text:
Oak View Group Launches COVID-19 Task Force To Lead Reopening Of Live Entertainment Venues
June 24, 2020 /in Featured2 /by Lorie Hollabaugh
Oak View Group, a global venue development, advisory, and investment company for the sports and live entertainment industries, has launched a new task force mandated to safely reopen America's live entertainment and public facilities as soon as possible.
Comprised of leaders across all facets of facility operations, including sanitization technology and equipment, research and development, design, food and beverage, capital expenditure management, content, and more, OVG's task force will provide development and operation protocols, guidance, and ongoing auditing of all COVID-19 and other infectious disease mitigation to ensure Americans can return safely to live entertainment facilities.
The team will review and test more than 180 products, solutions, and technologies to see which can be used in the most effective manner to protect all who enter into the facilities. It will develop industry standards to properly sanitize facilities as well as provide guidance on safety protocols for all venues using the most current and medically proven products and procedures. The group will specialize in making sure venues — the seats, concourses, restrooms, concession stands, locker rooms, staffing spaces, and clubs — are sanitized according to government health recommendations. The team will also take extra steps to ensure employees are screened on an ongoing basis for COVID-19 symptoms when they come into work, as well as to implement other security upgrades.
"Our team will be influential in implementing the new standards that will be adopted across the entire live entertainment industry," said Tim Leiweke, CEO of Oak View Group. "The health and safety of fans, artists, and touring staff, during and after this evolving global pandemic, remains our top priority. By aligning ourselves with global companies, all which I consider the best in their respective fields, we are developing and executing the necessary standards to protect anyone who may enter live entertainment venues."
To ensure a safe return to live venues during the 2020 calendar year, the task force will focus its attention across five pillars: developing standards along with its newly announced partnership with Delos, the Pioneer of Wellness Real Estate; testing solutions and technology; advising working groups on said standards and technology; advising the working groups on capital expenditure spends; and an auditing system that will lead to certification.
The task force will then develop a program based on the standardization to implement operational standards for all live entertainment and public facilities to follow. To deliver and execute these standards, the task force has developed a three-phased approach for facilities to follow in order to reopen: the creation of minimum operating standards to sanitize and cleanse facilities that align with Delos, a robust multi-day comprehensive training program for all staff and service providers involved in the production of events at facilities, and the development and implementation of an annual audit program and certification process to ensure venues are implementing and directing best practices.
Lorie Hollabaugh https://musicrow.com/wp-content/uploads/2019/03/MusicRow-header-logo-Mar19B.png Lorie Hollabaugh2020-06-24 15:09:282020-10-08 11:52:48Oak View Group Launches COVID-19 Task Force To Lead Reopening Of Live Entertainment Venues


text:
\section{The Geometric Set-Up}
In the following ($\mathcal{E},V$) is an affine space over a field $F$.
This means that we are given a data of:
\begin{enumerate}
  \item $V$ is a vector space over the field $F$.
  \item An action of the group $V$ on the set $\mathcal{E}$
  \[\mathcal{E} \times V\longrightarrow \mathcal{E} \]
  satisfying (on top of the axioms for a group action):

  For each $A,B\in \mathcal{E}$, there is a unique $\alpha\in V$ such that
  \[A + \alpha = B \]

  ( this $\alpha$ is denoted by $\overrightarrow{AB}$ ).
\end{enumerate}
A function $f:\mathcal{E}\longrightarrow \mathcal{E}$ is called an affine map if there exists a linear map $\overline{f}:V\longrightarrow V$ such that for every $A,B\in \mathcal{E}$
\[\overrightarrow{f(A)f(B)}=\overline{f}(\overrightarrow{AB}) \]
The linear map $\overline{f}:V\longrightarrow V$ is called the linear part of $f$.
An affine map is uniquely determined by its linear part $\overline{f}$ and by its action on a single point $A\in \mathcal{E}$.

For a linear map $L:V\longrightarrow V$ there is a standard construction of the module $V^L$ over the ring $F[X]$.

The underlying additive group is the additive group of $V$. For each $P(X)\in F[X]$ and each $w\in V$, \[P(X)\cdot w:=P(L)(w). \]
Given a linear map $L:V\longrightarrow V$, a key ingredient of our discussion is the submodule $(X-1)\cdot V^L$. Clearly $(X-1)\cdot V^L=\text{im}(L-\text{id}_V)$.

We define now the trajectory coset of an affine map.

\begin{definition}
Let $f:\mathcal{E}\longrightarrow \mathcal{E}$ be an affine map. Define the trajectory coset of $f$ to be the set: \[\tc(f):=\left\{\overrightarrow{Af(A)}\;\mid\; A\in\mathcal{E} \right\} \]
\end{definition}
We have to prove that the set $\tc(f)$ is a coset of the submodule $(X-1)\cdot V^{\overline{f}}$.
\begin{claim}
For each $A,B\in\mathcal{E}$, $$\overrightarrow{Af(A)}-\overrightarrow{Bf(B)}=(\id_V-\overline{f})(\overrightarrow{AB})$$
\end{claim}
\begin{proof}
\begin{IEEEeqnarray*}{rCl}
\overrightarrow{Af(A)}-\overrightarrow{Bf(B)}& = &
\left ( \overrightarrow{Af(A)}+\overrightarrow{f(A)B}\right )-\left ( \overrightarrow{f(A)B}+\overrightarrow{Bf(B)}\right )\\
& = & \overrightarrow{AB}-\overrightarrow{f(A)f(B)}   \\
& = & \overrightarrow{AB}-\overline{f}\left (\overrightarrow{AB}\right ) \\
& = &\left (\id_V - \overline{f} \right )\left ( \overrightarrow{AB}\right )
\end{IEEEeqnarray*}
\end{proof}

\begin{claim}
For each $A,B\in \mathcal{E},$
\[ \overrightarrow{Af(A)}\equiv\overrightarrow{Bf(B)}\quad \left (\;\text{mod}\quad(X-1)\cdot V^{\overline{f}} \right )\]
\qed
\end{claim}

\begin{claim}
Let $A\in\mathcal{E}$ and let $\gamma\in\overrightarrow{Af(A)}+(X-1)\cdot V^{\overline{f}}.$\\
Then there exists a point $B\in \mathcal{E}$, such that $\gamma=\overrightarrow{Bf(B)}$.
\end{claim}
\begin{proof}
Find $\alpha\in V$ such that: $\gamma =\overrightarrow{A f(A)}+\left( \overline{f}-\id_V\right )(\alpha)$.

Define $B$ by $B:=A+\alpha$.
\begin{IEEEeqnarray*}{rCl}
\overrightarrow{AB}& = & \alpha\\
\overline{Af(A)}-\overrightarrow{Bf(B)}& = &\left ( \text{id}_V-\overline{f}\right )(\overrightarrow{AB})\\
\overline{Af(A)}-\overrightarrow{Bf(B)}& = &\left ( \text{id}_V-\overline{f}\right )(\alpha)
\end{IEEEeqnarray*}
\begin{IEEEeqnarray*}{rCl}
\gamma & = &\overrightarrow{Af(A)} - \left ( \text{id}_V-\overline{f}\right )(\alpha)\\
& = & \overrightarrow{Af(A)}-\left ( \overrightarrow{Af(A)}-\overrightarrow{Bf(B)}\right ) = \overrightarrow{Bf(B)}.
\end{IEEEeqnarray*}
\end{proof}

\begin{claim}
Let $f:\mathcal{E}\longrightarrow\mathcal{E}$ be an affine map.
Fix a point $A\in\mathcal{E}$. Then:
\[\tc(f)=\overrightarrow{Af(A)}+(x-1)\cdot V^{\overline{f}}\].
\end{claim}
\noindent\textit{Proof:}
This claim follows from the previous two claims.

\section{Similarity of Affine Maps}
\begin{definition}
Let $f,g\in\mathcal{E}\longrightarrow\mathcal{E}$ be affine maps.

We say that $f$ and $g$ are similar (and write $f\sim g$) if there exists an affine isomorphism, $h:\mathcal{E}\longrightarrow\mathcal{E}$, such that $h\circ f\circ h^{-1}=g$
\end{definition}

\begin{claim}
For every affine map $h:\mathcal{E}\longrightarrow\mathcal{E}$, $h$ is an affine isomorphism, if and only if, the linear map $\overline{h}:V\longrightarrow V$ is a linear isomorphism.
\qed
\end{claim}

\begin{theorem}
Let $f,g:\mathcal{E}\longrightarrow\mathcal{E}$ be affine maps. If $h:\mathcal{E}\longrightarrow\mathcal{E}$ is an affine isomorphism such that $h\circ f\circ h^{-1}=g$, then the linear isomorphism $\overline{h}:V\longrightarrow V$ is an isomorphism of modules over $F[X]$, $\overline{h}: V^{\overline{f}}\longrightarrow V^{\overline{g}}$, it takes the submodule $(X-1)\cdot V^{\overline{f}}$ to the submodule $(X-1)\cdot V^{\overline{g}}$, and the coset $\tc(f)$ to the coset $\tc(g)$.
\end{theorem}

\begin{proof}
\begin{IEEEeqnarray*}{rCl}
h\circ f\circ h^{-1}&=&g\\
h \circ f&=&g \circ h\\
\overline{h\circ f}&=&\overline{g\circ h}\\
\overline{h}\circ\overline{f}&=&\overline{g}\circ\overline{h}
\end{IEEEeqnarray*}

On the left hand side of this equation, $\overline{f}$ stands for the action of $X$ on $V^{\overline{f}}$, while on the right hand side, $\overline{g}$ stands for the action of $X$ on $V^{\overline{g}}$.

Therefore the equation states that $\overline{h}\cdot X=X\cdot \overline{h}$.

Therefore $\overline{h}$ commutes with every polynomial in $F[X]$.

Since $\overline{h}$ commutes with $X-1$, it takes the submodule $(X-1)\cdot V^{\overline{f}}$ to the submodule $(X-1)\cdot V^{\overline{g}}$, \\
and a coset of $(X-1)\cdot V^{\overline{f}}$ to a coset of $(X-1)\cdot V^{\overline{g}}$ .

It remains to prove that the particular coset $\tc(f)$ is taken by $\overline{h}$ to the particular coset $\tc(\overline{g})$.

This follows because for each $A\in\mathcal{E}$,\\ $\overline{h}\left( \overrightarrow{Af(A)}\right)= \overrightarrow{h(A)h\left(f(A)\right)}=\overrightarrow{h(A)g\left(h(A)\right)}.$

So $\overline{h}$ takes the element $\overrightarrow{Af(A)}$ of $\tc(f)$ to the element $\overrightarrow{h(A)g\left( h(A)\right)}$ of $\tc(g)$.
\end{proof}

Now the converse is also true:
\begin{theorem}
Let $f,g:\mathcal{E}\longrightarrow\mathcal{E}$ be affine maps.\\
Assume $T:V\longrightarrow V$ is a linear isomorphism satisfying $T\circ\overline{f}\circ T^{-1}=\overline{g}$ (Otherwise put $T\circ \overline{f}=\overline{g}\circ T$ or $T:V^{\overline{f}}\longrightarrow V^{\overline{g}}$ is an isomorphism of modules over the ring $F[X]$).\\
Assume there exist $A,B\in\mathcal{E}$, such that $T\left(\overrightarrow{Af(A)}\right)=\overrightarrow{Bg(B)}$\\
Then:\\
There exists an affine isomorphism $h:\mathcal{E}\longrightarrow \mathcal{E}$ such that $h\circ f\circ h^{-1}=g$ and $\overline{h}=T$.\\
In particular $f\sim g$.
\end{theorem}

\begin{proof}
Remember that an affine map is determined by its linear part and by its action on a single point. We can therefore define an affine map $h:\mathcal{E}\longrightarrow\mathcal{E}$, by stipulating that $\overline{h}=T$ and that $h(A)=B$.\\
The affine map $h:\mathcal{E}\longrightarrow\mathcal{E}$ is an affine isomorphism because its linear part $\overline{h}:V\longrightarrow V$ is a linear isomorphism.\\
We assume that:
\begin{IEEEeqnarray*}{rCl}
T\left(\overrightarrow{Af(A)}\right)  &=& \overrightarrow{Bg(B)}\\
 \overline{h}\left(\overrightarrow{Af(A)}\right) &=& \overrightarrow{Bg(B)} \\
 \overrightarrow{h(A)h\left(f(A)\right)} &=& \overrightarrow{Bg(B)}\\
 \overrightarrow{Bh\left(f(A)\right)} &=& \overrightarrow{Bg(B)}
\end{IEEEeqnarray*}
From this we conclude that $h\left(f(A)\right)=g(B)$\\
that is: $h\left(f(A)\right)=g\left(h(A)\right)$\\
$h\circ f(A)=g\circ h(A)$.

Now look at the two affine maps $h\circ f\;,\;g\circ h:\mathcal{E}\longrightarrow \mathcal{E}$.\\
Their linear parts are equal:
\begin{IEEEeqnarray*}{rCl}
\overline{h\circ f}&=& \overline{h}\circ\overline{f}=T\circ\overline{f}=\overline{g}\circ T = \\
 &=& \overline{g}\circ\overline{h} = \overline{g\circ h}.
\end{IEEEeqnarray*}
Also they operate in the same way on the single point $A$:
\[h\circ f(A) = g\circ h(A)\]
From this we conclude that
\[h\circ f = g\circ h\]
that is $h\circ f\circ h^{-1}=g$
\end{proof}

Let us rephrase the last two theorems:
\begin{theorem}{\label{theo:a}}
Let $f,g:\mathcal{E}\longrightarrow\mathcal{E}$ be affine maps. Then $f\sim g$, if and only if, there is a linear isomorphism $T:V\longrightarrow V$ such that
\[T\circ \overline{f}\circ T^{-1} = \overline{g} \]
(Otherwise put:
$T:V^{\overline{f}}\longrightarrow V^{\overline{g}}$ is an isomorphism of modules over the ring $F[X]$) and $T$ takes the coset $\tc(f)$ to the coset $\tc(g)$
\end{theorem}

In our next section we shall introduce an invariant of cosets of the submodule $(X-1)\cdot V^{\overline{f}}$ which determines which coset of $(X-1)\cdot V^{\overline{f}}$
can be taken to which coset of $(X-1)\cdot V^{\overline{g}}$.

\section{The Algebraic Set-Up:}

Let $V$ be a finite dimensional vector space over a field $F$. Let $L:V\longrightarrow V$ be a linear map. We shall study the $F[X]-\text{module} \; V^L$, and in particular, the cosets of the submodule $(X-1)\cdot V^L$. We shall study an invariant of cosets, called $\tau$, introduced for the first time in \cite{TartarYakir}.

Let $\text{min}_L(X)\in F[X]$ be the minimal (monic) polynomial of $L$.

Write
\[ \text{min}_L(X)=P_1(X)^{n_1}\cdot P_2(X)^{n_2}\cdot P_3(X)^{n_3}\cdot\ldots\cdot P_\ell (X)^{n_\ell},
\]
where for each $i\quad(1\leq i\leq \ell)$ the polynomial $P_i(X)$ is prime (and monic) and for each $i,j\quad (i\neq j)$
\[\text{gcd}\left(P_i(X),\;P_j(X)\right)=1.
\]
For each $i\quad(1\leq i\leq \ell)$ write
\[V_i^L :=\text{ker}\left(P_i(L)^{n_i}\right).\]
Then $V^L=V_1^L\oplus V_2^L\oplus V_3^L\oplus \ldots\oplus V_\ell^L$ is the primary decomposition of $V^L$ as a module over the ring $F[X]$.
\begin{claim}
For each $i\quad (1\leq i\leq\ell)$ \\
if $X-1\neq P_i(X)$ then $V_i^L\subseteq (X-1)\cdot V^L$.
\end{claim}
\begin{proof}
Find $R(X), S(X)\in F[X]$, \\
such that $R(X)\cdot (X-1)+S(X)\cdot P_i(X)^{n_i}=1$.

For each $\alpha\in V$
\begin{IEEEeqnarray*}{rCl}
\left(R(X)\cdot(X-1)+S(X)\cdot P_i(X)^{n_i}\right)\cdot\alpha&=&1\cdot\alpha\\
R(X)\cdot(X-1)\cdot\alpha + S(X)\cdot P_i(X)^{n_i}\cdot\alpha &=&\alpha
\end{IEEEeqnarray*}
But if $\alpha\in V_i^L$ then $P_i(X)^{n_i}\cdot\alpha=\overline{0}$.

Hence: $$ R(X)\cdot(X-1)\cdot\alpha=\alpha,$$
so $$ \alpha\in (X-1)\cdot V^L.$$
\end{proof}

The following claim  is the first in an effort to find "simple" representatives of cosets of $(X-1)\cdot V^L$ in $V^L$.
\begin{claim}\label{claim:a}
If the polynomial $X-1$ does not divide the polynomial $\min\hbox{}_L(X)$, then $(X-1)\cdot V^L=V^L$.

If, however, $X-1\;\mid\;\min\hbox{}_L(X)$, say $X-1=P_1(X)$, then for each $\alpha\in V^L$ there exists an $\alpha '\in V_1^L$ such that
$$\alpha\equiv\alpha '\quad (\modp \; (X-1)\cdot V^L) $$
\end{claim}
\begin{proof}
If $X-1 \nmid \min\hbox{}_L(X)$, then for each $i\quad(1\leq i\leq\ell)$
\begin{IEEEeqnarray*}{lrCl}
          & V_i^L& \subseteq & (X-1)\cdot V^L,\\
\text{so}\quad & V^L& \subseteq & (X-1)\cdot V^L,\\
\text{so}\quad & V^L& = & (X-1)\cdot V^L.
\end{IEEEeqnarray*}
If $X-1\mid \min\hbox{}_L(X)$, say $X-1=P_1(X)$, then write $\alpha = \sum_{i=1}^{\ell}\alpha_i$ where for each $i\quad(1\leq i\leq\ell)\quad \alpha_i\in V^L_i$.\\
Write $\alpha':=\alpha_1$ and $\alpha'':=\sum_{i=2}^{\ell}\alpha_i$.\\
It follows that $\alpha=\alpha'+\alpha''$, $\alpha'\in V_1^L$ and $\alpha''\in(X-1)\cdot V^L$.
\end{proof}

We are going to introduce now the main tool in our study, the invariant $\tau(\alpha,L)$, an invariant which first appeared in \cite{TartarYakir}.
\begin{definition}
Let $V$ be a finite dimensional vector space over a field $F$.\\
Let $L:V\longrightarrow V$ be a linear map.\\
For each $\alpha\in V$ define $\tau(\alpha,L)$ to be the smallest $k\in\mathbb{N}$, such that
$$\alpha\in\im(L-\id_V)+\ker\left(\left(L-\id_V\right)^k\right). $$
$(\mathbb{N}:=\{0,\,1,\,2,\,3,\ldots\})$
\end{definition}
\begin{remark}
\begin{enumerate}
    \item For each $\alpha\in V$,\\
    $\tau(\alpha,L)=0$, if and only if, $\alpha\in (X-1)\cdot V^L$.
    \item If $X-1\; \nmid \;\min_L(X)$, then for each $\alpha\in V$, $\tau(\alpha,L)=0$
    \item If $X-1 \;\mid \;\min_L(X)$,\\
    say $X-1=P_1(X)$ (where $\min_L(X)=P_1(X)^{n_1}\cdot P_2(X)^{n_2}\cdot\ldots\cdot P_\ell(X)^{n_\ell}$)\\
    then for each $\alpha\in V,\quad \tau(\alpha,L)\leq n_1$.
    \item For each $\alpha,\;\alpha'\in V$, if $\alpha\equiv \alpha'\quad\left(\mod (X-1)\cdot V^L\right)$,\\
    then $\tau(\alpha, L)=\tau(\alpha', L)$.
\end{enumerate}
\end{remark}
\begin{definition}
For each $\alpha\in V^L$, write $\langle\alpha\rangle:=\Sp_{F[X]}(\alpha)$.
In other words:
$$\langle\alpha\rangle:=\left\{R(X)\cdot\alpha\;\mid\;R(X)\in F[X]\right\}. $$
\end{definition}
\begin{claim}\label{claim:b}
Assume $X-1=P_1(X)$ and that $\alpha\in V_1^L$ and $\beta\in\langle\alpha\rangle$. \\
Then either $\beta\in(X-1)\cdot V^L$ or $\langle\beta\rangle=\langle\alpha\rangle$.
\end{claim}
\begin{proof}
Put $\Ann(\alpha):=\left\{R(X)\in F[X]\;\mid\;R(X)\cdot\alpha=\overline{0}\right\}$.

Since $\alpha\in V_1^L$ it follows that $\left\langle P_1(X)^{n_1}\right\rangle\;\subseteq\;\Ann(\alpha)$.

This means that there is a $k\in\mathbb{N},\; k\leq n_1$, such that $\Ann(\alpha)=\left\langle P_1(X)^k\right\rangle$.

Since $\beta\in\langle\alpha\rangle$ there is a polynomial $Q(X)\in F[X]$, such that $\beta =Q(X)\cdot\alpha$.

If $P_1(X)\;\mid\;Q(X)$ then $\beta\in (X-1)\cdot V^L$.

If $P_1(X)\;\nmid\; Q(X)$ then $\gcd\left(Q(X) \, ,\, P_1(X)^k\right)=1$.

Find $R(X), S(X)\in F[X]$, such that $R(X)\cdot Q(X) + S(X)\cdot P_1(X)^k=1$
\begin{IEEEeqnarray*}{rCl}
\left(R(X)\cdot Q(X)+S(X)\cdot P_1(X)^k\right)\cdot\alpha &=& \alpha\\
R(X)\cdot Q(X) \cdot\alpha + S(X)\cdot P_1(X)^k\cdot\alpha&=&\alpha
\end{IEEEeqnarray*}
\begin{IEEEeqnarray*}{lrCl}
\text{But}\quad &P_1(X)^k\cdot\alpha & = & \overline{0}\\
\text{So}\quad & R(X)\cdot Q(X)\cdot\alpha & = & \alpha\\
 &R(X)\cdot\beta & = &\alpha \\
 \text{So}\quad & &\alpha\in\langle\beta\rangle&
\end{IEEEeqnarray*}\
and we conclude that $\langle\alpha\rangle=\langle\beta\rangle$.
\end{proof}
\begin{claim}\label{claim:c}
Assume $X-1=P_1(X)$ and fix $V_1^L=\langle\gamma_1\rangle\oplus\langle\gamma_2\rangle\oplus\langle\gamma_3\rangle\oplus\cdots\oplus\langle\gamma_t\rangle$ any decomposition of $V_1^L$ into a direct sum of cyclic submodules.

Then, for each $\alpha\in V^L$ there exist $\alpha_1,\alpha_2,\alpha_3,\cdots,\alpha_t\in V_1^L$, such that
$$\alpha\equiv\sum_{i=1}^{t}\alpha_i\quad(\modp (X-1)\cdot V^L)  $$
and for each index $i\;(1\leq i \leq t)$ either $\alpha_i =\overline{0}$ or $\langle\alpha_i\rangle = \langle\gamma_i\rangle$.
\end{claim}
\begin{proof}
Let $\alpha\in V^L$. \\
Use first Claim \ref{claim:a} to find $\alpha'\in V_1^L$ such that $\alpha\equiv\alpha'\quad (\modp (X-1)\cdot V^L) $.

Write then $\alpha'$ as: $\alpha'=\alpha_1'+\alpha_2'+\alpha_3'+\ldots+\alpha_t'$ \\
where for each index $i\quad(1\leq i\leq t)\quad \alpha_i'\in\langle\gamma_i\rangle$.

Now, for each index $i\quad(1\leq i\leq t)$ define:
\begin{equation*}
\alpha_i :=
\begin{cases}
\overline{0} & \alpha_i'\in (X-1)\cdot V^L\\
\alpha_i' & \alpha_i'\notin (X-1)\cdot V^L
\end{cases}
\end{equation*}
Clearly $\sum_{i=1}^{k}\alpha_i\equiv\sum_{i=1}^{k}\alpha'_i\quad(\modp (X-1)\cdot V^L)$.

It is also clear (Claim \ref{claim:b}) that for each index $i\;(1\leq i\leq t)$ if $\alpha'_i\notin (X-1)\cdot V^L$, then $\langle\alpha_i\rangle=\langle\alpha'_i\rangle=\langle\gamma_i\rangle$.
\end{proof}

\begin{claim}{\label{claim:d}}
Let $R$ be a ring, and let $W$ be a module over $R$.\\
Let $\varepsilon_1,\,\varepsilon_2,\,\varepsilon_3,\ldots ,\varepsilon_m \in W$.

Assume that $W=\langle\varepsilon_1\rangle\oplus\langle\varepsilon_2\rangle\oplus\langle\varepsilon_3\rangle\oplus\ldots\oplus\langle\varepsilon_m\rangle$.

Assume further that for each $i\;(1\leq i\leq m-1)$
$$\Ann(\varepsilon_i)\supseteq\Ann(\varepsilon_m) $$
Write $\varepsilon'_m:=\varepsilon_1 +\varepsilon_2 +\varepsilon_3 +\ldots +\varepsilon_m$

Then $\Ann(\varepsilon'_m)=\Ann(\varepsilon_m)$

and $W=\langle\varepsilon_1\rangle\oplus\langle\varepsilon_2\rangle\oplus\ldots\oplus\langle\varepsilon_{m-1}\rangle\oplus\langle\varepsilon'_m\rangle$.

Otherwise put: $\varepsilon'_m$ can be used instead of $\varepsilon_m$.
\end{claim}
\begin{proof}
$$\Ann(\varepsilon'_m)=\bigcap_{i=1}^{m}\Ann(\varepsilon_i)=\Ann(\varepsilon_m). $$
As to the second assertion, clearly
$$\varepsilon_1,\,\varepsilon_2,\ldots ,\varepsilon_{m-1},\,\varepsilon_m\,\in\, \langle\varepsilon_1\rangle +\langle\varepsilon_2\rangle +\dots +\langle\varepsilon_{m-1}\rangle +\langle\varepsilon'_m\rangle .$$
This implies that
$$W\;\subseteq\; \langle\varepsilon_1\rangle +\langle\varepsilon_2\rangle +\dots +\langle\varepsilon_{m-1}\rangle +\langle\varepsilon'_m\rangle$$
So
$$W\;=\; \langle\varepsilon_1\rangle +\langle\varepsilon_2\rangle +\dots +\langle\varepsilon_{m-1}\rangle +\langle\varepsilon'_m\rangle$$
It remains to show that the sum of submodules on the right, is
a direct sum.

Let $r_1,\, r_2,\ldots ,\,r_{m-1},\, s\in R$, such that $r_1\cdot\varepsilon_1 +r_2\cdot\varepsilon_2 +\ldots +r_{m-1}\cdot\varepsilon_{m-1} +s\cdot\varepsilon'_m =0 $.

$(r_1+s)\cdot\varepsilon_1+(r_2+s)\cdot\varepsilon_2+\ldots +(r_{m-1}+s)\cdot\varepsilon_{m-1}+s\cdot\varepsilon_m=\overline{0}$.

Hence, $s\cdot\varepsilon_m=\overline{0}$, \\
so $s\in\Ann(\varepsilon_m)$, \\
so for each $i\;(1\leq i\leq m-1)\; s\in\Ann(\varepsilon_i)$.

Also, for each $i\;(1\leq i\leq m-1)\;(r_i+s)\cdot\varepsilon_i=\overline{0}$, \\
so $r_i+s\in\Ann(\varepsilon_i)$, \\
so $r_i\in\Ann(\varepsilon_i)$.

To conclude: For each $i\;(1\leq i\leq m-1)\;r_i\in\Ann(\varepsilon_i)$ \\
so $r_i\cdot\varepsilon_i=\overline{0}$, \\
and also $s\in\Ann(\varepsilon_m)$ \\
(but $\Ann(\varepsilon_m)=\Ann(\varepsilon'_m)$) \\
so $s\cdot\varepsilon'_m=\overline{0}$.

This shows that the sum of submodules is indeed direct.
\end{proof}

The following claim will be the last in our quest of "simple" representatives $(\modp\;(X-1)\cdot V^L)$.

\begin{claim}{\label{claim:e}}
Assume $X-1=P_1(X)$. For each $\alpha\in V^L$, there exists an $\alpha'\in V_1^L$, such that the cyclic submodule $\langle\alpha'\rangle$ is complemented in $V_1^L$, and such that
$$\alpha\equiv\alpha'\quad(\modp (X-1)\cdot V^L). $$
\end{claim}
\begin{proof}
Start with a fixed decomposition $V_1^L=\langle\gamma_1\rangle\oplus\langle\gamma_2\rangle\oplus\langle\gamma_3\rangle\oplus\dots\oplus\langle\gamma_t\rangle$ of the module $V_1^L$ into a direct sum of cyclic submodules.

Now, let $\alpha\in V^L$.

According to Claim \ref{claim:c} there exist $\alpha_1,\,\alpha_2,\ldots,\alpha_t\in V_1^L$, such that
$$\alpha\equiv\sum_{i=1}^{t}\alpha_i\quad\left(\modp\, (X-1)\cdot V_1^L\right) $$
and such that for each index $i\;(1\leq i\leq t)$ either $\alpha_i=\overline{0}$, or $\langle\alpha_i\rangle=\langle\gamma_i\rangle$.

Write $$J:=\left\{ i\;\mid \; 1\leq i\leq t,\;\alpha_i\neq\overline{0}\right\}$$
(If $J=\emptyset$ then $\alpha'=\overline{0}$ and we are done).

Write $$\alpha':=\sum_{i\in J}\alpha_i$$
(It is also true that $\alpha'=\sum_{i=1}^{t}\alpha_i$).

For each $i\in J$, $\alpha_i\in V_1^L$ so $$P_1(X)^{n_1}\in\Ann (\alpha_i).$$
Thus there exists a $d_i\;(1\leq d_i\leq n_1)$ such that $$\Ann(\alpha_i)=\left\langle P_1(X)^{d_i}\right\rangle.$$
Fix an index $m\in J$ such that for each $i\in J,\; d_m\geq d_i$.
Thus for each $i\in J$, $$\Ann(\alpha_m)\subseteq\Ann (\alpha_i).$$
We plan to use Claim \ref{claim:d}.
Let $$W:=\bigoplus_{i\in J}\langle \alpha_i\rangle.$$
Write $$J':=J\smallsetminus\{m\}.$$
According to Claim \ref{claim:d}, the decomposition $$W=\bigoplus_{i\in J}\langle \alpha_i\rangle$$ can be replaced with another decomposition $$W=\left(\bigoplus_{i\in J'}\langle\alpha_i\rangle\right)\oplus\langle\alpha'\rangle.$$
Now, the submodule $\langle\alpha'\rangle$ is complemented in $W$, and $W$ is complemented in $V_1^L$, so $\langle\alpha'\rangle$ is complemented in $V_1^L$.
\end{proof}
\begin{definition}
Assume $X-1=P_1(X)$. Let $\alpha\in V^L$. According to Claim \ref{claim:e} there is an $\alpha'\in V_1^L$, such that $\alpha\equiv\alpha'\;\left(\mod\;(X-1)\cdot V^L\right)$ and such that the cyclic submodule $\langle\alpha'\rangle$ is complemented in $V_1^L$. Call $\alpha'$ a simple representative of $\alpha$.
\end{definition}
We shall make now a short excursion into the study of cyclic submodules of $V^L$.

We have (as usual) $V$, a finite dimensional vector space over $F$. \\
Let $L:V\longrightarrow V$ be a linear map. For each $\alpha\in V$, denote by $\langle\alpha\rangle$ the cyclic submodule of $V^L$ generated by $\alpha$. Thus
$$\langle\alpha\rangle := \left\{R(X)\cdot\alpha\,\mid\,R(X)\in F[X]\right\}. $$
\begin{claim}\label{claim:f}
Let $L:V\longrightarrow V$ be a linear map. Fix a polynomial $P(X)\in F[X]$, such that $\deg\left(P(X)\right)=1$. Then:
\begin{enumerate}
    \item For every $F-$subspace $U$ of $V$, $U$ is invariant under $L$, if and only if, $U$ is invariant under $P(L)$.
    \item For every $\alpha\in V^L$, the cyclic submodule $\langle\alpha\rangle$ is spanned (as a vector space over $F$) by the infinite sequence
    $$\alpha,\,P(X)\cdot\alpha,\,P(X)^2\cdot\alpha,\,P(X)^3\cdot\alpha,\,P(X)^4\cdot\alpha,\ldots $$
    \item Let $P(X)^d\cdot\alpha$ be the first non-pivotal element in the sequence above. (That is the first element which is spanned over $F$ by previous elements). Then the linear dependence (over $F$):
    $$\sum_{i=0}^{d-1}a_i\left(P(X)^i\cdot\alpha\right)=P(X)^d\cdot\alpha , $$
    when multiplied by $P(X)^j$, shows that for each $j\;(j\in\mathbb{N})$ the element $P(X)^{d+j}\cdot\alpha$ is non-pivotal, so $\dim\left(\langle\alpha\rangle\right)=d$.
    \item Denote by $L'$, the restriction of $L$ to $\langle\alpha\rangle$, that is
    $$L':=L\upharpoonright\langle\alpha\rangle . $$
    Then the linear dependence above gives rise to $\min_{L'}(X)$, the minimal polynomial of $L'$.
    $$\min\hbox{}_{L'}(X)=\sum_{i=0}^{d-1}(-a_i)\cdot P(X)^i + P(X)^d$$
\end{enumerate}
\end{claim}
\begin{proof}
An easy exercise.
\end{proof}
\begin{claim}{\label{claim:g}}
Assume $X-1=P_1(X)$. Let $\gamma\in V_1^L$ and assume that the cyclic submodule $\langle\gamma\rangle$ is complemented in $V_1^L$. Then
$$\Ann(\gamma)=\left\langle P_1(X)^{\tau (\gamma ,L)}\right\rangle . $$
\end{claim}
\begin{proof}
Since $\gamma\in V_1^L ,\, P_1(X)^{n_1}\cdot\gamma=\overline{0}$, so there exists $d\in\mathbb{N},\, d\leq n_1$, such that
$$\Ann(\gamma)=\langle P_1(X)^d\rangle . $$
\end{proof}
From Claim \ref{claim:f} it follows that the sequence
$$\gamma,\,P_1(X)\cdot\gamma,\,P_1(X)^2\cdot\gamma, \ P_1(X)^3\cdot\gamma,\ldots,P_1(X)^{d-1}\cdot\gamma $$
is a basis for the cyclic submodule $\langle\gamma\rangle$ (regarded as a vector space over $F$).

We regard the map
$$P_1(L)\upharpoonright\langle\gamma\rangle\; :\;\langle\gamma\rangle\longrightarrow\langle\gamma\rangle $$
Let us study its kernel and its image.

Clearly
$$P_1(X)^{d-1}\cdot\gamma\in\ker(P_1(L)\upharpoonright\langle\gamma\rangle) $$
(so $\dim\left(\ker\left(P_1(L)\upharpoonright\langle\gamma\rangle\right)\right)\geq 1$) and clearly
$$P_1(X)\cdot\gamma,\,P_1(X)^2\cdot\gamma, \ P_1(X)^3\cdot\gamma,\ldots,P_1(X)^{d-1}\cdot\gamma\in\im\left(P_1(L)\upharpoonright\langle\gamma\rangle\right) $$
so $$\dim\left(\im\left(P_1(L)\upharpoonright\langle\gamma\rangle\right)\right) \geq d-1 .$$
But since
$$\dim\left(\ker\left(P_1(L)\upharpoonright\langle\gamma\rangle\right)\right) + \dim\left(\im\left(P_1(L)\upharpoonright\langle\gamma\rangle\right)\right) =d ,$$
it follows that
\begin{IEEEeqnarray*}{rCll}
\dim\left(\ker\left(P_1(L)\upharpoonright\langle\gamma\rangle\right)\right) & = & 1 & \text{and}\\
\dim\left(\im\left(P_1(L)\upharpoonright\langle\gamma\rangle\right)\right) & = & d-1 &
\end{IEEEeqnarray*}
In particular
$$\im\left(P_1(L)\upharpoonright\langle\gamma\rangle\right) = \Sp\left(P_1(X)\cdot\gamma,\,P_1(X)^2\cdot\gamma,\, P_1(X)^3\cdot\gamma,\ldots,P_1(X)^{d-1}\cdot\gamma\right). $$
Our assumption is that the cyclic submodule $\langle\gamma\rangle$ is complemented in $V_1^L$. Remember that the primary submodule $V_1^L$ is complemented in the module $V^L$. Therefore $\langle\gamma\rangle$ is complemented in $V^L$. There exists therefore an $F[X]-$submodule $W$ of $V^L$, such that $$\langle\gamma\rangle\oplus W= V^L .$$
It follows that
$$\im\left(P_1(L)\right)\subseteq \Sp\left(P_1(X)\cdot\gamma,\,P_1(X)^2\cdot\gamma,\, P_1(X)^3\cdot\gamma,\ldots,P_1(X)^{d-1}\cdot\gamma\right)+W $$
Also:
$$\ker\left(P_1(L)^{d-1}\right)\subseteq \Sp\left(P_1(X)\cdot\alpha,\,P_1(X)^2\cdot\alpha,\, P_1(X)^3\cdot\alpha\ldots,P_1(X)^{d-1}\cdot\gamma\right)+W$$
So $\gamma\notin\im\left(P_1(L)\right)+\ker\left(P_1(L)^{d-1}\right)$ and therefore $\tau(\gamma,L)>d-1$.

But
$$\Ann(\gamma)=\left\langle P_1(X)^d\right\rangle, $$
so
$$\gamma\in\ker\left(P_1(L)^d\right) $$
Thus
$$\tau(\gamma,L)=d . $$
Therefore
$$\Ann(\gamma)=\left\langle P_1(X)^d\right\rangle = \left\langle P_1(X)^{\tau(\gamma,L)}\right\rangle .$$

And now to our main algebraic result:
\begin{theorem}{\label{theo:b}}
Let $V$ be finite dimensional vector space over a field $F$.
Let $L,M:V\longrightarrow V$ two linear maps. Assume that $L\sim M$ (L is similar to M) which means of course that the $F[X]-$modules $V^L$ and $V^M$ are isomorphic.

Then, for every $\alpha,\beta\in V$, \\
there exists an $F[X]-$isomorphism $T:V^L\longrightarrow V^M$,\\
taking the coset $\alpha+(X-1)\cdot V^L$ to the coset $\beta+(X-1)\cdot V^M$, \\
if and only if $\tau(\alpha,L)=\tau(\beta,M)$.
\end{theorem}
\begin{proof}
\begin{enumerate}
    \item Assume first that there exists an $F[X]-$isomorphism\\
    $T:V^L\longrightarrow V^M$ which takes the coset $\alpha + (X-1)\cdot V^L$ to the coset $\beta +(X-1)\cdot V^M $.
    For every $k\in\mathbb{N}$, if $\tau(\alpha,L)\leq k$ then
    \begin{IEEEeqnarray*}{rCl}
    \alpha & \in & (X-1)\cdot V^L+\ker\left(\left(L-\id_V\right)^k\right),\\
    T(\alpha) & \in & (X-1)\cdot V^M+\ker\left(\left(M-\id_V\right)^k\right),\\
    \beta & \in & (X-1)\cdot V^M+\ker\left(\left(M-\id_V\right)^k\right)
    \end{IEEEeqnarray*}
    (because $T(\alpha)\equiv\beta \quad(\modp(X-1)\cdot V^M)$).

    So $\tau(\beta,M)\leq k$.

    This shows that $\tau(\beta,M)\leq\tau(\alpha,L)$.\\
    The same proof using $T^{-1}:V^M\longrightarrow V^L$, \\
    will show that $\tau(\alpha,L)\leq\tau(\beta, M).$
    \item Conversely, assume that $\tau(\alpha,L)=\tau(\beta, M)$. We assume of course that $V^L\cong V^M$. Now, if $\tau(\alpha, L)=0$ (and then $\tau(\beta, M)=0$) then
    $$ \alpha\in(X-1)\cdot V^L\;,\;\beta\in(X-1)\cdot V^M.$$
    Every $F[X]-$isomorphism $T:V^L\longrightarrow V^M$,\\
    takes the submodule $(X-1)\cdot V^L$ to the submodule $(X-1)\cdot V^M$.

    Assume now that $\tau(\alpha,L)\geq 1$. Choose $\alpha'\in V_1^L$, a simple representative of $\alpha$. That is
    $$\alpha\equiv\alpha'\quad(\modp(X-1)\cdot V^L) $$
    and $\langle\alpha'\rangle$ is a complemented submodule in $V_1^L$.

    In the same manner, choose $\beta'\in V_1^M$, a simple representative of $\beta$. That is
    $$\beta\equiv\beta'\quad(\modp (X-1)\cdot V^M) $$
    and $\langle\beta'\rangle$ is a complemented submodule in $V_1^M$.

    Now (using Claim \ref{claim:g})
    \begin {IEEEeqnarray*}{rCl}
    \Ann(\alpha') & = &\left\langle P_1(X)^{\tau(\alpha',L)}\right\rangle\\
     & = & \left\langle P_1(X)^{\tau(\alpha, L)} \right\rangle.
    \end{IEEEeqnarray*}
    Also
    \begin {IEEEeqnarray*}{rCl}
    \Ann(\beta') & = &\left\langle P_1(X)^{\tau(\beta',M)}\right\rangle\\
     & = & \left\langle P_1(X)^{\tau(\beta, M)} \right\rangle.
    \end{IEEEeqnarray*}
    Our assumption is that
    $$\tau(\alpha,L)=\tau(\beta, M)\quad\text{so}\quad
    \Ann(\alpha')=\Ann(\beta').$$

    We shall construct an isomorphism $T:V^L\longrightarrow V^M$ \\
    such that $T(\alpha')=\beta'$.

    How shall we do it?

    Decompose $V^L$ into a direct sum of primary components. (This is a unique decomposition).

    Decompose then each primary component into a direct sum of cyclic submodules (this is not unique) taking care that one of the cyclic submodules of $V_1^L$ will be $\langle\alpha'\rangle$. This can be done because the submodule $\langle\alpha'\rangle$ is complemented in the primary submodule $V_1^L$.

    In the same manner decompose the module $V^M$ into a direct sum of primary components and then decompose each primary component into a direct sum of cyclic submodules, taking care that one of the cyclic submodules of $V_1^M$ will be $\langle\beta'\rangle$. This can be done because the submodule $\langle\beta'\rangle$ is complemented in the primary submodule $V_1^M$.

    All in all we have a decomposition
    \begin{IEEEeqnarray*}{rCl}
     V^L & = & \bigoplus_{i\in Y}\langle\alpha_i\rangle\\
     V^M & = & \bigoplus_{i\in Y}\langle\beta_i\rangle
    \end{IEEEeqnarray*}
    taking care that for each $i\in Y$,
    $$\Ann(\alpha_i)=\Ann(\beta_i) $$
    and taking care also that for the same index $i_0\in Y$,
    $$\alpha_{i_0}=\alpha'\; ,\; \beta_{i_0}=\beta' .$$
    We then define $T:V^L\longrightarrow V^M$, by stipulating that for every $i\in Y$,
    $$T(\alpha_i):=\beta_i $$
    (and in particular $T(\alpha_{i_0})=\beta_{i_0}$ that is $T(\alpha')=\beta'$).
\end{enumerate}
\end{proof}

\section{Back to Geometry}

Let $(\mathcal{E},V)$ be an affine space over a field $F$. We assume that $V$ is finite dimensional.

For an affine map $f:\mathcal{E}\longrightarrow\mathcal{E}$, we defined the trajectory coset of $f$ to be
$$\tc(f)=\left\{\overrightarrow{Af(A)}\;\mid\; A\in \mathcal{E}\right\}. $$
We proved that $\tc(f)$ is indeed a coset of the submodule $(X-1)\cdot V^{\overline{f}}$ of $V^{\overline{f}}$.

\begin{definition}
Let $f:\mathcal{E}\longrightarrow\mathcal{E}$ be an affine map. Choose $A\in \mathcal{E}$ and define
$$\tau(f):=\tau\left(\overrightarrow{Af(A)},\overline{f}\right). $$
\end{definition}
Notice that the definition does not depend on the choice of the point $A\in\mathcal{E}$. If $B\in\mathcal{E}$ then
$$\overrightarrow{Af(A)}\equiv\overrightarrow{Bf(B)}\;(\mod (X-1)\cdot V^{\overline{f}}) . $$
Notice also that $\tau(f)$ is the smallest $k\in\mathbb{N}$, such that
$$\overrightarrow{Af(A)}\in (X-1)\cdot V^{\overline{f}} + \ker\left(\left(\overline{f}-\id_V\right)^k\right) . $$
Our main theorem is:
\begin{theorem}
Let $(\mathcal{E},V)$ be an affine space over a field $F$. Assume $V$ is finite dimensional.

Then, for every affine maps $f,g:\mathcal{E\longrightarrow\mathcal{E}}$, $f\sim g$ if and only if, the linear maps $\overline{f},\overline{g}:V\longrightarrow V$ are similar (we write it $\overline{f}\sim\overline{g}$) and $\tau(f)=\tau(g)$.
\end{theorem}
\begin{proof}
From Theorem \ref{theo:a} it follows that $f\sim g$, if and only if, there is a linear isomorphism (of $F[X]-$modules)
$$T:V^{\overline{f}}\longrightarrow V^{\overline{g}} $$
which takes the coset $\tc(f)$ to the coset $\tc(g)$.

According to Theorem \ref{theo:b}, such a linear isomorphism exists, if and only if,
$$\tau\left(\overrightarrow{Af(A)},\overline{f}\right)=\tau\left(\overrightarrow{Bg(B)},\overline{g}\right) $$
(where $A,B$ are any points of $\mathcal{E}$)
\end{proof}

\section{Invariant Flats}

We shall study now invariant flats of affine maps, in order to give $\tau$ another meaning of a geometrical nature.

\begin{claim}{\label{claim:h}}
Let $(\mathcal{E},V)$ be an affine space over a field $F$.
Let $f:\mathcal{E}\longrightarrow\mathcal{E}$ be an affine map.

Then for every $A\in\mathcal{E}$ and every $W$ a subspace of $V$, the flat $A+W$ is invariant under $f$, if and only if, $\overrightarrow{Af(A)}\in W$ and $W$ is invariant under $\overline{f}$.
\end{claim}
\begin{proof}
Assume first that the flat $A+W$ is invariant under $f$.

In particular, $A\in A+W$ so $f(A)\in A+W$ so $\overrightarrow{Af(A)}\in W$.

Also, $A+W=f(A)+W$.

Now let $\alpha\in W$. Since $A+\alpha\in A+W$, it follows that $f(A+\alpha)\in A+W$.

So $f(A+\alpha)\in f(A)+W$.

That is $f(A)+\overline{f}(\alpha)\in f(A)+W$.

This implies that $\overline{f}(\alpha)\in W$.

Conversely, assume that $\overrightarrow{Af(A)}\in W$ and that $W$ is invariant under $\overline{f}$. Let $B\in A+W$. We have to show that $f(B)\in A+W$.

Our assumption that $B\in A+W$, implies that $\overrightarrow{AB}\in W$. Thus $\overline{f}\left(\overrightarrow{AB}\right)\in W$ (because $W$ is invariant under $\overline{f}$).
Now,
\begin{IEEEeqnarray*}{rCl}
\overrightarrow{Af(B)} &=& \overrightarrow{Af(A)} + \overrightarrow{f(A)f(B)}\\
 &=&\overrightarrow{Af(A)}+\overline{f}\left(\overrightarrow{AB}\right).
\end{IEEEeqnarray*}
$\overrightarrow{Af(A)}, \overline{f}\left(\overrightarrow{AB}\right)\in W$, so
$\overrightarrow{Af(A)}+ \overline{f}\left(\overrightarrow{AB}\right)\in W$, so
$\overrightarrow{Af(B)}\in W$, so
$f(B)\in A+W$
\end{proof}
\begin{definition}
Let $f:\mathcal{E}\longrightarrow\mathcal{E}$ be an affine map and let $A\in\mathcal{E}$.

Denote by $\Flat(A,f)$ the flat of $\mathcal{E}$ which passes throught the point $A$ and whose direction is $\left\langle\overrightarrow{Af(A)}\right\rangle$, the cyclic $F[X]-$submodule of $V^{\overline{f}}$, generated by the vector $\overrightarrow{Af(A)}$.

In other words:
$$\Flat(A,f):=A+\left\langle\overrightarrow{Af(A)}\right\rangle . $$
\end{definition}
\begin{claim}
Let $f:\mathcal{E}\longrightarrow\mathcal{E}$ be an affine map. Let $A\in\mathcal{E}$. Then:
\begin{enumerate}
    \item $A\in\Flat(A,F)$
    \item The flat $\Flat(A,f)$ is invariant under $f$.
    \item If $\mathcal{F}$ is a flat in $\mathcal{E}$ which is invariant under $f$ and which passes throught the point $A$, then $$\Flat(A,f)\subseteq\mathcal{F}. $$
    \item Denote by $\overline{f}'$ the restriction of $\overline{f}$ to $\left\langle\overrightarrow{Af(A)}\right\rangle$.
    That is:
    $$\overline{f}'=\overline{f}\upharpoonright
    \left\langle\overrightarrow{AfA}\right\rangle .$$
    Then:
    $$\dim\left(\Flat(A,f)\right)=\deg\left(\min\hbox{}_{\overline{f}'}(X)\right). $$
    That is, the dimension of the flat $\Flat(A,f)$ is the degree of the minimal polynomial of the linear map $\overline{f}':\left\langle\overrightarrow{Af(A)}\right\rangle\longrightarrow\left\langle\overrightarrow{Af(A)}\right\rangle$.
\end{enumerate}
\end{claim}
\begin{proof}
\begin{enumerate}
    \item Follows from the definition
    \item Claim \ref{claim:h}
    \item Claim \ref{claim:h}
    \item Follows from part 4. of Claim \ref{claim:f}.
\end{enumerate}
\end{proof}
\begin{theorem}{\label{theo:d}}
Let $(\mathcal{E},V)$ be an affine space over a field $F$. We assume that $V$ is finite dimensional.
Let $f:\mathcal{E}\longrightarrow\mathcal{E}$ be an affine map.

From the coset $\tc(f)$ (a coset of the submodule $(X-1)\cdot V^{\overline{f}}$ in the module $V^{\overline{f}}$) choose a simple representative $\alpha$. By this I mean:\\
If $\tau(f)=0$ (so $\tc(f)=(X-1)\cdot V^{\overline{f}}$) choose $\alpha:=\overline{0}$. \\
If $\tau(f)\geq 1$, choose an $\alpha\in V_1^{\overline{f}}$ such that the cyclic submodule $\left\langle\alpha\right\rangle$ is complemented in $V_1^{\overline{f}}$ (see Claim \ref{claim:e}).

After you chose the simple representative $\alpha$ from the coset $\tc(f)$, choose a point $A\in\mathcal{E}$, such that $\alpha=\overrightarrow{Af(A)}$. Then:
$$\dim\left(\Flat(A,f)\right)=\tau(f), $$
so we found a flat of $\mathcal{E}$ which is invariant under $f$, passes through the point $A$, and is of dimension $\tau(f)$. Its direction is a complemented submodule in $V_1^{\overline{f}}$
\end{theorem}
\begin{proof}
If $\tau(f)=0$ then $\alpha=\overline{0}$ and $\overrightarrow{Af(A)}=\overline{0}$ so $f(A)=A$.

$\Flat(A,f)=\{A\}$ and indeed in this case $\Flat(A,f)=A+\langle\overline{0}\rangle$.

Assume now that $\tau(f)\geq 1$. Write $k:=\tau(f)$. So we chose $\alpha\in\tc(f)$ (and $A\in\mathcal{E}$ such that $\overrightarrow{Af(A)}=\alpha$) such that $\alpha\in V_1^{\overline{f}}$, and the cyclic submodule $\langle\alpha\rangle$ is complemented in $V_1^{\overline{f}}$.

It follows from Claim \ref{claim:g} that $\Ann(\alpha)=\left\langle P_1(X)^{\tau(\alpha,\overline{f})}\right\rangle$.

That is $\Ann(\alpha)=\left\langle(X-1)^k\right\rangle$.

So, again, if we denote by $\overline{f}'$ the restriction of $\overline{f}$ to $\langle\alpha\rangle$, then
$$\min\hbox{}_{\overline{f}'}(X)=(X-1)^k.$$
Thus $\dim\left(\langle\alpha\rangle\right)=k$ (part~4 of Claim~\ref{claim:f}).

Remember that $\Flat(A,f)=A+\left\langle\overrightarrow{Af(A)}\right\rangle = A+\langle\alpha\rangle$.

So $\dim(\Flat(A,f))=k$.
\end{proof}
\begin{theorem}{\label{theo:e}}
Let $f:\mathcal{E}\longrightarrow\mathcal{E}$ be an affine map.

Then any flat of $\mathcal{E}$ which is invariant under $f$, has dimension $\geq\tau(f)$.
\end{theorem}
\begin{proof}
It suffices to prove our assertion for flats of the form $\Flat(A,f)$ (because every flat which is invariant under $f$ contains such a flat).

But $\Flat(A,f)=A+\left\langle\overrightarrow{Af(A)}\right\rangle$. \\
It suffices therefore to prove that for every $A\in\mathcal{E}$,
$$\dim\left(\left\langle\overrightarrow{Af(A)}\right\rangle\right) \geq \tau(f).$$
Assume (using the notation at the Algebraic Set-Up) that
$$\min\hbox{}_{\overline{f}}(X)=P_1(X)^{n_1}\cdot P_2(X)^{n_2}\cdot P_3(X)^{n_3}\cdot\ldots\cdot P_\ell(X)^{n_\ell},$$
where $P_1(X)=X-1$. (Remember that if $X-1\nmid\min_{\overline{f}}(X)$) then $\tau(f)=0$ and we are done).

For each $i\;(1\leq i\leq\ell)$ write $$V_i^{\overline{f}}:=\ker\left(P_i(\overline{f})^{n_i}\right) $$
and remember that
$$V^{\overline{f}}=V_1^{\overline{f}}\oplus V_2^{\overline{f}}\oplus V_3^{\overline{f}}\oplus \ldots\oplus V_\ell^{\overline{f}}. $$
Now let $A\in\mathcal{E}$. Write $\alpha:=\overrightarrow{Af(A)}$.
We have to show that
$$\dim\left(\langle\alpha\rangle\right)\geq\tau(f). $$
Write $\alpha=\beta+\gamma$ where $\beta\in V_1^{\overline{f}}$ and $\gamma\in\oplus_{i=2}^{\ell}V_i^{\overline{f}}$.
$$ \Ann(\alpha)=\Ann(\beta)\cap\Ann(\gamma)$$
So if
\begin{IEEEeqnarray*}{rCl}
\overline{f}' & := & \overline{f}\upharpoonright \langle\alpha\rangle\\
\overline{f}'' & := & \overline{f}\upharpoonright \langle\beta\rangle\\
\overline{f}'''& := & \overline{f}\upharpoonright \langle\gamma\rangle,
\end{IEEEeqnarray*}
then
\begin{IEEEeqnarray*}{rCl}
\min\hbox{}_{\overline{f}'}(X) & = & \lcm\left(\min\hbox{}_{\overline{f}''}(X)\, ,\,\min\hbox{}_{\overline{f}'''}(X)\right)\\
 & = & \min\hbox{}_{\overline{f}''}(X) \cdot\min\hbox{}_{\overline{f}'''}(X).
\end{IEEEeqnarray*}
Notice that $\gamma\in(X-1)\cdot V^{\overline{f}}$\\
so $\alpha\equiv\beta\;(\mod(X-1)\cdot V^{\overline{f}})$\\
so $\tau\left(\alpha,\overline{f}\right)=\tau\left(\beta,\overline{f}\right)$.\\
Now, since $\beta\in V_1^{\overline{f}}$, it follows that
$$\min\hbox{}_{\overline{f}''}(X)=P_1(X)^d $$
where
$$\tau\left(\beta,\overline{f}\right)\leq d\leq n_1 $$
But
$$ \tau\left(\beta,\overline{f}\right)=\tau\left(\alpha,\overline{f}\right)$$
so
$$\tau\left(\alpha,\overline{f}\right)\leq d .$$
\begin{IEEEeqnarray*}{rCl}
\min\hbox{}_{\overline{f}'}(X) & = & \min\hbox{}_{\overline{f}''}(X)\cdot\min\hbox{}_{\overline{f}'''}(X)\\
  & = & P_1(X)^d\cdot\min\hbox{}_{\overline{f}'''}(X)
\end{IEEEeqnarray*}
Thus
$$\deg\left(\min\hbox{}_{\overline{f}'}(X)\right)\geq d $$
So
$$\deg\left(\min\hbox{}_{\overline{f}'}(X)\right)\geq \tau\left(\alpha,\overline{f}\right). $$
Which means $$\dim\left(\langle\alpha\rangle\right)\geq\tau(f). $$
\end{proof}

The Invariance Level:

In \cite{Tarrida} the notion of invariance level of an affine map $f$ is defined and is denoted by $\rho(f)$.
\begin{definition}
Let $f:\mathcal{E}\longrightarrow\mathcal{E}$ be an affine map. If $\mathcal{E}$ has a flat, which is invariant under $f$, of dimension $k$, but not one of smaller dimension, then we say that the invariance level of $f$ is $k$. We write $\rho(f)=k$.
\end{definition}

Our Theorem \ref{theo:d} and Theorem \ref{theo:e} show:
\begin{theorem}{\label{theo:f}}
For every affine map
$$f:\mathcal{E}\longrightarrow\mathcal{E},\quad\rho(f)=\tau(f). $$
\qed
\end{theorem}
We can do better than Theorem~\ref{theo:e}. In Theorem~\ref{theo:e} we show that every flat which is invariant under $f$, is of dimension $\geq\tau(f)$.

One can actually show that if $\mathcal{F}$ is a flat which is invariant under $f$, and $\dim(\mathcal{F})=\tau(f)$, then $\dir(\mathcal{F})\subseteq V_1^{\overline{f}}$.

By looking closely at the decomposition $\alpha=\beta + \gamma$ in the proof of Theorem~\ref{theo:e}, one can show that not only
$\dim\left(\right\langle\alpha\rangle)\geq\dim\left(\langle\beta\rangle\right)$, but also that $\langle\alpha\rangle\supseteq\langle\beta\rangle$ (and $\langle\alpha\rangle\supseteq\langle\gamma\rangle$). We state it as:
\begin{theorem}{\label{theo:g}}
Assume $X-1\,\mid\,\min\hbox{}_{\overline{f}}(X)$, say $X-1=P_1(X)$.

Let $\mathcal{F}$ be a flat of $\mathcal{E}$ which is invariant under $f$.

If $\dim(\mathcal{F})=\tau(f)$, then $\dir(\mathcal{F})\subseteq V_1^{\overline{f}}$.
\end{theorem}
\begin{proof}
Let
\[ \min\hbox{}_{\overline{f}}(X)=P_1(X)^{n_1}\cdot P_2(X)^{n_2}\cdot P_3(X)^{n_3}\cdot\ldots\cdot P_\ell(X)^{n_\ell}\]
where for each index $i\;(1\leq i\leq\ell)$ the polynomial $P_i(X)$ is prime and (monic) and for each $i,j\;(i\neq j)$
\[\gcd\left(P_i(X),P_j(X)\right)=1\]
Assume $P_1(X)=X-1$.

Let $Q_1(X),\,Q_2(X),\,\ldots ,Q_\ell(X)$ be the sequence of "Lagrange interpolants" associated with the sequence
$P_1(X)^{n_1}\cdot P_2(X)^{n_2}\cdot P_3(X)^{n_3}\cdot\ldots\cdot P_\ell(X)^{n_\ell}$.

Remember that in the ring $F[X]$:
\[Q_j(X)\equiv 1\quad(\modp P_j(X)^{n_j}).\]
and if $i\neq j$ then
\[Q_j(X)\equiv 0\quad(\modp P_i(X)^{n_i}).\]
\end{proof}
The sequence $Q_1(X),\,Q_2(X),\,\ldots ,Q_\ell(X)$ \\
is uniquely determined $\left(\modp\;\min\hbox{}_{\overline{f}}(X)\right)$.

Remember also that
\[Q_1(X)+Q_2(X)+\ldots +Q_\ell(X)\equiv 1\quad \left(\modp\;\min\hbox{}_{\overline{f}}(X)\right).\]
Now, let $\mathcal{F}$ be a flat of $\mathcal{E}$ which is invariant under $f$.

Assume \[\dim(\mathcal{F})=\tau(f).\]

Clearly, there is a point $A\in\mathcal{E}$, such that $\mathcal{F}=\Flat(A,f).$

Write $\alpha := \overrightarrow{Af(A)}.$
\[\mathcal{F}=A+\langle\alpha\rangle .\]
\begin{IEEEeqnarray*}{rCl}
\alpha &=& 1\cdot \alpha = \left(Q_1(X)+Q_2(X)+\ldots +Q_\ell(X)\right)\cdot\alpha\\
\alpha &=& Q_1(X)\cdot\alpha +Q_2(X)\cdot\alpha +\ldots +Q_{\ell}(X)\cdot\alpha.
\end{IEEEeqnarray*}
For each index $i\;(1\leq i\leq\ell)$
\[Q_i(X)\cdot\alpha\in V_i^{\overline{f}}. \]
Write $\beta := Q_1(X)\cdot\alpha$\\
and $\gamma := Q_2(X)\cdot\alpha + Q_3(X)\cdot\alpha + \ldots + Q_\ell(X)\cdot \alpha$.\\
Thus $\alpha=\beta+\gamma$.\\
Since $\gamma\in (X-1)\cdot V^{\overline{f}}$ it follows that
\[\alpha\equiv \beta\quad\left(\modp\;(X-1)\cdot V^{\overline{f}}\right) \]
and therefore $\beta\in \tc(f)$.

There exists a point $B\in\mathcal{E}$ such that $\beta =\overline{Bf(B)}$.

Notice $\Flat(B,f)=B+\langle\beta\rangle$.

The flat $B+\langle\beta\rangle$ is invariant under $f$.

Therefore $\dim\left(\langle\beta\rangle\right)
\geq\tau(f)$.

But $\beta=Q_1(X)\cdot\alpha$\\
so $\langle\beta\rangle\subseteq\langle\alpha\rangle$.

From the assumption that $\dim\left(\langle\alpha\rangle\right)=\tau(f)$, \\
it follows that
\[\langle\beta\rangle=\langle\alpha\rangle\]
Thus
\[\dir(\mathcal{F})=\langle\beta\rangle\]
and
\[\langle\beta\rangle\subseteq V_1^{\overline{f}}\]
\qed

The moral of our journey, in conclusion, is that the study of $(X-1)\cdot V_1^{\overline{f}}$, is crucial for understanding questions such as similarity and invariance level.

\begin{remark}
In an independant work \cite{Hou} Xiang-dong Hou finds a system of distinct representatives of the conjugacy class of $AGL(m,F)$
\end{remark}

\printbibliography

\end{document}

redpajama_set_name: RedPajamaArXiv


text:
Ribe domkyrka i Ribe i Danmark är en av Danmarks största domkyrkor. Domkyrkan kallas även Vor Frue Kirke efter skyddshelgonet jungfru Maria.

Byggnadsbeskrivning och historia
Ribe domkyrka är en byggnad uppförd av tuffsten i romansk stil åren 1125-1160. Därefter har senare tillbyggnader av tegel uppförts på 1300-talet i tegelgotik – såsom det stora sydtornet och en rad tidigare sidokapell, både på norra och södra sidorna. Kyrkan restaurerades även fullständigt under åren 1883-1904 av arkitekten H.C. Amberg. Vid restaureringen återuppbyggdes av granit och tuff i sydväst sydtornet, kallat Mariatornet, som jämnats med jorden på 1700-talet; det är 50 meter högt. I det tornet hänger också kyrkans kyrkklockor.

Ribe torn eller Borgartornet
Det fyrkantiga stora tegeltornet Ribe tårn eller Borgertårnet i nordväst är 46,5 meter högt och byggdes på 1300-talet upp som stormklocketorn och vakttorn för tornvakten, med milsvid utsikt över marsklandet utanför Ribe. Det stod färdigt 1333 och uppfördes eftersom det gamla hade störtat samman på julmorgonen 1283. Ursprungligen hade detta torn en tornspira, men när en del av tornet störtade ner 1594 och tornet reparerades 1595, fanns inte pengar till att reparera tornspiran. Därför fick tornet då platt tak och blev 12 meter lägre.

Måttuppgifter
Kyrkan är 61,5 meter lång (varav koret 20,4) och 33,5 meter bred (med tvärskeppet). Mittvalvet över koret vid absiden har en höjd på 21 meter. Kyrkan är femskeppig och till mittskeppet sluter sig alltså fyra sidoskepp – inklusive de tidigare sidokapellen i tegel som har inkluderats i kyrkan). Dessa åtskiljs av fyra rader stenpelare, varigenom katedralens ursprungliga korsform nästan gått förlorad.

Bildgalleri

Referenser och externa länkar

Källor

Hemsidan https://web.archive.org/web/20070909171010/http://www.ribetourist.dk/ för turister till Ribe, med följande delar:
En sida om Ribes kyrkor och kloster, som länkar vidare till webbplatsen för Ribe domkyrka
En karta över centrala Ribe

Not

Domkyrkor i Danmark
Mariakyrkor
Ribe
Kyrkobyggnader i Ribe stift
Kröningskyrkor
Kyrkobyggnader i Esbjergs kommun
Kyrkobyggnader invigda under 1100-talet

redpajama_set_name: RedPajamaWikipedia


text:
Experts at the Belfer Center focus on a broad range of environmental and climate change issues.
Views on the Economy and the World (12)
Environment & Climate Change--Climate change policy (1646)
Environment & Climate Change--Climate agreements (440)
Environment & Climate Change--Air pollution (181)
Energy--Renewable energy (146)
Energy--China energy policy (141)
Energy--Coal, Carbon Capture, & Storage (118)
Energy--International cooperation (105)
Energy--Energy security (104)
Economics & Global Affairs--Trade (102)
Energy--Energy conservation (100)
Environment & Climate Change--Arctic (23)
Environment & Climate Change--Green chemistry (11)
(-) Energy--Biofuels (37)
(-) Economics & Global Affairs--International Finance (28)
Flickr Creative Commons
The Navy builds strength by saving energy
| July 5, 2016
Read more about The Navy builds strength by saving energy
Senior Fellow of the Future of Diplomacy Project, David Ignatius examines the Navy's energy diet that began seven years ago with an edict from newly appointed Secretary Ray Mabus, who issued five goals for radically changing how the Navy bought and consumed fuel.
Blog Post - Views on the Economy and the World
Fiscal Education for the G-7
Read more about Fiscal Education for the G-7
As the G-7 Leaders gather in Ise-Shima, Japan, on May 26-27, the still fragile global economy is on their minds. They would like a road map to address stagnant growth. Their approach should be to talk less about currency wars and more about fiscal policy.Fiscal policy vs. monetary policyUnder the conditions that have prevailed in most major countries over the last ten years, we have reason to think that fiscal policy is a more powerful tool for affecting the level of economic activity, as compared to monetary policy.
Talk on trade: TPP & Trump
Read more about Talk on trade: TPP & Trump
The ITC Wednesday released its mandated report on the economic effects estimated to result from the TransPacific Partnership. As is usual in standard trade models, the estimated welfare gains may sound small: on the order of ¼ % of income. But that would still be way worth doing. Furthermore the ITC study, by design, leaves out a lot. For example, the Petri-Plummer study from the Peterson Institute estimates income gains from TPP that are twice as large, in part because it takes into account Melitz-style opportunities for more productive firms to expand.
Journal Article - Climatic Change
Expert Views — and Disagreements — About the Potential of Energy Technology R&D
Erin Baker
Valentina Bosetti
Lara Aleluia Reis
| June 2016
Read more about Expert Views — and Disagreements — About the Potential of Energy Technology R&D
In order to make R&D funding decisions to meet particular goals, such as mitigating climate change or improving energy security, or to estimate the social returns to R&D, policy makers need to combine the information provided in this study on cost reduction potentials with an analysis of the macroeconomic implications of these technological changes. The authors conclude with recommendations for future directions on energy expert elicitations.
The Paris Agreement on Climate Change, C'est Bon
Read more about The Paris Agreement on Climate Change, C'est Bon
How should one evaluate the agreement reached in Paris December 12 by the 21st Conference of Parties to the UN Framework Convention on Climate Change (UNFCCC)? Some avid environmentalists may have been disappointed in the outcome. The reason is that the negotiators did not commit to limiting global warming to 1 ½ degrees centigrade by 2050, nor will the new agreement directly achieve the 2 degree limit.But such commitments would not have been credible. What came out of Paris was in fact better, because the negotiators were able to agree on meaningful practical near-term steps.
Paul Lowry, Creative Commons
Newspaper Article - The Wall Street Journal
What Will the U.S. Energy Industry Look Like Over the Next Five Years?
Read more about What Will the U.S. Energy Industry Look Like Over the Next Five Years?
Professor Meghan O'Sullivan was interviewed on November 15th, 2015 for a Wall Street Journal special section on energy, discussing the rapid transformation of the American energy sector in light of low fuel prices, new climate policies and other factors.
TPP Critics' Nighttime Fears Fade by Light of Day
Read more about TPP Critics' Nighttime Fears Fade by Light of Day
The TPP (Trans Pacific Partnership) that was finally agreed among trade negotiators of 12 Pacific countries on October 5 came as a triumph over long odds. Tremendous political obstacles, domestic and international, had to be overcome over the last five years. Now each country has to decide whether to ratify the agreement.Many of the issues are commonly framed as "Left" versus "Right." The unremitting hostility to the negotiations up until now from the Left – often in protest at being kept in the dark regarding the text of the agreement -- has carried two dangers.
Book - Oxford University Press
A Liberal Actor in a Realist World: The European Union Regulatory State and the Global Political Economy of Energy
| October 2015
Read more about A Liberal Actor in a Realist World: The European Union Regulatory State and the Global Political Economy of Energy
A Liberal Actor in a Realist World assesses the changing nature of the global political economy of energy and the European Union's response, and the external dimension of the regulatory state. The book concludes that the EU's soft power has a hard edge, which is derived primarily from its regulatory power.
Thomas Kohler, MCC/ZEW
Newsletter Article - Belfer Center for Science and International Affairs, Harvard Kennedy School Belfer Center Newsletter
Climate Change Agreement Takes Center Stage
| Fall/Winter 2014 - 15
Read more about Climate Change Agreement Takes Center Stage
The international agreement on greenhouse gas emissions and climate change to be determined in Paris in December 2015 is "the greatest opportunity the world has had in 20 years to make meaningful progress on this exceptionally challenging issue," Harvard Project on Climate Agreements (HPCA) Director Robert Stavins said in a Boston Globe op-ed in September. Stavins was in New York City during the week of the United Nations Climate Summit, which included numerous side events and a march that attracted several hundred thousand Americans calling for serious climate actions.

redpajama_set_name: RedPajamaCommonCrawl


Good quality DIY tutorials are the perfect way to attempt your own bridal hair or make up if you're not sure about hiring a professional. Be warned though, you must practice, practice and practice again if you're going to attempt it for your big day. If you're too scared to attempt it for the wedding day itself, what about your bridal shower, engagement party or your bachelorette? Rustic White Photography and hair and make up artist Claudia Mejerle have created the perfect organic looking half up half down 'do' for fine art brides and it's so simple we even think we could try this at home!
This hairstyle looks best and stays in place well when the hair has some texture. So if you have some natural wave or curl, don't fight it! If you have straight hair, wrap different sized sections of hair around 1 ¼" curling iron to get some movement through the hair. The waves don't have to be perfect.
Next, work some surf spray and through the hair, root to tip. Spray dry shampoo over the hair to give it even more texture and grip to make pinning the hair easier and help the waves to stay full.
Take a section of hair from the left side of the head, starting above the ear, and then twist it back toward the center of the head. Secure the twisted section with a couple of bobby pins.
Repeat Step three on the right side. Make sure this section of hair meets the first, left section of hair you pinned back in the center of the head.
Alternate repeating Steps three and four until all hair is pinned back, working down toward the base of the head.
You can stop at this point and leave the style a bit looser, or move on to the next step to finish the style to be more of a ponytail.
Take a small section of hair from underneath and wrap it around the hair that is left down. Depending on the length of you hair, you can wrap the small section around a couple of times. Then pin it underneath so you can hide the end of the wrapped section as well as the bobby pin used to do so.
You can rough up the pony with you fingers to give the hair a fuller, more "lived in" feel. A little more surf spray and a dusting of soft hold hairspray is a great way to finish this style.


text:
Woodland biodiversity, palaeo-human ecology and some implications for conservation management
Tipping R, Buchanan J, Davies A & Tisdall E (1999) Woodland biodiversity, palaeo-human ecology and some implications for conservation management. Journal of Biogeography, 26 (1), pp. 33-43. https://doi.org/10.1046/j.1365-2699.1999.00282.x
This paper appraises the role of prehistoric human societies in modifying the natural woodland biodiversity of several regions throughout Scotland. It draws attention to the likelihood that, contrary to popular belief, biodiversity was enhanced by interactions between agricultural communities and woodlands. The possible purposefulness of such interactions is discussed. Mindful of the present concern to re-establish 'native' woodlands in many parts of northern Britain, we then review three approaches to the identification of native tree types in Scotland, and conclude that insufficient attention in reafforestation schemes is being paid to (a) the palaeoecological record as a record of native woodland or (b) the likely former high taxonomic diversity of woodlands. Both these failings will lead to the creation of new woodlands lacking in species diversity.
'Native' woodland; palaeoecological record; Scotland; woodland biodiversity
Journal of Biogeography: Volume 26, Issue 1
Tipping, Richard; Buchanan, James; Davies, Althea; Tisdall, Eileen
Dr Eileen Tisdall
Lecturer, Biological and Environmental Sciences


text:
Rusty Anderson
born on 20/1/1959 in La Habra, CA, United States
Rusty Anderson is an American guitarist and singer-songwriter. He is known as the lead guitarist for Sir Paul McCartney, though he has worked with an extensive list of artists in addition to his own solo career.[1]
Anderson became a professional musician at age 14 with his first band, Eulogy. Eulogy won a sizeable local following in Los Angeles, playing with bands such as The Police, Van Halen, The Runaways, Quiet Riot, Rick Derringer, Eddie Money and The Motels.[2][3] Later, Anderson formed The Living Daylights, which shared the stage with Fishbone and OMD. The Living Daylights was a psychedelic/progressive rock band and Anderson's first as primary songwriter. The Living Daylights put out a record on the Greenworld Label, but subsequently disbanded when Anderson signed a solo development deal at Columbia Records.[4]
In 1993, along with Scott Cutler and Anne Preven, Carla Azar and Paul Bushnell, Anderson formed Ednaswap. Ednaswap shared the stage with Sublime and Social Distortion, and toured as opener for No Doubt and Weezer. The band released four major label records on East West and Island Records, and penned the worldwide Number One hit song "Torn".[5][6]
For his debut solo album Undressing Underwater, Anderson invited Paul McCartney to play bass and the rest of the McCartney band to sing backup vocals on the lead track, "Hurt Myself". Friend and former band mate Stewart Copeland also came in to play drums on the song "Catbox Beach". The album was self-produced by Anderson, along with David Kahne, Mudrock and Parthenon Huxley. Undressing Underwater was originally released in late 2003 by his own Oxide Records, and re-released in September 2005 on Surfdog Records, iTunes and Amazon.Anderson's second album Born on Earth was released October 2009.[7] His third release was a compilation album titled Until We Meet Again, and released in the spring of 2012 with pre-release copies available during the winter 2011 leg of Paul McCartney's On The Run tour.[8]
Rusty Anderson Afternoon
Joining with his longtime collaborator, Todd O'Keefe, Anderson formed his new band "Rusty Anderson Afternoon" in 2013. O'Keefe is a Los Angeles musician known for his work with Ray Davies, Black Francis, and The 88.[9] O'Keefe and Anderson had played together for several years in Anderson's solo group, introduced variously as "Rusty Anderson Band" and "Rusty Anderson and Friends". The band performed in cities throughout the United States, The UK and Iceland. Once Anderson and O'Keefe began sharing songwriting and lead vocal duties it was time to pick a name for the new band. A contest was held online in 2013. Fans voted for their favorite band name and the winner was "Rusty Anderson Afternoon."[10] The brand new band was showcased in 2013 at the Gibson Beverly Hills showroom debut of the Rusty Anderson Gibson ES-335 signature guitar, in coordination with the Natural Resources Defense Council, an environmental charity.[11]
Now officially a band, Rusty Anderson Afternoon released their first single, "Effortless", on April 30, 2013. "Effortless" made it to #1 at Sirius Radio's "20 on 20".[11] The band released Rusty Anderson Afternoon II on the Oxide Records label on July 31, 2014, produced by Anderson and Paul Ratazcjack, whom Anderson worked with years earlier with the bands Soldier and The Living Daylights. The band's second single, a version of the Stewart and Gouldman song "I'm Not in Love" began in rotation on "The Blend," a Sirius radio station, in August 2014. Rusty Anderson Afternoon was a featured act at SXSW 2014. In addition to a club performance, the band was featured at SXSW's tribute to Jimi Hendrix. Anderson and O'Keefe performed Hendrix's "Little Wing" and "Spanish Castle Magic" for the debut of the United States Postal Service Jimi Hendrix Forever Stamp in Butler Park in Austin, Texas. The band also performed at the Leo Fender Museum Center in Fullerton, California in 2014[11] and the Hard Rock Casino in Lake Tahoe, Nevada, in 2015.[12]
In April 2016 the band premiered its next single, "Evidence" on the syndicated worldwide show "Passport Approved", airing on over 30 alternative radio stations.[13] Their first full album, RAA, produced by Anderson and Ratazcjack on Anderson's Oxide label, was released shortly thereafter in May 2016[14] and debuted in the Top Ten on iTunes Alternative New Releases chart the week of its release.[15] Early media responses to the record described it as "most accomplished and brilliant.[16] During 2016-2017 the band performed several shows in California, including San Juan Capistrano[17] and at Pepperdine University in Malibu.[18][19]
In 2001, Anderson, along with drummer Abe Laboriel Jr., got a call from producers Mo Jackson and David Kahne to play and sing on Paul McCartney's studio album Driving Rain.[20] This became the birth of the widely praised Paul McCartney Band, which Variety called "as tight a unit as they come".[21] After finishing the record and performing at The Concert for New York City, the band began the Driving World Tour in April 2002, playing well over 100 shows in the US, Japan, and Mexico.The Driving World Tour was Billboard's Top Tour of 2002.[22] The Back in the World tour followed in 2003. Subsequent high profile performances included the Roman Colosseum, Live 8 UK and Red Square in Moscow. Following the success of this tour three Multi Platinum DVDs were released, all featuring Anderson on guitar: Back in the U.S., Paul McCartney in Red Square, and The Space Within US. Anderson also performed with McCartney at the Super Bowl twice, in 2002[23] and 2005.[24]
Anderson played guitar on McCartney's Chaos and Creation in the Backyard in 2005 and contributed guitar and vocals to the 2007 release, Memory Almost Full.[20] He also took part in Good Evening New York City, playing lead guitar on its Grammy winning song "Helter Skelter",[25] and Good Evening Europe. Anderson continued performing during the Summer Live '09 Tour and the Up and Coming Tour. In 2010, the McCartney band performed at the White House for President Barack Obama and First Lady Michelle Obama for the PBS series In Performance at the White House.[26] Several prominent performances in 2012 included the Olympics opening ceremony,[27] the Queen's Diamond Jubilee Concert[28] and the Grammy Awards.[29]
McCartney released New in 2013 on which Anderson played guitar and sang backing vocals.[30] McCartney's Out There tour also began in 2013 in South America and continued through 2015.[31] Anderson won critical praise during the tour for the many highlights[32] he contributed to the musical arrangements. In 2015, Anderson performed with McCartney on the Saturday Night Live 40th Anniversary Special on NBC.[33] The 2016-2017 worldwide One on One tour[34] was described in Salon as "perfect and life-affirming — well-sequenced, well-paced and incredibly satisfying".[35] Argentina's La Nación said during the tour "Anderson is the virtuoso of the six strings, the greatest rocker of all".[36] The Detroit Free Press noted that the band is "brimming with class, chops and energetic focal points in guitarist Rusty Anderson."[37]
After the 2018 announcements of McCartney's Egypt Station album and Freshen Up Tour, Anderson performed with McCartney at "secret" gigs at Abbey Road Studio[38] in London, The Cavern Club in Liverpool,[39][40] and New York City's Grand Central Station.[41]
Other collaborations and performances
Anderson's recordings include music by Elton John, Little Richard, New Radicals, Willie Nelson, Tim McGraw, Faith Hill, Michael Bublé, Sinéad O'Connor, The Wallflowers, Jewel, Santana, Stevie Nicks, Ronan Keating, Joe Cocker, and the surf riff in Ricky Martin's "Livin' La Vida Loca".[42] Also, The Bangles, Lisa Marie Presley, Matthew Sweet, Neil Diamond, Animal Logic (with Stewart Copeland and Stanley Clarke), Belinda Carlisle, Susanna Hoffs, Renée Fleming, k.d. lang, Parthenon Huxley, Carole King, Gwen Stefani, Perry Farrell, Steven Tyler, Regina Spektor, Gaby Moreno, Dido, Aimee Mann, Cat Stevens, Miley Cyrus, Eros Ramazzotti, Nelly Furtado, Lana Del Rey, Enrique Iglesias, OPM, Mandy Moore, and Robi Draco Rosa.[43] Anderson also played guitar for the multi-artist tribute project The Art of McCartney, which featured such musicians as Bob Dylan, Barry Gibb, B.B. King, Chrissie Hynde, Billy Joel, and Brian Wilson.[44]
He has performed live with such notables as Ringo Starr, Stevie Wonder, Neil Young, Doug Pinnick, Ron Wood, Rihanna, Robbie Krieger, Bruce Springsteen, Joe Walsh, Dave Grohl,[45] Slash,[46] Steve Lukather, Joe Sumner, Mary Bridget Davies, Bob Weir, Brittany Howard, Diana Krall, Krist Novoselic, and Wayne Kramer; Anderson has also played "The Star Spangled Banner" live on solo guitar at Dodger Stadium.[47]
In addition to Anderson's solo songwriting career, he has written with and produced acts such as Chris Shaffer of The Why Store, Jordan Lawhead, and Emma Burgess, resulting in song uses in TV shows Cold Case, Men In Trees, Kyle XY, and Wildfire.[48] Other soundtrack credits include Highlander, Highlander II, and Beyond Westworld.[49] The Rusty Anderson Afternoon song "Points of Interest" was featured on the hit CBS comedy The Big Bang Theory in 2014.[50]
Undressing Underwater (2005)
Born on Earth (2009)
Rusty Anderson Afternoon II (2014)
RAA (2016)
^ Perone, Pierre (August 5, 2010). "Maybe I'm amazed at being McCartney's chosen guitarist". The Independent. Retrieved 18 November 2017.
^ Blackett, Matt (June 3, 2013). "Spotlight Then and Now: Rusty Anderson". Guitar Player. Retrieved 18 November 2017.
^ "Rusty Anderson Interview Part One". Photos.mccartney.net. 2004-04-29. Retrieved 2017-03-08.
^ "Rusty Anderson Paul McCartney's #1 Band Leader". BlogTalkRadio. Retrieved 2016-07-19.
^ "Natalie Imbruglia — Torn — Listen, watch, download and discover music for free at". Last.fm. Retrieved 2017-03-08.
^ Deming, Mark. "Rusty Anderson Biography". Allmusic. Retrieved 18 November 2017.
^ "Music Review: Rusty Anderson – Born On Earth". Blogcritics.org. 2009-11-17. Retrieved 2017-03-08.
^ "Until We Meet Again". Amazon.co.uk. Retrieved 2017-03-08.
^ Todd O'Keefe. "Todd O'Keefe | Credits". AllMusic. Retrieved 2017-03-08.
^ "paulmccartney.com :: Index". Maccaboard.paulmccartney.com. Retrieved 2017-03-08.
^ a b c "Paul McCartney Lead Guitarist Issues New EP 'Rusty Anderson Afternoon II' @ Top40-Charts.com - New Songs & Videos from 49 Top 20 & Top 40 Music Charts from 30 Countries". Top40-charts.com. 2014-08-25. Retrieved 2017-03-08.
^ "Rusty Anderson Afternoon's night to remember in Tahoe". Tahoeonstage.com. Retrieved 2017-03-08.
^ "with Sat Bisla". Passport Approved. 2014-06-07. Retrieved 2017-03-08.
^ "rustyandersonmusic". Facebook. Retrieved 2017-03-08.
^ "iTunes charts - New Releases Alternative". Artofthemix.org. 2016-01-08. Retrieved 2017-03-08.
^ "https://soundcloud.com/howtofixthemusicbusiness/episode-4-an-afternoon-with
^ Madison Desler (2016-03-01). "Rusty Anderson Afternoon Playing Guitar For Home Fans". Concert Guide Live.com. Retrieved 2017-03-08.
^ "Rusty Anderson & Friends to Bring Melodies to Pepperdine's Smothers Theatre". Broadwayworld.com. 2017-01-14. Retrieved 2017-03-08.
^ "Tickets now on sale for Rusty Anderson show at Pepperdine". Malibu Surfside News. 2016-12-31. Retrieved 2017-03-08.
^ a b "Music – Music News, New Songs, Videos, Music Shows and Playlists from". MTV. 2016-09-30. Retrieved 2017-03-08.
^ Steve Chagollan (2014-08-11). "Paul McCartney Concert Review: Dodger Stadium". Variety. Retrieved 2017-03-08.
^ "Billboard". Books.google.com. 2002-12-28. p. 60. Retrieved 2017-03-08.
^ "Super Bowl XXXVI Pregame Ceremonies". PaulMcCartney.com. 2002-02-03. Retrieved 2017-03-08.
^ "5. Paul McCartney - Super Bowl XXXIX (2005) | The 10 Best Super Bowl Halftime Shows". Rolling Stone. 2013-01-31. Retrieved 2017-03-08.
^ "Paul McCartney wins Grammy for Helter Skelter". Reuters. 2011-02-13. Retrieved 2017-03-08.
^ "Honoring Sir Paul McCartney". Whitehouse.gov. 2010-06-03. Retrieved 2017-03-08.
^ "Paul McCartney, Olympics: Legend Closes London 2012's Opening Ceremony With 'Hey Jude'". The Huffington Post. 2012-07-27. Retrieved 2017-03-08.
^ Sutherland, Mark (2012-06-05). "Paul McCartney, Elton John Honor Queen at Diamond Jubilee Concert". Rolling Stone. Retrieved 2017-03-08.
^ Bosso, Joe. "Grammy guitar jam: Paul McCartney, Dave Grohl, Bruce Springsteen, Joe Walsh". MusicRadar. Retrieved 2017-03-08.
^ 00:00:00. "PaulMcCartney.com". PaulMcCartney.com. Retrieved 2017-03-08.
^ 00:00:00. "Tours". PaulMcCartney.com. Retrieved 2017-03-08.
^ Keith Spera. "Paul McCartney's career-spanning New Orleans concert celebrated the old and the 'New'". NOLA.com. Retrieved 2017-03-08.
^ Blistein, Jon (2015-02-16). "Paul McCartney, Miley Cyrus, Paul Simon Captivate at 'SNL 40'". Rolling Stone. Retrieved 2017-03-08.
^ "First Dates Of New 'One On One' Tour Confirmed". PaulMcCartney.com. Retrieved 2017-03-08.
^ "My dad's first Paul McCartney show: Why his concerts are the perfect intergenerational family outing". Salon.com. Retrieved 2017-03-08.
^ "McCartney: la respuesta a todas las preguntas - 19.05.2016". La Nacion. 2016-05-19. Retrieved 2017-03-08.
^ http://www.freep.com/story/entertainment/music/2017/10/02/paul-mccartney-detroit-little-caesars-arena-review/721624001/
^ https://www.thetimes.co.uk/article/macca-gets-back-for-secret-abbey-road-gig-vdsrhfvx6
^ https://www.bbc.co.uk/news/entertainment-arts-44964710
^ https://www.billboard.com/articles/news/8467312/paul-mccartney-cavern-club-liverpool-concert-recap
^ https://www.cbsnews.com/news/paul-mccartney-egypt-station-grand-station-terminal-secret-concert-new-york-city-today-2018-09-07/
^ "Music – Music News, New Songs, Videos, Music Shows and Playlists from". MTV. 2016-09-30. Retrieved 2017-03-08.
^ "Rusty Anderson - Relationships". MusicBrainz. Retrieved 2017-03-08.
^ Grow, Kory (2014-11-12). "Stream 'The Art of McCartney,' Featuring Bob Dylan, Brian Wilson, Kiss". Rolling Stone. Retrieved 2017-03-08.
^ "Bruce Springsteen, Joe Walsh, Rusty Anderson, Paul McCartney, Dave Grohl, And Brian Ray". GRAMMY.com. Retrieved 2017-03-08.
^ "SXSW 2014, Part Two | South By Southwest | Austin, TX | Southern Music Scene". Thesouthlandmusicline.com. 2014-03-26. Retrieved 2017-03-08.
^ "Rusty Anderson rocks anthem | MLB.com". M.mlb.com. Retrieved 2017-03-08.
^ "Rusty Anderson on GTR-B-Q". Musiciansradio.com. Retrieved 2009-01-19.
^ "Rusty Anderson". IMDb. Retrieved 2016-07-19.
^ Marinucci, Steve (2014-11-03). "McCartney's guitarist Rusty Anderson has music featured in 'The Big Bang Theory'". AXS. Retrieved 2017-03-08.
maccareport page of Rusty
Rusty Anderson at AllMusic
Rusty Anderson Facebook
Rusty on Instagram
@rustyanderson1 on Twitter
This article uses material from the article Rusty Anderson from the free encyclopedia Wikipedia and it is licensed under the GNU Free Documentation License.


text:
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Xenophobia: South Africa to release victims' names
The South African government said on Monday that the names of nationals killed in the attacks on foreigners by its citizens would soon be released.
The South African Presidency, in a statement, a copy of which was made available to one of our correspondents on Tuesday, said it had received media enquiries about the names of foreign nationals killed during the violence that broke out in KwaZulu-Natal.
"Government is still in the process of verifying the information and the names will be released as soon as there has been conclusive positive identification of the victims.
"President Zuma released the names of the three South Africans who were killed in KwaZulu-Natal and Mr. Manuel Jossias of Mozambique, who was killed in Alexandra township, Johannesburg," the South African Presidency said.
It added that Zuma once again extended condolences to all the affected families on behalf of the government and the people of South Africa.
The South African government, the statement said, was in touch with the embassies of affected countries to ensure careful management of "this difficult and painful process."
The government stated further that its police had been working round the clock to apprehend the perpetrators of the violence.
"President Zuma requests all who have information about the incidents in which the victims were killed to come forward so that those responsible would be brought to book," it said.
Meanwhile, the Deputy Chairman, House of Representatives Committee on Water Resources, Mr. Lasun Yusuf, has urged the Federal Government to take more drastic step against the government of South Africa to protest the xenophobic attack on Nigerians and other foreigners in the country.
Yusuf said this in a telephone interview with one of our correspondents in Osogbo, Osun State, on Monday.
The lawmaker, who is representing the Osogbo/Olorunda/ Irepodun/Orolu Federal Constituency in the House, urged the Federal Government to sever all bilateral agreements with South Africa.
He said the South African government and its people had been taking Nigeria for granted for too long, saying the time had come to show that Nigeria would no longer tolerate attacks on its citizens anywhere in the world.


text:
import passport from 'passport';
import { Strategy } from 'passport-local';
import UserModel from './model/user';

export function initialize() {
    passport.use(new Strategy({
        usernameField: 'email',
        passwordField: 'password'
    }, (email, password, done) => {
        UserModel.findOne({ email, password })
            .select('name email created').exec()
            .then(user => {
                return done(null, user);
            }, () => {
                return done({
                    message: 'Invalid email or password!'
                });
            });
    }));

    passport.serializeUser((user, done) => {
        done(null, user._id);
    });

    passport.deserializeUser((id, done) => {
        UserModel.findOne({ _id: id })
            .select('name email created').exec()
            .then(user => {
                done(null, user);
            });
    });
}

export function authenticatedMiddleware(req, res, next) {
    if (req.isAuthenticated()) {
        next();
    } else {
        res.redirect('/');
    }
}

redpajama_set_name: RedPajamaGithub


[Couldn't wait…A couple days pass] PREGNANT!
You see January 1st of 2016 I had pelvic pain due to my ovaries sticking to intestines. Thank you endo! I was prescribed Percocets to aliviate the pain. Mid January I began to feel he pain again so I took a Percocet. Sunday morning I felt an inkling to take a pregnancy test. Well, for someone who has been trying for so long, I had spare tests tucked away. Sure enough, I got the results! Now I'm not the type to get the pregnancy test that you have to figure out if there are 2 pink lines or guessing if that 2nd line is even there. I get the ones that say"PREGNANT" or "NOT PREGNANT" or "⏳" if it got an error for some reason. 🙄 At the time of this pregnancy, I was working full time and we were foster parents to 2 girls 2 and 4 years old. The thought of being pregnant was a thought that came and went.
So I wake up my husband with the test. I cried because I had just took a percocet the day before and because my cycle wasn't late yet, I had no clue how far along I was or if the pain killers affected my unborn child. I scheduled an appointment with the nurse practitioner a couple of days later. In the ultrasound, nothing was detected but my hcg levels were high and the pregnancy test at the NP office was a positive.
LOL!!! All joking to the side, I really wasn't sure what was going on. The NP explained that the fertilized egg was still making its way down to where it needed to be so she wanted to set up another appointment to see me in another week.
Week later and I got to see my fuzzy jelly bean nestled in the position needed to survive!! I cried all over again so grateful God saw fit to bless us again with another chance to naturally conceive. However, my heart remained guarded as I wasn't in the safe zone just yet. Only 6 weeks along and my mind kept getting flashbacks of my last pregnancy that came to a screeching halt at 7 weeks.
We get to 8 weeks and began to spotting a brownish color. Really trying not to panic but to still take precautionary measures, I go to the ER. When checked, they said baby was fine. The brown spotting was just old blod from the last cycle that just never fully went away but they still wanted to check me in another 2 weeks if no other bleeding happened. The bleeding never got heavier, thankfully, and actually went away. At my 10 week appointment, I was given a due date a week earlier than I was given based off of my LMP (last menstrual period). 1st due date was Sept. 29th, 2nd was Sept. 24th and I vaguely remember being given a 3rd due date but I honestly cannot recall what that date was.
Throughout the pregnancy, the baby was measuring closer to the Sept. 24th date so that's the date I went off of. I was so sure we were going to have a boy!! After all, my husband was the oldest and my oldest sibling was a boy and his first born was a boy so BOY is the guess I stuck with! Up until 2 weeks into my 2nd trimester, avocados and anything too sweet got me nauseous. This baby was determined to keep me healthy!! HA!
We planned to do a gender reveal at our place, inviting the few closest to us and going "live" on social media so friends and family further away could be apart of it. We bought a rainbow piñata because hey, we are in El Paso, which borders Mexico, let's get festive!😉The day arrives for my 20 week sonogram. We tell the technician not to reveal the gender and to place the gender reveal into an envelope we brought and she taped it up good. Now we didn't want anybody we knew knowing before us so we had two of my best friends go to a party store and hand the cashier the envelope and instruct the cashier to fill the piñata up with blue candy for a boy and red candy for a girl. Once done, tape the envelope back up, tape up the piñata and my friends were off. The colors were blue, for obvious reasons but red because that was my favorite color and pink was just so played out with gender reveals. As I finished getting ready, my husband set up the piñata in front of the house, my cousin helped set up the live feed and people began to trickle in and log into the live feed. Being so confident in my guess, I wore a white dress with a blue belt and blue sandals. My husband stood neutral with just white. Now i thought the piñata would've came with a stick to swing with but it didn't so my husband grabbed the metal bat we kept near the front door.
The surprise and excitement on both of our faces was priceless!!!
So from calling baby "him", we were now saying "her"! Yes, I was that sure. Lol.
I was the only girl and not a girly girl by any means so the thought of having a girl was scary for me. I wanted to be the best Mami I could be but would I?
We were planning a homebirth for babygirl and I was looking into more holistic ways of bringing her into this world. Never would I have thought that one needs to look into EVERY. SINGLE. DETAIL. when having a baby! I was supported by 2 amazing midwives that were incredibly loving, knowledgable and just absolutely a blessing to have had in this journey!! I saw a doctor for lab results but I saw my midwife for wellness appointments. I learned more from them than I did the doctor I would see.
Besides back pain and swollen feet, I was loving my pregnancy!
We prepped what we could till baby girl arrived. Now my mom was set to arrive a few days before the 24th. My mother in law also came down for a surprise visit the same day my mom's flight came in. A week came and went and no sign of baby. My best sister-friend, who was my photgrapher, had a planned trip to TN, my MIL flew back home and I had my mom another week. We kept talking to baby girl and let her know we were patiently awaiting her arrival but grandma leaves soon so she kind of needed to make her grand entrance asap. Haha! Boy did we have a good laugh on that one because my mother ended up having to extend her stay another week! She was due to leave Saturday, before extending, and I went into labor Sunday!
Got the contractions down to 7 minutes, affirmations posted all over my loving room, text the midwife and she instructed us to inflate the birthing tub! Now I thought I needed to look cute in pictures, so I proceeded to breath through contractions and put on lipstick. Yes this lasted all of about 2 hours or so before back labor kicked in and let me tell you, was it rough! Before going into labor, I was so excited that I would be able to eat while in labor but let me tell you, that was the last thing on my mind!!
The most comfortable position was squatting and the worst was laying on my back. After 24 hours of labor, exhaustion hit everyone. My husband's hands were shaking from applying pressure on my back, my midwife was tired and my mom would switch places with my husband to give him a break. I nibbled here and there but nothing solid. Drank water because I had to and my favorite thing to ingest were honey sticks! After 40 hours of labor, my midwife called it and said we had to go to the ER because I lost my stretngth and almost passed out. I bawled like a baby because I really don't like hospitals.
At the hospital, they took my vitals and then placed me in a room. Not long after being in the room that baby's heart rate kept dropping so they immediately got paperwork started for an emergency c section. Mind you, the worst position for me during contractions was laying down so at this point my hands are trembling during each contraction I got and you want me staying COMPLETELY still for an epidural?! Prayer goes a long way and I had the sweetest nurse holding my hands as they inserted the needle and I received instant relief from my contractions. The surgery team prepped everything and my husband was by my side. Because they have to have the operating rooms cold for equipment my body was shaking how cold it was.


Sorry for the tardiness of my reply. You're right. The mockup doesn't show all possibilities and it's not very acurate. I focused on the left side and therefore I simply forgot the right one. The main purpose for me doing this mockup was to check how the visual style of the indicators would fit to the rest of the graphics, to get a feeling for how many indicators would be visible and how I might be able to implement them (creatures will be shown on top of the indicators, so I can't just use my GUI library to do the job). This way I also noticed that it would help to see a preview of the car's position and orientation. BTW: You're idea of only showing indicators for a certain speed sounds reasonable.
Regarding driving cars in reverse: They will definitely need to hit zero before moving into negative speeds.
Regarding sub-turn rendering: I'm not sure about this one. I don't want to avoid it just because of its complexity but it needs to pay off so I will only go for it if the other approaches don't work.
Thanks again for your feedback. I wish several other forums I am using would be so energetic.
Aww... player cant flip a 180 and switch to reverse?
Hmmm... interesting point. I didn't think of that. Maybe there will be some exceptions (this could be some skill you need to unlock first or something like that).
Right now it's only purpose is to test the car movement ideas we were discussing in this thread.
You can move by using the W, A, S, D keys and enter the vehicle by pressing E. I don't want to tell you more about the contols because I would like to see if you can find it out on your own, i.e. if it's somewhat intuitive.
It requires Windows and the movement isn't ready yet but I thought getting critiques as soon as possible is a good thing.
BTW: Which operating systems are you using? I might be able to create a Linux version, too.
I made the walls look better and added doors. It's still very rough and still a very, very, very early version.
Are you afraid of me uploading malicious software? Should I move this topic to the thread "Early Development Feedback"? Can't you "play" it because you are using a different OS? Did you have problems running this app? A friend of mine said he couldn't run it because of a missing DirectX DLL. I now changed my app to use OpenGL by default instead. So this problem shouldn't happen anymore.
I know that this isn't much but I want to get feedback as early as possible in my development process.
EDIT: The video was done with software I didn't really know how to handle or they were crappy (or both), so the video runs a bit faster than the real recording was.
Often I have more time to simply comment on things and less time to actually work at something. I tried downloading and running your prototype last night but received an error regarding some DLL (which sounded related to Ogre). I'll give you an exact copy of the error when I can.
Looking at the video, however, I REALLY like the aesthetic of the car movement design. It would appear to be fluid enough to use without slowing down the player significantly.
Some of the controls appear limited, however, at least based on what you use in the video: changing velocity appears to be key-based while selecting a direction appears to be mouse-based. Ideally you'll want to be able to control all movement in the car with either or.
I also noticed that there are a rather large number of options to move, which could potentially make key-based movement rather convoluted. At the most, I was expecting seven distinct motions per velocity: forward, lean left/right, soft-turn left/right, and hard-turn left/right (where soft-turn is the full motion and soft-turn is for roads), which the exception of moving backward instead of forward when the car is in reverse. For some reason there's always two squares for forward, which isn't necessary if speed changes are handled separately; there are also some rather sharp leans, which I wouldn't expect a car to handle very easily; for soft/hard turns at speed 1 there's no noticeable difference, but when including leans, you have spaces that act differently depending on the orientation of the car. All of this depends on how you want the cars to move, though, so take it only as a consideration.
I'll try to give better feedback once I can get the prototype to run myself.
2013 7DRL Challenge blind-runs too!
I do roguelike LPs, usually blind. I'm always looking for criticism!
The car movement isn't final. I agree that the amount of movement options is way too high. Right now all these options are calculated and therefore there are some inaccuracies. For the final draft I probably will create the options by hand and of course I will add the ability to select an option with the keyboard, too.
Sounds reasonable. I will try this. By "lean left" you mean keeping the orientation but still putting the car a bit to the left?
I guess here's a typo? What degree/angle should a hard turn have in your opinion?
I really appreciate your help! Thanks in heaps!
RenderSystem_GL. System Error: The specified module could not be found.
in DynLib::load at C:\Programming\ogre\OgreMain\src\OgreDynLib.cpp (line 91)I installed the OGRE source just in case it was necessary, but it didn't help. Attached are the bnb/ogre logs as well.
By "lean left" you mean keeping the orientation but still putting the car a bit to the left?
Leans are one space to the left/right, X spaces forward, with a maintained directional orientation.
Soft turns are are X spaces left/right, X spaces forward, with a 90-degree rotational change in orientation.
Some of this could be scaled (perhaps once you're at speed 4-5 you'll lean two spaces over) but this is the general idea. The player should understand that faster cars have less control and get used to the idea of slowing down when maneuvering sharply: given the smoothness of the GUI, I don't think it'll end up clunky at all.
20:20:44: SceneManagerFactory for type 'DefaultSceneManager' registered.
20:20:44: MovableObjectFactory for type 'ParticleSystem' registered.
20:20:44: OverlayElementFactory for type Panel registered.
20:20:44: OverlayElementFactory for type BorderPanel registered.
20:20:44: OverlayElementFactory for type TextArea registered.
20:20:44: ArchiveFactory for archive type FileSystem registered.
20:20:44: ArchiveFactory for archive type Zip registered.
20:20:44: MovableObjectFactory for type 'Entity' registered.
20:20:44: MovableObjectFactory for type 'Light' registered.
20:20:44: MovableObjectFactory for type 'BillboardSet' registered.
20:20:44: MovableObjectFactory for type 'ManualObject' registered.
20:20:44: MovableObjectFactory for type 'BillboardChain' registered.
20:20:44: MovableObjectFactory for type 'RibbonTrail' registered.
20:20:44: OGRE EXCEPTION(7:InternalErrorException): Could not load dynamic library RenderSystem_GL.
Thank you very much Game Hunter. I can't put in words how valuable your help, your input and your patience is to me.
The RenderSystem_GL error took me by surprise. I didn't expect OpenGL to cause errors and I am still not sure why it can't load the lib on your system. Right now I just can make some guesses. Maybe some OpenGL libs are missing on your system, so the RenderSystem_GL can't be loaded properly. I also checked how Ogre is loading Plugins and it seems that the LoadLibraryEx function is called with the parameter LOAD_WITH_ALTERED_SEARCH_PATH whos behaviour is undefined if the filename contains a relative path (which is strange because it's Ogre's default).
Actually this looks really good. The amount of options seems straightforward and manageable. I will definitely give this a try.
Thanks again Game Hunter. I really appreciate your help.


EnviroLift also has many applications to match the needs of the commercial and industrial markets. Whether it be property managers looking to correct safety and operational hazards such as sidewalk trip hazards, cracked uneven warehouse floors, sunken community hall floors, settled interior floors or leaking parkades. Sidewalks being repaired can be walked on during application, interior floors in buildings can stay in use while being fixed and even driveways/parking lots can be driven on immediately after application.