text:
\section{Introduction}
\label{sdsdsdsd}

Diffeomorphism symmetry is the fundamental gauge symmetry which led Einstein to his unreasonably successful
theory of general relativity, and is the key symmetry which guides the quantization of this theory known as
loop quantum gravity \cite{thiemann2007, rovelli2004, BackLA}.  In order to make contact between loop quantum gravity and observational predictions, as well as
to test the classical limit of the theory in a simplified context, the framework of loop quantum cosmology was developed. There one starts with the homogeneous and (optionally) isotropic phase space of general relativity and quantizes using methods
as close as possible to those used in loop quantum gravity.  When homogeneity is imposed, almost all diffeomorphism symmetry
is thereby automatically fixed, except for a three parameter family of \textit{residual diffeomorphisms}.  When
isotropy is additionally imposed, the residual diffeomorphisms are further reduced to the one parameter group of isotropic dilations.

In the quantization of the homogeneous models, a key technical subtlety is encountered:
Because all dynamical fields become constant in space,
the action integral, and hence the symplectic structure and Hamiltonian derived from it, diverges. As a consequence,
it is necessary to fix a finite cell over which to integrate in order to have a well-defined framework.  This choice of a cell artificially
breaks the residual diffeomorphism symmetry further.  In particular, only \textit{volume preserving} residual diffeomorphisms
preserve the resulting classical Poisson algebra, and hence only these preserve the resulting quantum algebra of basic operators.

In  the work of Ashtekar and Campiglia \cite{UniquOfKinLQC}, invariance under residual diffeomorphisms is used to select uniquely
a representation of the reduced quantum holonomy-flux $^*$-algebra, in analogy to what is done in the full theory using
the full diffeomorphism group \cite{LOST}. Because only volume preserving residual diffeomorphisms
have a well defined action on the reduced algebra,
and such diffeomorphisms are non-trivial only in the non-isotropic case,
their analysis is limited to a homogeneous, non-isotropic case -- specifically Bianch I.

However, it is important to extend their uniqueness result to the isotropic case for two reasons (1.) the isotropic model is the
one that is actually being used to make contact with potential observations \cite{ABreview, AAN2013,  BCGM2014, BCT2011},
and (2.) thus far it is only in the isotropic
case that the more general framework introduced by Fleischhack is available \cite{CSL}.
The Fleischhack framework improves upon the standard framework in the sense that it is more faithful to the quantization
procedure used in the full theory, in that the configuration algebra includes holonomies along \textit{all} analytic curves and not just those along straight ones.

In the prior work \cite{JM}, uniqueness was achieved for the isotropic case by focussing attention on selecting only
the \textit{measure} on the quantum configuration space, and hence the inner product, using
invariance under residual diffeomorphisms, because there is no obstacle to defining the action of dilations
on the configuration space alone.  In this paper we propose a different solution to the problem which will allow for
a stronger uniqueness result, namely uniqueness of the full representation of the quantum algebra, similar to what
is done by Ashtekar and Campiglia for the homogeneous case and by the authors of \cite{LOST} for the full theory. Specifically, we address the problem at its source: By allowing dilations to act on the fiducial cell, covariance is recovered, and an action of dilations which is well-defined on the quantum algebra is achieved. We show that there is
only one state on both the standard algebra $\aA_s$ and on the Fleischhack algebra $\aA$ that is invariant under this action of dilations. We find that in the standard case, the corresponding GNS-representation is unitarily equivalent to the standard representation $\varrho_s$ used in LQC up until now. In the Fleischhack case, we have a canonical embedding $\iota\colon \aA_s\rightarrow \aA$,
where $\aA= \iota(\aA_s)\oplus\ID_0$ holds for $\ID_0$ the ideal generated by the additional elements coming from parallel transports along non-straight curves. Then, for $\varrho$ the GNS-representation that corresponds to the invariant state on $\aA$, (up to unitary equivalence) we have
\begin{align*}

	\varrho\cp \iota=\varrho_s\qquad\text{and}\qquad\varrho|_{\ID_0}=0,
\end{align*}
i.e., $\varrho$ is actually just given by its restriction to the standard holonomy flux $^*$- algebra of homogeneous isotropic LQC and equals the standard representation thereon.

\section{SYMMETRY REDUCTION, CANONICAL ANALYSIS, AND DILATIONS}
\label{sec:classical}

In this section, we derive the symmetry reduction of the Holst action and perform the Legendre transform
in order to obtain the form of the canonical variables in the spatially flat, homogeneous isotropic case.  As far as we are aware,
this is the first time the reduced canonical framework has been derived in this way. This derivation serves to fix conventions
as well as to carefully derive from first principles the various factors involved.

Loop quantum gravity is based on the Holst action of gravity \cite{holst1995}, in which the basic variables
are an $\SO(1,3)$ connection $\omega_\alpha^{IJ}$ and a co-tetrad $e^I_\alpha$. Here,
$I,J = 0, \dots, 3$ denote internal indices which are raised and lowered using $\eta_{IJ}:= \mathrm{diag}(-1,1,1,1)$.
To impose homogeneity and isotropy, one fixes a global foliation of space-time $\mathcal{M}$ into spatial 3-slices.
One then chooses one of the possible isometry groups of homogeneity and isotropy in three dimensions,
and fixes some action of this group on $\mathcal{M}$ via diffeomorphisms preserving each slice.
In this paper we choose our symmetry group to be the one so far suggested by observation --- the Euclidean group
$\mathcal{E}= \RR^3 \rtimes \SO(3)$ (i.e., the $k=0$, spatially flat case)
with multiplication $(u,r)\cdot(u',r') = (u+ru', rr')$, so that the 3-slices of $\mathcal{M}$ are diffeomorphic to $\RR^3$.
Moreover, we introduce a global chart $(x^\mathbf{\alpha}) = (x^\mathbf{0}, x^\mathbf{a})$ such that $x^\mathbf{0}$ is constant on each spatial slice, and the action $\mathcal{E} \ni (u,r) \mapsto \Phi_{(u,r)} \in \mathrm{Diff}(\mathcal{M})$
takes the standard form
\begin{align*}
	\Phi_{(u,r)}((x^\mathbf{0}, x^\mathbf{a}))=(x^\mathbf{0}, u^\mathbf{a}+r^\mathbf{a}{}_\mathbf{b}\hspace{0.5pt} x^\mathbf{b}).
\end{align*}
If one requires invariance of the basic variables $(\omega_\alpha^{IJ}, e_\alpha^I)$ under an action of $\mathcal{E}$ via diffeomorphisms alone \footnote{The basic fields in this section are naturally understood to be associated to an $\mathrm{SO}(1,3)$ principal fiber bundle $P$ over $\mathcal{M}$ which we assume to be trivial, and for which we fix a global trivialization $P = \mathcal{M} \times \mathrm{SO}(1,3)$ .  By ``diffeomorphisms alone'', we mean lifts of elements of $\mathrm{Diff}(\mathcal{M})$ to $P$ which act only on the $\mathcal{M}$-factor.}, the resulting solution space is empty; as a consequence, in order to have non-trivial invariant basic variables, it is necessary for $\mathcal{E}$ to act via gauge rotations as well.
Specifically, for each $r \in \SO(3)$ define
\begin{align*}
\Lambda(r)^I{}_J :=  \left( \begin{array}{cc} 1 & \vec{0}^T \\ \vec{0} & r^i{}_j \end{array} \right) \in \SO(1,3),
\end{align*}
and let each element $(u,r) \in \mathcal{E}$ act on the basic variables via \cite{engle2007, BojoKa}
$\Lambda(r) \circ \Phi_{(u,r)}$. Here, gauge transformations and diffeomorphisms are defined to act on the basic variables
in such a way that their action is a left action.
Let $\III$ denote the space of all pairs $(\omega,e)$ invariant under this action. They are
exactly of the form
\begin{align}
\label{invform}
\begin{aligned}
\omega_{\mathbf{a}}^{ij} = w\hspace{1pt} \epsilon_{\mathbf{a}ij},
\quad \omega_\mathbf{a}^{0i} = - \frac{c}{\gamma}\hspace{1pt}\delta^i_\mathbf{a},
\quad \omega_\mathbf{0}^{ij} = 0, \quad \omega^{0i}_\mathbf{0} = 0, \\
e^0_\mathbf{a} = 0, \quad  e^0_\mathbf{0} = N, \quad  e^i_\mathbf{0} = 0, \quad
e^i_\mathbf{a} = v \hspace{1pt} \delta^i_\mathbf{a},\quad\:\:
\end{aligned}
\end{align}
for some smooth real valued functions $w, c , v, N$, depending on $x^\mathbf{0}$ only.
At this point, let us additionally assume $N > 0$ (the time orientation part of the time gauge).

The (dynamically oriented) Holst action is given by
\begin{align*}
S_H =
\frac{1}{4k}\! \underset{(\mathcal{M}, [e])}{\int} \!\!\! \left( \epsilon_{IJKL}\: e^I  \wedge  e^J \wedge  \Omega^{KL}
\! - \! \frac{2}{\gamma}\: e^I  \wedge  e^J \wedge  \Omega_{IJ} \right)
\end{align*}
where $k:= 8\pi G$, $\gamma$ is the Barbero-Immirzi parameter \cite{barbero1995, immirzi1995}, $\Omega^{IJ}$
is the curvature of $\omega^{IJ}$, the four dimensional alternating tensor is defined such that
$\epsilon^{0123} = 1$ so that
$\epsilon_{0123}=-1$, and
where $(\mathcal{M}, [e])$ denotes $\mathcal{M}$ with the orientation defined by the co-tetrad $e$.
The Legendre transform of this action yields the canonical variables
$A^i_a = \Gamma^i_a + \gamma K^i_a$, the Ashtekar connection, and
$E^a_i = |\det e^j_b|\hspace{1pt} e^a_i$. Here, $\Gamma_a^i$ denotes the spin connection determined by $e_a^i$,
and $K^i_a := \omega^{i0}_a$.
Note in particular that the definition of the orientation of the action integral used here
makes it independent of any background orientation.
More importantly, this choice of orientation is necessary to ensure that the variable conjugate to $A^i_a$ is
$E^a_i$ defined with absolute value around the determinant of the triad, which ensures equivalence of the framework with the generalized ADM framework \cite{thiemann2007, BackLA}.

Let $\mathcal{L}_{H}$ denote
the integrand of this action, i.e., the Holst Lagrangian density.
Evaluated on $\III$,  this Lagrangian density reduces to
\begin{align*}
\mathcal{L}_{H} = \frac{3 |v|v }{k\gamma}\:\frac{\mathrm{d}}{\mathrm{d}x^{\mathbf{0}}}(c-w)
-\frac{3N|v|}{k\gamma^2}(c^2 - \gamma^2 w^2 - 2cw).
\end{align*}
As $\mathcal{L}_{H}$ is constant in space, the corresponding Lagrangian $L_{H}$ must be defined by integrating over a finite cell
$\mathcal{C}$ in space.  Let $V_0[\mathcal{C}]$ denote the unsigned coordinate volume of the cell $\mathcal{C}$ in the fixed
coordinates $x^{\mathbf{a}}$, so that
\begin{align*}
L_{H} \! = \! \frac{3 V_0[\mathcal{C}] |v|v }{k\gamma}\frac{\mathrm{d}}{\mathrm{d}x^{\mathbf{0}}}(c\!-\!w)
-\frac{3N V_0[\mathcal{C}] |v|}{k\gamma^2}(c^2\! - \! \gamma^2 w^2 \! - \! 2cw).
\end{align*}
The symplectic term in this Lagrangian tells us that the dynamical variable is $c-w$,
with canonically conjugate momentum
\begin{align}
\label{pdef}
p:= \frac{3 V_0[\mathcal{C}] |v|v }{k\gamma},
\end{align}
which motivates a change of variables
from $(c,w)$ to
\begin{align*}
c_{\pm} := c \pm w;
\end{align*}
yielding
\begin{align*}
 L_{H} = p\hspace{1pt}\dot{c}_- \hspace{1pt}
+\: &   N \sqrt{\frac{3 V_0[\mathcal{C}]}{16 k\gamma^3}} \,\,  |p|^{1/2}
\big( (1+\gamma^2) \hspace{1pt} c_+^2  \\
& \qquad \quad - (3-\gamma^2)\hspace{1pt} c_-^2 - 2(1+\gamma^2)\hspace{1pt} c_+ c_-\big).
\end{align*}
Here, $N$ and $c_+$ have no time derivatives in the Lagrangian, and so are Lagrange multipliers.
The constraint obtained by varying $c_+$ is
\begin{align}
\label{secclass}
c_+- c_- = 0.
\end{align}
In the Dirac constraint analysis of this system \cite{dirac1964}, the above constraint turns out to be second class,
for reasons similar to that used in \cite{holst1995}.
Specifically, note that the canonical momentum $\Pi_{c_+}$ conjugate to $c_+$ is trivially constrained to be zero
$\Pi_{c_+} \approx 0$. The canonical Poisson brackets then yield
$\{c_+-c_-, \Pi_{c_+}\} = 1$, so that both $c_+-c_- \approx 0$ and $\Pi_{c_+} \approx 0$ are second class.
Thus both are substituted into the Lagrangian prior to quantization.
Doing this, and writing the Lagrangian in terms of the original variables, we have
\begin{align*}
 L_{H} = p\hspace{1pt}\dot{c}
- N \sqrt{\frac{3 V_0[\mathcal{C}]}{k\gamma^3}} \,  |p|^{1/2} c^2,
\end{align*}
from which one can read off the conjugate variables $c$ and $p$, as well as the usual Hamiltonian constraint
of isotropic LQC \cite{as2011a}.
The Ashtekar connection $A^i_a$, and densitized triad $E^a_i$, when evaluated on $\III$, in terms of $c$ and $p$,
are of the form
\footnote{These hold only in the fixed coordinate system $x^{\mathbf{a}}$.
In order to have expressions valid in any coordinate system,
it is sufficient to recast them in a form covariant under coordinate transformation.
This can be done by introducing a fiducial triad and co-triad via
$\mathring{e}^a_i := \left(\frac{\partial}{\partial x^{\mathbf{i}}}\right)^a$ and
$\mathring{e}^i_a := (dx^{\mathbf{i}})_a$.
$V_0[\mathcal{C}]$ is then the volume of $\mathcal{C}$ with respect to $\mathring{q}_{ab}:=\mathring{e}^i_a\mathring{e}_{bi}$,
and the expressions for $A^i_a$ and $E^a_i$ take the covariant forms
$A^i_a = c \hspace{1pt}\mathring{e}^i_a$ and
$E^a_i :=  \frac{k\gamma}{3 V_0[\mathcal{C}]} \hspace{1pt} p \hspace{1pt} \det(\mathring{e}^j_b)\hspace{1pt} \mathring{e}^a_i$.
}
\begin{align}
\label{AErelation}
 A^i_\mathbf{a} = c \hspace{1pt}\delta^i_\mathbf{a}, \qquad
E^\mathbf{a}_i = \frac{k\gamma}{3 V_0[\mathcal{C}]} \hspace{1pt} p \hspace{1pt} \delta^\mathbf{a}_i .
\end{align}

\subsection*{THE ACTION OF DILATIONS}

Since the theory is restricted to a cell, and dilations do not preserve this cell, dilations are in fact not defined in the theory, \textit{unless we let the dilations act on the cell as well}.
This is the key new ingredient which will let dilations have a well-defined action on the quantum algebra.
We begin by deriving the action of a dilation on $c$ and $v$. For this, we let $\Phi_\lambda$ denote the dilation which expands space from the origin by a factor of $\lambda\in \RR_{>0}$, as measured in the
fixed coordinate system $x^{\mathbf{a}}$.
Then, letting $\Phi_\lambda$ act on the cell $\mathcal{C}$ transforms
$V_0[\mathcal{C}]$ to $|\lambda|^3 V_0[\mathcal{C}]$. Moreover, the $1$-forms $\w$ and $e$ transform via push forward by $\Phi_\lambda$ (pull-back by $\Phi_{\lambda^{-1}}$) \footnote{Since $\Phi_\lambda$ acts on curves via $\gamma\mapsto\Phi_\lambda\cp \gamma$, it acts on vector-fields via push-forward by $\Phi_\lambda$ (recall the definition of a tangent vector as a equivalence class of curves).
Since $\Phi_\lambda$ also acts on scalar fields $\phi$ via push-forward $[\Phi_\lambda]_*(\phi)= \phi\cp\Phi_{\lambda^{-1}}$, and since the natural pairing of a vector-field with a $1$-form is a scalar field, $1$-forms must transform via push-forward by $\Phi_\lambda$ as well. Moreover, it is with this action that standard constructions such as form integrals are diffeomorphism covariant.}, whereby
\begin{align*}
	[\Phi_\lambda]_*\colon \III&\rightarrow \III\\
	 (\w,e)&\mapsto ([\Phi_\lambda]_*(\w),[\Phi_\lambda]_*(e))
\end{align*}
is easily seen to map $\III$ to $\III$ (recall that the quantities in \eqref{invform} only depend on $x^{\mathbf{0}}$). More concretely, we have that
 $(v,c)$ transforms to $(\lambda^{-1} v, \lambda^{-1} c)$ \footnote{To see this
quickly, let $\ell$ denote the straight spatial curve $x^{\mathbf{a}}(t) := t\hspace{1pt} \ell^{\mathbf{a}}$, $t\in (0,1)$, so that
$\int_\ell \omega^{0i} = - (c/\gamma)\hspace{1pt} \delta^i_\mathbf{a} \ell^\mathbf{a}$ and
$\int_\ell e^i = v \hspace{1pt}\delta^i_\mathbf{a} \ell^\mathbf{a}$.
The diffeomorphism covariance of the form integral implies that both of these integrals remain unchanged when the dilation $\Phi_\lambda$ acts on $\omega^{0i}$, $e^i$, and $\ell$ simultaneously.
But under this dilation $x^{\mathbf{a}}$ transforms to $\lambda\hspace{1pt} x^{\mathbf{a}}$, so that
$c$ and $v$ transform as claimed.
}, so that, by equation (\ref{pdef}), $p$ transforms
to $\lambda \hspace{1pt} p$. That is, under the dilation $\Phi_{\lambda}$, we have
\begin{align}
\label{transfff}
	(p,c) \mapsto (\lambda\hspace{1pt} p,\lambda^{-1}c).
\end{align}

Now,
the elementary variables of the quantum theory are
$p$ and (a suitable dense subalgebra of) the complex $C^*$-algebra $\cC$ generated by parallel transports of the Ashtekar connection,
as determined by $c$ via (\ref{AErelation}), along a certain class of curves. $\cC$ is referred to as the configuration algebra.
Depending on whether the curves are restricted to be straight (standard LQC \cite{MathStrucLQC}) or include all embedded analytic curves (Fleischhack approach \cite{CSL}), we have
\begin{equation}
  \label{eq:Ksiiii}
  \cC =
  \begin{cases}
    \hspace{38.8pt} \CAP &\:\:\mbox{ standard LQC}\\
    C_0(\RR)\oplus \CAP &\:\:\mbox{ Fleischhack approach}.
  \end{cases}
\end{equation}
Here, $C_0(\RR)$ denotes the set of all continuous functions on $\RR$ that vanish at infinity, and $\CAP$ the space of  almost periodic functions on $\RR$, i.e.,  the $C^*$-subalgebra of the bounded functions on $\RR$ that is generated by the characters
 \begin{align*}
 	\chi_ \lambda\colon \RR\mapsto \CC,\quad t\mapsto\ee^{\I\lambda t}\qquad\forall\: \lambda\in \RR.
 \end{align*}
The action of dilations on these elementary variables is, from \eqref{transfff},
\begin{align}
\label{fdifdsioiu}
	\textstyle(p,\vp)\mapsto (\lambda\cdot p, \vp_\lambda)\quad\:\:\text{for}\quad\:\: \vp_\lambda\colon t\mapsto \vp(\lambda^{-1}\cdot t).
\end{align}
\section{QUANTIZATION}
\label{quanti}
The elementary variables of homogeneous isotropic LQC are the
elements of a suitable dense subalgebra $\DIDE$ of the configuration algebra \eqref{eq:Ksiiii}
 together with the momentum $p$. More precisely, the classical Poisson algebra is the  complex vector space $\CC\cdot p \times \DIDE$ with Lie bracket
\begin{align}
\label{dsopspod}
\{(z p,\vp),(z'p,\vp')\}=\{0,z'\dot\vp-z\dot\vp'\},
\end{align}
where $\dot\varphi$ denotes the derivative of $\vp$ with respect to its argument.
In the standard case, we choose $\DIDE$ to be the largest subalgebra  of $\cC$ for which \eqref{dsopspod} makes sense, namely,
\begin{align*}
	\DIDEZ:=\{\vp\in \CAP\cap C^\infty(\RR)\:|\: \vp^{(n)}\in \CAP\:\:\forall\in \NN\}.
\end{align*}
In the Fleischhack case, we define $\DIDE:=\DIDEE\oplus \DIDEZ$ for
\begin{align*}
	\DIDEE:=\{\vp\in C_0(\RR)\cap C^\infty(\RR)\:|\: \vp^{(n)}\in C_0(\RR)\:\:\forall\in \NN\}.
\end{align*}
In both cases, $\DIDE$ is dense in $\cC$. This is clear in the standard case, because
\begin{align}
\label{uoioioi}
	\VDZ:=\spann_\CC(\{\chi_\lambda\}_{\lambda\in \RR})\subseteq \DIDEZ
\end{align}
is dense in $\CAP$ by definition. In the Fleischhack case, we define $C_0(\RR)\ni\gauss:=\gauss_0$, with
\begin{align}
\label{gaussss}
\gauss_\epsilon\colon \RR\rightarrow \RR,\quad t\mapsto \ee^{-(\epsilon+t)^2}
\end{align}
for $\epsilon\in \RR$,
and observe that the products $\{\chi_\lambda\cdot \gauss\}_{\lambda\in \RR}$ separate the points in $\RR$; thus, generate a dense $^*$-subalgebra of $C_0(\RR)$ by the Stone-Weierstrass theorem as $\gauss$ vanishes nowhere.

\subsection*{THE QUANTUM ALGEBRA}
We now define the reduced quantum holonomy-flux $^*$-algebra $\aA$ over the basic variables $\p$ and $\wh{\vp}$ for $\varphi\in \DIDE$, that, setting $\hbar=1$, we want to obey the commutation relations
\begin{align}
\label{basiccomm}
\textstyle[\p, \widehat{\varphi}] = - \I \cdot \widehat{\dot\varphi}\qquad\forall\:\varphi\in \DIDE;
\end{align}
the quantum analogue to \eqref{dsopspod} for $p\equiv(p,0)$ and $\vp\equiv(0,\vp)$. In addition to that, we want to implement
\begin{align*}
	\wh{\vp}'\wh{\vp}=\wh{\vp'\vp}\qquad\forall\: \vp,\vp'\in \DIDE.
\end{align*}
For this, let $\cA$ denote the (mixed) complex tensor algebra over the complex vector spaces $\CC\cdot p$ and $\DIDE$. We define the involution $*\colon \cA\rightarrow\cA$ by antilinear extension of
\begin{align*}
	(o_1\ot{\dots}\ot o_n)^*:=o^*_n\ot{\dots}\ot o^*_1
\end{align*}
with $\vp^*:=\ovl{\vp}$ for each $\vp\in \DIDE$, $(z p)^*:=\ovl{z} p$ for each $z\in \CC$, as well as $\ovl{1}_\cA:=1_\cA$. Then, the
reduced quantum holonomy-flux $^*$-algebra $\aA$ is defined to be the quotient of
$\cA$ by the both sided $^*$-ideal $\ID\subseteq \cA$ that is generated by the elements
\begin{align*}
	 p\ot\varphi-\varphi&\ot p+\I\cdot\dot{\varphi}\\[2pt]
	 	\varphi_1\ot\varphi_2 &- \varphi_1\varphi_2\\[2pt]
	 	\chi_0 &- 1_\cA
\end{align*}
for all $\varphi,\varphi_1,\varphi_2\in \DIDE$. We then define $\p:=[ p]$, $\E:=[ \chi_0]$, as well as $\wh{\vp}:=[ \vp]$ for each $\vp\in \DIDE$.
\vspace{6pt}

\noindent
Finally, we impose the action of residual diffeomorphisms
on $\aA$ as follows. For each $\lambda\in \RR_{> 0}$, we define
$\scal_\lambda\colon \cA\rightarrow \cA$ by linear extension of
\begin{align*}
	\scal_\lambda\colon o_1\ot{\dots}\ot o_n\mapsto \scal_\lambda(o_1)\ot{\dots}\ot \scal_\lambda(o_n),
\end{align*}
with $\scal_\lambda(z p):=\lambda\cdot z p$ for each $z\in \CC$,
\begin{align*}
	\scal_\lambda(\varphi)\colon t\mapsto \vp(\lambda^{-1}\cdot t)\quad \forall\:\varphi\in \DIDE,
\end{align*}
as well as $\scal_\lambda(1_\cA):=1_\cA$, cf.\ \eqref{fdifdsioiu}.
Since $\scal_\lambda$ preserves the algebraic structure of $\cA$ as well as the ideal $\ID$, it carries over to a well-defined homomorphism $\Scal_\lambda\colon \aA\rightarrow \aA$.
\subsection*{STATES: THE GNS CONSTRUCTION}
Given a unital $^*$-algebra $\aA$, a {\bf state} (on $\aA$) is a linear functional $\w\colon \aA\rightarrow \CC$  with $\w(a^*a)\geq 0$ and $\w(\E)=1$. Here, we automatically have the Cauchy-Schwarz inequality
\begin{align}
\label{CASC}
	|\w(a^*b)|^2&\leq \w(a^*a)\cdot \w(b^*b)
\end{align}
for each $a,b\in \aA$. The corresponding Gel'fand (left) ideal is defined by $\ID_\w:=\{a\in \aA\:|\: \w(a^*a)=0\}$ and provides us with the pre-Hilbert space $(\Hil_\w,\langle\cdot,\cdot\rangle_\w)$, for $\Hil_\w$ the quotient of $\aA$ by $\ID_\w$, and
\begin{align*}
	\langle \psi_a,\psi_b\rangle_\w:=\w(a^*b)\qquad\forall\:\psi_a,\psi_b\in \Hil
\end{align*}
 ($\psi_a$ denotes the class of $a\in \aA$ in $\Hil_\w$). In addition to that, we have the $^*$-representation
	$\varrho_\w\colon \aA\rightarrow \Borel(\Hil_\w)$, defined by $\varrho_\w(a)(\psi_b):=\psi_{ab}$. Here, and in the following, $\Borel(\Hil)$ denotes the space of adjointable operators on the pre-Hilbert space $\Hil$.

	An important feature of the GNS-construction is that $\psi_\E$ is  cyclic, i.e., we have $\varrho_\w(\aA)(\psi_\E)=\Hil$. Moreover, if $(\varrho,\Hil,\langle\cdot,\cdot\rangle)$ is a further $^*$-representation of $\aA$ on a pre-Hilbert space $\Hil$, with
\begin{align*}
	\w(a)= \langle\psi,\varrho(a)(\psi)\rangle\qquad \forall\: a\in \aA
\end{align*}
for a cyclic vector $\psi\in \Hil$, then $(\varrho_\w,\Hil_\w,\langle\cdot,\cdot\rangle_\w)$ is unitarily equivalent to
$(\varrho,\Hil,\langle\cdot,\cdot\rangle)$ via the intertwiner
\begin{align*}
		 \Hil_\w\rightarrow \Hil,\quad \psi_a\mapsto \varrho(a)(\psi).
\end{align*}

\section{UNIQUENESS OF THE INVARIANT STATE}
\label{jndshjsd}
In the following, let $\aA$ denote the quantum holonomy-flux $^*$-algebra  defined in the first subsection of Sect.\ \ref{quanti}. We first show a continuity property of states on $\aA$, and then prove uniqueness and existence of the dilation-invariant state on $\aA$. Then, we show that the GNS-representation of this state is unitarily equivalent to the standard representation of homogeneous isotropic LQC in both the standard and the Fleischhack case. More precisely, we find that the GNS-Hilbert space is the same in both situations, and that in the Fleischhack case, the additional algebra elements are just mapped to zero operators.

\subsection*{CONTINUITY}
Let $\w\colon \aA\rightarrow \CC$ be a state, and let $\B=\DIDEZ$ in the standard case, as well as $\B=\DIDEE$ or $\B=\DIDEZ$ in the Fleischhack one. Moreover, denote by $\oB$ the closure of $\B$ in $\cC$, i.e.,
\begin{equation*}
  \oB =
  \begin{cases}
    \CAP &\mbox{if }\: \B=\DIDEZ\\
    \hspace{7.4pt}C_0(\RR)  & \mbox{if }\: \B=\DIDEE.
  \end{cases}
\end{equation*}
We claim that the linear functional
\begin{align}
\label{fdgdfgg}
	\w_{\B}\colon \B \rightarrow \CC,\quad \vp\mapsto \w(\wh{\vp})
\end{align}
is continuous w.r.t.\ $\|\cdot\|_\infty$, i.e., that it extends to a continuous linear functional $\w_{\oB}\colon \oB\rightarrow \CC$.
To see this, first observe that the smooth function \footnote{As figured out in \cite{CSL2}, in the previous versions of this article  it had been overseen that, instead of $\nu\cp \vp\in C_0(\RR)$, we have $\nu\cp \vp\in \spann_{\CC}(\chi_0) + C_0(\RR)$ for $\vp\in C_0(\RR)$ and $\nu\colon (-1,1)\ni t\mapsto \sqrt{1+t}$.
In the current version, we have fixed this issue by adding ``$-1$'' in the definition of $\nu$.
We then have $\sqrt{1+\phi} \in \DIDE$ (instead of $\sqrt{1+\phi} \in \B$, as wrongly stated in the previous versions of this article) -- This, however, makes no difference for our conclusion that $1+\w(\vp_\nu)\geq 0$ holds, as $\w$ is defined on full $\DIDE$.}
\begin{align*}
	\nu\colon (-1,1)\ni t\mapsto \sqrt{1+t}-1
\end{align*}
(and each of its derivatives) can be represented by a power series.
Thus, by completeness of $\oB$, for
$\vp\in \oB$ real valued with $\|\vp\|_\infty<1$, we have $\nu\cp \vp\in \oB$
as well as
\begin{align*}
	\nu^{(n)}\cp \vp\in \spann_{\CC}(\chi_0) + \oB\qquad\forall\: n\geq 1.
\end{align*}
Then, for $\vp\in \B$, the chain rule gives
\begin{align*}
	\textstyle\partial_t (\nu\cp \vp)=(\dot\nu\cp\vp)\cdot \dot{\vp}\in \ovl{\B}.
\end{align*}
Now, the right hand side is differentiable; and, applying the same arguments inductively, we find that $(\nu\cp\vp)^{(n)}\in \ovl{\B}$ holds for each $n\in \NN$, hence
\begin{align*}
	\nu\cp \vp=\sqrt{1+\vp}-\chi_0\in \B.
\end{align*}
We thus have $\vp_\nu:= \sqrt{1+\vp} \in \DIDE$ with
\begin{align*}
	1+\w_\B(\vp)&= \w(\E+\wh{\vp})=\w(\wh{\vp}_\nu^*\wh{\vp}_\nu)\geq 0.
\end{align*}
Then, for each non-zero and real-valued $\vp\in \B$, we have
\begin{align*}
	\textstyle 1\pm \frac{\epsilon}{\|\vp\|_\infty}\cdot \w_\B(\vp)\geq 0\quad&\textstyle\Longrightarrow\quad \pm\phantom{|}\w_\B(\vp)\phantom{|}\geq -\textstyle\frac{1}{\epsilon}\cdot \|\vp\|_\infty\\
	&\textstyle\Longrightarrow\quad \phantom{\pm}|\w_\B(\vp)|\leq \textstyle\phantom{-}\frac{1}{\epsilon}\cdot \|\vp\|_\infty
\end{align*}
for each $0<\epsilon<1$.
This shows $|\w_\B(\vp)|\leq \|\vp\|_\infty$ for each real-valued $\vp\in \B$. Then, for $\vp\in\B$ arbitrary, we conclude from $\|\Re(\vp)\|_\infty,\|\Im(\vp)\|_\infty\leq \|\vp\|_\infty$ that
\begin{align*}
	|\w_\B(\vp)|\leq |\w_\B(\Re(\vp))| + |\w_\B(\Im(\vp))|\leq 2\cdot\|\vp\|_\infty.
\end{align*}
\subsection*{UNIQUENESS}
Suppose that $\w$ is invariant under dilations. Then,
\begin{align*}
	\w(\p^*\p)=\w(\p^2)=\Scal_\lambda(\w)(\p^2)=\lambda^2\cdot \w(\p^*\p)
\end{align*}
for $0<\lambda\neq 1$, implies $\w(\p^*\p)=0$; hence, $\p\in \ID_\w$. Thus, we have
\begin{align}
\label{odpdpdd}
	\w|_{\aA\cdot \p}=0=\w|_{\p\cdot \aA}
\end{align}
by \eqref{CASC}, as $\p^*=\p$.
This implies that $\w$ is uniquely determined by its values on $\{\wh{\vp}\:|\: \vp\in \DIDE\}$, because, applying \eqref{basiccomm} successively, we see that each element in $\aA$ can be written as a sum of elements of $\{\wh{\vp}\:|\: \vp\in \DIDE\}$ and $\aA\cdot \p$. In addition to that, \eqref{odpdpdd} shows
	$\w(a\cdot \p)=0=\w(\p\cdot a)$ for each $a\in \aA$;
hence, $\w_\B\cp \partial_t=0$ as
\begin{align*}
	-\I\cdot \w(\hspace{1pt}\wh{\dot\vp}\hspace{1pt})=\w([\p,\wh{\vp}])=0\quad\forall\:\vp\in \DIDE.
\end{align*}
In particular, we have
\begin{align}
\label{gfhghg}
	\textstyle\w_{\DIDEZ}(\chi_\lambda)=-\frac{\I}{\lambda}\cdot\big(\w_{\DIDEZ}\cp \partial_t\big)(\chi_\lambda)=0\quad\forall\:\lambda\neq 0,
\end{align}
as well as $\w_{\DIDEZ}(\chi_0)=\w(\E)=1$. By denseness of $\VDZ$ in $\DIDEZ$ and by continuity of $\w_{\DIDEZ}$, this completely determines $\w_{\oB_\AP}$. Thus, uniqueness is clear in the standard case.

 Furthermore, in the Fleischhack case, we must have $\w_{\oB_0}=0$, so that it reduces to the standard case.
 To see this, let us first observe that $\w_{\oB_0}$ is positive, because for $C_0(\RR)\ni\vp\geq 0$, we have $\sqrt{\vp}\in C_0(\RR)$; hence,
\begin{align*}
	\w_{\oB_0}(\vp)=\textstyle\lim_n \w_{\oB_0}(\ovl{\vp}_n \vp_n)=\lim_n \w(\wh{\vp}_n^* \wh{\vp}_n)\geq 0
\end{align*}
for $\{\vp_n\}_{n\in \NN}\subseteq \B_0$ a sequence with $\lim_n\vp_n=\sqrt{\vp}$. Thus,
\begin{align*}
	\textstyle\w_{\oB_0}\colon \vp\mapsto \int \vp\: \dd\mu
\end{align*}
holds for some unique finite Radon measure $\mu$ on $\RR$, by the Riesz-Markov theorem. Then, dilation-invariance of $\w$ implies
$\mu=s\cdot \mu_\delta$ for some $s\geq 0$, with
\begin{equation*}
  \mu_\delta(A) :=
  \begin{cases}
    1 &\mbox{if } 0\in A\\
    0  & \mbox{if } 0\notin A
  \end{cases}
\end{equation*}
for each $A\in \Borel(\RR)$, cf.\ \cite{JM}. Then,
\begin{align*}
	0=(\w_{\DIDEE}\cp\partial_t)(\gauss_\epsilon)=\w_{\DIDEE}(\dot\gauss_\epsilon)=s\cdot \dot\gauss_\epsilon(0)=-2s\epsilon\cdot \gauss(\epsilon)
\end{align*}
shows $s=0$; hence, the claim. Here, $\gauss_\epsilon$ is defined by \eqref{gaussss}.

\subsection*{THE STANDARD FUNCTIONAL}
For existence of the dilation-invariant state, we will need the linear functional
\begin{align}
\label{hggffhjhjkj}
	\textstyle L\colon \cC\rightarrow \CC,\quad \vp\mapsto \lim_n \frac{1}{2n}\int_{-n}^n \vp(t)\:\dd t,
\end{align}
whose well-definedness follows easily from
\begin{align}
\label{hjhjgfhj}
	L(\chi_0)=1\qquad\text{and}\qquad L(\chi_\lambda)=0\quad\forall\:\lambda\neq 0,
\end{align}
from denseness of $\VDZ$ in $\CAP$, as well as from the fact that $L|_{C_0(\RR)}=0$ holds in the Fleischhack case.
$L$ is furthermore directly seen to be continuous. From this, and the fact that
$\scal_\lambda$ is an isometry for each $\lambda> 0$, we have
\begin{align}
\label{ssddssdsaaa}
	L\cp\scal_\lambda = L\qquad\forall\:\lambda\in \RR_{> 0}
\end{align}
as this clearly holds on $\VDZ$.
Moreover, it is clear that $\langle\vp_1,\vp_2 \rangle=\ovl{\langle\vp_2,\vp_1 \rangle}$ holds for
\begin{align}
\label{fddfgfsss}
	\langle\vp_1,\vp_2 \rangle:= L(\ovl{\vp}_1\vp_2)\qquad \forall\: \vp_1,\vp_2\in \cC;
\end{align}
and by continuity of $L$, we have $\langle\vp,\vp \rangle>0$ for each $0\neq\vp\in \CAP$, cf.\ Appendix A.
Thus, $\Hil:= \DIDEZ$ is a pre-Hilbert space w.r.t.\ $\langle\cdot,\cdot\rangle$.

\subsection*{EXISTENCE}
We now show existence of the dilation-invariant state for both the standard and the Fleischhack case simultaneously.
For this, we let  $\pap\colon \cC\rightarrow \CAP$ denote the identity on $\CAP$ in the standard case, as well as the projection onto $\CAP$ in the Fleischhack one. We define the operators
\begin{align*}
	\wt{\vp}\colon \Hil\rightarrow\Hil,&\quad \vp'\mapsto \pap(\vp)\cdot \vp'\\
	\wt{z p}\colon\Hil\rightarrow\Hil,&\quad \vp'\mapsto -z\I\cdot \dot\vp'
\end{align*}
for each $\vp\in \DIDE$ and $z\in \CC$, as well as $\wt{z1}_\cA:=z\cdot \id_\Hil$ for each each $z\in \CC$. Obviously, then
\begin{align}
\label{posdopsdpo}
	[\wt{p},\wt{\vp}]=-\I\cdot\wt{\dot\vp}\qquad\forall\:\vp\in \DIDE
\end{align}
holds. In addition to that, we have
\begin{align*}
	\langle\hspace{1pt} \wt{\ovl{\vp}}(\cdot),\cdot\rangle &= \langle \cdot,\wt{\vp}(\cdot)\rangle\qquad \forall\: \vp\in \DIDE\\
	\langle\wt{p}(\cdot),\cdot\rangle &= \langle \cdot,\wt{p}(\cdot)\rangle;
\end{align*}
whereby the first line is obvious, and the second one is due to partial integration.
 From this, it follows inductively that
$\Omega\colon  \cA\rightarrow \CC$ defined by linear extension of
\begin{align}
\label{pdisisad}
\Omega(o_1\ot{\dots}\ot o_n):=\langle \chi_0,(\wt{o}_1\cp{\dots}\cp \wt{o}_n)(\chi_0)\rangle
\end{align}
is positive. Moreover, it is straightforward from \eqref{posdopsdpo}
that $\Omega$ is zero on $\ID$; thus, defines a positive linear functional $\w$ on $\aA$, for which obviously $\w(\p^*\p)=0=\w(\p)$ holds. In addition to that, it is immediate from \eqref{ssddssdsaaa}, the definition of $\wt{p}$, and the chain rule that $\Omega$ is $\scal_\lambda$-invariant, i.e., that $\w$ is $\Scal_\lambda$-invariant.

\subsection*{THE GNS REPRESENTATION}
We define the $^*$-representation $\rho\colon \cA\rightarrow \Borel(\Hil)$ in analogy to \eqref{pdisisad}, by linear extension of
\begin{align}
\label{odood}
\rho(o_1\ot{\dots}\ot o_n):= \wt{o}_1\cp{\dots}\cp \wt{o}_n.
\end{align}
 By the same reasons as above, $\rho$ is zero on $\ID$; thus, defines a $^*$-representation $\varrho\colon \aA\rightarrow \Borel(\Hil)$. Obviously, $\chi_0\in \Hil$ is cyclic, and we have
\begin{align*}
	\w(a)=\langle\chi_0, \varrho(a)(\chi_0)\rangle\quad\forall\: a\in \aA
\end{align*}
by construction. Thus, $\varrho$ is unitarily equivalent to the GNS-representation that corresponds to $\w$, just by what we have discussed in the last part of Sect.\ \ref{quanti}.

Let us finally clarify the connection between the representations in the standard and the Fleishhack case. For this, let $\cA, \aA, \ID, \w, \rho, \varrho$ be defined as in the second case, and denote by $\cA_s, \aA_s, \ID_s,\w_s,\rho_s, \varrho_s$ the respective quantities in the standard one.
It is then obvious from the definition of $\w$ that $\ID_0\subseteq \ID_\w$ holds, for $\ID_0\subseteq \aA$ the both-sided ideal generated by $\wh{\B}_0:=\{\wh{\vp}_0\:|\: \vp_0\in \B_0\}$ \footnote{Alternatively, this can also be obtained from the properties of a dilation-invariant state that we have derived in the second part of Sect.\ \ref{jndshjsd}.}. This means that in the GNS-representation of $\w$, the elements of $\ID_0$ correspond to zero operators, as well as to zero-Hilbert states; which is the reason why we could use the same pre-Hilbert space $\Hil$ for the definition of $\varrho$ and $\varrho_s$.
More precisely, the canonical inclusion $\inn\colon \cA_s\rightarrow \cA$ carries over to an injection $\iota\colon \aA_s\rightarrow \aA$, and then we have
\begin{align*}
	\aA= \iota(\aA_s)\oplus\ID_0\quad\text{with}\quad
	\varrho\cp \iota=\varrho_s\quad\text{and}\quad\varrho|_{\ID_0}=0,
\end{align*}
cf.\ Appendix B. Thus, the GNS-representation in the Fleischhack case is actually just given by the standard one.

\section{EXTENSION: THE BIANCHI I CASE}

The key new elements in this paper, which have allowed all residual diffeomorphisms to have a well-defined action in the quantum
cosmological model, are (1.) to let residual diffeomorphisms act not only on the dynamical variables,
but also on the fiducial cell, and (2.) the use of the \textit{canonical}
momenta conjugate to the connection variable in defining the quantum algebra.
Here we apply these same two elements to the case of Bianchi I quantum cosmology.
We show that again one obtains
an action of \textit{all} residual diffeomorphisms on the quantum algebra,
and not just an action of volume preserving ones.
Furthermore, we show how invariance, under this action, of a GNS state on the algebra,
yields the unique state corresponding to the standard representation of Bianchi I defined in Ashtekar and Wilson-Ewing \cite{aw2009}.
Thus, we demonstrate that the approach of this paper leads to
the same conclusions as those in Ashtekar and Campiglia \cite{UniquOfKinLQC},
whence they may be viewed as generalizing the results of \cite{UniquOfKinLQC}
in such a way that the isotropic case --- as well as the Fleischhack case --- can also be handled.

The Bianchi I cosmological model is obtained by starting with the same full theory framework reviewed in
Sect.\ \ref{sec:classical}, but then imposing invariance under only the translation subgroup of the Euclidean group.
If one additionally uses diffeomorphism and gauge rotation freedom to impose that the triad be diagonal in the fixed coordinate system ---
$e^i_\mathbf{a} = v_i \delta^i_\mathbf{a}$, the so-called diagonal gauge --- then all terms in the Lagrangian involving
time derivatives reduce to
\begin{align}
\label{syltermBI}
&\sum_{i=1}^3 \frac{V_0[\mathcal{C}]}{k\gamma}\frac{|v_1 v_2 v_3|}{v_i} \frac{\mathrm{d}}{\mathrm{d}x^{\mathbf{0}}}
\left( \gamma\he\omega_{\mathbf{i}}^{i0} - \frac{1}{2}\epsilon^{ijk} \omega_{\mathbf{i}jk}\right)
 \end{align}
where the unbold and bold indices $i$ and $\mathbf{i}$ have the same numerical value,
the bold $\mathbf{i}$ denotes component with respect to $x^\mathbf{i}$,
and all fields are independent of the spatial variables $x^\mathbf{a}$.
Thus,
\begin{align}
\label{BIcanvars}
c^i := \gamma\he\omega_\mathbf{i}^{i0}-\frac{1}{2} \epsilon^{ijk} \omega_{\mathbf{i}jk},
\qquad
p_i := \frac{V_0[\mathcal{C}]}{k \gamma}\frac{|v_1 v_2 v_3|}{v_i}
\end{align}
are the dynamical variables and their canonical conjugates, with Poisson brackets
$\{c^i, p_j\}= \delta^i_j$.
The only components of $\omega_\alpha^{IJ}$ appearing in
the above symplectic term (\ref{syltermBI}) are $c^i$, so that the rest of its components are Lagrange multipliers.
Varying the action with respect to these Lagrange multipliers implies, in particular, the constraint that the Ashtekar connection
\begin{align*}
A^i_{\mathbf{a}} := -\left(\frac{1}{2}\epsilon_{ijk}\, \omega_\mathbf{a}^{jk} + \gamma \,  \omega_\mathbf{a}^{0i}\right)
\end{align*}
be diagonal, i.e., $A^i_\mathbf{a} = 0$ for $\mathbf{a} \neq i$.
This constraint is second class,
as we also trivially have the constraint on the conjugate momentum $\Pi_{A^i_{\mathbf{a}}} = 0$ for $\mathbf{a} \neq i$,
and $\{A^i_\mathbf{a}, \Pi_{A^i_{\mathbf{a}}} \} = 1$. Hence it must be solved prior to quantization,
yielding an Ashtekar connection of the form
\begin{align*}
A^i_\mathbf{a} = c^i \delta^i_\mathbf{a} .
\end{align*}

The ``residual diffeomorphisms'' in this Bianchi I framework consist in the following three dimensional group of
(possibly) \textit{anisotropic} dilations.
For each $\vec{\lambda} \in (\mathbb{R}_{>0})^3$,
let $\Phi_{\vec{\lambda}}$ denote the diffeomorphism which expands the $x^{\mathbf{a}}$ direction by a factor of
$\lambda^{\mathbf{a}}$ for $\mathbf{a} = 1,2,3$, as measured in the fixed coordinate system $x^{\mathbf{a}}$.
Letting $\Phi_{\vec{\lambda}}$ act on the cell $\mathcal{C}$ maps
$V_0[\mathcal{C}]$ to $|\lambda_1 \lambda_2 \lambda_3| V_0[\mathcal{C}]$.
Furthermore, $\Phi_{\vec{\lambda}}$ acts on the co-vectors
$e^i_{\mathbf{a}} = v^i \delta^i_{\mathbf{a}}$ and $A^i_{\mathbf{a}} = c^i \delta^i_\mathbf{a}$
via pull-back by $\Phi_{\vec{\lambda}}^{-1}$, inducing the action
$(v^i, c^i) \mapsto (\lambda_i^{-1} v^i, \lambda_i^{-1} c^i)$, which, via (\ref{BIcanvars}), translates into the action
\begin{align}
\label{canvaraction}
(p_i, c^i) \mapsto (\lambda_i \, p^i, \lambda_i^{-1} c^i)
\end{align}
of the dilations on the basic variables $(p_i, c^i)$.

In the Bianchi I approach of \cite{UniquOfKinLQC},
 holonomies of the Ashtekar connection $A^i_\mathbf{a} = c^i \delta^i_\mathbf{a}$ along straight curves parallel to one of the axes of the fixed coordinate system are considered in order to define the configuration algebra $\cC$, yielding $\CAPP$.
This is generated by the functions $\chi_\lambda\cp\prr_i$ for $\lambda\in \RR$ and $i=1,2,3$,
 where $\prr_i$ denotes projection to the $i^{\text{th}}$ component in $\mathbb{R}^3$.
 We let $\VDZ$ denote the span of these functions, in analogy to \eqref{uoioioi}.

The classical Poisson algebra is then the complex vector space $\CC\cdot p_1\times \CC\cdot p_2\times\CC\cdot p_3 \times \DIDE$
with bracket
\begin{align}
\label{dsopspodfgfgf}
\{(z p_i,\vp),(z'p_j,\vp')\}=\{0,z'\partial_j\vp-z\partial_i\vp'\}
\end{align}
for $1\leq i,j\leq 3$: here, $\DIDE\subseteq \cC$ is chosen to be the largest subalgebra of $\cC$ for which \eqref{dsopspodfgfgf} makes sense; namely that consisting of all $\vp\in \cC\cap C^\infty(\RR^3)$ such that
\begin{align*}
	(\partial_{k_1}\cp\dots\cp \partial_{k_n})(\vp)\in \cC\quad\text{ for }\quad 1\leq k_1,\dots,k_n \leq 3
\end{align*}
for each $n\in \NN$.
This is dense in $\cC$ as clearly $\VDZ\subseteq \DIDE$ holds.
Already if one additionally takes holonomies along straight curves not-parallel to the axes of the fixed coordinate system into account, existence of such dense $\DIDE$ closed under partial derivatives is by far not clear. For this reason, and because the Fleischhack algebra has not been calculated for Bianchi I so far, we do not consider the analogue of the Fleischhack generalization for the Bianchi I case.

The quantum holonomy-flux $^*$-algebra $\aA$ is the quotient of the (mixed) complex tensor algebra $\cA$,
over the vector spaces $\DIDE$ and $\CC\cdot p_i$ for $i=1,2,3$, by the both sided $^*$-ideal that is generated by
\begin{align*}
	 p_j\ot\varphi-\varphi&\ot p_j+\I\cdot\partial_j\varphi\\[2pt]
	 \varphi_1\ot\varphi_2- \varphi_1\varphi_2&\quad\:\: p_i\ot p_j-p_j\ot p_i\\[2pt]
	 	\chi_0 &- 1_\cA \hspace{48pt}
\end{align*}
for $\varphi,\varphi_1,\varphi_2\in \DIDE$ and $i,j=1,2,3$.
We let $\p_i:=[ p_i]$ for $i=1,2,3$, $\E:=[ \chi_0]$, and $\wh{\vp}:=[ \vp]$ for $\vp\in \DIDE$,
so that $[\p_i, \p_j] =0$ for $1\leq i,j\leq 3$ as well as
\begin{align*}
\textstyle[\p_i, \widehat{\varphi}] = - \I \cdot \widehat{\partial_i\varphi}\qquad\forall\:\varphi\in \DIDE.
\end{align*}
The action of dilations (\ref{canvaraction}) induces an action $\scal_{\vec{\lambda}}: \cA \rightarrow \cA$
for each $\vec{\lambda} = (\lambda_1, \lambda_2, \lambda_3)$ via
$\scal_{\vec{\lambda}}(z p_i):=\lambda_i\cdot z p_i$ for $z\in \CC$ and $i=1,2,3$, $\scal_{\vec{\lambda}}(1_\cA):=1_\cA$, as well as
\begin{align*}
	\scal_{\vec{\lambda}}(\varphi)\colon (x_1,x_2,x_3)\mapsto \vp( x_1/\lambda_1,x_2/\lambda_2,x_3/\lambda_3)

\end{align*}
for each $\vp\in \DIDE$. This action lifts to an action $\Scal_{\vec{\lambda}}\colon \aA\rightarrow\aA$
on the quantum algebra.
If $\w\colon \aA\rightarrow \cC$ is a $\Scal_{\vec{\lambda}}$-invariant state for all
$\lambda_1, \lambda_2, \lambda_3 > 0$, then the same arguments as in the homogeneous isotropic case show that it is uniquely determined by the continuous linear functional $\w_{\DIDEZ}\colon \DIDEZ\rightarrow \CC$,\: $\vp\mapsto \w(\wh{\vp})$
for $\DIDEZ:=\DIDE$ as we have $\wh{p}_i\in\ID_\w$ (the Gel'fand ideal) for $i=1,2,3$.
Moreover, in manner similar to (\ref{gfhghg}),
we find that $\w_{\DIDEZ}(\E)=1$ as well as $\w_{\DIDEZ}(\chi_\lambda\cp\prr_i)=0$ holds for $\lambda\neq 0$ and $i=1,2,3$. Uniqueness is then clear from continuity of $\w_{\DIDEZ}$,
which follows from the same arguments as in Sect.\ \ref{jndshjsd}.
For existence, we can just substitute \eqref{hggffhjhjkj} by
\begin{align*}
	\textstyle L\colon \cC\rightarrow \CC,\quad \vp\mapsto \lim_n \frac{1}{(2n)^3}\int_{[-n,n]^3}\vp(x)\:\dd x,
\end{align*}
and define $\langle\cdot,\cdot\rangle$ as in \eqref{fddfgfsss} as well as $\Hil:=\DIDEZ=\DIDE$ just as in the homogeneous isotropic case \footnote{Again, we have $\langle\vp,\vp \rangle>0$ for each $0\neq\vp\in \CAPP$: this just follows as in Appendix A if we replace $\RB\equiv\Spec(\CAP)$ by $\RBB\equiv\Spec(\CAPP)$ in the argumentation there.}.
We define the operators $\wt{z1}_\cA:=z\cdot \id_\Hil$ for each $z\in \CC$, as well as
\begin{align*}
	\wt{\vp}\colon \Hil\rightarrow\Hil,&\quad \vp'\mapsto \vp\cdot\vp'\\
	\wt{z p_j}\colon\Hil\rightarrow\Hil,&\quad \vp'\mapsto -z\I\cdot \partial_j\vp'
\end{align*}
for $j=1,2,3$, each $\vp\in \DIDE$ and $z\in \CC$. Then, $\Omega: \cA \rightarrow \mathbb{C}$
defined in exact analogy to \eqref{pdisisad}
carries over to a state on $\aA$ that has the desired properties.
The corresponding GNS-representation then is unitarily equivalent to that defined  by expression \eqref{odood},
which one sees to be the standard representation of Bianchi I
introduced by Ashtekar and Wilson-Ewing \cite{aw2009}.

\section*{ACKNOWLEDGMENTS}
We thank an anonymous referee for suggesting that we include an analysis of the Bianchi I case. We furthermore thank Christian Fleischhack for communicating to us \cite{CSL2} the technical oversight we have made in the continuity part in Sect.\ \ref{jndshjsd} in the  earlier versions of this article.
This work has been supported in part by the Alexander von Humboldt foundation of Germany and NSF Grant PHY-1505490.

\section*{APPENDIX A}
Let us first observe that
\begin{align*}
	\textstyle L(\vp)= \int \Gel(\vp)\:\dd\muRB \quad\forall\:\vp\in \CAP
\end{align*}
holds for $\muRB$ the Haar measure on the Bohr compactification $\RB\equiv\Spec(\CAP)$ of $\RR$, and
\begin{align*}
	 \Gel\colon \CAP&\rightarrow  C(\RB)\\
	 	\vp&\mapsto [\wh{\vp}\colon \alpha\mapsto \alpha(\vp)]
\end{align*}
 the corresponding Gel'fand transform. This
 is immediate for each $\vp\in \VDZ$ from general theory of Haar measures on compact abelian groups; and thus, by continuity, clear for all $\vp\in \CAP$.
 Consequently, we have
 \begin{align*}
 \textstyle 0=\langle \vp,\vp\rangle=\int |\wh{\vp}|^2\:\dd\muRB\quad&\Longrightarrow\quad \wh{\vp}=0\\
 &\Longrightarrow\quad \vp=0.
 \end{align*}
 This is because $\wh{\vp}$ is continuous, and because $\muRB(O)>0$ holds for each non-empty open subset $O\subseteq \RB$.

\section*{APPENDIX B}
First observe that $\inn\colon \cA_s\rightarrow \cA$ carries over to a map $\iota\colon \aA_s\rightarrow \aA$ because $\inn(\ID_s)\subseteq \ID$ holds, so that
\begin{align*}
	\rho\cp \inn=\rho_s\quad\Longrightarrow\quad \varrho\cp \iota=\varrho_s.
\end{align*}
Then, to verify injectivity of $\iota$, we define $\pr_s\colon \cA\rightarrow \cA_s$ by linear extension of
\begin{align*}
	\pr_s(o_1\ot{\dots}\ot o_n):= \pr(o_1)\cp{\dots}\cp \pr(o_n)
\end{align*}
with $\pr(\vp):=\pap(\vp)$ for each $\vp\in \DIDE$, and
\begin{align*}
	 \pr(z p):=zp\hspace{1pt},\quad \pr(z1_\cA):=z1_\cA\qquad\forall\: z\in \CC.
\end{align*}
Since $\pr_s(\ID)=\ID_s$ holds, $\pr_s$ carries over to a map $\pi\colon \aA\rightarrow \aA_s$; and then
\begin{align*}
	\pr_s\cp\inn=\id_{\cA_s}\quad\Longrightarrow\quad\pi\cp\iota=\id_{\aA_s}
\end{align*}
shows that $\iota$ is injective, and that
$\inn\cp \pr_s$
is a projection. We thus have
\begin{align*}
	\cA=(\inn\cp \pr_s)(\cA)\oplus \ker[\inn\cp \pr_s]=\inn(\cA_s)\oplus \ker[\pr_s],
\end{align*}
whereby $\ker[\pr_s]$ is easily seen to be the both-sided ideal $J_0\subseteq \cA$ generated by $\B_0$. Then, applying the quotient map $\quot\colon \cA\rightarrow \aA$ to both sides, we find that
\begin{align*}
	\aA= (\quot\cp \inn)(\cA_s)\oplus \quot(J_0)=\iota(\aA_s)\oplus \ID_0
\end{align*}
holds. Here, in the first step, the direct sum property is preserved, because
\begin{align*}
	a\in (\quot\cp \inn)(\cA_s)\cap \quot(J_0)\quad&\Longrightarrow\quad \quot(\inn(a_s))=a=\quot(a_0)\\
	&\Longrightarrow\quad \inn(a_s)-a_0\in \ID\\
	&\Longrightarrow\quad a_s=(\pr_s\cp\inn)(a_s)\in \ID_s\\
	&\Longrightarrow\quad \inn(a_s)\in \ID\\
	&\Longrightarrow\quad a=0,
\end{align*}
for some $a_s\in \cA_s$ and $a_0\in J_0=\ker[\pr_s]$.

\vfill

redpajama_set_name: RedPajamaArXiv


This entry was posted in #DoNoHarm, #ResistanceMatters #LetLoveLeadtheWay, arts, children, Common Core, Creative Writing, defending public school students, education activism, education revolution, equity, freedom, inhumanity, Jamaal Bowman, Mark Naison, Oppression, Poetry, Social Justice, student voice, Uncategorized. Bookmark the permalink.
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Survey participants are more than just judges!
When considering the participant's role in traditional survey research, you could think he or she takes up the role of a 'judge'. In this role, consumers rate you on your performance and help you improve. Nevertheless, the extent to which we allow participants help brands improve, is rather limited in surveys. Survey research mainly focuses on consumers rating your brand, without taking into account the other collaboration roles they could take up in research.
The role of peer, where consumers help brands immerse in the consumers' daily lives and develop fresh consumer insights. This is where consumers help us get a deeper understanding of their own daily lives.
The role of colleague, where consumers help brands generate new ideas, co-decide and review their strategies and plans. This is where consumers become part of the team.
The role of advocate, where consumers are activated by your brand and drive positive conversations. These consumers are your real brand advocates.
Judges, peers, colleagues and advocates… so what do we really need in survey research?
We believe the value of the future of surveys lies in a second survey dimension which we call 'The Village'. This village concept taps into the 3 learning dimensions we tackled in our previous blog post, namely engagement, conversation and consumer context, while also taking into account the different roles consumers can take in research.
The village is an additional survey element that participants, who have completed the regular survey, can access to participate in other fun task-based elements. This second survey dimension is optional, as we believe that only topic-, experience- and brand-engaged participants will bring more valuable learnings through richer content.
As the name reveals, the platform visually resembles an actual village with different buildings. Behind each building lies a specific task which focuses on a different consumer collaboration role.
A central building in the village is 'The Lounge', which is the social corner of the village in which participants can connect with other participants, researchers and brands. In this building participants can give additional feedback and advice to a brand. This is where consumers take up the role of both judge and peer.
'The Photo store' is a building where we ask consumers to upload pictures and reflect upon them. This is where consumers clearly take up the role of peer as they help us immerse in their daily lives.
In the 'Ideation building', participants are asked to think along with brands and provide ideas on different topic areas. It is a brainstorm tool where participants help brands to capture the solutions space and build recommendations. This task-based element allows consumers to take up the collaboration role of colleague.


Jazz up your holiday like a celebrity chef.
At this point, we assume you've noticed that the holidays come around every single year without fail. And while that's often exciting, it can also get monotonous. The same decorations, the same traditions, the same food. Traditions can be great — especially ones you do with your family — but sometimes it's just as much fun to shake things up, even just a little.
That's why, with the help of celebrity chef Chuck Hughes, we've pulled together some small ways to spice up your holiday dinners (in some cases, literally). Chuck was in town for his Holiday Takeover of Pusateri's in Toronto and he walked us through some of his favourite non-traditional holiday sides — the perfect way to shake up your meal without sending your oldest, most traditional relatives into cardiac arrest.
🎄Holiday Takeover🌲 . @pusaterisfoods✖ @chefchuckhughes ✖ @cxbo_chocolates – Link in bio!
This might sound like we're getting crazy right out of the gate, but Chuck swears that this is one of the best variations on traditional stuffing he's ever made.
"Oyster stuffing is traditional on the East Coast, Louisiana, where oysters are plentiful, and they use it to add some saltiness, brininess to [the stuffing]," Chuck said.
Find the recipe here (it's inside the turducken).
As for if you should be stuffing your bird or cooking your dressing in a pan, Chuck says the answer is BOTH.
"The way it works in my house is there's never enough stuffing and I think when you put it in the bird, it's good because it takes all that flavour from the juice but my favourite part is really the crispy edges which you don't get as much in the bird," he said.
Brussels sprouts may be highly controversial at your dinner table, but you can change anyone's mind about these little green leaf balls with a good helping of spicy maple dressing — hey, the less they look like veggies, the more likely it is your picky father-in-law will eat them, right? In his collab with Pusateri's, Chuck suggests glazing your Brussels sprouts in a dressing of maple syrup and his own pepper spread (but you can use whatever kind of spice you like, of course) before baking them to get those delicious charred edges.
Mashed potatoes are obviously a staple on every holiday table but leave it to a chef to elevate that humble dish to something so much more than just mushed up spuds. We're sure you already have a secret family recipe, but Chuck's tip is to simply swap in brown butter for a deeper, nuttier flavour.
If you're looking for a complete alternative to mashed potatoes — or just to put an additional mashed veg on the table — Chuck suggests mashed turnips either in the potatoes or on their own.
"In Quebec, my mom, when she does the mashed potatoes for Thanksgiving and Christmas, she adds turnip and carrots to the mashed potatoes so you have the fluffy mash and the crunchy bits of turnip and carrot and it's really good," Chuck shared.
Use your regular mashed potato recipe and substitute turnips or try this garlicy recipe.
Ready for something a little more ambitious? Chuck suggests adding a French tourtiere to your holiday menu. It may be a classic in French-Canadian homes, but meat pastries at Christmas dinner are probably pretty new territory for the rest of the country. Even better than the pie itself, though, is the tomato fruit jam that goes on top.
Get Chuck's tourtiere recipe here and a sweet and savoury tomato jam recipe here.
If you want to shake things up this holiday, why stop at the main course? Dessert is a great place to add in new favourites because no one would dare complain that there's too much dessert. One dish Chuck would strongly suggest you add to your holiday pie lineup is the maple pecan pie — it's like a regular pecan pie, but with maple syrup. Mmmm.
He assures us that if the pie turns out "leaky" you made it right.


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Category: Local News
NH Schools Closed On Monday
Posted on December 23, 2015 December 25, 2015 by Richard Hall
Police in Nashua, N.H. said that they have received threats regarding city's high schools. Keeping in mind the violent threats, schools were closed on Monday. Lately, on Sunday, the police kept investigating every possible chance of danger to the city. The Superintendent of Police said students are most probably not turning up to school till Tuesday. It is one of the largest districts of New Hampshire with around 11,500 students. Risking the life of many young children is not something feasible. Schools are closed, and children are assured full safety. As mentioned by the officials, the threat was very violent.
It promised to harm the local children as well as the staffs of the high school. The threat included many schools in the district, and not one. Many important details regarding the risks were clearly stated on the district website. Police remained close to every school in the district. Though the threat was imposed for children, many others had to suffer. The reason being obvious that a child's safety is the most important for everyone. Many working parents remained at home for the sake of their children. High ranking police officials recommended them not to leave their children as it will help them stay brave.
Both, federal and state agencies worked over the investigation. USA's second largest district Los Angeles also had to cancel classes on last Friday. Nearly 6,50,000 students stayed at home, as the officials were threatened through an email. A generic email threat was sent to not just one school in New York or elsewhere. Some people thought it to be a usual prank. However, New York City police commissioner denied it of being a student's prank. Officials were very cautious about the lives of school going children. It was soothing to see their support. However, people should stay strong when such threats are given for innocent children. Terrorism is something that should never stop education. It is very essential for every child. Make your child strong to resist such a situation as it is nothing for such a huge gift, "education".
Read also Mario Williams to Face Pay Cut Next Season?
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Dozens Of Pedestrians Hit In An Accident In Las Vegas
Posted on October 23, 2015 February 9, 2018 by Richard Hall
In an intentional act, a driver lurched into a crowded sidewalk. The driver did it several times on the Las Vegas strip. This act injured more than two dozen people and even killed one person. Officials said out of the surviving people seven are still in a critical condition. The accused driver was taken for several tests and is kept in custody. Though it is probably an intentional event, authorities have not declared it officially yet. It is still a doubt whether it was intentional or an accident. Thanks to the Police Department that they reached on time to the right place. Timely arrival of police cars and ambulances helped save the lives of many people.
When the incident took place, the scene turned really chaotic as people fled. Everyone came in a hassle situation, and they did not know where to go. A witness described as he was running in panic as he saw a car moving towards him. It looked as of the driver did not even wish to stop the car seeing many people in panic. It was a complete heartless incident. Whatever many be the reason, but such acts are never pardonable. People almost scrambled to control the driver, but it did not happen. It was a huge chaos, and that is all one can say to describe.
The entire incident was very disturbing. Men ran behind the car shouting to stop, but the driver did not listen to anyone. Near the Paris intersection, the car came to a stop, and people were punching its windows. Not wasting any time, the driver accelerated the car again mowing more people down. The traffic diverted people towards the Paris Las Vegas Hotel. The strip was shut down at this point. This incident grabbed the attention and anger of almost everyone. What is to happen with the accused is still awaited?
Read also Wrong Winner Announced By The Host Of Miss Universe Beauty Pageant
Wrong Winner Announced By The Host Of Miss Universe Beauty Pageant
Posted on September 23, 2015 December 25, 2015 by Richard Hall
Making a blunder on a Live TV show is something very awkward. The same kind of blunder happened at the Miss Universe Beauty Pageant 2015. Host Steve Harvey somehow mistakenly announced winner's name wrong. The event was going on in Las Vegas where the incident was witnessed. Harvey read out Miss Columbia's name as the winner, but actually she was the first runner-up. Originally, the winner to be declared was Miss Philippines. After the mistake was recognized, there were loud whispers among the audience. Harvey tried to take control of the mess. She said on stage "Folks let me take control of this".
She also showed the card with the name of the winner on camera. However, the event is getting a lot of criticism as why was the event allowed to continue after the announcement of the wrong winner. Nearly for two minutes, the ceremony continued wrongly. Miss Columbia was even crowned knowing that this was not to happen. After those 2 minutes, Harvey corrected and announced the right winner. Things are unclear, why things happened like that. You can watch video of stunned Miss Philippines after the correction was made. Everyone was like "what is this happening?"
The show aired in West Coast was unedited one. Production officials said the show will not be edited for airing it on international television where the show does not go live. It was very disappointing for Miss Columbia. A statement issued by the Miss Universe Organization said that the event was highly alluring, and a live event can have human errors. They apologized in their statement to the two beauty queens, their families and fans. It was highly controversial and is still continuing to be on the air. The Miss Universe organization was acquired by WME- IMG earlier this year. In the history, it was one of the very first Miss Universe events organized without Trump. On Twitter, Trump posted a comment which said "We need you Mr. Trump".
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iOS widget shows the message "Debes abrir My Data Manager después de cambiar la tarjeta SIM para mantener el widget actualizado" (message in spanish that translated means "You should open My Fata Manager after change the SIM card for maintaining the widget updating"), but doesn't show any data or graphic. The app is opened already but appears this message. I have tried to setup again the monthly plan, and I have even uninstalled and installed again the app, but it doesn't fix it.
Same problem for me with SIM swap. Not resolved by latest update today.
Same problem here ever updated to latest version.
If the widget doesn't work as a feature then just remove it :-). When it works then add it back to the app and release it.
update 5.3.6 , was not FIX the issue whit the widget , maybe is better to be removed , and when the problem is resolved for A to be added again. . Iphone 6+ , Ios 8.1 .
We have a new version of My Data Manager (v5.3.6) that should fix the widget issue for you.
If this still doesn't work, you may need to first uninstall the widget, then reboot your device, and finally reinstall the widget.
Also, please note that this issue was introduced due to an incompatibility with Apple's recent 8.3 iOS release affecting a small subset of users (the widget works fine on iOS 8.2 or earlier versions). We had a limited amount of time to test the widget running iOS 8.3, however we never encountered the bug prior to the widget going live. We are a very small team, so we really appreciate all your support on this community wiki as we work to fix issues as quickly as we can.
the problem affect all users . for example i use IoS 8.1 , my girlfriend 8.2 , we both have the problem whit the widget . I think u must admit the problem is more serious for what you try to say.
and don't fix this problem.
I've just uninstalled my data manager, rebooted my iPhone and reinstalled the app. It still gives the message on the widget in the notification centreabout needing to open the app before the widget will function due to SIM change.
Ok - we're seeing that several of you are still experiencing the widget issue on v5.3.6.
We are urgently trying to fix this issue and have started investigating a couple new hypotheses, however, we still haven't identified the specific set of circumstances leading to the failed SIM-card detection (which is responsible for that message appearing in the widget instead of one's actual data-usage).
We really appreciate your help and patience in regards to the widget still not functioning for many of you.
Just sent you the information requested by email. Hope it helps to figure it out somehow.
Besides the widget issue your app rocks. Please keep it up the good work.
It's not like the widget is core functionality. Without it the app is still plenty useful.
Thanks for a great app (and I will send in the info you requested as soon as possible).
I'm just sent you information requested by mail and wish the information can help to fix the issue soon. Thanks.
A big thank you to everyone that has written in - your information really helped us figure out how to reproduce the issue in-house (finally), and we are now are working towards a fix that we hope will work for virtually everyone that's been affected.
If all goes well, we'll have an updated version to submit to iTunes by end of day tomorrow (after that, the timing is going to depend on how quickly Apple reviews our app).
We'll keep you posted on our progress - thanks again for your help!
That's great! Thanks. Sorry I didn't email the information but I'm glad you have what you need. Look forward to update.
Just wanted to let you all know that we just submitted our latest fix for the widget issue to iTunes, so we're now waiting on Apple to review our submission and approve it to go live. Thanks again for you help!
You maybe droppped the ball on this widget, but you are surely quickly trying to pick it up again! Thanks for the work you are doing!
still no updates concerning the issue?
Wait .,.wait ....wait ..... !!!!!!!
Same here. Did all the above and removed the app and reinstalled. Widget still shows a SIM that was changed. I'll disable the widget for now. Thanks in any case for the attempts!
I'm running iOS 8.4 Beta 3 and just so the developers know. It doesn't work for me either and I have updated. Just thought you should know.
Mine works now and I didn't need to reboot or reinstall widget thanks so much.
Hello I updated the app 03/06 , but nothing I served , I restarted the phone, also to try to install the app again, still the same problem with the #widget not work, I have an iPhone 5c #ios 8.3 please alguma reply.
I had a new sim in my phone when the app updated and it worked so There must be a link with the specific sim that was changed previously.
A response would be nice Mobidia!
Sorry for the late reply - we've been working on another fix (submitted to iTunes already). We tried a different approach this time, and got a fixed version of our build out to a number of you that were experiencing the problem. We can confirm that this latest fix has worked for everyone that's used it so far, so we're hoping this will fix it for all of you as well.
This has been quite a challenging bug to pin-point and fix, so we really do appreciate your patience with us on this.
Just downloaded the app, but widget is not working, already tried the uninstall widget - reboot fix, but still doesn't work for me.
Data manager app doesn't work for me either.
How about that update?... It's been 2 weeks already..
Hard reset before even launching the updated app.
Then launch the app and then added the widget.
Works for me too and I'm on iOS 9. Thank you!!!!!
work and for me !!


Deion Sanders helped pioneer the modern idea of commercial endorsements back in the 90's. So it comes as no surprise that as part of his induction to the Hall of Fame last weekend, he thanked his corporate sponsor Under Armour in his Hall of Fame speech (to the dismay of some), wore a UA lapel and for his first official party as Hall of Famer, had not one, but six corporate sponsors: Grey Goose vodka, Sean John, Gigaré Collection, Kilwin's Chocolates, ASEA Water, and Signazon.com.
The result? A sponsorship and an adhesive-free flute design that included a profile of Sanders in uniform and his legendary number "21." All in all, this weekend proving Neon Deion may have retired, but he hasn't missed a beat at the sponsorship game. Check out his speech below.
Proceeds from Deion Sanders Hall of Fame party benefited Sanders' non-profit organization, The PrimeTime Association, a mentoring program for youth.


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Posted on June 21, 2007 by Kelp HQ
The Detroit Cobras Make Everything Clear
Music, Previews, Reviews
Alright, I know this is a little out of character for me lately, but I want to go way out on a limb here and recommend that you all leave your little huts next Wednesday, June 27th , and scamper down to the Wellfleet Beachcomber to see a new, tough little band called the Detroit Cobras. For any right-thinking American, leaving the house nowadays has to be regarded as a potentially hazardous proposition; yet I truly believe that these guys -once described as the Ronettes crossed with the Rolling Stones in Motown (!! -I mean, how cool would /that/ be?) – might be, uh, fun.
I use the term "guys" rather loosely here, as two of these guys, namely co-leaders Rachel Nagy on vocals and Mary Ramirez on guitar, clearly aren't. There's plenty of bad girl attitude to their work, but it's nicely understated (and, unlike Cape Cod's original, beloved bad girl group, the Extremes -and when will that reunion finally happen? – endearingly under-dressed, as the band uniform seems to lean heavily toward whatever-I-had-on-in-the-van wear.)
These folks play with a serious kick, loud and proud and relatively unadorned, and the music is a sort of rocked-up soul thing, but with a minimum of guitar effects and arena rock hooey. The guitars sound like guitars, careless and clangy, maybe not all that far from Joe Strummer-ville, and the drummer kicks. And the clear aim is to make you dance and go nuts, except for the occasional ballads, which were made for melting. Most of their songs sound like they were recorded in about half an hour, and there's a feeling of abandon to it that's just swell. Some of it's cave man stuff, but it's smart cave-man stuff (or, if you prefer, cave-person.)
One of the most distinctive and delightful things about the band is that they more or less steadfastly refuse to write their own songs, to which I can only say, thank you. The number of awful songwriters on tap lately has truly mushroomed out of control, and I'd like to thank the Detroit Cobras for at least trying to stem the tide. Plus, the material they pick is masterful, and obviously born of large record collections and dogged pursuit, alternating between obscure songs that you can't believe someone else knew about and obscure songs that you didn't know about. It's record collector's heaven: Ruby Johnson, the Cookies, the Blossoms, the Staple Singers, Sugar Pie DeSanto, Tammi Terrell, etc. (plus, the goils are really cute, and look like they might be on something.)
Even with all these other assets, though, the thing that really cinches the deal is singer Rachel Nagy, who had all sorts of interesting things to say -including "it's ironic that I'm a singer, 'cause it has always been my view that white women should not sing." From that point on, we got along famously.
Rachel's a great singer, not just a good one. She understands that the first focus of a singer shouldn't be showing off or belting or giving it all you got or putting in ten more doo-dads per syllable; it should be about singing the damn song, period, in a way that lets you know how the songwriter felt. For someone working a genre that sometimes leans toward histrionics, she has a restraint that's very intriguing.
More than anyone, Nagy sounds like Irma Thomas, most obviously on the Irma covers, "It's Raining", "Breakaway", and especially a just plain perfect version of "Cry On." There's a little of that bluesy feeling throughout, though: beat up, resigned, but sweet. She calls Irma "my godhead, my hero. I saw her live at her club in New Orleans [since washed away in Katrina, alas] and fell on the ground, screamed and cried. I could've washed her feet with my tears and hair." She got it.
Plus, she just turns out to be one of those rare people with pipes, an instantly satisfying and unadulterated sound so compelling that… well, I don't really buy that whole reading-the-phonebook analogy, but she's got pipes, and pipes is good. You don't have to show it off, everybody knows from note one.
Ms. Nagy seemed to share my feelings on the tragic lack of smoke in bars nowadays ("I mean, it's New York in the winter and you're supposed to stand outside and smoke -that's retarded!"), coddling of modern chillens through excessive monitoring of various health issues ("They're gonna turn 25 and a butterfly's gonna land on their head and they're gonna have an aneurysm") (and how'd we ever get going on that, anyway?), hatred of the singing of Janis Joplin (otherwise Janis seemed ok), and of people who get all bent out out of shape when they lose children (especially people who have multiples to begin with. After all, as Rachel pointed out, "you can always have more kids.")
At one point, she did imply that the band was born joyfully out of the record collection of its founder, since-departed guitarist Steve Shaw. She said she never meant to be a singer, and that she was plied with drinks early on, when the band played for the best reason, which is wanting to make those noises enough. "We sort of pulled a fast one -we said we were gonna break up, and then as soon as [Steve] left, Mary and I giggled and started up again. The only reason we keep boys around is as archivists; the girls job is to dance and sing."
I like a girl with her feet on the ground.
The Cobras' five records have their ups and downs, but the ups are pretty wonderful; their best album might still be 2001's "Life, Love, and Leaving", though their latest, "Tied and True", gives it a run for its money; the one Cobras' original, the fabulous, hilarious, "Hot Dog", is on 2005's "Baby". They're a good downloading project, as you might not need whole albums, but there's songs here that will rock your world.
And they've never been here before, and who knows when they'll pass this way again? Those who must rock should be in attendance.
This entry was tagged Detroit Cobras, Extremes, Irma Thomas, Mary Ramirez, Rachel Nagy. Bookmark the permalink.
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I am developing a project with various tables so I can have better understanding of model relation by communicating with the experts. Can anyone look at my code and see if I have designed my models based on the following conditions or not (only the relation in the model)?
Manufacturer can create Agency and Agents. Agency can create Agents. Agents can create their own account and Agency or Manufacturers can search and assign that agent under their ownership.
Browse other questions tagged python django or ask your own question.


1. Egypt was likely the first civilization to tax beer. Queen Cleopatra imposed a tax on beer in order, she claimed, to discourage public drunkenness, though it is believed that the tax was used to raise money to fund a war with Rome.
2. Beer is the most popular alcoholic beverage in the United States. According to a 2017 Gallup poll, 40% of Americans who drink alcohol say they prefer beer, down a few percentage points from last year. In 2016, the federal government collected $3.6 billion in excise taxes on domestic and imported beer alone.
3. In 1695, Great Britain raised taxes on beer, making gin the cheapest beverage in England. Gin was taxed at 2d (about 2 pennies) per gallon, while beer was taxed at 4 shillings 9d (about 57 pennies) per gallon. The difference in price is considered the root of a serious drinking problem in the country in the 18th century, especially among the poor.
4. In the United States, taxes on the production, distribution, and sale typically eat up 40% of the retail price of beer. That amount includes all taxes imposed on beer. In previous years, the federal excise tax was about 5 cents per drink (the nickel comes from the assumption that the average beer has an alcohol content of 4.5%).
5. If you thought tax reform only lowered income taxes, you'd be mistaken. The new law reduced the federal excise tax on beer according to output. Those rates were reduced to $3.50 per barrel on the first 60,000 barrels for domestic brewers producing fewer than 2 million barrels annually and $16 per barrel on the first 6 million barrels for all other brewers and all beer importers; the excise tax remains at $18 per barrel rate for those producing over 6 million. If those rates sound familiar, they closely mirror previously proposed legislation, including the BEER Act of 2013. There is one downside: Under tax reform, most corporate changes under tax reform are permanent and most individual changes are effective through 2025, the changes affecting the beer market will expire in 2020.
6. German beers are often labeled "Gebraut nach dem Bayerischen Reinheitsgebot von 1516" which translates roughly to "brewed according to the Bavarian Purity Law of 1516." The law originally limited the ingredients which can be used to make beer in Germany (barley malt, hops, yeast, and water) and allowed the government to tax beer. The Reinheitsgebot became an official part of the German tax code in 1919 but was largely gutted when Germany became part of the European Union.
7. To help pay for the Civil War, Congress imposed an excise tax on beer. The Revenue Act of 1862, signed into law by President Lincoln, included a tax on "all beer, lager beer, ale, porter, and other similar fermented liquors, by whatever name such liquors may be called." It may not be popular but taxing beer wasn't a bad idea from an economic standpoint, as it generates billions in revenue each year.
8. Arthur Guinness II – the father of Guinness stout – altered the family beer recipe to include unmalted roasted barley instead of black malt. The unmalted barley wasn't subject to extra taxes (more on those here) which made it affordable for the Guinness family – it also made the beer's taste distinctive. By the end of the 19th century, Guinness was the largest brewery in Europe.
9. According to the Beer Institute, last year, directly and indirectly, the beer industry employed nearly 2.23 million Americans, providing more than $103 billion in wages and benefits. The industry pays nearly $63 billion in business, personal and consumption taxes.
10. In 1991, President George H.W. Bush signed a bill which raised taxes on luxuries such as furs, yachts, private jets, jewelry and expensive cars (despite the "no new taxes" pledge) – that same bill nearly doubled the tax on beer. Bush called for the repeal of the tax just two years later, and while most of the taxes included in the bill were eventually repealed, the tax on beer remained in place and is still there today.
11. The most expensive state to buy a beer may be Tennessee where state excise taxes reach a whopping $1.29 per gallon, plus sales tax, making it the highest in the country. The cheapest state to buy a beer? Wyoming, where the excise tax is just $.02 per gallon.
13. According to Infogroup, with 1.54 beer-related businesses per capita, Bend, Oregon is the most favorable U.S. city for beer lovers. Oregon boasts two cities on the list. It probably helps that Oregon has no state sales tax to boost prices.
14. Sales of craft beer increased 8%, up to $26.0 billion, according to the Brewers Association, and now account for more than 23% of the beer market. California leads the way, boasting 2.2 breweries per 100,000 adults over the age of 21. You can see how your state ranks here.
15. In addition to sales of beer, "beer tourism" is a real thing. Sites like BrewTrail.com help consumers plan road trips and vacations around visits to breweries, which bring additional travel tax dollars. States and regions have gotten into the spirit, offering info on their own "ale trail" recommendations. One example? Just last week, my husband and I left our stamp on my hometown economy via the Wilmington Ale Trail (one of our favorites was Flytrap Brewing).
16. Cenosillicaphobia is the fear of an empty beer glass. Okay, that's not a tax fact, of course, just a fact. Don't live in fear: go, get a beer.


Make To Escape – Mansion Map is a puzzle map created by CrazyCowMM. This map is a dive into custom crafting recipes. You are stuck in a mansion and your mission is to solve puzzles by crafting items. Some crafting recipes will be given to you via the recipe book, while others will not. There is no in-game hint system however a walkthrough is provided. Default crafting recipes have been disabled, so don't worry about crafting the wrong thing. Download the map from the link below and enjoy.


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Q: line break in excel datatable How can I define a line break in a datatable excel when it is exploited?
https://jsfiddle.net/pa9q14mb/6/
At the time of exporting the excel the box shows it
-TEXT TEXT TEXT -TEXT TEXT TEXT -TEXT TEXT TEXT -TEXT TEXT TEXT

But I would like it this way in the excel
   -TEXT TEXT TEXT
   -TEXT TEXT TEXT
   -TEXT TEXT TEXT
   -TEXT TEXT TEXT

I would like to know if there is the possibility of doing that and I was  looking at some examples with the exportOptions function, but without results. regards

A: Where you're already setting your export options, you need to do something like this:
return column === 5 ?
       data.replace(/
/g,String.fromCharCode(10)) :
       data;

 appears in your DataTables cell, so you need to convert that to a carriage return (ascii code 10), which is what Excel expects for a new line.

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text:
Pete Rose, 1985
Screenprint on Lenox Museum Board
Pete Rose 360B was created by Andy Warhol in 1985, the year that Pete Rose broke the all-time hits record. Commissioned by the Cincinnati Art Museum, Warhol created an 4...
Pete Rose 360B was created by Andy Warhol in 1985, the year that Pete Rose broke the all-time hits record. Commissioned by the Cincinnati Art Museum, Warhol created an 4 image, acrylic on canvas screen print that was based off a photo by Cincinnati photographer, Gordon Baer. True to Pop Art style, the screen print mimics the style of a classic baseball card. Warhol produced 50 impressions of color screen prints of one of the four images. Andy used colors that exemplified who Pete was: a colorful character. Warhol died two years after this commission and Pete Rose was banned from baseball for placing illegal bets.
One of his late-period celebrity portraits, Warhol used acrylic paint on a canvas with four silk-screen images of Rose in a right-handed batting stance. Each image has a different color and the work is notable for being in the style of a baseball card. "It was so brilliant of Andy to make it into a baseball card," said Carl Soloway, the Cincinnati Art Dealer who approached Warhol about creating the portrait. "And that's so interesting because baseball cards are collectible and negotiable. So it was a statement about the commercialization of art, just like his soup cans are about the commercialization of branding."


Oxidative phosphorylation (OXPHOS) and glycolysis are two cellular metabolic pathways that play a crucial role in the functions of biological systems. Currently, magnetic resonance spectroscopy (MRS) (13C, 31P, and 1H) and positron emission tomography (PET) methods are used to investigate changes in these pathways that result from metabolic dysfunction. However, MRS methods are limited by low resolution and long acquisition times. While 18F-fluoro-2-deoxy-D-glucose (18F-FDG) PET is a widely used clinical modality, it requires the use of radioactive ligands. Thus, there is an unmet need for techniques to image these metabolic processes noninvasively, and with higher resolution in vivo. In this dissertation, we exploited the chemical exchange saturation (CEST) phenomenon to develop and optimize endogenous CEST magnetic resonance imaging (MRI) methods to measure OXPHOS and glycolysis, and demonstrated application of those techniques to study impaired metabolism in vivo. These CEST methods offer several orders of magnitude higher sensitivity compared to traditional spectroscopic techniques. Recently developed CEST imaging of free creatine (CrCEST) was targeted as a means of measuring OXPHOS. We optimized and validated this technique in healthy human skeletal muscle, showing that CrCEST imaging in dynamic exercise studies provides a measure of the mitochondrial rate of OXPHOS. CrCEST imaging was then implemented in a cohort of subjects affected by genetic disorders of the mitochondria. The results of these studies demonstrate that CrCEST has the capability to distinguish between healthy and impaired OXPHOS in muscle. In some diseases with altered metabolism, like cancer, aerobic glycolysis dominates, leading to increased lactate production. Existing methods for imaging lactate in vivo involve expensive, radiolabeled tracers. In this work, we demonstrated the feasibility of imaging lactate with CEST ("LATEST") in phantoms with physiological concentrations. Then, we validated the method dynamically in vivo by measuring lactate production and clearance in intensely exercised human skeletal muscle, which utilizes anaerobic glycolysis. Finally, we infused rats bearing lymphoma tumors with non-labeled pyruvate and demonstrated the ability of LATEST MRI to image tumors and measure dynamic lactate changes over time. Together, these studies demonstrate that metabolic processes can be monitored in vivo using CEST MRI, with potential for widespread clinical applications.
Debrosse, Catherine, "Chemical Exchange Saturation Transfer Imaging Of Endogenous Metabolites For Monitoring Oxidative Phosphorylation And Glycolysis In Vivo" (2017). Publicly Accessible Penn Dissertations. 2251.


THE WTACH is the brainchild of its founder and CEO, Chris Flynn, a former director for the Pacific region at the Pacific Asia Travel Association, a role he held for 15 years.
"WTACH works with destinations to provide development strategies and policy framework recommendations to avoid the kind of tourism meltdown we are seeing at Angkor Wat, Phi Phi Island and Mt Everest," says Flynn.
WTACH's position is that tourism needs to respect host communities and their cultural and heritage assets by adhering to a framework that has the host community at its heart.
"It's time for the tourism industry to take step back and look at the long term impact of its decision making," says Flynn.
Social media and mobile devices aren't helping. Carolyn Childs, CEO of MyTravelResearch.com, and a member of the WTACH advisory specialising in analysing data and trends, says it is no coincidence that WTACH is being born at a time when 'selfie' culture and the promotion of 'Instagramable' travel is sweeping the world.
"A unique image can 'create' a destination in moments - often leaving it unprepared or wrong-footed," says Childs. "This is particularly true if the image runs counter to cultural values. It risks tourism losing its 'social licence' with host communities. Ironically, these 'instadestinations' risk destroying the very thing travellers are seeking," she warns.
The desire for 'authenticity' in travel is also problematic. Childs cites an AirBnB survey which found that over 80% of millennial travellers (and 93% of Chinese millennials) seek a "unique" experience and want to "live like locals" while on holiday.
"The pressure on destinations and tour operators to find and monetise 'unique' and 'authentic' experiences will only increase as both millennial and mature travellers work through their 'been there done that' bucket lists," she says.
"Having the right frameworks in place help communities and tourists. They build a more sustainable destination that delivers truly rewarding experiences," she says.
The new association is deeply concerned that Turkey, for example, has decided to expand tourism arrivals from 40 million in 2018 to 70 million by 2023 - less than four years away.
"What interpretive and cultural safeguards have been put in place?" asks Flynn. "Have local communities been consulted? Is there an actual plan that involves a holistic government approach and key stakeholder and community engagement?"
Says Flynn: "At WTACH we know there's a better way. We are now seeking like-minded organisations and individuals to help us advance responsible tourism in culturally sensitive host communities."
The World Tourism Association for Culture and Heritage (WTACH) was formed to establish clear goals, objectives and strategies for the protection of cultural heritage through responsible and sustainable tourism practices. Working in collaboration with public, private and specialist academic sector organisations, WTACH determines best practice ethical principles and standards in line with the most robust global research available. A community of like-minded organisations and individuals, WTACH seeks to ensure that the unique attributes, history and cultural values of the world's communities are celebrated and preserved for generations to come.
The WTACH logo: We wanted to create a symbol that represents humanity. An image that characterised unity and togetherness. Although it may look like an ethnic design, the logo is in fact the molecular structure of a human DNA molecule. The colours used are also symbolic as they represent the first known pigments ever used by mankind. Earth pigments born out of rock and soil and used in the earliest known rock paintings. The high level of iron and limonite creating ochres that produced rich umber, sienna and yellows that were the pallet of humanity's first artists.


The following is a comprehensive overview of contracts.
Permanent members in the ETB sector are employed under memo V7.
Teachers on fixed term pro rata contracts are employed for the full year, meaning they are paid a regular salary until August 31st. They are also paid during holidays.
If you are a fully qualified part-time teacher (i.e. with fewer than 22 teaching hours per week) and have been engaged to teach for a specified number of hours per week over the course of the full school year with a start date not later than the first day of term following the mid-term break) you are entitled to a pro-rata contract.
A fixed term contract is a contract of employment for a specified period – usually 1st September to 31st August – and must be signed by both the employer and the employee.
You are entitled to a fixed term contract in written form.
The standard contract is appended to Circular Letters PPT 19/03 (Community and Comprehensive schools) and PPT 20/03 (VECs).
To be entitled to a pro-rata contract you must have been appointed following a formal selection process. This process involves advertisement of the position in the national press and competitive interview by a properly constituted interview board.
If you otherwise satisfy the conditions for award of a pro-rata contract (i.e. if you are qualified and have specified hours over the course of the school year) but have not been appointed following a formal selection process, you should contact your TUI Workplace Representative and, assisted by her/him, ask your employer to advertise the position and make an appointment following a formal selection process. There is a risk in this that you might not be appointed but the cost to you of not pursuing the matter is considerable.
If you have a fixed term pro-rata contract your salary for the school year is based on your weekly hours as a proportion of the weekly wholetime hours (22). To calculate your annual salary, you apply that proportion to the sum of your point of the CBS plus your qualifications allowances.
Full salary details are available on the TUI website or in the relevant section of the TUI diary/handbook.
A fully qualified part-time teacher who is employed to teach for a period in excess of 150 hours during the school year but for less than the course of a full school year is classified as a "non-casual part-time teacher" and is entitled to a non-casual part-time teaching contract. Also, a fully qualified part-time teacher who is employed to teach for a period of at least 22 successive weeks during the school year but for less than the course of a full school year is also classified as a non-casual part time teacher, even if the sum of the hours to be worked in that period does not exceed 150.
As such a teacher, you will be paid at an hourly rate calculated by dividing the sum of your point on the CBS and your qualifications allowances by 735 (i.e. the number of teaching hours in the school year).
If you are employed to replace a teacher on maternity leave or extended sick leave you may fall within this category.
If it is anticipated that a teacher will work no more than 150 hours and/or for fewer than 22 weeks in the course of the school year, the hours worked, to a maximum of 150, will be paid at a fixed hourly rate.
If, as such a teacher, you work for more than 150 hours in the school year each additional hour in excess of 150 will be paid at an hourly rate calculated by dividing the sum of your point on the CBS and your qualifications allowances by 735 (i.e. the number of teaching hours in the school year).
This type of contract applies in cases in which the work arises only on an intermittent basis and where the employer may reasonably expect that the hours worked will not exceed 150 in the course of the school year. (Please note that if you are employed to work specified hours for 22 weeks or more you are entitled to a non-casual part-time contract as opposed to a casual part time contract).
Following vigorous campaigning by TUI and the other teacher unions, 'expert groups' were established as part of the Haddington Road Agreement to investigate and tackle the crisis of casualisation in teaching and lecturing. At second level, the expert group reported in September 2014 – the Ward report – and details on the implementation of its seven key recommendations are set out in circular letter 24/2015, published in March 2015.
This circular letter applies to all TUI members contracted as teachers in settings including post-primary schools, colleges of further education, Youthreach and adult education settings.
A teacher will qualify for a Contract of Indefinite Duration (CID) after a period of continuous employment in excess of two years – ie upon commencement of a third year of continuous employment, subject to certain conditions. This is significantly better than the statutory provision and improves the initial position negotiated under the HRA. (Under fixed-term worker legislation the qualifying period is four years and under the HRA it had been 3 years for teachers).
Hours held by a fixed-term teacher who is covering for a teacher on career break or for a teacher on secondment will now be counted towards their CID. The teacher will be placed on a redeployment panel immediately prior to the return of the teacher on career break / secondment. In such circumstances the teacher will remain in receipt of his/her income/salary.
The hours of the CID will be the hours worked in the full school year prior to the issuing of the CID (i.e. the qualifying year), regardless of the source of the hours.
A teacher who has qualified for a CID after two years will be liable for redeployment if their subjects become surplus to the curricular requirements of their school.
Where a qualified teacher holds a part-time CID and is working additional hours under a separate fixed-term contract, they will receive a CID for these hours after a continuous period of employment (in the hours) in excess of one year, if the hours continue to be viable and available under the allocation.
All teachers on their first fixed-term contract will have their positions terminated at the end of year 1. The position will automatically be re-advertised and a new recruitment process undertaken. Therefore, the teacher will need to apply for and interview for the position at the end of year 1.
Entitlement to a CID - for all or some of the hours worked - does not accrue in circumstances where there are legitimate 'objective grounds' for the fixed term contract. Under circular letter 24/15, covering for a teacher on career break or secondment is no longer considered to be an objective ground for not awarding a CID.
If a teacher is of the view that s/he has an entitlement to a CID, in view of the above information, and has been refused a CID by management, there is right of appeal to the Adjudicator under the terms of CL 30/2018 . There is also a right of appeal to the Adjudicator if you have received a lesser number of hours in your CID than the number of hours held in the third year. There is a four week period from the date of refusal/receipt of the CID to appeal to the Adjudicator. All appeals should be processed through TUI Head Office. There is also an option of having refusal of a CID dealt with by a Rights Commissioner who will consider whether the legal entitlement to a CID under the terms of the Protection of Employees (Fixed Term) Act 2003 has been denied. There is a six month time frame for lodging such complaints. These complaints should be processed through TUI Head Office. It should be noted that under the Act entitlement to a CID arises on the fourth anniversary of appointment.


Here you can find all info about William's Wonder from Super Sativa Seed Club. If you are searching for information about William's Wonder from Super Sativa Seed Club, check out our Basic Infos, Lineage / Genealogy, Hybrids / Crossbreeds or User Comments for this cannabis variety here at this page and follow the links to get even more information - or list all William's Wonder Strains (±2) to find a different version. If you have any personal experiences with growing or consuming this cannabis variety, please use the upload links to add them to the database!
William's Wonder is an indica variety from S.S.S.C. and can be cultivated indoors. S.S.S.C.s William's Wonder is/was never available as feminized seeds.
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We found 25 direct descendants from S.S.S.C.s William's Wonder in the seedfinder strain database, here a short overview. To see all hybrids and their descendants, visit our William's Wonder Genealogy Page and check out all the direct crosses as like as the following generations.
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