Factorizable $R$-Matrices for Small Quantum Groups

Factorizable $\boldsymbol{R}$-Matrices for Small Quantum Groups

Simon LENTNER and Tobias OHRMANN S. Lentner and T. Ohrmann Fachbereich
Mathematik, University of Hamburg,
Bundesstraße 55, 20146 Hamburg, Germany<EMAIL_ADDRESS><EMAIL_ADDRESS>
Received January 16, 2017, in final form September 15, 2017; Published online
September 25, 2017

Representations of small quantum groups $u_{q}({\mathfrak{g}})$ at a root of
unity and their extensions provide interesting tensor categories, that appear
in different areas of algebra and mathematical physics. There is an ansatz by
Lusztig to endow these categories with the structure of a braided tensor
category. In this article we determine all solutions to this ansatz that lead
to a non-degenerate braiding. Particularly interesting are cases where the
order of $q$ has common divisors with root lengths. In this way we produce
familiar and unfamiliar series of (non-semisimple) modular tensor categories.
In the degenerate cases we determine the group of so-called transparent
objects for further use.

factorizable; $R$-matrix; quantum group; modular tensor category; transparent
object

17B37; 20G42; 81R50; 18D10

##  1 Introduction

Hopf algebras with $R$-matrices, so called _quasitriangular Hopf algebras_ ,
give rise to tensor categories with a braiding $c\colon V\otimes
W\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}W\otimes V$. Of particular
interest are braided tensor categories where the braiding fulfills a certain
non-degeneracy condition, see Definition 5.1, which is equivalent to the fact
that there are no _transparent objects_ $V$, i.e., no objects where the
double-braiding $c^{2}\colon V\otimes
W\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}V\otimes W$ is the identity
for all $W$. A $\mathbb{C}$-linear tensor category with a nondegenerate
braiding, as well as finiteness conditions and another natural transformation
$\theta\colon V\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}V$ (twist), is
called a  _modular tensor category_. Note that we do not require the category
to be semisimple.

Modular tensor categories have many interesting applications: They give rise
to topological invariants and mapping class group actions [7, 19]. For
example, the standard generators $T$, $S$ of the mapping class group of the
torus $\mathrm{SL}_{2}(\mathbb{Z})$ are constructed from $\theta$ and $c^{2}$,
respectively. A different source for modular tensor categories in mathematical
physics are vertex algebras. There are only few example classes of modular
tensor categories, in particular non-semisimple ones.

The aim of the present article is to provide modular tensor categories from
_small quantum groups_ $u_{q}(\mathfrak{g})$ at a primitive $\ell$-th root of
unity $q$ for a finite-dimensional simple complex Lie algebra $\mathfrak{g}$.
Lusztig [12] has constructed these finite-dimensional Hopf algebras and
provided an ansatz for an $R$-matrix $R_{0}\bar{\Theta}$, where the fixed
element $\bar{\Theta}\in u_{q}(\mathfrak{g})^{-}\otimes
u_{q}(\mathfrak{g})^{+}$ is constructed from a dual basis of PBW generators,
while $R_{0}\in u_{q}(\mathfrak{g})^{0}\otimes u_{q}(\mathfrak{g})^{0}$ is a
free parameter subject to some constraints. He gives one canonical solution
for $R_{0}$ whenever $\ell$ has no common divisors with root lengths,
otherwise there are cases where no $R$-matrix exists [8] and the quantum group
becomes more interesting [9], involving, e.g., the dual root system. Of
particular interest in conformal field theory [4, 5, 6] is the most extreme
case where all root lengths $(\alpha,\alpha)$ divide $\ell$. In particular our
article addresses the question, which modular tensor category appear in these
cases. We find indeed, e.g., in Lemma 4.7 that these extremal cases give
especially nice $R$-matrices; although in general they are not factorizable
and will require modularization (see for example [2]) to match the CFT side.

But even if there are no common divisors with the root length, the resulting
braided tensor categories may not fulfill the non-degeneracy condition and
hence provides no modular tensor category.

Both obstacles (existence and non-degeneracy) can be be resolved by extending
the Cartan part of the quantum group by a choice of a lattice
$\Lambda_{R}\subseteq\Lambda\subseteq\Lambda_{W}$ between root- and weight-
lattice, respectively a choice of a subgroup of the fundamental group
$\pi_{1}:=\Lambda_{W}/\Lambda_{R}$, corresponding to a choice of a Lie group
between adjoint and simply-connected form. These extensions are already
present in [12] as the choice of two lattices $X$, $Y$ with pairing $X\times
Y\to\mathbb{C}^{\times}$ (root datum). In this way the number of possible $R$
matrices increases and the purpose of our paper is to study them all.

In a previous article [11] we have already constructed some solutions $R_{0}$
in this spirit (under some additional assumptions). As it turns out, the
solutions can be parametrized by subgroups $H_{1},H_{2}\subset\pi_{1}$ and
group pairings between $H_{1}$, $H_{2}$, and the set of solutions depends on
the common divisors of $\ell$ not only with root lengths, but also divisors of
the Cartan matrix. Some cases admit no braided structure, while others have
multiple in-equivalent solutions. An interesting occurrence was for example
that $B_{n}$ behaves differently for $n$ odd or even, and that $D_{2n}$ with
non-cyclic fundamental group allows several more solutions with non-symmetric
$R_{0}$.

In the present article we conclude this effort: First we introduce more
systematical techniques that allow us to compute a list of all quasitriangular
structures (without additional assumptions, so we find more solutions). Then
our new techniques allow us to determine, which of these choices fulfill the
non-degeneracy condition. We also determine which cases have a ribbon
structure. A main role in the first part is played by a natural pairing
$a_{\ell}$ on the fundamental group $\pi_{1}$ which depends only on the common
divisors of $\ell$ with the fundamental group and encapsulates the essential
$\ell$-dependence. Then the non-degeneracy of the braiding turns out to depend
only on the $2$-torsion of the abelian group in question.

Our result produces a list of modular tensor categories for representations of
quantum groups. Moreover we use our methods to explicitly describe the group
of transparent objects if the category is not modular, which is for example a
prerequisite for modularization.

We now discuss our methods and results in more detail:

In Section 2 we briefly recall the Lie theory and Hopf algebra preliminaries:
For every finite-dimensional (semi-)simple complex Lie algebra $\mathfrak{g}$
and a primitive $\ell$-th root of unity $q$ Lusztig has introduced in [12] the
_small quantum group_ $u_{q}(\mathfrak{g})$ which has a triangular
decomposition $u_{q}^{+}u_{q}^{0}u_{q}^{-}$ where the (exponentiated) _Cartan
algebra_ $u_{q}^{0}$ is the groupring of the root lattice $\Lambda_{R}$ modulo
some suitable sublattice and $u_{q}^{\pm}$ are generated by _simple root
vectors_ $E_{\alpha_{i}}$, $F_{\alpha_{i}}$ fulfilling $q$-deformed Serre
relations. In [13, Section 32] he gives an ansatz for an $R$-matrix in the
form $R_{0}\bar{\Theta}$, where $\bar{\Theta}$ consists of dual PBW basis’ and
$R_{0}\in u_{q}^{0}\otimes u_{q}^{0}$ is an arbitrary element in the Cartan
part that has to fulfill certain relations.

Our goal is to study the existence and non-degeneracy of $R$-matrices of this
form for the quantum group $u_{q}(\mathfrak{g},\Lambda,\Lambda^{\prime})$ with
any choice of lattice between root- and weight-lattice
$\Lambda_{R}\subseteq\Lambda\subseteq\Lambda_{W}$ and any possible choice of
quotient by a subgroup $\Lambda^{\prime}\subseteq\Lambda$ in the Cartan part
$u^{0}=\mathbb{C}[\Lambda/\Lambda^{\prime}]$. Later, we prove that
$\Lambda^{\prime}$ is in fact unique if we want a quasitriagular structure
(Corollary 3.6).

The $R_{0}$-matrix has the following interpretation: It is an $R$-matrix for
the groupring $\mathbb{C}[\Lambda/\Lambda^{\prime}]$ and it appears as the
braiding between highest-weight vectors in our $u_{q}(\mathfrak{g})$-modules.
Thus the previous theorem clarifies which choices for an $R$-matrix for the
group ring lift to the quantum group.

In Section 3 we address the question of constructing quasi-triangular
$R$-matrices. First we briefly recall the following general combinatorial
result in [11]:

######  Theorem 3.3.

The $R_{0}$-matrix is necessarily of the form

$\displaystyle
f(\mu,\nu)=\frac{1}{d}q^{-(\mu,\nu)}g(\bar{\mu},\bar{\nu})\delta_{\bar{\mu}\in
H_{1}}\delta_{\bar{\nu}\in H_{2}},$

where $H_{1}$, $H_{2}$ are subgroups of $\Lambda/\Lambda_{R}\subseteq\pi_{1}$
with $|H_{1}|=|H_{2}|=:d$ $($not necessarily isomorphic!$)$ and $g\colon
H_{1}\times H_{2}\to\mathbb{C}^{\times}$ is a pairing of groups.

Then we proceed differently than in the previous article: Using the previous
result, we prove in Lemma 3.5 that the quasitriangularity of $R$ is equivalent
to the assertion that the group pairing
$\hat{f}:=|\Lambda/\Lambda^{\prime}|\cdot f$ between the preimages
$G_{i}:=\Lambda_{i}/\Lambda^{\prime}$ of the groups $H_{i}$ is _non-
degenerate_ (which is no surprise). In particular we show that this condition
fixes $\Lambda^{\prime}$ uniquely. In later applications we often encounter
$\hat{f}$ as a natural identification of $G_{1}$ and the dual $\hat{G}_{2}$,
e.g., when studying representation theory.

To find all solutions $f$ with this property we develop a machinery to push
$\hat{f}$ into the fundamental group $\pi_{1}$, which encapsulates all the
$\ell$-dependence: In Definition 3.8 we give an abstract characterization of a
_centralizer transfer map_

$\displaystyle A_{\ell}\colon\
\Lambda/\Lambda_{R}\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}\operatorname{Cent}_{\Lambda}^{\ell}(\Lambda_{R})/\operatorname{Cent}_{\Lambda_{R}}^{\ell}(\Lambda)$

(without proving that it always exists). In a generic case this is just
multiplication by $\ell$, but it severely depends on common divisors of $\ell$
with root length and divisors of the Cartan matrix. With this matrix we can
transfer $q^{-(\mu,\nu)}$ to a natural form $a_{g}^{\ell}$ on the fundamental
group. We prove that $\hat{f}$ is non-degenerate iff
$a_{g}^{\ell}(\mu,\nu)=q^{-(\mu,A_{\ell}(\nu))}\cdot g(\mu,A_{\ell}(\nu))$ is
non-degenerate. This explains why the set of solutions, say for fundamental
group $\mathbb{Z}_{n}$ always looks like the subset of invertible elements
$\mathbb{Z}_{n}^{\times}$ but it is shifted (namely by $A_{\ell}$) depending
on $\ell$ and the root system in question.

In Section 4 the remaining computational work is done for quasitriagularity:
We calculate a list containing $a_{g}^{\ell}$ for all simple $\mathfrak{g}$,
depending on common divisors of $\ell$ with root length and divisors. We thus
write down all solutions for $f$ and hence $R$-matrices. The calculation
starts with the Smith normal form for the Cartan matrix in question and uses
three cases: For $\Lambda=\Lambda_{W}$ we have a generic construction, the
cases $A_{n}$ with their large fundamental group $\mathbb{Z}_{n+1}$ is treated
by hand, as is $D_{2n}$ with non-cyclic fundamental group, which has the only
cases allowing $\Lambda_{1}\neq\Lambda_{2}$.

In Section 5 we address our main issue of factorizability with our new tools:

In Section 5.1 we introduce factorizability. Then we calculate the monodromy
matrix $R_{21}R$ for an arbitrary choice of $R$-matrix in terms the
$R_{0}$-part. This gives a purely lattice theoretic problem equivalent to the
factorizability of such an $R$-matrix. Then we prove in the main Theorem 5.9
that factorizability is equivalent to the non-degeneracy of a symmetrization
$\operatorname{Sym}_{G}\big{(}\hat{f}\big{)}$ of $\hat{f}$. As will turn out
later, the radical of this form is isomorphic to the group of transparent
objects.

In Section 5.2 we restrict ourselves to the _symmetric case_ where
$H_{1}=H_{2}$ and $f$, $g$ are symmetric. Other cases appear only in some of
the non-cyclic $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-extension for type
$\mathfrak{g}=D_{2n}$ and are dealt with in Section 5.3 and give surprising
new solutions.

The main result for the symmetric case is that the radical of the form
$\operatorname{Sym}_{G}\big{(}\hat{f}\big{)}$ is in this case simply the
$2$-torsion of $\Lambda/\Lambda^{\prime}$ (Example 5.11) and that this is non-
degenerate precisely for odd $\ell$ and odd $\Lambda/\Lambda_{R}$ as well as
for $\mathfrak{g}=B_{n}$, $\Lambda=\Lambda_{R}$, $\ell\equiv 2$ ${\rm
mod}~{}4$ including $A_{1}$.

In Section 5.4 we prove the following result:

######  Lemma 5.19.

The transparent objects in the category of representations of the Hopf algebra
$u_{q}(\mathfrak{g},\Lambda)$ with $R$-matrix given by Lusztig’s ansatz are
$1$-dimensional objects $\mathbb{C}_{\chi}$ and are the $f$-transformed of the
radical of $\operatorname{Sym}_{G}\big{(}\hat{f}\big{)}$:

$\displaystyle\chi(\mu)=f(\mu,\xi),\qquad\xi\in\operatorname{Rad}\big{(}\operatorname{Sym}_{G}\big{(}\hat{f}\big{)}\big{)}.$

In the following we summarize our results by a table containing all
quasitriangular quantum groups $u_{q}(\mathfrak{g},\Lambda)$ with their group
of transparent objects. In Section 6 we show that all our quasitriangular
quantum groups admit a ribbon element. The factorizable solutions and thus
modular tensor categories are $\ell$ odd, $\Lambda=\Lambda_{R}$ and the
following new factorizable cases:

($\ell$ odd, $E_{6}$, $\Lambda=\Lambda_{W}$) and ($\ell\equiv 2$ ${\rm
mod}~{}4$, $\mathfrak{g}=B_{n}$, $\Lambda=\Lambda_{R}$) (including $A_{1}$)
and ($\ell$ odd, $\mathfrak{g}=D_{2n}$, $\Lambda_{1}\neq\Lambda_{2}$). All
other cases can be modularized as discussed in Question 7.3.

The columns of the following table are labeled by

  1. 1)

the finite-dimensional simple complex Lie algebra $\mathfrak{g}$,

  2. 2)

the natural number $\ell$, determining the root of unity
$q=\exp\big{(}{\frac{2\pi i}{\ell}}\big{)}$,

  3. 3)

the number of possible $R$-matrices for the Lusztig ansatz,

  4. 4)

the subgroups $H_{i}\subseteq H=\Lambda/\Lambda_{R}$ introduced in Theorem
3.3,

  5. 5)

the subgroups $H_{i}$ in terms of generators given by multiples of fundamental
dominant weights $\lambda_{i}\in\Lambda_{W}$,

  6. 6)

the group pairing $g\colon H_{1}\times H_{2}\to\mathbb{C}^{\times}$ determined
by its values on generators,

  7. 7)

the group of transparent objects $T\subseteq\Lambda/\Lambda^{\prime}$
introduced in Lemma 5.19.

Table 1: Solutions for $R_{0}$-matrices. $\mathfrak{g}$ | $\ell$ | # | $H_{i}\cong$ | $H_{i}\,{\scriptstyle(i=1,2)}$ | $g$ | $T\subseteq\Lambda/\Lambda^{\prime}$
---|---|---|---|---|---|---
all | $\ell$ odd | 1 | $\mathbb{Z}_{1}$ | $\langle 0\rangle$ | $g=1$ | 0
$\ell\equiv 0$ mod $4$ | 1 | $\mathbb{Z}_{2}^{n}$
|  | $\infty$ |  |  |  | $\mathbb{Z}_{2}^{n-1}$, $2\nmid x$
| $\mathbb{Z}_{d}$ | $\langle\hat{d}\lambda_{n}\rangle$ | $g(\hat{d}\lambda_{n},\hat{d}\lambda_{n})=\exp\big{(}\frac{2\pi ik}{d}\big{)}$
$A_{n\geq 1}$ |  | $\mathbb{Z}_{2}^{n}$, $2\,|\,x$
$\pi_{1}=\mathbb{Z}_{n+1}$ |  | $d\,|\,n+1$ | $\hat{d}=\frac{n+1}{d}$ | $\operatorname{gcd}\big{(}d,\frac{k\ell-\hat{d}n}{\operatorname{gcd}(\ell,\hat{d})}\big{)}=1$
|  | $x=\frac{d\ell}{\operatorname{gcd}(\ell,\hat{d})}$
| $\ell\equiv 2$ mod $4$ | 1 | $\mathbb{Z}_{1}$ | $\langle 0\rangle$ | $g=1$ | 0
$\ell\equiv 2$ mod $4$ | 2 | $\mathbb{Z}_{2}$ | $\langle\lambda_{n}\rangle$ | $g(\lambda_{n},\lambda_{n})=\pm 1$ | $\mathbb{Z}_{2}$
$B_{n\geq 2}$ | $\ell\equiv 0$ mod $4$ | 2 | $\mathbb{Z}_{2}$ | $\langle\lambda_{n}\rangle$ | $g(\lambda_{n},\lambda_{n})=\pm 1$ | $\mathbb{Z}^{n}_{2}$
$\pi_{1}=\mathbb{Z}_{2}$
| $\ell$ odd | 1 | $\mathbb{Z}_{2}$ | $\langle\lambda_{n}\rangle$ | $g(\lambda_{n},\lambda_{n})=(-1)^{n+1}$ | $\mathbb{Z}_{2}$
| $\ell\equiv 2\text{ mod }4$ | 1 | $\mathbb{Z}_{1}$ | $\langle 0\rangle$ | $g=1$ | $\mathbb{Z}_{2}^{n-2}$
$\ell\equiv 2\text{ mod }4$ | 1 | $\mathbb{Z}_{2}$ | $\langle\lambda_{n}\rangle$ | $g(\lambda_{n},\lambda_{n})=1$ | $\mathbb{Z}_{2}^{n-1}$
$C_{n\geq 3}$ | $\ell\equiv 0\text{ mod }4$ | 2 | $g(\lambda_{n},\lambda_{n})=\pm 1$ | $\mathbb{Z}_{2}^{n}$
$\pi_{1}=\mathbb{Z}_{2}$
| $\ell$ odd | 1 | $g(\lambda_{n},\lambda_{n})=-1$ | $\mathbb{Z}_{2}$
| $\ell\equiv 2\text{ mod }4$ | 1 | $\mathbb{Z}_{1}$ | $\langle 0\rangle$ | $g=1$ | $\mathbb{Z}_{2}^{2(n-1)}$
$\ell\equiv 2\text{ mod }4$ | 1 | $\mathbb{Z}_{2}$ | $H_{1}\cong\langle\lambda_{2n-1}\rangle$ | $g(\lambda_{2n-1},\lambda_{2n})=(-1)^{n}$ | $\mathbb{Z}_{2}^{2n}$
$\ell\equiv 0\text{ mod }4$ | $2\delta_{2\,|\,n}$ | $g(\lambda_{2n-1},\lambda_{2n})=\pm 1$, $n$ even
$H_{2}\cong\langle\lambda_{2n}\rangle$
$\ell$ odd | 1 | $g(\lambda_{2n-1},\lambda_{2n})=-1$ | 0
$D_{2n\geq 4}$
$\pi_{1}=\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ | $\ell\equiv 2\text{ mod }4$ | $1$ | $\mathbb{Z}_{2}$ | $\langle\lambda_{2n}\rangle$ | $g(\lambda_{2n},\lambda_{2n})=(-1)^{n+1}$ | $\mathbb{Z}_{2}^{2n-1}$
| $\ell\equiv 0\text{ mod }4$ | $2\delta_{2\nmid n}$ | $g(\lambda_{2n},\lambda_{2n})=\pm 1$, $n$ odd | $\mathbb{Z}_{2}^{2n}$
| $\ell$ odd | 1 | $g(\lambda_{2n},\lambda_{2n})=-1$ | $\mathbb{Z}_{2}$
| $\ell$ even | $16$ | $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ | $\langle\lambda_{2n},\lambda_{2n+1}\rangle$ | $g(\lambda_{2(n-1)+i},\lambda_{2(n-1)+j})=\pm 1$ | $\mathbb{Z}_{2}^{2n}$
| $\ell$ odd | $6$ | $\det(K)=K_{12}+K_{21}=0$ mod $2$ | $\mathbb{Z}_{2}$
| $\det(K)=K_{12}+K_{21}=1$ mod $2$ | $\mathbb{Z}_{2}^{2}$
| $\ell\equiv 2\text{ mod }4$ | 1 | $\mathbb{Z}_{1}$ | $\langle 0\rangle$ | $g=1$ | $\mathbb{Z}_{2}^{2n}$
$\ell\equiv 2\text{ mod }4$ | 1 | $\mathbb{Z}_{2}$ | $\langle 2\lambda_{2n+1}\rangle$ | $g(2\lambda_{2n+1},2\lambda_{2n+1})=1$ | $\mathbb{Z}^{2n+1}_{2}$
$\ell\equiv 0\text{ mod }4$ | 2 | $g(2\lambda_{2n+1},2\lambda_{2n+1})=\pm 1$
$D_{2n+1\geq 5}$ | $\ell$ odd | 1 | $g(2\lambda_{2n+1},2\lambda_{2n+1})=-1$ | $\mathbb{Z}_{2}$
$\pi_{1}=\mathbb{Z}_{4}$
| $\ell$ even | 4 | $\mathbb{Z}_{4}$ | $\langle\lambda_{2n+1}\rangle$ | $g(\lambda_{2n+1},\lambda_{2n+1})=c$, $c^{4}=1$ | $\mathbb{Z}^{2n+1}_{2}$
| $\ell$ odd | 2 | $g(\lambda_{2n+1},\lambda_{2n+1})=\pm 1$ | $\mathbb{Z}_{2}$
| $\ell\equiv 2\text{ mod }4$ | 1 | $\mathbb{Z}_{1}$ | $\langle 0\rangle$ | $g=1$ | $\mathbb{Z}_{2}^{6}$
$\ell\equiv 0$ mod $3$ | 3 | $\mathbb{Z}_{3}$ | $\langle\lambda_{n}\rangle$ | $g(\lambda_{n},\lambda_{n})=c$, $c^{3}=1$ | $\mathbb{Z}_{2}^{6}$, $2\,|\,\ell$
$E_{6}$ | $\ell\equiv 1$ mod $3$ | 2 | $g(\lambda_{n},\lambda_{n})=1,\exp\big{(}\frac{2\pi i2}{3}\big{)}$
$\pi_{1}=\mathbb{Z}_{3}$ | 0, $2\nmid\ell$
| $\ell\equiv 2$ mod $3$ | 2 | $g(\lambda_{n},\lambda_{n})=1,\exp\big{(}\frac{2\pi i}{3}\big{)}$
| $\ell\equiv 2\text{ mod }4$ | 1 | $\mathbb{Z}_{1}$ | $\langle 0\rangle$ | $g=1$ | $\mathbb{Z}_{2}^{6}$
$\ell$ even | 2 | $\mathbb{Z}_{2}$ | $\langle\lambda_{n}\rangle$ | $g(\lambda_{n},\lambda_{n})=\pm 1$ | $\mathbb{Z}_{2}^{n}$
$E_{7}$
$\pi_{1}=\mathbb{Z}_{2}$ | $\ell$ odd | 1 | $g(\lambda_{n},\lambda_{n})=1$ | $\mathbb{Z}_{2}$
$E_{8}$ | $\ell\equiv 2\text{ mod }4$ | 1 | $\mathbb{Z}_{1}$ | $\langle 0\rangle$ | $g=1$ | $\mathbb{Z}_{2}^{8}$
$F_{4}$ | $\ell\equiv 2\text{ mod }4$ | 1 | $\mathbb{Z}_{1}$ | $\langle 0\rangle$ | $g=1$ | $\mathbb{Z}_{2}^{2}$
$G_{2}$ | $\ell\equiv 2\text{ mod }4$ | 1 | $\mathbb{Z}_{1}$ | $\langle 0\rangle$ | $g=1$ | $\mathbb{Z}_{2}^{2}$

In the case $D_{2n}$, $\Lambda=\Lambda_{W}$, $g$ is uniquely defined by a
$(2\times 2)$-matrix $K\in\mathfrak{gl}(2,\mathbb{F}_{2})$, s.t.
$g(\lambda_{2(n-1)+i},\lambda_{2(n-1)+j})=\exp\big{(}\frac{2\pi
iK^{g}_{ij}}{2}\big{)}$ for $i,j\in\\{1,2\\}$.

##  2 Preliminaries

###  2.1 Lie theory

Throughout this article, $\mathfrak{g}$ denotes a finite-dimensional simple
complex Lie algebra. We fix a choice of simple roots
$\Delta=\\{\alpha_{i}\,|\,i\in I\\}$, so that the Cartan matrix $C$ is given
by $C_{ij}=2\frac{(\alpha_{i},\alpha_{j})}{(\alpha_{i},\alpha_{i})}$, where
$(\,,\,)$ denotes the normalized Killing form. For a root $\alpha$, we define
$d_{\alpha}:=\frac{(\alpha,\alpha)}{2}$ and set $d_{i}=d_{\alpha_{i}}$. By
$\Lambda_{R}:=\mathbb{Z}[\Delta]$ and
$\Lambda_{R}^{\vee}:=\mathbb{Z}[\Delta^{\vee}]$ we denote the (co)root lattice
of $\mathfrak{g}$.

By $\Lambda_{W}$, we denote the _weight lattice_ spanned by fundamental
dominant weights $\lambda_{i}$, which are defined by the equation
$(\lambda_{i},\alpha_{j})=\delta_{i,j}d_{i}$. Finally, we define the _co-
weight lattice_ $\Lambda_{W}^{\vee}$ as the $\mathbb{Z}$-span of the elements
$\lambda_{i}^{\vee}:=\frac{\lambda_{i}}{d_{i}}$. The quotient
$\pi_{1}:=\Lambda_{W}/\Lambda_{R}$ is called the _fundamental group_ of
$\mathfrak{g}$.

One can easily see that the Killing form restricts to a perfect pairing
$(\,,\,)\colon\Lambda^{\vee}_{W}\times\Lambda_{R}\to\mathbb{Z}$ and that we
get a string of inclusions
$\Lambda_{R}\subseteq\Lambda_{R}^{\vee}\subseteq\Lambda_{W}\subseteq\Lambda_{W}^{\vee}$.

###  2.2 Lusztig’s ansatz for $\boldsymbol{R}$-matrices

The starting point for our work [11] was Lusztig’s ansatz in [13, Section
32.1] for a universal $R$-matrix of $U_{q}(\mathfrak{g})$. Namely, for a
specific element $\bar{\Theta}\in U_{q}^{\geq 0}\otimes U_{q}^{\leq 0}$ from a
dual basis and a suitable (not further specified) element in the coradical
$R_{0}\in U_{q}^{0}\otimes U_{q}^{0}$ we are looking for $R$-matrices of the
form

$\displaystyle R=R_{0}\bar{\Theta}.$

We remark that there is no claim that all possible $R$-matrices are of this
form. However they are an interesting source of examples, motivated by the
interpretation of $u_{q}(\mathfrak{g})$ as a quotient of a Drinfeld double and
thus well-behaved with respect to the triangular decomposition. This ansatz
has been successfully generalized to general diagonal Nichols algebras in [1].
In our more general setting $U_{q}(\mathfrak{g},\Lambda,\Lambda^{\prime})$, we
have

$\displaystyle
R_{0}\in\mathbb{C}[\Lambda/\Lambda^{\prime}]\otimes\mathbb{C}[\Lambda/\Lambda^{\prime}].$

This ansatz has been worked out by Müller in his dissertation [14, 15, 16] for
small quantum groups $u_{q}(\mathfrak{g})$ which we will use in the following,
leading to a system of quadratic equation on $R_{0}$ that are equivalent to
$R$ being an $R$-matrix:

######  Theorem 2.1 (cf. [16, Theorem 8.2]).

Let $u:=u_{q}(\mathfrak{g})$.

  1. $(a)$

There is a unique family of elements $\Theta_{\beta}\in u_{\beta}^{-}\otimes
u_{\beta}^{+}$, $\beta\in\Lambda_{R}$, such that $\Theta_{0}=1\otimes 1$ and
$\Theta=\sum_{\beta}\Theta_{\beta}\in u\otimes u$ satisfies
$\Delta(x)\Theta=\Theta\bar{\Delta}(x)$ for all $x\in\ u$.

  2. $(b)$

Let $B$ be a vector space-basis of $u^{-}$, such that $B_{\beta}=B\cap
u^{-}_{\beta}$ is a basis of $u^{-}_{\beta}$ for all $\beta$. Here,
$u_{\beta}^{-}$ refers to the natural $\Lambda_{R}$-grading of $u^{-}$. Let
$\\{b^{*}\,|\,b\in B_{\beta}\\}$ be the basis of $u_{\beta}^{+}$ dual to
$B_{\beta}$ under the non-degenerate bilinear form
$(\,\cdot\,,\,\cdot\,)\colon u^{-}\otimes u^{+}\to\mathbb{C}$. We have

$\displaystyle\Theta_{\beta}=(-1)^{{\rm tr}\beta}q_{\beta}\sum_{b\in
B_{\beta}}b^{-}\otimes b^{*+}\in u_{\beta}^{-}\otimes u_{\beta}^{+}.$

######  Theorem 2.2 (cf. [16, Theorem 8.11]).

Let $\Lambda^{\prime}\subset\\{\mu\in\Lambda\,|\,K_{\mu}\text{ central in
}u_{q}(\mathfrak{g},\Lambda)\\}$ a subgroup of $\Lambda$, and $G_{1}$, $G_{2}$
subgroups of $G:=\Lambda/\Lambda^{\prime}$, containing
$\Lambda_{R}/\Lambda^{\prime}$. In the following, $\mu,\mu_{1},\mu_{2}\in
G_{1}$ and $\nu,\nu_{1},\nu_{2}\in G_{2}$.

The element $R=R_{0}\bar{\Theta}$ with an arbitrary
$R_{0}=\\!\sum\limits_{\mu,\nu}f(\mu,\nu)K_{\mu}\otimes K_{\nu}$ is a
$R$-matrix for $u_{q}(\mathfrak{g},\Lambda,\Lambda^{\prime})$, if and only if
for all $\alpha\in\Lambda_{R}$ and $\mu$, $\nu$ the following holds:

$\displaystyle f(\mu+\alpha,\nu)=q^{-(\nu,\alpha)}f(\mu,\nu),\qquad
f(\mu,\nu+\alpha)=q^{-(\mu,\alpha)}f(\mu,\nu),$ (2.1)
$\displaystyle\sum_{\nu_{1}+\nu_{2}=\nu}f(\mu_{1},\nu_{1})f(\mu_{2},\nu_{2})=\delta_{\mu_{1},\mu_{2}}f(\mu_{1},\nu),\qquad\sum_{\mu_{1}+\mu_{2}=\mu}f(\mu_{1},\nu_{1})f(\mu_{2},\nu_{2})=\delta_{\nu_{1},\nu_{2}}f(\mu,\nu_{1}),$
$\displaystyle\sum_{\mu}f(\mu,\nu)=\delta_{\nu,0},\qquad\sum_{\nu}f(\mu,\nu)=\delta_{\mu,0}.$

##  3 Conditions for the existence of $\boldsymbol{R}$-matrices

###  3.1 A first set of conditions on $\boldsymbol{\Lambda/\Lambda^{\prime}}$

The target of our efforts is a Hopf algebra called small quantum group
$u_{q}(\mathfrak{g},\Lambda,\Lambda^{\prime})$ with Cartan part
$u_{q}^{0}=\mathbb{C}[\Lambda/\Lambda^{\prime}]$. It is defined, e.g., in [11]
and depends on lattices $\Lambda$, $\Lambda^{\prime}$ defined below. For
$\Lambda=\Lambda_{R}$ the root lattice and this is the usual small quantum
group; the choice of $\Lambda^{\prime}$ differs in literature.

In the previous section we have discussed an $R=R_{0}\bar{\Theta}$-matrix for
the quantum group $u_{q}(\mathfrak{g},\Lambda,\Lambda^{\prime})$ can be
obtained from an $R_{0}$-matrix of the form

$\displaystyle R_{0}=\sum_{\mu,\nu\in\Lambda}f(\mu,\nu)K_{\mu}\otimes
K_{\nu}\in\mathbb{C}[\Lambda/\Lambda^{\prime}]\otimes\mathbb{C}[\Lambda/\Lambda^{\prime}].$

In the following we collect necessary and sufficient conditions for
$R=R_{0}\bar{\Theta}$ to be an $R$-matrix.

######  Definition 3.1.

We fix once-and-for-all a finite-dimensional simple complex Lie algebra
$\mathfrak{g}$ and a lattice $\Lambda$ between root- and weight-lattice

$\displaystyle\Lambda_{R}\subseteq\Lambda\subseteq\Lambda_{W}.$

These choices have a nice geometric interpretation as quantum groups
associated to different Lie groups associated to the Lie algebra
$\mathfrak{g}$.

Another interesting choice is
$\Lambda_{R}\subseteq\Lambda\subseteq\Lambda_{W}^{\vee}\cong\Lambda_{R}^{*}$,
which would below pose no additional complications and may produce further
interesting factorizable $R$-matrices.

######  Definition 3.2.

We fix once-and-for-all a primitive $\ell$-th root of unity $q$. For
$\Lambda_{1},\Lambda_{2}\subseteq\Lambda_{W}^{\vee}$ we define the sublattice

$\displaystyle\operatorname{Cent}_{\Lambda_{1}}(\Lambda_{2}):=\\{\,\nu\in\Lambda_{1}\,|\,(\nu,\mu)\in\ell\cdot\mathbb{Z}\
\forall\,\mu\in\Lambda_{2}\\}.$

Informally, this is the centralizer with respect to the braiding
$q^{-(\nu,\mu)}$.

Contrary to [11] we do not fix $\Lambda^{\prime}$ but we prove later Corollary
3.6 that there is a necessary choice for $\Lambda^{\prime}$. In this way, we
get more solutions than in [11]. The only condition necessary to ensure that
the Hopf algebra $u_{q}(\mathfrak{g},\Lambda,\Lambda^{\prime})$ is well-
defined is
$\Lambda^{\prime}\subseteq\operatorname{Cent}_{\Lambda_{R}}(\Lambda_{R})$.

######  Theorem 3.3 (cf. [11, Theorem 3.4]).

The $R_{0}$-matrix is necessarily of the form

$\displaystyle f(\mu,\nu)=\frac{1}{d|\Lambda_{R}/\Lambda^{\prime}|}\cdot
q^{-(\mu,\nu)}g(\bar{\mu},\bar{\nu})\delta_{\bar{\mu}\in
H_{1}}\delta_{\bar{\nu}\in H_{2}},$

where $H_{1}$, $H_{2}$ are subgroups of
$H:=\Lambda/\Lambda_{R}\subseteq\pi_{1}$ with equal cardinality
$|H_{1}|=|H_{2}|=:d$ $($not necessarily isomorphic!$)$ and $g\colon
H_{1}\times H_{2}\to\mathbb{C}^{\times}$ is a pairing of groups.

The necessity of this form (in particular that the support of $f$ is indeed a
subgroup!) amounts to a combinatorial problem of its own interest, which we
solved for $\pi_{1}$ cyclic in [10] and for
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ by hand; a closed proof for all abelian
groups would be interesting.

######  Definition 3.4.

Let $g\colon G\times H\to\mathbb{C}^{\times}$ be a finite group pairing, then
the _left radical_ is defined as

$\displaystyle\operatorname{Rad}_{L}(g):=\\{\lambda\in
G\,|\,g(\lambda,\eta)=1\,\forall\,\eta\in H\\}.$

Similarly, the right radical is defined as

$\displaystyle\operatorname{Rad}_{R}(g):=\\{\eta\in
H\,|\,g(\lambda,\eta)=1\,\forall\,\lambda\in G\\}.$

The pairing $g$ is called _non-degenerate_ if $\operatorname{Rad}_{L}(g)=0$.
If in addition $\operatorname{Rad}_{R}(g)=0$, $g$ is called _perfect_.

For an $R_{0}$-matrix of this form, a sufficient condition is that they
fulfill the so-called _diamond-equations_ (see [11, Definition 2.7]) for each
element
$0\neq\zeta\in(\operatorname{Cent}(\Lambda_{R})\cap\Lambda)/\Lambda^{\prime}$.

However, we will now go into a different, more systematic direction that makes
use of the following observation:

######  Lemma 3.5.

An $R_{0}$-matrix of the form given in Theorem 3.3 is a solution to the
equations in Theorem 2.2, and hence produces an $R$-matrix $R_{0}\bar{\Theta}$
iff the restriction to the support

$\displaystyle\hat{f}:=d|\Lambda_{R}/\Lambda^{\prime}|\cdot f\colon\
G_{1}\times G_{2}\to\mathbb{C}^{\times}$

is a _perfect_ group pairing, where
$G_{i}:=\Lambda_{i}/\Lambda^{\prime}\subseteq\Lambda/\Lambda^{\prime}=:G$.

###### Proof.

We first show that a solution with restriction to the support a nondegenerate
pairing solves the equation.

The first equations are obviously fulfilled for the form assumed

$\displaystyle f(\mu+\alpha,\nu)=q^{-(\nu,\alpha)}f(\mu,\nu),\qquad
f(\mu,\nu+\alpha)=q^{-(\mu,\alpha)}f(\mu,\nu).$

For the other equations the sums get only contributions in the support
$\Lambda_{1}/\Lambda^{\prime}\times\Lambda_{2}/\Lambda^{\prime}$. The
quantities $f(\mu,\nu)\cdot d|\Lambda_{R}/\Lambda^{\prime}|$ for fixed $\nu$
(or $\mu$) are characters on the respective support, and by the assumed non-
degeneracy all $\nu\neq 0$ give rise to different nontrivial characters. Then
the second and third relations follows from orthogonality of characters. Note
that since $d|\Lambda_{R}/\Lambda^{\prime}|=|G_{1}|=|G_{2}|$ (equality of the
latter was an assumption!) we were able to chose the right normalization.

For the other direction assume a solution of the given form to the equations.
Then already the third equation shows that no $f(-,\nu)$ may be the trivial
character and hence the form on the support is nondegenerate and hence perfect
by $|G_{1}|=|G_{2}|$. ∎

######  Corollary 3.6.

A first consequence of the perfectness of $\hat{f}$ $($i.e., a necessary
condition for quasi-triangularity$)$ is

$\displaystyle\operatorname{Cent}_{\Lambda_{R}}(\Lambda_{1})=\operatorname{Cent}_{\Lambda_{R}}(\Lambda_{2})=\Lambda^{\prime}.$

This fixes $\Lambda^{\prime}$ uniquely. Moreover in cases
$\Lambda_{1}\neq\Lambda_{2}$, which can only happen for $\mathfrak{g}=D_{2n}$,
where $\pi_{1}$ is noncyclic, we get an additional constraint relating
$\Lambda_{1}$, $\Lambda_{2}$.

In our case, the only possibility for $\Lambda_{1}\neq\Lambda_{2}$, s.t.
$G_{1}\cong G_{2}$ is $\mathfrak{g}=D_{2n}$. In this case, we have
$\operatorname{Cent}_{\Lambda_{R}}(\Lambda_{W})=\operatorname{Cent}_{\Lambda_{R}}(\Lambda_{R})$
and thus the above condition is always fulfilled.

Our main goal for the new approach on quasitriangularity as well as the later
modularity is to reduce this non-degeneracy condition for $\hat{f}$ to a non-
degeneracy condition for $g$ on $H_{1},H_{2}\subset\pi_{1}$ that can be
checked explicitly.

###  3.2 A natural form on the fundamental group

We now define for each triple $(\Lambda,\Lambda_{1},\Lambda_{2})$ and each
$\ell$th root of unity $q$ a natural pairing $a_{\ell}$ on the subgroups
$H_{i}:=\Lambda_{i}/\Lambda_{R}$ of the fundamental group
$\pi_{1}:=\Lambda_{W}/\Lambda_{R}$. The simplest example is $a_{\ell}=e^{-2\pi
i(\mu,\nu)}$. In general it is a transportation of the natural form
$q^{-(\mu,\nu)}$ (which does not factorize over $\Lambda_{R}$) to $H_{i}$ by a
suitable isomorphism $A_{\ell}$.

This isomorphism $A_{\ell}$ will encapsulate the crucial dependence on the
common divisors of $\ell$, $|H|$ and the root lengths $d_{i}$; moreover, for
different $H$ these forms are _not_ simply restrictions of one another.

Then, we can moreover transport any given pairing $g$ together with
$q^{-(\mu,\nu)}$ along the isomorphism $A_{\ell}$ to the $H_{i}$ and thus
define forms $a_{\ell}^{g}$ on $H$. The main result of this section is in
Theorem 3.13 that the non-degeneracy condition in Lemma 3.5 for $R_{0}(f)$
depending on $H_{i}$, $g$ is equivalent to $a_{\ell}^{g}$ being non-
degenerate.

######  Definition 3.7.

Let $\Lambda\subseteq\Lambda_{W}^{\vee}$ be a sublattice, s.t.
$\Lambda_{R}\subseteq\Lambda$. By $\hat{\Lambda}\subset\Lambda_{W}^{\vee}$ we
denote the unique sublattice, s.t. the symmetric bilinear form
$(\,\cdot\,,\,\cdot\,)\colon\Lambda_{W}^{\vee}\times\Lambda_{W}^{\vee}\to\mathbb{Q}$
induces a commuting diagram

${\Lambda_{R}}$${\hat{\Lambda}}$${\Lambda_{W}^{\vee}}$${{\Lambda_{W}^{\vee}}^{*}}$${\Lambda^{*}}$${\Lambda_{R}^{*},}$$\scriptstyle{\cong}$$\scriptstyle{\cong}$$\scriptstyle{\cong}$

where $\Lambda^{*}:=\operatorname{Hom}_{\mathbb{Z}}(\Lambda,\mathbb{Z})$. In
particular, we have $\hat{\Lambda}_{R}=\Lambda_{W}^{\vee}$ and
$\hat{\Lambda}_{W}^{\vee}=\Lambda_{R}$.

######  Definition 3.8.

A _centralizer transfer map_ is an group endomorphism
$A_{\ell}\in\operatorname{End}_{\mathbb{Z}}(\Lambda)$, s.t.

  1. 1)

$A_{\ell}(\Lambda)\stackrel{{\scriptstyle!}}{{=}}\Lambda\cap\ell\cdot\hat{\Lambda}_{R}=\operatorname{Cent}_{\Lambda}^{\ell}(\Lambda_{R})$,

  2. 2)

$A_{\ell}(\Lambda_{R})\stackrel{{\scriptstyle!}}{{=}}\Lambda_{R}\cap\ell\cdot\hat{\Lambda}=\operatorname{Cent}_{\Lambda_{R}}^{\ell}(\Lambda)$.

Such a $A_{\ell}$ induces a group isomorphism

$\displaystyle\Lambda/\Lambda_{R}\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}\operatorname{Cent}_{\Lambda}^{\ell}(\Lambda_{R})/\operatorname{Cent}_{\Lambda_{R}}^{\ell}(\Lambda).$

Of course $A_{\ell}$ is not unique.

######  Question 3.9.

Are there abstract arguments for the existence of these isomorphism and for
its explicit form?

We will calculate explicit expressions for $A_{\ell}$ depending on the cases
in the next section. At this point we give the generic answers:

######  Example 3.10.

For $\Lambda=\Lambda_{W}^{\vee}$ we have
$A_{\ell}=\ell\cdot\operatorname{id}$.

For $\Lambda=\Lambda_{R}$ the two conditions are equivalent, so existence is
trivial (resp. obviously the two trivial groups are isomorphic) and we may
simply take for $A_{\ell}$ any base change between left and right side. The
expression may however be nontrivial.

######  Lemma 3.11.

Assume $\gcd(\ell,|\Lambda_{W}^{\vee}/\Lambda|)=1$, then
$A_{\ell}=\ell\cdot\operatorname{id}$. In particular this is the case if
$\ell$ is prime to all root lengths and all divisors of the Cartan matrix.

Moreover if $\ell=\ell_{1}\ell_{2}$ with
$\gcd(\ell_{1},|\Lambda_{W}^{\vee}/\Lambda|)=1$, then $A_{\ell}=\ell_{1}\cdot
A_{\ell_{2}}$.

This means we only have to calculate $A_{\ell}$ for all divisors $\ell$ of
$|\Lambda_{W}^{\vee}/\Lambda|$, which is a subset of all divisors of root
lengths times divisors of the Cartan matrix.

###### Proof.

For the first condition we need to show for any $\lambda\in\Lambda_{W}^{\vee}$
that $\ell\lambda\in\Lambda$ already implies $\lambda\in\Lambda$. But if by
assumption the order of the quotient group $\Lambda_{W}^{\vee}/\Lambda$ is
prime to $\ell$, then $\ell\cdot$ is an isomorphism on this abelian group,
hence follows the assertion. For the second condition applies the same
argument noting that
$|\hat{\Lambda}/\Lambda_{R}|=|\Lambda_{W}^{\vee}/\Lambda|$.

For the second claim we simply consider the inclusion chains

$\displaystyle
A_{\ell}(\Lambda)\subset\Lambda\cap\ell_{2}\cdot\hat{\Lambda}_{R}\subset\Lambda\cap\ell\cdot\hat{\Lambda}_{R},$
$\displaystyle
A_{\ell}(\Lambda_{R})\subset\Lambda\cap\ell_{2}\cdot\hat{\Lambda}\subset\Lambda_{R}\cap\ell\cdot\hat{\Lambda},$

where a first isomorphism is given by $A_{\ell_{2}}$ and again $\ell_{1}\cdot$
is a second isomorphism because it is prime to the index. ∎

Our main result of this chapter is the following:

######  Theorem 3.12.

Let $\Lambda_{R}\subseteq\Lambda_{1}$, $\Lambda_{2}\subseteq\Lambda_{W}$ be
intermediate lattices, s.t. the condition in Corollary 3.6 is fulfilled, i.e.,
$\operatorname{Cent}_{\Lambda_{R}}(\Lambda_{1})=\operatorname{Cent}_{\Lambda_{R}}(\Lambda_{2})=\Lambda^{\prime}$.
Assume we have a centralizer transfer map $A_{\ell}$.

  1. $1.$

The following form is well defined on the quotients:

$\displaystyle a_{g}^{\ell}\colon\ $
$\displaystyle\Lambda_{1}/\Lambda_{R}\times\Lambda_{2}/\Lambda_{R}\longrightarrow\mathbb{C}^{\times},$
$\displaystyle(\bar{\lambda},\bar{\mu})\longmapsto
q^{-(\lambda,A_{\ell}(\mu))}\cdot g(\lambda,A_{\ell}(\mu)).$

  2. $2.$

Let

$\displaystyle\operatorname{Cent}_{\Lambda_{1}}^{g}(\Lambda_{2}):=\big{\\{}\lambda\in\Lambda_{1}\,|\,q^{(\lambda,\mu)}=g(\lambda,\mu)\;\forall\,\mu\in\Lambda_{2}\big{\\}}.$

Then the inclusion
$\operatorname{Cent}_{\Lambda_{1}}^{g}({\Lambda_{2}})\hookrightarrow\Lambda_{1}$
induces an isomorphism

$\displaystyle\operatorname{Cent}_{\Lambda_{1}}^{g}({\Lambda_{2}})/\Lambda^{\prime}\cong\operatorname{Rad}\big{(}a_{g}^{\ell}\big{)}.$

######  Corollary 3.13.

The quasitriangularity conditions for a choice $R_{0}$ are by Lemma 3.5
equivalent to the non-degeneracy of the group pairing on
$\Lambda_{1}/\Lambda^{\prime}\times\Lambda_{2}/\Lambda^{\prime}$:

$\displaystyle\hat{f}(\lambda,\mu)=q^{-(\lambda,\mu)}g(\lambda,\mu).$

By the previous theorem this condition is now equivalent to the nondegeneracy
of $a_{g}^{\ell}$.

This condition on the fundamental group, which is a finite abelian group and
mostly cyclic, can be checked explicitly once $a_{g}^{\ell}$ has been
calculated.

###### Proof of Theorem 3.12.

The first part of the theorem is a direct consequence of the definition of the
centralizer transfer matrix $A_{\ell}$. For the second part, we first notice
that by assumption we have a commutative diagram of finite abelian groups

${\Lambda_{R}/\Lambda^{\prime}}$${\Lambda_{1}/\Lambda^{\prime}}$${\Lambda_{1}/\Lambda_{R}}$${\left(\Lambda_{2}/\operatorname{Cent}_{\Lambda_{2}}(\Lambda_{R})\right)^{\wedge}}$${(\Lambda_{2}/\Lambda^{\prime})^{\wedge}}$${\left(\operatorname{Cent}_{\Lambda_{2}}(\Lambda_{R})/\Lambda^{\prime}\right)^{\wedge},}$$\scriptstyle{q^{-(\cdot,\cdot)}}$$\scriptstyle{\hat{f}}$$\scriptstyle{\hat{f}^{\prime}}$

where $G^{\wedge}$ denotes the dual group of a group $G$.

Now, by the five lemma we know that $\hat{f}$ is an isomorphism if and only if
the induced map $\hat{f}^{\prime}$ is an isomorphism. Post-composing this map
with the dualized centralizer transfer matrix
$A_{\ell}^{\wedge}\colon(\operatorname{Cent}_{\Lambda_{2}}(\Lambda_{R})/\Lambda^{\prime})^{\wedge}\cong(\Lambda_{2}/\Lambda_{R})^{\wedge}$
gives $a_{g}^{\ell}$. ∎

##  4 Explicit calculation for every $\boldsymbol{\mathfrak{g}}$

In the following, we want to compute the endomorphism
$A_{\ell}\in\operatorname{End}_{\mathbb{Z}}(\Lambda)$ and the pairing
$a_{\ell}$ on the fundamental group explicitly in terms of the Cartan matrices
and the common divisors of $\ell$ with root lengths and divisors of the Cartan
matrix. We will finally give a list for all $\mathfrak{g}$.

###  4.1 Technical tools

We choose the basis of simple roots $\alpha_{i}$ for $\Lambda_{R}$ and the
dual basis of fundamental coweights $\lambda_{i}^{\vee}$ for the dual lattice
$\Lambda_{W}^{\vee}$ with $(\alpha_{i},\lambda_{j}^{\vee})=\delta_{i,j}$.

For any choice $\Lambda\subset\Lambda_{W}\subset\Lambda_{W}^{\vee}$, let
$A_{\Lambda}$ be a _basis matrix_ , i.e., any $\mathbb{Z}$-linear isomorphism
$\Lambda_{W}^{\vee}\to\Lambda$ sending the basis $\lambda_{i}^{\vee}$ of
$\Lambda_{W}^{\vee}$ to some basis $\mu_{i}$ of $\Lambda$. It is unique up to
pre-composition of a unimodular matrix $U\in\mathrm{SL}_{n}(\mathbb{Z})$.

The dual basis $A_{\hat{\Lambda}}$ of $\hat{\Lambda}$ is defined by

$\displaystyle\big{(}A_{\hat{\Lambda}}\big{(}\lambda_{i}^{\vee}\big{)},A_{\Lambda}\big{(}\lambda_{j}^{\vee}\big{)}\big{)}=\delta_{ij}.$

Explicitly, $A_{\hat{\Lambda}}$ is given by
$A_{\hat{\Lambda}}=\big{(}A_{\Lambda}^{-1}A_{R}\big{)}^{T},$ where
$(A_{R})_{ij}=(\alpha_{i},\alpha_{j})$. Now, let
$A_{\Lambda}=P_{\Lambda}S_{\Lambda}Q_{\Lambda}$ be the unique Smith
decomposition of $A_{\Lambda}$, which means: $P_{\Lambda}$, $Q_{\Lambda}$ are
unimodular and $S_{\Lambda}$ is diagonal with diagonal entries
$(S_{\Lambda})_{ii}=:d^{\Lambda}_{i}$, such that
$d^{\Lambda}_{i}\,|\,d^{\Lambda}_{j}$ for $i<j$.

######  Example 4.1.

For the root lattice the $d^{\Lambda_{R}}_{i}$ are the divisors of scalar
product matrix $(\alpha_{i},\alpha_{j})$. Their product is

$\displaystyle\prod_{i}d^{\Lambda_{R}}_{i}=\big{|}\Lambda_{W}^{\vee}/\Lambda_{R}\big{|}=\Big{(}\prod_{i}d_{i}\Big{)}\cdot|\pi_{1}|,\qquad
d_{i}=\frac{(\alpha_{i},\alpha_{i})}{2}.$

For the coweight lattice all $d^{\Lambda_{W}^{\vee}}_{i}=1$. For the weight
lattice we recover the familiar $d^{\Lambda_{W}}_{i}=d_{i}$.

Without loss of generality, we will assume the basis matrices $A_{\Lambda}$ to
be symmetric, i.e., $Q_{\Lambda}=P_{\Lambda}^{T}$. We then have the following
lemma:

######  Lemma 4.2.

Let $\Lambda_{R}\subseteq\Lambda\subseteq\Lambda_{W}^{\vee}$ be a lattice. We
define lattices

$\displaystyle A_{\rm
Cent}:=\big{(}P_{\Lambda}^{T}\big{)}^{-1}D_{\ell}P_{\Lambda}^{-1},\qquad
D_{\ell}:=\operatorname{Diag}\left(\frac{\ell}{\operatorname{gcd}\big{(}\ell,d_{i}^{\Lambda}\big{)}}\right).$

Then,

$\displaystyle\operatorname{Cent}_{\Lambda_{R}}(\Lambda)=A_{R}A_{\rm
Cent}\Lambda_{W}^{\vee},\qquad\operatorname{Cent}_{\Lambda}(\Lambda_{R})=A_{\Lambda}A_{\rm
Cent}\Lambda_{W}^{\vee}.$

###### Proof.

We compute explicitly,

$\displaystyle\operatorname{Cent}_{\Lambda_{R}}(\Lambda)=\Lambda_{R}\cap\ell\cdot\hat{\Lambda}=A_{R}\Lambda_{W}^{\vee}\cap\big{(}A_{\Lambda}^{-1}A_{R}\big{)}^{T}\ell\Lambda_{W}^{\vee}$
$\displaystyle\hphantom{\operatorname{Cent}_{\Lambda_{R}}(\Lambda)}{}=\big{(}A_{\Lambda}^{-1}A_{R}\big{)}^{T}\big{(}\big{(}\big{(}A_{\Lambda}^{-1}A_{R}\big{)}^{T}\big{)}^{-1}A_{R}\Lambda_{W}^{\vee}\cap\ell\Lambda_{W}^{\vee}\big{)}=A_{R}A_{\Lambda}^{-1}\big{(}A_{\Lambda}\cap\ell\Lambda_{W}^{\vee}\big{)}$
$\displaystyle\hphantom{\operatorname{Cent}_{\Lambda_{R}}(\Lambda)}{}=A_{R}\big{(}P_{\Lambda}S_{\Lambda}P_{\Lambda}^{T}\big{)}^{-1}\big{(}P_{\Lambda}S_{\Lambda}P_{\Lambda}^{T}\Lambda_{W}^{\vee}\cap\ell\Lambda_{W}^{\vee}\big{)}=A_{R}\big{(}P_{\Lambda}^{T}\big{)}^{-1}S_{\Lambda}^{-1}\big{(}S_{\Lambda}\Lambda_{W}^{\vee}\cap\ell\Lambda_{W}^{\vee}\big{)}$
$\displaystyle\hphantom{\operatorname{Cent}_{\Lambda_{R}}(\Lambda)}{}=A_{R}\big{(}P_{\Lambda}^{T}\big{)}^{-1}S_{\Lambda}^{-1}\operatorname{Diag}(\operatorname{lcm}(S_{\Lambda_{ii}},\ell))\Lambda_{W}^{\vee}=A_{R}\big{(}P_{\Lambda}^{T}\big{)}^{-1}D_{\ell}\Lambda_{W}^{\vee}=A_{R}A_{\rm
Cent}\Lambda_{W}^{\vee}.$

On the other hand,

$\displaystyle\operatorname{Cent}_{\Lambda}(\Lambda_{R})=\Lambda\cup\ell\hat{\Lambda}_{R}=\Lambda\cup\ell\Lambda_{W}^{\vee}=A_{\Lambda}\Lambda_{W}^{\vee}\cup\ell\Lambda_{W}^{\vee}=P_{\Lambda}S_{\Lambda}P_{\Lambda}^{T}\Lambda_{W}^{\vee}\cup\ell\Lambda_{W}^{\vee}$
$\displaystyle\hphantom{\operatorname{Cent}_{\Lambda}(\Lambda_{R})}{}=P_{\Lambda}\big{(}S_{\Lambda}\Lambda_{W}^{\vee}\cup\ell\Lambda_{W}^{\vee}\big{)}=P_{\Lambda}S_{\Lambda}D_{\ell}\Lambda_{W}^{\vee}=A_{\Lambda}\big{(}P_{\Lambda}^{T}\big{)}^{-1}D_{\ell}\Lambda_{W}^{\vee}=A_{\Lambda}A_{\rm
Cent}\Lambda_{W}^{\vee}.$

In particular, this means
$A_{\hat{\Lambda}}\operatorname{Cent}_{\Lambda}(\Lambda_{R})=\operatorname{Cent}_{\Lambda_{R}}(\Lambda)$.
∎

###  4.2 Case $\boldsymbol{\Lambda=\Lambda_{W}}$

In order to exhaust all cases that appear in our setting, we continue with
$\Lambda=\Lambda_{W}$:

######  Lemma 4.3.

In the case $\Lambda=\Lambda_{W}$, the centralizer transfer matrix $A_{\ell}$
is of the following form:

$\displaystyle A_{\ell}=\begin{cases}A_{\Lambda_{W}}A_{\rm
Cent}Q_{C}^{T}P_{C}^{-1}A_{\Lambda_{W}}^{-1},&\operatorname{gcd}(\ell,|\pi_{1}|)\neq
1,\\\ \ell\cdot\operatorname{id},&\text{else}.\end{cases}$

Here, $C=P_{C}S_{C}Q_{C}$ denotes the Smith decomposition of the Cartan matrix
of $\mathfrak{g}$.

###### Proof.

As we noted in Example 4.1, we have
$A_{\Lambda_{W}}=\operatorname{Diag}(d_{i})$, for $d_{i}$ being the $i$th root
length. Since $d_{i}\in\\{1,p\\}$ for some prime number $p$, up to a
permutation $A_{\Lambda_{W}}$ is already in Smith normal form: this means that
$P_{\Lambda_{W}}$ is a permutation matrix of the form
$(P_{\Lambda_{W}})_{ij}=\delta_{j,\sigma(i)}$ for some $\sigma\in S_{n}$, s.t.
$d_{\sigma(1)}\leq\cdots\leq d_{\sigma(n)}$. It follows that $A_{\rm
Cent}=\operatorname{Diag}\big{(}\frac{\ell}{\gcd(\ell,d_{i})}\big{)}$.

Using the definition $C_{ij}=\frac{(\alpha_{i},\alpha_{j})}{d_{i}}$, in the
case $\operatorname{gcd}(\ell,|\pi_{1}|)\neq 1$ we obtain

$\displaystyle A_{\rm Cent}C^{T}=CA_{\rm Cent}.$

Thus,

$\displaystyle A_{\ell}A_{R}=A_{\Lambda_{W}}A_{\rm
Cent}Q_{C}^{T}P_{C}^{-1}A_{\Lambda_{W}}^{-1}A_{R}=A_{R}C^{-1}A_{\rm
Cent}Q_{C}^{T}P_{C}^{-1}C$
$\displaystyle\hphantom{A_{\ell}A_{R}}{}=A_{R}A_{\rm
Cent}\big{(}C^{T}\big{)}^{-1}Q_{C}^{T}P_{C}^{-1}C=A_{R}A_{\rm Cent}.$

By the previous lemma, this proves the first condition for $A_{\ell}$. The
second condition follows immediately from the previous lemma.

The case $\operatorname{gcd}(\ell,|\pi_{1}|)=1$ follows from Lemma 3.11 and
the fact that $|\pi_{1}|=|\Lambda_{W}^{\vee}/\Lambda_{R}^{\vee}|$. ∎

###  4.3 Case $\boldsymbol{A_{n}}$

In the following example, we treat the case $\mathfrak{g}=A_{n}$ with
fundamental group $\Lambda_{W}/\Lambda_{R}=\mathbb{Z}_{n+1}$ for general
intermediate lattices $\Lambda_{R}\subseteq\Lambda\subseteq\Lambda_{W}$.

######  Example 4.4.

In order to compute the centralizer transfer map $A_{\ell}$, we first compute
the Smith decomposition of $A_{R}$:

$\displaystyle A_{R}=\begin{pmatrix}2&-1&0&\dots&&0\\\ -1&2&-1&0&&0\\\
0&-1&2&\ddots&\ddots&\vdots\\\ 0&0&\ddots&\ddots&-1&0\\\
\vdots&&\ddots&-1&2&-1\\\ 0&\dots&&0&-1&2\end{pmatrix}$
$\displaystyle\hphantom{A_{R}}{}=\begin{pmatrix}-1&0&0&\dots&&0\\\
2&-1&0&&&0\\\ 0&2&-1&\ddots&&\vdots\\\ 0&0&\ddots&\ddots&&0\\\
\vdots&&\ddots&2&-1&0\\\
0&\dots&&0&2&1\end{pmatrix}\\!\begin{pmatrix}1&0&0&\dots&&0\\\ 0&1&0&&&0\\\
0&0&1&\ddots&&\vdots\\\ \vdots&&\ddots&\ddots&&\\\ &&\ddots&&1&0\\\
0&\dots&&&0&n+1\end{pmatrix}\\!\begin{pmatrix}-2&1&0&\dots&&0\\\
-3&0&1&\ddots&&0\\\ -4&0&0&\ddots&&\vdots\\\ \vdots&\vdots&&\ddots&1&0\\\
-n&&&&0&1\\\ 1&0&\dots&&0&0\end{pmatrix}.$

A sublattice $\Lambda_{R}\subsetneq\Lambda\subsetneq\Lambda_{W}$ is uniquely
determined by a divisor $d\,|\,n+1$, so that
$\Lambda/\Lambda_{R}\cong\mathbb{Z}_{d}$ and is generated by the multiple
$\hat{d}\lambda_{n}$, where $\hat{d}:=\frac{n+1}{d}$. Then

$\displaystyle d_{i}^{\Lambda}=\begin{cases}1,&i<n,\\\ d,&i=n.\end{cases}$

Since $A_{n}$ is simply laced with cyclic fundamental group, the formula
$A_{\Lambda}=P_{R}S_{\Lambda}P_{R}^{T}$ gives us symmetric basis matrices of
sublattices $\Lambda_{R}\subseteq\Lambda\subseteq\Lambda_{W}$. We also
substitute the above basis matrix of the root lattice $A_{R}$ by
$A_{R}(Q_{R})^{-1}P_{R}^{T}$. It is then easy to see that the definition
$A_{\ell}:=P_{R}D_{\ell}P_{R}^{T}$ gives a centralizer transfer matrix. We
calculate it explicitly

$\displaystyle(A_{\ell})_{ij}=\big{(}P_{R}D_{\ell}P_{R}^{-1}\big{)}_{ij}=\begin{cases}\delta_{ij},&i<n,\\\
\displaystyle(n+1-j)\left(\frac{\ell}{\operatorname{gcd}(\ell,d)}-1\right),&i=n\text{
and }j<n,\\\
\displaystyle\frac{\ell}{\operatorname{gcd}(\ell,d)},&i=j=n.\end{cases}$

Now a form $g$ is uniquely determined by a $d$th root of unity
$g(\chi,\chi)=\exp\big{(}\frac{2\pi i\cdot k}{d}\big{)}=\zeta_{d}^{k}$ with
some $k$. Then we calculate the form $a_{g}^{\ell}$ on the generator

$\displaystyle
a_{g}^{\ell}(\chi,\chi)=q^{-(\chi,A_{\ell}(\chi))}g(\chi,A_{\ell}(\chi))=q^{-\frac{(n+1)^{2}\cdot\ell}{d^{2}\operatorname{gcd}(\ell,\hat{d})}(\lambda_{n}^{\vee},\lambda_{n}^{\vee})}\cdot
g(\chi,\chi)^{\frac{\ell}{\operatorname{gcd}(\ell,\hat{d})}}$
$\displaystyle\hphantom{a_{g}^{\ell}(\chi,\chi)}{}=\exp\left(\frac{2\pi
i\cdot(k\ell-\hat{d}n)}{d\cdot\operatorname{gcd}(\ell,\hat{d})}\right).$

For example the trivial $g$ (i.e., $k=0$) gives an $R$-matrix for all lattices
$\Lambda$ (defined by $\hat{d}d=n+1$) iff
$\frac{\hat{d}}{\operatorname{gcd}(\ell,\hat{d})}$ is coprime to $d$. For
$\ell$ coprime to the divisor $n+1$ this amounts to all lattices associated to
decompositions of $n+1$ into two coprime factors.

###  4.4 Case $\boldsymbol{D_{n}}$

Finally, we consider the root lattice $D_{n}$. Since we have
$\pi_{1}(D_{2n\geq 4})\cong\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ and
$\pi_{1}(D_{2n+1\geq 5})\cong\mathbb{Z}_{4}$, it is appropriate to split this
investigation in two steps. We start with $D_{2n\geq 4}$. In order to compute
the respective Smith decompositions, we used the software Wolfram Mathematica.

######  Example 4.5.

In the case $D_{2n\geq 4}$, we have three different possibilities for the
lattices $\Lambda_{R}\subseteq\Lambda_{1},\Lambda_{2}\subseteq\Lambda_{W}$:

1\. $\Lambda_{1}\neq\Lambda_{2}$, $H_{1}\cong H_{2}\cong\mathbb{Z}_{2}$: In
this case, the subgroups $\Lambda_{i}/\Lambda_{R}\subseteq\Lambda_{R}$ are
spanned by the fundamental weights $\lambda_{2(n-1)+i}$. As in the case
$A_{n}$, we define the centralizer transfer map
$A_{\ell}:=P_{R}D_{\ell}P_{R}^{-1}$ on $H_{2}$. This is possible since the
symmetric basis matrix $A_{\Lambda_{2}}=P_{R}S_{\Lambda_{2}}P_{R}^{T}$ of
$\Lambda_{2}$ is already in Smith normal form. Using the software Wolfram
Mathematica in order to compute $P_{R}$, we obtain
$A_{\ell}(\lambda_{2n})=\frac{\ell}{\operatorname{gcd}(2,\ell)}$. Combining
this with $(\lambda_{2n-1},\lambda_{2n})=\frac{n-1}{2}$, we get

$\displaystyle a_{g}^{\ell}(\lambda_{2n-1},\lambda_{2n})=\exp\left(\frac{2\pi
i\cdot(kl-2(n-1))}{2\cdot\operatorname{gcd}(2,\ell)}\right)$

for $g(\lambda_{2n-1},\lambda_{2n})=\exp\big{(}\frac{2\pi ik}{2}\big{)}$.

2\. $\Lambda_{1}=\Lambda_{2}$, $H_{i}\cong\mathbb{Z}_{2}$: Without
restrictions and in order to use the same definition for $A_{\ell}$ as above,
we choose $\Lambda_{i}$, s.t. the group $\Lambda_{i}/\Lambda_{R}$ is spanned
by $\lambda_{2n}$. Combining the above result
$A_{\ell}(\lambda_{2n})=\frac{\ell}{\operatorname{gcd}(2,\ell)}$ with
$(\lambda_{2n},\lambda_{2n})=\frac{n}{2}$, we obtain

$\displaystyle a_{g}^{\ell}(\lambda_{2n},\lambda_{2n})=\exp\left(\frac{2\pi
i\cdot(kl-2n)}{2\cdot\operatorname{gcd}(2,\ell)}\right)$

for $g(\lambda_{2n},\lambda_{2n})=\exp\big{(}\frac{2\pi ik}{2}\big{)}$.

3\. $\Lambda_{1}=\Lambda_{2}=\Lambda_{W}$,
$H\cong\mathbb{Z}_{2}\times\mathbb{Z}_{2}$: A group pairing
$g\colon(\mathbb{Z}_{2}\times\mathbb{Z}_{2})\times(\mathbb{Z}_{2}\times\mathbb{Z}_{2})\to\mathbb{C}^{\times}$
is uniquely defined by a matrix $K\in\mathfrak{gl}(2,\mathbb{F}_{2})$, so that

$\displaystyle g(\lambda_{2(n-1)+i},\lambda_{2(n-1)+j})=\exp\left(\frac{2\pi
iK_{ij}}{2}\right).$

Since $D_{n}$ is simply-laced, we have $A_{\ell}=\ell\cdot\operatorname{id}$.
Using $(\lambda_{2(n-1)+i},\lambda_{2(n-1)+j})\text{ mod
}2=\delta_{i+j\text{odd}}$, we obtain

$\displaystyle
a_{\ell}^{g}(\lambda_{2(n-1)+i},\lambda_{2(n-1)+j})=\exp\left(\frac{2\pi
i\cdot K_{ij}\ell}{2}\right)(-1)^{i+j}.$

The last step is the case $D_{2n+1\geq 5}$:

######  Example 4.6.

Since it it is simply-laced and its fundamental group is cyclic, the case
$D_{2n+1\geq 5}$ can be treated very similar to $A_{n}$. We distinguish two
cases:

1\. $\Lambda_{1}=\Lambda_{2}$, $H_{i}=\langle
2\lambda_{2n+1}\rangle\cong\mathbb{Z}_{2}$. As in the case $A_{n}$, we define
the centralizer transfer map $A_{\ell}:=P_{R}D_{\ell}P_{R}^{-1}$ on $H_{2}$.
Using $(\lambda_{2n+1},\lambda_{2n+1})=\frac{2n+1}{4}$, we obtain

$\displaystyle
a_{g}^{\ell}(2\lambda_{2n+1},2\lambda_{2n+1})=\exp\left(\frac{2\pi
i\cdot(k\ell-2(2n+1))}{2\cdot\operatorname{gcd}(2,\ell)}\right).$

for $g(2\lambda_{2n+1},2\lambda_{2n+1})=\exp\big{(}\frac{2\pi ik}{2}\big{)}$.

2\. $\Lambda_{1}=\Lambda_{2}=\Lambda_{W}$,
$H=\langle\lambda_{2n+1}\rangle\cong\mathbb{Z}_{4}$. By an analogous argument
as above, we obtain

$\displaystyle
a_{g}^{\ell}(\lambda_{2n+1},\lambda_{2n+1})=\exp\left(\frac{2\pi
i\cdot(k\ell-(2n+1))}{4}\right)$

for $g(\lambda_{2n+1},\lambda_{2n+1})=\exp\big{(}\frac{2\pi ik}{4}\big{)}$.

###  4.5 Table of all quasitriangular quantum groups

In the following table, we list all simple Lie algebras and check for which
non-trivial choices of $\Lambda$, $\Lambda_{i}$, $\ell$ and $g$ the element
$R_{0}\bar{\Theta}$ is an $R$-matrix. As before, we define
$H_{i}:=\Lambda_{i}/\Lambda_{R}$ and $H:=\Lambda/\Lambda_{R}$. In the cyclic
case, if $x_{i}$ are generators of the $H_{i}$, then the pairing is uniquely
defined by an element $1\leq k\leq|H_{i}|$, s.t.
$g(x_{1},x_{2})=\exp\big{(}\frac{2\pi ik}{|H_{i}|}\big{)}$. In the case
$D_{2n}$, $\Lambda=\Lambda_{W}$, $g$ is uniquely defined by a $2\times
2$-matrix $K\in\mathfrak{gl}(2,\mathbb{F}_{2})$, s.t.
$g(\lambda_{2(n-1)+i},\lambda_{2(n-1)+j})=\exp\big{(}\frac{2\pi
iK^{g}_{ij}}{2}\big{)}$ for $i,j\in\\{1,2\\}$.

The columns of the following table are labeled by

  1. 1)

the finite-dimensional simple complex Lie algebra $\mathfrak{g}$,

  2. 2)

the natural number $\ell$, determining the root of unity
$q=\exp\big{(}{\frac{2\pi i}{\ell}}\big{)}$,

  3. 3)

the number of possible $R$-matrices for the Lusztig ansatz,

  4. 4)

the subgroups $H_{i}\subseteq H=\Lambda/\Lambda_{R}$ introduced in Theorem
3.3,

  5. 5)

the subgroups $H_{i}$ in terms of generators given by multiples of fundamental
dominant weights $\lambda_{i}\in\Lambda_{W}$,

  6. 6)

the group pairing $g\colon H_{1}\times H_{2}\to\mathbb{C}^{\times}$ determined
by its values on generators,

  7. 7)

the group pairing $a_{g}^{\ell}\subseteq\Lambda/\Lambda^{\prime}$ introduced
in Theorem 3.12 determined by its values on generators.

Table 2: Solutions for $R_{0}$-matrices. $\mathfrak{g}$ | $\ell$ | # | $H_{i}\cong$ | $H_{i}\,{\scriptstyle(i=1,2)}$ | $g$ | $a_{g}^{\ell}$
---|---|---|---|---|---|---
all |  | 1 | $\mathbb{Z}_{1}$ | $\langle 0\rangle$ | $g=1$ | $1$
|  | $\infty$ | $\mathbb{Z}_{d}$ | $\langle\hat{d}\lambda_{n}\rangle$ | $g(\hat{d}\lambda_{n},\hat{d}\lambda_{n})=\exp\big{(}\frac{2\pi ik}{d}\big{)}$ | $\exp\big{(}\frac{2\pi i\cdot(k\ell-\hat{d}n)}{d\cdot\operatorname{gcd}(\ell,\hat{d})}\big{)}$
$A_{n\geq 1}$ |
$\pi_{1}=\mathbb{Z}_{n+1}$ |  | $d\,|\,n+1$ | $\hat{d}=\frac{n+1}{d}$ | $\operatorname{gcd}\big{(}d,\frac{k\ell-\hat{d}n}{\operatorname{gcd}(\ell,\hat{d})}\big{)}=1$
| $\ell$ even | 2 | $\mathbb{Z}_{2}$ | $\langle\lambda_{n}\rangle$ | $g(\lambda_{n},\lambda_{n})=\pm 1$ | $-1$
$B_{n\geq 2}$
$\pi_{1}=\mathbb{Z}_{2}$ | $\ell$ odd | 1 | $g(\lambda_{n},\lambda_{n})=(-1)^{n+1}$ | $\exp\big{(}\frac{2\pi i\cdot(k\ell-n)}{2}\big{)}$
| $\ell\equiv 2\text{ mod }4$ | 1 | $\mathbb{Z}_{2}$ | $\langle\lambda_{n}\rangle$ | $g(\lambda_{n},\lambda_{n})=1$ | $\exp\big{(}\frac{2\pi i\cdot(k\frac{\ell}{2}+1)}{2}\big{)}$
$C_{n\geq 3}$ | $\ell\equiv 0\text{ mod }4$ | 2 | $g(\lambda_{n},\lambda_{n})=\pm 1$ | $-1$
$\pi_{1}=\mathbb{Z}_{2}$
| $\ell$ odd | 1 | $g(\lambda_{n},\lambda_{n})=-1$ | $\exp\big{(}\frac{2\pi i\cdot(k\ell-2n)}{2}\big{)}$
| $\ell\equiv 2\text{ mod }4$ | 1 | $\mathbb{Z}_{2}$ | $H_{1}\cong\langle\lambda_{2n-1}\rangle$ | $g(\lambda_{2n-1},\lambda_{2n})=(-1)^{n}$ | $\exp\big{(}\frac{2\pi i\cdot(k\frac{\ell}{2}-n+1))}{2}\big{)}$
$\ell\equiv 0\text{ mod }4$ | $2\delta_{2\,|\,n}$ | $g(\lambda_{2n-1},\lambda_{2n})=\pm 1$, $n$ even
$H_{2}\cong\langle\lambda_{2n}\rangle$
$\ell$ odd | 1 | $g(\lambda_{2n-1},\lambda_{2n})=-1$ | $\exp\big{(}\frac{2\pi i\cdot(k\ell-2(n-1))}{2}\big{)}$
$D_{2n\geq 4}$
$\pi_{1}=\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ | $\ell\equiv 2\text{ mod }4$ | $1$ | $\mathbb{Z}_{2}$ | $\langle\lambda_{2n}\rangle$ | $g(\lambda_{2n},\lambda_{2n})=(-1)^{n+1}$ | $\exp\big{(}\frac{2\pi i(k\frac{\ell}{2}-n)}{2}\big{)}$
| $\ell\equiv 0\text{ mod }4$ | $2\delta_{2\nmid n}$ | $g(\lambda_{2n},\lambda_{2n})=\pm 1$, $n$ odd
| $\ell$ odd | 1 | $g(\lambda_{2n},\lambda_{2n})=-1$ | $\exp\big{(}\frac{2\pi i(k\ell-2n)}{2}\big{)}$
| $\ell$ even | $16$ | $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ | $\langle\lambda_{2n},\lambda_{2n+1}\rangle$ | $g(\lambda_{2(n-1)+i},\lambda_{2(n-1)+j})=\pm 1$ | $\exp\big{(}\frac{2\pi i\cdot K_{ij}\ell}{2}\big{)}(-1)^{i+j}$
| $\ell$ odd | $6$ | $\det(K)=K_{12}+K_{12}$ mod $2$
| $\ell\equiv 2\text{ mod }4$ | 1 | $\mathbb{Z}_{2}$ | $\langle 2\lambda_{2n+1}\rangle$ | $g(2\lambda_{2n+1},2\lambda_{2n+1})=1$ | $\exp\big{(}\frac{2\pi i\cdot(k\frac{\ell}{2}-2n-1)}{2}\big{)}$
$\ell\equiv 0\text{ mod }4$ | 2 | $g(2\lambda_{2n+1},2\lambda_{2n+1})=\pm 1$
$D_{2n+1\geq 5}$ | $\ell$ odd | 1 | $g(2\lambda_{2n+1},2\lambda_{2n+1})=-1$ | $\exp\big{(}\frac{2\pi i\cdot(k\ell-2(2n+1))}{2}\big{)}$
$\pi_{1}=\mathbb{Z}_{4}$
| $\ell$ even | 4 | $\mathbb{Z}_{4}$ | $\langle\lambda_{2n+1}\rangle$ | $g(\lambda_{2n+1},\lambda_{2n+1})=c$, $c^{4}=1$ | $\exp\big{(}\frac{2\pi i\cdot(k\ell-(2n+1))}{4}\big{)}$
| $\ell$ odd | 2 | $g(\lambda_{2n+1},\lambda_{2n+1})=\pm 1$
| $\ell\equiv 0$ mod $3$ | 3 | $\mathbb{Z}_{3}$ | $\langle\lambda_{n}\rangle$ | $g(\lambda_{n},\lambda_{n})=c$, $c^{3}=1$ | $\exp\big{(}\frac{2\pi i\cdot(k\ell-1)}{3}\big{)}$
$E_{6}$ | $\ell\equiv 1$ mod $3$ | 2 | $g(\lambda_{n},\lambda_{n})=1,\exp\big{(}\frac{2\pi i2}{3}\big{)}$
$\pi_{1}=\mathbb{Z}_{3}$
| $\ell\equiv 2$ mod $3$ | 2 | $g(\lambda_{n},\lambda_{n})=1,\exp\big{(}\frac{2\pi i}{3}\big{)}$
| $\ell$ even | 2 | $\mathbb{Z}_{2}$ | $\langle\lambda_{n}\rangle$ | $g(\lambda_{n},\lambda_{n})=\pm 1$ | $\exp\big{(}\frac{2\pi i\cdot(k\ell-1)}{2}\big{)}$
$E_{7}$
$\pi_{1}=\mathbb{Z}_{2}$ | $\ell$ odd | 1 | $g(\lambda_{n},\lambda_{n})=1$

The Lie algebras $E_{8}$, $F_{4}$ and $G_{2}$ have trivial fundamental groups
and thus have no non-trivial solution. We want to emphasize once more that the
choice $\Lambda_{i}=\Lambda_{R}$ always leads to a quasitriangular quantum
group.

The following lemma connects our results with Lusztig’s original result:

######  Lemma 4.7.

In Lusztig’s definition of a quantum group he uses the quotient

$\displaystyle\Lambda^{\prime}_{\rm
Lusz}=2\operatorname{Cent}_{\Lambda_{R}}(2\Lambda_{W}).$

This coincide with our choice
$\Lambda^{\prime}=\operatorname{Cent}_{\Lambda_{R}}(\Lambda_{1}+\Lambda_{2})$,
if and only if

$\displaystyle
2\gcd\big{(}\ell,d_{i}^{\Lambda}\big{)}=\gcd\big{(}\ell,2d_{i}^{W}\big{)},$
(4.1)

where the $d_{i}^{\Lambda}$ denote the invariant factors of
$\Lambda_{W}^{\vee}/\Lambda$ and the $d_{i}^{W}$ denote the invariant factors
of $\Lambda_{W}^{\vee}/\Lambda_{W}$ $($i.e., ordered root lengths$)$.

In particular, for $\ell$ odd these choices never coincide. For
$\Lambda=\Lambda_{W}$, $\Lambda^{\prime}=\Lambda^{\prime}_{\rm Lusz}$ holds if
and only if $2d_{i}\,|\,\ell$. This is the most extreme case of divisibility
and it is precisely the case appearing in logarithmic conformal field
theories.

###### Proof.

We first note that in our cases,
$\Lambda^{\prime}=\operatorname{Cent}_{\Lambda_{R}}(\Lambda_{1}+\Lambda_{2})=\operatorname{Cent}_{\Lambda_{R}}(\Lambda)$.
We have

$\displaystyle
2\operatorname{Cent}_{\Lambda_{R}}(2\Lambda_{W})=2\big{(}\Lambda_{R}\cap\widehat{2\Lambda_{W}}\big{)}$
$\displaystyle\hphantom{2\operatorname{Cent}_{\Lambda_{R}}(2\Lambda_{W})}{}=A_{R}2\left(\Lambda_{W}^{\vee}\cap
A_{W}^{-1}\frac{\ell}{2}\Lambda_{W}^{\vee}\right)=A_{R}\operatorname{Diag}\left(\frac{2\ell}{\gcd(\ell,2d_{i}^{W})}\right)\Lambda_{W}^{\vee}.$

By Lemma 4.2, this coincides with $\Lambda^{\prime}$ if and only if equation
(4.1) holds. ∎

##  5 Factorizability of quantum group $\boldsymbol{R}$-matrices

We first recall the definition of factorizable braided tensor categories and
factorizable Hopf algebras, respectively.

######  Definition 5.1 ([3]).

A braided tensor category $\mathcal{C}$ is _factorizable_ if the canonical
braided tensor functor
$G\colon\mathcal{C}\boxtimes\mathcal{C}^{\text{op}}\to\mathcal{Z}(\mathcal{C})$
is an equivalence of categories.

In [17], Schneider gave a different characterization of factorizable Hopf
algebras in terms of its Drinfeld double, leading to the following theorem:

######  Definition 5.2.

A finite-dimensional quasitriangular Hopf algebra $(H,R)$ is called
_factorizable_ if its _monodromy matrix_ $M:=R_{21}\cdot R\in H\otimes H$ is
non-degenerate, i.e., the following linear map is bijective

$\displaystyle H^{*}\to H,\qquad\phi\mapsto(\operatorname{id}\otimes\phi)(M).$

Equivalently, this means we can write $M=\sum_{i}R_{1}^{i}\otimes R_{2}^{i}$
for two basis’ $R_{1}^{i},R_{2}^{j}\in H$.

######  Theorem 5.3.

Let $(H,R)$ be a finite-dimensional quasitriangular Hopf algebra. Then the
category of finite-dimensional $H$-modules $H-\mathsf{mod}_{fd}$ is
factorizable if and only if $(H,R)$ is a factorizable Hopf algebra.

Shimizu [18] has recently proven a number of equivalent characterizations of
factorizability for arbitrary (in particular non-semisimple) braided tensor
categories. Besides the two previous characterizations (equivalence to
Drinfeld center and nondegeneracy of the monodromy matrix), factorizability is
equivalent to the fact that the so-called transparent objects are all trivial,
see Theorem 5.16 below, which will become visible during our analysis later.

###  5.1 Monodromy matrix in terms of $\boldsymbol{R_{0}}$

In order to obtain conditions for the factorizability of the quasitriangular
small quantum groups
$(u_{q}(\mathfrak{g},\Lambda,\Lambda^{\prime}),R_{0}(f)\bar{\Theta})$ as in
Theorem 2.2 in terms of $\mathfrak{g}$, $q$, $\Lambda$ and $f$, we start by
calculating the monodromy matrix $M:=R_{21}\cdot R\in
u_{q}(\mathfrak{g},\Lambda,\Lambda^{\prime})\otimes
u_{q}(\mathfrak{g},\Lambda,\Lambda^{\prime})$ in general as far as possible:

######  Lemma 5.4.

For $R=R_{0}(f)\bar{\Theta}$ as in Theorem 2.2, the factorizability of $R$ is
equivalent to the invertibility of the following complex-valued matrix $m$
with entries indexed by elements in $\mu,\nu\in\Lambda/\Lambda^{\prime}$:

$\displaystyle
m_{\mu,\nu}:=\sum_{\mu^{\prime},\nu^{\prime}\in\Lambda/\Lambda^{\prime}}f(\mu-\mu^{\prime},\nu-\nu^{\prime})f(\nu^{\prime},\mu^{\prime}).$

###### Proof.

We first plug in the expressions for $R_{0}$ from Theorem 3.3 and
$\bar{\Theta}$ from Theorem 2.2 and simplify:

$\displaystyle M:=R_{21}\cdot R=(R_{0})_{21}\cdot\bar{\Theta}_{21}\cdot
R_{0}\cdot\bar{\Theta}$
$\displaystyle\hphantom{M}{}=\left(\sum_{\mu_{1},\nu_{1}\in\Lambda}f(\mu_{1},\nu_{1})K_{\nu_{1}}\otimes
K_{\mu_{1}}\right)\left(\sum_{\beta_{1}\in\Lambda_{R}^{+}}(-1)^{{\rm
tr}\beta_{1}}q_{\beta_{1}}\sum_{b_{1}\in B_{\beta_{2}}}b_{1}^{*+}\otimes
b_{1}^{-}\right)$
$\displaystyle\hphantom{M=}{}\times\left(\sum_{\mu_{2},\nu_{2}\in\Lambda}f(\mu_{2},\nu_{2})K_{\mu_{2}}\otimes
K_{\nu_{2}}\right)\left(\sum_{\beta_{2}\in\Lambda_{R}^{+}}(-1)^{{\rm
tr}\beta_{2}}q_{\beta_{2}}\sum_{b_{2}\in B_{\beta_{2}}}b_{2}^{-}\otimes
b_{2}^{*+}\right)$
$\displaystyle\hphantom{M}{}=\sum_{\beta_{1},\beta_{2}\in\Lambda_{R}^{+}}\\!(-1)^{{\rm
tr}\beta_{1}+\beta_{2}}q_{\beta_{1}}q_{\beta_{2}}\\!\left(\sum_{\mu_{1},\mu_{2},\nu_{1},\nu_{2}\in\Lambda}\\!f(\mu_{1},\nu_{1})f(\mu_{2},\nu_{2})q^{\beta_{1}(\nu_{2}-\mu_{2})}K_{\nu_{1}+\mu_{2}}\otimes
K_{\mu_{1}+\nu_{2}}\right)\\!$
$\displaystyle\hphantom{M=}{}\times\left(\sum_{b_{1}\in B_{\beta_{1}},b_{2}\in
B_{\beta_{2}}}b_{1}^{*+}b_{2}^{-}\otimes b_{1}^{-}b_{2}^{*+}\right),$

where $\Lambda_{R}^{+}=\mathbb{N}_{0}[\Delta]$. The last equation holds since
$b_{1}^{-}\in u_{\beta_{1}}^{-}$ and hence fulfills
$K_{\nu_{2}}b_{1}^{-}=q^{-\beta_{1}\nu_{2}}b_{1}^{-}K_{\nu_{2}}$ and similarly
for $b_{1}^{*+}$. We have two triangular decompositions

$\displaystyle u_{q}=u_{q}^{0}u_{q}^{-}u_{q}^{+},\qquad
u_{q}=u_{q}^{0}u_{q}^{+}u_{q}^{-},$

and the $\Lambda_{R}^{+}$-gradation on $u_{q}^{\pm}$ induces a gradation

$\displaystyle u_{q}\otimes
u_{q}\cong\bigoplus_{\beta_{1},\beta_{2}}\big{(}u^{0}\otimes
u^{0}\big{)}\big{(}{u_{q}^{+}}_{\beta_{1}}{u_{q}^{-}}_{\beta_{2}}\otimes{u_{q}^{-}}_{\beta_{1}}{u_{q}^{+}}_{\beta_{2}}\big{)}.$

The factorizability of $R$ is equivalent to the invertibility of $M$
interpreted as a matrix indexed by the PBW basis. The grading implies a block
matrix form of $M$, so the invertibility $M$ is equivalent to the
invertibility of $M^{\beta_{1},\beta_{2}}\in(u_{q}\otimes
u_{q})_{(\beta_{1},\beta_{2})}$ for every
$\beta_{1},\beta_{2}\in\Lambda_{R}^{+}$ as follows

$\displaystyle
M^{\beta_{1},\beta_{2}}:=\left(\sum_{\mu_{1},\mu_{2},\nu_{1},\nu_{2}\in\Lambda}f(\mu_{1},\nu_{1})f(\mu_{2},\nu_{2})q^{\beta_{1}(\nu_{2}-\mu_{2})}K_{\nu_{1}+\mu_{2}}\otimes
K_{\mu_{1}+\nu_{2}}\right)$
$\displaystyle\hphantom{M^{\beta_{1},\beta_{2}}:=}{}\times\left(\sum_{b_{1}\in
B_{\beta_{1}},b_{2}\in B_{\beta_{2}}}b_{1}^{*+}b_{2}^{-}\otimes
b_{1}^{-}b_{2}^{*+}\right).$

Since the second sum in $M^{\beta_{1},\beta_{2}}$ runs over a basis in
${u_{q}^{+}}_{\beta_{1}}{u_{q}^{-}}_{\beta_{2}}\otimes{u_{q}^{-}}_{\beta_{1}}{u_{q}^{+}}_{\beta_{2}}$,
the invertibility of $M$ is equivalent to the invertibility for all
$\beta_{1}\in\Lambda_{R}^{+}$ the following element:

$\displaystyle
M_{0}^{\beta_{1}}:=\sum_{\mu_{1},\mu_{2},\nu_{1},\nu_{2}\in\Lambda/\Lambda^{\prime}}q^{\beta_{1}(\nu_{2}-\mu_{2})}f(\mu_{1},\nu_{1})f(\mu_{2},\nu_{2})K_{\nu_{1}+\mu_{2}}\otimes
K_{\mu_{1}+\nu_{2}}$
$\displaystyle\hphantom{M_{0}^{\beta_{1}}}{}=\sum_{\mu,\nu\in\Lambda/\Lambda^{\prime}}K_{\nu}\otimes
K_{\mu}\cdot\left(\sum_{\mu^{\prime},\nu^{\prime}\in\Lambda/\Lambda^{\prime}}q^{\beta_{1}(\mu^{\prime}-\nu^{\prime})}f(\mu-\mu^{\prime},\nu-\nu^{\prime})f(\nu^{\prime},\mu^{\prime})\right).$

Since $K_{\nu}\otimes K_{\mu}$ is a vector space basis of $u_{q}^{0}\otimes
u_{q}^{0}=\mathbb{C}[\Lambda/\Lambda^{\prime}]\otimes\mathbb{C}[\Lambda/\Lambda^{\prime}]$,
this in turn is equivalent to the invertibility of the following family of
matrices $m^{\beta_{1}}$ for all $\beta_{1}\in\Lambda_{R}^{+}$ with
rows/columns indexed by elements in $\mu,\nu\in\Lambda/\Lambda^{\prime}$:

$\displaystyle
m^{\beta_{1}}_{\mu,\nu}:=\sum_{\mu^{\prime},\nu^{\prime}\in\Lambda/\Lambda^{\prime}}f(\mu-\mu^{\prime},\nu-\nu^{\prime})f(\nu^{\prime},\mu^{\prime})q^{\beta_{1}(\mu^{\prime}-\nu^{\prime})}.$

We now use the fact that $R$ was indeed an $R$-matrix. By property (2.1) in
Theorem 2.2 we have

$\displaystyle
m^{\beta_{1}}_{\mu,\nu}=\sum_{\mu^{\prime},\nu^{\prime}\in\Lambda/\Lambda^{\prime}}f(\mu-\mu^{\prime},\nu-\nu^{\prime})f(\nu^{\prime}+\beta_{1},\mu^{\prime})q^{-\beta_{1}\nu^{\prime}}.$

Since the invertibility of a matrix $m_{\mu,\nu}$ is equivalent to the
invertibility of any matrix $m_{\mu,\nu+\beta_{1}}$, we may substitute
$\nu^{\prime}\mapsto\nu^{\prime}+\beta_{1}$, $\nu\mapsto\nu+\beta_{1}$, pull
the constant factor $q^{-\beta_{1}^{2}}$ in front (which also does not affect
invertibility) and hence eliminate the first $\beta_{1}$ from the condition.
Hence the invertibility of $R$ is equivalent to the invertibility of the
following family of matrices $\tilde{m}^{\beta_{1}}$ for all
$\beta_{1}\in\Lambda_{R}^{+}$:

$\displaystyle\tilde{m}^{\beta_{1}}_{\mu,\nu}:=\sum_{\mu^{\prime},\nu^{\prime}\in\Lambda/\Lambda^{\prime}}f(\mu-\mu^{\prime},\nu-\nu^{\prime})f(\nu^{\prime},\mu^{\prime})q^{-\beta_{1}\nu^{\prime}}.$

We may now use the same procedure to eliminate the second $\beta_{1}$, hence
the invertibility of $R$ is equivalent to the invertibility of the following
matrix with rows/columns induced by elements in
$\mu,\nu\in\Lambda/\Lambda^{\prime}$:

$\displaystyle
m_{\mu,\nu}:=\sum_{\mu^{\prime},\nu^{\prime}\in\Lambda/\Lambda^{\prime}}f(\mu-\mu^{\prime},\nu-\nu^{\prime})f(\nu^{\prime},\mu^{\prime}).$

This was the assertion we wanted to prove. ∎

######  Definition 5.5.

Let $g\colon G_{1}\times G_{2}\to\mathbb{C}^{\times}$ be a group pairing. It
induces a symmetric form on the product $G_{1}\times G_{2}$ we denote by
$\operatorname{Sym}(g)$:

$\displaystyle\operatorname{Sym}(g)\colon\ $ $\displaystyle(G_{1}\times
G_{2})^{\times 2}\longrightarrow\mathbb{C}^{\times},$
$\displaystyle((\mu_{1},\mu_{2}),(\nu_{1},\nu_{2}))\longmapsto
g(\mu_{1},\nu_{2})g(\nu_{1},\mu_{2}).$

######  Lemma 5.6.

If $g\colon G_{1}\times G_{2}\to\mathbb{C}^{\times}$ is a perfect pairing of
abelian groups, then the symmetric form $\operatorname{Sym}(g)$ is perfect.

###### Proof.

By assumption, $g\times g$ defines an isomorphism between $G_{1}\times G_{2}$
to $\widehat{G_{2}}\times\widehat{G_{1}}$. The symmetric form
$\operatorname{Sym}(g)$ is given by the composition of this isomorphism with
the canonical isomorphism
$\widehat{G_{2}}\times\widehat{G_{1}}\cong\widehat{G_{1}\times G_{2}}$. This
proves the claim. ∎

Consider for a finite abelian group $G$ and subgroups $G_{1},G_{2}\leq G$ the
canonical exact sequence

$\displaystyle 0\to G_{1}\cap G_{2}\to G_{1}\times G_{2}\to G_{1}+G_{2}\to 0.$
(5.1)

For $\mu\in G_{1}+G_{2}$, we denote its fiber by

$\displaystyle(G_{1}\times G_{2})_{\mu}:=\\{(\mu_{1},\mu_{2})\in G_{1}\times
G_{2}\,|\,\mu_{1}+\mu_{2}=\mu\\}.$

Moreover, we define

$\displaystyle\operatorname{Rad}:=\big{\\{}(\mu_{1},\mu_{2})\in G_{1}\times
G_{2}\,|\,\operatorname{Sym}\big{(}\hat{f}\big{)}((\mu_{1},\mu_{2}),x)=1\;\forall\,x\in(G_{1}\times
G_{2})_{0}\big{\\}},$
$\displaystyle\operatorname{Rad}_{\mu}:=\operatorname{Rad}\cap(G_{1}\times
G_{2})_{\mu},$
$\displaystyle\operatorname{Rad}_{0}^{\perp}:=\\{\mu_{1}+\mu_{2}\in
G\,|\,(\mu_{1},\mu_{2})\in\operatorname{Rad}\\}.$

######  Lemma 5.7.

We have two split exact sequences:

$\displaystyle
0\to\operatorname{Rad}_{0}\to\operatorname{Rad}\to\operatorname{Rad}_{0}^{\perp}\to
0,$ $\displaystyle 0\to\operatorname{Rad}_{0}^{\perp}\to
G\to\operatorname{Rad}_{0}\to 0.$

###### Proof.

The first sequence is exact by definition of the three groups. Moreover, we
know

$\displaystyle\operatorname{Rad}=\ker\big{(}\hat{\iota}\circ\operatorname{Sym}\big{(}\hat{f}\big{)}\big{)}\cong\ker(\hat{\iota})=\text{im}(\hat{\pi})\cong\hat{G}\cong
G,$

where $\hat{\iota}$, $\hat{\pi}$ denote the duals of the inclusion and
projection in (5.1). In Example 5.11 we will see that in the case
$G_{1}=G_{2}=G$, $\hat{f}$ symmetric, $\operatorname{Rad}_{0}$ is the
$2$-torsion subgroup of $G$, and the second map in the second exact sequence
is just the projection, hence both diagrams split in this case. If $\hat{f}$
is asymmetric, we will see in Section 5.3 that $\operatorname{Rad}_{0}$ is
isomorphic to $\mathbb{Z}_{2}^{k}$ for some $k\geq 2$, thus

$\displaystyle\operatorname{Rad}_{0}^{\perp}\longrightarrow\operatorname{Rad},\qquad
x\longmapsto\sum_{\tilde{x}\in\operatorname{Rad}_{x}}\tilde{x}$

is a section of the first exact sequence. Here we used that the sum over all
elements in $\mathbb{Z}_{2}^{k}$ vanishes. Again, it follows that both
diagrams split. Finally, if $G_{1}\neq G_{2}$ (i.e., in the case $D_{2n}$),
then $\hat{f}=q^{-(\cdot,\cdot)}$ on $G_{1}\cap G_{2}$. By the same argument
as in Example 5.11, $\operatorname{Rad}_{0}$ is the $2$-torsion subgroup of
$G_{1}\cap G_{2}$. But we have $G\cong G_{1}\cap G_{2}\times\pi_{1}$ in this
case, hence both sequences split. ∎

######  Corollary 5.8.

Using the projection $\alpha\colon G\to\operatorname{Rad}_{0}^{\perp}$ and the
inclusion $\beta\colon\operatorname{Rad}_{0}^{\perp}\to\operatorname{Rad}$
from the above lemma, we can define a symmetric form on $G$:

$\displaystyle\operatorname{Sym}_{G}\big{(}\hat{f}\big{)}\colon\ $
$\displaystyle G\times G\longrightarrow\mathbb{C}^{\times},$
$\displaystyle(\mu,\nu)\longmapsto\operatorname{Sym}\big{(}\hat{f}\big{)}(\beta\circ\alpha(\mu),\beta\circ\alpha(\nu)).$

Moreover, we have
$\operatorname{Rad}\big{(}\operatorname{Sym}_{G}\big{(}\hat{f}\big{)}\big{)}\cong\operatorname{Rad}_{0}$.

######  Theorem 5.9.

We have shown in Theorem 2.2 and Lemma 3.5 that the assumption that
$R=R_{0}(f)\bar{\Theta}$ is an $R$-matrix is equivalent to the existence of
subgroups $G_{1},G_{2}\subset\Lambda/\Lambda^{\prime}$ of same order some
$d|\Lambda_{R}/\Lambda^{\prime}|$ and $f$ restricting up to a scalar to a non-
degenerate pairing $\hat{f}\colon G_{1}\times G_{2}\to\mathbb{C}^{\times}$ and
$f$ vanishes otherwise.

In this notation the matrix $m$ as defined in the previous lemma can be
rewritten as

$\displaystyle
m_{\mu,\nu}=\frac{1}{d^{2}|\Lambda_{R}/\Lambda^{\prime}|^{2}}\sum_{\begin{subarray}{c}\tilde{\mu}\in(G_{1}\times
G_{2})_{\mu}\\\ \tilde{\nu}\in(G_{1}\times
G_{2})_{\nu}\end{subarray}}\operatorname{Sym}\big{(}\hat{f}\big{)}(\tilde{\mu},\tilde{\nu}).$

It is invertible if and only if $\operatorname{Rad}_{0}=0$. In this case,

$\displaystyle m_{\mu,\nu}=\frac{|G_{1}\cap
G_{2}|}{d^{2}|\Lambda_{R}/\Lambda^{\prime}|^{2}}\operatorname{Sym}_{G}\big{(}\hat{f}\big{)}.$

We first note that $\operatorname{Rad}_{0}=0$ implies
$\operatorname{Rad}_{0}^{\perp}=G$ and thus $G=G_{1}+G_{2}$. Together with
Corollary 3.6 this implies

######  Corollary 5.10.

$\displaystyle\Lambda^{\prime}=\operatorname{Cent}_{\Lambda_{R}}(\Lambda).$

Before we proof the theorem, we first give a simple example:

######  Example 5.11.

Let $G_{1}=G_{2}=G$ (correspondingly $\Lambda_{1}=\Lambda_{2}=\Lambda$) and
assume $\hat{f}$ is symmetric non-degenerate, then the radical measures
$2$-torsion:

$\displaystyle\operatorname{Rad}\big{(}\operatorname{Sym}_{G}\big{(}\hat{f}\big{)}\big{)}\cong\operatorname{Rad}_{0}=\\{\mu\in
G\,|\,2\mu=0\\}.$

Again, this is the only case appearing for cyclic fundamental groups. Hence in
all cases except $\mathfrak{g}=D_{2n}$ factorizability is equivalent to
$|\Lambda/\Lambda^{\prime}|$ being odd.

###### Proof of Theorem 5.9.

The first part of the theorem follows by applying Lemma 3.5 to the matrix $m$
as given in the previous lemma. Now, assume that $m$ is invertible. We must
have $G=G_{1}+G_{2}$, otherwise the matrix has zero-columns and rows,
differently formulated: the fibers $(G_{1}\times G_{2})_{\mu}$ in the short
exact sequence must be non-empty for all $\mu\in G$. If on the other hand,
$\operatorname{Rad}_{0}=0$, then $\operatorname{Rad}_{0}^{\perp}=G$ and thus
$G_{1}+G_{2}=G$ must also hold, thus we assume this from now on. By the short
exact sequence the fiber $(G_{1}\times G_{2})_{0}\cong G_{1}\cap G_{2}$, other
fibers are of the explicit form $\tilde{\mu}+(G_{1}\times G_{2})_{0}$ for some
choice of representative $\tilde{\mu}$. Therefore,

$\displaystyle
m_{\mu,\nu}=\frac{1}{d^{2}|\Lambda_{R}/\Lambda^{\prime}|^{2}}\sum_{\begin{subarray}{c}\tilde{\mu}\in(G_{1}\times
G_{2})_{\mu}\\\ \tilde{\nu}\in(G_{1}\times
G_{2})_{\nu}\end{subarray}}\operatorname{Sym}\big{(}\hat{f}\big{)}(\tilde{\mu},\tilde{\nu})$
$\displaystyle\hphantom{m_{\mu,\nu}}{}=\frac{1}{d^{2}|\Lambda_{R}/\Lambda^{\prime}|^{2}}\sum_{\tilde{\nu}\in(G_{1}\times
G_{2})_{\nu}}\operatorname{Sym}\big{(}\hat{f}\big{)}(\tilde{\mu},\tilde{\nu})\sum_{\tilde{\eta}\in(G_{1}\times
G_{2})_{0}}\operatorname{Sym}\big{(}\hat{f}\big{)}(\tilde{\eta},\tilde{\nu})$
$\displaystyle\hphantom{m_{\mu,\nu}}{}=\frac{|G_{1}\cap
G_{2}|}{d^{2}|\Lambda_{R}/\Lambda^{\prime}|^{2}}\sum_{\tilde{\nu}\in(G_{1}\times
G_{2})_{\nu}}\operatorname{Sym}\big{(}\hat{f}\big{)}(\tilde{\mu},\tilde{\nu})\cdot\delta_{\operatorname{Sym}(f)(\tilde{\nu},\\_)|_{G_{1}\cap
G_{2}}=1}=(*).$

Fix as above a representative $\tilde{\nu}$ of the fiber of $\nu$, i.e.,
$\tilde{\nu}\in(G_{1}\times G_{2})_{\nu}$ such that
$\operatorname{Sym}(f)(\tilde{\nu},\\_)|_{G_{1}\cap G_{2}}$ $=1$ holds. Two
elements fulfilling this property differ by an element in the subgroup
$\operatorname{Rad}_{0}\leq G_{1}\cap G_{2}$, thus

$\displaystyle(*)=\frac{|G_{1}\cap
G_{2}|}{d^{2}|\Lambda_{R}/\Lambda^{\prime}|^{2}}\operatorname{Sym}\big{(}\hat{f}\big{)}(\tilde{\mu},\tilde{\nu})\sum_{\tilde{\xi}\in\operatorname{Rad}_{0}}\operatorname{Sym}\big{(}\hat{f}\big{)}(\tilde{\xi},\tilde{\nu})\cdot\delta_{\operatorname{Sym}(f)(\tilde{\nu},\\_)|_{G_{1}\cap
G_{2}}=1}$ $\displaystyle\hphantom{(*)}{}=\frac{|G_{1}\cap
G_{2}||\operatorname{Rad}_{0}|}{d^{2}|\Lambda_{R}/\Lambda^{\prime}|^{2}}\operatorname{Sym}\big{(}\hat{f}\big{)}(\tilde{\mu},\tilde{\nu})\cdot\delta_{\operatorname{Sym}(\hat{f})(\tilde{\nu},\\_)|_{G_{1}\cap
G_{2}}=1}\,\delta_{\operatorname{Sym}(\hat{f})(\tilde{\mu},\\_)|_{\operatorname{Rad}_{0}}=1}.$

Since $m$ is symmetric, we have

$\displaystyle m_{\mu,\nu}=\frac{|G_{1}\cap
G_{2}||\operatorname{Rad}_{0}|}{d^{2}|\Lambda_{R}/\Lambda^{\prime}|^{2}}\operatorname{Sym}\big{(}\hat{f}\big{)}(\tilde{\mu},\tilde{\nu})\cdot\delta_{\operatorname{Sym}(\hat{f})(\tilde{\nu},\\_)|_{G_{1}\cap
G_{2}}=1}\delta_{\operatorname{Sym}(\hat{f})(\tilde{\mu},\\_)|_{G_{1}\cap
G_{2}}=1}$ $\displaystyle\hphantom{m_{\mu,\nu}}{}=\frac{|G_{1}\cap
G_{2}||\operatorname{Rad}_{0}|}{d^{2}|\Lambda_{R}/\Lambda^{\prime}|^{2}}\operatorname{Sym}_{G}\big{(}\hat{f}\big{)}(\mu,\nu)\delta_{\operatorname{Rad}_{\mu}\neq\varnothing}\delta_{\operatorname{Rad}_{\nu}\neq\varnothing}$

and this is invertible if an only if
$\operatorname{Rad}_{0}\cong\operatorname{Rad}\big{(}\operatorname{Sym}_{G}\big{(}\hat{f}\big{)}\big{)}=0$.
∎

###  5.2 Factorizability for symmetric $\boldsymbol{R_{0}(f)}$

For $R_{0}=\sum_{\mu,\nu}f(\mu,\nu)K_{\mu}\otimes K_{\nu}$ being the Cartan
part of an $R$-matrix, assume that $\hat{f}=|G|f$ on $G$ is symmetric. We have
shown in Example 5.11 that factorizability is equivalent to $|G|$ being odd.

We now want to give a necessary and sufficient condition for this:

######  Lemma 5.12.

Let $\Lambda_{R}\subseteq\Lambda\subseteq\Lambda_{W}$ be an arbitrary
intermediate lattice for a certain irreducible root system. Then the order of
the group $G=\Lambda/\operatorname{Cent}_{\Lambda_{R}}(\Lambda)$ is odd if and
only if both of the following conditions are satisfied:

  1. $1)$

$|\Lambda/\Lambda_{R}|$ is odd,

  2. $2)$

$\ell$ is either odd or $(\ell\equiv 2$ mod $4$, $\mathfrak{g}=B_{n}$,
$\Lambda=\Lambda_{R})$ including $A_{1}$.

###### Proof.

We saw that in all our cases, there exists an isomorphism

$\displaystyle\Lambda/\Lambda_{R}\cong\operatorname{Cent}_{\Lambda}(\Lambda_{R})/\operatorname{Cent}_{\Lambda_{R}}(\Lambda).$

Moreover, from Lemma 4.2 we know that
$|\Lambda/\operatorname{Cent}_{\Lambda}(\Lambda_{R})|=\det(D_{\ell})$, where
$D_{\ell}$ was the diagonal matrix
$\operatorname{Diag}\big{(}\frac{\ell}{\operatorname{gcd}(\ell,d_{i}^{\Lambda})}\big{)})$
with $d_{i}^{\Lambda}$ being the invariant factors of the lattice $\Lambda$
(i.e., the diagonal entries of the Smith normal form of a basis matrix of
$\Lambda$). Thus,

$\displaystyle|G|=|\Lambda/\operatorname{Cent}_{\Lambda_{R}}(\Lambda)|=|\Lambda/\operatorname{Cent}_{\Lambda}(\Lambda_{R})||\operatorname{Cent}_{\Lambda}(\Lambda_{R})/\operatorname{Cent}_{\Lambda_{R}}(\Lambda)|$
$\displaystyle\hphantom{|G|}{}=|\Lambda/\operatorname{Cent}_{\Lambda}(\Lambda_{R})||\Lambda/\Lambda_{R}|=\det(D_{\ell})|\Lambda/\Lambda_{R}|=\prod_{i=1}^{n}\frac{\ell}{\operatorname{gcd}(\ell,d_{i}^{\Lambda})}|\Lambda/\Lambda_{R}|.$

Clearly, this term is odd if $\ell$ and $|\Lambda/\Lambda_{R}|$ are odd. In
the case ($\ell\equiv 2$ mod $4$, $\mathfrak{g}=B_{n}$,
$\Lambda=\Lambda_{R}$), the Smith normal form $S_{R}$ of the basis matrix
$A_{R}$ is given by $2\cdot\operatorname{id}$. Thus, $|G|$ is odd in this
case. On the other hand, let $|G|$ be odd:

We first consider the case $\ell$ even. A necessary condition for
$|\Lambda/\Lambda^{\prime}|$ odd is that the multiplicity $m_{\ell}$ of the
prime $2$ in
$\prod\limits_{i=1}^{n}\frac{\ell}{\operatorname{gcd}(\ell,d_{i}^{\Lambda})}$
is at most the multiplicity $m_{\pi_{1}}$ of the prime $2$ in $|\pi_{1}|$. We
check this condition for rank $n>1$:

  * •

For $\mathfrak{g}$ simply-laced (or triply-laced $\mathfrak{g}=G_{2}$) we have
all $d_{i}=1$, hence $n\,|\,m_{\ell}$ (equality for $\ell=2$ ${\rm mod}~{}4$).
The cases $D_{n}$ with $m_{\pi_{1}}=2$ have rank $n\geq 4$, all others except
$A_{n}$ have $m_{\pi_{1}}=0,1$, so the necessary condition $m_{\ell}\leq
m_{\pi_{1}}$ is never fulfilled. The cases $A_{n}$ have
$2^{m_{\pi_{1}}}|(n+1)\leq(m_{\ell}+1)\stackrel{{\scriptstyle!}}{{\leq}}(m_{\pi_{1}}+1)$
which can only be true in rank $n=1$ treated above.

  * •

For $\mathfrak{g}$ doubly-laced of rank $n>1$, we always have always
$m_{\pi_{1}}=0,1$ but $m_{\ell}$ can be considerably smaller than above,
namely for $\ell=2$ ${\rm mod}~{}4$ equal to the number of short simple roots
$d_{\alpha_{i}}=1$ (otherwise $m_{\ell}$ again increases by $n$ for every
factor $2$ in $\ell$), hence the necessary condition $m_{\ell}\leq
m_{\pi_{1}}$ can be fulfilled only for $B_{n}$ (which would also include
$A_{1}$ above for $n=1$). More precisely, since $m_{\ell}=m_{\pi_{1}}$ and the
decomposition for $\Lambda/\Lambda^{\prime}$ has an additional factor
$|\Lambda/\Lambda_{R}|$, it can only be odd for $\Lambda=\Lambda_{R}$.

On the other hand, if $\ell$ is odd, then the whole product term is odd. But
since $|G|$ was assumed to be odd, also $|\Lambda/\Lambda^{\prime}|$ must be
odd. ∎

######  Corollary 5.13.

Let $\Lambda=\Lambda_{R}$. In the previous section we have seen that
$\hat{f}=q^{-(\cdot,\cdot)}$ gives always an $R$-matrix in this case. By the
proof of the previous lemma, we have

$\displaystyle\operatorname{Rad}_{0}\cong\prod_{i=1}^{n}\mathbb{Z}_{\gcd\big{(}2,\frac{\ell}{\gcd(\ell,d_{i}^{R})}\big{)}},$

where the $d_{i}^{R}$ denote the invariant factors of
$\Lambda_{W}^{\vee}/\Lambda_{R}$.

###  5.3 Factorizability for $\boldsymbol{D_{2n}}$, $\boldsymbol{R_{0}}$
antisymmetric

The split case $\mathfrak{g}=D_{2n}$, $G=G_{1}\times G_{2}$ is clearly
factorizable, so the only remaining case for which we have to check
factorizabilty is $\mathfrak{g}=D_{2n}$, $\Lambda=\Lambda_{W}$ for $\hat{f}$
being not symmetric. We know that in this case, the corresponding form $g$ on
$\Lambda/\Lambda_{R}$ is uniquely defined by a $2\times 2$-matrix
$K\in\mathfrak{gl}(2,\mathbb{F}_{2})$, s.t.
$g(\lambda_{2(n-1)+i},\lambda_{2(n-1)+j})=\exp\big{(}\frac{2\pi
iK_{ij}}{2}\big{)}$ for $i,j\in\\{1,2\\}$. From this we see that if $g$ is not
symmetric, it must be antisymmetric, i.e., $g(\mu,\nu)=g(\nu,\mu)^{-1}$. Thus,
the following lemma applies in this case, and hence there are no factorizable
$R$-matrices for $D_{2n}$, $\Lambda=\Lambda_{W}$.

######  Lemma 5.14.

For $\mathfrak{g}$ simply-laced and $\Lambda=\Lambda_{W}$, let
$\hat{f}=q^{-(\cdot,\cdot)}g\colon G\times G\to\mathbb{C}^{\times}$ be a non-
degenerate form as in Theorem 3.3 and Lemma 3.5, s.t. the form
$g\colon\pi_{1}\times\pi_{1}\to\mathbb{C}^{\times}$ is asymmetric. Then,

$\displaystyle\operatorname{Rad}_{0}\cong\bigoplus_{i=1}^{n}\mathbb{Z}_{\gcd(2,\ell
d_{i}^{R})},$

where the $d_{i}^{R}$ denote the invariant factors of $\pi_{1}$. In
particular, $\operatorname{Rad}_{0}=0$ holds if and only if
$\gcd(2,\ell|\pi_{1}|)=1$.

###### Proof.

We recall the definition of
$\operatorname{Rad}_{0}\big{(}\operatorname{Sym}_{G}\big{(}\hat{f}\big{)}\big{)}$
in this case:

$\displaystyle\operatorname{Rad}_{0}(\operatorname{Sym}_{G}(\hat{f}))=\big{\\{}\mu\in
G\,|\,f(\nu,\mu)^{-1}=f(\mu,\nu)\ \forall\,\nu\in G\big{\\}}$
$\displaystyle\hphantom{\operatorname{Rad}_{0}(\operatorname{Sym}_{G}(\hat{f}))}{}=\big{\\{}\mu\in
G\,|\,q^{(\nu,\mu)}g(\nu,\mu)^{-1}=q^{-(\mu,\nu)}g(\mu,\nu)\ \forall\,\nu\in
G\big{\\}}$
$\displaystyle\hphantom{\operatorname{Rad}_{0}(\operatorname{Sym}_{G}(\hat{f}))}{}=\big{\\{}\mu\in
G\,|\,q^{(\nu,\mu)}=q^{-(\mu,\nu)}\ \forall\,\nu\in G\big{\\}}$
$\displaystyle\hphantom{\operatorname{Rad}_{0}(\operatorname{Sym}_{G}(\hat{f}))}{}=\big{\\{}\mu\in
G\,|\,q^{(2\mu,\nu)}=1\ \forall\,\nu\in G\big{\\}}$
$\displaystyle\hphantom{\operatorname{Rad}_{0}(\operatorname{Sym}_{G}(\hat{f}))}{}=\big{\\{}\mu\in
G\,|\,2\mu\in\operatorname{Cent}_{2\Lambda_{W}}(\Lambda_{W})/2\operatorname{Cent}_{\Lambda_{R}}(\Lambda_{W})\big{\\}}=(\ast).$

For $\mathfrak{g}$ is simply-laced, we have $\Lambda_{W}=\Lambda_{W}^{\vee}$,
thus

$\displaystyle(\ast)\cong\operatorname{Cent}_{2\Lambda_{W}}(\Lambda_{W})/2\operatorname{Cent}_{\Lambda_{R}}(\Lambda_{W})=(2\Lambda_{W}\cap\ell
A_{R}\Lambda_{W})/2\ell A_{R}\Lambda_{W}$
$\displaystyle\hphantom{(\ast)}{}=P_{R}\operatorname{Diag}\big{(}\operatorname{lcm}\big{(}2,\ell
d_{i}^{R}\big{)}\big{)}\Lambda_{W}/P_{R}2\ell
S_{R}\Lambda_{W}=\Lambda_{W}/\operatorname{Diag}\big{(}\gcd\big{(}2,\ell
d_{i}^{R}\big{)}\big{)}\Lambda_{W}.$

This proves the claim. ∎

###  5.4 Transparent objects in non-factorizable cases

In this section, we determine the transparent objects in the representation
category of $u_{q}(\mathfrak{g},\Lambda)$ with our $R$-matrix given by
$R_{0}\bar{\Theta}$ and
$R_{0}=\frac{1}{|\Lambda/\Lambda^{\prime}|}\sum\limits_{\mu,\nu\in\Lambda/\Lambda^{\prime}}\hat{f}$
with $\hat{f}$ a group pairing
$\Lambda_{1}/\Lambda^{\prime}\times\Lambda_{2}/\Lambda^{\prime}\to\mathbb{C}^{\times}$.

######  Definition 5.15.

Let $\mathcal{C}$ be a braided monoidal category with braiding $c$. An object
$V\in\mathcal{C}$ is called _transparent_ if the double braiding $c_{W,V}\circ
c_{V,W}$ is the identity on $V\otimes W$ for all $W\in\mathcal{C}$.

The following theorem by Shimizu gives a very important characterization of
factorizable categories:

######  Theorem 5.16 ([18, Theorem 1.1]).

A braided finite tensor category is factorizable if and only if the
transparent objects are direct sums of finitely many copies of the unit
object.

######  Corollary 5.17.

In particular, for a Hopf algebra $H$ the representation category
$H-\mathsf{mod}_{fd}$ is factorizable if and only if the transparent objects
are multiples of the trivial representation and vice versa.

Since in our cases $\Lambda_{1}\neq\Lambda_{2}$ can only appear in $D_{2n}$,
and we know those are factorizable, we shall in the following restrict
ourselves to the case $\Lambda_{1}=\Lambda_{2}=\Lambda$. The proof below works
also in the more general case, but requires more notation. As usual we first
reduce the Hopf algebra question to the group ring and then solve the group
theoretical problem.

######  Lemma 5.18.

If a $u_{q}(\mathfrak{g})$-module $V$, with a highest-weight vector $v$ and
$K_{\mu}v=\chi(K_{\mu})v$, is a transparent object, then necessarily the
$1$-dimensional $\Lambda/\Lambda^{\prime}$-module $\mathbb{C}_{\chi}$ is a
transparent object over the Hopf algebra
$\mathbb{C}[\Lambda/\Lambda^{\prime}]$ with $R$-matrix $R_{0}$. If $V$ is
$1$-dimensional, then $V$ is transparent if and only if $\mathbb{C}_{\chi}$
is.

###### Proof.

Let $V$ be transparent. For every
$\psi\colon\Lambda/\Lambda^{\prime}\to\mathbb{C}^{\times}$ we have another
finite-dimensional module
$W:=u_{q}(\mathfrak{g})\otimes_{u_{q}(\mathfrak{g})^{+}}\mathbb{C}_{\psi}$
with highest weight vector $w=1\otimes 1_{\psi}$ which we can test this
assumption against

$\displaystyle c^{2}\colon\ V\otimes W\to W\otimes V\to V\otimes W.$

We calculate the effect of $c^{2}$ on the highest-weight vectors $v\otimes w$:

$\displaystyle c^{2}(v\otimes w)=\tau_{W\otimes
V}R_{0}\bar{\Theta}\tau_{V\otimes W}R_{0}\bar{\Theta}(v\otimes w).$

Because $v$, $w$ were assumed highest-weight vectors, the $\bar{\Theta}$ act
trivially. Hence follows that $\mathbb{C}_{\chi}$, $\mathbb{C}_{\psi}$ have a
trivial double braiding over the Hopf algebra
$\mathbb{C}[\Lambda/\Lambda^{\prime}]$ with $R$-matrix $R_{0}$. Because we
could achieve this result for any $\psi$ this means that $\mathbb{C}_{\chi}$
is transparent as asserted.

Now, let $V=\mathbb{C}_{\chi}$ be $1$-dimensional over $u_{q}(\mathfrak{g})$
and transparent over $\mathbb{C}[\Lambda/\Lambda^{\prime}]$, and let $w$ be
any element in any module $W$, then again the two $\Theta$ act trivially, one
time because $v=1_{\chi}$ is a highest weight vector, and one time because it
is also a lowest weight vector. But if the double-braiding of $v=1_{\chi}$
with any element $w$ is trivial, then $V=\mathbb{C}_{\chi}$ is already
transparent over $u_{q}(\mathfrak{g})$. ∎

######  Lemma 5.19.

$\mathbb{C}_{\chi}$ is a transparent object over the Hopf algebra
$\mathbb{C}[\Lambda/\Lambda^{\prime}]$ with $R$-matrix $R_{0}$ iff it is an
$f$-transformed of the radical of
$\operatorname{Sym}_{G}\big{(}\hat{f}\big{)}$, i.e.,

$\displaystyle\chi(\mu)=f(\mu,\xi),\qquad\xi\in\operatorname{Rad}_{0}.$

###### Proof.

Since $f$ is nondegenerate, we can assume $\chi(\mu)=f(\mu,\xi)$ and wish to
prove $\mathbb{C}_{\chi}$ is transparent iff $\xi\in\operatorname{Rad}_{0}$.
We test transparency against any module $\mathbb{C}_{\psi}$ and also write
$\psi(\mu)=f(\lambda,\mu)$ (note the order of the argument). We evaluate the
double-braiding on $1_{\chi}\otimes 1_{\psi}$ and get the following scalar
factor, which needs to be $=1$ for all $\psi$ in order to make
$\mathbb{C}_{\chi}$ transparent:

$\displaystyle\frac{1}{|G|^{2}}\sum_{\mu,\nu}\chi(\mu)\psi(\nu)\sum_{\begin{subarray}{c}\mu_{1}+\mu_{2}=\mu\\\
\nu_{1}+\nu_{2}=\mu\end{subarray}}\operatorname{Sym}\big{(}\hat{f}\big{)}((\mu_{1},\mu_{2}),(\nu_{1},\nu_{2}))$
$\displaystyle\qquad{}=\frac{1}{|G|^{2}}\sum_{\mu,\nu}f(\mu,\xi)f(\lambda,\nu)\sum_{\begin{subarray}{c}\mu_{1}+\mu_{2}=\mu\\\
\nu_{1}+\nu_{2}=\mu\end{subarray}}f(\mu_{1},\nu_{1})f(\nu_{2},\mu_{2})$
$\displaystyle\qquad{}=\frac{1}{|G|^{2}}\sum_{\mu,\nu}f(\mu,\xi)f(\lambda,\nu)\sum_{\nu_{1},\mu_{1}}f(\mu_{1},\nu_{1})f(\nu,\mu)f^{-1}(\nu_{1},\mu)f^{-1}(\nu,\mu_{1})f(\nu_{1},\mu_{1})$
$\displaystyle\qquad{}=\frac{1}{|G|}\sum_{\nu}f(\lambda,\nu)\sum_{\nu_{1},\mu_{1}}f(\mu_{1},\nu_{1})\delta_{\xi=-\nu+\nu_{1}}f^{-1}(\nu,\mu_{1})f(\nu_{1},\mu_{1})$
$\displaystyle\qquad{}=\frac{1}{|G|}\sum_{\nu}f(\lambda,\nu)\sum_{\mu_{1}}f(\mu_{1},\xi+\nu)f(\xi,\mu_{1})=f^{-1}(\lambda,\xi)f^{-1}(\xi,\lambda)=\operatorname{Sym}_{G}\big{(}\hat{f}\big{)}(\lambda,\xi).$

This scalar factor of the double braiding is equal $+1$ for all $\lambda$ (and
hence all $\mathbb{C}_{\psi}$) iff $\xi\in\operatorname{Rad}_{0}$ as asserted.
∎

The previous two lemmas combined imply that any irreducible transparent
$u_{q}(\mathfrak{g})$-module has necessarily the characters
$\chi(\mu)=f(\mu,\xi)$, $\xi\in\operatorname{Rad}_{0}$ as highest-weights, and
conversely if such a character $\chi$ gives rise to $1$-dimensional
$u_{q}(\mathfrak{g})$-modules (i.e., $\chi|_{2\Lambda_{R}}=1$), then these are
guaranteed transparent objects. Hence the final step is to give more closed
expressions for the $f$-transformed characters $\chi$ of the radical depending
on the case and check the $1$-dimensionality condition.

In all cases where $f$ is symmetric we have seen in Example 5.11 that
$\operatorname{Rad}_{0}\big{(}\operatorname{Sym}_{G}\big{(}\hat{f}\big{)}\big{)}$
is the $2$-torsion subgroup of $\Lambda/\Lambda^{\prime}$, so in these cases
$\chi$ gives rise to a $1$-dimensional object.

######  Corollary 5.20.

If $f$ is symmetric $($true for all cases except $D_{2n})$ then the
transparent objects are all $1$-dimensional $\mathbb{C}_{\chi}$ where the
characters $\chi$ are the $f$-transformed of the elements in the radical of
the bimultiplicative form $\operatorname{Sym}\big{(}\hat{f}\big{)}|_{G}$ on
$G=\Lambda/\Lambda^{\prime}$. In particular the group of transparent objects
is isomorphic to this radical as an abelian group.

######  Corollary 5.21.

In the case of symmetric $f$ $($all cases except $D_{2n})$ the fact that
$\operatorname{Rad}_{0}$ is the $2$-torsion of $\Lambda/\Lambda^{\prime}$ and
$f$-transformation is a group isomorphism shows:

The group $T$ of transparent objects consists of $\mathbb{C}_{\chi}$ where
$\chi|_{2\Lambda}=1$, i.e., the two-torsion of the character group.

The remaining case in $D_{2n}$ with $f$ nonsymmetric and has been done by hand
in Lemma 5.14.

##  6 Quantum groups with a ribbon structure

In [16, Theorem 8.23], the existence of ribbon structures for
$u_{q}(\mathfrak{g},\Lambda)$ is proven. In this section we construct a ribbon
structure for all cases. In the proof, we use several auxiliary results from
[16].

######  Theorem 6.1.

Let $u_{q}(\mathfrak{g},\Lambda)$ be quasitriangular Hopf algebra, with an
$R$-matrix satisfying the conditions in Theorem 2.2 and let
$u:=S(R_{(2)})R_{(1)}$. Then $v:=K_{\nu_{0}}^{-1}u$ is a ribbon element in
$u_{q}(\mathfrak{g},\Lambda)$.

###### Proof.

We consider the natural $\mathbb{N}_{0}[\alpha_{i}\,|\,i\in I]$-grading on the
Borel parts $u^{\pm}:=u_{q}(\mathfrak{g},\Lambda)^{\pm}$ [13]. Since $u^{\pm}$
is finite-dimensional, there exists a maximal
$\nu_{0}\in\mathbb{N}_{0}[\alpha_{i}\,|\,i\in I]$, s.t. the homogeneous
component $u_{\nu_{0}}^{\pm}$ is non-trivial. More explicitly $\nu_{0}$ is of
the form

$\displaystyle\nu_{0}=\sum_{\alpha\in\Phi^{+}}(\ell_{\alpha}-1)\alpha,$

where $\ell_{\alpha}:=\frac{\ell}{\operatorname{gcd}(\ell,2d_{\alpha})}$.

Using the formulas $u=\big{(}\sum f(\mu,\nu)K_{\mu+\nu}\big{)}^{-1}\vartheta$
and $S(u)=\big{(}\sum f(\mu,\nu)K_{\mu+\nu}\big{)}^{-1}S(\vartheta)$, where
$\vartheta=\sum\bar{\Theta}^{(2)}S^{-1}\big{(}\bar{\Theta}^{(2)}\big{)}$,
Müller proves the formula $K_{-\nu_{0}}^{2}=u^{-1}S(u)$. Using the fact that
$u$ commutes with all grouplike elements, this implies $v^{2}=uS(u)$. In order
to show that $v$ is central, we first show that $K_{\nu_{0}+2\rho}^{-1}$ is a
central element. By the $K,E$-relations, this is equivalent to

$\displaystyle\nu_{0}+2\rho\in\operatorname{Cent}_{\Lambda}(\Lambda_{R}),$

where $\rho=\frac{1}{2}\sum\limits_{\alpha\in\Phi^{+}}\,\alpha$ is the Weyl
vector.

We calculate directly that this is always the case:

$\displaystyle(\nu_{0}+2\rho,\beta)=q^{\sum\limits_{\alpha\in\Phi^{+}}(\ell_{\alpha}-1+1)(\alpha,\beta)}=q^{\ell\sum\limits_{\alpha\in\Phi^{+}}\frac{1}{\operatorname{gcd}(\ell,2d_{\alpha})}\cdot
2d_{\alpha}(\alpha^{\vee},\beta)}=1.$

Since $K_{2\rho}ux=xK_{2\rho}u$ holds for all $x\in
u_{q}(\mathfrak{g},\Lambda)$ (see [16, Lemmas 8.22 and 8.19]), we have

$\displaystyle
vx=K_{\nu_{0}}^{-1}ux=K^{-1}_{\nu_{0}+2\rho}K_{2\rho}ux=K^{-1}_{\nu_{0}+2\rho}xK_{2\rho}u=xK^{-1}_{\nu_{0}+2\rho}K_{2\rho}u=xv,$

hence $v$ is central. ∎

##  7 Open questions

######  Question 7.1.

It was surprising to us that the case $D_{2n}=\mathfrak{so}_{4n}(\mathbb{C})$
has so many more solutions that the other cases, in particular with non-
symmetric $R_{0}$, due to the non-cyclic fundamental group. Do these
additional modular tensor categories appear elsewhere? Does the non-symmetry
have interesting implications on the category?

######  Question 7.2.

Our procedure would be similarly possible for any diagonal Nichols algebra.
The Lusztig ansatz can in these cases be found in [1].

######  Question 7.3.

In each case where $u_{q}(\mathfrak{g},\Lambda),R$ is not factorizable, we can
modularize $($see [2]) the corresponding representation category and get a
modular tensor category, which should be representations over some “quasi-
quantum group” $u_{q}(\mathfrak{g},\tilde{\Lambda},\omega),R$ which is a
quasi-Hopf algebra where the group ring $\mathbb{C}[\tilde{\Lambda}]$ is
deformed by a $3$-group-cocycle $\omega$. Can we describe this quasi-Hopf
algebra in a closed form? Moreover, is every factorizable quasi-quantum group
the modularization of a quasi-triangular quantum group from our list?

More technically:

######  Question 7.4.

The centralizer transfer map $A_{\ell}$ in Definition 3.8 $($and
correspondingly the form $a_{\ell})$ had a very general characterization, but
we could only prove existence by a construction using the classification of
simple Lie algebras $($and distinguishing three cases$)$. We strongly suspect
that these maps exist under rather general assumptions.

Also the result Theorem 3.3 from our previous article [11] has only been
proven there for cyclic groups $($and by hand for
$\mathbb{Z}_{2}\times\mathbb{Z}_{2})$ although we strongly suspect it holds
for every abelian group.

### Acknowledgements

Both authors thank Christoph Schweigert for helpful discussions and support.
They also thank the referees, who gave a relevant contribution to improve the
article with their comments. The first author was supported by the DAAD
P.R.I.M.E program funded by the German BMBF and the EU Marie Curie Actions as
well as the Graduiertenkolleg RTG 1670 at the University of Hamburg. The
second author was supported by the Collaborative Research Center SFB 676 at
the University of Hamburg.

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