Modular Group Representations in Combinatorial Quantization

Modular Group Representations in Combinatorial
Quantization with Non-Semisimple Hopf Algebras

Matthieu FAITG

M. Faitg

IMAG, Univ Montpellier, CNRS, Montpellier, France<EMAIL_ADDRESS>
Received February 02, 2019, in final form September 24, 2019; Published online
October 03, 2019

Let $\Sigma_{g,n}$ be a compact oriented surface of genus $g$ with $n$ open
disks removed. The algebra $\mathcal{L}_{g,n}(H)$ was introduced by
Alekseev–Grosse–Schomerus and Buffenoir–Roche and is a combinatorial
quantization of the moduli space of flat connections on $\Sigma_{g,n}$. Here
we focus on the two building blocks $\mathcal{L}_{0,1}(H)$ and
$\mathcal{L}_{1,0}(H)$ under the assumption that the gauge Hopf algebra $H$ is
finite-dimensional, factorizable and ribbon, but not necessarily semisimple.
We construct a projective representation of $\mathrm{SL}_{2}(\mathbb{Z})$, the
mapping class group of the torus, based on $\mathcal{L}_{1,0}(H)$ and we study
it explicitly for $H=\overline{U}_{q}(\mathfrak{sl}(2))$. We also show that it
is equivalent to the representation constructed by Lyubashenko and Majid.

combinatorial quantization; factorizable Hopf algebra; modular group;
restricted quantum group

16T05; 81R05

##  1 Introduction

Let $\Sigma_{g,n}$ be a compact oriented surface of genus $g$ with $n$ open
disks removed and let $G$ be a connected, simply-connected Lie group. Fock and
Rosly defined a Poisson structure on the space of flat $G$-connections on
$\Sigma_{g,n}{\setminus}D$ (where $D$ is an open disk) in a combinatorial way,
by using a description of the surface as a ribbon graph, see [25, 26]. This
construction is another formulation of the Atiyah–Bott–Goldman Poisson
structure on the character variety [8, 28]. The algebra $\mathcal{L}_{g,n}$ is
an associative non-commutative algebra which is a combinatorial quantization
of the algebra of functions on the space of flat $G$-connections, by
deformation of the Fock–Rosly Poisson structure.

These algebras were introduced and studied by Alekseev–Grosse–Schomerus [2, 3,
4, 5] and Buffenoir–Roche [14, 15]. They replaced the Lie group $G$ by the
quantum group $U_{q}(\mathfrak{g})$, with $\mathfrak{g}=\mathrm{Lie}(G)$, and
defined non-commutative relations between functions on connections (also
called gauge fields) via matrix equations involving the $R$-matrix of
$U_{q}(\mathfrak{g})$. More precisely, in the Fock–Rosly graph description, a
flat connection is described by the collection of its holonomies along the
edges of the graph. For instance, with the canonical graph shown in Fig. 1, a
flat $G$-connection on $\Sigma_{1,0}{\setminus}D$ is represented by
$(x_{b},x_{a})\in G\times G$. The gauge group $G$ acts on connections by the
adjoint action and dually on gauge fields (on the right). The algebra of gauge
fields is generated by matrix coefficients $\overset{V}{A}{{}^{i}_{j}}$,
$\overset{V}{B}{{}^{i}_{j}}$ defined by
$\overset{V}{A}{{}^{i}_{j}}(x_{b},x_{a})=e^{i}(x_{a}\cdot e_{j})$ and
$\overset{V}{B}{{}^{i}_{j}}(x_{b},x_{a})=e^{i}(x_{b}\cdot e_{j})$, where $V$
is a $G$-module with basis $(e_{i})$ and dual basis $(e^{i})$. Then the
algebra $\mathcal{L}_{1,0}$ is generated by the coefficients of the matrices
$\overset{I}{A}$, $\overset{I}{B}$, where $I$ is now a representation of
$U_{q}(\mathfrak{g})$, modulo certain matrix relations involving the
$R$-matrix, see Definition 4.1 (also see Remark 4.9 for an explicit formula).
A connection is now $x_{b}\otimes x_{a}\in U_{q}(\mathfrak{g})^{\otimes 2}$.
The gauge algebra $U_{q}(\mathfrak{g})$ acts on gauge fields on the right by
the adjoint action. These relations give a non-commutative
$U_{q}(\mathfrak{g})$-equivariant multiplication between gauge fields.

$x_{m}$$m$

$b$$a$$x_{b}$$x_{a}$

Figure 1: Surfaces $\Sigma_{0,1}{\setminus}D$ and $\Sigma_{1,0}{\setminus}D$,
with canonical curves and discrete connections.

These quantized algebras of functions $\mathcal{L}_{g,n}$ and their
generalizations appear in various works of mathematics and mathematical
physics. Let us indicate papers which are related to this work. In [16], they
introduce multitangles (which are diagrams encoding transformations of the
graph and of the discrete connections) and use them to define a
comultiplication on the discrete connections (dual to the product in
$\mathcal{L}_{g,n}$), the holonomy of a path in the graph and the Wilson
loops. In [34], the axiomatic formulation of lattice gauge theory with Hopf
algebras is given and it generalizes the foundational works cited above. Their
formalism does not require semi-simplicity of the gauge algebra (except for
some properties about the holonomy); however, contrarily to the present work,
they do not use matrix coefficients, they work with an arbitrary graph filling
the surface and the purpose of their paper is different (they do not discuss
representations of mapping class groups). In [10], the algebras
$\mathcal{L}_{g,n}$ are recovered in a categorical setting based on
factorization homology; the case of the punctured torus is studied in [12],
recovering a version of $\mathcal{L}_{1,0}$. In [1], they study
$\mathcal{L}_{1,0}$ with the super Hopf algebra $\mathfrak{gl}(1|1)$ as well
as the associated $\mathrm{SL}_{2}(\mathbb{Z})$ action and they conjecture
that our result on the equivalence with the Lyubashenko–Majid representation
(Theorem 5.12) remains true with super Hopf algebras. [17] also discuss
related subjects.

The definition of $\mathcal{L}_{g,n}$ being purely algebraic, we can replace
the gauge algebra $U_{q}(\mathfrak{g})$ by any ribbon Hopf algebra $H$, thus
obtaining an algebra $\mathcal{L}_{g,n}(H)$. In the foundational papers on
combinatorial quantization, it is always assumed that $H$ is either
$U_{q}(\mathfrak{g})$ with $q$ generic or a semisimple truncation of
$U_{q}(\mathfrak{g})$ at a root of unity (defined in the context of weak
quasi-Hopf algebras). The moduli algebra, introduced in [3, 4, 5], is only
defined when $H$ is modular since its definition uses the particular
properties of the $S$-matrix in this case. Moreover, in [5], the
representation theory of $\mathcal{L}_{g,n}(H)$ is investigated and a
projective representation of the mapping class group of $\Sigma_{g,n}$ on the
moduli algebra is constructed, under the assumption that $H$ is a modular Hopf
algebra (and in particular semisimple).

In this paper we assume that $H$ is a finite-dimensional, factorizable, ribbon
Hopf algebra which is not necessarily semisimple, the guiding example being
the restricted quantum group $H=\overline{U}_{q}(\mathfrak{sl}(2))$. We
consider $\mathcal{L}_{0,1}(H)$, $\mathcal{L}_{1,0}(H)$ and we generalize to
this setting the projective representation of the mapping class group of [5];
in this way we obtain a projective representation of
$\mathrm{SL}_{2}(\mathbb{Z})$. The algebras $\mathcal{L}_{0,1}(H)$ and
$\mathcal{L}_{1,0}(H)$ deserve particular interest because they are the
building blocks of the theory, thanks to the Alekseev isomorphism,

$\mathcal{L}_{g,n}(H)\cong\mathcal{L}_{1,0}(H)^{\otimes
g}\otimes\mathcal{L}_{0,1}(H)^{\otimes n}.$

This isomorphism was stated in [2] for $H=U_{q}(\mathfrak{g})$, $q$ generic,
but it can be generalized to our assumptions on $H$ (see [22, Proposition
3.5]).

After having recalled the definition of $\mathcal{L}_{0,1}(H)$ and
$\mathcal{L}_{1,0}(H)$ and the $H$-action on them, we prove that
$\mathcal{L}_{0,1}(H)\cong H$ (Theorem 3.7) and that $\mathcal{L}_{1,0}(H)$ is
isomorphic to the Heisenberg double of the dual Hopf algebra $\mathcal{O}(H)$
(Theorem 4.8). In particular, it implies that $\mathcal{L}_{1,0}(H)$ is
(isomorphic to) a matrix algebra. We then define in Section 4.3 a
representation of the algebra of invariants,
$\mathcal{L}^{\mathrm{inv}}_{1,0}(H)$, on the space of symmetric linear forms
$\operatorname{SLF}(H)$ (Theorem 4.10). We will use this representation of the
invariant elements to define a projective representation of
$\mathrm{SL}_{2}(\mathbb{Z})$ on $\operatorname{SLF}(H)$.

The main ingredient of the construction of the mapping class group
representation is the natural action of the mapping class group
$\mathrm{MCG}(\Sigma_{1,0}{\setminus}D)$ on $\mathcal{L}_{1,0}(H)$, obtained
by considering the action of $\mathrm{MCG}(\Sigma_{1,0}{\setminus}D)$ on
$\pi_{1}(\Sigma_{1,0}{\setminus}D)$ and by replacing the loops representing
elements in $\pi_{1}(\Sigma_{1,0}{\setminus}D)$ by the corresponding matrices
in $\mathcal{L}_{1,0}(H)$, up to some normalization [5]. It turns out that
$\mathrm{MCG}(\Sigma_{1,0}{\setminus}D)$ acts by automorphisms on
$\mathcal{L}_{1,0}(H)$, and thus, since it is a matrix algebra, we get
elements which implements these automorphisms by conjugation. Representing
these elements on $\operatorname{SLF}(H)$, we get a projective representation
of $\mathrm{SL}_{2}(\mathbb{Z})$; this is explained in Section 5, with
detailed proofs under our assumptions on $H$.

We show (Theorem 5.12) that this projective representation is equivalent to
the Lyubashenko–Majid representation [33]. This gives a natural and
geometrical interpretation of the latter, which was constructed by categorical
methods.

Section 6 is devoted to the example of
$\overline{U}_{q}=\overline{U}_{q}(\mathfrak{sl}(2))$ as gauge algebra. All
the preliminary facts about $\overline{U}_{q}$ and the GTA basis, which is a
suitable basis of $\operatorname{SLF}\big{(}\overline{U}_{q}\big{)}$, are
available in [21]. In Theorem 6.5 we give the explicit formulas for the action
of $\mathrm{SL}_{2}(\mathbb{Z})$ on the GTA basis of
$\operatorname{SLF}\big{(}\overline{U}_{q}\big{)}$. The multiplication
formulas in this basis (see [21, Section 5], [27]), are the crucial tool to
obtain the result. The structure of
$\operatorname{SLF}\big{(}\overline{U}_{q}\big{)}$ under the action of
$\mathrm{SL}_{2}(\mathbb{Z})$ is determined. Thanks to the equivalence with
the Lyubashenko–Majid representation and the work of [24], the representation
obtained in Section 6.3 is equivalent to the one studied in [24], which comes
from logarithmic conformal field theory. Finally, in Section 6.4, we formulate
a conjecture about the structure of
$\operatorname{SLF}\big{(}\overline{U}_{q}\big{)}$ as a
$\mathcal{L}^{\mathrm{inv}}_{1,0}\big{(}\overline{U}_{q}\big{)}$-module.

To sum up, the main results of this paper are:

  * –

the construction of a projective representation of $\text{SL}_{2}(\mathbb{Z})$
(the mapping class group of the torus) on the space of symmetric linear forms
on $H$ (Theorem 5.8),

  * –

the equivalence of this representation to the one found by Lyubashenko and
Majid in [33] (Theorem 5.12),

  * –

in the case of $H=\overline{U}_{q}(\mathfrak{sl}(2))$, an explicit realization
of this representation and its structure (Theorem 6.5 and Theorem 6.8).

In [22], we generalize Theorems 5.8 and 5.12 to higher genus. The projective
representation of the mapping class group is shown to be equivalent to that
constructed by Lyubashenko using the coend of a ribbon category [32]. The
advantage of combinatorial quantization is that the construction of the
projective representation of the mapping class group is very explicit and that
the particular features of the algebra $\mathcal{L}_{g,n}(H)$ are helpful to
compute the formulas of the representation for a given $H$ (see Section 6.3).
In particular, for $H=\overline{U}_{q}(\mathfrak{sl}(2))$, one can hope that
the correspondence with logarithmic conformal field theory mentioned aboved
still exists in higher genus. Let us also mention that the subalgebra of
invariant elements $\mathcal{L}_{g,n}^{\mathrm{inv}}(H)$ is known to be
related to skein theory [16]; in fact, combinatorial quantization provides new
representations of skein algebras (work in progress).

Notations. If $A$ is an algebra, $V$ is a finite-dimensional $A$-module and
$x\in A$, we denote by $\overset{V}{x}\in\operatorname{End}_{\mathbb{C}}(V)$
the representation of $x$ on the module $V$. More generally, if $X\in
A^{\otimes n}$ and if $V_{1},\dots,V_{n}$ are $A$-modules, we denote by
$\overset{V_{1}\dots V_{n}}{X}$ the representation of $X$ on
$V_{1}\otimes\dots\otimes V_{n}$. As in [18], we will use the abbreviation PIM
for principal indecomposable module. Here we consider only finite-dimensional
representations.

Let $M\in\mathrm{Mat}_{m}(\mathbb{C})\otimes A=\mathrm{Mat}_{m}(A)$, namely
$M$ is a matrix with coefficients in $A$. It can be written as
$M=\sum_{i,j}E^{i}_{j}\otimes M^{i}_{j}$, where $E^{i}_{j}$ is the matrix with
$1$ at the intersection of the $i$-th row and the $j$-th column and $0$
elsewhere. More generally, every
$L\in\mathrm{Mat}_{m_{1}}(\mathbb{C})\otimes\dots\otimes\mathrm{Mat}_{m_{l}}(\mathbb{C})\otimes
A$ can be written as

$L=\sum_{i_{1},j_{1},\dots,i_{l},j_{l}}E^{i_{1}}_{j_{1}}\otimes\dots\otimes
E^{i_{l}}_{j_{l}}\otimes L^{i_{1}\dots i_{l}}_{j_{1}\dots j_{l}}$

and the elements $L^{i_{1}\dots i_{l}}_{j_{1}\dots j_{l}}\in A$ are called the
coefficients of $L$. If $f\colon A\to B$ is a morphism of algebras, then we
define
$f(L)\in\mathrm{Mat}_{m_{1}}(\mathbb{C})\otimes\dots\otimes\mathrm{Mat}_{m_{l}}(\mathbb{C})\otimes
B$ by

$f(L)=\sum_{i_{1},j_{1},\dots,i_{l},j_{l}}E^{i_{1}}_{j_{1}}\otimes\dots\otimes
E^{i_{l}}_{j_{l}}\otimes f\bigl{(}L^{i_{1}\dots i_{l}}_{j_{1}\dots
j_{l}}\bigr{)}$

or equivalently $f(L)^{i_{1}\dots i_{l}}_{j_{1}\dots
j_{l}}=f\bigl{(}L^{i_{1}\dots i_{l}}_{j_{1}\dots j_{l}}\bigr{)}$.

Let $M\in\mathrm{Mat}_{m}(\mathbb{C})\otimes A$,
$N\in\mathrm{Mat}_{n}(\mathbb{C})\otimes A$. We embed $M$, $N$ in
$\mathrm{Mat}_{m}(\mathbb{C})\otimes\mathrm{Mat}_{n}(\mathbb{C})\otimes A$ by

$M_{1}=\sum_{i,j}E^{i}_{j}\otimes\mathbb{I}_{n}\otimes M^{i}_{j},\qquad
N_{2}=\sum_{i,j}\mathbb{I}_{m}\otimes E^{i}_{j}\otimes N^{i}_{j},$

where $\mathbb{I}_{k}$ is the identity matrix of size $k$. In other words,
$M_{1}=M\otimes\mathbb{I}_{n}$, $N_{2}=\mathbb{I}_{m}\otimes N$, $\otimes$
being the Kronecker product. The coefficients of $M_{1}$ and $N_{2}$ are
respectively $(M_{1})^{ik}_{jl}=M^{i}_{j}\delta^{k}_{l}$,
$(N_{2})^{ik}_{jl}=\delta^{i}_{j}N^{k}_{l}$ where $\delta^{a}_{b}$ is the
Kronecker delta. Observe that $M_{1}N_{2}$ (resp. $N_{2}M_{1}$) contains all
the possible products of coefficients of $M$ (resp. of $N$) by coefficients of
$N$ (resp. of $M$): $(M_{1}N_{2})^{ik}_{j\ell}=M^{i}_{j}N^{k}_{\ell}$ (resp.
$(N_{2}M_{1})^{ik}_{j\ell}=N^{k}_{\ell}M^{i}_{j}$). In particular,
$M_{1}N_{2}=N_{2}M_{1}$ if and only if the coefficients of $M$ commute with
those of $N$. Now let
$Q\in\mathrm{Mat}_{m}(\mathbb{C})\otimes\mathrm{Mat}_{n}(\mathbb{C})\otimes
A$; then we denote $Q_{12}=Q$ and

$Q_{21}=\sum_{i,j,k,l}E^{k}_{l}\otimes E^{i}_{j}\otimes
Q^{ik}_{jl}\in\mathrm{Mat}_{n}(\mathbb{C})\otimes\mathrm{Mat}_{m}(\mathbb{C})\otimes
A.$

so that $(Q_{12})^{ac}_{bd}=Q^{ac}_{bd}$, $(Q_{21})^{ac}_{bd}=Q^{ca}_{db}$.
Such notations are obviously generalized to bigger tensors.

In order to simplify notations, we will use implicit summations. First, we use
Einstein’s notation for the computations involving indices: when an index
variable appears twice, one time in upper position and one time in lower
position, it implicitly means summation over all the values of the index. For
instance if $M\in\mathrm{Mat}_{m}(\mathbb{C})\otimes A$ and
$Q\in\mathrm{Mat}_{m}(\mathbb{C})\otimes\mathrm{Mat}_{n}(\mathbb{C})\otimes
A$, then $(M_{1}Q_{12})^{ac}_{bd}=M^{a}_{i}Q^{ic}_{bd}$; note that the
repeated subscript $1$ corresponds to matrix multiplication in the first
space. Second, we use Sweedler’s notation (see [30, Notation III.1.6]) without
summation sign for the coproducts, that is we write

$\Delta(x)=x^{\prime}\otimes
x^{\prime\prime},\qquad(\Delta\otimes\mathrm{id})\circ\Delta(x)=(\mathrm{id}\otimes\Delta)\circ\Delta(x)=x^{\prime}\otimes
x^{\prime\prime}\otimes x^{\prime\prime\prime},\qquad\text{and so on.}$

Finally, we write the components of the $R$-matrix as $R=a_{i}\otimes b_{i}$
with implicit summation on $i$, and we define $R^{\prime}=b_{i}\otimes a_{i}$.

For $q\in\mathbb{C}{\setminus}\\{-1,0,1\\}$, we define the $q$-integer $[n]$
(with $n\in\mathbb{Z})$ by

$[n]=\frac{q^{n}-q^{-n}}{q-q^{-1}}.$

We will denote $\hat{q}=q-q^{-1}$ to shorten formulas. Observe that if $q$ is
a $2p$-root of unity, then $[p]=0$ and $[p-n]=[n]$.

As usual $\delta_{s,t}$ is the Kronecker symbol and $\mathbb{I}_{n}$ is the
identity matrix of size $n$.

##  2 Some basic facts

We refer to [18, Chapters IV and VIII] for background material about
representation theory.

###  2.1 Dual of a finite-dimensional algebra

Let $A$ be a finite-dimensional $\mathbb{C}$-algebra, $V$ be a finite-
dimensional $A$-module and $\overset{V}{\rho}\colon
A\to\mathrm{End}_{\mathbb{C}}(V)$ be the representation associated to $V$. In
other words, $\overset{V}{\rho}$ is an element of
$\mathrm{End}_{\mathbb{C}}(V)\otimes A^{*}$ and if we choose a basis of $V$,
$\overset{V}{\rho}$ becomes an element, denoted $\overset{V}{T}$, of
$\mathrm{Mat}_{n}(\mathbb{C})\otimes A^{*}=\mathrm{Mat}_{n}(A^{*})$. In other
words $\overset{V}{T}$ is a matrix whose coefficients are linear forms on $A$.
The linear form $\overset{V}{T}{{}^{i}_{j}}$ is called a matrix coefficient of
$V$. We will always consider the representation
$\overset{V}{x}\in\mathrm{End}_{\mathbb{C}}(V)$ of $x$ on $V$ as the matrix
$\overset{V}{T}(x)\in\mathrm{Mat}_{\dim(V)}(\mathbb{C})$:
$\overset{V}{x}{{}^{i}_{j}}=\overset{V}{T}{{}^{i}_{j}}(x)$. Note that if a
$A$-module $V$ is a submodule or a quotient of a $A$-module $W$, then the
matrix coefficients of $W$ contain those of $V$ since $\overset{V}{T}$ is a
submatrix of $\overset{W}{T}$.

Since $A^{*}$ is finite-dimensional, it is generated as a vector space by the
matrix coefficients of the regular representation ${}_{A}A$. Indeed, let
$\\{x_{1},\dots,x_{n}\\}$ be a basis of $A$ with $x_{1}=1$ and let
$\big{\\{}x^{1},\dots,x^{n}\big{\\}}\subset A^{*}$ be the dual basis; it is
readily seen that $\overset{{}_{A}A}{T}{{}^{i}_{1}}(x_{j})=\delta_{i,j}$ and
thus $\overset{{}_{A}A}{T}{{}^{i}_{1}}=x^{i}$. This implies that in the sequel
one might restrict everywhere to the regular representation; however, when
studying examples it is very relevant to consider smaller representations in
order to greatly reduce the number of generators for the algebras that we will
consider in the sequel (like the fundamental representation
$\mathcal{X}^{+}(2)$ for $A=\overline{U}_{q}(\mathfrak{sl}(2))$, see Section
6).

Recall that the PIMs of $A$ are the indecomposable projective $A$-modules.
They are isomorphic to the direct summands of the regular representation
${}_{A}A$. In particular, the matrix coefficients of ${}_{A}A$ are exactly
those of the PIMs and it follows that the matrix coefficients of the PIMs span
$A^{*}$. Note however that the matrix coefficients of the PIMs (or
equivalently of ${}_{A}A$) do not form a basis of $A^{*}$ in general. Indeed,
even if we fix a family $(P_{\alpha})$ representing each isomorphism class of
PIMs, it is possible for $P_{\alpha}$ and $P_{\beta}$ to have a composition
factor $S$ in common. In this case, both $\overset{P_{\alpha}}{T}$ and
$\overset{P_{\beta}}{T}$ contain $\overset{S}{T}$ as submatrix. This is what
happens for $H=\overline{U}_{q}(\mathfrak{sl}(2))$, see, e.g., [21, Section
3]. In the semisimple case this phenomenon does not occur.

###  2.2 Braided Hopf algebras, factorizability, ribbon element

Let $H$ be a braided Hopf algebra with universal $R$-matrix $R=a_{i}\otimes
b_{i}$ (see, e.g., [30, Chapter VIII]). Recall that

$\displaystyle(\Delta\otimes\mathrm{id})(R)=R_{13}R_{23},\qquad(\mathrm{id}\otimes\Delta)(R)=R_{13}R_{12},$
(2.1) $\displaystyle(S\otimes\mathrm{id})(R)=\big{(}\mathrm{id}\otimes
S^{-1}\big{)}(R)=R^{-1},\qquad(S\otimes S)(R)=R,$ (2.2) $\displaystyle
R_{12}R_{13}R_{23}=R_{23}R_{13}R_{12}$ (2.3)

with $R_{12}=a_{i}\otimes b_{i}\otimes 1$, $R_{13}=a_{i}\otimes 1\otimes
b_{i}$, $R_{23}=1\otimes a_{i}\otimes b_{i}\in H^{\otimes 3}$. The relation
(2.3) is called the (quantum) Yang–Baxter equation.

Consider

$\begin{array}[]{crll}\Psi\colon&H^{*}&\rightarrow&H,\\\
&\beta&\mapsto&(\beta\otimes\mathrm{id})(RR^{\prime}),\end{array}$

where $R^{\prime}=b_{i}\otimes a_{i}$. $\Psi$ is called the
Reshetikhin–Semenov-Tian-Shansky map [37], or the Drinfeld map [19]. We say
that $H$ is factorizable if $\Psi$ is an isomorphism of vector spaces. By the
remarks above, we can restrict $\beta$ to be a matrix coefficient of some $I$.

Define $R^{(+)}=R$, $R^{(-)}=(R^{\prime})^{-1}$, and let

$\displaystyle\overset{I}{L}{{}^{(\pm)}}=\big{(}\overset{I}{T}\otimes\mathrm{id}\big{)}\big{(}R^{(\pm)}\big{)}=\bigl{(}\overset{I}{a_{i}^{(\pm)}}\bigr{)}b_{i}^{(\pm)}\in\mathrm{Mat}_{\dim(I)}(H)$
(2.4)

with $R^{(\pm)}=a_{i}^{(\pm)}\otimes b_{i}^{(\pm)}$ (note that
$\bigl{(}\overset{I}{a_{i}^{(\pm)}}\bigr{)}b_{i}^{(\pm)}$ is the matrix
obtained by multiplying each coefficient (which is a scalar) of the matrix
$\bigl{(}\overset{I}{a_{i}^{(\pm)}}\bigr{)}$ by the element $b_{i}\in H$).
Recall that $R^{(-)}$ is also a universal $R$-matrix and in particular it
satisfies the properties (2.1)–(2.3) above. We use the letters $I,J,\dots$ for
modules over Hopf algebras. If $H$ is factorizable, the coefficients of the
matrices $\overset{I}{L}\\!^{(\pm)}$ generate $H$ as an algebra. These
matrices satisfy nice relations which are consequences of (2.1) and (2.3):

$\displaystyle\overset{\\!\\!\\!\\!\\!I\otimes
J}{L^{(\epsilon)}_{12}}=\overset{I}{L}{{}^{(\epsilon)}_{1}}\overset{J}{L}{{}^{(\epsilon)}_{2}},$
$\displaystyle\overset{IJ}{R}{{}^{(\epsilon)}_{12}}\overset{I}{L}{{}^{(\epsilon)}_{1}}\overset{J}{L}{{}^{(\sigma)}_{2}}=\overset{J}{L}{{}^{(\sigma)}_{2}}\overset{I}{L}{{}^{(\epsilon)}_{1}}\overset{IJ}{R}{{}^{(\epsilon)}_{12}}\qquad\forall\,\epsilon,\sigma\in\\{\pm\\},$
$\displaystyle\overset{IJ}{R}{{}^{(\epsilon)}_{12}}\overset{I}{L}{{}^{(\sigma)}_{1}}\overset{J}{L}{{}^{(\sigma)}_{2}}=\overset{J}{L}{{}^{(\sigma)}_{2}}\overset{I}{L}{{}^{(\sigma)}_{1}}\overset{IJ}{R}{{}^{(\epsilon)}_{12}}\qquad\forall\,\epsilon,\sigma\in\\{\pm\\},$
$\displaystyle\Delta(\overset{I}{L}{{}^{(\epsilon)}\\!{}^{a}_{b}})=\overset{I}{L}{{}^{(\epsilon)}\\!{}^{i}_{b}}\otimes\overset{I}{L}{{}^{(\epsilon)}\\!{}^{a}_{i}}.$
(2.5)

For instance, here is a proof of the first equality with $\epsilon=+$:

$\overset{\\!\\!\\!\\!\\!I\otimes J}{L^{(+)}_{12}}=\bigl{(}\overset{I\otimes
J}{a_{i}}\bigr{)}_{\\!12}b_{i}=\bigl{(}\overset{I}{a_{i}}\otimes\overset{J}{a_{j}}\bigr{)}_{\\!12}b_{i}b_{j}=\bigl{(}\overset{I}{a_{i}}\bigr{)}_{\\!1}\bigl{(}\overset{J}{a_{j}}\bigr{)}_{\\!2}b_{i}b_{j}=\bigl{(}\overset{I}{a_{i}}\bigr{)}_{\\!1}b_{i}\bigl{(}\overset{J}{a_{j}}\bigr{)}_{\\!2}b_{j}=\overset{I}{L}\\!^{(+)}_{1}\overset{J}{L}\\!^{(+)}_{2},$

where we used (2.1); note that these are equalities between matrices in

$\mathrm{Mat}_{\dim(I)}(\mathbb{C})\otimes\mathrm{Mat}_{\dim(J)}(\mathbb{C})\otimes
H,$

which imply equalities among the coefficients. If the representations $I$ and
$J$ are fixed and arbitrary, we will simply write these relations as

$L^{(\epsilon)}_{12}=L^{(\epsilon)}_{1}L^{(\epsilon)}_{2},\qquad
R^{(\epsilon)}_{12}L^{(\epsilon)}_{1}L^{(\sigma)}_{2}=L^{(\sigma)}_{2}L^{(\epsilon)}_{1}R^{(\epsilon)}_{12},\qquad
R^{(\epsilon)}_{12}L^{(\sigma)}_{1}L\\!^{(\sigma)}_{2}=L^{(\sigma)}_{2}L^{(\sigma)}_{1}R^{(\epsilon)}_{12},$

the subscript $1$ (resp. $2$) corresponding implicitly to evaluation in the
representation $I$ (resp. $J$).

Recall that the Drinfeld element $u$ and its inverse are

$\displaystyle u=S(b_{i})a_{i}=b_{i}S^{-1}(a_{i})\qquad\text{and}\qquad
u^{-1}=S^{-2}(b_{i})a_{i}=S^{-1}(b_{i})S(a_{i})=b_{i}S^{2}(a_{i}).$ (2.6)

We assume that $H$ contains a ribbon element $v$. It satisfies

$\displaystyle v\text{ is central and
invertible},\qquad\Delta(v)=(R^{\prime}R)^{-1}v\otimes v,$ $\displaystyle
S(v)=v,\qquad\varepsilon(v)=1,\qquad v^{2}=uS(u).$ (2.7)

The two last equalities can be deduced easily from the others. A ribbon
element is in general not unique. A ribbon Hopf algebra $(H,R,v)$ is a braided
Hopf algebra $(H,R)$ together with a ribbon element $v$.

We say that $g\in H$ is a pivotal element if

$\displaystyle\Delta(g)=g\otimes g\qquad\text{ and }\qquad\forall\,x\in
H,\qquad S^{2}(x)=gxg^{-1}.$ (2.8)

Note that $g$ is invertible because it is grouplike, and hence $S(g)=g^{-1}$.
A pivotal element is in general not unique. But in a ribbon Hopf algebra
$(H,R,v)$ there is a canonical choice

$\displaystyle g=uv^{-1}.$ (2.9)

We will always take this canonical pivotal element $g$ in the sequel.

###  2.3 Dual Hopf algebra $\boldsymbol{\mathcal{O}(H)}$

The canonical Hopf algebra structure on $H^{*}$ is defined by

$\displaystyle(\psi\cdot\varphi)(x)=(\psi\otimes\varphi)(\Delta(x)),\qquad\eta(1)=\varepsilon,$
$\displaystyle\Delta(\psi)(x\otimes
y)=\psi(xy),\qquad\varepsilon(\psi)=\psi(1),\qquad S(\psi)=\psi\circ S$

with $\psi,\varphi\in H^{*}$ and $x,y\in H$. When it is endowed with this
structure, $H^{*}$ is called dual Hopf algebra, and is denoted by
$\mathcal{O}(H)$ in the sequel. In terms of matrix coefficients, we have

$\displaystyle\overset{I\otimes
J}{T}\\!\\!\\!_{12}=\overset{I}{T_{1}}\overset{J}{T_{2}},\qquad\eta(1)=\overset{\mathbb{C}}{T},\qquad\Delta(\overset{I}{T^{a}_{b}})=\overset{I}{T^{a}_{i}}\otimes\overset{I}{T^{i}_{b}},$
$\displaystyle\varepsilon(\overset{I}{T})=\mathbb{I}_{\dim(I)},\qquad
S(\overset{I}{T})=\overset{I}{T}{{}^{-1}},$ (2.10)

where $\mathbb{C}$ is the trivial representation. For instance, here are
proofs of the first and last equalities (with $h\in H$):

$\displaystyle\bigl{(}\overset{I\otimes
J}{T}\\!_{\\!\\!12}\bigr{)}(h)=\bigl{(}\overset{I\otimes
J}{h}\bigr{)}_{12}=\bigl{(}\overset{I}{h^{\prime}}\otimes\overset{J}{h^{\prime\prime}}\bigr{)}_{12}=\bigl{(}\overset{I}{h^{\prime}}\bigr{)}_{1}\bigl{(}\overset{J}{h^{\prime\prime}}\bigr{)}_{2}=\overset{I}{T_{1}}(h^{\prime})\overset{J}{T_{2}}(h^{\prime\prime})=\bigl{(}\overset{I}{T_{1}}\overset{J}{T_{2}}\bigr{)}(h),$
$\displaystyle\bigl{(}S(\overset{I}{T})\overset{I}{T}\bigr{)}(h)=S(\overset{I}{T})(h^{\prime})\overset{I}{T}(h^{\prime\prime})=\overset{I}{T}\bigl{(}S(h^{\prime})\bigr{)}\overset{I}{T}(h^{\prime\prime})=\overset{I}{T}\bigl{(}S(h^{\prime})h^{\prime\prime}\bigr{)}=\varepsilon(h)\overset{I}{T}(1)=\varepsilon(h)\mathbb{I}_{\dim(I)}.$

By definition of the action on the dual $I^{*}$, it holds

$\displaystyle S\big{(}\overset{I}{T}\big{)}={}^{t}\overset{\>I^{*}}{T},$
(2.11)

where t is the transpose. Also recall the well-known exchange relation

$\displaystyle\overset{IJ}{R}_{12}\overset{I}{T_{1}}\overset{J}{T_{2}}=\overset{J}{T_{2}}\overset{I}{T_{1}}\overset{IJ}{R}_{12}.$
(2.12)

As before, if the representations $I$ and $J$ are fixed and arbitrary, we will
simply write $T_{12}=T_{1}T_{2}$ and $R_{12}T_{1}T_{2}=T_{2}T_{1}R_{12}$.

##  3 The loop algebra $\boldsymbol{\mathcal{L}_{0,1}(H)}$

We assume that $H$ is a finite-dimensional factorizable Hopf algebra. The
ribbon assumption is not needed in this section.

###  3.1 Definition of $\boldsymbol{\mathcal{L}_{0,1}(H)}$ and
$\boldsymbol{H}$-module-algebra structure

Let $\mathrm{T}(H^{*})$ be the tensor algebra of $H^{*}$, which by definition
is linearly spanned by all the formal products $\varphi_{1}\cdots\varphi_{n}$
(with $n\geq 0$ and $\varphi_{i}\in H^{*}$) modulo the obvious multilinear
relations. There is a canonical injection $j\colon H^{*}\to\mathrm{T}(H^{*})$
and we let
$\overset{I}{M}\in\mathrm{Mat}_{\dim(I)}\bigl{(}\mathrm{T}(H^{*})\bigr{)}$ be
the matrix defined by $\overset{I}{M}=j\bigl{(}\overset{I}{T}\bigr{)}$, that
is $\overset{I}{M}{{}^{a}_{b}}=j\bigl{(}\overset{I}{T}{{}^{a}_{b}}\bigr{)}$.

######  Definition 3.1.

The loop algebra $\mathcal{L}_{0,1}(H)$ is the quotient of $\mathrm{T}(H^{*})$
by the following fusion relations

$\overset{I\otimes
J}{M}\\!_{12}=\overset{I}{M}_{1}\overset{IJ}{(R^{\prime})}_{12}\overset{J}{M}_{2}\overset{IJ}{\big{(}R^{\prime-1}\big{)}}_{12}$

for all finite-dimensional $H$-modules $I$, $J$.

The right hand-side of the fusion relation in Definition 3.1 is the one of
[13, Definition 1]; the one of [4, Definition 12] and [5, equation (3.11)] is
different, due to different choices of the action of $H$ on
$\mathcal{L}_{0,1}(H)$ and to particular normalization of Clebsch–Gordan
operators. Moreover in the papers [4, 5, 13] the matrix $\overset{I\otimes
J}{M}$ did not appeared, instead it was always decomposed as a sum of the
matrices of the irreducible direct summands of $I\otimes J$ thanks to
Clebsch–Gordan operators, which is relevant in the semisimple case only; in
[38], $I$ and $J$ are restricted to the regular representation and the matrix
$\overset{I\otimes J}{M}$ is thus denoted by $\Delta_{a}(M)$. In the
semisimple setting, the algebras resulting from each of these definitions are
isomorphic.

Note that the fusion relation is a relation between matrices in
$\mathrm{Mat}_{\dim(I)}(\mathbb{C})\otimes\mathrm{Mat}_{\dim(J)}(\mathbb{C})\otimes\mathcal{L}_{0,1}(H)$
(for all finite-dimensional $I$, $J$) which implies relations among elements
of $\mathcal{L}_{0,1}(H)$ (the coefficients of these matrices). Explicitly, in
terms of matrix coefficients it is written as

$\forall\,I,J,a,b,c,d,\qquad\overset{I\otimes
J}{M}{{}^{ac}_{bd}}=\overset{I}{M}{{}^{a}_{i}}\overset{IJ}{(R^{\prime})}{{}^{ic}_{jk}}\overset{J}{M}{{}^{k}_{l}}\overset{IJ}{(R^{\prime-1})}{{}^{jl}_{bd}},$

see the definition of the subscripts $1$ and $2$ in the Notations at p. 1. If
the two representations $I$ and $J$ are fixed and arbitrary, we will simply
write

$\displaystyle M_{12}=M_{1}R_{21}M_{2}R_{21}^{-1}$ (3.1)

the subscript $1$ (resp. $2$) corresponding implicitly to evaluation in the
representation $I$ (resp. $J$). Finally, note that if $f\colon I\to J$ is a
morphism, it holds

$\displaystyle f\overset{I}{M}=\overset{J}{M}f,$ (3.2)

where we identify $f$ with its matrix. Indeed,
$f\overset{I}{M}=f\bar{j}\bigl{(}\overset{I}{T}\bigr{)}=\bar{j}\bigl{(}f\overset{I}{T}\bigr{)}=\bar{j}\bigl{(}\overset{J}{T}f\bigr{)}=\bar{j}\bigl{(}\overset{J}{T}\bigr{)}f=\overset{J}{M}f$
where $\bar{j}$ is the linear map
$H^{*}\to\mathrm{T}(H^{*})\to\mathcal{L}_{0,1}(H)$.

######  Remark 3.2.

One can check that $\overset{(I\otimes J)\otimes
K}{M}=\overset{I\otimes(J\otimes K)}{M}$ holds thanks to the Yang–Baxter
equation.

We have an useful analogue of relation (2.12).

######  Proposition 3.3.

The following exchange relations hold in $\mathcal{L}_{0,1}(H)$:

$\overset{IJ}{R}_{12}\overset{I}{M}_{1}\overset{IJ}{(R^{\prime})}_{12}\overset{J}{M}_{2}=\overset{J}{M}_{2}\overset{IJ}{R}_{12}\overset{I}{M}_{1}\overset{IJ}{(R^{\prime})}_{12}.$

This relation is called the reflection equation. It can be written in a
shortened way if the representations $I$ and $J$ are fixed and arbitrary:

$\displaystyle R_{12}M_{1}R_{21}M_{2}=M_{2}R_{12}M_{1}R_{21}.$ (3.3)

###### Proof.

Let $P_{I,J}\colon I\otimes J\to J\otimes I$ be the flip map $P_{IJ}(v\otimes
w)=w\otimes v$. We have the isomorphism $P_{IJ}\overset{IJ}{R}\colon I\otimes
J\to J\otimes I$, hence $P_{IJ}\overset{IJ}{R}\overset{I\otimes
J}{M}=\overset{J\otimes I}{M}P_{IJ}\overset{IJ}{R}$ thanks to (3.2). This
gives

$\displaystyle(P_{IJ})_{12}\overset{IJ}{R}_{12}\overset{I}{M}_{1}\overset{IJ}{(R^{\prime})}_{12}\overset{J}{M}_{2}\overset{IJ}{(R^{\prime-1})}_{12}$
$\displaystyle=\overset{J}{M}_{1}\overset{JI}{(R^{\prime})}_{12}\overset{I}{M}_{2}\overset{JI}{(R^{\prime-1})}_{12}(P_{IJ})_{12}\overset{IJ}{R}_{12}$
$\displaystyle=(P_{IJ})_{12}\overset{J}{M}_{2}\overset{JI}{(R^{\prime})}_{21}\overset{I}{M}_{1}\overset{JI}{(R^{\prime-1})}_{21}\overset{IJ}{R}_{12}=(P_{IJ})_{12}\overset{J}{M}_{2}\overset{IJ}{R}_{12}\overset{I}{M}_{1}$

as desired. ∎

Consider the following right action $\cdot$ of $H$ on $\mathcal{L}_{0,1}(H)$,
which is the analogue of the right action of the gauge group on the gauge
fields

$\displaystyle\overset{I}{M}\cdot
h=\overset{I}{h^{\prime}}\overset{I}{M}\overset{I}{S(h^{\prime\prime})}$ (3.4)

or more explicitly $\overset{I}{M}{{}^{a}_{b}}\cdot
h=(\overset{I}{h^{\prime}}){{}^{a}_{i}}\overset{I}{M}{{}^{i}_{j}}\overset{I}{S(h^{\prime\prime})}{{}^{j}_{b}}$.
As in [14], one can equivalently work with the corresponding left coaction
$\Omega\colon\mathcal{L}_{0,1}(H)\to\mathcal{O}(H)\otimes\mathcal{L}_{0,1}(H)$
defined by

$\Omega(\overset{I}{M}{{}^{a}_{b}})=\overset{I}{T}{{}^{a}_{i}}S\big{(}\overset{I}{T}{{}^{j}_{b}}\big{)}\otimes\overset{I}{M}{{}^{i}_{j}},$

so that we recover $\cdot$ by evaluation: $x\cdot
h=(\langle?,h\rangle\otimes\mathrm{id})\circ\Omega(x)$. If we view
$\mathcal{O}(H)$ and $\mathcal{L}_{0,1}(H)$ as subalgebras of
$\mathcal{O}(H)\otimes\mathcal{L}_{0,1}(H)$ in the canonical way, then
$\Omega$ is simply written as
$\Omega(\overset{I}{M})=\overset{I}{T}\overset{I}{M}S(\overset{I}{T})$.

######  Proposition 3.4.

The right action $\cdot$ is a $H$-module-algebra structure on
$\mathcal{L}_{0,1}(H)$. Equivalently, $\Omega$ is a left
$\mathcal{O}(H)$-comodule-algebra structure on $\mathcal{L}_{0,1}(H)$.

###### Proof.

One must show for instance that $\Omega$ is a morphism of algebras, as in
[14]. For the sake of completeness we display the computation, where we use
the shortened notation explained before:

$\displaystyle\Omega(M)_{12}$ $\displaystyle=T_{12}M_{12}S(T_{12})$
(definition)
$\displaystyle=T_{1}T_{2}M_{1}R_{21}M_{2}R^{-1}_{21}S(T)_{2}S(T)_{1}$
(equations (2.10) and (3.1))
$\displaystyle=T_{1}M_{1}T_{2}R_{21}M_{2}R^{-1}_{21}S(T)_{2}S(T)_{1}$
(commuting coefficients)
$\displaystyle=T_{1}M_{1}T_{2}R_{21}M_{2}S(T)_{1}S(T)_{2}R_{21}^{-1}$
(equation (2.12))
$\displaystyle=T_{1}M_{1}T_{2}R_{21}S(T)_{1}M_{2}S(T)_{2}R_{21}^{-1}$
(commuting coefficients)
$\displaystyle=T_{1}M_{1}S(T)_{1}R_{21}T_{2}M_{2}S(T)_{2}R_{21}^{-1}$
(equation (2.12)) $\displaystyle=\Omega(M)_{1}R_{21}\Omega(M)_{2}R_{21}^{-1}$
$\displaystyle\quad\text{(definition)}.$ ∎

We say that an element $x\in\mathcal{L}_{0,1}(H)$ is invariant if for all
$h\in H$, $x\cdot h=\varepsilon(h)x$ (or equivalently,
$\Omega(x)=\varepsilon\otimes x$). For instance, for every
$\Phi\in\operatorname{End}_{H}(I)$, the element
$\mathrm{tr}(\overset{I}{g}\Phi\overset{I}{M})$ is invariant:

$\mathrm{tr}\big{(}\overset{I}{g}\Phi\overset{I}{M}\big{)}\cdot
h=\mathrm{tr}\big{(}\overset{I}{g}\Phi\overset{I}{h^{\prime}}\overset{I}{M}\overset{I}{S(h^{\prime\prime})}\big{)}=\mathrm{tr}\big{(}\overset{I}{g}\Phi\overset{I}{S^{-1}(h^{\prime\prime})}\overset{I}{h^{\prime}}\overset{I}{M}\big{)}=\varepsilon(h)\mathrm{tr}\big{(}\overset{I}{g}\Phi\overset{I}{M}\big{)}.$

We denote by $\mathcal{L}_{0,1}^{\text{inv}}(H)$ the subalgebra of invariants
elements of $\mathcal{L}_{0,1}(H)$.

###  3.2 Isomorphism $\boldsymbol{\mathcal{L}_{0,1}(H)\cong H}$

Recall the right adjoint action of $H$ on itself defined by $a\cdot
h=S(h^{\prime})ah^{\prime\prime}$ with $a,h\in H$, whose invariant elements
are the central elements of $H$.

######  Proposition 3.5.

If we endow $H$ with the right adjoint action, the following map is a morphism
of $($right$)$ $H$-module-algebras

$\begin{array}[]{crll}\Psi_{0,1}\colon&\mathcal{L}_{0,1}(H)&\rightarrow&H,\\\
&\overset{I}{M}&\mapsto&\big{(}\overset{I}{T}\otimes\text{\rm
id}\big{)}(RR^{\prime})=\overset{I}{L}{{}^{(+)}}\overset{I}{L}{{}^{(-)-1}}.\end{array}$

Hence, $\Psi_{0,1}$ brings invariant elements to central elements.

We will call $\Psi_{0,1}$ the Reshetikhin–Semenov-Tian-Shansky–Drinfeld
morphism (RSD morphism for short). The difference with the morphism $\Psi$ of
Section 2.2 is that the source spaces are different.

###### Proof.

Thanks to the relations of (2.5), we check that $\Psi_{0,1}$ preserves the
relation of Definition 3.1:

$\displaystyle\Psi_{0,1}(M)_{1}R_{21}\Psi_{0,1}(M)_{2}R^{-1}_{21}$
$\displaystyle=L^{(+)}_{1}L^{(-)-1}_{1}R_{21}L^{(+)}_{2}L^{(-)-1}_{2}R_{21}^{-1}$
$\displaystyle=L^{(+)}_{1}L^{(+)}_{2}R_{21}L^{(-)-1}_{1}L^{(-)-1}_{2}R_{21}^{-1}=L^{(+)}_{1}L^{(+)}_{2}L^{(-)-1}_{2}L^{(-)-1}_{1}$
$\displaystyle=L^{(+)}_{12}L^{(-)-1}_{12}=\Psi_{0,1}(M)_{12}.$

For the $H$-linearity

$\displaystyle\Psi_{0,1}\big{(}\overset{I}{h^{\prime}}\overset{I}{M}\overset{I}{S(h^{\prime\prime})}\big{)}$
$\displaystyle=\big{(}\overset{I}{T}\otimes\mathrm{id}\big{)}(h^{\prime}\otimes
1RR^{\prime}S(h^{\prime\prime})\otimes 1)$
$\displaystyle=\big{(}\overset{I}{T}\otimes\mathrm{id}\big{)}(h^{\prime}\otimes
1RR^{\prime}S(h)^{\prime\prime\prime}\otimes
S(h)^{\prime\prime}h^{\prime\prime\prime\prime})$
$\displaystyle=\big{(}\overset{I}{T}\otimes\mathrm{id}\big{)}(h^{\prime}S(h)^{\prime\prime\prime}\otimes
S(h)^{\prime\prime}RR^{\prime}1\otimes h^{\prime\prime\prime\prime})$
$\displaystyle=\big{(}\overset{I}{T}\otimes\mathrm{id}\big{)}(1\otimes
S(h^{\prime})RR^{\prime}1\otimes
h^{\prime\prime})=S(h^{\prime})\Psi_{0,1}(\overset{I}{M})h^{\prime\prime}.$

We used the basic properties of $S$ and the fact that
$\Delta^{\mathrm{op}}R=R\Delta$, with
$\Delta^{\mathrm{op}}(h)=h^{\prime\prime}\otimes h^{\prime}$. ∎

Write $\mathrm{T}(H^{*})=\bigoplus_{n\in\mathbb{N}}\mathrm{T}_{n}(H^{*})$,
where $\mathrm{T}_{n}(H^{*})$ is the subspace generated by all the products
$\psi_{1}\cdots\psi_{n}$, with $\psi_{i}\in H^{*}$ for each $i$.

######  Lemma 3.6.

Each element of $\mathrm{T}(H^{*})$ is equivalent modulo the fusion relation
of $\mathcal{L}_{0,1}(H)$ to an element of $\mathrm{T}_{1}(H^{*})$. It follows
that $\dim(\mathcal{L}_{0,1}(H))\leq\dim(H^{*})$.

###### Proof.

It suffices to show that the product of two elements of $T_{1}(H^{*})$ is
equivalent to a linear combination of elements of $\mathrm{T}_{1}(H^{*})$, and
the result follows by induction. We can restrict to matrix coefficients since
they linearly span $H^{*}$. If we write $R=a_{i}\otimes b_{i}$, then
$R^{\prime}=b_{i}\otimes a_{i}$ and the fusion relation is rewritten as

$\overset{I\otimes
J}{M}\\!\\!_{12}(\overset{IJ}{R^{\prime}})_{12}=\overset{I}{M}\\!_{1}(\overset{IJ}{R^{\prime}})_{12}\overset{J}{M}\\!_{2}=\overset{I}{M}\\!_{1}\big{(}\overset{I}{b_{i}}\big{)}_{1}\big{(}\overset{J}{a_{i}}\big{)}_{2}\overset{J}{M}\\!_{2}=\big{(}\overset{J}{a_{i}}\big{)}_{2}\overset{I}{M}\\!_{1}\overset{J}{M}\\!_{2}\big{(}\overset{I}{b_{i}}\big{)}_{1}.$

Using $S^{-1}(a_{j})a_{i}\otimes b_{i}b_{j}=1\otimes 1$, we get

$\displaystyle\overset{I}{M}\\!_{1}\overset{J}{M}\\!_{2}=\overset{J}{S^{-1}(a_{i})}_{2}\overset{I\otimes
J}{M}\\!\\!_{12}\big{(}\overset{IJ}{R^{\prime}}\big{)}_{12}\big{(}\overset{I}{b_{i}}\big{)}_{1}$
(3.5)

and this give the result since $\overset{I}{M}_{1}\overset{J}{M}_{2}$ contains
all the possible products between the coefficients of $\overset{I}{M}$ and
those of $\overset{J}{M}$. ∎

######  Theorem 3.7.

Recall that we assume that $H$ is a finite-dimensional factorizable Hopf
algebra. Then the RSD morphism $\Psi_{0,1}$ gives an isomorphism of
$H$-module-algebras $\mathcal{L}_{0,1}(H)\cong H$. It follows that
$\mathcal{L}_{0,1}^{\mathrm{inv}}(H)\cong\mathcal{Z}(H)$.

###### Proof.

Since $H$ is factorizable, $\Psi_{0,1}$ is surjective. Hence
$\dim(\mathcal{L}_{0,1}(H))\geq\dim(H)$. But by Lemma 3.6,
$\dim(\mathcal{L}_{0,1}(H))\leq\dim(H^{*})=\dim(H)$. Thus
$\dim(\mathcal{L}_{0,1}(H))=\dim(H)$. ∎

Let us point out obvious consequences. First, by comparing the dimensions, we
see that the canonical map
$H^{*}\hookrightarrow\mathrm{T}(H^{*})\twoheadrightarrow\mathcal{L}_{0,1}(H)$
is an isomorphism of vector spaces. Second, this shows that the matrices
$\overset{I}{M}$ are invertible since $RR^{\prime}$ is invertible. More
importantly, this theorem allows us to identify $\mathcal{L}_{0,1}(H)$ with
$H$ via $\overset{I}{M}=\overset{I}{L}\\!^{(+)}\overset{I}{L}\\!^{(-)-1}$,
where the matrices $L^{(\pm)}$ are defined in $\eqref{L}$. We will always work
with this identification in the sequel.

######  Remark 3.8.

Thanks to the isomorphism of vector spaces $\overset{I}{T}\to\overset{I}{M}$
and the relation (3.5), we can see $\mathcal{L}_{0,1}(H)$ as $H^{*}$ endowed
with a product $\ast$ defined by

$\displaystyle\varphi\ast\psi=\varphi\\!\left(?b_{j}b_{i}\right)\psi\big{(}S^{-1}(a_{i})?a_{j}\big{)},\qquad
x_{m}\mapsto\varphi(x_{m}^{\prime}b_{j}b_{i})\psi\big{(}S^{-1}(a_{i})x_{m}^{\prime\prime}a_{j}\big{)}.$
(3.6)

This is the product of the gauge fields $\varphi,\psi$ and its evaluation on
the connection which assigns $x_{m}\in H$ to the loop $m$, see Fig. 1. The
action of the gauge algebra $H$ on a gauge field $\varphi$ is $\varphi\cdot
h=\varphi(h^{\prime}?S(h^{\prime\prime}))$ and $\ast$ is $H$-equivariant.

We denote by $\operatorname{SLF}(H)$ the space of symmetric linear forms on
$H$:

$\operatorname{SLF}(H)=\\{\psi\in H^{*}\,|\,\forall\,x,y\in
H,\,\psi(xy)=\psi(yx)\\}.$

$\operatorname{SLF}(H)$ is obviously a subalgebra of $\mathcal{O}(H)$.
Consider the following variant of the map $\Psi$ of Section 2.2, which will be
useful in what follows

$\displaystyle\begin{array}[]{crll}\mathcal{D}\colon&\mathcal{O}(H)&\rightarrow&H,\\\
&\psi&\mapsto&\left(\psi\otimes\text{id}\right)\left((g\otimes
1)RR^{\prime}\right),\end{array}$ (3.9)

where $g$ is the pivotal element (2.9). Since $H$ is factorizable,
$\mathcal{D}$ is an isomorphism of vector spaces. A computation similar to
that of the proof of Proposition 3.5 shows that $\mathcal{D}$ brings symmetric
linear forms to central elements. Moreover, it is not difficult to show that
it induces an isomorphism of algebras
$\operatorname{SLF}(H)\cong\mathcal{Z}(H)=\mathcal{L}^{\mathrm{inv}}_{0,1}(H)$.

Let us fix a notation. Every $\psi\in H^{*}$ can be written as
$\psi=\sum_{i,j,I}\lambda_{ij}^{I}\overset{I}{T}\\!^{i}_{j}$ with
$\lambda_{ij}^{I}\in\mathbb{C}$. In order to avoid the indices, define for
each $I$ a matrix $\Lambda_{I}\in\mathrm{Mat}_{\dim(I)}(\mathbb{C})$ by
$(\Lambda_{I})^{i}_{j}=\lambda^{I}_{ji}$. Then $\psi$ can be expressed as
$\psi=\sum_{I}\mathrm{tr}\big{(}\Lambda_{I}\overset{I}{T}\big{)}$. We record
these observations as a lemma.

######  Lemma 3.9.

Every $x\in\mathcal{L}_{0,1}(H)$ can be expressed as

$x=\sum_{I}\mathrm{tr}\big{(}\Lambda_{I}\overset{I}{g}\overset{I}{M}\big{)}$

such that
$\mathcal{D}^{-1}(x)=\sum_{I}\mathrm{tr}\big{(}\Lambda_{I}\overset{I}{T}\big{)}$.
Moreover, if $x\in\mathcal{L}^{\mathrm{inv}}_{0,1}(H)$, then
$\mathcal{D}^{-1}(x)\in\operatorname{SLF}(H)$.

######  Remark 3.10.

Let us stress that, due to non-semi-simplicity, this way of writing elements
of $\mathcal{L}_{0,1}(H)$ and of $\operatorname{SLF}(H)$ is in general not
unique, see the comments in Section 2.1.

##  4 The handle algebra $\boldsymbol{\mathcal{L}_{1,0}(H)}$

We assume that $H$ is a finite-dimensional factorizable ribbon Hopf algebra.
Note however that the ribbon assumption is not needed in Sections 4.1 and 4.2.

###  4.1 Definition of $\boldsymbol{\mathcal{L}_{1,0}(H)}$ and
$\boldsymbol{H}$-module-algebra structure

Consider the free product $\mathcal{L}_{0,1}(H)\ast\mathcal{L}_{0,1}(H)$, and
let $j_{1}$ (resp. $j_{2}$) be the canonical injection in the first (resp.
second) copy of $\mathcal{L}_{0,1}(H)$. We define
$\overset{I}{B}=j_{1}(\overset{I}{M})$ and
$\overset{I}{A}=j_{2}(\overset{I}{M})$, that is
$\overset{I}{B}{{}^{a}_{b}}=j_{1}(\overset{I}{M}{{}^{a}_{b}})$,
$\overset{I}{A}{{}^{a}_{b}}=j_{2}(\overset{I}{M}{{}^{a}_{b}})$.

######  Definition 4.1.

The handle algebra $\mathcal{L}_{1,0}(H)$ is the quotient of
$\mathcal{L}_{0,1}(H)\ast\mathcal{L}_{0,1}(H)$ by the following exchange
relations:

$\overset{IJ}{R}_{12}\overset{I}{B}_{1}\overset{IJ}{(R^{\prime})}_{12}\overset{J}{A}_{2}=\overset{J}{A}_{2}\overset{IJ}{R}_{12}\overset{I}{B}_{1}\overset{IJ}{R}{{}^{-1}_{12}}$

for all finite-dimensional $H$-modules $I$, $J$.

The exchange relation above is the same as in [13, Definition 1] except that
$A$ and $B$ are switched; the one of [4, Definition 12] and [5, equation
(3.14)] is different, due to a different choice of the action of $H$ on
$\mathcal{L}_{1,0}(H)$. In the semisimple setting, the algebras resulting from
each of these definitions are isomorphic. This is a relation between matrices
in
$\mathrm{Mat}_{\dim(I)}(\mathbb{C})\otimes\mathrm{Mat}_{\dim(J)}(\mathbb{C})\otimes\mathcal{L}_{1,0}(H)$
(for all finite-dimensional $I$, $J$) which implies relations among elements
of $\mathcal{L}_{1,0}(H)$, namely

$\forall\,I,J,a,b,c,d,\qquad\overset{IJ}{R}{{}^{ac}_{ij}}\overset{I}{B}{{}^{i}_{k}}(\overset{IJ}{R^{\prime}})^{kj}_{bl}\overset{J}{A}{{}^{l}_{d}}=\overset{J}{A}{{}^{c}_{i}}\overset{IJ}{R}{{}^{ai}_{jk}}\overset{I}{B}{{}^{j}_{l}}\big{(}\overset{IJ}{R}{{}^{-1}}\big{)}^{lk}_{bd}.$

Like the other relations before, we can write the
$\mathcal{L}_{1,0}(H)$-exchange relation more simply as:

$\displaystyle R_{12}B_{1}R_{21}A_{2}=A_{2}R_{12}B_{1}R_{12}^{-1}.$ (4.1)

It immediately follows from (3.2) that if $f\colon I\to J$ is a morphism, it
holds

$f\overset{I}{B}=\overset{J}{B}f,\qquad f\overset{I}{A}=\overset{J}{A}f,$

where we identify $f$ with its matrix.

Similarly to $\mathcal{L}_{0,1}(H)$, consider the following right action of
$H$ on $\mathcal{L}_{1,0}(H)$, which is the analogue of the action of the
gauge group on the gauge fields:

$\displaystyle\overset{I}{B}\cdot
h=\overset{I}{h^{\prime}}\overset{I}{B}\overset{I}{S(h^{\prime\prime})},\qquad\overset{I}{A}\cdot
h=\overset{I}{h^{\prime}}\overset{I}{A}\overset{I}{S(h^{\prime\prime})}.$
(4.2)

As above, it is equivalent to work with the corresponding left coaction
$\Omega\colon\mathcal{L}_{1,0}(H)\to\mathcal{O}(H)\otimes\mathcal{L}_{1,0}(H)$
defined by

$\Omega(\overset{I}{B})=\overset{I}{T}\overset{I}{B}S(\overset{I}{T}),\qquad\Omega(\overset{I}{A})=\overset{I}{T}\overset{I}{A}S(\overset{I}{T}).$

######  Proposition 4.2.

The right action $\cdot$ is a $H$-module-algebra structure on
$\mathcal{L}_{1,0}(H)$. Equivalently, $\Omega$ is a left
$\mathcal{O}(H)$-comodule-algebra structure on $\mathcal{L}_{1,0}(H)$.

###### Proof.

One must show that $\Omega$ is an algebra morphism, as in [14]. This amounts
to check that $\Omega$ is compatible with the exchange relation, which is
similar to the proof of Proposition 3.4. ∎

We denote by $\mathcal{L}_{1,0}^{\text{inv}}(H)$ the subalgebra of invariant
elements of $\mathcal{L}_{1,0}(H)$. For instance, the elements

$\displaystyle\mathrm{tr}_{12}\Big{(}\overset{I\otimes
J}{g}\\!\\!\\!_{12}\Phi\overset{I}{A}_{1}\overset{IJ}{(R^{\prime})}_{12}\overset{J}{B}_{2}\overset{IJ}{R}_{12}\Big{)}$
(4.3)

with $\Phi\in\operatorname{End}_{H}(I\otimes J)$ and
$\mathrm{tr}_{12}=\mathrm{tr}\otimes\mathrm{tr}$, are invariant.

Finally, we describe a wide family of maps
$\mathcal{L}_{0,1}(H)\to\mathcal{L}_{1,0}(H)$. For
$w\in\mathcal{L}^{\mathrm{inv}}_{0,1}(H)=\mathcal{Z}(H)$ and
$m_{1},n_{1},\dots,m_{k},n_{k}\in\mathbb{Z}$, define

$\begin{array}[]{crll}j_{wB^{m_{1}}A^{n_{1}}\dots
B^{m_{k}}A^{n_{k}}}\colon&\mathcal{L}_{0,1}(H)&\rightarrow&\mathcal{L}_{1,0}(H),\\\
&\overset{I}{M}&\mapsto&\overset{I}{w}\overset{I}{B}{{}^{m_{1}}}\overset{I}{A}{{}^{n_{1}}}\cdots\overset{I}{B}{{}^{m_{k}}}\overset{I}{A}{{}^{n_{k}}}.\end{array}$

It is clear that these maps are morphisms of $H$-modules, but not of algebras
in general. Hence the restriction satisfies $j_{wB^{m_{1}}A^{n_{1}}\dots
B^{m_{k}}A^{n_{k}}}\colon\mathcal{L}^{\mathrm{inv}}_{0,1}(H)\to\mathcal{L}^{\mathrm{inv}}_{1,0}(H)$.
This gives a particular type of invariants in $\mathcal{L}_{1,0}(H)$. We will
more shortly write

$\displaystyle x_{wB^{m_{1}}A^{n_{1}}\dots
B^{m_{k}}A^{n_{k}}}=j_{wB^{m_{1}}A^{n_{1}}\dots B^{m_{k}}A^{n_{k}}}(x).$ (4.4)

We also use this notation for $x\in H$, thanks to the identification
$H=\mathcal{L}_{0,1}(H)$.

######  Remark 4.3.

Recall from Remark 3.10 that the matrix coefficients do not form a basis of
$\mathcal{L}_{0,1}(H)$. They just linearly span this space. However, the maps
$j_{wA^{m_{1}}B^{n_{1}}\dots A^{m_{k}}B^{n_{k}}}$ are well-defined. Indeed,
first observe that

$\displaystyle j_{B}\colon\
\mathcal{L}_{0,1}(H)\overset{j_{1}}{\longhookrightarrow}\mathcal{L}_{0,1}(H)\ast\mathcal{L}_{0,1}(H)\overset{\pi}{\longrightarrow\\!\\!\\!\\!\\!\rightarrow}\mathcal{L}_{1,0}(H),$
$\displaystyle j_{A}\colon\
\mathcal{L}_{0,1}(H)\overset{j_{2}}{\longhookrightarrow}\mathcal{L}_{0,1}(H)\ast\mathcal{L}_{0,1}(H)\overset{\pi}{\longrightarrow\\!\\!\\!\\!\\!\rightarrow}\mathcal{L}_{1,0}(H)$

are well-defined. Let us show for instance that the map $j_{A^{-1}B^{-1}A}$ is
well-defined. Assume that $\lambda^{a}_{b}\overset{I}{T}{{}^{b}_{a}}=0$.
Applying the coproduct in $\mathcal{O}(H)$ twice and tensoring with
$\mathrm{id}_{H}$, we get

$\lambda^{a}_{b}\overset{I}{T}{{}^{b}_{k}}\otimes\mathrm{id}_{H}\otimes\overset{I}{T}{{}^{k}_{l}}\otimes\mathrm{id}_{H}\otimes\overset{I}{T}{{}^{l}_{a}}\otimes\mathrm{id}_{H}=0.$

We evaluate this on
$\left(RR^{\prime}\right)^{-1}\otimes\left(RR^{\prime}\right)^{-1}\otimes
RR^{\prime}$:

$\lambda^{a}_{b}\big{(}\overset{I}{M}{{}^{-1}}\big{)}{{}^{b}_{k}}\otimes\big{(}\overset{I}{M}{{}^{-1}}\big{)}{{}^{k}_{l}}\otimes\overset{I}{M}{{}^{l}_{a}}=0.$

Finally, we apply the map $j_{A}\otimes j_{B}\otimes j_{A}$ and multiplication
in $\mathcal{L}_{1,0}(H)$:

$\lambda^{a}_{b}\big{(}\overset{I}{A}{{}^{-1}}\overset{I}{B}{{}^{-1}}\overset{I}{A}\big{)}{{}^{b}_{a}}=0$

as desired. A similar proof can be used to show that all the other maps
defined by means of matrix coefficients (like $\Psi_{1,0}$ or $\alpha$,
$\beta$ below etc..) are well-defined.

###  4.2 Isomorphism
$\boldsymbol{\mathcal{L}_{1,0}(H)\cong\mathcal{H}(\mathcal{O}(H))}$

Let us begin by recalling the following definition (see for instance [35,
Example 4.1.10]).

######  Definition 4.4.

Let $H$ be a Hopf algebra. The Heisenberg double of $\mathcal{O}(H)$,
$\mathcal{H}(\mathcal{O}(H))$, is the vector space $\mathcal{O}(H)\otimes H$
endowed with the algebra structure defined by the following multiplication
rules:

  * •

The canonical injections $H,\mathcal{O}(H)\to\mathcal{O}(H)\otimes H$ are
algebra morphisms. Thus we identify $\mathcal{O}(H)$ (resp. $H$) with
$\mathcal{O}(H)\otimes 1\subset\mathcal{H}(\mathcal{O}(H))$ (resp. with
$1\otimes H\subset\mathcal{H}(\mathcal{O}(H))$).

  * •

Under this identification, we have the exchange relation

$\forall\,\psi\in\mathcal{O}(H),\qquad\forall\,h\in H,\qquad
h\psi=\psi(?h^{\prime})h^{\prime\prime}=\psi^{\prime\prime}(h^{\prime})\psi^{\prime}h^{\prime\prime}.$

There is a representation of $\mathcal{H}(\mathcal{O}(H))$ on $\mathcal{O}(H)$
defined by

$\displaystyle
h\triangleright\psi=\psi(?h)=\psi^{\prime\prime}(h)\psi^{\prime},\qquad\varphi\triangleright\psi=\varphi\psi\qquad
h\in H,\qquad\psi,\varphi\in\mathcal{O}(H).$ (4.5)

This representation is faithful (see [35, Lemma 9.4.2]). Hence, if $H$ is
finite-dimensional, it follows that $\mathcal{H}(\mathcal{O}(H))$ is a matrix
algebra

$\displaystyle\mathcal{H}(\mathcal{O}(H))\cong\operatorname{End}_{\mathbb{C}}(H^{*}).$
(4.6)

Under our assumptions on $H$, a natural set of generators for
$\mathcal{H}(\mathcal{O}(H))$ consists of the matrix coefficients of
$\overset{I}{T}$ and of $\overset{I}{L}\\!^{(\pm)}$.

######  Lemma 4.5.

With these generators, the exchange relation of $\mathcal{H}(\mathcal{O}(H))$
is

$\overset{I}{L}\\!^{(\pm)}_{1}\overset{J}{T_{2}}=\overset{J}{T_{2}}\overset{I}{L}\\!^{(\pm)}_{1}\overset{IJ}{R}\\!^{(\pm)}_{12}$

and the representation $\triangleright$ is

$\overset{I}{L}\\!^{(\pm)}_{1}\triangleright\overset{J}{T_{2}}=\overset{J}{T_{2}}\overset{IJ}{R}\\!^{(\pm)}_{12},\qquad\overset{I}{T_{1}}\triangleright\overset{J}{T_{2}}=\overset{I\otimes
J}{T}\\!\\!\\!_{12}.$

###### Proof.

For the exchange relation, we apply the defining relation of
$\mathcal{H}(\mathcal{O}(H))$ together with the definition (2.4) of
$\overset{I}{L}{{}^{(\pm)}}$ and (2.1):

$\displaystyle\overset{I}{L}{{}^{(+)}_{1}}\overset{J}{T_{2}}$
$\displaystyle=\big{(}\overset{I}{a_{i}}\big{)}_{1}b_{i}\overset{J}{T_{2}}=\bigl{(}\overset{I}{a_{i}}\bigr{)}_{1}\overset{J}{T}(?b^{\prime}_{i})_{2}b_{i}^{\prime\prime}=\bigl{(}\overset{I}{a_{i}}\bigr{)}_{1}\overset{J}{T_{2}}\bigl{(}\overset{J}{b_{i}^{\prime}}\bigr{)}_{2}b_{i}^{\prime\prime}=\bigl{(}\overset{I}{a_{i}}\overset{I}{a_{j}}\bigr{)}_{1}\overset{J}{T_{2}}\bigl{(}\overset{J}{b_{j}}\bigr{)}_{2}b_{i}$
$\displaystyle=\overset{J}{T_{2}}\bigl{(}\overset{I}{a_{i}}\bigr{)}_{1}b_{i}\bigl{(}\overset{I}{a_{j}}\bigr{)}_{1}\bigl{(}\overset{J}{b_{j}}\bigr{)}_{2}=\overset{J}{T_{2}}\overset{I}{L}{{}^{(+)}_{1}}\overset{IJ}{R}_{12}.$

We used that $\overset{J}{T}(?x)=\overset{J}{T}\overset{J}{x}$ (which just
follows from the fact that $\overset{J}{T}$ is a morphism from $H$ to
$\mathrm{Mat}_{\dim(J)}(\mathbb{C})$). The proof for
$\overset{I}{L}{{}^{(-)}}$ is exactly the same since $R^{(-)}$ is also a
universal $R$-matrix. For the second formula we just apply the definition of
$\triangleright$:

$\overset{I}{L}{{}^{(+)}_{1}}\triangleright\overset{J}{T_{2}}=\bigl{(}\overset{I}{a_{i}}\bigr{)}_{1}b_{i}\triangleright\overset{J}{T_{2}}=\bigl{(}\overset{I}{a_{i}}\bigr{)}_{1}\overset{J}{T}(?b_{i})_{2}=\bigl{(}\overset{I}{a_{i}}\bigr{)}_{1}\overset{J}{T_{2}}\bigl{(}\overset{J}{b_{i}}\bigr{)}_{2}=\overset{J}{T_{2}}\bigl{(}\overset{I}{a_{i}}\bigr{)}_{1}\bigl{(}\overset{J}{b_{i}}\bigr{)}_{2}=\overset{J}{T_{2}}\overset{IJ}{R}_{12}.$

The proof for $\overset{I}{L}{{}^{(-)}}$ is exactly the same. The last formula
follows from (2.10). ∎

Thanks to [2], we know that there is a morphism from $\mathcal{L}_{1,0}(H)$ to
$\mathcal{H}(\mathcal{O}(H))$:

######  Proposition 4.6.

The following map is a morphism of algebras:

$\begin{array}[]{@{}crll}\Psi_{1,0}\colon&\mathcal{L}_{1,0}(H)&\rightarrow&\mathcal{H}(\mathcal{O}(H)),\\\
&\overset{I}{B}&\mapsto&\overset{I}{L}\\!^{(+)}\overset{I}{T}\overset{I}{L}\\!^{(-)-1},\\\
&\overset{I}{A}&\mapsto&\overset{I}{L}\\!^{(+)}\overset{I}{L}\\!^{(-)-1}.\end{array}$

###### Proof.

One must check that the fusion and exchange relations are compatible with
$\Psi_{1,0}$. Observe that the restriction of $\Psi_{1,0}$ to the first copy
of $\mathcal{L}_{0,1}(H)\subset\mathcal{L}_{1,0}(H)$ is just the RSD morphism
$\Psi_{0,1}$, thus $\Psi_{1,0}$ is compatible with the fusion relation over
$A$. For the fusion relation over $B$, we have

$\displaystyle\Psi_{1,0}(B)_{12}$
$\displaystyle=L^{(+)}_{12}T_{12}L^{(-)-1}_{12}$ (definition)
$\displaystyle=L^{(+)}_{1}L^{(+)}_{2}T_{1}T_{2}L^{(-)-1}_{2}L^{(-)-1}_{1}$
(equations (2.5) and (2.10))
$\displaystyle=L^{(+)}_{1}T_{1}L^{(+)}_{2}R_{21}T_{2}L^{(-)-1}_{2}L^{(-)-1}_{1}$
(Lemma 4.5)
$\displaystyle=L^{(+)}_{1}T_{1}L^{(+)}_{2}R_{21}T_{2}R_{21}L^{(-)-1}_{1}L^{(-)-1}_{2}R_{21}^{-1}$
(equation (2.5))
$\displaystyle=L^{(+)}_{1}T_{1}L^{(+)}_{2}R_{21}L^{(-)-1}_{1}T_{2}L^{(-)-1}_{2}R_{21}^{-1}$
(Lemma 4.5)
$\displaystyle=L^{(+)}_{1}T_{1}L^{(-)-1}_{1}R_{21}L^{(+)}_{2}T_{2}L^{(-)-1}_{2}R_{21}^{-1}$
(equation (2.5))
$\displaystyle=\Psi_{1,0}(B)_{1}R_{21}\Psi_{1,0}(B)_{2}R_{21}^{-1}$
$\displaystyle\quad\text{(definition)}.$

The same kind of computation allows one to show that $\Psi_{1,0}$ is
compatible with the $\mathcal{L}_{1,0}$-exchange relation. ∎

We will now show that $\Psi_{1,0}$ is an isomorphism under our assumptions on
$H$.

######  Lemma 4.7.

Every element in $\mathcal{L}_{1,0}(H)$ can be written as
$\sum_{i}(x_{i})_{B}(y_{i})_{A}$ with $x_{i},y_{i}\in\mathcal{L}_{0,1}(H)$. It
follows that
$\dim(\mathcal{L}_{1,0}(H))\leq\dim(\mathcal{L}_{0,1}(H))^{2}=\dim(H)^{2}$.

###### Proof.

This is the same proof as in Lemma 3.6. It suffices to show that an element
like $y_{A}x_{B}$ can be expressed as $\sum_{i}(x_{i})_{B}(y_{i})_{A}$. The
exchange relation can be rewritten as

$\displaystyle\overset{I}{A}_{1}\overset{J}{B}_{2}=\overset{I}{S^{-1}(a_{i})}_{2}\overset{IJ}{(R^{\prime})}_{12}\overset{J}{B}_{2}\overset{IJ}{R}_{12}\overset{I}{A}_{1}\overset{IJ}{(R^{\prime})}_{12}\overset{J}{(b_{i})}_{1},$
(4.7)

and the result follows since $\overset{I}{A}_{1}\overset{J}{B}_{2}$ contains
all the possible products between the coefficients of $\overset{I}{A}$ and
those of $\overset{J}{B}$. ∎

######  Theorem 4.8.

Recall that we assume that $H$ is a finite-dimensional factorizable Hopf
algebra. $\Psi_{1,0}$ gives an isomorphism of algebras
$\mathcal{L}_{1,0}(H)\cong\mathcal{H}(\mathcal{O}(H))$. It follows that
$\mathcal{L}_{1,0}(H)$ is a matrix algebra:
$\mathcal{L}_{1,0}(H)\cong\mathrm{Mat}_{\dim(H)}(\mathbb{C})$.

###### Proof.

Observe that $\Psi_{1,0}\circ j_{A}=i_{H}\circ\Psi_{0,1}$ where $i_{H}\colon
H\to\mathcal{H}(\mathcal{O}(H))$ is the canonical inclusion. Since
$\Psi_{0,1}$ is an isomorphism, there exist matrices
$\overset{I}{A}\\!^{(\pm)}$ such that

$\Psi_{1,0}\big{(}\overset{I}{A}\\!^{(\pm)}\big{)}=\overset{I}{L}\\!^{(\pm)}\in\mathrm{Mat}_{\dim(I)}(\mathcal{H}(\mathcal{O}(H))).$

Moreover, we have

$\Psi_{1,0}\big{(}\overset{I}{A}\\!^{(+)-1}\overset{I}{B}\overset{I}{A}\\!^{(-)}\big{)}=\overset{I}{T}\in\mathrm{Mat}_{\dim(I)}(\mathcal{H}(\mathcal{O}(H))).$

Thus $\Psi_{1,0}$ is surjective, and hence
$\dim(\mathcal{L}_{1,0}(H))\geq\dim(\mathcal{H}(\mathcal{O}(H)))=\dim(H)^{2}$.
This together with Lemma 4.7 gives
$\dim(\mathcal{L}_{1,0}(H))=\dim(\mathcal{H}(\mathcal{O}(H)))$. The last claim
is a general fact, see (4.6). ∎

######  Remark 4.9.

Due to Theorem 4.8, there is an isomorphism of vector spaces
$\mathcal{L}_{1,0}(H)\to H^{*}\otimes H^{*}$ given by
$\overset{I}{B}{{}^{i}_{j}}\overset{J}{A}{{}^{k}_{l}}\mapsto\overset{I}{T}{{}^{i}_{j}}\otimes\overset{J}{T}{{}^{k}_{l}}$.
This defines a product $\ast$ on $H^{*}\otimes H^{*}$; due to (4.7), we get
that it satisfies

$(\varepsilon\otimes\psi)\ast(\varphi\otimes\varepsilon)=\varphi\big{(}S^{-1}(a_{i})a_{j}?b_{k}a_{l}\big{)}\otimes\psi\big{(}\vphantom{S^{-1}}b_{j}a_{k}?b_{l}b_{i}\big{)}.$

Combining this with (3.6), we obtain the general formula

$\displaystyle(\varphi_{1}\otimes\psi_{1})\ast(\varphi_{2}\otimes\psi_{2})=\varphi_{1}(?b_{m}b_{n})\varphi_{2}\big{(}S^{-1}(a_{i})a_{j}S^{-1}(a_{n})?a_{m}b_{k}a_{l}\big{)}$
$\displaystyle\hphantom{(\varphi_{1}\otimes\psi_{1})\ast(\varphi_{2}\otimes\psi_{2})=}{}\otimes\psi_{1}\big{(}\vphantom{S^{-1}}b_{j}a_{k}?b_{o}b_{p}b_{l}b_{i}\big{)}\psi_{2}\big{(}S^{-1}(a_{p})?a_{o}\big{)},$
$\displaystyle x_{b}\otimes
x_{a}\mapsto\varphi_{1}(x_{b}^{\prime}b_{m}b_{n})\varphi_{2}\big{(}S^{-1}(a_{i})a_{j}S^{-1}(a_{n})x_{b}^{\prime\prime}a_{m}b_{k}a_{l}\big{)}$
$\displaystyle\hphantom{x_{b}\otimes
x_{a}\mapsto}{}\times\psi_{1}\big{(}\vphantom{S^{-1}}b_{j}a_{k}x_{a}^{\prime}b_{o}b_{p}b_{l}b_{i}\big{)}\psi_{2}\big{(}S^{-1}(a_{p})x_{a}^{\prime\prime}a_{o}\big{)}.$

This is the product of the gauge fields $\varphi_{1}\otimes\psi_{1}$,
$\varphi_{2}\otimes\psi_{2}$ and its evaluation on the connection which
assigns $x_{b}$ to the loop $b$ and $x_{a}$ to the loop $a$, see Fig. 1. The
action of the gauge algebra $H$ on a gauge field $\varphi\otimes\psi$ is
$(\varphi\otimes\psi)\cdot
h=\varphi(h^{\prime}?S(h^{\prime\prime}))\otimes\psi(h^{\prime\prime\prime}?S(h^{\prime\prime\prime\prime}))$
and $\ast$ is $H$-equivariant.

###  4.3 Representation of $\boldsymbol{\mathcal{L}_{1,0}^{\mathrm{inv}}(H)}$
on $\operatorname{SLF}(H)$

Recall from (4.5) that there is a faithful representation $\triangleright$ of
$\mathcal{H}(\mathcal{O}(H))$ on $\mathcal{O}(H)$. Using the isomorphism
$\Psi_{1,0}$, we get a representation of $\mathcal{L}_{1,0}(H)$ on
$\mathcal{O}(H)$, still denoted $\triangleright$:

$\forall\,x\in\mathcal{L}_{1,0}(H),\qquad\forall\,\psi\in\mathcal{O}(H),\qquad
x\triangleright\psi=\Psi_{1,0}(x)\triangleright\psi.$

Thanks to Lemma 4.5, it is easy to get

$\displaystyle\overset{I}{A}_{1}\triangleright\overset{J}{T}\\!_{2}=\overset{J}{T}\\!_{2}\overset{IJ}{(RR^{\prime})}_{12},$
$\displaystyle\overset{I}{B}_{1}\triangleright\overset{J}{T}\\!_{2}=\overset{I}{(a_{i})}_{1}\overset{I\otimes
J}{T}\\!\\!\\!_{12}\overset{I\otimes
J}{(b_{i})}_{12}\overset{IJ}{(R^{\prime})}_{12}=\overset{I}{(a_{i}a_{j})}_{1}\overset{I\otimes
J}{T}\\!\\!\\!_{12}\overset{I}{(b_{j})}_{1}\overset{J}{(b_{i})}_{2}\overset{IJ}{(R^{\prime})}_{12},$
(4.8)

where as usual $R=a_{i}\otimes b_{i}$ and the last equality is obtained using
(2.1).

In [2, Theorem 5] (which is stated in the case of $H=U_{q}(\mathfrak{g})$, $q$
generic), there is a representation of $\mathcal{L}_{1,0}^{\text{inv}}(H)$ on
a subspace of invariants in $H^{*}$. This can be generalized to our
assumptions.

######  Theorem 4.10.

The restriction of $\triangleright$ to $\mathcal{L}_{1,0}^{\text{\em inv}}(H)$
leaves the subspace $\operatorname{SLF}(H)\subset H^{*}$ stable

$\forall\,x\in\mathcal{L}^{\text{\em
inv}}_{1,0}(H),\qquad\forall\,\psi\in\operatorname{SLF}(H),\qquad
x\triangleright\psi\in\operatorname{SLF}(H).$

Hence, we have a representation of $\mathcal{L}_{1,0}^{\text{\rm inv}}(H)$ on
$\operatorname{SLF}(H)$. We denote it $\rho_{\mathrm{SLF}}$.

###### Proof.

For $h\in H$, define $\widetilde{h}\in\mathcal{H}(\mathcal{O}(H))$ by
$\widetilde{h}\triangleright\varphi=\varphi\bigl{(}S^{-1}(h)?\bigr{)}$ for all
$\varphi\in H^{*}$ (since the representation $\triangleright$ is faithful,
this entirely defines $\widetilde{h}$). It is easy to see that

$\forall\,g\in
H,\qquad\forall\,\psi\in\mathcal{O}(H),\qquad\widetilde{g}\widetilde{h}=\widetilde{gh},\qquad
g\widetilde{h}=\widetilde{h}g,\qquad\widetilde{h}\psi=\psi\big{(}S^{-1}(h^{\prime\prime})?\big{)}\widetilde{h^{\prime}}.$

We define matrices

$\displaystyle\overset{I}{\widetilde{L}}{{}^{(\pm)}}=\bigl{(}\overset{I}{a_{i}^{(\pm)}}\bigr{)}\widetilde{b_{i}^{(\pm)}}\in\mathrm{Mat}_{\dim(I)}\bigl{(}\mathcal{H}(\mathcal{O}(H))\bigr{)}$

with $R^{(\pm)}=a_{i}^{(\pm)}\otimes b_{i}^{(\pm)}$. By definition and (2.2),
they satisfy
$\overset{I}{\widetilde{L}}{{}^{(\pm)}_{1}}\triangleright\overset{J}{T}_{2}=\overset{IJ}{R}{{}^{(\pm)-1}_{12}}\overset{J}{T}_{2}$.
It is not difficult to show the following commutation rules

$\displaystyle\overset{I}{\widetilde{L}}{{}^{(\epsilon)}_{1}}\overset{J}{\widetilde{L}}{{}^{(\epsilon)}_{2}}=\overset{I\otimes
J}{\widetilde{L}}\\!\\!{{}^{(\epsilon)}_{12}},\qquad\overset{I}{\widetilde{L}}{{}^{(\epsilon)}_{1}}\overset{J}{L}{{}^{(\sigma)}_{2}}=\overset{J}{L}{{}^{(\sigma)}_{2}}\overset{I}{\widetilde{L}}{{}^{(\epsilon)}_{1}},\qquad\overset{IJ}{R}{{}^{(\epsilon)}_{12}}\overset{I}{\widetilde{L}}{{}^{(\epsilon)}_{1}}\overset{J}{T}_{2}=\overset{J}{T}_{2}\overset{I}{\widetilde{L}}{{}^{(\epsilon)}_{1}},$
(4.9)
$\displaystyle\overset{IJ}{R}\\!^{(\epsilon)}_{12}\overset{I}{\widetilde{L}}\\!^{(\epsilon)}_{1}\overset{J}{\widetilde{L}}\\!^{(\sigma)}_{2}=\overset{J}{\widetilde{L}}\\!^{(\sigma)}_{2}\overset{I}{\widetilde{L}}\\!^{(\epsilon)}_{1}\overset{IJ}{R}\\!^{(\epsilon)}_{12}\\!\qquad\forall\,\epsilon,\sigma\in\\{\pm\\},\qquad\overset{IJ}{R}\\!^{(\epsilon)}_{12}\overset{I}{\widetilde{L}}\\!^{(\sigma)}_{1}\overset{J}{\widetilde{L}}\\!^{(\sigma)}_{2}=\overset{J}{\widetilde{L}}\\!^{(\sigma)}_{2}\overset{I}{\widetilde{L}}\\!^{(\sigma)}_{1}\overset{IJ}{R}\\!^{(\epsilon)}_{12}\\!\qquad\forall\,\epsilon,\sigma\in\\{\pm\\}.$

For instance, here is a proof of the third equality with $\epsilon=+$:

$\displaystyle\overset{IJ}{R}{{}^{(+)}_{12}}\overset{I}{\widetilde{L}}{{}^{(+)}_{1}}\overset{J}{T}_{2}$
$\displaystyle=\overset{IJ}{R}{{}^{(+)}_{12}}\bigl{(}\overset{I}{a_{i}}\bigr{)}_{1}\widetilde{b_{i}}\overset{J}{T_{2}}=\overset{IJ}{R}{{}^{(+)}_{12}}\bigl{(}\overset{I}{a_{i}}\bigr{)}_{1}\overset{J}{T}\bigl{(}S^{-1}(b_{i}^{\prime\prime})?\bigr{)}_{2}\widetilde{b_{i}^{\prime}}=\overset{IJ}{R}{{}^{(+)}_{12}}\bigl{(}\overset{I}{a_{i}}\overset{I}{a_{j}}\bigr{)}_{1}\bigl{(}\overset{J}{S^{-1}(b_{i})}\bigr{)}_{2}\overset{J}{T_{2}}\widetilde{b_{j}}$
$\displaystyle=\overset{IJ}{R}{{}^{(+)}_{12}}\bigl{(}\overset{I}{a_{i}}\bigr{)}_{1}\bigl{(}\overset{J}{S^{-1}(b_{i})}\bigr{)}_{2}\overset{J}{T_{2}}\bigl{(}\overset{I}{a_{j}}\bigr{)}\widetilde{b_{j}}=\overset{J}{T_{2}}\overset{I}{\widetilde{L}}{{}^{(+)}_{1}},$

since $a_{i}\otimes S^{-1}(b_{i})=R^{(+)-1}$. Now, let

$\overset{I}{C}{{}^{(\pm)}}=\Psi_{1,0}^{-1}\big{(}\overset{I}{L}{{}^{(\pm)}}\overset{I}{\widetilde{L}}{{}^{(\pm)}}\big{)}\in\mathrm{Mat}_{\dim(I)}\\!\left(\mathcal{L}_{1,0}(H)\right).$

Thanks to (2.5) and (4.9), it is easy to see that $\overset{I\otimes
J}{C}\\!{{}^{(\pm)}_{12}}=\overset{I}{C}{{}^{(\pm)}_{1}}\overset{J}{C}{{}^{(\pm)}_{2}}$.

######  Lemma 4.11.

  1. $1)$

An element $x\in\mathcal{L}_{1,0}(H)$ is invariant under the right action of
$H$ if, and only if, $x\overset{I}{C}{{}^{(\pm)}}=\overset{I}{C}{{}^{(\pm)}}x$
for all $I$.

  2. $2)$

A linear form $\psi\in H^{*}$ is symmetric if, and only if,
$\overset{I}{C}{{}^{(\pm)}}\triangleright\psi=\psi\mathbb{I}_{\dim(I)}$ for
all $I$.

###### Proof.

1) Let $U=A$ or $B$, then by (2.4), (4.2) and (2.1) we get

$\displaystyle\overset{J}{U}_{2}\cdot
S^{-1}\bigl{(}\overset{I}{L}{{}^{(\pm)}_{1}}\bigr{)}$
$\displaystyle=\overset{J}{U}_{2}\cdot
S^{-1}\big{(}b_{i}^{(\pm)}\big{)}\bigl{(}\overset{I}{a_{i}^{(\pm)}}\bigr{)}_{1}=\overset{J}{S^{-1}\big{(}\left.b_{i}^{(\pm)}\right.^{\prime\prime}\big{)}}_{2}\overset{J}{U}_{2}\bigl{(}\overset{J}{\left.b_{i}^{(\pm)}\right.^{\prime}}\bigr{)}_{2}\bigl{(}\overset{I}{a_{i}^{(\pm)}}\bigr{)}_{1}$
$\displaystyle=\overset{J}{S^{-1}\bigl{(}b_{i}^{(\pm)}\bigr{)}}_{2}\overset{J}{U}_{2}\bigl{(}\overset{J}{b_{j}^{(\pm)}}\bigr{)}_{2}\bigl{(}\overset{I}{a_{i}^{(\pm)}}\overset{I}{a_{j}^{(\pm)}}\bigr{)}_{1}=\bigl{(}\overset{I}{a_{i}^{(\pm)}}\bigr{)}_{1}\overset{J}{S^{-1}\bigl{(}b_{i}^{(\pm)}\bigr{)}}_{2}\overset{J}{U}_{2}\bigl{(}\overset{I}{a_{j}^{(\pm)}}\bigr{)}_{1}\bigl{(}\overset{J}{b_{j}^{(\pm)}}\bigr{)}_{2}$
$\displaystyle=\overset{IJ}{R}{{}^{(\pm)-1}_{12}}\overset{J}{U}_{2}\overset{IJ}{R}{{}^{(\pm)}_{12}}$

with $R^{(\pm)}=a_{i}^{(\pm)}\otimes b_{i}^{(\pm)}$. Second, using (2.5) and
(4.9) we get

$\overset{I}{C}{{}^{(\pm)}_{1}}\overset{J}{U}_{2}\overset{I}{C}{{}^{(\pm)-1}_{1}}=\overset{IJ}{R}{{}^{(\pm)-1}_{12}}\overset{J}{U}_{2}\overset{IJ}{R}{{}^{(\pm)}_{12}}.$

For instance (with the shortened notation)

$\displaystyle\Psi_{1,0}\big{(}C^{(+)}_{1}A_{2}C^{(+)-1}_{1}\big{)}$
$\displaystyle=L^{(+)}_{1}\widetilde{L}^{(+)}_{1}L^{(+)}_{2}L^{(-)-1}_{2}\widetilde{L}^{(+)-1}_{1}L^{(+)-1}_{1}=L^{(+)}_{1}L^{(+)}_{2}L^{(-)-1}_{2}L^{(+)-1}_{1}$
$\displaystyle=R_{12}^{-1}L^{(+)}_{2}L^{(+)}_{1}R_{12}L^{(-)-1}_{2}L^{(+)-1}_{1}=R_{12}^{-1}L^{(+)}_{2}L^{(-)-1}_{2}R_{12}$
$\displaystyle=\Psi_{1,0}\big{(}R_{12}^{-1}A_{2}R_{12}\big{)}$

and we have equality since $\Psi_{1,0}$ is an isomorphism; the others cases
are similar. It follows that $\overset{J}{U}_{2}\cdot
S^{-1}\bigl{(}\overset{I}{L}{{}^{(\pm)}_{1}}\bigr{)}=\overset{I}{C}{{}^{(\pm)}_{1}}\overset{J}{U}_{2}\overset{I}{C}{{}^{(\pm)-1}_{1}}$
or in other words $\overset{J}{U}{{}^{c}_{d}}\cdot
S^{-1}\bigl{(}\overset{I}{L}{{}^{(\pm)}}{{}^{a}_{b}}\bigr{)}=\overset{I}{C}{{}^{(\pm)}}{{}^{a}_{i}}\overset{J}{U}{{}^{c}_{d}}\overset{I}{C}{{}^{(\pm)-1}}{{}^{i}_{b}}$,
which means that $\overset{J}{U}{{}^{c}_{d}}$ is invariant under the action of
$S^{-1}\bigl{(}\overset{I}{L}{{}^{(\pm)}}{{}^{a}_{b}}\bigr{)}$ if, and only
if, it commutes with $\overset{I}{C}{{}^{(\pm)}}{{}^{a}_{b}}$. Since the
elements $\overset{J}{U}{{}^{c}_{d}}$ (resp.
$S^{-1}\bigl{(}\overset{I}{L}{{}^{(\pm)}}{{}^{a}_{b}}\bigr{)}$) generate
$\mathcal{L}_{1,0}(H)$ (resp. $H$) as an algebra, we get that an element is
invariant if, and only if, it commutes with all the coefficients of the
matrices $\overset{I}{C}{{}^{(\pm)}}$, as desired.

2) Consider the left action $\diamond$ of $H$ on $H^{*}$ given by
$h\diamond\psi=\psi\big{(}S^{-1}(h^{\prime})?h^{\prime\prime}\big{)}$. It is
easy to see that $\psi$ is symmetric if, and only if, it is invariant under
$\diamond$ (namely $h\diamond\psi=\varepsilon(h)\psi$ for all $h\in H$). The
definitions and (2.1) yields

$\displaystyle\overset{I}{C}{{}^{(\pm)}}\triangleright\psi$
$\displaystyle=\bigl{(}\overset{I}{a_{i}^{(\pm)}}\overset{I}{a_{j}^{(\pm)}}\bigr{)}b_{i}\widetilde{b_{j}}\triangleright\psi=\bigl{(}\overset{I}{a_{i}^{(\pm)}}\overset{I}{a_{j}^{(\pm)}}\bigr{)}\psi\bigl{(}S^{-1}(b_{j})?b_{i}\bigr{)}$
$\displaystyle=\bigl{(}\overset{I}{a_{i}^{(\pm)}}\bigr{)}\psi\bigl{(}S^{-1}(b_{i}^{\prime})?b_{i}^{\prime\prime}\bigr{)}=\overset{I}{L}{{}^{(\pm)}}\diamond\psi.$

Since the coefficients of the matrices $\overset{I}{L}{{}^{(\pm)}}$ generate
$H$ as an algebra, we get the result. ∎

End of the proof of Theorem 4.10. Let $x\in\mathcal{L}_{1,0}^{\text{inv}}(H)$
and $\psi\in\operatorname{SLF}(H)$. We apply the previous lemma

$\overset{J}{C}{{}^{(\pm)}}\triangleright(x\triangleright\psi)=\big{(}\overset{J}{C}{{}^{(\pm)}}x\big{)}\triangleright\psi=\big{(}x\overset{J}{C}{{}^{(\pm)}}\big{)}\triangleright\psi=x\triangleright\big{(}\overset{J}{C}{{}^{(\pm)}}\triangleright\psi\big{)}=(x\triangleright\psi)\mathbb{I}_{\dim(J)}.$

Then $x\triangleright\psi\in\operatorname{SLF}(H)$, as desired. ∎

######  Remark 4.12.

Let
$\overset{I}{C}=\overset{I}{v^{2}}\overset{I}{B}\overset{I}{A}{{}^{-1}}\overset{I}{B}{{}^{-1}}\overset{I}{A}$.
It can be shown that $\overset{I}{C}$ satisfies the fusion relation of
$\mathcal{L}_{0,1}(H)$, that
$\overset{I}{C}=\overset{I}{C}{{}^{(+)}}\overset{I}{C}{{}^{(-)-1}}$, that
$x\in\mathcal{L}_{1,0}(H)$ is invariant if, and only if,
$x\overset{I}{C}=\overset{I}{C}x$ and that $\psi\in H^{*}$ is symmetric if,
and only if, $\overset{I}{C}\triangleright\psi=\psi\mathbb{I}_{\dim(I)}$. The
details in our general setting are given in [22] for arbitrary genus. Observe
that geometrically, $\overset{I}{C}$ is the boundary of the surface
$\Sigma_{1,0}{\setminus}D$, see Fig. 1.

We now need to determine explicit formulas for the representation of
particular types of invariants that will appear in the proof of the modular
identities in Section 5. If $\psi\in H^{*}$ and $a\in H$, we define

$\psi^{a}=\psi(a?),$

where $\psi(a?)\colon x\mapsto\psi(ax)$. This defines a right representation
of $H$ on $H^{*}$. Obviously, if $z\in\mathcal{Z}(H)$ and
$\psi\in\operatorname{SLF}(H)$ then $\psi^{z}\in\operatorname{SLF}(H)$.

Recall that $z_{A}=j_{A}(z)$ (resp. $z_{B}=j_{B}(z)$) is the image of
$z\in\mathcal{L}_{0,1}(H)$ by the map $j_{A}(\overset{I}{M})=\overset{I}{A}$
(resp. $j_{B}(\overset{I}{M})=\overset{I}{B}$). See (4.4) for the general
definition.

######  Proposition 4.13.

Let $z\in\mathcal{L}_{0,1}^{\mathrm{inv}}(H)=\mathcal{Z}(H)$ and let
$\psi\in\mathrm{SLF}(H)$. Then

$z_{A}\triangleright\psi=\psi^{z}\qquad\text{and}\qquad
z_{B}\triangleright\psi=\big{(}\mathcal{D}^{-1}(z)\psi^{v}\big{)}^{v^{-1}},$

where $\mathcal{D}$ is the isomorphism defined in (3.9).

###### Proof.

The first relation is obvious. For the second formula, we write
$z_{B}=\sum_{I}\mathrm{tr}\big{(}\Lambda_{I}\overset{I}{g}\overset{I}{B}\big{)}$
with
$\mathcal{D}^{-1}(z)=\sum_{I}\mathrm{tr}\big{(}\Lambda_{I}\overset{I}{T}\big{)}\in\operatorname{SLF}(H)$
by Lemma 3.9. We also write
$\psi=\sum_{J}\mathrm{tr}\big{(}\Theta_{J}\overset{J}{T}\big{)}$. Then, thanks
to (4.8):

$\displaystyle z_{B}\triangleright\psi$
$\displaystyle=\sum_{I,J}\mathrm{tr}_{12}\big{(}(\Lambda_{I})_{1}(\Theta_{J})_{2}\overset{I}{g}_{1}\overset{I}{B}_{1}\triangleright\overset{J}{T}_{2}\big{)}$
$\displaystyle=\sum_{I,J}\mathrm{tr}_{12}\big{(}(\Lambda_{I})_{1}(\Theta_{J})_{2}\overset{I}{g}_{1}\overset{I}{(a_{i}a_{j})}_{1}\overset{I\otimes
J}{T}\\!\\!\\!_{12}\overset{I}{(b_{j})}_{1}\overset{J}{(b_{i})}_{2}\overset{IJ}{(R^{\prime})}_{12}\big{)}$
$\displaystyle=\mathcal{D}^{-1}(z)(ga_{i}a_{j}?b_{j}b_{k})\psi(?b_{i}a_{k})=\mathcal{D}^{-1}(z)\big{(}?b_{j}b_{k}S^{2}(a_{i}a_{j})g\big{)}\psi(?b_{i}a_{k})$

with $\mathrm{tr}_{12}=\mathrm{tr}\otimes\mathrm{tr}$, $R=a_{i}\otimes b_{i}$.
Thanks to the Yang–Baxter equation, we have

$b_{j}b_{k}\otimes a_{i}a_{j}\otimes
b_{i}a_{k}=R_{23}R_{21}R_{31}=R_{31}R_{21}R_{23}=b_{i}b_{j}\otimes
a_{j}a_{k}\otimes a_{i}b_{k}.$

It follows that

$\displaystyle z_{B}\triangleright\psi$
$\displaystyle=\mathcal{D}^{-1}(z)\big{(}?b_{i}b_{j}S^{2}(a_{j}a_{k})g\big{)}\psi(?a_{i}b_{k})$
$\displaystyle=\mathcal{D}^{-1}(z)\big{(}?v^{-1}b_{i}a_{k}\big{)}\psi(?a_{i}b_{k})=\mathcal{D}^{-1}(z)\big{(}?\big{(}v^{-1}\big{)}^{\prime}\big{)}\psi\big{(}?v\big{(}v^{-1}\big{)}^{\prime\prime}\big{)},$

where we used (2.6), (2.9) and (2.2). Hence for $x\in H$:

$\displaystyle(z_{B}\triangleright\psi)(x)=\mathcal{D}^{-1}(z)\big{(}\big{(}v^{-1}\big{)}^{\prime}x^{\prime}\big{)}\psi\big{(}v\big{(}v^{-1}\big{)}^{\prime\prime}x^{\prime\prime}\big{)}=\big{(}\mathcal{D}^{-1}(z)\psi^{v}\big{)}\big{(}v^{-1}x\big{)}=\big{(}\mathcal{D}^{-1}(z)\psi^{v}\big{)}^{v^{-1}}(x)$

as desired. ∎

######  Lemma 4.14.

Let $z\in\mathcal{L}_{0,1}^{\mathrm{inv}}(H)=\mathcal{Z}(H)$ and let
$\psi\in\operatorname{SLF}(H)$. Then

$z_{B^{-1}}\triangleright\psi=\big{(}S\big{(}\mathcal{D}^{-1}(z)\big{)}\psi^{v}\big{)}^{v^{-1}}.$

It follows that if $S(\psi)=\psi$ for all $\psi\in\operatorname{SLF}(H)$, then
$\rho_{\mathrm{SLF}}(z_{B^{-1}})=\rho_{\mathrm{SLF}}(z_{B})$.

###### Proof.

This proof is quite similar to that of the previous proposition. Due to the
fact that
$\Psi_{1,0}\big{(}\overset{I}{B}\\!^{-1}\big{)}=\overset{I}{L}\\!^{(-)}S(\overset{I}{T})\overset{I}{L}\\!^{(+)-1}$
together with Lemma 4.5 and formulas (2.11), (2.2) and (2.1), it is not too
difficult to show that

$\overset{I}{B}\\!^{-1}_{1}\triangleright\overset{J}{T}\\!_{2}={\vphantom{\big{(}\overset{I^{*}}{(a_{i})}_{1}\overset{I^{*}\otimes
J}{T}\\!\\!\\!\\!_{12}\overset{I^{*}}{\big{(}a_{j}S^{-2}(b_{j}b_{k})\big{)}}_{1}\overset{J}{(a_{k}b_{i})}_{2}\big{)}}}^{t\otimes\mathrm{id}\\!\\!}\big{(}\overset{I^{*}}{(a_{i})}_{1}\overset{I^{*}\otimes
J}{T}\\!\\!\\!\\!_{12}\overset{I^{*}}{\big{(}a_{j}S^{-2}(b_{j}b_{k})\big{)}}_{1}\overset{J}{(a_{k}b_{i})}_{2}\big{)},$

where t⊗id means transpose on the first tensorand. Write
$z_{B^{-1}}=\sum_{I}\mathrm{tr}\big{(}\Lambda_{I}\overset{I}{g}\overset{I}{B}\\!^{-1}\big{)}$
with
$\mathcal{D}^{-1}(z)=\sum_{I}\mathrm{tr}\big{(}\Lambda_{I}\overset{I}{T}\big{)}\in\operatorname{SLF}(H)$
by Lemma 3.9, and
$\psi=\sum_{J}\mathrm{tr}\big{(}\Theta_{J}\overset{J}{T}\big{)}$. Thanks to
(2.11), observe that

$S\big{(}\mathcal{D}^{-1}(z)\big{)}=\sum_{I}\mathrm{tr}\big{(}\Lambda_{I}S(\overset{I}{T})\big{)}=\sum_{I}\mathrm{tr}\big{(}{\vphantom{\Lambda_{I}}}^{t}\Lambda_{I}\overset{I^{*}}{T}\big{)}.$

Using the fact that $S(g)=g^{-1}$ and (2.11), we thus get

$\displaystyle z_{B^{-1}}\triangleright\psi$
$\displaystyle=\sum_{I,J}\mathrm{tr}_{12}\big{(}(\Lambda_{I}\overset{I}{g})_{1}(\Theta_{J})_{2}{\vphantom{\big{(}\overset{I^{*}}{(a_{i})}_{1}\overset{I^{*}\otimes
J}{T}\\!\\!\\!\\!_{12}\overset{I^{*}}{(a_{j}S^{-2}(b_{j}b_{k}))}_{1}\overset{J}{(a_{k}b_{i})}_{2}\big{)}}}^{t\otimes\mathrm{id}\\!\\!}\big{(}\overset{I^{*}}{(a_{i})}_{1}\overset{I^{*}\otimes
J}{T}\\!\\!\\!\\!_{12}\overset{I^{*}}{(a_{j}S^{-2}(b_{j}b_{k}))}_{1}\overset{J}{(a_{k}b_{i})}_{2}\big{)}\big{)}$
$\displaystyle=\sum_{I,J}\mathrm{tr}_{12}\big{(}({\vphantom{\Lambda_{I}}}^{t}\Lambda_{I})_{1}(\Theta_{J})_{2}\overset{I^{*}}{(a_{i})}_{1}\overset{I^{*}\otimes
J}{T}\\!\\!\\!\\!_{12}\overset{I^{*}}{(a_{j}S^{-2}(b_{j}b_{k})g^{-1})}_{1}\overset{J}{(a_{k}b_{i})}_{2}\big{)}$
$\displaystyle=S\big{(}\mathcal{D}^{-1}(z)\big{)}\big{(}a_{i}?a_{j}S^{-2}(b_{j}b_{k})g^{-1}\big{)}\psi(?a_{k}b_{i})$
$\displaystyle=S\big{(}\mathcal{D}^{-1}(z)\big{)}\big{(}?a_{j}S^{-2}(b_{j}b_{k})g^{-1}a_{i}\big{)}\psi(?a_{k}b_{i})=S\big{(}\mathcal{D}^{-1}(z)\big{)}\big{(}?\big{(}v^{-1}\big{)}^{\prime}\big{)}\psi\big{(}?v\big{(}v^{-1}\big{)}^{\prime\prime}\big{)}.$

For the last equality we used (2.6), (2.9) and (2.2). Hence we get as in the
previous proof
$z_{B^{-1}}\triangleright\psi=\big{(}S\big{(}\mathcal{D}^{-1}(z)\big{)}\psi^{v}\big{)}^{v^{-1}}$.
∎

##  5 Projective representation of $\boldsymbol{\mathrm{SL}_{2}(\mathbb{Z})}$

As previously, $H$ is a finite-dimensional factorizable ribbon Hopf algebra.

###  5.1 Mapping class group of the torus

Let $\Sigma_{g,n}$ be a compact oriented surface of genus $g$ with $n$ open
disks removed. Recall that the mapping class group
$\mathrm{MCG}(\Sigma_{g,n})$ is the group of all isotopy classes of
orientation-preserving homeomorphisms which fix the boundary pointwise, see
[23].

Let us put the base point of $\pi_{1}(\Sigma_{g,n}{\setminus}D)$ on the
boundary circle $c$. Since $c$ is pointwise fixed, we can consider the action
of $\mathrm{MCG}(\Sigma_{g,n}{\setminus}D)$ on
$\pi_{1}(\Sigma_{g,n}{\setminus}D)$, obviously defined by

$\forall\,[f]\in\mathrm{MCG}(\Sigma_{g,n}),\qquad\forall\,[\gamma]\in\pi_{1}(\Sigma_{g,n}{\setminus}D),\qquad[f]\cdot[\gamma]=f_{*}([\gamma])=[f\circ\gamma].$

Until now, we identify $f$ with its isotopy class $[f]$ and $\gamma$ with its
homotopy class $[\gamma]$.

Here we focus on the torus $\Sigma_{1,0}=S^{1}\times S^{1}$. Consider
$\Sigma_{1,0}{\setminus}D$, where $D$ is an embedded open disk. The surface
$\Sigma_{1,0}{\setminus}D$ is represented as a ribbon graph in Fig. 1 together
with the canonical curves $a$ and $b$. The groups
$\mathrm{MCG}(\Sigma_{1,0}{\setminus}D)$ and $\mathrm{MCG}(\Sigma_{1,0})$ are
generated by the Dehn twists $\tau_{a}$, $\tau_{b}$ about the free homotopy
class of the curves $a$ and $b$. It is well-known (see [23]) that

$\mathrm{MCG}(\Sigma_{1,0})=\mathrm{SL}_{2}(\mathbb{Z})=\big{\langle}\tau_{a},\tau_{b}\,|\,\tau_{a}\tau_{b}\tau_{a}=\tau_{b}\tau_{a}\tau_{b},\,(\tau_{a}\tau_{b})^{6}=1\big{\rangle}.$

This presentation is not the usual one of $\mathrm{SL}_{2}(\mathbb{Z})$, which
is

$\mathrm{SL}_{2}(\mathbb{Z})=\big{\langle}s,t\,|\,(st)^{3}=s^{2},\,s^{4}=1\big{\rangle}.$

The link between the two presentations is
$s=\tau_{a}^{-1}\tau_{b}^{-1}\tau_{a}^{-1}$, $t=\tau_{a}$.

The action of the Dehn twists $\tau_{a}$ and $\tau_{b}$ on
$\pi_{1}(\Sigma_{1,0}{\setminus}D)$ is given by

$\displaystyle(\tau_{a})_{*}(a)=a,\qquad(\tau_{a})_{*}(b)=ba\qquad\text{and}\qquad(\tau_{b})_{*}(a)=b^{-1}a,\qquad(\tau_{b})_{*}(b)=b.$
(5.1)

For instance, the action of $\tau_{a}$ on $b$ is depicted by

$b$ |  | $\>ba$
---|---|---

###  5.2 Automorphisms $\boldsymbol{\alpha}$ and $\boldsymbol{\beta}$

The fundamental idea, proposed in [5] and [6], is to mimic the action of the
Dehn twists of $\Sigma_{g,n}{\setminus}D$ on
$\pi_{1}(\Sigma_{g,n}{\setminus}D)$ at the level of the algebra
$\mathcal{L}_{g,n}(H)$. Let us be more precise. We focus on the case
$(g,n)=(1,0)$.

In $\pi_{1}(\Sigma_{1,0}{\setminus}D)$ we have the two canonical curves $a$
and $b$, while in $\mathcal{L}_{1,0}(H)$ we have the matrices $\overset{I}{A}$
and $\overset{I}{B}$. In view of (5.1), let us try to define two morphisms
$\widetilde{\tau}_{a},\widetilde{\tau}_{b}\colon\mathcal{L}_{1,0}(H)\to\mathcal{L}_{1,0}(H)$
by the same formulas

$\displaystyle\widetilde{\tau}_{a}\big{(}\overset{I}{A}\big{)}=\overset{I}{A},\qquad$
$\displaystyle\widetilde{\tau}_{a}\big{(}\overset{I}{B}\big{)}=\overset{I}{B}\overset{I}{A},$
$\displaystyle\widetilde{\tau}_{b}\big{(}\overset{I}{A}\big{)}=\overset{I}{B}\\!^{-1}\overset{I}{A},\qquad$
$\displaystyle\widetilde{\tau}_{b}\big{(}\overset{I}{B}\big{)}=\overset{I}{B}.$

Let us see the behavior of these mappings under the fusion and exchange
relations. For the exchange relation, no problem arises

$\displaystyle
R_{12}\widetilde{\tau}_{a}(B)_{1}R_{21}\widetilde{\tau}_{a}(A)_{2}$
$\displaystyle=R_{12}B_{1}A_{1}R_{21}A_{2}$ (definition)
$\displaystyle=R_{12}B_{1}R_{21}A_{2}R_{12}A_{1}R_{12}^{-1}$ (equation (3.3))
$\displaystyle=A_{2}R_{12}B_{1}A_{1}R_{12}^{-1}$ (equation(4.1))
$\displaystyle=\widetilde{\tau}_{a}(A)_{2}R_{12}\widetilde{\tau}_{a}(B)_{1}R_{12}^{-1}$
$\displaystyle\quad\text{(definition)},$

and a similar computation holds for $\widetilde{\tau}_{b}$. The fusion
relation is almost satisfied

$\displaystyle\widetilde{\tau}_{a}(B)_{12}$ $\displaystyle=B_{12}A_{12}$
(definition)
$\displaystyle=\Delta(v)_{12}B_{12}v_{1}^{-1}v_{2}^{-1}R_{21}R_{12}A_{12}$
(trick based on (2.2))
$\displaystyle=\Delta(v)_{12}B_{1}R_{21}B_{2}R_{21}^{-1}v_{1}^{-1}v_{2}^{-1}R_{21}R_{12}A_{1}R_{21}A_{2}R_{21}^{-1}$
(equation (3.1))
$\displaystyle=\Delta(v)_{12}v_{1}^{-1}v_{2}^{-1}B_{1}R_{21}B_{2}R_{12}A_{1}R_{21}A_{2}R_{21}^{-1}$
($v$ is central)
$\displaystyle=\Delta(v)_{12}v_{1}^{-1}v_{2}^{-1}B_{1}A_{1}R_{21}B_{2}A_{2}R_{21}^{-1}$
(equation (4.1))
$\displaystyle=\Delta(v)_{12}v_{1}^{-1}v_{2}^{-1}\widetilde{\tau}_{a}(B)_{1}R_{21}\widetilde{\tau}_{a}(B)_{2}R_{21}^{-1}$
$\displaystyle\quad\text{(definition)},$

and we get similarly

$\widetilde{\tau}_{b}(A)_{12}=B^{-1}_{12}A_{12}=\Delta\big{(}v^{-1}\big{)}_{12}v_{1}v_{2}\widetilde{\tau}_{b}(A)_{1}R_{21}\widetilde{\tau}_{b}(A)_{2}R_{21}^{-1}.$

From this we conclude that the elements
$\overset{I}{v}\\!^{-1}\overset{I}{B}\overset{I}{A}$ and
$\overset{I}{v}\overset{I}{B}\\!^{-1}\overset{I}{A}$ satisfy the relation
(3.1). Since $v$ is central, we see that the exchange relation still holds
with these elements. We thus have found the morphisms which mimic $\tau_{a}$
and $\tau_{b}$. We denote them by $\alpha$ and $\beta$ respectively, and we
have the following proposition (the maps $\alpha,\beta$ and the fact that they
are automorphisms were already in [5, Lemma 2] and [6, equations (4.1) and
(4.2)]).

######  Proposition 5.1.

We have two automorphisms $\alpha$, $\beta$ of $\mathcal{L}_{1,0}(H)$ defined
by

$\displaystyle\alpha(\overset{I}{A})=\overset{I}{A},\qquad$
$\displaystyle\alpha(\overset{I}{B})=\overset{I}{v}\\!^{-1}\overset{I}{B}\overset{I}{A},$
$\displaystyle\beta(\overset{I}{A})=\overset{I}{v}\overset{I}{B}\\!^{-1}\overset{I}{A},\qquad$
$\displaystyle\beta(\overset{I}{B})=\overset{I}{B}.$

Moreover, these automorphisms are inner: there exist
$\widehat{\alpha},\widehat{\beta}\in\mathcal{L}_{1,0}(H)$ unique up to scalar
such that

$\forall\,x\in\mathcal{L}_{1,0}(H),\qquad\alpha(x)=\widehat{\alpha}x\widehat{\alpha}^{-1},\qquad\beta(x)=\widehat{\beta}x\widehat{\beta}^{-1}.$

###### Proof.

By Theorem 4.8, $\mathcal{L}_{1,0}(H)$ is a matrix algebra. Hence, by the
Skolem–Noether theorem, every automorphism of $\mathcal{L}_{1,0}(H)$ is inner.
∎

A natural question is then to find explicitly the elements $\widehat{\alpha}$,
$\widehat{\beta}$. The answer is amazingly simple (it has been provided in [5]
and [38] for the semisimple case). Recall the notation (4.4).

######  Theorem 5.2.

Up to scalar, $\widehat{\alpha}=v_{A}^{-1}$ and $\widehat{\beta}=v_{B}^{-1}$.

###### Proof.

We must show that

$v^{-1}_{A}\overset{I}{A}=\overset{I}{A}v^{-1}_{A},\qquad
v^{-1}_{A}\overset{I}{B}=\overset{I}{v}\\!^{-1}\overset{I}{B}\overset{I}{A}v^{-1}_{A}\qquad\text{and}\qquad
v^{-1}_{B}\overset{I}{A}=\overset{I}{v}\overset{I}{B}\\!^{-1}\overset{I}{A}v^{-1}_{B},\qquad
v^{-1}_{B}\overset{I}{B}=\overset{I}{B}v^{-1}_{B}.$

It is obvious that $v_{A}^{-1}$ (resp. $v_{B}^{-1}$) commutes with the
matrices $\overset{I}{A}$ (resp. $\overset{I}{B}$) since it is central in
$j_{A}(\mathcal{L}_{0,1}(H))$ (resp. in $j_{B}(\mathcal{L}_{0,1}(H))$). Let us
show the other commutation relation for $v_{A}^{-1}$. We use the isomorphism
$\Psi_{1,0}$. Observe that $\Psi_{1,0}(x_{A})=x$ for all $x\in H$. Hence,
using the exchange relation of Definition 4.4 and (2.2), we have

$\displaystyle\Psi_{1,0}\big{(}v_{A}^{-1}\overset{I}{B}\big{)}$
$\displaystyle=\overset{I}{L}\\!^{(+)}v^{-1}\overset{I}{T}\overset{I}{L}\\!^{(-)-1}=\overset{I}{L}\\!^{(+)}\overset{I}{T}\big{(}?v^{\prime-1}\big{)}v^{\prime\prime-1}\overset{I}{L}\\!^{(-)-1}=\overset{I}{L}\\!^{(+)}\overset{I}{T}\overset{I}{(v^{\prime})}{{}^{-1}}v^{\prime\prime-1}\overset{I}{L}\\!^{(-)-1}$
$\displaystyle=\overset{I}{L}\\!^{(+)}\overset{I}{T}\overset{I}{v}{{}^{-1}}\overset{I}{b_{i}}\overset{I}{a_{j}}v^{-1}a_{i}b_{j}\overset{I}{L}\\!^{(-)-1}=\overset{I}{v}{{}^{-1}}\overset{I}{L}\\!^{(+)}\overset{I}{T}\overset{I}{b_{i}}a_{i}\overset{I}{a_{j}}b_{j}\overset{I}{L}\\!^{(-)-1}v^{-1}$
$\displaystyle=\overset{I}{v}{{}^{-1}}\overset{I}{L}\\!^{(+)}\overset{I}{T}\overset{I}{L}\\!^{(-)-1}\overset{I}{L}\\!^{(+)}\overset{I}{L}\\!^{(-)-1}v^{-1}=\Psi_{1,0}\big{(}\overset{I}{v}\\!^{-1}\overset{I}{B}\overset{I}{A}v^{-1}_{A}\big{)}$

as desired. We now apply the morphism $\alpha^{-1}\circ\beta^{-1}$ to the
equality
$v_{A}^{-1}\overset{I}{B}=\overset{I}{v}\\!^{-1}\overset{I}{B}\overset{I}{A}v_{A}^{-1}$:

$\alpha^{-1}\circ\beta^{-1}\big{(}v_{A}^{-1}\overset{I}{B}\big{)}=\overset{I}{v}v_{B}^{-1}\overset{I}{B}\overset{I}{A}{{}^{-1}}=\alpha^{-1}\circ\beta^{-1}\big{(}\overset{I}{v}^{-1}\overset{I}{B}\overset{I}{A}v_{A}^{-1}\big{)}=\overset{I}{B}\overset{I}{A}{{}^{-1}}\overset{I}{B}v_{B}^{-1}.$

Using that $v_{B}$ and $\overset{I}{B}$ commute, we easily get the desired
equality. ∎

###  5.3 Projective representation of
$\boldsymbol{\mathrm{SL}_{2}(\mathbb{Z})}$ on
$\boldsymbol{\operatorname{SLF}(H)}$

We now show that the elements $v_{A}^{-1}$, $v_{B}^{-1}$ give rise to a
projective representation of $\mathrm{MCG}(\Sigma_{1,0})$ on
$\operatorname{SLF}(H)$ via the following assignment

$\tau_{a}\mapsto\rho_{\mathrm{SLF}}\big{(}v_{A}^{-1}\big{)},\qquad\tau_{b}\mapsto\rho_{\mathrm{SLF}}\big{(}v_{B}^{-1}\big{)}.$

We must then check that

$\rho_{\mathrm{SLF}}\big{(}v_{A}^{-1}v_{B}^{-1}v_{A}^{-1}\big{)}\sim\rho_{\mathrm{SLF}}\big{(}v_{B}^{-1}v_{A}^{-1}v_{B}^{-1}\big{)},\qquad\rho_{\mathrm{SLF}}\big{(}v_{A}^{-1}v_{B}^{-1}\big{)}^{6}\sim
1,$

where $\sim$ means equality up to scalar. As we will see, it turns out that
the braid relation holds in the algebra $\mathcal{L}_{1,0}(H)$ itself (the
scalar being $1$), while the relation
$\big{(}v_{A}^{-1}v_{B}^{-1}\big{)}^{6}\sim 1$ only holds in the
representation $\operatorname{SLF}(H)$. This is because the relation
$(\tau_{a})_{*}(\tau_{b})_{*}(\tau_{a})_{*}=(\tau_{b})_{*}(\tau_{a})_{*}(\tau_{b})_{*}$
holds on $\pi_{1}(\Sigma_{1,0}{\setminus}D)$, while the relation
$((\tau_{a})_{*}(\tau_{b})_{*})^{6}=1$ only holds on $\pi_{1}(\Sigma_{1,0})$.
As pointed out in Remark 4.12, the matrices $\overset{I}{C}$ corresponding to
the boundary circle vanishes on $\mathrm{SLF}(H)$. Thus applying
$\rho_{\mathrm{SLF}}$ amounts to gluing back the disk $D$. See [22] for the
generalization of this fact to higher genus.

Integrals on $H$ will play an important role. Recall that a left integral
(resp. right integral) is a non-zero linear form $\mu^{l}$ (resp. $\mu^{r}$)
on $H$ which satisfies

$\displaystyle\forall\,x\in
H,\qquad\big{(}\mathrm{id}\otimes\mu^{l}\big{)}\circ\Delta(x)=\mu^{l}(x)1\qquad\big{(}\text{resp.}\
\big{(}\mu^{r}\otimes\mathrm{id}\big{)}\circ\Delta(x)=\mu^{r}(x)1\big{)}.$
(5.2)

Since $H$ is finite-dimensional, this is equivalent to:

$\displaystyle\forall\,\psi\in\mathcal{O}(H),\qquad\psi\mu^{l}=\varepsilon(\psi)\mu^{l}\qquad\big{(}\text{resp.}\
\mu^{r}\psi=\varepsilon(\psi)\mu^{r}\big{)}.$ (5.3)

It is well-known that left and right integrals always exist if $H$ is finite-
dimensional. Moreover, they are unique up to scalar. We fix $\mu^{l}$. Then
$\mu^{l}\circ S^{-1}$ is a right integral, and we choose

$\displaystyle\mu^{r}=\mu^{l}\circ S^{-1}.$ (5.4)

The following proposition is important for the sequel (points 2 and 3 are
well-known and can be deduced from results of Radford, see [36]).

######  Proposition 5.3.

Let $\varphi_{v},\varphi_{v^{-1}}\in H^{*}$ defined by

$\varphi_{v}=\mu^{l}\big{(}v^{-1}\big{)}^{-1}\mu^{l}\big{(}g^{-1}v^{-1}?\big{)},\qquad\varphi_{v^{-1}}=\mu^{l}(v)^{-1}\mu^{l}\big{(}g^{-1}v?\big{)}.$

Then

$\mathcal{D}(\varphi_{v})=v,\qquad\mathcal{D}\big{(}\varphi_{v^{-1}}\big{)}=v^{-1}.$

It follows that

  1. $1)$

$\varphi_{v},\varphi_{v^{-1}}\in\operatorname{SLF}(H)$,

  2. $2)$

$\mu^{l}\big{(}g^{-1}?\big{)},\mu^{r}(g?)\in\operatorname{SLF}(H)$,

  3. $3)$

$\forall\,x,y\in H$, $\mu^{l}(xy)=\mu^{l}\big{(}yS^{2}(x)\big{)}$,
$\mu^{r}(xy)=\mu^{r}\big{(}S^{2}(y)x\big{)}$.

###### Proof.

Consider the following computation, where we use (2.2) and (5.2):

$\displaystyle\mathcal{D}\big{(}\mu^{l}\big{(}g^{-1}v^{-1}?\big{)}\big{)}$
$\displaystyle=\big{\langle}\mu^{l}\big{(}g^{-1}v^{-1}?\big{)}\otimes\mathrm{id},(g\otimes
1)RR^{\prime}\big{\rangle}$
$\displaystyle=\big{\langle}\mu^{l}\big{(}g^{-1}v^{-1}?\big{)}\otimes\mathrm{id},gv\big{(}v^{-1}\big{)}^{\prime\prime}\otimes
v\big{(}v^{-1}\big{)}^{\prime}\big{\rangle}$
$\displaystyle=\mu^{l}\big{(}\big{(}v^{-1}\big{)}^{\prime\prime}\big{)}v\big{(}v^{-1}\big{)}^{\prime}=\mu^{l}\big{(}v^{-1}\big{)}v.$

Since $H$ is factorizable, the map $\mathcal{D}$ is an isomorphism of vector
spaces. The left integral $\mu^{l}$ is non-zero, so
$\mu^{l}\big{(}g^{-1}v^{-1}?\big{)}$ is non-zero either. Since $\mathcal{D}$
is an isomorphism, it follows that
$\mathcal{D}\big{(}\mu^{l}\big{(}g^{-1}v^{-1}?\big{)}\big{)}=\mu^{l}\big{(}v^{-1}\big{)}v\neq
0$, and thus $\mu^{l}\big{(}v^{-1}\big{)}\neq 0$. Hence the formula for
$\varphi_{v}$ is well defined. Moreover, we have the restriction
$\mathcal{D}\colon\operatorname{SLF}(H)\overset{\sim}{\rightarrow}\mathcal{Z}(H)$,
so since $v\in\mathcal{Z}(H)$, we get that
$\varphi_{v}\in\operatorname{SLF}(H)$. This allows us to deduce the properties
stated about $\mu^{l}$. Using (5.4), we obtain the properties 1), 2) and 3)
for $\mu^{r}$. We can now proceed with the computation for $\varphi_{v^{-1}}$:

$\displaystyle\mathcal{D}\big{(}\mu^{l}\big{(}g^{-1}v?\big{)}\big{)}$
$\displaystyle=\big{\langle}\mu^{l}\big{(}g^{-1}v?\big{)}\otimes\mathrm{id},(g\otimes
1)RR^{\prime}\big{\rangle}=\mu^{l}(va_{i}b_{j})b_{i}a_{j}$
$\displaystyle=\mu^{r}(vS(b_{j})S(a_{i}))b_{i}a_{j}=\mu^{r}\big{(}vS(a_{i})S^{-1}(b_{j})\big{)}b_{i}a_{j}$
$\displaystyle=\big{\langle}\mu^{r}\otimes\mathrm{id},(v\otimes
1)(R^{\prime}R)^{-1}\big{\rangle}=\mu^{r}(v^{\prime})v^{\prime\prime}v^{-1}=\mu^{r}(v)v^{-1}=\mu^{l}(v)v^{-1},$

where we used (5.4), the property 3) previously shown and (2.2). We conclude
as before. ∎

Since $\mathcal{D}$ is an isomorphism of algebras, we have
$\varphi_{v^{-1}}^{-1}=\varphi_{v}$, and thus

$\displaystyle\varphi_{v^{-1}}\varphi_{v^{-1}}^{v^{-2}}=\frac{\mu^{l}\big{(}v^{-1}\big{)}}{\mu^{l}(v)}\varepsilon.$
(5.5)

By Proposition 4.13, the actions of $v_{A}^{-1}$ and $v_{B}^{-1}$ on
$\operatorname{SLF}(H)$ are

$\displaystyle\forall\,\psi\in\operatorname{SLF}(H),\qquad
v_{A}^{-1}\triangleright\psi=\psi^{v^{-1}}=\psi\big{(}v^{-1}?\big{)}\qquad\text{and}\qquad
v_{B}^{-1}\triangleright\psi=\big{(}\varphi_{v^{-1}}\psi^{v}\big{)}^{v^{-1}}.$
(5.6)

######  Lemma 5.4.

$\varphi_{v^{-1}}\varphi_{v^{-1}}^{v^{-1}}=\varphi_{v^{-1}}^{v^{-1}}$.

###### Proof.

For $x\in H$

$\displaystyle\big{\langle}\varphi_{v^{-1}}\varphi_{v^{-1}}^{v^{-1}},x\big{\rangle}$
$\displaystyle=\mu^{l}(v)^{-2}\mu^{l}\big{(}vg^{-1}x^{\prime}\big{)}\mu^{l}\big{(}g^{-1}x^{\prime\prime}\big{)}=\mu^{l}(v)^{-2}\big{\langle}\mu^{l}(v?)\mu^{l},g^{-1}x\big{\rangle}$
$\displaystyle=\mu^{l}(v)^{-1}\mu^{l}(g^{-1}x)=\varphi_{v^{-1}}^{v^{-1}}(x).$

We simply used (5.3). ∎

This lemma has an important consequence.

######  Proposition 5.5.

The following braid relation holds in $\mathcal{L}_{1,0}(H)$:

$v_{A}^{-1}v_{B}^{-1}v_{A}^{-1}=v_{B}^{-1}v_{A}^{-1}v_{B}^{-1}.$

###### Proof.

The morphisms $\alpha$ and $\beta$ satisfy the braid relation
$\alpha\beta\alpha=\beta\alpha\beta$. Hence by Theorem 5.2 and since
$\mathcal{Z}(\mathcal{L}_{1,0}(H))\cong\mathbb{C}$, we have $\lambda
v_{A}^{-1}v_{B}^{-1}v_{A}^{-1}=v_{B}^{-1}v_{A}^{-1}v_{B}^{-1}$ for some
$\lambda\in\mathbb{C}$. We evaluate on the counit

$\displaystyle\lambda
v_{A}^{-1}v_{B}^{-1}v_{A}^{-1}\triangleright\varepsilon=\lambda
v_{A}^{-1}v_{B}^{-1}\triangleright\varepsilon=\lambda
v_{A}^{-1}\triangleright\varphi_{v^{-1}}^{v^{-1}}=\lambda\varphi_{v^{-1}}^{v^{-2}},$
$\displaystyle
v_{B}^{-1}v_{A}^{-1}v_{B}^{-1}\triangleright\varepsilon=v_{B}^{-1}v_{A}^{-1}\triangleright\varphi_{v^{-1}}^{v^{-1}}=v_{B}^{-1}\triangleright\varphi_{v^{-1}}^{v^{-2}}=\big{(}\varphi_{v^{-1}}\varphi_{v^{-1}}^{v^{-1}}\big{)}^{v^{-1}}=\big{(}\varphi_{v^{-1}}^{v^{-1}}\big{)}^{v^{-1}}=\varphi_{v^{-1}}^{v^{-2}}.$

We used
$\varepsilon\big{(}v^{-1}?\big{)}=\varepsilon\big{(}v^{-1}\big{)}\varepsilon=\varepsilon$
and Lemma 5.4. It follows that $\lambda=1$. ∎

Observe that $(\alpha\beta)^{6}\neq\mathrm{id}$, thus the other relation of
$\mathrm{MCG}(\Sigma_{1,0})$ does not hold in $\mathcal{L}_{1,0}(H)$. In order
to show it in the representation, we begin with a technical lemma, in which we
use the notation of (4.4).

######  Lemma 5.6.

For all $z\in\mathcal{L}_{0,1}^{\mathrm{inv}}(H)=\mathcal{Z}(H)$, we have
$z_{v^{2}A^{-1}B^{-1}A}=z_{B^{-1}}$.

###### Proof.

Write as usual
$z=\sum_{I}\mathrm{tr}\big{(}\Lambda_{I}\overset{I}{g}\overset{I}{M}\big{)}$
(Lemma 3.9). We first show that $z_{v^{-1}AB^{-1}}=z_{vB^{-1}A}$ in the
Heisenberg double. We have

$\Psi_{1,0}\big{(}z_{v^{-1}AB^{-1}}\big{)}=\sum_{I}\mathrm{tr}\big{(}\Lambda_{I}\overset{I}{g}\overset{I}{v}\\!^{-1}\overset{I}{L}\\!^{(+)}\overset{I}{T}\\!^{-1}\overset{I}{L}\\!^{(+)-1}\big{)}=\sum_{I}\mathrm{tr}\big{(}\Lambda_{I}\overset{I}{g}\overset{I}{v}\\!^{-1}\overset{I}{a_{i}}b_{i}S(\overset{I}{T})\overset{I}{S(a_{j})}b_{j}\big{)}.$

Using the defining relation of $\mathcal{H}(\mathcal{O}(H))$ together with
(2.1) and (2.6), we get

$\overset{I}{a_{i}}b_{i}S\big{(}\overset{I}{T}\big{)}=\big{(}\overset{I}{a_{i}}\overset{I}{a_{k}}\overset{I}{S(b_{k})}\big{)}S\big{(}\overset{I}{T}\big{)}b_{i}=\big{(}\overset{I}{a_{i}}\overset{I}{g}{{}^{-1}}\overset{I}{v}\big{)}S\big{(}\overset{I}{T}\big{)}b_{i}.$

It follows that

$\displaystyle\Psi_{1,0}(z_{v^{-1}AB^{-1}})$
$\displaystyle=\sum_{I}\mathrm{tr}\big{(}\Lambda_{I}\overset{I}{S^{2}(a_{i})}S(\overset{I}{T})\overset{I}{S(a_{j})}b_{i}b_{j}\big{)}=\mathcal{D}^{-1}(z)\big{(}S^{2}(a_{i})S(?)S(a_{j})\big{)}b_{i}b_{j}$
$\displaystyle=S\big{(}\mathcal{D}^{-1}(z)\big{)}(S(a_{i})a_{j}?)b_{i}b_{j}=S\big{(}\mathcal{D}^{-1}(z)\big{)}\in\mathcal{O}(H)\otimes
1.$

A similar computation shows that
$\Psi_{1,0}z_{vB^{-1}A})=S\big{(}\mathcal{D}^{-1}(z)\big{)}$. Hence
$z_{v^{-1}AB^{-1}}=z_{vB^{-1}A}$. Applying the morphism $\alpha$ to this
equality, we find

$\alpha(z_{v^{-1}AB^{-1}})=z_{B^{-1}}=\alpha(z_{vB^{-1}A})=z_{v^{2}A^{-1}B^{-1}A}$

as desired. ∎

Consider $\widehat{\omega}=(v_{A}v_{B}v_{A})^{-1}\in\mathcal{L}_{1,0}(H)$,
which implements the automorphism $\omega=\alpha\beta\alpha$:
$\omega(x)=\widehat{\omega}x\widehat{\omega}^{-1}$. The key observation is the
following lemma.

######  Lemma 5.7.

For all $\psi\in\operatorname{SLF}(H)$:

$\widehat{\omega}^{2}\triangleright\psi=\frac{\mu^{l}\big{(}v^{-1}\big{)}}{\mu^{l}(v)}S(\psi).$

###### Proof.

First, we show the formula for $\psi=\varepsilon$:

$\displaystyle\widehat{\omega}^{2}\triangleright\varepsilon$
$\displaystyle=\big{(}v_{A}^{-1}v_{B}^{-1}v_{A}^{-1}\big{)}^{2}\triangleright\varepsilon=v_{A}^{-1}v_{B}^{-1}v_{A}^{-1}\triangleright\varphi_{v^{-1}}^{v^{-2}}=\bigl{(}\varphi_{v^{-1}}\varphi_{v^{-1}}^{v^{-2}}\bigr{)}^{v^{-2}}$
$\displaystyle=\frac{\mu^{l}\big{(}v^{-1}\big{)}}{\mu^{l}(v)}\varepsilon^{v^{-2}}=\frac{\mu^{l}\big{(}v^{-1}\big{)}}{\mu^{l}(v)}\varepsilon,$

where we used (5.6) and (5.5). Second, note that
$\omega(\overset{I}{A})=\overset{I}{v}{{}^{2}}\overset{I}{A}{{}^{-1}}\overset{I}{B}{{}^{-1}}\overset{I}{A}$
and $\omega(\overset{I}{B})=\overset{I}{A}$. Then, thanks to Lemma 5.6, it
follows that for every
$z\in\mathcal{Z}(H)=\mathcal{L}^{\mathrm{inv}}_{0,1}(H)$,

$\omega^{2}(z_{B})=\omega(z_{A})=z_{v^{2}A^{-1}B^{-1}A}=z_{B^{-1}}.$

Hence $\widehat{\omega}^{2}z_{B}=z_{B^{-1}}\widehat{\omega}^{2}$. Finally,
observe that by Proposition 4.13

$\forall\,\psi\in\operatorname{SLF}(H),\qquad\mathcal{D}\big{(}\psi^{v}\big{)}_{B}\triangleright\varepsilon=\psi.$

These three facts together with Lemma 4.14 yield

$\displaystyle\widehat{\omega}^{2}\triangleright\psi$
$\displaystyle=\widehat{\omega}^{2}\mathcal{D}\big{(}\psi^{v}\big{)}_{B}\triangleright\varepsilon=\mathcal{D}\big{(}\psi^{v}\big{)}_{B^{-1}}\widehat{\omega}^{2}\triangleright\varepsilon=\frac{\mu^{l}\big{(}v^{-1}\big{)}}{\mu^{l}(v)}\mathcal{D}\big{(}\psi^{v}\big{)}_{B^{-1}}\triangleright\varepsilon$
$\displaystyle=\frac{\mu^{l}\big{(}v^{-1}\big{)}}{\mu^{l}(v)}S\big{(}\psi^{v}\big{)}^{v^{-1}}=\frac{\mu^{l}\big{(}v^{-1}\big{)}}{\mu^{l}(v)}S(\psi)$

as desired. ∎

Recall that
$\mathrm{PSL}_{2}(\mathbb{Z})=\mathrm{SL}_{2}(\mathbb{Z})/\\{\pm\mathbb{I}_{2}\\}$
admits the following presentations

$\mathrm{PSL}_{2}(\mathbb{Z})=\big{\langle}\tau_{a},\tau_{b}\,|\,\tau_{a}\tau_{b}\tau_{a}=\tau_{b}\tau_{a}\tau_{b},\,(\tau_{a}\tau_{b})^{3}=1\big{\rangle}=\big{\langle}s,t\,|\,(st)^{3}=1,\,s^{2}=1\big{\rangle}.$

######  Theorem 5.8.

Recall that we assume that $H$ is a finite-dimensional factorizable ribbon
Hopf algebra. The following assignment defines a projective representation of
$\mathrm{MCG}(\Sigma_{1,0})=\mathrm{SL}_{2}(\mathbb{Z})$ on
$\operatorname{SLF}(H)$:

$\tau_{a}\mapsto\rho_{\mathrm{SLF}}\big{(}v_{A}^{-1}\big{)},\qquad\tau_{b}\mapsto\rho_{\mathrm{SLF}}\big{(}v_{B}^{-1}\big{)}.$

If moreover $S(\psi)=\psi$ for all $\psi\in\operatorname{SLF}(H)$, then this
defines actually a projective representation of
$\mathrm{PSL}_{2}(\mathbb{Z})$.

We denote this projective representation by $\theta_{\mathrm{SLF}}$.

###### Proof.

By Proposition 5.5, we know that the braid relation is satisfied in
$\mathcal{L}_{1,0}(H)$. By Lemma 5.7, we have

$(v_{A}^{-1}v_{B}^{-1})^{3}\triangleright\psi=\widehat{\omega}^{2}\triangleright\psi=\frac{\mu^{l}\big{(}v^{-1}\big{)}}{\mu^{l}(v)}S(\psi).$

If $S_{|\mathrm{SLF}(H)}=\mathrm{id}$, then

$\rho_{\mathrm{SLF}}\big{(}v_{A}^{-1}v_{B}^{-1}\big{)}^{3}=\frac{\mu^{l}\big{(}v^{-1}\big{)}}{\mu^{l}(v)}\mathrm{id}.$

Otherwise,

$\displaystyle\big{(}v_{A}^{-1}v_{B}^{-1}\big{)}^{6}\triangleright\psi$
$\displaystyle=\frac{\mu^{l}\big{(}v^{-1}\big{)}}{\mu^{l}(v)}\widehat{\omega}^{2}\triangleright
S(\psi)=\frac{\mu^{l}\big{(}v^{-1}\big{)}^{2}}{\mu^{l}(v)^{2}}S^{2}(\psi)$
$\displaystyle=\frac{\mu^{l}\big{(}v^{-1}\big{)}^{2}}{\mu^{l}(v)^{2}}\psi\big{(}g?g^{-1}\big{)}=\frac{\mu^{l}\big{(}v^{-1}\big{)}^{2}}{\mu^{l}(v)^{2}}\psi.$
∎

Observe that the quantity $\frac{\mu^{l}(v^{-1})}{\mu^{l}(v)}$ does not depend
on the choice up to scalar of $\mu^{l}$.

###  5.4 Equivalence with the Lyubashenko–Majid representation

Recall that $H$ is a finite-dimensional factorizable ribbon Hopf algebra.
Under this assumption, two operators $\mathcal{S},\mathcal{T}\colon H\to H$
are defined in [33]:

$\mathcal{S}(x)=\big{(}\mathrm{id}\otimes\mu^{l}\big{)}\big{(}R^{-1}(1\otimes
x)R^{\prime-1}\big{)},\qquad\mathcal{T}(x)=v^{-1}x.$

It is shown that they are invertible and satisfy
$(\mathcal{S}\mathcal{T})^{3}=\lambda\mathcal{S}^{2}$,
$\mathcal{S}^{2}=S^{-1}$, with $\lambda\in\mathbb{C}{\setminus}\\{0\\}$. We
warn the reader that in [33], they consider the inverse of the ribbon element
(see the bottom of the third page of their paper). That is why there is
$v^{-1}$ in the formula for $\mathcal{T}$.

Now we introduce two maps. The first is

$\begin{array}[]{crll}\boldsymbol{\chi}\colon&H^{*}&\rightarrow&H,\\\
&\beta&\mapsto&(\beta\otimes\mathrm{id})(R^{\prime}R),\end{array}$

while the second is

$\begin{array}[]{crll}\boldsymbol{\gamma}\colon&H&\rightarrow&H^{*},\\\
&x&\mapsto&\mu^{r}(S(x)?).\end{array}$

The map $\boldsymbol{\chi}$ is another variant of the map $\Psi$ of Section
2.2 and is called Drinfeld morphism in [24]. The map $\boldsymbol{\gamma}$ is
denoted $\widehat{\boldsymbol{\phi}}{{}^{-1}}$ in [24]. Under our assumptions,
the inverse of $\boldsymbol{\gamma}$ exists (this is a result of Radford
[36]), but we do not use it. Consider the space of left $q$-characters

$\mathrm{Ch}^{l}(H)=\big{\\{}\beta\in H^{*}\,|\,\forall\,x,y\in
H,\,\beta(xy)=\beta\big{(}S^{2}(y)x\big{)}\big{\\}}.$

These maps satisfy the following restrictions

$\boldsymbol{\chi}\colon\
\mathrm{Ch}^{l}(H)\longrightarrow\mathcal{Z}(H),\qquad\boldsymbol{\gamma}\colon\
\mathcal{Z}(H)\longrightarrow\mathrm{Ch}^{l}(H).$

This is due to the fact that they intertwine the adjoint and the coadjoint
actions (for the first the computation is analogous to that of the proof of
Proposition 3.5, while the second is immediate by Proposition 5.3).

It is not too difficult to show (see, e.g., [29, Remark IV.1.2]) that

$\forall\,z\in\mathcal{Z}(H),\qquad\mathcal{S}(z)=\boldsymbol{\chi}\circ\boldsymbol{\gamma}(z).$

It follows that $\mathcal{Z}(H)$ is stable under $\mathcal{S}$ and
$\mathcal{T}$. But since $S^{2}$ is inner, we have
$\mathcal{S}^{4}(z)=S^{-2}(z)=z$ for each $z\in\mathcal{Z}(H)$. Thus there
exists a projective representation $\theta_{\mathrm{LM}}$ of
$\mathrm{SL}_{2}(\mathbb{Z})$ on $\mathcal{Z}(H)$, defined by

$\theta_{\mathrm{LM}}(s)=\mathcal{S}_{|\mathcal{Z}(H)},\qquad\theta_{\mathrm{LM}}(t)=\mathcal{T}_{|\mathcal{Z}(H)}.$

We will need the following relation between left and right integrals.

######  Lemma 5.9.

Under our assumptions $H$ is unibalanced, which means that
$\mu^{l}=\mu^{r}\big{(}g^{2}?\big{)}$.

###### Proof.

The terminology “unibalanced” is picked from [11], where some facts about
integrals and cointegrals are recalled. Recall (see, e.g., [20, Proposition
8.10.10]) that a finite-dimensional factorizable Hopf algebra is unimodular,
which means that there exists $c\in H$, called two-sided cointegral, such that
$xc=cx=\varepsilon(x)c$ for all $x\in H$. Let $a\in H$ be the comodulus of
$\mu^{r}$: $\psi\mu^{r}=\psi(a)\mu^{r}$ for all $\psi\in\mathcal{O}(H)$ (see,
e.g., [11, equation (4.9)]). By a result of Drinfeld (see [35, Proposition
10.1.14], but be aware that in this book the notations and conventions for $a$
and $g$ are different from those we use), we know that

$uS(u)^{-1}=a(\mathfrak{a}\otimes\mathrm{id})(R),$

where $\mathfrak{a}\in H^{*}$ is the modulus of the left cointegral $c^{l}$ of
$H$. Here, since $c=c^{l}$ is two-sided, we have $\mathfrak{a}=\varepsilon$.
Thus $g^{2}=u^{2}v^{-2}=uS(u)^{-1}=a$ thanks to (2.2) and (2.9). We deduce
that

$\mu^{l}=\mu^{r}\circ S=\mu^{r}(a?)=\mu^{r}\big{(}g^{2}?\big{)},$

where the second equality is [11, Proposition 4.7]. ∎

The left $q$-characters are just shifted symmetric linear forms. More
precisely, we have an isomorphism of algebras:

$\begin{array}[]{crll}\big{(}g^{-1}\big{)}^{*}\colon&\operatorname{SLF}(H)&\rightarrow&\mathrm{Ch}^{l}(H),\\\
&\psi&\mapsto&\psi\big{(}g^{-1}?\big{)}.\end{array}$

Let us define shifted versions of $\boldsymbol{\chi}$ and of
$\boldsymbol{\gamma}$:

$\boldsymbol{\chi}_{g^{-1}}=\boldsymbol{\chi}\circ\left(g^{-1}\right)^{*}\colon\
\operatorname{SLF}(H)\overset{\cong}{\longrightarrow}\mathcal{Z}(H),\qquad\boldsymbol{\gamma}_{g}=g^{*}\circ\boldsymbol{\gamma}\colon\
\mathcal{Z}(H)\overset{\cong}{\longrightarrow}\operatorname{SLF}(H).$

The equality
$\mathcal{S}=\boldsymbol{\chi}_{g^{-1}}\circ\boldsymbol{\gamma}_{g}$ still
holds, but we have now $\operatorname{SLF}(H)$ instead of
$\mathrm{Ch}^{l}(H)$.

In order to show the equivalence of $\theta_{\mathrm{SLF}}$ and
$\theta_{\mathrm{LM}}$, we begin with two technical lemmas.

######  Lemma 5.10.

  1. $1.$

$\big{(}v^{-1}\big{)}^{\prime}\otimes
S^{-1}\big{(}\big{(}v^{-1}\big{)}^{\prime\prime}\big{)}=S\big{(}\big{(}v^{-1}\big{)}^{\prime\prime}\big{)}\otimes\big{(}v^{-1}\big{)}^{\prime}$.

  2. $2.$

$\forall\,\psi\in\mathcal{O}(H)$, $\forall\,h\in H$,
$\mu^{r}(h?)\psi=\mu^{r}(h^{\prime}?)\psi\big{(}S^{-1}(h^{\prime\prime})\big{)}$.

###### Proof.

The first equality follows from $S\big{(}v^{-1}\big{)}=v^{-1}$. For the second
one

$\displaystyle\big{\langle}\mu^{r}(h?)\psi,x\big{\rangle}$
$\displaystyle=\mu^{r}(hx^{\prime})\psi(x^{\prime\prime})=\mu^{r}(h^{\prime}x^{\prime})\psi\big{(}S^{-1}(h^{\prime\prime\prime})h^{\prime\prime}x^{\prime\prime}\big{)}=\psi\big{(}S^{-1}(h^{\prime\prime\prime})\mu^{r}(h^{\prime}x^{\prime})h^{\prime\prime}x^{\prime\prime}\big{)}$
$\displaystyle=\psi\big{(}S^{-1}(h^{\prime\prime})\mu^{r}((h^{\prime}x)^{\prime})(h^{\prime}x)^{\prime\prime}\big{)}=\mu^{r}(h^{\prime}x)\psi\big{(}S^{-1}(h^{\prime\prime})\big{)},$

where we simply used the defining property (5.2) of $\mu^{r}$. ∎

We will employ an immediate consequence of Lemma 5.9, namely (see Proposition
5.3)

$\displaystyle\varphi_{v}=\mu^{l}\big{(}v^{-1}\big{)}^{-1}\mu^{r}\big{(}gv^{-1}?\big{)}.$
(5.7)

######  Lemma 5.11.

It holds

$\rho_{\mathrm{SLF}}\big{(}v_{A}^{2}v_{B}\big{)}=\mu^{l}\big{(}v^{-1}\big{)}^{-1}\boldsymbol{\gamma}_{g}\circ\boldsymbol{\chi}_{g^{-1}}.$

###### Proof.

We compute each side of the equality. On the one hand:

$\boldsymbol{\gamma}_{g}\circ\boldsymbol{\chi}_{g^{-1}}(\psi)=\boldsymbol{\gamma}_{g}\big{(}(\psi\otimes\mathrm{id})\big{(}g^{-1}\big{(}v^{-1}\big{)}^{\prime}v\otimes\big{(}v^{-1}\big{)}^{\prime\prime}v\big{)}\big{)}=\psi\big{(}g^{-1}\big{(}v^{-1}\big{)}^{\prime}v\big{)}\mu^{r}\big{(}gS\big{(}\big{(}v^{-1}\big{)}^{\prime\prime}\big{)}v?\big{)},$

whereas on the other hand

$\displaystyle v_{A}^{2}v_{B}\triangleright\psi$
$\displaystyle=\bigl{(}\varphi_{v}\psi^{v}\bigr{)}^{v}=\mu^{l}\big{(}v^{-1}\big{)}^{-1}\big{(}\mu^{r}\big{(}gv^{-1}?\big{)}\psi^{v}\big{)}^{v}$
$\displaystyle=\mu^{l}\big{(}v^{-1}\big{)}^{-1}\big{[}\mu^{r}\big{(}g\big{(}v^{-1}\big{)}^{\prime}?\big{)}\psi\big{(}vg^{-1}S^{-1}\big{(}\big{(}v^{-1}\big{)}^{\prime\prime}\big{)}\big{)}\big{]}^{v}$
$\displaystyle=\mu^{l}\big{(}v^{-1}\big{)}^{-1}\mu^{r}\big{(}gvS\big{(}\big{(}v^{-1}\big{)}^{\prime\prime}\big{)}?\big{)}\psi\big{(}vg^{-1}\big{(}v^{-1}\big{)}^{\prime}\big{)}$

as desired. We used that $v_{A}\triangleright\psi=\psi^{v}$,
$v_{B}\triangleright\psi=\big{(}\varphi_{v}\psi^{v}\big{)}^{v^{-1}}$, (5.7)
and Lemma 5.10. ∎

The link between the two presentations of $\mathrm{SL}_{2}(\mathbb{Z})$ is
$s=\tau_{a}^{-1}\tau_{b}^{-1}\tau_{a}^{-1}$, $t=\tau_{a}$. Hence we define two
operators
$\mathcal{S}^{\prime},\mathcal{T}^{\prime}\colon\operatorname{SLF}(H)\to\operatorname{SLF}(H)$
by

$\mathcal{S}^{\prime}=\theta_{\mathrm{SLF}}\big{(}\tau_{a}^{-1}\tau_{b}^{-1}\tau_{a}^{-1}\big{)}=\rho_{\mathrm{SLF}}(v_{A}v_{B}v_{A}),\qquad\mathcal{T}^{\prime}=\theta_{\mathrm{SLF}}(\tau_{a})=\rho_{\mathrm{SLF}}\big{(}v_{A}^{-1}\big{)}.$

######  Theorem 5.12.

Recall that we assume that $H$ is a finite-dimensional factorizable ribbon
Hopf algebra. Then the projective representation $\theta_{\mathrm{SLF}}$ of
Theorem 5.8 is equivalent to $\theta_{\mathrm{LM}}$.

###### Proof.

Consider the following isomorphism of vector spaces

$\begin{array}[]{crll}f=\rho_{\mathrm{SLF}}\big{(}v_{A}^{-1}\big{)}\circ\boldsymbol{\gamma}_{g}\colon&\mathcal{Z}(H)&\rightarrow&\operatorname{SLF}(H),\\\
&z&\mapsto&\boldsymbol{\gamma}_{g}(z)^{v^{-1}}=\mu^{r}\big{(}gv^{-1}S(z)?\big{)}.\end{array}$

By Lemma 5.11,

$\mathcal{S}^{\prime}=\mu^{l}\big{(}v^{-1}\big{)}^{-1}\rho_{\mathrm{SLF}}\big{(}v_{A}^{-1}\big{)}\circ\boldsymbol{\gamma}_{g}\circ\boldsymbol{\chi}_{g^{-1}}\circ\rho_{\mathrm{SLF}}(v_{A}).$

Thus

$f\circ\mathcal{S}=\rho_{\mathrm{SLF}}\big{(}v_{A}^{-1}\big{)}\circ\boldsymbol{\gamma}_{g}\circ\boldsymbol{\chi}_{g^{-1}}\circ\boldsymbol{\gamma}_{g}=\mu^{l}\big{(}v^{-1}\big{)}\mathcal{S}^{\prime}\circ
f.$

Next,

$f\circ\mathcal{T}(z)=f\big{(}v^{-1}z\big{)}=\boldsymbol{\gamma}_{g}(z)^{v^{-2}}=\rho_{\mathrm{SLF}}(v_{A}^{-1})\big{(}\boldsymbol{\gamma}_{g}(z)^{v^{-1}}\big{)}=\mathcal{T}^{\prime}\circ
f(z).$

Then $f$ is an intertwiner of projective representations. ∎

##  6 The example of $\boldsymbol{H=\overline{U}_{q}(\mathfrak{sl}(2))}$

Let $q$ be a primitive root of unity of order $2p$, with $p\geq 2$. We now
work in some detail the case of $H=\overline{U}_{q}(\mathfrak{sl}(2))$, the
restricted quantum group associated to $\mathfrak{sl}(2,\mathbb{C})$, which
will be denoted $\overline{U}_{q}$ in the sequel. For the definitions and main
properties about $\overline{U}_{q}$,
$\mathcal{Z}\big{(}\overline{U}_{q}\big{)}$,
$\operatorname{SLF}\big{(}\overline{U}_{q}\big{)}$, their canonical bases and
the $\overline{U}_{q}$-modules, see [7, 21, 24, 27, 31, 39]. Here we take back
the notations and conventions of [21].

To explicitly describe the representation of $\mathrm{SL}_{2}(\mathbb{Z})$, we
use the GTA basis of $\operatorname{SLF}\big{(}\overline{U}_{q}\big{)}$ which
is studied in detail in [21], and which has been introduced in [27] and [7].

Even though $\overline{U}_{q}$ is not braided, we consider this example
because this Hopf algebra is well-studied and is related to certain models in
logarithmic conformal field theory [24]. As we shall recall below, it is very
close from being braided (it suffices to add a square root of the generator
$K$) and its important properties for our purposes are that it is factorizable
(by abuse of terminology since it is not braided) and that it contains a
ribbon element. It will follow that the defining relations of
$\mathcal{L}_{0,1}(H)$ and $\mathcal{L}_{1,0}(H)$ still make sense in this
context and that all the previous results remain true.

###  6.1 The braided extension of $\boldsymbol{\overline{U}_{q}}$

Recall that $\overline{U}_{q}$ is not braided [31, Proposition 3.7.3]. But its
extension by a square root of $K$ is braided, as shown in [24]. Let
$\overline{U}_{q}{{}^{\\!1/2}}$ be this extension and
$R\in\overline{U}_{q}{{}^{\\!1/2}}\otimes\overline{U}_{q}{{}^{\\!1/2}}$ be the
universal $R$-matrix, given by

$R=q^{H\otimes
H/2}\sum_{m=0}^{p-1}\frac{\hat{q}^{m}}{[m]!}q^{m(m-1)/2}E^{m}\otimes F^{m},$

with

$q^{H\otimes H/2}=\frac{1}{4p}\sum_{n,j=0}^{4p-1}q^{-nj/2}K^{n/2}\otimes
K^{j/2},$

where $q^{1/2}$ is a fixed square root of $q$. We use the notation
$q^{H\otimes H/2}$ because $q^{H\otimes H/2}v\otimes w=q^{ab/2}$ if
$K^{1/2}v=q^{a/2}v$ and $K^{1/2}w=q^{b/2}w$; also recall the notation
$\hat{q}=q-q^{-1}$. There is a ribbon element $v$ associated to this
$R$-matrix (we choose $g=K^{p+1}$ as pivotal element, and by (2.9) this fixes
the choice of $v$). The formulas for $RR^{\prime}$ and $v$ are in [24]; it is
important to note that $K^{1/2}$ does not appear in the expression of these
elements

$RR^{\prime}\in\overline{U}_{q}\otimes\overline{U}_{q}\qquad\text{and}\qquad
v\in\overline{U}_{q}.$

Moreover, thanks to the formula for $RR^{\prime}$, it is obvious that the map

$\begin{array}[]{crll}\Psi\colon&\overline{U}_{q}^{*}&\rightarrow&\overline{U}_{q},\\\
&\beta&\mapsto&(\beta\otimes\mathrm{id})(RR^{\prime})\end{array}$

is an isomorphism of vector spaces. Hence, even if $\overline{U}_{q}$ is not
braided, it can be thought of as a factorizable Hopf algebra.

Let $I$ be a $\overline{U}_{q}{{}^{\\!1/2}}$-module. Since
$\overline{U}_{q}\subset\overline{U}_{q}{{}^{\\!1/2}}$, $I$ determines a
$\overline{U}_{q}$-module, which we denote $I_{|\overline{U}_{q}}$. We say
that a $\overline{U}_{q}$-module $J$ is liftable if there exists a
$\overline{U}_{q}{{}^{\\!1/2}}$-module $\widetilde{J}$ such that
$\widetilde{J}_{|\overline{U}_{q}}=J$. Not every $\overline{U}_{q}$-module is
liftable, see [31]. But the simple modules and the PIMs are liftable, which is
enough for us. Indeed, it suffices to define the action of $K^{1/2}$ on these
modules. Take back the notations of [21] for the canonical basis of modules.
For the simple module $\mathcal{X}^{\epsilon}(s)$ ($\epsilon\in\\{\pm\\})$,
there are two choices for $\epsilon^{1/2}$, and so the two possible liftings
are defined by

$K^{1/2}v_{j}=\epsilon^{1/2}q^{(s-1-2j)/2}v_{j}$

and the action of $E$ and $F$ is unchanged. The two possible liftings of the
PIM $\mathcal{P}^{\epsilon}(s)$ are defined by

$\displaystyle K^{1/2}b_{0}=\epsilon^{1/2}q^{(s-1)/2}b_{0},\qquad$
$\displaystyle
K^{1/2}x_{0}=\big{(}\epsilon^{1/2}q^{p/2}\big{)}q^{(p-s-1)/2}x_{0},$
$\displaystyle
K^{1/2}y_{0}=\big{(}{-}\epsilon^{1/2}q^{p/2}\big{)}q^{(p-s-1)/2}y_{0},\qquad$
$\displaystyle K^{1/2}a_{0}=\epsilon^{1/2}q^{(s-1)/2}a_{0}$

and the action of $E$ and $F$ is unchanged.

Let $\widetilde{\mathbb{C}}^{-}$ be the $1$-dimensional
$\overline{U}_{q}{{}^{\\!1/2}}$-module with basis $v$ defined by $Ev=Fv=0$,
$K^{1/2}v=-v$ (which is a lifting of $\mathcal{X}^{+}(1)=\mathbb{C}$). If
$\widetilde{I}$ is a lifting of a simple module or a PIM $I$, then we have
seen that the only possible liftings of $I$ are
$\widetilde{I}^{+}=\widetilde{I}$ and
$\widetilde{I}^{-}=\widetilde{I}\otimes\widetilde{\mathbb{C}}^{-}$. Moreover,
using (2.1), we get equalities which will be used in the next section

$\displaystyle\bigl{(}\overset{\widetilde{I}^{-}\widetilde{J}}{R}\bigr{)}_{\\!12}=\bigl{(}\overset{\widetilde{I}^{+}\widetilde{J}}{R}\bigr{)}_{\\!12}\bigl{(}\overset{\widetilde{J}}{K^{p}}\bigr{)}_{2},\qquad$
$\displaystyle\bigl{(}\overset{\widetilde{I}\widetilde{J}^{-}}{\\!R\>}\bigr{)}_{12}=\bigl{(}\overset{\widetilde{I}\widetilde{J}^{+}}{\\!R\>}\bigr{)}_{\\!12}\bigl{(}\overset{\widetilde{I}}{K^{p}}\bigr{)}_{1},$
$\displaystyle\bigl{(}\overset{\widetilde{I}^{-}\widetilde{J}}{R^{\prime}}\bigr{)}_{\\!12}=\bigl{(}\overset{\widetilde{I}^{+}\widetilde{J}}{R^{\prime}}\bigr{)}_{\\!12}\bigl{(}\overset{\widetilde{J}}{K^{p}}\bigr{)}_{2},\qquad$
$\displaystyle\bigl{(}\overset{\widetilde{I}\widetilde{J}^{-}}{R^{\prime}}\bigr{)}_{\\!12}=\bigl{(}\overset{\widetilde{I}\widetilde{J}^{+}}{R^{\prime}}\bigr{)}_{\\!12}\bigl{(}\overset{\widetilde{I}}{K^{p}}\bigr{)}_{1}.$
(6.1)

###  6.2 $\boldsymbol{\mathcal{L}_{0,1}\big{(}\overline{U}_{q}\big{)}}$ and
$\boldsymbol{\mathcal{L}_{1,0}\big{(}\overline{U}_{q}\big{)}}$

We define $\mathcal{L}_{0,1}\big{(}\overline{U}_{q}\big{)}$ as the quotient of
$\mathrm{T}(\overline{U}_{q}{{}^{\\!*}})$ by the fusion relation

$\overset{I\otimes
J}{M}\\!_{12}=\overset{I}{M}_{1}\overset{\widetilde{I}\widetilde{J}}{(R^{\prime})}_{12}\overset{J}{M}_{2}\overset{\widetilde{I}\widetilde{J}}{\big{(}R^{\prime-1}\big{)}}_{12},$

where $I$, $J$ are simple modules or PIMs and $\widetilde{I}$, $\widetilde{J}$
are liftings of $I$ and $J$. From (6.1) and the fact that $K^{p}$ is central,
we see that this does not depend on the choice of $\widetilde{I}$ and
$\widetilde{J}$. As we saw in Section 2.1, the matrix coefficients of the PIMs
linearly span $\mathcal{L}_{0,1}(H)$, thus we can restrict to them in the
definition. However, the simple modules are added for convenience. All the
results of Section 3 remain true for
$\mathcal{L}_{0,1}\big{(}\overline{U}_{q}\big{)}$. In particular, $\Psi_{0,1}$
is an isomorphism since $\overline{U}_{q}$ is factorizable (in the generalized
sense explained in Section 6.1).

We now describe $\mathcal{L}_{0,1}\big{(}\overline{U}_{q}\big{)}$ by
generators and relations. Let

$M=\overset{\mathcal{X}^{+}(2)}{M}=\begin{pmatrix}a&b\\\
c&d\end{pmatrix}\qquad\text{and}\qquad\widetilde{R}=\overset{\widetilde{\mathcal{X}}^{+}(2)\widetilde{\mathcal{X}}^{+}(2)}{R}=q^{-1/2}\begin{pmatrix}q&0&0&0\\\
0&1&\hat{q}&0\\\ 0&0&1&0\\\ 0&0&0&q\end{pmatrix},$

where $\widetilde{\mathcal{X}}^{+}(2)$ is the lifting of $\mathcal{X}^{+}(2)$
defined by $K^{1/2}v_{0}=q^{1/2}v_{0}$. By the decomposition rules of tensor
products (see [39] and also [31]), every PIM (and every simple module) is a
direct summand of some tensor power $\mathcal{X}^{+}(2)^{\otimes n}$. Thus
every matrix coefficient of a PIM is a matrix coefficient of some
$\mathcal{X}^{+}(2)^{\otimes n}$ (with $n\geq p$). It follows from the fusion
relation that $a$, $b$, $c$, $d$ generate
$\mathcal{L}_{0,1}\big{(}\overline{U}_{q}\big{)}$. Let us determine relations
between these elements. First, we have the reflection equation:

$\widetilde{R}_{12}M_{1}\widetilde{R}_{21}M_{2}=M_{2}\widetilde{R}_{12}M_{1}\widetilde{R}_{21}.$

This equation is equivalent to the following exchange relations

$\displaystyle da=ad,\qquad db=q^{2}bd,\qquad dc=q^{-2}cd,$ $\displaystyle
ba=ab+q^{-1}\hat{q}bd,\qquad cb=bc+q^{-1}\hat{q}\big{(}da-d^{2}\big{)},\qquad
ca=ac-q^{-1}\hat{q}dc.$

Second, since $\mathcal{X}^{+}(2)^{\otimes
2}\cong\mathcal{X}^{+}(1)\oplus\mathcal{X}^{+}(3)$,111This decomposition does
not hold if $p=2$: in that case, we have $\mathcal{X}^{+}(2)^{\otimes
2}\cong\mathcal{P}^{+}(1)$. The morphism $\Phi$ remains valid and corresponds
to sending $\mathbb{C}=\mathcal{X}^{+}(1)$ in
$\mathrm{Soc}\bigl{(}\mathcal{P}^{+}(1)\bigr{)}$. there exists a unique (up to
scalar) morphism $\Phi\colon\mathcal{X}^{+}(1)\to\mathcal{X}^{+}(2)^{\otimes
2}$. It is easily computed

$\Phi(1)=qv_{0}\otimes v_{1}-v_{1}\otimes v_{0}.$

By fusion, we have

$\overset{\mathcal{X}^{+}(2)^{\otimes
2}}{M}\\!\\!\\!\\!\\!\\!_{12}\Phi=M_{1}\widetilde{R}_{21}M_{2}\widetilde{R}_{21}^{-1}\Phi=\Phi.$

This gives just one new relation, which is the analogue of the quantum
determinant

$ad-q^{2}bc=1.$

Finally, let us compute the RSD isomorphism on $M$

$\displaystyle\Psi_{0,1}\begin{pmatrix}a&b\\\ c&d\end{pmatrix}$
$\displaystyle=\overset{\widetilde{\mathcal{X}}^{+}(2)}{L^{(+)}}\overset{\widetilde{\mathcal{X}}^{+}(2)}{L^{(-)-1}}=\begin{pmatrix}K^{1/2}&\hat{q}K^{1/2}F\\\
0&K^{-1/2}\end{pmatrix}\begin{pmatrix}K^{1/2}&0\\\
\hat{q}K^{-1/2}E&K^{-1/2}\end{pmatrix}$
$\displaystyle=\begin{pmatrix}K+q^{-1}\hat{q}^{2}FE&q^{-1}\hat{q}F\\\
\hat{q}K^{-1}E&K^{-1}\end{pmatrix}.$

We deduce the relations $b^{p}=c^{p}=0$ and $d^{2p}=1$ from the defining
relations of $\overline{U}_{q}$.

######  Theorem 6.1.

$\mathcal{L}_{0,1}\big{(}\overline{U}_{q}\big{)}$ admits the following
presentation

$\left\langle
a,b,c,d\,\Bigg{|}\,\begin{matrix}da=ad,&db=q^{2}bd,&dc=q^{-2}cd\\\
ba=ab+q^{-1}\hat{q}bd,&cb=bc+q^{-1}\hat{q}\big{(}da-d^{2}\big{)},&ca=ac-q^{-1}\hat{q}dc\\\
ad-q^{2}bc=1,&b^{p}=c^{p}=0,&d^{2p}=1\end{matrix}\right\rangle.$

A basis is given by the monomials $b^{i}c^{j}d^{k}$ with $0\leq i,j\leq p-1$,
$0\leq k\leq 2p-1$.

###### Proof.

Let $A$ be the algebra defined by this presentation. It is readily seen that
$a=d^{-1}+q^{2}bcd^{-1}$ and that the monomials $b^{i}c^{j}d^{k}$ with $0\leq
i,j\leq p-1$, $0\leq k\leq 2p-1$ linearly span $A$. Thus $\dim(A)\leq 2p^{3}$.
But we know that
$2p^{3}=\dim\big{(}\overline{U}_{q}\big{)}=\dim\big{(}\mathcal{L}_{0,1}\big{(}\overline{U}_{q}\big{)}\big{)}$
since the monomials $E^{i}F^{j}K^{\ell}$ with $0\leq i,j\leq p-1$, $0\leq
k\leq 2p-1$ form the PBW basis of $\overline{U}_{q}$. It follows that
$\dim(A)\leq\dim\big{(}\mathcal{L}_{0,1}\big{(}\overline{U}_{q}\big{)}\big{)}$.
Since these relations are satisfied in
$\mathcal{L}_{0,1}\big{(}\overline{U}_{q}\big{)}$, there exists a surjection
$p\colon A\to\mathcal{L}_{0,1}\big{(}\overline{U}_{q}\big{)}$. Thus
$\dim(A)\geq\dim\big{(}\mathcal{L}_{0,1}\big{(}\overline{U}_{q}\big{)}\big{)}$,
and the theorem is proved. ∎

######  Remark 6.2.

A consequence of this theorem is that
$\mathcal{L}_{0,1}\big{(}\overline{U}_{q}\big{)}$ is a restricted version
(i.e., a finite-dimensional quotient by monomial central elements) of
$\mathcal{L}_{0,1}(U_{q})^{\mathrm{spe}}$, the specialization at our root of
unity $q$ of the algebra $\mathcal{L}_{0,1}(U_{q})$. A complete study of the
algebra $\mathcal{L}_{0,1}(U_{q})^{\mathrm{spe}}$ will appear in [9].

Applying the isomorphism of algebras $\mathcal{D}$ defined in (3.9) to the GTA
basis of $\operatorname{SLF}\big{(}\overline{U}_{q}\big{)}$, we get a basis of
$\mathcal{Z}\big{(}\overline{U}_{q}\big{)}$. We introduce notations for these
basis elements

$\displaystyle\overset{\mathcal{X}^{\epsilon}(s)}{W}=\mathcal{D}(\chi^{\epsilon}_{s}),\qquad
H^{s^{\prime}}=\mathcal{D}(G_{s^{\prime}})$ (6.2)

with $1\leq s\leq p$, $\epsilon\in\\{\pm\\}$ and $1\leq s^{\prime}\leq p-1$.
They satisfy the same multiplication rules than the elements of the GTA basis,
see [21, Section 5] or [27] (the elements $\boldsymbol{\chi}(s)$ defined in
[27] correspond to $[s]H^{s}$ here). Let us mention that under the
identification
$\mathcal{L}_{0,1}\big{(}\overline{U}_{q}\big{)}=\overline{U}_{q}$ via
$\Psi_{0,1}$, it holds by definition
$\overset{\mathcal{X}^{\epsilon}(s)}{W}=\mathrm{tr}(\overset{\mathcal{X}^{\epsilon}(s)}{K^{p+1}}\overset{\mathcal{X}^{\epsilon}(s)}{M})$,
since we choose $K^{p+1}$ as pivotal element. In particular,

$\displaystyle\overset{\mathcal{X}^{+}(2)}{W}=-qa-q^{-1}d=-\hat{q}^{2}FE-
qK-q^{-1}K^{-1}=-\hat{q}^{2}C,$ (6.3)

where $C$ is the standard Casimir element of $\overline{U}_{q}$.

Similarly, we define $\mathcal{L}_{1,0}\big{(}\overline{U}_{q}\big{)}$ as the
quotient of
$\mathcal{L}_{0,1}\big{(}\overline{U}_{q}\big{)}\ast\mathcal{L}_{0,1}\big{(}\overline{U}_{q}\big{)}$
by the exchange relations

$\displaystyle\overset{\widetilde{I}\widetilde{J}}{R}_{12}\overset{I}{B}_{1}\overset{\widetilde{I}\widetilde{J}}{(R^{\prime})}_{12}\overset{J}{A}_{2}=\overset{J}{A}_{2}\overset{\widetilde{I}\widetilde{J}}{R}_{12}\overset{I}{B}_{1}\overset{\widetilde{I}\widetilde{J}}{(R^{-1})}_{12},$

where $I$, $J$ are simple modules or PIMs and $\widetilde{I}$, $\widetilde{J}$
are liftings of $I$ and $J$. From (6.1), we see again that this does not
depend on the choice of $\widetilde{I}$ and $\widetilde{J}$. The coefficients
of $\overset{\mathcal{X}^{+}(2)}{A}$ and of $\overset{\mathcal{X}^{+}(2)}{B}$
generate $\mathcal{L}_{1,0}\big{(}\overline{U}_{q}\big{)}$. The morphism
$\Psi_{1,0}\colon\mathcal{L}_{1,0}\big{(}\overline{U}_{q}\big{)}\to\mathcal{H}\big{(}\mathcal{O}\big{(}\overline{U}_{q}\big{)}\big{)}$
is well-defined (the square root of $K$ does not appear). Indeed, the matrix
$\Psi_{1,0}\bigl{(}\overset{\mathcal{X}^{+}(2)}{A}\bigr{)}$ belongs to
$1\otimes\overline{U}_{q}\subset\mathcal{H}\big{(}\mathcal{O}\big{(}\overline{U}_{q}\big{)}\big{)}$
and is the same as the image of the matrix $M$ under the morphism $\Psi_{0,1}$
above; moreover, thanks to the commutation relations of the Heisenberg double
we get

$\displaystyle\Psi_{1,0}\bigl{(}\overset{\mathcal{X}^{+}(2)}{B}\bigr{)}=\overset{\widetilde{\mathcal{X}}^{+}(2)}{L^{(+)}}\overset{\widetilde{\mathcal{X}}^{+}(2)}{T}\overset{\widetilde{\mathcal{X}}^{+}(2)}{L^{(-)-1}}=\begin{pmatrix}K^{1/2}&\hat{q}K^{1/2}F\\\
0&K^{-1/2}\end{pmatrix}\begin{pmatrix}\alpha&\beta\\\
\gamma&\delta\end{pmatrix}\begin{pmatrix}K^{1/2}&0\\\
\hat{q}K^{-1/2}E&K^{-1/2}\end{pmatrix}$
$\displaystyle=\begin{pmatrix}q^{1/2}\alpha K+q^{-1/2}\hat{q}\delta
K+q^{1/2}\hat{q}\gamma KF+q^{-1/2}\hat{q}\beta E+q^{-1/2}\hat{q}^{2}\delta
FE&q^{-1/2}\beta+q^{-1/2}\hat{q}\delta F\\\
q^{-1/2}\gamma+q^{1/2}\hat{q}\delta K^{-1}E&q^{1/2}\delta
K^{-1}\end{pmatrix}.$

From these formulas we see that $\Psi_{1,0}$ is surjective: indeed, thanks to
the coefficients of
$\Psi_{1,0}\bigl{(}\\!\overset{\mathcal{X}^{+}(2)}{A}\\!\bigr{)}$ we obtain
$E$, $F$, $K$ and then thanks to the coefficients of
$\Psi_{1,0}\bigl{(}\overset{\mathcal{X}^{+}(2)}{B}\bigr{)}$ we obtain
$\alpha$, $\beta$, $\gamma$, $\delta$. Moreover, by the exchange relation in
$\mathcal{L}_{1,0}\big{(}\overline{U}_{q}\big{)}$ and the fact that
$\mathcal{L}_{0,1}\big{(}\overline{U}_{q}\big{)}\cong\overline{U}_{q}$, we
know that
$\dim\bigl{(}\mathcal{L}_{1,0}\big{(}\overline{U}_{q}\big{)}\bigr{)}\leq\dim\big{(}\overline{U}_{q}\big{)}^{2}=\dim\bigl{(}\mathcal{H}\big{(}\mathcal{O}\big{(}\overline{U}_{q}\big{)}\big{)}\bigr{)}$,
and hence $\Psi_{1,0}$ is an isomorphism of algebras. In order to get a
presentation of $\mathcal{L}_{1,0}\big{(}\overline{U}_{q}\big{)}$, one can
again restrict to $I=J=\mathcal{X}^{+}(2)$ and write down the corresponding
exchange relations. We do not give this presentation of
$\mathcal{L}_{1,0}\big{(}\overline{U}_{q}\big{)}$ since we will not use it in
this work.

With these definitions of $\mathcal{L}_{0,1}\big{(}\overline{U}_{q}\big{)}$
and $\mathcal{L}_{1,0}\big{(}\overline{U}_{q}\big{)}$, all the computations
made under the general assumptions before remain valid with $\bar{U}_{q}$ and
we have a projective $\mathrm{SL}_{2}(\mathbb{Z})$-representation on
$\mathrm{SLF}\big{(}\overline{U}_{q}\big{)}$.

###  6.3 Explicit description of the
$\boldsymbol{\mathrm{SL}_{2}(\mathbb{Z})}$-projective representation

Note that it can be shown directly that $\overline{U}_{q}$ is unimodular and
unibalanced, see for instance [29, Corollary II.2.8] (also note that in [11]
it is shown that all the simply laced restricted quantum groups at roots of
unity are unibalanced).

######  Proposition 6.3.

For all $z\in\mathcal{Z}\big{(}\overline{U}_{q}\big{)}$, $S(z)=z$ and for all
$\psi\in\operatorname{SLF}\big{(}\overline{U}_{q}\big{)}$, $S(\psi)=\psi$. It
follows that in the case of $\overline{U}_{q}$, $\theta_{\operatorname{SLF}}$
is in fact a projective representation of $\mathrm{PSL}_{2}(\mathbb{Z})$.

###### Proof.

By [24, Appendix D], the canonical central elements are expressed as
$e_{s}=P_{s}(C)$, $w^{\pm}_{s}=\pi^{\pm}_{s}Q_{s}(C)$ where $P_{s}$ and
$Q_{s}$ are polynomials, $C$ is the Casimir element and $\pi_{s}^{\pm}$ are
discrete Fourier transforms of $\big{(}K^{j}\big{)}_{0\leq j\leq 2p-1}$. It is
easy to check that $S(C)=C$ and that
$S\big{(}\pi^{\pm}_{s}\big{)}=\pi^{\pm}_{s}$, thus $S(e_{s})=e_{s}$ and
$S\big{(}w^{\pm}_{s}\big{)}=w^{\pm}_{s}$. Next, let
$\psi\in\operatorname{SLF}\big{(}\overline{U}_{q}\big{)}$. Since
$\boldsymbol{\gamma}_{g}$ is an isomorphism, we can write
$\psi=\boldsymbol{\gamma}_{g}(z)=\mu^{r}(gS(z)?)$ with
$z\in\mathcal{Z}\big{(}\overline{U}_{q}\big{)}$. Then

$S(\psi)=S\big{(}\mu^{r}(gz?)\big{)}=\mu^{r}\circ
S\big{(}?zg^{-1}\big{)}=\mu^{l}\big{(}g^{-1}z?\big{)}=\mu^{r}(gz?)=\psi.$

We used that $S(z)=z$, Proposition 5.3 and the fact that $\overline{U}_{q}$ is
unibalanced. ∎

Recall that the action of the elements $v_{A}^{-1}$, $v_{B}^{-1}$ on
$\mathrm{SLF}(H)$ implements the $\mathrm{SL}_{2}(\mathbb{Z})$-representation.
We now determine this action on the GTA basis. Some preliminaries are in
order; first recall (see [24]) that the expression of the ribbon element $v$
in the canonical basis $\big{(}e_{s},w_{s}^{\pm}\big{)}$ of
$\mathcal{Z}\big{(}\overline{U}_{q}\big{)}$ is

$\displaystyle
v=\sum_{s=0}^{p}v_{\mathcal{X}^{+}(s)}e_{s}+\hat{q}\sum_{s=1}^{p-1}v_{\mathcal{X}^{+}(s)}\left(\frac{p-s}{[s]}w^{+}_{s}-\frac{s}{[s]}w^{-}_{s}\right)$
(6.4)

with $\hat{q}=q-q^{-1}$ and

$v_{\mathcal{X}^{+}(s)}=v_{\mathcal{X}^{-}(p-s)}=(-1)^{s-1}q^{\frac{-(s^{2}-1)}{2}}.$

Note that
$\overset{\mathcal{X}^{\epsilon}(s)}{v}=v_{\mathcal{X}^{\epsilon}(s)}\mathrm{id}$
and $v_{\mathcal{X}^{+}(0)}$ is just a notation for $v_{\mathcal{X}^{-}(p)}$
used to unify the formula. Expressing $v^{-1}$ in this basis is obvious.
Second, it is easy to see that the action of
$\mathcal{Z}\big{(}\overline{U}_{q}\big{)}$ on
$\operatorname{SLF}\big{(}\overline{U}_{q}\big{)}$ is

$\displaystyle\big{(}\chi^{+}_{s}\big{)}^{e_{t}}=\delta_{s,t}\chi^{+}_{s},\qquad$
$\displaystyle\big{(}\chi^{-}_{s}\big{)}^{e_{t}}=\delta_{p-s,t}\chi^{-}_{s},\qquad$
$\displaystyle G_{s}^{e_{t}}=\delta_{s,t}G_{s}$
$\displaystyle\big{(}\chi^{+}_{s}\big{)}^{w^{\pm}_{t}}=0,\qquad$
$\displaystyle\big{(}\chi^{-}_{s}\big{)}^{w^{\pm}_{t}}=0,\qquad$
$\displaystyle G_{s}^{w^{+}_{t}}=\delta_{s,t}\chi^{+}_{s},\qquad$
$\displaystyle G_{s}^{w^{-}_{t}}=\delta_{s,t}\chi^{-}_{p-s},$ (6.5)

where $\varphi^{z}=\varphi(z?)$. Finally, we have the following lemma.

######  Lemma 6.4.

Let $z\in\mathcal{L}_{0,1}^{\mathrm{inv}}(H)=\mathcal{Z}(H)$ and let
$\psi\in\operatorname{SLF}(H)$. Then

$z_{vB^{-1}A}\triangleright\psi=S\big{(}\mathcal{D}^{-1}(z)\big{)}\psi.$

###### Proof.

The proof is analogous to those of the two similar results in Section 4.3 and
is thus left to the reader. Note that this lemma is not specific to the case
of $\overline{U}_{q}$. ∎

######  Theorem 6.5.

The actions of $v_{A}^{-1}$ and $v_{B}^{-1}$ on the GTA basis are given by

$v_{A}^{-1}\triangleright\chi^{\epsilon}_{s}=v^{-1}_{\mathcal{X}^{\epsilon}(s)}\chi^{\epsilon}_{s},\qquad
v_{A}^{-1}\triangleright
G_{s^{\prime}}=v^{-1}_{\mathcal{X}^{+}(s^{\prime})}G_{s^{\prime}}-v_{\mathcal{X}^{+}(s^{\prime})}^{-1}\hat{q}\left(\frac{p-s^{\prime}}{[s^{\prime}]}\chi^{+}_{s^{\prime}}-\frac{s^{\prime}}{[s^{\prime}]}\chi^{-}_{p-s^{\prime}}\right)$

and

$\displaystyle
v_{B}^{-1}\triangleright\chi^{\epsilon}_{s}=\xi\epsilon(-\epsilon)^{p-1}sq^{-(s^{2}-1)}\left(\sum_{\ell=1}^{p-1}(-1)^{s}(-\epsilon)^{p-\ell}\big{(}q^{\ell
s}+q^{-\ell
s}\big{)}\big{(}\chi^{+}_{\ell}+\chi^{-}_{p-\ell}\big{)}+\chi^{+}_{p}\right.$
$\displaystyle\left.\hphantom{v_{B}^{-1}\triangleright\chi^{\epsilon}_{s}=}{}+(-\epsilon)^{p}(-1)^{s}\chi^{-}_{p}\vphantom{\sum_{\ell=1}^{p-1}}\right)+\xi\epsilon(-1)^{s}q^{-(s^{2}-1)}\sum_{j=1}^{p-1}(-\epsilon)^{j+1}[j][js]G_{j},$
$\displaystyle v_{B}^{-1}\triangleright
G_{s^{\prime}}=\xi(-1)^{s^{\prime}}q^{-(s^{\prime
2}-1)}\frac{\hat{q}p}{[s^{\prime}]}\sum_{j=1}^{p-1}(-1)^{j+1}[j][js^{\prime}]\left(2G_{j}-\hat{q}\frac{p-j}{[j]}\chi^{+}_{j}+\hat{q}\frac{j}{[j]}\chi^{-}_{p-j}\right),$