text:
\section*{\bf{Introduction}}
Let $\mathscr{C}$ be a  braided fusion category  $\mathscr{C}$, that is,
a fusion category equipped with a braiding. Denote by $\mathscr{Z}(\mathscr{C})$ the \emph{Drinfeld center} of $\mathscr{C}$. The braided autoequivalences of $\mathscr{Z}(\mathscr{C})$ play important
roles in the study of braided fusion categories, see \cite{DN12,DGNO10,ENO10}.
For example, auto-equivalences  were used  to classify $G$-extensions of a given fusion category, see \cite{ENO10}. In order to
classify $G$-extensions of a given fusion category $\mathscr{C}$ using the
classical homotopy theory,  P. Etingof, D. Nikshych and V. Ostrik
introduced in \cite{ENO10} a 3-groupoid $\underline{\BP}(\mathscr{C})$, called the Brauer-Picard groupoid  of $\mathscr{C}$. This 3-groupoid can be truncated in the usual way into the Brauer-Picard group $\BP(\mathscr{C})$ of
$\mathscr{C}$, i.e. the group of the equivalence classes of invertible $\mathscr{C}$-bimodule categories. It turns out that there is a natural
 group isomorphism \cite[Thm 1.1]{ENO10}:
$$
\BP(\mathscr{C}) \cong  \Aut^{br}(\mathscr{Z}(\mathscr{C})),
$$
where  $\Aut^{br}(\mathscr{Z}(\mathscr{C}))$ is the group of isomorphism classes of braided autoequivalences of $\mathscr{Z}(\mathscr{C})$.
The name "Brauer-Picard group" speaks for itself that the group $\BP(\mathscr{C})$ has a close relation with the Brauer group $Br(\mathscr{C})$ of the category $\mathscr{C}$ which classifies the Azumaya algebras in $\mathscr{C}$, see \cite{VZ98}. In fact, every Azumaya algebra in $\mathscr{C}$ defines an invertible $\mathscr{C}$-bimodule category, so that $\Br(\mathscr{C})$ forms a subgroup of $\BP(\mathscr{C})$. The characterization of the Brauer group $\Br(\mathscr{C})$ in the group $\Aut^{br}(\mathscr{Z}(\mathscr{C}))$ has been done by A. Davydov and D. Nikshych in \cite{DN12}, where the braided autoequivalences corresponding to the Azumaya algebras are those trivializable on the base category $\mathscr{C}$, that is, $\Br(\baseC)\cong \Aut^{br}(\mathscr{Z}(\baseC),\baseC)$.

Now we look at braided fusion categories  from the angle of weak
Hopf algebras. Let $k$ be an algebraically closed field. It is well known that a braided fusion category $\baseC$ is equivalent to the category $_H\mathscr{M}^{fd}$ of finite dimensional modules over some finite dimensional quasitriangular semisimple weak Hopf algebra $(H,R)$ over $k$, see
\cite{ENO05,NTV03,O03}. When the weak Hopf algebra $H$ happens to be
a Hopf algebra, we know that the Brauer group of $\baseC$ is the
Brauer group $\BM(H,R)$ of $(H,R)$ consisting of Azumaya $H$-module
algebras, see \cite{VZ98}.  In this case, the Brauer group  $\BM(H,R)$ can be characterized by the quantum commutative Galois objects over the braided Hopf algebra $_RH$,
 the transmutation of the quasitriangular Hopf algebra $(H,R)$, see \cite{Zh04}. In fact, we have the following general exact sequence
of groups:
$$
1 \longrightarrow \Br(k) \longrightarrow \BM(H,R)
\longrightarrow \Gal^{qc}(_RH),
$$
where $\Gal^{qc} ({}_RH)$ is the group of quantum commutative
bi-Galois objects over ${}_RH$, and $k$ does not need to be algebraically closed. Now the question is whether the
group $\Gal^{qc}(_RH)$ is isomorphic to $\autoequivtri$, where $\baseC={}_H\mathscr{M}$. The answer
is positive, see \cite{DZ13}. The proof is based on the fact that an
autoequivalence of the comodule category over a Hopf algebra $H$ is
defined by a bi-Galois object over $H$. We don't know whether this
fact still holds for a weak Hopf algebra. However, one direction is
always true, that is, a bi-Galois object over a weak Hopf algebra
$H$ defines an autoequivalence of the comodule category over $H$.
In case $H$ is semisimple over an algebraically closed field, i.e. the braided category $_H\mathscr{M}^{fd}$ is a fusion category, we can show that both groups
$\Gal^{qc}(_RH)$ and $\autoequivtri$ are isomorphic to the Brauer
group $\BM(H,R)$, see \cite{ZZ2}. To obtain the isomorphisms, we
first construct a braided Hopf algebra $_RH$ from a quasitriangular weak Hopf algebra $(H,R)$. Unlike the Hopf algebra case, the original algebra $H$ can not be deformed into a Hopf algebra in the category of $H$-modules using Majid's transmutation theory. Here our braided Hopf algebra ${}_RH$ is nested on some centralizer subalgebra of $H$,  see \cite{LZ12}.

The next step is to use the braided Hopf algebra ${}_RH$ to describe the
Drinfeld center of the category of left $H$-modules using the category
of left ${}_RH$-comodules. Our result is the following (see Theorem 2.5).

\textbf{Theorem 1} \emph{Let $(H, R)$  be a quasitriangular weak Hopf algebra over a field $k$. Then the category  of Yetter-Drinfeld modules over $H$ is
equivalent to the category of  left  comodules over the braided Hopf algebra ${}_RH$ as a braided monoidal category.}

Following \cite[Thm 5.2]{Sc05} we know that a braided bi-Galois object $A$ over a braided Hopf algebra $\mathcal{H}$ in a braided monoidal category $\baseC$ defines an autoequivalence of the category $\baseC^\mathcal{H}$ of comodules over $\mathcal{H}$. Now we can apply this result to the braided Hopf algebra $_RH$ in the braided monoidal category $_H\mathscr{M}$ of a weak quasitriangular Hopf algebra $(H,R)$. Following Theorem 1, we know that the
category of  left comodules over ${}_RH$ is braided.  Thus a natural question arises: when is the autoequivalence defined by a braided bi-Galois object $A$ over $_RH$  a braided autoequivalence? Our answer is as follows ( see Theorem 3.6):

\textbf{Theorem 2} \emph{Let $(H, R )$  be a  quasi-triangular weak Hopf algebra over a field $k$. Assume that $A$ is a  braided bi-Galois object. Then the functor $A\Box-$ defines a braided autoequivalence of the category  of
Yetter-Drinfeld modules if and only if $A$ is quantum commutative.}

As a consequence, we obtain the following result:

\textbf{Theorem 3} \emph{Let $\mathscr{C}$ be a braided fusion category. Then the Drinfeld center of $\mathscr{C}$ is equivalent to the category of
finite dimensional left comodules over some braided Hopf algebra
${}_RH_\mathscr{C}$. If $A$ is a  braided bi-Galois object
over ${}_RH_\mathscr{C}$, then the functor $A\Box-$ defines a braided
autoequivalence of the Drinfeld center of  $\mathscr{C}$ trivializable on $\baseC$ if and only if $A$ is quantum commutative.}

The paper is organized as follows. In Section 1, we recall some
necessary definitions such as a weak Hopf algebra, a Yetter-Drinfeld
module and the Drinfeld center of a monoidal category.  In Section
2, we show that the category of Yetter-Drinfeld modules over a
quasitriangular weak Hopf algebra $(H,R)$ is equivalent to the
category of left comodules over the braided Hopf algebra $_RH$. In
Section 3, we show that a braided bi-Galois object $A$ over $_RH$
defines a braided autoequivalence of the category  of
Yetter-Drinfeld modules if and only if $A$ is quantum commutative.
Such a braided autoequivalence is trivializable on the base category $_H\mathscr{M}$. In case $(H,R)$ is semisimple and $k$ is algebraically closed, then every braided auto-equivalence of $^H_H\mathscr{YD}$ trivializable on $_H\mathscr{M}$ is given by a quantum commutative Galois object over $_RH$. The proof will be given in the forthcoming paper \cite{ZZ2} as it is a consequence of the exact sequence of the Brauer group. In the last section, we compute the braided Hopf algebras $_RH$ of the face algebras defined by Hayashi in \cite{Ha99} and the quantum
commutative Galois objects over $_RH$.

\section{\bf{Preliminaries}}

Throughout this paper $k$ is a fixed field.
 Unless otherwise stated, unadorned tensor products will be over $k$. For a coalgebra over $k$, the coproduct will be denoted by $\D $. We adopt
 Sweedler's notation for the comultiplication  in \cite{Sw69}, e.g., $\D (a)=a_{1}\o a_{2}$.

 We assume that the reader is familiar with the notions of a
(braided) monoidal category, a ribbon or a  modular category (see
 \cite{Ka95})  as well as a braided fusion category  in \cite{ENO05}.
Moreover, we make free use of the notions of algebras, bialgebras
and Hopf algebras in a braided monoidal category, see \cite{Ma95}.

\subsection{Weak Hopf algebras}
We first recall the notion of a weak Hopf algebra. For more detail
on weak Hopf algebras, the reader is referred to  \cite{BNS99}. A
\emph{weak Hopf algebra} $H$ is a $k$-algebra $(H, m, \mu )$
 and a $k$-coalgebra $(H, \D , \varepsilon )$
 such that the following axioms hold:
\begin{enumerate}
\item[(i)] \quad $\Delta(hk)=\Delta(h)\Delta(k)$,
\item[(ii)] \quad $\Delta^{2}(1)=1_{1}\o 1_{2}1_{1^{'}}\o
 1_{2^{'}}=1_{1}\o 1_{1^{'}}1_{2}\o 1_{2^{'}}$,
\item[(iii)] \quad $\varepsilon(hkl)=\varepsilon(hk_{1})\varepsilon(k_{2}l)
 =\varepsilon(hk_{2})\varepsilon(k_{1}l)$,
\item[(iv)] \quad \mbox {There exists a $k$-linear map $S: H\lr H$,
 called the \emph{antipode}, satisfying}\nonumber \\
 \quad \quad \quad $h_{1}S(h_{2})=\v (1_{1}h)1_{2}, \quad \quad
 S(h_{1})h_{2}=1_{1}\v (h1_{2}),
 \quad \quad S(h)=S(h_{1})h_{2}S(h_{3})$,
\end{enumerate} for all $h, k, l\in H$.
We have two idempotent linear maps $\varepsilon_{t},\varepsilon_{s}$:
$H\longrightarrow H$ defined respectively by
\begin{eqnarray*}
\varepsilon_{t}(h)=\varepsilon(1_{1}h)1_{2}, \quad \quad
 \varepsilon_{s}(h)=1_{1}\varepsilon(h1_{2}),
 \end{eqnarray*}
called the \emph{target map}  and the \emph{source map} respectively. Their images $H_{t}$ and $H_{s}$ are called the target space and the source space respectively.  In fact,  $H_{t}$ and $H_{s}$ are Frobenius-separable subalgebras of $H$. Moreover,  the following equations hold:
\begin{eqnarray}
&& h_{1}\o h_{2}S(h_{3})=1_{1}h\o 1_{2},\\
&& S(h_{1})h_{2}\o h_{3}=1_{1}\o h1_{2},\\
&&  h_{1}\o S(h_{2})h_{3}=h1_{1}\o S(1_{2}),
\end{eqnarray}
\begin{eqnarray}
&& h_{1}S(h_{2})\o h_{3}=S(1_{1})\o 1_{2}h,\\
&& \v (g\v _t (h))=\v (gh)=\v (\v _s (g)h),\\
&& y1_{1}\o S(1_{2})=1_{1}\o S(1_{2})y,\\
&& zS(1_{1})\o 1_{2}=S(1_{1})\o 1_{2}z,
\end{eqnarray}
for $g,h\in H, y\in H_s$ and $ z\in H_{t}$.

\begin{rem}
\begin{enumerate}
\item[(i)] A weak Hopf algebra $H$ is an ordinary Hopf
algebra if and only if $\Delta (1)=1\o 1$ if and only if $\v $ is a
homomorphism if and only if $H_t=H_s=k1_H$.
\item[(ii)] The antipode $S$ is an anti-algebra isomorphism between $H_{t} $ and $H_{s}$.
\item[(iii)]  A weak Hopf algebra $H$ is called \emph{regular} if
 $S^2 (x)=x$ for all $x\in H_tH_s$.
\item[(iv)] Every weak Hopf algebra can be obtained by twisting
the comultiplication of a regular weak Hopf algebra and keeping the same algebra structure, see \cite{Ni02}.
\end{enumerate}
\end{rem}

In what follows, a weak Hopf algebra always means a regular weak
Hopf algebra. We recall the definition of a quasitriangular weak Hopf algebra from \cite{BNS99,NTV03}.

\begin{defn}
  \emph{Let $H$ be a weak Hopf algebra with a bijective
antipode $S$.  A \emph{quasi-triangular weak Hopf algebra} is a
 pair $(H,R)$, where
 $$R=R^1\otimes
 R^2\in\Delta^{cop}(1)(H\otimes_{k}H)\Delta(1),$$ satisfies the
 following conditions:
\begin{eqnarray}
&&(id\otimes \Delta)R=R_{13}R_{12},\ \  \\
&&(\Delta\otimes id)R=R_{13}R_{23},\ \ \\
&& \Delta^{cop}(h)R=R\Delta(h),\ \ \ \
\end{eqnarray}
where $h\in H$, $R_{12}=R\otimes 1$, $R_{23}=1\otimes R$, etc. Moreover,
there exists an element $\overline{R}\in
\Delta(1)(H\otimes_{k}H)\Delta^{cop}(1)$ such that $R\overline{R}=\Delta^{op}(1)$ and $\overline{R}R=\Delta(1)$. Such an element  $R$ is often called an \emph{$R$-matrix.}  In particular, $(H,R)$ is
called a \emph{triangular weak Hopf algebra} if
$\overline{R}=R^2\otimes R^1.$}
\end{defn}

 For any $y\in H_s$ and $z\in H_t$, the  following equations hold:
\begin{eqnarray}
&&(1\o z)R=R(z\o 1),\quad \quad\quad~(y\o1)R=R(1\o y),\\
&&(z\o1)R=(1\o S(z))R,\quad \quad(1\o y)R=(S(y)\o 1)R,\\
&&R(y\o1)=R(1\o S(y)),\quad \quad R(1\o z)=R(S(z)\o1),\\
&&(\v_s\o id)(R)=\D(1),\quad\quad\quad\quad(id\o\v_s)(R)=(S\o id)\D^{cop}(1),\\
&&(\v_t\o id)(R)=\D^{cop}(1),\quad\quad\quad(id\o\v_t)(R)=(S\o
id)\D(1).
\end{eqnarray}

\subsection{Modules over weak Hopf algebras}
Let $H$ be a weak Hopf algebra. Denote by ${}_H\mathscr{M}$
the category of left   $H$-modules. Then ${}_H\mathscr{M}$ forms a monoidal category $({}_H\mathscr{M},  \o_{t}, H_{t}, a, l, r)$ as follows:
\begin{enumerate}
 \item[(i)] for any two objects $M $ and $N$ in  $
  {}_H\mathscr {M}$,  $$M\o_{t}N=\{\sum m_i\o n_i \in M\o N |\sum  \Delta(1)(m_i\o n_i)=\sum m_i\o n_i\}.$$  Clearly, $M\o_{t}N = \Delta(1)(M\o N) \subseteq M\o N;$
\item[(ii)] for any two objects $M $ and $N$ in  $
  {}_H\mathscr{M}  $, the $H$-module structure on $M\o_t N$ is as follows: \ \ $h\c (m\o_t n)= h_1\c m\o_t h_2\c n$  for all $h\in H$ and  $m\in M $ and
$n\in N $;
\item[(iii)] $H_t$ is the unit object with $H$-action $h\c z=\v_t(hz)$, where
$h\in H, z\in H_t$, and the  $k$-linear maps $l_M $, $r_M $ and
their inverses are given by
\begin{eqnarray*}
&&l_M (1_1 \c z \o 1_2\c m)=z\c m,\ \  l^{-1}_M ( m)=1_1\c 1_H \o
1_2\c m\\
&&r_M (1_1 \c m \o 1_2\c z)=S(z)\c m,\ \  r^{-1}_M ( m)=1_1\c m \o
1_2,
\end{eqnarray*}
 for any $z\in H_t$ and $m\in M$, where   $M $ is an object in  $
 {}_H\mathscr{M} $.
\end{enumerate}

If $(H,R)$ is a  quasi-triangular weak Hopf algebra, then  the category  $
 {}_H\mathscr{M} $ can be equipped with a braiding
$C$ as  follows \cite[Prop. 5.2]{NTV03}:
$$
C_{M,N}(m\o_t n)=R^2\c n\o_t  R^1\c m,\  \mbox {for all }m\in M\
\mbox {and } n\in N,
$$
where    $M $  and $N$ are any two objects in ${}_H\mathscr{M} $.

\subsection{ Yetter-Drinfeld modules and the Drinfeld center}
\begin{defn} \emph{Let $H$ be a  weak Hopf algebra. A left $H$-module $M$ is called  a left \emph{Yetter-Drinfeld module}
 if $(M, \r^L)$ is a left $H$-comodule such that the following two conditions:
\begin{eqnarray*}
&\mathrm{(i)}& \r ^L(m)=m_{[-1]}\o m_{[0]}\in H\o
 _{t} V,\\
&\mathrm{(ii)}&  (h\c m) _{[-1]} \o (h\c m) _{[0]}=h_1m _{[-1]}S(h_3) \o
h_2 \c m _{[0]},
\end{eqnarray*}
 are satisfied for all $h\in H$ and $m\in M$. For a Yetter-Drinfeld module $M$, we have the identity:
 \begin{eqnarray}
m_{[-1]} \o m _{[0]}= m _{[-1]}S(1_2) \o 1_1 \c m _{[0]}, \  \mathrm{for}\ m\in M.
\end{eqnarray}}
\end{defn}

Denote by  ${}^{H}_{H}\mathscr{YD}$ the category of  left Yetter-Drinfeld
 modules. A Yetter-Drinfeld morphism is both left $H$-linear and left
 $H$-colinear. If the antipode $S$ is bijective, then ${}^{H}_{H}\mathscr{YD} $ is a braided monoidal category with the braiding given by
\begin{eqnarray*}
C_{V,W}(v\o w)=v_{[-1]}\c w\o v_{[0]},
\end{eqnarray*}
where $v\in V \in {}^{H}_{H}\mathscr{YD}$ and $w\in  W\in
{}^{H}_{H}\mathscr{YD}$. In particular, if  $(H, R)$ is a
quasi-triangular weak Hopf algebra, then every left $H$-module $M$
is automatically a  left Yetter-Drinfeld module with the following left
coaction:
$$
 \r^L (m)= R^2 \o R^1\c m, \ \forall m\ \in M.
$$
It is easy to see that the category ${}_{H}\mathscr{M}$ is a braided
monoidal subcategory of  ${}^{H}_{H}\mathscr{YD}$.

\begin{defn}  \emph{Let $H$ be a weak Hopf algebra with a bijective antipode $S$. An algebra $A$ in $ {}^{H}_{H}\mathscr { YD}$  is called
\emph{quantum commutative} if the following equation:
\begin{eqnarray*}
xy=(x_{[-1]}\c y  )x_{[0]}
\end{eqnarray*}
holds for all $x,y \in A$.} \end{defn}

\begin{defn}  \emph{Let $H$ be a weak Hopf algebra with a bijective antipode.  The left\emph{ Drinfeld center} $\mathscr{Z}_l
({}_{H}\mathscr{M})$ of  the monoidal category ${}_{H}\mathscr{M}$  is
the category, whose objects are pairs $(U, \nu_{U,-})$, where $U$ is
an object of $ {}_{H}\mathscr{M}$ and $\nu_{U,-}$ is a natural
family of isomorphisms,  called \emph{half-braidings}:
$$
\nu_{U,V}:  U\o V  \longrightarrow V\o U, \ \forall \  V\in
{}_{H}\mathscr{M}
$$
 satisfying the Hexagon Axiom.  Similarly, one can define the right
Drinfeld center of ${}_{H}\mathscr{M}$.} \end{defn}

\begin{lem}  \emph{\cite[Thm 2.6]{CWY05}} Let $H$ be a weak Hopf
algebra with bijective antipode.  Then $\mathscr{Z}_l
({}_{H}\mathscr{M})$ is equivalent to ${}^{H}_{H}\mathscr{YD}$ as a
braided monoidal category. \end{lem}

\section{\bf{The Drinfeld center of a quasi-triangular weak Hopf algebra}}

Let $H$ be a  quasi-triangular weak Hopf algebra. In this section,
we show  that there is a braided monoidal equivalence between
the Drinfeld center of the category of left $H$-modules and   the
category of left comodules over some braided Hopf algebra.

Denote by $C_H(H_s) $ the centralizer subalgebra of $H_s$ in $H$.
Clearly,\  $ C_H(H_s) = \{1_1 h S(1_2)|  \ \forall\ h\in H\}. $ The
algebra $C_H(H_s) $ is a left $H$-module algebra with the
adjoint action: $ \ \ h\c x =h_1xS(h_2) $ for all $ h\in H$ and $x\in
C_H(H_s) . $

Now we need Majid's transmutation theory in the case of a
quasi-triangular weak Hopf algebra. Recall Theorem 3.11 from \cite{LZ12}.

\begin{lem}\label{lem2.1}  Let $(H, R)$ be a
quasi-triangular weak Hopf algebra. Then $C_H(H_s)$ is a Hopf
algebra in the braided monoidal category ${}_{H}\mathscr{M}$ with
the following structures:

\emph{(i)}   the multiplication $\overline{\mu}$ and the unit $
\overline{\eta}$ are defined by:
\begin{eqnarray*}
&& \overline{\mu}: C_H(H_s)\o_t C_H(H_s) \longrightarrow C_H(H_s),\
\ a\o_t b
 \longmapsto  (1_1\c  a) (1_2 \c b),\\
 && \ \ \ \ \quad\quad \quad   \ \overline{\eta}= Id_{H_t}: H_t  \longrightarrow C_H(H_s), \ \ x \longmapsto
 x.
\end{eqnarray*}

\emph{(ii)}  The comultiplication $\overline{\D} $ and the counit $
\overline{\v}$ are given by:
\begin{eqnarray*}
&&\overline{\D}: C_H(H_s) \longrightarrow C_H(H_s) \o_t C_H(H_s),\ \
x \longmapsto
x_1S(R^2) \o R^1\c x_2,   \\
&& \quad \quad \quad  \overline{\v}= \v_t:  C_H(H_s) \longrightarrow
H_t,\ \ \ \ x \longmapsto \v_t (x).
\end{eqnarray*}

\emph{(iii)} The antipode is $\overline{S}$ defined by
\begin{eqnarray*}
\overline{S}: C_H(H_s)\longrightarrow C_H(H_s),\ \ x\longmapsto R^2
R'^2S(R^1xS(R'^1)).
\end{eqnarray*}
Moreover, ${}_RH$ is  cocommutative cocentral in the sense of
\emph{\cite{Sc10}}. \end{lem}

A Hopf algebra in a braided monoidal category is usually called a braided Hopf algebra in case the category does not need to be mentioned. In the sequel, we shall call the Hopf algebra  $C_H(H_s)$ in  ${}_{H}\mathscr{M}$  a \emph{ braided Hopf algebra} and denote it by ${}_RH$.

\begin{defn} \emph{\cite{Ma95}} \emph{Let $H$ be a  quasitriangular
weak Hopf algebra. Let $M$ be a left $H$-module. We call $(M, \r^l)$
a \emph{left $_RH$-comodule} in the category $\HM$ if
 $(M, \r^l)$ is a left $_RH$-comodule such that $\r^l $ is left
$H$-linear, i.e.,
 $$\r^l( h\c m) = h_1\c  m_{(-1)} \o h_2\c  m_{(0)},\ \forall h\in H, \ m\in M.$$}
\end{defn}
Similarly, one can  define a right  $_RH$-comodule and an $_RH$-bicomodule in the category $\HM$. For convenience, in the sequel,  {\it a left (right, bi-) $_RH$-comodule in the category $\HM$  will be called a left (right, bi-) $_RH$-comodule for short}.

Let $(M, \r^l)$ and $(N, \r^l)$ be two left ${}_R H$-comdules.
The tensor product $M\o_t N$ is a left ${}_R H$-comdule with the following
comodule structure:
$$
h\c (m\o n)=h_1\c m\o h_2\c n,\ \ \ \r^l (m\o n) = (\overline{\mu}\o
1\o 1)(1 \o C \o 1)(\r^l \o \r^l) (m\o n),
$$
where  $m\in M$, $n\in N$, $h\in H$ and $C$ is the braiding  in
${}_{H}\mathscr{M}$.

Denote  by  ${} ^{_RH} (_{H}\mathscr{M})$ the category of left
$_RH$-comodules. Note that a morphism in  ${} ^{_RH} (_{H}\mathscr{M})$ is both left $H$-linear and left  $_RH$-colinear. It is easy to see that the category ${}^{_RH} (_{H}\mathscr{M})$ is a monoidal category with the unit object given by $H_t$.

Now we discuss  the relation between the category  ${} ^{_RH}
(_{H}\mathscr{M})$ and the category of left Yetter-Drinfeld $H$-modules.

\begin{lem}\label{lem2.3}  Let $H$ be a  quasitriangular weak Hopf algebra. If  $(M, \r^l  )$ is a left $_RH$-comodule,
then $M$ is a left Yetter-Drinfeld $H$-module with the following
$H$-comodule structure:
\begin{eqnarray*}
&& \r^L ( m)=  m_{(-1)}R^2\o  R^1 \c m_{(0)}\in H\o M,
\end{eqnarray*}
where $\r^l (m) =m_{(-1)} \o m_{(0)}$ for all $m\in M.$ \end{lem}
\proof For any $m\in M,$ we first have
$$
 1_1m_{(-1)}R^2\o  1_2R^1 \c m_{(0)} = m_{(-1)}1_1R^2\o  1_2R^1 \c
 m_{(0)}=m_{(-1)}R^2\o R^1 \c m_{(0)}.
$$
 So  $ \r^L( M)  \in H\o_t M$. Namely, $\r^L$ is well-defined.

Next we verify   that $(M, \r^L)$ is a left $H$-comodule. For the
coassociativity, we have:
\begin{eqnarray*}
(1\o \r^L)\r^L &=&(1\o \r^L)(m_{(-1)}R^2\o R^1 \c m_{(0)})\\
&=&m_{(-1)}R^2\o  (R^1 \c m_{(0)})_{(-1)}q^2  \o  q^1\c (R^1 \c m_{(0)})_{(0)}\\
&=&m_{(-1)}R^2\o  (R^1_1 \c m_{(0)_{(-1)}})q^2  \o  q^1\c (R^1_2 \c m_{(0)})\\
&=&m_{(-1)_1}S(r^2)R^2\o  (R^1_1r^1 \c m_{(-1)_2 })  q^2  \o  q^1  R^1_2 \c m_{(0)}\\
&\stackrel{(9)}{=}&m_{(-1)_1}S(r^2)p^2R^2\o  (p^1r^1 \c m_{(-1)_2 })  q^2  \o  q^1  R^1 \c m_{(0)}\\
&\stackrel{(8)}{=}&m_{(-1)_1}\v_s(r^2) R^2\o  (r^1 \c m_{(-1)_2 })  q^2  \o  q^1  R^1 \c m_{(0)}\\
&\stackrel{(14)}{=}&m_{(-1)_1}1_1 R^2\o  (S(1_2) \c m_{(-1)_2 })
q^2  \o  q^1  R^1 \c m_{(0)}\\
&=&m_{(-1)_1}1_1 R^2\o  (  m_{(-1)_2 }S^2(1_2))  q^2  \o  q^1  R^1 \c m_{(0)}\\
&=&m_{(-1)_1}1_1 R^2\o  (  m_{(-1)_2 }1_2)  q^2  \o  q^1  R^1 \c m_{(0)}\\
&=&m_{(-1)_1} R^2\o  m_{(-1)_2 }  q^2  \o  q^1  R^1 \c m_{(0)}\\
&=&(m_{(-1)_1}R^2_1\o (m_{(-1)_2}R^2_2 \o R^1 \c m_{(0)})\\
&=&(\D\o 1)(m_{(-1)}R^2\o R^1 \c m_{(0)})=(\D\o 1)\r^L (m).
\end{eqnarray*}
The counit axiom holds as well because we have:
\begin{eqnarray*}
(\v\o 1)\r^L (m) &=&\v (m_{(-1)}R^2) (R^1 \c m_{(0)})
\stackrel{(5)}{=}\v (m_{(-1)}\v_t (R^2)) (R^1 \c m_{(0)})\\
&\stackrel{(15)}{=}&\v (m_{(-1)}1_2) (S (1_1) \c m_{(0)})=\v (m_{(-1)}S(1_1)) (1_2 \c m_{(0)})\\
&=&\v ( m_{(-1)}1_1) (1_2 \c m_{(0)})=\v ( 1_1m_{(-1)}) (1_2 \c
m_{(0)})=m,
\end{eqnarray*}
where the last equality follows from the counit of a left
$_RH$-comodule, namely,
$$
l\circ(\v_t\o 1)( m_{(-1)}\o  m_{(0)})=\v_t (m_{(-1)}) \c m_{(0)}=m.
$$

Finally, the compatible condition holds since
\begin{eqnarray*}
 h_1( m_{(-1)}R^2)\o h_2\c [R^1 \c m_{(0)}]
&=&h_11_1 m_{(-1)}S(1_2)R^2\o  h_2R^1 \c m_{(0)}\\
&\stackrel{(3)}{=} & h_1 m_{(-1)}S(h_2)h_3R^2\o  h_4R^1 \c m_{(0)}\\
&\stackrel{(10)}{=}& h_1 m_{(-1)}S(h_2)R^2h_4\o  R^1h_3 \c m_{(0)}\\
&=&(h_1\c  m)_{(-1)}R^2 h_2\o  R^1\c (h_1\c  m)_{(0)}.
\end{eqnarray*}
for all $m\in M$ and $h\in H.$   \qed

The following lemma says that the converse of Lemma 2.3 is also
true.

\begin{lem} \label{lem2.4}
Let $H$ be a  quasitriangular weak Hopf algebra
with an antipode $S$. If $(N, \r^L  )$ is a left Yetter-Drinfeld
module,  then $N$ is a left $_RH$-comodule with the following
structure:
\begin{eqnarray*}
&&  \r^l (n) = n _{[-1]}S(R^2) \o R^1\c n _{[0]},
\end{eqnarray*}
where $\r^L (n)= n _{[-1]} \o n _{[0]} $ for all $n\in N$. \end{lem}
\proof First of all, we need to  check that $\r^l$ is well-defined.
For any $n\in N$,
\begin{eqnarray*}
 1_1 [n _{[-1]}S(R^2)S(1_2) \o R^1\c n _{[0]}
&=&1_1n _{[-1]}S(1_2R^2) \o R^1\c n _{[0]}\\
& \stackrel{(11)}{=}&1_1n _{[-1]}S(R^2) \o R^11_2\c n _{[0]}\\
&=&1_1n _{[-1]}S(R^2) \o R^1\c (1_2\c n _{[0]})\\
&=&n _{[-1]}S(R^2) \o R^1\c  n _{[0]};\\
\ \ \ 1_1\c  [n _{[-1]}S(R^2)] \o 1_2R^1\c n _{[0]}&=& [n _{[-1]}S(R^2)]S(1_1) \o 1_2R^1\c n _{[0]}\\
&=& [n _{[-1]}S(1_1R^2)] \o 1_2R^1\c n _{[0]}\\
&=& [n _{[-1]}S(R^2)] \o R^1\c n _{[0]}.
\end{eqnarray*}
So $\r^l (N)\subset {}_RH\o_t N$. The $H$-linearity of  the map
$\r^l$ follows from the equations below:
\begin{eqnarray*}
h_1\c  [n _{[-1]}S(R^2)] \o h_2R^1\c n _{[0]}
&=& h_1n _{[-1]}S(R^2)S(h_2) \o h_3R^1\c n_{[0]}\\
&=&h_1n _{[-1]}S(h_2R^2) \o h_3R^1\c n _{[0]}\\
&\stackrel{(10)}{=}&h_1n _{[-1]}S(R^2h_3) \o R^1h_2\c n _{[0]}\\
&=&(h_1n _{[-1]}S(h_3))S(R^2) \o R^1\c (h_2\c n _{[0]})\\
&=&(h\c n) _{[-1]}S(R^2) \o R^1\c (h\c n) _{[0]}=\r^l(h\c n),
\end{eqnarray*}
for all $h\in H$. Now we show that $(N, \r^l)$ is a left $_RH$-comodule. For any $n\in
N,$
\begin{eqnarray*}
(1\o \r^l) \r^l (n)
&=& n _{[-1]}S(R^2) \o (R^1\c n _{[0]})_{[-1]} S(r^2)\o r^1\c (R^1\c n _{[0]})_{[0]}\\
&=& n _{[-1]}S(R^2) \o R^1_1 n _{[0]_{[-1]}} S(R^1_3) S(r^2)\o r^1\c (R^1_2\c n_{[0]_{[0]}})\\
&=& n _{[-1]_1}S(R^2) \o R^1_1 n _{[-1]_{2}} S(r^2R^1_3) \o r^1R^1_2\c n _{[0]}\\
& \stackrel{(10)}{=}& n_{[-1]_1}S(R^2) \o R^1_1 n _{[-1]_{2}} S(R^1_2r^2) \o R^1_3r^1\c n _{[0]}\\
& \stackrel{(9)}{=}& n _{[-1]_1}S(R^2q^2) \o R^1\c  [n _{[-1]_{2}} S(r^2)] \o q^1r^1\c n _{[0]}\\
& \stackrel{(8)}{=}& [n _{[-1]_1}S(r^2_2)]S(R^2) \o R^1\c  [n _{[-1]_{2}} S(r^2_1)] \o r^1\c n _{[0]}\\
&=& [n _{[-1]}S(r^2)]_1S(R^2) \o R^1\c  [n _{[-1]} S(r^2)]_2 \o
r^1\c n _{[0]} \end{eqnarray*}
\begin{eqnarray*}
&=& \overline{\D}[n _{[-1]}S(r^2)]\o r^1\c n _{[0]}=(\overline{\D}\o
1) \r^l (n).
\end{eqnarray*}
Hence the coassociativity holds. Finally, we verify that $\v_t$ satisfies the counit axiom:
\begin{eqnarray*}
&& \v_t (n _{[-1]}S(R^2))\c  (R^1\c n _{[0]})\\
&=&(\v_t (n _{[-1]}S(R^2))R^1)\c n _{[0]}
 \stackrel{(9)}{=}(1_2R^1)\c n _{[0]}\v (1_1n _{[-1]}\v_t [S(R^2)])\\
&=&(1_2R^1)\c n _{[0]}\v (1_1n _{[-1]}S[\v_s (R^2)])\stackrel{(14)}{=}(1_2S(1'_2))\c n _{[0]}\v (1_1n _{[-1]}S(1'_1))\\
&=&(1_21'_1)\c n _{[0]}\v (1_1n _{[-1]}1'_2)=(1_21'_1)\c n _{[0]}\v (1_1n _{[-1]}S(1'_2))\\
&=&1_2\c n _{[0]}\v (1_1n _{[-1]}S(1_3))=n.
\end{eqnarray*}

Therefore,   $(N, \r^l)$ is a left $_RH$-comodule. \qed

Combining Lemma \ref{lem2.3} and Lemma \ref{lem2.4}, we obtain the following theorem.

 \begin{thm} \label{thm2.5}
 Let $(H, R )$  be a  quasitriangular weak Hopf algebra.   Then
there is a monoidal equivalence  $\mathcal{F}$ from the category ${}
^{_RH} (_{H}\mathscr{M})$ of left $_RH$-comodules to the category
${}^{H}_{H}\mathscr{YD}$ of left Yetter-Drinfeld modules:
$$
\mathcal{F} : \  {} ^{_RH} (_{H}\mathscr{M}) \longrightarrow
{}^{H}_{H}{\mathscr{Y}}{\mathscr{D}},  \ (M,\r^l)  \longmapsto  (M,  \r^L),
$$
where  $\r^L$ is defined in Lemma \ref{lem2.3}. The quasi-inverse of  $\mathcal{F}$ is
$$
\mathcal{G} : \ {}^{H}_{H}\mathscr{{YD}} \longrightarrow  {} ^{_RH}
(_{H}\mathscr{M}),  \ (N, \r^L) \longmapsto  (N, \r^l ),
$$
where $\r^l$ is defined in Lemma \ref{lem2.4}. \end{thm}

\proof  We show first that $ \mathcal{G}  \mathcal{F}(M)=M$ for any
object $M$ in ${} ^{_RH} (_{H}\mathscr{M})$.  It is enough to verify
that $\r^l (m)= m_{(-1)}\o m_{(0)}$ for all $m\in M.$  Indeed,
\begin{eqnarray*}
\r^l(m)&=&  m _{[-1]}S(R^2) \o R^1\c m
_{[0]}\\
&=&  m_{(-1)}r^2  S(R^2) \o R^1\c [r^1 \c m_{(0)}]=  m_{(-1)}r^2  S(R^2) \o (R^1r^1) \c m_{(0)}\\
&\stackrel{(8)}{=}&  m_{(-1)}\v_t(R^2) \o R^1 \c m_{(0)}\stackrel{(15)}{=} m_{(-1)}1_2 \o S(1_1) \c m_{(0)}\\
&=&  S^{-1} (1_2)\c m_{(-1)} \o S(1_1) \c m_{(0)}=  S  (1_2)\c m_{(-1)} \o S(1_1) \c m_{(0)}\\
&=&  1_1\c m_{(-1)} \o 1_2 \c m_{(0)}=   m_{(-1)} \o  m_{(0)}.
\end{eqnarray*}

Next we show that  $ \mathcal{F}\mathcal{G}(N)  =N$  for any object
of  ${}^{H}_{H}\mathscr{YD}$. For all $n\in N,$
\begin{eqnarray*}
\r^L (n)&=& n_{(-1)}R^2\o  R^1 \c n_{(0)}= n_{ [-1]} S(r^2) R^2\o  R^1 \c (r^1\c n_{ [0]})\\
&=& n_{ [-1]} S(r^2) R^2\o  (R^1r^1) \c  n_{ [0]}\stackrel{(8)}{=} n_{ [-1]} \v_s( R^2) \o  R^1 \c   n_{ [0]} \\
&\stackrel{(14)}{=}& n_{ [-1]} 1_1 \o  S(1_2) \c   n_{ [0]} =n_{ [-1]} S(1_2) \o  1_1 \c   n_{ [0]} \\
&=& 1'_1n_{ [-1]} S(1_2) \o  1_1 \c ( 1'_2\c  n_{ [0]}) =1_1n_{ [-1]} S(1_3) \o  1_2 \c  n_{ [0]} \\
&=& n_{ [-1]} \o   n_{ [0]}.
\end{eqnarray*}

Finally, we verify that the triple $(\mathcal{G}, Id, Id)$ is
monoidal. It is clear that $\mathcal{G}(H_t)=H_t.$   For any two
left Yetter-Drinfeld modules $U$ and $V$,  the left ${_RH}$-comodule
structure  on $\mathcal{G}(U)\o \mathcal{G}(V)$ is as follows:
\begin{eqnarray*}
&\ \ \ & (\mu\o 1\o 1)(1 \o C \o 1)(\r^L \o \r^L) (u\o v) \\
 &=& (\mu\o 1\o 1)(1 \o C \o 1) (u_{(-1)} \o u_{(0)}\o n_{(-1)}\o
 v_{(0)})\\
  &=& (\mu\o 1\o 1) (u_{(-1)} \o R^2\c v_{(-1)}\o R^1\c u_{(0)} \o
  v_{(0)})\\
  &=& u_{(-1)}(R^2\c v_{(-1)})\o R^1\c u_{(0)} \o
 v_{(0)},
\end{eqnarray*}
where $u\in U$ and $v\in V.$ Now we have
\begin{eqnarray*}
&&  u_{(-1)}(R^2\c v_{(-1)})\o R^1\c u_{(0)} \o
 v_{(0)}\\
&=&(u _{[-1]}S(p^2))R^2{}_1 (v _{[-1]}S(q^2)) S(R^2{}_2) \o R^1 \c (p^1\c u_{[0]}) \o  q^1\c v_{[0]}\\
&\stackrel{(8)}{=}&(u _{[-1]}S(p^2))r^2 (v _{[-1]}S(q^2)) S(R^2) \o (R^1r^1 p^1)\c u_{[0]}) \o  q^1\c v_{[0]}\\
&\stackrel{(8)}{=}&u _{[-1]}\v_s(r^2) (v _{[-1]}S(q^2)) S(R^2) \o (R^1r^1 )\c u_{[0]}) \o  q^1\c v_{[0]}\\
&\stackrel{(14)}{=}&u _{[-1]}S(1_2) (v _{[-1]}S(q^2)) S(R^2) \o (R^11_1 )\c u_{[0]}) \o  q^1\c v_{[0]}\\
&=&u _{[-1]}S(1_2) (v _{[-1]}S(q^2)) S(R^2) \o R^1\c (1_1 \c u_{[0]}) \o  q^1\c v_{[0]}\\
&=&u _{[-1]} (v _{[-1]}S(q^2)) S(R^2) \o R^1\c  u_{[0]} \o  q^1\c v_{[0]}\\
&=&(u _{[-1]}v _{[-1]})S(R^2q^2) \o R^1\c  u_{[0]} \o  q^1\c v_{[0]}\\
&\stackrel{(9)}{=}& (u _{[-1]}v _{[-1]})S(R^2) \o R^1\c  (u_{[0]} \o   v_{[0]})\\
&=&(u\o_t v) _{[-1]}S(R^2) \o R^1\c (u\o_t v) _{[0]}=\r^l(u\o_t v).
\end{eqnarray*}
Hence, $ \mathcal{G}(U\o V) =\mathcal{G}(U)\o \mathcal{G}(V). $ The
verification of the other axioms for a monoidal functor are obvious.  \qed

Since the category of Yetter-Dinfeld modules is braided, the
equivalence $\mathcal{G}$ in Theorem 2.5 induces a braiding in the category of
left ${}_RH$-comodules such that the equivalence becomes braided.

\begin{cor}\label{cor2.6}  Let $(H, R )$  be a  quasitriangular weak
Hopf algebra.  Then the category of left
$_RH$-comodules is a braided monoidal category with a braiding
$\widetilde{C}$ given by
\begin{equation}\label{newbraiding}
 \widetilde{C }(u\o v) = u _{(-1)}R^2\c  v \o R^1\c u _{(0)}, \ \forall u\in U, \forall v\in V,
\end{equation}
where $U$ and $V $ are any two left ${}_RH$-comodules. The inverse
of $\widetilde{C }$ is given by
\begin{eqnarray*}
&& \widetilde{C}^{-1} ( v\o u)= R^1\c u_{(0)}\o S^{-1}(u_{(-1)}R^2)
\c v.
\end{eqnarray*}
Moreover,  the functor $\mathcal{G}$ in Theorem 2.5 gives a braided
monoidal equivalence. \end{cor}

\proof  Consider the following commutative diagram of isomorphisms:
$$
\begin{picture}(250,60)
\thinlines \put(38,45){\shortstack{$\mathcal{G}(U)\o _t
\mathcal{G}(V)
  $}}
\put(110,47){\vector(1,0){100}}
\put(220,45){\shortstack{$\mathcal{G}(U\o_t V))$}}
\put(65,42){\vector(0,-1){30}}
 \put(70,26){\shortstack{${C}_{
\mathcal{G}(U), \mathcal{G}(V)}$}} \put(200,26){\shortstack{$
\mathcal{G}(C_{U,V}) $}} \put(38,0){\shortstack{$\mathcal{G}(V) \o_t
\mathcal{G}(U)$}} \put(210,2){\vector(-1,0){100}}
\put(220,0){\shortstack{$\mathcal{G}(V \o_t U),$}}
\put(247,42){\vector(0,-1){30}}
\end{picture}\\
$$
where the horizontal isomorphisms  are given by $Id:  \mathcal{G}(X)\o \mathcal{G}(Y)\cong \mathcal{G}(Y\o X)$. Thus, the braiding
$\widetilde{C }$ is just the composition $Id^{-1} \circ C_{U,V} \circ
Id$. In fact, we have
\begin{eqnarray*}
  \widetilde{C}_{U, V} (u\o v)
   &=&Id  \circ  C_{U,V} \circ Id (u\o v)\\
   &=&Id  \circ  C_{U,V}  (u\o v)\\
   &=&Id   (u_{[-1]}\c v\o u_{[0]})\\
   &=&  u_{(-1)}R^2\c v\o R^1\c u_{(0)}.
\end{eqnarray*}
Similarly, one can obtain the inverse of $\widetilde{C}$. \qed

By Lemma 1.6 and Corollary 2.6 we obtain the following corollary.

\begin{cor} Let $(H, R )$  be a  quasitriangular weak Hopf algebra.
Then the Drinfeld center $\mathscr{Z}_l(_{H}\mathscr{M})$ of left
$H$-modules is equivalent to the category ${} ^{_RH}
(_{H}\mathscr{M})$  of left $_RH$-comodules as a braided monoidal
category.
\end{cor}

 As a special case,  we have the following corollary on a quasitriangular Hopf algebra:

\begin{cor} \label{Bos} Let $(H, R )$  be a quasitriangular Hopf algebra.  Then
the Drinfeld center of left $H$-modules is equivalent to the
category of left $_RH$-comodules as a braided monoidal category.
\end{cor}

\begin{rem} \emph{(i) when  $H$ is a finite dimensional quasitriangular
Hopf algebra, the functor $\mathcal{G}$ was first proved in
\cite{Zh04} to have a right adjoint.}

\emph{(ii)  Let $H$ be a finite dimensional quasitriangular Hopf
algebra. Following \cite[Prop 4.1]{Maj93} the quantum double $D(H)$
is isomorphic to a semidirect product $A\rtimes H$, where $A=H^*$ is
a braided Hopf algebra. By Corollary \ref{Bos} we may choose $A$ as the dual braided Hopf algebra $(_RH)^*$. Thus we have the following equivalences of braided monoidal categories:
$$
{} _{(_RH)^*\rtimes H}\mathscr{M} \cong  {}_{D(H)}\mathscr{M} \cong ^{H}
_{H}\mathscr{YD}\cong  \mathscr{Z}_l(_{H}\mathscr{M}).
$$
In case $H$ is infinite dimensional, we  have neither the usual quantum double $D(H)$ nor the dual braided Hopf algebra $(_RH)^*$. But Corollary \ref{Bos} always holds for any (finite or
infinite dimensional) quasitriangular  Hopf algebra over any field
(or even over a commutative ring). In particular, the Drinfeld center
is naturally equivalent to the category of comodules over
$BU_q(\mathfrak{g})$ studied in \cite{Maj93}.}
\end{rem}

\section{\bf{Quantum commutative Galois objects}}
In this section we study (braided) Galois objects over the Braided Hopf algebra $_RH$ of a  finite dimensional quasitriangular weak Hopf algebra $(H, R)$. We shall construct  braided autoequivalences of the Drinfeld center of ${}_H \mathscr{M}$  from braided bi-Galois objects.
For the details about braided Galois objects over a braided Hopf
algebra one is referred to \cite{Sc05, Sc10}.

Let $(H, R)$ be a finite dimensional quasitriangular weak Hopf algebra.  An object $X$ in ${}_H \mathscr{M}$ is flat if tensoring with $X$ preserves equalizers. A flat object $X$ is called \emph{faithfully flat} if tensoring with $X$ reflects isomorphisms. It is not hard to see that ${}_RH$ is
flat in the category ${}_H \mathscr{M}$ since ${}_RH$ is finite and
has a dual object.

\begin{defn} \emph{\cite{Sc05}} \emph{An algebra $A$ in   ${}_H
\mathscr{M}$   is called a \emph{left ${}_RH$-comodule algebra} if
$A$ is a left ${}_RH$-comodule such that the left comodule map $\rho^l$
satifies:
 $$
\r^l(ab)= a_{(-1)}(R^2\c b_{(-1)}) \o  (R^1\c a_{(0)})b_{(0)},
$$
for all  $a,b\in A$, where   $\r^l (a)=a_{(-1)}\o a_{(0)}$ . Namely,
$\r^l$ is an algebra map in ${}_H \mathscr{M}$. }
\end{defn}
 Similarly, an
algebra $A$ in   ${}_H \mathscr{M}$ is called a \emph{right
${}_RH$-comodule algebra} if $A$  with a right ${}_RH$-coaction
$\r^r$ is a right ${}_RH$-comodule such that
 $$
\r^r(ab)= a_{(0)} (R^2\c b_{(0)}) \o  (R^1\c a_{(1)})b_{(1)}, $$
where $a,b\in A$ and  $\r^r (a)=a_{(0)}\o a_{(1)}$. An
\emph{${}_RH$-bicomodule algebra} is both a left and a right $_RH$-comodule algebra such that the left and the right coations commute.

Now let $A$  be a right ${}_RH$-comodule algebra. The subalgebra
$$A_0= \{a\in A| \r^r (a)=a \o_t 1 =1_1a\otimes 1_2 \}
$$
is called the \emph{coinvariant subalgebra}. Similarly, one can define the  coinvariant subalgebra of a left ${}_RH$-comodule algebra. An $_RH$-coinvariant subalgebra $A_0$ is said to be \emph{trivial} if $A_0=H_t$.

\begin{defn} \emph{ \cite[Defn 2.1]{Sc10}  Let $A$  be a right
${}_RH$-comodule algebra.  $A$ is called a \emph{right braided
${}_RH$-Galois object} if $ A$ is  faithfully flat and   the
morphism
$$
\b : A \o _t A \longrightarrow A \o_t {}_RH, \ a\o_t b \longmapsto
ab_{(0)} \o_t b_{(1)}
$$
is an isomorphism.  Similarly, one  can define a \emph{left braided
${}_RH$-Galois object} and a \emph{braided bi-Galois object}.
}\end{defn}

The coinvariant subalgebra $A_0$ of a  right ${}_RH$-Galois object $A$ is trivial. So is the coinvariant subalgebra
 of a  left ${}_RH$-Galois object $A$.  Moreover, it is not hard to see that $({}_R H, \t_{{}_R H, -})$ is an object in
the Drinfeld center   $ \mathscr{Z}_l({}_H\mathscr{M})$, where
$\t_{{}_R H, -}$ is a half-braiding
\begin{eqnarray*}
  && \t_{{}_R H, M} :  {}_R H\o  M  \longrightarrow   M \o {}_R H, \  h\o m\longmapsto  r^2 R^1\c m \o r^1 h
  R^2.
  \end{eqnarray*}
Since ${}_RH$ is cocommutative cocentral, for any left
${}_RH$-comodule  $ (M, \r^l)$,  by   \cite{Sc10} there exists a natural right comodule structure induced by the half-braiding $\t_{{}_R H,
M}:$
$$
\r^r= \t_{{}_R H, M} \circ \r^l: M \longrightarrow {}_RH \o M
\longrightarrow M \o {}_RH,
$$
so that $(M,\r^l, \r^r )$ becomes an ${}_RH$-bicomodule. By \cite{Sc10} we
call $M$ \emph{cocommutative} if the right ${}_RH$-comodule is
induced by the left ${}_RH$-comodule as above.

\begin{defn} \emph{A cocommutative braided bi-Galois object $A$
is called a \emph{quantum commutative Galois object }if $A$ is
quantum commutative as an algebra in  ${}^{H}_{H}\mathscr{YD}$.}
\end{defn}

By Theorem 2.5 and Corollary 2.6, a left Yetter-Drinfeld
module is an $_RH$-bicomodule in $_H\mathscr{M}$. Thus we can consider the contensor product $M \Box_{_RH} N$, or $M\Box N$ for convenience, for two left Yetter-Drinfeld modules $M$ and $N$:
$$M\Box N=\{m\o_t n\in M \o_t N| \r^r (m) \o_t n = m \o_t \r^l (n)\},$$
or precisely,
\begin{eqnarray}
M\Box N= \{m\o n \in M\o_t N| r^2 \c m _{[0]}  \o r^1m _{[-1]} \o
n= m\o n _{[-1]}S(R^2) \o R^1\c n
 _{[0]} \}.
\end{eqnarray}
If $A$ is a braided ${}_RH$-bi-Galois object,   by \cite{Sc05} we have  an
isomorphism:
$$
\xi: (A\Box M )\o_t  (A\Box N ) \cong A \Box (M\o_t N),
$$
given by $\xi ((a\o m ) \o (b\o n) ) = a   (R^2\c b) \o R^1\c m \o n$,
for all $a,b \in A$, $m\in M$ and $b\in N$. Following \cite{Sc10}
the cotensor functor $A\Box-$ is a monoidal autoequivalence of  ${}
^{_RH} (_{H}\mathscr{M})$.

\begin{lem}\label{lem3.4}  Let $(H, R )$  be a  finite dimensional
quasitriangular weak Hopf algebra. If $A$ is a quantum commutative
Galois object, then the functor $A\Box-$ is a braided
autoequivalence of  ${} ^{_RH} (_{H}\mathscr {M}).$ \end{lem}

\proof  Let $A$ be a quantum commutative Galois object. By Theorem
2.5 and \cite{Ka95} it suffices to verify that the following diagram
is commutative:
$$
\begin{picture}(250,60)
\thinlines \put(15,45){\shortstack{$(A\Box M)\o _t (A\Box N)
  $}}
\put(110,47){\vector(1,0){100}} \put(220,45){\shortstack{$A\Box
(M\o_t N))$}} \put(65,42){\vector(0,-1){30}}
 \put(70,26){\shortstack{$\widetilde{C }_{
A\Box M, A\Box N}$}} \put(200,26){\shortstack{$ A\Box \widetilde{C
}_{M,N} $}} \put(320,26){\shortstack{$ (*) $}}
\put(15,0){\shortstack{$(A\Box N) \o_t (A\Box M)$}}
\put(110,2){\vector(1,0){100}} \put(220,0){\shortstack{$A\Box (N\o_t
M)$}} \put(247,42){\vector(0,-1){30}}
\end{picture}\\
$$
Indeed, on the one hand, for any $a\o m \in A\Box M$ and $b\o
n\in A\Box N$, we have:
\begin{eqnarray*}
&& [ \xi \circ (\widetilde{C }_{
A\Box M, A\Box N})][(a\o m ) \o (b\o n) ]\\
&= &  \xi [(a\o m )_{(-1)} r^2\c (b\o n)
\o  r^1 \c (a\o m )_{(0)}]\\
&= &  \xi [a_{(-1)} r^2\c  (b\o n)
\o  r^1 \c (a_{(0)}\o m )]\\
&=  &  \xi [a_{(-1)_1} r^2_1\c b \o a_{(-1)_2} r^2_2\c n
\o  r^1_1 \c a_{(0)}\o r^1_2 \c m )]\\
&=  &  [a_{(-1)_1} r^2_1\c b] [ R^2r^1_1 \c a_{(0)}]\o R^1a_{(-1)_2} r^2_2\c n \o r^1_2 \c m \\
&=  &  [a _{[-1]_1}S(q^2_2)] r^2_1\c b] [ R^2r^1_1 \c [q^1\c a _{[0]}]]\o R^1[a _{[-1]_2}S(q^2_1)] r^2_2\c n \o r^1_2 \c m \\
&=  &  [a _{[-1]_1}[S(q^2)r^2]_1\c b] [ R^2r^1_1 q^1\c a _{[0]}]]\o R^1[a _{[-1]_2}[S(q^2)r^2]_2] \c n \o r^1_2 \c m \\
&\stackrel{(9)}{=}&  [a _{[-1]_1}[S(q^2)r^2p^2]_1\c b] [ R^2r^1 q^1\c a _{[0]}]]\o R^1[a _{[-1]_2}[S(q^2)r^2p^2]_2] \c n \o p^1 \c m \\
&\stackrel{(8)}{=}&  [a _{[-1]_1}[\v_s(r^2)p^2]_1\c b] [ R^2r^1 \c a _{[0]}]]\o R^1[a _{[-1]_2}[\v_s(r^2)p^2]_2] \c n \o p^1 \c m \\
&\stackrel{(14)}{=}&  [a _{[-1]_1}[1_1p^2]_1\c b] [ R^2S(1_2) \c a _{[0]}]]\o R^1[a _{[-1]_2}[1_1p^2]_2] \c n \o p^1 \c m \\
&=&  [a _{[-1]_1}p^2_1\c b] [ R^2S(1_2) \c a _{[0]}]]\o R^1[a _{[-1]_2}1_1p^2_2] \c n \o p^1 \c m \\
&\stackrel{(16)}{=} &  [a _{[-1]_1}p^2_1\c b] [ R^2\c a _{[0]}]\o
R^1[a _{[-1]_2}p^2_2] \c n \o p^1 \c m ,
\end{eqnarray*}
where   Corollary 2.6  and Lemma 2.4 were used in the first and
fifth equality, respectively. On the other hand, we have:
\begin{eqnarray*}
&& (1 \o \widetilde{C }) \circ \xi [(a\o m ) \o (b\o n)]
\end{eqnarray*}
\begin{eqnarray*}
 &= &  a
(r^2\c b) \o \widetilde{C } (r^1\c m
\o n )\\
&= &  a   (r^2\c b) \o (r^1\c m)_{(-1)}W^2 \c n \o
W^1\c (r^1\c m)_{(0)}\\
&= &  a   (r^2\c b) \o (r^1_1\c m_{(-1)})W^2 \c n \o
W^1r^1_2\c  m_{(0)}\\
&= &  a   (r^2\c b) \o r^1_1 m_{(-1)}S(r^1_2)W^2 \c n \o
W^1r^1_3\c  m_{(0)}\\
&= &  a   (r^2\c b) \o r^1_1 m_{[-1]}S(R^2)S(r^1_2)W^2 \c n \o
W^1r^1_3R^1\c  m_{[0]}\\
&= &  a   (r^2\c b) \o r^1_1 m_{[-1]}S(r^1_2R^2)W^2 \c n \o
W^1r^1_3R^1\c  m_{[0]}\\
&= &  a   (r^2\c b) \o r^1_1 m_{[-1]}S(r^1_3)S(R^2)W^2 \c n \o
W^1R^1r^1_2\c  m_{[0]}\\
&= &  a   (r^2\c b) \o r^1_1 m_{[-1]}S(r^1_3)  \c n \o
r^1_2\c m_{[0]}\\
&= &  a   (r^2\c b) \o r^1_1 m_{[-1]}S(1_2)S(r^1_3)  \c n \o
r^1_21_1\c m_{[0]}\\
&\stackrel{(14)}{=}&  a   (r^2\c b) \o r^1_1
m_{[-1]}S(R^2)p^2S(r^1_3) \c n \o
r^1_2p^1R^1\c m_{[0]}\\
&\stackrel{(18)}{=}&   (R^2\c a_{[0]})   (r^2\c b) \o
r^1_1R^1a_{[-1]}p^2S(r^1_3) \c n \o
r^1_2p^1\c m\\
&= &  [(R^2\c a_{[0]})_{[-1]}  \c  (r^2\c b)](R^2\c a_{[0]})_{[0]}
\o r^1_1R^1a_{[-1]}p^2S(r^1_3) \c n \o
r^1_2p^1\c m\\
&= &  [(R^2_1 a_{[-1]_2} S(R^2_3)r^2\c b)] (R^2_2\c a_{[0]})\o
r^1_1R^1a_{[-1]_1}p^2S(r^1_3) \c n \o
r^1_2p^1\c m\\
&\stackrel{(8)}{=}&   [(R^2 a_{[-1]_2} S(Q^2_2)r^2\c b)] (Q^2_1\c
a_{[0]})\o r^1_1Q^1R^1a_{[-1]_1}p^2S(r^1_3) \c n \o
r^1_2p^1\c m\\
&\stackrel{(10)}{=}&   [(a_{[-1]_1}R^2  S(Q^2_2)r^2\c b)] (Q^2_1\c
a_{[0]})\o r^1_1Q^1a_{[-1]_2}R^1p^2S(r^1_3) \c n \o
r^1_2p^1\c m\\
&\stackrel{(9)}{=}&   [(a_{[-1]_1}R^2  S(U^2)V^2r^2\c b)] (Q^2\c
a_{[0]})\o V^1U^1Q^1a_{[-1]_2}R^1p^2S(r^1_2) \c n \o
r^1_1p^1\c m\\
&\stackrel{(14)}{=}&   [(a_{[-1]_1}R^2  1_1r^2\c b)] (Q^2\c
a_{[0]})\o S(1_2)Q^1a_{[-1]_2}R^1p^2S(r^1_2) \c n \o
r^1_1p^1\c m\\
&\stackrel{(11)}{=}&   [(a_{[-1]_1}R^2  r^2\c b)] (Q^2S(1_2)\c
a_{[0]})\o Q^1a_{[-1]_2}1_1R^1p^2S(r^1_2) \c n \o
r^1_1p^1\c m\\
&\stackrel{(16)}{=}&  [(a_{[-1]_1}R^2 r^2\c b)] (Q^2\c a_{[0]})\o
Q^1a_{[-1]_1}R^1p^2S(r^1_2) \c n \o
r^1_1p^1\c m\\
&\stackrel{(9)}{=}&  [(a_{[-1]_1}R^2 \c b)] (Q^2\c a_{[0]})\o
Q^1a_{[-1]_1}R^1_1p^2S(R^1_3) \c n \o
R^1_2p^1\c m\\
&\stackrel{(10)}{=}&  [(a_{[-1]_1}R^2 \c b)] (Q^2\c a_{[0]})\o
Q^1a_{[-1]_1}p^2R^1_2S(R^1_3) \c n \o
p^1R^1_1\c m\\
&= &  [(a_{[-1]_1}R^2 \c b)] (Q^2\c a_{[0]})\o Q^1a_{[-1]_1}p^21_2
\c n \o p^11_1R^1\c m\\
 &= &  [(a_{[-1]_1}R^2 \c b)] (Q^2\c a_{[0]})\o
Q^1a_{[-1]_1}p^2 \c n \o p^1R^1\c m,
\end{eqnarray*}
where  Corollary \ref{cor2.6}  and Lemma \ref{lem2.4} were used in the second and
fifth equality, respectively; the twelfth and the thirteenth
equations stemmed from the compatible condition and the quantum
commutativity respectively. Thus $$ \xi \circ (\widetilde{C }_{
A\Box M, A\Box N}) =(1 \o \widetilde{C }) \circ \xi.$$

Therefore, $A\Box -$ is a braided autoequivalence of ${} ^{_RH} (_{H}\mathscr {M})$.  \qed

\begin{lem}\label{lem3.5} Let $(H, R )$  be a  finite dimensional
quasitriangular weak Hopf algebra. Assume that  $A$ is a braided
bi-Galois object. If the functor $A\Box-$ defines a braided
autoequivalence of   ${} ^{_RH} (_{H}\mathscr{M})$, then $A$ is
quantum commutative. \end{lem}

\proof Assume that  the functor $A\Box-$ defines a braided
autoequivalence. We  have  the commutative diagram $(*)$. Let
$M$ and $N$ be two left ${}_RH$-comodules.  Following the proof of Lemma
\ref{lem3.4} we obtain the following equation:
\begin{eqnarray}\label{eq3.2}
 && a_{(0)}   (r^2\c b) \o r^1_1 a_{(1)}p^2S(r^1_3)\c n  \o r^1_2p^1\c
 m  \nonumber\\
&=&[a_{(-1)_1} r^2_1\c b] [ R^2r^1_1 \c a_{(0)}]\o R^1a_{(-1)_2}
r^2_2\c n \o r^1_2 \c m,
\end{eqnarray}
for all $a\o m \in A\Box M$ and $b\o n\in A\Box N$. Now let $M={}_RH$. Since $a_{(0)}\o a_{(1)}, b_{(0)}\o b_{(1)}
\in A\Box {}_RH $, we may substitute them for the elements $a\o m$ and $b\o n$ in the above equation  and obtain the following equation:
\begin{eqnarray*}
&& a_{(0)}   (r^2\c b_{(0)}) \o (r^1\c a_{(1)})_{[-1]} \c b_{(1)} \o
(r^1\c a_{(1)})_{[0]}\\
&=& [a_{(-1)_1} r^2_1\c b_{(0)}] [ R^2r^1_1 \c a_{(0)}]\o
R^1a_{(-1)_2} r^2_2\c b_{(1)} \o r^1_2 \c a_{(1)}.
\end{eqnarray*}
 Now we apply the map $1\o \v_t \o \v_t$ to the foregoing equality and obtain the following:
\begin{eqnarray*}
&& [a_{(-1)_1} r^2_1\c b_{(0)}] [ R^2r^1_1 \c a_{(0)}]\o  \v_t (
R^1a_{(-1)_2}
r^2_2\c b_{(1)}) \v_t (r^1_2 \c a_{(1)})\\
&=& a_{(0)}   (r^2\c b_{(0)}) \o  \v_t [(r^1\c a_{(1)})_{[-1]} \c
b_{(1)} ] \v_t [(r^1\c a_{(1)})_{[0]}].
\end{eqnarray*}
Since $\v_t$ is an algebra map in the category  ${}_H \mathscr{M}$
and $A$ is a right ${}_RH$-comodule algebra, we have
\begin{eqnarray*}
[a_{(-1)} r^2\c b] [ r^1 \c a_{(0)}]=ab,
\end{eqnarray*}
 which is equivalent to $$ ab = (a_{[-1]} \c
b)a_{[0]}.$$ Thus $A$ is quantum commutative.

Now we show that $A$ is cocommutative. Namely, we need to verify
that the right coaction $\r^r$ on $A$ is induced by its left
coaction $\r^l$ and the half-braiding. Note that the regular left
$H$-module $H$ has an induced Yetter-Drinfeld module structure,
where the comodule structure is given by
$$
\r^L (h)=R^2 \o R^1 h:=h_{[-1]} \o h_{[0]}.
$$
 By Lemma \ref{lem2.4} we have a left ${}_RH$-comodule structure on $H$, where  $\r^l (h) = 1\o_t h$ for any $h\in H.$  Namely, $(H, \r^l)$ is a trivial left
$_RH$-comodule. Now consider $A\Box {}_RH$ and $A\Box H$. Note that
$1_A \o_t 1_H \in A\Box H$ and $a_{(0)} \o a_{(1)}  \in A\Box
{}_RH$. Using Equation \eqref{eq3.2} we easily get:
\begin{eqnarray*}
 &&
 a_{(0)}   (r^2\c 1) \o r^1_1 a_{(1)}p^2S(r^1_3)  \o r^1_2p^1\c
a_{(2)}\\
&=& [a_{(-1)_1} r^2_1\c 1] [ R^2r^1_1 \c a_{(0)}]\o R^1a_{(-1)_2}
r^2_2\o r^1_2 \c a_{(1)}.
\end{eqnarray*}
Now on the one hand, we have:
\begin{eqnarray*}
 && a_{(0)}   (r^2\c 1_A) \o r^1_1 a_{(1)}p^2S(r^1_3)  \o
r^1_2p^1\c a_{(2)}\\
&=& a_{(0)}   (\v_t (r^2)\c 1_A) \o r^1_1 a_{(1)}p^2S(r^1_3)  \o
r^1_2p^1\c a_{(2)}\\
&\stackrel{(15)}{=}&  a_{(0)}   (1_2'\c 1_A) \o S(1'_1)1_1
a_{(1)}p^2S(1_3)  \o
1_2p^1\c a_{(2)}\\
&=& 1_1'\c a_{(0)}  \o  1'_21_1 a_{(1)}p^2S(1_3)  \o
1_2p^1\c a_{(2)}\\
&=& a_{(0)}  \o 1_1 a_{(1)}p^2S(1_3)  \o
1_2p^1\c a_{(2)}\\
&=& a_{(0)}  \o a_{(1)}p^2S(1_2)  \o
1_1p^1\c a_{(2)}\\
&\stackrel{(11)}{=}& a_{(0)}  \o a_{(1)}p^2   \o
 p^1\c a_{(2)}.
\end{eqnarray*}
On the other hand, we have:
\begin{eqnarray*}
&& [a_{(-1)_1} r^2_1\c 1_A] [ R^2r^1_1 \c a_{(0)}]\o R^1a_{(-1)_2}
r^2_2\o r^1_2 \c a_{(1)}
\end{eqnarray*}
\begin{eqnarray*}
&=&  [\v_t (a_{(-1)_1} r^2_1)\c 1_A] [ R^2r^1_1 \c a_{(0)}]\o
R^1a_{(-1)_2}
r^2_2\o r^1_2 \c a_{(1)}\\
&=&  [\v_t (a_{(-1)_1} \v_t(r^2_1))\c 1_A] [ R^2r^1_1 \c a_{(0)}]\o
R^1a_{(-1)_2}
r^2_2\o r^1_2 \c a_{(1)}\\
&\stackrel{(4)}{=}&  [\v_t (a_{(-1)_1} S(1_1))\c 1_A] [ R^2r^1_1
\c a_{(0)}]\o R^1a_{(-1)_2}
1_2r^2\o r^1_2 \c a_{(1)}\\
&=&  [\v_t (a_{(-1)_1} )\c 1_A] [ R^2r^1_1 \c a_{(0)}]\o
R^1a_{(-1)_2}
r^2\o r^1_2 \c a_{(1)}\\
&=&  [1_1 \c 1_A] [ R^2r^1_1 \c a_{(0)}]\o R^11_2a_{(-1)}
r^2\o r^1_2 \c a_{(1)}\\
&=& S(1_1) R^2r^1_1 \c a_{(0)}]\o R^11_2a_{(-1)}
r^2\o r^1_2 \c a_{(1)}\\
&\stackrel{(11)}{=}& R^2r^1_1 \c a_{(0)}\o R^1a_{(-1)} r^2\o r^1_2
\c a_{(1)}.
\end{eqnarray*}
Thus, the following equation holds:
$$  a_{(0)}  \o a_{(1)}p^2   \o
 p^1\c a_{(2)}=R^2r^1_1 \c a_{(0)}\o R^1a_{(-1)} r^2\o r^1_2 \c a_{(1)}\in A \o   H \o  H. $$
Applying the map $ (1\o 1\o \v)$ to right side of the above equation, we obtain:
\begin{eqnarray*}
&&  (1\o 1\o \v)(R^2r^1_1 \c a_{(0)}\o R^1a_{(-1)} r^2\o r^1_2 \c a_{(1)})\\
&\stackrel{(5)}{=}&R^2r^1_1 \c a_{(0)}\o  [R^1a_{(-1)} r^2]  \v( \v_s (r^1_2) a_{(1)}  S(r^1_3) )\\
&\stackrel{(2)}{=}&R^2r^1_1 \c a_{(0)}\o  [R^1a_{(-1)} r^2]  \v(   a_{(1)}  S(r^1_2) )\\
&=&R^2r^1_1 \c a_{(0)}\o  [R^1a_{(-1)} r^2]  \v(   a_{(1)} \v_t( S(r^1_2) ))\\
&=&R^2r^1_1 \c a_{(0)}\o  [R^1a_{(-1)} r^2]  \v(   a_{(1)} S( \v_s(r^1_2) ))\\
&\stackrel{(3)}{=}&R^2r^11_1 \c a_{(0)}\o  [R^1a_{(-1)} r^2]  \v(   a_{(1)} S^2( 1_2) ))\\
&=&R^2r^11_1 \c a_{(0)}\o  [R^1a_{(-1)} r^2]  \v(   a_{(1)} S( 1_2) ))\\
&=&R^2r^11_1 \c a_{(0)}\o  [R^1a_{(-1)} r^2]  \v(  S( 1_2) a_{(1)}  ))\\
&=&R^2r^11_1 \c a_{(0)}\o  [R^1a_{(-1)} r^2]  \v(  1_2  a_{(1)}  )\\
&=&R^2r^1 S(\v_t ( a_{(1)} )) \c a_{(0)}\o   [R^1a_{(-1)} r^2] \\
&=&R^2r^1 \c a_{(0)}\o   [R^1a_{(-1)} r^2 ],
\end{eqnarray*}
where the counit of a right ${}_RH$-comodule $A$ was used in the
last equality.   Now we have
\begin{eqnarray*}
R^2r^1 \c a_{(0)}\o   [R^1a_{(-1)} r^2  ]&=&  (1\o 1\o \v)(a_{(0)}
\o a_{(1)}p^2 \o
 p^1\c a_{(2)})\\
 &=& a_{(0)}  \o a_{(1)}p^2 \v(
 p^1\c  a_{(2)})\\
 &=& a_{(0)}  \o a_{(1)}p^2 \v(
 \v_s(p^1_1) a_{(2)}  S(p_2^1)]\\
  &=& a_{(0)}  \o a_{(1)}p^2 \v(
  a_{(2)}  S(p^1)]\\
&=& a_{(0)}  \o a_{(1)}p^2 \v(
  a_{(2)}  S(\v_s (p^1))]\\
  &\stackrel{(14)}{=}& a_{(0)}  \o a_{(1)}1_2 \v(
  a_{(2)}  S(1_1)]\\
   &=& a_{(0)}  \o a_{(1)}1_2 \v(
   1_1a_{(2)} )\\
  &=& a_{(0)}  \o a_{(1)}  \v_t(
  a_{(2)})\\
  &=& a_{(0)}  \o a_{(1)},
\end{eqnarray*}
where  the counit on  ${}_RH$  was used in the last equality.  This
means that a right ${}_RH$-comodule structure on $A$ is indeed induced by
its left ${}_RH$-coaction. Therefore, $A$ is a quantum commutative
 Galois object. \qed

Summarizing the foregoing arguments, we obtain the main result of this section:

\begin{thm}\label{thm3.6} Let $(H, R )$  be a  finite dimensional
quasitriangular weak Hopf algebra.  Assume that  $A$ is a  braided
bi-Galois object.  Then the functor $A\Box-$ defines a braided
autoequivalence of the category ${}^H_{H}\mathscr{YD}$ of
Yetter-Drinfeld modules if and only if $A$ is quantum commutative.
\end{thm}

\proof   Assume that  $A$ is a   braided bi-Galois object. By
Lemma \ref{lem3.4} and Lemma \ref{lem3.5}, the functor $A\Box-$ defines a braided autoequivalence of ${} ^{_RH} (_{H}\mathscr{M})$ if and only if $A$
is quantum commutative. Since   ${}
^{_RH} (_{H}\mathscr{M}) \cong {}^H_{H}\mathscr{YD}$  as braided
monoidal categories,   the functor $A\Box-$ induces a braided
autoequivalence of ${}^H_{H}\mathscr{YD}$  if and only if $A$ is
quantum commutative. \qed

Recall that the Drinfeld center $\mathscr{Z}_l({}_H\mathscr{M})$ is tensor equivalent to the Yetter-Drinfeld module category $ {}^H_{H}\mathscr{YD}$. Thus the functor $A\Box-$ defines a braided autoequivalence of the Drinfeld center  if and only if $A$ is quantum commutative. This holds as well for any quasitriangular Hopf algebra.

In order to deal with the case of a braided fusion category, we need
to restrict ourself to the category of finite dimensional
representations.  Denote by ${}_H\mathscr{M}^{f.d}$ and ${}^H_H\mathscr{YD}^{f.d}$ the category of finite dimensional left $H$-modules and  the category of finite dimensional left Yetter-Drinfeld modules respectively. Then $\mathscr{Z}_l({}_H\mathscr{M}^{f.d}) \cong {}^H_H\mathscr{YD}^{f.d}$. Thus, Theorem \ref{thm3.6} applies to ${}_H\mathscr{M}^{f.d}$.

\begin{cor} Let $\mathscr{C}$ be a braided fusion
category. Then the Drinfeld center of $\mathscr{C}$ is equivalent to
the category of finite dimensional left comoduels
over some braided Hopf algebra ${}_RH_\mathscr{C}$. Moreover, if $A$
is a braided bi-Galois object over ${}_RH_\mathscr{C}$, then the
cotensor functor $A\Box-$ defines a braided autoequivalence of the
Drinfeld center of $\mathscr{C}$ if and only if $A$ is quantum
commutative. \end{cor}

\proof Suppose that  $\mathscr{C}$ is a braided fusion category.
By \cite{O03} there exists a semisimple connected weak Hopf algebra
$H_\mathscr{C}$ such that $\mathscr{C}$ is (tensor) equivalent to the
category ${}_{H_\mathscr{C}}\mathscr{M}^{f.d}$ of finite dimensional
left $H_\mathscr{C}$-modules. Similar to the proof of Corollary 2.6,
one can endow the category ${}_{H_\mathscr{C}}\mathscr{M}^{f.d}$
with a braiding $\Phi$ such that the equivalence between the two categories   preserves the braidings.
Following \cite[Prop 5.2]{NTV03} one can define a quasitriangular structure
$R$ on $H_\mathscr{C}$ so that the braiding $\Phi$ of ${}_{H_\mathscr{C}}\mathscr{M}^{f.d}$ is induced by the quaisi-triangular structure $R$ of ${}_{H_\mathscr{C}}$.  \qed

To end this section,  we show that the quantum commutative Galois
objects over $_RH$ form a subgroup of the group of braided bi-Galois
objects (see \cite{Sc05}). In the Hopf algebra case, this subgroup
was defined in \cite{Zh04}. In what follows, we fix a finite
dimensional quasitriangular weak Hopf algebra $(H,R)$. A Galois
object means a braided bi-Galois object over the braided Hopf
algebra $_RH$ in the category $_H\mathscr{M}$. It is easy to see
that $_RH\Box-$ defines the identity functor of
$^{_RH}(_H\mathscr{M})$. So $_RH$ is a quantum commutative Galois
object.

\begin{lem}\label{lem3.8}  If $A$ and $B$ are two quantum
commutative  Galois objects, so is $A\Box B$. \end{lem}

\proof Assume that $A$ and $B$ are quantum commutative Galois
objects. Then  $A\Box -$ and  $B\Box -$ are braided
autoequivalences. So is the composition $(A\Box B )  \Box - $. Thus
by Proposition 3.3  $A\Box B$ is quantum commutative. \qed

Let $A$ a bi-Galois object $A$.  One can define a braided bi-Galois
object $A^{-1}=: (_RH\o A)^{co{_RH}}\subset {}_RH\o A^{op}$  such
that $A\Box A^{-1} \cong {}_RH$ and $A^{-1}\Box A \cong {}_RH$. For
more detail on $A^{-1}$, one may refer to \cite{Sc05}.

\begin{lem}\label{lem3.9}   If $A$ is a quantum commutative
Galois object, so is $A^{-1}$. \end{lem}

\proof  Suppose that  $A$ is a quantum commutative Galois object.
The functor  $A\Box -$ is a braided autoequivalence functor. It is easy to
see that $A^{-1}\Box-$ gives the inverse of the functor  $A\Box -$. By Lemma \ref{lem3.5},  the Galois object $A^{-1}$ is quantum commutative. \qed

Denote by $Gal^{qc} ({}_RH)$ the set of isomorphism classes
of the quantum commutative  Galois objects. Let  $[A]$ denote the
isomorphism class of a quantum commutative Galois object $A$. By
Lemma \ref{lem3.8} and Lemma \ref{lem3.9} we obtain the following.

\begin{thm} The set $Gal^{qc} ({}_RH)$
forms a group. The multiplication is induced by the cotensor product $\Box$ over ${}_RH$, the identity is given by $[{}_RH]$ and
the inverse of an element $[A]$ is represented by $A^{-1}$. \end{thm}

It is well-known  that the category ${}_{H}\mathscr{M}$ is braided subcategory of the Yetter-Drinfeld module category ${}^H_{H}\mathscr{YD}$. If $M$ is a left $H$-module.  Then $M$ possesses  a left $H$-comodule structure:
$$
\r^L (m)=R^2 \o R^1 \c m:=m_{[-1]} \o m_{[0]},
$$
so that  $(M, \r^L)$ is a left Yetter-Drinfeld module. It follows from Lemma \ref{lem2.4} that the induced left ${}_RH$-comodule structure on $M$ is trivial, namely,  $\r^l (m) = 1\o_t m $ for all $m\in M.$    If $A$ is a braided bi-Galois object, then $A\Box M\cong M$. Thus the functor $A\Box-$
restricts to the identity functor on the category of left
$H$-modules.

Now we consider the image of the group  $Gal^{qc} ({}_RH)$ in the group $\Aut^{br}({}^H_{H}\mathscr{YD})$ of braided autoequivalences of the Yetter-Drinfeld module category.

\begin{defn} \emph{ \cite[Defn 2.1]{DN12}}\emph{ A braided
autoequivalence $F$ of ${}^H_{H}\mathscr{YD}$ is called
\emph{trivializable} on ${} _{H}\mathscr{M}$ if the restriction
$F|_{{} _{H}\mathscr{M}}$ is isomorphic to the identity functor as a
braided tensor functor.  } \end{defn}

Denote by $\Aut^{br}({}^H_{H}\mathscr{YD}, {} _{H}\mathscr{M} )$  the
group of isomorphism classes of braided autoequivalences  of
${}^H_{H}\mathscr{YD}$  trivializable on $ {} _{H}\mathscr{M}$.

\begin{cor} The group $Gal^{qc}({}_RH)$ is a subgroup of
 the group  $\Aut^{br}({}^H_{H}\mathscr{YD}, {}_{H}\mathscr{M})$.
 \end{cor}

We expect that the two groups are isomorphic for any finite
dimensional quasitriangular weak Hopf algebras $(H,R)$. This is the
case when $H$ is a Hopf algebra, see \cite{DZ13}. In case $H$ is
semisimple over an algebraically closed field, i.e. the fusion case,
the two groups are indeed isomorphic (to the Brauer group of the
braided fusion category), see \cite{ZZ2} or \cite{Zh12}.

\begin{eg}  \emph{Let $k$ be a field with $ch(k)\neq 2$. Let $H_4
$ be the Sweedler 4-dimensional Hopf algebra over $k$. Namely, $H_4$
is generated by two elements $g$ and $h$ satisfying
$$g^2=1, \ \ h^2=0,\ \ gh+hg=0.$$
The comultiplication, the counit and the antipode are given as
follows:
\begin{eqnarray*}
&& \D (g)= g\o g , \ \  \D (h)= 1\o h+h\o g\\
&& \v(g)=1,\ \ S(g)=g, \ \v(h)=0,\ \ S(h)=gh.
\end{eqnarray*}
It is known that $H_4$ has a quasitriangular structure $R_0$. All
quantum commutative Galois objects were computed in \cite {Zh04}.
Moreover, the group $Gal^{qc} (_{R_0}H)$ is isomorphic to $\Gamma
\rtimes Z_2$, where $\Gamma\cong k^+\times K^{\bullet }/K^{\bullet
2}.$ }\end{eg}

\section{\bf{Face algebras}}

In this section we compute the groups of quantum commutative Galois objects of a class of weak Hopf algebras, namely, the face algebras introduced by Hayashi in \cite{Ha99}.

Let $N\geq 2$ be an integer and $\mathbb{Z}_N$ the cyclic group
$\mathbb{Z}/N\mathbb{Z}.$ Let $\omega\in \mathbb{C}$ be a primitive $N^{th}$ root of unity. Let $H$  be the $\mathbb{C}$-linear span  of
$\{X^i_j (s) | i,j, s \in \mathbb{Z}_N \}$.  $H$ is  a
quasitriangular weak Hopf algebra equipped with the following
 structures:
\begin{eqnarray*}
&& \D (X^i_j (s))= \sum _{p+q=s} X^i_j (p)\o
X^{i+p}_{j+p} (q), \   \v(X^i_j (s))=\delta_{s,0},\\
&&X^i_j (p)X^k_l (q)=\delta_{j,k}\delta_{p,q}X^i_l (p),\ \   1=\sum_{i,p}X^i_i (p),\\
&& S(X^i_j (p))=X^{j+p}_{i+p} (-p),\\
&& R_1 \o R_2= \sum_{i,j,p}X^i_j (p) \o X^j_{j+p} (i-j)\omega^{-p(i-j)},\\
&&R'_1 \o R'_2= \sum_{i,j,p}  X^{j+p}_{i+p}(-p) \o X^{j}_{j+p}
(i-j)\omega^{-p(i-j)},
\end{eqnarray*}
 where  the target subalgebra $H_t$ of $H$ is  the $\mathbb{C}$-linear span  of $\{\sum_p
X^i_i(p)| i\in \mathbb{Z}_N\}$. Denote  by $1^i$ the sum $\sum _p
X^i_i (p)$ for all $i\in \mathbb{Z}_N.$ Then $H_t$
 is commutative and is equal to the direct sum $ \bigoplus_{i\in \mathbb{Z}_N } \mathbb{C}1^i$.

 Now we compute  the braided Hopf algebra ${}_RH $.

 \begin{lem}\label{lem4.1} The braided Hopf algebra ${}_RH $  is equal to the $\mathbb{C}$-linear span of $\{X^i_i (p)|i, p \in \mathbb{Z}_N\}$ equipped with the following structures:
\begin{eqnarray*}
&& \D' (X^k_k (s))= \sum_{w+q=s} X^k_k (w)\o
X^{k}_{k} (q),\ \  \v_t(X^i_i (s))=\delta_{s,0} \sum_p X^i_i (p),\\
&&X^i_i (p)X^k_k (q)=\delta_{i,k}\delta_{p,q} X^i_i (p) ,\
1=\sum_{i,p}X^i_i (p),\\
&& S (X^{k}_{k} (s)) =X^{k}_{k} (-s).
\end{eqnarray*}
 \end{lem}
\proof  Note that  $ \D(1_H) = \D(\sum_{i,s}X^i_i (s)) =
\sum_{i,s}\sum_{p+q=s} X^i_i (p) \o X^{i+p}_{i+p} (q).  $   We have
\begin{eqnarray*}
1_1 X^{m}_{n} (r) S(1_2)&=&  \sum_{i,s}\sum _{p+q=s}  X^i_i (p)
X^{m}_{n} (r)
 S( X^{i+p}_{i+p} (q) )\\
 &=&  \sum_{i,s}\sum_{p+q=s}  X^i_i (p)
X^{m}_{n} (r)
 X^{i+p+q}_{i+p+q} (-q)\\
 &=&  \sum_{i,s}\sum_{p+q=s}  \delta_{i,m}\delta_{n,i+p+q}  \delta_{p,r}\delta_{-q,r} X^i
_{i+p+q} (p)\\
&=&  \sum_{i} \delta_{i,m}\delta_{n,i} X^i _{i} (r)=\delta_{m,n} X^m
_{n} (r),
\end{eqnarray*}
for all $m,n,r\in \mathbb{Z}_N $.  So $_RH$ is the $\mathbb{C}$-linear
span of $\{X^i_i (p)| i, p \in V\}$.

Using the expression $ \D(R^1)\o R^2 = \sum_{i,j,p}\sum_{u+v=p} X^i_j(u) \o
X^{i+u}_{j+u}(v) \o X^j_{j+p}(i-j) \omega^{-p(i-j)}, $
  we compute the deformed comultiplication as follows:
\begin{eqnarray*}
&& \D' (X^k_k(s)) \\
&=& \sum_{w+q=s}   X^k_k(w)S(R^2) \o R^1 \c X^{k+w}_{k+w}(q)\\
&=& \sum_{w+q=s}   X^k_k(w)S(R^2) \o R^1_1
X^{k+w}_{k+w}(q)S(R^1_2)\\
&=& \sum_{w+q=s}\sum_{i,j,p}\sum_{u+v=p} X^k_k(w)S( X^j_{j+p}(i-j))
\o  X^i_j(u) X^{k+w}_{k+w}(q)S(X^{i+u}_{j+u}(v)) \omega^{-p(i-j)}\\
&=& \sum_{w+q=s}\sum_{i,j,p}\sum_{u+v=p} X^k_k(w) X_i^{i+p}(j-i)) \o
X^i_j(u)
X^{k+w}_{k+w}(q)X^{j+u+v}_{i+u+v}(-v)) \omega^{-p(i-j)} \\
&=& \sum_{w+q=s}\sum_{i,j,p}\sum_{u+v=p}\delta_{w,j-i}
\delta_{k,i+p} X^k_i(w) \o \delta_{u,q}\delta_{q,-v}
\delta_{j,k+w}\delta_{k+w,j+u+v}X^i_{i+u+v}(u)
 \omega^{-p(i-j)}\\
&=& \sum_{w+q=s}\sum_{i,j}\delta_{w,j-i} \delta_{k,i}
\delta_{j,k+w}X^k_i(w) \o X^i_{i}(q)\\
 &=& \sum_{w+q=s}\sum_{j}\delta_{w,j-k}
 \delta_{j,k+w}X^k_k(w) \o X^k_{k}\\
 &=&\sum_{w+q=s}
  X^k_k(w) \o X^k_{k}(q).
\end{eqnarray*}

By Lemma \ref{lem2.1} the antipode is given by\  $ \overline{S}(x) =
R^2R'^2S^2(R'^1)S(R^1x). $ For convenience,  we first compute
$R'^2S^2(R'^1)$. Indeed,
\begin{eqnarray*} R'^2S^2(R'^1)&=&
\sum_{i,j,p} X^j_{j+p}(i-j) S^2 (X^i_j (p)) \omega^{-p(i-j)}\\
&=&  \sum_{i,j,p} X^j_{j+p}(i-j) X^i_j (p) \omega^{-p(i-j)}\\
&=& \sum_{i,j,p} \delta_{j+p,i}  X^j_{j} (i-j)\omega^{-p(i-j)}.
\end{eqnarray*}
Now we have
\begin{eqnarray*}
\overline{S}(X^{k}_{k} (-s))&=&R^2 R'^2S(R^1X^{k}_{k} (s)S(R'^1))\\
&=& \sum_{i,j,p } \sum_{i',j',p' } \delta_{j'+p',i'}X^j_{j+p} (i-j)
X^{j'}_{j'} (i'-j')
 S(X^i_j (p) X^k_k (s) )\omega^{-[p(i-j)+p'(i'-j')]}\\
 &=& \sum_{i,j,p } \sum_{i',j',p' } \delta_{j'+p',i'}\delta_{j,k}\delta_{p,s}X^j_{j+p} (i-j)
X^{j'}_{j'} (i'-j')
 S(X^i_k (s) )\omega^{-[p(i-j)+p'(i'-j')]}\\
 &=& \sum_{i } \sum_{i',j',p' } \delta_{j'+p',i'}X^k_{k+s} (i-k)
X^{j'}_{j'} (i'-j')
 S(X^i_k (s) )\omega^{-[s(i-k)+p'(i'-j')]}\\
 &=& \sum_{i } \sum_{i',j',p' } \delta_{j'+p',i'}X^k_{k+s} (i-k)
X^{j'}_{j'} (i'-j')
 X^{k+s}_{i+s} (-s) \omega^{-[s(i-k)+p'(i'-j')]}\\
  &=& \sum_{i } \sum_{i',j',p' } \delta_{j'+p',i'}  \delta_{i-k,i'-j'}  \delta_{i'-j',-s}\delta_{k+s,j'}
 X^{k}_{i+s} (-s) \omega^{-[s(i-k)+p'(i'-j')]}\\
 \end{eqnarray*}
\begin{eqnarray*}
 &=& \sum_{i } \sum_{j',p' }    \delta_{i-k,p'}  \delta_{p',-s}\delta_{k+s,j'}
 X^{k}_{i+s} (-s) \omega^{-[s(i-k)+p'p']}\\
 &=& \sum_{i } \sum_{j' }    \delta_{i-k,-s} \delta_{k+s,j'}
 X^{k}_{i+s} (-s) \omega^{-[s(i-k)+(-s)(-s)]}\\
 &=& \sum_{i }  \delta_{i-k,-s}
 X^{k}_{i+s} (-s) \omega^{-[s(i-k)+(-s)(-s)]}\\
 &=&
 X^{k}_{k} (-s) \omega^{-[s(-s)+(-s)(-s)]}=X^{k}_{k} (-s).
 \end{eqnarray*}
Thus, the proof is completed. \qed

Take $i\in \mathbb{Z}_N$. Define $H^i$ to be the $\mathbb{C}$-linear
span of $\{ X^i_i (p)| p\in \mathbb{Z}_N\}$. It is obvious that $H^i$
is a subalgebra of ${}_RH$ with unity $ 1^i$. Moreover, ${}_RH $ is
the direct sum of all these $H^i$, i.e., $ {}_RH = \bigoplus_{i\in
\mathbb{Z}_N} H^i. $ We will show that  every $H^i$ is also an
ordinary Hopf algebra and so ${}_RH $ is actually the direct sum of
all these Hopf algebras. In order to verify that every $H^i$ can be
equipped with a coalgebra structure, we need to decompose the vector
space ${}_RH\o_t {}_RH. $

\begin{lem}\label{lem4.2}
${}_RH\o_t {}_RH = \bigoplus_{i\in \mathbb{Z}_N} (H^i \o H^i). $
 \end{lem}

\proof It is equivalent to show that
$$1_1\c X^a_a (b) \o 1_2\c X^u_u(w)=  \delta_{u,a}   X^a_a
  (b)    \o
X^{u}_{u}  (w),$$ for all $a,b,u,w \in \mathbb{Z}_N$. Indeed, we have
 \begin{eqnarray*}
 1_1\c X^a_a (b) \o 1_2\c X^u_u(w)
&=& \sum_{i,s} \sum_{p+q=s}  X^i_i (p) \c X^a_a (b)    \o
X^{i+p}_{i+p} (q) \c X^u_u(w)\\
&=& \sum_{i,s} \sum_{p+q=s}  \delta_{i,a} \delta_{p,0}  X^i_i
  (b)    \o   \delta_{i+p,u} \delta_{q,0}
X^{i+p}_{i+p}  (w)\\
&=& \sum_{i}   \delta_{i,a}   X^i_i
  (b)    \o   \delta_{i,u}
X^{i}_{i}  (w)=  \delta_{u,a}   X^a_a
  (b)    \o
X^{u}_{u}  (w),
 \end{eqnarray*}
for all $a,b,u,w \in \mathbb{Z}_N$. \qed

\begin{lem}\label{lem4.3} For all $i\in \mathbb{Z}_N$,
 $H^i$ is a
coalgebra over $\mathbb{C} 1^i $ with the following structures:
\begin{eqnarray*}
&& \D' (X^i_i (s))= \sum_{w+q=s} X^i_i (w)\o X^{i}_{i} (q),
\\
  &&\v_t(X^i_i (s))=\delta_{s,0} \sum_p
X^i_i (p).
\end{eqnarray*}
\end{lem}

\proof Follows from Lemma \ref{lem4.1} and Lemma \ref{lem4.2}. \qed

\begin{prop}\label{prop4.4} For all $i\in \mathbb{Z}_N $,  $H^i$ is a commutative and cocommutative Hopf
algebra over $\mathbb{C} 1^i $ equipped  with  the following
structures:
\begin{eqnarray*}
&&X^i_i (p)X^i_i (q)= \delta_{p,q} X^i_i (p) ,\ \ \ \ \
1_{H^i}=1^i,\\
&& \D' (X^i_i (s))= \sum_{w+q=s} X^i_i (w)\o
X^{i}_{i} (q), \  \\
&&\v_t(X^i_i (s))=\delta_{s,0} \sum_p X^i_i (p), \ \  S (X^{i}_{i}
(s)) =X^{i}_{i} (-s).
\end{eqnarray*}

\end{prop}

\proof Since we know already  that $H^i$ is both an algebra and a
coalgebra, it remains to be proved that    $\D'$ and $\v_t$ are
multiplicative, and that the axioms of   the antipode $S$ hold. We
first check that $\D'$ is multiplicative. Indeed,
 \begin{eqnarray*}
\D'(X^i_i (s)) \D''(X^i_i (t)) &=& [\sum_{p+q=s} X^i_i (p)\o
X^{i}_{i} (q)] [\sum_{p'+q'=t} X^i_i (p')\o
X^{i}_{i} (q')]\\
&=& \sum_{p+q=s}\sum_{p'+q'=t}
[X^i_i (p)X^i_i (p') \o X^{i}_{i} (q) X^{i}_{i} (q')]\\
&=& \sum_{p+q=s}\sum_{p'+q'=t} \delta_{p,p'}\delta_{q,q'}  [X^i_i
(p)
\o X^{i}_{i} (q)]\\
&=& \delta_{s,t} \sum_{p+q=s} X^i_i (p)\o  X^{i}_{i} (q) \\
&=&\D'(X^i_i (s) X^i_i (t)),
\end{eqnarray*}
for all $i,s,u,t\in \mathbb{Z}_N.$

Note that $\D' (1)=1 \o_t 1.$ It follows from Lemma \ref{lem4.2}  that
  $\D' (1^i)=   1^i \o 1^i$.

Next we verify that $\v_t$ is an algebra map.  For all $ s,t \in
\mathbb{Z}_N$, we have
\begin{eqnarray*}
\v_t (X^i_i (s)  ) \v_t (  X^i_i(t)) &= &
\delta_{s,0}\delta_{t,0}(\sum_p
X^i_i(p)) (\sum_q X^i_i(q))\\
&= & \delta_{s,0}\delta_{t,0} (\sum_p X^i_i(p)) =
\delta_{s,t}\delta_{s,0}\v_t (X^i_i (s))\\
&= &  \v_t (X^i_i (s) X^i_i(t)).
\end{eqnarray*}

Finally, we prove that the  antipode axioms  hold. Indeed,
\begin{eqnarray*}
m(1\o S) \D'' (X^i_i (s)) &=& \sum_{p+q=s} X^i_i(p)
S(X^i_i(q))= \sum_{p+q=s} X^i_i(p) X^i_i(-q)\\
&=& \delta_{p,-q}\sum_{p+q=s} X^i_i(p)= \delta_{s,0}\sum_{p\in
\mathbb{Z}_N} X^i_i(p)=\v_t (X^i_i (s)).
\end{eqnarray*}
for any $s\in  \mathbb{Z}_N.$ Similarly, we also have
\begin{eqnarray*}
 \sum_{w+q=s} S(X^i_i (w)X^{i}_{i} (q)&=&\sum_{w+q=s} X^i_i (-w)X^{i}_{i} (q))
 =\sum_{w+q=s} \delta_{-w,q} X^i_i (q)
  \end{eqnarray*}

\begin{eqnarray*}
&=&\sum_{q} \delta_{s,0} X^i_i (q)=\v_t (X^i_i (s)).
\end{eqnarray*}
Hence,  $H^i $ is an ordinary  Hopf algebra over $\mathbb{C} 1^i $.
\qed

In fact, $H^i$ is isomorphic to the dual Hopf algebra of the group Hopf algebra $k\mathbb{Z}_N$.

\begin{cor}\label{cor4.5}  The braided Hopf algebra $ {}_RH$ has a decomposition:
$$
_RH =\bigoplus_{i\in \mathbb{Z}_N} H^i,
$$
where $H^i$ is a Hopf algebra over $\mathbb{C} 1^i $ with unity
$1^i$.
 Moreover, there exists a Hopf algebra isomorphism from
$H^i$ to  $H^j$ defined by
$$
 \iota^j_i : X^i_i(p)\longmapsto X^j_j(p),
$$
for all $i,j,p\in \mathbb{Z}_N.$
\end{cor}

\proof Follows from  Proposition \ref{prop4.4}. \qed

Corollary \ref{cor4.5} indicates that braided bi-Galois objects over
${}_RH$ can be obtained from bi-Galois objects over a Hopf algebra
$H^i$.

Let the notations be as above.
Let $A$ be a quantum commutative Galois object over $_RH$.  Corollary \ref{cor4.5} implies that there is  a decomposition: $A =\bigoplus _{i\in
\mathbb{Z}_N} A^i,$ where $\r^r (A^i) \in A^i \o H^i$. Furthermore,
every $A^i$ is just a Galois object over $H^i$ (automatically a bi-Galois object as $H^i$ is cocommutative). Conversely, given a
Galois object $A'$ over  Hopf algebra $H^i$ for some $i\in
\mathbb{Z}_N$,  we can get a quantum commutative Galois object over
${}_R H$ as the direct sum $\bigoplus_{i\in \mathbb{Z}_N} A'^i$, where
every algebra $A'^i$ is a copy of $A'$.  Now we state the relation
between quantum commutative Galois object over ${}_R H$ and  Galois
object over $H^i$ as follows:

\begin{prop}
Let $A$ be a $\mathbb{C}$-algebra with unity. Then $A$ is   a
quantum commutative Galois object over $_RH$ if and only if $A $ is the direct
sum $\bigoplus_{i\in \mathbb{Z}_N}  A^i$, where every $A^i$ is an
$H^i$-Galois object. Moreover, there exists a group isomorphism
$$
\Omega:  Gal^{qc}({}_RH) \longrightarrow  Gal(H^i),\ \  A\longmapsto
A^i,
$$
for any fixed $i\in \mathbb{Z}_N$. The inverse of $\Omega$ is given as
follows: $$ \Omega':  Gal(H^i) \longrightarrow  Gal^{qc}({}_RH),\ \
A'\longmapsto \bigoplus_{i\in \mathbb{Z}_N} A'^i.
$$
\end{prop}

The detailed proof of the statement above is given in
\cite{Zh12} following a tedious and long computation. So the group
$Gal^{qc}({}_RH)$ can be obtained by computing the group $Gal(H^i)$ of Galois objects over  $H^i$. Since the Hopf algebra $H^i$ is commutative and cocommutative isomorphic to $k\mathbb{Z}_N$, we know that the group $Gal(H^i)$ is actually given by the second Galois cohomology group $H^2(\mathbb{Z}_N, k)$.

\section*{\bf{Acknowledgement}}

This work forms a part of the PhD thesis of the second named author at University of Hasselt. He would like to thank BOF of UHasselt for the financial support: BOF09-DOC009-R-1964.

redpajama_set_name: RedPajamaArXiv


text:
There are a handful of 'ghosts' in Walken's Filmography. Spooky? No. Sad.
Mostly, they are side-lined and forgotten tv and short-film projects never made available to the public except perhaps broadcast once on tv, like 40 years ago. And never seen again. Except on lists.
I've never seen any of these, so they won't get individual entries on my list, but let me give them some time right here. Perhaps as things surface online, and things get found, digitalized and uploaded, perhaps I will get the chance to watch some of these. Until then this will have to do.
Walken and Close in Sarah Plain and Tall
The Hallmark Hall of Fame series claims responsibility for several of these lost souls. In fact, Walken has had a long relationship with the HHoF. In addition to starring with Glenn Close in the SARAH PLAIN AND TALL trilogy (which I have seen and will cover properly) begun in the early 90s, Walken was in a handful of other less well-known HHoF episodes back as early as 1966.
FUN FACT: The HHoF is the longest-running primetime series in the history of television (1951-current). Take that, Simpsons!
In 1966, a 23 y.o. Christopher Walken appeared in an HHoF historical drama called BAREFOOT I N ATHENS about the trial of Socrates.
Walken plays Lamprocles, Socrates' eldest son.
In Walken's official filmography, BAREFOOT is the first entry and so it's sometimes regarded by some to be Walken's film debut. However, this is television. It has the look of theater that has been recorded, and it is a mere 76 minutes long. Not feature-length at all. It may be splitting hairs, but until it becomes available to watch, it remains a moot point.
Fun Fact: ME AND MY BROTHER is Walken's first film-film!
A few years later, Walken had a supporting role in a re-telling of the swashbuckling classic THE THREE MUSKETEERS. It was broadcast once in 1969 for Canadian television, and never seen again. Really.
Then in 1975 Walken did another 75 minute Hallmark HoF-episode called VALLEY FORGE, playing a minor character called The Hessian, in a dramatization of George Washington's strife during the Revolutionary War.
Young Walken in his first short film
There are a few other items of note in Walken's early years amidst a bunch of television. When he was just a lad of twelve or thirteen, he was in a short-film called THE BOY WHO SAW THROUGH (1956) about a child with x-ray vision. This is a live-action film directed by Mary Ellen Bute, who was one of the first female experimental filmmakers. Ever. She was a pioneer in animation, and abstract 'visual music'. She made this live-action short at the end of her film career, right before doing a Finnegan's Wake film. I'd love to see all of this.
In 1970, Walken had a nameless part in Michel Auder's CLEOPATRA, a kind of debauched, psychodelic, improvisational, Andy-Warhol-ian film. Auder never finished it, supposedly because of a fight with his producers, and for a while it was thought lost, and according to the director "only survives now as an uncut degraded copy of the original."
Jump ahead to 2001. In Walken's first and only attempt at directing, POPCORN SHRIMP is a 5 minute short starring Ralph "The Karate Kid" Macchio, and rapper/actor Master P. It was shown on Showtime once or twice, but can't seem to find it on youtube or anything. Drop me a line if you (hey you out there) find a clip, or a link, or whatever.
Walken supposedly said it was "awful", and I'm powerfully inclined to believe him, but then again Walken's never been good at judging his own work. If I ever get a chance to watch it, I'll let you know, NBD.
Then in 2002, Walken was the voice of a little girl's magical toy firetruck, in a 17 minute short film called ENGINE TROUBLE. It is part of a Showtime collection of other short films and documentaries focused on the 9/11 incident called Reflections From Ground Zero.
Lastly, there is a movie credit in 2011 for Walken's performance in an indie horror film known as EVIL CALLS: THE RAVEN (aka THE LEGEND OF HARROW WOODS). From what I can gather, Walken is never seen in the film. You hear him recite the famous Edgar Allan Poe poem The Raven, as he did for the audio CD Closed on Account of Rabies. It's good stuff, but it is unclear whether the director re-recorded Walken reading the poem or if he just used the same track that's on the CD.
Walken art by Martynas Juchnevicus
Either way, with the possible exception of Riki "Drop Dead Fred" Mayall,this film looks terrible, and the Walken-factor is redundant. If you want to hear Walken "Quoth the Raven" just get the Closed on Account… CD. It's full of great voices reading Poe's works.
But if you're still squawkin' for more Walken, check out this art that people from all over have created to celebrate the magnificence that is Walken.
Or go to my List of Walken Films and see if anything tickles your tits.
Something Something (todd)
Todd has a wonderful job in the evenings that not only allows him to write prodigiously during the day, but it also puts that trusty B.A. in Media (@USM) to regular use at night. He is a published writer, so that's good. And, can I tell you something? Something in private? The people. The people, they say he is *whispered* ... a lyrical maaadman! They say his words will set you on fire. Literally. So get friendly with the aloe, my friend. ... Actually, he IS required by the state of Maine & jolly old Uncle Sam to provide that warning written in the third person for your initial protection. He will not mention it again.
The Funeral (1996)
Basquiat (1996)
Celluloide (1996)

redpajama_set_name: RedPajamaCommonCrawl


OK, I know I said in a previous post that you should never pay for ads which just get "likes" for your page. And I said that because you should never pay for ads which just get people to like your page.
What does that mean exactly? For a quick recap in case you haven't been pouring over my blog posts... or indeed any reporting on any social media at all, like those people who have been living in mine shafts for example, Facebook have been systematically throttling the organic reach of Page posts meaning that all the people who who have opted to "like" your page, presumably because they are interested in your art and want info from you, don't. Or at least very few of them do.
I have noticed a massive drop in reach and engagement on posts over the last year. We were actually getting more interaction and reaching more people when we had about 2/5 of the following than we do now. There is obviously more competition now and there is only so much timeline space, that's also a factor but Facebook have been restricting who sees your stuff regardless of the fact people actually asked to do so.
I thought "sod it " and chucked a tenner at it and set it to the minimum spend of £1 per day with all the relevant targeting to make sure I was reaching likely candidates and also set it to only the UK and the USA as that's where we tour the most and let it do its thing. The least I would get is a few more fans and a blog post so you'll know exactly what you get for your money in case you were curious.
I got 40 new people liking my page so spent about 25p per person. If I share a page post I reach about 1000 people on average give or take, out of 27,000 so that 25p is worth 1/27 of a person I can now promote to on a post by post basis.
This tells me that there was no other activity other than the likes, that it showed the ad to over 1300 people that that I had a mediocre "relevance" score out of 10. This is odd because when I use almost identical demographics on my other Facebook ads I usually get a much higher relevance score around the 8 or even 9 mark sometimes.
In conclusion I won't be doing this again but it did satisfy my curiosity. I do know one person who has spent a considerable amount of money on Facebook to build up his audience, which he has, only to not be able to reach them.
So, if you are thinking about it, don't and then read my other blog post on what to do instead HERE.


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text:
Eligo Technology news round up 23/09/2016
Sep 23, 2016 by Eligo Technology, Industry News, Development & Testing...
Eligo Technology news round up, featuring the latest news and updates from the Tech industry- week ending 23/09/2016
Lenovo's Signature laptop refuses to install Linux
It has been confirmed that several of Lenovo's Yoga laptops are refusing to install Linux based operating systems. There is confusion surrounding this restriction as an employee posted that Linux was blocked because of an "agreement with Microsoft" but Lenovo denies enforcing a deliberate ban.
Brits are keen to start using fingerprint ID when making bank payments
Recent research from Visa has discovered that most Britons would trust banks with their fingerprint identification, but not the government. 60% of Britons would trust their banks to safely store biometric information such as fingerprint, iris and facial recognition. Whereas only 33% were happy for the government to have access to the same.
(City A.M.) 19/09/2016
Google launches its smart messaging app Allo
Allo, designed for Android and iOS is a "smart" messaging app that allows you to chat with friends and plan and find information. Its "smart" feature helps to keep conversations going by providing assistance such as phrases and words to say.
(Venture Beat) 20 /09 /2016
Microsoft called upon to compensate users for Windows 10 issues
UK consumer rights group Which? Has called for Microsoft to compensate users of Windows 10. The software upgrade received hundreds of complaints from printer problems to lost files with many users having to pay technicians to repair the issues.
(ZD Net) 22/09/2016
500m Yahoo user accounts have been compromised by a data breach
It has come to light that back in 2014 the dot-com giant Yahoo were hacked, compromising account information such as names, emails, telephone numbers and even passwords. The news of this data breach puts pressure on a current acquisition deal between Yahoo and telecommunications company Verizon.
(City.A.M) 22/09/2016


text:
import os
import sys
import copy
import time
import tarfile
import base64
sys.path.append('../../')

from externals.simple_oss import SimpleOss
from batchcompute import (
    Client, JobDescription, TaskDag, TaskDescription, ResourceDescription
)
import config as cfg

oss_clnt = SimpleOss(cfg.OSS_HOST, cfg.ID, cfg.KEY)

def upload_worker(bucket, local_dir, oss_path):
    '''
    A function to help upload worker package to oss.
    '''
    local_tarfile = 'worker.tar.gz'
    if os.path.exists(local_tarfile): os.remove(local_tarfile)

    def do_tar(worker_dir, tar_file):
        '''
        A function to tar worker package.
        '''
        tar = tarfile.open(tar_file, 'w:gz')
        cwd = os.getcwd()
        os.chdir(worker_dir)
        for root,dir,files in os.walk('.'):
            for file in files:
                tar.add(os.path.join(root, file))
        os.chdir(cwd)
        tar.close()

    do_tar(local_dir, local_tarfile)
    oss_clnt.upload(bucket, local_tarfile, oss_path)

def get_job_desc(package_path, verbose=True):
    # generate the worker command line string and encode it.
    argv = {}
    argv['-s'] = str(cfg.DATA_START)
    argv['-e'] = str(cfg.DATA_END)
    argv['-d'] = cfg.DATA_PATH
    argv['-o'] = cfg.OUTPUT_PATH
    argv['-b'] = cfg.OSS_BUCKET
    argv['-c'] = str(cfg.COUNT_TASK_NUM)
    argv['-l'] = cfg.LOCAL_DATA
    argv_str = ''
    for k, v in argv.iteritems():
        argv_str += ' %s %s'%(k, v)

    job_desc = JobDescription()
    find_task = TaskDescription()

    # Create find task.
    find_task.PackageUri = package_path
    find_task.ProgramName = 'count_prime_with_ossmapping_worker.py'
    find_task.ProgramType = 'python'
    find_task.ProgramArguments = argv_str
    find_task.ImageId = cfg.IMAGE_ID
    find_task.InstanceCount = cfg.COUNT_TASK_NUM
    find_task.EnvironmentVariables = {}
    find_task.StdoutRedirectPath = cfg.LOG_PATH
    find_task.StderrRedirectPath = cfg.LOG_PATH

    # Create count task.
    count_task = TaskDescription(find_task)
    count_task.InstanceCount = cfg.SUM_TASK_NUM
    # This mapping will mount a oss path to a local path in VM.
    count_task.OssMapping = {
        cfg.FULL_DATA: cfg.LOCAL_DATA
    }

    # Create task dag.
    task_dag = TaskDag()
    task_dag.add_task(task_name='Find', task=find_task)
    task_dag.add_task(task_name='Count', task=count_task)
    task_dag.Dependencies = {
        'Find': ['Count']
    }

    # count prime job description.
    job_desc.TaskDag = task_dag
    job_desc.JobName = 'find-prime'
    job_desc.Priority = 100
    return job_desc

def main():
    upload_worker(cfg.OSS_BUCKET, 'worker_package', cfg.PACKAGE_PATH)

    # Submit job to batch compute.
    clnt = Client(cfg.REGION, cfg.ID, cfg.KEY)
    job_desc = get_job_desc(cfg.FULL_PACKAGE)
    job = clnt.create_job(job_desc)
    t = 10
    print 'Sleep %s second, please wait.'%t
    time.sleep(t)

    # Wait for job Terminated.
    while(True):
        s = clnt.get_job(job)
        if s.State in ['Waiting', 'Running']:
            print('Job %s is now %s'%(job, s.State))
            time.sleep(3)
            continue
        else:
            # 'Failed', 'Stopped', 'Finished'
            print('Job %s is now %s'%(job, s.State))
            if s.State == 'Finished':
                result = oss_clnt.download_str(cfg.OSS_BUCKET, cfg.OUTPUT_PATH)
                # Print out the count of prime numbers from 0 to 10000.
                print('Total %s prime numbers from %s to %s.'%(
                    result, cfg.DATA_START, cfg.DATA_END))
            break
    # Release job from batchcompute.
    clnt.delete_job(job)

if __name__ == '__main__':
    main()

redpajama_set_name: RedPajamaGithub


Devoted to cooperation rather than competition, Aikido is a martial art and movement meditation created in the 20th century by Morihei Ueshiba, O'Sensei (1883-1969). His creation is Aikido, a defensive art form and a practice of mental and spiritual development rooted in the ancient Bushido tradition of Japan. Through training, practitioners cultivate self-awareness and resiliency, a powerful center and a calm spirit. We also develop what are first just techniques, but later become intuitive responses to resolving conflict in ourselves and in our environment.
Read more about What is Aikido?


text:
Lucidestea pallaryi is een slakkensoort uit de familie van de Rissoidae. De wetenschappelijke naam van de soort is voor het eerst geldig gepubliceerd in 1928 door Hornung & Mermod.

Rissoidae

redpajama_set_name: RedPajamaWikipedia


text:
Jacqueline Susann and Me
"Yeah, I think I'll be remembered as the voice of the 60's…Andy Warhol, the Beatles, and me!" – Jacqueline Susann.
I was first introduced to Jacqueline Susann by my father of all people. Oh, he didn't mean to. One summer he took my niece and me to the movies in Rosendale, the little town near our house upstate. Back then, movies took a long time before they played the cinema in Rosendale. The only movie playing was the movie version of the Susann novel Once is not Enough. Apparently my father didn't check the rating on this movie, so he had no idea that it was rated R. Not exactly the type of movie that you want to take a 10 or an 11 year old to but my niece and I were riveted at the story of January Wayne (played by Deborah Raffin) and her obsession/Electra complex with her father Mike Wayne (played Kurt Douglas). I'm not sure how much of it we understood, although I do remember Brenda Vaccaro talking about all the plastic surgery her character had had, and the love scene between Melina Mercouri and Alexis Smith which was tame by today's standards. After the movie was over, my father apologized profusely to us, he was so embarrassed.
A few years later, I caught Valley of the Dolls on late night TV, and fell in love all over again (Funnily enough Susann initially hated the movie version, she thought it was too campy). I had already discovered Sidney Sheldon and Harold Robbins thanks to cable and network miniseries. Now at the age of 14, I was ready to tackle Valley of the Dolls. I also read The Bell Jar that summer but it was Valley of the Dolls that stayed with me. The story of Anne Welles, Neely O'Hara and Jennifer North was filled with backstage gossip, sex, and drugs. Just the type of thing that a mother doesn't want her 14 year old reading about, particularly a 14 year old who had already decided that she would be an actress/writer/producer. Susann took the young women in New York theme first introduced by Rona Jaffe in THE BEST OF EVERYTHING, and ramped it up to 11. This novel is so enduring that two TV series have been made from it, one in the 1980's starring Catherine Hicks as Anne and Lisa Hartman Black as Neely O'Hara, and another version that was filmed and shown on late night television in 1994. There was also a sequel to the film written by none other than Roger Ebert, and directed Russ Meyer, that so incensed Jacqueline Susann that she spent the last years of her life suing the producers.
Although I enjoyed her books, I didn't know that much about Jacqueline Susann. I grew up in the era before the Internet and Google, an age where information was literally at your fingertips. Susann had died just as PEOPLE magazine arrived on the scene. It wasn't until years later that the world rediscovered Jacqueline Susann. Out of nowhere, two competing biopics came out about her life, no doubt fueled by Michael Korda's recollections of being her editor on her second novel THE LOVE MACHINE. And a biography LOVELY ME written by author Barbara Seaman was reissued in 1996.
Susann was one of the pioneers of the glitz and glamour novels that were so prevalent in the 1980's and 1990's. There would be no Judith Krantz or Jackie Collins, if Susann hadn't paved the way. While Harold Robbins and Sidney Sheldon also wrote books with similar themes, Susann wrote from the woman's POV. According to Wikipedia, "Valley of the Dolls was an instant success when it was first published and became the bestselling book of 1966. Since then it has sold more than 30 million copies, making it one of the bestselling books of all time. As the first roman à clef by a female author to achieve this level of sales in America, it led the way for other authors such as Jackie Collins to depict the private lives of the real-life rich and famous under a veneer of fiction."
Susann was in her forties when her first book 'Every Night, Josephine!' which was based on her life with her poodle, Josephine, was published in 1963. She'd spent years trying to make it as an actress, but never quite succeeding. Oh, she appeared on Broadway and on television but the parts never led to anything bigger. She was a has-been or a 'never quite been' to be more accurate. She'd won a beauty contest in Philadelphia when she was 18 (no doubt helped by the fact that her father was one of the judges). One of the prizes was a screen test with Warner Brothers. Unfortunately Jackie was told by both the make-up artist and the cameraman that she would never make it in film because she had 'big pores.'
Susann was born in Philadelphia on August 20, 1918. Her father, Robert, was a highly successful portrait painter. Her parents seemed to have been a miss-match, Robert was an unrepentant womanizer. They argued and made-up frequently. It was her mother who added the extra 'n' to the Sephardic Jewish family name while her father kept the original spelling. Young Jacqueline adored and idolized her handsome father who took her to the theatre and to the movies. She had a rockier relationship with her perfectionist school teacher mother Rose who thought Jackie should spend more time studying and less time daydreaming and reading movie magazines. Like me, Jackie was an indifferent student in school, although she had a high I.Q. No doubt she was bored, and already planning her escape to New York. Jackie auditioned repeatedly for a local radio show The Children's Hour until they finally gave in and let her occasionally do skits on the air which she wrote herself. Despite her mother's pleas, Jackie refused to even consider college. As soon she graduated high school, she was off to New York, living in Kenmore Hall, a women's residence, pounding the pavement, looking for work. It was at Kenmore Hall that Jackie met a young actress named Elfie who would later be the prototype for Nelly O'Hara in Valley of the Dolls.
She managed to score the role of a French maid in Clare Booth Luce's new play The Women, but she was fired during rehearsals. No matter, Jackie just hung around the theatre backstage during the run of the show until she eventually got hired for a bit part as a lingerie model, making her Broadway debut on June 2, 1937. She met her husband Irving Mansfield, a press agent, when she answered the phone at Walgreen's and it turned out to be for him. He was instantly smitten and used his clout to get her name mentioned in the theatrical and gossip columns in the New York papers. Although she liked Irving, she wasn't in love with him when they married in 1939. It was a practical decision, Irving had a good job, and she figured that he could do a lot for her career.
Despite Irving's devotion to Jackie, she was never faithful to him during their marriage. She had affairs with several men including the comedians Joe E. Lewis and Eddie Cantor. "Jackie was simply crazy for Jewish comics," her friend Maxine Stewart once said. Jackie often had crushes on women as well and may have been bisexual (there were rumors that Jackie made a pass at Ethel Merman and was rejected. Supposedly she later got back at Merman when she created the character of Helen Lawson in Valley of the Dolls). but sex was not a driving factor in her life. I think in some ways, she was replicating her parent's relationship, only with the roles reversed. In 1946, Jackie gave birth to her only child, a son named Guy. By the time he was three years old, he was diagnosed as autistic. Autism was still a relatively new diagnosis, doctors suggested that electroshock therapy might help, but it only made the little boy worse. Both Jackie and Irving felt that there only choice was to put him in an institution. Ashamed and embarrassed, they told people that he was at a special school in Arizona for children with severe asthma. Jackie's main goal, besides becoming famous, was to make enough money to ensure that her son was taken care of during his lifetime.
It was a diagnosis of breast cancer that spurred Jackie to take out her typewriter to try and write a novel. She'd already written two failed plays, and wrote most of the copy for her Schiffli Lace commercials as well as produced them. She had a mastectomy in 1962, but she kept the news of her illness the same way that she had kept Guy's a secret. After her diagnosis, she bargained with God, that if he would just give her ten more years, she would really make something of her life (her husband Irving once said that she treated God like the William Morris office.) And she took her new career as a novelist seriously. It took her a year and a half to write Valley of the Dolls. Each of her books from Valley of the Dolls through Once Is Not Enough went through at least 5 drafts (each one on different colored paper) before she turned into to her editor. She wrote every day from 10-5 pm. Even then, there was still more work to be done. Jackie surprised her editors by not being a prima donna when it came to making changes. She listened to their suggestions, some she agreed with, some she didn't. She knew that her strength lay in her ear for dialogue and her characters, not so much the plot, although she diagrammed her plots on a blackboard. After reading a Harold Robbins novel and tearing it apart, Jackie decided that secret formula was giving a set of characters a common denominator. In the Valley of the Dolls, it was pills. In The Love Machine, it was television and the main character Robin Stone, for whom all the female characters fell hopelessly in love despite the fact that he was a total bastard to them.
One of the most amazing things about Jackie Susann was her resilience. Life kept giving her hard knocks but she kept on going. And it was rough. Anyone who has a passion or a talent that they don't feel is appreciated knows what Jackie went through during her years as an actress. It was literally one step forward and eight steps back. But yet she kept on. She was one of the first to do confrontational interviews on television. Of course, Mike Wallace later became famous for his hard-nosed questions, but because Jackie was a woman, she never got enough credit for it.
She also didn't get enough credit during her lifetime for her writing. Jackie knew that no one would compare her writing to Flaubert or Philip Roth, but she knew what the average reader wanted after a hard day at some blue collar job. Although some people called her novels 'literary trash,' Jackie was a storyteller. She knew how to enthrall a reader. They wanted to be entertained; they wanted to feel as if their lives were ultimately better than the rich and famous. And Jackie provided that. She knew what people wanted because she had been that girl with her nose pressed against the glass for years, trying to get into the swanky parties, hob-nobbing with the rich and famous. Valley of the Dolls (dolls are slang for pills) in some ways was probably her most autobiographical novel. Both Jackie and Irving had taken pills for years, sleeping pills to get to sleep, amphetamines to wake-up, diet pills to stay slim. For Jackie, the pills were away to numb the pain. When things were going great, she would ease off the pills, a setback would send her straight back to them.
And she worked damn hard at her writing and ultimately at promoting her books. Jackie researched her books thoroughly, she made lists of the things that she didn't know that she needed to either look up or find out from friends who might know. During the writing of The Love Machine, she claimed to have interviewed men to make sure that her portrait of the main character Robin Stone was authentic. She kept detailed notes on everyone that she met while promoting her books, and send them handwritten notes. Now managing his wife's career full-time, Irving managed to find out the names of the 125 bookstores that The New York Times polled when compiling its best seller list. He recruited friends and acquaintances for a little strategic book-buying campaign, making sure that books were bought at every single one of the 125 bookstores. Nowadays, authors' giveaway books, tchotchkes, bookmarks, key chains etc. but no one was doing that in the 1960's before Jackie. When The Love Machine came out, she had gold ankhs made up, and gave them away to every one. During her publicity tour for 'Every Night Josephine,' she dressed herself and her dog in matching leopard-patterned pillbox hats and coats for TV appearances. Nothing and no one was too small for her to pay attention to when it came to selling her books. Susann had an unmistakable look, long dark hair augmented by falls, heavy eye make-up, and Pucci dresses. I still covet Pucci dresses because of her. Jackie knew that she was her own best pitch-woman for her books. But Jackie's books succeeded not just because of the publicity because she had empathy for the female emotional experience.
Although she was only paid $3,000 for the rights to Valley of the Dolls, the paperback rights were sold to Bantam for $200,000, the movie rights went for about the same amount. Not bad for a newbie writer! For The Love Machine, Jackie's advance from Simon & Schuster climbed to $200,000, with a $250,000 promotional budget. The Mansfields managed to forge separate agreement with Bantam for the paperback rights which gave them 100% royalties, unheard of at the time. When the book debuted, it toppled Philip Roth's Portnoy's Complaint from the top spot on the best seller list.
Jacqueline Susann was the first author ever to have 3 books catapult to the number one spot on the New York Times best-selling list. Not bad for a woman who only had a high-school education! I admire many things about Jacqueline Susann but I think it's her chutzpah and her ability to keep going, even through adversity. No matter how many doors slammed in her face, Jackie kept on knocking until one opened. She was determined to make her mark, whether through acting or writing, and she succeeded beyond her wildest dreams. One of the things that Jackie worried about when she was dying was that she would be forgotten, but she was so wrong. All you have to do is look at the list of people she influenced or references to Valley of the Dolls in pop culture on Wikipedia to see that. If Jacqueline Susann was alive today, not only would she have a Facebook page, but she would have taken to Twitter like a duck to water.
Unfortunately Susann didn't live to see her books reissued by Grove in 1997 or the two biopics that were made from her life story (suck on that Harold Robbins!) starring Bette Midler and Michelle Lee. In early 1973, Susann went into the hospital for a persistent cough. The news was not good, her cancer had returned, and this time it had spread to her lungs. Susann was only given months to live, but she persisted on promoting her final book Once Is Not Enough. She passed away on September 21st, 1973 at the age of 56.
Barbara Seaman Lovely Me: The Life of Jacqueline Susann, Seven Stories Press. 1996 (2nd Edition)
Anoka Tony said…
A sad early end, but Jaqueline Susann had a successful life.
I LOVED 'Valley of the Dolls' and 'Love Machine' Jacqueline Susann was a genius as far as I'm concerned :) Love her!