text:
\section{Introduction}
Let $(\mathcal{L}_+,\mathcal{L}_-,\mathcal{L}_0)$ be the standard
notation for the Conway triple of link diagrams.  The reduced Jones
polynomial $J(\mathcal{L};q)$ of a link (diagram) $\mathcal{L}$
 can be determined by the following skein relation
\begin{align}
q^2J(\mathcal{L}_+;q)-q^{-2}J(\mathcal{L}_-;q)=(q-q^{-1})J(\mathcal{L}_0;q),
\end{align}
and the normalization condition $J(U;q)=1$, where $U$ denotes the
trivial knot, i.e. the unknot throughout this paper. By induction on
the number of the components of the link $\mathcal{L}$. It is easy
to obtain, for an orient link $\mathcal{L}$ with $L$ components,
$J(\mathcal{L};q)\in \mathbb{Z}[q^{\pm 2}]$ \ for  $L$  odd, and
$qJ(\mathcal{L};q)\in \mathbb{Z}[q^{\pm 2}]$\ if $L$ even.
 In particular, for
a knot $\mathcal{K}$ (i.e. a link with only one component), we have
$ J(\mathcal{K};q)\in \mathbb{Z}[q^{\pm 2}].$ The Jones polynomial
$J(\mathcal{L};q)$ can be viewed as the $U_{q}(sl_2)$ quantum group
invariant with the fundamental representations \cite{RT}. More
generally, for a simple complex Lie algebra $\mathfrak{g}$,  T. Le
\cite{Le} showed that,  with a suitable normalization,  the quantum
group $U_{q}(\mathfrak{g})$ invariants with any irreducible
representations lie in the ring $\mathbb{Z}[q^{\pm 2}]$ which is
called the strong integrality (see Theorem 2.2 in \cite{Le}).

In this paper, we formulate a similar strong integrality theorem for
the reduced colored HOMFLYPT invariant defined by a basis in the
full HOMFLY skein of the annulus. We use the results from the HOMFLY
skein theory due to  Morton, Aiston, Lukac etc \cite{Ai,AM,Morton,L}
instead of using the quantum group theory by \cite{Lu,An}. So our
proof is completely different from Le's \cite{Le}.

Let $\mathcal{H}(\mathcal{L};q,a)$ be the framed HOMFLYPT invariant
of $\mathcal{L}$ determined by the local relations as showed in
Figure 1. Let $\mathcal{C}$ be the full HOMFLY skein of the annulus,
we will recall the definition in Section 2. In \cite{HM}, Hadji and
H. R. Morton constructed the basis elements $Q_{\lambda,\mu}$ in the
full skein $\mathcal{C}$. For a link $\mathcal{L}$ with $L$
components, we choose $L$ basis elements
$Q_{\lambda^\alpha,\mu^\alpha}$, $\alpha=1,...,L$ in the skein
$\mathcal{C}$. We construct the satellite link
$\mathcal{L}\star\otimes_{\alpha=1}^L
Q_{\lambda^\alpha,\mu^\alpha}$. Then the full colored HOMFLYPT
invariant for a link $\mathcal{L}$ is given by
\begin{align}
&W_{[\lambda^1,\mu^1],[\lambda^2,\mu^2],..,[\lambda^L,\mu^L]}(\mathcal{L};q,a)\\\nonumber
&=q^{-\sum_{\alpha=1}^L(\kappa_{\lambda^\alpha}+\kappa_{\mu^\alpha})w(\mathcal{K}_\alpha)}
a^{-\sum_{\alpha=1}^L(|\lambda^\alpha|+|\mu^\alpha|)w(\mathcal{K}_{\alpha})}
\mathcal{H}(\mathcal{L}\star\otimes_{\alpha=1}^L
Q_{\lambda^\alpha,\mu^\alpha};q,a).
\end{align}
In particular, when all $\mu^\alpha=\emptyset$ for $\alpha=1,..,L$,
$W_{[\lambda^1,\emptyset],[\lambda^2,\emptyset],..,[\lambda^L,\emptyset]}(\mathcal{L};q,a)$
is the ordinary colored HOMFLYPT invariant
$W_{\lambda^1,\lambda^2,..,\lambda^L}(\mathcal{L};q,a)$ studied in
\cite{Zhu,CLPZ}. We refer to Section 2 and 3 for a review of the
HOMFLY skein theory and the definition of the full colored HOMFLYPT
invariant for an oriented link. By the definition,
$W_{[\lambda^1,\mu^1],[\lambda^2,\mu^2],..,[\lambda^L,\mu^L]}(\mathcal{L};q,a)$
is not a polynomial of $q^{\pm 1}, a^{\pm 1}$ in general, because of
the factors $(q^k-q^{-k})$ in the denominator. In order to kill
these factors, we introduce the notion of reduced full colored
HOMFLYPT invariants.

For the case of a knot $\mathcal{K}$, it is natural to define the
reduced full colored HOMFLYPT invariant of $\mathcal{K}$ by
\begin{align}
P_{[\lambda,\mu]}(\mathcal{K};q,a)=\frac{W_{[\lambda,\mu]}(\mathcal{K};q,a)}{W_{[\lambda,\mu]}(U;q,a)}.
\end{align}
Then, we have
\begin{theorem}
For any knot $\mathcal{K}$, $P_{[\lambda,\mu]}(\mathcal{K};q,a)\in
\mathbb{Z}[q^{\pm 2},a^{\pm 2}]$.
\end{theorem}
In fact, Theorem 1.1 is a special case of the following Theorem 1.2.
Theorem 1.1 can also be proved directly by using the integrality
theorem \cite{Morton}(i.e Theorem 1 in \cite{Morton}) and the
Theorem 1.3 in the following for the case of knot.

It seems natural to generalize the definition of the reduced full
colored HOMFLYPT invariants for a general link $\mathcal{L}$ with
$L$ components as follow:
\begin{align}
P_{[\lambda^1,\mu^1],..,[\lambda^L,\mu^L]}(\mathcal{L};q,a)=
\frac{W_{[\lambda^1,\mu^1],..,[\lambda^L,\mu^L]}(\mathcal{L};q,a)}{\prod_{\alpha=1}^LW_{[\lambda^\alpha,\mu^\alpha]}(U;q,a)}.
\end{align}
Unfortunately,
$P_{[\lambda^1,\mu^1],..,[\lambda^L,\mu^L]}(\mathcal{L};q,a)$ does
not lie in the ring $\mathbb{Z}[q^{\pm 1},a^{\pm 1}]$ in general.
The recent works  by S. Nawata et al. \cite{NRZ,GNSSS} motivate us
to define the following  $[\lambda^\alpha,\mu^\alpha]$-reduced
HOMFLYPT invariants for a link $\mathcal{L}$:
\begin{align}
Q_{[\lambda^1,\mu^1],...,[\lambda^L,\mu^L]}^{[\mathcal{\lambda}^\alpha,\mathcal{\mu}^\alpha]}(\mathcal{L};q,a)
=\frac{W_{[\lambda^1,\mu_1],...,[\lambda^L,\mu^L]}(\mathcal{L};q,a)}
{W_{[\lambda^\alpha,\mu^\alpha]}(U;q,a)} \ \ \text{for}\ \
\alpha=1,...,L.
\end{align}
However,
$Q_{[\lambda^1,\mu^1],...,[\lambda^L,\mu^L]}^{[\mathcal{\lambda}^\alpha,\mathcal{\mu}^\alpha]}(\mathcal{L};q,a)$
still contains the factors of the forms $(q^k-q^{-k})$ in
denominator. In order to kill these factors, we add a factor to
$Q_{[\lambda^1,\mu^1],...,[\lambda^L,\mu^L]}^{[\mathcal{\lambda}^\alpha,\mathcal{\mu}^\alpha]}(\mathcal{L};q,a)$,
and introduce the notation of the normalized
$[\lambda^\alpha,\mu^\alpha]$-reduced invariant as
\begin{align}
P_{[\lambda^1,\mu^1],...,[\lambda^L,\mu^L]}^{[\lambda^\alpha,\mu^\alpha]}(\mathcal{L};q,a)&=\prod_{\beta=1,\beta\neq
\alpha}^{L}\prod_{\substack{\rho^\beta,\nu^\beta\\|\lambda^\beta|-|\rho^\beta|=
|\mu^\beta|-|\nu^\beta|}}\prod_{x\in \rho^\beta,\nu^\beta}(a\cdot
q^{cn(x)})(q^{hl(x)}-q^{-hl(x)})\\\nonumber &\cdot
Q_{[\lambda^1,\mu^1],...,[\lambda^L,\mu^L]}^{[\mathcal{\lambda}^\alpha,\mathcal{\mu}^\alpha]}(\mathcal{L};q,a).
\end{align}
See Section 3.1 for the definitions of $cn(x), hl(x)$ in formula
(1.6).

The main goal of this paper is to prove the following:
\begin{theorem}
For any $\alpha=1,...,L$, we have
\begin{align}
P_{[\lambda^1,\mu^1],...,[\lambda^L,\mu^L]}^{[\lambda^\alpha,\mu^\alpha]}(\mathcal{L};q,a)\in
\mathbb{Z}[q^{\pm 2}, a^{\pm 2}].
\end{align}
\end{theorem}
We give some special cases of Theorem 1.2.  For example, when all
the partitions $\mu^\beta= \emptyset$, we let
\begin{align}
P_{\lambda^1,...,\lambda^L}^{\lambda^\alpha}(\mathcal{L};q,a):&=
P_{[\lambda^1,\emptyset],...,[\lambda^L,\emptyset]}^{[\lambda^\alpha,\emptyset]}(\mathcal{L};q,a)\\\nonumber
&=\prod_{\beta=1,\beta\neq \alpha}^{L}\prod_{x\in
\lambda^\beta}(a\cdot
q^{cn(x)})(q^{hl(x)}-q^{-hl(x)})\frac{W_{\lambda^1,...,\lambda^L}(\mathcal{L};q,a)}
{W_{\lambda^\alpha}(U;q,a)}.
\end{align}
Moreover, if we take $\lambda^1=\lambda^2=\cdots =\lambda
^L=(r^\rho)$, by some straight calculations, we have
\begin{align}
\prod_{x\in
(r^\rho)}(q^{hl(x)}-q^{-hl(x)})=\prod_{i=0}^{\rho-1}\frac{\{r+i\}!}{\{i\}!},
\ \prod_{x\in (r^\rho)}(a\cdot
q^{cn(x)})=a^{r\rho}q^{\frac{1}{2}\rho r(r-\rho)},
\end{align}
where $\{n\}!=\prod_{i=1}^n\{i\}$, and $\{i\}=(q^i-q^{-i})$.

Hence,
\begin{align}
P_{(r^\rho),...,(r^\rho)}^{(r^\rho)}(\mathcal{L};q,a)=\left(a^{r\rho}q^{\frac{1}{2}\rho
r(r-\rho)}\prod_{i=0}^{\rho-1}\frac{\{r+i\}!}{\{i\}!}\right)^{L-1}\frac{W_{(r^\rho),...,(r^\rho)}(\mathcal{L};q,a)}
{W_{(r^\rho)}(U;q,a)}.
\end{align}
Theorem 1.2 implies
$P_{(r^\rho),...,(r^\rho)}^{(r^\rho)}(\mathcal{L};q,a)\in
\mathbb{Z}[q^{\pm 2},a^{\pm 2}]$. This is just a statement in
\cite{GNSSS}(see the last paragraph in page 42 in \cite{GNSSS}. In
fact, the definition of $P_{(r^\rho)}^{\text{fin}}(\mathcal{L};a,q)$
given by formula (3.2) in \cite{GNSSS} is equal to
 $(-1)^{r\rho
 (L-1)}P_{(r^\rho),...,(r^\rho)}^{(r^\rho)}(\mathcal{L};q,a)$ ).
According to the work \cite{GNSSS},
$P_{(r^\rho),...,(r^\rho)}^{(r^\rho)}(\mathcal{L};q,a)\in
\mathbb{Z}[q^{\pm 2},a^{\pm 2}]$ suggests that
$P_{(r^\rho),...,(r^\rho)}^{(r^\rho)}(\mathcal{L};q,a)$ may have a
categorification whose homology is finite dimensional. Similarly,
Theorem 1.2 provides us a fact that the categorification of the
general invariant
$P_{[\lambda^1,\mu^1],...,[\lambda^L,\mu^L]}^{[\lambda^\alpha,\mu^\alpha]}(\mathcal{L};q,a)$
has a finite dimensional homology.

In order to prove Theorem 1.2, firstly, we give a simple proof of
the following symmetries for the full colored HOMFLYPT invariants.
\begin{theorem}
For any link $\mathcal{L}$ with $L$-components, we have
\begin{align}
W_{[\lambda^1,\mu^1],..,[\lambda^L,\mu^L]}(\mathcal{L};q^{-1},a)&=(-1)^{\sum_{\alpha=1}^L(|\lambda^\alpha|+|\mu^\alpha|)}
W_{[(\lambda^1)^t,(\mu^1)^t],..,[(\lambda^L)^t,(\mu^L)^t]}(\mathcal{L};q,a),\\
W_{[\lambda^1,\mu^1],..,[\lambda^L,\mu^L]}(\mathcal{L};-q,a)&=(-1)^{\sum_{\alpha=1}^L(|\lambda^\alpha|+|\mu^\alpha|)}
W_{[\lambda^1,\mu^1],..,[\lambda^L,\mu^L]}(\mathcal{L};q,a),
\\
W_{[\lambda^1,\mu^1],..,[\lambda^L,\mu^L]}(\mathcal{L};q,-a)&=(-1)^{\sum_{\alpha=1}^L(|\lambda^\alpha|+|\mu^\alpha|)}
W_{[\lambda^1,\mu^1],..,[\lambda^L,\mu^L]}(\mathcal{L};q,a).
\end{align}
\end{theorem}
\begin{remark}
When all the $\mu^\alpha=\emptyset$, it is well-known that the above
symmetries hold for the ordinary colored HOMFLYPT invariants
$W_{\lambda^1,..,\lambda^L}(\mathcal{L};q,a)$ of a link
$\mathcal{L}$:
\begin{align}
W_{\vec{\lambda}}(\mathcal{L};q^{-1},t)&=(-1)^{\|\vec{\lambda}\|}W_{\vec{\lambda}^{t}}(\mathcal{L};q,t),\\
W_{\vec{\lambda}}(\mathcal{L};-q^{-1},t)&=W_{\vec{\lambda}^t}(\mathcal{L};q,t),\\
W_{\vec{\lambda}}(\mathcal{L};q,-t)&=(-1)^{\|\vec{\lambda}\|}W_{\vec{\lambda}}(\mathcal{L};q,t).
\end{align}
The first proof of these symmetries was given in the paper
\cite{LP1,LP2}, a different proof for (1.14) and (1.15) was given in
\cite{Zhu}. Then, in the joint paper with Q. Chen, K. Liu and P.
Peng \cite{CLPZ}, we provide a new simple proof for
(1.14),(1.15),(1.16). Recent,  D. Tubbenhauer, P. Vaz and P. Wedrich
\cite{TVW} also give a different proof for (1.14). In this paper, we
follow the method used in \cite{CLPZ} to prove Theorem 1.3 in
Section 4.
\end{remark}
Then, by using the integrality theorem in \cite{Morton} (i.e.
Theorem 1.1 in \cite{Morton}) with a slight modification,  we finish
the proof of Theorem 1.2  in Section 5.

\bigskip

{\bf Acknowledgements.} The author appreciate the collaborations
with Qingtao Chen, Kefeng Liu and Pan Peng in this area and many
valuable discussions with them within the past years.

\section{The Skein models}
Given a planar surface $F$, the framed HOMFLY skein $\mathcal{S}(F)$
of $F$ is the $\Lambda$-linear combination of orientated tangles in
$F$, modulo the two local relations as showed in figure \ref{Skein},
where $z=q-q^{-1}$,
\begin{figure}[!htb]
\begin{center}
\includegraphics[width=180 pt]{1.eps}\caption{\label{Skein}Two local relations}
\end{center}
\end{figure}
the coefficient ring $\Lambda=\mathbb{Z}[q^{\pm 1}, a^{\pm 1} ]$
with the elements $q^{k}-q^{-k}$ admitted as denominators for $k\geq
1$. The local relation showed in figure \ref{Removal}
\begin{figure}[!htb]
\begin{center}
\includegraphics[width=100 pt]{2.eps}\caption{\label{Removal}Removal of a null-homotopic closed curve}
\end{center}
\end{figure}
is a consequence of the above relations. It follows that the removal
of a null-homotopic closed curve without crossings is equivalent to
time a scalar $s=\frac{a-a^{-1}}{q-q^{-1}}$.

\subsection{The plane} When $F=\mathbb{R}^2$, it is easy to follow
that every element in $\mathcal{S}(F)$ can be represented as a
scalar in $\Lambda$. For a link $\mathcal{L}$ with a diagram
$D_{\mathcal{L}}$, the resulting scalar $\langle D_{\mathcal{L}}
\rangle \in \Lambda$ is the framed HOMFLYPT polynomial
 of the link $\mathcal{L}$. In the
following, we will also use the notation
$\mathcal{H}(\mathcal{L};q,a)$ to denote the $\langle
D_{\mathcal{L}}\rangle$ for simplicity. In particular, as to the
unknot $U$, we have $\mathcal{H}(U;q,a)=\frac{a-a^{-1}}{q-q^{-1}}$.

\subsection{The annulus}
Let $\mathcal{C}$ be the HOMFLY skein of the annulus, i.e.
$\mathcal{C}=\mathcal{S}(S^1\otimes I)$. $\mathcal{C}$ is a
commutative algebra with the product induced by placing one annulus
outside another. $\mathcal{C}$ is freely generated by the set
$\{A_m: m\in \mathbb{Z}\}$, $A_m$ for $m\neq 0$ is the closure of
the braid $\sigma_{|m|-1}\cdots \sigma_2\sigma_1$, and $A_0$ is the
empty diagram \cite{Turaev2}. It follows that $\mathcal{C}$ contains
two subalgebras $\mathcal{C}_{+}$ and $\mathcal{C}_{-}$ which are
generated by $\{A_m: m\in \mathbb{Z}, m\geq 0\}$ and $\{A_m:m\in
\mathbb{Z}, m\leq 0\}$.  The algebra $\mathcal{C}_+$ is spanned by
the subspace $\mathcal{C}_{n,0}$. There is a good basis
$\{Q_{\lambda}\}$ of $\mathcal{C}_+$ consisting of the closures of
certain idempotents of Hecke algebra $H_{n,0}$ \cite{L}.

In \cite{HM}, R. Hadji and H. Morton constructed the basis elements
$\{Q_{\lambda,\mu}\}$ explicitly for $\mathcal{C}$. We will review
this construction in next section.

\section{Full colored HOMFLYPT invariants}
\subsection{Partitions and symmetric functions}
A partition $\lambda$ is a finite sequence of positive integers
$(\lambda_1,\lambda_2,..)$ such that
$
\lambda_1\geq \lambda_2\geq\cdots
$
The length of $\lambda$ is the total number of parts in $\lambda$
and denoted by $l(\lambda)$. The degree of $\lambda$ is defined by
$
|\lambda|=\sum_{i=1}^{l(\lambda)}\lambda_i.
$
If $|\lambda|=d$, we say $\lambda$ is a partition of $d$ and denoted
as $\lambda\vdash d$. The automorphism group of $\lambda$, denoted
by Aut($\lambda$), contains all the permutations that permute parts
of $\lambda$ by keeping it as a partition. Obviously, Aut($\lambda$)
has the order
$
|\text{Aut}(\lambda)|=\prod_{i=1}^{l(\lambda)}m_i(\lambda)!
$
where $m_i(\lambda)$ denotes the number of times that $i$ occurs in
$\lambda$. We can also write a partition $\lambda$ as
$
\lambda=(1^{m_1(\lambda)}2^{m_2(\lambda)}\cdots).
$
We denote by $\mathcal{P}$ the
set of all the partitions.

Every partition can be identified as a Young diagram. The Young
diagram of $\lambda$ is a graph with $\lambda_i$ boxes on the $i$-th
row for $j=1,2,..,l(\lambda)$, where we have enumerate the rows from
top to bottom and the columns from left to right. The $j$-th box in
the $i$-th row has the coordinates $(i,j)$. The content $cn(x)$ of
the box $x=(i,j)$ is defined to be $j-i$.

Given a partition $\lambda$, we define the conjugate partition
$\lambda^t$ whose Young diagram is the transposed Young diagram of
$\lambda$ which is derived from the Young diagram of $\lambda$ by
reflection in the main diagonal. For the box $x=(i,j)\in \lambda$,
the hook length is defined to be
$hl(x)=\lambda_i+\lambda_j^{t}-i-j+1$.

The following numbers associated with a given partition $\lambda$
are used frequently in this paper: $
z_\lambda=\prod_{j=1}^{l(\lambda)}j^{m_{j}(\lambda)}m_j(\lambda)!, \
\kappa_{\lambda}=\sum_{j=1}^{l(\lambda)}\lambda_j(\lambda_j-2j+1). $
Obviously, $\kappa_\lambda$ is an even number and
$\kappa_\lambda=-\kappa_{\lambda^t}$.

The $m$-th complete symmetric function $h_m$ and elementary
symmetric function are defined by the following two generating
functions
\begin{align}
H(t)=\sum_{m\geq 0}h_mt^m=\prod_{i\geq 1}\frac{1}{(1-x_it)}, \
E(t)=\sum_{m\geq 0}e_mt^m=\prod_{i\geq 1}(1+x_it),
\end{align}
respectively.

The power sum symmetric function of infinite variables
$x=(x_1,..,x_N,..)$ is defined by $ p_{n}(x)=\sum_{i}x_i^n. $ Given
a partition $\lambda$, we define $
p_\lambda(x)=\prod_{j=1}^{l(\lambda)}p_{\lambda_j}(x). $ The Schur
function $s_{\lambda}(x)$ is determined by the Frobenius formula
\begin{align}
s_\lambda(x)=\sum_{|\mu|=|\lambda|}\frac{\chi_{\lambda}(C_\mu)}{z_\mu}p_\mu(x).
\end{align}
where $\chi_\lambda$ is the character of the irreducible
representation of the symmetric group $S_{|\mu|}$ corresponding to
$\lambda$. $C_\mu$ denotes the conjugate class of symmetric group
$S_{|\mu|}$ corresponding to partition $\mu$. The orthogonality of
character formula gives
\begin{align}
\sum_A\frac{\chi_A(C_\mu) \chi_A(C_\nu)}{z_\mu}=\delta_{\mu \nu}.
\end{align}

For $\lambda,\mu,\nu\in \mathcal{P}$, we define the
littlewood-Richardson coefficient $c_{\lambda,\mu}^{\nu}$ as
\begin{align}
s_{\lambda}(x)s_{\mu}(x)=\sum_{\nu}c_{\lambda,\mu}^{\nu}s_{\nu}(x).
\end{align}
It is easy to see that $c_{\lambda,\mu}^{\nu}$ can be expressed by
the characters of symmetric group by using the Frobenius formula
\begin{align}
c_{\lambda,\mu}^{\nu}=\sum_{\rho,\tau}\frac{\chi_{\lambda}(C_{\rho})}{z_{\rho}}\frac{\chi_{\lambda}(C_{\tau})}{z_{\tau}}\chi_{\nu}(C_{\rho\cup
\tau}).
\end{align}

\subsection{Basic elements in $\mathcal{C}$}
Given a permutation $\pi\in S_m$ with the length $l(\pi)$, let
$\omega_\pi$ be the positive permutation braid associated to $\pi$.
We have $l(\pi)=w(\omega_\pi)$, the writhe number of the braid
$\omega_{\pi}$.

 We define the quasi-idempotent element in Hecke alegbra $H_m$,
\begin{align}
a_m=\sum_{\pi\in S_m}q^{l(\pi)}\omega_{\pi}
\end{align}
Let element $h_m\in \mathcal{C}_{m,0}$ be the closure of the
elements $\frac{1}{\alpha_m}a_m\in H_m$, i.e
$h_m=\frac{1}{\alpha_m}\hat{a}_m$. Where $\alpha_m$ is determined by
the equation $a_ma_m=\alpha_m a_m$, it gives
$\alpha_m=q^{m(m-1)/2}\prod_{i=1}^m\frac{q^i-q^{-i}}{q-q^{-1}}$.

The skein $\mathcal{C}_+$ ($\mathcal{C}_-$) is spanned by the
monomials in $\{h_m\}_{m\geq 0}$ ($\{h_k^*\}_{k\geq 0}$). The whole
skein $\mathcal{C}$ is spanned by the monomials in $\{h_m,
h_k^*\}_{m,k\geq 0}$. $\mathcal{C}_+$ can be regarded as the ring of
symmetric functions in variables $x_1,..,x_N,..$ with the
coefficient ring $\Lambda$ \cite{L2}. In this situation,
$\mathcal{C}_{m,0}$ consists of the homogeneous functions of degree
$m$. The power sum $P_m=\sum x_i^m$ are symmetric functions which
can be represented in terms of the complete symmetric functions,
hence $P_m\in \mathcal{C}_{m,0}$. Moreover, the following identity
was first obtained in \cite{Ai}, and see \cite{Mor} for a different
proof:
\begin{align}
[m]P_m=X_m=\sum_{j=0}^{m-1}A_{m-1-j,j}.
\end{align}
where $[m]=\frac{q^m-q^{-m}}{q-q^{-1}}$ and $A_{i,j}$ is the closure
of the braid $\sigma_{i+j}\sigma_{i+j-1}\cdots
\sigma_{j+1}\sigma_{j}^{-1}\cdots \sigma_1^{-1}$. Given a partition
$\mu$, we define
\begin{align}
P_{\mu}=\prod_{i=1}^{l(\mu)}P_{\mu_i}.
\end{align}

Then, in \cite{HM}, Hadji and Morton constructed the basis
$Q_{\lambda,\mu}$ on the whole skein $\mathcal{C}$ as follow. Given
two partitions $\lambda, \mu$ with $l$ and $r$ parts. We first
construct a $(l+r)\times (l+r)$-matrix $M_{\lambda,\mu}$ with
entries in $\{h_m, h_k^*\}_{m,k\in \mathbb{Z}}$ as follows, where we
have let $h_{m}=0$, if $m<0$ and $h_{k}^*=0$ if $k <0$.

\begin{align}
M_{\lambda,\mu}=
\begin{pmatrix}
h_{\mu_{r}}^* & h_{\mu_{r}-1}^* & \cdots & h_{\mu_{r}-r-l+1}^* \\
h_{\mu_{r-1}+1}^* & h_{\mu_{r-1}}^* & \cdots & h_{\mu_{r-1}-r-l}^*\\
\cdot & \cdot & \cdots & \cdot \\
h_{\mu_1+(r-1)}^* & h_{\mu_1+(r-2)}^* & \cdots & h_{\mu_1-l}^*\\
h_{\lambda_1-r} & h_{\lambda_1-(r-1)} & \cdots & h_{\lambda_1+l-1}\\
\cdot & \cdot & \cdots & \cdot \\
h_{\lambda_l-l-r+1} & h_{\lambda_l-s-r+2} & \cdots & h_{\lambda_l}
\end{pmatrix}
\end{align}
It is easy to note that the subscripts of the diagonal entries in
the $h$-rows are the parts $\lambda_1,\lambda_2,...,\lambda_l$ of
$\lambda$ in order, while the subscripts of the diagonal entries in
the $h^*$-rows are the parts $\mu_1,\mu_2,..,\mu_r$ of $\mu$ in
reverse order.

Then, $Q_{\lambda,\mu}$ is defined as the determinant of the matrix
$M_{\lambda,\mu}$, i.e.
\begin{align}
Q_{\lambda,\mu}=\det M_{\lambda,\mu}.
\end{align}
Usually, we  write $Q_{\lambda}=Q_{\lambda,\emptyset}$ and
$Q_{\mu}^*=Q_{\emptyset,\mu}$, we have
\begin{align}
Q_{\lambda,\mu}=\sum_{\sigma,\rho,\nu}(-1)^{|\sigma|}c_{\sigma,\rho}^{\lambda}c_{\sigma^t,\nu}^{\mu}Q_{\rho}
Q_{\nu}^*.
\end{align}
The Frobenius formula (3.2) implies
\begin{align}
Q_{\lambda}=\sum_{\mu}\frac{\chi_{\lambda}(\mu)}{z_{\mu}}P_{\mu}.
\end{align}

\subsection{Full colored HOMFLYPT invariants}
Let $\mathcal{L}$ be a framed link with $L$ components with a fixed
numbering. For diagrams $Q_1,..,Q_L$ in the skein model of annulus
with the positive oriented core $\mathcal{C}$, we define the
decoration of $\mathcal{L}$ with $Q_1,..,Q_L$ as the link
\begin{align}
\mathcal{L}\star \otimes_{\alpha=1}^{L} Q_\alpha
\end{align}
which derived from $\mathcal{L}$ by replacing every annulus
$\mathcal{L}$ by the annulus with the diagram $Q_\alpha$ such that
the orientations of the cores match. Each $Q_\alpha$ has a small
backboard neighborhood in the annulus which makes the decorated link
$\mathcal{L}\otimes_{\alpha=1}^{L}Q_\alpha$ into a framed link.

In particular, when $Q_{\lambda^\alpha,\mu^\alpha}\in
\mathcal{C}_{d_\alpha,t_\alpha}$, where $\lambda^\alpha, \mu^\alpha$
are the partitions of positive integers $d_\alpha$ and $t_\alpha$
respectively, for $\alpha=1,..,L$.

\begin{definition}
The full colored HOMFLYPT invariant of $\mathcal{L}$ is defined as
follow:
\begin{align}
&W_{[\lambda^1,\mu^1],[\lambda^2,\mu^2],..,[\lambda^L,\mu^L]}(\mathcal{L};q,a)\\\nonumber
&=q^{-\sum_{\alpha=1}^L(\kappa_{\lambda^\alpha}+\kappa_{\mu^\alpha})w(\mathcal{K}_\alpha)}
t^{-\sum_{\alpha=1}^L(|\lambda^\alpha|+|\mu^\alpha|)w(\mathcal{K}_{\alpha})}
\mathcal{H}(\mathcal{L}\star\otimes_{\alpha=1}^L
Q_{\lambda^\alpha,\mu^\alpha};q,a).
\end{align}
$W_{[\lambda^1,\mu^1],[\lambda^2,\mu^2],..,[\lambda^L,\mu^L]}(\mathcal{L};q,a)$
is a framing independent invariant of link $\mathcal{L}$. In fact,
by the result in \cite{GT}, the framing factor for $Q_{\lambda,\mu}$
is $q^{\kappa_{\lambda}+\kappa_{\mu}} t^{|\lambda|+|\mu|}$.
\end{definition}

\section{The symmetries}
Before giving the proof of Theorem 1.3, we need an observation which
is an easy consequence of the formula (3.7). We introduce the
notation $\{m\}=q^{m}-q^{-m}$, and for a partition $\lambda\in
\mathcal{P}$, we let
$\{\lambda\}=\prod_{i=1}^{l(\lambda)}\{\lambda_i\}$.
\begin{lemma}
Given any two partitions $\lambda$ and $\mu$, for a knot
$\mathcal{K}$, we have
\begin{align}
\{\lambda\}\{\mu\}\mathcal{H}(\mathcal{K}\star
P_{\lambda}P_{\mu}^*;q,a)\in \mathbb{Z}[(q-q^{-1})^2,a^{\pm 1}].
\end{align}
\end{lemma}
\begin{proof}
By using the skein relations in Figure 1 for framed HOMFLYPT
invariant $\mathcal{H}(\mathcal{L};q,a)$, we have, for a link
$\mathcal{L}$ with $L$ components,
\begin{align}
(q-q^{-1})^L\mathcal{H}(\mathcal{L};q,a) \in
\mathbb{Z}[(q-q^{-1})^2,a^{\pm 1}].
\end{align}
The formula (3.7) gives:
\begin{align}
\{m\}P_m=(q-q^{-1})X_m=(q-q^{-1})\sum_{j=0}^{m-1}A_{m-1-j,j}.
\end{align}
Therefore, by using formula (4.2), we get
\begin{align}
\{m\}\mathcal{H}(\mathcal{K}\star
P_m;q,a)=(q-q^{-1})\sum_{j=0}^{m-1}\mathcal{H}(\mathcal{K}\star
A_{m-1-j,j};q,a)\in \mathbb{Z}[(q-q^{-1})^2,a^{\pm 1}],
\end{align}
since $\mathcal{K}\star A_{m-1-j,j}$ is an one-component link for
$j=0,..,m-1$. Similarly, it is straightforward to obtain the formula
(4.1) in the same way.
\end{proof}

By using the above integrality result, we prove the following
symmetries for the full colored HOMFLYPT invariants.
\begin{theorem}
For any link $\mathcal{L}$ with $L$-components,
\begin{align}
W_{[\lambda^1,\mu^1],..,[\lambda^L,\mu^L]}(\mathcal{L};q^{-1},a)&=(-1)^{\sum_{\alpha=1}^L(|\lambda^\alpha|+|\mu^\alpha|)}
W_{[(\lambda^1)^t,(\mu^1)^t],..,[(\lambda^L)^t,(\mu^L)^t]}(\mathcal{L};q,a),\\
W_{[\lambda^1,\mu^1],..,[\lambda^L,\mu^L]}(\mathcal{L};-q,a)&=(-1)^{\sum_{\alpha=1}^L(|\lambda^\alpha|+|\mu^\alpha|)}
W_{[\lambda^1,\mu^1],..,[\lambda^L,\mu^L]}(\mathcal{L};q,a),
\\
W_{[\lambda^1,\mu^1],..,[\lambda^L,\mu^L]}(\mathcal{L};q,-a)&=(-1)^{\sum_{\alpha=1}^L(|\lambda^\alpha|+|\mu^\alpha|)}
W_{[\lambda^1,\mu^1],..,[\lambda^L,\mu^L]}(\mathcal{L};q,a).
\end{align}
\end{theorem}
\begin{proof}
For convenience, we only provide the proof for the case of a knot
$\mathcal{K}$.  The method can be easily generalized to the case for
a general link. By Lemma 4.1, for any two partitions $\tau$ and
$\delta$, we have,
\begin{align}
\{\tau\}\{\delta\}\mathcal{H}(\mathcal{K}\star
P_{\tau}P_{\delta}^*;q,a)\in \mathbb{Z}[(q-q^{-1})^2,a^{\pm 1}].
\end{align}
Combing with the identity $(\{\tau\}\{\delta\})_{q\rightarrow
-q}=(-1)^{|\tau|+|\delta|}\{\tau\}\{\delta\}$, we obtain
\begin{align}
\mathcal{H}(\mathcal{K}\star
P_{\tau}P_{\delta}^*;-q,a)=(-1)^{|\tau|+|\delta|}\mathcal{H}(\mathcal{K}\star
P_{\tau}P_{\delta}^*;q,a).
\end{align}
by the  formula (3.12),
\begin{align}
Q_{\rho}=\sum_{|\tau|=|\rho|}\frac{\chi_{\rho}(\tau)}{z_{\tau}}P_{\tau},
\
Q_{\nu}^*=\sum_{|\delta|=|\nu|}\frac{\chi_{\nu}(\delta)}{z_{\delta}}P_{\delta}^*.
\end{align}
So we have
\begin{align}
\mathcal{H}(\mathcal{K}\star
Q_{\rho}Q_{\nu}^*;-q,a)=(-1)^{|\rho|+|\nu|}\mathcal{H}(\mathcal{K}\star
Q_{\rho}Q_{\nu}^*;q,a).
\end{align}
By the formula (3.11) for $Q_{\lambda,\mu}$,
\begin{align}
\mathcal{H}(\mathcal{K}\star
Q_{\lambda,\mu};-q,a)&=\sum_{\sigma,\rho,\nu}(-1)^{|\sigma|}c_{\sigma,\rho}^{\lambda}c_{\sigma^t,\nu}^{\mu}\mathcal{H}(\mathcal{K}\star
Q_{\rho}Q_{\nu}^*;-q,a)\\\nonumber
&=\sum_{\sigma,\rho,\nu}(-1)^{|\sigma|}c_{\sigma,\rho}^{\lambda}c_{\sigma^t,\nu}^{\mu}(-1)^{|\rho|+|\nu|}\mathcal{H}(\mathcal{K}\star
Q_{\rho}Q_{\nu}^*;q,a)\\\nonumber
&=(-1)^{|\lambda|+|\mu|}\sum_{\sigma,\rho,\nu}(-1)^{|\sigma|}c_{\sigma,\rho}^{\lambda}c_{\sigma^t,\nu}^{\mu}\mathcal{H}(\mathcal{K}\star
Q_{\rho}Q_{\nu}^*;q,a)\\\nonumber
&=(-1)^{|\lambda|+|\mu|}\mathcal{H}(\mathcal{K}\star
Q_{\lambda,\mu};q,a),
\end{align}
where the third ``=" is from the observation that
$c_{\sigma,\rho}^{\lambda}=0$ if $|\lambda|\neq |\sigma|+|\rho|$.
Therefore, by the definition of the full colored HOMFLYPT invariants
$W_{[\lambda,\mu]}(\mathcal{L};q,a)$ in formula (3.14), we obtain
\begin{align}
W_{[\lambda,\mu]}(\mathcal{K},-q,a)=(-1)^{|\lambda|+|\mu|}W_{[\lambda,\mu]}(\mathcal{K},q,a),
\end{align}
since $\kappa_{\lambda}$ is even for any partition $\lambda$.

Similarly, by formula (4.8), we have
\begin{align}
\mathcal{H}(\mathcal{K}\star
P_{\tau}P_{\delta}^*;q^{-1},a)=(-1)^{l(\tau)+l(\delta)}\mathcal{H}(\mathcal{K}\star
P_{\tau}P_{\delta}^*;q,a).
\end{align}
By using the identities,
\begin{align}
\chi_{\rho^t}(\tau)=(-1)^{|\rho|-l(\tau)}\chi_{\rho}(\tau), \
\chi_{\nu^t}(\delta)=(-1)^{|\nu|-l(\delta)}\chi_{\rho}(\delta),
\end{align}
we have
\begin{align}
\mathcal{H}(\mathcal{K}\star
Q_{\rho}Q_{\nu}^*;q^{-1},a)=(-1)^{|\rho|+|\nu|}\mathcal{H}(\mathcal{K}\star
Q_{\rho^t}Q_{\nu^t}^*;q,a).
\end{align}
By the formula (3.5) for $c_{\sigma,\rho}^{\lambda}$, it is easy to
get
\begin{align}
c_{\sigma,\rho}^{\lambda}=c_{\sigma^t,\rho^t}^{\lambda^t}, \
c_{\sigma^t,\nu}^{\mu}=c_{\sigma,\nu^t}^{\mu^t}.
\end{align}
Therefore,
\begin{align}
\mathcal{H}(\mathcal{K}\star
Q_{\lambda,\mu};q^{-1},a)=(-1)^{|\lambda|+|\mu|}\mathcal{H}(\mathcal{K}\star
Q_{\lambda,\mu};q,a).
\end{align}
i.e.
\begin{align}
W_{[\lambda,\mu]}(\mathcal{K},q^{-1},a)=(-1)^{|\lambda|+|\mu|}W_{[\lambda^t,\mu^t]}(\mathcal{K},q,a),
\end{align}
since $\kappa_\lambda=-\kappa_{\lambda^t}$ and
$\kappa_{\mu}=-\kappa_{\mu^t}$.

Finally, in order to show the last identity (4.7),  by using  the
Lemma 4.3 proved in \cite{CLPZ}, we have (see the formula (4.22) in
\cite{CLPZ}):
\begin{align}
\mathcal{H}(\mathcal{K}\star
P_{\tau}P_{\delta}^*;q,-a)=(-1)^{(|\tau|+|\delta|)(w(\mathcal{K})-1)}\mathcal{H}(\mathcal{K}\star
P_{\tau}P_{\delta}^*;q,a).
\end{align}

Then it is direct to show
\begin{align}
\mathcal{H}(\mathcal{K}\star
Q_{[\lambda,\mu]};q,-a)=(-1)^{(|\lambda|+|\mu|)(w(\mathcal{K})-1)}\mathcal{H}(\mathcal{K}\star
Q_{[\lambda,\mu]};q,a),
\end{align}
i.e.
\begin{align}
W_{[\lambda,\mu]}(\mathcal{K},q,-a)=(-1)^{|\lambda|+|\mu|}W_{[\lambda,\mu]}(\mathcal{K},q,a).
\end{align}

\end{proof}

\section{Proof of the strong integrality}
Let us first consider the case of a knot $\mathcal{K}$. Recall the
definition of the reduced full colored HOMFLYPT invariant
\begin{align}
P_{[\lambda,\mu]}(\mathcal{K};q,a)=\frac{W_{[\lambda,\mu]}(\mathcal{K};q,a)}{W_{[\lambda,\mu]}(U;q,a)}.
\end{align}
The main result in \cite{Morton} (see Theorem 1 in \cite{Morton})
can be written in the following form:
\begin{proposition}[Morton]
For any knot $\mathcal{K}$, $P_{[\lambda,\mu]}(\mathcal{K};q,a)$ is
a 2-variable integer Laurent polynomial, i.e.
$P_{[\lambda,\mu]}(\mathcal{K};q,a)\in \mathbb{Z}[q^{\pm 1},a^{\pm
1}]$.
\end{proposition}

By the formulas (4.6) and (4.7) in Theorem 4.2, we immediately
obtain:
\begin{align}
P_{[\lambda,\mu]}(\mathcal{K};-q,a)=P_{[\lambda,\mu]}(\mathcal{K};q,a),\
P_{[\lambda,\mu]}(\mathcal{K};q,-a)=P_{[\lambda,\mu]}(\mathcal{K};q,a).
\end{align}
Therefore, $P_{[\lambda,\mu]}(\mathcal{K};q,a)$ satisfies the
following strong integrality:
\begin{theorem}
For any knot $\mathcal{K}$, $P_{[\lambda,\mu]}(\mathcal{K};q,a)\in
\mathbb{Z}[q^{\pm 2},a^{\pm 2}]$.
\end{theorem}
In fact, Proposition 5.1 can be generalized slightly to the case of
any link as follow. For a link $\mathcal{L}$ with $L$ components
$\mathcal{K}_1,...,\mathcal{K}_L$, if we only decorate the
$\alpha$-th component $\mathcal{K}_\alpha$ with
$Q_{\lambda^\alpha,\mu^\alpha}$, then its full colored HOMFLYPY
invariant is denoted by
$W_{[\lambda^\alpha,\mu^\alpha]}(\mathcal{L};q,a)$.
 Let
\begin{align}
P_{[\lambda^\alpha,\mu^\alpha]}(\mathcal{L};q,a)=\frac{W_{[\lambda^\alpha,\mu^\alpha]}(\mathcal{L};q,a)}{W_{[\lambda^\alpha,\mu^\alpha]}(U;q,a)},
\end{align}
we also have
\begin{proposition}
$P_{[\lambda^\alpha,\mu^\alpha]}(\mathcal{L};q,a)\in
\mathbb{Z}[q^{\pm 1},a^{\pm 1}]$.
\end{proposition}
\begin{proof}
A slight modification for the proof of Proposition 5.1 presented in
\cite{Morton}( see page 333-334) will give the proof of Proposition
5.3. In fact, given a link $\mathcal{L}$ with $L$ components
$\mathcal{K}_1,...,\mathcal{K}_L$, we denote it by
$\mathcal{L}=\mathcal{K}_1\bigvee \mathcal{K}_2\bigvee \cdots
\bigvee \mathcal{K}_L$. We cut the component $\mathcal{K}_\alpha$
open and get a 1-tangle, so we can draw $\mathcal{L}$ in the annulus
as the closure of this 1-tangle. Decorating $\mathcal{K}_\alpha$
with a diagram $Q_{\alpha}$ gives a diagram $\mathcal{K}_1\bigvee
\cdots\bigvee \mathcal{K}_\alpha \star Q_{\alpha}\bigvee \cdots
\bigvee \mathcal{K}_L$ in $\mathcal{C}$, it induces a linear map
$T_{\mathcal{K}_\alpha}^{\mathcal{L}}: \mathcal{C}\rightarrow
\mathcal{C}$. Refer to the page 332 in \cite{Morton} for this
construction in the case of $\mathcal{K}$. If $Q_\alpha$ is an
eigenvector of $T_{\mathcal{K}_\alpha}^{\mathcal{L}}$ with
eigenvalue $a_{\mathcal{K}_\alpha}^{\mathcal{L}}$. Then
$\mathcal{K}_1\bigvee \cdots\bigvee \mathcal{K}_\alpha \star
Q_{\alpha}\bigvee \cdots \bigvee
\mathcal{K}_L=T_{\mathcal{K}_\alpha}^{\mathcal{L}}(Q_\alpha)=a_{\mathcal{K}_\alpha}^{\mathcal{L}}Q_\alpha=a_{\mathcal{K}_\alpha}^{\mathcal{L}}U\star
Q_\alpha$ implies that
\begin{align}
a_{\mathcal{K}_\alpha}^{\mathcal{L}}=\frac{\mathcal{H}(\mathcal{K}_1\bigvee
\cdots\bigvee \mathcal{K}_\alpha \star Q_{\alpha}\bigvee \cdots
\bigvee \mathcal{K}_L;q,a)}{\mathcal{H}(U\star Q_\alpha;q,a)}.
\end{align}
The eigenvectors of $T_{\mathcal{K}_\alpha}^{\mathcal{L}}$ are given
by $Q_{\lambda,\mu}$ since the map
$T_{\mathcal{K}_\alpha}^{\mathcal{L}}$ commutes with the meridian
map $\phi$, see \cite{HM}. We denote the eigenvalue of
$T_{\mathcal{K}_\alpha}^{\mathcal{L}}$ corresponding to eigenvector
by $a_{\mathcal{K}_\alpha}^{\mathcal{L}}(\lambda,\mu)$, the proof of
Theorem 1 in \cite{Morton}(see page 334) shows that
\begin{align}
a_{\mathcal{K}_\alpha}^{\mathcal{L}}(\lambda,\mu)\in
\mathbb{Z}[q^{\pm 1},a^{\pm 1}].
\end{align}
\end{proof}
Moreover, Proposition 5.3 immediately implies that
\begin{proposition}
For any link $\mathcal{L}$, the full colored HOMFLYPT invariants
$W_{[\lambda^1,\mu^1],...,[\lambda^L,\mu^L]}(\mathcal{L};q,a)$
contains a factor $W_{[\lambda^\alpha,\mu^\alpha]}(U;q,a)$ where
$\alpha$ can be chosen to be $1,...,L$.
\end{proposition}
This phenomenon has been showed in \cite{NRZ} after doing many
detailed calculations for colored HOMFLYPT invariants with symmetric
representations by Chern-Simons theory.

Now, we recall the definition of normalized
$[\lambda^\alpha,\mu^\alpha]$-reduced full colored HOMFLYPT
invariant in Section 1.
\begin{align}
P_{[\lambda^1,\mu^1],...,[\lambda^L,\mu^L]}^{[\lambda^\alpha,\mu^\alpha]}(\mathcal{L};q,a)&=\prod_{\beta=1,\beta\neq
\alpha}^{L}\prod_{\substack{\rho^\beta,\nu^\beta\\|\lambda^\beta|-|\rho^\beta|=
|\mu^\beta|-|\nu^\beta|}}\prod_{x\in \rho^\beta,\nu^\beta}(a\cdot
q^{cn(x)})(q^{hl(x)}-q^{-hl(x)})\\\nonumber &\cdot
Q_{[\lambda^1,\mu^1],...,[\lambda^L,\mu^L]}^{[\mathcal{\lambda}^\alpha,\mathcal{\mu}^\alpha]}(\mathcal{L};q,a).
\end{align}
where
\begin{align}
Q_{[\lambda^1,\mu^1],...,[\lambda^L,\mu^L]}^{[\mathcal{\lambda}^\alpha,\mathcal{\mu}^\alpha]}(\mathcal{L};q,a)
=\frac{W_{[\lambda^1,\mu_1],...,[\lambda^L,\mu^L]}(\mathcal{L};q,a)}
{W_{[\lambda^\alpha,\mu^\alpha]}(U;q,a)} \ \ \text{for}\ \
\alpha=1,...,L.
\end{align}
We have the following strong integrality:
\begin{theorem}
For $\alpha=1,...,L$, we have
\begin{align}
P_{[\lambda^1,\mu^1],...,[\lambda^L,\mu^L]}^{[\lambda^\alpha,\mu^\alpha]}(\mathcal{L};q,a)\in
\mathbb{Z}[q^{\pm 2}, a^{\pm 2}].
\end{align}
\end{theorem}

\begin{proof}

By formula (3.11) and (3.12),
\begin{align}
Q_{\lambda,\mu}&=\sum_{\sigma,\rho,\nu}(-1)^{|\sigma|}c_{\sigma,\rho}^{\lambda}c_{\sigma^t,\nu}^{\mu}Q_{\rho}
Q_{\nu}^*\\\nonumber
&=\sum_{\sigma,\rho,\nu,\tau,\delta}(-1)^{|\sigma|}c_{\sigma,\rho}^{\lambda}c_{\sigma^t,\nu}^{\mu}\frac{\chi_{\rho}(\nu)}{z_\tau}
\frac{\chi_{\nu}(\delta)}{z_\delta}\frac{1}{\{\tau\}\{\delta\}}X_{\tau}X_{\delta}^*.
\end{align}
Therefore,
\begin{align}
\mathcal{\mathcal{L}}\star \otimes_{\beta=1}^{L}
Q_{\lambda^\alpha,\mu^\alpha}& =\mathcal{K}_1\star
Q_{\lambda^1,\mu^1}\bigvee \cdots\bigvee \mathcal{K}_\alpha \star
Q_{\lambda^\alpha,\mu^\alpha}\bigvee \cdots \bigvee
\mathcal{K}_L\star Q_{\lambda^L,\mu^L}\\\nonumber
&=\sum_{\tau^\beta,\delta^\beta, \beta\neq
\alpha}C_{\tau^1,\delta^1,..,\hat{\tau^\alpha},\hat{\delta^\alpha},..,\tau^L,\delta^L}
\prod_{\beta=1,\beta\neq
\alpha}\frac{1}{\{\tau^\beta\}\{\delta^\beta\}}\\\nonumber
&\cdot\mathcal{K}_1\star X_{\tau^1}X^*_{\delta^1}\bigvee
\cdots\bigvee \mathcal{K}_\alpha \star
Q_{[\lambda^\alpha,\mu^\alpha]}\bigvee \cdots \bigvee
\mathcal{K}_L\star X_{\tau^L}X^{*}_{\delta^L},
\end{align}
where
\begin{align}
C_{\tau^1,\delta^1,..,\hat{\tau^\alpha},\hat{\delta^\alpha},..,\tau^L,\delta^L}=
\sum_{\sigma^\beta,\rho^\beta,\nu^\beta,\tau^\tau,\delta^\tau,\beta\neq
\alpha}\prod_{\beta,\beta\neq
\alpha}(-1)^{|\sigma^\beta|}c_{\sigma^\beta,\rho^\beta}^{\lambda^\beta}c_{(\sigma^\beta)^t,\nu^\beta}^{\mu^\beta}
\frac{\chi_{\rho^\beta}(\nu^\beta)}{z_{\tau^\beta}}
\frac{\chi_{\nu^\beta}(\delta^\beta)}{z_{\delta^\beta}},
\end{align}
and $\hat{\tau^\alpha}, \hat{\delta^\alpha}$ denote the indexes
$\tau^\alpha, \delta^\alpha$ do not appear in the summation.

We denote the link
$\mathcal{L}_{\tau^1,\delta^1,..,\hat{\tau^\alpha},\hat{\delta^\alpha},..,\tau^L,\delta^L}=\mathcal{K}_1\star
X_{\tau^1}X^*_{\delta^1}\bigvee \cdots\bigvee \mathcal{K}_\alpha
\bigvee \cdots \bigvee \mathcal{K}_L\star
X_{\tau^L}X^{*}_{\delta^L}$. By formula (5.5),
\begin{align}
a_{\mathcal{K}_\alpha}^{
\mathcal{L}_{\tau^1,\delta^1,..,\hat{\tau^\alpha},\hat{\delta^\alpha},..,\tau^L,\delta^L}}(\lambda,\mu)\in
\mathbb{Z}[q^{\pm 1},a^{\pm 1}],
\end{align}
and
\begin{align}
\mathcal{H}(\mathcal{\mathcal{L}}\star \otimes_{\beta=1}^{L}
Q_{[\lambda^\alpha,\mu^\alpha]};q,a)&=\sum_{\tau^\beta,\delta^\beta,
\beta\neq
\alpha}C_{\tau^1,\delta^1,..,\hat{\tau^\alpha},\hat{\delta^\alpha},..,\tau^L,\delta^L}
\prod_{\beta=1,\beta\neq
\alpha}\frac{1}{\{\tau^\beta\}\{\delta^\beta\}}\\\nonumber &\cdot
a_{\mathcal{K}_\alpha}^{
\mathcal{L}_{\tau^1,\delta^1,..,\hat{\tau^\alpha},\hat{\delta^\alpha},..,\tau^L,\delta^L}}(\lambda,\mu)
\mathcal{H}(U\star Q_{[\lambda^\alpha,\mu^\alpha]};q,a).
\end{align}

Recall that we have two expressions for the colored HOMFLYPT
invariant for unknot $U$, the first one is given by (see
\cite{LP1}),
\begin{align}
\mathcal{H}(U\star
Q_{\lambda};q,a)=\sum_{\mu}\frac{\chi_{\lambda}(\mu)}{z_\mu}\prod_{i=1}^{l(\mu)}\frac{a^{\mu_i}-a^{-\mu_i}}{q^{\mu_i}-q^{-\mu_i}}.
\end{align}
Another one is given by (see Lemma 3.6.1 in page 51 in \cite{L2} ),
\begin{align}
\mathcal{H}(U\star Q_{\lambda};q,a)=\prod_{x\in
\lambda}\frac{a^{-1}q^{cn(x)}-aq^{-cn(x)}}{q^{h(x)}-q^{-h(x)}}.
\end{align}
Therefore, we have
\begin{align}
\prod_{x\in
\lambda}(q^{h(x)}-q^{-h(x)})\sum_{\mu}\frac{\chi_{\lambda}(\mu)}{z_\mu}\prod_{i=1}^{l(\mu)}\frac{1}{q^{\mu_i}-q^{-\mu_i}}\in
\mathbb{Z}[q^{\pm 1},a^{\pm 1}].
\end{align}
So for any $\beta\neq \alpha$,
\begin{align}
\prod_{x\in \rho^\beta,
\nu^\beta}\{hl(x)\}\sum_{\rho^\beta,\nu^\beta}\frac{\chi_{\rho^\beta}(\tau^\beta)}{z_{\tau^\beta}}
\frac{\chi_{\nu^\beta}(\delta^\beta)}{z_{\delta^\beta}}\frac{1}{\{\tau^\beta\}\{\delta^\beta\}}\in
\mathbb{Z}[q^{\pm 1},a^{\pm 1}].
\end{align}
Therefore, we have
\begin{align}
\prod_{\beta=1,\beta\neq
\alpha}^{L}\prod_{\substack{\rho^\beta,\nu^\beta\\|\lambda^\beta|-|\rho^\beta|=
|\mu^\beta|-|\nu^\beta|}}\prod_{x\in
\rho^\beta,\nu^\beta}\{hl(x)\}\sum_{\tau^\beta,\delta^\beta,
\beta\neq
\alpha}C_{\tau^1,\delta^1,..,\hat{\tau^\alpha},\hat{\delta^\alpha},..,\tau^L,\delta^L}
\prod_{\beta=1,\beta\neq
\alpha}\frac{1}{\{\tau^\beta\}\{\delta^\beta\}}\in \mathbb{Z}[q^{\pm
1},a^{\pm 1}].
\end{align}
\begin{align}
\prod_{\beta=1,\beta\neq
\alpha}^{L}\prod_{\substack{\rho^\beta,\nu^\beta\\|\lambda^\beta|-|\rho^\beta|=
|\mu^\beta|-|\nu^\beta|}}\prod_{x\in
\rho^\beta,\nu^\beta}\{hl(x)\}Q_{[\lambda^1,\mu^1],...,[\lambda^L,\mu^L]}^{[\mathcal{\lambda}^\alpha,\mathcal{\mu}^\alpha]}(\mathcal{L};q,a)\in
\mathbb{Z}[q^{\pm 1},a^{\pm 1}].
\end{align}
Hence,
\begin{align}
P_{[\lambda^1,\mu^1],...,[\lambda^L,\mu^L]}^{[\lambda^\alpha,\mu^\alpha]}(\mathcal{L};
q,a)\in \mathbb{Z}[q^{\pm 1},a^{\pm 1}].
\end{align}

On the other hand, by formulas (4.6) and (4.7), we have
\begin{align}
Q_{[\lambda^1,\mu^1],...,[\lambda^L,\mu^L]}^{[\mathcal{\lambda}^\alpha,\mathcal{\mu}^\alpha]}(\mathcal{L};\pm
q,\mp a)=(-1)^{\sum_{\beta\neq
\alpha}(|\lambda^\beta|+|\mu^\beta|)}Q_{[\lambda^1,\mu^1],...,[\lambda^L,\mu^L]}^{[\mathcal{\lambda}^\alpha,\mathcal{\mu}^\alpha]}(\mathcal{L};
q, a).
\end{align}
Combing the two expressions (5.14) and (5.15) for colored HOMFLYPT
invariant of the unknot $U$, for any partition $\lambda$, we obtain
the identity:
\begin{align}
(-1)^{|\lambda|}=(-1)^{\sum_{x\in \lambda}(hl(x)+cn(x))}.
\end{align}
Thus
\begin{align}
P_{[\lambda^1,\mu^1],...,[\lambda^L,\mu^L]}^{[\lambda^\alpha,\mu^\alpha]}(\mathcal{L};\pm
q,\mp
a)=P_{[\lambda^1,\mu^1],...,[\lambda^L,\mu^L]}^{[\lambda^\alpha,\mu^\alpha]}(\mathcal{L};q,a).
\end{align}
Combing formulas (5.20) and (5.23), we obtain
\begin{align}
P_{[\lambda^1,\mu^1],...,[\lambda^L,\mu^L]}^{[\lambda^\alpha,\mu^\alpha]}(\mathcal{L};
q, a)\in \mathbb{Z}[q^{\pm 2},a^{\pm 2}].
\end{align}
\end{proof}

redpajama_set_name: RedPajamaArXiv


text:
A J'attendrai ("Várni foglak") egy dzsessz-sztenderddé vált francia sanzon. Tulajdonképpen a "Tornerai" című olasz dal fancia változata (zene: Dino Olivieri, szöveg: Nino Rastelli).

Eredendően Giacomo Puccini Pillangókisasszony című operájának zümmögő kórusa ihlette a dalt. A J'attendrai-t franciául először Rina Kettyvel vették lemezre 1938-ban.

A J'attendrai Lale Andersen Lili Marleenjének németországi párja. Vera Lynn We'll Meet Again című dala volt a hasonlóan sikeres Nagy-Britannia-i háborús sláger. A francia szöveget Louis Poterat írta, és a J'attendrai azonnali siker lett. Rina Ketty verzióját ugyanabban az évben követte a  belga énekesnő, Anne Clercy. Tino Rossi és Jean Sablon is lemezre vette 1939-ben. Amikor Franciaországot 1940-ben a németek megszállták, a J'attendrai a nagy francia háborús dallá vált.

Híres felvételek

Filmek

Jegyzetek

Források
 Songbook

Dalok
Francia nyelvű dalok

redpajama_set_name: RedPajamaWikipedia


text:
Q: Trouble calling a method from a class Can anybody help me with this?:
I'm trying call a method from a my class "numbers" to show that if the entered number is over 50, on button click a message box shows displaying "high" but if it's below 50 it displays "low".
I can't figure out what i'm doing wrong here.
This is the code from my class:
    private int number;
    private string getNumber(int num)
    {

        number = num;
        return number.ToString();
    }

        public int numProperty
    {
        get { return number; }
        set { number = value; }
    }

     public void isHighorlow()
        {
            if (number >=50)
            {

            }
            else
            {
                return;
            }

        }

Note: the int "number" is property that gets it's value from a text box too.
& here is the code from my form:
     numbers info = new numbers();
     private void Btn_Click(object sender, EventArgs e)
    {
      info.numProperty = Convert.ToInt32(numberBOX.Text);
      info.isHighorlow = Messagebox.Show = ("High");
    }

I know that I've not added the "low" bit yet because i'm still trying to see how this works.  Sorry if it seems confusing as i'm still learning c#.
I get the error message: cannot assign isHighorlow because it's part of a method group.
And I also realise it's much easier if I just do an if statement on the textbox, but I'm practising Classes and Methods so I'm trying to do it this way.
thanks.

A: I'm guessing you want something like this:
public string isHighorlow(int number)
{
    if (number >=50)
    {
         return "High";
    }
    else
    {
         return "Low";
    }
}

numbers info = new numbers();
private void Btn_Click(object sender, EventArgs e)
{
  Messagebox.Show(info.isHighorlow(Convert.ToInt32(numberBOX.Text)))
}

A: isHighorLow is a method, not a property.
MessageBox.Show is a method.
Not sure what you are trying to do but it should be :
 if(info.isHigh(Convert.ToInt32(numberBox.Text)))
   Messagebox.Show("High");
  else
    Messagebox.Show("Low");

Meaning you have a method isHigh like so:
public bool isHigh()
{
   return number>=50
}

A: Try to change your code this way:
private int _number;

private string GetNumber(int number)
{
    _number = number;
    return number .ToString();
}

public int Number
{
    get { return _number; }
    set { _number = value; }
}

public string IsHigh()
{
    get { if (number >= 50) return true; }
}

numbers info = new numbers();
private void Btn_Click(object sender, EventArgs e)
{
    info.Number = Convert.ToInt32(numberBOX.Text);
    MessageBox.Show(info.IsHigh ? "High" : "Low");
}

A: In your class you have defined void isHighorlow().
This means that you have a method that returns nothing.
Of course such method cannot be used on the left part of an expression like you have done.
Probably you want to write in your class
public bool isHighorlow()
{
    if (number >=50)
    {
        return true;
    }
    else
    {
        return false;
    }

}
in this way you declare a method that return True if the internal value is >= 50 or false otherwise.
Now, in your form you can use the method in this way
Messagebox.Show(info.isHighorlow() ? "High" : "Low");

However, if the requirement is simply to return a flag for true or false, it is better to use a read only property changing the class code in this way
public bool isHighorlow()
{
    get
    {
        return (number >=50 ? true : false);
    }
    // No set, read only
}

A: IsHighOrLow should be as follows
    public bool isHighorlow()
    {
        if (number >=50)
        {

           return true;
        }
        else
        {
            return false;
        }

    }

And in button click
if (info.isHighorlow){
//say high
} else
{
// say low
}

A: (disclaimer: double check boolean and associated constants per C#)
isHighOrLow doesn't do anything at all.  Perhaps this would be better:
     public boolean isHigh()
        {
            if (number >=50)
            {
                return true;
            }
            else
            {
                return false;
            }

        }

Or, more concisely:
     public boolean isHigh()
        {
            return number >=50;
        }

When you call it, this might be closer to what you need:
     numbers info = new numbers();
     private void Btn_Click(object sender, EventArgs e)
    {
      info.numProperty = Convert.ToInt32(numberBOX.Text);
      if (info.isHigh())
      {
        Messagebox.Show("High");
      }
      else
      {
        Messagebox.Show("Low");
      }
    }

redpajama_set_name: RedPajamaStackExchange


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Based on 2013 WHO expert panel review, admission criteria for SAM is MUAC <11.5cm and MAM >=11.5cms, discharge for both treatments is >=12.5cms. It is no longer recommended for SAM children to be transferred to MAM treatment when they reach 11.5cm on MUAC. Seeking clarification: what if SAM child defaults and comes back to the facility but now MUAC is >11.5cms. Is the child admitted as SAM or MAM?
This is a good question. Let me share what we do in our project in West Bengal India. We have no Community based management for SAM with RUTF so all cases of SAM are treated in the facility.We have many cases of defaulting from the facility like you mention- no complications, MUAC >11.5 <12.5. Mothers in most cases are unable to stay in the facility till such time that the child reaches a green MUAC.
We admit a child in such cases to the MAM programme because that is the best we can do - in addition to the MAM ration we think the major advantage is that with fortnightly contacts there is better monitoring of the child's nutritional status and opportunities to educate the mother on better care practices.
I am not sure where your project is situated and what is the mortality related to SAM but typically in our project area mortality with a RED MUAC is lower than reported from African contexts. In this situation, the approach employed above works for us.
I totally agree with doctor Charulatha.
I would treat this case as an interrupted treatment episode for SAM. This is a recovering SAM case and should be treated as such (i.e. re-admitted into the SAM program to complete the treatment episode). This is not a "MAM never SAM" case or a "SAM case spontaneously recovered to MAM" so entry into a MAM program is not indicated. This is looking at the issue from the perspective of individual cases.
We can look at this from a program coverage perspective. A returning defaulter is a good thing. We want to have no defaulters. Failing that, we want defaulters to come back to the program. I think that "punishing" the defaulter by putting them on a second tier program (i.e. less appetising food with lower energy density than RUTF, less clinical supervision, no systemic antimicrobials, &c.) may not be the best message to be giving out if you want defaulters to return.
the approach employed above works for us.
This is unlikely to be true as one of the main reasons for using MUAC in CMAM programs is that mortality risk at different MUAC thresholds is similar regardless of location (so much so that it is reasonable to assume that all studies measured the same underlying rate). See this Food and Nutrition Bulletin article for a review of evidence. It may be that you are mistaking case-fatality under treatment for case-fatality without treatment. Case-fatality under treatment will depend on factors such as early treatment seeking, timely case-finding and recruitment, and compliance of program and beneficiary which are known to vary from program to program.


text:
The Stars Beneath Our Feet
By David Barclay Moore
Narrated by Nile Bullock & David Barclay Moore / 6 hours 19 minutes
#9 in Bestselling Children's Audiobooks
"The right story at the right time. . . . It's not just a narrative; it's an experience. It's the novel we've been waiting for." —The New York Times
A boy tries to steer a safe path through the projects in Harlem in the wake of his brother's death in this outstanding debut novel that celebrates community and creativity.
** WINNER OF THE CORETTA SCOTT KING JOHN STEPTOE AWARD FOR NEW TALENT! **
MICHAEL B. JORDAN TO DIRECT MOVIE ADAPTATION!
SIX STARRED REVIEWS!
It's Christmas Eve in Harlem, but twelve-year-old Lolly Rachpaul and his mom aren't celebrating. They're still reeling from his older brother's death in a gang-related shooting just a few months earlier. Then Lolly's mother's girlfriend brings him a gift that will change everything: two enormous bags filled with Legos. Lolly's always loved Legos, and he prides himself on following the kit instructions exactly. Now, faced with a pile of building blocks and no instructions, Lolly must find his own way forward.
His path isn't clear—and the pressure to join a "crew," as his brother did, is always there. When Lolly and his friend are beaten up and robbed, joining a crew almost seems like the safe choice. But building a fantastical Lego city at the community center provides Lolly with an escape—and an unexpected bridge back to the world.
David Barclay Moore paints a powerful portrait of a boy teetering on the edge—of adolescence, of grief, of violence—and shows how Lolly's inventive spirit helps him build a life with firm foundations and open doors.
MORE PRAISE FOR THE STARS BENEATH OUR FEET:
A Time Magazine Top 10 Children's Books of the Year
A Boston Globe Best Children's Book of the Year
A Publisher's Weekly Best Book of the Year
A Kirkus Reviews Best Book of the Year
A Shelf Awareness Best Book of the Year
A YALSA Quick Pick
An ALA Notable Book
"A fast and furious read in which we meet some amazing people, people that stay with us. David Barclay Moore is an exciting new voice. We definitely haven't heard the last of his brilliance." —Jacqueline Woodson, Newbery Honor and National Book Award–winning author of Brown Girl Dreaming
"The Stars Beneath Our Feet is about the weight of the world on the back of a child, and the creative tools necessary to alleviate that pressure. I found myself rooting for Lolly, and you will too." —Jason Reynolds, Coretta Scott King Honor Winner for As Brave As You
Black History Month Listens
Project LIT Middle Grade Book Club
TIME Top Books of 2017
Author David Barclay Moore
Narrator Nile Bullock & David Barclay Moore
Publisher Penguin Random House Audio Publishing Group
Libro.fm Rank #386 Overall
Genre Rank #9 in Children's

redpajama_set_name: RedPajamaCommonCrawl


The flair for derring-do persists – last weekend, the Eastern Province Rugby Union president escaped floodwaters through the sunroof of his luxury sedan near Port Elizabeth. All well and good, but can his Southern Kings defy the force of rugby gravity by surviving in the Super rugby firmament for longer than a season?
The odds are remote, and Watson has attacked the South African Rugby Union's refusal to guarantee the new Port Elizabeth franchise the adjustment periods given to the Lions and Cheetahs, who both enjoyed years of cosseted safety from relegation.
Saru aren't budging, so Watson, CEO Anele Pamba and head coach Alan Solomons have a double-barrelled problem.
They must build a viable squad from scratch, while simultaneously fulfilling the urgent political mandate that justified their promotion – to trigger the deep transformation of South African rugby from the capital of its black heartland. Up in the presidential suite at Nelson Mandela Stadium, Watson is in his element – surrounded by old "strugby" comrades, many of them now luminaries of Port Elizabeth's business and political elite.
"In this region, we've always been fighters," he says.
"Black, white, coloured – ever since the colonial wars. We'll take this hand we've been dealt, and make the best of it. Look at Leonidas with his 300 Spartans, or King David with his army of drunkards and ragamuffins. They won wars. That's where we're at.
"We will only rest when the team is representative of all the demographics of South Africa. And when we hold the Super rugby trophy aloft," he says.
It's hard to tell which mission will prove tougher. At the final whistle of the Eastern Province Kings' hard-fought Currie Cup victory over the Falcons, there was only one South African player of colour on the pitch for the Kings, the promising centre Siyanda Grey.
Former Springbok team manager Zola Yeye, once Watson's wing partner at Spring Rose club, has complained recently about the Kings' milky complexion.
Watson, Pamba and Solomons all plead for patience.
"This doesn't change overnight. Perhaps 80% of players of colour in the Super rugby franchises hail from this region. EP has been disadvantaged because it was never linked to a franchise, so there hasn't been an incentive for a top player to stay," says Watson.
Despite the lack of a big title sponsor to bankroll recruitment, Pamba and Solomons have been shopping hard. Pamba will announce eight new signings within the next fortnight.
"There are one or two current Boks among them, and some South Africans who have been based overseas," says Pamba.
An Argentinian import is also on the cards, while Western Province flyhalf Demetri Catrakilis is signed up. Of the current side, only skipper Luke Watson and Jaco Engels are proven Super rugby campaigners, while the likes of Grey and fellow centre Lonwabo "Foxy" Ntleki are rated as potentially up to the mark.
One option would be to buy or borrow players from the relegated Lions – but it's doubtful whether the Johannesburg franchise would be inclined to hand their aces to their usurpers – whom they will likely confront in a relegation-promotion playoff at the end of next Super season.
Solomons wants to reclaim seasoned Eastern Cape products, while protecting the laaities from poachers.
Solomons, who assisted Nick Mallett during his 16-game unbeaten Bok run and coached Ulster to a four-year unbeaten home run, is arguably the Kings' key signing to date. Born in Uitenhage and schooled at Grey, he brings a mix of partisan passion and progressive thinking.
Since leaving a position as the IRB's high-performance consultant in 2010, Solomons has sparked a rugby renaissance in his home town. His first priority was to build a non-commercial academy, headed by Robbie Kempson. It's already working: graduates include EP Kings forwards Thembelani Bholi, Lizo Nqoboka and Siya Mangaliso.
Solomons predicts a bumper crop of Bok-quality players.
Factor in Model C powerhouses such as Grey, Framesby and Brandwag and you have a potent base of elite youth rugby – and Pamba emphasises the Kings are looking for white talent and black.
The Kings will have a remarkable stadium on their side, and Solomons wants to forge in it the kind of fortress mentality that defined his Ulster side.
The Watsons' evangelical convictions about transformation have incensed the Afrikaans rugby establishment, where many still see the game as their cultural birthright and territory. It doesn't help that Luke doesn't do subtlety. In 2008, in a secretly taped speech, he ruined his international career by foolishly remarking that he suffered from a nauseous reaction to the Bok emblem.
Having abandoned a lucrative career with Bath, Watson Jnr now has a chance to make his case non-verbally by leading a representative side to success at rugby's second-highest level.
"Luke has been unfairly labelled because he comes from a family that fought for liberation. I had nothing to do with his selection to Boks, or with his return to EP. Alan negotiated with Luke. Where I have to commend Luke is taking a serious pay cut to come back and be part of this dream," says Cheeky.
Has the mind of white rugby opened a bit since the furore over Luke's call-up?
Twenty years after unity, the Wallaby starting line-up features more black players – five – than the Bok line-up – three. Clearly it's not just a question of working harder. We're doing it wrong.


text:
Tag Archives: Julian
The Later Roman Empire – Ammianus (trans. Walter Hamilton)
Ammianus Marcellinus was the last great Roman historian, and his writings rank alongside those of Livy and Tacitus. The Later Roman Empire chronicles a period of twenty-five years during Ammianus' lifetime, covering the reigns of Constantius, Julian, Jovian, Valentinian and Valens, and providing eyewitness accounts of significant military events including the Battle of Strasbourg and the Goths' Revolt. Portraying a time of rapid and dramatic change, Ammianus describes an Empire exhausted by excessive taxation, corruption, the financial ruin of the middle classes and the progressive decline in the morale of the army. In this magisterial depiction of the closing decades of the Roman Empire, we can see the seeds of the events that were to lead to the fall of the city, just twenty years after Ammianus' death. This selection includes the major parts of the surviving books of the history. Walter Hamilton's fine translation captures the stylish vigour of the original, while Andrew Wallace-Hadrill's introduction describes the life and works of Ammianus and places the history in the context of its times.
Ammianus Marcellinus was the last great Roman historian. He was not a professional man of letters but an army officer of Greek origin born at Antioch and contemporary with the events described in what remains of his work. He set himself the task of continuing the histories of Tacitus from AD 96 down to his own day. The first thirteen of his thirty-one books are lost: the remainder describe a period of only twenty-five years (AD 354-378) and the reigns of the emperors Constantius, Julian, Jovian, Valentinian and Valens, for which he is a prime authority. He was a pagan and an admirer of the apostate Julian, to whose career about half the surviving books are devoted. But his treatment of Christianity is free from prejudice and his imparitality and good judgement have been generally recognized. His style is sometimes bizarre, but in all the essential qualities of an historian he deserves the praise accorded to him by Gibbon and is well able to stand comparison with Livy and Tacitus.
Review of Ammianus' The Later Roman Empire (AD 354-378)
While reading The Later Roman Empire, I feel an affinity with the Roman legions in Tacitus who have wandered far from Italian soil and too far into Germany. The fog is everywhere, the marshlands extends forever, the sun no longer shines. They have reached the edge of the known world. While Tacitus' legions have wandered too far in space, I feel, while reading Ammianus, that I have wandered too far in time. This is no longer the Roman Empire of old. This is something else. Not quite the Middle Ages, not quite classical antiquity.
In Ammianus' time, the world is a smaller space. The Roman Empire is sick. This is not Livy's Rome, which was expanding. Nor is it Tacitus' Rome, which was consolidating it's powers. This is an old Rome, sick unto death. In Livy and Tacitus, temporary setbacks due to lavishness, foreign influences, and profligacy (the usual suspects) could be overcome by a return to the mos maiorum, "the way of our ancestors." No longer in the Late Empire. The corruption is now systemic. Even Valentinian's attempts at reform are due to cruelty and a desire to see others suffer than the good old mos maiorum doctrine.
In Ammianus' time, the Empire is split between east and west. The emperor, or the "Augustus" watches over one half, while the prince, or the "Caesar" watches over the other. The barbarians constantly push against the boundaries and the army is constantly pushing back. The Augustus and the Caesar are on a neverending campaign. They themselves ride into battle in a magisterial if perhaps misguided show of noblesse oblige. Perhaps, like the Empire, they are tired of being alive.
In Ammianus' history, Rome is far away. What is going on in Rome? Are the senators coming up with new policies? Do the masses have enough grain? Who knows. Compared with the hostile frontier, Rome is an insignificant speck. Germans, Gauls, and Persians are the focus of attention. Guys names "Gundomadus" and "Dagalaifus" run around in Ammianus' history. We are no longer in Kansas, Dorothy.
Despite the book blurb, in style and substance, Ammianus cannot compare with Tacitus and Livy. Part of this may be due to the loss of the all-important prologue, which is, for me, the most interesting part of the text. In the prologue, the historian justifies the writing of his history. He talks of his predecessors' failings and how he is filling in a gap. He talks up the period of history he concentrates on and why he, and only he, can capture accurately the story of that time. Ammianus' style isn't as weird as the introduction makes it out to be (e.g. I didn't notice instances of what Gibbon refers to as Ammianus' disorder, perplexity of narrative, false ornaments, and turgid metaphors). All in all, a fascinating look at how the Roman world had changed. If we take the Hannibalic War (218 BC) as the "coming of age" of Rome, and the period covered by Ammianus as Rome's twilight (AD 378), it looks like, for some reason, Rome maintained her position as the primary Mediterranean world power for close to 600 years. What an amazing achievement. They ask: "Why did Rome fall?" But perhaps the real question is: "Why did Rome dominate for so long?"
This entry was posted in Reading List - Books and tagged Ammianus, Julian, Roman history, Walter Hamilton on January 19, 2019 by Edwin Wong.


I have listed a new property at 10015 107 ST in MORINVILLE.
This 10 year old bi-level has been completely renovated from top to bottom. This home comes complete with 3 bedrooms up, 2 full bathrooms, a large living room with bay window and open kitchen with eating nook. Recent upgrades include all new hardwood, tile and carpet, baseboards, casings, light fixtures and paint. The spacious master bedroom has a walk in closet and 3 piece bath. Direct access to the double attached garage. Fully landscaped and fenced.


Are Radical Right Media Feeding Political Violence and Hate?
That's the question posed this past week at PBS' Bill Moyers Journal. The program is a hard hitting examination of the impact of radical right talk radio, books, and TV shows not just on the nature of political discourse but also their link to violent actions against elected officials and fellow citizens.
When Glenn Beck says: "I'm thinking about killing Michael Moore and I'm wondering if I could kill him myself, or if I would need to hire somebody to do it," is it just harmless rhetoric or something more?
When Michael Reagan says that 9/11 deniers should be taken out and shot, is that harmless?
When Michael Savage refers to liberals as psychotic, as traitors, and as a virus on America, how does that shape audience perceptions and even actions against liberals?
What about when Rush Limbaugh tells his listeners to launch Operation Chaos against the Democrats with an explicit goal of generating violence at the Democratic Convention. As Limbaugh told his audience: "This is about chaos, this is why it is called Operation Chaos[...]the dream end, if people say what is your exit strategy. The dream end is this keeps up to the convention. And that we have a replay of Chicago 1968, with burning cars, protests, fires, literal riots and all of that. That's the objective here."
What do readers think? Could radical right media actually incite violence against liberals and or various out groups in society?


I'm so sorry Auntie that is total pants.
As long as you asked for an oral hearing you have a decent chance of getting your award back.
Did you get rid of your old blue badge auntie? Step by step is progress at whatever speed even with pauses.


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text:
Budget for the Organization You Want to Be
Home Finance, StrategyBudget for the Organization You Want to Be
The goal of the proposed ordinance is to create safe venues for customers and to discourage illegal activities. The governing officials of the City of Oxford will consider the following ordinance at the upcoming meeting(s) of the Board of Aldermen.
The bodies were lying in the streets un-buried. All railroads and vessels carrying food and such things into the great city had ceased runnings and mobs of the hungry poor were pillaging the stores and warehouses. Murder and robbery and drunkenness were everywhere. Already the people had fled from the city by millions—at first the rich, in their private motor-cars and dirigibles, and then the great mass of the population, on foot, carrying the plague with them, themselves starving and pillaging the farmers and all the towns and villages on the way.
Safe venues for citizens
"The man who sent this news, the wireless operator, was alone with his instrument on the top of a lofty building. The people remaining in the city—he estimated them at several hundred thousand—had gone mad from fear and drink, and on all sides of him great fires were raging. He was a hero, that man who staid by his post—an obscure newspaperman, most likely.
"For twenty-four hours, he said, no transatlantic airships had arrived, and no more messages were coming from England. He did state, though, that a message from Berlin—that's in Germany—announced that Hoffmeyer, a bacteriologist of the Metchnikoff School, had discovered the serum for the plague. That was the last word, to this day, that we of America ever received from Europe. If Hoffmeyer discovered the serum, it was too late, or otherwise, long ere this, explorers from Europe would have come looking for us. We can only conclude that what happened in America happened in Europe, and that, at the best, some several score may have survived the Scarlet Death on that whole continent.
"For one day longer the despatches continued to come from New York. Then they, too, ceased. The man who had sent them, perched in his lofty building, had either died of the plague or been consumed in the great conflagrations he had described as raging around him. And what had occurred in New York had been duplicated in all the other cities. It was the same in San Francisco, and Oakland, and Berkeley. By Thursday the people were dying so rapidly that their corpses could not be handled, and dead bodies lay everywhere. Thursday night the panic outrush for the country began. Imagine, my grandsons, people, thicker than the salmon-run you have seen on the Sacramento river, pouring out of the cities by millions, madly over the country, in vain attempt to escape the ubiquitous death. You see, they carried the germs with them. Even the airships of the rich, fleeing for mountain and desert fastnesses, carried the germs.
"Hundreds of these airships escaped to Hawaii, and not only did they bring the plague with them, but they found the plague already there before them. This we learned, by the despatches, until all order in San Francisco vanished, and there were no operators left at their posts to receive or send. It was amazing, astounding, this loss of communication with the world. It was exactly as if the world had ceased, been blotted out. For sixty years that world has no longer existed for me. I know there must be such places as New York, Europe, Asia, and Africa; but not one word has been heard of them—not in sixty years. With the coming of the Scarlet Death the world fell apart, absolutely, irretrievably. Ten thousand years of culture and civilization passed in the twinkling of an eye, 'lapsed like foam.'
"I was telling about the airships of the rich. They carried the plague with them and no matter where they fled, they died. I never encountered but one survivor of any of them—Mungerson. He was afterwards a Santa Rosan, and he married my eldest daughter. He came into the tribe eight years after the plague. He was then nineteen years old, and he was compelled to wait twelve years more before he could marry. You see, there were no unmarried women, and some of the older daughters of the Santa Rosans were already bespoken. So he was forced to wait until my Mary had grown to sixteen years. It was his son, Gimp-Leg, who was killed last year by the mountain lion.
Last modified: September 16, 2018
dannci says:
And then the battle\'s not so bad? You\'ve killed me! Oh, you\'ve killed me! And so we say goodbye to our beloved pet, Nibbler, who\'s gone to a place where I, too, hope one day to go. The toilet. Bender, being God isn\'t easy. If you do too much, people get dependent on you, and if you do nothing, they lose hope. You have to use a light touch. Like a safecracker, or a pickpocket.
Hey, tell me something. You\'ve got all this money. How come you always dress like you\'re doing your laundry? Come, Comrade Bender! We must take to the streets! Oh, you\'re a dollar naughtier than most. We need rest. The spirit is willing, but the flesh is spongy and bruised.
For the last time, I don\'t like lilacs! Your \'first\' wife was the one who liked lilacs! Goodbye, cruel world. Goodbye, cruel lamp. Goodbye, cruel velvet drapes, lined with what would appear to be some sort of cruel muslin and the cute little pom-pom curtain pull cords. Cruel though they may be…


text:
Apple TV 4K review
The Apple TV 4K offers an attractive, customizable interface and affordable UHD content, but design flaws and a subpar remote hold it back.
By Marshall Honorof published 8 February 21
Fantastic 4K HDR content
Intuitive home screen
Simple iOS and macOS integration
Fiddly remote
No digital audio or USB ports
Tom's Guide Verdict
Today's best Apple TV 4K deals
Editor's note: The Apple TV 4K (2017) has been upgraded. Check out our Apple TV 4K (2021) review
The fourth-generation Apple TV was a competent, if unexciting streaming box, buoyed by strong Siri navigation and hamstrung by navigational oddities and a high price tag. Now that Apple has had two years to improve, the company has debuted the Apple TV 4K ($179 to $199). This set is everything we loved (and hated) about Apple's streaming box, but now with 4K and HDR capabilities making the best streaming device ... for Apple users, at least.
But as one of the best Apple HomeKit products, you'll still find an attractive and customizable interface, strong Apple ecosystem integration, and powerful voice search via Siri. You'll also still find unintuitive navigation options, a remote with a learning curve and an asking price that borders on prohibitive. If you live, breathe and dream Apple and need a way to watch iTunes content in 4K, the Apple TV 4K may be worth a look. Otherwise, this set doesn't offer many advantages over its cheaper competitors.
Apple 4K TV (64GB Black) at very.co.uk for £189
Apple TV 4K review: Price and release date
The Apple TV 4K starts at $179 on Amazon for the 32GB model. Its 64GB counterpart is $199.
The Apple TV 4K came out on September 22, 2017.
Apple TV 4K specs
Video: 4K, HDR, Dolby Vision
Audio: Dolby Atmos
Processor: A10X Fusion
Dimensions: 3.9 x 3.9 x 1.4 inches
In the box: Apple TV 4K, Siri remote, Lightning to USB cable
Smart assistant support: Siri
Apple TV 4K review: Design
Apple has generally adopted an "if it ain't broke, don't fix it" approach to the Apple TV, which has evolved from a small black box into a slightly more elaborate black box. Measuring 4.0 x 4.0 x 1.3 inches, it's neither small enough to hide behind your TV nor big enough to affect the overall appearance of your entertainment center. It looks unremarkable yet elegant.
The back of the device is similarly uncluttered. There's a power port, an HDMI port and an Ethernet port. (The device also works just fine via Wi-Fi.) The lack of a digital audio port could be a pretty significant oversight, because that port can enhance your audio experience significantly. Presenting 4K HDR videos without a high-fidelity audio option makes this set feel incomplete.
Apple TV 4K review: Interface
The Apple TV 4K's interface is something of a paradox. Visually, it's one of the cleanest, most attractive, most customizable menu systems around. Navigationally, it could still use a lot of work, particularly for those times when you're trying to switch among multiple services.
First and foremost, Apple TV's home screen and menus are simply gorgeous. Unlike the tedious Amazon Fire TV or the jumbled Nvidia Shield TV, the Apple TV makes use of large, clean, movable icons for each of your apps. There is absolutely no guesswork in figuring out where to go for your favorite content. Every app you have is available right from the home screen, and you can drag and drop their icons into whichever order is most convenient for you, or even make folders to cluster similar apps together. The addition of folders, in fact, puts the Apple TV slightly ahead of the Roku, long the gold standard in streaming-box organization.
Away from the home screen, however, things are a little less straightforward. Apple has always favored surface-level aesthetics over deep-cut functionality in the Apple TV, but after two years, barely anything has changed. Unless what you're looking for is on the front page, the App Store is a mess (more on that later). Apple's own iTunes apps are difficult to parse, separating movies, TV shows and music into three different stores for — well, no real reason, as far as I can tell.
The TV app still is not any easier than just loading up an app directly, unless you are watching so much stuff that you can't remember where you are in any given series.
Arguably, the biggest navigational problem is the TV app, which concatenates streaming content from a wide variety of sources. Although the app debuted last year, it still feels half-baked. In theory, the TV app is kind of a one-stop shop for everything you're watching, drawing in content from apps like HBO Go, Hulu and ABC and even keeping track of which episodes you're watching on which service. Because the TV app makes use of only those apps to which you're already subscribed, it can theoretically show you a lot of cool content that won't cost you any additional money.
In practice, though, it's still not any easier than just loading up an app directly, unless you are watching so much stuff that you can't remember where you are in any given series. The recommendations are not tailored for individual users, instead just showing what's popular in any given service. While it's good to know what you can watch without paying extra (aside from a small row of new releases available for a charge through iTunes, which we'll let slide), it's not worth giving up the algorithms that tailor recommendations to your taste within individual apps.
Apple TV 4K review: Remote
The Siri remote that comes with the Apple TV 4K is unlike most other streaming remotes in that you must rely on a square touchpad to get around, rather than directional buttons. You'll tap and slide your fingers around to navigate through content instead of physically pressing anything.
Whether I was scrolling through shows or trying to control a character in a game, the touchpad was overly sensitive and imprecise.
Save for a white circle around the Menu button, it's exactly the same as Apple's remote from 2015, complete with the touchpad, a home button, a play/pause button and a volume control that syncs directly with your TV.
In theory, touch controls mean you can swipe to your favorite content much faster; in practice, it means that you're going to overshoot your destination — at least until you've really spent a lot of time with the remote. Whether I was scrolling through shows, inputting a password or trying to control a character in a game, the touchpad was overly sensitive and imprecise.
The one handy feature of the remote is that you can hit a button to bring up the Siri search, which is, by and large, still excellent. Granted, this feature is a lot more common among streaming remotes now than it was when the 4th-gen Apple TV debuted, but it's still a pretty darn useful button.
Apple TV Plus review: Search
The Apple TV 4K is unparalleled is in its search features. Accessing the same technology that goes into Siri, you can use voice or text to search for just about any piece of content imaginable. If it's on iTunes — or one of more than 60 compatible apps, including Netflix and Hulu — the search will find it. You can search for episodes of Star Trek with obscure guest stars; you can search for 4K movies starring Hugh Jackman; you can search for poorly reviewed movies from the '90s. Apple TV 4K's search handled everything that I threw at it, at least in terms of movies, TV and music.
However, the Siri search was considerably less adept at picking out apps. It was unable to find the Disney Channel, Sling TV or FX Now — three fairly common streaming services. This isn't helped at all by the fact that the App Store is still, to put it charitably, a mess. If the app or game you want is popular enough to make the front page, that app is easy enough to find. Otherwise, the menus are both haphazard and incomplete.
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There's no "A-Z" listing, meaning that your browsing will always be restricted to what's popular, or what the Apple devs want you to see. If you don't know the exact name of your app and don't find it within the fairly anemic lists in the App Store, you'll probably never locate it at all. Having a streamlined store is convenient, perhaps, but it does take all the joy out of finding that one specialized app you didn't even know you needed.
Apple TV 4K review: Content and Apps
The Apple TV 4K has just about everything the average streaming enthusiast needs. In addition to iTunes, you can get content via Netflix, Hulu, Crackle, Spotify and Pandora, as well as a plethora of network-specific apps from Fox, ABC, CBS, FX, HBO, Starz and others.
You've got your choice of cable TV alternatives, with access to Sling TV and YouTube TV (among others), and you can even specify your TV provider during the initial system setup. This saves you from a lot of tedious logins later on.
Apple TV 4K review: Apple Arcade
While Apple Arcade debuted on iPhones, it's now available on iPads, Macs and, of course, Apple TVs. This subscription services costs $5 per month, and lets you download as many premium mobile games as you want from a library of dozens. From the gorgeous high-octane action of Sayonara Wild Hearts to the cyberpunk dialogue-driven adventures in Neo Cab, there's something for everyone.
Tom's Guide has not yet had a chance to test Apple Arcade on an Apple TV, but the selection alone is a feather in the device's cap, as far as games go.
Apple TV 4K review: Other features
Back when the fourth-gen Apple TV came out, the company was very quick to tout its gaming features. With high-budget, innovative titles such as Transistor and Beat Sports, it was easy to believe that the Apple TV might be the second streaming box to make an honest go of gaming. (The Shield TV was the first, obviously.)
Since then, the Apple TV has gained a lot of good games and even more cut-rate, sloppy, ignorable ones. As far as I could tell, the system doesn't have any new exclusive titles, and highly anticipated releases such as Jenova Chen's Sky aren't available yet.
Apple no longer insists that developers make their games compatible with the remote; games can require a controller instead. There still aren't that many Apple TV controllers, although the good news is that the SteelSeries Nimbus is a quite good one. Still, be warned that if you play games with an Apple TV remote, the controls tend to be extremely imprecise and floaty. This can be especially damaging in a platformer like Momoka: An Interplanetary Adventure, in which I couldn't finish the first dungeon, because it was too difficult to jump, aim and shoot at the same time.
Lots of streaming boxes offer games; comparatively few offer smart home options. The Apple TV 4K allows you to control smart home devices, provided they run on the Apple HomeKit protocol. This includes a wide variety of HomeKit devices, including thermostats from Honeywell, security cameras from D-Link and light bulbs from Philips. Once you've gone through a rather cumbersome and clumsy setup process, involving an iOS device and a plethora of settings menus, you can use your Apple TV as a smart home hub. This lets you control your gadgets, even when you're away from your home Wi-Fi network. The Apple TV is pretty much the only HomeKit hub on the market, so that alone might make it worth a look.
One advantage for existing Apple customers is AirPlay, a screen-mirroring capability that resembles Miracast and Google Cast. Using AirPlay, users can either broadcast their iTunes libraries up on the big screen or mirror their screens directly. This can be useful if you want to show a presentation on a TV screen or get some productivity work done in your living room; otherwise, the feature is much less necessary than it used to be. Broadcasting iTunes content from your iDevice or Mac isn't really necessary, because your iTunes content is already available on the Apple TV. Even your own photos and videos are just a click away in the built-in Photo app.
Most of these boxes have USB ports for side-loaded content; the Apple TV does not. You can still set up a media server and stream your personal content from programs like Plex or VLC, but you can't just plug in a thumb drive and watch your own personal content — and you can't charge your remote, unless you have a wall adapter or a USB port on a powered device handy.
Apple TV 4K review: Performance
The biggest thing that sets apart the Apple TV 4K from its predecessor is the new device's ability to display videos in 4K and HDR, and this box's A10X Fusion chip is more than up to the task. The 4K resolution is self-explanatory; the HDR comes in both Dolby Vision and HDR 10 flavors, ensuring that you'll get a vibrant color palette, no matter which studio your movie or TV show comes from.
You'll need a pretty powerful Internet connection to stream 4K HDR content (you need at least 25 Mbps down, which is more than what we got on a standard home Wi-Fi network), but content loads quickly and smoothly. Streams usually took just a few seconds to buffer before reaching full 1080p HD, and perhaps an additional 5 seconds before 4K HDR kicked in. This will vary depending on the strength of your internet connection, but if you have the requisite speed, the Apple TV 4K will leverage it.
Unsurprisingly, movies and TV look gorgeous in 4K HDR as well. In The Lego Batman Movie, the reds and oranges of explosions played off the shiny, black Batmobile, while the eerie blue-green sky ensured that the vibrant heroes and villains looked as real as the toys that inspired them.
Apple has also indulged in a bit of generosity by offering viewers 4K movies at the same price as their full-HD counterparts, with free upgrades on movies customers have already purchased. Although studios ultimately set the price of their movies, this offer applies to every piece of content in the iTunes library.
The only disappointing part of the visuals is that the upscaling doesn't do much for older content. Rather than switching back and forth between different screen resolutions (this can be annoying, but it at least gives you the proper picture), the Apple TV simply upscales SD and HD shows to mimic a higher resolution. The scaling worked better on some shows than others. The Handmaid's Tale on Hulu looked great, but Penn & Teller: Fool Us on iTunes appeared grainy and oversaturated. If you go for older shows on Netflix and Hulu, the jump from SD could be even more jarring.
Otherwise, navigation through the menus feels snappy and immediate, and the 4K menus are a nice touch. Most of Apple's competitors still use 1080p menus, even if they're capable of displaying 4K movies and shows.
Apple TV 4K vs. the Competition
Since the Apple TV 4K is considerably more expensive than most of its competitors, it's worth seeing how they all line up in terms of functionality.
The Google Chromecast Ultra is a 4K HDR device that relies on your phone, tablet or computer to work. It has no visible operating system or remote control; just choose a piece of content on your own device, then cast that media to your TV. (The protocol is not the same thing as Miracast, because your phone is not actually providing content, just telling the Chromecast what to do.) The picture quality is comparable to that of the Apple TV 4K, and the Google device costs nearly a full $100 less, but you won't have access to many games or smart home perks.
The Amazon Fire TV costs $100 and offers 4K features, but no HDR. The biggest advantage of the Fire TV over the Apple TV 4K (aside from the price) is that it ties into your Amazon account and offers you personalized recommendations based on your Amazon Video and Music history. (Even when the Apple TV 4K gets an Amazon app, it won't do this right from your home screen.) You can also control smart home gadgets if you have Amazon's Alexa gadget. The Fire TV one's good if you have an Amazon Prime subscription, but it's nothing special without one.
The Roku Ultra ($100) is probably the most complete streaming device for the money, in terms of both content and features. Not only do you get the largest app selection on the market, but you also get 4K HDR picture and a remote control with a headphone jack. (Apple TV requires Bluetooth headphones for private listening.) The game selection isn't as good as with Apple TV 4K, however, and you don't get any smart home features.
The most comparable system to the Apple TV, in terms of both price and functionality, is the Nvidia Shield TV ($200). Like with the Apple TV, you get 4K HDR content and smart home features; unlike on the Apple TV, you can also access some of the latest and greatest big-budget games through the Nvidia GeForce Now streaming service. With some extra accessories, you can even stream live TV from an HD antenna.
Apple TV 4K review: Bottom Line
In "Through the Looking-Glass and what Alice Found There" by Lewis Carroll, the Red Queen says to Alice, "Now, here, you see, it takes all the running you can do to keep in the same place." Apple has run very fast indeed to address some of the biggest problems with the 4th-gen Apple TV — and yet, given the pace of its competition, Apple has mostly stayed in the same spot.
To give credit where it's due, Apple TV 4K hosts gorgeous content in vivid colors. The home screen is second-to-none, and the set's iTunes integration alone might be worth the cost of admission for viewers who are already invested in the Apple ecosystem.
On the other hand, the new Apple TV doesn't do much more than streaming boxes that cost almost $100 less — and in some cases, it doesn't even do as much. In addition, as slick as the touchpad remote is, it's less precise than remotes with physical buttons.
Ultimately, the Apple TV 4K is gorgeous and intuitive, but it's not ambitious enough for the high price. If you've got a 4K TV and a lot of money tied up in the Apple ecosystem, this box is worth a look. Otherwise, you could probably invest the extra $80 in some excellent 4K movies instead.
Credit: Shaun Lucas/Tom's Guide
Marshall Honorof
Marshall Honorof is a senior editor for Tom's Guide, overseeing the site's coverage of gaming hardware and software. He comes from a science writing background, having studied paleomammalogy, biological anthropology, and the history of science and technology. After hours, you can find him practicing taekwondo or doing deep dives on classic sci-fi.
1 Comment Comment from the forums
LazarXD
There's no mention that unlike the others, the Apple TV also functions as a shopping device with the apps you can download for stores that are integrated with Siri. The stores are on a bit of the boutique side, but that's the market that Apple has always favored.
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Players with Texas Roots
51 Greatest Super Bowl Moments
Texas Roots
Players' indelible marks
50 years in, Super Bowl lore owes a lot to participants with Texas roots
On Sunday, the NFL's modern-day championship game will begin its second half-century in a state whose native sons have played a considerable role in shaping the first 50 years of Super Sundays.
Duane Thomas: His own man
Duane Thomas always flowed to the beat of a different drummer. Some running backs looked for the hole. To Thomas, the entire field was the hole. To some, running plays were diagrams. To him, they were art.
Forrest Gregg: Leader of the Pack
If not for the exploits of Forrest Gregg and his Green Bay Packers teammates, chances are the NFL might be handing out the Tom Landry Trophy at Super Bowl LI.
Cliff Branch: A sprinter's delight
Patience has always been a virtue for Cliff Branch. He waited until the fourth round of the 1972 NFL draft to hear his name called, but his combination of speed and receiving skills was ideal for the Raiders.
Lester Hayes: His game had substance
Lester Hayes lives in NFL lore as the Jedi Master of stickum, the goop that helped him corral 18 interceptions during the regular season and playoffs en route to the Oakland Raiders' Super Bowl XV win.
Dexter Manley: Determined to get into the big game
Dexter Manley made it to three Super Bowls playing for Washington, celebrating two championships as the unofficial "secretary of defense." His first Super Bowl, though, was in his hometown of Houston.
Coaches: Title winners Landry, Johnson, Kubiak share Texas ties
Six Texans have led their teams into Super Bowls as head coaches, and three walked away with the Lombardi Trophy. The Dallas Cowboys' Tom Landry accounted for five of those appearances.
Houston-area high schools with the most Super Bowl players
No Texas high school has produced more Super Bowl players than them. Those four schools have each had five alumni move on to play in the NFL and eventually make it to a Super Bowl.
Bob Lilly: First a toss, then the ring
Bob Lilly holds two Super Bowl records, albeit unofficial, that are likely to stand the test of time. The one Lilly most enjoys remembering came in Super Bowl VI.
Joe Greene: Man behind the Curtain
"Mean" Joe Greene is in good health and is still receiving accolades for a career that took him from segregated Central Texas to the highest rungs of the NFL.
Kenneth Davis and Thurman Thomas: Something in common
Four times Kenneth Davis and Thurman Thomas played in a Super Bowl with the Buffalo Bills, and four times they lost. But don't waste your time sympathizing with the former Texas high school standouts.
Santana Dotson and Kevin Smith: Excellent prep work
For a high school football player in Texas, nothing can beat winning a state championship - unless, of course, it's making it to the NFL and winning a Super Bowl ring. Ah, but what about winning both?
The 1968 Jets: Oh, how they soared
Pete Lammons, from Jacksonville in East Texas, was one of 11 Texans - nine of whom played high school football in the state - on the Jets' 1968 roster.
Drew Brees: The arm of a champion
Fifty years into the NFL's Super Bowl era, Drew Brees stands alone among the more than 300 former Texas high school players who have played for the Lombardi Trophy.
© 2018 Hearst Newspapers, LLC.


A delicious 3 course 'dark' dinner including tea/coffee, surrounded in Gothic splendour with background music in a soft Gothic/Alternative 80's style.
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Numbers are limited and it is advisable to book early to avoid disappointment. All tickets must be pre-booked with a £20 deposit per person taken at the time of booking, with full payment and menu selections required 1 week prior to the event.


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Q: How to draw graph stick with rotated rectangle properly in matplotlib? I am trying to figure out how to draw the following properly.
Lets say I have two nodes, i.e. 2 points in R^2, namely src=(x1,y1) and dst=(x2,y2).
I want to draw an edge/line between them, but starting with a box.
This is an example I created with Tikz:

src and dest are the centers of these nodes. Note that they can also be on different heights. The problem is when I try to build the box (using patches.Rectangle), I have to rotate it and then it changes its size.
Thanks in advance

A: If you're happy to use a filled rectange for the graph with a rotated rectangle as stick, this can be emulated by a simple line, which is as thick as a scatter point at the graph edges.
import matplotlib.pyplot as plt
import numpy as np

def plot_spec_line(src, dst, length=0.3, srctext="v", dsttext="w",
                   ax=None, color="k", textcolor="k"):
    if not ax: ax=plt.gca()
    lend=src+(dst-src)*length
    ax.plot([src[0],lend[0]], [src[1],lend[1]], lw=24,solid_capstyle="butt", zorder=1, color=color )
    ax.plot([src[0],dst[0]], [src[1],dst[1]], lw=2,solid_capstyle="butt", zorder=0, color=color)
    ax.scatter([src[0],dst[0]], [src[1],dst[1]], s=24**2, marker="o", lw=2, edgecolors=color, c="w", zorder=2)
    ax.text(src[0],src[1], srctext, fontsize=12, ha="center", va="center", zorder=3, color=textcolor)
    ax.text(dst[0],dst[1], dsttext, fontsize=12, ha="center", va="center", zorder=3, color=textcolor)

s = np.array([1,1])
d = np.array([3,2])
plot_spec_line(s, d, length=0.3, srctext="v", dsttext="w")

s = np.array([1.5,0.9])
d = np.array([2.8,1.2])
plot_spec_line(s, d, length=0.2, srctext="a", dsttext="b", color="gray")

s = np.array([1,2])
d = np.array([2,1.9])
plot_spec_line(s, d, length=0.7, srctext="X", dsttext="Y", textcolor="limegreen", color="limegreen")

plt.margins(0.2)
plt.show()

To obtain a rectangle, which can be filled or not and more imporatantly, has an edge, you can use a Rectangle.
You can rotate a rectangle by setting an appropriate transform.
It then makes sense to use a plot of equal aspect ratio, such that the Rectangle and circles are not skewed.

import matplotlib.pyplot as plt
import matplotlib.transforms
import numpy as np

def plot_spec_line(src, dst, length=0.3, radius=0.1, srctext="v", dsttext="w",
                   ax=None, color="k", textcolor="k", reccolor="w", lw=1 ):
    if not ax: ax=plt.gca()
    lend=np.sqrt(np.sum(((dst-src)*length)**2))
    s = dst-src
    angle = np.rad2deg(np.arctan2(s[1],s[0]))
    delta = np.array([0,radius])
    tr = matplotlib.transforms.Affine2D().rotate_deg_around(src[0],src[1], angle)
    t = tr + ax.transData
    rec = plt.Rectangle(src-delta, width=lend, height=radius*2, ec=color,facecolor=reccolor, transform=t, linewidth=lw)
    ax.add_patch(rec)
    ax.plot([src[0],dst[0]], [src[1],dst[1]], lw=lw,solid_capstyle="butt", zorder=0, color=color)
    circ1= plt.Circle(src, radius=radius, fill=True, facecolor="w", edgecolor=color, lw=lw)
    circ2= plt.Circle(dst, radius=radius, fill=True, facecolor="w", edgecolor=color, lw=lw)
    ax.add_patch(circ1)
    ax.add_patch(circ2)
    ax.text(src[0],src[1], srctext, fontsize=12, ha="center", va="center", zorder=3, color=textcolor)
    ax.text(dst[0],dst[1], dsttext, fontsize=12, ha="center", va="center", zorder=3, color=textcolor)

s = np.array([1,1])
d = np.array([3,2])
plot_spec_line(s, d, length=0.3, srctext="v", dsttext="w", radius=0.06,
               reccolor="plum", )

s = np.array([1.5,0.9])
d = np.array([2.8,1.2])
plot_spec_line(s, d, length=0.2, srctext="a", dsttext="b", color="gray", lw=2.5)

s = np.array([1,2])
d = np.array([2,1.9])
plot_spec_line(s, d, length=0.7, srctext="X", dsttext="Y", textcolor="limegreen",
               reccolor="palegreen", color="limegreen")

plt.margins(0.2)
plt.gca().set_aspect("equal")
plt.show()