serre functor $\mathbb{S}$ and $\mathscr{X}$ a non-zero contravariantly finite rigid subcategory of $\mathscr{C}$. Then $\mathscr{X}$ is cluster tilting if and only if the quotient category $\mathscr{C}/\mathscr{X}$ is abelian and $\mathbb{S}(\mathscr{X})=\mathscr{X}[2]$.

2 Preliminaries

We recall the definition of Auslander-Reiten triangle.

Definition 2.1**.**

[Ha, Definition 4.1]* Let $\mathscr{C}$ be a triangulated category. A triangle*

[TABLE]

in $\mathscr{C}$ is called an Auslander-Reiten triangle if

• •

$A,B$* are indecomposable;*

• •

$w\neq 0$**

• •

If $f\colon X\to C$ is not a split epimorphism, then there exists a morphism $f^{\prime}\colon X\to B$ such that $f^{\prime}v=f$.

In this case, we call $u$ is a source morphism and $v$ is a sink morphism. The notions of a source morphism and a sink morphism are also known as minimal left almost split morphism and minimal right split morphism, respectively.

Assume that $\mathscr{C}$ is a $k$-linear Hom-fnite triangulatedcategory where $k$ is a field. Recall from [BK] a Serre functor $\mathbb{S}\colon\mathscr{C}\to\mathscr{C}$ is a $k$-linear equivalence with bifunctorial isomorphisms

[TABLE]

for any $A,B\in\mathscr{C}$, where $D$ is the duality over $k$. Reiten and Van den Bergh [RV] proved that if $\mathscr{C}$ admits a Serre functor $\mathbb{S}$, then $\mathscr{C}$ has Auslander-Reiten triangles. Moreover, if $\tau$ is the Auslander-Reiten translation in $\mathscr{C}$, then $\mathbb{S}\simeq\tau[1]$. We say that a triangulated category $\mathscr{C}$ is $2$-Calabi-Yau if $\mathbb{S}\simeq[2]$.

Let $\mathscr{C}$ be an additive category and $\mathscr{X}$ be a subcategory of $\mathscr{C}$. We denote by $\mathscr{C}/\mathscr{X}$ the category whose objects are objects of $\mathscr{C}$ and whose morphisms are elements of $\mbox{Hom}_{\mathscr{C}}(A,B)/\mathscr{X}(A,B)$ for $A,B\in\mathscr{C}$, where $\mathscr{X}(A,B)$ the subgroup of $\mbox{Hom}_{\mathscr{C}}(A,B)$ consisting of morphisms which factor through an object in $\mathscr{X}$. Such category is called the quotient category of $\mathscr{C}$ by $\mathscr{X}$. For any morphism $f\colon A\to B$ in $\mathscr{C}$, we denote by $\overline{f}$ the image of $f$ under the natural quotient functor $\mathscr{C}\to\mathscr{C}/\mathscr{X}$.

3 Main result

In order to prove our main result, we need the following some lemmas.

Lemma 3.1**.**

[BM, Theorem 3.1]* and [J, Theorem 2.2] Let $\mathscr{C}$ be a triangulated category and $\mathscr{X}$ a contravariantly finite subcategory of $\mathscr{C}$. Then the quotient category $\mathscr{C}/\mathscr{X}$ is a left triangulated category with respect to the following loop functor and left triangles:*

• (a)

For any object $C\in\mathscr{C}$, we take a triangle

[TABLE]

where $\beta$ is a right $\mathscr{X}$-approximation of $C$. Then $\Omega$ gives a well-defined functor of $\mathscr{C}/\mathscr{X}$, which is the loop functor of $\mathscr{C}/\mathscr{X}$.

• (b)

A left triangle in $\mathscr{C}/\mathscr{X}$ is by definition a diagram which is isomorphic in $\mathscr{C}/\mathscr{X}$ to a diagram

[TABLE]

arising by forming the homotopy pull-back in $\mathscr{C}$

[TABLE]

of a morphism $w\colon B\to C$ along the triangle $\Omega C\xrightarrow{~{}~{}}X\xrightarrow{~{}g~{}}C\xrightarrow{~{}~{}}\Omega C[1]$ where $\beta$ is a right $\mathscr{X}$-approximation of $C$. Equivalently it is easy to see that the left triangles in $\mathscr{C}/\mathscr{X}$ are the diagrams which are isomorphic in $\mathscr{C}/\mathscr{X}$ to diagrams (3.1), arising from triangles $A\xrightarrow{~{}~{}}B\xrightarrow{~{}~{}}C\xrightarrow{~{}~{}}A[1]$ in $\mathscr{C}$, where the morphism $\emph{\mbox{Hom}}_{\mathscr{C}}(\mathscr{X},B)\to\emph{\mbox{Hom}}_{\mathscr{C}}(\mathscr{X},C)$ is an epimorphism.

Lemma 3.2**.**

Let $\mathscr{C}$ be a triangulated category and $\mathscr{X}$ a subcategory of $\mathscr{C}$. If

[TABLE]

is the Auslander-Reiten triangle in $\mathscr{C}$ and $A\notin\mathscr{X}$, then we have the following exact sequence in the quotient category $\overline{\mathscr{C}}:=\mathscr{C}/\mathscr{X}$:

[TABLE]

where $M\in\overline{\mathscr{C}}$.

Proof.

Since $gf=0$ and then $\overline{f}\circ\overline{g}=0$. Thus we have ${\rm Im}(\overline{g})\subseteq{\rm Ker}(\overline{f})$.

Conversely, let $\overline{u}\colon B\to M$ be any morphism in $\overline{\mathscr{C}}$ such that $\overline{u}\circ\overline{f}=0$. Then there exists morphisms $s\colon A\to X$ and $t\colon X\to M$ such that $uf=ts$. Since $A\in\notin\mathscr{X}$, we have that $s\colon A\to X$ is not split monomorphism. So there exist a morphism $x\colon B\to X$ such that $s=xf$ and then

[TABLE]

Thus there exists a morphism $y\colon C\to M$ such that $u-tx=yg$. It follows that $\overline{u}=\overline{y}\circ\overline{g}$.

This shows that ${\rm Ker}(\overline{f})\subseteq{\rm Im}(\overline{g})$. ∎

Now we prove the main result of this article. The proof is not too far from Beligiannis case, compare with [Be, Theorem 7.3].

Theorem 3.3**.**

Let $\mathscr{C}$ be a triangulated category with a Serre functor $\mathbb{S}$ and $\mathscr{X}$ a non-zero contravariantly finite rigid subcategory of $\mathscr{C}$. Then $\mathscr{X}$ is cluster tilting if and only if the quotient category $\mathscr{C}/\mathscr{X}$ is abelian and $\mathbb{S}(\mathscr{X})=\mathscr{X}[2]$.

Proof.

We first show the ‘only if’ part.

By Theorem 1.3, we know that the quotient category $\mathscr{C}/\mathscr{X}$ is an abelian category.

Now we show that $\mathbb{S}(\mathscr{X})=\mathscr{X}[2]$.

Since $\mbox{Hom}_{\mathscr{C}}(\mathscr{X},\mathbb{S}\mathscr{X}[-2][1])=\mbox{Hom}_{\mathscr{C}}(\mathscr{X},\mathbb{S}\mathscr{X}[-1])\simeq D\mbox{Hom}_{\mathscr{C}}(\mathscr{X}[-1],\mathscr{X})=0$ and $\mathscr{X}$ is cluster tilting, we have $\mathbb{S}(\mathscr{X})\subseteq\mathscr{X}[2]$.

Conversely, note that $\mbox{Hom}_{\mathscr{C}}(\mathbb{S}^{-1}\mathscr{X}[2],\mathscr{X}[1])\simeq D\mbox{Hom}_{\mathscr{C}}(\mathscr{X}[1],\mathscr{X}[2])=0$, thus for any $M\in\mathbb{S}^{-1}\mathscr{X}[2]$, we have $\mbox{Hom}_{\mathscr{C}}(M,\mathscr{X}[1])=0$.

Since $\mathscr{X}$ is contravariantly finite in $\mathscr{C}$, we can take a triangle

[TABLE]

with $g$ is a right $\mathscr{X}$-approximation of $M$. Apply the functor $\mbox{Hom}_{\mathscr{C}}(\mathscr{X},-)$ to the above triangle, we have the following exact sequence:

[TABLE]

Since $g$ is a right $\mathscr{X}$-approximation of $M$, we have that $\mbox{Hom}_{\mathscr{C}}(\mathscr{X},g)$ is an epimorphism. It follows that $\mbox{Hom}_{\mathscr{C}}(\mathscr{X},N[1])=0$ implies $N\in\mathscr{X}$ since $\mathscr{X}$ is cluster tilting. We obtain that $h=0$. Thus the triangle (3.2) splits. It follows that $M$ is a direct summand of $X_{0}$ and then $M\in\mathscr{X}$. Hence $M\in\mathbb{S}^{-1}\mathscr{X}[2]\subseteq\mathscr{X}$ and then $\mathscr{X}[2]\subseteq\mathbb{S}\mathscr{X}$.

To prove the ‘if’ part.

Since $\mathscr{X}$ is rigid, we have that $\mathscr{X}\subseteq\{M\in\mathscr{C}~{}|~{}\mbox{Hom}_{\mathscr{C}}(\mathscr{X},M[1])=0\}$.

Suppose that $M$ is an indecomposable object satisfying $\mbox{Hom}_{\mathscr{C}}(\mathscr{X},M[1])=0$.

Now we assume $M\notin\mathscr{X}$. Since

[TABLE]

we have $\mbox{Hom}_{\mathscr{C}}(M,\mathscr{X}[1])=0$. Then $\tau M\notin\mathscr{X}$ since $\mbox{Hom}_{\mathscr{C}}(M,\tau M[1])\simeq D\mbox{Hom}_{\mathscr{C}}(M,M)\neq 0$. Let

[TABLE]

be the Auslander-Reiten triangle ending at $M$.

Since $\mathscr{X}$ is contravariantly finite in $\mathscr{C}$, then for the object $M$ there exists a triangle

[TABLE]

where $y$ is a right $\mathscr{X}$-approximation of $M$. Since $M\notin\mathscr{X}$, we have that $y$ is not a split epimorphism. It follows that $y$ factors through $g$ and there exists a morphism of triangles

[TABLE]

which by Lemma 3.1 induces the following left triangle in the left triangulated category $\mathscr{C}/\mathscr{X}$:

[TABLE]

If $\overline{g}=0$, then the above left triangle gives a direct sum decomposition $\tau M\simeq\Omega M\oplus C$ in $\mathscr{C}/\mathscr{X}$. Since $M$ is indecomposable, we have that $\tau M$ is indecomposable. Thus we have $C=0$ in $\mathscr{C}/\mathscr{X}$ or else $\Omega M=0$ in $\mathscr{C}/\mathscr{X}$, that is to say, either $C\in\mathscr{X}$ or else $\Omega M\in\mathscr{X}$. If $\Omega M\in\mathscr{X}$, we have $z\in\mbox{Hom}_{\mathscr{C}}(M,\mathscr{X}[1])=0$. Then $y$ is a split epimorphism, namely $M$ lies in $\mathscr{X}$ as a direct summand of $X_{0}$ and this is a contradiction to $M\notin\mathscr{X}$. We infer that $C\in\mathscr{X}$.

Since $\overline{g}=0$, there exist morphisms $s\colon C\to X_{1}$ and $t\colon X_{1}\to M$ such that $g=ts$ where $X_{1}\in\mathscr{X}$. Since $M\notin\mathscr{X}$, we have that $t$ is not a split epimorphism. So there exists a morphism $k\colon X_{1}\to C$ such that $t=gk$ and then $g(1-ks)=g-gks=g-ts=0$. Thus there exists a morphism $m\colon C\to\tau M$ such that $1-ks=fm$ and then $\overline{f}\circ\overline{m}=\overline{1}$. We obtain $\overline{f}$ is a split epimorphism in $\mathscr{C}/\mathscr{X}$. Then $C$ is a direct summand of $\tau M$. Since $\tau M\notin\mathscr{X}$, we have $C\notin\mathscr{X}$. This is a contradiction to $C\in\mathscr{X}$.

So we get $\overline{g}\neq 0$. Let

[TABLE]

be the Auslander-Reiten triangle starting at $M$. Then $\overline{v}\neq 0$ since $\overline{g}\neq 0$.

We claim that $\overline{v}$ is an epimorphism $\mathscr{C}/\mathscr{X}$. In fact, let $\overline{\alpha}\colon\tau^{-1}M\to L$ be any morphism in $\mathscr{C}/\mathscr{X}$ such that $\overline{\alpha}\circ\overline{v}=0$. Since $\mathscr{C}/\mathscr{X}$ is abelian, we assume that $\overline{\beta}\colon\tau^{-1}C\to L$ is the cokernel of $\overline{u}$. Then there exists a morphism $\overline{\omega}\colon L\to\tau^{-1}M$ such that $\overline{v}=\overline{\omega}\circ\overline{\beta}$ since $\overline{v}\circ\overline{u}=0$. Note that $\overline{\beta}\circ\overline{u}=0$, by Lemma 3.2, there exists a morphism $\overline{\delta}\colon\tau^{-1}M\to L$ such that $\overline{\beta}=\overline{\delta}\circ\overline{v}$. It follows that

[TABLE]

Since $\overline{\beta}$ is an epimorphism in $\mathscr{C}/\mathscr{X}$, we have $\overline{\delta}\circ\overline{\omega}=\overline{1}$ and $\overline{\omega}\circ\overline{\delta}$ is an idempotent in ${\rm End}_{\mathscr{C}/\mathscr{X}}(\tau^{-1}M)$. Since $\overline{v}\neq 0$, we have $\overline{\omega}\circ\overline{\delta}\neq 0$. Thus $\overline{\omega}\circ\overline{\delta}$ is not nilpotent. Since ${\rm End}_{\mathscr{C}/\mathscr{X}}(\tau^{-1}M)$ is local, we have $\overline{\omega}\circ\overline{\delta}$ is an isomorphism. Hence $\overline{\omega}$ is an isomorphism in $\mathscr{C}/\mathscr{X}$ which implies that $\overline{v}$ is an epimorphism in $\mathscr{C}/\mathscr{X}$.

Note that $\overline{w}=0$ since $\overline{v}$ is an epimorphism. Thus there exists morphisms $s^{\prime}\colon\tau^{-1}M\to X_{2}$ and $t^{\prime}\colon X_{2}\to M[1]$ such that $w=t^{\prime}s^{\prime}$ where $X_{2}\in\mathscr{X}$. Note that $t^{\prime}\in\mbox{Hom}_{\mathscr{C}}(X_{2},M[1])=0$, we have $w=0$. Therefore the Auslander-Reiten triangle (3.3) splits, a contradiction. So we get $M\in\mathscr{X}$.

This shows that $\{M\in\mathscr{C}~{}|~{}\mbox{Hom}_{\mathscr{C}}(\mathscr{X},M[1])=0\}\subseteq\mathscr{X}$ and hence $\mathscr{X}$ is cluster tilting. ∎

This theorem immediately yields the following important conclusion.

Corollary 3.4**.**

Let $\mathscr{C}$ be a $2$-Calabi-Yau triangulated category and $\mathscr{X}$ a non-zero contravariantly finite rigid subcategory of $\mathscr{C}$. Then $\mathscr{X}$ is cluster tilting if and only if the quotient category $\mathscr{C}/\mathscr{X}$ is abelian.