Some algebras having relations like those for the 4-dimensional Sklyanin algebras

TL;DR

This paper investigates algebras similar to 4-dimensional Sklyanin algebras, disproves a conjecture about a related family of algebras, and explores their structural features and classifications.

Contribution

It refutes a conjecture linking certain algebras to Sklyanin algebras and introduces a new classification approach using projective space graph structures.

Findings

Most Cho-Hong-Lau algebras are characterized by a bijection graph in P^3

The studied algebras share features with Sklyanin algebras but also differ in key ways

A new family of 4-generator 6-relator algebras includes Sklyanin and Cho-Hong-Lau algebras

Abstract

The 4-dimensional Sklyanin algebras are a well-studied 2-parameter family of non-commutative graded algebras, often denoted A(E,tau), that depend on a quartic elliptic curve E in P^3 and a translation automorphism tau of E. They are graded algebras generated by four degree-one elements subject to six quadratic relations and in many important ways they behave like the polynomial ring on four indeterminates apart from the minor difference that they are not commutative. They are elliptic analogues of the enveloping algebra of sl(2,C) and the quantized enveloping algebras U_q(gl_2). Recently, Cho, Hong, and Lau, conjectured that a certain 2-parameter family of algebras arising in their work on homological mirror symmetry consists of 4-dimensional Sklyanin algebras. This paper shows their conjecture is false in the generality they make it. On the positive side, we show their algebras exhibit features that are similar to, and differ from, analogous features of the 4-dimensional Sklyanin algebras in interesting ways. We show that most of the Cho-Hong-Lau algebras determine, and are determined by the graph of a bijection between two 20-point subsets of the projective space P^3. The paper also examines a 3-parameter family of 4-generator 6-relator algebras admitting presentations analogous to those of the 4-dimensional Sklyanin algebras. This class includes the 4-dimensional Sklyanin algebras and most of the Cho-Hong-Lau algebras.

Tables

Table 1. Table 1. Automorphisms ψ 1 , ψ 2 , ψ 3 subscript 𝜓 1 subscript 𝜓 2 subscript 𝜓 3 \psi_{1},\psi_{2},\psi_{3} .

	$x_0$	$x_1$	$x_2$	$x_3$
${\psi }_1$	$bc{x}_1$	$-i{x}_0$	$-ib{x}_3$	$-c{x}_2$
${\psi }_2$	$ac{x}_2$	$-a{x}_3$	$-i{x}_0$	$-ic{x}_1$
${\psi }_3$	$ab{x}_3$	$-ia{x}_2$	$-b{x}_1$	$-i{x}_0$

Table 2. Table 2 . The action of ℤ 2 × ℤ 2 subscript ℤ 2 subscript ℤ 2 {\mathbb{Z}}_{2}\times{\mathbb{Z}}_{2} as automorphisms of A 𝐴 A .

	$x_0$	$x_1$	$x_2$	$x_3$
${\gamma }_1$	$x_0$	$x_1$	$-x_2$	$-x_3$
${\gamma }_2$	$x_0$	$-x_1$	$x_2$	$-x_3$
${\gamma }_3$	$x_0$	$-x_1$	$-x_2$	$x_3$

	$x_0$	$x_i$	$x_j$	$x_k$
$\psi$	${\lambda }_0{x}_i$	${\lambda }_i{x}_0$	${\lambda }_j{x}_k$	${\lambda }_k{x}_j$

Table 4. Table 3. The points in 𝔓 𝔓 {\mathfrak{P}}

${𝔓}_{\infty }$	${𝔓}_0$	${𝔓}_1$	${𝔓}_2$	${𝔓}_3$
$(1,0,0,0)$	$(abc,a,b,c)$	$(a,-ia,-i,-1)$	$(b,-1,-ib,-i)$	$(c,-i,-1,-ic)$
$(0,1,0,0)$	$(abc,a,-b,-c)$	$(a,-ia,i,1)$	$(b,-1,ib,i)$	$(c,-i,1,ic)$	${\gamma }_1$
$(0,0,1,0)$	$(abc,-a,b,-c)$	$(a,ia,-i,1)$	$(b,1,-ib,i)$	$(c,i,-1,ic)$	${\gamma }_2$
$(0,0,0,1)$	$(abc,-a,-b,c)$	$(a,ia,i,-1)$	$(b,1,ib,-i)$	$(c,i,1,-ic)$	${\gamma }_3$

Table 5. Table 4. Contragredient action of H 4 subscript 𝐻 4 H_{4} on V ∗ superscript 𝑉 V^{*} .

	$x_0^{\ast }$	$x_1^{\ast }$	$x_2^{\ast }$	$x_3^{\ast }$
${\psi }_1$	$i{x}_1^{\ast }$	${\left(bc\right)}^{-1}{x}_0^{\ast }$	$-c^{-1}{x}_3^{\ast }$	$i{b}^{-1}{x}_2^{\ast }$
${\psi }_2$	$i{x}_2^{\ast }$	$i{c}^{-1}{x}_3^{\ast }$	${\left(ac\right)}^{-1}{x}_0^{\ast }$	$-a^{-1}{x}_1^{\ast }$
${\psi }_3$	$i{x}_3^{\ast }$	$-i{b}^{-1}{x}_2^{\ast }$	$i{a}^{-1}{x}_1^{\ast }$	${\left(ab\right)}^{-1}{x}_0^{\ast }$

Equations

$$\{\begin{array}{ll}{x}_0{x}_1-x_1{x}_0& =\alpha (x_2{x}_3+x_3{x}_2)x_0{x}_1+x_1{x}_0=x_2{x}_3-x_3{x}_2\\ x_0{x}_2-x_2{x}_0& =\beta (x_3{x}_1+x_1{x}_3)x_0{x}_2+x_2{x}_0=x_3{x}_1-x_1{x}_3\\ x_0{x}_3-x_3{x}_0& =\gamma (x_1{x}_2+x_2{x}_1)x_0{x}_3+x_3{x}_0=x_1{x}_2-x_2{x}_1.\end{array}$$

$$\alpha +\beta +\gamma +\alpha \beta \gamma =0\text{ and}\{\alpha ,\beta ,\gamma \}\cap \{0,\pm 1\}=\varnothing ,$$

$$[x_0,x_i]={\alpha }_i\{x_j,x_k\},\{x_0,x_i\}=[x_j,x_k],(i,j,k)a\; cyclic\; permutation\; of(1,2,3).$$

$$a{x}_4{x}_3+b{x}_3{x}_4+c{x}_3{x}_2+d{x}_4{x}_1=0,$$

$$a{x}_3{x}_2+b{x}_2{x}_3+c{x}_4{x}_3+d{x}_1{x}_2=0,$$

$$a{x}_2{x}_1+b{x}_1{x}_2+c{x}_1{x}_4+d{x}_2{x}_3=0,$$

$$a{x}_1{x}_4+b{x}_4{x}_1+c{x}_2{x}_1+d{x}_3{x}_4=0,$$

$$a{x}_3{x}_1-a{x}_1{x}_3+c{x}_4^2-c{x}_2^2=0,$$

$$b{x}_4{x}_2-b{x}_2{x}_4+d{x}_3^2-d{x}_1^2=0.$$

$${\psi }_i(x_0)=a_j{a}_k{x}_i,{\psi }_i(x_i)=-i{x}_0,{\psi }_i(x_j)=-i{a}_j{x}_k,and{\psi }_i(x_k)=-a_k{x}_j.$$

$$H_4:=⟨{\epsilon }_1,{\epsilon }_2,\delta |{\epsilon }_1^4={\epsilon }_2^4={\delta }^4=1,\delta {\epsilon }_1={\epsilon }_1\delta ,{\epsilon }_2\delta =\delta {\epsilon }_2,{\epsilon }_1{\epsilon }_2=\delta {\epsilon }_2{\epsilon }_1⟩.$$

$${\epsilon }_1\mapsto {\nu }_1^{-1}{\psi }_1,{\epsilon }_2\mapsto {\nu }_2^{-1}{\psi }_2,{\epsilon }_3\mapsto {\nu }_3^{-1}{\psi }_3.$$

$$\frac{{\lambda }_0{\lambda }_i}{{\lambda }_j{\lambda }_k}=-1,\frac{{\lambda }_0{\lambda }_j}{{\lambda }_k{\lambda }_i}=-{\alpha }_j,and\frac{{\lambda }_0{\lambda }_k}{{\lambda }_i{\lambda }_j}={\alpha }_k.$$

$$\frac{A}{([A,A])}=\frac{\mathbb{k}[x_0,x_1,x_2,x_3]}{(x_1,x_2,x_3)\cap (x_0,x_2,x_3)\cap (x_0,x_1,x_3)\cap (x_0,x_1,x_2)}.$$

$$\operatorname{Proj}\bigg{(}\frac{A}{([A,A])}\bigg{)}\;=\;\{e_{0}=(1,0,0,0),e_{1}=(0,1,0,0),e_{2}=(0,0,1,0),e_{3}=(0,0,0,1)\}.$$

$$A(\alpha ,\beta ,\gamma )$$

$$\cong A(-\alpha ,-\gamma ,-\beta )\cong A(-\beta ,-\alpha ,-\gamma )\cong A(-\gamma ,-\beta ,-\alpha ).$$

$$[x_0,-x_1]=-\alpha \{x_3,x_2\},[x_0,x_3]=-\gamma \{x_2,-x_1\},[x_0,x_2]=-\beta \{-x_1,x_3\},$$

$$\{x_0,-x_1\}=[x_3,x_2],\{x_0,x_3\}=[x_2,-x_1],\{x_0,x_2\}=[-x_1,x_3],$$

$$⟨p,q,r,s{⟩}_A:=({\mu }_1{\nu }_1,{\mu }_2{\nu }_2,{\mu }_3{\nu }_3)\in {\mathbb{k}}^3$$

$$[x_p,x_q]$$

$$[x_p,x_r]$$

$$[x_p,x_s]$$

$$\{\begin{array}{ll}⟨0,1,2,3{⟩}_A=⟨1,0,3,2{⟩}_A=⟨2,3,0,1{⟩}_A=⟨3,2,1,0{⟩}_A=({\alpha }_1,{\alpha }_2,{\alpha }_3)& \text{and}\\ ⟨0,1,3,2{⟩}_A=⟨1,2,3,0{⟩}_A=⟨2,1,0,3{⟩}_A=⟨3,1,2,0{⟩}_A=(-{\alpha }_1,-{\alpha }_3,-{\alpha }_2).& \end{array}$$

$${\lambda }_0{\lambda }_1\{x_p^{\prime },x_q^{\prime }\}={\lambda }_2{\lambda }_3[x_r^{\prime },x_s^{\prime }],{\lambda }_0{\lambda }_2\{x_p^{\prime },x_r^{\prime }\}={\lambda }_3{\lambda }_1[x_s^{\prime },x_q^{\prime }],{\lambda }_0{\lambda }_3\{x_p^{\prime },x_s^{\prime }\}={\lambda }_1{\lambda }_2[x_q^{\prime },x_r^{\prime }].$$

$${\lambda }_0{\lambda }_1[x_p^{\prime },x_q^{\prime }]={\alpha }_1{\lambda }_2{\lambda }_3\{x_r^{\prime },x_s^{\prime }\},{\lambda }_0{\lambda }_2[x_p^{\prime },x_r^{\prime }]={\alpha }_2{\lambda }_3{\lambda }_1\{x_s^{\prime },x_q^{\prime }\},{\lambda }_0{\lambda }_3[x_p^{\prime },x_s^{\prime }]={\alpha }_3{\lambda }_1{\lambda }_2\{x_q^{\prime },x_r^{\prime }\}.$$

$$[x_p^{\prime },x_q^{\prime }]={\alpha }_1{\lambda }_0^{-1}{\lambda }_1^{-1}{\lambda }_2{\lambda }_3\{x_r^{\prime },x_s^{\prime }\},[x_r^{\prime },x_s^{\prime }]={\lambda }_0{\lambda }_1{\lambda }_2^{-1}{\lambda }_3^{-1}\{x_p^{\prime },x_q^{\prime }\},$$

$$[x_p^{\prime },x_r^{\prime }]={\alpha }_2{\lambda }_0^{-1}{\lambda }_2^{-1}{\lambda }_3{\lambda }_1\{x_s^{\prime },x_q^{\prime }\},[x_s^{\prime },x_q^{\prime }]={\lambda }_0{\lambda }_2{\lambda }_3^{-1}{\lambda }_1^{-1}\{x_p^{\prime },x_r^{\prime }\},$$

$$[x_p^{\prime },x_s^{\prime }]={\alpha }_3{\lambda }_0^{-1}{\lambda }_3^{-1}{\lambda }_1{\lambda }_2\{x_q^{\prime },x_r^{\prime }\},[x_q^{\prime },x_r^{\prime }]={\lambda }_0{\lambda }_3{\lambda }_1^{-1}{\lambda }_2^{-1}\{x_p^{\prime },x_s^{\prime }\}.$$

$$M:=\left(\begin{array}{cccc}-x_1& x_0& -\alpha x_3& -\alpha x_2\\ -x_2& -\beta x_3& x_0& -\beta x_1\\ -x_3& -\gamma x_2& -\gamma x_1& x_0\\ -x_3& -x_2& x_1& -x_0\\ -x_1& -x_0& -x_3& x_2\\ -x_2& x_3& -x_0& -x_1\end{array}\right).$$

$$M^{\prime }:=\left(\begin{array}{cccccc}-x_1& -x_2& -x_3& -x_3& -x_1& -x_2\\ x_0& \beta x_3& \gamma x_2& x_2& -x_0& -x_3\\ \alpha x_3& x_0& \gamma x_1& -x_1& x_3& -x_0\\ \alpha x_2& \beta x_1& x_0& -x_0& -x_2& x_1\end{array}\right).$$

$$x_0{x}_1-{\lambda }^2{x}_2{x}_3=0,x_0{x}_2-{\mu }^2{x}_1{x}_3=0,x_0{x}_3-{\nu }^2{x}_1{x}_2=0,$$

$$(0,1,0,0),(0,0,1,0),(\lambda \mu \nu ,\lambda ,\mu ,\nu ),(\lambda \mu \nu ,-\lambda ,-\mu ,\nu ),$$

$$(1,0,0,0),(0,0,0,1),(\lambda \mu \nu ,-\lambda ,\mu ,-\nu ),(\lambda \mu \nu ,\lambda ,-\mu ,-\nu ).$$

$$x_0{x}_1-{\lambda }^2{x}_2{x}_3=(x_0-\lambda \mu x_3)x_1+(x_1-\lambda {\mu }^{-1}{x}_2)\lambda \mu x_3$$

Full text

Recently, Cho, Hong, and Lau, conjectured that a certain 2-parameter family of algebras arising in their work on homological mirror symmetry consists of 4-dimensional Sklyanin algebras. This paper shows their conjecture is false in the generality they make it. On the positive side, we show their algebras exhibit features that are similar to, and differ from, analogous features of the 4-dimensional Sklyanin algebras in interesting ways. We show that most of the Cho-Hong-Lau algebras determine, and are determined by the graph of a bijection between two 20-point subsets of the projective space ${\mathbb{P}}^{3}$.

The paper also examines a class of 4-generator 6-relator algebras admitting presentations analogous to those of the 4-dimensional Sklyanin algebras. This class includes the 4-dimensional Sklyanin algebras and most of the Cho-Hong-Lau algebras.

1. 1 Introduction
2. 2 Algebras $A(\alpha,\beta,\gamma)$ with a Sklyanin-like presentation
3. 3 The zero locus of the relations for $A(\alpha,\beta,\gamma)$
4. 4 Point schemes, graphs and flat families
5. 5 The algebras $R(a,b,c,d)$ of Cho, Hong, and Lau
6. 6 Central elements

1. Introduction

1.1.

This paper examines three families of graded algebras with four generators and six quadratic relations. The only commutative algebra in these families is the polynomial ring on 4 variables. All algebras in these families are, like the polynomial ring on 4 variables, generated by 4 elements subject to 6 homogeneous quadratic relations.

The members of the first of these families are denoted by $A(\alpha,\beta,\gamma)$, depending on a parameter $(\alpha,\beta,\gamma)\in\Bbbk^{3}$ where $\Bbbk$ is a field that will be fixed throughout the paper. They are generated by $x_{0},x_{1},x_{2},x_{3}$ subject to the relations

[TABLE]

Among these algebras, those for which

[TABLE]

are so starkly different from the rest that we consider them as a separate family. These constitute the second of our three families and are called non-degenerate 4-dimensional Sklyanin algebras. Algebras in the third family are denoted by $R(a,b,c,d)$, depending on a parameter $(a,b,c,d)$ that is required to lie on the quadric $\{ad+bc=0\}$ in the projective space ${\mathbb{P}}^{3}$. They are defined in 1.7.

The algebras $R(a,b,c,d)$ were discovered by Cho, Hong, and Lau in their work on mirror symmetry [8], and the motivation for this paper is their conjecture that these are 4-dimensional Sklyanin algebras. We prove the conjecture false in the generality in which it is made, but on the positive side

1. (1)

for a Zariski-dense open subset of points on the quadric $\{ad+bc=0\}$, $R(a,b,c,d)$ is isomorphic to $A(\alpha,\beta,\gamma)$ for some $(\alpha,\beta,\gamma)$, but $(\alpha,\beta,\gamma)$ does not always satisfy the condition $\alpha+\beta+\gamma+\alpha\beta\gamma=0$; 2. (2)

there are two lines $\ell_{1},\ell_{2}\subseteq\{ad+bc=0\}$ such that $R(a,b,c,d)$ is isomorphic to $A(\alpha,1,-1)$ for all $(a,b,c,d)\in\ell_{1}\cup\ell_{2}-\{\text{12 points}\}$; 3. (3)

the automorphism group of almost all $R(a,b,c,d)$ has a subgroup isomorphic to the Heisenberg group of order $4^{3}$.

The non-degenerate Sklyanin algebras may be parametrized by pairs $(E,\tau)$ consisting of an elliptic curve $E$ and a translation automorphism $\tau:E\to E$. We write $A(E,\tau)$ for the Sklyanin algebra corresponding to this data. It is striking that the translation automorphism for those $R(a,b,c,d)$ that are non-degenerate Sklyanin algebras has order 4; i.e., if $(a,b,c,d)\in\ell_{1}\cup\ell_{2}-\{\text{12 points}\}$, then $R(a,b,c,d)\cong A(E,\tau)$ for some elliptic curve $E$ and some $\tau$ having order 4 (Propositions 5.2 and 5.3).

1.2.

A striking feature of the algebras $R(a,b,c,d)$ is that almost all of them determine, and are determined by, a set of 20 points in the product ${\mathbb{P}}^{3}\times{\mathbb{P}}^{3}$ of two copies of the three-dimensional projective space.

1.3.

Because Sklyanin algebras, appearing first in [18, 19], have played such a large role in the development of non-commutative algebra and algebraic geometry over the past thirty years (see [21, 15, 22, 20, 25] for example), it is sensible to examine the larger class of algebras $A(\alpha,\beta,\gamma)$ defined by the “same” relations minus the constraint $\alpha+\beta+\gamma+\alpha\beta\gamma=0$.

We do not undertake an exhaustive study of the algebras $A(\alpha,\beta,\gamma)$ when $\alpha+\beta+\gamma+\alpha\beta\gamma\neq 0$ but it appears to us that there are interesting questions about them that might be fruitfully pursued. We mention some of these questions in 1.10.

1.4.

We use the notation $[x,y]=xy-yx$ and $\{x,y\}=xy+yx$.

1.5. The algebras $A(\alpha,\beta,\gamma)$

Let $\Bbbk$ be an arbitrary field and $\alpha_{1},\alpha_{2},\alpha_{3}\in\Bbbk$. Define $A(\alpha_{1},\alpha_{2},\alpha_{3})$, or simply $A$, to be the free algebra $\Bbbk\langle x_{0},x_{1},x_{2},x_{3}\rangle$ modulo the six relations

[TABLE]

We always consider $A$ as an ${\mathbb{N}}$-graded $\Bbbk$-algebra with ${\rm deg}\{x_{0},x_{1},x_{2},x_{3}\}=1$. Thus, $A$ is the quotient of the free algebra $TV/(R)=\Bbbk\langle x_{0},x_{1},x_{2},x_{3}\rangle/(R)$ where $V={\rm span}\{x_{0},x_{1},x_{2},x_{3}\}$ and $R\subseteq V^{\otimes 2}$ is the linear span of the six elements in $V^{\otimes 2}$ corresponding to the relations 1-3.

1.6. Degenerate and non-degenerate 4-dimensional Sklyanin algebras

Suppose $\alpha+\beta+\gamma+\alpha\beta\gamma=0$. We call $A(\alpha,\beta,\gamma)$ a 4-dimensional Sklyanin algebra in this case. If, in addition, $\{\alpha,\beta,\gamma\}\cap\{0,\pm 1\}=\varnothing$ we call $A(\alpha,\beta,\gamma)$ a non-degenerate 4-dimensional Sklyanin algebra. If $\alpha+\beta+\gamma+\alpha\beta\gamma=0$ and $\{\alpha,\beta,\gamma\}\cap\{0,\pm 1\}\neq\varnothing$ we call $A(\alpha,\beta,\gamma)$ a degenerate 4-dimensional Sklyanin algebra.

By [21], non-degenerate 4-dimensional Sklyanin algebras are noetherian domains having the same Hilbert series as the polynomial ring in 4 variables. By [21] and [15], they have excellent homological properties. The representation theory is intimately related to the geometry of $(E\subseteq{\mathbb{P}}^{3},\tau)$.

Some degenerate 4-dimensional Sklyanin algebras are closely related to better known algebras. For instance, the algebra $A=A(0,0,0)$ has a degree-one central element, $z$, such that $A/(z-1)\cong A[z^{-1}]_{0}\cong U({\mathfrak{so}}(3,\Bbbk))$, the enveloping algebra of the Lie algebra ${\mathfrak{so}}(3,\Bbbk)$. Similarly, if $\Bbbk={\mathbb{C}}$ and $\beta\neq 0$, then $A=A(0,\beta,-\beta)$ has a degree-two central element $\Omega$ such that $A[\Omega^{-1}]_{0}\cong U_{q}({\mathfrak{s}}{\mathfrak{l}}(2,{\mathbb{C}}))$, a quantized enveloping algebra of ${\mathfrak{s}}{\mathfrak{l}}(2,{\mathbb{C}})$.

1.7. The algebras of Cho, Hong, and Lau

Let $(a,b,c,d)\in\Bbbk^{4}$. We write $R(a,b,c,d)$, or simply $R$, for the free algebra $\Bbbk\langle x_{1},x_{2},x_{3},x_{4}\rangle$ modulo the relations

[TABLE]

Since $a$, $b$, $c$, and $d$, enter into the relations in a homogeneous way, the algebra $R(a,b,c,d)$ depends only on $(a,b,c,d)$ as a point in ${\mathbb{P}}^{3}$. If we impose the condition that $ac+bd=0$, we obtain a 2-dimensional family of algebras $R(a,b,c,d)$ parametrized by a quadric (isomorphic to ${\mathbb{P}}^{1}\times{\mathbb{P}}^{1}$) in ${\mathbb{P}}^{3}$. Cho, Hong, and Lau conjecture that, when $ac+bd=0$, $R$ is a 4-dimensional Sklyanin algebra [8, Conj. 8.11].

Although only a 1-parameter family of the $R(a,b,c,d)$ are Sklyanin algebras, we find it remarkable that almost all of them (Zariski-densely many, that is) have the “same” relations as the 4-dimensional Sklyanin algebras. We do not understand the deeper reason for this; our proof is just a calculation. We also find it remarkable that the translation automorphism for those that are Sklyanin algebras has order 4—the only translation automorphisms of a degree-four elliptic curve in ${\mathbb{P}}^{3}$ that extend to automorphisms of the ambient ${\mathbb{P}}^{3}$ are the translations by the points in its 4-torsion subgroup. We do not know in what way, in the context of the work of Cho-Hong-Lau, those $R(a,b,c,d)$ that are Sklyanin algebras are special.

1.8. Results about $A(\alpha,\beta,\gamma)$

Suppose $\alpha\beta\gamma\neq 0$. In Section 2 we show that the Heisenberg group of order $4^{3}$ acts as automorphisms of $A(\alpha,\beta,\gamma)$. In Proposition 2.4, we determine exactly when two of these algebras are isomorphic to each other.

In Section 3 we give a geometric interpretation of the relations defining $A(\alpha,\beta,\gamma)$. To do this we first write $A(\alpha,\beta,\gamma)$ as $TV/(R)$, the quotient of the tensor algebra $TV$ on a 4-dimensional vector space $V$ by the ideal generated by a 6-dimensional subspace $R$ of $V^{\otimes 2}$. We then consider elements in $V\otimes V$ as forms of bi-degree $(1,1)$ on the product ${\mathbb{P}}(V^{*})\times{\mathbb{P}}(V^{*})$ of two copies of projective 3-space. We now define the closed subscheme $\Gamma\subseteq{\mathbb{P}}(V^{*})\times{\mathbb{P}}(V^{*})$ to be the vanishing locus of the elements in $R$. Proposition 3.3 shows that $\Gamma$ is finite if and only if $\alpha\beta\gamma\neq 0$ and $\alpha+\beta+\gamma+\alpha\beta\gamma\neq 0$. Propositions 3.3 and 3.4 show that in that case

1. (1)

$\Gamma$ consists of 20 distinct points; 2. (2)

$\Gamma$ is the graph of a bijection between 20-point subsets of ${\mathbb{P}}(V^{*})$; 3. (3)

$R=\{f\in V^{\otimes 2}\;|\;f|_{\Gamma}=0\}$.

1.9. Centers

[18, Thm. 2] states that two explicitly given degree-two homogeneous elements, which are denoted there by $K_{0}$ and $K_{1}$, belong to the center of the 4-dimensional Sklyanin algebra. Although Sklyanin writes that it is “straightforward” to prove these elements central, the details are left to the reader. We and others have found the calculations less than straightforward.111The problem of showing that $K_{0}$ and $K_{1}$ are central is mentioned in a talk given by Tom Koornwinder at Nijmegen on 12 November 2012—see https://staff.science.uva.nl/t.h.koornwinder/art/sheets/SklyaninAlgebra1.pdf, retrieved on 01-20-2017. Koornwinder says that part of the proof is “straightforward” and appeals to the Mathematica package NCAlgebra 4.0.4 at http://www.math.ucsd.edu/~ncalg/ for the remainder of the proof. Sklyanin says that an alternative proof can be given by using a lemma in his paper [14] with Kulish. Presumably, the relevant lemma is equation (5.7) in [14]. However, “due to space limitations [they] do not present [t]here the complete proof of (5.7)”. We have been unable to find a complete proof of [18, Thm. 2] in the literature so give a direct proof that $K_{0}$ and $K_{1}$ are central in Proposition 6.1 below. We do not use the same notation as Sklyanin, so in this introduction we label the elements in Proposition 6.1 $\Omega_{0}$ and $\Omega_{1}$.

For most 4-dimensional Sklyanin algebras $K_{0}$ and $K_{1}$, or equivalently $\Omega_{0}$ and $\Omega_{1}$, generate the center of the algebra. In sharp contrast, when $\alpha\beta\gamma\neq 0$ and $\alpha+\beta+\gamma+\alpha\beta\gamma\neq 0$ the elements $x_{0}^{2}$, $x_{1}^{2}$, $x_{2}^{2}$, and $x_{3}^{2}$ belong to the center of $A(\alpha,\beta,\gamma)$ (Proposition 6.2).

Cho, Hong, and Lau, write down two degree-two elements in $R(a,b,c,d)$ that they conjecture belong to the center of $R(a,b,c,d)$. We verify their conjecture in Proposition 6.3 and Corollary 6.5.

It is interesting to compare the proof of these results about the centers to the proof that the Casimir elements in the enveloping algebras $U({\mathfrak{s}}{\mathfrak{l}}_{2})$ and $U_{q}({\mathfrak{s}}{\mathfrak{l}}_{2})$ belong to the center. The latter proofs are absolutely straightforward, whereas the computations involved in describing the centers of $A(\alpha,\beta,\gamma)$ are far less routine because these algebras do not have a PBW basis (or, apparently, any basis that makes computation routine). See however, the notion of an $I$-algebra in [23].

1.10. Some questions about $A(\alpha,\beta,\gamma)$

Computer calculations by Frank Moore suggest that the dimensions of the homogeneous components $A(\alpha,\beta,\gamma)_{n}$ are $1,4,10,16,19,20,20,20,\ldots$ when $\alpha+\beta+\gamma+\alpha\beta\gamma\neq 0$. Is this true? If so, then for a generic linear combination $\Omega$ of the central elements $x_{0}^{2},x_{1}^{2},x_{2}^{2},x_{3}^{2}\in A(\alpha,\beta,\gamma)_{2}$ the localization $A[\Omega^{-1}]_{0}$ is a finite dimensional algebra having dimension 20. We expect the representation theory of these finite dimensional algebras is interesting.

We do not know if $A(\alpha,\beta,\gamma)$ is a Koszul algebra (Sklyanin algebras are) but whether it is or not its quadratic dual $A(\alpha,\beta,\gamma)^{!}$ deserves investigation.

We show in Section 3, when $\alpha+\beta+\gamma+\alpha\beta\gamma\neq 0$ and $\alpha\beta\gamma\neq 0$, that the algebra $A(\alpha,\beta,\gamma)$ determines and is determined by a configuration of 20 points in ${\mathbb{P}}^{3}\times\mathbb{P}^{3}$ that is the graph of a bijection between two 20-point elements of $\mathbb{P}^{3}$. We do not understand this configuration but the representation theory of $A(\alpha,\beta,\gamma)$ and these related algebras is governed by it. The details of this are likely to be interesting and novel.

It would be interesting to understand how the configuration of 20 points relates to the features of $R(a,b,c,d)$ that are relevant to the work of Cho, Hong, and Lau.

1.11. Acknowledgements

We wish to thank Frank Moore whose computer calculations involving the algebras $A(\alpha,\beta,\gamma)$ defined in 1-3 were of great assistance to us at an early stage of this project. His calculations showed that over certain finite fields the elements $x_{0}^{2},\ldots,x_{3}^{2}$ belong to the center of $A(\alpha,\beta,\gamma)$ when $\alpha\beta\gamma\neq 0$ and $\alpha+\beta+\gamma+\alpha\beta\gamma\neq 0$. Based on those calculations we then proved the centrality of those elements over all fields (Proposition 6.2).

The work of the first author was partially supported by NSF grant DMS-1565226.

2. Algebras $A(\alpha,\beta,\gamma)$ with a Sklyanin-like presentation

2.1. Notation

Throughout this paper $\Bbbk$ denotes a field whose characteristic is not 2, and $i$ denotes a fixed square root of $-1$.

Whenever we use parameters $\alpha,\beta,\gamma\in\Bbbk$ we will assume they have square roots $a,b,c\in\Bbbk$.

We fix a 4-dimensional $\Bbbk$-vector space $V$. Always, $x_{0},x_{1},x_{2},x_{3}$ will denote a basis for $V$.

2.1.1.

We write $TV$ for the tensor algebra on $V$. Thus $TV$ is the free algebra $\Bbbk\langle x_{0},x_{1},x_{2},x_{3}\rangle$. We always consider $TV$ as an ${\mathbb{Z}}$-graded $\Bbbk$-algebra with $\deg(V)=1$. All the algebras in this paper are of the form $A=TV/(R)$ for various 6-dimensional subspaces $R$ of $V^{\otimes 2}$.

2.1.2.

Let $\alpha,\beta,\gamma\in\Bbbk$. The algebra $A(\alpha,\beta,\gamma)$ is the free algebra $TV$ modulo the relations in 1-1.

2.1.3.

We will often write $(\alpha_{1},\alpha_{2},\alpha_{3})=(\alpha,\beta,\gamma)$. In Section 2 and Section 3, $a,b,c$ will denote fixed square roots of $\alpha,\beta,\gamma$. We will often write $(a_{1},a_{2},a_{3})=(a,b,c)$.

2.1.4.

Let $A$ be a ${\mathbb{Z}}$-graded $\Bbbk$-algebra. We write $\operatorname{Aut}_{{\sf gr}}(A)$ for the group of graded $\Bbbk$-algebra automorphisms of $A$.

If $\lambda\in\Bbbk^{\times}$, $\phi_{\lambda}$ denotes the automorphism of $A$ that is multiplication by $\lambda^{n}$ on $A_{n}$. The map $\Bbbk^{\times}\to\operatorname{Aut}_{{\sf gr}}(A)$, $\lambda\mapsto\phi_{\lambda}$, is an injective group homomorphism whose image lies in the center. We will often identify $\lambda$ with $\phi_{\lambda}$. If $\psi\in\operatorname{Aut}(A)$, we will write $\psi^{m}=\lambda$ if $\psi^{m}=\phi_{\lambda}$ and $\lambda\psi$ for $\phi_{\lambda}\psi$.

2.1.5.

Suppose $a,b,c\in\Bbbk^{\times}$. We define $\psi_{1},\psi_{2},\psi_{3}\in\operatorname{GL}(V)$ by declaring that $\psi_{i}(x_{j})$ is the entry in row $\psi_{i}$ and column $x_{j}$ in Table 1.

In the notation of 2.1.3, if $(i,j,k)$ is a cyclic permutation of $(1,2,3)$, then

[TABLE]

2.1.6. The Heisenberg group of order $4^{3}$

The Heisenberg group of order $4^{3}$ is

[TABLE]

2.2.

By [11] and [22, pp. 64-65], for example, the Heisenberg group $H_{4}$ acts as graded $\Bbbk$-algebra automorphisms of the 4-dimensional Sklyanin algebras when $\Bbbk={\mathbb{C}}$. The next result records the fact that $H_{4}$ acts as graded $\Bbbk$-algebra automorphisms of $A(\alpha,\beta,\gamma)$ whenever $\alpha\beta\gamma\neq 0$ and $\Bbbk$ is a field having square roots of $\alpha$, $\beta$, $\gamma$, and $-1$.

Proposition 2.1**.**

Suppose $\alpha\beta\gamma\neq 0$. Fix $\nu_{1},\nu_{2},\nu_{3}\in\Bbbk^{\times}$ such that $a\nu_{1}^{2}=b\nu_{2}^{2}=c\nu_{3}^{2}=-iabc$.

1. (1)

The maps $\psi_{1},\psi_{2},\psi_{3}:V\to V$ in Table 1 extend to $\Bbbk$-algebra automorphisms of $A(\alpha,\beta,\gamma)$. 2. (2)

There is an injective homomorphism $H_{4}\to\operatorname{Aut}_{\sf gr}(A)$ given by

[TABLE]

Under this map, $\delta\mapsto\phi_{i}$, the automorphism that is multiplication by $i^{n}$ on $A(\alpha,\beta,\gamma)_{n}$. 3. (3)

The subgroup of $\operatorname{Aut}_{\sf gr}(A)$ generated by $\gamma_{1}:=\varepsilon_{1}^{2}$, $\gamma_{2}:=\varepsilon_{2}^{2}$, and $\gamma_{3}:=\varepsilon_{3}^{2}$ is isomorphic to ${\mathbb{Z}}_{2}\times{\mathbb{Z}}_{2}$. The value of $\gamma_{i}(x_{j})$ is the entry in row $\gamma_{i}$ and column $x_{j}$ of Table Table 2 below.

Proof.

Let $(i,j,k)$ be a cyclic permutation of $(1,2,3)$ and let $\lambda_{0},\lambda_{i},\lambda_{j},\lambda_{k}\in\Bbbk^{\times}$. In [19, Prop. 4], Sklyanin observed that the linear map $\psi:V\to V$ acting on $x_{0},x_{i},x_{j},x_{k}$ as

extends to an automorphism of the Sklyanin algebra if and only if

[TABLE]

A straightforward calculation shows that $\psi$ extends to an automorphism of $A(\alpha,\beta,\gamma)$ without any restriction on $\alpha,\beta,\gamma$ other than $\alpha\beta\gamma\neq 0$ if and only if 2-1 holds. The maps $\psi_{1}$, $\psi_{2}$, and $\psi_{3}$, satisfy these conditions so extend to graded $\Bbbk$-algebra automorphisms of $A$.

It is easy to check that $\psi_{1}\psi_{2}=\delta\psi_{2}\psi_{1}$, $\psi_{2}\psi_{3}=\delta\psi_{3}\psi_{2}$, and $\psi_{3}\psi_{1}=\delta\psi_{1}\psi_{3}$. It follows that $\varepsilon_{1}\varepsilon_{2}=\delta\varepsilon_{2}\varepsilon_{1}$, $\varepsilon_{2}\varepsilon_{3}=\delta\varepsilon_{3}\varepsilon_{2}$ ,and $\varepsilon_{3}\varepsilon_{1}=\delta\varepsilon_{1}\varepsilon_{3}$.

It is easy to check that $\gamma_{1},\gamma_{2},\gamma_{3}$ act on $x_{0},x_{1},x_{2},x_{3}$ as in Table 2. Hence $\varepsilon_{1}^{4}=\varepsilon_{2}^{4}=\varepsilon_{3}^{4}=1$. Simple calculations show that $\psi_{1}^{2}=-ibc\gamma_{1}$, $\psi_{2}^{2}=-iac\gamma_{2}$, and $\psi_{3}^{2}=-iab\gamma_{3}$, where $\gamma_{1}$, $\gamma_{2}$, and $\gamma_{3}$, are the automorphisms in Table 2. We leave the rest of the proof to the reader. ∎

2.2.1.

The maps $\gamma_{i}\in\operatorname{GL}(V)$ given by Table 2 extend to graded $\Bbbk$-algebra automorphisms of $A(\alpha,\beta,\gamma)$ for all $\alpha,\beta,\gamma\in\Bbbk$.

2.3.

In the next result, whose proof we omit, $([A,A])$ denotes the ideal in $A$ generated by all commutators $ab-ba$, $a,b\in A$. Thus, $A/([A,A])$ is the largest commutative quotient of $A$.

Proposition 2.2**.**

Suppose $\alpha\beta\gamma\neq 0$. Let $A=A(\alpha,\beta,\gamma)$.

1. (1)

As a quotient of the polynomial ring $\Bbbk[x_{0},x_{1},x_{2},x_{3}]$,

[TABLE] 2. (2)

As a subscheme of ${\mathbb{P}}(V^{*})$,

[TABLE] 3. (3)

$A$* has exactly four graded quotients that are polynomial rings in one variable, namely the quotients by the ideals $(x_{1},x_{2},x_{3})$, $(x_{0},x_{2},x_{3})$, $(x_{0},x_{1},x_{3})$, and $(x_{0},x_{1},x_{2})$.*

Lemma 2.3**.**

There are algebra isomorphisms

[TABLE]

Proof.

There is an isomorphism $A(\alpha,\beta,\gamma)\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}A(\beta,\gamma,\alpha)$ given by $x_{0}\mapsto x_{0}$ and $x_{i}\mapsto x_{i+1}$ for $i\in\{1,2,3\}={\mathbb{Z}}/3$. Similarly, $A(\beta,\gamma,\alpha)\cong A(\gamma,\alpha,\beta)$. Since

[TABLE]

and

[TABLE]

there is an isomorphism $A(\alpha,\beta,\gamma)\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}A(-\alpha,-\gamma,-\beta)$ given by $x_{0}\mapsto x_{0}$, $x_{1}\mapsto-x_{1}$, $x_{2}\mapsto x_{3}$, and $x_{3}\mapsto-x_{2}$. ∎

Proposition 2.4**.**

Suppose $\alpha\beta\gamma\neq 0$ and $\alpha^{\prime}\beta^{\prime}\gamma^{\prime}\neq 0$. Then $A(\alpha,\beta,\gamma)\cong A(\alpha^{\prime},\beta^{\prime},\gamma^{\prime})$ as graded $\Bbbk$-algebras if and only if $(\alpha^{\prime},\beta^{\prime},\gamma^{\prime})$ is a cyclic permutation of either $(\alpha,\beta,\gamma)$ or $(-\alpha,-\beta,-\gamma)$.

Proof.

($\Leftarrow$) This is the content of Lemma 2.3.

($\Rightarrow$) Before starting the proof we introduce some notation. If $(p,q,r,s)$ is a permutation of $(0,1,2,3)$ we define

[TABLE]

where $\mu_{1},\mu_{2},\mu_{3},\nu_{1},\nu_{2},\nu_{3}$ are the unique scalars such that

[TABLE]

in $A$. It is easy to see that

[TABLE]

If $\langle p,q,r,s\rangle_{A}=(\lambda_{1},\lambda_{2},\lambda_{3})$, then $\langle p,r,s,q\rangle_{A}=(\lambda_{2},\,\lambda_{3},\lambda_{1})$. Using this and the equalities in 2-2, it is easy to compute $\langle p,q,r,s\rangle_{A}$ for all permutations $(p,q,r,s)$ of $(0,1,2,3)$.

Let’s write $A=A(\alpha,\beta,\gamma)$ and $B=A(\alpha^{\prime},\beta^{\prime},\gamma^{\prime})$. To distinguish the presentation of $A$ from that for $B$ we will write $x_{0},x_{1},x_{2},x_{3}$ for the generators of $A$, as in 1-3, and write $x_{0}^{\prime},x_{1}^{\prime},x_{2}^{\prime},x_{3}^{\prime}$ for the generators of $B$. Thus, if $(\beta_{1},\beta_{2},\beta_{3})=(\alpha^{\prime},\beta^{\prime},\gamma^{\prime})$, then $[x_{0}^{\prime},x_{i}^{\prime}]=\beta_{i}\{x_{j}^{\prime},x_{k}^{\prime}\}$ and $\{x_{0}^{\prime},x_{i}^{\prime}\}=[x^{\prime}_{j},x^{\prime}_{k}]$ for each cyclic permutation $(i,j,k)$ of $(1,2,3)$.

Suppose $\Phi:A\to B$ is an isomorphism of graded $\Bbbk$-algebras. The restriction of $\Phi$ to $A_{1}$ is a vector space isomorphism $A_{1}\to B_{1}$. It induces an isomorphism $\varphi:{\mathbb{P}}(B_{1}^{*})\to{\mathbb{P}}(A_{1}^{*})$. Let’s denote the points $(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)\in{\mathbb{P}}(B_{1}^{*})$ by $e_{0}^{\prime},e_{1}^{\prime},e_{2}^{\prime},e_{3}^{\prime}$ respectively. Since $\Phi$ induces an isomorphism $A/([A,A])\to B/([B,B])$, $\varphi$ restricts to an isomorphism $\operatorname{Proj}(B/([B,B])\to\operatorname{Proj}(A/([A,A])$. Therefore $\varphi(\{e_{0}^{\prime},e_{1}^{\prime},e_{2}^{\prime},e_{3}^{\prime}\})=\{e_{0},e_{1},e_{2},e_{3}\}$. Since each $x_{m}$ vanishes at exactly 3 points in $\{e_{0},e_{1},e_{2},e_{3}\}$, $\Phi(x_{m})$ vanishes at exactly 3 points in $\{e_{0}^{\prime},e_{1}^{\prime},e_{2}^{\prime},e_{3}^{\prime}\}$. It follows that there are non-zero scalars $\lambda_{0},\lambda_{1},\lambda_{2},\lambda_{3}$ and a permutation $(p,q,r,s)$ of $(0,1,2,3)$ such that $\Phi(x_{0})=\lambda_{0}x_{p}^{\prime}$, $\Phi(x_{1})=\lambda_{1}x_{q}^{\prime}$, $\Phi(x_{2})=\lambda_{2}x_{r}^{\prime}$, and $\Phi(x_{3})=\lambda_{3}x_{s}^{\prime}$.

Since $\{x_{0},x_{i}\}=[x_{j},x_{k}]$ for every cyclic permutation $(i,j,k)$ of $(1,2,3)$,

[TABLE]

Since $[x_{0},x_{i}]=\alpha_{i}\{x_{j},x_{k}\}$ for every cyclic permutation $(i,j,k)$ of $(1,2,3)$,

[TABLE]

It follows that

[TABLE]

[TABLE]

[TABLE]

Therefore $\langle p,q,r,s\rangle_{B}=(\alpha,\beta,\gamma)=\langle x_{0},x_{1},x_{2},x_{3}\rangle$. It now follows from 2-2 and the sentence after it that $\langle 0,1,2,3\rangle_{B}$ is a cyclic permutation of either $(\alpha,\beta,\gamma)$ or $(-\alpha,-\beta,-\gamma)$; since $\langle 0,1,2,3\rangle_{B}=(\alpha^{\prime},\beta^{\prime},\gamma^{\prime})$, the proof is complete. ∎

3. The zero locus of the relations for $A(\alpha,\beta,\gamma)$

The ideas in 3.1 apply to all graded algebras defined by 4 generators and 6 quadratic relations, i.e., to all algebras $A$ of the form $TV/(R)$ where $V$ and $R$ are as in the next paragraph.

3.1. Quadratic algebras on 4 generators with 6 relations

Let $V$ be a 4-dimensional vector space over $\Bbbk$, $R$ a 6-dimensional subspace of $V^{\otimes 2}$. Let ${\mathbb{P}}={\mathbb{P}}(V^{*})\cong{\mathbb{P}}^{3}$. Let $\Gamma\subseteq{\mathbb{P}}\times{\mathbb{P}}$ be the scheme-theoretic zero locus of $R$ (viewed as forms of bi-degree $(1,1)$). For example, if $A$ is the polynomial ring, then $R$ consists of the skew-symmetric tensors and $\Gamma$ is the diagonal.

Since $\dim_{\Bbbk}(R)=6=\dim({\mathbb{P}}^{3}\times{\mathbb{P}}^{3})$, $\Gamma\neq\varnothing$.

Proposition 3.1**.**

Suppose $\dim(\Gamma)=0$. Then

1. (1)

$\Gamma$* consists of 20 points counted with multiplicity, and* 2. (2)

the subspace of $V\otimes V$ that vanishes on $\Gamma$ is $R$ **[17, Thm. 4.1]**.

Proof.

(1) The Chow ring of ${\mathbb{P}}^{3}$ is isomorphic to ${\mathbb{Z}}[t]/(t^{4})$ with $t$ the class of a hyperplane. The Chow ring of ${\mathbb{P}}^{3}\times{\mathbb{P}}^{3}$ is isomorphic to ${\mathbb{Z}}[s,t]/(s^{4},t^{4})$ and the class of the zero locus of a non-zero element in $V\otimes V$ is equal to $s+t$. If $\dim(\Gamma)=0$, then the class of $\Gamma$ is $(s+t)^{6}$ since $\dim(R)=6$. But $(s+t)^{6}=20s^{3}t^{3}$ so the cardinality of $\Gamma$ is 20 when its points are counted with multiplicity.

(2) This is [17, Thm. 4.1]. ∎

3.2.

We now explain our strategy for computing $\Gamma$ for $A(\alpha,\beta,\gamma)$.

Let ${\bf x}$ denote the row vector $(x_{0},x_{1},x_{2},x_{3})$ over $\Bbbk\langle x_{0},x_{1},x_{2},x_{3}\rangle$ and let ${\bf x}^{\sf T}$ denote its transpose.

The relations defining $A(\alpha,\beta,\gamma)$ can be written as a single matrix equation, $M{\bf x}^{\sf T}=0$, over $\Bbbk\langle x_{0},x_{1},x_{2},x_{3}\rangle$ where

[TABLE]

The relations can also be written as ${\bf x}M^{\prime}=0$ where

[TABLE]

Consider the entries in $M$ (resp., ${\bf x}$) as linear forms on the left-hand (resp., right-hand) factor of ${\mathbb{P}}^{3}\times{\mathbb{P}}^{3}={\mathbb{P}}(V^{*})\times{\mathbb{P}}(V^{*})$. Then $\Gamma$ is the scheme-theoretic zero locus of the six entries in $M{\bf x}^{\sf T}$ when those entries are viewed as bi-homogeneous elements in $\Bbbk[x_{0},x_{1},x_{2},x_{3}]\otimes\Bbbk[x_{0},x_{1},x_{2},x_{3}]$.

Let ${\operatorname{pr}}_{1}:\Gamma\to{\mathbb{P}}^{3}$ and ${\operatorname{pr}}_{2}:\Gamma\to{\mathbb{P}}^{3}$ be the projections ${\operatorname{pr}}_{1}(p,p^{\prime})=p$ and ${\operatorname{pr}}_{2}(p,p^{\prime})=p^{\prime}$.

If $p\in{\mathbb{P}}^{3}$, then $p\in{\operatorname{pr}}_{1}(\Gamma)$ if and only if there is a point $p^{\prime}\in{\mathbb{P}}^{3}$ such that $M(p){\bf x}^{\sf T}(p^{\prime})=0$; i.e., if and only if $\operatorname{rank}(M(p))<4$. Thus, ${\operatorname{pr}}_{1}(\Gamma)$ is the scheme-theoretic zero locus of the $4\times 4$ minors of $M$. Similarly, ${\operatorname{pr}}_{2}(\Gamma)$ is the scheme-theoretic zero locus of the $4\times 4$ minors of $M^{\prime}$.

Lemma 3.2**.**

If $\lambda,\mu,\nu$ are non-zero scalars, then the intersection of the three quadrics

[TABLE]

consists of the eight points

[TABLE]

Proof.

The line $x_{0}-\lambda\mu x_{3}=x_{1}-\lambda\mu^{-1}x_{2}=0$ lies on the quadric $x_{0}x_{1}-\lambda^{2}x_{2}x_{3}=0$ because

[TABLE]

and on the quadric $x_{0}x_{2}-\mu^{2}x_{1}x_{3}=0$ because

[TABLE]

Continuing in this vein, the lines $x_{0}=x_{3}=0$, $x_{1}=x_{2}=0$, $x_{0}-\lambda\mu x_{3}=x_{1}-\lambda\mu^{-1}x_{2}=0$, and $x_{0}+\lambda\mu x_{3}=x_{1}+\lambda\mu^{-1}x_{2}=0$, lie on the quadrics $x_{0}x_{1}-\lambda^{2}x_{2}x_{3}=0$ and $x_{0}x_{2}-\mu^{2}x_{1}x_{3}=0$. By Bézout’s theorem, the intersection of these two quadrics is a curve of degree 4 in ${\mathbb{P}}^{3}$ so is the union of these four lines.

The quadric $x_{0}x_{3}-\nu^{2}x_{1}x_{2}=0$ meets the line $x_{0}=x_{3}=0$ at $(0,1,0,0)$ and $(0,0,1,0)$; the line $x_{1}=x_{2}=0$ at $(1,0,0,0)$ and $(0,0,0,1)$; the line $x_{0}-\lambda\mu x_{3}=x_{1}-\lambda\mu^{-1}x_{2}=0$ at $(\lambda\mu\nu,\lambda,\mu,\nu)$ and $(\lambda\mu\nu,-\lambda,-\mu,\nu)$; and the line $x_{0}+\lambda\mu x_{3}=x_{1}+\lambda\mu^{-1}x_{2}=0$ at $(\lambda\mu\nu,-\lambda,\mu,-\nu)$ and $(\lambda\mu\nu,\lambda,-\mu,-\nu)$. The proof is complete. ∎

Proposition 3.3**.**

The scheme $\Gamma$ associated to the algebra $TV/(R)=A(\alpha,\beta,\gamma)$ is finite if and only if $\alpha\beta\gamma\neq 0$ and $\alpha+\beta+\gamma+\alpha\beta\gamma\neq 0$.

Proof.

Before starting the proof we introduce some notation.

We label the following four polynomials in the symmetric algebra $SV$:

[TABLE]

We write $h_{ij}$ for the $4\times 4$ minor of $M$ obtained by deleting rows $i$ and $j$. Up to non-zero scalar multiples,

[TABLE]

These are the “same” expressions as those in the proof of [21, Prop. 2.4].222The phrase “making frequent use of (0.2.1)” in the second sentence in the proof of [21, Prop. 2.4] should be deleted in order to make that sentence true.

We write $g_{ij}$ for the $4\times 4$ minor of $M^{\prime}$ obtained by deleting columns $i$ and $j$.

If $p=(\lambda_{0},\lambda_{1},\lambda_{2},\lambda_{3})\in{\mathbb{P}}^{3}$ we write $\ominus p$ for the point $(-\lambda_{0},\lambda_{1},\lambda_{2},\lambda_{3})$ and $M^{\prime}(\ominus p)$ for the matrix $M^{\prime}$ evaluated at $\ominus p$. The matrices $M^{\prime}(\ominus p)$ and $-M(p)^{\sf T}$ are almost the same: the only difference is that the top row of $M^{\prime}(\ominus p)$ is the negative of the top row of $-M(p)^{\sf T}$. This observation makes it easy to compute the $4\times 4$ minors of $M^{\prime}$ from the $4\times 4$ minors of $M$. Doing that, up to non-zero scalar multiples we obtain

[TABLE]

In particular, up to non-zero scalar multiples, $g_{ij}(x_{0},x_{1},x_{2},x_{3})=h_{ij}(-x_{0},x_{1},x_{2},x_{3})$ for all $i$ and $j$. Hence ${\operatorname{pr}}_{2}(\Gamma)=\ominus{\operatorname{pr}}_{1}(\Gamma)$.

It follows that ${\operatorname{pr}}_{1}(\Gamma)$ is finite if and only if ${\operatorname{pr}}_{2}(\Gamma)$ is finite if and only if $\Gamma$ is finite.

($\Leftarrow$) Suppose $\alpha\beta\gamma\neq 0$ and $\alpha+\beta+\gamma+\alpha\beta\gamma\neq 0$.

Let $C$ be an irreducible component of ${\operatorname{pr}}_{1}(\Gamma)$. Since $h_{12}$, $h_{13}$, and $h_{23}$, vanish on $C$, either $q$ vanishes on $C$ or $C$ is in the zero locus of the other factors of $h_{12}$, $h_{13}$, and $h_{23}$; i.e., in the common zero locus of $x_{0}x_{1}-\alpha x_{2}x_{3}$, $x_{0}x_{2}-\beta x_{1}x_{3}$, and $x_{0}x_{3}-\gamma x_{1}x_{2}$; but that common zero locus is finite by Lemma 3.2 so either $C$ is finite or $q$ vanishes on $C$. Likewise, if $q_{j}\in\{q_{1},q_{2},q_{3}\}$, either $C$ is finite or $q_{j}$ vanishes on $C$. Thus, either $C$ is finite or all four of $q$, $q_{1}$, $q_{2}$, and $q_{3}$, vanish on $C$. However, the set $\{q,q_{1},q_{2},q_{3}\}$ is linearly independent because the determinant

[TABLE]

is non-zero so the common zero-locus of $q$, $q_{1}$, $q_{2}$, and $q_{3}$, is empty. We conclude that $C$ is finite. It follows that ${\operatorname{pr}}_{1}(\Gamma)$, and hence $\Gamma$, is finite.

($\Rightarrow$) Suppose $\Gamma$ is finite.

If $\alpha+\beta+\gamma+\alpha\beta\gamma=0$, then ${\rm span}\{q,q_{1}\}={\rm span}\{q,q_{2}\}={\rm span}\{q,q_{3}\}$. It follows that all $h_{ij}$ vanish on $\{q=q_{1}=0\}$ whence $\{q=q_{1}=0\}\subseteq{\operatorname{pr}}_{1}(\Gamma)$. But this is ridiculous because $\{q=q_{1}=0\}$ is a curve, hence infinite, so we conclude that $\alpha+\beta+\gamma+\alpha\beta\gamma\neq 0$.

If $\alpha=0$, then all $h_{ij}$ vanish on the line $x_{0}=x_{1}=0$; i.e., $\{x_{0}=x_{1}=0\}\subseteq{\operatorname{pr}}_{1}(\Gamma)$; this is not the case because $\Gamma$ is finite so we conclude that $\alpha\neq 0$. If $\beta=0$, then $\{x_{0}=x_{2}=0\}\subseteq{\operatorname{pr}}_{1}(\Gamma)$; this is not the case so we conclude that $\beta\neq 0$. If $\gamma=0$, then $\{x_{0}=x_{3}=0\}\subseteq{\operatorname{pr}}_{1}(\Gamma)$; this is not the case so we conclude that $\gamma\neq 0$. Thus, $\alpha\beta\gamma\neq 0$. ∎

3.3.

Suppose $\alpha\beta\gamma\neq 0$. Let ${\mathfrak{P}}\subseteq{\mathbb{P}}^{3}$ denote the set of points in the following table.

Let $\ominus:{\mathbb{P}}(V^{*})\to{\mathbb{P}}(V^{*})$ and $\theta:{\mathfrak{P}}\to\ominus{\mathfrak{P}}$ be the maps $\ominus(\xi_{0},\xi_{1},\xi_{2},\xi_{3})=(-\xi_{0},\xi_{1},\xi_{2},\xi_{3})$ and

[TABLE]

Proposition 3.4**.**

Suppose the subscheme $\Gamma\subseteq{\mathbb{P}}^{3}\times{\mathbb{P}}^{3}$ determined by the relations for $A(\alpha,\beta,\gamma)$ is finite. Then $\Gamma$ is the graph of the bijection $\theta:{\mathfrak{P}}\to\ominus{\mathfrak{P}}$ and consists of 20 distinct points.

Proof.

Since $\Gamma$ is finite both $\alpha\beta\gamma$ and $\alpha+\beta+\gamma+\alpha\beta\gamma$ are non-zero. Since $\alpha\beta\gamma\neq 0$, each column of Table 3 consists of four distinct points. It is easy to see that

[TABLE]

If $(a,\xi_{1},\xi_{2},\xi_{3})\in{\mathfrak{P}}_{0}$, then $\xi_{1}=a\xi_{2}\xi_{3}$; if $(a,\xi_{1},\xi_{2},\xi_{3})\in{\mathfrak{P}}_{1}$, then $\xi_{1}=-a\xi_{2}\xi_{3}$; hence ${\mathfrak{P}}_{0}\cap\,{\mathfrak{P}}_{1}=\varnothing$. Similarly, if $(b,\xi_{1},\xi_{2},\xi_{3})\in{\mathfrak{P}}_{0}$, then $\xi_{2}=b\xi_{1}\xi_{3}$ whereas if $(b,\xi_{1},\xi_{2},\xi_{3})\in{\mathfrak{P}}_{2}$, then $\xi_{2}=-b\xi_{1}\xi_{3}$ so ${\mathfrak{P}}_{0}\cap\,{\mathfrak{P}}_{2}=\varnothing$. The same sort of argument shows that ${\mathfrak{P}}_{0}\cap\,{\mathfrak{P}}_{3}=\varnothing$. Thus, ${\mathfrak{P}}$ is the disjoint union of five sets each of which consists of four distinct points. Hence ${\mathfrak{P}}$ consists of 20 distinct points.

Let $\Gamma_{\theta}$ denote the graph of $\theta:{\mathfrak{P}}\to\ominus{\mathfrak{P}}$.

To complete the proof we must show that the vanishing locus in ${\mathbb{P}}\times{\mathbb{P}}$ of the polynomials

[TABLE]

is exactly $\Gamma_{\theta}$.

Clearly, if $p\in{\mathfrak{P}}_{\infty}$, then all six polynomials vanish at $(p,p)=(p,\theta(p))$.

Suppose $p\in{\mathfrak{P}}_{0}$. Let $(i,j,k)$ be a cyclic permutation of $(1,2,3)$. Since $(p,\theta(p))=(p,\ominus p)$, $x_{0}\otimes x_{i}+x_{i}\otimes x_{0}$ and $x_{j}\otimes x_{k}-x_{k}\otimes x_{j}$ vanish at $(p,\theta(p))$; the three polynomials in the second column of 3-3 therefore vanish at $(p,\theta(p))$. On the other hand, $x_{0}\otimes x_{i}-x_{i}\otimes x_{0}\,-\,\alpha_{i}(x_{j}\otimes x_{k}+x_{k}\otimes x_{j})$ vanishes at $(p,\theta(p))$ if and only if $2x_{0}\otimes x_{i}\,-\,2\alpha_{i}x_{j}\otimes x_{k}$ does. This vanishes at $(p,\ominus p)$ because $2x_{0}x_{i}\,-\,2\alpha_{i}x_{j}x_{k}$ vanishes at $p$.

Let $(i,j,k)$ be a cyclic permutation of $(1,2,3)$. Suppose $p=(\xi_{0},\xi_{1},\xi_{2},\xi_{3})\in{\mathfrak{P}}_{i}$. Then $\theta(p)=(\xi_{0}^{\prime},\xi_{1}^{\prime},\xi_{2}^{\prime},\xi_{3}^{\prime})$ where $\xi_{i}^{\prime}=-\xi_{i}$ and $\xi_{\ell}^{\prime}=\xi_{\ell}$ if $\ell\in\{0,1,2,3\}-\{i\}$. It follows that

[TABLE]

[TABLE]

[TABLE]

A case-by-case inspection shows that these three expressions are zero; thus, three of the polynomials in 3-3 vanish at $(p,\theta(p))$. The other three polynomials in 3-3 also vanish at $(p,\theta(p))$ because

[TABLE]

[TABLE]

vanish at $(p,\theta(p))$.

We have shown that the polynomials in 3-3 vanish on $\Gamma_{\theta}$. This completes the proof that $\Gamma_{\theta}\subseteq\Gamma$. In particular, ${\mathfrak{P}}\subseteq{\operatorname{pr}}_{1}(\Gamma)$. To complete the proof of the proposition we must show the polynomials in 3-3 do not vanish outside $\Gamma_{\theta}$ or, equivalently, that ${\operatorname{pr}}_{1}(\Gamma)={\mathfrak{P}}$.

With this goal in mind let $p\in{\operatorname{pr}}_{1}(\Gamma)$. We observed in the proof of Proposition 3.3, that

[TABLE]

If $q$ does not vanish at $p$, then Lemma 3.2 implies that $p\in{\mathfrak{P}}_{\infty}\cup{\mathfrak{P}}_{0}$. Likewise, if $q_{j}\in\{q_{1},q_{2},q_{3}\}$ and does not vanish at $p$, then Lemma 3.2 tells us that $p\in{\mathfrak{P}}_{\infty}\cup{\mathfrak{P}}_{j}$. We conclude that $p\in{\mathfrak{P}}$. ∎

Corollary 3.5**.**

If $\alpha\beta\gamma\neq 0$ and $\alpha+\beta+\gamma+\alpha\beta\gamma\neq 0$, then $A(\alpha,\beta,\gamma)$ is isomorphic to $TV/(R)$ where $R\subseteq V^{\otimes 2}$ is the subspace consisting of those $(1,1)$ forms that vanish on the graph of the bijection $\theta:{\mathfrak{P}}\to\ominus{\mathfrak{P}}$.

The deeper meaning of the data $({\mathfrak{P}},\theta)$ eludes us.

3.4. Remarks

In 3.4 we assume that $\alpha\beta\gamma\neq 0$ but do not make any assumption on $\alpha+\beta+\gamma+\alpha\beta\gamma$.

3.4.1.

Let $\psi_{1},\psi_{2},\psi_{3}\in\operatorname{GL}(V)$ be the maps defined in 2.1.5. Let $\gamma_{1},\gamma_{2},\gamma_{3}\in\operatorname{GL}(V)$ be the maps defined in 2.2.

Let $x_{0}^{*},x_{1}^{*},x_{2}^{*},x_{3}^{*}$ be the dual basis to $x_{0},x_{1},x_{2},x_{3}$. The contragredient action of the maps $\psi_{j}$ acting on $V^{*}$ is given by Table 4. The subgroup of $\operatorname{GL}(V^{*})$ generated by $\psi_{1},\psi_{2},\psi_{3}$ is isomorphic to $H_{4}$. The center of $H_{4}$ acts trivially on ${\mathbb{P}}(V^{*})$ so we obtain an action of ${\mathbb{Z}}_{4}\times{\mathbb{Z}}_{4}$ on ${\mathbb{P}}(V^{*})$.

It is easy to see that $\psi_{j}({\mathfrak{P}})={\mathfrak{P}}$ for all $j$. We note that

[TABLE]

It follows rather easily that ${\mathfrak{P}}_{0}\cup{\mathfrak{P}}_{1}\cup{\mathfrak{P}}_{2}\cup{\mathfrak{P}}_{3}$ is a single orbit under the action of $H_{4}$ and therefore a single ${\mathbb{Z}}_{4}\times{\mathbb{Z}}_{4}$-orbit. The subgroup $\{\mathrm{id},\gamma_{1},\gamma_{2},\gamma_{3}\}$ of ${\mathbb{Z}}_{4}\times{\mathbb{Z}}_{4}$ is an essential subgroup and, as is easy to see, none of $\gamma_{1},\gamma_{2},\gamma_{3}$ fixes any point in ${\mathfrak{P}}_{0}\cup{\mathfrak{P}}_{1}\cup{\mathfrak{P}}_{2}\cup{\mathfrak{P}}_{3}$ so the homomorphism

[TABLE]

is injective. It follows that ${\mathfrak{P}}_{0}\cup{\mathfrak{P}}_{1}\cup{\mathfrak{P}}_{2}\cup{\mathfrak{P}}_{3}$ consists of 16 distinct points. Hence ${\mathfrak{P}}$ consists of 20 distinct points (even without the hypothesis $\alpha+\beta+\gamma+\alpha\beta\gamma\neq 0$).

3.4.2.

The points in ${\mathfrak{P}}_{\infty}$ are fixed by the action of ${\mathbb{Z}}_{2}\times{\mathbb{Z}}_{2}$ given by Table 2.

3.4.3.

If $i\neq\infty$ and $p$ is the topmost point in the column ${\mathfrak{P}}_{i}$, then the other points in that column are $\gamma_{1}(p)$, $\gamma_{2}(p)$, and $\gamma_{3}(p)$, in that order from top to bottom. Thus, when $j\neq\infty$, ${\mathfrak{P}}_{j}$ is a single ${\mathbb{Z}}_{2}\times{\mathbb{Z}}_{2}$-orbit.

3.5. The scheme $\Gamma$ for the 4-dimensional Sklyanin algebras

We review the Sklyanin algebra case (see [21] and [15]). Let $A=A(\alpha,\beta,\gamma)$ be a non-degenerate Sklyanin algebra.

Then $\Gamma$ is the graph of an automorphism of $E\cup\{e_{0},e_{1},e_{2},e_{3}\}$ where $E\subseteq{\mathbb{P}}^{3}$ is the quartic elliptic curve cut out by the equations

[TABLE]

and $e_{i}$ is the vanishing locus of $\{x_{0},x_{1},x_{2},x_{3}\}-\{x_{i}\}$. The points $e_{i}$ are the vertices of the four singular quadrics that contain $E$. The automorphism fixes each $e_{i}$ and its restriction to $E$ is a translation automorphism. Furthermore, $R=\{f\in V\otimes V\;|\;f|_{\Gamma}=0\}$. Thus, $R$ and $\Gamma$ determine each other.

We fix a point $o\in E\cap\{x_{0}=0\}$ and impose a group law on $E$ such that $o$ is the identity and four points on $E$ are collinear if and only if their sum is $o$. The 2-torsion subgroup $E[2]$ is $E\cap\{x_{0}=0\}$. We write $\oplus$ for the group law and $\ominus$ for subtraction, i.e., $p\oplus q=r$ if and only if $p=r\ominus q$.

If $p=(\xi_{0},\xi_{1},\xi_{2},\xi_{3})\in E$, then $\ominus p=(-\xi_{0},\xi_{1},\xi_{2},\xi_{3})$.

See [4, §7] for a longer explanation that uses the same notation as here.

Proposition 3.6**.**

If $A$ is a non-degenerate 4-dimensional Sklyanin algebra, then

1. (1)

${\mathfrak{P}}_{0}\cup{\mathfrak{P}}_{1}\cup{\mathfrak{P}}_{2}\cup{\mathfrak{P}}_{3}\subseteq E$* where $E$ is the elliptic curve given by the equations in 3-4;* 2. (2)

${\mathfrak{P}}_{0}=\tau^{\prime}\oplus E[2]$* where $\tau^{\prime}=(abc,a,b,c)$;* 3. (3)

${\mathfrak{P}}_{i}=\varepsilon_{i}\oplus\tau^{\prime}\oplus E[2]$* and $E[2]=\{o,2\varepsilon_{1},2\varepsilon_{2},2\varepsilon_{3}\}$ for suitable $\varepsilon_{1},\varepsilon_{2},\varepsilon_{3}\in E[4]$.*

4. Point schemes, graphs and flat families

Consider the family of algebras $A(\alpha_{1},\alpha_{2},\alpha_{3})$ as the parameters $\alpha_{i}$ are allowed to vary. This section is devoted to studying the behavior of the scheme $\Gamma$ introduced in 3.1 as the fiber of a family over the parameter space consisting of the points $(\alpha_{1},\alpha_{2},\alpha_{3})$, or in fact more generally over the family of six-dimensional relation spaces for four-generator algebras.

4.1.

Throughout Section 4, $V$ denotes a fixed four-dimensional space of generators for our quadratic algebras with a fixed basis consisting of the generators $x_{i}$, $0\leq i\leq 3$, $\mathbb{G}=\mathbb{G}(6,V^{\otimes 2})$ is the Grassmannian of $6$-planes in $V^{\otimes 2}$, and we regard the points of $\mathbb{G}$ as spaces of relations for four-generator-six-relator connected graded algebras, so that $\mathbb{G}$ will be the parameter space for the algebras in question. We encode a relation space $R\in\mathbb{G}$ as either a $6\times 4$ matrix $M$ or a $4\times 6$ matrix $M^{\prime}$ with entries in $V$ analogous to 5.2 and respectively 3-2, so that for ${\bf x}=(x_{i})$ the relations $R$ read either $M{\bf x}^{\sf T}=0$ or ${\bf x}M^{\prime}=0$.

As mentioned above, we denote by $\Gamma$ the family $\pi:\Gamma\to\mathbb{G}$ whose $R$-fiber $\Gamma_{R}$ for $R\in\mathbb{G}$ is by definition the subscheme of $\mathbb{P}(V^{*})\times\mathbb{P}(V^{*})$ consisting of the pairs of points $(p,p^{\prime})$ as in the discussion from 3.1, whose defining property is $M(p){\bf x}^{\sf T}(p^{\prime})=0$.

We further set

[TABLE]

Let $U=\{R\in\mathbb{G}\;|\;\dim(\mathbb{X}_{R})=0\}$ and define $U^{\prime}$ similarly for $\mathbb{X}^{\prime}$. Finally, given a family $\pi:\mathbb{S}\to\mathbb{T}$ and an open subset $W\subseteq\mathbb{T}$, we denote the restricted family $\pi^{-1}(W)$ by $\mathbb{S}_{W}$ by a slight abuse of notation.

When the algebra $A_{R}$ corresponding to $R$ is Artin-Schelter regular and has some other good properties that we will not specify here, the scheme $\mathbb{X}_{R}$ is (one incarnation of) the point scheme of $A_{R}$ (i.e. the scheme of classes of point modules in $\operatorname{\sf QGr}(A_{R})$). Moreover, in many cases of interest $\mathbb{X}_{R}$ equals $\mathbb{X}^{\prime}_{R}$, and $\Gamma_{R}$ is the graph of an automorphism of this scheme (e.g. for non-degenerate $4$-dimensional Sklyanin algebras [21], for $3$-dimensional AS-regular algebras [1], etc.). Moreover, we saw above in Proposition 3.4 that even when $\mathbb{X}_{R}\neq\mathbb{X}^{\prime}_{R}$, the scheme $\Gamma_{R}$ is often the graph of an isomorphism.

For these reasons, we regard $\Gamma$ and its projections $\mathbb{X}$ and $\mathbb{X}^{\prime}$ as stand-ins for the point scheme even when we lack the requisite regularity properties for this to be literally the case.

We first prove a statement analogous to [5, Theorem 2.6]. That result says, essentially, that the line schemes of connected graded algebras with four generators and six quadratic relations form a flat family provided they have minimal dimension. We prove here that the family $\Gamma\to\mathbb{G}$ is similarly well behaved.

First, we have the following analogue of [5, Proposition 2.1].

Proposition 4.1**.**

The subset $U,U^{\prime}\subset\mathbb{G}$ are open and dense.

Proof.

This is entirely analogous to the proof of [5, Proposition 2.1]. We focus on the case of $U$, to fix ideas.

First, Van den Bergh’s result [24] that, generically, four-generator-six-relator algebras have twenty point modules ensures that $U$ contains a dense open subset of $\mathbb{G}$. Let $\mathbb{X}_{i}$ be the irreducible components of $\mathbb{X}$, and $\pi_{i}:(\mathbb{X}_{i})_{\mathrm{red}}\to\pi(\mathbb{X}_{i})_{\mathrm{red}}$ the restriction and corestriction of $\pi$ to $(\mathbb{X}_{i})_{\mathrm{red}}$.

Each $\pi_{i}$ is projective and hence closed. By [12, Exercise II.3.22(d)] applied to each $\pi_{i}$ individually, the complement of $U$, being the image of the closed subset

[TABLE]

of $\mathbb{X}$, is closed in $\mathbb{G}$. ∎

In conclusion, we get

Corollary 4.2**.**

The locus $W\subset\mathbb{G}$ over which the family $\Gamma$ has zero-dimensional fibers is open and dense.

Proof.

The fiber $\Gamma_{R}$ is zero-dimensional if and only if its two projections $\mathbb{X}_{R}$ and $\mathbb{X}^{\prime}_{R}$ are, so $W$ is simply the intersection $U\cap U^{\prime}$. The conclusion now follows from Proposition 4.1. ∎

We next turn, as hinted above, to proving certain regularity properties for the families $\mathbb{X}$, $\mathbb{X}^{\prime}$ and $\Gamma$ over the good open loci of $\mathbb{G}$ identified in Propositions 4.1 and 4.2.

Lemma 4.3**.**

The schemes $\mathbb{X}_{U}$ and $\mathbb{X}^{\prime}_{U^{\prime}}$ are Cohen-Macaulay.

Proof.

Once more, we focus on the case of $\mathbb{X}_{U}$ without loss of generality.

Locally on $U$, the equations that define $\mathbb{X}_{U}$ as a $U$-subscheme of the relative projective space $\mathbb{P}(V^{*})_{U}=\mathbb{P}(V^{*})\times U$ are given by the $4\times 4$ minors of a $6\times 4$ matrix. Moreover, over $U$, $X_{U}$ has codimension $3$ in $\mathbb{P}(V^{*})_{U}$.

In general, the quotient by the ideal $I$ generated by the $r\times r$ minors of a $p\times q$ matrix in a Cohen-Macaulay ring is again Cohen-Macaulay, provided $I$ has maximal codimension $(p-r+1)(q-r+1)$ (see e.g. the discussion in [10, $\S$ 18.5] on determinantal rings and [2] for a proof). In our case $p=6$, $q=4$ and $r=4$, hence the critical codimension is precisely $(6-4+1)(4-4+1)=3$. This concludes the proof. ∎

Theorem 4.4**.**

The families $\mathbb{X}_{U}\to U$, $\mathbb{X}^{\prime}_{U^{\prime}}\to U^{\prime}$ and $\Gamma_{W}\to W$ are flat.

Proof.

We divide the argument into two parts.

(1): $\mathbb{X}$ and $\mathbb{X}^{\prime}$. Symmetry allows us to once again focus on $\mathbb{X}$. Given the Cohen-Macaulay property for $\mathbb{X}_{U}$, the proof of the theorem mimics that of [5, Theorem 2.6] verbatim.

Let $x\in\mathbb{X}_{U}$ be a point, and set $B=\mathcal{O}_{\pi(x),U}$ and $A=\mathcal{O}_{x,\mathbb{X}_{U}}$. In order to show that $A$ is flat as a $B$-module, denote by ${\mathfrak{p}}$ the maximal ideal of $B$. Since the fiber $(\mathbb{X}_{U})_{\pi(x)}$ is of minimal dimension [math], we have the equality

[TABLE]

This implies the flatness of $A$ over $B$ via [10, Theorem 18.16 (b)], given that $A$ is Cohen-Macaulay by Lemma 4.3 and $B$ is regular.

(2): $\Gamma$. The result for $\Gamma_{W}\to W$ follows from part (1) and the observation that over $W$ the projection ${\operatorname{pr}}_{1}:\Gamma\to\mathbb{X}$ is an isomorphism. ∎

Finally, as an application of the flatness results just proven, we provide an alternate argument for the fact that the 20 points in Table 3 exhaust the “point scheme” of $A(\alpha_{1},\alpha_{2},\alpha_{3})$ under certain non-degeneracy conditions on the parameters $\alpha_{i}$. We begin with

Corollary 4.5**.**

For every $R\in U$, the scheme $\mathbb{X}_{R}$ consists of 20 points counted with multiplicity. Similarly for $\mathbb{X}^{\prime}_{U^{\prime}}$ and $\Gamma_{W}$.

Proof.

We can embed the family $\mathbb{X}\to\mathbb{G}$ into the relative projective space $\mathbb{P}(V^{*})_{\mathbb{G}}=\mathbb{P}(V^{*})\times\mathbb{G}$ in the usual fashion.

The flatness of Theorem 4.4 ensures that all fibers $\mathbb{X}_{R}$ have the same degree in $\mathbb{P}(V^{*})$ so long as $R\in U$, i.e. when $\dim(\mathbb{X}_{R})=0$. But we know there are six-dimensional relation spaces $R$ where the degree is $20$ (e.g. the algebras in [24, 26, 16, 4, 3]).

The case of $\mathbb{X}^{\prime}_{U^{\prime}}$ is analogous, while that of $\Gamma_{W}$ follows as in the proof of Theorem 4.4, using the fact that ${\operatorname{pr}}_{1}:\Gamma\to\mathbb{X}$ is an isomorphism over $W$. ∎

Proposition 4.6**.**

If $\sum\alpha_{i}+\prod\alpha_{i}\neq 0$ and $\prod\alpha_{i}\neq 0$ and $A(\alpha_{1},\alpha_{2},\alpha_{3})=TV/(R)$, then $\mathbb{X}_{R}$ consists of the 20 points in Table 3.

Proof.

If we show that every closed point of $\mathbb{X}_{R}$ is one of the points in Table 3, then $\dim(\mathbb{X}_{R})=0$ so, by Corollary 4.5 above, $\mathbb{X}_{R}$ will consist of 20 points counted with multiplicity.

Let $p$ be a closed point in $\mathbb{X}_{R}$. Consider the four quadratic polynomials appearing as the right hand factors of $h_{12}$, $h_{14}$, $h_{25}$, and $h_{36}$, which are

[TABLE]

Either all of them vanish at $p$, or at least one does not.

Not all of them can vanish at $p$ because the determinant

[TABLE]

is non-zero by hypothesis.

Thus, at least one of the four quadratic polynomials does not vanish at $p$. Note that the four quadratic polynomials are permuted up to scaling by the $H_{4}$-action discussed above (well-defined by our assumption $\alpha\beta\gamma\neq 0$), so we may as well assume that $\sum x_{i}^{2}$ does not vanish at $p$. But then, examining the minors $h_{12}$, $h_{13}$ and $h_{23}$, we see that $p$ is one of the finitely many points in

[TABLE]

These points belong to ${\mathfrak{P}}$ so $p\in{\mathfrak{P}}$. However, it is easy to see that every $h_{ij}$ vanishes on ${\mathfrak{P}}$ so $\mathbb{X}_{R}={\mathfrak{P}}$. ∎

5. The algebras $R(a,b,c,d)$ of Cho, Hong, and Lau

In this section $a,b,c$ do not denote square roots of $\alpha,\beta,\gamma$.

5.1. The definition

Let $(a,b,c,d)\in\Bbbk^{4}-\{0\}$ and define $R(a,b,c,d)$, or simply $R$, to be the free algebra $TV=\Bbbk\langle x_{1},x_{2},x_{3},x_{4}\rangle$ modulo the relations

[TABLE]

For example, $R(1,-1,0,0)$ is the commutative polynomial ring on 4 generators.

Since $R(a,b,c,d)$ depends only on $(a,b,c,d)$ as a point in ${\mathbb{P}}^{3}$, the family of algebras $R(a,b,c,d)$ is parametrized by ${\mathbb{P}}^{3}$. Proposition 5.2 concerns those algebras $R(a,b,c,d)$ parametrized by the points on the quadric $ac+bd=0$. That quadric is isomorphic to ${\mathbb{P}}^{1}\times{\mathbb{P}}^{1}$.

The lines

[TABLE]

on that quadric and their open subsets

[TABLE]

play a distinguished role.

5.2.

At [8, Conj. 8.11], Cho, Hong, and Lau conjecture that when $ac+bd=0$, $R$ is isomorphic to a 4-dimensional Sklyanin algebra, i.e., isomorphic to $A(\alpha,\beta,\gamma)$ for some $\alpha,\beta,\gamma\in\Bbbk$ such that $\alpha+\beta+\gamma+\alpha\beta\gamma=0$. Proposition 5.2 shows that $R$ is isomorphic to a 4-dimensional Sklyanin algebra if and only if $(a,b,c,d)\in\ell_{1}\cup\ell_{2}$. Nevertheless, $R$ is always isomorphic to $A(\alpha,\beta,\gamma)$ for some $\alpha,\beta,\gamma$.

Proposition 5.1**.**

Let $z_{0}=\frac{1}{2}(x_{2}+x_{4})$, $z_{1}=\frac{1}{2}(x_{1}+x_{3})$, $z_{2}=\frac{1}{2}(x_{1}-x_{3})$, and $z_{3}=\frac{1}{2}(x_{2}-x_{4})$. The algebra $R(a,b,c,d)$ is equal to $\Bbbk\langle z_{0},z_{1},z_{2},z_{3}\rangle$ modulo the relations

[TABLE]

Proof.

Since $x_{1}=z_{1}+z_{2}$, $x_{2}=z_{0}+z_{3}$, $x_{3}=z_{1}-z_{2}$, and $x_{4}=z_{0}-z_{3}$, the relations (R1)–(R4) can be replaced by the four relations

[TABLE]

[TABLE]

It follows that $R$ is equal to $\Bbbk\langle z_{0},z_{1},z_{2},z_{3}\rangle$ modulo the relations

[TABLE]

Rearranging these gives the presentation in the statement of the proposition. ∎

Proposition 5.2**.**

Let $a,b,c,d\in\Bbbk$ and define $p:=a+b$, $q:=a-b$, $r:=c+d$, and $s:=c-d$. Suppose that $ac+bd=0$ and

[TABLE]

1. (1)

$R(a,b,c,d)\cong A(\alpha,\beta,\gamma)$* where*

[TABLE] 2. (2)

$\alpha+\beta+\gamma+\alpha\beta\gamma=0$* if and only if $(a,b,c,d)\in\ell_{1}\cup\ell_{2}$.* 3. (3)

If $(a,b,c,d)\in\ell_{1}\cup\ell_{2}$, then $R(a,b,c,d)$ is isomorphic to the Sklyanin algebra $A(\alpha,1,-1)$ where

[TABLE]

and is generated by homogeneous degree-one elements $Y_{\pm},K,K^{\prime}$ such that

[TABLE]

Proof.

(1) Condition 5-1 ensures that the denominators in the expressions for $\alpha,\beta,\gamma,$ are non-zero.

Let $z_{0},z_{1},z_{2},z_{3}$ be as in Proposition 5.1. Condition 5-1 ensures that the denominators in the following expressions are non-zero, so $R$ is defined by the following relations:

[TABLE]

For brevity, let’s write these relations as

[TABLE]

Define $v_{0}:=z_{0}$, $v_{1}:=\sqrt{\nu_{2}\nu_{3}}\,z_{1}$, $v_{2}:=\sqrt{\nu_{3}\nu_{1}}\,z_{2}$, and $v_{3}:=\sqrt{\nu_{1}\nu_{2}}\,z_{3}$. Thus, $R$ is the free algebra $\Bbbk\langle v_{0},v_{1},v_{2},v_{3}\rangle$ modulo the relations

[TABLE]

where

[TABLE]

(2) The expression (q^{2}-s^{2})(r^{2}-q^{2})ab\big{(}\alpha+\beta+\gamma+\alpha\beta\gamma) is equal to

[TABLE]

But $p^{2}+q^{2}=2(a^{2}+b^{2})$ and $r^{2}+s^{2}=2(c^{2}+d^{2})$ so

[TABLE]

Therefore (q^{2}-s^{2})(r^{2}-q^{2})ab\big{(}\alpha+\beta+\gamma+\alpha\beta\gamma)=2(ab+cd)(p^{2}q^{2}-r^{2}s^{2}). Hence $\alpha+\beta+\gamma+\alpha\beta\gamma=0$ if and only if $(ab+cd)(p^{2}q^{2}-r^{2}s^{2})=0$; i.e., if and only if $(a^{2}-b^{2}-c^{2}+d^{2})(a^{2}-b^{2}+c^{2}-d^{2})(ab+cd)=0$.

The locus $ac+bd=(a^{2}-b^{2}-c^{2}+d^{2})(a^{2}-b^{2}+c^{2}-d^{2})(ab+cd)=0$ consists of 8 lines: the locus $ac+bd=a^{2}-b^{2}-c^{2}+d^{2}=0$ is the union of the four lines

[TABLE]

the locus $ac+bd=a^{2}-b^{2}+c^{2}-d^{2}=0$ is the union of the four lines

[TABLE]

the locus $ac+bd=ab+cd=0$ is the union of the four lines

[TABLE]

If $(a,b,c,d)$ is on the line $a-b=c+d=0$ or on the line $a-d=b+c=0$, then $0=a-b-c-d=q-r$; hypothesis 5-1 excludes this possibility so $(a,b,c,d)$ is not on either of those lines. If $(a,b,c,d)$ is on the line $a=d=0$ or on the line $b=c=0$, then $abcd=0$; hypothesis 5-1 excludes this possibility so $(a,b,c,d)$ is not on either of those lines. If $(a,b,c,d)$ is on the line $a+d=b-c=0$ or on the line $a+b=c-d=0$, then $0=a+b-c+d=p-s$; hypothesis 5-1 excludes this possibility so $(a,b,c,d)$ is not on either of those lines. Thus, $\alpha+\beta+\gamma+\alpha\beta\gamma=0$ if and only if $(a,b,c,d)$ is on the union of the lines $a+id=c+ib=0$ and $a-id=c-ib=0$; i.e., $(a,b,c,d)\in\ell_{1}\cup\ell_{2}$.

Not every $(a,b,c,d)\in\ell_{1}\cup\ell_{2}$ satisfies 5-1. The points $(a,b,c,d)$ on the line $a+id=c+ib=0$ that do not satisfy 5-1 are

[TABLE]

The points $(a,b,c,d)$ on the line $a-id=c-ib=0$ that do not satisfy 5-1 are

[TABLE]

(3) Suppose $a+id=c+ib=0$. Then $(\alpha,\beta,\gamma)=(-1,\beta,1)$ where $\beta=(b+d+ib-id)^{2}/(b+d-ib+id)^{2}$. As $(a,b,c,d)$ runs over the points in 5-2, $\beta$ takes on the values $-1$, $-1$, $1$, $1$, [math], $\infty$, respectively. As $(a,b,c,d)$ runs over the other points on the line $a+id=c+ib=0$, $\beta$ takes on every value in $\Bbbk-\{0,\pm 1\}$.

Suppose $a-id=c-ib=0$. Then $(\alpha,\beta,\gamma)=(-1,\beta,1)$ where $\beta=(b+d-ib+id)^{2}/(b+d+ib-id)^{2}$. As $(a,b,c,d)$ runs over the points in 5-3, $\beta$ takes on the values $-1$, $-1$, $1$, $1$, $\infty$, [math]. As $(a,b,c,d)$ runs over the other points on the line $a-id=c-ib=0$, $\beta$ takes on every value in $\Bbbk-\{0,\pm 1\}$.

By Lemma 2.3, $A(\beta,1,-1)\cong A(-1,\beta,1)\cong A(1,-1,\beta)$ so to prove (3) it suffices to show that $A(\alpha,1,-1)$ has generators $Y_{\pm},K,K^{\prime}$ satisfying the stated relations. We do this in Proposition 5.3 below. ∎

Proposition 5.3**.**

Suppose $\alpha\in\Bbbk-\{0,\pm 1\}$. Let $i$ be a square root of $-1$.

1. (1)

$A(\alpha,1,-1)$* is a Sklyanin algebra $A(E,\tau)$ with $\tau$ being translation by a 4-torsion point.* 2. (2)

There is a basis $Y_{\pm},K,K^{\prime}$ for $A(\alpha,1,-1)_{1}$ such that

[TABLE]

[TABLE]

Proof.

Let $A=A(\alpha,1,-1)$. By Lemma 2.3, $A(\alpha,1,-1)\cong A(1,-1,\alpha)\cong A(-1,\alpha,1)$. Algebras of the form $A(1,-1,\alpha)$ are those identified in equation (1.9.4) of [21] so the results in [21] apply to $A$. The zero locus of the $4\times 4$ minors in the proof of [21, Prop. 2.4] is the curve $E$ given by the equations

[TABLE]

The restrictions on $\alpha$ imply that the Jacobian matrix has rank 2 at all points of $E$ so $E$ is an elliptic curve. (The description of $E$, in particular, the formula for the polynomial $g_{2}$, in [21, Prop. 2.4] does not make sense when $\beta=1$.) The formula for the automorphism $\sigma:E\to E$ in [21, Cor. 2.8] is

[TABLE]

where the last equality uses the fact that $x_{0}^{2}-x_{1}^{2}+\alpha x_{2}^{2}-\alpha x_{3}^{2}=0$ on $E$. The formula for $\sigma$ can also be verified by observing that

[TABLE]

for all $(x_{0},x_{1},x_{2},x_{3})\in E$; the $6\times 4$ matrix in the previous equation is the $6\times 4$ matrix in 5.2

Corollary 2.11 in [7] involves elements $a,b,c\in\Bbbk$ such that $a^{2}=\alpha$, $b^{2}=1$, and $c^{2}=-1$; let $(a,b,c)=(a,1,i)$; [7, Cor. 2.11] then says there is a 4-torsion point $\varepsilon_{1}\in E$ such that if $p=(x_{0},x_{1},x_{2},x_{3})\in E$, then

[TABLE]

by [7, §2.6], there is a 2-torsion point $\gamma_{2}\in E$ such that $p+\varepsilon_{1}+\gamma_{2}\;=\;(x_{1},x_{0},x_{3},-x_{2})$; thus

[TABLE]

Calculations like those in [21, §1.2] show that $Y_{\pm}:=x_{0}\pm x_{1}$, $K:=x_{0}+x_{1}$, $K^{\prime}:=x_{0}-x_{1}$, satisfy the relations in (2). ∎

Part (2) of Proposition 5.3 remains true when $A=A(0,1,-1)$ and in that case, $A(0,1,-1)$ is a homogenization of the quantized enveloping algebra $U_{q}({\mathfrak{s}}{\mathfrak{l}}_{2})$ with $q=-i$. See [6, §2.4] for details.

5.3. Parameter spaces and modular curves

After Proposition 5.3(1), it is natural to ask which pairs $(E,\tau)$ have the property that $A(E,\tau)\cong R(a,b,c,d)$ for some point $(a,b,c,d)$ in the “Sklyanin locus” $\ell_{1}^{\circ}\cup\ell_{2}^{\circ}$. Similarly, we can ask how much redundancy there is in this parametrization: how many $(a,b,c,d)\in\ell_{1}^{\circ}\cup\ell_{2}^{\circ}$ lead to the same pair $(E,\tau)$?

Since the transformation

[TABLE]

interchanges $\ell_{1}$ and $\ell_{2}$ and intertwines the respective transformations

[TABLE]

it suffices to consider what happens for $\ell_{1}$.

Note first that the map

[TABLE]

recovering the parameter $\alpha$ of the Sklyanin algebra $A(\alpha,1,-1)$ from $(a,b,c,d)\in\ell_{1}^{\circ}$ is a two-fold cover of

[TABLE]

Now, to each $\alpha\in X$ associate the elliptic curve $E_{\alpha}$ of point modules of $A(\alpha,1,-1)$, defined by 5-4. Since furthermore the point $(1,1,i,i)$ belongs to all $E_{\alpha}$, the $\alpha$-indexed family $E\to X$ of elliptic curves $E_{\alpha}$ over $X$ has a section.

Finally, 5-5 defines an automorphism of order $4$ of the family $E\to X$. Since the section $(1,1,i,i)$ puts on $E$ a unique structure of an abelian curve over $X$ [13, Theorem 2.1.2], we can identify said automorphism with a point of $E$ of order (precisely) $4$. In other words, we obtain a family of abelian curves over $X$ with marked order-$4$ points. This moduli problem is represented by the modular curve $Y_{1}(4)$ classifying such data (see e.g. [9, Theorem 8.2.1] and references therein), and hence we obtain a morphism

[TABLE]

The following results give the full picture of the parametrization of the Cho-Hong-Lau algebras.

Proposition 5.4**.**

The map $\psi:X=\mathbb{P}^{1}-\{\pm 1,0,\infty\}\longrightarrow Y_{1}(4)$ defined above as

[TABLE]

is a two-fold cover, identifying $\pm\alpha$.

Proof.

Note first that the automorphism

[TABLE]

of $\mathbb{P}^{3}$ interchanges the elliptic curves $E_{\alpha}$ and $E_{-\alpha}$, and moreover it intertwines their respective order-$4$ automorphisms defining them as points of $Y_{1}(4)$. This implies that $\psi$ factors through a morphism

[TABLE]

Since $Y_{1}(4)$ is known to have three cusps and the left-hand side is a thrice-punctured projective line, $\psi^{\prime}$ extends to a endomorphism $\overline{\psi^{\prime}}$ of $\mathbb{P}^{1}$. It follows from the fact that three distinct points have singleton preimages that $\overline{\psi^{\prime}}$ is an isomorphism, and hence so is $\psi^{\prime}$. ∎

In conclusion, we have

Corollary 5.5**.**

The maps $\ell^{\circ}_{i}\longrightarrow Y_{1}(4)$, $i=1,2$, that send $(a,b,c,d)$ to the underlying elliptic curve and automorphism of the Sklyanin algebra $R(a,b,c,d)$ are fourfold covers.

Proof.

Simply compose $\psi$ and its analogue for $\ell_{2}$ (which are double covers by Proposition 5.4) with the two-fold covers of the form 5-6. ∎

6. Central elements

6.1. Central elements in $A(\alpha,\beta,\gamma)$

The next result is often asserted but we could not find a proof in the literature so we include one here.

Proposition 6.1**.**

Let $\Bbbk$ be any field. If $\{\alpha,\beta,\gamma\}\cap\{0,\pm 1\}=\varnothing$ and $\alpha+\beta+\gamma+\alpha\beta\gamma=0$, then $-x_{0}^{2}+x_{1}^{2}+x_{2}^{2}+x_{3}^{2}$ and $x_{0}^{2}+\beta\gamma x_{1}^{2}-\gamma x_{2}^{2}+\beta x_{3}^{2}$ belong to the center of $A(\alpha,\beta,\gamma)$.

Proof.

Let’s simplify the notation by omitting the $x$’s and just retaining the subscripts, so $kji$ denotes $x_{k}x_{j}x_{i}$, $ii0$ denotes $x_{i}x_{i}x_{0}$, and so on. We also write $\{i,j\}$ for $\{x_{i},x_{j}\}$, $[i,j]$ for $[x_{i},x_{j}]$, etc.

For each cyclic permutation $(i,j,k)$ of $(1,2,3)$, we define

[TABLE]

Straightforward computations in the free algebra $\Bbbk\langle x_{0},x_{1},x_{2},x_{3}\rangle$ show that

[TABLE]

When $\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{1}\alpha_{2}\alpha_{3}=0$, error-prone calculations333In carrying out these calculations one should not attempt to “simplify” the expressions $ijk+kji$ and $ikj+jki$. show that

[TABLE]

equals $[x_{0},x_{1}^{2}+x_{2}^{2}+x_{3}^{2}]$. Hence $[x_{0},-x_{0}^{2}+x_{1}^{2}+x_{2}^{2}+x_{3}^{2}]=0$ in $A(\alpha_{1},\alpha_{2}.a_{3})$.

A similar calculation shows that

[TABLE]

Hence $[x_{1},-x_{0}^{2}+x_{1}^{2}+x_{2}^{2}+x_{3}^{2}]=0$. The transformation $x_{0}\mapsto x_{0}$, $x_{i}\mapsto x_{i+1}$ for $i=1,2,3$, and $\alpha_{i}\mapsto\alpha_{i+1}$ for $i=1,2,3$, leaves $-x_{0}^{2}+x_{1}^{2}+x_{2}^{2}+x_{3}^{2}$ fixed; it follows that $[x_{2},-x_{0}^{2}+x_{1}^{2}+x_{2}^{2}+x_{3}^{2}]=0$ and then that $[x_{3},-x_{0}^{2}+x_{1}^{2}+x_{2}^{2}+x_{3}^{2}]=0$. This completes the proof that $-x_{0}^{2}+x_{1}^{2}+x_{2}^{2}+x_{3}^{2}$ belongs to the center of $A(\alpha_{1},\alpha_{2},\alpha_{3})$ when $\{\alpha_{1},\alpha_{2},\alpha_{3}\}\cap\{0,\pm 1\}=\varnothing$ and $\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{1}\alpha_{2}\alpha_{3}=0$.

The automorphism $\psi_{1}$ in Table 1 sends $-x_{0}^{2}+x_{1}^{2}+x_{2}^{2}+x_{3}^{2}$ to $x_{0}^{2}+\alpha_{2}\alpha_{3}x_{1}^{2}-\alpha_{3}x_{2}^{2}+\alpha_{2}x_{3}^{2}$ so the latter also belongs to the center of $A(\alpha_{1},\alpha_{2},\alpha_{3})$. ∎

Proposition 6.2**.**

Let $\Bbbk$ be any field. If $\alpha\beta\gamma\neq 0$ and $\alpha+\beta+\gamma+\alpha\beta\gamma\neq 0$, then $x_{0}^{2}$, $x_{1}^{2}$, $x_{2}^{2}$, and $x_{3}^{2}$, belong to the center of $A(\alpha,\beta,\gamma)$.

Proof.

We use the same notation as that in the proof of Proposition 6.1. Calculations show that if $(i,j,k)$ is a cyclic permutation of $(1,2,3)$, then

[TABLE]

and

[TABLE]

and

[TABLE]

and

[TABLE]

Since the images of $a_{i}$ and $c_{i}$ in $A(\alpha_{1},\alpha_{2},\alpha_{3})$ are zero, if $\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{1}\alpha_{2}\alpha_{3}\neq 0$, then $[x_{0},x_{1}^{2}]=[x_{0},x_{2}^{2}]=[x_{0},x_{3}^{2}]=0$ and $[x_{1},x_{3}^{2}]=[x_{2},x_{1}^{2}]=[x_{3},x_{2}^{2}]=0$ and $[x_{1},x_{2}^{2}]=[x_{2},x_{3}^{2}]=[x_{3},x_{1}^{2}]=0$ in $A(\alpha_{1},\alpha_{2},\alpha_{3})$. ∎

6.2. Degree-two central elements in $R(a,b,c,d)$

It is conjectured at [8, p. 47] that the elements

[TABLE]

and

[TABLE]

generate the center of $R(a,b,c,d)$ (we have suppressed an irrelevant scaling constant from the original expression of $C_{2}$ in [8]). In order to have a little more symmetry, and to emphasize the parallels with the Sklyanin algebras, we will replace $C_{1}$ by

[TABLE]

which is equal to $2C_{1}$, and replace $C_{2}$ by the element $Z_{2}$ in Corollary 6.5 below, and show that $Z_{1}$ and $Z_{2}$ belong to the center of $R(a,b,c,d)$.

Proposition 6.3**.**

For all $a,b,c,d\in\Bbbk$, the element $Z_{1}$ is central in $R(a,b,c,d)$.

Proof.

In terms of the generators $z_{i}$ in Proposition 5.1,

[TABLE]

Using the expression for $Z_{1}$ in 6-1, we get

[TABLE]

We now label the relations in the statement of Proposition 5.1 (in the form LHS $-$ RHS) according to which commutator or anticommutator involving $z_{0}$ they contain. For example, the first and third relations in Proposition 5.1 are

[TABLE]

and

[TABLE]

With this in place, we leave the reader to check that 6-2 equals

[TABLE]

which obviously belongs to the ideal generated by the relations $c_{i}$ and $a_{i}$. Thus $[z_{0},Z_{1}]=0$.

We now prove that $[Z_{1},z_{i}]=0$ for $i=1,2,3$ by changing the labels of the $z_{i}$ and the structure constants $a$, $b$, etc. so that both $Z_{1}$ and the space of relations in Proposition 5.1 are preserved. The transformation

[TABLE]

is such a relabeling so the fact that $[Z_{1},z_{0}]=0$ implies $[Z_{1},z_{1}]=0$. The transformation

[TABLE]

(while $c$ and $d$ are fixed) is another such transformation, so the fact that $[Z_{1},z_{0}]=0$ implies $[Z_{1},z_{3}]=0$. Finally, composing the two transformations will prove that $[Z_{1},z_{2}]=0$. ∎

Proposition 6.4**.**

Let

[TABLE]

Assume that the denominators in the expressions for $\rho_{2}$ and $\rho_{3}$ are non-zero. Fix $q_{2}$ and $q_{3}$ such that $q_{2}^{4}=\rho_{2}$ and $q_{3}^{4}=\rho_{3}$, and define

[TABLE]

The linear map $\psi:R_{1}\to R_{1}$ given by the formula

[TABLE]

extends to an algebra automorphism of $R(a,b,c,d)$.

Proof.

By Proposition 5.1, $R$ is $\Bbbk\langle z_{0},z_{1},z_{2},z_{3}\rangle$ modulo the relations

[TABLE]

where $(i,j,k)$ runs over the cyclic permutation of $(1,2,3)$ and

[TABLE]

Furthermore,

[TABLE]

Since

[TABLE]

we have

[TABLE]

Hence $\psi$ extends to an algebra automorphism, as claimed.

Since $\psi^{2}(z_{0})=\tau_{0}\tau_{1}z_{0}=-z_{0}$, $\psi^{2}\neq\mathrm{id}_{R}$. Since $(\tau_{0}\tau_{1})^{2}=(\tau_{2}\tau_{3})^{2}=1$, $\psi^{4}=\mathrm{id}_{R}$. ∎

Corollary 6.5**.**

With the notation and hypotheses in Proposition 6.4, The element

[TABLE]

belongs to the center of $R$.

Proof.

Let $\psi$ be the automorphism in Proposition 6.4. Since Z_{2}=\psi\big{(}\frac{1}{2}Z_{1}\big{)}, the result follows from Proposition 6.3. ∎