Explicit representations of Sklyanin algebras at a point of order 2

# Explicit Representations of 3-dimensional Sklyanin algebras associated to a point of order 2

Daniel J. Reich and Chelsea Walton Reich: Department of Mathematics, North Carolina State University, Raleigh, North Carolina 27695, USA Walton: Department of Mathematics, Temple University, Philadelphia, Pennsylvania 19122, USA
###### Abstract.

The representation theory of a 3-dimensional Sklyanin algebra depends on its (noncommutative projective algebro-) geometric data: an elliptic curve in , and an automorphism of given by translation by a point. Indeed, by a result of Artin-Tate-van den Bergh, we have that is module-finite over its center if and only if has finite order. In this case, all irreducible representations of are finite-dimensional and of at most dimension .

In this work, we provide an algorithm in Maple to directly compute all irreducible representations of associated to of order 2, up to equivalence. Using this algorithm, we compute and list these representations. To illustrate how the algorithm developed in this paper can be applied to other algebras, we use it to recover well-known results about irreducible representations of the skew polynomial ring .

###### Key words and phrases:
Azumaya locus, irreducible representation, Maple algorithm, three-dimensional Sklyanin algebra
###### 2010 Mathematics Subject Classification:
16S38, 16G99, 16Z05

## 1. Introduction

We work over the ground field . The motivation of this work is to study, up to equivalence, irreducible finite-dimensional representations (irreps) of Sklyanin algebras of global dimension 3 [Definition 1.2]. Past work on this problem include results on bounds on the dimension of irreps of [Walton], and on a geometric parametrization of (trace-preserving) irreps of [DeLaetLeBruyn]. The focus of this paper is to determine, for a class of Sklyanin algebras, all explicit irreps up to equivalence. Namely, we compute irreducible matrix solutions to the defining equations of , up to an action of a general linear group. A geometric parametrization of the set of irreps of is also presented, as this is the typical approach to understanding aspects of Sklyanin algebras.

###### Remark 1.1.

We directly compute the irreps via a Maple algorithm. A more conceptual technique, using noncommutative projective algebraic geometry (and Clifford theory for these particular Sklyanin algebras), can be used to solve this problem. We nevertheless hold to the computational approach because it can be adapted (much more easily in some cases) to other algebras; for further discussion of the complexity of this approach, see Remarks 1.7 and 1.8.

To begin, let us define the algebra under investigation.

###### Definition 1.2.

[ATV1] The 3-dimensional Sklyanin algebra over is generated by three non-commuting variables , , subject to the following relations:

 (1.1) ayz+bzy+cx2 = azx+bxz+cy2 = axy+byx+cz2 = 0.

Here, , with and .

This algebra is rather resistant to noncommutative Gröbner basis methods, that is, it is difficult to write down a -vector space basis of (consisting of monomials in ); see, for instance, [Rogalski, Exercise 1.7]. (The reader may also be interested in recent work of Iyudu-Shrakin [IyuduShkarin].) In fact, it is common practice to consider the geometric data of in the context of Noncommutative Projective Algebraic Geometry [ATV1, Rogalski, StaffordvandenBergh] to analyze its ring-theoretic behavior. By [ATV1, Equations 1.6 and 1.7], the geometric data of consists of an elliptic curve defined the equation,

 (1.2) Ea,b,c:  (a3+b3+c3)(uvw)−(abc)(u3+v3+w3)=0,

and an automorphism of this elliptic curve given by

 (1.3) σa,b,c([u:v:w])=[acv2−b2uw : bcu2−a2vw : abw2−c2uv].

Here, the automorphism is given by translation of the point , where is the origin of . The order of , denoted , is the smallest such that . If no such exists, then . Consider the following terminology.

###### Definition 1.3.

We say that a Sklyanin algebra is associated to a point () of order  if the automorphism has order .

The role of this geometric data for our work will be explained towards the end of this section.

Now let us recall some basic representation theory terminology. Take to be a positive integer. An -dimensional representation of is an algebra homomorphism where is a -vector space of dimension . Since is isomorphic to , there is a one-to-one correspondence between the -dimensional representations of and the matrix solutions to the system of equations (1.1). Here, , , and .

Next, we discuss irreducibility. Given a representation , a subspace of is called -stable if , for all , . Such a subspace yields a sub-representation of , given as . We say that is irreducible if the only -stable subspaces of are and itself, that is, if there are no proper sub-representations of . Similarly, there is a notion of irreducibility for a matrix solution to equations (1.1); see Lemma 2.1.

Now we recall when two representations/ matrix solutions of are equivalent. We say that -dimensional representations are equivalent if there exists a matrix so that , for all . Likewise, two matrix solutions and to (1.1) are equivalent if there exists such that , , and . Note that two equivalent representations/ matrix solutions are either both irreducible or both reducible.

As the reader can imagine, studying explicit finite-dimensional representations of the algebras is difficult computationally. Now by [Walton, Theorem 1.3], we only have non-trivial finite-dimensional representations of when the automorphism of (1.3) has finite order. So, we refine our goal: we study the irreps of associated to a point of order 2. (Note that the order 1 case is precisely the case when is commutative [Lemma 2.4].)

###### Lemma 1.4 (Lemma 2.5).

A Sklyanin algebra is associated to a point of order 2 if and only if .

In this case, we assume that by rescaling. Therefore, our goal is to study the representation theory of the 3-dimensional Sklyanin algebra , where by Definition 1.2, . By Lemma 2.6, all 1-dimensional irreps of are trivial, and all irreps of are finite-dimensional, of at most dimension 2. Thus, we only need to compute the irreps of dimension 2; we achieve this as follows.

###### Theorem 1.5.

The non-trivial explicit irreps (or matrix solutions) of the 3-dimensional Sklyanin algebra are of dimension 2. They are classified up to equivalence; the representatives of equivalence classes of irreps of are provided in Tables 3 and 4 in Sections 5 and 6, respectively.

In Section 2, we provide background material and some preliminary results. In Section 3, we give an outline (Steps 0-2, 3a, 3b) of our algorithm to prove Theorem 1.5. The algorithm then begins in Section 4, where we determine all of the 2-dimensional representations of , and exclude ‘families’ of reducible representations; this is Steps 0-2 of the algorithm. In Sections 5 and 6, we determine representatives of equivalence classes of 2-dimensional irreps of ; this is Steps 3a and 3b of the algorithm.

The study of the irreps of ends in Section 7, where for completion, we discuss a geometric parametrization of equivalence classes of irreps of (e.g., we illustrate the Azumaya locus of over the center of ). Namely we have the result below.

###### Theorem 1.6 (Theorem 7.1).

The set of equivalence classes of irreps of is in bijective correspondence with the points of the 3-dimensional affine variety:

 Xc:=V(g2−c2(u31+u32+u33)−(c3−4)u1u2u3)⊆C4{u1,u2,u3,g}.

In particular, is the Azumaya locus of over its center (i.e., points of correspond to 2-dimensional irreps of ), and the origin of corresponds to the trivial representation of .

###### Remark 1.7.

We would like to point out that one can adjust our algorithm to prove Theorem 1.5 to examine equivalence classes of irreps of other algebras with generators and relations, especially those that are module-finite over their center. Although, the run-time and complexity of the output of the algorithm is in direct correlation with the number of generators and relations of the algebra, along with the algebra’s polynomial identity degree (if applicable).

We illustrate the remark above in Section 8, where we tailor our algorithm to examine irreps of the following skew polynomial ring:

 C−1[x,y]:=C⟨x,y⟩/(xy+yx).

Like , it is well-known that all irreps of are finite-dimensional, of dimension at most 2 [Lemma 8.1(c)]. See Proposition 8.2 and Corollary 8.3 for the results on the representation theory of .

Unless stated otherwise, computational results in this work are performed with the computer algebra system Maple™(version 16). All code will be presented in  typewriter typeface, and are available on the authors’ websites. 1

###### Remark 1.8.

Part of the novelty of this work is that we obtain noncommutative algebraic/ representation theoreric results with Maple, which is a computer algebra system that is used typically for commutative computations. We hope that in the future the task of determining equivalence classes of irreps of noncommutative algebras (presented by generators and relations) can be achieved easily using a computer algebra system that handles noncommutative Gröbner bases, such as GAP [GAP-GBNP].

## 2. Preliminaries

We begin with a result on the irreducibility of a representation/ matrix solution of a Sklyanin algebra . This result is well-known, and we will use it often without mention.

###### Lemma 2.1.

Let be an -dimensional representation of , with corresponding matrix solution to the system of equations (1.1). Then, the following are equivalent:

1. is irreducible;

2. the corresponding -module (where acts on via ) is simple;

3. is surjective;

4. generates as a -algebra; and

5. every matrix in can be expressed as a noncommutative polynomial in over . ∎

If any of the above conditions hold, we say that the matrix solution is irreducible.

On the other hand, we can determine when a matrix solution of is reducible by using Lemma 2.1.

###### Corollary 2.2.

An matrix solution to (1.1) (corresponding to a representation of ) is reducible if and only if there exists a subspace of of dimension with for all . Here, we embed into so that is given by matrix multiplication. ∎

If is a Sklyanin algebra associated to a point of infinite order, then by [Walton, Theorem 1.3(i)], we have that all finite-dimensional irreps of are trivial. On the other hand, Sklyanin algebras associated to points of finite order have an interesting representation theory, due to the following result.

###### Proposition 2.3.

Let be a Sklyanin algebra associated to a point of finite order. Then, all irreducible representations of are finite-dimensional, of at most dimension .

###### Proof.

In this case, we have that the Sklyanin algebra is module-finite over its center by [ATV2, Theorem 7.1]. Further, has PI degree  by [Walton, Proposition 1.6]. Hence, the irreducible representations of are all finite-dimensional by [MR, Theorem 13.10.3(a)], of dimension at most  by [BG, Proposition 3.1]. ∎

Now we analyze parameters so that the automorphism from (1.3) has finite order. Recall that two projective points , are equal if and only if if and only if for all , for some nonzero . Omitting the conditions on parameters for now, it is worth noting the following the result.

###### Lemma 2.4.

The automorphism from (1.3) has order 1 if and only if . In this case, is the commutative polynomial ring .

###### Proof.

If has order 1, then we obtain that . Therefore, , which (by taking the coefficient of ) implies that . Without loss of generality, take . Now, , and we must have that since . Therefore, the forward direction holds. For the converse, note that , so has order 1. The last statement is clear. ∎

Consider the following preliminary results about Sklyanin algebras associated to a point of order 2.

###### Lemma 2.5.

Take to be a 3-dimensional Sklyanin algebra associated to the automorphism of (1.3). Then, if and only if .

###### Proof.

Without loss of generality, take . Consider the following routine with comments.

Let , for .

f1:=c*v^2-b^2*u*w:        g1:=b*c*u^2-v*w:
h1:=b*w^2-c^2*u*v:
f2:=c*g1^2-b^2*f1*h1:     g2:=b*c*f1^2-g1*h1:
h2:=b*h1^2-c^2*f1*g1:


We want , or equivalently, we need that =. Hence, we want the expressions below to be simultaneously zero for some and .

v1:=u*g2-f2*v:    v2:=u*h2-f2*w:       v3:=v*h2-g2*w:


By Definition 1.2, we exclude . Now we extract the coefficients of and solve for .

var:=[u,v,w];
Coeffs:=[coeffs(collect(v1,var,’distributed’),var),
coeffs(collect(v2,var,’distributed’),var),
coeffs(collect(v3,var,’distributed’),var)];
solve([op(Coeffs),b<>0,c<>0],{b,c});
>     {b = 1, c = c}


Hence, and there are no conditions on (other than those in Definition 1.2).

The converse is clear by the computation above, but we can verify this directly. If , then . So,

 σ21,1,c([u:v:w])=[c(cu2−vw)2−(cv2−uw)(w2−c2uv):c(cv2−uw)2−(cu2−vw)(w2−c2uv):(w2−c2uv)2−c2(cv2−uw)(cu2−vw)]=[u(c3u3+c3v3+w3−3c2uvw):v(c3u3+c3v3+w3−3c2uvw):w(c3u3+c3v3+w3−3c2uvw)]=[u:v:w],

as desired. ∎

Hence, to work with Sklyanin algebras associated to a point of order 2, we take .

###### Lemma 2.6.

We have the following statements for the Sklyanin algebra .

1. The only -dimensional representation of is the trivial representation.

2. All irreducible representations of are finite-dimensional, of at most dimension equal to 2.

###### Proof.

(a) One can compute this directly, or by using the short routine below:

solve([x*y+y*x+c*z^2,y*z+z*y+c*x^2, z*x+x*z+c*y^2], [x,y,z]);
>     [[x = 0, y = 0, z = 0]]


(b) This follows from Proposition 2.3 and Lemma 2.5. ∎

## 3. Methodology and Terminology

In this section, we provide an outline of the algorithm used to prove Theorem 1.5; see Sections 4-6 for the full details. The goal is to obtain irreducible representative families of as defined below.

###### Definition 3.1.

We say that a set of matrix solutions of the defining equations of (or of equations (1.1) with ) is a representative family of matrix solutions, if no two members within the set are equivalent. Further, we call this set an irreducible representative family if all of its members are irreducible matrix solutions of .

Note that we aim to have the parameter of free.

So due to Maple’s default alpha ordering, we refer to as zc in the code below.

First, we make the following simplification.

Step 0: Assume that the matrix is in Jordan form. Due to Lemma 2.6 we know that all non-trivial irreps of are of dimension . Hence, we only study matrix solutions of (1.1) with . Initially, the entries of are for . We further simplify the problem by assuming that is in Jordan form. This simplification is made because we wish to classify the irreps up to equivalence, and equivalence is determined by simultaneous conjugation by an invertible matrix. So, we take to be either a single Jordan block or diagonal so that we have 3 or 2 less unknowns, resp. We consider these cases separately.

Step 1: Find all families of matrix solutions. Now, we solve (1.1) with for matrices . The output consists of 2-dimensional (matrix solution) families of . The solutions are grouped according to the default behavior of Maple. We refer to these groups as Families.

Step 2: Eliminate reducible matrix solutions. We run this step now to cut down on the run-time of the algorithm and the complexity of its output. Given a family of matrix solutions, we use Corollary 2.2 to determine if all members of this family are reducible. Namely, we let w =<<p,q>> be a basis of a 1-dimensional subspace of . Note that if and , for and , then precisely when . We examine when is stable under the action of ; namely, we need to be a scalar multiple of . So, we solve for , subject to the following conditions:

• is not the zero subspace p*conjugate(p)+q*conjugate(q)<>0

• p*Xw[2][1]-q*Xw[1][1] = 0

• p*Yw[2][1]-q*Yw[1][1] = 0

• p*Zw[2][1]-q*Zw[1][1] = 0

• conditions on .

If there is a solution, then this implies that all members of the specified family are reducible. We remove such families from further computations by forming a list NonRedFams consisting of families for which there is no satisfying the conditions above.

Steps 3a and 3b are independent of each other, and either can be run after Step 2.

Step 3a: Account for equivalence between families. For the remaining families of matrix solutions, we determine conditions when members of one family NonRedFams[i] is equivalent to members of another family NonRedFams[j]. These conditions are collected in the list BetweenFams.

We do so as follows. First, we force variables of NonRedFams[i] to be in terms of instead of , for ; this is executed with eval(NonRedFams[...],ChangeVars). Next, we conjugate the relabeled matrices simultaneously by a matrix Q to form Xconj, Yconj, Zconj. Then, we solve for variables subject to the following conditions:

• Xconj is equal to the -matrix Xj of NonRedFams[j] Equiv1 = 0

• Yconj is equal to the -matrix Yj of NonRedFams[j]Equiv2 = 0

• Zconj is equal to the -matrix Zj of NonRedFams[j]Equiv3 = 0

• conditions on and invertibility of Q.

The output is [i, j, {conditions on } ], which we interpret as follows.

Interpretation: We can eliminate NonRedFams[i] from our consideration if all of its members are equivalent to members of NonRedFams[j] for some . This occurs if we get an output

 [i, j,...{ each of uℓ,vℓ,wℓ is free }...] for i

We obtain that NonRedFams[i] forms a representative family if we get output

[i, i,...{ restrictions on }...]

under one of the following conditions:

(i) each of , , is free and   (ii) each of , , is free, or depends only on , , ; or

(i) each of , , is free and      (ii) each of , , is free, or depends only on , , .

In either case above, we set the free variables in (ii) equal to 1 to obtain representative families. Otherwise, a careful examination is needed.

Conditions may depend on entries of the matrix Q. In this case, we can conclude that such variables are free as long as this does not violate invertibility of Q.

Step 3b: Check for full irreducibility conditions. Here, we run the same code as in Step 2 except that we solve for along with all variables . The conditions are collected in a list called IrConditions. If the output for NonRedFams[i] is [i] (or empty), then all members of NonRedFams[i] are irreducible.

## 4. Families of non-reducible representations of S(1,1,c)

Here, we execute Steps 0-2 of the algorithm discussed in the previous section. Namely, we find all 2-dimensional representations of by determining matrix solutions to (1.1) with . Here, is in Jordan form, either one Jordan block or two Jordan blocks (diagonal). Moreover, we eliminate the families of solutions for which all of its members are reducible.

Steps 0 and 1  We set up the defining equations.

restart;                            with(LinearAlgebra):


For 1 Jordan block, uncomment #. For 2 Jordan blocks, uncomment ##.

#  X:= <<x1, 0|1, x1>>:  Y:= <<y1, y3|y2, y4>>:  Z:= <<z1, z3|z2, z4>>:
## X:= <<x1, 0|0, x4>>:  Y:= <<y1, y3|y2, y4>>:  Z:= <<z1, z3|z2, z4>>:


Continue by entering the following. Again, we refer to by in the code below.

XX:= Multiply(X,X):                 XY:= Multiply(X,Y):                 XZ:= Multiply(X,Z):
YX:= Multiply(Y,X):                 YY:= Multiply(Y,Y):                 YZ:= Multiply(Y,Z):
ZX:= Multiply(Z,X):                 ZY:= Multiply(Z,Y):                 ZZ:= Multiply(Z,Z):
Eq1:= convert(YZ+ZY+zc*XX,list):    Eq2:= convert(XZ+ZX+zc*YY,list):    Eq3:= convert(XY+YX+zc*ZZ,list):


We enter conditions on and solve for , , , subject to these conditions, to get all 2-dimensional representations of . For 1 Jordan block, uncomment #. For 2 Jordan blocks, uncomment ##.

Conditions:= [zc<>0,zc^3<>-8,zc^3<>1]:
# Vars:= {zc,x1,y1,y2,y3,y4,z1,z2,z3,z4}:
## Vars:= {zc,x1,x4,y1,y2,y3,y4,z1,z2,z3,z4}:
M:= solve([op(Eq1),op(Eq2),op(Eq3),op(Conditions)],Vars):


We need to work with all values of roots in expressions for , call it .

L:= []:
for i from 1 to nops([M]) do        T:=map(allvalues,{M[i]}):
for j from 1 to nops(T) do          L:=[op(L),T[j]]:
end do:       end do:


We build solution families from the list .

Families:=[]:
for i from 1 to nops(L) do          Families:=[op(Families),[eval(X,L[i]),eval(Y,L[i]),eval(Z,L[i])]]:
end do:


Step 2  We now remove families whose members are all reducible.

w:=<<p,q>>:                         NonRedFams:=[]:
for i from 1 to nops(Families) do
Xw:=Multiply(Families[i][1],w):     Yw:=Multiply(Families[i][2],w):     Zw:=Multiply(Families[i][3],w):
NonRed:=solve([p*conjugate(p)+q*conjugate(q)<>0,
p*Xw[2][1]-q*Xw[1][1], p*Yw[2][1]-q*Yw[1][1], p*Zw[2][1]-q*Zw[1][1], op(Conditions)],[p,q]):
Ψif NonRed=[] then                   NonRedFams:=[op(NonRedFams),[Families[i]]]:
Ψend if:  end do:


The output of Steps 0-2 can be viewed by entering the following:

for i from 1 to nops(NonRedFams) do    print(NonRedFams[i]):     end do:


Now, we obtain the results below.

 α=c2z2z3z4+c2z34+(3c4z22z23z24+3c4z2z3z44+c4z64−cz33+c4z33z32)1/2β=−c2z2z3z4−c2z34+(3c4z22z23z24+3c4z2z3z44+c4z64−cz33+c4z33z32)1/2

Table 1. Output of Steps 0-2: NonRedFams for is one Jordan block case

 (1)X=⎛⎜ ⎜⎝cz242y400−cz242y4⎞⎟ ⎟⎠Y=⎛⎝−y4−y34−z34y4y3y3y4⎞⎠Z=⎛⎜⎝−z4−z4(8y34+c3z34)4y24y30z4⎞⎟⎠\inner@par(2)X=(−x400x4)Y=(00y30)Z=⎛⎝0−cx24y300⎞⎠\inner@par(3)X=⎛⎜ ⎜⎝cy242z400−cy242z4⎞⎟ ⎟⎠Y=⎛⎜⎝−y4−y4(8z34+c3y34)4z3z240y4⎞⎟⎠Z=⎛⎝−z4y34−z34z4z3z3z4⎞⎠\inner@par(4)X=(−x400x4)Y=⎛⎝0−cx24z300⎞⎠Z=(00z30)\inner@par(5)X=⎛⎜⎝γc2y3z300−γc2y3z3⎞⎟⎠Y=⎛⎜⎝−y4−2z4γc2y3z3+cy24cy3y3y4⎞⎟⎠Z=⎛⎜⎝−z4−−2γy4c2y3z3+cz24cz3z3z4⎞⎟⎠\inner@par(6)X=⎛⎜⎝−δc2y3z300δc2y3z3⎞⎟⎠Y=⎛⎜ ⎜⎝−y4−2z4δc2y3z3+cy24cy3y3y4⎞⎟ ⎟⎠Z=⎛⎜⎝−z4−2δy4c2y3z3+cz24cz3z3z4⎞⎟⎠
 γ=−z23z4−y23y4+(z43z24+2z23z4y23y4+y43y24+c3y3z33y24+c3y33z3z24−2c3y23z23y4z4)1/2δ=z23z4+y23y4+(z43z24+2z23z4y23y4+y43y24+c3y3z33y24+c3y33z3z24−2c3y23z23y4z4)1/2

Table 2. Output of Steps 0-2: NonRedFams for is two Jordan block case

## 5. Equivalence and Irreducibility: one Jordan block case

We wish to classify the matrix solutions from Steps 0-2 (in the previous section) up to equivalence and extract the irreducible equivalence classes. So, we would like to know under what conditions is a matrix solution equivalent to a member of the same/different solution family. We then specify conditions for which the representative of an equivalence class of matrix solutions is irreducible. This achieved with Steps 3a and 3b, respectively, as described in Section 3. In this section, we continue the algorithm of Section 4 in the case when is one Jordan block.

Step 3a  To execute Step 3a, as described in Section 3, enter the following:

BetweenFams:=[]:                    ChangeVars:=[x1=u1,x4=u4,y1=v1,y2=v2,y3=v3,y4=v4,z1=w1,z2=w2,z3=w3,z4=w4]:
Q:=<<q1,q3|q2,q4>>:                 Qi:=MatrixInverse(Q):
for i from 1 to nops(NonRedFams) do
ΨXconj:=Multiply(Q,Multiply(eval(NonRedFams[i][1][1],ChangeVars),Qi)):
ΨYconj:=Multiply(Q,Multiply(eval(NonRedFams[i][1][2],ChangeVars),Qi)):
ΨZconj:=Multiply(Q,Multiply(eval(NonRedFams[i][1][3],ChangeVars),Qi)):
Ψfor j from i to nops(NonRedFams) do
ΨΨXj:=NonRedFams[j][1][1]:            Yj:=NonRedFams[j][1][2]:             Zj:=NonRedFams[j][1][3]:
ΨΨEquiv1:= convert(Xj-Xconj,list):    Equiv2:= convert(Yj-Yconj,list):     Equiv3:= convert(Zj-Zconj,list):
ΨΨConditions:= [zc<>0,zc^3<>-8,zc^3<>1,q1*q4-q2*q3<>0]:
ΨΨEquiv:= solve([op(Equiv1),op(Equiv2),op(Equiv3),op(Conditions)]):
ΨΨBetweenFams:=[op(BetweenFams),[i,j,Equiv]]:
Ψend do:     end do:


The output of Steps 0-3a can be viewed by entering the following:

for i from 1 to nops(BetweenFams) do      print(BetweenFams[i]):         end do:


Interpretation  Consider the snippets of output:

{Verbatim}

[baselinestretch=0.7] [1, 2, q1 = q1, q2 = q2, q3 = 0, q4 = q1,

3 3 1/2 -4 w4 + v4 - v4 v4 = v4, w4 = w4, y4 = ———————–, z4 = -w4, zc = - —————-, 2 1/2 2 1/2 v4 - 4 3 2 zc q2 w4 [1, 5, q1 = q1, q2 = q2, q3 = 0, q4 = q1, v4 = 0, w4 = w4, y4 = ———, z4 = w4, zc = zc] q1

In the first snippet, one sees that with a choice of and , the parameter can be considered free without violating the invertibility of . We can also conclude that any member of NonRedFams[1] is equivalent to a member of NonRedFams[2], except when , or equivalently when or . From the second snippet of output, we see that any member of NonRedFams[1] is equivalent to a member of NonRedFams[5] when . Moreover by Table 1, we have that in NonRedFams[1] (identified with ) cannot be .

So, we exclude NonRedFams[1] from further computation.

Now consider another two snippets of output:

{Verbatim}

[baselinestretch=0.7] [2, 4, q1 = q1, q2 = q2, q3 = 0, q4 = q1, v4 = v4, w4 = w4, z2 = z2,

2 2 (2 RootOf(_Z + 1 + _Z) w4 q2 - q1 z2) q1 RootOf(_Z + 1 + _Z) w4 q2 - q1 z2 z3 = —————————————–, z4 = - ———————————-, 2 q2 q2 2 2 (2 RootOf(_Z + 1 + _Z) w4 q2 - q1 z2) q1 zc = - ——————————————-] 2 4 3 1/2 2 (v4 + (v4 - 8 w4 v4) ) q2 2 zc q2 w4 [2, 5, q1 = q1, q2 = q2, q3 = 0, q4 = q1, v4 = 0, w4 = w4, y4 = ———, z4 = w4, zc = zc] q1

Through a choice of and , we consider to be free in [2,4,.... We conclude that any member of NonRedFams[2] is equivalent to a member of NonRedFams[4] for all values of and except when , or equivalently when or . From the second snippet of output, we see that if , any member of NonRedFams[2] is equivalent to a member of NonRedFams[5]. From Table 1, we see that (identified with ) in NonRedFams[2] cannot be .

So, we exclude NonRedFams[2] from further computation.

Now take into account the following snippets of output:

{Verbatim}

[baselinestretch=0.7] [3, 4, q1 = q1, q2 = q2, q3 = 0, q4 = q1, w2 = w2, w3 = w3, w4 = w4,

2 2 2 q1 w4 q2 + w2 q1 - w3 q2 -w3 q2 + q1 w4 z2 = —————————-, z3 = w3, z4 = ————–, zc = zc] 2 q1 q1

[4, 5]

This implies that NonRedFams[3] is equivalent to NonRedFams[4].

So, we exclude NonRedFams[3] from further computation.

Further, no member of NonRedFams[4] is equivalent to a member of NonRedFams[5].

Finally, we determine when the remaining families are representative families. Consider:

{Verbatim}

[baselinestretch=0.7] [4, 4, q1 = q1, q2 = q2, q3 = 0, q4 = q1, w2 = w2, w3 = w3, w4 = w4,

2 2 2 q1 w4 q2 + w2 q1 - w3 q2 -w3 q2 + q1 w4 z2 = —————————-, z3 = w3, z4 = ————–, zc = zc] 2 q1 q1

q1 (-y4 + v4) [5, 5, q1 = q1, q2 = - ————-, q3 = 0, q4 = q1, v4 = v4, w4 = w4, y4 = y4, z4 = w4, zc = zc] 2 zc w4

We get that a member of NonRedFams[5] is equivalent to another member of this family for any value of . Without loss of generality, set .

So, NonRedFams[5] is a representative family with .

In NonRedFams[4], we obtain any value for , say , by setting . (Note that by Table 1, , identified by , is not equal to 0.) This choice of does not violate the invertibility of . Further, it is easy to check that in this case, . Thus, without loss of generality, set

So, NonRedFams[4] is a representative family with .

Step 3b  Given the results above, we only need to execute this step for NonRedFams[4] and NonRedFams[5], but we complete this for the whole list NonRedFams as follows:

IrConditions:=[]:
for i from 1 to nops(NonRedFams) do
Xw:=Multiply(NonRedFams[i][1][1],w): Yw:=Multiply(NonRedFams[i][1][2],w): Zw:=Multiply(NonRedFams[i][1][3],w):
Ir:=solve([p*conjugate(p)+q*conjugate(q)<>0,
p*Xw[2][1]-q*Xw[1][1],p*Yw[2][1]-q*Yw[1][1],p*Zw[2][1]-q*Zw[1][1], zc<>0,zc^3<>1,zc^3<>-8]):
IrConditions:=[op(IrConditions),[i,Ir]]:
end do:


To see the output, enter:

for i from 1 to nops(IrConditions) do      print(IrConditions[i]):       end do:


One gets that, for each , all members of NonRedFams[i] are irreducible matrix solutions of .

Conclusion  By entering eval(NonRedFams[4],[z4=1]); and eval(NonRedFams[5],[y4=1]);, one obtains the representatives of equivalence classes of irreducible matrix solutions of equations (1.1), where is assumed to be one Jordan block. The output is listed in the table below.

 X=(0100)Y=⎛⎜ ⎜⎝−βcz3−−cz22z3−cz2+2βcz3z3−cz2z3−cβcz3⎞⎟ ⎟⎠Z=(−1z2z31)X=(0100)Y=⎛⎝−11cz24−cz241⎞⎠Z=(−z42cz40z4)
 β=−c2z2z3−c2+(3c4z22z23+3c4z2z3+c4−cz33+c4z33z32)1/2

Table 3. Representatives of equivalences classes of 2-dimensional irreps of , when is one Jordan block

## 6. Equivalence and Irreducibility: two Jordan block case

As in the one Jordan block case, we wish to classify the matrix solutions from Steps 0-2 (in Section 4) up to equivalence and extract the irreducible equivalence classes. So, we would like to know under what conditions is a matrix solution equivalent to a member of the same/different solution family. We then specify conditions for which the representative of an equivalence class of matrix solutions is irreducible. This achieved with Steps 3a and 3b, respectively, as described in Section 3. In this section, we continue the algorithm of Section 4 in the case when is two Jordan blocks.

Step 3a  To execute Step 3a, as described in Section 3, enter the code for Step 3a provided in Section 5. (The memory and time for this operation was 27068.0 MB and 523.78 seconds, respectively.) The output of Steps 0-3a can be viewed by entering the following:

for i from 1 to nops(BetweenFams) do      print(BetweenFams[i]):         end do:


Interpretation  Consider the following snippet of output:

{Verbatim}

[baselinestretch=0.7] v3 q4 [1, 1, q1 = —–, q2 = 0, q3 = 0, q4 = q4, v3 = v3, v4 = y4, w4 = z4, y3 = y3, y4 = y4, z4 = z4, zc = zc y3

Note that in NonRedFams[1] by Table 2.

So, NonRedFams[1] is a representative family with (identified with ) is 1 without loss of generality.

Now take:

{Verbatim}

[baselinestretch=0.7] v3 q4 [2, 2, q1 = —–, q2 = 0, q3 = 0, q4 = q4, u4 = x4, v3 = v3, x4 = x4, y3 = y3, zc = zc] y3

Note that in NonRedFams[2] by Table 2.

So, NonRedFams[2] is a representative family with (identified with ) is 1 without loss of generality.

Consider the output:

{Verbatim}

[baselinestretch=0.7] w3 q4 [3, 3, q1 = —–, q2 = 0, q3 = 0, q4 = q4, v4 = y4, w3 = w3, w4 = z4, y4 = y4, z3 = z3, z4 = z4, zc = zc z3

Note that in NonRedFams[3] by Table 2.

So, NonRedFams[3] is a representative family with (identified with ) is 1 without loss of generality.

Next, consider the snippet of output below:

{Verbatim}

[baselinestretch=0.7] 2 zc x4 q3 [2, 4, q1 = 0, q2 = - ———, q3 = q3, q4 = 0, u4 = -x4, v3 = v3, x4 = x4, z3 = z3, zc = zc] z3 v3

By Table 2, we have that for NonRedFams[4]. So by the output above, we get that any member of NonRedFams[4] is equivalent to a member NonRedFams[2].

We exclude NonRedFams[4] from further computation.

Consider the output:

{Verbatim}

[baselinestretch=0.7] v3 q4 z3 v3 [5, 5, q1 = —–, q2 = 0, q3 = 0, q4 = q4, v3 = v3, v4 = y4, w3 = —–, w4 = z4, y3 y3

y3 = y3, y4 = y4, z3 = z3, z4 = z4, zc = zc

We have that in NonRedFams[5] by Table 2. Without loss of generality, we can take (identified with ) to be 1. In this case, .

So, NonRedFams[5] is a representative family with .

Now let us take:

{Verbatim}

[baselinestretch=0.7] v3 q4 z3 v3 [5, 6, q1 = —–, q2 = 0, q3 = 0, q4 = q4, v3 = v3, v4 = y4, w3 = —–, w4 = z4, y3 y3

y3 = y3, y4 = y4, z3 = z3, z4 = z4, zc = zc]

Note that by Table 2, we have for NonRedFams[6]. So by the output above, we get that any member of NonRedFams[6] is equivalent to a member NonRedFams[5].

We exclude NonRedFams[6] from further computation.

But we still need to analyze the equivalence between members of NonRedFams[1], NonRedFams[2], NonRedFams[3], NonRedFams[5]. In this case, the output is easier to interpret if we run Step 3b before Step 3a again.

Step 3b  Given the results above, we only need to execute this step for NonRedFams[1], NonRedFams[2], NonRedFams[3], NonRedFams[5], but we complete this for the whole list NonRedFams by entering the code for Step 3b provided in Section 5. Consider the snippets:

{Verbatim}

[baselinestretch=0.7] y4 q [1, p = - —-, q = q, y3 = y3, y4 = y4, z4 = 0, zc = zc] y3

[2, p = 0, q = q, x4 = 0, y3 = y3, zc = zc]

z4 q [3, p = - —-, q = q, y4 = 0, z3 = z3, z4 = z4, zc = zc] z3

2 [5, p = 0, q = q, y3 = 0, y4 = RootOf(_Z + 1 + _Z) z4, z3 = z3, z4 = z4, zc = zc,

z4 q z4 y3 p = - —-, q = q, y3 = y3, y4 = —–, z3 = z3, z4 = z4, zc = zc] z3 z3

We obtain that:

members of NonRedFams[1], NonRedFams[2], NonRedFams[3] are irreducible precisely when , , , respectively, and

members of NonRedFams[5] are irreducible precisely when {, } or {}.

Step 3a again  We execute Step 3a again for

{Verbatim}

NewNonRedFams[1]:=eval(NonRedFams[1],[y3=1]): NewNonRedFams[2]:=eval(NonRedFams[2],[y3=1]): NewNonRedFams[3]:=eval(NonRedFams[3],[z3=1]): NewNonRedFams[4]:=eval(NonRedFams[5],[y3=1]):

The code and output is:

{Verbatim}

NewBetweenFams:=[]: ChangeVars:=[x1=u1,x4=u4,y1=v1,y2=v2,y3=v3,y4=v4,z1=w1,z2=w2,z3=w3,z4=w4]: Q:=¡¡q1,q3—q2,q4¿¿: Qi:=MatrixInverse(Q): for i from 1 to 4 do Xconj:=Multiply(Q,Multiply(eval(NewNonRedFams[i][1][1],ChangeVars),Qi)): Yconj:=Multiply(Q,Multiply(eval(NewNonRedFams[i][1][2],ChangeVars),Qi)): Zconj:=Multiply(Q,Multiply(eval(NewNonRedFams[i][1][3],ChangeVars),Qi)): for j from i+1 to 4 do Xj:=NewNonRedFams[j][1][1]: Yj:=NewNonRedFams[j][1][2]: Zj:=NewNonRedFams[j][1][3]: Equiv1:= convert(Xj-Xconj,list): Equiv2:= convert(Yj-Yconj,list): Equiv3:= convert(Zj-Zconj,list): Conditions:= [zc¡¿0,zc^3¡¿-8,zc^3¡¿1,q1*q4-q2*q3¡¿0]: Equiv:= solve([op(Equiv1),op(Equiv2),op(Equiv3),op(Conditions)]): NewBetweenFams:=[op(NewBetweenFams),[i,j,Equiv]]: end do: end do:

for i from 1 to nops(NewBetweenFams) do print(NewBetweenFams[i]): end do:

{Verbatim}

[baselinestretch=0.7] ¿ [1, 2, q1 = q1, q2 = q1 v4, q3 = 0, q4 = q1, v4 = v4, w4 = 0, x4 = 0, zc = zc,

2 q1 = -q3 v4, q2 = -v4 q3, q3 = q3, q4 = 0, v4 = v4, w4 = 0, x4 = 0, zc = zc,

q1 (q1 - q4) q2 q1 = q1, q2 = q2, q3 = - ————, q4 = q4, v4 = —-, w4 = 0, x4 = 0, zc = zc] q2 q1

[1, 3, q1 = 0, q2 = q2, q3 = q3, q4 = 0, v4 = -y4, w4 = -y4, y4 = y4, z4 = y4,

3 2 2 zc = RootOf(q3 _Z y4 + 8 q3 y4 + 4 q2),

2 q1 = 0, q2 = q2, q3 = q3, q4 = 0, v4 = -y4, w4 = -RootOf(_Z + 1 + _Z) y4, y4 = y4,

2 3 2 2 2 z4 = RootOf(_Z + 1 + _Z) y4, zc = RootOf(q3 _Z y4 + 8 q3 y4 - 4 q2 RootOf(_Z + 1 + _Z) - 4 q2)]

[1, 4] [2, 3] [2, 4] [3, 4]

We obtain that in NewNonRedFams[1] precisely when any member of NewNonRedFams[1] is equivalent to a member of NewNonRedFams[2]. On the other hand, we have that in NewNonRedFams[2] precisely when any member of NewNonRedFams[2] is equivalent to a member of NewNonRedFams[1]. But members of NewNonRedFams[1] and NewNonRedFams[2] are reducible when and , respectively.

Now by a choice of , , we can consider to be free in [1, 3, ...]. So, we get that for in NewNonRedFams[3] precisely when any member of NewNonRedFams[3] is equivalent to a member of NewNonRedFams[1].

Putting this together we conclude that:

NewNonRedFams[1]=eval(NonRedFams[1],[y3=1]) is an irreducible representative family when .

NewNonRedFams[2]=eval(NonRedFams[2],[y3=1]) is an irreducible representative family when .

NewNonRedFams[3]=eval(NonRedFams[3],[z3=1]) is an irreducible representative family when , and there is no overlap with NewNonRedFams[1] when for .

NewNonRedFams[4]=eval(NonRedFams[5],[y3=1]) is an irreducible representative family when .

Conclusion  We obtain the following representatives of equivalence classes of irreducible matrix solutions of equations (1.1), where is assumed to be two Jordan blocks.

 X=⎛⎜ ⎜⎝cz242y400−cz242y4⎞⎟ ⎟⎠Y=⎛⎝−y4−y34−z34y41y4⎞⎠Z=⎛⎜⎝−z4−z4(8y34+c3z34)4y240z4⎞⎟⎠ for z4≠0\inner@parX=(−x400x4)Y=(0010)Z=(0−c2x2400) for x4≠0\inner@parX=⎛⎜ ⎜⎝cy242z400−cy242z4⎞⎟ ⎟⎠Y=⎛⎜⎝−y4−y4(8z34+c3y34)4z240y4⎞⎟⎠Z=⎛⎝−z4y34−z34z41z4⎞⎠ for \begin{tabular}[]{l}y4≠0\ z4≠ζy4, ζ3=1\end{tabular}\inner@parX=⎛⎜⎝γc2z300−γc2z3⎞⎟⎠Y=⎛⎜⎝−y4−2z4γc2z3+cy24c1y4⎞⎟⎠Z=⎛⎜⎝−z4−−2γy4c2z3+cz24cz3z3z4⎞⎟⎠ for \begin{tabular}[]{l}y4≠e±2πi/3z4\ z4≠y4z3\end{tabular}
 γ=−z23z4−y4+(z43z24+2z23z4y4+y24+c3z33y24+c3z3z24−2c3z23y4z4)1/2

Table 4. Representatives of equivalences classes of 2-dimensional irreps of , when is two Jordan blocks

## 7. Geometric parametrization of irreducible representations of S(1,1,c)

Since the Sklyanin algebra is module finite over its center, we can use the center of to provide a geometric parametrization of the set of equivalence classes of irreducible representations of . (Recall by Definition 1.2, , .) Namely, we depict the Azumaya locus of over its center [BGbook, III.1.7]. We refer the reader to [Smith] for an introduction to affine varieties.

###### Theorem 7.1.

Let be the center of the Sklyanin algebra .

1. We have that