text:
\section{Introduction}
Toroidal algebras are an important class of algebras in the theory of the extended affine Lie algebras. The first realization of a toroidal algebra appeared as
a vertex representation of the affinized Kac-Moody algebra \cite{F}. The
loop algebraic presentation of toroidal Lie algebras was given by Moody et al.~\cite{MRY},
which shows the similarity with the affine Kac-Moody algebras.
Quantum toroidal algebras appeared in the study of the Langlands reciprocity for algebraic surfaces \cite{GKV}
and was constructed geometrically by Nakajima's quiver varieties \cite{N}. More general quantum toroidal
 Kac-Moody algebras were constructed in \cite{J2} using Drinfeld presentation and vertex representations
 where the quantum Serre relations were found to be closely connected with some nontrivial relations
 of Hall-Littlewood symmetric functions.
The quantum toroidal algebras have been studied
in various contexts: toroidal Schur-Weyl duality
 \cite{VV}, its general representation theory \cite{M},
 vertex representations \cite{S}, McKay correspondence
 \cite{FJW}, toroidal actions on level one
representations \cite{STU}, higher level analogs for quantum affine algebras \cite{TU},
fusion products \cite{H1} and an excellent survey can be found in \cite{H2}.
Recently quantum toroidal algebras have found more interesting rich structures and applications
in \cite{FJM1, FJM2, FJM3}.

Like the theory of affine algebras, quantum toroidal algebras also have twisted analogs. Based on the quantum
general linear algebra action on
the quantum toroidal algebra \cite{GJ}, a quantum Tits-Kantor-Kocher (TKK) algebra was constructed \cite{GJ2} using homogeneous $q$-deformed vertex operators in connection with special unitary Lie
algebras, where the Serre relation was also found to be equivalent to
some combinatorial identity of Hall-Littlewood polynomials.
More recently, a new twisted quantum toroidal algebra of type $A_1$ has also been constructed
as an analog of the quantum TKK algebra \cite{JL}.
On the other hand, a principal quantum affine algebra was constructed \cite{J3}
by deforming the Kac-Moody algebras with an involution. However that algebra is larger than its classical analog
as the Serre relations were not known. In the original classical situation the principal realization of affine Lie
algebras \cite{LW} has played an important role in the theory of vertex operator algebra \cite{FLM}.
In light of quantum homogeneous constructions \cite{FJ, J1}, the quantum principal realization was also expected.
In this paper, we construct a principally graded quantum Kac-Moody algebras and obtain
the associated Serre relations, which will guarantee the finiteness of the representation theory.

The content is organized as follows. Section 2 prepares some background information  regarding twisted quantum toroidal algebras. Section three introduces a twisted quantum Heisenberg algebra associated to the root lattice of the Kac-Moody algebra and then its Fock space representation is constructed. After that, Section four is concerned with proving
the quantum Serre relations of $(-1)$-twisted quantum toroidal algebras.
 \\[0.1cm]

\section{$(-1)$-Twisted quantum toroidal algebras}
We start by recalling the construction
of the principally graded quantum algebras constructed in \cite{J2} and
then state the newly found Serre relations. Let $\mathfrak g$ be the complex finite dimensional simple Lie algebra of a simply laced type. Let $\alpha_{1}, \cdots,\alpha_{l}$ be the simple roots. The normalized invariant bilinear form $(\cdot|\cdot)$ on $\mathfrak g(A)$ satisfies the property that $(\alpha_{i}|\alpha_{j})=a_{ij}$, where $A=(a_{ij})$
is the associated Cartan matrix of $\mathfrak g$.

The twisted quantum toroidal algebra $T_q^-({\mathfrak g})$ is the complex unital associative algebra generated by
\begin{equation*}
 q^{c},\quad h_{im}
\quad x_{in}^{\pm}, ~m\in2\mathbb{Z}+1, ~n\in \mathbb{Z},
\quad i=0,\ldots, l,
\end{equation*}
subject to the relations written in terms of generating series.

Let
\begin{equation}
 x_i^{\pm}(z)=\sum_{n\in \mathbb{Z}}x_{in}^{\pm} z^{-n},
\end{equation}
\begin{equation}
\phi_{i}(z)=\exp\big\{(q^{-1}-q) 2\sum_{m\in2\mathbb{N}-1}^{\infty} h_{i, -m}z^{m}\big\}=\sum_{n\geqslant 0}\phi_{i,-n}z^{n},
\end{equation}
\begin{equation}
\psi_{i}(z)=\exp\big\{(q- q^{-1}) 2\sum_{m\in2\mathbb{N}-1}^{\infty}h_{i, m}z^{-m}\big\}=\sum_{n\geqslant 0}\psi_{i,n}z^{-n},
\end{equation}
where $\mathbb{N}$ is the set of natural numbers $1, 2, \cdots$. Then the relations of the twisted toroidal algebra are the following:
\begin{equation}
[q^c, \phi_{i}(z)]=[q^c, \psi_{i}(z)]=[q^c, x_{i}^{\pm}(z)]=0,
\end{equation}
\begin{equation}
[x_{i}^{\pm}(z), x_{j}^{\pm}(w)]= [x_{i}^{+}(z), x_{j}^{-}(w)]=0, ~~{if}~(\alpha_{i}|\alpha_{j})=0,
\end{equation}
\begin{equation}
(z+w)[x_{i}^{+}(z), x_{j}^{-}(w)]=0, ~~{if}~(\alpha_{i}|\alpha_{j})=-1,
\end{equation}
\begin{equation}
[x_{i}^{+}(z), x_{i}^{-}(w)]= \frac{2(q+q^{-1})}{q-q^{-1}}\Big(\psi_{i}(q^{-c/2}z)\delta(\frac{w}{z}q^{c})
-\phi_{i}(q^{c/2}z)\delta(\frac{w}{z}q^{-c})\Big).
\end{equation}
The Serre relations are given as follows:
 \begin{eqnarray}
\begin{split}
Sym_{z_{1},z_{2}}\Big\{(z_{1}+q^{\mp2}z_{2})(z_{2}-q^{\mp2}z_{1})\Big(z_{2}x_i ^{\pm}(z_1 )x_i ^{\pm}(z_2 )x_j^{\pm} (w)-(z_{1}+z_{2})\\\cdot x_i ^{\pm}(z_1 )x_j^{\pm} (w)x_i ^{\pm}(z_2 )+z_{1}x_j^{\pm} (w)x_i ^{\pm}(z_1 )x_i ^{\pm}(z_2 )\Big)\Big\}=0, \quad if ~(\alpha_{i}|\alpha_{j})= -1,
\end{split}
\end{eqnarray}
\begin{eqnarray}
\begin{split}
\sum_{r=0,\sigma \in \mathfrak{S}_{k+1}}^{k+1}
\sigma .\Big(\prod_{m<n}(z_{m}+q^{\mp2}z_{n})(z_{n}-q^{\mp2}z_{m})x_i ^{\pm} (z_1 )x_i ^{\pm}(z_2 )
\cdots x_i ^{\pm}(z_{r})\\ \cdot x_j^{\pm} (w)x_i ^{\pm}(z_{r+1})\cdots x_i ^{\pm}(z_{k+1} )\Big)=0,~if~(\alpha_{i}|\alpha_{j})= -k, ~k\in \mathbb N,~k\geqslant2.
\end{split}
\end{eqnarray}
 Let
 \begin{equation*}
 \displaystyle G_{ij}(x)=\sum_{n=0}^{\infty}G_{n}x^n
 \end{equation*}
 be the Taylor series at $x=0$ of the following functions
 \begin{align}
 G_{ij}(x) &=\frac{q^{(\alpha_{i}|\alpha_{j})}x-1}{q^{(\alpha_{i}|\alpha_{j})}x+1}\frac{x+q^{(\alpha_{i}|\alpha_{j})}}{x-q^{(\alpha_{i}|\alpha_{j})}},\nonumber
\end{align}
 then the relations  of the twisted quantum {toroidal} algebra are expressed in terms of generating series:
\begin{eqnarray}
\phi_{i}(z)\psi_{j}(w)=\psi_{j}(w)\phi_{i}(z)G_{ij}(q^{-c}\frac{z}{w})/G_{ij}(q^{c}\frac{z}{w}),
\end{eqnarray}
\begin{equation}
[\phi_{i}(z),\phi_{j}(w)]=[\psi_{i}(z),\psi_{j}(w)]=0,
\end{equation}
\begin{align}\label{commutator1}
\phi_i(z)x_j^{\pm}(w)\phi_i(z)^{-1}&=x_j^{\pm}(w)G_{ij}(\frac{z}{w}q^{\mp c/2})^{\pm 1},\\ \label{commutator2}
\psi_i(z)x_j^{\pm}(w)\psi_i(z)^{-1}&=x_j^{\pm}(w)G_{ij}(\frac{w}{z}q^{\mp c/2})^{\pm 1},
\end{align}
where $\delta(z)=\sum_{n\in \mathbb{Z}}z^{n}$ is the formal $\delta$-function.\\[-0.1cm]

\section{Fock space representations}

The twisted quantum Heisenberg algebra $U_{q}(\widetilde{h})$ is the associative algebra generated by $a_{i}(m)$ $(m\in  2\mathbb{Z}+1)$ and the central element $\gamma=q^c$ subject to the following relations:
\begin{equation}
[a_{i}(m),a_{j}(n)]=\delta_{m,-n}\frac{[(\alpha_{i}|\alpha_{j})m]}{2m}\frac{\gamma^{m}-\gamma^{-m}}{q-q^{-1}},
\end{equation}
\begin{equation}
[a_{i}(m),\gamma]=[a_{i}(m),\gamma^{-1}]=0.
\end{equation}

Let $Q$ be the root lattice of the simply finite dimensional Lie algebra $\mathfrak g$ of the simply laced type with the standard bilinear form given by $(\alpha_{i}|\alpha_{j})=a_{ij}$.  Then let $\hat{Q}$ be the central extension of the root lattice $Q$ such that
\begin{equation}
1\longrightarrow \mathbb{Z}_{2} \longrightarrow \hat{Q}\longrightarrow Q\longrightarrow 1
\end{equation}
with the commutator
\begin{equation}
aba^{-1}b^{-1}=(-1)^{(\alpha|\beta)},
\end{equation}
where $a$ and $b$ are the preimages of $\alpha$ and $\beta$ respectively.

Let
\begin{equation}
S(\mathcal{H^{-}})=\mathbb{C}\left[a_{i}(n):1 \leqslant i \leqslant {l}, n\in -(2\mathbb{N}-1)\right]
\end{equation}
be the symmetric algebra generated by $a_{i}(n),  1 \leqslant i \leqslant {l},  n \in -(2\mathbb{N}-1)$.
Then $S(\mathcal{H^{-}})$ is an
$\mathcal{H}$-module under the action that
$a_{i}(n)$ acts as a differential operator for $n \in 2\mathbb{N}-1$,
and $a_{i}(n)$ acts as a multiplication operator for $n \in -(2\mathbb{N}-1)$.

Denote the preimage of $\alpha_{i}$ by $a_{i}$, and let $T$ be the $\hat{Q}$-module such that
\begin{equation}
a_{i}a_{j}=(-1)^{(\alpha_{i}|\alpha_{j})}a_{j}a_{i},
\end{equation}
the Fock space is defined as
\begin{equation}
V_{Q}=S(\mathcal{H}^{-})\otimes T.
\end{equation}

Introduce the twisted vertex operators acting on $V_{Q}$ as follows:
\begin{equation}
E_{-}^{\pm}(\alpha_{i},z)=\exp\Big(\pm\sum_{n=1, odd}^{\infty}\frac{2q^{\mp n/2}}{[n]}a_{i}(-n)z^{n}\Big),
\end{equation}
\begin{equation}
E_{+}^{\pm}(\alpha_{i},z)=\exp\Big(\mp\sum_{n=1, odd}^{\infty}\frac{2q^{\mp n/2}}{[n]}a_{i}(n)z^{-n}\Big),
\end{equation}
\begin{equation}
X_{i}^{\pm}(z)=E_{-}^{\pm}(\alpha_{i},z)E_{+}^{\pm}(\alpha_{i},z)a_{i}^{\pm1}=\sum_{n\in \mathbb{Z}}X_{i}^{\pm}(n)z^{-n}.
\end{equation}
Then we have the following result.
\begin{thm}
The space $F$ is a level one module for
the twisted quantum toroidal algebra $T_q^-({\mathfrak g})$ under the action defined by $\gamma\mapsto q$, $h_{im}~\mapsto$ $a_{i}(m)$, and $x^{\pm}_{i,n}$ $\mapsto$ $X_{i}^{\pm}(n)$.
\end{thm}

We will prove the theorem in this and the next section. To compute the operator product expansion we need the following $q$-analogs of series $(z-w)^{r}$ introduced in \cite{J1, J2}. For $r\in\mathbb{C}$, we call
the following $q$-analogs:
\begin{equation}
(a; q)_{\infty}=\Pi_{n=0}^{\infty}(1-aq^{n}),
\end{equation}
\begin{equation}
(1-z)_{q^{2}}^{r}=\frac{(q^{-r+1}z; q^{2})_{\infty}}{(q^{r+1}z; q^{2})_{\infty}}=exp(-\sum_{n=1}^{\infty}\frac{[rn]}{n[n]}z^{n}).
\end{equation}
Similarly the twisted q-analog is defined by \cite{J3}:
\begin{equation}
(\frac{1-z}{1+z})_{q^{2}}^{r}=\frac{(1-z)_{q^{2}}^{r}}{(1+z)_{q^{2}}^{r}}=exp(-\sum_{n\in 2\mathbb{N}-1}\frac{2[rn]}{n[n]}z^{n}).
\end{equation}

The operator product expansions (OPE) for $X_{i}^{\pm}$(z) are given by
\begin{eqnarray}\label{A}
X_{i}^{\pm}(z)X_{j}^{\pm}(w)=\!\!\!\!&&\!\!\!\!:X_{i}^{\pm}(z)X_{j}^{\pm}(w):
\left(\frac{1-q^{\mp1}w/z}{1+q^{\mp1}w/z}\right)_{q^{2}}^{(\alpha_{i}|\alpha_{j})},
\\X_{i}^{\pm}(z)X_{j}^{\mp}(w)=\!\!\!\!&&\!\!\!\!:X_{i}^{\pm}(z)X_{j}^{\mp}(w):
\left(\frac{1+w/z}{1-w/z}\right)_{q^{2}}^{(\alpha_{i}|\alpha_{j})}.
\end{eqnarray}

In particular, when $(\alpha_{i}|\alpha_{j})=-1$,
\begin{eqnarray}
X_{i}^{\pm}(z)X_{j}^{\pm}(w)=\!\!\!\!&&\!\!\!\!:X_{i}^{\pm}(z)X_{j}^{\pm}(w):
\frac{z+q^{\mp1}w}{z-q^{\mp1}w},
\\X_{i}^{\pm}(z)X_{j}^{\mp}(w)=\!\!\!\!&&\!\!\!\!:X_{i}^{\pm}(z)X_{j}^{\mp}(w):
\frac{z-w}{z+w} .
\end{eqnarray}

\begin{lem}\cite{J3} The operators ~$\phi_{i}(zq^{-1/2})=:X_{i}^{+}(zq^{-1})X_{i}^{-}(z):$  ~and ~$\psi_{i}(zq^{1/2})=\\:X_{i}^{+}(zq)X_{i}^{-}(z):$  are given by
\begin{equation}
\phi_{i}(z)=\exp\big\{(q^{-1}-q) 2\sum_{m\in 2\mathbb{N}-1} h_{i, -m}z^{m}\big\}=\sum_{n\geqslant 0}\phi_{i,-n}z^{n},
\end{equation}
\begin{equation}
\psi_{i}(z)=\exp\big\{(q- q^{-1}) 2\sum_{m\in 2\mathbb{N}-1}h_{i, m}z^{-m}\big\}=\sum_{n\geqslant 0}\psi_{i,n}z^{-n}.
\end{equation}
\end{lem}

\section{Twisted quantum Serre relations}

Now we use the vertex representation and quantum vertex operators to prove that the Serre relations are
satisfied by our representation. In the following we treat the ``$+$''-case, as the ``$-$''-case can be proved similarly.

\begin{pro}\label{21}If $(\alpha_{i}|\alpha_{j})= -1$, the ``$+$''-Serre relation can be written as:
\begin{eqnarray}\label{marker}
\begin{split}
Sym_{z_{1},z_{2}}\Big\{(z_{1}+q^{-2}z_{2})(z_{2}-q^{-2}z_{1})\Big(z_{2}X_i^{+} (z_1 )X_i^{+}(z_2 )X_j^{+}(w)\\-(z_{1}+z_{2})X_i^{+}(z_1 )X_j^{+}(w)X_i^{+}(z_2 )+z_{1}X_j^{+}(w)X_i^{+}(z_1 )X_i^{+}(z_2 )\Big)\Big\}=0.
\end{split}
\end{eqnarray}
\end{pro}

\begin{proof}
When $(\alpha_{i}|\alpha_{j})= -1$, the OPEs are
\begin{align*}
X_{i}^{+}(z_{1})X_{j}^{+}(w)&=:X_{i}^{+}(z_{1})X_{j}^{+}(w):\frac{z+q^{-1}w}{z-q^{-1}w},\\
X_{i}^{+}(z_{1})X_{i}^{+}(z_{2})&=:X_{i}^{+}(z_{1})X_{i}^{+}(z_{2}):\frac{z_{1}-z_{2}}{z_{1}+z_{2}}
\cdot\frac{z_{1}-q^{-2}z_{2}}{z_{1}+q^{-2}z_{2}}.
\end{align*}
Thus the bracket inside the left-hand side of Eq. (\ref{marker}) is simplified as
\begin{eqnarray*}
(z_{1}+\!\!\!\!&&\!\!\!\!\!\!q^{-2}z_{2})(z_{2}-q^{-2}z_{1})\Big(z_{2}X_i^{+} (z_1 )X_i^{+} (z_2 )X_j^{+} (w)
\\-(z_{1}+\!\!\!\!\!\!&&\!\!\!\!z_{2})X_i^{+} (z_1 )X_j^{+} (w)X_i^{+} (z_2 )+z_{1}X_j^{+} (w)X_i^{+} (z_1 )X_i^{+}(z_2 )\Big)
\\=\!\!\!\!&&\!\!\!\!:X_i^{+} (z_1 )X_i^{+} (z_2 )X_j^{+} (w):\frac{(z_{1}-z_{2})(z_{1}-q^{-2}z_{2})(z_{2}-q^{-2}z_{1})}{z_{1}+z_{2}}
\\\!\!\!\!&&\!\!\!\!\Big( z_{2}\cdot\frac{z_{1}+q^{-1}w}{z_{1}-q^{-1}w}\cdot\frac{z_{2}+q^{-1}w}{z_{2}-q^{-1}w}
+(z_{1}+z_{2})
\\\!\!\!\!&&\!\!\!\!\cdot \frac{z_{1}+q^{-1}w}{z_{1}-q^{-1}w}\cdot\frac{w+q^{-1}z_{2}}{w-q^{-1}z_{2}}+z_{1}\frac{w+q^{-1}z_{1}}{w-q^{-1}z_{1}}\cdot\frac{w+q^{-1}z_{2}}{w-q^{-1}z_{2}}\Big)
\\=\!\!\!\!&&\!\!\!\!:X_i^{+} (z_1 )X_i^{+} (z_2 )X_j^{+} (w):\frac{z_{1}-z_{2}}{z_{1}+z_{2}}\prod_{i=1}^{2}\frac{(w-q^{-1}z_{i})^{-1}}{(z_{i}-q^{-1}w)}
\\\!\!\!\!&&\!\!\!\!\cdot \Big\{z_{2}(z_{1}+q^{-1}w)(z_{2}+q^{-1}w)(w-q^{-1}z_{1})(w-q^{-1}z_{2})
\\\!\!\!\!&&\!\!\!\!+(z_{1}+z_{2})(z_{1}+q^{-1}w)(z_{2}-q^{-1}w)(w-q^{-1}z_{1})(w+q^{-1}z_{2})
\\\!\!\!\!&&\!\!\!\!+z_{1}(z_{1}-q^{-1}w)(z_{2}-q^{-1}w)(w+q^{-1}z_{1})(w+q^{-1}z_{2})\Big\}.
\end{eqnarray*}
The proposition is proved if the following lemma holds. \end{proof}
\begin{lem}
Let $\mathfrak{S}_{2}$ act on $z_{1},z_{2}~via~\sigma.z_{i}=z_{\sigma(i)}$. Then
\begin{eqnarray*}
\sum_{\sigma\in\mathfrak{S}_{2}}\sigma.\Bigg\{(z_{1}-z_{2})\Big(z_{2}(z_{1}+q^{-1}w)(z_{2}+q^{-1}w)(w-q^{-1}z_{1})(w-q^{-1}z_{2})
\\+(z_{1}+z_{2})(z_{1}+q^{-1}w)(z_{2}-q^{-1}w)(w-q^{-1}z_{1})(w+q^{-1}z_{2})
\\+z_{1}(z_{1}-q^{-1}w)(z_{2}-q^{-1}w)(w+q^{-1}z_{1})(w+q^{-1}z_{2})\Big)\Bigg\}=0.
\end{eqnarray*}
\end{lem}
\begin{proof}
Considering the left-hand side as a polynomial in $w$, we extract the constant term:
\begin{equation*}
\sum \limits_{\sigma  \in \mathfrak{S}_2 }\sigma.(z_{1}-z_{2})\Big(z_{2}q^{-2}z_{1}^{2}z_{2}^{2}+
(z_{1}+z_{2})(-q^{-2}z_{1}^{2}z_{2}^{2})+z_{1}q^{-2}z_{1}^{2}z_{2}^{2}\Big)=0.
\end{equation*}
Similarly the highest coefficient of $w^{4}$ is seen to be zero.
The coefficients of $w$ and $w^3$ are respectively computed as follows.
\begin{align*}
\sum \limits_{\sigma  \in \mathfrak{S}_3 }&\sigma.
(z_{1}-z_{2})\big(z_{2}(z_{1}+z_{2})+(z_{1}+z_{2})(z_{1}-z_{2})-z_{1}(z_{1}+z_{2})
\big)=0,\\
\sum \limits_{\sigma  \in \mathfrak{S}_3 }&\sigma.
(z_{1}-z_{2})\big[(z_{1}+z_{2})((q^{2}+q^{-2})z_{1}z_{2}-(z_{1}+z_{2})^{2})
\\&\qquad+(z_{1}+z_{2})((q^{2}+q^{-2})z_{1}z_{2}+(z_{1}-z_{2})^{2})
\big]=0.
\end{align*}
Thus Lemma 1 is proved.
\end{proof}

Until now, we have discussed the Serre relation
with $(\alpha_{i}|\alpha_{j})= -1$. Next we consider the general situation when $(\alpha_{i}|\alpha_{j})= -k,~k\in \mathbb{N}, ~k\geqslant2$. The twisted quantum Serre relations are proved generally
as follows.

\begin{pro}\label{32} If $(\alpha_{i}|\alpha_{j})= -k, ~k\in \mathbb{N},~k\geqslant2$, then
\begin{eqnarray}\label{B}
\begin{split}
\sum_{r=0,\sigma \in \mathfrak{S}_{k+1}}^{k+1}\sigma .\Big(\prod_{m<n}(z_{m}+q^{-2}z_{n})(z_{n}-q^{-2}z_{m})X_i^{+} (z_1 )X_i^{+}(z_2 )
\cdots \\ \cdot X_{i}^{+}(z_{r}) X_j^{+} (w)X_i^{+} (z_{r+1})\cdots X_i^{+} (z_{k+1} )\Big)=0,
\end{split}
\end{eqnarray}
where the symmetric group $\mathfrak{S}_{k+1}$ acts on the variables $z_1, \cdots, z_{k+1}$.
\end{pro}
\begin{proof} For $(\alpha_{i}|\alpha_{j})= -k, ~k\in \mathbb{N},~k\geqslant2$, from (\ref{A}) it follows that
\begin{equation*}
X_{i}^{+}(z_{1})X_{i}^{+}(z_{2})=:X_{i}^{+}(z_{1})X_{i}^{+}(z_{2}):
\frac{z_{1}-z_{2}}{z_{1}+z_{2}}\cdot\frac{z_{1}-q^{-2}z_{2}}{z_{1}+q^{-2}z_{2}},
\end{equation*}
\begin{equation*}
X_{i}^{+}(z)X_{j}^{+}(w)=:X_{i}^{+}(z)X_{j}^{+}(w):\frac{z+q^{-k}w}{z-q^{-k}w}
\cdot\frac{z+q^{-k+2}w}{z-q^{-k+2}w}\cdots\frac{z+q^{k-2}w}{z-q^{k-2}w}.
\end{equation*}
Let
\begin{equation*}
[z, w; k]_{q^{2}}=(z-w)(z-wq^{2})\cdots(z-wq^{2(k-1)}).
\end{equation*}
Then the left-hand side of (\ref{B}) can be simplified as
\begin{align*}
&\sum_{r=0,\sigma \in \mathfrak{S}_{k+1}}^{k+1}\sigma .\Big(\prod_{m<n}(z_{m}+q^{-2}z_{n})
(z_{n}-q^{-2}z_{m})X_i^{+} (z_1 )\cdots X_{i}^{+}(z_{r})\\
&\qquad\quad X_j^{+} (w)\cdot X_i^{+} (z_{r+1})\cdots X_i^{+} (z_{k+1} )\Big)
\\=&:X_j^{+} (w )\prod_{r=1}^{k+1}X_i^{+} (z_r):
\sum_{\sigma \in \mathfrak{S}_{k+1}}\sigma .\big(\frac{\prod_{m<n}(z_{m}-z_{n})}{\prod_{l=1}^{k+1}(w-q^{-1}z_{l})(z_{l}-q^{-1}w)}\cdot f(z_{1},z_{2},\cdots,z_{k+1})\big)\\
=&:X_j^{+} (w )\prod_{r=1}^{k+1}X_i^{+} (z_r):\frac{\prod_{m<n}(z_{m}-z_{n})}{\prod_{l=1}^{k+1}(w-q^{-1}z_{l})(z_{l}-q^{-1}w)}
\sum_{\sigma \in \mathfrak{S}_{k+1}}sgn(\sigma)\sigma.f(z_{1},z_{2},\cdots,z_{k+1}),
\end{align*}
where we have put (for $k$ even)
\begin{align*}
f&(z_{1},z_{2},\cdots,z_{k+1})\\
&=[w,-z_{1}q^{-k}]_{q^{2}}\cdots[w,-z_{k+1}q^{-k}]_{q^{2}}[z_{1},wq^{-k}]_{q^{2}}\cdots[z_{k+1},wq^{-k}]_{q^{2}}
\\&\quad+[z_{1},-wq^{-k}]_{q^{2}}[w,-z_{2}q^{-k}]_{q^{2}}
\cdots[w,-z_{k+1}q^{-k}]_{q^{2}}\\
&\qquad\qquad\cdot[w,z_{1}q^{-k}]_{q^{2}}[z_{2},wq^{-k}]_{q^{2}}\cdots[z_{k+1},wq^{-k}]_{q^{2}}+\cdots\\
&\quad+[z_{1},-wq^{-k}]_{q^{2}}\cdots[z_{k+1},-wq^{-k}]_{q^{2}}[w,z_{1}q^{-k}]_{q^{2}}\cdots[w,z_{k+1}q^{-k}]_{q^{2}}.
\end{align*}
Here we briefly write
$[z, w; k]_{q^2}$ as $[z, w]_{q^2}$.
Observe that each summand in $f(z_1, \cdots, z_{k+1})$ has at least one symmetry under a transposition.
For example
$$[z_{1},-wq^{-k}]_{q^{2}}[w,-z_{2}q^{-k}]_{q^{2}}\cdots[w,-z_{k+1}q^{-k}]_{q^{2}}
[w,z_{1}q^{-k}]_{q^{2}}[z_{2},wq^{-k}]_{q^{2}}\cdots[z_{k+1},wq^{-k}]_{q^{2}}$$
 is invariant under
switching $z_2$ by $z_3$ when $k\geq 2$. Therefore the antisymmetrizer of this
summand under $\mathfrak S_{k+1}$ is zero. Subsequently $\sum_{\sigma \in \mathfrak{S}_{k+1}}sgn(\sigma)\sigma .f(z_{1},z_{2},\cdots,z_{k+1})=0$, and the Serre relation is proved.
\end{proof}

\bigskip

\centerline{\bf Acknowledgments}

 NJ gratefully acknowledges the support of
Humboldt Foundation, Simons Foundation
grant 198129, and NSFC grant 11271138 during this work.

\bigskip

\bibliographystyle{amsalpha}


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Apple developed HFS (Hierarchical File System) as Macintosh file system format. The format is also used in iPods. It does not have the 4GB file limit but the usage of HFS+ is very limited to Apple device. It does not work on Windows computer natively nor almost all consumer electronics.
NTFS stands for New Technology File System. It is a proprietary file system developed by Microsoft but it has a high compatibility to consumer electronics next to FAT32. It also does not have the 4GB file limit. Macintosh can read NTFS volume but cannot write to it without third party software. There are several TV manufacturers licensed NTFS as one of the file system recognized through the USB port.
FAT32 was an old file system architect that has gain wide popularity. But it becomes obsoleted with device with demanding storage capacity. Back in Windows 95 age, almost no files will be bigger than 4GB. However, with the popularity of videos and HD quality videos, files can easily be larger than 4GB. FAT64 or exFat (Extended File Allocation Table) is developed to replace FAT32. It breaks through the 4GB file limit and is gaining popularity on consumer electronic devices. However, it is still not as universal as FAT32 that is compatible in almost all devices.
Which Format to Use for USB Flash Drive?
Basically, for maximum consumer electronic playback compatibility, FAT32 is the must use format. Therefore, if you have video files over 4GB in size, it is better to break them down into chapters. It also makes it easier to navigate small file instead of fast forwarding in a long video.
If you REALLY cannot break down the file to chapter, NTFS wiil be the next best choice. However, if you are using Macintosh, you might want to consider adding third party software such as NTFS for Mac by Paragon.
Which Format to Use as most compatible to both PC and Mac?
If you need to format a USB drive (or portable USB harddrive) for PC or Mac read/write, exFat would be the best choice as exFat is cross platform. However, older operating system like Windows XP natively will not read exFat formatted without update to SP2. In that case, format the drive in NTFS for Windows only access. However, Mac OS can only READ NTFS files but cannot write to it without adding third party softare. There were also reports on exFAT drive formatted on PC not using the "Default Allocation Unit" value that Mac OS will not mount the drive too.
A TV with a USB port is just a regularly TV with a USB Port. They are not SmartTV. More and more TV now can playback videos on their USB port. However, it is up to the TV manufacturers to add that capability or not. Therefore, you cannot assume your client can view the film on the USB drive foramtted in FAT32 on their big screen TV. There're still many TVs (even new TV sets) still won't playback video files from their USB port.
Our latest invention TakyBox HDMI TV Player powered by TakyBox HTML5 Menu System is a breakthrough method to watch films on any TV with HDMI Port 100% compatible. With the remote control, it can completely replace the need for DVD and BluRay for clients to watch their films on TV.


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'Putin sent tame oligarchs on deniable missions to win bid'
With Russia spying on it, England's 2018 bid team hired its own intelligence experts, who came to suspect foul play. Insight reports
November 30 2014, 12:01am, The Sunday Times
ENGLAND'S bid to host the 2018 World Cup gathered intelligence about suspected corruption in the rival Russian camp in a two-year campaign, a dossier published by parliament has revealed.
Details of dispatches on the activities of Russia's 2018 bid by a network of private intelligence operatives are revealed for the first time in evidence sent to parliament by The Sunday Times.
It reports how England's bid officials were warned that Vladimir Putin, then the Russian prime minister, had "cranked up . . . the operation" months before the secret ballot and was alleged to have tasked a trusted cabal of oligarchs with doing "deniable" deals to win the support of Fifa voters.
The intelligence was gathered from a network of British embassies and private intelligence…

redpajama_set_name: RedPajamaCommonCrawl


"In this instructable i'll show you how to build a very inexpensive wireless switch for Phillips Hue Lights.
These lights needs permanent power supply, wall switches have to be always on.
If you go to bed and turn off the wall switch the light will never shine again if the hue bridge try to switch on the light, for example in the morning to wake up you softly with a "warm light".
You have to buy the Hue Tap or dimmer switches, which are very expensive, especially if you need one for each room.
The solution is the using of an ESP8266. These tiny controllers with inbuild usb adapter are available for less than 3$. For the power supply you only need 2 pcs. AAA battery, by using the DeepSleep modus of the Esp8266 the battery work for a long time.


Anas (radi Allahu anhu) passed by some children and greeted them (by saying Assalamu Alaykum to them). Then he (radi Allahu anhu) said: The Prophet (sal Allahu alaihi wa sallam) used to do the same.
To greet (saying Assalamu Alaykum) small children is an excellent expression of humbleness.
Similarly, a believer is also recommended to greet his wife and children while entering his home, and this will be a sign of his humbleness as well.
The believer is also supposed to greet those who may be less in worldly status than himself (such as a servant etc.). Not doing this may be an indication of pride and arrogance and shows a deviation from the practice of the Prophet (sal Allahu alaihi wa sallam) which is enough to incur the displeasure of Allah.


Members of the Del Ray Business Association present Pork Barrel BBQ's Bill Blackburn, fourth from right, with last year's Heart of Del Ray Award. Online voting is open through Feb. 7 for the 2019 award.
Use your bow to hunt down animals in the best https://games.lol/hunting/">free hunting games for kids online! We got the most thrilling deer, bear, duck and animal video hunting games online for PC!
This is really a high quality content. Very helpful & informative.


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Tom Jones: "Last year was a bloody wake-up call. Without an audience, I can't do anything."
Posted bypaulkirkley 14th May 2021 14th May 2021 Posted inInterviewsTags:Surrounded By Time, Tom Jones
Five years ago, Sir Tom Jones wondered if he would ever sing again. His wife Linda – the childhood sweetheart he'd married when they were both 16 – was dying of lung cancer, and wanted to know what her husband was going to do next.
"I said, 'I don't think I can sing – the bloody words will get stuck in my throat,'" Tom recalls. "And she said, 'Well you must sing, you've got to. Don't fall with me. I'm sorry, I've got to go. I don't want to, but I've got to. But don't let it pull you down. Promise me you won't do that.'
"So I tried," he adds. "I got a rhythm section together and tried a few songs I didn't think I'd be able to do. A therapist that I saw in Los Angeles said, 'What do you think will be the hardest song to sing?' and I said, 'There's a Bob Dylan song, What Good Am I?', which I'd recorded on my Praise and Blame album. She said, 'Well, that's the first one you should do. That's the road to recovery.'"
Fast forward to 2021, and Tom has kept his promise with a new album, Surrounded By Time, and its highly-charged opening track – a cover of Bernice Johnson Reagon's gospel classic I Won't Crumble With You If You Fall that stands as a moving tribute to the love of Tom's life (even if, famously, there has been no shortage of other lovers along the way).
But the record is also a defiant two fingers to the doctors who, when Tom got ill and had to cancel his upcoming tour dates a couple of years ago, advised him it was probably time to pack it in. Red rag, meet bull…
"He said, 'You're 78 years old, you don't really need to do these shows'," says Tom, sounding fighting fit and raring to go after receiving both Covid vaccine jabs. "I said, 'Fuck yeah I do, I've still got an audience, for Christ's sake. I want to go out as long as I possibly can.' I had a urinary infection, and they put me on a drip, when really they could have just given me a fistful of antibiotics and I would have carried on with the tour. It was just a little glitch. No big deal."
By 2019, he was back on the road. "We were sold out all over Europe, and were already selling out for 2020," he says. "I was thinking, 'I'm really looking forward to 2020. I'm alive and well, and singing well, and I turn 80 in June… what could stop me now?'" He gives a throaty, rueful laugh. "It was a bloody wake-up call," he says. "Without an audience, I can't do anything…"
Instead, he's spent most of the past year in his new London flat overlooking the Thames, having recently returned from the States after 40 years. "Linda and I were looking for somewhere to move back to, because that's where our family is – my son and daughter-in-law and my grandkids. And what did we have to stay in LA for? Been there, done that. So I came home. But sadly it was too late for Linda."
Surrounded By Time is the fourth album in a fertile collaboration with producer Ethan Johns that has seen the 100 million-selling Welsh superstar pivoting away from belting out big, dramatic showstoppers like Delilah and What's New Pussycat? in favour a stripped-back, more vulnerable sound inspired by his lifelong love of American gospel, blues and folk. (It is also, incredibly, the first record he's ever made in Wales.)
One track in particular – the late Bobby Cole's I'm Growing Old – is a song Tom's been keeping in his back pocket for half a century. "Bobby came to see me in Vegas and said, 'I've got this song I've just written, what do you think?'" he remembers. "I said, "Well I don't think I'm old enough' – I was about 32, 33 at the time – 'hopefully if I get to 80 I'll do it.' And that's exactly what I did. It's right for now. I am growing old and some of the things in that song do happen. You feel the cold a bit more, you walk a bit slower… Also, I love that line, 'I no longer ponder life'. Because when you're young, you're thinking, 'I wonder what's this and I wonder what's that…' But when you get older, hopefully you've worked out what's happening."
So, at 80, he's got this whole 'life' business figured out? "Yes. I've gone through it. I've been through all the ups and downs of showbusiness. I've had slim years – in the 70s. And then, thank God, I started making some noise again…"
Linda said, 'You must sing, you've got to. Don't let it pull you down. Promise me you won't do that…'
It's nearly 60 years since Pontypridd coal miner's son Thomas Woodward first emerged from the clubs of South Wales and began making some noise with his debut hit, It's Not Unusual. In the years that followed, his baritone voice graced everything from a Bond theme (Thunderball) to his own top-rated US TV show and four decades of Vegas residencies, and he's successfully reinvented himself several times over, finding new audiences collaborating with the likes of The Art of Noise (on a cover of Prince's Kiss) and Mousse T (on 2000 Euro smash Sex Bomb). He's performed for queens and presidents, won Grammies, BRITS and Golden Globes and has a star on the Hollywood Walk of Fame. But the highlight of his career, he says, was receiving his knighthood in 2006.
"There's nothing to touch that," he says. "My parents, they would have loved it. Well, my father not so much, he wasn't a royalist. But my mother would have. There's a picture, which my sister still has, of my grandfather. His name was Albert Rees Jones, he was in the Royal Engineers, and he died in the First World War. So when I got my OBE, the first thing I thought of was my grandfather, and his medals. And then of course to get a knighthood, a miner's son… My god, that's unbelievable.
"I love the Queen," he adds. "I saw her being crowned, and I'm glad I've lived at the same time as her. Looking back over my life, the thing I think more than anything else is: 'My god, I lived in the same time as Muhammad Ali, I lived in the same time as Frank Sinatra, Elvis Presley. These people, these one-offs."
Of course, Tom hasn't just shared the same earth as these people – he moved in their orbit. "Yes, I knew them personally. Muhammad Ali came to see me in Philadelphia, and when he walked into my dressing room he was going 'Tom Jones, Tom Jones'. I said, 'Why are you repeating my name?' and he said 'It's my daughter, she watches your TV show and that's what she says.'"
Elvis became a good friend. Was that… weird? "It got to feel more normal the more I knew him, because he was a normal person," he says. "But when I first met him in '65, I was in Hollywood, doing The Ed Sullivan Show, and he was walking towards me, singing With These Hands, which was my recording. I thought, 'my god, if the boys back home could see me now'. Because when I used to sing in the pubs in Wales, I'd do a lot of Elvis Presley, and I had a feeling I would meet him one day, which no-one believed. So as he was walking towards me, I could see all those fellas in my mind.
"Elvis was envious of me because I'd married my childhood sweetheart," he adds. "Linda used to come up with me to his suite. He gave her a beautiful necklace. He said, 'Tom, you're lucky'. He told me he'd fallen in love when he was young, but it hadn't worked out. It bothered him."
Another regular face back in the day was Donald Trump, whose rise to the White House Tom tackles on recent single Talking Reality Television Blues. "He used to come and see me in Atlantic City," Tom recalls. "He owned three of the hotels I used to sing at, and when he'd come to the shows he'd want the spotlight on him. I'd say, 'the hotel owner is here tonight, ladies and gentlemen, Donald Trump'. I just looked at him as being this playboy business guy. I didn't even know he was interested in politics."
Some would say he still isn't… "Well, there you go. It's debatable."
A rumbling, swampy, six-minute spoken-word cover of Todd Snider's folk-country song, Talking Reality Television Blues follows the evolution of TV up to its natural end point, of "reality killed by a reality star".
"I was there at the beginning of TV," says Tom. "I was ill in bed with TB from '52 to '54 – from when I was 12 until I was 14 – and my mother and father got me a television set, and we watched the Coronation. I watched the moon landing live from trailer on the east coast of America. All those things." (And now, of course, as a judge on The Voice, he's a reality TV star himself.)
Surrounded By Time closes with Tom's take on Terry Callier's Lazarus Man. Does he feel like a man who's risen again? "Definitely, because I've survived," he says. "We did talk about finishing the album with I'm Growing Old, but that that felt a bit negative. So we decided on Lazarus Man, because that's like a new beginning – a man who kicks off his sheets and walks out of his tomb."
So this isn't the final chapter of the Tom Jones story? "No, no," he says. "If push came to shove, if this was the last one, then so be it – it represents me well. But I've still got the fire in my belly. Sir Geraint Evans, the great Welsh bass-baritone, once said to me, 'I see you still have the fire, Tom'. And that was 30 years ago. As long as the fire is in you, you've got to do it."
THIS IS TOM JONES:
Despite being one of Wales' most famous exports, Sir Tom is actually three-quarters English, with family roots in Gloucestershire, Somerset and Wiltshire.
His manager Gordon Mills gave him the stage name Tom Jones to cash in on the success of the 1963 film adaptation of Henry Fielding's novel, starring Albert Finney.
Tom played in Las Vegas for at least one week every year from 1967 to 2011.
In 1969, he was nominated for a Golden Globe Best Actor award – for playing himself in his TV show, This Is Tom Jones. He has also appeared in The Simpsons and Tim Burton's Mars Attacks.
He celebrated his 80th birthday last June with a family gathering at his grandson's house in Henley-on-Thames. "I had a few drinks and it was lovely."
Surrounded by Time is out now.
A version of this article was first published in Waitrose Weekend on April 22, 2021.
Don't leave me this way: the Reverend Richard Coles on the madness of grief
Stephen Mangan: "What is comedy? It's laughing at someone else's pain."


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This award was generously made possible by Anonymous.
In the words of The New Yorker, "With his winning of the Pulitzer Prize for the little match girl passion (one of the most original and moving scores of recent years), Lang, once a postminimalist enfant terrible, has solidified his standing as an American master."Recent works include the concerto man made for the ensemble So Percussion and a consortium of orchestras, including the BBC Symphony and the Los Angeles Philharmonic; mountain for the Cincinnati Symphony; death speaks for Shara Nova, Bryce Dessner, Nico Muhly, and Owen Pallett, at Carnegie Hall; and the whisper opera for International Contemporary Ensemble, which premiered at the Museum of Contemporary Art in Chicago and at the Mostly Mozart Festival at Lincoln Center.Upcoming highlights include the world premiere staging of his opera anatomy theater, written in collaboration with visualist Mark Dion, at Los Angeles Opera in June 2016; the North American release in December of Academy Award-winning director Paolo Sorrentino's film Youth, for which Lang composed the score; and Cantaloupe Music's release of the world premiere recordings by the Los Angeles Master Chorale of Lang's the national anthems and the choral version of the little match girl passion.Lang is a Professor of Music Composition at the Yale School of Music and is co-founder and co-artistic director of New York's legendary music festival Bang on a Can.
Portrait photo by Peter Serling.
davidlangmusic.com


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Viewing Mendelssohn, Viewing Elijah
April 29 - May 1, 2009 | ASU Tempe campus
From child prodigy to the most celebrated composer of his time: Felix Mendelssohn Bartholdy was hailed as a genius, reviled as a sentimentalist, beloved as a model of assimilated thinking and attacked for his Jewish heritage.
In honor of the 200th anniversary of the birth of one of the greatest composers— and one of the most paradoxical figures—of the Romantic age, Jewish Studies, the Herberger College of Fine Arts School of Music, and Faculty of Religious Studies in the School of Historical, Philosophical and Religious Studies at Arizona State University present "Viewing Mendelssohn, Viewing Elijah: assimilation, interpretation and culture." Mendelssohn and Elijah have been viewed through a variety of cultural and ideological lenses, with often contradictory outcomes. This conference explores this variety of interpretive perspectives, and features a performance of Mendelssohn's oratorio, Elijah, presented by the Arizona State University Symphony Orchestra and Choirs.
download event program booklet
Celebratory events beyond the conference:
Mendelssohn Birthday Concert Faculty Chamber Music Recital
Katzin, 7:30 p.m.
ASU School of Music faculty members Robert Barefield, Nancy Buck, Andrew Campbell, Carole FitzPatrick, Thomas Landschoot, and Robert Spring join guest pianist Laura Melton for a varied program of songs and instrumental chamber works by Mendelssohn.
ASU Chamber Singers with pianists Dian Baker and Eckart Sellheim
Katzin Concert Hall, 7:30 p.m.
A Valentine's Day concert featuring part-songs by Mendelssohn and the Brahms Liebeslieder Walzer (Lovesong Waltzes)
Trio Soleil in Recital
Featuring Mendelssohn's Piano Trio No. 2 in C Minor, opus 66
Tuesday Morning Music and Tea
Kerr Cultural Center, 10:30 a.m.
A program of chamber music by Mendelssohn, featuring his masterpiece, the Octet in E-flat Major, opus 20. Performers include Carole FitzPatrick, Russell Ryan, and Robert Spring.
Vocal Master Class with Joel Revzen of Arizona Opera
Recital Hall, 7:30 p.m.
Lecture: "Elijah in Sight and Sound"
Phoenix Art Museum, 2 p.m.
Free with museum admission. David Schildkret and Celina Chiarello speak about the artworks to be shown during the performance of Elijah later this month.
Latin Dance Pachanga, Marimba Maderas de Comitan
School of Music Courtyard, 6 p.m.
Featuring selections from the "Songs Without Words" arranged as boleros and waltzes.
ASU Gammage, 7:30 p.m.
Pre-concert talk by Joel Revzen of AZ Opera at 6:30 p.m.
ASU Chamber Singers, Symphonic Chorale, Choral Union, and Symphony Orchestra with faculty soloists Carole FitzPatrick, soprano; Judy May, mezzo-soprano; Randall Black, tenor; and Robert Barefield, baritone. David Schildkret, conductor
Mendelssohn's beloved oratorio performed by faculty and students of the ASU School of Music. During the performance, artworks illustrating the story will be projected near the stage in an exhibit curated by the Education Department of the Phoenix Art Museum.
Joel Gereboff
David Schildkret
Hava Tirosh-Samuelson
Sponsors and support
This conference was made free and publicly accessible, thanks to the following:
ASU Jewish Studies
Herberger Institute for Design and the Arts, School of Music, Dance and Theatre
Religious Studies in the School of Historical, Philosophical & Religious Studies
This conference is made possible in part by the generous support of an anonymous donor.


So as of last week, it's official. They had like 10 suit and ties come down to the break room last week with like 4 shifts of mechanics STUFFED in the same room. The announcment was this: 100 mechanics are gone. Even with 3 weeks for older mechanics to op out an early retirement, the chances are at least 80% that I won't have a job anymore.
So, things are probably the worst they have been since high school. Maybe even worse. This might be as low as my life can get. Scratch that, maybe lower depending on how long my unemployment checks last from Uncle Sam.
Needless to say, FA is the last thing on my mind, as well as being social in any form. I'm sorry if I just don't chat with anyone anymore. This is the hardest time in my life, and I'm scared and confused. I might just have to vanish for a while, or not. I don't know. At least you all know why if I do.
I'm open to any help or kind words. Weather it be a job referral, or just an "I'm sorry" or gift art, or donation or whatever. This is my darkest hour, so any love our way would help.


Kokula Krishna Hari K an Indian born in Pondicherry, South India in 1980's is a pure veteran with acquired knowledge in Business Administration, Computer Research and Entrepreneurship. Has a wide range of contacts through various pipelines and channels across the globe either in Academic Network or Industrial Network or Research Network. Carried away with a greater commitment to the consoled projects, he has helped many people to achieve greater laurels.


You lost your Apple TV remote. Maybe it is broken. What should you do now? You could buy a new one, but they're expensive if you have a newer Apple TV with a Siri remote, and would not you rather spend that money on a movie? You may not have a working remote, but you probably already have the next best one in your pocket.
1; you are now a professional.
Setting Up the Remote  If you do not meet these two requirements, go to Option 2. If you have already set up and already set up your iPhone to control your Apple TV, the Apple TV remote will be visible in the Control Center after you've upgraded to iOS 12. To access the Control Center, swipe down from the top right of the screen (iPhone X or later) or swipe up from the bottom of the screen (all other iPhones).
You do not see the control as an option? You probably never set up your iPhone to control your Apple TV, but it's not too late. Open your Settings app, tap Control Center, and choose Customize Controls. On the next page, go to Other Controls and tap (+) next to "Apple TV Remote."
Now swipe down or up the control center depending on your iPhone model. The Apple TV Remote icon appears as a new option. Tap to start. You will be asked to select your Apple TV from the list.
If your Apple TV does not appear, make sure your Apple TV is on, the same Wi-Fi network as your iPhone is displayed on your TV screen and on the computer. If you have turned off an Apple TV, you can turn it back on by unplugging and reconnecting the power cord.
If you selected your Apple TV from the list, look for a four-digit passcode on the Apple TV interface on your TV screen. Enter this number in the prompt of your iPhone. And that's all it takes – your iPhone is now paired with your Apple TV.
You can now use this control on your iPhone as you would with your traditional Siri remote. The "Menu", "Play / Pause" and "Siri" buttons work just like your old Siri remote, as does the Apple TV App / Home button. The space above these buttons is like the "Glass Touch" area of your old remote.
No volume buttons appear on the screen, but the volume is controlled by the hardware buttons on your iPhone. Of course, as with the old Siri remote control, the volume can only be adjusted in this way when it comes directly from the Apple TV, for example when Bluetooth speakers are connected. Otherwise, if you are routing via HDMI, you will need to use your TV, soundbar, etc. to change the volume unless the connected audio / video receiver supports HDMI-CEC.
Siri remote control activates If you select a text box, you can tap your iPhone on the spot without having to scroll through a long list of characters. The remote UI even changes if text selection is available, as you can see from the small keyboard icon in the third image below.
Use the following link to view the Apple TV Remote to install your iPhone from the App Store. After installation, open the app, tap Add Apple TV, and then select your Apple TV from the list. If you do not see your Apple TV here, make sure your Apple TV is turned on, on your TV screen, and on the same Wi-Fi network as your iPhone. If the Apple TV is off, you can turn it back on by unplugging the power cord and reconnecting it.
Your iPhone displays a four-digit passcode. On your Apple TV, use your original remote to go to Settings -> General -> Remote controls, and then select your iPhone at the bottom of the page. If you do not have a working original remote control, some have succeeded in connecting Apple TV to iTunes, erasing it and restoring it with the computer keyboard, but we have not tried it.
"Remotes" menu may take a minute for your Apple TV to recognize your device. Enter the four-digit number displayed on the screen. If the code is not displayed, it probably means that your iPhone is already paired. So try controlling your Apple TV. If this does not work, the pairing process will be repeated.
Once all settings have been made, you can control your Apple TV with some extra features as expected, so it's worth it. The buttons "Menu", "Play / Pause", "Siri" and "Apple TV App / Home" are all there. There are also the new buttons "Forward Jump" and "Backward". The distance above these buttons is just like the "Glass Touch" surface on your Siri remote, if you have a new model of Apple TV.
When you send audio to a Bluetooth speaker with Apple TV Audio, you can control the volume using the physical volume buttons on your iPhone's side of the device. When audio is routed via HDMI, you must adjust the volume with the unit's remote controls, unless the connected audio / video receiver supports HDMI-CEC .
The Apple TV Remote app also allows typing directly on the screen instead of using the clumsy Apple TV interface, which is the main reason for using this app. If you're running both iOS 10.3 or later and tvOS 10.2 or later, you'll be able to access the lyrics and playlists on the Now Playing screen, as well as select chapters, audio, and captions.
If you or a friend opt for an Apple Remote or Siri decide remotely, you do not have to use his lame writing scheme. You can set up your Apple TV to prompt you for input on your iPhone to save time and patience.
As long as your iPhone is connected to the same iCloud account as your Apple TV, you can enter it from your iPhone. Make sure both your iPhone and your Apple TV are on the same Wi-Fi network and that Bluetooth is enabled on your iPhone. Then just select a text box on your Apple TV. Your iPhone will receive a push notification that will allow you to tap on your Apple TV after opening.
<img src = "https://img.gadgethacks.com/img/51/79/63679597759764/0/control-your-apple-tv-with- just-your-iphone.w1456.jpg "alt =" How To Control Your Apple TV With Your iPhone  While Apple does not confirm this on its website anywhere, it seems that this notification knit is only available on models from Apple TV 4K and the fourth 3rd generation and below will not respond when entering text fields, even if paired with the same iCloud account.
This article became Movies & TV on Mobile during Gadget Hacks' annual coverage Read all movies and TV shows on mobile devices.


Rich Whitney, Green Party candidate for Governor of Illinois did well in the heartland of Illinois. In Peoria county, home of the giant multinational corporation Caterpillar, Whtney was awarded 15% of a usually strong union Democratic vote. While just across the river in the industrial towns of Pekin and East Peoria 15% of those voters also approved of Mr. Whitney. In Woodford County, a traditional Republican stronghold, 16% of the voters there turned out for Green Party candidate Whitney. Just north of Woodford County sits Marshall County, a rural farming area of the state where 16% of the votes went to Mr. Whitney. For a Green party candidate to do so well in the heartland of Illinois in such disparate political arenas is an indication of the level of dissatisfaction ordinary Americans have with their government.
In a 2002 Columbia Law School Survey on the public perception of attorneys the only profession thought less honest than lawyers were politicians. It would seem Illinois politicians take pride in that fact. Three of the previous seven governors; George Ryan, Dan Walker and Otto Kerner have all been convicted of white-collar crime. Two of the previous seven were indicted (but not convicted) and the present governor is under investigation. He was reelected. If you are wondering how that could happen you need look no further than his opponent – Judy Topinka. She was a close associate of George Ryan, the previous governor who is now serving time in the Illinois State penitentiary. I think voters have had enough. I know I have.


It can be incredibly easy to drink too much caffeine. One morning cup of coffee turns into two, and then there's the mid-morning coffee break, and maybe one in the afternoon just to get through the work day. Maybe an energy drink before gym or before going out with friends. Then we wonder why it can be so hard to fall asleep.
Which isn't to say coffee isn't a glorious, fantastic gift from the gods, but more than four cups of coffee a day can lead to not only insomnia, but increased nervousness and anxiety, migraines, and irritability. Which is one reason so many people are saving the coffee, and starting their mornings with a mug of fragrant tea instead.
Believe it or not, tea is the most widely consumed beverage in the world after water. Said to have been discovered in China in the 3rd century, tea has been traded and sold around the world ever since. It is a cultural tradition everywhere from Turkey to Ireland, and there are a thousand different ways to drink it. From cakes of strong, fermented black tea to the sweet, chilled, orange blossom-spiced Thai tea, the flavors of tea can vary widely. But its popularity never wanes, and there are good reasons for that.
First of all, let's just get this out of the way, tea is cheap. Even if you, like the average Turkish citizen, drink over 6lbs a year of the stuff, it will still never make the dent on your wallet that a year's worth of six dollar lattes can. But more importantly, it's healthier for you! Not only does tea help keep you hydrated, because it's mostly water after all, it's full of powerful antioxidants that can help you body stay strong against cancer, illness, and early aging. Green tea is also great for your metabolism. And a little bit of honey as a sweetener is way less calories than the cream and sugar so many of us pour in our coffee.
Don't worry, there's still some caffeine in tea. No one is asking you to wake up completely on your own. But with on average half the caffeine of a regular cup of coffee in black tea, you will find yourself overall consuming much less caffeine, which means sleeping better and feeling a lot less restless by the end of the day. In an age, where we're all combatting the daily modern stress of media and work, a little less caffeine can help you stay mindful and in control of your anxiety levels.
If you're a regular strong coffee drinker, try weaning yourself off with a strong morning black tea blend. We also love making jug of iced green tea to keep in the fridge for those mid-day pick-me-ups. You can even add some blueberry or pomegranate juice, for extra youth-preserving antioxidants. It makes a great pre-run or pre-workout drink as well.
The good news, once you start reducing caffeine in your daily routine, your tolerance will adjust. So you can save the coffee for emergency wake-up needs, because it will be extra effective.


slicedbread is the company, sharedo is the product and we combine the best people, technology and ideas to provide our clients with innovative solutions designed to disrupt the norm and positively impact their operations.
We are a privately funded company that is owned by its employees. We have a passion for perfection and work closely with clients to form strong bonds that lead to greater understanding, relationships and smarter solutions.
Founder of slicedbread. 15+ years experience designing and delivering case management solutions across industries including legal, insurance, care and debt.
Founder of slicedbread. 20+ years of software implementation, account management and employment in early phase of software product organisations.
15+ years in technology leadership. Significant experience in architecting and engineering case management solutions.
20+ years of technology implementation experience. Recently VP at Hitachi Consulting with responsibility for delivering £3M services PA.


Minneapolis, MN – Inc. Magazine announces that Scrimmage has made its 37th Annual Inc. 5000 list, the most prestigious ranking of the nation's fastest-growing private companies and a hallmark of entrepreneurial success. The list represents a unique look at the most successful companies within the American economy's most dynamic segment of independent businesses. Microsoft, Dell, Pandora, Timberland, LinkedIn, Yelp, Zillow, and many other well-known names gained their first national exposure, as honorees on the Inc. 5000.
Not only have the companies on the 2018 Inc. 5000 been very competitive within their markets, but the list as a whole shows staggering growth compared with prior lists. The 2018 Inc. 5000 achieved an astounding three-year median growth rate of 171.8 percent. The Inc. 5000's aggregate revenue was $206.1B in 2017, accounting for 664,095 jobs over the past three years.
The annual Inc. 5000 event honoring the companies on the list will be held October 17th-19th, 2018, at the JW Marriott San Antonio Hill Country Resort, in San Antonio, Texas. As always, speakers include some of the greatest innovators and business leaders of our generation. Joel McHale is set to host the Awards Ceremony and Gala Event on Friday evening, where the honorees will be celebrated.
For more information on Inc. and the Inc. 5000 Conference, visit: http://conference.inc.com/.
The 2018 Inc. 5000 is ranked according to percentage revenue growth when comparing 2014 and 2017. To qualify, companies must have been founded and generating revenue by March 31, 2014. They had to be U.S.-based, privately held, for profit, and independent—not subsidiaries or divisions of other companies—as of December 31, 2017. (Since then, a number of companies on the list have gone public or been acquired.) The minimum revenue required for 2014 is $100,000; the minimum for 2017 is $2 million. As always, Inc. reserves the right to decline applicants for subjective reasons.
Founded in 1979 and acquired in 2005 by Mansueto Ventures, Inc. is the only major brand dedicated exclusively to owners and managers of growing private companies, with the aim to deliver real solutions for today's innovative company builders. Inc. took home the National Magazine Award for General Excellence in both 2014 and 2012. Meanwhile, the total monthly audience reach for the brand has been growing significantly, from 2,000,000 in 2010 to more than 18,000,000 today. For more information, visit www.inc.com.


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Music Monday: Mildly Musing Melodies
By Leo Dillinger
Students, Alumni, Faculty, lend me your ears! Yes, it's Monday again. The start of a new week may be tough to get through, but don't be scared. Music Monday is here to make your life a little bit easier. These six tracks have been meticulously chosen for your listening pleasure. So turn on, tune in, and drop out as you let the music sink into the caverns of your mind.
1.) 50 years ago, Beatlemania invaded the United States and changed music forever. This past weekend, fans across the country celebrated the anniversary of the band's first visit to the U.S. by listening to some of the Fab Four's greatest hits from 1964. Ladies and gentlemen, The Beatles!
2.) Today marks the 10th anniversary of Kanye West's album, The College Dropout. One of the most captivating songs on the album for me is "All Falls Down." From the instrumental, to Kanye's narrative-style lyrics, and even the hook sung by Syleena Johnson are the key elements for a hip hop masterpiece. Kanye was also added to this list to hype fans up for when he comes to the Bryce Jordan Center this Thursday.
3.) ICYMI: At the Winter Olympic opening ceremony in Sochi, a Russian Police choir covered Daft Punk's hit single "Get Lucky" and it might be one of the most entertaining things from this past week.
4.) On his final episode of Late Night, Jimmy Fallon sang "The Weight" by The Band as a farewell anthem with accompaniment by the Muppets. Whether or not you are a fan of Fallon, this is a magical cover that is sure to make you smile.
5.) Toronto Mayor Rob Ford declared February 6 "Bob Marley Day" in order to commemorate the life of the reggae icon. Although Marley's birthday was last week, here's some heart-warming reggae vibes to counteract the cold weather outside.
6.) The 1975 is an alternative rock band based out of Manchester, U.K. that has grown in popularity over the past year with the release of their self-titled debut album. Here is a rendition of their most notable single "Chocolate."
Leo Dillinger
Penn State Junior, Print Journalism Major, Minors in English and Sociology, Writer of Arts, Entertainment, News, Tomfoolery and Opinion.
Other posts by Leo
Being and Becoming: Leo Dillinger's Senior Column
Leo Dillinger reflects on his time spent as a Penn State student.
Movin' On 2014 Exceeds Expectations
Hardwell and The Allure of EDM
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Music Monday: Beatles, Kanye West, Bob Marley, and More!
https://onwardstate.com/2014/02/10/music-monday-mildly-musing-melodies/


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You can download my full CV. You can also download my Teaching Statement.
Current and Previous Positions
2018-present: University of St. Gallen, assistant professor at the School of Finance (since August 2019 also executive director of the PhD Program in Finance).
2015-2019: Vilnius University, researcher at the Faculty of Economics and Business Administration (part-time).
2015-2018: Bank of Lithuania, researcher at the Center for Excellence in Finance and Economic Research (until July 2017 Senior Economist, thereafter Principal Economist).
2011-2015: University of Amsterdam and Tinbergen Institute, PhD in economics (thesis title: Behavioral Economics and the Public Sector; advisor: Arthur Schram).
2009-2011: Tinbergen Institute, MPhil in economics.
2003-2009: University of Freiburg, Diplom in mathematics (BSc+MSc equivalent; specialization in mathematical and applied statistics).
2021: Vernon Smith Young Talent Award by the Society for Experimental Finance (for "Intertemporal Prospect Theory", with I. Lampe).
2021: HSG Latsis Prize by Fondation Latsis Internationale (awarded to a researcher in any field below the age of 40 at the University of St. Gallen for significant research contributions).
2019: Vladas Jurgutis Award for significant research works by the Bank of Lithuania (for "The Non-Equivalence of Labour Market Taxes: A Real-Effort Experiment," with A. Schram).
2018: Teaching recognition Best CEMS Course based on student evaluations.
2018: Award for the best publication in the social sciences and humanities at Vilnius University in 2017 (for "The Non-Equivalence of Labour Market Taxes: A Real-Effort Experiment," joint work with A. Schram).
2016: Finalist (top-3) of the Joop Hartog Dissertation Prize for the best PhD thesis defended at the Amsterdam School of Economics in the years 2014 and 2015.
2020-2020: HSG Basic Research Grant (project on decision making; ca. CHF 30 000).
2019-2022: HSG Basic Research Grant (project on learning in social networks, joint with C. Aymanns; ca. CHF 46 000).
2018-2021: HSG Basic Research Grant (project in behavioral finance; ca. CHF 10 000).
2018: HSG FT-50 Success Fee (CHF 10 000).
2017: Funding for laboratory experiments from the Bank of Lithuania and the ERC IBSEN Horizon 2020 project, joint with A. Kopányi-Peuker (€ 10 000).
2012-2015: Research Talent Grant of The Netherlands' Organisation for Scientific Research, joint with A. Schram (ca. € 170 000).
2011-2014: Funding for laboratory experiments from the Research Priority Area Behavioral Economics of the University of Amsterdam (ca. € 20 000).
2010-2011: Tinbergen Institute Scholarship (ca. € 12 000).


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Countries, States, and Cities
Where is Martinique located?
Q: Where is Martinique located?
What Continent is Martinique in?
Martinique is an island.. it isn't on a continent. it's located in the Caribbeans, near South America.
What is the capital of martinique location?
Martinique is an island located in the Caribbean Sea. It is an overseas region of France. The inhabitants of Martinique are French citizens. Paris, the capital city in France, is Martinique's capital city.
The location of mt pelee?
it's located in martinique.
Where is martinique?
Martinique is an island located in the eastern Caribbean Sea, and its population is 402,000 (historical population 2008 estimate)
Where is Martinique located what language do they speak?
french and creole are the languages that they speak in martinique. considering i am doing a project on it i would know!
Where is Fort-de-France located?
There is a Fort De France located on the island of Martinique.
Where is the island of Martinique located?
Martinique is an island in the Lesser Antilles in the eastern Caribbean Sea. It is directly north of St. Lucia and south of Dominica. Martinique is one of the twenty-seven regions of France.
What is the capital of la martique?
Which continent is Martinique located on?
West Indies Martinique is an island. It is in the Caribbean, close to South America. It is part of France. So it's in Europe
What continent is martinque?
Martinique is a French overseas department. It is located in North America.
What is martinques continent?
Martinique is an overseas French department. It is located in North America.
What cities in France begin with the letter F?
Martinique is located in the Caribbean Sea. It is an overseas of France and part of the French Republic. Fort-de-France is a city in Martinique.
Where is mt pelee located?
Northern tip of the french overseas department of Martinique in the Caribbean.
How do you spell Martinique in French?
What is the slogan of Martinique?
La Martinique, fleur des Caraibes (Martinique, flower of the Caribbean).
What island is in the West Indies and is located at 14 degrees north and 61 degrees west?
What island in the West Indies is located at 14 degrees north and 61 degrees west?
The Answer is Martinique
What is the size of Martinique?
The area of Martinique is 1,128 square kilometers. The population of Martinique is 436,000 people.
What is the proper adjective for Martinique?
Martinican is the proper adjective for Martinique.
Is martinique a volcanic island?
Yes. Martinique is volcanic.
Why is Dutch spoken in Martinique?
Martinique does not speak Dutch. For more information about the languages of Martinique, click here.
What is the hotel martinique?
The Hotel Martinique is a company that provides Hotel Listings, Car Rentals and Cruises for Martinique. Martinique is an island in the Caribbean. On the website, you will find everything you need to set up a whole vacation in Martinique.
What is the state flower of Martinique?
The state flower of Martinique is... Anguilla
What is the currency used in Martinique?
Euro is the currency used in Martinique
What is the main imports and exports of Martinique?
what are martinique imports and exports