text:
\section{Introduction}

The formal construction of the homogeneous coordinate ring of a
toric variety, that I.M. Musson in \cite{M2}, D.A. Cox in
\cite{Cox} and others discovered in the early 1990s, takes a fan
$\Delta $ and creates a torus action on an open subset of an
affine space whose quotient is the toric variety of $\Delta $. We
reverse this process in \cite{R}. To be more specific, let $k$ be
an algebraically closed field of characteristic $0$. We take a
torus action on the affine space $X=k^r\times (k^{\times})^s$ and
create various fans whose toric varieties are the quotients under
the action of the torus of an open subset of $X$.  This "reverse
engineering" has appeared in other places, see for instance
Chapter 10 of Miller and Sturmfels book \cite{MS} where they take
an algebraic torus times a finite abelian group acting on $k^n$.

In this paper, we extend the construction of the fan associated to
the action given in \cite{R} to cover also Miller and Sturmfels
situation but we focus on making the process computational. Let
$H$ be an algebraic torus times a finite abelian group, we give an
algorithm to obtain a set of fans associated to the diagonal
action of $H$ on $X$. Using these fans, we will be able to give a
computational description of the graded components of Laurent
polynomial rings.

Let $S$ be the Laurent polynomial ring
$k[x_1,\ldots,x_{r},x_{r+1}^{\pm 1},\ldots ,x_n^{\pm 1}]$, with
$n=r+s$. The action of $H$ on $X$ extends to an action of $H$ on
$S$. As in the polynomial ring case, this action determines a
multigrading of $S$ by the finitely generated abelian group
$A=Hom(H,k^{\times})$. Multigradings of polynomial rings are
treated in detail in \cite{MS}, Chapter 8.  In section
\ref{multigrading}, we extend the definition of multigrading to
Laurent polynomial rings and in section \ref{compFan}, we provide
an algorithm to obtain a set of fans compatible with the
multigrading of $S$ by $A$. This fans allow us to see graded
components of $S$ as polyhedra. Each graded component $S_a$, $a\in
A$ is a module over the ring of differential invariants $\mathcal{D}
=k[x_1,\ldots,x_{r},x_{r+1}^{\pm 1},\ldots ,x_n^{\pm 1},
\partial_1 ,\ldots ,\partial_n]^H$ where $\partial_i=\partial/\partial
x_i$, $i=1,\ldots ,n$. In \cite{MR}, we studied graded components
of $S$ from this point of view . The connection between group
actions and finite fans allows a computational description of this
family of $\mathcal{D}$-modules that we present in this paper.

When computing the graded components of $S$, it is important to
decide whether they are finite dimensional as $k$ vector spaces or
not. Let $S_0$ be the $k$ vector space of homogeneous polynomials
of degree $0$. As in the polynomial ring case, either all the
graded components are finite or infinite dimensional as we
conclude in the last section of the paper. In section
~\ref{special}, we give an algorithm to check whether $S_0$ is
finite dimensional.

Given a finite fan $\varDelta$, a shift with nonempty intersection
of the dual cones of the cones of $\varDelta$  gives a polyhedron.
Furthermore, such a polyhedron is a polytope if and only if the
fan is not contained in a half-space, as we prove in section
~\ref{fans}. In section ~\ref{polytopes}, given a fan compatible
with the multigrading of $S$ by $A$, we use polyhedra obtained in
this manner to give a set of generators for each $S_a$ as a module
over $S_0$. If the graded components of $S$ are finite dimensional
we compute a $k$-basis of $S_a$ using the lattice points of a
polytope. We provide a set of linear inequalities determining this
polytopes so that the number of lattice points can be counted with
LattE, \cite{LTY}. When $S_0$ is infinite dimensional,  we
determine a finite set of generators $f_1,\ldots ,f_h$,
corresponding to the Hilbert basis of a cone, such that
$S_0=k[f_1,\ldots ,f_h]$. To determine the infinite dimensional
$S_a$, we compute a finite set of generators of $S_a$ as an
$S_0$-module corresponding to the lattice points of a polytope.

We begin our paper describing in section ~\ref{action} the group
$H$ and the diagonal action of $H$ on $X$. Throughout this paper,
a particular action will be specified by a matrix $L$ whose
columns are the weights of the action. In some special cases, we
can replace the matrix $L$ by another one with the special form
given in section ~\ref{specialmatrix} which gives the same action.
Most of the algorithms given in this paper will take the matrix
$L$ or its special form as an input.

\section{Description of the group and the action}\label{action}

We consider a diagonal action of $H$ on $X=k^r\times
(k^{\times})^s\subseteq k^n$ with $n=r+s$. This is an action that
extends to a diagonal action on $k^n$. Such an action is given by
an embedding of $H$ into the group $T$ of diagonal matrices in
$GL_n(\mathbb{Z})$. Details about this action in the algebraic torus
case are given in \cite{MR}, \S 2.1. The construction developed
there applies to the more general case treated here.

Identify $H$ with $G\times\mathbb{F}$ where $G=(k^{\times})^p$ and
$\mathbb{F}$ is the product of cyclic groups of orders $d_1,\ldots
,d_t$. We can identify the group of characters of $H$,
$A=Hom(H,k^{\times})$ with
\begin{displaymath}
A=\mathbb{Z}^p\oplus \mathbb{Z}/d_1\mathbb{Z} \oplus\ldots\oplus\mathbb{Z}/d_t\mathbb{Z}.
\end{displaymath}
We think of $A$ as a space of column vectors with integer entries.

There exist $\eta_1,\ldots ,\eta_n\in A$ such that {\em H acts on
$X$ with weights $\eta_{1},\ldots ,\eta_n$ .} Let $m=p+t$. We
denote by $L$ the $m\times n$ matrix with $i$-th column vector
$\eta_i$, $i=1,\ldots ,n$. We say that {\em H acts on $X$ by the
matrix $L$}. We may and we will assume that $H$ acts faithfully on
$X$. Therefore $L$ has rank $m$.

Let us also denote by $D$ the diagonal matrix with entries
$d_1,\ldots ,d_t$. This $t\times t$ matrix will be used in several
computations throughout this paper.

\section{Finite fans}\label{fans}

In this section we state some results about finite fans that will
be needed in the rest of the paper. As far as possible we follow
the notation of \cite{F} , Chapter 1.

Let $N= \mathbb{Z}^l$ be the $l$-dimensional lattice. Let $\varDelta $
be a fan in $N$, which is a collection of strongly convex rational
polyhedral cones in $N_{\mathbb{R}}=N\otimes _{\mathbb{Z}} \mathbb{R}\equiv\mathbb{R}^l$.
Through this paper, cones will be strongly convex, rational and
polyhedral except in a few occasions where we will say convex cone
to mean convex rational polyhedral cone.

Let $M=Hom_{\mathbb{Z}}(N,\mathbb{Z})$ and $\langle \_\;,\_\rangle : M\times
N\rightarrow \mathbb{Z}$ the natural bilinear pairing. For each
$\sigma\in\varDelta$, let
\begin{equation}\label{lambda}
\Lambda_{\sigma}=M\cap\sigma^{\vee}=\{u\in M|\langle u,v\rangle\geq
0\mbox{ for all }v\in\sigma\}.
\end{equation}

We review some terminology introduced in \cite{R}, \S 1.2. Denote
by $\varDelta (1)$ the set of cones of $\varDelta $ with dimension
one. Given $v\in N$, let $\tau_v=\mathbb{R}_+ v$ be the ray generated by
$v\in N$. Given $\sigma\in\varDelta$ we define $[\sigma
]=\{i\in\{1,\ldots ,r\} \mid \tau_{v_i}\mbox{ is a face of }\sigma
\}$. Given $u\in M_{\mathbb{R}}=M\otimes_{\mathbb{Z}}\mathbb{R}\equiv\mathbb{R}^l$, a
subset of the form $H_{u}=\{ v\in N_{\mathbb{R}} \mid \langle
u,v\rangle\geq 0\}$ with $u\neq 0$ is called a half-space in
$N_{\mathbb{R}}$. We say that the fan $\varDelta $ is {\em contained in
a half-space} if we can find $0\neq u\in M_{\mathbb{R}}$ such that
$\sigma\subseteq H_{u}$ for all $\sigma\in\varDelta$.

Given a fan $\varDelta$, let $\{v_1,\ldots ,v_r\}$ be a set of
vectors in $N$ generating the one dimensional cones of
$\varDelta$. Fix $\varphi=(\varphi_1,\ldots
,\varphi_n)\in\mathbb{N}^r\times\mathbb{Z}^s$ and define the sets
\begin{equation}\label{gamma-sigma-m}
\gamma_{\sigma ,\varphi}=\{u\in M_{\mathbb{R}} \mid \langle
u,v_{i}\rangle\geq -\varphi_i \mbox{ for all } i\in [\sigma] \}
\end{equation}
for each $\sigma\in\varDelta$. Consider the polyhedron
\begin{equation}\label{polytope}
\mathcal{P}_{\varphi}=\displaystyle{\cap_{\sigma\in\varDelta}}\;\gamma_{\sigma
,\varphi}= \{u\in M_{\mathbb{R}} \mid \langle u,v_i\rangle \geq
-\varphi_i\mbox{ for all }i=1,\ldots ,r\}.
\end{equation}
The purpose of the following two lemmas is to prove that
$\mathcal{P}_{\varphi}$ is bounded if and only if the fan $\varDelta$ is not
contained in a half-space.

\begin{lem}\label{continuous}
Fix an integer $a\geq 0$ and $v\in\mathbb{Z}^l$ and consider the set
$\Omega =\{u\in\mathbb{R}^{l} \mid \langle u,v\rangle\geq -a \} $. Given
$u\in S=\{u\in \mathbb{R}^{l}||u|=1\}$, let $l_{u}=sup \{\mu\in\mathbb{R}^{+}
| \mu u\in \Omega\}$. Then the map $f:S\rightarrow \mathbb{R}^{l}$
defined by $f(u)=l_{u}$ is continuous in its domain.
\end{lem}
\begin{proof}
It can be easily seen that if $l_{u}<\infty$, then
$[0,l_{u}]=\{\mu\in\mathbb{R}^{+} | \mu u\in \Omega\}$. In fact, $\langle
l_{u}u,v\rangle =-a$, i.e. if $v=(v_1,\ldots ,v_l )$ then $l_{u}u$
belongs to the hyperplane of $\mathbb{R}^{l}$ given by the equation
$v_{1}x_{1}+\ldots+v_{l}x_{l}=-a$. On the other hand $l_{u}u$ also
belongs to the line in $\mathbb{R}^{l}$ through the origin in the
direction of $u=(u_1,\ldots ,u_l )$. Therefore $l_u u$ is solution
of the matrix equation $M_{u}X=B$, where
\begin{displaymath}
M_u=\left[\begin{array}{cccc}
v_1 & v_2 &\ldots &v_l\\
u_2-u_1 &\; &\; &\; \\
\; &u_3-u_2 &\; &\; \\
\; &\; &\ddots &\;\\
\; &\; &\; &u_l-u_{l-1}
\end{array}\right],
\indent B=\left[\begin{array}{clcr}-a\\0\\
\vdots \\0\end{array}\right]
\end{displaymath}
and $X$ is the column vector of unknowns.

Let $M_{u}^{i}$ be the matrix obtained exchanging the $i$-th
column of $M_{u}$ by $B$. Using Cramer's Rule, and the fact
$l_{u}=|l_{u} u|$, we obtain the following expression for $f(u)$
\begin{displaymath}
f(u)=l_u= {\left( \sum_{i=1}^{r} {\left(
\frac{detM_{u}^{i}}{detM_{u}}\right)}^{2} \right)}^{\frac{1}{2}}
\end{displaymath}
This proves the result.
\end{proof}

\begin{lem}\label{bounded}
The set $\mathcal{P}_{\varphi}$ is bounded if and only if there does not
exist $u\in M_{\mathbb{R}}$, $u\neq 0$ such that $\mathbb{R}^{+}u\subset
\mathcal{P}_{\varphi}$.
\end{lem}
\begin{proof}
Obviously if $\mathcal{P}_{\varphi}$ is bounded the conclusion is clear. Let
us prove the other direction. Consider the sets
\begin{displaymath}
\Omega_{i}=\{u\in M_{\mathbb{R}} \mid \langle u,v_{i} \rangle\geq
-\varphi_i \}
\end{displaymath}
for all $i\in\cup_{\sigma\in\varDelta}\; [\sigma]=\{1,\ldots
,r\}$. For each $\sigma\in\varDelta$ we have $\gamma_{\sigma
,\varphi}=\cap_{i\in [\sigma ]}\;\Omega_{i}$. Then
$\mathcal{P}_{\varphi}=\cap \Omega_{i}$ where the intersection is taken
over all the $i\in\{1,\ldots ,r\}$. Let
\begin{equation}\label{lui}
l_{u}^{i}=sup\{\mu\in\mathbb{R}^{+} \mid \mu u\in \Omega_{i}\}
\end{equation}
for $u\in \mathbb{R}^{l}$ and $i\in\{1,\ldots ,r\}$. Then for each
$i\in\{1,\ldots ,r\}$ we define maps $f_{i}:S\longrightarrow
\mathbb{R}^{+}$ where $f_{i}(u)=l_{u}^{i}$ for every $u\in S$. Each
$f_{i}$ is continuous in its domain by Lemma ~\ref{continuous}.
Now we define the map $F:S\longrightarrow\mathbb{R}^{+}$ where
\begin{equation}\label{F}
F(u)=\mbox{inf}\{ l_{u}^{i} \mid i\in \{1,\ldots ,r\}\}
\end{equation}
for all $u\in S$. By \cite{Mu}, Chapter 2, Section 18, Exercise 8,
$F$ is continuous in its domain. By hypothesis,  there is no $u\in
M_{\mathbb{R}}$, $u\neq 0$ such that $\mathbb{R}^{+} u\subseteq
\mathcal{P}_{\varphi}$. Therefore the  domain of $F$ is $S$. By \cite{Co},
Chp. II 5.8, $F(S^{r-1} )$ is compact in $\mathbb{R}^{+}$ and by the
Heine-Borel theorem it is also bounded. Therefore we can find a
positive integer $N$ such that $F(u)\leq N$ for all $u\in S$. If
$y\in \mathcal{P}_{\varphi}$, then $y=\mu u$ for some $u\in S$ and by
~\eqref{F} and ~\eqref{lui}, $\mu\leq F(u)\leq N$. Therefore
$\mathcal{P}_{\varphi}$ in included in the sphere of radius $N$.
\end{proof}

\begin{prop}\label{half-space}
 The set $\mathcal{P}_{\varphi}$ is a polytope
 if and only if the fan $\varDelta $ is not
contained in a half-space.
\end{prop}
\begin{proof}
Suppose there exists $u\in M_{\mathbb{R}}$, $u\neq 0$ such that the fan
$\varDelta$ is contained in the half-space $H_{u}$. Then given
$i\in\{1,\ldots ,r\}$ and for any $\mu\in\mathbb{R}^{+}$ it holds that
$\langle\mu u,v_{i}\rangle\geq 0\geq -\varphi_i$. Thus
$\mathbb{R}^{+}u\subseteq \mathcal{P}_{\varphi}$ and hence $\mathcal{P}_{\varphi}$ is
not bounded.

Suppose now that $\mathcal{P}_{\varphi}$ is not bounded. Then by Proposition
~\ref{bounded} there exists $u\in M_{\mathbb{R}}$, $u\neq 0$ such that
$\mathbb{R}^{+}u\subseteq \mathcal{P}_{\varphi}$. By ~\eqref{gamma-sigma-m}, we
have $\mu \langle u,v_{i}\rangle\;\geq -\varphi_i$ for every
$i\in\{1,\ldots ,r\}$ and every $\mu\in\mathbb{R} ^{+}$. Then $\langle
u,v_{i}\rangle\geq\; 0$ for all $i\in\{1,\ldots ,r\}$. Therefore the
fan $\varDelta$ is contained in the half-space $H_{u}$.
\end{proof}

\begin{rem}\label{f}
A fan $\varDelta$ is not contained in a half-space if and only if
the intersection of the dual cones is zero,
$\cap_{\sigma\in\varDelta} \sigma^{\vee} =0$. Moreover, given a
set of generators $\{v_1,\ldots ,v_r\}$ of cones in $\varDelta
(1)$ and $0\neq u\in M_{\mathbb{R}}$, the following statements are
equivalent:
\begin{enumerate}
\item $\varDelta$ is contained in the half-space $H_u$.

\item The vector $u$ is a solution in $\mathbb{R}^l$ of the system of
linear inequalities $\{ \langle x, v_i\rangle \geq 0, i=1,\ldots
r\}$.

\item The set $\mathbb{R}_{\geq 0} \{v_1,\ldots ,v_r\}$ is a convex cone
contained in the half-space $H_u$.
\end{enumerate}
\end{rem}

There are well known algorithms to find a nonzero solution of a
system of linear inequalities $\{ \langle x, v_i\rangle \geq 0,
i=1,\ldots , r\}$. Because we need to use it in section
\ref{special}, we give next an algorithm to find such a solution
in the language of half-spaces. The algorithm is based on the
following lemma.

\begin{lem}\label{addvector}
Let $\sigma$ be a cone in $N_{\mathbb{R}}$ and let $v\in N_{\mathbb{R}}$. Then
$\mathcal{C} = \mathbb{R}_{\geq 0} (\sigma\cup \{v\})$ is a convex cone if and
only if $\mathcal{C}\subseteq H_u$ for some $u\in M_{\mathbb{R}}$ such that
$\langle u, v\rangle =0$.
\end{lem}
\begin{proof}
Let us suppose that $ \{v_1,\ldots ,v_r\}$ is a set of generators
of $\sigma$. Then  $ \{v_1,\ldots ,v_r, v\}$ is a set of
generators of $\mathcal{C}$. If $\mathcal{C}$ is a cone, by \cite{Sch}, Theorem
7.1 it is the intersection of the half-spaces $H_u$, $u\in
M_{\mathbb{R}}$ containing $ \{v_1,\ldots ,v_r, v\}$ and such that $\{
v\in N_{\mathbb{R}} \mid \langle u,v\rangle = 0\}$ is spanned by $l-1$
linearly independent vectors from $ \{v_1,\ldots ,v_r, v\}$. Let
$H_{u_0}$ be one the this half-spaces such that $\langle
u_0,v\rangle = 0$. This proves the result.
\end{proof}

\begin{al}\label{hsp1}
{\sc hsp1}($\{w_{1},\ldots ,w_t\}$ ,$\sigma$ , $u$) {\sf Given} a
convex cone $\sigma$ contained in the half-space $H_u$ with $0\neq
u\in M_{\mathbb{R}}$ and a set of vectors $\{w_{1},\ldots ,w_t\}$ in
$N_{\mathbb{R}}$, the algorithm {\sf decides }  whether $\mathcal{C}=\mathbb{R}_{\geq
0}(\sigma\cup \{w_{1},\ldots ,w_t\})$ is contained in a half-space
and if the answer is affirmative it {\sf returns } a nonzero
vector $u_0$ such that $\mathcal{C}$ is contained in $H_{u_0}$.
\begin{enumerate}
\item $i:=1$, $\mathcal{C}:=\sigma$.

\item While $i\leq t$ do
\begin{itemize}
\item[2.1] If $w_i\in\mathcal{C}$ then i:=i+1

\item[2.2] else take $u\in M_{\mathbb{R}}$ such that $\langle u,
w_i\rangle =0$. If $\mathcal{C}\subseteq H_{\pm u}$ then $\mathcal{C}:=\mathbb{R}_{\geq
0} (\mathcal{C}\cup \{w_i\})$, $i:=i+1$, $u_0:=\pm u$ else return {\sf
''Not contained in a half-space"}.
\end{itemize}
\item Return $u_0$ and {\sf ''Contained in a half-space"}.
\end{enumerate}
\end{al}

Let $\mathcal{V}=\{v_1,\ldots ,v_r\}$ be a set of generators of a fan
$\varDelta$ in $N$. If $\mathcal{V}$ has rank less than $l$ then
$\varDelta$ is contained in a half-space. Otherwise $\mathcal{V}$ contains
a subset $\mathcal{W}$ with $l$ linearly independent vectors. By
\cite{Sch}, Theorem 7.1, $\mathbb{R}_{\geq 0} \mathcal{W}$ is a cone contained
in the half-space $H_u$ for some $u\in M_{\mathbb{R}}$ such that
$\mathcal{W}\subseteq H_u$ and  $\{ v\in N_{\mathbb{R}} \mid \langle u,v\rangle
= 0\}$ is spanned by $l-1$ vectors of $\mathcal{W}$. Algorithm ~\ref{hsp1}
applied to ($\mathcal{V}\setminus\mathcal{W} $, $\mathbb{R}_{\geq 0} \mathcal{W}$, $u$) decides
whether $\varDelta $ is contained in a half-space.

\section{Multigradings}\label{multigrading}

The Laurent polynomial ring $S$ is the ring of regular functions on
$X$, $\mathcal{O} (X)$. We consider the action of $H$ on $\mathcal{O} (T)$ (or $\mathcal{O}
(X)$) given by right translation, see \cite{MR}, (12). This
convention implies that $x_i$ has weight $\eta_i$.

We say that a Laurent polynomial ring $S$ is {\em multigraded} by
$A$ when it has been endowed with a degree map $deg
:\mathbb{Z}^n\rightarrow A$. Multigradings of polynomial rings are
defined in \cite{MS}, \S 8.1 in the same manner.

Using the matrix $L$ that gives the action of $H$ on $X$, we
define a degree map as follows. The restriction map
$\mathbb{X}(T)=Hom(T,k^{\times})\longrightarrow \mathbb{X} (H)$ is given by
left multiplication by $L$. If we identify $\mathbb{X}(T)$ with $\mathbb{Z}^n$
we have an exact sequence (see lemma 4.1)
\begin{equation}\label{multigrading_sequence}
0\longleftarrow A\longleftarrow \mathbb{Z}^n\longleftarrow
\FRAK{K}\longleftarrow 0
\end{equation}
which determines a multigrading of S by $A$. Given $\lambda
=(\lambda_1,\ldots ,\lambda_n)$, we write
$x^{\lambda}=x_1^{\lambda_1}\cdot\ldots\cdot x_n^{\lambda_n}$. The
middle semigroup homomorphism $deg :\mathbb{Z}^n\rightarrow A$ gives a
multigrading of $S$ by $A$ sending each monomial $x^\lambda$ to
its {\em degree} $deg(\lambda )= L\lambda $. We will say that the
{\em multigrading of $S$ by $A$ is given by the matrix $L$}. Let
$Q=deg(\mathbb{N}^r\times\mathbb{Z}^s)$ denote the semigroup of $A$ generated
by $deg(x_1),\ldots ,deg(x_r), \pm deg(x_{r+1}),\ldots ,\pm
deg(x_n)$.

The next lemma proves that ~\eqref{multigrading_sequence} is an
exact sequence. Let $l=n-p$.

\begin{lem}\label{deg}
The degree map $deg$ is surjective and there exists an $n\times l$
matrix $K$ whose columns are a $\mathbb{Z}$-basis of $\FRAK{K}$.
\end{lem}
\begin{proof}
By ~\cite{A}, Theorem 12.4.3, there exist matrices $U\in GL_m(\mathbb{Z})$
and $V\in GL_n(\mathbb{Z})$ such that
\begin{equation}\label{smith}
L'=ULV=\left[\begin{array}{ccccccc}
f_1   &0  &\ldots &0     &0&\ldots&0\\
0     &f_2&\;     &0     &\;& \;  &\;\\
\vdots&\; &\ddots&\vdots &\vdots&\;&\vdots\\
0  &\ldots&\;     &f_m   &0&\ldots &0
\end{array}\right],
\end{equation}
with $f_i\neq 0$ for all $i=1,\ldots ,m$. Consider the $n\times l$
matrix
\begin{equation}
E=\left[\begin{array}{cc}
0&0\\
0&D\\
I_{l'}&0
\end{array}\right]
\end{equation}
with $D$ as in section \ref{action}, the integer $l'=l-t$ and
$I_{l'}$ the $l'\times l'$ identity matrix.

The columns of $E$ are a $\mathbb{Z}$-basis of the kernel of the group
homomorphism  $\mathbb{Z}^n\longrightarrow A$ given by right
multiplication by $L'$. Let $K:=VE$, then $LK=0$. Therefore the
columns of $K$ are a $\mathbb{Z}$-basis of the kernel of $L$. We can write
$\FRAK{K}$ as $K\mathbb{Z}^l$. Then $K$ is a presentation of
$deg(\mathbb{Z}^n)$. By \cite{A}, Proposition 5.12 we also have a
presentation $E=V^{-1}K$ of $deg(\mathbb{Z}^n)$. We conclude that
$deg(\mathbb{Z}^n)=A$.
\end{proof}

We are ready to introduce the graded components of $S$ that we are
computing in this paper. For $a\in A$, let $S_a$ denote the vector
space of homogeneous polynomials having degree $a$ in the
$A$-grading, that is

\begin{equation}
S_a=span\{x^\lambda\in S\mid L\lambda =a\}.
\end{equation}
Therefore $S=\bigoplus_{a\in Q} S_a$ since $S_a$ is empty if
$a\notin Q$ and $S_a\cdot S_b = S_{a+b}$, $a,b\in Q$. Note that
the subring $S^H$ of $S$ of invariants under the action of $H$
equals the semigroup ring
\begin{equation}\label{ceroComp}
S_0=k[\FRAK{K}\cap (\mathbb{N}^r\times\mathbb{Z}^s)].
\end{equation}

\section{Construction of fans compatible with the multigrading}\label{compFan}

The concept of fan compatible with the multigrading of a
polynomial ring was given in \cite{MS}, section 10.3. Next, we
extent the definition so that it applies to multigradings of
Laurent polynomial rings. In this section, we give an algorithm to
construct a fan compatible with a multigrading from the matrix $L$
describing the action of $H$ on $X$.

Applying the contravariant functor $Hom(\_, \mathbb{Z})$ to the sequence
~\eqref{multigrading_sequence} we obtain the exact sequence
\begin{equation}\label{dual}
0\longrightarrow Hom(A,\mathbb{Z})\longrightarrow \mathbb{Z}^n\longrightarrow
\FRAK{K}^{\vee}\longrightarrow Ext^1(A,\mathbb{Z})\longrightarrow 0.
\end{equation}

By the previous lemma ~\ref{deg}, $\FRAK{K}$ is a free abelian
group of dimension l. Then we identify $\FRAK{K}$ and
$\FRAK{K}^{\vee}= Hom(\FRAK{K},\mathbb{Z})$ with $\mathbb{Z}^l$. We fix the
columns of the matrix $K$ given by lemma ~\ref{deg} as a
$\mathbb{Z}$-basis of $\FRAK{K}$.

The middle morphism $\varphi : \mathbb{Z}^n\longrightarrow
\FRAK{K}^{\vee}\equiv \mathbb{Z}^l$ of the sequence ~\eqref{dual} is
given by right multiplication by the matrix $K$. Let $e_i$ be the
$i$-th standard basis vector of $\mathbb{Z}^n$ and call $v_i =\varphi
(e_i)$ the $i$-th row vector of $K$, $1\leq i\leq n$. In
particular, $\varphi$ is onto when $H$ is an algebraic torus, see
\cite{R}, paragraph following Lemma 2.2. We call $\{v_1,\ldots,
v_r\}$ a {\em set of vectors associated to the action} of $H$ on
$X$.

We call a fan $\varDelta $ in $N$ {\em compatible} with the
multigrading of $S$ by $A$ if its set of one dimensional cones
equals $\varDelta (1) =\{\tau_{v_i} \mid i=1,\ldots ,r\}$. The cones
of $\varDelta$ are subcones of $\mathcal{C}=\mathbb{R}_{\geq 0}\{v_1,\ldots,
v_r\}$.

The proof of lemma ~\ref{deg} provides an algorithm to compute a set
of vectors associated to the action of $H$ on $X$.

\begin{al}\label{compatible} {\sc associated-vectors}($L$,$D$)
{\sf Given} the $m\times n$ matrix $L$ associated to the action of
$H$ on $X=k^r\times (k^{\times})^s$ and the $t\times t$ matrix $D$
given in section \ref{action}, the algorithm {\sf returns} a set
of vectors $\{v_1,\ldots ,v_r\}$ associated to the action of $H$
on $X$.
\begin{enumerate}
\item Compute the Smith normal form of $L$; that is, compute
invertible matrices $U$ and $V$ such that $U\cdot L\cdot V$ is the
concatenation of a diagonal matrix and a zero matrix.

\item Let $l':=n-m$, and
\begin{equation} E:=\left[\begin{array}{cc}
0&0\\
0&D\\
I_{l'}&0
\end{array}\right].
\end{equation}

\item $K:=V\cdot E$.

\item Return the set $\{v_1,\ldots ,v_r\}$ of the first $r$ row
vectors of $K$.
\end{enumerate}
\end{al}

\section{Special form of the matrix $L$}\label{specialmatrix}

For some of the constructions in the next sections we need the
matrix $L$ (introduced in section \ref{action}) to have the
special form presented in the next lemma. Now suppose $H$ acts on
$X$ via the matrix $L$ and set
\begin{equation}
\Sigma_L=\{\lambda \in\mathbb{N}^r\times\mathbb{Z}^s \mid deg(\lambda )=0_A\},
\end{equation}
where $0_A$ is the class of zero in $A$. We show that in some
cases we can obtain a matrix $L'$ equivalent to $L$ with the
special form and giving an action of $H$ on $X$ with the same set
of invariants.

Given a matrix $M$ we will write $\textsl{subm}(M,r_1\ldots r_2,
c_1\ldots c_2)$ to denote the submatrix of $M$ obtained selecting
rows $r_1$ through $r_2$ and columns $c_1$ through $c_2$ of $M$.

\begin{lem}
If $\eta_{r+1},\ldots ,\eta _n$ are linearly independent, there
exist matrices $\Gamma\in GL_m (\mathbb{Z})$, $\Delta\in GL_n (\mathbb{Z})$ such
that
\begin{enumerate}
\item  $L'=\Gamma\cdot L\cdot\Delta$ has the block matrix form
\begin{equation}\label{newL}
\left[\begin{array}{cc}L_1&dI_p\\L_3&L_4\end{array}\right],
\end{equation}
where $I_p$ is the $p\times p$ identity matrix and $d$ is a nonzero
positive integer.

\item $\Sigma_L$ is isomorphic to $\Sigma_L'$.
\end{enumerate}
\end{lem}
\begin{proof}
\begin{enumerate}
\item Since $L$ has rank $m=p+t$ there exists a matrix $\Delta \in
GL_n(\mathbb{Z} )$ such that the last $p$ columns of $L\cdot\Delta$ are
linearly independent and such that $\Delta x\in
\mathbb{N}^r\times\mathbb{Z}^s$ for any $x\in\mathbb{N}^r\times\mathbb{Z}^s$. To be more
precise, if $p\leq s$ then $\Delta$ is the identity matrix since
$\eta_{r+1},\ldots ,\eta _n$ are linearly independent whereas if
$p>s$ then $\Delta$ is a permutation of the first $r$ columns of
$L$.

Furthermore, we can find a matrix $\Gamma_1\in GL_m(Z)$ permuting
the rows of $L\cdot\Delta$ such that $L_2= subm(\Gamma_1\cdot
L\cdot\Delta, 1\ldots p, l+1\ldots n)$ has rank $p$. Let $d=det(
L_2)$. Then $\Gamma_1=d L_2^{-1}$ is matrix in $GL_p (\mathbb{Z} )$. Let
\begin{displaymath}
\Gamma = \left[\begin{array}{cc}\Gamma_1&0\\0&D\end{array}\right]
\end{displaymath}
then $\Gamma\cdot L\cdot \Delta$ has the desired form
~\eqref{newL}.

\item Since $\Delta$ permutes only the first $r$ columns of $L$
and $\Gamma 0_A=0_A$ we can define the following isomorphism
\begin{equation}
\Sigma_L\longrightarrow \Sigma_{L'}
\end{equation}
given by $x\mapsto \Delta^{-1}x$.
\end{enumerate}
\end{proof}

From the proof of statement 1. in the previous lemma we derive the
following algorithm.

\begin{al} {\sc special-matrix}($L$,$D$)
{\sf Given} the $m\times n$ matrix $L$ associated to the action of
$H$ on $X=k^r\times (k^{\times})^s$, where the columns
$\eta_{r+1},\ldots ,\eta _n$ are linearly independent, and the
$t\times t$ matrix $D$ given in section \ref{action}, the
algorithm {\sf returns} an $m\times n$ matrix $L'$ with the
special from ~\eqref{newL}.
\begin{enumerate}
\item $p:=m-t$, $l:=n-p$.

\item If $p> s$ then permute the first $r$ columns of $L$ to
obtain $L_c$ so that\\ $rank(subm(L_c,1\ldots n,l+1\ldots
r))=p-s$.

\item $L:=L_c$. Permute the rows of $L$ to obtain $L_r$ so that
$rank(subm(L_r,1\ldots p, l+1\cdots n))= p$.

\item $L:=L_r$. $L_2:=subm(L, 1\ldots p, l+1\ldots n)$, $d:=det(
L_2)$, $\Gamma_2:=dL_2^{-1}$.

\item $\Gamma :=
\left[\begin{array}{cc}\Gamma_2&0\\0&D\end{array}\right]$.

\item Return $L':=\Gamma\cdot L$.
\end{enumerate}
\end{al}

\section{Determining the special case $S_0=k$}\label{special}

To describe the graded components of $S$ we need to decide in the
first place whether they are finite or infinite dimensional. For
that matter, it is enough to decide whether $S_0$ is finite
dimensional as we will see in section \ref{polytopes}. In this
section, we give an algorithm to decide whether $S_0 =k$ or
equivalently whether $S_0$ is finite dimensional.

We define next an isomorphism allowing us to go from points in the
lattice $M\equiv\mathbb{Z}^l$ to points in the lattice $\FRAK
{K}\subset\mathbb{Z}^n$ and we use it to prove the next proposition. There
is a natural bilinear pairing $(\;\;,\;):\mathbb{X}(T)\times
\mathbb{Y}(T)\longrightarrow \mathbb{Z}$ defined by the requirement that
$(a\circ b)(\lambda )=\lambda^{(a,b)}$ for all $a\in \mathbb{X}(T) $,
$b\in \mathbb{Y}(T)= Hom(k^{\times }, T)$ and $\lambda\in k^{\times }$. We
have an isomorphism $\omega :M\rightarrow \FRAK {K}$ given by
\begin{equation}\label{w}
\langle u,\varphi (b)\rangle =(\omega (u),b)
\end{equation}
for all $u\in M$, $b\in \FRAK {K}$.

\begin{thm}\label{fanS0}
The following statements are equivalent.
\begin{enumerate}
\item The only polynomials of degree 0 are the constants; i.e.
$S_0=k$.

\item The $k$-vector space $S_0$ is finite dimensional.

\item Any fan $\varDelta$ compatible with the multigrading of $S$
given by $A$ is not contained in a half-space.
\end{enumerate}
\end{thm}
\begin{proof}
The proof is analogous to \cite{R}, Lemma 4.2. Let $\varDelta$ be
a fan compatible with the multigrading of $S$ given by $A$. Let
\begin{align}\label{phi}
\phi _{\sigma } &=\{\lambda\in \FRAK{K} \mid (\lambda ,e_{i})\geq 0
\mbox{ for all }i \in [\sigma ]\}.
\end{align}
Then by ~\eqref{ceroComp},
$S_0=k[\cap_{\sigma\in\varDelta}\phi_{\sigma }]$. Therefore $S_0$
is finite dimensional if and only if
$\cap_{\sigma\in\varDelta}\phi_{\sigma }$ is a finite set.
Furthermore $\omega (\cap_{\sigma\in\varDelta} \Lambda_{\sigma
})=\cap_{\sigma\in\varDelta}\phi_{\sigma }$ and
$\cap_{\sigma\in\varDelta} \Lambda_{\sigma }$ is a finite set if
and only if $\cap_{\sigma\in\varDelta} \sigma^{\vee}$ is bounded.
This happens if and only if $\cap_{\sigma\in\varDelta}
\sigma^{\vee}=0$ which is equivalent to statements 1. and 3..
\end{proof}

An $A$-grading of $S$ verifying any of the equivalent conditions
given in the previous proposition is called {\em positive}.

\begin{prop}
The following are necessary conditions for the grading of $S$ by $A$
to be positive.
\begin{enumerate}
\item $p>s$.

\item $\eta_{r+1},\ldots ,\eta_n$ are linearly independent.
\end{enumerate}
\end{prop}
\begin{proof}
\begin{enumerate}
\item Let $\{v_1, \ldots ,v_r\}$ be a set of vectors associated to
the action of $H$ on $X$. If $p\leq s$ or equivalently $l\geq r$
then $\mathcal{C} =\mathbb{R}_{\geq 0}\{v_1, \ldots ,v_r\}$ is a cone and
equivalently $\varDelta$ is contained in a half-space. By theorem
~\ref{fanS0} the result follows.

\item Let $\hat{G}$ be the m dimensional algebraic torus, then
$H\subset\hat{G}$. Consider the action of $\hat{G}$ on $X$ given by
the matrix $L$, then $S^{\hat{G}}\subseteq S^H$. If
$\eta_{r+1},\ldots ,\eta_n$ are linearly dependent, ~\cite{MR},
Lemma 4.1(1) implies that $S^{\hat{G}}$ is not equal to $k$ and
therefore $S_0$ is not finite dimensional.
\end{enumerate}
\end{proof}

Let us suppose that $\eta_{r+1},\ldots ,\eta_n$ are linearly
independent. Then we can assume that $L$ has the special form
~\eqref{newL}. Let $\mathcal{V}=\{v_1,\ldots ,v_r\}$ be the output of
algorithm ~\ref{compatible} applied to $L_1$ and $D$, and let
$\varDelta$ be a fan compatible with the multigrading of $S$ by
$A$ given by $L$.

For $1\leq i\leq l$, let $\rho_i$ be the restriction of $\eta_i$ to
the subtorus $G$ of $H$. These characters can be thought of as the
columns of $L_1$.

\begin{lem}
If $S_0=k$ then $\rho_i\neq 0$ for all $i=1,\ldots ,l$.
\end{lem}
\begin{proof}
By \cite{R}, Lemma 4.2 if $\rho_i=0$ for some $i=1,\ldots ,l$ then
$\varDelta$ is contained in a half-space and by proposition
~\ref{fanS0} the result follows.
\end{proof}

The converse of the previous lemma does not hold but we can modify
the matrix $L$ to get an action of $H$ on $X$ whose only
invariants are the constant polynomials whenever $\rho_i\neq 0$
for all $i=1,\ldots ,l$. Given a set $I\subseteq \{1,\ldots ,r\}$
define
\begin{equation}
v_i^I=\left\{\begin{array}{c}
-v_i\mbox{ if }i\in I\\
v_i\mbox{ if }i\notin I
\end{array}\right., i=1,\ldots ,r.
\end{equation}
Let $\varDelta_I$ be a fan in $N$ with $\varDelta_I
(1)=\{\tau_{v_i^I} |i=1,\ldots ,r\}$. For $1\leq i\leq n$, set
\begin{equation}\label{LI}
\varsigma_i=\left\{ \begin{array}{c}
-\eta_i\mbox{ if }i\in I\\
\eta_i\mbox{ if }i\notin I
\end{array}\right. .
\end{equation}
Let $L_I$ be the matrix with columns $\varsigma_1,\ldots
,\varsigma_n$. Then $H_I$ denotes the group $H$ acting on $X$ by the
matrix $L_I$. Then the fan $\varDelta_I$ is compatible with the
multigrading of $S$ by $A$ given by $L_I$.

We explain next how to obtain a set $I$ so that $S^{H_I}=k$
whenever $\rho_i\neq 0$ for all $i=1,\ldots ,l$. Let us call
$b_{i,j}$ the entries of $L_1$, $i=1,\ldots ,p$, $j=1,\ldots ,l$.

\begin{lem}\label{lin}
When the matrix $L$ is of the special kind ~\eqref{newL}, then
$v_1,\ldots ,v_l$ are linearly independent.
\end{lem}
\begin{proof}
By Lemma ~\ref{deg}, $LK=0$. Then for $j=l+1,\ldots ,n$
\begin{equation}
dv_j=-\sum_{i=1}^l b_{j-l,i}v_i.
\end{equation}
Thus $v_{l+1},\ldots ,v_n$ belong to the $\mathbb{R}$-span of
$v_1,\ldots ,v_l$. By lemma ~\ref{deg} $K$ has rank $l$ and the
result follows.
\end{proof}

By Lemma ~\ref{lin}, $\mathcal{B}=\{v_1,\ldots ,v_l\}$ is a basis of
$N_{\mathbb{R}}$ and  respect to $\mathcal{B}$ the vectors $v_{l+1},\ldots ,v_n$
have coordinates
\begin{equation}\label{coordinates}
v_j=-\frac{1}{d}(b_{j-l,1},\ldots ,b_{j-l,l}),\indent j=l+1,\ldots
,n.
\end{equation}
Let $v_1^*,\ldots ,v_l^*$ be the dual basis of $\mathcal{B}$. Then
\begin{equation}\label{coordinates2}
\langle v_i^*, v_j\rangle =\frac{-1}{d}b_{j-l,i}
\end{equation}
for $i=1,\ldots ,l$, $j=l+1,\ldots ,r$.

Let us suppose that $p>s$ or equivalently $l<r$. Given
$j\in\{l+1,\ldots ,r\}$, let
\begin{equation}\label{Iplus}
I_j^+=\{i\in\{1,\ldots ,l\} \mid \frac{-1}{d}b_{j-l,i} \;>0\},
\end{equation}
\begin{equation}\label{Iminus}
I_j^-=\{i\in\{1,\ldots ,l\} \mid \frac{-1}{d}b_{j-l,i} \;<0\}.
\end{equation}

\begin{lem}
If $\rho_i\neq 0$ for all $i=1,\ldots ,l$ then there exists
$J\in\{l+1,\ldots ,r\}$ such that
\begin{equation}\label{J}
\cup_{j=1}^J (I_j^+\cup I_j^-)=\{1,\ldots ,l\}.
\end{equation}
\end{lem}
\begin{proof}
Otherwise there exists $i\in \{1,\ldots ,l\}$ such that
\begin{equation*}
\frac{-1}{d}b_{j-l,i}=0 ,\; j=l+1,\ldots ,r,
\end{equation*}
then $\rho_i=0$.
\end{proof}

This lemma ensures that the next algorithm terminates.

\begin{al}\label{I} {\sc positivity-set}($L_1$, $d$)
{\sf Given} the matrix $L_1$, where the columns $\rho_i\neq 0$ for
all $i=1,\ldots ,l$, and the positive integer $d$ in the special
form of the matrix $L$ giving the action of $H$ on $X$, the
algorithm {\sf returns} a chain of subsets
$\mathcal{I}_{l+1}\subseteq\ldots\subseteq\mathcal{I}_J$ of $\{1,\ldots ,l\}$ such
that the grading of $S$ by $A$ given by $L_{\mathcal{I}_J}$ is positive.
\begin{enumerate}
\item $J:=l+1$.

\item Compute $I_J^-$ and $I_J^+$ using ~\eqref{Iplus} and ~\eqref{Iminus}.

\item $\mathcal{I}_J:=I_J^+$.

\item While $\cup_{j=l+1}^J (I_j^-\cup I_j^+)\neq \{ 1,\ldots
,l\}$ then
\begin{enumerate}
\item[4.1] $I_J^0:=\{ 1,\ldots ,l\}\setminus I_J$.

\item[4.2] Compute $I_{J+1}^-$ and $I_{J+1}^+$.

\item[4.3] $\mathcal{I}_{J+1}:=\mathcal{I}_J\cup((\cap_{j=l+1}^{J}I_j^0)\cap
I_{J+1}^+)$.

\item[4.4] $J:=J+1$.
\end{enumerate}
\item Return $\mathcal{I}_{l+1},\ldots ,\mathcal{I}_J$.
\end{enumerate}
\end{al}

The next proposition shows that the grading of $S$ by $A$ given by
$L_{\mathcal{I}_J}$, with $\mathcal{I}_J$ as in the output of algorithm ~\ref{I}, is
positive.

\begin{prop}\label{not in half-space}
Let us suppose that $\rho_i\neq 0$ for all $i=1\ldots l$ and let
$\mathcal{I}_J$ be the subset of $\{1,\ldots ,l\}$ obtained by {\sc
positivity-set}($L_1$,$d$), then the grading of $S$ by $A$ given
by $L_{\mathcal{I}_J}$ is positive.
\end{prop}
\begin{proof}
It can be proved as in \cite{R}, Proposition 4.5 that
$\varDelta_\mathcal{I} $ is not contained in a half-space. The result
follows by theorem ~\ref{fanS0}.
\end{proof}

If $\mathcal{I}_J=\emptyset$ then the previous proposition ensures that
$\varDelta = \varDelta_{\emptyset}$ is not contained in a
half-space. On the other hand, if $\mathcal{I}_J\neq\emptyset$ the fan
$\varDelta$ might still be contained in a half-space. We give next
an algorithm to check whether $\varDelta$ is contained in a
half-space using the output set of algorithm ~\ref{I}. The
algorithm is based on the following fact.

\begin{lem}
Let $\{\mathcal{I}_{l+1} ,\ldots ,\mathcal{I}_J\}$ be the output of {\sc
positivity-set}($L_1$, $d$). Given $k\in \{1,\ldots ,J-(l+1)\}$,
if $\mathcal{I}_{J-k}=\emptyset$ and $\mathcal{I}_{J-k+1}\neq \emptyset$ then
$\mathcal{C}= \mathbb{R}_ {\leq 0}\{v_1,\ldots ,v_{J-k+1}\}$ is contained in
$H_{v_i^*}$ for all $i\in \mathcal{I}_{J-k+1}$.
\end{lem}
\begin{proof}
Given $i\in \mathcal{I}_{J-j+1}$ then $i\in (\cap_{j=l+1}^{J-k}I_j^0)\cap
I_{J-k+1}^+$. Thus by ~\eqref{coordinates2}, ~\eqref{Iplus},
~\eqref{Iminus} and 4.1 in algorithm ~\ref{I} we have $\langle
v_i^*, v_j\rangle\geq 0$ for all $j=l+1,\ldots ,J-k+1$. This
proves the result.
\end{proof}

\begin{al}\label{hsp2}
{\sc hsp2}($\{\mathcal{I}_{l+1} ,\ldots ,\mathcal{I}_J\}$, $\mathcal{V}$) {\sf Given}
$\{\mathcal{I}_{l+1} ,\ldots ,\mathcal{I}_J\}$ the output set of algorithm
~\ref{I} and $\mathcal{V}=\{v_1,\ldots ,v_r\}$ a set of vectors associated
to the action of $H$ on $X$, the algorithm {\sf decides} whether
$\mathcal{C}=\mathbb{R}_{\geq 0} \mathcal{V}$ is contained in a half-space and if the
answer is affirmative it {\sf returns} a vector $0\neq u\in
M_{\mathbb{R}}$ such that $\mathcal{C}$ is contained in $H_u$.
\begin{enumerate}
\item $\mathcal{C}:=\mathbb{R}_{\geq 0}\{v_1,\ldots ,v_l\}$.

\item If $\mathcal{I}_{l+1},\ldots ,\mathcal{I}_J$ are nonempty sets then
\begin{itemize}
\item[2.1] $\mathcal{C}:=\mathbb{R}_{\geq 0}(\mathcal{C}\cup\{v_{l+1}\})$, $u:=v_i^*$ with $i\in
\mathcal{I}_{l+1}$.

\item[2.2] If $J=l+1$ return $u$ and {\sf '' Contained in a half-space"}
else return {\sc hsp1}($\{v_{l+2},\ldots ,v_J\}$, $\mathcal{C}$ , u).
\end{itemize}

\item If $\mathcal{I}_J=\emptyset$ then return  {\sf '' Not contained in a half-space"}.

\item Let $1\leq i\leq J-(l+1)$ be the smallest such $\mathcal{I}_{J-i}=\emptyset$ and $\mathcal{I}_{J-i+1}\neq\emptyset$.
\begin{itemize}
\item[4.1] $\mathcal{C}:=\mathbb{R}_{\geq 0}(\mathcal{C}\cup\{v_{l+1},\ldots
,v_{J-i+1}\})$ and $u:=v_i^*$ with $i\in \mathcal{I}_{J-i+1}$.

\item[4.2] If $J-i+1=J$ return $u$ and {\sf '' Contained in a half-space"}.

\item[4.3] Return {\sc hsp1}($\{v_{J-i+2},\ldots ,v_J\}$, $\mathcal{C}$ ,
u).
\end{itemize}

\end{enumerate}
\end{al}

From the previous results we derive the following algorithm to
test whether an $A$-grading is positive.

\begin{al} {\sc positivity-test}($L$)
{\sf Given} the $m\times n$ matrix $L$ associated to the action of
$H=G\times \mathbb{F}$ on $X=k^r\times (k^{\times})^s$ where $G$ is a
torus of dimension $p$, the algorithm {\sf decides} whether the
$A$-grading of $S$ by $L$ is positive and if the answer is
negative it {\sf returns} $\mathcal{I}\subset\{1,\ldots l\}$ such that
$A$-grading of $S$ by $L_{\mathcal{I}}$ is positive.
\begin{enumerate}
\item If $p>s$ or $\eta_{r+1},\ldots ,\eta_n$ are linearly
dependent return {\sf ''The grading is not positive"}.

\item
$\left[\begin{array}{cc}L_1&dI_p\\L_3&L_4\end{array}\right]:=${\sc
special-matrix}($L$,$D$).

\item If any of the columns of $L_1$ is $0$ then return {\sf ''The
grading is not positive"}.

\item $\{\mathcal{I}_{l+1},\ldots ,\mathcal{I}_J\}:=${\sc
positivity-set}($L_1$,$d$).

\item If {\sc hsp2}($\mathcal{I}_{l+1} ,\ldots ,\mathcal{I}_J$) returns {\sf ''Not
contained in a half-space"} then return {\sf''The grading is
positive"} else return $\mathcal{I}_J$ and {\sf ''The grading is not
positive"}.
\end{enumerate}
\end{al}

\section{Polyhedral description of graded components of S}\label{polytopes}

In this section we describe the graded components of $S$ in terms
of polyhedra. We distinguish to main cases depending on the
dimension.

Let $\varDelta$ be a fan compatible with the multigrading of $S$
by $A$ given by $L$. Given $a\in Q$, there exists
$\varphi=(\varphi_1,\ldots ,\varphi_n)\in\mathbb{N}^r\times\mathbb{Z}^s$ such
that $L\varphi=a$. Consider the polyhedron $\mathcal{P}_{\varphi}$ defined
by ~\eqref{polytope}.

\begin{lem}\label{Sa}
Let $a\in Q$ then for any $\varphi=(\varphi_1,\ldots
,\varphi_n)\in\mathbb{N}^r\times\mathbb{Z}^s$ such that $L\varphi=a$ it holds
\begin{equation}\label{Sa2}
S_a=x^{\varphi}k[\omega (\mathcal{P}_{\varphi}\cap\mathbb{Z}^l)],
\end{equation}
with $\omega$ as in ~\eqref{w}.
\end{lem}
\begin{proof}
For each $a\in Q$ we have $S_a=k[(\varphi + \FRAK{K})\cap
(\mathbb{N}^r\times\mathbb{Z}^s)]$. Given $\sigma\in\varDelta$ let
\begin{align}\label{phim}
\phi _{\sigma ,a} &=\{\varphi +\mu\in \varphi +\FRAK{K} \mid (\mu
,e_{i})\geq -\varphi_i\mbox{ for all } i\in [\sigma ]\}.
\end{align}
Then $S_a=k[\displaystyle{ \cap_{\sigma\in\varDelta}} \phi
_{\sigma ,a}]$. Define the sets
\begin{equation}
\psi_{\sigma ,a}=\{\lambda\in \FRAK{K} \mid (\lambda ,e_{i})\geq
-\varphi_i, \mbox{ for all }i\in [\sigma ]\}.
\end{equation}
Thus $\phi_{\sigma ,a}=\varphi +\psi_{\sigma ,a}$ and
$k[\phi_{\sigma ,a} ]= x^{\varphi } k [\psi_{\sigma ,a} ]$.
Therefore $S_a=x^{\varphi} k[ \cap_{\sigma} \psi_{\sigma ,a} ]$.
Let
\begin{equation}
\Lambda_{\sigma ,a}=\{x\in M\mid \langle x,v_{i}\rangle\geq
-\varphi_i\;\mbox{for all}\; i\in [\sigma ]\}.
\end{equation}
Then $\psi_{\sigma ,a}=\omega (\Lambda_{\sigma ,a})$ and $\omega
(\cap_{\sigma\in\varDelta} \Lambda_{\sigma ,a })= \cap_{\sigma}
\psi_{\sigma ,a}$. Finally $\cap_{\sigma\in\varDelta}
\Lambda_{\sigma ,a }=\mathcal{P}_{\varphi}\cap \mathbb{Z}^l$.
\end{proof}

The following theorem gives two more properties of positive
gradings.

\begin{thm}
The following statements are equivalent:
\begin{enumerate}
\item There exits $a\in A$ such that the $k$-vector space $S_a$ is
finite dimensional.

\item For all $a\in A$, the $k$-vector space $S_a$ is finite
dimensional.

\item Any fan compatible with the multigrading of $S$ given by $A$
is not contained in a half-space.
\end{enumerate}
\end{thm}
\begin{proof}
Let $\varDelta$ be a fan compatible with the multigrading of $S$
by $A$. Given $\varphi\in\mathbb{N}^r\times\mathbb{Z}^s$, by proposition
~\ref{half-space} the set $\mathcal{P}_{\varphi}\cap \mathbb{Z}^l$ is finite if
and only if $\varDelta$ is not contained in a half-space. Given
$a\in Q$ and $\varphi\in\mathbb{N}^r\times\mathbb{Z}^s$ such that $L\varphi
=a$ by the previous lemma $S_a$ is finite dimensional if and only
if $\mathcal{P}_{\varphi}\cap \mathbb{Z}^l$ is finite. This proves the result.
\end{proof}

\subsection{Finite dimensional case}

At this point we can determine the graded components in the finite
dimensional case.

\begin{thm}\label{thmFin}
Let us suppose that the grading of $S$ by $A$ given by $L$ is
positive. Given $a\in Q$, then the dimension of $S_a$ equals the
number of lattice points inside the polytope $\mathcal{P}_{\varphi}$ for any
$\varphi$ such that $L\varphi =a$.
\end{thm}
\begin{proof}
By proposition  ~\ref{half-space}, the polyhedron $\mathcal{P}_{\varphi }$
is bounded. By lemma ~\ref{Sa} the graded component $S_a$ is
spanned over $k$ by the finite set of monomials $x^u$ such that
\begin{displaymath}
u\in \varphi+\omega (\mathcal{P}_{\varphi}\cap \mathbb{Z}^l).
\end{displaymath}
This proves the result
\end{proof}

\begin{rem}
{\sc polyhedral description} The proof of theorem ~\ref{thmFin}
provides an algorithm to determine $S_a$ using a polytope.
\end{rem}

Let us suppose that the grading of $S$ by $A$ given by $L$ is
positive, that is $S_0=k$. Then by \cite{MR}, Lemma 4.1(1) the
vectors $\eta_{r+1},\ldots ,\eta_n$ are linearly independent and
we can assume that $L$ is of the special kind ~\eqref{newL}. Let
$\{v_1,\ldots ,v_r\}$ be a set of vectors associated to the action
of $H$ on $X$. By lemma ~\ref{lin}, the set
$\bar{\mathcal{B}}=\{-v_1\ldots ,-v_l\}$ is an $N_{\mathbb{R}}$ basis. Let us
consider the $r\times l$ matrix $P$ whose $i$-th row is the row
vector of the coordinates of $v_i$ in the $N_\mathbb{R}$ basis
$\bar{\mathcal{B}}$, $i=1\ldots r$. In the dual basis of $\bar{\mathcal{B}}$ the
polytope $\mathcal{P}_{\varphi}$ equals the set
\begin{equation}\label{poli}
\{x\in \mathbb{R}^l| Px\leq \varphi_i, i=1,\ldots ,r\}.
\end{equation}

\subsection{Infinite dimensional case}

We describe next the graded components of $S$ for the infinite
dimensional case. Let us suppose that $S_0$ is not finite
dimensional. Let $\mathcal{V} =\{v_1,\ldots ,v_r\}$ be a set of vectors
associated to the action of $H$ on $X$ and let $\varDelta$ be a
fan compatible with the multigrading of $S$ by $A$. By theorem
~\ref{fanS0}, $\varDelta$ is contained in a half-space and by
remark ~\ref{f}, the set $\mathcal{C} =\mathbb{R}_{\geq 0} \{v_1,\ldots ,v_r\}$
is a cone. Let $\mathcal{C}^{\vee}$ be the dual cone of $\mathcal{C}$. By
\cite{Sch}, Theorem 16.7 there exists a Hilbert basis of
$\mathcal{C}^{\vee}$.
\begin{prop}
Let $S_0$ be infinite dimensional and let $\mathcal{H}=\{w_1,\ldots
,w_h\}$ be a Hilbert basis of $\mathcal{C}^{\vee}$. Then
\begin{displaymath}
S_0 = k[x^{\omega (w_1)},\ldots ,x^{\omega (w_h)}].
\end{displaymath}
\end{prop}
\begin{proof}
 For each
$\sigma\in\varDelta$, let $\phi _{\sigma }$ be defined as in
~\eqref{phi} and $\Lambda_{\sigma}$ as in ~eq\ref{lambda}. Then
$\cap_{\sigma\in\varDelta} \phi_{\sigma} = \FRAK{K}\cap
(\mathbb{N}^r\times\mathbb{Z}^s)$,  $\cap_{\sigma\in\varDelta}
\Delta_{\sigma}= \mathcal{C}^{\vee}\cap M$ and $\omega
(\cap_{\sigma\in\varDelta} \Delta_{\sigma}) =
\cap_{\sigma\in\varDelta} \phi_{\sigma}$. Thus $\{ \omega
(w_1),\ldots ,\omega (w_h)\}$ is a minimal generating set of the
partially ordered set $\FRAK{K}\cap (\mathbb{N}^r\times\mathbb{Z}^s)$ and this
proves the result.
\end{proof}

Every graded component of $S$ is infinite dimensional. Given $a\in
Q$ let $\varphi=(\varphi_1,\ldots
,\varphi_n)\in\mathbb{N}^r\times\mathbb{Z}^s$ such that $L\varphi=a$. The next
result will be used to determine $S_a$.

By proposition ~\ref{half-space}, the polyhedron $\mathcal{P}_{\varphi }$
is not bounded. By \cite{Sch}, Corollary 7.1b then $\mathcal{P}_{\varphi
}=\mathcal{P} +\mathcal{C}^{\vee}$ for some polytope $\mathcal{P}$.

\begin{lem}\label{descomposition}
Let $\mathcal{H}=\{w_1,\ldots ,w_h\}$ be a Hilbert basis of $\mathcal{C}^{\vee}$,
let $B =\{ \sum_{i=1}^h \alpha_i w_i \mid 0\leq \alpha_i \leq 1\}$
and let $\mathcal{P}$ be a polytope such that $\mathcal{P}_{\varphi } = \mathcal{P}
+\mathcal{C}^{\vee}$. Then

\begin{equation}
\mathcal{P}_{\varphi }\cap\mathbb{Z}^l = ((\mathcal{P}
+B)\cap\mathbb{Z}^l)+(\mathcal{C}^{\vee}\cap\mathbb{Z}^l)
\end{equation}

\end{lem}
\begin{proof}
We prove the nontrivial inclusion $\mathcal{P}_{\varphi }\cap\mathbb{Z}^l
\subseteq ((\mathcal{P} +B)\cap\mathbb{Z}^l)+(\mathcal{C}^{\vee}\cap\mathbb{Z}^l)$. Given
$p\in \mathcal{P}_{\varphi }\cap\mathbb{Z}^l$ then $p=q+c$ with $q\in Q$ and
$c\in \mathcal{C}^{\vee}$. We can write $c=\sum \alpha_i w_i$ with
$\alpha_i\geq 0$. Consider the integral vector
 $c'=\sum \lfloor \alpha_i\rfloor w_i$, thus $b=c-c'\in B$. We
 can write $p=q+b+c'$ with $q+b=p-c'\in(\mathcal{P} +B)\cap\mathbb{Z}^l$ and $c'\in \mathcal{C}^{\vee}\cap\mathbb{Z}^l$.
\end{proof}

Given a set of generators of a cone, we can use the algorithm given
in \cite{H}, section 5.5 to compute a Hilbert basis of the dual
cone.

\begin{thm}\label{thmInf}
For all $a\in Q$, $S_a$ is finitely generated as an $S_0$-module.
\end{thm}
\begin{proof}
Let us consider the cone $\bar{\mathcal{C}}$ generated by vectors
$(v_i,\varphi_i)\in\mathbb{Z}^{l+1}$, $i=1,\ldots ,r$ and
$(0_{\mathbb{Z}^l},1)\in \mathbb{Z}^{l+1}$ with $0_{\mathbb{Z}^l}$ the zero vector
in $\mathbb{Z}^l$. By \cite{Sch}, Theorem 16.7 there exists a Hilbert
basis $\mathcal{H}$ of the dual cone $\bar{\mathcal{C}}^{\vee}$ of $\bar{\mathcal{C}}$
which equals
\begin{displaymath}
\{(x,\lambda)\in \mathbb{R}^l\times\mathbb{R}_{\geq 0}\mid \langle
x,-v_i\rangle -\lambda\varphi_i\leq 0\}
\end{displaymath}
The set $\mathcal{H}_0=\{(x,\lambda)\in\mathcal{H}\mid \lambda=0\}$ is a Hilbert
basis of $\mathcal{C}^{\vee}$. Let $\mathcal{P}$ be the convex hull of the
$x\in\mathbb{Z}^l$ with $(x,1)\in \mathcal{H}$. Now $x\in \mathcal{P}_{\varphi}$ if and
only if $(x,1)\in \bar{\mathcal{C}}^{\vee}$. Then $\mathcal{P}_{\varphi } = \mathcal{P}
+\mathcal{C}^{\vee}$.

By lemma ~\ref{Sa} and lemma  ~\ref{descomposition} we conclude
that $S_a$ is generated as an $S_0$-module by the elements $x^u$
such that $u\in \varphi+\omega ((\mathcal{P} +B)\cap\mathbb{Z}^l)$.
\end{proof}

\begin{rem}
{\sc polyhedral description} The proof of theorem ~\ref{thmInf}
provides an algorithm to determine $S_a$ using polyhedra.
\end{rem}

redpajama_set_name: RedPajamaArXiv


text:
Streaming TVShould you get Disney Plus in Ireland?
Should you get Disney Plus in Ireland?
If you love a feel good Disney classic or a piece of Marvel action then Disney Plus could be for you. Here's what you need to know about the next big streaming service.
What is Disney Plus?
It's an on demand streaming service that's free of adverts and brought to us by The Walt Disney Company.
It includes content from these Wonderful Worlds:
With its huge range of family entertainment, Disney+ rivals the likes of Netflix, Amazon Prime Video and Apple TV+.
Subscribe to Disney+ Ireland
Pay €89.90 upfront for the year or €8.99 per month to start streaming.
Disney+ Ireland costs €89.90 a year or €8.99 a month so it's cheaper to pay upfront if you can.
You'll save €17.98 by paying annually, which is the equivalent of getting two months free.
What's included in the price?
You'll get access to over 1,000 movies, series, and documentaries from: Disney, Pixar, Marvel, Star Wars, National Geographic and Star.
You'll also be able to watch exclusive Disney+ originals including feature length films and short form content.
If the whole family want to watch content at the same time, you can stream any series or movie on up to four screens at once.
What movies and series are available?
There will be classics like The Lion King and Grey's Anatomy at your fingertips as well as a range of new content like:
The Mandalorian and The Bad Batch from Star Wars
Raya the Last Dragon, Frozen II and Mulan from Disney
The Falcon, Loki, Avengers Endgame and Wanda Vision from Marvel
Luca, Monsters at Work and Soul from Pixar
The world according to Jeff Goldblum and Gordon Ramsey: Unchartered from National Geographic
Family Guy, The Walking Dead and Nomadland from Star
What is Star?
Star is the latest channel to join Disney Plus, and it brings you TV series, movies and new originals.
Its content is geared towards adults, allowing you to binge on series like 24, Prison Break and The X Files, and movies like Die Hard and Grosse Pointe Blank.
Due to its more adult themed content, you'll need to check you have the right parental controls set up for your kids, and ensure you can access the higher rated shows.
What do you need to get started?
To start watching and downloading content, you'll need:
A reliable broadband service with a speed of around 3Mbps
Enough data and storage for downloads - that you can watch without an internet connection
To find out whether your broadband is suitable for streaming, read our guide on the best broadband for streaming.
What happens when you subscribe to Disney Plus?
Once you've subscribed, you'll be able to watch content from most internet enabled devices that stream video, for example:
Smart phones, tablets and laptops
Watch from up to four devices at once
Do unlimited downloads on up to ten devices
Create up to seven account profiles for the family
You'll have access to over 200 avatars to choose from to personalise each family member's profile.
How does it compare to other streaming services?
There's no free trial for Disney+ currently, but paying upfront for the year at €89.90 works out at €7.49 a month.
An Entertainment Membership with NOW formerly Now TV is currently €15.00 per month. There's more flexibility, with a cancel anytime policy and the choice of TV and movies is wider.
Amazon Prime Video costs £5.99 (around €6.99) a month and our dedicated Amazon guide outlines what it includes to help you compare the two services.
You can also check out our TV streaming guide which looks at the best streaming services available in Ireland.
Is Disney+ worth getting?
How good value Disney+ is, depends on how much of the content you and your family will watch.
Even if you're not the biggest Disney fan, there's loads of content to enjoy on the other channels, including Star, which brings a welcomed boost of content for the adult members of the family.
At just €89.90 for the year, it offers quite a low cost family entertainment option, and gives you access to exciting new content sooner.
So, if you're looking for top quality entertainment for the whole family in one place, it's certainly worth checking out the Disney+ streaming service.
Be aware that your Disney+ subscription will automatically renew at the end of the billing period if you don't cancel.
Disney Plus FAQs
Can I pay for Disney Plus monthly?
Yes, it costs €8.99 a month which works out more expensive for the year by €17.98, compared with paying €89.90 upfront.
Can I watch Disney+ on my NOW TV smart stick?
Yes. You'll need to buy Disney+ direct from Disney first. Then, you can download the app from the NOW TV device App Store and sign into your Disney+ account to start watching.
Disney+ isn't included in any NOW TV Passes, you have to subscribe to Disney directly.
How do I cancel my Disney Plus subscription?
You can cancel any time before the end of the billing period and you'll continue to receive the service until then too. You won't get a refund for any unused days in a billing period.
If you don't cancel before the end of the billing period, your subscription will automatically renew and payment will be taken from the card you have registered.
Here are full cancellation terms including how to cancel.
What will happen to my DisneyLife subscription when Disney Plus launches?
Disney Life is being rebranded as Disney+ so if you're currently a Disney Life customer, you'll get communication from Disney about how to move your subscription over.
When did Disney Plus launch in Ireland?
Disney+ launched in Ireland on 24 March 2020.
It was also rolled out in the UK, France, Germany, Italy, Spain, Austria and Switzerland on the same day.

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: How to manipulate data in an Excel Spreadsheet with Google Apps Script I have stored a Excel spreadsheet in my Google Drive. I am able to open it with `Excel 2003 and with Google sheets. The operating system is Windows 10.
My goal is to update the spreadsheet with data from a Google Form and use the spreadsheet as an input to a windows program.
I have tried to open the spreadsheet with methods openById(id) and openByUrl(url) without any success.
Is there any way I can manipulate the data in this spreadsheet with Google Apps Script?

redpajama_set_name: RedPajamaStackExchange


I assume your organisation has already started the digitalization journey, you have gained some hands-on experience and it is time to scale results for larger benefits. For example, technical experiment on AI or IoT may have taken place. You'd like to proceed without creating an overwhelming change load or future cost chaos in the impacted organisations. Planning rules may still be a bit confusing and thus, for example, company wide and product level development projects may be partially overlapping.
Gartner's CIO agenda 2019 estimates that 33% of organisations have reached digital scale, digital business is maturing from tentative experiment to massive scale. Scaling means that organisations are making long-lasting business model changes and improving customer experience. Accordingly, also external risks are increasing in areas of digital service performance, customer trust and cybersecurity. Internal challenges in leadership and technology remain in the agenda.
To make it easier for you to plan next transformation initiatives and to avoid known problems, I have combined some practical tips that a business leader should take into account when scaling digitalization. The tips can be applied at company or lower levels, for example in one product family portfolio.
Digital transformation should be seen as a journey that consists of agile experiments and programs. During the journey and as an outcome of each transformation program, you start harvesting benefits from a more digital business. Each program in the journey is organised in modern way, it combines several development sub-streams and integrates a solution that solves a business case, ensures that the new solution is fitting the existing ICT and organisational environment around it, and finally deploys the solution to market and operational use. It is important to understand that content of each program includes changes that impact in minimum people, technologies, data and way of working. In case your benefit targets include new business, the scope impacts business environment outside your current organisation. A key challenge is to manage this digital intensive change as an integrated solution. Following agile development principles, each program release should be able to synchronise development streams frequently and demonstrate draft integrated solutions several times during the program execution. In your digital transformation programs, have you demonstrated new way of working side by side new digital applications, frequently?
Digital transformation planning is in a crucial role to ensure success when scaling out. Program leadership is another major capability any organisation has to master. Even though the default digital development approach is agile, combining changes in customer interface, business rules, people roles and technical elements call for senior leadership. No matter how small steps you take at a time, digitalization remains a complex transformation problem.
We at Midagon know the key challenges of digital transformation. They include specific failure factors you want to avoid. And success factors you want to include. More information about these critical factors is available in a fresh Midagon whitepaper, supported by academic research (https://www.midagon.com/insights/).
Make sure that every digital transformation program is properly started.
In context of digitalization, these steps require experimental-minded people. Leaders are to demonstrate strong top management support, experienced transformation managers are to lead pragmatic program releases and innovative team members implement incremental results. If the start of a transformation program is well supported, there is a good chance that business benefits can be harvested at the end.
If an organisation tries to shortcut in implementing the above mentioned basic practices, scaling of digitalization remains very difficult. It may be that you are underestimating the transformation challenge ahead. If digital innovation flow is not impacting core strategic portfolio, how can you ensure that top management support is in place? A cross-functional transformation needs top priority, funding and people. It is a well-known fact that agile transformation steps can't flow quickly through the organisation without dedicated resources and skilled transformation management.
Good news is that you can start improving the planning system in your own portfolio, for example at product family level. Companies have planning processes in place, so implementation of digital intensive components is hopefully a tuning issue rather than a major change in way of working. And in pragmatic way, a program can be scoped and started properly by an experienced program manager.
Let's continue closing the scaling gap. What are your next steps?
For more information, contact tero.lindholm@midagon.com. Mobile +358 50 484 5277.


Sign Up and Book your FREE introductory session to discover how TeleLingo works. This is a chance for you and your TeleTutor / TeleMentor to review your goals and define the learning strategy just for you. Take advantage of this goal setting and skills evaluation sample session.
Basic exercises, breathing techniques, scales, music reading, classic/rock/pop/folk/blues songs.
Are you preparing for a saxophone certification? These sessions will help you prepare and practice.
Are you a young learner (over 6 years old) wanting to play saxophone like a pro?
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Long-term learning commitments require daily practice with immediate feedback to inform and motivate you over the long haul. Virtual tools enable your progress and provide feedback to you and your TeleTutor / TeleMentor.
"Practice makes Perfect" is our motto. Aside from personalized sessions with TeleTutors / TeleMentors, we provide hands-on experience and a wide range of practice exercises we recommend.
TeleTutors / TeleMentors provide regular feedback and conduct tests. They can assign real-life tests as well as practice and evaluation of TeleStudent recordings.


There's nothing funny about depression. Or is there?
On this week's episode Julie talks to Jessica Holmes, a Canadian comedian, actress and author who just released a new book called Depression the Comedy. It's a very funny, honest and uplifting memoir that describes how Jessica eventually came to terms with, and learned to live alongside, clinical depression.
Jessica is best known for her work on CBC's Royal Canadian Air Farce – a comedy sketch show that used to be on CBC television. She has opened for Jerry Seinfeld, Ellen DeGeneres, Leslie Neilson and Russel Peters. Oh, and a former talk show host you may have heard of – Oprah Winfrey.
For more information about the show, visit mindofherownpodcast.com. You can also join the Facebook Group at facebook.com/groups/mindofherownpodcast and connect with Julie on Twitter at @julietrites.


A new leaf is being turned over in the "Lounge Du Soleil" compilation series - presenting the latest and nicest music productions of the Lemongrassmusic label artists.
1. Glam Sam And His Combo - "Those Sweet and Funny Nights"
2. Gushi & Raffunk - "Letter"
3. Alexandra Hampton - "Distant Shore"
4. Derrick - "Owner Of Trust"
5. Mirage Of Deep - "Ave Maria (Sights From The Mediterranean)"
6. Mark Oakland - "Simple Debris"
7. T2'n - "3 Minutes Avant Le Grand Saut"
8. Dee C'rell - "Departure"
9. Weathertunes - "That's Right"
10.Dub Mars - "Bask In The Sun"
11.Newton feat. Twila.too - "Parler Aux Étoiles"
13.Placid Larry - "Wishing Well"
14.MoShang - "Inscribed In Jade"
15.The Angel Flies In The Sky - "Wellness"


i was wondering if ne1 has any reference pics for a realistic human body (women). Im trying a good at an image in my head. Seeing what i can come up with, but im new to organic modelling so i thought it would be easier to try some references first then build on from that. So any help would be appreciated. Thanks in advance hope someone can help me out.
They seem to be down at the moment, but try again in a couple of days. They have alot of great human referanses and resources!
Don't be put off by the pay-site www.3d.sk (also a great resource, but more directed towards 3d artists), because they have enormous resources there for free.


Matthew Rasanow, DDS is a graduate of North Ridgeville High School and The Cleveland State University, as well as the University of Detroit Mercy School of Dentistry. When not working at Leidgenheimer Dental Group in Elyria, Ohio, Doctor Rasanow enjoys hiking, biking, fishing, archery and woodworking. Matthew is also a volunteer college-level tutor and attends Avon Baptist where he serves as pianist. Dr. Rasanow was recently married in 2018 and resides in Olmsted TWP.


Thank you so much for reaching out to us here at the Disney Parks Moms Panel with your great planning questions about the Move It! Shake It! Dance & Play It! Street Party and Once Upon a Time.
You are correct that sometimes the times for certain repeating events don't get posted until closer to the date itself. I can't guarantee that the events that you mentioned will be occurring, but I also haven't heard anything to the contrary either. I would keep an eye out for updated times as we do get closer! Sorry that I can't be more helpful in this answer.


text:
template<typename PixelType>
class ofxColorPalette_ {

public:
    /// \brief The Channel which should be the modifier for some Generators
    /// generateMonoChromatic, generateComplementary,
    enum ColorChannel{
        SATURATION,
        BRIGHTNESS
    };
    ofxColorPalette_<PixelType>();
    ofxColorPalette_<PixelType>(ofColor_<PixelType> baseColor);
    ~ofxColorPalette_<PixelType>();

    // Sets the Base Color for the Generators
    void setBaseColor(ofColor_<PixelType> col);

    //### Generators
    void inline initGen();

    /// \brief generates random colors
    void generateRandom(int numColors=4);

    /// \brief generates a monochromatic color palette
    /// for now the base color is the brightest in the palette
    void generateMonoChromatic(ColorChannel channel = SATURATION, int numColors = 4);

    /// \brief generates a palette with a fade to the Complementary Color
    ///
    void generateComplementary(ColorChannel channel = SATURATION, int numColors = 4);

    /// \brief generates 3 colors evenly distributed in hue space
    void generateTriad();

    /// \brief generates a palette with the complementary triad
    /// generates a palette of 3 colors
    /// see http://en.wikipedia.org/wiki/Color_scheme#Split-Complementary for description
    void generateComplementaryTriad(float spread = 0.083);

    /// \brief generates a palette of analogous colors
    /// \param numColors the number of Colors which should be generated
    /// \param spread the area (between 0-1) in which the hue is expanded
    void generateAnalogous(int numColors = 5, float spread = 0.2);

    /// \brief get a reference to a random Color which is IN the palette
    ofColor_<PixelType> & getRandom();

    //### Operator overloading
    const ofColor_<PixelType> & operator [] (int n) const;
    ofColor_<PixelType> & operator [] (int n);

    /// \brief returns the size of the palette
    int size();

private:
    ofColor_<PixelType> _baseColor;
    vector<ofColor_<PixelType> > _palette;

    // used to "wrap around" values to PixelType.limit() (1.27 becomes 0.27, -0.3 becomes 0.7)
    float normalizeValue(float val);
};

typedef ofxColorPalette_<unsigned char> ofxColorPalette;
typedef ofxColorPalette_<unsigned short> ofxShortColorPalette;
typedef ofxColorPalette_<float> ofxFloatColorPalette;

redpajama_set_name: RedPajamaGithub


text:
Register / Enter
Entry Guidelines
Winner's Gallery
Judging Overview
About Creativity Awards
Competition History
Creativity Partners
Traveling Winner Exhibit
Official Creativity Partner
Sakol Mongkolkasetarin
Born in Bangkok, Thailand, Sakol always had a passion for art. From there, he started his advertising career in 1990 after graduating with a BFA in Visual Communications Design from California State University, Long Beach. Over the years, he has worked at Wieden & Kennedy, Hal Riney & Partners, Leagas Delaney, BBDO West and even started his own advertising agency, Acme Advertising. The work he's done has been recognized in The One Show, Communication Arts, New York Art Directors Club and many other national and international creative and design competitions. He likes to think of himself as a thinker. A collaborator. A crafter of print ideas. TV ideas. Virtual ideas. Viral ideas. All ideas. Derived from insight. Make it unique. Make it compelling. His motto: Always think, explore, learn, create and communicate. And most of all, have fun along the way.


Celebrate Mother's Day by making yours a "Mum in a Million"
Tiverton Town Centre Partnership along with Tiverton Pannier Market has launched a competition to find Tiverton's 'Mum in a Million'.
• Plus much, much more.
The competition requires candidates to be nominated by their families. If anyone would like to nominate their mum, they need to write a few lines describing why their mum is special to them and deserves to be Tiverton's 'Mum in a Million'.
The lucky winner will be awarded the title 'Mum in a Million for Tiverton 2019' and will be showered with gifts presented at the market by MP Neil Parish, on Saturday, March 30.
Businesses throughout the town will be supporting the occasion and local schools have also been involved by creating Mothers' Day cards, which will be on display at the market.
Alan Ottey, Tiverton's Town and Market Manager said: "We're looking to receive nominations from next of kin explaining why their mother is special to them. We've arranged a small committee, chaired by Town Mayor Councillor Janet Rendle, who'll help judge the nominations to find our 'Mum in a Million' winner.
"We encourage people to take part by nominating their mums ahead of Mothers' Day on Sunday, 31 March, in recognition for their unconditional love, support and forgiving nature that holds them so dear to our hearts.
The closing date for nominations is Wednesday, March 27.


FabulousSonya . LalaNeko. AlmaGrey. AlmaGrey.
EvaPierce4uTonnyMazurMatureClaraLydiaMae .soulbrosexytara92KeyQueenBradNMelody .PetiteDreamsUniquemiaCHADMAXdianablackgirl .JERICHOTBelatriXKristenLaneDirtySlu1t .latinblondieniceNicoleParadisebayoletandguysEasternyGirl .LilianRedHotLuxloveKeyQueenScarlletCute .AriaSunLadyGraceVegasTWOPASSIONSEXLilianRedHot .TonnyMazurSweetTTTFantasyBarbie1RobertoLuy .funPlaysPamellaNixonVarellyEvaPierce4u .KeyQueenCaressaSweetEvaPierce4uTSonTOP69 .PetiteDreamsEttaHazeCHADMAXUniqueDaisy .
PreciousJolie1PornoPrincessAlmaGreyKeyQueen .HardCoupleBoysxDelaylaSummerLydiaMaeBradNMelody .latinblondieniceCuteAngelBellaUniquemiawildtyra .TonnyMazurPamellaNixonTWOPASSIONSEXkathyalvarez .PreciousJolieMatureClaraVarellyAniaGreen .MelaneeLalaNekoFantasyBarbie1CHADMAX .SweetEmmaEmaFriendlyAnneliseFruitflyDelaylaSummer .NatZionFantasyBarbie1EttaHazeCumBandit .EasternyGirlPapiAlexFitbayoletandguysCaressaSweet .dianablackgirlKeyQueenCaressaSweetSexy007Yumi .Sexy007YumiCuteEdiTonnyMazurAronHotx .


100 exceptional toilets installed across 100 sites, enabling lightsource to continue delivering 100% renewable energy.
Lightsource have over 100 solar farms spread across Britain in remote locations without any access to water.
They have security guards on each of their sites and needed low maintenance great value toilets.
The KL1 is the ideal option for remote locations such as solar farms. They're great value, easy to use waterless toilets that are extremely environmentally friendly.
Find out how they work?


text:
El Ramal F23 pertenece al Ferrocarril General Belgrano, Argentina. Está enteramente dentro de la ciudad de Santa Fe de la Vera Cruz.

Ubicación
Se halla enteramente dentro de la ciudad de Santa Fe de la Vera Cruz, en la Provincia de Santa Fe, dentro del departamento La Capital.

Características
Es un ramal secundario de la red de vía estrecha del Ferrocarril General Belgrano, que sirve de empalme entre el Ramal C y el F1, de la misma línea.

Véase también
Ferrocarril General Belgrano
Ferrocarril Provincial de Santa Fe

Referencias

F23
Ferrocarril en la provincia de Santa Fe

redpajama_set_name: RedPajamaWikipedia


text:
A38 är en motorväg i centrala Tyskland.

Trafikplatser

|}

Motorvägar i Tyskland
Vägar i Niedersachsen
Vägar i Hessen
Vägar i Thüringen
Vägar i Sachsen-Anhalt
Vägar i Sachsen
Listor över avfartsnummer


At Sandbar Powersports, in addition to our word-class selection of boat rentals and ATV tours, we offer luxurious RV rentals in Lake Havasu City, AZ. Our new travel trailer rentals have the latest amenities, and we provide a convenient delivery service to RV campgrounds near the lake. Request a quote online now or call (928) 854-4242.
Whether it's your first time renting an RV or you're a seasoned pro at RVing, we know that setting up camp takes away from the initial thrill of starting your vacation. At Sandbar Powersports, we only offer delivery service for our travel trailers so you don't have to worry about picking up your RV and getting everything in place. We'll have your RV rental ready at the campground of your choice near Lake Havasu.
For more ideas, view our comprehensive list of campgrounds in the Lake Havasu area.
As a family-owned and operated business in Arizona, our friendly staff makes the rental experience a breeze so your group can enjoy a camping adventure of a lifetime. We keep our RV rental fleet clean and well-maintained for our customers. When we deliver a travel trailer, we'll supply you with the keys along with helpful instructions for getting started.
For seasonal activities in the area, check out our boats and jet ski rentals to add extra excitement to your vacation. Contact us at (928) 854-4242 to learn more about our quality RV rentals and powersports rentals. Sandbar Powersports serves Lake Havasu City, Parker, and the surrounding areas.


Although these books should be required to read at schools, everyone needs to read them!
Reading at a young age can influence your thinking and personality. Books are not just stories to read for fun. They can shape our thinking and make us see the world from different point of views. They can also help us understand other cultures and people's nature so we get to understand how the world works. Students who read develop a great deal of imagination and confidence to state their opinions.
This series taught me to appreciate friendship and family. It also helped me understand that we cannot change our past, but we can work on our future. Harry Potter is a journey full of happy and sad memories. It is also one of the main reasons that got kids to read.
This book is everything to me. I read it when I was an adult but I wish I read it at a younger age. It is about childhood and how children try to understand the adult world. It has one of my favorite characters ever who is the father in the book. He works as a lawyer and sets such a great example for fathers and how they should understand their children.
I read this book when I was nine years old. It is an amazing adventure about a girl who goes through a hole to another world. It shows the strength of children imagination.
I read this book over the past two years. I will study this book as a part of my class for children literature. It taught me that even if you are poor, you can achieve your dreams if you behave in the right way.
This is my favorite book ever. It taught me that words can change the world and save it. It is such a heart-breaking novel that captured my breath while reading it. It shows the importance of books and words during World War II. Shakespeare has mentioned once that words are the mightiest in this world; that they can never be destroyed. This novel proves that.
What do I know about Africa? I have always asked myself this question. I found the answer in this book. This book helped me to shape Africa in my mind. Unity and love for others is good for society to grow and humans to improve.
This book is described as a tool to transform girls into women. It is a classic British novel that sets in the 18th century. Every feminist need to read this book to see how women can be effective and strong in their society.
This book is about animals rebelling against their owner and how they will rule the farm after him. It is a powerful short book. Orwell managed to portray animals as humans. The book gave me chills at the end and made me rethink about politics. It shows that it is easier to control uneducated people than educated ones. The final thing I learnt is that power is corrupting.


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Home /AG Barr appoints new BOP director amid Epstein controversy
AG Barr appoints new BOP director amid Epstein controversy
Attorney General William Barr on Monday announced the appointment of a new director for the Bureau of Prisons amid the scandal over Jeffrey Epstein's suicide while in federal custody.
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Dr. Kathleen Hawk Sawyer, who previously served in the position for over a decade, between 1992-2003, will take over the role from acting director Hugh Hurwitz, who has been temporarily serving in the position since May of last year.
According to a Department of Justice press release, Hurwitz will remain in the BOP as Assistant Director of the bureau's Reentry Services Division.
Attorney General William Barr speaks at a press conference in New Orleans, Aug. 12, 2019.
Barr also announced the appointment of Dr. Thomas R. Kane as the Deputy Director of BOP.
The leadership announcement comes at a moment of crisis for the BOP, following the suicide death of Epstein in the Manhattan's Metropolitan Correctional Center more than a week ago.
There are at least five ongoing federal investigations into the matter, after Barr last week singled out what he described as "serious irregularities" that investigators had already found surrounding the death of one of the government's most high-profile inmates.
New York State Division of Criminal Justice/EPA-EFE/REX via Shutterstock, FILE
An undated handout photo made available by New York State Division of Criminal Justice showing Jeffrey Epstein, issued 25 July 2019.
Last Tuesday, Barr announced that two guards who had been responsible for observing Epstein had been placed on administrative leave, and that a new acting warden would serve at MCC until the conclusion of the investigations.
The move marks the second time that Barr has appointed Sawyer as the nation's BOP director. During his first time serving as Attorney General under President George H.W. Bush, Barr appointed Dr. Sawyer to the position, where she served until her retirement in 2003.
According to her bio on the BOP website, Dr. Sawyer started her career in 1976 as a psychologist at FCI Morgantown, West Virginia, and eventually became the leader of the prison system.
Thomas Kane, served as Deputy Director and then Acting Director during the Obama Administration and through the Trump administration and has had an over 40 year career at the Bureau of Prisons.
Like Dr. Sawyer, Dr. Kane is a psychologist.


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news / PechaKucha Night: Waterways bringing together thought leaders from around the globe for inspirational evening on the Cuyahoga
PechaKucha Night: Waterways bringing together thought leaders from around the globe for inspirational evening on the Cuyahoga
Free community event on June 20 to feature artists, activists, and special presentation from the Stokes family
Release Date: 5.29.2019
CLEVELAND – Taking place on the banks of the Cuyahoga almost 50 years to the day of the river's last burning, PechaKucha Night: Waterways will highlight clean water issues in a free, exciting format for anyone who loves creativity, clean water and Cleveland.
PechaKucha Night Cleveland (PKN Cleveland) and LAND studio will host the groundbreaking event, presented by the Cleveland Foundation, on Thursday, June 20 at Jacobs Pavilion at Nautica. Doors will open at 7 p.m. and the event will officially kick off at 8 p.m. The event is free and open to the public, but advanced registration is requested at http://globalpkn.com.
"We believe gathering diverse voices to explore our relationship to water creates a powerful space to reflect on the Cuyahoga River's recovery and foster a global dialogue about our shared freshwater future," said Stephen Love, Cleveland Foundation Program Officer for Environment.
"PechaKucha is a simple presentation format that is fast-paced and engaging, with each presentation consisting of 20 slides timed to 20 seconds each," said Michael Christoff, co-founder of PKN Cleveland. "At PechaKucha Night: Waterways, architects, activists and artists from around the world will leverage the presentation format to share their unique perspectives on our most critical natural resource."
The crowd-pleasing format was devised by Astrid Klein and Mark Dytham of Klein Dytham architecture in Tokyo in the early 2000s. The first PechaKucha Night was held in Tokyo in 2003. PechaKucha Nights, informal and fun gatherings where creative people get together and share their ideas in the PechaKucha 20×20 format, are now held in more than 1,000 cities worldwide.
The event will be the 34th PechaKucha Night to take place in Cleveland and the first to feature nationally and internationally renowned experts in addition to local presenters. Additionally, PechaKucha co-founder, Mark Dytham, will help emcee the event.
Presenters include:
Pat Conway and Saul Kliorys from Cleveland, Ohio
Conway (founder) and Kliorys (sustainability manager) of Great Lakes Brewing Company, a principle-centered, environmentally respectful and socially conscious company committed to crafting fresh, flavorful, high-quality beer.
Adid Dada from Beirut, Lebanon
Dada is the founder and lead architect at theOtherDada, an architecture lab aimed at invoking new relationships between climate, landscape, and inhabitants.
Sharon Day from Minneapolis, Minnesota
Day is an Ojibwe elder and water walker who is carrying forward a new tradition to honor and protect our shared waters. As part of the Cleveland Foundation's Creative Fusion: Waterways to Waterways Edition, Day will lead a water walk which will begin at the Cuyahoga's headwaters in northern Ohio and end four days later; she will carry the sacred river water to the mouth of the Cuyahoga where it empties into Lake Erie. At PechaKucha Night: Waterways, she'll speak about her journey.
Kizito Johnson from Kampala, Uganda
Johnson serves as the Director of Works for Drink Local. Drink Tap., Inc., an organization that inspires individuals to recognize and solve our water issues through creative education, events, and providing safe water access to people in need.
Sadhu Johnston from Vancouver, British Columbia
Johnston helped found the Cleveland Green Building Coalition in the early 2000s. Today, Johnston is the city manager of Vancouver, where his Greenest City Action Plan hopes to make the British Columbia seaport into the most eco-friendly in the world by 2020.
Monica Lewis-Patrick from Detroit, Michigan
Lewis-Patrick is Co-Founder & President of We the People of Detroit. As an educator, entrepreneur, and human rights activist/advocate, she is one of the leaders at the forefront of the water rights struggle in Detroit.
Dr. Taraneh Meshkani from Cleveland, Ohio
Meshkani is an urbanist, architect, and educator who currently serves as an Assistant Professor at Kent State University's College of Architecture & Environmental Design.
Cordell Stokes and Chuck Stokes from Cleveland, Ohio
Cordell Stokes, the son of Cleveland Mayor Carl Stokes, is the chairman and CEO of CLC Stokes Consulting Group in Cleveland and senior advisor for Journey Steel Inc. in Cincinnati, OH. Chuck Stokes, the son of Rep. Louis Stokes, is the editorial and public affairs director for WXYZ-TV/Channel 7 in Detroit, MI.
Michael Tsegaye from Addis Ababa, Ethiopia
Tsegaye is an artist and photographer. Much of his work presents a glimpse of life in contemporary Ethiopia, although an extended catalog of his images come from his travels abroad. As part of the current Creative Fusion cohort, Tsegaye will create an exhibition of aerial photographs of the Cuyahoga River and its crooked bend from a helicopter.
Creative Fusion: Waterways to Waterways Edition, including the PechaKucha Night: Waterways event, is a part of Cleveland's broader Cuyahoga50 initiative (www.cuyahoga50.org) that will bring people together throughout 2019 to celebrate clean water and ignite future action through storytelling, discussion, and debate.
About PechaKucha Night Cleveland (PKN Cleveland): PechaKucha Nights are fun and informal gatherings where creative people get together and share their ideas, works, thoughts, project — just about anything, really — in the PechaKucha 20×20 format. PechaKucha Nights were pioneered in Tokyo in the early 2000s and quickly spread to over 1,000 cities worldwide, including here in Cleveland. Volunteer members of the PKN Cleveland Creative Team organize three to four PechaKucha Nights at venues around the city each year in an effort to showcase Cleveland talent. For more information, visit https://www.pechakucha.com/cities/cleveland and follow us on Facebook.
About the Cleveland Foundation: Established in 1914, the Cleveland Foundation is the world's first community foundation and one of the largest today, with assets of $2.5 billion and 2018 grants of more than $100 million. Through the generosity of donors, the foundation improves the lives of residents of Cuyahoga, Lake and Geauga counties by building community endowment, addressing needs through grantmaking and providing leadership on vital issues. The foundation's program areas include: arts and culture, economic and workforce development, education, environment, neighborhoods, and youth development and social services. For more information, visit www.ClevelandFoundation.org and follow us on Facebook, Twitter and Instagram.
About Creative Fusion: Since 2008, the Cleveland Foundation has brought more than 90 accomplished or rapidly rising artists from around the world to Cleveland as part of an international arts residency program. Each year, artists from Creative Fusion cohorts are hosted by local cultural institutions during a three-month residency. Creative Fusion artists collaborate with local Cleveland artists on a commissioned project during their stay, and engage directly with the community, making strong connections and exchanges. Beginning in 2016, the program strengthened collaboration between local Cleveland artists and the visiting artists by focusing each Creative Fusion cohort along a theme, those have included Creative Fusion: Data Edition, Creative Fusion: Composer Edition, The Madison Residencies as part for FRONT International: Cleveland Triennial for Contemporary Art, Creative Fusion: Cuba Edition, and Creative Fusion: Street Art Edition. More information about Creative Fusion: Waterways to Waterways can be found at www.clevelandfoundation.org/creative-fusion/, and by following #CreativeFusionCLE across Cleveland Foundation and host organization social media properties.


Have some festive fun with your brownie baking. Cut these decadent chocolate brownies into holiday shapes using on-hand cookie cutters for a simple twist.
Preheat oven to 350°F. Microwave chocolate and butter in large microwavable bowl on MEDIUM 2 minutes or until butter is melted; stir until chocolate is completely melted.
Add granulated sugar, eggs and vanilla; stir until well blended. Add flour; mix well. Spread into greased 13x9-inch baking pan.
Bake 30 to 35 minutes or until toothpick inserted in centre comes out with fudgy crumbs. (Do not overbake.) Cool in pan. Cut into 24 brownies using holiday cookie cutters. Garnish with icing sugar just before serving, if desired.
This is an indulgent recipe to enjoy on special occasions.


From beginners to elite athletes, we have a training program just for you.
Each class starts with a coach-led warm-up, specifically designed to prepare you for the WOD (Workout of the Day). Then, depending on which level of programming you select, your coach will guide you through either a STRENGTH, SKILL, or MOBILITY session, followed by the WOD.
Romeo Athletics is the home of CrossFit Revelation, where you'll experience CrossFit programming at its best. Perform constantly varied functional movements at a high intensity to develop overall fitness. Recommended for all fitness levels.
Basic CrossFit movements, minus the barbell. It's all about intensity, and you will get your heart rate up! Recommended for beginners.
You will have a custom day-by-day plan developed by one of our experienced coaches who will provide expert guidance.
Recommended for anyone, from first-timers to athletes, who wants tailored programming, nutrition consulting, and accountability.
Whether your goal is to improve body composition, get stronger, focus on a specific skill, or beat your best 5K time, we can customize a workout plan just for you.
Recommended for anyone of any skill level who wants to reach their goals faster and with more focus.
If you are feeling beat up, sore, or just not at 100%, book a session with Jill—our on-staff Physical Therapist—to see how she can help!
Recommended for anyone looking to improve recovery and feel better between workouts.
Designed for the athlete who is looking for workouts focused on improving on-field performance and fostering athleticism.
The Athletic Development Program incorporates dynamic movement prep, intensity sprints, barbell lifts, bodyweight movements, volume sprinting, and conditioning.
Recommended for anyone looking to increase Athletic Performance. .
Develop full-body strength using Olympic weightlifting. Learn to become more proficient and competitive in the fast lifts.
Recommended for both beginning and competitive weightlifters or athletes looking for an edge.
Endurance training utilizes high-intensity speed work, which is constantly varied by length, time domain, and movement. You will train running, cycling, rowing, and swimming as a skill.
Recommended for anyone who wants to increase endurance capacity, to competitors getting ready for a local 5K, adventure race, triathlon, half-marathon, marathon, or even the toughest mud race or ultra-marathon.


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