text:
\section{\normalsize{\bf INTRODUCTION}}
Quantized universal enveloping algebras of Lie superalgebras
were introduced in the late 1980s \cite{Bracken, Chaichian}
to describe the type of supersymmetries exhibited
by some $2$ - dimensional statistical mechanics models \cite{Perk}.
Since then these quantum superalgebras have been intensively
studied, leading to the development of an extensive theory
on both the structure and representations.
We mention in particular that the quasi - triangular Hopf superalgebraic
structure of the quantum superalgebras was investigated in \cite{Double};
the representation theory of the type I
quantum superalgebras,  the $gl(m|n)$  super Yangians
and the quantum affine superalgebras with symmetrizable
Cartan matrices were developed in
\cite{II}. The theory of quantum superalgebras
had significant impact on
a range of areas of physics and mathematics.
Its applications to two dimensional
integrable models in statistical mechanics and quantum
field theory were extensively explored in
\cite{Bracken, Delius} and many other publications.
The  application to knot theory and $3$ - manifolds
\cite{Links}\cite{Manifolds} has
yielded many new topological invariants, notably, the
multi - parameter generalizations of Alexander - Conway polynomials.

The associated quantum supergroups are in contrast less studied
in the literature.  So far only the quantum supergroup
$GL_q(m|n)$ has been systematically investigated \cite{Zhang}.
In \cite{Zhang}, the structure and representation theories
of $GL_q(m|n)$ were developed.  The irreducible
covariant and contravariant tensorial  representations were
studied in detail within the framework of parabolic induction,
resulting in a quantum Borel - Weil theorem for
these representations. The aim of this paper is to
treat the $\osp$  series of quantum supergroups at generic $q$.

The $\osp$ - series of Lie superalgebras, $\mbox{osp}(1|32)$ in
particular, featured prominently in recent developments
of string theory.  An Inonu - Wigner contraction of
$\mbox{osp}(1|32)$ yields the
11 - dimensional Poincar\'e superalgebra with two and five form
central charges, which is the underlying symmetry of M - theory;
the superalgebra $\mbox{osp}(1|32)$ itself also plays an important
role in the theory of supermembranes \cite{Townsend}.
From a mathematical point of view,
$\osp$ is also rather exceptional amongst all the finite
dimensional simple Lie superalgebras
in that its Cartan matrix is symmetrizable, and the structure
of its finite dimensional representations   is completely understood.
In particular, it is known that all finite dimensional
representations are completely reducible.

Many properties of $\osp$ carry over to the quantum case when $q$
is generic.  It is particularly useful to recall that the Drinfeld
version of $\mbox{U}_q(\osp)$ is, algebraically,
a trivial deformation of $\mbox{U}(\osp)$\footnote{This fact is
known to experts, and may be easily inferred from results
of \cite{Scheunert}.}
in the sense of Gerstenhaber.  Therefore,
{\em finite dimensional representations of} $\mbox{U}_q(\osp)$
{\em are also completely reducible}.
This remains true for the Jimbo version of $\mbox{U}_q(\osp)$
at generic $q$.
One way to see this is through the specialization of
the indeterminate of the Drinfeld algebra to a generic
complex parameter; the other is through the
isomorphism between $\mbox{U}_q(\osp)$ and $\mbox{U}_{-q}(so(2n+1))$
established by a kind of Bose - Fermi
transmutation \cite{I}.  There is also an interesting
connection between the representation theory of
$\mbox{U}_q(\osp)$  and quantum para - statistics, details on
which can be found in \cite{Palev}.

This paper will study structural and representation theoretical
properties of the quantum supergroup $\OSPq$,  and also investigate
its underlying geometries.  This quantum supergroup will be defined
by its superalgebra of functions, which is the $\Z_2$ - graded Hopf
algebra generated by the matrix elements of the vector
representation of $\mbox{U}_q(\osp)$.  Two major results in the structure
theory are presented, namely,
the existence of a left and right integral,
which will be called a quantum Haar functional,
and a quantum Peter - Weyl theorem.

Corresponding to each reductive subalgebra $\Uk$  of $\mbox{U}_q(\osp)$,
we introduce a quantum homogeneous superspace, which is
defined by specifying its superalgebra of functions $\Eq$.
A quantum homogeneous supervector bundle over the
quantum homogeneous superspace is induced from any given
finite dimensional $\Uk$ - module. We shall show that
the space of sections $\Hq(V)$ of this bundle is projective
and is of finite type both as a left and a right module
over $\Eq$. Therefore our definition of
quantum homogeneous supervector bundles is consistent
with the general definition of noncommutative vector bundles
in Connes' theory \cite{Connes}.

Quantum homogeneous supervector bundles will be applied to
develop a theory of induced representations for $\OSPq$.
Amongst the results obtained are quantum versions of
Frobenius reciprocity and the Borel - Weil theorem.
The latter provides a concrete realization of finite
dimensional irreducible $\OSPq$ representations in terms
of quantum analogs of `holomorphic' sections of quantum
homogeneous supervector bundles.

We wish to point out that in the context of Lie supergroups
at the classical level, the mathematical theories of
homogeneous superspaces and
homogeneous supervector bundles were studied in
\cite{Manin, Penkov}.  The development of
a Bott - Borel - Weil theory was also initiated and extensively
investigated by Penkov and co - workers\cite{Penkov}.
However, complications arising from supermanifold  geometry
render these subjects very difficult to study.  So far as
we are aware, many aspects of the subjects remain to be fully
developed.  It seems that the Hopf algebraic approach
developed here and in \cite{Zhang} is also worth exploring
at the classical level, and is likely to
provide a new method complementary to the geometric approach
of \cite{Manin, Penkov}.

The organization of the paper is as follows.
Section 2 reviews some known facts about $\mbox{U}_q(\osp)$,
which will be needed later. Section 3 studies the quantum
supergroup $\OSPq$.  Section 4 investigates the quantum
homogeneous superspaces and quantum homogeneous
supervector bundles determined by this quantum supergroup,
while the last section applies results of section 4 to
study the representation theory of $\OSPq$.

\section{\normalsize{$\mbox{U}_q(\mbox{osp}(1|2n))$}}
This section reviews some known results on the
quantized universal enveloping algebra $\mbox{U}_q(\mbox{osp}(1|2n))$.
Let $E$ be the $n$ - dimensional Euclidean space spanned by
the vectors $\epsilon_i$, with the inner product $(\, ,\, )$
defined by $(\epsilon_i,\, \epsilon_j)=\delta_{i j}$.
We can express the simple roots of the Lie superalgebra
$\osp$ in terms of the $\epsilon$'s as
\ban
\alpha_i& =&\epsilon_i -\epsilon_{i+1}, \quad i=1, 2, ..., n-1, \\
\alpha_n&=&\epsilon_n,
\ee
where $\alpha_n$ is the odd simple root.
The Cartan matrix $A=\left(a_{i j}\right)_{i, j=1}^n$
of $\osp$ is then given by
$a_{ i j}={{2(\alpha_i,\,  \alpha_j)}\over{(\alpha_i,\,  \alpha_i)} }$.
An element $\mu\in E$ will be called integral if
\ban
l_i&=&{{2(\mu,\,  \alpha_i)}\over {(\alpha_i,\,  \alpha_i)} }\in\Z,
\quad \forall i<n, \\
l_n&=&{{(\mu,\,  \alpha_n)}\over {(\alpha_n,\,  \alpha_n)} }\in\Z,
\ee
and the set of all integral elements will be denoted by $\cP$.
(Note the unusual form of $l_n$.)
Set $\cP_+ =$  $\{\mu\in\cP\, |\, l_i, l_n\in\Z_+\}$. Elements of
$\cP_+$ will be called integral dominant.

The Jimbo version of the quantum superalgebra
$\mbox{U}_q(\osp)$ is a $\Z_2$ - graded complex associative
algebra generated by $\{  k_i^{\pm 1}, \ e_i, \ f_i,
\ i\in\Nn\}$, $\Nn=\{1, 2, ..., n\}$,
subject to the relations
\bea
k_i k^{-1}_i=1, &   k_i k_j = k_j k_i,  \nonumber \\
k_i e_j = q^{(\alpha_i, \ \alpha_j)} e_j k_i,
&k_i f_j= q^{-(\alpha_i, \ \alpha_j)} f_j k_i,\nonumber \\
{[}e_i, f_j\} = \delta_{i j}{{k_i - k_i}\over
{q - q }},
&\forall i, j\in I, \nonumber \\
(Ad e_i)^{1-a_{i j}}(e_j) = 0, &(Ad f_i)^{1-a_{i j}}(f_j)=0,
\ \   \forall i\ne j.
\label{definition}
\na
All the generators are chosen to be homogeneous,
with  $k^{\pm 1}_i, \ \forall i$, and $e_j$, $f_j$, $j<n$,
being even, and $e_n$, $f_n$ being odd.
For a homogeneous element $x$, we define $[x]=0$ if $x$ is even,
and $[x]=1$ when odd.  The graded commutator $[ .\  ,\ . \}$
represents the usual commutator when any one of the two arguments
is even, and the anti commutator when both arguments are odd.
The adjoint operation $Ad$ is defined by
\begin{eqnarray*}
Ad e_i(x)&=& e_i x - (-1)^{[e_i][x]} k_i x k_i^{-1} e_i, \\
Ad f_i(x)&=& f_i x - (-1)^{[f_i][x]} k_i x k_i^{-1} f_i.
\end{eqnarray*}
For $x$ being a monomial in $e_j$'s or $f_j$'s, it carries a
definite weight $\omega(x)\in H^*$. Then
$Ad e_i(x)$ $=$ $e_i x $ $-$
$(-1)^{[e_i][x]} q^{(\alpha_i, \ \omega(x))} x e_i$,
and similarly for $Ad f_i(x)$.  For convenience, we will
use the notation $\frak g$ to denote $\osp$, and
$\Uq$ to denote $\mbox{U}_q(\osp)$.
As is well known, this algebra has the structures of a ${\bf Z}_2$
graded Hopf algebra. We will denote the co-multiplication by
$\Delta$, the co-unit by $\epsilon$ and the antipode by $S$.

The representation theory of $\Uq$ was developed in \cite{I}.
For any finite dimensional $\Uq$ - module, there exists
a homogeneous basis relative to which  the $k_i$ are represented
by diagonal matrices. Here we will only consider such finite
dimensional $\Uq$ - modules
that the eigenvalues of the $k_i$ tend to $1$ as $q$ approaches $1$.
We will denote the set of all such $\Uq$ - modules by $\Mod$.
Recall that all objects of $\Mod$ are semi-simple.

If $W(\lambda)$ is a simple object of $\Mod$,
then there exists the unique (up to scalar multiples)
highest weight  vector $v_+$, such that
\begin{eqnarray*}
e_i v_+ =0, & \quad  k_i v_+ = q^{(\lambda, \ \alpha_i)} v_+,
& \quad \lambda\in\cP_+,
\end{eqnarray*}
and the module $W(\lambda)$ is uniquely determined by
the highest weight $\lambda$.
We will denote the lowest weight of
$W(\lambda)$ by $\bar\lambda$, and define
$\lambda^\dagger = -\bar\lambda$.
The dual module of $W(\lambda)$ has highest weight $\lambda^\dagger$.

The irreducible $\Uq$ - module with highest weight $\epsilon_1$
plays a special role in the representation theory of $\Uq$.
We denote this module by  $\E$, and refer to it as the vector module.
Let us now examine this module in some detail.
Denote by $w_1$ the highest weight vector of $\E$, which is assumed to
be even. Define
\ban
w_i=f_{i-1} w_{i-1}, &\quad 1<i\le n\\
w_0=f_n w_n, &\quad w_{-n} = f_n w_n, \\
w_{-j}= f_j w_{-j-1}, &\quad n>j\ge 1.
\ee
Then $\{w_\mu\, |\, \mu=0, \pm 1,\pm 2, ...,\pm n\}$
forms a weight basis of $\E$. We will denote by $t$ the
{\irrep } relative to this basis. The matrix elements of the
$e_i$, $f_i$ and $k_i$ can be immediately written down.
We have
\ban
t(e_i)_{\mu \nu} &=& \delta_{\mu i}\,  \delta_{\nu,  i+1}
                  + \delta_{\mu, -i-1}\,  \delta_{\nu,  -i}, \\
t(f_i)_{\mu \nu} &=& \delta_{\mu,  i+1}\,  \delta_{\nu i}
                  + \delta_{\mu, -i}\,  \delta_{\nu,  -i-1},  \quad i<n, \\
t(e_n)_{\mu \nu} &=& \delta_{\mu n}\,  \delta_{\nu 0}
                  - \delta_{\mu 0}\,  \delta_{\nu,  -n}, \\
t(f_n)_{\mu \nu} &=& \delta_{\mu 0}\,  \delta_{\nu n}
                  + \delta_{\mu, -n}\,  \delta_{\nu 0},\\
t(k_j)_{\mu \nu} &=& \delta_{\mu \nu}\,
                q^{(\alpha_j, \, \epsilon_\mu)},  \quad  \quad 1\le j\le n,
\ee
where $\epsilon_0=0$, and $\epsilon_{-i}=-\epsilon_i$.

Let $\{w^*_\mu\}$
be the basis of $\E^*$ defined by $w^*_\mu(w_\nu)=\delta_{\mu\nu}$.
$\E^*$ has a natural $\Uq$ - module structure with
the $\Uq$ action given by
\bea x w^*_\mu &=&\sum_\nu  (-1)^{[x]\delta_{\mu 0}}
     t(S(x))_{\mu \nu} w^*_\nu.  \label{duality}\na
The lowest weight of $\E$ is $-\epsilon_1$.
Thus the module $\E$ is self dual.
This implies that there exists a
$\Uq$ - module isomorphism $M: \E \rightarrow \E^*$, which
is unique up to scaler multiples.
The $w^*_{-1}$, being the highest weight vector of $\E^*$,
will be identified with $w_1$ so that this arbitrariness in
$M$ can be removed. Now let
\ban
w^*_\mu &=&\sum_\nu w_\nu M_{\nu \mu}.
\ee
Then
\bea
M_{\mu \nu} &=& m_\mu \delta_{\mu+\nu\, 0},\nonumber\\
m_\mu &=&\left\{ \begin{array}{l l}
                  (-q)^{\mu-1},  &\mu>0,\\
                  (-q)^n,        &\mu=0,\\
                  (-q)^{2n+\mu}, &\mu<0.
                 \end{array}\right. \label{M}
\na

It follows from earlier discussions that
repeated tensor products of $\E$ are completely reducible.
Furthermore, every finite dimensional irreducible $\Uq$
module is embedded in some $\E^{\otimes k}$ for at least one $k\ge 0$.

For later use, we consider two classes of $\Z_2$ - graded Hopf
subalgebras of $\Uq$.  Corresponding to  any subset $\bf\Theta$ of $\Nn$,
we introduce
\ban
{\cal S}_k &=&\{ k_i^{\pm 1}, i\in \Nn;\
               \ e_j, \ f_j, \ j \in {\bf\Theta}\};\\
{\cal S}_{p}&=& {\cal S}_k \cup \{ e_j,
      j \in \Nn\backslash{\bf\Theta}\}.
\ee
The elements of each set generate a $\Z_2$ - graded Hopf subalgebra of
$\Uq$.  The subalgebra generated by the elements of ${\cal S}_k$ will
be denoted by $\Uk$, and
called a reductive subalgebra of $\Uq$, while that generated by
the elements of ${\cal S}_p$ will be denoted by $\Up$ and called
a parabolic subalgebra. Note that $\Uk$ is a
$\Z_2$ - graded Hopf subalgebra of $\Up$.
If we replace $e_i$ by $f_i$ and vice versa in ${\cal S}_{p}$,
we obtain  another set, which will generate a
$\Z_2$ - graded Hopf subalgebra of $\Uq$ having similar
properties as $\Up$.  Results presented in the remainder of the paper can
also be formulated using such algebras.

Observe that there are two types of
reductive subalgebras, depending on whether $\bf\Theta$ contains $n$.
The first type arises when $n\not\in \bf\Theta$, and in this case,
$\Uk$ is the direct product of
quantized universal enveloping algebras associated with
a series of ordinary (i.e., non - graded) Lie algebras of
type $A$ supplemented by the algebra generated by some
$k_i^{\pm 1}$.  The second type arises when
$n\in\bf\Theta$.  This time, $\Uk$ is the direct product of
the first type with a  $\mbox{U}_q(osp(1|2m))$ for some $m<n$.
In both cases, the finite dimensional representations of
$\Uk$ are completely reducible.
This fact will be of great importance to the main
subject of the paper.

Let $V_\mu$ be a finite dimensional irreducible $\Uk$-module.  Then
$V_\mu$ is of highest weight type.  Let $\mu$ be the highest weight
and $\tilde\mu$ the lowest weight of $V_\mu$ respectively.  We can
extend $V_\mu$ in a unique fashion to a $\Up$-module, which is still
denoted by $V_\mu$, such that the elements of ${\cal S}_{p}\backslash
{\cal S}_k$ act by zero.  It is not difficult to see that all finite
dimensional irreducible $\Up$-modules are of this kind.

Consider a finite dimensional irreducible $\Uq$-module
$W(\lambda)$, with highest weight $\lambda$  and lowest weight
$\bar\lambda$.  $W(\lambda)$ can be restricted  in a natural way
to a $\Up$-module, which
is always indecomposable, but not irreducible in general.
It can be readily shown  that
\ban
\dim_{\C} \mbox{Hom}_{\Up} ( W(\lambda), \ V_\mu) &=&
    \left\{ \begin{array}{l l}
           1, &  \bar\lambda={\tilde\mu}, \\
           0, &  \bar\lambda\ne {\tilde\mu}.
     \end{array}\right.  \label{Hom}
\ee

\section{\normalsize {\bf THE QUANTUM SUPERGROUP} $\mbox{OSP}_q(1|2n)$}
There exist well established methods
for quantizing ordinary Lie groups in
the nonsupersymmetric setting (See \cite{Chari} and references therein.).
These methods can also be extended
to construct $\OSPq$, and this will be done here. However,
we should point out that it is in general much more difficult
to study quantum supergroups.  See \cite{Zhang} for details
on $\mbox{GL}_q(m|n)$.

We will show that the quantum supergroup $\OSPq$ admits a
quantum Haar functional, and also  a Peter - Weyl basis.
This, however, is an exception rather than the rule.
It is known that the finite dimensional representations of
all the quantum superalgebras but $\mbox{U}_q(\osp)$ are
not completely reducible.  This fact renders it
impossible to construct Peter - Weyl bases for
the corresponding quantum supergroups (which are yet to be
defined except $\mbox{GL}_q(m|n)$.).
It also forbids the existence of quantum Haar functionals
on the quantum supergroups in view of the dual Maschke theorem.

Let us recall some general results about $\Z_2$ - graded Hopf algebras.
Let $A$ be a $\Z_2$ - graded Hopf algebra with co - multiplication
$\Delta$, co - unit $\epsilon$ and antipode $S$. We define the finite
dual $A^0$ of $A$ to be a subspace of $A^*$ such that
for any $f\in A^0$, $Ker f$ contains a two-sided ideal $\cal I$
of $A$ which is of finite co-dimension, i.e.,
$\dim A/{\cal I}<\infty$. Of course in the most general
situation, there is no guarantee that $A^0$ will not be zero.
But when $A^0$ is nontrivial, then it is
also a $\Z_2$ - graded Hopf algebra with a structure
dualizing that of $A$. More explicitly, the multiplication
is defined, for $f,\, g\in A^0$, $a,\, b\in A$,  by
\ban
\langle f g,\ a\rangle &=& \langle f\otimes g, \ \Delta(a)\rangle\\
                &=&\sum_{(a)} (-1)^{[g][a_{(1)}]}
 \langle f,\ a_{(1)} \rangle \langle g, \ a_{(2)} \rangle.
\ee
It is easy to see that the unit of $A^0$ is $\epsilon$.
Denote the co - multiplication, the co - unit and the
antipode of $A^0$ respectively by $\Delta_0$, $\epsilon^0$ and $S_0$.
Then
\ban
\langle \Delta_0(f),\  a\otimes b\rangle &=&
        \sum_{(f)} (-1)^{[f_{(1)}][f_{(2)}]} \langle f_{(1)}, \ a\rangle
                   \langle f_{(2)},\ b\rangle \\
&=&\langle f, \  a b\rangle,\\
\langle S_0(f),\ a\rangle &=& \langle f,\ S(a)\rangle, \\
\epsilon^0(f) &=& \langle f,\ \hbox{\sf 1\kern-.4em\hbox{1}}_A \rangle.
\ee

Now we come back to the quantum supergroup $\mbox{OSP}_q(1|2n)$.
As is well known, we can not define the quantum supergroup
directly. Instead, we need to find the algebra of functions
on it. Introduce $t_{\mu \nu}\in (\Uq)^*$,
$\mu, \ \nu=0, \pm 1, \pm 2, ..., \pm n$, defined by
\ban
t_{\mu \nu}(x)&=& t(x)_{\mu \nu}, \quad \forall x\in\Uq,
\ee
where $t$ is the vector representation of $\Uq$. We call
the $t_{\mu \nu}$ the matrix elements of $t$.
Finite dimensionality of $\E$ implies that $t_{\mu \nu}\in(\Uq)^0$,
$\forall \mu,\ \nu$.

We define the superalgebra $\Tq$ of functions
on $\OSPq$  to be the
$\Z_2$-graded  subalgebra
of $(\Uq)^0$ generated by the matrix elements of the vector
representation of $\Uq$, i.e., $t_{\mu \nu}$,
$\mu, \ \nu=0, \pm 1, \pm 2, ..., \pm n$.  Then
\begin{theorem}
\begin{enumerate}
\item $\Tq$ is a $\Z_2$ - graded Hopf algebra.
\item Let $t^{(\lambda)}$ be the \irrep \  of $\Uq$ with
highest weight $\lambda\in\cP_+$, and let $t^{(\lambda)}_{i j}$,
$i,\, j=1, 2, ..., d_\lambda$ ($d_\lambda= \mbox{dim}t^{(\lambda)})$,
be the matrix
elements of $t^{(\lambda)}$. Then
\bea
\Tq&=&\bigoplus_{\lambda\in\cP_+} \bigoplus_{i, j=1}^{d_\lambda}
 \C t^{(\lambda)}_{i j}.
\na
\end{enumerate}
\end{theorem}
\begin{proof} {Proof}
The $\Z_2$ - graded bi - algebra structure of
$\Tq$ is obvious, and the existence of the antipode
follows from the self duality of the vector module $\E$ over $\Uq$.
Part 2) immediately follows from the complete
reducibility of finite dimensional
representations of $\Uq$.
\end{proof}

Let us now work out the explicit forms of the
co - multiplication and the antipode.  The co - multiplication is
given by
\ban
    \Delta_0(t_{\mu \nu}) &=& \sum_{\sigma} (-1)^{(\delta_{\mu 0}
+\delta_{\sigma 0})(\delta_{\nu 0} + \delta_{\sigma 0})}
     t_{\mu \sigma}\otimes t_{\sigma \nu}.
\ee
The antipode can be constructed from (\ref{duality}) by using the
$\Uq$ - module isomorphism $M$. We have
\ban
S_0(t_{\mu \nu})&=& (-1)^{(\delta_{\mu 0} + \delta_{\nu 0})\delta_{\mu 0}}
                     (M^{-1} t M)_{\nu \mu}\\
                &=& (-1)^{(\delta_{\mu 0} + \delta_{\nu 0})\delta_{\mu 0}}
                 \ { {m_{-\mu}\ t_{-\nu,\,  -\mu} }\over{m_{-\nu}} },
\ee
where $m_\mu$ is given by (\ref{M}).

Here we introduce more notations for later use.
Let $\{w^{(\lambda)}_i | i=1, 2, ..., d_{\lambda}\}$ be
the homogeneous basis of $W(\lambda)$ with respect to which
the representation $t^{(\lambda)}$
is defined.  We denote by $\{{\tilde w}^{(\lambda)}_i
| i=1, 2, ..., d_{\lambda}\}$
the basis of $W(\lambda)^*=W(\lambda^\dagger)$ such that
${\tilde w}^{(\lambda)}_i(w_j)=\delta_{i j}.$
The $\Uq$ - module structure of $W(\lambda)^*$
enables us to define
${\tilde t}^{(\lambda)}_{i j}\in\Tq$ by
\begin{eqnarray*}
x {\tilde w}^{(\lambda)}_i &=&\sum_{j} {\tilde t}^{(\lambda)}_{j i}(x)
{\tilde w}^{(\lambda)}_j,
\quad \forall x\in\Uq.
\end{eqnarray*}
Then
\begin{eqnarray*}
{\tilde t}^{(\lambda)}_{j i}
&=& (-1)^{[i]([i]+[j])} S_0(t^{(\lambda)}_{i j}),
\end{eqnarray*}
where $[i]=0$ or $1$ depending on whether $w_i$ is even or odd.
Cleary the ${\tilde t}^{(\lambda)}_{j i}$ are linear combinations
of $t^{(\lambda^\dagger)}_{i j}$. Furthermore,
the ${\tilde t}^{(\lambda)}_{j i}$, $\forall\lambda\in{\cal P}_+$,
also form a basis of $\Tq$.

From here on, we will omit the subscript $_0$ from $\Delta_0$
and $S_0$.

Let us now turn to the discussion of
a Haar functional on the quantum supergroup $\Tq$.
But before embarking on this task, we first consider the notion
of an integral on an arbitrary  $\Z_2$ - graded Hopf algebra $A$.
Let $A^*$ be its dual, which has a natural
$\Z_2$ - graded algebraic structure induced by the co-algebraic
structure of $A$.
An even homogeneous element $\int^l\in A^*$ is called a
left integral on $A$ if
\begin{eqnarray*}
f\cdot \int^l &=& \langle f, \hbox{\sf 1\kern-.4em\hbox{1}}_A\rangle\ \int^l, \ \ \ \forall f\in A^*.
\end{eqnarray*}
Similarly,  an even homogeneous element
$\int^r\in A^*$ is called a right integral on $A$ if
\begin{eqnarray*}
\int^r\cdot f &=& \langle f, \hbox{\sf 1\kern-.4em\hbox{1}}_A\rangle \ \int^r, \ \ \ \forall f\in A^*.
\end{eqnarray*}
A straightforward calculation shows that the defining properties of
the integrals are equivalent to the following requirements
\begin{eqnarray}
(\id\otimes \int^l)\Delta(x)=\int^l x, \quad
(\int^r\otimes \id)\Delta(x)=\int^r x,\quad  \forall x\in A.
\end{eqnarray}
where ${\rm id}$ is the identity map on $A$.

A Haar functional $\int\in A^*$ on $A$ is an integral on $A$ which is
both left and right, and sends $\hbox{\sf 1\kern-.4em\hbox{1}}_A$ to $1$, i.e.,
\begin{eqnarray}
&(i).&  (\int\otimes \id) \Delta(x) = (\id\otimes \int)\Delta(x)
       = \int x, \quad \forall x\in A,\nonumber\\
&(ii).&  \int \hbox{\sf 1\kern-.4em\hbox{1}}_A  = 1. \label{integral}
\end{eqnarray}

In the case of $\Tq$,  it is an entirely straightforward
matter to show that
\begin{theorem}
The element $\int\in (\cT)^*$ defined by
\begin{eqnarray}
\int \hbox{\sf 1\kern-.4em\hbox{1}}_{\cT}=1; \quad  \quad \int t^{(\lambda)}_{i j}=0,
\quad  0\ne \lambda\in \cP_+, \nonumber
\label{Haar}
\end{eqnarray}
gives rise to a Haar functional on $\cT$.
\end{theorem}

Denote by $2\rho$ the sum of the positive roots of $\frak g$.
Let $K_{2\rho}$ be the product of powers of $k_i^{\pm 1}$'s such that
\begin{eqnarray*}
K_{2\rho} e_i K_{2\rho}^{-1} &=& q^{(2\rho, \ \alpha_i)} e_i,
\quad \forall i.\end{eqnarray*}
Then it can be easily shown  that
\begin{eqnarray*}
S^2(x) &=& K_{2\rho} x K_{2\rho}^{-1}, \quad \forall x\in\Uq.\end{eqnarray*}
We define the quantum super-dimension of the
irreducible $\Uq$-module  $W(\lambda)$ by
\begin{eqnarray*}
\mbox{SD}_q(\lambda)&:=& Str\{t^{(\lambda)}(K_{2\rho})\}.
\end{eqnarray*}
For quantum superalgebras other than the $\osp$ series,
there exists a class of finite dimensional irreducible representations,
the typicals, of which the super-dimensions vanish identically.
Again, $\mbox{U}_q(\osp)$ is an exception, and we have the following
important property: for any irreducible $\mbox{U}_q(\osp)$ -
module $W(\lambda)$ with highest weight $\lambda\in\cP_+$,
\begin{eqnarray*}
\mbox{SD}_q(\lambda)&\ne& 0.
\end{eqnarray*}
Now the Haar functional $\int$ satisfies the following properties.
\begin{lemma}\label{tresult}
\begin{eqnarray}
\int t^{(\lambda)}_{i j} {\tilde t}^{(\mu)}_{r s}
(-1)^{[j] [r] + [i]+[j]}
&=& \delta_{i r} \delta_{\lambda  \mu}
{{t^{(\lambda)}_{s j}(K_{2\rho})}\over{\mbox{SD}_q(\lambda)} }, \nonumber \\
\int {\tilde t}^{(\lambda)}_{i j} t^{(\mu)}_{r s}
(-1)^{[j][r]}&=&
\delta_{j s} \delta_{\lambda  \mu}
{{{\tilde t}^{(\lambda)}_{ i r}(K_{2\rho})}\over{\mbox{SD}_q(\lambda)} }.
\end{eqnarray}
\end{lemma}
\begin{proof} {Proof}
Consider the first equation.
The $\lambda \ne \mu$ case is easy to prove:
the integral vanishes because
the tensor product $W(\lambda)\otimes W(\mu^\dagger)$
does not contain the trivial $\Uq$-module.
When   $\lambda = \mu$,  we introduce the notations
\begin{eqnarray*}
\phi_{i r; s j} = \int t^{(\lambda)}_{i j}
                    {\tilde t}^{(\lambda)}_{r s}
                  (-1)^{[j] [r] + [i]+[j]};
&\quad \Phi[s, j] = \left( \phi_{i r; s j}\right)_{i, r=1}^{d_\lambda};
&\quad \Psi[i, r]= \left( \phi_{i r; s j}\right)_{s, j=1}^{d_\lambda}.
\end{eqnarray*}
It is clearly true that
$Str\left(\Psi[i, r]\right)$ $=$ $\delta_{i r}$.

Note that corresponding to each $x\in\Uq$, there exists an
${\tilde x}\in (\cT)^*$ defined by
${\tilde x}(a) = \langle a, \ x\rangle$, $\forall a\in\cT$.
The left integral property of $\int$ leads to
\begin{eqnarray*}
\epsilon(x)\phi_{i r; s j} &=& ({\tilde x}.\int)t^{(\lambda)}_{i j}
{\tilde t}^{(\lambda)}_{r s} (-1)^{[j] [r] + [i]+[j]}\\
&=& \sum_{(x)} \sum_{i', r'} t^{(\lambda)}_{i i'}(x_{(1)})
t^{(\lambda)}_{ r' r}(S(x_{(2)})) \phi_{i' r'; s j} (-1)^{[x]([i]+[j])
+ [x_{(2)}]([j]+[s])},\\
\mbox{i.e.}\quad  \epsilon(x) \Phi[s, j]
&=& \sum_{(x)} t^{(\lambda)}(x_{(1)}) \Phi[s, j] t^{(\lambda)}(S(x_{(2)}))
(-1)^{[x_{(2)}]([j]+[s])},
\quad \forall x\in\Uq.
\end{eqnarray*}
Schur's lemma forces $\Phi[s, j]$ to be proportional to the identity
matrix, and we have
\begin{eqnarray*}
\Psi[i, r]&=&\delta_{i r} \psi,
\end{eqnarray*}
for some $d_\lambda \times d_\lambda$ matrix $\psi$.
The right integral property of $\int$ leads to
\begin{eqnarray*}
\epsilon(y) \psi &=& \sum_{(y)} t^{(\lambda)}(K_{2\rho})
t^{(\lambda)}(y_{(1)}) t^{(\lambda)}(K_{2\rho}^{-1}) \psi
t^{(\lambda)}(S(y_{(2)})).
\end{eqnarray*}
Again by using Schur's lemma we conclude that $\psi$ is proportional
to $t^{(\lambda)}(K_{2\rho})$. Since its supertrace is $1$, we have
\begin{eqnarray*} \psi &=& { {t^{(\lambda)}(K_{2\rho})}\over{\mbox{SD}_q(\lambda)} }. \end{eqnarray*}
This completes the proof of the first equation of the lemma.
The second equation can be shown in exactly the same way.
\end{proof}

It is worth observing that this Lemma and part 2) of Theorem 1
provide a quantum analog of the Peter - Weyl
theorem for $\OSPq$.

\section{\normalsize{\bf QUANTUM HOMOGENEOUS SUPERVECTOR BUNDLES}}
In this section we will investigate the quantum homogeneous
superspaces and quantum homogeneous supervector bundles
arising from the quantum supergroup $\OSPq$ by adapting the
methods and techniques of \cite{Zhang, Gover} to the
present context.
Let us start by introducing two types of actions of $\Uq$ on $\cT$.
The first action will be denoted by $\circ$, which
corresponds to the right translation in the classical theory
of Lie groups.  It is defined by
\begin{eqnarray}
x\circ f&=&\sum_{(f)} (-1)^{[f_{(1)}][f_{(2)}]} \
         f_{(1)} \ \langle f_{(2)}, \ x\rangle,
\quad x\in\Uq, \ f\in\cT,  \label{circ}
\end{eqnarray}
Straightforward calculations show that
\begin{eqnarray*} x\circ (y \circ f ) &=& (x y)\circ f;\\
(x\circ f) ( y ) &=& f(y x), \\
(id_{\cT}\otimes x\circ) \Delta(f) &=& \Delta( x\circ f ).  \end{eqnarray*}
The other action, which corresponds to the
left translation in the classical  Lie group theory,
will be denoted by $\cdot$. It is defined by
\begin{eqnarray}
x\cdot f &=&\sum_{(f)} \langle f_{(1)}, \ S^{-1}(x) \rangle f_{(2)}.
\end{eqnarray}
It can be easily shown that
\begin{eqnarray*}
(x\cdot f)(y)&=& (-1)^{[x][y]} f( S^{-1}(x) y ),\\
x\cdot(y\cdot f)&=& (x y)\cdot f, \quad  x,\ y\in \Uq, \ f\in \cT.
\end{eqnarray*}
Furthermore, the two actions graded - commute in the following sense
\begin{eqnarray*}
x\circ(y\cdot f)&=&(-1)^{[x][y]} y\cdot(x\circ f),
\quad  x, y\in\Uq, \ f\in\cT.
\end{eqnarray*}

Let $V$ be a finite dimensional module over $\Uk$.
We extend the actions $\circ$ and $\cdot$ trivially to
$V\otimes \Tq$: for any
$\zeta= \sum v_i \otimes f_i  \in V\otimes\Tq$,
\begin{eqnarray*}
x\cdot\zeta &=& \sum (-1)^{[x][v_i]} v_i \otimes x\cdot f_i, \\
x\circ\zeta&=& \sum (-1)^{[x][v_i]} v_i \otimes x\circ f_i,
\ \ \ \  x\in\Uq.
\end{eqnarray*}

We now introduce two important definitions:
\begin{eqnarray}
\Eq&:=&\left\{ f\in\Tq \ | \ x\circ f =\epsilon(x) f,
\quad \forall x\in\Uk \right\};   \label{space}\\
\Hq(V)&:=& \left\{ \zeta\in V\otimes \Tq \
| \ x\circ\zeta = (S(x)\otimes id_{\mbox{${\cal A}_q$}}) \zeta, \ \forall x\in\Uk\right\}.
\label{section}
\end{eqnarray}

The remainder of this section is devoted to studying the properties
of these objects. Let us first prove the following
\begin{proposition}\label{plenty}
1). $\Eq$ is an infinite dimensional subalgebra of $\Tq$. \\
2). $\Hq(V)$ is an infinite dimensional supervector space
if the weight of any vector of $V$ is $\Uq$-integral, and is
zero otherwise.
\end{proposition}
\begin{proof} {Proof}
We first show that
$\Eq$ is a subalgebra of $\Tq$.
Since $\Uk$ is  a Hopf subalgebra of $\Uq$, for any $x\in\Uk$,
$\Delta(x)=\sum_{(x)}x_{(1)}\otimes x_{(2)}$ $\in\Uq\otimes\Uq$.
Hence
\begin{eqnarray*}
x\circ ( a b )&=&
\sum_{(x)} (-1)^{[x_{(2)}][a]}  \{x_{(1)}\circ a\} \{ x_{(2)}\circ b\}
   =\epsilon(x) ab,
\end{eqnarray*}
that is, $a b\in\Eq$.

Since the finite dimensional representations of $\Uk$ are completely
reducible, the study of properties of $\Hq(V)$  reduces to the case
when $V$ is irreducible. Let $V_\mu$ be a
finite dimensional irreducible $\Uk$ -module with highest weight $\mu$
and lowest weight $\tilde\mu$.
Any element $\zeta$ $\in$  $\Hq(V_\mu)$
can be expressed in  the form
\begin{eqnarray*}
\zeta &=& \sum_{\lambda\in\cP_+}
\sum_{i, j} v_{i j}^{(\lambda)}\otimes {\tilde t}_{i j}^{(\lambda)},
\end{eqnarray*}
for some $v_{ i j}^{(\lambda)}\in V_\mu$.
Fix an arbitrary $\lambda\in\cP_+$. For any nonvanishing
$w\in W(\lambda)$, the following linear map is clearly surjective:
\begin{eqnarray*}
\mbox{Hom}_{\C}(W(\lambda), \ V_\mu)\otimes w &\rightarrow& V_\mu,\\
\phi\otimes w&\mapsto& \phi(w).
\end{eqnarray*}
Thus there exist $\phi^{(\lambda)}_i$
$\in$ $\mbox{Hom}_{\C}(W(\lambda), \ V_\mu)$ such that
$v_{ i j}^{(\lambda)} = \phi^{(\lambda)}_i ( w_j^{(\lambda)} )$,
where $\{w_i^{(\lambda)}\}$ is the basis of $W(\lambda)$ discussed
before. Therefore,  we can rewrite $\zeta$ as
\begin{eqnarray*}
\zeta &=& \sum_{\lambda\in\cP_+}
\sum_{i, j} \phi_i^{(\lambda)}(w_j^{(\lambda)})
\otimes {\tilde t}_{i j}^{(\lambda)}.
\end{eqnarray*}
The defining property of $\Hq(V_\mu)$ states that
\begin{eqnarray*}
\ell\circ\zeta&=& ( id_{\cT}\otimes S(\ell)) \zeta, \quad \forall \ell\in\Uk .
\end{eqnarray*}
Thus  we have
\begin{eqnarray*}
\sum_{\lambda\in\cP_+} \sum_{i, j, k}
t_{j k }^{(\lambda)}(S(\ell)) \phi_i^{(\lambda)}(w_j^{(\lambda)})
\otimes (-1)^{[\ell] [\phi_i^{(\lambda)}]} {\tilde t}_{i k}^{(\lambda)}
&=& \sum_{\lambda\in\cP_+} \sum_{i, j} S(\ell)
\phi_i^{(\lambda)}(w_j^{(\lambda)}) \otimes {\tilde t}_{i j}^{(\lambda)}.
\end{eqnarray*}
Recalling that the ${\tilde t}_{k i}^{(\lambda)}$ are linearly
independent. So the above  is equivalent to
$$
\ell \phi_i^{(\lambda)}(w_j^\lambda)
= (-1)^{[\ell] [\phi_i^{(\lambda)}]}
\phi_i^{(\lambda)}(\ell w_j^\lambda), ~~~ \forall \ell\in\Uk .
$$
This equation is precisely the statement that the $\phi_i^{(\lambda)}$
be $\Uk$-module homomorphisms of degrees $[\phi_i^{(\lambda)}]$,
\begin{eqnarray*}
\phi_i^{(\lambda)}&\in&\mbox{Hom}_{\Uk}\left(W(\lambda), V_\mu\right)
\subset \mbox{Hom}_{\C}\left(W(\lambda), V_\mu\right),
\quad \forall i.
\end{eqnarray*}
Thus finding sections in $\Hq(V_\mu)$ is equivalent to finding,
for all $\lambda\in\cP_+$, the homomorphisms $\phi^{(\lambda)}$ $\in$
$\mbox{Hom}_{\Uk}\left(W(\lambda), V_\mu\right)$.
Note that each such homomorphism $\phi^{(\lambda)}$
determines $d_\lambda$ linearly independent sections
$$
\zeta^{(\lambda)}_i =  \sum_{j}
\phi^{(\lambda)}(w_j^{(\lambda)})\otimes {\tilde t}_{i j}^{(\lambda)}.
$$

However, when $\mu$ is not integral with respect to $\Uq$,
$\mbox{Hom}_{\Uk}\left(W(\lambda), V_\mu\right)=0$, and hence
$\Hq(V_\mu)$ vanishes in this case.

Now consider the case with $\mu=0$, we have $\Hq(V_{\mu=0})=\Eq$
as supervector spaces.
There is a homomorphism from the trivial representation of $\Uq$,
$W(0)={\bf C}$, onto $V_0={\bf C}$. This gives the constant sections of
$\Eq$.  Let $\gamma$ be the highest root of $\frak g$.
Recall that in the classical situation,
${\frak k}$ is reductive with
$N=r-|{\bf\Theta}|$ independent central elements.
This,  transcribed to the quantum case,
implies the existence of $N$ linearly independent $\Uk$-homomorphisms
$W(\gamma)\to {\bf C}$.
As mentioned above each of these corresponds
to $d = \mbox{dim} ({\frak g})$  linearly independent sections.
So the representation
$W(\gamma)$ determines $ N d$
linearly independent sections.
Further linearly independent sections can  be obtained using the
following lemma
\begin{lemma} \label{prodhom}
Suppose there are non-trivial $\Uk$-homomorphisms
$W(\lambda_1)\to V_{\mu_1}$
and $W(\lambda_2)\to V_{\mu_2}$. Then there is an induced non-trivial
$\Uk$-homomorphism
$$
W(\lambda_1+\lambda_2)\to V_{\mu_1+\mu_2}.
$$
\end{lemma}
For example, for any positive integer $m$, there exist $(m|N)$
(partition of $m$ into $\leq N$ parts) linearly independent
homomorphisms $W(m\gamma)\to {\bf C}$.  Thus we have proved that
the algebra $\Eq$ is infinite dimensional.

Now let us consider the case with $0\neq\mu\in \cP$.
It is an elementary exercise to verify that $V_\mu$ is
$\Uk$-isomorphic to a $\Uk$-irreducible part of $W(\lambda')$,
where $\lambda'$ is the dominant weight in the Weyl group orbit
of $\mu$. Thus there is a non-trivial $\Uk$-homomorphism
$$ W(\lambda') \to V_\mu , $$
and this  determines at least  $d_{\lambda'}$ linearly independent
sections in $\Hq(V_\mu)$. Further linearly independent sections
can be constructed explicitly using lemma
\ref{prodhom} which promises a family of homomorphisms
$$
W(\lambda'+m\gamma)\to V_\mu ~~~~~m\in {\bf N}_+.
$$
This establishes that $\Hq(V_\mu)$ is infinite dimensional.
\end{proof}

$\Eq$ may be regarded as the quantum analog of the algebra of
functions over the superspace $\mbox{OSP}(1|2n)/K$, where $K$
is the subgroup of $\mbox{OSP}(1|2n)$ with Lie superalgebra ${\frak k}$.
Such homogeneous superspaces were studied in the work of Manin
\cite{Manin}, Penkov \cite{Penkov} and others.
Here we wish to make some investigations into their quantum analogs.

As is well known,  one can not define a noncommutative (in the
$\Z_2$ - graded sense) space directly in geometrical
terms. Instead,  such a space  has to be defined by specifying
its algebra of functions.  We will take $\Eq$ as
the algebra of functions over the quantum homogeneous superspace
which corresponds to $\mbox{OSP}(1|2n)/K$ in the classical situation.
Let us now study properties of $\Hq(V)$. First observe that
\begin{theorem}
$\Hq(V)$ furnishes  a two-sided  $\Eq$ module
     under the multiplication of $\Tq$.
\end{theorem}
\begin{proof} {Proof}
The left and right actions of $\Eq$
on $\Hq(V)$ are respectively defined by
\begin{eqnarray*}
a\zeta&=&\sum_r (-1)^{[a][v_i]} v_i\otimes a f_i,\\
\zeta a&=&\sum_r v_i\otimes f_i a,
\end{eqnarray*}
where $a\in\Eq$ and $\zeta=$ $\sum_i v_i\otimes f_i$ $\in\Hq(V)$.
Now for $p\in\Uk$,
\begin{eqnarray*}
p\circ(a\zeta) &=& \sum_{(p)} (-1)^{[p_{(2)}][a]}
                 \{ p_{(1)}\circ a \} \{p_{(2)}\circ \zeta\}\\
               &=& (-1)^{[p][a]} a\{p\circ \zeta\}
                = (S(p)\otimes id_{\mbox{${\cal A}_q$}}) a\zeta;\\
p\circ(\zeta a)&=& \sum_{(p)} (-1)^{[p_{(2)}][\zeta]}
                \{p_{(1)}\circ \zeta\} \{ p_{(2)}\circ a \}\\
               &=&\{p\circ \zeta\}a= (S(p)\otimes id_{\mbox{${\cal A}_q$}}) \zeta a.
\end{eqnarray*}
This completes the proof.
\end{proof}

When $V$ is actually a $\Uq$ - module,
the $\Eq$ - module $\Hq(V)$ has a particularly simple structure.
\begin{proposition}\label{projective}
Let $W$ be a finite dimensional left $\Uq$-module, which we
regard as a left $\Uk$-module by restriction.
Then $\Hq(W)$ is isomorphic to
$W\otimes \Eq$ either as a left or right $\Eq$-module.
\end{proposition}
\begin{proof} {Proof}
We first construct the right $\Eq$-module
isomorphism.
Being a  left $\Uq$-module, $W$ carries a natural
right $\mbox{${\cal A}_q$}$  co-module structure with the co-module action
$\delta: W \rightarrow W\otimes\cT$
defined by
\begin{eqnarray}
\delta(w)(x) &=& x w, \quad  x\in\Uq, \ \ w\in W.
\label{comodule}
\end{eqnarray}
(Here the notation requires some clarification. If we express
$\delta(w)=\sum_{(w)} w_{(1)}\otimes w_{(2)}$,
then $\delta(w)(x) = \sum_{(w)} (-1)^{[x][w_{(1)}]}
w_{(1)} \langle w_{(2)}, \, x\rangle$.)
Define $\eta: W\otimes\Tq \longrightarrow W\otimes\Tq$
by the composition of maps
\begin{eqnarray*}
W\otimes\Tq  \stackrel{\delta\otimes id}{\longrightarrow}
W\otimes \Tq\otimes\cT
\longrightarrow W\otimes\Tq,
\end{eqnarray*}
where the last map is the multiplication of $\mbox{${\cal A}_q$}$.
Then $\eta$ defines a right $\Eq$-module isomorphism,
with the inverse map given by the composition
\begin{eqnarray*}
W\otimes\Tq \stackrel{\delta\otimes id}{\longrightarrow}
W\otimes\Tq\otimes\Tq
\stackrel{(id\otimes S\otimes id)}{\longrightarrow}
W\otimes \cT\otimes \Tq\longrightarrow W\otimes \Tq,
\end{eqnarray*}
where the last map is again  the multiplication of $\mbox{${\cal A}_q$}$.
It is not difficult to show that
\begin{eqnarray*}
x\circ\eta(\zeta)&=&\eta\left(\sum_{(x)}(x_{(1)}\otimes  id_{\mbox{${\cal A}_q$}})
                     x_{(2)}\circ \zeta\right),\\
x\circ\eta^{-1}(\zeta)&=&\eta^{-1}\left(\sum_{(x)}
(S(x_{(1)})\otimes  id_{\mbox{${\cal A}_q$}}) x_{(2)}\circ\zeta\right),
\quad \forall \zeta\in\Tq\otimes W, \ \  x\in\Uq.
\end{eqnarray*}
Consider $\zeta\in\Hq(W)$. We have
\begin{eqnarray*}
p\circ\eta(\zeta)&=&
  \eta\left(\sum_{(p)} (p_{(1)}\otimes
   id_{\mbox{${\cal A}_q$}})p_{(2)}\circ\zeta\right)\\
&=&\eta\left(\sum_{(p)}  (p_{(1)} S(p_{(2)})\otimes
   id_{\mbox{${\cal A}_q$}})\zeta\right)\\
&=&\epsilon(p)\eta(\zeta), \quad  \forall \ p\in\Uk.
\end{eqnarray*}
Hence $\eta(\Hq(W))\subset W\otimes\Eq$.
Conversely, given any $\xi\in  W\otimes \Eq$, we have
\begin{eqnarray*}
p\circ\eta^{-1}(\xi)&=&
\eta^{-1}\left(\sum_{(p)} (S(p_{(1)})\otimes  id_{\mbox{${\cal A}_q$}})
p_{(2)}\circ \xi\right)\\
&=&\eta^{-1}\left(\sum_{(p)}(S(p_{(1)})\epsilon(p_{(2)})
\otimes  id_{\mbox{${\cal A}_q$}})\xi\right)\\
&=& (S(p)\otimes  id_{\mbox{${\cal A}_q$}}) \eta^{-1}(\xi),  \quad  \forall \ p\in\Uk.
\end{eqnarray*}
Thus $\eta^{-1}(W\otimes \Eq)\subset\Hq(W)$.
Therefore the restriction of $\eta$ to $\Hq(W)$  provides
the desired right $\Eq$-module isomorphism.

The left module isomorphism is given by the restriction to $\Hq(W)$
of the linear map
$\kappa: W\otimes \Tq$ $\rightarrow$
$W\otimes \Tq$, which is defined by the following composition of maps
\begin{eqnarray*}
W\otimes \Tq \stackrel{\delta\otimes id}{\longrightarrow}
W\otimes\Tq\otimes\Tq
\stackrel{id\otimes P(S^2\otimes id)}{\longrightarrow}
W\otimes\Tq\otimes \Tq\longrightarrow W\otimes \Tq,
\end{eqnarray*}
where
\begin{eqnarray}
P: \Tq\otimes\Tq&\rightarrow &\Tq\otimes \Tq, \nonumber \\
   a\otimes b &\mapsto& (-1)^{[a][b]}\,  b \otimes a. \label{P}
\end{eqnarray}
The inverse map $\kappa^{-1}$ is given by
\begin{eqnarray*}
W\otimes \Tq \stackrel{\delta\otimes id}{\longrightarrow}
W\otimes\Tq\otimes\Tq
\stackrel{id\otimes P(S\otimes id)}{\longrightarrow}
W\otimes\Tq\otimes \Tq\longrightarrow W\otimes \Tq.
\end{eqnarray*}
\end{proof}

With the help of this Proposition, we can now prove the following
important result.
\begin{theorem}\label{proj}
$\Hq(V)$ is projective and of finite type
both as a left and right module over the superalgebra $\Eq$ of
functions on the quantum homogeneous superspace.
\end{theorem}
\begin{proof} {Proof}
Since $\Uk$ is a reductive subalgebra of $\Uq$,
all finite dimensional representations of $\Uk$
are completely reducible.
Let $V_s$, $s=1, 2, ..., K\le\infty$, be the irreducible
direct summands of $V$ such that their weights are all integral
with respect to $\Uq$. Then $\Hq(V)=\bigoplus_s \Hq(V_s)$.
Consider any $V_s$, and denote its highest weight by $\mu_s$.
There exists such a ${\hat\mu}_s$ in the Weyl group orbit of
${\frak g}$ that is integral dominant with respect to $\frak g$.
Let $W({\hat\mu}_s)$ be the irreducible $\Uq$ - module
with highest weight ${\hat\mu}_s$, which can be regarded
as a $\Uk$ - module in the natural way. There always exists a
$\Uk$ - module $V_s^\bot$ such that $W({\hat\mu}_s)
= V_s\oplus V_s^\bot$.
Write $V^\bot=\oplus_s V_s^\bot$, and $W=\oplus_s W({\hat\mu}_s)$.
We have
\begin{eqnarray*}
\Hq(V) \oplus \Hq(V^\bot) &=&\Hq(W)\\
&\cong& W\otimes\Eq,
\end{eqnarray*}
where the last step follows from Proposition \ref{projective}.
\end{proof}

Recall that in classical differential geometry,
the space $\cal H$ of sections of
a vector bundle over a compact manifold $M$ furnishes a module over
the algebra ${\cal A}(M)$ of functions on $M$.
It then follows from Swan's
theorem that this module must be projective and is of finite type.
Conversely, any projective module of finite type over ${\cal A}(M)$
is isomorphic to the space of sections of some vector bundle over $M$.
This result is taken
as the starting point for studying vector bundles in noncommutative
geometry: one defines a vector bundle over a noncommutative space in
terms of the space of sections which is required to be a finite type
project module over the noncommutative algebra
of functions on the virtual noncommutative space.
Therefore, $\Hq(V)$ will be called the space of
sections of a quantum supervector bundle over the
quantum homogeneous superspace associated with $\Eq$.

Homogeneous supervector bundles at the classical level
were studied in \cite{Manin, Penkov}. We will not enter
the discussion of the subject, but refer to
the original papers for detail. It is worth mentioning
that the supergeometry of these objects are very interesting,
and many aspects of it remain to be understood.
The Hopf algebra approach to quantum homogeneous supervector bundles
adopted here works equally well at the classical level,
and should provide a new approach to the supergeometry of
homogeneous supervector bundles.

Follwoing the classical terminology, we will call a
quantum supervector bundle trivial if the sections form
a free module over the superalgebra of functions on the
quantum superspace. It immediately follows from
Proposition \ref{projective} that
\begin{proposition}
If the $\Uk$ - module $V$ is in fact a finite dimensional left $\Uq$
- module, then the quantum homogeneous supervector bundle with the
space of sections $\Hq(V)$ is trivial.
\end{proposition}

\section{\normalsize{\bf INDUCED REPRESENTATIONS}}
In this section we will investigate induced representations of the
quantum supergroup $\OSPq$ by using results of the last section.
The following Proposition explains how quantum homogeneous
supervector bundles enter representation theory.
\begin{proposition}
$\Gamq(V)$ furnishes a left $\Uq$ - module under the $\cdot$ action,
and also a right $\Tq$ co - module under the action
$\omega=\mbox{id}_V\otimes (\mbox{id}_{\Tq}\otimes S^{-1})\Delta$.
\end{proposition}
\begin{proof} {Proof}
For $p\in\Uk$, $x\in\Uq$, and $\zeta\in\Hq(V)$, we have
\begin{eqnarray*}
p\circ  ( x\cdot\zeta )&=& (-1)^{[p][x]} x\cdot(p\circ\zeta)\\
&=&(S(p)\otimes id_{\mbox{${\cal A}_q$}}) ( x\cdot\zeta ).
\end{eqnarray*}
Thus $\Gamq(V)$ indeed furnishes a left $\Uq$ - module under
the $\cdot$ action.  The $\Tq$ - co-action $\omega$ is
just the dual of this left $\Uq$ -  action.
\end{proof}

We call $\Hq(V)$ an induced $\Uq$ module, and also
an induced $\Tq$ co-module.
For such induced modules, we have the following
quantum analog of Frobenius reciprocity.
\begin{theorem} Let $W$ be a $\Uq$ module,
the restriction of which furnishes a $\Uk$ module in a natural way.
Then there exists a canonical isomorphism
\begin{eqnarray}
\mbox{Hom}_{\Uq} ( W, \ \Hq(V)  )
&\cong& \mbox{Hom}_{\Uk} ( W, \  V),
\end{eqnarray}
where $\Uq$ acts on the left module $\Hq(V)$ via the $\cdot$ action.
\end{theorem}
\begin{proof} {Proof}
  We prove the Proposition by explicitly constructing
the isomorphism, which we  claim to be  the linear map
\begin{eqnarray*}
F: \mbox{Hom}_{\Uq} ( W, \ \Hq( V ) ) &\rightarrow&
   \mbox{Hom}_{\Uk} ( W, \ V), \\
    \psi &\mapsto& \psi(1_{\Uq}),
\end{eqnarray*}
with the inverse map
\begin{eqnarray*}
\bar{F}: \mbox{Hom}_{\Uk} ( W, \ V) &\rightarrow&
        \mbox{Hom}_{\Uq} ( W, \ \Hq( V ) ), \\
       \phi &\mapsto& \bar{\phi}=(\phi \otimes S) \delta,
\end{eqnarray*}
where $\delta: W\rightarrow W\otimes\Tq$ is
the right $\mbox{${\cal A}_q$}$ co-module action defined by (\ref{comodule}).

To verify our claim, we first need to demonstrate that
the  image of $F$  is contained in
$\mbox{Hom}_{\Uk} ( W, \ V)$.  Consider $\psi$ $\in$
$\mbox{Hom}_{\Uq} ( W, \ \Hq( V ) )$.  For any $p$ $\in$ $\Uk$ and
$w$ $\in$ $W$,  we have
\begin{eqnarray*}
p ( F \psi (w) ) &=& (S^{-1}(p)\circ\psi (w)) (\hbox{\sf 1\kern-.4em\hbox{1}}_{\Uq}),
\end{eqnarray*}
where we have used the defining property of $\Hq( V )$.
Note that
\begin{eqnarray*}
(S^{-1}(p)\circ\psi (w)) (\hbox{\sf 1\kern-.4em\hbox{1}}_{\Uq})&=& (p\cdot\psi (w)) (\hbox{\sf 1\kern-.4em\hbox{1}}_{\Uq}).
\end{eqnarray*}
The $\Uq$-module structure of $\Hq( V )$ and the given condition
that $\psi$ is a $\Uq$-module homomorphism immediately leads to
\begin{eqnarray*}
p ( F \psi (w) ) &=& (-1)^{[\psi][p]}\,  \psi(p w) (\hbox{\sf 1\kern-.4em\hbox{1}}_{\Uq}) \\
                 &=&(-1)^{[\psi][p]}\,  F \psi (p w),
\quad   p\in \Uk, \  w\in W.
\end{eqnarray*}

Now consider ${\bar F}$.
We first show that the image $Im({\bar F})$
of ${\bar F}$ is contained in
$\mbox{Hom}_{\Uq} ( W, \ \Hq( V ) )$.
Note that $Im({\bar F})$ $\subset$  $\mbox{Hom}_{\C}
(W, \  V\otimes \Tq)$.
Some relatively simple manipulations lead to
\begin{eqnarray*}
(x\cdot{\bar\phi} (w) )  &=& {\bar\phi} ( x w ),\\
( p\circ {\bar\phi} (w) ) &=& (S(p)\otimes id_{\mbox{${\cal A}_q$}})
                           {\bar\phi}(w), \quad
x\in\Uq, \ \ p\in\Uk, \ \ w\in W.
\end{eqnarray*}
Therefore,  $Im({\bar F})$ $\subset$
$\mbox{Hom}_{\Uq} ( W, \ \Hq( V ) )$.
Now we show that $F$ and ${\bar F}$ are inverse to each other.
For $\psi\in$ $\mbox{Hom}_{\Uq} ( W, \ \Hq( V ) )$, and
$\phi$ $\in$ $\mbox{Hom}_{\Uk} ( W, \ V)$, we have
\begin{eqnarray*}
(F \bar{F}  \phi)(w) &=& (\bar{F} \phi)(w) (1_{\Uq})\\
                             &=& \phi(w), \\
(\bar{F} F \psi)(w)(x)&=& (-1)^{[x]([w]+1)}\  ( F \psi) (S(x) w)\\
        &=&(-1)^{[x]([w]+1)}\   \psi (S(x) w) (1_{\Uq})\\
        &=&(-1)^{[x]([w]+[\psi]+ 1)}\    (S(x)\cdot \psi (w)) (1_{\Uq})\\
        &=& \psi (w) (x), \ \ \ \ x\in\Uq, \ \ w\in W.
\end{eqnarray*}
This completes the proof of the Proposition.
\end{proof}

Let $V_\mu$ be a finite dimensional irreducible $\Up$-module with
highest weight $\mu$ and lowest weight $\tilde\mu$.
Since $V_\mu $ is a $\Up$-module the following is a well defined
subspace of $\Hq(V_\mu)$,
\begin{eqnarray*} \mbox{$\cal O$}_q(V_\mu) & :=& \left\{ \zeta\in \Hq(V_\mu) \ |
\ p\circ \zeta =
(S(p) \otimes id_{\mbox{${\cal A}_q$}})\zeta , \ \forall p\in\Up\right\}.
\end{eqnarray*}
We may regard $\mbox{$\cal O$}_q(V_\mu)$
as the quantum analog of the space of `holomorphic sections'.
Recall that the notation $W(\lambda)$ denotes
the irreducible $\Uq$ module with highest weight $\lambda$.  We have
the following result.
\begin{theorem} \label{BW}
There exists the following $\Uq$ module isomorphism
\begin{eqnarray}
\mbox{$\cal O$}_q(V_\mu) &\cong&\left\{\begin{array}{ l l }
       W((-{\tilde \mu})^\dagger),& -{\tilde \mu}\in\cP_+,\\
       0,& \mbox{otherwise}.
       \end{array} \right.
\end{eqnarray}
\end{theorem}
\begin{proof}{Proof}
Each  $\zeta$ $\in$ $\mbox{$\cal O$}_q(V_\mu)$
can be expressed in the form
\begin{eqnarray*}
\zeta &=& \sum_{\lambda\in\cP_+}
\sum_{i, j} v_{i j}^{(\lambda)}\otimes  {\tilde t}_{i j}^{(\lambda)},
\end{eqnarray*}
for some $v_{ i j}^{(\lambda)}\in V_\mu$ ($i,j=1,\cdots d_\lambda$).
Arguing as in the proof of proposition \ref{plenty}
one concludes, for each $\lambda\in\cP_+$, that
there exist $\phi^{(\lambda)}_i$ $\in$ $\mbox{Hom}_{\C}(W(\lambda), \
V_\mu)$ such that $v_{ i j}^{(\lambda)} = \phi^{(\lambda)}_i (
w_j^{(\lambda)} )$, where $\{w_i^{(\lambda)}\}$ is the basis of
$W(\lambda)$, relative to which the \irrep \  $t^{(\lambda)}$ of $\Uq$ is
defined.  Thus we can rewrite $\zeta$ as
\begin{eqnarray}
\zeta &=& \sum_{\lambda\in\cP_+} \sum_{i, j}
  \phi_i^{(\lambda)}(w_j^{(\lambda)})
\otimes {\tilde t}_{i j}^{(\lambda)}.
\end{eqnarray*}
Similar reasoning as in the proof of proposition \ref{plenty}
shows that
the $\phi_i^{(\lambda)}$ must be $\Up$-module homomorphisms of
degree $[\phi_i^{(\lambda)}]$.
It immediately follows from (\ref{Hom}) that
\begin{eqnarray*}  \phi^{(\lambda)}_i &=& c_i \, \phi^{(\lambda)},\quad
c_i\in \C, \end{eqnarray*}
and $\phi^{(\lambda)}$ may be nonzero only when
\begin{eqnarray*}
{\bar \lambda}&=&\tilde\mu.
\end{eqnarray*}
Hence, if $-\tilde\mu\not\in\cP_+$, we have $\mbox{$\cal O$}_q(V_\mu)=0$.
When $-\tilde\mu\in\cP_+$,  we set $$\nu=(-\tilde\mu)^\dagger.$$
Then, we may conclude that $\mbox{$\cal O$}_q(V_\mu)$ is spanned by
\begin{eqnarray}
\zeta_i &=& \sum_{j}
\phi^{(\nu)} (w_j^{(\nu)})\otimes {\tilde t}_{i j}^{(\lambda)},
\label{irrep}
\end{eqnarray}
which are  obviously linearly independent. Furthermore,
\begin{eqnarray*} x\cdot\zeta_i &=&
(-1)^{[x][\phi^{(\nu)}]}\  \sum_{j} t^{(\nu)}_{j i} (x)\  \zeta_j,
\ \ \ \ \ x\in\Uq.    \end{eqnarray*}
Thus $\mbox{$\cal O$}_q (V_\mu)\cong W(\nu)$.  More explicitly,
the isomorphism is given by
\begin{eqnarray}
W(\nu)
\stackrel{(id\otimes S)\delta}{\longrightarrow}
\mbox{$\cal O$}_q( W(\nu))
\stackrel{\phi^{(\nu)}\otimes id }{\longrightarrow}
\mbox{$\cal O$}_q( V_\mu).
\end{eqnarray}
This completes the proof of the theorem.
\end{proof}

This result provides an analog of
the celebrated Borel - Weil theorem for the quantum
supergroup $\OSPq$.
For the classical Lie supergroups, the program of developing
a  Bott - Borel - Weil theory was extensively
investigated by Penkov and co - workers\cite{Penkov},
although it has not yet been completed so  far as we are aware.
Also,  a quantum Borel-Weil theorem for the covariant and
contravariant tensor representations of quantum $GL(m|n)$
was obtained in \cite{Zhang}.

When $\mu=0$, the theorem implies that
\begin{eqnarray*}
\left\{ f\in\Tq | p\circ f = \epsilon(p) f, \ \ \forall p\in\Up\right\}
& =& \C \epsilon.
\end{eqnarray*}
Combining this result with  with Proposition \ref{projective},
we obtain

\bigskip
\noindent{\bf Corollary }:
{\em Let $W$ be any finite dimensional $\Uq$-module.
Then, as $\Uq$-modules,
\begin{eqnarray*}
\mbox{$\cal O$}_q(W) &\cong & \epsilon\otimes W.
\end{eqnarray*}}

\bigskip
\noindent {\bf Acknowledgement}:  This work is supported
by the National Science Council (ROC) Grant NSC-87-2112-M008-002.
Zhang wishes to thank the Department of Physics and
the Center for Complex Systems at the National Central University
for the hospitality extended to him during a visit
from late 1997 to early 1998, when this work was largely
completed.

\pagebreak
\small

redpajama_set_name: RedPajamaArXiv


From our enticing range of formal and casual dining venues to our lively bars and lounges and even our action packed Krazy Kids Playland, Smithfield RSL can keep the entire family entertained for hours!
Come on down, bring your friends and family, and enjoy the exciting atmosphere that is Smithfield RSL. There truly is something for everyone.


text:
Q: Error: Actions must be plain objects. Use custom middleware for async actions. ToDo app I checked a lot of blogs but none got me this error solved. Can someone please explain why i am getting this error and the remedy.
I am new to redux. I was following all the commands given by a youtube video. In that video he is not using dispatch anywhere. Can you also tell me where to use dispatch.
src/index.js
import React from 'react';
import ReactDOM from 'react-dom';
import './index.css';
import {Provider} from 'react-redux'
import {createStore} from 'redux';
import reducers from './reducer/'
import App from './App';
import * as serviceWorker from './serviceWorker';

const store=createStore(reducers);

ReactDOM.render(
<Provider store={store}>
  <App />
</Provider>
  ,
  document.getElementById('root'));
serviceWorker.unregister();

action/index.js
const addTask=(task)=>{
  return{
    type:'ADD_TASK',
    payload:task
  }
}
const deleteTask=(taskId)=>{
  return{
    type:'DELETE_TASK',
    payload:taskId
  }
}
export  {addTask,deleteTask};

task/index.js
import React from 'react'
import {connect} from 'react-redux'
import {bindActionCreators} from 'redux'
import {addTask} from '../../actions/'
import {deleteTask} from '../../actions/'

class Task extends React.Component{
  render(){
    return(

      {this.props.task1}

      <button onCLick={()=> this.props.dispatch(deleteTask(this.props.id))}>delete</button>

    )
  }
}
function mapDispatchToProps(dispatch){
  return (bindActionCreators({deleteTask},dispatch))
}
export default connect(()=>{},mapDispatchToProps)(Task)

taskbar/index.js
import React from 'react';
import {connect} from 'react-redux';
import {bindActionCreators} from 'redux'
import {addTask} from '../../actions/index'
class Taskbar extends React.Component{
  render(){
    return(

      <input type="text" ref="task" placeholder="add your task" />
      <button onClick={()=>this.props.dispatch(addTask(this.refs.task.value))}> ADD TASK</button>

    );
  }
}
function mapDispatchToProps(dispatch){
  return bindActionCreators({addTask},dispatch);
}
export default connect(()=>{},bindActionCreators)(Taskbar) ;

tasklist/index.js
import React from 'react';
import {connect} from 'react-redux';
import Task from '../task';
class Tasklist extends React.Component{
  render(){
    return(

      Tasks
      Actions

      {this.props.tasks.map((t,i)=> <Task key={i} task1={t} />)}

    );
  }
}
function mapStateToProps (state){
return{  tasks: state.tasks
};
}
export default connect(mapStateToProps)(Tasklist);

A: The error info you are facing:

Use custom middleware for async actions.

Follow it and add applyMiddleware to your createStore.
import { composeWithDevTools } from 'redux-devtools-extension';
import { createStore, applyMiddleware } from 'redux';
import thunk from 'redux-thunk';

(async () => {
  const preloadStore = {
  };

  const store = createStore(
    reducer,
    preloadStore,
    composeWithDevTools(
      applyMiddleware(thunk)
    )
  );
..

redpajama_set_name: RedPajamaStackExchange


text:
Q: how to check which tables and fields take more space? I have one database that has a bigger size than expected, how can I query it to retrieve the fields that take more space.
let's say I have 2 only tables
Table1.FieldOne
Table1.FieldTwo
Table2.FieldOne
Table2.FieldTwo

Let's say 99.9% of DB data is in Table2.FieldTwo, how do I query the db?
Of course I have many many tables.
The db is on sqlserver 2000

A: I can't think of an easy way to do this off the top of my head.
One way would be do dynamically generate the TSQL to list the total size of each column in each table which could be done using something like this:
DECLARE @SQL VARCHAR(MAX)

SELECT @SQL = COALESCE(@SQL + CHAR(10) + 'UNION ALL' + CHAR(10), '') +
    'SELECT ''' + QUOTENAME(t.name) + ''' AS TableName, ''' + QUOTENAME(c.name) + ''' AS ColumnName, SUM(DATALENGTH(' + QUOTENAME(c.name) + ')) AS TotalSizeBytes
    FROM ' + QUOTENAME(t.name)
FROM sys.columns c
    JOIN sys.tables t ON c.object_id = t.object_id
WHERE t.name = 'VoterAccount'

SET @SQL = 'SELECT TableName, ColumnName, TotalSizeBytes FROM (' + @SQL + ') x ORDER BY TotalSizeBytes DESC'
PRINT @SQL -- Just PRINT the SQL out. You can then copy it, and execute manually

Which would return e.g.
TableName    ColumnName    TotalSizeBytes
Table1       Column1       123456789
Table1       Column3       34356464
Table2       ColumnA       33987987

However, my concern is this is likely to be a pretty hefty query to run so I'd think carefully about it first. If there are no better suggestions and you try this, it might be worth restricting it to only look at a subset of columns - i.e. ignore any columns that you know from the start are small in size, focus on the columns you know are likely to be the bigger ones.
The ideal solution would be if this information was already held in system tables/accessible via DMVs somehow - unfortunately I don't know if it is, I've never come across it if it is available.


Insurance law issues can come into play in many different contexts, and firm practitioners have the ability to help clients with all manner of insurance law issues.
For businesses, consultation with counsel is often prudent at the insurance purchasing stage when proposals are being evaluated from multiple insurance providers and cost must be balanced with the scope of coverage being provided. Our attorneys can review complicated policy language and help clients understand the scope and limitations of coverage proposals, a subject which may not be fully disclosed or understood by insurance agents.
For both businesses and individuals, insurance issues arise when losses occur and sources of insurance coverage may not be readily apparent. We can help identify policies that may provide coverage for losses. At other times, the scope of provided coverage is in question. Our lawyers can help analyze whether insurance coverage exists under policies that the client has purchased – so-called first party claims – or whether insurance recoveries may be available from others who may be responsible for losses, known as third party claims. In either context, a proper understanding of insurance law is vital before making strategic decisions about how to best move forward in order to maximize insurance recoveries and minimize losses.
For significant business claims, the assistance of counsel starting early in the claims process is advisable to make sure that the insured complies with technical policy requirements to preserve insurance coverage, and that insurers honor their duty to pay claims or their duty to defend lawsuits filed against their insured.
When insurers wrongfully refuse to pay first party claims, Georgia law provides statutory penalties that can be used to motivate insurers to comply with their policy obligations. When first party coverage disputes with insurers cannot be resolved prior to suit, our attorneys have experience in prosecuting claims against insurers to obtain coverage benefits.
When insurers wrongfully refuse to pay or defend against third party claims, Georgia bad faith law can cause insurers to be responsible for the entire loss regardless of the insurer's underlying policy limits. This potential penalty means that bad faith law can be sued as potent tool in negotiations with insurers to ensure that third party claims are fairly evaluated by insurers, and that payment is promptly paid when due.
Our attorneys are adept at handling insurance litigation which may arise when insurance companies seek a declaratory judgment from the court that the insurance policy at issue does not provide coverage for a particular loss. Our attorneys have represented both insurers and putative insureds in such litigation which requires a thorough understanding of insurance law and litigation procedure.
When analyzing personal injury claims or losses, there are often multiple policies of insurance that provide coverage to injured parties. This will likely include a mixture of first party coverages and third party insurance policies. Our attorneys are skilled in finding all of the insurance policies that may provide coverage for an injury or loss, and evaluating: how those policies interact with each other; whether policies may be aggregated or "stacked," and; whether policies provide primary coverage or excess coverage.
Injured parties may also be eligible to recover benefits under their own auto policies when involved in collisions with uninsured or underinsured motorists. Our attorneys have experience in the recovery of uninsured motorist ("UM") benefits and the many technical requirements related thereto. There may be multiple UM polices applicable to any given claim, and the amount payable under each policy may require a complex analysis of the type of UM policy involved and its relation to both liability policies and other UM policies. Statutory penalties are available when insurers wrongfully refuse to pay UM benefits, but the nature of UM claims is such that these claims are often litigated. Consequently, it is imperative that claimants have counsel who is well versed in Georgia UM law.
A related consideration for personal injury and other claimants is defending against claims by first party insurers, including UM and group health insurers, for reimbursement or subrogation where the insurer seeks to have the injured pay back the insurer for benefit payments previously paid on the insured's behalf when a third party is responsible for the loss. Group health insurers have become increasingly active in pursuing such claims against injured parties, even when the insurer's claim may not be legally enforceable. Our attorneys will analyze reimbursement or subrogation claims early in the claims process and structure settlements to either avoid such claims or minimize their effect.
Whatever type of insurance policy is in question, we can help. Areas of practice include general liability insurance, automobile insurance, homeowner's insurance, professional liability insurance, errors and omissions, directors' and officers' liability insurance, garage owner's liability insurance, accident and disability insurance and life insurance. (See our separate web page on title insurance).
David L. Turner has practiced in the area of insurance law throughout his career.


On this week's episode of Sneak Attack!, the gang continues their fight through the enchanting and mysterious sky temple. After clearing the first room, they move into the second, eager to peel back its secrets and discover what lies beneath. Will they be able to work together to solve the puzzles? Find out right now!


text:
Worksite Education for Employees
State Employees and Healthcare Professionals
White Plains Branch
Long Island Branch
Syracuse Branch
Florida Branch
Madison in the News
Everyone has a story.
Survey Details Back-to-School Spending for 2019
Submitted by Madison Planning Group on August 29th, 2019
The National Retail Federation (NRF) and Proper Insights & Analytics recently released a survey detailing back-to-school spending for the 2019 school year. The survey looks at how American families plan to shop for clothing, supplies, and other items for the school year.
Record spending expected for 2019
As students prepare to go back to school and college, families are spending more than ever on school supplies. "Consumers are in a strong position given the nation's growing economy, and we see this reflected in what they say they will spend on back-to-class items this year," NRF president and CEO Matthew Shay said.
According to the NRF, families with children in elementary school through high school plan to spend an average of $696.70, up from $684.79 last year and topping the previous record of $688.62 set in 2012. Families with college students are expected to spend even more — an average of $976.78, which is up from $942.17 last year and exceeds the previous record of $969.88 set in 2017.
The survey also revealed that total combined spending for K-12 and college is projected to reach $80.7 billion. This figure is down from last year's $82.8 billion, but is attributed to the decreased number of households surveyed with children in K-12 or attending college.
Spending trends
Clothing and accessories are expected to top K-12 families' expenses at an average of $239.82, followed by electronics such as computers, calculators, and phones ($203.44); and shoes ($135.96) and supplies such as notebooks, pencils, backpacks, and lunch boxes ($117.49). K-12 families plan to do most of their shopping at department stores (53%), discount stores (50%), online (49%), clothing stores (45%), and office supply stores (31%).
College shoppers plan to spend the most on electronics ($234.69), followed by clothing and accessories ($148.54), dorm and apartment furnishings ($120.19), and food items ($98.72). They plan to do most of their shopping online (45%), followed by department stores (39%), discount stores (36%), college bookstores (32%), and office supply stores (29%).
The survey shows that among K-12 shoppers, teens are expected to spend an average of $36.71 of their own money, up from $30.88 ten years ago. Pre-teens plan on spending $26.40, up from $11.94 ten years ago. According to Shay, "Members of Generation Z are clearly becoming more involved with back-to-school purchasing decisions rather than leaving the choices up to mom and dad."
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Disclosures and Privacy
bschwartz@madisonplanning.com
Securities offered through Lion Street Financial, LLC (512.776.8400), member FINRA, SIPC. Investment advisory products and services offered through Madison Advisory Services, Inc. an investment advisor registered with the SEC. LSF is not an affiliated company. Neither Lion Street Financial, LLC nor Madison Advisory Services, Inc. provide tax or legal advice. Representatives may transact business, which includes offering products and services and/or responding to inquiries, only in state(s) in which they are properly registered and/or licensed. This site is published for residents of the United States only. Representatives may only conduct business with residents of the states and jurisdictions in which they are properly registered. Therefore, a response to a request for information may be delayed until appropriate registration is obtained or exemption from registration is determined. Not all of services referenced on this site are available in every state and through every advisor listed. For additional information, please contact Benjamin Schwartz at 800.222.2091.
*Third party awards, rankings, and recognitions are no guarantee of future investment success and do not ensure that a client or prospective client will experience a higher level of performance or results. These ratings should not be construed as an endorsement of the advisor by any client not are they representative of any one client's evaluation.
© 2022 Madison Planning Group. All rights reserved.

redpajama_set_name: RedPajamaCommonCrawl


text:
Cheat River Review
Table of Contents Archive
Statement of Ethics
Spring 2021 Contributing Artist Statements
Post author By cheatriverreview
Tiffany Babb
"As a queer, mixed-race writer, I am interested in moments of uncovering what has been buried or ignored or forgotten. I think of my writing as an attempt to reach into the fog and solidify it. My writing is driven by the attempt to slow the ever increasing unknown. Robert Frost wrote that poetry "a momentary stay against confusion." He didn't expect poetry to be the answer to all questions or even a step on the path to great enlightenment. It was a momentary stay, not a cure. Though I recognize that I construct an identity out of a history that may not be true, I can still create a foothold into my history as I understand it."
Tiffany Babb is a poet, essayist, and cultural critic. She's a regular contributor to The AV Club's Comic Panel and the Eisner Award winning PanelxPanel Magazine. You can find her poetry in Rust + Moth, Third Wednesday Magazine, and Cardiff Review. You can follow her on twitter @explodingarrow and sign up for her monthly newsletter at tiffanybabb.com/puttingittogether.
Pages: 1 2 3 4 5 6 7 8 9 10 11
← An Interview with Vince Granata → Issue 15 – Spring 2021
© 2022	Cheat River Review


text:
Gun used in Chicago hospital shooting was stolen from Indiana: Officials
No one was hurt in the Monday shooting at Chicago's Jesse Brown VA Hospital.
Emily Shapiro
Man arrested after opening fire at Chicago VA hospitalThe unidentified man allegedly entered the Jesse Brown VA Medical Center in Chicago with an assault rifle at around 2 p.m on Monday. Antonio Perez/TNS
A convicted felon who allegedly opened fire at a Chicago hospital brought a rifle that had been stolen from a federal firearms licensee in a neighboring state, according to prosecutors.
Bernard Harvey Jr., a 40-year-old from Indianapolis, is accused of storming Chicago's Jesse Brown VA Hospital with a semi-automatic rifle on Monday, prosecutors said.
Shots were fired inside and outside of the building, but no one was hurt, authorities said. Within 30 seconds, authorities responded and Harvey was taken into custody, said FBI special agent in charge Jeffrey Sallet.
Police officers enter Jesse Brown VA Medical Center after reports of possible shots fired inside the hospital, Aug. 12, 2019 in Chicago.
Police officers enter Jesse Brown VA Medical Center after reports of possible shots fired inside the hospital, Aug. 12, 2019 in Chicago.Antonio Perez/TNS
(MORE: 'We avoided tragedy': Man arrested after opening fire at Chicago VA hospital, police say)
The gun used at the hospital had been reported stolen from a federal firearms licensee in Indiana, according to federal prosecutors.
The weapon was among 31 guns swiped by thieves who smashed display cases at Element Armaments, a firearm shop in New Whiteland on July 27, reported ABC Chicago station WLS.
The five thieves made their escape with 13 long guns, 20 handguns and two silencers, WLS reported.
(MORE: Fingerprinting gun owners approved by Illinois lawmakers)
It's not clear if Harvey was involved in the theft or obtained the gun afterward, according to WLS.
A man is detained by police outside of Jesse Brown VA Medical Center after reports of possible shots fired inside the hospital, Aug. 12, 2019 in Chicago.
A man is detained by police outside of Jesse Brown VA Medical Center after reports of possible shots fired inside the hospital, Aug. 12, 2019 in Chicago.Antonio Perez/Chicago Tribune via Newscom
Three of the five suspected thieves have been arrested, New Whiteland police told WLS.
Officials with New Whiteland police and Element Armaments could not immediately be reached by ABC News.
Ahead of Monday's shooting, Harvey had been convicted of multiple felonies, including gun offenses, prosecutors said. He was charged this week with one count of illegal possession of a firearm by a convicted felon, prosecutors said.
Harvey has not entered a plea and is set to return to court on Friday.
Aug 14, 8:31 PM


You have a unique medical history. Therefore, it is essential to talk with your doctor about your personal risk factors or experience with anemia. By talking openly and regularly with your doctor, you can take an active role in your care.
What is my risk of getting anemia?
Am I getting adequate nutrition in my diet?
Is my diet appropriate for my weight and lifestyle?
Should I be checked for anemia?
Is a transfusion necessary, or is there a safer alternative?
If I need medicine, how long will I need to take it?
Will the medicine interact with anything else I am taking?
How often should I check back with you?
Can I treat my anemia by making changes in what I eat?
Do I need to alter my diet or other lifestyle habits?
Are there any activities I should avoid?
Will anemia affect my quality of life? Will this be short or long term?
How long will I have to live with limits?
Getting the most out of your doctor appointment. Family Doctor—American Academy of Family Physicians website. Available at: https://familydoctor.org/tips-for-talking-to-your-doctor. Updated January 19, 2018. Accessed October 15, 2018.


Peanut, or Liam as we should call him since his is growing up so quickly, is becoming one of my favorite characters in this or just about any series.
Liam is the only a few months old, but as the first dragon since Dragos was created at the beginning of creation, no one is certain what to expect as far as powers or his unusual growth. He seems to have a blend of his parents personalities and abilities, but he keep surprising us all with new abilities of his very own. He even managed to sneak up on his Dad without his father sensing him, which his quite the feat.
On his first day of school, he is picked on by a few of the older boys as well as his teacher. He handles both instances with just the right mixture of intelligent thought, bravery and a little intimidation of his own. I am sure he will shine as a future leader in this series.
I can't wait to see what else Liam's unusual character will bring to the series as far as his adventures and his abilities. I just hope he doesn't grow up too fast. I am not sure how I would feel about him growing up and getting his own love story when he is only, say two calendar years old.
The tricky thing about using a cloaking spell is that cloaking spells are tricky.
Liam snickered to himself as he tiptoed around the patio furniture and changed into his dragon form. His dragon had grown to the size of an adult lion, and he had to be careful not to knock over furniture as he shapeshifted.
He also managed to hold on to his cloaking spell, which was a bit fat win. Dad had said Liam's cloaking ability was one of the best he'd ever seen, although it was difficult to stay hidden while changing forms.
But since his dad could do it, Liam felt he would be able to it also. Eventually. Sometimes. If he kept practicing, pretty soon he should be able to stay cloaked all the time if he needed to.
Liam was playing his favorite game, Spy Wyr, which he had totally made up himself. When he grew up, he was going to be a secret sentinal. Uncle Graydon would send him out on missions, and when he returned after saving somebody, or maybe even after saving everybody, Mom and Dad would be really proud of him.
Of course, because it was undercover, Dad would have to give him medals in secret. Sometimes they might be silver or bronze ones, or when he did something amazing, they might be gold. Or maybe when he did something really outstanding, Dad would give him a sparkly medal with diamonds on it. Then Liam would have to find a super-secret place to hide them.
His dragon side like the sound of that. It made him feel growly and fine.
By day, Liam would be, oh, maybe a basketball player. Basketball players traveled a lot, so it would be a good cover, and besides it would be fun to play ball all the time, so that would be big fat win.
Hi, my name if Cuelebre, Liam Cuelebre. My code name is Double Oh Peanut, but you can call me Rock Star for short.
Excited to read this installment. I am amazed at how much the Elder Races series keeps surprising me and always keeps me guessing.


For over 25 years Larry Dressler has designed and facilitated high-stakes conversations and advised leaders who seek to weave authentic engagement and collaboration into the fabric of their organizations. He has facilitated such meetings in settings ranging from the offices of Fortune 100 companies to a solar-powered chocolate factory in the Amazon. Larry has served as a trusted consultant to leaders at New Belgium Brewing, First Bank of Colorado, Facebook, Nike, and The Denver Foundation. In addition to being a skilled practitioner, Larry is the author of Consensus Through Conversation: How to Achieve High Commitment Decisions. His second book, Standing in the Fire: Leading High- Heat Meetings with Clarity, Calm, and Courage has been published in the US, China, and Japan.


Our high-quality, handcrafted lampshades have made us a go-to resource for retailers and home decor companies throughout the country. If you stock and re-sale lampshades, we would love to work with you as a stocking wholesaler.
Complete the application below and submit it via email or fax 704-376-0268. In addition to wholesale discounts, we offer unique bulk purchasing options on our site. Once you are registered as a wholesaler, you will be able to access the wholesale purchase link on our site, so you can buy shades with a click of a button. We also offer custom and COM shade capabilities to wholesale customers.
If you are an OEM or have a large commercial, hospitality or private label job please email us.


text:
Contact/Mailing List
ELLIOTT JAMAL ROBBINS
"Snow White Clapping"
OPENING RECEPTION: JUNE 3rd, 2018 (6-8pm)
June 3rd - July 22nd 2018
"It is a peculiar sensation, this double-consciousness, this sense of always looking at one's self through the eyes of others, of measuring one by the tape of a world that looks on in amused contempt and pity. One ever feels his two-ness,—an American, a Negro... two thoughts, two unreconciled strivings; two warring ideals in one dark body, whose dogged strength alone keeps it from being torn asunder" – W.E.B Dubois, The Souls of Black Folk
"Is there not a concurrent regression to and fixation at pregenital levels of sexual development? Self-castration? (The Negro is taken as a terrifying penis.) Passivity justifying itself by the recognition of the superiority of the black man in terms of sexual capacity? It is obvious what a variety of questions it would be interesting to raise." Frantz Fanon, Black Skin, White Masks
Kai Matsumiya presents "Snow White Clapping," a solo debut by Elliott Jamal Robbins.
White strangers at Western themed locales. Snow White eclipsed in black. Oblique propositions: Robbins's protagonist is the embodiment of a boy, circumscribed by the trappings of representation, queerness, and race. In the film from which the exhibition's title heralds, Elliott Robbins fuses a series of frames from Disney's Snow White with the hand-drawn, virtual armature of a black body. In the gallery's back room, a figure claps in silence: perhaps at an audience, or perhaps at a spectator always already embodied by the modality of projection. The ambiguity here is one between subject and identity, but also between viewer and gaze.
In the four series comprising the exhibition, Robbins' maneuvers intersectionality as if it were a minefield-- rather than an already established discursive territory from which one could nominate a Self. "A Black Aesthetic," is comprised of graphite drawings; "Cartoons Series," are works on paper executed with black ink, there are also his "collages"; and hypnotic loops of animation film. His ink drawings are executed with a slapstick ethos and are juxtaposed with harrowing exercises in new media and work in graphite. A sense of silence graces the visual space of "Black Aesthetic," a large-scale graphite drawing in which the protagonist stares at the words "Dead Means Legally Black/ Black Means Legally Dead." By making direct the relationship between the eye and the hand, Robbins is able to instrumentalize the very languages that serve to mediate the black body.
The figure in play is only referred to as "Brother"-- always "him," or "he," but never "I," and never "we." Our so-called "hero" is a dysmorphic personage anchored in globular shoes. He occupies the picture plane like one might a personal pronoun. It's a mediation that aims as much to conceal its own violence as it does to conceal its own origin, and it's from a position of both autonomy and confusion that Robbins' contorts, expands—only to contort again— the horizons of representation against which the black body is composed.
Robbins writes of his work,
"Through the use of appropriated and self-generated imagery and text, as well as the inclusion of the black male cartooned figure, the viewer is presented with a disjointed narrative. The narrative in question is an exploration of the intersection of societal readings of a black body, as well as a subjective experience, and the dichotomies to be found between. By collaging together bits of text, and the inclusion of a black figure, my work examines experience, with no concrete conclusions to be drawn or clear distinctions that can delineate the personal and the political. This body of work is as much a personal narrative, as it is an interrogation of the performative nature of blackness and masculinity."
We often see what we want to see measured against what we want to see; we become who we are in the confrontation of what is there. What's at stake is a world, and the identity in question, a state of affairs. The line between individual and constituent body is a confused adolescent casually cruising. In every instance, Robbins encourages us to not to find answers, but rather to explore the inherent complexities of the condition of the unnamed character.
Robbins is a graduate of the University of Arizona (MFA 2017). He lives and works in Tucson, AZ.
From Cartoon Series
Sumi Ink on Paper
11" x 8,5"
From Collages
Black Geppetto
153 ½ Stanton Street
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