2. Preliminaries

2.1. Right $(d+2)$ -angulated categories

Let $\mathscr{C}$ be an additive category with an endofunctor $\Sigma^d\colon\mathscr{C}\rightarrow\mathscr{C}$ ( $\Sigma^d$ is called the d-suspension functor), and d be an integer greater than or equal to one. A $(d+2)-\Sigma^d-sequence$ in $\mathscr{C}$ is a sequence of morphisms

\begin{equation*}A_0\xrightarrow{f_0}A_1\xrightarrow{f_1}A_2\xrightarrow{f_2}\cdots\xrightarrow{f_{d-1}}A_d\xrightarrow{f_d}A_{d+1}\xrightarrow{f_{d+1}}\Sigma^d A_0.\end{equation*}

Its left rotation is the $(d+2)-\Sigma^d$ -sequence

\begin{equation*}A_1\xrightarrow{f_1}A_2\xrightarrow{f_2}A_3\xrightarrow{f_3}\cdots\xrightarrow{f_{d}}A_{d+1}\xrightarrow{f_{d+1}}\Sigma^d A_0\xrightarrow{(\!-\!1)^{d}\Sigma^d f_0}\Sigma^d A_1.\end{equation*}

A morphism of $(d+2)-\Sigma^d$ -sequences is a sequence of morphisms $\varphi=(\varphi_0,\varphi_1,\cdots,\varphi_{d+1})$ such that the following diagram

commutes, where each row is a $(d+2)-\Sigma^d$ -sequence. It is an isomorphism if $\varphi_0, \varphi_1, \varphi_2, \cdots, \varphi_{d+1}$ are all isomorphisms in $\mathscr{C}$ , and a weak isomorphism if $\varphi_i$ and $\varphi_{i+1}$ are isomorphisms for some $0\leq i\leq d+1$ (with $\varphi_{d+2}:=\Sigma\varphi_0$ ). Note that the composition of two weak isomorphisms need not be a weak isomorphism.

We recall the notion of a right $(d+2)$ -angulated category from [Reference Z.13, Definition 2.1]. The axiom (N4) presented below is based on the higher octahedral axiom introduced in [Reference E., P. and M.1].

Definition 2.1 A right $(d+2)$ -angulated category is a triple $(\mathscr{C}, \Sigma^d, \Theta)$ , where $\mathscr{C}$ is an additive category, $\Sigma^d$ is the d-suspension functor of $\mathscr{C}$ , and $\Theta$ is a class of $(d+2)-\Sigma^d$ -sequences (whose elements are called right $(d+2)$ -angles), which satisfies the following axioms:

1. (N1) (a) The class $\Theta$ is closed under isomorphisms, direct sums and direct summands.

2. (b) For each object $A\in\mathscr{C}$ the trivial sequence

   \begin{equation*}A\xrightarrow{\textrm{Id}_A} A\rightarrow 0\rightarrow 0\rightarrow\cdots\rightarrow 0\rightarrow \Sigma^dA,\end{equation*}
   belongs to $\Theta$ .

3. (c) Each morphism $f_0\colon A_0\rightarrow A_1$ in $\mathscr{C}$ can be extended to a right $(d+2)$ -angle:

   \begin{equation*}A_0\xrightarrow{f_0}A_1\xrightarrow{f_1}A_2\xrightarrow{f_2}\cdots\xrightarrow{f_{d-1}}A_d\xrightarrow{f_d}A_{d+1}\xrightarrow{f_{d+1}}\Sigma^d A_0.\end{equation*}

4. (N2) If a $(d+2)-\Sigma^d$ -sequence belongs to $\Theta$ , then its left rotation belongs to $\Theta$ .

5. (N3) Each solid commutative diagram

   

   with rows in $\Theta$ can be completed to a morphism of $(d+2)-\Sigma^d$ -sequences.

6. (N4) Given the solid part of the diagram

   

   with commuting squares and with rows in $\Theta$ , the dotted morphisms exist such that each square commutes, and the $(d+2)-\Sigma^d$ -sequence

   \begin{equation*}A_2\xrightarrow{\left( \begin{matrix} f_2 \\ \varphi_2 \\ \end{matrix} \right)} A_3\oplus B_2\xrightarrow{\left( \begin{matrix} -f_3 & 0 \\ \varphi_3 & -g_2 \\ \phi_3 & \psi_2 \\ \end{matrix} \right)} A_4\oplus B_3\oplus C_2\xrightarrow{\left( \begin{matrix} -f_4 & 0 & 0 \\ -\varphi_4 & -g_3 & 0 \\ \phi_4 & \psi_3 & h_2 \\ \end{matrix} \right)}A_5\oplus B_4\oplus C_3\end{equation*}
   \begin{equation*}\xrightarrow{\left( \begin{matrix} -f_5 & 0 & 0 \\ \varphi_5 & -g_4 & 0 \\ \phi_5 & \psi_4 & h_3 \\ \end{matrix} \right)}\cdots\xrightarrow{\scriptsize\left(\begin{matrix} -f_{d} & 0 & 0 \\ (\!-\!1)^{d+1}\varphi_{d} & -g_{d-1} & 0 \\ \phi_{d} & \psi_{d-1} & h_{d-2} \\ \end{matrix} \right)}A_{d+1}\oplus B_{d}\oplus C_{d-1}\end{equation*}
   \begin{equation*}\xrightarrow{\left( \begin{matrix} (\!-\!1)^{d+1}\varphi_{d+1} &-g_{d} &0 \\ \phi_{d+1}& \psi_{d}& h_{d-1} \\ \end{matrix} \right)}B_{d+1}\oplus C_{d}\xrightarrow{(\psi_{d+1},\ h_{d})}C_{d+1}\xrightarrow{\Sigma^d f_1\circ h_{d+1}}\Sigma^d A_2 \end{equation*}
   belongs to $\Theta$ .

The notion of a left $(d+2)$ -angulated category is defined dually.

If $\Sigma^d$ is an automorphism, it is easy to see that the converse of an axiom (N2) also holds, thus the right $(d+2)$ -angulated category $(\mathscr{C},\Sigma^d, \Theta)$ is a $(d+2)$ -angulated category in the sense of Geiss–Keller–Oppermann [Reference C., B. and S.8, Definition 1.1] and in the sense of Bergh-Thaule [Reference P. and M.5, Theorem 4.4]. If $(\mathscr{C},\Sigma^d, \Theta)$ is a right $(d+2)$ -angulated category, $(\mathscr{C},\Omega, \Phi)$ is a left $(d+2)$ -angulated category, $\Omega$ is a quasi-inverse of $\Sigma^d$ and $\Theta=\Phi$ , then $(\mathscr{C},\Sigma^d, \Theta)$ is a $(d+2)$ -angulated category.

Suppose that $\Sigma^d$ is an automorphism. Now modify axiom (N1) into a new axiom (N1 $^\ast$ ).

1. (N1)* (a) The class $\Theta$ is closed under weak isomorphisms.

2. (b) For each object $A\in\mathscr{C}$ the trivial sequence

   \begin{equation*}A\xrightarrow{\textrm{Id}_A} A\rightarrow 0\rightarrow 0\rightarrow\cdots\rightarrow 0\rightarrow \Sigma^dA,\end{equation*}
   belongs to $\Theta$ .

3. (c) Each morphism $f_0\colon A_0\rightarrow A_1$ in $\mathscr{C}$ can be extended to a $(d+2)$ -angle:

   \begin{equation*}A_0\xrightarrow{f_0}A_1\xrightarrow{f_1}A_2\xrightarrow{f_2}\cdots\xrightarrow{f_{d-1}}A_d\xrightarrow{f_d}A_{d+1}\xrightarrow{f_{d+1}}\Sigma^d A_0.\end{equation*}

Remark 2.2 [Reference P. and M.5, Theorem 3.2] If $\Theta$ is a collection of $(d+2)-\Sigma^d$ -sequences satisfying the axioms (N2) and (N3), and $\Sigma^d$ is an automorphism. Then the following are equivalent:

1. (1) $\Theta$ satisfies (N1);

2. (2) $\Theta$ satisfies (N1).

Let $\mathscr{C}$ be an additive category and $f\colon A\rightarrow B$ a morphism in $\mathscr{C}$ . A weak cokernel of f is a morphism $g\colon B\rightarrow C$ such that for any $X\in\mathscr{C}$ the sequence of abelian groups

\begin{equation*}\mbox{Hom}_\mathscr{C}(C,X)\xrightarrow{\ g^\ast\ }\mbox{Hom}_\mathscr{C}(B,X)\xrightarrow{\ f^\ast\ }\mbox{Hom}_\mathscr{C}(A,X),\end{equation*}

is exact. Equivalently, g is a weak cokernel of f if $gf=0$ and for each morphism $h\colon B\rightarrow X$ such that $hf=0$ there exists a (not necessarily unique) morphism $p\colon C\rightarrow X$ such that $h=pg$ . These properties are subsumed in the following commutative diagram

Clearly, a weak cokernel g of f is a cokernel of f if and only if g is an epimorphism. The concept of a weak kernel is defined dually.

Lemma 2.3 Let $(\mathscr{C}, \Sigma^d, \Theta)$ be a $(d+2)$ -angulated category and

\begin{equation*}A_0\xrightarrow{f_0}A_1\xrightarrow{f_1}A_2\xrightarrow{f_2}\cdots\xrightarrow{f_{d-1}}A_d\xrightarrow{f_d}A_{d+1}\xrightarrow{f_{d+1}}\Sigma^d A_0.\end{equation*}

be a $(d+2)$ -angle. Then $f_{i}\circ f_{i-1}=0, \ i=1,2,\cdots,d+1$ . That is to say, any composition of two consecutive morphisms in a $(d+2)$ -angle vanishes.

Proof. This is [Reference C., B. and S.8, Prposition 2.5 (a)]. See also [Reference P. and M.5, Lemma 3.1].

Lemma 2.4 Let $(\mathscr{C}, \Sigma^d, \Theta)$ be a $(d+2)$ -angulated category and

\begin{equation*}A_0\xrightarrow{f_0}A_1\xrightarrow{f_1}A_2\xrightarrow{f_2}\cdots\xrightarrow{f_{d-1}}A_d\xrightarrow{f_d}A_{d+1}\xrightarrow{f_{d+1}}\Sigma^d A_0.\end{equation*}

be a $(d+2)$ -angle. Then $f_{i}$ is a weak cokernel of $f_{i-1}$ and $f_{i-1}$ is a weak kernel of $f_{i}$ , $i=1,2,\cdots,d+1$ .

Proof. This follows from [Reference C., B. and S.8, Prposition 2.5 (a) and Remark 2.2 (c)]. See also [Reference Z.12, Lemma 2.4].

2.2. Auslander–Reiten $(\textbf{d}+2)$ -angles

Let $\mathscr{C}$ be a $(d+2)$ -angulated category. We denote by $\textrm{rad}_{\mathscr{C}}$ the Jacobson radical of $\mathscr{C}$ . Namely, it is given by the formula

\begin{equation*}\textrm{rad}_{\mathscr{C}}(X,Y)=\{g\colon X\to Y\ |\ \textrm{Id}_{X}-hg\ \mbox{is invertible for any}\ h\colon Y\to X\},\end{equation*}

for all objects X and Y in $\mathscr{C}$ .

Next, we recall from [Reference M. and I.2] some terminology for the Auslander–Reiten theory. Let $f\colon X \to Y$ be a morphism. One says that f is right almost split if f is not a split epimorphism and every non-split epimorphism $g\colon M \to Y$ factors through f. In a dual manner, one defines f to be left almost split.

Definition 2.5 [Reference O. and Y.9, Definition 3.8], [Reference F.7, Definition 5.1] Let $\mathscr{C}$ be a $(d+2)$ -angulated category. A $(d+2)$ -angle

\begin{equation*}A_{\bullet}:\ \ {A_0 \xrightarrow{\ \alpha_0\ }A_1 \xrightarrow{\ \alpha_1\ } A_2 \xrightarrow{\ \alpha_2\ } \cdots \xrightarrow{\ \alpha_{d - 1}\ } A_d \xrightarrow{\ \alpha_{d}\ } A_{d+1}\xrightarrow{\ \alpha_{d+1}\ } \Sigma^d A_0},\end{equation*}

in $\mathscr{C}$ is called an Auslander-Reiten $(d+2)$ -angle if $\alpha_0$ is left almost split, $\alpha_d$ is right almost split and when $d\geq 2$ , also $\alpha_1,\alpha_2,\cdots,\alpha_{d-1}$ are in $\mbox{rad}_{\mathscr{C}}$ .

Remark 2.6 [Reference F.7, Remark 5.2] Assume $A_{\bullet}$ as in the above definition is an Auslander–Reiten $(d+2)$ -angle. Since $\alpha_0$ is left almost split which implies that $\mbox{End}(A_0)$ is local and hence $A_0$ is indecomposable. Similarly, since $\alpha_d$ is right almost split, then $\mbox{End}(A_{d+1})$ is local and hence $A_{d+1}$ is indecomposable. Moreover, when $d=1$ , we have $\alpha_0$ and $\alpha_d$ in $\mbox{rad}_{\mathscr{C}}$ , so that $\alpha_d$ is right minimal and $\alpha_0$ is left minimal. When $d\geq 2$ , since $\alpha_{d-1}\in\mbox{rad}_{\mathscr{C}}$ , we have that $\alpha_d$ is right minimal and similarly $\alpha_0$ is left minimal.

Remark 2.7 [Reference F.7, Lemma 5.3] Let $\mathscr{C}$ be a $(d+2)$ -angulated category and

\begin{equation*}A_{\bullet}:\ \ {A_0 \xrightarrow{\ \alpha_0\ }A_1 \xrightarrow{\ \alpha_1\ } A_2 \xrightarrow{\ \alpha_2\ } \cdots \xrightarrow{\ \alpha_{d - 1}\ } A_d \xrightarrow{\ \alpha_{d}\ } A_{d+1}\xrightarrow{\ \alpha_{d+1}\ } \Sigma^d A_0},\end{equation*}

be a $(d+2)$ -angle in $\mathscr{C}$ . Then the following statements are equivalent:

1. (1) $A_{\bullet}$ is an Auslander-Reiten $(d+2)$ -angle;

2. (2) $\alpha_0,\alpha_1,\cdots,\alpha_{d-1}$ are in $\textrm{rad}_{\mathscr{C}}$ and $\alpha_d$ is right almost split;

3. (3) $\alpha_1,\alpha_2,\cdots,\alpha_{d}$ are in $\textrm{rad}_{\mathscr{C}}$ and $\alpha_0$ is left almost split.

Lemma 2.8 [Reference F.7, Lemma 5.4] Let $\mathscr{C}$ be a $(d+2)$ -angulated category and

\begin{equation*}A_{\bullet}:\ \ {A_0 \xrightarrow{\ \alpha_0\ }A_1 \xrightarrow{\ \alpha_1\ } A_2 \xrightarrow{\ \alpha_2\ } \cdots \xrightarrow{\ \alpha_{d - 1}\ } A_d \xrightarrow{\ \alpha_{d}\ } A_{d+1}\xrightarrow{\ \alpha_{d+1}\ } \Sigma^d A_0},\end{equation*}

be a $(d+2)$ -angle in $\mathscr{C}$ . Assume that $\alpha_d$ is right almost split and if $d\geq2$ , also that $\alpha_1,\alpha_2,\cdots,\alpha_{d-1}$ are in $\mbox{rad}_{\mathscr{C}}$ . Then the following are equivalent:

1. (1) $A_{\bullet}$ is an Auslander-Reiten $(d+2)$ -angle;

2. (2) $\mbox{End}(A_0)$ is local;

3. (3) $\alpha_{d+1}$ is left minimal;

4. (4) $\alpha_{0}$ is in $\mbox{rad}_{\mathscr{C}}$ .

In the case $d=1$ , so in the case of a triangulated category, a morphism can be extended to a triangle in a unique way up to isomorphism. On the other hand, for $d\geq 2$ , a morphism can be extendend to a $(d+2)$ -angle in different non-isomorphic ways. However, we still have a unique “minimal” $(d+2)$ -angle extending any given morphism.

Lemma 2.9 [Reference S. and H.14, Lemma 5.18], [Reference F.7, Lemma 3.14] Let $d\geq 2$ and $h\colon A_{d+1}\to \Sigma^d A_0$ be any morphism in a $(d+2)$ -angulated category $\mathscr{C}$ . Then, up to isomorphism, there exists a unique $(d+2)$ -angle of the form

\begin{equation*} {A_0 \xrightarrow{\ \alpha_0\ }A_1 \xrightarrow{\ \alpha_1\ } A_2 \xrightarrow{\ \alpha_2\ } \cdots \xrightarrow{\ \alpha_{d-2}\ } A_{d-1} \xrightarrow{\ \alpha_{d - 1}\ } A_d \xrightarrow{\ \alpha_{d}\ } A_{d+1}\xrightarrow{\ h\ } \Sigma^d A_0},\end{equation*}

with $\alpha_1,\alpha_2,\cdots,\alpha_{d-1}$ in $\textrm{rad}_{{\mathscr{C}}}$ .

3. Every Serre functor is a $(d+2)$ -angulated functor

Let $\mathscr{C}$ be a k-linear Hom-finite additive category. When there is no danger of confusion, we will sometimes instead of

\begin{equation*}\mbox{Hom}_\mathscr{C}(X,A)\xrightarrow{\ \mbox{Hom}_\mathscr{C}(X,\ f)\ }\mbox{Hom}_\mathscr{C}(X,B),\end{equation*}

write one of the following simplified forms:

\begin{equation*}\mbox{Hom}(X,A)\xrightarrow{\ \mbox{Hom}(X,\ f)\ }\mbox{Hom}(X,B),\end{equation*}

\begin{equation*}(X,A)\xrightarrow{\ (X,\ f)\ }(X,B).\end{equation*}

A right Serre functor is a k-linear additive functor $\mathbb{S}\colon\mathscr{C}\to \mathscr{C}$ together with isomorphisms

\begin{equation*}\eta_{A,B}\colon\mbox{Hom}(A,B)\to \mbox{Hom}(B,\mathbb{S} A)^\ast,\end{equation*}

for any $A,B\in\mathscr{C}$ which are natural in A and B, where $(-)^{\ast}:=\mbox{Hom}_{k}(-,k)$ .

A left Serre functor is a k-linear additive functor $\mathbb{S}\colon\mathscr{C}\to\mathscr{C}$ together with isomorphisms

\begin{equation*}\zeta_{A,B}\colon \mbox{Hom}(A,B)\to \mbox{Hom}(\mathbb{S} B,A)^\ast,\end{equation*}

for any $A,B\in\mathscr{C}$ which are natural in A and B.

Let $\eta_A\colon\mbox{Hom}(A,\mathbb{S} A)\to k$ be given by $\eta_{A,A}(\textrm{Id}_A)$ and let $f\in\mbox{Hom}(A, B)$ . Looking at the following commutative diagram (which follows from the naturality of $\eta_{A,B}$ in B)

we find for $g\in\mbox{Hom}(B,\mathbb{S} A)$

\begin{equation*}\eta_{A,B}(f)(g)=\eta_A (g f).\end{equation*}

Similarly by the naturality of $\eta_{A,B}$ in A we obtain a commutative diagram

This yields for $g\in \mbox{Hom}(B,\mathbb{S} A)$ the formula

(3.1) \begin{equation}\begin{array}{l}\eta_{A,B}(f)(g)=\eta_B(\mathbb{S}(f) g).\end{array}\end{equation}

Bondal and Kapranov [Reference A. and M.4, Proposition 3.3] prove that every Serre functor is a triangulated functor in a triangulated category. Van den Bergh has since provided a different proof, see [Reference R.3, Theorem A.4.4]. The following result shows that the Serre functor is a $(d+2)$ -angulated functor in a $(d+2)$ -angulated category. This generalizes the work by Bondal and Kapranov. Our proof is an adaptation of the proof of Van den Bergh. For more details, see also [Reference P.16, Theorem 10.5.1].

Proof. Assume that $(\mathbb{S},\eta_{X,Y})$ is a right Serre functor. Put

\begin{equation*}\eta_{X}:=\eta_{X,X}(\textrm{Id}_X)\in\mbox{Hom}(X,\mathbb{S} X)^{\ast},\ \mbox{for all}\ X\in\mathscr{C}.\end{equation*}

Step 1: We claim that there exists a natural isomorphism $\zeta\colon \mathbb{S}\Sigma^d\to \Sigma^d\mathbb{S}$ such that

(3.2) \begin{equation}\begin{array}{l}\eta_{X}(\Sigma^{-d}\zeta_X\circ\Sigma^{-d}f)=(\!-\!1)^d\eta_{\Sigma^dX}(f),\ \mbox{for all}\ f\in\mbox{Hom}(\Sigma^dX,\mathbb{S}(\Sigma^dX)) \ \textrm{and}\ X\in\mathscr{C}.\end{array}\end{equation}

By the following two isomorphisms

\begin{equation*}\mbox{Hom}(\Sigma^dX,\mathbb{S}(\Sigma^dX))^{\ast}\xrightarrow{\ \ \sim\ \ }\mbox{Hom}(X,\Sigma^{-d}\mathbb{S}(\Sigma^dX))^{\ast}\end{equation*}

\begin{equation*}\eta^{\ast}_{X,\ \Sigma^{-d}\mathbb{S}(\Sigma^dX)}\colon\mbox{Hom}(X,\Sigma^{-d}\mathbb{S}(\Sigma^dX))^{\ast}\xleftarrow{\ \ \sim\ \ }\mbox{Hom}(\Sigma^{-d}\mathbb{S}(\Sigma^dX),\mathbb{S} X),\end{equation*}

and $(\!-\!1)^d\eta_{\Sigma^dX}\in\mbox{Hom}(\Sigma^dX,\mathbb{S}(\Sigma^dX))^{\ast}$ , there exists $\zeta_X\in\mbox{Hom}(\mathbb{S}(\Sigma^dX),\Sigma^d(\mathbb{S} X))$ such that

(3.3) \begin{equation}\begin{array}{l}\eta^{\ast}_{X,\ \mathbb{S}\Sigma^{-d}(\Sigma^dX)}(\Sigma^{-d}\zeta_X)(g)=(\!-\!1)^d\eta_{\Sigma^dX}(\Sigma^dg),\ \mbox{for all}\ g\colon X\to\mathbb{S}\Sigma^{-d}(\Sigma^dX).\end{array}\end{equation}

Using Serre duality, the above equality (3.3) is equivalent to

(3.4) \begin{equation}\begin{array}{l}\eta_{X,\ \Sigma^{-d}\mathbb{S}(\Sigma^dX)}(g)(\Sigma^{-d}\zeta_X)=(\!-\!1)^d\eta_{\Sigma^dX}(\Sigma^dg).\end{array}\end{equation}

By the equality (3.1), the above equality (3.4) is identified with

\begin{equation*}\eta_{X}(\Sigma^{-d}\zeta_X\circ \Sigma^{-d}f)=(\!-\!1)^d\eta_{\Sigma^dX}(f),\ \mbox{for all}\ f\in\mbox{Hom}(\Sigma^dX,\mathbb{S}(\Sigma^dX)),\end{equation*}

as desired. That is to say, the equality (3.2) holds.

It remains to show that $\zeta$ is a natural isomorphism.

By the following two isomorphisms

\begin{equation*}\eta^{\ast}_{\Sigma^dX,\ \Sigma^d(\mathbb{S} X)}\colon\mbox{Hom}(\Sigma^d(\mathbb{S} X),\mathbb{S}(\Sigma^{d}X))\xrightarrow{\ \ \sim\ \ }\mbox{Hom}(\Sigma^{d}X,\Sigma^d(\mathbb{S} X))^{\ast}\end{equation*}

\begin{equation*}\mbox{Hom}(\Sigma^dX,\Sigma^d(\mathbb{S} X))^{\ast}\xleftarrow{\ \ \sim\ \ }\mbox{Hom}(X,\mathbb{S} X)^{\ast}\end{equation*}

and $(\!-\!1)^d\eta_{X}\in\mbox{Hom}(X,\mathbb{S} X)^{\ast}$ , there exists $\theta_X\in\mbox{Hom}(\Sigma^d(\mathbb{S} X),\mathbb{S}(\Sigma^dX))$ such that

(3.5) \begin{equation}\begin{array}{l}\eta^{\ast}_{\Sigma^dX,\ \Sigma^d(\mathbb{S} X)}(\theta_X)(g)=(\!-\!1)^d\eta_{X}(\Sigma^{-d}g),\ \mbox{for all}\ g\colon\Sigma^dX\to\Sigma^d(\mathbb{S} X).\end{array}\end{equation}

Using Serre duality, the above equality (3.5) is equivalent to

(3.6) \begin{equation}\begin{array}{l}\eta^{\ast}_{\Sigma^dX,\ \Sigma^d(\mathbb{S} X)}(g)(\theta_X)=(\!-\!1)^d\eta_X(\Sigma^{-d}g).\end{array}\end{equation}

By the equality (3.1), the above equality (3.6) is identified with

(3.7) \begin{equation}\begin{array}{l}\eta_{\Sigma^dX}(\theta_X\Sigma^dg)=(\!-\!1)^d\eta_{X}(g),\ \mbox{for all}\ g\in\mbox{Hom}(X,\mathbb{S} X).\end{array}\end{equation}

We now show that $\zeta$ is an isomorphism. For any morphism $g\colon X\to \mathbb{S} X$ , we have

By the bilinear map $(-,-)\colon\mbox{Hom}(X,\mathbb{S} X)\times \mbox{Hom}(\mathbb{S} X,\mathbb{S} X)\longrightarrow k$ is non-degenerate, we obtain $\Sigma^{-d}(\zeta_X\theta_X)=\textrm{Id}_{\mathbb{S} X}$ and then $\zeta_X\theta_X=\textrm{Id}_{\Sigma^d(\mathbb{S} X)}$ .

Similarly, for any morphism $g\colon \Sigma^dX\to \mathbb{S}(\Sigma^dX)$ , we have

So $\theta_X\zeta_X=\textrm{Id}_{\mathbb{S}(\Sigma^dX)}$ . This shows that $\zeta_X$ is an isomorphism.

We now show that $\zeta$ is a natural transformation. Namely, for any morphism $f\colon X\to Y$ , we have the following commutative diagram:

Note that $\Sigma^{-d}\zeta_X\in\mbox{Hom}(\Sigma^{-d}\mathbb{S}(\Sigma^dX),\mathbb{S} X)$ and $\Sigma^{-d}\zeta_Y\in\mbox{Hom}(\Sigma^{-d}\mathbb{S}(\Sigma^dY),\mathbb{S} Y )$ . For any morphism $h\in\mbox{Hom}(Y,\Sigma^{-d}\mathbb{S}(\Sigma^dX))$ , we have

Consider the following commutative diagram:

It follows that $\mbox{Hom}(-,\mathbb{S} f)(\Sigma^{-d}\zeta_X)=\mbox{Hom}(\Sigma^{-d}\mathbb{S}(\Sigma^df),\ -))(\Sigma^{-d}\zeta_Y)$ and then

\begin{equation*}\mathbb{S}(f)\Sigma^{-d}\zeta_X=\Sigma^{-d}\zeta_Y\Sigma^{-d}\mathbb{S}(\Sigma^df).\end{equation*}

Hence $\Sigma^{d}(\mathbb{S} f)\zeta_X=\zeta_Y\mathbb{S}(\Sigma^df)$ as required.

Step 2: Assume that

\begin{equation*}A_0\xrightarrow{\alpha_0}A_1\xrightarrow{\alpha_1}A_2\xrightarrow{\alpha_2}\cdots\xrightarrow{\alpha_{d-1}}A_d\xrightarrow{\alpha_d}A_{d+1}\xrightarrow{\alpha_{d+1}}\Sigma^dA_{0}\end{equation*}

is any $(d+2)$ -angle in $\mathscr{C}$ . The morphism $\mathbb{S}\alpha_0$ can be embedded in a $(d+2)$ -angle

\begin{equation*}\mathbb{S} A_0\xrightarrow{\mathbb{S}\alpha_0}\mathbb{S} A_1\xrightarrow{\beta_1}B_2\xrightarrow{\beta_2}\cdots\xrightarrow{\beta_{d-1}}B_d\xrightarrow{\beta_d}B_{d+1}\xrightarrow{\beta_{d+1}}\Sigma^d\mathbb{S} A_{0}.\end{equation*}

In order to prove that $(\mathbb{S},\zeta)$ is a $(d+2)$ -angulated functor, we only need to show that there are morphisms $\omega_i\colon B_i\to\mathbb{S} A_i,\ i=2,3,\cdots,d+1$ such that the following diagram commutes:

Note that the class of $(d+2)$ -angles is closed under weak isomorphisms, then

\begin{equation*}\mathbb{S} A_0\xrightarrow{\mathbb{S}\alpha_0}\mathbb{S} A_1\xrightarrow{\mathbb{S}\alpha_1}\mathbb{S} A_2\xrightarrow{\mathbb{S}\alpha_2}\cdots\xrightarrow{\mathbb{S}\alpha_{d-1}}\mathbb{S} A_d\xrightarrow{\mathbb{S}\alpha_d}\mathbb{S} A_{d+1}\xrightarrow{\zeta_{A_0}\mathbb{S}\alpha_{d+1}}\Sigma^d\mathbb{S} A_{0}\end{equation*}

is also a $(d+2)$ -angle since the first row is a $(d+2)$ -angle. Then we are done.

In fact, since $\beta_1$ is a weak cokernel of $\mathbb{S}\alpha_1$ and $\mathbb{S}\alpha_1\circ\mathbb{S}\alpha_0=0$ , there exists a morphism $\omega_2\colon B_2\to\mathbb{S} A_2$ such that $\mathbb{S}\alpha_1=\omega_2\beta_1$ . Since $\beta_2$ is a weak cokernel of $\beta_1$ and $\mathbb{S}\alpha_2\circ\omega_2\beta_1=\mathbb{S}\alpha_2\circ\mathbb{S}\alpha_1=0$ , there exists a morphism $\omega_3\colon B_3\to\mathbb{S} A_3$ such that $\mathbb{S}\alpha_2\circ\omega_2=\omega_3\beta_2$ . Continue this process, there are morphisms $\omega_i\colon B_i\to \mathbb{S} A_i,\ i=4,5,\cdots,d$ such that $\mathbb{S}\alpha_{i-1}\circ\omega_{i-1}=\omega_{i}\beta_{i-1}$ .

It remains to show that there exists a morphism $\omega_{d+1}\colon B_{d+1}\to\mathbb{S} A_{d+1}$ such that

(3.8) \begin{equation}\begin{array}{l}\zeta_{A_0}\mathbb{S}\alpha_{d+1}\circ\omega_{d+1}=\beta_{d+1},\end{array}\end{equation}

(3.9) \begin{equation}\begin{array}{l}\omega_{d+1}\beta_d=\mathbb{S}\alpha_d\circ\omega_d,\end{array}\end{equation}

Claim I: The equality (3.8) holds if and only if

(3.10) \begin{equation}\begin{array}{l}\eta_{A_{d+1}}(\omega_{d+1}\circ\Sigma^d\varphi\circ\alpha_{d+1})=(\!-\!1)^d\eta_{A_0}(\Sigma^{-d}\beta_{d+1}\circ\varphi),\ \mbox{for all}\ \varphi\in\mbox{Hom}(A_0,\Sigma^{-d}B_{d+1}).\end{array}\end{equation}

The equality (3.9) holds if and only if

(3.11) \begin{equation}\begin{array}{l}\eta_{A_{d+1}}(\omega_{d+1}\beta_d\phi)=\eta_{A_d}(\omega_d\phi\alpha_d),\ \mbox{for all}\ \phi\in\mbox{Hom}(A_{d+1},B_d).\end{array}\end{equation}

We first show that the equality (3.8) holds if and only if the equality (3.10) holds.

Suppose that the equality (3.8) holds. For any morphism $\varphi\in\mbox{Hom}(A_0,\Sigma^{-d}B_{d+1})$ , by the equality (3.8), we have

\begin{equation*}\Sigma^{-d}(\zeta_{A_0}\mathbb{S}\alpha_{d+1}\circ\omega_{d+1})\circ\varphi=\Sigma^{-d}\beta_{d+1}\circ\varphi\colon A_0\to\mathbb{S} A_0.\end{equation*}

It follows that

This shows that the equality (3.10) holds.

Conversely, assume that the equality (3.10) holds. For any morphism $\varphi\in\mbox{Hom}(A_0,\Sigma^{-d}B_{d+1})$ , by the equalities (3.10), (3.1) and (3.2), we have

\begin{equation*}\eta_{A_0}(\Sigma^{-d}\beta_{d+1}\circ\varphi)=\eta_{A_0}(\Sigma^{-d}(\zeta_{A_0}\mathbb{S}\alpha_{d+1}\circ\omega_{d+1})\circ\varphi).\end{equation*}

As

\begin{equation*}\mbox{Hom}(A_0,\Sigma^{-d}B_{d+1})\times \mbox{Hom}(\Sigma^{-d}B_{d+1}),\mathbb{S} A_0)\longrightarrow k,\ \ (\varphi,g)\longmapsto\eta_{A_0}(g\varphi),\end{equation*}

is bilinear and non-degenerate, we have $\zeta_{A_0}\mathbb{S}\alpha_{d+1}\circ\omega_{d+1}=\beta_{d+1}$ . Namely, the equality (3.8) holds.

Now we show that the equality (3.9) holds if and only if the equality (3.11) holds.

Assume that the equality (3.9) holds. For any morphism $\phi\in\mbox{Hom}(A_{d+1},B_d)$ , by the equality (3.9), we have

\begin{equation*}\omega_{d+1}\beta_{d}\phi=\mathbb{S}\alpha_{d}\circ\omega_d\phi\colon A_{d+1}\to \mathbb{S} A_{d+1}.\end{equation*}

It follows that

This shows that the equality (3.11) holds.

Conversely, assume that the equality (3.11) holds. For any morphism $\phi\in\mbox{Hom}(A_{d+1},B_d)$ , we have

By the bilinear map

\begin{equation*}\mbox{Hom}(A_{d+1},B_d)\times \mbox{Hom}(B_d,\mathbb{S} A_{d+1})\longrightarrow k,\ \ (\phi,h)\longmapsto\eta_{A_{d+1}}(h\phi),\end{equation*}

is non-degenerate, we obtain $\omega_{d+1}\beta_n=\mathbb{S}\alpha_d\circ\omega_d$ . That is to say, the equality (3.9) holds.

Claim II: The following statements are equivalent.

$(\clubsuit)$ The equalities (3.10) and (3.11) hold.

$(\spadesuit)$ If there are morphisms $\varphi\colon A_0\to \Sigma^{-d}B_{d+1}$ and $\phi\colon A_{d+1}\to B_d$ such that $\Sigma^d\varphi\circ\alpha_{d+1}=\beta_d\phi$ , then

(3.12) \begin{equation}\begin{array}{l}\eta_{A_{d}}(\omega_{d}\phi\alpha_d)=(\!-\!1)^d\eta_{A_0}(\Sigma^{-d}\beta_{d+1}\circ\varphi).\end{array}\end{equation}

If $(\clubsuit)$ hold, it is obvious that the equality (3.12) also holds.

Conversely, assume that $(\spadesuit)$ holds. Put

\begin{equation*}V_1:=\{\Sigma^d\varphi\circ\alpha_{d+1}\ |\ \varphi\in\mbox{Hom}(A_0,\Sigma^{-d}B_{d+1})\}\subseteq\mbox{Hom}(A_{d+1},B_{d+1}),\end{equation*}

\begin{equation*}V_2:=\{\beta_d\phi\ |\ \phi\in\mbox{Hom}(A_{d+1},B_d)\}\subseteq\mbox{Hom}(A_{d+1},B_{d+1}).\end{equation*}

Take a set of basis elements $a_1,a_2,\cdots,a_m$ of $V_1\cap V_2$ , and expand them to a set of basis elements

\begin{equation*}a_1,\cdots,a_m,a_{m+1},\cdots,a_{m+t},\end{equation*}

of $V_1$ . At the same time, it is also extended to a set of basis elements

\begin{equation*}b_1=a_1,\cdots,b_m=a_m,b_{m+1},\cdots,b_{m+s},\end{equation*}

of $V_2$ . Take a set of basis elements $c_1,\cdots,c_n$ of $\mbox{Hom}(B_{n+1},\mathbb{S} A_{n+1})$ . Write

\begin{equation*}\eta_{A_{n+1}}(c_ja_i)=a_{ij},\ i=1,\cdots,m+t;\ j=1,\cdots,n,\end{equation*}

\begin{equation*}\eta_{A_{n+1}}(c_jbi)=b_{ij},\ i=1,\cdots,m+s;\ j=1,\cdots,n,\end{equation*}

\begin{equation*}M=(a_{ij})_{(m+t)\times n},\ N=(b_{ij})_{(m+s)\times n}.\end{equation*}

Thus the first m rows of the matrices of M and N are equal. Since $a_1,\cdots,a_m,a_{m+1},\cdots,a_{m+t}$ is a set of basis elements of $V_1$ , there are morphisms

\begin{equation*}\varphi_1,\cdots,\varphi_m,\varphi_{m+1},\cdots,\varphi_{m+t}\in\mbox{Hom}(A_0,\Sigma^{-d}B_{d+1}),\end{equation*}

such that

\begin{align*}a_1& =\Sigma^{d}\varphi_1\circ\alpha_{d+1},\cdots,a_m=\Sigma^{d}\varphi_m\circ\alpha_{d+1},a_{m+1}=\Sigma^{d}\varphi_{m+1}\circ\alpha_{d+1},\cdots,\\[3pt] a_{m+t}&=\Sigma^{d}\varphi_{m+t}\circ\alpha_{d+1}.\end{align*}

Put

\begin{equation*}d_i=(\!-\!1)^d\eta_{A_0}(\Sigma^{-d}\beta_{d+1}\circ\varphi),\ \ i=1,\cdots,m+t.\end{equation*}

Similarly, there are morphisms

\begin{equation*}\phi_1,\cdots,\phi_m,\phi_{m+1},\cdots,\phi_{m+t},\end{equation*}

such that

\begin{equation*}b_1=\beta_d\phi_1,\cdots,b_m=\beta_d\phi_m,\cdots,b_{m+1}=\beta_d\phi_{m+1},\cdots,b_{m+s}=\beta_d\phi_{m+s}.\end{equation*}

Put

\begin{equation*}e_i=\eta_{A_n}(\omega_d\phi_i\alpha_d),\ \ i=1,\cdots,m+s.\end{equation*}

Since $\beta_d\phi_i=b_i=a_i=\Sigma^d\varphi_i\circ\alpha_{d+1},\ \ i=1,\cdots,m.$ By the equality (3.12), we have

\begin{equation*}e_i=\eta_{A_d}(\omega_d\phi_i\alpha_d)=(\!-\!1)^d(\Sigma^{-d}\beta_{d+1}\circ\varphi_i)=d_i,\ \ i=1,\cdots,m.\end{equation*}

It follows that there exists a morphism $\omega_{d+1}\colon B_{d+1}\to \mathbb{S} A_d$ such that the equalities (3.10) and (3.11) hold if and only if the following two linear equations have solutions

\begin{equation*}M\left(\begin{matrix}x_1\\\vdots\\x_n \end{matrix} \right)=\left(\begin{matrix}d_1\\\vdots\\d_{m+t} \end{matrix} \right),\end{equation*}

\begin{equation*}N\left(\begin{matrix}x_1\\\vdots\\x_n \end{matrix} \right)=\left(\begin{matrix}e_1\\\vdots\\e_{m+s} \end{matrix} \right).\end{equation*}

Note that the first m rows of the matrices M and N are equal. Let $L=\left(\begin{matrix}M\\N'\end{matrix}\right)$ where N’ consists of the last s rows of N. Hence, the necessary and sufficient conditions for the two linear equations stated above to have a solution is that the following linear equations

(3.13) \begin{equation}\begin{array}{l}L\left(\begin{matrix}x_1\\\vdots\\x_n \end{matrix} \right)=\left(\begin{matrix}d_1\\\vdots\\d_m\\d_{m+1}\\ \vdots\\d_{m+t}\\e_{m+1}\\\vdots\\e_{m+s} \end{matrix} \right).\end{array}\end{equation}

have a solution. Since the bilinear map

\begin{equation*}\mbox{Hom}(A_{d+1},B_d)\times \mbox{Hom}(B_d,\mathbb{S} A_{d+1})\longrightarrow k,\ \ (u,v)\longmapsto\eta_{A_{d+1}}(vu)\end{equation*}

is non-degenerate, we get that L has row full rank, and thus, that (3.13) has a solution.

Step 3: We show that the equality (3.12) holds.

Assume that there are morphisms $\varphi\colon A_0\to \Sigma^{-d}B_{d+1}$ and $\phi\colon A_{d+1}\to B_d$ such that $\Sigma^d\varphi\circ\alpha_{d+1}=\beta_d\phi.$ Consider the following morphisms of $(d+2)$ -angles in $\mathscr{C}$ :

where $\delta_1,\delta_2,\cdots,\delta_d$ exist by the axiom (N3). It follows that

as we wished. This completes the proof.

4. Connection between Auslander–Reiten $(d+2)$ -angles and Serre functors

In this section, we build a link between Auslander–Reiten $(d+2)$ -angles and Serre functors.

Lemma 4.1 Let $\mathscr{C}$ be a $(d+2)$ -angulated category,

\begin{equation*}A_{\bullet}:\ \ {A_0 \xrightarrow{\ \alpha_0\ }A_1 \xrightarrow{\ \alpha_1\ } A_2 \xrightarrow{\ \alpha_2\ } \cdots \xrightarrow{\ \alpha_{d - 1}\ } A_d \xrightarrow{\ \alpha_{d}\ } A_{d+1}\xrightarrow{\ \alpha_{d+1}\ } \Sigma^d A_0},\end{equation*}

and

\begin{equation*}B_{\bullet}:\ \ {B_0 \xrightarrow{\ \beta_0\ }B_1 \xrightarrow{\ \beta_1\ } B_2 \xrightarrow{\ \beta_2\ } \cdots\xrightarrow{\ \beta_{d - 1}\ } B_d \xrightarrow{\ \beta_{d}\ } B_{d+1}\xrightarrow{\ \beta_{d+1}\ } \Sigma^d B_0},\end{equation*}

be two Auslander–Reiten $(d+2)$ -angles in $\mathscr{C}$ . Then $A_0\simeq B_0$ if and only if $A_{d+1}\simeq B_{d+1}$ .

Proof. Assume that $\varphi_{d+1}\colon A_{d+1}\to B_{d+1}$ is an isomorphism. Then $\varphi_{d+1}\alpha_d$ is not split epimorphism, otherwise, $\alpha_d$ is a split epimorphism and this is impossible. Since $\alpha_d$ is right almost split, there exists a morphism $\varphi_d\colon A_d\to B_d$ such that $\varphi_{d+1}\alpha_d=\beta_d\varphi_d$ . Thus we have the following commutative diagram

of $(d+2)$ -angles in $\mathscr{C}$ .

Since $A_{\bullet}$ and $B_{\bullet}$ are Auslander–Reiten $(d+2)$ -angles, by Remark 2.6, we have that $A_0$ and $B_0$ are indecomposable objects. If $\varphi_0$ is not an isomorphism, we obtain that $\varphi_0$ is not a split monomorphism since $A_0$ and $B_0$ are indecomposable. Because $\alpha_0$ is left almost split, there exists a morphism $h\colon A_1\to B_0$ such that $\varphi_0=h\alpha_0$ . Hence $\beta_{d+1}\varphi_{d+1}=(\Sigma^d\varphi_0)\alpha_{d+1}=\Sigma^dh(\Sigma^d\alpha_0)\alpha_{d+1}=0$ which implies $\beta_{d+1}=0$ . This is a contradiction since $B_{\bullet}$ is an Auslander-Reiten $(d+2)$ -angles in $\mathscr{C}$ . Hence $\varphi_0\colon A_0\to B_0$ is an isomorphism.

Remark 4.2 Let $\mathscr{C}$ be a $(d+2)$ -angulated category, and

\begin{equation*} {A_0 \xrightarrow{\ \alpha_0\ }A_1 \xrightarrow{\ \alpha_1\ } A_2 \xrightarrow{\ \alpha_2\ } \cdots \xrightarrow{\ \alpha_{d - 1}\ } A_d \xrightarrow{\ \alpha_{d}\ } A_{d+1}\xrightarrow{\ \alpha_{d+1}\ } \Sigma^d A_0},\end{equation*}

be an Auslander-Reiten $(d+2)$ -angle in $\mathscr{C}$ . By Lemma 4.1, we know that $A_0$ is uniquely determined up to isomorphism. In this case, we write $A_0=\tau_dA_{d+1}$ .

In order to prove the main result of this section, we need the following lemma.

Lemma 4.3 Let $\mathscr{C}$ be a $(d+2)$ -angulated category. Suppose that

\begin{equation*} {\tau_dC \xrightarrow{\ \alpha_0\ }C_1 \xrightarrow{\ \alpha_1\ } C_2 \xrightarrow{\ \alpha_2\ } \cdots \xrightarrow{\ \alpha_{d - 1}\ } C_d \xrightarrow{\ \alpha_{d}\ } C\xrightarrow{\ \alpha_{d+1}\ } \Sigma^d\tau_dC,}\end{equation*}

is an Auslander–Reiten $(d+2)$ -angle in $\mathscr{C}$ and B is a indecomposable object in $\mathscr{C}$ . Then the following hold:

1. (1) For any non-zero $g\in \mbox{Hom}_{\mathscr{C}}(B,\tau_d\Sigma^dC)$ , there is an $f\in Hom_{\mathscr{C}}(C,B)$ such that $\alpha_{d+1}=gf$ .

2. (2) For any non-zero $f\in \mbox{Hom}_{\mathscr{C}}(C,B)$ , there is a $g\in \mbox{Hom}_{\mathscr{C}}(B,\tau_d\Sigma^dC)$ such that $\alpha_{d+1}=gf$ .

Proof.

1. (1) For any non-zero morphism $g\colon B\to\tau_d\Sigma^dC$ , there exists a $(d+2)$ -angle

   \begin{equation*} {\tau_dC \xrightarrow{\ \beta_0\ }B_1 \xrightarrow{\ \beta_1\ } B_2 \xrightarrow{\ \beta_2\ } \cdots \xrightarrow{\ \beta_{d - 1}\ } B_d \xrightarrow{\ \beta_{d}\ } B\xrightarrow{\ g\ } \Sigma^d\tau_dC,}\end{equation*}
   in $\mathscr{C}$ . Since g is non-zero, then $\beta_0$ is not split monomorphism. It follows that there exists a morphism $h\colon C_1\to B_1$ such that $\beta_0=h\alpha_0$ . Thus we have the following commutative diagram
   

   of $(d+2)$ -angles in $\mathscr{C}$ . We obtain $\alpha_{d+1}=gf$ .

2. (2) For any non-zero morphism $f\colon C\to B$ , there exists a $(d+2)$ -angle

   \begin{equation*} {A_0 \xrightarrow{\ \gamma_0\ }A_1 \xrightarrow{\ \gamma_1\ } \cdots \xrightarrow{\ \gamma_{d-2}\ }A_{d-1}\xrightarrow{\ \gamma_{d-1}\ } C \xrightarrow{\ f\ } B\xrightarrow{\ \gamma\ } \Sigma^d A_0},\end{equation*}
   in $\mathscr{C}$ . Since $f\gamma_{d-1}=0$ and $f\neq 0$ , we have that $\gamma_{d-1}$ is not split epimorphism. Since $\alpha_d$ is right almost split, there exists a morphism $u\colon A_{d-1}\to C_d$ such that $\alpha_du=\gamma_{d-1}$ and then
   \begin{equation*}\alpha_{d+1}\gamma_{d-1}=\alpha_{d+1}\alpha_du=0.\end{equation*}

So there exists a morphism $g\colon B\to \Sigma^d\tau_dC$ such that $\alpha_{d+1}=gf$ .

The following lemma can be found in [Reference I. and M.15, Proposition I.1.4].

Lemma 4.4 Let $\mathscr{C}$ be an additive category and $\mathbb{S}\colon \mathscr{C}\to\mathscr{C}$ be an additive functor. If there exists a k-linear map $\eta_X\colon\textrm{Hom}(X,\mathbb{S} X)\longrightarrow k$ such that the bilinear map

\begin{equation*}(-,-)\colon\textrm{Hom}(X,Y)\times\textrm{Hom}(Y,\mathbb{S} X)\longrightarrow k,\ \ (f,g)=\eta_X(gf),\end{equation*}

is a non-degenerate pairing for all $X,Y\in\mathscr{C}$ , then $\mathbb{S}$ can be viewed as a right Serre functor, where

\begin{equation*}\eta_{X,Y}\colon\textrm{Hom}(X,Y)\longrightarrow \textrm{Hom}(Y,\mathbb{S} Y)^{\ast},\end{equation*}

given by

\begin{equation*}\eta_{X,Y}(f)(g)=\eta_X(gf),\ \mbox{for all}\ f \in \mbox{Hom}(X,Y), g\in \mbox{Hom}(Y,\mathbb{S} X).\end{equation*}

Now we state and prove our second main result. This generalizes the work by Reiten and Van den Bergh [Reference I. and M.15, Theorem I.2.4] for triangulated categories.

If either of these properties holds, then the action of the Serre functor on objects coincides with $\tau_d\Sigma^d$ , namely $\tau_d=\mathbb{S}\Sigma^{-d}$ . In this case, $\tau_d$ is called the d-Auslander-Reiten translation.

Proof. We first show the “only if” part. Suppose that $\mathscr{C}$ has Auslander–Reiten $(d+2)$ -angles. For any indecomposable object $C\in\mathscr{C}$ , put $\mathbb{S} C=\tau_d \Sigma^d C$ . Then there exists an Auslander-Reiten $(d+2)$ -angle

\begin{equation*} {\tau_dC \xrightarrow{\ \alpha_0\ }C_1 \xrightarrow{\ \alpha_1\ } C_2 \xrightarrow{\ \alpha_2\ } \cdots \xrightarrow{\ \alpha_{d - 1}\ } C_d \xrightarrow{\ \alpha_{d}\ } C\xrightarrow{\ \alpha_{d+1}\ } \Sigma^d\tau_dC.}\end{equation*}

Thus $\alpha_{d+1}\neq 0$ . For the indecomposable object C, take a k-linear map

\begin{equation*}\eta_{C}\colon \mbox{Hom}(C,\mathbb{S} C)\to k,\end{equation*}

such that $\eta_{C}(\alpha_{d+1})\neq0$ . By Lemma 4.3, for any indecomposable object $B\in\mathscr{C}$ , the pairing

\begin{equation*}\mbox{Hom}(C,B)\times\mbox{Hom}(B,\mathbb{S} C)\to k \ \textrm{given by}\ (f,g)=\eta_{C}(gf),\end{equation*}

is a non-degenerate k-bilinear map. By Lemma 4.4, $\mathbb{S}$ can be viewed as a right Serre functor. Dually, we can prove that $\mathscr{C}$ has a left Serre functor.

We now prove the other implication. Assume that $\mathscr{C}$ has a Serre functor $\mathbb{S}$ . Then for any indecomposable object C, there exists a k-linear isomorphism

\begin{equation*}(\eta^{\ast}_{C,C})^{-1}\colon \mbox{End}(C)^{\ast}\to \mbox{Hom}(C,\mathbb{S} C).\end{equation*}

Since $\mbox{End}(C)$ is local and k is an algebraically closed field, we have $\mbox{End}(C)/\mbox{rad}\mbox{End}(C)\simeq k$ . Then $\theta_C\colon \mbox{End}(C)\to \mbox{End}(C)/\mbox{rad}\mbox{End}(C)\simeq k$ is a natural projection with $\mbox{Ker}\ \theta_C=\mbox{rad}\mbox{End}(C)$ . For

\begin{equation*}\delta=(\eta^{\ast}_{C,C})^{-1}(\theta_C)\in\mbox{Hom}(C,\mathbb{S} C),\end{equation*}

we have $\delta\neq0$ and so $\alpha_d$ is not a split epimorphism. Consider a $(d+2)$ -angle extending $\delta$ :

\begin{equation*} {\Sigma^{-d}\mathbb{S} C \xrightarrow{\ \alpha_0\ }C_1 \xrightarrow{\ \alpha_1\ } C_2 \xrightarrow{\ \alpha_2\ } \cdots \xrightarrow{\ \alpha_{d-2}\ } C_{d-1}\xrightarrow{\ \alpha_{d-1}\ } C_d \xrightarrow{\ \alpha_d\ } C\xrightarrow{\ \delta\ } \mathbb{S} C}\end{equation*}

where when $d\geq2$ , we can choose $\alpha_1,\alpha_2,\cdots,\alpha_{d-1}$ in $\mbox{rad}_{\mathscr{C}}$ by Lemma 2.9.

We claim that

\begin{equation*} {\Sigma^{-d}\mathbb{S} C \xrightarrow{\ \alpha_0\ }C_1 \xrightarrow{\ \alpha_1\ } C_2 \xrightarrow{\ \alpha_2\ } \cdots \xrightarrow{\ \alpha_{d-2}\ } C_{d-1}\xrightarrow{\ \alpha_{d-1}\ } C_d \xrightarrow{\ \alpha_d\ } C\xrightarrow{\ \delta\ } \mathbb{S} C,}\end{equation*}

is an Auslander-Reiten $(d+2)$ -angle in $\mathscr{C}$ . We first show that $\alpha_d$ is a right almost split morphism. Let $\beta\colon B\to C$ not be a split epimorphism in $\mathscr{C}$ . Then the composition

\begin{equation*}\mbox{Hom}(C,B)\xrightarrow{\mbox{Hom}(C,\ \beta)}\mbox{End}(C)\xrightarrow{\ \theta_Z\ }k,\end{equation*}

is zero. Consider the following commutative diagram

we can obtain

\begin{equation*}\delta\beta=(\eta^{\ast}_{C,C})^{-1}(\theta_Z)\beta=(\eta^{\ast}_{C,B})^{-1}\mbox{Hom}(C, \beta )^\ast(\theta_C)=(\eta^{\ast}_{C,B})^{-1}(\theta_C)\mbox{Hom}(C, \beta)=0.\end{equation*}

So there exists a morphism $h\colon C\to C_d$ such that $\beta=\alpha_dh$ . This shows that $\alpha_d$ is a right almost split morphism.

Since C is an indecomposable object, we have that $\Sigma^{-d}\mathbb{S} C$ is also indecomposable which implies that $\mbox{End}(\Sigma^{-d}\mathbb{S} C)$ is local. By Lemma 2.8, we know that

\begin{equation*} {\Sigma^{-d}\mathbb{S} C \xrightarrow{\ \alpha_0\ }C_1 \xrightarrow{\ \alpha_1\ } C_2 \xrightarrow{\ \alpha_2\ } \cdots \xrightarrow{\ \alpha_{d-2}\ } C_{d-1}\xrightarrow{\ \alpha_{d-1}\ } C_d \xrightarrow{\ \alpha_d\ } C\xrightarrow{\ \delta\ } \mathbb{S} C,}\end{equation*}

is an Auslander–Reiten $(d+2)$ -angle in $\mathscr{C}$ .

Dually, we can show that for any indecomposable object $C\in\mathscr{C}$ , there exists an Auslander–Reiten $(d+2)$ -angle $ {C \xrightarrow{\ \beta_0\ }C_1 \xrightarrow{\ \beta_1\ } C_2 \xrightarrow{\ \beta_2\ } \cdots \xrightarrow{\ \beta_{d-1}\ } C_d \xrightarrow{\ \beta_d\ } \mathbb{S}^{-1}\Sigma^dC\xrightarrow{\ \beta_{d+1}\ } \Sigma^dC}$ .

Now we give an example illustrating our main result in this section.

Example 4.6 We first recall the standard construction of $(d+2)$ -angulated categories given by Geiss–Keller–Oppermann [Reference C., B. and S.8, Theorem 1]. Let $\mathscr{C}$ be a triangulated category and $\mathcal{T}$ a d-cluster tilting subcategory which is closed under $\Sigma^{d}$ , where $\Sigma$ is the shift functor of $\mathscr{C}$ . Then $(\mathcal{T},\Sigma^{d},\Theta)$ is a $(d+2)$ -angulated category, where $\Theta$ is the class of all sequences

\begin{equation*}A_0\xrightarrow{f_0}A_1\xrightarrow{f_1}A_2\xrightarrow{f_2}\cdots\xrightarrow{f_{d-1}}A_d\xrightarrow{f_d}A_{d+1}\xrightarrow{f_{d+1}}\Sigma^{d} A_0,\end{equation*}

such that there exists a diagram

with $A_i\in\mathcal{T}$ for all $i\in\mathbb{Z}$ , such that all oriented triangles are triangles in $\mathscr{C}$ , all non-oriented triangles commute, and $f_{d+1}$ is the composition along the lower edge of the diagram.

Assume that $\mathscr{C}$ has a Serre functor $\mathbb{S}$ . By [Reference O. and Y.9, Proposition 3.4], we know that $\mathcal{T}=\mathbb{S}\Sigma^{-d}\mathcal T=\Sigma^d\mathbb{S}^{-1}\mathcal T$ . By hypothesis, $\Sigma^d\mathcal T=\mathcal T$ . It follows that $\mathbb{S}\mathcal T=\mathcal T=\mathbb{S}^{-1}\mathcal T$ . In particular, $\mathcal T$ has an auto-equivalence functor $\mathbb{S}$ . Hence $(\mathcal{T},\Sigma^{d},\Theta)$ has a Serre functor $\mathbb{S}$ . By Theorem 4.5, we know that $\mathcal T$ has Auslander-Reiten $(d+2)$ -angles and $\tau_d=\mathbb{S}\Sigma^{-d}$ .

5. $(\textbf{d}+2)$ -angulated quotient categories

Let $\mathscr{C}$ be an additive category, and $\mathscr{X}$ be a subcategory of $\mathscr{C}$ . Recall that we say a morphism $f\colon A \to B$ in $\mathscr{C}$ is an $\mathscr{X}$ -monic if

\begin{equation*}\mbox{Hom}_{\mathscr{C}}(f,X)\colon \mbox{Hom}_{\mathscr{C}}(B,X) \to \mbox{Hom}_{\mathscr{C}}(A,X),\end{equation*}

is an epimorphism for all $X\in\mathscr{X}$ . We say that f is an $\mathscr{X}$ -epic if

\begin{equation*}\mbox{Hom}_{\mathscr{C}}(X,f)\colon \mbox{Hom}_{\mathscr{C}}(X,A) \to \mbox{Hom}_{\mathscr{C}}(X,B),\end{equation*}

is an epimorphism for all $X\in\mathscr{X}$ . Furthermore, we say that f is a left $\mathscr{X}$ -approximation of A if f is an $\mathscr{X}$ -monoic and $B\in\mathscr{X}$ . We say that f is a right $\mathscr{X}$ -approximation of B if f is an $\mathscr{X}$ -epic and $A\in\mathscr{X}$ .

A subcategory $\mathscr{X}$ is called contravariantly finite if any object in $\mathscr{C}$ admits a right $\mathscr{X}$ -approximation. Dually we can define covariantly finite subcategory.

Definition 5.1 Let $\mathscr{C}$ be a $(d+2)$ -angulated category. A subcategory $\mathscr{X}$ of $\mathscr{C}$ is called strongly contravariantly finite, if for any object $C\in\mathscr{C}$ , there exists a $(d+2)$ -angle

\begin{equation*}B\xrightarrow{}X_1\xrightarrow{}X_2\xrightarrow{}\cdots\xrightarrow{}X_{d-1}\xrightarrow{}X_{d}\xrightarrow{\ g\ }C\xrightarrow{\ \ }\Sigma^dB,\end{equation*}

where g is a right $\mathscr{X}$ -approximation of C and $X_1,\cdots,X_d\in\mathscr{X}$ . Dually we can define a strongly covariantly finite subcategory.

A strongly contravariantly finite and strongly covariantly finite subcategory is called strongly functorially finite.

Proof. Since the proof is similar to [Reference Z.12, Theorem 3.7], we omit it. See also [Reference Z.13, Remark 3.8].

The following is just the dual statement of Lemma 5.2.

The notion of mutation pairs of subcategories in a $(d+2)$ -angulated category was defined by Lin [Reference Z.12, Definition 3.1]. We recall the definition here.

Let $\mathscr{C}$ be a $(d+2)$ -angulated category. Recall that a subcategory $\mathcal{A}$ of $\mathscr{C}$ is called extension closed if for any morphism $\alpha_{d+1}\colon A_{d+1}\to\Sigma A_0$ with $A_0,A_{d+1}\in\mathcal{A}$ , there exists a $(d+2)$ -angle

\begin{equation*}A_0\xrightarrow{\alpha_0}A_1\xrightarrow{\alpha_1}A_2\xrightarrow{\alpha_2}\cdots\xrightarrow{\alpha_{d-2}}A_{d-1}\xrightarrow{\alpha_{d-1}}A_d\xrightarrow{\alpha_d}A_{d+1}\xrightarrow{\alpha_{d+1}}\Sigma^d A_0.\end{equation*}

where each $A_i\in\mathcal{A}$ . It is clear that $\mathscr{C}$ is extension closed in $\mathscr{C}$ .

Lemma 5.6 Let $\mathscr{C}$ be a $(d+2)$ -angulated category with a Serre functor $\mathbb{S}$ , $\mathscr{X}$ be a subcategory of $\mathscr{C}$ and let

(5.1) \begin{equation}\begin{array}{l} {A_0 \xrightarrow{\ \alpha_0\ }A_1 \xrightarrow{\ \alpha_1\ } A_2 \xrightarrow{\ \alpha_2\ } \cdots \xrightarrow{\ \alpha_{d - 1}\ } A_d \xrightarrow{\ \alpha_{d}\ } A_{d+1}\xrightarrow{\ \alpha_{d+1}\ } \Sigma^d A_0,}\end{array}\end{equation}

be a $(d+2)$ -angle. Assume that $\tau_d\mathscr{X}=\mathscr{X}$ . Then $\alpha_0$ is an $\mathscr{X}$ -monic if and only if $\alpha_d$ is an $\mathscr{X}$ -epic.

Proof. Apply the functor $\mbox{Hom}(X,-)$ with $X\in\mathscr{X}$ to the $(d+2)$ -angle (5.1), we have the following long exact sequence:

\begin{equation*}\mbox{Hom}(X,A_d)\xrightarrow{\ }\mbox{Hom}(X,A_{d+1})\xrightarrow{\ }\mbox{Hom}(X,\Sigma^dA_0)\xrightarrow{\ }\mbox{Hom}(X,\Sigma^dA_1),\end{equation*}

For $\alpha_d\colon A_d\to A_{d+1}$ to be an $\mathscr{X}$ -epic is the same as for the first arrow in the long exact sequence always to be epimorphism. This is the same as for the second arrow always to be zero, which is again the same as for the third arrow always to be injective.

Using Serre duality, the third arrow can be identified with

\begin{equation*}\mbox{Hom}(A_0,\Sigma^{-d}\mathbb{S}X)^{\ast}\longrightarrow\mbox{Hom}(A_1,\Sigma^{-d}\mathbb{S}X)^{\ast},\end{equation*}

which is monomorphism if and only if

is epimorphism. For this always to be an epimorphism is the same as for $\alpha_0\colon A_0\to A_1$ to be a $(\tau\mathscr{X})$ -monic, that is, an $\mathscr{X}$ -monic.

Proof.

1. (1) $\Longrightarrow$ (2). By Lemma 5.5, we have that $\mathscr{C}/\mathscr{X}$ is a $(d+2)$ -angulated category.

2. (2) $\Longrightarrow$ (1). Since $\mathscr{X}$ is strongly covariantly finite, for any object $A\in\mathscr{C}$ , by Lemma 5.2, there exists a $(d+2)$ -angle in $\mathscr{C}$ :

   \begin{equation*} {A\xrightarrow{\ \alpha_0\ }X_1 \xrightarrow{\ \alpha_1\ } X_2 \xrightarrow{\ \alpha_2\ } \cdots \xrightarrow{\ \alpha_{d - 1}\ } X_d \xrightarrow{\ \alpha_{d}\ }\mathbb{H}A\xrightarrow{\ \alpha_{d+1}\ } \Sigma^d A,}\end{equation*}

   where $X_1,\cdots,X_d\in\mathscr{X}$ and $\alpha_0$ is a left $\mathscr{X}$ -approximation of A. Since $\mathscr{X}$ is strongly contravariantly finite, for the object $\mathbb{H}A\in\mathscr{C}$ , by Lemma 5.3, there exists a $(d+2)$ -angle in $\mathscr{C}$ :

   \begin{equation*} {\mathbb{L}\mathbb{H}A\xrightarrow{\ \beta_0\ }Y_1 \xrightarrow{\ \beta_1\ } Y_2 \xrightarrow{\ \beta_2\ } \cdots \xrightarrow{\ \beta_{d - 1}\ } Y_d \xrightarrow{\ \beta_{d}\ }\mathbb{H}A\xrightarrow{\ \beta_{d+1}\ } \Sigma^d \mathbb{L}\mathbb{H}A,}\end{equation*}
   where $Y_1,\cdots,Y_d\in\mathscr{X}$ and $\beta_d$ is a right $\mathscr{X}$ -approximation of $\mathbb{H}A$ .

Since $\beta_d$ is a right $\mathscr{X}$ -approximation of $\mathbb{H}A$ and $X_d\in\mathscr{X}$ , then there exists a morphism $\varphi_d\colon X_d\to Y_d$ such that $\beta_d\varphi_d=\alpha_d$ . Thus we have the following morphisms of $(d+2)$ -angles

Since $\mathscr{C}/\mathscr{X}$ is a $(d+2)$ -angulated category, then the morphism $\overline{\varphi}\colon A\to\mathbb{L}\mathbb{H}A$ is invertible in $\mathscr{C}/\mathscr{X}$ . It follows that there exists a morphism $\varphi'\colon\mathbb{L}\mathbb{H}A\to A$ such that $\overline{\varphi'}\circ\overline{\varphi}=\textrm{Id}_A$ and then $\textrm{Id}_A-\varphi'\varphi$ factorizes through an object in $\mathscr{X}$ . So it factorizes through the left $\mathscr{X}$ -approximation $\alpha_0\colon A\to X_1$ of A. Namely there exists a morphism $u\colon X_1\to A$ such that $\textrm{Id}_A-\varphi'\varphi=u\alpha_0$ . Note that $(\textrm{Id}_A-\varphi'\varphi)\circ\Sigma^{-d}\alpha_{d+1}=u(\alpha_0\circ\Sigma^{-d}\alpha_{d+1})=0$ and then

\begin{equation*}\Sigma^{-d}\alpha_{d+1}=(\varphi'\varphi)\circ\Sigma^{-d}\alpha_{d+1}=\varphi'\circ\Sigma^{-d}(\beta_{d+1}).\end{equation*}

Hence $\alpha_{d+1}=\Sigma^d\varphi'\circ\beta_{d+1}$ and we also have the following morphisms of $(d+2)$ -angles

Now we show that $\alpha_d$ is a right $\mathscr{X}$ -approximation of $\mathbb{H}A$ . Indeed, for any morphism $a\colon X\to \mathbb{H}A$ with $X\in\mathscr{X}$ , since $\beta_d$ is a right $\mathscr{X}$ -approximation of $\mathbb{H}A$ , there exists a morphism $b\colon X\to Y_d$ such that $\beta_db=a$ . It follows that $a=\beta_db=\alpha_d(\phi_db)$ . This shows that $\alpha_d$ is a right $\mathscr{X}$ -approximation of $\mathbb{H}A$ .

Similarly, one can show that for any object $C\in\mathscr{C}$ , there exists a $(d+2)$ -angle in $\mathscr{C}$ :

\begin{equation*}{D\xrightarrow{\ \gamma_0\ }Z_1 \xrightarrow{\ \gamma_1\ } Z_2 \xrightarrow{\ \gamma_2\ } \cdots \xrightarrow{\ \gamma_{d - 1}\ } Z_d \xrightarrow{\ \gamma_{d}\ }C\xrightarrow{\ \gamma_{d+1}\ } \Sigma^d A,}\end{equation*}

where $Z_1,\cdots,Z_d\in\mathscr{X}$ , $\gamma_0$ is a left $\mathscr{X}$ -approximation and $\alpha_d$ is a right $\mathscr{X}$ -approximation.

This shows that $(\mathscr{C},\mathscr{C})$ is an $\mathscr{X}$ -mutation pair.

(1) $\Longrightarrow$ (3). It suffices to show that $\mathscr{X}\subseteq\mathbb{S}\Sigma^{-d}\mathscr{X}$ and $\Sigma^d\mathbb{S}^{-1}\subseteq\mathscr{X}$ . We only show that $\mathscr{X}\subseteq\mathbb{S}\Sigma^{-d}\mathscr{X}$ , dually, we can show that $\Sigma^d\mathbb{S}^{-1}\subseteq\mathscr{X}$ .

For any indecomposable object $X\in\mathscr{X}$ , then $\Sigma^d\mathbb{S}^{-1}X$ is also indecomposable object. Since $\mathscr{C}$ has a Serre functor $\mathbb{S}$ , by Theorem 4.5, there exists an Auslander-Reiten $(d+2)$ -angle

\begin{equation*} {X\xrightarrow{\ \alpha_0\ }A_1 \xrightarrow{\ \alpha_1\ } \cdots \xrightarrow{\ \alpha_{d-2}\ }A_{d-1}\xrightarrow{\ \alpha_{d-1}\ }A_d \xrightarrow{\ \alpha_d\ }\Sigma^d\mathbb{S}^{-1}X\xrightarrow{\ \alpha_{d+1}\ } \Sigma^d X,}\end{equation*}

Since $(\mathscr{C},\mathscr{C})$ is an $\mathscr{X}$ -mutation pair, then for the object $\Sigma^d\mathbb{S}^{-1}X$ , there exists a $(d+2)$ -angle

\begin{equation*} {B\xrightarrow{\ \beta_0\ }X_1 \xrightarrow{\ \beta_1\ } X_2 \xrightarrow{\ \beta_2\ } \cdots \xrightarrow{\ \beta_{d - 1}\ } X_d \xrightarrow{\ \beta_{d}\ }\Sigma^d\mathbb{S}^{-1}X\xrightarrow{\ \beta_{d+1}\ } \Sigma^d B,}\end{equation*}

where $X_1,\cdots,X_d\in\mathscr{X}$ , $\beta_0$ is a left $\mathscr{X}$ -approximation and $\beta_d$ is a right $\mathscr{X}$ -approximation.

If $\Sigma^d\mathbb{S}^{-1}X\notin\mathscr{X}$ , then $\alpha_d\colon X_d\to\Sigma^d\mathbb{S}^{-1}X$ is not a split epimorphism. Thus there exists a morphism $u\colon X_n\to A_n$ such that $\beta_d=\alpha_du$ . It follows that $\alpha_{d+1}\beta_d=\alpha_{d+1}\alpha_du=0$ . So there exists a morphism $v\colon \Sigma^dB\to\Sigma^dX$ such that $v\beta_{d+1}=\alpha_{d+1}$ . Since $\beta_0$ is a left $\mathscr{X}$ -approximation and $X\in\mathscr{X}$ , there exists a morphism $w\colon X_1\to X$ such that $w\beta_0=\Sigma^{-d}v$ and then $v=\Sigma^dw\circ\Sigma^d\beta_0$ . Hence $\alpha_{d+1}=v\beta_{d+1}=\Sigma^dw\circ(\Sigma^d\beta_0\circ\beta_{d+1})=0$ . This is a contradiction.

Thus we obtain $\Sigma^d\mathbb{S}^{-1}X\in\mathscr{X}$ and $X\in\mathbb{S}\Sigma^{-d}\mathscr{X}$ . Since $\mathscr{C}$ is a Krull-Schmidt category we infer that $\mathscr{X}\subseteq\mathbb{S}\Sigma^{-d}\mathscr{X}$ .

(3) $\Longrightarrow$ (1). It follows from Lemma 5.6.

In Theorem 5.7, if $d=1$ , we have the following result.

Now we give an example illustrating our main result in this section.

We describe the Auslander–Reiten quiver of $\mathcal{T}$ in the following:

It is straightforward to verify that $\mathscr{C}:=\mbox{add}(S_3\oplus P_1\oplus S_1)$ is a 2-cluster tilting subcategory of $\mathcal{T}$ . Moreover, $\mathscr{C}[2]=\mathscr{C}$ . By [Reference C., B. and S.8, Theorem 1], we know that $(\mathscr{C},[2])$ is a 4-angulated category with a Serre functor $\mathbb{S}=[2]$ . Let $\mathscr{X}=\mbox{add}(S_3\oplus S_1)$ . Then the 4-angle

\begin{equation*}P_1\xrightarrow{\ \ }S_1\xrightarrow{\ \ }S_3\xrightarrow{\ \ }P_1\xrightarrow{\ \ }P_1[2],\end{equation*}

shows that $\mathscr{X}$ is strongly functorially finite subcategory of $\mathscr{C}$ and $(\mathscr{C},\mathscr{C})$ is an $\mathscr{X}$ -mutation pair. By Theorem 5.7, we obtain that $\mathscr{C}/\mathscr{X}$ is 4-angulated category and $\tau_2\mathscr{X}=\mathscr{X}$ .