Classification of Quantum Groups via Galois Cohomology

Abstract

The first example of a quantum group was introduced by P. Kulish and N. Reshetikhin. In the paper Kulish et al. (J Soviet Math 23:2435–2441, 1983), they found a new algebra which was later called $U_q (\mathfrak {sl}(2))$. Their example was developed independently by V. Drinfeld and M. Jimbo, which resulted in the general notion of quantum group. Later, a complimentary approach to quantum groups was developed by L. Faddeev, N. Reshetikhin, and L. Takhtajan in Faddeev et al. (Leningr Math J 1:193–225, 1990). Recently, the so-called Belavin–Drinfeld cohomology (twisted and non-twisted) have been introduced in the literature to study and classify certain families of quantum groups and Lie bialgebras. Later, the last two authors interpreted non-twisted Belavin–Drinfeld cohomology in terms of non-abelian Galois cohomology $H^1({\mathbb {F}}, {\mathbf {H}})$ for a suitable algebraic ${\mathbb {F}}$-group ${\mathbf {H}}$. Here ${\mathbb {F}}$ is an arbitrary field of zero characteristic. The non-twisted case is thus fully understood in terms of Galois cohomology. The twisted case has only been studied using Galois cohomology for the so-called (“standard”) Drinfeld–Jimbo structure. The aim of the present paper is to extend these results to all twisted Belavin–Drinfeld cohomology and thus, to present classification of quantum groups in terms of Galois cohomology and the so-called orders. Low dimensional cases $\mathfrak {sl}(2)$ and $\mathfrak {sl}(3)$ are considered in more details using a theory of cubic rings developed by B. N. Delone and D. K. Faddeev in Delone and Faddeev (The theory of irrationalities of the third degree. Translations of mathematical monographs, vol 10. AMS, Providence, 1964). Our results show that there exist yet unknown quantum groups for Lie algebras of the types $A_n, D_{2n+1}, E_6$, not mentioned in Etingof et al. (J Am Math Soc 13:595–609, 2000).

Appendices

Double Cosets and Orders (by Juliusz Brzezinski and A. Stolin)

1.1 Double Cosets and Orders in ${\mathbb {K}}^n$

In this subsection, we consider ${\mathbb {K}}^n$ as a ${\mathbb {K}}$-algebra with ${\mathbb {K}}$ embedded diagonally into ${\mathbb {K}}^n$. Our purpose is to describe the double cosets which we discussed in the preceding section in terms of ${\mathbb {O}}$-orders in the algebra ${\mathbb {K}}^n$.

Definition A.1

An ${\mathbb {O}}$-module $M\subset {\mathbb {K}}^n$ is called a lattice on ${\mathbb {K}}^n$ if its rank over ${\mathbb {O}}$ is equal to n.

Clearly, ${\mathbf {GL}}(n, {\mathbb {K}})$ acts transitively on the set of lattices in ${\mathbb {K}}^n$ because any lattice has a form $M=G\cdot {\mathbb {O}}^n$ for some $G \in {\mathbf {GL}}(n, {\mathbb {K}})$. Hence,

$$\begin{aligned} {\mathbf {GL}}(n, {\mathbb {K}})/{\mathbf {GL}}(n, {\mathbb {O}})\cong \{\mathrm {lattices}\ \mathrm {in}\ {\mathbb {K}}^n\}. \end{aligned}$$

Definition A.2

An order in ${\mathbb {K}}^n$ is a subring $\Lambda$ of ${\mathbb {K}}^n$ containing ${\mathbb {O}}$, finitely generated as an ${\mathbb {O}}$-module and such that $\Lambda {\mathbb {K}}= {\mathbb {K}}^n$.

Definition A.3

Let $M\subset {\mathbb {K}}^n$ be a lattice. Then $I(M)=\{x\in {\mathbb {K}}^n:\ xM\subset M \}$ is called the set of multipliers of M.

The following lemma is well known.

Lemma A.4

• 1) Any order $\Lambda$ is contained in ${\mathbb {O}}^n$.

• 2) For any order $\Lambda$, we have $I(\Lambda )=\Lambda$.

• 3) For any lattice M, I(M) is an order. $\quad \square$

Proposition A.5

There is a canonical surjection

$$\begin{aligned}&{\mathbf {GL}}(n, {\mathbb {O}})\backslash {\mathbf {GL}}(n, {\mathbb {K}})/\mathbf {Diag}(n, {\mathbb {K}})\cong \mathbf {Diag}(n, {\mathbb {K}})\backslash {\mathbf {GL}}(n, {\mathbb {K}})/{\mathbf {GL}}(n, {\mathbb {O}})\rightarrow \\&\quad \{\mathrm {orders}\ \mathrm {in}\ {\mathbb {K}}^n \}. \end{aligned}$$

Proof

Consider two lattices in ${\mathbb {K}}^n$, $M_1=G\cdot {\mathbb {O}}^n$ and $M_2=H\cdot G\cdot {\mathbb {O}}^n$, where $H\in \mathbf {Diag}(n, {\mathbb {K}})$. Clearly, multiplication by $H=\mathrm {diag}\,(a_1,\ldots ,a_n)$ coincides with multiplication by $h=(a_1,\ldots ,a_n)\in {\mathbb {K}}^n$. Let $g\in I(M_1)$. Since the ring ${\mathbb {K}}^n$ is commutative, it follows that $g\in I(M_2)$ and so, by symmetry we have $I(M_1)=I(M_2)$. Thus, the correspondence $G\mapsto I(M_1)$ defines the required map

$$\begin{aligned} \omega _n: \mathbf {Diag}(n, {\mathbb {K}})\backslash {\mathbf {GL}}(n, {\mathbb {K}})/{\mathbf {GL}}(n, {\mathbb {O}})\rightarrow \{\mathrm {orders}\ \mathrm {in}\ {\mathbb {K}}^n \}. \end{aligned}$$

It is a surjection because for any order $\Lambda$ we have $I(\Lambda )=\Lambda$. $\quad \square$

Generally speaking, the map defined above is not injective. Let us define its kernel in the sense of sets. More exactly, for any order $\Lambda$ we will find the subset $\omega _n^{-1} (\Lambda )$.

Definition A.6

Given an order $\Lambda$, we say that a lattice Mbelongs to $\Lambda$ if $\Lambda =I(M)$.

It is clear that M and $h\cdot M$, $h\in {\mathbb {K}}^n$, belong to the same order $\Lambda =I(M)$.

Definition A.7

We say that two lattices $M_1$ and $M_2$ are in the same lattice class of $\Lambda$ if $M_1=hM_2$ for some $h\in {\mathbb {K}}^n$.

Let us consider a canonical map $\omega : \{\mathrm {lattices}\ \mathrm {in}\ {\mathbb {K}}^n\}\rightarrow \{\mathrm {orders}\ \mathrm {in}\ {\mathbb {K}}^n\}$ defined as $M\mapsto I(M)$. The following proposition is obvious.

Proposition A.8

$\omega (M_1)=\omega (M_2)$ if $M_1$ and $M_2$ belong to the same lattice class. $\quad \square$

Remark A.9

We remind readers that ${\mathbf {GL}}(n, {\mathbb {K}})/{\mathbf {GL}}(n, {\mathbb {O}})\cong \{\mathrm {lattices}\ \mathrm {in}\ {\mathbb {K}}^n\}$. Therefore, we can define a map

$$\begin{aligned} d: \{\mathrm {lattices}\ \mathrm {in}\ {\mathbb {K}}^n\}\rightarrow \mathbf {Diag}(n, {\mathbb {K}})\backslash {\mathbf {GL}}(n, {\mathbb {K}})/{\mathbf {GL}}(n, {\mathbb {O}}) \end{aligned}$$

and it is easy to see that $\omega (M)=\omega _n (d(M))$. Moreover,

$$\begin{aligned} \mathbf {Diag}(n, {\mathbb {K}})\backslash {\mathbf {GL}}(n, {\mathbb {K}})/{\mathbf {GL}}(n, {\mathbb {O}})\cong \bigcup _{\Lambda \subset {\mathbb {O}}^n} \ \{\mathrm {lattice\ classes}\ \mathrm {belonging}\ \mathrm {to}\ \Lambda \}. \end{aligned}$$

Let us fix an order $\Lambda$ and consider the set of lattices $L(\Lambda )$ belonging to $\Lambda$. If $M\in L(\Lambda )$, then $hM\in L(\Lambda )$. Therefore, $L(\Lambda )$ is a disjoint union of lattice classes. Let us denote the number of such classes by $\mathrm {lc}(\Lambda )$. The number $\mathrm {lc}(\Lambda )$ is finite because $\Lambda$ is finitely generated over ${\mathbb {O}}$, which is a discrete valuation ring. The following proposition is a direct consequence of the remark above.

Proposition A.10

$\omega _n^{-1}(\Lambda ) =\{\mathrm {lattice\ classes}\ \mathrm {belonging}\ \mathrm {to}\ \Lambda \}$ and hence, $\omega _n^{-1}(\Lambda )$ consists of $\mathrm {lc}(\Lambda )$ elements. $\quad \square$

The result below was proved by Brzezinski in [4].

Theorem A.11

$\mathrm {lc}(\Lambda )=1$ if and only if $\Lambda$ is a Gorenstein ring. $\quad \square$

1.2 Quantum Groups Over $\mathfrak {sl}(2)$

We begin with a corollary to Brzezinski’s theorem.

Corollary A.12

The map $\omega _2$ is a bijection.

Proof

Let $\Lambda$ be an order in ${\mathbb {O}}^2$. Then it is of the form $\Lambda ={\mathbb {O}}[y]$, where y satisfies a quadratic equation $y^2+ay+b=0$ with $a,b\in {\mathbb {O}}$. It is known that such a ring is Gorenstein. Therefore, $\mathrm {lc}(\Lambda )=1$ and $\omega _2$ is a bijection. $\quad \square$

Proposition A.13

Any order $\Lambda \subset {\mathbb {O}}^2$ is a free ${\mathbb {O}}$-module $\Lambda _n$ with a basis $\{(1,1), (t^n,0)\}$, $n=0,1,\ldots$. The orders $\Lambda _{n_1}$ and $\Lambda _{n_2}$ are not isomorphic if $n_1\ne n_2$ and hence, quantum groups of non-twisted type over $\mathfrak {sl}(2)$ are indexed by non-negative integers.

Proof

Let $\Lambda$ have a basis $\{(1,1), (a,b)\}$ with $a,b\in {\mathbb {O}}$. Then $\{(1,1), (a-b,0)\}$ is also a basis. Let $a-b=xt^n$, where $x\in {\mathbb {O}}$ is invertible and n is a non-negative integer number. Therefore, $\{(1,1), (t^n,0)\}$ is a basis. The rest is clear. $\quad \square$

Let us also discuss the twisted case, in other words the double cosets

$$\begin{aligned} {\mathbf {D}}_2\backslash J^{-1}{\mathbf {GL}}(2, {\mathbb {K}})J /J^{-1}{\mathbf {GL}}(2, {\mathbb {O}})J, \end{aligned}$$

where ${\mathbf {D}}_2=\{\mathrm {diag}\,(d,\gamma _1 (d)): \ d\in {\mathbb {L}}\}$.

The lemma below is straightforward.

Lemma A.14

$J^{-1}{\mathbf {GL}}(2, {\mathbb {K}})J ={\mathbf {U}}(1,1)$. $\quad \square$

Here, in an analogy with the real numbers, we denote by ${\mathbf {U}}(1,1)$ the group which consists of matrices of the form

$$\begin{aligned} P=\left( \begin{array}{cc} x &{} y\\ \gamma _1 (y) &{} \gamma _1 (x)\\ \end{array}\right) \end{aligned}$$

with $x,y\in {\mathbb {L}}$.

The group ${\mathbf {U}}(1,1)$ acts naturally on ${\mathbb {L}}$ via the formula $Pd=xd+y\gamma _1 (d)$. In fact, this action comes from the natural action of ${\mathbf {U}}(1,1)$ on ${\mathbb {L}}^2$ and the embedding ${\mathbb {L}}\rightarrow {\mathbb {L}}^2$, $d\mapsto (d,\gamma _1 (d))$.

Now we can repeat the non-twisted considerations above.

Definition A.15

$M\subset {\mathbb {L}}$ is a lattice in ${\mathbb {L}}$ if it is an ${\mathbb {O}}$-submodule of ${\mathbb {L}}$ of rank 2.

It is not difficult to show that

$$\begin{aligned} J^{-1}{\mathbf {GL}}(2, {\mathbb {K}})J /J^{-1}{\mathbf {GL}}(2, {\mathbb {O}})J\cong \{\mathrm {lattices}\ \mathrm {in}\ {\mathbb {L}}\}. \end{aligned}$$

Definition A.16

$\Lambda \subset {\mathbb {L}}$ is an order in ${\mathbb {L}}$ if it is a lattice and a sub-ring of ${\mathbb {L}}$ which contains the unit of ${\mathbb {L}}$.

Remark A.17

One can show that in fact $\Lambda \subset {\mathbb {O}}_{\mathbb {L}}={\mathbb {O}}[j]$.

Using the result by Brzezinski [4], we deduce the final classification of the twisted quantum groups for $\mathfrak {sl}(2)$.

Theorem A.18

There is a canonical bijection

$$\begin{aligned} \rho : {\mathbf {D}}_2\backslash J^{-1}{\mathbf {GL}}(2, {\mathbb {K}})J /J^{-1}{\mathbf {GL}}(2, {\mathbb {O}})J\rightarrow \{\mathrm {orders}\ \mathrm {in}\ {\mathbb {L}}\} = \{{\mathbb {O}}[t^{n+\frac{1}{2} }] :\ n\in {{\mathbb {Z}}}_+\}. \end{aligned}$$

Proof

As in the non-twisted case, we have to show that any order $\Lambda$ in ${\mathbb {L}}$ is a Gorenstein ring, which is clear because it can be easily shown that in this case $\Lambda ={\mathbb {O}}[t^{n+\frac{1}{2}}]$, $n\in {{\mathbb {Z}}}_+$. $\quad \square$

Corollary A.19

Quantum groups such that their classical limit is $\mathfrak {sl}(2)$ are in a one-to-one correspondence with the set of orders in separable quadratic rings, i.e. ${\mathbb {O}}^2$ and ${\mathbb {O}}_{\mathbb {L}}$. The corresponding orders were described above. $\quad \square$

1.3 Quantum Groups Over $\mathfrak {sl} (3)$ and Orders in Cubic Rings

The aim of this subsection is to classify quantum groups such that their classical limit is $\mathfrak {sl}(3)$ with the Lie bialgebra structure defined by $r_{\mathrm{DJ}}$ and $jr_{\mathrm{DJ}}$ in terms of cubic rings.

Our considerations are based on results about orders in cubic rings obtained in [5] and [11], see also [2] and [13]. We begin with the non-twisted case. If $n=3$, the bijection

$$\begin{aligned} \mathbf {Diag}(3, {\mathbb {K}})\backslash {\mathbf {GL}}(3, {\mathbb {K}})/{\mathbf {GL}}(3, {\mathbb {O}})\cong \bigcup _{\Lambda \subset {\mathbb {O}}^3} \ \{\mathrm {lattice\ classes}\ \mathrm {belonging}\ \mathrm {to}\ \Lambda \}. \end{aligned}$$

has been already constructed.

Let us turn to the twisted case. We have

$$\begin{aligned} J=J_{3}=\left( \begin{array}{ccc} 1 &{} 0 &{} 1\\ 0 &{} 1 &{} 0 \\ -j &{} 0 &{} j \\ \end{array}\right) \end{aligned}$$

(see [15]). Because of this particular form of $J_3$, our treatment of the case $n=3$ is very similar to the case $n=2$.

Let us present an element of ${\mathbb {L}}\oplus {\mathbb {K}}$ in the form $(x,a,\gamma _1 (x))$, where $x\in {\mathbb {L}}$, $a\in {\mathbb {K}}$. Then, it is clear that there is a bijection of sets

$$\begin{aligned} J_3^{-1}{\mathbf {GL}}(3, {\mathbb {K}})J_3 /J_3^{-1}{\mathbf {GL}}(3, {\mathbb {O}})J_3 \cong \{\mathrm{lattices}\ \mathrm{in}\ {\mathbb {L}}\oplus {\mathbb {K}}\}. \end{aligned}$$

Let us define ${\mathbf {D}}_3=\{\mathrm {diag}\,(d,a,\gamma _1 (d)): d\in {\mathbb {L}},a\in {\mathbb {K}}\}$.

Let N be a lattice in ${\mathbb {L}}\oplus {\mathbb {K}}$. Let us define the ring of multipliers of N as $I(N)=\{x\in {\mathbf {D}}_3: \ xN\subset N\}$. Clearly, $I(N)\subset {\mathbb {O}}_{\mathbb {L}}\oplus {\mathbb {O}}$ is an order. The following result takes place.

Theorem A.20

1) Quantum groups of the twisted type which quantize the Lie bialgebra structure on $\mathfrak {sl}(3)$ defined by $jr_\mathrm {DJ}$ are parameterized by

$$\begin{aligned} {\mathbf {D}}_3\backslash J_3^{-1}{\mathbf {GL}}(3, {\mathbb {K}})J_3 /J_3^{-1}{\mathbf {GL}}(3, {\mathbb {O}})J_3 \end{aligned}$$

2) There is a natural surjection

$$\begin{aligned} \rho _3: {\mathbf {D}}_3\backslash J_3^{-1}{\mathbf {GL}}(3, {\mathbb {K}})J_3 /J_3^{-1}{\mathbf {GL}}(3, {\mathbb {O}})J_3 \rightarrow \{\mathrm {orders}\ \mathrm {in}\ {\mathbb {O}}_{{\mathbb {L}}}\oplus {\mathbb {O}}\}. \end{aligned}$$

$\square$

Given an order $\Lambda \subset {\mathbb {O}}_{\mathbb {L}}\oplus {\mathbb {O}}$, we say that a lattice Nbelongs to $\Lambda$ if $I(N)=\Lambda$. Further, we say that two lattices $N_1$ and $N_2$ are in the same lattice class if $N_2=xN_1$ for some $x\in {\mathbb {L}}\oplus {\mathbb {K}}$. Clearly, $I(N_1)=I(N_2)$ and the set of lattices belonging to $\Lambda$ is a disjoint union of lattice classes.

Corollary A.21

$$\begin{aligned}&{\mathbf {D}}_3\backslash J_3^{-1}{\mathbf {GL}}(3, {\mathbb {K}})J_3 /J_3^{-1}{\mathbf {GL}}(3, {\mathbb {O}})J_3\\&\quad \cong \bigcup _{\Lambda \subset {\mathbb {O}}_{\mathbb {L}}\oplus {\mathbb {O}}} \ \{\mathrm {lattice\ classes}\ \mathrm {belonging}\ \mathrm {to}\ \Lambda \}. \end{aligned}$$

$\square$

Further, we need to study orders in cubic rings ${\mathbb {K}}^3$ and ${\mathbb {L}}\oplus {\mathbb {K}}$. In the next two subsections, we give two approaches to this description.

1.4 Classification of Cubic Orders Contained in Separable Cubic Algebras I

We begin with a general construction of cubic rings following [5], see also [2, 13]. Let R be a discrete valuation ring (e.g., $R={\mathbb {O}}$) and K its quotient field. Assume that any quadratic field extension of K is generated by an element of R whose square equals a generator of the maximal ideal of R. Let A be a cubic separable K-algebra. For every R-order $\Lambda$ in A, write $\Lambda = R+R\omega +R\theta$. Translating $\omega$ and $\theta$ by appropriate elements of R, we can achieve that $\omega \theta =n\in R$. Such a basis we will call normal. So, we got the following multiplication table:

$$\begin{aligned} \omega \theta =n, \quad \omega ^2= m +b\omega -a\theta , \quad \theta ^2 =l+d\omega -c\theta , \end{aligned}$$

where $a,b,c,d,l,m,n\in R$. One can show that the associative law implies that $(n,m,l)=(-ad,-ac,-bd)$, i.e., we get

$$\begin{aligned} \omega \theta = -ad, \quad \omega ^2= -ac +b\omega -a\theta , \quad \theta ^2 =-bd + d\omega -c\theta . \end{aligned}$$

(A.22)

Now let us consider the index form$f(x,y)=ax^3+bx^2y+cxy^2+dy^3$ of $\Lambda$. Notice that the index form f determines $\Lambda =\Lambda (f)=\Lambda _{abcd}$ uniquely up to an isomorphism.

Let $P_\omega (X)=X^3 -bX^2 + acX- a^2d$ and $P_\theta (X)= X^3+cX^2+bdX+ad^2$.

Lemma A.23

$P_\theta (\theta )=0$ and $P_\omega (\omega )=0$.

Proof

To derive the first equation, we multiply both sides of the third relation in the multiplication table above by $\theta$ and take into account that $\omega \theta =n=-ad$. We get the second equation similarly. $\quad \square$

Remark A.24

If $ad\ne 0$, then $P_{\theta }(-ad/X)=(ad^2 /X^3)P_{\omega }(X)$. If $a=1$, then $P_{\omega } (X)=f(X,-1)$.

Theorem A.25

If $A=K\Lambda (f)$, then A is a field if and only if $P_\omega (X)$ is irreducible over K.

Proof

Let A be a field. Since $\omega \in A {\setminus } K$ is a zero of the polynomial $P_\omega (X)$ of degree 3, this polynomial is minimal for $\omega$ over K. Thus, it is irreducible over K. Conversely, if $P_\omega (X)$ is irreducible over K,  then $K(\omega )$ is a field extension of degree 3 over K, so $K(\omega )=A$. $\quad \square$

Remark A.26

Clearly, if $P_\omega (X)$ is irreducible, then $P_\theta (X)$ is also irreducible because irreducibility of $P_\omega (X)$ implies that $ad\ne 0$, and then we can use Remark A.24.

As we know, if A is a separable algebra of degree 3 over K, then A is either a (separable) field extension of K, or A is isomorphic to a product of a quadratic (separable) field extension L of K by K, or A is isomorphic to $K^3$. If $A=K\Lambda (f)$, then we already know that A is a field if and only if $P_\omega (X)$ (and $P_\theta (X)$) are irreducible. Moreover, the algebra $A=K\Lambda (f)$ is separable if and only if the discriminant

$$\begin{aligned} \Delta (f) = 18abcd + b^2 c^2-4ac^3-4db^3-27a^2 d^2 \ne 0. \end{aligned}$$

Now we want to distinguish between the two remaining cases using the index form f(X, Y).

We need the following auxiliary result:

Lemma A.27

The elements $1,\omega ,\omega ^2$ form a basis of A over K if and only if $a \ne 0$, while $1,\theta ,\theta ^2$ form a basis of A if and only if $d \ne 0$.

Proof

Follows immediately from the relations (A.22) taking into account that $1,\omega , \theta$ is a basis of A. $\quad \square$

Proposition A.28

If $a \ne 0$ and the polynomial $P_\omega (X)$ is reducible over K, then

• (a) A is isomorphic to $L\oplus K$ if $P_\omega (X)$ has only one zero in K,

• (b) A is isomorphic to $K^3$ if $P_\omega (X)$ has three (different) zeros in K.

The same is true when $d \ne 0$ and $P_\omega (X)$ is replaced by $P_\theta (X)$.

Proof

If $a \ne 0$, then by the lemma above, the elements $1,\omega ,\omega ^2$ generate A, which implies that $A \cong K[X]/(P_\omega (X))$ and both (a) and (b) are evident. The same arguments work when $d \ne 0$ and $P_\omega (X)$ is replaced by $P_\theta (X)$. $\quad \square$

It remains the case when $a=d=0$. The multiplication rules (A.22) reduce then to

$$\begin{aligned} \omega \theta =0, \quad \omega ^2= b\omega , \quad \theta ^2 =-c\theta . \end{aligned}$$

Notice that $\Delta (f)=b^2c^2 \ne 0$, since A is separable.

Proposition A.29

If $a=d=0$ and $A=K\Lambda (f)$ is a separable algebra, then $A \cong K^3$.

Proof

It is easy to see that $A \cong K[X,Y]/(X^2-bX,Y^2+cY,XY)$. Since A is separable, we have to exclude a possibility that A contains a quadratic field extension of K. In our case, such a quadratic field extension is generated by an element of A whose square equals a generator t of the maximal ideal of R. A general element of $K[X,Y]/(X^2-bX,Y^2+cY,XY)$ has the form $\alpha +\beta x+\gamma y$, where $\alpha , \beta , \gamma \in K$ and $x^2=bx, y^2=-cy,xy =0$. Thus $(\alpha +\beta x+\gamma y)^2 =t$ implies that $\alpha ^2=t$, where $\alpha \in K$, which is impossible. $\quad \square$

Notice that in the case $a=d=0$ the polynomials $P_\omega (X), P_\theta (X)$ have all their zeros in K. Thus, we have

Corollary A.30

• (a) The separable algebra $A=K \Lambda ( f)$ is isomorphic to $K^3$ if and only if both polynomials $P_{\omega }(X)$ and $P_{\theta }(X)$ have all their zeros in K.

• (b) The separable algebra $A=K \Lambda (f)$ is isomorphic to $L\oplus K$ if and only if at least one of the polynomials $P_{\omega }(X)$ or $P_{\theta }(X)$ has a root in L.

Now return to the case $R={\mathbb {O}}$. We are almost ready to complete our description of double cosets (and therefore, our classification of the corresponding quantum groups) in terms of quadruples (a, b, c, d).

First, we define an action of ${\mathbf {GL}}(2,{\mathbb {O}})$ on the set of index forms and hence, on the set of quadruples (a, b, c, d). Let $g\in {\mathbf {GL}}(2,{\mathbb {O}})$. The action is defined as follows:

$$\begin{aligned} f(u,v)\mapsto g\cdot f(u,v)=\frac{1}{\mathrm {det}(g)}f((u,v)g). \end{aligned}$$

Here, we consider (u, v) as a row.

The result below was proved in [13], see also [2] and [5].

Proposition A.31

Let S be either a local ring or a principal ideal domain. Then there is a bijection between the set of orbits of the action of ${\mathbf {GL}}(2,S)$ on the set of index forms (and hence, on the set of quadruples (a, b, c, d)) and the set of isomorphism classes of cubic rings over S. $\quad \square$

Let us make the following observation:

Lemma A.32

Let $r:\Lambda \rightarrow \Lambda '$ be an ${\mathbb {O}}$-algebra isomorphism. Then we can extend r to a ${\mathbb {K}}$-isomorphism $r': {\mathbb {K}}\Lambda \rightarrow {\mathbb {K}}\Lambda '$ of the corresponding enveloping algebras (and therefore, they are isomorphic).

Proof

Clearly, r can be extended to $r': {\mathbb {K}}\Lambda \rightarrow {\mathbb {K}}\Lambda '$ as $r' (a\otimes k)=r(a)\otimes k$. $\quad \square$

Let us denote the set of quadruples (a, b, c, d) such that the corresponding cubic order is contained in ${\mathbb {K}}^3$ (resp. ${\mathbb {L}}\oplus {\mathbb {K}}$) by ${{\mathcal {P}}}$ (resp. ${{\mathcal {Q}}}$).

Corollary A.33

The sets ${{\mathcal {P}}}$, ${{\mathcal {Q}}}$ are invariant under the action of ${\mathbf {GL}}(2,{\mathbb {O}})$. $\quad \square$

Let $\mathrm {Aut}_{\mathbb {K}}({\mathbb {K}}\Lambda )$ be the group of ${\mathbb {K}}$-automorphisms of the enveloping algebra ${\mathbb {K}}\Lambda$.

Corollary A.34

There are two bijections of sets

$$\begin{aligned}&\mathrm {Aut}_{\mathbb {K}}({\mathbb {L}}\oplus {\mathbb {K}})\backslash \{\mathrm {orders}\ \mathrm {in}\ {\mathbb {O}}_{\mathbb {L}}\oplus {\mathbb {O}}\}\cong {\mathbf {GL}}( 2,{\mathbb {O}})\backslash {{{\mathcal {Q}}}}, \\&\mathrm {Aut}_{\mathbb {K}}({\mathbb {K}}^3)\backslash \{\mathrm {orders}\ \mathrm {in}\ {\mathbb {O}}^3\}\cong {\mathbf {GL}}(2,{\mathbb {O}})\backslash {{{\mathcal {P}}}}. \end{aligned}$$

Proof

It is sufficient to notice that any ${\mathbb {K}}$-automorphism of the enveloping algebra preserves the corresponding maximal order, ${\mathbb {O}}^3$ or ${\mathbb {O}}_{\mathbb {L}}\oplus {\mathbb {O}}$. $\quad \square$

Remark A.35

It is easy to show that $\mathrm {Aut}_{\mathbb {K}}({\mathbb {L}}\oplus {\mathbb {K}})\cong \mathrm {Aut}_{\mathbb {K}}({\mathbb {L}})\cong {\mathbb {Z}}/2{\mathbb {Z}}$ and $\mathrm {Aut}_{\mathbb {K}}({\mathbb {K}}^3)\cong S_3$, the symmetric group.

Now, we can describe the set of quantum groups related to the orders contained in ${\mathbb {K}}^3$ as follows.

• Choose a representative (a, b, c, d) in ${\mathbf {GL}}(2,{\mathbb {O}})\backslash {{{\mathcal {P}}}}$.

• Construct $\Lambda _{abcd}\subset {\mathbb {K}}^3$.

• Quantum groups corresponding to the orbit of the quadruple (a, b, c, d) are in a one-to-one correspondence with lattices in ${\mathbb {K}}^3$ such that their ring of multipliers is $\gamma (\Lambda _{abcd})$, where $\gamma$ is an automorphism of ${\mathbb {K}}^3$.

The set of quantum groups related to the orders contained in ${\mathbb {K}}\oplus {\mathbb {L}}$ has an almost identical description.

Example A.36

Assume that $ad\ne 0$ and $(a,b,c,d)\in {{{\mathcal {P}}}}$. The equation $P_{\theta }(x)=0$ has three roots $x_1,x_2,x_3\in {\mathbb {O}}$ and we can set $\theta =(x_1,x_2,x_3)\in {\mathbb {K}}^3$. Then $\omega =(-ad/x_1,-ad/x_2,-ad/x_3)$ and $\mathrm {Aut}_{\mathbb {K}}({\mathbb {K}}^3)=S_3$ acts on $\Lambda _{abcd}$ as a permutation group. It is not necessary that all six orders $\gamma (\Lambda _{abcd}),\ \gamma \in S_3$ are distinct. It might happen that some of them coincide.

In order to complete our description of quantum groups in terms of quadruples, we have to describe the set of lattice classes belonging to an order $\Lambda$ in terms of a, b, c, d.

The result below is a consequence of general results of [11] applied to the ring ${\mathbb {O}}$.

Theorem A.37

If $a,b,c,d\in t{\mathbb {O}}$, then $\mathrm {lc}(\Lambda _{abcd})=2$. Otherwise, $\mathrm {lc}(\Lambda _{abcd})=1$. $\quad \square$

Remark A.38

Notice that, according to Theorem A.11, if $a,b,c,d\in t{\mathbb {O}}$, then $\Lambda _{abcd}$ is not Gorenstein. Otherwise, $\Lambda _{abcd}$ is Gorenstein.

1.5 Classification of Cubic Orders Contained in Separable Cubic Algebras II

Here, we give a different approach to the classification problem of cubic orders. Again, let R be a discrete valuation ring (e.g., $R={\mathbb {O}}$) and K its quotient field. Denote by t a generator of the maximal ideal of R. If $\Lambda ' \subset \Lambda$ are two R-orders in a K-algebra A, then the product of the invariant factors (see [19, (4.14)]) of this pair (of R-lattices) is a power of the ideal (t). We write $[\Lambda :\Lambda ']=t^k$ if this product of the invariant factors is $(t^k)$ and we call $t^k$ or simply k for the index of $\Lambda '$ in $\Lambda$.

Description of all R -orders in the K -algebra $K^3$ .

We consider the field K as diagonally embedded into $K^3$. The maximal order in this algebra is $\Lambda =R^3$. Choose as a basis of $R^3$ the following elements: $e_1=1 = (1,1,1)$, $e_2=(0,1,0)$ and $e_3=(0,0,1)$. Of course, we have $e_2^2=e_2, e_3^2=e_3$ and $e_2e_3=0$. Let $\Lambda ' \subset \Lambda$ be any R-suborder of $\Lambda$. Let $1,f_2,f_3$ be an R-basis of $\Lambda '$. It is clear that 1 always can be chosen as a part of such a basis since $\Lambda '/R$ is torsion-free and $\Lambda '$ is R-projective (even free). This means that we can choose $f_2 = \alpha e_2+\beta e_3, f_3=\gamma e_2+\delta e_3$, where $\alpha ,\beta ,\gamma ,\delta \in R$.

Assume now that $\Lambda '$ is a Gorenstein order, that is, $\alpha ,\beta ,\gamma ,\delta$ are relatively prime. Otherwise, we have $\Lambda ' = R + t\Lambda ''$, where $\Lambda ''$ is a suborder of $\Lambda$. When $\Lambda '$ is Gorenstein, at least one of $\alpha ,\beta ,\gamma ,\delta$ is invertible in R, say $\alpha$, and we can assume that $\alpha =1$. Thus, we may choose $\gamma =0$, so that $f_3=\delta e_3$. Further, we may assume that $\delta =t^k$ for a nonnegative integer k. Since $\Lambda '$ is an order, we have $f_2^2, f_3^2, f_2f_3 \in \Lambda '$. Only the first condition puts some restrictions on $\beta$:

$$\begin{aligned} f_2^2= e_2+ \beta ^2e_3 \end{aligned}$$

implies that there exist $k,l \in R$ such that $e_2+ \beta ^2e_3 =k(e_2+\beta e_3)+lt^ke_3$. Hence, we get $k=1$ and $\beta ^2=\beta +lt^k$. The second equation shows that $\beta \equiv 0,1 \pmod {t^k}$. Thus, we get two possibilities: $f_2=e_2, f_3= t^ke_3$ or $f_2=e_2+e_3, f_3=t^ke_3$. It is easy to check that the orders $\Lambda _k =R+Re_2+Rt^ke_3$ and $\Lambda '_k=R + R(e_2+e_3)+Rt^ke_3$ are Gorenstein and different if only $k > 0$ (if $k=0$, we get the maximal order $\Lambda$). Thus, we have proved the following

Theorem A.39

For every index $[\Lambda :\Lambda ']=t^k$, where $k >0$, we have exactly two Gorenstein suborders of $\Lambda =R^3$, namely $\Lambda _k$ and $\Lambda '_k$. All other proper suborders of $\Lambda$ are not Gorenstein and are $\Lambda _{k,l}=R+t^l\Lambda _k$ and $\Lambda '_{k,l}=R+t^l\Lambda '_k$, where $k >0$, $l >0$. The number of all suborders of $\Lambda$ of given index $n=k+2l$ equals $\left[ \frac{n}{2} \right] +1$, $n \ge 0$. $\quad \square$

Description of all R -orders in the K -algebra $K \oplus L$ , where L is a quadratic field over K .

Let $L=K(j)$, where $j^2=t$. We consider the field K as embedded diagonally into $K \oplus L$. The maximal order in this algebra is $\Lambda = R \oplus S$, where S is the maximal R-order in L. Choose as a basis of $R \oplus S$ the following elements: $e_1=1 = (1,1)$, $e_2=(0,1)$ and $e_3=(0,j)$. Of course, we have $e_2^2=e_2, e_3^2=te_2$ and $e_2e_3=e_3$. Let $\Lambda ' \subset \Lambda$ be any R-suborder of $\Lambda = R \oplus S$. Let $1,f_2,f_3$ be an R-basis of $\Lambda '$. It is clear that 1 always can be chosen as a part of basis of $\Lambda '$ for the same reasons as in the case of $\Lambda = R^3$. This means that we can choose $f_2 = \alpha e_2+\beta e_3, f_3=\gamma e_2+\delta e_3$, where $\alpha ,\beta ,\gamma ,\delta \in R$.

Assume now that $\Lambda '$ is a Gorenstein order, that is, $\alpha ,\beta ,\gamma ,\delta$ are relatively prime. Otherwise, we have $\Lambda ' = R + t\Lambda ''$, where $\Lambda ''$ is a suborder of $\Lambda$. When $\Lambda '$ is Gorenstein, at least one of $\alpha ,\beta ,\gamma ,\delta$ is invertible in R.

Case I. If one of $\alpha ,\gamma$ is invertible in R, then without loss of generality we can assume that $\alpha =1$. Thus, we may choose $\gamma =0$, so that $f_3=\delta e_3$. Further, we may assume that $\delta =t^k$ for a nonnegative integer k. Since $\Lambda '$ is a suborder, we have $f_2^2, f_3^2, f_2f_3 \in \Lambda '$. As before, only the first condition puts some restrictions on $\beta$:

$$\begin{aligned} f_2^2= e_2+2\beta e_3+ \beta ^2te_2=(1+\beta ^2)te_2+2\beta e_3 \end{aligned}$$

implies that there exist $k,l \in R$ such that $e_2+2\beta e_3+ \beta ^2te_2=k(e_2+\beta e_3)+lt^ke_3$. Hence, we get $k=1+\beta ^2t$ and $2\beta = k\beta +lt^k$, which gives $2\beta =(1+\beta ^2t)\beta +lt^k$. Thus, we have $\beta \equiv 0 \pmod {t^k}$. As a consequence, we get that $\Lambda _k=R+Re_2+Rt^ke_3$ is the only Gorenstein suborder of $\Lambda$ of index $[\Lambda :\Lambda _k]=t^k$. All other proper suborders of $\Lambda$ are not Gorenstein and are $\Lambda _{k,l}=R+t^l\Lambda _k$, where $k >0$, $l >0$. Observe also that in this case the number of all suborders of $\Lambda$ of given index $n=k+2l$ equals $\left[ \frac{n}{2} \right] +1$, $n \ge 0$.

Case II. If t divides both $\alpha , \gamma$ and one of $\beta ,\delta$ is invertible in R, then without loss of generality we can assume that $\beta =1$. Thus, we may choose $\delta =0$, so that $f_2=\alpha e_2+e_3$ and $f_3=\gamma e_2$. As before, since $\Lambda '$ is a suborder, we have $f_2^2, f_3^2, f_2f_3 \in \Lambda '$. We easily check that also this time only the first condition puts some restrictions on the coefficients (this time $\alpha ,\gamma$):

$$\begin{aligned} f_2^2= \alpha ^2e_2+2\alpha e_3+ te_2=(\alpha ^2+t)e_2+2\alpha e_3 \end{aligned}$$

implies that there exist $k,l \in R$ such that $(\alpha ^2+t)e_2+2\alpha e_3 =k(\alpha e_2+e_3)+l\gamma e_2$. Hence, we get $k=2\alpha$ and $\alpha ^2+t=k\alpha +l\gamma$, which implies that $l\gamma =t-\alpha ^2$. Since $t \mid \gamma$ and $t^2 \mid \alpha ^2$, we get $l \in R$ only if $t^2 \not \mid \gamma$. Hence, we can choose $f_2=e_3$ and $f_3= te_2$, so $\Lambda '=R+Rte_2+Re_3$ is the only Gorenstein suborder of $\Lambda$ in this case. The order $\Lambda '_k = R + t^k\Lambda '$ for integer $k >0$ is not Gorenstein and has index $[\Lambda :\Lambda '_k]=t^{2k+1}$.

To summarize, we get the following

Theorem A.40

The maximal order $\Lambda = R \oplus S = R+Re_2+Re_3$, where $e_2^2=e_2, e_3^2=te_2$ and $e_2e_3=e_3$ in $K \oplus L$ contains exactly one Gorenstein suborder $\Lambda _k =R+Re_2+Rt^ke_3$ of every index $k > 1$, while for $k=1$, there are two Gorenstein suborders of index 1, $\Lambda _1 =R+Re_2+Rte_3$ and $\Lambda '_1=R+Rte_2+Re_3$. All non-Gorensteins suborders of $\Lambda$ are $\Lambda _{k,l}=R+t^l\Lambda _k$, where $k>0,l >0$ (of index $k+2l$) and $\Lambda '_k = R + t^k\Lambda '_1$, where $k > 0$ (of index $2k+1$). The total number of suborders of $\Lambda$ of given index n is equal $\left[ \frac{n}{2} \right] +1$ for even n and $\left[ \frac{n}{2} \right] +2$ for odd n. $\quad \square$

Remark A.41

At this point we would like remind the reader that in the case $R={\mathbb {O}}$ we have one quantum group corresponding to a Gorenstein order and two quantum groups which correspond to a non-Gorenstein order.

Our results are quite unexpected: there are “too many” quantum groups which are not isomorphic as Hopf algebras over ${\mathbb {O}}$. However, we make a conjecture that after tensoring by ${\mathbb {K}}$ there will be only two Hopf algebras over ${\mathbb {K}}$ related to non-twisted and twisted Belavin–Drinfeld cohomology.

Belavin–Drinfeld Cohomology for Exceptional Simple Lie Algebras (by E. Karolinsky and Aleksandra Pirogova)

In this appendix we discuss Belavin–Drinfeld cohomology for exceptional simple Lie algebras. We keep notation introduced in Sect. 6. Let ${\mathbf {G}}$ be a split simple simply connected (i.e., $X=P$) algebraic group of exceptional type. If ${\mathbf {G}}$ is of type $G_2$, $F_4$, or $E_8$, then $P=Q$, i.e., ${\mathbf {G}}$ is of adjoint type, and therefore, by Proposition 6.1, the centralizer ${\mathbf {C}}({\mathbf {G}}, r_{\mathrm{BD}})$ is connected for any Belavin–Drinfeld r-matrix $r_{\mathrm{BD}}$. The remaining cases are $E_6$ and $E_7$. In the $E_6$ case, $\Gamma =\{\alpha _1,\ldots ,\alpha _5,\alpha _6\}$ is enumerated in a way that $\{\alpha _1,\ldots ,\alpha _5\}$ is the simple root system of type $A_5$ (with the standard enumeration).

Theorem B.1

• 1) In the $E_6$ case, the centralizer ${\mathbf {C}}({\mathbf {G}}, r_{\mathrm{BD}})$ is not connected if and only if one of the following (mutually non-exclusive) conditions hold: either $\alpha _1$ and $\alpha _2$ are in the same string and $\alpha _4$ and $\alpha _5$ are also in the same string, or $\alpha _1$ and $\alpha _5$ are in the same string and $\alpha _2$ and $\alpha _4$ are also in the same string. In these cases ${\mathbf {C}}({\mathbf {G}}, r_{\mathrm{BD}}) = {\mathbf {T}}\times \mathbf {\mu }_3$, where ${\mathbf {T}}$ is a split torus and $\mathbf {\mu }_3$ is the group of cubic roots of unity.

• 2) In the $E_7$ case, the centralizer ${\mathbf {C}}({\mathbf {G}}, r_{\mathrm{BD}})$ is connected for any Belavin–Drinfeld r-matrix $r_{\mathrm{BD}}$.

Proof

The proof is via brute force aided by a computer. Namely, first, using a program written in C++, we list all possible admissible triples and compute the corresponding strings. Then, using Wolfram Mathematica, in each case we solve the corresponding system of equations (6.3) and compute the centralizer. $\quad \square$

Applying [18, Remark 4.11 and Corollary 4.13], we get

Corollary B.2

Let the base field ${\mathbb {F}}$ be of cohomological dimension 1. Let $r_{\mathrm{BD}}$ be a Belavin–Drinfeld r-matrix with $r_0\in {\mathfrak {h}} \otimes _{\mathbb {F}}{\mathfrak {h}}$.

• 1) In the $E_6$ case, $H({\mathbf {G}}, r_{\mathrm{BD}})={\mathbb {F}}^\times /({\mathbb {F}}^\times )^3$ in the cases when ${\mathbf {C}}({\mathbf {G}}, r_{\mathrm{BD}}) = {\mathbf {T}}\times \mathbf {\mu }_3$. Otherwise, $H({\mathbf {G}}, r_{\mathrm{BD}})=\{1\}$.

• 2) In the $G_2$, $F_4$, $E_7$, and $E_8$ cases, $H({\mathbf {G}}, r_{\mathrm{BD}})=\{1\}$. $\quad \square$

For the $E_6$ case, totally there are $406=203\times 2$ admissible triples (with non-empty $\Gamma _1$ and $\Gamma _2$). Among these, $70=35\times 2$ triples satisfy the condition of Theorem B.1. They are listed below (up to interchanging $\Gamma _1$ and $\Gamma _2$). First, we list the corresponding strings, and then the admissible triples having the given string structure.

• $\{\alpha _1, \alpha _2\}$, $\{\alpha _4, \alpha _5\}$

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _4\}$, $\Gamma _2=\{\alpha _2, \alpha _5\}$, $\tau (\alpha _1)=\alpha _2$, $\tau (\alpha _4)=\alpha _5$;

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _5\}$, $\Gamma _2=\{\alpha _2, \alpha _4\}$, $\tau (\alpha _1)=\alpha _2$, $\tau (\alpha _5)=\alpha _4$.

• $\{\alpha _1, \alpha _5\}$, $\{\alpha _2, \alpha _4\}$

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _2\}$, $\Gamma _2=\{\alpha _5, \alpha _4\}$, $\tau (\alpha _1)=\alpha _5$, $\tau (\alpha _2)=\alpha _4$;

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _4\}$, $\Gamma _2=\{\alpha _5, \alpha _2\}$, $\tau (\alpha _1)=\alpha _5$, $\tau (\alpha _4)=\alpha _2$.

• $\{\alpha _1, \alpha _2\}$, $\{\alpha _3, \alpha _4, \alpha _5\}$

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _3, \alpha _4\}$, $\Gamma _2=\{\alpha _2, \alpha _4, \alpha _5\}$, $\tau (\alpha _1)=\alpha _2$, $\tau (\alpha _3)=\alpha _4$, $\tau (\alpha _4)=\alpha _5$.

• $\{\alpha _1, \alpha _2, \alpha _3\}$, $\{\alpha _4, \alpha _5\}$

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _2, \alpha _4\}$, $\Gamma _2=\{\alpha _2, \alpha _3, \alpha _5\}$, $\tau (\alpha _1)=\alpha _2$, $\tau (\alpha _2)=\alpha _3$, $\tau (\alpha _4)=\alpha _5$.

• $\{\alpha _1, \alpha _5\}$, $\{\alpha _2, \alpha _3, \alpha _4\}$

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _3, \alpha _4\}$, $\Gamma _2=\{\alpha _5, \alpha _2, \alpha _3\}$, $\tau (\alpha _1)=\alpha _5$, $\tau (\alpha _3)=\alpha _2$, $\tau (\alpha _4)=\alpha _3$.

• $\{\alpha _1, \alpha _3, \alpha _5\}$, $\{\alpha _2, \alpha _4\}$

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _2, \alpha _3\}$, $\Gamma _2=\{\alpha _3, \alpha _4, \alpha _5\}$, $\tau (\alpha _1)=\alpha _3$, $\tau (\alpha _2)=\alpha _4$, $\tau (\alpha _3)=\alpha _5$;

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _2, \alpha _5\}$, $\Gamma _2=\{\alpha _3, \alpha _4, \alpha _1\}$, $\tau (\alpha _1)=\alpha _3$, $\tau (\alpha _2)=\alpha _4$, $\tau (\alpha _5)=\alpha _1$;

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _4, \alpha _5\}$, $\Gamma _2=\{\alpha _5, \alpha _2, \alpha _3\}$, $\tau (\alpha _1)=\alpha _5$, $\tau (\alpha _4)=\alpha _2$, $\tau (\alpha _5)=\alpha _3$.

• $\{\alpha _1, \alpha _2\}$, $\{\alpha _4, \alpha _5, \alpha _6\}$

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _4, \alpha _6\}$, $\Gamma _2=\{\alpha _2, \alpha _6, \alpha _5\}$, $\tau (\alpha _1)=\alpha _2$, $\tau (\alpha _4)=\alpha _6$, $\tau (\alpha _6)=\alpha _5$;

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _5, \alpha _6\}$, $\Gamma _2=\{\alpha _2, \alpha _6, \alpha _4\}$, $\tau (\alpha _1)=\alpha _2$, $\tau (\alpha _5)=\alpha _6$, $\tau (\alpha _6)=\alpha _4$.

• $\{\alpha _1, \alpha _2, \alpha _6\}$, $\{\alpha _4, \alpha _5\}$

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _4, \alpha _6\}$, $\Gamma _2=\{\alpha _6, \alpha _5, \alpha _2\}$, $\tau (\alpha _1)=\alpha _6$, $\tau (\alpha _4)=\alpha _5$, $\tau (\alpha _6)=\alpha _2$;

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _5, \alpha _6\}$, $\Gamma _2=\{\alpha _6, \alpha _4, \alpha _2\}$, $\tau (\alpha _1)=\alpha _6$, $\tau (\alpha _5)=\alpha _4$, $\tau (\alpha _6)=\alpha _2$.

• $\{\alpha _1, \alpha _5\}$, $\{\alpha _2, \alpha _4, \alpha _6\}$

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _2, \alpha _4\}$, $\Gamma _2=\{\alpha _5, \alpha _4, \alpha _6\}$, $\tau (\alpha _1)=\alpha _5$, $\tau (\alpha _2)=\alpha _4$, $\tau (\alpha _4)=\alpha _6$;

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _2, \alpha _6\}$, $\Gamma _2=\{\alpha _5, \alpha _4, \alpha _2\}$, $\tau (\alpha _1)=\alpha _5$, $\tau (\alpha _2)=\alpha _4$, $\tau (\alpha _6)=\alpha _2$;

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _4, \alpha _6\}$, $\Gamma _2=\{\alpha _5, \alpha _6, \alpha _2\}$, $\tau (\alpha _1)=\alpha _5$, $\tau (\alpha _4)=\alpha _6$, $\tau (\alpha _6)=\alpha _2$.

• $\{\alpha _1, \alpha _5, \alpha _6\}$, $\{\alpha _2, \alpha _4\}$

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _2, \alpha _5\}$, $\Gamma _2=\{\alpha _5, \alpha _4, \alpha _6\}$, $\tau (\alpha _1)=\alpha _5$, $\tau (\alpha _2)=\alpha _4$, $\tau (\alpha _5)=\alpha _6$;

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _2, \alpha _6\}$, $\Gamma _2=\{\alpha _5, \alpha _4, \alpha _1\}$, $\tau (\alpha _1)=\alpha _5$, $\tau (\alpha _2)=\alpha _4$, $\tau (\alpha _6)=\alpha _1$;

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _4, \alpha _6\}$, $\Gamma _2=\{\alpha _6, \alpha _2, \alpha _5\}$, $\tau (\alpha _1)=\alpha _6$, $\tau (\alpha _4)=\alpha _2$, $\tau (\alpha _6)=\alpha _5$.

• $\{\alpha _1, \alpha _2, \alpha _4, \alpha _5\}$

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _2, \alpha _4\}$, $\Gamma _2=\{\alpha _4, \alpha _5, \alpha _2\}$, $\tau (\alpha _1)=\alpha _4$, $\tau (\alpha _2)=\alpha _5$, $\tau (\alpha _4)=\alpha _2$;

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _2, \alpha _4\}$, $\Gamma _2=\{\alpha _5, \alpha _4, \alpha _1\}$, $\tau (\alpha _1)=\alpha _5$, $\tau (\alpha _2)=\alpha _4$, $\tau (\alpha _4)=\alpha _1$;

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _2, \alpha _5\}$, $\Gamma _2=\{\alpha _4, \alpha _5, \alpha _1\}$, $\tau (\alpha _1)=\alpha _4$, $\tau (\alpha _2)=\alpha _5$, $\tau (\alpha _5)=\alpha _1$;

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _2, \alpha _5\}$, $\Gamma _2=\{\alpha _5, \alpha _4, \alpha _2\}$, $\tau (\alpha _1)=\alpha _5$, $\tau (\alpha _2)=\alpha _4$, $\tau (\alpha _5)=\alpha _2$.

• $\{\alpha _1, \alpha _2\}$, $\{\alpha _3, \alpha _6\}$, $\{\alpha _4, \alpha _5\}$

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _3, \alpha _5\}$, $\Gamma _2=\{\alpha _2, \alpha _6, \alpha _4\}$, $\tau (\alpha _1)=\alpha _2$, $\tau (\alpha _3)=\alpha _6$, $\tau (\alpha _5)=\alpha _4$.

• $\{\alpha _1, \alpha _5, \alpha _6\}$, $\{\alpha _2, \alpha _3, \alpha _4\}$

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _2, \alpha _3, \alpha _5\}$, $\Gamma _2=\{\alpha _6, \alpha _3, \alpha _4, \alpha _1\}$, $\tau (\alpha _1)=\alpha _6$, $\tau (\alpha _2)=\alpha _3$, $\tau (\alpha _3)=\alpha _4$, $\tau (\alpha _5)=\alpha _1$;

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _2, \alpha _3, \alpha _6\}$, $\Gamma _2=\{\alpha _6, \alpha _3, \alpha _4, \alpha _5\}$, $\tau (\alpha _1)=\alpha _6$, $\tau (\alpha _2)=\alpha _3$, $\tau (\alpha _3)=\alpha _4$, $\tau (\alpha _6)=\alpha _5$;

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _3, \alpha _4, \alpha _5\}$, $\Gamma _2=\{\alpha _5, \alpha _2, \alpha _3, \alpha _6\}$, $\tau (\alpha _1)=\alpha _5$, $\tau (\alpha _3)=\alpha _2$, $\tau (\alpha _4)=\alpha _3$, $\tau (\alpha _5)=\alpha _6$.

• $\{\alpha _1, \alpha _2\}$, $\{\alpha _3, \alpha _4, \alpha _5, \alpha _6\}$

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _3, \alpha _5, \alpha _6\}$, $\Gamma _2=\{\alpha _2, \alpha _5, \alpha _6, \alpha _4\}$, $\tau (\alpha _1)=\alpha _2$, $\tau (\alpha _3)=\alpha _5$, $\tau (\alpha _5)=\alpha _6$, $\tau (\alpha _6)=\alpha _4$.

• $\{\alpha _1, \alpha _2, \alpha _3, \alpha _6\}$, $\{\alpha _4, \alpha _5\}$

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _2, \alpha _4, \alpha _6\}$, $\Gamma _2=\{\alpha _3, \alpha _6, \alpha _5, \alpha _1\}$, $\tau (\alpha _1)=\alpha _3$, $\tau (\alpha _2)=\alpha _6$, $\tau (\alpha _4)=\alpha _5$, $\tau (\alpha _6)=\alpha _1$.

• $\{\alpha _1, \alpha _2, \alpha _3, \alpha _4, \alpha _5\}$

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _2, \alpha _3, \alpha _4\}$, $\Gamma _2=\{\alpha _2, \alpha _3, \alpha _4, \alpha _5\}$, $\tau (\alpha _1)=\alpha _2$, $\tau (\alpha _2)=\alpha _3$, $\tau (\alpha _3)=\alpha _4$, $\tau (\alpha _4)=\alpha _5$.

• $\{\alpha _1, \alpha _2, \alpha _4, \alpha _5, \alpha _6\}$

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _2, \alpha _4, \alpha _6\}$, $\Gamma _2=\{\alpha _4, \alpha _5, \alpha _6, \alpha _2\}$, $\tau (\alpha _1)=\alpha _4$, $\tau (\alpha _2)=\alpha _5$, $\tau (\alpha _4)=\alpha _6$, $\tau (\alpha _6)=\alpha _2$;

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _2, \alpha _4, \alpha _6\}$, $\Gamma _2=\{\alpha _5, \alpha _4, \alpha _6, \alpha _1\}$, $\tau (\alpha _1)=\alpha _5$, $\tau (\alpha _2)=\alpha _4$, $\tau (\alpha _4)=\alpha _6$, $\tau (\alpha _6)=\alpha _1$;

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _2, \alpha _5, \alpha _6\}$, $\Gamma _2=\{\alpha _4, \alpha _5, \alpha _6, \alpha _1\}$, $\tau (\alpha _1)=\alpha _4$, $\tau (\alpha _2)=\alpha _5$, $\tau (\alpha _5)=\alpha _6$, $\tau (\alpha _6)=\alpha _1$;

  $*$:

  $\Gamma _1=\{\alpha _1, \alpha _2, \alpha _5, \alpha _6\}$, $\Gamma _2=\{\alpha _5, \alpha _4, \alpha _6, \alpha _2\}$, $\tau (\alpha _1)=\alpha _5$, $\tau (\alpha _2)=\alpha _4$, $\tau (\alpha _5)=\alpha _6$, $\tau (\alpha _6)=\alpha _2$.

We also list the admissible triples that satisfy the conclusions of Proposition 4.10. There are $40=20\times 2$ such triples (with non-empty $\Gamma _1$ and $\Gamma _2$). Their list (up to interchanging $\Gamma _1$ and $\Gamma _2$) is given below.

• $\Gamma _1=\{\alpha _1\}$, $\Gamma _2=\{\alpha _5\}$, $\tau (\alpha _1)=\alpha _5$;

• $\Gamma _1=\{\alpha _2\}$, $\Gamma _2=\{\alpha _4\}$, $\tau (\alpha _2)=\alpha _4$;

• $\Gamma _1=\{\alpha _1, \alpha _2\}$, $\Gamma _2=\{\alpha _4, \alpha _5\}$, $\tau (\alpha _1)=\alpha _4$, $\tau (\alpha _2)=\alpha _5$;

• $\Gamma _1=\{\alpha _1, \alpha _2\}$, $\Gamma _2=\{\alpha _5, \alpha _4\}$, $\tau (\alpha _1)=\alpha _5$, $\tau (\alpha _2)=\alpha _4$;

• $\Gamma _1=\{\alpha _1, \alpha _3\}$, $\Gamma _2=\{\alpha _3, \alpha _5\}$, $\tau (\alpha _1)=\alpha _3$, $\tau (\alpha _3)=\alpha _5$;

• $\Gamma _1=\{\alpha _1, \alpha _4\}$, $\Gamma _2=\{\alpha _2, \alpha _5\}$, $\tau (\alpha _1)=\alpha _2$, $\tau (\alpha _4)=\alpha _5$;

• $\Gamma _1=\{\alpha _1, \alpha _4\}$, $\Gamma _2=\{\alpha _5, \alpha _2\}$, $\tau (\alpha _1)=\alpha _5$, $\tau (\alpha _4)=\alpha _2$;

• $\Gamma _1=\{\alpha _1, \alpha _6\}$, $\Gamma _2=\{\alpha _6, \alpha _5\}$, $\tau (\alpha _1)=\alpha _6$, $\tau (\alpha _6)=\alpha _5$;

• $\Gamma _1=\{\alpha _2, \alpha _3\}$, $\Gamma _2=\{\alpha _3, \alpha _4\}$, $\tau (\alpha _2)=\alpha _3$, $\tau (\alpha _3)=\alpha _4$;

• $\Gamma _1=\{\alpha _2, \alpha _6\}$, $\Gamma _2=\{\alpha _6, \alpha _4\}$, $\tau (\alpha _2)=\alpha _6$, $\tau (\alpha _6)=\alpha _4$;

• $\Gamma _1=\{\alpha _1, \alpha _2, \alpha _3\}$, $\Gamma _2=\{\alpha _3, \alpha _4, \alpha _5\}$, $\tau (\alpha _1)=\alpha _3$, $\tau (\alpha _2)=\alpha _4$, $\tau (\alpha _3)=\alpha _5$;

• $\Gamma _1=\{\alpha _1, \alpha _2, \alpha _4\}$, $\Gamma _2=\{\alpha _4, \alpha _5, \alpha _2\}$, $\tau (\alpha _1)=\alpha _4$, $\tau (\alpha _2)=\alpha _5$, $\tau (\alpha _4)=\alpha _2$;

• $\Gamma _1=\{\alpha _1, \alpha _2, \alpha _5\}$, $\Gamma _2=\{\alpha _4, \alpha _5, \alpha _1\}$, $\tau (\alpha _1)=\alpha _4$, $\tau (\alpha _2)=\alpha _5$, $\tau (\alpha _5)=\alpha _1$;

• $\Gamma _1=\{\alpha _1, \alpha _3, \alpha _4\}$, $\Gamma _2=\{\alpha _5, \alpha _2, \alpha _3\}$, $\tau (\alpha _1)=\alpha _5$, $\tau (\alpha _3)=\alpha _2$, $\tau (\alpha _4)=\alpha _3$;

• $\Gamma _1=\{\alpha _1, \alpha _4, \alpha _6\}$, $\Gamma _2=\{\alpha _5, \alpha _6, \alpha _2\}$, $\tau (\alpha _1)=\alpha _5$, $\tau (\alpha _4)=\alpha _6$, $\tau (\alpha _6)=\alpha _2$;

• $\Gamma _1=\{\alpha _1, \alpha _4, \alpha _6\}$, $\Gamma _2=\{\alpha _6, \alpha _2, \alpha _5\}$, $\tau (\alpha _1)=\alpha _6$, $\tau (\alpha _4)=\alpha _2$, $\tau (\alpha _6)=\alpha _5$;

• $\Gamma _1=\{\alpha _1, \alpha _2, \alpha _3, \alpha _4\}$, $\Gamma _2=\{\alpha _2, \alpha _3, \alpha _4, \alpha _5\}$, $\tau (\alpha _1)=\alpha _2$, $\tau (\alpha _2)=\alpha _3$, $\tau (\alpha _3)=\alpha _4$, $\tau (\alpha _4)=\alpha _5$;

• $\Gamma _1=\{\alpha _1, \alpha _2, \alpha _3, \alpha _6\}$, $\Gamma _2=\{\alpha _6, \alpha _3, \alpha _4, \alpha _5\}$, $\tau (\alpha _1)=\alpha _6$, $\tau (\alpha _2)=\alpha _3$, $\tau (\alpha _3)=\alpha _4$, $\tau (\alpha _6)=\alpha _5$;

• $\Gamma _1=\{\alpha _1, \alpha _2, \alpha _4, \alpha _6\}$, $\Gamma _2=\{\alpha _4, \alpha _5, \alpha _6, \alpha _2\}$, $\tau (\alpha _1)=\alpha _4$, $\tau (\alpha _2)=\alpha _5$, $\tau (\alpha _4)=\alpha _6$, $\tau (\alpha _6)=\alpha _2$;

• $\Gamma _1=\{\alpha _1, \alpha _2, \alpha _5, \alpha _6\}$, $\Gamma _2=\{\alpha _4, \alpha _5, \alpha _6, \alpha _1\}$, $\tau (\alpha _1)=\alpha _4$, $\tau (\alpha _2)=\alpha _5$, $\tau (\alpha _5)=\alpha _6$, $\tau (\alpha _6)=\alpha _1$.