Twisted algebras of geometric algebras

Abstract

A twisting system is one of the major tools to study graded algebras; however, it is often difficult to construct a (nonalgebraic) twisting system if a graded algebra is given by generators and relations. In this paper, we show that a twisted algebra of a geometric algebra is determined by a certain automorphism of its point variety. As an application, we classify twisted algebras of three-dimensional geometric Artin–Schelter regular algebras up to graded algebra isomorphism.

1 Introduction

The notion of twisting system was introduced by Zhang in [14]. If there is a twisting system $\theta =\{ \theta _n \}_{n \in \mathbb {Z}}$ for a graded algebra A, then we can define a new graded algebra $A^{\theta }$ , called a twisted algebra. Zhang gave a necessary and sufficient algebraic condition via a twisting system when two categories of graded right modules are equivalent [14, Theorem 3.5]. Although a twisting system plays an important role to study a graded algebra, it is often difficult to construct a twisting system on a graded algebra if it is given by generators and relations.

Mori introduced the notion of geometric algebra $\mathcal {A}(E,\sigma )$ , which is determined by geometric data which consist of a projective variety E, called the point variety, and its automorphism $\sigma $ . For these algebras, Mori gave a necessary and sufficient geometric condition when two categories of graded right modules are equivalent [11, Theorem 4.7]. By using this geometric condition, we can easily construct a twisting system.

Cooney and Grabowski defined a groupoid whose objects are geometric noncommutative projective spaces and whose morphisms are isomorphisms between them. By studying a correspondence between the morphisms of this groupoid and a twisting system, they showed that the morphisms of this groupoid are parametrized by a set of certain automorphisms of the point variety [2, Theorem 28].

In this paper, we focus on studying a twisted algebra of a geometric algebra $A=\mathcal {A}(E,\sigma )$ . For a twisting system $\theta $ on A, we set $\Phi (\theta ):=\overline {(\theta _{1}|_{A_{1}})^{\ast }} \in \mathrm {Aut}_{k}\,\mathbb {P}(A_1^{\ast })$ by dualization and projectivization. We find a subset $M(E,\sigma )$ of $\mathrm {Aut}_{k}\,\mathbb {P}(A_{1}^{\ast })$ parametrizing twisted algebras of A up to isomorphism. As an application to three-dimensional geometric Artin–Schelter regular algebras, we will compute $M(E,\sigma )$ (see Theorems 4.8 and 4.11), which completes the classification of twisted algebras of three-dimensional geometric Artin–Schelter regular algebras.

Itaba and the author showed that for any three-dimensional quadratic Artin–Schelter regular algebra B, there are a three-dimensional quadratic Calabi–Yau Artin–Schelter regular algebra S and a twisting system $\theta $ such that $B \cong S^{\theta }$ [8, Theorem 4.4]. Except for one case, a twisting system $\theta $ can be induced by a graded algebra automorphism of S. By using $M(E,\sigma )$ , we can recover this fact in the case that B is geometric (see Corollary 4.12).

2 Preliminary

Throughout this paper, we fix an algebraically closed field k of characteristic zero and assume that a graded algebra is an $\mathbb {N}$ -graded algebra $A=\bigoplus _{i \in \mathbb {N}} A_i$ over k. A graded algebra $A=\bigoplus _{i \in \mathbb {N}} A_i$ is called connected if $A_0=k$ . Let $\mathrm {GrAut}_{k}\,A$ denote the group of graded algebra automorphisms of A. We denote by $\mathrm {GrMod} A$ the category of graded right A-modules. We say that two graded algebras A and $A'$ are graded Morita equivalent if two categories $\mathrm {GrMod} A$ and $\mathrm {GrMod} A'$ are equivalent.

2.1 Twisting systems and twisted algebras

Definition 2.1 Let A be a graded algebra. A set of graded k-linear automorphisms $\theta =\{ \theta _n \}_{n \in \mathbb {Z}}$ of A is called a twisting system on A if

$$ \begin{align*} \theta_n(a\theta_m(b))=\theta_n(a)\theta_{n+m}(b) \end{align*} $$

for any $l,m,n \in \mathbb {Z}$ and $a \in A_m$ , $b \in A_l$ . The twisted algebra of A by $\theta $ , denoted by $A^{\theta }$ , is a graded algebra A with a new multiplication $\ast $ defined by

$$ \begin{align*}a \ast b =a\theta_m(b)\end{align*} $$

for any $a \in A_m$ , $b \in A_l$ . A twisting system $\theta =\{ \theta _n \}_{n \in \mathbb {Z}}$ is called algebraic if $\theta _{m+n}=\theta _m \circ \theta _n$ for every $m,n \in \mathbb {Z}$ .

We denote by $\mathrm {TS}^{\mathbb {Z}}(A)$ the set of twisting systems on A. Zhang [14] found a necessary and sufficient algebraic condition for $\mathrm {GrMod} A \cong \mathrm {GrMod} A'$ .

Theorem 2.2 [14, Theorem 3.5]

Let A and $A'$ be graded algebras finitely generated in degree $1$ over k. Then $\mathrm {GrMod} A \cong \mathrm {GrMod} A'$ if and only if $A'$ is isomorphic to a twisted algebra $A^{\theta }$ by a twisting system $\theta \in \mathrm {TS}^{\mathbb {Z}}(A)$ .

Definition 2.3 For a graded algebra A, we define

$$ \begin{align*} &\mathrm{TS}^{\mathbb{Z}}_{0}(A):=\{ \theta \in \mathrm{TS}^{\mathbb{Z}}(A) \mid \theta_0=\mathrm{id}_A \}, \\ &\mathrm{TS}_{\mathrm{alg}}^{\mathbb{Z}}(A):=\{ \theta \in \mathrm{TS}^{\mathbb{Z}}_{0}(A) \mid \theta \text{ is algebraic} \}, \\ &\mathrm{Twist}(A):=\{ A^{\theta} \mid \theta \in \mathrm{TS}^{\mathbb{Z}}(A) \}/_{\cong},\\ &\mathrm{Twist}_{\mathrm{alg}}(A):=\{ A^{\theta} \mid \theta \in \mathrm{TS}^{\mathbb{Z}}_{\mathrm{alg}}(A) \}/_{\cong}. \end{align*} $$

Lemma 2.4 [14, Proposition 2.4]

Let A be a graded algebra. For every $\theta \in \mathrm {TS}^{\mathbb {Z}}(A)$ , there exists $\theta ' \in \mathrm {TS}^{\mathbb {Z}}_{0}(A)$ such that $A^{\theta } \cong A^{\theta '}$ .

It follows from Lemma 2.4 that $\mathrm {Twist}(A)=\{ A^{\theta } \mid \theta \in \mathrm {TS}^{\mathbb {Z}}_{0}(A) \}/_{\cong }$ , so we may assume that $\theta \in \mathrm {TS}^{\mathbb {Z}}_{0}(A)$ to study $\mathrm {Twist}(A)$ . By the definition of twisting system, it follows that $\theta \in \mathrm {TS}^{\mathbb {Z}}_{\mathrm {alg}}(A)$ if and only if $\theta _n=\theta _{1}^{n}$ for every $n \in \mathbb {Z}$ and $\theta _{1} \in \mathrm {GrAut}_{k}\,A$ , so

$$ \begin{align*}\mathrm{Twist}_{\mathrm{alg}}(A)=\{ A^{\phi} \mid \phi \in \mathrm{GrAut}_{k}\,A \}/_{\cong},\end{align*} $$

where $A^{\phi }$ means the twisted algebra of A by $\{ \phi ^{n} \}_{n \in \mathbb {Z}}$ .

2.2 Geometric algebra

Let V be a finite-dimensional k-vector space, and let $A=T(V)/(R)$ be a quadratic algebra where $T(V)$ is a tensor algebra over k and $R \subset V \otimes V$ . Since an element of R defines a multilinear function on $V^{\ast }\times V^{\ast }$ , we can define a zero set associated with R by

$$ \begin{align*}\mathcal{V}(R)=\{ (p,q) \in \mathbb{P}(V^{\ast}) \times \mathbb{P}(V^{\ast}) \mid g(p,q)=0 \text{ for any } g \in R \}.\end{align*} $$

Definition 2.5 Let $A=T(V)/(R)$ be a quadratic algebra. A geometric pair $(E,\sigma )$ consists of a projective variety $E \subset \mathbb {P}(V^{\ast })$ and $\sigma \in \mathrm {Aut}_{k} E$ .

1. (1) We say that A satisfies (G1) if there exists a geometric pair $(E,\sigma )$ such that

   $$ \begin{align*}\mathcal{V}(R)=\{ (p,\sigma(p)) \in \mathbb{P}(V^{\ast}) \times \mathbb{P}(V^{\ast}) \mid p \in E \}.\end{align*} $$
   In this case, we write $\mathcal {P}(A)=(E,\sigma )$ , and call E the point variety of A.

2. (2) We say that A satisfies (G2) if there exists a geometric pair $(E,\sigma )$ such that

   $$ \begin{align*}R=\{ g \in V \otimes V \mid g(p,\sigma(p))=0 \text{ for all } p \in E \}.\end{align*} $$
   In this case, we write $A=\mathcal {A}(E,\sigma )$ .

3. (3) We say that A is a geometric algebra if it satisfies both (G1) and (G2) with $A=\mathcal {A}(\mathcal {P}(A))$ .

For geometric algebras, Mori [11] found a necessary and sufficient geometric condition for $\mathrm {GrMod} A \cong \mathrm {GrMod} A'$ .

Theorem 2.6 [11, Theorem 4.7]

Let $A=\mathcal {A}(E,\sigma )$ and $A'=\mathcal {A}(E',\sigma ')$ be geometric algebras. Then $\mathrm {GrMod} A \cong \mathrm {GrMod} A'$ if and only if there exists a sequence of automorphisms $\{ \tau _n \}_{n \in \mathbb {Z}}$ of $\mathbb {P}(V^{\ast })$ for $n \in \mathbb {Z}$ , each of which sends E isomorphically onto $E'$ , such that the diagram

commutes for every $n \in \mathbb {Z}$ .

2.3 Artin–Schelter regular algebras and Calabi–Yau algebras

Definition 2.7 A connected graded algebra A is called a d-dimensional Artin–Schelter regular algebra (simply AS-regular algebra) if A satisfies the following conditions:

1. (1) $\mathrm {gldim} A=d<\infty $ ,

2. (2) $\mathrm {GKdim} A:=\inf \{ \alpha \in \mathbb {R} \mid \dim _{k}(\sum _{i=0}^{n}A_i) \leq n^{\alpha } \text { for all } n\gg 0 \} <\infty $ , and

3. (3) $\mathrm {Ext}^{i}_{A}(k,A)= \begin {cases} k, \,\,\,\text { if } i=d, \\ 0, \,\,\,\text { if } i \neq d. \end {cases}$

Artin and Schelter proved that a three-dimensional AS-regular algebra A finitely generated in degree $1$ is isomorphic to one of the following forms:

$$ \begin{align*}k\langle x,y,z \rangle/(f_1,f_2,f_3) \text{ or } k\langle x,y \rangle/(g_1,g_2),\end{align*} $$

where $f_i$ are homogeneous polynomials of degree $2$ (the quadratic case) and $g_j$ are homogeneous polynomials of degree $3$ (the cubic case; see [1, Theorem 1.5]).

We recall the definition of Calabi–Yau algebra introduced by [4].

Definition 2.8 [4]

A k-algebra S is called d-dimensional Calabi–Yau if S satisfies the following conditions:

1. (1) $\mathrm {pd}_{S^e}\,S=d<\infty $ , and

2. (2) $\mathrm {Ext}^i_{S^e}(S,S^e) \cong \begin {cases} S, \,\,\,\text { if } i=d \\ 0, \,\,\,\text { if } i \neq d \end {cases}$ (as right $S^e$ -modules),

where $S^e=S^{\mathrm {op}} \otimes _{k} S$ is the enveloping algebra of S.

The following theorem tells us that we may assume that three-dimensional quadratic AS-regular algebra is Calabi–Yau up to graded Morita equivalence.

Theorem 2.9 [8, Theorem 4.4]

For every three-dimensional quadratic AS-regular algebra A, there exists a three-dimensional quadratic Calabi–Yau AS-regular algebra S such that $\mathrm {GrMod}\,A \cong \mathrm {GrMod}\,S$ .

Lemma 2.10 ([6, Lemma 2.8], [7, Theorem 3.2], and [12, Lemma 3.8])

Every three-dimensional geometric AS-regular algebra A is graded Morita equivalent to $S=k \langle x,y,z \rangle /(f_1,f_2,f_3)=\mathcal {A}(E,\sigma )$ in Table 1 .

Table 1 Defining relations and geometric pairs.

Remark 2.11 The original definition of geometric algebra given by Mori [11] is different from our definition. In the sense of Definition 2.5, there exists a three-dimensional quadratic AS-regular algebra which is not a geometric algebra. Strictly speaking, a three-dimensional quadratic AS-regular algebra is a geometric algebra in our sense if and only if the “point scheme” is reduced.

3 Twisted algebras of geometric algebras

In this section, we study twisted algebras of geometric algebras. Let $E \subset \mathbb {P}(V^{\ast })$ be a projective variety where V is a finite-dimensional k-vector space. We use the following notations introduced in [2].

Definition 3.1 Let $E \subset \mathbb {P}(V^{\ast })$ be a projective variety and $\sigma \in \mathrm {Aut}_{k}\,E$ . We define

$$ \begin{align*} &\mathrm{Aut}_{k}(E \uparrow \mathbb{P}(V^{\ast})):= \{ \tau \in \mathrm{Aut}_{k}\,E \mid \tau=\overline{\tau}|_{E} \text{ for some } \overline{\tau} \in \mathrm{Aut}_{k}\,\mathbb{P}(V^{\ast}) \}, \\ &\mathrm{Aut}_{k}(\mathbb{P}(V^{\ast}) \downarrow E):= \{ \tau \in \mathrm{Aut}_{k}\,\mathbb{P}(V^{\ast}) \mid \tau|_{E} \in \mathrm{Aut}_{k}\,E \}, \\ &Z(E,\sigma):=\{ \tau \in \mathrm{Aut}_{k}(\mathbb{P}(V^{\ast}) \downarrow E) \mid \sigma\tau|_{E}\sigma^{-1}=\tau|_{E}\}, \\ &M(E,\sigma):=\{ \tau \in \mathrm{Aut}_{k}(\mathbb{P}(V^{\ast}) \downarrow E) \mid (\tau|_{E}\sigma)^{i}\sigma^{-i} \in \mathrm{Aut}_{k}(E \uparrow \mathbb{P}(V^{\ast}))\,\,\, \forall i\in\mathbb{Z} \}, \\ &N(E,\sigma):=\{ \tau \in \mathrm{Aut}_{k}(\mathbb{P}(V^{\ast}) \downarrow E) \mid \sigma\tau|_{E}\sigma^{-1} \in \mathrm{Aut}_{k}(E \uparrow \mathbb{P}(V^{\ast})) \}. \end{align*} $$

Note that $ Z(E,\sigma ) \subset M(E,\sigma ) \subset N(E,\sigma ) \subset \mathrm {Aut}_{k}(\mathbb {P}(V^{\ast }) \downarrow E)$ , and $Z(E,\sigma )$ , $N(E,\sigma )$ are subgroups of $\mathrm {Aut}_{k}(\mathbb {P}(V^{\ast }) \downarrow E)$ .

Lemma 3.2 Let $E \subset \mathbb {P}(V^{\ast })$ be a projective variety and $\sigma \in \mathrm {Aut}_{k}\,E$ . If $\sigma \mathrm {Aut}_{k}(E \uparrow \mathbb {P}(V^{\ast }))=\mathrm {Aut}_{k}(E \uparrow \mathbb {P}(V^{\ast }))\sigma $ , then $M(E,\sigma )=N(E,\sigma )=\mathrm {Aut}_{k}(\mathbb {P}(V^{\ast }) \downarrow E)$ .

Proof Since $M(E,\sigma ) \subset N(E,\sigma ) \subset \mathrm {Aut}_{k}(\mathbb {P}(V^{\ast }) \downarrow E)$ in general, it is enough to show that $\mathrm {Aut}_{k}(\mathbb {P}(V^{\ast }) \downarrow E) \subset M(E,\sigma )$ . We will show that for any $\tau \in \mathrm {Aut}_{k}(\mathbb {P}(V^{\ast })\downarrow E)$ , $(\tau |_E\sigma )^{i}\sigma ^{-i} \in \mathrm {Aut}_{k}(E \uparrow \mathbb {P}(V^{\ast }))$ for every $i \in \mathbb {Z}$ by induction so that $\tau \in M(E,\sigma )$ . The claim is trivial for $i=0$ . If $(\tau |_E\sigma )^{i}\sigma ^{-i} \in \mathrm {Aut}_{k}(E \uparrow \mathbb {P}(V^{\ast }))$ for some $i \ge 0$ , then $(\tau |_E\sigma )^{i+1}\sigma ^{-i-1}= \tau |_E\sigma ((\tau |_E\sigma )^i\sigma ^{-i})\sigma ^{-1} \in \mathrm {Aut}_{k}(E \uparrow \mathbb {P}(V^{\ast }))$ . If $(\tau |_E\sigma )^{-i}\sigma ^i \in \mathrm {Aut}_{k}(E \uparrow \mathbb {P}(V^{\ast }))$ for some $i \ge 0$ , then

$$ \begin{align*}(\tau|_E\sigma)^{-(i+1)}\sigma^{i+1}=\sigma^{-1}\tau|_E^{-1}((\tau|_E\sigma)^{-i}\sigma^i)\sigma \in \mathrm{Aut}_{k}(E \uparrow \mathbb{P}(V^{\ast})).\\[-34pt]\end{align*} $$

Let $A=\mathcal {A}(E,\sigma )$ be a geometric algebra. The map $\Phi : \mathrm {TS}^{\mathbb {Z}}_{0}(A) \to \mathrm {Aut}_{k}\,\mathbb {P}(A_1^{\ast })$ is defined by $\Phi (\theta ):=\overline {(\theta _{1}|_{A_{1}})^{\ast }}$ .

Lemma 3.3 Let $A=\mathcal {A}(E,\sigma )$ be a geometric algebra. Then

$$ \begin{align*}\Phi(\mathrm{TS}^{\mathbb{Z}}_{0}(A))=M(E,\sigma).\end{align*} $$

Proof Let $\theta \in \mathrm {TS}^{\mathbb {Z}}_{0}(A)$ . We set $V:=A_1=(A^{\theta })_1$ . Then $\theta _n$ is also a graded k-linear isomorphism from $A^{\theta }$ to A and satisfies $\theta _n(a \ast b)=\theta _n(a)\theta _{n+m}(b)$ for every $n, m, l\in \mathbb {Z}$ and $a \in A^{\theta }_m$ , $b \in A^{\theta }_l$ . Let $\tau _n : \mathbb {P}(V^{\ast }) \to \mathbb {P}(V^{\ast })$ be automorphisms induced by the duals of $\theta _n|_V : V \to V$ . By [2, Remark 15], $\tau _n \in \mathrm {Aut}_{k}(\mathbb {P}(V^{\ast }) \downarrow E)$ and the diagram of automorphisms

commutes for every $n \in \mathbb {Z}$ . Then $(\tau _{1}|_E\sigma )^{n}\sigma ^{-n}=\tau _n|_E \in \mathrm {Aut}_{k}(E \uparrow \mathbb {P}(V^{\ast }))$ for every $n \in \mathbb {Z}$ , so it holds that $\Phi (\theta )=\tau _{1} \in M(E,\sigma )$ .

Conversely, let $\tau \in M(E,\sigma )$ . Since $(\tau |_E\sigma )^{n}\sigma ^{-n} \in \mathrm {Aut}_{k}(E \uparrow \mathbb {P}(V^{\ast }))$ , there is an automorphism $\tau _n \in \mathrm {Aut}_{k}\,\mathbb {P}(V^{\ast })$ such that $\tau _n|_E=(\tau |_E\sigma )^{n}\sigma ^{-n}$ for every $n \in \mathbb {Z}$ . By [2, Remark 15], there exists $\theta \in \mathrm {TS}^{\mathbb {Z}}_{0}(A)$ such that $\overline {(\theta _{n}|_{A_{1}})^{\ast }}=\tau _n$ for every $n \in \mathbb {Z}$ . Hence, it follows that $\Phi (\theta )=\overline {(\theta _{1}|_{A_{1}})^{\ast }}=\tau $ .

Let $A=T(V)/I$ be a connected graded algebra. Let $\Psi : \mathrm {GrAut}_{k}\,A \to \mathrm {PGL}(V)$ be a group homomorphism defined by $\Psi (\phi )=\overline {\phi |_V}$ . We set

$$ \begin{align*}\mathrm{PGrAut}_{k}\,A:=\mathrm{GrAut}_{k}\,A/\mathrm{Ker\,} \Psi.\end{align*} $$

Lemma 3.4 Let $A=\mathcal {A}(E,\sigma )$ be a geometric algebra.

1. (1) $\Phi (\mathrm {TS}^{\mathbb {Z}}_{\mathrm {alg}}(A))=Z(E,\sigma )$ .

2. (2) $\mathrm {PGrAut}_{k}\,A \cong Z(E,\sigma )^{\mathrm {op}}.$

Proof (1) Let $\theta =\{ \theta _{1}^{n} \}_{n \in \mathbb {Z}} \in \mathrm {TS}^{\mathbb {Z}}_{\mathrm {alg}}(A)$ . We set $V:=A_1=(A^{\theta })_1$ . Let $\tau _n : \mathbb {P}(V^{\ast }) \to \mathbb {P}(V^{\ast })$ be automorphisms induced by the duals of $\theta _{1}^n|_V : V \to V$ . Then we can write $\tau _n=\tau _{1}^n$ for every $n \in \mathbb {Z}$ . By the proof of Lemma 3.3, it follows that $(\tau _{1}|_E)^n=(\tau _1|_E\sigma )^n\sigma ^{-n}$ for every $n \in \mathbb {Z}$ . If $n=2$ , then $\tau _{1}|_E\sigma =\sigma \tau _{1}|_E$ , so $\Phi (\theta )=\tau _{1} \in Z(E,\sigma )$ .

Conversely, let $\tau \in Z(E,\sigma )$ . Since $\tau |_E\sigma =\sigma \tau |_E$ , $(\tau |_E\sigma )^{n}\sigma ^{-n}=(\tau |_E)^{n}$ for every $n \in \mathbb {Z}$ . By [2, Remark 15], there exists $\theta =\{ \theta _{1}^n \}_{n \in \mathbb {Z}} \in \mathrm {TS}^{\mathbb {Z}}_{\mathrm {alg}}(A)$ such that $\overline {(\theta _{1}|_{V})^{\ast }}^n=\tau ^n$ for every $n \in \mathbb {Z}$ . Hence, it follows that $\Phi (\theta )=\tau $ .

(2) By the following commutative diagram

it follows that $\mathrm {PGrAut}_{k}\,A \cong \Phi (\mathrm {TS}^{\mathbb {Z}}_{\mathrm {alg}}(A))=Z(E,\sigma )^{\mathrm {op}}$ .

Theorem 3.5 Let $A=\mathcal {A}(E,\sigma )$ be a geometric algebra.

1. (1) $\mathrm {Twist}(A)=\{ \mathcal {A}(E,\tau |_E\sigma ) \mid \tau \in M(E,\sigma ) \}/_{\cong }$ .

2. (2) $\mathrm {Twist}_{\mathrm {alg}}(A)=\{ \mathcal {A}(E,\tau |_E\sigma ) \mid \tau \in Z(E,\sigma ) \}/_{\cong }$ .

Proof By [2, Proposition 13], for every $\theta \in \mathrm {TS}^{\mathbb {Z}}_{0}(A)$ , $A^{\theta } \cong \mathcal {A}(E,\Phi (\theta )|_E\sigma )$ . By Lemma 3.3, it follows that

$$ \begin{align*} \mathrm{Twist}(A)&:=\{ A^{\theta} \mid \theta \in \mathrm{TS}^{\mathbb{Z}}_{0}(A) \}/_{\cong} \\ &=\{ \mathcal{A}(E,\Phi(\theta)|_E\sigma) \mid \theta \in \mathrm{TS}^{\mathbb{Z}}_{0}(A) \}/_{\cong} \\ &=\{ \mathcal{A}(E,\tau|_E\sigma) \mid \tau \in \Phi(\mathrm{TS}^{\mathbb{Z}}_{0}(A)) \}/_{\cong} \\ &=\{ \mathcal{A}(E,\tau|_E\sigma) \mid \tau \in M(E,\sigma) \}/_{\cong}. \end{align*} $$

By Lemma 3.4, we can similarly show that

$$ \begin{align*}\mathrm{Twist}_{\mathrm{alg}}(A)= \{ \mathcal{A}(E,\tau|_E\sigma) \mid \tau \in Z(E,\sigma) \}/_{\cong}.\\[-34pt]\end{align*} $$

4 Twisted algebras of three-dimensional geometric AS-regular algebras

In this section, we classify twisted algebras of three-dimensional geometric AS-regular algebras. We recall that for connected graded algebras A and $A'$ generated in degree 1, $\mathrm {GrMod}\,A \cong \mathrm {GrMod}\,A'$ if and only if $A' \in \mathrm {Twist}(A)$ , so

$$ \begin{align*}\mathrm{Twist}(A)=\{ A' \mid \mathrm{GrMod}\,A' \cong \mathrm{GrMod}\,A \}/_{\cong}. \end{align*} $$

By Lemma 2.10, we may assume that A is a three-dimensional geometric Calabi–Yau AS-regular algebra in Table 1 to compute $\mathrm {Twist}(A)$ . The algebras in Table 1 are called standard in this paper. For three-dimensional standard AS-regular algebras, we will compute the subsets $Z(E,\sigma )$ and $M(E,\sigma )$ of $\mathrm {Aut}_{k}(\mathbb {P}^2 \downarrow E)$ . We remark that some of the computations were given in [2, Section 4].

For a three-dimensional geometric AS-regular algebra $\mathcal {A}(E,\sigma )$ , the map

$$ \begin{align*}\mathrm{Aut}_{k}(\mathbb{P}^{2} \downarrow E) \to \mathrm{Aut}_{k}(E \uparrow \mathbb{P}^{2}); \tau \mapsto \tau|_E\end{align*} $$

is a bijection, so we identify $\tau \in \mathrm {Aut}_{k}(\mathbb {P}^2 \downarrow E)$ with $\tau |_E \in \mathrm {Aut}_{k}(E \uparrow \mathbb {P}^2)$ if there is no potential confusion.

Let E be an elliptic curve in $\mathbb {P}^2$ . We use a Hesse form

$$ \begin{align*}E=\mathcal{V}(x^3+y^3+z^3-3 \lambda xyz),\end{align*} $$

where $\lambda \in k$ with $\lambda ^3 \neq 1$ . It is known that every elliptic curve in $\mathbb {P}^2$ can be written in this form (see [3, Corollary 2.18]). The j-invariant of a Hesse form E is given by $j(E)=\frac {27\lambda ^3(\lambda ^3+8)^3}{(\lambda ^3-1)}$ (see [3, Proposition 2.16]). The j-invariant $j(E)$ classifies elliptic curves in $\mathbb {P}^2$ up to projective equivalence (see [5, Theorem IV.4.1(b)]). We fix the group structure on E with the zero element $o:=(1,-1,0) \in E$ (see [3, Theorem 2.11]). For a point $p \in E$ , a translation by p, denoted by $\sigma _{p}$ , is an automorphism of E defined by $\sigma _{p}(q)=p+q$ for every $q \in E$ . We define $\mathrm {Aut}_{k}(E,o):=\{ \sigma \in \mathrm {Aut}_{k}\,E \mid \sigma (o)=o \}$ . It is known that $\mathrm {Aut}_{k}(E,o)$ is a finite cyclic subgroup of $\mathrm {Aut}_{k}\,E$ (see [5, Corollary IV.4.7]).

Lemma 4.1 [7, Theorem 4.6]

A generator of $\mathrm {Aut}_{k}(E,o)$ is given by:

1. (1) $\tau _E(a,b,c):=(b,a,c)$ if $j(E) \neq 0,12^3$ ,

2. (2) $\tau _E(a,b,c):=(b,a,\varepsilon c)$ if $\lambda =0$ (so that $j(E)=0$ ),

3. (3) $\tau _E(a,b,c):=(\varepsilon ^2a+\varepsilon b+c,\varepsilon a+\varepsilon ^2b+c,a+b+c)$ if $\lambda =1+\sqrt {3}$ (so that $j(E)=12^3$ ),

where $\varepsilon $ is a primitive third root of unity. In particular, $\mathrm {Aut}_{k}(E,o)$ is a subgroup of $\mathrm {Aut}_{k}(E \uparrow \mathbb {P}^{2})=\mathrm {Aut}_{k}(\mathbb {P}^{2} \downarrow E)$ .

Remark 4.2 When $j(E)=0, 12^3$ , we may fix $\lambda =0,1+\sqrt {3}$ , respectively, as in Lemma 4.1, because if two elliptic curves E and $E'$ in $\mathbb {P}^2$ are projectively equivalent, then for every $\mathcal {A}(E,\sigma )$ , there exists an automorphism $\sigma ' \in \mathrm {Aut}_{k}\,E'$ such that $\mathcal {A}(E,\sigma ) \cong \mathcal {A}(E',\sigma ')$ (see [12, Lemma 2.6]).

It follows from [7, Proposition 4.5] that every automorphism $\sigma \in \mathrm {Aut}_{k}\,E$ can be written as $\sigma =\sigma _{p}\tau _E^i$ where $\sigma _{p}$ is a translation by a point $p \in E$ , $\tau _E$ is a generator of $\mathrm {Aut}_{k}(E,o)$ , and $i \in \mathbb {Z}_{|\tau _E|}$ . For any $n \ge 1$ , we call a point $p \in E$ n-torsion if $np=o$ . We set $E[n]:=\{ p \in E \mid np=o \}$ and $T[n]:=\{ \sigma _{p} \in \mathrm {Aut}_{k}\,E \mid p \in E[n] \}$ .

If $A=\mathcal {A}(E,\sigma )$ is a three-dimensional standard AS-regular algebra, we write

$$ \begin{align*}\mathrm{Aut}_{k}(\mathbb{P}^{2} \downarrow E)=Z(E) \rtimes G(E)\end{align*} $$

as in Table 2 where $Z(E) \leq \mathrm {Aut}_{k}(\mathbb {P}^{2} \downarrow E)$ with $Z(E) \subset Z(E,\sigma )$ and $G(E) \leq \mathrm {Aut}_{k}(\mathbb {P}^{2} \downarrow E)$ so that

$$ \begin{align*}N(E,\sigma)=Z(E) \rtimes (G(E) \cap N(E,\sigma)).\end{align*} $$

Table 2 The list of Z(E) and G(E).

Table 2 can be checked by the following three steps:

• • Step $1$ : Calculate $\mathrm {Aut}_{k}(\mathbb {P}^{2} \downarrow E)$ .

• • Step $2$ : Find $Z(E) \leq \mathrm {Aut}_{k}(\mathbb {P}^{2} \downarrow E)$ with $Z(E) \subset Z(E,\sigma ) \cong (\mathrm {PGrAut}_{k}\,A)^{\mathrm {op}}$ (see Lemma 3.4(2)).

• • Step $3$ : Find $G(E) \leq \mathrm {Aut}_{k}(\mathbb {P}^{2} \downarrow E)$ .

$\mathrm {Aut}_{k}(\mathbb {P}^{2} \downarrow E)$ and $\mathrm {PGrAut}_{k}\,A$ were computed in [13]. We explain these steps for Type S. By Lemma 2.10, $E=\mathcal {V}(x) \cup \mathcal {V}(y) \cup \mathcal {V}(z)$ and

$$ \begin{align*}\begin{cases} \sigma(0,b,c)=(0,b,\alpha c), \\ \sigma(a,0,c)=(\alpha a,0,c), \\ \sigma(a,b,0)=(a,\alpha b,0), \end{cases}\end{align*} $$

where $\alpha ^3 \neq 0,1$ . By [13, Lemma 3.2.1],

$$ \begin{align*}\mathrm{Aut}_{k}(\mathbb{P}^{2} \downarrow E)=\left\{ \begin{pmatrix} 1 & 0 & 0 \\ 0 & e & 0 \\ 0 & 0 & i \end{pmatrix} \middle| ei \neq 0 \right\} \rtimes \left\langle \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \right\rangle.\end{align*} $$

By [13, Theorem 3.3.1],

$$ \begin{align*} Z(E,\sigma)&=(\mathrm{PGrAut}_{k}\,A)^{\mathrm{op}} \\ &= \begin{cases} \left\{ \begin{pmatrix} 1 & 0 & 0 \\ 0 & e & 0 \\ 0 & 0 & i \end{pmatrix} \middle| ei \neq 0 \right\} \rtimes \left\langle \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix} \right\rangle, &\text{if } \sigma^2 \neq \mathrm{id}, \\ \mathrm{Aut}_{k}(\mathbb{P}^{2} \downarrow E), &\text{if } \sigma^2=\mathrm{id}, \end{cases} \end{align*} $$

so we may take

$$ \begin{align*}Z(E)=\left \{ \begin {pmatrix} 1 & 0 & 0 \\ 0 & e & 0 \\ 0 & 0 & i \end {pmatrix} \middle | ei \neq 0 \right \} \rtimes \left \langle \begin {pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end {pmatrix} \right \rangle\ \mathrm{and}\ G(E)=\left \langle \begin {pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end {pmatrix} \right \rangle. \end{align*} $$

Remark 4.3 By Table 2:

1. (1) $|G(E)|<\infty $ if and only if A is of Types P, S, S’, NC, and EC, and, in this case, there exists $\tau _E \in \mathrm {Aut}_{k}(\mathbb {P}^{2} \downarrow E)$ such that $G(E)=\langle \tau _E \rangle $ is a finite cyclic group.

2. (2) $|G(E)|<\infty $ but $|G(E)| \neq 2$ if and only if A is of Type P ( $|G(E)|=1$ ), or Type EC with $j(E)=0$ ( $|G(E)=6|$ ), or Type EC with $j(E)=12^3$ ( $|G(E)|=4$ ).

Theorem 4.4 If $A=\mathcal {A}(E,\sigma )$ is a three-dimensional quadratic AS-regular algebra of Types T, T’, and CC (so that $|\sigma |=\infty $ (cf. [9])), then $Z(E,\sigma )=M(E,\sigma )=N(E,\sigma )$ .

Proof Writing $\mathrm {Aut}_{k}(\mathbb {P}^{2} \downarrow E)=Z(E) \rtimes G(E)$ as in Table 2, it is enough to show that $G(E) \cap N(E,\sigma )=\{ \mathrm {id} \}$ .

Type T: For every $\tau =\begin {pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & i \end {pmatrix} \in G(E)$ ,

$$ \begin{align*}\begin{cases} \sigma\tau|_E\sigma^{-1}(a,-a,c)=(a,-a,(1-i)a+ic), \\ \sigma\tau|_E\sigma^{-1}(a,-\varepsilon a,c)=(a,-\varepsilon a,\varepsilon^2(1-i)a+ic), \\ \sigma\tau|_E\sigma^{-1}(a,-\varepsilon^2 a,c)=(a,-\varepsilon^2a,\varepsilon(1-i)a+ic). \end{cases}\end{align*} $$

If $\tau \in N(E,\sigma )$ , then there exists $\overline {\tau } \in \mathrm {Aut}_{k}\,\mathbb {P}^2$ such that $\sigma \tau |_E\sigma ^{-1}=\overline {\tau }|_E$ . Then $1-i=\varepsilon ^2(1-i)=\varepsilon (1-i)$ , so $i=1$ .

Type T $'$ : For every $\tau = \begin {pmatrix} 1 & 0 & 0 \\ 0 & e & 0 \\ 0 & 0 & e^2 \end {pmatrix} \in G(E)$ ,

$$ \begin{align*}\begin{cases} \sigma\tau|_E\sigma^{-1}(0,b,c)=(0,eb,e(1-e)b+e^2c), \\ \sigma\tau|_E\sigma^{-1}(a,b,c)=(a,(e-1)a+eb,(e-1)^2a+2e(e-1)b+e^2c). \end{cases}\end{align*} $$

If $\tau \in N(E,\sigma )$ , then there exists $\overline {\tau } \in \mathrm {Aut}_{k}\,\mathbb {P}^2$ such that $\sigma \tau |_E\sigma ^{-1}=\overline {\tau }|_E$ . Then $e(1-e)=2e(e-1)$ , so $e=1$ .

Type CC: For every $\tau = \begin {pmatrix} 1 & 0 & 0 \\ 0 & e & 0 \\ 0 & 0 & e^2 \end {pmatrix} \in G(E)=\mathrm {Aut}_{k}(\mathbb {P}^{2} \downarrow E)$ ,

$$ \begin{align*} &\sigma\tau|_E\sigma^{-1}(a,b,c) \\ &= \left( a+(1-e)b,eb,-3\frac{(a+b)^2}{eb}+3(a+b)-eb+e^{-2}\left( 3\frac{a^2}{b}+3a+b+c \right) \right). \end{align*} $$

If $\tau \in N(E,\sigma )$ , then there exists $\overline {\tau } \in \mathrm {Aut}_{k}\,\mathbb {P}^2$ such that $\sigma \tau |_E\sigma ^{-1}=\overline {\tau }|_E$ . Then there exists $0 \neq e' \in k$ such that

$$ \begin{align*}(1+(1-e)b,eb)=(1,e'b) \text{ in } \mathbb{P}^1\end{align*} $$

for every $(1,b,c) \in E$ , so $e'=e=1$ .

Lemma 4.5 Let $A=\mathcal {A}(E,\sigma )$ be a three-dimensional standard AS-regular algebra of Type S, S’, NC, or EC. For every $i \ge 1$ , $\sigma ^{i} \in \mathrm {Aut}_{k}(E \uparrow \mathbb {P}^{2})=\mathrm {Aut}_{k}(\mathbb {P}^{2} \downarrow E)$ if and only if $\sigma ^{3i}=\mathrm {id}$ .

Proof This lemma follows from [9, Theorem 3.4].

Lemma 4.6 Let $A=\mathcal {A}(E,\sigma )$ be a three-dimensional standard AS-regular algebra. If $\sigma \tau \sigma ^{-1}, \sigma ^{-1}\tau \sigma \in \mathrm {Aut}_{k}(E \uparrow \mathbb {P}^{2})$ for every $\tau \in G(E)$ , then $M(E,\sigma )=N(E,\sigma )=\mathrm {Aut}_{k}(\mathbb {P}^{2} \downarrow E)$ .

Proof Every $\tau \in \mathrm {Aut}_{k}(\mathbb {P}^{2} \downarrow E)$ can be written as $\tau =\tau _1\tau _2$ where $\tau _1 \in Z(E)$ , $\tau _2 \in G(E)$ . Since $\sigma \tau \sigma ^{-1}=\sigma \tau _1\tau _2\sigma ^{-1}=\tau _1\sigma \tau _2\sigma ^{-1}$ , it holds that $\sigma \tau \sigma ^{-1} \in \mathrm {Aut}_{k}(E \uparrow \mathbb {P}^{2})$ . Similarly, every $\tau \in \mathrm {Aut}_{k}(\mathbb {P}^{2} \downarrow E)$ can be written as $\tau =\tau _2\tau _1$ where $\tau _1 \in Z(E)$ , $\tau _2 \in G(E)$ , so $\sigma ^{-1}\tau \sigma =\sigma ^{-1}\tau _2\tau _1\sigma = \sigma ^{-1}\tau _2\sigma \tau _{1}$ , and hence $\sigma ^{-1}\tau \sigma \in \mathrm {Aut}_{k}(E \uparrow \mathbb {P}^{2})$ . The result now follows from Lemma 3.2.

Theorem 4.7 Let $A=\mathcal {A}(E,\sigma )$ be a three-dimensional standard AS-regular algebra such that $|G(E)|=2$ .

1. (1) If $\sigma ^2=\mathrm {id}$ , then $Z(E,\sigma )=M(E,\sigma )=N(E,\sigma )=\mathrm {Aut}_{k}(\mathbb {P}^{2} \downarrow E)$ .

2. (2) If $\sigma ^{6}=\mathrm {id}$ , then $M(E,\sigma )=N(E,\sigma )=\mathrm {Aut}_{k}(\mathbb {P}^{2} \downarrow E)$ .

3. (3) If $\sigma ^{6} \neq \mathrm {id}$ , then $Z(E)=Z(E,\sigma )=M(E,\sigma )=N(E,\sigma )$ .

Proof (1) We will give a proof for Type S’. The other types are proved similarly. By Lemma 2.10 and Table 2, $E=\mathcal {V}(x) \cup \mathcal {V}(x^2-\lambda yz)$ ,

$$ \begin{align*}\begin{cases} \sigma(0,b,c)=(0,b,\alpha c), \\ \sigma(a,b,c)=(a,\alpha b,\alpha^{-1}c), \end{cases}\end{align*} $$

where $\lambda =\frac {\alpha ^3-1}{\alpha }$ and $\alpha ^3 \neq 0,1$ , and $\tau _E= \begin {pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end {pmatrix}$ . In general, $Z(E) \subset Z(E,\sigma ) \subset M(E,\sigma ) \subset N(E,\sigma ) \subset \mathrm {Aut}_{k}(\mathbb {P}^{2} \downarrow E)$ . In this case, $\sigma ^2=\mathrm {id}$ if and only if $\alpha ^{2}=1$ . Since

$$ \begin{align*}\begin{cases} \sigma\tau_E\sigma^{-1}(0,b,c)=(0,c,\alpha^2 b), \\ \sigma\tau_E\sigma^{-1}(a,b,c)=(a,\alpha^{2}c,\alpha^{-2}b), \end{cases}\end{align*} $$

if $\sigma ^{2}=\mathrm {id}$ , then $\sigma \tau _E\sigma ^{-1}=\tau _E$ . Since $\tau _E \in Z(E,\sigma )$ , $Z(E,\sigma )=\mathrm {Aut}_{k}(\mathbb {P}^{2} \downarrow E)$ .

(2) By direct calculations, $(\tau _E\sigma )^2=\mathrm {id}$ , so $\sigma \tau _E\sigma ^{-1}\tau _E^{-1}=\sigma ^2=\tau _E^{-1}\sigma ^{-1}\tau _E\sigma $ . By Lemma 4.5, $\sigma \tau _E\sigma ^{-1}\tau _E^{-1}, \tau _E^{-1}\sigma ^{-1}\tau _E\sigma \in \mathrm {Aut}_{k}(\mathbb {P}^{2} \downarrow E)$ if and only if $\sigma ^6=\mathrm {id}$ . In particular, if $\sigma ^6=\mathrm {id}$ , then $\sigma \tau _E\sigma ^{-1}, \sigma ^{-1}\tau _E\sigma \in \mathrm {Aut}_{k}(\mathbb {P}^{2} \downarrow E)$ . By Lemma 4.6, $M(E,\sigma )=N(E,\sigma )=\mathrm {Aut}_{k}(\mathbb {P}^{2} \downarrow E)$ , and hence (2) holds.

(3) If $\sigma ^6 \neq \mathrm {id}$ , then $\tau _E \notin N(E,\sigma )$ . Since $G(E) \cap N(E,\sigma )=\{ \mathrm {id} \}$ , $N(E,\sigma )=Z(E)$ , and hence (3) holds.

Theorem 4.8 Let $A=\mathcal {A}(E,\sigma )$ be a three-dimensional standard AS-regular algebra except for Type EC. Then Table 3 gives $Z(E,\sigma )$ and $M(E,\sigma )$ for each type.

Proof By Theorems 4.4 and 4.7, the result holds.

Table 3 $Z(E,\sigma)$ and $M(E,\sigma)$ except for Type EC.

Definition 4.9 Let $E=\mathcal {V}(x^3+y^3+z^3) \subset \mathbb {P}^2$ so that $j(E)=0$ , and define

$$ \begin{align*}\mathcal{E}:=\{ (a,b,c) \in E \mid a^9=b^9=c^9 \} \subset E[9] \setminus E[6].\end{align*} $$

In this paper, we say that a three-dimensional quadratic AS-regular algebra is exceptional if it is graded Morita equivalent to $\mathcal {A}(E,\sigma _{p})$ for some $p \in \mathcal {E}$ .

Lemma 4.10 Let $A=\mathcal {A}(E,\sigma _{p})$ be a three-dimensional standard AS-regular algebra of Type EC and $\sigma _{q}\tau _{E}^{i} \in \mathrm {Aut}_{k}(\mathbb {P}^{2} \downarrow E)$ where $q \in E[3]$ , $i \in \mathbb {Z}_{|\tau _E|}$ . Then:

1. (1) $\sigma _{q}\tau _{E}^i \in Z(E,\sigma _{p})$ if and only if $p-\tau _{E}^i(p)=o$ ,

2. (2) $\sigma _{q}\tau _{E}^i \in N(E,\sigma _{p})$ if and only if $p-\tau _{E}^i(p) \in E[3]$ , and

3. (3) $M(E,\sigma _{p})=N(E,\sigma _{p})$ .

Proof (1) Since $\sigma _{p}(\sigma _{q}\tau _{E}^i)\sigma _{p}^{-1}=\sigma _{q+p-\tau _{E}^{i}(p)}\tau _{E}^i$ , $\sigma _{q}\tau _{E}^{i} \in Z(E,\sigma _{p})$ if and only if $p-\tau _{E}^{i}(p)=o$ .

(2) Since $\sigma _{p}(\sigma _{q}\tau _{E}^i)\sigma _{p}^{-1}=\sigma _{q+p-\tau _{E}^{i}(p)}\tau _{E}^i$ , by [11, Lemma 5.3], $\sigma _{q}\tau _{E}^{i} \in N(E,\sigma _{p})$ if and only if $p-\tau _{E}^{i}(p) \in E[3]$ .

(3) In general, $M(E,\sigma _{p}) \subset N(E,\sigma _{p})$ , so it is enough to show that $N(E,\sigma _{p}) \subset M(E,\sigma _{p})$ . Let $\sigma _{q}\tau _E^{i} \in N(E,\sigma _{p}) \subset \mathrm {Aut}_{k}(\mathbb {P}^{2} \downarrow E)$ where $q \in E[3]$ and $i \in \mathbb {Z}_{|\tau _E|}$ . Since $\sigma _{p}(\sigma _{q}\tau _E^{i})\sigma _{p}^{-1}=\sigma _{q+p-\tau _E^{i}(p)}\tau _E^{i} \in \mathrm {Aut}_{k}(\mathbb {P}^{2} \downarrow E)$ , $p-\tau _E^{i}(p) \in E[3]$ . For any $j \ge 1$ , we can write

$$ \begin{align*}(\sigma_{q}\tau_E^{i}\sigma_{p})^{j}(\sigma_{p})^{-j}=\sigma_{r_{j}}\tau_E^{ji},\end{align*} $$

where $r_{j}=\sum _{l=0}^{j-1}\tau _E^{li}(q)+\sum _{l=1}^{j}\tau _E^{li}(p-\tau _E^{(j-l)i}(p))$ , and

$$ \begin{align*}(\sigma_{q}\tau_E^{i}\sigma_{p})^{-j}(\sigma_{p})^{j}=\sigma_{s_{j}}\tau_E^{-ji},\end{align*} $$

where $s_{j}=\sum _{l=1}^{j}(-\tau _E^{-li}(q))+\sum _{l=0}^{j-1}\tau _E^{-li}(p-\tau _E^{(l-j)i}(p))$ . By [7, Lemma 4.19], for any $j \ge 1$ , $r_{j}, s_{j} \in E[3]$ , so

$$ \begin{align*}(\sigma_{q}\tau_E^{i}\sigma_{p})^{j}(\sigma_{p})^{-j}, (\sigma_{q}\tau_E^{i}\sigma_{p})^{-j}(\sigma_{p})^{j} \in \mathrm{Aut}_{k}(\mathbb{P}^{2} \downarrow E),\end{align*} $$

and hence (3) holds.

Theorem 4.11 Let $A=\mathcal {A}(E,\sigma _{p})$ be a three-dimensional standard AS-regular algebra of Type EC. Then Table 4 gives $Z(E,\sigma _{p})$ and $M(E,\sigma _{p})$ .

Table 4 $Z(E,\sigma)$ and $M(E,\sigma)$ for Type EC.

Proof By Lemma 4.10(3), it is enough to calculate $Z(E,\sigma _{p})$ and $N(E,\sigma _{p})$ . By Lemma 3.4(2), $Z(E,\sigma _{p})$ was given in [10, Proposition 4.7]. The set of points satisfying $p-\tau _{E}^i(p) \in E[3]$ was given in [10, Theorem 3.8]. By Lemma 4.10(1) and (2), the result follows.

Corollary 4.12 shows that in most cases a twisting system can be replaced by an automorphism to compute a twisted algebra.

Corollary 4.12 Let $A=\mathcal {A}(E,\sigma )$ be a three-dimensional nonexceptional standard AS-regular algebra. If $\sigma ^6 \neq \mathrm {id}$ or $\sigma ^2=\mathrm {id}$ , then $Z(E,\sigma )=M(E,\sigma )$ , so $\mathrm {Twist}_{\mathrm {alg}}(A)=\mathrm {Twist}(A)$ .

Proof By Theorems 4.4 and 4.7, it is enough to show the case that $A=\mathcal {A}(E,\sigma _{p})$ is of Type EC such that $j(E)=0$ , $p \notin \mathcal {E}$ , or $j(E)=12^3$ .

(1) $j(E)=0$ , $p \notin \mathcal {E}$ : Let $E=\mathcal {V}(x^3+y^3+z^3) \subset \mathbb {P}^2$ . By Theorem 4.11, if $p \in E[2]$ or $p \notin E[6]$ , then $N(E,\sigma _{p})=M(E,\sigma _{p})=Z(E,\sigma _{p})$ .

(2) $j(E)=12^3$ : Let $E=\mathcal {V}(x^3+y^3+z^3-3\lambda xyz) \subset \mathbb {P}^2$ where $\lambda =1+\sqrt {3}$ . By Theorem 4.11, if $p \in E[2]$ or $p \notin E[6]$ , then $N(E,\sigma _{p})=M(E,\sigma _{p})=Z(E,\sigma _{p})$ .

Let $E \subset \mathbb {P}^2$ be a projective variety. For $\tau \in \mathrm {Aut}_{k}\,E$ , we define

$$ \begin{align*}||\tau||:=\inf\{ i \in \mathbb{N}^{+} \mid \tau^i \in \mathrm{Aut}_{k}(E \uparrow \mathbb{P}^{2}) \}.\end{align*} $$

Corollary 4.13 For every three-dimensional nonexceptional geometric AS-regular algebra B, there exists a three-dimensional standard AS-regular algebra S such that $\mathrm {Twist}(B)=\mathrm {Twist}_{\mathrm {alg}}(S)$ .

Proof By Lemma 2.10, there exists a three-dimensional nonexceptional standard AS-regular algebra $A=\mathcal {A}(E,\sigma )$ such that $\mathrm {GrMod}\,B \cong \mathrm {GrMod}\,A$ . If $\sigma ^6 \neq \mathrm {id}$ , then $\mathrm {Twist}(B)=\mathrm {Twist}(A)=\mathrm {Twist}_{\mathrm {alg}}(A)$ by Corollary 4.12, so we assume that $\sigma ^6=\mathrm {id}$ . Set $\tau :=\sigma ^2 \in \mathrm {Aut}_{k}\,E$ and $S:=\mathcal {A}(E,\sigma ^3)$ . Since $||\tau ||=||\sigma ^2||=|\sigma ^6|=1$ by [9, Theorem 3.4], $\tau \in \mathrm {Aut}_{k}(E \uparrow \mathbb {P}^{2})$ . Since $\tau ^{i+1}\sigma =\sigma ^{2i+3}=\sigma ^3\tau ^i$ for every $i \in \mathbb {Z}$ , $\mathrm {GrMod}\,A \cong \mathrm {GrMod}\,S$ by Theorem 2.6. Since $(\sigma ^{3})^2=\mathrm {id}$ , $\mathrm {Twist}(B)=\mathrm {Twist}(A)=\mathrm {Twist}(S)=\mathrm {Twist}_{\mathrm {alg}}(S)$ by Corollary 4.12.

Example 4.14 shows that even if $B \cong S^{\theta }$ for some three-dimensional quadratic Calabi–Yau AS-regular algebra S, there may be no $\phi \in \mathrm {GrAut}_{k}\,S$ such that $B \cong S^{\phi }$ . We need to carefully choose S in order that $B \cong S^{\phi }$ for some $\phi \in \mathrm {GrAut}_{k}\,S$ .

Example 4.14 Let $E \subset \mathbb {P}^2$ be an elliptic curve. Assume that $j(E) \neq 0,12^3$ . We set three geometric algebras of Type EC; $B:=\mathcal {A}(E,\tau _{E}\sigma _{p})$ , $A:=\mathcal {A}(E,\sigma _{p})$ , and $S:=\mathcal {A}(E,\sigma _{3p})$ , where $p \in E[6] \setminus (E[2] \cup E[3])$ . By [8, Theorem 4.3], these algebras are three-dimensional quadratic AS-regular algebras. Moreover, A and S are standard. By [7, Theorem 4.20], $\mathrm {GrMod}\,B \cong \mathrm {GrMod}\,A \cong \mathrm {GrMod}\,S$ . Since $|\sigma _{p}|=6$ and $|\sigma _{3p}|=2$ , $M(E,\sigma _{3p})=Z(E,\sigma _{3p}) \neq Z(E,\sigma _{p})$ by Table 4, so $\mathrm {Twist}(B)=\mathrm {Twist}_{\mathrm {alg}}(S) \neq \mathrm {Twist}_{\mathrm {alg}}(A)$ .