# Sklyanin algebras and a cubic root of 1

Natalia Iyudu and Stanislav Shkarin

###### Abstract

We consider Sklyanin algebras $S$ with 3 generators, which are quadratic
algebras over a field ${\mathbb{K}}$ with $3$ generators $x,y,z$ given by $3$
relations $pxy+qyx+rzz=0$, $pyz+qzy+rxx=0$ and $pzx+qxz+ryy=0$, where
$p,q,r\in{\mathbb{K}}$. This class of algebras has enjoyed much attention. In
particular, using tools from algebraic geometry Artin, Tate and Van Den Berg
[4] showed that if at least two of the parameters $p$, $q$ and $r$ are non-
zero and at least two of three numbers $p^{3}$, $q^{3}$ and $r^{3}$ are
distinct, then $S$ is Artin–Schelter regular. More specifically, $S$ is Koszul
and has the same Hilbert series as the algebra of commutative polynomials in 3
indeterminates. It has became commonly accepted that it is impossible to
achieve the same objective by purely algebraic and combinatorial means like
the Gröbner basis technique. The authors have previously dispelled this
belief. However our previous proof was no less complicated than the one based
on algebraic geometry. It used a construcion of a Gröbner basis in a suitable
one-sided module over $S$ and had quite a number of cases to consider. In this
paper we exhibit a linear substitution after which it becomes possible to
determine the leading monomials of a reduced Gröbner basis for the ideal of
relations of $S$ itself (without passing to a module). We also find out
explicitly (in terms of parameters) which Sklyanin algebras are isomorphic.
The only drawback of the new technique is that it fails if the characteristic
of the ground field equals 3.

MSC: 17A45, 16A22

Keywords: Sklyanin algebras, Quadratic algebras, Koszul algebras, Hilbert
series, Gröbner bases, PBW-algebras, PHS-algebras

##  1 Introduction

Throughout this paper ${\mathbb{K}}$ is an arbitrary field of characteristic
different from $3$. If $B$ is a graded algebra, the symbol $B_{m}$ stands for
the $m^{\rm th}$ graded component of the algebra $B$. If $V$ is an
$n$-dimensional vector space over ${\mathbb{K}}$, then $F=F(V)$ is the tensor
algebra of $V$. For any choice of a basis $x_{1},\dots,x_{n}$ in $V$, $F$ is
naturally identified with the free ${\mathbb{K}}$-algebra with the generators
$x_{1},\dots,x_{n}$. For subsets $P_{1},\dots,P_{k}$ of an algebra $B$,
$P_{1}\dots P_{k}$ stands for the linear span of all products $p_{1}\dots
p_{k}$ with $p_{j}\in P_{j}$. We consider a degree grading on the free algebra
$F$: the $m^{\rm th}$ graded component of $F$ is $V^{m}$. If $R$ is a subspace
of the $n^{2}$-dimensional space $V\otimes V$, then the quotient of $F$ by the
ideal $I$ generated by $R$ is called a quadratic algebra and denoted $A(V,R)$.
For any choice of bases $x_{1},\dots,x_{n}$ in $V$ and $g_{1},\dots,g_{k}$ in
$R$, $A(V,R)$ is the algebra given by generators $x_{1},\dots,x_{n}$ and the
relations $g_{1},\dots,g_{k}$ ($g_{j}$ are linear combinations of monomials
$x_{i}x_{s}$ for $1\leqslant i,s\leqslant n$). Since each quadratic algebra
$A$ is degree graded, we can consider its Hilbert series

$H_{A}(t)=\sum_{j=0}^{\infty}{\rm dim}_{{\mathbb{K}}}A_{j}\,\,t^{j}.$

Quadratic algebras whose Hilbert series is the same as for the algebra
${\mathbb{K}}[x_{1},\dots,x_{n}]$ of commutative polynomials play a
particularly important role in physics. We say that $A$ is a PHS-algebra (for
polynomial Hilbert series) if

$H_{A}(t)=H_{{\mathbb{K}}[x_{1},\dots,x_{n}]}(t)=(1-t)^{-n}.$

Following the notation from the Polishchuk, Positselski book [19], we say that
a quadratic algebra $A=A(V,R)$ is a PBW-algebra (Poincare, Birkhoff, Witt) if
there are linear bases $x_{1},\dots,x_{n}$ and $g_{1},\dots,g_{m}$ in $V$ and
$R$ respectively such that with respect to some compatible with multiplication
well-ordering on the monomials in $x_{1},\dots,x_{n}$, $g_{1},\dots,g_{m}$ is
a (non-commutative) Gröbner basis of the ideal $I_{A}$ generated by $R$. In
this case, $x_{1},\dots,x_{n}$ is called a PBW-basis of $A$, while
$g_{1},\dots,g_{m}$ are called the PBW-generators of $I_{A}$. In order to
avoid confusion, we would like to stress from the start that Odesskii [17] as
well as some other authors use the term PBW-algebra for what we have already
dubbed PHS. Since we deal with both concepts, we could not possibly call them
the same and we opted to follow the notation from [19].

Another concept playing an important role in this paper is Koszulity. For a
quadratic algebra $A=A(V,R)$, the augmentation map $A\to{\mathbb{K}}$ equips
${\mathbb{K}}$ with the structure of a commutative graded $A$-bimodule. The
algebra $A$ is called Koszul if ${\mathbb{K}}$ as a graded right $A$-module
has a free resolution $\dots\to M_{m}\to\dots\to M_{1}\to A\to{\mathbb{K}}\to
0$ with the second last arrow being the augmentation map and with each $M_{m}$
generated in degree $m$. The last property is the same as the condition that
the matrices of the above maps $M_{m}\to M_{m-1}$ with respect to some free
bases consist of elements of $V$ (=are homogeneous of degree $1$).

The notion of a noncommutative potential was introduced in [15]. We make use
of an equivalent definition from [5]. An element $F$ of ${\mathbb{K}}\langle
x_{1},\dots,x_{n}\rangle$ is called cyclicly invariant if $F$ is invariant for
the linear map $C:{\mathbb{K}}\langle
x_{1},\dots,x_{n}\rangle\to{\mathbb{K}}\langle x_{1},\dots,x_{n}\rangle$
defined by its action on monomials as follows: $C(1)=1$ and $C(x_{j}u)=ux_{j}$
for every $j$ and every monomial $u$. The symbol ${\mathbb{K}}^{\rm
cyc}\langle x_{1},\dots,x_{n}\rangle$ stands for the vector space of all
cyclicly invariant elements of ${\mathbb{K}}\langle x_{1},\dots,x_{n}\rangle$.
We also consider the linear maps $\frac{\delta}{\delta
x_{j}}:{\mathbb{K}}\langle x_{1},\dots,x_{n}\rangle\to{\mathbb{K}}\langle
x_{1},\dots,x_{n}\rangle$ defined by their action on monomials $u$ as follows:
$\frac{\delta u}{\delta x_{j}}=0$ if $u$ does not start with $x_{j}$ and
$\frac{\delta u}{\delta x_{j}}=v$ if $u=x_{j}v$. A potential algebra $A_{F}$
defined by a potential $F\in{\mathbb{K}}^{\rm cyc}\langle
x_{1},\dots,x_{n}\rangle$ is a ${\mathbb{K}}$-algebra given by the generators
$x_{1},\dots,x_{n}$ and the relations $\frac{\delta F}{\delta x_{j}}=0$ for
$1\leqslant j\leqslant n$. For the sake of convenience, we consider the onto
linear map $G\mapsto G^{\rcirclearrowleft}$ from ${\mathbb{K}}\langle
x_{1},\dots,x_{n}\rangle$ onto ${\mathbb{K}}^{\rm cyc}\langle
x_{1},\dots,x_{n}\rangle$ defined by its action on homogeneous elements by
$u^{\rcirclearrowleft}=C(u)+{\dots}+C^{d}u$, where $d$ is the degree of $u$.
Foe example, ${x^{4}}^{\rcirclearrowleft}=4x^{4}$ and
${x^{2}y}^{\rcirclearrowleft}=x^{2}y+xyx+yx^{2}$.

Prime examples of a potential algebras are Sklyanin algebras. Recall that if
$(p,q,r)\in{\mathbb{K}}^{3}$, the Sklyanin algebra $Q^{p,q,r}$ is the
quadratic algebra over ${\mathbb{K}}$ with generators $x,y,z$ given by $3$
relations

$\text{$pyz+qzy+rxx=0$},\quad\text{$pzx+qxz+ryy=0$},\quad\text{$pxy+qyx+rzz=0$}.$

It is easy to observe that $Q^{p,q,r}$ is a potential algebra given by the
potential
$r(x^{3}+y^{3}+z^{3})+pxyz^{\rcirclearrowleft}+qxzy^{\rcirclearrowleft}$.

It is worth mentioning that our use of potentiality of Sklyanin algebras is
just a matter of convenience. Namely, it makes the job of parforming a linear
substitution much easier. Instead of making the sub in each of the three
relations, one can perform it with the potential and then compute the
’derivatives’ $\frac{\delta}{\delta x_{j}}$ yielding the same result.

Odesskii [17] proved that in the case ${\mathbb{K}}={\mathbb{C}}$, a generic
Sklyanin algebra is a PHS-algebra. That is,

$\textstyle
H_{Q^{p,q,r}}(t)=\sum\limits_{j=0}^{\infty}\frac{(j+2)(j+1)}{2}\,t^{j}\ \
\text{for generic $(p,q,r)\in{\mathbb{C}}^{3}$,}$

where generic means outside the union of countably many algebraic varieties in
${\mathbb{C}}^{3}$ (different from ${\mathbb{C}}^{3}$). In particular, the
equality above holds for almost all $(p,q,r)\in{\mathbb{C}}^{3}$ with respect
to the Lebesgue measure. Polishchuk and Positselski [19] showed in the same
setting and with the same meaning of the word ’generic’, that for generic
$(p,q,r)\in{\mathbb{C}}^{2}$, the algebra $Q^{p,q,r}$ is Koszul but is not a
PBW-algebra. The same results are implicitly contained in the Artin, Shelter
paper [2].

Artin, Tate and Van den Berg [4, 3], and Feigin, Odesskii [18], considered
certain family of infinite dimensional representations of Sklyanin algebra.
Namely, they used representations, where variables are represented by infinite
matrices with one nonzero upper diagonal. Equivalently, they considered graded
modules with all graded components being one-dimensional. The geometric
interpretation of the space of such modules is in the core for most of their
arguments. Artin, Tate and Van den Berg showed that if at least two of the
parameters $p$, $q$ and $r$ are non-zero and the equality $p^{3}=q^{3}=r^{3}$
fails, then $Q^{p,q,r}$ is Artin–Shelter regular. More specifically,
$Q^{p,q,r}$ is Koszul and has the same Hilbert series as the algebra of
commutative polynomials in three variables.

It became commonly accepted that it is impossible to obtain the same results
by purely algebraic and combinatorial means like the Gröbner basis technique,
see, for instance, comments in [17, 26]. We have dispelled this notion in
[13]. However our previous proof is rather complicated. It uses a construcion
of a Gröbner basis in a suitable one-sided module over $Q^{p,q,r}$ and has
quite a number of cases to consider. In this paper we exhibit a linear
substitution after which it becomes possible to determine the leading
monomials of a reduced Gröbner basis for the ideal of relations of $S$ itself
(without passing to a module). We also find out explicitly (in terms of
parameters) which Sklyanin algebras are isomorphic. The only drawback of the
new technique is that it fails if the characteristic of the ground field
equals 3.

######  Theorem 1.1.

The algebra $Q^{p,q,r}$ is Koszul for any $(p,q,r)\in{\mathbb{K}}^{3}$. The
algebra $Q^{p,q,r}$ is PHS if and only if at least two of $p$, $q$ and $r$ are
non-zero and the equality $p^{3}=q^{3}=r^{3}$ fails.

We stress again that the above theorem is essentially one of the main results
in [4] with a different proof provided in [13]. Our new proof here is however
much easier and shorter, since we found clever change of variables which makes
the Gröbner basis regular (which is a generalisation of finiteness).

##  2 General background

We shall use the following well-known facts, all of which can be found in
[19]. Every monomial quadratic algebra $A=A(V,R)$ (=there are linear bases
$x_{1},\dots,x_{n}$ and $g_{1},\dots,g_{m}$ in $V$ and $R$ respectively, such
that each $g_{j}$ is a monomial in $x_{1},\dots,x_{n}$) is a PBW-algebra.
Next, if we pick a basis $x_{1},\dots,x_{n}$ in $V$, we get a bilinear form
$b$ on the free algebra $F=F(V)$ defined by $b(u,v)=\delta_{u,v}$ for every
monomials $u$ and $v$ in the variables $x_{1},\dots,x_{n}$. The algebra
$A^{!}=A(V,R^{\perp})$, where $R^{\perp}=\\{u\in V^{2}:b(r,u)=0\ \text{for
each}\ r\in R\\}$, is called the dual algebra of $A$. Clearly, $A^{!}$ is a
quadratic algebra in its own right. Recall also that there is a specific
complex of free right $A$-modules, called the Koszul complex, whose exactness
is equivalent to the Koszulity of $A$:

$\dots\mathop{\longrightarrow}^{d_{k+1}}(A^{!}_{k})^{*}\otimes
A\mathop{\longrightarrow}^{d_{k}}(A^{!}_{k-1})^{*}\otimes
A\mathop{\longrightarrow}^{d_{k-1}}\dots\mathop{\longrightarrow}^{d_{1}}(A^{!}_{0})^{*}\otimes
A=A\longrightarrow{\mathbb{K}}\to 0,$ (2.1)

where the tensor products are over ${\mathbb{K}}$, the second last arrow is
the augmentation map, each tensor product carries the natural structure of a
free right $A$-module and $d_{k}$ are given by $d_{k}(\varphi\otimes
u)=\sum\limits_{j=1}^{n}\varphi_{j}\otimes x_{j}u$, where
$\varphi_{j}\in(A^{!}_{k-1})^{*}$, $\varphi_{j}(v)=\varphi(x_{j}v)$. Although
$A^{!}$ and the Koszul complex seem to depend on the choice of a basis in $V$,
it is not really the case up to the natural equivalence [19]. We recall that

every PBW-algebra is Koszul; $\displaystyle\text{$A$ is Koszul $\iff$ $A^{!}$
is Koszul};$ $\displaystyle\text{if $A$ is Koszul, then
$H_{A}(-t)H_{A^{!}}(t)=1$}.$ (2.2)

Note that the Koszul complex (2.1) of any quadratic algebra is exact at its
last $3$ terms: ${\mathbb{K}}$, $(A^{!}_{0})^{*}\otimes A=A$ and
$(A^{!}_{1})^{*}\otimes A$. The following lemma proved in [13] allows us to
prove Koszulity of Sklyain algebras once we have computed their Hilbert
series. We say that $u\in A=A(V,R)$ is a right annihilator if $Vu=\\{0\\}$ in
$A$. A right annihilator $u$ is non-trivial if $u\neq 0$.

######  Lemma 2.1.

Let $A=A(V,R)$ be a quadratic algebra such that $A^{!}_{4}=\\{0\\}$,
$A^{!}_{3}$ is one-dimensional and $wA_{2}^{!}\neq\\{0\\}$ for every non-zero
$w\in A_{1}^{!}$. Then the following statements are equivalent$:$

  * (2.1.1)

$A$ is Koszul;

  * (2.1.2)

$A$ has no non-trivial right annihilators and $H_{A}(-t)H_{A^{!}}(t)=1$.

Our next observation is that neither Koszulity nor the Hilbert series of a
quadratic algebra $A=A(V,R)$ is sensitive to changing the ground field.

######  Remark 2.2.

Fix the bases $x_{1},\dots,x_{n}$ and $r_{1},\dots,r_{m}$ in $V$ and $R$
respectively. Then $A=A(V,R)$ is given by the generators $x_{1},\dots,x_{n}$
and the relations $r_{1},\dots,r_{m}$. Let ${\mathbb{K}}_{0}$ be the subfield
of ${\mathbb{K}}$ generated by the coefficients in the relations
$r_{1},\dots,r_{m}$ and $B$ be the ${\mathbb{K}}_{0}$-algebra defined by the
exact same generators $x_{1},\dots,x_{n}$ and the exact same relations
$r_{1},\dots,r_{m}$. Then $A$ is Koszul if and only if $B$ is Koszul (see, for
instance, [19]) and the Hilbert series of $A$ and of $B$ coincide. The latter
follows from the fact that the Hilbert series depends only on the set of
leading monomials of the Gröbner basis. Now the Gröbner basis construction
algorithm for $A$ and for $B$ produces exactly the same result. Thus if a
quadratic algebra given by generators and relations makes sense over 2 fields
of the same characteristic, then the choice of the field does not effect its
Hilbert series or its Koszulity. In particular, replacing the original field
${\mathbb{K}}$ by its algebraic closure or by an even bigger field does not
change the Hilbert series or Koszulity of $A$. On the other hand, the PBW-
property is sensitive to changing the ground field [19].

##  3 Elementary observations

The following elementary facts are proved in [13]. The next lemma is proved by
a direct computation of the reduced Gröbner basis for the ideal of relations
of the duals of Sklyanin algebras (using the usual left-to-right degree
lexicographical ordering on monomials assuming $x>y>z$. The last statement is
verified directly since knowing the Gröner basis we know the multiplication
table in the relevant finite dimensional (in this case) algebra $A^{!}$.

######  Lemma 3.1.

Let $(p,q,r)\in{\mathbb{K}}^{3}$ and $A=Q^{p,q,r}$. Then the Hilbert series of
$A^{!}$ is given by

$H_{A^{!}}(t)=\left\\{\begin{array}[]{ll}1+3t&\text{if $p=q=r=0$;}\\\
\textstyle\frac{1+2t}{1-t}&\text{if $p^{3}=q^{3}=r^{3}\neq 0$ or exactly two
of $p$, $q$ and $r$ equal $0$;}\\\
(1+t)^{3}&\text{otherwise}.\end{array}\right.$ (3.1)

Moreover, $A_{2}^{!}w\neq\\{0\\}$ for each non-zero $w\in A_{1}^{!}$ provided
$H_{A^{!}}(t)=(1+t)^{3}$.

Note [19] that for every quadratic algebra $A=A(V,R)$ (Koszul or otherwise),
the power series $H_{A}(t)H_{A^{!}}(-t)-1$ starts with $t^{k}$ with
$k\geqslant 4$. This allows to determine ${\rm dim}\,A_{3}$ provided we know
${\rm dim}\,A^{!}_{j}$ for $j\leqslant 3$. Applying this observation together
with (3.1), we immediately obtain the following fact.

######  Corollary 3.2.

Let $(p,q,r)\in{\mathbb{K}}^{3}$ and $A=Q^{p,q,r}$. Then

${\rm dim}\,A_{3}=\left\\{\begin{array}[]{ll}27&\text{if $p=q=r=0;$}\\\
12&\text{if $p^{3}=q^{3}=r^{3}\neq 0$ or exactly two of $p$, $q$ and $r$ equal
$0;$}\\\ 10&\text{otherwise}.\end{array}\right.$ (3.2)

Obviously, multiplying $(p,q,r)\in{\mathbb{K}}^{3}$ by a non-zero scalar does
not change the algebra $Q^{p,q,r}$. It turns out that there are non-
proportional triples of parameters, which lead to isomorphic (as graded
algebras) Sklyanin algebras. The following three lemmas can be found in [13].
Since their proof is very short, we present it for the sake of reader’s
convenience.

######  Lemma 3.3.

Assume that $(p,q,r)\in{\mathbb{K}}^{3}$ and $\theta\in{\mathbb{K}}$ is such
that $\theta^{3}=1$ and $\theta\neq 1$. Then the graded algebras $Q^{p,q,r}$
and $Q^{p,q,\theta r}$ are isomorphic.

###### Proof.

The relations of $Q^{p,q,r}$ in the variables $u$, $v$, $w$ given by $x=u$,
$y=v$ and $z=\theta^{2}w$ read $puv+qvu+\theta rww=0$, $pwu+quw+\theta rvv=0$
and $pvw+qwv+\theta ruu=0$. Thus this change of variables provides an
isomorphism between $Q^{p,q,r}$ and $Q^{p,q,\theta r}$. ∎

######  Lemma 3.4.

Assume that $(p,q,r)\in{\mathbb{K}}^{3}$ and $\theta\in{\mathbb{K}}$ is such
that $\theta^{3}=1$ and $\theta\neq 1$. Then the graded algebras $Q^{p,q,r}$
and $Q^{p^{\prime},q^{\prime},r^{\prime}}$ are isomorphic, where
$p^{\prime}=\theta^{2}p+\theta q+r$, $q^{\prime}=\theta p+\theta^{2}q+r$ and
$r^{\prime}=p+q+r$.

###### Proof.

A direct computation shows that the space of the quadratic relations of
$Q^{p,q,r}$ in the variables $u$, $v$, $w$ given by $x=u+v+w$, $y=u+\theta
v+\theta^{2}w$ and $z=u+\theta^{2}v+\theta w$ (the matrix of this change of
variables is non-degenerate) is spanned by
$p^{\prime}uv+q^{\prime}vu+r^{\prime}ww=0$,
$p^{\prime}wu+q^{\prime}uw+r^{\prime}vv=0$ and
$p^{\prime}vw+q^{\prime}wv+r^{\prime}uu=0$. Thus $Q^{p,q,r}$ and
$Q^{p^{\prime},q^{\prime},r^{\prime}}$ are isomorphic. ∎

######  Lemma 3.5.

The Sklyanin algebra $A=Q^{p,q,r}$ is PBW and therefore is Koszul if $r=0$, or
if $p=q=0$, or if $p^{3}=q^{3}=r^{3}$. Moreover, $H_{A}(t)=(1-3t)^{-1}$ if
$p=q=r=0$, $H_{A}(t)=\frac{1+t}{1-2t}$ if exactly two of $p$, $q$ and $r$ are
$0$ or if $p^{3}=q^{3}=r^{3}\neq 0$ and $H_{A}=(1-t)^{-3}$ if $r=0$ and
$pq\neq 0$. Furthermore, $A$ is Koszul and satisfies $H_{A}=(1-t)^{-3}$ if
$(p+q)^{3}=-r^{3}\neq 0$ and $p^{3}=q^{3}=r^{3}$ fails.

###### Proof.

First, if $p=q=r=0$, then $Q^{p,q,r}$ is the free algebra and therefore
$A=Q^{p,q,r}$ is PBW and therefore Koszul and has the Hilbert series
$H_{A}(t)=(1-3t)^{-1}$. If exactly two of $p$, $q$ and $r$ are $0$, then $A$
is monomial and therefore is PBW and therefore Koszul. In this case
$H_{A}(t)=\frac{1+t}{1-2t}$. If $p^{3}=q^{3}=r^{3}\neq 0$, one easily checks
that the defining relations of $A$ form a Gröbner basis in the ideal they
generate. Hence $A$ is PBW and therefore Koszul. Furthermore, the Hilbert
series of $A$ is the same as for the monomial algebra given by the leading
monomials $xx$, $xy$ and $xz$ of the relations of $A$. It follows that again
$H_{A}(t)=\frac{1+t}{1-2t}$. If $r=0$ and $pq\neq 0$, the defining relations
form a Gröbner basis again. This time the leading monomials are $xy$, $xz$ and
$yz$, the same as for the algebra ${\mathbb{K}}[x,y,z]$ of commutative
polynomials. Hence $A$ is a PBW (and therefore Koszul) PHS-algebra. The latter
means that $H_{A}=(1-t)^{-3}$. As a matter of fact, $A$ in this case is an
algebra of quantum polynomials on three variables. Finally, assume that
$(p+q)^{3}=-r^{3}\neq 0$ and $p^{3}=q^{3}=r^{3}$ fails. By Remark 2.2, by
passing to a field extension, we can, without loss of generality, assume that
${\mathbb{K}}$ is algebraically closed. Since $(p+q)^{3}=-r^{3}\neq 0$, Lemma
3.3 implies that $A=Q^{p,q,r}$ is isomorphic to $Q^{p,q,-p-q}$. Thus without
loss of generality $r=-p-q$. Since ${\rm char}\,{\mathbb{K}}\neq 3$ and
${\mathbb{K}}$ is algebraically closed, we can find $\theta\in{\mathbb{K}}$
such that $\theta^{3}=1\neq\theta$. By Lemma 3.4, $A$ is isomorphic to
$Q^{p^{\prime},q^{\prime},r^{\prime}}$, where $p^{\prime}=\theta^{2}p+\theta
q+r=(\theta^{2}-1)p+(\theta-1)q$, $q^{\prime}=\theta
p+\theta^{2}q+r=(\theta-1)p+(\theta^{2}-1)q$ and $r^{\prime}=p+q+r=0$. If
$p^{\prime}q^{\prime}\neq 0$, we are back to quantum polynomials and the
result follows. Observe that $p^{\prime}+q^{\prime}=-3(p+q)$. Thus the case
$p^{\prime}+q^{\prime}=0$ is impossible since otherwise $p+q=0$, which
conradicts $(p+q)^{3}=-r^{3}\neq 0$. In particular, it is impossible to have
$p^{\prime}=q^{\prime}=0$. Thus it remains to consider the case when exactly
one of $p^{\prime}$ or $q^{\prime}$ is zero. By Corollary 3.2, we then have
${\rm dim}\,A_{3}={\rm dim}\,Q^{p^{\prime},q^{\prime},r^{\prime}}=12$. Since
condition $(p+q)^{3}=-r^{3}\neq 0$ implies that at least two of $p,q,r$ are
non-zero, the same Corollary 3.2 yields $p^{3}=q^{3}=r^{3}\neq 0$, which
contradicts the assumptions. ∎

######  Lemma 3.6.

The Sklyanin algebra $A=Q^{p,q,r}$ is Koszul and satisfies $H_{A}=(1-t)^{-3}$
if $p^{3}=q^{3}\neq 0$.

###### Proof.

If $r=0$, the result follows from Lemma 3.5. Thus we can assume that $r\neq
0$. By normalizing, we can without loss of generality assume $p=1$. Then
$q^{3}=1$ and the defining relations take form
$xx=-\frac{1}{r}yz-\frac{q}{r}zy$, $xy=-qyx-rzz$ and
$xz=-\frac{1}{q}zx-\frac{r}{q}yy$. A direct computation shows that the reduced
Gröber basis of the ideal of relations comprises the defining relations
$rxx+yz+qzy$, $xy+qyx+rzz$, $qxz+ryy+zx$ together with $yyz-q^{2}zyy$ and
$yzz-q^{2}zzy$ (the basis is finite; one has to use the equality $q^{3}=1$).
Now the normal words (the words, which do not contain any of the leading
monomials $xx$, $xy$, $xz$, $yyz$, $yzz$ of the basis as submonomials) are
exactly $z^{k}(yz)^{m}y^{l}x^{\varepsilon}$, where $k,m,l$ are non-negative
integers and $\varepsilon\in\\{0,1\\}$. As there are precisely
$\frac{(n+1)(n+2)}{2}$ normal words of degree $n$, we have $H_{A}=(1-t)^{-3}$.
Since the set of normal words is closed under multiplication by $z$ on the
left, the map $u\mapsto zu$ from $A$ to itself is injective. In particular,
$A$ has no non-trivial right annihilators. By Lemma 3.1 $H_{A^{!}}=(1+t)^{3}$
and $wA_{2}^{!}\neq\\{0\\}$ for every non-zero $w\in A_{1}^{!}$. Now Lemma 3.2
implies that $A$ is Koszul. ∎

We need another observation made in [13].

######  Lemma 3.7.

For every $p,q,r\in{\mathbb{K}}$, the Sklyanin algebra $A=Q^{p,q,r}$ satisfies
${\rm dim}\,A_{n}\geqslant\frac{(n+1)(n+2)}{2}$ for every
$n\in{\mathbb{Z}}_{+}$.

##  4 Proof of Theorem 1.1

Throughout this section $p,q,r\in{\mathbb{K}}$ and $A=Q^{p,q,r}$. By Remark
2.2, by passing to a field extension, we can, without loss of generality,
assume that ${\mathbb{K}}$ is algebraically closed. Since ${\rm
char}\,{\mathbb{K}}\neq 3$ and ${\mathbb{K}}$ is algebraically closed, we can
find $\theta\in{\mathbb{K}}$ such that $\theta^{3}=1\neq\theta$. If
$p^{3}=q^{3}=r^{3}$ or $p=q=0$ or $r=0$ or $(p+q)^{3}+r^{3}=0$ or
$p^{3}=q^{3}$, the conclusion of Theorem 1.1 follows from Lemma 3.5. Thus for
the rest of the proof, we can assume that none of these equalities holds.
Moreover, if $p+q=0$, one easily sees that a substitution from Lemma 3.4 to
breaks this equality. Thus we can additionally assume that $p+q\neq 0$.

Since $r\neq 0$, by scaling the relations, we can without loss of generality
assume that $r=1$. Then $(p,q)\neq(0,0)$, $(p^{3}-1,q^{3}-1)\neq(0,0)$,
$p+q\neq 0$ and $(p+q)^{3}+1\neq 0$.

Our proof hinges on finding a convenient linear substitution. The main
objective is to make specific monomials (namely, $xx$, $xy$ and $yz$) into
leading monomials of defining relations with respect to the standard left-to-
right degree-lexicographical ordering assuming $x>y>z$. We perform the
substitution in a number of steps: each of the steps is a linear sub itself
and the resultipng substitution is their composition. We keep the same letters
$x,y,z$ for both old and new variables. We introduce a substitution by showing
by which linear conmbination of (new) $x,y,z$ must the (old) variables be
replaced. For example, if we write $x\to x+y+z$, $y\to z-y$ and $z\to 7z$,
this means that all occurances of $x$ (in the relations, potential etc.) are
replaced by $x+y+z$, all occurances of $y$ are replaced by $z-y$, while $z$ is
swapped for $7z$.

Note that our Sklyanin algebra $Q^{p,q,1}$ is potential with the potential

$F_{p,q}=x^{3}+y^{3}+z^{3}+pxyz^{\rcirclearrowleft}+qxzy^{\rcirclearrowleft}.$

First, we perform the sub $x\to-\frac{x}{p+q}$, $y\to y$ and $z\to z$. As a
result, we see that $A=Q^{p,q,1}$ is isomorphic to the potential algebra with
the potential

$\textstyle
F^{\prime}_{p,q}=-\frac{(p+q)^{3}+1}{(p+q)^{3}}x^{3}+(x^{3}+y^{3}+z^{3})-\frac{p}{p+q}xyz^{\rcirclearrowleft}-\frac{q}{p+q}xzy^{\rcirclearrowleft}.$

Note that we have used the condition $p+q\neq 0$. Next, we do the sub $x\to
x+y+z$, $y\to x+\theta^{2}y+\theta z$ and $z\to x+\theta y+\theta^{z}$. As a
result, we see that $A=Q^{p,q,1}$ is isomorphic to the potential algebra with
the potential

$\textstyle
F^{\prime\prime}_{p,q}=-\frac{(p+q)^{3}+1}{(p+q)^{3}}(x+y+z)^{3}+\frac{3((1-\theta)p+(1-\theta^{2})q)}{p+q}xyz^{\rcirclearrowleft}+\frac{3((1-\theta^{2})p+(1-\theta)q)}{p+q}xzy^{\rcirclearrowleft}.$

Note that the $(x+y+z)^{3}$ coefficient in $F^{\prime\prime}_{p,q}$ is non-
zero since $(p+q)^{3}+1\neq 0$. Since scaling multiplying by a potential by a
non-zero constant has no effect on the corresponding potential algebra, $A$ is
isomorphic to the potential algebra with the potential

$G_{a,b}=(x+y+z)^{3}+axyz^{\rcirclearrowleft}+bxzy^{\rcirclearrowleft},$

where

$\textstyle
a=\frac{3(p+q)^{2}((\theta-1)p+(\theta^{2}-1)q)}{(p+q)^{3}+1},\quad
b=\frac{3(p+q)^{2}((\theta^{2}-1)p+(\theta-1)q)}{(p+q)^{3}+1}.$

Note that $a=0\iff p=\theta^{2}q$, $b=0\iff p=\theta q$ and $a=b\iff p=q$.
Since $p^{3}\neq q^{3}$, we have $ab(a-b)\neq 0$. Furthermore,
$a+b=-\frac{9(p+q)^{3}}{(p+q)^{3}+1}$ and therefore $a+b\neq 0$ as well.

Now we perform the sub $x\to\frac{a}{a+b}x$, $y\to\frac{b}{a+b}x+y-z$ and
$z\to z$. Note that we need the fact that $ab(a+b)\neq 0$ for this sub to be
non-degenerate. As a result, we see that $A=Q^{p,q,1}$ is isomorphic to the
potential algebra with the potential

$G^{\prime}_{a,b}=(x+y)^{3}+\frac{ab}{a+b}xxz^{\rcirclearrowleft}+\frac{a^{2}}{a+b}xyz^{\rcirclearrowleft}+\frac{ab}{a+b}xzy^{\rcirclearrowleft}-axzz^{\rcirclearrowleft}.$

By this point we have already reached our main objective. One easily sees that
the leading monomials of the defining relations of the potential algebra with
the potential $G^{\prime}_{a,b}$ are indeed $xx$, $xy$ and $yz$. However, we
shall perform one final sub (with a triangular matrix) in order to simplify
and/or kill some of the coefficients. Namely, we use the sub $x\to
x-\frac{a-b}{a}y+\frac{(a+b)^{2}+a^{2}b}{(a-b)^{3}}z$,
$y\to\frac{a-b}{a}y-\frac{(a+b)^{2}+ab^{2}}{(a-b)^{3}}z$ and
$z\to-\frac{a+b}{(a-b)^{2}}z$. Note that the sub is non-degenerate since
$ab(a+b)(a-b)\neq 0$. As a result, we see that $A$ is isomorphic to the
potential algebra with the potential

$P_{\alpha,\gamma}=x^{3}-xyz^{\rcirclearrowleft}+\alpha
yyz^{\rcirclearrowleft}-(\gamma-\alpha^{2})z^{3},$

where

$\displaystyle\textstyle\alpha$
$\displaystyle=-\frac{(a+b)^{3}+ab(a^{2}+b^{2})}{(a-b)^{4}}$
$\displaystyle\textstyle\gamma$
$\displaystyle=-\frac{(a+b)^{4}(a^{2}-ab+b^{2})+ab(a+b)^{3}(2a^{2}+2b^{2}-3ab)+a^{2}b^{2}(a^{4}+b^{4}+a^{2}b^{2}-a^{3}b-ab^{3})}{(a-b)^{8}}$

######  Remark 4.1.

There is no black magic to this string of substitutions. We have tried this
because there is no visible pattern to the leading monomials of the Gröbner
basis of the ideal of relations of the Sklyanin algebras in their original
form. We had to change something and the most drastic change ensues from
altering the leading monomials of the defining relations. However for generic
Sklyanin, it is impossible to get rid of the square $xx$ of the largest
generator. It is equally impossible to lose both $xy$ and $xz$, so we might as
well keep $xy$. The only freedom we have now is to find a sub, which will
change the smallest leading term $xz$ and this can not possibly go below $yz$.
So we set on changing it into $yz$. Next, we used elementary linear algebra to
find necessary and sufficient conditions on a potential for the corresponding
potential algebra to have $xy$, $xz$ and $yz$ as the leading terms of defining
relations. These have the form of equations on the coefficients of the
potential. Each of the first few of the above subs were designed to alter the
potential in such a way that one equation is satisfied without spoiling the
previously obtained ones. The last sub is there for the sake of neatness.

The case $\alpha=\gamma=0$ does not occur (does not come from a Sklyanin
algebra). One can see it directly using the above formulas for $\alpha$ and
$\gamma$. However there is an indirect way. The potential algebra with the
potential $P_{0,0}$ enjoys a rank 1 quadratic relation $xy=yy$ (we mean the
usual rank in $V\otimes V$). On the other hand, it is elementary to see that
no such thing exists for $Q^{p,q,r}$ provided there are at least two non-zero
numbers among $p$, $q$ and $r$ and $p^{3}=q^{3}=r^{3}$ fails. We shall discuss
the potential $P_{0,0}$ later since it is peculiar indeed.

Thus we have that $A$ is isomorphic to the potential algebra $B$ difined by
the potential $P_{\alpha,\gamma}=x^{3}-xyz^{\rcirclearrowleft}+\alpha
yyz^{\rcirclearrowleft}-(\gamma-\alpha^{2})z^{3}$ with
$\alpha,\gamma\in{\mathbb{K}}$. We know that $(\alpha,\gamma)\neq(0,0)$. Since
$A$ and $B$ are isomorphic, the proof will be complete if we show that Then
$B$ is Koszul and $H_{B}=(1-t)^{-3}$. A direct computation shows that $B$ is
presented by generators $x,y,z$ and the relations

$xx-zx+zy+\alpha zz=0,\ \ xy-yy-\alpha zx+\gamma zz=0,\ \ yz-zx+zy+\alpha
zz=0$ (4.1)

(observe the leading monomials $xx$, $xy$ and $yz$). Let $I$ be the right
ideal in $B$ generated by $y$ and $z$: $I=yB+zB$. For $f,g\in B$, we write
$f=g\,({\rm mod}\,I)$ if $f-g\in I$. For each $k\in{\mathbb{Z}}_{+}$ consider
the following property:

  * ($\Pi_{k}$)

there exist $a_{k},b_{k}\in{\mathbb{K}}$ such that
$xz^{k}x=a_{k}xz^{k+1}\,({\rm mod}\,I)$ and $xz^{k}y=b_{k}xz^{k+1}\,({\rm
mod}\,I)$.

Note that according to (4.1), $\Pi_{0}$ is satisfied with $a_{0}=b_{0}=0$. We
shall prove the following Claim.

Claim: If $k\in{\mathbb{Z}}_{+}$ and $\Pi_{k}$ is satisfied, then either
$\Pi_{k+1}$ is satisfied or

  * ($\Sigma_{k+1}$)

$xz^{k+1}x=xz^{k+1}y\,({\rm mod}\,I)$ and $xz^{k+2}=0\,({\rm mod}\,I)$.

###### Proof of Claim.

Assume that $\Pi_{k}$ is satisfied. Then there are
$a_{k},b_{k}\in{\mathbb{K}}$ such that $xz^{k}x=a_{k}xz^{k+1}\,({\rm mod}\,I)$
and $xz^{k}y=b_{k}xz^{k+1}\,({\rm mod}\,I)$. Using (4.1), we see that
$xz^{k}yz=xz^{k+1}x-xz^{k+1}y-\alpha xz^{k+2}$. On the other hand,
$xz^{k}yz=b_{k}xz^{k+2}\,({\rm mod}\,I)$. These equalities yield

$-xz^{k+1}x+xz^{k+1}y+(b_{k}+\alpha)xz^{k+2}=0\,({\rm mod}\,I).$

By (4.1), $xz^{k}xy=xz^{k}yy+\alpha xz^{k+1}x-\gamma xz^{k+2}$. Using
($\Pi_{k}$), we then have $xz^{k}xy=b_{k}xz^{k+1}y+\alpha xz^{k+1}x-\gamma
xz^{k+2}\,({\rm mod}\,I)$. Directly from ($\Pi_{k}$), we have
$xz^{k}xy=a_{k}xz^{k+1}y\,({\rm mod}\,I)$. These two equalities yield

$\alpha xz^{k+1}x+(b_{k}-a_{k})xz^{k+1}y-\gamma xz^{k+2}=0\,({\rm mod}\,I).$

Again, by (4.1), $xz^{k}xx=xz^{k+1}x-xz^{k+1}y-\alpha xz^{k+2}$. Using
($\Pi_{k}$), we have $xz^{k}xx=a_{k}xz^{k+1}x\,({\rm mod}\,I)$. These two
equalities yield

$(a_{k}-1)xz^{k+1}x+xz^{k+1}y+\alpha xz^{k+2}=0\,({\rm mod}\,I).$

In the matrix form, the equations in the above three displays read

$M\left(\begin{array}[]{c}xz^{k+1}x\\\ xz^{k+1}y\\\
xz^{k+2}\end{array}\right)=\left(\begin{array}[]{c}0\\\ 0\\\
0\end{array}\right)\ ({\rm mod}\,I),\ \ \text{where}\ \
M=\left(\begin{array}[]{ccc}-1&1&b_{k}+\alpha\\\ \alpha&b_{k}-a_{k}&-\gamma\\\
a_{k}-1&1&\alpha\end{array}\right).$ (4.2)

If the first two columns of matrix $M$ are linearly independent, $\Pi_{k+1}$
follows straight away.

It remains to consider the case when the first two columns of matrix $M$ are
proportional. In this case $a_{k}=0$ and $b_{k}=-\alpha$. Thus in this case
$M$ has the form

$M=\left(\begin{array}[]{ccc}-1&1&0\\\ \alpha&-\alpha&-\gamma\\\
-1&1&\alpha\end{array}\right).$

Since $(\alpha,\gamma)\neq(0,0)$, (4.2) is now equivalent to
$xz^{k+1}x=xz^{k+1}y\,({\rm mod}\,I)$ and $xz^{k+2}=0\,({\rm mod}\,I)$, which
is $\Sigma_{k+1}$. This concludes the proof of the claim. ∎

According to the above claim we have two options: either $\Pi_{k}$ holds for
all $k\in{\mathbb{Z}}_{+}$ or for some $k\in{\mathbb{Z}}_{+}$,
$\Pi_{0},\dots,\Pi_{k}$ and $\Sigma_{k+1}$ are satisfied. Note that the first
option is what happens for generic $(\alpha,\gamma)\in{\mathbb{K}}^{2}$, while
the validity of $\Pi_{0},\dots,\Pi_{k}$ and $\Sigma_{k+1}$ occurs for
$(\alpha,\gamma)$ from a non-trivial algebraic variety.

Case 1: $\Pi_{k}$ holds for all $k\in{\mathbb{Z}}_{+}$. The only monomials,
which do not contain any of $yz$, $xz^{k}x$ and $xz^{k}y$ for
$k\in{\mathbb{Z}}_{+}$ are $z^{k}y^{m}$ and $z^{k}y^{m}xz^{j}$ for
$k,m,j\in{\mathbb{Z}}_{+}$. Denote this set of monomials $N$. The number of
monomials of degree $n$ in $N$ is exactly $\frac{(n+1)(n+2)}{2}$. Since
$\Pi_{k}$ is saisfied for each $k$, $B$ is the linear span of $N$. It follows
that ${\rm dim}\,B_{n}\leqslant\frac{(n+1)(n+2)}{2}$ for each $n$ and these
inequalities turn into equalities precisely when monomials from $N$ are
linearly independent in $B$. On the other hand, by Lemma 3.7, ${\rm
dim}\,A_{n}={\rm dim}\,B_{n}\geqslant\frac{(n+1)(n+2)}{2}$ for each $n$. Hence
${\rm dim}\,B_{n}=\frac{(n+1)(n+2)}{2}$ for every $n$, that is,
$H_{B}=(1-t)^{-3}$, and monomials from $N$ are linearly independent in $B$.
Equalities in $\Pi_{k}$ can be written as the equalities
$xz^{k}x-a_{k}xz^{k+1}+f_{k}$ and $xz^{k}y-b_{k}xz^{k+1}+g_{k}$ in $B$ with
$f_{k},g_{k}$ being homogeneous elements of $I$ of degree $k+2$. According to
the above observations, the equality $H_{B}=(1-t)^{-3}$ implies that $yz-
zx+zy+\alpha zz$ together with $xz^{k}x-a_{k}xz^{k+1}+f_{k}$ and
$xz^{k}y-b_{k}xz^{k+1}+g_{k}$ for $k\in{\mathbb{Z}}_{+}$ form a reduced
Gröbner basis in the ideal of relations of $B$ and that $N$ is the set of
corresponding normal words. Since $N$ is closed under multiplication by $z$ on
the left, the map $u\mapsto zu$ from $B$ to itself is injective. Thus $B$ has
no non-trivial right annihilators. Since $A$ and $B$ are isomorphic
$H_{A}=(1-t)^{-3}$ and $A$ has non non-trivial right annihilators. By Lemma
3.1 $H_{A^{!}}=(1+t)^{3}$ and $wA_{2}^{!}\neq\\{0\\}$ for every non-zero $w\in
A_{1}^{!}$. Now Lemma 3.2 implies that $A$ is Koszul.

Case 2: there is $k\in{\mathbb{Z}}_{+}$ such that $\Pi_{0},\dots,\Pi_{k}$ and
$\Sigma_{k+1}$ hold. By $\Sigma_{k+1}$, $xz^{k+1}x=xz^{k+1}y\,({\rm mod}\,I)$
and $xz^{k+2}=0\,({\rm mod}\,I)$. Using these equalities and (4.1), we have
$xz^{k+1}xx=xz^{k+1}yx\,({\rm mod}\,I)$ and
$xz^{k+1}xx=xz^{k+2}x-xz^{k+2}y-\alpha xz^{k+3}=0\,({\rm mod}\,I)$. Hence

$xz^{k+1}yx=0\,({\rm mod}\,I).$

The only monomials, which do not contain any of $yz$, $xz^{j}x$ for
$0\leqslant j\leqslant k+1$, $xz^{j}y$ for $0\leqslant j\leqslant k$,
$xz^{k+2}$ and $xz^{k+1}yx$ are the words of the form $z^{j}y^{m}w$, where
$m,j\in{\mathbb{Z}}_{+}$ and $w$ is an initial subword (empty allowed) of the
infinite word $xz^{k+1}yyyy\dots$. Denote this set of monomials $N$. As
before, the number of monomials of degree $n$ in $N$ is exactly
$\frac{(n+1)(n+2)}{2}$. Now exactly the same argument as in Case 1 yields
$H_{A}=H_{B}=(1-t)^{-3}$ and shows that $A$ is Koszul. Note that this time the
Gröbner basis in the ideal of relations of $B$ turns out to be finite: it
consisits of $2k+6$ elements. This concludes the proof of Theorem 1.1.

##  5 Isomorphic Sklyanin algebras

The question when two different Sklyanin algebras are isomorphic as graded
algebras is touched in [4]. The authors associate to each Sklyanin algebra
some geometric data (an elliptic curve and its automorphism) and observe that
Sklyanin algeras are isomorphic precisely, when the same happens to geometric
data associated to them. This certainly may be useful on many occasions,
however this characterization of isomorphic Sklyanin algebras is hard to use
when the only data we have are just the triples of parameters (possible, of
course, but rather inconvenient). In this section we find out explicitly in
terms of just the values of parameters which Sklyanin algebras are isomorphic
to each other.

One may wonder whether the answer depends on whether we allow all algebra
isomorphisms instead of just graded (=linear in our case) ones. Well, it does
not according to the following very general observation. It must be somewhere
in the literature, however we could not come up with a reference. For this
reason we include its proof. We banish the said proof to the next section
devoted to remarks in order to keep the flow of arguments intact.

######  Proposition 5.1.

Let $A=A(V_{1},R_{1})$ and $B=A(V_{2},R_{2})$ be two quadratic algebras over
the same ground field ${\mathbb{K}}$ $($any field of any characteristic is
allowed here$)$. Assume also that $A$ and $B$ are isomorphic as algebras. Then
they are also isomorphic as graded algebras. In other words, there is a linear
change of variables turning $A$ into $B$.

Keeping the above proposition in mind, we shall just deal with isomorphisms in
the category of graded algebras. Recall that we still assume that $\rm
char\,{\mathbb{K}}\neq 3$. Throughout this section we shall also assume that
there is $\theta\in{\mathbb{K}}$ such that $\theta^{3}=1\neq\theta$ and use
$\theta$ for this element without further reference. Note that the absence of
a non-trivial cubic root of $1$ does effect the results below both directly
(when $\theta$ features in a statement) and indirectly.

First, we remove the degenerate Sklyanin algebras from the picture. We say
that a Sklyanin algebra $Q^{p,q,r}$ is degenerate if $(pq,pr,qr)=(0,0,0)$
(there are at least two zeros among $p$, $q$ and $r$) or $p^{3}=q^{3}=r^{3}$.
Note that according to Theorem 1.1, degenerate Sklyanin algebras are precisely
the Sklyanin algebras whose Hilbert series differs from $(1-t)^{-3}$.

######  Lemma 5.2.

Let $A=Q^{p,q,r}$ be a degenerate Sklyanin algebra. Then

  * •

$A={\mathbb{K}}\langle x,y,z\rangle\iff p=q=r=0;$

  * •

$A$ is isomorphic to the monomial algebra $Q^{1,0,0}$ given by the relations
$xy=zx=yz=0$ if and only if either $r=0$, $pq=0$ and $(p,q)\neq(0,0)$ or
$p^{3}=q^{3}=r^{3}\neq 0$ and $p\neq q;$

  * •

$A$ is to the monomial algebra $Q^{0,0,1}$ given by the relations $xx=yy=zz=0$
if and only if either $p=q=0$ and $r\neq 0$ or $p=q\neq 0$ and $p^{3}=r^{3};$

Furthermore, the algebras ${\mathbb{K}}\langle x,y,z\rangle$, $Q^{1,0,0}$ and
$Q^{0,0,1}$ are pairwise non-isomorphic and a degenerate Sklyanin algebra can
not be isomorphic to a non-degenerate one.

###### Proof.

The statement $A={\mathbb{K}}\langle x,y,z\rangle\iff p=q=r=0$ is obvious. If
exactly two of $p$, $q$ and $r$ are zero, then up to scaling we have three
options for $(p,q,r)$: $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$. Note that swapping
$x$ and $y$, while leaving $z$ as is, provides an isomorphism betweem
$Q^{1,0,0}$ and $Q^{0,1,0}$. It remains to deal with the case
$p^{3}=q^{3}=r^{3}\neq 0$. If $p=q$, then by scaling we can make $p=q=1$. Then
$r^{3}=1$. By Lemma 3.3, $Q^{p,q,r}$ is isomorphic to $Q^{1,1,1}$. By Lemma
3.4, $Q^{1,1,1}$ is isomorphic to $Q^{p^{\prime},q^{\prime},r^{\prime}}$,
where $p^{\prime}=\theta^{2}+\theta+1=0$, $q^{\prime}=\theta+\theta^{2}+1=0$
and $r^{\prime}=1+1+1=3\neq 0$. Hence $Q^{p,q,r}$ is isomorphic to
$Q^{0,0,1}$. It remains to deal with the situation $p^{3}=q^{3}=r^{3}$ and
$p\neq r$. By scaling, we can make $p=1$. Then $r^{3}=q^{3}=1$ and $q\neq 1$.
By Lemma 3.3, $Q^{p,q,r}$ is isomorphic to $Q^{1,q,q^{2}}$. By Lemma 3.4,
$Q^{1,q,q^{2}}$ is isomorphic to $Q^{p^{\prime},q^{\prime},r^{\prime}}$, where
$p^{\prime}=q+q^{3}+q^{2}=0$, $q^{\prime}=q^{2}+q^{2}+q^{2}=3q^{2}\neq 0$ and
$r^{\prime}=1+q+q^{2}=0$. Thus $Q^{p,q,r}$ is isomorphic to $Q^{1,0,0}$.

Clearly, ${\mathbb{K}}\langle x,y,z\rangle$ is not isomorphic to any of
$Q^{1,0,0}$ and $Q^{0,0,1}$ (for example, ${\mathbb{K}}\langle x,y,z\rangle$
has no non-trivial zero divisors, while $Q^{1,0,0}$ and $Q^{0,0,1}$ have non-
trivial zero divisors provided by the defining relations). The algebras
$Q^{1,0,0}$ and $Q^{0,0,1}$ are non-isomorphic since the only one-dimensional
representation of $Q^{0,0,1}$ is the augmentation map, while the map sending
$x$ to $1$ and $y$ and $z$ to $0$ extends to a one-dimensional representation
of $Q^{1,0,0}$ ($Q^{0,0,1}$ has only one one-dimensional representation, while
$Q^{1,0,0}$ has more than one; as a matter of fact it has infinitely many of
those if ${\mathbb{K}}$ is infinite).

Finally, by Theorem 1.1 and Lemma 3.5, every degenerate Sklyanin algebra has
exponential growth, while every non-degenerate Sklyanin algebra has polynomial
growth. Hence a degenerate Sklyanin algebra can not be $\cal A$-isomorphic to
a non-degenerate one. ∎

Note that monomiality of degenerate Sklyanin algebras was first noticed by
Smith [21]. Lemma 5.2 effectively takes degenerate algebras out of the
picture, so we can concentrate on the non-degenerate ones. That is we have to
deal with $Q^{p,q,r}$ for

$(p,q,r)\in M_{0}=\\{(p,q,r)\in{\mathbb{K}}^{3}:(pq,pr,qr)\neq(0,0,0)\ \
\text{and}\ \ (p_{3}-q^{3},p^{3}-r^{3},q^{3}-r^{3})\neq(0,0,0)\\}.$

We split $M_{0}$ into two disjoint subsets $M_{0}=M_{1}\cup M_{2}$, where

$M_{1}=\\{(p,q,r)\in{\mathbb{K}}^{3}:r\neq 0,\ (p,q)\neq(0,0),\
(p+q)^{3}+r^{3}\neq 0\ \ \text{and}\ \
(p_{3}-q^{3},p^{3}-r^{3},q^{3}-r^{3})\neq(0,0,0)\\}$

and $M_{2}=M_{0}\setminus M_{1}$. Just as we have removed degenerates, we
shall do the same with $M_{2}$, which will leave us free to concentrate on
$M_{2}$ afterwards.

######  Lemma 5.3.

Let $(p,q,r)\in{\mathbb{K}}^{3}$ be such that $r\neq 0$,
$(p_{3}-q^{3},p^{3}-r^{3},q^{3}-r^{3})\neq(0,0,0)$ and $(p+q)^{3}+r^{3}=0$.
Then $Q^{p,q,r}$ is isomorphic to the quantum polynomial algebra
$Q^{1,-\alpha,0}$ given by the relations $xy=\alpha yx$, $zx=\alpha xz$ and
$yz=\alpha zy$, where $\alpha=\theta\frac{p-\theta^{2}q}{p-\theta
q}\in{\mathbb{K}}^{*}$.

###### Proof.

Clearly, $Q^{p,q,r}=Q^{s,t,1}$, where $s=\frac{p}{r}$ and $t=\frac{q}{r}$.
Since $(p+q)^{3}+r^{3}=0$, we have $(-s-t)^{3}=1$. By Lemma 3.3, $Q^{s,t,1}$
is isomorphic to $Q^{s,t,-s-t}$. By Lemma 3.4, $Q^{s,t,-s-t}$ is isomorphic to
$Q^{p^{\prime},q^{\prime},r^{\prime}}$, where
$p^{\prime}=(\theta^{2}-1)s+(\theta-1)t$,
$q^{\prime}=(\theta-1)s+(\theta^{2}-1)t$ and $r^{\prime}=0$. Note that the
assumptions yield that $Q^{p,q,r}$ is non-degenerate and therefore so is
$Q^{p^{\prime},q^{\prime},r^{\prime}}$. Hence $p^{\prime}q^{\prime}\neq 0$.
Then $Q^{p^{\prime},q^{\prime},r^{\prime}}=Q^{1,-\alpha,0}$ with
$\alpha=-\frac{(\theta-1)s+(\theta^{2}-1)t}{(\theta^{2}-1)s+(\theta-1)t}$.
After easy simplifications, one gets
$\alpha=\theta\frac{p-\theta^{2}q}{p-\theta q}\in{\mathbb{K}}^{*}$. ∎

######  Lemma 5.4.

For every $(p,q,r)\in M_{2}$, the only one-dimensional representation of
$Q^{p,q,r}$ is the augmentation map.

###### Proof.

Let $A=Q^{p,q,r}$ with $(p,q,r)\in M_{1}$. We shall verify that $A$ has no
one-dimensional representations other than the augmentation map. Let
$\varphi:A\to{\mathbb{K}}$ be a representaion (=an algebra homomorphism) and
let $a=\varphi(x)$, $b=\varphi(y)$ and $c=\varphi(z)$. We have to prove that
$a=b=c=0$. From the defining relations of $A$ it follows that
$(p+q)ab=-rc^{2}$, $(p+q)bc=-ra^{2}$ and $(p+q)ac=-rb^{2}$. Multiplying these
equalities, we get $a^{2}b^{2}c^{2}((p+q)^{3}+r^{3})=0$. Since $(p,q,r)\in
M_{1}$, $(p+q)^{3}+r^{3}\neq 0$. Hence $abc=0$. Without loss of generality, we
may assume that $a=0$. From $(p+q)ab=-rc^{2}$ and $(p+q)ac=-rb^{2}$ we now
have $rb=rc=0$. Since $(p,q,r)\in M_{1}$, $r\neq 0$ and therefore $a=b=c=0$. ∎

######  Lemma 5.5.

For every $(p,q,r)\in M_{2}$, $Q^{p,q,r}$ is isomorphic to a quantum
polynomial algebra $B^{\alpha}=Q^{1,-\alpha,0}$ given by the relations
$xy=\alpha yx$, $zx=\alpha xz$ and $yz=\alpha zy$ for some
$\alpha\in{\mathbb{K}}^{*}$. Next, $B^{\alpha}$ and $B^{\beta}$ are isomorphic
if and only if either $\alpha=\beta$ of $\alpha\beta=1$. Furthermore,
$Q^{p,q,r}$ and $Q^{p^{\prime},q^{\prime},r^{\prime}}$ are non-isomorphic if
$(p,q,r)\in M_{2}$ and $(p^{\prime},q^{\prime},r^{\prime})\in M_{1}$.

###### Proof.

If $(p,q,r)\in M_{2}$, from definition of $M_{2}$ it follows that either $r=0$
and $pq\neq 0$ or $r\neq 0$,
$(p_{3}-q^{3},p^{3}-r^{3},q^{3}-r^{3})\neq(0,0,0)$ and $(p+q)^{3}+r^{3}=0$. If
$r=0$ and $pq\neq 0$, then $Q^{p,q,r}$ is isomorphic to $B^{\alpha}$ with
$\alpha=-\frac{q}{p}\in{\mathbb{K}}^{*}$. Otherwise, Lemma 5.3 yields an
isomorphism of $Q^{p,q,r}$ and $B^{\alpha}$ with
$\alpha=\theta\frac{p-\theta^{2}q}{p-\theta q}\in{\mathbb{K}}^{*}$. In any
case, $Q^{p,q,r}$ is isomorphic to $B^{\alpha}$ with
$\alpha\in{\mathbb{K}}^{*}$.

If $(p^{\prime},q^{\prime},r^{\prime})\in M_{1}$, by Lemma 5.4,
$Q^{p^{\prime},q^{\prime},r^{\prime}}$ has just one one-dimensional
representation, the augmentation map. On the other hand each $B^{\alpha}$ has
plenty one-dimensional representations. For instance, the map $x\mapsto 1$,
$y\mapsto 0$ and $z\mapsto 0$ extends to a one-dimensional representation for
each $B^{\alpha}$. Since $Q^{p,q,r}$ for $(p,q,r)\in M_{2}$ is isomorphic to a
$B^{\alpha}$, we see that $Q^{p,q,r}$ and
$Q^{p^{\prime},q^{\prime},r^{\prime}}$ are non-isomorphic if $(p,q,r)\in
M_{2}$ and $(p^{\prime},q^{\prime},r^{\prime})\in M_{1}$.

It remains to prove that $B^{\alpha}$ and $B^{\beta}$ are isomorphic if and
only if either $\alpha=\beta$ of $\alpha\beta=1$. Since
$B^{1}={\mathbb{K}}[x,y,z]$ is the only commutative algebra among
$B^{\alpha}$, it is non-isomorphic to any $B^{\alpha}$ with $\alpha\neq 1$.
Thus it remains to deal with the case $\alpha\neq 1$ and $\beta\neq 1$. First,
observe that swapping $x$ and $y$, while leaving $z$ as is, provides a $\cal
G$-isomorphism between $B^{\alpha}$ and $B^{\alpha^{-1}}$. Now let
$\alpha,\beta\in{\mathbb{K}}^{*}$, $\alpha\neq 1$, $\beta\neq 1$,
$\alpha\neq\beta$ and $\alpha\beta\neq 1$. The proof will be complete if we
show that $B^{\alpha}$ and $B^{\beta}$ are non-isomorphic. Assume the
contrary: there is a graded algebra isomorphism $\varphi:B^{\alpha}\to
B^{\beta}$. Let $x$, $y$, $z$ be the usual generators of $B^{\beta}$, while
$u$, $v$ and $w$ be the images under $\varphi$ of the usual generators of
$B^{\alpha}$. Then $u$, $v$ and $w$ are homogeneous degree one elements of
$B^{\beta}$, they form a linear basis in the degree $1$ component
$B_{1}^{\beta}$ of $B^{\beta}$ and they satisfy $uv=\alpha vu$, $wu=\alpha uw$
and $vw=\alpha wv$. Furthermore, the linear span of $uv-\alpha vu$, $wu-\alpha
uw$ and $vw-\alpha wv$ treated as elements of ${\mathbb{K}}\langle
x,y,z\rangle$ must coincide with the linear span of $xy-\beta yx$, $zx-\beta
xz$ and $yz-\beta zy$. In particular, $uv-\alpha vu$, $wu-\alpha uw$ and
$vw-\alpha wv$ must be ’square-free’: when written in terms of $x$, $y$ and
$z$ they should not contain any of $xx$, $yy$ and $zz$. Since $\alpha\neq 1$,
it immediately follows that each of $u$, $v$ and $w$, when written in terms of
$x$, $y$ and $z$ must be a scalar multiple of a single element of
$\\{x,y,z\\}$. That is, the matrix of our linear substitution $\varphi$ is a
product of a diagonal matrix and a permutation matrix (of size $3\times 3$).
If the permutation in question is even, one easily sees that $uv-\alpha vu$,
$wu-\alpha uw$ and $vw-\alpha wv$ are scalar multiples of $xy-\alpha yx$,
$zx-\alpha xz$ and $yz-\alpha zy$ (in some order), while if the permutation is
odd $uv-\alpha vu$, $wu-\alpha uw$ and $vw-\alpha wv$ are scalar multiples of
$\alpha xy-yx$, $\alpha zx-xz$ and $\alpha yz-\alpha zy$ (in some order).
Thus, the spans of the triples $uv-\alpha vu$, $wu-\alpha uw$ and $vw-\alpha
wv$ and $xy-\beta yx$, $zx-\beta xz$ and $yz-\beta zy$ can coincide only if
$\alpha=\beta$ or $\alpha\beta=1$. This contradiction completes the proof. ∎

Next lemma clinches description of isomorphic Sklyanin algebras. Since every
$(p,q,r)\in M_{1}$ satisfies $r\neq 0$, by dividing by $r$, $(p,q,r)$ can be
turned into a unique triple $(a,b,1)$ with

$(a,b)\in M,\ \ \text{where}\ \ M=\\{(a,b)\in{\mathbb{K}}^{2}:(a,b)\neq(0,0),\
(a+b)^{3}+1\neq 0,\ (a^{3}-1,b^{3}-1)\neq(0,0)\\}.$

Since scaling the triple of parameters does not change the Sklyanin algebra,
in order to describe which Sklyanins with $(p,q,r)\in M_{1}$ are isomorphic,
it suffices to do so in the case $(p,q,r)=(a,b,1)$ with $(a,b)\in M$.

######  Lemma 5.6.

If both $(a,b)$ and $(a^{\prime},b^{\prime})$ belong to $M$, then the Sklyanin
algebras $Q^{a,b,1}$ and $Q^{a^{\prime},b^{\prime},1}$ are isomorphic if and
only if $(a,b)$ and $(a^{\prime},b^{\prime})$ are in the same orbit of the
group action on $M$ generated by two maps $(a,b)\mapsto(\theta a,\theta b)$
and $(a,b)\mapsto\left(\frac{\theta
a+\theta^{2}b+1}{a+b+1},\frac{\theta^{2}a+\theta b+1}{a+b+1}\right)$. This
group is finite, consists of $24$ elements $($thus if ${\mathbb{K}}$ is
infinite, for generic $(a,b)\in M$, there are exactly $23$ other elements of
$M$ giving rise to an isomorphic Sklyanin algebra$)$ and is isomorphic to
$SL_{2}({\mathbb{Z}}_{3})$.

The complete list of pairs $(a^{\prime},b^{\prime})\in M$ such that for a
given $(a,b)\in M$, $Q^{a,b,1}$ and $Q^{a^{\prime},b^{\prime},1}$ are
isomorphic is as follows$:$

  * •

$(\theta^{j}a,\theta^{j}b)$ with $j\in\\{0,1,2\\};$

  * •

$(\theta^{j}b,\theta^{j}a)$ with $j\in\\{0,1,2\\};$

  * •

$\left(\frac{\theta^{j}a+\theta^{k}b+\theta^{m}}{a+b+\theta^{n}},\frac{\theta^{k}a+\theta^{j}b+\theta^{m}}{a+b+\theta^{n}}\right)$
with $n\in\\{0,1,2\\}$ and $\\{j,k,m\\}=\\{0,1,2\\}.$

The last condition means that $(j,k,m)$ is a permutation of $(0,1,2)$. Each of
the first two lines in the above list provide $3$ pairs, while the last line
yields $18$.

###### Proof.

Assume that $(a,b)$ and $(a^{\prime},b^{\prime})$ belong to $M$, then the
Sklyanin algebras $A=Q^{a,b,1}=A(V,R)$ and
$B=Q^{a^{\prime},b^{\prime},1}=A(V,R^{\prime})$ are isomorphic. By Theorem
1.1, ${\rm dim}\,A_{3}={\rm dim}\,B_{3}=10$. Since ${\rm dim}\,V^{3}=27$, we
have ${\rm dim}\,(VR+RV)={\rm dim}\,(VR^{\prime}+R^{\prime}V)=27-10=17$. Since
${\rm dim}\,VR={\rm dim}\,RV={\rm dim}\,VR^{\prime}={\rm dim}\,R^{\prime}V=9$,
it follows that ${\rm dim}\,(RV\cap VR)={\rm dim}\,(R^{\prime}V\cap
V^{\prime}R)=1$. On the other hand, obviously, the potentials
$F=x^{3}+y^{3}+z^{3}+axyz^{\rcirclearrowleft}+bxzy^{\rcirclearrowleft}$ and
$F^{\prime}=x^{3}+y^{3}+z^{3}+a^{\prime}xyz^{\rcirclearrowleft}+b^{\prime}xzy^{\rcirclearrowleft}$
for $A$ and $B$ respectively satisfy $F\in RV\cap VR$ and $F^{\prime}\in
R^{\prime}V\cap VR^{\prime}$. Since a linear substitution $T\in
GL_{3}({\mathbb{K}})$ facilitating the graded isomorphism of $A$ and $B$ must
send $RV\cap VR$ to $R^{\prime}V\cap VR^{\prime}$, it transforms $F$ to
$F^{\prime}$ up to a scalar multiple. Hence $T$ must transform the
abelianization of $F$ $G=x^{3}+y^{3}+z^{3}+3(a+b)xyz\in{\mathbb{K}}[x,y,z]$
(the image of $F$ under the canonical map from ${\mathbb{K}}\langle
x,y,z\rangle$ to ${\mathbb{K}}[x,y,z]$) into the abelianization
$G^{\prime}=x^{3}+y^{3}+z^{3}+3(a^{\prime}+b^{\prime})xyz\in{\mathbb{K}}[x,y,z]$
of $F^{\prime}$ up to a scalar multiple. Hence, $T$ provides an isomorphism
between the elliptic (projective) curves $C$ given by $G=0$ and $C^{\prime}$
given by $G^{\prime}=0$.

Since $(a,b)$ and $(a^{\prime},b^{\prime})$ belong to $M$, we have
$(a+b)^{3}+1\neq 0$ and $(a^{\prime}+b^{\prime})^{3}+1\neq 0$ and therefore
the curves $C$ and $C^{\prime}$ are regular. Note that the each of the curves
$C$ and $C^{\prime}$ have the same exactly collection of nine inflection
points which are the nine lines $L_{k}$ for $1\leqslant k\leqslant 9$ spanned
by $(1,-1,0)$, $(1,-\theta,0)$, $(1,-\theta^{2},0)$, $(1,0,-1)$,
$(1,0,-\theta)$, $(1,0,-\theta^{2})$, $(0,1,-1)$, $(0,1,-\theta)$ and
$(0,1,-\theta^{2})$ respectively. Since $T$ is an isomorphism between
$C^{p,q}$ and $C^{p^{\prime},q^{\prime}}$, $T$ must leave the union of $L_{j}$
invariant. It is a routine exercise to verify that the subgroup $G$ of
$GL_{3}({\mathbb{K}})$ leaving the union of $L_{j}$ invariant is generated by

$\left(\begin{array}[]{ccc}\lambda&0&0\\\ 0&\lambda&0\\\
0&0&\lambda\end{array}\right)\ \ (\lambda\in{\mathbb{K}}^{*}),\ \
\left(\begin{array}[]{ccc}0&1&0\\\ 0&0&1\\\ 1&0&0\end{array}\right),\ \
\left(\begin{array}[]{ccc}1&0&0\\\ 0&1&0\\\ 0&0&\theta\end{array}\right)\ \
\text{and}\ \ \left(\begin{array}[]{ccc}\theta&\theta^{2}&1\\\
\theta^{2}&\theta&1\\\ 1&1&1\end{array}\right).$ (5.1)

One way to see it is to notice that, first of all, each linear map in the
above display leaves the union of $L_{j}$ invariant. Next, one easily checks
that the group $K$ generated by matrices in the above display acts
transitively on the set of triples of $L_{j}$ that span ${\mathbb{K}}^{3}$.
This fact together with an (easily verifiable) observation that $K$ contains
all permutation matrices and all scalar matrices yields that $K$ contains all
elements of $G$ and therefore $K=G$. Thus $T\in G$.

Now the linear transformations given by the first two matrices in the above
display provide automorphisms of each $Q^{p,q,1}$, while the linear
transformation given by the third matrix in the above display facilitates an
isomorphism between $Q^{p,q,1}$ and $Q^{\theta p,\theta q,1}$ for each
$(p,q)\in M$. Finally, a direct computation shows that the linear
transformations given by the last matrix in the above display provides an
isomorphism of $A^{p,q}$ and $A^{p^{\prime},q^{\prime}}$ for every $(p,q)\in
M$, where $(p^{\prime},q^{\prime})=\bigl{(}\frac{\theta
p+\theta^{2}q+1}{p+q+1},\frac{\theta^{2}p+\theta q+1}{p+q+1}\bigr{)}$. This
completes the proof of the isomorphism statement. Thus the existence of an
isomorphism between $A$ and $B$ is equaivalent to $A$ and $B$ being in the
same orbit of the group action on $M$ generated by two maps
$(a,b)\mapsto(\theta a,\theta b)$ and $(a,b)\mapsto\left(\frac{\theta
a+\theta^{2}b+1}{a+b+1},\frac{\theta^{2}a+\theta b+1}{a+b+1}\right)$.

A direct computation shows that this group consists of the maps
$(a,b)\mapsto(\theta^{j}a,\theta^{j}b)$ with $j\in\\{0,1,2\\}$
$(a,b)\mapsto(\theta^{j}b,\theta^{j}a)$ with $j\in\\{0,1,2\\}$ and
$(a,b)\mapsto\left(\frac{\theta^{j}a+\theta^{k}b+\theta^{m}}{a+b+\theta^{n}},\frac{\theta^{k}a+\theta^{j}b+\theta^{m}}{a+b+\theta^{n}}\right)$
with $n\in\\{0,1,2\\}$ and $\\{j,k,m\\}=\\{0,1,2\\}$. Hence the group has $24$
elements. As for this group being isomorphic to $SL_{2}({\mathbb{Z}}_{3})$,
this can be done by computing enough features of this group (for instance, it
has a two-element centre, maximal order of an element in it is $6$ etc.) to be
able to identify it in the well-known list of $24$-element groups. ∎

##  6 Appendix

We start this section by proving Proposition 5.1.

###  6.1 Proof of Proposition 5.1

Let $A=A(V_{1},R_{1})$ and $B=A(V_{2},R_{2})$ be quadratic algebras over the
same ground field ${\mathbb{K}}$. Let $x_{1},\dots,x_{n}$ be a linear basis in
$V_{1}$ and $y_{1},\dots,y_{m}$ be a linear basis in $V_{2}$. Assume also that
$f_{1},\dots,f_{s}$ is a linear basis in $R_{1}$, while $g_{1},\dots,g_{t}$ is
a linear basis in $R_{2}$. Assume also that there is an algebra isomorphism
$\varphi:A\to B$. The proof will be complete if we construct another algebra
isomorphism $\psi:A\to B$ such that its restriction to $V_{1}$ is a linear
isomorphism from $V_{1}$ onto $V_{2}$ (equivalently, $\psi$ is a graded
algebra isomorphism). Since $f_{j}\in V_{1}^{2}$,
$f_{j}=\sum\limits_{1\leqslant k,p\leqslant n}a^{(j)}_{k,p}x_{k}x_{p}$ with
$a^{(j)}_{k,p}\in{\mathbb{K}}$. Let $u^{(j)}=\varphi(x_{j})$ for $1\leqslant
j\leqslant n$. As usual, for an element $u$ of a graded algebra, $u_{j}$
stands for degree $j$ homogeneous component of $u$. Since $\varphi$ is an
algebra isomomorphism, $u^{(j)}$ must generate $B$. Since the first component
of every element of $B$ in the subalgebra generated by $u^{(j)}$ is in the
linear span of $u^{(j)}_{1}$, the said span must coincide with $V_{2}$. In
particular, $n\geqslant m$. The same argument with the role of $A$ and $B$
switched yields $m\geqslant n$. Hence $m=n$ and $u^{(j)}_{1}$ for $1\leqslant
j\leqslant n$ form a linear basis in $V_{2}$. Since $\varphi$ is an algebra
homomorphism, we have $\varphi(f_{j})=0$ in $B$, that is,

$\sum\limits_{1\leqslant k,p\leqslant n}a^{(j)}_{k,p}u^{(k)}u^{(p)}=0\ \
\text{in $B$ for $1\leqslant j\leqslant s$.}$ (6.1)

The zero component of the above equation reads

$\sum\limits_{1\leqslant k,p\leqslant n}a^{(j)}_{k,p}u_{0}^{(k)}u_{0}^{(p)}=0\
\ \text{in ${\mathbb{K}}$ for $1\leqslant j\leqslant s$.}$ (6.2)

The degree $1$ component of (6.1) looks like

$\sum\limits_{1\leqslant k,p\leqslant
n}a^{(j)}_{k,p}(u_{0}^{(k)}u_{1}^{(p)}+u_{0}^{(p)}u_{1}^{(k)})=0\ \ \text{in
$V_{2}$ for $1\leqslant j\leqslant s$.}$ (6.3)

Since $u^{(j)}_{1}$ are linearly independent, the left-hand sides in (6.3)
must be zero as polynomials in $u^{(j)}_{1}$. Hence,

$\sum\limits_{1\leqslant k,p\leqslant
n}a^{(j)}_{k,p}(u_{0}^{(k)}u^{(p)}+u_{0}^{(p)}u^{(k)})=0\ \ \text{in $B$ for
$1\leqslant j\leqslant s$.}$ (6.4)

Denote $v^{(j)}=u^{(j)}-u^{(j)}_{0}$. Combining (6.1), (6.2) and (6.4) we get

$\sum\limits_{1\leqslant k,p\leqslant
n}a^{(j)}_{k,p}(u^{(k)}-u^{(k)}_{0})(u^{(p)}-u^{(p)}_{0})=0\ \ \text{in $B$
for $1\leqslant j\leqslant s$.}$ (6.5)

The second degree component of (6.5) now is

$\sum\limits_{1\leqslant k,p\leqslant n}a^{(j)}_{k,p}u^{(k)}_{1}u^{(p)}_{1}=0\
\ \text{in $B$ for $1\leqslant j\leqslant s$.}$ (6.6)

Since $f_{j}$ are linearly independent and $u^{(j)}_{1}$ are linearly
independent, (6.6) provides $s$ linearly independent quadratic relations in
$B$. In particular, $s\geqslant t$. The same argument with the roles of $A$
and $B$ reversed, gives $t\geqslant s$. Hence $s=t$ and the linear span of the
left-hand sides of (6.5) must coincide with $R_{2}$. It follows that the map
$x_{i}\mapsto u^{(i)}_{1}$ extends to an algebra isomorphism of $A$ and $B$.
Clearly, this is the graded isomorphism, we were after.

###  6.2 Some remarks

Observe that our results on Sklyanin algebras collapse if the characteristic
of the ground field equals $3$. Indeed, a cubic root of 1 plays an essential
role in the substitution we construct as well as in the description of
isomorphic Sklyanin algebras.

As for the potential

$P_{0,0}=x^{3}-xyz^{\rcirclearrowleft},$

which does not correspond to a Sklyanin algebra, the algebra $W$ it does
generate is peculiar indeed. The defining relations of $W$ are $xx-zx+zy=0$,
$xy-yy=0$ and $yz-zx+zy=0$. A direct computation shows that the reduced
Gröbner basis in the ideal of relations of $W$ is finite. It comprises $xx-
zx+zy$, $xy-yy$, $yz-zx+zy=0$, $yyy$, $xzx-xzy+zyx-zzx+zzy$ and $xzyx$, which
allows us to compute the Hilbert series of $W:$
$H_{W}=\frac{(1+t)(1+t^{2})(1+t+t^{2})}{1-t-t^{3}-2t^{4}}$. On the other hand,
the dual $W^{!}$ is given by the relations $xx+yz+zx$, $xy+yy$, $xz$, $yx$,
$zx+zy$ and $zz$. Those together with $yyy$, $zyz$, $yyz-zyy$ and $yzy-zyy$
form the reduced Gröbner basis in the ideal of relations of $W^{!}$. The
corresponding normal words are $1$, $x$, $y$, $z$, $yy$, $yz$, $zy$ and $zyy$,
yielding $H_{W^{!}}=(1+t)^{3}$. Clearly, the the duality relation
$H_{W}(t)H_{W^{!}}(-t)=1$ fails and therefore $W$ is non-Koszul. Thus $W$
provides an example of a non-Koszul quadratic potential algebra on three
generators.

We did not bother to find out which Sklyanin algebras are isomorphic in two
cases: when ${\rm char}\,{\mathbb{K}}=3$ and when ${\rm
char}\,{\mathbb{K}}\neq 3$ but ${\mathbb{K}}$ possesses no nontrivial cubic
roots of $1$.

Acknowledgements. We are grateful to IHES and MPIM for hospitality, support,
and excellent research atmosphere. This work is funded by the ERC grant
320974, the ESC Grant N9038, and EPSRC grant EP/T029455/1.

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Natalia Iyudu

Department of Mathematics and Statistics

Lancaster University, Lancaster, LA1 4YF

E-mail address<EMAIL_ADDRESS><EMAIL_ADDRESS>
<EMAIL_ADDRESS><EMAIL_ADDRESS>

Stanislav Shkarin

Queens’s University Belfast

Department of Pure Mathematics

University road, Belfast, BT7 1NN, UK

E-mail addresses: and<EMAIL_ADDRESS>