text:
The Florida Project is a touching ode to the magic of childhood
Posted on November 22, 2017 November 22, 2017 Author Néil RogersPosted in UncategorizedLeave a Reply
Director: Sean Baker Starring: Brooklyn Prince, Bria Vinaite, Willem Dafoe, Christopher Rivera, Caleb Landry Jones Runtime: 115 Minutes
For decades, Disney World Florida has served as a cornerstone of childhoods and imaginations across the globe. Every year millions of visitors flood its ticket booths in search of the infamous "magic" that Disney has to offer. Within the park's walls, guests are welcomed to a world unlike their own: castles stretch to dizzying heights, magnificent firework displays light up the night sky, and fairytales come to life before your very eyes.
Beyond its boundaries, however, the same magic and wonder is harder to find. In the theme park's shadows, strip malls, run-down hotels and sun-drenched swampland stretch for miles. The families in the encompassing areas struggle to hold onto what little they have. It is in this economic wasteland that the modern face of homelessness shows itself. Families move desperately from hotel to hotel in search of permanent accommodation. Rising rents determine what little luxuries they can afford. Jobs are hard to come by.
It is this world, however, that 6-year old Moonee and her friends embrace as their own private wonderland. On the margins of Disneyland and, indeed, of life, the young children search for magic around every corner, oblivious to the harsh reality of their circumstances.
The Florida Project follows the story of Moonee (Prince) and her young mother Halley (Vinaite) throughout the course of a summer they spend at a budget hotel named 'The Magic Castle'. The hotel, managed by the strict yet compassionate Bobby (Dafoe), caters to those families who are down on their luck. Despite her surroundings, a defiant Moonee goes about her summer in search of the wonderful world she knows exists. As Moonee and her friends explore the world around them, the severity of their situation deepens as Halley struggles to provide for her family and turns to dangerous means in order to make ends meet.
Similarly to his previous project Tangerine, director Sean Baker has cast his film in an unusual way. Instead of the multitude of actors that would have lined up to be in his next project, Baker opted for relatively unknown actors and real people to embody the roles he had created. Bria Vinaite, Moonee's mother, was discovered via an Instagram post that one of Baker's friends shared with him, while many of the child actors had never previously stared in films.
This daring tactic truly speaks to the type of filmmaker Baker is and the world he creates for his movies. There is an authenticity to The Florida Project that is difficult to come by in other films, where stories don't just unfold. Instead, you live with the characters, get to know them and ultimately endure their heartbreak with them. Baker's ability to command such fantastic performances from his cast again speaks to his talents as a director.
However, Vinaite and Prince seem like natural stars. The wonderful on-screen chemistry they share allows for the audience sit with them through their moments of elation and joy, as well as the devastating circumstances under which they spend their summer at 'The Magic Castle'. This relationship truly offers a window through which the audience can peer into the lives of those experiencing struggles similar to Halley and Moonee's. Willem Dafoe (obviously not one of those unknown actors!) is remarkable as the hotel manager, Bobby. On the outside, Bobby is a stern, strict worker looking to impress his boss and run the hotel as effectively as possible. However, beneath this exterior he is a compassionate, genuine and sweet man who guides and protects the children throughout the summer. Dafoe demonstrates the various layers of his character impressively and easily gives one of the best performances of the year and of his long, remarkable career.
The Florida Project has a vibrancy and energy that, considering the subject matter, almost feels like it shouldn't. However, Baker's film is almost always shown through the eyes of Moonee and her friends, all of whom average around 6 years of age. As such, the reality of their situation evades them as they run amok on the hotel's grounds making new friends and enjoying the summer sunshine. Baker attempts to communicate the resilient power of a child's imagination, and succeeds in doing so. In a touching scene, Moonee brings her friend to a nearby field in order to stare at cows, remarking "I took you on a safari" – a simple, yet wonderful note on the innocence and imagination of the children.
As we view most of the film through their eyes, the camera's attention is drawn to the striking imagery that surrounds the children. The hotels are painted in an array of vibrant colours, with names such as 'The Magic Castle' and 'Futureland'. The nearby sweetshop is shaped like an ice-cream cone, while a giant, preposterous-looking wizard sits atop the local gift shop. While the reality is that these have been designed to attract Disney World guests, the children accept them as part of their everyday surroundings, as if giant wizards and ice-cream cones were magical, everyday structures. Cinematographer Alexis Zabe impressively captures these sites, and more, in the mesmerising Florida sunlight, aiding in Baker's efforts to craft an authentic world for his story.
While we spend the majority of our time on these wondrous adventures, The Florida Project never attempts to sensationalise the issue of homelessness or the problems that these families face. By telling the story through the children's perspectives, one of the main messages of the film is simply and effectively communicated – reality can corrupt the innocence of childhood. The film is sympathetic to its characters' struggles and how the children are affected. As Moonee remarks in the film, she is able to see the issues surrounding her, but perhaps not fully understand them: "I can always tell when adults are about to cry."
As helicopters transporting important guests fly over head, Halley and Moonee attempt to earn their living on the fringe of Disney World, flogging perfume and stollen tickets to the theme park's visitors. In a poignant scene, Moonee, her mother and her friend stare above them as the famous Disney World fireworks light up the night sky. While the sky has been rented out for the enjoyment of others, the fireworks strangely seem to face away from them, as they stare in awe of the magic that they are not able to experience first-hand.
The Florida Project is an absolutely mesmerising film. This colourful and vibrant masterpiece is an ode to the innocence and magic of childhood in the face of difficult realities. To call it an important film is an understatement – the new face of homelessness (being experienced across the world) is explored in vivid detail and with true compassion for the struggles that accompany it. As the old saying goes: you'll laugh, you'll cry, you'll absolutely adore every second of it.
The Florida Project is a wonderful, wonderful film.
About Néil Rogers
Originally hailing from Galway, Film In Dublin kindly adopted Néil to cover film on the other side of the country. With previous experience contributing to FilmFixx.com and Flirt FM, Néil is a dedicated cinema fan, who believes the only thing better than watching film, is talking about it!
View all posts by Néil Rogers →
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text:
\section{Introduction}

Let $k$ be any perfect field and $R=k[x_1,...,x_d]$ its polynomial ring in $d$ variables. In this case it is known \cite[IV, Th\'eor\`eme 16.11.2]{EGAIV} that the ring $\DD$ of $k$--linear differential operators on $R$ is the $R$-algebra (which we take here as a definition)
\[
\DD:=R \left\langle D_{x_i,t} \mid i=1,\ldots,d \mbox{ and } t\geq 1 \right\rangle\subseteq\mathrm{End}_{k}(R),
\]
generated by the operators $D_{x_i,t}$, defined as
\[
D_{x_i,t}(x_j^s)=\begin{cases}
\binom{s}{t}x_i^{s-t},\text{ if }i=j\text{ and }s\geq t,\\
0,\text{ otherwise }.\end{cases}
\]
For a non-zero $f\in R$, the natural action of $\DD$ on $R$ extends to $R_f$ in such a way that $R_f=\DD\frac{1}{f^m}$, for some $m\geq 1$. Whilst there are examples of $m>1$ in characteristic $0$ (e.g. \cite[Example 23.13]{Twentyfourhours}), it is $m=1$ in positive characteristic (\cite[Theorem 3.7 and Corollary 3.8]{AlvarezBlickleLyubeznik2005}). This is shown by proving the existence of a differential operator $\delta\in\DD$ such that $\delta(1/f)=1/f^p$, i.e., $\delta$ acts as Frobenius on $1/f$. We will suppose that $k=\mathbb{F}_p$ and fix an algebraic closure $\overline{k}$ of $k$ from now on.

For an integer $e\geq 0$, let $R^{p^e}\subseteq R$ be the subring of all the $p^e$ powers of all the elements of $R$ and
set $\DD^{(e)}:=\End_{R^{p^e}} (R)$, the ring of $R^{p^e}$-linear ring-endomorphism of $R$. Since $R$ is a finitely generated $R^p$-module, by \cite[1.4.8 and 1.4.9]{Yekutiely1992}, it is
\[
\DD=\bigcup_{e\geq 0}\DD^{(e)}.
\label{filter0}
\]
Therefore, for $\delta\in\DD$, there exists $e\geq 0$ such that $\delta\in \DD^{(e)}$ but $\delta\not\in \DD^{(e')}$ for any $e'<e$. Such number $e$ is called the level of $f$.

The level of a polynomial has been studied in \cite{AlvarezBlickleLyubeznik2005} and \cite{BoixDeStefaniVanzo2015}. In \cite{BoixDeStefaniVanzo2015}, an algorithm is given to compute the level and a good number of examples are exhibited. Moreover, if $f$ is a cubic smooth homogeneous polynomial defining an elliptic curve $\mathcal{C}=V(f)=\{(x:y:z)\in \mathbb{P}^2_k: f(x,y,z)=0\}$, the level of $f$ can be used to characterise the supersingularity of $\mathcal{C}$ in the following way:

\begin{teo}(\cite[Theorem 1.1]{BoixDeStefaniVanzo2015})
Let $f\in R$ be a cubic homogeneous polynomial such that $\mathcal{C}=V(f)$ is an elliptic curve over $k$. Denote by $e$ the level of $f$. Then
\begin{enumerate}[(i)]
\item $\mathcal{C}$ is ordinary if and only if $e=1$.
\item $\mathcal{C}$ is supersingular if and only if $e=2$.
\end{enumerate}
\label{intro1}
\end{teo}

The goal of the present work is to extend the results of \cite{BoixDeStefaniVanzo2015} to hyperelliptic curves of genus $g\geq 2$ defined over $k$. Such a curve $\mathcal{C}$ is birationally equivalent to the vanishing locus $V(f):=\{(x:y:z)\in \mathbb{P}^2_k: f(x,y,z)=0\}$, where $f$ is a homogeneous polynomial of degree $2g+1$ defined over $k$. We will write $\mathcal{C}\cong V(f)$. Notice that $V(f)$ is singular at the infinity point.

Denote by $\mathrm{Jac}(\mathcal{C})$ its Jacobian. It is well known that for
an integer $n>0$ then
\begin{align*}
    \mathrm{Jac}(\mathcal C)[n](\overline k)&=(\mathbb{Z}/n
    \mathbb{Z})^{2g}\quad \text{if $\mathrm{char}\, k  \not|\ n$,}\\
    \mathrm{Jac}(\mathcal C)[p^m](\overline
    k)&=(\mathbb{Z}/p^m\mathbb{Z})^i\quad \text{if $p=\mathrm{char}\, k,\
    m>0$},
\end{align*}
where $i$ can take every value in the range $0\le i \le g$, and is called the
$p$-rank of $\mathcal C$.

For the convenience of the reader, we recall here the following standard terminology, which will be used in this work

\begin{df}The curve $\mathcal{C}$ is said to be ordinary if its
$p$-rank is maximal. The curve $\mathcal{C}$ is said to be supersingular (resp.\
superspecial) if $\mathrm{Jac}(\mathcal{C})$ is isogenous (resp.\ isomorphic)
over $\overline{k}$ to the product of $g$ supersingular elliptic curves.
\end{df}
Our generalisation of Theorem \ref{intro1} is as follows:

\begin{teo}Let $f\in R$ be a homogeneous polynomial of degree $2g+1$ such that $\mathcal{C}\cong V(f)$ defines a hyperelliptic curve over $\overline{k}$ of genus $g$ and denote by $e$ the level of $f$. Assume $p>2g^2-1$. Then
\begin{itemize}
        \item[(i)] $e=2$ if $\mathcal C$ is ordinary,
        \item[(ii)] $e>2$ if $\mathcal C$ is supersingular but not
          superspecial.
    \end{itemize}
\label{intro2}
\end{teo}
The remaining cases are more complicated and require further study, in particular it would be very interesting to find a relation (if such a relation exists), between the level, the $p$-rank, and the $\alpha_p$-number. We give some examples showing that, beyond the ordinary and supersingular cases, the level is not enough to determine the $p$-rank, namely, we exhibit: a) hyperelliptic curves of genus $2$ having $p$-rank $1$ and level $e=2$, b) hyperelliptic
curves of genus $2$ having $p$-rank $1$ and level $3$. Hence the level cannot be used to characterise the $p$-rank, being then $e=2$ a strict necessary condition for $\mathcal{C}$ to be ordinary.

The superspecial case also remains open, although we exhibit c) two infinite families of superspecial hyperelliptic curves of arbitrary genus $g\geq 2$ such that, for a fixed genus $g$, there exist an infinite family of primes for which the curve has level $e>2$. Due to these explicit results plus computational evidence, we make the following

\begin{con}Let $f\in R$ be a homogeneous polynomial of degree $2g+1$ such that $\mathcal{C}\cong V(f)$ is a hyperelliptic curve over $\overline{k}$ of genus $g\geq 2$. Denote by $e$ the level of $f$. If $\mathcal{C}$ is superspecial, then $e>2$.
\end{con}

Our article is structured as follows: in Section \ref{preliminaries section} we review some definitions and properties of the level, following \cite{BoixDeStefaniVanzo2015}, as well as useful characterisations of the $p$-rank in terms of the Cartier-Manin matrix, following \cite{Yui1978}. We prove that the level is invariant under homogeneous change of coordinates, however, as we show via an example, the level is not invariant under birational transformation, and we will define the level of a hyperelliptic curve as the level of the polynomial defining its imaginary model . In Section \ref{results on the level} first, we prove that on the contrary to the elliptic case, the level of a hyperelliptic curve of genus $g\geq 2$ must be at least $2$. Then, we prove some technical results (Propositions 3.2 and 3.3), and use them to prove our main result, labelled above as Theorem \ref{intro2} and labelled as Theorem \ref{main result in the ordinary case} in Section \ref{results on the level}. Theorem \ref{main result in the supersingular case} relates the maximal non-vanishing power of the Cartier-Manin matrix with the level, which allows us to conclude the following:
\begin{cor}Let $\mathcal{C}$ be a hyperelliptic curve of genus $g\geq 2$ defined by a homogeneous polynomial $f$ of degree $2g+1$. Denote by $e$ the level of $f$. Assume $\mathcal{C}$ is supersingular but not superspecial. Then, $e>2$.
\end{cor}
Finally, in Section \ref{some examples} we provide several numerical examples of hyperelliptic curves of genus 2 with $p$-rank 1 and different level (Subsection \ref{non ordinary examples}) and exhibit two infinite families of supersingular hyperelliptic curves of genus $g\geq 2$ with level bigger than $2$. On the grounds of this evidence, we close our work by posing a conjecture:

\begin{con}Let $\mathcal{C}$ be a superspecial hyperelliptic curve of genus $g$ over $k$. Then, the level is strictly bigger than $2$.
\end{con}

The present work extends the research already carried out by the authors for genus $2$ and presented in the 13th International Conference on Finite Fields and Applications, held in Gaeta, in June 2017. We take the occasion to thank the organisers for having allowed us to present our preliminary version.

Likewise, the authors are indebted to Gary McGuire for regular discussions on this topic as well as for carefully reading our manuscript and useful comments which have improved our work.

The following notation will be used throughout the article: for $\mathbf{\alpha}=(a_1,\ldots ,a_d)\in\N^d$ we shall denote $\mathbf{x}^{\mathbf{\alpha}}:=x_1^{a_1}\cdots x_d^{a_d}.$ and set $\norm{\mathbf{x}^{\mathbf{\alpha}}}:=\max\{a_1,\ldots ,a_d\}$, and sometimes, even $\norm{\mathbf{\alpha}}$ instead of $\norm{\mathbf{x}^{\mathbf{\alpha}}}$. For any polynomial $g\in k [x_1,\ldots ,x_d]$, we define
\[
\norm{g}:=\max_{\mathbf{x}^{\mathbf{\alpha}}\in\supp (g)} \norm{\mathbf{x}^{\mathbf{\alpha}}},
\]
where if $g=\sum_{\mathbf{\alpha}\in\N^d} g_{\mathbf{\alpha}} \mathbf{x}^{\mathbf{\alpha}}$, $\supp (g):=\left\{x^{\mathbf{\alpha}}\in R\mid\ g_{\mathbf{\alpha}}\neq 0\right\}$. Finally, the total degree of $g$ is
\[
\\mathrm{deg} (g):=\max_{\mathbf{x}^{\mathbf{\alpha}}\in\supp (g)} \mathrm{deg} (\mathbf{x}^{\mathbf{\alpha}}),\text{ where }\mathrm{deg}(\mathbf{x}^{\mathbf{\alpha}}):=
a_1+\ldots +a_d.
\]

\section{Preliminaries}\label{preliminaries section}

\subsection{Reminders about differential operators in prime characteristic}

First, let us denote $\DD^n:=R \left\langle D_{x_i,t} \mid i=1,\ldots,d \mbox{ and } 1\leq t\leq n \right\rangle$ so that one has
\begin{equation}
\DD=\bigcup_{e\geq 0}\DD^n.
\label{filter1}
\end{equation}
From \cite[1.4.8 and 1.4.9]{Yekutiely1992}, it is $\DD^{p^e-1}=\DD^{(e)}$ for each $e\geq 0$, and hence, by \ref{filter1}, it is also
\begin{equation}
\DD=\bigcup_{e\geq 0}\DD^{p^e-1},
\label{filter2}
\end{equation}
making $\DD$ a filtered ring.

\subsection{The ideal of $p^e$-th roots} For any ideal $J \subseteq R$, and for any integer $e\geq 1$, denote by $J^{[p^e]}$ the ideal generated (over $R$) by all the $p^e$-th powers of elements in $J$, or equivalently, by the $p^e$-th powers of any set of generators of $J$.

\begin{df}\cite[page 465]{AlvarezBlickleLyubeznik2005} and \cite[Definition 2.2]{BlickleMustataSmith2008}
Given $g\in R$, for any integer $e\geq 1$, we set $I_e(g)$ to be the smallest ideal $J\subseteq R$ such that $g\in J^{[p^e]}$.
\end{df}
\begin{rk}Since $R$ is a free $R^{p^e}$-module for each $e\in \N$, with basis given by the monomials $\{\mathbf{x}^{\mathbf{\alpha}} \mid \norm{\alpha} \leq p^e-1\}$, for $g\in R$ such that
\[
g=\sum_{0\leq\norm{\mathbf{\alpha}}\leq p^{e}-1}g_{\mathbf{\alpha}}^{p^e} \mathbf{x}^{\alpha},
\]
then $I_e (g)$ is the ideal of $R$ generated by elements $g_{\mathbf{\alpha}}$ \cite[Proposition 2.5]{BlickleMustataSmith2008}.
\label{remarkI}
\end{rk}

\begin{rk} \label{I_e homog} Notice that, if $g$ is a homogeneous polynomial, then, for all $e \in \N$, $I_e(g)$ is a homogeneous ideal. Indeed, if we write $g=\sum_{0\leq\norm{\mathbf{\alpha}}\leq p^{e}-1}g_{\mathbf{\alpha}}^{p^e} \mathbf{x}^{\alpha}$, then we can assume without loss of generality that every $g_\alpha^{p^e} \mathbf{x}^{\alpha}$ has degree equal to $\mathrm{deg}(g)$. Moreover, $g_\alpha$ is homogeneous of degree
\[
\mathrm{deg}(g_\alpha) = \frac{\mathrm{deg}(g)-\mathrm{deg}(\mathbf{x}^{\alpha})}{p^e}.
\]
\label{remarkdeg}
\end{rk}

The relation between ideals of $p^e$--th roots and differential operators is as follows:

\begin{lm}\cite[Lemma 3.1]{AlvarezBlickleLyubeznik2005}
For any integer $e\geq 0$, it is $I_e (g)^{[p^e]}=\DD^{(e)}\cdot g$, where
$\DD^{(e)}\cdot g:=\{\delta (g):\ \delta\in\DD^{(e)}\}.$
\label{some piece of Frobenius descent}
\end{lm}

\subsection{The level of a polynomial}As pointed out in the introduction, the level of a non-zero polynomial $f\in R$ is defined as the minimal integer $e\geq 1$ such that there exist $\delta\in \DD^{p^e-1}$ with $\delta(1/f)=1/f^p$. Here we make this idea more precise and recall another characterization, following \cite[Definition 2.6]{BoixDeStefaniVanzo2015}. First, we observe that as an immediate consequence of \cite[Lemma 3.4]{AlvarezBlickleLyubeznik2005}, there is a decreasing chain of ideals
\begin{equation}\label{cadena incordio}
R=I_0 (f^{p^0-1})\supseteq I_1 (f^{p-1})\supseteq I_2(f^{p^2-1})\supseteq I_3 (f^{p^3-1})\supseteq\ldots
\end{equation}
The level can be characterised as the minimal step where this chain stabilises. More precisely:

\begin{teo}\cite[Proposition 3.5, and Theorem 3.7]{AlvarezBlickleLyubeznik2005}
Define
\[
e:=\inf\left\{s\geq 1\mid\ I_{s-1}\left(f^{p^{s-1}-1}\right)=I_s\left(f^{p^s-1}\right)\right\}.
\]
Then, the following assertions hold.
\begin{enumerate}[(i)]

\item The chain of ideals \eqref{cadena incordio} stabilizes rigidly, that is $e < \infty$ and $I_{e-1}\left(f^{p^{e-1}-1}\right)=I_{e+s} \left(f^{p^{e+s}-1}\right)$ for any $s\geq 0$.

\item One has
\[
e=\min\left\{s\geq 1\mid\ f^{p^s-p}\in I_s\left(f^{p^s-1}\right)^{[p^s]}\right\}.
\]

\item There exists $\delta\in\DD^{(e)}$ such that $\delta(f^{p^e-1}) = f^{p^e-p}$, or equivalently such that $\delta (1/f)=1/f^p$.

\item There is no $\delta '\in\DD^{(e')}$, with $e'<e$, such that $\delta ' (1/f)=1/f^p$.
\end{enumerate}
\label{first step of the algorithm}
\end{teo}

\begin{df}\label{the level for any polynomial}(\cite[Definition 2.6]{BoixDeStefaniVanzo2015})
For a non-zero polynomial $f \in R$, we call the integer $e$ defined in Theorem \ref{first step of the algorithm} the level of $f$. Also, we will say that $\delta \in \DD^{(e)}$ such that $\delta(f^{p^e-1}) = f^{p^e-p}$, or equivalently such that $\delta(1/f) = 1/f^p$, is a differential operator associated with $f$.
\end{df}

\subsection{The level of a hypersurface}Next, we prove that the level of a non-zero homogeneous polynomial $f\in R$ is an invariant under change of homogeneous coordinates, hence depending just on the projective hypersurface defined by $f$.

Let $k$ be (for the moment) any perfect field of prime characteristic $p$, let $R=k[x_1,\ldots ,x_d],$ standardly $\Z$--graded (meaning $\mathrm{deg} (x_i)=1$ for any $1\leq i\leq d$). Denote $G:=\operatorname{GL}_d (k)$ and observe that $R$ has a right action of $G$ defined by $(f|A)(x_1,...,x_d):=f(y_1,...,y_d),$ where
\[
\begin{pmatrix} y_1\\ \vdots\\ y_d\end{pmatrix}=A\cdot\begin{pmatrix} x_1\\ \vdots\\ x_d\end{pmatrix},
\]
for $A\in G$.
Observe as well that a matrix $A\in G$ induces an isomorphism $\phi_A$ of graded $k$--algebras $\xymatrix@1{R\ar[r]^-{\phi_A}& R}$ defined by $\phi_A(f)=f|A$. This is clear, since $A$ being invertible, both sets $\{\mathbf{x}^{\alpha}:\ \alpha\in\mathbb{Z}_{\ge0}^d\}$ and
$\{\mathbf{y}^{\beta}:\ \beta\in\mathbb{Z}_{\ge0}^d\}$ are $k$--basis of $R$ and for matrices $A,B\in G$ and $f\in R$, it is easy to check that $(f|A)|B=f|AB$.

\begin{df}
Given homogeneous $f,g\in R$, we say that $f$ and $g$ are $G$--equivalent if there is $A\in G$ such that $\phi_A(f)=g$.
\end{df}

\begin{lm}\label{homogeneous change of coordinates give p basis}
Notations as before, let $y_1,\ldots ,y_d\in R$ be homogeneous elements of degree $1$ such that
\[
\begin{pmatrix} y_1\\ \vdots\\ y_d\end{pmatrix}=A\cdot\begin{pmatrix} x_1\\ \vdots\\ x_d\end{pmatrix},
\]
for some $A\in G$ Then, for any $e\geq 1$ the set
\[
\mathcal{B}:=\{\mathbf{y}^{\alpha}:=y_1^{a_1}\cdots y_d^{a_d}:\ \alpha =(a_1,\ldots ,a_d)\in\mathbb{Z}_{\ge0}^d,\ 0\leq a_i\leq p^e-1\text{ for any }1\leq i\leq d\}
\]
is a basis of $R$ as $R^{p^e}$--module.
\end{lm}

\begin{proof}
Since $G$ is invertible, both $\{\mathbf{x}^{\alpha}:\ \alpha\in\mathbb{Z}_{\ge0}^d\}$ and
$\{\mathbf{y}^{\beta}:\ \beta\in\mathbb{Z}_{\ge0}^d\}$ are $k$--basis of $R$, hence it is enough to check that
\[
\mathcal{B}:=\{\mathbf{x}^{\alpha}:=x_1^{a_1}\cdots x_d^{a_d}:\ \alpha =(a_1,\ldots ,a_d)\in\mathbb{Z}_{\ge0}^d,\ 0\leq a_i\leq p^e-1\text{ for any }1\leq i\leq d\}
\]
is a basis of $R$ as $R^{p^e}$--module.

Indeed, let $g\in R,$ and write
\[
g=\sum_{\beta\in\mathbb{Z}_{\ge0}^d}g_{\beta}\mathbf{x}^{\beta},
\]
where $g_{\beta}\in k$ and $g_{\beta}=0$ up to a finite number of
terms; in this way, given $\beta=(b_1,\ldots ,b_d)\in\mathbb{Z}_{\ge0}^d$ so that
$\mathbf{x}^{\beta}\in\supp (g),$ after doing the Euclidean division
we can write in a unique way $b_i=q_i p^e+a_i,$ where both $q_i$
and $a_i$ are non--negative integers, and $0\leq a_i\leq p^e-1.$ Therefore, one
has that
\[
g=\sum_{\alpha\in\mathcal{B}}g_{\alpha}^{p^e}\mathbf{x}^{\alpha},
\]
for some $g_{\alpha}\in R$.
\end{proof}

We can now prove that the level is $G$--invariant, what is immediate after the following result.

\begin{teo}
Let $f,g\in R$ be homogeneous $G$--equivalent polynomials via $A\in G$. Then, for any $e\geq 0$, it is
$\phi_A (I_e (f))=I_e(g)$ and $\phi_A^{-1} (I_e (g))=I_e(f),$ where, given any ideal $J\subseteq R$ and a ring endomorphism $\xymatrix@1{R\ar[r]^-{\varphi}& R},$ $\varphi (J)$ denotes the ideal generated by the images of elements of $J$ under $\varphi.$
\label{the level is a geometric invariant}
\end{teo}

\begin{proof}
Setting
\[
\begin{pmatrix} y_1\\ \vdots\\ y_d\end{pmatrix}=A\cdot\begin{pmatrix} x_1\\ \vdots\\ x_d\end{pmatrix},
\]
and applying Lemma \ref{homogeneous change of coordinates give p basis} we see that for any $e\geq 1,$ the set
\[
\{\mathbf{y}^{\alpha}:=y_1^{a_1}\cdots y_d^{a_d}:\ \alpha =(a_1,\ldots ,a_d)\in\mathbb{Z}_{\ge0}^d,\ 0\leq a_i\leq p^e-1\text{ for any }1\leq i\leq d\}
\]
is a basis of $R$ as $R^{p^e}$--module.

Now, for $e\geq 0,$ and write
\[
f=\sum_{0\leq\norm{\alpha}\leq p^e-1}f_{\alpha}^{p^e}\mathbf{x}^{\alpha},
\]
for $f_{\alpha}\in R$. Then
\[
g=\phi_A(f)=\sum_{0\leq\norm{\alpha}\leq p^e-1}\phi_A(f_{\alpha})^{p^e}\mathbf{y}^{\alpha},
\]
which shows that $\phi_A (I_e (f))\subseteq I_e (g)$. Equality holds because
$\phi_A$ is an isomorphism.
\end{proof}

Motivated by Theorem \ref{the level is a geometric invariant}, we introduce the following

\begin{df}\label{the level of a homogeneous hypersurface}
Let $H\subseteq\mathbb{P}_{k}^n$ be a projective hypersurface of positive degree defined
over a perfect field $k$ of prime characteristic $p$. We define the level of $H$ as
the level of any element in the below set:
\[
\{f\in K[x_0,\ldots ,x_n] :\ f\text{ is homogeneous, } \mathrm{deg} (f)=h,\  V(f)=H\}/\sim,
\]
where $\sim$ denotes $\operatorname{GL}_{n+1} (k)$--equivalence.
\end{df}

\subsection{Elliptic curves}
From now on, let $k=\mathbb{F}_p$ ($p>2$)and $R=k[x,y,z]$ from now on. Let $E\subseteq\mathbb{P}_{k}^2$ be an elliptic curve
defined over $k$ by a homogeneous cubic $f\in R$. For a prime $l\geq 2$, $E[l](\overline{k})$ is the kernel of the multiplication-by-$l$ isogeny $[l]$, which is a subgroup of $E(\overline{k})$. As mentioned in the introduction, for $l\neq p$, it is $|E[l](\overline{k})|=p^2$ whilst $|E[p](\overline{k})|=p^i$ (for $i=0,1$). If $|E[p](\overline{k})|=p$, then $E$ is said to be ordinary and otherwise supersingular.

Writing  $f^{p-1}=h\cdot (xyz)^{p-1}+\ldots,$ for some $h\in k$, it is also well known that $E$ is supersingular if and only if $h=0$ \cite[V.4.1]{Silverman1986}. The following characterization is given in \cite[Theorem 6.9]{BoixDeStefaniVanzo2015}.

\begin{teo}\label{starting point elliptic setting}
For $E$, $f$ and $p\geq 2$ as above, the following assertions hold.
\begin{enumerate}[(i)]

\item $E$ is ordinary if and only if $f$ has level one.

\item $E$ is supersingular if and only if $f$ has level two.

\end{enumerate}
\end{teo}

Being ordinary or supersingular is related with the level via the Cartier differential operator
\[
\Delta:=\frac{1}{(p-1)!^3}\cdot\frac{\partial^{p-1}}{\partial z^{p-1}}
\frac{\partial^{p-1}}{\partial y^{p-1}}\frac{\partial^{p-1}}{\partial x^{p-1}}.
\]
If $E$ is ordinary, then $h\neq 0$ and therefore one has that
\[
\left(\frac{1}{h}\cdot\Delta\right)(f^{p-1})=1,
\]
which is to say that $f$ has level one. If $h=0$, then the best one can say is that there is a (not necessarily unique) differential operator $\delta$ of level two with the property that $\delta (f^{p^2-1})=f^{p^2-p}$. This operator can be expressed as a certain $R$-multiple (see \cite[Proposition 6.2]{BoixDeStefaniVanzo2015}
for some examples) of the operator
\[
\frac{1}{(p^2-1)!^3}\cdot\frac{\partial^{p^2-1}}{\partial z^{p^2-1}}
\frac{\partial^{p^2-1}}{\partial y^{p^2-1}}\frac{\partial^{p^2-1}}{\partial x^{p^2-1}}.
\]

\subsection{Hyperelliptic curves} Let $\mathcal{C}$ be a hyperelliptic curve of genus $g\geq 2$ defined over $k$ by an affine equation $f(x,y)=0$, with $f(x,y)=y^2-h(x)$, and $h(x)\in k[x]$ a polynomial with no multiple roots and with  degree $2g+2$ (real hyperelliptic curve) or $2g+1$ (imaginary hyperelliptic curve). Notice that the homogeneized  polynomial $\tilde{f}$ defines a plane curve which is singular at infinity, and it is only birationally equivalent to the hyperelliptic curve $\mathcal{C}$, which is smooth in $\mathbb{P}^3(k)$. Moreover, every hyperelliptic curve of genus $g$ is birationally equivalent to an imaginary model via the transformation
\begin{equation}
(x,y)\mapsto \left(\frac{1}{x-a},\frac{y}{(x-a)^{g+1}}\right),
\label{moebiusreal}
\end{equation}
where $a$ is a root of $h$.

As we see in the next example, in general the level is not invariant under this transformation.

\begin{ex}Consider the hyperelliptic curve with imaginary model given by the homogeneous polynomial $f(x,y,z)=y^2z^3-x^5-2x^3z^2-2x^2z^3-xz^4-2z^5$ over $\mathbb{F}_{13}$. The level of this polynomial is $3$. By applying ( \ref{moebiusreal} ) with $a=-1$, we obtain the homogeneous polynomial $h(x,y,z)=y^2z^4-2x^6+2x^4z^2-8x^3z^3+x^2z^4-6xz^5-8z^6$, which has level $2$.
\end{ex}

\begin{df}The level of a hyperelliptic curve of genus $g$ is the level of the polynomial defining its imaginary model, up to change of homogeneous coordinates.
\end{df}

Denote by $\mathrm{Jac}(\mathcal{C})$ the Jacobian of $\mathcal{C}$, which is an abelian variety defined over $k$ of dimension $g$.

The $p$-rank of $\mathcal{C}$ is defined to be the $p$-rank of $\mathrm{Jac}(\mathcal{C})$. Likewise, we say that $\mathcal{C}$ is ordinary (respectively, supersingular, superspecial) if so is $\mathrm{Jac}(\mathcal{C})$.

The ordinary/supersingular/superspecial character of $\mathcal{C}$ can be easily described by the Cartier operator, which we recall next.

\begin{df}\label{the cartier manin matrix in general}
For a positive integer $k\geq 1$, write
\[
h(x,1)^{(p^k-1)/2}=\sum_{j=0}^{N_k} c^{(k)}_j x^j,
\]
where $N_k:=((p^k-1)/2)(2g+1)$. It is convenient to define the $g\times g$ matrix with elements on $k$
\[
C_k:=\begin{pmatrix} c^{(k)}_{p^k-1} & c^{(k)}_{p^k-2} & \ldots & c^{(k)}_{p^k-g}\\
c^{(k)}_{2p^k-1}& c^{(k)}_{2p^k-2}& \ldots & c^{(k)}_{2p^k-g}\\
\vdots & \vdots & \ddots & \vdots\\
c^{(k)}_{gp^k-1}& c^{(k)}_{gp^k-2}& \ldots & c^{(k)}_{gp^k-g}\end{pmatrix}.
\]
Set $c_j:=c^{1}_j$ and notice that $C:=C_1$ is the Cartier-Manin matrix of $\mathcal{C}$. It corresponds to the Cartier operator acting on $H^0(\mathcal{C},\Omega_{\mathcal{C}})$, where $\Omega_{\mathcal{C}}$ stands for the sheaf of regular differentials on $\mathcal{C}$. Notice that since our curve is defined over $k$, the Cartier operator is linear (cf. \cite{achter} for details and a relevant recent discussion on the Cartier-Manin and Hasse-Witt matrices).
\label{numerosc}
\end{df}

We will also make use of the following property.

\begin{lm}\label{high hassewitt} For $k\ge 1$, it is
$$
C_k=C^k.
$$
\end{lm}
\begin{proof}
The case $k=1$ is trivial. Since
$$
h(x,1)^{(p^k-1)/2}=(f(x,1)^{(p^{k-1}-1)/2})^ph(x,1)^{(p-1)/2},$$
it is
$$
C_k=C_{k-1}^pC,
$$
and since the coefficients of $f$ belong to $k$, they are invariant under the $p$-Frobenius and, by induction, the result holds.
\end{proof}

The matrices $C_k$ are related with the ordinary/supersingular/superspecial character of $\mathcal{C}$ in the following way.

\begin{teo}\cite[Theorem 3.1]{Yui1978}
Let $\mathcal{C}$ be a hyperelliptic curve of genus $g\geq 2$ defined over $k$  where $p\geq 7$. Then  $\mathcal{C}$ is ordinary if and only if $\rank (C)=g.$
\label{yuith}
\label{conditions and the Cartier Manin matrix}
\end{teo}

In the other extreme, Nygaard proved \cite[Theorems 4.1]{Nygaard1981} that $\mathcal{C}$ is superspecial if and only if $C$ is identically zero.

As for the supersingular case, the following result holds.

\begin{teo}\cite[Theorem 2.1]{Nygaard1983}
Let $\mathcal{C}$ be a hyperelliptic curve of genus $g\geq 2$ defined over $k$. Then, $\mathcal{C}$ is supersingular if and only if, for $1\leq i,j\leq g$ , it is

\[
c^{(g+2(n-1))}_{ip^{g+2(n-1)-j},g+(n-1)}\equiv 0\pmod{p^n},
\]

where $1\leq n\leq G$, with
\[
G=\begin{cases}
\frac{g^2+1}{2}-(g-1),\text{ if }g\text{ is odd ,}\\
\frac{g^2+1}{2}-\frac{3(g-1)}{2}\text{ if }g\text{ is even}.\end{cases}
\]
\label{conditions and the Cartier Manin matrix2}
\end{teo}

Notice that for $g=2$, Lemma \ref{high hassewitt} and Theorem \ref{conditions and the Cartier Manin matrix2} imply that $\mathcal{C}$ is supersingular if and only iff $C^2$ is identically zero. Moreover, this is equivalent to the $p$-rank being equally $0$.

\section{Results on the level of hyperelliptic curves}\label{results on the level}

Hereafter, unless otherwise is specified, assume  $p\geq 3$ and let $f(x,y,z)=y^2z^{2g+1}-h(x,z)$ with $h(x,z)\in k[x,z]$ a homogeneous polynomial of degree $2g+1$. For any field extension $K/k$, integers $r,n\geq 1$ and any matrix $G\in M_{r\times r}(K)$, denote by $G^{[n]}$ the matrix obtained by raising each entry of $G$ to the $n$-th power. Since the coefficients of $f$ belong to $k$, so are the numbers $c_i$ in Definition \ref{numerosc}, and hence, for the Cartier-Manin matrix $C$ of the curve defined by $f$, it is $C^{[p^r]}=C$ for each $r\geq 1$.

Unlike the elliptic case, the following holds:

\begin{prop}\label{level not one for higher genus: first try}Let $\mathcal{C}$ be a hyperelliptic curve over $k$ of genus $g\geq 2$ Then, if $p\geq 2g-1$, the level of $\mathcal{C}$ is strictly bigger than $1$.
\end{prop}
\begin{proof}
Let $\mathcal{C}$ be defined by a homogeneous polynomial $f\in R$ of degree $d=2g+1$. It is enough to check that the level of $f$ is strictly bigger than $1$. Indeed, if it were not the case, we would have $R=I_1(f^{p-1})$ and so, $1\in I_1(f^{p-1})$. But if we write $f^{p-1}=\displaystyle \sum_{||\alpha||\leq p-1} f_{\alpha}^px_{\alpha}$, for $\{x_{\alpha}\}_{||\alpha||\leq 3(p-1)}$ a basis of $R$ as $R^p$-module and since all the terms $f_{\alpha}^px_{\alpha}$ are distinct, it follows that
$$p\mathrm{deg}(f_{\alpha})+\mathrm{deg}(x_{\alpha})=(2g+1)(p-1),$$
hence $\mathrm{deg}(f_{\alpha})\geq (2g-2)(1-1/p)$, so that $\mathrm{deg}(f_{\alpha})\geq 2g-2$ if $p\geq 2g-1$. Since, as seen in  \ref{remarkI}, $I_1(f^{p-1})$ is generated by the elements $f_{\alpha}$, this contradicts that $1\in I_1(f^{p-1})$.
\end{proof}

\subsection{Preliminary calculations}
Our main result will rely on the computations which we provide here. We will denote for now on $f(x,y,z)=y^2z^{2g-1}-h(x,z)$ with $h(x,z)\in k[x,z]$ a homogeneous polynomial of degree $2g+1$. For $r\geq 1$ write
\begin{equation}\label{binomial expansion of curve}
f^{p^r-1}=\sum_{j=0}^{p^r-1}\binom{p^r-1}{j}y^{2(p^r-1-j)}z^{(2g-1)(p^r-1-j)}h(x,z)^j,
\end{equation}
and
\begin{equation}\label{Rp expansion of curve}
f^{p^r-1}=\sum_{||\alpha||\leq p^r-1}f_{\alpha}^{p^r}x^{\alpha},
\end{equation}
where the elements $\{x^{\alpha}\}_{||\alpha||\leq p^r-1}$ are an $R^{p^r}$-basis of $R$.

For a polynomial $h(x,y,z)\in R$, $\mathrm{deg}_x(h)$ stands for the degree of $h$ as a polynomial in $x$ with coefficients in the ring $k[y,z]$ and analogously for $\mathrm{deg}_y(h)$ and $\mathrm{deg}_z(h)$.

\begin{prop}Let $g\geq 2$ and $r\ge 1$ be an integer. Then the following assertions hold.
\begin{enumerate}
	\item[(i)] An $R$-multiple of $y^2$ cannot be an element of a generating set for $I_r$,
    \item[(ii)] Assume $p^r>2g^2-1$. Then, no monomial of the form $x^ayz^b$ can be an element of the generating set of $I_r$ if $a+b<2g-2$.
    \item[(iii)] Assume $p^r>2g^2-1$. Then, the elements $\{z^ix^{2g-2-i}\}_{i=0}^{g-2}$ cannot be generators of $I_r$.

    \end{enumerate}
   \label{lemaideal1}
\end{prop}
\begin{proof}Since the degree in $y$ of each monomial in $f^{p^r-1}$ is between $0$ and $2(p^r-1)$, which is smaller than
  $2p^r$, assertion (i) holds.

For assertion (ii), assume that $x^ayz^b=f_{\alpha}$ for some $||\alpha||\leq p^r-1$ and $a+b<2g-2$. since $\mathrm{deg}_z(f_{\alpha}x^{\alpha})=bp^r+\mathrm{deg}_z(x^{\alpha})$, it must be $\mathrm{deg}_z(f_{\alpha}x^{\alpha})<(2g-2)p^r$. Hence, $a+b=2g-3$, for if it were smaller, then
$$
\mathrm{deg}(x^{\alpha})=(2g+1)(p^r-1)-(a+b+1)p^r>2p^r-2g+1>p^r,
$$
which is a contradiction.

Write $\mathrm{deg}(f_{\alpha}x^{\alpha})=(2g-2)p^r+(2g+1)(p^r-1)-(2g-2)p^r=(2g-2)p^r+3p^r-(2g+1)$. This forces $\mathrm{deg}(x^{\alpha})=3p^r-(2g+1)$.

Setting $x^{\alpha}=x^{p^r-a_1}y^{p^r-a_2}z^{p^r-a_3}$ so that $\mathrm{deg}_y(f_{\alpha}^{p^r}x^{\alpha})=p^r-a_2$, it is $2g+1\geq 2p^r-\mathrm{deg}_y(f_{\alpha}^{p^r}x^{\alpha})$. On the other hand, from (\ref{binomial expansion of curve}) we see that
$$
\mathrm{deg}_z(f_{\alpha}^{p^r}x^{\alpha})\geq \frac{2g-1}{2}\mathrm{deg}_y(f_{\alpha}^{p^r}x^{\alpha}),
$$
hence
$$
2g+1\geq 2p^r-\mathrm{deg}_y(f_{\alpha}^{p^r}x^{\alpha})\geq 2p^r-\frac{2}{2g-1}\mathrm{deg}_z(f_{\alpha}^{p^r}x^{\alpha})>\frac{2p^r}{2g-1}.
$$
This would yield $p^r\leq 2g^2-1$, which is a contradiction.

For (iii), if any monomial in the list were a generator $f_{\alpha}$, then $\mathrm{deg}_y(f_{\alpha}^{p^r}x^{\alpha})=p^r-a_2$ with $a_2\leq 2g+1$, as in (ii). Then, it would be
$$
\mathrm{deg}_z(f_{\alpha}^{p^r}x^{\alpha})\geq\frac{2g-1}{2}\mathrm{deg}_y(f_{\alpha}^{p^r}x^{\alpha})\geq\frac{(2g-1)(p^r-2g-1)}{2}=(g-1)p^r+\frac{p^r-4g^2+1}{2},
$$
a contradiction with the fact that the $z$-degrees of monomials are less than or equal $g+1$.
\end{proof}

\begin{prop}Assume $p^r>2g^2-1$. Then the ideal $I_r$ is contained in the ideal
$$
M=(\{z^ax^b|a+b=2g-2,a\geq g-1,b\geq 0\}\cup\{z^ax^b|a+b=2g-1,0\leq a<g-1\}).
$$
\end{prop}\label{lemaideal2}
\begin{proof}First, we observe that for $p^r>2g+1$, it must be $2g-2 \leq \mathrm{deg}(f_{\alpha})\leq 2g$. From Proposition \ref{lemaideal1} (i), the generators $f_{\alpha}$ cannot be multiple of $y^2$, so they belong to two categories: a) monomials of the form $x^ayz^b$ with $a+b\geq 2g-2$ (due to Proposition \ref{lemaideal1} (ii)) or b) monomials of the form $x^az^b$ with $a+b\leq 2g$.

For the elements of the form $x^az^b$, it must be $a+b\geq 2g-2$, so we have two cases:

\begin{itemize}
\item[a)] $a+b=2g-2$. This provides the list $\{x^iz^{2g-2-i}\}_{i=0}^{g-1}$, because of Proposition \ref{lemaideal1} (iii).
\item[b)] $a+b\geq 2g-1$. For $a+b=2g-1$, the monomials $z^ax^{2g-1-a}$ with $g-1\leq a\leq 2g-1$ are multiples of the elements in the list of case a), so this case provides the extra terms $\{z^ix^{2g-1-i}\}_{i=0}^{g-2}$. The monomials with $a+b=2g$ are multiples of cases a) or b).
\end{itemize}

Finally, the elements $x^ayz^b$ with $a+b\geq 2g-2$ are multiples of the elements without $y$.
\end{proof}

\begin{df}The ideal $M=(\{z^ax^b|a+b=2g-2,a\geq g-1,b\geq 0\}\cup\{z^ax^b|a+b=2g-1,0\leq a<g-1\})$ will be called the relevant ideal of $\mathcal{C}$.
\end{df}

\subsection{Ordinary hyperelliptic curves of genus $g\geq 2$}
We use here our former results to prove that the level of hyperelliptic curves of genus $g\geq 2$ is $2$.

\begin{teo}\label{main result in the ordinary case}Let $\mathcal{C}$ be a hyperelliptic curve over $k$ defined by a homogeneous polynomial of degree $2g+1$. Assume $p>2g^2-1$. If $\mathcal{C}$ is ordinary then
$$I_1=I_2=M.$$
Therefore the level of the $\mathcal{C}$ is $2$.
\end{teo}
\begin{proof}
First, for $r=1,2$, notice that for each $1\leq j\leq g$, the expansion of $f^{p^r-1}$ contains the linear combination
$$
\lambda(c_{p^r-j}z^{2g-2}+c_{2p^r-j}z^{2g-3}x+...+c_{gp^r-j}z^{g-1}x^{g-1})^{p^r})x^{p^r-j}y^{p^r-1}z^{p^r-(2g-j)},
$$
with $\lambda=(-1)^{\frac{p^r-1}{2}}{{p^r-1}\choose{\frac{p^r-1}{2}}}\neq 0$, hence for $1\leq j\leq g$,
$$c_{p^r-j}z^{2g-2}+c_{2p^r-j}z^{2g-3}x+...+c_{gp^r-j}z^{g-1}x^{g-1}\in I_r.$$

Since both $\mathcal{C}$ and $\mathcal{C}^2$ are invertible (because of Theorem \ref{yuith}), it follows that the first $g$ generators of $M$, namely $\{z^ax^b\}_{a=g-1}^{2g-2}$, belong to $I_r$.

We need now to check that the the rest of the generators of $M$, i.e. $\{z^ax^b\}_{a=0}^{g-1}$ also belong to $I_r$. Let us start showing that $x^{g-1}z^{g-2}\in I_r$.

Set $u:=\left\lfloor\frac{(g-1)p^r}{2g-1}\right\rfloor-1$ and consider
$$
h_{u,1}(x,y,z)={{p^r-1}\choose{u}}(y^2z^{2g-1})^u(-x^{2g+1})^{p^r-1-u}.
$$
it is
$$
(g-1)p^r-2(2g-1)\leq \mathrm{deg}_z(h_{u,1})\leq (g-1)p^r-(2g-1),
$$
and
$$
(g+1)p^r+\frac{p^r}{2g-1}\leq \mathrm{deg}_x(h_{u,1})\leq (g+1)p^r-(2g+1).
$$

Since $p^r>2g^2-1$, it follows that $(g-2)p^r\leq \mathrm{deg}_z(h_{u,1})\leq {(g-2)p^r+(p^r-1)}$ and $(g+1)p^r\leq \mathrm{deg}_x(h_{u,1})\leq (g+1)p^r+(p^r-1)$, so the term $x^{g+1}z^{g-2}$ is a summand in some of the elements $f_{\alpha}$, generators of $I_r$.

Since $u<p^r-1$, $h_{u,1}$ only appears in the $u$-th term of the binomial sum (\ref{binomial expansion of curve}):
\begin{equation}
(y^2z^{2g-1})^u(g(x,z))^{p^r-1-u}=\mu h_{u,1}(x,y,z)+y^{2u}z^{(2g-1)u}[x^{(2g+1)(p^2-2-u)}+...]
\label{eqnaux}
\end{equation}
with $\mu\neq 0$. Setting $h_{u,1}(x,y,z)=x^{g+1}z^{g-2}x^{}\alpha$, with $x^{\alpha}$ an element in the basis of $R$ as $R^{p^r}$-module, and observing in Equation \ref{eqnaux} that the degree in $z$ of right hand summand is  bigger than or equal to $g-1$, the polynomial $f_{\alpha}$ containing the monomial $x^{g+1}z^{g-2}$ consists in the sum of a) this monomial (homogeneous of degree $2g-1$, b) a homogeneous component of degree $2g-2$ with degree in $z$ bigger than or equal to $g-1$ and no $y$ and c) a sum of homogeneous polynomials of smaller degrees.

Since $I_r$ is a homogeneous ideal, it follows that $f_{\alpha}$ contains, in its expansion, the term
$$
\rho x^{g+1}z^{g-2}+\sum_{i=g-1}^{2g-2}a_iz^ix^{2g-2-i}\in I_r$$
with $\rho\neq 0$. And since the terms $z^ix^{2g-2-i}$ belong to $I_r$ for $g-1\leq i\leq 2g-2$, also $x^{g+1}z^{g-2}\in I_r$.

The rest of the elements $x^{g+k}z^{g-k-1}$ can be now inductively proved to belong to $I_r$ for $2\leq k\leq g-1$: set
$$
u_k:=\left\lfloor\frac{(g-k)p^r}{2g-1}\right\rfloor-1
$$
and
$$
h_{u_k,k}(x,y,z)={{p^r-1}\choose{u_k}}(y^2z^{2g-1})^u(-x^{2g+1})^{p^r-1-u_k}.
$$
With the same argument as for $x^{g+1}z^{g-2}$, we show that $x^{g+k}z^{g-1-k}$ appears in the expansion of some generator $f_{\alpha_k}$ of $I_r$ and the the previous homogeneity argument, we conclude that $f_{\alpha_k}$ contains, in its expansion, the term
$$
\rho_k x^{g+k}z^{g-1-k}+\sum_{i=1}^{k-1}\rho_i x^{g+i}z^{g-1-i}+\sum_{i=g-1}^{2g-2}a_iz^ix^{2g-2-i}\in I_r
$$
with $\rho_i\neq 0$, hence $x^{g+k}z^{g-1-k}\in I_r$.

Finally, to finish the proof, we have to see that $x^{2g-1}\in I_r$, take a non-negative integer $0\leq u_r<2g+1$ with $2gp^r\equiv u_r \pmod{2g+1}$, and consider, in the binomial expansion (\ref{binomial expansion of curve}), the term
$$
(y^2x^{2g-1})^{\frac{p^r+u_r-(2g+1)}{2g+1}}(-x^{2g+1})^{\frac{2gp^r-u_r}{2g+1}},
$$
which appears with non-zero coefficient and can be written as
$$
\pm (x^{2g-1})^{p^r}x^{p^r-u_r}y^{\frac{2(p^r+u_r-(2g+1))}{2g+1}}z^{ \frac{(2g-1)(p^r+u_r-(2g+1))}{2g+1} }.
$$
\end{proof}

\subsection{Some partial results in the non-ordinary case}

\begin{lm}\label{ideal} For $r\ge 1$, there are elements generators $h_{r,1},...,h_{r,g}$ of $I_r$ of the form
$$\left[\begin{matrix}
	h_{r,1} & h_{r,2} & \cdots & h_{r,g}
	\end{matrix}\right]=\left[\begin{matrix}
	z^{2g-2} & xz^{2g-3} & \cdots & x^{g-1}z^{g-1}
	\end{matrix}\right]C_r.
$$
\end{lm}
\begin{proof}
	It is enough to check that the expansion of $(y^2z^{2g-1})^{(p^r-1)/2}g(x,z)^{(p^r-1)/2}$ contains a sum of the form
	$$
	\sum_{j=1}^g\sum_{i=1}^gc^{(r)}_{ip^r-j}z^{(2g-1-i)p^r}x^{(i-1)p^r}x_{\alpha_{i,j}},
	$$
for $x_{\alpha_{i,j}}$ elements in the basis of $R$ as $R^{p^r}$-module. Notice that since the degree of $y$ is between $0$ and $2(p^r-1)$, the power $y^{p^r-1}$ do not appear anywhere else in the expansion of $f(x,y,z)^{p^r-1}$ than in the considered term so than the sum cannot cancel with expansions corresponding to other powers.

But, this is clear:
$$
y^{p^r-1}z^{\frac{(2g-1)(p^r-1)}{2}}c^{(r)}_{ip^r-j}x^{ip^r-j}z^{\frac{(2g+1)(p^r-1)}{2}-ip+j}=c^{(r)}_{ip^r-j}g_{r,i}^{p^r}x_{\alpha_{i,j}},
$$
with $x_{\alpha_{i,j}}=x^{p^r-j}y^{p^r-1}z^{p^r-(2g-j)}$.
\end{proof}

\begin{prop}
For $r\geq 1$, the $k$-linear combinations of the set $\{z^{2g-2-z}x^i \: | \: 0\le i \le g-1\}$ which belong to $I_1$ are generated by  the entries of the following row-vector:
 $$[\begin{matrix}z^{2g-2} & \cdots & z^{g-1}x^{g-1} \end{matrix}]\cdot [C_g^r \: | \: C_g^{r-1} H_g],$$
where $H_g\in M_{g\times s_g}(k)$ with $s_g\geq 1$.
\end{prop}
\begin{proof}
For $r=1$, write
$$
f^{p-1}=...+\sum_{i=0}^{g-1}a_i(x^iz^{2g-2-i})^p\sum_{\alpha,\beta,\gamma}x^{p-\alpha}y^{p-\beta}z^{p-\gamma}+...
$$
with $\alpha+\beta+\gamma=2g+1$. The value $\beta=1$ yields the product
$$
[\begin{matrix}z^{2g-2} & \cdots & z^{g-1}x^{g-1} \end{matrix}]\cdot [C_g],
$$
while odd values $3\leq \beta\leq 2g+1$ yield the columns
$$
[\begin{matrix}z^{2g-2} & \cdots & z^{g-1}x^{g-1} \end{matrix}]\cdot [H_g].
$$

For $r>1$, the $R^{p^r}$-basis terms in the expansion $f^{p^r-1}$ can be expressed as
$$
x^{p^r-\alpha} y^{p^r-\beta} z^{p^r-\gamma}=(x^{p^{r-1}-a}y^{p^{r-1}-1}z^{p^{r-1}-b})^px^{ap-\alpha} y^{p-\beta} z^{bp-\gamma},
$$
where $a+1+b=2g+1$. Again, collecting the $R^{p}$-coefficients for $\beta=1$ yields
$$
[\begin{matrix}z^{2g-2} & \cdots & z^{g-1}x^{g-1} \end{matrix}]\cdot C_g^r,
$$
and the odd terms $3\leq \beta\leq 2g+1$ yield
$$
[\begin{matrix}z^{2g-2} & \cdots & z^{g-1}x^{g-1} \end{matrix}]\cdot C_g^{r-1}H_g.
$$
\end{proof}

\begin{rk}By choosing a suitable ordering of the columns in the matrix $H_g$, we can assume that its first column is

$$
\left[\begin{array}{c}
c_{p-(g+1)}\\
c_{2p-(g+1)}\\
\vdots\\
c_{gp-(p+1)}
\end{array}
\right],
$$
i.e., it corresponds to the sum $\sum_{i=1}^{g}c_{ij-(g+1)}x^{(2g-1-i)p}z^{(i-1)p}x^{p-(g+1)}y^{p-1}z^{p-(2g-j-1)}$ in the expansion of $g(x,z)^{\frac{p-1}{2}}$.
\label{remH}
\end{rk}

\begin{teo}\label{main result in the supersingular case}
	If $C_g^r\ne 0$ but $C_g^{r+1}=0$, then the level is strictly greater than $r+1$.
\end{teo}
\begin{proof}Since $C_g^r\ne 0$, there exits a non-zero linear combination of $\{z^{2g-2-z}x^i \: | \: 0\le i \le g-1\}$ in $I_r$. Since $C_g^{r+1}=C_g^{r+2}=0$, there is no non-zero linear combination of $\{z^{2g-2-z}x^i \: | \: 0\le i \le g-1\}$ in $I_{r+2}$. Therefore, $I_r\ne I_{r+2}$ and hence $I_r\ne I_{r+1}$ and $e>r+1$.
\end{proof}

\begin{cor}Let $\mathcal{C}$ be a supersingular but not superspecial hyperelliptic curve of genus $g\geq 2$. Then, the level is strictly greater than $2$.
\end{cor}
\begin{proof}Since $\mathcal{C}$ is not superspecial, by Theorem 4.1 in \cite{Nygaard1981}, it is $C_g\neq 0$. Since $\mathcal{C}$ is supersingular, by Theorem \ref{conditions and the Cartier Manin matrix2}, it is also $C_g^g=0$, hence there exists $1\geq r<g$ such that $C_g^r\neq 0$ but $C_g^{r+1}=0$ and hence $e>r+1\geq 2$.
\end{proof}

\section{Some examples}\label{some examples}
So far, we have studied the ordinary and supersingular (but not superspecial) cases. In this last section we give computational and theoretical evidence which leads us to pose the following conjecture.

\begin{con}Let $\mathcal{C}$ be a superspecial hyperelliptic curve of genus $g$ over $k$. Then, the level of $\mathcal{C}$ is strictly bigger than $2$.
\end{con}

All the computer calculations appearing here were done with Macaulay2 \cite{M2} and Magma \cite{Magma}.

\subsection{The non-ordinary and non-superspecial cases: some examples}\label{non ordinary examples}

First, we show by three numerical examples that the level cannot be used in general to characterize the $p$-rank of hyperelliptic curves.

\begin{ex}\label{intermediate with also level 2}
Consider the curve defined by the quintic $f=y^2z^3-x^5-2z^5$, for
$p=11$, $13$ and $17$.

For $p=11$, the $p$-rank is $2$ (i.e. the curve is ordinary), and as predicted by Theorem 3.5, the algebra package gives level $2$, the chain of ideals of $p^e$th roots being $R \supset (z^2,xz,x^3).$

For $p=13$, the level of $f$ is $4$ and the $p$-rank is $0$. Finally, for $p=17$, the level of $f$ is $3$ and the $p$-rank is also $0$.
\end{ex}

\subsection{The superspecial case: some examples and a conjecture}
The authors do not know if, in general, the level of a superspecial curve is always bigger than or equal $3$. Indeed, denote
$$
\tilde{C_g}=[C_g| H_g].
$$
In the superspecial case, $C_g=0$, but if $\tilde{C_g}\neq 0$, it would be still true that there is a non-zero linear combination of the terms $\{z^{2g-1-i}z^i\}_{i=1}^{g}$ in $I_1$, while, again, there is none in $I_2$, which would yield $e>2$. It is not clear for the moment how to control the numbers $c_i$ from the coefficients of $f$ and the multinomial coefficients, so it is not clear whether in general $\tilde{C_g}=[C_g| H_g]$ vanishes or not. Our numerical examples seem to show that this is the case.

However, this is so for the following two infinite families.

\begin{ex}\label{first superspecial family}
Let $\mathcal{C}_p$ be the curve defined by $f=y^2z^{2g-1}-x^{2g+1}-\mu xz^{2g}$ with non zero $\mu\in k$. By Theorem 2 in \cite{Val1995}, $\mathcal{C}_p$ is superspecial if an only if $p\equiv 2g+1\pmod{4g}$ or $p\equiv -1\pmod{4g}$. In the first case, $c_{p-g-1}\neq 0$, and in the second case, $c_{gp-g-1}\neq 0$, so, by Remark \ref{remH}, it is $\tilde{C_g}\neq 0$ and hence $e>2$.
\end{ex}

Another family:

\begin{ex}\label{second superspecial family}
Let $\mathcal{C}_p$ be the curve defined by $f=y^2z^{2g-1}-x^{2g+1}-\mu z^{2g+1}$ with non zero $\mu\in k$. Again, by Theorem 2 in \cite{Val1995}, $\mathcal{C}_p$ is superspecial if an only if $p\equiv 2g+1\pmod{2g+1}$, what implies that $c_{gp-g-1}\neq 0$ and again $e>2$.
\end{ex}

Computational evidence together with these examples lead us to formulate the following conjecture, with which we conclude our work.

\begin{con}Let $\mathcal{C}$ be a superspecial hyperelliptic curve of genus $g$ over $k$. Then, the level is strictly bigger than $2$.
\end{con}

\bibliographystyle{alpha}

redpajama_set_name: RedPajamaArXiv


The original idea to adopt graves of the American liberators came up in February of 1945. To this effect the "Burger Comité Margraten" (Citizens Committee Margraten) was formed. The committee's goal was to support the set-up of the American Cemetery with an extensive adoption campaign. The adopters were supposed to regularly visit the adopted grave and, in case this was appreciated, keep in touch with the next of kin in the U.S.
The campaign gained massive support. At the first Memorial Day in 1945 every grave was decorated with flowers. At the second Memorial Day one year later all graves (at the time an incredible amount of 18,764) had been adopted. Captain Shomon, the founder of the American Cemetery, praised the members of the committee for all the work they had done.
The "Citizens Committee Margraten" is partially responsible for the correspondence that developed between the adopters and the descendants of the fallen soldiers. It often resulted in connections and friendships that have survived to this day. Due to the ageing of the members of the Citizens Committee and to the arrival of new means of information and communications, the committee's work was taken over by the specially formed "Foundation for Adopting Graves at the American Cemetery in Margraten". The new foundation is mainly staffed by persons who are part of Margraten-Eijsden's social and economic network. Its intention is to carry on the activities of the former citizens committee in the best possible way.
For an extensive historiography of the phenomenon of adoption in Margraten we kindly refer to a recent publication by Peter Schrijvers named "The Margraten Boys", How a village refused to forget its American liberators. The book can be ordered from our webshop.
Take a look at our webshop for more publications on this subject.


Gorgeous 18th Century Silver Oil And Vinegar Cruet Holder By 'Maître Orfèvre' Jean Louis OUTREBON.
This beautiful piece comes complete with the hand blown Crystal cruets.
This lovely cruet holder is worked in the Louis XVI Style, with the two cruet holders worked with silver seashells and vine garlands.
The two receptacles can be unscrewed to enable better cleaning of the piece.
The oval base is decorated with acanthus leaves scrolls and grape clusters, and has two round holders to accommodate the toppers of the crystal cruets in order not to have to put these on the tables while serving yourself!
The stand reste on four silver feet done with acanthus scrolls.
The Crystal cruets are perfectly adapted to the piece with an Intaglio grape décor, and wheel cut necks.
The toppers are done in silver with the same grape clusters that decorate the stand.
As usual the work is very refined, and beautifully executed.
It is indeed rare to find an 18th century piece in this condition!
The height given is that of the cruet stands with the Crystal cruets.
The Marks are 18th Century Marks.
The Year Hallmark is a Crowned A used by Fermiers Generaux (Tax collectors in essence..) Jean Baptiste FOUACHE & Dominique COMPANT in Paris from November 18th 1774 until July 12th 1780.
We also have (not pictured) the discharge mark of a monkey's head used during the same period of 1774.
We have the JLO Mark with a crowned fleur de Lys registered by Jean Louis OUTREBON on July 15th 1772, in Paris and used until 1789.
We can peg this piece as being from 1774.


More 1965 alumni from Pueblo County HS have posted profiles on Classmates.com®. Click here to register for free at Classmates.com® and view other 1965 alumni.
If you are trying to find people that graduated in '65 at Pueblo County in Pueblo, CO, check the alumni list below that shows the class of 1965.


text:
Other Voices: Fruitful advice for persimmon thieves
Written by Dave Izant
Courtesy of Dave Izant
A security camera captures a suspected persimmon thief.
We live in north Los Altos and have a very prolific persimmon tree. After we used what we could and gave away what we could, my industrious wife decided to sell the rest. She offered them at a very good price and sold approximately 50 before trouble arrived at our doorstep.
Her procedure was to set them up on a table in the garage and when customers showed up, she would let them select. She would monitor our camera when folks arrived so she could go meet them.
While she was watching for her next customer, two people (we think female, but not totally sure) showed up in hoodies and appeared on our camera. One of them ran into the garage and stole the entire tray of persimmons. My wife ran out and shouted at them (fortunately, we are old and don't run very fast) to stop and return our persimmons.
Even though they were already a block or so away, my wife's loud shout seemed to cause the person running with the tray with 28 persimmons to drop them and to keep running. We picked up the persimmons and the tray. The persimmons were still quite hard, so no damage done.
I will now address the thieves and hope that you are reading this. Others who are contemplating similar activities would also benefit from this. Several years ago, the city of New York instituted a "broken windows" policy to reduce crime. The idea was to arrest people for relatively low-level crimes and give them some modest punishment, the hope being that this would deter them from going on to bigger things. It was deemed to be a very successful program.
Stealing persimmons is about as low-level crime as one would imagine, but it is a start and would be considered a crime, which may stay on your record and come to haunt you later on. Also, remember that there are cameras everywhere.
That, however, is not the point. Honesty is what you do when no one is looking. I hope the thieves who took our persimmons will realize that this time they dodged a bullet and may not be so lucky next time. Time to start living an honest life!
Dave Izant is a Los Altos resident.
The Town Crier welcomes letters to the editor on current events pertinent to Los Altos, Los Altos Hills and Mountain View. Write to us at 138 Main St., Los Altos 94022, Attn: Editor, or email editor Bruce Barton at bruceb@latc.com. Because editorial space is limited, please confine letters to no more than 200 words. Include a phone number for verification purposes. Anonymous letters will not be printed.
You can also have your say right here at losaltosonline.com – scroll to the bottom of any story to add a comment.

redpajama_set_name: RedPajamaCommonCrawl


Racelogic's Inertial Measurement Unit (RLVBIMU04-V2) provides highly accurate measurements of pitch, roll, and yaw rate using three rate gyros, as well as x, y, z acceleration via three accelerometers. The CAN based unit is temperature compensated and has improved calibration and stability.
The RLVBIMU04 is designed for use either as a standalone sensor with simple connection and configuration via the CAN bus interface, or for use with VBOX GPS data loggers.
When used in conjunction with VBOX 3i, data from the IMU can be seamlessly integrated with GPS to produce pitch and roll angle accurate to 0.06° (RMS) as well as smoother velocity data. This ensures GPS data even when satellite reception is interrupted.
The IMU is constructed with a splash proof casing, which is rated to a limited ingress IP rating of IP67*, making it ideal for use on boats or in harsh environments, as well as automotive testing.
Using synchronous 16 bit sampling for each of the internal sensors provides a high degree of accuracy, with yaw rate resolution of 0.0137 degrees per second and acceleration resolution down to 1.5 mg.
Please note: Inertial measurement sensors are highly sensitive mechanical systems. Their performance and life span can be impacted by severe vibration or heavy knocks, and we can only guarantee the gyroscope and accelerometer specifications for a maximum of 2 years from the date of purchase.
The new IMU04 uses higher grade inertial sensors which have less measurement noise and a wider range. This has enabled the Kalman Filter in the VBOX3i to accurately track Pitch, Roll and Yaw angles.
The easiest method of mounting the IMU is by placing it directly on the vehicle roof, co-located with the GPS antenna.
A specially designed IMU Roof Mount (RLACS216) allows for an IMU04 (only) to be securely fastened within the machined enclosure while it's magnetic base ensures that it stays safely in place.
The antenna can then be placed directly above the IMU allowing for the data sources to be measured at the same point.
By co-locating the antenna and IMU on the vehicle's roof the distance between the two does not have to be measured, as you do in a standard setup, leaving only the required translation to – typically – the vehicle's centre of gravity. This method is easier and isn't prone to human measurement error. Co-locating the antenna with the IMU also greatly improves the performance of the Kalman filter.
When using a vehicle with a non-ferrous roof such as aluminium, carbon fibre or glass (sun roof), the magnetic based IMU Roof Mount will not hold and a suction mount is required along with the magnetic base.
With its three points of contact the Vacuum Tripod Mount (RLACS256) ensures a very stable fixture and reduces IMU vibration that might be transmitted by the vehicle's roof.
The adjustable lever arms allow for the levelling of the IMU on a non-flat roof whilst extending the mounting options on the vehicle. For example, you can fix the mount partially or fully on a coupé's glass roof when trying to achieve a greater antenna separation in a dual antenna setup.
If you are testing with a VBOX 3i Dual Antenna and are looking for a maximum antenna separation* the use of a Roof Mounting Pole (RLACS171) in combination with the Pole Mount with Ferrous Plate (RLACS257) will allow for a quick dual antenna alignment.
An additional pair of pole suction arms is supplied to stabilise the IMU on the roof.
If roof mounting the IMU is not possible the IMU can be fixed inside the vehicle. A flexible way to fix the IMU rigidly within the vehicle is by using the Racelogic mounting arm. The three part telescopic handle is fully adjustable to any length between 70 and 150 cm to which another 20 cm can be added by extending a third section using the compression lever.
Both ends are fixed to an 8 x 13 cm plate which sits on a joint to accommodate for uneven surfaces. Pressed against the IMU on the floor and the vehicle's ceiling, the mounting prop ensures that the IMU is fixed tightly.
It is also possible to fix the IMU firmly to the body of the vehicle. Make sure it is mounted in the direction of travel - as shown in the image below. It is also important to mount the sensor so that it is level with the ground.
For best results, mount the IMU and GPS antenna as close to each other as possible. For example: Bolt the IMU to the seat rails and place the GPS antenna on the roof directly above.
If you are using an IMU04 or IMU03, you must measure the relative position of the antenna in relation to the IMU to within +/- 5 cm. These distances must then be entered into the VBOX either via VBOX Tools > VBOX Setup or using a VBOX Manager.
Survey-grade high-speed GPS is the most accurate way to measure velocity - as long as view to the sky is clear.
Problems arise when buildings or tall trees obstruct the testing ground. If the GPS signal is interrupted, dropouts cause spikes in data, which is not ideal when you are relying on a clean velocity signal.
In order to keep high GPS accuracy even where sky visibility is less than perfect, Racelogic couples data from an IMU and the VBOX 3i GPS data logger.
By blending GPS with data from an Inertial Measurement Unit, housing three gyros and three accelerometers, smoother, more reliable data is produced. The solution can deal with GPS dropouts, maintaining high accuracy. IMU integration realises the high accuracies that VBOX 3i can achieve even when external conditions are compromised, meaning that data has now become more reliable and easier to interpret.
How does IMU Integration improve the GPS data channels?
IMU integration decreases the noise for a more accurate reading, and maintains a precise velocity measurement throughout the GPS drop outs.
The heading reading here is very consistent, providing more accurate results than the GPS only data, which exhibits some noise, even in low speeds.
When mapping vehicle position along a heavily tree lined road, the GPS signal incurs some drop outs and reflections, producing noisy position readings. However, IMU integration corrects the position measurements, creating smoother, more accurate data.
IMU Integration data provides a smoother, more accurate representation of longitudinal acceleration (measured in G-force) during an ABS brake stop.
Conducting brake tests on tall vehicles with long suspension travel can result in a speed overshoot of the velocity data, due to the measurements being taken at the high roof position of the GPS antenna. As the brakes are initially applied, there is a higher rate of change in velocity at the roof than there is at the vehicle's centre of gravity (COG).
However, the integration of an IMU04 or IMU03 with a suitably upgraded VBOX 3i can be used to counteract this 'lever-arm' effect by placing the IMU at the COG, which measures the vehicle pitch as it brakes. This data, when combined with that from GPS, provides a compensation for the overshoot and allows for consistent brake stop testing.
In this example of a high-dynamic brake stop, the blue trace (GPS Speed) overshoots at the initial point of brake application, and then exhibits a damped oscillation as the deceleration continues. The IMU-corrected data (red trace) accurately records the brake stop from the vehicles centre of gravity.
Counteracting the lever-arm effect will also aid test engineers when conducting high-dynamic manoeuvres other than brake stops. In slip angle measurements the speed overshoot can occur if the antenna is moving through a greater arc of travel than that of the vehicle's centre of gravity as it corners. Procedures such as lane change manoeuvres can therefore benefit from IMU integration and lever-arm compensation.
The traces between the green and red vertical lines are of a car going over a speed hump. Note how the GPS speed alters as the vehicle roof moves independently of the COG as it goes over the hump. The integrated speed logs the correct speed of the vehicle.
All VBOX units purchased before Jan 2014 need to be upgraded to enable IMU04 integration.
Works with any VBOX unit.
RL Module Mode works with all VBOX units, IMU03 Integration only with VBOX 3i units.
The overall performance of the Kalman Filter (IMU03 Integration) is not as good as with IMU04. IMU03 Integration will not provide lever arm correction if the firmware is not upgraded to version 2.1+, nor body attitude (pitch & roll angles).
*Inertial measurement sensors are highly sensitive mechanical systems. Their performance and life span can be impacted by severe vibration or heavy knocks, and we can only guarantee the gyroscope and accelerometer specifications for a maximum of 2 years from the date of purchase.


text:
Q: Creating new list on Etsy I am trying to create a new listing on Etsy.

*

*I used oauth to authenticate and got
OAUTH_CONSUMER_KEY
and
OAUTH_CONSUMER_SECRET

*I check it with this code and I got return og all the seller data, so everything is ok with the OAuth.
$oauth = new OAuth(OAUTH_CONSUMER_KEY, OAUTH_CONSUMER_SECRET, OAUTH_SIG_METHOD_HMACSHA1, OAUTH_AUTH_TYPE_URI);
$oauth->setToken("key","secret");

try {
 $data = $oauth->fetch("http://openapi.etsy.com/v2/users/__SELF__", null, OAUTH_HTTP_METHOD_GET);
 $json = $oauth->getLastResponse();
 print_r(json_decode($json, true));

} catch (OAuthException $e) {
 error_log($e->getMessage());
 error_log(print_r($oauth->getLastResponse(), true));
 error_log(print_r($oauth->getLastResponseInfo(), true));
 exit;
}

*I am trying to crate a new listings. First i managed to create a new listing through the api browser on the production. Now, i want to create a new listing through PHP. This is what i did, and it return my error:
This is my code:
$oauth = new OAuth(OAUTH_CONSUMER_KEY, OAUTH_CONSUMER_SECRET,OAUTH_SIG_METHOD_HMACSHA1, OAUTH_AUTH_TYPE_URI);

$oauth->setToken("key","secret");

try {
 $url = "http://openapi.etsy.com/v2/listings";

 $params = array('description' => 'thisisdesc','materials' => 'yes','price'=>"5.99"
 ,'quantity'=>"2",'shipping_template_id'=>"52299",'shop_section_id'=>"1"
 ,'title'=>"thisistitle",'category_id'=>"i_did",'who_made'=>"5.99"
 ,'is_supply'=>"1",'when_made'=>"2010_2012");

 $oauth->fetch($url, $params, OAUTH_HTTP_METHOD_POST);
 print_r(json_decode($json, true));

} catch (OAuthException $e) {

 print_r($e);

 error_log($e->getMessage());
 error_log(print_r($oauth->getLastResponse(), true));
 error_log(print_r($oauth->getLastResponseInfo(), true));
 exit;
}

I get the response of:
Invalid auth/bad request (got a 403, expected HTTP/1.1 20X or a redirect)
This method not accessible to restricted API key.

A: If you are still developing your application you can create listings without gaining full API access, so long as the shop is on your account. Make sure you parse in listings_w as a permission scope when making the first OAUTH request. I have altered the example provided by ETSY in the code below.
    // instantiate the OAuth object
    // OAUTH_CONSUMER_KEY and OAUTH_CONSUMER_SECRET are constants holding your key and   secret
    // and are always used when instantiating the OAuth object
    $oauth = new OAuth(OAUTH_CONSUMER_KEY, OAUTH_CONSUMER_SECRET);

    // make an API request for your temporary credentials
    $req_token = $oauth->getRequestToken("https://openapi.etsy.com/v2/oaut/request_token?scope=email_r%20listings_r%20listings_w", 'oob');

    print $req_token['login_url']."\n";`

Notice the scope=email_r%20listings_r%20listings_w;
Hope this helps

A: Aha, here's the answer. From an Etsy developer:

Your API was not yet approved for full API access.  I've fixed that, so you should be able to use those methods now.

Hence, get in touch with the firm, and ask for your key to be approved.

redpajama_set_name: RedPajamaStackExchange


text:
The Tipping Point PDF Book by Malcolm Gladwell
Click here to Download The Tipping Point PDF Book by Malcolm Gladwell English having PDF Size 3.2 MB and No of Pages 287.
It made no sense to them that shoes that were so obviously out of fashion could make a comeback. "We were told that Isaac Mizrahi was wearing the shoes himself," Lewis says. "I think it's fair to say thai at the time we had no idea who Isaac Mizrahi was." By the fall of 1995, things began to happen in a rush. first the designer John Bartlctt called.
Name of Book The Tipping Point
Author Malcolm Gladwell
About Book – The Tipping Point PDF Book
He wanted to use I lush Puppies in his spring collection. Then another Man hattan designer, Anna Sui, called, wanting shoes for her show as well. In Los Angeles, the designer Joel Fitzgerald put a twenty-five-foot inflatable basset hound — the symbol of the Hush Puppies brand — on the roof of his Hollywood store and gutted an adjoining art gallery to turn it into a Hush Puppies boutique.
While he was still painting and putting up shelves, the actor Pee-wee Herman walked in and asked for a couple of pairs. "It was total word of mouth," Fitzgerald remembers. In 1995, the company sold 450,000 pairs of the classic Hush Puppies, and the next year it sold lour times that, and the year after that still more, until Hush Puppies were once again a staple of the wardrobe of the young American male.
Click here to Download The Tipping Point PDF Book
In 1996, Hush Puppies won the prize for best accessory at the Council of Fashion Designers awards dinner at Lincoln Center, and the president of the firm stood up On the stage with Calvin Klein and Donna Karan and accepted an award for an achievement that — as he would be the first to admit — his company had almost nothing to do with.
Hush Puppies had suddenly exploded, and it all started with a handful of kids in the East Village and Soho. How did that happen? The Tipping Point is the biography of an idea, and the idea is very simple. It is that the best way to understand the emergence of fashion trends, the ebb and How of crime waves, or, for that matter, the transformation ot unknown books into bestsellers, or the rise of teenage smoking.
War Doctor Surgery on the Front Line PDF
Seeds of Change PDF
I Will Teach You to Be Rich PDF
Kafka on the Shore PDF
Where the Crawdads Sing PDF
Brief Answers to the Big Questions PDF
The Hating Game PDF
Or the phenomena of word of mouth, or any number of the other mysterious changes that mark everyday life is to think of them as epidemics. Ideas and products and messages and behaviors spread just like viruses do. The rise of Hush Puppies and the fall of New York's crime rate are textbook examples of epidemics in action.
Although they may sound as if they don't have very much in common, they share a basic, underlying pattern. First of all, they are clear examples of contagious behavior. No one took out an advertisement and told people that the traditional Hush Puppies were cool and they should start wear ing them.
Those kids simply wore the shoes when they went to clubs or cafes or walked the streets of downtown New York, and in so doing exposed other people to their fashion sense. They infected them with the Hush Puppies "virus." The crime decline in New York surely happened the same way. It wasn't that some huge percentage ol wouldbe murderers suddenly sat up in 1993 and decided not to commit any more crimes. The Tipping Point PDF Book
When we sav that a handful of East Village kids started the Hush Puppies epidemic, or that the scattering of the residents of a few housing projects was sufficient to start Baltimore's syphilis epidemic, what we are really saying is that in a given process or system some people matter more than others. This is not, on the face of it, a particularly radical notion.
Economists often talk about the 80/20 Principle, which is the idea that in any situation roughly 80 percent of the "work" will be done by 20 percent of the participants. In most societies, 20 percent of criminals commit 80 percent of crimes. Twenty percent of motorists cause 80 percent of all accidents.
Twenty percent of beer drinkers drink 80 percent of all beer. When it comes to epidemics, though, this disproportionality becomes even more extreme: a tiny percentage of people do the majority of the work. Potterat, for example, once did an analysis of a gonorrhea epidemic in Colorado Springs, Colorado. The Tipping Point PDF Book
Looking at everyone who came to a public health clinic for treatment of the disease over the space of six months. He found that about half of all the cases came, essentially, from four neighborhoods representing about 6 percent of the geographic area of the city. Half of those in that 6 percent, in turn, were socializing in the same six bars.
Every time someone in Baltimore comes to a public clinic for treatment of syphilis or gonorrhea, John Zenilman plugs his or her address into his computer, so that the case shows up as a little black star on a map of the city. It's rather like a medical version of the maps police departments put up on their walls, with pins marking where crimes have occurred.
On Zenilman's map the neighborhoods of East and West Baltimore, on either side of the downtown core, tend to be thick with black stars. From those two spots, the cases radiate outward along the two central roadways that happen to cut through both neighborhoods. In the summer, when the incidence of sexually transmitted disease is highest. The Tipping Point PDF Book
The clusters of black stars on the roads leading out of East and West Baltimore become thick with cases. The disease is on the move. But in the winter months, the map changes. When the weather turns cold, and the people of East and West Baltimore are much more likely to stay at home, away from the bars and clubs and street corners where sexual transactions are made, the stars in each neighborhood fade away.
The seasonal effect on the number of cases is so strong that it is not hard to imagine that a long, hard winter in Baltimore could be enough to slow or lessen substantially — at least for the season — the growth of the syphilis epidemic. Epidemics, Zenilman's map demonstrates, are strongly influenced by their situation — by the circumstances and conditions and particulars of the environments in which they operate.
It wasn't that his connections hadn't helped him. It was that he didn't think of his people collection as a business strategy. He just thought of it as something he did. It was who he was. Horthow has an instinctive and natural gift for making social connections. He's not aggressive about it. The Tipping Point PDF Book Download
He's not one of those overly social, back-slapping types for whom the process of acquiring acquaintances is obvious and self-serving. He's more an observer, with the dry, knowing manner of someone who likes to remain a little bit on the outside. He simply likes people, in a genuine and powerful way.
And he finds the patterns of acquaintanceship and interaction in which people arrange themselves to be endlessly fascinating. When I met with Horchow, he explained to me how he won the rights to revive the Gershwin musical Girl Crazy as Crazy for You. The full story took twenty minutes. This is just a portion.
If it seems at all calculating, it shouldn't. Horchow told this story with a gentle, selfmocking air. He was, I think, deliberately playing up the idiosyncrasies of his personality. But as a portrait of how his mind works — and of what makes someone a Connector — I think it's perfectly accurate: I have a friend named Mickey Shannon, who lives in New York. The Tipping Point PDF Book Download
He said, I know you love Gershwin. I have met George Gershwin's old girlfriend. Her name is Emily Paley. She was also the sister of Ira Gershwin's wife, Lenore. Then he stopped himself, as if he realized what he was saying, and burst out laughing. "Look, you can save a whole dollar! In a year's time I could probably save enough for a whole bottle of wine."
Alpert is almost pathologically helpful. He can't help himself. "A Maven is someone who wants to solve other people's problems, generally by solving his own," Alpert said, which is true, although what I suspect is that the opposite is also true, that a Maven is someone who solves his own problems — his own emotional needs — by solving other people's problems.
Something in Alpert was fulfilled in knowing that I would thereafter buy a television or a car or rent a hotel room in New York armed with the knowledge he had given me. "Mark Alpert is a wonderfully unselfish man," Leigh MacAllister, a colleague of his at the University of Texas, told me. "I would say he saved me fifteen thousand dollars when I first came to Austin. The Tipping Point PDF Book Download
He helped me negotiate the purchase of a house, because he understands the real estate game. I needed to get a washer and dryer. He got me a deal. I needed to get a car. I wanted to get a Volvo because I wanted to be just like Mark. Then he showed me an on-line service that had the prices of Volvos all over the State of Texas and went with me to buy the car.
He helped me through the maze of all the retirement plans at the University of Texas. He simplified everything. He has everything processed. That's Mark Alpert. That's a Market Maven. God bless him. He's what makes the American system great." Here is another example of the subtleties of persuasion.
A large group of students were recruited for what they were told was a market research study by a company making high-tech headphones. They were each given a headset and told that the company wanted to test to sec how well they worked when the listener was in motion — dancing up and down, say, or moving his or her head. The Tipping Point PDF Book Download
All of the students listened to songs by Linda Ronstadt and the Eagles, and then heard a radio editorial arguing that tuition at their university should be raised from its present level of $587 to $750. A third were told that while they listened to the taped radio editorial they should nod their heads vigorously up and down.
The next third were told to shake their heads from side to side. The final third were the control group. They were told to keep their heads still. When they were finished, all the students were given a short questionnaire, asking them questions about the quality of the songs and the effect of the shaking.
Slipped in at the end was the question the experimenters really wanted an answer to: "What do vou feel would be an appropriate dollar amount for undergraduate tuition per year?" The answers to that question are just as difficult to believe as the answers to the newscasters poll. The students who kept their heads still were unmoved by the editorial. The Tipping Point PDF Book Free
The tuition amount that they guessed was appropriate was $582—or just about where tuition was already. We normally think of the expressions on our face as the reflection of an inner state. I feel happy, so I smile. I feel sad, so I frown. Emotion goes inside-out. Emotional contagion, though, suggests that the opposite is also true. If I can make you smile, I can make you happy.
If I can make you frown, I can make you sad. Emotion, in this sense, goes outside-in. If we think about emotion this way — as outside-in, not inside-out — it is possible to understand how some people can have an enormous amount of influence over others. Some of us, after all, are very good at expressing emotions and feelings, which means that we are far more emotionally contagious than the rest of us.
Psychologists call these people "senders." Senders have special personalities. They are also physiologically different. Scientists who have studied faces, for example, report that there are huge differences among people in the location of facial muscles, in their form, and also — surprisingly — even in their prevalence. The Tipping Point PDF Book Free
"It is a situation not unlike in medicine," says Cacioppo. "There are carriers, people who are very expressive, and there are people who are especially susceptible. It's not that emotional contagion is a disease. But the mechanism is the same." Howard Friedman, a psychologist at the University of California at Riverside.
Has developed what he calls the Affective Communication Test to measure this ability to send emotion, to be contagious. There was no ambiguity in his presentation. Everything was written on his face. I could not see my own face, of course, but my guess is that it was a close mirror of his. It is interesting, in this context, to think back on the experiment with the nodding and the headphones.
There was an example of someone persuaded from the outside-in, of an external gesture affecting an internal decision. Was I nodding when Tom Gau nodded? And shaking my head when Gau shook his head? Later, I called Gau up and asked him to take Howard Friedman's charisma test. As we went through the list, question by question, he started chuckling. By question 11 —" I am terrible at pantomime, as in games like charades" — he was laughing out loud. "I'm great at that! The Tipping Point PDF Book Free
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War Doctor Surgery on the Front Line PDF Book by David Nott
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Quarterbacks, Receiver, Defense Backs, Running Backs og Linebacker SkulderbeskytterThe Zena 25 Series is designed for the "mid sized" player to the larger Quarterback and Back position.The "patented" breast plate assembly allows a fit for all players, and are adjustable up and down. The anatomically designed pads specifically contour to fit a woman's body.
Designet til Offensive Line, Defensive Line, Linebackers, Tight Ends, Defensive Ends, Centers and Snappers. The Zena Ms. D is still the pad of choice for Offensive and Defensive line, as well as, Tight Ends, Defensive Ends, Centers, and Long Snappers. Its low profile, streamlined design allows greater range of motion and increased vision, allowing the athlete to not be restricted. The anatomically designed pads specifically contour to fit a woman's body. It is important that women athletes understand they now have a "choice" in innovative shoulder pad protection other than those available in the industry.


Caelan Pascoe from Keebra Park State High School was officially sworn in today as the new City of Gold Coast Junior Mayor.
Mayor Tom Tate congratulated Caelan and said his community spirit was inspiring.
"Our Junior Council program has helped hundreds of young people like Caelan develop their leadership skills and learn about local government," Mayor Tate said.
In an innovative first for the program, the swearing-in ceremony was livestreamed.
Mayor Tate said live streaming would expand its reach.
"Live streaming allows us to tap into all 29 participating schools and reach an anticipated audience of 30,000 students," he said.
Caelan's strengths in leadership and teamwork will give him an excellent foundation on which to build his skills in the role of Junior Mayor.
"With 113 students from 29 schools across the Gold Coast, our Junior Council continues to be the largest Junior Council program in Queensland," Mayor Tate said.
The Junior Council provides an opportunity for young people to 'have a voice', raise issues, exchange ideas, discuss community issues and provide input into city planning, programs and services. It helps develop key skills such as communication, consultation, leadership and program development.


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Erickson: Week 6 fantasy football rankings overview
Green Bay Packers running back Aaron Jones (33) against the Atlanta Falcons on Monday, Oct. 5, 2020, at Lambeau Field in Green Bay, Wis. Apc Packers Vs Falcons 2993 100520 Wag
By Andrew Erickson
In this week's preview of the NFL slate, I'm highlighting key insights on the seven games that I didn't review in PFF's huge NFL Week 6 Preview.
I'll identify sleepers, busts, DFS picks and waiver-wire options with consideration for injuries and other fantasy-relevant news and notes. It's the behind the scenes look into my weekly fantasy rankings.
I'll also add some suggested potential replacements for players from the Atlanta Falcons and Minnesota Vikings — the NFL announced on Thursday that the Falcons' facility was shut down in light of multiple positive test results. We just learned Friday morning that the Indianapolis Colts have closed their facility after positive tests as well.
This has become a weekly theme in 2020, so we need to be prepared with contingency plans.
Houston Texans @ Tennessee Titans
The Houston Texans have allowed the second-highest touchdown rate (12.1%) and quarterback rating (143.4) to wide receivers from the slot this season. That doesn't impact A.J. Brown, who is already in a great spot against the Texans — they've allowed the fourth-most receiving yards to WRs over the past two weeks — but it makes Titans slot receivers Adam Humphries and Kalif Raymond extremely enticing from a DFS perspective at just $4,000 and $3,200, respectively.
Brown is going to be shadowed by Texans cornerback Bradley Roby, but that is a matchup I expect the Titans' receiver to win. Last season, Brown went over 100 receiving yards in both games he played against the Texans.
In coverage versus just Robey in one of those matchups, Brown compiled three catches, 27 yards and a touchdown receptions on six targets.
Corey Davis is still on the COVID/reserve list, so if he misses a second straight week it's hard to imagine building a cash lineup in DFS without Brown at his $5,600 price point.
Both starting running backs in this matchup find themselves inside my top 16 this week. Derrick Henry should feast on the Texans, who have allowed the most rushing yards to running backs this season, and David Johnson should be solid versus the Titans and the whopping 5.6 yards per carry they've allowed this season to opposing running backs.
Tennessee has also allowed the highest rate of rushing yards after contact per attempt (3.8) this season.
Johnson has seen 18-plus touches the past two weeks even with Duke Johnson back in the mix, so the volume and matchup should prove to be a favorable fantasy outing.
Will Fuller also remains as a top-option for me this week sliding inside my top 12 wide receivers. The Titans have allowed the second-most fantasy points over the past four weeks to wide receivers and Fuller has averaged 19.8 fantasy points per game in his four healthy games this season. Brandin Cooks is ranked much lower because of his inconsistent play so far this year but is a viable WR3/FLEX option in the solid matchup.
Both receivers should see plenty of cornerback Malcolm Butler, who Deshaun Watson has torched in years past. Since 2018, Watson has the highest passer rating (118.5) when targeting a receiver that Butler is covering.
This year, Butler ranks 44th out of 50 qualifying cornerbacks in PFF's coverage grade.
New York Jets @ Miami Dolphins


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Insurers say no flood defences, no cover as gardaí warn of bogus tradesmen trying to capitalise on stormy weather
Fermoy successfully deployed its €30m flood protection scheme which saw the erection of giant barriers along Kent Bridge
Elaine Loughlin and Cormac O'Keeffe
Upwards of 500 houses have been evacuated since Storm Desmond hit, with more than 270 families still unable to return to their homes.
It comes as disagreement brews between the Government and insurers, who say they would not sign up to a scheme to help flood-hit families and businesses until significant flood defences are built.
Gardaí are also warning homeowners about bogus tradesmen calling to their doors seeking to fix any damage caused to their property by the stormy weather.
John Barry, chair of the national co-ordination group, yesterday confirmed at least 270 homes remain evacuated. However, the group is still awaiting figures from a number of local authorities, so this number is expected to rise.
A further 200 families who had to leave their flooded homes have now returned.
Mr Barry said: "Other information has come forward on businesses. We have a figure of 388 businesses who have been affected by flooding."
The Dáil has cleared time to let the Taoiseach, Tánaiste, and other deputies discuss the flooding crisis when it returns next week.
Although Met Éireann has forecast drier weather over the weekend, apart from some heavy showers along coastal areas, it will be many weeks before floods subside.
The Red Cross said it has already paid out around €450,000 to 166 flood-hit businesses that were unable to obtain insurance. This figure is also expected to increase as more applications are processed. Colette Morris of the Red Cross said: "Next week assessors will be contacting people and they will be for applicants who are looking for claims that are in excess of €5,000."
Mr Barry warned there was still significant flooding on roads in many parts of the country, and motorists are advised to heed all diversions and not to follow their satnavs.
He also warned the public not to take unnecessary trips in boats across flood waters and said there had been some reports of "flood tourism", with eager photographers putting themselves in danger to take pictures of waterlogged areas.
Gardaí have also warned bogus outfits might seek to capitalise on damage caused by the adverse weather. Sgt Kelvin Courtney of the National Crime Prevention Office said: "It tends to be February, into March, that the bogus tradesmen come out of the woodwork, with the spring cleaning."
But with the recent damage caused by high winds and driving rain to properties, not just those in flooded areas, the bogus tradesmen may start approaching homes sooner.
"It is an opportunity for people now, if they want to be unscrupulous, to take advantage of people in these vulnerable situations," said Sgt Courtney.
He urged people who need to get urgent jobs done on their home, such as roof, boundary fencing or door repairs, to take the time to find a reputable tradesman.
"People can be desperate to get a job done quick, but if you act in haste you might regret it," he said.
"They might tell you it's a few quid for the job, but then they come back down and say it's a much bigger job and demand a lot more."
The flood barriers protecting the town in Mallow, Co Cork. Authorities there estimate defences protected the centre from flooding five times last month. Picture Dan Linehan
He advised people not to take on anyone they are suspicious of and to fit door chains, door limiters, or door viewers on their front door — and if concerned to ring local gardaí.
Insurance Ireland, the representative body of Ireland's insurers, yesterday said it would be unwilling to agree to any special scheme for flood-prone areas until significant flood defences are put in place by Government.
The Taoiseach is due to meet with insurers on Tuesday to discuss the issue.
However, Michael Horan of Insurance Ireland blamed flooding on inadequate investment by the Government and said it would not be logical to insure areas which are going to flood again.
"The flood problem has been caused by inadequate investment in flood defences over the years combined with development on flood plains," he said.
- Daniel McConnell
Taoiseach Enda Kenny might wait until March before calling the general election because of the floods crisis, senior Fine Gael sources have said.
The growing consensus has landed on Friday, February 26, being polling day, but in light of the floods crisis, it has been suggestedthat Mr Kenny might seek to hold on to create some separation in the minds of voters.
A second senior party figure confirmed the idea of a March election was now "doing the rounds", saying it was 50:50 that the election would be in February.
Hundreds still waiting for flood relief funding
Most in favour of state-back insurance fund to help weather victims
€106m set aside to fix roads damaged in recent floods
Government announces plan for new group to combat River Shannon flooding
FloodsfloodingTOPIC: Winter floods


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PREMIER ZILLE ANNOUNCES GARDEN ROUTE REBUILD INITIATIVE
Western Cape Premier Helen Zille has announced the establishment of a Garden Route Rebuild Initiative.
The Garden Route Rebuild Initiative will, in partnership with Garden Route Municipalities, community members, and private sector donors, develop a rebuilding strategy for the area and identify where public and private funds can best be spent.
The Garden Route was badly affected by the fires in June. People lost their lives, homes were destroyed in the greater Knysna area and Plettenberg Bay, and the environment was badly damaged. Many jobs have been lost temporarily, especially in the tourism service industry.
Officials from all three spheres of government and countless volunteers assisted in the immediate aftermath of the disaster. "Now it is time to rebuild. I sent a team to the region a week ago to assess how best the Province can assist further. It is clear that the Garden Route requires significant support in the rebuilding process," said Premier Zille.
The Rebuild Initiative will provide immediate support to Municipalities for building approvals and coordination of humanitarian efforts and will launch skills initiatives to enable local residents to participate in the rebuilding process.
Premier Zille said: "I am establishing a Garden Route Rebuild Initiative, which will be responsible for coordinating all of the public and private initiatives in the Garden Route region. The initiative will be led by Minister Alan Winde, and coordinated by Dr. Hildegarde Fast, who was previously Head of Department: Local Government who coordinated the disaster response to the Eden drought in 2009/10".
The provincial team will draw in additional national and provincial departments and resources. Knysna Municipality has established workstreams to deal with the challenges, and the provincial team will support the work that is already under way.
"In recovering from this terrible disaster, we have an opportunity to build a resilient economy, create more sustainable jobs, and create a more united community. We can only do this if the government, civil society, and the private sector join forces to create a future for the Garden Route that is more prosperous than before," said Zille.
July 6, 2017 /by Bomber Webb
https://i0.wp.com/www.edgenews.co.za/wp-content/uploads/2017/07/Helen-Zille.jpg?fit=495%2C400&ssl=1 400 495 Bomber Webb https://www.edgenews.co.za/wp-content/uploads/2016/06/blank_340w.png Bomber Webb2017-07-06 14:18:002017-07-06 14:18:26PREMIER ZILLE ANNOUNCES GARDEN ROUTE REBUILD INITIATIVE
GARDEN ROUTE BIOSPHERE DECLARED FESTIVE OYSTERFEST