If you've ever taken a long camping trip or found your home without electricity during wild weather, you'll understand just how terrifying it can be to live without power. We rely on our phones, laptops, lighting, medical equipment, and refrigeration to keep us safe and healthy, and when they're no longer available it can cause quite a few problems.
The Yeti 1250 Solar Generator is the solution to all of these problems, with a huge range of added extras that just can't be found with traditional generators. Whether you need to run a fridge or medical supplies, this renewable source of energy can get you through without the gases or noise of standard power supplies.
This truly unique generator can offer you enough energy to keep your campsite or home running, by using free energy straight from the sun. Without relying on dangerous gases and non-renewable power sources to work and a whisper quiet performance you can ensure your safety and comfort with the Yeti 1250 Solar Generator.
1250 WH for ultimate power - This device will allow you to power tools, CPAP, display monitors, refrigerators, and more so that you don't need to live without power when you need it most.
Three power source options - With the ability to charge your Yeti 1250 three different ways, you'll never have to live without portable power again. Recharge from the sun's energy, plug into a power outlet or use your car's battery to recharge.
Cordless and portable - Comes with a rolling cart so you can move it around with ease and with no cords, this product is safe and comfortable to use anywhere.
Priced at about $1,200 this product costs around the same as standard generator but with so many more benefits that make it worthwhile. If you're used to hiring a generator for camping trips or relying on an emergency power source, this is a great investment to make.
Although advertised for days of use, this product could really get you around 12 hours of energy for larger appliances such as a refrigerator. However, when used smartly this should be all you need for emergency situations.
Priced far more affordable than standard backup generators, it makes sense to invest in the Yeti 1250 if you usually rely on backup power. While not in use, you can power it with energy from the sun so running costs are virtually free, other than the additional purchases for solar panels.
This product should only be used to power emergency appliances or provide your home with enough energy to keep smartphones and laptops charging or lighting around the home. If you try to power up too many large appliances you'll need to constantly recharge the battery.
For a smart and comfortable way to provide your home with emergency power, the Yeti 1250 Solar Generator is the right product for you. This amazing generator signals the future of backup generators in an environmentally and people friendly device.


text:
\section{Intoduction}
In three dimensional classical Einstein gravity, every two solutions are locally isometric and there is no propagating degrees of freedom. In 3d gravity with negative cosmological constant (AdS gravity), the existence of the BTZ black hole \cite{Banados:1992wn, Banados:1992gq} makes the theory more interesting to consider this theory as a toy model to understand the higher dimensional gravity \cite{Deser:1983tn}-\cite{Grumiller:2019ygj}.\\

 Three dimensional AdS gravity has asymptotic Virasoro symmetry. During the quantization of this theory, Brown-Henneaux showed that the theory has left and right moving Virasoro algebras which are  part of the structure of the conformal field theory \cite{Brown:1986nw}. The corresponding central charge is $c=\frac{3l}{2g}$ (where $l$ is the length of the AdS space).  This shows the existence of the boundary conformal field theory. This duality is an example of the AdS/CFT correspondence, which  is a correspondence between a bulk gravity, and boundary CFT in the lower dimension \cite{Maldacena:1997re}-\cite{Witten:1998qj}.

 Using the AdS/CFT dictionary,  one can obtain useful informations about the bulk theory by studying the boundary CFT. Solvability of the 3d gravity and the AdS/CFT correspondence, makes this theory more powerful to reveal some fundamental aspects of the quantum gravity\cite{Afkhami-Jeddi:2019zci}-\cite{Harvey:2020jvu}.

We  investigate the minimal theory of 3d gravity in this paper. It is believed that the 3d pure gravity is dual to the holomorphically factorizable extremal 2d CFT on the boundary, where there are some constraints on the central charges of the CFT \cite{witten}.  These constraints on the dual CFT come from the equivalence between 3d Einstein gravity and Chern-Simons gauge theory \cite{achucarro1986, witten1988cs}.

In three dimension, the Einstein-Hilbert action with negative cosmological constant can be expressed in terms of the gauge invariant action as follows:
\begin{align}\label{E5}
 I=\frac{k}{4\pi}\int  {\rm Tr}^*\,(A\wedge dA+\frac{2}{3}A \wedge A \wedge A),
 \end{align}
where the gauge field $A$ is built from $SO(2,1)$ gauge field $w$(spin connection), and verbin $e$. The gauge group of this action is $SO(2,2)$. The gauge group $SO(2,2)$ is locally equivalent to $SO(2,1)\times SO(2,1)$. Therefore, in the oriented space-time the action can be written as:
\begin{align}\label{E9}
I&=\frac{k_L}{4 \pi}\int {\rm Tr}\, (A_L \wedge dA_L + \frac{2}{3}A_L \wedge A_L \wedge A_L) - \frac{k_R}{4 \pi} \int {\rm Tr}\,(A_R\wedge dA_R + \frac{2}{3} A_R\wedge A_R \wedge A_R)\\\nonumber
&=k_L I_L + k_RI_R.
\end{align}
where $k_L$ and $k_R$, are  Chern-Simons couplings.  The allowed values of the  $k_L$ and $k_R$, can be calculated from the quantization of the Chern-Simons couplings. Since the fundamental group of the gauge group $SO(2,1)\times SO(2,1)$ is $U(1)\times U(1)$, every diagonal covering groups of  this group can be considered as the gauge group of the Chern-Simons theory. For an n-fold diagonal  cover of $SO(2,1)\times SO(2,1)$, the quantization condition of the $k_L$ and $k_R$ is obtained as follows
\begin{equation}\label{klquanta}
k_L \in \bigg \{ \begin{array}{ll}
n^{-1} \mathbb{Z} ~~~~~~n \in 2\mathbb{Z}+1\\
(2n)^{-1}\mathbb{Z}~~~~~~n \in 2 \mathbb{Z}
\end{array}
\end{equation}
\begin{equation}\label{kdifquanta}
k_L-k_R \in \mathbb{Z}
\end{equation}
Using the AdS/CFT dictionary, equation \eqref{E9} shows  the partition function of the dual CFT should be holomorphically factorizable:
\begin{equation}
Z(\tau,\bar{\tau})=Z(\tau)\bar{Z}(\bar{\tau})
\end{equation}
Where $Z(\tau)$ and $\bar{Z}(\bar{\tau})$ are chiral and anti chiral characters of the partition function which we called them as chiral and anti chiral functions respectively. For $c_L$ ( $c_R$ )integer multiple of 24, the chiral function  (anti-chiral function)  itself is a modular invariant partition function, and it can be  considered a corresponding purely chiral conformal field theory. Otherwise chiral function  is just the characters of the partition function \cite{flm1}.

The quantization condition \eqref{klquanta} leads to the following quantization condition on the left and right central charges:
\begin{equation}
(c_L,c_R)=(24k_L, 24k_R)
\end{equation}
 Quantization condition \eqref{kdifquanta} is equivalent to the $T$ invariant constraint  of the partition function of the dual CFT \cite{witten}.

  The purity of the 3d gravity  for the dual CFT  means that the primary fields of low dimensions should come from the identity. Therefore, the scaling dimension of the lowest primary fields excluding identity should be $k+1$. This class of CFTs is called extremal CFT \cite{hon1,hon2}. The vacuum state of the extremal CFT corresponds to the AdS space and the other primary fields  correspond to the BTZ black hole. \\
  Now, we are ready to solve the pure gravity. In order to achieve this goal, we need to find the dual CFT. First step toward obtaining the partition function of the extremal CFT is  determining the gauge group of the Chern-Simons gravity.  As   \eqref{klquanta} shows, the simplest gauge group is $SO(2,1)\times SO(2,1)$. For this gauge group, $k_L$ and $k_R$ are integer numbers and the dual extremal CFT satisfies all of the constraints. The left and right central charges are integer multiple of 24 and the chiral and anti-chiral  functions are modular invariant.

Modular invariance  is a powerful tool that reveals interesting aspects of the conformal field theory \cite{heller1}-\cite{Cheng:2020srs}. Modularity determines the partition function precisely. Every modular function can be expressed in terms of the polynomial  of the Klein J function:
\begin{align}\label{wit4}
Z(q)=\sum_{r=0}^{k}f_rJ^r.
\end{align}
 For $c_L=24,48$, and $72,$  the partition function of the extremal CFT is calculated as follows:
\begin{align}\label{wit5}
Z_1(q)&=J(q)=q^{-1}+196884q+\cdots\\\nonumber
Z_2(q)&=J(q)^2-393767\\\nonumber
&=q^{-2}+1+42987520 q+\cdots\\\nonumber
Z_3(q)&=J(q)^3-590651J(q)-64481279\\\nonumber
&=q^{-3}+q^{-1}+1+2593096794 q+\cdots.
\end{align}
The closed-form of the extremal partition function is derived using the Hecke operators
 \begin{align}\label{He2}
T_nf(\tau)=\sum_{d|n}\sum_{b=0}^{d-1}f\left(\frac{n\tau +bd}{d^2}\right).
\end{align}
The Hecke operators map the modular functions to the modular functions and for $f(\tau)=q^{-1}+\mathcal{O}(q)$;
 \begin{align}\label{He2}
T_nf(\tau)=q^{-n}+\mathcal{O}(q).
\end{align}
Therefore, the partition function of external CFT is obtained as follows \cite{appostal}\footnote{One can also write the closed-form of the partition function in terms of the unique modular function $J_m(\tau)$ which only has an order-m pole at $q=0$ \cite{McGady:2018rmo}:
\begin{align}
Z_k(\tau) := J_k(\tau) + \sum_{m = 0}^{k-1}  \{ p(k-m) - p(k-m-1) \}J_m(\tau),
\end{align}
where $p(m)$ is the partition number.}:

\begin{align}\label{wit6}
Z_k(\tau)=\sum_{r=0	}^{k}a_{-r}T_rJ(\tau).
\end{align}
Where,  $a_{-r}$ are the coefficients of the low states of the vacuum. For $c_L=24$, it is believe that  there exist $71$ holomporphic CFT \cite{flm1}.  70 of these holomorphic CFTs have  Kac-Moody or current algebra symmetry. Therefore, they can not be a candidate for pure gravity. The partition function $Z_1$ is one of the 71 holomorphic CFTs with no Kac-Moody symmetry. This model first was constructed by Frenkel, Lepowsky, and Meurman and its uniqueness was conjectured  \cite{ flm2}. For $c>24$ it is not know whether such CFTs exist or not. Although  the existence of extremal CFT were investigated for large value of $k$, but its existence  is still an open question \cite{gaber1, Gaberdiel:2008pr}. If such CFT's exist, they are good candidates for pure gravity.

In \cite{Maloney:2007ud, Keller:2014xba}, the authors attempted to compute the partition function of the pure gravity form different point of view by summing over the known saddles in the Euclidean gravitational path integral. In \cite{Alday:2019vdr}, using the Rademacher expansiosn  the partition function of the pure gravity have been calculated. The resulting partition function is an interesting modification of the partition function in \cite{Maloney:2007ud, Keller:2014xba}. Their results showed some unphysical features  (e.g., the negativity of the density of states at special values of the primary fields). Despite the attempts to solve these issues the  partition function of the pure gravity is still unknown \cite{Benjamin:2019stq}-\cite{Maxfield:2020ale}.

In \cite{witten}, Witten considered the $SO(2,1)\times SO(2,1)$ gauge group, now the question arise: ``Does any covering group of the gauge group $SO(2,1)\times SO(2,1)$ exists which satisfies all constraints on the dual CFT?''

In order to address this question, we  study the holomorphically  factorizable CFT  in this paper. We show that modular invariance  of    the holomorphically factorizable CFT  is  necessary and sufficient condition for deriving the allowed values of the covering group. In section \ref{s2}, we  study the modularity of the holomorphically factorizable partition functions and we  calculate the allowed covering group. From modular invariance of the holomorphically factorizable partition function we conclude that  the chiral  and the anti-chiral  functions are modular covariant. We explicitly detail the list of holomorphic and anti-holomorphic functions that serve as candidates for chiral and anti-chiral partition functions and note that modular covariance is only consistent with such functions when the left (resp. right) central charge is an integer multiple of 8, $c\in 8\mathbb{N}$.  Since the chiral and the anti-chiral  functions are not modular invariant, we can not use the Hecke operators. In section \ref{sec3}, we  introduce the generalized Hecke operators which map modular covariant functions into the modular covariant functions. We also investigate its Fourier expansion. In section \ref{sec4}, we  find related constraints on the symmetry group of the corresponding topological, Chern-Simons, theory in the bulk of AdS. We show that the symmetry group of the theory can be one of two choices: either $SO(2; 1) \times SO(2; 1)$ or its three-fold diagonal cover. In this section, we study the dual CFT for the case where the gauge group is three-fold diagonal cover of the group $SO(2,1)\times SO(2,1)$. we obtain conjectural partition functions for extremal CFT2s, and the corresponding microcanonical entropies, when the chiral central charges are multiples of eight.
 \section{Holomorphically Factorizable Partition Function}\label{s2}
 \subsection{Partition Function}\label{sub21}
 The partition function of the unitary 2d CFT in the upper half plane $\tau=\tau_r+i\tau_i$ ($\bar{\tau}=\tau_r-i\tau_i$), is defined as
  \begin{equation}
    \label{z1}
    Z(\tau,\bar{\tau})=q^{\frac{-c_L}{24}}{\bar q}^{\frac{-c_R}{24}}\sum_{h,\bar{h}=0}\rho(h,\bar{h})\,q^{h}{\bar q}^{\bar{h}},
    \end{equation}
  where, $c_L$ and $c_R$ are the left and right central charges and $\rho(h,\bar{h})$ is the density of the state.

The holomorphically factorizable partition functions can be written as the multiplications of the chiral and anti-chiral  functions as follows:
\begin{equation}
Z(\t,\bar{\tau})=Z(\t)\bar{Z}(\bar{\tau}),
\end{equation}
where the chiral function $Z(\tau)$,  and the anti-chiral  function $\bar{Z}(\bar{\tau})$ are defined  as follows
\begin{equation}\label{HF2}
Z(\t)=\sum_{h=0}\rho(h)e^{2\pi i\t(h-\frac{c_L}{24})},
\end{equation}
\begin{equation}\label{HF3}
\bar{Z}(\bar{\tau})=\sum_{\bar{h}=0}\rho(\bar{h})e^{-2\pi i\bar{\tau}(\bar{h}-\frac{c_R}{24})}.
\end{equation}
\subsection{Modularity of Partition Function}\label{sub22}
The modular covariance of  the chiral and anti-chiral  functions is the necessary and sufficient condition for the  modular invariance of the partition function $Z(\t ,\bar{\tau})$. Modular covariant  means that the chiral and anti-chiral  functions take  an overall phase under  the $S$ and $T$ transformations.

Modular covariance of $Z(\t)$ under the $S$ transformation demands:
\begin{equation}\label{HF4}
Z\left(\frac{-1}{\t}\right)=e^{i\beta}Z(\t).
\end{equation}
The identity $S^2=1,$ shows  the phase $\beta$ should be $\pi$ or $2\pi$.

For  $\beta=\pi$, we called the corresponding  function $Z^{-}(\t)$. The Fourier expansion of $Z^{-}(\t)$ under $S$ transformation at the self dual point $\t=i$ yields:
\begin{equation}\label{HF5}
SZ^{-}(\t)\big\vert_{\t=i}=\sum_{h=0} \rho(h) e^{-2\pi(h-\frac{c_L}{24})}.
\end{equation}
Using \eqref{HF4} for $\beta=\pi$, and  \eqref{HF5} show that $Z^{-}(\t=i)$ is equal to  zero. In Eq. \eqref{HF5} all phases are positive, so some of the density of states should be negative. Therefore, $Z^{-}(\t)$ is not a  physical function.

Covariance of $Z(\t)$ under $T$ transformation requires:
\begin{equation}\label{HF6}
TZ(\t)=e^{-2\pi i\alpha}Z(\t).
\end{equation}
Plugging \eqref{HF2} into \eqref{HF6}, for $\tau_r=0$ yields
\begin{equation}\label{HF7}
\sum_{h=0}\rho(h){e}^{-2\pi\tau_i(h-\frac{c_L}{24})}\left(1-\cos2\pi (h-\frac{c_L}{24}+\alpha)\right)=0,
\end{equation}
\begin{equation}\label{HF8}
\sum_{h=0}\rho(h){e}^{-2\pi\tau_i(h-\frac{c_L}{24})}\sin2\pi \left(h-\frac{c_L}{24}+\alpha\right)=0.
\end{equation}
The summands  \eqref{HF7} and \eqref{HF8} are non-negative. Therefore, $\left(h-\frac{c_L}{24}+\alpha\right)$ should be integer. The vacuum state $(h=0)$, requires that
\begin{equation}\label{alpha}
 \alpha=\frac{c_L}{24},
 \end{equation}
Therefore,
\begin{equation}\label{h}
h\in\mathbb{N}.
\end{equation}
Using $(ST)^3=1$ and  invariance of $Z(\t)$ under $S$ transformation one can obtain:
\begin{equation}\label{HF10}
e^{-2\pi i\frac{c_L}{8}}=1.
\end{equation}
Consequently;
\begin{equation}\label{c8}
c_L=8m_L,\quad m_L\in\mathbb{N}
\end{equation}
 Similarly, for $\bar{Z}(\bar{\tau})$ we have:
\begin{eqnarray}\label{HFC2}
\bar{h}\in\mathbb{N}\quad\quad c_R=8m_R,\quad m_R\in\mathbb{N}.
\end{eqnarray}
 For $m_L, m_R \notin3\mathbb{N}$ modular invariance of the partition function $Z(\t,\bar{\tau})$, enforces that $m_L=m_R=k$. In this case, the partition function $Z(\t,\bar{\tau})$ automatically becomes real. For $m_L, m_R \in3\mathbb{N}$,  if we put the weak condition of the reality of the partition function, this constrains leads to the equality of the right and left central charges.
\subsection{The Basis for the Modular Covariant  Functions}\label{sub32}
In this section we  derive the basis for $Z(\t)$. For $m_L$ integer multiple of  three, i.e. $c_L\in 24\mathbb Z$, $Z(\t)$ is modular invariant.  Therefore, it is a polynomial  in terms of the Klein function $J(\t)$ \cite{appostal}:
\begin{equation}
      Z(\tau)=\sum_{r=0}^{k}h_r J^{r}.
     \end{equation}
    with some coefficients $h_r$.  The Klein function has the following Fourier expansion:
    \begin{eqnarray}
    \label{g1}
    J(\t)&=&\mathfrak{j}(\t)^3\\
    \label{j11}
    &=&q^{-1}+744+196884\, q+\cdots,
    \end{eqnarray}
    where, $\mathfrak{j}$ has expansion in terms of the Jacobi Theta functions and Eta function as follows
\begin{eqnarray}
    \label{g11}
    \mathfrak{j} (\t)&:=&\frac{1}{2}\left[ \left(\sqrt{\frac{\theta_2(\t)}{\eta(\t)}}\right)^{16}+\left(\sqrt{\frac{\theta_3(\t)}{\eta(\t)}}\right)^{16}+\left(\sqrt{\frac{\theta_4(\t)}{\eta(\t)}}\right)^{16}\right] \nonumber\\
     &=&q^{\frac{-1}{3}}\left(1+248\,q+\cdots\right).
      \end{eqnarray}
   In order to obtain the bases for $Z(\t)$, we use the lemma  in \cite{Ashrafi-16}.
    \begin{Lemma}\label{t41}
 The $S$-invariant function $f^{(r)}\left(\lbrace a^{(r)}\rbrace,\t\right)$  with  Fourier expansion
    \begin{align}\label{g31}
    &f^{(r)}\left(\lbrace a^{(r)}\rbrace,\t\right)=q^{\frac{-p}{3}}\left[\sum _{n=-r}^{0} a_n^{(r)} q^n+\sum _{n=1}^{\infty } a_n^{(r)} q^{n}\right],&p\in\{0,1,2\}.
    \end{align}
    on the upper half  $\tau$-plane, is a polynomial in $\mathfrak{j}$.
    \end{Lemma}
       \begin{proof}
  From Eq.\eqref{j11} and Eq.\eqref{g11} we conclude  that there exist $\lbrace a^{(r-1)}\rbrace$ such that
    \begin{equation}\label{g311}
    q^{\frac{-p}{3}}\sum _{n=-r}^{\infty} a_n^{(r)} q^n=a_{-r}^{(r)}\,\mathfrak{j}^p\, J^r+q^{\frac{-p}{3}}\sum _{n=-r+1}^{\infty} a_n^{(r-1)} q^n.
    \end{equation}
    Therefore,
    \begin{equation}\label{g312}
    f^{(r)}\left(\lbrace a^{(r)}\rbrace,\t\right)=a_{-r}^{(r)}\,\mathfrak{j}^p\, J^{r}+f^{(r-1)}\left(\lbrace a^{(r-1)}\rbrace,\t\right).
    \end{equation}
    Since $f^{(r)}\left(\lbrace a^{(r)}\rbrace,\t\right)$, $\mathfrak{j}$ and $J$ are  $S$-invariant, so $f^{(r-1)}\left(\lbrace   a^{(r-1)}\rbrace,\t\right)$ is also $S$-invariant.
   The order of the pole of $f^{(r)}\left(\lbrace a^{(r)}\rbrace,\t\right)$ and $f^{(r-1)}\left(\lbrace a^{(r-1)}\rbrace,\t\right)$ are  $r$ and $r-1$ respectively. By iteration one can obtain
    \begin{equation}\label{g313}
    f^{(r)}\left(\lbrace a^{(r)}\rbrace,\t\right)=\mathfrak{j}^p\left[a_{-r}^{(r)}J^{r}+a_{-(r-1)}^{(r-1)}J^{r-1}+\cdots +a_{0}^{(0)}\right]+f^{(-1)}\left(\lbrace a^{(-1)}\rbrace,\t\right),
    \end{equation}
    where
    \begin{equation}\label{g32}
    f^{(-1)}\left(\lbrace a^{(-1)}\rbrace,\t\right)=q^{1-\frac{p}{3}}\sum_{m\ge1} a^{(-1)}_m q^{m-1}.
    \end{equation}
   Since the function $f^{(r)}\left(\lbrace a^{(r)}\rbrace,\t\right)$ is $T^3$ invariant for all values of $r$,  therefore  the function $\left[f^{(-1)}\left(\lbrace a^{(-1)}\rbrace,\t\right)\right]^3$ is modular invariant. It has no pole on the upper half plane and is zero at $\tau=i\infty$. Thus, it is zero on the upper half plane.
    \end{proof}
  \begin{Corollary}\label{C3}
    The  chiral  function  is a polynomial in function $\mathfrak{j} $ as follows:
     \begin{align}
     \label{ch1}
 Z(\t)=\mathfrak{j}^{k}\sum_{r=0}^{[k/3]}n_r J^{-r}, \quad &n_r\in {\mathbb N}.
    \end{align}
    \end{Corollary}

\section{The Hecke Operators}\label{sec3}
 Let us define the subgroup of the modular group with $S$ and $T^3$ generators, which we call this group  $\Gamma_3$.

The modular covariant function $f_3(\tau)$ with the Fourier expansion:
 \begin{equation}
 f_3(\tau)=\sum_{m=-k}^{\infty}a\left(m-\frac{1}{3}\right)q^{m-\frac{1}{3}},
 \end{equation}
is invariant under the group $\Gamma_3$.

The Hecke operators are  linear operators which map modular form space $M_k$, onto itself and are defined as follows:
\begin{equation}\label{h0}
T_nf(\tau)=n^{k-1}\sum_{d|n}d^{-k}\sum_{b=0}^{d-1}f\left(\frac{n\tau +bd}{d^2}\right).
\end{equation}
The Hecke operators map the Modular functions, onto the modular functions. \\

  In this section, we  generalize  the definition of the Hecke operators, 3-Hecke operators for the group $\Gamma_3$ , which map the modular covariant functions onto the modular covariant functions. \\
 \begin{Definition}\label{def1}
 For positive integer values of $n$ and $n\neq 3\mathbb{N}$, the operator $T^{(3)}_n$ on the modular covariant function $f_3(\tau)$ is defined as follows
   \begin{equation}\label{he3}
 T^{(3)}_nf_3(\tau)=\sum_{d|n}\sum_{b=0}^{d-1}f_3\left(\frac{n\tau +3bd}{d^2}\right),\quad\quad  \end{equation}
\end{Definition}
We called the operators $T^{(3)}_n$, the 3-Hecke operators. We show that the 3-Hecke operators map the modular covariant functions  $f_3(\tau)$,  onto the modular covariant functions.
First, we study the Fourier expansion  of  $T^{(3)}_nf_3(\tau)$.
\begin{Theorem}
If $f_3(\t)$ has the  Fourier expansion
 \begin{equation}\label{fourierf3}
 f_3(\tau)=\sum_{m=-k}^{\infty}a\left(m-\frac{1}{3}\right)q^{m-\frac{1}{3}}.
 \end{equation}
 then, $T^{(3)}_nf_3(\tau) $ has the  Fourier expansion:
\begin{equation}\label{fourierheck}
T^{(3)}_nf_3(\tau)=\sum_{m=-k}^{\infty} \gamma_n \left(m-\frac{1}{3}\right) q^{m-\frac{1}{3}},
\end{equation}
where
\begin{equation}\label{h2}
\gamma_n (m) =\sum_{d|(n,3m)}\frac{n}{d}a\left(\frac{nm}{d^2}\right).
\end{equation}
\end{Theorem}
\begin{proof}
By putting the Fourier expansion of the function $f_3(\tau)$\eqref{fourierf3} into \eqref{he3} we have
\begin{equation}\label{fourier1}
T^{(3)}_nf_3(\tau)=\sum_{m=-k}^{\infty}a\left(m-\frac{1}{3}\right)\sum_{d|n}e^{\frac{2\pi i n\tau}{d^2}\left(m-\frac{1}{3}\right)}\sum_{b=0}^{d-1}e^{\frac{2\pi ib(3m-1)}{d}}.
\end{equation}
The last sum in \eqref{fourier1} is zero for $d\nmid 3m-1$ and is equal to $d$ for $d\mid 3m-1$:
\begin{equation}\label{fourier2}
T^{(3)}_nf_3(\tau)=\sum_{m=-k}^{\infty}a\left(m-\frac{1}{3}\right)\sum_{d|n, d\mid 3m-1}de^{\frac{2\pi i n\tau}{d^2}(m-\frac{1}{3})}.
\end{equation}
Since, $d\mid 3m-1$; writing  $3m-1=pd$ and replacing $\frac{n}{d}$ with $d$ (because $d\mid n$) yeilds
\begin{equation}\label{fourier3}
T^{(3)}_nf_3(\tau)=\sum_{p=-3k-1}^{\infty}\sum_{d|n}a\left(\frac{pn}{3d}\right)\frac{n}{d}e^{\frac{2\pi i \tau dp }{3}}.
\end{equation}
The last term in the sum has the form $q^({\frac{pd}{3})}$. For all  terms which ${\frac{pd}{3}}$ is constant ${\frac{pd}{3}}=m-\frac{1}{3}$ one can obtain:
\begin{equation}\label{fourierheck1}
T^{(3)}_n f_3(\tau)=\sum_{m=-k}^{\infty}\sum_{d|(n,3m-1)}\frac{n}{d}a\left((m-\frac{1}{3})\frac{n}{d^2}\right) e^{2\pi i\tau(m-\frac{1}{3})}.
\end{equation}
\end{proof}
\subsection{The Order $n$ Transformations}\label{sub31}
\indent For positive integer $n$, the order $n$ transformation $\Gamma(n)$, is defined as follows
\begin{equation}
\tau\rightarrow A\tau=\frac{a\tau+b}{c\tau+d},\quad\quad ad-bc=n,
\end{equation}
 where  $a,b,c$, and $d$ are integers. The $\Gamma (1)=\Gamma$ transformations  correspond to the modular transformations.\\
 \indent The transformations $A_1$ and $A_2$ in  $\Gamma (n)$ are called equivalent if there exist a modular transformation $V\in\Gamma$, such that
 \begin{equation}
A_2\sim  A_1\quad \mbox{if}\quad A_2=VA_1.
 \end{equation}
 It is clear that the relation $\sim$ is an equivalence relation. So, the transformations $\Gamma (n)$  can be divided into the equivalence classes. Two element of $\Gamma (n)$ are in the same class, if and only if, they are equivalent.\\

\begin{Lemma}\label{lem2}
For every equivalence class of $\Gamma (n)$, there is a triangular representation $A_3$:
 \begin{equation}
 A_3=\left(\begin{array}{cc} a & 3b \\ 0 & d \\ \end{array} \right),
\end{equation}
where $n=3p+i$$(i=1,2)$ and $p\in\mathbb{N}$.
\end{Lemma}
\begin{proof}
As shown in \cite{appostal}, in every equivalence class of $\Gamma(n)$ there is a representation of triangular form
 \begin{equation}
 A_1=\left(\begin{array}{cc} a_1 & b_1 \\ 0 & d _1\\ \end{array} \right)\quad\quad d_1>0.
\end{equation}
 For $A_1$ and $A_2$ (two equivalent elements in $\Gamma (n)$), there is $V=\left(\begin{array}{cc} 1 & q \\ 0 & 1 \\ \end{array} \right)\in\Gamma$
 Such that
\begin{equation}
 A_2=VA_1=\left(\begin{array}{cc} a_1 & qd_1+b_1 \\ 0 & d_1 \\ \end{array} \right).
 \end{equation}
In order to prove this theorem, it  is necessary to show that $b_2$ is multiple of three.
\begin{equation}\label{1}
b_2=qd_1+b_1.
 \end{equation}
 Since $a_1d_1=n$ and $n\neq 3\mathbb{N}$, so $d_1$ can not be multiple integer of three and takes   $3s+1$ or $3s+2$ values. For fixed value of d, $b_1$  takes  $3r, 3r+1$ and $3r+2$  values. By substituting these values to \eqref{1}, one can show $b_2$ can be multiple of three (by choosing  appropriate values of $q$).
\end{proof}
\begin{Theorem}\label{theorem4}
A complete system of nonequivalent elements of  $\Gamma (n)$ is given by the set of triangular transformations of the form:
 \begin{equation}\label{2}
\left(\begin{array}{cc} a & 3b \\ 0 & d \\ \end{array} \right),
\end{equation}
 where $d$ runs through the positive divisors of n and for fixed values of $d$, $a=\frac{n}{d}$ and $b$ runs through a complete residue system of modulo $d$.\\
\end{Theorem}
\begin{proof}
The  lemma \eqref{lem2} shows every element of $\Gamma (n)$ is equivalent to one of the transformations in \eqref{2}. So, we should show  two  transformations $A_1$ and  $A_2$  are equivalent, if and only if
  \begin{equation}\label{pr3}
  a_1=a_2, \quad d_1=d_2, \quad \mbox{and}\quad  b_2=b_1+q'd.
  \end{equation}
  where,
   \begin{equation}
 A_i=\left(\begin{array}{cc} a_i & 3b_i \\ 0 & d_i \\ \end{array} \right) \quad\quad i=1,2.
\end{equation}
 First, we show if \eqref{pr3} holds, then $A_1\sim A_2$. For some integer $q$, if we consider $V$ as follows
 \begin{equation}
 V=\left(\begin{array}{cc} 1 & q \\ 0 & 1 \\ \end{array} \right)\in\Gamma.
 \end{equation}
 where $q=3q'$, then, $VA_1=A_2$, so $A_1\sim A_2$.\\
 Conversely, if $A_1\sim A_2$ there exists $V\in \Gamma$
 \begin{equation}
V=\left(\begin{array}{cc} p & q \\ r & s \\ \end{array} \right),
 \end{equation}
 such that
 \begin{equation}
\left(\begin{array}{cc} a_2 & 3b_2 \\ 0 & d_2 \\ \end{array} \right)=\left(\begin{array}{cc} p & q \\ r & s \\ \end{array} \right) \left(\begin{array}{cc} a_1 & 3b_1 \\ 0 & d_1 \\ \end{array} \right)= \left(\begin{array}{cc} pa_1 & 3pb_1+qd_1 \\ ra_1 & 3rb_1+sd_1 \\ \end{array} \right).
 \end{equation}
 The above equality shows $r=0$ (since $a_1\neq 0$). From $ps-qr=1$, we can conclude  $ps=1$, so $p=s=1$ or $p=q=-1$. Let us consider $p=s=1$ (for the other case we replace $V$ by $-V$).  By equating the entries in the above equation, we have
  \begin{equation}\label{3}
  a_1=a_2,\quad  d_1=d_2, \quad and\quad  3b_2=3b_1+qd.
  \end{equation}
  Since, $a_1d_1=n$ and $n$ is not integer multiple of three, therefore, $q=3q'$:
    \begin{equation}\label{3}
 b_2=b_1+q'd.
  \end{equation}
\end{proof}

 \begin{Lemma}\label{lem5}
 For $A_1\in \Gamma (n)$, $V_1\in \Gamma$ there exists transformation $A_2\in\Gamma(n)$ and $V_1\in \Gamma$ such that
 \begin{equation}
A_1V_1=V_2A_2.
\end{equation}
where
  \begin{equation}
 A_i=\left(\begin{array}{cc} a_i & 3b_i \\ 0 & d_i \\ \end{array} \right), \quad and \quad\quad V_i=\left(\begin{array}{cc} \alpha _i & \beta _i \\ \gamma _i & \delta _i \\ \end{array} \right).
 \end{equation}
 \end{Lemma}
\begin{proof}
$det(A_1V_1)=detA_1detV_1 =n$, so $A_1V_1\in \Gamma (n)$. According to lemma \ref{lem2}, there exists  $A_2\in \Gamma (n)$ and $V_2\in \Gamma$ such that
 \begin{equation}\label{4}
A_1V_1=V_2A_2.
\end{equation}
\end{proof}

 Now, by using \eqref{4} for $V_1=S=\left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \\ \end{array} \right)$ transformation, we derive  the elements of $A_2$ and $V_2$ in terms of the element of $A_1$. By equating the entries in \eqref{4}, we have
 \begin{eqnarray}\label{5}
 a_2&=&\frac{d_1}{\gamma_2},\\\nonumber
  3b_2&=&-a_1\delta_2, \\\nonumber
   d_2&=&a_1\gamma_2,\\\nonumber
  \alpha_2 &=&\frac{3b_1\gamma_2}{d_1},\\\nonumber
   \beta_2 &=&\frac{3b_1\delta_2}{d_1}-\frac{1}{\gamma_2}.\\\nonumber
 \end{eqnarray}
From \eqref{5}, we recognize that  $V_2$ has two independent entries $\delta_2$ and $\gamma_2$. Since; $n=a_1d_1\neq 3\mathbb{N}$ , the second and the forth equation in \eqref{5} show  $\delta_2$ and $\alpha_2$ are multiples of three. \\
\indent We already know  $S$ and $T=\left(\begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} \right)$ transformations are generators of the modular group and each elements of the modular group can be written in the below form
\begin{equation}
ST^{n_1}ST^{n_2}.......
\end{equation}
 Since $V_2$ has two independent entries, one can write it as follows
  \begin{equation}\label{6}
V_2=ST^{n_1}ST^{n_1}ST^{n_2}.
 \end{equation}
 From \eqref{6} we have
 \begin{eqnarray}\label{7}
  \alpha_2 &=&\frac{3b_1\gamma_2}{d_1}=-n_1,\\\nonumber
   \beta_2 &=&\frac{3b_1\delta_2}{d_1}-\frac{1}{\gamma_2}=1-n_1n_2,\\\nonumber
   \gamma _2&=&n_1^2-1,\\\nonumber
   \delta_2 &=& n_{2}(n_1^2-1)-n_1.
 \end{eqnarray}
 For the case where $n$ is not multiple of three, from \eqref{5} and \eqref{7} we conclude  $n_1$ and $n_2$ are multiple of three.\\

\begin{Theorem}
 For  an integer value of  $n$ which is not multiple of three, if $f_3(\tau)$ is modular covariant then, $T^{(3)}_n f_3(\tau)$ is covariant under modular transformations. Hence, $T^{(3)}_n f_3(\tau)$ maps the $\Gamma_3$ invariant function $f_3(\t)$ onto the $\Gamma_3$ invariant function.
 \end{Theorem}
\begin{proof}
 Since $d|n$ one can  rewrite the 3-Hecke operator as follow
 \begin{equation}\label{he1}
T^{(3)}_nf_3(\tau)=\sum_{a\geq 1, ad=n}\sum_{b=0}^{d-1}f_3(A\t),
\end{equation}
where $A$ is an element of $\Gamma (n)$:
\begin{equation}
A\tau=\frac{a\tau+3b}{d}.
\end{equation}
 From \eqref{he1} we have:
  \begin{equation}\label{he01}
T^{(3)}_n f_3(S\tau)=\sum_{a_1\geq 1, a_1d=n}\sum_{b=0}^{d-1}f_3(A_1S\t),
\end{equation}
Using lemma \eqref{lem5}, we have
  \begin{equation}\label{he21}
f_3(A_1S\t)=f_3(ST^{n_1}ST^{n_1}ST^{n_2}A_2\t)=e^{\frac{-2(2n_1+n_2)\pi i}{3}}f_3(A_2\t).
\end{equation}
 As we showed earlier,  $n_1$ and $n_2$ are multiple of three, therefore:
  \begin{equation}\label{heck44}
f_3(A_1S\tau)=f_3(A_2\tau).
 \end{equation}
 Substituting \eqref{heck44} to \eqref{he1} yields to:
  \begin{equation}\label{he12}
T^{(3)}_n f_3(S\tau)=\sum_{a_2\geq 1, a_2d=n}\sum_{b=0}^{d-1}f_3(A_2\t)=T^{(3)}_n f_3(\tau).
\end{equation}
\eqref{he12} shows that the 3-Hecke operators are invariant under $S$ transformation. The Fourier expansion \eqref{fourierheck}, shows that the 3-Hecke operators are modular covariant under $T$ transformation and are invariant under $T^3$ transformation. Since every elements of the group of $\Gamma_3$ are built from the multiplication of the $S$ and $T^3$ generators, we conclude that the 3-Hecke operators map the $\Gamma_3$ invariant functions $f_3(\t)$ into the $\Gamma_3$ invariant functions.
\end{proof}
\section{Three Dimensional Gravity}\label{sec4}
Our focus in this section is solving the pure quantum gravity in the sense of  finding the dual boundary CFT. As  it is shown in \cite{witten}, the dual CFT is extremal which means that the lowest dimension of the primary fields excluding the identity, is $k+1$ for $c=24k$, and the partition function should be holomorphically factorizable. The allowed values of the left  and right central charges can be obtained from the symmetry group of the Chern-Simons gauge theory. The symmetry group can be the group $SO(2,1)\times SO(2,1)$ and its $n$th diagonal cover:
\begin{equation}\label{klkr2}
k_L \in \bigg \{ \begin{array}{ll}
n^{-1} \mathbb{Z} ~~~~~~n \in 2\mathbb{Z}+1\\
(2n)^{-1}\mathbb{Z}~~~~~~n \in 2 \mathbb{Z}
\end{array}
\end{equation}
where,  $k_L$ and $k_R$ are the Chern-Simons couplings. From the AdS/CFT dictionary, the corresponding central charges are obtained as follows
\begin{equation}
(c_L,c_R)=(24k_L, 24k_R)
\end{equation}
Therefore, the first step to solve the quantum gravity is determining  the covering group. The holomorphically factorizable and modular invariance of the partition function is necesary and sufficient condition for determining the symmetry group.

In section \eqref{s2} we show  for the modular invariant holomorphically factorizable partition function, the scaling dimensions of the chiral and anti-chiral  functions should be an integer number and the left and right central charges are integer multiples of eight:
\begin{equation}\label{clcr}
c_L=8m_L, \quad\quad c_R=8m_R,\quad\quad\quad\quad m_L,m_R\in\mathbb{N}.
\end{equation}
From \eqref{klkr2} and \eqref{clcr}, we conclude that the allowed values of  $n$ should be $1$ or $3$. So,  the symmetry group of the Chern-Simons gauge theory should be $SO(2,1)\times SO(2,1)$ and its three-fold diagonal cover.

For pure gravity with the asymptotic space-time  AdS$_3$, the  vacuum state is the trivial state where its chiral  function is obtained as follows
\begin{equation}\label{zvac}
Z^{vac}(\tau)=q^{-\frac{c_L}{24}}\prod_{n=2}^{\infty}\frac{1}{1-q^n},
\end{equation}
The  vacuum state   corresponds to the Anti de Sitter Space, classically. Since, the vacuum partition function \eqref{zvac}, is not modular covariant, there should be other states in the theory. This is in the agreement with the existence of the BTZ black hole in the theory. The mass and the angular momentum of the classical BTZ black hole in terms of the Virasoro generators $L_0$ and $\bar{L}_0$ are obtained as follows
\begin{eqnarray}\label{mass}
M&=&\frac{1}{l}(L_0+\bar{L}_0),\\\nonumber
J&=&(L_0-\bar{L}_0).
\end{eqnarray}
where
\begin{equation}\label{mj}
Ml\geqslant |J|.
\end{equation}
and the  entropy is
\begin{equation}\label{entropy0}
S=4\pi\left(\sqrt{\frac{c_L}{24}L_0}+\sqrt{\frac{c_R}{24}\bar{L}_0}\right).
\end{equation}
From \eqref{mj} and \eqref{entropy0}, we conclude that $L_0\geqslant1$. Hence, the full chiral function has the following form:
\begin{equation}\label{ztot}
Z(\tau)=q^{-\frac{c_L}{24}}\prod_{n=2}^{\infty}\frac{1}{1-q^n}+{\cal O}(q).
\end{equation}
The modular covariant constraint determines the partition function uniquely. For $k_L, k_R\in\mathbb{Z}$, the chiral and anti-chiral  functions are modular functions. In \cite{witten}, the partition function and the entropy are investigated. In this section we  study the case where  $k_L, k_R\in\frac{\mathbb{Z}}{3}$. We showed in section \eqref{s2}, modular invariant of $Z(\t,\bar{\tau})$ demands $c_L=c_R=8k, k\in\mathbb{Z}$.

From corollary \eqref{C3}, the chiral  function is obtained as follows
 \begin{align}
     \label{ch11}
 Z(\t)=\mathfrak{j}^{k}\sum_{r=0}^{[k/3]}n_r J^{-r},\quad &n_r\in {\mathbb N}.
    \end{align}
    where the $n_r$  coefficients are determined from the fact that the density of the low dimensional state, should be equal to the density of the corresponding low dimensional state of the vacuum. It is worth noting that for $k=3m$ for some $m\geq 1$, these candidate partition functions are exactly the candidate torus path integral  introduced in \cite{witten} . For $k=1$ to $k=11$, the chiral  functions is obtained in \cite{Avramis:2007gx}. Here are some examples:
    \begin{align}\label{wit14}
Z_8(\tau)=\mathfrak{j}(\tau)=q^{\frac{-1}{3}}+248q^{\frac{2}{3}}+4125q^{\frac{5}{3}}+\cdots,
\end{align}
\begin{align}\label{wit15}
Z_{16}(\tau)=\mathfrak{j}^{2}(\tau)=q^{\frac{-2}{3}}+496q^{\frac{1}{3}}+69752q^{\frac{4}{3}}\cdots,
\end{align}
\begin{align}\label{wit16}
Z_{32}(\tau)&=\mathfrak{j}^{4}(\tau)-992\mathfrak{j}(\tau)\\\nonumber
&=q^{\frac{-4}{3}}+139504q^{\frac{1}{3}}+69332992q^{\frac{5}{3}}\cdots,
\end{align}
\begin{align}\label{wit17}
Z_{40}(\tau)&=\mathfrak{j}^{5}(\tau)-1240\mathfrak{j}^{2}(\tau)\\\nonumber
&=q^{\frac{-5}{3}}+20620q^{\frac{1}{3}}+86666240q^{\frac{4}{3}}\cdots.
\end{align}
The chiral function \ref{ch11} is unitary modular covariance and has the positive density of state. For $c=8$ and $c=16$, the chiral functions $Z_8$ and $Z_{16}$ are well known and is believed that they are unique. The chiral functions $Z_8$ and $Z_{16}$ are the vacuum character of the level 1 affine $\hat{E}_8$ and the level 1 affine $\hat{E}_8\times\hat{E}_8$ theory respectively. For $c=32$  and $c=40$ the chiral function has been identified with $\mathbb{Z}_2$ orbifolds of theories defined on even unimodular lattices of the respective rank possessing no vectors of squared length 2. For $c>40$, the existence of these CFTs are not known and is an open question \cite{hon1, hon2}. If these CFT's exist they are good candidates for the pure gravity.

The closed-form of the chiral function is obtained by using the generalize 3-Hecke operator \eqref{he3}. The Fourier expansion \eqref{fourierf3} and \eqref{fourierheck} shows for
   \begin{align}\label{j1}
\mathfrak{j}(\tau)=q^{\frac{-1}{3}}+{\cal O}(q).
\end{align}
and
\begin{align}\label{j2}
\mathfrak{j}^{2}(\tau)=q^{\frac{-2}{3}}+{\cal O}(q).
\end{align}
The 3-Hecke operators have  the following expansion
   \begin{align}\label{he3j1}
T^{(3)}_{n}\mathfrak{j}(\tau)=q^{\frac{-n}{3}}+{\cal O}(q).
\end{align}
and
\begin{align}\label{he3j2}
T^{(3)}_{n}\mathfrak{j}^{2}(\tau)=q^{\frac{-2n}{3}}+{\cal O}(q).
\end{align}
Therefore, for $c=8k$   the chiral  function is obtained as follows
\begin{equation}\label{z2k+1}
Z_{k}(\tau)=\sum_{r=0}^{k}a_{-r}T^{(3)}_r\mathfrak{j}^{i}(\tau),\quad\quad i=1,2.
\end{equation}
where $i=1$ is for $k=\mbox{odd}$  and $i=2$ corresponds with $k=\mbox{even}$. The  $a_r$ coefficients, are the low state density of the vacuum:
\begin{equation}
Z^{vac}(\t)=\sum_{r=-k}^{\infty}a_rq^r.
\end{equation}
In order to determine the entropy, let us write the chiral function as:
\begin{equation}\label{partition11}
Z_k(\t)=\sum_{m=-k}^{\infty}b_{k,m}q^m.
\end{equation}
Using \eqref{z2k+1} and \eqref{fourierheck}, the  $b_{k,m}$ coefficients are  obtained as follows
\begin{equation}\label{bkm}
b_{k,m}=\sum_{r=0}^{k}a_{-r}\sum_{d|(r,3m)}\frac{r}{d}c_i\left(\frac{rm}{d^2}\right),\quad i=1,2.
\end{equation}
where $c_1(m)$, and $c_2(m)$ are  the $\mathfrak{j}$  and $\mathfrak{j}^2$ Fourier expansion's coefficients, respectively. The  $c_i$  coefficients are obtained as follows ( up to the exponentially  suppressed terms)
 \cite{Loran:2010bd}, \cite{Alday:2019vdr}:
\begin{eqnarray}\label{cm7}
c_i(m)=2\pi\sqrt{\frac{i/3}{m-i/3}}I_1\left(4\pi\sqrt{\frac{i}{3}(m-\frac{i}{3})}\right),
\end{eqnarray}
and
\begin{equation}
a_r=P(r)-P(r-1).
\end{equation}
The partition numbers $P(r)$ are obtained from Peterson-Rademacher expansion:
\begin{equation}
P(r)=2\pi\left(\frac{1/24}{r-1/24}\right)^{3/4}\sum_{k=1}^{\infty}\frac{1}{k}\mbox{Kl}\left(r-\frac{1}{24},-\frac{1}{24};k\right) I_{3/2}\left(\frac{4\pi}{k}\sqrt{\frac{1}{24}(r-\frac{1}{24})}\right).
\end{equation}
where, Kl(a,b;k) is the Kloosterman sum.
Using \eqref{bkm}, the microcanonical entropy obtained as follows
\begin{eqnarray}\label{entropy1}
S(k,m)&=&\ln{b_{k,m}}
\end{eqnarray}
We obtain the entropy in the semiclassical limit. In  \eqref{bkm}, the leading term in the large $km$ limit is $d=1$:
\begin{equation}\label{log}
b_{k,m}=kc(km)+(k-1)c((k-1)m)+(k-2)c((k-2)m)+\cdots,
\end{equation}
Now, we compare the first and the second terms in the Eq.\eqref{log}. Using \eqref{cm7} and the asymptotic behavior of the Bessel Function:
\begin{equation}
I_1(z)\sim\frac{e^z}{\sqrt{2\pi z}},\quad\quad z\gg 1,
\end{equation}
 we have
\begin{eqnarray}\label{cd23}
\frac{I_1\left(4\pi\sqrt{\frac{k}{3}(m-\frac{1}{3})}\right)}{I_1\left(4\pi\sqrt{\frac{(k-1)}{3}(m-\frac{1}{3})}\right)}\sim e^{2\pi\sqrt{{(m-1/3)}{k}}}.
\end{eqnarray}
As \eqref{cd23} shows the first term in \eqref{log} is the dominant term and the other terms are exponentially small in $\sqrt{\frac{m-1/3}{k}}$, which are important for large values of $\frac{m-1/3}{k}$.

 In the large $m$  limit, the  $c_i(m)$ coefficients are obtained as follows
  \begin{equation}\label{cm}
\ln c_i(m)=4\pi\sqrt{\frac{i}{3}(m-\frac{i}{3})}-\frac{3}{4}\ln(m-\frac{i}{3})+\frac{1}{4}\ln\frac{i}{3}-\frac{1}{2}\ln 2\quad\quad i=1,2
\end{equation}
 In the large $k$  and $m$,  where $\frac{m}{k}$ is constant, the leading terms in the entropy  comes from $r=k$. Therefore, in the semiclassical limit the entropy is obtained as follows
\begin{eqnarray}\label{entropy}
S(k,m)&=&\ln{b_{k,m}}=\ln{kc_i\left(k(m-\frac{1}{3})\right)}+\cdots\\\nonumber
&=&4\pi\sqrt{\frac{i}{3}(k(m-\frac{i}{3}))}-\frac{3}{4}\ln(m-\frac{i}{3})+\frac{1}{4}\ln k+\frac{1}{4}\ln\frac{i}{3}-\frac{1}{2}\ln 2+\cdots
\end{eqnarray}
the first  term is well known Bekenstein-Hawking entropy which is proportional with the area of the BTZ black hole.  The other terms determine the logarithmic corrections to the entropy which were calculated before\cite{Strominger:1997eq}-\cite{Govindarajan:2001ee}. These logarithmic correction to the entropy  typically appears in the microcanonical entropy.
Eq \eqref{cd23} shows the portion of  $r=k$ and $r=k-1$ terms. As this equation shows, for large values of $\frac{m-1/3}{k}$, which means the size of the BTZ black  hole is in the order of AdS scale, these terms are exponentially suppressed. The physical interpretation of these terms would be interesting to investigate.

\section{Summary}
In this study we have investigated the 3d quantum gravity and its corresponding CFT.  The equivalence between 3d gravity and Chern-Simons gauge theory, shows that the boundary CFT should be extremal and the partition function  should be holomorphically factorizable and the left and right central charges are $(24k_L, 24k_R)$. Existence of extremal CFT for $k_L, k_R>1$ is still an open problem. There are two approaches to view this subject. In the positive approach we view the current failure of attempts to non- existence proof as an      indication that extremal CFT exist as a mathematical object. While in the negative approach, the lack of current constructed extremal CFTs is viewed as an indication that they do not exist. Additional investigation is required to clarify matter. The values of the gauge couplings have been obtained from the gauge group of the Chern-Simons theory.

We studied the holomorphic factorized CFT.  From modular invariant of partition function we showed that the chiral and anti-chiral  functions are $S$ invariant and $T$ covariant. Therefore,  the chiral and anti-chiral  functions are modular covariant. For chiral CFT, scaling dimension of the primary fields are integer numbers and the central charges are integer multiple of 8 $(c_L=24k_L, c_R=24k_R$ for $k_L, k_R\in\frac{\mathbb{Z}}{3})$  . We have obtained  the bases for the chiral  functions in-terms of the Klein function and $\mathfrak{j}=J^{\frac{1}{3}}$. It is believed that for $k\in3\mathbb{Z}$ where the chiral  function is modular invariant (therefore the chiral function itself is a modular invariant partition function and can be consider as a corresponding purely chiral CFT ), the chiral extremal CFT is holomorphically dual with chiral gravity  \cite{Li:2008dq, Maloney:2009ck}.

The gauge group of the Chern-Simons theory can be $SO(2,1)\times SO(2,1)$ and its n-fold diagonal cover. From modular invariance of the holomorphically factorizable partition function, we have shown the symmetry group of the theory can be one of two choices: either $SO(2; 1) \times SO(2; 1)$ or its three-fold diagonal cover. Furthermore, we have introduced the generalized Hecke operators (3-Hecke operators) which map the modular covariant functions to modular covariant functions. We have studied the 3d pure gravity and corresponding boundary CFT for  the case where the gauge group of the Chern-Simons theory is 3-fold diagonal cover (i.e.  left and right central charges are integer multiple of eight ). For the case where the central charges are multiple of 24, it has been tried to calculate the partition function of the pure gravity by different methods including summing over the known classical geometries contributions to the partition function, including quantum corrections \cite{Maloney:2007ud, Keller:2014xba} and the Rademacher expansions \cite{Alday:2019vdr}. The resulting partition functions have some unknown physical properties including the negative norm sates. In this case the chiral partition function is modular invariant.  The calculation of the  partition function of the pure gravity for central charges multiple of 8 (where the chiral  function is modular covariant), using the same method as in\cite{Maloney:2007ud}, would be an interesting matter which can be studied in the future.

 Using the 3-Hecke operators we have obtained the closed-form for the conjectural partition functions for extremal CFT2s, and the corresponding microcanonical entropies, when the chiral central charges are multiples of eight. We have computed subleading corrections to the Beckenstein-Hawking entropy in the bulk gravitational theory with these conjectural partition functions. We showed  the  microcanonical  entropy is equal to the  Bekenstein-Hawking entropy, the logarithmic corrections and some subleading terms which, are important when the size of the BTZ black hole is of the order of the AdS scale. The logarithmic corrections were obtained before for the BTZ black hole, but the subleading terms are new terms which studying their physical interpretation would be interesting

 \subsection*{Acknowledgments}
    I am grateful to Steven Carlip, David Mc Gady, and Farhang Loran for reading the manuscript and Stevan Carlip, David Mc Gady, Nathan Benjamin and especially referee for their useful comments.

redpajama_set_name: RedPajamaArXiv


Cheapest kamagra oral jelly uk famvir 250 mg price famvir 750 mg price viagra us price viagra brand canada. Kamagra onlineapotheke de cheap generic kamagra online buy amoxicillin 500 generika apotheke kamagra kamagra generic viagra cheap generic kamagra uk. Seriöse online apotheke kamagra famvir 500mg price best price for famvir famvir 500mg price kamagra kaufen in der apotheke stada sildenafil al ohne rezept. Viagra us cost famvir uk price famvir 250 mg price best price for viagra in us famvir price walmart online apotheke schweiz kamagra famvir generic cost. Price for famvir sildenafil von stada ohne rezept famvir generic price sildenafil abz ohne rezept price of viagra in the us kamagra oral jelly birmingham uk. Lamictal or depakote for anxiety kamagra oral jelly 100mg online bestellen us price of viagra kamagra online apotheke gutschein sildenafil ratiopharm ohne rezept. Cost of viagra in usa kamagra oral jelly order online kamagra oral jelly online pharmacy kamagra rezeptfrei in der apotheke. Buy kamagra 100mg generic viagra famvir price famvir once price kamagra oral jelly uk delivery buy amoxicillin 875 mg amoxicillin 250 mg to buy famvir 750 mg price. How much does viagra cost in the us kamagra super online-apotheke famvir once price amoxicillin 875 mg buy online. Kamagra oral jelly online kaufen sildenafil ratiopharm 100mg ohne rezept viagra cialis levitra canada kamagra oral jelly expired sildenafil 1a pharma 50mg ohne rezept. Kamagra oral jelly vendita online kamagra-onlineapotheke.to erfahrung famvir price australia famvir price australia famvir price walmart kamagra cheap generic. Viagra canada mastercard online apotheke österreich kamagra cheap generic viagra co uk kamagra oral jelly 100mg famvir price ireland buy amoxicillin 250 kamagra oral jelly effet secondaire. Famvir tablets price price for famvir kamagra oral jelly kaufen apotheke famvir 125 mg cost generic viagra canada pharmacy. Kamagra oral jelly 100mg uk famvir price south africa famvir price canada sildenafil stada 100mg filmtabletten ohne rezept. Famvir price south africa lamictal weight loss or gain how much does viagra cost in usa famvir price in india kamagra generic viagra uk. Famvir price uk sildenafil pfizer ohne rezept deutsche online apotheke kamagra buy amoxicillin 250 mg price of pfizer viagra in us.
Generic kamagra oral jelly viagra generic vs pfizer cheap generic viagra co uk kamagra tablets coupon code for generic viagra levaquin generic kamagra cheap generic. Unisom gel caps vs tablets kamagra 100mg generic viagra unisom sleep tabs vs sleepgels kamagra uk fast delivery free seroquel xr coupon. Cheap generic kamagra seroquel xr manufacturer coupons kamagra online cheap viagra generic equivalent kamagra jelly cheap uk kamagra cheapest online where to buy generic zoloft. Unisom gel vs tabs cheapest kamagra gel is there a generic version of viagra cheap kamagra now uk kamagra jelly uk fast delivery. Viagra brand and generic names viagra generic brand has a generic version of viagra been approved generic brand for viagra. Kamagra for cheap kamagra cheap next day delivery unisom sleep tabs vs ambien seroquel coupons astrazeneca seroquel coupon cheap kamagra jelly online. Where can i buy generic zoloft kamagra generic viagra uk seroquel xr coupon card generic form of viagra levaquin generic cost walgreens levaquin generic coupons. Seroquel xr 300 mg coupon buy generic zoloft online unisom sleep tabs vs gels mildronate buy online australia levaquin generic coupon. Generic viagra brand cheap generic kamagra uk seroquel xr coupon buy kamagra in new zealand levaquin generic name generic viagra vs pfizer viagra cheapest kamagra oral jelly. Buy generic zoloft kamagra oral jelly wholesalers cheapest kamagra online uk kamagra cheap online buy zoloft generic online. Generic levaquin prices are there generic versions of viagra Kamagra 30 Pills 100mg $121 - $4.03 Per pill seroquel xr discount coupon seroquel xr free trial coupon. Kamagra oral jelly nz mildronate online mildronate buy online generic sildenafil or viagra viagra generic vs viagra brand price generic levaquin 500 mg. Kamagra jelly cheapest buy kamagra new zealand generic zoloft buy buy kamagra uk cheap seroquel coupon 2015 sildenafil generic for viagra unisom sleep tabs vs zzzquil. Cost of ventolin inhaler in australia kamagra vs generic viagra seroquel xr coupons online generic kamagra uk zoloft generic buy mildronate online buy. Is a generic version of viagra available cheap generic viagra co uk french kamagra viagra brand and generic printable seroquel xr coupons.
Kamagra oral jelly birmingham uk kamagra oral jelly uk shop strattera drug testing suhagra 50 mg online purchase. Order ventolin online buy flagyl online australia adderall and strattera drug interactions buy kamagra uk cheap. Strattera risperdal drug interactions buy flagyl online fast delivery kamagra 100mg tabletten kaufen strattera fail drug test online shopping suhagra. Buy flagyl metronidazole online strattera abuse drug tricor medication generic cheap kamagra oral jelly uk kamagra menge kaufen tricor 48 mg generic. Kamagra fast shipping usa does strattera test positive drug screen cheap kamagra now uk kamagra oral jelly from uk strattera in drug test. Kamagra fast discount code ventolin online order kamagra cheap generic kamagra oral jelly order online ventolin evohaler order online drug interactions strattera wellbutrin. Buy flagyl pills online information about the drug strattera is strattera a scheduled drug can strattera cause a positive drug test. Cheapest kamagra gel ventolin inhaler order online buy flagyl online usa strattera zyrtec drug interactions strattera and drug test kamagra cheap uk buy flagyl online cheap. Buy liquid flagyl online suhagra 100 buy online buy flagyl er online buy online suhagra suhagra 50 mg online order ventolin inhaler online strattera military drug test. Buy flagyl suppositories online order flagyl 500mg online adhd drug strattera new warnings buy generic tricor. Buy flagyl antibiotics online drug test for strattera kamagra oral jelly.rs kamagra kaufen nl drug interactions strattera adderall. Azithromycin buy online usa ventolin hfa order online kamagra kaufen nrw order ventolin hfa online Buy meldonium online uk order flagyl online overnight. Buy flagyl online ireland order ventolin online canada online degree programs for pharmacy technician.
Hoodia p57 buy where can you buy pure hoodia cymbalta online coupon cymbalta 60 mg coupon hoodia buy australia. kamagra fast uk delivery valtrex film tablet 500 mg 42 tb Kamagra oral jelly 5 pack cymbalta coupons online kamagra oral jelly for sale in uk. Where can you buy unique hoodia accutane usa buy hoodia where to buy generic viagra pills online generic valtrex tablet canada generic viagra online valtrex tablets australia. Kamagra for sale sydney kamagra for sale uk kamagra for sale dublin kamagra oral jelly for sale kamagra oral gel 100 mg. Generic viagra online cheapest cymbalta discount coupon card kamagra for sale in dublin cymbalta 20 mg coupon kamagra oral jelly generic4all Kamagra 30 Pills 100mg $121 - $4.03 Per pill. Valtrex tabletten kaufen valtrex film tablet 500 mg 10 tb can you buy unique hoodia in stores valtrex tablets 500mg dosage buy accutane online usa. Online prescription for generic viagra safest generic viagra online buy accutane us generic viagra in canada online. Coupon for cymbalta 30 mg hoodia pills where to buy generic viagra online bestellen buy fluoxetine online uk best rx online net viagra generic. Unique hoodia where to buy best online pharmacy for generic viagra Colchicine tablets over the counter uk kamagra oral jelly sale cymbalta 30 mg coupon reputable online pharmacy generic viagra. Buy canadian generic viagra online where can i buy 100 pure hoodia cheap kamagra for sale uk valtrex tablet 500 mg kamagra uk oral jelly. Valtrex tabletten rezeptfrei online pharmacy for pain meds best online source for generic viagra cymbalta discount coupon. Online pharmacy for generic viagra best online generic viagra site where can you buy hoodia pills.


text:
Portable Telemedicine Device for Medics
Home » Featured Projects » Portable Telemedicine Device for Medics
A robust portable device for monitoring vital signs and providing communications for medics developed with the support of ESA offers a lifeline even in the remotest areas on Earth via satcoms.
The Tempus Pro combines the diagnostic facilities found in standard hospital vital signs monitors with extensive two-way communications, packaged in a compact, robust, highly portable unit that can be tailored to user needs.
The unit has been developed by Remote Diagnostic Technologies (RDT) in the UK, with funding from ESA's Advanced Research in Telecommunications Systems programme.
It has GSM (3G), GPS, wi-fi, bluetooth and ethernet connectivity, and can use available VSAT facilities to exchange voice, video, medical data and GPS positioning.
Various external devices can be connected such as a digital stethoscope, video laryngoscope, contac t temperature sensors and electrocardiogram leads and USB ultrasound probe.
Teaming up with International SOS, a leading company for medical and travel security services, meant that RDT could prove the device works in real-life situations, under rugged conditions, over a six-week period in Algeria and Nigeria.
International SOS provides medical and security services for businesses and large organisations through an extensive network of offices across 35 countries, staffed with clinical personnel. They provide direct medical care, supported by a network of assistance centres staffed by doctors, nurses and specialist call handlers.
"Initially, participants were slightly sceptical of the idea of telemedicine," observed Dr Arnaud Derossi of International SoS. "However, they were quickly won over by the Tempus Pro. By the end of the six-week period, they were extremely positive about the possibilities it offers.
"The key to the unit's success is that it is a fully functional, hospital-grade vital signs monitor. On top of that, it offers very useful communications facilities."
International SoS is active typically in locations served by poor roads with security challenges, and medical evacuations by air can be costly and problematic to arrange.
"Often it is simply a matter of a medical practitioner in the field wanting to get a second opinion from a colleague. At times, a remote consult can even save an unnecessary medical evacuation," added Derossi.
The RDT team has observed that it takes less than an hour for an experienced medical professional to learn the basic functions.
For non-medical users, RDT offers a less advanced model, the Tempus IC, which was developed during an earlier ARTES project that ran from 2006 to 2008. "The Tempus IC has been a very successful product," says RDT's project manager, Mark Williams. "We've sold more than a thousand units."
The Tempus IC is typically bought by airlines and cruise ships, where it can be used in medical emergencies by non-medical staff. According to Williams, there is great interest in the new unit, which is designed expressly for medical professionals, both in the civilian and in the military world.
"What made the Amazon project remarkable," says ESA's Technical Officer, Arnaud Runge, "was that it didn't concern just the development of a prototype – the Tempus Pro is a fully-certified medical device which has been validated with an end-to-end operational service.
"ESA is highly active in health and life sciences, supporting some such 160 projects during the past 10 years, easily half of which have been telemedicine applications."
AMAZON Project Web Page
Video: Portable Telemedicine Device
Remote Diagnostic Technologies
11 March 2014 - Last updated at 11 March 2014 - 09:49

redpajama_set_name: RedPajamaCommonCrawl


Close iMovie and in the Finder, open the iMovie project folder. Inside it is the project.mov, a movie with the same name as your project, with the suffix ".mov". Duplicate that file (in the finder you can control click copy - control click paste). Double-click the duplicate to open it in QT Player. (From now on, we'll refer to this as the video layer.)... Extend it to 60 seconds or something longer. Then fill in the credits with all your stuff. Then fill in the credits with all your stuff. The key is to do each line as TAB, TAB, then text.


One shot of gluta drip can remove toxins from your body. As we are exposed to radicals on daily basis therefore a single shot will be able to boost your immune system and make you glow which will define your beauty.
There are several gluta drip manila clinics and that's what we're going to shed some light on today and make you aware of the facts and information related to these gluta drip manila clinics so let's begin.
Having a rating above 4.5 on all social networking sites, this clinic is said to provide affordable and effective services. Their customer's gluta drip manila reviews are always happy with the ambiance, products and services and love to go back there again for more.
the strength of these ingredients that are combined with other antioxidants and nutrients that help in skin perfecting make sure that health benefits and beauty are absorbed quickly when you need them the most.
The properties of these ingredients is the same as above.
It consists of concentrated L-Carnitine for a firmer body and improvement in metabolism.
It is an ideal combination of premium Glutathione, potency of Multivitamins and benefits of vitamin C which are sure to grant a lucent and healthy glow to your skin.
The robust anti-aging properties of Placenta make sure to increase the production of collagen and skin elasticity perfect for that youthful and bright glow.
For more products relating to Gluta manufacturing kindly visit this LINK . You can also chat with us on our social media here in FACEBOOK.
Gluta drip manila includes this spa which is recommended by many people and they claim that the products they use are absolutely safe and effective. Their staff is very accommodating and have smiles on their faces all the time. Moreover, the place is also very clean and homey. All of their staff is approachable and very professional.
Another gluta drip manila offers a variety of services including acne management, thread lift, hangover drip, memory boost and cognitive drip and much more. With a rating of over 4.8 on social media, their customers are highly satisfied with their services.
You can see many gluta drip before and after pictures and decide for yourself if you want to use this service. There are many gluta drip deals and gluta drip promo available in the clinics that you can avail who also offer gluta drip for sale. Many people have given awesome and positive gluta drip reviews that you can read anywhere. Gluta drip clinics are now developing in the manila economy rapidly. Not only manila but also gluta drip Makati is getting famous.
If you are planning to have a beauty product business? Bioskin Is The Best Beaut Product Manufacturer In The Philippines Right Away!
I am planning to open IV drip vitamins business and looking for supplier. Do you supplies IV drips?
Hi Maam, we don't have IV drips po.


text:
Sprint Contests 2001-2002
Top Moments
Rally Car Museum
Cabinet for Rent
Trafic Park for the Kids
Rallying Events
JM Fan Merchandise
Rally Car's Spare Parts
Rally Ireland 30 January – 1 February 2009
My aim was to drive for the position between 1 – 4 and to get a good start of the season but things started to go wrong already a week before the race. I was in a traffic accident during the tests when a car coming towards me hit me heavily. I was able to yield so much that our cars only hit sideways. Luckily there were no casualties although the cars were badly wrecked. The start of the rally was a dream-come-true when I started the race in heavy rain with so called snow tyres without studs while the other drivers used tyres meant for driving on tarmac. I was the fastest on stage 1 beating the second, Urmo Aava, by 17 seconds. On stage 2 everything went wrong when the car hit a puddle, started to slide and I had to straighten it. There was a corner and the car went too much in hitting a rock which broke the left front tyre and suspension. I finished the stage with an empty tyre but on the finish line the left front suspension broke totally. Miikka and I tried to fix it all we could but we were unable to get it strong enough so we had to retire. We were so very disappointed but that's the name of the game! Due to SupeRally we could continue the race on Saturday and Sunday getting more experience and feeling of driving on tarmac. The conditions were very difficult because of heavy rain on Saturday but Sunday was drier. My stage times varied because I tried to drive without any unnecessary risks but I got a few good times, too, which made me feel happy. We finished the rally 14 th and got two manufacturer points.
Invalid Displayed Gallery
JM-Rally Parc Fermé


I wished a brief refresher simply because i'm task searching and this occasionally comes up on interviews because of my profession. I havent needed to use this kind of math for particularly some time although. I dont disagree in spite of the fact that with a few reviewers who've complained that the cloth will never be written for novices. I needed to fight with a lot of the publication and needed to cross over yes chapters greater than as soon as. additionally, there's not approximately adequate clarification of definite key techniques. you're anticipated, for instance, to just accept complex equation is an equation with an influence no more than 2 within which the equation is decided to equivalent 0. but the accompanying workbook used to be a superb aid; in reality, there's barely enough rationalization of every proposal in every one bankruptcy of the workbook that i may virtually suggest paying for the workbook in addition to the most textbook. after all, notwithstanding it used to be a fight every now and then, i used to be capable of train myself algebra from this publication, and used to be capable of take my placement examination with a passing grade. Like another commenters. i purchased this booklet simply because i'm trying to find a brand new activity 2 years after leaving my past one, and was once prompt that I may be able to resolution questions about it. i used to be panicked, simply because i used to be by no means a robust math pupil and have not taken a math classification in additional than 20 years.
This publication offers an creation to the interaction among linear algebra and dynamical platforms in non-stop time and in discrete time. It first stories the independent case for one matrix A through caused dynamical platforms in ℝd and on Grassmannian manifolds. Then the most nonautonomous ways are awarded for which the time dependency of A(t) is given through skew-product flows utilizing periodicity, or topological (chain recurrence) or ergodic houses (invariant measures).
Imperative closure has performed a task in quantity thought and algebraic geometry because the 19th century, yet a contemporary formula of the idea that for beliefs possibly all started with the paintings of Krull and Zariski within the Thirties. It has constructed right into a device for the research of many algebraic and geometric difficulties.
Realizing why smaller or fewer is better Why is better than ? Most people can relate better to smaller numbers. You can picture 4 out of 5 things in your mind more easily than you can picture 48 out of 60 — refer to Figure 3-1 if you don't believe me. indd 41 4/16/10 11:04 AM 42 Part I: Starting Off with the Basics Figure 3-1: Forty-eight of 60, or 4 of 5? You decide. A few more situations may help you get this: ✓ A survey found that 162 out of 198 people preferred Bix Peanut Butter. reduces to , which offers clearer information as far as The fraction the preference for the peanut butter.
Could be . You do the same thing on the top and bottom of the fraction, so you really just divided by 1, which doesn't change the value — just how it looks. Multiplying by one Any number multiplied by 1 equals that number: For any real number n, n × 1 = n. Like division, you can multiply by 1 and change how a fraction looks without changing its value. That is, 4×1=4 –8 · 1 = –8 In the case of fractions, instead of actually using 1, a fraction equal to 1 is used: Using the fractional value for the number 1 allows you to change how fractions look without changing their value.
Story problems are the whole point of doing algebra. Why do algebra unless there's a good reason? Oh, I'm sorry — you may just like to solve algebra equations for the fun alone. " Going through each step and using each tool to play this game is entirely possible. Simplify, factor, solve, check. That's good! Lucky you. It's time to dig in! indd 18 4/16/10 11:01 AM Chapter 2 Assigning Signs: Positive and Negative Numbers In This Chapter ▶ Signing up signed numbers ▶ Using operations you find outside the box ▶ Noting the properties of nothing ▶ Doing algebraic operations on signed numbers ▶ Looking at associative and commutative properties N umbers have many characteristics: They can be big, little, even, odd, whole, fraction, positive, negative, and sometimes cold and indifferent.


Are you looking for other service companies that treat you honestly and with integrity like we do? We here at The Plumbing Detectives have our "Circle Of Trust" that contains companies that we've used ourselves and have a great relationship with. Our customers deserve the best and these folks, like us, deliver.


I have a program on my computer consisting of the exe files,dlls etc is there any way I can convert this into a bootable cd?
I imagine it'd explain in the help files how todo this.
You must also make sure that in your bios the cdrom is selected in the boot sequence before your harddrive.
Which files are they? Why you want to 'convert' them into bootable CD? Or you want to just make any bootable CD with those files burned on it?
Please use MagicISO to [URL=***********.magiciso.com/tutorials/miso-make-bootable-cd.htm]make bootable CD[/URL] from .exe .dll and other type files.
A straightforward Win98 bootdisk or any other that loads real mode CD and mouse drivers etc and Nero is all you need for a basic bootable CD.
If you need to access NTFS partitions there are several applications that are freely available. ReadNTFS springs to mind.
You could go a step further if you like and use Winimage etc to create a 2.88M boot floppy or even further still and create a hard drive boot image. The only limitation with the latter being that the drive must be a couple of percent smaller than the blank CD/DVD.
If you have a floppyless system then [url=********chitchat.at.infoseek.co.jp/vmware/vfd20-041003-2.zip]VFD[/url] will create a virtual floppy of up to 2.88M that you can use.
MagicISO looks remarkably similar to UltraISO. Is it a ripoff?


text:
Das in englischsprachiger Literatur als rite of the single shoe ("Ritual des einzelnen Schuhs", ) bezeichnete Ritual gehörte im 15. und 16. Jahrhundert zum Einführungsritual irischer Clan-Anführer und Häuptlinge.

Überlieferung
In einer anderen Form – und vermutlich ohne Zusammenhang mit dem späteren Ritual – wurde ein einzelner Schuh bereits im frühen Mittelalter bei der Amtseinführung des Königs von Leinster verwendet. In einer im 12. Jahrhundert entstandenen Vita Maedog von Ferns werden die Abgaben der Könige von Leinster an das Kloster Ferns aufgezählt, unter ihnen ein am Tag der Amtseinführung des Königs zu übergebender mit Silber gefüllter Schuh.

Aus dem 15. und 16. Jahrhundert ist die Verwendung eines einzelnen Schuhs in den Amtseinführungsritualen der Clans der Uí Chonchobhair von Síol Muireadaigh, der Uí Néill und der Méig Uidhir bekannt:
 Bei den Einführungen von Feidhlimidh Fionn Ó Conchobhair im Jahr 1461 und von Feidhlimidh Ó Conchobhair Ruadh im Jahr 1488 zog der Anführer des Clans Mac Diarmada als "Königsmacher" dem neuen Herrscher einen Schuh an.
 Die Annalen der vier Meister berichten, dass Conchobhar Ruadh Mag Uidhir (Connor Roe Maguire, † 1625) 1589 in der Auseinandersetzung mit Aodh Mag Uidhir (Hugh Maguire) um die Führung des Clans einen Schuh in Sgiath Gabhra, dem traditionellen Krönungsort, deponierte und so seinen Führungsanspruch öffentlich machte, sich damit aber nicht durchsetzen konnte.
 Während der für das 15. und 16. Jahrhundert belegten Zeremonie der Uí Néill in Tulach Óg warf der Führer des Clans Uí Catháin als wichtigster Vasall eine goldene Sandale über den Ó Néill als neuen Herrscher von Tyrone. Anschließend zog der Führer der Uí Ágáin als Wächter von Tulach Óg dem Herrscher die andere Sandale an. Das Ritual in Tulach Óg wurde 1593 zum letzten Mal durchgeführt.

Die Zeremonie in Tulach Óg wurde im späten 16. Jahrhundert von John Perrot in seiner Chronicle of Ireland festgehalten. Eine zeichnerische Darstellung findet sich auf einer um 1602 entstandenen Landkarte von Ulster von der Hand eines unbekannten Zeichners, vermutlich eine Kopie nach Richard Bartlett. Darauf hält der unmittelbar rechts des Herrschers dargestellte Ó Catháin die Sandale über dessen Kopf. Ein für die Zeremonie verwendeter "Krönungsschuh" aus Messing soll noch in der Mitte des 19. Jahrhunderts existiert haben.

Deutung
Das Ritual wird als Zeichen der engen Bindung zwischen dem jeweiligen Häuptling und seinem wichtigsten Vasallen gedeutet. Sowohl für die Position des Ó Conchobhair als auch für die des Ó Néill liegen zeitgenössische bischöfliche Schreiben vor, die betonen, dass eine Amtseinführung nur unter Beteiligung des jeweiligen Vasallen Rechtskraft erlangte. Damit hing die Legitimität des neuen Anführers maßgeblich von der Stärke seines mächtigsten Gefolgsmannes ab.

Auf einer tieferen Ebene verknüpfte das Ritual die am Übergang vom Mittelalter zur Neuzeit regierenden Häuptlinge mit den halbgöttlichen Königen der irischen Mythologie. In alt- und mittelirischen Texten ist der einzelne Schuh ein Erkennungszeichen des rechtmäßigen Königs. Auch übermenschliche Helden, übernatürliche Besucher in der Welt der Sterblichen, Zauberer und Heilige werden durch ihn offenbart.

Literatur

Anmerkungen

Einzelnachweise

Irland im Mittelalter
Geschichte Irlands in der Frühen Neuzeit
Ritual

redpajama_set_name: RedPajamaWikipedia


text:
Wer bist Du? är musikgruppen Megaherz andra musikalbum, utgivet den 12 juli 1997.

Låtförteckning
 Gott sein – 4:15
 Wer bist Du? – 3:07
 Schlag' zurück – 4:01
 Das Leben – 3:59
 Finsternis – 1:19
 Licht – 3:51
 Negativ – 3:44
 Kopf durch die Wand – 4:31
 Müde – 4:39
 Krone der Schöpfung – 4:05
 Tanzen gehen – 3:23
 Die Gedanken sind frei – 3:59
 Hänschenklein Siebenundneunzig – 2:52
 Wer bist Du? (Noel Pix Mix) – 4:31

Musikalbum 1997
Tyska musikalbum


text:
Nevada Tourism grants $504,230 to rural areas for tourism promotion
CARSON CITY — The Nevada Commission on Tourism (NCOT) has awarded $504,230 in grants to rural Nevada tourism organizations for marketing efforts, Lt. Gov. Mark Hutchison announced Wednesday. Among the awardees are five first-time applicants, as well as cultural and recreational events that drive overnight stays in Nevada communities. "The NCOT grant program supports rural Nevada in its efforts to increase overnight visitation, a vital economic component in so many of our state's communities," Hutchison, NCOT chairman, said. "Programs that drive overnight visitation bring room tax revenue to these rural communities as well as to the state, and NCOT strongly supports rural Nevada in those efforts." NCOT awarded funding to 89 marketing projects designed to increase overnight stays, boost brand awareness and expand marketing programs for communities in rural Nevada. Grants are awarded twice a year, and this recent distribution is the second in fiscal 2017. A total of $1.4 million was awarded in fiscal 2017. "Nevada's rural tourism community continues to demonstrate ingenuity and sound strategy in its efforts to attract overnight visitors, and we're pleased to see an increase in the applicant pool during this latest round of grants," Claudia Vecchio, Nevada Department of Tourism and Cultural Affairs (NDTCA) director, said. NDTCA oversees the Nevada Division of Tourism — popularly known as Travel Nevada — whose advisory body is NCOT. Among the funded projects:
The National Historic California Emigrant Trails Interpretive Center Foundation was awarded $3,500 to help promote Trails Days. The event, set for June 3-4, 2017, will take place at the California Trail Interpretive Center in Elko. Trails Days raises awareness of the vast American migration to the west in the mid-19th century; event promotion will include display ads, videos and brochures.
The Ely Shoshone Tribe, a first-time applicant to the grant program, received $2,500 to promote the Ely Shoshone Tribe Pow Wow, a cultural event that is open to the public and set to run in July. Funds will be spent on creative design for the event's promotional campaign, as well as posters, print advertising and social media.
The Pahrump Valley Museum and Historical Society, another first-time applicant to the grant program, was awarded $2,500 to create a seven- to 10-minute video about the museum that will be shown on Public Broadcasting Service channels in Clark and Washoe counties as well as in southern California and in the Sacramento area.
The White Pine Historic Railroad Foundation received $20,000 to promote the Nevada Northern Railway to a target market of railroad enthusiasts. The historical railway in Ely offers several experiential programs, including the opportunity to work on the railway for a week, that are of interest to this market. Grant funds will be used to promote those programs in such publications as Classic Trains magazine.
The Lake Tahoe Visitors Authority was awarded $6,000 for public relations efforts around the AMGEN professional cycling race; the agency will host the first two days of the 2017 women's race. This high-profile event raises awareness of the Lake Tahoe area as a world-class destination for recreation and elite cycling. To amplify that message, grant funds will be directed toward digital advertising, social media and public relations.
The Tahoe Rim Trail Association was awarded $7,500 to construct the group's new website. The Tahoe Rim Trail, a 165-mile rail around Lake Tahoe, is one of the premiere outdoor destinations in the region and a key contributor to Nevada's outdoor recreation economy — a $14.9 billion business, according to the Outdoor Industry Association. A modern, updated website will help ensure the trail's continued success.
Grants are given to nonprofit entities for tourism marketing projects that will result in overnight stays, ultimately increasing room tax revenue for the state. Grant recipients must provide a 50-50 match in funds or volunteer hours unless a waiver is approved. NCOT distributes the grants in the form of reimbursements to the recipients after the projects are completed and labor and funding details are documented. Funding for the Rural Grants Program does not come from the state's general fund. Travel Nevada receives three-eighths of 1 percent of room tax revenue as its operations budget. A portion of those funds is used to offer the Rural Grants Program. For a complete listing of grant recipients by territory, click here. The Nevada Commission on Tourism (NCOT) advises the Nevada Division of Tourism, which is part of the Nevada Department of Tourism and Cultural Affairs. NCOT is responsible for promoting and marketing Nevada as a travel destination to domestic and international travelers. For more, visit www.travelnevada.biz.


I'm starting to think that it's going to be a while before DCU fans are able to sleep at night with both eyes closed. While the likely upcoming success of James Wan's Aquaman and David F. Sandberg's Shazam! is bound to go a long way in helping ease the tension, it's hard not to wonder how many well-received films it will take before people react to relatively minor information as if the sky is falling. Case in point: the Shazam! reshoots.
I get it. There have been several films in recent years that had highly reported reshoots and ended up being seen as messes when the final product hit theaters. Two of those were DCU films. However the thing that I think most fans often miss is that reshoots typically happen all the time in the film industry. Heck, speaking from my own personal experience on an indie filmmaker level, I can only think of a handful of projects I've worked on where our team didn't opt to invest some extra time and effort in doing reshoots.
I suppose if you wanted to indulge in conspiracy theories, you might argue that Sandberg is lying or that he's contractually obligated to defend the reshoots. It's admittedly a possibility, albeit a ridiculously slim one. So I think the healthiest option for everyone is to take it at face value that these are just regular reshoots like for any other film.
How are you feeling about Shazam! these days? Let us know!


I think Australia might be trying to kill me. I am not sure yet. Sometimes, it seems like Australia loves me and never wants me to leave. Like when the sun shines, and the waves crash and the guy at the counter calls me Sheila (that hasn't actually happened but I trust that it will at some point). But much of the time I think Australia wants me dead.
Why? Because She has waged a campaign of biological warfare against me and my kin. Over the last two weeks, my girls and I have been ravaged by a flu/cold from hell. Fortunately, Mike has been spared. Good thing too. He works for an investment group. Not sure he would survive a physical assault as well. This is cold/flu number 15 since we have arrived.
My theory may be wrong. Maybe Australia is just trying to bolster my wimpy immune system, because it cares; like a drill Sargent trying to beat the weakness out of me. If this is the case, the World Youth Day Pilgrimage to Sydney was like the boot camp finale. The Pilgrims brought with them love, prayers, and germs from around the world. So it could be that this hellish flu is not Australian at all. Maybe this humdinger is Latvian. Hard to say at this point. It is more likely an Asian flu, because it has brought out Chaylee's legendary kung fu skills.
You see, when Chaylee weaned (see Suddenly Seymour) she may have stopped suckling, but her fascination with the human mouth did not cease. She continues to rely upon sticking her hand in my mouth when she is in need of comfort or rest.
Therefore: Chaylee + Illness = root canal for mommy.
Example: Last night I was up from 2am till 4am being worked on. She used some of her traditional moves like the inner cheek scratch and gum claw. I used some of my own defense techniques like the lullaby method and snuggle hold. When these defenses faltered I attempted to hold a pillow in my mouth in hopes that she could not enter . The pillow was no match for her misery and immense need for comfort. She pulled a side maneuver and entered through the gap between fabric and cheek. Exhaustion and empathy prevailed.
It was like negotiating with a little terrorist. Okay, okay, you can put your hand in my mouth, just stay away from my frenulum…and no scratching.
Now that my hair is longer, she has incorporated some new comfort seeking techniques. Namely, twirling my hair and then jamming it into my mouth. This is a favorite of mine. Another meditative strategy she now employs involves a flat hand and a swift jabbing motion aimed directly at the uvula. Any attempts to thwart said strategy is met with rivers of tears and agonizing cries of "momma", cough, cough, sneeze, cough, "momma". How can you say no to a sickly baby dentist. I can't. Especially one as lovely as she.
Word of the Day: Oldies: Parents.
I refuse to use this in a sentence.
Poll: Does anyone else have a child or know of a child that is comforted by the inside of his/her mother's mouth?
Family Fun Fact: Chaylee Pasley DDS is an amazing little girl…it would take ten blogs…even more…to capture her loveliness. She is a delight not just a dentist. Just wanted to let the world know.
Surgeon General Notice: By the time guests arrive in Sydney once again, our family should be completely immune to germs worldwide.


What does 'real time' mean?
All units of measurement have a precise scientific calibration and time is no different. It's probably an attempt to keep order in a world full of chaos.
A second is defined as the time interval required for 9,192,631,770 transitions between two energy states of the cesium atom to take place.
But we don't measure seconds like that; not a lot of us anyway.
For fairly obvious reasons, to measure our seconds or even, minutes, hours and days by how we have filled them, what we have achieved and how that makes us feel. We've also been around long enough now to fully appreciate that time is actually our most precious commodity, so we must spend it wisely.
We're all working longer and harder, some of us chose to and some of us have to. Modern technology allows us to work anywhere and at anytime. We're contactable 24 hours a day and the temptation is to just keep working, dipping into the e-mails, writing up another report, catching up on correspondence we didn't have time to do when we were actually 'working'.
But what's that actually doing to us? We're more stressed, have less time for our families, spend less time outside and are less fit and healthy as a society than we were even 30 years ago.
Everything is fast now; trains, cars, broadband, food. Most things can happen 'while you wait', but you don't wait – because you're still working, because you can.
What would happen if you left your smart phone at home or, even better, turned it off and stepped outside and traded your virtual office for the real world? What if, once you were outside, you walked and found yourself on a bustling street, or a hiking trail going uphill; vibrant and alive with people or in nature - peaceful and alive with birdsong? What if you stopped and took in the view, spoke to someone you've never met before or bought a cup of coffee and sat down to really appreciate it? What if you took your children, family or friends with you to do this?
The truth is that nothing would happen. The world would not fall apart. Deadlines would still be met, work would still get done. The mundane drudgery associated with living would still all take place.
The truth is that you would feel better.
Better for the time you spent away from the pressure of everyday life and work. Better for the time you spent with yourself or with others. Better aligned to cope with going back to it all.
Time is precious, so make it real. Don't be filled with regrets when your allotted quota nears its expiry.


text:
Black Business News | Directory of Black-Owned Businesses
News and more about African American entrepreneurs and Black-owned businesses. Also features a directory of events, publications, organizations, and more.
News, Resources & More For and About Black-Owned Businesses
Home News/ Blog Events Publications Organizations Grants/ Funding
Meet the First African American Entrepreneur to Develop His Own 4G LTE Smart Phone
Freddie Figgers, CEO and founder of Figgers Communications, owns the first and only Black-owned telecommunications firm competing head-on with AT&T, Verizon and T-Mobile
Freddie Figgers is an inventor, computer programmer, software engineer, and entrepreneur. He is the youngest person in history to hold a FCC license, which allows him to operate his own cell phone company. (Watch his amazing story on video!)
Founded in Quincy, Florida, his company, Figgers Communications, has already developed it's soon to be released Figgers F2 4G LTE smart phone. Competing head-on with the new iPhone and other popular smart phones, this elegant device has a 5.7-inch (18:9) JDI full lamination screen that is thin, bright and offers stunning visual effects. The silky technology innovation brings super-fast screen response speed, and has zero night time glare with blue light filter technology that is easy on your eyes.
It has dual camera technology with 13.0 megapixel capacity so that users can get high-quality photos. It also has a 5.0MP assistant camera that can create a sense of depth to your photos.
Even more, the phone's long battery life comes from the power-saving combination of the Super AMOLED display. The device comes with 5V2A quick charge technology, meaning that you can get 80% battery after just 12 minutes of charging.
Not just a phone
Freddie's company does a lot more than just develop and sell smart phones. As a telecommunications company that is directly competing with other major networks, he also offers competitive plans.
The most basic plan is for Seniors and only costs $15/month, but the plan for unlimited talk, text and 4G data is $60/month.
Freddie's background
Freddie was abandoned at birth, but was adopted by a loving family when he was just 2-days old. At the age of 9, he received a broken computer that he was able to fix within a week. When his father developed dementia and would often wonder off and get lost, Freddie designed a smart shoe for his dad with GPS tracking and a two-way communicator.
At the age of 12, he landed his first job as a computer technician working for the city of Quincy, Florida. At the age of 15, he started his own cloud computing services company.
By the time Freddie had turned 24-years old, he had over 80 custom software programs designed, built, and executed. Even more, he had sold every one of them.
All of this led up to him eventually launching Figgers Communications, the only Black-owned (and minority-owned) telecommunications firm in the United States that is competing with major networks like AT&T, Verizon, and T-Mobile. His company owns 4 utility patents and 2 design patents in the industry.
For more details about Figgers Communications, visit www.Figgers.com
For more details about Freddie and/or to watch a 2-minute video about his story, visit www.Figgersinc.com
For media interviews, call his assistant at (786) 716-1883.
Mga etiketa:
Find a Black-Owned Business:
Apparel & Footwear (99)
Beauty & Makeup (19)
Books & Publishing (55)
Construction & Real Estate (55)
Design & Manufacturing (59)
Education & Training (28)
Entertainment & Film (73)
Finance & Investing (138)
Food & Beverages (242)
Hair & Skin Care (109)
Marketing and PR (21)
Professional Services (77)
Technology & Inventions (144)
Tourism & Travel (22)
Transportation & Logistics (33)
Most Popular Stories:
Meet the 25-Year Old Black Lawyer Who Runs Her Own Law Firm
Black-Owned Animation Studio Secures $500K Deal on Shark Tank With Kevin Hart and Mark Cuban
This Black-Owned Salon Does Not Service Women With Straight Hair
Steve Harvey Obtains Licensing Rights to Bring 'Family Feud' to Ghana and South Africa
This Couple Helps Black-Owned Brands Get Their Products into Grocery Stores
Black Entrepreneur Revolutionizes Haircare with Sensitive Skin Friendly Silk and Satin-Lined Headwear
Introducing the First Black Woman-Owned Skateboard Company
Single Black Mom of 2 Buys Her Kids a Vending Machine to Teach Them About Business
Don Cheadle to Produce and Star in Movie About the First Black Millionaire
Woman Who Was Once Evicted From Her Home Just Made $1M in Sales in 40 Minutes
HIGHLY RECOMMENDED:
Send your company's story to all the African American newspapers, magazines, TV and radio stations! Visit BlackPR.com
Get Black Business News In Your Email:
Contact us at: (866) 910-6277/ info@dantelee.com or view our privacy policy
Copyright © 2001-2021 Dante Lee International | All Rights Reserved


This is orange cashew chicken that we eat often around here... I basically use this orange chicken recipe and tweak it for my diet. I sub coconut aminos for the soy sauce and I use tapioca starch instead of cornstarch. I will often leave out the sesame seeds and add raw cashews to saute with the chicken once it is browned. I add the cashews before the sauce. I also add honey instead of brown sugar most of the time, too. This is a very flexible recipe and you can add or leave out any of the spicy ingredients to get the heat or lack of heat you prefer. I serve it over rice and add a vegetable or blend of veggies that I can stir fry to go with it.
I have taken to making my own "takeout" so I can control what goes into it.
Showing off a new tablecloth I picked up at Walmart and a table runner I crocheted that I shared the other day with you guys.
This was inspired and very close to the recipe found at Julia's Album website. It was very Autumn festive and delicious! I think this is going to happen again for Thanksgiving.
My hubby is the grillmaster so I took photos of these delicious pork chops he grilled. I marinated them in a balsamic honey vinaigrette I made. I will do a post on vinaigrettes sometime for you.
I made these gluten free pumpkin donuts for us for a Saturday morning breakfast not too long ago. The recipe is not mine... I got it here at Food By Teva (and yes, I used her version so I could use the Bob's Red Mill Gluten free 1to1 Flour) No one missed the gluten and they did not last long!
I tweaked this to fit my diet but here is my homemade Pumpkin Spice Latte inspired by this site's chart. I used almond milk and I have also used cashew milk and left out the cocoa from this one above. I am super crazy good at shipping almond and cashew milk into a froth but you need to warm the milk slightly and use a good whisk. Coconut milk does really well, too.
This might be one of my favorite things in the world. I made some Paleo Coconut Shrimp, grits and Brussel sprouts on a weekend when I was home all day Saturday by myself. I was playing with food and this just happened! LOL Coconut shrimp is AMAZING and even more so with grits!
Another "takeout" dish I make regularly.... sweet and sour chicken over rice. You can probably see little bits of broccoli in there, too. I love a one bowl meal!
Made a stop this past weekend at my favorite grocery market called Sprouts (yes, even more so than Whole Foods because until the grand opening tomorrow that is all we had in the "health food" realm of full markets) I had a special helper with me.
We just recently got one on the east side of town. There is one on the west side, too. I had not visited the east side one until this past weekend. Their produce is beautiful and always reasonably priced compared to just about everyone in town.
So, there are some highlights from my October cooking adventures.
What are you up to in the kitchen this month?
Do you have any special things you like to make in October?
Do you have a favorite place to shop for foodie stuff?
I'm still eating away on that spaghetti squash/mushroom tetrazinni. Realized I need to not just inventory the freezer, but start really eating out of it. I was digging in the "fish" section (lower shelf, right side in back), only to find I'd been tucking other stuff in that area because there was room. So lamb chops it is for dinner tonight, rather than salmon! Seriously love lamb almost better than a good steak. It's a rare treat, though.
I have some ground lamb in my freezer right now that were bought to make meatballs. I am glad you mentioned lamb. I am pulling that out to make meatballs this weekend. It is a rare treat... I happened to get this on sale or I might not have bought it. I have been clearing out my freezer to make room for new turkeys and maybe a duck soon with the holidays coming. I find all kinds of crazy stuff I forgot I had when I do that. I also store a lot of things like rice and certain grains in there to fight weevils. Weevils are a big thing here. I really need to do a full defrost soon, too.