---
abstract: 'In noncommutative algebraic geometry, an Artin-Schelter regular (AS-regular) algebra is one of the main interests and every $3$-dimensional quadratic AS-regular algebra is a geometric algebra introduced by Mori whose point scheme is either $\mathbb{P}^{2}$ or a cubic curve in $\mathbb{P}^{2}$ by Artin-Tate-Van den Bergh.  In the preceding paper by the authors, we determined the all possible defining relations for these geometric algebras.  However, the authors did not check the AS-regularity of these geometric algebras.  In this paper, by using twisted superpotentials and twists of superpotentials in the sense of Mori-Smith, we check the AS-regularity of geometric algebras whose point schemes are not elliptic curves.  For geometric algebras whose point schemes are elliptic curves, we give a simple condition to be $3$-dimensional quadratic AS-regular algebras.  As an application, we show that every $3$-dimensional quadratic AS-regular algebra is graded Morita equivalent to a Calabi-Yau AS-regular algebra.' 
address:
- |
     Department of Mathematics, Faculty of Science, Tokyo University of Science\
    1-3 Kagurazaka, Shinjyuku, Tokyo, 162-8601, Japan
- |
     Graduate School of Integrated Science and Technology, Shizuoka University\
    Ohya 836, Shizuoka 422-8529, Japan
author:
- Ayako Itaba
- Masaki Matsuno
title: 'AS-regularity of geometric algebras of plane cubic curves'
---

Introduction
============

In noncommutative algebraic geometry, an Artin-Schelter regular (AS-regular) algebra introduced by Artin-Schelter [@AS] is one of the main interests.  Artin-Tate-Van den Bergh [@ATV1] proved that there exists a one-to-one correspondence between $3$-dimensional AS-regular algebras and regular geometric pairs.  This work convinced us that algebraic geometry is very useful to study even noncommutative algebras.  Dubois-Violette [@D] and Bocklandt-Schedler-Wemyss [@BSW] showed that every $3$-dimensional quadratic AS-regular algebra $A$ is isomorphic to a derivation-quotient algebra $\mathcal{D}(w)$ of a twisted superpotential $w$, and Mori-Smith [@MS1] showed that such $w$ is unique up to non-zero scalar multiples.  So it is interesting to study AS-regular algebras using both algebraic geometry and twisted superpotentials.  In fact, Mori-Smith [@MS2] classified $3$-dimensional quadratic Calabi-Yau AS-regular algebras by using superpotentials. 

In the preceding paper [@IM] by the authors, in terms of geometric algebras defined by Mori [@Mo], we determined all possible defining relations for geometric algebras whose point schemes are either $\mathbb{P}^{2}$ or cubic curves in $\mathbb{P}^{2}$ and classified them up to graded algebra isomorphism and up to graded Morita equivalence.  However, in [@IM], the authors did not check the AS-regularity of these classified geometric algebras.  So, one of the aims of this paper is to check the AS-regularity of them.  Note that, Iyudu-Shkarin [@IS] recently gave a list of defining relations of $3$-dimensional AS-regular algebras by using twisted superpotentials, but no proof of AS-regularity of these algebras was given in their paper.  For geometric algebras listed in [@IM Theorem 3.1], we give a list of candidates of twisted superpotentials to serve our purposes (see Proposition \[TSP\]).  By using this list, we give a complete list of superpotentials whose derivation-quotient algebras are $3$-dimensional quadratic Calabi-Yau AS-regular algebras whose point schemes are not elliptic curves (see Theorem \[SP\]).  By using [*a twist of a superpotential*]{} (in the sense of [@MS1]), we showed that potentials listed in Proposition \[TSP\] are in fact twisted superpotentials and derivation-quotient algebras of them are $3$-dimensional quadratic AS-regular algebras (see Theorems \[prop-3\] and \[thm-RTSP\]).  For a geometric algebra $A$ whose point scheme is an elliptic curve in $\mathbb{P}^{2}$, we give a simple condition that $A$ is AS-regular (see Theorem \[Main\]).  As an application of Corollary \[cor\_conj\] and Theorem \[Main\], we prove the following theorem (see Theorem \[Main2\]). 

\[conj\] For every $3$-dimensional quadratic AS-regular algebra $A$, there exists a Calabi-Yau AS-regular algebra $S$ such that $A$ and $S$ are graded Morita equivalent. 

Theorem \[conj\] tells us that, for a $3$-dimensional quadratic AS-regular algebra $A$, to study the noncommutative projective scheme ${\rm Proj}_{\rm nc}A$ of $A$ in the sense of Artin-Zhang [@AZ] is reduced to study ${\rm Proj}_{\rm nc}S$ for the Calabi-Yau AS-regular algebra $S$. Note that [@Ueyama2 Example 14] gave one example of a $3$-dimensional cubic AS-regular algebra which is not graded Morita equivalent to any Calabi-Yau AS-regular algebra. 

This paper is organized as follows: In Section 2, we recall the definition of an AS-regular algebra defined by Artin-Schelter [@AS], a Calabi-Yau algebra by Ginzburg [@G], a twisted superpotential and a twist of a superpotential in the sense of [@MS1].  Also, we recall Zhang’s twist and twisted algebras from [@Z] and some lemmas which are needed to show our Theorem \[conj\].  Moreover, we recall the definitions of a geometric algebra for quadratic algebras introduced by Mori [@Mo], and the result of our preceding paper [@IM].  In Section 3, we prove Theorem \[conj\] for geometric algebras whose point schemes are not elliptic curves.  Finally, in Section 4, we prove Theorem \[conj\] for geometric algebras whose point schemes are elliptic curves in $\mathbb{P}^{2}$.

Preliminaries
=============

Throughout this paper, let $k$ be an algebraically closed field of characteristic $0$. A graded $k$-algebra means an $\mathbb{N}$-graded algebra $A=\bigoplus_{i\in \mathbb{N}}A_{i}$. A [*connected graded*]{} $k$-algebra $A$ is a graded $k$-algebra such that $A_{0}=k$. We denote by ${\rm GrMod}\,A$ the category of graded right $A$-modules and graded right $A$-module homomorphisms and we say that two graded $k$-algebras $A$ and $B$ are graded Morita equivalent if two categories ${\rm GrMod}\,A$ and ${\rm GrMod}\,B$ are equivalent. 

AS-regular algebras and Calabi-Yau algebras 
--------------------------------------------

Let $A$ be a graded $k$-algebra.  We recall that $${\rm GKdim}\,A:=
 {\rm inf}\{\alpha\in \mathbb{R} \mid {\rm dim}_{k}(\sum_{i=0}^{n}A_{i})\leq n^{\alpha}
 \text{ for all }n \gg 0\}$$ is called the [*Gelfand-Kirillov dimension*]{} of $A$. 

\[AS-reg\] A connected graded $k$-algebra $A$ is called a [*$d$-dimensional Artin-Schelter regular [(]{}AS-regular[)]{} algebra*]{} if $A$ satisfies the following conditions:

(i) ${\rm gldim}\,A=d<\infty$,

(ii) ${\rm GKdim}\,A<\infty$,

(iii) ([*Gorenstein condition*]{}) ${\rm Ext}_{A}^{i}(k,A)\cong \left\{
                                             \begin{array}{ll}
                                             k   &\quad (i=d), \\
                                             0   &\quad (i\neq d).  
                                             \end{array}
                                             \right.$

Any $3$-dimensional AS-regular algebra $A$ finitely generated in degree $1$ is a graded algebra isomorphic to an algebra of the form $$k\langle x, y, z \rangle/(f_{1},f_{2},f_{3})\text{ (quadratic case) } ,
\ \text{ or }\ k\langle x, y\rangle/(g_{1}, g_{2})\text{ (cubic case) }$$ where $f_{i}$ are homogeneous polynomials of degree $2$ and $g_{i}$ are homogeneous polynomials of degree $3$ ([@AS Theorem 1.5 (i)]).  In this paper, we focus on $3$-dimensional quadratic AS-regular algebras.  Let $V$ be a $3$-dimensional $k$-vector space and $T(V)$ the tensor algebra of $V$. We choose a basis $\{x_{1},x_{2},x_{3}\}$ of $V$. Also, for an algebra $T(V)/(R)$, we choose a basis $\{f_{1},f_{2},f_{3}\}$ of $R \subset V^{\otimes 2}$. We set $\bm{x}:=(x_{1},x_{2},x_{3})^{t}$ and $\bm{f}:=(f_{1},f_{2},f_{3})^{t}$, where, for a matrix $N$, $N^{t}$ means the transpose of $N$. There is a unique $3\times 3$ matrix $M$ with entries in $V$ such that $\bm{f}=M\bm{x}$ (see [@AS]).  From [@ATV1 p.34], $T(V)/(R)$ is called [*standard*]{} if there are bases for $V$ and $R$ such that the entries in $\bm{x}^{t}M$ are also a basis for $R$.

\[ATV1\_thm1\] Let $V$ be a $3$-dimensional $k$-vector space and $R$ a $3$-dimensional subspace of $V^{\otimes 2}$. Then $T(V)/(R)$ is a $3$-dimensional AS-regular algebra if and only if $T(V)/(R)$ is standard and the common zero locus in $\mathbb{P}^{2}$ of the $2\times 2$ minors of the matrix $M$ in the above is empty. 

Here, we recall the definition of a Calabi-Yau algebra introduced by [@G]. 

A $k$-algebra $S$ is called [*$d$-dimensional Calabi-Yau*]{} if $S$ satisfies the following conditions:

(i) $\mathrm{pd}_{S^{\rm e}}S=d<\infty$,

(ii) $
           \mathrm{Ext}_{S^{\rm e}}^{i}(S,S^{\rm {e}})\cong
           \begin{cases}
            S & \text{ if } i=d, \\
            0 & \text{ if } i \neq d
           \end{cases}
           $ (as right $S^{\rm e}$-modules)

where $S^{\rm e}=S^{{\rm op}} \otimes_{k} S$ is the enveloping algebra of $S$.

For example, it is known that an $n$-th polynomial ring $k[x_{1},x_{2},\ldots,x_{n}]$ is $n$-dimensional Calabi-Yau.  Note that, for a $3$-dimensional quadratic AS-regular algebra $A$, the quadratic dual $A^{!}$ of $A$ is a Frobenius algebra by [@S Proposition 5.10].  Hence, we can consider the Nakayama automorphism $\nu_{A^{!}}$ of $A^{!}$. By using the following consequence proved by Reyes-Rogalski-Zhang [@RRZ], we can determine whether these algebras $A$ are Calabi-Yau algebras or not. 

\[thmRRZ\] Let $A$ be a $3$-dimensional quadratic AS-regular algebra.  Then $A$ is Calabi-Yau if and only if the Nakayama automorphism $\nu_{A^{!}}$ of $A^{!}$ is the identity [(]{}that is, $A^{!}$ is symmetric[)]{}. 

Twisted algebras
----------------

In this subsection, we recall the notion of twisting system and twisted algebra introduced by Zhang [@Z]. 

A set of graded $k$-linear automorphisms of $A$, say $\theta=\{\theta_{i}\mid i \in \mathbb{N}\}$, is called a [*twisting system*]{} of $A$ if $$\theta_{n}(a\theta_{l}(b))=\theta_{n}(a)\theta_{n+l}(b)$$ for all $l,m,n \in \mathbb{N}$ and all $a \in A_{l} , b\in A_{m}$ ([@Z Definition 2.1]).  Let $\theta=\{\theta_{i}\mid i \in \mathbb{N}\}$ be a twisting system of $A$.  Then, a new graded and associative multiplication $\ast$ on the underlying graded $k$-vector space $\bigoplus_{i\in \mathbb{N}}A_{i}$ is defined by $$a\ast b:=a \theta_{l}(b)\quad\text{for all }a \in A_{l}, b\in A_{m}.$$ We denote by $1_{\theta}$ the identity with respect to $\ast$. The graded $k$-algebra $(\bigoplus_{i\in \mathbb{N}}A_{i},\ast,1_{\theta})$ is called the [*twisted algebra*]{} of $A$ by $\theta$ and is denoted by $A^{\theta}$ ([@Z Definition 2.3]).  Any graded algebra automorphism $\theta \in {\rm Aut}\,A$ defines a twisting system of $A$ by $\{ \theta^{i} \}_{i \in \mathbb{N} }$.  The twist of $A$ by this twisting system is denoted by $A^{\theta}$ instead of $A^{\{\theta^{i} \}_{i \in \mathbb{N}}}$ which is called the [*twist*]{} of $A$ by $\theta$.

\[twistingequi\] Let $A$ and $A'$ be two connected graded $k$-algebras with $A_{1}\neq 0$. Then $A'$ is isomorphic to a twisted algebra of $A$ if and only if ${\rm GrMod}\,A$ and ${\rm GrMod}\,A'$ are equivalent. 

\[twist\_inv\] Let $A$ be a connected graded $k$-algebra and $\theta$ be a twisting system of $A$.  Then $A$ is a $3$-dimensional quadratic AS-regular algebra if and only if the twist $A^{\theta}$ of $A$ by $\theta$ is also a $3$-dimensional quadratic AS-regular algebra. 

Derivation-quotient algebras
----------------------------

Now, we recall the definitions of superpotentials, twisted superpotentials and derivation-quotient algebras from [@BSW] and [@MS1].  Also, we recall the definition of a [*twist of a superpotential*]{} due to [@MS1] (see Definition \[MStwist\]). 

Fix a basis $\{ x_{1},x_{2},x_{3} \}$ for $V$. For $w\in V^{\otimes 3}$, there exist unique $w_{i}\in V^{\otimes 2}$ such that $w=\sum_{i=1}^{3}x_{i}\otimes w_{i}$. Then the [*partial derivative*]{} of $w$ with respect to $x_{i}$ ($i=1,2,3$) is $
\partial_{x_{i}}(w):=w_{i},
$ and the [*derivation-quotient algebra*]{} of $w$ is $$\mathcal{D}(w):=T(V)/(\partial_{x_{1}}w,\partial_{x_{2}}w,\partial_{x_{3}}w).$$ Note that we call an element $w \in V^{\otimes 3}$ a [*potential*]{} in this paper.  We define the $k$-linear map $\varphi$: $V^{\otimes 3}\longrightarrow V^{\otimes 3}$ by $
\varphi(v_{1}\otimes v_{2}\otimes v_{3}):=v_{3}\otimes v_{1}\otimes v_{2}
$.  We write ${\rm GL}(V)$ for the general linear group of $V$.

\[def\_sp\] Let $w \in V^{\otimes 3}$ be a potential. 

1.   If $\varphi(w)=w$, then $w$ is called a [*superpotential*]{}. 

2.   If there exists $\theta\in {\rm GL}(V)$ such that $$(\theta\otimes {\rm id}\otimes {\rm id})\varphi(w)=w,$$ then $w$ is called a [*twisted superpotential*]{}. 

By Dubois-Violette [@D] and Bocklandt-Schedler-Wemyss [@BSW], every $3$-dimensional quadratic AS-regular algebra $A$ is isomorphic to a derivation-quotient algebra $\mathcal{D}(w)$ of a twisted superpotential $w$, and by Mori-Smith [@MS1], such $w$ is unique up to non-zero scalar multiples. 

\[MStwist\] For a superpotential $w\in V^{\otimes 3}$ and $\theta \in {\rm GL}(V)$, $$w^{\theta}:=(\theta^{2}\otimes \theta\otimes {\rm id})(w)$$is called a [*Mori-Smith twist*]{} ([*MS twist*]{}) of $w$ by $\theta$.

For a potential $w \in V^{\otimes 3}$, we set $${\rm Aut}\,(w):=
\{\theta \in {\rm GL}(V) \mid  
(\theta^{\otimes 3})(w)
=\lambda w,\,\exists \lambda \in k \setminus \{0\}\}.$$ For any $w \in V^{\otimes 3}$, it follows from [@MS1 Lemma 3.1] that ${\rm Aut}\,(w) \subset {\rm Aut}\,\mathcal{D}(w)$.

\[[[@MS1 Proposition 5.2]]{}\] \[twist\] For a superpotential $w\in V^{\otimes 3}$ and $\theta \in {\rm Aut}\,(w)$, we have that $\mathcal{D}(w^{\theta}) \cong \mathcal{D}(w)^{\theta}$.

\[lem-1\] If $w \in V^{\otimes 3}$ is a superpotential and $
\theta \in
{\rm Aut}\,(w)
$, then the MS twist $w^{\theta}$ of a superpotential $w$ by $\theta$ is a twisted superpotential. 

Let $w \in V^{\otimes 3}$ be a superpotential and $\theta \in {\rm Aut}\,(w)$. By definition, there exists $\lambda \in k \setminus \{0\}$ such that $(\theta^{\otimes 3})(w)=\lambda w$. We set $\theta':=\lambda^{-1}\theta^{3} \in {\rm GL}(V)$. Since $w$ is a superpotential, $$\begin{aligned}
(\theta' \otimes {\rm id} \otimes {\rm id})(\varphi(w^{\theta}))
            &=(\theta' \otimes {\rm id} \otimes {\rm id})({\rm id} \otimes \theta^{2} \otimes \theta)(\varphi(w)) \\
            &=\lambda^{-1}(\theta^{3} \otimes \theta^{2} \otimes \theta)(w) 
            =\lambda^{-1}(\theta^{2} \otimes \theta \otimes {\rm id})(\lambda w)
            =w^{\theta},\end{aligned}$$ so the MS twist $w^{\theta}$ is a twisted superpotential. 

If $\theta \in {\rm GL}(V)\setminus{\rm Aut}\,(w)$, then the MS twist $w^{\theta}$ of a superpotential $w\in V^{\otimes 3}$ by $\theta$ may not be a twisted superpotential.  Indeed, let $w:=x^{3}\in V^{\otimes 3}$. Since $\varphi(w)=w$, we see that $w$ is a superpotential.  Take $
\theta:=\left(
\begin{array}{ccc}
0 &1 & 0 \\
1 & 0& 0 \\
0 & 0& 1
\end{array}
\right) \in {\rm GL}_{3}(k)
$.  Then $
w^{\theta}=(\theta^{2}\otimes \theta \otimes {\rm id})(w)=xyx.  
$ Since, for any $\theta' \in {\rm GL}(V)$, $
(\theta'\otimes {\rm id}\otimes {\rm id})(\varphi(w^{\theta}))
=\theta'(x)xy
\neq w^{\theta}
$, the MS twist $w^{\theta}$ is not a twisted superpotential.  Note that we have $\theta \not\in {\rm Aut}\,(w)$ by $
(\theta^{\otimes 3})(w)=y^{3}\neq w
$. 

\[de-pot\] Let $w \in V^{\otimes 3}$ be a potential. 

1.   A potential $w$ is called [*regular*]{} if the derivation-quotient algebra $\mathcal{D}(w)$ is a $3$-dimensional quadratic AS-regular algebra. 

2.   A potential $w$ is called [*Calabi-Yau*]{} if the derivation-quotient algebra $\mathcal{D}(w)$ is a $3$-dimensional Calabi-Yau AS-regular algebra. 

\[Bockdland\] By Bockdlandt [@B], every $3$-dimensional quadratic Calabi-Yau AS-regular algebra is isomorphic to a derivation-quotient algebra $\mathcal{D}(w)$ of a superpotential $w$ ([@B Theorem 3.1]). 

\[CY-superpotential\] Let $w \in V^{\otimes 3}$ be regular.  Then $w$ is Calabi-Yau if and only if it is a superpotential. 

\[lem-2\] If $w$ is a Calabi-Yau superpotential and $\theta \in {\rm Aut}\,(w)$, then the MS twist $w^{\theta}$ of a superpotential $w$ by $\theta$ is a regular twisted superpotential. 

Let $w \in V^{\otimes 3}$ be a Calabi-Yau superpotential and $\theta\in {\rm Aut}\,(w)$. By Lemma \[lem-1\], the MS twist $w^{\theta}$ of $w$ by $\theta$ is a twisted superpotential, so it is sufficient to show that $w^{\theta}$ is regular.  Since $w$ is Calabi-Yau, $\mathcal{D}(w)$ is Calabi-Yau AS-regular, and since $\theta \in {\rm Aut}\,(w)$, by Lemma \[twist\], we have that $\mathcal{D}(w^{\theta}) \cong \mathcal{D}(w)^{\theta}$. Since it holds from Lemma \[twist\_inv\] that AS-regularity is presented by twisting, $\mathcal{D}(w^{\theta})$ is AS-regular, that is, the MS twist $w^{\theta}$ is regular. 

\[ex\_S1\] We set $
w:=(xyz+yzx+zxy)-(zyx+yxz+xzy)\in V^{\otimes 3}
$.  Then, we see that $\varphi(w)=w
$.  So, $w$ is a superpotential.  The derivation-quotient algebra of $w$ is $$\begin{aligned}
\mathcal{D}(w)
&=k\langle x,y,z\rangle/(\partial_{x}w,\partial_{y}w,\partial_{z}w)\\
&=k\langle x,y,z\rangle/(yz-zy,zx-xz,xy-yx)=k[x,y,z].  \end{aligned}$$ It is known that a polynomial ring $k[x,y,z]$ is Calabi-Yau AS-regular.  So, by Definition \[de-pot\] (2), $w$ is a Calabi-Yau superpotential.  Take $
\theta:=\left(
\begin{array}{ccc}
\alpha &0 & 0 \\
0 & \beta& 0 \\
0 & 0& \gamma
\end{array}
\right)\in {\rm GL}_{3}(k)
$.  Calculating the MS twist $w^{\theta}$ of the superpotential $w$ by $\theta$, $$\begin{aligned}
w^{\theta}&=(\theta^{2}\otimes \theta\otimes {\rm id})(w)\\
&=(\alpha^{2}\beta xyz+\beta^{2}\gamma yzx+\alpha\gamma^{2}zxy)
-(\beta\gamma^{2}zyx+\alpha\beta^{2}yxz+\alpha^{2}\gamma xzy).  \end{aligned}$$ Therefore, the derivation-quotient algebra of $w^{\theta}$ is as follows: $$\begin{aligned}
\mathcal{D}(w^{\theta})
&=k\langle x,y,z\rangle/(\partial_{x}w,\partial_{y}w,\partial_{z}w)\\
&=k\langle x,y,z\rangle/
(\alpha^{2}\beta yz-\alpha^{2}\gamma zy,
\beta^{2}\gamma zx-\alpha\beta^{2}xz,
\alpha\gamma^{2}xy-\beta\gamma^{2}yx)\\
&=k\langle x,y,z\rangle/
(\beta yz-\gamma zy,
\gamma zx-\alpha xz,
\alpha xy-\beta yx).  \end{aligned}$$ Since ${\rm Aut}\,(w)={\rm GL}\,(V)$, we see that $\theta \in {\rm Aut}\,(w)$. By Lemma \[lem-2\], $w^{\theta}$ is a regular twisted superpotential, so, $\mathcal{D}(w^{\theta})$ is an AS-regular algebra. 

Geometric algebras
------------------

Let $(E,\mathcal{O}_{E})$ be a scheme where $\mathcal{O}_{E}$ is the structure sheaf on $E$. An [*invertible sheaf*]{} on $E$ is defined to be a locally free $\mathcal{O}_{E}$-module of rank $1$. For a quadratic algebra $A=T(V)/(R)$, we set $$\mathcal{V}(R):=\{(p,q)\in \mathbb{P}(V^{\ast})\times \mathbb{P}(V^{\ast})
\mid f(p,q)=0 \text{ for all } f\in R\}.$$ Let $E \subset \mathbb{P}(V^{\ast})$ be a closed $k$-subscheme and $\sigma$ an automorphism of $E$. For the rest of the paper, we fix

(a) $\pi: E \rightarrow \mathbb{P}(V^{\ast})$ is the embedding,

(b) $\mathcal{L}:=\pi^{\ast}(\mathcal{O}_{\mathbb{P}(V^{\ast})}(1))$.

In this case, $\mathcal{L}$ becomes an invertible sheaf on $E$. The map $$\mu: {\rm H}^{0}(E,\mathcal{L}) \otimes {\rm H}^{0}(E,\mathcal{L})
\rightarrow
{\rm H}^{0}(E,\mathcal{L}) \otimes {\rm H}^{0}(E,\mathcal{L}^{\sigma})
\rightarrow
{\rm H}^{0}(E,\mathcal{L}\otimes_{\mathcal{O}_{E}} \mathcal{L}^{\sigma})$$ of $k$-vector spaces is defined by $v \otimes w \mapsto v \otimes w^{\sigma}$ where $\mathcal{L}^{\sigma}=\sigma^{\ast}\mathcal{L}$ and $w^{\sigma}=w \circ \sigma$.

For a quadratic algebra, a [*geometric algebra*]{} was introduced by Mori [@Mo]. 

A quadratic algebra $A=T(V)/(R)$ is called [*geometric*]{} if there is a pair $(E,\sigma)$ where $E \subset \mathbb{P}(V^{\ast})$ is a closed $k$-subscheme, and $\sigma$ is a $k$-automorphism of $E$ such that

-   (G1): $\mathcal{V}(R)=\{
            (p,\sigma(p)) \in \mathbb{P}(V^{\ast}) \times \mathbb{P}(V^{\ast}) \mid p \in E
            \}$, and

-   (G2): $R={\rm ker} \mu$ with the identification $${\rm H}^{0}(E,\mathcal{L})
            ={\rm H}^{0}(\mathbb{P}(V^{\ast}),\mathcal{O}_{\mathbb{P}(V^{\ast})}(1))
                    =V\text{ as $k$-vector spaces.}$$

When $A$ satisfies the condition (G2), we write $A=\mathcal{A}(E,\sigma)$.

Let $A=T(V)/(R)$ be a quadratic algebra.  If $A=\mathcal{A}(E,\sigma)$ is a geometric algebra, then $E$ is called the point scheme of $A$.  If $E$ is reduced, then the condition (G2) is equivalent to the condition (G2’): $R=\{
f \in V \otimes V \mid f|_{\mathcal{V}(R)}=0
\}$ (see [@Mo]). 

\[ATV1\] Let $A$ be a quadratic algebra.  Then $A$ is a $3$-dimensional AS-regular algebra if and only if $A$ is isomorphic to a geometric algebra $\mathcal{A}(E,\sigma)$ which satisfies one of the following conditions[:]{}

1.   $E=\mathbb{P}^{2}$ and $\sigma \in {\rm Aut}_{k}\,\mathbb{P}^{2}$.

2.   $E$ is a cubic curve in $\mathbb{P}^{2}$ and $\sigma \in {\rm Aut}_{k}\,E$ such that $\sigma^{\ast}\mathcal{L} \not\cong \mathcal{L}$ and $$(\sigma^{2})^{\ast}\mathcal{L} \otimes_{\mathcal{O}_{E}} \mathcal{L}
            \cong \sigma^{\ast}\mathcal{L} \otimes_{\mathcal{O}_{E}} \sigma^{\ast}\mathcal{L}.$$

The types of $(E,\sigma)$ of $3$-dimensional quadratic AS-regular algebras are defined in [@MU1] which are slightly modified from the original types defined in [@AS] and [@ATV1].  We extend the types defined in [@MU1] as follows (see [@IM Subsection 2.3]):

[(1) Type P]{}

:   $E$ is $\mathbb{P}^{2}$, and $\sigma \in \mathrm{Aut}_{k}\mathbb{P}^{2}=\mathrm{PGL}_{3}(k)$ (Type $\mathbb{P}^{2}$ is divided into Type P$_{i}$ ($i=1,2,3$) in terms of the Jordan canonical form of $\sigma$). 

[(2-1) Type S$_{1}$]{}

:   $E$ is a triangle, and $\sigma$ stabilizes each component. 

[(2-2) Type S$_{2}$]{}

:   $E$ is a triangle, and $\sigma$ interchanges two of its components. 

[(2-3) Type S$_{3}$]{}

:   $E$ is a triangle, and $\sigma$ circulates three components. 

[(3-1) Type S$'_{1}$]{}

:   $E$ is a union of a line and a conic meeting at two points, and $\sigma$ stabilizes each component and two intersection points. 

[(3-2) Type S$'_{2}$]{}

:   $E$ is a union of a line and a conic meeting at two points, and $\sigma$ stabilizes each component and interchanges two intersection points. 

[(4-1) Type T$_{1}$]{}

:   $E$ is a union of three lines meeting at one point, and $\sigma$ stabilizes each component. 

[(4-2) Type T$_{2}$]{}

:   $E$ is a union of three lines meeting at one point, and $\sigma$ interchanges two of its components. 

[(4-3) Type T$_{3}$]{}

:   $E$ is a union of three lines meeting at one point, and $\sigma$ circulates three components. 

[(5) Type T$'$]{}

:   $E$ is a union of a line and a conic meeting at one point, and $\sigma$ stabilizes each component. 

[(6) Type CC]{}

:   $E$ is a cuspidal cubic curve. 

[(7) Type NC]{}

:   $E$ is a nodal cubic curve (Type NC is divided into Type NC$_{i}$ $(i=1,2)$). 

[(8) Type WL]{}

:   $E$ is a union of a double line and a line (Type WL is divided into Type WL$_{i}$ $(i=1,2,3)$). 

[(9) Type TL]{}

:   $E$ is a triple line (Type TL is divided into Type TL$_{i}$ $(i=1,2,3,4)$). 

[(10) Type EC]{}

:   $E$ is an elliptic curve. 

\[rem\_IM1\] All possible defining relations of $3$-dimensional quadratic AS-regular algebras are listed in each type up to isomorphism from (1) through (9) in [@IM Theorem 3.1], and (10) in [@IM Theorem 4.9]. 

Classifications of twisted superpotentials
==========================================

In this section, we will give complete lists of superpotentials and twisted superpotentials whose derivation-quotient algebras are $3$-dimensional quadratic AS-regular algebras except for Type EC, by using the following three steps:

Step I

:   (Proposition \[TSP\]) Find the candidates of regular twisted superpotentials corresponding to defining relations listed in [@IM Theorem 3.1]. 

Step II

:   (Theorem \[SP\]) Find all superpotentials among the above candidates and show that they are Calabi-Yau superpotentials. 

Step III

:   (Theorem \[prop-3\], Theorem \[thm-RTSP\]) Show that all above candidates can be written as MS twists of Calabi-Yau superpotentials and that they are in fact regular twisted superpotentials. 

As a byproduct, we will prove that, for any $3$-dimensional quadratic AS-regular algebra $A$ except for Type EC, there exist a Calabi-Yau AS-regular algebra $S$ and $\theta \in {\rm Aut}\,S$ such that $A$ is isomorphic to $S^{\theta}$ as graded $k$-algebras.  This result is needed to prove our main result Theorem \[conj\]. 

\[TSP\] Every $3$-dimensional quadratic AS-regular algebra except for Type EC is isomorphic to $\mathcal{D}(w)$ of a potential $w$ in Table $1$.

[|p[0.6cm]{}|@p[2.9cm]{}|p[1.2cm]{}|p[3.2cm]{}|p[2.9cm]{}|]{}\
&    potential $w$ & Cond. &  $\partial_{x}w$, $\partial_{y}w$, $\partial_{z}w$ & $\nu_{A^{!}}$\
$\rm{P}_1$ & $\begin{array}{l}
             \alpha^2\beta xyz+\beta^2\gamma yzx \\ +\gamma^2\alpha zxy-\alpha^2\gamma xzy
             \\ -\gamma^2\beta zyx-\beta^2\alpha yxz
\end{array}$ & $\alpha\beta\gamma\neq 0$ & $\left\{
              \begin{array}{ll}
               \alpha^2\beta yz-\alpha^2\gamma zy, \\
               \beta^2\gamma zx-\beta^2\alpha xz, \\
               \gamma^2\alpha xy-\gamma^2\beta yx
              \end{array}
             \right. 
             $ &

------------------------------------------------------------------------

$
             \left(
              \begin{array}{ccc}
              \dfrac{\alpha^{2}}{\beta\gamma} & 0& 0 \\
              0 & \dfrac{\beta^{2}}{\alpha\gamma}& 0 \\
              0 & 0& \dfrac{\gamma^{2}}{\alpha\beta}
             \end{array}
             \right)
             $\
$\rm{P}_2$ & $\begin{array}{l}
             xyz+\alpha yzx+ \!  \alpha^{2}zxy
             \\ -\alpha xzy-\alpha^{2}zyx \\ -yxz
             +y^{2}z \! - \!  2\alpha yzy \\ +\alpha^{2}zy^{2}
\end{array}$ & $\alpha\neq 0$ & $\left\{ \!\!  
              \begin{array}{ll}
                yz-\alpha zy, \\
                yz-2\alpha zy+\alpha zx \\ \hfill -xz,\  \\
                \alpha^{2}y^{2}+\alpha^{2}xy-\alpha^{2}yx
              \end{array}
             \right. 
             $ &

------------------------------------------------------------------------

$
             \left(
              \begin{array}{ccc}
              \dfrac{1}{\alpha} & \dfrac{3}{\alpha}& 0 \\
              0 & \dfrac{1}{\alpha}& 0 \\
              0 & 0& \alpha^{2}
             \end{array}
             \right)
             $\
$\rm{P}_3$ & $\begin{array}{l}
             -xyz-yzx-zxy \\ +xzy+zyx+yxz \\
             -z^{2}x+2zxz-xz^{2} \\ -zy^{2}+zyz+z^{2}y \\ 
             -y^{2}z+2yzy-2yz^{2} \\ -z^{3}
\end{array}$ & nothing &

------------------------------------------------------------------------

$\left\{ \!\!  
              \begin{array}{ll}
              zy-yz-z^{2}, \\
              xz-yz-2z^{2}\\
              \hfill -zx+2zy,\  \\
              -xy+yx-y^{2}-zx\\
              \hfill +2xz+yz+zy \\ \hfill -z^{2} 
              \end{array}
             \right. 
             $ & $
             \left(
              \begin{array}{ccc}
              1 & 3& 3 \\
              0 & 1& 3 \\
              0 & 0& 1
             \end{array}
             \right)
             $\

[|p[0.6cm]{}|@p[2.9cm]{}|p[1.2cm]{}|p[3.2cm]{}|p[2.9cm]{}|]{}$\rm{S}_1$ & $\begin{array}{l}
             \beta xyz+\gamma yzx \!  + \!  \alpha zxy \\ -\alpha\beta xzy
             -\alpha\gamma zyx \\ -\beta\gamma yxz
\end{array}$ & $\alpha\beta\gamma\neq 0,1$ & $\left\{
              \begin{array}{ll}
               \beta yz-\alpha\beta zy, \\
               \gamma zx-\beta\gamma xz, \\
               \alpha xy-\alpha\gamma yx
              \end{array}
             \right. 
             $ &

------------------------------------------------------------------------

$
             \left(
              \begin{array}{ccc}
              \dfrac{\beta}{\gamma} & 0& 0 \\
              0 & \dfrac{\gamma}{\alpha}& 0 \\
              0 & 0& \dfrac{\alpha}{\beta}
             \end{array}
             \right)
             $\
$\rm{S}_2$ & $\begin{array}{l}
             -yzx-xzy+ \!  \dfrac{1}{\beta}x^{2}z \\ +\dfrac{1}{\alpha}zx^{2}+\alpha y^{2}z
             \\ +\beta zy^{2}
\end{array}$ & $\alpha\beta\neq 0$ & $\left\{
              \begin{array}{ll}
              \dfrac{1}{\beta}xz-zy, \\
              \alpha yz-zx, \\
              \dfrac{1}{\alpha}x^{2}+\beta y^{2}
              \end{array}
             \right. 
             $ &

------------------------------------------------------------------------

$
             \left(
              \begin{array}{ccc}
              0& -\alpha& 0 \\
              -\dfrac{1}{\beta} & 0& 0 \\
              0 & 0& \dfrac{\beta}{\alpha}
             \end{array}
             \right)
             $\
$\rm{S}_3$ & $\begin{array}{l}
             -xzy-zyx-yxz
             \\ +\beta x^{3}+\gamma y^{3}+\alpha z^{3}
\end{array}$ & $\alpha\beta\gamma\neq 0,1$ & $\left\{
              \begin{array}{ll}
               \beta x^{2}-zy, \\
               \gamma y^{2}-xz, \\
               \alpha z^{2}-yx
              \end{array}
             \right. 
             $ &

------------------------------------------------------------------------

$
             \left(
              \begin{array}{ccc}
              1& 0& 0 \\
              0& 1& 0 \\
              0 & 0& 1
             \end{array}
             \right)
             $\
$\rm{S'}_1$ & $\begin{array}{l}
               \beta xyz+ \!  \beta yzx \!  + \!  \alpha zxy
               \\ -\alpha\beta xzy-\alpha\beta zyx \\ -\beta^{2}yxz+\beta x^{3}
\end{array}$ & $\alpha\beta^{2}\neq 0,1$ & $\left\{
              \begin{array}{ll}
              \beta x^{2}+\beta yz-\alpha\beta zy, \\
              \beta zx-\beta^{2}xz, \\
              \alpha xy-\alpha\beta yx
              \end{array}
             \right. 
             $ &

------------------------------------------------------------------------

$
              \left(
              \begin{array}{ccc}
              1& 0& 0 \\
              0& \dfrac{\beta}{\alpha}& 0 \\
              0 & 0& \dfrac{\alpha}{\beta}
              \end{array}
              \right)
              $\
$\rm{S'}_2$ & $\begin{array}{l}
              -zxy-yxz+xy^{2} \\ +y^{2}x+xz^{2}+z^{2}x
              \\ +x^{3}
\end{array}$ & nothing & $\left\{
              \begin{array}{ll}
               x^{2}+y^{2}+z^{2}, \\
               yx-xz, \\
               zx-xy
              \end{array}
             \right. 
             $ &

------------------------------------------------------------------------

$
              \left(
              \begin{array}{ccc}
              1& 0& 0 \\
              0& 0& -1 \\
              0& -1& 0
              \end{array}
              \right)
              $\
$\rm{T_1}$ & $\begin{array}{l}
             \rule{0pt}{12pt}
             \beta x^2y \\ +(\alpha-\beta+\gamma)xyx \\ 
             +(\alpha-\beta-\gamma)yxy \\ -\alpha y^2x-yxz+yzx
             \\ +\beta yx^2+xyz-xzy \\ -\alpha xy^2+zxy-zyx
\end{array}$ & $\alpha+\beta+\gamma\neq 0$ & $\left\{ \!\! 
              \begin{array}{ll}
              \beta xy+(\alpha \! - \!  \beta+\gamma)yx\\
              \hfill +yz-zy-\alpha y^2, \\
              (\alpha \! - \!  \beta \! - \!  \gamma)xy-\alpha yx\\
              \hfill -xz+zx+\beta x^2, \\
              xy-yx
              \end{array}
             \right. 
             $ & $
              \left(
              \begin{array}{ccc}
              1& 0& 0 \\
              0& 1& 0 \\              
              \delta& \varepsilon& 1
                            \end{array}
              \right)
              $

------------------------------------------------------------------------

$
              \delta:=-\alpha+2\beta-\gamma
              $,

$
              \varepsilon:=2\alpha-\beta-\gamma
              $\
$\rm{T_2}$ & $\begin{array}{l}
\rule{0pt}{12pt}
              (1-\beta-\gamma)x^{3} \\ -(\alpha+2\gamma)yx^{2}+zx^{2}
              \\ -xy^{2}
              +\gamma y^{3}-zy^{2} \\ -x^{2}z+xzy+\beta x^{2}y
              \\ -y^{2}z+yzx+\alpha y^{2}x
\end{array}$ & $\alpha+\beta+\gamma\neq 0$ & $\left\{
              \begin{array}{ll}
              (1-\beta-\gamma)x^{2}-y^{2}\\
              \hfill -xz + zy+\beta xy, \\
             -(\alpha+2\gamma)x^{2}+\gamma y^{2}\\
             \hfill -yz+zx+\alpha yx,\\
             x^{2}-y^{2}
             \end{array}
             \right. 
             $ & $
              \left(
              \begin{array}{ccc}
              0& -1& 0 \\
              -1& 0& 0 \\
              \delta & \varepsilon &  -1
              \end{array}
              \right)
             $

------------------------------------------------------------------------

$\delta:=-\beta+\gamma$

$\varepsilon:= -\alpha+\gamma$\
$\rm{T_3}$ & $\begin{array}{l}
\rule{0pt}{12pt}
             -x^3+y^3+x^2y \\ +xyx +yx^2-xy^2 \\ -yxy  -y^2x+x^2z \\ 
             +xzx 
             +zx^2+
             zy^2 \\ +yzy  +y^2z-xyz \\ -yzx  -zxy
\end{array}$ & nothing & $
             \left\{
             \begin{array}{ll}
             -x^{2}+xy+yx-y^{2} \\
             \quad +xz+zx-yz,\\
             y^{2}+x^{2}-xy-yx\\
             \quad+zy+yz-zx, \\
             x^{2}+y^{2}-xy
             \end{array}
             \right. 
             $ & $
              \left(
              \begin{array}{ccc}
              1& 0& 0 \\
              0& 1& 0 \\
              0& 0& 1
              \end{array}
              \right)
             $\
$\rm{T'}$ & $\begin{array}{l}
\rule{0pt}{12pt}
            \alpha x^2y- \! ( \alpha-2\beta)xyx \\ 
            +(\beta^2-\alpha\beta)xy^2 \\
            +xyz -xzy+\alpha yx^2 \\ -yxz  +yzx-\alpha yzy
            \\ +\alpha \beta^2 y^3 \\ -(\beta^2-\alpha \beta)y^2x \\
            -\beta y^2z+zxy-zyx \\ -\beta zy^2
\end{array}$ & $\alpha+2\beta\neq 0$ & $
             \left\{
             \begin{array}{ll}
            \alpha xy-(\alpha-2\beta)yx\\
            \hfill +(\beta^{2}-\alpha\beta)y^{2} \\
            \hfill +yz-zy, \\
            \alpha x^{2}-xz+zx \\ 
            \hfill -\alpha zy+\alpha \beta^{2} y^{2}\\
            \hfill -(\beta^2-\alpha \beta)yx \\ 
            \hfill -\beta yz, \\
            xy-yx-\beta y^{2}
             \end{array}
             \right. 
             $ & $
              \left(
              \begin{array}{ccc}
              1& \delta& 0 \\
              0& 1& 0 \\
              2 \delta & \delta^{2}& 1
              \end{array}
              \right)
              $

------------------------------------------------------------------------

$\delta:=\alpha-\beta$\

[|p[0.6cm]{}|@p[2.9cm]{}|p[1.2cm]{}|p[3.2cm]{}|p[2.9cm]{}|]{}$\rm{CC}$ & $\begin{array}{l}
\rule{0pt}{12pt}
            -3x^{3}-y^{2}x-yxy \\ -xy^{2}+y^{2}z+yzy \\ +zy^{2}
            -xyz-yzx \\ -zxy+xzy+zyx \\ +yxz
\end{array}$ & nothing & $
             \left\{ \!\! \!  
             \begin{array}{ll}
            -3x^{2}-y^{2}-yz+zy, \\
            -yx-xy+yz+zy\\
            \hfill -zx+xz, \\
            y^{2}-xy+yx
             \end{array}
             \right. 
             $ & $
            \left(
            \begin{array}{ccc}
            1& 0& 0 \\
            0& 1& 0 \\
            0& 0& 1
            \end{array}
            \right)
            $\
$\rm{NC}_1$ & $\begin{array}{l}
              \dfrac{1 \! -  \alpha^{3}}{\alpha} x^{3} \!  + \!  
              \dfrac{1 \!  -\alpha^{3}}{\alpha}y^{3} \\
              +xyz+yzx+zxy \\ -\alpha(xzy+zyx \\ \hfill +yxz)
\end{array}$ & $\alpha^{3}\neq 0,1$ &

------------------------------------------------------------------------

$
             \left\{
             \begin{array}{ll}
              \dfrac{1-\alpha^{3}}{\alpha} x^{2}+yz \\
              \hfill -\alpha zy, \\
              \dfrac{1-\alpha^{3}}{\alpha}y^{2}+zx \\
              \hfill -\alpha xz, \\
              xy-\alpha yx
             \end{array}
             \right. 
             $ & $
              \left(
              \begin{array}{ccc}
              1& 0& 0 \\
              0& 1& 0 \\
              0& 0& 1
              \end{array}
              \right)
              $\
$\rm{NC}_2$ & $\begin{array}{l}
\rule{0pt}{12pt}
              -2xyx+x^{2}z+ \!  zx^{2} \\ -2yxy+y^{2}z+zy^{2} \\ +yzx+xzy
\end{array}$ & nothing & $
             \left\{
             \begin{array}{ll}
              -2yx+xz+zy, \\
              -2xy+yz+zx,\\
              x^{2}+y^{2}
             \end{array}
             \right. 
             $ & $
              \left(
              \begin{array}{ccc}
              0& 1& 0 \\
              1& 0& 0 \\
              0& 0& 1
              \end{array}
              \right)
              $\

$\rm{WL}_1$ & $\begin{array}{l}
 \rule{0pt}{12pt}
              -(1+\gamma)y^2x \\ +\alpha(1+2\gamma)yxy \\
              -\alpha^2(1+\gamma)xy^2 \\ +\alpha^2xyz+yzx \\
              +\alpha zxy-\alpha^2 xzy \\ -zyx-\alpha yxz
\end{array}$ & $\alpha\neq 0,1$ & $
             \left\{ \!\!\! 
             \begin{array}{ll}
              -\alpha^{2}(1+\gamma)y^{2} \\
              \hfill +\alpha^{2}yz-\alpha^{2}zy, \\
              -(1+\gamma)yx \\ 
              \  +\alpha(1+2\gamma)xy +zx \\ 
              \hfill -\alpha xz, \\
              \alpha xy-yx
             \end{array}
             \right. 
             $ & $
              \left(
              \begin{array}{ccc}
              \alpha^{2}& 0& 0 \\
              0& \dfrac{1}{\alpha}& 0 \\
              0& \delta& \dfrac{1}{\alpha}
              \end{array}
              \right)
              $

$\delta:=\dfrac{1}{\alpha}(2+3\gamma)$\
$\rm{WL}_2$ & $\begin{array}{l}
              -(1+\gamma)y^{2}x \\
              +(1+2\gamma)yxy \\ 
              -(1+\gamma)xy^{2}+xyz \\ +yzx+zxy
              -xzy \\ -zyx-yxz
\end{array}$ & nothing &

------------------------------------------------------------------------

$
             \left\{
             \begin{array}{ll}
              -(1+\gamma)y^{2}+yz \\
              \hfill -zy, \\
              -(1+\gamma)yx \\
              \ \  +(1+2\gamma)xy +zx \\
              \hfill -xz, \\
              xy-yx
             \end{array}
             \right. 
             $ & $
              \left(
              \begin{array}{ccc}
              1& 0& 0 \\
              0& 1& 0 \\
              0& 2+3\gamma& 1
              \end{array}
              \right)
              $\
$\rm{WL}_3$ & $\begin{array}{l}
 \rule{0pt}{12pt}
              x^{2}y-2xyx+yx^{2} \\ -(1+\gamma)y^{2}x \\ +(1+2\gamma)yxy
              \\ -(1+\gamma)xy^{2}+xyz \\ +yzx+zxy-xzy \\ -zyx-yxz
\end{array}$ & nothing & $
             \left\{ \!\!\! 
             \begin{array}{ll}
              xy-2yx-(1+\gamma)y^{2} \\
              \hfill +yz-zy, \\
              x^{2}-(1+\gamma)yx \\
              \hfill +(1+2\gamma)xy +zx \\
              \hfill -xz, \\
              xy-yx
             \end{array}
             \right. 
             $ & $
              \left(
              \begin{array}{ccc}
              1& 0& 0 \\
              0& 1& 0 \\
              3& 2+3\gamma& 1
              \end{array}
              \right)
              $\
$\rm{TL}_1$& $\begin{array}{l}
 \rule{0pt}{18pt}
             -\dfrac{1}{\alpha^{2}}zxy+\dfrac{1}{\alpha}zyx \\[8pt] +\alpha^{2}yxz
             \\ -\alpha yzx+\dfrac{1}{\alpha}xzy \\[8pt] -\alpha xyz -x^{3}
\end{array}$ & $\alpha\neq 0$ & $
             \left\{
             \begin{array}{ll}
             \dfrac{1}{\alpha}zy-\alpha yz -x^{2}, \\
             \alpha^{2}xz-\alpha zx, \\
             -\dfrac{1}{\alpha^{2}}xy+\dfrac{1}{\alpha}yx
             \end{array}
             \right. 
             $ & $
             \left(
             \begin{array}{ccc}
             1& 0& 0 \\
             0& \alpha^{3}& 0 \\
             0& 0& \dfrac{1}{\alpha^{3}}
             \end{array}
             \right)
             $\
$\rm{TL}_2$& $\begin{array}{l}
             \beta x^2y+\beta xyx \\ 
             +(-\beta^2 \! - \!  1)x^3 \! \!  + \!  2yxy \\ 
             -y^2x \! - \!  2\beta yx^2+zxy
             \\ -zyx \! - \!  \beta zx^2 \!  -\beta x^2z \\ 
             +2\beta xzx \! - \!  yxz \!  + \!  yzx 
             \\ -xzy +xyz -xy^2
\end{array}$ & nothing &

------------------------------------------------------------------------

$
             \left\{ \!\! 
             \begin{array}{ll}
             \beta xy+\beta yx \\
             +(-\beta^2-1)x^{2}  -  \beta xz \\
             \hfill +2\beta zx-zy+yz  \\
             \hfill -y^{2}, \\
             2xy-yx-2\beta x^{2} \\
             \hfill -xz +zx, \\
             xy-yx-\beta x^{2}
             \end{array}
             \right. 
             $ & $
             \left(
             \begin{array}{ccc}
             1& 0& 0 \\
             3\beta& 1& 0 \\
             3\beta& 3& 1
             \end{array}
             \right)
             $\

[|p[0.6cm]{}|@p[2.9cm]{}|p[1.2cm]{}|p[3.2cm]{}|p[2.9cm]{}|]{}$\rm{TL}_3$& $\begin {array}{l}
             -2yxy-y^2x+zxy \\ +zyx-yxz-yzx \\ +xzy-xyz
             - x^3 \\
              -  xy^2
\end{array}$ & nothing &

------------------------------------------------------------------------

$
             \left\{ \! 
             \begin{array}{ll}
             zy-yz-x^{2}-y^{2}, \\
             -2xy-yx-xz \\
             \hfill -zx, \\
             xy+yx
             \end{array}
             \right. 
             $ & $
             \left(
             \begin{array}{ccc}
             1& 0& 0 \\
             0& -1& 0 \\
             0& 3& -1
             \end{array}
             \right)
             $\
$\rm{TL}_4$& $\begin{array}{l}
 \rule{0pt}{12pt}
             -x^3+yx^2+x^2y \\ -2xyx+zxy+xyz \\ +yzx-zyx-yxz \\ -xzy
\end{array}$ & nothing & $
             \left\{
             \begin{array}{ll}
             -x^{2}+xy-2yx \\
             \hfill +yz-zy, \\
             x^{2}+zx-xz,\\
             xy-yx
             \end{array}
             \right. 
             $ & $
             \left(
             \begin{array}{ccc}
             1& 0& 0 \\
             0& 1& 0 \\
             3& 0& 1
             \end{array}
             \right)
             $\

All possible defining relations $f_{1},f_{2},f_{3}$ of $3$-dimensional quadratic AS-regular algebras except for Type EC were given in [@IM Theorem 3.1].  In each type, it is enough to find $w$ such that $(\partial_{x}{w},\partial_{y}{w},\partial_{z}{w})=(f_{1},f_{2},f_{3})$.

We will give a proof for Type T$_{2}$ algebras.  For the other types, the proofs are similar.  From [@IM Theorem 3.1], Type T$_{2}$ algebras are given as $A=k\langle x,y,z \rangle/(f_{1},f_{2},f_{3})$: $$\begin{cases}
f_{1}&=x^{2}-y^{2}, \\
f_{2}&=xz-zx-\beta xy+(\beta+\gamma)y^{2},\\
f_{3}&=yz-zx-\alpha yx+(\alpha+\gamma)x^{2}, 
\end{cases}$$ where $\alpha+\beta+\gamma\neq 0$.

Taking a potential $$\begin{array}{l}
w=(1-\beta-\gamma)x^{3}-(\alpha+2\gamma)yx^{2}+zx^{2}
     -xy^{2}+\gamma y^{3}-zy^{2}\\
      \quad\quad -x^{2}z+xzy+\beta x^{2}y-y^{2}z+yzx+\alpha y^{2}x
\end{array}$$ as in Table 1, we have $$\begin{cases}
\partial_{x}w&=(1-\beta-\gamma)x^{2}-y^{2}-xz+zy+\beta xy,\\
\partial_{y}w&=-(\alpha+2\gamma)x^{2}+\gamma y^{2}-yz+zx+\alpha yx,\\
\partial_{z}w&=x^{2}-y^{2}.  
\end{cases}$$ Since $\partial_{z}w=x^{2}-y^{2}=f_{1}$, $\partial_{x}w=f_{1}-f_{2}$ and $\partial_{y}w=-\gamma f_{1}-f_{3}$, it follows that $
\mathcal{D}(w)
=k\langle x,y,z\rangle/(\partial_{x}w, \partial_{y}w,\partial_{z}w)
=k\langle x,y,z\rangle/(f_{1}, f_{2}, f_{3})=A
$. 

For an algebra $A=k \langle x,y,z \rangle/(f_{1},f_{2},f_{3})$ of any Type, we can take a potential $w=xf_{1}+yf_{2}+zf_{3}$ such that $A=\mathcal{D}(w)$. But it is difficult to check that this potential $w$ is a regular twisted superpotential and, in many cases, it is not so.  The potentials $w$ listed in Proposition \[TSP\] were chosen so that they are candidates of regular twisted superpotentials. 

Next, we give a complete list of Calabi-Yau superpotentials as follows:

\[SP\] For every type except for Type EC, the following table is a complete list of Calabi-Yau superpotentials $w_{0}$.

[|c|p[4.5cm]{}|p[1.3cm]{}|p[4.6cm]{}|]{}\
& CY Superpotential $w_{0}$ & Cond. &  $\partial_{x}w_{0}$, $\partial_{y}w_{0}$, $\partial_{z}w_{0}$\
$\rm{P}_1$ & $\begin{array}{l}
             xyz+yzx+zxy \\ -\alpha(xzy+zyx+yxz)
\end{array}$ & $\alpha^{3}=1$ &

------------------------------------------------------------------------

$\left\{
             \begin{array}{ll}
             yz-\alpha zy,\\
             zx-\alpha xz,\\
             xy-\alpha yx
             \end{array}
             \right. 
             $\
$\rm{S}_1$ & $\begin{array}{l}
              xyz+ yzx+ zxy \\ -\alpha (xzy+zyx+yxz)
\end{array}$ & $\alpha^{3}\neq 0,1$ &

------------------------------------------------------------------------

$\left\{
             \begin{array}{ll}
              yz-\alpha zy,\\
              zx-\alpha xz,\\
              xy-\alpha yx
             \end{array}
             \right. 
             $\
$\rm{S}_3$ & $
             xzy+zyx+yxz-\alpha(x^{3}+y^{3}+z^{3})
             $ & $\alpha\neq 0,1$ &

------------------------------------------------------------------------

$\left\{
             \begin{array}{ll}
             zy-\alpha x^{2},\\
             xz-\alpha y^{2},\\
             yx-\alpha z^{2}
             \end{array}
             \right. 
             $\
$\rm{S'}_1$ & $\begin{array}{l}
                xyz+ yzx+ zxy \\
               -\alpha (xzy+zyx+yxz)+x^{3}
\end{array}$ & $\alpha^{3}\neq 0,1$ &

------------------------------------------------------------------------

$\left\{
             \begin{array}{ll}
               yz-\alpha zy+x^{2}, \\
               zx-\alpha xz, \\
               xy-\alpha yx
             \end{array}
             \right. 
             $\
$\rm{T_1}$ & $\begin{array}{l}
             xyz+yzx+zxy \\ -(xzy+zyx+yxz) \\ 
             +(x^2y+xyx+yx^2) \\
             -(y^2x+yxy+xy^2) 
\end{array}$ & nothing & $\left\{
             \begin{array}{ll}
             yz-zy+xy+yx-y^{2}, \\
             zx-xz+x^{2}-yx-xy, \\
             xy-yx
             \end{array}
             \right. 
             $\
$\rm{T_3}$ & $\begin{array}{l}
             xyz+yzx+zxy \\ -(x^{2}y+xyx+yx^{2}) \\ 
             +(xy^{2}+yxy+y^{2}x) \\ -(x^{2}z+xzx+zx^{2}) \\ -
             (zy^{2}+yzy+y^{2}z)+x^{3}-y^{3}
\end{array}$ & nothing &

------------------------------------------------------------------------

$\left\{
             \begin{array}{ll}
             yz-xy-yx+y^{2}-xz-zx \\
             \hfill +x^{2}, \\
             zx-x^{2}+xy+yx-zy-yz \\ 
             \hfill -y^{2}, \\
             xy-x^{2}-y^{2}
             \end{array}
             \right. 
             $\
$\rm{T'}$ & $\begin{array}{l}
             xyz+yzx+zxy \\ -(xzy+zyx+yxz) \\
            +(x^{2}y+xyx+yx^{2}) \\
            -(y^{2}z+yzy+zy^{2})+y^{3}
\end{array}$ & nothing & $\left\{
             \begin{array}{ll}
            yz-zy+xy+yx, \\
            zx-xz+x^{2}-yz-zy+y^{2}, \\
            xy-yx-y^{2}
             \end{array}
             \right. 
             $\
$\rm{CC}$ & $\begin{array}{l}
            xyz+yzx+zxy \\ -(xzy+zyx+yxz) \\ 
            +(y^{2}x+yxy+xy^{2}) \\ -(y^{2}z+yzy+zy^{2})+3x^{3}
\end{array}$ & nothing & $\left\{
             \begin{array}{ll}
            yz-zy+y^{2}+3x^{2}, \\
            zx-xz+yx+xy-yz-zy, \\
            xy-yx-y^{2}
             \end{array}
             \right. 
             $\
$\rm{NC}_1$ & $\begin{array}{l}
                            xyz+yzx+zxy \\ -\alpha(xzy+zyx+yxz)
              +x^{3} {\!}  + {\!}  y^{3}
\end{array}$ & $\alpha^{3}\neq 0,1$ &

------------------------------------------------------------------------

$\left\{
             \begin{array}{ll}
               yz-\alpha zy+x^{2}, \\
               zx-\alpha xz+y^{2}, \\
               xy-\alpha yx
             \end{array}
             \right. 
             $\

$\rm{WL}_2$ & $\begin{array}{l}
                            xyz+yzx+zxy \\ -(xzy+zyx+yxz) \\
              -\dfrac{1}{3}(y^{2}x+yxy+xy^{2})
\end{array}$ & nothing &

------------------------------------------------------------------------

$\left\{
             \begin{array}{ll}
             yz-zy-\dfrac{1}{3}y^{2},
             \vspace{0.5em} \\
             zx-xz-\dfrac{1}{3}(yx+xy), \\
             xy-yx
             \end{array}
             \right. 
             $\
$\rm{TL}_1$& $\begin{array}{l}
             xyz+yzx+zxy \\
             -\alpha(xzy+zyx+yxz)
              -x^{3}
\end{array}$ & $\alpha^{3}=1$ &

------------------------------------------------------------------------

$\left\{
             \begin{array}{ll}
             yz-\alpha zy -x^{2},\\
             zx-\alpha xz,\\
             xy-\alpha yx
             \end{array}
             \right. 
             $\

Let $A$ be a $3$-dimensional quadratic AS-regular algebra except for Type EC.  By Lemma \[thmRRZ\], $A$ is Calabi-Yau if and only if the Nakayama automorphism $\nu_{A^{!}}$ is identity, where $A^{!}$ is the quadratic dual of $A$.  Considering the condition that $\nu_{A^{!}}$ in Table 1 is the identity, we have a superpotential, so it is sufficient to show that $w_{0}$ is regular, that is, $\mathcal{D}(w_{0})=k\langle x,y,z \rangle /(\partial_{x}w_{0},\partial_{y}w_{0},\partial_{z}w_{0})$ is a $3$-dimensional quadratic AS-regular algebra.  In fact, if $w$ is regular, then by Lemma \[CY-superpotential\], it is Calabi-Yau.  In order to prove AS-regularity of $\mathcal{D}(w_{0})$, we will check that $\mathcal{D}(w_{0})$ satisfies the conditions of Theorem \[ATV1\_thm1\].  Note that if $w$ is a superpotential, then the derivation-quotient algebra $\mathcal{D}(w)$ is standard if and only if the partial derivatives $\partial_{x}x$, $\partial_{y}w$, $\partial_{z}w$ are linearly independent (for example, see [@MS2 Proposition 2.6]). 

We will give a proof for Type T$_{1}$ algebras.  For the other types, the proofs are similar.  Let $w_{0}=xyz+yzx+zxy-(xzy+zyx+yxz) 
+(x^2y+xyx+yx^2)
-(y^2x+yxy+xy^2)$. It is easy to check that $\partial_{x}w_{0},\partial_{y}w_{0},\partial_{z}w_{0}$ are linearly independent.  For the potential $w_{0}$, we have the unique $3 \times 3$ matrix

$
M:=
\left(
\begin{array}{ccc}
y & x-y-z & y \\
x-y+z & -x & -x \\
-y & x & 0
\end{array}
\right)
$ such that $\left(
\begin{array}{c}
\partial_{x}w_{0} \\
\partial_{y}w_{0} \\
\partial_{z}w_{0}
\end{array}
\right)=M\left(
\begin{array}{c}
x \\
y \\
z
\end{array}
\right)$.

By calculation, we have $
\left(
\begin{array}{ccc}
x & y & z
\end{array}
\right)
M=
\left(
\begin{array}{ccc}
\partial_{x}w_{0} &
\partial_{y}w_{0} &
\partial_{z}w_{0}
\end{array}
\right)
$.  Hence, $\mathcal{D}(w_{0})$ is standard.  We denote by $\Delta_{ij}$ the $(i,j)$-th $2\times 2$ minors of the matrix $M$ ($1\leq i,j \leq 3$).  Since $\Delta_{11}=-x^{2}$, $\Delta_{22}=y^{2}$ and $\Delta_{33}=-x^{2}+xy-y^{2}+z^{2}$, we have that $\mathcal{V}(\{\Delta_{ij}\mid 1\leq i,j \leq 3\})=\emptyset$.

Therefore, by Theorem \[ATV1\_thm1\], $\mathcal{D}(w_{0})$ is a $3$-dimensional quadratic AS-regular algebra, that is, $w_{0}$ is a Calabi-Yau superpotential. 

\[prop-3\] For a potential $w$ in Table $1$, there exist a Calabi-Yau superpotential $w_{0}$ in Table $2$ and $\theta \in {\rm Aut}\,(w_{0})$ such that $w=(w_{0})^{\theta}$ as in Table $3$.

[|p[0.8cm]{}|p[5cm]{}|p[5cm]{}|]{}\
& $w_{0}$ & $\theta$\
P$_{1}$ & $xyz+yzx+zxy-(xzy+zyx+yxz)$ &

------------------------------------------------------------------------

$
                     \left(
                     \begin{array}{ccc}
                     \alpha & 0& 0 \\
                     0 & \beta & 0 \\
                    0 & 0& \gamma
                    \end{array}
                     \right)
                    $\
P$_{2}$ & $xyz+yzx+zxy-(xzy+zyx+yxz)$ &

------------------------------------------------------------------------

$
                        \left(
                        \begin{array}{ccc}
                        1 & 1& 0 \\
                        0 & 1 & 0 \\
                        0 & 0& \alpha
                        \end{array}
                        \right)
                        $\
P$_{3}$ & $xyz+yzx+zxy-(xzy+zyx+yxz)$ &

------------------------------------------------------------------------

$
                      \left(
                      \begin{array}{ccc}
                      1 & 1& 0 \\
                      0 & 1 & 1 \\
                      0 & 0& 1
                     \end{array}
                     \right)
                   $\
S$_{1}$ & $(xyz+yzx+zxy)-\sqrt[3]{\alpha\beta\gamma}(xzy+yxz+zyx)$ &

------------------------------------------------------------------------

$
                          \left(
                          \begin{array}{c@{}c@{}c}
                          \sqrt[3]{\beta}{\gamma^{-1}} & 0& 0 \\
                           0 & \sqrt[3]{\gamma\alpha^{-1}} & 0 \\
                          0 & 0& \sqrt[3]{\alpha\beta^{-1}}
                          \end{array}
                          \right)
                           $\
S$_{2}$ & $(xyz+yzx+zxy)+(xzy+yxz+zyx)$ &

------------------------------------------------------------------------

$
                         \left(
                         \begin{array}{c@{}c@{}c}
                         0 & -\sqrt[3]{\alpha^{2}\beta}& 0 \\
                         -\frac{1}{\sqrt[3]{\alpha\beta^{2}}} & 0 & 0 \\
                        0 & 0& \sqrt[3]{\beta\alpha^{-1}}
                        \end{array}
                        \right)
                        $\

[|p[0.8cm]{}|p[5cm]{}|p[5cm]{}|]{} S$_{3}$ & $(xzy+yxz+zyx)-\alpha(x^{3}+y^{3}+z^{3})$ &

------------------------------------------------------------------------

$
                         \left(
                         \begin{array}{ccc}
                          1 & 0& 0 \\
                          0 & 1 & 0 \\
                          0 & 0& 1
                          \end{array}
                           \right)
                          $\
S’$_{1}$ & $(xyz+yzx+zxy)-\sqrt[3]{\alpha\beta^{2}}(xzy+yxz+zyx)+x^{3}$ &

------------------------------------------------------------------------

$
                           \left(
                           \begin{array}{ccc}
                            1 & 0& 0 \\
                            0 & \sqrt[3]{\beta\alpha^{-1}} & 0 \\
                           0 & 0& \sqrt[3]{\alpha\beta^{-1}}
                           \end{array}
                            \right)
                           $\
S’$_{2}$ & $(xyz+yzx+zxy)+(xzy+yxz+zyx)+x^{3}$ &

------------------------------------------------------------------------

$
                      \left(
                      \begin{array}{ccc}
                      1 & 0& 0 \\
                      0 & 0 & 1 \\
                      0 & 1& 0
                      \end{array}
                      \right)
                       $\
T$_{1}$ & $(xyz+yzx+zxy)-(xzy+yxz+zyx)+(x^{2}y+xyx+yx^{2})-(xy^{2}+yxy+y^{2}x)$ &

------------------------------------------------------------------------

$
                          \left(
                          \begin{array}{ccc}
                            1 & 0& 0 \\
                            0 & 1 & 0 \\
                           \lambda\nu^{-1} & \mu\nu^{-1}& 1
                          \end{array}
                          \right)
                         $

------------------------------------------------------------------------

$\lambda:=\frac{1}{3}(-\alpha+2\beta-\gamma)$,

------------------------------------------------------------------------

$\mu:=\frac{1}{3}(2\alpha-\beta-\gamma)$,

------------------------------------------------------------------------

$\nu:=\frac{1}{3}(\alpha+\beta+\gamma)$\
T$_{2}$ & $(xyz+yzx+zxy)-(xzy+yxz+zyx)+(x^{2}y+xyx+yx^{2})-(xy^{2}+yxy+y^{2}x)$ &

------------------------------------------------------------------------

$
                         \left(
                         \begin{array}{ccc}
                         0 & -1& 0 \\
                         -1 & 0 & 0 \\
                        -\lambda\nu^{-1} & -\mu\nu^{-1}& -1
                        \end{array}
                        \right)
                        $

------------------------------------------------------------------------

$\lambda:=\frac{1}{3}(-\alpha+2\beta-\gamma)$,

------------------------------------------------------------------------

$\mu:=\frac{1}{3}(2\alpha-\beta-\gamma)$,

------------------------------------------------------------------------

$\nu:=\frac{1}{3}(\alpha+\beta+\gamma)$\
T$_{3}$ &

------------------------------------------------------------------------

$\begin{array}{@{}l}(xyz+yzx+zxy)-(x^{2}y+xyx \\
+yx^{2})+(xy^{2}+yxy+y^{2}x) \\ -(x^{2}z+xzx+zx^{2})-(zy^{2} \\
+yzy+y^{2}z)+x^{3}-y^{3}\end{array}$ & $
                        \left(
                        \begin{array}{ccc}
                        1 & 0& 0 \\
                        0 & 1 & 0 \\
                        0 & 0& 1
                        \end{array}
                          \right)$\
T’ & $(xyz+yzx+zxy)-(xzy+yxz+zyx)+(x^{2}y+xyx+yx^{2})-(y^{2}z+yzy+zy^{2})+y^{3}$ &

------------------------------------------------------------------------

$
                 \left(
                 \begin{array}{ccc}
                 1 & \lambda^{-1}\mu& 0 \\
                 0 & 1 & 0 \\
                 2\lambda^{-1}\mu & \lambda^{-2}\mu^{2}& 1
                 \end{array}
                 \right)
                 $

------------------------------------------------------------------------

$\lambda:=\frac{1}{3}(\alpha+2\beta)$,

------------------------------------------------------------------------

$\mu:=\frac{1}{3}(\alpha-\beta)$\
CC & $\begin{array}{@{}l}(xyz+yzx+zxy)-(xzy+yxz \\
+zyx)+(y^{2}x+yxy+xy^{2}) \\
-(y^{2}z+yzy+zy^{2})+3x^{3}\end{array}$ &

------------------------------------------------------------------------

$
               \left(
               \begin{array}{ccc}
                1 & 0& 0 \\
               0 & 1 & 0 \\
               0 & 0& 1
               \end{array}
               \right)
               $\
NC$_{1}$ & $(xyz+yzx+zxy)-\alpha(xzy+yxz+zyx)+x^{3}+y^{3}$ &

------------------------------------------------------------------------

$
                           \left(
                           \begin{array}{ccc}
                            1 & 0& 0 \\
                            0 & 1 & 0 \\
                            0 & 0& 1
                           \end{array}
                          \right)
                          $\
NC$_{2}$ & $(xyz+yzx+zxy)+(xzy+yxz+zyx)+x^{3}+y^{3}$ &

------------------------------------------------------------------------

$
                        \left(
                        \begin{array}{ccc}
                        0 & 1& 0 \\
                        1 & 0 & 0 \\
                        0 & 0& 1
                      \end{array}
                       \right)
                      $\

[|p[0.8cm]{}|p[5cm]{}|p[5cm]{}|]{} WL$_{1}$ & $\begin{array}{@{}l}xyz+yzx+zxy-(xzy+zyx \\
+yxz)-\dfrac{1}{3}(y^{2}x+yxy+xy^{2})\end{array}$ &

------------------------------------------------------------------------

$
                          \left(
                          \begin{array}{ccc}
                          \alpha & 0& 0 \\
                           0 & 1 & 0 \\
                           0 & \dfrac{2}{3}+\gamma & 1
                          \end{array}
                          \right)
                        $\
WL$_{2}$ & $\begin{array}{@{}l}xyz+yzx+zxy-(xzy+zyx \\
+yxz)-\dfrac{1}{3}(y^{2}x+yxy+xy^{2})\end{array}$ &

------------------------------------------------------------------------

$
                          \left(
                         \begin{array}{ccc}
                         1 & 0& 0 \\
                         0 & 1 & 0 \\
                        0 & \dfrac{2}{3}+\gamma & 1
                       \end{array}
                       \right)
                      $\
WL$_{3}$ & $\begin{array}{@{}l}xyz+yzx+zxy-(xzy+zyx \\
+yxz)-\dfrac{1}{3}(y^{2}x+yxy+xy^{2})\end{array}$ &

------------------------------------------------------------------------

$
                         \left(
                        \begin{array}{ccc}
                         1 & 0& 0 \\
                         0 & 1 & 0 \\
                          1 & \dfrac{2}{3}+\gamma& 1
                         \end{array}
                         \right)
                        $\

TL$_{1}$ & $xyz+yzx+zxy-(xzy+zyx+yxz)-x^{3}$ &

------------------------------------------------------------------------

$
                          \left(
                          \begin{array}{ccc}
                          1 & 0& 0 \\
                          0 & \gamma & 0 \\
                          0 & 0& \gamma^{-1}
                         \end{array}
                        \right)
                        $\
TL$_{2}$ & $xyz+yzx+zxy-(xzy+zyx+yxz)-x^{3}$ &

------------------------------------------------------------------------

$
                         \left(
                         \begin{array}{ccc}
                         1 & 0& 0 \\
                         \beta & 1 & 0 \\
                         0 & 1& 1
                        \end{array}
                        \right)
                        $\
TL$_{3}$ & $xyz+yzx+zxy-(xzy+zyx+yxz)-x^{3}$ &

------------------------------------------------------------------------

$
                          \left(
                          \begin{array}{ccc}
                          1 & 0& 0 \\
                          0 & -1 & 0 \\
                          0 & 1& -1
                          \end{array}
                          \right)
                           $\
TL$_{4}$ & $xyz+yzx+zxy-(xzy+zyx+yxz)-x^{3}$ &

------------------------------------------------------------------------

$
                       \left(
                       \begin{array}{ccc}
                       1 & 0& 0 \\
                       0 & 1 & 0 \\
                       1 & 0& 1
                       \end{array}
                       \right)
                       $\

By direct computation, for a potential $w$ in Table 1, we find a Calabi-Yau superpotential $w_{0}$ in Table 2 and $\theta \in {\rm Aut}\,(w_{0})$ such that $w=(w_{0})^{\theta}$ as in Table 3. 

We will give a proof for Type T$_{1}$ algebra.  For the other types, the proofs are similar.  Let $w$ be a potential of Type T$_{1}$ in Table $1$. We take a superpotential $w_{0}=xyz+yzx+zxy-(xzy+yxz+zyx)+(x^{2}y+xyx+yx^{2})-(xy^{2}+yxy+y^{2}x)$ and $\theta=
    \left(
    \begin{array}{ccc}
    1 & 0& 0 \\
    0 & 1 & 0 \\
    \lambda\nu^{-1} & \mu\nu^{-1}& 1
    \end{array}
    \right)
    $ where $\lambda:=\frac{1}{3}(-\alpha+2\beta-\gamma)$, $\mu:=\frac{1}{3}(2\alpha-\beta-\gamma)$ and $\nu:=\frac{1}{3}(\alpha+\beta+\gamma)$. Since $(\theta^{\otimes 3})(w_{0})=w_{0}$, $\theta \in {\rm Aut}\,(w_{0})$. By calculation, we have that $(w_{0})^{\theta}=(xyz+yzx+zxy)-(xzy+yxz+zyx)
    +\nu^{-1}(\beta x^{2}y+(\alpha-\beta+\gamma)xyx+\beta yx^{2})
    -\nu^{-1}(\alpha y^{2}x+(-\alpha+\beta+\gamma)yxy+\alpha xy^{2})$. By taking $\theta'=
    \left(
    \begin{array}{ccc}
    1 & 0& 0 \\
    0 & 1 & 0 \\
    0 & 0 & \nu^{-1}
    \end{array}
    \right)
    \in {\rm GL}_{3}(k)$, it follows that $(\theta'^{\otimes^{3}})((w_{0})^{\theta})=\nu^{-1}w$, so $\mathcal{D}((w_{0})^{\theta}) \cong \mathcal{D}(w)$.

By Lemma \[lem-1\], Lemma \[lem-2\], Proposition \[TSP\], Theorem \[SP\] and Theorem \[prop-3\], the following theorem immediately holds. 

\[thm-RTSP\] Any potential $w$ in Table $1$ is a regular twisted superpotential. 

\[rem\_reg\] It turns out from Theorem \[thm-RTSP\] that the defining relations listed in [@IM Theorem 3.1] are in fact those of $3$-dimensional quadratic AS-regular algebras (see also Remark \[rem\_IM1\]). 

By Theorem \[prop-3\], for a $3$-dimensional AS-regular algebra $A$ except for Type EC, there exist a Calabi-Yau superpotential $w_{0}$ and $\theta \in {\rm Aut}(w_{0})$ in Table $3$ such that $A \cong \mathcal{D}((w_{0})^{\theta})$. Since $\theta \in {\rm Aut}(w_{0})$, we have that $\mathcal{D}((w_{0})^{\theta}) \cong \mathcal{D}(w_{0})^{\theta}$ by Lemma \[twist\].  Since $\mathcal{D}(w_{0})$ is Calabi-Yau AS-regular, we have the following corollary:

\[cor\_conj\] For a $3$-dimensional quadratic AS-regular algebra $A$ except for Type EC, there exist a Calabi-Yau AS-regular algebra $S$ and $\theta \in {\rm Aut}\,S$ such that $A$ is isomorphic to $S^{\theta}$ as graded $k$-algebras. 

Geometric algebras of Type EC
=============================

We say that a geometric algebra $A=\mathcal{A}(E,\sigma)$ is [*of Type EC*]{} if $E$ is an elliptic curve in $\mathbb{P}^{2}$. In this section, we give a criterion when a geometric algebra of Type EC is a $3$-dimensional quadratic AS-regular algebra. 

Divisors on curves and Hesse forms
----------------------------------

Let $E$ be a projective smooth curve over $k$. The [*Picard group*]{} of $E$, denoted by $({\rm Pic}\,E,[\mathcal{O}_{E}],\otimes)$, is the group of isomorphism classes of invertible sheaves on $E$ under the operation $\otimes$ (see [@H p.143]). 

A [*divisor*]{} on $E$ is an element of the free abelian group $${\rm Div}\,E:=\left\{ \sum_{p \in E}n_{p}\cdot p\,\middle |\,n_{p} \in \mathbb{Z} \right\}$$ where only finitely many $n_{p}$ are different from zero.  We write the group of divisors $({\rm Div}\,E,0,+)$ where $0$ is the zero divisor, that is, $n_{p}=0$ for any $p \in E$. For any divisor $D \in {\rm Div}\,E$, there exists an invertible sheaf on $E$, denoted by $\mathcal{O}_{E}(D)$, and the map $D \mapsto \mathcal{O}_{E}(D)$ gives a surjective homomorphism from $({\rm Div}\,E,0,+)$ to $({\rm Pic}\,E,[\mathcal{O}_{E}],\otimes)$ (see [@H Proposition II 6.13] and [@H Corollary II 6.16]), that is, for any $[\mathcal{M}] \in {\rm Pic}\,E$ there exists a divisor $D \in {\rm Div}\,E$ such that $[\mathcal{M}]=[\mathcal{O}_{E}(D)]$. The zero divisor $0$ maps to the isomorphism class $[\mathcal{O}_{E}] \in {\rm Pic}\,E$.

For $\sigma \in {\rm Aut}_{k}\,E$, we define a map $\tilde{\sigma}:\operatorname{Div}E \rightarrow \operatorname{Div}E$ by $$\tilde{\sigma}\left( \sum_{p \in E}n_{p} \cdot p \right)
=\sum_{p \in E}n_{p} \cdot \sigma^{-1}(p).$$ This map $\tilde{\sigma}$ is a group automorphism of $(\operatorname{Div}E,0,+)$. On the other hand, for $\sigma \in {\rm Aut}_{k}\,E$, the rule $\mathcal{M} \mapsto \sigma^{\ast}\mathcal{M}$ where $\mathcal{M}$ is an invertible sheaf on $E$ induces a group automorphism of the Picard group $\sigma^{\ast}:\operatorname{Pic}E \rightarrow \operatorname{Pic}E$. It follows from [@H II Ex.  6.8] that, if $D \in \operatorname{Div}E$, then $$\sigma^{\ast}(\mathcal{O}_{E}(D)) \cong \mathcal{O}_{E}(\tilde{\sigma}D).$$

Let $E$ be an elliptic curve in $\mathbb{P}^{2}$. It is well-known that the [*$j$-invariant*]{} $j(E)$ classifies elliptic curves up to isomorphism, that is, two elliptic curves $E$ and $E'$ in $\mathbb{P}^{2}$ are isomorphic if and only if $j(E)=j(E')$ (see [@H Theorem IV 4.1(b)]).  For $p \in E$, we define ${\rm Aut}_{k}\,(E,p):=\{ \sigma \in {\rm Aut}_{k}\,E \mid \sigma(p)=p \}$. It follows from [@H Corollary IV 4.7] that, for every point $p \in E$, ${\rm Aut}_{k}\,(E,p)$ becomes a cyclic group of order $$|{\rm Aut}_{k}\,(E,p)|
=\begin{cases}
2 &\quad\quad \text{if } j(E) \neq 0, 12^{3}, \\
6 &\quad\quad \text{if } j(E)=0, \\
4 &\quad\quad \text{if } j(E)=12^{3}. 
\end{cases}$$ For each point $o \in E$, we can define an addition $\oplus$ on $E$ so that $(E,o,\oplus)$ is an abelian group with the zero element $o$ and, for $p \in E$, the map $\sigma_{p}$ defined by $\sigma_{p}(q):=p \oplus q$ is a scheme automorphism of $E$, called the [*translation*]{} by a point $p$.

Type EC
-------

Throughout this subsection, for an elliptic curve $E$ in $\mathbb{P}^{2}$, we use a [*Hesse form*]{} $E=\mathcal{V}(x^{3}+y^{3}+z^{3}-3{\lambda}xyz)$ where ${\lambda}\in k$ with ${\lambda}^{3} \neq 1$. The [*$j$-invariant*]{} of a Hesse form is given by the following formula (see [@F Proposition 2.16]): $$j(E)=\frac{27{\lambda}^{3}({\lambda}^{3}+8)^{3}}{({\lambda}^{3}-1)^{3}}.$$

We fix the group structure on $E$ with the zero element $o_{E}:=(1:-1:0) \in E$. Every automorphism $\sigma \in {\rm Aut}_{k}\,E$ can be written as $\sigma=\sigma_{p}\tau^{i}$ where $p \in E$, $\tau$ is a generator of ${\rm Aut}_{k}\,(E,o_{E})$ and $i \in \mathbb{Z}_{|\tau|}$ ([@IM Proposition 4.5 and Theorem 4.6]). 

We call a point $p \in E$ [*$3$-torsion*]{} if $p \oplus p \oplus p=o_{E}$. We set $E[3]:=\{ p \in E \mid p \oplus p \oplus p=o_{E} \}$. For $p\in E$ and $i\in \mathbb{Z}$, $A=\mathcal{A}(E,\sigma_{p}\tau^{i})$ is of Type EC if and only if $p \in E \setminus E[3]$ ([@IM Lemma 4.14]). 

The map $p \mapsto [\mathcal{O}_{E}(p-o_{E})]$ is an injective homomorphism from $(E,o_{E},\oplus)$ to $(\operatorname{Pic}E,[\mathcal{O}_{E}],\otimes)$ (see [@H Example IV 1.3.7]).  For $p \in E$ and $n \in \mathbb{Z}$, we use a notation $[n]p:=\underbrace{p \oplus \cdots \oplus p}_{n}$. It is easy to check the following lemma. 

\[sum\] Let $(E,o_{E},\oplus)$ be an elliptic curve in $\mathbb{P}^{2}$, $p \in E$ and $n \in \mathbb{Z}$. Then $$[\mathcal{O}_{E}([n]p-o_{E})]=[\mathcal{O}_{E}(n(p-o_{E}))].$$

Since the zero element $o_{E}=(1:-1:0)$ is an inflection point of $E$, it follows that $\mathcal{L} \cong \mathcal{O}_{E}(3o_{E})$ where $\mathcal{L}=\pi^{\ast}(\mathcal{O}_{\mathbb{P}^{2}}(1))$.

\[Lemma1\] Let $\pi:E \rightarrow \mathbb{P}^{2}$ be the embedding and $\mathcal{L}=\pi^{\ast}\mathcal{O}_{\mathbb{P}^{2}}(1)$. Then an automorphism $\sigma \in {\rm Aut}_{k}E$ can be extended to an automorphism of $\mathbb{P}^{2}$ if and only if $\sigma^{\ast}\mathcal{L} \cong \mathcal{L}$.

A $3$-dimensional [*Sklyanin algebra*]{} is defined by $\mathcal{A}(E,\sigma_{p})$ where $p=(a:b:c)\in E \setminus E[3]$ and the defining relations are given as follows: $$\begin{cases}
ayz+bzy+cx^{2}, \\
azx+bxz+cy^{2}, \\
axy+byx+cz^{2}.  
\end{cases}$$ A $3$-dimensional Sklyanin algebra $\mathcal{A}(E,\sigma_{p})$ is a $3$-dimensional quadratic AS-regular algebra by [@ATV1 Section 1].  It follows from [@IM Theorem 4.12 (1)] that $\sigma_{p}\tau^{i}$ is not extended to an automorphism of $\mathbb{P}^{2}$, so, by Lemma \[Lemma1\], $$(\sigma_{p}\tau^{i})^{\ast}\mathcal{L} \not\cong \mathcal{L}.$$

For a geometric algebra $A=\mathcal{A}(E,\sigma)$ of Type EC, we give a criterion when $A$ is AS-regular. 

\[Main\] Let $A=\mathcal{A}(E,\sigma)$ be a geometric algebra of Type EC where $\sigma=\sigma_{p}\tau^{i}$, $p \in E \backslash E[3]$ and $i \in \mathbb{Z}_{|\tau|}$. Then the following are equivalent[:]{}

1.   $A$ is a $3$-dimensional quadratic AS-regular algebra. 

2.   $p \ominus \tau^{i}(p) \in E[3]$.

3.   $A$ is graded Morita equivalent to a $3$-dimensional Sklyanin algebra $\mathcal{A}(E,\sigma_{p})$.

\(1) $\Rightarrow$ (2): Assume that $A=\mathcal{A}(E,\sigma)$ is a $3$-dimensional quadratic AS-regular algebra.  By Theorem \[ATV1\], it holds that $$(\sigma^{2})^{\ast}\mathcal{L} \otimes_{\mathcal{O}_{E}} \mathcal{L} \cong
    \sigma^{\ast}\mathcal{L} \otimes_{\mathcal{O}_{E}} \sigma^{\ast}\mathcal{L}.$$ Since $\mathcal{L} \cong \mathcal{O}_{E}(3o_{E})$, $$\begin{aligned}
    &\sigma^{\ast}\mathcal{L} \cong \sigma^{\ast}(\mathcal{O}_{E}(3o_{E}))
    \cong \mathcal{O}_{E}(\tilde{\sigma}(3o_{E}))=\mathcal{O}_{E}(3q), \\
    &(\sigma^{2})^{\ast}\mathcal{L} \cong (\sigma^{2})^{\ast}(\mathcal{O}_{E}(3o_{E}))
    \cong \mathcal{O}_{E}(\tilde{\sigma^{2}}(3o_{E}))
    =\mathcal{O}_{E}(3r),
    \end{aligned}$$ where $q:=\sigma^{-1}(o_{E})=\ominus\tau^{-i}(p)$ and $r:=\sigma^{-2}(o_{E})=\sigma^{-1}(q)=q \oplus \tau^{-i}(q)$, so $$\begin{aligned}
    &(\sigma^{2})^{\ast}\mathcal{L} \otimes_{\mathcal{O}_{E}} \mathcal{L}
    \cong \mathcal{O}_{E}(3r) \otimes_{\mathcal{O}_{E}} \mathcal{O}_{E}(3o_{E})
    \cong \mathcal{O}_{E}(3(r+o_{E})), \\
    &\sigma^{\ast}\mathcal{L} \otimes_{\mathcal{O}_{E}} \sigma^{\ast}\mathcal{L}
    \cong \mathcal{O}_{E}(3q) \otimes_{\mathcal{O}_{E}} \mathcal{O}_{E}(3q)
    \cong \mathcal{O}_{E}(6q). 
    \end{aligned}$$ Therefore, $$\begin{aligned}
    (\sigma^{2})^{\ast}\mathcal{L} \otimes_{\mathcal{O}_{E}} \mathcal{L} \cong
    \sigma^{\ast}\mathcal{L} \otimes_{\mathcal{O}_{E}} \sigma^{\ast}\mathcal{L}
    &\Longrightarrow
    \mathcal{O}_{E}(3(r+o_{E})) \cong \mathcal{O}_{E}(6q) \\
    &\Longrightarrow
    \mathcal{O}_{E}(3(r-o_{E})) \cong \mathcal{O}_{E}(6(q-o_{E})) \\
    &\Longrightarrow
    \mathcal{O}_{E}([3]r-o_{E}) \cong \mathcal{O}_{E}([6]q-o_{E}) \\
    &\Longrightarrow
    [3]r=[6]q \\
    &\Longrightarrow
    [3](q \ominus \tau^{-i}(q))=o_{E} \\
    &\Longrightarrow
    q \ominus \tau^{-i}(q) \in E[3]. 
    \end{aligned}$$ Since $q=\ominus \tau^{-i}(p)$, $
    q \ominus \tau^{-i}(q)=\ominus \tau^{-i}(p) \oplus \tau^{-2i}(p)=\tau^{-2i}(p \ominus \tau^{i}(p))
    $.  Hence we have $
    p \ominus \tau^{i}(p) \in E[3]
    $. 

\(2) $\Rightarrow$ (3): Assume that $p \ominus \tau^{i}(p) \in E[3]$. By [@IM Theorem 4.20], $A=\mathcal{A}(E,\sigma_{p}\tau^{i})$ and $\mathcal{A}(E,\sigma_{p})$ are graded Morita equivalent. 

\(3) $\Rightarrow$ (1): Assume that $A=\mathcal{A}(E,\sigma_{p}\tau^{i})$ is graded Morita equivalent to a $3$-dimensional Sklyanin algebra $\mathcal{A}(E,\sigma_{p})$. By Lemma \[twistingequi\], $A$ is isomorphic to a twisted algebra of $\mathcal{A}(E,\sigma_{p})$. Since being AS-regular is invariant under twisting system by Lemma \[twist\_inv\], a twisted algebra of $\mathcal{A}(E,\sigma_{p})$ is a $3$-dimensional quadratic AS-regular algebra.  Therefore, $A=\mathcal{A}(E,\sigma_{p}\tau^{i})$ is a $3$-dimensional quadratic AS-regular algebra. 

Now, we are ready to prove Theorem \[conj\] in Introduction. 

\[Main2\] For every $3$-dimensional quadratic AS-regular algebra $A$, there exists a Calabi-Yau AS-regular algebra $S$ such that $A$ and $S$ are graded Morita equivalent. 

: By Lemma \[twistingequi\] and Corollary \[cor\_conj\], the statement holds. 

: Let $A=\mathcal{A}(E,\sigma_{p})=\mathcal{D}(w)$ be a $3$-dimensional Sklyanin algebra where $p=(a:b:c) \in E \setminus E[3]$ and $w=a(xyz+yzx+zxy)+b(xzy+zyx+yxz)+c(x^{3}+y^{3}+z^{3})$. Since $w$ is a superpotential, by \[CY-superpotential\], $\mathcal{D}(w)$ is Calabi-Yau AS-regular.  By Theorem \[Main\], the statement holds. 

In Section 3, Corollary \[cor\_conj\] tells us that, for a $3$-dimensional AS-regular algebra except for Type EC, there exist a Calabi-Yau AS-regular algebra $S$ and $\theta \in {\rm Aut}\,S$ such that $A$ is isomorphic to $S^{\theta}$ as graded $k$-algebras.  We prove this by using Theorem \[prop-3\], that is, for a potential $w$ in Theorem \[TSP\], there exist a superpotential $w_{0}$ in Theorem \[SP\] and $\theta \in {\rm Aut}\,(w_{0})$ such that $w=(w_{0})^{\theta}$. On the other hand, it follows from [@IM Theorem 4.9] that, for a potential $w$ of a geometric algebra of Type EC, there exist a Calabi-Yau superpotential $w_{0}$ and $\theta \in {\rm GL}\,(V)$ induced by $\tau^{i}\in {\rm Aut}_{k}\,E$ such that $w=(w_{0})^{\theta}$. But this $\theta$ is not necessary in ${\rm Aut}\,(w_{0})$, so $w=(w_{0})^{\theta}$ may not be a twisted superpotential nor regular (see Example \[examEC\]).  This means that we do not know whether Theorem \[prop-3\] holds or not for Type EC.  So, we need to divide the proof of Theorem \[conj\] into two cases of non Type EC and Type EC. 

Let $E=\mathcal{V}(x^{3}+y^{3}+z^{3}-3 {\lambda}xyz)$ be an elliptic curve in $\mathbb{P}^{2}$ with $j(E) \neq 0,12^{3}$ and $A=\mathcal{A}(E,\sigma_{p}\tau)$ where $p \in E \setminus E[3]$. In this case, we have that $\tau(p)=\ominus p$. By Theorem \[Main\], $A$ is a $3$-dimensional quadratic AS-regular algebra if and only if $[2]p \in E[3]$, that is, $p \in E[6]$ where $E[6]:=\{ q \in E \mid [6]q=o_{E} \}$.

\[examEC\] In general, it is not true that if $w$ is a regular superpotential and $\theta \in {\rm GL}(V)$, then the MS twist $w^{\theta}$ is regular.  Let $E=\mathcal{V}(x^{3}+y^{3}+z^{3}-3\lambda xyz)$ be an elliptic curve with $j(E) \neq 0,12^{3}$ and $A=\mathcal{A}(E,\sigma_{p}\tau)$ where $p=(a:b:c) \in E \setminus E[6]$. By Theorem \[Main\], $A$ is not AS-regular.  By [@IM Theorem 4.6] and [@IM Theorem 4.9], we have that $A=\mathcal{A}(E,\sigma_{p}\tau)=\mathcal{D}(w^{\theta})$ where $w=a(xyz+yzx+zxy)+b(xzy+yxz+zyx)+c(x^{3}+y^{3}+z^{3})$ and $\theta= \left(
    \begin{array}{ccc}
    0& 1& 0 \\
    1& 0& 0 \\
    0& 0& 1
    \end{array}
    \right) \in {\rm GL}_{3}(k)$. Note that if $a \neq b$, then $\theta \notin {\rm Aut}\,(w)$. Since $A$ is not a $3$-dimensional quadratic AS-regular algebra, $w^{\theta}$ is not regular. 

The first author was supported by Grants-in-Aid for Young Scientific Research 18K13397 Japan Society for the Promotion of Science. 

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